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Those too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Un des symboles suivants apparaitra sur la dernidre image de cheque microfiche, selon le cas: le symbols ~^> signifie "A SUIVRE", le symbols V signifie "FIN". Les cartes, planches, tableaux, etc., peuvent dtre film^s 6 des taux de reduction diffdrents. Lorsque le document est trop grand pour dtre reproduit en un seul cliche, il est film6 6 partir de Tangle sup6rieur gauche, de gauche 6 droite, et de haut en bas, en prenant le nombre d'images ndcessaire. Les diagrammes suivants illustrent la mdthode. 1 2 3 1 2 3 4 5 6 J THEORY OF STRUCTURES AND STRENGTH OF MATERIALS. WITH DIAGRAMS, ILLUSTRATIONS, AND EXAMPLES. BY HENRY T. BOVEY, M.A., D.C.L., F.R.S.C, PROFESSOR OP CIVIL ENGINEEKING AND APPLIED MECHANICS, m'giI.L UNIVBKSITV, MONTREAL. MBMBBR OP THE INSTITUTION OP CIVIL BNGINBBRS ; MEMHBK OF THB INSTITU- TION OF MECHANICAL ENGINEERS; LATE FELLOW OF queens' COLLEGE, CAMBRIDGE (ENG.). .f -. NEW YORK: JOHN WILEY & SONS, 68 East Tenth Street. 1893. v^ '■\V\;W\'?, ■■Hi •. U' '- Copyright, 1893, BV HENRY T. BOVEY. Febrir Bbo*., Prinitr*, tX, Pearl Street, New York. ROBSBT DBCWIOHD, jClwtrotvper, 411 & i4S Pearl Btieetf Mew York. TyEDICATED TO WClHsm €. mc-BonaW, WHOSE BENEFACVION^ TO M'gILL UNIVERSITY HAVE DONE SO MUCH :o ADVANCE THE CAUSE OF SCIENTIFIC EDUCATION. «W M( aul int by ovv: Mo of ' Dai Car and We ] who man I stud plet( PREFACE. The present work treats of that portion of Applied Mechanics which has to do with the Design of Structures. Free reference has been made to the works of other authors, yet a considerable amount of new matter has been introduced, as, for example, the Articles on " Surface Loading" by Carus-Wilson, " The Flexure of Columns " by Findlay, and " The Efficiency of Riveted Joints " by Nicolson ; also my own Articles on "Maximum Shearing Forces and Bending Moments," " The Flexure of Long Columns," " The Theorem of Three Moments," etc. I am much indebted to Messrs. C. F. Findlay and W. B. Dawson for valuable information respecting the treatment of Cantilever Bridges, Arched Ribs, and the Live Loads on Bridges. To Messrs. J. M. Wilson, P. A. Peterson, C. Macdonald, and others, many thanks are due for data respecting the Dead Weights of Bridges. I am under deep obligation to my friend Prof. Chandler, who has kindly revised the proof-sheets, and who has made many important suggestions. I have endeavored so to arrange the matter that the student may omit the advanced portions and obtain a com- plete elementary course in natural sequence. At the end of each chapter, a number of Examples, Vi PREFACE. selected for the most part from my own experience, are arranged with a view to illustrating the subject-matter — an important feature, as it is admitted that the student who care- fully works out examples obtains a mastery of the subject which is otherwise impossible. The various Tables in the volume have been prepared from the most recent and reliable results. A few years ago I published a work on " Applied Me- chanics," consisting mainly of a collection of notes intended for the use of my own students. The present volume may be considered as a second edition of that work, but the subject- matter has been so much added to and rearranged as to make it almost a new book. I venture to hope that this volume may prove acceptable not only to students, but to the profes- sion at large. Henry T. Bovey. McGiLL College, Montreal, November, 1892. CONTENTS. CHAPTER I. Frameu Structures. PACK Definitions i Frames of Two or More Members 2 Funicular Polygon 3 Polygon of Forces 4 Line of Loads 5 Mansard Roof 6 Non-closing Polygons 7 Funicular Curve 10 Centre of Gravity 11 Moment of Inertia 12 Cranes, Jib 13 " Derrick 16 " Composite 31 Shear Legs 17 Bridge Trusses > 17 Roof Trusses iq King-po&t Truss 21 Incomplete Frames 27 Queen-post Truss.,. 31 Composite Frames 32 Roof-weights 37 Wind-pressure 38 Distribution of Loads 39 Examples of Roof Trusses 41 Examples of Bridge Trusses (Fink, Bollman, Howe, Bowstring, Single- intersection, etc) 52 Method of Sections 62 Piers 65 vii ':S1 VIM CONTENTS. PACK Tables of Roof -weights and Wind-pressures 67 Examples 70 CHAPTER II. Shearing Forces ano Bending Moments. Equilibrium of Beams 93 Shearing Force 95 Bending Moment 96 Examples of Shearing Force and Bending Moment 97 Relation between Shearing Force and Bending Moment 108 Eftt:;t of Live (or Rolling) Load iil Graphical Representation of Moment of Forces with Respect to a Point.. . 116 Relation between Bending Moments and Funicular Polygon 118 Maximum Shear and Maximum Bending Moment at any Point of an Arbi- trarily Loaded Girder 121 Hinged Girders 136 Examples 131 CHAPTER IIL General Pkincii'LEs, etc. Definitions 140 Stress, Simple 140 " Compound 140 Hooke's Law 141 Coefficient of Elasticity 141 Poisson's Ratio 143 Effect of Temperature 142 Specific Weight. 143 Limit of Elasticity 143 Breaking Stress 147 Dead and Live Loads 143 Repeated Stress Effect 145 WOhler's Lxperimenis 145 Testing of Metals 147 Launhardt's Formula 159 Wey ranch's Formula 1 53 Unwin's Formula 159 Flow of Solids 162 Work, Internal and External 168 Energy, Kinetic and Potential 167 Oblique Resistance 169 Values of k 174 Momentum. Impulse 176 Angular Momentum 177 CONTENTS. IX PACK Useful Work. Waste Work 178 Centrifugal Force 181 Impact 184 Extension of a Prismatic Bar 189 Oscillatory Motion of a Weight at the End of a Vertical Elastic Rod 190 Inertia 198 Balancing 198 Curves of Piston Velocity 205 Linear Diagrams of Velocity 206 Curves of Crank-eflfort. . . 207 Curves of Energy 207 Fluctuation of Energy '. . . 207 Tables of Strengths, Elasticities, and Weights of Materials 210 Tables of the Breaking Weights and Coefficients of Bending Strengths of Beams . 713 Table of the Weights and Crushing Weights of Rocks, etc 4 Table of Expansions of Solids .!I3 Examples 2t6 CHAPTER IV. Stresses, Strains, Earthwork, and RtfAiNiNO \Valls. Internal St: as 2'^e Simple Strain 235 Compound Strain 236 Principal Stresses 240 Curves of Maximum Shear and Normal Intensity 240 Combined Bending and Twisting Stresses 244 Combined Longitudinal and Twisting Stresses 247 Conjugate Stresses 247 Relation between Principal and Conjugate Stresses 247 Ratio of Conjugate Stresses 250 Relation between Stress and Strain 231 Rankine's Earthwork Theory 255 Pressure against a Veriiv:al Plane 257 Earth Foundations 258 Retaining Walls 260 Retaininij Walls. Conditions of Equilibrium 260 Rankine 3 Earthwork Theory applied to Retaining Walls 264 Line of Rupctire 265 Practical Rules respecting Retaining Walls 267 Reservoir Walls 271 General Case of Reservoir Walls 275 General Equations of Stress 276 Ellipsoid of Stress a8i Stress-strain Equations 281 fl ^:iiS m CONTENTS. 1 1 Isotropic Bodies 283 Relation between A, \, and G 285 Traction 287 To.sion 268 Woric done in the Small Strain of a Body (Clapeyron) 292 Examples 294. CHAPTER V. . i , . Friction. Friction 300 Laws of Friction 300 Inclined Plane 301 We /edge 302 Screws 306 Endless Screw 309 Rolling Friction 310 Journal Friction 312 Pivot 316 Cylindrical Pivot 316 Wear 318 Conical Pivot 319 Schiele's Pivot {anti-friction) 320 Belts and Ropes 321 Brakes 323 Effective Tension of a Belt 324 Effect of High Speed 325 Slip of Belts 326 Prony's Dynamometer 327 StifTness of Belts and Ropes 327 Wheel and Axle 329 Toothed Gearing 331 Be vel- wheels 335 Efficiency of Mechanisms. . . 335 Table of Coefficients of Journal Friction 336 Examples 337 CHAPTER VI. Transverse Strength of Beams. Elastic Moment 340 Moment of Resistance 340 Neutral Axis 340 Transverse Deformation 344 Coefficient of Bending Strength 344 CONTENTS. . » FAGE Equalization of Stress 349 Surface Loading 350 Effect of Bending Moment in a Plane which is not a Principal Plane 354 Springs 355 Beams of Uniform Strength 358 Flanged Girders 365 Classification of Flanged Girders 365 Equilibrium of Flanged Girders 366 Moments of Inertia of I and other Sections 371 Design of a Girder of I-section 381 Deflection of Girders 384 Camber , 3S7 Stiffness 389 Distribution of Shearing Stress 391 Beam acted upon by Forces Oblicue to its Direction 396 Similar Girders 401 Allowance to be made for Weight of Bcim 405 Examples 407 CHAPTER VII. Transverse SxRENfiTH ok Beams— {Co«//w«l * J i\ . 1 « THEORY OF STRUCTURES. 5. To show that the Intersection of the First and Last Sides of the Funicular Polygon (i.e., the Point G) is a Point on the Actual Resultant of the System of Forces P,, P, , Pj, . . . — First consider two points T^, , /*,, MNP being the force and ABCD the funicular polygon. Let yJZ>, DC, the first and last sides of the latter, be pro- FiG. la. duced to meet in g^^ also let Z?C produced meet the line of action of /*, in H. Produce t^/'and MN to meet in K. Let the lines of action of /", and /*, meet in L. By similar triangles, H P. Hence or KP KN HC ~ HL' KN KO ~ HB HC KO KM~ HB' KP KN KO KN KO KM ~ HC HB Hg, ' HL HC HB ' KP KM~ ' HV pnd therefore, since the angle H is equal to the angle K, the line PM\s parallel to the line Lg^. FRAMES LOADED A T THE JOINTS. But PM represents in magnitude the resultant of the forces P,, P^, and is parallel to it in direction. Therefore Lg^ is also parallel to the direction of the re- sultant. But L is evidently a point on the actual resultant of P^ , P, . Hence ^, must be a point on this resultaiit. 7\V.i7, let there be three forces, P^, P^, P^. Replace P, , P^ by their resultant X acting in the direction Lg^. The force and funicular polygons for the forces A' and P, are evidently MPQ and Ag^DE, respectively ; and g^ , the point of intersection of Ag^ and ED produced, is, as already proved, a point on the actual resultant of X and P^ , i.e., of P, , P, , and P3 . Hence the Jirst and /ast sides, AB, ED, of the funicular polygon ABCDE of the forces P^, P^, P^ , with respect to the pole O, intersect in a point which is on the actual resultant of the given forces. The proof may be similarly extended to four, five, and any number of forces. If the forces are all parallel, the force pul\-gon of the two forces P^ , Pj becomes a straight line, MNQ. Draw the funicular Fig. 13. polygon ^5C/? as before, and through ^if,, the intersection of the y?rj/and /^/j/ sides, dravv^'-, Kparallel tOiT/(2. and cutting Z?C" in Y. By similar triangles, ON ~ ON ~ BY' and ON ~ ON ~ CY ' P. CY^ BY' I! 10 THEORY OF STRUCTURES. ' ^1 :, Hence Vg^, which is parallel to the direction of the forces /*,, /*,, divides the distance between their lines of action into seg- ments which are inversely proportional to the forces, and must therefore be the line of action of their resultant. The proof may be extended to any number of forces, as in the preceding. Funicular Curve. — Let the weights upon a beam AB become infinite in number, and let the distances between the weights diminish indefinitely. The load then becomes continuous, and the funicular poly- gon is a curve, called the funicular curve. The equation to this curve may be found as follows : Let the tangents at two consecutive points /*and Q meet in R. This point is on the vertical through the centre of gravity of the load upon the portion MN of the beam. A X Mt/JN B Fig. 14. Fig. 15. Let 17/ be the hne of loads, and let OS, OT ho. the radial lines from O, the pole, parallel to the tangents at P and Q. Take ^ as the origin. Let ^ be the inclination of the tangent at P to the beam, and let the polar distance OV = p. wdx = the load upon the portion JAV. 'riicn wc^x = ST=S1'- TV =p tan ^ - / tan Kj^ + d^) == — pdti, approximately. de d'y ''=^dv = Kix- since ^ = dx' Integrating twice, c^ and i*, being constants of integration. CENTRES OF GRAVITY. II If the intensity, w, of the load is constant, py ^ - -^ -\- c,x -^ c^, and the curve is a parabola. 6. Centres of Gravity. — Let it be required to determine the centre of gravity of any plane area symmetrical with re- spect to an axis- 2 ' / 2 + ...= 2/ " The sum 2{ax'') is the moment of inertia, /, of the plane area with respect to gG. Hence, A =—. or / = 2Ap, 2p h ' MOMENT OF INERTIA. n The moment of inertia /^ of the area, with respect to a parallel axis at distance ^^ from gG, is given by the equation \vhere5= ^, + /4, + . . . I et the new axis intersect Cyg and kg in the points q and r. The triangles qgr and Om are similar. qr I « S " y~ P ~ / ' and, therefore, the area A' of the triangle qgr 'Im Hence ly = 2pA -\-2pA' = 2p{A 4- A'). I W y4 iVi?^^. — If /> be made = — = — , 2 2 and 7=^' and 5y,'rr^^', .-. /, = .4(/i+^';- The angle lOn is also evidently a right angle. 8. Cranes. — (a). Jib-crane. — Fig. 19 is a skeleton diagram of an ordinary jib-crane. OA is the post fixed in the ground at 0\ OB is the jib ; AB is the tie. The jib, tie, and gearing are suspended from the top of the post by a cross-head, which admits of a free rotation round the axis of the post. Let the crane lift a weight W. :l |i| t in 14 THEORY OF STRUCTURES. Three forces in equilibrium meet at B\ viz., W, the tension T in the tie, and the thrust C along the jib. Fig. ao. Draw the reciprocal figure 55,5, of B, S^S^ representing W. T SS„ AB and W~ S,S,~ Aa W~ S,S,'~ Ad' The load is not suspended directly from ^, but is carried by a chain passing over pulleys to a chain-barrel usually fixed to the crane-post. The stress 5 in the chain depends upon the W system of pulleys, and is, e.g., — , if « is the number of falls of chain from B and if friction is 7teglected. In order to obtain the true values of 7"and 6" this tension'5must be compounded with W. Draw 5,/t parallel to the direction in which the chain passes ' om B to the chain-barrel, and take S^k to represent 5 in CRANES. 15 magnitude. The line S,k evidently represents the resultant force at B due to IV and S. Draw ki parallel to AB. The tension in the tie and the thrust in the jib are now evidently represented by tk, tSi , respectively. Generally the effect of chain-tension is to ditiiinish the ten- sion of the tie and to increase the thrust on the jib. , ^ . ^BD „.BD Thevertical component of T, viz., B—rj; mitted through the post. \V- AO' is trans- Thc total resultant pressure along the post at O Tsin BAB -\-C sin BOF=-lV^^-\-lV-^ — -J^ ""^ ivli^^^±^=jv. The pull upon the tie tends to upset the crane, and its moment with respect to O is r cos BAD XA0= W^^AO = WAD = WOF, AU AB OF being the horizontal projection of AB. OF is often called the radius or throw of the crane. If the post revolves about its axis (as in //V-cranes), the jib and gearing are bolted to it, and the whole turns on a pivot at the toe G. In this case, the frame, as a whole, is kept in equilibrium by the weight W, the horizontal reaction // of the web-plate at O, and the reaction R at G. The first two forces meet in F and, therefore, the reaction at G must also pass through F. Hence, since OFG may be taken to represent the triangle of forces. H= W OF OG and R = W- GF OG' In a portable crane the tendency to upset is counteracted by means of a weight placed upon a horizontal platform OL attached to the post and supported by the tie AL. The horizontal projection fm of fk represents the horizontal i 'i i i6 THEORY OF STRUCTURES. Fig. 31. pull at A, and if tn be drawn parallel to AL, the intercept vin cut off on the vertical through m by the lines tin and tn repre- sents the counter-weight required at L. {b) Derrick-crane, — The figure shows a combination of a der- B rick and crane, called a derrick- crane. It is distinguished from the jib-crane by having two back-stays, AD, AE. One end of the jib is hinged at or near the foot of the post, and the other is held by a chain which passes over pulleys to a winch on the post, so that the jib may be raised or lowered as required. The derrick-crane is gener- ally of wood, is simple in con- struction, is easily erected, has a vertical as well as a lateral motion, and a range equal to a circle of from lo to 60 feet radius. It is therefore useful for temporary works, setting masonry, etc. The stresses in the jib and tie are calculated as in the jib- crane, and those in the back-stays and post may be obtained as follows : Let the plane of the tie and jib intersect the plane DAE of the two back-stays in the line AF, and suppose the back-stays replaced by a single tie AF. Take OF to represent the hori- zontal pull at A. The pull on the " imaginary" stay A F is then represented by AF and is evidently the resultant pull on the two back-stays. Completing the parallelogram EG AH, AH will represent the pull on the back-stay AF., and AG that upon AD, their horizontal components being OK, OL, respectively. The figure OKFL is also a parallelogram. If the back-stays lie in planes at right angles to each other, OL = OF cos B — T sin a cos Q, and is a max. when ^ = 0°, and OK =^ OF sin 6 = T sin <>e sin 6, and is a max. when 6 90°, BRIDGE AND ROOF TRUSSES. 17 B being the angle FOL, and a the inclination of the tie to the vertical. Hence the stress in a back-stay is a maximum when the plane of the back-stay and post coincides with that of the jib and tie. Again, let (i be the inclination of the back-slays lu Lhe ver- tical. The vertical components of the back-stay stresses are Tsin a cos 6* cot ft and T sin n sin (9 cot ft\ and, therefore, the corresponding stress along the post is T sin a cot /i (cos B -\- sin if), which is a maximum when B = 45°. 9. Shear Legs (or Shears) and Tripods (or Gns) are ow Pig. 23. often employed when heavy weights are to be lifted. The former consists of two struts, AD, AE. united at A and sup- ported by a tie A C, which may be made adjustable so as to admit of being lengthened or shortened. The \vei^\^ i" 1 y -^^pX . t ^ •; y^ \. ^ i To Fig. as- Fig. 24. Take the vertical line 5,5", to represent P, the weight at O. Draw 55i parallel to OC, and SS^ to OD. Draw the horizontal SH, and let the angle AOC =:^ a. The thrust along OC = S^S = S^H cosec a = — cosec a. P The tension along OA = SH = S,H cot a = — cot a. The horizontal and vertical thrusts upon the masonry at C P P (or D) are — cot a and — , respectively. If the girder is uniformly loaded, P is one half of the whole load. II. In the figure a straining cill, EF, is introduced, and the girder is supported at two intermediate points. 1 f Fig. as. Fig. s6. Let Pht the weight at each of the points E and F. Draw the reciprocal SS,Ho[ the point E, ^i//" representing P. ill BRIDGE AND ROOF TRUSSES. 19 The thrust in EC (or FD) = SS, = p:^% = P^. and the o,/7 AL horizontal thrust in the straining piece If a load is uniformly distributed over AB, it may be assumed that each strut carries one half of the load upon AF (or BE), and that each abutment carries one half of the load upon AE (or BF). By means of straining cills the girders may be supported at several points, i, 2, . . . , and the weight concentrated at each may be assumed to be one half of the load between the two adjacent points of support. The calcula- tions for the stresses in the struts, etc., are made precisely as above. If the struts are very long they are liable to bend, and counterbraces, AM, BN, are added to counteract this tendency. 12. The triangle is the only geometrical figure of which the form cannot be changed without varying the lengths of the sides. For this reason, all compound trusses for bridges, roofs, etc., are made up of triangular frames. Fig. 28 represents the simplest form of roof-truss. AC, BC a.rQ rafters of equal length inclined to the horizontal at an angle 01, and each carries a uniformly distributed load iV. Fig. a?. H ■ i > . \i ,11 Fig. 28. The rafters react horizontally upon each other at C, and their feet are kept in position by the tie-beam AB. Consider the rafter A C. The resultant of the load upon AC, i.e., IF, acts through the middle point D. I 90 THEORY OF STRUCTURES. Let it meet the horizontal thrust // of BC upon AC \t\ F. For equilibrium, the resultant thrust at A must also act through F. The sides of the triangle AFE evidently represent the three forces. Hence „ ,,,AE WAR W , "— ^iTr- — — 77P = — cot a ; hb 2 DE 2' R = wf^=W. -\-£F' V ^^ = V-^i(^2'=V- + cot a The thrust R produces a tension H in the tie-beam, and a vertical pressure W upon the support. Also, if y is the angle FAE, FF DF ^ If the rafters AC, BC are unequal, let or,, or, be their in- clinations to A, B, respectively. Let W^ be the uniformly distributed load upon AC, W, that upon BC. Px^M F/'^i Fic. 39. Let the direction of the mutual thrust P zt C make an angle /? with the vertical, so that if CO is drawn perpendicular ROOF TRUSSES. ai to i^C, the angle COB = fi\ the angle ACF=go° - ACO = QO'^ - 1^ - <^). Draw AM perpendicular to the direction of P, and consider the rafter AC. As before, the thrust A', at A, the resultant weight lV^ at the middle point of AC, and the thrust P aX. C mecL ill the point F. Take moments about A. Then P.AM= W,AE. But AM = AC sin A CM = AC cos {/3 — or,), and ^£ = — - cos or,. 4W P = ^^. cos o'. 2 cos(/^ — ar,)' Similarly, by considering the rafter BC, P = W., cos a„ W. cos or. 2 sin(/S + «, — 90°) 2 cos(/S4-«,)' Hence cos a, w. 2 cos{fi — a,) cos a„ 2 cosi/H -\- a;)' and therefore tan /? = PF,+ ^, W^ tan «, — W^3 tan a The horizontal thrust of each rafter = Psin /?. The vertical thrust upon the support A = W^ — P cos fi. » The vertical thrust upon the support B = W^-\- P cos /?. 13. King-post Truss. — The simple triangular truss may be modified by introducing a C king-post CO, which carries a portion of the weight of the beam AB, and transfers it through the rafters so as to act p^^ upon the tie in the form of a tensile stress. w ■Til I;.: ;i a2 THEORY OF STRUCTURES. Let P be the weight borne by the king-post ; represent it by CO. Draw OD parallel to BC, and DE parallel to AB. DC = CE sin u cosec a is the thrust in CA due to P, and is of course equal to DO, i.e., the thrust along CB. P DE = CE cot 01 = — cot a is the horizontal thrust on each rafter, and is also the tension in the tie due to P. Let W be the uniformly distributed load upon each rafter. cot ^f The total horizontal thrust upon each rafter = ( W-\- P)- The total vertical pressure upon each support = W -\- P If the apex C is not vertically over the centre of the tic- beam take COy as before, to represent the weight P borne by the king-post ; draw (^Z) parallel to BC, and DE parallel to AB. The weight P produces a B thrust CD along CA, DO along fe^S CB, and a horizontal thrust DE ^'^^- 3i- upon each rafter. CE is the portion of P supported at A, and EG that sup- ported at B. DE, and therefore the tension in the tie AB, diminishes with AO, being zero when AC is Vertical. Sometimes it is expedient to support the centre of the tie- beam upon a column or wall, the king-post being a pillar against which the heads of the rafters rest. Consider the rafter AC. The normal reaction R' of CO upon AC, the resultant KiG, 3a. weight W at the middle point D, and the thrust R dX A meet in the point F. In ' ROOF TRUSSES. lake moments about A. Then n w R'AC= W.AE, or R' = — cos a. Thus the total thrust transmitted through CO to the sup- W port at is 2 — cos oe . cos a = W cos' a. The horizontal thrust upon each rafter W . W . = — cos a sm a = — sm 2(x. 2 4 14. If the rafters are inconveniently long, or if they are in danger of bending or breaking transversely, the centres may be supported by struts OD, OE. A portion of the weight upon Fio. 34- the rafters is then transmitted through the struts to the vertical tie (king-post cr rod) CO, which again transmits it through the rafters to act partly as a vertical pressure upon the supports, and partly as a tension on the tie-beam. The main duty, indeed, of struts and tics is to transform transverse into longitudinal stresses. This king-post truss is the simplest and most economical frame for spans of less than thirty feet. In larger spans two or more suspenders may be introduced, or the truss otherwise modified. mi: l|! H THEORY OF STRUCTURES. % % iHii Let there be a load 2 W uniformly distributed over the rafters ACy BC, and assume it to be concentrated at the joints A, D, C, E, B, m the proportion — , — ■, — , — , — . 42224 Also, let the load (including a pol-tion of the weight upon the tie-beam AB, and the weights of the members OD, OE, OC) borne directly at O be P. P The total reaction at each support; is W -\ — , and acts in W ' an opposite direction to the weight — there concentrated. 4 Hence the resultant reaction at a support is ^W-\-\P. Thus, the weights at the points of support A and B are taken up by the abutments, and need not be considered in de- termining the stresses in the several members of the frame. Draw the reciprocal S^HS^ of ^. Then xW P SyH = 1 — ; //"i^, = tension in AO; 42 5g5i = compression in AD. Draw the reciprocal S^S^S3S^ of D. Then 5,5, = compression in OB; S^S^ = compression in DC; 5,5, = — = weight at D. 2 Draw the reciprocal S^S^S^S^S^ of C, Then 5,5, = tension in CO = — -f- /*; 5,5, = compression in CE; W S^S^ = —• — weight at C. CE\ ROOF TRUSSES. Draw the reciprocal 5,5,5,5,5^ of E. Then 35 S^S^ = compression in OE ; 5,5, = compression in BE; W 0,0, = weight at E. Draw S^K horizontally. Then S,KS\S, is evidently the reciprocal of B ; A'5, = f W-\- ^P, being \he reaction at B, and S^K the tension in the tie BO. The reciprocal of O is also the figure SjHKStS^S,S^ , and //A!" =: P. 15. Collar-beams (/?£■), queen-posts (DF, EG), braces, etc., may bo employed to prevent the deflection of the rafters. The complexity of the truss necessarily increases with the span and with the weight to be borne. A'-" sr^^ Pig. 35 Fig. 36. With a single collar-beam and a uniformly distributed load, SJ7S, is the reciprocal of A, and 5,5,^5,5,5, the reciprocal of B ; 5,// being the reaction at A, and 5^5, the weight at B. D /^\ E \^ ^\B t\ - c ^ U- Fig. 37. Pig. 38, With a collar-beam DE, two king-posts DF, EG, and a uniformly xlistributed load, the stresses at the joints D and E 1 in >"* r !ii[ 1 I ^i 26 THEORY OF STRUCTURES. become indeterminate. To render them determinate it is sometimes assumed that the components of the weights at D and E, normal to the rafters, are taken up by the collar-beam and corresponding king-post. Thus S^HS^ is the reciprocal of A, and ^,5,535<5j the reciprocal of D, S^H being the reaction at Ay 6"j,.S, the weight at D\ SJS., is the normal component of the weight, and the components of S^S^, viz., SJS^ horizontal and Sj^., vertical, represent the stresses borne by DE and DP, re- spectively. This frame belongs to the incomplete (Art. 1 8) class, and if it has to support an unequally distributed load, braces must be introduced from D to G and from E to F. l6. The truss ABC, Fig. 40, having the rafters supported at two intermediate points, maybe employed for spans of from 30 to 50 feet. Suppose that these intermediate points of sup- port trisect the rafters, and let each mfter carry a uniformly dis- tributed load W. Fig. 39- Fig. 40. Then a weight may be considered as concentrated at eacli W of the joints //, D, C, E, K. This wei'dit = — . 3 Let P be the weight directly supported at each of the joints F, G. The resultant reaction Vit A — l\V-\- P. S^//S\, is the reciprocal of A, S^H representing ^W-{- P. S^SjS^S^S^ is the reciprocal of //. S^SJ/KS,S, is the reciprocal of P, HK representing P, the weight directly borne at P. 6^,5,5,5,5,53 is the reciprocal of />, S^S, representing the weight at D, S,S^ the thrust along HD, S,S, the tension in DP, S,S, the thrust along El), and S^S, the thrust along CV. INCOMPLETE FRAMES. V ; each )f the 5. the As in the preceding case, this truss will be found incomplete if the load is unevenly distributed, and the reciprocals of D and E will not close. In practice, however, the friction at the joints, the stiffness of the several members, and the mode of construc- tion render the truss sufificiently strong to meet the ordinary variations of load. 17. General Remarks. — In the trusses described in Arts. n and 14 the vertical members are ties, i.e., are in tension, and the inclined members are struts, i.e., are in compression. By inverting the respective figures another type of truss is obtained in which the verticals are struts while the inclined members are ties. Both systems are widely used, and the method of calcu- lating the stresses is precisely the same in each. In designing any particular member, allowance must be made for every kind of stress to which it may be subjected. The collar-beam DE, for example, must be treated as a pillar subjected to a thrust in the direction of its length at each end ; if it carry a transverse load, its strength as a beam, supported at the points D and Ey must also be determined. Similarly, the rafters AC, BC, etc., must be designed to carry transverse loads and to act as pillars. But it must be remembered that struts and queen-posts provide additional points of support over which the rafters are continuous, and it is practically suf- ficient to assume thart the rafters are divided into a number of short lengths, each of which carries one half oi the load between the two adjacent supports. When a tie-beam is so long as to require to be spliced, allowance must be made for the weakening effect of the splice. 18. Incomplete Frames. — The frames discussed in the preceding articles (excepting those referred to in Art. 15) will support, without change of form, any load consistent with strength, and the stresses in the several members can be found in terms of the load. It sometimes happens, however, that a frame is incomplete, so that it tends to change form under every distribution of load. An example of this class is the simple trapezoidal truss, consisting of the two horizontal members AB, DE, and the two equal inclined members AD, JUi, Fig. 41. First, let there be a weight W at each of the points D, E. 1 , ill'-! 38 THEORY OF STRUCTURES. The triangles of forces for the joints D and E, viz., SS^H and SSyH, can be drawn, and hence it follows that there must be Fig. 41. Fig. 42. equilibrium. This is also evident from the symmetrical char- acter of the loading. The same triangles represent the forces at the points of support A, B. .-. reaction at .r^ — S,I/= W= S^H = reaction at B. Next, let iherc e << W,) at E. \;«;ight fF, at D and a weight W, Fig. 43. It will now be found that the diagram of forces will not close, so that there cannot be equilibrium. The joint B will be pushed in and the frame distorted. The distortion may be prevented by introducing a brace irom A to E or from B to D. In the latter case S^mSS^S^ represents the stress-diagram, the triangle S^HS being the reciprocal of the joint E, and the quad- rilateral Sm S ^ f/ thiit of the joint D. Drawing the horizontal ;«//, the triangle mnS^ and the quadrilateral mSS^n are evidently the reciprocals of A and B, respectively. .*. nS\ = reaction at A and «S, = reaction at B. INCOMPLE TE FRA MES. 29 In practice the loads are usually transmitted toZ? and E by means of two vertical queenposis {queen-rods or queens) DF, EG. I D 1 ^ A/ \ \b 1 = G Fig. 45. If there are no diagonal braces DG, EF, the distortion of the frame under an unevenly distributed load can only be pre- vented by the friction at the joints, the stiffness of the mem- bers, and by the queens being rigidly fixed to AB at /'"and G. Let W^ be the load at F transmitted through the queen FD to D. Let W^ (< W^ be the load at G transmitted through the queen GE to E. If the frame is rigid, the reactions R^ at A and R^ at B, which will balance these weights, can easily be found by taking moments about B and A, successively. Thus, and W W W IV RJ = -:r{l - c) + -^{l -V c), where AB = I and FG = c. Draw the triangle of forces SHS^ for the joint A, 5//" rep- resenting R^ . The triangle SS,X is the reciprocal of the joint at D, and the tension in FD should, therefore, be XS^ = SH =. 7?, . But the ';ii M M r I 1 i ' i:^l 3© THEORY OF STRUCTURES tension in FD is actually W^ , so that there is an unbalanced force, = W, - R, W, - W , l-c 2 ' / ' acting along FD. To take up this unbalanced force and render the frame rigid the diagonal DG is introduced, and the stress for which it should be designed is evidently W — W I — c s ( W, - R^) sec FDG = --^ ^' '—^ - , s being the length of the diagonal and a? the depth of the truss. The complete stress diagram is as shown in Fig. 46. Cor. I. The manner in which distortion is prevented by the stiffness of AB may be shown as follows: Let X be the force of resistance which AB, by its stiffness, can exert at F or 6^ against any load which tends to make it deviate from the horizontal. If W\s the load at F, the actual downward pull upon D is W— X ; this must necessarily produce an equal upward pull at E, which must be balanced by the force of resistance x at G, '. W — X = X, and W X = 'i 1:: I i Thus the beam AB will be acted upon by an upward pull W — at F and an equal downward pull at G, forming a couple IV of moment — c, and showing that equilibrium is impossible. The upward reaction R^ at A is ' ~" 7v T ~2 ~ "2 2~"/ ~ ~J1 I COMPOSITE FRAMES. %\ = downward reaction at B^ and the moment at F (or G) 2/2 Cor. 2. Let a weight IF be supported at the joint D of anj quadrilateral frame ADEB. Draw the reciprocal SS^S^ of D, Fig. 47- Fig. 48. 5,5, representing W. Draw ^S, parallel to EB and intersecting the vertical 5,5, produced in 53 . The weight which can be borne at E consi.stent with equilibrium is represented b}' 5,53 . 19. Composite Frames or Trusses (i.e., frames made up of two or more simple frames). — An example of this class has already been given in the case of the king-post roof (Art. 13). Bent Crane. — Fig. 49 shows a convenient form of crane when much head-room is required near the post. The crane is merely a semi-girder, and may be tubular with plate-webs if the loads are heavy, or its flanges may be braced together as in the figure for loads of less than ten tons. The flanges may be kept at the same distance apart throughout, or the distance may be gradually diminished from the base towards the peak. Let the nuqpbers in Fig. 50 denote the stresses in the cor- responding members. Three forces, 5, , C, , and W, act through the point (l), so that 5, and (7, may be obtained in terms of W\ three forces, S^, S^, T,, act through (2), so that 5, and 7", may be obtained in terms of 5, and therefore of IV; four forces, 5j, (T,, 5,, (7<, act through (3), and the values of 5, , Ci 32 THEORY OF STRUCTURES. ! 1 being known, those of vS,, C^ may be determined. Proceed- ing in this way, it is found that of the forces at each succeed- ing joint only two are unknown, and the values of these are consequently determinate. Fio. so. The calculations may be checked by the method of mo7nents. and by the stress diagram (Fig. 50). E.g., let W= 10 tons. T? :e moments about the point (7). Then r,(.r7) = 10(^7) or r, = ^ = 26 tons = (68) in Fig. 50. I. -5 No other forces enter- into the equation of moments, as the portion of the crane above a plane intersecting (68) and passing through (7) is kept in equilibrium by the weight of 10 tons and the stresses T^, S^, C^; the moments of 5, and C, about (7) are evidently zero. In the stress diagram (Fig. 50) PaQ is the reciprocal of the point I, adQ of the point 2, PcdQ of 3, Qdai q^ 4, and so on. Other examples of composite roof and bridge frames will now be given. 20. Roof-trusses. — A roof consists of a covering and of trusses (or frames) by which it is supported. The covering is generally laid upon a number of common rafters which rest. ROOF TRUSSES. 33 upon horizontal beams (or purlins), the latter being carried y-y trusses spaced at intervals varying with the type of con- struction but averaging about lO ft. The truss rafters arc called principal rafters, and the trusses themselves are often designated as principals. In roofs of small span the trusses and purlins are sometimes dispensed with. Types of Trtiss. — A roof-truss may be constructed of tim- ber, of iron or steel, or of these materials combined. Timber is almost invariably employed for small spans, but in the longer spans it has been largely superseded by iron, in con- sequence of the combined lightness, strength, and durability of the latter. Attempts have been made to classify roofs according to the mode of construction, but the variety of form is so great as to render it impracticable to make any further distinction than that which may be drawn between those in which the reac- tions of the supports are vertical and those in which they are inclined. Fig. 51. Fig. S3. Fig. S3. Fig. 54. Fig. ss. Fig. s6. Fig. 57. Fig. 58. 11; '(,[1? Fig. 59. \\ Fig. 51 is a simple form of truss for spans of less that 30 ft. Fig. 52 is a superior framing for spans of from 30 to 40 ft.; it may be still further strengthened by the introduction of struts, Figs. 53 and 54, and with such modification has been employed to span openings of 90 ft. It is safer, however, to limit the use of the type shown by Fig. 53 to spans of less than It 34 THEORY OF STRUCTURES. 60 ft. Figs. 55, 56, 57, 58, and 59 are forms of truss suitable for spans of from 60 to 100 ft. and upwards. Arched roofs, Figs. 58 and 59, admit of a great variety of treatrrent. They have a pleasing appearance, and cover wide spans without intermediate supports. The flatness of the arch is limited by the requirement of a minimum thrust at the abutments. The thrust may be resisted either by thickening the abutments or by introducing a tie. If the only load upon a roof-truss were its own weight, an arch in the form of an inverted catenary, with a shallow rib, might' be used. But the action of the wind induces oblique and transverse stresses, so that a considerable depth of rib is generally needed. If the depth exceed 12 in., it is better to connect the two flanges by braces than by a solid web. Roofs of wide span are occasion- ally carried by ordinary lattice-girders. Principals, Purlins, etc. — The principal rafters m Figs. 51 to 57 are straight, abut against each other at the peik, and are prevented by tie-rods from spreading at the heels. When made of iron, tee (T), rail, and channel (both single 1 — 1 and double ][ ) bars, bulb-tee (T) and rolled (I) iron beams, are all excellent forms. Timber rafters are rectangular in section, and for the sake of economy and appearance, are often made to taper uniformly from heel t() peak. The heel is fitted into a suitable cast-iron skew-back, or is fixed between wrought-iron angle-brackets (Figs. 60, 61, 62), and rests either directly upon the wall or upon a wall-plate. Fig. 61. Fig. 62. When the span exceeds 60 ft., allowance should be made for alterations of length due to changes of temperature. This ROOF TRUSSES. 35 may be effected by interposing a set of rollers between the skew-back and wall-plate at one heel, or by fixing one heel to the wall and allowing the opposite skew-back to slide freely over a wall-plate. The junction at the peak is made by means of a casting or w rought-iron plates (Figs. 63, 64, 65). Fig. 63. Fig. 64. Fig. 65. Light iron and timber beams as well as angle-irons are em- ployed as purlins. They are fixed' to tire top or sides of the rafters by brackets, or lie between them in cast-iron shoes (Figs. 66 to 71), and are usually held in place by rows of tie- FlG. 70. Fig. 71. Fig. 72. rods, spaced at 6 or 8 ft. intervals between peak and heel, running the whole length of the roof. The sheathing boards and final metal or slate covering are fastened upon the purlins. The nature of the coverin'^^ regu- lates the spacing of the purlins, and the size of the purlins is governed by the distance between the main rafters, which may ",i m 36 THEORY OF STRUCTURES. vary from 4 ft. to upwards of 25 ft. But when the interval between the rafters is so great as to cause an undue deflection of the purlins, the latter should be trussed. Each purlin may be trussed, or a light beam may be placed midway between the main rafters so as to form a supplementary rafter, and trussed as in Fig. 72. Struts are made of timber or iron. Timber struts are rectangular in section. Wrought-iron struts may consist of L-irons, T-bars, or light columns, while cast-iron may be em- ployed for work of a more ornamental character. The strut- heads are attached to the rafters by means of cast caps, wrought-iron straps, brackets, etc. (Figs, "Jt^ to "jQ), and the strut-feet are easily designed both for pin and screw connec- tions (Figs. Ty to 80). Fio. 7J. Fig. 74- Fig. 7s. Fig. 76. Fig. 77. Fig. 78. Fig 79. Fig. 80. 1; .rl! Tics may be of flat or round bars attached cither by eyes and pins or by screw ends, and occasionally by rivets. The greatest care is necessary in properly proportioning the dimen- sions of the eyes and pins to the stresses that come upon them. To obtain greater security, each of the end panels of a roof may be provided with lateral braces, and wind-ties are often made to run the whole length of the structure through the feet of the main struts. tn ROOF WEIGHTS. 37 Due allowance must be made in all cases for changes of temperature. 21. Roof-weights. — In calculating the stresses in the different members of a roof-truss two kinds oi load have to be dealt with, the one permanent and the other accidental. The permanent "load consists of the covering, the framing, and ac- cumulations of snozv. Tables at the end of the chapter show the weights of various coverings and framings. The weight of freshly fallen snow may vary from 5 to 20 lbs. per cubic foot. English and European engineers consider an allowance of 6 lbs. per square foot sufficient for sno'.v, but in cold climates, similar to that of North America, it is probably unsafe to estimate this weight at less than 12 lbs. per square foot. The accidental or live load upon a roof is the wind-pressure, the maximum force of which has been estimated to vary from 40 to 50 lbs. per square foot of surface perpendicular to the direction of blozc. Ordinary gales blow with a force of from 20 to 25 lbs., which may sometimes rise to 34 or 35 lbs., and even to upwards of 50 lbs. during storms of great severity. Press- ures much greater than 50 lbs. have been recorded, but they are wholly untrustworthy. Up to the present time, indeed, all wind-pressure data are most unreliable, and to this fact may be attributed the frequent wide divergence of opinion as to the necessary wind allowance in any particular case. The great differences that exist in all recorded wind-pressures are pri- marily due to the unphilosophic, unscientific, and unpractical character of the anemometers which give no correct informa- tion either as to pressure or velocity. The inertia of the mov- ing parts, the transformation of velocities into pressures, and the injudicious placing of the anemometer, which renders it subject to local currents, all tend to vitiate the results. It would be practically absurd to base calculations upon the violence of a wind-gust, a tornado, or other similar phe- nomena, as it is almost absolutely certain that a structure would not lie within its range. In fact, it maj' be assumed that i'i wind-pressure of 40 lbs. per square foot upon a surface v'^B I i III 38 THEORY OF STRUCTURES. perpendicular to the direction of blow is an ample and perfectly safe allowance, especially when it is remembered that a greater pressure than this would cause the overthrow of nearly all the existing towers, chimneys, etc. 22. Wind-pressure upon Inclined Surfaces. — The press- ure upon an inclined surface may be obtained from the follow- ing formula, which was experimentally deduced by Hutton, viz, : M p„ = p sin a'.84cosa-,. (A> p being the intensity of the wind-pressure in pounds per square foot upon a surface perpendicular to the direction of blow, and />„ being the normal intensity upon a surface inclined at an angle a to the direction of blow. Let />/, , /j, be the components of p„ , parallel and perpen- dicular, respectively, to the direction of blow. Ph'= Pn sin «, and p^ = /„ cos a. W Hence, if the inclined surface is a roof, and if the wind blows horizontally, a is the roof's pitch. Again, let v be the velocity of a fluid current in feet per second, and be that due to a head of Ji feet. Let w be the weight of the fluid in pounds per cubic foot. Let p be the pressure of the current in pounds per square foot upon a surface perpendicular to its direction. If the fluid, after striking the surface, is free to escape at right angles to its original direction, V p = 2]nv = — zv. Hence for ordinary atmospheric air, since w = .08 lb., approx- imately, p =z f' = 32 \, 20, (B) ■ I III I DISTRIBUTION OF LOADS. 39 When the wind impinges upon a surface oblique to its . Iv sin (i\ direction, the intensity of the pressure is 1 — -^ ) , v being the absolute impinging velocity, and yS being the angle between the direction of blow and the surface impinged upon. (Sec chapter on Bridges.) Taoles prepared from formulae A and B are given at the end of the chapter. 23. Distribution of Loads. — Engineers have been accus- tomed to assume that the accidental load is uniformly dis- tributed over the whole of the roof, and that it varies from 30 to 35 lbs. per square foot of covered surface for short spans, and from 35 to 40 lbs. for spans of more than 6.") ft. But the wind may blow on one side only, and although its direction is usually horizontal, it may occasionally be inclined at a con- siderable angle, and be even normal to a roof of high pitch. It is therefore evident that the horizontal component (/»/,) of the normal pressure (/„) should not be neglected, and it may cause a complete reversal of stress in members of the truss, especially if it is of the arched or braced type. If P„ ^s the total normal wind-pressure on the side of a roof of pitch a, its horizontal component P^ sin a will tend to push the roof horizontally over its supports. This tendency must be resisted by the reactions at the supports. In roofs of small .span, the foot of each rafter is usually^avrt^ to its support, and it may be assumed that each support exerts P sin (y the same reaction, which should therefore be equal to —" — . 2 In roofs of large span the foot of one rafter is fi.vcd, while that of the other rests upon rollers. The latter is not suited to with- stand a horizontal force, and the whole of the horizontal com- ponent of the wind-pressure must be borne at the fixed end, where the reaction should be assumed to be equal to P„ sin ci. In designing a roof-truss it is assumed that the wind blows on one side only, and that the total load is concentrated at the joints (or points of support) of the principal rafters. E.g., let the rafters AB, AC oi a truss be each supported at 11 11 f 1 ■ ;j- I : ;<(! . It «-v... .. ' 'If m 40 THEORY OF STRUCTURES. two intermediate points (or joints), D, E and F, G, respectively, and let the wind blow on the side AB. Fig. 8i. ;i 1' Take BD = CF=l,, DE = FG = l^, EA = GA-l^\ and let /, -f-^'j + /a = /; .". BC — 2/ cos or, a being the angle ABC. Let W be the permanent (or dead) load per square foot of roof-surface. Let pn be the normal wind-pressure per square foot of roof- surface. Let d be the horizontal distance in feet from centre to centre of trusses. The total normal live load concentrated at^^A'^^; at Z> = p^d l.-Vh. at E — p^d l. + l. atA=p„dj. \ The total vertical dead load concentrated at D and F = j^,^i±/? ; at £ and C; = wd^^^'- ; at ^ = wdl,, 2 2 • Let /?, , R^ be the resultant vertical reactions at B and C, respectively (i.e., the total vertical reactions less the dead eights \ivd-\ concentrated at these points). w DISTRIBUTION OF LOADS. 41 Take moments about C. :. R^2l cos a = sum of moments of live loads about C-\- sum of moments of dead loads about C, ^ moment of resultant wind-pressure about C -f- moment of resultant dead load about C, = pjdi- + / cos 2a j + wd{l^ -\- 2/, -|- 2/,)/ cos or, -where — f-/cos 2a is the perpendicular from C upon the line of action of the resultant wind-pressure which bisects AB normally. (N.B. The moment of the horizontal reaction at ^ or C about C is evidently nil.) R^ may be found by taking moments about B, To determine the stresses in the various members of a roof- truss two methods may be pursued : {x) A single stress diagram may be drawn to represent the combined effect of the live and dead loads. This will be found to be the quickest and most useful method. {y) The normal wind-pressure (/>„) may be resolved into its vertical {p^) and horizontal {p^ components ; p^ may then be combined with the dead load W, and a stress diagram drawn for the vertical loads only. A second diagram may be drawn for the horizontal loads. The resultant stresses will be the algebraic sum of the corresponding stresses in the two dia- grams. A third method will be referred to in a subsequent article. 24. Ex. I. Method {x) applied to the roof-truss ^IBC, Fig. 82. The dead load = w/(^ concentrated at A. Id The live loads = p,~- acting at each of the points A and B, normally to AB. The vertical reaction at B rvld pjd 1 1 cos 2a ' 2 cos «\4 ' 2 )• M| li i ii 42 THEORY OF STRUCTURES. Let rollers be placed underneath C. The total horizontal reaction = pjd sin a, and is wholly borne at B» Fig. 83. ■Si bciiu At given. Fig. 83. At B there arejivc forces in equilibrium, of which three are known, and the reciprocal of B may be thus described : Draw 5,5, to represent the normal wind-pressure (Ai"^) at B ; S,S^ to represent R, ; S,S^ to represent the horizontal reaction {p„M sin a) ; S^S^ parallel to BD ; ^'j^^ parallel to AB. The closed figure S^S^S^S^S^S^ is the reciprocal required, and the stresses in BD, AB, at B, are represented by 5,5^, S^S^, respectively, being a tension and a thrust. At D there are three forces in equilibrium, of which the tension in DB has been found. Drawing 5,5„ horizontally and 5j5, parallel to AD, the triangle S^S,S^ is evidently the reciprocal of D, the stresses in DA, BE being represented by 5j5g, 6",6\, respectively, and being both tensions. , ROOF- TRUSSES. 45 by Again, the triangle S^S^S^ is the reciprocal of E, the stressc s in EC\ EA being represented by 5,5, , 5,5, , respectively, and being both tensions. At A there are six forces in equilibrium, of which two, viz., / ld\ the normal pressure, [pn"^], and the dead weight, {wld), arc given, while the stresses in AB, AD, AE have been found. Id Draw S^S^ to represent /„ — , and S^S^ to represent wld. Five of the forces at A are therefore represented by the following lines, taken in order : S,S.,, S^S, , S,S^, S,S,, S^S, . Hence the closing line 5,5^ must necessarily represent in direction and magnitude the force in AC at A, and it is a thrust. Also, 5,5^5, must be the reciprocal of C, and therefore S^S^ represents the reaction at C. T/ie resultant reaction at B is represented in direction and magnitude by S^S^ . The line SJS^ must pass through the point S^ , as 535, , the horizontal reaction, is merely the horizontal projection of S^S^ , the total wind-pressure. The dotted lines show the altered stresses if rollers are under /j, the end C being fixed. The stress in each member is diminished, and as the truss should be designed to meet tlie most unfavorable case, the stresses should be calculated on the assumption that the rollers are on the leeward side. This may be considered an invariable rule for roof-trusses. Ex. 2. Method (x) applied to the roof-truss ABC, Fig. 84. The vertical uead load = at each of the pomts E, A, G. The live load, acting normally to AB, = pjd at each of the points B and A, and pJd at F. The vertical reaction R^ at B = \zvld -{- :77^— (cos 2» 4" i)* 2 cos a liKjfF*^" r '1- >1. v\\ Iti 44 THEORY OF STRUCTURES. I 1 The horizontal reaction at ^ = pjd sin a, rollers being under C as before. Fig. 85. Describe the stress diagram in precisely the same manner as in Ex. i. Taking S^S^ to represent the normal wind-pressure at jff, 5,5, " " vertical reaction R^ at B, " horizontal reaction at B, .-. S,S^S,S,S,S, is the reciprocal of B, SAS.S,S,S, " F, S,S,S,S,S, " D, 0,0,0,0,„0,,0,aO, " ^, 5,,>j,o>Ji3'Ju5,, " G^, '^9'^*'^18'^10«^» " j&, 545,40,3 « C. 5,5^ is the horizontal projection of S^St + SiS, -\- S^S^^ , i.e., of the total normal wind-pressure, and therefore the vertical through 5„ must pass through 5« . J1:: ROOF TRUSSES. 45 The dotted lines show the altered stresses if rollers are under B. The resultant reaction at B is represented in direction and nia;4iiitLide by S^S^. Ex. 3. Method {x) applied to the truss represented by Fig. 86. Fig. 86. Data. —Pitch = 30° ; AD - BD - AE - CE = 21 ft. ; trusses 13 ft., centre to centre ; dead weight = 8 lbs. per square foot of roof-surface ; wind-pressure on one side of roof (say AE) normal to roof-surface = 28 lbs. per square foot; DF=DH = EG — EK\ DF and EG are vertical ; rollers under one end, say C\ span =: 79 ft. ; AF=.BH = 21 ft., nearly ; FN = 3^ ft, nearly. Total live load and = Total dead load = 4459 lbs. f= 13 -28] at each of the points F, H, 3822 lbs.( = — .13.28] at each of the points A, B. 1274 lbs.(=:-^*. 13.8) at each of the points F, //, K, G, and = 2184 lbs. ( = 21 . 13 . 8) at the point A. I :. I lit .■ '1 .1. ' :'i . \ m i ii; 46 THEORY OF STRUCTURES. «Jl t/3/ i III -:.M «l if CO S 2 2; ^ CO OT ot«"ot wT ^ ROOF TRUSSES. 47 Resultant vertical reaction at B = K4 X 1274 + 2184) + ^^- = 13201.8 lbs. Horizontal reaction at B = 16562 sin 30° = 8281 lbs. Let I inch represent 16,000 lbs., and on this scale draw 5,5j = 3822 lbs., the normal wind-pressure at B ; S^S^ — 13201.8 lbs., the vertical reaction at B ; S.,S, = 8281 lbs., the horizontal reaction at B ; StS^ parallel to BD, and s^s^ parallel to BA. The figure S^S^S,S^S^S^ is the reciprocal of B. The stress diagram can now be easily completed, the recip- rocals of the points H, F, D, A, G, K, E, and C being 5,e5„5„5.,5.„ , S,,S,^S,,S,^S,,, and S,^S,^S,S,,, respectively. St , as before, is in the vertical line S^^S^^ produced. On the assumed scale, '■'dm , t i \ t tfil 5«5j = the tension at BD ; S,A,= " " " AD; 5,5.,= " " " BE; s,s,= thrust inBH SA = ki " HF 5g>Jii = (< " AF s.s,= << " DH s,s,= (( " DF. These are the maximum stresses to which the members of me half of the truss can be subjected, and for which they should be designed. It is also usual, except in special cases, to make the two halves symmetrical. 5,5, is the resultant reaction at B. If the end C is fixed and rollers placed under B, the reduced stresses may be shown by dotted lines as in Exs. i and 2. ■if! "-^M 48 THEORY OF STRUCTURES. Exs. 4 and 5. Method {x) applied to the trusses repre- sented by Figs. 88 and 90. • It is assumed, as before, that there is a normal wind-press- ure upon ABy and that rollers are under C. Figs. 89 and 91 are the maximum stress diagrams corre- sponding to Figs. 88 and 90, respectively, and are drawn in pre- cisely the i-ame manner as described in the preceding examples. Remark on Fig. 88. — The stresses at the joints F and D are indeterminate, and it is assumed that the stress in FL Fig. 88. Fig. 89. is equal to that in FH. The reciprocal of F thus becon 6",„5„5e5g5,5„5,35"„, 5„5,3(= S,S,) being the stress in FD. Tl truss is an example of a frame with redundant bars, in which the stresses can only be determined when the relative yield of the bars is known. ii im ROOF TRUSSES. 49 Remark on Fig. 90. — The stress-diagram, Fig. 91, for each of the joints in the horizontal BC (Fig. 90) is closed by the return of one side upon another. Thus at D the stress diagram is S,S,StS^S, , the closing line 5,6", (the tension in D£) returning 3^ f^ u [X, c ^^^ D E '\1 ^ ^ ^ ^- Fig. gi. upon 5^5", (the tension in DB). The total stress in AE is evidently represented by S^^S^^, the reciprocal of A being Ex. 6. A truss with curved upper and lower chords, the portions, however, between consecutive joints being assumed straight. Under a uniformly distributed load the truss (Fig. 92) is evidently incomplete, and the stress diagrams at the joints in the lower chord will not close, so that equilibrium is impossible. T'l'' frame is made complete and the stresses determinate by inuouucing ties as in Fig. 93, the corresponding stress diagram for OP" half the truss being shown by Fig. 94. N t, let there be a wind-pressure on the side AB of the truss. In order to prevent a reversal of stress in the diagonal ties on the side AC (Fig. 93), additional ties DE, FG, called coimter-braus, arc introduced as in Fig. 95. Fig. 96 gives the .!»! 1 ■m .,.-,. a giSfSM L 'i 50 THEORY OF STRUCTURES. stress diagram due to wind-pressure only, it being assumed that the end C rests upon rollers and that B is fixed. Pic. 93. Fig. 93. Fig. 94. Fig. 95. Fig. q6. Note. — 21 = wind-press, at .5 = f wind-press, upon BM, 23 = 34 = 45 = 56 = 67 = "M = '^31=-. ^' = " 6> = . " A = BM, MO, MO, OA, OA. m \K = vertical reaction at B, HK = horizontal reaction at B. Ex. 7. A single example will serve to illustrate method (7). Take the truss represented by Fig. 97. Fig. 98 is the stress diagram due to the vertical load upon the roof, viz., the dead weight -\- vertical component of wind- pressure, />g is the vertical reaction at B and is 4 4 ROOF-TRUSSES, qni is the weight at F and is p„ cos a-\- w BH = weight at //" = inn. 51 Fig. 97. Fig. 98. Fig. 99 is the stress-diagram due to horizontal component of wind-pressure, rollers being placed under B and the end C beincr fixed. p'o = downward reaction at B pjd sin" n 4 cos tx , , , . , , , . , ,, A, CD DF / w/ <§/ 8 / FH / / AC / / < r \ c^ \ EG \ - ^ w\ <^\ HK km\ \ w \ MQ w QT \ Fig. III. which they are subjected. The directions of these stresses at the joints, and hence also their character, are easily determined by following in order the sides of the reciprocals. The verti- cals are evidently all ties and the diagonals all stmts. If the load is unevenly distributed, the stresses in different members may be reversed. For example, Fig. 113. Let the truss carry a single weight P at any point D. The reciprocal of D is S^S.,S^S,S^S, (Fig. 112), 5,5, represent^ H> t i ! \\ 1. * «o THEORY OF STRUCTURES. ; ! ing P, and the arrow-heads showing the directions of the forces now acting at D. Thus the force in DE at Z>, represented by 5,5^ , acts from D towards E, and is, therefore, a tension. Hence, in order that DE may not be subjected to a tensile force, counterbraces CF, EH are introduced so that the por- tion of P borne on the support at T may be transmitted through the system CFEH to H and from H to T through the regular system HGKLMNQRT. The reciprocal of D is now ,S,5,S,5, (Fig. 1 1 3), and the reciprocal of C the figure HS^S^S^S^H, the arrow-heads showing the directions of the forces at C. It will be at once observed that FC must be a strut. In order to make provision for a varying load, as when a train passes over a bridge, counterbraces are introduced in the panels on both sides of the centre, and although they may not be necessary in every panel, they will give increased stiffness to the truss. Note. — Generally speaking, a panel is that portion of the bridge-truss between two consecutive verticals, and the ends of the verticals are called panel-points. Ex. 7. Fig. 1 14 represents a Pratt truss, and is merely an inverted Howe truss. The diagonals become ties and the t II' Pig. 1x4. verticals struts. Counterbraces are introduced to resist the action of a varying load, precisely as described in Ex. 6. Ex. 8. The bowstring truss in its simplest form is repre- FlG. lis. sented by Fig. 115. Assuming that the portions of the upper chord between consecutive joints are straight, the stress dia- BOWSTRING TRUSS. 6i gram for a uniformly distributed load and for one half the truss is Fig. Il6. The panels, however, are incomplete frames, and if the truss Ss , S, Fig. ii6. Fig. 117. has to carry an unequally distributed load, ties similar to that shown by the dotted line MNviwxsX. be introduced in the several panels in order to prevent distortion. For example, let there be a single load P at the joint N, and let there be no brace NM. The stress in the first vertical is evidently nil. The reciprocal of iV is S^S^S^S^S^S^ , Fig. 117, S^S^ representing P. The reciprocal of L is HSJS^S^H, and the arrow-heads show the directions of the forces at H. Thus the force in 6^Z, which is represented by S^S, , acts from towards Z, and is, therefore, a compression. But, under a uniformly distributed load, the diagonals are all ties, and NM is introduced to take up that portion of P which would be otherwise transmitted through LO in the form of a compression. In this case the reciprocal of L is HS^S^H, since the stress in LO due to P is assumed to be nil. Also the reciprocal of N is S,S.,S^S^S^S^S^. The stress in NM, represented by ^g^, , acts from A^to J/ and is a tension. Hence the diagonals NM are also ties, and the portion of the weight P borne at L is carried to Q through the system NAIOQ. Ex. 9. Fig. 118 is a bowstring truss with isosceles bracing. Under an arbitrary load Fig. 119 is the stress diagram, the loads at a, b, c, d, e,f, g being 12, 23, 34, 45, 56, 67, 78, respect- ively. As in the Warren girder, the diagonals may, under the action of a varying load, be subjected to both tensile and com- % li f I \ \:'\\ hi i ^2 THEORY OF STRUCTURES. pressive stresses. They mut.f, therefore, be designed to bear such reversal of stress. ah c a e J (/ Fig. ii8. p Fig. 119, It is assumed, as before, that the portions of the upper chord between consecutive joints are straight. Note. — The design of bridge-trusses will be further con- sidered in a subsequent chapter. 26. Method of Sections. — It often happens that the stresses in the members of a frame may be easily obtained by the method of sections. This method depends upon the following principle : If a frame is divided by a plane section into two parts, and if each part is considered separately, the stresses in the bars (or members) intersected by the secant plane must balance the external forces upon the part in question. Hence the algebraic sums of the horizontal components, '2{X\ of the vertical components, "^{Y), and of the moments of the forces with respect to any point, 2{M), are severally zero ; i.e., analytically, 2{X) = o, 2(y) = o, and 2{M) = o. These equations are solvable, and the stresses therefore determinate, if the secant plane does not cut more than three members. EXAMPLES, 63 Ex. I. ABC is a roof-truss of 60 ft. span and 30° pitch. Q N 4KTOtl8 Fig. 120 The strut DF = GH = i, ft.; the angle FDA =90°. Also AF=FB = AG= GC. The vertical reaction at ^ = 5 tons. The weight concen- trated at Z> = 4J- tons. Let the angle /i^F= a. AB — 30 sec 30° — 20 ^^3 ; cot ioi/3 a = = 2V3, sin a — V13 cos a z= 2i^ VY3' If the portion of the truss on the right of a secant plane A/N be removed, the forces C, 7", , T^ in the members AD,AF, FG must balance the external forces 5 tons and 4^ tons in order that the equilibrium of the remainder of the truss may be pre- served. Hence, revolving horizontally and vertically, r, + r, cos (« + 30°) - C sin 60° = o ; 7; sin {a + 30°) - C cos 60° + 5 - 4^ = o. Taking moments about F, Cs- SBF cos {t,o'= - a) -f ^IDF sin 30° = c. Rut 3^3 cos(«-f30°) = -4=-, sin («4-30°)= ^-7^. cos (30° - a) =—^-. 2 1^13 2 ^13 '2 Vii ;!<; 'M ' m ii lr:[ »^u m f I . Ill r li }■ /; :i: \ 9 I jjl !|1 > }■, . " ' i 1'-' ' 1 <; i ^ ■ i ,1 ■ ^4 THEORY OF STRUCTURES. BF— BD sec a = 5 ^Ti, and DF = 5 tt. 2 V13 2 T-^.l^J-^-.^ + i^zo. 2 ^13 <^". 5 - 5 . 5 v/13 . -4-- + 4i. 5 •i = O. 2 V13 Hence C= 15^ tons, Z, = 9.89 tons, and 7!, ::= 6.35 tons. Ex. 2. The figure represents a portion of a bridge-truss cut off by a plane A/N and supported at the abutment at A. Tiic vertical reaction at A = 409,400 lbs. ^ "'f"^ The weight at Z? = 49,500 lbs. F,c. .... ^ «' " " C: = 38,700 lbs. 409,400 11)S. p, I* ]| AB = BC= 24 ft. ; BD = 24 ft.; {7/: = 29^ ft. The forces C, D\ T in the members met by MN must balance the external forces at A, B, C. Revolving horizontally and vertically, T-\- D' cos a — C cos /8 — o; D' sin a 4- C"' sin ft — 409400 + 49500 +38700 = o ; a and ft being the inclinations to the horizon of EF, DEy respectively. Taking moments about E, — T X 29^ 4- 409400 X 48 — 49500 X 24 = o. PIERS. 2g\ II 6^ But tan a = — ''-" = 24 9 and tan /S = ^- = -. 24 9 .•. sin a = II 9 cos « ^ ~ 1/202 Hence T = 62g,42y^j lbs. ; 981450O ^/-- 1090500 , sin/3=— ^, cos /? = ~2_ 4/202 1^85 ■v'ss c ^r:f-*'«5 = --^fp.^s lbs.; 1994600 /; zz: ■ r 202 lbs. 27. Piers. — 1 o determine the stresses in the members of the braced piers (Fig. 122) supporting a deck bridge. 40 tons 20 tons ^2 tons// 9«! B f / I ' E/ i y^ — 1— / s G H Fig. 132. Fic. i»3. Z>rf/rt.— Height of pier = 50 ft. ; of truss — 30 ft. Width >f pier at top :== 17 ft. ; at bottom -- ii\ ft. T .1 1 ' oo THEORY OF STRUCTURES. t ! i The bridge w hen most heavily loaded throws a weight of l(X) tons on eaeh of the points A and B. Weight of half-pier = 30 tons. The increased weight at each of the points C, D and E, F, from the portions AD and CF oi the pier = 5 tons. Resultant horizontal wind-pressure on train = 40 tons at 87^ feet above base. Resultant horizontal wind-pressure on truss = 20 tons at 65 feet above base. Resultant horizontal wind-pressure on pier = 2| tons at each of the points C and E. With the wind-pressure acting as in the figure, the diagonals CB, ED, and GF are requireil. When the wind blows on the other side, the \agonals D to A, F to C, and // to E are brought into plaj-. The moment of the couple tending to ove' turn the pier = 40 X 87I 4- 20 X O5 -1- 4 X 25 = 4900 ton-feet. 33l The moment of stability = {2(X) -f- 30) X 3871! ft.-tons. Thus the difference, = 4900 — 3871^ = 1028^ ft.-tons, must be provided for in the anchorage. The pull on a vertical 10281^ anchorage-tie at 6 = — ,r- = SOA'r tons. ^ 331'"' Again, if // be the horizontal force upon the pier at A due to wind-pressure, //■ X 50 = 40 X 87A- 4- 20 X 65 = 4800 ; // — 96 tons. The stress diagram can now be easily drawn. The reciprocals of the points A, /), C, D, E, F are 4321. 2561, 11-10-4169, 65789, 13-12-11-98 14, and 87-1 5-16- 14, respec- tively. In the stress diagram 43 = 96 tons, 32 = 25 = 100 tons, 57 = 7-15 = 4-IO =: 1 1-12 = 6 tons, and 10-11=12-13 = 2^ tons. The stress in EG is of an opposite kind to the stresses in AC, CE. Note. braced \ maxim u pressure structure of the st posts of the bridt Dcscripi Hoarding (| Hoarding ai Cast-iron pi Copper, Corru. ,itf: I'clt.a.^i I, ■ Ffl; and gr.i Inilvanized i l.allis and pi I'.intiles. . . . SiicL't lead.. SlR-t--t-zinc.. Slieet-iron (c Slicet-iron (1 Sliingler, (i6- " (lull .Sl)eathing (i- " (cli (asl Slates (ordin; *^l'ites (large) ■'^iates and in Thatch Tiles Tiles and mo Timbering ol roofs (addit WEIGHT OF KOOF-COVERIXGS. 67 Note. — In computin;4 the stresses in the leeward posts of a l^raced pier, it is usual in American practice to assume that the maximum load is upon the bricl<^e and that the wind exerts a pressure of 30 lbs. per sq. ft. upon the surfaces of the train and structure, or a pressure of 50 lbs. per sq. ft. upon the surface of the structure alone. The negative stresses in the windward posts of the pier are determined when the minimum load is on the bridge, the wind-pressure remaining the same. TABLE OF WEIGHTS OF ROOF-COVERINGS. Description of Coveririii. lioar'Jing (J-inch) Hoarding aiiiJ sheet-iron.. Cast-iron plates (,§-inch). . Copper Corru .ito-l Ik ,1 and laths. Felt, a.-.; li :*'i ,d Kelt and gravel dalvanized iron Laths and plaster i'.intiles Sheet lead Sheet-zinc Sheet-iron (corrugated). . . Sheet-iron (16 W.G.) and laths Shingles (i6inch) (h'lig) Sheathing (i-inch pine) " (chestnut and maple) (ash, hickory, oak) Slates (ordinary) tites (large). Weight of Covering in lbs. per sq. ft. of Covered Area. Slates and iron laths Thatch Tiles Tiles and mortar Timbering of tiled and slate roofs (additional) Dead Weight of Roof in lbs. per sq. ft. of Covered Area. 2.5 to 3 6.5 15 .8 to 1.25 5-5 .3 10 .4 8 to 10 I to 3 9 to ID 6 to 10 5 to 8 1.25 to 2 3 4 3-4 5 2 3 3 4 5 5 to () 9 to II 10 6.5 7 to 20 25 to 3c 55 to 6.5 S without boards and 11 with boards for spans u|) to 75 ft. 12 without boards and 15 with boards for spans from 75 to 150 ft. ID on laths for spans up to 75 ft. 14 on laths for spans from 75 to 150 ft. 13 without boards or on laths and 16 on i^-in. boards for s[)ans up to 75 ft. 17 without boards or on laths and 20 on li-in. boards for spans from 75 to 150 ft. ;i;i .^: J! 68 THEORY OF STRUCTURES. WEIGHTS OF VARIOUS ROOF-FRAMINGS. TABLE O |i I - Description of Roof. Pent Common Truss. ^ Timber rafters and struts, iron ties Common Truss. Location. j Liverpool I 1 Dociis ( j Liverpool » I Docks S Bowstring .... j Manchester " Lime Street " Birmingham Arched Strasburg Paris Dublin Derby iSvdenham Cover- ing. Felt Zinc Zinc Zinc Slates St. Pancras Cremorne Span Width of Bays.} Weight in ]bs. per sq. It. of Covered Area. Fram- Cover- ing. I ing. ft. in. [ft. 15 o' 37 o 40 50 o 53 3 54 o 55 o 72 o 62 o 76 o 79 o 80 8 90 2 84 100 o ,130 o ' 50 o 1154 o 211 97 153 41 81 120 72 o 240 o 45 5 = 40, AS DETERMINED BY THE FORMULA Pn = P . sin a>84 cosa-i. Pitch of Roof. Pn Pv Pk 5° 5.0 4.9 •4 10' 9-7 9.6 1-7 20" 18. 1 17.0 6.2 3°: 26.4 22.8 132 40^ 33-3 25-5 21.4 50° 38.1 24-5 29.2 60' 40,0 20.0 34-0 70° 41. c 14.0 38.5 80' 40.4 7.0 39-8 90^ 40.0 0.0 40.0 TABLE PREPARED FROM THE FORMULA / = ©■ Vc!ociiier> in Velocities in Pressurv; in feet per second. miles per hour. lbs. per sq. ft. 10 6.8 •25 20 13.0 I.OC 40 27.2 4.CX3 60 40.8 g.cx) 70 47.6 12.25 80 54-4 16.00 90 6r.2 20.25 100 68.0 25.00 IIO 74. 3 30-25 120 81.6 36.00 130 88.4 42.25 150 102.0 56.25 \0 'i i I'. 70 THEORY OF STRUCTURES. EXAMPLES. 1. Show that the locus of the poles of the funicular polygons oi which the first and last sides pass through two fixed points on the clus ing line, is a straigat line parallel to the clcjsing line. 2. The first and last sides of a funicular polygon of a system of forces intersect the closing line in two fixed points. Show that for any position of the pole each side of the polygon will pass through a fixed point on the closing line. 3. Four bars of equal weight and length, freely articulated at the extremities, form a square ABCD. Tiie system rests in a vertical plane, the joint A being fixed, and the form of tlie square is preserved bv means of a horizontal string connecting the joints B and D. If \V be the weight of each bar, sliow (ii) tiiat the stress at C is horizontal and W rVs = — -, (d) that the stress on BC at B is IV12. and makes with tiie ver- 2 2 tical an angle tan -'^, (r) that the stress on A/> at B is ll'J-lA and makes with the vertical an angle tan"'ii, (d) that the stress upon AB at A is | PK (c) that the tension of the string is 2 IF. 4. Five bars of equal length and weight, freely articulated at tiie extremities, form a regular pentagon ABCDli. The system rests in a vertical plane, the bar CD being fixed in a horizontal position, and the form of the pentagon being preserved by meani of a string connecting the joints B and E. If the weight of each bar be \V, sliow that the \V tension of the string is — (tan 54"" + 3 tan iS°), and find the niagni- 2 tudes and directions of the stresses at the joints. 5. Six bars (jf equal length and weight ( = W), freely articulated at the extremities, form a regidar hexagon /IBCHEF. First, if the system hang in a vertical plane, the bar AB being fixed in a horizontal positicjn, and the form of tlu- hexagon being preser\'c 1 by means of a string connecting the middle points of AB and DI-1, siiow W and that (li) the tension of the string is 3 \V, (b) the stress at C is -^3 horizontal, (r) the stress at D is W if ^^. and makes with the vertical an angle cot ■' 2 4/3. Second, if the system rest in a vertical plane, t'^i' bar DF. being fixed in a horizontal position, aiul tlie form of tlie hi;xagoii being ijreserved by mean tii>: tensi makes w and niak Thira and the f necling . each of t AD is 2 / at the joi from the 6. She liorizonta which are 7. If t that the < spect to s which is i; 8. A s> fixed poinl the joints, tangents are in ariti 9. Ifar librium in angles of i iiig from tl 10. Thi The systeii same horiz the stress i II. Thr tern is kept mine the st Wiiat rt ■' m EXAMPLES. 71 hy means of a string connecting the joints C and F, show that ( Then stress in BD — 2 sin (60° -I- a) 18. If ii whole of tl stress diagr 19. A tr beam BC, a from A by 1 each rafter in) the stres What {b) wi liobeam be tilted again.' pressure bet Ans. {a) Si (*) . ■)' .SAMPLES. n i6. The boom AB of tne accompanying truss is supported at fiv'e intermediate points dividing the lengt'i into six segments each lo ft. long, depth of the truss = lo ft. ]_. aw diagrams for the following cases : The tress ib:;ofirf4f5/B Fig. J24. (rt) A weight of 100 lbs. at each intermediate point of support. {b) Weights of 100, 200, 300, 400, 500 lbs. in order at these points. Ans. (a) Stress in « = 375 ; ^ = 325 ; c— 375 ; h = 450 ; ;« = 125 1^13: n- soVS' <}- 50^5; p = 2i Vii lbs. (J)) Stress in « = 875 ; /^ = 825 ; F=4i6f; AD = 502 lbs. Stresses due to 300 lbs. in BE = 1012^ ; EA = 1800; ^1/^ = 2812^; BD=3847i; AD = 22S0 ; DC = 2160 ; AC = 2700 ; DF = o. 18. If it be assumed in the first part of the last question that the whole of the weight is concentrated at the points E and F, draw the stress diagram. 19. A triangular truss consists of two equal rafters AB, AC Sind a tie- beam BC, all of white pine; the centre D of the tie-beam is supported from A by a wrought-iron rod AD; the uniformly distributed load upon each rafter is 8400 lbs., and upon the tie-beam is 36000 lbs. ; determine (ii) the stresses in the different members, /?C being 40 ft. and /ID 20 ft. What {b) will be the effect upon the several members if the centre of the lie-beam be supported upon a wall, and if for the rod a post be substi- tuted against which the heads of the rafters can rest.' Assume that the pressure between the rafter .ind post acts at right angles to the rafter. Ans. (a) Stresses in BD = 13200; AD — 18000; AB = 13200 4/2 lbs. {b) " " = 4200; " = 8400; " = 6300 V'2~lbs. 'U; 1 41 H p 74 THEORY OF STRUCTURES. . 1 I lit? ' ll 20. A triangular truss of white pine consists of a rafter AC, a vertical post A/>\ and a horizontal tie-beam BC; the load upon the rafter is 300 lbs. per lineal foot; ^!C= 30 ft., AB = 6 ft. Find the resultant pressure at ' '. Ans. 4409 lbs. Find the stresses in the several members when the centre D of the rafter is also supported by a strut from B. Ans. Stress in BC = 4500 y6\ CD = 22500; DB = 11250; J)A = 1 1250; .IB — 2250 lbs, 21. The rafters y//>', AC oi a roof-truss are 20 ft. long, and are sup- ported at the centres by the struts Uh', DF\ the centre D of the tie- beam BC is supported by a tie-rod AD, 10 ft. long; the uniformly dis- tributed load upon AB is Sooo lbs., and upon AC is 2400 lbs. Determine the stresses in all the members. What will be the effect upon the several members ii AB be subjected to a horizontal pressure of 156 lbs. per lineal foot? A/is. {ii) Stress in BD — 4600 ^^3 ; BE = 9200; EA = 5200; £D = 4000 ; AD — 2600 ; DE = 1200 ; AE = 5200 ; CE = 6400 ; CD = 3200 |''3. (d) Tens, in BE = 520 v'3 ; -ID — 260 V'3 ; compres. in ED = 520 i/3"; AC = 520 4/3 ; DC= 780. No stresses in BD, AE. 22. Determine the stresses in all the members of the truss in the preceding question, assuming the tie-beam to be also loaded with a weight of 600 lbs. per lineal foot. Ans. Stress in ^IB iiicicased by 6000 4/3 lbs. ; in BC by 9000 lbs.; in AD by 6000 4/3 lbs. 23. A horizontal beam is trussed and supported by a vertical strut at its middle point. If a loaded wheel roll across tiie beam, show that the stress in each member increases proportionately with the distance of the wheel from the end. IV ^Lns. Stress in tie-beam (hor.) = -j--^' ^ot ^! on tie = JVx I sin 0' on strut = W -2X. 24. A frame is composed of a horizontal top-beam 40 ft. long, two vertical struts 3 ft. long, and three tie-rods of which the middle one is horizontal and 15 ft. long. Find the stresses produced in the several members when a single load of 6000 lbs. is concentrated at the head ot each strut. Ans. Stress in horizontal members = 50000 lbs. " " sloping " = 51420 " " " struts = 12000 " U: " EXAMPLES. 7S 25. If a wheel loaded with i2ocx) lbs. travel over the top-beam in the last question, what membeis must be introduced to prevent distortion? What are the, niaximum stresses to which these members will be sub- jected ? Ans. 191 22 lbs. 26. A beam of 30 ft. span is supported by an inverted queen-truss, tiie queens beinij; each 3 ft. long and the bottom horizontal member 10 ft. ioni;. Find the stresses in the several members due to a weic,'ht /P at the head of a queen, introducing the diagonal required to prevent distortion. Also find the stresses due to a weight JT'at centre ot beam Ans. I. Stress in AB = ~°ir- AE = 2.32 IF; EF = =°]V; 9 9 BE=: -II 3 CF = HF:=\.\61V 1)F= i.\6lV. BC . 10 , IV\ 2. Stress in AB =^ "W; 3 EF = ^ H-'. AE^ \.7\IV; BE=. IV 27. A roof-truss of 20 ft. span and 8 ft. rise is composed of two rafters and a horizontal tie-rod between the feet. The load upon the truss = 500 lbs. per foot of span. Find the pull on the tie. What would the pull be if the rod were raised 4 ft.? A/is. 3125 lbs.; 6250 lbs. 28. The rafters AB, AC of a roof are unequal in length and are in- clined at angles (x, fi to the vertical ; the uniformly distributed load upon AJi = IVi , upon ^IC = Jl'-2 . Find the tension on the tie-beam. JVi + 11 ^ sin (I sin^i 2 sin {(I + fJY Alls. 29. In the last question, if the span = 10 ft., a = 60' and /i = 45 \ find the tension on the tie, the rafters being spaced 2^ ft. centre to centre, and the roof-load being 20 lbs. per square foot. Alts. 198 lbs. 30. The equal rafters AB, AC for a roof of 10 ft. span and 2^ ft. rise are spaced 2^ ft. centre to centre ; the weight of the rcjof-covering, etc. = 20 lbs. per square foot. Find the vertical pressure auvl outward thrust at the foot of a rafter. Alls. Total vertical pressure = 125^/5 lbs. — horizontal thrust. 31. The lengths of the tie-beam and two rafters of a roof-truss are in the ratios of 5 : 4 : 3. Find the stresses in the several members when tlie load upon each raftt^r is uniformly distributed and equal to 100 lbs. .Ins. Stress in tie = 48 lbs. ; in one nifter = 60 lbs.; in otlier = 80 lbs. *>. \f- v^, IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 IM 1125 145 Ma !- IM |||||2J^ " «- IIIIM 1-4 IIIIII.6 1 t./f ^ o p /2 ^ /}. Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y 14580 (716) 872-4503 ,\ iV 4t>^ :\ \ % .V 6^ ^S^ % ^^ ^ -^ i^ ^^^ ». i I I ,1 : I '^ » I 1 w* ill ^ .,. . ■ . ■'1 76 THEORY OF STRUCTURES. 32. In a triangular truss the rafters each slope at 30° ; the load upon the apex = 100 lbs. Find the thrust of the roof and the stress in each rafter. Ans. 100 lbs.; 86.6 lbs. 33. A roof-truss is composed of two equal rafters and a tie-beam, and the span = 4 times the rise; the load at the apex = 4000 lbs. Find the stresses in the several members. Secondly, if a man of 150 lbs. stands at the middle of a rafter, by how much will the stress in the tie-beam be increased ? Ans. — 1. Stress in tie = 4000 lbs. ; in each rafter = 2000 1/5 lbs. 2. 75 lbs. 34. A king-post truss for a roof of 30 ft. span and 7i ft. rise is com- posed of two equal rafters AB, AC, the horizontal tie-beam BC, the vertical tie AD, and the struts DE, DF from the middle point D of the tie-beam to the middle points of the rafters ; the roof-load = 20 lbs. per square foot of roof-surface, and the rafters are spaced 10 ft. centre to centre. Find the stresses in the several members. Second, find the altered stresses when a man of 150 lbs. weight stands ■on the ridge. Third, find the altered stresses when the tie-beam supports a ceiling weighing 12 lbs. per square foot. Ans. — I. Stress in BE = 56250 lbs. ; BD —. 2250 i/Jlbs.; AE = '6875 lbs. ; DE = 9375 lbs.; AD = 7 500 \/J lbs. 2. Stresses in BD, BE, AE increased by 150 lbs., 75 4/5 lbs., and 75 ^5 lbs., respectively ; other stresses un- changed. 3. Stresses in AD, tie-beam, and rafters increased by 1800, 1800, and 900 -/s lbs., respectively ; other stresses unchanged. 35. The platform of a bridge for a clear span of 60 ft. is carried by two queen-trusses 15 ft. deep; the upper horizontal member of the truss is 20 ft. long; the load upon the bridge = 50 lbs. per square foot of plat- form, which is 12 ft. wide. Find the stresses in the several members. Ans. Stress in vertical --= 6000 lbs.; in each sloping member = loooo lbs. ; in each horizontal member = 8000 lbs. 36. If a single load of 6000 lbs. pass over the bridge in the last ques- tion, and if its effect is equally divided between the trusses, find (a) the greatest stress in the members of the truss, and also {l>) in the members which must be introduced to prevent distortion. Also find (c) the stresses when one half the bridge carries an additional load of 50 lbs. per square foot of platform. EXAMPLES. 77 Am. — {a) In sloping end strut = 3333^^ lbs.; horizontal tie = 26668^ lbs.; horizontal strut = 1333^ lbs. {c) In sloping end strut = 6250 lbs.; horizontal tie = 5000 lbs.; horizontal strut = 3000 lbs. {b) In case (a) = i666f lbs.; in case {c) = 2500 lbs. 37. A roof-truss consists of two equal rafters AB, AC inclined at 60° to the vertical, of a horizontal tie-beam BC of length /, of a collar-beam DK of length — , and of queen-posts DF, EG at each end of the collar-beam ; the truss is loaded with a weighs of 2600 lbs. at the vertex, a weight of 4000 lbs. at one collar-beam joint, a weight of 1200 lbs. at the other, and a weight of 1 500 lbs. at the foot of each queen ; the diagonal DG is inserted to provide for the unequal distribution of load. Find the stresses in all menioers. Alts. Stress in BD = i i733i ; BF = 5866^ ^3 ; DF= 1^,00; DA = 2600 ; DE = 3633i f^ ; /;(; =.- 1S66S ; GC = 4933i Vi ; OE = 2433^ ; CE = 9866^ ; AE = 2600 lbs. 38. The rafters AB, AC are supported at the centres by the struts DE, BE; the centre of the tie-beam is supported by the tie AD; BC = 30 ft., AD= 7i ft. ; the load upon AB is 4000 lbs., that upon AC 1600 lbs. r.nd the stresses in all the members. By an accident the strut /?£■ was torn away; how were the stresses in the other members affected ? Ans. — Case i ; Stress in BE = 2400 ^5 ; ^D = 4800 ; DE = 1000 v'5 : AE = 1400 v^5 ; AE = 1400 1/5 ; DE = 400 1^5 ; FC = 1800 j/J"; DC = 3600 lbs. Case 2 : Stress in BA — 1400 \/$ ; BD = 2800 ; AD = 400 ; AE = 1400 4/5 ; EC =: 1800 {/J; DE = 400 4/5"; DC = 3600. 39. The platform of a bridge for a clear span of 60 ft. is carried by two trusses 15 ft. deep, of the type shown by the c n f aiconipanying diagram ; the load upon the - bridge is 50 lbs. per square foot of platform, which is 12 ft. wide. Fuul the stresses in the several members. ^'°- "'• Afis. Stress in BE = 1 3500 ; BG = 6750 \/J; EG = 40CO ; ED = 13500; GD= 2250 I 7; GA =4500 V'J; AD = 9000 lbs. ;t 'ii'. ^'\\ 78 THEORY OF STRUCTURES. \ '\ ^ i\ 40. If a single weight of 2000 lbs. pass ove-- a truss similar to that shown in the preceding question, find the stresses in the several members when the load is (i) at E, (2) at D. Ans. — Lase i: Stress in BG = 1500 |/J; BE= 3000; EG = 2000 ; ED = 3000 ; GD = 1000 1^5 ; AG = 500 ^5 ; AH- soo\/J; DN=o; F// = o; DF = 1000 ; FC = 1000 ; C// = 500 1/5" lbs. Case 2 : Stress in BA and CA = 1000 1/5 ; BD and DC — 2000 ; AD = 2000 lbs., and in other members = o. 41. A white-pine triangular truss consists ot two rafters AB, AC, of unequal length, and a tie-beam BC. A vertical wrought-iron rod from A, 10 ft. long, supports the tie-beam at a point D, dividing its length into the segments BD = 10 ft. and CD = 20 it. The load upon each rafter is 300 lbs. per lineal ft. ; the load upon the tie-beam is 18,000 lbs., uniformly distributed. Determine the stresses in the several members. Ans. In AB = 9650 V2 lbs. ; AC= 4825 Vj lbs. ; BD = CD = 9650 lbs. 42. The post of a jib-crane is 10 ft. ; the weight lifted = IV; the jib is inclined at 30°, and the tie at 60°, to the vertical. Find (a) the stresses in the jib and tie, and also the B. M. at the foot of the post. How (/>) will these stresses be modified if the chain has/our falls, and if it passes to the chain-barrel in a direction bisecting the angle between the jib and tie? Ans.— {a) Stress in tie = IV; in jib = fV ^J. g, m _ j |/Jft, tons. id) " " =.87 IV; •' =i.S7lV. 43. An ordinary jib-crane is required to lift a weight of 10 tons at a horizontal distance of 9 ft. from the axis of the post. The hanging part of tile chain is in four falls; the jib is 15 ft. long, and the top of the post is i6i ft. above ground. Find the stresses in the jib and tie when the chain passes (i) along the jib, (2) along the tie. The post turns round a vertical axis. Find the direction and magni- mde of the pressure at the toe, which is 3 ft. below ground. Ans, — (i) Stress in tie = 32jL5 tons; in jib = iij| tons. II (2) If '€= 20 ft.; BDz=2oit.\ the weight lilted = 25 tons; r'^^j:;;;^^^ AC slopes at 45° ; the cliain hangs in four falls and passes ^^ "' from A to D. Find the stresses in all the members and c o the upward pull at D. F'o- »»«• Ans. Stress in BC = 26 ; AC = 47.6 ; AB = 28.4 ; CD = 32.8 tons. Vertical pull at Z> = 31.3 tons. 45. The figure represents the framing of an hydraulic crane. AB=BD = IJF = y-Y; = //A' = 5 ft.; A'(J = BC = 2i ft. Find the "«--X^r^ Stresses in the members of the crane when the weight K g'f dV' a (I ton) lifted is {a) at A ; {/>) at B ; (c) at /). Also Uf) find Fio. la?. the stresses when there is an additional weight of ^ ton at each of the points B, D, F, and G. Ans— {a) Stress in tons in AB - BD ='2 ; DP = FG = -" ; KG = L'/ ; AC= )/J; CE = - vT; 20 9 i^//=^=^=Vr;C/;=5_^- 9009 9 99 _ 143 NG = ^^; BC=EF = o. 26 (S) Stress in tons inAB = oz= AC; BC— i; CE=^^ \'2 ; HE = ^ V7; Z)F= /-(; = — ; GA' = -^; 13 u 2 CZ?=7V5'; DE= ^V'JT?; 9 99 28 II ^ V^3"T7; 7 i^r EC = -^~ V i\7 ; HG ^ -. 143 26 (f) Stress in ^^ = o = AC = BC = DC = BD ^ CE\ DF=^-^^ = FG; DE = ^^ \'^t, ^^=.f3^^7''''^=2l*'5"; HE 20 ,- = - |/2 ; KG = 2i. 13 !iC ■i I' 4: * mm m \'' •} * i ^ fee 80 THEORY OF STRUCTURES. id) Stress in tons in AB = = AC; BC=-= EF; 21 62 5 \/z FD= - = FG, KC=-\ CE=-^ I' 13 9 ^^ = 3ov:2 ;/?,? = J3 8 99 ^317; Fig. ia8. 46. The inclined bars of the trape- zoidal truss represented by the figure make angles of 45° with the vertical ; a load of 10 tons is applied at the top joint of the left rafter in a direction of 45° with the vertical. Assuming the reaction at tlie right to be vertical, fintl the stresses in all the pieces of the frame. Ans. Vert, reaction at Z> = - t/2 ; stress in DE = ■— \'2\ •J J ' DB = 6i; BE = 6^; BA = "~ ^2 ; AE = - s/z; AC CE = - V2 tons. 10 ; CA = 3 3 47. The post of a derrick-crane is 30 ft. high ; the horizontal traces of the two back-stays are at riglit angles to each other, and are 15 ft. and 25 ft. in length. Show that tiie angle between the shorter trace and the plane of the jib and tie, when the stress in the post is a maximum, is 30' 58'. Also find the greatest stresses in the different members of the crane when the jib, which is 50 ft. long and is hinged at the foot of the post, is inclined at 45° to the vertical, the weight lifted being 4000 lbs. Ans. Stress in jib = 66661} lbs.; in tie = 4768.4 lbs.; max. thrust along post = 10991.5 lbs. ; max. stress on Ions.; back-stay = 7362.7 lbs.; on short back-stay = 10539 'bs. 48. A queen-truss for a roof consists of two horizontal members, the lower 48 ft. long, the upper 16 ft. long; two inclined members AB, DC, and two queens BE, CF, each 8 ft. long; the points if, /"divide .,4Z> into three equal segments ; the load upon the members AB, BC, CD is 120 lbs. per lineal foot. Find (a) tlie stresses in the several members. How {b) will these stresses be modified if struts are introduced from the m EXAMPLES. 8l feet of the queens to the middle points G, H of the inclined members ? In this latter case also, determine (r) the stresses due to a wind-pressure of 1 20 lbs. per lineal ft. normal to AB, assuming that the horizontal re- action is equally divided between the two supports at A and D. Ans.—{a) Stress in lbs. in AE = 4066.56 = EF = DE = BC\ AB = 4546. yb^CD; BE = 2033. 28 = CF. (b) Stress in lbs. in AE= 5' 39- ^4 = E>E\ BC = 4066. 56 = AE\ AG = 5746. 56 rr- n//; BG = 4546.56 = C7/; EG - 1 200 = EH; BE = 536.64 ~ CE. (c) Additional stress in AG = 1040 f'J ; BG = 680 4,^5 ; GE= 600 v^5 ; AE=2^2o ; BE = 600 ; BC — 400 \^$ i BE= 400 4/5 ; CE = 400 ; CB = 400 {/$ ; EE = 1 1 20 ; /^Z>=320. (In case (c) the brace BE is introduced to prevent distortion.) 49. A pair of shear-leg.s, each 25 ft. long, with the point of suspension 20 ft. vertically above the ground surface, is sinnnorted by a tie 100 ft. long; distance between feet of legs = 10 4/5 ft. Find the thrusts along the legs and the tension in the tie when a weight of 2 tons is being lifted. Ans. Tension in tie = 1. 137 tons; compn. in each leg = 1.87 tons. 50. In the crane ABC, the vertical post AB = 15', the jib AC= 23', iir;! the angle BAC = 30". Find (a) the stresses in the jib and tie, and also the bending moment at the foot of the post when the crane lifts a weight of 4 tons. The throw is increased by adding two horizontal members CE, BD and an inclined member DE, the figure BE being a parallelogram and the diagonal CZ> coincident in direction with CA. Find {b) the stresses in the several members of the crane as thus modified, the weight lifted being the same. In the latter case show {c) how the stresses in the members are affected when the chain, which is in four falls, passes from E to // and then down the post. Ans. — (rt) Tension in tie = 3! tons; thrust in jib = 6/j tons; (b) Stress in CE = 934; in ED = 10.16; in CB =■ 13.49; in CD = 6.15; in DA — 10.7 ; in BD — 7 tons. {c) Stress in CE — 8,9; in ED = 10.7; in CB = 12.9; in CD = 5.8 ; in DA = 10.7 ; in BD = 7.4 tons. 51. The horizontal traces of the two back-stays of a derrick-crane are .1 and j feet in length, and the angle between them is ft Show that cos (fi — 0) X the stress in the post is a maximum when — :::,.,; ^ •"=-.» being the cos angle between the trace x and the plane of the jib and tie. 1 . iii»' m iir iiiJ if 'i 83 THEORY OF STRUCTURES. 1 ., li 1 )M ' i 3 t - m ■4 . ;■ 1 ^ '' ! . i ■ 52. The two back-stays of a derrick-crane are each 38' long, and the angle between their horizontal traces 2 tan"\'j; height of the crane- post = 32'; the length of the jib = 40'; the throw of the crane = 20'; the weight lifted = 4 tons. Determine the stresses in the several mem- bers and the upward pull at the foot of each back-stay when the plane of tlie jib and post (a) bisects the angle between the horizontal traces of the back-stays, (d) passes through a back-stay. Ans. In jib = 5 ; in tie = 2.52 tons ; in back-stay in (a) = 2.56, in (d) = 4.7 tons. 53. Find the stresses in the members of the crane represented E 35' B ^y ^''^ figure ; also find balance-weight at C. Ans. Stress in BE = 25 ; DE = 26.9; DB = 21.08; ZJA = 26.08; BA = .24; BC = 18.12 tons. A ~ Counterweight at C = 15.14 Fig. I2Q. tons. 54. Draw tlie stress diagram for the truss represented by the figure, the load at each of the points B and C being 500 lbs. angle EA£> = 3o\) 00 V Fig. 130. Also, if the rafter AB is subjected to a nominal wind-pressure of 100 lbs. per lineal ft., introduce the additional member required to prevent deformation, and state in lbs. the stress it should be designed to bear. Draw the stress diagram of the modified truss, assuming that the foot A is fixed, and that there are rollers at />. (AB = AE =z is' : BC = 10' ; angle BAD = 45 55. The post AB of a jib-crane is b 20 ft. ; the jib AC is inclined at 30° and 'lie tie BC at 45° to the vertical ; the weight lifted is 5 tons. Find the stresses in the jib and tie when the chain passes (a) along the jib, {d) along the tie, (c) horizontally from C to the post. The chain lias /zoo falls. Fig. i-ji. 56. In a mansard roof of 12 ft. rise, the upper triangular portion (of 4 ft. rise) has its rafters inclined at 60° to the vertical. The rafters of the "A" /B0\ A--Xii_/^jy c- & tons ^^55^^ EXAMPLES. «3 lower portion are inclined at 30° to the vertical. If there is a load of 1000 lbs. at the ridge, find the load at each intermediate joint necessary for equilibrium, and the thrust of tiie roof. A load of 2000 lbs. is concentrated at each of the intermediate joints and a brace is inserted between these joints. Find the stress in the brace. Alls. 1000 lbs. ; thrust = 500 4/3 lbs.; 333^ 4/ J lbs. 57. The horizontal boom CD is divided into eight segments, each 8 ft. long, by seven intermediate supports ; the depth of the truss at each end = 16 ft.; a weight of i ton is concentrated at C and at D. and a weight of 2 tons at each of the points of division. Determine the stresses ill tiie several members. 58. The figure is a skeleton diagram of a roof-truss of 72 ft. span and 12 ft. deep ; G, K, L, O, H are respectively the middle points of AE, EL, EF, LF, FB\ AE=EL^ LF= FB = 20 ft.; thetrussesare 12 ft. centre to centre ; thedead weight of the roof = 12 lbs. per sq. ft.; the normal wind-pressure upon AE may be taken = 30 lbs. per sq. ft.; the end A is fixed and Show by dotted lines how the Fig. 13a. ^>l F G H [I*' Fig. 133. B is on rollers. Draw a stress diagram stress diagram is modified with rollers under A, B being fixed. 59. The platform of a bridge of 84 ft. span bode nnd 9 ft. deep is carried by a pair of trusses of the type shown in the figure. If the load borne by each truss is 300 lbs. per lineal ft., find the Fig. 134. stresses in all the members. Ans. Stress in AB — 6000 ; AC = 1200 4/73 ; AD = 3600 |/i7 ; BC = 4800 ; CD = 14400 ; DE = 28800. Stress in horiz. chord = 288000 ; in each vertical = 3600 lbs. 60. The figure represents the shore portion of one of the trusses for a cantilever higliway bridge. The depth of truss over pier = 51 ft.; the length of each pane! = 17 ft.; the load at A (from weight of centre span) =z 16800 lbs.; the width of road- way = 15 ft.; the load per sq. ft. of roadway P'°* ^35- = 80 lbs. Find the stresses in all the mem- bers, assuming the reaction at the pier F to be vertical. Ans. /i = /j = 28000 ; /a = 36500 ; ^4 = 45000 ; h = 53500 ; /a = 55200; A =48400; /e = 41600 = /» ; ci = 5600 4/34; f, = 7300 V34 ; 7^1 = 10200 ; V3 = 15300; Vt = 20400 ; Afi t« h ti ti tn tr ti toB I'J.WW lbs, 'I 84 THEORY OF STRUCTURES. V\ = 25500; Vt = 45900; V, = 20400; 7/, = 15300; v» = 10200 ; ca = 9000 v^34 ; r, = 10700 t/34 ; Ci = \2\00 \^2>-^\ re — 77500: ri =69000; r» = 60500; Ca = 52000 ; t/i = 1700 V'34 ; (ft = 1700 V6i ; r/i = 1700 4/106 ; «'< = 22100; (it. = 1700 4/97 ; rt'o = 3400 4/13 ; ^A = 8500 lbs. 61. The inner flange of a bent crane forms a quadrant of a circle of 20 ft. radius, and is divided into /(3;/r equal bays. Tlie outer flange forms the segment of a circle of 23 ft. radius. The two flanges are 5 ft. apart at the foot, and are struck from centres in the same horizontal line. The bracing consists of a series of isosceles triangles, of which the bases are the equal bays of the inner flange. The crane is required to lift a weight of 10 tons. Determine the stresses in all the members. 62. A braced semi-arch is 10 ft. deep at the wall and projects 40 ft. The upper flange is horizontal, is divided into/(3Kr equal bays, and carries a uniformly distributed load of 40 tons. The lower flange forms the segment of a circle of 104 ft. radius. The bracing consists of a series of isosceles triangles of which the bases are the equal bays of the upper flange. Determine the stresses in all the members. 63. The domed roof of a gas-holder for a clear span of 80 ft. is strength- ened by secondary and primary trussing as in the figure. The points />' and C are connected by the tie BPC passing beneath the central strut AP,vih'\ch is 15 ft. long, and is also common to all the primary trvisses ; the rise of A above the horizontal is 5 ft.; the secondary truss ABEF consists of the equal bays A//, HG, GB, the ties BE, EF, FA, of which BE is horizon- tal, and the stu- .'.s GE, FH, which are each 2 ft. 6 in. long and are par- allel to the radius to the centre of GH\ the secondary truss ACLK is similar to ABEF; when the holder is empty the weight supported by the truss is 36000 lbs., which may be assumed to be concentrated at G. H, A, M, N, in the proportions 8000, 4000, 1000, 4000, and 8000 lbs., re- spectively. Determine the stresses in the different members of the truss. 64. The flgure is the skeleton diagram of a cantilever for a viaduct in 4S tgu 48 tdu 411 Ufa JD tou Fig. 137. EXAMPLES. 85 India. Determine ^<,'-m/)///(((i//>' the stresses in the various members un- der ilie loading indicated. f)5. In the accompanying roof-truss AB = AC-= 30 ft., and the struts are ail normal to tlie rafters. Find the stresses in all the members, the load at each of the joints in tlic rafters being 2 tons (angle AliC = 30' and angle DEC = 10 ). How will the stresses be mod- ified if theie is a force of 2 tons acting at each of the points of support between // and /)' at right angles to the rafter, and a force of i ton at A, assum- ing tiiat the end // is fixed and that C rests upon rollers? 66. The figure represents a portion of a Warren girder cut off by the 1^ plane iJAV and supported upon the abutment at j A. The reaction at A = 20 tons ; the load con- centrated at each of the points B = 4. tons. Find the stresses in each of the members met by AfN. Ans. Stress in tension chord = — 1/3 tons ; Fir.. 138. Fig. 139. 16 .7 toni in compression chord = 32 4/3 tons; compression in diagonal = — 4/3 tons. 67. The figure represents a portion of a roof-truss cut off by a plane MX and supported at A. The strut DC is vertical ; AD = 23 ft., and the distance of D from AC= 7^ ft.; the angle between AC and the horizontal r= cos"'^ ; the vertical reac- tion at ^ = 7 tons ; the horizontal reaction at A = 2i tons ; at each of the points B and Ca weight of 4 tons is concentrated. Find the stresses in the members met by MN. {AD and Ti make equal angles with the rafter.) Ans. Ci = 13.2 tons ; Tj = 2.1 tons ; Ti = 10.8 tons. 68. The feet of the equal roof-rafters AB, AC are tied by rods BD, CD which meet under the vertex and are joined to it by a rod AD. If H'l , IVi are the uniformly distributed loads in pounds upon AB, AC, respectively, and if .S' is the span of the roof in feet, find the weight of metal (wrought- iron) in the ties. 5 IVx + IV^ Pig. 140. Ans. — S cot /3, / being inch-stress in pounds, and 6 / fi the angle ABD. {a) If AB = AC = 20 ft.. AD — 5 ft., the angle BAD = 60°, find '■i^M \ ' m Li ' n f u fli mf THEORY OF STRUCTURES. the stresses in the several members when a weight of 3500 lbs. is con- centrated at the vertex. Ans. 7000 lbs.; 6309.8 lbs.; 3500 lbs. (d) The roof in (a) is loaded with 10 lbs. per square foot on one side and 33 lbs. per square foot on the other; the trusses being 13 ft. centre to centre. Determine (a) the stresses ;n the several members. Examine (/>) the effect of a horizontal pressure of 14 lbs. per square fool on the most heavily loaded side, assuming that the reaction is equally divided between th^ two supports. A/is. (a) 1 1 180 lbs.; 10077.65 lbs.; 5590 lbs. In the truss represented in the accompanying figure, the load on An = IV, . on AC = IV^ ; the angle ABD = H; AD z= BD= AE = CE. Find the total weight of metal (wroiiglit-iron) in the tie-rods. 69. Fig. 141. Ans. - 6 5 /F. + W; f Sco\. /?; 5 being the span and /the inch stress. (a) If the stress in BD or EC is equal to the stress in DE, show that a /3 = 60° ; a being the angle ABC. 3 {b) The trusses are 12 ft. centre to centre; the span is 40 ft.; the hori- zontal tie is 16 ft. long; the rafters are inclined at 60° to the vertical; the dead weight of the roof, including snow, is estimated at 10 lbs. per sq. ft. of roof-surface. Determine the stress in each member when a wind blows on one side with a force of 30 lbs. per sq. ft. normal to the roof- surface, assuming that the horizontal reaction is equally divided between the supports. Ans. Stress in AB = 8956.8 lbs. ; BD = 10015.2 lbs. = EC; AD = 2503.8 lbs. = AE; DE = 8196 lbs.; AC = 11356.8 lbs. 70. In the truss represented by the accompanying figure, the load upon AB = JV, , upon AC = fF, ; the angle ABD=z /3 ; the span BC = S ; the ties AD, BD, AE, CE are equal ; E and G are the middle points of the rafters. Find the amount of metal in the tie-rods (wrought- iron). 5 ^ / Ans. - — Fig. 142. IVi +(JVi + JV,) cos' /3 sin /3 cos /i (a) The struts DE and EG are each 5 ft.; the angle ABC = 30° ; the dead weight of the roof, including snow. Is 9 lbs. per square foot of roof- surface, and the trusses are 12 ft. centre to centre. Determine the stresses in the several members wlien a wind blows with a force of 30 EXAMPLES. 87 lbs. per square foot of roof-surface normal to the side AB. The span — 60 ft., and the (mkI C rests upon rollers. Secondly, dctermirie the stresses produced in the members of the truss in the preceding question when a single weight of 3000 lbs. is sus- pended from G. //;«.— (I) Stresses in DD\ DA; DE\ EA ; EC; 31238.55; 19852.35; 12633.6; 8113.5; 24379.43; JiF; FA; FD; CG; GA; GE. 29620.44; 28685. 16 ; 7855.2 ; 22420.44; 21485.16; 1620 lbs. (2) Stresses in BD; 375 s'y) ; FA; 2625; BE; 2625; DA; DE; EA; EC; 4/^- loooVT; 8751/39; "25*^39: CG; GA; GE. _ 7875; 6375; 15001^3 lbs. '25 V39 FD; o; (b) The rafters AB, AC are of unequal length and make angles of 60° and 45°, respectively, with the vertical; the strut DF = 7^ ft.; the tie DE is horizontal j the dead load upon each rafter = 100 lbs. per lineal foot; the wind-pressure normal to iB = 300 lbs. per lineal foot; rollers are placed at C. Find the stresses in all the members. The rafter AB = 45 ft. Show by dotted lines how tlie stress diagram will be modified: (1) If the rollers aie placed at B. (2) If the strut DF is omitted. (3) If a single weight of 500 lbs. is concentrated at D. (c) If it is assumed that the horizontal .eaction is equally divided be- tween B and C, show that the stress in DE due to a horizontal wind- pressure upon AB is nil ; the angle ABC hc'xn^ 30°. ((/) In a given roof, the rafters are of pitch-pine, the tie-rods of wrought-iron ; the span is 60 ft.; the trusses are 12 ft. centre to centre; DF= 5 lx..=EG; the angle ABC = 30"; the dead weight of the roof, in- cluding snow, is 9 lbs. per sq. ft. of r'^'~ '-surface ; rollers are placed at C; a single weight of 3000 lbs. is suspended from F, and the roof is also designed to resist a normal wind-pressure of 26.4 lbs. per sq. ft. of roof- surface on one side AB. Determine the stresses in the several members. 71. In the truss represented in the accompanying figure, the struts DF. DH, EG, EK are equal, and the ties BD, AD, a E.l, EC are also equal; the load upon AB is ff 1 , . h^^^^k and upon AC is IV-, . Find the weight of metal -^^i^*^ ^"^ (wrought-iron) in the ties. Ans. -2- 5 5 4^V^+j(lV, + Fig. 143. fF,)cos'/S 18 / cos fi sin fi (a) AD = AE — BD =1 EC = 2-^ ft.; the angle ABC = 30° ; the span = 79 ft.; the trusses are 13 ft. centre to centre; the heel B is free to i.t i 88 THEORY OF STRUCTURES. slide on a smooth wall-plate; the dead weight of the roof, including snow, is 8 lbs. per square foot of roof-surface. Determine the stress to which each member is subjected when the wind blows horizontally with a force of 40 lbs. per square foot of vertical surface (i) upon the side AB, (2) upon the side AC. Ans. See Ex. 3, Art. 24. ib) The rafters AB, AC are inclined at 60° to the vertical and are each 40 ft. in length. The foot Crests on rollers, and the foot B is fixed. The strut DF is vertical, is 10 ft. long, and is equal to the strut DE in length. Also AF = HF = 10 ft. The dead load carried by the rafters is 120 lbs. per lineal foot. Provision has also to be made for a normal wind-pressure upon AB of 300 lbs. per lineal foot. Draw the stress diagram, and show how it will be modified if the strut DF is re- P'°- '«• moved. Ans. Vertical reaction at B = 10528 lbs. both before and after DF is removed. Horizontal reaction at ^ = 6000 lbs. The dotted lines show the modified stresses for one half of the truss. 72. The load upon a roof-truss of the accompanying type is 1000 lbs. at each joint ; the span 100 ft.; the rise = 25 ft. Find the stresses in Fig, 145. the different members. How will the stresses be affected by an addi- tional load of 250 lbs. at each of the joints between the foot and ridge on one side ? Ans. Stress in BD =5500 Vj; DF = 5000 \/J; FH = 4500 4/5 ; HL — 4000 4/5"; LN= 3500 |/s"; NA = 3000 4/5 ; DE = o ; /"G = 500 ; HK = 1000; LM= 1 500 ; NO — 2000 ; AP = 5000 ; BE = I rooo = EG ; G/v = loooo ; KM = 9000 ; MO = 8000 ; OP = 7000 ; DG = 500 y'J; FK = 1000 4/2 ; HM = 500 1/13 ; LO — 1000 ^J; NP — 500 4/29 lbs. EXAMPLES. 89 73. The dead load upon a roof-truss of accompanying type consists of 1000 lbs. at F, 1000 lbs. at K, and 500 lbs. at G\ the wind-pressure is a normal force of 30 lbs. per square foot of roof-surface upon AB; the span = 90 ft.; the rise = 25 ft.; the trusses are 25 ft. centre to centre. Find the stresses in the several members when rollers :i229if lbs.; vertical reaction horizontal reaction at i? = Fig. 146. are {a) at C, (b) at B. Ans. — {a) Reaction (vertical) at C- at B = 22>9SH lbs.; 18750 lbs. Tension in BD = 48625 ; DL = 34475 ; LE = 21675 '< EC =2212$; DN=7S6i^; AL = 15888^; /^E = 250 lbs. Compression in BF=^666i y'loe; F>Y=2788| |/io6; //A = 1977I |/io6 ; AA!' = 2325 4/106 ; KG = 24o8J^ 4/106 ; GC = 2458^ 4/106 ; DF = 1572I 4/106; LH- 1505^ 4/181 ; ZA' ^ S^i 4/T81 ; EG=So i^ToS lbs. (6) Only alteration in stresses is that each stress in the different sections of the horizontal tie is diminished , by 18750 lbs.; all the remaining stresses are un- changed. 74. In the accompanying roof-truss, angle ABC = 30° ; the span = goj^ ft.; DF = EG = loj ft.; each rafter is divided into four equal segments by the points of support ; the trusses are 20 ft. centre to centre ; the weight of a bay of the roof = 24416 lbs. Determine the Fig. 147. stress in each member. Also determine the stresses due to a wind-pressure of 30 lbs. per square foot of roof-surface acting normally to AB, when rollers are under (a) C, {f>) B. 75. The figure represents a bowstring truss of 80 ft. span, cut off by the plane MN and supported at O. The upper flange OCDE is an arc of a circle of 85 ft. radius ; OA = AB = etc. = 10 ft. ; the rise of the truss = 10 ft. ; a load of 1 5 tons is concentrated at each of the points A and B\ the reaction at O = 45 tuns. Find the stresses in the members cut by the plane MN. ^ IM Pio. 148. '%^' m i 1 . : ■ \\ . r'-i ..\ If .^;,l:„::.i^ r.ij. \ I s \ ■'-:l:i! ■'Ml ■' V i ] i|l'i T 53,800 lbs, 90 THEORY OF STRUCTURES. 76. The figure is a portion of a bridge-truss cut off by the plane MX and supported upon the abutment at A ; AC = CE- I4,V ft.: the depth BC = DE = 17^ ft. ; in the third panel the compression in the upper -<6i,6oo lbs. choi 1 is 64,600 lbs.; the tension in the lower chord is 53,800 lbs. Find the reaction at^^, the equal weights supported at C and E, and the diagonal stress T. Ans.' Reaction = 19,474 lbs.; weight at C and at £■ = 9737 lbs. ; T= 17,977 lbs. 77. The top beam of a roof for a clear span of 96 ft. consists of six bars AB, BC, CD, DE, EF, EG, equal in length and so placed that A, B, C, D, E, F, G are on circle of 80 ft. radius; the lower boom also consists of six equal rods AH, HK, KL, LM, MN, NG, the points H, K, L, M, and N being on a circle of 148 ft. radius; B is connected with H, C with K, D with L, E with M, and F with N \ the opposite corners of the bays are connected by cross-braces ; the end A is fixed to its sup- port, G being allowed to slide freely over a smooth bed-plate. Determine graphically the stresses in the various members when there is a normal wind-pressure per lineal foot of 460 lbs. upon AB, 340 lbs. upon BC, and 60 lbs. upon CD. 78. A bowstring roof-truss, with vertical and diagonal bracing, of 50 ft. rise, and five panels, is to be designed to resist a wind blowing horizontally with a pressure of 40 lbs. per square foot. The depth of the truss at the centre is 10 ft. Determine, ^;'rt'////ra//)', the stresses in the several members of the truss, assuming that the roof rests on rollers at the windward support. 79. Determine the chord, vertical and diagonal stresses in a Howe truss of So ft. span, 8 ft. depth, and ten panels, due to a load of 40 tons (a) concentrated at the centre ; (^) concentrated at the third panel point ; {c) uniformly distributed ; {d) distributed so that 5 tons is at first panel point, 10 tons at second, and 25 tons at third. Ans. Panel stresses in tension chord : I I a 1st 2d 3d 4th 5th 6th 7th 8th 9th loth 20 40 60 80 IOC 100 80 60 40 20 h 28 56 84 72 60 60 48 36 24 12 c 18 32 42 48 50 50 48 42 32 IS d 30 55 70 60 50 50 40 30 20 10 EXAMPLES. Panel stresses in compression chord : 91 a 20 40 60 80 80 60 40 20 b 28 56 84 72 48 36 24 12 c IS 32 42 48 4S 42 32 18 1 d 30 55 70 60 40 30 20 10 Stresses in verticals : a 20 20 20 20 40 20 20 20 20 b 28 28 28 12 12 12 12 12 12 c 18 14 10 6 4 6 10 14 18 d 30 25 15 ID 10 10 10 10 10 Diagonal stresses : a 20 ^2 tons in each diago nal. b 28V'2 2iV2 284/2 124/2 124/2 124^2 124^2 124/2 124/2 124/2 c 184/2 I4I/2 104/2 64/2 24/2 24^2 64/^ 104^2 1W2 1SV2 d 3oV'2 254/2 rsV^ 104^2 104^2 104^2 104^2 ro4/2 104/2 104/2 80. A Warren girder of 60 ft. span, composed of six equilateral triangles, carries upon its lower chord a weight of 2 tons at the first and second joints, 15 tons at the centre joint, and 7^ tons at the fourth and fifth joints. Find the stresses in all the members. Am. Stresses in tension chord: ist bay = Y 4/J; 2d = ',"/ 1^3 ; 3d = s,V' Vy^ 4th = =^,v Vy, 5th = V V3 ; 6th = i^ 4' 3. Stresses in com pr. chord : ist bay = '/ 4/3; 2d = V 4/3; 3d = V- 4/3J 4th = V i' J: 5th = V V3. Diag. stresses ist and 2d bays = V 4/3 ; 3d and 4th = -\?- \/'2- 5th and 6th = V- 4/3 ; 7th and Sth = V- Vy, 9th and loth = V V/3 ; nth and 12th = -',," 4''3'; 81. Determine the stresses in the members of a Fink truss of 240 ft. span and sixteen panels; depth of truss = 30 ft. ; uniformly distributed load = IV. Is; ■' : '. t !i .Mi :m h I W9 In liiii:- \ *m '.in J' ■ ^ I , ; ■-It 92 THEORY OF STRUCTURES. -^VHD- Ans.— a! l ft n o p q r s' Stress in BA, BM, DM, DO, FOy FQ, HQ, HS, same and — B C D E F Q H Fig. 150. 64 in IV CA, CO, GO, GS same and = —7- 4/2 ; 10 IV - W , m EA, ES same and = -5- y'c ; in AK = — r 17 ; in BL, o ■' 4- ff W DN, FP, HR, same and = — 7- ; in Ol/, Cg same and = -5- ; w w in £0 = — - ; \n KS= — ; in AM, MO, OQ, OS same and 64 82. Determine the stresses in the members of a Bollman truss 100 ft. long and \2\ ft. deep, under a uniformly distributed load of 200 tons, to- gether with a single load of 10 tons concentrated at 25 ft. from one end. Ans. Stress \n AB = ^\/\; ^Z = if 4/2 ; /f Z? = ^ 4/5 ; DL = ^- \/yj ; AF = i^ i/Fo"; FL = ^ 4/26; AH - ^ i/vj = HL; in BC = 2$ = FG = HK = etc.; DE= 50 tons; compression along /iZ = 193I tons. A''.?/^.— Questions 53, 54, 57-59, 61, 66, 67, 70, 71, 73, and 74 can be easily solved graphically. 83. Determine the stresses in the several members when the throw of the crane in Question 55 is increased by the introduction of the new members, shown by the dotted lines. I'll rfi ' CHAPTER II. SHEARING FORCES AND BENDING MOMENTS. Note. — In this chapter it is assumed that all forces act in one and the same plane, and that the deformations are so small as to make no sensible alteration either in the forces or in their relative positions. I. Equilibrium of Beams. — A beam is a bar of somewhat considerable scantling, supported at two points and acted upon by forces perpendicular or oblique to the direction of its length. Case I. AB is a beam resting upon two supports in the same horizontal plane. The reactions A^, and R^ at the points of support are vertical, and the resultant P of the remaining external forces must also "^ ^? act vertically in an opposite direction ^'°' '^'' at some point C. According to the principle of the lever, iA _c r^ ^. = ^z^. R. P^ AB' and R,-^R^ = P. Case II. AB is a beam supported ox fixed at one end. Such a support tends to prevent any deviation from the straight in that portion of the beam, II t p and the less the deviation the more ^ ' 1T ! ^^ perfect is the fixture. ''■q " ^? The ends may be fixed by means of ^'°' '"■ two props (Fig. 153), or by allowing it ^ , 3 to rest upon one prop and preventing jp upward motion by a ledge (Fig. 154), or Fig. 154. by building it into a wall (Fig. 155). In any case it may be assumed that 1 ' R Ac -[M I i B the effect of the fixture, whether perfect >ifp or imperfect, is to develop two unequal ^'°' '55- forces, Q and R, acting in opposite di- rections at points M and N, These two forces are equivalent 93 \' ■ ' r ' ( . .^i r' !i ill I I , rn^ 1 It^ •■i 94 THEORY OF STRUCTURES. to a left-handed couple (0, — Q), the moment of which is Q.MN, and to a single force R— Q dX N. Hence R — Q must = P. Case III. AB is an inclined beam supported at A and resting upon a smooth vertical surface ati?. The vertical weight P, acting at the point C, is the resultant load upon AB. Let the direction of P meet the hori- zontal line of reaction at B in the point D. The beam is kept in equilibrium by the weight P, the reaction R^dX A, and the reaction 7?, at B. Now the two forces R^ and P meet at D, so that the force R^ must also pass through D. Fic. 156. Hence R, = P- I cos ADC and R, = P tan ^27(7. A*!?/^. — The same principles hold if the beam in Cases I and n is inclined, and also whatever may be the directions of the forces P and R^ in Case HI, Case IV. In general, let the beam AB be in equilibrium under the action of any number of forces /*,,/*,, P, , ... , !2i > (2(1 > G3 . • • • > of which the magnitudes and points of appli- At Fig. 157. cation are given, and which act at right angles to the length of the beair. Suppose the beam to be divided into two segments by an imaginary plane MN. Since the whole beam is in equi- librium, each of the segments must also be in equilibrium. Consider the segment AMN. ? ^^ ^'^ ?' f 1 N i 1 V V Qa Q2 EQUILIBRIUM OF BEAMS. 95 It is kept in equilibrium by the forces P^, P^, P^, . . . and by the reaction of the segment BAIN upon the segment AMN at the plane MN\ call this reaction E^. The forces/*,,/*,, I\, . . . are equivalent to a single resultant R^ acting at a point distant r, from AIN. Also, without affecting the equilibrium, two forces, each equal and parallel to /v, , but opposite to one another in direction, may be applied to the segment AMN at the plane MN, and the three equal forces are then equivalent to a single force R^ at MN, and a couple {R^ , — R^) of which the moment is /?,r, . Ru ^jT'^-^Ru^R.'^'^ 33 I V_-ii^ Fig. 158. Thus the external forces upon AMN are reducible to a sincflo force /?, at AIN, and a couple {R^ , — R^). These must bo balanced by E^ , and therefore E^ is equivalent to a single force — R, at A/N and a couple {— R,, /?,). In the same manner the external forces upon the segment BMN are reducible to a single force /?, at AIN, and a couple (A\, — R^ of which the moment is R^r^. These again must be balanced by E^, the reaction of the segment AMN upon the segment BMN. Now /", and E^ evidently neutralize each other, so that the force /?, and the couple (/l, , — /?,) must neutralize the force A\ and the couple {R^, —-^0- Hence the force /?, and the couple {R^ , — R,) are respectively equal but opposite in effect to the force R, and the couple {R^, — R,) ; i.e., .■'§1' J \\ < r< 7?, = Rj and Rr^ ^ R^r^ ; r, = r,. The force R^ tends to make the segment AMN slide over the segment BMN aX. the plane AIN, and is called the Shearing 53 . I- I ft ^ J 1 II ^il 96 THEORY OF STRUCTURES. Force with respect to that plane. It is equal to the algebraic sum of the forces on the left of MN, = p^ + p^-p^j^... = :s{P). So ^, = (2, — ^3 — S3 + • • = ^(0 is the algebraic sum of the forces on the right of MN, and is the force which tends to make the segment BAIN slide over the segment AMN at the plane J/A''. ^3 is therefore the Shearing Force -wXih. respect to MN, and is equal to R^ in magnitude, but acts in an opposite direction. Again, let/, , A . A » • • • . ^1 » ^» . ^3 . • • • . be respectively the distances of the points of application of /*,,/!,, P, ,.. . ^Q^,Q^, <2, , . . . from MN. Then R^r^ , = the algebraic sum of the moments about MN of all the forces on the /eft of AfN, = P.P. + P.P. - /'3 A + • . . = ^{Pp\ is the moment of the couple iR^, — /?,). This couple tends to bend the beam at the plane MN, and its moment is called the Bending Moment with respect to MN of all the forces on the left of MN. So /v',;-, , := the algebraic sum of the moments about il/A^ of all the forces on the right of MN, = Q.Q. - Q.^. = ^{QqI is the Bending Moment, with respect to MN, of all the forces on the right of MN, and is equal but opposite in effect to R^r^ . It is seen that the Shearing Force and Bending Moment change sign on passing from one side of MN to the other, so that to define them absolutely it is necessary to specify the seg- ment under consideration. Remark. — The reaction E^ has been shown to be equivalent to the force — R^ and the couple (— R^ , /?,). The Moment of this couple may be called the Elastic Moment, the Moment of Resistance, or the Moment of Inflexibility, and is equal in magnitude, but opposite in effect, to the corresponding Bend- ing Moment due to the external forces. 2. E: ments.- example of length Ex. is fixed /'at O. The j point of stant and Upon equal or between ; shearing i Again distant x Upon tional to point of moment a Ex. 2. take AB e( distance be resents the Again, t is nil at (9, ^C equal o SHEARING FORCES AND BENDING MOMENTS. 97 Fig. 139. 2. Examples of Shearing Forces and Bending Mo- ments. — In each of the following examples the beam is horizontal and of length /. Ex. I. The beam OA, Fig. 159, is fixed at A and carries a weight /'at O. The Shearing Force {S) at every point of the beam is evidently con- stant and equal to P. Upon the verticals through A and O take AB and (9Ceach equal or proportional to P\ join BC. The vertical distance between any point of the beam and the line BC represents the shearing force at that point. Again, the Bendbig Moment (M) at any point of the beam distant x from O is Px ; it is nil at O, and P/ at A. Upon the vertical through A take AD equal or propor- tional to P/; join DO. The vertical distance between any point of the beam and the line DO represents the bending moment at that point. Ex. 2. The beam OA, Fig. 160, is fixed at A, and carries a uniformly distributed load, of in- tensity zu per unit of length. The resultant force on the right of a vertical plane J/iVdistant x from O is wx and acts half-way between O and MNi The SJiearing Force (S) at MN is therefore zvx ; it is nil at O, and ivl at A. Upon the vertical through A take AB equal or proportional to wl\ join BO. The vertical distance between any point of the beam and the line BO rep- resents the shearing force at that point. X tax' Again, the Bending Moment (M) at A/N is tux- — ; it is nil at O, and — -at A. Upon the vertical through A take zvi' J C equal or proportional to . ^w.x Fig. 160. ■^ mu\ U:l*Ll THEORY OF STRUCTURES. I s ■: The bending moment at any point of the beam is repre- sented by the vertical distance between that point and a pa- rabola CO having its vertex at and its axis vertical. Ex. 3. The beam OA, Fig. 161, is fixed ', and A take OC, BE, each equal or Pb [)roportional to -v-, and BF, AD, each equal or proportional to - y ; join CE and DF. The shearing force at any point of the beam is represented by the vertical distance between that point and the broken line CEFD. t 'is •<4-! WMil 1 1 i ri i: li'l 1 » lilMi F ~ iJiii- '■ f. lOO THEORY OF STRUCTURES. It IS Again, the Bending Moment (M) at any point between O and ^ distant x from 6> is R^x = P-.x\ it is nil at O, and P^* at /i. The Bending ^lament (Af) at any point between B and A distant x from 6> is R,x — /'(;«; — a) = P-A/ — x) P-. at B, and nil at y?. Upon the vertical through B take BG equal or proportional to P—f\ join 6^6^ and AG. The bending moment at any point of the beam is represented by the vertical distance be twecn that point and the line OGA. P tor. — If /*bc at the centre of the beam, 5 = -, and M at the centre = PI Ex. 5. The beam OA, Fig. 163, rests upon two supports p at C? and A, and carries a uni- formly distributed load of inten- sity zv per unit of length. The reactions at and A are wl each equal to — The resultant force between and a plane MN distant x from Fig. 163- is wx, and acts half-way between and MN. The Shearing Force (S) at MN is theiefoie wl wl zvx; it is at O, nil at the middle point B, and at A. Upon the verticals through O and A take OC wl and ^Z?, each equal or proportional to - — ; join CJy. The shearing force at any point of the beam is represented by the vertical distance between that point and the line CD. SHEA KING FOJtCES AND BENDING MOMENTS. lOI Again, the Bending Moment (Af) at MN is w/ X ti'i Tta' — X — wx- = — X 2.22 2 it is nil at O and ^t A \ it is a maximum and equal to --— at o the middle point B. Upon the vertical through B take BE equal or proportional to - -. The bending moment at any point of the beam is represented b\' the vertical distance be- tween that point and a parabola OEA having its vertex at £ and its axis vertical. Cor. I. The shearing force is a minimum and zero at the centre, a. maximum and — at the ends, and increases uni- 2 formly with the distance from the centre. Cor. 2. The bending moment is a minimum and zero at the ends, a maximum and ■wT 8 at the centre, and diminishes as the distance from the centre increases. Ex. 6. The beam OA, Fig. 164, rests upon two supports at and A, and carries a weight P at a point B, together with a uniformly distributed load of intensity u> per unit of length. Let the lengths of the segments OB, BA be a and b, respectively. The reactions /?, at 0, and R^ at A, are vertical, and according to the principle of the lever, R. and Fig. 164. ' Prt wl -, at ^. / 2 Upon the verticals through O, B, and y4 take OC equal or . , Pb , wl „^ proportional to —,- -\ , BD equal or proportional to Pb , 7vl „^ , . ^ ivl Pa _| ^(i^ }jj^ equal or proportional to ^ wa, wl Pa and /i/^ equal or proportional to j ; join CD and EF. The shearing force at any point of the beam is rep- resented by the vertical distance between that point and the broken line CDEF. wl Pa \i — > ~r + ^^^. BE is positiv^e, and therefore E is ver tically above B. Again, the Bending Moment (M) at any point between O and B is b wl\ wx^ H+x ;^ it is nil at O, and ( ^b , wl\ wa* \p-^--]a-~^,B. and SHEARING FORCES AND BENDING MOMENTS. 103 The bending moment (J/) at any point between B and A distant x from O is (b , tvl\ zva' r^ , ., . It IS \P-, -\ \a at ^, and ml at A. Upon the vertical through B take BG equal or propor- / b tul\ wa^ tional to \P-.-\---\a . The bending moment at any point of the beam between O and B is represented by the vertical distance between that point and a parabola OGH having its axis HT vertical and its vertex at a point //, where and 0T= --LP-7 + - • I / ^ wiy HT is equal or proportional to -—[P-i + — J • The bending moment at any point between B and A is represented by the vertical distance between that point and a parabola AGK having its axis A'F vertical and its vertex at a point A", where w\ / ' 2 /' and KV is equal or proportional to 2ZV Pi + tU""- Cor. — If the weight /* is at the centre, P PI UfP S = — , and M at the centre = 1 — ^r-. 2 ' 48 Note. — The ordinates of the lines CD and EF in Ex. 6 are equal to the algebraic sum of the corresponding ordinates of the .! 'i n 8.. .-iii-l-t i* 104 THEORY OF STRUCTURES. ' lines CE, FD in Ex. 4, and the line CD in Ex. 5. Also, the ordinates of the curves OG, AG are equal to the algebraic sum of the corresponding ordinates of the lines OG, AG in Ex. 4, and the curve OEA in Ex. 5. Hence the same conclusions as in Ex. 6 are arrived at by treating the weight P and the load ivl independently, and then superposing the respective results. Ex. 7, In fine, a beam, however loaded, may be similarly treated, remembering that if the load changes abruptly at dif- ferent points, the portions of the beam between the points of discontinuity are to be dealt with separately. For example, the beam OA, Fig. 165, rests upon two supports at and A, and carries three weights /*, , P^, P^ at points C, D, E, of which the distances from O are />,, />, , p^, respectively. A point B divides OA into segments OB = a and BA = b, which are i I F r^G /^ ^■^T?"^--^^^ \ / IC 1 1 Mk \ l« iD^"~^v. E \ A '?^' f~ i y Pi P« N 1 1 N K ^. P3 ^ T Fig 165. uniformly loaded with weights of intensities w, and w, per unit of length, respectively. The reactions 7?, and R^ at O and A are vertical, and according to the principle of the lever, R,l = />.(/ - /.) + Pil - A) + Pit - A) and Kl = P. P. + P.P. + P^^ + ^- + tvb[- 4- 4 SHEARING FORCES AND BENDING MOMENTS. 105 To represent graphically the Shearing Force at different points of the beam : Upon the verticals through O, C, B, D,E, A, take OF, CG, CH, BK, DL, DM, EN, EV, and ^r, respectively equal or proportional to ^f^ltHI ^. R, — w,p„ R - 7v,p, - P,,R,~ 7i\a R. — zc\a — P, — 7i',{f, — a), R. - Zi\a, — P, — w,(A - (i)- P.„ . A'. - w.a — P, — 7i\ip, — (I) - P„ A', - n -P,- xi'lp, -a)-P,- P„ A, - ZV,.. - P, - ZVJ) - P,-P,:= R,. P., ind Join FG, HK, KL, MN, and VT. The shearing force at any point of the beam is represented by the vertical distance between that point and the broken line FGHKLMNVT. To represent graphically the Bending Moment {M) at dif- ferent points of the beam : Mat = 0; Mat C= Rj>, — w^p^ 7U rt;' MatB- R,a '- P,{a - p,) ; M atD = R,p, - w,a[p, - I) --,^^'-/'.(A-A). .'"i ti t h \ 1 1' I 1. I * i I I^J io6 THEORY OF STRUCTURES. MB.tE = R,p, - 2va[p, - I) — w. ^.0>,-A)-^.(A-A); and Mat A= o. Upon the verticals through C, B, D, and E take Ci, ^2, Z?3, and £'4, respectiv^-ly equal or proportional to the bending moments at these points. The bending moment at any point of the beam is repre- sented by the vertical distance between that point and the parabolic arcs Oi, 12, 23, 34, and 4A. The axes of these pa- rabolas are vertical, and the positions of the vertices may be easily found from the several equations. Ex. 8. A beam OA, Fig. 166, of which the weight may be neglected, is 15 ft. long, is fixed at O, and carries a weight of 80 lbs. at A. Determine the bend- ing moment at a point distant 10 ft. from the free end. Also illus- trate the shearing force and bend ing moment at different points of the beam graphically. The re- quired bending moment is 80 X 10 = 800 Ib.-ft. The shearing force is the same at every point of the beam, and eq-ial to 80 lbs. Choose a vertical scale of measurement so that half an inch represents 160 lbs. Upon OA describe a rectangle OABC, in which OC = AB = ^". The ordinate from every point of BC to AO is i", or 80 lbs., and is therefore the shearing force at the foot of such ordinate. Again, the bending moment at O is 80 X 15 = 1200 Ib.-ft. Choose a vertical scale of measurement so that i inch repre- sents 1200 Ib.-ft. Upon the vertical through O take OD ^ Fig. 166. SHEARING FORCES AND BENDING MOMENTS. \0J I inch ; join DA. The ordinate from any point of DA to OA is the bending moment at its foot. For example, at \\\ ft. from the ordinate is i", or 300 Ib.-ft., and this is equal to 80 X 3$', i-e., the bending moment. Ex. 9. A beam OA, Fig. 167, of which the weight may be neglected, rests upon tw^ supports at and A, 30 ft. apart, and carries a uniformly distributed load of 200 lbs. per lineal foot, together with a single weight of 600 lbs. at a point B dividing the beam into segments OB, BA, of which the lengths are 10 and 20 ft. respectively. Determine the shearing force and bending mo- ment at the points C and D, distant 5 ft. from the nearest end. Also, illustrate graphically the shearing force and bending moment at differ- ent points of the beam. Let R^ , R^ be the reactions at O and A, respectively. Then .^, . 30 = 600 . 20 + 200 . 30 . 1 5 = 102000 ; .•. R, = 3400 lbs., and R^ = 200 . 30 + 600 — /?, = 3200 lbs. The Shearing Force ?Lt C = 3400 — 200 . 5 = 2400 lbs. " " " " Z>= 3400— 200. 25— 600= — 2200 lbs. The Bending Moment at C = 3400 . 5 — 200 . 5 . - = 14,500 Ib.-ft. 25 The Bending Moment at Z) = 3400. 25 — 200 . 25 . — — 600. 1 5 = 13,500 Ib.-ft. Next, considering the segment OB, the shearing force at O is 3400 lbs., and at B 1400 lbs. Considering the segment BA, the shearing force at A is — 3200 lbs., and at B 800 lbs. Choose a vertical scale of measurement so that i inch repre- sents 3000 lbs. Upon the verticals through 0, B, A take OE = lyV'. ^^= iV. -^^ = tV'» and AH=. i^-g" ; join £Fand Fig. 167. ' 1' i^ 1 '! i 1 I ' 1 1 1, I I I S iP, io8 THEORY OF STRUCTURES. GH. The ordinate from any point of the broken line EFGH to OA is the Shearing Force at its foot. For example, the ordinate at Z> is — \\" , or — 2200 lbs. Again, the bending moment at B is 3400. 10— 200. 10. 5 = 24,ocx3 Ib.-ft. Choose a vertical scale of measurement so that I inch represents 24,000 Ib.-ft. Upon the vertical through />' take BK=^ I inch. Draw the parabolas OK, AK, with their vertices at points determined as in Example (6). The ordi- nate from any point of the curves OK, AK is the bending moment at its foot. For example, at a point 14 ft. from O the curve ordinate is 1^1^"^ or 25,600 Ib.-ft., and this is the Bending Moment at the same point, being also the greatest for the segment BA. The vertex of AK is, therefore, vertically above the point of which the horizontal distance from 6> is 14 ft. 3. Relation between Shearing Force and Bending Moment. — Let a beam AB be arbitrarily loaded with weights w, , zo^, w^, . . . concentrated at the points i, 2, 3, . . . The^ i/,, at A M, " I = J/, " 2 = i/„ " n = Hence ginning an shearing fi tcrval. Let J J bending m r~i r-H I .11 , Fig. 168. Let a,, a,, a^, . . .he the lengths of the segments Ai, 12, 23, ... , respectively. Let M^i , Mb be the moments at A and B. These moments are of course m/ if the beam merely rests upon supports at its ends. The reaction R at A is given by the equation Rl = wil - a,) + wll -a,-a,)-\-...-^MB-\-M^, 1 being the length of the beam. The shearing force S, between A and i = 7? ; 5, " I " 2 = ^ - 7f , ; .S", " 2 " 3 = /? - zv, - w, ; This re the nuntbc? number an Thus, i: or, the shea of the bcndi The abc shearing fo) at the corr curve. The she 5„ " n-i " n = R-'2{w)\ ^(w) denoting die sum of the first (« — i) weights. .^ Li-; SHEARING ^ORCE AND BENDING MOMENT. IO9 The bending moment Mj at A=M^\ J/, " 2 = R{a, + a,) - w,a, + J/, = i^, + S,^;, ; Hence ///f" difference betzveen the bending moments at the be- ginning and end of any interval is equal to the product of the shearing force (5) for that interval by the length (a) of the in- terval. Let AM denote the difference between any two consecutive bending monnents ; then Mf = Sa. This result has been deduced without any assumption as to the number of the loads. They may therefore be infinite in number and in the limit form a continuous load. Thus, if 5 be the shearing force at a distance x from /3, d_M_ dx = 3; ' or, the shearing force at any point is eqtial to the rate of increase of the bending moment per unit of length. The above results may also be expressed as follows : The shearing force at any point is measured by the tangent of the slope at the corresponding point of the b ending-moment polygon or curve. The shearing force is positive^ zero, or negative according as R = ^{w) ; < wT't ^■\ » 'J ilt fiix^. i 1 :.iS' •■. .;,i. 1 < i ) V \ 'if \\ ' \ I I mi ■I 11 ■t/ HI '\\\V i 111 I 1 '' 1 ■■ no THEORY OF STRUCTURES. ^(w) being the sum of the weights up to the point under con- sideration. In the case of a continuous load, of intensity cb, f H = f wdx. Thus the bending moment iJ/at the same point is a maxi- mutn (or a minimum in certain special cases) when the shear- ing force changes sign, i.e., when S — o. Again, with an arbitrarily distributed load and with a continuous load Thus the difference between the ordinates of the bending- moment diagram at any point and A is proportional to the area of the shearing-force diagram between the same points. From this result an important deduction may at once be made. The bending moment M^ at any point between r and r + i distant x from r is M., = R{a,-\-a^-\- . . . -\-a,-\-x)—wSa^-\-a^-{- . . . -{•ar-\-x)-\- . . . -WrX-\-M^ = My -\- x{R — zv^ — w^ — . . . ~ Wr) Now My^.^ = My -\- «!,.+i5,.+i, and therefore 5,.+, is zero if Mr^^ = My, and also M^ = My = My^^. Thus, the bending moment is the same at every point between r and r + i . (i^^d the case is one of simple bending without shear, as, e.g., with a carriage-axle. 4. T( Let a sin OA of 1 supports The ■ at B dist and is tl points be IV accorc or 0. U take OD ( force at a; the latter tical dista Also, t R, - W -- distance b to OD. Again, Fi tween the { and, its vert Note.—", on both sid with one ha Cor. I. point are m, For exai i I EFFECT OF A ROLLING LOAD. Ill 4. To Discuss the Effect of a Rolling Load. — Case I. Let a single wciglit W travel from left to right over a girder OA of length /, resting upon two supports at O and A, The reaction /?, at O, when W is at B distant x from 0, is W — -. — Fig. 169. and is the Shearing Force for all points between O and B\ it is nil or \V according as the weight is at A or 0. Upon the vertical through O take OD equal or proportional to W\ join DA. The shearing force at any point of tlie beam between O and the weight, as the latter travels from A towards O, is represented by the ver- tical distance between that point and the line AD. Also, the shearing force at any point between B and A is X R^— W = — W-., and is equal or proportional to the vertical distance between that point and the line OE where AE is equal to OD. Again, the Bending Moment at B, when W is at B, is —X ; It IS ml at O and at A : W- it is a maximum and = Wl at the Fig. 170. middle point D. The bending mo- ment at any point of the beam when the weight is at that point is repre- sented by the vertical distance be- tween the point and the parabola OEA, having its axis vertical ]Vi and, its vertex at E, where DE is equal or proportional to . Note. — The shearing and bending actions are symmetrical on both sides of the centre, and it is therefore sufficient to deal with one half of the girder only. Cor. I. The shearing force and bending moment at any point are maxima at the instant the weight passes that point. For example, the shearing force at B for the segment OB, h « ' ' Mi fii 1 it! ■ II iw 112 THEORY OF STRUCTURES. i \ % I I when the weight is at B, is equal or proportional to BC (Fi^r 169), wliich is evidently greater than GH, representing the shearing force at />, when the weight is at any other point G. Again, the bending moment at B (Fig. 170), when W is at /— X B, is W — -. — X. If IV is at any other point G distant a from O, the bendmg moment at B is wa — -. — or Wx — j— , according as rz < or > x, and in either case is greatest when a =. x, i.e., when the weight is at B. Cor. 2. In addition to the rolling load, let the girder carry a permanent weight W at the centre. Consider one half of the girder only, and, for convenience trace the shearing-force and bending-moment diagrams for W below OA. The compound diagram for maximum shearing forces is W DTLFD (Fig. 171), where KT is equal or proportional to — , and KL = OF is equal or proportional to W The maximum shearing force at a point distant x from the centre is represented hy X Y — -—[ — [- .1^) -| . V ^""^^^ ■^T 1 !k \ 1 i 1 1 ! !. F Y L c A / / 1 / 1 D A 1 Y\ Fig. 171. Fig. 17a. Again, the compound diagram for maximum bending mo- ments is OEFO (Fig. 172), where DF is equal or proportional W'l to , and OF is a straight line. EFFECT OF A ROLLING LOAD. 113 u; i The maximum bending moment at a point distant x from the centre is represented by WIP _ I V ~ XY = + 'f(^4 Cor. 3. Theoretically, the total volume of material required in the web of the girder in Cor. 2 is equal or proportional to 2 X ar&ixDTLF 3 Wl I W'l /, being the web unit stress. So, if d be the effective depth of the girder, and / the unit stress in one of the flanges, the total volume of metal in that flange is equal or proportional to 2 X area OEFO 2 WP , WT i Wl' . i W'l^ + + « fd ~ lAfd ^ %fd~6fd^Sfd' Case II. Let a train weighing zv per unit of length travel over the girder from right to left, and let the total length of the train be not less than that of d the girder. The reaction at A, Fig. 173, when the front of the train is at . wx' , B distant x from O, is — j , and is the shearing force for all points between A and B. Upon the ver- F'o- '73. ticals through A and O take AD and OE each equal or pro- wl portional to — . Thus between A and B the shearing force at any point is represented by the vertical distance between that point and a parabola having its axis vertical and its vertex at 0. After the end of the train has passed O, the shearing force at any point of the uncovered portion of the girder is evidently represented by the vertical distance between that point and the parabola AFE, having its axis vertical and its vertex at A. Again, as the train moves from O towards B, the reaction lii ; t ■ I t'l fi ' ii 1 ! ! -; 114 THEORY OF STRUCTURES. \\ \\ i \ :\ I at Af and consequently the bending moment at B^ continually increase. On passing B^ the reaction at A still increases, and the bending moment at B when the train covers a length a of the girder is -^{l-x) - -{a - xf = -^a{2l-a) - -— . This expression is evidently a maximum when a{^l — a) is a maximum, i.e., when a — I. Hence the bending moment, and therefore the flange stresses, at any point are greatest when the moving load covers the whole girder. Cor. I. The shearing force at any point 5 is a maximum when the train covers the longest segment OB. This is evidently the case until the train arrives at B, for the reaction at A^ and therefore the shearing force at B, will continually increase up to this point. When the train passes B and covers a length a(^x) of the girder, the shearing force at B IS — -. zv{a — x). 1VX But this is < — T- , the shearing force at B when OB is 2/ ' d — x^ covered, if — —, — -~->. i^?m > ^G $ y^ ^ y^ -. Fig. 174. ■"I:' EFFECT OF A ROLLING LOAD. nS The maximum shearing force at a point distant x from the centre is represented by XY and is equal to f;(^+')"+ W X. Again, the maximum flange-stresses are obtained by assum- ing the total load upon the girder to be zf 4- w' per unit of length. - ... Ex. The two main girders of a single-track bridge are 80 ft. in the clear and 10 ft. deep. The dead load upon the bridge is 2500 lbs, per lineal foot. If the bridge is traversed by a uniformly distributed live load of 3000 lbs. per lineal foot, determine the maximum bending moment and shearing force at a point of the girder distant 10 ft. from one end. The bending moment at any point is a maximum when the train covers the whole of the bridge, in which case the total distributed load is 5500 lbs. per lineal foot, of which each girder carries one half. Thus the reaction at each support = — . 80. = i io,ccX) lbs., and t\\Q bending moment at the given point = iiooooX 10 — 10 X 2750 X 5 = 962,500 Ib.-ft. The shearing force at the given point due to the dead load =^ I loooo — 10x2750 = 82,500 lbs. Tlie shearing force due to the live load is a maximum, when the live load covers the 70 ft. segment, and its value is then 1500x70 ' 2x80 = 45.937ilbs. Hence the total maximum shearing force = 82,500 + 45, 937i = 128,437^ lbs. u\\ , f ,, i^ ;%■ ■Ail 1 ■"■HRin? t i: ' 1 1 ii6 THEORY OF STRUCTURES. 5. Moments of Forces with respect to a given Point Q. — First, consider a single force P,. Describe the force and fu- nicular polygons, i.e., the line 5,5„ and the lines AB, BC. Through the point Q draw a line parallel to S^S^, cutting the lines AB and CB produced in x and J. Drop the perpendiculars EM and ON upon yx and 5,5, produced. Then BM ~ 0N~ ON' .: P,BM =xy.ON. But BM is equal to the length of the perpendicular from Q to the line of action of /*, , and the product xy . ON is, there- fore, equal to the moment of P^ with respect to Q. Hence, if a scale is so chosen that ON =. unity, this moment becomes equal to xy ; i.e., it is the intercept cut off by the two sides of the funicular polygon on a line drawn through the given point parallel to the givcfi force. Next, let there be two forces, P P . Describe the force and fu- nicular polygons 5j5,5, and ABCD. Let the first and last sides {AB and DC) be produced to meet in G, and let a line through the given point Q parallel to the line 5,5, intersect these lines in X and y. Draw GM perpendicular to xy, and ON perpendicular to 5,5^. Then xy SjS, _ resultant of /*, and P, Gll^ ON ^ "ON ' Fig. 176. II' r'F" MOMENTS OF FORCES. 117 and henc6 (the resultant of P, and P^ X GM — xy . ON. But GM is equal to the length of the perpendicular from Q ipon the resultant of /*, and P^, which is parallel to S^S, and must necessarily pass through G. Hence, if a scale is so chosen that ON =^ unity, xy is equal to the moment of the forces with respect to Q ; i.e., it is the intercept cut off by the first and last sides of the funicular polygon on a line drawn through the given point parallel to the resultaftt force. A third force P, may be compounded with /*, and /*, , and the proof may be extended to three, four, or any number of forces. The result is precisely the same if the forces are parallel. The force polygon of the n parallel forces P^, P^, . . . P„ Fio. 177. becomes the straight line S^S^S^ . . . 5„ . Let the first and last sides of the funicular polygon meet in G. Drop the perpen- iliculars GM, ON upon xy and S,S„, xy, as before, being the intercept cut off on a line through the given point Q parallel to S^S„ . Then xy . ON = GM. S^S„ . Hence, etc. Thus the moment of any number of forces in one and the same plane with respect to a given point may be represented by the intercept cut off by the first and last sides of the funicu- i 1 >' f V ) \'i i i. t -s 'i I r .!'■ f H: ■ ii8 THEORY OF STRUCTURES. lar polygon on a line drawn through the given point parallel to the resultant of the given forces. 6. Bending Moments. — Stationary Loads. — Let a hori- zontal beam AB, supported at A and B, carry a number of weights -P,, /*,, /',,... at the points iV,, N,, N,, ... - Fronr Fig. 178. The force polygon is a vertical line 1234 . . . n, where 12 = /',, 23 = P„etc. Take any pole O and describe the funicular polygon A,A^A^ . . . Let the ^rst and /ast sides of this polygon be produced to meet in G and to cut the verticals through A and B in the points Cand D. - Join CD, Let the vertical through G cut AB in L and CD in AT; LG is the line of action of the resultant. Draw OH parallel to CD. From the similar triangles OvH and GCK, ■ ' 0H~ CK' R. , K be But L Henci Thus //«r CD d one is equi Let it point Mo side of M In the correspont sides of th OH, and y these sides the forces bending m from O up Hence, unity, the b ccpt 071 the CD and tht 7- Mov action of m of a railwa) determine t loads. It I wheels whi( apart. At any \ BENDING MOMENTS. From the similar triangles OnH and GDK, nH GK 119 OH ~ DK \H DK nH BL R, CK ~ AL r: R, , R, being the reactions at A and B, respectively. But iH-\- nH = m ='P, -\- P^-\- . . . = R^-^ R^. Hence iH=R, and nH=R^. Thus the line drawn throtigh the pole parallel to the closing line CD divides the line of loads into two segments, of which the one is equal to the reaction at A and the other to that at B. Let it now be required to find the bending moment at any point M of the beam, i.e., the moment of all the forces on one side of M with respect to M. In the figure these forces are R^, P^, P^, P^, P^, P^, and the corresponding force polygon is //i 23456. The first and last sides of the funicular polygon of the forces are CD parallel to OH, and A^A^ parallel to 06. If the vertical through J/ meet these sides in x and y, then, as shown in Art. 5, the moment of the forces R^, /*, , P^, P^, P,, P^ with respect to M, i.e., the bending moment at M, = ON.xy, ON being the perpendicular from O upon iH produced. Hence, if a scale is chosen so that the polar distance ON is ?aiity, the bending moment at any point of the beam is the inter- cept on the vertical through that point cut off by the closing line CD and the opposite bounding line of the funicular polygon. 7. Moving Loads. — Beams are often subjected to the action of moving loads, as, e.g., in the case of the main girders of a railway bridge, and it becomes a matter of importance to determine the bending moments for different positions of the loads. It may be assumed that the loads are concentrated on wheels which travel across the bridge at invariable distances apart. At any given moment, let the figure represent a beam 1 1 * ! 1 "h 'I .1 M ^ rf ' kHA ^ i '!l A .1! ill 120 THEORY OF STRUCTURES. under the loads Z', , /*, , -P, . . . Describe the corresponding funicular polygon CC'C" . . . D, the closing line being CD. Let the loads now travel from right to left. The result will be precisely the same if the loads remain stationary and if the supports 1 1 are made to travel from left to right. Thus, if the loads successively move through the distances ^^r-?^ .-'>''!./ Fig. 179. 12, 23, 34, . . . to the left, the result will be the same if the loads are kept stationary and if the supports are successively moved to the right into the positions 22, 33, 44, . . . Tli? new funicular polygons are evidently C'C" . . . D' , C"C' ' , . . D' , C"C"" . . . D'", ... the new closing lines being CD', C"D", CD'", . . . The bending moment at any point M is measured by xy for the first distribution, x'y' for the second, x"y" for the third, etc., the position of M for the successive distributions being de- fined by MM' - 12, M'M" = 23, M"M"' = 34, . . . Similarly, if the loads move from left to right, the result will be the same if the loads are kept stationary and if the sup- ports are made to move from right to left. It is evident that the envelope for the closing line CD for all distributions of the loads is a certain curve, called the enve- lope of moments. The intercept on the vertical through any point of the beam cut off by this curve and the opposite bound- MAXIMUM SHEAR AND BENDING MOMENT. 121 ary of the funicular polygon is the greatest possible bending moment at that point to which the girder can be subjected. Example. Loads of 12 and 9 tons are concentrated upon a horizontal beam of 12 ft. span at distances of 3 and 9 ft. from the right-hand support. Find (a) the B. M. at the middle point 12' 9' 6' 3' I » tons Tl8ton8-^| y I ?^ Fig. i8a of the beam, and also (d) the max. B. M. produced at the same point when the loads travel over the beam at the fixed dis- tances of 6 ft. apart. Sca/es for lengths, ^ in. = I ft. ; for forces, ^ in. = i ton. Take polar distance = f in. = ic tons. Case a. B.M. = ;r>/ X 10 = 3.15 X 10 tons = 31^ ton-ft. Case b. B. M. —x'y' X 10 = 3.6 X 10 tons = 36 ton-ft. 8. Analytical Method of Determining the Maximum Shear and Bending Moment at any Point of an Arbitrarily Loaded Girder AB. — At any given moment let the load con- sist of a number of weights w, , w, , . . . w„, concentrated at points distant a, , rr, , . . . a„ , respectively, from B. The corresponding reaction R^ at a is given by R,l — %v,a, + w^a^ + . . . + w^a^, /being the length of the girder. Let W^ = zv,-\-'^i-\' • ' ' + ^« > the sum of the n weights. " Wr — 'w^-\-%v^-\-...-{-Wr^ the sum of the first r w'ts. The shear at a point P between the rth and the (r -\- i)th weights is ' 5, = /?, — w, — M/, — . . . — w^ = /?, — W'^ . li t ' twwww I u il'1 li^l' f :-: :. 122 THEORY OF STRUCTURES. Let all the weights now move towards A through a distance X, and let / of the weights move off the girder, q of the weights be transferred from one side of P to the other, and s new weights, viz., w„+, , w„+, , . . . w„+, , advance upon the girder, their distances from B being a;„+, , ^„+, , . . . «„+,, respectively. Let L-= w^-^-w^-^- . . . -\-Wp, the total weight leaving the girder. Let T = w4, + Wyj^ 4" • • • + "^r+q , the total weight trans, ferred from one side of P to the other. Let Rpl = w,a, -]- w^a,-{- ...-}- '^p^p . " RJ = W„+,rt«+, + Zf«+:^„+, -f . . . + W„+.«„+,. Thus Rp, R^, R, are the icart'' ns a' •' due, respectively, to the weight which leaves the girder, the weight which is trans- ferred, and the new weight which advances upon the girder. The reaction R, at A with the new distribution of the loads is given by RJ = w^+,Kn -\-^) + •^ph{^p+2 + ^) + . . . + wXar + x) . + zi>„^,a„+, = RJ - Rpl +x{W„-L) + RJ, and hence (R, -R,)l= (/?. - RpY +x{W^^ L). Also, the corresponding s/iear at P is S, = R,- (m/^, + m'^+, -f . . . + w, + w^, + • • • + Wh^) = 7?,-(IK-i^+n , . Hence the s/iear at P with the first distribution of weights is greater or less than the s/iear at the same point with the second distribution according as or R,~W,>R,-W,-\-L-T, or .. T-L^R,- R,, M. or Note.— R,,Rp, an according j i.e., accord! through w/i ivcight on t Again, The bn weights is i]/, = RH - = Ril- The hem tribution is M, = Rll- = Rli - Hence tl of weights ii same point ^ or HI- -2)- MAXIMUM SHEAR AND BENDING MOMENT. 123, or T - L%R,- R.^'^i^W^- L). (A). Note. — When no weights leave or advance upon the girder, R,, Rp, and L are severally nil, and hence according as c > c •^1 < 'jj » , i.e., according as the weight transferred divided by the distance through which it is transferred is greater or less than the total zvcight on the girder divided by the span. Again, let z be the distance of P from B, and let ' • R^l = zv,a^ -\- w^a^ -{-... ■\- w^a^ . The bending moment at P with the first distribution of weights is il/, = R,{1 — Z) — W^a^ — 2) — Ws(^a — 2) — . . . — W^ttr — Z) = Ril-z)-RJ-[-zW,. The bending moment at the same point with the second dis- tribution is M, = R,{1 —s)— w^,(a^. -\-x-z)- w^+3^+, -\-x — z)—... — w^itr -\-x — z) — . . . — %v^^^{arJ,^ -\-x — z) = Rll-z) - {RJ-Rpl+RJ) -{x- z){Wr -L+T). Hence the bending moment at P with the first distribution of weights is greater or less than the bending moment at the same point with the second distribution according as M,>M,, or Ril-.z)-Rrl-\-zWr%Rll-z)-.{R,-Rp + R,)l r^;i :i; M ' H ' , ii I. 1 !!' -'if' (Ii ». . ' I ' » .' I i ,'> t- .i-'l'lfi Ii •. ! 124 or THEORY OF STRUCTURES. zW,-{R,-R^l^{x-z){Wr^L-\-T)\{R,--R:){J'-z) or z{L -T+R.- R,)^l{R, - R.) + x{Wr -L+T) >^M-^){IV„-L) (B) NoU. — If no weights leave or advance upon the girder R,, Rp and L are severally nil, and M,%M,, according as . -zT^lR,^x{Wr^T^-fl-z)W^. , If also the point P coincide with the rth weight, and the distance of transfer, x^ = a,. - ^,.^, , then RJ, = WrJ^^a^^ , T = w,.^., , and 2 = Ur. Hence J/, ^ J/, , according as or l-a<'l ' i.e., according as t/te sum of the first r weights divided by the length of the corresponding segment is greater or less than the total weight upon the girder divided by the span. If the weights are concentrated at the panel point3 of a truss, the last relation may be expressed in the form first (r) weights ^ total weight r panels "^ total number of panels' Example. A series of loads of 3000, 23,600, 20,100, 21,700, MA 22,900, 1 8,5 over a truss Let Ap panel poin compare th lbs. has rea( weights hav R.= 9 Hence 5", ^ , 300 and Let the v Hence 5, ^ 5 23600 — o^ and Hence the weight of 300 Again, let moment at / when the wei towards A. First. 2 = gjiiiii; MAXIMUM SHEAR AND BENDING MOMENT. 125 22,900, 18,550, 18,000, 18,000, and 18,000 lbs. travel, in order, over a truss of 240 ft. span and ten panels. Let Ap^p^ . . . B be the truss, p^, P^, Pt, . . . being the panel points. Let the loads travel from B towards A, and compare the shear in the panel/,/, when the weight of 3000 lbs. has reached />, with the shear in the same panel when the weights have advanced another 24 ft. i?. = i- . 18550 = 1855 lbs., Rp = o, ^ = --, W„ = 91300 lbs., Z = o, T = 3000 lbs. Hence 5, ^ Sj, according as (see A) 3000 - o>r,i855 + —(91300 - o)^ 10985, and • • Oj ^^ o* • Let the weights again advance 24 ft. I X \ R, = — . 18000 = 1800 lbs., Rp = o, - — — , 10 / 10 IF„ = 109,300 lbs,, Z = o, T = 23,600 lbs. Hence 5", ^ 5, , according as (see A) 23600 — O^ 1800 — o + —(109300 — o), or 23600^ 12730, and • • 0| ^^ Oq • Hence the shear in the panel p^p^ is a maximum when the weight of 3000 lbs. is at /», . Again, let the 3000 lbs. be at/< , and compare the bending moment at p^ with the bending moment at the same point when the weights have advanced first 24 ft. and then 48 ft. towards^. First. z=. \20 ft., Z = o, T — 22,900 lbs., R,l= 18000X24, mmm'WwTr^ 1 ||||:f ' V ■ lim m I a ■ II ii 126 THEORY OF STRUCTURES. Rp = 0, i?/ = 22900 X 96, ;jr = 24 ft., PT^ = 68400 lbs., W„ = 145,850 lbs. . Hence M^'^M^, according as (see B) .120(0 — 22900 + 1800 — o) -f 22900 X 96 — 18000 X 24 4- 24(68400 — o + 22900) ^ -—(240 — 120X145850 — 0), 240 or and 1425600^ 1750200, .-. M^, is a maximum when the weight of 3000 lbs. is at /, , i.e., when all the panel points are loaded. 9. Hinged Girders. — Any point of a girder at which the bending moment is nil is termed a point of contrary flexure, and on passing such a point the bending moment must neces- sarily change sign. Consider a horizontal girder resting upon supports at A, B, C, D, and hinged at the points E and F in the side spans. In order that there may be no distortion by the turning of I ;the hinges, the latter must not be subject to any bending] action ; i.e., they must be points of contrary flexure. HINGED GIRDERS. 127 Let AE = a,EB^ 3, BC - c, CF — e, DF = d. Let W„ U\, 1V„ W„ W, be the loads upon AE, EB, BC DF, FC, respectively, and let ^r, , x^, x^, x^, x^ be the several distances of the corresponding centres of gravity from the points E, B, C, F, C. Fig. 181. The two portions AE and DF zx& evidently in precisely the same condition as two independent girders of the same lengths, carrying the same loiids and supported at the ends. EF may also be treated as an independent girder supported at B and C, carrying the weights W^, W^, W^, and loaded at the cantilever ends E and F with weights equal to the reactions at E and F for the portions AE, Z^/^ assumed to be independent girders. Let R^, R^, R^, R^ be the reactions at A, B, C, D, respec- tively. Then ?nd R,a= lV,x„ R,d=W,x,. Hence, since R^ and R^ are always positive, there can be no upward pull either at A or D, and no anchorage will be needed at these points. Next, taking EF as an independent girder, the load atE=lV,-R,= W^.(i - ^) ; « " F= W,-R,= W^i - ^. ■■■m .: ^^.: pi it ! Ill ' mm Iff i >'. 128 THEORY OF STRUCTURES Take moments about C and B. Then -{W,- R,){b -\-c)- Wlx, + c) + R,c - W,x, + W,x, + ( W, - R,)e = o. and -{W,- R,)b - IV^x, + WXc - X,) - R,c -f- ^^^ + ^) + {K- ^0(^ + e) = o; two equations giving /?, and i?,, since ^, and R^ have been already determined. The pier moments P, at B and /*, at C are P, = - ( ^F. - /?,)* - W^, = W,-{a - X,) - W^, and />,=:_ (J^, - R)e - U\x, = - W,-^{d - X,) - W,x, ; their values depending solely upon the loads on the spans contain- ing the hinges. The bending moment at any point in BC distant x from B = R^x-{W,- R,){b 4-x)- Wlx, -\-x)-M = P^^x{R,-{-R,- W,- W,)-M; M being the bending moment due to the load upon the length x. The shearing-force and bending-moment diagrams for the whole girder can now be easily drawn. For any given loads upon the side spans, let AEH and DFL be the bending-moment curves for the portions AB, CD ; BH and CL representing the pier moments at B and C, respec- tively. The bending moments for the least and greatest loads upon BC will be represented by two curves HKL, IIK'L, and the distances TT', VV through which the points of contrary flexure must move, indicate those portions of the girder which are to be designed to resist bending actions of opposite signs. HINGED GIRDERS. 129 Again, let the two hinges be in the intermediate span. Let AB = a,BE=i b, EF = c, FC = e, CD = d. Let W,, W„ W„ JV,, W, be the loads upon AB, BE, EF, CD, CF, rtively, and let ,r, , x^, x,, x^, x^ be the several distances ^i the corresponding centres of gravity from the points B, B, F, C, C. Fig. i8a. EF evi(^ 'tly may be treated as an independent girder sup- ported at wo ends and carrying a load W^. ^^ anu ^^'^ may be treated as independent girders carry- ing the loads W^ , W^ and W^, W^, respectively, and also loaded at the cantilever ends E and F with weights equal to the reac- tions at E and F due to the load W^ upon girder EF, which is assumed to be independent. Thus the load at ^ = PF,- ; : " = '■'.(■ - 7)- The pier moments P, at B and P, at C are and P.= W,x,+ wii-'^)e; their values depending solely upon the loads on the span contain- ing the hinges. 130 THEORY OF STRUCTURES. V Let R,, Rj, Rg, R, be the reactions at A, B, C, D, respec- tively, and take moments about the points B, A, D, C. Then R,a - W,x, + [w,x, + W,x!^ = o = ^.^ - W,x, + P, ; - R,a + W,{a - X,) + Wla + a',) + VV,x rt + 3 = o; R,d -w{i- p){e + d)- IVXx, + a') - W,{a -x,) = o: - R,d -wii- ^).- - W,x, + W,x, = o = - R,d + W,x, - P,. R^ and R, are always positive; A', \s/>ositivi' or ncgaiivc according as W,^:, ^ P, ; and R, " " " " W,x,->P,. Thus there will be a downward pressure or an upward pull at each end according as the moment of the load upon the ad- joining span is greater or less than the corresponding pier mo- ment. The ends must therefore be anchored down or they will rise off their supports. ' The shearing-force and bending-moment diagrams for the whole girder can now be easily drawn. Let HEFL be the bending moment curve for any given load upon the span BC, BH and CL being the pier moments at B and C, respectively. The bending-moment curves for the least and greatest loads on the side spans may be represented by curves A TH, ATM and DVL, DV'L, and the distances TT, VV through which the points of contrarj' flexure move indicate those por- tions of the girder which are to be designed to resist bending actions of opposite signs. Reverse strains may, however, be entirely avoided by making the length of EF sufficiently great as compared with the lengths of the side spans. The preceding examples serve to illustrate the mechanical principles governing the stresses in cantilever bridges. I. A be upon a suj liDrizontal segment. Sliow til 150 lbs. is s iiig-force ai 2. A ma anci the cigl if the man i 3. Atiml: reartion at t 4. Two h beam suppor of the bolls V weight whicl of maximum same weight 5. In the load which tl Draw the 6. A unifc 1,'roiind and tl at Oo° to the bending men: uniformly disi bLiim should 2000 lbs. may Draw curv EXAMPLES. in EXAMPLES. 1. A beam 20 ft. long and weighing 20 lbs. per lineal foot is placed upon a support dividing it into segments of 16 and 4 ft., and is kept horizontal by a downward force /' at the middle point of the smaller segment. Find the value of /' and the reaction at the support. Show that the required force /' will be doubled if a single weight of 150 lbs. is suspended from the end of the longer segment. Draw shear- ing-force and bending moment diagrams in both cases. Arts. 1200 lbs. ; 1600 lbs. 2. A man and eight boys carry a stick of timber, the man at the end and the eight boys at a common point. Find the position of this point, if the man is to carry twice as much as each boy. Ans, Distance between supports = ^ length of beam. 3. A timber beam is supported at the end and at one other point; the rcaftion at the latter is double that at the end. Find its position. Ans. Distance between supports = ^ length of beam, 4. Two beams ABC, BCD are bolted at /> and C so as to act as one beam supported at/? and D\ AB — 12 ft., BC — 4 ft., CD = 16 ft. ; each of liie bolls will bear a bending moment of 100 lb. -ft. Find the greatest weight which can be concentrated o 1 the portion BC. Draw diagrams of nia-ximum shearing force and bending moment when a wheel of the same weight rolls over the beam. Ans. \\^^ lbs. 5. In the preceding question find the greatest uniformly distributed load which the beam will bear. Draw the shearing-force and bcr>ding-moment diagrams. Ans. ilW lbs. 6. A uniform beam 20 1/3 ft. in length rests with one end on the s,'round and the other against a smooth vertical wall ; the beam is inclined at 60° to the vertical and has a joint in the middle which can bear a bending moment of 30,000 lb. ft. Find the greatest load which may be uniformly distributed over the beam. Also find how far the foot of the biani should be moved towards the wall in order that an additional :ooo lbs. may be concentrated at the joint. Draw curves of shearing force and bending moment in each case. Ans. 8000 lbs. ; distance = 10 ft. \ >i\. 1 J fifiiititrt (ili. ;'•■ ''« -\- -r '!> i^ "■ \ ! I mm ', i "■I'la I m 'J' 132 THEORY OF STRUCTURES. 7. A man of weight W ascends a ladder of length / which rests against a smooth wall and the ground and is inclined to the vertical at an angle a. The ladder has « rounds. Find the bending moment ;ii the nil round from the foot when the man is on the/th round from the foot. (Neglect weight of ladder.) Ans. ^'pl ,.. . ->i sm a. C« + I)' 8. A regular prism of weight W and length a is laid upon a beam of length 2/(>rt). If the prism is so stiff as to bear at its ends only, show- that the bending action on the beam is less than if the bearing were con- tinuous from end to end oi the prism. Ans. — 1st. Max. B.M. 2d. 9. A railway girder, 50 ft. in the clear and 6 ft. deep, carries a uni- formly distributed load of 50 tons. F'ind the maximum shearing stress at 20 ft. from one end, when a train weighing ij tons per lineal foot crosses the girder. Also find the minimum theoretic thickness of the web at a supper. 4 tons being the safe shearing inch-stress of the metal. Ans. i6Jtons; .195 in. 10. A beam is supported at one end and at a second point dividing its length into the segments ;// and ;/. Find the two reactions. Also finrl the ratio of m to n which will make the maximum positive moment equal to the maximum negative moment. w — ,— «'), -_— (/« + nf , m : n :: I + v 3 : r 2. w Ans. — (;«' 2m 2>n II. One of the supports of a horizontal uniformly loaded beam is at the end. Find the position of the other support so that the straining of the beam may be a minimum. length Ans. Distance from end support — V2 12. A rolled joist 17 ft. long is supported at one end and at a point 13 ft. distant from* that end. Two wagon-wheels 5 ft. apart and eacli carrying a load of 1300 lbs. pass over the joist. Find the maxiniiini positive and negative moments due to these weights, and also the corre- sponding reactions. Ans. Max. positive B. M. = 5512J Ib.-ft. ; reactions = 1550 and 1050 lbs. Max. negative B. M. = 5200 lb. -ft. ; reactions = 1700 lbs. and — 400 lbs. or = 2900 lbs. and — 300 lbs. Denotin diagram for 13. A u supports so 14. Two su^ ed u disic it X fr each /ft. in X must lie. 15. A uni supports at given point OP:OQ::C Draw cur points of the 16. A bea mW and «^ supports. S according as 17- A whe tlie wheel in and find its v 18. Two \ n\V x.ox\% m b the bending i whose distant / < a[^ + |/. '9- Find tl ai the two enc one end to w EXAMPLES. 133 Denoting the distance from a support by x, the max. positive B. M. diagram for each half of the 13-ft. span is given by Mx = 100(21 — 2x)x. 13. A uniformly loaded beam rests upon two supports. Place the supports so that the straining of the beam may be a minimum. Ans, Distance cf each support from centre = /( i ]. \ V2/ 14. Two bars AC, CB in the same horizontal line are jointed at C and suy ed upon two props, the one at A, the other at some point in CB disu it X from C. The joint C will safely bear n Ib.-ft. ; the bars are each /ft. in length and w lbs. in weight, Find the limits within which X must lie. Wl ± 2« Ans. I ywlT 2«' 15. A uniform load PQ moves along a horizontal beam resting upon supports at its ends A and B. Prove that the bending moment at a given point O is a maximum when PQ occupies such a position that OP:OQ:\OA : OB. Draw curves of maximum shearing force and bending moment for all points of the beam. 16. A beam is supported at the ends and loaded with two weights rnlV and nW at points distant a, b, respectively, from the consecutive supports. Show that the bending action is greatest at in IV or nW A- fnyb accordmg as - ^ — . 17. A wheel supporting 10 tons rolls over a beam of 20 ft. span. Place the wheel in such a position as to give the maximum bending moment, and find its value. Ans. At the centre ; 50 ton-ft. 18. Two wheels a ft. apart support, the one vilV tons, the other n ir tons, m being > «, and roll over a beam of / ft. span. Show that the bending moment is an absolute maximum at the centre or at a point whose distance from the nearest support is na 2{in + n) according as / ^ (M I 4- V 1. and find its value in each > \ ^ m + nj case. m Wl , VI +«,,,(, na ) ' , Ans. ton-ft. ; 7- W \l \ ton-ft. 4 4/ ( ;« -I- « \ 19. Find the max. B. M. on a horizontal beam of length / supported ui the two ends and carrying a load which varies in intensity from w at one end to w + px at the other. ' I ill I ; If 134 THEORY OF STRUCTURES. ').■ S. 20. Four wheels each carrying 5 tons travel over a girder of 24 ft. clear span at equal distances 4 ft. apart. Determine, graphically, the ma.x B. M. at 8 ft. from a support, and also the absolute max. B. M. on the girder. Ans. ^\^ ton-ft. ; 80 ton-ft. 21. Two wheels each supporting 7 tons roll over a beam of ^\ ft. span. Find the maximum bending moment for the whole span, and also the curve of the maximum bending moment at each point when the wheels are 4 ft. apart. Ans. Abs. max. B. M. = Yo'- ton-ft. at wheel at 2\ ft. from one ' end. Denoting the distance from support by x, the max. B. M. curve for the first 3^ ft. is given by ^/:« = ij(ii — 2x)x, and for the remaining 4 ft. by M. = \\{n-x)x. 22. Two wheels supporting, the one 11 tons, the other 7 tons, travel over a beam of 1 2\ ft. span. Find the maximum bending moment for the whole span, and also the curves of the max. shearing force (both positive and negative) and maximum bending moment at each point when the wheels are 6 ft. apart. Ans. Abs. max. B. M. = 37.2 ton-ft. I The max. positive shearing force at each point is given by the equations S,= 183— iS .r and S.t = 7(12^ -.r) I2i ' The max. negative %\\e&x'\n^ force at each point is giv^en by the equations s --2^ 5,= 45i iS.r S. = - 42 + i8.r S.= I2i' - 12i ' - I2i The max. B. M. curve is given by the equations 183— i8a- , ,, ii(i2i ^ x and Afx = - JXX -V) 124 I2i N.B. — In the above cases x is measured from the support to the nearest load. 23. In the preceding questio.i show that the maximum negative ?,\\ta.r at 4t^ ft. from a support, when the 7-ton wheel only is on the beam, is the same as the maximum negative shear at the same point when botli of the wheels maximum the ii-ton beam, and 24. Sol 1250 lbs. (: Ans. 25. Thre roll over a b give the ma: 26. Place any point b( its value. give the sam value. 27. Four roll over a be give the max Am 28. All tL B- M. at the c that for a par the beam, wheels. EXAMPLES. 135 the wheels are on the beam, and find its value. Also show that the maximum negative shear at ^ ft. from a support is the same when only the ii-ton wheel is on the beam as when the two wheels are on the beam, and find its value. Ans. If J tons ; -\V^ tons. 24. Solve question 22 when the beam carries an additional load ot 1250 lbs. (= % ton) per lineal foot. / 7418^ \ Ans. Abs. max. B. M. is at 5.284 ft. I = ft. 1 from support. Max. positive shearing-force diagram is given by Sx = 18.54625— 2.065^: from -«r==otojir=6ift., and 5^=14.90625— 1.505.^ from X — 6\ to ;r= 12^ ft. The max. negative shtaring-force diagram is given by 5« =: — .56^: from x = to X = \i\ ft.; = 3.64 — 1.44^ from x = 4/^ tox = 6i ft. ; = '/.54625 — 2.065^ from .r = 6J to jt = 6^ ft.; "^ 3.90625 — 1.505X from -r = 6J to;r = 9jit.; = 9.18625 — 2.065.V from jr = 9f to X = I2| ft. Max. B. M. curve is given by j]/x = (18.54625 — 1.7525^).^, and Mx = (14.90625 — i.i925.r).r. 25. Three wheels, each loaded with a weight IV and spaced 5 ft. apart, roll over a beam of 12 ft. span. Place the wheels in such a position as to give the maximum bending moment, and find its value. Ans. Middle weight at centre of beam ; 4 IV. 26. Place (rt) the v/heels in the preceding question so that B.M. at any point between the two hindmost wheels may be constant, and find its value. Also (d) determine all the positions of the wheels which will give the same bending moment at 6 and 12 ft. from one end, and find its value. Am: — (a) ist wheel at i ft. from support ; B. M. = 7 IV. (d) When distance between end wheel and sup- port is ^ 2 ft. and ^ 5 ft.; B. M. = 7 IV. 27. Four wheels each loaded with a weight IV and spaced 5 ft. apart roll over a beam of 18 ft. span. Place the wheels in such a position as to give the maximum bending moment, and find its value. Ans. One wheel off the beam and middle wheel of remaining three at the centre ; max. B. M. = 8J IV. If all wheels are on beam, max. B. M. = 8 IV. 28. All the wheels in the preceding question being on the beam, the B. M. at the centre for a certain range of travel is constant and equal to that for a particular distribution of the wheels when only three are on the beam. Find the range, the B. M., and the position of the three wiieels. d k_ mti ;--*^*iriaAii ^l*''^^ 1^ 136 THEORY OF STRUCTURES. Ans. While the end wheel travels 3 ft. from the support ; 8 W; first wheel 5 ft. from the support. 29. A span of / ft. is crossed by two cantilevers fixed at the ends and hinged at the centre. Draw diagrams of shearing force and bending moment (i) for a single weight W 2X the hinge, (2) for a uniformly dis- tributed load of intensity iv. Ans. Taking hinge for origin, the shearing-force and bending- nioment diagrams are given by (I) M:c-=- An (3) Sx = iDX ; Mx ivx' ,@ i •, 30. A beam for a span of 100 ft. is fixed at the ends. Hinges are in- troduced at points 30 ft. from each end. Draw curves of shearing force and bending moment (i) when a weight of 5 tons is concentrated on each hinge ; (2) when a uniformly distributed load of \ ton per linear foot covers (a) the centre length, {b) the two side lengths, ic) the whole span. Ans. Take a hinge as origin ; the diagrams are given by — (i) For each side span Sx =■ I, Mx=- — 'ix; for centre span Sx = o, Mx = o. (2) — (a) For each side span Sx = 5 5 5 Af.r=--x: for centre span Sx = — x 8' Mx=^^x- i6" 15 x"" ip) For each side span 5jr = „-, Mx - for centre span Sx = o, Mx = o. t. X 2^ x^ (c) For each side span Sx = — -t- V . ^x = x + -> ; for centre soan ■^■'-2 8 Mx = —x 2 • 2 i6 31. If the load on each of the wheels in question 27 is 5 tons, and if the beam also carries a uniformly distributed load of 20 tons, and two loads of 2 and 3 tons concentrated at points distant 5 and 9 ft., respec- tively, from one end, find the maximum shearing force (both positive and negative) and the maximum bendmg moment for the whole span; also tind the loci for the maximum shearing force and bending moment at each point. 32. A rol carries a uni apart, the on joist. Find t ft. from each 33- A bea of rn IV lbs. a from the righ the weights i How will the 34- A roll< carries the fou end. Find th both positive j Ans. 35- Solve I lineal foot is wheels. 36. The loa a beam of 6c 16,900, 16,900, EXAMPLES. 137 Ans. Denoting the distance from support by x, the max. positive shearing-force diagram is given by Sx = ^ — ^x from ^ = o to JT = 3 ; 5i- = V" — \x from jr = 3 to .r = 8 ; 5j: = -i^/ — \x from ;ir = 8 to 5-^ to x=\Z ft. The ^ = 13 ; "^-^ = 5 — 78 ^'■o'" •*' = '3 max. negative shearing-force diagram is given by Sx = — -^^x from X -=0 to x = 5 ; 5j: = f f — |jr from ;r = 5 to .1' = 10 ; ^.i- = 'i.' — 1-^ from -r = 10 to x = 1 5 ; 50 20.r 5j: = -7- -g- from .r = 15 to;r = 18, Max. positive / shear = -'/ tons ; max. negative shear = ^ tons ; max. bending-moment curve is given by Mx = --^-x — ^§-x'^ from X = o to .r = 3 ; ^/.^ = ^^x — 4f.r* from .r = 3 to ;>r = 5 ; Mx^ H^x — ^'.r' — 1 5 from .r = 5 to jr = 8 ; .^/!r = ^^i-x — \\x''-\- 12 from jr = 8 to JT = 9 ; abs. max. B. M. = 142 ton. -ft. 32. A rolled joist weighing 150 lbs. per lineal foot and 20 ft. long carries a uniformly distributed load of 6000 lbs., and two wheels 5 ft, apart, the one bearing 5000 lbs. and the other 3000 lbs., roll over the joist. Find the maximum shears at the supports, at the centre, and at 5 ft. from each end. Ans. 10,250 lbs. ; 9750 lbs. ; 3250 lbs. ; 6750 lbs.; 6250 lbs. 33. A beam lit. long and weighing w lbs. per lineal foot has a load of m W lbs. at a ft. from the left end and a load oi nW lbs. at b ft. from the right end. Find the shearing forces and bending moments at the weights and at the middle of the beam, a and b being each < — . / How will the result be affected '\{b> — ? 'J. 34. A rolled joist weighing 450 lbs. pe*" lineal foot and 20 ft. long carries the four wheels of a locomotive at 3, 8, 13, and 18 ft. from one end. Find the maximum bending moment and the maximum shears, both positive and negative, the load on each wheel being 10,000 lbs. Ans. Max. B. M. = 102,000 lb. -ft. ; max. shears = 19,000 lbs. and 21,000 lbs. 35. Solve the preceding question when a live load of 2\ tons per lineal foot is substituted for the four concentrated weights on the wheels. 36. The loads on the wheels of a locomotive and tender passing over a beam of 60 ft. span are 14,180, 14,180, 21,260, 21,260, 21,260, 21,260, 16,900, 16,900, 16,900, 16,900 lbs., counting in order from the front, the M , ) '^ 4^ -u f r .1 ' ^ Hi 5 ' J [ i I f V ¥ ■ i^ i i- ■" if III': : i; ;, . t _ _ , 1 11 138 THEORY OF STRUCTURES. intervals being 5, 5^, 5, 5, 5, 8|, 5, 4, 5 ft. Place the wheels in such a position as to give the maximum bending moment, and find its value. Also find the maximum bending moments for spans of 30, 20, and 16 feet. Ans. For 60-ft. span, max. B. M. is at 5th wheel and = 1,559,925.4 Ib.-ft. when ist wheel is 7.95 ft. from support. For 30-ft. span, max. B, M. at 5th wheel when 2d wheel is .596 ft. from support and = 436,761.4 Ib.-ft. For 2o-ft. span, max. B. M. at centre when 3d wheel is 2^ ft. from support and = 212,600 Ib.-ft. = max. B.M. at same point when 4th wheel is 5 ft. from support. For_i6-ft. span, max. B. M. is at 5th wheel and = 132,875 Ib.-ft. when 4th wheel is 5 ft. from sup- port. 37. If the 60-ft. beam in the preceding question also carries a uni- formly distributed load of 60,000 lbs., find the curves of maximum shearing force and bending moment at each point. 38. If a beam is supported at the ends and arbitrarily loaded, show that the ordinate at the point of maximum moment divides the area of the curve of loads into two parts which are equal to the supporting forces. If a and b are the distances of the centres of gravity of the parts from the ends of the beam, and if W is the total weight on the beam, I show that the maximum bending moment is W ■ & 39. A span of / ft. is crossed by a beam in two half-lengths, sup- ported at the centre by a pier whose width may be neglected. The suc- cessive weightson the wheels of a locomotive and tender passing overthe beam are 14.000, 22,000, 22,000, 22,000, 22,000, 14,000, 14,000, 14,000, 14,000 lbs., the intervals being T\, \\, 4J, \\, loj, 5, 5, 5 ft. Place the wheels in such a position as to throw the greatest possible weight upon the centre pier, and find the magnitude of this weight for spans of (1) 50 ft.; (2) 25 ft.; (3) 20 ft.; (4) 18 ft. 40. Loads of 3f, 6, 6, 6, and 6 tons follow each other in order over a ten-panel truss at distances of 8, 5J, 4*, and \\ ft. apart. Appiy the results of Art. 8 to determine the position of the loads which will give the maximum diagonal and flange stresses in the third and fourth panels. 41. A truss of 240 ft. span and ten panels, has loads of 12^, 10, 12, 11, 9, 9, 9, 9, and 9 tons concentrated at the panel points. Find, by scale measurement, the bending moments at the four panel points which are the most heavily loaded, and determine by Art. 8 whether these are the greatest be travel over 42. Loa( beam of 2 respectively moment at the given di 43- A be the intermei load upon tl 10 tons unif( FC, 30 tons at the midd FC = 10 h.; points of infl 44- Four girder of 24 1 left support, centre of th( distances apj jected. 45- Three are placed up 'lie left abut left abutment weights trave which the bet Ans. EXAMPLES. 13^ greatest bending moments to which the truss is subjected as the weights travel over the truss at the panel distances apart. 42. Loads of 7i, 12, 12, 12, 12 tons are concentrated upon a horizontal beam of 25 ft. span at distances of 18, 108, 164, 216, and 272 in., respectively, from the left support. Find, graphically, the bcndiny; moment at the centre of the span. If the loads travel over the truss at the given distances apart, find the maximum B. M. at the same section. 43. A beam ABCD is supported at four points A, B, C, and D, and the intermediate span ^C is hinged at the two points E and F. The load upon the beam consists of 15 tons uniformly distributed over AB, 10 tons uniformly distributed over BE, 5 tons uniformly distributed over FC, 30 tons uniformly distributed over CD, and a single weight of 5 tons at the middle point of EF. AB = 1$ ft.; BE = ^ ft.; EF= is ft.; FC = 10 it. ; CZ> = 25 ft. Draw curves of B, M. and S. F., and tind points of inflexion. 44. Four wheels loaded with 4, 4, 8, and 8 tons are placed upon a girder of 24 ft. span at distances of 3 in., 6^ ft., 8J ft., and 9 ft. from the left support. Find by scale measurement the bending moment at the centre of the girder. If the wheels travel over the girder at the given distances apart, find the maximum B. M. to which the girder is sub- jected. 45. Three wheels loaded with 8, 9, and 10 tonsand spaced 5 ft. apart, are placed upon a beam of 15 ft. span, the 8-ton wheel being 3 ft. from the left abutment. Determine graphically the B. M. at 6 ft. from the left abutment. Also find the greatest B. M. at the same point when the weights travel over the beam, and the ads. max. bending moment to which the beam is subjected. Ans. 47| ton-ft.; 53I ton-ft.; abs. max, B. M. = 56^^ ton-ft. at 2d wheel when ist is 2^ ft. from support. '1, I .1 Mil' ui . ! ? 1 > ■ i! I CHAPTER III. DEFINITIONS AND GENERAL PRINCIPLES. I. Definitions. — The science relating to the strength of materials is partly theoretical, partly practical. Its primary object is to investigate the forces developed within a body, and to determine the most economical dimensions and form, con- sistent with stability, of that body. Certain hypotheses have to be made, but they are of such a nature as always to be in accord with the results of direct observation. The materials in ordinary use for structural purposes may be termed, generally, so/td bodies, i.e., bodies which offer an ap- preciable resistance to a change of form. A body acted upon by external forces is said to be strained or deformed, and the straining or deformation induces stress amongst the particles of the body. The state of strain is simple when the stress acts in one direction only, and the strain itself is measured by the ratio of the deformation to the original length. The state of strain is compound when two (or more^ stresses act simultaneously in different directions. A strained body tends to assume its natural state when the straining forces are removed : this tendency is called its elas- ticity. A thorough knowledge of the laws of elasticity, i.e., of the laws which connect the external forces with the internal stresses, is absolutely necessary for the proper comprehension of the strength of materials. This property of elasticity is not possessed to the same degree by all bodies. It may be almost absolute, or almost zero, but in the majority of cases it has a mean value. Hence it naturally follows that solid bodies may be classified between two extreme, though ideal, states, viz., 140 a perfectly elastic bod forms exa( fectly soft resistance t Bodies ( tioii under 2. Strej jected to fi\ {a) A lo {/>) A lo (e) A sh stress tendi which it is i (d) A tr (e) A tw Under a elastic defoi formation, c 3. Resrj Let a straig stretched or distributed the line of a let / be the < formation. If the tn pared with lim'ts, the fo and to the a: these quantil where ^ is a and is called dently the fc elastic bar of ELEMENTARY PRINCIPLES OF ELASTICITY. 141 a perfectly elastic state and a perfectly soft state. Perfectly elastic bodies which have been strained resume their original forms exactly, when the straining forces are removed. Per- fectly soft bodies are wholly devoid of elasticity and ofTer no resistance to a change of form. Bodies capable of undergoing an indefinitely large deforma- tion under stress are said to ht plastic. 2. Stresses and Strains. — Every body may be sub- jected to five distinct kinds of stresses, viz. : {a) A longitudinal pull, or tension. {b) A longitudinal thrust, or compression'. [c) A shear, or tangential stress, which may be defined as a stress tending to make one surface slide over another with which it is in contact. {(i) A transverse stress. {c) A twist or torsion. Under any one of these stresses a body may suffer either an elastic deformation, of a temporary character, or a plastic de- formation, of a permanent character. 3. Resistance of Bars to Tension and Compression. — Let a straight bar of homogeneous material and length L be stretched or compressed longitudinally by a force P uniformly distributed over the constant cross-section A of the bar ; let the line of action of P coincide with the axis of the bar, and let / be the consequent extension or compression, i.e., the de- formation. . If the transverse dimensions of the bar are small as com- pared with the length, experiment shows that, xvitliin certain limits, the force P is directly proportional to the deformation / and to the area A, and inversely proportional to the length L, these quantities being connected by the relation where ^ is a constant dependent upon the material of the bar and is called the coefficient or modnhts of elasticity. It is evi- dently the force which will double the length of a perfectly IrilMP ' "[, elastic bar of unit section. Denoting the unit stress i-jj by/", \ !ii ! '■■ *. I- » jS S40 THEORY OF STRUCTURES. and the strain per unit of length (-yj by A, the above equation . may be written or the unit stress = E times the unit strain. Thus the equation is the analytical expression of Hooke's law, that for a body in a state of simple strain the strain is pro- portional to the stress. The longitudinal strain is accompanied by an alteration in the transverse dimensions, the lateral unit strain being — — , m where ;;/ is a coefficient which usually varies from 3 to 4 for solid bodies and is approximately 4 for the metals of construc- tion. In the case of india-rubber, if the deformation is small, in is about 2. Generally the deformation may be calculated per unit of original length without sensible error, but for india-rubber it i-; more accurate to make the calculation per unit of stretcJud length (= -7-^)- lateral strain \-\-M The ratio — = m longitudinal strain is called Poisson's ratio. If the transverse dimensions of a bar under compression are small as compared with the length Z, a slight disturbing force will cause the bar to bend sideways, and the bar will be subject -d to a bending action in addition to the compression. If the bar is to be capable of resisting a direct thrust only, the ratio of L to its least transverse dimension should not exceed a certain limit depending upon the nature of the ma- terial. For example, experiment indicates that this limit should be about 5 for cast-iron, 10 for wrought-iron, 7 for steel, and 20 for dry timber. If the temperature of the bar is raised /°, the consequent strain is at, a being the coefficient of linear dilatation ; and a stress EoLf will be developed if a change of length is pre- vented. 4ii4'«i>ii'fi;j SPECIFIC WEIGHT.— COEFFICIE^TT OF ELASTICITY. 143 4. Specific Weight ; Coefficient of Elasticity ; Limit of Elasticity ; Breaking Stress. — Before the strength of a body can be fully known, certain physical constants, whose values depend upon the material, must be determined. {(i) Specific Weight. — The specific weight is the weight of a unit of volume. The specific weights of most of the materials of construction have been carefully found and tabulated. If the specific weight of any new material is required, a conven- ient approximate method is to prepare from it a number of regular solids of determinate volume and weigh them in an ordinary pair of scales. The ratio of the total weight of these solids to their total volume is the specific weight. It must be remembered that the weight may vary considerably with time, etc.; thus a sample of greenheart weighed 69.75 lbs. per cubic foot when first cut out of the log, and only 57 lbs. per cubic foot at the end of six months. When the strength of a timber is being determined, it is important to note the amount of water present in the test-piece, since this appears to have a great influence upon the results. The straining of a structure is generally largely due to its own weight. The total load upon a structure includes all the external forces applied to it, and in practice is designated dead {perma- nent) or live {rolling), according as the forces are gradually ap- plied and steady, or suddenly applied and accompanied with vibrations. For example, the weight of a bridge is a dead load, while a train passing over it is a live load ; the weight of a roof, together with the weight of any snow which may have acc'inuL. upon it, is a dead load ; 7t tenacity of Wohler fou but alternat repetitions LIMIT OF ELASTICITY. 145 The coefficients of direct elasticity for the different metals and timbers are sometimes obtained by subjecting bars of the material to forces of extension or compression, or by observing the deflections of beams loaded transversely. The coefficients for blocks of stone and masonry might also be found by trans- verse loading ; they are of little, if any, practical use, as, on account of the inherent stiffness of masonry structures, their deformations, or settlings, are due rather to defective work- manship than to the natural play of elastic forces. The torsional coQ^cxcnt of elasticity, i.e., the coefficient of elastic resistance to torsion, has been shown by experiment to vary from two fifths to three eighths of the coefficient of direct elasticity. {e) Limit of Elasticity. — Whe'i the forces which strain a body fall below a certain limit, the body, on the removal of the forces, will resume its original form and dimensions without sensible change (disregarding any effects due to the develop- ment of heat) and may be treated as perfectly elastic. But if the forces exceed this limit, the body will receive a permanent deformation, or, as it is termed, a set. Such a limit is called a /////// of elasticity, and is the greatest stress that can be applied 1:0 a body without producing in it an appreciable and permanent deformation. rhis is an unsatisfactory definition, as a body passes from the elastic to the non-elastic itate by such imperceptible degrees that it is impossible to fix any exact line of demarca- tion between the two states. Fairbairn defines the limit more correctly, as the stress below which the deformation is approxi- mately proportional to the load which produces it, and beyond which the deformation increases much more rapidly than the load. In fact, both the elastic and ultimate strengths of a ma- terial depend upon the nature of the stresses to which they are subjected and upon X.\\c frequency of their application. For ex- ample, in experimenting upon bars of iron having an ultimate tenacity of 46,794 lbs. per sq. in. and a ductility of 20 %, Wohler found that with repeated stresses of equal intensity, but alternately tensile and compressive, a bar failed after 56,430 repetitions when the intensity was 33,000 lbs. per sq. in. ; a «l lll.^iLt iii 146 THEORY OF STRUCTURES. second bar failed only after 19,187,000 repetitions when the intensity was 18,700 lbs. per sq. in. ; while a third bar remained intact after more than 132,000,000 repetitions when the inten- sity was 16,690 lbs. per sq. in. These experiments therefore indicated that the limit of elasticity ior the iron in question, under repeated stresses of equal intensity, but alternately tensile and compressive, lay between 16,000 and 17,000 lbs. per sq. in., which is much less than the limit under a steadily ap- plied stress. Similar results have been shown to follow when the stresses fluctuate from a maximum stress to a minimum stress of the same kind. Generally speaking, then, the limit of elasticity of a ma- terial subjected to repeated stresses, is a certain maximum stress below which the condition of the body remains unim- paired. Bauschinger's experiments indicate that the application to a body of any stress, however small, produces a plastic or permanent deformation. This, perhaps, is sometimes due to a want of uniformity in the material, or to the bar being not quite straight initially. In any case, the deformations under loads which are less than the elastic limit, are so slight as to be of no practical account and may be safely disregarded. The main object, then, of the theory of the strength of materials, is to determine whether the stresses developed in any particular member of a structure exceed the limit of elasticity. As soon as they do so, that member is permanently deformed, its strength is impaired, it becomes predisposed to rupture, and the safety of the whole structure is threatened. Still, it must be borne in mind that it is not absolutely true that a material is always weakened by being subjected to forces superior to this limit. In the manufacture of iron bars, for instance, eacli of the processes through which the metal passes changes its elasticity and increases its strength. Such a material is to be treated as being in a new state and as possessing new properties. The strength of a material is governed by its tenacity and rigidity, and the essential requirement of practice is a tougli material with a high elastic limit. This is especially necessary for bridges and all structures BREAKING STRESS. 147 liable to constantly repeated loads, for it is found that these repetitions lower the elastic limit and diminish the strength. In the majority of cases, experience has fixed a practical limit for the stresses, much below the limit of elasticity. This insures greater safety and provides against unforeseen and accidental loads, which may exceed the practical limit, but which do no harm unless they pass beyond the elastic limit. Certain operations have the effect of raising the limit of elasticity : a wrought-iron bar steadily strained almost to the point of its ultimate strength and then released from strain and allowed to rest, experiences an elevation both of tenacity and of the elastic limit. If the bar is stretched until it breaks, the tensile strength of the broken pieces is greater than that of the bar. A similar result follows in the various processes employed in the manu- facture of iron and steel bars and wires : the wire has a greater ultimate strength than the bar from which it was drawn. Again, iron and steel bars, subjected to long-continued com- pression or extension, have their resistance increased, mainly because time is allowed for the molecules of the metal to as- sume such positions as will enable them to oiTer the maximum resistance ; the increase is not attended by any ap- preciable change of density. Under an increasing stress a brittle material will be fractured without any great deformation, while a tough material will become plastic and undergo a large deformation. (^) Breaking Stress. — When the load upon a material increases indefinitely, the material may merely suffer an increasing deformation, but generally a limit is reached at which fracture suddenly takes place. Cast-iron is perhaps the most doubtful of all materials, and the greatest care should be observed in its employment. It possesses little tenacity or elasticity, is very hard and brittle, and may fail sud- ^'°' '^*' deiily under a shock or an extreme variation of temperature. Unequal cooling may predispose the metal to rupture, and its w ill 11 ill ll.. If ■? 1 1 i ■ 148 THEORY OF STRUCTURES. strength may be still further diminished by the presence of air-holes. Cast-iron and similar materials receive a sensible set even under a small load, and the set increases with the load. Thus at no point will the stress-strain curve be absolutely straight, and the point of fracture will be reached without any great change in the slope of the curve and without the development of much plasticity. Wroiight-iron and steel are far more uniform in their be- havior, and obey with tolerable regularity certain theoretical laws. They are tenacious, ductile, have great compressive strength, and are most reliable for structural purposes. Their strength and elasticity may be considerably reduced by high temperatures or severe cold. When a bar of such material is tested, the stress-strain curve (/ = ± ETC), as has already been pointed out, is almost absolutely straight within the elastic limit, e.g., from O lo A in tension and from O to B in com- pression. As the load increases beyond the elastic limit, the in- creasing deformation becomes plastic and permanent, and the stress-strain diagram takes an ap- preciable curvature between the limits A and B and the points D and E corresponding to the maxi- mum loads. In tension, as soon as the point D is reached, the bar rapidly elongates and is no longer able to sustain the maximum load, its sectional area rapidly dimin- ishes, and fracture ultimately takes place under a load much less than the maximum load. The point of fracture is represented in the figure by the point F the ordinate of F being the actual Jiltiuiatc final load on the bar intensity of stress — — ■- -j—. — j -■ — . •^ -^ area of fractured section Fig. iB^. vt* BREAKING STRESS. The exact form of the stress-strain curve-between D and F is unknown, as no definite relation has been found to exist be- tween the stress and strain during the elongation from D to F. Ordinarily, the hrtaking tensile stress has been defined to be the maximum lead applied divided by the initial sectional area of the bar; but this, although convenient, is manifestly in- correct. It is important to note that, as the deformation gradually increases under the increasing load, the molecules of the ma- terial require greater or less time to adjust themselves to the new condition. During the tensile test of a ductile material there is, at some point beyond the elastic limit, an abrupt break GH in the continuity of the stress-strain curve, the curve again becom- ing continuous from H to D. The point G has been called the Yield Point or the Breaking-down Point, and the deformation from H onward is almost wholly plastic or permanent. In compression there is no local stretch as in tension, and there is consequently A no considerable change in the curvature of Fig. ise. the compression stress-strain curve up to the point of fracture. Timber is usually tested by being subjected to the action of tensile, compressive, or transverse loads. Other character- istics, however, must be known before a full conception of the strength of the wood can be obtained. Thus the specific weight must be found ; the amount of water present, the loss in drying, and the corresponding shrinkage should be deter- mined ; the structural d'x^QV&ncQS of the several specimens, the rate of growth, etc., should be observed. The chief object of experiments upon masonry and orick- icork is to discover their resistance to compression, i.e., their crushing strength. In fact, their stiffness is so great that they may be compressed up to the point of fracture without sensible change of form, and it is therefore very difficult, if not impos- sible, to observe the limit of elasticity. iTTir ii I ! * .i I 150 THEORY OF STRUCTURES. The cement or mortar uniting the stones and bricks is most irregular in quality. In every important work it should be an invariable rule to prepare specimens for testing. The crushing strength of cement and of mortar is much greater than the ten- sile strength, the latter being often exceedingly small. Hence it is advisable to avoid tensile stresses within a mass of masonry, as they tend to open the joints and separate the stones from one another. Attempts are frequently made to strengthen masonry and brickwork walls by inserting in the joints tarred and sanded strips of hoop-iron. Their utility is doubtful, for, unless well protected from the atmosphere, they oxidize, to the detriment of the surrounding material, and, besides this, they prevent an equable cistribution of pressure. They are, how- ever, far preferable to bond-timbers. The working load (or stress, or strength) is the maximum stress which a material can safely bear in ordinary practice, and depends both upon the character (see Art. 5, below) of the stress and upon the ultimate strength of the material, the ratio of the ultimate or breaking stress to the working stress being usually called a factor of safety. For example, the factor is about 3 for long-span iron bridges, or bridges having great weight as compared with the live load (a moving train). 4 for ordinary iron bridges. 5 for ordinary metal shafting. 8, 10, and even more for long struts and members subjected to repeated stresses of varying magnitude. 10 is also generally taken to be the factor of safety for timber. Under a steady, or a merely statical load, even as great as |- of the breaking stress, a member of a structure may prob- ably not be unsafe. 5. Wohler's Law. — It is now generally admitted that variable forces, constantly repeated loads, and continued vibra- tions diminish the strength of a material, whether they pro- duce stresses approximating to the elastic limit, or exceedingly small stresses occurring with great rapidity. Indeed many engineers design structures in such a manner, that the several members £ of the evil Although 1 tacitly ack first to giv vation and " That of a bar, tl peated strt differences affected in produce Ixi This • la ments the great rapid resulting st to the rapi stress, and subject for The exf repetitions ( rapid than t depends boi cnce or flue, The efife nately tensii in Art. 4. Bars of t bore 31,132 moved betw wlien the stt The tabl ments on ste The axle- when subjecl site in kind, ; were of the shearing stre VVOHLER'S LAW. «5i members arc strained in one way only, so convinced are they of tlie evil effect of alternating tensile and compressive stresses. Although the fact of a variable ultimate strength had thus been tacitly acknowledged and often allowed for, Wohler was the first to give formal expression to it, and, as a result of obser- vation and experiment, enunciated the following law : "That if a stre^.s t, due to a static load, cau' : the fracture of a bar, the bar may also be fractured by a series of often-re- peated stresses, each of v.'hich is less than t\ and that, as the differences of stress increase, the cohesion of the material is affected in such a manner that the minimum stress required to produce fracture is diminished." This -law is manifestly incomplete. In Wohler's experi- ments the applications of the load followed each other with great rapidity, yet a certain length of time was required for the resulting stresses to attain their full intensity ; the influence due to the rapidity of application, to the rate of increase of the stress, and to the duration of individual strains still remains a subject for investigation. The experiments, however, show that the rate of increase of repetitions of stress required to produce fracture, is much more rapid than the rate of decrease of the stresses themselves, and depends both upon the maximum stress and upon the differ- ence ox fluctuation of stress. The effect of repeated stresses of equal intensity, but alter- nately tensile and compressive, has been already pointed out in Art. 4. Bars of the same material repeatedly bent in one direction, bore 31,132 lbs. per square inch when the load was wholly re- moved between each bending, and 45,734 lbs. per square inch when the stress fluctuated between 45,733 lbs. and 24,941 lbs. The table on page 152 gives the results of similar experi- ments on steel. The axle-steel was found to bear 22,830 lbs. per square inch, when subjected to repeated shears of equal intensity but oppo- site in kind, and 29,440 lbs. per square inch, when the shears were of the same kind. It would therefore appear that the shearing strengths of the metal in the two cases are about ^ 152 THEORY OF STRUCTURES. of the strengths of the same metal under alternate bending and under bending in one direction, respectively. Character of Fluctuation. Alternating stresses of equal intensity . . . . Complete relief from stress between each bending Partial relief from stress between each bending Maximum Resistance to Repeated Stresses in lbs. per square inch. Axle-steel. 29,000,-29,000 49,890, O 83,110, 36,380 .S|prinf;-steel (un- hardened). 52,000, 93,500, 62,240 From torsion experiments with various qualities of steel, the important result was deduced, that the maximum resistance of the steel to alternate twisting was f of the maximum resist- ance of the same steel to alternate bending. ' Wohler proposed 2 as a factor of safety, and considered that the maximum permissible working stresses should be in the ratios of 1:2:3, according as members are subjected to alternate tensions and compressions (alternate bending), to tensions alternating with entire relief, or to a steady load. The weakening of metal by repeated stresses has been called / If 3 is the factor of safety, 3= 1100(1 +-^0) (6) Example i. — The stresses upon a bar of Phoenix axle-iron, normal to its cross-section, vary from a maximum tension of 50000* to a minimum tension of 20000*. Determine the admis- sible stress per cent." and the necessary sectional area. By (4), / , I 20OOO\ „ , ^ =:. 70o( I + - — -- j = 840* per cent.', and 2 50000 _ 50000 50000 .*. F = ; — = „ = 59.52 sq. centimetres. 840 Let/ be the dead load and q the total load, per lineal unit of length, upon the flanges of roof and bridge trusses. :_ili.-. -I ( ]■ ^ t i \u 156 • THEORY OF STRUCTURES. P .-. = - , and equations (4) and (6) respectively become b= 700(1 +i^) (7) *=uoo(,+f,|).. ..... (8) Ex. 2. — Determine the limiting stress per cent." for the flanges of a wrought-iron lattice girder when the ratio of the dead load to the greatest total load is By (7). ZV ^ = 700(1+^-1) = 800*. 7. Weyrauch's Formula. — Let a bar of a unit sectional area be subjected to stresses which are alternately different in kind, and which vary from an absolute numerical maximum a {= max. £) of the one kind to a maximum a" (= max. B') of the other kind. f Let a' -^ a" = d = the maximum numerical difference of stress. = 0'. Let a max. ±( a! ~ max. B If a" = 0, a' — d —u. If a" == s, d a = s = —. 2 By Wohler's Law ' a' cxd =/d, (9) / being an unknown coefficient of which the value remains to be determined. If a' =■ u, If a' = s, WEYRAUCH'S FORMULA. 157 Weyrauch's assumption, viz., / -, satisfies 2u — s — a' these extreme conditions, the most reliable results of the few experiments yet recorded, and also Wohler's deduction that a' diminishes as d increases and vice versa. Hence (9) becomes n — s a' = 2n — s — a id = n — s 211 — s — a -{a' + a"\ and / u — sa"\ I u — s ,\ •.a' = u[i -j = u[i 0'). (10) This is Weyrauch's formula, and it may be always applied to those cases in which a member is subjected to stresses alter- nating between tension and compression, or due to shearing actions in opposite directions. In the Phoenix iron experiments already referred to it was found that s = 1 1 70* per cent.' ; u u 7_ 15 Taking 7^ = 2100* as before, and making » w ^_ I ~ 2 (10) becomes «' = 2100(^1 -j. ..... (11) If 3 is the factor of safety, = 700I I I (12) Weyrauch considers 3 to be the proper factor of safety for bridges and similar structures. It is also a suitable factor for the parts of machines subjected to determinate straining actions. A larger factor will be required when other con- tingencies have to be provided against. I .fi ^3 IlM 158 THEORY OF STRUCTURES. i. In the steel experiments, Wohler found that s — 2050* per cent.' ; u — s u _5_ 12 Taking u = 3300* and s = 1800*, u — s 5 u II and (10) becomes a' = 3300(1 - j\(f' } (13) If 3 is the factor of safety, d = 1100(1 -^j \ m wwW m • Ui 160 THEORY OF STRUCTURES. which fluctuate between a tension of 14 tons and a compression of 6 tons. Find the statical strength of the iron. a „ i4-(- -6) 20 b — 3 ' 3 d = fluctuation of stress = t = 10.17 tons per sq. in. 9. Remarks upon che Values of t, u, s, and b. — As yet the value of u in compression ' is not been satisfactorily detcr- nsined, and for the present its value may be assumed to be the same both in tens'on and compression. If, as Wohler states, "repeated stresses" are detrimental to the strength of a material, then the values of u and s diminish as the repetitions increase in number, and are minima in struc- tures designed for a practically unlimited life. Only a very few of Wohler's experiments give the values of ^, u, s, and a, so ihat Launhardt's and Weyrauch's assumptions for the value of /must be regarded as tentative only, and re- quire to be verified by further experiments. The close agree- ment of Wohler\> results from tests upon untempered cast-steel (Krupp), with those given by Launhardt's formula, may be seen from the following : For /= 1 100 centners* per sq. zoll, Wohler found that u = 500 centners per sq. zoll. Thus (2) becomes «, 500 I + 5 ^^, ). and .*. a' — 500^?, — 6oo\ (17) V and in being certain coefifiicients which depend upon the nature of the material and also upon the manner of the loading. Consider three cases, the material in each case being wrought- iron : {a) Let the stresses vary between a maximum tension and an equal maximum compression ; then and 0=1, ,•. b = 700(1 — i) = 350* per cent.'. {l>) Let the material be subjected to stresses which are either tensile or compressive, and let it always return to the original unstrained condition ; then min. B = o, or max. B' = o, and .•. = 0. .'. b = 700(1 ± o) = 700* per cent.'. {(-) Let the material be continually subjected to the same dead load ; then min. B = max. B, I' i '! : i 1; \i ■f 3: h ; Itllf ■fi \i i i ' 162 and THEORY OF STRUCTURES. yoo{i -f- i) = 1050* per cent.' = 14,934 lbs. per sq. in., which is one third of the ultimate breaking strength, viz.. 1050* per cent.'. Thus in these three cases the admissible stresses are in the ratios of 1:2:3, ratios which have been already adopted in ma- Chine construction as the result of experience. VVohler, from his experiments upon untempered cast-steel (Krupp), concluded that for alternations between an unloaded condition and either a tension or a compression, b = 1 100, and for alternations between equal compressive and tensile stresses, d = 580. In America it has often been the practice to take max. B -\- max. B' a' -f- a" 700 7CX) for stresses alternately tensile and compressive, it being as- sumed that if the stresses are tensile only, their admissible values may vary from o* to 700* per cent.". Since z= —r, .-.a = a 700F a' 700 ,,and.-.^ = -p = ^-^-,. (US) i. h I I, 560, 467, 400, 350, 612, 525. 437. 350. 1 + Comparing this with (12), for 0' = o, ; (18) gives d — 700, and (12) gives fi = 700, 10. Flow of Solids. — When a ductile body is strained beyond the elastic limit, it approaches a purely plastic con- dition in which a sufificiently great force will deform the body indefinitely. Under such a force, the elasticity disappears and the material is said to be in a ^uu/ state, behaving preciscl)- like a fluid. For example, it flows through orifices and shows a contracted section. The stress developed in the material is called \.\\Q flvid pressure or coefficient of fluidity. The general principle of the flow of solids, deduced by Trcsca, may be enunciated as follows: A pres motion of This g in materia ing, rolling ably it als certain ext Rails u have acqui due to the the rails an of solids ar and in the of fluidity disappear a In punc at first com CDmmences brought un( illustration I 75 inches, length of th flow must h; shearing thi really shorn iiuasure A t area and th( bserved by -le fractured j,nven them a 'ie tound cur '•ottom to th 'ating planes and rolling th In experi lates, one ab '65, with a h( lead was alwa u mi FLOW OF SOLIDS. J63 A pressure upon a solid body creates a tendency to the relative motion of the particles in the direction of least resistance. This gives an explanation of the various effects produced in materials by the operations of wire-drawing, punching, shear- ing, rolling, etc., and in the manufacture of lead pipes. Prob- ably it also explains the anomalous behavior of solids under certain extreme conditions. Rails which have been in use for some time are found to have acquired an elongated lip at the edge. This is doubtless due to the flow of the metal under the great pressures to which the rails are continually subjected. Other examples of the flow of solids are to be observed in the contraction of stretched bars and in the swelling of blocks under compression. The period of fluidity is greater for the more ductile materials, and may disappear altogether for certain vitreous and brittle substances. In punching a piece of wrought-iron or steel, the metal is at first compressed dSidi flows inwards, while the shearing only commences when the opposite surface begins to open. A case brought under the notice of the author may be mentioned in illustration of this. The thickness of a cold-punched nut was 1.75 inches, the nut-hole was .3125 inch in diameter, and the length of the piece punched out was only .75 inch. Thus the flow must have taken place through a depth of I inch, and the ^lu.iring through a depth of .75 inch. Hence the surface iLilly shorn was n X .3 '25 X .75 =■ .736 sq. in. in area, and a niLiisurt A the shearing action is the product of this surface area and the fluid pressure. The nature of the flow may be bserved by splitting a cold-punched nut in half and treating iiu' fractured surfaces with acid, after having planed them and Ljivrn them a bright polish. The metal bordering the core will i)c lound curved downwards, the curvature increasing from the Dottom to the top, and well-defined curves will mark the sepa- rating planes of the plates which were originally used in piling and rolling the iron. In experimenting upon l-^ad. Tresca placed a number of >lates, one above the other, in a -trong cylinder. Fig. 188, page i6>, with a hole in the bottom. Upon applying pressure the lead was always f uund to flow when the coefficient of fiuidity m \mm ■ 1 1 164 THEORY OF STRUCTURES. I was about 2844 lbs. per sq. in., the difference of stress beini; double this amount. The separating planes assumed curved forms analogous to the corresponding surfaces of flow when water is substituted in the cylinder for the lead. The flow of ductile metals, e.g., copper, lead, wrought-iron, and soft steel, commences as soon as the elastic limit is ex- ceeded, and in order that the flow may be continuous the dis- torting stress must constantly increase. On the other hand, in the case of truly plastic bodies, flow commences and con- tinues under the same constant stress. It evidently depends upon the hardness of the material, and has been called the co- efficient of hardness. The /(3;/^rr the stress acts the greater is the deformation, which gradually increases indefinitely or at .1 diminishing rate. Experiment shows that there is very little alteration in the density of a ductile body during its plastic deformation, anil Tresca's analytical investigations are based on the assumption that the body is deformed without sensible change of volume. Consider a prismatic bar undergoing plastic deformation. Let L be the length and A the section of the bar at com- mencement of deformation. Let Z -|- .r be the length and a the section of the bar at a subsequent period. Let/ be the intensity of the fluid pressure. Since the volume remains unchanged, LA={L± x)a, (i) the positive or negative sign being taken according as the bar is in tension or compression. Let /^, be initial force on bar. Let P be force on bar when its length \s L ±. x. Then and the fc -L- y CD of a cyl D. A hole oi the fac( flows under by a piston, pressed to the correspc First, as the mass rei If dx be DO correspc length of th Integrati .)' = o when P.=pA. P=pa, and hence Pa L P, ~ A" L±x . . (2) Hence P{L ± x) = P,L = a constant, . . . . (3) Second, a (d/r transfer '-^ L'N'lindrical FLOW OF SOLIDS. 165 V and the force diminishes a > the bar stretches and increases as the bar contracts under pressure. If equation (3) be referred to rect- angular axes, the ordinates repre- senting different values of P and the abscissc-E the corresponding values of x, the stress-strain dia- '■% grams, tt in tension and cc in corri- pression, are hyperbolic curves, having as asymptotes the axis of X, XOX, and a line parallel to the axis of _;v at a distance from it equal to the length L of the bar. Next consider a metallic mass (e.g., lead) resting upon the end CD of a cylinder of radius R, and filling up a space of depth D. A hole of radius r is made at the centre of the face CD, through which the mass flows under the pressure of fluidity exerted by a piston. When the mass has been com- pressed to the thickness DO =■ x, let y be the corresponding length KE of the "jet." First, assume that the specific weight of the mass remains constant. If dx be the diminution in the thickness DO corresponding to an increase dy in the length of the jet, then Fig. 187. nICdx -\- Ttr^dy = 0. (0 Integrating eq. I, and remembering that 1' = o when X =. D, R\D — x)- ry ^ o. (2) Second, assume that the cylindrical portion EFGH \% gradti- ally transformed into NMPLKQN, of which the part PMNQ is cylindrical, while the diameter of the part PLKQ gradually otic too THEORY OF STRUCTURES. increases from the face of the cylinder to KL ( = EF\ at tlie end of the jet. Then n{R*-r')dx amount of meta! which flows into the central cylinder = 27trdrx, (3) dr being the depth to which the metal penetrates. Third, assume that the diminution of the diameter of the cylindrical portion PMNQ is directly proportional to the said diameter. Then, if 2 be the radius of the cylinder PQNM, dr dz r ~ z ' By eqs. (3) and (4), X 2 Integrating, {R'-r')\og,x = 2r'\og,z-{-c, c being constant of integration. When X = D, z — r, .'. (^"-01og,-^- = 2rMog,|, (4) or /f»-r» 2 _ lx\ ='■' 7~ \DI (5) I 1 WORK. By eqs, (2) and (5), ^' 187 (6) which is the equation to the profile PL or QK. Note. — If K' = sr", eq, (6) represents a straight line. '' F^ = 2r\ " " " parabola. II. Work. — Work must be done to overcome a resistance. Thus bodies, or systems of bodies, which have their parts suit- ably arranged to overcome resistances are capable of doing work and are said to possess energy. This energy is termed kinetic ox potential according as it is due to motion or to posi- tion. A pile-driver falling from a height upon the head of a pile drives the pile into the soil, doing work in virtue of its motion. Examples of potential energy, or energy at rest, are afforded by a bent spring, which does work when allowed to resume its natural form ; a raised weight, which can do work by falling to a lower level ; gunpowder and dynamite, which do work by exploding ; a Leydcnjar charged with electricity, which does work by being discharged ; coal, storage batteries, a head of water, etc. It is also evident that this potential energy must be converted into kinetic energy before work can be done. A familiar example of this transformation may be seen in the action of a common pendulum. At the end of the swing it is at rest for a moment and all its energy is potential. When, under the action of gravity, it has reached the lowest point, it can do no more work in virtue of its position. It has acquired, however, a certain velocity, and in virtue of this velocity it does work which enables it to rise on the other side of the swing. At intermediate points its energy is partly kinetic and partly potential. A measure of energy, or of the capacity for doing work, is the zvork done. The energy is exactly equivalent to the actual work done in the following cases : {a) If the effort exerted and the resistance have a common point of application. H If If'' ill! Iffi^ i ! i ; l\ \i r: li i68 YHEORY OF STRUCTURES. {b) If the points of application are different but are rigidly connected. {c) If the energy is transmitted from member to member, provided the members do not change form under stress, and that no energy is absorbed by frictional resistance or restraint at the connections. Generally speaking, work is of two kinds, viz., internal work, or work done against the mutual forces exerted between the molecules of a body or system of bodies, and external zvork, or work done by or against the external forces to which the bod\- or bodies are subjected. In cases {a), {b), {c) above, the inter- nal work is necessarily nil. As a matter of fact, every body yields to some extent under stress, and work must be done to produce the deformation. Frictional resistances tend to oppose the relative motions of members and must also absorb energy. If, however, the work of deformation and the work absorbed by frictional resistance are included in the term work done, the relation still holds that Energy = work done. A measure of work done is the product of the resistance by the distance through which it is overcome. When a man raises a weight of one pound one foot against the action of gravity he does a certain amount of work. To raise it two feet he must do twice as much work, and ten times as much to raise it ten feet. The amount of work must therefore be propor- tional to the number oi feet through which the weight is raised. Again, to raise two pounds one foot requires twice as much work as to raise one pound through the same distance ; while five times as much work would be required to raise five pounds, and ten times as much to raise ten pounds. Thus the amount of work must also be proportional to the weight raised. Hence a measure of the work done is the product of the number of pounds by the number of feet through which they are raised, the resulting number being designated foot-pounds. Any other units, e.g., a pound and an inch, a ton and an inch, a kilogramme and a metre, etc., may be chosen, and the work done represented in inch-pounds, inch-tons, kilogram- = A' cos 6 arc \n. OBLIQUE RESISTANCE. 169 metres, etc. This standard of measurement is applicable to all classes of machinery, since every machine might be worked by means of a pulley driven by a falling weight. 12. Oblique Resistance. — Let a body move against a resistance R inclined at an angle to the direction of motion |Fig. 189). No work is done against the normal component R sin 0, as there is ^~ no movement of the point of applica- \ tion at right angles to the direction of motion. This component is, there- fore, merely a pressure. The work done against the tangential component R . cos 8 between two consecutive ^"^' '^' points M and A'' of the path of the body is R cos B . MN. Hence the total work done between any two points A and B of the path = 2{R cos e . MN) = f'R cos dds, s being the length of AB. If AB is a straight line (Fig. 190), and if R is constant in direction and magnitude, the total work = R cos . AB = R.AC, .-iC" being the projection of the displacement upon the line of action of the resistance. Let the path be the arc of a circle Rsiae R Fig. 190. Fig, 191. iFig. 191) subtending an angle a at the centre. If R and 6 re- main constant, the work done from A to B = K cos e arc AB = R cos . OA . a = R . OM cosB.a = Rpa = Ma, ■i Is 1 w II ll ■■■: '>■ ; J "•!f" m I? M fjllli ■-'I';' IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I - m 22 1.8 1.25 1.4 1.6 -m 6" — ► Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N. IT. 14580 (716) 872-4503 1^ ^"^'m f/j «^ m 170 THEORY OF STRUCTURES. p being the perpendicular from O upon the direction of /?, and M = Rp being the moment of resistance to rotation. If there are more resistances than one, they may be treated separately and their several effects superposed. In such case, M will be the total moment of resistance and will be equal to the algebraic sum of the separate moments. The normal component R sin 6 produces a pressure. 13. Graphical Method. — Let a body describe a path AB B (Fig. I92)against a variable resistance of such a character that its magnitude in the direction of motion may be repre- sented at any point M by an ordinate MN to the curve CD. Let the path AB be subdivided into a number of parts, each part MP being so small that the resistance from M to P may be considered uniform. The mean r ,. . MN-^PQ J ^ value of this resistance = , and the work done in 2 MN-\-PQ overcoming it = . MP — the area MNQP in the limit. Hence the total work done from A to B — the area bounded by the curves AB, CD and the ordinatcs AC, BD. 14. Kinetic Energy. — The velocity v acquired by a body of weight tt> and mass m in falling freely from rest through the vertical distance // is V = ^2gh ; , W V V* .*. wli = = m—, g2 2 Thus an amount of work zvh is done, and the body possesses the kinetic energy ;;/— . Again, let v' be the velocity of the body after falling through a further distance x, measured vertically. Then w{h -f- ^) = ;//?' and Ktl^Eftt EMEkdV. .*. wx = --\v — tr). 171 Thus the work done in falling through the vertical distance x is zvx, and is equal to the corresponding change of kinetic energy. 15. Example i. Let it be required to determine the 7uork done in stretching or compressing a bar of length L and sec- tional area A by an amount /. Suppose that the force applied to ihe bar gradually in- creases from o until it attains the value /*; its mean value is P P -, and the work done is therefore -/. 2 2 , / But P = EAj ; E being the coefTicient of elasticity. ■i . h % 4 ';■,«!? .*. the work done 2 L J IPVAL E\AI 2 ' This formula is only true for small values of the ratio j. In the case of a compressive force it is assumed that the bar does not bend. P . A suddenly applied force, - , will do as much work as a steady force which increases uniformly from o to /", and hence it follows that a bar requires twice the strength to resist with safety the sudden application of a given load than is necessary when the same load is gradually applied. If /is tho proof stress or elastic limit per unit of sectional / area, -A is the corresponding /rtfn, , . . . iu„ being the masses of the particles. The sum between the brackets is called the moment of in- ertia of the system of particles about the axis and is usually denoted by /. .*. the total kinetic energy = A' I ' I ii b. ,1 Again, it appears from the definition that every moment of inertia is the product of a mass and the square of a length. This length is called the radius of gyration and is usually designated by the symbol k. If M be the total mass of the system, and W the total weight, W S ,',. . i^Mf^y W {Akf , and the total kmetic energy = M = ~. the re- &/ 2 g "2- ,^ M r S r ? >74, THEORY Of structures. suit being the same as if the particles were collected in a ring of radius k, sometimes called the equivalent ring or fly-wheel. Let Ig be the moment of inertia of the system with respect to a parallel axis through the centre of gravity, and let // be the distance between the two axes. Then I, = mih - x,Y + m,{h -x,Y +...-]- mlh - ^„)» = ie'2(in) — 2h'2{mx) + 2(w4r'). Since the new axis passes through the centre of gravity, "^mx = MA. Also, 2{m) = M and 2{mx') = /; .-. /^ = M/i' 4- / - 2M/1' ; .-. I^I,-\-Mk\ So, if /' is the moment of inertia about another parallel axis at the distance //' from the centre of gravity. I' =I,-\-Mir .\i-Mw = r ■ Mh'\ i i , iil' Hence, if the positions of two parallel axes relatively to the centre of gravity are known, and if the moment of inertia about one is given, the moment of inertia about the other can be obtained by means of the last formula. Note. — Nothing has been said as to the number of the par- ticles. They may be infinite in number and infinitely near each other, forming in fact a solid body. The summation 2{mx') is then best effected by integration. 16. Values of A;*. 1. For a rectangular plate of depth d with re- spect to an axis through the centre _ perpendicular to the side d. ^' = — . 2. For a circular plate of radius r with re- , spect to a diameter k* = —, VALUES OF k\ 175 3. For an annulus of external radius r, and internal radius r, with respect to a diameter ^' Note. — If r,— r, = t, and the breadth t of the annulus is small as compared with the radius r, , then _ K 4- r: k' — ' ' ^ ' — '- — -^^ — '-, approx., 4 2 and the area = n{r^ — r/) = 2nr^t, approx. 4. For the plates in (2) and (3) with respect to an axis through the centre perpen- dicular to the plates, the numerators remain the same but the denominator is in each case 2. '5. For a sphere of radius r with respect to a diametCi k^ 6. For a solid cylinder of radius r with re- spect to its axis k^ 7. For an elliptic plate of which the major and minor axes are 2b and id respec- tively : With respect to the major axis /•' With respect to the minor axis k^ 8. For a triangular plate of height // with re- spect to an axis coinciding with the base . k^ 5 r' 2 4 b' 1^ 6" .ill ' 'r I \m \W\ 1" Tll^ • ' I i IH. >■ lj6 THEORY OF STRUCTURES. 17. Momentum— Impulse. — A moving body of weight it' and mass m acted upon in the direction of motion for a time / by a force F will acquire a velocity v which is directly propor- tional to /^and to t, and inversely proportional to w. Hence V — n- Ft w n being some coefificient. If F= tf, the velocity generated in one second is^. and or ••• g=n. Ft Ft .'. v=g— = — , mv = Ft. This is the analytical statement of Newton's Second Law of Motion, which has been expressed by Clerk Maxwell in the following form : " The change of momentum (i.e., the product of the mass and velocity) is numerically equal to the impulse (i.e., the product of the force and the time during which it acts) xvhicU produces it, and is in the same direction." Again, let p be the perpendicular from a fixed axis O upon the direction of motion of the body, and let r be the radius OP to the body. Then imp = Ftp — Fpt = Mt, where M = Fp\ or the change of the moment of momentutn, i.e., of the angular momentum, is equal to the moment of impulse. The above results are also true for two or more bodies or systems of bodies severally acted upon by extraneous forces, and the equations may be written 2mv = 2Ft, 2mvp = 2Fpt = 2Mt. In words, the total change of momentum in any assigned direction is equal to the algebraic sum of the impulses in the same direction. MOMENTUM— IMP ULSE. 177 /'. y^\ y T and the total change of angular momentum is equal to the alge- braic sum of the moments of the impulses. Hence it follows that if two or more bodies or systems of bodies mutually attract or repel each other, and if there are no extraneous forces, the total momentum in «, yy any assigned direction is constant (the principle of the conservation of linear momentum), and the angular momentum / x about a given axis is constant (the prin- ciple of the conservation of angular momentum). /,''' Suppose that the velocity of the body 0*^':" r^. of weight zv and mass m changes from "•^v„ / t', to %\ in the time / under the action of "\ /' a couple of moment M, and let/,,/, be F'g- '93- the corresponding values of p, and r, , r, those of r, Fig. 193. .-. ;«(t'j>, - v^^ = Mt- ' or if w, , w^ are the components of i\, f, in directions perpen- dicular to r, , r,, respectively, m{u\r^ — ii\r^ = Mt. For example, a weight W of water passing through a turbine of external radius r, and internal radius r, has its angular mo- W W mentum changed from — w,r, to — 't^'3^3, w, , ^v.^ being the tangential components of the velocity with which the water enters and leaves the wheel. The water, therefore, exerts W upon the wheel a couple of moment — (tf,r, — w,rj, and if the wheel rotates with an angular velocity A, the work done upon the wheel by the water W W = --^(ic'.r, - w^r^ = — (w,«, - w,u,), u, and «, being the circumferential velocities corresponding to ?•, and r,, respectively. m .,v •' 'viF viiliy' '♦f'' if't i m I I I 178 THEORY OF STRUCTURES. I i8. Useful Work— Waste Work. — Let a body of mass in and weight w pass over the distance s under the action of a force F acting in the direction of motion for a time /, and let the velocity of the body change from ?', to 7',. Assume / to be so small that, for the interval in question, the velocity may be regarded as constant and of the average value -^— - — - ; s = V, + V, t. But Ft = {mv^ — mv^. m Fts=^-{v:-v,y, or tn Fs = -iv: - v,% t Thus Fs, the work done, is equal to the change of kinetic energy in the given interval. If the body is a material particle of a connected system, a similar relation hcilds for every other particle of the system, and the total work done = \{'2mv.^ — ^viv^). A part of thio work may be expended in doing what is called effective work, i.e., in overcoming (i) an external resist- ance, or in doing useful work, and (2) frictional resistance, or in doing xvasted work. Denoting the total effective work by T, and the total motive work by T^ , tne last equation may be written 7;- T, = \{'2mv:-:2mv^\ and the difference between the total motive work and the total effective work is equal to the total change of kinetic energy. In the case of a machine working at a normal speed the velocities of the different parts are periodic, being the same at the beginning and end of any period or number of periods. For any such interval, therefore, v^ = v,, and .•. T„ = T,, so GENERAL CASE. 179 that there is an equality between the motive work and the effective work. 19. General Case.— Let x,, ^, , x;, be the co-ordinates of tlie C. of G. of a moving body of mass A/ with respect to three itctangular axes at any given instant. Let x\, >', , xr, be the co-ordinates of ilic same point after a, unit of time. Let ^, , J, , ^, be the co-ordinates of any particle of mass m at the given instant. Let ^, , j,j, ^, be the co-ordinates of the sdme particle after a unit of time. .-. Mx, = ^Xw;jr,), M^, = ^{my^), M7, = 2{ms,) ; Mx, = 2{mx,), Mj>^ = 2{>'iy^, M'z^ = 2(»ic,)', ••• ^(-i-, - ^'.) = 2w{x, - .r,), M{j', - J,) = 2m{}', - 7,), or Mu =^ 2»iu, Mv = 2f>iv, Mw = 2mw, iiy V, w being the component velocities of the C. of G. at the given instant with respect to the three axes, and u, v, w the component velocities of the particle ;« at the same instant. From these last equations, Mil = 2muu, Mv = 2mz'v, Mza = 2mww. .'. M{h -\- V -\-w)=- 2m{mi -\-vv-{- ivw), which may be written in the form M(u +V 4- ^') + ^>«K« - "^y -\-{v- vY + (w - wf\ MU' + 2m V = 2viv\ or !•! '.i'» n i; ,3 .'I m •■A I - . If--; til i8o TliKOKV or SlKUCrVh'ES V hv'\\\\^ tho resultant vcKnity of the (.'. of li. ; ;•, that of tl\c parliclr ; antl /', that of tlu> paiiide ichitivcly to the C. of ll. The hist equation may he written J/T' i';///" i'w;-' Thus the energy of tiie total mass eolleeteil at the centre oi j^ravity, toj^olher with the enei^jy relatively to the ceiitn- of j;ravity. is equal to the total enerj^y of motion. If the botly revolves arouml an axis throu[;h its C. of G. with an angular veloeitj- ./. the seeoml term of the last e(|ua- tioi\ becomes ~:£mr\r V A* - 2.titr ~ — /, 2 2 ' Total energy of explosion = energ^y of shot -|- energy of recoil ; Energy of shot = energy of translation -|- energy of rotation _ 50 (3020V 500 I f!^-li Jo_2oy /3i535\' ~ 32.- ' 2 "^ 32:2 ■ 2 • V -iV ' ■ AO.\%i \ 12 I = 31680124.2+97758.6 = 31777882.S ft.-lbs. ; ., 36 X 2240 (16JV , , ,^ Emrgy of ncoil = ^~^:;^~ . — ,— = 330652.1 ft.-lbs. 0"~ ' m CENTKIFUGAI. lOS'CE. l8l ilciicc, if Cbe the energy of i lb. of powder, ' C. 300 = 3i777«H2.8 4- 330652.1 = 32108534.9 ft.-lbs., iiul hence ^ C= 107028.45 ft.-lbs — 47.7 ft.-tons. Ex. 2. Let W^be the weight of a fly-wheel in lbs., .ind let its max. antl min. anj^ular velocities be //,,//,, respectively. The motion beinj; one of rotation only, the enerj^y stored up when the velocity rises from //, to .',, or ^ivin out when it falls from /J, to /i,, is 1 IV IV -^ {a; - a:) = -^^n^i: - a,-) = ^^; {v; - v^), il PI * ;. "11 '^\ 1 t', , ?', being the linear velocities corresponding to A,, /!,, and I' being taken equal to the mean radius of the wheel. It is usual to specify that the variation of velocity is not to exceed a certain fractional part of the mean velocity. Let V be the mean velocity, and - the fraction. Then V V, — t\ = -- ; also v,-{-v,~2V; v; - v: V* w V^ Hence the work stored or s;tven out = , 21. Centrifugal Force. — A body constrained to move in a plane curve exerts upon the body which constrains it, a force \1 ~ . i 182 THEORY OF STRUCTURES. •0 Pig. 194. called centrifugal force, which is equal and opposite to the de- viating (or centripetal^ force exerted by the constraining body upon the revolving body. Let a particle of mass m move from a point Z' to a consecutive point Q (Fig. 1 94) of its path during an interval of time t under the action of a normal deviating force. Let the normals at P and Q meet in O ; PQ may be con- sidered as the indefinitely small arc of a circle with its centre at 0. If there were no constraining force, the body would move along the tangent at Z' to a point T ^uch that PT = vt, v being the linear velocity at P. Under the deviating force the body is pulled towards O through a distance /W= ^/V',/ being the normal accelera- tion, and QN being drawn perpendicular to OP, Also, in the limit, PQ = PT = QN = vt. But QN' = PN. 2OP', .'. v'f - \ff2R, R being the radius 0P\ and hence f f=-^ = A'R, A being the angular velocity. Hence the deviating force of the mass m = mf= fft p = fnA*r, and is equal and opposite to the centrifugal force. Again, if a solid body of mass M revolve with an angular velocity A about an axis passing through its C. of G., the total '.V- CENTRIFUGAL FORCE. 183 centrifugal force will be nil, provided the axis of rotation is an axis of symmetry, or is one of the principal axes of inertia at the C. of G. If the axis of rotation is parallel to one of these axes, but at a distance R from the C. of G., the centrifu- gal force W ..^ \ = ^mrA* = A'2mr = A'MR = — A'R, r being the distance of a particle of mass m from the axis, and W^the weight of the body. Thus the centrifugal force is the same as if the whole mass were concentrated at the C. of G. If the axis of rotation is inclined at an angle 6* to the prin- cipal axis, the body will be con- stantly subjected to the action of a couple of moment 2E tan 6^, E being the actual energy of the body. Example. — A ring of radius r rotates with angular velocity A about its centre 0. Let p be the weight of the ring per unit of length of periph- ery. Consider any half-ring AFB. The centrifugal force of any element CC^^-^A^r. g The component of this force parallel to AB, is balanced by an equal and opposite force at C" , the angle COB being = the angle CO A. Thus the total centrifugal force parallel to AOB\% nil. The component of the force at (7, perpendicular to AB, ./.^A\ sin COD=^^—-A'r cos C'CE g g Fig. 19s. g CC '^ g W -.!■! \\\ ?4| rfl i ^ 1: I '■iil I I I \' u 1411 I '1 I !* ■■ : 184 THEORY OF STRUCTURES. ' Hence, the total centrifugal force perpendicular to AB If T is the force developed in the material at each of the points A and B^ 2 7'= 2^A'r\ g since the direction of T is evidently- perpendicular to AB. .-. T = -A'r' = ^v\ g g V being the circumferential velocity. Let / be the intensity of stress at A and B, and w the specific weight of the material. Assuming that T is distributed uniformly over the sectional areas at A and B, XV g if-, K Thus, the stress is independent of the radius for a given value of V, and the result is applicable to every point of a flex- ible element, whatever may be the form of the surfaces over which it is stretched. 22. Impact. — When a body strikes a structure, or member of a structure, the energy of the blow is expended in (i) overcoming the resistance to motion of the body struck ; (2) deforming the body struck ; (3) the kinetic energy of either or of both of the bodies after impact, if the motion is sensible ; (4) deforming the striking body; (5) producing vibrations. IMPACT. 185 Generally speaking, the energy represented by (5) is very small and may be disregarded. Also, if the striking body is very hard, the energy (4), absorbed in its deformation, is inap- preciable and may be neglected. First, let a body of weight P fall through a vertical dis- tance h and strike a second body, the point of application moving in the direction of the blow through a distance x against a mean resistance R'. Then Piji -\- x)=z work done = R'x. Let Fbe the velocity of the striking body at the moment of impact. Then P V energy of blow = = Rx = P{Ji -f- x). S 2 The actual resistance is directly proportional to the dis- tance through which the point of application moves, so long as the limit of elasticity is not exceeded. Its initial value is nil, and if R is its max. value, the mean value '\s R' = — . r 1 i f 1 NHi ] ! > )» ^ •p;vi. t'li *!h •' I », . |.| * .8 i 5 I ; f 1 ,if PV' Rx .-. -7-= Y=^(^^+^)- \i h = o, R = 2P, or the sudden application of a load P from rest, produces a pressure equal to twice the load, pro- vided the limit of elasticity is not exceeded. Example. A i-oz. bullet moving with a velocity of 800 ft. per sec. strikes a target and is stopped dead in the space of -^(j inch (^ = 32). Then i.-iV.^.(8oo)' = /?'./^.-iV; > II "itmBMM-l :?r:.: ^i 1 86 THEORY OF STRUCTURES. .'. R', the mean resistance overcome by the bullet, = 5000 lbs. The time in which the bullet is brought to rest momentum iV • tjV • 800 force 5000 3200 sec. Ni'xf, let a body of weight IV, moving in a given direction with a velocity 7\ strike a body of weight W, moving in the same direction with a velocity ?', . After impact let the bodies continue to move in the same direction with a common ve- locity V, w vv — ^i\ 4" — ^^'a = momentum before impact momentum after impact W Wy or rr.r, + w,v, = {w,+ w,)v. Energy before impact = — ' -'-H -' — . A'" 2 I'" 2 after = ( yy. + n \ w g /2 Energy lost by impact -(^F..-'+^F,TO--(^F, + rr,) If ei must b( energy EXA liead of l; round pile beii Let : " / " ^ « V Px Also, ( and then Again, Finally m IMPACT. 187 If either of the bodies is subjected to any constraint, energy iTuist be expended to overcome such constraint, and the loss of energy by impact will be less. EXAMI'LIC I. Let a weight of W^ tons fall h ft. upon the head of a pile weighing W^ tons and drive it a ft. into the i;round against a mean resistance of R tons, the head of the pile being crushed for an appreciable length x ft. Let V be the velocity of the weight when it strikes the pile ; mean force of the blow ; distance through which pile moves during ac- tion of blow ; duration of the blow in seconds ; common velocity of the pile and weight during action of blow ; ' XT " " distance through which pile moves after the blow. Px -f- Ry = work done in crushing the pile -f- work done in overcoming ground-resistance in time t = energy dissipated by blow y <• «' t " " = W./i IV, + w, r (I) Also, considering the change of momentum first of weight and then of pile, W W Pt = — '(z; - F) = y?^ 4- — ' V. s s (2) Again, Rz = work done after blow = — ^~ — ? — . . (2) g 2 ^^' l=£/(£. If the weight is made to descend to a point C, and is then left free to return to its state of equilibrium, it must necessarily describe a series of vertical oscilla- tions about B as centre. Take B as the origin, and at any time t let the ^veight be at i^ distant x from B; also let BC = c. Two cases may be considered. First, suppose the end of the rod to be gradually forced duwu to C and then suddenly released. M Fig. 196. Acc( or and henc f,t Now above B Again and integi and the os When. and the tir OSCILLATORY MOTION. I9I According to the principle of the conservation of energy, ( — r ) = the work done between C and M £ 2 \dtl g _ EA le x\ or PildxV Pi and hence V, the velocity of the weight at M, = a/S^Cc^ _ x^)i . Now V is zero when x = ± c, so that the weight wili rise above B to a. point C^ where BC^ , — s =z BC. Again, from the last equation, ^^\j I -{e- x'y and integrating between the limits o and x, /y/f = sin-^, and the oscillations are therefore isochronous. When x =. c, It IT and the time of a complete oscillation is \ i ! '■ I, i !i .'^ Hf? I ■\ m^i- ir 192 THEORY OF STRUCTURES. Next, suppose the oscillatory motion to be caused by a weiffht /' falliii|:j without friction from a point D, and beini; suddenly checked and held by a catch at the lower end of the rod. Take the same origin and data as before, and let AD = h. The elastic resistance of the rod at the time t is EA so tha oscillat Cor and the equation of motion of the weight is and hen Pd\v l-\-x /r de L = />-y(/+,r)=-P7; d\v g Integrating, ./.r\^ // j = — "4.1-" -\- c^, c^ being a constant of integration. dx g But -J- is zero when x = c, and c^ = rc'. Hence ri)'=f(--) = This is precisely the same equation as was obtained in the first case, and between the limits o and x a/t = ''"'-' 7- or Cor. : the rod oscilJatioi Cor. 3 Z + /, wl to its elas motion. lias reachi spending distance a corresponi force agai may be co however si intervals, t but the inc tiiiually. picx on ac( ^fluctiiatio than the st mm 1 VII OSCILLA TOR Y A/0 TION. •93 SO that the motion is isochronous, and the time of a complete oscillation is Cor. I. When x = — I. and hence "vl- ri)"=-". I (C» - /■') = 2gh, or Cor. 2. If h = o, i.e., if the weight is merely placed upon tlic rod at the end A, c = ± /, and the amplitude of the oscillation is twice the statical elongation due to /-*. Cor. 3. The rod may be safely stretched until its length is L-\- /, while a further elongation c might prove most injurious to its elasticity, which shows the detrimental effect of vibratory motion. If a small downward force Q is applied to /'when it has reached the end of its vibration, it will produce a corre- sponding descent, and the weight P will then ascend an equal distance above its neutral position. At the end of the interval corresponding to P's natural period of vibration, apply the force again, and P will descend still further. This process may be continued indefinitely, until at last rupture takes place, however small Pa.nd (2 may be. If Q is applied at irregular intervals, the amplitude of the oscillations will still be increased, but the increase will be followed by a decrease, and so on con- tinually. In practice the problem becomes much more com- plex on account of local conditions, but experience shows that 7i fluctuation of stress is always more injurious to a structure than the stress due to the maximum load, and that the injury 194 THEORY OF STRUCTURES. is aggravated as the periods of fluctuation and of vibration of the structure become more nearly synchronous. An example of a fluctuating load is a procession marching in time across a suspension-bridge, which may strain it far more severely than a much greater dead load, and may set up a synchronous vibration which may prove absolutely dangerous. In fact, a bridge has been known to fail from this cause. Cor, 4. The coefficient of elasticity of the rod may be ap- proximately found by means of the formula "VJ T being the time of a complete oscillation. For suppose that the rod emits a musical note of n vibrations per second, then V ^ 2« is the time of travel from C to C, ; m liv « sS ; li .i t ,', I = —5—5 , and hence E = ^it n A g Cor. 5. Sujjpose that the weight is perfectly free to slide along the rod. When it returns to A, it will leave the end of the rod and rise with a certain initial velocity. This velocity is evidently V2gh, and the weight accordingly ascends to D, then falls again, repeats the former operation, and so on. The equations of motion are in this case only true for values of x between x = -\- c and x =^ — I. 25. On the Oscillatory Motion of a Weight at the End of a Vertical Elastic Rod of Appreciable Mass. — Suppose the mass of the rod to be taken into account, and assume : {a) That all the particles of the rod move in directions par- allel to the axis of the rod. Hence OSCILLATORY MOTION. 195 ~ ^° + /V^^^A' + /> 4- p.gAx = o. (^) That all the particles, which at any instant are in a plane perpendicular to the axis, remain in that plane at all times. As before, the rod O/l of natural length L and sectional area A is fixed at O and carries a weight /\ at A. Take O as the origin, and let OX be the axis of the rod. Let S, S -{- 1 '. \'-lh But if the thickness dS of the slice MM' is indefinitely diminished, P is evidently the elastic reaction, and its value is dB,—dx t: A l^^ \ ^^—dr- = ^'^\^x-'V Hence - P„ + ^pAXdH + EA {~^_ -1)4- p„gAx = 9. ! 1 ' TH 'T" n": 196 THEORY OF STRUCTUREH. Differentiating with respect to x. dB, d'S, pAX-r- + £A-r^ + p,gA = o. dx dx" a, I' \\\ But pdB = p^dx, or .'. p,AX-\- EA j^ + p,gA = o, Also, p„AXdx is the resistance to acceleration arising from the inertia of the slice, and is therefore equal to -p^Adx-^, so that X=- d^ 'df ' Hence d'B Ecea df ~ p„ dx" ■\-g- (I) To solve this equation. — In the state of equilibrium, -(J!-) is the tension in the section of which the distance from OSCILLATORY MOTION. 197 is X, and counterbalances the weight P^ and the weight ft^A{l — x)g of the portion AMN of the red. .-. £^ (^ - l) = />, + p^g(l - x), vf f i % or Integrating, S='+^+¥('--)- 5 = .+ A,+P^(,,_£). ... (3) There is no constant of integration, as x and $ vanish together. This value of ^ is a particular solution of (i), and is inde- pendent of /. Put ^ = * + Ei- + ¥('--T)+-- ■ W, I ,^ wm ''' Mil • ■ ! ■ "?' ^^ being a new function of x and /. Then d^a dx' - B^'^dx'' ^""^ de ~ dt" Hence, from eq. (i). d^z _ E (£z de ~ p,dx^ d^s P.' The integral of this equation is of the form s = F{x^ v,t) +/(.r - v,t), ! jiiill 198 THEORY OF STRUCTURES. ^^(=vf)^"'"^ ig the velocity of propagation of the vibrations. The full solution of (i) is therefore of the form 26. Inertia — Balancing. — Newton's First Law of Motioiv called also i\\c Law of Inertia, states that "a body will continue in a state of rest or of uniform motion in a straight line unless it is made to change that state by external forces." This property of resisting a change of state is termed inertia, and in dynamics is always employed to measure the quantity of matter contained in a body, i.e., its mass, to which the inertia must be necessarily proportional. Thus, to induce motion in a body, energy must be expended, and must again be absorbed before it can be brought to rest. The inertia of the reciprocating parts of a machine may therefore heavily strain the framework, which should be bolted to a firm foun- dation, or must be sufificiently massive to counteract by its weight the otherwise unbalanced forces. Example i. Consider the case of a direct-acting horizontal steam-engine. Fig. 198. At any given instant let the crank OP and the connecting-rod CP make angles B and 0, respectively, with the line of stroke AB. Let V be ':he velocity of the crank-pin centre P, and let u be the corresponding piston velocity, which must evidently be the same as that of the end C of the connecti.ig-rod. Let CP produced meet the vertical through C in /. At the moment under consideration, the points C and Pare turning about /as an instantaneous centre. Fig. 198. U V IC IP sin (^+0) cos (J> INER TIA —BALA NCING, 199 Let W be the weight of the reciprocating parts, i.e., the piston-head, piston-rod, cross-head (or motion-block), and a por- tion of the connecting-rod. Assume (i) that the motion of the crank-pin centre is uni- form; (2) that the obHquity of the connecting-rod may- be disregarded without sensible error, and .-, = 0. Draw PN perpendicular to AB, and let ON=x\ ON is equal to the distance of the piston from the centre of the stroke, corresponding to the position OP of the crank. The kinetic energy of the reciprocating parts :t id;. W^i^_W_ v" sin' e W v' g '^~ g 2 r being the radius OP. .". the change of kinetic energy, or work done, corresponding to the values x^ , x^ of x. ■mm W %!" (X^ - ^„»^ g r^^i- '^■lU Let R be the mean pressure which, acting during the same interval, would do the same work. Then W z>' x; - x," y 2 ' r' = R{x, — x^), and , Wv^x. + x, g 2 r' Hence, in the limit, when the interval is indefinitely small, x^ = .r, = X, and the pressure corresponding to x becomes R = -X. g r ,1 . 200 THEORY OF STRUCTURES. This is the pressure due to inertia, and may be written in the form R=C-, r C {— ) being the centrifugal force of W assumed con- centrated at the crank-pin centre. /? is a maximum and equal to C when x =^ r, i.e., ?.t the points A, B, and its value at intermediate points may be represented by the vertical ordi- nates to AB from the straight line EOF drawn so that A£ = BF = C In low-speed engines, C may be so small that the effect of inertia may be disregarded, but in quick-running engines, C may become very large and the inertia of the recip- rocating parts may give rise to excessive strains. Another force acting upon the crank-shaft is the centrifu- gal force of the crank, crank-pin, and of that portion of the con- necting-rod which may be supposed to rotate with the crank- pin. Let w be the weight of the mass concentrated at the crank- pin centre which will produce the same centrifugal force as these rotating pieces (i.e., wr = sum of products of the weights of the several pieces into the distances of their centres of gravity from O). The centrifugal force of w = . g r Thus the total maximum pressure on the crank-shaft WV^ V* = C+~j:=~{lV-{-zv) = r{W + zv)j A being the uniform angular velocity of the crank-pin. This pressure may be counteracted by placing a suitable balance-weight (or weights) in such a position as to develop in the opposite direction a centrifugal force of equal magnitude. Let or from whi Durir represent reciproca triangle I when the Durin reciprocal the piston direction. InAE of the rec ^?, AE rep Draw E'C During E'O' reprt celerate retarded The ca N.B.— follows : u = V Wd_ g d %: INER TIA—BA LA NCING. 201 Let W^ be such a weight and R its distance from 0. Then or RW, = r{W+w), m ^^rfll H Hffli ? ,r ■ ''!;|j s i si t. 1 f ■ r 1 from which, if R is given, W^ may be obtained. During the first half of the stroke an amount of energy represented by the triangle AEO is absorbed in accelerating the reciprocating parts, and the same amount, represented by the triangle BOF, is given out during the second half of the stroke when the reciprocating parts are being retarded. During the up-stroke of a vertical engine the weights of the reciprocating parts act in a direction opposite to the motion of the piston, while during the down-stroke they act in the same direction. \r\AE produced (Fig. 199) take EE' to represent the weight of the reciprocating parts on the same scale g- as AE represents the pressure due to inertia. E Draw E'O'F' parallel to EOF. A During the up-stroke the orrlinates of E'O' represent the pressures required to ac- celerate the reciprocating parts, the pressures while they are retarded being represented by the ordinates of O'F' . The case is exactly reversed in the down-stroke. X N.B. — The formula R = C- may be easily deduced as T follows : r. , , . ^« ^d9 v" « = t; sin 6', the acceleration = -y- =1 v cos 6 —- = -x \ at at r W du .: — '77 — accelerating force = force due to inertia -!Z^- - rf — ^ r'^~ r' ml ^''IH 202 THEORY OF STRUCTURES. Ex. 2. Consider a double-cylinder engine with two cranks at right angles, and let d be the distance between the centre lines o^ the cylinders (Fig. 200). f C.COS Q .(%ntre line of Cylr. Y Centre line iC.Blne o'Cylr. Fig. 200. The pressures due to inertia transmitted to the crank-pins when one of the cranks makes an angle B with the line of stroke are P,— C cos and P^= C sin B. These are equivalent to a single alternating force P = C(cos B ± sin B) acting half-way between the lines of stroke, together with a couple of moment M=P-= C-(cos B ± sin B). 2 2 The force and couple are twice reversed in each revolution, and their maximum values are Cd ^ P.na.. = CV2 and M^^, = -J ^• ...-■]' i .i"o\d the evils that might result from the action of the ' ( couple at high speeds, suitable weights are introduced in such positions that the centrifugal forces due to their ro For ex; upon thi ing-whei Let ; diametri crank-pii from the Let e the balar Thee and this f Fig. forces eacl the angle 1 that betwe ant couple the axes o to the line Q and < 2F CO'. and Fe sin IliiJ INER TIA—BA LA NCING. 203 their rotation tend to balance both the force and the couple. For example, the weights may be placed upon the fly-wheels, or again, upon the driv- ing-wheels of a locomotive. Let a balance-weight Q be placed nearly diametrically opposite to the centre of each crank-pin (Fig. 201), and let R be the distance from the axis to the centre of gravity of Q. Let e be the horizontal distance between the balance-weights. The centrifugal force 7^ due to the rotation of Q Fig. 201. ™ ^Wm H ■ R \ K 1 ' M' ;mp, >\k 1 ^ p , -til iVt ' jiiiiiiiii 'rrrfTr % if e (velocity of Qf g R and this force F is equivalent to a single force F acting half-way between the weights and to a couple of moment > \ 90' A 90°; \ '\y Let be the angle between the radius Fig. 202. 2 to a balance-weight, and the common bisector of the angle between the two cranks (Fig. 202). Since there are two weights Q, there will g be two couples each of moment F - , and two forces each equal to F acting half-way between the weights, the angle between the axes of the couples being 1 80°— 20, and that between the forces being 20. The moment of the result- ant couple is Fe sin 0, and its axis bisects the angle between the axes of the separate couples ; the resultant force parallel to the line of stroke = 2/^ cos 0. Q and may now be chosen so that 2F cos = maximum alternating force = C V2, and Fe sin = maximum alternating couple = — V2. d .'. tan = - , 111 ■f :4r ' i;:: :! ' 1 . ri hi I 304 and or and THEORY OF STRUCTURES. gr" g r ey e'-\-d* ^' + ^' Ex. 3. Again, the pressure d7 at a dead point may be balanced by a weight Q diametrically opposite. If R is the radius of the weight-circle, then g r g r" and ,'.Q^W R ' e-i-d The weight Q may be replaced by a weight Q—~— on the 2e e — d near and a weight Q—z — on the far wheel. Thus, since the cranks are at right angles, there will be two weights 90° apart e -^ d e d on each wheel, viz., Q in line with the crank and Q . These two weights, again, may be replaced by a single weight B whose centrifugal force is the resultant of the centrifugal forces of the two weights. Thus IB 7ry_ (Q ej±_d vy iq e^-d v_y ^g R^~^g 2e Rl'^^g 2e Rl' v' being the linear velocity at the circumference of the weight- circle. .-. ^ = (2' 2^' or If a 1 Note.- nil, and B 27. Ct Ex. I. Let CP in OP take The pis centre are c If the veloc by OP, then velocity u. CURVES OF PISTON VELOCITY. 205 or B Q A' 4- cV If a is the angle between the radius to the greater weight e + d Q and the crank radius, 2e Q e-d v^l g 2c R e — d Qej±_dv'^~e^d' g 2e R Note. — In outside-cylinder engines e — dh approximately nil, and ^ = (2 = W^. 27. Curves of Piston Velocity.— Consider the engine in Ex. I. s Fig. 203. Let CP produced intersect the vertical through O in T', and in CPtake OT' = OT. The piston velocity u and the velocity v of the crank-pin centre are connected by the relation u V sin {0 H- 0) _0T _ OT^ "OF cos OP . . . (I) If the velocity v is assumed constant, and if it is represented by OP, then on the same scale OT' will represent the piston velocity «. Drawing similar lines to represent the value of u I i ^K I f :i 1 1 ill a \ \\\ 1 'i ') ' If If . ! ;i 206 THEORY OF STKUCTUKES. for every position of tlic crank, the locus of 'f will be lountl to consist of two closed curves OGS, OUT, caWcd X\\iz polar curies of piston velocity. They pass through the point O and through the ends 5 and T of the vertical diameter. On the side towards the cylinder they lie outside the circles having OS and OT as diameters, while on the side away from the cylinder they lie inside the circles. If the connecting-rod is so long that its -obliquity may be disregarded, = and u = t> sin 6, and the curves coincide with the circles. A rectangular dxdigrdim of velocity may be drawn as follows : C M Upon the vertical through C, Fig. 204, take CL = OT; the locus of L is the curve required for one stroke. A similar curve may be drawn for the return stroke either below A/N ov upon the prolongation A^R {= MN) of MN. If the obliquity of th«^ connecting-rod is neglected, the curves evidently coincide with the semicircles upon AIN and AT?, MN (r= NR) defining the extreme positions of C. Tiic obliquity, however, causes the actual curve to fall above the semicircle during the first half of the stroke, and below during the second half. Again, let the connecting-rod (/) = n cranks {f). Then n. f ' a X au ^\ / • /J I sin '9 cos 6* \ , ^ // = V (sm p -X- cos V tan 0) = 7' ( sin v A — ). (2) \ Vn' - sin" ej sin ^ / sin "" r and by eq. I, If th( tan and 28. C j)ositii)ii ( If the resented t resent tht already dr to reprcsei If tlie pre usually the radius, rej. of P. Aft take OP' ii of the crai lesponding directly oh parallel to sent the rec may be dra 29. Cur curve of en position 01 being equal n { CURVE OF CRANK-EFFORT— CURVES OF ENERGY. 20? If the obliquity is very small, tan = sin = SI in e n , approximately, and ( ■ ft ^ sin^cos ft] i . n , sin 2^ 28. Curve of Crank-effort. — The crank-cjfort F for any position OP o{ the crank is tlie component along the tangent at P oi the thrust along the connecting-rod. This thrust = cos ^_^sin (^y + 0) cos If the pressure /* upon the piston is constant, and if it is rep- resented by OP, then, on the same scale, OT', Fig. 203, will rep- resent the crank-effort. Thus, the curves of piston velocity already drawn may also be taken to represent curves of crank-effort. If the pressure P is variable, as is usually the case, let OP, the crank radius, represent the initial value of P. After expansion has begun, take OP' in OP, for any position OP of the crank, to represent the cor- responding pressure which may be ilirectly obtained from the indicator-diagram. Draw P'T' parallel to PT, and take OT" ^ OT. Then OT" will repre- sent the required crank-ef?ort,and the linear and polar diagrams may be drawn as already described. 29. Curves of Energy— Fluctuation of Energy. — In the curve of crank-efYort as usually drawn, the crank-effort for any position OP of the crank is the ordinate S'H, the abscissa DH being equal to the arc AP, i.e., to the distance traversed by the F"IG. 205. Hi m H : : ,iM ^^B.') ■ t ' ■: ■ 1 1' ' t ■ " :4- 1 ■- -i, ■ ^^H I ■ u[ ■ f ^- ■ '< . V v»;it? Mm (■ . tft m i 208 THEORY OF STRUCTURES. ii point of application of the crank-effort. Thus, DSE andEVG being tlic curves, DE = EG = semi-circumference of crank-circle = nr. If the obliquity is neglected, the curves of crank-effort arc the two curves of sines shown by the dotted lines. The area DS H also evidently represents the zvork done as the crank moves from OA to OP, and the total work done is represented by the area DSE in the forward and by E VG in the return stroke. Let F^ be the mean crank-effort. Then F, X 2nr— 2PX 2r, assuming Pto be constant. /^,= 2P 7t ' 2P Draw the horizontal line 1234567 at the distance — from DEG, and intersecting the verticals through D, E, and ^7 in i, 4, and 7, and the OJrves in 2, 3, 5, and 6. The engine ma be supposed to work against a constant resistance R equal and opposite to the mean crank-effort F„ . From Z> to 2, 7? > crank-effort, and the speed must there- fore continually diminish. From 2 to 3, i? < crank-effort, and the speed must contin- ually increase. Thus 2 is a point of min. velocity, and therefore also of min. kinetic energy. From 3 to £', ^ > crank-effort, and the speed must contin- ually diminish. are necessari "W^ CUFVES OF ENERGY— FLUCTUATION OF ENERGY. 209 Thus 3 is a point of max. velocity, and therefore also of max. kinetic energy. Similarly, in the return stroke, 5 and 6 are pnnts of min. and max. velocity, respectively. The change or fluctuation of kinetic energy from 2 to 3 = area 283, bounded by the curve and by 23. The fluctuation from 3 to 5 = area 3^5, bounded by 35 and by the curve. F Fr Again, since -j^=l— ^ the ordinatcs of the curves may be taken to represent the moments of crank-effort, and the abscissae arc then the corresponding values of ^. The work done between A and any other position P of the crank-pin db Jo Jo \ Vn^- sin' 0) = Pr{i — cos ft -\- n — V'«" - sin""^). If there are two or more cranks, the ordinatcs of the crank- effort curve will be equal to the algebraic sums of the several crank-efTorts. For example, if the two cranks are at right angles, and if F, , F^ are the crank-efforts when one of the cranks (/^,) makes an angle with the line of stroke, ^, = 4i„. + ?i'if^) and =H cos 6 /?__}_/?= P(sin ^ + cos 6^ sin 2^ 2n /■ combined crank-effort. P being supposed constant. Note. — In the case of the polar curves of crank-effort, if a circle is described with O as centre and a radius = mean crank- 2P effort = — , it will intersect the curves in four points, which are necessarily points of max. and min. velocity. m (■ ' :i m 210 THEORY OF STRUCTURES. w w H a < ?: O OS o H O Q < u H < t/1 H o z W Oi, H J3 O u • is u'^.s rf O O C >S5 - a fl. « O u O k. "= il 5 U K O 8 8 8 8 8 8 i/> »o m in lo W1 ■8 "8 'S'S'8^ 888i to u> in c 1 ^ m en in in H ^ 00 00 00 00 00 m -r ■* o a c c o o o N rn S» aacucQac oooooooo m^t \%. ^1 K c t/1 ^ I ■a s s m in (>co oo o6 ^ ^ «1 8^ SiifeS * . 2 Jo .•a uJS »l o it '5. u Be 5° si (« "Q - o T3 O •a a rt > c u c M JO tic bcu c ^ rt j^ _ w ^ 2 a* 3 bg c u •a c rt □ 9 o u e rt rt Wrt •S c-a s o 21 a AM a u 1^ IF 3 01 1 (A Acacia. . ■Mder. ... Apple Ash, Canadi Asli, Enelis Heech..... Mirch Hux Hluegiiui. Cedar Clicstnut. . Kbony .lil.T:, Canadi TABLES. THE STRENGTHS, ELASTICITIES, AND WEIGHTS OF VARIOUS ALLOYS. ETC. SII Mutcriul. Aluminum Urass " (li.immcred). Urass wire Coppt-r-platc, liammercd,. . . " annealed (Upper wire (iiiiimelal Lead Lead wire rliosplior-bronze I'm Zinc Leather Max. Load on OriKinal Area in lbs. per aq. in. Tension. Com- Shcar- pression. "iK. aS.Soo 17,600 10,380 53,000 3o,o 58,000 fKl,,CX-HJ 1,850 7,100 .i.>'» 57.000 4,y8o 7.500 4,000 Young's Modulus, E (in lbs.). 9,600,000 0,IfX>,000 I .;,ooo,oo,ofX3 7ro,(x>o 9.^0, 000 14,000,000 5,69o, 255 5' 146 03 1.63 to 2.86 Weight in lbs. per cu. ft. 34 to 37 58 to 72 47J 42 to 63 32 to 38 41 to 83 53 35 49 57 to 58 61 61 49 to 58 36 34 41 to 58 34 34 32 40 23 to 26 29 to 32 .36 to 43 41 to 52 38 to 57 24 to 35 * The results for these timbers are deduced from experiments carried out by Bauschinger Lanza, and others, on comparatively large specimens. THE BREAKING WEIGHTS AND COEFFICIENTS OF BENDING STRENGTH IN TONS (of 2240 lhs.) OF VARIOUS RECT- ANGULAR BEAMS, THE WEIGHTS HAVING BEEN UNIFORMLY DISTRIBUTED. Material. Yellow pine (Quebec), 1 joists.. 2 beams. Fir (Baltic), a beams " " II joists " (Swedish), a ioists Pine (Baltic), 2 beams Baltic redwood deal (Wyberg), 2 joists . . . Spruce deals (St. John), 3 pairs with bridg- ing pieces Clear Span Breadth between in Supports inches. in inches. 14a 3, 142 ^k 126 >4 126 14 142 3* 142 3 ia6 •3* 14a 3 14a 3 Depth in inches. 15 »4 II 9 '3* 9 Mean Breaking Weight of each Joist or Beam. 5.66 7.89 60.97 46.6 8.29 5'7 58.43 5.75 6.81 Coef- ficient of Bending Strength. 2.48 t.aS 1.83 1.6 2.0B a. 49 2.24 2.5a a.qB THE BF STRE Matei Yell ow pin Pitch pine. Baltic pine. American ei Greenheart. Red pine. A'.iS.-The ments carried i pool. ■Wfl TABLES. 213 THE BREAKING WEIGHTS AND COEFFICIENTS OF BENDING STRENGTH IN TONS (of 2240 lbs.) OF VARIOUS RECT- ANGULAR BEAMS LOADED AT THE CENTRE. Material. Clear Span between Supports in inches. Yellow pine . Pitch pine. Baltic pine. American elm. ■Greenheart. Red pine. 129 129 45 45 45 45 45 45 129 129 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 139 147 147 Breadth in inches. 14 14 5 5 5 5 2i 2i 14 14 5 5 5 5 2i 2J 5 S 2i 2i 5 5 2i 2* 5 2i 3 2i 2i 9 6 6 Depth in inches. 15 15 7 7 5 5 3i 3i 15 15 7 7 7 7 ■:k 3i 7 7 3i 3i 7 7 3i 34 5 7 3 3i 3i 8 12 12 Breaking Weight in tons. Coef- ficient of Bending Strength. 38.15 34 9 7 I 05 •925 I -075 59-25 60.25 7.8 9-75 10.65 II 1.6 1-35 7 8.5 1. 125 1.2 14.9 15-6 2.65 2.6 14 "•45 3-85 4.00 3-55 24.5 7-5 8.45 Remarks. 2.34 2.og 1.62 1-57 1.67 1.64 2.04 2.37 3-64 3-7 2.14 2.68 2.92 3-03 3-52 2.97 1. 91 2.34 2.48 2.64 4.1 4.29 5-84 5-73 7-56 6.11 9.62-, 8.81 7.82 8.87 1.91 2.15 Old timber Old timber Old timber A''.5.— The results contained in the last two tables are mainly deduced from experi- ments carried out under the supervision of W. Le Mesurier, M.Inst.C.E., Dock Yard, Liver- pool. TJ m ■m t I \ ' . 1H I I ir -•r U [sl ii - r H fl w I' it r II IHI if 1 1 1 jU ■1 lil 214 THEORY OF STRUCTURES. THE WEIGHTS AND CRUSHING WEIGHTS OF ROCKS ETC. Material. Weight per cu. ft. in lbs. Asphalt Basalt, Scotch u " Greenstone Welsh Beton Brick, common " stock (Eng.) " Sydney, N. S " yellow-faced (Eng.)., " Staffordshire blue. . . " fire " pressed (best) Brickwork Cement, Portland Roman Clay Concrete, ordinary " in cement. . . . .. Earth Firestone Freestone Glass, flint " crown " common green " plate Granite, Aberdeen gray.. . , red " Cornish Sorrel Irish U. S. (Quincy)..., Argyll Gneiss Limestone , Lime, quick Mortar , " (average) Masonry, common brick.., " in cement rubble Marble, statuary " miscellaneous.... OSlite, Portland stone Bath stone Sand, quartz " river " pit " fine Sandstone, red (Eng.) " Derby grit. . . •• paving (Eng.). " Scotch U. S Shingle Slate, Anglesea " Cornish , " Welsh Trap 156 184 181 172 100 to J35 150 112 86 to 94 100 119 119 137 77 to 125 112 192 157 153 172 163 165 i66 167 96 to 175 154 to 162 53 86 to 119 106 116 to 144 170 168 to 170 151 i?3 177 "7 100 95 133 150 15(J lo 157 153 to 155 83 179) 157 - 180) 170 Crushing Weight in lbs. per sq. in. 8,300 17,200 16,800 800 to 1,400 550 to 800 2,250 2,200 1,440 7,200 1,700 10, 200 1,700 to 6,000' 460 to 775 19,600 3,000 to 3, SCO' 27,500 31,000 31,000 10,800 14,000 12,800 10,450 15,000 10,900 19,600 7,500 lo 9,c oo- 120 10 240 500 to 800 760 y^ of cut stone- 3,200 8,000 to 9,700' 4,100 5.700 3,100 5,700 to O.ooo' 5,300 to 7,800. 5.300 10,000 10 24,000 Ti M( Ml Brass Bronze.. .. Cast-iron,, Copper. . . , Fir Glass Gold Gun-metai. . Iron wire . . Lead Oak Platinum... Silver Steel, unhardi " hardene Tin Wrought-iron tt Zinc, cast... hammer '1 i kj TABLES. ACTORS OF SAFETY. Good Ordinary Work. Timber 4 to 5 for dead load, 8 to 10 for live load. Metals 3 " " " 6 Masonry 4 8 EXPANSIONS OF SOLIDS. 21$ (( tt Materials. Brass Bronze Cast-iron Copper Fir Glass Gold Gun-metal Iron wire Lead Oak Platinum Silver Steel, unhardened " hardened Tin Wrought-iron (bar) " (for smith-work), Zinc, cast " hammered Linear Expansion per Unit of Length. From 3a° F. to 212° F. .00x868 .00182 .001075 ,001718 .00352 .00861 .001466 .00181 = .00144 . 0002848 . 000746 .000884 .001909 .001079 .00124 .002173 .001235 .001182 .00^2941 .003108 ffiff laVr TlVf itK itVt til = Bi BtST Tfftl From 3 Expansion in Built. 12° F. to From 32° F. ° F. to 212° F. 572° F. .001883 = ,fj .001468 = tIt .0065 .0054 .0033 .0055 .0027 .0057 .0036 .0066 .0036 .0058 m m i'-i' w <> ! til ' 11 I ■in ' ill Hii iilil i 1\^ • '.' ; ' '.ii ^^y 2l6 THEORY OF STRUCTURES. Ill EXAMPLES. r. How many square inches are there in the cross-section of an iron rail weighing 30 lbs. per lineal yard? How many in a yellow-pine beam of the same lineal weight? Ans. 3 sq. in.; 45 sq. in. 2. A vertical wrought-iron bar 60 ft. long and i in. in diameter is fixed at the upper end and carries a weight of 2000 lbs. at the lower end. Find the factors of safety for both ends, the ultimate strength of the iron being 50,000 lbs. per sq. in. Ans. 191^^; i^fi-^. 3. A vertical rod fixed at both tads ii- w jighted with a load iv at an intermediate point. How is the load distributed in the tension of the upper and compression of the lowc. portion of the rod? w, . Inversely as the lengths. 4. Find the length of a steel bar of sp. gr. 7.8 which, when suspended vertically, would break by its own weight, the ultimate strength of the metal being 60,000 lbs. per sq. in. Ans, 17,723 ft. 5. The iron composing the links of a chain is \ in. in diameter; the chain is broken under a pull of 10,000 lbs. What is the corresponding tenacity per sq. in. ? Ans, 57,272j\ lbs. 6. A vertical iron suspension-rod 90 ft. long carries a load of 20,000 lbs. at its lower end ; the rod is made up of three equal lengths square in section. Find the sectional area of each length, the ultimate tenacity of th ~ 'ron being 50,000 lbs. per sq. in., and 5 a factor of safety. Ans. V^ sq. in. ; Vff^ sq. in. ; ^^%^ sq. in. 7. If the rod in the previous question is of a conical form, what should be the area of the upper end ? Also find the intensities of the tension at 30 and 60 ft. from the lower end. Ans. 2.0407 sq. in.; 9999.612 lbs., 9999.605 lbs. per sq. in. 8. The dead load of a bridge is 5 tons and the live load 10 tons per panel, the corresponding factors of safety being 3 and 6. If the two loads are taken together, making 9 tons per panel, what factor of safety would you use ? Ans. 5. 9. The end of a beam 10 in. broad rests on a wall of masonry. If it be loaded with 10 tons, what length of bearing surface is necessary, the safe crushing stress for stone being 150 lbs. per sq. in. ? Ans. \i\ in. 10. Find diameter of bearing surface at the base of a column loaded with 20 tons, the same stress being allowed as in the preceding question. Ans, 1/380. 1 2. II. In alternatel} the chain the metal 12 .12 in Ar und 13- A J cable stret respond inj 14. A 1 1,200,000 1 under a pi 15. A 5oooolbs. 16. A V pull of 52,5 E being 21; 17. A A between t E = 29,oo< the tempei 18. Wh pressure is surface an< 19. A f done in sti 20. A r subjected elongation 21. An to a tensic show by a within the 22. A t 30 ft. from what thicl pillar and ' EXAMPLES. 217 11. In the chain of a suspension-bridge five flat links dovetail with four alternately, and a cylindrical pin passes through the eyes. The pull on the chain is 200 tons. Find the area of the pin, the bearing strength of the metal being 6 tons per sq. in. Ans. \\ sq. in. 12. An iron bar of uniform section and 10 ft. in length stretches .12 in. under a unit stress of 25,000 lbs. Find E. Ans. 25,000,000 lbs. 13. A ship at the end of a 600-ft. cable and one at the end of a 500-ft. cable stretch the cables 3 in. and 2\ in., respectively. What are the cor- responding strains.' Ans. 5^5. 14. A rectangular timber tie is 12 in. deep and 40 ft. long. If ^ = 1,200,000 lbs., find the proper thickness of the tie so that its elongation under a pull of 270,000 lbs. may not exceed 1.2 in. Ans. 7^ in. 15. A wrought-iron bar 60 ft. long is stretched 5 in. by a pull of 5ooo^bs. Find its diameter, E being 25,000,000 lbs. Ans. .59 in. 16. A wrought-iron rod 984 ft. long alternately exerts a thrust and a pull of 52,910 lbs. ; its cross-section is 9.3 sq. in. Find the loss of stroke, E being 29,000,000 lbs. Atis. 4.632 in. 17. A wrought-iron bar 2 sq. in. in sectional area has its ends fixed between two immovable blocks when the temperature is at 32° F. If E = 29,000,000 lbs., what pressure will be exerted upon the blocks when the temperature is 100° F. ? Ans. 27388I lbs. 18. What should be the diameter of the stays of a boiler in which the pressure is 30 lbs. per sq. in., allowing one stay to each i^ sq. ft. of surface and a stress of 3500 lbs. per sq. in. of section of iron ? Ans. li in. 19. A force of 10 lbs. stretches a spiral spring 2 in. Find the work done in stretching it successively i in., 2 in., 3 in., up to 6 in. Ans. |, V, V-. ¥-. -F. ^^ in. -lbs. 20. A roof tie-rod 142 ft. in length and 4 sq. in. in sectional area is subjected to a stress of 80,000 lbs. U E = 30,000,000 lbs., find the elongation of the rod and the corresponding work. Ans. 1. 136 in.; 3786I ft.-lbs. 21. An iron wire i in. in diameter and 250 ft. in length is subjected to a tension of 600 lbs., the consequent strain being j^. Find E, and show by a diagram the amount of work done in stretching the wire within the limits of elasticity. Ans. 1 4,661, 8 iS^^ lbs. 22. A timber pillar 30 ft. in length has to support a beam at a point 30 ft. from the ground. If the greatest safe strai.. of the timber is 3-J^, what thickness of wedge should be driven between the head of the pillar and the beam ? Ans. -^ ft. I- , .« -■•■-'» .. .ir 2l8 THEORY OF STRUCTURES. 23. An hydraulic hoist-rod 50 ft. in length and i in. in diameter is attached to a plunger 4 in. in diameter, upon which the pressure is 800 lbs. per sq. in. Determine the altered length of the rod, E being 30,000,000 lbs. Ans. .0213 ft. 24. A short cast-iron post is to sustain a thrust of 1000 lbs., the ul- timate crushing strength of the iron being 80,000 lbs. per sq. in. and ic a factor of safety. Find the dimensions of the post, which is rectangular in section with the sides in the ratio of 2 to i. Ans. 4 in.; 2 in. 25. The length of a cast-iron pillar is diminished from 20 ft. to 19.97 ft. under a given load. Find the strain and the compressive unit stress, E being 17,000,000 lbs. Ans. .0015 ; 25,500 lbs. per sq. in. 26. A rectangular timber strut 24 sq. in. in sectional area and 6 ft. in length is subjected to a compression of 14,400 lbs. Determine the diminution of the length, E being 1,200,000 lbs. Ans. .003 ft. 27. Find the height from which a weight of 200 lbs. may be dropped so that the maximum admissible stress produced in a bar of i sq. in. section and 5 ft. long may not exceed 20,000 lbs. per sq. in., the co- efficient of elasticity being 27,000,000 lbs. Ans. ^j ft., or, more accurately, j^o ^^• 28. Find the H. P. required to raise a weight of 10 tons up a grade of I in 12 at a speed of 6 miles per hour against a resistance of 9 lbs. per ton. Ans. 31.3. 29. A square steel bar 10 ft. long has one end fixed ; a sudden pull of 40,000 lbs. is exerted at the other end. Find the sectional area of the bar consistent with the condition that the strain is not to exceed j\^. E = 30,000,000 lbs. Find the resilience of the bar, Ans. 2 sq. in. ; 533^ ft.-lbs. _j,Q 30, How much work is done in subjecting a cube of 125 cu. in. uf iron to a tensile stress of jooo lbs. per sq. in. } Ans. 11^ ft.-lbs. 31. A signal-wire 2000 ft. in length and ^ in. in diameter is subjected to a steady stress of 300 lbs. The lever is suddenly pulled back, and the corresponding end of the wire moves through a distance of 4 in. De- termine the instantaneous increase of stress. Ans. S^\jI lbs. 32. If the total back-weight is 350 lbs., what is the range of the sig- nal end of the wire? Ans. -rifh ft. 33. A steel rod of length L and sectional area A has its upper end fixed and hangs vertically. The rod is tested by means of a ring weigh- ing 60 lbs. which slides along the rod and is checked by a collar screwed to the lower end. A scale is marked upon the rod with the zero at the fixed end. If the strain in the steel is not to exceed 7-J-5, what is the reading from which the weight is to be dropped } What should be the reading of the collar } E — 35,000,000 lbs. Ans. Distance from point of suspension = (|§J — l^A)L ; ^^^L. 34- A io pending roc 2 sq. in., le duced. 35- If th 2000 lbs. is J 36. Steal upon a pis length and done upon t 37. What proof of .001 38. A boi diameter har ft. of rope w; block being ■ lbs. What w 39- The b it was again 1 40. What rope is not to 41. The st versed. Shov 42. A weig weight is furth to oscillate, equilibrium. 43- If a spr period of oscil 44. Show ti a longitudinal and ml when * 45. A steel 1 20,000 lbs. per per cubic inch. EXAMPLES. 219 34. A load of 1000 lbs. falls i in. before commencing to stretch a sus- pending rod by which it is carried. If the sectional area of the rod is 2 sq. in., length 100 in., and E = 30,000,000 lbs., find the stress pro- duced. Ans. 17,828 lbs. per sq. in. 35. If the rod carries a load of 5000 lbs., and an additional load of 2000 lbs. is suddenly applied, what is the stress produced ? Ans. 4.500 lbs. per sq. in. 36. Steam at a pressure of 50 lbs. per sq. in. is suddenly admitted upon a piston 32 in. in diameter. The steel piston-rod is 48 in. in length and 2 in. in diameter, E being 35,000,000 lbs. Find the work done upon the rod. Ans. 117.69 ft. -lbs. 37. What should be the pressure of admission to strain the rod to a proof of .001 } Ans. 68ff lbs. per sq. in. 38. A boulder-grappler is raised and lowered by a wire rope i in. in diameter hanging in double sheaves. On one occasion a length of 1 50 ft. of rope was in operation, the distance from the winch to the upper block being 30 ft. The grappler laid hold of a boulder weighing 20,000 lbs. What was the extension of the rope, E being 15,000,000 lbs. ? Ans. -rhr ft. 39. The boulder suddenly slipped and fell a distance of 6 in. before it was again held. Find the maximum stress upon the rope. Ans. 50,452^ lbs. per sq. in. 40. What weight of boulder may be lifted if the proof-stress in the rope is not to exceed 25,000 lbs. per sq. in. ol gross sectional area? Atts. 78,5711 lbs. 41. The steady thrust or pull upon a prismatic bar is suddenly re- versed. Show that its effect is trebled. 42. A weight W is suspended by a spring, which it stretches. The weight is further depressed i ft., when it is suddenly released and allowed to oscillate. Find its velocity at a distance x from the position of equilibrium. Ans. y 10(1 — 10.1-^)-"—. 43. If a spring deflects .001 ft. under a load of i lb., what will be the period of oscillation of a weight of 14 lbs. upon the spring? 44. Show that the change of a unit of volume of a solid body under a longitudinal stress is a( i ), which becomes — if /// = 4, as in metals, \ ml 2 and «//when tn = 2, as in india-rubber (page 142). 45. A steel bar stretches TjVif'h of its original length under a stress of 20,000 lbs. per sq. in. Find the change of volume and the work done per cubic inch. Ans. ^■^^\.\\ ; | ft. -lb. per cu, in. . I ' '•■« ,j .1 ? 1 t". 1 'I ii 1 I 'm t i[- ^ ! '. I , » 220 THEORY OF STRUCTURES. m iii 46. During the plastic deformation of a prismatic bar, show that the change in sectional area is proportional to the deformation calculated on the altered length of the bar. 47. A prismatic bar of volume V changes in length from L to L ± x under the " fluid pressure"/. Find the corresponding work. Ans. pV\oge{L ± x). 48. Show that the total work done in raising a number of weights through to a given level is the product of the sum of the weights and the vertical displacement of their centre of gravity. 49. An engine has to raise 4000 lbs. 1000 ft. in 5 minutes. What is its H. P. ? How long will the engine take to raise 10,000 lbs. 100 ft. ? Ans. 24/j H. P. ; ij min. 50. How many men will do the same work as the engine in the pre- ceding question, assuming that a man can do 900,000 ft. -lbs. of work in a day of 9 hours ? Ans. 480 men. 51. Determine the H. P. which will be required to drag a heavy rock weighing 10 tons at the rate of 10 miles an hour on a level road, the coefficient of friction being 0.8. What will be the speed up a gradient of I in 50, the same power being exerted ? Ans. 47711 ; ^l\ miles per hour. 52. Two horses draw a load of 4000 lbs. up an incline of i in 25 and 1000 ft. long. Determine the work done. Ans. 160,000 ft.-lbs. 53. At what speed do the horses walk if each horse does 16,000 ft.- lbs. of work per minute ? Ans. 2r(\ miles per hour. 54. A wrought-iron rod 25 ft. in length and i sq. in. in sectional area is subjected to a steady stress of 5000 lbs. What amount of live load will instantaneously elongate the rod by i in., E being 30,000,000 lbs. .' Ans. 6250 lbs. 55. Determine the shortest length of a metal bar a sq. in. in sec- tional area that will safely resist the shock of a weight of JV lbs. falling a distance of ^ ft. Apply the result to the case of a steel bar i sq. in. in sectional area, the weight being 50 lbs., the distance 16 ft., the proof- strain 7-57. and E = 35,000,000 lbs. 2EJVak Ans. ;.,/ being the safe unit stress ; ^ Jlg " ft. ay^-2£lV/ 56. A shock of TV ft.-lbs. is safely borne by a bar / ft. in length and a sq. in. in sectional area. Determine the increased shock which the bar will bear when the sectional area of the last mthoi its length is increased Ans. Nil 1 . \ m rmj 57. The bar in Example 12 is i sq. in. in section. Determine the work stored up in the rod in foot-pounds and compare it with the work which would be st( to 4 in. 58. If 25 resilience fo 59. A st< E = 3S,ooo,c area of the the weight o determine tl /j in., succei An. 60. A hm 32 F. Dete the work stoi 61. A wro stretches .000 E = 29,000,0c How may ening walls tl 62. Find 1 weight of a s number of, slal 63. How r plates, each 3 strong as the wrought-iron I 64. A horiz area S, has its /F suspended 2ontal. Show f being the inl ficient of elasti 65, A heav) ends fixed in a weight. Find throughout). EXAMPLES. 221 would be stored up if for half its length the rod has its section increased to 4 in. Ans. 125 ft.-lbs.; ^ of 125 ft.-lbs. 58. If 25,000 lbs. per sq. in. is the proof-stress, find the modulus of resilience for the i-in. rod. Ans. 25 in in. -lb. units. 59. A steel rod 100 ft. in length has to bear a weight of 4000 lbs. If E = 35,000,000 lbs., and if the safe strain is .0005, determine the sectional area of the rod (i) when the weight of the rod is neglected ; (2) when the weight of the rod is taken into account. Also in the former case, determine the work done in stretching tlie rod -jV '"•• tV '"- fV '" {^ in., successively. Ans. ^g sq. in. ; //yV sq- in. ; 33i, 133^, 300, ... 1200 in. -lbs. 60. A line of rails is 10 miles in length when the temperature is at 32 F. Determine the length when the temperature is at 100° F.. and the work stored up in the rails, E being 30,000,000 lbs. Ans. 10.008 miles ; 10.24 H. P. 61. A wrought-iron bar 25 ft. in length and i sq. in. in sectional area stretches .0001745 ft. for each increase of 1° F. in the temperature. If E = 29,000,000 lbs., determine the work done by an increase of 20° F. How may this property of extension under heat be utilized in straight- ening walls that have fallen out of plumb ? Ans. 7.064 ft.-lbs. 62. Find the work done in raising a Venetian blind, iv being the weight of a slat, a the distance between consecutive slats, and n the number of slats. , n{n + i) Ans. wa . 63. How many |-in. rivets must be used to join two wrought-iron plates, each 36 in. wide and \ in. thick, so that the rivets may be as strong as the riveted plates, the tensile and shearing strength of wrought-iron being in the ratio of 10 to 9? Atis. 17 rivets (16.3). 64. A horizontal string, without weight, of length 2a and sectional area S, has its two ends fixed in the same horizontal plane. A weight /K suspended from its centre draws the string slightly out of the hori- zontal. Show that, approximately. '=K^')* '""* '^^'^l^)* m t being the intensity of the tension, d the depression, and E the coet- ficient of elasticity. 65. A heavy wire of length 2a, sectional area 5, and weight W has its ends fixed in a horizontal plane and is allowed to deflect under its own weight. Find the deflection d and the tenacity / (assumed uniform throughout). •f!s 222 THEORY OF STRUCTURES. \\i\ •> 66. A length 270 ft. of wire i sq. in. in section and of sp. gr. 7.8 is subjected to tlie above conditions. Find the tenacity of the wire and the deflection, the coefficient of elasticity, E, being 25,300,000 lbs. 67. A brick wall 2 ft. thick, 12 ft. high, and weighing 112 lbs. per cu. ft. is supported upon solid pitch-pine columns 9 in. in diameter. 10 ft. in length, and spaced 12 ft. centre to centre. Find the compress- ive unit stress in the columns (1) at the head; (2) at the base. The tim- ber weighs 50 lbs. per cu. ft. Ans. 507.03 lbs. ; 510.5 lbs. 68. If the crushing stress of pitch-pine is 5300 lbs. per sq. in. and the factor of safety 10, find the height to which the wall may be built. Ans. 12.46 ft. 69. Determine the diameter of the wrought-iron columns which might be substituted for the timber columns in question 67, allowing a working stress in the metal of 7500 lbs. per sq. in. Ans. 2.36 in. 70. Find the greatest length of an iron suspension-ro^i which will carry its own weight, the stress being limited to 4 tons per sq. in. What will be the extension under this load, E being 12,500 tons.' Ans. 2700 ft. ; .864 ft, 71. A horizontal cast-iron bar i ft. long exactly fits between two verti- cal plates of iron. How much should its temperature be raised so that it might remain supported between the plates by the friction, the coef- ficient of friction being j^ ? Ans. -^'' F. 72. The fly-wheel of a 40 H. P. engine, making 50 revolutions per minute, is 20 ft. in diameter and weighs 12,000 lbs. What is its kinetic energy? If the wheel gives out work equivalent to that done in raising 5000 lbs. through a height of 4 ft., how much velocity does it lose? Tho axle of the fly-wheel is 12 in. in diameter. What proportion of the H. P. is required to turn the wheel, the coefficient of friction being .08? If the fly-wheel is disconnected from the engine when it is making 50 revolutions per minute, how many revolutions will it make before it comes to rest ? Ans. 51 1,260.4 ft.-lbs. ; 1.04 ft. per sec. ; |ths ; 169.4. 73. The velocity of flow of water in service-pipe 48 ft. long is 64 ft. per sec. If the stop-valve is closed in \ of a sec, find the increase of pressure near the valve. Ans. 375 lbs. per sq. in. 74. Work equivalent to 50 ft.-lbs. is done upon a bar of constant sectional area, and produces in it a uniform tensile stress of 10,000 lbs. per sq. in. Find the cubic content of the bar, E being 30,000,000. Ans. 360 cu. in. 75. A fly-wheel weighs 20 tons and its radius of gyration is 5 ft. How much H per min 76. '. 30,000 fi was con same six n. / uniform fulfil ? 78. E 1,650,000 nules pel 79- A miles an weight ol 80. A long, agai H. P. of 1 H. P., is . 81. Th tional are metal is n pillar mig 82. A area carrit termine tl 35,000.000 83. A strained to the rod, E 84. Wh nail is driv of the mass 85. A h 10 ft. per s( force on th 86. A hi a nail 5 in. the rnomen steady press •\ir n "WZi m EXAMPLES. 223 much work is given out while the speed fails from 60 to 50 revolutiono per minute ? ^hts. y+iVA ft. -tons 76. The resilience of an iron bar i sq. in. in section and 20 ft. long is 30,000 ft.-l' Vhat would be the resilience if for 19 ft. of its length it was compc of iron 2 sq. in. in section, the remaining foot being the same size as before ? Ans. 8625 ft.-lbs. 77. A particle under the action of a number of forces moves with a uniform velocity in a straight line. What condition must the forces fulfil } Ans. Equilibrium. 78. Determine the constant effort exerted by a horse which does 1,650,000 ft.-lbs. ol work in one hi)ur when walking at the rate of 2^ miles [ler hour. Ans. 125 lbs. 79. A train is drawn by a locomotive of 160 H. P. at the rate of 60 miles an hour against a resistance of 20 lbs. per ton. What is tlie gross weight of the train ? Av.s. 50 tons. 80. A train of 292J tons is drawn up an incline of i in 75, 5^ miles long, against a resistance of 10 lbs. per ton, in ten minutes. Find the H. P. of the ' 'ine. The speed on the level, the engine exerting 769.42 H. P., is 43. "S per hour. Wiiat is the resistance in pounds per ton ? Ans. 1027 H. P. ; 22.7 lbs. per ton. 81. The dead load upon a short hollow cast-iron pillar with a sec- tional area of 20 sq. in. is 50 tons (of 2000 lbs.). If the strain in the metal is not to exceed .0015, find the greatest live load to which the pillar might be subjected, E being 17,000,000 lbs. Ans. 255,000 lbs. 82. A steel suspension-rod 30 ft. in length and ^ sq. in. in sectional area carries 3500 lbs. of the roadway and 3000 lbs. of the live load. De- termine the gross load and also the extension of the rod, E being ',5,000.000 lbs. ^i"^- ^Ih ft- 83. A steel rod 10 ft. in length and ^ sq. in. in sectional area is strained to the proof by a tension of 25, 00 lbs. Find the resilience of the rod, E being 35,000,000 lbs. Ans. 178^ ft.-lbs. 84. What form does the useful work done by a hammer take when a nail is driven into any material ? What becomes of the rest of the energy of the mass of the hammer after striking the blow } 85. A hammer weighing 2 lbs. strikes a steel plate with a velocity of 10 ft. per sec, and is brought to rest in .0001 sec. What is the average force on the steet.' Ans. 6250 lbs. 86. A hammer weighing 10 lbs. strikes a blow of 10 ft.-lbs. and drives a nail 5 in. into a piece of timber. Find the velocity of the hammer at the moment of contact, and the mean resistance to entr3\ Also find the steady pressure that will produce the same effect as the hammer. Ans. 8 ft. per sec. ; 240 lbs. ; 480 lbs. :W 5< M 224 THEORY OF STRUCTURES. 87. When :i nail is driven into wood, why do the blows seem to have little if any effect unless the wood is backed up by a piece of metal or stone ? 88. In Question 86, takinj^ the weight of the nail to be 4 oz. and the weight of tiie piece of timber to be 100 lbs., find the depth and time ol the penetration (ii) when the timber is fixed ; (b) when the timber is free to move. Also in case {p) find tlie distance through which the timber moves. Ans. — (d) j',' in.; b'b sec. (/') .44245 ill.; .0009448 sec. ; .04113 in. Si). Show that the greater part of the eiicigy of impact is expended in local damage at high velocities, and in straining the impinging bodies a.s a whole at low velocities. 90. A pile-driver of 300 lbs. falls 20 ft., and is stopped in ^ sec. What is the average force exerted on the pile ? Ans, 3344 lbs. yi. A weight falls 16 ft. and docs 2560 ft.-lbs. of work upon a pile which it drives 4 in. against a uniform resistance. Find the weight of the ram, and the resistance. Ans. 160 lbs. ; 7680 li)s. 92. A pitch-pine pile 14 in. square is 20 ft. above ground, and is being driven by a falling weight of 112 lbs. \{ E = 1,500,000 lbs., find the fall so that the inch-stress at the head of the pile may be less than 800 11)S. Supposing that the pile sinks 2 in. into the ground, by how much would it be safe to increase the fall.' Ans. 7.456 ft. ; 1 16.5 ft. 93. A weight of W\ tons falls // ft., and by n successive instantaneous blows drives an inelastic pile weighing \Vi tons a ft. into the ground. Assuming the pile and weight to be inelastic, lind ( resistance of tlie ground. If the ground-resistance increases directly as the depth of penetration, find (/') how far the pile will sink under tlie /th blow. If the head of tin* pile is crushed for a length of .1 ft., x being very small as compared with the depth — of penetration, find (i) the mean thrust, during the blow. between the weight and hammer ; (2) the time of penetrating the ground : (3) the time during which the blow arts. Ans,-(ii) ri) w «// • ' 1 (2) \Vx 4- \W ) proportional to tlie dcptii ol jieiietration. If tiie resistance is uniform, how lon^ (t) docs cacii move- iiuMit of the pile last? How many blows (d) are re(|iiirefl to drive tiie l)ilc the first lialf of tiic depth, viz., \\ ft., tiie ground-resistance iieing 7168 lbs. ? How far {c) does the pile sink under the last IjIow } Alls, (a) 14,336 lbs. ; {/>) 28,672 lbs. ; (c) .0107 sec. ; (d) 30; (c) .016 in. 95. A steamer of 8000 tons displacement sailing due east at 16 knots an hour collides witii a steamer of 5000 tons displacement sailing at 10 knots an hour. Find the energy of collision if the latter at the moment of collision is going (1) due west; (2) north-west; (3) north-east. t)G. A hammer weighing 2 lbs. strikes a nail with a velocity of 15 ft. per sec, driving it in A in. What is the mean jjressure overcome tjy the nail ? A/IS. 673 iljs. 97. A beam will safely carry i ton witli a deflection of 1 in. Fr(jm what height may a weight of 100 lbs. drop without injuring it, neglecting the elfect of inertia ? //w. 11.2 in. 98. A rifie-l)iillet .45 in. in diameter weighs 1 oz. ; the cliarge of pow- der wei^lis 85 grains; the mnzzle-velocity is 1350ft. per sec; the weight of the rifle is 9 lbs. Nc}:;lcctin}r the twist determine the energy of 1 lb. of powder. If the bullet loses \ of its velocity in its passage tlirougii the air, find the average force of llu; blow on the target into wiiich tiie bullet binks I, in. If tliere is a twist of i in 20 in., find the charge to give the same muzzle- velocity, the length of tl;': barrel being 33 in. 99. A leather belt runs at 2400 ft. per minute. Find how much its tension is increased by centrifugal action, the weight of leather being taken at 60 lbs, per cubic foot, , Ans. 20% lbs. 100. Find the centrifugal force arising from a cylindrical crank pin 6 in. long and 3^ in. in diameter, the axis of the pin being 12 in. fnmi the axis of the engine-shaft, \.!Mch makes 200 revolutions per minute. How would you balance such a pin } ^Ins. 55.02 lbs. 101. The i)ull on one of the tension-bars of a lattice girder fluctuates Irom 12.8 tons to 4 tons. If 24 tons is the statical breaking strength ot ilie metal, 15 tons the primitive strength, determine the sectional area of the bar, 3 being a factor of safety. Ans. 2.15 sq. in. (Launhardt) ; 1.87 sq. in. (Unwin). 102. The stress in a diagonal of a steel bowstring girder fluctuates from a tension of 15.15 ^""s to a compression of 7.65 tons. If the primitive strength of the metal is 24 tons and the vibration strength 12 H I; rl 226 THEORY OF STRUCTURES. tons, find the proper sectional area of the diagonal, 3 being a factor of safety. Ans. 2.53 sq. in. (Weyrauch) ; 1.7 sq. in. (Unvvin), 40 tons per sq. in. being statical strength. 103. A wrought-iron screw-shaft is driven by a pair of cranks set at light angles. Neglecting the obliquity of the connecting-rods, and assuming that the pull on the crank-pin is constant, compare the coef- ficients of strength (ti' and /) to be used in calculating the diameter of I he shaft. How is the result affected by the stopping of the engine.'' Ans. a = .82/; ) VVith in 100.5 (<■■) If tlu- (0 ON the in( () With the same. H. P. what will be the speed up a gradient of i in 100? (c) If the steam is shut ofT, how far will the train run before Stopping (i) on the incline; (2) on the level.'' ((/) If the draw-bar suddenly breaks, in what distance would the carriages (100 tons in weight) be stopped if the brakes are applied im- mediately the fracture occurs, the weight of the brake-van being 20 tons and the coefficient of friction .2 .^ ■H I' 'ii4 '4 1 ■ \:l '•-♦ •'t- -If , '■i Mm ,udMm 228 THEORY OF STRUCTURES. (e) If the engine (weight = 60 tons) continued to exert the same power after the fracture, what would be its ultimate speed ? (/) What resistance would be required to stop the whole train after steam is shut off, in 1000 yards on the level ? Ans. {a) 128; (d) g^j\ miles per hour; (c) (i) 199.2 ft., (2) 6776 ft. ; (flf) 680.3 ft. on the level, 52.9 ft. on the incline; (e) 80 miles an hour on the level, 24.6 miles on the incline ; (/) 22. 5S lbs. per ton. 117. A 4-in. X 3-in. diameter crank-pin is to be balanced by two weights on the same side of the crank ; the length of the crank is 12 in. ; the engine makes 100 revolutions per minute; the distance of the C. of G of each weight from the axis of the shaft is 6 in. Find the weights. 118. A shaft is worked with cranks at 120°. Assuming the pressure on the crank-pin to be horizontal and constant in amcjunt, compare the coelBcients of actual and ultimate strength to be used in calculating the diameter of the shaft. Ans. a' = .507/. 119. In a horizontal marine engine with two cranks at riglit angles distant 8 ft. from one another, weight of reciprocating parts attached to each crank is lo tons, revolutions 75 per minute, stroke 4 ft. Find the alternating force and couple due to inertia. Afis. 54.2 tons; 216.8 ft.-tons. 120. An inside-cylinder locomotive is running at 50 miles an hour; the driving-wheels are 6 ft. in diameter; the distance between the centre- lines of the cylinders is 30 in., the stroke 24 in,, the weight of one piston and rod 300 lbs., and the horizontal distance between the balance- weights 4^ ft. ; the diameter of the weight-circle is 4^ ft. Find the alternating force and couple, and also the magnitude and position of suitable balance- weights. Ans. 7871 lbs; 9839 ft.-lbs. ; 106.5 lbs. ; 27f°. 121. The pressure equivalent to the weight of the reciprocating parts of an engine is 3 lbs. per sq. in.; the stroke is 36 in. ; the number of revolutions per minute is 45; the back-pressure is 2 lbs. per sq. in.; ihe absolute initial steam- pressure is 60 lbs. per sq. in.; the rale of expansion is 3. Find the pressure necessary to start the piston, and also the effec- tive pressure at each ^ of the stroke. 122. An engine with a 24-in. cylinder and a connecting-rod = six cranks = 6 ft., makes 60 revolutions per minute. Show that the pressure re- quired to start and stop the engine at the dead-points = ^^ of the weight of reciprocating parts. 123. Find the ratio of hrust at cross-head to tangential effort on crank-pin when the crank is 45° from the line of stroke, the connecting- rod being = four cranks. if EXAMPLES. 229 124. Draw the linear diagram of cranic-effort in tlie case of single crank, the connecting-rod being = four cranks. Assume the resistance uniform and a constant pressure of 9000 lbs. on the piston, the stroke oeing 4 ft. and the number of revolutions per minute 55. Also find the fluctuation of energy in ft.-lbs. for one revolution. 125. An engine with a connecting-rod = six cranks = 6 ft. receives steam at 70 lbs. pressure per sq. in., and cuts oft at one-quarter stroke. Find the crank-effort when the piston has travelled one third of its for- ward stroke. Diameter of piston = 2 ft. Also find the position of the piston where its velocity is a maximum. 126. Data: Stroke = 3 ft. ; number of revolutions per minute = 60; cut-off at one-half stroke; initial pressure = 56 lbs. persq. in. absolute; diameter Oi piston = 10 in. ; weight of reciprocating parts = 550 lbs. ; back-pressur.^ = \\ lbs. per sq. in. absolute. Find the effective pressure at each fourtl: of the stroke, taking account of the inertia of the piston. Also find the pressure equivalent to inertia at commencement of stroke. 127. A pair of 250 H. P. engines, with cranks at 90°, and working against a uniform resistance and under a uniform steam-pressure, are running at 60 revolutions per minute. Assuming an indefinitely long connecting-rod, find the maximum and minimum moments of crank- effort, the fluctuation of energy, and the coefficient of energy. 128. An inside-cylinder locomotive runs at 25 miles per hour; its drivers are 60 in. in diameter; the stroke is 24 in. ; the distance between the centre-lines of the cylinders = 30 in.; weight of reciprocating parts = 500 lbs. ; horizontal distance between balance-weights = 59 in. ; diameter of weight-circle = 42 in. Find the alternating force, alternat- ing couple, and the magnitude and position of suitable balance-weights. Ans. 226.8 lbs.; 41 13.8 ft.-lbs. ; = 26^ 129. Draw a diagram of crank-effort for a single crank, the connect- ing-rod being equal to four cranks, the stroke 4 ft., and the number of revolutions per minute 55. Assume a uniform resistance and a constant pressure of 9000 lbs. on the piston. 130. A vertical prismatic bar of weight ffi , sectional area .^, and length L has its upper end fixed, and carries a weight W% at the lower end. Find the amount and work of the elongation. ea\- + ^^^j •' "'^^'^ = ea[-J 131. A right cone of weight W and height h rests upon its base of radius r. Find the amount and work of the compression. Wh I W^ Am. Ext. + lVtlV, + lVi^] Ans, Comp. = — 27TEr' work = 8 rrE'r 132. A tower of height A, in the form of a solid of revolution about a vertical axis, carries a given surc/iartfc. If the specific weight of the n fm '■ m ■?>'!' 'U T i- V^ 230 THEORY OF STRUCTURES. material of the tower is w, and the radius of the base a, determine the curve of the generating line so that the stress at every point of the tower may be/. If the surcharge is zero and the heiglit of the tower becomes infinite, show that its volume remains finite. Ans. y = ac wx vol. of tower of infinite height i=—na w VIX T. 133. Determine the generating curve when the tower in the last question is hollow, the hollow part being in the form of a right cylinder upon a circular base of given radius R. Ans. y — R* - (rt" - R')e 134. A iieavy vertical bar of length / and specific weight w is fixed at its upper end and carries a given weight IV at the lower end. Deter- mine the form of the bar so that the horizontal sections may be pro- portionate to the stress/ to which they- are subjected. {Note. — Such a bar is a bar of uniform strength.) Ans. Sectional area at distance .«• from origin = — e 135. Find the upper and lower sectional areas of a steel shaft of uni- form strength, 200 ft. in length, which will safely sustain its own weij^ht and 100 tons, 7 tons per sq. in. being tlie woricing stress. Ans. 14.3 sq. in. ; 17.8 sq. in. 136. A vertical elastic rod of natural length L and of which the mass may be neglected, is fixed at its upper end and carries a weight W\ at the lower end. A weight Wt falls from a height h upon W\ . Find the velocity and extension of the rod at any time /. Wi -f- lV-\ L ) \dil X being measured from mean position of {IVi + IVi), 137. Determine the functions F kind/ in Art. 24 wlien Pi is zero, ami also when the rod is perfectly free ; i.e., when. A = o and A = o. 138. An elastic trapezoidal lamina ^y5CA of natural length / and thickness unity, has its upper edge AB {2a) fixed and hangs verti- cally. If a weight IV is suspended from the lower edge CD {2/A, show th.1t, neglecting the weight of the lamina, the consequent elongation = ^ -fzr ■■ , log,—. If an adcitional weight is placed upon IF iu\i\ then suddenly removed, show that the oscillation set up is isochronous r fF/log, and that the time of a complete oscillation = n^ — pr — Examine the case when a = b. f>) i Ans. Ext. I IVl --" ; time of oscillation = itA/ JU- 139. J w, find li work of e Work 140. A its base A its elonga = unity, a 141- A its upper e striking th (middle C of the rod 142. Di of 6 ft. radi second. E measured a 143- A ( second abo normal to t 144- In i of water of denly check E being the coefficient o pipe circumf 145- A he a horizontal vertical dept If a unif( position. Also find i EXAMPLES. 23' 139. If the specific weight of the lamina in the preceding question is w, find how mucli it will stretch under its own weij^ht, and also the work of extension. Determine the result when a = b. I 'uib'T b wra+b wP Alls. —- :— log - + --7 - — -, ; -TT- 2t {a — bf ^ a i,E a — b zE Work = ■wP \E{a — by \-"'' b" 2 b^)-b^W^,~ 140. An elastic lamina in the form of an isosceles triangle ABC has its base AB (= la) fixed and hangs vertically. If its weight is W, fmd its elongation. Take coefficient of elasticity = is, thickness of lamina = unity, and L the distance of C from AB. ^ \VL •' Ans. . 6,ah 141. A metal rod \ sq. in. in area and 5 ft. long hangs vertically with its upper end fixed and carries a weight of 18 lbs. at the lower end. On striking the rod it emitted a musical note of 264 vibrations per second (middle C of piano-forte). Find the coefficient of elasticity, the weight of the rod being neglected. Ans. 30,979,160 lbs. 142. Diameter of a pipe is 18 in. ; at one point it is curved to an arc of 6 ft. radius. Water flows round the curve with a velocity of 6 ft. per second. Determine the centrifugal force per foot of length of elbow measured along the axis. Ans. 124.3 'bs. 143. A disk of weight fFand area A sq. ft. makes n revolutions per second about an axis through its centre, inclined at an angle Q to the normal to the plane of the disk. Find the centrifugal couple. WAii" Ans. 5.12 tan ft. -lbs. 144. In a circular pipe of internal radius r and thickness t, a column of water of length I, flowing with a velocity due to the head //, is sud- denly checked. Show that gh^ EtX^ |' + 2K'+l)+$^[' E being the coefficient of elasticity of the material of the pipe, E^ the coefficient of compressibility of the water, and /I the extension of the pipe circumference corresponding to E. 145. A heavy ball attached by a string to a fixed point O revolves in a horizontal circle with a given uniform angular velocity gj. Find the vertical depth of the centre of the ball below the point of attachment. If a uniform rod be substituted for the ball and string, find its position. Also find the position when the ball is attached to the fixed point by ' • Mum ■ 1i ' 1 f u •lt::;il rw% 1 1 232 THEORY OF STRUCTURES. a uniform rod ; r being the ratio of the weight of the rod to the weight of the ball. Ans. h = = .C. . = 3£ GJ- // = ^^-;/.=^ H I + - g 2 2 W' O) ' n I + - 3 Ans. 146. The deflection of a truss of /ft. span is / x .001 under a station- ar\' load W. What will be the increased pressure due to centrifugal force when If crosses the bridge at the rate of 60 miles an hour? 242 W 725 "7"' 147. A fly-wheel 20 ft. in diameter revolves at 30 revolutions per minute. Assuming weigiit of iron 450 lbs. per cu. ft., find the intensity of the stress on the transverse section of the rim, assuming it unaffected by the arms. Ans. 96 lbs. per sq. in. 148. Assuming 15,000 lbs. per sq. in. as the tensile strength of cast- iron, and taking 5 as a factor of safety, find the maximum working speed and the bursting speed for a cast-iron fly-wheel of 20 ft. mean diameter and weighing 24,000 lbs., the section of the rim being 160 sq. in. 149. A 60-in. driving-wheel weighs 3^ tons, and its C. of G. is i in. out of centre. Find the greatest and the least pressure on the rails. 150. A wheel of weight W, radius of gyration k, and making n revolutions per second on an axle of radius /?, comes to rest after having made N revolutions. Find the coefficient of friction. ■ Ans. sm = — - — , and coefi. of fric. = tan . Ng 151. A train starts from a station at A and runs on a level to a station at B, /ft. away. If the speed is not to exceed v ft. per sec, show that the time between the two stations is l_ W_v_ P + B v"^ g 2 {P-R){B + R)' W being the gross weight of the train, P the mean uniform pull exerted by the engine, R the road resistance, and B the retarding effect of the brakes. Also, if the speed ['• not limited, show that the least time in which the train can run between the specified points is V 2/ IV P + B g iP-R)(B + R) and that the maximum speed attained is - sec, -R)( B + R) P + B ft. per sec. 152. A a 24-in. si between t locomotiv lbs.); the the wheel; friction be distance i attained ; 1 wheels. A 153. If find (a) th< is brought '54. If lbs. per sq hauled bet lbs. per toi the train u Also fii drivers, th attained. Ans.- 155- Th locomotive diameter o quired to s 156. Tw lbs.), run b an average the other s fuel in the per hour. 157. If 1 the wind, w gale will dr ton. 158. A 1 ried by the ( ton, what n^ %£ EXAMPLES. 233 152. A locomotive capable of exerting a uniform pull of 2 tons, with a 24-in. stroke, 20-in. cylinder, and 60-in. driving-wheels, hauls a train between two stations 3 miles apart. The gross weight of the train and locomotive = 200 tons; the road resistance = 12 lbs. per ton (of 2000 lbs.); the brakes, when applied, press with two thirds of the weight on the wheels of the engine and brake-van, viz., 90 tons, the coefficient of friction being .18. Find (a) the least time between the stations ; {b) the distance in which the train is brought to rest; (c) the maximum speed attained ; (rf) the pressure of the steam ; (e) the weight upon the driving- wheels. Ans. — {a) 513.8 sec. ; (b) 990 ft.; {c) 42 miles per hour; {d) 25 lbs. per sq. in. ; (e) \\\ tons. 153. If the speed in the last question is limited to 30 miles an hour, find {a) the time between the stations; ( sin B. inn cosec u A The normal com- ponent of the intensity on m'n' = p sin' 6 = pj. Pig. 207,. 235 f'J *1 236 THEORY OF STRUCTURES. The tangential component or shear on m'n' = / sin 6* cos = //. So, if m"n" is an oblique plane perpendicular to m'n', the nor- mal component of the intensity on 7n''n" = p cos" B = p„". The tangential component or shear on m"n" =zp cos 6 sin 6 = p/'. ... p^' _|_ pj' = p and p/ = pr =ps\ne cos d = P_^}^, The shear is evidently a maximum when 2B = 90° or e ^ 45°. 3. Compound Strain. — {a) First consider an indefinitely small rectangular element OACB (Fig. 208) of a strained body, p p p p ^^Pt "' equilibrium by stresses ^^ \ \ \ acting as in the figure. ' ■ '" ' ip p is the intensity of stress on -"■''^ the faces OB, A C, and a its ob- -O'/ liquity. "' q is the intensity of stress on '^ '"^ the faces OA, EC, and /? its ob- FiG. 208. liquity. OB .p cos a, the total normal stress on OB, is balanced by AC.p cos a, the total normal stress on AC. OB .p sin a, the total shear on OB, is equal in magnitude but opposite ii direction to AC .p sin a, the total shear on AC. These two forces, therefore, form a couple of moment OB .p sin a.OA. Similarly, the total normal stresses on the faces OA, BC balance and the total shears form a couple of moment OA. g sin /3. OB. In order that equilibrium may be maintained the two couples must balance. or .-. OB .p sin a.OA = OA.q sin /?. OB, p sin a T= q sin ft = t, suppose. COMPOUND STRAIN. ^17 B Fig. 209. Hence, at any point of a strained body, the intensities of the shears on any two planes at right angles to each other are equal. (d) Next consider an indefinitely small triangular element OAB (Fig. 209) of the strained body, bounded by a plane AB and two planes OA, OB at right angles to each other. Let p be the intensity of stress on OB, a its obliquity. Let g be the intensity of stress on OA, li its obliquity. Let t be the intensity of shear on each of the planes OA, OB. Then ^ = / sin « = ^ sin ft. p„ , the normal component of/, = / cos a. q„ , " " " " ^. = ? cos /3. Produce OA and take OC = p„ . OB -}- t . OA = the total force on OB in the direction of OA. Produce OB, and take OD = q„. OA -\- t . OB = the total force on OA in the direction of OB. Complete the rectangle CD. OE represents in direction and magnitude the resultant of the two forces OC, OD, and must therefore be equal in magni- tude and opposite in direction to the total stress on AB. Let/,, be the intensity of stress on AB. Then (/, . AB^' - OE' OC + OD' = {p„.OB + t. OA)" + ( cos y — sin '/» sin y Px—Pi. Px— P-i — sin ;^ cos y = /. sin ONR 2pr sih 2;/. /,- 2/, -^ 2 .-. sin 6>^V^ = sin 2;/, or 0NR — 2y. . . (13) Let A''A' (Fig- 214) produced in both directions meet OA in F and OB in G. , The angle OFN = 180° - ONR - NOF = 180° — 2y - (90° -y) = go° — y = FON. .-. NF = NO; so, NG = NO = NF. ■ Also and .•. A'' is the middle point of FG. RF= FN -NR=ON- NR=p, RG = RN-\- NG = AW+ ON = p,. i ! N. v5.— The shear at O P.- P. cos {2y — 90") = (/, — p^ sin ;/ cos ^. ..^:n THEORY OF STRUCTURES. 8. Maximum Shear. — ON has no component along AB. Hence, the shear on AB \s NR cos (angle between A^R and AB)y and is evidently a maximum when the angle is nil. Its value is then NR, or ^'"J"^' . 9. Application to Shafting. — At any point in a plane sec- tion of a strained solid, let r be the intensity of stress, and its obliquity. At the same point in a second plane let s be tlie intensity of stress, and B' its obliquity. By Art. 6, r and s are the resultants of two constant stresses and A±A ^„j Aj-A, ... {^)'= f-4-^-=)'+ .- - KA + A) cos e (,4) (^-^f = (^&)'+ .■ - .(A + A) cos »'. (.5) and Subtracting one equation from the other, v =A +A r' — r r cos 6^ — 5 COS . . . (16) First. Consider the case of combined torsion and bending, as when a length of shafting bears a heavy pulley at some point .„ between the bearings. Let p be the intensity of stress (compression or tension) due to the bending moment J/4. ■ ft APPLICATION TO SHAFTING. 245 1 Let q be the intensity of shear due to the twisting nno- ment Mt . p and q act in planes at right angles to each other. ( .-. r cos Q — pt rs\nd — q-s, ar .' 6' — 90''. 4 .'•r'=f-\-Q- and s = q. Hence, by eq. (i6), \ P.^-P.^P\ (17) and by eq. (15), {P^P-^U+f; (,8) ••A = 5+Vt+^ (19) and = i-\/f+^ (^) The max. shear = A -A =/ + ?'; .... (21) also p=^ (Chap. VI.) and ? = ^' (Chap. IX.) for a shaft of radius r. -'-P. = ^AM,-^ V~M,^ + Mt\', • ' . (22) ll'fi' ! » f -it w.!>aWI£ 246 and THEORY OF STRUCTURES. A -P 2 Ttr * n~ / (2.3> M --LA Perhaps the irtost important example of the application of the above principle is the case of a shaft acted upon by a crank (Fig. 2 1 5). A force /'applied to the centre 6" of the crank-pin is resisted by an equal and opposite force at the bearing B, forming a couple of moment P. CB = M. This couple may be resolved into a F'G. 21S. bending couple of moment Mt = P. AB = P . EC cos S = M cos d, and a twisting couple of moment Mt = P . AC = P . BC sin 6 = M sin 6 ; d being the angle ABC. ■ .:p, = -^\Mcos6-\-M\ = ^cos^^; . . (24) ^j^^ Ttr nr and the max. shear = 2M 7ir . . . (25) If the working tensile or compressive stress (/,) and the working sKear stress [— -j are given, the corresponding values of r may be obtained from eqs. (22) and (23) or eqs. (24) and (25) ; the greater value being adopted for the radius of the shaft. Second. Consider the case of combined torsion and tension or compression. Let the tensile or compressive force be P. A the intensity of the tension or compression, = —5 ; shear 2M^ Ttr' S'Jj-i PRINCIPAL AND CONJUGATE STRESSES. 247 and •••A=^.j-V-+^'!- • • <^^^ li t* P. -p. 2itr y ' r' ^ ' 10. Conjugate Stresses. — Consider the equilibrium of an indefinitely small parallelopiped abed {¥\g. 216) of a strained body, the faces ab, cd being parallel to the plane XOX, and the faces ad, Y — -^ -0'- hc to the plane YO Y. Let the stresses on ab, cd act parallel to the plane YOY. The total stresses on ab and cd are '''°- "*• equal in amount, act at the centres of the faces, are parallel to YOY, and therefore neutralize one another. Hence the total stresses on ad diwd be must also neutralize one another. But they arc equal in amount, and act at the middle points of ad, be; they must therefore be parallel to XOX. , , Hence, if two planes traverse a point in a strained body, and if the stress on one of the planes is parallel to the other plane, then the stress on the latter is parallel to the first plane. Such planes are called planes of conjugate stress, and the stresses themselves are called conjugate stresses. Principal stresses are of course conjugate stresses as well. Conjugate stresses h.uc equal obliquities, each obliquity being the complement of the same angle. II. Relations between Principal and Conjugate Stresses (Fig. 217).— Take any line C^TV =^^^-^^ . 11 If''' . ^1 ■•■ 1 '. ii THEORY OF STRUCTURES. P — P With N as centre and a radius = — ^ , describe a semi- 2 circle. • Let Q be the common obliquity of a pair of conjugate stresses. Fig. ai7. Draw ORS, making an angle H with ON, and cutting the semicircle in the points R and S. Join NR, NS. OR and OS are evidently a pair of conjugate stresses. Draw NV perpendicular to RS and bisecting it in V. I Draw the tangent OT; join NT. ] Let OR = r, 0S = s. Then rs-OR.OS= OT* = ON' - NT* ' = (n4-M)"=AA, (.3) and r-\- s = 2OV = 2ON cos ^ = (A+A)cos 6. (29) The maximum value of the obliquity, i.e., of 6, is the angle TON. Call this angle 0. Then NT p,—p, , , sm0 = 7r-.7 = T-rT (30) ON A+A PRINCIPAL AND CONJUGATE STRESSES. Let OR, OR be a pair of con- jugate stresses (Fig. 218). Let OG, OH be the axes of r; ;4ieatest and least principal stress, respectively. Draw ON normal to OR'. Let the angle GOR = ^, RON = e, HON= GOR' = y, as before. Then 249 f = gO°-y-e; and by eqs. (8), ~ cot y = tan ip =cot {y-\-ff)', . cot(r + ^) _ A cot X A' sin0= sin 6 or A -A _ coty-cot(>^ + g) _ A+A cot >' + cot (;/ -f 6^) sin (2;/ 4-^)" sin ^ .•• sin (2y 4- p) = -^ — :r, ^ '^ ' ' sin >' = i{-* + M^)l <3') Hence, mgle ^W= 90° - y = ^{ i8o°+^-sin-(|^) } (32) and anglcffOR=y + d = 1 1 ^ + sin-(~^) | . . . (33) mi r \i 250 THEORY OF STRUCTURES. 12. Ratio of V to s. l_qR _ OV-R V _ OV -VNR'-N V* s ~ OS ~ OV i-RV ~ oy-L. {^NR' — NV _ ON cos a - \/NW^^~W^rn) ~ ON cos B '^^VnW^'WN^^^^ NR A -A A^T- . _.. . , 7TT7 = - , = TTiTT = Sin TON = sin 0. ON A+A <^^ r cos — l/sin' — sin" 6 " s ~ cDse+ i/sin'^ - sin' ^ But cos COS cos ^ — i/cos" 6^ — cos" COS ^ + ^cos' 6 — cos' • • • • (34) Let cos COS 8 = sin Of. Then I T cos a I ± cos or = tan a a ' - or = cot' - , (35) If ^ = 0, « = 90 — 0. .-. J = tan" \4S 1(0=0, a =z 90°. (45 -f) or =cot'(45-f). . (36) r _ s ~ (37) RELATION BETWEEN STRESS AND STRAIN. 251 13. Relation between Stress and Strain. — Let a solid body be strained uniformly, i.e., in such a manner that lines of particles which are parallel in the free state remain parallel in the strained state, their lengths being altered in a given ratio, which is practically very small. Lines of particles which are oblique to each other in the free state are generally inclined at different angles in the strained state, and their lengths are altered in different ratios. Let the straining of the body convert a rectangular portion ABCD (Fig. 219) into the rectangle AB'C'D', where AB' = (I + a)AB and AD' = (i + fi)AD. Now a and fi are very small, so that their joint effect may be considered to be equal to the sum of their separate effects. Hence : • ' First. Let a simple longitudinal strain in a direction paral- lel to AB convert the rectangle ABCD into the rectangle ABED, where BB' ^\ .j'X':..K_.C' = a.AB. A line OF will move into the posi- tion OF', where FF' = a . DF, and OF' - OF the strain along OF = FF' cos d a.BFcosff OF = a cos' 8, "b' Fig, aig. "" OF OF being the angle OFD. Also, the " distortion or deviation from rect angularity" = angle FOF' = FF'9,me a.DFsmO OF OF = a cos ^ sin ^. Seco7id. Let a simple longitudinal strain in a direction parallel to AD convert the rectangle ABCD into the rectangle ABKD', where DD' = fi.AD. The line OF will move into the portions O'F", where 00' = fi.AOdindF"F=DD' = ^.AD. the strain along OF = O'F' OF OF 4 "ife. '\ 252 THEORY OF STRUCTURES. Draw O'M parallel to OF. Then O'F"-- 0F= O'F"- 0'M= F"M sin fi = {F"F-FM) sin = {DV - 00') sin e = ti{AD - AO) sin = ft. on sin tf. . , ^ ^ ft.OD sin e ^ . . ^ .'. the strain along OF = ^jx = /? sin 0. RE Consid( /, and />, ///;, BC ai Chap, ill, The distortion = the anrrlc F"0'M F"I\f cos fi.ODcos ) ^ . , ; =■ p sni t) cos c'. 6>i'^ OF Hence, when the strains are simultaneous, the line OF yj'xW take the position OF '' between O'F" d^nd OF', and the total strain along OF =^ a cos' -\- fi sin' ^ ; the total distortion =.{a — fi) sin ^ cos 0, Again, draw a line OG perpendicular to OF. The angle OGA — 90° — (f, and hence, from the above, the total strain along OG — a sin' G -^ /3 cos^ 6, and the corresponding distortion — {a — /?) sin 6^ cos 6^. Denote the strain along OF by <•, , that along OG by ^, , and each of the equal distortions by t. Then e, 4- r, = « -f /y. Again, if OF, OG are the sides of a rectangle enclosed in the rectangle A BCD, the straining will convert the rectangle into an oblique figure with its opposite sides parallel. The lengths of adjacent sides are altered by the amounts f, and e, , and the angle 6' by 2t. The above results may also be consid- ered to hold true if the straining, instead of being uniform, varies continuously from point to point. and the str Uici: A/iCl If the s i.e., if the o A-- .'. a = Thus th so that the 1 Also, if point of AL ^. = and the dist( i? RELATION BETWEEN STRESS AND STRAIN. 253 Consider a unit cube A BCD subject to stresses of intensity /I, and /, upon the parallel faces ///;, nC and AB, DC. By Art. 3, [" Chap, ill, Di i -C P. A a ■=■ E mE ' ^ ~ ni/i ^ E ' >P, K Fig. and the strain perpendicular to the hicc A ncjj= -\-A-. ntE VI It If the stresses are of equal intensity but of opposite kind^ i.e., if the one is a tension and the other a compression, /. = — A=/. suppose. E^ .'. a = — /? — y.l 1 4- - ), and the third strain is ni/. Thus the volume of the strained solid = (l 4- «)(! — ^)(0 = I — «' = I, approximately, so that the volume is not sensibly changed. Also, if OGHF is an enclosed square, O being the middle point of AD, 6 = 45°, and e^ = ej= = o = strain along OF or OG, and the distortion .— change in angle O 2t = 2 a — p = " = |(-+i-)- 254 7'HKOKV (>/•■ s'rA'UcruA'/-:.s. This result may be at once cli;cliiccnt of a strained solitj tin shears along plan -s at right angles are of ecpial intcnsitj'. Tlu' elTect of such stresses is merely to pioduce a (distortion c/ _/fj;'"//;r, and generally without sensible change of volume. Thus, shears of intensity .v along the parallel faces of t'u' unit square A BCD will merely distort the square into a rhom bus yI^>C"/^' (Fig. 2j0. Denoting tlu- change of angle by 2/, S and assuming that the "stress is propui- ■ C' tional to the strain," s ~ G.2t, where G is a coefficient calleil the modulus of transverse elasticity, or the eoejfieient of rigidity, and ilepeiuls upon a e/iangr oj Fig. aji. forill. Consider a section along the diagonal BD. The stresses on the faces. /A'. A J), and on CB, CD, resolved parallel and perpendicular to BD, are evidently equivalent to nil and a normal force .v i 2, respectively. Thus, there is no sliding tendency along JU), but the two portions ABD and s \ 2 CBD exert upon each other a pull, or tension, of intensity -77% _ i^ — \ 2 Similarly it maybe shown that there is no tendency to slide .along AC, but that the two portions .IBC and ADC exert ,, 1 1 1 1 1 1 1 1 / A ^ -e" B , consistent with equi- librium. The limiting ratios of a pair of conjugate stresses in a mass of earthwork may also easily be determined. By eq- (34), the ratio = cos — i/cos" ti — cos'' cos B + |/cos" d — cos' (39- Hence the ratio cannot exceed cos B 4- j/cos" B — cos'^ _ __ ' , • cos B — V'cos" B — cos" nor can it be less than cos B — Vcos' B — cos" cos B -f- v'cos" B — cos' If 6* = o, the ratio becomes I T sin I ± sm (40) For example, let the ground-surface be horizontal. The pair of conjugate stresses become a vertical stress /, and a horizontal stress/, . " /, =^ r — sin 0' (41) or as in eq. Pressu tJie groun horizon at Consid Let s 1 at Z?. Let r b at /A This coi the ground Take D. DE DC r s Then w., magnitude t vertical plan ' ' v'olumc of Join C£. The intei Hence, t prism nC£ Again, column CD. S IS m «PH^^ ■iV^linpMfWf ill RANKINE'S EARTIfWORK THEORY. 257 or (42) /, > I — sin /, = I +sin 0' ' as in eq. (38). Pressure agaiftst a Vertical Plane. — Let ACB (Fig. 222), the ground-surface of a mass of earthwork, be inclined to the horizon at an angle 0. Consider a particle at a vertical depth CD = x below C. Let s be the vertical intensity of pressure on the particle at Z*. Let r be the conjugate intensity of pressure on the particle at /A This conjugate pressure acts in the direction ED parallel to the ground-surface, and its obliquity is 6. Take DE so that DE DC r s cos 6 — 4 cos' 8 — cos' cos -\- Vcos'' — cos' Then w . ED represents in direction and magnitude the intensity of pressure on the vertical plane DC at the point D, w being the weight of a unit of volume of the earthwork. Join CE. The intensity of pressure at any other point m is evi- dently w. inn, mn being drawn parallel to DE.. Hence, the total pressure on the plane /?C — weight of prism DCE w.DC.DE ,^ zv.DC'r — cos V = cos r/ 2 2 S wx' „„ ..cos — Vcos 6 — cos = cos n =^ ~- — , 2 cos /^ -)- V cos" & — cos' (43) Again, s is the pressure du> to the weight of the vertical column CD. .•. s =r tt.*:r COS tf, (44) "-'KSl; 258 and THEORY OF STRUCTURES. r z=. wx cos ^ cos B — Vcos" B — cos" cos -\- y cos' B — cos (45^ By means of this last equation the total pressure on CD may be easily deduced as follows : The pressure on an element dx at a depth x =1 rdx = zvx cos B cos B — •i/cos" B — cos' dx. cos B -\- ^coi' B — cos" .*. total pressure = / "rdx = etc. The total resultant pressure is parallel in direction to the ground-surface, and its point of application is evidently at tivo thirds of the total depth CD. 15. Earth Foundations. — Case I. Let the weight of the superstructure be uniformly distributed over the base, and let p^ be the intensity of the pressure produced by it. If //, is the maximum horizontal intensity of pressure cor- responding to/„, A_ < I +sin ph ~ I — sin ' In the natural ground, let p^ be the maximum vertical in- tensity of pressure corresponding to the horizontal intensity //, • Then Ph_ < I + si" /„ ~ I — sin " Hence A < (^ +sin0 /, = \i —sin If X is the depth of the foundation, and w the weight of a cubic foot of the earth, p^ =-wx\ . A, . ivx — \ ombining (48) and (49), (49) /, MG in Case I. Hn^ Take MH — f, and the triangle X. MHK = R (Fig. 226). The pressure on the bed is now represented by the triangle MHK. 'V. .M O Fig. jafi. K N R = \MH. MK= \f. MK. The ordinate through the 262 THEORY OF STRUCTURES. centre of gravity 01 the triangle MHK parallel to HM cuts MN in the centre of pressure F. qt = 0F= OM- MF=z-- — . 2 3 But MK=Y\ t 1 R •*• ^^ = T 7 ; and hence 2 3 / I 2i? , . .^ , I (J — - y and IS evidently > 6' (53) Case III. Let the maximum intensity/ < MG in Case I. Take ML =f, and the trapezoid MLSN = R (Fig. 227). The pressure on the bed is now represented by the trape- zoid MLSN. G .'.R = ^{ML-\-NS)ALY = l{f-\-NS)t. 1 ^i-^ — n=---, and M o Fig. 347. a 2R ML-\-NS=-~ ^s=lf-. The ordinate through the centre of gravity of the trapezoid parallel to Zy>/cuts MN in the centre of pressure F. Draw ST parallel to NM. The moment of MLSN with respect to = moment of MTSN with respect to O -j- moment of ZST^with respect to O, or TS' ^ML + NS)MN .OF = o^^LT-^ :.--jtqt = l{ML-NS)-^^ =-[2f.-^)-^, Hence, q — -A-^ — i), and is evidently < ^. . . . (54) RE TAINING- WALLS. 263 Now W must be ;i function of x, the vertical depth of N below B ; P also may be a function of x. Hence if /is given, and the corresponding value of q from (53) o"" (54) substituted in (52), x may be found. When (53) is employed, the value of x found must make q>\' When (54) is employed, the value of x found must make q<\' Example. The rectangular wall in [a), the safe crushing strength of the material being 10,000 lbs. per square foot(=/). By (53), Substituting in (52), Hence, X ^ 9.03 ft. Again, ^ > '■ — > .4248, and is h fortiori > ^. R = W= : 500;jr Q I ~ 2 X 120* If (54) is employed, Hence, by (52), 1/80 \ ^ =321— -ij. < i ^^ ^ I , » By trial x is found to lie between 12 and 13 ; each of these values makes ^ > ^, which is contrary to (54). The first is therefore the correct substitution. (c) The angle between the directions of the resultant pres- sure and a normal to the bed must be less than the angle of friction. ■.:--ll-:r. .^% t -<» .o. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 M 2.0 1.8 1-4 11.6 'm m /A <^^ (? / ^« -^ Photographic Sciences Corporation 33 WEST MAIN STREET WEBSTER, NY 14580 (716) 872-<503 i 264 THEORY OF STRUCTURES. Let be the angle of friction, R the mutual normal pressure. Resolving along the bed and perpendicular to it, Pcos a-\- fi — W^sin a W{s\n cos a+cos 0sin a), or /* cos /? + a 4- +a) | t W{qt T rt) cos a. (56) The other conditions of equilibrium may be discussed as in Art. 16. i8. Line of Rupture. — Another expression for the press- ure on AB may be obtained as follows : If the whole mass in front of AB (Fig. 229) were suddenly removed, some of the earthwork behind AB would fall away. Suppose that the volume ABC would slip along the plane CB. The stability of ABC is maintained by the reaction P ow AB, the weight W oi ABC, and the frictional resistance along BC. Let the direction of P make an angle /3 with the horizon. Let the angle CBA = i. Let R be the mutual pressure on the plane BC. Resolving along and perpendicular to BC, Fig. 339. V:m.- and -/'cos (90° -»■ — /?)+ H^cos« = ^tan0; P sin (90° -/-/?)-(- W^sin i - R. im t .-. - P sin {/3 + t) -f- Wcos i = tan | Pcos {/3 + t) -f fTsin t\ , and P= W cos i — sin i tan = W-r cos {i + 0) sin {fi + + cos (/« + tan sin {ti-\-i-\- 0)' iv t Ml' »i'l tr 266 Also THEORY OF STRUCTURES. BA.BC . . W= w sin { 7vx^ COS d sin t ~2~ cos {e -\- iy \P- zvx' COS ^ sin t cos {t -\- 0) 2 cos {tf-{-t) sin {p-\-i-{-ZAr = 90° = / 4- + / = 2? 4- 0. ^ = 45 - 19. Practical Rules. — When the surface of the earthwork is horizontal and the face of the wall against which it abuts ver- tical, the pressure on the wall according to Rankine's theory is P=--^ sin 2 I + sin ' and the direction of P is horizontal. This result is also identical with that obtained in Art. 18, on the assumption of Coulomb's wedge of maximum pressure (Poncelet's Theory). Pit »4i::nl :.t,. J 268 THEORY OF STRUCTURES. Experience has conclusively proved that the theoretical value of P given above is very much greater than its real value, so that the thickness of a wall designed in accordance with theory would be in excess of what is required in practice. In the deduction of the formula, indeed, the altogether inadmissible assumption is made that there is no friction between the earth- work and the face of the wall. This is equivalent to the sup- position that the face is perfectly smooth and that therefore the pressure acts normally to it. Boussinesque, Levy, and St. Venant have demonstrated that the hypothesis of a normal pressure only holds true, either, first, if the ground surface is horizontal and the wall- face inclined at an angle of 45° to the vertical, or, second, if the wall-face is vertical and the ground-surface inclined at an angle to the horizon. When the surface of the ground is horizontal and the face of the wall vertical, and when = 45°, the above formula gives the correct magnitude of P. Its direction, however, is not hori- zontal, but makes an angle with the vertical equal to the r ngle of friction between the earth and the wall. The wall-face is gen- erally sufficiently rough to hold fast a layer of earth, and in all probability Boussinesque's assumption that the friction between the wall and the earth is equal to that inherent in the earth is a near approximation to the truth. The direction of P will thus be considerably modified, leading to a smaller moment of stability and a corresponding diminution in the necessary thick- ness of the wall. In practice the thrust P may always be made small by carrying up the backing in well-punned horizontal layers. In order to neutralize the very great thrust often induced by alternate freezing and thawing and the consequent swelling, a most effective expedient is to give a batter of about i in i to the rear line of the wall extending below the line to which frost penetrates. The greatest difficulty in formulating a table of earth-thrusts arises from the fact that there is an infinite variety of earth- work. As an example of this, Airy states that he has found PRACTICAL RULES. 269 the cohesive power of clay to vary from 168 to 800 pounds per square foot, the corresponding coefficients of friction varying from 1. 1 5 to .36, and that even this wide range is less than might be found in practice. A correct theory for the design of retaining-walls is as yet wanting. According to Baker, experience has shown that with good backing and a good foundation the stability of a wall will be insured by making its thickness one-fourth the height, and giving it a front batter of i or 2 in. per foot, and that under no conditions of ordinary surcharge or heavy backing need its thickness exceed one-half the height. Baker's usual practice in ground of average character is to make the thickness one-third the height from the top of the footings, and if any material is taken out to form a face panel, three-fourths of it is put back in the form of a pilaster. General Fanshawe's rule for brick walls of rectangular section retaining ordinary material is to make the thickness 24^ of the height for a batter of i in 5 ; 25" •• (( " I in 6 ; 26" « " I in 8 ; 27" << " I in 10; 28" (( " I in 12 ; 30" l( " I in 24; 32" li for a vertical wall The thickness at the footing adopted by Vauban for walls with a front batter of i in 5 or i in 6 and plumb at the rear, is approximately given by the empirical formula thickness = .19// + 4 f^., H being the height of the wall above the footing. Counter- forts were introduced at intervals of 15 feet for walls above 35 feet in height, and at intervals of 12 feet for walls of less height. The counterfort projects from the wall a distance of -r "i~ 3 ^'• approximately, and the approximate width of the counterfort is - + 3 ft., diminishing to -- + 2 ft. If,!;' f1 i.*<'l ' ! : m I I ti I )«'( 270 THEORY OF STRUCTURES. ' ■ II a? J i>- * i' ^ Brunei curved the face of the wall and made its thickness one-fifth or one-sixth the height. Counterforts 2 ft. 6 in. in thickness were introduced at intervals of 10 ft. The vast importance of the foundation will be better appre- ciated by bearing in mind that the great majority of failures have been due to defective foundations. If water can percolate to the foundation, a softening action begins and a consequent settlement takes place, which is most rapid in the region sub- jected to the greatest pressure, viz., the toe. In order to coun- teract this tendency to settle, the toe may be supported by rak- ing piles, the rake being given to diminish the bending action of the thrust on the piles. It is also advisable to distribute the weight as uniformly as possible over the base, a condition which is not compatible with large front batters and deep offsets, as they tend to concentrate weight on isolated points. In the case of dock-walls, too, a large front batter will keep a ship farther away from the coping and will necessitate thicker fenders, as well as cranes with wider throws. As an objection to offsets Bernays urges that, in settling, the backing is liable to hang upon them, forming large holes underneath. He there- fore favors the substitution of 'a batter for the offsets. On the other hand, if water stands on both sides of the walls, the hydrostatic pressure on the offsets will greatly increase its stability. Dock-walls are liable to far greater variations of thrust than ordinary retaining-walls. The water in a dock with an im- permeable bottom may stand at a much higher level than the water at the back of the wall, and its pressure may thus even more than neutralize the thrust due to the backing. With a porous bottom the stability of a wall may be greatly dimin- ished by an upward pressure on the base. The experience of dock-wall failures has led to the conclusion that a large moment of stability is not of so much importance as " weight with a good grip on the ground." Many authorities, both practical and theoretical, have urged the great advantages in economy and strength attending the employment of counterforts. The use of Portland cement, or cement concrete, will guard against the breaking away of the counterforts from the main body of RESERVOIR WALLS. 271 B the wall, as has often happened in the case of the older walls. But a uniform distribution of pressure as well as of weight is important, .ind it therefore seems more desirable to introduce the extra weight of the counterforts into the main wall. Be- sides, the building of the counterforts entails of itself an in- creased expense. 20. Reservoir Walls. — Let /be the maximum safe press- ure per square foot of horizontal base, at inner face of a full reservoir, at outer face when empty. Let w be the wciglit of a cubic foot of the masonry. Assume that the wall is to be of uniform strength, i.e., that the section of the wall is of such form that in passing from any horizontal section to the consecutive one below, the ratio of the in- crement of the weight to the increment of the surface is constant and equal to/. Let AB, Fig. 231, be the top of the wall. Take any point as origin, and the vertical through O as the axis of x. Let OA = t„ 0B = t^, and let T = t, + t, = AB. For the profile AP consider a layer of thickness dx at a depth X. Then tvydx C/itN G (/j Fig. 231. dy = /. (I) or w y c being a constant of integration. When .a; = o, / = /, ; - / o = — log* 'i + <^» w X- ' *• % t I .r^-J ■>: I y is' 11; I If II 272 and hence THEORY OF STRUCTURES. f y ■ (2> which is the equation to ^/*and is the logarithmic curve. It may be similarly shown that the equation to BQ is - = ^-log.f w • • • . . (3) Equations (2) and (3) may also be written in the forms y - t,ef and TO —-X y - V^ (4) (5) Corresponding points on the profiles, e.g., /*and Q, have a common subtan^ent of the constant value — , for NT = PNt.nNPT[=y^)={^ (6) Area PNOA = />. = /,(/-> - ^) = £(V^ - /.), (7) where PN- F,. Area QNOB = J^'^ydx where QN = V,. .-. Area QPAB = £-(F.+ K, -7^^) = ^(7" - T), . (9) where PQ = F, + F, = r. = ~(I^,-a (8^ RESERVOIR WALLS. 273 Thus the area of the portion under consideration is equal to the product of the subtangent and the diflference of thick- ness at top and bottom. Lines of resistance with reservoir empty. Let g^ be the point in which the vertical through the C. of G. of the portion UAPN intersects PN. Then ii Mm ,rH'l li'if' Ng, X area OAPN — f ydx^- ; .*. Ngl^ y^ - A)- = r -I ydy = 7 - ( Y,' - /,») ; .•.^^, = -^4^' So if ^, be the point in which the vertical through the C. of G. of the portion C^^QA'' intersects QN, Ng,=. v^ + f. Let G be the point in which the vertical through the C. of G. of the whole uiass ABQP intersects PQ. Then NG X area ABQP = Ng, X area A ONP- Ng^ X area BONQ, or NG^-f^ f; - /. + y. - = 7 -( y" - A') - - --( y: - /,'). w i t.SfJR' 56S: 274 THEORY OF STRUCTURES. The horizontal distance between G and a vertical t'lrough the middle point of AB =NG-\{t-Q= ( K.-/.y-(K-0' _ (K.~0-(K.-0 4(K.-/.+ K-0- 4 = one half of the horizontal distance between the verticals through the middle points of AB and CD. The locus of G can therefore be easily plotted. Lines of Resistance zvith Reservoir Full. — Let R be the centre of resistance in PQ (Fig. 232). Draw the vertical QS, and consider the equilibrium of the mass QSAPQ. Let w' = weight of a cubic foot of water. Si BOA w'x' X Fig. 333. or w'x' - = moment of water-pressure against QS about R = moment of weight of QBS about R -f- moment of weight of QPAB about R, f. , — moment of QBS about R-\--\T' ~ T)w.GR. The first term on the right-hand side of this equation is generally very small and may be disregarded, the error being on the safe side. In such case ^„ I vj' x'' GR = ^ 6 / T'- T' Also the mean intensity of the vertical pressure _ _ TV X area APQB _ J _T\ —A— PQ —jy T'J' Is m RESERVOIR WALLS. and the maximum intensity of the vertical pressure 275 = A = 2R {i-Z9)T ^ = 1/ I — 2q or = y7(l-f-6y)=/(l+6^)[l -y,j. General Case. — Let the profile be of any form, and consi' er .-iny portion ABQP, Fig. 233. Take the vertical through Q as -"i the axis of x, and the horizontal line coincident .\/l. top of wallas the axis of J/. The horizontal distance {y) be- tween the axis of x and the vertical through the C. of G. of the portion under consideration is given by the equation ■^'/'^^-^ ^ / V-^' QL-i-_-_- / being the width, dx the thickness, ^^'^^ '33- and y the horizontal distance from OQ of the C. of G. of any layer AIN at a depth x from the top. When the reservoir is empty, the deviation of the centre of resistance from the centre of base = gT=Y--y<-^ When the reservoir is full, let q'T be the deviation of the centre of resistance from the centre of the base, and disregard the moment of the weight of the water between OQ and the profile BQ. Then - 1 ,11 276 THE'.RY OF STRUCTURES. ,^ moment of water pr. ± moment of wt. of ABQP g T= rflZTf.-T-TTjTTT^ T y = ±6 w'x' 'i"^^-^ weight of ABQP ±y^ Y. w Hence {g± Fig. 234. 21. General Equations of Stress. — Let x, y, s be the co-ordinates with respect to three rectanguhir axes of any point O in a strained body. Consider the equilibrium of an element of the body in the form of an indefinitely small parallelo- piped with its edges 0A{= dx), OB{--=.dy), OC{=dz) parallel to the axes of x, y, z. It is assumed X that the faces of the element arc sufficiently small to allow of tlio distribution of stress over them being regarded as uniform. TIk" resultant force on each face will therefore be a single force acting at its middle point. Let X^, F, , Z^ be the components parallel to the axes x, y, z of the resultant force per unit area, on the face EC. " X,, Y^, Z^ be the corresponding components for the face^C. " X^ , y, , Z, be the corresponding components for the face ^5. These components are functions' of x, y, z, and therefore become -(^.+f4-(''.+'M-(^.+f4 for the adjacent face AIJ\ ui^ mm GENERAL EQUATIONS OF STRESS. dy"''r V ' Ty 277 -(-.+f4-(n+f4-(^.+f4 for the adjacent face BD\ for the adjacent face DC. Hence, the total stress parallel to the axis of 4r = X.dyds - [x, + '^-^^d^dydz + X,dzdx - [x^ + "^-^^-d^dzdx J^X,dxdy- \X,^'-j^dz]dxdy IdX. , dX. , dX\ , , , dy Similarly, the total stress parallel to the axis of j^ (dV, ,dY, dY\ ^~\'dx'^-dy'^'dz¥''^y'^'^ and the total stress parallel to the axis of z IdZ, . dZ, , dZX . , , Let p be the density of the mass at O, and let P,, Py, P, be the components parallel to the axes of x, y, z of the external force, per unit mass, at O. pdxdydzP^ is the component parallel to the axis of x of the external force on the element ; pdxdydzPy'xs the component parallel to the axis of _;/ of the external force on the element ; pdxdydzP, is the component parallel to the axis of z of the external force on the element. » . 'I 278 THEORY OF STRUCTURES. The element is in equilibrium. dx dV, , dV, , dV, dx ^ dy ' dz dZ^ dZ.^ dx dy pPy\ dz '^ '' (I) These are the general equations of stress. Again, take moments about axes through the centre of the element parallel to the axes of co-ordinates, and neglect terms involving {dxydydz, dx{dyYdz, dxdy{dz)^. V, = Z,, Z, = X,, and X,= F,. (--) Adopting Lamp's notation, i.e., taking iV, , iVj , iV, as the normal intensities of stress at on planes perpendicular to the axes of x, y, z ; T^ as the tangential intensity of stress at O on a plane perpendicular to the axis of x if due to a stress parallel to the axis of y, or on a plane perpen- dicular to the axis of y if due to a stress parallel to the axis of x ; and T^ , 7", similarly, — equa- tions (i) become dx"^ dy"^ dz -^P^-^' y ' dx ^ dy ^ dz ~P ' dx "^ dy ~^ dz ~ f ' . i3) GENERAL EQUATIONS OF STRESS. 279 Next consider the equilibrium of a tetrahedral element having three of its laces parallel to the co-ordinate planes. Let /, w, n be the direction-cosines of the normal to the fourth face. Also, let X, V, Z be the compo- nents parallel to the axes of x, y, z of the intensity of stress R on the fourth face. X=lN,-\- w. r, + « r, + \pPJdx. But the last term disappears in i^'g- »3s. the limit when the tetrahedron is indefinitely small, and hence Zz=lT,-\-mT, -\-nN,. (4) These three equations define R in direction and magnitude when the stresses on the three rectangular planes are known. Let it be required to determine the planes upon which the stress is wholly normal. We have X = IR, Y= inR, Z = nR. . (5) Substituting these values of X, Y, Z in eqs. (4) and eliminat- ing /, m, n, we obtain -{N,N,N, - N,T: - N,T:-N,T:-\-2T,T,T,)=0', (6) a cubic equation giving three real values for R, and therefore three sets of values for /, in, and «, showing that there are three planes at O on each of which the intensity of stress is wholly normal. These planes are at right angles to each other and are c^WqA principal planes, the corresponding stresses hemg prin- cipal stresses. They are the principal planes of the quadric, N,x' + N,f + A^,j' + 2 T,yz + 2 T,2x + 2 T,xy = c. (7) I \ \ i. ! " ^iiii lilST H im: ;l^- 280 THEORY OF STRUCTURES. For, the equation to the tangent plane at the extremity of a radius r whose direction-cosines are /, ;«, n is Xrx ■\- Yry -\- Zrs = c, (8) and the equation of the parallel diametral plane is Xx -\- Yy ■\- Zz — o (9) The direction-cosines of the perpendicular to this plane are X__ K_ f_ so that the resultant stress R must act in the direction of this perpendicular. Hence the intensities of stress on the planes perpendicular to the axes of the quadric (7) are wholly normal. Refer the quadric to its principal planes as planes of refer- ence. All the 7"s vanish and its equation becomes Also, the general equations (3) become (10) dy *^ '' dN, de = pP,- • • • . . . (II) Again, (ir+(0+©'='-+"''+''-= I. . . (12) M^ IV iittx RELATION BETWEEN STRESS AND STRAIN. 281 Consider X, Y, Z as the co-ordinates of the extremity of the straight line representing R in direction and magnitude. Equation (12) is then the equation to an ellipsoid whose semi- axes are iV, , iV, , TV, . As a plane at O turns round (? as a fixed centre, the extremity of a line representing the intensity of stress/? on the plane will trace out an ellipsoid. This ellipsoid is called the ellipsoid of stress. Note I. The coefficients in the cubic equation (6) are in- variants. Thus, N^-\- N^-\- N, is constant, or the sum of three normal intensities of stress on three planes placed at right angles at any point of a strained body is the same for all positions of the three planes. Note 2. The perpendicular/ from O on the tangent plane, equation (8), = Wr=^- pr ^ ^' Note 3. Let the stress be the same for all positions of the plane at 0. Then N,—N^ = N,, and the ellipsoid (12) be- comes a sphere. The stress is therefore everywhere normal, and the body must be a perfect fluid. Conversely, if the stress is everywhere normal, the body must be a perfect fluid, the ellipsoid becomes a sphere, and therefore iV, = iV, = TV, . 22. Relation between Stress and Strain.— In Art. 13 it was shown that when the size and figure of a body are altered in two dimensions, there is an ellipse of strain analogous to the ellipse of stress. If the alteration takes place in three dimen- sions, it may be similarly shown that every state of strain may be represented by an ellipsoid of strain analogous to the ellip- sold of stress. The axes of the ellipsoid are the principal axes of strain, and every strain may be resolved into three simple strains parallel to these axes. i ■n I if 11' W ml R il W 11 :-r; \m I;-!,";'! ■If !• •;• 282 . THEORY OF STRUCTURES. It is assumed that the strains remain very small, that the jj stresses developed are proportional S' to the corresponding strains, and that their effects may be superposed. Consider an element of the uii- '_ [_.Q' strained body in the form of a rcct- "* angular parallelopiped, having its edges PQ (= h), PR (= k\ PS {= I) parallel to the axes of co-ordinates. When the body is strained, the ^'"' '^*' element becomes distorted, the new edges being P'Q', P'R', P'S'. Let X, y, z be the co-ordinates of P. Let X -\- u, y -\- V, z -\- w ho. the co-ordinates of P'. By Taylor's Theorem the co-ordinates with respect to /*' of „, ,/ , dti\ ,dv .dtv ay \ ay I ay t > • (14) Hence, strain parallel to axis of j»r = y = z = p'Q'-P Q du:\ PQ ~ ~dx' P'R' -P R dv ^ PR ' ~dy' P'S' -P S dw PS ~dz' (•5) ISOTROPIC BODIES. 285 Again, cos QP'R' du dxidy (■+S)^+('+-)*+ dw dw dy I dx ' dy dx [l-+(Sv©v©'H(^p"+(.-4;)vgf)'l]* In the limit, this reduces to • I • • (16) cos G'/"/e' = ^ + $^^ dy dx Similarly, cos QP'S' = ~ ^~\V C0SR'P'S'=:^ + '^. ay dz ^ Volume of unstrained element = hkl\ Volume of distorted element = ^^'^A ^+77")\^+3~)( ^+7") multiplied by the cosines of small angles in the limit. Difference of volume dii , dv . dtv , , ^•^ . . . 3:::^ L ^ ^ ^ (17) " Vol. of unstrained element dx dy dz^ ' ' ' = the volume or cubic strain. 23. Icc-*Topic Bodies, i.e., bodies possessing the same elas- tic properties in all directions. A nornja) stress of intensity TV, parallel to the axis of x N produces a simple longitudinal strain — i, and two simple .V ^ lateral strains, each = ^, parallel to the axes of y and s, nth '*•, •i.!| ' t it Lis > I . - x. m i L'i ii & i^ iiH^lli 384 THEORY OF STRUCTURES. E being the ordinary modulus of elasticity and — , Poisson's /« ratio (Art. 3, Chap. III). Normal stresses N^ , N^ parallel to the axes o{ y and 2 may be similarly treated. Let the three normal stresses act simultaneously and super- pose the results. Then total strain parallel to axis of jr = -' E II. II II fi II II y = inE dx ' TV, N,-\-N, dv z = -=^ E inE dy N, N,-\-N, _div mE ~ dz : )- (18) lIP' V ill i The form in which these equations are given is due to Grashof. Solving for iV;, TV,, -/V,, N. K N, m{m — \)E du + ; mE {m -f- i)(w— 2) dx "^ (;« + i)('«— 2)\dy Idv dw\ ^ m{m — i)E dv + - mE idw , du (w-f- i){m— 2)dy '^(;«+ ^){^— 2)\d2 ^dxr Y (19) _ m{m — i)E dw vr + mE Idu . dv {m + i)(w— 2) dz ' (w 4- iXw— 2)\dx ' dy ^dvl' The last equations may be written dx \iy dz ] ' .dv , yidw , du\ dy .dw = <^ + H.T + ^)^J . . (20) ISOTROPIC BODIES. 285 where \ = mE (;« + iX''^— 2) , is the coefficient of dilatation, and A _ in{m—\)E ~ {m-\- \){tH— 2) Again, the straining changes the angle RPS by an amount '-1^ 4- — . producing two tangential stresses, each equal to dy as Gl -- + "T )> parallel to the axes of ^ and s. Similarly, ^' = ^('1 + ll"') ' r" i") ^■ = ) o y 1 « »••*»•# 1 • > • • • • • (40) aiul by cq. ( uS\ (,|h n = f/i'ff satisfies this hist equation ami also rq. {^6), if f^ (4.') Ajjaitu the algebraic sum of the moments T, , 7", with it spcct to the axis of x. • . . (43^ t^rrfrfrirfOf^x agt The tulttl iiiuinriil {/if «if llir cMUjilr prudiicing turilun f>' 4- / /[,y [-f>wus s= r;/' fr/'' ..I < ami flio torsiiiiml iljjiility /]/ nf> ,,,■ J =^ " /,.-q:- !•• I • ■ • « • ■ • (44) (t) TnrNlmi of a bur nf rrMniifiiiliU" ^rrtlon. As In case (A), ff nnist; '•mtiflfy the n(|iiiitioti -^ o. • •««••• (45) AUo, the iMUmtinii^ of ('(»ii(li(i(»n coiirspniKliii^,' 1(» r«j, (jH) arr -4-. «- ^7 *- wlipn V -*^ :h (^ (4''') ,iii-,o when /rrn J:r; ... (47) .'/' iiiul 2r ((^ < f) bcinf^ the sidcR of thr rcctanfrlc, Thr total Dininoiit of torsion, v\/.,,^/'{7\j> ■-- T^ayfS is then found lo bo M r *''" 1 , , Ian // (2n I l) , 3 rr* r '^" If // ex If, i.e., if the section ib a square, eq. (48) bi-comei il/a= .843462 /67y (4(>) /( lb*) bcinpj the momrnt of inertia with respect to the axes. lSf<- Cha|). IX). if ■f ;c| Li !i I 292 THEORY OF STRUCTURES. k' • • . , - " .....'. If - is very small, eq. (48) becomes M=xU^cGe{^^-,2V>)., ~ . . . (50) The torsional rg'dity of a rectangular section is sometimes ex- pressed by the formula .... . M _ 5 dY ' ; - . ^ (^ - isd' + c' • • ^50 For the further treatment of this subject, the student is re- ferred to St. Venant's edition of Clebsch, and to Thomson and Tait's Natural Philosophy. 3. Wor& done in the small strain of a body (Clapeyron's Theorem). — M ultiply eqs. (3 ) hyudxdy dz, v dx dy dz, ivdxdydz, and find the triple integral of theif sum throughout the whole of the solid. The terms involving the components P^^P^^ P^ may be dis- regarded, as the deformations due to their action are generally inappreciable. Also, ' 'SSf-d^'''^^'^y^' ■ =J'f^NJu, - NJ'uJ')dydz -fffNf^dxdydz, NJ, NJ' being the values of N^ at the two points in which the line parallel to the axis of x cuts the surface of the body, aiul uj, uj' the corresponding values of u. Let dS, dS' be the elementary areas of the surface at these points, and /', /" the cosines of the angles between the normals to these elements and the axis of x. The double integral on the right-hand side of the last equa- tion then becomes _ . . , • ■ ■ . //{NJl'uJdS - NJ'V'uJ'dS) = :S{NJudS). APPLICATIOITZ 293 Treating the other terms similarly, = 2K^/+ T,in + T.,n)u + (7,/+ N,m + T,n)v + ( r/ + r. w + -N,n)w \ dS df^ . ^ (dv dzv\ Idw du\ Idu . dv\] + ^' UJ +^"J + ^' i^ + ^; + ^' l^ + ^J f Hence, the ivork done = ^2{Xu -\- Yv -\- Zw)dS = \Sffdxdydz I c~^^^{N. + ^, + ^,r • = ,fffd.dyd.{^^^^^ -. E being the ordinary modulus of elasticity. ]• t» m tlie ami ;hcsc mals :qua- MH: *v 294 THEORY OF STRUCTURES. EXAMPLES. 1. At a point within q strained solid there are two conjugate stresses viz., a tension of 200 lbs. and a thrust of 150 lbs. per square inch, tlir common obliquity being 30". Find (a) the principal stresses; {b) \.\\\i maximum shear and the direction and magnitude of the correspond itij^ resultant stress ; (c) the resultant stress upon a plane inclined at 30' to the axis of greatest principal stress. Am. — {li) A tension of 204.65 lbs. and a thrust of 146 95 lbs. per sq in. (b) 175.8 lbs. per sq. in.; 173.2 lbs. in a direction making an angle of 40° 13' with the axis of greatest principal stress. (0 163.3 lbs. per sq. in. 2. A wall with a plumb rear face is to be 30 ft. high and 4 ft. wide at the top ; the earth slopes up from the inner edge at the angle of 20°, 30° being the angle of repose. Assuming Rankine's theory, determine the proper width of the base, the masonry weighing 144 lbs. per cubic foot, and tiie eartli no lbs. 3. A wall 6 ft. wide at the bottom, plumb at the rear, and with a front batter of i in 12, retains water level with the top. Find {a) the limiting position of the centre of pressure at the base so that the stress may be nov/here negative. How {h) high may the wall be built when subjected to this condition.' (a cubic foot of masonry = 125 lbs.). Ans, {a) 12 in. from middle poMit of base; (*) height = 8.9 ft. 4. A wall is built up in layers, the water face being plumb and the rear stepped. If t be the thickness of the «th layer and y the depth o( water above its lower face, show that width of layer x thickness of layer = V4''''' + (>Atz 4- ;;//('' — 2A ; .,'/ being the sectional area of the wall above the layer in question, s the horizontal distance between the water face and the Ime of action of the resultant weight above the layer, / the layer's thickness, and ;// the ratio of the specific weights of the water and masonry. 5. At a point within a strained solid, the stresses on two planes at right angles to each other are a thrust of 30 1^2 lbs. and a tension of 60 lbs. per square inch, the obliquities being 45° and 30° respectively. Determine (<») the principal stresses; (/') tiie ellipse of stress; (<) ilie intensity of stress upon a plane inclined at 60° to the major axis. Ans. — Or) A thrust of 61.76 lbs. and a tension of 39.80 lbs. {c\ A thrust of 66.5 lbs. y k EXAMPLES, 295 •I ' -\\- 6. If the principles of the ellipse of stress arc applicable within a mass of earth, and if at any point of the mass the stress upon a plane is double its conjugate stress, the angle between the two stresses being 20° 28', show that the angle of repose of the earth is 28'. i. 7. The total stress at a point O upon a plane ///> is 60 lbs per square in'''\ and its obliquity is 30" ; the normal component upon a plane CD at the point O is 4.0 lbs, per square inch ; CD is perpendicular to AH. Find ((«) the total stress upon CD, and also its obliquity ; {h) the princi- pal stresses at O ; {c) the equal conjugate stresses at O. Ans. — (rt)tan-'(}); 50 lbs. {!)) 76.57 lbs. and 15.39 lbs. (r) 34.23 lbs, ; obliquity = 41° 42'. 8. Assuming Rankine's theory, find the pressure on the vertical face of a retaining-wall, 30 ft. high, which retains earth sloping up from the top at the angle of repose, viz., 30'. (Weight of masonry = 128 lbs. per cubic foot.; weight of earth = 120 lbs. per cubic foot.) Ans. 46,764 lbs. 9. At a point within a strained solid the stress on one plane is a ten- sion of 50 lbs. per square inch with an obliquity of 30", and upon a second plane is a compression of 150 !bs. per square inch with an ob- liciuity of 45". Find (a) the principal stresses ; {b) the angle between the two planes ; (c) the plane upon which the resultant stress is a shear, and the amount of the shear. Ans. — (a) p\ r- 153.8 lbs. (comp.) ; /j = — 20 lbs. (tens.) (d) 71" 55'. () the plane upon which the stress is wholly a shear; (t) the planes of principal stress. Ans. — (a) 1 1' 38'. (/>} 64.6 lbs ; y = 3" 26'. (c) />i = 106 46 (tens );/„ = — 39.26 (compr.). 1 1. In the preceding question find the conjugate stresses at the given point having the common obliquity 45°. Ans. Impossible. 12. At a point within a strained mass the principal stresses at a given point are in the ratio of 3 to i. Find the ratio of the conjugate stresses at the same poit\t having the common obliquity 30". Also find the in- clination of the axis of greatest principal stress to the horizontal. Ans. Equal ; 60". 13. A wall 3 feet thick, of rectangular section and weighing 125 lbs. \)v\- cubic foot, is subjected to a horizontal thrust of 800 lbs. per foot run 296 THEOKY OF STRUCTURES. il ill. f«i ■ at its t(jp. What sliould be llic liciglit of the wall in order that all the joints above the base may be frictionally stable? Coefficient of friction = unity. Ans. 12 ft. 14. A wall 12 ft. high, 2 ft. wide at the top, and 3 ft. wide at the bot- tom, is constructed of masonry weighing 120 lbs. per cubic foot. The overturning force on the rear face of the wall, which is plumb, is a hori- zontal force P acting at 4 ft. from tiie base. Find P so that the devia- tion of the centre of pressure in the base may not exceed \ ft. The centre of pressure being fi.xed at 2 in. from the middle of the base, show that I of the section may be removed without altering its stability, ami find the increase in the inclination of the resultant pressure on the b;ise to the vertical, consequent on the removal. Ans. 360 lbs.; tangents of angles are in ratio of 5 to 3. 15. A reservoir wall is 4 ft. wide at top, has a front batter of i in 12. ;i. rear batter of 2 in 12, and is constructed of masonry weighing 125 lbs. per cubic foot ; the maximum compression is not to exceed 12,800 lbs. per square foot. Find the limiting height of the wall. Ans. 24 ft., q being %% 16. A dock-wall, plumb at the rear and having a face with a batter of I in 24, is 20 ft. high and 9 ft. wide at the base. Counterforts are built at intervals of 12 ft., projecting 3 ft. from the rear and 6 ft. wide. Determine the thickness of an equally strong wall without counterforts, with the same face-batter and also plumb in the rear. Ans. 10.95 ^'^• 17. If the walls in the preceding question are founded in earth weigh- ing 112 lbs. per square foot and having an angle of repose of 32', find the least depth of foundation in each case, the masonry weighing 125 lbs. per cubic foot. Ans. 2.72 ft. ; 2.71 ft. 18. A vertical retaining-wall is strengthened by means of vertical rectangular anchor-plates having their upper and lower edges 18 and 22 ft., respectively, below the surface. Find the holding power per foot of width, the earth weighing 130 lbs. per cubic foot and having an angle of repose of 30°. Ans. 27,733^ lbs. 19. Determine the limiting depths of foundation for {a) a wall of rectangular section 20 ft. high ; (U) for a wall of trapezoidal section hav- ing plumb rear and front faces 4 and 20 ft. high respectively. Angle of '■epose of earth =30°; weight of earth = 112 lbs. per cubic foot; of masonry = 140 lbs. Ans. (a) 3.22 ft.; ((5) 1.93 ft. 20. A wall 20 ft. high and 6 ft. thick retains earth on one side level with the top, and on the other the earth rises up the wall at its natural iope, viz., 45°, to the height of 5 ft. Will the wall stand or fall ? (Weight of masonry per cubic foot = 130 lbs.; of earth = 120 lbs.) Fmd the locus of the centres of pressure of successive layers. m EXAMPLES. 297 Ans. Overturning moment = 4128 ft. -lbs ; moment of stability = 93600^ + 750 (Y — (yq) = 3691 2j ft.-lbs \[q =.\, The wall is stable. 21. The upper half of the section of a masonry wall is a rectangle 4 ft. wide, and the lower half a rectangle 6 ft. wide, one face being plumb. Find the height of the wall so that the stress on the base may nowhere exceed 10,000 lbs. per square foot when the wall retams water (a) on the plumb face, (/') on the stepped face. (Masonry weighs 125 lbs. per cubic foot.) Ans. {a) 13.08 ft. ; {h) 9.8 ft. 22. A masonry dam h ft. high is a right-angled triangle ABC in sec- tion, and retains water on the vertical (AceA/J. Show that the thickness i of the base BC is given bv (^ — — -, at bemg the deviation of the 5(6^^ + 1) centre of pressure in the base from the middle point. Also show that the thickness will be given by /" = ^Ji^ if the 3(6$' + I) rock upon which the wall is built is seamy, and if it is assumed that the communication between the water in the seams, and that in the reservoir produces an upward pressure upon the base BC, varying uniformly from that equivalent to the head at B to nil at C. If ^ = \, show that, in order that the wall may slide, the coefficient of friction must be less than 67 per cent in the first and 81 per cent in the second case. (Weight of a cubic foot of masonry = 2| x weight of cubic foot of water.) 23. A wall 30 ft. high is of triangular section ABC, the face AB being plumb, and water being retained on the side AC level with the top of the wall; the masonry weighs 125 lbs. per cubic foot. Find the thickness of the base.5C (a) when q =^\\ (p) when stress in masonry is not to exceed 10,000 lbs. per square foot ; {c) when q =^\ and the wall also retains earth on the side AB level with the top, the angle of repose being 30°. Ans. (ii) 17.69 ft.; {b) 13.19 ft.; {c) 17 ft. 24. A wall 4 ft. wide at the top, with a front batter of i in 8, and a rear batter of i in 12, is 30 ft. high. Will the wall be stable or unstable (i) when it retains water level with the top ; (2) when it retains earth ? (Weight of masonry per cubic foot = 125 lbs. ; of earth =112 lbs. ; angle of repose = 30° ,- and q — f .) Ans. (i) Moment of wt. = 128,863 ft.-lbs. ; overturning moment = 281,250 ft.-lbs., artd wall is therefore unstable. (2) Moment of wt. = 148,251 lbs.; overturning moment = 168,000 lbs., and wall is therefore unstable. 25. The faces of a reservoir wall 4 ft. wide at top and 40 ft. high have the same batter, and water rises on one side to within 6 ft. of the top. Find the batter, assuming {a) that tb« pressure on the horizontal base is B!i I 298 THEORY OF STRUCTURES. to be nowhere negative ; {b) that the pressure varies uniformly and at no point exceeds 10,000 lbs. per square foot. (Weight of masonry = 125 lbs. per cubic foot.) Ans. (a) 35.8 ft.; (d) 30 ft. 26. The faces /i A -^C of a wall are parabolas of equal parameters hav- ing their vertices at B and C; water rises on one side to the top of the wall. Determine the thickness of the horizontal base BC, {a) for a wail 50 ft. high; (6) for a wall 100 ft. high, so that the pressure on the base may at no point exceed 10,000 lbs. per square foot. Also (c) compare the volume of such wall with the volume of an equally strong wall of the same height, but with a section in the form of an isosceles triangle with its vertex at A. (Weight of masonry = 125 lbs. per cubic foot.) Ans. (rt) 32.44 ft.; (d) 119.17 ft. (c) in case (a) ratio = 7 : |/i 18 ; (d) •' = ^Ts6 : 21. 27. The water-face AC of a wall has a batter of i in 10 ; the width of the wall AD at the top is 6 ft. ; the rear of the wail DEF has two slopes, DE, having a batter of 2 in 10, and EF, a batter of 78 in 100; the masonry weighs 125 lbs. per cubic foot, and the maximum compression must not exceed 85 lbs. per square inch. Find the safe heights of the two portions AE and EC. 28. The section ABCD of a retaining-wall for a reservoir has a verti- cal face ^Cand a parabolic water-face AD, with the vertex at D. The width of the base DC = 4 x width of the top AB. If AB = 6 ft., find the height of the wall, and trace the curves of resistance {a) when the reservoir is full ; (i) when empty. (Cubic foot of masonry = 2 x cubic foot of water.) Ans. 32 ft. if ^ = ^, and then max. compn. = 8000 lbs. per sq. ft. 29. The figure represents the section of the upper portion of a masonry dam which has to retain water level with the top of the dam. The face AC is plumb for a depth of 7} ft. The width of the section is constant and = 22^ ft. for a depth AB = 40 ft. F Find the maximum stress in the masonry at the \ horizontal bed BF. With the same maximum stress, \ what should be the width of the horizontal bed CO, \ FG being straight ? G (Masonry weighs 130 lbs. per cubic foot.) ^■°- '^T- ' Ans. 20,720 lbs. per sq. ft. 30. A wall of an isosceles triangular section with a base 36 ft. wide has to retain water level with its top. How high may such a wail be built consistent with the condition that the stress in the masonry is nowhere to exceed 10.546I lbs. per square foot ? (Weight of masonry per cubic foot = 125 lbs.) Ans. 54 ft., and q = j'ff. B till,; n EXAMPLES. 299 31. When a cylindrical bar is twisted, show that it is subjected to shears along transverse and radial longitudinal sections, or to tensions and compressions on helices at 45° to the axis. 32. Find the work done in gradually and uniformly compressing a body of volume V\ to the volume Fa, p being the final intensity of pressure and k the modulus of compression. Also show that the intensity of stress is constant throughout the body. Ans. t^. 33. A bar is stretched under a force of intensity/. If the bar is pre- vented from contracting, find the lateral stress; also find the extension. P P '"^ Ans. j)t — 2 m — I iL m {in — i ) 34. Taking the value of the coefficient of elasticity {E) and the co- eiiicient of rigidity (C) to be 15,000 and 5750 tons for steel, 13,950 and 5450 tons for vvrouglit-iron, and 9500 and 3750 tons for cast-iron, find the coeflicient of elasticity of volume (A'), and also the values of the direct elasiicity {A) and the lateral elasticity (A), assuming the metals to be isotropic. m K A 3f i2777i ^G 3ll io559f W^ 3f 6785^ ^G 35. A body is distorted without compression or expansion ; find the work done. Ans, Steel Wrought-iron. , Cast-iron A. ^G WG %G Ans. i,f^ Nx' + AV + N»' + 2(T,' + Ti' + T,')ldS. 36. Find the work required to twist a hollow cylinder of external radius /?i, internal radius /?a, and length /through an angle a, Ans. ;u!^V.* - ^a*). 41 Prove that torsion is equivalent to a shear at each point. 37. Show thai a simple elongation is equivalent to a cubical dilation and a pair of shearing or distorting stresses. 38. Find the resultant shearing stress at any point in the surface of the transverse section of an elliptic cylinder. (Art. 24, Case d.) nG ^V Ans. 2O- i,/ being the perpendicular from the centre P6' + c' upon the tangent to the ellipse at the given point, and 2d, 2c the major and minor axes. 39. A cylinder undergoes torsion round its axis. Show that the curves of no traction are concentric circles. ai...: ■ 1 , ( : , i V. CHAPTER V. FRICTION. I. Sliding Friction. — Friction is the resistance to motion which is always developed when two substances, whether solid, liquid, or gaseous, are pressed together and are compelled to move the one over the other. If /* is the mutual pressure, and if /'" is the force which must act tangentially at the point of •contact to produce motion, the ratio oi F to P is called the co- efficient of friction and "may be denoted by/". The value of/ does not depend upon the nature of any single substance, but upon the nature and condition of the surfaces of contact of a pair of substances. It is not the same, e.g., for iron upon iron as for iron upon bronze or upon wood ; neither is it the same when the surfaces are dry as when lubricated. The laws of friction as enunciated by Coulomb are : (i) That / is independent of the velocity of rubbing ; (2) that /is independent of the extent of surface in contact; (3) that /"depends only on the nature of the surfaces in contact. The friction between two surfaces at rest is greater than when they are in motion, but a slight vibration is often suffi- cient to change the friction of rest to that of motion. Morin's elaborate friction experiments completely verified these laws within certain limits of pressure (from | lb. to 128 lbs. per square inch) and velocity (the maximum velocity being 10 ft. per second), and under the conditions in which they were made. A few of his more important results are given in the follow- ing table : 300 SLIDING FRICTION. 3OT Material. Wood on wood Metal on wood i< II ■• Metal on metal Mclal and wood on each other or each on itself State of Surfaces. dry dry wet dry wet slightly oily occasionully lubricated as usual constantly lubricated Coelticient of Kriciiun. 5 6 26 2 .35 to .2 " 22 " •15 " ■3 •15 .07 to .08 .05 Tlic appar.itus employed in carrying out these experiments consisted of a box whicli could be loaded at pleasure, and which was made to slide along a horizontal bed by means of a cord passing over a pulley and carrying a weight at the end. The contact-surfaces of the bed and bo.x were formed of the materials to be experimented upon. The pull was meas- ured and recorded by a spring dynainometer. More recent experiments, however, have shown that Coulomb's laws cannot be regarded as universally applicable, but that / depends upon the velocity, the pressure, and the temperature. At very low velocities Morin's results have been verified (Fleeming Jenkin). At high velocities / rap- idly diminishes as the velocity increases. Franke, having carefully examined the results of various series of experi- ments, especially those of Poiree, Bochet, and Galton, has suggested the formula / = /„-«^ V being the velocity and/,, a, coefficients depending upon the nature and condition of the rubbing surfaces. For example, /, = .29 and a = .04 for cast-iron on steel with dry sur- faces. /„ = .29 and a = .02 for wrought-iron on wrought-iron with dry surfaces. .; . » /„ = .24 and a = .0285 for wrought-iron on wrought-iron with slightly damp surfaces. Ball has shown that at very low pressures / increases as a 30« THEORY OF STRUCTURES, the pressure diminishes, while Rcnnie's experiments indicate that at very high pressures y rapidly increases with the press- ure, and this is perhaps partly due to a depression, or to an abrasion of the rubbing surfaces. 2. Inclined Plane. — Let a body of weight P slide uni- formly up an inclined plane under a force Q inclined at an angle (i to the plane. Let /^ be the friction resisting the mo. tion, R the pressure on the plane, and a the plane's inclination. The two equations of equilibrium are F ■= Q cos /? — /* sin « and R— — Q sin /3-\-P cos a. Qcosfi — Psina „. ,,. . = coefficient of friction =/. K — Q sin fi -}- P cos a I . ■? :^1 Let the resultant of F and R make an angle (p with the normal to the plane. Then n I si F Q cos ft — P?,in n tan = -5- = R~ - Qs\r\ ft^Pcosa' or -f; = Q_ sin {a -f 0) P cos (ft — (p)' (/) is called the afi_§-/e of friction. It has also been called the angle of repose, since a body will rem..; in at rest on an inclined plane so long as its inclination does rot exceed the angle of friction. If a = o = /?, then ^ = tan = /. The work done in traversing a distance x =■ Q cos /S.jtr. If Q is variable, the work done —. I Q cos ft . dx. 3. Wedge. — The wedge, or key, is often employed to con- nect members of a structure, and is generally driven into posi- WEDGE. 303 tion by the blow of a hammer. It is also employed to force out moisture from materials by induc- '"o '^ pressure thereon. The figure represents a wedge de- scending vertically under a continuous pressure P, thus producing a lateral motion in the horizontal member C, which must therefore exert a pressure upon the vertical face AB. The member // is fixed, and it is assumed that the motion of the machine is uniform, so that the wedge and 67 are in a state of relative equilibrium. Let K^ , R, be the reactions at the faces DE, DF, respec- tively, their directions making an angle 0, equal to the angle of friction, with the normals to the corresponding faces. Let a be the angle between DE and the vertical, a' the angle between DF and the vertical. Consider the wedge, and neglect its weight, which is usually inappreciable as compared with P. Resolving vertically, R, cos (90° - a + 0) + /?, cos (90° - «'-f-0) = P=7?,sin(a-f 0) + /?, sin(a'-;-0). . (i) Resolving horizontally, R, sin (90" — a + 0) — R^ sin (90° - a' + 0) = o, or R^ cos (or -|- 0) = R^ cos (a' -|- 0) (2) Consider the member C, and neglect its weight. Resolving horizontally, R, cos (« -f 0) = e = ^, cos {a' + 0) (3) Assuming the wedge isosceles, as is usually the case, a = a', and hence, by eq. (2), R, = R^, and by eq. (i), 2R, sin (a -|- 0) = P. (4) ) I-, •. ii I. Ij. i.tl ^ 3CM nu-.oKv 01- s7'A'rc/'CA'/i.s. Hence, by cqs. (^^) ami (4), Q cot (fir -f 0) _ cxtcnial icslstancc overcome (A\A\ — This ralii) 90. the ratio .y is negative, which is impossible, while if «>--}- = 90°, .y is zero, and in order to WHDGH, 305 ovcuomo C.'. Iinvvcvcr Hiniill il iiiif;lil be, /'would rccjuiic tu he iiifmitcly giciit, llciicc, » |- (/» imiHt be < yo", .111(1 hrlnw this limit ;, diniinlHhcH aH Incrcnscs. Similiirly. il may be hIiovvii fiom ctj. (7) that when Q is the rffnrt .md /' ihr icHistaiicc, must be < or, Q and that bcU)W this limit ., incrcaHCH with 0. Elftcicmy. iJuiin^^ llic unifoiiii mution of the machine, let any point a descend veitieally to the point h. The correspond- ing hoii/onlal di^piatemcrtl in evidently .iln . The motive work - - /'. ah ; " useful work Q.ibc, Q'^fx: Q Hence, the emcicncy --: ,, , — ,, . 2 tana I .all r = tan (V cot [(y |- (/'), by ecj. (5). This is a maximum for a ^dven value of «/» when a = 45 2' ;uul the max. eflficiency -- tan (45" — ) (d (45" -f- -j \ _ I — sin (J> ]~ I -{- sin 0" I - tan -^ ' 2 I -f tan For the reverse motion, the efficiency P.ah Q.2ln ' = cot a tan (« — 0). Fin « : |.-4 -1 H 'I } 306 THEORY OF STRUCTURES. This is a maximum when a ■=. 45° -j — . Thus the max. efficiency (45" +f) cot ^45° + t) tan (45' ♦ 0\ I — sin I -\- sin 0' 4. Screws. — A screw is usually designed to produce a hnear motion or to overcome a resistance in the direction of its length. It is set in motion by means of a couple acting in a plane perpendicular to its axis. A reaction is produced be- tween the screw and nut which must necessarily be equivalent to the couple and resistance, the vioiion being steady. Take the case of a .y^?^rt:rr * -threaded screw. It may be assumed that the reaction is concentrated along a helical line, whose diameter, d, is a mean between the external and internal diameters of the thread, and that iT:s distribution along this line is uniform. It will also be supposed that the axes of the couple and screw are coincident, so that there will be no lateral pressure on the nut. Let M be the driving couple. " Q '* " axial resistance to be over- come. " r " " reaction at any point a of the helical line, and let be angle between its direction and the normal at « ; is the angle of friction. " a " " angle between the tangent at a and the horizontal ; a Fig. 340. is called the pitch-angle. Since the reaction between the screw and nut must be equivalent to M and Q, then *Squarethreaded screws work more accurately than those with a V-thread, but the efficiency of the latter has be --n shown to be very little less than that of the former (Poncclet). On the other hand, the V-thread is the stronger, much less metal lieing removed in cutting it than is the case with a square thread. Again, with a V-thread there is a tendency to burst the nut, which does not obtain in a screw with a square thread. H< IF- •WPP f^ .I^J" li!' « SC/HE WS. 307 1; ' Q = algebraic sum of vertical components of the reac- tions at all points of the line of contact, = 2lr cos {a -|- 0)] = cos {a -{- (p)2{r), (0 and M = algebraic sum of the moments with respect to the axis of the horizontal components of the reactions at all points of the line of contact, = ^ r sin (a + 0) ^ = -sin (or -\- (p)2{r). (2) Let the couple consist of two equal and opposite forces, P, acting at the ends of a lever of length/, so that M = Pp. Hence, by eqs, (1) and (2), J7 = 7^=^^°'(^+^)' and the mechanical advantage (3) Q s/ If = o, p- == ~7 cot a, and the effect of friction may be allowed for, by assuming the screw frictionless, but with a pitch-angle equal to a -\- 0. Again, let the figure represent one complete turn of the thread developed in the plane of the ' paper. CD is the corresponding length of the thread ; DE the circumference nd\ CE, parallel to the axis, the pitch jj" h\ and CDE i\\Q pitch-angle or. Fig. 541. The motive work in one revolution = M. 27t = Pp . 2n, The useful work done in one revolution = Qli. Hence, the efficiency = 7, = ~ cot (« -f- 0) h — -, cot {a -|- 0) = tan a cot {a -\- 0). (4) : I i ! ', I 308 THEORY OF STRUCTURES. .. This is a maximum when a = 45° , its value then being tan I -\- tan In practice, however, a is generally much smaller, efficiency being sacrificed to secure a large mechanical advantage, which, according to eq. (3), increases as a diminishes. If «f -f- — 90°, -^ = o, so that to overcome Q, however small it may be, would require an infinite effort P. I : .-. ff -|- < 90°. Suppose the pitch-angle sufficiently coarse to allow of the screw being reversed. Q now becomes the effort and P the resistance. The direction of r falls on the other side of the normal, and the relation between P and Q is the same as above, — being substituted for 0. Thus, ^ = -^ cot (a - 0), and therefore the mechanical advantage = ^ = T^t^"(^-'^)' P If a = 0. If a < 0, reversal of motion is impossible, and the screw then possesses the property, so important in practice, of serv- ing to fasten securely together different structural parts, or of locking machines. m 11 !0«l? !i!T;: ENDLESS SCREWi. 309 Again, it may be necessary to take into account the friction between the nut and its seat, as well as the friction at the end of the screw. The corresponding moments of friction with respect to the axis are (Art. 8) / id:-d: and 3 / being the coefificient of friction, d^ , d^ the external and inter- nal diameters of the seat, and d' the diameter of the end of the screw. 5. Endless Screws (Fig. 242). — A screw is often made to work with a toothed wheel, as, for ex- ample, in raising sluice-gates, when the screw is also made sufficiently fine to prevent, by friction alone, the gates from falling back under their own weight. The theory is very similar to the preceding. Let the screw drive. A tooth rises on the thread, and the wheel turns against a tangential resistance Q, which is approximately parallel to the axis of the screw. Let Fig. 243 represent one complete turn of the thread developed in the plane of the paper, a being the pitch-angle as before. Consider a tooth. It is acted upon by (2 in a direction parallel to the axis, and by the reaction R between the thread and tooth, making an angle (the angle of friction) with the normal to the thread CD. Fig. 242. Fig. 343. .'. Q= R cos (a + 0). Again, the horizontal component of R, viz., R sin (a -f- 0), d has a moment R sin (« -(- 0) - with respect to the axis of the 3IO TIlkORY OF STRUCTURES. screw, and this must be equivalent to the moment of the driv- ing-couple, viz., Pp (Art. 4). .-. Pp = R- sin {a + 0). Thus the relation between P and Q is the same as in the pre- ceding article. Similarly if the wheel acts as the driver. ^ = -tan(«-0). 6. Rolling i" "cti-.-i.- -The friction between a rolling body and the surface over which it rolls is called rolling friction. Prof. Osborne Reynolds has given the true explanation of the resistance to rolling in the case of elastic bodies. The roller produces a deformation of the surfaces in contact, so that the distance rolled over is greater than the actual distance between the terminal points. This he verified by experiment, and con- cluded that the resistance to rolling was due to the sliding of one surface over the other, and that it would naturally increase or diminish with the deformation. In proof of this he found, for example, that the resistance to an iron roller on india- rubber is Un times as great as the resistance when the roller is on an iron surface. Hence the harder and smoother the sur- faces, the less is the rolling friction. The resistance is not sensibly affected by the use of lubricants, as the advantage of a smaller coefficient of friction is largely counteracted by the increased tendency to slip. Other experiments are yet re- quired to show how far the resistance is modified by the speed. Generally, as in the case of ordinary roadways, the resist- ance is chiefly governed by the amount of the deformation of the surface and by the extent to which its material is crushed. Let a roller of weight rr(Fig. 244) be on the point of motion under the action of a horizontal pull R. ROLLING FRICTIOA^. 311 The resultant reaction between the surfaces in contact must pass through the point of intersection of R and W. Let it also cut the surface in the point B. Let d be the horizontal distance between B and W. " p " vertical " " B " R. Taking moments about B, Rp = Wd, or R< — R = the resistance = W-. P Coulomb and Morin inferred, as the results of a series of ex- periments, that d is independent of the load upon the roller as well as of its diameter,* but is dependent upon the nature of the surfaces in contact. *Dupuit's experiments led him to the conclusion that d is proportional to the square root of the diameter, but this requires further verification. Let n be the coefficient of sliding friction. The resistance of the roller to sliding is /< IV, and " rolling " will be insured d if R < /< W, i.e., if - < tan 0, which is generally the case so long as the direc- P lion of R does not fall below the centre of the roller. Assume that R is applied at the centre. The radius r may be substituted for/, since d'\s very small, and hence R = ivi r An equation of the same form applies to a wheel rolling on a hard roadway over obstacles of small height, and also when rolling'on soft ground. In the latter case, the resistance is proportional to the product of the weight upon the wheel into the depth of the rut, and the depth for a small arc is inversely pro- portional to the radius. Experiments on the tractional resistance to vehicles on ordinary roads are few in number and incomplete, so that it is impossible to draw therefrom any general conclusion. From the experiments carried out by Easton and Anderson, it would appear that the value of d in inches varies from 1.6 to 2.6 for wagons on soft ground, and that the resistance is not sc ibiy affected by the use of springs. Upon a hard road, in fair condition, the resistance was found to be from i to J of that on the soft ground, the average value of d being \ mch, and was very sensibly diminished by the use of springs. ■ IIH! " "' !* m ^■ti.^M^iiUIiihiwLn ' ^MHHPi' h\V ' '1'- iS^^^^myu ; 1 ;..'■' 1 1 L i; KH p.:-''^, •:i^J 312 THEORY OF STRUCTURES. 7. Journal-friction. — Experiments indicate that / is not the same for curved as for plane surfaces, and in the ordinary cases of journals turninjr in weii- lubricated bearings the value of/ is probably governed by a combina- tion of the laws of fluid friction and of the sliding friction of solids. The bearing part of the journal is generally truly cylindrical and is terminated by shoulders resting against the ends of the step in which the journal turns. Consider a journal in a semicircular bearing with the cap removed. When the cap is screwed on, the load upon the journal will be increased by an amount approximately equal to the tension of the bolts. Let Pbe the load. Assume that the line of action of the load is vertical and that it intersects the axis of the shaft. This load is balanced by the reaction at the surface of contact, but much uncertainty exists as to the manner in which this reaction is distributed. There are two extremes, the one corresponding to a normal pressure of constant intensity at every point of contact, the other to a normal pressure of an intensity varying from a maximum at the lowest point /^ to a minimum at the edge of the bearing B. Let / be the length of the bearing, and consider a small element AS at any point C, the radius OC {= r) making an angle with the vertical OA. First. Let / be the constant normal intensity of pressure. P - 2{pAS cos ^. /) = pl-^ipU) = 2plr. Frictional resistance =2{//>ASl)= /p/2{AS)=/p/7rr=/P-. The frictional resistance probably approximates to this limit when the journal is new. '' tl > JO URN A L-FRICTION. 313 Second. Let / = /„ cos ^, so that the intensity is now proportional to the depth CD and varies from a maximum /„ at A to nil ^t B. This, perhaps, represents more accurately the pressure at different points when the journal is worn. ' -Is; .: P = 2(/ J5 cos e. I) = 2[p,AS cos' d. I) = 2pjr, f- cos' 6* . ^6^ = pjr\ 2 and i\ie frictional resistance = 2{/pJS/) = 2/pJr = fP-. ... 7r 4 Hence, the frictional resistance lies between fP~ and /"/*— . 2 TT It may be represented by fxP, ix being a coefficient of friction to be determined in each case by experiment. The total uiomcnt of frictional resistance must necessarily be equal and opposite to the moment M of the couple twisting the shaft ; i.e., M = fxPr. Thus, the total reaction at the surface of contact is equiva- lent to a single force P tangential to a circle of radius ^r having its centre at and called the friction-circle. The work absorbed by axle-friction per revolution = M.27t = 2)xnPr. The work absorbed by axle-friction per minute = 2}ntPrN — piPv, N being the number of revolutions and v the velocity per minute. 314 THEORY OF STRUCTURES. ' The work absorbed by frictional resistance produces an equivalent amount of heat, which should be dissipated at once in order to prevent the journal from becoming too hot. This may be done by giving the journal sufificient bearing surface (an area equal to the product of the diameter and the len<^th of the bearing), and by the employment of a suitable unguent. Suppose that h units of heat per square inch of bearing surface {Id) are dissipated per minute. Let / inches be the length and d inches the diameter of the journal. hdl = heat-units dissipated = heat-units equivalent to frictional resistance 12/ ~I^' J being Joule's equivalent, or 778 ft.-lbs. 12/ h PN , 12//1 Pv and ixn I fx ~ Id ' Let jz =p = pressure per square inch of bearing surface. Id \2jh . pv = = a constant. M In Morin's experiments ^^ varied from 2 to 4 in., P from 330 lbs. to 2 tons, and v did not exceed 30 ft. per minute," so that/z' was < 5CX)0, and the coefificient of friction for the given limits was found to be the same as for sliding friction. Much greater values oi pv occur in modern practice. Rankine g\vesp{v -\- 20) = 44800 as applicable to locomo- tives. Thurston gives ^z^ = 60000 as applicable to marine engines and to stationary steam-engines. Frictional wear prevents the diminution of /below a certain JOURNAL FRICTION. 315 limit at which the pressure per unit of bearing surface exceeds a value/ given by the formula. where P=pld= pkd'' ; / k = d In practice /i = ^ for slow-moving journals (e.g., joint-pins), and varies from i^ to 3 for journals in continuous motion. The best practice makes the length of the journal equal to four diameters (i.e., k = 4) for mill-shafting. Again, if the journal is considered a beam supported at the ends, CPl qcP 32" -^, q being the maximum permissible stress per square inch, and C a coefficient depending upon the method of support and upon the manner of the loading. .*. d"^ Of. — . For a given value of /*, d diminishes as q increases. Also, it has been shown that the work absorbed by friction is directly proportional to d. Hence, for both reasons, d should be a minimum and the shaft should be made of the strongest and most durable material. In practice the pressure per square inch of bearing surface may be taken at about 2 tons per square inch for cast- iron, 3J tons per square inch for wrought-iron, and 6^ tons per square inch for cast-steel. It would appear, however, from the recent experiments of Tower and others, that the nature of the material might become of minor importance, while that of a suitable lubricant would be of paramount importance. They show that the friction of properly lubricated journals follows the laws of fluid friction much more closely than those of solid friction, and that the 3l6 THEORY OF STKUCrURES. lubrication might be made so perfect as to prevent any ab- solute contact between the journal and its bearing. The journal would therefore float in the lubricant, so that there would be no metallic friction. The loss of power due to fric- tional resistance, as well as the consequent wear and tear, would be very considerably diminished, while the load upon the journal might be increased to almost any extent. Tower's experiments also indicate that the friction dimin- ishes as the temperature rises, a result which had already been experimentally determined by Him. It was also inferred bv Hirn that, if the temperature were kept uniform, the friction would be approximately proportional to sU\ and Thurston has enunciated the law that, with a cool bearing, the friction is approximately proportional to Vv for all speeds exceedinj^ 100 ft. per minute. With a speed of 150 ft. per minute and with pressures vary- ing from 100 to 750 lbs. per square inch, Thurston found ex- perimentally that /"varied inversely as the square root of the intensity of the pressure. The same law, but without any limitations as to speed or pressure, had been previously stated by Hirn. 8. Pivots. — Pivots are usually cylindrical, with the circular edge of the base removed and sometimes with the whole of the base rounded. Conical pivots are employed in special machines in which, e.g., it is important to keep the axis of the shaft in an invariable position. Spherical pivots are often used for shafts subject to sudden shocks or to a lateral move- ment. {a) Cylindrical Pivots, — If the shafts are to run slowly, the intensity of pressure (/>) on the step should not be so great as to squeeze out the lubricant. Reuleaux gives the following rules: The maximum value of p in lbs. per square inch should be 700 for wrought-iron on gun-metal, 470 for cast-iron on gun- metal, and 1400 for wrought-iron on lignum-vitae. For rapidly-moving shafts, d—cVPn, H ;/ being ' H to be det( ™ and/^th( Suppc divided i rin^rs be h In on( by the fri Hence th( where and d^,d^i face in con If the T work absorl Again, t and the tota If^.= o. Thus, in ; times the mo til PIVOTS. 3<7 II id be gun- n being the number of revolutions per minute, c a coefficient to he determined by experiment ( = .0045), ,111(1 /'the load upon the pivot. Suppose the surface of the step to be tiiviiled into rings, and let one of these riiv^^s be bounded by the radii x, x -(- dx. In one revolution the work absorbed by the friction of this ring = /t . 2nx . dx . 2nx. Hence the total v^oxV absorbed in one revolution Pic. 346. = / ^iipit^x^dx I^P^^ /jz jj\ 2 d* — d^ where P=pjL^d:-d:), and d^ , d^ are the external and internal diameters of the sur- face in contact. If the zvholc of the surface is in contact, d^ = O, and the work absorbed = ^;A7rPd^. Again, the moment of friction for the ring = fxp.2nx .dx.x = 2/u7r/>x^ . d. ; and the total moment ~7^» d' — d* 2ixnpx'dx = \f^7tp — in 8 12 ^ ' ^' 3 d^' — d^ uP If d^ r= o, the moment — ^-—d^. 3 Thus, in both cases, the work absorbed by friction = 2n times the moment of friction. 3i8 'j7//-:oA'y o/'- si'A'L'crc/A'/is. Let D be the wmn diameter of the surface in contact . 2 ' Let 2y be the width of the surface in contact = d^ — d^. Then work absorbed = /<^/^fZ?+ ~-j. Some»-imes shafts have to run at high speeds and to bear heavy pressures, as, e.g., in screw-propellers and turbines. In order that there may be as little vibration as possible, / nuist be as small as practicable, and this is to some extent insured by using a collar-journal. Let N be the number of collars, and let 7T^ , ,, ,,, „ /' (i.^ — r/,' , - . . = ^^~- {d^ — /)/: sin a . 27rx /■'" =:: 2n/>x I dx =■ 2fr/>x{x, ■»',). /'/' Hence total moment of friction = - -; — (.r, f-.tO. 2 SM) tr^ ' ' '' ( a sin 6\ and hence if/> is constant, y />i/fi;\ Any small element JUi' (= rfs) of the belt is acted upon by a pull T tan- ^/ {fcntial to the pulley at B, a pull T— dT tau'^^ential to the pulley at B' , and by a iciction equivalent to a normal force Rds It the middle point of /)'/)", and a tan- gential force, or frictional resistance, Let the angle COD — 8, and the an- gle BOB' = de. P'<-- '49. Resolving normally, {T-\T- dT) s\n^ - Rds = o. . . . (i) % mif m I I . H"; I ja^. • . THEORY OF STRUCTURES. Resolving tangentially, dH {T — T — dT) cos — — ixRds = o, ... (2) /i being the coefficient of friction. . . .. , . . „ . dO . . , dB Now ad being very small, sin — is approximately — . dd ' cos — is approximately unity, and small quantities of the second order may be disregarded. Hence, eqs. (i) and (2) may be written TdB-Rds-o (3j !] and dT — jxRds = o. dT .-. dT — jxTdB, or -^p — t^dd. . . . Integrating, log,T=M(^+C, C being a constant of integration. When (9 = 0, T = T^, and hence log, 7", = (7. (4' (5> I 1 .-. log,-:™ = fxO, or -;f = ^^ • (6) When e = cr, T - T„ and hence (7) t- being the number 2.71828, i.e., the base of the Naperian system of logarithms. . (6) . (7) iperi ;in BELTS AND ROPES. 323 Fig. 250. If « is increased by /?, the new ratio of tensions will be c^^ times the old ratio ; so that if a increases in arithmetical progression, the ratio of tensions will increase in geometrical progression. This rapid increase in the ratio of the tensions, corresponding to a comparatively small increase in the arc of contact, is utilized in " brakes" for the purpose of absorbing surplus energy. For example : A flexible brake consisting of an iron .or steel strap, or, again, of a chain, or of a series of iron bars faced with wood and jointed together, embraces about three-fourths of the cir- cumference of an iron or wooden drum. One end of the brake is secured to a fixed point and the other to the end B o( a. lever A OB turning about a fulcrum at O. A force applied at .i will cause the brake to clasp the drum and so produce fric- tion which will gradually bring the drum to rest. Let a) be the angular velocity of the drum before the brake is applied. Let /be the moment of inertia of the drum with respect to its axis. /a)* The kinetic energy of the drum = . When the brake is applied, the motion being in the direc- tion of the arrow, let the greater and less tensions at its ends be r, , T^, respectively. Let n be the number of revolutions in which the drum is brought to rest. Then i/oj' = {T, - T,)ndn, (8) ./being the diameter of the drum. Also, if Pis the force applied at A, and if p and q are the perpendicular distances of O from the directions of P and 7",, rcsp(,'ctively. Pp - T^f' (9) ;:( 'I'l Ss'j'tl II' i ■ n I i i !■ m IH'lp ?ii ' A ■: 1 u , .,1 ' ! ; ! .' ■ 1 1' THEORY OF STRUCTURES. Again, y, = T^c^", (lo) a being the angle subtended at the centre by the arc of contact. Hence, by eqs. (8), (9), (lO), , n = qJoa^ 2Pp{c>"'— \)nd' (n) If the motion of the drum were in the opposite direction, q would be the perpendicular distance of from the direction of Z, , and then Pp = T^q. Proceeding as before, , n = ^/coV"" 2Pp{e*"-— \)niV and therefore the number of turns '.1 the second case, before the drum comes to rest, is e^*^ times the number in the first, which is consequently the preferable arrangement. The coefficient of friction fi varies from .12 for greasy shop belts on iron pulleys to .5 for new belts and hempen ropes on wooden drums. In ordinary practice, an average value of /< for dry belts on iron pulleys is .28, and for wire ropes .24; if the belts are wet, }x is about .38. Formulae (6) and (7) are also true for non-circular pulleys. 10. Effective Tension. — The pull available for the trans- mission of power = r, — 7", = 6". Let HP be the horse- power transmitted, v the speed of transmission in feet per sec- ond, a the sectional area of the rope or belt, and s the stress per square inch in the advancing portion of the belt. ThcK, if T', and 7", are in pounds, HP (7-, - T,)v Sv = — , and 550 550' as. The working tensile stress per square inch usually adopted for leather belts varies from 285 lbs. (Morin) to 355 lbs. (Claudcl), EFFECT OF HIGH SPEED. 325 "■"Wli! \m liiMl * i ■■■I f m RfHpRr ' i ^ \ T- 1 ^ r an average value being 300 lbs. In wire ropes, 8500 lbs. per square inch may be considered an average working tension. Hempen ropes fpr the transmission of power generally vary from 4^ to 6^ in. in circumference. II. Effect of High Speed. — When the speed of trans- mission is great, tlie effect of centrifugal force must be taken into account. wads v' The centrifugal force or the element ds = , zu being the specific weight of the belt or rope, and r the radius of the pulley. Eq. (3) above now becomes Tdd — Rds = o, or ^ ,„ wadd , „ , Tdd —~v* — Rds = o\ and hence, by eq. (4), dT T— — v' S = ^de. Integrating, T- wa log* g ■v^ = txe g since T — T^ when Q — o. Also, T—T^ when 6^ = a, and therefore wa . T. g -V T, wa ~g z=ei"', -v^ or wa 7; = 7;^« ■y'(>"- 1). 1 THEORY OF STRUCTURES, The work transmitted per second = (r.-r>=:(r,z;-'^z.')(.^--i), which is a maximum and equal to f 7',(r'*» — i) when and the two tensions are then in the ratio of V = A / , a 2^^" + I to 3. The speed for which no work is transmitted, i.e., the Hniit- ing speed, is given by T.v -v' o, or V zva 12. Slip of Belts.— A length / of the belt (or rope) becomes /(i + -^jon the advancing side and /(i -\- —j on the s/acJt side, T T^ where />, = -' and />, = --, B being the coefficient of elasticity. Thus, the advancing pulley draws on a greater length than is given off to the driven pulley, and its speed must therefore exceed that of the latter by an amount given by the equation ; i 4 1 reduction of speed, or slip speed of driving pulley A -A The slip or creep of the belt measures the loss of work. In ordinary practice the loss with leather belting does not ex- ceed 2 per cent, while with wire ropes it is so small that it may be disregarded. PKON Y'S D YNAMOME TEli. 327 13. Prony's Dynamometer. — This dynamometer is one of tiie commonest forms of friction-brake. The motor whose power is to be measured turns a wheel E which revolves be- tween the wood block B and a band of wood blocks A. To A^? Fig. 351. the lower block is attached a lever of radius p carrying a weight P at the free end. By means of the screws C, D the blocks may be tightened around the circumference until the unknown moment of frictional resistance FR is equal to the known moment Pp. The weight /*, which rests upon the ground when the screws are slack, is now just balanced. The work absorbed by friction per minute = znRFn = mPpn, II being the number of revolutions per minute. 14. Stiffness of Belts and Ropes. — The belt on reaching the pulley is bent to the curvature of the periphery, and is straightened again when it leaves the pulley. Thus, an amount of work, increasing with the stiffness of the belt, must be ex- pended to overcome the resistance to bending. As the result of experiment, this resistance has been expressed in the form i>R' T being the tension of the belt, a its sectional area, R the radius of the pulley, and 6 a coefificient to be determined. According to Redtenbacher, /f = 2.36 in. for hempen ropes. t( n a ^ = 1.67 " " " " « •• Reuleaux, b = 3.4 " " leather belts. 1 r. !'|! 1 1 1 J . ( h ,.;'!.. -'in lif f 1 328 THEORY OF STRUCTURES. Let the figure represent a sheave in a pulley-block turning in the direction of the arrow about a journal of radius r. Let T, be the effort, 7!, the re- sistance. The resistance due to the stiff- ness of the belt may be allowed for aT aTs by adding -ttb to the force 71. The Fig. 2s». frictional resistance at the journal- surface is Psin or//*, P being the resultant of 7,, 7,. The motion being steady, taking moments about the centre, T,R=.\T^-\- Tr )R+fPr, or T,-T,-]r ^^ -\rf pP- If 7, and 7, are parallel, 7* = 7, -|- 7,, and the last equa- tion becomes ,' T.= T,-^YR'^f'^{T,-\-T^. Let the pulley turn through a small angle B. The counter-efficiency of the sheave motive work 7,# 7, 2/r a ~ useful work ~ TJ^~ T~^ '^ R - fr^ b R — fr' In the case of an endless belt connecting a pair of pulleys of radius R^, R^, the resistance due to stiffness m?y be taken equal to -r-y-p + d/- ^bemg the mean tension ^= ^ The resistance due to journal-friction — /rP\j^ ~^'rJ' The useful resistance = T, — T, = S. WHEEL AND AXLE. Hence, the counter-efficiency 329 = ^ + (1 + ^)11^+ "■^'■?)- \R, ' Rj\bS In wire ropes the stress due to bending may be calculated as follows : * Let X be the radius of a wire. The radius of its axis is sensibly the same as the radius R of the pulley. The outer layers of the wire will be stretched, and the inner shortened, while the axis will remain unchanged in length. Hence, X change of length of outer or inner strands unit stress R ~ length of axis ~ E ' and the unit stress due to bending = E-jc . 15. Wheel and Axle. — Let the figure represent a wheel of radius p turning on an axle of radius r, under the action of the two tangential forces P and Q, in- clined to each other at an angle d. The resultant R oi P and Q must equilibrate the resultant reaction be- tween the wheel and axle at the sur- face of contact. Let the directions of P and Q meet in T. If there were no friction, the re- sultant reaction and the resultant R would necessarily pass through O and T. Taking friction into account, the direction of R will be inclined to TO. Let its direction intersect the circumference of the axle in the point A. The angle between TA and the noxmaX AO dA A, the motion being steady, is equal to the angle of friction ; call it 0. •} i, ; ' ( ■ 1 ■ Fig. 2S3- 1 J I e 3. ii iii^ ri I! 330 THEORY OF STRUCTURES. Taking moments about ^, Pp — Qp — Rr s\n (p^o (i) Also, R' - P" -\- Q ^ 2PQ cos 6 (2) Let/= sin = A* Eq. (i) may now be written , jj. being the coefficient of friction. Pp- Qp- fRr = O. . . If Pand Q are parallel in direction, ^ = O and R = P-l-Q. Let the figure represent a wheel and axle. (3) Let P be the effort and Q the weight lifted, the directir r. of Pand Q being parallel. Let ^Fbe the weight of the " wheel and axle." Let /?, and R^ be the vertical reactions at the bi. iigs. Let/ be the radius of the wheel. Let q " " " axle. Let r " " " bearings. Take moments about the axis. Then Pp — Qq — R,r sin cp — R^r sin = o. (4) TOOTHED GEARING. 331 But Hence, R,-\-R,= W+P+Q (5; Pp - Qg = (JV-{-PJr Q)r sin = (W+PJr Q)/r, or P{p-/r) = Q{g + /r)+/lVr Efficiency. — In turning through an angle 6', motive work = PpB, useful work = Qq^, Qq^ Qq (6> .'. efficiency = Q Ppe - Pp ' and the ratio j,- is given by eq. (6). 16. Toothed Gearing. — In toothed gearing the friction is partly rolling and partly sliding, but the former will be disre- garded, as it is small as compared with the latter. Fig, ass- Let the pitch-circles of a pair of teeth in contact at the point B ' ^uch at the point A ; and consider the action before reaching the line of centres (9,6>, , i.e., along the arc of approach. ^it;?!-;. ili -y I 11 l> i I ^sf^sm •- ■ 332 THEORY OF STRUCTURES, The line /i/)' is nonnal to the surfaces in contact at the point /)' Let R be the resultai^t reaction at //. Its direction, the motion being steady, makes an angle 0, equal to the angle of friction, with /!/>. Let ^ be the angle between (7,C7, and AB, Let the motive force and force of resistance be respective!)- equivalent to a force /* tangential to the pitch-circle C?, ,anil to a force Q tangential to the pitch-circle (>, . Let ;•, , i\ be the radii of the two wheels. The work absorbed by friction in turning through the small arc d& -=^{P-Q)ds (1) Consider the wheel (?, , and take moments about the centre, Pi\ =: K\r, sin (^ - ) + x sin 0(, . . . (j) where /IB = r. Similarly, from the wheel O^ Qr^ — R\r^ sin {H — ) — X sin ) - - sin 'P~ X "' sin {0 — 0) -f- - sin • • • • (4) and therefore P-Q=Q (- "I — \v sin {^ ?\v\ (^ — 0) sin Hence, the work absorbed by friction in the arc ds (^+;-)-»^sin0flJf = <2 . . . (f'l sin {0 -' 0) - '- sin In 1 leaving and t'^c Th e 1 arc respc f^iven by the arc o A[Tain i.e., if and this c Simpl( ci'liciency H diffe with r, a urcd from Hence TOOTHED GEARING. 333 111 precisely the same manner it can be shown that, after leaving the line of centres , i.e., in the arc of recess. Q sin (^ -J- 0) sin sin {H-\- 0)-f- - sin «/» and t^c work absorbed by friction in the arc ds (7) Q . (8) sin {(i -{- ^(^+^)/' 03) ' 1 ' a and the counter-efficiency '1 ' j' - =:l+;i;r(~ + i) (15) This last equation shovvs that the efficiency increases with the number of teeth. \Mi ' EFFICIENCY OF MECHANISMS. 335 If the follower is an annular wheel, — — — must be substi- Thus, with an an- tuted for — \-— in the above equations. nular wheel the counter-efficiency is diminished and the efficiency, therefore, increased. It has been assumed that R and Q are constant, as their variation from a constant value is probably small. It has also been assumed that only one pair of teeth are in contact. The theory, however, holds good when more than one pair are in contact, an effort and resistance, corresponding to P and Q, being supposed to act for each pair. 17. Bevel-wheels. — Let I A, IB represent the develop- ments of the axes of the pitch- circles //, , //, of a pair of bevel- wheels when the pitch-cones are spread out flat, 0, , O^ being the corresponding centres. The preceding formulae will ap- ply to bevel-wheels, the radii being OJ, OJ, and the pitch being meas- ured on the circumferences I A, IB. 18. Efificiency of Mechanisms. — Generally speaking, the ratio of the effort P to the resistance Q in a. mechanism may be expressed as r function of the coefificient of fric- tion fx. Thus, Q = Hm)' If, now, the mechanism is moved so that the points of application of P and Q traverse small distances Jx, Ay in the directions of the forces, \ (1 the efficiency = PAx I Ay Jd^Ax' 3S^ THEORY OF STRUCTURES. Ay But the ratio -j-^ depends only upon the geometrical rcLi- tions between the different parts of the mechanism, and will therefore remain the same if it is assumed that i-i is zero. In such a case the efficiency would be perfect, or the motive vvoik {PAx) would be equal to the useful work {QAy), and therefore I. In ^ i I = Hence, the efficiency I Ay F{o)A^' 2?1; TABLE OF COEFFICIENTS OF AXLE-FRICTION. i '. ii Bell-met.il on bell-raetal Brass on brass Br.iss on cast-iron Cast-iron on bell-metal C.ist-iron on brass Cast-iron on cast-iron Cast-iron on liRnum-vitae Lignum-vitae on cast-iron Litfnum-s'itoe on lignum-vitae... Wroutrht-iron on bell-mclal WroU(jht-iron on cast-iron Wrou({ht-iron on liffnum-vit.ie. . c •a a 'fi : .S.S >y £^ ■a 3 0-^ Q c .097 .079 .072 •'94 .161 .075 .194 .075 •»37 .075 .185 . I .116 .251 .189 ■075 .075 .187 "5 c w o 3 u C 3 ° J u .040 .054 • 054 .054 .092 ■ '7 .07 ,054 •"54 I U 3 .065 .09 ^3 .14 • 14 EXAMPLES. 337 EXAMPLES. 1. In a pair of four-sheaved blocks, it is found that it requires a force P to raise a \veifi;ht 5/", and a force 5/'' to raise a weigiit 15/^'. Show that the j^enerid relation between the force P and the weight W to be raised is given by P = -IV- P'. Find the efficiency when raising the weights 5/" and 15/". 2. Find tlie mechanical advantage v/hen nn inch bolt is screwed up byaiS-in. spanner, the elTective diameter of the nut being I'i in., the diameter at the base of the thread .84 in., £ iid . 1 5 being the coclFicient of friction. 3. A belt, embracing one-half the circumference of a pulley, transmits 10 H. P. ; the pulley makes 30 revolutions per minute and is 7 ft. iri diameter. Neglecting slip, find T\ and T-i\ i-i being .125. 4. A A-in. rope passes over a 6-in. pulley, tlie diameter of the axis bointf i in. ; the load upon the axis = 2 x the rope tension. Find tlie elhcieiicv ol tiic pulley, the coefficient of axle-friction being .08 and the coefficient for slilTnoss .47. Hence also deduce the efficiency of a pair of three-sheaved blocks. 5. If the pulleys are 50 ft. c. to c. and if the tight is three times the slack tension, find the length of the belt, the coefficient of friction being iand the diameter of one of the pulleys 12 in. 6. Show that the work transmitted by a belt passing over a pulley will be a ma.ximum when it travels at the rate of 4/ _ ft. per sec, T% being the slack tension and tn the mass of a unit of length of the belt. The tight tension on a 20-in. belt, embracing one -half the circum- ference of the pulley, is 1200 lbs. Find the maximum work the belt will transmit, the thickness of the belt being .2 in. and its weight .0325 lb. per cubic inch. (Coefficient of friction = .28.^ 7. In an endless belt passing over two pulleys, the lea.st tension is 150 lbs., the coeflicient of friction .28, and the angle subtended by the arc of contact 148". Find the greatest tension. The diameter of the larger wheel is 7S in., of the smaller 10 in., of the bearings 3 in. Find the efficiency. A lightening-pulley is mndc to press on the slack side of the belt. Assuming that the working tension is to the coefficient of elasticity in the ratio of i to 80, find the increment of the arc of contact I'fT'' 'PI IP ii.>mj. ^ uf i Ill 'ill. p' Fig. 2SJ, Fig. 258. CHAPTER VI. ON THE TRANSVERSE STRENGTH OF BEAMS. I. To determine the Elastic Moment. — Let the plane of R the paper be a plane of symmetry Avith respect to the beam PQRS. If the beam is subjected to the action of external forces in this plane, PQRS is bent and as- sumes a curved form P'Q'R'S'. The upper layer of fibres, Q'R', is extended, the lower layer, P'S', is compressed, while of the layers within the beam, those nearer P'S' are compressed and those nearer Q'R' are extended. Hence, there must be a layer M'N' between P'S ' and Q'R' which is neither compressed nor extended. It is called the neutral surface (or cylinder), and its axis is perpendicular to the plane of flexure. In the present treatise it is proposed to deal with flexure in one plane only, and, in general, it will be found more convenient to refer to M'N' as the neutral line (or axis), a term only used in refer- ence to a transverse section. If a force act upon the beam in the direction of its length, the lower layer P'S' , instead of being compressed, may be stretched. In such a case there is no neutral surface within tthe beam, but theoretically it still exists some- where ivithout the beam. Let ABCD be an indefinitely small rect- angular element of the unstrained beam, and let its length be s. Let A'B'C'D', Fig. 260, be the element after deformation by the external forces. 340 / B '1 p p D Q C Fig. 359. %\ Ilia THE ELASTIC MOMENT. P'Q', the neutral line, being neither com- 341 A'. pressed nor extended, is unchanged in length p! and equal to PQ — s. Let the normals at P' and Q' to the neutral line meet in the point \ O h the centre of curvature of P'Q'. Also, as the flexure of the element is very small, the normal planes through OP' and OQ' may be assumed to be perpendicular to all the layers which traverse the corresponding sec- tions of the beam, so that they must coincide with the planes A'D' and B'C, respectively. The assumptions made in the above are : (^2:) That the beam is symmetrical with respect to a certain plane. [b) That the material of the beam is homo- geneous. (r) That sections which are plane before bending remain plane after bending. {d) That the ratio of longitudinal stress to the correspond- ing strain is the ordinary (i.e., Young's) modulus of elasticity notwithstanding the lateral connection of the elementary layers. {c) That these elementary layers expand and contract freely under tensile and compressive forces. Consider an elementary layer p'q', of length s', sectional area a^ , and distant j, from the neutral surface. -LctOP' =R= OQ'. From the similar figures OP'Q' and Op'q' , Op' p'q' OP' ~ P'Q' or R±_y. R = — , and therefore ^ = R s' — s Also, if /, is the stress along the layer /»'^', j ■ \il ^ S' —S r- f^ ^ 342 THEORY OF STRUCTURES. n > E being the coefificient of elasticity of the material of_the beam. So, il f,, a^, y^, it, a^, jt, . . . are respectively the stress, sectional area, and distance from the neutral surface, of the several layers of the element, _E _E The total stress along the beam is the algebraic sum of a these elementary stresses, E E R> R Again, the moment of /, about P' = t^y^ — — a^y^ ; i E it U «« / " « / ^ ^ .. 3 . *« — *a /a — r> "a J'a > R E u u n 4 «« " = / 1/ n V ' • and so on. Thus, the Elastic Moment for the section A' D' — the alc^e. braic sum of the moments of all the elementary stresses in the different layers about P' , = ^ijV. + t^y^ + t,y^ 4- . . . = T,-(«, y^ + «,/»' + •• •) R E R = %^iccf)' I'l THE ELASTIC MOMENT. 343 Now, 2 («y) is tbe tnoment of inertia of the section of the beam through A'D\ with respect to a straight Hne passing through the neutral Hne and perpendicular to the plane of flexure, i.e., the plane of the paper. It is usually denoted by / or Ak'\ A being the sectional area, and k the radius of gyration. Thus, E E the elastic moment = - /= -^Ak^, But the elastic moment is equal and opposite to the bending moment {M) due to the external forces, at the same section. Hence ~I=.~Ak^ = M. the alc;e- es in the Note. — It is necessary in the above to use the term alge- braic, as the elementary stresses change in character, and therefore in sign, on passing from one side of the neutral sur- face to the other. Cor. I. Bearing in mind assumption () = o, or 2{a}') = o, showing that P' must be the centre of gravity of the section through A'D'. Hence, when the external forces produce no longitudinal stress in the beam, the neutral line is the locus of the centres of gravity of all the sections perpendicular to the length of the beam. Cor. 3. If /, a, y be, respectively, the stress, sectional area, and distance of a fibre from the neutral line, then -^ay = t, or -^ jj/ = - = intensity of stress =-fy, suppose, fyE.E /=M==^'-/. Example i. A timber beam, 6 in. square and 20 ft. long, rests upon two supports, and is uniformly loaded with a weight of 1000 lbs. per lineal foot. 'Determine the stress at the centre at a point distant 2 in. from the neutral line. Also find the central curvature, E being 1,200,000 lbs. I—-^— — 108, M =^ looa X 10— 1000 X 5= 5000 ft.-lbs. 12 = 60,000 inch-lbs., and y ■= 2 in. and the stress in the metal is necessarily greatest at the central section. W-\- 5200 M, at the centre, = — . 30. 12 inch-lbs. ; 8 / ''2 1 /, = 2 X 2240 lbs., and - = ?rrV = — . 15'.. -. ^HasMiii 'Jtic !i i|'".'f INTERNAL STRESSES. 345 Hence from the above equations, I20OO00 „ ^ /y „ — 73— X 108 = 60000 = —108. K 2 Thus R = 2160 in. = 180 ft., and ^ = 11 ii-^ lbs. per sq. in. Ex. 2. A standpipe section, 33 ft. in length and weighing 5720 lbs., is placed upon two supports in the same horizontal plane, 30 ft. apart. The internal diameter of the pipe is 30 in., and its thickness .V inch. Determine the additional uniformly distributed load which the pipe can carry between the bearings, so that the stress in the metal may nowhere ex- ceed 2 tons per square inch. Let W be the required load in pounds. 30 The weight of the pipe between the bearings = — . 5720 — 5200 lbs. Thus, the total distributed weight between the bearings = (rr+ 5200 lbs.) Now c W-\- 5200 22 , I ~ . 30. 12 = 2 . 2240 .— . 15'. - = 72000 X 22, o 72 and hence W ^= 30,000 lbs. ■}■. i- ii 346 THEORY OF STRUCTURES. Cor. 4. The beam is strained ♦"o the limit of safety when either of the extreme layers A' B\ D'C is strained to the limit of elasticity. In such a case, the least of the values of — for y the extreme \s.ycrs A' B' , D' C is the greatest consistent with the strength of the beam ; and \i f^ and c are the corresponding intensity of stress, and distance from the neutral axis, R c EXAMTLE. — Compare the strengths of two similarly loaded beams of the same mai°rial, of equal lengths and equal sectional areas, the one being round and the other square. Let r be the radius of the round beam ; /,., the intensity of the okin stress. Let a be a side of the square beam; _/,, the intensity of the skin st.'ess. Then TTr' = rt' ; /, for round bar, = , and for square bar = --. 4 ' 12 Also, since the beams are similarly loaded, the bending moments at corresponding points are equal. prcs l^ut stros shea diffc with so that " r ^ M a 12' 2 fr 2 a' 2 /. 3 ^^" 3 V /22 /88 7 ~ V 63 Thus, under the same load, the round beam is strained to a greater extent than the square beam, and the latter is the stronger in the ratio of i 88 to V63. B/i EA KIA'G WEIGH TS. 347 Cor. 5. The neutral surface is neither stretched nor com- pressed, so that it is not subjected to any longitudinal stress, liut it by no means follows that this surface is wholly free from stress, and it will be subsequently seen that the effect of a shearing force, when it exists, is to stretch and compress the different particles in diagonal directions making angles of 45° with the surface. Cor. 6. For a rectangular beam, / = 12 and c ■=■ c -7 = T-bir a 12 6 If the beam is fixed at one end and loaded at the other with a weight IV, the maximum bending moment = IV/. If the beam is fixed at one ei:d and loaded uniformly with a weight wl = W, the maximum bending moment 2 m 2 If the beam rests upon two supports and carries a weight W IVl at the centre, the maximum bending moment =:: — — . If the beam rests upon two supports and carries a uniformly distributed load of ivl = W, the maximum bending moment - '8 ~ 8"* f 1. jt Hence, in the first case, W = y —j~\ " " second " W 6^ r " " third " JF=^4y; " '• fourth " ]V= i^'T- 6 I (!' I I yi 348 In general, THEORY OF STRUCTURES. ^" il. q being some coefificient depending upon the manner of the loading. Now, if the laws of elasticity held true up to the point of rupture, these equations would give the breaking weights (JF), corresponding to different ultimate unit stresses (/), but tlic values thus derived differ widely from the results of experi- mert. It is usual to determine the breaking weight {W) of a rectangul beam from the formula W ^^ ^ ~7~' where C is a constant which depends both upon the manner of the loading and the nature of tlie material, and is called the coefficient of rupture. The modulus of rupture is the value of/" in the ordinary bcnding-moment formula (J/ — -/) when the load on the beam is its breaking load. The preceding equations, however, may be evidently em- ployed to determine the breaking weights in the several cases by making "^(7 = C. In this case /is no longer the real stress, but may be called the coefficient of bending strength. The values of C for iron, steel, and timber beams, supported at the two ends and loaded in the centre, are given in the Tables at the end of Chapter III. The corresponding value of / is obtained from the equation or f-.--\C. Example. — Determine the central breaking weight of a * \\i EQUALIZATION OF STRESS. 349 red-pine beam, lo in. deep, 6 in. wide, and resting upon two supports 20 ft. apart. The value of C for red pine is about 5700. Hence, the breaking weight = IV = 5700 6.10' 20 X 12 - = 14,250 lbs. 2. Equalization of Stress. — The stress at any point of a beam under a transverse load is proportional to its distance from the neutral plane so long as the elastic limit is not ex- ceeded. At this limit materials which have no ductility give way. In materials possessing ductility, the stress may go on iiicieasing for some distance beyond the elastic limit witliout prixl.'cing rupture, but the stress is no longer proportional to C iistance from the neutral plane, its variation being much slower. This is due to the fact that the portion in compres- sion acquires increased rigidity and so exerts a continually increasing resistance (Chap. Ill) almost if not quite up to the point of rupture, while in the stretched portion a flow of metal occurs and an appro.ximately constant resistance to the stress is developed. Thus, there will be a more or less perfect equalization of stress throughout the section, accompanied by an increase of the elastic limit and of the apparent strength, the increase depending both upon the form of section and the ductility. For example, if the tensile elastic limit is the same as the compressive, the shaded portion of Fig. 262 gives a graphical "W Fig. 26a. Ban - ' I Fig. 263. Fig. 264. representation of the total stress in a beam of rectangular section when the straining is within the elastic limit. Beyond this limit, it may be represented as in Fig. 263, and will be it' J' ) n If I' 350 THEORY OF STRUCTURES. intermediate between Fig. 262 and the shaded rectangle of Fig. 264 which corresponds to a state of perfect equaliza- tion. 3. Surface Loading.* — It may be well to draw attention to another important assumption upon which is based the matlie- matical treatment of the problem of Beam Flexure. It has been assumed that the external forces acting on a beam can be so applied that they may be considered as dis- tributed uniformly over the whole section. Thus when a beam encastr^ is loaded at the free end, Fig. 265, the load P is as- !b Fig. 2*5. sumed to be uniformly distributed over the section ab^ i.e., each element in the section is supposed to experience the same amount of strain due to the load, and the reaction of the wall is also supposed to be uniformly distributed over each element in the section cd. It is clear that such suppositions must be far from the truth. In practice, the load P must be hung by some means from the beam, say by a stirrup passing over the top. The whole load is then concentrated at the line of contact of the stirrup with the beam, and it is obviously untrue to say that every * This aiiicle was kindly written by Professor Carus-Wilson and is an abstract ol a Paper presented by hiin to the Ptiysiiai Society, SURFACE LOADIXG. 351 element in the section ab is equally strained. But more than this. It has been assumed that, taking the effect of the load as distributed uniformly over the section ab, and a certain deflection thereby produced, the effect of P on each element of the section ab may be disregarded in com- parison with the strains involved in the deflection whicla P produces. It will probably be difificult at first to grasp the fact that certain measurable effects have been actually neglected, but that this is so may be seen by supposing the beam in question to be a pine beam, and the stirrup of iron. Experience proves that with a very moderate load the beam will be indented at a. But the theory shows that the longitudinal tension at a is zero and increases to a maximum at d. Thus, so far from the squeezing effect of the load being distributed uniformly over the section ab, it is concentrated at rt, and hence it is impossible to neglect it- Engineers have always recognized the existence of this "surface-loading" effect in practice, and where possible, have provided a good " bearing" in order to a^^oid such local strains : but this cannot always be done — as, for instance, in the case of rollers under bridge ends. The theory of flex- ure is therefore manifestly incomplete if it cannot take into account the actual manner in which the loads are and must be applied. ■( t Fig. 266. It can be shown that the effect of placing a pressure of p tons per inch run, say in the form of a loaded roller, on a beam resting upon a flat surface, as in Fig. 266, to prevent it from ! I I y ! i\ 352 THEORY OF STRUCTURES. bending, is to compress every element say along ab with an intensity given approximately by the equation r_2p[\ _X\ where /is the pressure at a distance x from a^ the point of con- tact, and h = ab. This is the equation to a curve be which is approximately an hyperbola. When a beam is bent by the application of external forces, a very close approximation to the true condition may be obtained by superposing this surface-loading effect on that found for bending. Take the case of a beam supported at the ends and loaded at the centre, and let it be required to find the condition along ab, Fig. 267. Fig. 267. The effect of the bending is to produce compression above and tension below the point f, and these effects may be repre- sented by a right line dc passing through c. The surface-loading effect may be represented by an hyper- bola giving the compression at any point along ab due to tlic load. The hyperbola and straight line will intersect in two S UKFA CE 1. 0.1 DING. 3ii poin :s h and/, which shows that at two points h' ,f' along ^'(5 the vertical squeeze produced by the load is of equal intensity to the horizontal squeeze produced by the bending; hence aii cl -inent at each of these points is subject to cubical compres- Mon only. From a to/' the beam is squeezed vertically, from / to li' it is squeezed horizontally, and from/;' to b it is stretched horizontally. The intensities are given at every point by the difference between the ordinates of the line of bending dc and the curve of loading. It will appear that one effect of surface- loading is to make the neutral axis rise up under the load and pass through the point h\ for there is neither compression nor tension at that point. This can be verified by examining the condition of a bent glass beam by polarized light. The neutral axis is pushed up under the load and there is a black ring passing through the point /'. If the span i diminished and the load kept constant, it is clear that ae will become less, while the curve of loading remains the same, until the line dec ceases to cut the curve ; every clement along ah will then be subjected to horizontal stretch, and the stretch is greatest at a ; the result obtained by neglect- ing the surface loading is that only elements from c to b are stretched, the greatest stretch being at b. The position of the " neutral points " is given by the equation h 4 i6 m where y is the distance from the top edge, h equals the depth ah. 111 = —. 4, and a = one-half of the span. For all elements in ab to be stretched, the ratio of span to depth, viz., ^^, must be equal to or less than 4.25. In other words, for any beam, and any load, if the span is less than 4^ times the depth, every element in the normal under the load is ^.tretched horizontally. *••*, 354 THEORY OF STRUCTURES. 4. Beam acted upon by a Bending Moment in a Plane which is not a Principal Plane. Let XOX, YO V be the principal axes of the plane section of the beam. Fig. 368. Let the axis MOM of the bending moment M make an angle a with OX. M may be resolved into two components, viz., M cos a z=i X and M sin « =r F. These components may be dealt with separately and the results superposed. Thus, the total stress,/, at any point (x,y) = stress due to .^ -|- stress due to F = - - -f- -7- = /, I., , ly being the moments of inertia with respect to the axes XOX, YO Y, respectively. If the point {xy) is on the neutral axis, then Xy . Yx ■'■X -"jl or tan/8 = -y _ fi being the angle between the neutral axis and XOX. ifli^^ SPRINGS. 355 Also see Art. 6, Chap. VIII. In this article ^ is the angle between the neutral axis and the axis of the couple, i.e.. d- (i — a. 5. Springs. — {a) Flat Springs. — If two forces, each equal to P but acting in opposite directions in the same straight line, are applied to the ends of a straight uniform strip of flat steel spring, the spring will assume one of the forms shown below, known as the clastic ctirvc. This curve is also the form of the linear arch best suited to withstand a fluid pressure, Chap. XIII. Consider a point B of the spring distant y from the line of action of P. Then FlI Py — bending moment at ^ = -zy- , r"Tr 356 THEORY OF STRUCTURES. Ill' , i) S '.■: : R being the radius of curvature at B, and / the moment of inertia of the section. If E and /are both constant, Rf a constant is the equation to the elastic curve. {d) Spiral Springs (as, e.g., in a watch). — Let the figure rep- resent a spiral spring fixed at C and to an arbor at A, ami subjected at every point of its length to a bending. action only. Q ] I Consider the equilibrium of any 4^^J portion AB of the spring. A' ^^-Q The forces at A are equivalent to a B ^p couple of moment M, and to a force P ^'°- '7^- acting in some direction AD. This couple and force must balance the elastic moment at.^. .-. M-{- Py = £1 X change of curvature at B, y being the distance of B from the line of action of P, or M- R^ being the radius of curvature at B before winding, and R that after winding. Let ds be an elementary length of the spring at B. Then, for the whole spring, 2(/./+ Py)ds = £/:s(f - ~) = E!2{dd - de;), 6r M2ds -{- P2yds —E/X total change of curvature between A and C ; , .'.Ms-YPsy^EI{B- 6/„), SPJilJVGS. 357 s being the length of the spring, _y the distance of its C. of G. from AD, B the angle through vvhicii the spring is wound up, and B^ the " unwinding" due to the fixture at C. With a large number of coils the distance between the C. of G. and A may i)e assumed to be nil and then y = o. Also, if the spring is so secured that there is no change of direction relatively to the barrel, 6', — o, and Ms = Eld. Let the winding-up be effected by a couple of moment Qq = M, Q being a tangential force at the circumference of a circle of radius q. The distance through which Q moves (or deflection of Q) = ge = %s, since M = -L ^ ch c /being the skin stress, and c the distance of the neutral axis of the spring from the skin. Thus, if b is the width of a spring of circular or rectangular section, c =■ — . and hence ' 2 "zqf the deflection = -rr-S' bE The work done = -Q x deflection = g6 = 2 2 q^ 2 2 Ec' ~ 2 Ec' ~ 2E C 5" > /' being the square of the radius of gyration, A the sectional area of the spring, and V its volume. k' I In case of spring of rectangular section -p = - . " circular ^ c' I 4 .V^a' IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 ^m m ,.. 1^ |||22 1 36 '""=== i e iiiio 1.8 U IIIIII.6 i^. V] <5> ^ //. ^•y% 'el. ? '^i Photographic Sdences Corporation 33 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 V U.x V. \ IH i;ii 'ikm IT t! !■ at ii 358 THEORY OF STRUCTURES. Again, the spiral spring in Fig. 277 is wholly subjected tf> a bending action by means of a twisting couple of moment M z=i Qq'xn a plane perpendicular to the axis of the spring. Any torsion in the spring itself is now due to the coils not being perfectly flat. (( t< "• Fig. J77. Let /?, = radius of a coil before the couple is applied. ^ being che angle of twist ; or . • , . Qqs Ms s s ... .: . -^ = ^ = ^- - Z = (^- ^•)2'^' JV being the number of coils before the couple is applied, and X << << << << after " " " " The distance through which Q acts, i.e., the " deflection,'* N, " " =^''=^ and the work done "2 2E c"' I rv — >- —p~ for spring of rectangular section, I rv - 2, E " circular 6. Beams of Uniform Strength. — A beam having the same maximum unit stress (/) at every section is said to be a beam of uniform strength. s BEAMS OF UNIFORM STRENGTH. 359 At any section of a beam An{=:l) denote the bending moment by M, the depth of the beam by y, and its breadth by b. Then ll= M= ^Ak\ c c c being the distance of the skin from the neutral axis, and A the area of the section. Evidently c and k are each proportional toj/, and A to by. or nfbf = M, J".il' « being a coefficient whose value depends upon the form of section. Four cases will be considered. Case a. Assume that the breadth b is constant, and let nfb = -. Then or y — ± s/JM. Thus AB may be either the lower edge of the beam, the ordinates of the upper edge being the different values of y, or it may be a line of symmetry with respect to tlic profile, in y which case the ordinates are the different values of ± -. Example i. A cantilever AB loaded at the free end zvith a wcig/it \V,. At a distance x from A, f = pM = p W\.x, Theoretically, therefore, the beam, in elevation, is the area "^ ACD. the curve CAD being a Pic. 178. '.' I <. ! \'M ' I'li ^liSt'i; u i . I i i:i t I! hi. 360 THEORY OF STRUCTURES. parabola with its vertex at A and having a parameter The max. depth = 2CB = CD= VpWJ. The form of this beam is very similar to tliat adopted for cranks and for the cast-iron beams of engines. In the latter, the material is usually concentrated in the flanges, a rib being reserved along the neutral axis for purposes of connection. Again, geometrical conditions of transmission require the teeth of wheels to be of approximately uniform strength. A cantilever of approximately uniform strength may be ob- tained by taking the tangents CE, DF as the upper and lower edges of the beam instead of the curves CA, DA. The depth of the beam at A is then EF = \CD = ^ S'pW). Although, theoretically, the depth at A is nil, practicall}' the beam nui.st have sufficient sectional area at A to bear the shear due to Jf, , and the depth ^ V/JF,/ will be found ample for this purpose. Note. — The dotted lines show the beams of uniform strength, when the lower edge is the horizontal line AB. Ex. 2. A cantilever AB carrying a uniformly distributed load W,. At a distance x from A, /=pM=^x' 2l or y = -/-^-. Fig, 279. The beam, in elevation, is there fore the area ACD, AC, AD being two straight lines, and the maximum depth being • CD = 2BC=l^''-^^ = sJ The sectional area at A is nil, as both the bending moment and shear at that point are zero. BEAMS OF UXIFOKM STKENGTH. 361 Xotc. — The (Jotted lines show the cantilever of uniform strength when AB is the lower edge. Ex. 3. A canlilcvcr AB carrying a zvcight U\ at the free end A and also a uniformly distributed load l\\. IG. a8o. At the distance x from A, f = pM .= p[w,x + V/f^). Thjs equation may be written in the form / w \ y pjvii 2]]\ — I. Theoretically, therefore, the beam, in elevation, is the area ACD, the curve CAD being an hyperbola having its centre at / IV \ H i where AH — W^Ji and semi-axes equal to ■~l and The maximum depth CD = \/ p[wj -\- wA = 2BC, « ' 362 THEORY OF STRUCTURES. !■ H ; A cantilever of approximately uniform strength may be ob- tained by taking the tangents CE, DF as the upper and low er edges of the beam instead of the curves CA^ DA. It may be W easily shown that the depth of this beam at A is ^ ' rr^^r), and this will give sufficient sectional area at A to bear tlic shear due to W^. . « Note. — The dotted lines show the cantilever of uniform strength, when the lower edge is the line ^^. , . Ex. 4. A beam AB supported at A and B, and carrying a loq^d W^ at the middle point 0. At a distance x from 0, f = pM = p W, (^4 2 \2 Theoretically, therefore, tlie beam, in elevation, is the area ACBD, the curves CAD, CBD being two equal parabolas. having their vertices at A and B, respectively, and having parameters equal to \p \\\ . The maximum depth = CD — 2CO = ^ V])WJ. A beam of approximately uniform strength may be ob- tained by taking the tangents CE, CG as the upper edges instead of the curves CA, CB, and the tangents DF, DH as the lower edges instead of the curves DA, DB. The depth of the beam at A and B is now EF = GH = — , 2 and this depth will give a sectional area at the ends of the beam sufficient to bear the shears at these point, viz., — - . Note. — The dotted lines show the beam of uniform strength when the line AB is the lower edge. Ex. 5. A beam AB supported at A and B, and carrying a uniformly distributed load W^. i !. BEAMS OF UN /FORM STRENGTH. 363 At a distance x from the middle point O, This equation may be writ- ten in the form 4 pwi 8 = I. Theoretically, therefore, the beam, in elevation, is an ellipse ACBD, having its centre at O and axes AB = l and CD hw^i The maximum deptii is of course the axis CD ■= 2CO. Practically, the beam must have a certain depth at A and B in order to bear the shears due to the reactions at these W points, viz., — ^. If the horizontal tangents at Cand at D are substituted for the curves, the volume of the new beam is to the volume of the elliptic beam in the ratio of 4 to rr. Note. — The dotted line shows the beam of uniform strength when its lower edge is the line AB. Ex. 6. A beam AB supported at A ayid B, and carrying a load JV, at the middle point O and also a uniformly distributed load ir,. At a distance x from O, / =pM = p\--\l-x) + -^'-[i - V) f- This equation may be written in the form -f + 2 IK -**^- + i j|! ,14 H\ IV, I- iv: + w-.') g^;(4 w, IK + IK' + rf 7) = I. ' I-: ' 'i'i :iJi 364 THEORY OF STRUCTURES. Theoretically, therefore, the beam, in elevation, is the area ACBD, the curves CAD and CBD being the arcs of ellipses having the centres at the points K and L, respectively, where 0K= 0L = 2w: The maxinnum depth CD =20C = P^ \ — ^~ -\- WJ , WJ 8 }*■ A beam of approximately uniform strength maybe obtained by taking as the upper edge the tangents to the curves at C. and as the lower edge the tangents to the curves at D. It may be easily shown that the depth at the cxxdsA and /)' \y _|_ w is now ^^— 117— i — T>7» and this depth will make allowance for W -\- W the shear — ^— ' ? at these points. Note, — The dotted lines show the beam of uniform strength when the lower edge is the line AB. Case b. Assume that the ratio of the breadth (^) to the depth (j) is constant, i.e., that transverse sections are similar. y on b f^','j ( Iffin ('■ '■''■ ■ HR ■ -l^l [I \ ' ' -i ' i < 1 ' ■ (■■ i.t 15 1 . ^■'■'1' I 'h»| ' '' iB V r 61 »'' r T i f I 1 . H ■ ' ! ' t .■ •I v' % ■ T 366 THEORY OF STRUCTURES. I. Girders with Horizontal Flanges. — In these the flan^'es can only convey horizontal stresses, and the shearing force, which is vertical, must be wholly transmitted to the flanges through the medium of the web. If the web is open, or lattice-work, the flange stresses are transmitted throuj^h the lattices. If the web is continuous, the distribution of stress, arising from the transmission of the shearing force, is indeterminate, and may lie in certain curves; but the stress at every point is resolvable into vertical and horizontal components. Thus, the portion of the web adjoining the flanges bears a part of the horizontal stresses, and aids the flanges to an extent depend- ent upon its thickness. With a thin web this aid is so trifling in amount that it may be disregarded without serious error. II. Girders with one or both Flanges Curved. — In these the shearing stress is borne in part by the flanges, so that the web has less duty to perform and requires a proportionately less sectional area. Equilibrium of Flanged Girders. — AB is a girder in equi- librium under the action of external forces, and has its upper flange com- pressed and its lower flange cx- ten.ded. Suppose the girder to be divided into two segments by an iN 'L. jL. B Fig. 291. imaginary vertical plane MN. Consider the segment AM^'. It is kept in equilibrium by the external forces on the left of jSIN, by the compressive flange stress at iV ( = C), by the tensile flange stress at J/ ( = T), and by the vertical and horizontal web stresses along MN. The horizontal web stresses may be neglected if the web is thin, while the vertical web stresses pass through M and jV, and consequently have no moments about these points. Let d be the effective depth of the girder, i.e., the distance between the points of application of the resultant flange stresses in the plane MN. ,., ^ ,_ Take moments about J/ and //successively. Then Cd = the algebraic sum of the moments about M of ssscs are FLANGED GIRDERS, ETC. 367 the external forces upon AMN —\}ci^ bending moment at Mis' = M. So, Td = M; .'. Cd=M^ Td, and C = T. Hence, the flange stresses at anv vertical section of a girder are equal in magnitude but opposite in kind. The flange stress, whether compressive or tensile, will be denoted by F. Example. — A flanged girder, of which the effective depth is 10 ft., rests upon two supports 80 ft. apart, and carries a uni- formly distributed load of 25CK) lbs. per lineal foot. Determine the flange stress at 10 ft. from the end, and find the area of the flange at this point, so that the unit stress in the metal may not exceed 10,000 lbs. per square inch. The vertical reaction at each support 80 X 2500 {1 wKk 111 ■i. 1 i '^ ,!' i 1 1 H ,1' )'• = 100,000 lbs. .-. F.\o = M= looooo X 10 — 2500 X 10 X 5 = 875,000 ft.-lbs. .•.F= 87,500 lbs. !:i in. ^, . , 87500 The required area = = 8.75 sq. 1 ^ lOOOO ^ Cor. I. Fd=M=^I=-^^I. K y Cor. 2. At any vertical section of a girder, let '^. + 2/', + /0 + A(/^ + /0 yjyj^ - 2A GG,-- -A,{A,+/i,) + A,{/t, + /i,) 2A nn - A,{/i, + /i,) + AX/i,-\-2/i, + /i,) 2A //, A,{h,+7ji,\-h:)+A,{K+ h:) , h //, /., ^,= 6^6^3+7 = ^ -^-i-~2-~2 _h (/^,+/,.)(^,+yj,4-^.)-^ , (//.4-2//,+//.)-/JA+/0 2 2A h __ AlK + /Q - ^.(/''. + /Q - ^IK - ^) ' ~ 2 2A So, y, = GG, + ^- = etc Again, /, with respect to G, = :fl2!Lj^A,.G,G'-^^^^^ + A,.G,G'-^A,^+A,.G,G' 12 ' ' ' ' 12 ' ' ' 12 ' AA' + AA' + AJi,* ' /, being equal to Hence, A / = /. + ^ i ^,(/^ + ^) + ^,(/^ H- 2/^ + /^) }' + ^]^ A + 2//, + A,) + ^,(>^, + /l,)\\ EXAMPLES OF MOMENTS OF INERTIA. m =^-+:M-i = /.+ 4^' = /,4- = /, 4^' + A,A,\h, + A.)' + A,A:{h, + 2//. + V 4- (^3^^' + ^sM.XA + 2//, + /O' L+ 2^,^,^3(7/, 4- 2/r, + f^.W^, + Z-', + /^, + /O /^,.>i3(//. + //,)M + A,Alh, + /^M ] -A,A,Alh,-^2K + h,r + (/J./i,' + ^3M,)(//. + 2/., + /0' + 2^A^s(/';. + 2//, + /0' Hence, finally, AX^AX + AA I- 12 4- ^\A,Alh, 4- //;/ + A^Alh, 4- h,r 4- ^,^3(/.. 4- 2//, 4- h,y\. Cor. I. If //, and /*, are small compared with //,, put /:,- = h' Then y = AM'+A^{/.' + '^) ,^ ,, 2A 2A, + A , -^ , nearly, m'-^m ;ii: i( i 'i 'I 1 I < Vr. 37« and / = THEORY OF STRUCTURES. /f/// + A(//'-'^^^')' + /M.' 12 ( 12 "^ ~4// ) Note. — If A^ is also very small, as in the case of an open web, then A A A j'^ = /{'!- and /= /<'"—-—-', approximately. Cor. 2. Let )',,.. I'a be the distances of G from the uppei aiul lower edges, respectively ; let /], , //, be the corresponding maximum working; unit stresses. From the preceding corollary, /* = j, = — '~T~~'' or ^. + A + ^. y.^-yk 2.-i, + W. 2^» Jb ^j/b Jb ^Jb Hence, /. ^. + tV = ^, EXAMl'LES Oh MOMENTS OF INENTIA. %n I 12 "•" 4A , . } ' m ■ _j^,i 4A ,A , + a: f- ^,A, 4- \2A,A, \ ' \ 12A ) 12 12 yl ^•'(^/:") + '^■^'(" TT^I + 7;(^^.^.+ 12/^.') Lij:ltt2A 4- yi 1 J Fifth. T-section. Let the area of the flange be /J, , and its depth //,. Let the area of the web be A^ , and its depth /;,. Let A,'^- A, = A, and h,-\-h^ = h. Let G be the centre of gravity of the sec- tion, G^ of the flange, and C, of the web. G 8, Flo. 295. Let J, be the distance of G from foot of the web. Then (/t.H-w,)(//.+ //,) ^ ax- A A U ! u- und TJ/HOA'V OF S/A'UCrUAH^i. •^' 2 "^ 2(//,-f /i.) 2 "^ 2^ ^.(;=-' 4- //.->'. = ^3^ and ^,^=;..--'=_L-. Hence /, with respect to a liorizontal line through G^ whicli reduces to Ajt: + Ajt: A.Aji^ Cor. I. If /l^ is very small as compared with //, , put then }\{A,+A:) = AJi' + /J,(^^'- - y = [a, -f ^-') A', nearly, or 2K A )■ and AA' + A,{/,'-^^)' A,A,[/i' + ^^J 12 44 or \I2 4^ / Cor. 2. Let j'a be the distance of the compressed, or upper, side from the neutral axis. TO DUSIGN A UIKDKK VF UNll-ONM STRENGTH. 38 1 Let J'/, be tlic distance of the stretched, or lower, side from tlic neutral axis. Lct/a be the crushing unit stress, /» the tensile unit stress. I'rom the preceding, y^ — ' "^ ' J — -' ; but h' =7,. -\-y^ ; 2 A 2 /1. 4-^', J., =='------—, J, and A,=:A/~'^-'^=:A/' ^' 2jft • 2// IIciicc, / becomes _ /r 2/. -/, /. +/,_y^ + y, _ //' _ /I - ^ anu — - — . AWr. — Although the preceding approximate methods arc often useful, they can only be regarded as tentative and shoukl always be checked by an accurate determination of the moment of inertia and of the position of the neutral axis. 8. To design a Girder of Uniform Strength, of an I-section with equal Flange Areas, to carry a Given Load. Let J be the depth of the girder at a distance x from its middle point. Let A be the sectional area of each flange at a distance x from its middle point. Let A' be the sectional area of the web at a distance x from its middle point. Lot M be the bending moment at a distance x from its mid tile point. Let 5 be the shearing force at a distance x from its middle point. Then . ,, • f(A+'~)y = M, /being the safe unit stress in tension or compression. :i.; 382 THEORY OF STRUCTUKES. Web. — Assume tli.it the web transmits the xv/toic of tlic shearing force. This is not strictly correct if the flanj;;^ is curved, as tlie flange then bears a portion of the shearing force. The error, however, is on tlie safe side. Theoretically, the web should contain no more material than is absolutely necessary. Let/, be the safe unit stress in shear. Then /J' = I: /.' and the sectional area is, therefore, independent of the depth. The tJiickncss of the web = A' 5 7./ but this is often too small to be of any practical use. Experience indicates that the minimum thickness of a plate which has to stand ordinary wear and tear is about \ or -^^ in., while if subjected to saline influence its thickness should be I or ^ in. Thus, the weight of the web rapidly increases with the depth, and the greatest economy will be realized for a cer- tain definite ratio of the depth to the span. The thickness of the web in a cast-iron girder usually varies from i to 2 in. In the case of riveted girders with plate webs of medium size, all practical requirements are effectively met by specifyiiiif that the shearing stress is not to exceed one-half o{ the flaiii,fe tensile stress, and that stiffeners are to be introduced at inter- vals not exceeding txvice the depth of the girder when tlie thickness of the web is less than one-eightieth of the depth. Again, it is a common practical rule to stiffen the web of a plate girder at intervals approximately equal to the depth of the girder, whenever the shearing stress in pounds per square inch exceeds 12000-^ (i -\ ), H being the ratio of the depth of the web to its thickness. • , ' ■ Flanges. — First. Assume that the flanges have the same sectional area from end to end of girder. TO DESIGN A GIKDEK OF UN 1 1- ON M STRENGTH. 383 If the effect of the vvcb is nef^lected, _ M ^ ~ /A' and the depth of the beam at any point is proportional to the ordinate of the bendin^-moinent curve at the same point. l""or example, let the U)ad be uniformly distributed and of intensity w\ and let / l)e the span. Then M = 2(4-4 o Kn.. ay6. and the beam in elevation is the parabola ACH, havinj.' its vertex at C and a central depth CO = i. i"> The depths thus determined are a little greater than the depths more correctly given by the equation • M /(''+^:-) Second. Assume that the depth j of the girder is constant. Then and, neglecting the effect of the web, the area of the flange at any point is proportional to the ordinate of the curve of bend- ini,^ moments at the same point. Let the load be uniformly distributed and of intensity w, also, let the flange be of the same uniform width ^ throughout. gjLc C ff =4. A 1 2 3 S S 1 B Fiii. j'iT. The flange, in elevation, is then the parabola ACB, having its vertex at C and its central thickness CO = xtt.- Such 8/yo w:% II -.LiM 384 THEORY OF STRUCTURES. beams are usually of wrought-iron or steel, and are built up hy means of plates. It is impracticable to cut these plates in sucii a manner as to make the curved boundary of the flange a true parabola (or any other curve). Hence, the flange is generally constructed as follows: Draw the curve of bending moments to any given scale. By altering the scale, the ordinates of the same curve will represent the flange thicknesses. Divide the span into seg- ments of suitable lengths. From A to i and B to 7 the thickness of the flange is \a = 7/; from 1 to 2 and 7 to 6 the thickness is 2b =^ 6e; from 2 to 3 and 6 to 5 the thickness is y = ^d\ and from 3 to 5 the thickness is CO. M A'\ . : -2' I IS somewhat less than that now determined, but the error is on the safe side. Again, at any section, The more correct value of A E R 2f and hence R oc y, the depth. Thus the curvature diminishes as the depth increases, so that a girder with horizontal flanges is superior in point of stiffness to one of the parabolic form. The amount of mct.il in the web of the latter is much less than in that of the former. If great flexibili!:y is requircil, as in certain dyna- mometers, the parabolic form is of course the best. 9. Deflection of Girders.--The principles of economy and strength require a girder to be designed in such a manner that every part of it is proportioned to the greatest stress to which it rnay be subjected. When such a girder is acted upon by external forces, it is uniformly strained throughout, and in benuing, the neutral axis must necessarily assume the form nt an arc of a circle, provided the limit of elasticity is not ex- ceeded. It might be supposed that the curve of deflection is dependent upon the character of the web, and this is doubtless the case, but experiments indicate that so long as the flani;e iM , liM i I uilt up by es in svicii :loe a true generally iven scale. curve will into seg- e flange is = 6^ ; from I 3 to 5 the somewhat 3n the sate ncreases, so in point of t of mct.il hat of the tain dyna- n onomy ^mJ anner th;it ss to which d upon by lut, and ill the form "t is not ex- eflection is s doubtle -.s the flange DEFLECTION OF GIRDERS. 38; unit stresses are unaltered in amount, the influence of the web ma}' be disregarded witliout sensible error. Let y" be the unit stress in the beam at a distance y from the neutral axis; let d be the depth of the beam. Then / M E - = y = ^ = a constant, y I K assuming that the neutral axis is an arc of a circle of radius R. But y a d, and I = Ak' o^ Ad\ Hence/a jv a d; and if the depth is constant, / is also con- stant and the beam is of uniform strength. If the area A is constant, d a ^/ j7. Example i. A cantilever bent under the action of exter- nal forces, so that its neutral axis AB assumes the form of an arc of a circle having its centre ^E^^^!r\~f at 0. pl""*^ Draw the verticals OA, BF, and ihe horizon- j ' ^ tals BR, FA. \ The vertical deviation of B from the hori- | zontal, viz., BF, is the maximum deflection. Denote it by D. Let radius of circle = R. '"^' "^^^ Since the deflection is very small, BE is approximately equal to AB ( = /), the length of the cantilever. .-. /-■ = BE' = AE{2R -AE)= iRD ~ IT = 2RD, a.s 17 may be disregarded without much error. Also, the deflection at any point distant x from A is evi- "J^'itly :^. If /is the stress in the material at a distance y from the neutral axis, / E 2DE ^ 7DEy v = R = --7^ °'' ^=-^- v ii <" ii . .1! ,' . ' ! '. • -■■ % *■ i 1 ■ i i ■ ''^il ■m 1|: 386 THEORY OF STRUCTURES. if M X\ Ex. 2. A girder resting upon two supports at A and B is bent under the action of external forces so that its neutral axis ACB assumes the form of an arc of a circle having its centre at O. Draw the vertical OC, meeting the vB horizontal AB in F. CF is the maximum deflection ; denote it by D. Since D is very small, its square may be disregarded and the horizontal AB may be supposed equal to the length ACB { = /) of the girder, without much error. Then ^' AF' = FC{2R - FC) = 2RD - D' = 2RD. Fig. 299. 4 Hence, A, • / ^ Also, smce — = ^, y K D = SR' / = %DEy jj The deflection at a distance x from F =. D ^. 2R Ex. 3. A timber beam of 20 ft. span, is 12 in. deep and 6 in. wide : what uniformly distributed load ( W) will deflect the beam i in., E being 1,200,000 lbs.? By Ex. 2, • Also, . I< = 7200 m. W.20 E^ S--''-R^- 1200000 d(^' 1200000 6. 12' 7200 12 7200 12 .•. W = 4800 lbs. nwrfmrr 1 DEFLECTION OF GIRDERS. 387 and B jxternal is Acn a circle :ting the ;flection ; ts square supposed 3ut much ID. Ex. 4. Let s^, f^, d^, and s^, f^, d^, respectively, be the length, unit stress, and distance fronti the neutral axis of the stretched and compressed outside fibres in Examples (i) and (2). Let d^-\- d^ = d = the total depth of the girder. Hence, from similar figures, s, R + d, ^ s, R-d, 1 = -R- ^''^ 1 = -R- Also, . -y, — s ^ _ d, -|- d^ d " I ~ R -R' /.. ?,-/ d, /. V - s, d^ E~ I ~R ^""^ E" I ~'R' E ~ I ~ R ~ R' ''■>nHI '■ i 'j ■ i ■J i leep and 6 deflect the 12 Ex. 5. A truss of span 120 ft. and 15 ft. deep is. strained so that the flange tensile and compressive unit stresses are 10,000 and 8000 lbs., respectively. Find the deflection, and difference of length between the extreme fibres. lOOOO+Sooo _ .y, — .f, _ 15 30000000 120 .*. .y, — J, = .864 in., and R = 25,000 ft. Hence also D = (120)' 8 X 25000 = .864 in. ID. Camber. — Owing to the play at the joints, a bridge- truss, when first erected, will deflect to a much greater extent ':! 388 THEORY OF STRUCTURES, than is indicated by theory, and the material of the truss will receive a permanent set, which, however, will not prove detri- mental to the stability of the structure, unless it is increased by subsequent loads. If the chords were made straight, they would curve down- wards, and, although it does not necessarily follow that tlic strength of the truss would be sensibly impaired, the appear- ance would not be pleasing. In practice it is usual to specify that the truss is to have such a camber, or upward convexity, that under ordinary loads the grade line will be true and straight. The camber may be given to the truss by lengthening the upper or shortening the lower chord, and the difference of length should be equally divided amongst all the panels. The lengths of the web members in a cambered truss are not the same as if the chords were horizontal, and must be carefully calculated, otherwise the several parts will not fit accurately together. To find an approximate value for the camber, etc. : Let d be the depth of the truss. . , Let J, , J, be the lengths of the upper and lower chords, respectively. Let /", , f be the unit stresses in the upper and lower chords, respectively. Let d^ , d, be the distances of the neutral axis from the upper and lower chords, respectively. Let R be the radius of curvature of the neutral axis. Let / be the span of the truss. 6 R -I E = ^J and -'- ^ / A R — — -. — = -p , approximately, the chords being assumed to be circular arcs. Hence, the excess in length, of the upper over the lower chord, ij Si- S, = -gr(/, f /,) = /^- = J/. STIFFNESS. 389 Let x^ , X, be the cambers of the upper and lower chords, respectively. R -\- d^ and R — d^ are the radii of the upper and lower chords, respectively. By similar figures, the horizontal distance between the ends of the upper chord = — - — 'l, and the horizontal distance be- tween the ends of the lower chord Hence, A-^/„ R ^ J —^~n = ^i ■ 2{R + ^1). approximately, and /i R-d,X ,„ ^ . , I -^ — /I = x^ . 2(/c — 0' B' Fig. 300. tions AB, A'B', transverse to the horizontal neutral axis 00'. Let the abscissae of these sections with respect to an origin ! I f 392 THEORY OF STRUCTURES. V: n J ; in tlic neutral axis be x and x -\-dx, so that the thickness of the shce is dx. In the Hmit, since dx is indefinitely small, corresponding linear dimensions in the two sections are the same. Let / be the moment of inertia of the section AB (or A' B' in the Hmit) with respect to the neutral axis. Let c be the distance of A (or A' in the limit) from the neutral axis. Let/, , /, be the unit stresses at A and A', respectively. Consider the portion ACC A' of the slice, CC being parallel to and at a distance Y from the neutral axis. Since it is in equilibrium, the algebraic sum of the horizontal forces acting upon it must be nil. These forces are: The total horizontal force upon ACC, " A'C'C\^x\^ " " " shear along the surface CC . The horizontal force upon an element PQ of thickness dy and at a distance y from the neutral axis ^f.^^dy, z being the width PQ,. Thus the total horizontal force upon ACC Ifel = :^[^^y::dy) = ^^-^{yzdy) =~ J]'^dy = ^Ay, A being the area of ACC, and,;/ the distance of the centre of gravity of this area from 00'. Similarly, the total horizontal force upon A'C'C = -cj^y'^y = fiA- -L-.A y- '/mi . r DISTRIBUTION OF SHEARING STRESS. 393 kncss of iponding (or A'E from the lively. C being is. Since \tal forces ickness dy [force upon Hence, (— ' — ~\Ay = difference of the horizontal forces ^'^ '' upon A CCiiud A' C'C, = horizontal shear along CC, = qwdx ; <] being the intensity of this shear, and tv the width of the section at CC. Let il/ and M — r/J/ be the bending moments at the two consecutive sections AB, A'B'. Then M=--I and M-dM=^--I, c c and therefore dM = f^' - ^)l. Hence, dM .- , — - Ay = qwdx, or since dMAy S .- ^^^ ^ 'dx T ^ 1^'^ -—- = shearing force at the section AB = S. dx " Example i. So/id rectangular section of zuidth b. -Ay, 12 e centre of or . = liy - y). B Fig. 302. ./ / and the intensity of the shear at any point of AB may be rep- resented by the horizontal distance of the point from the 3^" parabola A VB, having its vertex at V, where OV = 4 be ii nliU a !l »! rr r i 1 394 THEORY OF STRUCTURES. The inaximum intensity of shear is at and its value is _ 3 5 The value of the average intensity is 5 Qav = b . 2C • • Qmax. • Yati. • • 3 • ^* Ex. 2. /I hollow rectangular section ; B and 2c being the external and B' and 2c' the internal xvidth and depth. At the neutriil axis, q{B-B') = j I ^(.'-o+^^' ic' - n I . Thus, as in Ex. I, the intensity of shear is again greatest al the neutral plane, i.e., when Y = o. Ex. 3. Solid circular section of radius c. Ay =^2^ 1V -/dy = K^' - Vi, w = 2(c' ~ V')\ and / = inc* nc' and the intensity of the shear at any point of AB may be rep- resented by the horizontal distance of the point from the pa- AS rabola AVB, where OV = 37tc' Also, g„,a^, = -^—^ and ^^,. = — •,. ^TtC 7tC .--qa 'max. • fav 4: 3. DISTRIBUTION OF SHEARING STRESS. 395 Ex. 4. A double-flanged section, each of the flanges con- sisting oi five 8-in. X i-in. plates riveted to a 24-in. X i-in. web by tii'o 3-in. X 3-in. X i-in. angles. To find the intensity of shear at the surface of contact between the angles and the flange : ^^ = 20 X I3i = 265 ; w = 6i in. ; / = 8975J, neglecting the effect of the rivet-holes in the tension flange. Hence ^f! q = S 2120 466739 Let S = 49 tons. Then q = .2226 ton per square inch. Let the rivets have a pitch of 4 in., then 6i the total shear on each rivet = -^ x 4 X .2226 = 2.8938 tons. Let the coefificient of shearing strength be 4 tons per square inch, and suppose that the surfaces of the angle-irons and of the flange are close together ; then c • ^ 2.8938 area of rivet = — ^^- = 4 7234 sq. in., and its diameter = .96 in. If the surfaces arc not close together, so that the rivet may be subjected to a bending action, then, by Ex. 3, the average intensity of shear in a section =1.4 = 3 tons per sq. in., and hence area , , ^ 2.8938 ^ ^ of rivet = — ^^ = .9646 sq. in. ; 3 its diameter is i.i in. 396 THEORY OF STRUCTURES. 13. Beam acted upon by Forces Oblique to its Direc tion, but lying in a Plane of Symmetry.— In discussing the equilibrium of such a beam the forces may be resolved into components parallel and perpendicular to the beam, and their respective effects superposed. Fig. 304. Let AB he the beam, P,,Pj,P^, ... the forces, and a, , a-, , a^ their respective inclinations to the neutral axis. Divide the beam into any two segments by an imaginary plane MN perpendicular to the beam, and consider the seg- ment AjVN. It is kept in equilibrium by the external forces on the left of MN and by the elastic reaction of the segment BA/N upon the segment AMN at the plane AIN. The resultant force along the beam is the algebraic sum of the components in that direction, of /*, , P^ , P, , . . . , = /*, cos a, + P, cos a^-\- , . . = ^{P cos a). It may be assumed that this force acts a'ong the neutral axis, and is uniformly distributed over tht^ sejtion A/N. Thus, if A is the area of the section, -— — ^ is the in- A tensity of stress due to this force. Again, the components of /*,,/*,, P, , ... , perpendicular to the beam, are equivalent to a single force and a couple at MN. The single force at AIN is the Shearing Force, is per- pendicular to the beam, and is the algebraic sum of P^ sin or,, /!, sin a, , . . . , = P, sin Of, -\- P, sin a, + = 2{P sin a). BEAM ACTED UPOX BY OBLIQUE FORCES. 397 This force develops a mean iixttgcutial unit stress of 1- — '- ' in MN. and deforms the bca;ii, but so sliglitly as to A be of little account. The moment of the couple is the algebraic sum of the moments with respect to MN oi P, sin «, , /-', sin or, , . . . , = /*, sin a^/>^ -\- /\ sin nr,/), -f- . . . = 2{P/> sin a), />,,/>,,... being respectively the distances of the points of application oi P, , J\ , . . . from MN. Now, ^{P/> sin a) is the resultant moment of all the external forces on the left of MN, for the resultant moment of the com- ponents along the beam is evidently ni/. Hence, ^{P/> sin a) = A/=^-^/= -n. E R and /,. = j:E{Pp?.\n a), is the unit stress in the material of the beam at a distance y from the neutral axis due to the bending action at MN oi the external forces on the segment AMN. Hence, also, the total \xmt stress in the material in the plane MN at a distance y from the neutral axis is ± ^(jP cos a) , ^ ■S'(/^co sry) y ^, ^^^ . , ., -^^ ±fy= ± -^ ± -j^{Pp sm a) =/;, the signs depending upon the kind of stress. It will be observed that this formula is composed of tivo intensities, the one due to a direct pull or thrust, the other due to a bending action. The latter is proportional to the distance of the unit area under consideration from the neutral axis. It is sometimes assumed that the same law of variation of stress holds true over the real or imaginary joints of masonry and brickwork structures, e.g., in piers, chimney-stacks, walls, arches, etc. In such cases the loci of the centres of pressure correspond to the neutral axis of a beam, and the maximum m 59« THEORY OF STRUCTURES, III; and minimum values of the intensity occur at the edges of the joint. Example i. A horizontal beam of length /, depth d, and sectional area A is supported at the ends, and carries a weight W d.X. its middle point. It is also subjected to the action of a force H acting in the direction of its length. First. Let the line of action of // coincide with the axis of the beam. The intensity of the stress in the skin at the centre But coid, and / = Ak" a Ad\ - . ^^ = A, sm a = — cos a. ' 2 , jp; Ji j i * 400 THEORY OF STRUCTURES. ■|.' The component of zvx along the beam = wx sin a. The component of w-ir perpendicular to the beam = wxco?, ix. Hence, »r,^ tv/ cos'' a , the total compression at NM = : -4- wx sm a- = t •, ^ 2 sin a ' the shearing force at MN tvl = — cos a wx COS a = .S\ ; \ zul x the bending moment at MN ■=■ x — cos a — - wx cos a = J/^; and Jy — A J ^ , 1 \ flii 5;-: Ml These expressions may be interpreted graphically as already described, C".,- , ^\ being represented by the ordinates of straigiit lines, and J/j. ,_/^ by the ordinates of parabolas. fy , for example, consists of two parts which may be treated independently. Draw 0£ and J/' perpendicular to OA, and respectively equal or proportional to wl cos" a w/ cos' a xvl . — : — ; and — i —. + — r sni a. 2 A sm a 2 A sm a A Join EF. The unit stress at any point of tlie beam due to direct com- pression is represented by the ordinate (drawn parallel to (V; or AF) from that point to EF. Upon ♦^he line GG' drawn through the middle point B per- pendicular to Ox-i, take BG = BG\ equal or proportional lo ywl* _--— - cos a. According as the stress due to the bending action J o at any point of the beam is compresswe or tensile, it is repre- sented by the ordinate (drawn parallel to OE and AF) from that point to the parabola OGA or OG'A ; G and G', respec- tively; being the vertices, and GG' a common axis. m. iili ■i nt-ii SIMILAR GIRDERS,— PRINCIPAL PROPERTIES. 401 By superposing these results, the parabolas EHF, EH F are obtained, the ordinates of which are respectively proportional to the values of fy for the compressed and stretched parts of the beam, i.e., for the parts above and below the neutral surface. 14. Similar Girders. — Two girders are said to be similar when the linear dimensions of the one bear the same constant proportion to the corresponding linear dimensions of the other. Thus, if fi, li' , rf, 6', .\, A' are corresponding breadths, depths, and lengths of two similar girders, (i 8 \ •-J7 = — , =; -r-, = a constant = ^, suppose. pox 15, To Deduce the Principal Properties of Similar Girders. — {u) The weight of a girder is proportional to the product of an area and length, i.e., to the cube of a linear dimension. Hence, the weights of similar girders vary directly as the cubes of their linear dimensions. Hence, too, the unit stresses iiuist vary directly as their linear dimensions. \b) The Breaking Weight of a girder is calculated from a ad formula of the form W — ^~r, ^ being an area, rtfa depth, and /a leiigtii. d Now J is constant for similar girders, so that W is propor- tional to a, i.e., to the square of a linear dimension. Hence, //tf Breaking Weights of similar girders vary directly as the squares of their linear dimensions. Example. — A girder resting upon two supports 80 ft. apart is 10 ft. deep and weighs 6 tons. Determine the length and depth of a similar girder weighing 48 tons. /lengthy^ /dq)thy^ 48 ^ ^ Hence, the length = 160 ft. and the depth = 20 ft. Also, the unit stresses are in the ratio of 10 to 20, and the breaking weights in the ratio of 10' to 20'. I 402 THEORY OF STRUCTURES. l; \ 16. To Discuss the Relations between the Correspond- ing Sectional Areas, Moments of Inertia, Weights, Bend- ing Moments, etc., of two Girders which have the same Sectional Form and are thus related : The forces upon the one being P^ , P^ , I\ , . . . with absciss.e x^,x^,x^, . . . those upon the other are nP^, iiP,, nP^, . . . with abscissiK /».r, , /.v, , /.r, . The spans and corresponding lengths are in the constant ratio /. Corresponding sectional breadths are in the constant ratio ^. Corresponding sectional depths are in the constant ratio r. Let A, A' be corresponding sectional areas ; /,/' ti «« moments of inertia ; Q,Q', <( l( weights ; S,S' _ Hence, b Wb IV B W W,-B, IV.- b: IV - />; Example. — Apply the preceding results to a cast-iron girder of rectangular section resting upon two supports 30 ft. apart. The girder is 12 in. deep and carries a uniformly dis- tributed load of 30,000 lbs. Take 4 as a factor of safety ; b. is given by 120000 C b^ where C Hence, 30,000 lbs., d=- \2 in., and / = 360 in. ; .-.*,= 5 in. « 5 X 12 , „ B, = ——— X 30 X 450 = 5625 lbs. ; 144 W, — B, = 30000 — 5625 = 24,375 lbs. ; m 30000.5 ^ „ , b = = 6A in. ; 24375 ^^ ^= ^^ ^ 5625 = 6923tV lbs. ; 24375 W= M^. + ^ = 36,923tV lbs. : ri ■r, Ml ':'. 1 4 aJ EXAMPLES. bar is bent tli( of ircle of ft. di Iiameter: the coefficient of elasticity is 30,000,000 lbs. Find the moment uf resist- ance of a section of the bar and the maximum intensity of stress iji the metal, {a) when the bar is round and i in. in diameter, {d) when the bar is square having a side of i inch. If the metal is not to be strained above lo.ooo lbs. per sq. in., find (c) the diameter of the smallest circle into which the bar can be bent. /Ins. — («)^|4?r in. -lbs.; sooolbs. (/>) 8334- in. -lbs.; 5000 lbs. (c) 250 ft. 2. A piece of timber 10 ft. long, 12 in. deep, 8 in. wide, and having a working strength of 1000 lbs. per sq. in., carries a load, including its own weight, of lu lbs. per lineal foot. Find the value of 7f, (a) when the timber acts as a cantilever ; {i>) when it acts as a beam supported at the ends. Find (<.) stress in material 3 in. from neutral axis at fixed end of cantilever and at middle of beam. A/is. — (a) 320 lbs.; (d) 1280 lbs.; (c) 500 lbs. per sq. in. 3. Is it safe for a man weighing 160 lbs. to stand at the ceiitre of a spruce plank 10 ft. long, 2 in. wide, and 2 in. thick, supported by vertical ropes at the ends.' The safe working strength of the timber is 1200 lbs. per sq. in. Alts. No ; the maximum safe weight at the centre is 53^ lbs. 4. Compare the uniformly distributed loads which can be borne by two beams of rectangular section, the several linear dimensions of the one being « times the corresponding dimensions of the other. Also compare the moments of resistance of corresponding sections. Ans. «' ; «'. 5. A cast-iron beam of rectangular section, 12 in. deep, 6 in. wide, and 16 ft. long, carries, in addition to its own weight, a single load P ; the coefhcient of working strength is 2000 lbs. per sq. in. Find the value of P when it is placed (a) at the middle point; {d) at 2i ft. from one end. Ans.— (a) 847 s\hs.; {/>) 11,300 lbs. 6. A round and a square beam of equal length and equally loaded are to be of equal strength. Find the ratio of the diameter to the side of the square. Ans. |/s6 : ^33. 407 , y.-t .M ■:.■:•;«* i 1 li jf'' • ! : 1 Us MR' '■f. i . II 408 THEORY OF STRUCTURES. 7. Compare the relative strengilis of two beamsof the same length and material ( Ans. £= 55826821V; 13.44 inch-tons; H lb., J^ lb., \Y 'b. 22. The effective length of the Conway tubular bridge is 412 ft. ; the effective depths of a tube at the centre and quarter spans are 23.7 ft. and 32.25 ^^■' respectively ; the sectional areas of the top and bottom flanges are respectively 645 sq. in. and 536 sq. in. at the centre and 506 sq. in. and 461 sq. in. at the quarter spans ; the corresponding sectional areas of the web are 257 sq. in. and 241 sq. in. Assume the total load upon a tube to he equivalent to 3 tons per lineal foot, and that thecontinuity of tiie web compensates for the weakeniiif; of the tension flange by the rivet-holes. Find the flange stresses and the deflection at the centre and quarter spans, E being 24,000,000 lbs. V/hat will be the increase in the ■ w rl^fJ ri- i If <4 • i i ■ :| • 'i pi|i| 'iH ; a ?u; 15:.* ^ 410 THEORY OF STRUCTURES. central Range stresses under a uniformly distributed live load of J ton per lineal foot? / Ans. At centre — - = 181485 ; /j = 4-3799 tons per sq. in. ; '+4 /. = 3.9326 Deflection = 8.33 in. At quarter span -~ = 132774 ; ft = 1.414 " " '44 /,= 1.256 " " Deflection = 6.25 in. The stresses and deflections are increased by the live load in the ratio of 5 to 4. 23. A plate girder of 64 ft. span and 8 ft. deep carries a dead load of 2 tons per lineal foot. At any section the two flanges are of equal area, and their joint area is equal to that of the web. Find the sectional area at the centre of the girder, so that the intensity of stress in tlie metal may not exceed 3 tons per sq. in. The deflection of the girckr is ? in. at the centre. Find J£ and the radius of curvature. Ans. 128 sq. in. ; 15,360 ft.; 25,804,800 lbs. 24. Taking the coefficient of direct elasticity at 15,000 tons, the coelTi- cient of latt-ral elasticity at 60,000 tons, and the liiiiit of elasticity at 10 tons, determine the greatest deviation from the straight lineof a wrought- iron girder of breadth b and depth d. b"^ Ans. - ,. 24000^ 25. Find the stress at the skin and also at a point 4 in from the neu- tral axis in a piece of 10" x8" oak, {a) with the 10" side vertical; (b) with the 8" side vertical. The oak rests upon supports 3 ft. apart and carries a load of 4900 lbs. at its middle point. Also compare (r) the strength of the beam with its strength when a diagonal is horizontal. Ans. — (ii) 330J; 132/jlbs. per sq. in. {c) 4 : 4/41 or 5 : /^li^. 26. Find the uniformly distributed load which can be borne by a rolled T-iron beam, 6" X4" x i", 10 ft. long, fixed at one end and free at the other, the coefficient of strength being 10,000 lbs. per sq. in. Ans. 438 lbs. 27. One of the tubes of the Britannia bridge has an efTective length of 470 ft., depth of 27^ ft., and deflects 12 in. at the centre unuer a uniformly distributed load of 1587 tons. Find ^and the central flange stresses, the sectional areas of the top flange, bottom flange, and web being 648 sq.i".. 585 sq. in., and 302 sq. in., respectively. Ans. E = 22,910,496 lbs.; /( = 5.37 tons per. sq. in.; • . . ■. /c=4.8i ' EXAMPLES. 411 :8. Find the moment of resistance to bending, of a steel I-beam, each flange consisting of a pair of 3-in. x 3 in. x i in. angle-irons, riveted to a 12 in. X I in. web. the coefficient of strength being 5 tons per sq. in. .1 load will the beam < arry at 5 ft. from one end, its span being 20 .t. .'' Find the central deflection, and also the deflection at tlie loaded point, E being I5,cxx) tons. Ans. 287 J J in. -tons ; 6j2J tons disregarding weight of beam, or l^\ii tons if weight of beam is talien into account ; deflection at centre — \ in., at loadeJ point = ^^ in. 29. A shaft 5i in. deep x 5 in. wide x 98 in. lonq; has one end abso- luteiv ti.xed, while at the other a wheel turns at t!ie rate of 270 revolutions per minute ; a weight (jf 200 lbs. is concentrated in the rim, its C. of G. being 2{ ft. from the axis of the shaft. Find the maximum stress in the material of the shaft, and also find the maximum deviation of the shaft from the straight, E bring 27,000,000 lbs. 30. The square of the radius of gyration of the equal-flanged sect ui of a wrought-iron i^'irder of depth d is i^//"; the area of the sei uon — IcC ; tiie span = 50 ft. In addition to its own weight it carries a uiii- ily distributed load of Ij'y lbs. per lineal foot; the maximum intensity ess = 10,000 lbs. per sq. in. Find the depth. Also determine the ji.^/^«j. A" being 25,000,000 lbs. ^Itis. 3* in.; -,fj. 31. The central section of a cast-iron girder is lo^ in. deep; its web area isy^Tv times the area of the topflangc, and the moment of resistance of the section is 360,000 in. -lbs.; the tensile and compressive intensities of stress are 3000 and 7500 lbs. per sq. in., respectively. Find the span and load so that the girder may have a stiffness — .001, E being 17,000,000 lbs. Ans. rti = la-j^T- sq. in.; a-i — \\\\ sq. in.; a^ + at = 97{|f sq. in.; span = 136 ft.; uniformly distributed load = 1764}* lbs. 32. A double-flanged cast 'ron girder 14 in. deep and 20 ft. between sup- ports carries a uniformly distributed load of 20 tons. Find suitable di- mensions for the section, the tensile and compressive inch-stresses being 2 tons and 5 tons, respectively. Also f.nd the s/iffness of the beam, E being 8000 tons. Ans. Let thickness of web = i in.; rti = 22|| sq. in.; at = 4tVj sq. in.; stiffness = .001875. 33. The deflection of a uniformly loaded horizontal beam supported at the ends is not to exceed i in. in 50 feet of span, and the stress in tlie ma- terial is not to exceed 400 lbs. per sq. in. Find the ratio of span to depth. £■ being 1,200,000 lbs. per sq. in., and the neutral axis being at half the depth of the beam. Ans. 20. 34. Two equal weights are placed symmetrically at the points of tri- scction of a beam of uniform section supported at the ends. These weights 412 TllF.Oh'V or STKUCTirUKS. are then rnudvcd aiul oilier two cinial wri^lits .ire placed at the qiiaitcr spans, Kind llio ratio of the two sets of wci^litH stb of a wroujjjht iron cylindrical beam 4 in, in diameter, tiic modulus of rupture being 42,ocx) lbs. What nniforinlv distributed lo;id will bre.ik a cylindrical beam of the same material :o li, lon^ and 4 in. in diameter ? Ans. 64.8 ft. ; 8800 lbs. 39. ,'\ red-pine beam irl ft. lotiR has to support a weight of 10,000 Ihs at the centre. Tiie section is rectangular and tlie dc|Hh is twice tin- breadth. Kind the tiansverse dimensions, the modulus of rupture bcin^ S500 lbs., and to being a factor of safety. (Neglect the weight of tin; be.im.) Ans. h = 9.84 in. ; d =■ 19,68 in, 40. A round Oak cantilever 10 ft. long is just broken by a load of Cifx) lbs. suspended from the free end. Kind itsdiameter, the moduiusof rup- ture being 10,000 lbs. (Neglect the weight of the Oc.im.) Alls. 4.185 in. 41. Determine the breaking weight at the centre of a cast-iron beam of 6 ft. s|»au and 4 in. square, the coefTicicnt of rupture being 30,txx)li)s. Ans. r.6,66r)|| lbs. 42. The flooring of a corn warehouse is supported upon ycllow-piiic joists 20 ft. in the clear, S in. wide, 10 in. deep, and spaced 3 ft. centre to centre. Kind tlie height to wliich corn weighing 48^ lbs. per cii. ft. may be heaped upon the floor, 10 being a factor of safety and yMo lbs. the coeflicient of rupture. Ans. .68 ft. 43. A yellow-pine beam 14 in. wide, 15 in. deep, and resting upon siip- port'j 126 in. apart, broke down under a uniformly distributed load of •60.97 tons. Find the coelficient of rupture. Ans. 2731.456. 1 1 : ■ .mi:^\ = 11 1 V 'Ah !■ ■;i»lr :i K.XAMl'LES. 413 .1). Kitirl tlir brcakin^r wrij^'lit iit llir rnifir f)f a ("iiniidiiin ;isfi liriim \\ ill wiilc, 3i in. (Ircp. iitid of 45 in. spjin, tlir crjciririciit of iii|i(,iirc l)i 'w<^ 7250. //"'. 4'Al4!r'ii- 41;. A limber iH'uin <'> in.(ir("|i. 3 in. vvidi', |torl'«, ;iii'! wci^liin)^ 50 11)H. per en, ft., iiiiiki- down imdcf a W(!i^lit. f)f if>,(x)o Ih, ;it ilic (M-nlrc. Find llic t:(ndli(i(Mil of inpUiic. Ans. 8911^. 4(1. A wrouKlil-iron bur 2 in. wide, 4 in. deep, nnd r44 in. between Hiip- |i(iiis, carries a unifonniy diHiribiiled load VV in ndfblion to its own wci^lit- I'ind W'. 4 beiii^ u factor of safely and 5o,o(K) jbn. the roefii- ( iciil of nipiiirc. Aifi. 5235^ ll)H, .(; I'iii in. lon^f, 2I in. de(|), and 7 in. wide. What is iIk; breakinj{ weiylil of a tiMith, the cocllii i<-iit of nipliiii' btMnj.; 5000 ll)s. ? Ans. 50,625 lbs. 41;. A wroiiKht-iron bar 4 in. dee|), 3 in- wide, and rij^idly fixed at one iiui j^ave way at 32 in. from the load when loarled witii 1568 lbs. at the (ice (lid. I''ind the coefTu'ient of rupture. fills. 4iKi^. 50. A east-iron ijeiim \2 in. wide rests upon suppfirts 18 fl. a|)art, ;uk1 carries a i2-in. brieU wall which is I2i ft. in liei).(ht anri w in. ; (//) 6i in. ; (r) 14.23 in. 51. A cast-iron girder 27^ in. deep, rests upon support.s 2G ft. apfirt. lis l)(»itom llaiigi^ has an area of 48 sq. in. and is 3 in. thick. Find the breaking weight at the centre, the ultimate tensile strength of the iron hcini; i5,(wo lbs. per stj. in. (Neglect the cflect of the web.) All.'!. 253,846 ,«,, lbs. 52. A bcatn of rectangular section, of breadth /i and depth r/, is acu-d upon by a couple in a plane inclined at 45 to llu; axis of the secli"ii. C(iiiii)are the moment of rcsi.stance to bending with that about either a.xis. ^^^^ 24/2//^ . 2^/^H " 2//' +f/"' '26' + a''' 53. A 2-in. wrought-iron bar 10 ft. long is held at the ends and is whirled about a parallel axis at the rate t>f 50 revolutions per minute. If the distance between the axis of the bar and the axis of rotation is JO ft., find the maximum stress to which the material is subjected. Ans. 17148.5 lbs. per sq. in. m 414 THEORY OF STRUCTURES. \ \ 1 W ■ i 54. A block of ice 3 in. wide and 4 in. deep has its ends resting upon supports 30 in. apart and carries a uniformly distributed load of 4800 lbs. An increase of pressure 10 the extent of 1 125 lbs. per sq. in. lowers the freezing point i" F. Assuming tliat the ordinary theory (jf ilexure holds good, find the temperature of the ice. Ans. 30' F. 55. Find the limiting length of a cantilever of uniform transverse section,/ being the coefficient of strength, k the ratio of length to deptii, and w ihe specific weight of the material. Ans. — ~y^t 7/ being a coefficient depending upon the form of the section. 56. If the beam in the preceding question is to be supported at its two I I 52/« wk ends, what will its limiting length be ? Ans. 57. Find the limiting length of a cedar cantilever of rectangular sec- tion, k being 40, w = 36 lbs. per cu. ft., and/= 1800 lbs. per sq. in. Ans. 60 ft. 58. A steel cantilever 2 in. square has an elastic strength of 15 tons per sq. in. What must its limiting length be so that there may be no seif Ans. 23.4 ft. 59. Find the limiting length of a wrought-iron beam of circular sec- tion, k being 64 and tlie elastic strength 8 tons per sq. in. What will iliis length be if a beam of I-section, having, equal flange areas and a web area equal to the joint area of the flanges, is substituted for the circular section ? Ans. 84 ft. ; 224 fi. 60. A rectangular cast-iron beam having its length, depth, and breadth in the ratio of 60 to 4 to i, rests upon supports at the two ends, Find the dimensions of the beam so that the intensity of stress under its own weight may nowhere exceed 4500 lbs. per sq. in. Ans. / = 128 ft. ; <■/ = SiV ft. ; b= 2-,=^ ft. 61. A beam supported at the ends can just bear its own weight //'to W gether with a single weight — - at the centre. What load may be placed at the centre of a beam whose transverse section is similar but vr as great, its length being n limesas great ? If the beam could support only its own weight, what would be the relation between m and « } I vi^ in'\ a Ans. IV ; m — \n 2 j 2 62. The flanges of a rolled joist are each 4 in. wide by \ in. thick; the web is 8 in. deep by i inch thick. Find the position of tlie neutral axis, the maximum intensities of stress per square inch being kjooo lbs. in tension and 8000 lbs. in compression. Ans. h\ = 3J ; //g = 4I. ii^. ff^^ fiiiil EXAMPLES. 415 Ling upon i of 4800 in. lowers of liexure . 30' ^''• transverse 1 10 depth, le form of ;d at its two wk ingular sec- sq. in. Ans. 60 ft. h of I 5 tons I may be no u. 23.4 ft- circular st'C- /hai will this s and a web the circular t. ; 224 ft. depth, and Ihe two ends, ess under its \b = 2f,T ft. veight fTto |ay be placeri ir but in- ai Isupportonly //i = ^ ■ \ in. thick; tlie neii trill |being 10000 ; hi = 4ft' 63. A continuous lattice-girder is supported at four points, each of the side spans being 140 ft. 11 in. in length, 22 ft. 3 in. in depth, and weigh- ing .68 ton per lineal foot. On one occasion an excessive load lifted the end of one of the side spans off the abutment. Find the consequent in- tensity of stress in the bottom flange at the pier, wliere its sectional area is 127 sq. in, Ans, 2.3893 tons per sq. in. 64. A railway girder is 101.2 ft. long, 22.25 ft. deep, and weighs 3764 lbs, per lineal foot. Find the maximum shearing force and flange stresses at 25 ft. from one end when a live load of 2500 lbs. per lineal foot crosses the girder. 65. A floor with superimposed load weighs 160 lbs. per sq. ft. and is carried by tubular girders 17 ft. c. to c. and 42 ft. between be; ' ings. Find the depth of the girders (neglecting effect of web), the safe inch- stress in the metal being 9000 lbs., and the sectional area of the tension flange at the centre 32 sq. in. Ans. 24.99 '"• 66. Design a timber cantilever of fl//r<7,rmrt/J 8- »l-3^ ^ 10 Fig. 307. V io^ * Fig. 3og. Also find the diameters of the rivets R in Fig. i, neglecting the weak- ening effect of the rivet-holes in the bottom flange. What is the ratio of the maximum tensile and compressive stresses in each section } m 4i5 THEORY OF STRUCTURES. \.i,'\ (X) A trapezoidal section, the top side, bottom side, and depth // (inches) being in the ratio of i to 2 to 4. Ans. ^h from top side ; 7^5^' in. -tons. (XI) A section in the form of a rhombus of depth 2c and with a hori- zontal diagonal of length 2b. Ans. \hc'^ ; ^^bc ; 9 to 8. (XII) An angle-iron 2 in. x 2 in. x ^ in. .Ins. Neutral axis divides depth into segments of jf in. and \\ in. ; WW in.-tons; ^§^^ ton; 1334 to 1369. (XIII) A hollow circular section of external radius C and interna! radius C. Ans. .99 112 C' 33 C'-C 4 C» + CC + €•■' 3 C • 4 C'» + CC + C" • 3 C' + C" (XIV) A cruciform section made up of aflat steel bar 10 in. by \ in. and four steel angles, each 4 in. by 4 in. by i in., all riveted together, (Neglect weakening eflfect of rivet-holes.) 75. A girder of 21 ft. span has a section composed of two equal flanges each consisting of two 3i-in. x 5-in. x \-\n. angles riveted to a 39-in. X |-in. web ; the cover-plates on the flanges are each 12 in. x | in., and the rivets in the covers alternate with those connecting the angles and web ; the pitch of the rivets is 3^ in. Find the diameter and also firui the maximum flange stresses, {a) disregarding the we '.!;ening eflfect of thu rivet-holes in the tension flange ; (b) taking this eflfect into account The load upon the girder is a uniformly distributed load of 20,800 lbs. (including weight of girder) and a load of 50,000 lbs. concen trated at each of the points distant 4^ ft. from the middle pomt of the girder. Ans. Diam. of rivets = .48 in. if tight, = .54 in. if subject to flexure. {a) /i —f-i = 7762 lbs. per sq. in. (b) /, = 8248 lbs. per sq. in., /a = 7847 lbs. per sq. in. 76. A beam of triangular section 12 in. deep and with its base hori- zontal can bear a total shear of 100 tons. If the safe maximum intensity of shear is 4 tons per sq. in., find tlie width of the base. Ans. 6J in, 77. Assuming that the web and flanges of a rolled beam are rectangular in section, determine the ratio of the maximum to the average intensity n of shear in a section from the following data : the total tiep/A is - times the breadth of each hange, « times the thickness of each flange, and m times the thickness of the web. Show also that this ratio is ^ or V according as the area of the web is equal to the joint area of the two flanges or is equal to the area of each flange. How much of the shear in^- force is borne by the web ? How much by the flange ? 3(«' -)- I2« — I2)(« + 6) Ans. ratio 2(«' -f- i8«' — 36« 4- 34) ' 7o)« ; 85^. :t to flexure. EXAMPLES. 419 78. In a rolled beam with equal flanges, the area of the web is propor- tional to the «ih power of the depth. Find the most economical distribu- tion of metal between the flanges and web, and the moment of resistance to bending of the section thus designed. Also find the ratio of the aver- age to the maximum intensity of shear. Ans. Area of each flange : web area :: 2« — i : 6 ; n 2 n—J^y' B. M. = - > / being the coefficient of strength, 5 the total area of section, and y the depth. Max. intensity of shear : av. intensity :: (« -f \)(\ii -(- i) : 6«. 79. Find the moment of resistance to bending, the resistance to shear, and the ratio of maximum to the average intensity of a shear in the case of a section consisting of two equal flanges, each composed of a pairof 5-in. X 3i-in. x f-in. angle-irons riveted to a 3ii-ia x f-in. web, tiie 5-in. sides of the angles being horizontal, and 4^ tons per sq. in. being the coefficient of strength. Ans. 1501.06 in. -tons ; 22.36 tons ; 4.916. 80. The floor-beam for a single-track bridge is 15 ft. between bearings, and each of its flanges is composed of a pairof 2|-in. x 2^-in, x f-in. angle-irons riveted to a 30-in. x f-in. web. The uniformly distributed load (including weight of beam) upon the beam is 4200 lbs., and a weight of 1600 lbs. is concentrated at each of the rail-crossings, 2\ ft. from the centre. Find ia) the maximum flange stress, (I)) the ratio of the maxiinitin and average intensities of shear ; (f) the stiffness, E being 27,000,000 lbs. Ans. (rt) 6523.4 lbs ; (b) 2.037; (c) .00033. 24512.59 / = 1024 neglecting effect of rivet-holes. 81. A beam 36 ft. betv\'een bearings is a hollow tube of rectangular sec- tion and consists of a 24ii.. x ^-in. lop plate, a 24-in. x ,}-in. bottom plate, and two side plates each 35 in. x ^ in. The plates are riveted together at the angles of the interior rectangle by means of four 6-in. X 4-in. x^-in. angle-irons, the 6-in. side being horizontal. Determine — {a) The intensity of shear at the surface between the angle-irons and the upper and lower'plates. {b) The diameter of the rivets, the pitch being 4 in. and assuming an effective width of 5^ in. in shear per rivet. ( at a point K distant x from O. M^ M+tIM / \ ' K V K'* IV L I L' Fig. 3to. Consider the conditions of equilibrium of a slice of the girder bounded by the vertical planes KL, K'L', of which the abscissae are x, x -{• dx, respectively. The load between these planes may, without sensible error, be supposed to be uniformly distributed, and its resultant pdx therefore acts along the centre line W . The forces acting upon the slice at the plane KL are equiv- alent to an upivard shearing force S, and a right-handed couple of which the moment is M, while the forces acting upon the slice at the plane K'L' are equivalent to a dowmvard shearing force S -\- dS, and a left-handed couple of which the moment \i>M^dM. Since there is to be equilibrium, , S--{^S-\-dS)—pdx=^ the algebraic sum of the vertical forces =o. dS dx -\-p = O ((?) 42S GENERAL EQUATIONS. 429 And, M-{M-^ dM) + S— + (5 + ^5) — = the alge- braic sum of the moments of the forces with respect to V or V = o. dM „ , — 5 = 0. dx (^) f ^ 7 The term — '- — is disregarded, being indefinitely small as compared with the remaining terms. Equations {a) and {b) are the general equations applicable to girders carrying loads of which the intensity is constant or varies continuously. Their integration is easy, and introduces two arbitrary constants which are to be determined in each particular case. Cor. I. From equations {ci) and {b\ dx' ~ dx~ ^' Let p = wf{x\ w being a constant, and f{x) some function of X. Then and M = f, + c,x - tv r rf{x)dx\ |l! !■ f, and r.j being the constants of integration, and o and x the limits. Example. — Let the girder rest upon two supports and cany a uniformly distributed load of intensity w^. Then dM am px J — — - = c^ — I iVydx = t, — w^x, dx v% and M= c^-\- c,x — w, 430 THEORY OF STRUCTURES. But M is zero when ^ = o and also when x = /. Hence Therefore, and f , = o and r, W.l TV. „ M= -~-x -x\ 2 2 ' ^ dM zv.l S — -y = — ZV.X. dx 2 ' ik_._ il ii Cor. 2. The bending moment is a maximum at the point defined by -y- = o = S, i.e., at a point at which the shearing dx force vanishes. In tlie preceding example, the position of tlie maximum bendmg moment is given by .S = o = -— 7i'^x, or x = -, and its correspondmg value is — — — = —^-. ^ 22248 The shearing force is greatest and equal to — when.T=o. Cor. 3. Suppose that the load, instead of varying contin- ., M , , uously, consists of a number of finite wci'fhts at isolated points. By reason of the discontinuity of the load- Fig. 311. jpg^ ^.j^g general equations can only be inte- grated between consecutive points. Let Nr, ^r+i » be any two such points, of abscissae x,. , x^+,, respectively. Between these points equations {a) and (d) become dS , dM ^ ^ = 0, and ^; = 5. .'. S = a constant = S,., suppose, between N^ and N,.+, ■ i f i- GENERAL EQUATIONS. 43 1 dM Hence, '^ = S^, and M= S^x -\- c, between N^ and ilX Nr^, , c being a constant of integration. Let M — Mr wlicn x = x^- Then C = Mr — SrXr , and M = Sr{x — Xr) + Mr . maximum Also, if M — Mr+, when x — Xr^, , M..+ , =-- S,.(^.+ , - Xr) + i/;. The terminal conditions will give additional equations, by mc.'uis of which the solution may be completed. Example. — The girder OA, of length /, rests upon two supports at O, A, and carries weights /', , P, , at points B, C, 4 R> R, Tx Fig. 312. dividing the girder into three segments, OB, BC, CA, of which the lengths are r, s, t, respectively. The reaction R, a.\. O = ^^^^+J) + ^^\ The reaction i?, at /J = ^^ + ^'^(^ + .y) ^ Between and B, S is constant = 5,. suppose, = /?, , .-. M = SrX, there being no constant of integration, as M= o when x = o. Also, when x = r, M = SrT. % H H! ^ IT m J i !. \ =^JP^i» i^ii;; I 432 THEORY OF STRUCTURES. Between B and C, S is constant = S, suppose, = /?, — /*,. c' being the constant of integration. But yl/= S^T when x — /-. .'. c' — {Sr — S^r, and M = S,x -{- {S^ — S,)r. Also, v.'hen x ■■= r -\- s, M = S^ -f- S^r. Betwcien 6" and A, S is constant = S/ suppose,=y?,— /*,— P„ and hence M=S,x-\-c", c" being tha constant of integration. But M = S,s + •S,f wlien x = r -{- s. .'. c" = S,s -{- S,r — St[r -f- s), and S'W M = V + S,s 4- 5,7- - S^{r -\- s). Hence, zt A, o = 5/ + 5^^ + S^r Cor, 4. The equation dx S indicates that the shearing force at a vertical section of a girder is the increment of the bending moment at that section per unit of length, and is an important relation in calculating the number of rivets required for flange and web connections. 2. On the Interpretation of the General Equations — The bending moment M at any transverse section of a girder FI may be obtained from the equation M= ~,R being the H -J DEFLECTION OF BEAMS. 433 /?, - Pv :.)r. ^'P-P. nent of the h, and is an ets required Equations.— of a girder V being the radius of curvature of the neutral axis at the section under consideration. Let OAy in Figs. 313 and 314, represent a portion of the neutral axis of a bent girder. 0.= 1 ^ X ■wX Fig. 313 Fig. 314. Take O as the origin, the horizontal line OX as the axis of .r, and the line OY drawn vertically downwards as the axis o{ y. Let x,y be the co-ordinates of any point P in the neutral axis. If R is the radius of curvature at P, then R=^ "dx" 1-H©T ..dd = ±cos6^^; the sign being -f- or — according as the girder is bent as in Fig- 313 or as in Fig. 314, and B being the angle between the tangent at P and OX. dy Now, -^ is the tangent of the angle which the tangent line at /'to the neutral axis makes with OX, and the angle is always ^m dy , • „ ■ very small. Thus, ■— is also very small, and squares and he shearing ■ -^ ' dx j' > m dy higher powers of -j- may be disregarded without serious error. Hence, • . ^=±^i-, approxnnately, and the bending-moment equation becomes d^v JllBfigJt, ii,:. hh 1^ 434 TI/EOKV OF STRUCrUKKS. The integration of this equation introihiccs two arhitran- constants, of whiclj the v.ihicsarc to he dctrrniincd from j,'iv('ii conditions. At the point or points of support, for exain[)lc, the neutral axis may bo horixA)ntal or may sU)pe at a j^ivcii an^rlc. Let he the slope at /'. Since is <;enerally very small. aiul hence W = tan (K api^roxiniately, or try dB M dx ^ EI dx* (A) From this last equation e = ± j^/Mdx, and the c/untj^i' of slope between any given limits is represented by the corresponding area of the bending-moment curve. Also, since "- = ^, y ~ J ^''''*'» and the deflection is measured by the area of a curve repre- senting the slope at each point. Again, by Art. i, d'M dS dx* ~ dx = -p. (B) Comparing eqs. (A) and (B), it will be observed that y, a, and TjT^., i.e., the deflection, slope, and bending moment, are connected with one another in precisely the same manner NEVTh'AI. AXIS 01' A tOADED liEAM. 435 s as My 5, and p, i.e., the liciidiiifj moment, shearin^^ force, and lo.id. Tltiis, llu: mutual relations hi-tween curves drawn to iipri'si-nt tlie tifjlrction, slope, .umI /uiu/inx mouicut must he the same, mutatis mutandis, as those between the turves of bendin^f inoMient, slu-arin^ force, and load. l''or example, ilivide the ijD'cctivc bending-moment area hito a number of elrmentary areas by drawing vertical lines at con- venient distances apart, and su|)pose these elementary areas to irprc.'sent weights. Two reciprocal figures connecting j/, ^, and ;)/ may now be drawn exactly as described in Chap. I, and it at once follows that — (rt) Any two sides of the funicular polygon, or, in the limit (when the widths of the elementary areas are indefim'tely (liniinished), any two tangents to the funicular or dcjlection curve, meet in a point which is vertically bilow the centre of gravity of the corresponding ijj'cctivf moment area. (/;) The segments !//, ////into which the line of weights is divided hy drawing OH parallel to the closing line CD, give the slopes (= ^Mdx) at the supports. N.B. — In the casti of a semi-girder, the last side of a polygon is the closing line, and \n gives the total change of >l()pe. {c) If the polar distance is made equal to EI, the intercept between the closing line and the funicular or deflection curve measures the deflection. 3. Examples of the Form assumed by the Neutral Axis of a Loaded Beam. ExAMi'i.K I. A semi-girder fixed at one end O so that the neutral axis at that point is horizontal carries a weight W at tlie other end A. At any point {x, y) of the neutral axis -^^Elf^, = W{l~x). . . (A) Integrating, ^, being a constant of integration. But ^'°' 3«s- tiic girder is fixed at O, so that the inclination of the neutral ! I I 1: '• I 436 THEORY OF STRUCTURES, axis to the horizon at this point is zero, and thus, when ;r = 0, dy ■J- is o, and therefore c.=^o. ax ' Hence, ' EI%^W{^.-^ (B) Integrating, £/,= (f(/?'-^;")+.„ f, being a constant of integration. But y = when x — 0, and therefore c, = o. Hence, E^y = iv[i'^ -^) (C) dy Equation (B) gives the value of -^ , i.e., the slope, at any point of which the abscissa is ;f. Equation (C) defines the curve assumed by the neutral axis, and gives the value of y, i.e., the deflection, corresponding to any abscissa x. Let or, be the slope, and d^ the deflection at A. From (B), I wr and from (C), d, = 3 £ r NEUTRAL AXIS OF A LOADED BEAM. 437 Ex. 2, A semi-girder fixeJ at one end carries a uniformly distributed load of intensity xv. At any point P {x, y) of the j neutral axis, ,(Vy XV A-EI-.-;^^ M-xf dx' 2 W . = - (l' -2/x+x'). . (A) 2 ^ ' '' ^ ' Fio. 3i«. . Integrating, t-, being a constant of integration. dy But -,— = o when x = o, and therefore f, = O. Hence, ■^^%=jh-'^'+i) m Integrating, 6', being a constant of integration. But ^ = o when x = o, and therefore f, = o. Hence, ('T-4+f^) (^) W I X -^ 2 \ 2 3 12 Let ff, be the slope and d^ the deflection at A. Hence, from iB), and from (C), I wP tana, = g^; _ I wl* i ■' 438 THEORY OF STRUCTURES. Ex. 3. A semi-girder fixed at one end carries a uniformly distributed load of intensity w, and also a single weight WdX the free end. Tiiis is merely a combination of Examples I and 2, and the resulting equations are : El''^^W{l-x)Jr'^-(l-x)' (A) Also, if A is the slope and D the deflection at the free end, — from (B), tan A = Yl\~r "^ ~^) "^ *^" "' + *^" "' ' and from (C), I (wr , w/*\ £l\ 3 Cor. — The slope (a) and deflection (d) of an arbitrarily loaded semi-girder may be determined in the manner de- scribed in Art. 2. Let F be the area of the bending-moment curve. Its centre of gravity is at the same horizontal distance .*• from the vertical through A as the point T in which the tangent at A intersects OX. .'. -F>= a = angle A TX = -. In Ex. 3, e.g., £1 2 ' 3 2 ' and Fx wr2 wP% i+-^-i=Erd. 2364 NEUTRAL AXIS OF A LOADED BEAM. 439 Note. — If the semi-girder in the three preceding examples is on\y partially fixed at O, so tiiat the neutral axis, instead of being horizontal at the support, slopes at an angle ^, then dy when X ^=o, -.— = tan &, and the constant of integration, c, , is also ^/tan ^. Thus, the left-hand side of eqs. (B) and (C), respectively, become £f[j- - tan ^j and £/(j - x tan ^). Ex. 4. The girder OA rests upon two supports at O, A, and carries a weight W at the centre. The neutral axis is evidently symmetrical with respect to the middle point C, and at any point P {x, y) between (? and C, B F^ p -f w Fig. 317. Integrating, dx 2 dx 4 ' * .... (A) £, being a constant of integration. But the tangent to the neutral axis at C must be horizontal, so that when x — -, -y- = O, and therefore c. = 2~ • 2 dx ' 16 Hence, Integrating, — EI-T- = —x^ ^ dx 4 ID (B) vr w , ivr ^ Ely =-x --^^ + ,., c, being a constant of integration. * : I. ';m 440 THEORY OF STA'UCTUJfES. But J = o when x = o, and therefore f, = o. Hence - Efy w , wr 12 16 (C) Cor. — Let «, be the slope at 0, and d, the doflection at the centre. Then, I wr I WP from (B), tan a, = j^ -^ ; and from (C), d, = ~ ~pj. B P ~c" Fig. 318. ^ Ex. 5. The girder OA rests ' '^ upon supports at O, A, and carries a uni'"ormly distributed load of in- tensify tu. Ai any point /' {x, j>) of the neutral axis, integrating, „ - dy xvl TVX' EI~ — — X dx 2 2 dy wl zvx^ , dx 4 6 ' " (A) I- t: m c^ being a constant of integration. dj I But -7- = o when x — -, and therefore tr, = — dx 2 ' Hence, 24 ^j^ w/ , wx^ wP — EI -J- = — ,r' ^ •. . , . (B) dx A 6 24 • • V / Integrating, 24 wx^ wr wl ^ ..^ Ely = —A-' X 4- <:,, 12 24 24 ' • c, being a constant of integration. But y -—Q when x = o, and therefore r, = o. Hence - Ely = —.1-' ^^ - -~x (C) 12 24 24 NEUTRAL AXIS OF A LOADED BEAM. 441 Let <*, be the slope at 0, and d.^ the deflection ;it the centre, Tlien, I ivl^ c wl^ from (B), tan ix. = ^,--7- ; and from (C), d. z=l — - ----. ^ ' 24,hl \ /' 1 2,^4 hi Ex. 6. A girder rests upon two supports and carries a uni- formly distributed load of intensity w, together witii a single woif^iit VV at the centre. This is merely a combination of Examples 4 and 5, and the resulting equations are : nl'r.1 = — X H •^ dx 2 ' 2 2 (A) and dv W W wl wx^ wr -El-f =-x' - ~-P J^—x-' - -^ -^^ , . (B) dx 4 16 ' 4 6 24 ^ ' W W - Ely = -x' - -J'x + -x' X. (C) •^ 12 16 ' 12 2/ "^ -^ 7C'/ . Tt'Jir* w/' 24 24 Also, if A is the slope at the origin, and D the central deflection, we have, from (B), tan/J = -^l^-^ + -— -j = tan a,+ tan a,; and from (C), ^=ii{'^+h"'') = <+<- Cor. — The slope and deflection of an arbitrarily loaded t,nrdcr resting upon two supports may be determined in the manner described in Art. 2. Let C be the lowest point of the deflection curve. The tan<^cnts at C and will intersect in a point T which is ver- tically below the centre of gravity of the bending-moment area corresponding to OC. Denote this area by F and the hoi izontal distance of centre 442 THEORY OF STRUCTURES. of gravity from OY by ^. Let ix be the angle between 01' and CT produced. Then EI a ^ being the maximum deflection. In Ex. 6, e.g., the girder being synnnetrically loaded, I. f 242'382 i6 32' 24 82 m7\' Ex. 7. Suppose that the end O of the girder in Ex. 5 _ „ is fixed. The fixture introdiicfs ^R, Ra I 4 -'"A P Fig. 319. neutral axis, a Icft-liaiidcd couple at ; let its ■X moment be J/,. Let the reactions at O and A bo R^ , A', , respectively. At any point /' {x, y) of the - £7-7-4 = ^,'f ^, (I) '*) But M, i.e., — £I-j^> is zero when x = /, \hi 7x>r ... o = A,/ J/., 1 ^ 1 Integrating eq. (i), ^rdy „x^ wx' dy (2) (3) There is no constant of integration, as -,- = o when x = 0. Integrating eq. (3), - E/y = R, x* zvx* .,x* 24 M,-. (4) There is no constant of integration, as x and y vanish to- gether. m cd, I 2 ~ :zEId. in Ex. 5 introduces 0; let its \.0 aiul /) .y) of the • . (1) NEUTRAL AXIS OF A LOADFD BEAM. But 7 also vanishes when ;r = /, so that o^' '^i' n^f = R,2- MJ. . . . , '6 24 ' Hence, by cqs. (2) and (5), 445 (5> M, = g-, A', == |7f/, and so R^ -^ |w/. . (6) Thus, the bcnding-momcnt , slope, and deflection equations arc, respectively, 7i' ,, IV r 2' --jt M, Mr 16 W 7VP 6' - ^^'^' 5 w/ ■^ 48 24 7VP 16 ,-.r (7) (8) (9) C(3r. I. The bending moment is «//at points given by 5 , «^ . w^ i.e., when x ^= - or I. Take (?/^ = — . K 4 4 \ Since y-7 = o, /^ is a point of inflexion. V ,:^' ^A If the girder is cut through at this point, and a hinge introduced sufficiently fi*" strong to transmit the shear {= ^ivl), Fio. 320. tlic s, ability of the girder will not be impaired. Hence, the girder may be considered as made up of two independent portions, viz. : / {a) A cantilever OF oi length — , carrying a uniformly dis- 4 trihuted load of intensity w, together with a weight \zul at /*". 444 THEORY OF STRUCTURE S. The maximum bending moment on OF is at 0, and is 3 . I . tvl I xvP ^s^^i + Ts^X" ii (/;) A girder FA of length -, carrying a uniformly distrib- uted load of intensity re'. The maximum bending moment on FA is at the middle 8 W >' ~ 128' This result may also be obtained from eq. (7) by putting -— - ■= o. Whence dx point D, and is — - l ' / ) = O = ^zvl — zvx, or X = §/, and therefore M., tI¥«'^° The shearing force and bending moment at different points of the girder may be represented graphical!) as follows : The shearing force at any point of which the abscissa is x is 5 = %xvl — tvx. Take OB and AC, respectively equal or proi)crtional to i;ri' and fW; join BC The line BC cuts OA in B, where OD — |r The shearing force at any point is represented by the ordinate between that point and the line BC. The bending moment at the point {x,jy) is 5 7CI ivl* M = k'^vIx x'' 5-. 828 Take OG, DE, and OF, respectively equal or proportional W' 9 ,, , ^ 0-. . ,• to -5—, ^uu\ and -. Ine benduig moment at any point is o 128 4 represented by the ordinate between that point and the parab- ola passing through G, F, and A, having its vertex at A .nii' its axis vertical. NEUTRAL AXIS OF A LOADED BEAM. dy 44S Cor. 2. The deflection is a maximum wiien -.- = o, i.e., when ax wl' 5 7C i6 6 8 or at the point ^iven by jr = 5 ('5 — ^^ii)- Substituting this value of x in cq. (9), the corresponding value of y may lie obtained. Ex. 8. If both ends of the girder in eq. (7) are fixed, the wl reaction at eacli support is evidently — , and the equation of moments becomes — LI--/:, — '-X M, dx' 2 2 ' (I) Integrating, - EI% ^ '^x' - Ix^ dx 4 6 M.x. dy . . . (2> No constant of integration is required, as -7- — o when x =. o. ; - is also zero when x r= I (also when x = -]. dx \ 2 / wl* wl* 4 6 dud hence M,= xvl* 12 Integrating eq. (2), wl J, zv wP — Ely = — X X' X' (.3) (4) 24 24 There is no constant of integration, as x and j' vanish together. 1 he central deflection I i.e., when x = -- j = — , -y^' \ 2 / 384 El 446 THEORY OF STRUCTURES If tlie load, instead of being uniformly distributed, is a weight (F concentrated at the centre, then, {ox one //^z//" of the girder, dx' 2 ' (5' N- ^(4 ;:, Integrating, dy W EI-^= —x-'-M.x. dx 4 ' dy (6, There is no constant of integration, as -t- = o when x — O). dy dx 7 is also evidently zero when x = -, and hence i6 '2 '8 • (7) Integrating eq. (6), W x" ETv = —X' - M, 12 ' w Wl . . (8) 2 12 l6 ' There is no constant of integration, as x and y vanisli together. I Wi The central deflection = 192 EI' 4. Supports not in same Horizontal Plane. — In the preceding examples it has been assumed that the ends of the girder are in the same horizontal plane. Suppose that one end, e.g., A, falls below by an amount jk, , y^ being small as com- pared with /. The abscissae of points in the neutral axis are not sensibly changed, but the conditions of integration are altered. Con- sider Ex. 4. Between and C, (>) ited, is a alf of the . . (5' . . (6' tien ;f = 0. (7> . . . (8> |sh together. le.— In the lends of the lliat one end, lall as com- lot sensibly lered. Con- . ('^ 447 ^/^= ---V +<^> (2) NEUTRAL AXIS 01- A LOADED BEAM. Integrating -eA = 4 (\ being a co istant of integration. Intcgr.'i'^.ng again, W - l^h = — -'t-' + c,x (3) There is no constant duo to the last integration, as x and y vanisli together. Ihtzvct'H C and A, Integrating twice, dv - El-^y- dx W {l-x)'-\-c. and W - Ely = — (/ - xy + c,x + (4) (5) (6) c,, c, being constants of integration. The tangent at C is no longer horizontal, but makes a dcfi- dy nite ancrle f^ with the horizon, so that -,- is now tan when "^ dx • I dy X ■=. -. Also, the values of -7- and I' at 6, viz., tan d and d, as 2 dx ' given by eqs. (2) and (3), must be identical with those given by eqs. (5) and (6), while the value jj', at A, as given by eq. (6) when x ■=. l,'\?, equal to y^ . Therefore W W A' ^c, = -EI tan e = - ^P + c, , 16 ' ' 16 ' ^ and / IV I go 2 96 2 - Ely, = cj -jr c,. I ill i h 448 Hence, THEORY OF STRUCTURES. V W c — ~ EI— — --/' c / 16 fully defining both halves of the neutral axis. V W W dy Again, in Ex. 6 it is no longer true that ->- = o when ax I X ■=.—, but the conditions of integration are j = o when x = 0, dy and y = j, when x = I. These, together with ^^ = o when X =^ o, are also the conditions in Ex. 7. Other cases may be similarly treated. 5. To Discuss the Form assumed by the Neutral Axis of a Girder OA which rests upon Supports at o and A, and carries a Weight P at a Point JB, distant r from o. Let OBA be the neutral axis of the deflected girder. r , The reactions at O and A arc .,,—' ^ I — r r P — -, — and P-., respectively. Y ' Let BC, the deflection at C F'G. 3"- = d. Let a be the slope of the neutral axis at B. The portions OB, BA must be treated separately, as the weight at B causes discontinuity in the equation of moments. First, at any point {x, y) of OB, ---iB -EI dy dx' -X, (I) Integrating, ^4=^^ rx + ^„ r, being a constant of integration. NEUTRAL AXIS OF A LOADED BEAM. 449 But -^ = tan Of when x = r, and therefore ax I -rr' — EI tan a = P — J- - + <^i • Hence, hitcgrating, / — r fx" /•' \ - E/{y - X tan a) = P —^ [-^ - -xj. (2) (3) There is no constant of integration, as x and j/ vanish to- gether. Also, f = d when x = r. EI{d - r tan a) = - P I -rr' I l ' - (4) In the same manner, if A is taken as the origin, and AB treated as above, equations similar to (i), (2), (3), and (4) will be obtained, and may be at once written down by sub- v-.^ / — r stituting in these equations n — a for oi, P-. for P — -. — , I— r for r, and r for I — r. Thus, the equation corresponding to (4) is -EI\d-{l - r) tan(;r-a)| = - P I 3 Subtracting (5) from (4), P Ell tan a = -r{l- r){l and from (4), £/d = - P r\l-ry 3 / • (5) — 2r) ; . . . . (6) (7) ■'.^' ',1' 11 ■ii ; 1 j i ■ ) : I : i^ t ii i, a h 'i f Jl:- 450 THEORY OF STRUCTURES. Thus, eqs. (2) and (3) become and ^dy PI -r ^ Pr ,, ^^y =6-7-^^°-67(^-'')(2^-''K- • (9) the latter being the equation to the portion OB of the neutral axis, and the former giving its slope at any point. Next, at any point {x,y) of BA, ipy I — r -EI-:ii,=iP—y-x-P(x-r) (10) dx' I Integrating, ax / 2 2^ J ^ t> c^ being a constant of integration. But -7- = tan a when ;f = r. dx I -rr" Pr I 2 -\-c^= — £/ tan a = li/ — ^)i^ — 2r), 3 * and Hence, Pl-r 6 <:.= .-^— -r(2/-r). ^4 = fV-'-^(^-'')'-?7(^-''X^^-^)-^") Integrating, ^, being a constant of integration. But y — d when x = r. 4> NEUTRAL AXIS OF A LOADED BEAM. 45 X P I — r P r .-. g- -^ r' - -g- -j(/ - r)(2/ - r> + r, = - Eld = - Pr\l-ry and c^ = o. Hence, P I — r P P r -EIy = -^ —j-x' - -Ax -rf -p- -j{l - r){2l - r)x, ( 1 2) / 6 r which is the equation to the portion BA of the neutral axis, eq. (11) giving its slope at any point. In the figure r < — , and the maximum deflection of the girder will evidently lie between B and ^, at a point given by dy putting -j~=^ o in eq. (12), which easily reduces to — 2lx-\- 2/' + = 0, and therefore --4 r -r' 3 IS the abscissa of the most deflected point. The corresponding deflection is found by substituting this value of x in eq. (12). If r > — , the maximum deflection lies between O and B, at 2 ' dy a point determined by putting -3- = o in eq. (8), which then easily reduces to r(2/ - r) = X' from which . = ^fc> IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I ;:f ilM ^ iilM 1116 1^ IlM 2.0 1.8 1.25 1.4 1.6 ^ 6" — ► V] W e: 0>* VI 6> '^f A 1^ ^% s op /A Photographic Sciences Corporation A ^^' % # :\ ^ ^> ^V \ ^ m » n. 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 ^#»'*' ^ ) C/j ' f , t 1 t ! i ': 1 I it ) > 1 ; U- m it u t m 452 •• THEORY OF STRUCTURES. Substituting this value of x in eq. (9), J n ' P l-r(r{2l- r)\l the maximum deflection = Tp'T~l~ \ "/ " Example. — P = 15,000 lbs., / = 100 ft., r = 90 ft. The distance of most deflected point from O sj'^- _./?oX^=57^ft., and the maximum deflection 15000 10 ^ 90 X I io\* X 10 ^ 90 X I ip y _ 100 \ 3 / ~ 500000, ,, 6. To Discuss the Form of the Neutral Axis of a Girder OA which rests upon Supports at O and A and carries several Weights i*, , P, , P, , . . . , at points i, 2, 3, . . . , of which the Distances from O are r, , r, , r, , . . . , respectively. :^X X Fig, 333, It may be assumed that the total effect of all the weights is the sum of the effects of the separate weights, and thus each may be treated independently, as in the preceding article. Let or, , or, , or, , ... be the slopes at the points I, 3, 3, . . . of the neutral axis. Considering P^ , the equation to 0\ is l :.'. - Ely = -y ^^x^ ~ ^ 7'(/ - r ){2l-r,)x; NEUTRAL AXIS OF A GIRDER OA. 4S31 and to lA, ^ Ely = § ^^V - ^{x - r.)' - § "jil - r.)(2/- r,)x. Considering P^ , the equation to O2 is - Efy = § ^V - § ^(/ - r,)(.2/ - r,)A: ; and to 2A, - £/;' = § ^-=^V - ^(x - r.)' - § "jil - r,)(2/ - 0;r ; and so on for P, , P, , etc. The total deflection V at any point {x, Y) is the sum of the deflections due to the several loads. Take, e.g., a point between 3 and 4, and let d?, , - 5(^ - r.)' ; -£M = § ^-^V - § ^^'(Z - r.)(2/ - r> - ^\x - r,y ; -£M = ^* ^V - § ^X/ - 0(2/ - r> ; and so on. Hence, - EIY= - EI{d, + ■ I *■ iUf •*-.. t ^1' I' i i i !■' ^ i. i li '■ .- L 458 THEORY OF STRUCTURES. Case c. Spring of constant width but parabolic in elevation. Let d, be the depth at a distance x from the fixed end. Then IW r Pic. 396. {d.\_ l-x \d]- I ' and /at the same point = ' bd:_ _ bjT il - Ar\t ~2 ~ I2A / /• Integrating twice, y = 12W I* ~'Erbd' \^{l-x)^-2l\l-x)^\l% and hence £ bd'~ lEd' (7) Also, rF-^ = ^ ^JL^lLMl=-f-W. 61 ' lEd <) E lE ' .: V = The work done WA _ i/'V 2 ~6 E ' 9. Girder Encastrfe at the Ends.— The girder BCDEFG rests upon supports at the ends, is held in position by blocks forced between the ends and the abutments, and carries a uni- formly distributed load of intensity w. GIRDER ENCASTR£ AT THE ENDS. AS9 It is required to determine the pressure that must be devel- oped between the blocks and the girder so that the straight portion between vertical sections at points O and A of the R I D 1 "^^^m -t— i i 01 ^^pTa 1 1 ^^ — ■ t' : i i i !. ■,■.■1 )i ■■\\ \:,t I. . Fig. 327. neutral axis may be in the same condition as if the girder were Jixcd at these sections. Let / be the length of OA. Let R be the reaction at the surface BC, and r its distance from O. Let H be the reaction between the block and the end CD, and k its distance from O. Let P be the weight of the segment on the left of the ver- tical section O, and/ its distance from 0. Then for the equilibrium of the segment on the left of the section at O, wl tvl^ R4---P=o, and Rr - Pp - Hh • =0. '2 ^ 12 .■.R = P--. and H = I wis _ tvl* h = the required pressure. #1 Again, take O as the origin, OA as the axis of x, and a vertical through O as the axis of y. At any point (;r, y) of the neutral axis, d^y wl WX* wl* EI~, = ~x- — - — dx^ 2 2 12 (See Ex. 8.) I'll I 1 ■ ! !iti n u '•'! 1 .' ) 460 THEORY OF STRUCTURES. 10. On the Work done in bending: a Beam. — Let ^ A'B'C'D' be an originally rectangular ele- ^' ment of a beam strained under the action of Q' external forces. Let the surfaces A' D' , B'C meet in 0\ O is the centre of curvature of the arc P'Q of the neutral axis. Let OF = R= OQ'. Let the length of the arc P'Q' = dx. Consider any elementary fibre /»y, of length dx', of sectional area a, and distant y from tlie neutral axis. Let / be the stress \np'q'. The work done in stretching/'^' =z ^~-[dx' - dx^. But dx ~ P'Q' R-\-y , ^ dx' — dx ^ y and t — Ea ; = Ea j,. R dx \E The work done in stretching/'^' = — -w^dxay'', and the work done in deforming the prism A'B'C'D' = ^[-^^^dxay') = '-^,dx2iaf) ^\^. dx. Hence, the total work between two sections of abscissx A' \ET EI C* I M But -^ — -py; therefore the work between the given limits EI /*"' '^^%) =2EI m^m TRANSVERSE VIBRATIONS. 461 This expression is necessarily equal to the work of the ex- ternal forces between the same limits, and is also the semi vis- viva acquired by the beam in changing from its natural state of equilibrium. Ccr. — If the proof load P is concentrated at one point of a P beam, and if d is the proof Jeflection, the resilience = —d. If a proof load of intensity zv is uniformly distributed over the beam, and if y is the deflection at any point, the resili- ence = ~ / wydx, the integration extending throughout the whole length of the beam. The case of the single weight, however, is the most useful in practice. II. On the Transverse Vibrations of a Beam resting upon Two Supports in the same Horizontal Plane. It is assumed — (a) That the beam is homogeneous and of uniform sectional area. {b) That the axis {neutral) remains unaltered in length. {c) That the vibrations are small. [d) That the particles of the beam vibrate in the vertical planes in which they are primarily situated. In reality, these particles have a slight angular motion about the horizontal axis throup[h the centre of gravity of the section, but for the sake of simplicity the effect of this motion is disregarded. dx A-X Ci;C' i s+as vodx fY Fig. 329. Let OA be the beam. Take O as the origin, the neutral line OA as the axis of x, and the vertical C>Fas the axis o[ y. Consider an element of the beam, bounded by the vertical SbHI^H :eay ^^^^^H KflP tf 1 w [ 1* 1 L t X ■ ■j ■ ,t- ; ■ n- ^^ "■El' I? ■■^ FUL. 1 ;■ I, I! 409 THEORY OF STRUCTURES. planes BC, B'C\ of which the abscissae are x and x -\- dx, respectively. Let w be the intensity of the load per unit of length ; hence wdx is the load upon the given element, and acts vertically through its centre. Let S be the shearing force at .5 ; S -{• dS the shearing force at B'. Let M be the bending moment at B\ M-\-dM the bend- ing moment at B'. Also, the resistance of the element to acceleration = - ~ -. Hence, at any time t. w d^v -dx -^ + 5 - (5+ ^5) - wdx = o, O or dy g dS df w dx -g=o. (I) Again, taking moments about the middle point of BB' or CC, M.il 4' :;!' M-{M + dM) + S~ + {S-{-dS)~ = o, or ^dy dM 'dx = S. But iW = — £/-T=i . Therefore dx 2 • « • • < (2) dx' dS dy •^=-^^^' ^"^ di=-^^d-^^ Hence, from (i). 5+^£/~^-^=o. (3) dt ' IV dx* ^ CON^TINUOUS GTRDERS. 463 This equation docs not admit of a finite integration, but may be integrated in the form of a partial differential equation. 12. Continuous Girders. — When a girder overhangs its bearings, or is supported at more than two points, it assumes a wavy form and is said to be continuous. The convex portions are in the same condition as a loaded girder resting upon a single support, the upper layers of the girder being extended and the lower compressed. The concave portions are in the same condition as a loaded girder supported at two points, the upper layers being compressed and the lower extended. At certain points, called points of contrary flexure, ox points ?f in- flexion, the curvature changes sign and the flange stresses ai'- necessarily zero. Hence, apart from other practical considera- tions, the flanges might be wholly severed at these point vith- oiit endangering the stability of the girder. 13. The Theoreii of Three Moments. — It is required tw determine a relation between the bending tuoments at any • >t, consecjiiix'C ^ Jnts of support of a loaded continuous girder of several spans. Rr-l A Rr iX V Rr+1 Fig 330. Let O, X, V be the (r — i)th, rth, and (r+ i)th supports, respectively. Let OX=l^,XV-lr^r' Case A, Let w^ be the load per unit of length on OX, tcv+i the load per unit of length on XV. Let Rr-i, Rr, Rr + i bc the reactions at 0,X, V, respectively. Let Mr-,, Mr, Mr^i be the bending moments at O, X, V, respectively. Let a be the angle which the tangent to the girder at X makes with V. Consider the segment OX, and refer it to the rectangular axes Ox, Oy. \ i M ;{' u Mil 464 THEORY OF STRUCTURES. The equation of moments at any point {x, y) is - EI-^. = Rr-,X - w, y + M,., =AI. . . (I) At X, X = I, . and M — M, . /' . ^) c; imilarly, the segment A^F gives ^,.^,/, + ,-7C,+ ,-^+J/, + , =il/,. . . (3) Combining (2) and (3), = MXlr + ^r + ■) (4) Integrating (1), -T^y EI d. V X X ,- = R,., V ~ ^- "0 + ^r-i^ + <^, . . (5) c being a constant of integration. dv When X — I, , -',-- — tan a. dx Ir ly .'. — EI tan ix — AV_, -- — iv,. — + ^r-Jr + ^>, • • (^) Integrating (5), X* V* x^ - Ely = Rr_, -^ -Wr '-- 4- J/,., -- + ex. . (7» There is no constant of integration, as x and y vanish together. Also, >' = o when x = l, . ■■■o = xJ^-w/£ + mJ-^ + J„ or .=:-/?,., ^ + «V^-i^/.-i'. ... (8) -t Ip '^ ii ; 1^1 iL.a i ,L ^^F"W^P ™"l mT* « THE TIIEOKEM OF THREE MOMENTS. Substituting this value of c in cq. (6), - EI tan a = /?,., -^ - w, " + M,., ~. Similarly, the segment A'f^ gives 465 (9) lif tan (;r - ..) - R, , . ^^ - w,. , / '^^ -f ^r!'-^ . do) Adtling cqs. (9) and (lo), transposing, and simplifying, = IWJ; + |«', + A , . - IMr.Jr - IMJ,,.,. (I I) Finally, combining cqs. (4) and (11), M,_J,. + 2ArXlr + Ir,) + ^/m/h. = - aw: -f 7t/. + , /V,). (12) If the girder is sujjported at « points, there are n — 2 equa- tions connecting the corresponding bending moments, and two additional equations result from the conditions of su])port at the enils. For example, if the ends merely rest upon the sup- (iy ports !/■, '— O and M„ := o ; if an end is fixc^d, -,- ::= o at that point. The point of maximum bending moment, the points of inflexion, and the point of maximum deflection in any span are ,. dM - . , + <^. = ^r-,-r- + ^/.-. A + ^, , ^■■PWfWpf Tin: THEOREM OT THREE MOMENTS. 467 Integrating, dy x'' P Kjf + ^/.-.^ + c,p, = R.J^- + M^Jf -\-c,p,+c,, and so that and o = AV_,|- - ^'(4 - /.)' + ^/.-.-;- + cj, + ^. ; f, = 0, C, = f. /' p /. 6+'6fy'-^^'>^-^^-i" ' ' (7) Let a be the slope at A'; then, by eqs. (5) and (7), - £/tan a = i?,./^ - ^^(/. _^,)X24 +/.) + .»/.-. ^-^ (8) Similarly, the segment XV gives - £/ tan {tt - a) = kJ-^ + i1/.+,^". (9) il ' I Al ! r f i 468 THEORY OF STRUCTURES. Adding eqs. (8) and (9), and transposing, K_,K^ + Rr^rr^. = -%lr - P.)\2lr +/.) ^1 -|J/,.,/,-|J/,+,/,+,. (10, Again, taking moments about X, whence R^ _,/,» + /?.+./%+, = M^{1, + /,^,) - M^_J^ -Jf,+,/^, + P./.(4-A). (12) and finally, by eqs. (lo) and (12), JC/. + 2J/.(/. + /.+.) + ^/.+/.+. = -Z'. 7(4' -A')- (13) The effect of each weight may be discussed in the same manner, and hence the relation between i1/,._, , J/,., and i^/^^, may be expressed in the form M, Ppr ;_ /, + 2Mllr + 4+:) + i^/.4..4+: = — ^-,-(/.' - P , '■r -:^^(/%+,-^'). (14) Cor. I. The relation between M^_i, M^, M^^^ tor a uni- formly distributed load may be easily deduced from eq. (14), For example^ let a uniformly distributed load of intensity:;' cover a length 2a {A' being the rth span, let OBX be the curve Fig. 331. of bending moments, supposing OX an independent girder, i.e., cut at and X. On the same scale as this curve is drawn, take the verticals OE and XF to represent Mr-^ and M^ , re- ■ I, t •Mi m m i i 470 THEORY OF STRUCTURES, ■I r spectively, and join EF. The curve OBX corresponds to the portion \-j x — MA of the above equation, and the line EF to the remainder, i.e., — ^'1 /,. — x\-\- -fx. The actual bend- in^ moment at any point of OX is represented by the algebraic sum of the ordinates of the curve and line at the same point. which will be the intercept between them, since they represent bending moments of opposite kinds. Let A be the effective moment area, or the algebraic sum of the areas for the load and for the moments at O and X, and let X be the horizontal distance of its centre of gravity from O. Let Ar be the area for the load, i.e., the area of the curve OBX, and let s^ be the horizontal distance of its centre of gravity from O. Then Ax = A,z^ + mJ^ + ; (^. - ^.-:)/; ^ 3 = ^^, + iJ/,..4' + iJ/,/A This result will be referred to in a subsequent article, 14. Applications. — Example i. Swing-bridges of two spans revolving about a single support at the pivot pier. This is a case of a girder of two spans, 0X{= /,), XV {— /^i, resting upon supports at O and V, and continuous over a pier at X. The bending moments at and Fare X V both ;///. ^ Let M be the bending moment at X. For a uniformly distributed load. T — ■ T" Fig. 33a 2il/(/. + /,) = - i(«;/.' + «.,//), or J/=- I W,l^ + li\ll 8 /. + /. ' w, being the intensity of the load on OX, w, that on XV. SWING-BRIDGES, For an arbitrarily distributed load. 471 2 J/(A + /,) = - yi - />', or M=~\^^^^, where ^ = ^'^.^(/.' - /) and B = ^^!^ {l^ - q'). Let K^, A\, y?., be the reactions at O, X, V, respectively. For a 2iniforiuly distributed load, ^. = -:^+-, - 8/AA + /,) For an arbitrarily distributed load, J. __^ P{^.-P) I M _ P{l,-p ) I A±B_^ '" A "'^/;" /, 2 /,(/, + /,)• If 7^, = o, or if P and hence A = o, then A, is negative. So if zy, = o, or if Q and ^ence i) = o, then A'^ is nega- tive. Hence, if either of the spans is unloaded, the reaction at the abutment end of the unloaded span is negative and that end is subjected to a liammcriug action. This evil may be obviated : (rt) By loading the spans sufficiently to make A, and R^ zero or positive. This result is attained for A, \{izvj:'-\-A'ii;Kk>w,i:, or if :s^^^^)>i^±^^, and for 7?, if 4«'//,' + iw^l: > ^A', or if :s^l^^ > 1 ^^^_"^^-^. n i; I' ' 'ii:t n 472 THEORY OF STRUCTURES. {b) By using a latching apparatus to keep the ends from rising. (<■) Ry employing suitable machinery to exert an upward pressure, at least equal to the corresponding negative reaction upon each end, which is thus wholly prevented from leaving its seat. Cor. I. When the load is uniformly distributed, the di>. tance x of the point of inflection in OX from O is given by M ^♦^'l-^ , , r 2/?, O = K.x , and therefore x = '2 w, Similarly, the distance of the point of inflection in XV from 2R V= ' Ui If /, = /, = /, then And if w, = «', = ti', then 7w, w. 16 -V, K = 16 /. M=- iwP, R, = iwl =R,, R,= 2wl -R,-R,^ fzc/. 2/?, -, 2R. In the latter case = |/ , and thus a hinge may be introduced in each span at*a distance from the centre pier equal to one fourth of the span, without impairing the stability of the girder. Hence, also, the continuous girder of two equal spans may be considered as consisting of two independent girders, each of length |/, resting upon end supports, and of / two cantilevers each of length — . Ex. 2. Swing-bridges with fwo points of support at the Ri R« R» R4 h Wi W^' r Fig. 333. pivot pier, as, e.g., when they are carried upon rollers running: in a circular path. lii_ :! I'M A P PLICA TIONS. 473 This is a case of a continuous girder of three spans. Let /,,/,, /, be the lengths of the spaces, 7f , , tc, , w, the corresponding intensities of the loads, which are assumed to be uniformly distributed. Let R^y A',, R^, R^ be the reactions at the supports; ,t/,, M^, J/j, M^ the corresponding bending moments. Then Ml + 2 J///, + 4) + ^./, = - \(zv,i: + T.',/,') ; . ( I ) i^u + 2iJ/3(/, + A) + ^y, - - iKC + ^^//)- • (2) Let the ends of the girders rest upon the supports, and assume, as is usually the case in practice, that the centre span is unloaded, i.e., that 7f, = o. Then M,—Q and M^ = o. From (i) and (2), 2Mll,^i:)^MJ, = -\wj: (3) J/,/, + 2yJ/3(/, + A) = - M/3' (4) - 2wj:{k + h) + ^^,/oV, and Hence, '^'-4(4//. + 3C-h4/./3 + 4V,)' and M^ = zvJ:1, - 2zv,/,V, + /,) Taking moments about the second support, RA = ^ + M, - -^e>.(6/.V, + 6/.V3 + 6/X + 8/.Vy,) + ri'JX 4(4//, + 3/,' + 4//. + M) Taking moments about the third support, _ w,{6/X + 6 /.V. + 6/, V,' + 8/3V/,) + w,/X 4(4/,/, + 3/;' + M + 4/y,) (5) (6) (7) (8) : i i 474 THEORY OF STRUCTURES. Thus /?, and R^ are both positive for all uniform distribti tions of load over the side spans, and no hammering action can take place at tiie ends. Again, if the span on the left is unloaded, i.e., if w, = o, 7^/, is positive and J/, negative ; and if the span on the riglit is unloaded, i.e., if tc, = o, iJ/, is negative and M, positive. Thus, at the piers, the flanges of the gink r will be sub- jected to stresses which are alternately tensile and compressive, and must be designed accordingly. The same result is also true for arbitrarily distributed loails. Ex. 3. The weights on the wheels of a locomotive passin ^1.-2 > ^n-3 . • • • being given by the law, a„-i = 4^n-a — «»-3 ; m ATPUCA TIONS. 477 «. = 4^. — rt, = 209 ; <». = 4«, -<^= 15; «. = 4^. = 4 ; a, = I. Commencing with equations w — 3 and « — 2, and proceed- iiig as before, -_^_^ an-l^, + ^,) •- "— 3(^1 + «^a) + an-Sw^ + W,) — . . . ± 1 5(w«_4 + w«-3) T 4(«^»-3 + "^n-^ ± (^«-» 4- w„_,) [ , (2) the upper or lower sign being taken for the terms within the brackets according as n is odd or even. Solving the two equations j and j, - (a„_,rt„_3— rt'„_,^„.^ — 3)zy3 + . . . :f(3««-. + <^»-4 — ««-3)^«-a and ±w»-i(«'«-, — = /' ( 4 ^ - (3«„_, + ««-4 - «»-3)«'. T (^^-.««-3 — «»-ia«-4 - 3)z«'«-2 Hence, since w, ,«/,,... w^ are positive integers, the value of ;;/„ will be greatest when «/, , it', , w^ , 7t\ , w, , . . . are greatest and Ti'j, zfc'j, a;, , . . . are least ; and the value of »/,.., will be greatest when w„ , ze'„_, , w„_3, w„_^, . . . are greatest, and 7i;'„_, , :i'„_^, w„_<,, . . . are least. In other words, the bending monr ents at the 1st and (« — i)th intermediate supports have their maxi- f it i: 1^ lit. I Iritl' ' [ I 1l 1- i ^> :jJ I III i .! .,i if 478 THEORY OF STRUCTURES. mum values when the two spans adjacent to the support in question, and then every alternate span, are loaded, and the re- maining spans unloaded. w, ,;«,,... w„_j may now be easily determined. Thus, by eq. (i), P 4 /' ( 4 • + (a!„_,«„_, — 3 are each less than unity, and or I'L or I' I, or the co- COiV TIN UO US G/h'DKRS. 481 It may now easily be shown that Ac — /> and lib — A arc each positive. Hence, ///,._, and w,. are both of tiie same sii,ni. The bendin^Lj moment ;//,, at any intermediate suijport on the left of /- — 1 is given by III -^ -j -III,., if q and r are the one even and the other odd, or ///, — itr-l ;;/,.., if q and r are both even or both odd. Thus the bending moment at the ^th support is increased in the former case and diminished in the latter. If (/is on the right of r, III, or ««- _|_ ^m±Lj,i^ if q and r are both even or both odd, a tt-r+i n, III,— — n-q+t ih,-r '--in^ if q and r are the one even and the other odd, +2 and the bending moment on the ^th support is increased in the foiniLM- case and diminished in the latter. Thus, the general principle may be enunciated, that, "in a hori/oiital continuous girder of n equal spans, with its ends rcstiii<^ u[)on two abutments, the bending moment at an inter- nuuhate support is greatest when the two spans adjacent to such support, and the alternate spans counting in both direc- tions, are loaded, the remainder of the spans being unloaded." Case III. The same general principle still holds true when the two end spans are of different lengths. E.g., let the length of the first span be kl, k being a numerical coefficient, and let 2(1 -\- IS) = x, Eq. (i) now becomes m^x -(- w, = o. : '■::*!# ^%\ ]■■' 4 f :; 482 THEORY OF STRUCTURES. Proceeding as before, Ill - + '"' - /// 7;, -' = + ..., the coefficients /;, , /^j, /',, . . . being given by the same law as before, viz., /', rz: I ; 1>,=X', l\ = 4K - -/\ = 4x-i; i\ = aK - -K=-- 1 5-f - 4 ; • • • -/;, ^ 56,t-- 15; The two sets of coefficients {a) and {b) are identical when A- = 4; and when x > 4, all the coefficients b except the first (^, = i) arc numericallj' increased. Hence, the same general results will follow. A'./>. — The equations giving »i^ are simple and easily ap- plicable in practice. They may be written and ;//. in, (la I^ ^t' .r . y \ r r . = ± — — —. — - — - It ^ IS on the left of r, ' "^ a,-, be- I ^ ' = ± "«-i~i7~Z — " ^ '^ °" '"^ ""'ght of r. If there are several weights on the rth span, A = 2 -J?(/' - /) and B = ^7v ^(/ - p){2l - p). Example. — The viaduct over the Osse consists of two end spans, each of 94 ft., and five intermediate spans, each of 126 ft. The platform is carried by two main girders which arc con- MAXlMUAf BENDING MOMENTS. 483 2;//,(/&+l)+'«, = -^V'.ii + i); 4 m, -f 4;//, 4- ;;/. = - -(i| + i^) ; w, + 4w. + m. /' di + i); *«6 + 4"^ + "^ = 4 • f«, + 2W,(/t+ l) 4 Hut ^' = r^ = f , very nearly. ^' 23 7w, + 2;«, = - - . g- (0 timious from end to end. The total dead load upon the {girders may l)e taken at one ton (of 2000 lbs.) per lineal foot. Denote the supports, taken in order, by the letters a, b, c, d, <'./•.'.''> /'' •^"'^' ''-"•^ 't '^^' recpiired to find the maximum bendinfj nioiueut at d when the bridge is subjected to an additional proc ' load of l^ tons [K-r lineal foot. The spans ab, of, dc, /V of each girder carry i^ tons per lineal f()ot. Tile spans be, ef, i^li of each girder carry \ ton per lineal foot. Denoting the bending moments at a, h, c, d, c, f,g, h, re- spectively, by vi^ , ///, , . . . w„, the intermediate spans by /, the end spans by kl, and remembering that /«, = o = ?«, , we have in. + 4"'3 + '«4 = - 7 /' 7 4 2 (2) ." :: i 484 TIfKOKV OF STh'UCTUKKS. ^' 5 w, + 4m, 4- w, = - - . J ; m^ + 4"^ + w. = - w, -f 4/;/, I- ;//, = - 2w. -1- 7;;/, From cqs. (i), (2), (3), 97///, -\- 26m ^ = From cqs. (4), (5), (6). 4 2' ^_: 7 4'2' /• '3 4 ■ 4 ■ (3) • (4,1 ^ 347 4- 8 • 26/;/, -|- gym, = r 279 4" 4 ' Hence, m^, the niaxiimiin required, /• 19151 •4 ■ 8 X 8733 = — 605.5 ft.-tons. 16. General Theorem of Three Moments. — The most general form of theorem of three moments may be deduced as follows : O, Oj 0, T^ I Fig. 334. Let 0, X, V, the (/• — i)th, rth, and (r -f- i)th supports of a continuous girder of several spans, be depressed the vertical distances U, {= 0,0), r/, {=0,X), and ^/, (= 0,V), respectively, below the proper level 0,0^0, of the girder. ai-.iXi-.NAi. riii'.OKi'.M oi- riiNEE momi:.\'is. 485 ,1^ , re beinj^ 60 in., whih the web consists of tv/o ^f'\n. plaus, j6 in. clet'i) and 18 in. apart. V glectin^' the elfect of the anj^le- iroris uniting tin: web plates to the llanges, determine tlie tiuiinent of resistance. The (,'irder has to carry a uniformly distributed dead loail of 56 tons, a uniformly distributed live load <'f 54 tons, and a local load at the j^'ivin section of 100 t(jns. What are the corresponding (lange stresses per s(|uare inch } How many J-in. rivets are required at the given .section to unite the angle-irons to the llanges? Ans. 238.13 X coeff. of strength ; 3.3186 tons ; 3.896 tons. 8. A yellow-pine beam, 14 in. wide and 15 in. deep, was placed upon supports 10 ft. 9 in. apart, and deflected j in. under a load of 20 tons at the centre. Find K, neglecting the weight of the beam. .Ills. 11 = 1,272,112 lbs. 9. What were the intensities of the normal and tangential stresses at : ft from a sujjport and 2.i in. from neutral jjlane, upon a plane inclined at 30" to the axis of the beam in the preceding question.-' A/is. 132.83 and 218.91 lbs. 10. A beam is supported at the ends and bends under its own weight. Shoa- that the upward force at the centre which will exactly neutralize llic bending action is equal to ^ or J of the weight of the beam (u'), according as the ends avi^/rtY ox fixed. Find the neutralizing forces at the quarter spans. Ans. Ends free ^'^y,io at eaeli or ,'."f,7i/ at one of the points of divisif)n. Ends fixed -^^70 at eue/i or ^70 at one of the points of division. 11. A beam 8 in. wide and weighing 50 lbs. per cubic foot rests upon supports 30 ft. apart. Find its depth so that it may deflect | in. unrier its own weight. (E— i ,200,000 lbs.) Ans. 9.185 in. 12. A rectangular girder of given length (/) and breadth (/>) rests upon two supports and carries a weight /' at the centre. Find its depth so that the elongation of the lowest fibres may be ,5'^nj "^ ^'"^ original length. /2100/'/ Ans. \/ . oh 13. A yellow-pine beam, 14 in. wide, 15 in. deep, and weighing 32 lbs. per cubic foot, was placed upon supports 10 ft. 6 in. apart. Under uniformly distributed loads of 59,734 lbs. and of 127,606 lbs. the central V-.}- 492 THEORY or STKUCTURF.S. I f dencctions were respectively .uS in. .iiul .29 in. FMnrl the mean value of /•;. Also (letcrinine the additional weii^lit at the cx-ntre which will increase the lirst dcHection by ,'0 of an inch. Atis. 2,552,980 lbs.; 24,121 lbs. 14. In tile preceding question lind for the load of 59.734 lbs. ilie inaximuni intensities of thrust, tension, and shear at a point half wav between the neutral axis and the outside skin in a transverse set tioii ,ii one of the points of trisection of the beam. Also find the inclinaliniis of the planes of principal stress at the point. Alts. 1O09.255, 165.562, 1 19.364 lbs. ; =: 3" 48f. 15. A pitch pine beam, 14 in. wide, 15 in. deep, and weighing 45 lbs. per cul)ic foot, is jjlaced upon supports 10 ft. 9 in. apart, and can ies ,1 load of 20 tons at the centre. Find the deflection and curvature, /; being 1,270,000 lbs. What slitTness does this give } Wliat amount of uniformly distributed load will produce the same dellection ? Ans, ^\^\ 32 tons. 16. In the preceding question find the maximum intensities of thrust. tension, and shear at points (a) half-way between the neutral a.\is and the outside skin, (/') at one third of the depth of the beam, in a lians- vcrse section at one of the quarter spans. Also lind the inclinations of the planes of principal stress at tiiese points. Ans. — {a') 951. S53, 292.969, 329.442 lbs.; 0=9° 34*'. {!>) 65S.774, 171.108, 243. S33 lbs.; 0^ r5 50'!'. 17. A piece of greenheart, 142 in. between supports, 9 in. dcei). ,mil 5 in. wide, was tested by being loaded at two points, distant 23 in. from the centre, with equal weights. Under weights at each point of 44^10 lbs., 11,200 lbs., and 17,920 lbs. the central deflections were . 13 in.. .37 in., .67 in., respectively. Find the mean coelFicient of elasticity. The beam broke under a load of 32,368 lbs. at each point. Find the coefHcient of bending strength. 18. A sample cast-iron girder for the Waterloo Corn Warehouses. Liverpool, 20 ft. J\ in. in length and 21 in. in depth (total) at the centre, was placed upon supports iS ft. \\ in. apart, and tested under a uniformly distributed load. The to|) flange was 5 in. x i^ m., the bottom tlange was 18 in. x 2 in., and the web was \\ in. thick. The girder deflected .15 in., .2 in., .25 in., and .28 in. under loads (including' weight of girder) of 63,76311)3., 88,571 lbs., 107,468 lbs., and 119,746 lbs., respectively, and broke during a sharp frost untler a load of 390,282 li)s. Find the mean coelTicient of elasticity and the central flange stresses at the moment of rupture. Am. 7 = 3309.122; £■ = 17,427,327 lbs.; 20,121 lbs., 47,168 lbs. 19. A steel rectangidar girder, 2 in. wide, 4 in. deep, is placed upon EXAA/J'LEU, 493 supports 20 ft. apart. If K is 35,ckx),ooo Ihs., fiiul the woij4ht wliicli, il placed at tlic centre, will rau,se the beam to dellect i in. j-lns. I stX^sV 't^s. 20. A timber joist weif^iiinfj 48 lbs per cubic foot, 2 in. wide x 12 in, deep X 14 ft. long, deflected .825 in. iirder a load of 887 ll)s. at tiic cciiire. Find A'. Ans. 3(^7,880 lbs. 21. A beam of span / is uniformly Jdaded. Cnmpare its strenglii and sliflness (rt) when merely resting upon supports at the ends; (/') vvlu'ii fixed at one end and resting upon a support at the otlur; (i) when tix(;(l at both ends. In case ic) two liingi.'s an; introfluced at points distatit y (roin the centre ; show that the s/rcfii^/Zi oi the beam is economized to tlie best ellect when/ — - . and Uiat the stiJTness is a maximum when 252 y = - very nearly. 4 Ans. Cases {a) and (h). iitx : nix :: i : i ; D\ . Di :: i : .416. Cases (a) and (< ). ni\ : Wa :: 3 : 2 ; l)\ : Ih '•'• 10 : 3. Case ((•). Max, economy, Wi : im :: 2 : i : 1), . /A :: 5 : 2 y'2. Max, stiffness, Wi : ///;, :: 4 : 3 ; /^i : Dt, :: 15:4 (approx,). 22. A beam AB o[ span /, carrying a uniformly distributed load of intensity w, rests upon a support at // and is imperfectly fixed at „■/, so I 7.'/' tliat the neutral axis at ,/ has a slojjc of tlian . / f)y an amount I ivl' 48 A7 ■ Find the reactions, The end // is lower How much must li Fmri Ans ■ -joi, — 3A/ 3A/ loo — 2o.r'' + .r'j. (b) First. Loaded span between support and weight. )' ;oo / „ \ox\ Loaded span between weight and hinge, y = -—f 28i25o.r - 703125 j. Unloaded side span horizontal ; centre span straight between liinges. 5000000 5ooo.r/ x2 Second. Side span, v = - >rr 5-»' — -7- hl \ 6 2500.V/ .f2\ centre b^an,/ = ->y-(25 ] + \L^ jntre span EXAMPLES. 495 30. A uniformly loaded beam, with both ends absolutely fixed, is hinged at a point dividing the span into segments a and b. Draw curves of siiearing force and bending moment, and compare the strength and stirtness of the beam when the hinge is {a) at the middle point ; {b) at a point of trisection ; yc) at a quarter-span. Also, determine the slope of the segments of these points. IV ^a" + 8rt^' + ib^ Ans. A'l = <«• + b' K: loa a* + 4"':: 6.25 : 3.29 : 2.66. 1/2 Slopes in {a) = — — - ; in (^) = — - — for segment a, E c b h ct>\ / / 23 and = — — --- for segment b ; t f 162 in (c) =. — — -~ for segment a, and ^ '^ E c 176 ^ / / 92 » i: — ^ — for segment b. E c 2,g\ ^ 31. A horizontal beam rests upon two supports and is loaded with a weight fF at a point dividing the span into segments a and b. Find the deflection at this point and the worli done in bending the beam. Ans. IV a-'b* IV a'b' 3 E/[a + b) ' bEI a + b — — X deflection 32. A wrought-iron beam of rectangular settion and 20 ft. span is 16 in. deep, 4 in. wide, and is loaded with o proof load at the centre. If the proof strength is 7 tons per square inch, find the proof deflection and the resilience, E being 12,000 tons. Arts. .029 ft. ; 650 ft. -lbs. 33. Design a wooden cantilever 12 ft. long, of circular section and uniiorm strengtli, to carry a uniformly distributed load of 2 tons, the (oetficitMit of working strength being i ton per square inch. Alst), find the deflection of the free end. Ans. Taking fixed end as origin and 3 being radius in inches at distance x ft. from origin, then i ic'' = 141 12 — .r)'. _ „ . , 697.6 Deflection at end = m. E 34. A girder fixed at both ends carries (2« 4- i) weights ff concen- trated at points dividing the length of the girder into 2/1 + 2 equal divisions. Find the total central deflection. , « + i IV/* Ans. 192 EI 496 THEORY OF STRUCTURES. !.i '.) i: 1 35. A girder 30 m. long has both ends rixed and carries a uniformly distributed load of 5800 k. per lineal metre. Find the deflection and the work of flexure. ^„,. &21^ooo ^_^^ 36. A steel beam of circular section is to cross a span of 15 ft. and to carry a load of 10 tons at 5 ft. from one end. Find its diameter, the stiffness being such that the ratio of viaximutn deflection to span is .00125. £= 1 3,000 tons. Ans. 10.3 ui. 37. Determine the dimensions of a beam of rectangular section which might be substituted for tiie round beam in the preceding ques- tion, the stiffness remaining the same and the coefficient of working strength being 7i tons per square inch. Ans. lui^ — 320. 38. The flange of a girder consists of a pair of angle-irons and of a plate which extends over the middle portion of the girder for a cci tain required distance. Show that the greatest economy of materia! is secured when the length of the plate is t of the span and the sectional areas of the plate and angle-irons are as 4 to 5 (the girder being uniformly loaded). 39. The flange of a uniformly loaded girder is to consist of two plates, each of which extends over the middle portion of the girder fur a certain required distance, and of a pair of angle-irons. Show tiiat the greatest economy of material is realized when the lengths of tlie plates and angle-irons are in the ratio of 12:18: 23, and when the areas of the plates are in the ratio of 4 : 5. What should be the relative lengths of the plates if they are of equal sectional area? Ans. i •.\'lUS2-\- \). 40. An elastic beam rests upon supports at its ends, and a weight placed at a point A produces a certain deflection (d) at a point B. Show that if the weight is transferred to B the same deflecti(jii (V/i is produced at A. 41. A uniform beam is supported by four equidistant props, of which two are terminal. Show that the two points of inflexion in the middle segment are in the same horizontal plane as the props. 42. Find the slope and deflection at the free end of the following cantilevers when bending under their own weight, / being the length, 2b the depth at the fi.xed end, 10 the specific weight, and £■ the coeMicient of elasticity : {a) Of constant thickness / and with profile in the form of a trapezoid with the non-parallel sides equal and of depth 2a at the free end. {b) Of circular section and with profile in the form of an isosceles triangle. nifornily lion and — km. 15 11. and leter, the 3 span is 10.3 in. xr section [ins; ques- f working " = 3-0- 5 and of a r a certain material is e sectional irder beini; isist of !\\o J girder t'C Show tiiat ;ths of liie en the areas ire of equal |nd a weight a point />'. Iction ((/> i^ js. of wliicli the middle lie following the Icniitli, |e coefficient a trapezoid end. Lin i#oscc les EXAMPLES. 497 ((•; Of constant thickness and with profile in the form of a parabola symmetrical with respect to the ;ixis and having its vertex at the fiee end. z^'P 3?c'/ \ AflS.— (0- rr-r; -.-.r ""si lOg , f I -UJi,' '^^rr,. '£' ('-■) b 6(5' 43. Deduce the slope and deflection at the free end — ((/) When the depth 2ii in {a) of the preceding question is «//, i.e., when the profile is an isosceles triangle. (ij Due to a unifcjrmly distributed load of intensity p over the cantilever ((?). Hence, also, deduce the slope and defiection when the depth 2ii is /!/'/. {/ \ Due to a wcitiht H'at the free end of {a). (,!,'i Due to a uniformly distributed load of intensity p upon the cantilever 0). U'P UT AHs.—((i) (^) 2/3'' .\Eb'' 3 pi' \ , /> (ib - a)(b - a) 4 EtKb — ay' ( ' a 2b'' pP \ 4 £/(b ■ 3 /{l 4 A7^' ■ _;-.,3.'l<.g-^~ + a (2^+ lab - a'')(b - a) ) (g) p P Eb't ' 3 _frp 2 E/ib- 10 aW (1 ^ ) log- — r. "\ a 2b'' jb — a 2b' \ 90" 44.. A cantilever of length /, specific weight w, and square in section, aside of tiie section being 2b at the fixed and 2a at the free end, bends under its own weitiht. Find the slope and deflection of the neutral axis at the free end. Hence, also, deduce corresponding results when the cantilever is a regular pyramid. (/; + i^yi>P_ _ {b + 2 a\n number of equal sides, show that the neutral axis is a parabola with its vertex at the point of fixture. 4^^'. The section of a cantilever of length / is an ellipse, the major axis (verihii/) being twice the minor axis. Find the deflectfon at the end a m iti \ 498 r/f/':oA'y or sjw^uctu/^ks. ntiflor II hiiij^Ii" wciglit //',/ Ijciii^; llu" cociricicnt nf wotkiiiii; stic iii;ili and A' tin' liU-HuioiU of cliisticity. / -M7 /V'' \\ ■■'"'■ [7000 /-ir)- 47. A cast-iioii IxMin i>f an iiivottoii 'rseclioii rests upon sii|iiinits 22 ft. apart ; tlu- wcl) is i in. iliicU and 20 in. doep; tlu- tlan^i- is 1,: m. tliirk and u in. widi-; ilic Ix'ain laiiii-s a unifoindy (iislrihuti'd Icjad di 99,ixx) lbs. I'liid tlu' nia.xiiniim dutk-clion, A' i)i'inj; ij.ijjo.ckxj lbs. , ht.s. .Mj J in. ( / — KjdS.o^i. 4S. I'ind the ni.ixiinnni dcfk-ction of a casi-iion canliliAcr 2 1.1. wkIc X 3 in. deep x 1:0 in. long under its own wriglit. A," being 17,920.000 llis. . h/s. \l in. 4'). A girder of uniform stmii^t/iy of length /. breadth /'. aii variable. yh/s. 4 a;/' 50. A seiui-girder of uitifonn s/n-ii^i^t/i, of length /, breadlii /-, and depth f/, carries a weight //'at the free end which produces an inch- stress of /"lbs. at every point of the material. Prove that the ma.ximiiin 4 ( A/|3/ /> \» detlection is ^r.-L ,,% wiien /' is constant and ti variable, and that 3 A \6/// it is twice as great as it would be if the section were uniform tliioii^hout and e(]nal to that at the supjiori. What would be the ma.xinumi deflection if tlie semi-girder were subjected to a uniformly distributed load of w lbs. per unit of length .' .Ins. 51. The neutral a.xis of a symmetrically loaded girder, who.^e .luiiiiciii of inertia is constant, assumes the form of an elliptic or circular arc. Show that the bending moment at any point of the deflected girder is inverselv proportional to the cube of the vertical distance between llu' point and the centre of the ellipse or circle. 52. A vertical row of water-tight sheet piling, 12 ft. Iii!.;li, is supported by a series of uprights placed 6 ft. centre to centre .ind securely ti.\ed at the base. Find the greatest deviation of an ui)rii,'lit from the vertical when the water rises to the iop of the piling. What will the maxiiiuim deviation be when the water is 6 ft. from the top.-' Am. 7.'/'//' 30/:/ 3 1 1 0400 7i'0 , ^^ 7('/>c- , ^ 21S72O LI i'^EI 24A./ tl KXAMPIES. 499 ,(•;• sni>|i<>rls is 1 2 in. 1 Inad oi 1)S. 2 ill. wiiio :o,(H)u llis. ■. \\ ill. :1 (IcpUi ,/, ,(l.)l7.'lllS. VflV l'i>i"i /•; \ ,v-''i I ,/ IS CDIl- e .iioiuciit lircular arc. |il irinler is ^'iwoi'ii iIk' lii^li. is Icentie and ,aii upriii lit "K- What the top : ; 1 8720 53. A vi'iiit:iil ri)W of water tiidit sli'-ct piling, 30 ft. hi)^!', is siipiiorlcfl liv a scries of iipti^;iil.i placed .S It. c(!iilr(! to cetilrc .infl securely (i.xed at tin; liasc, while the upper ends are kept in the vertical by struts slo[)inj3; ill 45 ■ If the water ris<;s t(j the to|) of the pilitij^, liiid («) the thrust on a strut; (b)\.\\M. niaxiinuin intensity of stress in an iipri(.,dit; (c) the amount and position of the inaxiuiiim deviation of an u|)ri}^lit from the vertical. Alls. 45000^/2 lt)S.; max. H. M. — -, and max. intensity of '54/5 I ( 7f'//' r .S7.'//« I ^ . aticss -^ , ' — 1 V — ; deflection is a max. when ./ I .0 / ,5 ^'5 ) .r = 30 , and lis amount 7.V/' .32 4/5 \'% '■' 7S^Vs' v( The pilin;.,' in the preccdinf^ example is strenj^thciicd by a seconfl s.rics of struts sloi)in).( at 45" from tiie points of maximum deviation. Imiu! the normal reactions upon an upright and tiie bcndinj; moment at I'S llli )l. What will be the reactions and bcndinj; moment if the second row of siiiiis starts from the middle of the uprights? .his. .007547.'//-'; .13771'/''; .920277.'//»; ,V()W/^'; ^JJw/^'; Ulw/i\ jy A continuous girder of three spans, the outside spans being wjual. is uniformly loaded. What must be the ratio of the lengths of ilif centre and a side span so that the neutral axis maybe hori/ontal over t lie intermediate supjjorts.' /^«5. 4^3 • ^ ^■ 56. What should the ratio bo if the centre span is hinged (n) at the centre; (/;) at the points of trisection ? /ifts. (a) 1^2 : i ; (/;) 3 : 2 \^2. 57. Four weights, each of 6 tons, follow each other at fixed distances of 5 ft. over a continuous girder of two spans, each equal to 50 ft. If the second and third supports are i in. and 1^ in., respectively, vertically iiclow the first support, find the maximum H. M. at the intermediate sup[K)rt. Ans. (.9855 - ~ ft. -tons. 40000/ 58. A continuous girder of two equal 50-ft. spans is fixed at one of the end supports. The girder carries a uniformly distributed load of 1000 lbs. per lineal foot. Find the reactions and bending moments at the points of support. How much must the intermediate support be lowered so that it may bear none of the load ? How much should the free end be i/ten lowered to bring upon the supports the same loads as at the first ? Ans. Reactions- !3,2[4f. 57,1425, 19,6425 lbs. ; Bending moments — 178,571!, 267,857} ft.-lbs; coo T//EOUV OF SrKUCTU/^ES. m; \t f '> m^' 59. Four loads, each of 12 tons ami spaced 5. 4, and 5 ft. apart, travel in oi'dcr over acotuinuousi^irdur of two spans, tlio otinof 30 and the oilKr of 20 ft. I'lace the wheels so as to throw a maximum B. M. upon the centre support, and tind the correspondinii; reactions. Draw a diagram of B. M., and find the maximum deflection of each span. 60. The loads upon the wheels of a truck, locomotive, and tender, counting in order from the front, are 7, 7, 10, 10, 10, 10, 8, 8, 8, 8 tons. the intervals beinj,' 5, 5, 5, 5, 5, 9, 5. 4, 5 ft. Tl c loads travel over a continuous girder of two 50-ft. spans AB, DC. Place the locomotive, etc., ((0 on the span AB so as to give a maximum B. M. at B\ (/'i so as to give an absolute tiiaxiinum B. M. at B. 61. A continuous girder of two spans /i^, .flC has its two ends,/ and C fixed to the abutments. The load upon AB is a weight /' disiaiu / from A, an'C = /a. The bending moments at A, B, C arc J/i , M~. A/3, respectively. The areas of the bending-moment curves for liie spans AB, BC assumed to be independent girders arcAi, A-i, respect- ively. Show that MJy + M^ih + /a) + M:,h = - 2{Ai + At), and Mt{h + A) = — 2(^,/> + A^q). If /, = /a = /, show that il/a is a tnaxiinuin if 62. A continuous girder of two spans AB, BC rests upon supports at A, B. A uniformly distributed load £"/-' travels over the girder. (/, is the centre of gravity of the portion BE upon AB, and 6^3 iliat of the ^ox\\o\\BF upon BC. If the bending moment at B is a maxtiiiiiin.^hiiw that AE.EB AG, CF . FB ~ CO', ' 63. An eight-wheel locomotive travels over a continuous girder (^f two loo-ft. spans ; the truck-wheels are 6 ft. centre to centre, the loaii upon each pair being 8000 lbs. ; the driving-wheels are 8;^ ft. centie to centre, the load upon each pair being 16.000 lbs. ; the distance centre tu centre between ths front drivers and the nearest truck-wheels is alsn ^\ ft. Place the locomotive so as to throw a maximur"! B. M. upon the centre support, and find the corresponding reactions. 64. If an end of a continuous girder of any number of spans is fixeH. show that the relation between the moment of fixture (J/i) and the bending moment (il/a) at the consecutive support, is 2.I/1 -f J/a = — or 2;l/i -I- J/a = — .j2[ /'/(/ — /)( 2/ — /)], according as the load iip! If ■ ; ! EXAMl'LES. 501 e load upon the span (/) between the fixed end and the const ciiiivc support is of uniform intensity or consists of a number of weights /';. Ft, Pi, . . . concentrated at points distant A, A, A, . . , from tho lixcd end. 65. A continuous girder of two spans AH, BC, carrying a load of tuiilorm intensity, has one end A Jixfti, and the other end rests upon the support at C. If the bending moments at A and B are equal, show that tiie s|)ans are in the ratio of 1^3 to \^2, and find the reactions at the supports, Wi being the load upon ^IB, and W^a that upon BC. Ans. At .-/ reaction = A /f, . " B " = ur, + %ii\. " c " =hv.. 66. A viaduct over the Garonne at Rordeaux consists of seven spans, viz., two end spans, each of 57.371; ni., and five intermediate spans, each of 77.06 ni. ; the main girders are continuous from end to end, and are t;icii subjected to a dead load of 3050 k. per lineal metre. Determine the absolute maximum bending moment at the third support from one end. .\ls ) find the corresponding reactions, the points of inflexion, and the liia.xinuim deflection in the first and second spans. 67. A continuous girder consists of two spans, each 50 ft. in length; the effective ilepth of the girder is 8 ft. If one of the end bearings settles to the extent of i in., find the maximum increase in the fiange and siiearing stress caused thereby, and show by a diagram the change in the distribution of the stresses throughout the girder. (.Assume the section of the girder to be uniform, and take /i = 25,000,000 lbs.) Afis. Increase of maximum B. M. = V/'W' ^ — i )■ \2!67£' / " " shearing force = H^- 70 being weight per unit of length, and /the moment of inertia. 68. A girder carrying a uniformly distributed load is continuous over four supports, and consists of a centre span (/j) and two equal side spans (A). Find the ratio of A to A , so that the neutral axis at the intermediate supports may be horizontal. Also find the value of the laiiowhena hinge is introduced (a) at the middle point of the centre span ; (/;) at the points of trisection of the centre span ; {c) at the middle points of the half lengths of the centre span, 69. In a certain Howe truss bridge of eight panels, the timber cross- tie< are directly supported by the lower chords, and are placed sufii- cientlv close to distribute the load in an approximately uniform inanner over the whole length of these chords, thus producing an additional stress due to flexure. Assuming that the chords may be regiarded as girders supported at the ends and continuous over seven intermediate 1 rnT-r-r- ?1 t i^ I (i I Hm 502 TIfEOKY OF STRUCTURES. supports coincident with the pane! points, and that these panel puint- are in a truly horizontal line, determine (a) the bending moments and reactions at the panel points ; {b) the maximum intermediate bending moments; and (c) the points of inflection, corresponding to a load of r^' per unit of length, / being the length of a panel. Ans. — (a) At ist support ; 2d support; 3d support; B. M. = o ; -sVf-f/'; -3V'.'''^^ Reaction = /?, = )i\\iul\ Ri = i\livl\ /?s = WVwl ; At 4th sup[)(jrt ; 5th support. B. M. = - aVVc//^ 3Ss' Reaction = /^t = iJssw/; /\\ = ats''-'^- (fi) Maximum intermediate B. M = 6600.5 1 1704.5 „. -—riiJi in 1st span; (388,1' ^ 5104.5 ,, . , 6600.5 .„ . = —Jc. M ill 2d ; = — ,:„-^w/' n 3d ; (388)-' (388)' -^ 6208.5 = — T^TT-n'vl'^ in 4th. (' A', = 350 A'o = 55 H- I El Second, i*?! = 1 55 — <. ■^ : A = 35° + R, 55- I EI ' 2 P ' I EI 4 I'' I AY 4 i^' I EI 8 /' £7 /3 Where 18^5 1 1 232 72. Two tracks, 6 ft. apart, cross the Torksey IJridge, and are sup- ported by single-webbed plate cross-girders 25 ft. long and 14 in. deep. If the whole of the weiglit upon a pair of drivers, viz., 10 tons, be directly transmitted to one of these cross-girders, draw the corresponding shear- ing force and bending-moment diagrams (i) if the ends of the cross- girder are yf.itv?' to the bottom flanges of the main girders ; (2) if they merely rest on the said flanges. Find the iiiaxiimim rleflection of the cross-girder and the luorlc done in bending it, in each case. Ans. (I) 7088.45 - at 13.208 ft. from one end. _ , , , „ 67161 6 Total work of flexure = .- ft -tons. 73. A swing-bridge consists of the tail end ///>', and of a span BC of length I ft., the pivot being at />. The balIast-bo.\ oi weight ff extends over a length AD (= zc ft.), and the weight of the bridge from D to />' is Ti' tons per lineal foot. If DB = x. if / is the cost per ton of tiie uridine, and if g is the cost per ton of the ballast, show that the total cost is a minimum when x -f- c _(g( C-c') \^ \ 2/ , and that the corresponding weight of the ballast is wxi— — i J 4- -wr. \9 I Q . ,_....,^s*.L.. M 504 THEORY OF STRUCTURES. 74 Co:npa'C graphically, the shcariiifj forces and bending moments along the span BC ot the bridge in the preceding question when the bridge is closed, with their values when the bridgf is open. What pro- vision should be made to meet the change in the kind oi stress? 75. Each of the main girders of a railway bridge resting upon two end supports and five intermediate supports is fixed at the centre sup- port, is 3 ft. deep throughout, and is designed X.o carry a uniformly dis- tributed dcad\oi\A of \ ton and a live load of A ton per lineal foot. The end spans are each 51 ft. 8 in. and the intermediate spans each 50 ft in the clear. Find the reactions at the supports. 'I'he girders are sini^it;- webbed and double-flanged ; the flanges are 12 in. wide and equal ;n sectional area, the areas ff)r the intermediate spans i)eing 13 sq. in. luid 17 sq. in. at the centre and piers respectively. Find the correspond in;,' moments of resistance and flange stresses, the web being f in. iliick. Ans. Reaction at island 7th supports = iSsonSill '< ^^ 2d and iiii supports = 43'^J5(iJ" ; ''^ 3^ ^"^^ 5^^ supports =3;,';;;:; at 4th support = 38"///^*', tons. Ax. piers— = 693 and flange stresses are 3.59 tons per sq. in. .at 2d support, 2.45 at 3d, and 2.£3 at 4th. At Centre --= 549 and flange stresses in istspan 3.- tons per sq. in., in 2d = 1.3, and in 3d = 1.78. 76. A continuous beam of four equal spans carries a uniformly dis- tributed load of Ti' intensity per unit of length. The second support is depressed a certain distance nsistsof three spans, .//>', />C, fZ?. The two main girders are continuous and rest upon the abutments at ./ and D and upon piers at />' and C. The effective length of each of the spans .IB, CD is 20S ft. 6 in., and of the centre span y)'C'243 ft. The permanent load upon a main girder is 1277 lbs. per lineal foot, and the proof loafl is 2688 lbs. per lineal foot. Find the reactions at the sup- ports (I) when the pr'; (2) when the proof load covers the span BC ; (3) when the proof load cover the spans AB and liC; (4) when tlie proof load covers the whole girder. Draw shearing-force and bending-moment diagrams for each case. (b) At the piers i.ie web is i in. thick and 18 ft. in depth, and each flange is made up of four plates A in. thick and 3 ft. wide. Determine the flange stresses for cases (i j and (3). 1(1 The angle-irons connecting the flanges with the web at the pier are riveted to the former with (J-in. rivets and to the latter with i-in. rivets. How many of each kind are required in one line per lineal foot on both sides of the pier at B, Sooo lbs. per square inch being safe shearing stress? (d) The effective height of the pier at B is 41 ft., its mean thickness is 14 ft. 9 in., its width is 42 ft. 9 in., and it weighs 125 lbs. per cubic foot. If there is no surcharge on the bridge, and if the coefficient of friction between the sliding surfaces at the top of the pier is taken at .15, show that the overturning moment due to the dilatation of the girders is about y'j of the amount of stability of the pier. 5o6 THEORY OF STRUCTURES. « ,T *■ 5 • i< « tt ' is lowered by an amount —r-r.- Find the reactions. How mucii nui.st I> 16A/ be lowered so that the whole of the weight may be borne at .-/? I lel* Ans. {\iifl at A, {^wl at B ; --rr.- EXAMPLES. 507 83. Solve the preceding question supposing the fixture at A to be imperfect, the neutral axis making with the horizontal an angle whose I wl* 7 wl^ tangent is — „ -^. Ans. 84. A wrought-iron girder of I-section, 2 ft. deep, with flanges of equal area and having their joint area equal to that of the web, viz., 48 sq. in., carries \ ton per lineal foot, is loo ft. long, consists of five equal spans, and is continuous over six supports. Find the reactions when the third support is lowered \ in. How much must this support be lowered so that the reaction may be nil at () Ul '"■: W 2 j^ in. ; (d) i/3»rin.; (e) 2,,f^in.: (/) 6i in. 85. If the three supports of any two equal ccjnsecutive spans of a continuous girder of any number of spans are depressed below the horizontal, show that tiie relation between the three bending moments at the supports will be unafTected if the depression of the centre support is a mean between the depressions of the other two supports. 86. A girder consists of two spans AB, BC, each of length /, and is continuous over a centre pier B. A uniform load of length 2a{< / ) and of intensity w travels over ^/i')'. Find the reactions at tiie supports for any given position of the load, and show that the bending moment at , ... , , (I'^l I ''' ' , the centre pifir is a maximum and equal to -[i — -jr,\ when the 3 V'3 V ^7 87. A continuous girder rests upon three supports and consists of two unequal spans AB (= /,), BC (= A). A uniform load of intensity w travels over AB, and at a given instant covers a length .ID (= r) of the span. \l Ri, Ri are the reactions at .1 and C. respectively, show that Rxh' + RJ^' = wrU' - K-/, + ^^ Draw a diagram showing the shearing force in front of the nuwing load as it crosses the girder. 88. If the live load in the preceding question may cover both spans, show that the shearing force at any point Z* is a maximum when .ID and BC are loaded and BD unloaded. Illustrate this iatze graphically, taking into account the dead load upon the girder. centre of the load is at a distance 3 V'3 from /i. 11 ^^%.4|^;5jj I Vn .1 ■n i 508 THEORY OF STRUCTUKES. 89. A continuous-girder bridge has a centre span of 300 ft. and two side spans, each of 200 ft. The dead load upon each of the main girders is 1250 lbs. per lineal foot. In one of the side spans there is also an additional load of 2500 lbs. per lineal foot upon each girder. Finfl ilie reactions and points of inflexion. How much must the third suppcjii from the loaded end be lowered so that the pressure upon it may be jusi zero } Ans. Let IV = weight on loaded span = 750,000 lbs. i^. = iVA ^ lbs. ; A', = i §f I IV lbs.; ^. = iVW ^^ lbs. ; A'. = iVs'e ^^ lbs. Af, = - i!,V/ ^y ft. -lbs.; A/» = - V/fi' ^^ ft. .s. Distance of point of inflexion in loaded span from nearest end support = i62i5',' ft. Distance of point of inflexion in unloaded end span from nearest end support = 145^? ft. Distance of point of inflexion in intermediate span from end support in unloaded span is the value of .r in the equation .r* — '|','".i- + ^5^5"^^ = o. , , , ,. 56350000 fF 3d support must be lowered a distance = —. 87 E/ 90. A continuous girder AC consists of two equal spans AB, BC, eacii of length /, and carries a uniformly distributed load of intensity icn upon AB, and of intensity 7£'a upon BC. Determine the bending moments at the supports, the maximum intermediate bending moments, and the re- actions {a) when both ends of the girder are fixed; {b) when one end ./ is fixed and the other free. Ans. Denoting tiie reactions and bending moments at A, B, C by Bi , Af, , A'a , Afi , A's , Ma , respectively : /' /' {a) Afi = -- (— 574y, + Wa) ". ^J/a = (wi + Wj); 48 24 /" /? ' Ml =-i(wi — 5Wa); Af,„ax. in AB = -— + Afi , in BC 48 27l>i = — + Af, 2Wi ^'-76^- /?, = -(9W1 10 Wt); Bi= -(wj+wj); ■7V: + gWa). (6) Mi = -(30/1 — Wa); Afi = 28 28 {wi + 2Wi); Aft = o ; ATmax. in AB = — -V Afx, in BC = — ; 27£/i 2W, / / /?, = -—(16a/, — 3Wa); A', = -r-(l3Wi + l9Wa); 28 28 /?, = —A—W, + I27£/a). 20 EXAMPLES. 509 91. In the preceding question, if lOx = Wq = «/, find the points of in- flexion and tlie maximum deflection in each case and for each span. Ahs. — (a) Points of inflexion for //// or BC are given by e.v" - 6.1V + /» = o. Max. deflection for AJ>' or J>'C is given by , -E/y= (a/.v-.r'-/"). 24 in wliich the value of .v is found from !.r' - 3/.r + / = o. (d) Points of inflexion in AB are given by 14.1" — i3.r/ + 2/" = o, and ;n BC by .r = ^^/, Max. deflection for ^IB is given ijy -y^/v. 7t7.t-' 168 , (i3/.r- 6/' -7.r'), and 28a-' — 39/.r + 12/" = o. Max. deflection for BC is given by 4/'), and 28^' -33-1-' 4- 4/' = o. 92. A continuous girder AC consists of two equal spans AB, BC of 15 m. each. Determine the bending moments at the supports, the maxi- mum intermediate bending moments, and the reactions (a) wiien the load upon each span is 3000 it. per metre ; {b) wh(;n the load per metre is 3000 iv. upon y//j' and 1000 1<. upon BC. Call J/i , .I/a, J/s the bending moments and Ri , A'a , A'a the reactions at A, /i, C, respectively, and con- sider three cases, viz., when both ends of the girder arc free, when both ends are fixed, and when one end is free and the other fixed. Alts. —Case I : (a) Ml =0 = Mi ; Mi = — 84375 km. ; M„,ax. in AB or BC = 47460.9375 km. /?, = A'a = 16875X' ; A\ = 56250/(-. {b) Mt — o = J/a ; Afj ~ — 56250 km. ; Mmax. in AB = 58593.75 km., in /)'t' = 7031.25 km. /?, = 18750 k. ; A'a = 37500 k.; A\ — 3750 k. Case II; (a) M: = M, = M, ■■ = 28125 km. A" — 56250 km.; Mmax. in AB or BC /?, = - = A'a = 22500 k. III H i M "I;' h I 510 THEORY OF STRUCTURES. 'n • {b) M\ = — 65625 km. ; Af, = — 37500 km. \ Ms = — 9375 km. Mmax. in .4// = 33398.4375 km., in BC = 6445.3125 km.; Ri = 24375 k.; A'a = 30000 k.; R3 = 5625 k. Case III : {a) Ml = — 48214? km.; JAj = — 72321^ km.; J/:, = o. M,„a.v. ill An -— 24537^5^ km., in BC = 52088}^'^ l' if the two ends are fixed. Find the corresponding bending moments at the central pier. PI 2 Ans. ---; -PI. 6V3 =7 95. A girder with both ends fixed carries two equal loads W^'at points dividing the girder into segments a, b, c. Determin'e the reactions and bending moments at the supports. Ans, 2rt' + (iii'b 4- yib"^ 4- <^' + 6rtV 4- dabc + 3<^V R-, = IV Ml = IV M-, = W- {a + b + o* 2rtV 4- 2abc 4- be'' 4- ab\ {a + b + c-y ' lac' 4- 2abc 4- (I'b 4- b'^c () under the same dead load together with an additional prcjof load of 2000 k. per lineal metre on one span. Tlie depth of the girder = 3.228 m., and /= .093929232444. Ans. — (ti) 14.85 m.from the abutments ; S3308.5 kilogrammetres (km.); 46,8613'j km.; 1.4315 k.persq. mm. {/>) 16.18 m. from abutment on loaded side; 11.876 m. from abutment on unloaded side;. 132313.5 km.; 101991. 65625 km.; 2.27356 k. per sq. mm. 99. The Estressol viaduct consists of four spans of 25 m. ; the main girders are continuous and tiieir ends rest upon abutments; the dead load upon each girder is 1700 k. per lineal metre. Determine the position of points of inflexion in each span, the reactions and bending moments at the supports when an additional load of 2000 k. per lineal metre crosses (n) the ist span; (/^) the 1st and 2d spans; (c) all the spans. Also, find the absolute maximum bending moments at the inter- mediate supports. Ans, Call .ti , Xi .va, .w the distances of points of inflexion in 1st, 2d, 3d, and 4th spans from the ist, 2d, 4th, and 5th supports, respectively; R\, R-i, A'a, R^, Rt, the reactions; J/i ('=0). J/a, Mt, Ah, J/s (=0) the bending moments. (a) ,r, = 20.72 m. ; xj is given by 1700X2" — 4240174x3 + 395089^ = o; x% by 1700.1-3" — 47767f.r.i 4- 238839^ =0 ; .1-4 = 19.38 m. Rx = 38348V'4 k-; ^i = 8n6of k. ; /?» = 34107+ k.; Rt = 49910?^ k. ; Rt = i6473y\ ^■ M.,= i97544i-^~ km. ; Ah = 535717 km. ; Mt= 4776^4 km. ; 1 : i mm lid U-,H 11 ! II I 512 THEORY OF STRUCTURES. (d) .1-1 = 19.556 m. ; 3700.1,' — losooo.ra + 503571!,' - o; I700.r3''' — 41071^.13 4- 205357I = o; .r. = 20. 1(6 111. A', = 36178T k. ; A\ = 107821^^ k.; A'3 = 629645 k.; A'* = 45^92? k.; A\ = 17142! k. Mi= 258928f km. ; M3 = 120535^ km.; Mt— 102678^ km. (fi) xi = 19.64 m. ; .t'a and .vj are given by i4-i'-375-' + i«75 = o. J/j= A/i = 247767^ km. ; J/3 = 165178;} km. Abs. max. B. M. at 2(1 support (= max. H. M. at 4th sup- port) occurs when 1st, 2(J, and 4tli spans are loaded, and — 264508}! km. Abs. max. B. M. at 3d support occurs when 2d and 3d spans are loaded and = 2098211' Urn. 100. In the preceding question find the iibsohite maximum Hani^'c unit stress at the piers, / being .093929232444. ^/is. 4.5 k. per sq. mm. 101. The Osse iron viaduct consists of seven spans, viz., two ciui spans of 28.8 m. and five intermediate spans of 38 m. ; each main girder is continuous and carries a dead load of 1450 k. per lineal metre. Find the bending moments at the supports when a proof load of 2250 k. per line;d metre for each girder covers all the spans; and also find the absolute maximum bending moment at the fourth support. Is the following section of sufficient strength ? — two equal flanges, each com- posed of a 6oo-mm. x 8-mm. plate riveted by means of two loo-nim. X loo-mm. X 12-mm. angles to a 6oo-mm. x lo-mm. vertical web plate and two 80-mm. x 8o-mm. x ii-mm. angles riveted to each horizontal plate with the ends of the horizontal arms 15 mm. from the edges of the plates; the whole depth of the section being 4.016 m., and the dis- tance between the web plates, which is open, being 2.8 m. If insuf- ficient, how would you strengthen it.-" Ans. J/a = 416,518 km.; J/s = 452,790 km.; Mk = 443.722 km. Max. B. M. = 542,199 km. / = .14074440467. .•. — = .07009183, and max. fiance stress = -^ = 7.73 k. per sq. mm. I This is much too large. The section maybe strengthened by adding two 6oo-mm. x 8-mm. plates to each flan.i;e. / is thus increased by .0783425536, and the flange unit stress becomes 5 k. per sq. mm. ':1P CHAPTER VIII. PILLARS. 1. Classification. — The manner in which a material fails under pressure depends not merely upon its nature but also upon its dimensions and form. A short pillar, e.g., a cubical block, will bear a weight that will almost crush it into powder, while a thin plank or a metal coin subjected to enormous com- pression will be only condensed thereby. In designing struts or posts for bridges and other structures, it must be borne in mind that such members have to resist buckling and bending in addition to a direct pressure, and that the tendency to buckle or bend increases with the ratio of the length of a pillar to its least transverse dimension. Hodgkinson, guided by the results of his experiments, divided all pillars with truly flat and firmly bedded ends into three classes, viz. : (A) Short Pillars, of which the ratio of the length to the diameter is less than 4 or 5 ; these fail under a direct pressure. (B) Medium Pillars, of which the ratio of the length to the diameter exceeds 5. and is less than 30 if of cast-iron or tim- ber, and less than 60 if of wrought-iron ; these fail partly by crushing and partly by flexure. (C) Long Pillars, of which the ratio of the length to the diameter exceeds 30 if of cast-iron or timber, and 60 if of \vroui,fht-iron ; these fail wholly by flexure. 2. Further Deductions from Hodgkinson's Experi- ments. — A pillar with both ends rough from the foundry so that a load can be applied only at a few isolated points, and a pillar with a rounded end so that the load can be applied only •■'■"-/ , .. --.::- • ■' ■■.■ 513 n: 1 { i ^ 514 THKORY OF STRUCTURES. along the axis, are each one-third of the strength of a pillar of class B, and from one-third to two-thirds of the strength of a pillar of class C, the pillars being of the same dimensions. The strength of a pillar with one end flat and the other round is an arithmetical mean between the strengths of two pillars of the same dimensions, the one having both ends flat and the other both ends round. Disks at the ends of pillars only slightly increase their strength, but facilitate the formation of connections. An enlargement of the middle section of a pillar sometimes increases its strength in a small degree, as in the case of solid cast-iron pillars with rounded ends which are made strongei- by about onc-scventh ; hollow cast-iron pillars are not affected. The strength of a disk-ended pillar is increased by about one- cis;hth or one-ninth when the middle diameter is lengthened by 50 per cent., but for slight enlargements the increase is imper- ceptible. The strength of hollow cast-iron pillars is not affected by a slight variation in the thickness of the metal, as a thin shell is much harder than a thick one. The excess above or deficiency below the average thickness should not exceed 25 per cent. 3. Form. — According to Hodgkinson, the relative strengths of long cast-iron pillars of equal weight and length may be tabulated as follows : (rt) Pillars with flat ends. The strength of a solid round pillar being lOO, " " " square " is 93; *' " *' triangular " is 1 10. (/;) Pillars, with round ends, i.e., ends for hinging or pin connections. The strength of a hollow cylindrical pillar being 100. •^ '• " " an H -shaped " is 74.6: *' " a 4-shaped " is 44.:: The strengths of a long solid round pillar with flat ends, and a long hollow cylindrical pillar v/ith round ends, are ap- proximately in the ratio of 2.3 to I. The stiffcst kind of wrought-iron strut is a built tube, the •1" : 1 ■>?'" . pillar of igth of ;i ions. the other IS of two 1 ends flat ease their sometimes ,se of suVul le stronger ot affected. ^' about ("'('• igthened by ^se is imper- iffccted by a thin shell is 1- deficiency per cent. ve strengths gth may be lOO, 93-. no. Iging or pin |r being lOO. is lA-^'- is 44'-- Ith flat eivis, [nds, are ap- THE FAILURE OF PILLARS. 515 m Fig. 335. section consisting of a cell or of cells, which may be circular, rectangular, triangular, or of any convenient form. In experimenting upon hollow tubes, Hodgkinson found that, other conditions remaining the same, the circular was the strongest, and was followed in order of strength by the square xwfour compartments i+1 ; the rectangle \x\ /wt? compartments, rr I : the rectangle, o ; and the square. The addition of a diaphragm across the middle of the rect- angle doubled its resistance to crippling. 4. Modes of Failure. — The manner in which the crush- ing of short pillars takes place depends upon the material, and the failure may be due to splitting, bulging, or buckling. (a) Splitting into fragments is characteristic of such crys- talline, fibrous, or granular substances as glass, timber, stone, brick, and cast-iron. The compressive strength of these substances is much greater than their tensile strength, and when they fail they do so suddenly. A hard vitreous material, e.g., glass or vitrified brick, splits into a number of prisms (Fig. 335). A fibrous material, e.g., timber, and granular materials, e.g., cast-iron and many kinds of stone and brick, shear or slide along planes oblique to the direction of the thrust, and form one or more wedges or pyramids (Figs. 336, 337, 338). Sometimes a granular or a crystalline substance will sud- denly give way and be reduced to powder. (/') Bulging, i.e., a lateral spreading out, is characteristic of blocks of fibrous materials, e.g., wrought iron, copper, lead, and timber, and fracture occurs in the form of longitudinal cracks. All substances, however, even the most crystalline, will bulge slightly before they fail, if they possess some degree of toughness. (() Buckling is characteristic of fibrous materials, and the resistance of a pillar to buckling is always less than its resist- ance to direct crushing, and is independent of length. Fig. 336. Fig. 337. Fig. 338. rlt tube, the H Thin malleable plates usually fail by the bending, pucker- - f. i' 5i6 THEORY OF STRUCTURES. 'Si ! ing, wrinkling, or crumpling up of the fibres, and the same phenomena may be observed in the case of timber and of long bars. >'-a ■ .- ■ ■ .-• "-• ,■ Long plate tubes, when compressed longitudinally, first bend and eventually fail by the buckling of a short length on the concave side. The ultimate resistance to buckling of a well-made and well-shaped tube is about 27,000 lbs. per square inch section of metal, which may be increased to 33,000 or 36,000 lbs. per square inch by dividing the tube into two or more compart ments. A rectangular wrought-iron or steel tube offers the greatest resistance to buckling when the mass of the material is con- centrated at the angles, while the sides consist of thin plates or lattice-work sufficiently strong to prevent the 'bending of the angles. Timber offers about twice the resistance to crushing when dry that it does when wet, as the presence of moisture dimin- ishes the lateral adhesion of the fibres. 5. Uniform Stress.— Let a short pillar be subjected to a pressure of IV lbs. uniformly distributed over its end and acting in the direction of its axis. Let .5) be the transverse sectional area of the pil- lar. W-' W Let/>= -^ be the intensity of stress per unit of area of any transverse section AB. Let A'B' be any other section of area S', in- clined to the axis at an angle 6. The intensity of stress per WW.,, /sin 6, which may be Fig. 339. unit of area oi A'B' = -— = sin 6^ S' S resolved into a component/ sin" 6 normal to A'B', and a com sin 2ff ponent/ sin cos 6, i.e.,/ -, parallel to A'£ . The last intensity is evidently a maximum when 6 = 45°, so that the plane along which the resistance to shearing is least, and there- fore along which the fracture of a homogeneous material would tend to take place, makes an angle of 45° with the axis. li^'i; UNIFORMLY VARYING STRESS, 517 None of the materials of construction are truly homo- geneous, and in the case of cast-iron the irregularity of the texture and the hardness of the skin cause the angle between the plane of shear and the direction of the thrust to vary rom 32° to 42°. Brick chimneys sometimes fail by the shear- ing of the mortar, the upper portion sliding over an oblique plane. ' ■ Hodgkinson's experiments upon blocks of different mate- rials led him to infer that the true crushing strength of a ma- terial is obtained when the ratio of length to diameter is at least i^; for a less ratio the resistance to compression is un- duly increased by the friction at the surfaces between which the block is crushed. 6. Uniformly Varying Stress. — The load upon a pillar is rarely, if ever, uniformly distrib- uted, but it is practically sufficient to assume that the pressure in any transverse section varies uni- jornily. Any variable external force ap- plied normally to a plane surface AA of area 5 may be graphically represented by a cylinder AABB, the end BB being the locus of the extremities of ordinates erected upon A A, each ordinate being pro- portional to the intensity of press- ure at the point on which it is erected. Let P be the total force upon A A, and let the line of its resultant intersect A A in C; Cis the centre of presstire of A A, and the ordinate CC necessarily passes through the centre of gravity of the cylinder. Again, the resultant internal stress developed in AA is P, and may of course be graphically represented by the same cylinder yi/4v95. , " # ' - Assume that the pressure upon A A varies uniformly ; the surface BB is then a plane inclined at a certain angle to AA. Fig. 340. k B,U ■■ riir *1' i 'i !»il 518 THEORY OF STRUCTURES. Take (?, the centre of figure of AA, as the origin, and AA as the plane o{ x^y. ' . Let Y, the axis of jy, be parallel to that line ££ of the plane BB which is parallel to the plane AA. Through EE draw a plane DD par- allel to AA, and form the cylinder A ADD. The two cylinders ^^i5^ and AADD are evidently equal in volume, and Of, the average ordinate, represents the mean pressure over AA ; let it be denoted byA- At any point R of the plane AA, erect the ordinate RQP, intersecting the planes DD, BB, in Q and P, respect- ively. ^"'' 341- Let x,y be the co-ordinates of R. The pressure at R = p = PR = PQ + QR = PQ-\.OF=ax+p^ > a being a constant depending upon the variation. Note. — The sign of x is negative for points on the left of 0, and the pressure at a point corresponding to R xs p^ — nx. Let x^, y^ be the co-ordinates of the centre of pressure 6. Let AS be an elementary area at any point R. Then pAS is the pressure upon AS, and 2{pJS} is the total pressure upon the surface AA, 2 being the symbol of summation. Hence, x,2{pAS) = 2(pxAS), and y,S(pAS)=^ ■ Butp=p^-^ax. ; • • .-. x,2{{p, + ax)AS\ = 2{(p,x + ax*)AS} o). and .y.^KA + ax)AS\ = 2\{p,j> + axy)AS\. UNIFORMLY VARYIiVG STRESS. 5'9 Now is the centre of figure of A A, and therefore "S^ixAS) and 'H^^yAS) are each zero. Also, ^X'^'^) = -^f ^X-*'"^-^) is the moment of inertia (/) of A A with respect to OY, and "^{xyAS) is the product 0/ inertia [K) about the axis OZ. .'. x„p,S= aI=x^P 0) and yJ,S=^aK = y,P. (2) Cor. I. In any symmetrical section y^ is zero, and x, is the deviation of the centre of pres- sure C from the centre of fig- ure 0. Let x^ be the distance from of the extreme points A of the section. |A A —9- X » "C Fig. 34a. The greatest stress in A A is/^ -+- ax^ =/, , suppose. Buta=^^,by eq. (1). • •AH 7 — —Pi* or A P. i + ^S (3) It is generally advisable, especially in masonry structures, to limit x„ by the condition that the stress shall be nowhere negati'i'e, i.e., a tension. Now the minimum stress is/, — ax^ , so that to fr '1 this condition, />f > or = ax, . But/, = ax, -\-p, ; .-. /, < or = 2/». ii ■'^^tiiujA^i^fc' a i! ij^ pillars ; /being the ultimate crushing strength in tons per square inch ; 6" being the sectional area of the pillar in square inches. Again, if the ends of a cast-iron pillar are rounded, the above formulae may be still employed to determine its strength, A being 14.9 for a solid and 13 for a hollow pillar. ( i 522 THEORY OF STRUCTURES. 'iK \\\ 8. Gordon's Formula for the Ultimate Strength of a Pillar. — The method discussed in the preceding articles, being practically very inconvenient, is not generally used, and the present article will treat of Professor Gordon's formula, which has a better theoretical basis and is easier of application. The effect of a weight f^upon a pillar of lengtli / and sectional area S may be divided into two parts : {a) A direct thrust, which produces a uniform com- . W pression of intensity -rr .— p^ . (^) A bending moment, which causes the pillar to yield in the direction of its least dimension (/;). Let J/ be the greatest deviation of the pillar from the vertical. The bending moment M at the point of maximum stress may be represented by Wy. Let /, be the stress in the extreme layers due to this bend- ing moment. Now M=^-^I = txpJ>h\ c being the distance of the layer under consideration from the neutral axis, /< a constant depending upon the sectional form, and b the dimension perpendicular to the plane of flexure. .*. yw////' = Wy, and p^ or < 28.5. er APPLICATIONS OF GORDON'S FORMULA. 525 ilts, and (is. ISI5 "s'li) soil ) Ti^M 3 ^5STi 3 SB^Oii JTi'il TTOS •ss'di! TTrtii Huff K) Ts\is XJ irb DO uJfi DO ISU 1 ends, 4(? d and one Strength /I ild-steel the corre- lurves are exceeds cast-iron lor wcakei 80,000 lbs, 67,300 lbs, ClOfO lbs, II. Application of Gordon's Formula to Pillars of other Sectional Forms. In any section whatever, the least transverse dimension for Ccilculation (i.e., h) is to be measured in the plane of greatest ricxnre. Thus, it may be taken as the least diameter of the rectant;le circumscribing ice (Fig. 345), chaiuiel (Y\^. 346\ and cruciform I Fig. 347) sections, and as the perpendicular from the angle to the opposite side of a triangle circumscribing rtz/^V^^l-'ig. 348) sections. 7i, i f-«-| •^ K iJ 'S l.< --, Fig. 345- Fig. 346. Fir, 3t7- Fig. 348. From a series of experiments upon wrought-iron pillars of I these sections, /"was found to be 42,e;oo lbs., and n, 900 In cast-iron struts of a cruciform section / — 8o,oCK) lbs. and a — inii 400 :ii ^26 THEORY OF STRUCTURES. Ni i These results are only approximately true, and apply to pillars fixed at both ends. 12. Rankine's Modification of Gordon's Formula.— The factor a in Gordon's formula is by no means constant, and not only varies with the nature of the material, with the length of the pillar, with the condition of its ends, etc., but also with tiic sectional form of the pillar. The variation due to this latter cause may be eliminated, and the formula rendered somewhat more exact, by introducing the least radius of gyration instead of the least transverse dimension. If k is the least radius of gyration, • mass nbh n ' m and n being constants which depend upon the sectional fo»-m, Thus, Gordon's formula for pillars with square ends may be written W = A = / /»» in which a^ is independent of the sectional form, all variations of the latter being included in k^. This modified form of Gordon's formula was first suggested by Rankine. 4^', is substituted for «, if the pillar has two pin ends, and -a^ or 2rt, is substituted for a, if the pillar has one pin end and one square end. Rankine gives for wrought iron, /= 36000 lbs., for cast-iron, '/= 80000 lbs., for dry timber, /= 7200 lbs.. I I = 36000; = 6400; = 3000; FAN KIKE'S MODIFICATION OF GORDON'S FORMULA. 527 In good American practice the safe working unit stress in bridge compression members is determined by the formula Safe vvorkiner unit stress = /' f being 8000 lbs. for wrought-iron and 10,000 lbs. for steel, and - being 40,000 for two square ends, 30,000 for one square and one pin end, and 20,000 for two pin ends. Another formula often employed is. Working stress in lbs. per sq. in. X (4 + ^) = i -\-xH' ' If bt-'ing the ratio of length to least breadth, where, in the case of wrought-iron, /' /' /' 38,500 lbs. and - 38,500 " " 37,800 " " 5820 for two square ends; 3000 " one square and one pin end. — igoo " two pin ends. Tht factor of safety, viz., 4 -j , increases with H, and par- tially provides for the corresponding decrease in the strength to resist side blows. Examples. — According to Rankine the ultimate compres- sive strength of wrought-iron struts, in pounds per square inch, is 36000 ■ 1 + I r 36000^ m I ill :.^i^.Jtf!iU-v luil • : m w 111 it i i V 528 THEORY OF STRUCTURES. h' If the section is a solid rectangle, 1^ — — , and hence 36000 A = ' 3000 /[' le If the section is a solid circle, k^ = -z, and hence 36000 A = 1 + I /'• 2250 /t" le If the section is a thin annulus, k^ = -^, nearly, and hence 36000 A = 1 + I /' 4500 A' I . Cor, — If T is small, W ■=■ fS. 1 f^h^ fT If ^ is large, {^= — .-=_. Comparing the last result with eq. (5), Case 4, Art. 16, I ^Eit' tty f ' which gives a theoretical value of a, , the actual value being somewhat different. 13. Values of h^ for Different Sections. / A' {a) Solid rectangle : 1^ ■=—■=. —, h being the least dimen- O 12 sion. {b) Hollow rectangle: k* = -^ = — f , . _ ,/./ ), b, h being the greatest and least outside dimensions, and b\ h' the great* est and least inside dimensions, respectively. m ;■' ! VALUES OF k* FOR DIFFERENT SECTIONS. Let / be the thickness of the metal. Then b' t=b — 2t and li = h — 2t, 529 and hence I bh' -{b- 2t){h - 2ff h' zb-\-h ^' ~ 12 bh-{b- 2t){h - 2t) ~ 12 b-\-/? approximately, when / is small compared with /i, i.e., for a i/u'n hollotv rectangle. For a square cell, k' = -^. (c) Solid triangle : k^ = -^, = — ^, k being the height. (d) Hollow triangle : k" = ^^ = -^ , . _ ,,., ,b, h being the base and height of the outside triangle, and b\ Ji! the base and b h height of the inside triangle, respectively. Also, -r, = -,-. : [ •i ' I Hence, for a t/tin triangular cell, k' = — . / li' (e) Solid cylinder : k' = - = —^, h being the diameter. / n (/) Hollow cylinder : k' = T. = ~fi(^'' "^ ^''' ")' ^^ '^"^ ^^' '^^'"g the external and internal diameters, respectively. Hence, for a t/nn cylindrical cell, k'' = -—, approximately. o Example. — Gordon's formula for hollow cylindrical cast- iron pillars is W = A = / / ^ ' 500 /:' ^ ' 4000 k* 'i.aci^f I .^ UtM^^ MiiiM lU !*M !! ^530 • "theory OF STRUCTU^AS. The relation /, = — 77 may be assumed to hold for ' 4000 k^ hollow square struts and also for struts of a cruciform section. Ex. I. For a hollow square having its diagonal equal to the internal diameter of the hollow cylinder, i.e., h', f 6 =77' ^"'^ f^=-~~iri" 1 + 1000 //'* 1 -fl ir Ex. 2. If the side of the square is equal to the external diameter, i.e., h, then '^^ ~ "6 ' ^"^ ^' / 1 + 3 /'• 2000 h^ (g) Cruciform section, the arms being equal : V b ^1 I Fig. 349. , M' , hb' b' ^ ,. ,, /_ L — ; 5 = 2bh — fV. 12 '12 12 I bh' + hb'-b' h' .*. ')' I • {k) H-iron, breadth of flanges being b, length of web k, and thickness of metal / : , b't , he b't , o z , r 7=2 — H = 2 — , nearly ; 5=2W + ^/. 12 ' 12 12 "^ ' •. /fe' = - 2(5/ \2 2bt-\-ht 12A + B' A being the area of the flanges, and B the area of the web. (/) Circular segment, of radius r and length rO : k* = r* 1 sm t' ^ 2 2 + "2^ 6*^^ Hence, for a semicircle, since 6 =z tt, ,&» = r« { - - ^ I = ^ , nearly. ^jSj li '■ <■ If I' V M ;|::!l 'Hi liti •i ■,! ': f if i I II 532 THEORY OF STRUCTURES. (;«) Barlcnv rail: k' =■ — , nearly. (») Two Barlow rails, riveted base to base : k* = 'ijlr\ nearly. 14. American Iron Columns. — In 1880 Mr. G. Bouscaren read before the American Society of Civil Engineers a paper containing the results of a series of experiments made for the Cincinnati Southern Railroad upon Keystone, square, Phoenix, and American Bridge Co.'s columns. KEYSTONE Fig. 351. AM. BRIDQE CO, SQUARE PHOENIX F>G- 352. Fio. 353. Fig. 354. These experiments show, as those of Hodgkinson and others have also shown, that the strength of iron and steel columns is not only dependent on the ratio of length to diam- eter, and on the form of the cross-section, but also on the proportions of parts, details of design and workmanship, and on the quality of the material of which the columns are con- structed. Further, they seem to lead to the conclusions that Gordon's formula is more correct as modified by Rankine, and that, in the case of columns hinged at both ends, Rankine's formula, with rt, assumed at double the value it has when the formula is applied to columns with flat ends, is practically correct. The subjoined table gives the values of the constants a^ and f as deduced from Bouscaren's experiments by Prof. W. H. Burr. In 1 88 1 Messrs. Clarke, Reeves & Co. presented to the American Society of Civil Engineers a paper containing the results of experiments upon twenty Phoenix columns, which appeared to show that neither Gordon's nor Rankine's formula expressed the true strength of a column of the Phoenix type. In the di.scussion that followed the reading of this paper, how- ever, it was demonstrated that, within the range of the experi- ments, the strength of intermediate lengths and sections of AMERICAN IRON COLUMNS. 533 = .393'-'. ouscaren a paper e for the Phoenix, g i. " • .'^ kM. BRIDQECO. FiC. 354- inson and and steel h to diam- Iso on the .nship, and IS are con- It Gordon's ind that, in s formula, formula is ;ect. constants s by Prof. ted to the gaining the ins, which ■e's formula Icenix type. ^aper, how- [the experi- Isections of For keystone columns with flat ends — swelled " " " " " —straight (open or closed) •' " " " " " —open (swelled straight) " " " " pin ends — swelled For square columns with flat ends " " pinends For Phoenix columns with flat ends " " " " round ends " pinends For American Bridge Co.'s columns with flat ends " " " round ends.... " " " " pinends / in lb>. 36,000 39.500 38,300 38.300 39,000 39,000 42,000 42,000 42,000 36,000 36,000 36,000 "I msoo issoo 1 mooo 1 ttooo 1 17000 \ (0000 1 11600 1 SITOO 1 4«000 1 11*00 1 titoo Phoenix columns can be obtained either from Rankine's for- mula by slightly changing the constants, or from very simple new formulae. I Mr. W. G. Bouscaren showed that by making «, = W lOOOOO and/= 38000, the calculated values of -^ agree very nearly with the actual experimental results. Mr. D. J. Whittemore gave the following (only applicable for lengths varying from 5 to 45 diameters) as expressing the probable ultimate strength of these columns : W^ lbs. = (1200 — H)io-\- 525000 //being the ratio of length to diameter. Mr. C. E. Emery stated that the ultimate strength in each case is approximately represented by the formula ^Tlbs. = 355063 + 30950// i^+6.175 //being the ratio of length to diameter. i r ii ; 1 In 1 II aia.:>^'';';: S^'ji^i S84t THEORY OF STRUCTURES. Taking the different values of // as abscissai, and of W as ordinates, this is the equation of an hyperbola. It agrees very accurately with the experimental results from 20 diameters upwards; at 15 diameters the calculated values of W are greater than those given by the exf oriments ; for a less num ber of diameters the experimental results are the higher, but the variations are slight, and are provided for in the factor of safety. The following very simple formuL-e, due to Prof. W. H. Burr, give results agreeing closely with those obtained in the experiments : / For values of -r ^ 30, the ultimate strength in pounds per square inch = 64700 — 4600, \fi- For values of -r between 30 and 140, the ultimate strength in pounds per square inch = 39640-46^, k being the radius of gyration. 15. Long Thin Pillar. — Let ACB be the bent axis of a ,B thin pillar of length /, having two pin ends and carry- ing a load W at B. Let d be the greatest deviation of the axis from the vertical. Then E Wd — bending moment = -h/, . • (0 — being the curvature of the pillar and /the moment of inertia of the most strained transverse section. This equation is only true on the assumptions that— (i) initially, the pillar is perfectly straight; (2) initially, the line of action of the load coincides with the axis of the pillar ; wm LONG THIN PILLAR. 535 (3) the material of the pillar is homogeneous. These assumptions cannot be fulfilled in practice, and varia- tions from theoretical accuracy may, perhaps, be provided for by supposing that the line of action of the load is at a small distance x from the axis of the pillar. The bending-moment equation then becomes ^n^+^) = |/=7A (2) /, being the skin stress due to bending at a distance c from the neutral axis. Again, assuming that the bent axis is in the form of an arc of a circle, I U . . R^-P (3) .'.W{d^x)=.%EIj^^^p, .... (4) and consequently Wx ^^-p~Zw ^5) where p='^ (6) If the line of action of the load W coincided with the axis of the pillar, then x would be nil. Hence, by eq. (5), so long as the load is less than /', d = o, and the failure of the pillar would be due to direct crush- ing. If the load is equal to P, d? would become indi terminate (=-) and the pillar would remain in a state of neutral equi- librium at any inclination to the vertical. It is impossible that W should exceed P, as d would then be negative ; and therefore a load greater than P would cause the pillar to bend over laterally until it broke. -' W ij» ^ VA] Mi f 1' 1 1 f n r' ^ ML. V 536 THEORY OF STRUCTURES. Thus, P — -yr must be the theoretical maximum compres- sive strength of the pilar. Again, let A be the area of the section under consideration ; " p be the total intensity of the skin stress at the section ; " f be the intensity of the direct stress due to W _W_ ~ A ' " /", be the intensity of the stress due to P __ P_ ~A' Then P=f±A = ^±W{d+x)j (7) the sign of /, being positive for the coinpressed side of the pillar and negative for the side in tension. .•./ = -^'(i±(«r+^)4^)=/(i±(^+^)|5), '. (8) k being the radius of gyration. Let h be the least transverse dimension of the section in the plane of flexure. Then c . i i |! ' i |; :1 ' 1 i. |lK H ;■ H i |»W i IH ■] 1 1 m ; H i 538 THEORY OF STRUCTURES. Also find the deviation (-r) of the line of action of a load of 20,000 lbs. from the axis of the pillar, sc \hat the maximum intensity of stress may not exceed 10,000 lbs. per square inch. By Gordon's formula and the table, page 524, the crushing load _ 672oo;r . | I _i_ 4 /izov 85292.3 lbs. Again, the theoretical maximum compressive strength P SEI 8 X 28000000 7r(3)* /' (120)' 64 = 61875 lbs. /. P- w-f,-f 6i_875 41875 99 67- Hence 20000/ , QQ I0000 = -^^^I + '^-\ 67 8 ^ X 3 )' ox X = .65 in. 16. 9 M L Long Columns of Uniform Section. ( Pluler's Theory.) Case I. Columns wiih both ends hinged. — The column OA of length / is bent under a thrust P and takes the curved form OMA. Take O as the origin, the vertical through O as the axis ^^i x. and the hoii zontal through O as the axis oi y. Consider a section at any point M {x,y). If there is equilibrium and if the line of action of P coincides with the -""-"-^yj ' axis of the column, the equation of mo- FiG. 356. ments at M is or 1? EI y = a*y. ^0 LONG COLUMNS OF UNIFORM SECTION. 539 dy Multiplying each side of the equation by -j- and integrating, (!)■ = «•{*•-/), (2) h being a constant of integration dy • \/b^ = adx. Integrating, sin"' ( -^j = ax -|- Cy or y ^= b s\n {ax -f- c), (3> c being a constant of integration. When X = o, y \s also o, and hence ^ = o or t = o. \[ b =^ o, y \s always o, and lateral flexure is impossible., Take c = o. Then y = b s\n ax. .... Also, when x — OA = OMA , nearly, = l,y z=o. .'. O = b sin al, {^1 ot nn = al =V^' and hence TT" P^n'EI-ji, (5) Now the least value of P evidently corresponds to « = i, and hence the minimum thrust which will bend the column laterally is P^EI~. wm- 540 THEORY OF STRUCTURES, Cor. I. If the column is made to pass through N point: dividing the vertical OA into N A,- \ equal divisions, then J/ = o when x = and therefore, by eq. (4), o = ^ sin or = nn, N-\-i' al N-\-i' N-\- I and hence .TC' Fig. 357. P=n'EIj,{N-\-i)\ As before, the least value of P corresponds to « = i, and P=EIj,{N-\-ir is the least force which will bend the column laterally. Hence, the strength of the column is increased in the ratio of 4, 9, 16, etc., by causing it to pass through points which divide its length into 2, 3, 4, etc., equal parts, respectively. Cor, 2. The value of b may be approximately determined as follows : Let ds = length of element at M. Let = inclination to vertical of tangent at M. Then dx pressure upon ds =: P cos 6 = P-f and the compression of ds = _^_.ds = ^dx, A being the sectional area of the column. Hence, the total diminution of the length of the column P I = ' ^^" EA I, II LONG COLUMNS OF UNIFORM SECTION. 541 ii-.gain, the length of the cclumn = I (i-{- (^-Jj dx= Mi + a'b' cos' ax)dx, = / ( 1 -j cos axjax, approximately, Hence, if L is the initial length of the column, i.e., the length before compression, and consequently 6' = 2 EIIL- I 'P\ I r 'a: M Case 2. Columns zvith one end fixed and the other constrained to lie in the same vertical. Assume that the lateral deviation is prevented x- by means of a horizontal force H at the top of a JA column. Then -EI^, = Py-H{l-x)., . . (I) A particular solution of this is Oz:^Py - H{1 - x\ Let^ =^' -(- «. dy__d\j_ ■*• lx~dx'' V Fig. 358. §': ^542 THEORY OF STRUCTURES. and eq. (i) becomes or = — au. . . (2^ The solution of the last equation is y — y =^ u = b sin {ax -{• c), . . . . b and c being constants of integration. .' . y = —{I — x) -\- b sin {ax -\- c). . . . But dy -p- = o when x =^o, ax and J = o when ^ = o and when x =■ I. H .•. O = — 7, -|- '''^ COS ^ ; O = 'p^'V ^sin^; O = ^sin {al-\-c^. Hence al-\-c — o and ^/ = — tan ^r = tan a/, and therefore which may be written in the form (3) (4' P— 2.045 ;r'' EI (5) LONG COLUMNS OF UNLFORM SECTION. It is sufficiently approximate to write Ef P ~ 2W-' /' 543 (6) y Fig. 359. Case 3. Columns zvith one end fixed and the other free. A rigid arm AB is connected with the x free end ^ of ^ column, and a vertical b a I force /^applied at B bends the column j ^ '^ laterally, until its axis assumes the curved "p form OMA. Let AB = q, AC = p, and let / be the leii^^th of the column, — OC, nearly. The inclination of AB to the horizon is so small that the difference in length between AB and its horizontal projection may be disregarded. The moment equa- tion at any point AI(x,jy) is £/^^. = ^^(/> +?-/), or dy Multiplying each side by 2-- and integrating, d being a constant of integration. dy But ^- = o when y = o, and hence ^ = o. (I) • • t • (2) or dy V2{p + g)y-y adx. S44 Integrating, THEORY OF STRUCTURES. COS - ^—~~ — - =ax-\-c, c being a constant of integration, or But y ■=■0 when ;jr = o, and hence c =.q. pJ^q-y p^q = COS ax. Also, y = p when x =■ I. (3) (4) .'. — ^ — = COSrt/. / + !/ (5) If ^ is very small or «//, the term - ■ . may be disregarded, and then o = COS al. .'. al ''~2 = ^yE'r •••••• (6) n being a whole odd number. The least value of /'corresponds tow = i, and the minimum pressure which will cause the column to bend laterally is • (/ SI; Cor. I. By eq. (5) the deviation of the top of the column from the vertical is AC=p=q I — cos a/ cos al (8) LONG COLUMNS OF UNIFORM SECTION. 545 Cor. 2. Let the force applied at B be oblique and let its vertical and horizontal components be P and H, respectively. The moment equation now becomes EIjl-^=P{p-^q-y)^H{l-x). ... (9) A particular solution of this is Let jj/ = J/' -(- ti. Substituting in eq. (9), (10) or d*u i = — a u. (II) dx The solution of this equation is 2t = 6 sin {ax -\-c) z=y — y\ ^and c being constants of integration. .'.y=p-^q^p{l-x)^bsm{ax-\-c).. . (12) dy When x = o,y and -j- are each = o ; and when x=zl,y = q. Hence, IT o=-p-Vq-\- -p/+3sin c\ P H 0= — -p-\- ab cos c ; O = P -\- b sxn {al -\- c) ', IMAGE EVALUATION TEST TARGET (MT-S) # .<'%* 1.0 I.I I4i »50 IM 32 li5 IM M ^ 1^ III 2.0 1.8 1.25 1.4 1.6 .4 6" — ► v] <% ''w^ ->i \> ^ M ^ y Photogi^phic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 Lfi ^^^X^^^, -^* 4^ #/ ■fA ! li 546 THEORY OF STRUCTURES. three equations giving b, c, and p, and therefore fully deter- mining y. Case 4. Column with both cuds fixed. Let i-i be the end moment of fixture. Then X d'v iix or '^^=-a'yJ^a'b = a\b-y), . . (i) 1/ , M where b — p. Kio. 300. Multiplying each side of the equation by 2,- and integrating, i^fj=a\2by-f)^d, d being a conr-*:ant of integration. But '^'- = o when y — o, and hence d = 0. dx ...(|)=.x%-/)/. . . (2) or dy V2by — y ^ = adx. Integrating, CCS' or b b — y — ax -\-c, = cos {ax -f- c), being a constant of integration. (3) LONG COLUMNS OF UNIFORM SECTION, But y —o when x = o and when x = I. Hence I = cos c and I = cos {al-\- c). and therefore t = o and al = inn, It being a whole number. Hence, 54; V^^Z "'"''' or P^ie ./^I it^ (4) The least value of P corresponds to w = i, and the mini- mum thrust which will cause the column to bend laterally is P = ^EI 71 (5) 17. Remarks. — From the preceding it appears that the maximum theoretical compressive strength of a column per unit of area may be expressed in the form / = r n' .n ^^A T' ~ ^^^'^'ji't k i)cing the radius of gyration, and A a coefficient whose value i- I, 2, ^, or 4, according as the column has tzvo hinged tfii/s, one end fixed and the other guided in the direction of thrust, one cud fixed and the other free, or two fixed ends. This formula is easy of application, but Hodgkinson's experiments show that the value of P as derived therefrom is DO large. This may be partly due to the assumption that the lasticity of the material is perfect. The factors of safety to be used with this formula vary from 4 to 8 for iron and steel and from 4 to 15 for timber. riie objection to the use of flat bars as compression mem- bers has sometimes been overestimated. Consider, e.g., the case of a flat bar hinged at both ends. ■i ! ■i i ''•'A 4> '•'• 11' n 8 548 T.IEORY OF STRUCTURES. Let the coefficient of elasticity of the material be 25,000,- 000 lbs. Let the working stress per square inch be 8000 lbs. The bar will not bend lat'^rally under pressure so long as the unit stress < Ek'j^, and n-d* 8000 < 2500(xxx)- ji , or 2 < 507. Hence, the length of a flat bar in compression seems to be comparatively limited. If, however, both ends are securely fixed, the strength is quadrupled and the admissible length of bar is doubled, while it may be still further increased by fixine; the bar at intermediate points as indicated in Corollary i, page 540. This shows the marked advantage to be gained by rivet- ing together the diagonals of lattice-girders at the points where they cross each other. P The value of /= -j (Art. 15) must not exceed the clastic limit. It is difficult to define with any degree of accuracy the elastic limit of cast-iron and timber. It is claimed, indeed, that the latter has no elastic limit, properly so called, but that a permanent set is produced by every elastic change of form. It may be assumed, however, that the elasticity of these materials is practically unaffected so long as they are not loaded to more than one half of the ultimate crushing load. Hence, taking E E E E E 29,000,000 lbs. and / = 20,000 lbs. for wrought-iron, 29,000,000 " " / z= 33,600 " " soft steel, 29,000,000 " " / = 56,000 '• " hard steel. = 17,000,000 — 1,500,000 / = 40,000 " " cast-iron, / = 3,600 " " dry timber. / / the pillars will not bend laterally unless the ratio of - , or — LONG COLUMNS OF UNIFORM SECTION. 549 yd being the shortest side of a rectangular section and r the radius of a circular section) exceeds the values given in the following table : Material. Value of -j. a Wruught-iron 34.5 Soft steel 26.6 Hard steel 20.3 Cast-iron 18.7 Dry timber 18.5 Wrought-iron 48.8 Soft steel 37.7 Hard steel 28.8 Cast-iron 26.4 Dry timber 26.1 Wrought-iron 17.2 Soft steel 13.3 Hard steel 10. 1 Cast-iron 9,3 Dry timber g.2 Wrought-iron 69 Soft steel 53.3 Hard steel 40.7 Cast-iron 374 Dry timber 37 Value of — . Formula. 29. 24 J7 10. 16 42 34. 25- 22. 22. 14 12 8. 8. 8 59- 48 35 32 32 ^ - A *^ I' P ifl /=3— ? Baker has deduced by experiment the following formulae for tlie strength of wrought-iron and steel pillars of from 10 to 30 diameters in length and with fixed ends, the tensile strength of the metals ranging from 20 to 60 tons (2240 lbs.) per square inch : Let t be the tensile .strength of the iron or steel, and H the ratio of length to diameter. Then the ultimate compressive resistance, in pounds per square inch, for solid round pillars for thin tubes for tubes with stiffening ribs for girder sections (.4 -.oo6//)(/+ 18); (.44 - .oo4/^)(^ -f 18); (.44-.oo2//)(^+ 18); (.4 - .oo4//)(/+ 18). ji! 550 THEOIiY OF STRUCTURES. 18. Weyrauch's Theory of the Resistance to Buckling. — In order to make allowance for buckling, Weyrauch piu- poses the two following methods : Method I. Let F^ be the necessary sectional area, and h. the admissible nnit stress for a strut subjected to loads vary- ing from a maximum compression /j, to a minimum com- pression /),. Let F' be the necessary sectional area, and b' the admissible unit stress for a strut subjected to loads which vary between a given maximum tension and a given maximum compression. B' being the numerically absolute maximum load, and B" tlic maximum load of the opposite kind. According to Art. 7, Chap. Ill, if there is no tendency to buckling, B. .'(1 + .4) (0 and F' - ^' - ^ - b' - B' I v'\\ JV \* * * * * * (2) If there is a tendency to buckling, let / be the length of the strut, F its required sectional area, and T the mean unit stress at the moment of buckling. Then, according to the theory of long struts, ^^ EI EI (3) 6 being a coefificient depending upon the method adopted foi securing the ends, E the coefificient of elasticity, and / tlu least moment of inertia of the section. Also, let / be the statical compressive strength of the ma- terial of the strut, and take t = ^T. Then - 1 -lE^ ^- T~ 6 EI' FP_ o7' (4) KESISTANCE TO BUCKLING— WE YRAUCir S THEORY. 55 1 where for an area F would be -., . /' If the strut under the pressure B is liable to buckling;, its required sectional area will be -— , since T is the mean unit stress at the moment of buckling. Let x be the unit stress iit the moment of buckling, for the area F. Assuming that the unit stresses in the two cases are in the same ratio as the required sectional areas, ihen B B B A* • . • • • Fit X = B_l B (6) The force which, when uniformly distributed over the area F, will produce this stress, is Fx = /'. Hence, allowance may be made for buckling by substitut- \\\g for the compressive forces in equations (i) and (2), their values multiplied by }x. Thus, equation (i) becomes «i'!. <,i1 txB, nB, and equation (2) becomes ^B' -/> - = f^f,, (7) ^''('-''''«y 7TT-, if B' is a compression, (8) : I I .1 ; i 552 and B' b' THEORY OF STRUCTURES. B' ,1 ,^B"\ — -d77\, if B" is a compression. (9) If yw < I, equations (i) and (2) give larger sectional areas than equations (7), (8), and (9), so that the latter are to be ap- plied only when ;< > i. Method II. General formula; applicable to all values of \x may be obtained by following the same line of reasoning as that adopted in the proof of Gordon's formula. It is there assumed that the total unit stress in the most strained fibre is pA\-\- a r-,j,Pt being the stress due to direct compression, and p^a jj that due to the bending action. So, instead of employing equations (i) and (2) when /< < i, and equations (7), (8), and (9) when /< > i, formulae including a/l cases may be obtained by substituting for the compressive forces in equations (i) and (2) their values multiplied by i -j-/<. Thus, equation (i) becomes F = ii+M)B, il+M)J'^, (10) and equation (2) becomes (I 4- ^^)B" F = v'[i m B' (i+A^)^ ) , if B' is a compression, (11) or F=: v'[i - B' — /.I \p/\ ' 'f ^" 's a compression. (12) m' Ki ) B' Equation? (7), (8), (9), respectively, give larger values of F than the corresponding equations (10), (11), and (12). RESISTAXCE TO BUCKLING— WEYRAUCH' S THEORY. 553 Dn. (9) ional areas i to be up- values of )X sasoning as It is there ined fibre is )ression, and when /i < I, ilae including ; compressive edby !+/*• . . . (10) Iression, (n) )ression. (I2) values of f 1 1 2). Note, — For wrought-iron bars it may be assumed, as in Arts. 5, 6, Chap. Ill, that x\ = v' = 700 k. per sq. cm., and w, = ;«' The value of o is given by formula (5), but is unreliable, and varies in practice from io,CXX) to 36,cxx) for struts with fixed ends. When the ends are fixed, d = 4^-', according to theory. Hence, a = 4-' J. Therefore, \{ E =■ 2,000,000 k. per sq. cm., and t = 3300 k. per sq. cm., 6 = 23,926, or in round numbers 23,900; 24,000 is the value usually adopted by Wcyrauch. Example. — The load upon a wrought-iron column 360 cm. long varies between a compression of 50,000 k. and a compres- sion of 25,000 k. Calculate the sectional area of the column, assuming it to ht first solid and j^rt?;/^/ hollow, allowance being made for buckling. First. By eq. (i), ^.= 50000 700(1 +ix Mm) 400 , / r being the radius of the section. Also, / = nr^ " / ~ r' ~ 50' Hence, by eq. (4), 360 X 360 fji = — X — = 1. 188. 24000 50 Thus ^ > I, and by eq. (7) the required sectional area is F, X 1. 188 = A^^ X 1. 188 = 67.9 sq. cm. Second. F, = ^^= n{r* - r,'), >", being the external and r, the internal radius of the section. 554 THEORY OF STRUCTURES, Let /", = 9 cm. and r, = 7.92 cm. Then 'rl^' - r^) - 57.43 sq. cm. Also, / = 71 (' K) I r;' + r,' 143-7^64 Hence, by eq. (4), h = 360 X 3^XD — X 4000 1437-64 = .15. cncy Thus, in the latter case, since /< < i, there is no tend to buckling. If the area is determined by equation (10), its value become^ I.I 5 X A 2 a — 65 sq. cm. 19. Flexure of Columns. — In Art. 16 the moment equa- tion has been expressed in the form and this is sufficiently accurate if the deviation of the axis of the (dyV strut from the vertical is so small that ( -f-1 may be neglected without sensible error. The more correct equation is P p being the radius of curvature. Consider, e.g., the strut in Art. 16, Case I. Then P \ dd dd , ^ •^ EI' li ds dy no tendency alue beconur [lomcnt cqua- FLEXURE OF COLUMNS. 555 '^bciny the inclination of the tangent at J/ to the axis of a", and ds an element of the bent strut at J/. lntc<;rating, .'. — d'ydy = sin f^Jff. d'y = cos — cos 6„ , 2 »' (I) K,^ being the value of ft at a strut end. Let sin — = u and sin - = u sin 0. Then 2 2 «y = 2yM'(l — sin''0), or 2/A y = — cos(p (2) Let Fbe the maximum deviation of the axis of the strut from the vertical, i.e., the value of y when ft = o or = o. Then F^ 2h a sin 4 a (3) Again, ds — pdft = —■ de ^ \ \ — jji' s\n' Hence, if /be the length of the strut, d(t'i 2 /*" (10 2 « / \ I- fu' sin'' « ^ (4) F„(0) being an elliptic integral of the Jirst kind. Let P' be the /m.y/ thrust which will make the strut bend. As shown in Art. 16, I im 5 $6 THEORY OF STRUCTURES. and, by eq. (4), the corresponding value of the modulus /< is given by PM) = \ (5) Let the actual thrust on the strut be P=n'F (6) «* being a coefficient > unity. The corresponding value of the modulus is given by j:,,^. I I~P I n FA ~^J, V\ - A'sin''0''^ ;-|^3^(i-;.'sin>K0-y^>^_^t,.„,^j n = ^!2£,(0j-/v(0)[; £^(0) being an elliptic integral of the second kind. Hence, the diminution in the length of the strut = L-X=l\F,{)-E,{ — total stress at the distance c from tii-j /:/ neutral axis, and/— stress due to direct thrust (=; —1, so ihat the stress due to /fending = /»—/. It is also assiuncd that the form of the axis of the column before it is acted upon by the thrust /', is a curve of sines defined by the equation TtX J', = J cos ~, . . (4' the origin being half-way between the ends of the strut, and J being the maximum initial deviation of the axis from the ver- tical, i.e., the value of j'„ when .r = o. dx' An* itx -F '^^^ T' and hence, by eq. (3), dx -.r = -a\y + d) — Aj, cos -j. . (5) A solution of this equation is cosa.r cos - cos TTX {6^ I — n' FLEXURE OF COLUMNS. 559 Now — is alwavs small for such values of / as would con- -titute a safe working load, and therefore al cos - = I 2 ,5 /J a I -, approximately, i'T or tliat eq. (6) becomes • -|- « = « cos a.v\\ ~\ ^j + A cos — r I i -| ,- j, approx. (7) Let Fbe the maximum value of j'. i.e., the value oi y when .1- z^ 0. Then y =<+-)+ '^■ (8i Hence, by eq. (3), the total maximum intensity of stress where /; = - - J — -I and c s \o It I d-^ A Eq. (9) is a quadratic from which /may be found in terms if/. As a first approximation, /> may be siibstitutetl for / in lie last term of the portion within brackets, tlie error beini; in he direction of safety. Fixed Ends.— hct J/, be the moment of fixture. Eq. (3) now becomes = -a{j' + ^+^)- (10) 5 III 560 THEORY OF STRUCTURES. Assuming again that the initial form of the axis is a curve of sines, the solution of the last equation is cos nx Initially, (n.i cos I — y^ = A cos -y, dy dy I I ' and -7- is equal to ~ when ;tr = — or = . dx ^ dx 22 Hence, -.7=-K . al 71 PI al cos — 2 a'P n or <^+f = (12) Again, the value of j at the point a- = o is Y PI 8 (13' Also, '^i Pi, Pi are the total maximum intensities of stress at the end and at the most deflected point, then ^^--^'('^+^)=etc., .... (.4^ and A-/ zE ' = -AY+ -+?■) etc. ; . (15* two equations from which / may be found as before. The following conclusions are drawn from the above inves- tigation : First. The actual strength of a column depends partly upon FLEXURE OF COLUMNS. 561 i IS a curve known facts as to dimensions, material, etc., and partly upon accidental circumstances. Second. Experiments upon the crippling or destruction of columns cannot be expected to give coherent results when applied to the determination of the constants in such an equa- tion as No. (9). Third. It is a question whether p should be made the dastic liuiit of the material and the working load a definite fraction of the corresponding value of / derived from eq. (9), or whether/ should be the allowable skin working stress, and the working stress,/ be found by means of the same equation. The former seeVns to be the more logical assumption. Fourth. It would appear that the strength of hinged col- umns is likely to be much more variable than the strength of columns with fixed ends, as it depends upon two variable elements d and J, while the end fixture eliminates d. Note, — The Tables on the following page give the numerical values of elliptic integrals of the first and second kind, and are useful in applying the results of Art. 18. ••t^=. rir ■■'p (.i- fi [U 1 1 1 ri i i i1 ! 562 THEORY OF STRUCTURES. FIRST ELLIPTIC INTEGRAL . FM). ♦ H = ji = .1 fi = .J M = -7. M = .4 M = .5 H = .6 M =.7 1 H =.8 ft = .9 »» = . 0° 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 5° 0.087 U.087 0.087 0.087 0.087 , 0.087 0,087 O.0S7 O.0S7 0.087 0,087 10 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 0.175 *s: 0.262 0.262 0.262 0.262 0.262 0.263 0.263 0.263 a. 264 0.264 0.265 20" o-34<) 0.349 0.349 0.350 0.3501 0.351 j 0.352 0.353 0.354 0.355 0.350 25 0.436 0.436 0.437 O.43S 1 0.439 0.440 , 0.441 0.443 0.445 0.448 0.451 30 0.524 0.524 0.525 0.526 0.527 1 0.529 0.532 0.536 0.539 0.544 549 35 0.6 n 0.61 1 0.612 0.614 0.617 1 0.620 0.624 0.630 0.636 0.644 0.653 4°: 0.698 0.699 0.700 1 0.703 0.707 0,712 0.718 727 0.736 0.748 0.763 45^ 0.785 0.786 0.789 0.792 0.798 0.804 0.814 0.826 0.839 0.858 o.cSi SO 0.873 0.874 0.877 0.882 0.889 0.898 0.911 O.92S 0.947 0.974 lOII 15" 0.960 0.961 0.965 0.972 0.981 0.993 1. 010 1.034 I 060 1.099 1.1:4 1.047 1.049 1.054 1.062 1.074 I.jf ::, and vi and n coefficients, show that n /{•' )■ 4- — — must be substituted for_j' in eq. (3). Ill r 38. In one of Christie's experiments an angle-bar 2 in. X2 in. x ,'j in., wall hinged ends, for which .- had the value 154, deflected .01 in. for an d A increase in the load of 3000 lbs. Show that + -^ = .01 in. o TT 39. A long column with pin ends is bent laterally until the angular deviation (5o) at the ends is 4°. Find the total maximum intensity of stress, the section of the column being {a) a circle; {h) a square. A"= 29,000,000 lbs., and the stress due to direct thrust = 1500 lbs. per square inch. Ans. — (a) 30,615 lbs.; {b) 26,715 lbs. 40. With the same vtaximuin stress as in the last question, find the angular deviation at the ends so that the stress due to direct thrust may be 10,000 lbs. per square inch. Ans. — {a) V 5'; {d) i 33'. 41. Show that the load required to produce an angular deviation of 14' at the two pin ends of a long column is on\y one per cent greater than that which just produces flexure. m, of sectional and that the le coefficient of le axis, ^ being CHAPTER IX. TORSION. I. Torsion is the force with which a thread, wire, or pris- matic bar tends to recover its original state after having been twisted, and is produced when the external forces which act upon the bar are reducible to two equal and opposite couples (the ends of the bar being free), or to a single couple (one end of the bar being fixed), in planes perpendicular to the axis of the bar. The effect upon the bar is to make any transverse section turn through an angle in its own plane, and to cause originally straight fibres, as DE, to assume helicoidal forms, as FG or DC. This induces longitudinal stresses in the fibres. Fig. 360. * and transverse sections become warped. It is found suf- ficiently accurate, however, in the case of cylindrical and regu- lar polygonal prisms, to assume that a transverse section which is plane before twisting remains plane while being twisted. In order that the bar may not be bent, its axis must coincide with the axis of the twisting couple. 2. Coulomb's Laws. — The angle turned through by one transverse section relatively to another at a unit distance from it, is called the Angle of Torsion, and Coulomb deduced from 568 111! TORSIONAL STRENGTH OF SHAFTS. 569 experiments upon wires, that this angle is directly proportional to the moment of the twisting couple, and inversely propor> tional to the fourth power of the diameter. Thus, if a force P, at the end of a lever of radius /, twists a cylindrical bar of length L and radius R^ and if Q is the drcUf lar measure of the angle of torsion, then Q oc Pp, and also a R" so that 6/ = C-^,, C being a constant depending only upon the nature of the material. ' - Let T be the total angle of torsion, in circular measure, i.e., the angle turned through by one end of the bar relatively to the other. Then L R*' 3. Torsional Strength of Shafts (see Art. 23, Chap. IV). — Consider a portion of the shaft bounded by the planes C£ and MN, Fig. 361. It is kept in equilibrium by the couple (P, —P), and by the elastic resistance at the section AfN. Hence, this elastic resistance must be equivalent to a couple equal and opposite to (P, —P). Fig. 36a. Let Fig. 362 be the transverse section at MN, on an en- enlarged scale, and let abb' a' be any elementary area (= ^A^ {P,—P) of the surface bounded by the radii OA, OB, and by the concentric arcs aa', bb'. Let :r, be the distance of ^A^ from O. It is assumed, and is approximately true, that the resist- ance of any element abb' a' to torsion is directly proportional to the angle of torsion (^), to its distance from the axis (;»:,), and to its area {^A,), and also that it acts at right angles to the radial line of the element, i.e., to OA or OB. Thus, the resistance of abb'a' to torsion = GBx^AA^ , G be- ing a constant to be determined by experiment. The corresponding moment of resistance about the axis = GBx^AA^. Similarly, if x^, x^, x^, ... are the distances f P.p %7o THEORY OF STRUCTURES. from the axis of any other elements, J^,, AA^, AA^,... respectively, the corresponding moments of resistance are Gdx^AA^, Gtix^AA^, . . . Hence, the Ma/ moment of resist- ance of the section = G(f{x,'JA,+x:dA,-{-...), ;. = G(^^{x'^A) = Gei, V , /being the moment of inertia with respect to the axis. But this moment of resistance {M) is equal and opposite to the moment of the couple (7^, — /') Hence, '''-/;'/": :"^^.:----^:'-^ M=Gei=z]p. ,'/,.'. '-';'',, The twisting moment will of course vary with a variable resistance, and the last equation gives its mean value. The shaft, however, must be designed (see Cor. 4) for the maximum couple to which it may be subjected, and the moment of this couple (= J/,) may be expressed in terms of the mean by the equation :^ « f4 being a coefficient to be determined in each case. In a series of experiments with different engines, Milton found that /.i varied from 1.3 to 2.1, but doubtless the variation is often be- tween still wider limits. Cor. 1. Let /be the stress at the point farthest from the axis. For a so/uil round shaft, of diameter D, / = — , and /= G^-. .'. M=Pp = GdI= —^flT = .196/Z)'. Let T" be the total torsion in degrees. Then ^~ L 180' TORSIONAL STRENGTH OF SHAFTS. 57I and hence •■ • ■ ■ i* » •'- ■^~ Z 180 2' or Taking the following mean values of G and/: Material. G , f Cast-iron 6,3CX),ooo 5,600 Wrought-iron 10,500,000 7,200 Steel 12,000,000 11,200 „ = 9.87'° for cast iron, = \2.jT° for wrought-iron, = 9.3 T" (or steel. Thus, the twist is 1° each 9.8 diameters in length for cast- iron, each 12.7 diameters in length for wrought-iron, and each 9.3 diameters in length for steel. This is often mucli too small, and in practice the twist is usually limited to ■^^° per lineal foot of length. For a hollow round shaft, D being the external and Z), the internal diameter, , I^j^{D^-D:), and f=Ge^. •■• ^ = ^/ = l6'^—D = -'^^-^ ~~D~^' If the thickness {T) of the hollow shaft is small compared with D, D'-D; = IT -{D- 2Ty = WT, approximately, and .; M=Pp=i.S7n'T. .1.;.' IM i ] i f'' r lU: r ill 572 THEORY OF STRUCTURES. The use of compressed steel admits of shafts being made hollow. For a solid square shaft, H being the side of thesquare, / = 6'' and /, the stress at the end of a diagonal, = GQ —- r ^ H 6 = -^/^' = .2i6/H\ and GI~ GH' <= 2/ 4/2/^ In these results it is assumed that Cr^l = -^ or = , . is constant at different points of the cross-section, which, how- ever, is only true for circular sections. In non-circular sections the stress is more generally greatest at points in the bounding surface which are nearest to the axis and least at those points which are farthest from the axis. St. Venant, who first called attention to this fact, gave the fol- lowing, amongst others, as the results of his investigations. Designating by unity the torsional rigidity I = -^-1 of a shaft with circular section, the torsional rigidity of a shaft of equal sectional area is .8863, .8863 X yiF /;• ac- 2n qr7» 7255. or cording as the section is a square, a rectangle with sides in the ratio « to i, an equilateral triangle, or an ellipse whose major and minor axes are 2a and 2b, respectively. Cor. 2. The torsional stress per unit of area at a distance x from the axis is G6x. > Hence, if ^ =:: 1 and x =^ 1, 6" is the force that will twist a . > ^m\ Hk TORSIONAL STRENGTH OF SHAFTS. 573 being made )f thesquare, /2 unit of area at a unit of distance from the axis through an angle unity. Cauchy found analytically that in an isotropic body G is two-fifths of the coefficient of direct elasticity. Experiments indicate that G is about three-eighths or one- third of the coefficient of direct elasticity. Cor, 3. For a solid cylinder, Pp = , R being the radius, Pp and therefore R* a -^^. If the shaft is to have a certain spea- Pp fed stiffness, i.e., if # is fixed, R^ a y,-, and for a given twisting ^ or = -^y- j n, which, how- lerally greatest rest to the axis Irom the axis, |t, gave the fol- investigations, l:^) of a shaft 1 shaft of equal IT> or ac- with sides in ellipse whose It a distance x It will twist a moment R* a ^. Now G is nearly the same for wrought-iron and steel, so that there is little if any advantage to be gained by the use of the latter. After passing the elastic limit, the stress varies much more slowly than as the distance from the axis, and there will be a partial equalization of stress, the apparent torsional strength being increased. Cor. 4, In any transverse ,>ection of a solid cylindrical shaft, the maximum unit stress / = rft D 16 M, 1/, being the moment of the maximum twisting couple. This relation is true so long as the stress does not exceed the elastic limit, and agrees with the practical rule that the diameter of a cylindrical shaft subjected to torsional forces is proportional to the cube root of the twisting couple. The rule is usually expressed in the form M, = KD\ so that K = fJL i6- Wohler's experiments show that the value of / depends, to some extent, upon its fluctuation under the variable twist- .: 'M 574 THEORY OF STRUCTURES. ing moment. Ordinarily it should not exceed 7200 lbs. per square inch for wrought-iron, in which case K =■ ^\%^ X ^ = 1414. {No/i\ — If P, is the torsional breaking weight, W" s SI I I \ I is the coefficient of torsional rupture.) Cor. 5. Let ^F'be the work transmitted to a shaft of D in. diameter, in foot-pounds per minute, N being the correspond- insi number of revolutions. Then 12 fF = inch-pounds transmitted = 2nMN = 2tt — 'iV /^ KD\, — 2n~ — N, M Since M = mean twisting moment = — . Hence, Let HP he the horse-power transmitted per minute. Then IV = 33000 HP. Also for u'roughi-iron K = i| g^ x ^^^ TT 490 HP _,, , .. Hence ix^ ^ = D\ and if yw = 1.43, V ^v ' If J I a formula agreeing with the best practice in the case of wrought-iron shafts subjected to torsional forces only. Such shafts should, therefore, carry no pulleys. Cor. 6. The resilience of a cylindrical axle is the product of one half of the greatest moment of torsion into the correspond- ing angle of torsion. ' Cor. 7. It often happens in practice that a shaft (or beam) is subjected to a bending as well as to a torsional action. DISTANCE liETlVF.EX THE BEARINGS OF SHAFTING. 575 The combined bctidiiip; and twisting moments are equiv- alent (Art. 8, Chap. IV) to the moment M, = M, + ^M: -t- M': = J/, (« + )/n' + I). 'vhere Mi, = nM,, M,, being the bending and Mi the twisting moment at the given section. Hence, remembering that the maximum twisting moment J/, is equal to yiMt , we have for a wrought-iron shaft, 4goIfP /'('^ + v^«'+0^^- = ^'. ■ V = ,36 -\- , this becomes Z? = 4i . /HP a formula agreeing with the best practice in the case of trans- mission with bending, as, e.g., in the crank-shafts of marine engines. It often happens that ;/ has a still larger value, as, e.g., in the case of head shafts properly supported against springing. The usual formula is then I) = s V HP 'N' corresponding to « = .72 -{- . 4. Distance between Bearings. — The distance between the bearings of a line of shafting is limited by the considera- tion that the stiffness of the shaft must be such as will enable it to resist excessive bending under its own weight and under any other loads (e.g., pulleys, wheels, etc.) applied to it. For this reason, the ratio of the maximum deviation of the axis of the shaft from the straight to the corresponding dis- tance between bearings should not exceed a certain fraction whose value has been variously estimated by different writers. Let / be the distance in feet between bearings, a the diameter of the shaft in inches, w the weight of the ma- ir :i 576 THEORY OF STRUCTURES. terial of the shaft per cubic foot, and let the applied load be equivalent to a load per lineal unit of length in times that of the shaft. Assume a stiffness of pgVo"' ^"*^ ^^^' ^^^ '^'^'s of the shaft is truly in line at the bearings. The maximum de- flection of the shaft is given by the formula (Art. 3, Ex. 8, Chap. VII) _^_ I {m 4- i)(\veight of shaft)/' . 1728 ^ - '385 EI 1 , .nd* I , 64 /• . 1728 z=-^(m4- l) ZVl — -jr =r » 384^ ' •' 4 144 ftd* £ D •*• / "" 100 4 144 I (w + i)w /* 2E d" or -i/ Ed' ^ozj{tn +0' ExAMPLii:. — For wrought-iron, E = 3o,ooo,ocx) lbs. and w = 480 lbs. » / d' .\l— 12.7a / — —-. If the applied load, instead of being uniformly distributed is concentrated at the centre, the maximum deflection = Z>in. i_ {m -\- ^)( weight of shaft)/' . 1728 192 EI and hence =/. Ed' ioozv{in + i) ' Example. — For wrought-iron /= 8.5A / — —^ . CYLINDRICAL SPIRAL SPRINGS. 577 5. Efficiency of Shafting.— Let it require the whole of the driving moment to overcome the friction in the case of a shaft of diameter d and length L. The efficiency of a shaft of / the same diameter and length / = i — j- . But fnd —^ = {Pf) = moment of friction wnd^ d /( L— lAwnd^ L, li) being the specific weight of the material of the shaft, and yu the coefficient of friction. Hence, I Zif and the efficiency = i — 2^-7-, 6. Cylindrical Spiral Spring. — Let the figure represent a cylindrical spiral spring of length s, supporting a weight W. Consider a section of the spring at any point B. At this point there is a shear W^and a torque Wy, y being the distance of B from the axis of the spring, i.e., the radius of the coil. The effect of W may generally be neglected as compared with the effect of the moment Wy, and it may be therefore assumed that the spring is under torsion at every point. Let there be n coils. Then ':, . Hence, 5= 2nyn, approx. Fig. 363. r being the radius of the spring. 578 THEORY OF STRUCTURES. The elongation of the spring . _ 2WfS Syf 2nfn/ Wy Wy'^S Pnr'S The work done = dS = ' ^ , = :p^ — . 2 GTrr 4G A weight hung at the lower end tends to turn as well as lengthen the spring, and this is due to a slight bending action. According to Hartnell, /= 60,000 lbs. per square inch for f-in. steel, / = 50,000 lbs. per square inch for ^-in. steel, and G varies from 13,000,000 lbs. for ^in. steel to 11,000,000 lbs. for ^-in. steel. Also for wire less than f in. in diameter, W = 1^°°°^' y and the deflection = Wny* 288ooor* Example. — A wrought-iron shaft in a rolling-mill makes 50 revolutions per minute and transmits 120 H. P., which is sup- plied from a waterfall by means of a turbine. Determine the diameter of the shaft (i) if the maximum stress in the metal is not to exceed 9000 lbs. per square inch ; (2) if the angle of torsion is not to exceed ^° per lineal foot. As a matter of fact, the diameter of the shaft is 3! in. at the bearings and 4 in. in the intermediate lengths. What are the corresponding maximum inch-stresses in the metal? Let the twisting couple be represented by force P at the end of an arm /. Then .\Pp P X 27rp X 95 = 120 X 33.000 ft.-lbs 120 X 33000 First, 2n X 95 126000 X i^ 126000 , ,. 126000 . ,, ft.-lbs. = X I2m.-lbs. 19 19 19 and hence P/. _ Z""-^' _ 9000 X 22 ^P - 16 - 16 X 7 ' D = 3.56 in 126000 Second, X 12 = Pp = '19 ^ n I EXAMPLE. 579 32 But d = ~g - X 77 X — ; take G = 10,500,000. Then 126000 X 12 _ 10500000 22 I I I 2r? 19 32 7 ISO 13 12 7 Hence, Z>*= 689.45, and Z)=5.i2in. Third, the maximum stresses in the real shaft at the bear- ings and in the intermediate lengths are respectively given by 126000 stress 22 , „^. X 12 = —^ X y X (3l)', and 19 126000 stress 22 , ^, ,9 X.2 = -^-XyX(4)-. From the former, the maximum stress = 7682 lbs. per sq. inch. " " latter, " ♦' " = 6330 " " " " t »■' ■;v'-i"T'-" Kl tail illi i r I s ' 580 THEORY OF STRUCTUKES, EXAMPLES. 1. A steel shaft 4 in. in diameter is subjected to a twisting couple which produces a circumferential stress of 15,000 lbs. What is the stress (shear) at a point i in. from the centre of the shaft } Determine the twisting couple. Ans. 7500 lbs.; 23,571^ lbs. 2. A weight of 2j tons at the end of a i-ft. lever twists asunder a steel shaft if in. in diameter. Find the breaking weight at the end of a 2-ft. lever, and also the modulus of rupture. Ans. i^tons; 23,510 lbs. 3. A couple of A'^ft.-tons twists asunder a shaft of diameter d. Find the couple which will twist asunder a shaft of the same material and diameter id. Ans. %N. 4. Compare the couples required to twist two shafts of the same material through the same angle, the one shaft being / ft. long and d in. in diameter, the other 2/ ft. long and 2d in. in diameter. Compare the couples, the diameter of the latter shaft being -. Ans. I to 8; 32 to i. 5. Ashaftisft. long and 4i in. in diameter is twisted through an angle of 2° under a couple of 2000 ft.-lbs. Find the couple which will twist a shaft of the same material 20 ft. long and T\ in. in diameter through an angle of 2^°. Ans. 12,288 ft.-lbs. 6. A round cast-iron shaft 15 ft. in length is acted upon by a weight of 2000 lbs. applied at the circumference of a wheel on the shaft ; the diameter of the wheel is 2 ft. Find the diameter of the shaft so that the total angle of torsion may not exceed 2°. Ans. 3.53 in. 7. A wrought-iron shaft is subjected to a twisting couple of 12,000 ft.- lbs. ; the length of the shaft between the sections at which the power is received and given off is 30 ft. ; the total admissible twist is 4°. Find the diameter of the shaft, fi (page 570) being \, and m 10,000,000 lbs, Ans. 7.74 in. 8. A wrought-iron shaft 20 ft. long and 5 in. in diameter is twisted through an angle of 2°. Find the maximum stress in the material, m being 10,500,000 ft.-lbs. Ans. 3819.2 lbs. per sq. in. i EXAMPLES. 9. A crane chain exerts a pull of 6000 lbs. tangentially to the drum upon which it is wrapped. Find the diameter of a wrought-iron axle which will transmit the resultinp; rmiplc, the effective radius of the drum being 7^ in. ; the safe working stress per square inch being 7200 lbs. Ans. 3.17 in. 10. Find the diameter and the total angle of torsion of a 12-ft. vvrought-iron shaft driven by a water-wheel of 20 H. P., making 25 revolutions per minute, m being 10,000,000 lbs., and the working stress 7200 lbs. per square inch. Ans. 5.6 in.; 2°. 2. 11. A turbine makes 114 revolutions per minute, and transmits 92 H. P. through the medium of a shaft 8 ft. 6 in. in length. What must be the diameter of the shaft so that the total angle of torsion may not ex- ceed — , m being 10,500,000 lbs. ? Ans. 4.7 in. Determine the side of a square pine shaft that might be substituted for the iron shaft. 12. A steel shaft 20 ft. in length and 3 in. in diameter makes 200 revolutions per minute and transmits 50 H. P. Through what angle is the shaft twisted ? A wrought-iron shaft of the same length is to do the same work at the same speed. Find its diameter so that the stress at the circumference may not exceed | of that at the circumference of the stee! shaft. Ans. 2°. 6; 3.556 in. 13. A vertical cast-iron axle in the Saltaire works makes 92 revolu- tions per minute and transmits 300 H. P.; its diameter is 10 in. Find the angle of torsion. Ans, .0144° per lineal foot. 14. In a spinning-mill a cast-iron shaft 8J in. in diameter makes 27 1° revolutions per minute; the angle of torsion is not to exceed — per lineal foot. Find the work transmitted. Ans. 62.19 H. P. 15. A square wooden shaft 8 ft. in length is acted upon by a force of 200 lbs., applied at the circumference of an 8 ft. -wheel on the shaft. Find the length of the side of the shaft, so that the total torsion may not exceed 2° {m = 400000). What should be the diameter of a round shaft of equal strength and of the same material ? Ans, 4.96 in.; 5.09in. 16. A shaft transmits a given H. P. at A'' revolutions per minute with- out bending. Find the weight of the shaft in pounds per lineal foot. /H.P.\! ^1 "1 r 582 THEORY OF STRUCTURES. 17. The working stress in a steel shaft subjected to a twisting couple of 1000 in.-tons is limited to 11,200 lbs. per square inch. Find its diam- eter; also find the diameter of the steel shaft which will transmit 5000 H. P. at 66 revolutions per minute, yu being f. Ans. 10 in. ; 6.88 in. 18. A wrought-iron shaft is twisted by a couple of 10 ft.-tons. Find its diameter {a) if the torsion is not to exceed 1° per lineal loo\.,{b) if the safe working stress is 7200 lbs. per square inch, in = 10,000,000 lbs. ^«5.— (tf) 3.7 in. ; (^) 5.7 in. 19. A steel shaft 2 in. in diameter makes 100 revolutions per minute and transmits 25 H. P. Find the maximum working stress and the tor- sion per lineal foot, jn being 10,000,000 lbs. Also find the diameter of ;i shaft of the same material which will transmit 100 H. P. with the same maximum working stress. Ans. io,o32^*'r ''^s. ; .0478°; 3.17 in. 20. Tlie crank of a horizontal engine is 3 ft. 6 in. and the connecting- rod 9 ft. long. At half-stroke the pressure in the connecting-rod is 500 lbs. What is the corresponding twisting moment on the crank-shaft ? Ans. 1716^ ft.-lbs. 21. If the horizontal pressure upon the piston end of the connecting rod in the previous question is constant, find the maximum twisting mo- ment on the crank-shaft. ^1 ■ A sin cos 6 \ . , . Ans. F[ sm + — _ — , bemg given by \ ^/w" — sin'O/ n" cos'O + w^sin" G cos'' — i) + «' sin* 6(1 + sin" 6) — sin » 6 = where n = J//" = V"- A^.B. — If sin'O is neglected as compared with «', „ . - / cos G^ the maximum moment = /'smG( i -+- . „/ cosG\ .mG(i+-^). G being very nearly 72°. 22. Show that a hollow shaft is both stiffer and stronger than a solid shaft of the same weight and length. 23. Find the percentage of weiglit saved by using a hollow instead of a solid shaft. Ans. If of equal stiffness = 200 w' -H I If of equal strength = 100 1" ^/ m\m^ - 1) ) m being the ratio of the external to the internal diameter f ! of hoilow shaft. 24. A hollow cast-iron shaft of 12 in. external diameter is twisted by a couple of 27,000 ft.-lbs. Find the proper thickness of the metal so that the stress may not exceed 5000 lbs. per square inch. Ans. .619 in. EXAMPLES. 5«3 25. The external diameter of a hollow shaft is/ times the internal. Compare its torsional strength with that of a solid shaft of the same ma- terial and weight. j^ ^^a _ j ■^"•f- --z/-^ • p' 4- I 26. If the solid shaft is 10 in. in diameter, and the internal diameter of the hollow shaft is 5 inches, find the external diameter and compare the torsional strengths. ' ^;^j, 5 ^j in. . ^^ to 3. 27. A hollow steel shaft has an external diameter d and an internal diameter — . Compare its torsional strength with that of {a) a solid steel shaft of diameter d\ {b) a solid wrought-iron shaft of diameter d; the safe working stresses of steel and iron being 5 tons and 3J tons re- spectively. Ans.—(a)-^^\{b)l\. 28. What twisting moment can be transmitted by a hollow steel shaft of 8 in. internal and 10 in. external diameter, the working stress being 5 tons per square inch ? Ans. 184* in.-tons. 29. If/i is the safe torsional working stress of a shaft, and /j is the safe working stress when the shaft acts as a beam, show that the tor- sional resistance of the shaft is to its bending resistance in the ratio of 2/, to/,. 30. The wrought-iron screw shaft of a steamship is driven by a pair of cranks set at right angles and 21.7 in. in length; the horizontal pull upon each crank-pin is 176,400 lbs., and the effective length of the shaft is 866 in. Find the diameter of the shaft so that (i) the circumferential stress may not exceed 9000 lbs. per square inch ; (2) the angle of torsion 1° may not exceed — - per lineal foot ; in being 10,000,000 lbs. The actual diameter of the shaft is 14.9 in. What is the actual \.ox%\ot\ ? Ans.—{\') 14.53 '"•; (2) 14.89 in.; (3) total torsion = 5°.545- 31. The ultimate tensile strength of the iron being 60,000 lbs. per, square inch, find the actual ultimate strength under unlimited repetitions of stress. Ans. 54,899 lbs. (Unwin's formula). 32. What is the torsion in the preceding question when one of the cranks passes a dead point ? 33. A steel shaft 300 feet in length makes 200 revolutions per minute and transmits 10 H, P. Determine its diameter so that ihe greatest stress in the material may be the same as the stress at tiie circumference of an iron shaft i in. in diameter and transmitting 500 ft. -lbs. Ans. .807 in. (= \ in.) 34. Determine the coefficient of torsional rupture for the shaft in Question 33, 10 being the factor of safety. 35. A wrought-iron shaft in a rolling-mill is 220 feet in length, makes 95 revolutions per minute, and transmits 120 H. P. to the rolls ; the main body of the shaft is 4 in. in diameter, and it revolves in gudgeons 3f in. 584 THEORY OF STRUCTURES. in diameter. Find the greatest shear stress in the shaft proper anri in the portion of the shaft at the gudgeons. Ans. 6330.2 lbs.; 7508 lbs. 36. Power is taken from a shaft by means of a pulley 24 inches in diameter which is i)e^;{c) [' - ^)"- EXAMPLES. 585 43. A steel shaft carries a 5-ft, pulley midway between the supports and nialtes 6 revolutions per minute, t'u tangential force on the pulley being 500 lbs. Taking the coefficient ot working strength at 11,200 lbs. per square inch, find the diameter of the shaft and the proper distance between tlie bearings. 44. A steel shaft 4 inches in diameter and weighing 490 lbs. per cubic (out makes 100 revolutions per minute. If the working stress in the metal is 11,200 lbs. per square inch, find the twisting couple and the dis- tance to which the work can be transmitted ; the coefficient of friction being .05, and the efficiency of the shaft f . Am. 140,800 in. -lbs. ; 8228^ ft. 45. If the shaft is of steel, and if the loss due to friction is 20 per cent, lind the distance to which work may be transmitted, /< being .05. Ans. 6582^ ft. 46. A wrought-iron shaft 220 ft. between bearings and 4 in. in diam- eter can safely transmit 120 H. P. at the rate of 95 revolutions per minute. What is the efficiency of the shaft ? (// = -^.~) Ans. .976. 47. The efficiency of a wrought-iron shaft is ^ ; the working stress in the metal is 7200 lbs. per square inch ; the coefficient of friction is. 125. How far can the work be transmitted .' Ans. 4320 ft. 48. A spring is formed of steel wire; the mean diameter of the coils IS I inch ; the working stress of the wire is 50,000 lbs. per square inch ; the elongation under a weight of iQjS^- lbs. is 2 inches; the coefficient of transverse elasticity is 12,000,000 lbs. Find the diameter of the wire and the number of coils. 49. Find the weight of a helical spring which is to bear a safe load of 6 tons with a deflection of 1 inch, G being 12,000,000 lbs., andy 60,000 lbs. 50. Find the time of oscillation of a spring, the normal displacement under a given load being A. /~Z Ans. ny — ■ 51. Find the deflection under the weight W^of a conical helical spring iiiiof circular section ; (l>) of rectangular section, the radii of the e.xtreme coils being yi andja , and the radial distance from the axis to a point of the spring at an angular distance

'' = o and v> = H; 6' + A' IV - S and A being the sides of the rectangular section. 52. The efficiency of an axle is i ; the working stress in the shaft is 9000 lbs. per square inch; the coefficient of friction is .lo. How far may work be transmitted ? 1 Iff I 585a THEORY OF STRUCTURES. B Fig. A shows the distortion produced by twisting a round >i-in. iron bar. Fig. B shows the distortion produced by twisting a square J^-in. iron bar. DISTORTION OF IRON BARS BY TWISTING. iZ^b The above figures show the distortion produced by twisting &\%yt J^" iron bar. :, .-it! ^f a' II 5-^ CHAPTER X. STRENGTH OF CYLINDRICAL AND SPHERICAL BOILERS. Fig. 364. I. Thin Hollow Cylinders ; Boilers ; Pipes. Let r be the radius of the cylinder. Let t be the thickness of the metal. Let p be the fluid pressure upon each unit of surface. Let/ be the tensile or compressive unit stress, according as / is an internal or exter- nal pressure. ' ' Assume (l) that the metal is homogeneous and free from initial strain ; . " (2) that t is small as compared with r ; f (3) that the pressures are uniformly distributed 1 'I over the internal and external surfaces; .V (4) that the ends are kept perfectly flat and rigid ; ;} (5) that the stress in the metal is uniformly dis- tributed over the thickne.ss. The last assumption is equivalent to supposing that it isj the mean circumferential stress which is governed by the] strength of the metal, while in reality it is the internal or maxi- mum circumferential stress which is so governed. The figure represents a cross-section of the cylinder ofj thickness unity. A section made by any diametral plane, as AB, must de- velop a total resistance of 2tf, and this must be equal and! opposite to the resultant of *^he fluid pressure upon each half| i.e., to ipr. Hence, 2tf = 2pr, or // = pr. (0 586 THIN HOLLOW CYLINDERS; BOILERS; PIPES. 587 This formula may be employed to determine the btirsting, [ froof, or working pressure in a cylindrical or approximately cj'lindrical boiler, provided that/", instead of beinj^ the tensile or compressive unit stress, is some suitable coefficient which has been determined by experiment. If rf is the efificiency of 1 a riveted joint, the formula may be employed to determine the working pressure in a cylin- I drical or approximately cylindrical boiler. In ordinary practice the values of // and /"are given by the I (ollowing table : Material. Joint, 1 /in lbs. per sq. in. ft'roughi-iron Sieel '. '.. Single-riveted Double-riveted Treble-riveted Single-riveted Double-riveted Treble-riveted •55 • 7 .8 to .85 •55 •7 . 8 to .85 8000 to 9000 (4 ( 1 12000 to 13000 1 ( ii tt For cast-iron cylinders the working value of /may be taken latabout 2000 lbs. per square inch. The total pressure upon each of the flat ends of the cylinder - - . . = 7rr*p. [The longitudinal tension in a thin hollow cylinder 2nrt 2/' ^ '' |iind is one half of the circumferential stress/. Cor. I. Let the cylinder be subjected to an external pressure 1^ as well as to an internal pressure p. Then /t=pr-p'r\ (3> r being the radius of the outside surface of the cylinder, /is. htension or a pressure according as pr %_p'r'. ^1 I ' 1 :'^' 588 THEORY OF STRUCTURES. Generally, r — r' is very small, and the relation (3) may be written 2. Thick Hollow Cylinder.— If t is large, the stress is no longer uniformly distributed over the thickness. Suppose that the assumptions (i) and (3) of Art. i still hold, also that the cylinder ends are free, and that the annulus forming the section of the cylinder is composed of an infinite number of concentric rings. Under these conditions the straining of the cylinder cannot affect its cylindrical form. Hence, right sections of the cylinder in the unstrained state remain planes after the strain- ing, so that the longitudinal strain at every point must be the same. Two methods will be discussed. First Method. — Let dx be the thickness of one of the rings of radius x, and let dq be th«, intensity of the circum- 1 ferential stress. pr — p'r' = difference between the total pressures from , within and without = total circumferential stress = / dg. If it be assumed that the thickness (= r' — r) remains imj changed under the pressure, then the circumferential extension j of each of the concentric rings must be equal to the same con- stant quantity \, and therefore dq = Edx 2nx' E being the coefficient of elasticity. Hence, , , E\ r'dx E\^ t' pr —p'r' = — / — = — log, -. r^ ^ Let / be the tensile unit stress. Then /= -£ if the elastic limit is not exceeded, and therefore pr-p'r'=^fr\og,-, ID V :| or THICK HOLLOW CYLINDER. 589 i'/i-n,allascomparedwi.h/; and hence/ ^^«^ fr ■^2\-~Jr~)- ■ . (5) m most cases which oen^r : &F ^ero in equation (5), ^ A' ^-27)' ' . P • • • •■ (6) cJ!!-rT'"''^ ^y- ^^P^^ence '^"'""^ °f strength ano J; i^etlti?'' '" ^'' ^^^^'^^ Mechanics, obtains by 'f/ be neglected. Hence, _ / ^ 2 7^ ' approximately, •'^■■"-aU as compared with/ and therefore k :: 590 THEOKY OF STh'UCTUKES. r r f\ ^ 2/r an equation identical with (6). Second Mf.thod. — Consider a ring bounded by tlic radii X, X -\- dx, at any point. Let q be the normal (i.e., radial) intensity of stress. Let /be the intensity of stress tangential to the ring. •' s " " " " " perpendicular to the plane of the ring. Let tx, /S, y be the corresponding strains. Let E ?nd /// ' he respectively the coefificients of direct and lateral elas Kxlj. Then, since E,/, s are principal stresses (Chap. IV), '^~ E mE ' ^ ~ ;£ iuE ' ^ ~ E mE ' ^'^ But Y is constant. Also, since the ends are free, the total pressure on a transverse section is nil, and hence it might be inferred that s is zero at every point. Adopting this value of s, Byeq. (I), ... f-\-q=. a constant =c (2) Again, d{qx) = fdx = xdq -\- qdx (3) By eqs. (2) and (3), xdq -\- iqdx = cdx. ' .•. d{x*q) = cxdx. Integrating, ^V = ~ + ^, . ..... (4) (f being a constant of integration. When X =■ r, the internal radius, q '=^ p. " X =■ r\ the external radius, q z= p\ _■'-.' Hence, by eq. (4), SPHERICAL SHELLS. 59' and therefore 2 /• ^^^—' -^ and c' ^^{tnjYr'* Hence, by eq. (4), q = -S -^ -.P-P r'r .»«/9 and, by eq. (2), r' _ *' » i — r — r ,/ a • • • (5) . (6) 3- Spherical Shells —I ^f ^i . T>.e »«i„„ „„j, ^ ^^^ "<= "- -me as before. «sta„cQ of 2^«/. Then ^ "'" """^ develop a total or W^pr. . Hence, a spherical «fi n • ' * * vU longest parts of ,a-.v./,.,/bol , t° t T'"'' ^^ "'^' 'he ^^■^^ I. Let the shf-M k ! ^"^" ^'"'^■'^- ''- weU a. to an inte'™,";::^^-';' ';,:„«'"-' P--ure 2 ^-f ~ ^fp ~ nr'y, ,\f{r' + r)t =, r'p -^ r' y - ' ^^ ' /' = ?(;»-./.> . . (3) 592 THEORY OF STRUCTURES. Cor. 2. For a thick hollow sphere, Rankine obtains P = 2fzr^ r" - r' r" + 2r' , approximately. • (4) 4. Practical Remarks. — A common rule requires that the working pressure in fresh-water boilers should not exceed one- sixth of the bursting pressure, and in the case of marine boilers that it should not exceed one-seventh. An Ens^lish Board of Trade rule is that the tensile workiii'^ stress in the boiler-plate is not to exceed 6000 lbs. per square inch of gross section, and French law fixes this limit at 4250 lbs. per square inch. The thickness to be given to the wrought-iron plates of a cylindrical boiler is, according to French law, / = .oo2)6nr -f- .1 in.; according to Prussian law, ;; / = (^•°°3" — \)r -\- I m. = .003«r + -i in-> approximately, r being the radius in inches, and n the excess of the internal above the external pressure in atmospheres. The thickness given to cast-iron cylindrical boiler-tubes is, according to French law, five times the thickness of equivalent wrought-iron tubes ; according to Prussian law, f _ ^^.oi« _ jy _|_ ^ in. — ,oinr -\- ^ in., approximately. Steam-boilers before being used should be subjected to a hydrostatic test varying from i^ to 3 times the pressure at which they are to be worked. Fairbairn conducted an extensive series of experiments upon the collapsing strength of riveted plate-iron flues, by enclosing the flues in larger cylinders and subjecting them to hydraulic pressure. From these experiments he deduced the following formula for a wrought-iron cylindrical flue or tube: Collapsing pressure in pounds per square inch of surface I =/ =403150-^- -^ = 403150 a' Ir' ^y riveting an^Je- or t • . ^ and =:°°'^''*^ + -°^-.«of.W..o„. ^a/;- or £.^./ hccordfng as the plate is m. , '^^'■?'-d]y fixed around\r ^^^^'"PP^'-^^d^'-ound the r'. I -responding deflections ^th^platar^ "^ ^^^-^^^• 61?;: and 1(7)*^' ittl E' 594 THEORY OF HTKUCTUKES, EXAMPLES. 1. What should be the thickness of the plates of a cylindrical boiler 6 ft. in diameter and worked to a pressure of 50 lbs. per square inch, in order that the working tensile stress may not exceed 1.67 tons per square inch of gross section ? Ans. .42 in. 2. A cylindrical boiler with hemispherical ends is 4 ft. in diameter and 22 ft. in length. Determine the thickness of the plates for a steam- pressure of 4 atmospheres. 3. What is the collapsing pressure of a flue 10 ft. long, 36 in. in diameter, and composed of i-in. plates? Also of a flue 30 ft. long, 48 in. in diameter, and j\ in. thick? Ans. 490.84 lbs.; 91.59 lbs. 4. Determine the thickness of a 2-in. locomotive fire-tube to support pn external pressure of 5 atmospheres. 5. A copper steam-pipe is 4 in. in diameter and \ in. thick. Find the working pressure, the safe coefficient of strength for copper being 1000 lbs. per square inch. Ans. 125 lbs. per square inch. 6. A 7-ft. boiler of -j'ls-in. plates was burst at a longitudinal double- riveted joint by a pressure of 310 lbs. per square inch. Find the coef- ficient of ultimate strength. Ans. 29,760 lbs. 7. A 50-in. cylindrical boiler of -^ in. plates is made of wrought- iron wliose safe coefficient of strength is 4000 lbs. per square inch. Find the working pressure. Ans, 50 lbs. per squaie inch. 8. A lo-in. cast-iron water-pipe is subjected to a pressure of 250 lbs. per square inch. Find its thickness, the coefficient of working strength being 2000 lbs. per square inch. Ans. \\ in. 9. A steel spherical shell 36 in. in diameter and f in. thick is sub- jected to an internal fluid pressure of 300 lbs. per square inch. Find its coefficient of strength. Ans. 7200 lbs. 10. A thin, hollow, spherical, elastic envelope, whose internal radius is R, was subjected to a fluid pressure which caused it to expand gradually until its radius became R\ . Determine the work done. 11. The plates of a cylindrical boiler 5 ft. in diameter are \ in. thick. Find to what pressure the boiler may be worked so that the tensile stress in the plates may not exceed \\ tons per square inch of gross section. '' M EXAMPLES. 595 12. Show that the assumption of a uniform distribution of stress in the thickness of a cylindrical or spherical boiler is only admissible when the thickness is very small. 13. A metal cylinder of internal radius r and external radius nr is sjbjected to an internal pressure of p tons per square inch. Show that the total work done in stretching the cylinder circumferentially is y- , _ - ft. -tons per square foot of surface, £ bemg the metal's co- etficient of elasticity. .... 14. The cast-iron cylinder of an hydraulic press has an external diameter twice the internal, and is subjected to an internal pressure of /tons per square inch. Find the principal stresses at the outer and inner circumferences. Also, if the pressure is 3 tons per square inch, and if the internal diameter is 10 in., find the work done in stretching the cylinder circumferentially, £ being 8000 lbs. Am. At inner circumference,'V =/. a thrust, and/= — |/,a tension. At outer circumference, g = o, and/ = — |/>, a tension. Work = 126 ft.-lbs. per square foot of surface. 15. The chamber of a 27-ton breech-loader has an external diameter of 40 in. and an internal diameter of 14 in. Under a powder pressure of 18 tons per square inch, find the principal stresses at the outer and inner circumferences, and also the work done ; £ being 13,000 lbs. Ans. At inner, y = 18 tons, compression ; at outer, ^ = o. At inner,/ = — 23^ tons, tension ; at outer,/ = — s^\ tons, tension. Work = li ft.-tons per sq. ft. of surface. 16. What should be the thickness of a 9-in. cylinder (a) which has to withstand a pressure of 800 lbs. per square inch, the maximum allow- able tensile stress being 24,000 lbs. per square inch ; (d) which has to withstand a pressure of 6000 lbs. per square inch ; the maximum allow- able tensile stress being 10,000 lbs. per square inch ? Afts. — (a) 1.86 in. ; (d) 4^ in. 17. Show that the radial (a) and hoop (/3) strains in thick hollow cylinders and spheres are connected by the relation a = — ; — . ax 18. Prove that the relation in Ex. 17 is satisfied by the values ob- tained for/ and g in the Second Method of Art. 2, Chap. X. 19. A thick hollow sphere of internal radius r and external radius 'ir is subjected to an internal pressure / and an external pressure p . Determine the principal stresses at a distance x from the centre. Ans. q =■- «°— I P + l_-I' «•— I / _p'n^-p p-p' n*r* Ml «• - I 2X' «' — I I if' m^': ill 596 THEORY OF STRUCTURES. 20. Assuming tliat the annulus forming the section of a cylindrical boiler is composed of a number of infinitely thin rings, show that the pressure at the circumference of a ring of radius r is —~ per unit of surface, and that the circumferential stress is - B -, A and B de- r mr'» + • noting arbitrary constants, and m being the coefficient of lateral con- traction. Find the values of A and B.po and pi being respectively the internal and external pressures. 21. Show that in the case of a spherical boiler the pressure and cir- cumferential stress are respectively-^j^jj^ and — + — : , . Find fi + m (fn - Or"* -3 A and B. 22. Solve Questions i, 2, 6, 7, 8, 9, and 11 on the supposition th?.c / is not small as compared with r. 23. Taking/ =4000 lbs. per square inch and £' = 30,000,000 lbs,, Find the thickness and deflection of the end plates of the boiler in Ques- tion 7. .-''■■■'■■ ■■'•■?■ ' "■■■■ I- Class eral classes, tilcver bridg (D) Arched I of bridges in I 2. Compi Flanges in 1 sometimes vJ claimed that J chord a slope, ^ucii a truss flanges and or bolic form is r nution in dept tion. Again, i tile lower part and the girder of several span- continuity isne "lost economic; uniform throug ^ange at any p plates. 3. Depth of ally varies from the span. It is 1 . CHAPTER XI. ' BRIDGES. I- Classification Wr,- 1 "lever bridges (A ,. ,„ fc) I '"'f'^""'^] girders; (H, |^". ID) Arched bridges (cti^-^f, rxT" ''"''^" '^'•'' ' ^ ' ')^ »' bridges in Classes A and H ol "'"'■■'" '^'"'P'" "■<--at=i 2- Comparative Advantages nf r Flanges in Girders for Brid^. „f *^r"' /"" Horizontal »met,n,es varied for the sake 'f '^'^ ^'^••- ^P'l- is *™ed that an economy otlteria,''^''";''""- ""^ " ''a'»o herd a slope, as, e.g., in'^.he «se "f ' h! %Tt ''^ ^'^f^ "- ^"cl, a truss is intermediate beta' ^"''" ^"''S': (Art. ,„). W c form ,s not well suited to „1„ ^ '=''™'' "i- Para- ""•-n in depth lessens the r "sS ! "-'™"ion, and a din, -• »»■ Again, if the bottom Ta' J °' ""^ ^'"^" '" Astor. *' lower part of the girder is ref, ■! !'"'"'■ "" bracing f„r ■»d the girder itself mL. b- ndep'nT ,"'''''" ""™" "™'-'° »' several spans any advantage wX,° I '° """ '" a bridge »"fnuity is necessarily lost oli^"' ""^'" ^' ''"'^able from ;;;t economical form'^f ^^e^ .S ''''\''"^- "'^ "est and ■"form throughout, and fn ' h U '" """'' "'^ "^'Pt" is 1 nge at any point f^ obtainld h '""''">- ">ickness of pia.es, ""'^'"^d by .ncreasing the number of f Arie?f?om^«2^,:,::/"-'^ (Class A).-The depth usu- *----sgenei;;r;:::--;(a-;v^^ give large girders 597 598 THEORY OF STRUCTURES. an increased depth, and they should, therefore, be designed to have a specified strength. If the span is more than twelve times the depth, the deflection becomes a serious consideration, and the girder should be designed to have a specified stiffness. The depth should not be more than about i^ times the width of the bridge, and is therefore limited to 24 ft. for a single and to 40 ft. for a double-track bridge. 4. Position of Platform. — The platform may be supported either at the top or bottom flanges, or in some intermediate position. In favor of the last it is claimed that the main girders may be braced together below the platform (Fig. 365), while the upper portions serve as parapets or guards, and also that the vibration communicated by a passing train is diminished. The position, however, is not conducive to rigidity, and a large amount of metal is required to form the connections. Fig. 365. Fig, 366. The method of supporting the platform on the top flanges (Fig. 366) renders the whole depth of the girder available for bracing, and is best adapted to girders of shallow depth. Heavy cross-girders may be entirely dispensed with in the case of a single-track bridge, and the load most effectively distrib- uted, by hying the rails directly upon the flanges and vertically above the neutral line. Provision may be made for side spaces by employing sufficiently long cross-girders, or by means of short cantilevers fixed to the flanges, the advantage of the POSITION OF PLATFORM. 599 former arrangement being that it increases the resistance to lateral flexure, and gives the platform more elasticity. Figs. 367, 368, 369 show the cross-girders attached to the bottom flanges, and the desirability of this mode of support increases with the depth of the main girders, of which the cen- tres of gravity should be as low as possible. If the cross-girders are suspended by hangers or bolts below the flanges (Fig. 369), the depth, and therefore the resistance to flexure, is increased. Fig. 36b. Fig. 369. In order to stiffen the main girders, brakes and verticals, consisting of ang'e- or teeiron, are introduced and connected with the cross-girders by gusset pieces, etc. ; also, for the same purpose, the cross-girders may be prolonged on each side, and the end joined to the top flanges by suitable bars. When the depth of the main girders is more than about 5 ft., the top flanges should be braced together. But the minimum clear headway over the rails is i6 ft., so that some other method should be adopted for the support of the plat- form when the depth of the main girders is more than $ ft. and less than 16 ft. Assume that the depth of the platform below the flanges is 3 ft., and that the depth of the transverse bracing at the top is I ft. ; the total limiting depths are 7 ft. and 19 ft., and if i to 8 is taken as a mean ratio of the depth to the span, the corre- sponding limiting spans are 56 ft. and 152 ft. 6oo THEORY OF STRUCTURES. 5. Comparative Advantages of Two, Three, and Four Main Girders.— A bridge is {jreneraily constructed with two main girders, but if it is crossed by a double track a tliird is occasionally added, and sometimes each track is carried by tw independent girders. The employment of four independent girders possesses the one great advantage of facilitating the maintenance of the bridge, as one-iialf may be closed for repairs without int pt- ing the traffic. On the other haiid, the rails at the appr^ ichcs must deviate from the main lines in order to enter the bridge, so that the width of the bridge is much increased, and far more material is required in its construction. Few, if any, reasons can be ur^ed in favor of the introduc- tion of a third intermediate girder, since it presents all tin objectionable features of the last system without any corre- sponding recommendation. The two-girder system is to be preferred, as the rails, hy such an arrangement, may be continued over the bridge with- out deviation at the approaches, and a large amount of ma- terial is economized, even taking into consideration the in- creased weight of long cross-girders. 6. Bridge Loads. — In order to determine the stresses in the different members of a bridge truss, or main girder, it is necessary to ascertain the amount and character of the load to which the bridge may be subjected. The load is partly dead, partly livt\ and depends upon the type of truss, the span, the number of tracks, and a variety of other conditions. The dead load increases with the span, and embraces the weight of the main girders (or trusses), cross-girders, platform. rails, ballast, and accumulations of snow. As to the live load see Art. 19. 7. Trellis or Lattice Girders. — The ordinary trellis or lattice girder consists of a pair of horizontal chords and two series of diagonals inclined in opposite directions (Fig. 370). The system of trellis is said to be single, double, or treble, ac- cording to the number of diagonals met by the same vertical section. TRELLIS OR I^-rir^r^ UR LATTICE GIRDERS. g^j p.^ f'o. 370. '.-I. ^^"■'^^X!^t^'tx,^Z:^^^^^ ■'"'-- f- the -^^'J plates, or of links and pir"' ""••" °' "'"■=<' ""d "^lir ■^inmiir The verticals and diagonal, maJ'hr , '"°- "- °"!" ^"""blc section, but the dht , "' ^" ^ ^- '■ H, u, or ' »">Kle system of .reUis, are ,„ fv "' 'k"''"^' '" ""^ "- "' «'l.c points of intersection ^ ^'' ''"^' "^'''^d together ,,, A" Objection . o this da. Of girder is ,e number Of .he •« t .nr :hia"r;n'; t^,:";t - ''"-ined on the ass„.p. Attributed be,v„een the diL,o„ T "'"'"' '«"™ ''^ equally "1'.ivalent .0 the subl.it , bn '^f' "" "^ "■-' section, Ihich '"'^tresses in the several bar" ""■"" =^'™^ '"' "■= differ. <• p - ;« » "e the permanent load concentrated at each ape. '=t."'^^=7i«'-4-=tL ''"="'■"« '"-« -t the tnis shearing force mu«f h ^ „, """^""^ '"-""-tted through the diag. onals. Hence, the stress in ab due to th. aue to the permanent load 4 ^^^ ^ = g zi'sec ft 602 THEORY OF STRUCTURES. Again, let w' be the live load concentrated at an apex. The greatest shear at mn due to the live load occurs when every apex between a and 7 is loaded. This shear = corresponding reaction at i = \%w' , and the stress in ab due to the live load = i X Ww' sec 6* = Ww' sec B. Hence, the total maximum stress in ab = (|w -f- ||w') sec d. The greatest stress of a kind opposite to that due to the dead load is produced in ab when the live load w' is concen- trated at every apex between I and b. The shear to be transmitted is then 2\w due to the dead load, and — ^w' due to the live load, and the resultant stress in ab = ^{2^w — \^tv') sec 6 = {^zv — ^^zv') sec 0. This stress may be negative, and must be provided for by introducing a counter-brace or by proportioning the bar to bear both the greatest tensile and the greatest compressive stress to which it may be subjected. The stress in any other bar may be obtained as above. The chord stresses are greatest when the live load covers the whole of the girder, and may be obtained by the method of moments, or in the manner described in the succeedhig articles. In the above it is assumed that the members of the girder are riveted together. If they are connected by pins, each of the diagonal systems may be treated as being independent. Thus, the system i 2^^34567 transmits to the supports the stresses due to loads at a, 3, and 5. The shear due to the dead load, transmitted through al), = reaction at i — load at a = -w — w = — . 2 2 , w Hence, *:he stress in ab due to the dead load = — sec 6. • 2 The sircss in ab due to the live load is greatest when tv' is concentrated at each of the points 3 and 5, The maj and the corr Hence, tl as compared sumption. 8. Warre horizontal ch. iinjjle triangu The princi girders are equ T'le cross-gi the apex of eac When the p sistance of the ; 'lorizontal braci bracing betweei VVlien the ] additional cross tile upper chord rigidity of the n Let zu be the " zu' " " " / " " " /k " " " $ n u " ^'-f I be WARREN GIRDER. 603 The maximum shear due to live load transmitted through ab = if w' = Jw', and the corresponding stress in ab = ^w sec B, Hence, the total maximum stress in ab (t+H = (~ + ^«;']sece. as compared with (Jzf/ + 1|^') sec B obtained on the first as- sumption. 8. Warren Girder. — The Warren girder consists of two horizontal chords and a series of diagonal braces forming a jinpfle triangulation, or zigzag, Fig. 375. n-l JS'+L N-1 2 4 /t N-2 Fig. 375. The principles which regulate the construction of trellis girders are equally applicable to those of the Warren type. The cross-girders (floor-beams) are spaced so as to occur at the apex of each triangle. When the platform is supported at the top chords, the re- sistance of the structure to lateral flexure may be increased by horizontal bracing between the cross-girders and by diagonal bracing between the main girders. When the platform is supported on the bottom chords, additional cross-girders may be suspended from the apices in the upper chords, which also have the effect of adding to the rigidity of the main girders. Let w be the dead load concentrated at an apex or joint. " span of the girder. " depth " " " " length of each diagonal brace. " inclination of each diagonal brace to the ver- tical. iV-|- 1 be the number of joints. w I k s 6 €o4 THEORY OF STRUCTURES. Then sec 5 = ^, tan 6* = ^. Two cases will be considered. Case \. All the joints loaded. Chord Stresses. — These stresses are greatest when the live load covers the whole of the girder. Let 5„ be the shearing force at a vertical section between the joints n and n-\- i. Let Hn be the horizontal chord stress between the joints « — I and n •\- \. The total load due to both dead and live loads = {w -\- w'){N - I). The reaction at each abutment due to this total load w -A- w' The shearing forces in the different bays are w -\- w' 5„ = {N — i), between o and i ; w A- v/ 5. = — ~-(^-3). 2; and 5. = -^(^-5). ' w A- w' 5. = —^-{N- 7), Sn= -^-{N - 2n - I). 2 " 3; 3 " 4; The corresponding diagonal stresses are Sa sec 0, S, sec 0, .... S^ sec 6. WARREN GIRDER. 605 The last stresses multiplied by sin B give the increments of the chord stresses at each joint. Thus, H^ = tension in 02 = 5„ tan B w + w' I _w-\-w' I N — \ H^ — compression in i 3 = 5, tan 8-j- S, tan _w-\- w' I N — \ w-Vw' I N — 3 ~ 2 li~~N~^ 2 li N _w-\-w' I 2{N— 2) ~ 2 ~k N ' H^ = tension in 2 4 = //, -|" "S^i tan ^ + 6", tan _ w-\- iv' I 2>{N- i) ,, \ " ~ 2 k n' ' ■':"-■'':"■■■',::■■-'.:■ H^ — compression in 3 5 = ^a + 5, tan ^ + 5, tan B _zv -\-w' I 4{N — 4) _ ~ 2 ^ N ' and //„ = horizontal stress in chord, between the joints w-\- w' / n(N— n) , . . , n — I and n-\- i = y tz — , being a tension for a ' 2 k N ' ^ bay in the bottom chord, and a compression for a bay in the top chord. Note. — The same results may be obtained by the method of moments ; e.g., find the chord stress between the joints "— I and n -{- I. Let a vertical plane divide the girder a little on the tight of n. The portion of the girder on the left of the secant plane is kept in equilibrium by the reaction at the left abutment, the liorizontal stresses in the chords, and the stress in the diagonal from « to « -j- I. Take moments about the joint n. Then i'i'il 6o6 THEORY OF STRUCTURES. 2N w 4- w' , N — n — ■ In- N .*. H,t = etc. Diagonal Stresses due to Dead Load. Let d„ be the stress in the diagonal n, n-\- i, due to the dead load. The shearing forces in the different bays due to the dead load are W IV -{N — i), between o and i ; -{N — 3), between i and 2; 2 2 W W -{N-s\ " 2 " 4; -(A^-7), " 3 " 4; w and —{N — 2n — i), between n and « -j- i. The corresponding diagonal stresses are a compression -(N— i) sec 6 =. -{N — iK = ^„ inoi 2 ■ ZV IV s a tension - (iV — 3) sec 9 = —{N — 3)t = df, in i 2 w w s . a compression —{N— 5) sec ^ = —{N — 5)r = a, m 23 2 2 ^ and the stress in the »th diagonal between n and n-\-iis s W S "i WARREN GIRDER. 607 being a tension or a compression according as the brace slopes down or up towards the centre. Diagonal Stresses due to Live Load. — The live load produces the greatest stress in any diagonal («, n-\- \), oi the same kind as that due to the dead load, when it covers the longer of the segments into which the diagonal divides the girder. Repre- sent this maximum stress by D„. The live load produces the greatest stress in any diagonal («, n -\- l), of a kind opposite to that due to the dead load, when it covers the shorter of the segments into which the diagonal divides the girder. Represent this maximum stress by /?„'. The shearing force at any section due to the live load, as it crosses the girder, is the reaction at the end of the unleaded segment, and the corresponding diagonal stress is the product s of this shearing lorce by sec B, or j- The values of the different diagonal stresses are : , : /?, = compression in o i when all the joints are loaded _ sw' N{N- i) ~ k 2 N ' H-' m ZJ, = tension in 1 2 when all the joints except one are loaded _ sw' jN- i){N-2) ~ k 2 N ' i Z), = compression in 2 3 when all the joints except i and 2 are s zv' {N - 2){N - 3) loaded = k 2 N M D^ = tension in 34 when all the joints except i, 2, and 3 are k 2 N D^ = stress in n, n-\- i when all the joints except i, 2, 3, , . . , , , s w' (N — n)(JV — n — 1) and n are loaded = r ^r ~ k 2 N A = stress in o i before the load comes upon the girder = o. ni 6o8 THEORY OF STKUVTUKES. S W Z?/= compression in i 2 when the joint i is loaded = t^i. s w DJ— tension in 2 3 when the joints i and 2 are loaded = -r \,x. ' k N^ /?,'= compression in 34 when the joints i, 2, and 3 are loaded /?„'= stress in «, « + I, when the joints 1,2,... and ware loaded s 7v' n{n-\- i) ~ kN 2""' The total maximum stress in the «th diagonal of the same kind as that due to the dead load = d^-\- D„. The resultant stress in the «th diagonal when the load covers the shorter segment =: d„— D„'. This resultant stress is of the same kind as that due to the dead load so long as d„ > D„', and need not be considered since d„ -\- D^ is the maximum stress of that kind. If D^ > d^, it is necessary to provide for a stress in the given diagonal of a kind opposite to that due to d„ + -^-o ^"<^ equal in amount to D^' — d„: This is effected by counterbracing or by proportioning the bar to bear both the stresses d„ -\- D„ and D„' — d„. Case II. Only joints denoted by even numbers loaded. 2-4 N-2 N Fig. 376. N-1 N-2 N N~T Fig. 377. Chord Stresses. — The stresses are greatest when the live load covers the whole of the girder. The total load d "e to both dead and J .Ve loads 609 The '•eaetton at each abutment zv -\- To find H, , take '"e to this totaUoad . - moments about i. Th en wi.".: To find /^r, take AT A __ ^ + ?f'' / moments about 2. r • i. »«... • v--/- To find A^„ take ^•^='-^'(^-.).^. moments about 3. T° find ^, take „o„,ems about 4. . / • 4 ^^-2)4-^^^^.^^,^^_/^ ,Jofind^., take moments aboutVand^.;,:t « be even. 4 ^^-2)«- "^"+"')^-K«~2)+.(«^ A^ and 4)4-...+6 + 44.2f 4 "~:j -j, 4 >^ iv^ ■• I 6io THEORY OF UTRUCTURES, Next, let n be odd. Then -(w + tt^)^{(» - a)-|-(« - 4) + . . . -f 5-f 3+ ij / . '. ^ ((i\^-2)« {n - ly ) and //- = w-\-w' I (N— 2)n - (« - I)' 4 k N Note. — If — is even, 2 ' w 4- w' / HiV, the stress in the middle bay, = — ^ — tN N If — is odd, "^ -L. 7^;' / ^' ^ /T-v, the stress in the middle bay, = v — t — tt — . Diagonal Stresses due to the Dead Load. — The shearing forces w in the different bays due to the d.ad load are -{N — 2) between w w O and 2, -(N — 6) between 2 and 4, ~{N — 10) between 4 and 4 4 6, etc. The corresponding diagonal stresses are S IV , ^, in o I = T -iN — 2) = ^, in I 2 ; J w ^. in 2 3 = ^ 7(^^- 4) = d, in 34; jr Tf*, d,mA^ = j-^{N- io)'^d,\n 56; etc.. etc., etc. The truss : combined. The chord! "lore parallel r 3"o\v iron susp ^^'gs. 379 and HOWE TRUSS. Thus the stresseq in f k a- ^** io'-; are e.ual -Ca^ • .::,t~ ^^^-H n.eet at an unloaded Diagonal Stresses ^ue to i^T r-^T^ '" '^'"^- - '" Case I. and '' ''' ^^^^ ^''-^•^Thesc are found iV ^^T'«<>dd. there is a single stress at'L r these coJumns. ' ^^' ^°°t of each of The maximum rp<5iilfo.,4. . 'oads /, obuined a/::?o': ' ^'"" ''"= "" ""h ^ead and „Ve ^•g-» the maximum resuJtanf of segment is loaded "^''"' ^^''"^ '" 3 4 when the longer = ^, + A = a; + /?„ and the maximum resulfnnf * -ent is loaded " "'"^^^"^ ^'^^ '" 34 when the shorter seg- ^-^ ^s generally 6oo, in which case ;= 3/ 9. Howe Truss.-Fiff ,73 ,« , ^ Howe truss. ^' ^^« '« a skeleton diagram of a ^''G- 378. 'lie chords f = Kmi, 7^ parallel members phc J™' r"?"'"'y""^«' of three .r ^ '>on s„spe„ders w.Wew'ei''''!,''''''''"" -P"t so" "J (F'fis. 379 and 380). '"'"'='' '"d^ '<> pass between them B^n 612 THEORY OF STRUCTURES. Each member is made up of a number of lengths scarfed or fished together (Figs. 381 and 382). The main braces, shown by the full diagonal lines in Fig. 378, are composed of two or more members. The counter-braces, which are introduced to withstj^id the effect of a live load, and are shown by the dotted diagonal lines in Fig. 378, are cither single or are composed of two or more members. They are set between the main braces, and are bolted to the latter at the points of intersection. . The main braces and counters abut against solid hard-wood or hollow cast-iron angle-blocks (Fig. 380). They are designed to withstand compressive forces only, and are kept in place by tightening up the nuts at the heads of the suspenders. ,-■ sA h'^ V ■■ ""' f- UV r^ ^^ I -I — T -TT- ^5^ Fig. 379- Pig. -do. Fig. 381. rr ?-* V r^ Fig. 38a, hi ZI jl ■ Lil ■ Fig. 383. Fit;. 384. The angle-blocks extend over the whole width of the chords; if they are made of iron, they may be strengthened by ribs. If the bottom chord is of iron, it may be constructed on the same principles as those employed for other iron girders. It often consists of a number of links, set on edge, and connected by pins (Figs. 383 and 3° |.)- I'l such a case the lower an;^le- blocks should hav^e grooves to receive the bars, so as to prevent lateral flexure. If the truss is made entirely of iron, the top chord may be formed of lengths of cast-iron provided with suitable flanges by which they can be bolted together. Angle-blocks may also be cast in the same p'oce with the chord. To determine the stresses in the different members, the HOWE TRUSS. 613 same data are assumed as for the Warren girder, except that A'' is now the number of parcls. Chord Stresses. — These stresses are greatest when the live load covers the whole of the girder. Let //„ be the chord stress in the «th panel. The total load due to both dead and live loads = {w-\- zv'){N — i). The reaction at each abutment due to this total load = !!i±i":(Ar_,). ■ ■ Let a plane MM' divide the truss as in Fig. 378. The por- tion of the truss on the loft of the secant plane is kept in equilibrium by the load upon that portion, the reaction at the left abutment, the chord stresses in the «th panels, and the tension in the «th suspender. First, let the load be on the top chord and take moments about the foot of the «th suspender. Then iv-\-tv' I I n{n — i) H„k = — 7— (A^ - i)«^ - {^v + zv')-^ 3 ^ or H^ zi -\- zv' n(N — n) w -\-iv' I n{N — n) 2~k N • NexJ, let the Joad be on the bottom chord and take moments about the head of the nth suspender. Then BJ zv + zc' n{N - n) ;; — / T^: , as before. Thus, //■„ is the same for corresponding panels, whether the load is on the top or bottom cliord. Diagonal Stresses due to the Dead Load. — Let VJ be the 6i4 THEORY OF STRUCTURES. shearing force in the «th panel, or the tension on the «th sus- pender due to the dead load. Fir&t, let the load be on the top chord. Then F;' = --(A^- \)-nw vi^-^ - «). Next, let the load be on the bottom chord. Then The corresponding diagonal stresses are and Diagonal Stresses due to the Live Load. — Let V^' be the shearing force in the «th panel, or tension on the «th suspen- der, when the live load covers the longer segment. First, let the load be on the top chord. The greatest stress in the «th brace, of the same kind as that produced by the dead load, occurs when all the panel points on the right of MM' are loaded. With such load, Vn\ the shearing force on the left of MM', = the reaction at o «'' , .N— n * and the corresponding diagonal stress, D„, N-n s k s w k 2 = tV:' ^T--^N-n- I)- N Hence, the resultant tension on the «th suspender due to both dead and live loads = F„ = fV + V^' lN~x \ , «;' ^ N-n N ' m HOWE TRUSS. 615 and the resultant maximum compression on the «th brace due to both dead and live loads The live load tends to produce the greatest stress in the «th counter when it covers the shorter segment up to and in- cluding the «th panel point. Even then there will be no stress in the counter unless the effect of the live load exceeds that of the dead load in the (« -f" 0^^^ brace. The shearing force on the right of MM' = the reaction at N w' n{n -f- i) ~2 W Hence, s w' n(n -f- i) D„ , the corresponding diagonal stress, = j ^—77 — -, *ii+i ind the resultant stress in the counter = D„' — Next, let the load be on the bottom chord. Then VJ' = -{N- n) N- « + i and /?« s w k 2 {N-n) N N- n-\-\ N ' Hence, K = K' + v." = .(^ - „) + ^ V _ „)^^=^. id ^. + Z?, = ^1 w[-j- - «J + -{N - n) ^ J . Also, the stress in the «th counter is d: s\iv' n-\-i w ^■-")[- 6i6 THEORY OF STRUCTURES. Note.— A common value of 6 is 45°, when sec 6* = -7 = 1414, and tan 6 = -ry. = i . iVife The end panels and posts, shown by the dotted lines in Fig. 378, may be omitted when the platform is suspended irom the lower chords. 10. Single and Double Intersection Trusses. — Fie,'. 385 represents the simplest form of single-intersection (or l\ \. / \^ y X \X / \ / J Fig. 385. Pratt) truss ; i.e., a truss in which a diagonal crosses one panel only. It may be constructed entirely of iron or steel, or may have the chords and verticals of wood. The verticals arc in compression and the diagonals in tension. The angle-blocks are therefore placed above the top and below the bottom chord. Counter-braces, shown by the dotted diagonals, are in- troduced to withstand the effect of a live load. If the truss is inverted it becomes one of the Howe t\pe. and the stresses in the several members of both trusses may be found in precisely the same manner. Fig. 386 represents a double-intersection (or Whipple! ^C2A'4 Cf 6 I 8 H 7 Fig. 386. truss, i.e., a truss in which a diagonal crosses fzvo panels. It may be constructed entirely of iron or steel. It is of tlie piii- connected type, and the two diagonal systems may be treated independently. Let B' be the inclination of AB to the vertical. " e " " " " A\, CD, . .. to the vertical. Chord Stresses. — These stresses are greatest when the live load covers the whole of the girder. The reaction at ^ from the system ABCD . . . = 4(w-|-ri' 1: " " /i " " " Al2i . . . =^{20^10 )\ SINGLE AND DOUBLE INTERSECTION TRUSSES. 617 M'^xv' being the dead and live loads concentrated at the panel points (7, 2, £, 4, . • . The shearing foces in the different bays are : • ..^ ^ ^ • ;' 4(zy -■}- w') in AC, from the system A BCD ;;",•'''-; ^" ■■ ^123... ABCD... A12 ^ . ,., ABCD ■ ^123 ... ABCD A I 2 ^ . . . ^{w -\- w') in AC, " 3(w + w') in C2, " " •' |(w + w') in 2^, " '" 2{w + w') in £4, " " |(w + w') in 46^, " i(w + w;') in (76, " " ^{w + M/') in 6/, The corresponding diagonal stresses are : 4{w -\- w') sec &' in AB ; 3i(zf + w') ^ec 6 in A i ', l{w -{- w') sec B in (7Z> ; 2|(z£' + ^'') sec in 23 ; etc. Hence, the top chord stresses are : C/m AC = i\{w -\- zv') tan 6' -\- il{zv + iv') tan 6-, C, in C2 = C,-\- T){w + tc') tan 6* = 4(w -|- 2£;') tan 6' -\- 6^{w -f- w') tan 6^ ; Cj in 2E = C^-{- 2^{w -\- w') tan 6 = 4{w -4- w') tan 6^' -f- 9(w -f- ze;') tan 6^ ; etc. The bottom chord stresses are : r, in j5i = 4(7€' 4- w') tan ^' ; 7; in iZ> = r, + i\(:cv + w') tan ^ = df{w -\- w') tan 0' -\- 3i(w + w') tan 6/ = C, . So, T, = C,, Tt = C^, etc., etc. Again, the stress in any diagonal 4 5 of the system A i 2 . . . due to the dead load = i^w sec 6*. , . • ' The live load produces the greatest stress in 4 5, of the same \'t >i i 6i8 THEORY OF STRUCTURES. kind as that due to the dead load, when it is concentrated at all panel points of the system ^ i 2 3 ... on the right of 4. The reaction at A is then ^w' , ^and the corresponding diagonal stress = ^w' sec Q. Hence the maximum resultant stress in 45 = (|ze' -}--'/«/') sec B. The live load tends to produce the greatest stress in any counter 5 8 when it is concentrated at all the panel points of the system ^ i 2 3 ... on the left of 8. The reaction at the right abutment is then \w\ and the corresponding stress in the counter = \w' sec B. Thus, the resultant stress in the counter = ^w' —\w) sec B, \w sec Q being the stress in 6 7 due to the dead load. Similarly, the stresses in any other diagonal and counter may be found. The Pratt truss composed entirely of iron and with some of the details of the Whipple truss is sometimes called a Murpliy-W hippie truss. The Linville truss is a Whipple truss made of wrought-iron, the verticals being tubular columns. II. Post and Quadrangular Trusses. — The peculiarity of the Post truss (Fig. 387) is that the struts are inclined at an angle of about 22° 30' to the vertical, with a view to an economy of material. ° with the vertical. Fig. 388. Fig. 387. The ties cross two panels at an angle of 45 In the quadrangular truss (Fig. 388) the bottom chord has additional points of support half- way between the panel points. The Bollman, Fink, and other bridge-trusses have been referred to in a previous chapter. 12. Bowstring Girder or Truss. — The bowstring girder in its simplest form is represented by Fig. 389, and is an excel- lent structure in point of strength and economy. The top chord is curved, and either springs from shoes (sockets) which are held together by a horizontal tie, or has its ends riveted to those of the tie. t : - / , :■ The strongest bow is one composed of iron or steel cyHn- entrated at ?ht of 4. rresponding ^ress in any J points of v\ and the Thus, the sec 6 being BOWSTRmc TRUSS. ' ^ drical tubes hnf o^ • , ^^ •erhc-i and diagonals. '"'"'" '" '^e attachment of "cal^blf S tr''''"fr'' '""" "-= bow by means o, ^ wnicn are usual v of an To »• ''^ '"eans of ver. ke greatest bread,!, transvefse so" s , "' '"" ^--^ ^=' '^"h lateral flexure. In large br;d;es,,e""T-* "" ■■^^■»-" 1'agonals may be lattice-work ""' "' ^"""1. and If the load upon the girri^r ; ;tat.o„ary, verticals only are renuLd"f°™'^ "'""'""«' ^"d ke neutral axis of the bow should be t ^"=^P"-°". and "ly distributed load, such a That d,! , ''^™''°'''- ^n irregu- » change the shape of the bow and d" " "T'"^ '^^'■"' '"^^ to resist this tendency. ' '' diagonals are introduced ';,e-,bethedeadloadper,i„ealfoot. " ''^^ ^ live " << „ .'. 1 '.' ',' "P"" °f the girder. greatest depth ^^ of .h,gr^^^ ^'^ord Stresses Thp ta covers the whole of",: gS:; "= «-^'«' -hen the live C^aWSi 620 THEORY OF STRUCTURES. Let Hht. the horizontal thrust at the crown. " 7* " . •' " tension in the tie. Imagine the girder to be cut by a vertical plane a little on the right of BD. The portion ABD is kept in equilibrium by the reaction dX A, the weight upon AD, and the forces H and T. '' Take moments about B and D. Then «;•::_ ~ and "w A- iv' \ , '• Let H' be the thrust along the chord at any point P. Let X be the horizontal distance of P from B. The portion PB is kept in equilibrium by the thrust //at B, the thrust H' at P, and the weight {w -\- w')x between P and B. Hence, . , • ^^. . , H'stc'i = H" = H'-{-{w-\-wYx\ i being the inclination of the tangent at P to the horizontal, and - : ,• the thrust at ^ = (-:^/)^^.+ ij . ..J Diagonal Stresses due to Live Load. — Assume that the load is concentrated at the panel points, and let it move from A towards C. If the diagonals slope as in Fig. 390, they are all ties, and the live load produces the greatest stress in any one of them, as QS, when all the panel points from A up to and including Q are loaded. BOWSTRING TRUSS. mt Let X, y be the horizontal and vertical co-ordinates, respec- . tively, of any point on the parabola with respect to B as origin. , u . ■.,, _ .■ . ■ . The equation of the parabola is y = 77^'. (I) Let the tangent at the apex P meet DB produced in L, and DC produced in £. Draw the horizontal line PM. - • --..-■ - ■ ^ From the properties of the parabola, LM = 2BM. - ' Let PM = X and BM = y. From the similar triangles LMP and LDE, LM MP' LD DE' or 21 X k+y • :'.QE=x" ■y 2y x+QE' P -4x' Sx ' Also , c. = e.-(£-.) = <'-=i^. CE ''QE I 2X l-\-2x' (2) Draw ^'F perpendicular to QS produced, and imagine the girder to be cut by a vertical plane a little on the right of PQ. The portion of the girder between PQ and C is kept in equilibrium by the reaction A' at C, the thrust in the bow at P, the tension in the tie at (2, and the stress in the diagonal QS. Denote the stress by D„ , and let the panel OQ be the «th. Let B be the inclination of QS to the horizontal. _ Take moments about E. Then D^EF^R X CE, ■:,ui.:.' i ;.. CE A = ^oE ^^^^^ ^' ^^^ * '■ 622 THEORY OF STRUCTURES. ^ , L.et iV be the total number of panels. Then > I I ■TT^ is a panel length, and tv'-jj is a panel weight. Also.. = 4- i, and hence • . r a • I — 2x CE _N — n R, the reaction at C when the n panel points preceding T are loaded, _ w' .n{n -f- i) \-\ Thus, equation (3) becomes w' N — n D„ = — /(« + i) — -r-^— cosec 6. N' (4) Again, by equation (i). .'.ST=ki I -4 n-\- 1 I -i)} = 4^' (N-n- !)(«+ I) A^= and cosec e=-^ = 5jr— . Hence, finally, '/ivr-.^+^u^-«-o(«+i)rJ_ ^^^ 8 /& N N— n— 1 This formula evidently applies to all the diagonals between D and C. V- } BOWSTRING TRUSS. 623 Similarly, it may be easily shown that the stress in any diagonal between D and A is given by an expression of pre- cisely the same form. Hence, the value of D^ in equation (5) is general for the whole girder. A load moving from C towards A requires diagonals in- clined in an opposite direction to those shown in Fig. 390. Stresses m the Verticals due to the Live Load. — Let F„ be the stress in the «th vertical PQ due to the live load. This stress is evidently a compression, and is a maximum when all the panel points from A up to and including ar^ loaded. Imagine the girder to be cut by a plane S' S" very near PO, Fig. 390. The portion of the girder between S S" and C is kept in equilibrium by the reaction R' at C, the thrust in the bow at P, the tension in the tie at O, and the compression V^ in the vertical. Take moments about E. Then V^QE = R'X CE, or F„ = R' N^ n n and R' , the reaction at C when the (« — i) panel points from A to and ii Hence, w' tiin i^ up to and including are loaded, = — /■ „ -- (6) a general formula for all the verticals. Let v„ be the tension in the «th vertical due to the dead load. The resultant stress in it when the live load covers AO isz'„ — V„, and if negative, this is the maximum compression to which PQ is subjected. If v„ — V„ is positive, the vertical PQ is never in compression. The maximum tension in a vertical occurs when the live / load covers the whole of the girder and = 7£'' -f- the tension due to the dead load. iVote. — The same results are obtained when JV is odd. ii)l I. f mn 624 THEORY OF S7KUCTUKES. 13. Bowstring Girder with Isosceles Bracing. I)itii;;<)nat StrtsSiS liiii' to the Dead Load. — Uiuicr a iload i.ui the bow is equilibrated aiul the tic is subjected to a unidim tensile stress equal in amount to the horizontal thrust at the crown. The braces merely serve to transmit the load to ilu: bow and are all ties. Lot Z, , T', be the tensile stresses in the two diagonals meeting at any panel point Q. Let <^, , ^^ be the inclinations of the diagonals to the horizontal. Let W be the panel wei^jht suspended from Q. --->*.£ Fig. 391. The stress in the tie on each side of Q is the same, and therefore T^, T^, and ^Fare necessarily in equilibrium. Hence, , T - IV ^°^ ^' A T - W—^^'— ^'~ "^^sinl^. + ^J' ''"'' ^'- ''^sin{H,-\-Hy Dia£■OHa/ Stresses due to the Live Load. — Let iV^be the num- ber of /;«//" panels. 2/ The length of a panel = -rj ; the weight at a panel point ' .2/' ^■-.7.: '>"-7 : = w N Let the load move from A towards C. All the braces in- clined like OP are ties, and all those inclined like QP are struts. The live load produces the greatest stress in OP when it covers the girder between A and O. Denote this stress by D„ ; OG is the «th half-panel. As before, D„EF=R X CE. . . . . . . . (i) ifc'tt BoiyarniNG girder with isosceles BRAChva. 625 The load upon AO = nxv'-rt, and hence R = -j^ 2]V~ ' The ratio of CJi to EF is denoted by the same expression as in the preceding article. Thus, , IK = xv' I n-\-2N 8 /fe « -I- 1 N " ^ — ^. (2) N — n — I The live load produces the greatest stress in OAf when it covers the girder up to and including /?. Denote the stress by D„' ; BG is now the «th half panel. Let K' be the reaction at C. As before, - . . CE DJ = R'-^cosccO, I) being the angle MOD. I The weight upon AD = (« — i)w'^, and hence "• (3) R' w' n* — I 2 "W^- III It may be easily shown, as in the preceding article, that CE _N_ 0E~ n n-\- I , and cosec = N [/■+J^i(^-»)»r]* 4«^(iV — n) .-./?' = — w„-,iv-«-,[''+^V-«>rJ % k n N N-n (4) Hence, when the load moves from A towards C, eq. (2) gives the diagonal stress when n is even, and eq. (4) gives the stress when n is odd. If the load moves from C towards A, the stresses are re- versed in kind, so that the braces have to be designed to act both as struts and ties. ,,, , - 626 THEORY OF STRUCTURES. Note. — By inverting Fig. 391, a bowstring girder is obtained with the horizontal chord in compression and the bow in tension. 14. Bowstring Suspension Bridp;e {Lenticular Truss). — This bridge is a combina;;ion of tlie ordinary and inverted bowstrings. The most important example is that erected at Saltash, Cornwall, which has a clear span of 445 feet. The bow is a wrought-iron tube of an elliptical section stiffened at intervals by diaphragms, and the tie is a pair of chains. A girder of this class may be made to resist the action of a passing load either by ':he stiffness of the bow or by diagonal bracing. In Fig. 392, let BD = k, B'D = k' , Let //"be the horizontal thrust at B, and T the horizontal pull at B' , when the live load covers the whole of the girder. Then _ w^iv' r _ First, let k — k' . Then w + w' /' //'= T = 16 k' which is one half of the corresponding stress in a bowstrini; girder of span / and depth /•. (>.ie half of the total load is supported by the bow and ono half is transmitted through the verticals to the tie. Hence, the stress in each vertical — -j;j{^v' -|- w"), zu" being the portion of the dead weight per lineal foot borne by the verticals, and iV"the number of panels. CANTILEVER TRUSSES. 627 The diagonals are strained only under a passing load. Let PP' be a vertical through E, the point of intersection of any two diagonals in the same panel, and let the load move from A towards 0. By drawing the tangent at P and proceeding as in Art. 13, the expression for the diagonal stress in QS becomes, as before. %v' n{n- \) l-2x Similarly, the stress in the vertical QQ' is (I) N N' l-^2X (2) Next, let k and k' be unequal. Let ^Fbe the weight of the bow, W the weight of the tie. Then, under these loads, ' _ TT TTf _ _ 8 k~ ~"8"y6' w k ~k" . (3) The verticals are not strained unless the platform is attached to them along the common chord ADO. In such a case, the weight of the platform is to be included in W' The tangents at /'and P' evidently meet AO produced in the same point O' , for EO' is independent of k or k' . Hence, the stresses in the verticals and diagonals due to the passing load may be obtained as before. 15. Cantilever Trusses. — A cantilever is a structure sup- ported at one end only, and a bridge of which such a structure forms part may be called a cantilever bridge. Two cantilevers 33- BRIDQEOVER ST. LAWRENCE AT NIAGARA. Fig. 393. may project from the supports so as to meet, or a gap may be left between them which may be bridged by an independent 628 THEORY OF STRUCTURES. girder resting upon or hinged to the ends of the cantilevers. The form of the cantilever is subject to considerable variation. ^^ SUKKL'R BRIDGE Fig. 394, FORTH BRIDGE. Fig. 395. Fig.s. 396 to 401 represent the simplest forms of a cantilever frame. If the membei AB has to support a uniformly dis- Fig. 396. Fig. 397. Fig. 398. N\KI\M\ A B Fig. 400. Pig. 401, Fig, 402. . /f^TTVfv?^. Fig. 403. Fio. 404. tributed load as well as a concentrated load at />', intermediate stays may be introduced as shown by the full or by the dt)ttod CA N TILE VKR Tk' USSES. 629 linos in Figs. 398 and 399. Should a live load travel over /i;V, each stay must be designed to bear with safety the maximum stress to which it may be subjected. Figs. 400 and 401 show cantilever trusses with parallel chords. If the truss is of the double-intersection type, Fig. 401, the stresses in the members terminating in Ji become in- determinate. They may be made deternn'nate by introducing a short link BD, Fig. 402. Thus, if, in DJi produced, BG be taken to represent the resultant stress along the link, and if the parallelogram HK be completed, J)K will represent the stress along JiJi, and JUI that along Jil''. This lini. device has been employed to equalize the pressure on the turn-table TT of a swing-bridge (Fig. 403). An " equal- izer" or " rocker-link" BD, Fig. 404, conveys the stresses trans- niit*^"' *^hrough the members of the truss terminating in J) to tht Ci'MVe posts J)T. '1 heoretically, therefore, the pressure over TT will be evenly distributed, whatever the loading may be, if the direction of HI) bisects the angle TB'f and if friction is neglected. The joint between the central span and the cantilever re- quires the most careful consideration and should fulfil the tollinving conditions: (a) The two cantilevers should be free to expand and con- tract under changes of temperature. {/>) Tlie central span should have a longitudinal support which will enable it to withstand the effect of the braking of a train or the pressure of a wind blowing longitudinally. (f) The wind-pressure on the central span should bear equally on the two cantilevers. {i/} The connections at both ends should have sufficient lateral rigidity to check undue lateral vibration. Conditions Ui) and {c) would be fulfilled by supporting the central span like an ordinary bridge-truss upon a rocker bolted down at one ciul and upon a rocker resting on expansion rollers at the other. This, however, would not satisfy condition (/;). It is preferable to support the span by means of rollers or links at botii ends, and to secure it to one cantilever only on the ccntr.d line of the bridge with a large vertical pin, adapted to 630 THEORY OF STRUCTURES. transmit all the lateral shearing force. A similar pin at the other end, free to move in an elongated hole, or some equiva- lent arrangement, as, e.g., a sleeve-joint bearing laterally and with rollers in the seat, is a satisfactory method of transmitting the shearing force at that end also. (If there is an end post, it may be made to act like a hinge so as to allow for expansion, etc.) The points of contrary flexure of the whole bridge under wind-pressure are thus fixed, and all uncertainty as to wind- stresses removed. Where other spans have to be built adjacent to a large can- tilever span, it should not be hastily assumed that it is neces- sarily best to counterbalance the cantilever by a contiguous cantilever in the opposite direction. If it is possible to obtain good foundations and if piers are not expensive, it might be cheaper to build a number of short independent side spans and to secure the cantilever to an independent anchorage. If this is done, care must be taken to give the abutiiunt sufficient sta- bility to take up the unbalanced thrust along the lower boom of the cantilever. Suppose that the cantilever is anchored back by means of a single back-stay. Let W =■ weight necessary to resist the pull of the back- stay ; h = depth of end post of cantilever ; z = horizontal distance between foot of post and anchorage ; M — bending moment at abutment = Wz. If it is now assumed that the sectional areas of the post and back-stay are proportioned to the stresses they have to bear (which is never the case in practice), the quantity of material in these members must be proportional to h hs which is a minimum when z = s/2h. If a horizontal member is introduced between the feet of CA .V I 'IL E I 'ER y 'A' i \SSE S. 6'M the back-stay and the post, the quantity of material becomes proportional to W"-±-~ + W/i + IV^- = 2M'' ^" k h zli which is a minimum when z = h, i.e., when the back-stay slopes at an an J'!' ' CAN TILE VER TR USSES. 633 Agali, the preceding remarks iiulicatc a method of finding the most economical cantilever length in any given case. Tak J, e.g., an opening spanned by two equal cantilevt-rs and t\\ int'jrmediatc girder. Having selected the type of bridge to be cuipioyed for the intermediate span, estimate, either from existing bridges or otherwise, the weights of independent budges of the same type and of different spans. Sketch a skeleton diagram of the cantilever, extending over one-half of the whole span, and apply to it the processes referred to in (A) and \\\\ If /.is the length of the cantilever and P that of a panel, the following table, in which the intermediate span increases by two panel lengths at a time, may be prepared : c _ 2S End lever nier- c U Can- for tons [i. Can- lie at rom diate Can- ue to ILive d its ight. £il 0.-* •0 ■3 B . u "-§s "-UH &s U^ u — 'J^ « v rt *i U C C I- c 3 s ^•^ 0. - X ^ IS ^ H L 1P L - 2P 4/" L -aP bl' L - 6P %P L - 8/" etc. etc. Weight in col. 3 = one-half oi the weight of the intermediate girder ■\- one-half of the live load it carries if uni- formly distributed. (The proportion will be greater than one-half for arbitrarily distributed loads, and may be easily de- termined in the usual manner.) Col. 5 gives the weights obtained as in A. weight on end of cantilever Col. = col. 5 X ■ . ^ 100 Col. 7 gives the weights obtained as in B, Col. 8 = col. 2 + col. 6 -|- col. 7. It is iirportant to bear in mind that an increase in the weight of the central span necessitates a corresponding increase in the 11 ^34 THEORY OF STRUCTURES. weights of the cantilevers. Hence, in order that the weight of the structure may be a minimum, the best material with the highest practicable working unit stress should be employed for the centre span. The table must of course be modified to meet the require- ments of different sites. Thus, if anchorage is needed, a column may be added for the weights of the back-stays, etc. i6. Curve of Cantilever Boom. — Consider a cantilever with one horizontal boom OA, and let x, y be the co-ordinates of any point P in the other boom, O being the origin of co-or- i — * f-3* 405- Fig. 406. dinates and A the abutment end of the cantilever. Let Whit the portion of the weight of an independent span supported at O. Let w be the intensity of the load at the vertical section through P. Assume (i) that there are no diagonal strains, and, hence, that the web consists of vertical members only; (2) that the stress H in the horizontal boom is constant, and therefore the bending moment ^tP=Hy\ (3) that the whole load is transmitted through the vertical members of the web. Let k be such a factor that kTl is the weight of a member of length /, subjected to a stress T. {Note. — If / is in feet and T in tons, then k for steel is about .0003, allowance being made for loss of section or increase of weight at connections.) w consists of two parts, viz., a constant part p, due to the weight of the platform, wind-bracing, etc., which is assumed to El cantilever co-ordinates :Tin of co-or- CURVE OF CANTILEVER BOOM. 635 be uniformly distributed ; and a variable part, due to the weight of the cantilever, which may be obtained as follows: Weight of element dx of horizontal boom = kHdx, " web corresponding to dx ~ kwydx. " element of curved boom corresponding to dx « =./.0^)'... Hence the variable intensity of weight ds' = kH+kwy + kH[£)\ and ds\' w =p-\- kH-\- kwy -(- kH \-j-\ Again, if M is the bending moment and S the shearing force at the vertical section through P, then d'M _dS_ _ dy dx' ~ dx ~ ~ dx' ' .'.H^j^,=p + kH-\-kwy + kH (~J dx' Integrating twice. Iiy = A-\-Bx + {p+2kH)^+kH^, A and B being constants of integration. When X — o, y = o, and H^ ~ W. .dy dx Thus, ^ = o and B= W. 636 Hence, THEORY OF STRUCTURES. Hy = Wx + (/. + 2kHf~ + kH^ is the equation to the curve of the boom, and represents an eUipse with its major axis vertical, and with the lengths of the two axes in a ratio equal to I — t^t j • The depth of the longest cantilever is determined by the vertical tangent at the end of the minor axis, and corrcsijoiids to the value of y given by making — - = o m the preceding equation, which gives y = — . For a given value of H the curve of the boom is independ- ent of the span. Again, for a given length of cantilever with a boom of this elliptic form, a value of H may be found which will make the total weight a minimum, and which will there- fore give the most economical depth. Such an investigation, however, can only be of interest to mathematicians, as the hypotheses are far from being even approximately true in practice, and the resulting depth would be obviously too great. Assumption (l) on page 634 no longer holds when a live load has to be considered. Diagonal bracings must then be introduced, which become heavier as the depth increases, in consequence of their increased length. The diagonal bracings are also largely affected by the length of the panels. If the panels are short, and if a great depth of cantilever, diminishing rapidly away from the abutment, is used, the angles of the diagonal bracing, near the abutment, will be unfavorable to economy. This difficulty may be avoided by adopting a double system of triangulation over the deeper part of the cantilever only, or even a treble system for some distance in a large span. The objections justly urged against multiple systems of triangulation in trusses lose most of their force in large cantilevers. In the first place, the method of erection by building out injures that each diagonal shall take its proper share of the dead load ; and in the second place, it should be mmw^ CURVE OF CANTILEVER BOOM. 637 remembered that only in large spans c(jLild a double system have anytliin;^ to reeommend it, and then only near the abut- ment where the stresses are greatest : in sueh cases the moving load only produces a small portion of the entire stress in the web. In practice, a compromise has to be made between dif- tcrent requirements, and the depth must be kept within such limits as will admit of reasonable proportions in other respects, while the diagonal ties or struts may be allowed to vary in in- clination, to some extent, from one panel to another. Again, in h.King the panel length, care must be taken that there is no undue excess of platform weight, as this will pro- duce a corresponding increase in the weight of the cantilever. An excessive depth of cantilever generally causes an in- crease in the cost of erection. Both theory and practice, however, indicate that it will be more advantageous to choose a greater depth for a cantilever than for an ordinary girder bridge. An ordinary proportion for a large girder bridge would be one-ninth to one seventh of the span, and if for the girder were substituted two cantilevers meeting in the middle of the span, the depth might with advantage be considerably increased beyond this proportion at the abutment, if it be reduced to )iil where the cantilevers meet. When a central span is introduced, resting upon the ends of the two cantilevers, the concentrated load on the end gives an additional reason for still further in- creasing the depth at the ■A\iw\.xt\Q\\X. proportionally to the Ungtii of the cantilever. The greatest economical depth has probably been reached in the Indus bridge, in which the depth at the abutment = ,54 X length of cantilever. Probably the propor- tion of one-third of the length of the cantilever would be ample, except where the anchorage causes a considerable part of the whole weight, but each case must be considered on its own merits. The reduction of deflection obtained by increas- 'vAs^ the depth is also an appreciable consideration. If a depth be chosen not widely different from that which makes the quantity of material a minimum, the weight will be only slightly increased, while it is possible that great structural advantages may be gained in other directions. In recommend- ii ., if IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I l^|2| 2.5 22 lis lllllio 1.8 - 1.25 1.4 1.6 "^ 6" — ► Hiotographic Sciences Corporation ^^ 4 V «^^^ \\ c ^ ^ 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 873-4503 ^ ^^ 'r /. /a J 638 THEORY OF STRUCTURES. ing a great depth for a cantilever at its abutment, it is assumed that the depth will be continuously reduced from the abutmcMit outwards. If the load were continuously distributed, it is by no means certain that a cantilever of uniform depth would re- quire more material than one of varying depth, but it has already been pointed out to what extent the weight of the structure itself necessarily varies, and if the concentrated load at the end were separately considered, the economical truss would be a simple triangular frame of very great depth. From economic considerations, it would be well to reduce the depth of the cantilever at the outer end to nil, but in many cases it is thought advisable to maintain a depth at this point equal to that at the end of the central span, so that the latter may be built out without false-works, under the same system of erection as is pursued in the case of the cantilever. The post at the ends of the central span and cantilever is sometimes hinged to allow for expansion. 17. Deflection. — A serious objection urged against can- tilever bridges is the excessive and irregular deflection to whicli they are sometimes subject. They usually deflect more than ordinary truss-bridges, and the deflection is proportionately increased under suddenly applied loads. In the endeavor to lecover its normal position, the cantilever springs back with increased force and, owing to the small resistance offered by the weight and stiffness at the outer end, there may result, especially in light bridges, a kicking movement. It must, how- ever, be borne in mind that the deflection, of which the impor- tance in connection with iron bridges has always been rcco — RJt — w,{h -\- 1 — a, — x) — w^h -\-l — a^ — x) — . . . — Wjjl -\-l—ar — x) — 7CV+,(/^ -\-l— (Ir^x — x) — ... — Wr+lh -\-l— a,+^ - x). It is assumed, for simplicity, that no weights leave or ad- vance upon the bridge. ' ' r,- ,,--. .-. A = A, LIVE LOAD. 641 according as RJi - will -f / - «^) ^ wih + / - ^,) - . . . -_ w,{h + l-a^) = RJi — w,{h + /—«, — ■*') — ^^>(''' + / — <7, — ^) — . . . — wXh + / — «r — ^) — Wr+,(/' + / — r+,{h + / - «.+,) + . ■ . + w,+,(A 4- / — a^^.,) and Hence, according as R,-R,=jW^. A I A. R:{l-^h)^x{W^-\-T)+W„jh. Take, e.g., the truss represented by the accompanying dia- ijrain (Sault Ste. Marie Bridge), the live load being that shown by Fig. 411, i.e., the loading from a Standard Consolidation engine with four drivers and one leading wheel. .lOrtJt \ cnite 1 X IM •sj'* •^ M^ M J ^ 'X 'J A"rr '-P,""" ' p,"* W','"" •P4"" 'p, i^ H ^ Pi Fig. 411. Span = 239 ft. Length of centre verticals = 40 ft.^ of end verticals = 27 ft. ■f » 642 THEORY OF STRUCTURES. Applying the principles referred to in the preceding it is found that the distributions of live load, concentrated at tin panel points, which will give the maximum stresses in the several members, may be tabulated as below : Distribu- tions, Case I . a . 3' 4 • 5- 6. 7 • 8. Dead weight End Reac- Load Load Load Load Load Load Load tion MA. at/,. at/j. at/,. at/4. at/,. at/,. at/,. •87990 49500 38700 459»5 43750 36225 36000 36000 162920 49500 38700 45925 43750 36225 36000 124230 6400 47200 40200 43400 45800 37100 9S030 6400 47200 40200 43400 45800 69410 6400 47200 40J00 43400 47400 6400 47200 40200 29100 6400 47200 15380 6400 121500 27000 27000 27000 27000 27000 27000 27000 Load at/,. 36000 36000 36000 37100 45800 43400 40200 47200 27000 Load at/,. 360C0 3g <^-')l u I "3 N-i.N-i u 3 !1 Bo S •= — «' s H '" 3 AT Q V + V. ct I a •a ■3 N.N— I ■il 52 > ! S ^ jj — T 5 - "< •S — N.N-iD 2 A^ fifP LIVE LOAD. 645 Col. I designates the several diagonals. Col. 2 gives the multiplier ^.V — r for different values of r. Col. 3 gives the maximum vertical shears due to E trans- mitted through the several diagonals. This shear for any given diagonal is the product of the corresponding multiplier in col. 2 and -jj . Cols. 4 and 5, g and 10 give similar quantities for the live and dead loads. Col. 6 gives the sums of the shears in cols. 3 and 5, i.e., it gives the total maximum vertical shears due to live load. Col. 8 gives the maximum diagonal stresses due to live load. For any specified diagonal it is the product of the cor- responding shear in col. 7 and the secant of the angle between the vertical and the diagonal in question. Col. 1 1 in like manner gives the maximum diagonal stress due to dead load. Col. 12 gives the total maximum diagonal stresses due to both live and dead loads. Another column might be added giving the sectional areas of the diagonals. In the above table the diagonal stresses due to the live and dead loads are separately determined, as different coefficients of strength are sometimes specified for the two kinds of load. With a suitable compound coefficient of strength, cols. 6, 8, and II may be replaced by a column giving the sums of the corresponding shears in cols. 3, 5, and 10. These sums, multi- plied by secant ^, give the maximum diagonal stresses. Stresses in the Verticals. — The maximum stress in any ver- tical, say at the rth panel point, is evidently the vertical com- ponent of the maximum diagonal stress in the r\.\\ panel, i.e., it is the maximum vertical shear in the rth panel. To be more accurate, this amount should be diminished by the portion of the weight of the lower chord borne at the foot of the vertical in question. Chord Stresses. — Take the load at each panel point t, i'll I i 1 646 THEORY OF STRUCTURES. TABLE II. (Compression Chord.) TABLE % ■8-6 II ■5 c I ■2 un fi c 1 5 t Vertical transi c (4 H Stress tr to Chor Diagon US "5 h 13 3 456 Col. I designates the chord panel length. Col. 3 gives the several vertical shears transmitted to the chords through the diagonals. They are the product of -j^ \j^ -j- ^ "f" -^j «^"^ ^^^ multipliers in col. 2. Col. 5 gives the chord stresses due to these shears, i.e., the product of the shears in col. 3 and the corresponding values of tan B in col. 4. Col. 6 gives the total chord stresses in the several panels. In any given panel the total chord stress is equal to the chord stress due to the shear in that panel //?/j the total chord stress in the preceding panel. Another column for the sectional areas of the several lengths of chord may be added if required, each length being designed as a strut, hinged or fixed at the ends, according to ^'iC method of construction. A precisely similar table may be prepared for the tension chord. Example i. An '° ■*''"■ «]^/i/-panelled deck-truss of 108 ft. span and 18 ft. deep, with a single diagonal system. Concentrated load E for each truss = 25,000 lbs. Train load 7" for each truss = 1600 lbs. per lineal ft. = 2 1 ,600 lbs. per panel. Bridge (dead) load D for each truss = 800 lbs. per lineal ft. = 10,800 lbs. per panel. sec = \, tan 6* = f . :« C k' y 7 6 5 4 3 It will mum posU lbs., the fo The result l/iat dtie t minterbrac in the sixth in order to TABLE Chord Si TABLE OF Member. 4(9 mwi LIVE LOAD. Un TABLE OF MAXIMUM DIAGON.-XL STRESSES. (See Table I.) "3 c a k 1 00 II li- 1 00 "k 1 N i II loo 567-X3 Toul Max. Shear due to Live Load. • c 1 •fsi 98218} t 1 » 28 Diag. S'tress due to Dead Load. Total Max. Uiag. Stress. ■^^ 7 21875 21 78575 ■ J 37800 4725" I454«l ,/, 6 18750 •5 40500 59250 li 74062* 30 27000 33750 107812* '^. 5 15625 10 270fX) 42625 5328it 12 16200 20250 7353'i ''f 4 12500 6 16200 28700 35875 4 5400 0750 42625 843l - 4 - 5400 - 6750 "5093} i* a 6250 I 2700 8950 11187* — 12 —16200 —20250 •ii I 3115 3"5 39o6i — JO —27000 -33750 It will be observed that in the fifth panel there is a maxi- mum positive shear of 17,47S 'bs. and a negative shear of 5400 lbs., the former due to the live and the latter to the dead load. The resultant shear of 12,075 lbs., which is opposite in kind to that due to the dead load, is provided for by means of the counterbrace ab. No counterbraces are theoretically required in the sixth and seventh panels, but they are often introduced in order to stiffen the truss. TABLE OF MAXIMUM STRESSES IK THE VERTICALS. ^. = 78575 + 37800 = 1 16.375 lbs. v^ — 59250 + 27000 = 86,250 " v^ = 27000 -|- 16200 = 43,200 " z/«=i62oo-j- 5400= 21,600 " Chord Stresses. — Load at each panel point i = g-+^+ ^ = 35.525 lbs. TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD. Member. 4(9 - ar). ?'f^ = 44401. Tan 9. Chord Stress due to Shear. Total Max. Chord Stress. Cl 28 20 12 4 124337! 888 1 2i 53287! 17762! 93253! 66609! 39965* 13321J 93253! 159862^ 1998281 213150 648 THEORY OF STRUCTURES. TABLE OF MAXIMUM STRESSES IN TENSION CHORD. Member. 4(9 - ar). "r'=4.4o.. Tan 9. Chord Stress due Total M.i» to Shear. Chord Stress t% 28 20 12 I24337i 888i2i 53287i J J 300 00 Total Max. Shear due to Live Load. u ,)S Ma». Diag. Stress due to Live Load. i "a m II Diag. Stress due to Dead Load. Tf.tal Max. ! Dia(j Stress. \ / 7 21875 21 56700 78575 98218} 28 37800 47350 MM68J 7S50 36300 I 45375 I2i 16875 21093} 6646s} di S 15625 3i 9450 25075 45>35 «♦ "'475 20655 65790 i* 4 12500 a* 6750 1925° 34650 4i 6075 >oy35 45585 '',* 3 9375 i 1350 10725 «■ 1930s i 675 • 2:5 20520 i* a 6250 i 1350 7600 « 13680 - 3+ - 47*5 - 8S05 5"75 i* k I56»* 1562* I 2812} - 7* — 10125 -18325 d-, k 1562* 1563* ^ •953* -Hi -»55aS -27945 The counterbrace cf is required to take up the resultant shear of 6250 — 4725 = 1525 lbs., which is opposite in kind to that due to the dead load. The first line in the table gives the maximum thrust alon^ the end post (/). It is made up of the stresses transmitted m through t at the firs TA The m when the : V, z Chords TABLE OF Member Multip 2S Note. — c^ : TABLE OF Membe Multi plier. Ex. 3. A t having the sam LIVE LOAD. 649 throiii^h the two systems of diagonals when the 2^,QOO lbs. is at tlie first panel point. TABLE OF MAXIMUM STRESSES IN VERTICALS. The maximum stress in an end vertical evidently occurs when the 25,000 lbs. is concentrated at its foot. 7', = 25000 -f- 10800 = 35800 lbs. (ten.sion) ; 7', = 19250-1- 6075 = 25325 " (compression); 7',= 10725 -j- 675 = 11400 " " V, = 6250— 4725 = 1525 " Chord Stresses. — Load at each panel point = |^-l-^-^-z) = 35525. It TABLE OF MAXIMUM STRESSES IN COMPRESSION CHORD. Member Multiplier. 3"'5 - A ^ ■ g = 44-.0,. Tan e. Chord Siress due tu Shear. Tola! Maximum Chord Stress. ( 28 ] +I2i ( + 8i 4i i 1243371 55507iJ 37745t^s 199821;! 2220 ,»« * i 93253i 4163011* 566171! • i9i5oiSi 29974 fl's 3330JI 22H76ji; 22jSO()J Note. — c^ is made up of the thrusts transmitted through TABLE OF MAXIMUM STRESSES IN TENSION CHORD. Member. Multi- plier. 35535 . 8 =444oi. TanO. Chord Stress due to Shear. Total Maximum Chord Stress. 28 8i I24337i 55507H 37745 A i i f 93253* 4i630Ji 5661 7|i 93253i l348S3«f Ex. 3. A through-bridge of the Warren type (Fig. 415) having the same span and loading as in Exs. i and 2. C\ Ct C3 C4 ti ti ti ti liiil Hi n :> h 1^^ i:- ■; 1 1^ w t I: 1 i Fig. 415. €$0 THEORY OF STRUCTURES. TABLE OF MAXIMUM STRESSES IN DIAGONALS. 2 8 ■i Mazi hear itted. E~. i a ?3 a 3 S 1 C4 1 00 a 21 II NO 00 N a, ■3 X II 1 " V) a las 9 c« HBi: 1-155 3 . x ■In d^=:d^ 7 2I87S 56700 28 37800 1 16375 1 344 14 d,=d. 6 .8750 15 J 0500 20 27000 86250 gi/)i() dt=.dt 5 15625 ro ■.'7000 12 16200 58825 <-'7i)43 d,=dt 4 12500 6 16200 4 5400 34100 3<)386 dt=dio 3 9375 3 8ioo - 4 - 5400 12075 I3')47 ^ii=dii 2 6250 I 2700 — 12 — 16200 dia=dn I 3125 —20 — 27000 The resultant stresses, d, = = aver. <"»"•)■ -^ 200, or ^ "'^'"'"' '«' = (velocity in „i,e, p", 200* According to Dines the ro....,3houId be 2000' , The Wind.Pressur*^ r^ . . I ,.™„,, '«ure Co„n,.ss,on (Eng.) recommended the /> = lOO' *o.t ii = 4 „ . , "''*"' «""d velocities should be ;- ««,« JpttS'o/rstr '""''^'' - --ponding fc tir tr' ''---) -id"her::ed°r ^'^^ =■- €§2 THEORY OF STRUCTURES. per hour would be 13.1 lbs. per square foot according to Smeaton's rule and only 9.18 lbs. according to Dines. Again, certain experiments at Greenwich indicated tliat the pressure was increased by the stiffness of the copper wire connecting the recording pencil with the pressure plate, and a flexible bra. s chain was therefore substituted for the wire. Thus modified, a pressure of 29 lbs. per square foot was regis- tered as corresponding to a velocity of 64 miles per hour. whereas with the copper wire a pressure of 49!- lbs. per .square foot had been registered with a velocity of only 53 mile.s jxr hour. These facts tend to show that the actual pressure is much less than that given by a recording instrument, and that the very high pressures, as, e.g., 80 lbs. per square foot and even more, must be due to gusts or squalls having a purely local effect. This opinion seems to be confirmed by Sir B. Baker's experiments at the Forth Bridge, which also indicate that the pressure per square foot diminishes as the area acted upon increases. No engineering structure could withstand a press- ure of 80 lbs. per square foot of surface, and a pressure of 28 lbs. to 32 lbs. would overturn carriages, drive trains from the track, and stop all traffic. It is, of course, well known that wind-forces sufficiently powerful to uproot huge trees and to demolish the strongest buildings are occasionally developed by whirlwinds, tornadoes, and cyclones, but these must be classed as acta Dei and can scarcely be considered by an engineer in his calculations. Numerous observations as to the effect of wind upon struc- tures in different localities must yet be made before any useful and reliable rules can be enunciated. In the case of existing bridges the elongation of the wind-braces during a storm can easily be measured within -^-^ of an inch. Investigations should be made as to the action of the wind upon surfaces of different forms and upon sheltered surfaces, as, e.g., upon the surfaces behind the windward face in bridge-trusses. Again, it is quite possible, if not probable, that many of the recorded upsets have been due to a combined lifting and side action, requiring a much less flank-pressure than would be necessary EMPIRICAL REGULATIONS. 653 if there were no upward force, and hence further light should be obtained on this point. Under any circumstances, the wind-stresses should be as small as possible, compatible with safety, seeing how largely they influence the sections of the several members, especially in bridges of long span.- 22. Empirical Regulations. Wind-Pressure Coinmission Rules. — For railway bridges and viaducts assume a maximum pressure of 56 lbs. per square foot upon an area to be estimated as follows : A. In close-%\xdi&x bridges or viaducts the area = area of windward face of girder -f- area of train surface above the top of the same gir- der. B. In o/f«-girder bridges or viaducts the area for the wind- zuard girder = area of windward face, assumed close, between rails and top of train -f- calculated area of windward surface above the top of the train -f- calctilated area of windward surface below the rails. For the leeward girder or girders the area = calculated area of surface of one girder above the top of the train and below the level of the rails, the pressure being 28, 42, or 56 lbs. per square foot, according as this area < f5, > \S and < J5, or >f 5, where 5' is the total area within the out- line of the girder. The assumed factor of safety is to be 4. American Specifications. — [ci) The lateral bracing iti the plane of the roadrvay is to be designed so as to bear a pressure of 30 lbs. per square foot upon the vertical surface of one truss and upon the surface of a train averaging 12 sq. ft., per lineal foot, i.e., 360 lbs. per lineal foot ; this latter is to be re- garded as a live load. The lateral 'tracing in the plane of the other chord is to be designed so as to bear a pressure of 50 lbs. per square foot upon tzoice the vertical surface of one truss. {b) The portal, vertical, and horizontal bracing is to be 654 THEORY OF STRUCTURES. proportioned for a pressure of 30 lbs. per square foot upon twice the vertical surface of one truss and upon the surface of a train averaging 10 sq. ft. per lineal foot, i.e., 300 lbs. per lineal foot, the latter being treated as a live load. (£•) Live load in plane of roadway due to wind-pressure = 300 lbs. per lineal foot. Fixed load in plane of roadway due to wind-pressure = 150 lbs. per lineal foot. Fixed load in plane of other chord due to wind-pressure = 150 lbs. per lineal foot. Lateral Bracing. — Consider a iruss-bridge with parallel chords and panels of length p. Let A be the area of the ver- tical surface of one truss. According to {a), the lateral bracing in the plane of the roadway is subjected to (i) a panel live load of 360/ lbs. and (2) a panel fixed load of 30^ lbs., while in the plane of the other chord it is subjected to a panel fixed load of 50 X 2^1 = \ooA lbs. Thus, if the figure represent the bracing in the plane of ihe roadway of a ten-panel truss, and if the wind blow upon the I I 8«5^^P 86j^p segP se^p 3«^P Mj^p Fig. 416. side AB, the maximum horizontal force for which any diagonal, e.g.- CD, is to be designed is = 4$A lbs. due to the horizontal force of 30^! lbs. at each panel point -f- 756/ lbs. due to the horizontal force of 360/' lbs. at each panel point between C and B. The dotted lines show the bracing required when the \yind blc"W3 on the opposite side. It is sometimes maintained that the wind-forces in the I aioKDs. 655 plane of the upper chords of a through-bridge or the lower chords of a deck-bridge arc transmitted to the floor-bracing through the posts. This can hardly be correct in the case of long posts, as they do not possess sufficient stiffness. It lias, however, been pointed out by Mr. W. B. Dawson that, in through-bridges, the cumulative effect of the wind-pressure at the ends of the bridge might produce a serious bending action ill the end posts. This action would have to be resisted by additional plating on the end posts below the portals, or by an increase of their sectional area. Under wind-pressure the floor-beams act as posts ; hence, if the wind-bracing is attached to the top or compression flange of a floor-beam, the flange's sectional area must be propor- tionately increased. If the bracing is attaclicc^ to the lower or tension flange, the stresses in the latter will be diminished. 22' Chords. — The wind-pressure transmitted through the floor-bracing increases the stresses in the several members, or panel lengths, of the leeward chord, the greatest increments being due to a horizontal force of (36q/> -|- 30.i) lbs. at each of the panel points in AB. The corresponding chord stresses in tile ten-panel truss-bridge referred to above are : K, = 0; C, = 4K360/' + 30/^) tan 6 lbs. ; k=C + 3K36o/+3O/0tan^ r. = C, 4- 2\{2>6op + zoA) tan ^ Y = C, + ik{l6op + 2,oA) tan 6 8(360/' + 30/i) tan B lbs. ; 101(360/ -f-- 30/i) tan ii lbs. ; 12(360/' + 30.i) tan ^ lbs. |50°— 6* being the angle between a diagonal and a chord. Again, the wind-pressure tends to capsize a train and throws l»n additional pressure'of P'^ lbs. per lineal foot upon the lee- pard rail, /"being the pressure in pounds per lineal foot on the pin surface, y the vertical distance between the line of action t/'and the top of the rails, and G the gauge of the rails. % Mi ii m iilil li ^5^ THEORY OF STRUCTURES. Thus, the total pressure on leeward rail and = (- -{- P 7^\ lbs. per lineal foot, the total pressure on windward rail (w v\ 2 ~ ^r) ^^^- P^*" lineal foot, w being the weight of the train in pounds per lineal foot. Hence, the total vertical pressure at a panel point of the leeward truss S-G 2S S being the distance between the trusses. 24. Stringers. — Each length of stringer between consecu- tive floor-beams may be regarded as an independent girder resting upon supports at the ends, and should be designed to bear with safety the absolute maximum bending moment to which it may be subjected by the live load. If the beams are not too far apart, the absolute maximum bending moment will be at the centre when a driver is at that point. Again, in the case of the Sault Stc. Marie Bridge, it may be easily shown that the maximum bending moment is produced when the four pairs of drivers- are between the floor-beams. Let J/ = distance of first driver from nearest point of support. The reaction at this support = J-|tr(824 - Ay) - ^(206 -y). The bending moment is evidently a maximum at the second or third driver, and at the second driver = J5^iL(2o6 — j)(56 -{-y) — 12000 X 56 ; MAXIMUM ALLOWABLE STRESS. 6$7 at the third driver = ■5-§i(2o6 — j/XioS +/) — 12000(52 + 108). In the first case it is an absobite maximum when y = 75" ; " " second " " " J' = 49 its value ill each case being 2,i88,i66f in. -lbs. Hence, the bending moment is an absolute maximum and tqual to 2,i88,i66f in.-lbs., at two points distant 75 in. from each point of support. Also, if /, is the moment of inertia of the section of the stringer at these points, t, the distance of the neutral axis from the outside skin, and /, the coefficient of strength, then 2 / -(2i88i66|) =f,~ for the inner stringer, 3 ^1 and -(2i88i66f) =/, - for the outer stringer. 3 ^i The continuity of the stringers adds considerably to their :5t length. 25. Maximum Allowable Stress. — Denoting by A and B, respectively, the numerically greatest and least stresses to 'A'hich a member is to be subjected, the following rules will give results which are in accordance with the best practice : I. Members subjected to Tensile Stresses only. For wrottght-iron, maximum stress per square inch f ^\ ( ^\ = loooo lbs. = 8000'^ I + ^ j lbs. =: ^3.81 4- '-9 J J *^°"s- For steel, maxin\um stress per square inch = 12000 lbs. = 10000(^1 + -rj lbs. = 1 5.08 -\~ 2.54-jj tons. 658 THEORY OF STRUCTURES, II. Members subjected both to Tensile and Compressive Stresses. For wronght-iron, maximum stress per square inch = 8000^1 - ^j lbs. = ^3 81 — 1-9^) tons. For steel, maximum stress per square inch = 10000(1 - ~j lbs. = [5.08 - 2.54 -^j tons. III. Members subjected to Compressive Stresses only. Denote the ratio of the length (/) to the least radius of gyration {k) by r. The maximum stress per square inch = / I + ar' lbs., /being 8000 lbs. for wrought-iron and to,ooo lbs. for steel, and - being 40,000, 30,000, or 20,000, according as the member has a two square (fixed) ends, one square and one pin end, or two pin ends. Again, the maximum stress per square inch for steel struts with two pin ends = (loooo— 6or)( i -|- -^j lbs.; " " square ends = (lOOOO— 4or)f i + ^j lbs.; « " pin ends = (s - ^)(i + ;^) tons; " " square ends = f 5 — ^]( 1 + -j-) tons. In the last two expressions r < 40. These expressions may be also employed in the case of alternating stresses, but the factor must then be changed to ( i -j ^j. CAMBER. 659 26. Camber. — Owing to the play at the joints, a girder or truss will deflect to a much greater extent than is indicated by tiicory, and the material will receive a permanent set, which, however, will not prove detrimental to the stability of the structure unless it is increased by subsequent loads. If the chords were initially made straight, they would curve down- wards ; and although it does not necessarily follow that the strength of the truss would be sensibly impaired, the appear- ance would not be pleasing. In practice it is often specified that the girder or truss is to nave such a camber or upward convexity that under ordinary loads the grade line will be true and straight ; or, again, that a camber shall be given to the span by making the panel lengths of the top chord greater than those of the bottom chord by .125 in. for every 10 ft. The lengths of the web members in a cambered truss are not the same as if the chords were horizontal, and must be care- fully calculated so as to insure that the several parts will fit together. To find an Approximate Value for the Camber, etc. Let d be the depth of the truss. Let J, , J, be the lengths of the upper and lower chords, re- spectively. Let /, , /, be thp unit stresses in upper and lower chords, respectively. Let H Fill. 490, in the inelineil bars. Let r' be the ih'stanee between the liius of action of these two resultants. The correspomliiif; btiidiiti; action upon the pin is that due to a couple of which tin- mo- ment is /";•. Let // be the distance between the lines of action of tiic equal resultants // of the horizontal stresses upon each side of the pin. The correspondinj; bending action upon the pin is that due to a couple of which the moment is ////. Hence, the niaximum bendin<; action is that due to a couple of which the moment is the resultant of the two moments Vv and ////, viz., Eyt'-bars. — In Enp[land it has been the practice to roll bars having enlarged ends, and to forge the eyes under hydraulic EYRHANS AXn I'/XS. U^i Fro. 431. prcssurr with siiil.ibl)' .li.ipi d dies. In America botli Ii;imiiit.*r- |cii|;ctl .111(1 liydi.iiilir forbid cyo |).irs .iic made, tlic lallfi htin^j i.illcd 7tvA//('.v.v rj'i-/'iifs, Canfid iiiallicnialit al and fxpcriinrn- t.il nivcsli^^.ilions liavi- l»rcn » an icd niit tn d( tcrniiiM: the |»ii'|)r|- (linuMisiotis of III! Iiid<-li(-.id and |)in, hnt owin;.; to lln \iiy iiiniplcx character of the stie.sses developed in t iu' nntal .tionnd I'm eye, ,\u accnratc inathein.ilii al sohition is nnpossihle. I-et // he tile widlli and / tlir tliiclvnesH of the slutiik of the eye-har represented in I'i};. ,|.M. I, el .S' he the widtii of tin metal at llie sides of tlio lyr, and // tlie width at end. Let I) be tlie diameter (if the pin. The proijortions of the head nc ^'overned iiy the ^^eiieral condition that each and every part sliouM i)e at least as stron;.; as the shank. When the bar is snbjected to a tensile stress the pin is ti;.;htly embraced, anil failure may arise from any one of tlic (ollowiii}^ causes : (<■/) The pin may fie shorn throui^h. Hence, if the pin is in double shear, its sectional area should be at least one-half i\\A\. of the shank. It may happen that the pin is bent, but that fracture is pre- vented by the closinff up of the pieces between the pin-head and nut ; the efficiency, however,of the connection is destroyed, as the bars are no lon}^'er free to turn on the pin. In practice, D for Hat bars varies from |c/ to \(l, but usually lies i)etvveen J^/and j|c/. The diameter of the pin for the end of a round bar is gen- erally made equal to i\ times the diameter of the bar. The pin should be turned so as tf) fit the eye accurately, but the best practice allows a difference of from ^^ to ^J^ of an inch in the diameters of the pin and eye. {h) The link may tear across MN. On account of the perforation of the head, the direct pull on the shank is bent out of the straijjht and distributed over Mil illlli ' t; m 664 T/mONy OF STNUCrVK/.S. the sectiotis X There is no rcasdii for the assuinpliMn ili, t the (h'sttihiilioii is uniform, ami it is oijviously pidhalilr iImi the intensity o{ stress is [greatest in the metal next tiie iuik. llcnce, the seetiomil area of the metal aeross JAV must he ,it least equal to that of the shank, ami in practice is alwa)' greater. 5 usually varies from .qi;// to .()2 5f/. The sectional area throu^^h the sides of the eye in the he, id ni a round bar varies from li limes to twice that of the bar. (( ) '/7/f /•/« tfutv /'(• torn f/irottx/t tlic luiut rhiOt(tiuu7\\ the sectional area of the metal across l\) should be cue half that of the shank. The met.d in fnnit dI the pin. h()\\ever, may be likened to .i uniformly loailed i.Mr(lcr with l)oth eiuls fixed, and is subjected to a bemliin; as well .is CO a shearinjT action. Hence, the itiiiiiiiiuni value of // h.i . Iieni fixed at l(t .uid if If is m;»de e(|u.d to (/. botli kinds ol aelioii will be amply jirovided {ow {d) The lu (III II i^ surface ntoy i'e if/SNj/reiei/f. If such be the case, the intensil)- of the pressure upon the bearini; surface is excessive, the eye becomes oval, the met.il is upset, and a fracture takes place. Or a^^ain, as tlu: hole elon- gates, the metal in the sections vS" n^'xt the hole w ill be diawii out, aiul a crack will commence, extending outwards until li,u- ture is produced. In practice, adequate bearing surf.ice m.i\- be obtained by thickening the head so as to confine the maximum intensity of the pressure within a given limit. (e) The head may he torn titrougli the shoulder at XV. Hence, A'l'is made equal to d. The radius of curvature A' of the shoulder varies from ihl to 7.6 u: hoU' clou- iU Uv (h.wvii Is \intil li.u- STK/iL EYEllARS. 665 Vahio Valiir Viilue of 5. ..( ,/. ..f I). WrI.llrin lla'jimrrrd IIkir. M.UB. t f)0 .f.7 '•5 '■33 1 . nc > .75 I.S '•33 1 . (II ) 1 . 1 If 1 1 .=; t.'^.O T .(Jll 1 .l^ 1 u « 50 I .no I n •7 1 . ' 1' ) I . f,n > H5 1. 67 1 III) ' • Tn S.Od 1.07 1 .on a ■ 01 1 ■.■.2'^ 'TS ■ Alio, 111 WplillffiM l);u«. // ; .S. ill li.iitimpicd IjHrs, II ~d. 29. Steel Eye-bars. — llydiiMilic-ffjr^'cd steel eye-bars are iinu heiiij.; l.nj^cly iiiiule. Tin: steel has an (liliinate teiiaeily of tniin 60,000 to 68, OCK) lbs. i)cr square inch, an elastic limit of not less than 50 i)er cent, aiul an elonj^ation of from 17 to 20 per cent ill a li'ii}.;th e(|iial to A;/ times the least transverse dimension. The riin-nix Hridf^e Co. and tlu' Va\<^^ki Moor Iron C! 4^ I ■y^% 3 8 2 2 J 54 I ■iSt 4 3,',i 9 a* 24 54 I ■},^% '! :Mli. i/'>.'tU 10 24 3i f-i r 33^ 5 3 :;. .»i°„ 1 1 3 2i ''4 1 33'^ 5 .1 >,. 5,"« 12 3 4 K S 33^ 5 5 A. 4 i 13 4 ^! '>4 \ 33:j; U 4 13 4 loi \ ysi 6 Si",,. '?,".,.■ 5 \ 14 5 4« Hi \ Zl% f) '',",T ''lA.'' 8 15 5 5ii 124 i 7,1% 7 ?i"a IS f> 5i 134 i 31% 7 f'lS. 7i"„. 7 lO () f'i 144 I 21% 7 • 7 7 511 ■ 54 1! 40% 8 'i/'fl 17 7 74 17 4"% 8 f>IJ.'>i8. 7i"„ iS 8 5! '7 I 40% 8 7U. Sil 1') 8 OS 18 I 4'i% 8 8i. •)& 20 <) 7A. 7U 20 ') 81). 8J 21 10 SB 22 10 8a, «jI 23 10 10, loi 24 In l)oth the Plueiiix and Edge Moor bars the thickness of the head is the same as that of the Ixidy of the bar, or does not exceed it by more than ,'j in. ittiii Zsmk' ■ 666 THEORY OF STRUCTURES, 30. Rivets. — A rivet is an iron or steel shank, sHqlitlv tapered at one end (tlie taW), and surmounted at the otlici by a cup ox pan-shaped head {Vx^, 422). It is used to join steel or iron plates, bars, etc. For this purpose the rivet is generally heated to a cherry-red, the shank or spindle is passed through ^ Fio. 4ia. Fig. 433. Fig. 434, ZZii Fig. 425. Ku.. 4j6. the hole prepared for it, and the tail is made into a button, or point. The hollow cup-tool gives to the point a nearly hemi- spherical shape, find forms what is called a snap-rivet (Fig. 423). Snap-rivets, partly for the sake of appearance, arc com- monly used in girder-work, but tiiey are not so tight as conical. pointed rivets (j/<(/!^-rivets), which are hammered into shape until almost cold (Fig. 424). When a smooth surface is required, the rivets are counter-sunk (Fig. 425). The counter-sinking is drilled and may e.xteiul through the plate, or a shoulder may be left at the inner eilgc. Cold-riveting is adopted for the small rivets in boiler work and also wherever heating is impracticable, but tightly-drivcii turned bolts are sometimes substituted for the rivets. In all such cases the material of the rivets or bolts should be of su- perior quality. Loose rivets are easily discovered by tapping, and. if very loose, should be at once replaced. It must be borne in niii)d, however, that expansions and contractions of a complicated character invariably accompanj' hot-riveting, and it cainiot be supposed that the rivets will bj perfectly tight. Intleed, it is doubtful whether a rivet has any hold in a straight drilled hole, except at the ends. Riveting is accomplished either by hand or machine, the latter being far the more effective. A machine will s(|uee/e a rivet, at almost any temperature, into a most irregular hole, but the exigencies of practical conditions often prevent its use, except for ordinary work, and its advantages can rarely be obtained Sl'KKArCTH OF PUNCHED AND DRILLED PLATES. 667 where the)- would be most appreciated, as, e.tj., in the rivetitiij; up of connections. 31. Dimensions of Rivets. — Tlie diameter (<-/) of a rivet in ordinary |jfirder-\vork varies from \ in. to i inch, and rarely exceeds 1^ in. The thickness {i) of a plate in ordinary tjirder-work should never be less than \ in., and a thickness of f in., or even -^^ in., is preferable. Let T be the total thickness through which a rivet passes. According to Fairbairn, When t <\ in., d should be about 2t. When t > \ in., d should be about \^t. According to Unwin, When / varies from \ in. to i in. and passes through tivo thicknesses of plate, d lies between \t -\- -^ and T <; When the rivets join several plates, d = (- -. 8 8 According to I'^ench practice, Diameter of head = i^d. Length of rivet from head = T -\- l^d. According to Rankine, Length of rivet from head —-T-\- 2^d. The rise of the heatl = %d. The diameter of the rivet-hole is made larger than that of the shank b\' from jV to -J in., so as to allow for the expansion of the latter when hot. There seems to be no objection to the use of long rivets, provided they are properly heated and secured. 32. Strength of Punched and Drilled Plates. — Experi- ment shows that the tenacity of iron and steel plates is con- siderably diminished by punching. This deterioration in tenacity seems to be due to a molecular change in a narrow aniiulus of the metal around the hole. The removal of the annulus largely neutralizes the effect of the punching, and, hence, the holes are sometimes punched ^ in. less in diameter than the rivets and are subsequently rimered or drilled out to the full size The original stren;,';th may also be almost t:l III '. ' -f )\' 'i T'J7a&<1B^ H i 'f 11^ I ' k 1 !> V • P k i: €68 THEORY OF STRUCTURES. entirely restored by annealing, and, generally, in steel work, either this process is adopted or the annulus referred to above is removed. Punching does not sensibly affect the strength of Landoie- Siemens unannealcd plates, and only slightly diminishes the strength of thin steel plates, but causes a considerable loss of tenacity in thick steel plates ; the loss, however, is less tlian for iron plates. The harder the material the greater is the loss of tenacity. Iron seems to suffer more from punching when the lioles are near the edge than when removed to sonic distance from it, while mild steel suffers less when the hole is one diameter from the edge than when it is so far that there is no bulging at the edge. The injury caused by punching may be avoided b}' drilling the holes. In important girder-work and whenever great accuracy of workmanship is required, a uniform pitch nia\' be insured and the full strength of the metal retained by the use of multiple drills. Drilling is a necessity for first-class work when the diameter of the holes is less than the thickness of the plate, and also when several plates are piled. It is impos- sible to punch plates, bars, angles, etc., in spite of all ex- pedients, in such a manner that the holes in any two exactly correspond, and the irregularity becomes intensified in a pile, the passage of the rivet often being completely blocked. A drift, or rimer, is then driven through the hole by main force, cracking and bending the plates in its passage, and separating them one from another. The holes may be punched for ordinary work, and in plates of which the thickness is less than the diameter of the rivets. Whenever the metal is of an inferior quality, the holes should be drilled. 33. Riveted Joints. — In lap joints (Figs. 427 and 430) the plates overlap and are riveted together by one or more rows of rivets which are said to be in single shear, as each rivet has to be sheared through one section only. In Jish (or butt) joints (Figs. 428 and 429) the rivets are in double shear, i.e., must be each sheared through two sections. RIVETED JOINTS. 669 Thus they are not subjected to the one-sided pull to which rivets in single shear are liable. IH Fig. 429. Fig. 430. \x\ fish pints the ends of the plates meet, and the plates are riveted to a single cover (Fig. 428), or to two covers (Fig. 429), by means of one or more rows of rivets on each side of the joint. A fish joint is properly termed a butt joint when the plates are in compression. The plates should butt evenly against one another, although they seldom do so in practice. Indeed, the mere process of riveting draws the plates slightly apart, leav- ing a gap which is often concealed by caulkmg. A much better method is to fill up the space with some such hard sub- stance as cast-zinc, but the best method, if the work will allow of the increased cost, is to form 7i jjwip joint, i.e., to plane the ejes of the plates carefully, and then bring them into close contact, when a short cover with one or two rows of rivets will suffice to hold them in position. The riveting is said to be single, donhh\ triple, etc., according as the joint is secured by one, two, three, or more rows of rivets. 000 000 000 000 000 000 o°o o o 0% CHAIN Fig. 431. ZIGZAG Fig. 432. Double, triple, etc., riveting may be chain (Fig. 431) or zig- zag (Fig. 432). In the former case the rivets form straight lines longitudinally and transversely, while in the latter the rivets in each row divide the space between the rivets in adja- cent rows. Experiments indicate that chain is somewhat stronger than zigzag riveting. I I it. 670 THEORY OF STRUCTURES. Figs. 433 to 435 show forms of joint usually adopted for bridge-work. In boiler-work the rivets are necessarily very close together, and if the strength of the solid plate be assumed to be 100, the strength of a single-riveted joint hardly exceeds 50, while double-riveting will only increase it to 60 or 70. Fair- Fig. 433. Fig. 434. Fig. 435. bairn proposed to make the joint and unpunched plate equally strong by increasing the thickness of the punched portion of the plate, but this is somewhat difficult in practice. The stre.sses developed in a riveted joint are of a most com- plex character and can hardly be subjected to exact mathe- matical analysi.s. ' For example, the distribution of stress will be necessarily irregular {a) if the pull upon the joint is one- sided ; {b) when local action exists, or the plates stretch, or in- ternal straiiis are in the metal before punching; {c) if there is a lack of symmetry in the arrangement of the rivets, so that one rivet is more severely strained than another; (d) when the workmanship is defective. The joint may fail in any one of the following ways : (i) The rivets may shear. (2) The rivets may be forced into and crush the plate. (3) The rivets may be torn out of the plate. (4) The plate may tear in a direction transverse to that of the stress. The resistance to rupture should be the same in each of the four cases, and always as great as possible. The shearing and tensile strengths of plate-iron are very nearly equal. Thus, iron with a tenacity of 20 tons per square iarh has a shearing strength of 18 to 20 tons per square inch. !vv:t-iron is usually somewhat stronger than plate-iron. \gain, the shearing strength of steel per square inch varies THEORETICAL DEDUCTIONS. 671 from . 30ut 24 tons for steel, with a tenacity of about 30 tons, to about 33 tons for steel, with a tenacity of about 50 tons ; an average value for rivet-steel with a tenacity of 30 tons being 24 tons. Hence, if 4 be a factor of safety, the working coefficients become T^ , ^ . ( 5 tons per square inch in shear, and Tor wroueht-iron -( -^ ^ ^ ^ ( 5 " " " " '• tension. ^ . (6 tons per square inch in r or steel ( 7Jr " " " " *' in shear, and tension Allowance, however, must be made for irregularity in the dis- tribution of stress and for defective workmanship, and in riveting wrought-iron plates together it is a common practice to make the aggregate section of the rivets at least equal to and sometimes 20 per cent greater than the net section of the plate through the rivet-holes. Hence, the working coefficients are reduced to and 4 or 4^ tons per square inch for wrought-iron, 5 or 5i " " " " " steel. according to the character of the joint. There is very little reliable information respecting the in- dentation of plates by rivets and bolts, and it is most uncertain to what extent the tenacity of the plates is affected by such indentation. Further experiments are required to show the effect of the crushing pressure upon the bearing area (i.e., tJie diameter of the rivet niultiplicd by the thickness of the plate], although a few indicate that the shearing strength of the rivet diminishes after the intensity of the bearing pressure exceeds a certain maximum limit. 34. Theoretical Deductions. Let S be the total stress at a riveted joint ; /,,/,,/,,/,, be the safe tensile, shearing, compressive, and bearing unit stresses, respectively ; t be the thickness of a plate, and w its width , liil ill! I .1 -1 'it- 672 THEORY OF STRUCTURES. iVbe the total number of rivets on one side of a joint; n be the total number of rivets in one row ; p be the pitch of the rivets, i.e., the distance centre to centre ; d be the diameter of the rivets ; X be the distance between the centre line of the nearest row of rivets and the edge of the plate. Value of x. — It has been found that the minimum safe value of X is d, and this in most cases gives a sufificient overlap {= 2x), while X = ^d is a maximum limit which amply pro- vides for the bending and shearing to which the joint may be subjected. Thus the overlap will vary from 2d to ^d. X may be supposed to consist of a length x,, to resist the shearing action, and a length x^ to resist the bending action. It is impossible to determine theoretically the exact value of x^, as the straining at the joint is very complex, but the metal in front of each rivet (the rivets at the ends of the joint ex- cepted) may be likened to a uniformly loaded beam of length d d, depth x^ , and breadth /, with both ends Jixed. Its (^> - t) ' f being the / moment of resistance is therefore yrt 6 maximum unit stress due to the bending. Also, if P is the load upon the rivet, the mean of the bending moments at the P end and centre is -^d. o Hence, approximately ^ i^^-ii - -=i^(-^)- It will be assumed that the shearing strength of the rivet is equal to the strength of a beam to resist cross-breaking. Single-riveted lap and single-cover joints (Figs. 427 and 428). nd' /. = {p-d)tA = dtf,) . . (I) THEORE TJCA L DED UCTIONS. 673 8 / (2) ^ ^ \ " 2/ 4. ■" .-. ;«;,= - + I / d' f (3) As already pointed out, these joints are weakened by the bending action developed, and possibly also by the concentra- tion of the stress towards the inner faces of the plates. Single-riveted double-cover Joints (Fig. 429). Ttd' ^ ^■4 ^' 2^//= = 2-—/,. {p-d)tf,=dtf,', . nd' (4) • • *^ \ Id ndl 4 t ' (5) ^/\* dV _ I na- ~'2l -2~A-^' ^ , I ' 2 ' 4 \/^r r (6) These joints are much stronger than joints with siiigle covers. Also, equation (4) shows that the bearing unit stress in a double-cover joint is twice as great {theoretically) as in a single-cover joint (eq. i), so that rivets of a larger diameter may be employed in the latter than is possible in the former, for corresponding values of -. m t^: ; 674 THEORY OF STRUCTURES. Chain-riveted joints (Fig. 431). fiw-nd)t = S = f,Ndt\ _ y.r'^d S = N — /^ when there is one cover only ; . 4 nd' S = N /, when there are two covers. . . . This class of joint is employed for the flanges of bridge- girders, the plates being piled as in Figs. 436, 437, 438, and ;/ being usually 3, 4, or 5. In Fig. 437 the plates are grouped so as to break joint, and opinions differ as to whether this arrangement is superior to the full butt shown in Fig. 438. The advantages of the latter /.^/~^^-^ r^ r^ r\ r> /-\ I.e., a (7) tucen Le (8) 13, • ■ Til dently (9) Let sile strt 2 2,3i ^ X 3 Fig. 436. I Ci Ci — Ci Q lO Ci Q Q_ X ru X Fig. 437. ^3- "C "O \J KJ C7 ^^~~0 — CT" Fig. 438. are that the plates may be cut in uniform lengths, and the flanges built up with a degree of accuracy which cannot be otherwise attained, while the short and awkward pieces accom- panying broken joints are dispen.sed with. A good practical rule, and one saving much labor and ex- pense, is to make the lengths of the plates, bars, etc., multiples of the pitch, and to design the covers, connections, etc., so as to interfere with the pitch as little as possible. The distance between two consecutive joints of a group (Fig. 437) is generally made equal to t^vice the pitch. An excellent plan for lap and single-cover joints is to arrange the rivets as shown in Figs. 431 to 435. The strength of the plate at the joint is only weakened by one rivet-hole, for the plate cannot tear at its weakest section, Assuir Hence above rek A, the ass 35. Co must not 1 a si//£-le CO if there an When happens th the greatei COVERS. 675 i.e., along the central row of rivets {an), until the rivets be- tween it and the edge are shorn in two. Let there be m rows of rivets, i 1,22, 33,... (Fig. 439). The total number of rivets dcntly IS evi- VI' Let /,,^,,^3,^, be the unit ten- sile stresses in the plate along the lines i i, 2 2, 3 3, respectively. Then nd' 3 2 I 3 2 Fio. 439. S = {w- d)tf^ = —- ;«y; , for the line i i ; 4 I i 11 4 2 2 ; 4 3 3i ill! Ttd^ = (w-4^)/^, = — -( m 6y, S = (w — d)tf, = {w — 2d) m m — I ;^?, = 44; t\ > are each less than Assume that/, = q^ . Then w =. {in' + i)^. Hence, by substituting this value of iv in the first of the above relations, - = — -9. Since g. , q, t II /, 33.^4 /, , the assumption is justifiable. 35. Covers. — In tension joints the strength of the covers must not be less than that of the plates to be united. Hence, a single cover should be at least as thick as a single plate ; and if there are two covers, each should be at least half as thick. When two covers are used in a tension pile it often happens that a joint occurs in the top or bottom plate, so that the greater portion of the stress in that plate may have to be p! 676 THEORY OF STRUCTURES. borne by the nearest cover. It is, therefore, considered advis- able to make its thickness five-eighths that of the plate. The number of the joints should be reduced to a minimum, as the introduction of covers adds a large percentage to the dead weight of the pile. Covers might be whollj' dispensed with in pcrfcct-jnmp joints, and a great econoni)'^ of material effected, if the dif- ficulty of forming such joints and the increased cost did not render them impracticable. Hence, it may be said that covers are required for all compressicn joints, and that they must be as strong as the plates ; for, unless the plates butt closely, the whole of the thrust will be transmitted through the covers. In some of the best examples of bridge construction the ten- sion and compression joints are identical. 36. Efficiency of Riveted Joints.* — The efficiency of a riveted joint is the ratio of the maximum stress which can be transmitted to the plates through the joint to the strength of the solid plates. Denote this maximum efficiency by r}. Let / be the pitch of the rivets ; d " diameter of the, rivets ; / " thickness of the plates ; ft " tenacity of the solid plate ; mft " " " " riveted plate ; f, " shearing strength of the rivets ; N " number of rivets in a pitch length ; e " ratio of the strength of a rivet in double shear to its strength in single shear. Then I/, = efficiency as regards the plates = — J. — - ^^P_=J), (I) »;, = efficiency as regards the rivets = eN-d'f. 4 Ptft ' * From an article by Professor Nicolson in the Engineer, Oct. 9, ifi (2) EFFICIENCY OF RIVETED JOINTS. 677 The eflficiency of the joint is, of course, the sr,aller of these two values ; and the joint is one of maximum efficiency when »/, = 1;, = V ; that is, when nt p — d __ 4_ ^ or .71 {p-d)tmf, = cN-dV. 4 (3) In this expression the quantities m ft, N, and c are con- stants for any given joint, being of necessity known, or having been fixed beforehand ; and the equation thus expresses one condition governing the relations of the three variables p, d, and t to each other. It is obvious, however, that, in order to determine the values of any two of these variables in terms of the third, another relation between them must be postulated. In short, in designing a joint, the value of one of the three ratios ~, -, and - must be fixed. at t ■ P Case I. Suppose that the ratio -, has a certain value. This is very frequently the quantity predetermined ; but it is most usually done by fixing the value of v> V very obviously P r I A volving -,', m fact Tf=m\i — -y. in d' ' ''\' pi Equation (3) may be written p ^ eN'^d' A j^ d, ^ 4 t mfi ' or P = d[eN~%^^-\-'^ (4) d If the ratio - be denoted by k, then mu d 4 '«/* (5) 678 THEORY OF STRUCTURES. _, ; ' m{p —■ d) ' ' But Since 7 = ^ -, P P _ m 1~ m-r, (^) Therefore, substituting in (5), and, ultimately, '^~ eNTt f/m-v ^^^ The process of designing a joint of maximum efficiency for a boiler of given diameter and pressure of steam, when /; for the ratio ^j is fixed, is then as follows : Settle the number of rivets per pitch (i.e., N) ; the value to be allowed for e (de- pending on the nature of the shearing stress on the rivets^ ; and the values of m,fif and /,. Then k is known from equa- tion (8). But / may be found from the relation, pressure X diameter = rf X 2t/t , or pressure X diameter / = ^^ • • • ... (9^ Hence, since ^ = — is known, d may be found ; and since -. = is known, p is also fixed. ": . d m — 1? -^ :.■■■■■■■'. P Case II. When -, the ratio of rivet pitch to plate thick- ness, is given, equation (5) must be otherwise manipulated. Mu have Putting For bre and solv Ther A, T, their valu term beii: Now, and since plate (/) : values of the knowi This n rational ol will remai of rivets t( the relativ Case must first ii'li ff- ( EFFICIENCY OF RIVETED JOINTS. 6;9 Multiplying it by -, and substituting for d its value kt, we P have . = £^'i*.A+i. 4 P I'ifi P Putting this in the form of a quadratic equation in k, (10) eNn f cNn f, t . . (II) For brevity, substituting A for —ry- , T for -j- , and R for -^ and solving the quadratic, AT I k=^-—±-\fAT-^AA TR. . . . (12) ^f'lii The method of designing the joint is, then, as follows : A, T, and R being known, k may be found by substituting their values in equation (12), the positive sign of the second term being taken. Now, ff = m{l -j) = m{i-^) = ^(i -|) ; and since both k and R are now known, the thickness of plate {() may be found, as in Case I, by equation (9), The values of the diameter and pitch of rivets follow at once from the known values of k and R. This method of designing a joint appears to be the most rational of the three. For the greatest pitch for which a joint will remain steam-tight depends mainly on the relation of pitch of rivets to thickness of plates ; although it is also affected by the relative size of rivets and of rivet-heads. d Case III. If — , or k, be predetermined, the value of 7 must first be obtained, in order that the plate thickness may ill. I 68o THEORY OF STRUCTURES. P ~ d be found by means of equation (9). Now, r} = m- may be put into the form / = tnd and if this value is substituted for/ in equation (4), md m — V \ A V From this is finally deduced rf =. m Vlft eNnkf )d eNnkf, + 4w// (13) The plate thickness may now be found by equation (9); the diameter of rivet from d = kt, and the pitch from p = . In the above investigations no account has been ^ m— rf ° taken of the effect of the bearing pressure on the rivets or plate. li fc be the allowable bearing pressure per projected square inch of rivet surface, the following relation must obtain : {p-d)tmf,^Ndtf, (14) This may be written / = {p — d)mft Nd ^ ' (IS) Then if/,, be estimated by this equation, and if it should be greater than 43 tons per square inch in a lap joint, or 45 to 50 tons in a butt joint, such joint will fail by the rivets shearing before the full strength of the plate is exerted, as Kennedy's experiments show that with these values of f the rivets do not attain their natural ultimate shearing strength (viz.,/,), but fail at shearing stresses much beL .v this. Again prelimina using the deduced fi {Unwin suj In desij any value should be r Note.- have been f bridges in q EFFICIENCY OF RIVETED JOINTS. d 68 1 Again, the maximum allowable ratio of - (i.e., K) as the preliminary datum for the design of a joint, may be fixed by using the expression deduced from the obvious relati'^n — similar to (14)^ (16) eN-d'f, = Ndtf,. 4 (Unwin suggests the relation d=:. ^ V^.) In designing the joint by any of the methods given above, any value obtained for k greater than that supplied by (16) should be rejected. * m Itii:. Note, — The following Tables of the Weights of Bridges have been prepared from data supplied by the engineers of the bridges in question. 'i H:i '^i t 682 THEORY OF STRUCTURES, tr ° o z o o u ^ Q O S < z C g ■< «« Si tC « O !« >-< 'i'*5 in — mr^.o^r^inO'^O'^o coin m I????. U'ti-o 00 O « wi^ IT) m m m m>o m*Na\t^ mw vo w' ox T (>. ^ - CO - - 00 00 aj ^; I ^-•p •?" 1 c^ Oqqcicvqooqci f>qci qC):t;Cic:iqqnoQ d Q qCia;0!Nk(;(5(; | u b £ Q J3 J3 ^ i' H HMHMHMiHHHH(4NHHC4HHNMHt4nHMH « M WWNHMNHMMl ^°2 1 H K «i •». *, ^ ^ ^ ^ fi w 5, ;f ^ ^ ^ . .V . ^" . O c^^o vo 8- •k «, *M Vi "in "ui*in'Vvo '^\r\\n 'V>^o lb m in invo« l^vo t^MH ] Q w ro fi i V?" ; _ ! 1^.00^ 00 X 00 00 9 \ SSP; 5 5 s-g, .? J St; :?R 00 >o o 2 ' 6>» i H „MMMHM1H«.1-I-HI-M«HMHH-MI-MHMH ■«• i/t >• H H B M B' n M •• , log. (rt J o c Oh o u = i w H X U XI < 1) (fa a. a 3 T3 E o ill in c u 5 •S-S ■^ '»;'»; O O C _ UUUwO§ Cf.CS-' tS???. S;C5K^.c;C;s; 1 Ct H H H H M « |vb« Vy: VVV TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 683 lilt u c m 03 >- < t/1 o 2 ^ w a. O l-J ^ (I- -1 5 u u -^ tt. "a. C g. 4-1 •a E o J' c (/) 10 •5)2 000 TJ •c — TS fid U U) o u :^ - o o 613 M . o He o o inn" O in S3 •3^2 . « u c w fco J- § •5, S U)I3 . 4:, K. ^ ^ -a jf " * 3 o a ^ o m C" M 00 CiCiQCi^^CiCiqCl^C^ Cl qo C^Q Ri C| 5; KSCi^-QCs (s, "SJ3 o « M N M N W w N (N M N N N 03 r*- r>. 00 *b V. Pi ^ ^ in m invo c^> (*. t-» ^ "n i: "o V Vo V. 00 "g "0 *« Vi& *in *b 0000^OO^ 0« d^O*OHOOOOOO*- »^ I H H M H H rO I 684 THEORY OF STRUCTURES, TABLl "O c V) n u ?: o ■< o S »-H H pa < Vi 7^ «/) o D m tc OS H H H W ^ fri u Ou o O < ^ z o OiS Q u U •a* o S s Z O b 2 o ^ O) H OS E O o o w u H U < o n ««; CU U) c IS (4 •a E 2 V. o . ») u 0) c u c WOQ Ji o o « SU c- 11 O - g •S."' " c 5"; " a i n o I 5 a V 03 CQ o o a a (Q (A W tf) 3 3 c c .2.2 u u u u « (fl V- u c o be be c o BO 'o" i a II II tn vi £i .£. . II o- II a J! 2j< :" u u Sc2c .£ .2 T" j" o J« 1> i i' Si; E £ Ji S ii ij [fl ?; « pi ui :/: g aj tat .2 -.2 c OS . « = (3 t ix" c .u ^ n O aj ° o ^J 5 c o U -=? c u oo=-h"5 = uUwc f- MM \^ 8 8 S^oS- hsMN vO00«»O VON vC«0 c:! qqf>, f-.C|C| f^ qc:)Cit,hC|C)t«,C)Ot:iqqq^,qt,K, q rt « H « W « H M M N l-tM«MM(4HMMMMHMC4MHHM M 6 V) » 03 H MM ^ MWM MM ^ CI «»- W « 8 :v'S' I H 00 M r^ Ot M r^Vi Vivb "V*"- Vvb vb "b V. O O ? ^ ? S ? ^ -«>C o ^ >^ •^ •^OO VO ■♦ ts 0\Wn^wC?'N'0 l°l n H m m (A o .c •a c z" ■< H w a, Q < Z o Q o .-. < Q S i -■ ^ c tr. o ^ (/3 O u h O < o K a: s •o II "5. c !S5 ' ^•o' (/] °13 6 J ■a !5h c ^ , A a tn u •D Q F o b. £=.S **— ■ MUft « 2t J" c uft. Nil TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 68$ o Q ~ '< 3C >< J3 C < s a, (/5 Z (« O a: w w a S o !/) H -r 3 f- U < O u cc < T3 C *-• E o I 8 II I o S .h: ^ T « XI en C II eg M .- r* U C 00 » O 0^»J 1. 'S o g o u S£ rt— §-, !"»:& »»! a V 3 3 O Q !' ->- 4> u. a> Was a o o o u ?= CQ •= c « 8 S O 1) O 4J o ;i 3 0.9; OS* •a Ed o dU u . , §.-0 o S ° ft.T3 S" Q ' J3 I O 3 u En KEnK ClQSN^t,t>,^(^^^ EnK {n, o ti n HMNM«W« », ft 4J O •a « r» ir. Q w \n t^«o ^oo inmoQONvom 1/5 i 686 THEORY OF STRUCTURES. TABLL Si 13 c CI) < s t/5 o Q i-i ■< tn z; o as OS m S Z a. u n a, _) O < z 2; o cd U u < Q ^ O ^ Jrt as" I £ tfl »J u CU roso .-■' .■9-OT3 ■a aa »2a u u 1^ a J c u B c^ = 00 ^ u = - O uvtfi 41 . > a . "zl .It COM II S) U t^ CI ;?■ rn CO - N 1^00 N O N M N ^ u-jvo CO CI ro cj r^ m QQOC; ^s, c^oqq RiOdCiq KClt^^ .T32 U 5 in M « « W « ■*vo VO N CT>o I/-, iriM 1/1 »/l^ *C M C 00 t^vO "* M vo *0 t^ p. ^ 00^ ^ VC M O^ 4J tic •a •c n ** »i 2 5 " r u c u S i ° c c " oi oi - ui S S u W « « N N N a u rt « c u u u c > tn N W M •*3 •e V--- i^ r u a>-^ •aapug KqAVBUB^ aSpijg sisdMsmJSnoj CM S c rt'3 •a n "^ rt U U 4, a! k> 1m ^ * j; u 11 u c u >• t> t. 3 •aSpua J3AI^ ssoiEqx 3 OS c ■ nt ^ . U z I o s j ? 2. tn 2 1 cfl o H H s; a. ja u - •Hii a. c < z ^ • Z ^^ ,"1 3 w < CS C)' . U u f Q . [1. -) = ^ O^ t/) •> -• <-> ^» !5S m u 5 ^ 2 a Q ►J « < jy :3'* H =*< •Sug U-" Mo-S •^.^ " -Z. ►JU^ b< a ci. ^ 3 U K • f ,2 0. < ■- H-^ 0) nt T3 E d o TABLE OF ACTUAL WEIGHTS OF MODERN BRIDGES. 687 si i i|t' |l \i 688 THEORY OF STRUCTURES. TABLE OF LOADS FOR HIGHWAY BRIDGES. Span in Feet. loo and under loo to 200 200 to 300 300 to 400 above 400 City and Suburban Bridges liable to Heavy Traffic. Bridi^es in Manu- facturing Districts. Ballasted Koad. Bridges in Country Districts, Unballasted Koads. 100 lbs. per sq. ft. 80 •• " 70 " " 60 " " 50 go lbs. per sq, 60 " " 50 " " 50 " " 50 " " ft. 70 lbs. per sq. ft. 60 " " 50 " " 45 " " 45 " " ^ EXAMPLES. 689 1^ EXAMPLES. I. A bridge of A^ equal spans crosses a span of Z ft.; the weights in tons per lineal foot of the main girders of the platform, permanent way, etc., and of the live load, are wi , Wa , 71/, , respectively. Show that LA w, = N- LB' where A = Wt^pk + ^) + iv%{pk + q) and B =pk ■\- r, k being the ratio of span to depth, and p, q, r numerical coefficients. Hence also determine the limiting span of a girder. If X is the cost of a pier and if Y is the cost per ton of the super- structure, find the va'.ue of iV which will make the total cost per lineal foot a maximum, and prove that this is approximately the case when the spans are so arranged that the cost of one span of the bridge structure is equal to the cost of a pier. Ans. A span < 77-7-^ Cost is a minimum when A'' — LB = Z4/__, pk -^ q. and the minimum cost of the span -(-f) X, approx. 2. A car of weight W for a gauge of 4 ft. 81 in. is 33 ft. long, 6 ft. deep, and its bottom is 2 ft. 6 in. above the rails. Find the additional weight thrown upon the leeward rail when the wind blows upon a side of the car with a pressure of 20 lbs. per square foot. Also find the minimum pressure that will blow the car over. Ans. 4625.84 lbs. ; .428 W. 3. A lattice-girder 200 ft. long and 20 ft. deep, with two systems of ri^'ht-angled triangles, carries a dead load of 800 lbs. per lineal foot. Determine the greatest stresses in the diagonals and chords of the fourth bay from one end when a live load of 1200 lbs. per lineal foot passes over the girder. Ans. \i riveted: Diagonal stress = 37,2004/2 lbs. ; Chord stress = 450,000 lbs. _ \{ pin-connected : Diagonal stress = 44,800 1^2 and 29,600 V2 lbs.; Chord stress =: 460,000 lbs. i 1:!^ I t 1 690 THEORY OF STRUCTURES. 4. A latticc-girdcr 80 ft. long and 8 ft. deep carries a uniformly dis- tributed load of 144,000 lbs. Find the flange inch-stresses at the centre, the sectional area of the top flange being 56i sq. in. gross, and of ilie bottom flange 45 sq. in. net. What should be the camber of the girder, and what extra length should be given to the top flange, so that the bottom flange of the loaded girder may be truly horizontal ? {E =. 29,000,000 lbs.) Ans, 3185.8 lbs. ; 4000 lbs. x\ = .29735 ; -t' = • 2987 ; J. - 5, = jYM- 5. A lattice-girder 80 ft. long and 10 ft. deep, with four systems of right-angled triangles, carries a dead load of 1000 lbs. per lineal foot. Determine the greatest stresses in the diagonals met by a verticul plane in the seventh bay from one end when a live load of 2500 lbs, per lineal foot passes over the girder. Design the flanges, which are to consist of plates riveted together. The lattice-bars are riveted to angle-irons. Find the number of |-in, rivets required to connect the angle-irons with the flanges in the first bay, 10,000 lbs. per square inch being the safe shearing strength of the rivets. Ans. \i riveted: Diagonal stress = io,664-,ij|/2 lbs. \{ pin-connected : " = 9062^1/2; 6250V2; 15,4684 V"2; ii,87sV2 lbs. 22 rivets (2i-i\). 6. The bracing of a lattice-girder consists of a single system of tri- angles in which one of the sides is a strut and the other a tie inclined to the horizontal at angles of a and fi respectively ; in order to give the strut sufficient rigidity its section is made i times that indicated by theory, the coefficient I' being > unity. Show that the amount of ma- terial in the struts and ties is a minimum when tan a = i tan /3. 7. A lattice-girder of 40 ft. span, 5 ft. depth, and with horizontal chords has a web composed of two systems of right-angled triangles and is designed to support a dead and a live load, each of i ton per lineal foot. Determine the maximum stresses in the members of the third bay from one end met by a vertical plane. Ans. If riveted: Diagonal stress = ^^2 tons ; Chord stress = 27 tons. If pin-connected : Diagonal stress = \\^2 and fjV2 tons ; Chord stress = 26 tons. 8. A lattice-truss of 100 ft. span and 10 ft. depth has a web composed of four systems of right-angled triangles. The maximum stress in the EXAMPLES. 691 II II (iiagonal joining the sixth apex in the upper chord to the fourth apex in The lower is 16 tons. F"ind the dead load, the live load bring i ton per lineal foot, assuming the truss to be («) riveted, (b) pin-connected. Ans. (a) .554 ton ; {6) 1.062 tons. 9. A lattice-girder of 40 ft. span has a web composed of t\/o systems of triangles (base = 10 ft.) and is designed to carry a live load of i6oo .us. per lineal foot and a dead load of 1200 lbs. y^er lineal foot. Defin- ing the stress-length of a member to bf the product of its length into tiie stress to which it is subjected find the depth of the truss so that its A'/.// stress-length may be a minimum. Ans. 10.19 ^t. 10. Determine the maximum stresses in the members of a lattice- truss of 40 ft. span a;id 4 ft. depth, with two systems of triangles (base = 8 ft.), (a) when riveted together; (d) when pin-connected. Dead load — i ton per lineal foot, live load = i ton per lineal foot. Ans. Bays — ist; ( h r ^ \;^V(\AAA/\/\/\A/ 277if lbs. per lin. ft «i <^ <:i M Fig. 440. Stresses in diagonals : t/? = rfj = 9 V3 ; 18 '53 '7' 1.0965 '^1 '7 63* 80^ 1.0965 rf. 16 56* 72* '■345 d. '5 75795 48* '353'5 211110 6'* '3985' 350961 '■345 di 14 70742 4'-* 118575 '893'7 53* 121659 310976 '■345 d. «3 65689 35* 99045 '64734 42* 96645 261379 '■345 d. 12 60636 30* 85095 •4573' 34* 78453 224184 '■345 '^ 11 24* 23* '■345 dn 10 2U|^ '5* ••345 d> 9 '.5* 4* '■345 «I0 8 40424 ■2* 34875 75299 - 3* - 7959 '•345 dn 7 3537' 8* 237'5 -M* - 32973 '•345 Max. Stress, 301528 90572 3512a 7'i = 139200 lbs. «'a = 35096' " ^t = 3'0976 " r'4 = 261379 lbs, 7'5 = 224184 '• ''« = '77377 " 7'i — 142972 lbs. vg = 98955 " 7', = 67340 27. Prepare a table showing the stresses in the several members (including counters) of a ten-panel double-track through railway bridge of 184^ ft. span and 34 ft. depth, the live and dead loads being respect- ively 2250 lbs. and iioo lbs. per lineal foot. (Tliamcsville Bridge.) 28. Determine the minimum stress-length (stress in a member multi- plied by its length) for a double-intersection Pratt truss of 1 54 ft. span and with eleven panels. The panel loads for engine = 44,000 lbs., for train = H It 1 696 THEORY OF STRUCTURES. 27,500 lbs., for bridge = 13,200 lbs. ; coefficientof working strengtli = 8000 lbs. per square inch for both compression and tension. 29. A six-panel single-intersection Pratt truss is uniformly loaded. Assuming the same coefficient of strength both for compression and tension, show that the economy of material will be greatest when tlic diagonals are inclined at 32° 25' to the vertical. 30. A double-intersection truss for a single-track through-bridge 01 204 ft. span is 29 ft. deep, 20 ft. wide, and has twelve panels. Find ihc stresses produced in the members of the leeward truss by a panel wind- pressure of 5000 lbs. acting 8 ft. above base of rails (5-ft. gauge). Ans, Sloping members : 1st = 27500 sec « ; 2d = 1 2708^ sec a ; 3d = 10208^ sec li\ 4th = 7708^ sec /i; 5th — 52o8i sec a ; 6th = 2708^ sec li\ 7th = 208^ sec li. Tension chord : ist panel = 27500 tan a = 2d ; 3d = 40208 J tan a; 4th = 40208^ tan « 4- io2o8i tan fi; 5th = 40208 jr tan tr 4. 17916! tan /i; 6ih = 40208^ tan nr + 23125 tan li. Compression chord : ist =:; 40208^ tan a 4- 10208^ tan (i ; 2d = 40208^ tan a 4- 17916^- tan (i ; 3d = 40208^- tan (f + 23125 tan ft : 4th = 402o8i tan a + 25833^^ tan ft ; 5th = 40208^ tan (t- -f- 26041 1 tan ft. Verticals: ist=5ono; 2d = 7708^; 3d = 5208!; 4th = 2708^ ; 5th = 2084^ lbs. tan 0: = ^; tan ft = f*. 31. In the preceding question find tiie maximum stresses in the members of the fourth panel met by a vertical plane ; engine panel Inad = 85,000 lbs., train panel load = 40,800 lbs., bridge panel load = 22,500 lbs. Ans. Stresses in tension chord = 456,430.45 lbs. ; in compression chord = 645,31 1.77 lbs. ; in sloping members = 206,242.5 lbs,; and 139,705.62 lbs. 32. Each of the two Pratt single-intersection five-panel trusses for a single-track bridge is 55 ft. centre to centre of end pins and 1 1 ft. 6 in. deep. Timber floor-beams are laid upon the upper chords 2^ ft. centre to centre ; the width between the chords = 10 ft. Find the proper scant- ling of the floor-beams for the loading given in Fig. 407, page 639. Also determine the maximum chord and diagonal stresses in the centre panel due to the same live load. 33. Prepare a table giving the stresses in the several members of a double-intersection deck-truss of 342 ft. span, 40 ft. depth, and with EXAMPLES. 697 i^ B nineteen panels. (Double-track bridge.) The panel engine, train (or live), iind dead loads are 96,000, 53,000, and 43,200 lbs., respectively. 34. Prepare a table giving the stresses in the several members of a deck-truss for a double-track bridge of 342 ft. span, 33 ft. depth, and with eighteen panels. The panel engine, live, and dead loads are 96,000, 54,000, and 36,000 lbs., respectively. 35. The two trusses for a 16 ft. roadway are each 100 ft. in the clear, 17 ft. 3 in. deep, and of the type repre- ^ ^ sented in the figure ; under a live load of 1120 lbs. per lineal foot the greatest total , stress in AB is 35,400 lbs. Determine the permanent load. Fig. 445. The diagonals and verticals are riveted to angle-irons forming part of the flanges. How many f-in. rivets are required for the con- nection of AB and BC at B? Also, how many are required between A and C to resist the tendency of the angle-irons to slip longitudinally .' \Vorking-shear stress = 10,000 lbs. per square inch. A/is. 708.9 lbs. ; 8 ; 4 ; II. 36. The compression chord of a bowstring truss is a circular arc of 80 ft. span and 10 ft. rise ; the bracing is of the isosceles type, the bases of the isosceles triangles dividing the tension chord into eight equal lengths. Determine the maximum stresses in the members met by a vertical plane 28 ft. from one end. The live and dead loads are each i ton per lineal foot. 37. Design a parabolic bowstring truss of 80 ft. span and 10 ft. rise for a dead load of i ton and a live load of i ton per lineal foot. The joints between the web and the tension chord are to divide the latter into eight equal divisions. 38. The compression chord of a bowstring truss is a circular arc. The depth of the truss is 14 ft. at the centre and 5 ft. at each end ; the span = 100 ft. ; the load upon the truss = 840 lbs. per lineal foot. Find tile stresses in all the members. Determine also the maximum stresses in the members met by a vertical 25 ft. from one end wlien a live load of 1000 lbs. per lineal foot crosses the girder. What counter-braces are required ? 39. A Pratt truss with sloping end posts has a length of 150 ft. centre to centre, and a height of 30 ft. centre to centre, with panels 15 ft. long; the dead load is 3000 lbs. per lineal foot, and the live load 1200 lbs. Determine the maximum stresses in the end posts, in the third post trom one end, in the middle of the bottom chord, and in the members of the third panel met by a vertical plane. 40. Design a cross-tie for a double-track open-web bridge, the ties i lit ' 698 THEORY OF STRUCTURES. being 18 ft. J in. centre to centre, nnd the live load for the floor system being 8000 lbs. per lineal foot. 41. A bowstring roof-truss of 50 ft. span, 15 ft. rise, and five panels is to be designed to resist a wind blowing horizontally with a pressure of 40 lbs. per square foot. The depth of the truss at the centre is 10 ft. Determine, graphically, ihe stresses in the several members of the truss, assuming that the roof rests on rollers at the windward support. 42. A bowstring truss of 120 ft. spai. and 15 ft. rise V: of the isosceles braced type, the bases of the isosceles triangles dividing the tension chord into twelve equal divisions ; the dead and live loads are \ ton and I ton per lineal foot, respectively. Find the maximum stresses in ilic members met by vertical planes immediately on the right of the second and fourth joints in the tension chord. 43. The figure is a si Sum. tan. Total Stress. Total Ma.x.' Stress. •^i lo 32700 - 3270 45 I 45 106650 1 106650 242730 f? 169911 169911 c-i — I - 3270 34 80580 773'° 54"7 224028 c\ — I - 3270 n 545'0 51240 35868 259896 c* — I - 3270 12 28440 25170 17619 277515 Ci — I - 3270 1 2370 - 900 — 630 276885 i'\ = 35i97o lbs. ; i/j = 90,210 ; v^ = 68,040 ; 1/4 = 47,470 ; v^ = 28,500 lbs. 48. Compare the relative amounts of iron required in the webs of a single- and a double-intersection Pratt deck-truss of loo ft. span and having eight panels. Panel live load = L, panel dead load = D. 49. The figure represents a pier, square in plan, supporting the ends of two deck-trusses, each 200 ft. long and 30 ft. deep. The height of tiie pier is 50 ft. and is made up of three panels, the upper and lower being each 17 ft. deep. Ten square feet of bridge surface and ten square fett of train surface per lineal foot are subjected to a wind-pressure of 40 lbs. per square foot. Tlie centre of pressure for the bridge is 68 ft., and for the train 86 ft., above the pier's base. The wind also produces a horizontal pressure of 4000 Fio. 450. lbs. at each of the intermediate panel points on the windward side of the pier. Width of pier = 17 ft. at top and 33$ ft. at bottom. The bridge load = 1600 lbs. per lineal foot, live load = 3000 lbs. per lineal foot. Determine— (rt) The overturning moment (3180 ft.-tons). (b) The horizontal force due to the wind at the top of the pier. (61.6 tons.) (c) The tension in the vertical anchorage ties at 5 and T. (Xil.) {d) The vertical and horizontal reactions at T. (275 and 65.6 tons.) Draw a diagram giving the wind-stresses in all the menibers, and in- dicate which are in tension and which in compression. Ascertain whether the wind-pressure of 40 lbs. per square foot upon a train of empty cars weighing 900 lbs. per lineal foot will produce a tension anywhere in the inclined posts. What will be the tension in the anchorage ties ? (20.75 tons.) Find the stresses in the traction bracing (i) when a loaded train trav- elling at 30 miles an hour is braked just as the engine is over the pier and brought to rest in a length of 300 ft. ; (2) when a loaded traih with t'le engine over the pier is started by a sudden admission into the cylin- ders of steam at 100 lbs. per square inch. Stroke of cylinder = 16 in., diameter of drivers = 5 ft. HI! EXAMPLES. 701 50. The figure represents one half of one of the piers of the Bouble Viaduct. The spans are crossed by two lattice-gir- "^^ ders, 14' 9" deep and having a deck platform. The heightof the pier is 183' 9" and is made up of eleven panels of equal depth. Width of pier at top = 13' li", at bottom = 67' 7". With wind-pressure at 55.3 lbs. per square foot, the total pressure on the jjirder, train, and pier have been calculated to be 20, 16.2, and 20 tons, acting at points 196.2, 210.3, «i"d 92.85 ft., respectively, above the base. The dead weight upon each half pier is 222J tons, of which 60 tons is weight of half span, 120 tons the weight of the half pier, and 42^ tons the weight of the train. Assuming that the wind-pressure on the pier is a horizontal force of 2 tons at each panel point on the windward side, and that the weight of the pier may be considered as a weight of 6 tons at each panel point, determine — (a) The overturning moment. Fig. 451. (b) The total horizontal force at the top of the pier due to the wind. ic) The tension in each of the vertical anchorage ties at 5 and 7" due to the wind-pressure. (d) The vertical and horizontal reactions at T. Show tliat the greatest compressive stress occurs in the member RT, and that it amounts to 422 tons. Draw a stress diagram giving the stresses in all the members, indi- caiin;^ which are in tension and which in compression. Width of pier at ^ = 20 ft., ax. B = 23^ ft., at C = 36^ ft. What will be the effect of braking the train when running at 30 miles an hour, so as to bring it to rest within a distance of 220 ft. ? Width of pier in direction of bridge = 9I ft. at top and = 20 ft. at bottom. Ans.—{a) 9188 ft.-tons; (b) 39.9 tons; {c) 24^ tons, {d) Hori- zontal reaction = 59.9 tons ; vertical reaction = 247 tons. 51. The accompanying figure represents a portion of a cantilever truss, the horizontal distances of the points A, B, C from the free end being h , h , h, respectively. The boom ABC is inclined at an angle a, and the boom A' VZ at an angle ft, to the horizon. Find the deflections at the end of the cantilever due to X Y 2. {a) an increase kAB in the length of AB; (2) an Fig. 45a. increase k^B K in the length of ^ F ; (3) a decrease kiXY in the length oi XY; (4) a decrease iiBX in the length of BX. k.hAB ^'"-^'^ BX^^nABX' M Si ij «li li!'L-':». 702 THEORY OF STRUCTURES. (3) ^•^sin i^^K . ( BX cos a , , „,,,, <4) >f'< i . , „ V. - />(cot BXY~ cot yiZ.'.Vi ( sill ABA ' In the preceding question, if ki = ^-a = /'a = ^'< = l\ and if ^/ W is parallel to ^5^, and AX to i? F. show that the angle between WX and ^K after deformation = 2/C-(cot ABX + cot /y F.V). Hence also, if the truss is of uniform deptli d, show that the " deviation " 2k of the boom per unit of length is constant and equal to ^• a 52. Six bars have to be arranged upon a steel pin ; each bar is i in. wide and is subjected to a stress of 64,000 lbs. Should the bars be ar- M.OOO lb8..4— ^ U.OOO-^ MjOOO-*- M,0004 81.000 ■<—^^~ W,000< ^ "t:^ — ^>-61,U001b». 2- ->- 64,000 'J > um Fig. 433.— Method 1. ;2s— »-M,ooo 7 >-64,000 J ■^ — ►.M.OOO Fig. 454. — Method 2. ranged according to method i or method 2 .' Why ? Determine the di- ameter of the pin. 53. The accompanying sketch represents one of the pin connections in a certain bridge which was recently overthrown. The two innermost bars are web members inclin-jd to the horizon at an angle whose cosine 'V^ra^ 48,000 lba<'p l>4'hnri 3'! | I ' t>^ — y ^8.100 lb8, «,«00 \\m t^V4- i I I I ' Fig. 455. is .815. The thickness of the bars and the maximum stresses to which they are severally subjected are shown on the diagram. Is the 3-in. wrought-iron pin sufficiently strong? CHAPTER XII. SUSPENSION-BRIDGES. 1. Cables. — The modern suspension-bridge consists of two or more cables from which the platform is suspended by iron or steel rods. The cables pass over lofty supports (piers), and are secured to anchorages upon which they exert a direct pull. Chain or link cables are the most common in England and Europe, and consist of iron or steel links set on edge and pinned together. Formerly the links were made by welding the heads to a flat bar, but they are now invariably rolled in one piece, and the proportional dimensions of the head, which in the old bridges are very imperfect, have been much im- proved. Hoop-iron cables have been u.sed in a few cases, but the practice is now abandoned, on account of the difficulty attend- ing the manufacture of endless hoop-iron. Wire-rope cables are the most common in America, and form the strongest ties in proportion to their weight. They consist of a number of parallel wire ropes or strands, compactly bound together in a cylindrical bundle by a wire wound round the outside. There are usually seven strands, one forming a core round which are placed the remaining six. It was found impossible to employ a seven-strand cable in the construction of the East River Bridge, New York, as the individual strands would have been far too bulky to manipulate. The same ob- jection held against a thirteen-strand cable (thirteen is the next number giving an approximately cylindrical shape), and it was finally decided to make the cable with nineteen strands. Seven of these are pressed together so as to form a centre core, around which are placed the remaining twelve, the whole being con- tinuously wrapped with wire. 703 \m 704 THEORY OF STRUCTURES. In laying up a cable yreat care is required to distribute the tension uniformly amon<,'st the wires. This ma\- be effected either by giving each wire the same deflection or by usiit^ straight wire, i.e., wire which when unrolled upon the floor from a coil remains straight and shows no tendency to spriii;' back. The distribution of stress is practically uniform in un- twisted wire ropes. Such ropes are spun from the wires and strands without giving any twist to individual wires. The back-stay is the portion of the cable extending from an anchorage to the nearest pier. The elevation of the cables should be sufficient to allow fur settling, which chiefly arises from the deflection due to the load and from changes of temperature. The cables may be protected from atmospheric influence by giving them a thorough coating of paint, oil, or varnish, but wherever they are subject to saline influence, zinc seems to be the only certain safeguard. 2. Anchorage, Anchorage Chains, Saddles. — The an- chorage, or abutment, is a heavy mass of masonry or natural rock to which the end of a cable is made fast, and which re- sists by its dead weight the pull upon the cable. Fig. 456. Fig. 457. Fig. 458. The cable traverses the anchorage as in Figs. 456 to 458, and passes through a strong, heavy cast-iron anchor-plate, and. if made of wire rope, has its end effectively secured by turning it round a dead-eye and splicing it to itself. Much care, how- ever, is required to prevent a wire-rope cable from rustin" on account of the great extent of its surface, and it is co; • advisable that the wire portion of the cable should alv .r- minate at the entrance to the anchorage and there be a. lied to a massive chain of bars, which is continued to the anclior- plate or plates and secured by bolts, wedges, or keys. ANCHORAGE, ANCHORAGE CHAINS, SADDLES. 705 In order to reduce as much as possible the depth to which it is necessary to sink the anchor-plates, the anchor-chains are frequently curved as in Fig. 458. This gives rise to an oblitiue force, and the masonry in the part of the abutment subjected to such force should be laid with its beds perpendicular to the line of thrust. The anchor-chains arc made of compound links consisting alternately of an odd and an even number of bars. The friction of the link-heads on the knuckle-plates considerably lessens the stress in a chain, and it is therefore usual to diminish its sectional area gradually from the entrance E to the anchor. This is effected in the Niagara Suspension Bridge by varying the section of the bars, and in the East River Bridge by vary- ing both the section and the number of the bars. The necessity of preserving the anchor-chains from rust is of such importance that many engineers consider it most essential that the passages and channels containing the chains and fastenings should be accessible for periodical examination, painting, and repairs. This is unnecessary if the chains are first chemically cleaned and then embedded in good hydraulic cement, as they will thus be perfectly protected from all at- mospheric influence. The direction of an anchor-chain is changed by means of a saddle or knuckle-plate, which should be capable of sliding to an extent sufificient to allow for the expansion and contraction of the chain. This may be accomplished without the aid of rollers by bedding the saddle upon a four- or five-inch thickness of asphalted felt. The chain, where it passes over the piers, rests on saddles, the object of which is to furnish bearings with easy vertical curves. '' it^her the saddle may be constructed a>> in Fig. 459, so as to allow the cable to slip over it with compara- ti' y little friction, or the chain may be secured to the saddle, and the saddle supported upon rollers which work over a per- fectly true and horizontal bed formed by a saddle-plate fixed to the pier. 1 ■a m 1. '■:: r 7o6 THEORY OF STRUCTURES. 3. Suspenders. — The suspenders are the vertical or in. clined rods which carry the platform. Fig. 460. Pig. 461. Fig. 463. Fig. 463. In Fig. 460 the suspender rests in the groove of a cast- iron yoke which straddles the cable. Fig. 461 shows the suspender bolted to a wrought-iron or steel ring which em- braces the cable. When there are more than two cables in the same vertical plane, various methods are adopted to insure the uniform distribution of the load amongst the set. In Fig. 462, for example, the suspender is fastened to the centre of a small wrought-iron lever PQ, and the ends of the lever are connected with the cables by the equally strained rods PR and QS. In the Chelsea bridge the distribution is made by means of an irregularly shaped plate (Fig. 463), one angle of which is supported by a joint-pin, while a pin also passes through another angle and rests upon one of the chains. The suspenders carry the ends of the cross-girders (floor- beams), and are spaced from 5 to 20 ft. apart. They should be provided with wrought-iron screw-boxes for purposes of adjustment. 4. Curve of Cable. — . ASEA. An arbitrarily loaded flexible cable takes the shape of one of the catenaries, but the true catenary is the curve in which a cable of uniform section and material hangs under its own v/eight only. Let A be the lowest point of the cable, and take the ver- tical through A as the axis oi y. Take the horizontal tlirough as the axis of x, the origin O being chosen so that p .AO = H = mp. (0 / being the weight of a unit of length of the cable, and //' the horizontal pull at A. tflil! ciblc true ami Ycr- [rigin the CURVE OF CABLE. 707 m OX AO \s the parameter, or modulus, of the catenary, and OG is the directrix. Let X, y be the co-ordinates of any point P, the length of the arc AP being s. Draw the tangent /'T'and the ordinate PN. The triangle PNT is evidently a triangle of forces for the portion AP^ PN representing the weight of AP (viz., ps), PT the tangential pull T at P, and NT the horizontal pull H at A. ,,±=:,,nPTN=g^=.^=^, ... (2) ax TN H m ^ which gives the differential equation to the catenary. It may be easily integrated as follows : ^s - I , fdyV / , s' I (3) or (is Vs' 4- m' dx m' ,\\og{s-\-Vs'-\-m') = w; £ being a constant of integration. When x = 0, s — o, and therefore log m = c. Hence, log J + V/ 4- vt^ X m m' 7o8 THEORY OF STRUCTURES. .# or s-\-^/s^ -\-m*=me"', m Again, and hence, j=__(^«_^-«.). (^y dy s I , * - 5. -j- = — — -U >" — e '«); ax m 2 ' 7 = -(^ " + ^ ") (5) The constant of integration is zero, since y = m when X =0. The last equation is the equation to the catenary, while eq. (4) gives the length of the arc AP. By equations (4) and (5), f = s' + m\ (6) Draw NM perpendicular to PT^ and let the angle PTN = PNM - e. Then PM = PN sinewy and MN = PNcos e=y '" - i^s' + in \=5, , . . (7) = w, . . . (8) since tan 6/ = -f- = _. ax in Thus, the triangle PMN possesses the property that the side PM is equal to the length of the arc AP, and the side J/iV is equal to the modulus ;«(= ^(9). The area APNO ydx = — (f •• — e~ m):=z ms = 2 X triangl gPMN. The PG bein^ At A, Again P. being tl These and constr the assuni] in practice density. Case I posed of a tween the Ic any given lii the horizonti CURVE OF CABLE, The radius of curvature, p, at P 709 1- + (I)T (£)■ y ^ d y y m dx' m PG being perpendicular to PT. At A, y =. m, and the radius of curvature is also m. (9) (10) Again, ps PT_ NP y cosec 6 = -, .'.T = py', (II) H = pm=pp,', (12) p^ being the radius of curvature at A. These catenary formul.'e are of little if any use in the design and construction of suspension-bridges, as they are based upon the assumption of a purely theoretical load which never occurs in practice, viz., the weight of a chain of uniform section and density. Case B. Let the platform be suspended from chains com- posed of a number of links, and let W be the whole weight be- tween the lowest point O of the chain and the upper end P oi any given link. Let the direction of this link intersect that of the horizontal pull {H) at O in E. Drop the perpendicular PN^, 7IO THEORY OF STRUCTURES. The triangle PNE is evidently a triangle of forces ; and if the angle PEN = 6, ^ PN W tan e and hence NE tan e (X W. H' Thus, by treating each link separately, commencing with the lowest, the exact curve of the chain may be easily traced. Generally speaking, the distribution of the load may be assumed to be approximately uniform per horizontal unit of length, the load being suspended from a number of points along each chain or cable by means of rods. The curve of the cable will then be a parabola. Let w be the intensity of the load per horizontal unit of length. Let X, y be the co-ordinates of any point P of the cable with respect to the horizontal OX and the vertical 6^ F as axes of X and y, respectively. Let ^ be the inclination of the tangent at /'to the horizon- tal. The portion OP of the cable is kept in equilibrium by the horizontal pull H at 0, by the tangential pull T at P, and by the load ivx upon OP, which acts vertically through the middle point E of ON, PN being the ordinate at P. Hence, the tangent at P must also pass through E, and PEN is a triangle of forces. Hence, X wx ~~ y' , 2H or x^ = — -y, (I) the equation to a parabola with its vertex at 0, its axis vertical, , 2H and its parameter equal to — -. Again, I T_PE^_ H~ EN~ cos e* and hen and the , that at t Also, The so that th 5- Par and B, res Let 01 By equ / Denote th< Also, PARAMETER, ETC. and hence Tcose = H = wx 2y' 7" (2) and the horizontal pull at every point of the cable is the same as that at the lowest point. Also, ^ wx „ WX' T — sec u = — 2y 2y /iH — ^-='wx K \-\ -, \J - V 4/ The radius of curvature at P 2y[ x'\~ Hi so that the radius at Ois or Po = H w' « H = wp^. w H 5. Parameter, etc. — Let h^, //, be the elevations of A and B, respectively, above the horizontal line COD, Fig. 465. Let OD = a, , OC — a^, and let a, + a, = a = CI). By equation (i), Art. 4. /2H _ a, _ y w ~ ^h\ ~ a. a,-\-a. a Vh, Vh, Vh, + Vh, Vh,-\-Vh, Denote the parameter by P. Then w Also, tan d \Vh, + VhJ' _'2y _ ZUX _ 24r _ Fy ~~x~ll~~P~^\l P i 'ill 712 THEORY OF STRUCTURES. If ^, , 6*, be the values of ^ at /i and B, respectively. '-sll tan 6^, Note.—\{ h.^K — hy and tan d. v/^ and hence a a" tan e, = — = tan ^, . 6. Length of Arc of Cable.— Let OP = s, Fig. 465. Since tan 6 = 7l> W sec' ^^^ = -jjdx =. -jyds cos 6^, or ds H dd w cos' 6' Hence, _Hf'de H 111 VQ w ^° cos' B 2W — I tan ^ sec 6* + log, (tan ^ + sec ^) j . Again, tan (y = -fyx, and /■ w sec^ = ^/i+— 4r». WEIGHT OF CABLE. 7ti Note. — An approximate value of the length of the arc may be obtainf^H as follows : ds^ = dx^^df = dx-^ I I + (g)'[ = dx^ (i + -^). / I 'W^X'^\ .', ds = dx\i -\ /TT J. approximately. Integrating between and P, s^OP = x+~ I IV^X* 6 //" ~ ^ 2>x' 7. Weight of Cable. — The ultimate tenacity of iron wire is 90,000 lbs. per square inch, while that of steel rises to 200,000 lbs., and even more. The strength and gauge of cable wire may be insured by specifying that the wire is to have a certain ultimate tenacity and elastic limit, and that a given number of lineal feet of wire is to weigh one pound. Each of the wires for the cables of the East River Bridge was to have an ultimate tenacity of 3400 lbs., an elastic limit of 1600 lbs., and 14 lineal feet of the wire were to weigh one pound. A very uniform wire, having a coefificient of elasticity of 29,000,000 lbs., has been the result, and the process of straightejiing has raised the ultimate tenacity and elastic limit nearly 8 per cent. Let W^ be the weight of a length «, (= OD) of a cable of sufficient sectional area to bear safely the horizontal tension H. Let fF, be the weight of the length J, ( = OA) of the cable of a sectional area sufficient to bear safely the tension 7", at .^. Let /be the safe inch-stress. Let q be the specific weight of the cable material. Then and W,= y a,q W^ = ^L^s^. f llfil '0 I /H THEORY OF STRUCTURES. •••'^.= "'.^"., = ^(. + f^)(.+f4-...), or ^F,= fr.(i+^-^), nearly. A saving may be effected by proportioning any given section to the pull across that section. At any point {x, y) the pull = H sec B, and the correspond. // sec /9 ing sectional area = 7 — . The weight per unit of length Hs&cd / q, and the total weight of the length j, (= OA) is But x' = -j-v. < J Hence, and also ■■■"■-'in Hq( . aK\ , The weight of a cubic inch of steel averages .283 lb. The weight of a cubic inch of wrought-iron averages .2781b. IT The volume in inches of the cable of weight W^,= \2afj . DEFLECTION OF CABLE. W, 1 2a, jj=. .283 lb. or .278 lb., 7 7»5 ":if according as the cable is made of steel or iron. Let the safe inch-stress of steel wire be taken at 33,960 lbs., of the best cable-iron at 14,958 lbs., and of the best chain-links at 9972 lbs. Then W, = Ha, X .283 X -^ = -—- for steel cables ; 339CX) lOOOO W, = Ha, X .278 X -^ = -~ for iron cables ; ' 14958 4500 W, = Ha, X .278 X = — -- for link cables. ' ' 9972 3000 NoU. — About one-eighth may be added to the net weight of a chain-cable for eyes and fastenings. 8. Deflection of a Cable due to an Elementary Change in its Length. By the corollary of Art. 6 the total length {S) of the cable AOB is Now a, and a^ are constant ; /i, — //, is also constant, and therefore d/i, = dh^. Hence, If the alteration in length is due to a change of t° in the temperature, dS = ctS, c being the coefificient of linear expansion and = 5 — -— — -^ per degree Fahr. for wrought-iron. m i Li 7i6 THEORY OF STRUCTURES. In England the effective range of temperature is about 60° Fahr., while in other countries it is usual to provide for a range of from 100° to 150° F, If the alteration is due to a pull of intensity /per unit of area, dS = jj S, E being the coefficient of elasticity of the cable material. If //,=./!, = //, a \G h a, = a, = — , and dS = d/t. ' 2' 3 rt 9. Curve of Cable from which the load is suspended by a series of sloping rods. /y T' E Fig. 466. Let be the lowest point of such a cable. Let the tangent at O, and a line through O parallel to the suspenders, be the axes of X and }>, respectively. Let tc' be the intensity of the oblique load. Consider a portion OP of the cable, and let the co-ordinates of /'with respect to OX, OVhe x and j. Draw the ordinate FN, and let the tangent at Pmeet ON in E. As before, PNE is a triangle of forces, and E is the middle point of ON. Then w'x PN 2y . 2H H ~ NE or w -y^ the equation to a parabola with its axis parallel to (9F and its 2// focus at a point S, where i\SO — — > . CURVE OF CABLE WITH OBLIQUE SUSPENDERS. /I/ Cor. I. Let the axis meet the tangent at O in T\ and let its inclination to OX be /'. Let A be the vertex, and ON' a perpendicular to the axis. Then SO = ST' = SA-i-AT' = SA + AN'. But 4AS . AN' = ON" = N' T" tzin' i= 4AN " tan' i. AS .:AS= AN' tan' t, and SO = AS(i 4- cot' t) = -.^r-- ^ ' ' sin z Hence, the parameter = dtAS = 4^6^ sin' /. Cor. 2. Let P be the oblique load upon the cable between and P. Let Q be the total thrust upon the platform at E. " w " " load per horizontal unit of length. " q " " rate of increase of thrust along platform. " t " •' length of PE. Then w . , and q ■=■ %v cot i ; w sm I * 3 w X H = — "- = 2w'. SO = 2AS4^-. = 2AS-^.; 2y sm t sm' i X y x'' f = y"^ -\- ^ — \- xy cos i. 4 Cor. 3. Let s be the length of OP, and let 6* be the inclina- tion of PE to (9 F. Then s = AP- AO = ^^^^' I tan (90° - ^) sec (90° - 0) + log, I tan (90° - ^) + sec (90° -ft)\- tan (90°- i) sec (90° -z) — log,{ tan (90°- /) 4- sec (90° - /) } | fffiin'ti . „ ,. .,, cot & 4- cosec \ = 7— < cot ^ cosec 6 — cot t cosec ? 4- log, — -^—. — . f . 2w i " cot ^ -|* cosec ? ) isi; '•■!?«- B' 7i8 THEORY OF STRUCTURES. It may be easily shown, as in the Note to Art. 6, that ap- proximately . , 2 y sin' i 3 ^+7 cos* 10. Pressure upon Piers, etc. Let T^ be the tension in the main cable at A. " T; " " " " " back-stay at A. " a, ft be the inclinations to the horizontal of the tangents at A to the main cable and back-stay, respectively. The total vertical pressure upon the pier at A = T', sin a -f- 7", sin /? = R. The total resultant horizontal force at A =x Tj cos « ~ T", cos ft — Q. If the cable is secured to a saddle which is free to move horizontally on the top of the pier (Fig. 467), Q — the frictional resistance to the tendency to motion, or Qt f^A Ml being the corresponding coefficient of friction. Fig. 467. Let D, Fig. 468, be the total height of the pier, and let IV be its weight. Let FG be the base of the pier, and K the limiting position of the centre of pressure. ^ AUXILIARY OR STIFFENING TRUSS. 719 Let /, q be the distance of P and W, respectively, from K. Then for stabiltty of position Q _ tt , and for stability of friction, when the pier is of masonry, ^ the coefficient of friction of the masonry. P+ W If jw, is sufficiently small to be disregarded, Q is approxi- mately nil, and Z', cos i* = T, cos fi =■ H. The pressure upon the pier is now wholly vertical and is = ^(tan a -\- tan (i). When the cable slides over smooth rounded saddles (Fig. 459), the tensions T^ and T^ are approximately the same. Thus, R = T^(s\x\ Of + sin /?) and Q = T^cos a — cos /3). U a = fi, Q = O, and the pressure upon the pier is wholly vertical, its amount being 2 7", sin a. The piers are made of timber, iron, steel, or masonry, and allow of great scope in architectural design. The cable should in no case be rigidly attached to the pier, unless the lower end of the latter is free to revolve throutrh a small angle about a horizontal axis. II. Auxiliary or Stiffening Truss. — The object of a stiff- ening truss (Fig. 469) is to distribute a passing load over the cable in such a manner that it cannot be ilistorted. The pull upon each suspender must therefore be the same, and this vir- tually assumes that the effect of the extensibility of the cable and suspenders upon the figure of the stiffening truss may be disregarded. -e*^* ^ Si'-r '■ 720 THEORY OF STRUCTURES. The ends'6> and A must be anchored, or held down by pir.s, but should be free to move horizontally. Let there be n suspenders dividing the span into {fi ^ i) equal segments of length a. Let P be the total weight transmitted to the cable, and z the distance of its centre of gravity from the vertical through 0. Let T be the pull upon each suspender. Taking moments about O, n(n -4- i^ nl Ps= r(« + 2a + 3« + . . . + nd) - Ta ' ^ ' = T—, 2 2 /being the length of OA. Also, if t is the intensiy of pull per unit of span, tl = nT, and hence P2 = t — . 2 Let there be a central suspender of length s. There will, 11 — I therefore, be suspenders on each side of the centre. The parameter of the parabola = ~t • Hence, the total length of all the suspenders ..+2|«-^.+4^.{/+2«+3-F...i-(^y]l ' / 24 \ ' 3 « -|~ J If there is no central suspender, i.e., if n is even, / h n \ the total length =z {n — \)\s -\ TT]' Denote the total length of suspenders by L. Then the strcss-leni^th = TL == -,PL. ^ nl 'il ;. i ■ AUXILIARY OR STIFFENING TRUSS. 721 Let tu be the uniform intensity of the dead load. Case I. T/u bridge partially loaded. Let w' be the maximum uniform intensity of the live load, and let this load advance from A and cover a length AB. Let OB = X, and let /?, , R^ be the pressures at O and A, respectively. For equilibrium, R, + R, + tl-wl-w'{l-x) = 0) . . . (I) /' /' za' R^lJ^t--w~--{l-x)^ =0. . . . (2) * g Also, since the whole of the weight is to be transmitted through the suspenders, i/z=wl-{- w\l — x). From eqs, (i), (2), and (3), (3) w X -R. = —j{l-x) = R,, (4) which shows that the reactions at and A are equal in mag- nitude but opposite in kind. They are evidently greatest when / X = -, i.e., when the live load covers half the bridge, and the w'l * common value is then -x- . The shearing force at any point between and B distant x' from = R^J^{t-7V).x' -^w'^-^[x' --^,. . . (5) i?i> Ei ''i'i- which becomes W X T7 (/ - X) R^ = /?, when x' equal x. Thus the shear at the head of the live load is equal in magni- tude to the reaction at each end, and is an absolute maximum III i! 722 THEORY OF STRUCTURES. when the live load covers half the bridge. The web of the w'l truss must therefore be designed to bear a shear of -^- at the o centre and ends. Again, the bending mome it at any point between and B distant x' from O „ , , t — w ,„ w' I — X R.x'A x" = — ' ' 2 2 —{x'^-xx'), . . (6) which is greatest when ;jr' z=— , i.e., at the centre of OB, its w' I — X value then being — -- — j-x''. Thus, the bending moment is d . 2 an absolute maximum when -r-ilx'' — x^) = o, i.e., when x = -/, dx y ' ' 2 T and its value is then P. 54 The bending moment at any point between B and A dis- tant x' from O t — IV w W X, = R,x' + -^-x'^ - -{x' -xy = - j{x' - x){l- x'\ (7) which is greatest when -r--,\{x' — x){l — .r')} = o, i.e., when l^x x' =: , or at the centre of AB, its value then being *w X q" -(/— xy. Thus, the bending moment is an absolute maxi- mum when -,-|;r(/ — xy\ = o, i.e., when x — -, and its value is ax 3 za' then + — /'. 54 Hence, t/ie maximum baiding moments of the unloaded and \ loaded divisions of the truss are equal in magnitude but opposite in direction, and occur a.' the points of trisection {D, C) of OA ; f%:k«ii •m'' I' i i I AUXILIARY OR STIFFENING TRUSS. 723 ■when the live load covers one-tliird ( 4C) and two-thirds {AD) of the bridge, respectively. Each chord must evidently be designed to resist both tension and compression, and in order to avoid unnecessary nicety of calculation, the section of the truss may be kept uni- form throughout the middle half of its length. Case II. A single concentrated load W at any point B of the truss. W^now takes the place of the live load of intensity w' . The remainder of the notation and the method of pro- cedure being precisely the same as before, the corresponding equations are R. + K 4- (^ -w)l-W = o. • • • • (!') lir / — w RJ+ —r-l' W{l-x) = o (2') W t-w=-j. (30 -R^ = ~{x~^=R, (40 which shows that the reactions at O and A are equal in mag- nitude but opposite in kind. They are greatest when x ~ o and when x = I, i.e., when IV is either at O or at A, and the common value is then — . 2 The shearing force at any point between O and B distant x' from O W I l\ ^R^^{t-W)x' = -j{x'-X^-;^, . 'vhich is a maximum when x' = x, and its value is then (50 W w The web must therefore be designed to bear a shear of throughout the whole length of the truss. 1' 1 . ; >«'\''' , ;''! 1 k i 1 724 THEORY OF STRUCTURES. Again, the bending moment at any point between and B distant x' from O W (-tr" ^R^,'^it-w)-^-j^-^x\--x)\. • (6' First, let w <—. The bending moment is positive and is a maximum when x' = x, its value then being W Next,\QX. X >— . The bending moment is then neg-ativc and I is a maximum when x' =^ x — —, its value then being W( '-I The bending moment at any point between B and A dis. tant x' from O r" W Ix' \ = R^^' + (^ _ ^)i_ _ W{x' -x)= -j{x' - l)[ -^ - .-), (7') which is a maximum when i.e., when x' =: x -{- -, and its value is then jlx 1 . Note. — The stiffening truss is most effective in its action, but adds considerably to the weight and cost of the whole struc- ture. Provision has to be made both for the extra truss and for the e.Ktra material required in the cable to carry this extra load. '! 1 AUXILIARY OR STIFFENING TRUSS. 725 Stiffening Truss hinged at the Centre. — Provision may be made for counteracting the straining due to changes of tem- perature by hinging the truss at the centre E. Let a live load of intensity tv' advance from A. First, let the live load cover a length AB =^ x [> — Let R^ , Rj be the pressures at O, A, respectively. The equations of equilibrium are ^. + ^, + (^ - w)/ -W'X = 0', . . (0 IV ^4 + ('-'")?-¥■" = °' (3) Eqs. (2) and (3) being obtained by taking moments about E. Hence, ■'■■■■■' zv' i -.w = — -jt{1^ - 4^^ + 2x') ; ... (4) I w R, = -^{r--4^x+3x^)', .... (5) I w' R.-^-jil-x)\ . ... (6) Next, let the live load cover the length BO ( < -j. Let AB = ;r as before, and let R,', R,', t' be the new values of R,, R,, t, respectively. The equations of equilibrium are now R,' + R,' + (/' - wy - zoy ^ x) = o; . (7) 7- 2 w K'\ -f it' - «') 8 ~ 7^^^ - •^) = o ; (8) 726 THEORY OF STJiUCTURES. i R/--\■{t'-^vY^^o^ and hence, w (9) t' -w=^ 2y,-(/ - ;r)' [= - (/ - w - w')-\ ; . (10) I w ^:=--^-'f{i'-Alx-\-lx^){=-Ry, . (II) r: = I w j{i-xy{=-R,) (12) Diagravi of Maximum Shearing Force. — The shear at any point distant z from A in the unloaded portion BO when the hve load covers AB = R.-V{t-w){l-z) (13) = - \R,'-\- it' -w- w'){l - z)\ = - {R,'-{-{t' - w){/-2) - w'{l-2)\ = minus the shear at the same point when AB is unloaded and the live load covers BO. For a given value of z the maximum shear, positive or negative, at any point of OB, is found by making (see eq. (13) ) dR,-\-{l-~2)d{t-w) = o, or w tif , _-(_ 2/+ IX) -jAl- 2){- 4/+ 4^) = O, or «f x = l 4^—2/ AZ-l' (14) and occur ^B^ and i "-«- ">' e,s. (,), (;,, (,3j_ (,^^^ 727 Fig. 4;^j. the ;«^4r.W^ ,/,,^^^ ^ i^^'^fljlif and mav hp r^r, ^ ~ x' ' ' 05) :Mrf/4'.'^,:~';/,,''rti.e ordinate .j; the shears are greatest when "=^ ^'' U u and their values ar. -^'ues are, respectively, ^^gain, the shear -.f = »«wtl,e shear at th. , »W.^ and the ,.Ve irarret'l^;^" ^^ '^ the shears I i!?',, increasing for a given v,l„„ , "-™u„ whe„ : :7'^'-;^ ^ With /_ ., ,„,, ,,^^^,^^_^^ ^ the maximum shear = i i^'/, ^--=:^^L-/rCt^-- -overs ^''nen It covers BO. It (16) 728 THEORY OF STRUCTURES. may be represented by the ordinate {^positive or negative) of the curve orsq. For example, at the points denned by z^x^l, \l, 1/, \l, \l, the maximum shears given by eq. (i8) are, respectively. Diagram of Maxinuun Bending Moment. — The bending moment at any point in BO distant z from A when the live load covers AB = Ril - z) -\-{t- w)- {i-zr (19) t' — 7t) — W/ '1^^ ~ (. = - I RXl - .) + {f - u^jtlJL _ Jl^ \ ■= minus the bending moment at the same point when the live load covers BO. Hence, by eqs. (4), (5), (19), the bending moment I lU I zv = ± - -yi/" - V'tr + 3,v')(/ -z) T - -j{l' - 4/-V + 2x\l -z) For a given value of z this is a maximum and equal to 2lz w' zl ~ zl — 2:. ^ T (/ - 2Z) when X = /-f- 2Z' Thus, the maximum bending moment may be represented by the ordinate {positive or negative) of a curve. For example, at the points defined by /, lA i/, |/. 2' AUXILIARY OR STIFFENING TRUSS. 729 the bending moments • x-c greatest when x = */; their values being, respectively, o, TM^w'r, T-.hf'l', ^sh^'P. o- -h \ ^- -■-' Fic. 471. The absolute maximum bending moment may be found as follows : For a given value of x the bending moment (see eq. (19) ) is a maximum when R^J^{t-w){l- z)^0, or I- Z-- R. t — w Hence, the maximum bending moment ^ 2 t — IV ^8 /' ~ 4/^ + ix" It will be an absolute maximum for a value of x found by put ting its differential with respect to x equal to nil. This differential easily reduces to ix' - gix^ + erx -r = 0. X = ^l is an approximate solution of this equation, and the cor- responding maximum bending moment := ^^f-j^t'V. The preceding calculations show f/iat at every point in its length the truss may be subjected to equal maximum shears and equal maximum bending moments of opposite signs. Again, it may be easily shown, in a similar manner, tiiat ill! \ Hi V: ■ \ il!i1»f*sii 730 THEORY OF STRUCTURES. when a single weight W travels over the truss, the maximum positive shear at a distance z from A W the maximum negative shear W either = -^(Z" — %lz-{- 4^') I W or = - -j{ll - 4^) ; and the maximum bending moment = ± Y2{/ - Z){1 — 2Z). 12. Suspension-bridge Loads. — The heaviest distributed load to which a highway bridge may be subjected is that due to a dense crowd of people, and is fixed by modern French practice at 82 lbs. per square foot. Probably, however, it is unsafe to estimate the load at less than fromicxD to 140 lbs. per square foot, while allowance has also to be made for the con- centration upon a single wheel of as much as 36,000 lbs., and perhaps more. A moderate force repeatedly applied will, if the interval between the blows corresponds to the vibration interval of the chain, rapidly produce an excessive oscillation (Chap. Ill, Cor. 2, Art. 24). Thus, a procession marching in step across a suspension-bridge may strain it far more intensely than a dead load, and will set up a synchronous vibration which may prove absolutely dangerous. For a like reason the wind usually sets up a wave-motion from end to end of a bridge. The fixctor of safety for the dead load of a suspension-biidge should not be less than 2^ or 3, and for the live load it is advisable to make it 6. With respect to this point it maybe remarked that the efficiency of a cable does not depend so much upon its ultimate strength as upon its limit of elasticity, MODIFICATIONS OF THE SIMPLE SUSPENSION-BHIDGE. 731 and so long as the latter is not exceeded the cable remains un- injured. For example, the breaking xveight of one of the 1 5-inch cables of the East River Bridge is estimated to be 12,000 tons, its limit of elasticity being 81 18 tons ; so that with ii^ only as a factor of safety, the stress would still fall below the elastic limit and have no injurious effect. The continual application of such a load would doubtless ultimately lead to the destruc- tion of the bridge. The dip of the cable of a suspension-bridge usually varies from iV *° tV °^ ^^^ span, and is rarely as much as ji-j, except for small spans. Although a greater ratio of dip to span would give increased economy and an increased limiting span, the passage of a live load would be accompanied by a greater dis- tortion of the chains and a larger oscillatory movement. Steadiness is therefore secured at the cost of economy by adopting a comparatively flat curve for the chains. 13. Modifications of the Simple Suspension-bridge. — The disadvantages connected with suspension-bridges are very great. The position of the platform is restricted, massive anchorages and piers are generally required, and any change in the distribution of the load produces a sensible deformation in the structure. Owing to the want of rigidity, a considerable vertical and horizontal oscillatory motion may be caused, and many efforts have been made to modify the bridge in such a manner as to neutralize the tendency to oscillation. {ci) The simplest improvement is that shown in Fig. 472, where the point of the cable most liable to deformation is attached to the piers by short straight chains AB. Fio. 472. (^) A series of inclined stays, or iron ropes, radiating from the pier-saddles, may be made to support the platform at a number of equidistant points (Fig. 473). Such ropes were used in the Niagara Bridge, and still more recently in the East River 1 i i I I ■s'i>^ ^^^%^ IMAGE EVALUATION TEST TARGET (MT-3) /. 4. J' C>, .7I O / /i Photographic Sciences Corporation s^ e^ %^ \ «,-' 6^ ^■^v^^ %" A "^ '^^^ 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 732 THEORY OF STRUCTURES. Bridge. The lower ends of the ropes are generally made fast to the top or bottom chord of the bridge-truss, so that the cor- responding chord stress is increased and the neutral axis pro- portionately displaced. To remedy this, it has been proposed to connect the ropes with a horizontal tie coincident in position with the neutral axis. Again, the cables of the Niagara and Fig. 473- East River bridges do not hang in vertical planes, but are in- clined inwards, the distance between them being greatest at the piers and least at the centre of the span. This drawing in adds greatly to the lateral stability, which may be still further increased by a series of horizontal ties. {c) In Fig. 474 two cables in the same vertical plane arc diagonally braced together. In principle this method is similar Fig. 474- to that adopted in the stiffening truss (discussed in Art. 1 1), but is probably less efficient on account of the flexible character of the cables, although a slight economy of material might doubt- less be realized. The braces act both as struts and ties, and the stresses to which they are subjected may be easily calcu- lated. {d) In Fig. 475 a single chain is diagonally braced to the platform. The weight of the bridge must be sufficient to insure Fig. 475. that no suspender will be subjected to a thrust, or the efficiency of the arrangement is destroyed. An objection to this as well II' MODIFICATIOXS OF THE SIMPLE SUSPENSIOX-BRIDGE. 733 as to the preceding method is that the variation in the curva- ture of the chain under changes of temperature tends to loosen and strain the joints. The principle has been adopted (Fig. 476) with greater per- fection in the construction of a foot-bridge at Frankfort. The Fig. 476. girder is cut at the centre, the chain is hinged, and the rigidity is obtained by m* ans of vertical and inclined braces which act both as struts and ties. {e) In Fig. 477 the girder is supported at several points by Fig. 477. straight chains running directly to the pier-saddles, and the chains are kept in place by being hung from a curved chain by vertical rods. (/) It has been proposed to employ a stiff inverted arched rib of wrought-iron instead of the flexible cable. All straining action may be eliminated by hinging the rib at the centre and piers, and the theory of the stresses developed in this tension rib is precisely similar to that of the arched rib, except that the stresses are reversed in kind. {g) The platform of every suspension-bridge should be braced horizontally. The floor-beams are sometimes laid on the skew in order that the two ends of a beam may be sus- pended from points which do not oscillate concordantly, and also to distribute the load over a greater length of cable. r-r% 734 THEORY OF STRUCTURES, EXAMPLES. 1. The span of a suspension-bridge is 200 ft., the dip of the chains is 80 ft., and the weight of the roadway is i ton per foot run. Find the ten- sions at the middle and ends of each chain. Ans. 31^ tons ; 58.94 tons. 2. Assuming that a steel rope (or a single wire) will bear a tension of 15 tons per square inch, sliow that it will safely bear its own weight over a span of about one mile, the dip being one-fourteenth of the span. Ans, Max. tension = 33,074 lbs. 3. Show that a steel rope of the best quality, with a dip of one-seventh of the span, will not break until the span exceeds 7 miles, the ultimate strength of the rope being 60 tons per square inch. Ans. Max. tension = 59.545 tons per square inch. 4. The river span of a suspension-bridge is 930 ft. and weighs 5976 tons, of which 1439 tons are borne by stays radiating from the summit of each pier, while the remaining weight is distributed between four 15-in. steel-wire cables, producing in each at the piers a tension of 2064 tons, rind the dip of the cables. Ans. 66.44 ft- The estimated maximum traffic upon the river span is 131 1 tons uniformly distributed. Determine the increased stress in the cables. Ans. 596.4 tons. To what extent might the traffic be safely increased, the limit of elasticity of a cable being 81 16 tons, and its breaking stress i:,3oo tons ? Ans. To 13,303 tons uniformly distributee!. 5. If the span = /, the total uniform load = W, and the dip = / \i show that the maximum tension = 1.58 ff'', the minimum tension = 1.5 \V, the length of the chain = 1.018/, and find the increase of dip corresponding to an elongation of i in. in the chain. 6. A cable weighing / lbs. per lineal foot of length is stretched be- tween supports in the same horizontal line and 20 ft. apart. If the ma.\- imum deflection is \ ft., determine tiie greatest and least tensions. Ans. Parameter ;// = 100 ft.; max. tension = 100^/; min. ten- sion = 100/. 7. A light suspension-bridge carries a foot-path 8 ft. wide over a river 90 ft. wide by means of eight equidistant suspending rods, the dip being 10 ft. Each cable consists of nine straight links. Find their several lengths. If the load upon the platform is 120 lbs. per square foot, and EXAMPLES. 735 if one-fourth of the load is borne by the piers, find the sectional areas of the several links, allowing 10,000 lbs. per square inch. Ans, Lengths in ft., 10; 10.049; 10.198; 10.44; 'o??- Tensions in lbs., 45000; 4500^^101 ; 45001/104; 4 500 if/ 1 09 ; 4500 1/116. Areas in sq. in., 4.5 ; 4.522 ; 4.59 ; 4.698 ; 4.847. 8. A suspension-bridge of 200 ft. span and 20 ft. dip has 48 sus- penders on each side; the dead weight = 3000 lbs. per lineal foot ; the live load = 2000 lbs. per lineal foot. Find the maximum pull on a sus- pender, the maximum bending moment and the maximum shear on the stiffening truss. Also, find the elongation in the chain due to the live load. Atts. Max. pull = 12,500 lbs.; max. shear = 30,000 lbs.; max. B.M. = i,o66,666S ft. -lbs. ; elongation = 89,600,000 -*- EA, A being sectional area of a cable, and E the coefficient of elas- ticity. 9. A foot-path 8 ft. wide is to be carried over a river 100 ft. wide by two cables of uniform sectional area and having a dip of 10 ft. Assum- ing the load on the platform to be 112 lbs. per square foot, find the greatest pull on tiie cables, tlieir sectional area, length, and weight. (Safe stress = 8960 lbs. per square inch ; specific weight of cable = 4S0 lbs. per cubic foot.) Ans. H = —:-- 7^=56,000 lbs.; area =6.73 sq. in.; V29 length = io2| ft.; weight = ■y.^oz.ti, lbs. 10. Find the depression in the cables in the last question due to an increment of length under a change of 60° F. from the mean temperature. (Coefficient of expansion = 1 -*- 144000.) Ans. .0S02 ft. 11. Each side of the platform of a suspension-bridge for a span of 100 ft. is carried by nine equidistant suspenders. Design a stitl'eiiing truss for a live load of 1000 lbs. per lineal foot, and determine the pull upon the suspenders due to the live load when the load produces (i) an ixhsoliitc ma.ximuin shear ; (2) an absolute tiiaximum beiidiit^ moment. Ans. Max. shear = 6250 lbs.; max. B.M. = 92, 592^ J ft. -lbs.; pull on suspender = (i) 2777J lbs., (2) = 18515? "'«• »•■ yT^ZW lbs. 12. In a suspension-bridge (recently blown down) each cable was de- signed to carry a total load of 84 tons (including its own weight). The distance between the piers = 1270 ft.; the deflection of the cable = 91 ft. Find (a) the length of the cable ; (/') the pull on the cable at the piers und at the lowest point ; (c) the amounts by which these pulls are changed by a variation of 40° F. from the mean temperature; (d) the tension in the back-stays, assuming them to be approximately straight and inclined to the vertical at the angle whose tangent is \. '% i ill \k. 1 736 THEORY OF STRUCTURES. Ans.—(a) 1287.4ft.; (6) //= = 1461^3 tons; (r) depression due 1.04 tochaiigeof lenip. = .936 ft. and amount of change in //= '- // y = li tons, in 7=1.45 tona ; ('/) 394-55 tons, neglecting pier friction, 13. The platform of the bridge in tiie preceding question was liun^^ from the cables by means of 480 suspenders (240 on each aide). Find the pull on each suspender and the total length of the suspenders, the lowest point of a cable being 14 ft. above tiie platform. Arts. .35 ton; 10,565^]', ft. 14. A suspension-bridge has a dip of 10 ft. and a span of 300 ft. Fimi the increase of dip due to a change of 100" F. from the mean tempera- ture, the coefficient of expansion being .00125 P*^*" '80° F, Aus. 1. 17 ft. Also, find the corresponding flange stress in the stifTeniiig truss. which is I2i ft. deep, the coeiricient of elasticity being 8000 tons. A/ts. 6.24 tons. 15. The ends of a cable are attached to saddles free to move hori/^on- tally. If Jrt is the horizontal movement of each saddle due to the ex- pansion of the cables in the side spans, and if ^S is the extension of the chain between the two saddles, show that the increment cf the dip (//) is approximately (3a _/r {8 A a 16. The platform of a suspension-bridge of 150 ft. span is suspended from the two cables by 88 vertical rods (44 on each side) ; the dip of the cables is 15 ft.; there are two stiffening trusses ; the dead weight is 2J40 lbs. per lineal foot, of which o/tf-Ad// is divided equally between the two piers. Find the stresses at tiie middle and ends of the cables when a uniformly distributed load of 78,750 lbs. covers one half of the bridge. .-\lso, find the maxiinun\ shears and bending moments to which the stid- ening trusses are subjected when a live load of 1050 lbs. per lineal foot crosses the bridge. Ans. Pull on suspender = 2741 J lbs. ; // = -_^ T= 203,4374 lbs. y29 Max. shear on each truss at centre and due to 78,750 lbs. = 9843^ lbs. = that due to 1050 lbs. per lineal foot. Max. H. M. due to 78,750 lbs. is at centre of loaded and un- loaded halves and = 184,570/^ ft. -lbs. Abs. max. B.M. due to 1050 lbs. per lineal foot is at points of trisection and = 218,750 ft. -lbs. 3 '' 16 A EXAMPLES. 717 17. Solve the preceding question when the trusses are hinged at the centre. Ans. Pull on suspender = 2741 S lbs. ; // = - _ 7^ = 1 54,2 18} lbs. Max. shear due to 78,750 lbs. = 9843J lbs. at centre of span and at end of loaded half of bridge ; max. shears due to 1050 lbs.' per lineal foot =13,125, 5906J, 4921 J, 8305 ,"2^, and 9843^ lbs. at ends of the half truss and at the points dividing the half span into four equal seg- ments. Max. H. M. due to 78,750 lbs. is at centre of half truss and = i84,57o,V ft.-lbs. Max. B. M. due to 1050 lbs. per lineal foot =176,1803 S3, 221,4841, and 1 53,808 J |! ft.-lbs. at points dividing the half truss into four equal segments. 18. Show that tiic total extension of a cable of uniform sectional area A under a uniformly distributed load ai inteiisiiy 7*/ is JoP_[ I + 6 ^'\ 3 rl' 16 3 / being the span and d the dip. 19. The dead weight of a suspension-bridge of 1600 ft. span is i ton per lineal foot; the dip = span 13 ' Find the greatest and least pulls upon one of the chains. The ends of the chains are attached to saddles on rollers on the top of piers 50 ft. higli, and the bacl< -stays are anchored 50 ft. from the foot of each pier. Find the load upon the pier j and the pull upon the anchorage. ^bis. 255 tons ; 243I tons ; 637^ tons ; 344.6 tons. 20. A bridge 444 ft. long consists of a central span of 180 ft. and two side spans each of 132 ft. ; each side of the platform is sus|)ended by vertical rods from two iron-wire cables; each pair of cables passes over two masonry abutments and two piers, the former being 24 ft. and the latter 39 ft. above the surface of the groand ; the lowest point of the cables in each span is 19 ft. above the ground surface ; at the abutments the cables are connected with straight wrought-iron chains, by means of which they arc attached to anchorages at a horizontal distance of 66 ft. from the foot of each abutment ; the dead weight of the bridge is 3500 lbs. per lineal foot, and the bridge is covered with a proof load of 4500 lbs. per lineal foot. Determine — {a) The stresses in the cables at the points of support and at the lowest points. {b) The dimensions and weights of the cables (i) if of uniform sec- I if 738 THEORY OF STRUCTURES. tion throughout ; (2) if each section is proportioned to the pull across it. (Unit stress = 14.958 lbs. per square inch.) () two side spans ; (c) the centre span. Ans. — {a) Pull on stays in centre span = 840.050 lbs. (^h) = double that in (a). (c) " ' sids span — 948,432 lbs. EXAMPLES. 739 22. A floating landing-stage is held in position by a number of 4|- lii. steel-wire cables anchored to the shore, a shoreward movement being prevented by rigid iron booms, pivoted at the ends and stretching from sh(\ f 1 1 " ,'\ 11 \ ! \ 1 "\ r :N! M CHAPTER XIII. ARCHES AND ARCHED RIBS. I. An arch may be constructed of masonry, brickwork, timber, or metal. Fig. 478. In the figure ABCD represents the profile of an arch. The under surface AD is called the soffit or intrados. The upper surface BC is sometimes improperly called the cxtrados. The highest point K o{ the soffit is the croton or key of the arch. The spriugings or skeivbacks are the surfaces AB^ DC from which the arch springs, and the haunches are the portions of the arch half-way between the springings and the crown. Upon each of the arch faces stands a spandril wall, and the space between these two external spandrils may be occupied by a series of internal spandrils spaced at definite distances apart, or may be filled up to a certain level with masonry (i.e.. backing) and above that with ordinary ballast or other rough m.aterial (\.e.,jiiling). A masonry arch consists of courses of wedge-shaped blocks with the bed-joints perpendicular, or nearly so, to the soffit. 740 EQUll.lHKATED POl.YGOX AXD I.IXE OF RESISTAXCE. 74^ Tlic blocks arc called voussoirs, and the voussoirs at tli' crown are the keystones of the arch. A Prick arch is usually built in a number of rings. Consider the portion of the arch houndetl by the vertical I)lane KE at the key and by tlie plane .//>'. It is kept in equilibrium by the reaction R at KE, the reac- tion A^, at Ali, .Mid the wei^dit K, of the portion under con- sideration and Its superincumbent load. Let ^' and T be the points of application of A*, and R, respectivel}'. Let the directions of A, and A intersect in a point. The direction of F, must also pass through the same point. Taking moments about S, />, and J', being the perpendicular distances of the directions of A and \\ from S, respectively. Similarly, the portion KECD of the arch gives the equation F, being the weight to which it is subjected, and p^, )\ the perpendicular distances of the directions of A and K, from the point of application K of the reaction at the plane DC. If the arch and the loading are symmetrical with respect to the plane KE, ^1 = ^!i' Ji =J's» ^n^ therefore /, =■ p^. Hence the direction of A will be horizontal, which might have been inferred by reason (^f the symmetry. The magnitudes of the reactions are indeterminate, as the positions of the pnintr. of application {S, T, V ) are arbitrary, and can only be fixed by a knowledge of the law of the varia- tion of the stress in the material at the bounding planes AB, KE. 2. Equilibrated Polygon and Line of Resistance. — Suppose an arch divided into a number of elementar}- portions r-- n 74^ rilEOKY Oh STRUCTURES. kc' , k'e" . . , Cc.g., tlic vousaoirs of a masonry arch) by a scries of ioints kc, k'c' . . . Fig. 479. Fig. 480. Let Wt, IVj, ... be the loads directly supported by the several portions. These loads generally consi.st of the wciijlu of a portion (e.g., ke') -f- the weiglit of the superiiicumbciit mass + the load upon the overlying roadway ; the lines of action of the loads are, therefore, nearly always vertical. Each elementary portion may be considered as acted upon and kept in equilibrium by t/inr forces, viz., the external load and the pressures at the joints. If the pressure and its point of application at any given joint have been determined, the pressures and the corresponding points of application at the other joints may also be found. For, let I 2 34 . . . be the line of loads, so that 12= W,, 2 3 = n\ , . . . Assume that the pressure P and its point of application r at any given joint ke are known. Draw o i to represent P in direction and magnitude. Then 02 evidently represents the resultant of /^and W, in direction and magnitude, and this resultant must be equal and opposite to the pressure /•, at the joint k'/. Hence, a line n'n drawn through «, tlie intersection of P and W,, parallel to 20, is the direction of the pressure P,, and intersects k'e' in the point of application r' of P, Again, o 3 represents the resultant of /' and IV^ in direc- tion and magnitude, and this resultant must be equal and opposite to the pressure /', at the joint k"e". The line «"«' drawn through ;/', the intersection of P^ and EQUILIBRATED POLYGOX AND LINE OF A'ES/.S lAXCE. 743 VV,„ parallel to 30, is the direction of the pressure /', and inter- sects k"i'" in the point of application r" of J\. Proceeding in this manner, a scries of points of application or ccntris of resistance r', r", r"\ . . . may be found, the corresponding pressures being represented by 02, 03,04, . . . T\\c polygon of pressures formed by the lines of action of P,Pi, F,, . , . \s termed an equilibrated poly^A 3 f.PQ 1+6^' PQ and in the limit when PO = — , i.e., when the intensity of stress varies uniformly from /at Pto «//at Q, 1 and iV: /•PQ (See Art. i6, Chap. IV.) Similarly, if Q is the most compressed edge, the limiting position of O, the centre of resistance or pressure, is at a point PQ a defined b> QO' = -^. Hence, as there should be no tendency on the part of the joints to open at either edge, it is inferred that PO or QO' PO should be > — -, i.e., that the point O should lie within the middle third of the joint. Experience, however, shows that the " middle-third " theory cannot be accepted as a solution of the problem of arch stability, and that its chief use is to indicate the proper dimensions of the abutments. Joint cracks are to be found in more than 90$^ of the arches actually constructed, and cases may be instanced in which the joints have opened so widely that the whole of the thrust is transmitted through the cds^es. In Telford's masonry arch over the Severn, of 150 ft. span, Baker discovered that there had been a settlement (15 in.) sufficient to induce a slight reverse curvature at the crown of the sofifit. Again, the position of the centre of pressure at a joint is indeterminate, and it is therefore impossible as welt as useless to make any calculations as to the maximum intensity of stress due to the pressure at the joint. What seems to happen 1 exceeds formly di curve of arch are been dec Baker ma Let : width of Let/ square foe An arc surfaces a sures, can advance w were possi compressil ments, ho reliance ca values. 4. Joini between w] the centre pressure m; different va! Let ^y horizontal t X is called t The an{ i if I JOINT OF RUPTURE. 747 happen in practice is, that the straining at the joints generally exceeds the limit of elasticity, and that the pressure is uni- formly distributed for a certain distance on each side of the curve of pressures. Thus, the proper dimensions of a stable arch are usually determined by empirical rules which have been deduced as the results of experience. For example, Baker makes the following statement : Let T be the thrust in tons or pounds per lineal foot of width of arch. Let / be the safe working stress in tons or pounds per square foot. An arch will be stable if an ideal arch, with its bounding surfaces at a minimum distance of — — from the curve of pres- sures, can be traced so as to lie within the actual arch. An advance would be made towards a more correct theory if it were possible to introduce into the question, the elasticity and compressibility of the materials of construction. These ele- ments, however, vary between such wide limits that no reliance can be placed upon the stresses derivable from their values. 4. Joint of Rupture. — Let i 2, 3 4 be the bounding surfaces between which the curve of pressures must lie, and let 4 be Fig. 483. the centre of pressure at the crown. A series of curves of pressure may be drawn for the same given load, but with different values of the horizontal thrust h. Let ^^y be that particular curve which for a value H o[ the horizontal thrust is tangent to the surface I 2 at 4r ; the joint at X is called \.\\^ joint of rupture. The angle which the joint of rupture mgkes with the i;ii^! m.» l!: 748 THEORY OF STRUCTURES. horizontal is about 30° in semicircular and 45° in elliptic arches. The position of the joint in any given arch may be tenta- tively found as follows : Let J be any joint in the surface i 2. Let ^Fbe the weight upon the arch between y and i. Let X be the horizontal distance between J and the centre of gravity of W. Let y be the vertical distance between J and 4. It will also be assumed that the thrust at 4 is horizontal. If the curve of pressure be now supposed to pass through y, the corresponding value of the horizontal thrust Ji is given by hY =z WX. By means of this equation, values of h may be calculated for a number of joints in the neighborhood of the haunch, and the greatest of these values will be the horizontal thrust H for the joint x. This is evident, as the curve of pressure for a smaller value of k must necessarily fall bcloiv 4x_y. When this happens, the joints will tend to open at the lower edge of the joint I 4 and at the upper edges of the joints at .V and at 2 3, so that the arch may sink at the crown and spread, unless the abutments and the lower portions of the arch are massive enough to counteract this tendency. If the curve of pressure fall aiove ^xy, an amount of back- ing sufificient to transmit the thrust to the abutments must be provided. The same result may be attained by a uniform in- crease in the thickness of the arch ring, or by a gradual increase from the crown to the abutments. For example, the upper sur- face (extrados) of the ring for an arch with a semicircular soffit A KB, having its centre at O, may be delineated in the following manner: Let X define the joint of rup- ture in the soffit ; then AOx = 30°. In Ox ness at the The an duced may ring, and tl xf to the ai 5. Mini resultant rectangular Let y b tion from Let // a components Let zv b in the abuti Let h be Let / be In order toe D, the to D ph than the mo the n lus thi This relai abutment mj which define; ment. MINIMUM THICKNESS OF ABUTMENT. 749 In Ox produced take xx' = 2 X KD, KD being the thick- ness at the crown. The arc Dx' of a circle struck from a centre in DO pro- duced may be taken as a part of the upper boundary of the ring, and the remainder may be completed by the tangent at x! to the arc Dx' . 5. Minimum Thickness of Abutment. — Let T be the resultant thrust at the horizontal joint BC of a rectangular abutment A BCD. Let y be the distance of its point of applica- tion from B. Let // and V be the horizontal and vertical components of T. Let w be the specific weight of the material in the abutment. Let Ji be the height AB of the abutment. Let t be the width AD of the abutment. F'c 485. In order that there may be no tendency to turn about the toe D, the moment of the weight of the abutment with respect to D plus the moment of V with respect to D must be greater than the moment of H with respect to D. Or, wht-^ V{t-y)> Hk, ■1 i' 1 %:\ ^ 1 m-i lit lllli!5 or / + z^V 2// , 2 F , F' — -r - — y -\ —. — • zv wh zv'h* This relation must hold good whatever the height of the abutment may be ; and if h is made equal to 00 , t> V w which defines a minimum limit for the thickness of the abut- ment. 750 THEORY OF STRUCTURES. 6. Empirical Formulae.— In practice the thickness t at the crown is often found in terms of s, the span, or in terms of p, the radius of curvature at the crown, from the formulae t ■=. c V'j, or / = Vcp, t, s, and p being all in feet, and c being a constant. According to Dupuit, i = .36 ^s for a full arch ; / = .27 i^s for a segmental arch. According to Rankine, = i^.i 2/3 for a single arch ; / = V.I 7/3 for an arch of a series. 7. Examples of Linear Arches, or Curves of Pressure. (a) Linear Arch in the Form of a Parabola, — Suppose that the cable in Art. 4, Chap. XII, Case B, is exactly inverted, and that it is stiffened in such a manner as to resist distortion. Suppose also that the load still remains a uniformly distributed weight of intensity w per horizontal unit of length A thrust will now be developed at every point of the inverted cable equal to the tension at the corresponding point of the original cable. Thus the inverted parabola is a linear arch suitable for a real arch which has to support a load of intensity zo per horizontal unit of length. The horizontal thrust at the crown ■=. H =■ wp, p being the radius of curvature at the crown. {l>) Linear Arch in the Form of a Catenary. Transformed Catenary. — If the cable in Art. 4, Chap. XU, Case A, is in- 1^ p Q K, verted and stiffened as before, a linear arch is obtained suitable for a real arch which has to support a load dis- tributed in such a manner that tiie weight upon any portion AP is pro- { , portional to the length of AP, and is in fact =/'^. The area OAPN =^ ins. Thus, a lamina of thickness unity and specific weight zv, bounded by the curve AP, the directrix ON, and the verticals AO, PN, weighs xojns, and may be taken Fig. 486. to repre; i.e., if th Th( the radii A dis catenary through sary that meet the may be o Upon horizonta generatin Cut tl quired an, solid will arch, and the arch v lamina art The pr length. The pi of the line; of the ang! Let X, formed cat, Let X, _ in the cate Let A'( Then he equp.tii EXAMPLES OF LINEAR ARCHES. 751 lis \s. y Ix In to represent the load upon the arch if %vms = ps, i.e., if zvm —P, i.e., if the weight of m units of the lamina is w. The horizontal thrust at the crown = // = wvi =■ zcp, the radius of curvature (p) at the crown being equal to ///. A disadvantage attached to a linear arch in the form of a catenary lies in the fact that only o/ie catenary can pass through tzvo given points, while, in practice, it is often neces- sary that an arch sliall pass through ^/ine points in order to meet the requirements of a given rise and span. This difificulty may be obviated by the use of the transfornied catenary. Upon the lamina PAPNN as base, erect a solid, with its horizontal sections all the same, and, for simplicity, with its generating line perpendicular to the base. Cut this solid by a plane through A'N inclined at any re- quired angle to the base. The intersection of the plane and solid will define a transformed catenary /''.<4'/'', or a new linear arch, and the shape of a new lamina P'A'P'NN, under which the arch will be balanced. This is evident, as the new arch and lamina are merely parallel projections of the original. The projections of horizontal lines will remain the same in length. The projections of vertical lines will be c times the lengths of the lines from which they are projected, c being the secant of the angle made by the cutting plane with the base. Let X, F be the co-ordinates of any point /'' of the trans- formed catenary. Let X, y be the co-ordinates of the corresponding point P in the catenary proper. Then y_P^_ _A'0 _M y~ FN ~^~ A0~'^« ^^' The equation to the catenary proper is m % ■P lit 2 \ (2) 752 THEORY OF STRUCTURES. Substituting in the last equation the value oi y given by eq. (i), and her \ (3) which is the equation to the transformed catenary. With this form of linear arch the depths M over the crown and Fover the springings, for a span 2x% may be assumed, and tlic corresponding value of m determined from eq. (3). It is convenient, in calculating ;«, to write eq. (3) in the form X m = iog.|j^+y^,-i|. . . . (4) The slope ?' at /" is given by tan i' ., dV Ml- -'-\ Ms ax 2)fn I m s being the length AP of the catenary proper, corresponding to the length A'P' of the transformed catenary. The area OA 'p'N= r '■*,,, Mm Ydx = - \e"' — e '") Ms. The triangle P' TN is a triangle of forces for the portion A'P'. Tlie triangle PTN is a triangle of forces for the portion AP. (The tangents at 7^ and P' must evidently intersect ON in the same point T.) Let H' be the horizontal thrust at A', //being that at A. Let P' be the weight upon A'P', /'being that upon AP. Let R' be the thrust ai P'. Then P' _ area OA'P'N _Ms _ M • ■ ~P ~ area OAPN ~~ nis ~~ m ' Thei and the r of the tra The t to a linea (c) Ch which has normal pn sity shouk Consid element L ■turned t( straight. Let the ant pressur Let CE CD. The an^ The hor This is < •'. the he Similarh pressure =^ and hence EXAMPLES OF LINEAR ARCHES. ^ M M .^ t^ = — /^= —wms = wMs ; tn m 753 y A in' H = F cot i' = wMs-^j- = wne = H\ Ms R'= H' sec i' =wrn'\/ i + /; w Vm* + APs' . m The radius of curvature p' at the crown = ir=^ M' H' = wMp' = H— wp, and the radius of the " catenary proper " is M times the radius of the transformed catenary. The term " equiHbrated arch " has generally been applied to a linear arch with a horizontal extrados. {c) Circular and Elliptic Linear Arches. — A linear arch which has to support an external normal pressure of uniform inten- sity should be circular. Consider an indefinitely small element CD, which meiy be as- sumed to be approximately straight. Let the direction of the result- ant pressure upon CD, viz., / . CD, make an angle 6 with OB. Let CE, DE be the vertical and horizontal projections of CD. The angle DCE = ft. The horizontal component of />. CD =/. CD cos =p. CE, This is distributed over the vertical projection CE. .'. the horizontal intensity of pressure = p . CE -^ CE ■==■ p. Similarly, it may be shown that the vertical intensity of pressure =/>. Fig. 487. li' w 1 1 \ 1 \ ['.' \ i i 'lilllliil 754 THEORY OF STRUCTURES. Thus, at any point of the arch, the horizontal intensity of pressure = vertical intensity = normal intensity =/. Again, the total horizontal pressure on one-half of the arch = 2(/ . CE) = p2{CE) =pr = H, and the total vertical pressure on one-half of the arch = 2(/> . DE) = p2{DE) =ipr = P. Hence, at any point of the arch the tangential thrust = pr. Next, upon the semicircle as base, erect a semi-cylinder. Cut the latter by an inclined plane drawn through a line in the Fig. 488. plane of the base parallel to OA, The intersection of the cut- ting plane and the semi-cylinder is the semi-ellipse B'AB', in which the vertical lines are unchanged in length, while the lengths of the horizontal lines are c times the lengths of tlie corresponding lines in the semicircle, c being the secant of the angle made by the cutting plane with the base. A semi- elliptic arch is thus obtained, and the forces to which it is sub- jected are parallel projections of the forces acting upon the semicircular arch. These new forces are in equilibrium (see Corollary). Let P" = the total vertical pressure upon one-half of the arch ; H' = the total horizontal pressure upon one-half of the arch ; EXAMPLES OF LINEAR ARCHES. F Py — vertical intensity of pressure = yrrr, ; pj = horizontal intensity of pressure = tt-j-. Then P' = P=I/ = pr] 755 0) p^ B' pr /_ ^' ~ 0B'~ c,OB~~ cr~ c' • • • • (2) H' = cH=cP=:cP'\ (3) H' c.H cpr ^' -OA'-aA-T-'^ w Hence, by eq. (3), H' OB' P' ~^ " oa'' or, the total horizontal and vertical thrusts are in the ratio of the axes to which they are respectively parallel, and, by eqs. (2) and (4), / 8 » OB or, the vertical and horizontal intensities of pressure are in the ratio of the squares of the axes to which they are respectively parallel. Any two rectangular axes OG, OK in the circle will project into a pair of conjugate radii OG', OK' in the ellipse. Let OG' =zr„ OK' = r,; Q = total thrust along elliptic arch at K; Then H r ^ H r " R ~r' I I I Hi r l! : • ( m in I 756 THEORY OF STRUCTURES. or, the total thrusts along an elliptic arch at the extremities suitable for a load distributed in such a manner that the vertical and horizontal intensities (eqs. (2) and (4)) at any point of tli' arch are unequal, but are uniform in direction and magnitude. Corollary. — It can be easily shown that the projected forces acting upon the elliptic arch are in equilibrium. The equations of equilibrium for the forces acting upon the circular arch may be written «'(r|)+Frf.=o; 7* being the thrust along the arch at the point xy, and X, Y the forces acting upon the arch parallel to the axes of x and y, respectively. If T\ X\ V be the corresponding projected forces, ^ = J, Xds = cX'ds', Yds = Y'ds'. as as Hence, the above equations may be written d[^,cdx'^-\-cX'cis' =0» d^-'dy) + FV/ = o; and or and i-%) + FV/=o. Hence, the forces T\ X\ and Y' are also in equilibrium. EXAMPLES OF LINEAR ARCHES. 757 ^<^o {d) Hydrostatic Arch. — Let the figure represent a portion of a linear arch suited to support a load ^ _ which will induce in it a normal pressure at every point. Tlie pressure beiii^ normal has no tangential component, and tlie thrust (7") along the arch must therefore be everywhere the same. Consider any indefinitely small element CD. It is kept in equilibrium by the equal thrusts (7") at the extremities C and D, and by the pressure p . CD. The intensity of pressure / being assumed unk^m for the element CD, the line of action of the pressure /TcZ? bisects C i right angles. Let the normals at C and D meet in (9, , the centre of curvature. Take 0,C = 0,D = p, and the angle CO,D = 2 JO. Resolving along the bisector of the angle CO^D, Fig. 489. or and hence, 2Tsir\ ^6= p. CD = pp. 2 Ad, 2TJ0=pp.2J0\ T = Pp = a constant. . . . (i) Thus, a series of curves may be obtained in which p varies inversely as/, and the hydrostatic arch is that curve for which the pressure p at any point is directly proportional to the depth of the point below a given horizontal plane. Denote the depth by_j/. and let iv be the specific weight of the substance to which the pressure/ is due. Then and p = wy, (2) T = Pp = wyp = a constant. • • • (3) The curve may be delineated by means of the equation yp = const. (4) !<■ h -t >} iir' i'! ■■,■«■ :)' 758 THEORY OF STRUCTURES. It may be shown, precisely as in Case {c\ that the horizontal intensity of pressure {J>^ = the vertical intensity (Py) =p. • (5) Take as the origin of co-ordinates the point O vertically above the crown of the arch, in the given horizontal plane. Let the horizontal line through O be the axis of x. " " vertical " " " " " " " j. Any portion AM oi the arch is kept in equilibrium by the equal thrusts {T) at A and A/, and by the resultant load P upon AM, which must necessarily act in a direction bisecting the angle ANM. Fio. 490. Complete the parallelogram AM, and take SIV = NM to represent T. The diagonal NL will therefore represent P. Let Q be the inclination of the tangent at M to the hori- zontal. The vertical load upon AM= vertical component of P = LK = Tsin 6 = pfJt sin S = wyp sin 6 = wy^p^ sin 6, . (6) J/,, p„ being the values of j, p, respectively, at A. The horizontal load upon A AT = horizontal component oi P = NK=SN-KS= T- Tcosd=2Tism^ 2f) 2tvyp sin — = zwj/^p, 4^ = 2pP (sin -j Again, the vertical load upon AM •• (^'" I) • (7) = / pi^-^ — "^^ J ^'^^ ~ "^'^'Po sin ^ ; (8) the horizontal load upon AM ^ X!^'^^ "^ ^X!^'^^ ~ ^^^' ~-^'*^ ~ ^^yoP, (sin -) . (9) EXAMPLES OF LINEAR ARCHES. 759 Equation (8) also shows that the area bounded by the curve AM, the verticals through M and A, and the horizontal through is equal to yj\ sin ^, and is tlierefore proportional to sin ^. At the points defined hy B — 90° the tangents to the arch are vertical, and the portion of the arch between these tangents is alone available for supporting a load. The vertical and horizontal loads upon one-half the arch are each equal to Corollary. — The relation given in eq. (i) holds true in any arch for elements upon which the pressure is wholly normal. This has been already proved for the parabola and catenary, in cases {a) and (p). At the point A' of the elliptic arch, c'r^ OA' = — = cr. Hence, the horizontal thrust at A' = p,p--f>-pcr- cH. {e) Geostatic Arch. — IhQ geostatic is a parallel projection of the hydrostatic arch. The vertical forces and the lengths of vertical lines are unchanged. ' _-— ^A The horizontal forces and lengths of hori- zontal lines are changed in a given ratio ^ to I. /B__yB: Let B'A be the half-geostatic curve de- Fig. 491. rived from the half-hydrostatic curve BA. The vertical load on AB' -P' = P= thrust along arch at B'. ... (i) The horizontal load on AB' — H' ~ cH = thrust along arch at A. . . , (2) The new vertical intensity = A' C.OB- c ~ c ^3i m % m \\ OB' \V ili M 76p THEORY OF STRUCTURES. The new horizontal intensity _ ' _ J^' _ £^ _ ~^' ~ 0A~ OA ~'^^^ cp. (4) Thus, the geostatic arch is suited to support a load so dis- tributed as to produce at any point a pair of conjugate press- ures ; pressures, in fact, similar to those developed according to the theory of earthwork. Let R^ , R^ be the radii of curvature of the geostatic arcli at the points A, B\ respectively, and let r, , r, be the radii of curvature at the corresponding points A, B o{ the hydrostatic arch. The load is wholly normal at A and B' . Thus, H' = p;R, = ^R, = cH=cpr,. . . . Also, .-. R, = c'r, P' =: p^'R, = cpR, ^P = pr, (6) (7) cR, = r. (8) {f) General Case. — Let the figure represent any linear P arch suited to support a load which is sym- metrically distributed with respect to the crown A, and which produces at every point of the arch a pair of conjugate pressures, the one horizontal and the other vertical. Take as the axis oi y the vertical through the crown, and as the axis of x the hori- FiG. 49a. zontal through an origin 6? at a given dis- tance from A. Any portion AM oi the arch is kept in equilibrium by the horizontal thrust H at A, the tangential thrust T at M, and the resultant load upon AM, which must necessarily act through the point of intersection A'^ of the lines of action of H and T. Since the load at A is wholly vertical, H is given by //■„ = / P« , (0 (6) (7) Irough hori- bn dis- the r, and Irough Id T. (0 EXAMPLES OF LINEAR ARCHES. 761 />„ and p„ being, respectively, the vertical intensity of pressure and the radius of curvature at A. Let MN= T, and take NS = H,. Complete the parallelogram SM\ the diagonal NL is the resultant load upon AM in direction and magnitude. The vertical {KL) and the horizontal {KN) projections of NL are, therefore, respectively, the vertical and horizontal loads upon AM. Denote the vertical load by V, the horizontal by H. Then 7" sin = KL — V— J pydx, (2) and H^ KN= SN- SK=H,- Fcot B, . . (3) 6 being the angle between MJV and the horizon. dV Py, the vertical intensity of pressure, = px, the horizontal intensity of pressure dx' dH d ,^_ ^, = -7-= - -r(^'cot^). dy d)r ' (4) (5) Example. — A semicircular arch of radius r, with a hori- zontal extrados at a vertical distance R from the centre. The angle between the radius to J/ and the vertical = 6. .'.x^rsxwB, y = R — r cos 0. . . . (i) ' dx =r cos 6d0, dy — r s,m Odd (2) p^ = wy = w{R — r cos (f), (3) w being the specific weight of the load. Hence, V= u'J^\j^ - r cos ey cos fidd = wr[R sm & - — — J. ... (4) IF 'i' 1 I I m it-.' t ■■ P 762 THEORY OF STRUCTURES. Equations (3) and (4) give H \ iox /, = wiji — r) , .... and hence H, = wr{R -r) Px , the horizontal intensity of pressure, r 6 — sin 6 cos 6 (5) (6) = -~{Vcote)=^.[R-l' sin — r cos B )• (7) Rankine gives the following method of determining whether a linear arch may be adopted as the intrados of a real arch. At the crown a of a linear arch ab measure on the normal a length ac, so that c may fall within the limits required for stability (e.g., within th^ middle third). At c two equal and opposite forces, of the same magnitude as the horizontal thrust H at a, and acting at right angles to ac, may be introduced without altering the equilibrium. Thus the thrust at a is replaced by an equal thrust at c, and a right-handed couple oi moment H . ac. Similarly, the tangential thrust T'at any point d of ab ma}' be replaced by an equal and parallel thrust at e, and a couple of moment T . de. The arch will be stable if the length of de, which is normal to ab at d, is fixed by the condition 7". de ^^ H . ac, and if the line which is the locus of e falls within a certain area (e.g., within the middle third of the arch ring. 8. Arched Ribs in Iron, Steel, or Timber. — In the fol- lowing articles, the term arched rib is applied to arches con- structed of iron, steel, or timber. The coefficients of elasticity are known quantities which are severally found to lie between certain not very wide limits, and their values maybe introduced into the calculations with the result of giving to them greater accuracy. There are other considerations, however, involved in the problem of the stability of arched ribs which still render its solution more or less indeterminate. It has been shown that the curve of pressure, or linear arch, m-e^. \ ARCHED RIB U.VDER A VERTICAL LOAD. 763 is a funicular polygon of the extraneous forces which act upon the real arch. It is, therefore, also the bending-moinent curve, drawn to a definite scale, for a similarly loaded Jiorizontal girder of the same span, whose axis is the springing line. When the arched rib carries a given symmetrically dis- tributed load, it will be assumed that the linear arch coincides with the axis of the rib, and that the thrust at any normal cross-section is axial and uniformly distributed. The total stress at any point is made up of a number of subsidiary stresses, of which the most important are : (l) a direct thrust ; (2) a stress due to flexure ; (3) a stress due to a change of temperature. Each of these may be investigated separately, and the results superposed. 9. Bending Moment (M) and Thrust (T) at any Point of an Arched Rib under a Vertical Load. — Let ABC be the axis of the rib. Let D and E be points on the same vertical line, E being 'D^ Fig. 493. on the axis of the rib and D on the linear arch for nny given distribution of load. Resolve the reaction at A into its vertical and horizontal components, and denote the latter by H, Since all the forces, excepting H, are vertical, the difference between the moments at D and E = H . DE. But moment at Z> = o. Hence, moment at i? = J/= H . DE. Let the normal at E meet the linear arch in D'. Then, if T is the thrust along the axis at E, J)'E T cos DED' — H = ^^p^ approximately, or ;il 1! ■ H.DE--^ T.D'E= M. 764 THEORY OF STRUCTURES. 10. Rib with Hinged Ends ; Invariability of Span.— Let ABC be the axis of a rib supported at the ends on pins or --.^.-^.-^8 Fig. 494. i M on cylindrical bearings. The resultant thrusts at A and C must necessarily pass through the centres of rotation. The vertical components of the thrusts are equal to the corre- sponding reactions at the ends of a girder of the same span and similarly loaded, and H is given by the last equation in the preceding article when BE has been found. Let ADC be the linear arch for any arbitrary distribution of the load, and let it intersect the axis of the rib at 6^ Tlie curvature of the more heavily loaded portion AES will be flattened, while that of the remainder will be sharpened. The bending moment at any point E of the axis tends to change the inclination of the rib at that point. Let the vertical through E intersect the linear arch in D and the horizontal through A in F. Let 6 be the inclination of the tangent at £ to the hori- zontal. Let / be the moment of inertia of the section of the rib at£. Let ds be an element of the axis at E. Mds H.DE.ds Change of inclination 3i^ E = dO — —=j EI If this change of curvature were effected by causing the whole curve on the left of E to turn about E through an angle dd, the horizontal displacement of A would be EF. dO = ?* . H.DE.EF.ds EI ARCHED RIB WITH HINGED ENDS. 765 This is evidently equal to the horizontal displacement of E, and the algebraic sum of the horizontal displacements of all points along the axis is M.DE.EF.ds =/^ DE.EF.ds EI = 0, (I) since the length AC '\s assumed to be invariable. Thus, the actual linear arch must fulfil the condition ex- pressed by eq. (i), which may be written J 'DE.EF.ds = 0, (2) since H and E are constant. If the rib is of uniform section, /is also constant, and eq. (2) becomes jDE.EF.ds=o (3) Also, since DE is the difference between DF and EF, C{pF - EF)EF. ds = o =fDF. EF . ds- J EF'ds (4) Remark. — Eq. i expresses the fact that the span remains invariable when a series of bending moments, H .DE, act at points along the rib. These, however, are accompanied by a thrust along the arch, and the axis of the rib varies in length with the variation of thrust. Let H^ be the horizontal thrust for that symmetrical loading which makes the linear arch coincide with the axis of the rib. Let T^ be the corresponding thrust along the rib at E. The shortening of the element ds at E of unit section T -ds. Example i. Let the axis of a rib of uniform section and hinged at both ends be a semicircle of radius r. Let a single weight W be placed at a point upon the rib whose horizontal distance from O, the centre of the span, is a. %\- jg 11 I 111 1.^' IH ( iAs.^a itii^-i ^(A THEORY OF STRUCTURES. The " linear arch " (or bending-moment curve) consists of two straight lines DA, DC. / \\ Draw any vertical line intersecting the axis, the linear arch, and the springing line AC in E\ D', F', respectively. Let OF' = X, and let dx be the horizontal projection upon AC ol the element ds at E'. Then ds T^,^r-, ^ ^ = cosec E'OF = -^j, , or E'F'ds = rdx (i) Applying condition (4), f^ D'F'rdx -f f D'F'rdx = f E'F'rdx, or y" D'F'dx + f D'F'dx = f E'F'dx, or area of triangle ADC — area of semicircle. And if z be the vertical distance of D from AC, nr' zr = ;f r or ARCHED RIB WITH HINGED ENDS, j6y nr z =■ — — one-half of length of rib. ... (2) I $: nr .: DE = DF- EF= ~- Vr* - a' 2 (3) Hence, if h be the horizontal thrust on the arch due to W, 3 3 h . DE = M = W^-^^^ (4) Similarly, if there are a number of weights W^, IV,, W^, . . . upon the rib, and if //,, ^,, ^3, . . . are the corresponding hori- zontal thrusts, the total horizontal thrust H will be the sum of these separate thrusts, i.e., /f=//. + /^,+ (5) It will be observed that the apices (Z>, , D^, D^, . . .) of the several linear arches (triangles) lie in a horizontal line at the nr vertical distance — from the springing line. Ex. 2. An ar'-.hcd rib hinged at the ends and loaded with weights W^, W,, W,, . . . \ \ \ ^i^^v, s r: Fig. 4g6. l^ Fig. 497. Let I 2 3 4 . . . « be the line of loads, W^ being represented by I 2, W^ by 2 3, W^ by 34, etc.. and let the segments \Xy i!|i; k: \' \ LIU \ 768 THEORY GF S'JRUCTURES. nx, respectively, represent the vertical reactions at A and C. Take the horizontal length xPto represent //, and draw the radial lines Pi, P2, Pi, . . . The equilibrium polygon ^^'-.i'-^, . . . must be the funicu- lar polygon of the forces with respect to the pole P, and there- fore the directions of the resultant thrusts from A to Z:',, £^ to /;,,, P^ to /ij, ... are respectively parallel to Pi, P2, P^, . . . The tangential (axial) thrust and shear at any point /> uf the rib, e.g., between Zi, and £,, may be easily found by draw- ing Pi parallel to the tangent at/, and 3/ perpendicular to J'/. The direct tangential thrust is evidently represented by P/, and the normal shear at the same point by 3/. The latter is borne by the web. If / is a point at which a weight is concentrated, e.g., £^, draw /V7" parallel to the tangent at £, and 5/', 6i" perpen- dicular to Pi'i". Pt' represents the axial thrust immediately on the left of ^5 , and 5/' the corresponding normal shear, while /^^" repre- sents the axial thrust immediately on the right of E^, and 6/" the corresponding normal shear. A vertical line through P can only meet the line of loads at infinity. Thus, it would require the loads at A and C to be infinitely great in order that the thrusts at these points might be vertical. Practically, no linear arch will even approximately coincide with the axis of a rib rising vertically at the springings, and hence neither a semicircular nor a semi-elliptical axis is to be recommended. Ex. 3. Let the axis of the rib be a circular arc of span 2/ and radius r, subtending an angle 201 at the centre N. Let the angles between the radii NE, NE' and the vertical be p and 0, respectively. The element ds at E' = rdO. Also, E'F' = r(cos d - co^ a); AF = /-r sin d- D'F' ^ r,—n - r sin B). ARCHED RIB WITH HINGED ENDS, 769 yo- N Fic. 498. Applying condition (5), r'(cos ^ — cos afrdO L =/ 7— — (/— r sin ^)r(cos — cos a)rd9 + X^-^/- r sin 6)r{cos S — cos a)rdO, which easily reduces to r{ar(cos 2a + 2) — f sin 2a\ = 7, i \ /'(sin oc — a cos «) + - (cos 2a — cos 2^) — /■/ cos a'(cos a — cos /5) — /«(sin /? — /? cos a) >• , an equation giving z or Z?/^ Also, DE = DF- EF, and the corresponding horizontal thrust may be found, as before, by the equation h,DE= W I f il. M L 770 THEOKY OF STRUCTURES. Note.—U a° = 90°, 5 _ 2" (^ ~ ^*] or £ ~ — as in Ex. I. Ex. 4. Let the axis be a parabola of span 2/ and rise Jt. (Fig. 498, Ex. 3), From the properties of the parabola, E'F' = /&(i-^). t ± a and or, approximately. ds* = «'.ir'(i + 4^^'), +4V). ^j = dx\ I Applying condition (5), which easily reduces to an equation giving 2 or Z>K iVc^/^. — If the arch is very flat, so that ds may be considered ARCHED RIH WITH ENDS ABSOLUTELY FIXED. 77 ^ /{•' as approximately equal to ix, the term 2j,x' in the above equation may be disregarded, and it may be easily shown that I /' + a' {■-a-q i6 'is' or kr • - - 5 5/' - a" II. Rib with Ends absolutely Fixed.— Let ABC be the axis of the rib. The fixture of the ends introduces two un- Fig. 499. known moments at these points, and since H is also unknown, three conditions must be satisfied before the strength of the rib can be calculated. Represent the linear arch by the dotted lines KL ; the points K, L may fall above or below the points A, C. Let a vertical line DEF intersect the linear arch in D, the axis of the rib in E, and the horizontal through A in F. r . ,. . ^ ,„ Mds As m Art. lo, change of mclination at h, or dB, = —f^. k.1 But the total change of inclination of the rib between A and C must be nil, as the ends are fixed. pMds 'J ~EI ° = / H.DE.ds EI ' (I) ■which may be written / -j-ds = O, (2) II tf since H and E are constant. ibiv ! U 772 THEORY OF STRUCTURE'S. If the section of the rib is uniform, / is constant and eq. (2) becomes . v*'- •" i'--- .-• • v • •• / DE .ds = 0. (3) Again, the total horizontal displacement between A and C will be nil if the abutments are immovable. If they yield, the amount of the yielding must be determined in each case, and may be denoted by an expression of the form f^H, M being some coefificient. .. ._"'■. ■ .: As in Art. 10, the total horizontal displacement l .' .; H.DE.EF.ds •••/ ~ J EI H.DE.EF.ds EI = o or = jxH. ... (4) But H and E are constant. >DE . EF. ds •■•/- / = o or = E' . (5) If the section of the rib is uniform, / is also constant, and hence / DE . EF .ds = o or E' ... (6) and since DE is the difference between DF and EF, this last may be written ^fDF.EF.ds '^J'EF\ds = o or =^I. ... (7) Again, the total vertical displacement between A and C must be nil. ._. The vertical displacement of E (see Art. 10) M.AF.ds -AF,dB^ Ml ARCHED RIB WITH ENDS ABSOLUTELY FIXED. 773 Hence, the total vertical displacement ^H.DE.AF f- EI ds = 0, (8) which may be written -DE . AF /ut. . At , J — as — o, (9) since //'and E are constant. If the section of the rib is also constant, fDE.AF.ds=o=fDF,AF.ds-fEF,AF.ds. (lo) t Eqs. (2), (5), and (9) are the three equations of condition. In eq. (9) AF must be measured from same abutment throughout the summation. The integration extends from A to C. Example i. Let the axis of the rib be a circular arc of span 2/, subtending an angle 2a at the centre N'. Let a weight IV , be concentrated on the rib at a point E ! y ^.2 ft / N Fig, 500. whose horizontal distance from the middle point of the span is a. Let the radius NE make an angle /? with the vertical. . The " linear arch " consists of two str-ight lines DA', DC. Let AA' — y,. DF = z, CC =)\ . I' I J 5 f< 774 \ THEORY OF STRUCTURES. Draw any ordinate E'F' intersecting the linear arch in Lf. Let the radius NE' make an angle ^ with the vertical. Then E'F' = r{cos 6 — cos a). AF' = /-rsme, and D'F' =={/ -r sine)'^—^-{.y^ if F' is on the left of F; AF' = I -\-r sine and D'F' = {/- r sin d)-j-^ J^ y^ l-a \{ F' is on the right of F. Also, ds = rdB. i Applying condition (i), + r {('-'• sin «)^|^ +>., I the "linear arch." Then, - fDE . EF. ds =zj*EF'ds = r' J\cos 6 - cos afdd = r'|a(2-f cos 2a) — |sin2a'|. Also, / = r sin a. ''•^iin^^'*^^+^°^^"^~*^'"^"^ ±26//= a Note, — If the axis is a semicircle, a = 90°, and .— ~±2etl:=:0. Second, let the rib h^ fixed at both ends. The " linear arch" is now a straight line A'C at a distance j?(= DF) !" ;-n .<4C given by the equation / DE .ds = o. .'. JdF . ds =fEF . ds, e ids = r' I (cos d — cos a)dB, az = r(sin a — a cos a). or or Also, J' BE . EF. ds ^f{DF. EF-^EPys = zjEFds -^fEF'ds — 2^rr'(sin or — « cos «)~ r'|«(2 + cos 2a) — | sin 2a\. .| 2^r'(sin a — or cos a) — r'|a(2 + cos 2a) — | sin 2a | ( ± 2etl = o, EI and / = r sin a. Ex. 2. Let the axis AEC of a rib of uniform section be a parabola of span 2/ and rise k. (See Fig. 501 in Ex. i.) / Fin The DE and hen Secon The from AC or .'. ^ or Also, fDE. EI =zfEF.i 779 EFFECT OF A CHANGE OF TEMPERATURE. First, let the rib be hinged at both ends. The straight line AC\s the linear arch. Then and hence, ~ k% -; 4- --— 77 ± 2€tl = O. EI V15 ' 105 /V or Second, let the rib ht fixed eA. both ends. The linear arch is the line A'C at a distance z {z=.DF) from A C given by the equation CdE ,ds = o = J{DF ~ EF)ds, DFCds^ fEF.ds. or Also, /"/?£ .EF.ds =fDF .EF.ds-^ J EF'ds i ii- ft 1 I li 78o THEORY OF STRUCTURES. Hence, Ht Akl { I .2k {.(.+-;f;)~f(.+f^)} ± 2etl = o. Remark. — The coefficient of expansion per degree of Fah. renheit is .0000062 and .0000067 for cast- and wrought-iron beams, respeictively. Hence, the corresponding total expansion or contraction in a length of 100 ft., for a range of 60° F. from the mean temperature, is .0372 ft. {— -^a") and .0402 ft. (= ^"). In practice the actual variation of length rarely exceeds (?«r- half oi these amounts, which is chiefly owing to structural con- straint. 13. Deflection of an Arched Rib. Fig. 502. Let the abutments be immovable. Let ABC be the axis of the rib in its normal position. Let ADC represent the position of the axis when the rib is loaded. Let BDF be the ordinate at the centre of the span ; join AB, AD. Then /arc A D\ ' DF' = AD' - AF' = AB'l ~\ - AF". \arc ABI But arc AB — a rc AD f arc AB ~E' /being the intensity of stress due to the change in the length of the axis. / .-. DF' = AB'[i - ^-1 - AF' = BF' - AB' | 2^ E \EI ^ i ELEMENTARY DEFORMATION OF AN ARCHED RIB. 78 1 /. AB'^ \^~e~^h) 1 "^ ^'^'~ ^^' ^ ^^^^ ~ DF){BF+ DF) = 2BP\BD), approximately. y—rj is also sufficiently small to be disregarded. Hence, AB' f k' -{-r f . , BD, the deflection, = -j^ jr — — 7 — ^ , approximately. 14. Elementary Deformation of an Arched Rib. X The arched rib represented by Fig. 503 springs from two abutments and is under a vertical load. The neutral axis PQ is the locus of the centres of gravity of all the cross-sections of the rib, and may be regarded as a linear arch, to which the conditions governing the equilibrium of the rib are equally ap- plicable. Let AA' be any cross-section of the rib. The segment AA'P is kept in equilibrium by the external forces which act upon it, and by the molecular action at AA'. The external forces are reducible to a single force at C and to a couple of which the moment M is the algebraic sum of the moments with respect to C of all the forces on the right of C. The single force at C may be resolved into a component T along the neutral axis, and a component 6" in the plane AA'. r\. (28) Hence, by equations (19) and (26), (A) i=:-^J^M,x^sf^+(^-^^-^)^y, . (29) and by equations (19) and (27), I i .^ . e-*' , f8/&^ \x* ^ (B) ,= - — |^„^H-.S.-+(-^-«;J^ -^{4r-(i-r)/{'|. (30) When X — l,i= /, = o, and therefore, by the last equation, : .r,,V o = ^ + ^-+i-75--«'J6-6-'''/'. . . . (31) S5!^ 1 ARCHED RIB OF UNIFORM DEPTH. dv 791 Again, let « = -7- . Then */ a t/ a ■dy , _ Hdv dy _ .dy dx ~ 1 dx dx ~ ^dx But t, = o, and -j^ = -y By the conditions of the problem, x' — x and ^' — ^ are each zero at Q. Hence, equations (20) and (21), respectively, become o=-^/'/'^'^^'- (^^^^)^; • • (33) o=- J^ idx. (34) Substitute in eqs. (33) and (34) the value of i given by eq. (30), and integrate between the limits o and /. Then and ° = -£7,i-2-+-6-+l— -«^)i5-«"'iir which may be written • ' "4 ' \ / ^20 20 ;li r ' ' (1 •1 i . i: !^ 792 and THEORY OF STRUCTURES, j o=J/. + 5.^ + (^-«;)^-z.V:^. . . . . . (36) Hence, by eqs. (31), (35), (36), , "^ * wl w l^r^ M, = 12 3 (-^) 4 '3 (37) (38) -fT; 8" 4 8 ' 4 y 4 ^ K'+^iiii-) . . (39) When x = l,M=M„ and 5 = S,, Hence, by eqs. (25) and (2y), and S, = S,-\- \^-ji w]l - w'rl, n^ ,^ . c, , /8/fe^ Y w'r'P Substituting in these equations the values of 5„, M^, given above, we have and 5,= -^-»V/(,-^ + l)+^, . . . (40) J/, = wTr' r + -) + -/&^. . . (41) To find the greatest intensity of stress, etc. — The intensity of . T H the stress due to direct compression = -r = -^ . ARCHED RIB OF UNIFORM DEPTH. 793 The intensity of the stress in the outside layers of the rib due to bending is the same as that in the outside layers of a horizontal beam of uniform section A^ acted upon by the same moments as act on the rib, for the deflections of the beam and rib are equal at every point (eq. (19) ). Also, since the rib is fixed at both ends, the bending moment due to that portion of the load which produces flexure is a maximum at the loaded end, i.e., at Q. Hence, the maximum intensity of stress (/,) •LJ occurs at Q, and/*, = ^- ± M^j, z^ being the distance of the layers from the neutral axis. H and J/, are both functions of r, and therefore /»i is an ab- solute maximum when But dr~^~Adr^I,dr ^42> dH iSw'l' r\i- rf and dr dM. 4 k 1+^ 45 /. ' 4 Ak^ Hence, /i is an absolute maximum when I I 2 kz\ Qz^wTr{\ -rf' ^(■+f.4=) ^7; (43) The roots of this equation are r = \ 45 /. and 0^+4 A,k' 2 A^zji ± I (45) ! I ■ I 1^ 794 THEORY OF STRUCTURES. r = I makes tt ^^^o, so that the maximum value of p. corresponds to one of the remaining roots. ■ Thus, "■' .■";' *••■ : . ■; , :■• Xht max. thrust = -j-\H -{-——■ Ma =p^' . . (46) and the max. tension = -j-\— ^ -\- ~j~^^-) —Px't (47) ■"1 •'1 the values of H and J/, being found by substituting in eqs. (39) and (41) r = 4 A,k' 5.-i^'- 2A,2^k or 2 ^ 4^,/^' 1 + 2 y4,2',/^ . (48) according as the stress is a thrust or a tension. If eq. (47) gives a negative result, there is no tension at any point of the rib. Note. — The moment of inertia may be expressed in the form q being a coefficient depending upon X}\q. form of the section. Hence, the maximum intensity of stress = -j\±. H -\ ' j . . (49) Corollary i. — If the depth of the rib is small as compared with k, the fraction t will be a small quantity, and the maxi- mum intensity of stress will approximately correspond to r = |^. ARCHED RIB OF UNIFORM STIFFNESS. ;95 The denominator in eq. (39) may be taken to be k, and it may be easily shown that the values of/,',//' are ^•-^. ( 8 U+T;fe'J^4-^-+^5-^|:- (50) Cor. 2. If the numerator in eqs. (48) is greater than the denominator, then r must be unity. Hence, by eq. (39) and _ r zv -\- w' /[S^eiEI^ 8 bk ^'^~blF'' (52) and by eqs. (38) and (41), M M ^' ( I /^ I - ^ , 1 5 ^fET, - 16' b ^k^^\'b~T (53) Thus, //,//' can be found by substituting these values of //"and J/, in eqs. (46) and (47). 19. Parabolic Rib of Uniform Stiffness, hinged at the Ends. Let the rib be similar to that of the preceding article. Since the ends are hinged, M^ — O — M^, while i is an un- determined constant. The following equations apply : (A) i=:i;+^___^J^. ^j^^ (B) S=iS,-\-\-j^ -w)x-w'\x-{\-r)l\', . (55) 796 THEORY OF STRUCTURES, t w'\ m But by (62), = 5. + ^L^i _ '^f.Vv, ^ ' vr ' ' m\ 2 / 2m Hence, J/', the maximum bending moment, P* ~ / , , ZkH\ 2\W -\-W jj— J As before, the greatest stress (a thrust) (71) • . • (7? and the value of r which makes // an absoiu, m? imum is dp' given by -^ = o. But by (71), M' involves r" in the numera- ARCHED KIB OF UNIFORM STIFFNESS. 799 dp' tor and r" in tlic denomhutor, so that -,— = O will be an dr equation involving r'\ One of its roots isr = i, which generally gives a niimuiinn value of /»,'. Dividing by r — l, tlie equation reduces to one of \\m thirteenth order, but is still far too complex for use. It is found, however, that ^ = i gives a close approximation \. \n « I 7V By (71), M'= , sU^^+tJ— -+4f w \n 7U 4- - — I w n '2 (74) (75) (76) (77) (78) iV(?/t'. — If the rib is merely supported at the ends but not fixed, the horizontal displacement of the loaded end may be 8oo m:>^:%: THEORY OF STRUCTURES. represented by jaH {Art. ii). Thus the term — fiH must be added to the right-hand side of eq, (15). 20. Parabolic Rib of Uniform Stiffness, hinged at the Crown and also at the Ends. — In this case M — o dX the crown, which introduces s. fourth equation of condition. By (57). which may be written o = S,^^-^-w)^-w'l[r^-r+'^. . . (79) Eliminating 5, between (79) and (62), Hence, By (79)» By (68), By (66), — — w = w {— 2r' -\- <\r — i). ff = :^^\w - w\2r' - 4r + i)\. . . . (80) P^S^=^^r-.r Zr. = wT 24EI, By (/o) and (82), '-X=: w'l, ., (81) (82) (1-4^ + 4^'-/) (83) iw'ir — \f 4 = "•••••••• v"4) By (71). w'l M' = — g-(r -•- i)' , (85) JILMJ-M... I'* 1 «-WSKS^ PARABOLIC RIB OF UNIFORM STIFFNESS. When r = i /7 — n w' ^=8i'r+-J. s.= 2 r li 8 ' "^^ ~ 8 ' '• - - 384 i:7. ' '/» and J/' = 64- 801 (86) These results agree with those of (73) to (78), if « = I. Lt general, when n =■ i, 4) !5) w + — (5r' — 5/ + 2O = w — w'(2r' — 4r -f- 1), by (65) and (80). Htnce, 2r' — 5r* + 9^' + Sr + 2 = o = {2r — i)(r - i)'(r' — 2), and the roots are r =^^, r — \, r ^= ± 1^2. Hence, ?i = l only renders the expressions in {'66) identical with the corresponding expressions of the preceding article when ;/ = ^ or I. Again, the intensity of thrust is greatest at the outer flange of the loaded and the inner flange of the unloaded half of the rib, and is . _ /' )Z,W' \l SA, U, S ' k \w^ -)!• The intensity of tension is greatest at the inner flange of the loaded and the outer flange of the unloadv;d half of the rib, and is /' ( ^'i If/' I / , w' 8^, (/, The greatest total horizontal thrust occurs when r = i, and its value is U (w -f- w'). 802 THEORY OF STRUCTURES. the deflection is an absolute maximum when "-\y' — y) '= o. 21. Maximum Deflection of an Arched Rib. — The deflec- tion must necessarily be a maximum at a point given by i =:^ o. Solve for x and substitute in (i6) to find the deflection j/' — y; d dr' The resulting equation involves r to a high power, and is too intricate to be of use. It has been found by trial, however, that in all ordinary cases the absolute maximum deflection occurs at the middle of the rib, when the live load covers its / whole length, i.e., when x = -, and r = i. Case I. Rib of An. i8. For convenience, put i + — Vra= J. ^ ' 4 Ak Then, by (39), (87) By (38) and (41), -M.= -{w + wi-—^:^l'-l^ = -M,., (88) 12 4 J k By (36) and (38), By (30), (38), (89), ■. (89) - ^^{^M,x - zM^ -{- 2M^. . . . (90) Ei: Hence, the maximum deflection - J, tdx = - -I \x-i-j^ 2j,)dx = - ^-^-- - r W -\- W' S — I 5 6//' ^ MAXIMUM DEFLECTION OF AN ARCHED RIB. 803 The central deflection d^ of a uniform straight horizontal beam of the same span, of the same section as the rib at the crown, and with its ends fixed, is '^'~ iH~Ejr' •••••• (92) Hence, neglecting the term involving the temperature, ^■ = ^^" (93) Case II. Rib of Art. 19. By (65), r- _ l^ W -\- W' u~~Ji (94) • By (66) and (62), ^»-2-i^/> + ^)-7r =Ill7- • • • (95) By (30), (94), and (95), EI,\i2 2 "^3/7 (96) Hence, the maximunr. deflection If the ends of th Seam in Case I are free, its central de- flection = 5 ^^±^1_,, 384 £f ~ ' ' •■•^■' = ^'^'' (98) Thus, the deflection of the arched rib in both cases is less than that of the beam. 8o4 THEORY OF STRUCTURES. 22. Arched Rib of Uniform Stiffness fixed at the Ends and connected at the Crown with a Horizontal Distribut- ing Girder. — The load is transmitted to the rib by vertical struts so that the vertical displacements of corresponding points of the rib and girder are the same. The horizontal thrust in the loaded is not necessarily equal to that in the un- loaded division of the rib, but the excess of the thrust in the loaded division vvill be borne by the distributing girder, if che rib and girder are connected in such a manner tiiat the hori- zontal displacement of each at the crown is the same. The formulae of Art. i8 are applicable in the present case with the modification that /, is to include the moment of inertia of the girder. The maximum thrust and tension in the rib are given by equations (64) and (65). Let z' be the depth of the girder, A' its sectional area. H M z' The greatest thrust in the girder = — — j — — -j — ~. (99) The greatest tension in the girder = A,^A' ' 2EI, 2EL . (100) H and J/, being given by equations (66) and {6y), respectively. The girder must have its ends so supported as to be capable of transmitting a thrust. 23. Stresses in Spandril Posts and Diagonals. — Fig. 505 represents an arch in which the spandril consists of a series of vertical posts and diagonal braces. Fig. 505. Let the axis of the curved rib be a parabola. The arch is then equilibrated under a uniformly distributed load, and the diagonals will be only called into play under a passing load. STJ^£:SS£S IN SPANDRIL POSTS AND DIAGONALS. 805 Let X, y be the co-ordinates of any point F of the parabola with respect to the vertex C. Then y = V^ • Let the tangent at F meet C£ in L, and the horizontal ££ in G. hGt BC=k'. Then BL = BC - CL = BC - CN = k' - y. J^et A^ be the total number of panels. Consider any diagonal FB between the nth and {n -\- i)th posts. Let za' be the greatest panel live load. The greatest compression in £D occurs '^•Lcn the passing load is concentrated at the first ?i — i panel points. Imagine a vertical section a little on the left of £F. The portion of the frame on the right of this section is kept in equilibrium by the reaction R at P, and by the stresses in the three members met by the secant plane. Taking moments about G, D.GF cose = R. AG, D being the stress in DE, and B the angle DEP, Now, R = w n\ •) is le Also, X -^ GB _ k' -\- y ^ GB ~ k' -y'' GB _ k'x — xy - ^7~' and hence, GE=.GB^x^^^-^^^^^-, and 2y Hence, _ w' n{n — i)ly -\- k'x t k'x — GA =- + 1^ — 2 2y xy D- — 2 xy N k'x -(- xy sec B. The stresses in the counter-braces (shown by dotted lines in the figure) may be obtained in the same manner. 8o6 THEORY OF STRUCTURES. I ■ The greatest thrust in EF = w' -|- w. The greatest tension in EF ~ Bcos — w, w being the dead load upon EF. If the last expression is negative, EF is never in tension. 24. Clerk Maxwell's Method of determining the Re- sultant Thrusts at the Supports of a Framed Arch. — Let As be the change in the length s of any member of the frame under the action of a force P, and let ci be the sectional area of the member. Then ± -jT-s = As, . Ea the sign depending upon the character of the stress. Assume that all the members except the one under con- sideration are perfectly rigid, and let A/ be the alteration in the span / corresponding to As. The ratio -- is equal to a constant m, which depends only upon the geometrical form of the frame. .-. A/ = in. As = ± mP-fT- . ha Again, P may be supposed to consist of two parts, viz.,/, due to a horizontal force H between the springings, and /, due to a vertical force F applied at one springing, while the other is firmly secured to keep the frame from turning. By the principle of virtual velocities, Al A Similarly, ~ is equal to some constant n, which depends only upon the form of the frame. .'. Al= ± {m'H ^ mnV)-^. ds CLERK MAXWELL'S METHOD. 807 Hence, the total change in / for all the members is If the abutments yield, let '2,Al = ^J^, /< being some co- efficient to be determined by experiment. Then /r= ± 2 [mn V- s Ea If the abutments are immovable, 2AI is zero, and ri =■ — (C) <''4J . . (D) Fis the same as the corresponding reaction at the end of a girder of the same span and similarly loaded. The required thrust is the resultant of H and V, and the stress in each member may be computed graphically or by the method of moments. In any particular case proceed as follows : (1) Prepare tables of the values of m and n for each member. (2) Assume a cross-section for each member, based on a probable assumed value for the resultant of V and H. (3) Prepare a table of the value of Pi^-pr for each member, and form the sum 2 ('"■iJ- (4) Determine, separately, the horizontal thrust between the springings due to the loads at the different joints. Thus, let v^ , 7', be the vertical reactions at the right and left supports due to any one of these loads. Form the sum ^y-ninV-jr-L using v^ for all the members on the right of the load and 7', for all those on its left. The corresponding thrust may then be 111 I 8o8 THEORY OF STRUCTURES. found by eq. (C) or eq. (D), and the total thrust H is the sum of tlie thrusts due to all the weights taken separately. (5) Repeat the process for each combination of live and dead load so as to find the maximum stresses to which any member may be subjected. (6) If the assumed cross-sections are not suited to thes^ maximum stresses, make fresh assumptions and repeat the whole calculation. The same method may be applied to determine the result- ant tensions at the supports of a framed suspension-bridge. Note. — The formulae for a parabolic rib may be applied without material error to a rib in the form of a segment of a circle. More exact formulae may be obtained for the latter in a manner precisely similar to that described in Arts. 18-22, but the integrations will be much simplified by using polar co- ordinates, the centre of the circle being the pole. EXAMPLES. 809 EXAMPLES. 1. Assuming that an arch may be divided into elementary portions by imaginary joint planes parallel to the direction of the load upon the arch, find the limiting span of an arch with a horizontal upper surface and a parabolic soffit (latus rectum = 40 ft.;, the deptii over the crown being 6 ft. and the specific weight of the load 120 lbs. per cubic foot; the thrust at the crown is horizontal (:::; P) and 4 ft. above the soffit. 2. A masonry arch of 90 ft. span and 30 ft. rise, with a parabolic in- trados and a horizontal extrados, springs from abutments with vertical faces and 10 ft. thick, the outside faces being carried up to meet the extrados. The depth of the keystone is 3 ft. The centre of resistance at the springing is the middle of the joint, and at the crown 12 in. below the extrados. Tlie specific weight of the masonry may be taken at 150 lbs. per cubic foot. Determine (a) the resultant pressure in the vertical joint at the crown ; (p) the resultant pressure in the horizontal joint at the springing ; (c) the maximum stress in the vertical joint aligning with the inside of an abutment. 3. The intrados of an arch of 100 ft. span and 20 ft. rise is the segment of a circle. Tlic arch ring has a uniform thickncs-^ of 3 ft. and weighs 140 lbs. per cubic foot ; the superincumbent load may be taken at 480 lbs. per lineal fcjot of the ring. Determine the mutual pressures at the key and springing, their points of application being 2 ft. and i^ ft., re- spectively, from the intrados. Also find the curve of the centres of pres- sure. 4. The soffit of an arch of 30 ft. span and 10 ft. rise is a transformed catenary. The masonry rises 10 ft. over the crown, and the specific weight of the load upon the arch may be taken at 120 lbs. per cubic foot. Determine the direction and amount of the thrust at the springing. 5. A concrete arch has a clear spring of 75 ft. and a rise of 7A ft. ; the heigiit of masonry over crown = 5 ft. ; the weight of the concrete = 144 lbs. per cubic foot. Determine the transformed catenary, the amount and direction of the thrust at the springing, and the curvatures at the crown and springing. Alls. ;;/ = 23.9; thrust = 91,354 lbs. ; slope at springing = 25!° ; radius of curvature = 114.2 ft. at crown and =248.7 ft. at springing. 6. Determine the transformed catenary for an arch of 60 ft. span and 15 ft. rise, the masonry rising 6 ft. over the crown and weighing 120 lbs. per cubic foot. Also find the amount and direction of the thrust at the abutments. 8io THEORY OF STRUCTURES. 7. Determine the transformed catenary for an arch of 30 ft. span and ']\ ft. rise, the height of masonry over the crown being \\ ft. ; weight of the masonry = 125 lbs. per cubic foot. Also find the thrust at the spring- ing and the curvature at the crown and the springing. 8. In a parabolic arch of 50 ft. span and 10 ft. rise, hinged at both ends, a weigiit of i ton is concentrated at a point whose horizontal dis- tance from the crown is 10 ft. Find the total thrust along the axis of the rib on each side of the given point, allowing for a change of 60 from the mean temperature (e = .0000694). 9. A parabolic arched rib of 100 ft. span and 20 ft. rise is fixed at the springings. The uniformly distributed load upon one-half of the arch is 100 tons, and upon the other 200 tons. Find the bending moment and shearing force at 25 ft. from each end. 10. An arched rib with parabolic axis, of 100 ft. span and i2i ft. rise, is loaded with i ton at the centre and i ton at 20 ft. from the centre, measured horizontally. Determine the thrusts and shears along the rib at the latter point, and show how they will be affected by a change of 100° F. from the mean; the coefficient of linear expansion being .00125 for 180° F. 11. A parabolic arched rib hinged at the ends, of 64 ft. span and 16 ft. rise, is loaded with i ton at each of the points of division of eight equal horizontal divisions. Find the horizontal thrust on the rib, allowing for a change of 60° F. from the mean temperature. Also find the maxi- mum flange stresses, the rib being of d(nible-tee section and 12 in. deep throughout. (Coefficient of linear expansion per 1° F. = /-^ 144000.) 12. The axis of an arched rib of 50 ft. span, 10 ft. rise, and hinged at both ends is a parabola. Draw the linear arch when the rib is loaded with two weights each equal to 2 tons concentrated at two points 10 ft. from the centre of the span. If the rib is of double-tee section and 24 in. deep, find the maximum flange stresses. If the arch is loaded so as to produce a stress of 10,000 lbs. per square inch in the metal, show that the rib will deflect .029 ft., E being 25,000,000 lbs. 13. A steel parabolic arched rib of 50 ft. span and 10 ft. rise is hinged at both ends and loaded at the centre with a weight of 12 tons. Find the horizontal thrust on the rib when the temperature varies 60° F. from the mean, and also find the maximum flange stresses, the rib being of double-tee section and 12 in. deep. 14. A semicircular rib, pivoted at the crown and springings, is loaded uniformly per horizontal unit of length. Determine the position and magnitude of the maximum bending moments, and show that the hori- zontal thrust on the rib is one-fourth of the total load. 15. Draw the linear arch for a semicircular rib of uniform section EXAMPLES. 8ii under a load uniformly distributed per horizontal unit of length (<;)when hin,<,'cd at botii ends; (h) when hinged at both ends and at the centre; (<) when tixed at both ends. !6. A semi-elliptic rib (axes 2rt and 2/') is pivoted at the springings. Find tlie position and magnitude of the maximum bending moment, the load being uniformly distributed per iiorizontal unit of length. How will the result be alTected if the nb is also pivoted at the crown ? 17. Draw the equilibrium polj'gon for a parabolic arch of 100 ft. span and 20 ft. rise when loaded with weights of 3, 2, 4, and 2 tons, re- spectively, at the end of the third, sixth, eighth, and ninth division from the left support, of ten equal horizontal divisions. (Neglect the weight of the rib.) If the rib consist of a web and of two flanges 2^ ft. from centre to centre, determine the maximum flange stress. 18. Find the flange stresses at the ends of the rib, in the preceding question, and also at the points at which the weights are concentrated, when both ends are absolutely fixed. 19. A semicircular rib of 28 ft. span carries a weight of \ ton at 4 ft. (measured horizontally) from the centre. Find the tlinist and shear at the centre of the rib and at the point at which the weiglit is concen- trated. 20. The axis of an arched rib hinged at both ends, for a span of 50 ft. and a rise of 10 ft., is a parabola. Draw the equilibrium polygon when the arch is loaded with two equal weiglus of 2 tons concentrated at two points 10 ft. from the centre of the span. Also flctermine the maximum flange stress in the rib, which is a double-tee section 2 ft. deep. 21. The load upon a parabolic rib of 50 ft. span and 15 ft. rise, hinged at both ends, consists of weights of i, 2, and 3 tons at points 15, 25, and 40 ft., respectively, from one end. Find the axial thrusts and the shears at these points. Ans. Horizontal thrust = 9.6 tons. Axial thrusts : above i ton = 9.3 tons ; below I " =97 " above 3 tons = 8.3 " below 3 " = lo.i " Shears : above i ton = 3. i tons ; below I " =2.2 " above 3 tons = 5 " below 3 " = 2.6 " 22. Draw the linear arch and determine the maximum flange stresses for an arched rib of 80 ft. span, 16 ft. rise, and loaded with live weights each of 2 tons at the end of the first, second, third, fourth, and fifth division, of eight equal horizontal divisions. The rib is of double-tee 8l2 THEORY OF STRUCTURES. section and 30 in. deep. Also find the shears and the axial thrusts at the fifth point of division. 23. A wrought-iron parabolic iib of 96 ft. span and i6 ft. rise is hinged at the two abutments; it is of a double-tee section uniform throughout, and 24 in. deep from centre to centre of tlie llange?. Ueter- minc the compression at the centre, and also the position and amount of the maximum bending moment {a) when a load of 48 tons is concen- trated at the centre ; {b) when a load of 96 tons is uniformly distributed per horizontal unit of length. Determine the deflection of the rib in each case. 24. Design a parabolic arched rib ol 100 ft. span and 20 it. rise, hinged at both ends and at the middle joint ; dead load =40 tons uniformly distributed per horizontal unit of length, and live load — i ton per hori- zontal foot. 25. Show how the calculations in the preceding question are affected when both ends are absolutely fixed. 26. In the framed arch represented by the figure, the span is 120 ft., the rise 12 ft., the depth of the truss at the crown 5 ft., the fixed load at each top joint Fig. 506. 10 tons, and the moving load 10 tons. De- termine the ma.\imum stress in each member with any distribution of load. Show that, approximately, the amount of metal required for the arch: the amount required for a bowstring lattice-girder of the same span and 17 ft. deep at the centre : the amount required for a girder of the same span and 12 ft. deep :: 100 : 155 : 175. 27. The steel parabolic ribs for one of the Harlem River bridges has a clear opening of 510 ft., a rise of 90 ft., a depth of 13 ft., and are spaced 14 ft. centre to centre. The dead weight per lineal foot is estimated at 33,000 lbs. and the live load at 8000 lbs. ; a variation in temperature of 75" F. from the mean is also to be allowed for. Determine the maxi- mum bending moment (assuming /constant), and the maximum deflec- tion. E = 26,000,000 lbs. Show how to deduce the play at the hinges. 28. A cast-iron arch (see figure) whose cross-sections are rectangular Do'o' -^ and uniformly 3 in. wide, has a straiglit horizon- tal extrados, and is hinged at the centre and at the abutments. Calculate the normal intensity of stress at the top and bf)ttom edges D, E of the Fig. 507. vertical section, distant 5 ft. from the centre of the span, due to a vertical load of 20 tons concentrated at a point dis- tant 5 ft. 4 in. horizontally from B. Also find the maximum intensity of the shearing stress on the same section, and state the point at which it occurs. {AB = 21 ft. 4 in.). INDEX. Allowance for the weight of a beam, 403. Allernaiing stresses, 152. American iron columns, 53a. Anchorage, 704. Angle of repose, 237. " " torsion, 568. Angular momentum, 177, Anti-friction curve, 320. " pivots, 320. Arch, 470. Arch abutment, maximum thickness of, Arch, conditions of equilibrium of, 745. '' formulx for thickness of, 750. " linear, 743, 750, 760. Arched ribs, 740, 762. " " deflection of, 780, 802. Arched ribs, effect of change of tem- perature oil, 770, 780. Arched ribs, elementary deformation of, 751. Arched ribs, general equations of equi- librium of, 784. Arched ribs, graphical determination of stresses in, 677. Arched rib of uniform stiffness, 7S8, -aij. 795. 800, 804. Arched ribs with fixed ends, 771. " with hinged ends, 764. Arched ribs with axis in form of circu- lar arc, 769, 773. Arched ribs with parabolic axis, 760, 775. Arched ribs with semicircular axis, 765, 775- Arches, middle-third theory of, 746. Auxiliary truss, 719. Back-stays, i6, 704. Baker's formulee for Strength of pillars, 549- Balancing, I98. Beam acted upon by oblique forces, 396. Beam, transverse strength of, 340, 429. Beam, transverse vibration of loaded, 461. Beams, equilibrium of, 93. " of uniform strength, 358-365. Bearing surface, 314, 315. Belts, 321. " effect of high speed in, 325. " effective tension of, 324. " slip of, 326. " stiffness of, 327. Bending moment, 96, ri8, 434. Bending moment In plane which is not a principal plane, 354. Bending moment, relation between, and shearing stress, loS. Bevel-wheels, 335. Boilers, 586. Bollman truss, 56, 618. Bowstring truss, 61, 618. Brace, i, 25. Brakes, 323. Breaking-down point, 149. Breaking stress, 147. weights, 343, 399. Breaking weights of iron girders, 369, 370. Breaking weiKhis, tables of, 212. Brickwork, 149. Bridge, bowstring suspension, 626. " loads, 600. " trusses, 17, 52. " ' '' chords of, 625. " " depth of, 597. Bridge trusses, maximum allowable stress in, 657. Bridge trusses, stiffness of, 598. stringers of, 656. Bridges, 597. " position of platform of, 598. Buckling of pillars, 513, 515. 813 8X4 INDEX. Cable with sloping suspenders, 717. Cables, 703. " curves of, 706. " deflection of, 714. " length of arc of, 712. " parameter of, 711, " weight of, 713. Camber, 3SS, 659. v • Cantilever, 365. " curve of boom of, 634. " deflection of, 638. " depth of 637. " weight of, 632. Cast-iron, 147. Catenary, 34, 706, 750. Cement, 150. Centres of gravity, 11. Centre of resistance, I, 743. Centrifugal force, iSi. Centripetal force, 1S2. Clapeyron's theorem, 292. Coefficient of cu> ic elasticity, 255. " " e' isticity, 141, 143. " " f uidity, 162. " " lardness, 164. " " i^'eral elasticity, 144. " " rigidity, 254, 285. " " rupture, 248. " " torsional rupture, 574. " " transverse elasticity, 285. Collar-beams, 25. Columns, see Pillars, 513, 53S. flexure of, 554, 557. Compound strain, 236. Compression, 141. Conjugate stresses, 247. Continuous girders, 463. Continuous girders, advantages and dis- advantages of, 486. Continuous girders, maximum bending moment in, 465. Coulomb's laws, 568. Counterbrace, 60. Counter-efficiency, 328, Counterforts, 270. Covers of riveted joints, 665. Cranes, 13. " bent, 31. " derrick, 16. ■ ' " jib, 13. pit, 15. Crank effort, 207. Cubic elasticity, 255. " strain, 283. Dead load, 143, 600. Deflection, curve of, 434. of girde.., 384-386, 638. Deformation, 140, 251, 254. Dock walls, 270. Dynamometer, Prony's, 327. Earth foundations, 258. Earthwork, 255. '' pressure of, 257. Earthwork, Rankine's theory applied to retaining walls, 264. Efficiency of mechanisms, 335. " of riveted joints, 606. Elastic curve, 355. " moment, 96, 340. Elasticity, 140. coefficient of, '.41, 143. '• " " cubic, 255. '■ " " lateral, 144. " 1. .< transverse, 285. " limit of, 145. Ellipse of stress, 241. Ellipsoid of stress, 281. Empirical rules for wind-pressure, 663. Encasire girders, 458. Energy, 207. " curves of, 207. " fluctuation of, 207. " kinetic, 167, 169, 170. " potential, 167. Envelope of moments, 121. Equalization of stress, 349. Equalizer, 629. Equilibrated polygon, 740, Equilibrium of bea.ms, 428. Equilibrium of beams, general equations of, 428. Equilibrium of flanged girders, 366. Euler's theory of the strength of pillars, 537- Examples, 69-92, 132-139, 21(1-234, 294-205, 337-339. 407-427. 4QO-5I2, '56?-567. 5S>o-5S5, 594-59^. (>89-702. 734-739. S09-812. Expansion of solids, 215. Extension of prismatic bar, 289. Extrados, 740. Eyebars, 661. " steel, 665. Factor of safety, 150. Fatigue, 152. Fink truss, 54. Flanged girders, 365. " " equilibrium of, 36b. " " stiffness of, 384. Flanges, 365, 597. " curved, 366. " horizontal. 366. Flexure of columns, see Pillars, 554, 557. Flow of solids, 162. INDEX. 815 Fluctuation of stress, 151. Fluid pressure, 162. Force polygon, 3, 7, 119. Foundations, earth, 258. " limiting depth of, 258. " of walls, 270. Fracture, 141. Framed arch, stresses in, 804. Framed arch. Clerk Maxwell's method of determining stresses in a, 806. Frames, i, 2. " incomplete, 27, 61. Friction, 300. angle of, 237. " coefficient of, 300, 313. " journal, 310. " rolling, 310. Funicular curve, 10. polygon, 3, 7, 117, 119. Gins, 17. Girder of uniform strength, 381. Gordon's formulfe for pillars, 522. Hinged girders, 127-131. Hodgkinson's formulae for the strength of pillars, 513, 517-521. Hooke's law, 142. Howe truss, 58, 6ir. Impact, 184. Impulse, 176. Incomplete frames, 27. Inertia, 198. " moment of, 12, 342, " pressure due to, 200. Inflection, point of, 453-463. Internal stress, 235. Isotropic bodies, 283. Joint of rupture, 747, Keystone, 741. Lateral bracing, 654. Lattice girder, 600. Launhardt's formula, 153. Lenticular truss, 626. Limit of elasticity, 145. Line of loads. 5. " " resistance, 273-276, 741, 750. " " rupture, 265. Linear arch, 743, 753-760. Loads, live, iii, 115, 119, 600,639,641, 730. Loads, stationary (dead), 118, 600, Long pillars, 535. Mansard roof, 6. Masonry, 149. Mechanical advantage, 294. Middle third theory of arches, 746. Modulus of (ilasticity, 141. " " rupture, 348. " " transverse elasticity, 254. Moment of forces, n6 " " inertia, 12, 342. " " " examples of, 371-81. Moment of inertia, variable section of, 455. Moment of inflexibility, 96. " " resistance, 96, Momentum, 176. Mortar, 150. Neutral axis, 340. " of a loaded beam, 435-454. " " surface, 340. Oblique resistance, 169. Oscillatory motion, 190, 195. Panel points, 52. Panels, 54. Piers, 65. Pillars, 513. " Euler's formulae for, 527. " failure of, 515. " flexure of, 515. " formulae for American, 527-532. " Gordon's formulae for, 522. Pillars, Hodgkinson's formulae for, 517- 521. Pillars, Rankine's formula for, 526. Pillars with stress uniformly distributed. 516. Pillars with uniformly varying stress, 517- Pins, 661. Piston velocity, curves of, 205. Pivots, 316. " conical, 319 " cylindrica. , 316. Schieie's (anti-friction), 320. Plasticity, 141. Poisson's ratio, 142. Pole, 7. Polygon of forces, 3, 7. " " pressure, 743. Pratt truss, 60. Primitive strength, 153, Principals, 33, 34. Prony's dynamometer, 327. Proof strain, 171. " stress, 171. Purchase 304. Purlins, 33, 34. Radius of gyration, 174, 528-531. Rale of twist, 289. 8i6 INDEX. Redundant bars, 48. Reservoir walls, 271. Resilience, 171, Retaining walls, 260. Retaining walls, conditions of equiib- rium of, 260. Retaining walls, Rankine's theory ap- plied 10, 264. Rivet connection bef.veen flange and web, 660. Riveted joints, 668 " " covers of, 675. •« " design of, 67S. " " efficiency oi', 679. " " failure of, 670. " " stresses in, 670. " " theory of, 671-675. Riveting, 666. RivetE, 666. " dimensions of, 667. Rocker-link, 629. Rollers, 35, 639. Roof trusses, 17. " " distributionof loadson, 39. " types of, 33. ' " weights of, 37. Ropes, 321. Saddles, 704. S-^hiele's pivots, 320. Screws, 306. " endless, 309. Sections, method of, 62 Set, 145. Shafting, distance between bearings of, 575- Shafting, efficiency of, 577. " internal stress in, 237. " stiffness of, 573. " torsional strength of, 288-291. Shear, 141. " maximum, 121, 237. Shearing force, 95. " " examples of, 97, 108. Shearing force and bending moment, relation between, 108. Shearing stress, 19S. " " distribution of, 391. Shear-legs, 17. Similar girders, 401-404. Skew-backs, 34, 740. Soffit, 740. Spandril, 740. Specific weight, 143. Spherical shells, 591. Spritigings, 740. Springs, 355, 45M5S. " simple rectangular, 456. Springs, spiral, 477. Springs of constant depth, but triangu- lar in plan, 457. Springs of constant width, but parabolic in elevation, 457. Statical breaking strength, 153. Steel, 148. Stiffening truss, 719, " hinged at centre, 725. Stiffness, 190, 387, 389. Strain, 140, 251. Straining cill, 18. Stress, 141, 251 " and strain, relation betw«en, 281. " general equations of, 277. " principal, 241. " " planes of, 237. Stresses, conjugate, 248, 250. Stress-strain curves, 147-149. " line, 144. Strut, r. St. Venant's torsion results, 572. Surface loading, 350. Suspenders, 706. Suspension-bridges, 703. " loads on, 730. Suspension-bridges, modifications of, 731- Suspension-bridges, pressure upon piers of, 718. Swing-bridges, 470-472. Tables of breaking weights and coeffi- cients of bending strength of limber, 212, 213. Table of coefficients of axle friction, 336. Table of coefficients in Gordon's for- mula, 524. Table of coefficients in Rankine's mod- ification of Gordon's formula, 526. Tables of diagonal and chord stresses, 644-650. Tables of efficiencies, 587. " " elliptic integrals, 562. " " expansion of solids, 215. " " eyebar dimensions, 665. " " factors of safety, 214. Tables of loads for highway bridges, 687. Tables of strengths, elasticities, and weights of iron and steel, 210. Table of strengths, elasticities, and weights of various alloys, 211, Tables of weights of modern bridges, 682-687. Table of weights and crushing strength of rocks, 214 Table of weights of roof coverings, 67. INDEX. 817 Table of weights of roof frames, 67. Tension, 141. Theorem of hree moments, 463-470. Thick hollou cylinder, 588. Timber, 149, Torsion, 141, 568. St. Venani's results, 572. Torsional coefficient of elasticity, 145. " resilience, 574. Torsional strength of shafts, 2S8, 280, 571, 572. Transverse strength, 141. Transverse vibration of a loaded beam 461. ' Trellis girder, 600. Tresca's theory of flow of solids, 162. Tripods, 17. Truss, 2. " composite, 31. king-post, 21. " queen-post, 25. " roof, 32. " triangular, 19, Trussed beams, 53, Twist, 141. Unwin's formula, 159. Values of /6*. 174, 528-531. Vibration strength, 153. Voussoir, 741, Warren truss, 57. Web thickness, 382. Wedge, 303 Weights of roof coverings, 67. " " frames, 67. Weyrauch's formula, 153. Weyrauch's theory of buckling of pillars. 550. Wheel and axle, 329. Whipple truss, 618. Wind pressure, 38, 67, 629, 651, 653. Wmd-pressure, American specifications of, 652. Wind-Pressure Commission rules, 653. Wind-pressure, empirical regulations, 653- Wohler's law, 150. Work, 167. " effective, 178. " external, 168. " internal, 168. " useful, 178. " waste, 178. Work done in bending a beam, 460. Work done in small deformation of a body, 292. Work of journal friction, 314. Working load, 150. strength, 150. " stress, 150, Wrought-iron, 148. Yield-point, 149.