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ON THE STRENGTH OF MATERIALS ; CONTAINING VAKIOUS ORIGINAL AND USEFUL FORMULA, SPECIALLY APPLIED TO TUBULAK BRIDGES, WROUGHT IRON AND CAST IRON BEAMS, ETC. THOMAS TATE ; AUTHOR OF " THE PRINCIPLES OF THE DIFFERENTIAL AND INTEGRAL CALCLXCS, " FACTORIAL, ANALYSIS," ETC. LONDON: PRINTED FOR LONGMAN, BROWN, GREEN, AND LONGMANS, PATERNOSTER-ROW. 1850. - LONDON ; SPOTTISWOODES and SHAW, New- street- Square. INTRODUCTION. THE railway system in this country has recently given rise to a new and important principle of construction, that of tubular beams ; and the facts elicited by the experiments made on this subject have placed in the hands of the mathematician new and valuable data relative to the strength of materials. Mathematicians had long known, that in order to attain the condition of maximum strength, with a given amount of material, in a beam subjected to transverse strain, the material in the section should be so dis- tributed, that when it is about to yield by the force of compression on the upper side of the beam, it should at the same time be upon the point of yielding to that of extension on the under side. If the resistance of the material to compression were in all cases equal to the resistance which it presents to extension, then it is obvious that the form answering to maximum strength would be that in which the material would be equally accumulated towards the upper and under sides of the beam. But this is not generally the case. At the commencement of the present century, Watt and others, being aware of the fact that cast iron presents a A 2 117177 VI INTRODUCTION. much greater resistance to compression than it does to extension, constructed cast-iron beams with flanges at the under side in order to compensate for the weakness at that part. Professor Hodgkinson afterwards deter- mined with more precision the relative dimensions of these top and bottom flanges, so as to have them equally strong. On the same principle, hollow beams were found to possess an eligible form of strength. Until very recently, however, the resistance of a combination of wrought-iron plates to transverse strain had not claimed the attention of experimentalists. When Mr. Stephenson first suggested the bold and original idea of a tube, composed of sheet iron, through which railway trains might pass, the practicability of the scheme had to be tested by experiment, and the true form and distribution of the material had to be de- termined so as to secure a maximum strength with a given section. By a series of inductive experiments, Mr. Fairbairn showed, that such a tube should have the form of a rectangular beam, having the plates at the upper side arranged in the form of cells, as in Fig. 10., to counteract the tendency which plates have to crumple when subjected to severe strain, and that the areas of the top and bottom parts should be in the ratio of about 12 to 11. His experiments, as well as those subse- quently conducted by Mr. Hodgkinson, also lead to the conclusion, that the plates composing the beam should not be less than about one-half inch in thickness. In estimating the transverse strength of beams where the material is accumulated in the upper arid under INTRODUCTION. Vll sides, the effect of the rib or plates, as the case may be, connecting these parts may be neglected without in- curring any sensible or practical amount of error. On this principle, Mr. Hodgkinson found by experiment, that the strength of cast-iron beams with double flanges varied nearly as the product of the sectional area of the bottom flange multiplied by the depth. Although this formula can scarcely be regarded as a mathematical deduction, yet from the wide range of experiments from which he derived the constant in the formula, it may be taken as sufficiently exact for all beams not widely different in form from those upon which he based his calculations. As applied to tubular beams, this for- mula is much less empirical. Here the depth of the beam is large as compared with the thickness of the plates or the depth of the bottom part ; and this cir- cumstance gives to the formula in this case more of the character and decided impress of a mathematical de- duction. (See Art. 59.) Notwithstanding, it must be conceded, that the strict mathematical principle upon which calculations of this kind should be made, is that of the properties of similar beams, as explained in Art. 34. of this work. These properties, in certain restricted cases, were investigated by me in Mr. Fairbairn's work on tubular bridges ; but I believe that they are now given, for the first time, in the following Treatise, in their most general and comprehensive form. Various new formula are also given throughout the work rela- tive to the strength of different forms of beams, &c. Isolated, as far as possible, from the influence of viii INTRODUCTION. private feelings and sympathies, it is principles not men which should form the subject of all philosophical dis- cussions. Accordingly, wherever I have presumed to differ from others in opinion, I have done so from a love of truth rather than controversy ; and with respect to the three celebrated men associated with the de- velopment of one of the boldest and most successful enterprises which has taken place in the history of modern engineering, I have only to state, that I en- tertain, in common with every one devoted to the study of practical science, a deep sense of the obligations they have conferred on society by their great perseverance and talent. T. TATE. Twickenham, July, 1850. TABLE OF CONTENTS. Page Preliminary Observations and Formulae - - - 1 Neutral axis of a Beam ..... -4 General Theorem relative to the Neutral Axis - - * ? Conditions of Rupture - - - - - -10 General Formulae of the Moment of Inertia - - - - 17 Centres of Compression and Extension - - - 18 To change the Axis of Movements - - 19 Deflection of Beams - * - -21 GENERAL FORMULAE RELATIVE TO SIMILAR BEAMS ' - Neutral Axis in similar Beams - - ; -: - - 24 Moment of Inertia of similar Beams - - - - - 28 Transverse Strength of similar Beams - - - - - 30 Deflections of similar Beams - - _ _ . - 37 General Formulae relative to Beams only in certain respects similar - 38 SRENGTH, ETC. or VARIOUS FORMS OF BEAMS - - - 41 Hollow rectangular Beams - - - - - - 42 Comparison of Strength of solid and hollow Beams - - - 43 Rectangular Cells, maximum Strength - - - - 44 To find the Moment of Inertia of Angle-Iron - - 46 Mr. Fairbairn's Model Tubular Beam - - - - - 47 Angle-Iron taken into the Calculation - - - - 49 Rivet-Holes, do. - - - - - -51 1. Approximate formula of Strength of Tubular Beams - - 52 Limits of Error in do. - - - - - 56 2. Approximate Formulae "-'<" " " " -57 Strength, &c. of Beams with Flanges - - - - 60 Mr. Hodgkinson's Experiments on cast-iron Beams with double Flanges 65 New Formula for cast-iron Beams - - - - - 66 Strongest Form of cast-iron Beams - - - - - 69 STRENGTH, ETC. OF CYLINDRICAL BEAMS - - - - 70 Hollow cylindrical Beams, &c. - - - - - 71 Comparative Strengths of cylindrical and square Beams - - 73 Comparative Strengths of circular and square Cells - - 74 a X CONTENTS. Page Observations relative to the best Form of the Cells in a tubular Beam - - 75 Why the cellular Structure exhibits such Strength - - 76 Moment Inertia of a Semicircle, &c. &c. - >< - - 77 Strength of a mixed Form of Beam - - 81 STRENGTH, ETC. or ELLIPTICAL BEAMS, ETC. - - 82 Hollow elliptical Beams - - - - - 83 Strength of a mixed Form of Beam - - - - - 85 Moments of Inertia,, c. of triangular and trapezoidal Surfaces - 86 STRENGTH, ETC. OF PARABOLIC BEAMS - - - - 91 Doctrine of similar Beams applied to a few particular Forms - - 94 ON THE STRENGTH OF MATERIALS, ON THE STRENGTH OF MATERIAL. PRELIMINARY OBSERVATIONS AND FORMULA. 1. THE particles of a rigid body are connected together by the force of cohesion, and this force must be completely overcome before the body can be broken or ruptured. A force acting upon a rigid body, first tends to change the relative position of its particles, and finally to separate them from each other. The property which bodies possess of taking a new form under the action of a force, and of resuming their original form when the force is withdrawn, is called elasticity. All bodies possess elasticity in a greater or less degree. 2. When the fibres of a beam are pressed together by a force acting longitudinally, the resistance which the material presents is called the resistance of compression. On the contrary, when the fibres of a beam are stretched by a force tending to pull them asunder, the resistance which the material presents is called the force of tension, or the resistance of extension. When the resistance of compression is equal to that of ex- tension, the material in this respect is said to be perfectly elastic, which is nearly the case in bars of wrought iron. But in most kinds of material these forces are different : thus in cast iron the compressive resistance is about 6^- times that of the tensile resistance. B 2 PRELIMINARY FORMULA. 3. A beam undergoes transverse strain or rupture when it is broken across, or transversely, as may be the case with joists and beams supporting mason work. Let us suppose that a beam rests upon two supports, as \&fig. 1. p. 5., and that it has a load placed upon its middle, then this weight causes the beam to bend, and thereby gives rise to a complex action in the fibres of the material: the fibres in the top of the beam become compressed, while those at the bottom part become extended ; and there is a certain part in the beam which is neither compressed nor extended; this portion is called the plane of the neutral axis. If a beam of timber be cut with a very fine saw about one half through, that is as far as its neutral axis, then it will be found that the strength of the beam is scarcely at all impaired. The position of this axis depends upon the form of the section of the beam, as well as upon the relative resistances of the material to compression and extension. If the transverse section of the beam is rectangular, and the material perfectly elastic, the neutral axis will ob- viously lie in the central axis of the beam. In general this neutral axis will lie towards the parts where the material presents the greatest resistance, so that the forces on each side of the axis may be duly balanced. Barlow was the first ex- perimentalist, who precisely defined the position of the neutral axis in beams undergoing transverse strain. Modulus of Elasticity. 4. When the elongation or compression of the fibres of a beam does not exceed a certain limit, they tend to return to the position which they at first had with a force proportional to the space through which they have been extended or compressed) as the case may be. But if this elongation or compression be carried beyond a certain point, called the limit of elasticity, then the fibres of the beam remain inactively in their new position. ELONGATION AND COMPRESSION. 3 The law of perfect elasticity, that the amount of extension or compression is proportional to the force, holds strictly in relation to gases, as expressed in Marriotte's law; but there seems ground for believing that it is only approximately true in solid bodies. However, all our theoretical calculations on the strength of materials are based upon the assumption that the law is applicable to the rigid bodies employed in con- struction. The modulus of elasticity is that force E which is necessary to elongate a uniform bar, one square inch section, to double its length (supposing such a thing possible) or to compress it to one-half its length. Elongation and Compression. 5. Let L be the length of a bar 1 square inch of section, / its elongation or compression with a force of p Ibs., then we have from the law of perfect elasticity : Force to produce an elongation of L in. = E, . _ E ,j )) 1 in. - y LI E/ that is, Now if the bar contains A square inches in the section, then the force P necessary to extend this bar I inches must ob- viously be A times p : B 2 NEUTRAL AXIS OF BEAMS. or if the value of I be required, we have In these expressions L and I are in the same linear unit. Example. If, according to Tredgold, the modulus of elas- ticity of wrought iron be 24900000 Ibs., what force will be required to extend a bar 10 ft. long and ^ in. section, through the space of ^ in. ? Here A = |, E = 24900000 Ibs., L = 10 x 12, / = ; hence by eq. (2.), 4 x 24900000 x A P 10x12 -- = 17 lbs< nea y< THE NEUTKAL AXIS OF A BEAM. 6. From the property of the neutral axis, explained in Art. 3., it follows that the sum of the adhesive forces of the fibres resisting rupture on the upper side of this axis must be equal to the sum of the adhesive forces of the fibres on the under side of it. Moreover, as the strain upon the fibres is in proportion to their distances from the plane of the neutral axis, therefore, by Art. 4., the resistances of these fibres to extension or compression, as the case may be, will also be in the same proportion* NEUTRAL AXIS OF BEAMS. 5 7. Let AB represent a beam (whose section is made up of rectangles) resting upon the supports A and B, and undergoing transverse strain from the weight w placed upon it. If n be the neutral axis of the section of rupture ad, then a J d the fibres on the upper side nad will undergo compression, while the fibres on the lower side will undergo exten- sion. Let ov be a line of fibres parallel to ad, or, what is the same thing, parallel to the neutral axis of the section; then the resistance of the fibres in ov will be to the resistance of the fibres in ad as no is to na } that is, as their distances from the neutral axis of the section. Let h, AJ r= the respective distances of the neutral axis n from the top and bottom of the beam ; b, b l = the respective breadths of the top and bottom parts ; S, s x = the compressive and tensile forces respectively exerted by a square inch of the material at the distances h and h l from the neutral axis ; s, s l = the compressive and tensile forces respectively exerted by a square inch of the material at the distance of unity from the neutral axis ; x = n o, a variable distance from the neutral axis. Compressive force per sq. in. at a unit from n = s ; a: units =sx. Area element of surface at v = b A x ; but it has been shown that sx is the resistance of a unit of surface at v, B 3 6 NEUTRAL AXIS OF BEAMS. .'. Compressive force of this element of surface = sx x b A x = sbx Ax ; .-. Sum of all the compressive forces between ad and ov h sb^ xAx, y^A = sbl xdx 9 %J x Similarly we have Sum of aU the tensile forces == ^ (V~*i*> Hence by Art. 6., (#-*)-!& (V-'tf (4.) Neutral axis of a solid rectangular beam. 8. In eq. (4.), let x = 0, ^ = 0, and b = ^ ; then, but, from the principle of elasticity, Art. 4., 8 therefore, by substitution in eq. (5.), 8^ = 8^! . . . . * . (6.), which expresses the relation of the distances of the neutral axis from the upper and under sides of a rectangular beam. If the elasticity of the material is perfect, then s = s l - } and, NEUTRAL AXIS OF BEAMS. 7 therefore, by eq. (5.), h=h l9 that is, in a beam of this kind, the neutral axis lies in the centre of the section. 9. The ratio of the moments of the material undergoing compres- sion and extension about the neutral axis is equal to the ratio of the forces of extension and compression. Thus if A be put for the area of the section above the neutral axis, G the distance of its centre of gravity from this axis, and A V G p the corresponding symbols for the parts below the neutral axis, then A . G s. Let SQ^SjQ^^p &c. represent the section of rupture, KG the neutral axis, Sm = Qw = h, gm = kn = e, SQ = gk = b, = Q l n = h v g l m=k l n = e l , B l Q l = g l k l = ^ ; also let a = area s k, g = the distance of the centre of gravity of sk from KG, A = the section of the whole material above the neu- | tral axis, G the distance of the centre of gravity of A from the neutral axis, and ct v g l9 A p and GJ the corresponding dimensions of the parts below the neutral axis : Then put- s, o, ting e for x, eq. (4.) will express the relation from which the position of the neutral axis is determined. By an easy reduc- tion this expression becomes but =g, b (h e)-= a, and so on; Now, if we regard s and ^k ly as elementary portions of the section of the beam, the sum of all the elementary moments B 4 8 NEUTRAL AXIS OF BEAMS. sag will be equal to the sum of all the elementary moments fc&^| . . (7.), but 2U<7 = A . G, and ^oc l o l = AJ . G p .'. 5 . A . G=S l . A l . G l . . . . (8.), ..... (9.) Q Q 10. Substituting for s, and-! for s v eq. (9.) becomes h where A and A p in this formula, are the distances of the upper and under edges of the section from the neutral line. 11. Let ,/and/j be put for the respective resistances of the material at the centres of gravity of the top and bottom parts of the section, then /= G . s, and/! = G 1 . Sj ; therefore, by eq. (8.), we have /A=/lA, ...... (11.) 12. When the material in a beam is perfectly elastic the neutral axis passes through the centre of gravity of the section of rupture. For in this case s = s v and then eq. (8.) becomes A. G = A I . G! ..... (12.); that is, taking the neutral axis as the axis of moments, the moment of the section on the upper side of it is equal to the moment of the section on the under side ; therefore, the neutral NEUTRAL AXIS OF BEAMS. 9 axis must pass through the centre of gravity of the whole section. To find the neutral axis when the elasticity of the material is perfect, that is, when the neutral axis passes through the centre of gravity of the section. Assume NO (see fig. 2. p. 7.) as any convenient axis of moments (as, for example, the centre of the vertical depth of the beam), and let KG be the neutral axis passing through the centre of gravity of the section. Let K = the area of the whole section, a, e^ the areas of the portions sk and SjAj respectively, and g, g l = the distances of their respective centres of gravity from NO. x the distance of NO from KG. The moment of s k = a.g, and Sj^ = 1 ^ 1 , .*. The sum of the moments of the portions above NO=^ay, sum of the moments below N o = 2$ a^. But, because KG passes through the centre of gravity of the section, the moment of the whole section is equal to K#. . . . (130 13. If , and G + Gr v be the same in two beams, then the A i neutral axis will have the same position, in the two beams, with respect to the centres of gravity of the upper and under parts of the sections. 10 CONDITIONS OF RUPTURE. For by eq. (12.), G, A * = = a constant, G A! G + G, = a constant ; but G + Gj = a constant. Therefore G and G x are constants, that is, they have the same values for both beams. 14. If be a constant ratio in two beams, the ratio of the A i resistances at the centres of gravity of the top and bottom parts will also be constant. For by eq. (11.) A A but by assumption = 9 CONDITIONS OF KUPTUEE. 15. When rupture is about to take place the beam turns upon the neutral axis n of the section of rupture, as a fulcrum (see^zy. L p. 5.); there are, therefore, two forces to be con- sidered, whose moments are in equilibrium with each other, viz., the weight w tending to rupture the beam, and the resistances of CONDITIONS OF RUPTURE. 11 the material on the upper and under sides of the neutral axis in the section of rupture. 16. The weight w (see fig. 1.) placed in the middle of the \V beam, will produce a pressure of -^ upon each of the supports A and B ; now, as the beam turns round on n as a fulcrum, this pressure acting at B will have a leverage of half the distance between the supports, or ; hence the moment of the weight w tending to rupture the beam will be the pressure acting at w 1 w Z B x half the distance between the supports, or x ^ =., which we shall represent by the symbol M. This is called the moment of rupture. 17. It has been explained, Art. 6., that the fibres in the section of rupture present different degrees of resistance as they are more or less distant from the neutral axis, and as the beam turns upon this axis the moment of the resistance of any fibre, or its efficacy to prevent rupture, will be its resistance multiplied by its distance from the neutral axis. The sum of all these moments of resistance of the fibres, above as well as below the neutral axis, will be equal to the moment of rupture M. Taking the figures, notation, &c. of Arts. 7. and 9., we have Compressive force of an element of surface = sbx&x; but this force acts with the leverage x from the neutral axis KG (see^. 2.), . . Moment of this element sbx^^Xy Sum of all the moments of the forces of compression between gk and SQ, or moment of the rectangle 12 CONDITIONS OP RUPTURE. = -(#-*). .,, > . . (14.) Similarly we have Moment rectangle s l Q l k^ =. \^ (hf e*\ Let KJ = area rectangle mSQn, K 2 =:area mgkn, T> 1 = sm = Sj m Q l n, d z ^=(/ l m=.Jt l n. Then the above expressions become Moment rectangle SQ kg -(K^T)^ K 2 D 2 2 ). . . (15.) Moment rectangle & l Q l k l g l =-( ^ d^ k 2 d). . . (16.) Now, if we regard s& and SjAj (Jig. 2.) as elementary por- tions of the section, the sum of all the moments on the upper side of the neutral axis added to those on the under side will be equal to the sum of the moments of the resistances to rup- ture, and therefore equal to the moment of w tending to produce rupture. . * . Sum of the moments of resistance of the material above KG, Or m = ^(K 1 D 1 2 -K 2 D 2 2 ) (17.) = 51, where I is put for ^ (K^^ K 2 D 2 2 ), or ^b (h? e 3 ), which is called THE MOMENT OF INERTIA of the section on the upper side of the neutral axis. Hence from eqs. (14.) and (15.), CONDITIONS OF RUPTURE. 13 The moment of inertia of the rectangle SQ% about KG as an axis = I (A - e 3 ) or * (K^\ - K 2 D 2 2 ). If e = 0, or s^ = CONDITIONS OF RUPTURE. 15 or gS^K . . . (26.); where K is put for the area of the whole section. Hence the moment of rupture in a rectangular beam varies as the area of the section x the resistance per sq. in. of the material at one edge of the beam x the distance of that edge from the neutral axis. When the beam is loaded in the middle, and supported at the wl extremities, M = , therefore eq. (26.) becomes wl 1 1 = ^SDiK, or gSiDiK; Now SiK is the direct tensile strength of the whole section of the beam, therefore the transverse strength of a rectangular beam, loaded in the middle and supported at the extremities, varies as the direct tensile strength, multiplied by the distance of the neutral axis from the under edge, divided by the distance between the supports. 20. If the elasticity of the material be perfect, then (Art. 3.), D! = J D, where r> is put for the whole depth of the beam ; and in this case we have from eq. (26.) M = |S. D.K ....... (28.), and from eq. (27.), w = 2 -l ....... (29 .) 16 CONDITIONS OF RUPTURE. If the material be incompressible, according to the hypothesis of Galileo and Leibnitz, then the neutral axis will lie at the upper edge of the beam, and therefore d l = D, hence eq. (26.) becomes M=^Sj . D . K (30.) In like manner, when the material does not admit of ex- tension, we have M = ^S . D . K (31.) The results in eqs. (28.), (30.), and (31.), have been derived on three distinct hypotheses ; but in all these expressions it will be observed that the moment of rupture varies as the area of the section multiplied by the depth. Mr. Barlow confirmed this law by experiment, and showed that the quantity s, called the modulus of rupture, is a constant for beams of the same material. This constant, or modulus of rupture, has been determined for all the kinds of material employed in construction (see Moseley's Engineering, p. 622.). It has been customary for practical men to assume this law as applicable to beams of all forms in the section. Mr. Hodgkinson showed, by a series of able experi- ments, that the law was approximately true in reference to cast iron beams with double flanges. Mr. Fairbairn, by an induc- tion of these facts, assumed that the same law would be appli- cable to the strength of tubular beams ; and he accordingly calculated the proportions of the Conway Tube on this as- sumption (see Mr. Fairbairn's letter to Mr. Stephenson, p. 66. of Mr. F.'s work). Under certain restrictions, the author of this work confirmed Mr. Fairbairn's views by mathematical analysis. A more exact and comprehensive investigation of this subject will be hereafter given in this work. MOMENT OF INERTIA. 17 GENERAL FORMULA OF THE MOMENT OF INERTIA. 21. It has been shown in Art. 17., that the moment of inertia of the sum of the rectangular spaces SQ% (see^y. 2.) is Now when e = o } this expression becomes the moment of inertia of the sum of all the rectangles SQnm ; that is, Let RNDB be the section of a beam, undergoing transverse strain; NOB the axis passing through the centre of gra- vity of the section ; s Q n m a small rect- B angular element. Put y = $m,x = om, and Ax-=mn } then eq. (32.) be- comes (32.) (33.), where the limits of integration remain to be assigned. Suppose the limits to be taken between x = 0, and x, then the moment of inertia of the space o R s m will be where y may be determined in terms of x from the equation to the curve RSN. This expression is equivalent to the following form of double integration, C 18 CENTRES OF COMPRESSION AND EXTENSION. r * where dx I y^dy is the moment of inertia of the element SQ?zwz, % " and the whole expression is the sum of the moments of these elements taken between OR and s?w, that is, the moment of the space ORSW. CENTRES OF COMPRESSION AND EXTENSION. 22. DEFINITION. The centre of compression is that point in the surface undergoing compression where the whole material may be collected without altering its moment of resistance to rupture. And similarly with respect to the centre of ex- tension. Let q, (^ = the distances of the centres of compression and extension respectively from the neutral axis ; m, m l the moments of resistance to rupture of the surfaces of compression and extension ; A, a = the surfaces of compression and extension, and so on to the notation before employed. By the definition m = SA<^ and m l = Hence, in beams of the same material and sectional area, the strength varies as the distances of the centres of compression and extension from the neutral axis. Now s = qs y m = SAg = but m = si (35.) TO CHANGE THE AXIS OF MOMENTS. 19 and similarly, (36.) Hence it appears from the property of the moment of inertia given in works on Mechanics, that I is the moment of inertia of the surface of compression, and that q is the distance of its centre of gyration from the neutral axis; and so on with respect to the centre of extension. If s = s ly then, putting Q = the distance of gyration of the whole section, K, of the beam, we have M = SKQ ..... (37.) and KQ 2 = i + T! = I O OTI X . . . . (38.), as the case may be. TO CHANGE THE AXIS OF MOMENTS. 23. Let (fig* 2.) KG be the neutral axis of the section passing through the centre of gravity, and NO any other axis parallel to KG at the distance x from it. Let I, Tj = the moments of inertia of the sections above and below NO ; 1^=1 +T 15 the moment of inertia of the whole section referred to the axis NO. I = the moment of inertia of the whole section referred to KG. The remaining notation being the same as in the preceding investigations. By eq. (20.) Art. 17., we have c 2 20 TO CHANGE THE AXIS OF MOMENTS. but, by reduction, observing that a=b (h e), and #= - we have | \(h + ^f-(e 4 *)} = | (*-) + = 1 + 2xA . By substituting -x, for a; &c., we have but i + ii = I j A + AJ = K, and by Art. 12., AG AjGj = 0, .-. I X = I O +?K: . . . . (39.) .... (40.); that is, the moment of inertia of the section taken with respect to the axis passing through the centre of gravity is equal to moment of inertia taken with respect to any other parallel axis, minus the square of the distance between these axes multiplied by the area of the whole section. 24. Now if i x be determined with respect to any convenient axis NO (seefy. 2.) at the distance x from the neutral axis KG, then the moment of inertia I about this neutral axis is given in eq. (40.). Therefore in cq. (24.) the value I is given in eq. DEFLECTION OF BEAMS. 21 (40.) ; and, moreover, if d be put for the distance of the axis NO from the upper edge of the beam, we have h=d+x. It is further worthy of observation that the value of x is determined in eq. (13.), Art. 12. Hence, w= 4 SI Q hi 4s (i x l(dx) . . . (41.) DEFLECTION OF BEAMS. 25. Let the annexed figure represent a beam fixed at one extremity, and bent by a weight w suspended from the extremity L; NK, a small portion of the neutral axis, and o its centre of curvature. Let C = NK = A, the length of a small portion of the neutral axis LNK, ! = B, the extension of the fibre Aa* Fig. 4. p=,O=ON, the radius of curvature of the neutral line NK, a = K A = N a, the distance of the upper side of the beam from the neutral axis, E=the modulus of elasticity. And so on to other notation as employed in the preceding investi- gations. c 3 22 DEFLECTION OF BEAMS. By eq. (1.) Art. 4., Force to produce an elongation c l} or p = - ; But the force elongating the fibres at A, or p = sa ; S CL i By the similar triangles ONK and Ba : N::NK : OK, that is, d : a : : c : p, c p substituting in eq. (42.) From eq. (23.), Art. 18., M=SI (42.) (43.), (44.) .'. M/>=sEI . . . Let HNL represent the curve of de- flection, then taking H as the origin of coordinates, Put x = H v, y = VN, 7 = HR the length L of the beam, and S = RL the greatest deflection. The moment of w considered in rela- tion to the point N of the beam is, (45.) y ' 5 ' DEFLECTION OF BEAMS. 23 MzrW X RV =w(/-*) therefore by eq. (45.) But - = - dx ] . See the Author's Calculus, p. 215. Now - is equal to the tangent of the angle which a line Ci X touching the curve at N makes with HR; but as the deflection of beams is always small, this angle must also be small, and dy* consequently -~-^ may be neglected. Comparing this with eq. (46.), we have Successively integrating between the limits x and 0, which is the equation of the curve of deflection.* * For a comprehensive view of the subject of Deflections, see Moseley's " Mechanics of Engineering." c 4 24 SIMILAR BEAMS. Now when y = RL = S, # = 7, w wZ 3 /A0 . - i; - = ^ w: which is the greatest deflection of a beam loaded at one ex- tremity and fixed at the other. When a beam Z between the supports is loaded in the middle (see fig. 2.) with a weight w, the deflection takes place at this w Z point as if a force -^ were applied at the distance ^ from it : hence the greatest deflection in this case will be found by w Z substituting -^ for w, and - for in eq. (48), that is . 48EI (49.) v } GENERAL FORMULA RELATIVE TO SIMILAR BEAMS. NEUTRAL AXIS IN SIMILAR BEAMS. 26. The neutral axis in similar beams divides the vertical axis proportionally. First, suppose the elasticity of the material to be perfect. Assume NO (Jig. 2. p. 7.) to pass through the centre of the vertical axis of the beam, and KG the neutral axis passing through the centre of gravity of the section. NEUTRAL AXIS IN SIMILAR BEAMS. 25 Let r = the linear ratio of the parts of two beams in all respects similar in section, ~x, a, &c. = the elements of the section of one beam as in Art. 12., x f i of, &c. = the corresponding elements in the other beam. By eq. (13.), Art, 12, K' but K' = r*K, a' = r^a, g' = rg, and so on, r 2 K 27. Second, suppose the resistance of the material to com- pression and extension to be unequal. In this case let KG (see fig. 2.) be the neutral axis, then adopting the notation of Art. 9., we have the following relations : A . G A 9 Supposing the ratios and to be the same for the two beams, then let us inquire what are the geometrical relations requisite for producing these conditions. 26 SIMILAR BEAMS. S S ' A A' By the assumption =.,, and = -, therefore by the two preceding equalities, .'. Sr^te*. .' (50.), G' G f G\ where G is put for G + GJ the distance between the centres of gravity of the surfaces of compression and extension, and si- milarly with respect to G'. Let h) A! = the distances of the upper and under edges of the beam from the neutral axis ; d = A + hi the whole depth of the beam. 5 Si fii Si h s """"ir^gi * hi ~h Similarly : 7 = S ' Vi h h r d h ht ,. N The formulae (50.) and (51.) express the geometrical relations involved in the assumptions. But besides these two inde- A A' pendent relations we have also = r Now it is obvious that A! A ' these three geometrical relations will be fulfilled when the NEUTRAL AXIS IN SIMILAR BEAMS. 27 sections of the beams are similar; and hence eq. (51.) shows that the depths of the beams are divided proportionally by the neutral axis. At the same time it is interesting to observe, that the conditions may not be restricted to exact similarity of form in the sections of the beams. 28. By a similar process of reasoning the first case may be established ; for we have by eq. (12.), AG , A'G' = 1, and . 7- = 1, AG AxG A A' A" = A T **! ^*- 1 2-21 G! "" G'x G G - = - G x G' which is the same relation as that given in eq. (50.). Since 5 or ^=7 = ^ /6 ^ simUarly which is the same relation as that given in eq. (51.). 28 SIMILAR BEAMS. MOMENTS OF INERTIA OF SIMILAR SURFACES. 29. The moments of inertia of two similar surfaces, about any axis which divides their vertical depth proportionally, are to each other as the fourth powers of their linear dimensions* Let NO (sQQjfig. 2. p. 7.) be assumed as any convenient axis of moments, and let KG be the neutral axis of the section passing through the centre of gravity. Suppose NO to divide the ver- tical depth of the two beams proportionally; then adopting the notation of the preceding formulae, we have by eq. (20.), Art. 17., where K 15 D! &c., are taken in reference to the axis NO. but K / 1 =r 2 K 1 , &c., D'iSsrDi, &c.; .-. K / 1 D / 1 2 =r 4 K 1 D I 2 , &c.; = r 4 ! ........ (52.) In like manner, I'x^r*!!. ... ....... (53.) Adding equations (52.) and (53.), MOMENTS OF INERTIA OF SIMILAR BEAMS. 29 but i x = I + ii, and i' x i' + i\, .'. l'=r*i x ..... (54.) which is the analytical expression of the Theorem. Transferring the moments to the axis passing through the centres of gravity, we have by eq. (40.), Art. 23. but i' x = r 4 i x , ~x f = rx, and K' = r 2 K, (56.); that is, the moments of inertia of two similar surfaces, about their centres of gravity as axes, are to each other as the fourth powers of their linear dimensions, 30. The distances of the centres of gyration of similar surfaces from their axes of rotation, similarly situated, are proportional to the linear dimensions of the surfaces. By eq. (38.), Art. 22., KQ 2 r 2 KQ /2 30 SIMILAR BEAMS. -$-> '-" - which is the analytical expression of the Theorem. TRANSVERSE STRENGTH OF SIMILAR BEAMS. 31. The moments of rupture of similar surfaces are to each other as the cubes of their linear dimensions. By eq. (17.) Art. 17. si si' but, Art. 26., since the neutral axes divide the surfaces propor tionally, l'r=r I, and h' = rh, , si (57.) In like manner, m / 1 = r 8 m 1 . ..... (58.) Adding equations (57.) and (58.), m f + m\ = r 3 (m TRANSVERSE STRENGTH OF SIMILAR BEAMS. 31 but M' = m' + m\, and M = m + m 19 .-. M' = r 3 M ..... (59.), which is the analytical expression of the Theorem. Or thus. By eq. (19.) Art. 17., . ,_BI MI " h f + ~K\ Sr 4 I 32. TAe breaking weights of similar beams are to each other as the squares of their linear dimensions. By eq. (22.) Art. 17., but by eq. (59,) Art. 31., M' = r 3 M, and I' = r/, (60.) Or thus. Supposing the elasticity of the material to be perfect, we have by eq. (41.) Art. 24., 32 SIMILAR BEAMS. l(dx) l'(d r *') But d' = rd; by eq. (54.) i^rzr 4 !^; by Art. 26., = r/; and K / = r 2 K; 33. Strength of the Conway Tube calculated by this formula. In the model Conway Tube w = 89'15 tons, 1 = 900, and in the Conway tube itself I = 4800 ; _ 4800 _ 16 = = "3' f = (Y)* X 89-15 = 2535 tons, which is the breaking weight of the Conway Tube, supposing it in all respects similar to the model tube. In the same manner the strength of any beam (whatever may be the form of its section or the nature of its material) may be calculated from the breaking weight of an experimental beam whose linear dimensions are similar. 34. The breahing weights of similar beams are to each other as their sectional areas. STRENGTH OF SIMILAR BEAMS. 33 By eq. (60.), Art. 32., but A / = r <2 A, or - = r\ w' A' -=-. ...... (61.) W A 35. Strength of the Conway Tube calculated by this formula. In the model tube W =: 89*15 tons, A, or area whole section of the material =. 54, A', or area section of the Conway tube = 1530, w' _1530 *' 89a5" ! ~64~' .-. w' = 2526 tons. 36. Strength of Mr. Cubitfs cast-iron beams with double flanges. In his best model beam, seej^. 11, CD = 5'9, CE= '84, AB = 272, AR = -83, depth AC = 17^, thickness rib nv = -68, area section =18, length =180, and w=16 tons. Let it be re- quired to find the breaking weight of a similar beam whose sec- tion is 6 sq. in. Here w = 16 tons, A = 18, A' = 6, 37. The breaking weights of similar beams are to each other as the areas of their bottom sections. D 34 STRENGTH OF SIMILAR BEAMS. Let a and a f be put for the areas of the bottom sections of two similar beams, then A'_^' therefore by eq. (61.) = - (62.) W a v J 38. In similar beams the cubes of the breaking weights are to each other as the squares of the weights of the beams. Let w and w' be put for the weights of the beams respec- tively, then /a W f T^ W 9 ' 7*6 w* ; but by eq. (60.) Art. 32. w 3 hence we have by equality, w 3 " (63.) Let L and I/ be put for the weights of the breaking loads of the two similar beams, then w = L + |, and w'= L 7 + , but w'= r 2 w, therefore by substitution STRENGTH OF SIMILAR BEAMS. 35 substituting r*w for w', and reducing, . . . (64.), which expresses the breaking load, in terms of the breaking load and weight of the experiment beam. 39. The breaking weight of beams is equal to the continued product of the sectional area, the depth, and a constant determined by experiment for the particular FORM of beam, divided by the distance between the supports. Suppose the breaking weight w to be determined by experi- ment for a beam whose length is /, depth d, and sectional area K. Assume > where c is a constant, which may be determined from this equation By eq. (60.) Art. 32. I r 2 K. rd . C rl That is the formula, ....'... (65.), holds true for all similar tubes. D 2 36 STRENGTH OF SIMILAR BEAMS. In like manner, if a be put for the sectional area of the bottom portion of the beam, and c a new constant, we have (66.) a constant, .-. K = ar l) by substituting in (65.) we have ad -j. adc where the new constant c is determined from c by the equality TT f = rG or ....... (67.) 40, Given the areas of the bottom portions of two similar beams, to find the linear ratio, r, of the other parts, &c. which is the expression for the depth, and so on to other parts. Conversely if the depths be given, we have a' r*a, and r =-7, 41. The breaking weights of learns of the same length and similar in their sections) are as the cubes of the linear ratio of the sections. By eq. (59.), M r = r 3 M. DEFLECTIONS OF SIMILAR BEAMS. 37 Now when the beams have the same length, w'Z wl .-. w'=r 3 w ....... (68.) 42. In beams having similar sections, but different lengths I and 4 corresponding to the breaking weights w and W 1} the relation is expressed by the equality r 3 W/ Vi=-j- ......... (69.) l i This result at once follows from eq. (68.), observing that the breaking weights are inversely as the lengths. Substituting A / 2l or A / for r, we have V a V K These two theorems are useful in comparing the strengths of beams having different lengths and sections. See Art. 78. DEFLECTIONS OF SIMILAR BEAMS. 43. The ultimate deflections of similar beams are to each other as their linear dimensions. By eq. (49.), Art. 25., ~48EI Z /3 w' ' 48 EI'' D 3 38 BEAMS IN CERTAIN RESPECTS SIMILAR. By eq. (56.) i' = r 4 i, and by eq. (60.), w' = r 2 w, and more- over I = rl, 48E . r 4 ! ' 48EI GENERAL FORMULA RELATIVE TO BEAMS ONLY IN CERTAIN RESPECTS SIMILAR. STRENGTH OF BEAMS. 44. Suppose the top and bottom parts of a beam to be com- posed of rectangular plates, and that the depth of those parts, as well as the thickness of the plates and the whole depth of the beam, to be similar to the corresponding parts of the experi- mental beam from which the constant c, &c., are determined ; to find the breaking weight, c., when the length is any quantity / 2 , and the breadth any quantity 2 , or the area of the section A 2 , the sides connecting the top and bottom parts being neglected. First, let the only variation from similarity in the two beams be in the length. By eq. (22.), BEAMS IN CERTAIN RESPECTS SIMILAR. 39 but because the two sections are similar, therefore by eq. (59.), Art. 31, M' = , .. . . . (72.) Or, substituting =- for w, this eq. becomes *,-*? ...... (73.) Second, let the breadth, as well as the length, of the beams differ from similarity. Now, the breaking weights of beams of this kind must evi- dently be in proportion to their breadths ; hence, if b' is put for the breadth of the beam in eq. (72.) and b 2 for the breadth of the beam whose breaking weight we shall call W 2 , then we have from eq. (72.), observing that 4 is changed to 7 2 for the sake of uniformity, but b f rb, From eq. (73.), changing ^ to / 2 and E/ to K 2 , _ K 2 ct G . W 2 i V'-> where K 2 is the sectional area, d f the depth, 7 2 the length, and c a constant having the same value as in eq. (65.). Hence the rule contained in Art 39. applies to beams of any length and breadth, provided that the depths and the structure B 4 40 BEAMS IN CERTAIN RESPECTS SIMILAR. of the top and bottom parts are similar. This gives a con- siderable flexibility to the application of the formula. It may perhaps be said, that the vertical plates in the top and bottom parts of a tubular beam would interfere with the principle assumed in the preceding investigation, viz. that the sectional area is proportional to the breadth. Now, so far as the strength or theoretical investigation of the general theorem is effected, the vertical plates may be distributed in any manner ; hence we may conceive them to be arranged so as to allow the principle assumed to hold strictly true. Hence the general theorem, Art. 39., admits of the following important modification. The breaking weight of a beam, composed of rectangular pieces, is equal to the continued product of the sectional area, the depth, and a constant determined from an experimental beam divided by the distance between the supports, provided that the vertical linear Dimensions of the beam are similar to those of the model beam. 45. Deflection of beams similar in section, but varying in length. By eq. (48.) Art. 25., 48Ei' By eq. (72.) Art. 44., w, = ^ r 3 w, 'i and since the sections of the beams are similar, therefore, by eq. (56.) Art. 29., i' = r 4 !, hence, by substitution, we have, STRENGTHS, ETC., OF BEAMS. 41 48Erl STRENGTHS, &c., OF VARIOUS FORMS OF BEAMS. 64. Let ABCD (seej^. 6.) be a solid rectangle in which th< neutral axis NO passes through the centre of the section; then by Art. 17., p. 13., the moment of inertia of the rectangle ABON is ^KjDj 2 ; but the moment of the solid rectangle ABCD is double this, (77.), which is the moment of inertia of a solid rectangle about an axis passing through its centre of gravity, and parallel to the upper or under edge of the section. Or, putting b for the breadth AB of the beam, this expression becomes (78.) A1 SI Also, M = =- 2 (79.) It must be observed that all the expressions given throughout 42 HOLLOW RECTANGULAR BEAMS. this work for M, or the moment of rupture, is equal to the moment of the breaking weight w (see Art. 16.). HOLLOW RECTANGULAR BEAMS. 47. Let ABCD be the transverse section of a hollow rect- angular beam of uniform thickness, NO the axis passing through the centre of the section, then AN = ^AD. -BJ Put d= AD, the depth of the beam, b = AB, the breadth, d l = ad, the interior depth, Z>j.= 6, the interior breadth, I = the moment of inertia of the whole section. By Art. 17., the moment inertia of the rectangle NOB A = ^ v27 == A^^ 3 - Similarly the moment inertia of the in- terior rectangle = -^ b l d^. Moment inertia section NOAB = moment inertia NOB A moment inertia noba = J^(bd 3 &i^i 3 ). But the inertia of the whole is double this expression, M! 8 ) (80.) When b = d, and b^ = d lf that is, when the section of the beam is square, this equality becomes 'o = Or, if K be put for the section of the material, we have, COMPARATIVE STRENGTH OF BEAMS. 43 = ^() where the strength varies as the area of the section multiplied by the depth. Comparison of Strength of solid and hollow Beams. 48. In order to compare the strength of a solid beam with a hollow one of the same depth and section of material, we have, by subtracting eq. (79.) from eq. (84.), * for the ex- cess of strength of the latter over the former. If the platos composing the hollow beam be thin, then d is nearly equal to d, and therefore, in this case, the hollow beam will have double the strength of the solid one. 44 RECTANGULAR CELLS. 49. To determine the moment of inertia, -c., of a square rectangular cell A BCD, about any axis NO parallel to DC, or AB. By eq. (82.), Art. 47., the moment of inertia of this cell about an axis KG passing through its centre of gravity is N K Fig. 7. 12 And by eq. (39.), Art. 23., = (* + rfi'+12?) . . . (86.), which is the expression required, x being the distance between KG and NO, or the distance of the centre of gravity of the cell from the axis NO. SO. To find the moment of inertia of a rectangular cell ABCD, about any axis NO parallel to DC or AB (see last figure). Here, substituting the value of I given in eq. (80.), we have Mi 3 )+*K. . . . (87.) This expression will also give moment of inertia of a series of cells ABSK (see^y. 10.) about KG ; where b = AB the exterior breadth, d = AR the exterior depth, &!= the whole breadth of the spaces in the cells, d = the interior depth of the cells, K = the area of the section of the material, and x = the distance of the centre of the cells from the axis KG. 51. To determine the conditions of maximum strength ; b, d, x, and K being constant. HOLLOW RECTANGULAR BEAMS. . 45 Here i xi and, consequently, M X , will be a maximum when bidf is a minimum, or putting K X for the area of the internal spaces, ^dj 3 = K^! 2 = a minimum, or d ls a minimum, that is, when the material is accumulated in the two horizontal plates AB and RS. However, in practice, it is necessary that the vertical plates should have a certain thickness to counteract the tendency which the horizontal plates have to crumple. 52. To find the strength, Sfc., of a hollow beam, when the ma- terial is not equally distributed on each side of the neutral axis. Let KG (see Jig. 6.) be drawn bisecting the vertical depth of the beam, and parallel to the neutral axis NO. Let K 19 K 2 = the areas of the external and internal rectangles above KG; D x and D 2 being the respective depths of these rectangles estimated from KG. k l9 hy d 1} d 2 the corresponding dimensions below KG. Then we have for the position of the neutral axis (see Art. 12.) . . (88.), 2K which expresses the distance of the neutral axis NO from the central line KG. By Art. 17., we have for the moment of inertia i x about the axis KG, Moreover, for the moment of inertia referred to the neutral axis, we have by eq. (40.), Art. 23., 46 MOMENT OF INERTIA OF ANGLE IRON. And, finally, si. (90.) (91.), where the value of x is given in eq. (88.), and I in eqs. (89.) and (90.). In these expressions D x - d : ~> or one-half the whole depth of the beam. S3. To find the moment of inertia of angle-iron. Let ABCD be the section of the angle-iron, and Gi the axis of moments. Let a B i = AG, the distance of the Fi 9- 8 - &9- edge A B from G I ; D c A B b = AB = BC, the length of the angle-iron ; =AF = DC, the thickness of j the angle-iron. G Then in^. 9., we have, by Art. 17., I = moment inertia rectangle AB I G moment inertia rectangle FEHG moment inertia rectangle DCIH. K I G HI Similarly inj^. 8., we have -$fl 8 . (93.) In eq. (92.) if a = b, or Ci = 0, then CONWAY MODEL BEAM. In eq. (93.) if a = 0, or AG = Bl = 0, then 47 . . . . (95.) TO DETERMINE THE STRENGTH, ETC. OF A TUBULAR BEAM ABDC, COMPOSED OF SQUARE CELLS, A H : 1 1. 1 : 7. 10. A S, IN THE TOP PART, AND OF THICK PLATES, CD, IN THE BOTTOM PART, SIMILAR TO THE MODEL TUBE EX- PERIMENTED UPON BY MR. FAIR- BAIRN. 54. To find the neutral axis, the elas- ticity of the material being perfect. Let a = the area of the material in the top, A s ; /3 = the area of the bottom plates. y = the area of the two side plates, RE and SF ; a l = the distance of the centre of gravity of the top cells from the lower edge, CD; ft = half the thickness of the plates in c D ; y! = the distance of the centre of gravity of the side plates from C D ; K = area of the whole section ABDC ; x n K c = G D, the distance of the neutral axis from c D. Then taking CD as the axis of moments, we have, Art. 12., ' -f L - __ a ! -f (96.) 48 CONWAY MODEL BEAM. 55. In the model tube, a = 24, /3 = 22, y = 7, ! = 507, & = -3, y : = 24, and K = 53, - __ 24x50- __ Now, as the whole depth of the beam is 54, we have the dis- tance of the neutral axis from the upper edge, viz. A K = 54 26*2 = 27 '8. Hence, it appears that the neutral axis passes nearly through the centre of the beam. 56. To find the moments of rupture, 8fc. Let &=AK:=:BG, the distance of the upper edge from the neutral axis K G ; ^ =r CK = DG, the distance of the lower edge from KG ; p =. the whole breadth of the spaces in the top cells ; e = AR =BS, the depth of the top cells; k = the thickness of the plates in the top cells ; b A B = CD, the breadth of the tube ; h = the thickness of the bottom plates ; & 2 the sum of the thicknesses of the side plates, KE and SF. I = moment ABGK moment space KGSR moment space in the cells AS + moment KGDC moment space KGFE. Now moment ABGK = &A 3 , moment space (/i-e) 3 , moment cells AS = ^p | (7* A) 3 - (h - e + kj | , ANGLE IRON TAKEN INTO THE CALCULATION. 49 moment KGDC = -bh* 3 moment space KGFE=^(6 & 2 ) (^ AJ 3 . Hence we have, by substitution and reduction, i- V) - (5 - A,) {(A- ) + (ft, - *,) } . (97.), which expresses the moment of inertia of the whole section. By substituting this value for I in eq. (24.), Art. 18., the expression for the breaking weight, &c. is determined. 57. If the angle-iron is to be taken into the calculation, the following method of investigation may be adopted. First, let us calculate the moment of inertia of the top-cells A s, about an axis, passing through the centre of the cells, parallel to AB or RS. Let e, e' = the exterior and interior depth of the top cells respectively, by Z/ = the exterior and interior breadth of the same (see observations, Art. SO.), a! = the distance of the centre of the top cells from KG, a = the area of the material in the top cells, r = the length of the angle iron, t = the thickness of the angle iron, n = the total number of angle irons in the top part, <7 = the moment of inertia of the top part about its axis, Q = the moment of the same about KG. 50 STKENGTH OF THE CONWAY TUBE, ETC. These symbols being accented for the corresponding dimen- sions below the neutral axis, and the other notation being the same as in the preceding investigations, we have, by eq. (92.) Art. 53., Moment of inertia of the angle iron in the top cells, about a line passing through their centre, . (98.) By eq. (80.) Art. 47., moment of inertia of the plates in the top cells (99.) Adding eq. s. (98.) and (99.), ^(be*-b'e'^. (100.) In order to refer this moment to the axis KG, we have, by eq. (39.), Art. 23., Q = 2 + a /2 a ...... (101.) Moment inertia side plates = \\ (hef ... i = Q + l 2 (/i-e) 3 . . . . (102.) Supposing the bottom part of the tube to be constructed with cells, as in the Conway and Britannia tubes, we similarly have Qi = ?i + "i /2 *i ... (103.) i l = Q l + i* 1 (V-O'- - ( 104 RIVET HOLES TAKEN INTO THE CALCULATION. 51 Hence we have for moment of rupture of the whole section, Art. 18., M = |(I + I I ) ..... (105.) Where, if s be given, M, as well as w, may be determined ; or if M = ^ be determined by experiment, then s may be found from eq. (105.) ; and the value of h may be found from eq. (96.), where ft must be taken for the area of the material in the bottom cells, and /5 X for the distance of their centre of gravity from KG. 58. It is desirable that an allowance should be made for the rivet holes of the angle iron ; for this purpose, let there be two rivet holes in each arm of the angle iron, and Let c = the diameter of the rivet holes, v, v' = the distances of the edges of the rivet holes, in the vertical plates, at the greatest distance from the central axis of the top cells ; q f = moment of inertia of all the rivet holes, in the top cells, about the central axis of the cells. Moment inertia rivet holes in the horizontal plates Moment inertia rivet holes in the vertical plates v*-(k+t) (v-c)* + (k+t) v'*-(Jc + t) (t/-c) 3 } (106.) E 2 52 APPROXIMATE FORMULA FOR TUBULAR BEAMS. If there is only one rivet in each arm of the angle iron, then Therefore, in the place of eq. (101.), we must in this case employ the following expression for Q, viz. Q =?-?' + ' 2 a ...... (107.) And so on similarly for the value of Q L . 1. Approximate Formula. 59. In tubular beams, as they are now usually constructed., the strength nearly varies as the area of the top or bottom part mul- tiplied by the depth divided by the distance between the supports. That is, if a, ! = the top and bottom areas respectively, d = the depth of the beam, w = the breaking weight, I = the distance between the supports, c, G! = constants determined from experiment on a tube con- structed on the" peculiar principle of tubular beams ; then adc a,dCi Let e = AR = BS, the depth of the top cells (see^. 10.) ; g == the distance of the centre of the top cells from the neutral axis KG; b the breadth of the material in the top cells, supposing it collected in vertical plates ; h the distance of the upper edge of the beam from the neutral axis ; and so on. APPROXIMATE FORMULA FOR TUBULAR BEAMS. 53 Without infringing upon the peculiarity of the structure, we may suppose that the material in the top and bottom cells is collected in vertical plates, then, neglecting the material in the side plates KR and GS, we have, Art. 17., eq. (14.), sb m=. -r neglecting y^ 2 as being very small, and also observing that be = a, we have m = In like manner, but by eq. (8.) Art. 9., sag = .'. M = sag(g + g^ or s^^, (g + q^ = sag or s^a-^g^ (108.) = sad x ~ or s.a^ X ^j hd E 3 54 APPROXIMATE FORMULA FOR TUBULAR BEAMS. taking jj and ~-^ as constants, each of them being, in all prac- II CL II \ CL tical cases, nearly equal to unity, we have (109.), where the strength varies as the product of the top or bottom areas multiplied by the depth. When the beam is supported at each end and loaded in the middle, wl = M = S a d or Si a l d 4s, a, d w = T or y 1 - adc a,dc* /,-, \ .-. w = por-^-. . . (110.), where the values of c and c x are determined as follows : 60. In the model Conway tube, Z=900, a v = 22-5, rf= and w = 89*15 tons, therefore from eq. (110.) __ 900 x 89-15 22-5 x 54 = 66 tons. Or making a deduction of 3 '5 inches for the rivet holes in the bottom area, 900 x 89-15 Ci= r^ - FT = 78 tons. 19 x 54 STRENGTH OF THE CONWAY TUBE, ETC. 55 Substituting the value of the first constant in eq. (110.), we have (111.), which expresses the breaking weight, in tons, of any tubular beam loaded in the middle and supported at the extremities, a^ being the area of the bottom part of the beam in inches, d the whole depth of the beam, and / the distance between the two supports expressed in the same units as d. Cor. From eq. (110.), we have, 4s = c, .-. s = c. 61. Strength of the Conway Tube calculated by formula (111.) Here a l = 517, d= 302, and I 4800, = 2146 tons. 66 x 517 x 302 4800 62. When the depth and distance between the supports are the same in two beams, the breaking weights are as the areas of the top or bottom parts. This property at once follows from eq. (110.) The truth of this theorem is confirmed by Mr. Fairbairn's experiments on the model tube. From experiments 1, 3, 4, and 10 on the model tubular beam, as given in his work on Tubular Bridges, we have the following table of results : No. of Experi- ments. Area, bottom. Breaking Weight. Proportional Area to Breaking Weight. 1 6-8 79,578 1 91 3 12-8 126,128 1 98 4 17-8 148,129 1 84 10 22-45 192,892 I 86 E 4 56 LIMITS OF ERROR OF THE FORMULA. Here the law almost exactly obtains in the two last experi- ments. In the two first cases the breaking weights are slightly in excess of that which would be indicated by the law ; but this might have been anticipated in conformity with the relation w == j , since the areas of the tops of these tubes are in excess of the proportion which they should have to the bottom areas. However, every candid person cannot fail of recognising the preceding formula as being a manifest induction from these ex- periments. Limits of Error involved in Formula (110.) e 1 63. First with respect to the quantity T^; in the model tube e = 6-5, by Art. 55., AK = 27-8, ... <,>27-3-2>23-8, Hence this portion of the formula which is rejected is less th part of that which is retained, so that the relation, M = sagG, or s 1 a l ff 1 G 9 is almost mathematically exact. Second, with respect to the quantity *[L_ in the model tube, /*! e = 6-5, e, = '6,d= 54, h, = 26'2, g l - 26-2 - ^ = 26 > G = 9 + 9\ = 51 > nearly, 26 51 25 Hence there is an error of only about y^th occasioned by the assumption of the formula (109.) A MOKE EXACT FORMULA. 57 64. In the tubular beams which Mr. Fairbairn has recently constructed, the depth of the beams are increased in ratio ; hence in such cases the error is less than that determined in the fore- going calculation. The friends of Mr. Stephenson, actuated by an intemperate zeal for their distinguished patron, have publicly declared that this important formula is empirical. It would certainly better serve their cause if they could sustain their declamation by an appeal to mathematical analysis. The fact is, the formula is as nearly true as any general formula of the kind can profess to be. It will be hereafter shown, that it is a nearer approximation to truth than Mr. Hodgkinson's celebrated formula relative to cast- iron beams. 2. Approximate Formula. 65. In beams where the material is properly distributed, we iould always have = we have, from eq. (8.), should always have =: , or s a = ^ a l ; in this case, therefore, CL\ S that is, the distances of the centres of gravity of the top and bottom parts of a beam, from the neutral axis, are equal to one another. Hence we have, from eq. (108.), . a . | G . G G + yj "By substituting g + - for h, and g l + for h l9 we obtain pre- '-* JL cisely the same expressions for m and m l as those given in Art. 59. : hence we also have, in this case, the relation of eq. (109.), the same quantities being neglected in the formula. Hence, also, the expression for the breaking weight given in eq. (110.) applies to beams having double flanges, with the same general formula of error. Application of eq. (110.) to cast-iron Beams. 72. In order to calculate the limit of error, we shall first assume, in accordance with eq. (119.), that the neutral axis lies in the middle of the vertical line joining the centres of the top and bottom flanges. In Mr. Hodgkinson's 12th experiment, p. 428., of his edition of Tredgold, e=-31, ^=-66, d= 5-125, .-. G=5-125-' 31 l'' 66 = 4-64, 9l = 2-32, and h, = 2-32 + -33 = 2-65, 2-32x4-64 11 h^" 2-65x5-125" 13 where there is an error of about J-th. Again taking the neutral axis at the lower edge of the upper flange, we have d= 5-125, G = 5-125 _ >66 + >31 = 4 . 64? ffi - 66 2 5-125--31 -^ = 4-48, * 1= = 5-125- -31 = 4-815, CAST-IRON BEAMS. 65 where there is an error of about ^th, as before. These calculations show that the formula, Art. 59., theo- retically considered, is not so applicable to beams of this kind, as to the tubular beams. However, although $- may not be equal to unity, yet it will very nearly be a constant for all beams which differ but little from similarity. We shall now inquire how far these theoretical deductions agree with the results of experiment. 73. Mr. Hodgkinsorfs experiments on cast-iron learns with double flanges, as given in his edition of Tredgold on the Strength of Material, Table 11., page 434. With the view of ascertaining how far the results of the preceding articles coincide with the deductions of experiment, the following table presents the proportion of the depths of the beams to their breaking weights. In experiments 10. and 11., the breaking weights are reduced to the same distance between the supports as in experiments 1, 2, 3, 4, and 5. No. of Experi- ments. Depth. Area Section. Breaking Weight. Breaking Weight re- duced to unity of Section. Proportion of Depth to re- duced breaking Weight. 1 4-1 6-54 13,543 2,070 1 505 2 5-2 6-94 15,129 2,180 1 420 3 6-0 7-08 15,129 2,130 1 355 4 6-93 7-67 22,185 2,890 1 417 5 6-98 7-40 19,049 2,570 1 368 10 10-25 7-83 36,800 4,700 1 459 11 10-25 9-10 41,400 4,550 1 444 Here it will be observed that where the beams approach to 66 NEW FORMULA APPLICABLE TO an identity of section, we have, as might be expected, a near approach to the equality of ratio, as for example, in the case of experiments 10. and 11., and also in 3. and 5. And where the beams approach to similarity of section, or to such modifications as are explained in Art. 44., we have, in accordance with the law of similar beams demonstrated in Art. 39., a near approach to an equality of ratio, as for example in the case of experi- ments 2. and 11. But where the beams are decidedly dissimilar we find a marked deviation from the law, as for example in experiments 1. and 5., and also in 3. and 10. ; in such cases the variation amounts to about jth as determined in Art. 72. New Formula relative to Cast-iron Beams. 74. For similar beams the true formula of strength is that given in Art. 39., and when the beams are not similar, formula (114.), Art. 65., determines the breaking weight with con- siderable precision, viz., 4sG 2 4s, 1 G 2 yy - - Qj* _ i _ i _ l(G + e) l(G + ej Where the constants s and SJL are determined from the equations _W/(G + e) 4G 2 W/(G + and s, = ; , 2 To determine S 19 we have from experiment 12., given by Mr. Hodgkinson (see Art. 72.), same thickness of plates, and the diameter of the one equal to the Side of the other. By eq. (82.), Art. 47., we have, for the square beam, And by eq. (136.), Art. 80., we have, for the cylindrical beam, M _4K but K = cP d^, and KJ (d 2 d^), 5.-1 KJ^TT' M 4 4 16 74 COMPARATIVE STRENGTHS OF CELLS. Hence the comparative strengths of the two beams are as follows : ^ , ,,. , , , strength sq. beam 16 For equal thickness or plates, - SL. / . strength cir. beam STT - , strength sq. beam 16 4 4 For equal areas, - ^rr ^ r = o s =-^ strength cir. beam STT TT 3 With the same material, the square beam has 1^ times the strength of the cylindrical one. COMPARATIVE STRENGTHS OF CIRCULAR AND SQUARE CELLS. 84. Let KEGF be a circular cell, and A BCD a square one, where the areas of the sections are the same, the diameter 1C 1) o I Fig. 13. N O Fig. 14. KG = the side AB = d, the lines KG passing through the centres of the two c^lls are at the same distance (.r) from the axis of moments, NO. By Art 49., Moment of rupture of the square cell, M, By Arts. 80. and 23,, eqs. (136.) and (39.), BEST FORM OF CELLS. 75 Moment of rupture of the circular cell, M 1? 4 where d% 2 = the square of the interior diameter = d 2 -- K. 7T Now M will be greater or less than M 1 according as ft but - K is greater than K, therefore M is greater than MJ ; that 7T is, the square cell is stronger than the circular one. OBSERVATIONS RELATIVE TO THE BEST FORM OF THE CELLS IN A TUBULAR BEAM. 85. It appears from the foregoing investigation, that with a given section of material the square form of cells is stronger than the circular form. Mr. H., however, found from ex- periment that the circular cells are stronger than the square ones when subjected to a simple crushing force exerted equally THE 76 STRENGTH OF CELLULAR STRUCTURE. over their sections ; arid he therefore infers that the former are better adapted to a tubular beam than the latter. This in- ference admits of dispute, for it is contrary to theory. The cells in a transverse strain undergo a different kind of strain to what they are subjected to in a simple crushing force equally distributed over the section of the tube. In this case all the parts of the section are equally compressed, and it is reasonable to conclude that the best form of the cell will be that in which the material is equally distant from the axis of pressure ; but the case of transverse strain is very different ; the upper edge undergoes the greatest strain, and of the other parts, that which is nearest the neutral axis of the beam under- goes the least : in this case, therefore, the material in the square cells is symmetrically distributed with respect to the axis of pressures the neutral axis of the beam. Be this as it may, we are not disposed to give up a principle established by theory until some direct experiments should prove the contrary. In a rectangular cell the weakest part is obviously at the corners or points of junction of the plates ; but if these points are strengthened by angle irons, as they are in the Conway Tube, the cells are not only rendered more perfect, but a new important auxiliary element is introduced into their structure. WHY THE CELLULAR STRUCTURE EXHIBITS SUCH STRENGTH. 86. Let AC and EG represent the sections of two beams undergoing transverse strain, in all A respects the same, excepting that in the former case the material at AB is composed of horizontal plates in contact with each other, whereas in the latter case the material is arranged in the form of cells; the Fig. 15. Fig. ie. horizontal plates being connected by vertical plates or ribs, a : b, c, d, &c. MOMENT OF INERTIA OF CIRCULAR SPACES, ETC. 77 When thin plates of wrought iron are subject to compression, they double or crumple up long before the material would be destroyed under ordinary circumstances by crushing. In the first case of construction (fig. 15.), no means are employed to counteract this tendency to double ; but in the latter case it is different. Here the horizontal plates, as well as the vertical ones, have a tendency to double up ; but the direction of this tendency in the former is in a plane at right angles to the plane in which the latter would take place ; that is to say, the horizontal plates EF and KJ tend to crumple vertically, while the vertical plates ab, cd, &c. tend to crumple horizontally. Again, the direction of greatest strength in the vertical plates is in the vertical line, and at the same time the direction of weak- ness of the horizontal plates is also in the vertical line, and vice versa: hence the horizontal plates EF and KJ are prevented from crumpling by the vertical plates ab, dc, &c., and vice versa. It is evident, that the horizontal plates EF and KJ could not crumple without exerting a vertical strain upon the vertical plates ab, dc, &c. : and, in like manner, the vertical plates could not crumple without exerting a horizontal strain upon the horizontal plates. It appears, therefore, that the object of the cellular structure is to counteract the tendency which thin plates, acted upon by a compressive force, have to crumple, and thus to cause the tubular beam to be subjected to the same laws of transverse strain as an ordinary beam. Hence it is not necessary that the bottom part of the tube should have a cellular structure. MOMENT OF INERTIA OF CIRCULAR SPACES, ETC. 87. To determine the moment a of inertia of a semicircle about an axis passing through its centre of gravity parallel to the diameter. Let ABR be a semicircle, KG the axis passing through its centre of gravity parallel to the diameter AB. (See^. 17.) 78 MOMENT OF INERTIA OF CIRCULAR SPACES, ETC. Put x the distance of KG from AB. By eq. (131.) the moment of inertia of a semicircle about its diameter is ^Trr 4 ; therefore, by eq. (40.), Art. 23., the moment about the axis through the centre of gravity is I = H 7T7* 4 area x x 2 o (139.); 4r but x = , (see the Author's Calculus, p. 226.), 37T = -llr 4 , nearly. . . . (140.) 88. To determine the moment of inertia of a semicircle about any axis NO parallel to the diameter AB. Let e = N A = OB, I =the moment of inertia of the semicircle about the axis KG, and i^the moment about NO; then I is given in eq. (139.); therefore, by eq. (39.) Art. 23., I X =I + areaARB x OG 2 Fig. 17. (141.); MOMENT OF INERTIA OF CIRCULAR SPACES, ETC. 79 which is the moment of inertia of a semicircle about any axis parallel to its diameter. Or thus by a direct investigation : Let Se be a small element of surface, c the centre of the circle ; put x Ce 'Dm, y Se, A # = the breadth of the elementary rectangle s^-=^(f* + 4>)'. . (142.) .*. moment semicircle lying below the diameter AB = moment whole circle moment semicircle ABR 80 MOMENTS OF INERTIA, ETC. If in this case e rr DR = e + r ; then we have moment inertia semicircle ABR placed in an inverted position (144.) 89. The moment of inertia of a circle about an axis forming a tangent to its circumference is Jive times the moment when the axis passes through the centre. In eq. (142.) let e = r, then moment inertia about the tangent = r=5 x moment inertia about the diameter, by eq. (131.) 90. The difference of the moments of inertia of the semicircle ABR and ABV about NO as an axis, is equal to the moment of inertia of the rectangle ABON about AN as an axis. (Seej^. 17.) Subtracting eq. (143.) from eq. (141.), we obtain the differ- ence x = 5 e (2 r) 3 = moment inertia rectangle ABON about o o the axis AN. 91. To determine the moment of inertia of the space BSRw about the axis NO. Seej^. 17. Let q be put for the moment of inertia of the space BSR/Z, then Moment inertia BSRTZ = moment square R n B c moment inertia quadrant BCR, 045.) MOMENTS OF INERTIA, ETC. 81 92. To determine the moment of inertia of the section ABDC of a beam with double flanges, having the quadrants (rs) of circles forming the interior angles. Let b, b l AB and CD respectively; h, ^ = the distances of AB and CD re- spectively from the neutral axis KG; K v h' = K R the distance of R S from K G ; k 2 ~ the thickness of the vertical n G D rib ; Fig. is. r =r o s = o r, the radius of the circles forming the quadrants rs, &c. ; e = vs, the distance of the centre of the circle from KG; Q=:the moment of inertia of the rectangular parts above KG; q rrthe moment of inertia of the circular portion rts above KG; and so on to these symbols accented for the parts below KG. By eq. (120.), Art, 70., j. . . (146.) And by eq. (145.), Art. 91., q. is given, .-. l = Q + 2g, and i! = 0^ + 2^; if the material is perfectly elastic, we have M=(Q + 2 7 + Q 1 + 2 !7l ) . . (147.); 82 STRENGTH, ETC. OF ELLIPTICAL BEAMS. and when the beam is supported at each extremity, w =-JT- JQ + 2<7 + Q! + 2q l | . . (148.); where it is to be observed that Q is given in eq. (146.), q in eq. (145.), and so on to Q { and q r The neutral axis is determined from eq. (13.). STRENGTH, &c. OF ELLIPTICAL BEAMS, &c. 93. Let NRBD (seejft?. 3.) represent the transverse section of an elliptical beam, ON the semi-minor diameter, and OR the semi-major diameter. Let Z> = ON = OB, arrOR, x=Om, y = sm ; then, by the equation to the curve, therefore by substitution in eq. (34.), Art. 21., (149.), which is the expression for the quadrant NOR of the ellipse. Hence the moment of inertia of the whole ellipse about its minor diameter will be (150.), HOLLOW ELLIPTICAL BEAMS. 83 but area section = 7rab = K, s 1 =^ . Ktt 2 = SKa 4a 4 (151.) where d is put for the whole depth of the beam or the major axis of the ellipse. 94. If the beam be supported at each end, s Kd /-,*o\ T -. . . . (152.); that is, the breaking iveight of a solid elliptical beam is equal to the continued product of the sectional area, the depth, and a con- Q stant (-) divided by the distance between the supports. It will be observed that this formula applies to all elliptical beams, whatever may be the ratio of the depth to the breadth. Hollow elliptical Beams. 9. Let a, and b l be put for the semi-axes of the interior ellipse; then from eq. (150.), 7T (*rf>-W) . . (153.); G 2 84 HOLLOW ELLIPTICAL BEAMS. x i a a, , taking =j } we find, U Ui SK . . -or- (154.) by taking -~ l = I, which it is very nearly when the plate forming the hollow beam is thin as compared with either n diameter. In experiment 19 , made by Mr. Fairbairn on elliptical tubes, ^ = -98. 96. When the beam is loaded in the middle and supported at the extremities, we have /1KK\ w-=-y- ..... (155.); the mean of the constant s, determined from Mr. Fairbairn's experiments on elliptical tubes, is 15 tons. Hence we have the following rule for finding the strength of hollow elliptical beams. In hollow elliptical beams formed of thin plates, the breaking weight in tons is equal to the continued product of the sectional ELLIPTICAL BEAMS. 85 area, the depth, and a constant (15.), divided by the distance be- tween the supports. These dimensions are all in inches. This result is analogous to that derived for rectangular and cylindrical beams. Arts. 39. and 82. 97. To determine the strength, fyc., of a beam having a rect- angular section, ABDC (seejft?. 18.), hollowed at the sides by the semi-ellipses, RUE and Let m, n = the major and minor diameters of the ellipse respectively. Here, by eq. (78.), the moment of inertia of the solid rect- angle, ABDC, about the axis, KG, passing through the middle of the beam, is TO^^ wnere # = AB = CD the breadth of the beam, and C?=:AC = BD, the depth. And the moment of inertia of the ellipse is given in eq. (150.) : which expresses the moment of rupture of the beam. If m = d, then M = s