KINETIC THEORY OF ENGINEERING STRUCTURES DEALING WITH STRESSES, DEFORMATIONS AND WORK FOR THE USE OF STUDENTS AND PRACTITIONERS IN CIVIL ENGINEERING BY DAVID A. MOLITOR, B.C.E., C.E. < AUTHOR "HYDRAULICS OF RIVERS, WEIRS AND SLUICES," ETC.; MEM. AM. soc. c. E.; MEM. DETROIT ENGINEERING SOC.; MEM. SOC. FOR THE PROM. ENG. EDU.J MEM. A. A. A. S., ETC.; FORMERLY DESIGNING ENGINEER ISTHMIAN CANAL COMMISSION; FROFESSOr, IN CIVIL ENGINEERING, CORNELL UNIVERSITY. McGRAW-HILL BOOK COMPANY 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E.G. 1911 Copyright, 1910, BY DAVID A. MOLITOR THE SCIENTIFIC PHES9 ROBERT ORUMMONO AND COMP BROOKLYN. N. V. PREFACE " ALL that mankind has done, thought, gained, or been; it is lying in magic preserva- tion in the pages of books"; and so in presenting this work for publication, the author hopes to preserve the results of many years of painstaking labors, study and exped- ience, to civil engineering, his chosen profession. The field of usefulness which it is proposed to supply is that of an advanced treatise on stresses and deformations in engineering structures, for the use of students and prac- titioners in civil engineering and particularly for specialists engaged in the design of so-called " higher structures " requiring more than the ordinary methods of statics for their analysis. The reader is supposed to possess a thorough knowledge of higher mathematics, including the calculus; also, of the elements of statics, embracing the composition and resolution of forces, force and equilibrium polygons, etc. All these are usually given in the two first years of a Civil Engineering course in connection with mathema- tics and mechanics. Details pertaining to the design of bridge members and connections are not dealt with. The present volume might advantageously be employed as a text-book in bridge stresses with the aim of giving a more thorough training in the analysis of stresses and deformations by economizing somewhat on the time at present devoted to detailing and shop practice, which is not really justified in a four-year course of theoretical preparation. In the opinion of the author this would be highly desirable, because details and shop practice can best be learned in the shop, where experience is the teacher. It devolves on the college to give to its students that thorough theoretical training which they cannot readily acquire in practice, a feature which must ultimately distinguish the college-bred engineer from the purely practical man. The present work, which is the gradual evolution of years of labor, thus represents an effort to place before the profession a treatise on the analysis of engineering struc- tures which is based on the most advanced theories and researches of the present time. To enhance the educational value of the book, each subject is prefaced by a few brief historical remarks and all important theorems bear the names of their originators, a practice which has been shamefully neglected by many modern writers. The author has made free use of any and all literature bearing on the various sub- jects treated and hereby expresses his grateful acknowledgment to all who have contrib- uted to the sum total of our present knowledge. Prominent among these should be vi PREFACE mentioned Hooke, Bernoulli, Lagrange, Coulomb, Navier, Lame, de Saint- Venant, Clapeyron, and Menabrea, as the founders of the theories of elasticity and virtual work. To Clerk Maxwell, Castigliano, Mohr, Fraenkel, Culmann, Winkler, Mueller-Breslau, Engesser, Weyrauch, Manderla, Asimont, Landsberg, Melan, Zimmerman, Ritter, Land, and Mehrtens, we are indebted for advancing these theories to their present state of per- fection and usefulness. Professor Otto Mohr enjoys the high distinction of being the foremost originator of novel principles in the analysis of engineering structures. A bibliography of the works used by the author will be found at the end of this volume. The author has avoided a sharp differentiation between analytic and graphic methods and uses both without discrimination, aiming always to secure the most practical solu- tion for any problem in hand. While this work was produced with very great care and diligence, hoping to eliminate typographical and other errors, it is certain that some have escaped detection. The author, therefore, invites the kind indulgence of the reader and will be truly thankful for having his or the publishers' attention called to any errors that may be discovered. Finally, as the love liberated in our work is a true manifestation of character, so may the present production reflect the character of, the man, the engineer, the author. ITHACA, N. Y., September 9, 1910. DEFINITIONS OF TERMS USED THROUGHOUT THIS WORK FREQUENTLY some of the letters are employed in a special connection other than here defined, but in such cases the definitions locally given will prevent misunderstandings. ART. 3. p = number of pin points of any frame. m = number of members of any frame. n = number of redundant conditions in any structure. n' = number of external redundants in any structure. n" = number of internal redundants in any structure. e = number of elements or simple frames in a structure. Sr = number of necessary reaction conditions for any structure. 2 = the summation sign. ART. 4. S = total stress in any member of a frame resulting from any loads P or other causes and + for tension. & = the stress produced in any member by applied loads P. /= the unit stress in any member, + for tension. Z = length of any member or span of a structure for the condition of no stress. JZ = change in length of a member due to any cause, positive for elongation. F cross-section of any member, prismatic in form. t=& uniform change in temperature in degrees, + for rise. = coefficient of expansion per degree temperature. E= Young's modulus of elasticity for direct stress. a = the angle which a member makes with the x axis of coordinates. /? = the angle which a member makes with the y axis of coordinates. m = any point of a structure chosen for illustration. p = l/EF = the extensibility of a member employed for brevity. .R = any reaction force, or condition, of a structure resulting from loads P. .ft = any reaction force, or condition, of a structure resulting from loads P. Jr = any displacement of the point of application of a force R in the direction of this force. P = any external load or force applied to a structure. P = a load or force identical in position and point of application with the force P but having a different intensity. ART. 5. J = any displacement of a pin point. = the change in length of the member a for condition Xb=l. d&a = the change in length of the member b for condition X a =l, etc. d at = the change in length of the member a resulting from a change in temperature of t, when the principal system is not otherwise loaded, hence X = 0, P = 0. dbt, $ct are similarly defined for members b and c respectively, ART. 13. f x , f y , / 2 = the unit normal stresses in the directions X, Y, and Z, respectively , for any isotropic body. T XV and r x:s = the unit tangential stresses in the YZ plane. r vx and r vz = the unit tangential stresses in the XZ plane. T ZX and T zv = the unit tangential stresses in the XY plane. G = the modulus of tangential stress. m = the Poisson number or ratio of lateral to longitudinal deformation. ART. 14. N = & normal thrust on any section. R = any oblique thrust. Q = a tangential force or shear. I = the moment of inertia of anv cross-section. DEFINITIONS OF TERMS USED THROUGHOUT THIS WORK ix t/ = an ordinate usually the distance from the neutral axis to the extreme fiber of a cross-section. h = height of a section. 6 = base or breadth of a section. /? = coefficient of shearing strain, Eq. (ML) and Eq. (14M). ART. 16. w = weight of a moving body. H = height of fall. Later used to represent the horizontal thrust of an arch, also, the pole distance of a force polygon. v = velocity at instant of impact. g = acceleration due to gravity. ART. 17. jj = any ordinate to an influence line. d = a panel length. p = uniform moving load per unit of length. ART. 18. m = a load point, being any one of the many possible points of application of a certain moving load. n = the point for which art influence line is drawn. Also, the location of a section under consideration. ART. 20. i = the load divide for positive and negative effects. For arches the load divide is sometimes designated by d. ART. 21. r = the lever arm of a member in the equation S = M/r. S a = ihe stress in any member S due to a reaction unity at A, when the load producing this reaction is to the right of the section. Si, = the stress in the same member due to a reaction unity at B, when the load producing this reaction is to the left of the section. ART. 28. e = the kernel point for the extreme fiber of the extrados of a solid web arch section. i = the kernel point for the extreme fiber of the intrados of the same arch section. ART. 35. w = an elastic load, representing an abstract number without any particular unit of measure. ART. 45. m + Zr, ........... (SB) then the structure is not in stable equilibrium and may be called statically insufficient. When 2p+ra-2/>=8+6-2X7=0. Eq. (3E) gives n' = 2r e2=86 2=0. Hence, both equations may serve to test the external redun- dancy, when the elements alone are considered, but internal redundancy requires count- ing all members and pin points in the entire structure. It is thus important to understand the exact interpretation of these two test equations. As a general rule it is best to count intersecting web members as four members instead of two, though the result is usually identical. Thus in Fig. SK, n = 2r+m 2p = 34-1720=3+2124=0. But in this example the structure is not composed of a succession of triangles and when pairs of members, meeting in a point, are removed, the structure will not reduce to a single triangle, showing that it is not stable. Fig. 3cj is another example of the toggle joint when the members at c are not connected. The following table was arranged to illustrate all these points with reference to the several cases represented by Figs. 3D to 3n. Fig. JY m p e n = ~ r + -m 2p n 1 = r-e-2 Remarks. 3D 3 1 2 1 3+ 1- 4 = 3-1-2=0 Determinate. 3E 6 1 2 1 6+ 1- 4 = 3 6-1-2=3 3 times ext. ind. 3r 6 1 2 1 6+ 1- 4 = 3 6-1-2=3 3 times ext. ind. 3G 3 8 5 1 3+ 8-10=1 3-1-2=0 Once int. ind. 3H 4 21 12 1 4+21-24 = 1 4-1-2=1 Once ext. ind. 3j 6 17 10 1 6+17-20 = 3 6-1-2=3 Thrice ext. ind. 3K 3 21 12 1 3+21-24 = 3-1-2=0 Not stable. 3L 5 45 24 1 5+45-48 = 2 5-1-2=2 Twice ext. ind. 3M 3 33 18 1 3+33-36 = 3-1-2=0 Determinate 3N 8 56 32 6 8+56-64 = 8-6-2=0 Determinate. 3o 4 10 7 2 4+10-14 = 4-2-2=0 Determinate. 3p 10 16 13 8 10+16-26 = 10-8-2 = Determinate. 3<> 3 9 6 1 3+ 9-12 = 3-1-2=0 Case of infinite stress. 3R 8 3 4 3 8+ 3- 8 = 3 8-3-2=3 Thrice ext. ind. CHAPTER II THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES ART. 4. ELASTIC DEFORMATIONS, FUNDAMENTAL EQUATIONS Elastic Deformations Let S = total stress in any member of a frame resulting from any cause, or causes, designating tension by + . I = length of any member when its S = 0. Al= change in I due to stress S, + for elongation. F = cross-section of any member, prismatic in form. t = a uniform change in temperature in degrees, + for rise. = coefficient of expansion per degree of temperature. E = modulus of elasticity. fS/Fumt stress in any member. JZ/Z=relative elongation. l/FE =|0=the extensibility, frequently employed for brevity. Then according to Hooke's law and within the elastic limit of the material, EF E from which SI (4A) This equation represents the elastic deformation for any member of any frame and is a fundamental elasticity condition. Referring now to Fig. 4A, let AB be any member of any frame which, as a result of elastic distortions of the frame, is made to undergo displacements Ji at A and J 2 at B and a change in its original length of Al. The new length of the member is then I + M and its new position may be shown as A\Bi. Since the displacements Ji and A% are very small compared with the length AB, the two lines AB and A\BI are assumed parallel for the present purpose. The member and its displacements are referred to rectangular axes in Fig. 4A. Then FIG. 4A. 11 12 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II which, differentiated, gives 2ldl=2(x b x n ) (dx b -dx a ) +2(y b -y a ) (dy b -dy a ) . Also from the figure x b x a =l cos a. and y b y a ^l cos /9, which values substituted in the previous equation, dividing through by 21, and replacing the differentials by small finite increments A, gives Al = (Ax b - Jx a ) cosa + (4y b - Ay a } cosp = -- + etl, ..... (4s) tir which is a fundamental elasticity condition for any frame. Hence, every frame will involve twice as many of the unknown, subscript-bear- ing displacements Ax and Ay, as there are pin points, while there will be as many equa- tions of the form Eq. (4e), as there are members in the frame. It will now be shown that any frame, whether involving redundancy or not, is capable of anatysis by proving the following: For any frame in stable equilibrium, there are as many possible condition equations as there are unknown quantities, provided the elasticity conditions Eq. (4s) are included. This will be true of a statically determinate frame without including the elasticity conditions. The ultimate analysis of any frame includes the determination of the reaction forces or conditions; the stresses in all the members; and the deformation of the frame as a result of these stresses. The deformation is considered solved when the displacements Ax and Ay, of all the pin points, are found. It is assumed that for any structure under consideration, the externally applied loads P, the changes in temperature t and the abutment displacements Ar (if any exist) are all known and that wherever movable connections or roller bearings occur, these are frictioriless. Let S\, 82, 83, etc., be the stresses in the several members meeting in some particular pin point m. ct\, (Xz, 3, etc., be the angles which these members make with the x axis of coordiantes. /?i, /?2, /?3, etc., be such angles with the y axis. P x =the sum of the components parallel to the x axis, of all the external forces P acting on the point m. P v =the sum of the components parallel to the y axis, of all these forces P. p, m and 2r as previously defined in Art. 3. Then (4c) cos ,5=0 because for a state of equilibrium, thje sum of the horizontal and vertical components of all the forces acting on any one pin point must respectively equal zero. Hence, for every pin point of a frame, two equations of the form of Eqs. (4c) may be written, expressing equilibrium of the internal and external forces for that pr'- ART. 4 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 13 Also, one equation of the form of Eq. (4s) may be written for each member of a frame. Besides these, there will be one equilibrium equation for each reaction condition. Therefore, every stable frame affords 2r+m+2p condition equations of the first degree, involving as many unknowns as there are equations, thus: 2p equations of the form (4c) involving ra unknowns S. m equations of the form (4B) involving 2p unknowns Ax and Ay. Zr equations for the reactions involving Zr unknowns R. Total 2p-fra + Zr equations, involving 2p+m + r unknowns. It follows then that the stresses S, the reactions R of known direction, and the Ax and Ay projections of the pin-point displacements, may all be represented as linear func- tions of the horizontal and vertical components of all the external forces P, plus similar functions of assumed temperature changes t, plus linear functions of the abutment displacements Ar, Jr 2 , Ar 3 , etc. Thus any unknown function of any frame may be expressed by an equation of the form Z=f(P l ,P 2 , P 3 , etc.)+/i(0+/ 2 (M,M, ^ 3 , etc.), ..... (4o) in which the coefficients are independent of the values P, t and Ar, but depend on the lengths and directions of the members also on E, e and the manner in which the frame is supported. The law of the summation of similar partial effects. In Eq. (4o), every set of causes or conditions P, t and Ar, produces a partial value Z r for Z and the ultimate total value of Z=Z' +Z" +Z'", etc., is the sum of all the partial values or effects resulting from the respective sets of independent causes or conditions. Thus each effect, such as a stress, whether due to loads, temperature or abutment displacements, may be ascertained or invest- igated by itself and the sum total effect Z will then be the sum of the several similar partial effects. This is the law of the summation of similar partial effects resulting from various causes and is fundamental to the analysis of all structures involving redundancy. The law of proportionality between cause and effect. Eq. (4o) , being true for any set of effects, would remain true for any multiple of these effects. Hence, if the loads are doubled the resulting stresses will likewise be doubled, etc. Therefore, the law of pro- portionality holds true between the causes and their effects. Thus, if a set of loads P produces stresses S in the members of a frame then another_set of parallel loads P, acting at the same points, would produce stresses S which are P/P times as great as^the stresses S. Or, if a single load unity, acting on a point m of any frame produces reactions Ri, and stresses Si, then a load P m acting at the same point will produce reactions R = P m Ri, and stresses S =P m Si. 14 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II ART. 5. GENERAL WORK EQUATIONS FOR ANY FRAME The general work equation, Clapeyron's law. A loaded frame is a machine in equilibrium and within the limits of proportional elasticity its deformation varies directly with the magnitude of the superimposed loads. The question of deformation does not enter into statics and hence a general and comprehensive treatment of the elastic frame must necessarily involve the principles of mechanics. , For example, given a frame with definite loading and supports, constituting a sys- tem in external equilibrium. The frame in turn will undergo certain deformations which will steadily increase in proportion to the internal stresses created in the members by the external forces as they are gradually applied. The final deformation will occur in the instant when the external forces are exactly balanced by the internal stresses, pass- ing into the condition known as static equilibrium. While this deformation is taking place the applied loads travel through certain distances which are the paths or displacements of the points of application of the loads. Thus a certain quantity of positive work of deformation is performed by the external forces. The internal stresses in the members must accommodate themselves to the deformed condition of the frame and in resisting this action must produce negative work of deformation. In the instant when static equilibrium is established between the loads and stresses the positive and negative work of deformation produced in the same interval of time must exactly balance. Let A e =the positive or externally applied work of deformation. .4;= the negative or internally overcome work of deformation. P=any externally applied loads including the reactions. $=the stress in. any member due to the loads P. M ihe change in length of any member due to the stress S. =the displacement of the point of application of any force P measured in the direction of this force. A positive amount of work is always produced when the force and its displacement act in the same direction. The product %Pd represents the actual work produced by a force gradually applied and increasing from its initial zero value to a certain end value P, thus exerting only its average intensity during the entire time of traversing the path to perform this work. Hence, for all the external forces acting on any frame, the total positive external work of deformation would be A e =^Pd. Similarly %SAl represents the actual work in any member subjected to a gradually increasing stress of end value S, with an average intensity S/2 during the entire time while producing a change in its length of JZ. Hence for all internal stresses in any frame the total negative work of deformation would be A t -=^2$JZ. By the " doctrine of the conservation of energy," the applied work must equal the work overcome, hence (5A) or ART. 5 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 15 which is Clapeyron's law (1833), and may be stated as follows: For any frame of constant temperature, and acted on by loads which are gradually applied, the actual work produced during deformation is independent of the manner in which these loads are created and is always half as great as the work otherwise produced by forces retaining their full end values during the entire act of deformation. By regarding any elastic body as composed of an infinite multiplicity of members, it is readily seen that Clapeyron's law applies equally to frames and solid web elastic structures when properly supported. Clapeyron's law finds extensive application in problems dealing with deformations of framed structures. Cases involving dynamic impact would imply a certain amount of kinetic energy in excess of the applied work of deformation. That is, the applied forces have some initial value, greater than zero, while the stresses are still zero. This would not be repre- sented by the work Eq. (OA), wherein A e Ai=Q, but would give rise to the following equation : Ae-Ai-jPf, ..... . ..... (OB) indicating a state of accelerated motion instead of one of equilibrium. For problems under this heading see Art. 16. The law of virtual work will now be considered. It is a general law, permitting of more varied application than does Clapeyron's law. The law of virtual work was first enunciated by Galilei and Stevin, and in its more general form by Joh. Bernoulli. Lagrange (1788) reduced the law to an algebraic expres- sion for which the following derivation may be given. According to the general law expressed by Eqs. (4c). all forces (and stresses) acting on a pin point of any structure in equilibrium will have components parallel to any axis and the sum of such a set of parallel components must be zero. Calling o: the angle which any force or stress P makes with the axis chosen, then for any pin point 2P cosa=0. Now if a displacement J, parallel to the axis, be arbitrarily assigned to this pin point and assuming equilibrium to continue, then the product of J with the sum of the com- ponents must still be zero. This is equivalent to multiplying the above equation by J to obtain 2(Pcosa)J=0. But the displacements A cos a=d are the projections of the displacements J on the directions of the several forces, hence 0, ............. (5c) wherein P signifies that the forces are in every sense independent of the displacements d. This is the law of virtual work, and is applicable to any point or group of points and hence to any frame or group of bodies. The displacements d may be any possible dis- placements whether or not subject to the law of elastic deformation, provided equilibrium exists. KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II Stated in words this law implies that for any point, frame or body acted on by loads P in an established state of equilibrium, the sum total work performed by these forces, in moving over any small arbitrary but possible displacements o, must always be equal to zero. If the time element is introduced into Eq. (oc) giving SPo/f =0, the law of virtual velocities is obtained. Professor Otto Mohr's work equations. Proceeding from Eq. (5c), Professor Mohr, in 1874, developed two equations which furnish the most comprehensive laws for framed struc- tures. These equations are now derived. Given any frame carrying the arbitrarily assumed loads PI, P 2 , P 3 , etc., which in combination with temperature changes produce stresses Si, *S 2 , 83, etc., in the several members of the frame and thus constituting a system of forces in static equilibrium. All the pin points of the frame are now supposed to be subjected to small, arbitrarily assigned displacements A\, J 2 , J 3 , etc., which may have been produced by some other system of actual loads PI, P 2 , P 3 , etc., entirely independent of the loads P. The J's will naturally vary in amount and direction for each pin point. According to the la\vs for static equilibrium, the sum of the components, taken in any fixed direction, of all forces acting on any pin point of a frame must equal zero. Choosing for the fixed direction the displacement J lr then for any particular pin point, Fig. SA, the follow- ing equation may be written: cos which multiplied through by JiP x cos 61 + Ji 5 cos 0=0; i gives cos 0=0. But from the figure, A\ cos #1 = for the two ends of the same member give the sum Jl, or actual change in the length of such member. Hence, the terms containing the stresses S will give the sum where the minus sign is due to the fact that the stresses applied to pin points as external forces are always opposite in direction with their respective JZ's. The final sum equation for all pin points then becomes or 2P3 = 2SM, ............ (5s) which may be said to represent a condition of elastic equilibrium as distinguished from static equilibrium. The law thus expresses the equality between the external and internal virtual work of deformation for real displacements and arbitrary cases of loading, provided elastic equilibrium exists. The external forces P necessarily include the reactions due to the applied loads P. However, the work of the reactions is always zero when the abutments are immovable. In the general case involving abutment displacements Jr in the directions of the reac- tions R produced by loads P, Eq. (OE) may be written .... (OF) It should be repeated that the displacements d, Ar and Al are actual and mutually dependent on the same causes, such as the actual loads P and temperature changes. The arbitrary loads P, reactions R and stresses S form a system in elastic equilibrium which is independent of the actual loads P, reactions R and stresses S, and hence inde- pendent of the actual displacements. In the special case where the forces P, S and R become identical with the forces P, S and R, Eq. (5r) becomes v _ v - y S2 * EF' which is in accordance with Clapeyron's law. Eq. (5r) is the fundamental law of framed structures and includes all conditions of equilibrium of the external forces; of the external forces and internal stresses; and of all relations existing between the stresses and the distortions of any frame. Professor Mohr's second work equation serves the purpose of determining any displace- ment d m produced by any case of loading. It follows from Eq. (5r) by allowing all the arbitrary loads to vanish and substituting therefor a single load unity and the stresses Si and reactions RI resulting from such unit load. The new work equation then becomes is KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II wherein the unit loading, henceforth called the conventional loading, may be a single force unity or a moment equal to unity. In the latter case the o will represent a rotation measured in arc. By dividing Eq. (5n) by unity the stresses Si become abstract num- bers and the equation will read The applications of this law, universally known as Mohr's icork equation, will be shown in the following articles 6 and 7. The work represented by the conventional loading will generally be designated by W. ART. 6. DISPLACEMENTS OF POINTS. STATICALLY DETERMINATE STRUCTURES BY PROFESSOR MOHR'S WORK EQUATION The relative displacement 8 m between any two pin points mi and m, of any frame Fig. GA, when Al is given for each member of the frame, may be found by applying Eq. (5i) . If the abutments undergo known displacements Ar, as a result of the given actual loading, this effect on the required d m must be included. Likewise temperature displacements may be considered, but the problem will first be treated by neglecting this effect. Now assume two unit forces, applied at the points mi and m respectively, and, acting in opposite directions along the line tn\ < m. The directions of these unit forces should be so chosen as to make the conventional work l-o m a positive quantity. This means that if d m is an elongation, then the unit forces must be so applied as to elongate the distance m\m. This rule will be universally applied to determinate structures as well as to indeterminate for redundant members or conditions. A negative result will indicate an erroneous assumption in the direction of the unit conventional load. The stresses Si and abutment reactions R i resulting from the two unit loads acting at mi and m are now determined from a Maxwell dia- gram or by computation. The work Eq. (5n) applied to this assumed or conventional loading then gives, after substitution of values Al=Sl/EF, excluding temperature effect, FIG. GA. (GA) The stresses S are those produced in the structure by any real case of loading, as loads P, temperature changes or abutment displacements, giving rise to the actual changes Al in the lengths of the members. ART. 7 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 19 Eq. (6A) then offers a solution to the above problem, because all the quantities except o, n are known or can be found from the given data. If temperature effects are to be included then Eq. (4A) gives the value for JZ, thus furnishing the final equation ~+eU)-ZR l Jr ..... \ . . (6 B ) &V / In a similar manner the relative displacement of any pair of points, the deflection of any point, or the angular change between any pair of lines of any frame, resulting from given changes M of all the members of the frame, may be found by assigning such con- ventional loads as will produce the conventional work 1 d on the frame at the point or points in question. See also Art. 9 for other examples of this class. ART. 7. INDETERMINATE STRUCTURES BY PROFESSOR MOHR'S WORK EQUATION A truss, according to Chapter I, may include more than the statically necessary number of members or reaction conditions, and is then called statically indeterminate. It is now proposed to show the manner in which Mohr's work equation may be employed to find the stresses and reactions in such a structure, loaded by any system of loads P concentrated at the several pin points of the frame. The structure must be so constituted that when all redundant members and reaction forces are removed, the remaining frame will represent a statically determinate structure, including the necessary conditions for proper support. This determinate frame will always be called the principal frame or system.. Now let X a , Xi, X c , etc., represent the stresses produced by the applied loads in any redundant members or supports, as the case may demand. When these stresses are applied to the principal system, together with the external loads, the resulting stresses in all the principal members will be identical with those produced in these same mem- bers by the original loading of the whole indeterminate frame. This will also be true of the deformations. It follows then that the loads P applied to the indeterminate frame, and the loads P and X applied to the principal frame produce identical stresses and deformations in the principal members. Also, the deformations are in each case definitely fixed by the elastic changes Al of the necessary members of the principal system. Hence, for applied loads only (without temperature effects) the work of the stresses in the principal frame is always greater when a redundant member is omitted. The work which a redundant member can do must necessarily lessen the work otherwise required of the principal system. If, then, the elastic displacements of the several pin points representing the points of application of the redundant forces X, are found for the principal system, these dis- placements will suffice to solve one elasticity equation for each unknown X. This may be done in two ways: (1) By applying Mohr's work equation to the indeterminate sys- tem, and (2) by finding the requisite elastic displacements of the principal system from Maxwell's law. The solution by Mohr's work equation will now be given. 20 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II The following definitions of terms will be strictly adhered to in all succeeding discussions. Let S =the stress in any member of the principal system. >S> =the stress in this member due to the loads P when the several redundants X and t are all zero, to be known as condition X=Q. a =the stress in this member of the principal system when no load other than X a = 1 is active. Condition X a = 1 . Sb=ihe same when no load other than Xi = l is active on the principal system. /S c =the same when no load other than X c = l is active on the principal system. j=the stress in the same member caused by a uniform change in temperature. R t =& reaction produced by a uniform change in temperature. M t = a moment produced by a uniform change in temperature. X a , Xb, X CJ etc., the stresses in the redundant members or reactions. R , R a , Rb R c , etc., are denned like the S's with like subscripts, but represent reaction forces instead of stresses. da, <>b, $c, are changes in the lengths of the redundant members X a , X b , X c , respectively. The forces X a , Xb, X c , etc., may be the stresses in redundant members or they may be redundant reactions as in the case of fixed arches or continuous girders, etc. Each of the above cases of conventional loading will be known as conditions. Thus condition X =0 will mean that all the redundant conditions are removed, while condi- tion X a = l will signify that this force alone is applied to the principal system and all other X's, S t , and P's are removed. The stress S, in any member of a frame involving redundant conditions, is a linear function of the loads P, X a , Xb, X c , etc., all treated as external forces applied to the principal system. This follows because all conditions of equilibrium are represented by linear equations. Hence, according to the law of the summation of effects expressed by Eq. (4o) , the general equation for stress in any member of a truss involving redundancy, would have the form wherein ax =the stress in any member due to the external force X a acting alone on the principal system. Similarly 5 6j .=the stress in the same member due to the external force Xb acting alone on the principal system, etc. But by the law of proportionality S ax =S a X a , Sb x =SbXb and S cx =S c X Ci etc. Hence, the general equation for any case of redundancy may be written : for the stress in any principal member /S=*So S a X a SbXb S C X C , etc., +/S< + for any reaction of the principal system R=R R a X a Rb Xb R c X c , etc., -\-Rt ...... (7 A) and for any moment on the principal system M=M -M a X a -M h X b -M c X c , etc.,+M ART. 7 THEOREMS, LAWS, AND FORMULAE FOR FRAMED STRUCTURES 21 In these equations the quantities S , R and M are all linear functions of the externally applied loads P, while all the stresses, reactions, and moments bearing subscripts a, b, c, etc., are constants due to conventional loadings and are absolutely independent of all external loads P and X. The work of any member or reaction is obtained from Eqs. (7 A) by multiplying both sides of these equations by the deformation which each sustains. Thus, neglecting temperature effects, the work equations are SM=[S -S a X a -S b X b -S c X c , etc.]JZ RJr=[R -R a X a -R b X b -R c X c , etc.]Jr The negative signs in Eqs. (7 A) and (7e) , indicate that the quantities involving the redundants X must always be of opposite sign to the stress S resulting from the loads P, for reasons above stated in this article. Hence the increment of work performed by the redundants X, as shown by Eqs. (7s), is always negative with respect to the work performed by the stress S because the forces X are classed with the external forces. The conventional unit loadings are always applied in the opposite direction to the forces they replace, and this in turn puts these (negative) unit forces in the same direc- tion as the displacements d which they accompany. Hence the work 1 o is always positive. This is exactly the same as the case illustrated in Art. 2, where the unit load was independent of all redundant conditions. Therefore, whenever in the following a conventional unit load represents a redundant condition, the direction of this unit load must be taken in a direction opposite to the force representing the stress in such member. This is equivalent to saying that when a member X elongates under stress, then the unit load X = l must be so applied as to move the adjacent pin points apart, and the converse when the member shortens under stress. Whenever the direction of a redundant force cannot be correctly foreseen, then some assumption is made which, if it be erroneous, will result in a negative value for such redundant X. The following illustration is given to make these points clear and to avoid confu- sion of ideas in all future problems. Figures 7A to 7E represent an indeterminate structure involving three redundant conditions as may be verified by applying Eq. (3c) to the problem in Fig. 7A. The truss is supported by hinged bearings at A and D and by columns at B and C, all of which may be subjected to certain displacements which may be estimated from the conditions of the problem. Figure 7s, then, represents the principal system carrying loads P, when all redundant conditions are removed and shows condition X =0. The other figures represent, respec- tively, the conventional loadings X a = 1 , X b = 1 and X c = 1 , the three X's being the redundant conditions. Note here the directions which the unit forces take. They are negative 22 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II with respect to the forces they replace, but in the same directions as the displacements which their points of application undergo as a result of the loads P. The reaction displacements are assumed to be downward by amounts A A, AB, AC, and AD for these supports respectively, and it is also assumed that A and D move apart by an amount AL. Temperature effects will be taken up later being omitted for the present. GIVEN CONDITIOM CONDITION X=O. FIG. 7A. CONDITION Xf I. FIG. 7c. FIG. 7o. CONDITION X c -l. FIG. ?E. The various reactions in the following table, given for the four conventional load- ings, are now found for the principal system, which in this case is a three-hinged arch. For Condition. A B C D H* V f\ 2P(L-e) D (\ 2Pe A n L-P(L-2e) Ao L .DO U o vJ D L H 2h ,, Uh-J:} A-a V, 1 ^o = l(L-d) B a = v R, n a=0 c', n L> a \J D h - ld fla- il H h ld - Afc 1 V 1 A b L A 1 ' d Dfe U Bn <-& i L D KL-rf) Hb 2h l.d A c 1 Ac ~ L c (J c U Dc L Hc " 2h * Obtained by taking moment equation about the middle hinge O, in a clockwise direction. ART. 7 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 23 Hence from Eq. (7A) for reactions, these values give: -e L-d d A=A - =0 C=0 and h k d d H = H H a X a H b X b H C X C =H X a ylXb FJT S & & a X a &bXi) S C X C . . . (7c) The work equation (GA), when applied to each of the three redundant forces X a , X b and X c and observing that the conventional work 1-d in each case is positive, gives = + 1 AL - + 1 AB (7D) The values for ^R Ar, in each of the Eqs. (?D) , may also be written out from the above tabulated quantities for the three conventional loadings X a = 1 , Xb = l and X c = I . They are: h A b AA-D b AD-H b AL L d . . d .^ d -A C AA-D C AD-H C AL d . L d . d . (7B) It should be observed that when the direction of a force is opposite to the direc- tion of the displacement over which the force travels, then the product which represents work is always negative. This is the case with all the work quantities in Eqs. (7E) . Hence, when these are substituted into Eqs. (7o) they become positive. The quantities 2>S a Al, HSbAl and ^S C AI still remain to be evaluated. Eq. (4A) gives M=Sp + ett, where S=S S a X a -S b X b -S c X c from Eq. (7 A). Then by substitution M=[S -S a X a -S b X b -S c X c ]p+dl, which gives the final value . . . (7c) 24 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IT Similar expressions follow for 2S&JZ and 2$ C JZ, and these substituted in Eqs. (7o) give for the values d: d a = . . (7n) d c = wherein the summations, except those of the reactions, include only the members of the principal system. Now d a , 5, and d c being the changes in the lengths of the three redundants, these may be evaluated from Eq. (4A) in terms of the lengths, areas and stresses of these members (or reactions) and become -A- , , , . (73) Hence all the terms in Eq. (7n) are now known except the three redundant forces X a , X b and X c , and having three elasticity equations involving only these unknowns, the latter may be found by solving Eqs. (7n) for simultaneous values of the X's, with the aid of Eqs. (?E) and (7j) . It is thus seen that the abutment displacements are of vital importance in determin- ing the magnitudes of the stresses in any statically indeterminate structure. When- ever these displacements cannot be determined with any degree of certainty, or when small displacements indicate large resulting stresses, then such structures should not be built. This applies particularly to fixed arches and continuous garders. Eqs. (7n), based on Professor Mohr's work equation, thus furnish a means for the analysis of any statically indeterminate structure. ART. 8. INDETERMINATE STRUCTURES BY MAXWELL'S LAW The application of Maxwell's law to the same analysis will now be given by express- ing the summations in Eqs. (7n) in terms of the elastic deformations of the structure instead of the stresses in the members. Leto ma =the displacement of the point of application m of any load P, n , in the direction of this load, when the principal svstem is loaded with only Xa = l. d mb =fi similar displacement of the same point m for the conventional load- ing X 6 = l. d nlc =ihe same for condition X c = l. daa =the change in length of the member a for condition X a = l. d^ =the change in length of the same member a for condition X b = l. doc =the similar change for condition X c = \. dba = the change in length of the member b for condition X a = 1 . dbb =the change in length of the member 6 for condition Xb = l. ART. 8 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 25 dbc = the change in length of the member 6 for condition X c = l. d ca =the change in length of the member c for condition X a = l. d c t, =the change in length of the member c for condition X& = 1. d cc =the change in length of the member c for condition X c l. o al =the change in length of the member a resulting from a change in tempera- ture t, when the principal system is otherwise not loaded, hence x=o, p=o. Ob<=the same for the member b. d ct =the same for the member c. For the sake of simplicity, the abutments will be assumed immovable, thus making all Jr=0. This part of the previous problem remains unchanged. The work Eq. (OE) for the condition X 0, (when the loads P only are acting on the principal system and producing the stresses S ) now becomes for the three cases of displacements > Similarly applying the work Eq. (GA) to the conditions X a = l, Xb = l and X e = l, the various displacements become 1 d aa = 2S a M u ; l-3 ab = 2S a M b and 1 3 ac = 2S a M c ; and by inserting values for Al aj 4l b and Al c from Eq. (4A) then and similarly for conditions X b = 1 and X c = 1 2 S c S a p =d c a] S S c Si,p =d c b', 2 S c S c p = d c (SB) Finally the work Eq. (6B), for temperature displacements for each of the conven- tional unit loadings, gives SS etf - 1 3 at ; 2S b dl = l-3 bt and SS c stf = 1 d ct ...... (8c) Substituting all these values from Eqs. (SA) , (SB) , and (8c) into Eqs. (?H) , the follow- ing important equations are obtained : Xcdac S# a Jr +d at Ob = ZPmdmb X a dba X b dbbX c db c T>Rb d c = 2P m 3 mc -X a d m -X h d cb -X c 3 cc - 2R c (SD) It should be noted that for the quantities on the right-hand side of Eqs. (So), the single subscripts and the second one of the double subscripts always refer to the con- 26 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II ventional loadings X a = l, X b = l and X c = l, while the first of the double subscripts always refers to a point or member of the structure. According to Maxwell's law (proven in Art. 9) the following equalities exist in Eqs. (So) : $ab = dba', dac=^ca and d bc = O cb . This means that the order of the subscripts may be interchanged at will without altering the equations, thus greatly simplifying the solution of problems. All of the displacements d, in Eqs. (So), may now be determined by four Williot- Mohr displacement diagrams (see Chapter VI), drawn respectively for the conditions X a = l, X b = l, X c = l and t = l acting on the principal system. Hence, the three redun- dant conditions X are again found by solving Eqs. (SD) for simultaneous values. It is thus seen that any indeterminate structure may be analyzed in either of two ways, by Eqs. (7n) or (SD), according to the choice of the designer or the nature of the problem. The redundants X being found by either of the above methods, all the stresses S and reactions R may now be determined from Eqs. (7 A) . In these the X's are the redun- dant forces from Eqs. (7n) or (So) and the stresses S , S a , S b and S c are those found for the conventional loadings on the principal system and are independent of the values X and of each other. The summations in all the previous equations include only the members of the principal frame. However, an equation of the form (7n) may be written to cover all members, including the redundant, and this form is frequently very useful. The work Eqs. (7o) when made to cover all members of an indeterminate frame become S# c Jr = SS c JZ ..... . (SE) These values inserted into Eq. (7o), and others of that form, give + 2S a ea 1 (8r) 2# c Ar = 2S c S p -X a ^S c Saf) -X b ZS c S bf ) - X c ^S wherein the summations extend over all the members including the redundant, and all the terms retain their previous significance. Therefore, when X a = l, then S =0, S a = l=X a , S b =0, and S c =0. The first terms of the right-hand side of each of the Eqs. (8r), according to Eqs. (SA), may be expressed in terms of the external loads as follows: 2iS S a p= 2P m $ mo ; ^S S b f>~^P m d mb ', 2)S *ST C jO=lIP m ^ 7nc , . . . (SG) and may be evaluated from Williot-Mohr displacement diagrams, Chapter VI. All the other summations may be determined once for all either from Maxwell stress diagrams, or in terms of displacements by employing Eqs. (SB) . Problems of the kind treated in Art. 6 can now be solved for any indeterminate structure by following precisely the same method there indicated except that the redundants ART. o THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 27 must first be found by one of the two methods just given, finally employing Eq. (?A), to ascertain the stresses. These stresses being found, the elastic changes in all members are computed. Then by applying a conventional unit load to the point for which the displacement is desired, determine the stresses in the principal system for this unit load and proceed as before by using Eq. (6A). It may be well to mention in closing this subject that there is usually a wide latitude n the choice of the redundant conditions. Thus, in Fig. ?A, one abutment might have >een placed on rollers, thus making the principal system a simple truss on two supports^ Then X a would have become the horizontal thrust instead of the stress in a member ART. 9. PROF. MAXWELL'S THEOREM (1864) This theorem, known as the law of reciprocal displacements, and previously men- tioned in discussing Eqs. (8 D ), establishes the mutual relation between the elastic dis- placements of a pair of points, or a pair of lines, whenever these displacements result from simultaneous conditions of loading ;_ provided that the arrangement of the membe remains unchanged and the supports are immovable. For the sake of simplicity it is assumed that the frame is in a condition of no stress and that there are no temperature changes. Clapeyron's law then applies and the equation for actual work becomes wherein the P's are concentrated loads, and the 3's are the deflections of the points on which the loads act in the respective directions of these loads. FIG. QA. FIG. OB. Each of the products \P m d m may now be regarded as a summation of work pro- duced by some group of loads such that the work of the group is exactly identical with the work rep- resented by %Pmd m . Figs. QA, 9s, and 9c will illustrate just what is meant by such groups of loads. If in Fig. OA, two equal and opposite . forces P m are applied at the points m l and m, then the resulting o m represents the relative displacement between these two points. We call this case the loading of a pair of points corresponding to the case illustrated in Fig. GA, where the loads P are each equal to unity. FIG. 9c. 28 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IT Fig. OB shows a pair of equal and opposite forces perpendicular to a line ik and producing a couple of moment 1-Pm. This represents the loading of a line. The result- ing d m represents the angular change in the line ik expressed in arc. This case may be designated as the loading of a line m and, whenP = l, it becomes the unit loading of this line. Fig. 9c represents the loading of a pair of lines m^ and m. The d m of this group of loads is the relative angular change between the two lines, or it is the change in the angle 6, which results from the conventional loading. The three groups of loadings are typical, and any particular load applied to a frame may become one load of such a group, but not of two or more groups simultaneously. For brevity, any such group of loads will be known as a loading and the corresponding o will be the path or elastic displacement of the loading. The several d's are always linear functions of the applied loads P and can be expressed as follows: d bm P m ; \ (9B> O m ~ I>M r a ~r rnb* b ~r O mln L n wherein the d's with the double subscripts are independent of the loads P, and for example o am represents the special value of d a when P m = 1 and all the other loads P are zero, all in accordance with the nomenclature employed in Art. 8. Now apply loads P m to any frame, producing stresses S m in the several members, and changes in their lengths Jl m =S m l/EF. Likewise for a system of loads P n , producing stresses S n , and changes Al n = S n l/EF. Let d' mn = the value of d m , for the point m, when certain loads P n only are active. Also, let d' nm = the special value of d n for the point nj when a certain set of loads P m only are active. Then note that d' mn = d mn when 2P n =l and o' nm = d nm when 2P TO =1. The work equation for the loads P m , stresses S m and displacements d' mn and Al n due the loads P n only, is according to Eq. (OE) S I 2jPmO mn = ^o wz Jt n == Lo^-pp. Similarly S I TiP n d' nm = 2}S n dl m ~ ^^n^W> therefore, (9c) which is Betti's law (1872) and which extends Maxwell's law to the summation of all members of a structure. If instead of systems of loads P m and P n as above, only the unit loads P m = 1 and P n = 1 are successively applied to the frame, then Eq. (9c) becomes d mn = 8 nm , ............. (9D) which is Maxwell's law (1864) and may be stated thus: ART. 10 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 29 1. The relative displacement d mn of a pair of points TOI and TO, resulting from a unit loading of another pair of points n\ and n, is equal to the relative displacement o nm of the pair of points n\ and n, caused by a unit loading of the pair of points m\ and TO. 2. The same equality exists between the relative angular changes of two pairs of lines, successively loaded with a unit loading. 3. The same equality also exists between the linear change in a pair of points result- ing from a unit loading of a pair of lines, and the angular change (expressed in arc) of the pair of lines, resulting from a unit loading of the pair of points. The practical value of Maxwell's law when applied to redundant conditions was shown in Art. 8. Its application to displacement influence lines is given in Chapter VIII. ART. 10. THEOREMS RELATING TO WORK OF DEFORMATION (a) Menabrea's law (1858), or theorem of least work. Given a statically indeterminate framework in an initial condition of no stress for which the temperature is known and remains constant; also, assuming that the supports are either rigidly fixed or permit frictionless movements, such that 2.RJr = for the entire structure, then the changes in the lengths of the members are expressed by the formula M = Sl/EF, and the work equations for the several statically indeterminate quantities X a , Xf,, X c , etc., are, accord- ing to Eqs. (SB) , EF etc., the value S in which is given by Eq. (7A.) as S = S S a X a SbXf, S C X C , etc. The partial differentiation of S with respect to X a gives likewise for the other X's, (10B) (lOc) The actual work of deformation for the entire frame, including redundant members as external forces, becomes by Clapeyron's law or Eq. (5A) (10D) 30 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II which when differentiated gives _ and by dividing through by ~dX a this becomes, after substitution of values from Eqs. (lOc) - - EF EF ' which by Eq. (10A) must equal zero. Hence, , . .. . dA . 9 A , . = and similarlv ^-^-=0 and ^^=0, .... (10E) which proves that the redundant or indeterminate conditions, reduce the actual work of defor- mation of the frame to a minimum. This is the theorem of least work. (b) Castigliano's law (1879), or derivative of the work equation. This law deals with the displacement of the point of application of a force. Any external load P m acting on a framework will, by Clapeyron's law, produce the actual applied work of deformation A=$P m d m , ......... . . (10F) when d m is the displacement of the point of application of P m in the direction of P m and within range of proportionality, or within the elastic limit of the material. Also, the actual internal work of deformation, for the entire frame, becomes, accord- ing to Eq. (10D) The partial differential derivative of A with respect to any certain external load P m is 'S 2 l\ .^ s dS_ J_ 3P m EF' wherein 3*S/3P OT is the derivative of the stress S in any member of a determinate or indeterminate structure. The general expression for S, from Eq. (lOs), by substituting for S its equivalent value in terms of external loads becomes , etc., +S m P m -S X a -S b X b -S c X c , etc., . . . (10j) in which Si is the stress in the member S for PI =1, while S a is that stress when X a = l, etc.; finally S is the stress in any particular member caused by the combined effects of all P's and Z's. The several values Si to S m are thus independent of the loads P and X. Also, the loads P and X are independent of each other. Hence, S may be partially differentiated with respect to P m or X and Eq. (10j) when so treated gives --=S m and '' == ~ s > etc ........ ( 10K ) ART. 11 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 31 Substituting this value in Eq. (lOn) then but, according to Mohr's law, Eq. (GA), when abutment displacements are zero, then (10M) In Eqs. (IOL) and (10M) the summation extends to all the members, including the redundant conditions, and hence 7)4 Ld m =-, ....... ........ (ION) which, expressed in words, means that the path o m of a load P m is equal to the partial differential derivitave of the actual work of the frame with respect to P m . Castigliano's law thus expressed is equivalent to Mohr's work equation, that is, leading to the identical result by a more circuitous process. In similar manner the same law may be deduced for girders with solid webs. Mohr's law thus offers the most direct method of finding any displacement which may result from any specific cause. Castigliano's law will lead to the same conclusions by a somewhat less direct method. ART. 11. TEMPERATURE STRESSES FOLLOW MENABREA'S AND CASTIGLIANO'S LAWS When a structure is subjected to a uniform change in temperature t then, from Eq. (4A), This value of M when substituted into Eq. (10D) for Sl/EF, gives ; ........ (llA) and the differential of A with respect to S is (llB) Similarly when the temperature effect is introduced into Eqs. (10A) then for immovable abutments as before y c cf/ 2S^l==+8 a <&-Q ......... (He) 32 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. II Also, Eqs. (lOc) apply equally when temperature effects are included, hence by divid- ing Eq. (lie) by 3A" a , and substituting values from Eqs. (lOc), then Eq. (11s) becomes i 9 A a ~ EF But, by Eq. (lie) this expression is equal to zero, hence, as before "dA 3A o^r- = and similarly ~^- =0 etc., ...... (HE) oX a oXb which proves Menabrea's law applicable to the general case of stress from temperature and applied loads, provided the abutments are rigid and immovable. The introduction of the temperature factor SefcS/ from Eq. (!!A) into Eqs. (10o), to (ION) will suffice to prove that Castigliano's law also applies to the general case includ- ing temperature effects. It is not deemed necessary to repeat the transformations here. ART. 12. STRESSES DUE TO ABUTMENT DISPLACEMENTS Abutment displacements produce stresses which follow Menabrea's and Castigliano's laws. This is readily seen when it is considered that the supporting elements may always be replaced by linked members which take up these displacements and undergo elastic deformations. The immovable supports are then outside of these connecting links as illustrated in Figs. ID, IE and IF, and the links are counted with the structural members. If then the abutment displacements are defined as distortions Jr in these several links, which are in every way equal to the former displacements, the resulting effect on the structure remains unchanged. Therefore, when the distortions Jr must be considered in the computation of the X's and d's, it is only necessary to extend the work equations to include these links. It is also clear that these links may have any desired sections, lengths or values of E, as may be required for metal or masonry supports. Thus, let r be the length of a supporting link; R the load which the link carries; Ar the change in length of this link; F r the cross-section of this link; e r the coefficient of expansion ; t the change in temperature from the normal; E r the modulus of elasticity for any material. then from Eq. (4 A) The general work Eq. (11 A) then becomes V027 ART. 12 THEOREMS, LAWS, AND FORMULA FOR FRAMED STRUCTURES 33 wherein Ar may be regarded as a constant and Menabrea's and Castigliano's laws again apply. For the case of one external indeterminate, R becomes X a and Ar becomes o a , then from Eq. (12n) By differentiating for 8 with respect to X a and imposing the condition for minimum this becomes s 3S 3 n9 v :t + da ~ Also, S=S S a X a and its differential is Substituting the value from Eq. (12o) into Eq. (12c) and solving for X a then the latter becomes 2S S a ~ + 2dlS a -3 a Xa= ~' ........ (12E) EF which might be used as the expression for the horizontal thrust of a two-hinged arch where X a is the redundant thrust. CHAPTER III THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS ART. 13. GENERAL WORK EQUATIONS In the previous chapters no consideration was given to solid web structures or other isotropic solid bodies only in so far as was necessary in demonstrating the general laws and theorems of framed structures. While it is true that the foregoing discussion is generally applicable to all isotropic solid bodies which are supported in any of the ways given in Chapter I, and stressed Avithin the elastic limit by externally applied loads, yet the formula? previously given will require some modifications to better adapt them to solid web and other massive and homogeneous structures. Also, the introduction of shearing stress now enters as a further complication. As was previously mentioned, all structures which are purely isotropic an involve only external redundancy. It is, therefore, desirable to take up this subject to the extent, at least, of giving the special formu- lae applicable to any solid in which the physical properties of the material are pre- sumably uniform in any and all directions. Such bodies are called isotropic solids. It will scarcely be necessary to prove at length all of the theorems and laws pre- viously given for frames, but a passing refer- ence at the proper time will be deemed suf- ficient proof of their general acceptance. The elastic deformation resulting from given stresses in an infinitesimally small particle of a body is a certain and definitely measurable quantity and is dependent only on the magnitude of the given stresses and the physical properties of the material. If then Fig. 13A represents a small parallelepiped of mass, referred to axes X, Y and Z, and having dimensions dx, dy and dz, coincident with these axes respectively, then the stress acting at the corner m is determined in magnitude and direction by nine components, viz., three normal stresses and six tangential stresses acting in the three coordinate planes and positive in the directions of the arrows. FIG. ISA. ART. 13 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 35 Let f x , fy and f z be the unit normal stresses (tension or compression) in the direc- tions X, Y and Z, respectively. Also, let - xy and r xz be the unit tangential stresses in the YZ plane; r yx and r yz those in the XZ plane; and - zx and ~ zy those in the XY plane. The first subscript indicating the normal stress to which they belong, and the second subscript referring to the axis. The two unit tangential stresses intersecting in any one point are equal, in each case, otherwise the body would rotate about its center of gravity, which is the intersec- tion of the three normal unit stresses. Also, because the three normal stresses and four tangential stresses, ~, fz , r xz , r, y and T ZX , cannot have any moment about such a gravity axis OZ', hence, the moment of the two remaining stresses -c yj . and r xy must equal zero. But, the latter having equal lever arms must be equal to each other. Therefore, The resulting stress at the point m is then determined by the three normal stresses f x ,fy,fz and the three tangential stresses r x , T U , and T Z in Eqs. (13A), and these in turn determine the deformation of the parallelepiped. Let Adx, M,y and Adz represent small elastic changes which the lengths dx, dy and dz undergo, then the forces represented by the unit normal stresses / will produce the following virtual work : =f x dydzJdx+f y dzdxJdy+f z dxdyJdz = f x + Also, let j- x , fy and f z be the tangents of small angular distortions of the parallelepiped such that +f x is an increase in the angle between the Y and Z axes; f v the change in the angle between the X and Z axes, etc. Then the virtual work of the tangential forces is found thus : For the distortion of the YZ plane, the path is f x dy and the force is r x dxdz, and since the vertical force r^jjan do no work in this direction then the work of the tangential stress for the plane XZ becomes r x dxdz^r x dy, and similarly for the other two planes, whence the total work becomes x ...... (13c) The expressions Adxjdx, 4dy/dy, and 4dz/dz in Eqs. (13B) are rates of elongation, or coefficients, for which the values a x , a v and a z may be substituted. Then by making dxdydz =dV, and combining Eqs. (13s) and (13c), the total internal virtual work is obtained. Also, since by Clapeyron's law, this must be equal to the virtual work of the externally applied forces, then the fundamental work equation for isotropic bodies becomes: W = This equation is applicable to any case of related displacements and elastic deforma- tions o dr, a, and 7-, so long as these are small in comparison with the dimensions of the structure. The external forces must include dead loads and all frictional resistances which may be active at points of support. 36. KIXETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill As was previously shown for indeterminate frames, so also the various elements in Eq. (13D) may now be expressed MS linear functions of the applied loads and any redundant conditions, thus: R=R R a X a R b X b , etc. J x = J xo J xa-A-a J xb-^-bj CtC. fy = fyofyaX a ~fyb^b, etc. fz =^fzo fiaX a fzbXb, etc. ?x = T ro ~raX a T xb X. b , etc. etc., etc. Also, the virtual work of the reactions for each of the conventional loadings becomes, (see Eqs. (?E)): (13E) xa f (13P) f 2R b Ar =J [f x b( etc. Then by writing Eq. (13D) for a load P a = l the following is obtained: -'ZR a Similar expressions result for 2/S^Jr, etc. The general work equation is now found by inserting for the actual distortions a and f their values in terms of stresses and the moduli of elasticity for direct and tan- gential stress. The length dx subjected to the unit stress f x and a rise in temperature of t degrees will be changed by an amount ART. 13 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 37 The two other unit stresses/, and f z will diminish the effect on a x by an amount /*+/. mK ' wherein m is the " Poisson number " (1829) which is given as 3.33 for structural steel and 3 5 for hi-h steel. Professor F. E. Turneaure gives 0.1 to 0.125 for concrete, quantity m is also denned as the ratio of lateral to longitudinal deformation an determined by experiment. Hence, a x and similarly a y and a,, also, T *> Y y and r* may be evaluated as follows. m and and (13J) and mE wherein G is the modulus of tangential stress or shear and has the value G 2(jn + 1) ' while m is the " Poisson number "just given. Inserting the values given by Eqs. (18,) into Eq. (13D), the fo.lowmg equat.on obtained for the actual work of an isotropic body : . . (13K) \lso by substititingthe values from Eqs. (13i) into Eq. (13n), reducing and integrat- in , tPS^/AT., etc., the following simple form is obtamed by msertmg the value from Eq. (13KJ, when temperature effect is negled VR^^- and similarly S^Jr = ^, etc. - - (13L) OA a When the abutments are immovable and no temperature effects exist then -0 and -0, etc., MCT ab<)ve fed P m , or redundant X m : ^ KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill The value of A in all the Eqs. (13L) to (13N) is that given by Eq. (13K). \Yhen the abutments are immovable the last term of Eq. (13N), becomes zero and Castigliano's law is again established. Proceeding from Eq. (13o), and writing same for two related conditions of unit loading, it is easily shown that Maxwell's law likewise applies here. ART. 14. WORK OF DEFORMATION DUE TO SHEARING STRESS For a beam of any section and any loading applied in the vertical plane YY, Figs. and 14n, the resultant of the external forces on one side of any section A A may be represented by a force R which may in turn be resolved into a normal force N and a tangential force Q acting at the point of application of R on the section A A. T,, FIG. 14A. Let M be the static moment of the normal force N about the Z axis and I z the moment of inertia of the section about this same axis. Then the unit stress/ at any point of the section is by Navier's law: N My f = J+~r (14.A) * Z If now this stress undergoes a small change df due to the differential change in R for a neighboring section, distant dx from the first, then the shear on the area 2zidx becomes 2T J * l dx=f(f+df)dF-ffdF=JdfdF. where r x is the unit shear on the differential area d-F. Assuming N constant in Eq. (HA) , and treating M as a variable, then by differentiation (14c) AKT. 14 THEOREMS, LAWS, AND FORMULAE FOR ISOTROPIC SOLIDS 39 Also, since the shear is the differential of the bending moment, then dM=Qdx (14D) Combining Eqs. (14B), (14c) and (14o), then CydMdF Qdx C , 1- x z-idx= I I ydF. 04E) J L z l z J and since J ydF is the static moment M s of the cross-section, this integration may be considered performed to obtain Return now to Fig. 14s, take any section pq perpendicular to the ?/ axis and let Q produce a unit shear r in any point i of this section. The line of this shear intersects the ?/ axis in a point H, which is determined by the tangent pH. This is a pure assumption, but from the nature of the case it is the most rational assumption to make. The shear r may now be resolved into components T U and T Z , while r x , for this sectional plane, must be zero. However, r y now has the same value as previously found for - x , and hence *-!&->' <-> Again, from Fig. MB, tan0 = = - from which r g = -r y . But tan d = , therefore, O7 i f) P P / * \ j. n ft A \ T a = Tj,( itan (14H) \ z i/ For 0=0, T 2 =0, which is the case for any surface point, the tangent to which is parallel to the y axis. From Eqs. (14r) and (14o) it follows that for a given loading and section, the shear- ing stresses depend only on M 8 and attain a maximum when the section coincides with a gravity axis parallel to the neutral axis. In practical cases it is usually sufficient to consider the shearing stress of a section as extending only over a unit length of the beam as given by Eq. (14F). The actual work produced by shearing stress alone is then found from Eq. (13K) as By substituting the value of r, from Eq. (14n), into Eq. (14i), the latter becomes /f (14K) 40 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill and by substituting for Ty its value from Eq. (14c), then Eq. (14K) may be solved for any special section. This gives in general for any section A=> and and (14L) In Eqs. (14L) ft is an involved function of the shape of the section, called by German authors the " distribution number " of the section, or " coefficient of shearing strain." This number @ is 1.2 for any solid square or rectangular section, and 1.111 for a solid circular or elliptic section. Other values will be given after illustrating the computation of this number by the use of Eq. (14L). bh 3 For a rectangle of height h and breadth b = 2zi, then 1,. = -^; F = bh; and dF = bdy, \m hence, Q and For any I section, Professor Mehrtens finds the general formula for /? as follows: {3 = M\T 2 dF = j2\--(8b 3 l2b 2 t t 3 + 6bt 2 )+(-^-\ -+t' 2 a) , . . (14ivi) wherein i= - and the special dimensions here used are shown in Fig. 14c. Professor L. von Tetmajer, in his " Elastizi- taets und Festigkeitslehre," 1905, p. 49, gives a table of values for /? for I beams ranging from 3i"-4 Ibs. to 19"-95 Ibs., 'for which /3=2.39 to 2.03 respectively. Also for riveted girders made of f" webs, 4 L's 3&"X3&"Xf" each, and 2 plates 8f" xf" on each chord. Then for _!.. __, , -Y 2.71 23$" 2.49 27 A" 2.35 The value is independent of the unit of length and is, therefore, the same for metric and U. S. measures. ART. 1.) THEOREMS, LAWS, AND FORMULAE FOR ISOTROPIC SOLIDS 41 ART. 16. WORK OF DEFORMATION FOR ANY INDETERMINATE STRAIGHT BEAM Direct ^tress. For axial or direct stress only, Eq. (13K) gives (ISA) Also, by observing that a x =f x /E, then Eq. (13n) becomes for direct stress alone, - ........ < 15B > Now ( f x dF=N, the normal direct stress, and, as dV =dFdx, then from Eq. (15A) C&HV- C N f* dx - C N f* F dx_ CN*dx J 2E a[ 'I ~2E~~J 2EF J 2EF ' and similarly from Eq. (15B) ff x df x dV_ CN-df x dx_^ (N-df x Fdx_ fNZNdx ( J E-dX a J EdX a J EF3X a J EFdXa The temperature effects in Eqs. (15A) and (15B) will now be determined on the supposition that the effect is not uniform, but as shown on Fig. 14A, where fo=the change in temperature from normal at the gravity axis of the section; J=the difference in temperature of the two extreme fibers; h = height of the section; t/=any ordinate; t=the temperature above normal, of any point of the section then (15B) Substituting this value for t into the temperature element of Eq. (15A), the latter becomes But ff x dF=N and r/ x ?/dF=M=moment of resistance of the section, therefore, dx ......... (15F) By differentiating Eq. (15?), and dividing through by 3X a , the value of the tem- perature element for Eq. (15u), is obtained thus: n . . (15G) 42 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, in The bending moment. From Eq. (HA), f x =My/I g and I z ==fy 2 dF. With these values and neglecting the temperature element, Eq. (!OA) becomes By differentiating Eq. (Ion) and dividing through by "dX a then 'M The general work equation, including all effects, may now be written out by collect- ing the several Eqs. (14L), (15A), (15c), (15n), and (lor) into the following, represent- ing the total actual work for any isotropic body by r r M Similarly the following differential expression of Eq. (!OK) is obtained from Eqs. (14L), (15s), (15D), (15.i) and (15o) or directly by differentiating Eq. (!OK) and divid- ing through by ~dX a , thus: - (loL) Equation (15L) being true for any force X a is, of course, true for a force P m , hence by substitution of the latter value and then inserting the value 3A/3P TO . thus obtained from Eq. (15L) into Eq. (13N) the following equation for elastic deflection is obtained: c Q ^Q , C t ^N j GFZP^ dx+ J t 3p- 3M The last three equations (15K), (15i.) and (15M) are the three fundamentals from which all cases of redundancy for isotropic bodies can be solved. There are always as many of these equations as there are redundant conditions X. In each of these the first term expresses the effect due to direct or normal stress; the second term that due to pure bending; the third term that due to shearing stress; the fourth term gives the effect due to a uniform rise in temperature to; and the fifth term that due to a difference At in the temperature of opposite extreme fibers. ART. 16 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 43 The object in presenting these rather long fundamental equations is not to make the subject appear complicated, but rather with a view to showing once for all combined effect from all causes, thus permitting the easy choice in combining any effects or in omitting such as may seem negligible in any specific problem. Equation (15K) may also be derived from Eq. (5n) for virtual work and offers very useful applications. Thus assuming a column subjected to a thrust N ttt and a bending moment M m the virtual work on the column from Eq. (OH) becomes where A r rt and M a are due to the usual conventional loadings. But A^te nd . MJx lioncG W . C l N m N a dx , r lM rnM a dx W = \-d m EF r l M m M a dx J El When the virtual work becomes the actual work A then Eq. (15N) becomes identical with Eq. (15K) term for term. ART. 16. WORK OF DEFORMATION DUE TO DYNAMIC IMPACT Problems involving the deflection or strength of a structure subjected to impact, are frequently met with, and their treatment is here considered as properly belonging to the subject of the present chapter. Let LO = weight of a moving body; H = height of a fall; v= velocity at instant of impact; g = acceleration due to gravity ; a=any elastic displacement produced by the moving body in some structure. Then the work expended by the moving mass is represented, either in terms of velocity or height of fall as follows: wherein v*/2g represents the velocity height or height through which a body falls in acquiring a velocity v. When the body moves with a velocity v along a horizontal path, thenv 2 /2g will be the height to which the body would raise itself in order to expend its energy and come to a state of rest. 44 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Ill The work thus produced by kinetic energy, when it is instantly taken up by a quiescent body or structure, is twice as great as the work of the same moving body gradually applied, and hence produces twice the internal work in the body struck as would result from an equivalent static load. Problem 1. The weight w falls on top of a column, stressing the column normally, what sectional area is required in order that the unit compressive stress does not exceed /? From Eq. (!OK) the work due to direct stress is given by the first term, where N is the total static load, and to balance the dynamic energy, twice this amount is taken; hence (16B) and since o = ~ this gives for F, m Problem 2. A weight w falls from some height striking a beam resting on two sup- ports. The weight strikes the center of the beam with a velocity v; what will be the stress / in the extreme fiber for a given beam section? Shear and bending resistances are to be considered. From Eq. (!OK) the actual internal work is Let P=an equivalent static load producing the same stress / in the given beam. Then, for any point of the beam of depth h and span I, P My Plh 8/7 -x; also /=--* = __ or P = -^- ..... (16s) Hence i , + JM 2 dx_ P 2 rt 2 _ P2/3 _2f 2 ll i_ ~2ET~4EI Jo X WEI~3Eh 2 (16r) p Also, for Q = the second term of Eq. (16D) becomes =dx = _ 2GF 4GFjo SGF GFlh 2 ....... 2 The sum of Eqs. (16p) and (16c) gives A, according to Eq. (16D), and this must equal a}V*/2g. Therefore, AKT. 16 THEOREMS, LAWS, AND FORMULA FOR ISOTROPIC SOLIDS 45 Her e G=modulus of shear = 1 =0.385 E for structural steel (see Eqs. (13j). The coefficient of shearing strain /? is given by Eqs. (14L) and (14M). By applying Eq. (16n) to a steel beam 2.5X2.5 inches by 78 inches long, it is found that when shear is included, the stress / is about f per cent smaller than when this term is omitted, showing that the internal resistance due to shear is very small and usually negligible. ' This is also true when computing deflections. ' In the same manner all problems involving impact may be solved. Suppose a ship weighing w tons, and moving with a velocity of v feet per second were to collide with a fixed structure, then the work which the ship is capable of expend- in- is represented by a>v*/2g ft.tons. This work may be expended in injuring the structure or the ship itself, a condition depending on the relative strength of the two bodies, which must be ascertained form the design and construction of each. CHAPTER IV INFLUENCE LINES AND AREAS FOR STATICALLY DETERMINATE STRUCTURES ART. 17. INTRODUCTORY Professors MOHR and WINKLER, in 1868, published simultaneously the first treatises on influence lines describing, at that early date, practically all the uses and applications of these lines known at the present time. Professor J. Weyrauch, in 1873, introduces the name influence line not used by Mohr and Winkler in their earlier work. Professor Mohr, in a series of articles published from 1870 to 1877, was the first to apply influence lines to deflections and to redundant conditions. Professor Geo. F. Swain, in 1887, gave the first treatise on the subject in English. An influence line is the graphic representation of some particular effect produced, at a certain point of a structure, by a single moving load occupying, successively, all possible positions over the entire span. The effect may be the shear, the bending moment or the deflection at any certain point of the structure; it may also represent any reaction force or the stress in any member. The single moving load is usually taken equal to unity, though in certain special cases it may be desirable to use any load P. An influence line represents a certain effect for a certain point or member of a structure and for any position of a moving load, while a shear or moment diagram represents effects due to a single position of the load or loads for all points of a structure. The ordinate to any influence line is thus an influence number or factor, usually designated by TJ when stresses are dealt with and sometimes by o when deflections are under consideration. As a matter of convention, all positive influence line ordinates will be laid off down- ward from the axis of abscissae. A load point is any particular one of the many possible positions of the moving load. A summation influence line is one which gives the total effect at some particular point due to a train of concentrated loads. The actual loads are here employed and each influence ordinate is made to represent the summation of influences of the same kind for a certain position of the train of loads. Usually this position is taken so that the first load is over the point for which the influence line is constructed. An influence area is the area included between the influence line, the axis of abseissce and the two end ordinates. The maximum effect is always produced when the load point coincides with the maximum ordinate of the influence area. Hence influence lines are eminently suited 46 ART . 17 INFLUENCE LINES AND AREAS 47 to the solution of all problems relating to position of moving loads for maximum and minimum effects and stresses. The load divide is such a load point, the ordinates to either side of which have opposite signs. Hence, the influence ordinate at a load divide passes through zero and the influence line intersects the axis of abscissae at such load divide. It is assumed that if a unit load produces some effect 77 at a certain point of a given structure, then a load P will produce the effect PTJ at this same point, so long as the material is not stressed beyond the elastic limit. This follows from the law of proportionality stated at the end of Art. 4. Hence, the total effect Z produced by a system of moving loads, PI, P 2 , etc., will be the sum of the effects Pi] of all the loads and the maximum and minimum value of Z will be determined by the position of the moving loads. The total effect is thus expressed by the following equation when the case of loading is simultaneous and within working limits : (17A) Therefore, having given the influence line for a certain effect on some structure, then the total effect, due to any system of loading, is easily found by a summation of the prod- ucts of loads into corresponding influence ordinates or numbers for any desired positions of the loads. For a uniform moving load p per foot of length, Eq. (!?A) becomes wherein the integral represents the area of the influence polygon between the end ordi- nates of the uniform load. The equation of any influence line may be written out by expressing the desired function for a particular point in question in terms of a moving load unity acting at any variable distance x from one end of the structure taken as origin. Since influence lines represent all possible effects it is readily seen that maximum and minimum stresses may be found from the same lines. Therefore, only one half of a symmetric structure requires analysis. The left half is usually treated, as a matter of conventional uniformity. Direct and indirect loading. In the above it was assumed that the loads were directly applied to the beam, which is rather the excep- tion. Usually they are taken up by the floor system and then transferred to the panel points as load concentrations. The former case of loading will be known as direct loading and the latter as indirect loading. The influence line between two successive panel points is always a straight line, regardless of the riG 1?A system of loading or the particular influence. Let Fig. 17A represent two successive floor beams of any truss and the loadP = IS KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV is transferred to the points a and b by the floor stringer ab. The load effects PI and Po, which are carried to the panel points a and b, are - and Now the total influence of P l and P 2 , on the structure as a whole, must be exactly equal to that produced by the resultant P. But the influence line a'b' '', extending over the panel ab, must give, by Eq. (!?A), , 17 x from which fd-: For any influence line, >?i, r) 2 and d are constants, hence Eq. (17c) represents a straight line, which was to be proven. ART. 18. INFLUENCE LINES FOR DIRECT LOADING The several influence lines for reactions, shears, moments and deflections for a beam with direct loading will now be given. They will always be known by the names indicated in Fig. ISA. Thus the A line is the influence line for the reaction A. The influence line for a reaction is of the first importance, because the other lines are generally derived from the reaction influence line. This is really seen from the circumstance that for the unloaded portion of a beam or truss, the shear is always equal to the end reaction and the moment is equal to this reaction into the distance from the section in question to the end of the beam. Hence, in drawing influence lines it is always best to consider the unloaded portion of the span to the right or left of the section as the case may be, beca-use the only external force on that side of the section will then be the reaction. In the following the moving load will always be assumed as coming on the span from the right end and the effect produced on the left half of the span only, is considered. (a) Reaction influence lines A and B. Using the dimensions indicated in Fig. ISA, the two reactions become Px' Px A= f- and B= ~T .......... For x' and x variable, both expressions represent equations of straight lines which are easily plotted. When P = l and x' =1 then A=l and for x' =0, A =0. Hence, the A line is drawn by laying off a distance unity down from A and joining this point with B. The reaction A for any load P acting at the load point m is then A m =Prj m . In like manner the B line is found, and the corresponding reaction B m for a load P at m becomes B m =Pj)' m , and i? m + fy' = l for every point of the span. ART. IS INFLUENCE LINES AND AREAS (b) Shear influence lines. The shear Q n for a certain point n to the left of the load point m is always equal to the reaction A. But for a load point to the left of n (not shown) the shear is Q n = A P = B, because P =A +B. The influence line for Q n is thus derived from the influence lines for A and B as indicated in Fig. ISA, and consists of the polygon An'n"R. The point n becomes the load divide for shears at n and hence there will always be a maximum and a minimum value for Q n depending on whether the positive or the negative influence area is fully loaded. The shear due to any single load P must change sign in passing the point n. (c) Moment influence lines. The moment for the point n is now found when the moving load is to the right or left of the section at n. Then for P.I m x>a, M n =Aa; and for M n =B(la). Thus the moment influence line is also derived from the reaction lines because the ordinates of the latter, when multiplied by a or (I a) as the case may be, give the ordinates to the moment line. Hence the bounding lines of the moment influence line are easily found. Since A and B are both unity, then the ordinate A A' --=a and the ordinate BB'=la and the lines AB' and A'B inclose the required moment influence line. Also, the ordinate at n is the ordinate of the intersection n' between the two bounding lines. Hence, if BE' should fall off the drawing, then the line ~AB' may be drawn from A to n' without finding B' '. In either case the moment influence line is then the polygon An'B with all ordinates positive so long as the point n is not outside the span, a case which will be given later. The maximum ordinate is always under n and has the value l-a(l a)/l, which offers still another construction for this influence line. The middle ordinate of the AB' \me=(l-a) which furnishes a convenient construction for this line when B' falls off the drawing. It is clearly seen that if a single load is placed at n over the maximum ordinate, then a maximum moment is produced. (d) Deflection influence lines. The deflection for a point n, produced by a single load at any point m, is given in terms of the moment of inertia of the beam section and the modulus of elasticity, as follows: FIG. ISA QEIl -* a*) P. (18s) 50 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV This is a cubic equation and when plotted for all values of x and making P = l, the influence line for deflections is obtained. In general, any deflection polygon, drawn for a load unity acting on a fixed point n of any structure, is the deflection influence line for the point n of that structure. This follows from Maxwell's law. As the subject of deflection influence lines is fully treated in Chapter VIII, no further consideration is given to it now. B ART. 19. INFLUENCE LINES FOR INDIRECT LOADING In the present case the loads are applied to a stringer and floor-beam system and are thus transferred to the beam at certain panel points as load concentrations. The beam is then indirectly loaded and since an influence line was shown to be a straight line between panel points the present case will require some slight modification of the influence lines just found for direct loading. See Fig. 19A. However, all that was said for cases of direct loading applies here and the modifications made necessary by the indirect loading occur only in the panel wherein the point n is located. As before, n is the point for which the influence line is drawn and m is any one of the possible load points for the moving loadP = l. (a) Reaction influence lines remain the same whether the loading is direct or indirect, since the point for which the influence is sought is alwavs at A or B, - 1 / which are also panel points. Furthermore, the reaction influence lines being straight over the length of the span will always be straight between successive panel points. (b) Shear influence lines. The Q line, outside of the panel ce and containing the point n, remains the same as for direct loading. But since the influence line within a panel must be a straight line, therefore, the points c' and e' must determine the influence line for the panel ce regardless of the location of the point n so long as this point is within the panel ce. Hence, the polygon Ac'e'B is the shear influence line for the panel ~ce and the point i is the load divide for this panel. The limiting values for shear are thus the same for any point n of the same panel. (c) Moment influence lines. Here again the influence line within the panel ~ce is all that requires modification in the case of indirect loading. FIG. 19x. ART. 20 INFLUENCE LINES AND AREAS 51 Hence for the same reasons just given to determine the modified Q line, the M line for the point n is now the polygon Ac'e'B, instead of the triangle An'B for direct loading. The point n is the center of moments. ART. 20. THE LOAD DIVIDE FOR A TRUSS In constructing influence lines for truss web members it frequently happens that one or the other end ordinate falls outside the limits of the drawing. There is a very simple way of locating the load divide i and thus facilitating the construction of any influence line. The method is given first and the proof follows: Let Fig. 20A represent a truss arranged for bottom chord loading, but no loads are shown. Required to find the load divide i for the panel fg necessary to determine the stress in the diagonal eg cut by the section tt. FIG. 20A. Construction. Prolong the unloaded chord member ek in the panel fg until it inter- sects the verticals through the end reactions in a and 6. Then the intersection i of the two lines af and bg will be the required load divide. Proof. Suppose the unit moving load is now located in the vertical through i. Then if i is the desired load divide, the unit load for the load point ii will produce zero stress in the member eg and the influence line ordinate for the point i must be zero. Let F and G be the panel concentrations in the points / and g due to the load P = 1 at i. Then the polygon abfg may be regarded as the equilibrium polygon for the forces A, B, F and G. Also, the resultant R of all forces on one side of the section tt must pass through the intersection n of the two included sides of the equilibrium polygon. But, the point n is the intersection of the two chords fg and ek, which point is also the center of moments for the diagonal eg. Hence the moment of this resultant R about n must be zero and cannot produce stress in the member eg, thus proving that the load P = 1 is actually located in the load divide. When the top chord is the loaded chord, a similar construction furnishes the point i', as the load divide for the member eg, according to the same proof above given. This construction applies to any structure even when one or both chords are straight. When the center of moments n falls inside the span then there is no load divide and all loads on the entire span will produce the same kind of stress in any particular web member. 52 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV ART. 21. STRESS INFLUENCE LINES FOR TRUSS MEMBERS According to Professor Ritter's well-known method of sections given in Art. 58, the stress in any member of a determinate truss may be expressed thus : (21 A) where M is the moment of all the external forces, including the reactions, on one side of the section and r is the lever arm, of the member in question, measured from a center of moments determined by the intersection of the two other members cut by the section. Hence, the influence line for the stress in an}^ member is the influence line for the corresponding moment for this member, the ordinates of which are divided by r. For any moment influence line, the ordinate at any point represents the moment due to a unit load at that point and, therefore, from Eq. (2lA), i)=M=rS. But the end ordinates for a moment influence line were formerly found to be a and a', therefore, the end ordinates for a stress influence line must be 1/r times as great, whence o' AA'=-, and BB"=- (21u) r' r Now l-a/r=$ =the stress in a member S due to a reaction unity at A, when the load producing this reaction is to the right of the section through S. Also I a' /r =${,= the stress in this same member S due to a reaction unity at B when the load producing this reaction is to the left of the section. Hence, the end ordinates for any stress influence line are equal to the unit stresses S a and Sb produced in the member by reactions unity at A and B respectively. Since these stresses are readily determined either by Ritter's method or by a Max- well diagram, all stress influence lines for the members in any determinate truss are easily found. In Fig. 2lA, the stress influence lines are drawn for a general case of truss design. The top chord is the loaded chord and the influence lines are drawn for the three mem- bers L, U and D of the panel cut by the section it. te =the stress in the member L for a reaction unity at A. This is evaluated by Ritter's method in terms of the lever arms ai and r t . Sib=ihe stress in the member L for a reaction unity at B and is expressed in terms of the lever arms a'i and r t . The stress influence line for the bottom chord L, is drawn in accordance with the method illustrated in Fig. 19A, observing, however, that the end ordinates now become the stresses S ta and Sn, respectively. The center of moments for this member being at E, which is vertically over E', the stress influence line for the member L becomes a triangle AE'B, determined by the end ordinates A A' = to and BB' =#. The influence line for the top chord, J7 is drawn in exactly the same manner, but in this instance the resulting triangle AF'B does not offer a straight line over the panel ART. 21 INFLUENCE LINES AND AREAS 53 Here F is PJG and hence the line E'G' must be drawn to complete the influence line, the center of moments for the member U. The influence line for the diagonal D is similarly drawn and the load divide %' ', found in the lower diagram, is seen to coincide with the point i found on the truss diagram by the method previously given. Also the intersection 0', between the limiting rays is verticallv under the center of moments for the AE' and G'B of the D member D. line. ISTRESS INFLUENCE LINE FOR L. STRESS |NFJLUE!NCE LINE FOR. u." ' ' " STRESS INKLUEJNCE LINE FOR o It should be observed that in all three influence lines, Fig. 21 A, the limiting rays intersect on the vertical through the center of moments f or the particular member. This then serves as a check on the diagram. Regarding the sign of the influence area the following rule should be observed: When the center of moments is located between the supports A and B then all ordinates of that particular stress influence line will be of the same sign; when the center of moments is off the span, then there exists a load divide and there will be positive and negative influence areas giving rise to stresses of opposite signs. 54 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, iv The criterion for position of a moving load for maximum and minimum stresses in any member is thus clearly shown by influence lines and the maximum effect of any load is always produced when that load is situated over the maximum influence ordinate. This subject is more fully treated in another article.. It is sometimes necessary to construct influence lines without the divisor r, and dividing the final stress thus found by r. Either method is employed as circumstances may demand, though it is always better to avoid the divisor and follow the method given in Fig. 21 A, unless there is some very good reason for doing otherwise. Attention is also called to the fact that moment and shear influence lines are not required when the stress influence lines are used. The former were given to render the treatment complete and may be employed to find stresses, but it is usually preferable to draw the stress influence lines directly. The final result of an analysis will be more satisfactory when this is done. Influence lines are seldom used for finding dead load stresses, though such a procedure may sometimes be warranted, and then the following points must be observed : For direct loading and for beams, it makes no difference whether the influence lines be used for dead or live loads, but in dealing with framed structures an error might be committed because then the influence lines are always drawn for a certain loaded chord while the dead loads act along both chords. This is easily remedied by drawing the influence lines for both chords loaded and applying the dead loads to the proper lines. These influence lines are identical except in the panel containing the member in question. Hence by observing this circum- stance, dead load stresses may be found from the same influence lines without difficulty. ART. 22. REACTION SUMMATION INFLUENCE LINES Most summation influence lines become rather complicated and, therefore, their use is practically restricted to a few cases for which the construction is simple. In general, any ordinate to a summation influence line is expressed by Eq. (17A), as and it is readily seen that when P and ij are both variables, such a line would ordinarily require much labor for its determination. However, the summation influence line for an end reaction of a simple truss is very easily constructed and serves a most valuable purpose in finding the end shears for a system of moving loads. This influence line will be called simply the sum A line to distinguish it from the ordinary A line in Figs. ISA and 19A. The general usefulness, of the sum A line originated by Professor Winkler, will be shown later; suffice it to say here that it affords a ready means of finding stresses in the web members of any truss, because, as previously shown, the shears and moments for any point of a truss are easily found when the end reaction for the particular loading is known. The sum A line for a system of concentrated wheel loads will now be demonstrated using but five loads for simplicity, though the method applies to any number of loads. ART. 22 INFLUENCE LINES AND AREAS 55 Given the train of moving loads PI, P 2 , etc., in kips of 1000 Ibs., and spaced as shown in Fig. 22A, to construct the sum A line for a span of 25 ft. Since a reaction influence line is the same both for direct and indirect loading, the sum A line is independent of the number and location of panel points. For standard position of loads on a truss, the train is always assumed as coming on the 'span from the right end and moving toward the left, but the above loads are placed in exactly the reverse order, with the first load PI over the support B. The reason for this will appear later. The loads are applied consecutively from PI up on the vertical through A, using any convenient scale. r POSITJON OF LOADS roft CONST *GcT ION or SUM A unr B FIG. 22A. The several rays drawn from B to the respective load points c, d, e, etc., form a force polygon with pole distance H =1. The sum A line is then drawn as the equilibrium polygon for this force polygon and the loads P, by making 1 -2 \\Bd, 2 -3 \\Bc, 3 -4 \\Bf, etc. It will now be shown that the equilibrium polygon so obtained is really the sum- mation influence line for the end reaction A . Referring again to Fig. 22A, the line Be is easily recognized as the influence line for A due to a moving load PI. Alsp^the influence line for A when the moving load is P 2 , may be represented by the line Bd when Be is the axis of the absciss. Similarly ~Be represents the A line for P 3 , and so on for any other loads. The summation of all odinrates should then represent the desired summation influence line. For the train coming on the span from right to left, the ordinate TJ, under the forward load PI, always represents the reaction A. Thus when the train has advanced 56 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV to the point n, Fig. 22A, the ordinate v? represents the sum of the ordinates t, and T}=A n =U+t 3 +t 2 +ti\vhkb. is readily seen by inspection. The ordinates t are the con- tributions to A from the several loads on the span. It is now seen why the train was first placed in reverse position for the construction of the equilibrium polygon. Hence the sum A line for direct or indirect loading is an equilibrium polygon drawn for a pole distance equal to the length of span and any system of train loads placed in reverse order on the span with the forward load at B. The reaction A for any position of loading is then the ordinate to the sum A line measured under the forward load when the train approaches from right to left. The greatest possible reaction A will always be the end ordinate A5, Fig. 22A. When extraordinary accuracy is required the end ordinate A 5 can be readily checked by computation. The sum A line for uniformly distributed loads may be constructed in precisely the same manner as above illustrated for concentrated loads, merely by laying off the load line Ag to represent the total load on the span =pl and dividing this into some con- venient number of equal parts. The greater this number the more accurate will be the result, and the polygon finally becomes a parabola. The application of the sum A line, to finding the shear at any point of a beam or truss, will now be presented. The shear at any point n of a beam, for a case of direct loading, is equal to the reaction A minus the loads between A and n. Hence, the sum A line gives a complete solution for this case. The subtraction of any loads to the left of n can be performed graphically on the diagram, Fig. 22 A, by taking^ off the proper ordinate less the loads between A and n. For indirect loading, the shear at n is equal to the end reaction A, provided the forward load is exactly over the panel point n. But when the load extends over into the panel, then the shear is equal to the end reaction A, minus the panel reaction a at the left hand pin point of the loaded panel. See Fig. 22ra. The values of o for all possible positions of loads in a single panel may be found by drawing a sum a line noe for one panel, using as many of the loads from the forward end of the train as may be placed into one panel. If this auxiliary sum a line is drawn on tracing paper, it will also serve_ a ready means for finding the position of the train for maximum shear in any panel mn. The method of finding the sum a line is exactly the same as for the sum A line only the pole distance for the former is the panel length d. When the forward wheel of the train is at n, then the shear is the ordinate rj n . When the forward wheel is at r the shear in the panel mn is Q r =rcre=ec where re = -n a the panel reaction a for loads PI and P 2 in the panel and load P 3 at n. Similarlv when the forward wheel is at s and the second wheel is at n the shear Q 8 =s/" os. The position of the train for max.Q in the panel mn is then easily found by selecting such a point s for which the ordinate O/=T? max. This maximum jj is easily found with a pair of dividers and will always occur when some one of the forward wheels is at the panel n. INFLUENCE LINES AND AREAS 57 If the sum a line is drawn on tracing cloth, it may be superimposed on any panel of the sum A line and TJ max. can be quickly determined. It sometimes happens that two positions of the train give the same maximum shear, in which case either may be used. In Fig. 22e this would be true if oe\\cf then ~ec=of and both Q r and Q s would thus be equal. Load potitiori for Qm<. in panel m-n FIG. 22s. ART. 23. POSITIONS OF A MOVING TRAIN FOR MAXIMUM AND MINIMUM MOMENTS (a) Point of greatest moment for direct loading. Given a span of certain length I, loaded with a train of loads, PI, Pa, PS, etc., to find the point n which is subjected to the greatest bending moment. It is assumed that all loads remain on the span. See Fig. 23 A. Let R = resultant of all loads P on the span. RI = resultant of all loads to the left of n. R 2 = resultant of all loads to the right of n. A and B are the end reactions for R. n =the point of maximum moments to be found. Then the moment about n is Rx (22A) 58 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV By differentiating and equating to zero, the abscissa for the point of maximum moments is found thus : dM R n a I r9 ., x -=- r (l-2x-a) =0 or x+-=-^ ....... UOB) ax t z z Eq. (23B) signifies that the maximum moment under any system of loads occurs for a point n when the center of gravity of the system and the point n are equidistant f rom the center of the beam. Introducing the value of x from Eq. (23 B) into Eq. (23A) the value of max. M is obtained as max.M^R^-R^. . . ........ (23c) l> In any special case it is necessary to find the particular load P n which must act at the point n to produce a real maximum. The moment influence line clearly indicates that one of the loads must fall at n to obtain the maximum moment. FIG. 23A. Usually the point n falls very near the center of the beam and in most practical problems it will suffice to place that load at n which falls nearest the center of gravity of the sj'stem of loads. When dealing with floor stringers it frequently happens that the stringer is long enough to carry three wheel loads but that the max. M so found is less than when only two of the loads are on the stringer. A few of these' special cases are here given. Case I. When there is only one load P on the span, then x=l/2 and a=0. Hence Eq. (23c) gives M Pl max.M= (23D ) Case H. When there are two equal loads P on the span, distant e from each other, then R=2P and |=|, hence z=--|=_- 1 which gives, from Eq. (23c), 2P/Z max. M =:-[ Equating Eqs. (23o) and (23c) to find when one or two loads give equal maxima, this results in the condition e=0.5858Z. Hence, when e>0.5858Z, then one load gives a greater moment than two loads, all loads being of same magnitude. ART. 23 INFLUENCE LINES AND AREAS 59 Case III. When there are three equal loads P on the span, distant e from each other, then R =3P and x =1/2 making a =-0. Also Ri=P and a t =e, hence ,, 3PI n n x max. M=. Pe=P( e}. 4 \4 (23F) From Eqs. (23E) and (23r) it is found that when e>0.4494Z, then two loads give a greater max. M than do three loads, all loads being equal. Case IV. For four equal loads on the span may. M occurs under the second load and Eq. (23c) gives max.M=P(l+-2e] ........ (23o) Here again it is found that when e>0.2679Z, then three loads give a greater max- imum than four loads. (b) Critical loading for maximum moments. Direct loading. In this case the moment influence line for any point n is a triangle and the discussion is, therefore, applicable to any beam or truss for which the moment influence line is a triangle. See Fig. 23B. :i n-~lD..O 01 n h n FIG. 23B. Let R = 2P =the total load, or resultant of loads on the span. RI =the resultant of all loads between A and n for a certain position of a train of loads. RZ= the resultant of all loads between n and B for the same position of the train of loads. Calling rji and rj 2 the influence ordinates for the load points of RI and R 2 respec- tively, then the moment for the point n may be written as 2RiXi tan ai +R 2 x 2 tan 2 . . . . . . (23n) Suppose now that the train is moved to the right by a small distance dx, producing the change +dx in Xi and the change dx in x 2 , then the change in M n is dM n and the differential coefficient thus obtained is equated to zero for maximum and minumum, thus: dM n dx =Ri tan ai R% tan a 2 =0. (23j) 60 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV But since tan i =^-. and tan a 2 = , Eq. (23j) gives, by substitution, a i a-2 ^1=??. (23K) ai a 2 ' By observing that a l +a 2 =Z and ^i +R 2 =R, then by composition, Eq. (23K) gives: _ a\ -f 02 a\ I Either Eq. (23K) or Eq. (23L) may serve as a criterion for the critical position of the train load on the span. The load P n , falling at n, is divided equally between 1^ and R-Z. In the first instance the criterion would be satisfied when the average unit loads on both sides of the point n are equal, a condition which is absolutely fulfilled by a uniformly distributed load. According to Eq. (23L) the criterion for max. M would be that the average unit load to the left of the point n must equal the average unit load on the whole span. Also, since the maximum ordinate y is under the load point n, no position of loads can give a real maximum unless one of the loads is directly over n. Hence it is readily understood that there are usually two or more maxima for any point n, depending on which load is placed over this point. Therefore, so long as the criterion is fulfilled, all positions giving such a max. M n must be used to determine the real maximum. In general for any moment influence line, Fig. 23s, the moment will be maximum for a point n when a load P is applied at n and when the angle ^ of the influence line exceeds 180. The moment at n will be minimum only when there is a load at some point B for which 0'<180, that is where the angle between two successive elements of the influence line is convex upward. Hence all angles in a moment influence line represent critical load points and correspond to maxima or minima accordingly as these angles are convex downward or upward, respectively. (c) Critical loading for maximum moments. Indirect loading. For any point w, other than a panel point, the moment influence line is a quadrilateral, hence all cases for which this is true are included here. Let .RI the resultant of all loads acting between A and c. R% =the resultant of all loads acting in the panel ce. R s =the resultant of all loads acting between e and B. Other notation, as shown in Fig. 23c, which represnts a beam or girder with four panels, each of length d. From Eq. (23H) the general expression for an influence line of any number of sides may be written as M = 2Rx tan a, ........... (23 M ) and the differential coefficient, equated to zero for maximum and minimum, becomes -- = 2R tan a =0 ...... ... . . (23u) ART. 23 INFLUENCE LINES AND AREAS 61 The various values of R tan a are now found from the special case in hand, using the notation and illustration in Fig. 23c. Three such terms are here required and these are evaluated as follows: jj V Tj c a\ d\ fje a 2 d 2 tana-i=--; tanas-; also ~~^~' a\ a 2 TJ a\ T} a 2 from which tana 2 = With #1, /2 2 and # 3 and these values of on, 2 and 3 , and observing that for dx to the right the value R 3 tan 3 must be negative, then Eq. (23N) becomes _ dx _ d (23o) FIG. 23c. which may be reduced to give the criterion (23p) Observing that R=Ri+R 2 +Rs as before; also that d=d l +d 2 and that l=( then by composition of Eq. (23?) and substitution of these values the second form of criterion is obtained as d\ _!l = :? (23Q) a\ I Eq (23r) expresses the criterion for max. M n when the average unit loads on both sides of the point n are equal, provided the resultant R 2 is distributed between R, and R* in the proportion di : d 2 . Eq (23q) stipulates that the average unit load over the distance 01 must be equal to the average unit load over the whole span, provided the portion R^/d is added to the resultant R\. . . The question of finding which particular load must be placed at either c < fulfill either of the above criteria may be solved, by trial, as for the case of direct loading. 62 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV ART. 24. POSITIONS OF A MOVING TRAIN FOR MAXIMUM AND MINIMUM SHEARS (a) Critical loading for maximum and minimum shear. Direct loading. The shear influence line is then as shown in Fig. ISA and the max. +Q n is obviously produced by a full loading of the entire span between B and the load divide n. The min.Q n is produced by a full loading over the distance An, which corresponds to the negative influence area. This is absolutely correct for uniformly distributed loads. For a concentrated load system, the train must cover the portion of span between n and B, with the first load just to the right of n so that its influence ordinate is nn" without any negative effect from the ordinate nn' . However, if the first load is small compared with the second load, it may necessitate placing the second wheel at n to attain the real max. +Q n =A Pi. Hence, it is clear that either PI or P 2 must be just to the right of n to obtain max. +Q, and whichever gives the greater value is then the real maximum. (b) Critical loading for maximum and minimum shear. Indirect loading. The influence line for shear, Fig. 19A, or the stress influence line for a web member, Fig. 21 A, are both included here. Hence the criterion for position of loads for a max. Q will also serve for maximum stress in a web member. For uniformly distributed loads the load divide, as found in Fig. 20A, will always establish the point at which the head end of the load must be located for max. +Q or min. Q. When dealing with concentrated load systems the position for maximum and minimum shear is most advantageously found by graphics, as illustrated in Fig. 22e, where the magnitude of these shears is at once determined, together with the critical position of loads. When analytic methods are employed, the following criterion may serve a useful purpose. Let Fig. 24A represent a truss with bottom chord loaded and i is the load divide for the diagonal D. The train of loads covers the span from B to i, making RI the resultant of the loads in the panel CF and R 2 the resultant of the loads between B and F. The line A'C'F'B' is the stress influence line for the member >. Then the stress in the member D is (24A) By shifting the train dx to the right, both Xi and x are diminished by this amount dx. Hence the stress now becomes ART. 24 INFLUENCE LINES AND AREAS 63 Subtracting Eq. (24s) from Eq. (24A) the differential change in S D is obtained dS =S D -S D ' =r t \ -^, +^===. | . (24c) thus From Eq. (24 c) the value dS/dx is found and equated to zero for maximum value of S D . This gives RI R 2 Ri+R 2 Ri R T,=l^^ or by composition __=_=_, . . . (24D ) where R is the resultant of all loads between B and i. This furnishes the criterion, provided no additional loads have entered the panel CF and no other loads have come on the span in making the shift. FIG. 24A. It is evident from the influence line, Fig. 24A, that any loads to the left of i would produce a negative influence on SD, so that the minimum value of SD may be found by loading the left portion Ai of the span, and the criterion then becomes RI R This criterion expresses exactly the same condition as previously found for moments in Eq. (23L), but the span now becomes the length a\ or a 2 , as the case may be, cover- ing only that portion of the stress influence line D, which has the same sign. As previously found, by the graphic method, it is always necessary to place a load at the panel point F for max. D and one at C for min. D, allowing as many loads inside the panel CF as may be required to satisfy the criterion. 64 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IV The above discussion of the various criteria for position of moving loads to produce maximum and minimum stresses in any member of a statically determinate structure is ample demonstration of the high value and practical use of influence lines. It was not deemed advisable to extend this discussion to cover the various types of special and composite structures, as the influence lines themselves, when used for the determination of stresses, will give complete and comprehensive answers in all cases without special analytic treatment. However, when such is desirable, the methods previously employed will indicate the procedure to be followed for any particular influence line. The subject of influence lines for statically indeterminate structures and for deflec- tions will be treated after presenting the general subject of distortions and deflection polygons of framed structures. CHAPTER V SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY DETERMINATE STRUCTURES ART. 25. DOUBLE INTERSECTION TRUSSES Usually trusses of this type are analyzed as two separate systems, dividing the loads between them in such manner as may seem most probable. The stresses found for the separate systems are then combined for the double system wherever the same member forms part of each system, provided simultaneous cases of loading were used. The method of influence lines has advantages especially in the distribution of simultaneous load effects, and the combined stress in a member belonging to both systems is at once found for the same position of loads. Double intersection trusses should always be designed with an even number of panels, thus retaining symmetry of both systems with respect to the center of the span. Otherwise the loading carried by one system in the left half span must go to the other system in the right half span, which is not desirable. Fig. 25A represents a double intersection truss for which stress influence lines will now be drawn. Bottom chord member DE. A section tt, passed through the panel DE, cuts the diagonals FE and GK, hence the chord DE forms part of two panels for which there are two centers of moments, F and G. Therefore, each system gives rise to a stress influence line for the member DE, shown in Fig. 25A, by the two dotted triangles A'C'B' and A'D'B'. The top chord members are regarded straight between alternate pin points for each system in question. Loads acting at panel points are supposed to be carried by the system to which each such point belongs except at the panel where the load is divided equally between the two systems. Hence a load at D goes entirely to one system and its influence is represented by the ordinate DD' ', of the influence line A'D'B'; while a load at E is carried by the other sys- tem and its influence ordinate becomes EE' in the influence line A'C'B'. This same reasoning applies to all panel points. Since influence lines between . successive panel points must be straight lines, the actual influence line for DE is represented by the shaded area A'O'C'D'E'K' to B' . The point 0' must lie midway between the two ordinates for because at this point the load is carried equally by both systems. This is not the case at M, where the entire load goes to one system. 65 60 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V Top chord member LF. The influence line for DE is like the influence line for LF, except that the end ordinates are slightly different. Here the lever arms are measured normally to the members LF and FG. Diagonal member FE. The method of reasoning just applied to chord members answers in every sense for the web members. The influence line for FE is constructed as indicated in the figure. The stresses in FE, due first to a unit load at A and then at B, are found for the system to which FE belongs. FIG. 25A. Since loads at the panel points of the other system cannot cause stress in FE, the influence line ordinates for these points must be zero. This means that the other influence line is the base line A'B'. Observing again that the required influence line must be straight between successive panel points the final line becomes A'O'C'D'E'K' to B'. The point 0' , as before, is midway between the ordinates under 0. The influence line indicates clearly that when loads are spaced 2d apart, then the entire web stress goes into one system, making the design verj'- uneconomical. This would easily happen for 50 ft. locomotives and 25 ft. panels. ART. 26 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 07 ART. 26. CANTILEVER BRIDGES Any statically determinate structure extending over more than two supports (piers or abutments), when continuous over r2 of these supports, may be regarded as a cantilever. If r is the total number of supports, then such a structure must be provided with r2 intermediate hinges such that the bending moments at these hinged points are zero for any system of loads. The reaction and typical stress influence lines for such a cantilever will now be drawn. Reaction influence lines for all simple structures are independent of the panels and may, therefore, be illustrated on a simple cantilever beam, Fig. 26A. FIG. 26A. The illustration represents an anchor arm A B with a downward reaction or anchorage at A and an upward pier support at B. The span EF is a simple beam on two supports E and F, with an overhanging cantilever arm at_each_ end. _A similar arm projects out from the anchor arm, making three cantilevers, BC, DE and FG, in the whole system. The two spans CI) and GH are known as intermediate spans and are supported on hinged ends which permit of horizontal displacements. Hence, there is only one roller bearing at E, and the anchorage at A also admits of slight horizontal motion. The whole system is, therefore, determinate and can have no temperature stresses, neither will slight displacements of the supports cause any internal stresses. The reaction influence lines thus involve no principles other than those already elucidated for simple beams on two supports. The central span WF is a suitable beginning, as it rests on two supports R 3 and #4 with a roller bearing at E. Hence the influence lines for these two reactions are drawn precisely as for a simple beam between the points E and F. But the two arms at the ends of this span must also be considered. Regarding first the influence line for R 3 , it is seen that a load at F has zero effect, 68 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V and as the load proceeds to the left its effect on R% is uniformly increasing and this continues to be the case until the load reaches the point D, where the effect becomes a maximum. The same is true for a load between F and G, hence the ordinate unity at E' determines the RS influence line from D^_io G' . A load at C or at H produces zero effect on R s and loads outside the span CH have no influence, hence the line C'D'G'H' is the influence line for R s . Further proof will scarcely be needed, though the several moment equations for the separate spans CD and GH will readily supply this. Similarly the line C"D"G"H" represents the influence line for R$, based on the unit ordinate at F". The influence line for Rs is the same as for a simple beam GH. Precisely the same reasoning applies to the reactions Ri&nd R-2, the influence lines for which are shown to the left in Fig. 26A. It should be noted with regard to RI that the negative influence area being greater than the positive, provision must be made for a downward reaction. When the upward reaction RI, due to live load from A to B, is greater than the downward reaction R'i, due to dead load from A to D, then the fastening at A must be designed to take stress in both up and down directions. Shear influence lines. These are readily deduced from the reaction influence lines, as may be seen by comparing Fig. 26s with Fig. 26A. Four typical panels are assumed. One each for the central span EF, the cantilever arms BC and FG, and the anchor arm A B, respectively. FIG. 26s. For the panel ef the shear influence line within the span is precisely as for any simple beam on two supports, while outside the span these lines are continued to the intersections D' and G' with the verticals through the hinges D and G, respectively. The shear influence line for the panel ef is thus found as C'D'e'f'G'H' . For the panel ab the same construction is applied as for a span on two supports A and B followed by a cantilever to the right of B exactly as for the previous case. The shear influence line for the panel ab is the polygon A'a'b'B'C"D" '. When the panel is located in one of the cantilever arms as for gh, then the circum- stances are slightly different, because the shear in any cantilever arm is always equal to the sum of the applied forces between the section and the outer end of the arm. ART. 26 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 69 Accordingly the shear must be composed of the end reaction from the intermediate Hence the influence line It should be noted that truss GH and the constant effect of loads between h and G. for the panel gh is easily found to be the polygon g'h'G"H' . loads to the left of g, even on the next span EF, can have no effect on the shear in the panel gh. The shear influence line for the panel cd is similarly found to be the polygon c'd'C"'D" '. Any panel located within either of the intermediate spans CD and GH, is treated precisely as for a span on two supports, since no load outside these spans can produce any effect on these trusses. Moment influence lines. These offer very little difficulty and follow directly from the methods previously given. In Fig. 26c the anchor arm was omitted, as the portion of the structure from C to H affords all the illustration required. Here, again, the central span EF is treated exactly like a simple beam and the negative effects of the adjoining cantilever arms are found by prolonging the lines so FIG. 26c. obtained to the intersections D' and G' with the verticals through the hinged points. The moment influence line for the point n, in the panel EF, thus becomes C'D'E'e'f'F'G'H' , and it is readily seen that a load on any portion of the structure CH affects the moment at n. For the section t, in the panel gh, the moment is the reaction at G into the lever arm a 3 , hence for a unit load, the ordinate at G must be l-a 3 , and observing that the influence line must be straight over the panel gh, the required moment influence line for the section t becomes F"g'h'G"H". Stress influence lines. The general method for stress influence lines, illustrated in Fig. 21 A, serves every purpose in connection with cantilevers. In every determinate structure the stress in any member may be expressed as S=M/r and from this it follows that the stress influence line for any truss member is similar to the moment influence line drawn for the center of moments pertaining to the member. The difference between the stress and moment influence lines is only the constant divisor r applied to the ordinates of the latter to obtain the ordinates for the stress influence line. For convenience, the central span CE, Fig. 26o, will be treated first, because this 70 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V portion of the structure is exactly like a simple truss on two supports, and all the rules previously given apply here. Regarding the web members it should be observed that the center of moments may fall inside or outside the limits of the span. In the former case there will be no load divide within the span. Both cases are illustrated in Fig. 26D, where the stress influence lines for members U, L, D, and D' are presented. The designations regarding the stresses in these members for unit reactions at C and E, will be as previously adopted in describing Fig. 2lA, viz., S uc =ihe stress in the member U for a reaction unity acting at C. FIG. 26o. Since these stresses are always necessary in constructing influence lines, it is best to find them for all members of the structure before proceeding to draw any influence .lines. These stresses are very easily found either by slide rule, using Ritter's moment method, or by drawing a Maxwell diagram. Stress influence line for U, central span. This being a top chord member, and in compression, its influence line is negative between C and E, which is also recognized by the stresses S^, and S ue found for this member for the reactions unity, acting first at C and then at E. These stresses are laid off from A'B', Fig. 26D, upward and on the verticals through ART. 20 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 71 C and E, respectively. The stress influence line is then easily drawn, observing that the intersection n' must fall under the center of moments n for the member U. Hence this is a check and may also be used as a shorter method of finding the U line, when only S llc is known. Between the panel points m and o the influence line must be straight, hence m'o' is drawn. To complete the influence line over the side spans, the U line, just found, is continued to intersections B" and F" with the verticals through the hinged points B and F and the final U line is then found to be A'B"m'o'F"G' . The portions below the base line A'G' are positive and the upper portion is negative. The positions of loads to produce max. U and min. U are thus clearly indicated and no other criterion is required. Stress influence line L, central span. This is constructed in precisely the same manner as the U line, but L being in tension, as found for C = l, the stress +Si c is laid off down- ward under C. The point m" . vertically under the center of moments m for the member L, is thus found arid may serve to complete the L line without using the stress Si e . The L line is then obtained as A'B'm"F'G' and is positive over the span CE and negative over the two cantilever spans AC and EG. Stress influence line D, central span. The center of moments for this member is at i, which is within the supports, and hence this member has no load divide. The stress influence line is constructed exactly as was done for the U line, by laying off the positive stresses S dc and S de to obtain the two limiting rays, with intersection i' under the point i. However, since the rate of increase in the stress D must be uniform for a load advanc- ing from E to o, the ray E' r7 i' must be continued to o ly finally drawing the line m^i to complete the polygon. Over the two side spans the D line is drawn as for the two previous cases, giving the complete line A'"B f "C f "mio l E'"F"'G"'. Stress influence line D', central span. The center of moments for D' is outside the span at k. hence the influence line must show a load divide. The stresses + S'd c and S'd e serve to construct the stress influence line for D' which is in all other respects like the D line. The limiting rays intersect in k' and the load divide is at q. Stress influence line U, cantilever arm, Fig. 26B. The effect of a moving load, coming on the span from left to right, begins when the load is just to the right of A and increases uniformly until the load reaches B. From B toward m the effect is uniformly decreas- ing and becomes zero when the load is over n, which is the center of momentsjor U. The stress in U for a load P = l at B is-S ub , and laying this off from the_base A'o', the negative U line is easily drawn. For the effect due to loads in the panel mo draw m'o' to complete the influence line. Loads to the right of o or to the left of A have no influ- ence on U. Stress influence line L, cantilever arm. The center of moments for L is at o and the stress in L for a load P = l at B is+S lb . Hence the influence line becomes A"B"o" and is positive. Stress influence line D', cantilever arm. The center of moments for this diagonal is at i, which is inside the cantilever arm. The rule for load divide is now reversed because loads on opposite sides of i produce opposite stresses in D', and a load at i cannot produce any stress. Hence i is the load divide for D'. 72 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V The stress in D' due to a unit load at F is S' d f, hence making the ordinate under F equal to this stress the influence line s'i'F'G' is obtained. Since loads to the left of h cannot affect D', and since the influence line is a straight line over a panel, the influence line is completed by drawing h's'. Stress influence line D, cantilever arm. The center of moments now falls off the arm at k and hence all loads produce the same kind of stress in D. Lay off the stress +S df on the vertical through F and complete the influence line as shown in Fig. 26E, obtaining s"v'F"G" for the D line. The intermediate spans A B and FG are simple trusses on two supports and require no further consideration. FIG. 26E. ART. 27. THREE-HINGED FRAMED ARCHES In taking up this subject it is necessary to show briefly the general relations between the reactions and external loads. Let Fig. 27A represent any three-hinged arch ACB, which may be framed or solid. The single load P may be applied at any point m. For all positions of P between A and C, the reaction R% will always have the direction BC and the two reactions RI and R% must be components of P. Similarly for all positions of the load P between C and B, these reactions will have the directions AC and eB, respectively. Hence for any position of the load P, the reactions at A and B are found by resolving P into two components as shown, and the points of application of all possible positions of P will lie on the prolongation of either AC or BC. If the reactions R and R 2 be resolved, each into a vertical component and one along AB, the two last-named components H' will be equal and opposite, while the two vertical components A and B will be precisely the same as for a simple beam on two supports A and B, because the forces acting at C are in equilibrium among themselves. Therefore, A+B=P; Pa 2 A -, and B= = . (27A) AKT. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 73 The horizontal component of H' is known as the horizontal thrust of the arch and is found from H=H'cosa (27B) If the reactions R\ and R 2 had been decomposed into components _4/, B' and H, then the former would no longer be the reactions for a simple beam AB. However, when a =0 and the end supports are on a horizontal line, then the vertical end reactions are always as for a beam on two supports. FIG. 27A. Taking moments about the hinge C, when P alone is acting, then for condition of equilibrium at C M=Al l -H'y=Al l -H'fcosa=Q, and substituting values from Eqs. (2?A> and (27s) this gives Hi or H= J-J-. . . (27c) When P = l is applied at C the value H c becomes the ordinate vertically under C for the influence line for horizontal thrust. Hence, ,-i, (27D) -_^ and when Zi=Z 2 =o, then making the H influence line a triangle with middle ordinate T? C under C, as given by Eq. (27D). The influence lines for the vertical end reactions are found exactly as for a simple beam AB. The stress influence line for any member of a three-hinged framed arch will now be developed. When the structure is symmetric, the half arch only requires analysis, but for unsymmetric arches the entire structure must be treated, though the method remains the same as would appear from the previous discussion. KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V In Fig. 2?B, the half arch is shown with a load P producing reactions R\ and R 2 , and since the latter must be transmitted to the point C, the right half of the span may be considered removed and its effect replaced by R 2 acting at C. The left half of the span is then held in equilibrium by the two reactions R : and R 2) the latter applied at C. The reaction R\ is now resolved into A and K and the reaction R 2 is similarly replaced by a vertical reaction C and a reaction K which is equal arid opposite to the first K. FIG. 27s. If the forces K were zero, then the half span would represent a simple truss on two supports, with vertical reactions A and C and the influence line for any member could be found by the method previously given. Thus, the influence line for any member- is easily drawn when the stresses, first for A=l, and then for 5 = 1 are known. Such an influence line is drawn for the diagonal D, and is shown as the polygon A'E'F'C' ',. Fig. 2?B, where the stress D a , due to A = 1 is positive, and the stress D c , resulting from C = l, is negative. In each case the half span is supposed to be held first at C and then at A in order to find these stresses. ART. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 75 The influence line thus found, must now be corrected so as to include the effect of the force K. Since K=Hsec{3, therefore K is proportional to H and the stress Dk, produced by the force K, must likewise be proportional to K and H. Hence, the influence line for the stress D k , is similar to a horizontal thrust line and must be a triangle whose middle ordinate under the hinge C is equal to the stress D k , produced by a unit load applied to the whole span at the point C. To find this stress D k for a unit load at C, substitute P--=l into Eqs. (27 A) and (27c) and obtain A=i and # = l'i- J If the stress in D is found first for A = 1/2 and then for // =l/4f } separately, then the combined effect will be the stress D k =%D a +D H , which is much more convenient than to find D k directly for the diagonal force K, The stress D H is the stress in D for the above force H = l-l/4f applied horizontally at A and supposing the half arch rigidly held at C. The stress D a was previously found, and hence the required stress D^ is readily obtained from (27B) In the present case D^ was found to be negative and the influence line for the member D due to the effect K is drawn as the triangle A"C"B" ' . The final D line is now obtained by combining the two influence lines just found- This is done in the last figure and gives the polygon A'"E"F"C'"B" r . The stress +D a is laid off down from A'" and the stresses D c and D^ being negative, are laid off upward from the base line A'"B'" and being of the same sign they are added. The polygon is then drawn as indicated in the figure and as a check, the point n" must fall vertically under the 'center of moments n for the member D. Still another check is found from the circumstance that a load at i produces zero stress in the member D and hence the influence line must intersect the base line vertically under i in the point i'. This last mentioned condition is always true even though the point i falls below C. The point i is the intersection between BC and An, where n is the center of moments for D. This offers a very valuable condition and obviates the necessity of finding the stress D c , whenever the center of moments n can be located within limits of the drawing. It is thus seen that when n and i are conveniently located, which is always the case for chord members, then the influence line can be determined when D a alone is given. However, as a check, it is always desirable to know D^, so that when the stresses in all the members are to be determined by influence lines, it is advisable to find S a and Sh for each member before proceeding to draw the lines. This can be done by drawing two Maxwell diagrams, one for A = 1 and the other for H=l/4f. A slide-rule computa- tion will also answer if the lever arms are carefully determined. In Fig. 27s the loads were supposedly applied to the top chord. Had they been applied to the bottom chord then the panel line E"F" would have become G"H". 76 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V Typical stress influence lines will now be given for the three-hinged, framed arch, Fig. 27c, treating the top chord as the loaded chord. The example will be taken up in such manner as to indicate a complete solution for all stresses in the several members by the method of influence lines. ~?B B' I UU INFLUENCE LINE. "^luiii]|p|i||]^ 90 LIN&THV The first step will be to determine the stresses S a and S k for all the members and this will be done analytically by Ritter's method of moments, being very careful about the sign of each stress. The same stresses could also be found in a very convenient way by drawing two Maxwell diagrams, one for A~l, supposing the half truss ~AC fixed at C; and the other ART. 27 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 77 for a load P = l acting vertically downward at C and producing the horizontal thrust // = l-Z/4/= 0.833, which is again applied at A, supposing the half truss fixed at C. The following table contains the stresses S a and S a thus obtained and the stresses S=S a +S are then easil found. MEMBER. STRESSES. MEMBER. STRESSES. -Sa SH 8t Sa S H Sh U,A -1.00 + 1.02 + 0.52 U& -0.45 + 0.52 + 0.30 f7,L, + 0.93 -0.57 -0.11 iW -1.37 + 1.08 + 0.40 u& -1.29 + 0.79 + 0.15 U 3 U t -3.11 + 1.97 + 0.42 U 2 L 2 + 1.32 -0.85 -0.19 AL t 0.00 -1.44 -1.44 U 3 L 2 -1.65 + 0.77 -0.05 L,L. + 0.85 -1.87 -1.45 U 3 L 3 + 2.19 -1.15 -0.06 L 2 L 3 + 2.08 -2.38 -1.34 U t L 3 -2.45 + 0.96 -0.26 L 3 C + 4.40 -3.35 -1.15 u t c + 3.05 -1.19 + 0.33 + for tension and for compression. The values in this table will suffice for the construction of all the stress influence lines, but only those for (7 2 t/ 3 , L 2 L 3 , U 2 L 2 and / 4 L 3 will be drawn, as shown in Fig. 27c. Stress influence line U 2 U 3 . The stress S a = 1.37 is laid off, to a convenient scale, upward from A' and the stress ^=+0.40 is laid off downward from A'B', vertically under C. The lines EC' and C'B' are then drawn. The center of moments for U 2 U 3 is at L 2 , hence find Z/ 2 vertically under L 2 and Finally since the influence line must be straight over the loaded panel As a check observe draw A'L 2 '. UjU~ 3 , draw U 2 'U S to complete the influence line A'U 2 UzC'B'. that the load divide i/ falls vertically under i', which latter is the intersection of AL 2 with BC produced. Still another check may be had by resolving a unit load at U 2 into components parallel to U 2 U S and U 2 L 2 as was done to the right of the truss diagram. The stress thus found for U 2 U 3 must equal the ordinate U 2 'F. Lastly, the ordinate S c is the stress in U 2 U 3 for a load unity at C when the half truss AC is fixed at A and the half span CB is removed. Stress influence line L 2 L 3 . The method is precisely as for the previous case, being careful to observe the signs of the stresses S a and S k in laying them off from the base line A 7 /?'. The point U' 3 , vertically under the center of moments for this member, again serves to complete the stress influence line, with all the checks just mentioned for the upper chord member. Stress influence lines U 2 L 2 and U 4 L 3 . The same method again applies and the two lines illustrate the effects of the signs of the stresses S a and S k . The center of moments n, for the member U 2 L 2 now falls outside the span, but the same relations hold good as before. Similarly for the center HI of the member U 4 L 3 . In this last case the load divide t' 3 falls below the middle hinge C and is no longer a real load divide, as the influence line for TT^Ls indicates, but all other relations continue to exist as before. KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V ART. 28. THREE-HINGED, SOLID-WEB AND MASONRY ARCHES Fig. 28A illustrates a three-hinged, solid-web arch, which may be built of masonry or steel, according to circumstances. If of masonry, then the section will be rectangular, and if of steel, then the section will resemble that of a plate girder. Whenever the depth of the section is small compared with the radius of the arch center line, then the effects of curvature are very minute and the stresses in the arch rib may always be found by the well known beam formula of Navier. This condition always prevails in bridges and all larger arches, but when dealing with small arched culverts and structures of that class, then the effects of curvature may require special consideration. For any girder section then, the stress on the extreme fiber is derived from N wherein M is the bending moment, A" the normal axial thrust, F the cross-section, / the moment of inertia of the section and ?/ the distance from the center of gravity of the section to the extreme fiber. M k is the total bending moment of the external forces about one of the outer kernel points e or i (Fig. 28A) and W is the moment of resistance of the section. Calling r the radius of gyration of the section, then 7 Fr 2 r 2 I W=-= =Fk where k=^-~ ...... (2SB) y y y Fy The quantity k is the distance from the gravity axis to a kernel point and the negative sign indicates that it is measured in the opposite direction from y for all sections. Hence, the influence lines for the kernel moments M e and Mi will serve to find the stresses f e and /; for the extreme fibers of extrados and intrados of any particular section tl, by using the multiplier /y. = l/W = l/Fk. For unsymmetric sections the two kernel distances are not equal so that {i e = l/Fk e for extrados and fi.i = \/Fki for intrados. For each section the stresses on the extreme fibers thus become ''-7S, and fi =~K ..... ' ' (28c) where M e =K ac e and Mi=K ac i, and K ac is the end reaction at A for a load unity at C found by resolving the vertical reaction A along AC and AB. See also Art. 49 on two- hinged arches. The moment influence lines M e and M; are now found by the method previously outlined for framed arches. Each of these moments becomes zero when the unit moving load passes through a load divide d. Hence the load divides are easily located for each case. Also since moments and not stresses are now considered, the end ordinate for the influence line at A is equal to x e or Xi as the case may be. ART. 28 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 79 Hence to draw the influence line Mi, lay off Xi at A and project the point d down to d', draw Gd'C' and after projecting down the center of moments i to find i', finally complete the influence line by drawing A'i' and C'B'. Now find as a check, that ~C'~D = x'i. \ B' FIG. 28A. The Af t - influence line becomes the/; influence line with a factor m=* The influence line M e is found in precisely the same manner. In both cases the ordinate C'C" is equal to M t or M e as given by Eq. (28c) for a unit load at C. The M e influence line becomes the / influence line with a factor 80 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, v The shear influence line Q OT is found by a similar process. Here the center of moments, for a web member at m, would be practically at an infinite distance because the chords are parallel or nearly so. Hence the line AdAj.t will intersect the chord C/I in the load divide d, which in this case is only imaginary, but serves the purpose as before. The equation for shear along the section it is easily found to be and this gives, for a unit reaction at A, when the half arch is fixed at C, the end ordinate A'G = l-cos . Also since the shear is constant between A and m, the limiting lines A'm' and Gd' must be parallel. Hence the shear influence line is easily constructed by drawing GC'd' , and then completing the polygon by drawing A'm' || Gd' and joining C' with B' ' . Since the loading is directly applied the line Wira', vertically under m, completes the Q 7> , line. Finally the ordinate C'C" =H[ - - ) wherein 7/ = l--^rfor a unit load at C. \ cos a ) fl When the loads are indirectly applied, then the condition that the influence line must be straight between load or panel points must be carefully observed. For masonry arches the section is always rectangular and 7 =/>D 3 /12, hence k = r' 2 /y = ^F^Z), where D is the thickness of the arch ring. The points e and i thus become the middle third points of the arch section and Eqs. (28c) become , GM e , f*~~TF and ^ = from which the stresses in a masonry arch section are readily found in terms of the moments about the kernel points of the section. The stress influence lines thus become moment influence lines with a factor 6/D 2 for each section. ART. 29. SKEW PLATE GIRDER FOR CURVED DOUBLE TRACK In general, influence lines for skew bridges are drawn in all respects like those for any simple truss or beam with the exception that at one end of the span the load always leaves the span before it is entirely outside of the limits of the longer truss. This merely calls for a slight correction in the end panel of the influence line as ordinarily drawn. At the opposite end of such truss some load comes on the end floor beam before the moving load is on the truss, but since the end floor beam transmits its load directly to the bridge support and not to the truss, this last named circumstance does not affect the influence line. These points will be brought out in connection with the following problem. It may seem inappropriate to apply influence lines to plate girders, but the practical designer who has dealt with problems of skew plate girders on curved double track will readily appreciate the advantages of the method there presented. KT. 29 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 81 MOMENT INFLUENCE LINE " ~ -- POINT 7, GIRDER NO.I.TRACK NO.2. FIG. 29x. 82 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.) This method was first suggested and used by Mr. T. B. Moenniche, C. E., in designing a skew double-track girder of 125 ft. span for the Virginian Ry. Co., in 1906. Thi author is indebted to Mr. Moenniche for the solution which follows. The principal advantage in the present application of influence lines consists ii reducing the solution of this rather complicated problem to a method instead of a time consuming process of experimentation. The principle of the method may be outlined as follows: The floor-beam reaction! are first determined for a unit load situated on track Xo. 1, and then for a unit load 01 track No. 2, which requires a careful computation of the coordinates or intersectioi points of each track on each floor beam. These floor beam reactions are then used a! influence or reduction factors for the ordinates to any particular influence line, becaus< the factors represent the proportion of a total load which can reach either girder a any panel point. This will give two influence lines for each panel point of a girder one separately for each track. In Fig. 29A, the upper diagram represents the structure in plan and the two sue ceeding figures show the floor beam reactions graphically. The last two diagram: are the moment influence lines for point 7, girder No. 1, for tracks No. 1 and 2, respectively The two reaction diagrams were drawn merely as a convenient way of showing the figures and in practice they would serve the same purpose, though they might b( dispensed with entirely. The reactions appearing as ordinates are plotted to sonn convenient scale making the sum of two reactions for each floor beam equal to unity except at the ends where the floor beams do not connect with both girders. The moment influence lines for point 7 are drawn in the usual manner by laying off a 2 vertically down from B and drawing A'T and 7'B' '. Since the load on track \o 1, when coming on from the right, begins to affect girder No. 1 as soon as it passes the abutment at E, therefore, the distance #10 must be treated as a panel arid the influence line is drawn straight from 10 to B'. The ordinates d of the influence line are now successively reduced by multiplying them with the corresponding ordinates or factors m of the reaction diagram to obtaii the real ordinates y of the required moment influence line. Thus ^6 =< Vn 6 and similar!} for the influence line point 7, girder 1, track No. 2, the ordinate Jjg=0 8 w 8 . With the use of these two influence lines, represented by shaded areas, the actual moment for point 7 may be found for any case of loading for each track and the sun: of the two gives the combined max. M 7 . Shear influence lines may be drawn in a similar manner, though they would scarcely serve a useful purpose except when dealing with a truss. In the above example onl} the reaction influence lines would be useful to determine the end shears. AKT. :-o INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES S3 ART. 30. TRUSSES WITH SUBDIVIDED PANELS From a point of economy, large trusses must have long panels, but with the increase in panel length the stringers of the floor system soon attain such proportions as to impose a practical limitation on the panel length. To overcome this difficulty, modern practice has developed a new type of truss especially adapted to long spans, wherein the stringers are supported by floor beams placed at the half-panel points. These intermediate floor beams are carried at the ends by a secondary set of truss members, which latter serve to carry the loads, thus locally applied, to the main truss system. This has given rise to the truss with subdivided panels. In the analysis of these truss types, it is not always a simple matter to determine the criterion for maximum stresses in the members, and it is believed that the following treatment by influence lines will serve to clear up the doubts usually encountered. Fig. 30A shows a truss of 200 ft. span with four different forms of subdivided panels. These are all combined in the one structure, though in practice only one form should be used in the same structure excepting in the first panels ALiL 2 , where the several compression members shown offer distinct advantages in stiffening the ends of the truss. Otherwise the panel L 2 L 3 , Case I, is the best and most economical type for adoption, as will appear later. In general, all primary members in the structure will receive stresses which are at least equal to those which would exist if the secondary members were all removed and in addition to these primary stresses nearly all of the members, excepting the verticals, will receive increased stresses due to local loads from the secondary members. This local loading is transmitted to the truss through the vertical M Z K which never acts as a primary member and gets no stress other than that due to a full panel load, besides the dead weight of bridge floor and bottom chords. It is the load, locally transferred to the points M , which enters as a disturbing element, and the four cases here treated will illustrate about all the practical points. To make the four cases directly comparable, the same panel L 2 L 3 was retained, but the diagonals were successively rearranged. The stress influence line for the primary stress in U 2 L S remains the same in each case, hence it is drawn only once and is then modified according to local conditions in the panel L 2 L 3 for each case. The Diagonal Member U ,L ,. Case I. The stress influence line for U 2 L 3 , when acting only as a principal truss member, is drawn by laying off the end ordinate A' A" = +0.568 which is the stress U 2 L 3 for a reaction unity at A. The ordinate B'B" (not shown) is 2.790 and is the stress in the same member for a unit reaction at B. By drawing A" ' B' and A'B" the points U' 2 and L' 3 are found and the line t/ 2 'L 3 ' completes the influence line A f U 2 L s 'B' with load divide at i. This gives the maximum stress for M 3 L 3 but not for U 2 M Zl because a load at K would produce additional stress in the latter member without affecting the former. Hence to find the influence line for U 2 M 3 , it is necessary to determine the stress 84 KINETIC THEORY OF ENGINEERING STRUCTURES i STRESS INFLUENCE LINE M 3 L 3 . FIG. 30A. ART. :so INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 85 in this member due to a local load unity at M 3 which gives the total ordinate under M 3 . A load at M z will be carried entirely by U 2 M 3 and M^U~ 3 , hence the stress due to a unit load is readily found by resolving the unit load into components parallel to these two members as indicated by the triangle of forces in the figure for Case I, drawn to half the scale of the influence line ordinates. The stress in U 2 M 3 is thus found to be +0.5, which is laid off from m to complete the influence line A'U 2 'M 3 L 3 B' for the member U 2 M 3 . It is seen that the point M' 3 does not fall on the line B'A" as usually assumed by authors on this subject. The error from this assumption is more apparent in Case II, and is not always on the safe side. Case II. The influence line for the primary stress in U 2 L 3 is as before, and again gives maximum for the lower portion M 3 L 3 . However, a unit load at M 3 is now carried to the members U 2 M 3 and L 2 M 3 , and these stresses are again found by a triangle of forces drawn in the figure for Case II. The stress thus produced in U 2 M 3 is now +0.62, from which the new influence line for U 2 M 3 is readily obtained as A'U 2 'M 3 L 3 B'. Case III. Here the primary stress holds good for the upper portion U 2 M 3 and the stress in the lower portion M 3 L 3 is diminished by the compression due to local loading at M 3 . A unit load at M 3 is now carried by M 3 L 3 and M 3 U 3 and the triangle of forces again gives the ordinate iM 3 , which is here negative. Case IV. This is like Case III, but a local load at M 3 is now carried by L 2 M 3 and M 3 L S equally. Hence the stress in M 3 L 3 due to a unit load at M 3 is found as 0.62, which gives the ordinate iM 3 . Members L 2 M 3 and M 3 U 3 , whenever they occur alone, can act only as secondary members to carry the portion of load locally applied at M 3 . When the panel is subject to counter stress as in panels L 4 L 5 and L 5 Le, then the local stress must be combined with the counter stress when a counter U^MQ is employed. However, it is usually preferable to omit the counter member and design the main diagonal to take both the direct and counter stresses. The verticals U 2 L 2 , U 3 L 3 , etc., may be treated entirely as primary members assuming the panels to be L 2 L 3} L 3 L 4 , etc. The vertical UjLi is merely a suspender for the double panel AL 2 and its maximum stress will occur when the span is fully loaded from A to L 2 . The member M 6 V, and others of similar character often employed in large bridges, are entirely secondary and perform no work as truss members. The sole purpose of these members is to stiffen certain long compression members. Bottom chord member L 2 L 3 . The influence line for this member is easily drawn. The center of moments is at U 2 and the end ordinate at A is the stress in the member for a unit reaction at A. The influence line thus becomes a triangle A'L 2 'B'. Case I. With the diagonals all as tension members the local loads at the middle points cannot affect the bottom chord stresses and the influence line just described gives maximum stress. Case II. Here the bottom chord stress is increased by the local load transmitted from M 3 to L 2 . This is shown in Fig. 30A. For a unit load at M 3 the stress in the 86 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. V bottom chord is the horizontal component in L 2 M 3 found in the small force diagram for Case II. This stress, equal to +0.36 i _must then be the additional plus ordinate under M 3 and the stress influence line for L 2 L 3 thus becomes A'L 2 K'L 3 B'. as shown. Case III is exactly like Case II, and the horizontal component of M 3 L 3 due to a unit load at M 3 will be the positive ordinate to be added to the ordinary influence line under K. Case IV. Here the two members L 2 M 3 and M 3 L 3 will transmit the unit load from M 3 after the manner of a three-hinged arch. The local stress in the oottom L 2 L 3 20 chord then becomes ===== = = +0.358 for a unit load at M 3 . 4M 3 K 4X13.95 Hence the positive ordinate under K must be increased by 0.358 to determine the proper influence ordinate under K. Top chord member U 2 U 3 has four cases as just described for the bottom chord. Case I. Here the local load at M 3 produces a compression in the top chord like a suspension s} r stem. The amount of this compression for a unit load at M 3 is readily found from the Eq. (27o) given for three-hinged arches and is H' =l\L 2 /fd cos a, where li +I 2 =d=pa,nel length, see the truss diagram Fig. 30A. This will give the negative stress ordinate to be added to the negative stress ordinate under M 3 for the final influence line. Cases II and III are similar and here the local effect in the top chord is the com- ponent in U 2 M 3 or M 3 U 3 parallel to U 2 U 3 . Case IV does not influence the top chord for local load at M 3 . It is similar to Case I for bottom chord. It is thus seen that the best criterion for trusses of this type is obtained from influence lines and the ease and clearness with which any case is solvable speaks greatly in favor of this method of analysis, where critical positions of loads and stresses are all given from the one solution. In conclusion it may be remarked that the type of panel illustrated in Case I is in all respects preferable to the others. The type Case II deserves second choice, and the others, especially Case IV, should be avoided whenever possible. The panel L\L 2 is in a certain sense indeterminate and should never be used except in end panels as here illustrated, where the several posts tend to steady the shock of trains coming on the structure. CHAPTER VI DISTORTION OF A STATICALLY DETERMINATE FRAME BY GRAPHICS ART. 31. INTRODUCTORY The displacement between any two points or pairs of lines of any determinate frame may be found analytically as shown in Arts. 6 and 9. However, many problems, to be considered later, require a complete solution for every pin point of a structure, and then the analytic method would become quite laborious. The graphic solution here given includes the two following problems: (1) To find the distortion, or change in form, of a frame resulting from changes in the lengths of its members, first published in 1877 by the French engineer Williot; and (2) to find the effect on this distortion produced by a yielding of the supports, or reaction displacements. This latter contribution came from Professor Otto Mohr in 1887, and without it the Williot diagram had little real value. Hence it is proper to call the complete diagram a Williot- Mohr displacement diagram, contrary to the position taken by Professor Mueller- Breslau, who erroneously credits the entire subject to Williot. Having given the changes in the lengths of all the members of a structure, by Eq. (4fi), when the stresses and cross-sections are known, let it be required to find the hori- zontal and vertical displacements of all the pin points. The Williot diagram, to be presented first, offers a partial solution of this problem, which is completed by Mohr's rotation diagram. ART. 32. DISTORTIONS DUE TO CHANGES IN THE LENGTHS OF THE MEMBERS BY A WILLIOT DIAGRAM Since any determinate frame is formed by a succession of triangles, this elementary frame is first examined. The process may then be extended to include any number of such triangles or a complete frame. In Fig. 32A suppose the point c to be connected with the points a and 6 by members 1 and 2, and that the lengths of these members also undergo changes Al and + J2, respectively, while the points a and b move to new positions a' and &'. It is now required to find the direction and amount of displacement resulting for the point c. Both members are first moved parallel to themselves until they occupy the positions a'c 2 and b'c i} respectively. The length of member 1 is now shortened by the amount Jl because this is a negative change; similarly the member 2 is lengthened by the amount J2, giving the new lengths a'c 4 and b'c 3 , respectively. 87 8S KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI With a' and 6' as centers and corrected lengths a'c 4 and b'c 3 as radii, describe arcs which may intersect in a point c' representing the new position of the point c. But because Al and Al are always very minute compared with the lengths of the members to which they belong, the arcs c 4 c' and c 3 c' are replaced by their tangents, drawn respec- tively perpendicular to the original directions of the members. The diagram shows these displacements several hundred^ times larger than they actually are compared with the real lengths of the members ac and be. It is clear that the point c' might have been determined in a separate figure, without including the members themselves, by dealing exclusively with the directions and changes in the lengths of these members. Thus, in Fig. 32B, the Jl's are plotted on a very large scale, affording greater accuracy in the results. If the point is regarded as a fixed point or pole, and if from this pole the dis- placements of the points a and 6 are drawn, making Oa' =aa' and Ob' =66', both in magnitude and direction; also, applying at a' the shortening Jl in the length of the member 1; and at b' the lengthening +J2 of the member 2; then the perpendiculars FIG. 32B. erected at the extremities of A\ and J2 will intersect in c' , which is the new position of c. The displacement of c is then represented in direction and amount by the ray Oc'. In laying off the values of Al, in Fig. 32s, the following rule must be observed with regard to signs: If a' is regarded as fixed, then Al representing the shortening of ac, it follows that c moves in the direction from c toward a, and hence Jl must be drawn from a' in the direction of c to a. Likewise, if 6' is regarded fixed, then since +J2 represents a lengthening of be, it follows that c moves away from 6 and hence +J2 must be drawn from 6' in the direction of 6 to c. Any succession of triangles, as in Fig. 32c, may be treated in the same manner. It is necessary, however, to assume that one of the members, as ab, retains its direction, and that some point of ab, as a, remains fixed. The changes Al in the several members being given, the construction is carried out as follows : The point 0, Fig. 32c, is the assumed pole, and since the point a is considered fixed it must coincide with 0. Also, since the member ab does not change in direction, then Jl is drawn parallel to ab and, being negative, it must be applied in the direction of 6 toward a. The displacement of the point 6 is thus given by Ob'. ART. 33 DISTORTION OF A STATICALLY DETERMINATE FRAME 89 The point c recedes from a by an amount +J3, and approaches b by an amount J2. If, therefore, J2 be drawn from b' in the direction of c toward b and parallel to cb, and if J3 be drawn from or a', in the direction of a toward c and parallel to ac, then the intersection c' of the perpendiculars, to the extremities of J2 and J3, will determine the new position of c. The displacement of c is thus given, both indirection and amount, by a ray Oc' not drawn in the figure. The point d is joined to a and c by the members ad and dc, Fig. 32c. Its displace- ment Od' is now found, in Fig. 32o, by drawing +J4 parallel to ad from and in the direction a to d; also by drawing J5 parallel to erf from c' in the direction d to c. The intersection d' , of the two perpendiculars to the extremities of J4 and J5, is the new position of d. Finally the new position e' of the point e is similarly found by applying the same method. I \ \ \ \ \ FIG. 32o. Fig. 32o, whose polar rays Ob', Oc' , Od' , and Oe' give the displacements of the several points b, c, d and e, both in direction and magnitude, is called the "Williot displace- ment diagram," for the frame abcde, giving credit to the name of its originator. ART. 33. ROTATION OF A RIGID FRAME ABOUT A FIXED POINT PROFESSOR MOHR'S ROTATION DIAGRAM In the Williot diagram it was assumed that the member ab did not alter its position, but merely changed its length. However, this is not always the case and more frequently this member ab is also subjected to a rotation about some instantaneous center. In the latter event, the elastic displacements, given by the Williot diagram, will require further changes to make them comply with the initial change in the position of the member ab. It is supposed, for the present, that the frame has undergone its elastic deformation and has passed into a rigid state. The solution of this phase of the problem was first proposed by Professor Mohr and is based on the following theorem in mechanics: The motion of a rigid body, at any given instant, may be defined as a rotation about a certain point called the instantaneous center of rotation, and the direction of motion of any point of this body, at the instant in 90 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI question, will be perpendicular to the line joining such point with the instantaneous center of rotation. Thus, if abcde, Fig. 33A, represents a rigid frame rotating about the instantaneous center P, each arrow, perpendicular to its respective ray, will then represent the direction of motion of the point to which the ray is drawn. Now, if from any pole 0, Fig. 33B, a series of radial rays be drawn respectively parallel to the arrows representing the motions of the several points, then these rays Oa", Ob", Oc", Od", and Oe" will in turn represent, both in magnitude and direction, the several displacements of the several points, a b, c, d and e. The figure a"b"c"d"e" will be the new position of the rigid frame abcde. For, a"0-LaP; b"0-LbP; c"0-LcP; etc., because the direction of motion of each such point of the rigid frame was perpendicular to the ray joining this point with the instantaneous center of rotation. Also. a"0:b"0:c"0, eic.=aP:bP:cP, etc., because the displacements of the several points a, b, c, etc., are respectively proportional to their velocities, and these in turn FIG. 33A. FIG. 33n. are proportional to the distances of the several points from the instantaneous center of rotation. From these conditions it. follows that: (a) If the points a", b", c" , etc., of the diagram Fig. 33B, are joined by straight lines, so that for every member ab of the frame, there will be a corresponding line a"b" in the displacement diagram, then these latter lines will produce a figure a"b"c"d"e" which will be similar to the rigid frame. (6) Also, the line joining any two points of the rigid frame, as ab, will be perpen- dicular to the corresponding line a"b" of the displacement diagram. Hence, if it is possible to determine the position of any two of the points a", b", c", etc., of the displacement diagram, then the figure a"b"c"d"e" can be drawn by- similarity with the figure abcde. The method of determining two such points a" ', b" will be illustrated in the examples. Therefore, the secondary displacements, due to rotation of the rigid frame, are found by inserting in the Williot displacement diagram a figure a"b"c"d"e" similar to the frame abcde, and having its sides respectively perpendicular to the sides of the ART. M DISTORTION OF A STATICALLY DETERMINATE FRAME 91 latter frame. The original displacements Ob', Oc', etc., are then combined with the displacements Ob", Oc", etc., due to rotation to produce the resultant displacements V'J', c r '7, etc. The complete solution for the displacement of the pin points of any framed structure, definitely supported, may then be outlined in the three following steps: (a) Assume any member and a point on the axis of this member as fixed and construct a Williot displacement diagram. (b) Then assume the frame as rigid and subjected to a rotation such as will conform to the actual conditions of the supports. (c) The displacements of the several pin points may then be found, in magnitude and direction, by scaling the distances between the m" and m' points. The direction of any displacement will always be from m" toward m' '. The complete diagram, combining the Williot displacement diagram with the Mohr rotation diagram, will hereafter be known as a Williot-Mohr diagram. The analytic solution of deflection problems is conducted with the aid of Professor Mohr's work, Eq. (GA), and may be advantageously employed when only one point of the structure is to be treated, or when great accuracy is required in the solution. This method is fully discussed in Art. 6 and, in combination with Maxwell's theorem given in Art. 9, offers the most accurate analytic solution of a great variety of problems dealing with displacements of points and lines. However, all problems requiring a complete solution for all the pin points of a structure, may be solved by a Williot-Mohr displacement diagram when extreme accuracy is not necessary. Attention is called to the fact that the value for E, in all deflection problems, should be chosen low rather than high, because there is always a slight amount of lost motion and permanent set, which follows the first loading of a new structure and which is not truly an elastic deformation. This can be compensated for in the calculations by assuming a small E. As a general rule, the changes Al, in the lengths of the members, are best figured on gross sections rather than on net sections. Experience indicates that this gives results nearer the truth. Also the stresses in the members must be simultaneous and not maximum. ART. 34. SPECIAL APPLICATIONS OF WILLIOT-MOHR DIAGRAMS (1) Distortion of a simple truss. Fig. 34A has been solved in three ways, for the purpose of illustrating that the displacements are not affected by the choice of the member and point which are regarded as fixed in position, and also in order to show how to select the most convenient of the possible forms of solution. The truss is supported on rollers at e and is fixed at a. The extensions (+) and contractions ( ) of the members are assumed to be as marked upon them. The solution given in diagram 6 is made under the supposition that the direction of the member af and the point a remain fixed. According to the method outlined in Art. 32, a displacement diagram is constructed 92 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI x DOUBLE *C/UE. / y /(d)\ OF DISPLACE MCNTS. i , , , i , . . , i ,,,,11 o i 2! FIG. 34A. ART. 34 DISTORTION OF A STATICALLY DETERMINATE FRAME J3 with pole at 0. Since a is now supposed to remain fixed, the point a' coincides with 0, and the displacements of the various points 6, c, d, e and / are obtained by scaling the distances Ob 7 , Oc r , Od' . . .Of. In reality, however, the direction of the member uf does not remain fixed, for the member revolves about the point a. Therefore, the displacements just found must be combined with those displacements 06", Oc", Od", etc., which are caused by revolving the rigid frame, or truss, abge about the point a until the point e shows a resulting displacement, e'e" ', parallel to the direction of motion of the roller bed at e. In other words, the resultant movement of the point e will be horizontal when the roller bed is horizontal. If the bed were inclined, as em, diagram a, then the point e would move parallel to em. The figure a"b"c"d"e"f"g"h", similar to the figure of the truss, can then be con- structed, making the sides respectively perpendicular to the members of the truss. This figure is definite and can be drawn when the points a" and e" are found. The point a" will coincide with a' or 0, since it remains fixed and a"e" will be perpendicular to ae. Also e'e" will be parallel to the roller bed, which is horizontal in the present case, but may be in any direction, as e'm" ', for a skew-back. The actual displacements of the points 6, c, d, etc., will then be given in direction and in amount by the distances &"&', c"c', d"d f , etc., but the horizontal and vertical projections of these displacements are more generally desired and may be scaled from the drawing. The deflection polygon of the bottom chord is found graphically by projecting the points a', a", 6', b", etc., on to the verticals through the panel points a, 6, etc. The points a", 6", c", etc., will be projected in a'", 6"', c'", etc., and will form a straight line a'"e'" . The points 6', c' , d' , etc., will fall in 6 , c , d , etc., and the ordinates 6 6"', C(,c'" , etc., will give the vertical deflections of the panel points 6, c, etc. In diagram c the direction cb is assumed and the point c is fixed. All that has been said regarding the first solution applies here. It will be seen that this solution gives exactly the same displacements as previously found, while it occupies only about half the space of the first diagram because the member be has a lesser angular motion than the member af. The third solution, diagram d, is the simplest of all, and its diagram covers the least area; for the member eg, which is now assumed as fixed, j-eally has no angular motion, but simply drops vertically. The relative displacement b'g' of any two points, 6 and g, may be scaled off directly. Although this displacement diagram was drawn on the assumption that the line eg and the point c remained fixed, nevertheless any two points may be directly compared. Hence, any point may be chosen as the origin of coordinates from which to scale the horizontal and vertical displacements of all other points relatively to this origin. Naturally the displacements desired would be those with respect to the point a, for this is the fixed point, giving for the point e a horizontal movement from a' to e'. All other points move to the right and down from their original positions by amounts which may be scaled from the diagram, taking a' for the origin. Hence it may be concluded that if a truss contains a member which will move 94 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI parallel to itself, or which has no angular motion, then this member is the best one to choose as the fixed member in constructing the displacement diagram. The rotation displacements are thereby eliminated. (2) A French roof truss. Fig. 34s. In this example the stresses were found from a Maxwell diagram, and the corresponding changes in the lengths of the members were obtained by means of the formula Jl=Sl/EF, taking E at 25,600,000 pounds per square inch for wrought iron. Changes of length, due to temperature, are neglected. The values of Al being very small, ten times these values were taken. All the data necessary in the solution of this truss are given in the following table: TABLE OF THE VALUES OF S, F, I, AND 10JZ IN FIG. 34n Member. Stress, Lbs. S Area, Sq. in. F Length, Inches, I 10UO, Inches. Member. Stress, Lbs. S Area, Sq. in. F Length, inches. I 10(40, Inches. 1 -21,800 6.82 83.53 -0.106 15 -20,970 6.82 83.53 -0.102 2 -20.040 6.82 83.53 -0.094 16 -19,210 6.82 83 . 53 -0.091 3 - 18,300 6.82 83.53 -0.087 17 -17,470 6.82 83.53 -0.083 4 -16.540 6.82 83.53 -0.079 18 -15,710 6.82 83.53 -0.075 5 + 19,730 2.02 96.53 + 0.370 19 + 15,330 2.02 96.53 + 0.288 6 + 16,430 2.02 96.53 + 0.307 20 + 13,570 2.02 96.53 + 0.256 7 + 9,040 1.86 100 . 08 + 0.189 8 + 9,260 1.86 96.53 + 0.189 22 + 6,200 1.86 96.53 + 0.126 9 + 12,560 1.86 96.53 + 0.256 23 + 7,960 1.86 96 . 53 + 0.161 10 - 3,300 1.40 48.07 -0.043 24 - 1,680 1.40 48.07 -0.024 11 + 3,300 0.78 96.53 + 0.162 25 + 1,680 0.78 96.53 + 0.087 12 - 6,600 2.48 96.53 -0.102 26 - 3,520 2.48 96.53 -0.055 13 + 3,300 0.78 96.53 + 0.162 27 + 1,680 0.78 96.53 + 0.087 14 - 3.300 1.40 48.07 -0.043 28^ - 1,680 1.40 48.07 -0.024 The truss is composed of the two frames aeh and sqh, designated as I and II. These are connected by the member as and by the pin at h. The displacement diagram of frame I is first drawn, assuming as fixed the direction of any member, as ab, and the position of a point, as a, of this member. The point a' then coincides with the pole 0, and the displacement Ob' , of the point b will be equal to J12. The points c' , d', e*, f, g' and h' are then found as directed in Art. 32. The displacement diagram of frame II is next drawn, assuming as fixed the direction of the member sk and the position of the point s. Having thus determined the points k' , I', m', q' , ri ', p' and h' in diagram II, the relative changes in the positions of the points h, s and q, parallel to the straight lines joiningjthe points h and s, h and q, and s and q, may be found. These changes are called Ahs, Ahq and Asq. Diagram I may now be completed by inserting the values Ahs, Ahq and Asq previously found from diagram II. The point s' is found from Al and Ahs, and the point q' is found from Ahq and Asq. Since q moves on a horizontal roller bed and since e is fixed, the figure c"d"b"g"h"f"a"s"q" can be drawn as in the preceding problem. This figure gives the displacements of all the points of frame I, and those of the points s and q. ART. 34 DISTORTION OF A STATICALLY DETERMINATE FRAME The displacement diagram of frame II may now be completed by transferring the displacements of the points h and q from diagram I into diagram II, thus determiniaf Lj_ loo \ DIAGRAM I. \ N DISPLACEMENTS. i DIAGRAM H. too zoo FIG. 34s. the points q" and h" from which the figure q"l"s"ri'h"m"k"p" can be drawn. As a check, it should be remembered that the line q"h" in diagram II must be 96 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, vi perpendicular to the line hq, and that the displacement s's" of the point s must be the same, both in direction and in amount, in both diagrams. Diagram II might have been dispensed with in the present case, as the values dhs t Ahq and Asq might have been found directly by summing the J/'s. However, the use of diagram II is general, and it becomes necessary when the points h, p, k, m and q, or h n and s, or /, s and q, are not in straight lines, as in the case of a curved top chord. Measurements from diagram I show that the point g undergoes the greatest displacement, having a horizontal movement of 1.93/10=0.193 inch to the right, a vertical downward movement of 2.33/10 =0.233 inch, or a resultant movement of oV 7 =3.02/10 = 0.302 inch. The horizontal movement of the point q is = q'q" =2.61/10 = FIG. 34c. 0.261 inch. The displacements given by the diagram are here divided by 10, because, as already stated, the changes Al were originally taken ten times too large. (3) A three-hinged arch. Fig. 34c. It is required to draw the displacement diagram. Independent diagrams for each of the elastic frames, I and II, are first drawn by assuming as fixed the direction of, and a point on, some member of each frame, as was done in the previous problem. The frames I and II are then regarded as rigid, and each frame is supposed to revolve in such a way as to satisfy the conditions imposed by the supports. The displacement diagrams are omitted and only the second part of the problem is solved. ART. 34 DISTORTION OF A STATICALLY DETERMINATE YRA ME 97 Supposing that the points a' and c' in Fig. 34c, diagram I, and the points b' and c', diagram II, have been found from the corresponding displacement diagrams (not shown) , let it be required to complete the solution by adding Mohr's rotation diagram. The following conditions must exist between the figures a"c" and b"c" , which are to be similar to the frames I and II respectively: a. The displacement of the point a is zero, hence a" will coincide with a'. b. Similarly b" will coincide with b'. c. The line a"c", in diagram I, must be perpendicular to the line ac. d. The line b"c" , in diagram II, must be perpendicular to the line be. e. Both diagrams, I and II, must give the same displacement c'c" for the point c. Hence a"c" is drawn through a', perpendicular to ac, also b"c" through b' perpen- dicular to be. Now, in diagram I, c'n is drawn parallel to ac, intersecting a"c" in n, and the projection of the required displacement c'c", parallel to ac, is thus obtained. Likewise, in diagram II, c'm is drawn parallel to cb, intersecting b"c" in m, giving c'm, the projection of c'c" parallel to cb. Diagram I may now be completed by transferring c'm from diagram II and erecting a perpendicular to c'm at m. This perpendicular will intersect na"c" in c", and the figure a"c" can now be drawn by similarity with the frame I, since the members in the two figures are respectively perpendicular. In like manner diagram II may be completed by transferring c'n from diagram I and drawing nc" perpendicular to c'n, thus determining c". As a check, the displace- ments c'c" must be equal and parallel in the two diagrams. (4) A cantilever bridge, similar in principle to the Memphis bridge, is represented in Fig. 34D. It is assumed, as in the two preceding problems, that separate displacement diagrams have been made for each of the elastic frames, I, II, III, and IV, and of these, the diagrams for I and III can be completely solved, as was done with the example in Fig. 34A, since each is a determinate framed structure on two supports. It is here deemed necessary only to complete the rotation diagrams of the frames II and IV. The displacements of these frames depend on those of the points c and m and upon that of the point s, respectively. Let it be assumed that the points d' and n' have been determined for the elastic frame II as in diagram II, and let it be required to find the figure d"n" which shall be similar to frame II. The members cd and ran are elongated by Jl and J2 respectively, and their elonga- tions must be applied to the^points d' and n' in diagram II, giving the points c' and m'. Now if the displacements ~c'c" and m'm" be drawn as found by diagrams I and III (not shown), the points c" and ra" are obtained. These represent the original positions of the points c and ra. The point n" must have originally occupied the same relative level with ra", also d" must have been_at the same relative level with c". Lastly, the line = .-= = =---. r r un r mh r The two new elastic loads, which were substituted for w, may now be evaluated from the following equations, without involving the lever arm r, thus: Wu =* and w n (36c) 104 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. Yii In most cases this simple relation proves of valuable assistance, because r is frequently outside the limits of the drawing. Regarding the signs of the two elastic loads iv u and w n it is discerned from the above that the signs must always be opposite and it is necessary to distinguish which of the two is the positive load and then call the other one negative. The sign of Al alone determines this without regard to the sign of the work S a Al and Professor Mohr gives the following simple rule, which will always identify the positive elastic load w. Calling all top chord members negative and all bottom chord members positive and giving the proper sign to Al for the web member in question, then the positive w is found on that side of the section or panel where the sign of Al coincides with that of the adjacent chord. The same result is obtained from the force polygon, Fig. 36s, drawn for the three loads w. This is also seen when the distortion in the quadrilateral ingun, due to the change Al in the member mn, is considered. Assuming the top chord mg immovable, then Al in mn will cause the point u to drop, hence w u is positive when the bottom chord is the loaded chord. Were the top chord the loaded chord, then nn would be considered immovable in applying this reasoning. When the center of moments o falls inside the span there is no load divide, whence the deflection influence line for a web member can have only a positive area. The maximum stress is then due to the total span loaded, the same as for maximum moments. However, this in no wise affects the previous discussion, but some proof of the identity of the effects of the substitute loads w u and w n with that of w must be furnished for this case. Thus the elastic load w=Al/r, acting at c, is in equilibrium with the reactions A and B resulting from w, and it is also the resultant of the two substitute loads w u and w n . Hence the sum of the moments of the forces, A, B, w u and w n about any point o in the plane of forces, must be zero. But the resultant w of w u and w n , always passes through the center of moments o, hence this resultant must be equal and opposite to the resultant of A and B. The method is thus general and applies equally to cases of o inside and outside the span. The elastic loads w. u and w n may also be expressed in terms of the moments of the external forces, as was previously done for the chords, and the following general expres- sions are thus obtained: Al Mp Al Mp w u = = and w n = = =F--; . . . . . . . (36n) U ^ r n r n r '~ '* also Al , Al Al Mp . 1,11 w ~= -*- = -f> hence -=- T - ...... (36 E ) v 'n ' > it r n This condition is easily seen directly from Fig. 36B. (c) The deflection polygon for the loaded chord. This deflection polygon may be found by applying the formulae (36B) and (36o) to any given frame. The elastic loads w are now computed for each chord and each web member and all those acting at the same pin point are algebraically added for total effects. The deflection polygon is then found by combining the loads w into a force polygon and drawing the moment or equilibrium polygon exactly as shown in Fig. 36A, following ART. 36 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 105 the description relating thereto. These moments may also be computed to obtain the deflections analytically. Before concluding this subject a few special cases will be illustrated. (1) When there is no end post, as in Fig. 36B, then Hitter's section cuts only two members 'ad and ac and some doubt may exist regarding the application of the formula for w. To overcome this difficulty, add a member ak\\cd and of any_convement length. Then the center of moments o l will be the_center for the member ad treated as a web member. The two imaginary members ak and kd are assumed rigid and Eqs. (36D> again apply. The load w thus found for the point a will have no effect, and the result is the same as would be found by applying Eq. (36n) to the center of moments at c with the lever r\. (2) When there is a vertical end post and the deflection polygon for the lower chord is required, then a load applied at a, Fig. 36c, will not produce stress in the verticals nor end post. The end chord section, and these members, therefore, do not influence the deflection polygon of the lower chord. t\ FIG. 36c. If the load is applied at the upper point c, then it is best to construct the deflection polvgon by regarding the Al=Q for the two end posts. Now since the whole top chord is lowered by amounts -M and -M, of these vertical end posts, it is merely necessary to raise the closing line of the deflection polygon by these amounts at the two respective ends. The deflection ordinates are thus corrected by the addition of the end pos (3) Frames with vertical posts, as in Figs. 36c and 36D. When the center of moments cannot be ascertained by a section through three members, as for the member w', Fig. 36D, the elastic load for the point a' cannot directly found. ,, To overcome this difficulty assume the disposition in Fig. 36E, where the small chord dl is interposed at a' and the vertical aa' is split into two posts such that the deflection remains the same as for the original case. Then the original solution is again applicable by passing two sections as indicated and cutting three members by each section The vertical post is now made up of two substitute posts for which the work mu* be equal to that of the original post, thus: SI \ 4-2-1 ^2l EF I 10G KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, vn Also for each substitute there will be a load w active at the center of moments o and the resultant of these two loads will be w a , Fig. 860. Hence, Jl Al In the ordinary case, Fig. 36c, the deflection polygons for top and bottom chords are alike when the web system is neglected. When the latter is included it is best to construct the deflection polygon of the chord which ends in the supports (the lower chord in this instance) and the deflection of the other chord is then found by simply correcting the first polygon by the Al of the verticals, being careful to regard the sign. (4) Trusses with parallel chords become very much simplified because the height of the truss and the inclination of the web members are then uniform, Fig. 36F. FIG. 36F. Hence for all chord members, Eq. (36s) gives Jl (36F) For the web members, the center of moments is at infinity, making the elastic loads equal and opposite for the same section. Each pair thus constitutes a moment which is balanced by an infinitesimally small force acting at an infinite distance. For this and other reasons it is better to determine the elastic loads for the web members directly from the shear. Calling the shear Q, Eq. (36o) then gives (36o) Here r u =r n for each panel and is the angle of the diagonal with the vertical. The figure shows the case where the w loads are desired for the bottom chord panel points. When = -. r * /.* Adding these expressions and solving for Aa\ + J2 = J gives Jxx Jl b / 7 ?- (37A) But J c cos i ={, cos 2 =/", hence Eq. (3?A) becomes .. (37B) ry b Now from Fig. 3?A, Jx_Jy_f f Mc_fc. Mb_fb. ^_ cot/? . and y._ cotr T~Y F J Z c ~' Z 6 ~' r~ *<*> r~ tr ' which values substituted into Eq. (3?B) give ^a = (/ a -/ 6 )cot r + (/ a -/ c )cot^. . _. , "."- \ . . (37c) The values EJ{3 and EAj- can be derived in the same way to obtain the following: E Aa - (f a -f b ) cot r - (f c -f a ) cot B 1 = (f b -f c ] cota-(/ a -/ 6 ) cot r \. . ."../. . (37n) = (f e -fJ cot B-(f h -ft cot a (3?E) wherein tensile stresses / are positive and compressive stresses are negative. Since the sum of the three angles of a triangle is always constant, therefore, whence the third value is easily found after having computed any two from Eqs. (37n). The above formula are extensively used in Art. 61 dealing with secondary stresses. When temperature changes are to be included then the unit stresses / must be corrected by dE. A positive t produces elongation in all members, hence dE must then be positively applied to all values/. The contrary is true for t. The change in any peripheral angle ^ of a frame is easily found by summing the changes Aa occurring in the several angles a whose apices meet in that angle 0. Then ^ = Sa and J^ = 2Ja, making EI>Aa ...... .... (37F) ART. 37 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 109 using the minus sign when 0=360 Ha which is the case when the vertices of the angles a are on the opposite side of the kinematic chain from the peripheral angle nij j. of the force polygon, then -r-=-fr. This value of ae inserted in Eq. (37c) gives ds ti Also from Fig. 37s, .Vi 2/2 =fe sin 4 :. The differential of Eq. (37j) gives Ji/i Ay 2 =A1 2 sin X 2 +1 2 cos k 2 d& 2 =i d 2 (37iv) which must equal o\ o 2 because 'both expressions represent the deflection of the point 2 with respect to the point 1. Also from the figure d 2 =1 2 cos X 2 which divided into Eq. (37K) gives and in like manner is found for + ^ , -- -j =-;- tan "3 h Making H = 1 and substituting these values into Eqs. (37n) then ~ tan A 2 +-r-^ tan A 3 ....... (37L) But 180->(2 + >l3 = 2 then - Also -r^=Tf and -7-^=^, whence Eq. (37i.) becomes 1 2 & /a Jb f 2 t&n A 2 +/ 3 tan 4 ....... (3?M) The general expression for any pin point n is then Ew n = E4 tyn -f n tan X n +/ + 1 tan 4 . + 1 ..... (3?N) in which Ed n is given by Eq. (37r) for top chords, using the sign for bottom chords. For any negative angle ^ (see Fig. 37s) tan A also becomes negative. Similarly the signs of the unit stresses / must be considered in substituting values in Eqs. (37r) and (37N). The equations as they stand are written for positive values in all cases. // the elastic loads w are multiplied by E and the force polygon be constructed to any ART. 37 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 111 convenient scale of forces with a pole H=E to scale of forces, when the scale of the truss diagram is l:a, then the deflections will be actual to the scale I: a. But if the deflections should appear m times actual when measured with the scale 1 : a, then the pole distance must be made E/m to the scale of forces. Finally the closing line A 'B' is found from the reaction displacements. If these are zero, then dA and d B are zero and the closing line must join the ends of the equilibrium polygon. Otherwise $A and d B must be ascertained from the conditions of the supports. When the web system does not include vertical members then it is preferable to choose these web members for the kinematic chain rather than a chain of chords, because the deflection polygon will then furnish the deflections of all the pin points instead of only those belonging to one chord. When the deflections in one direction only are required, then the above polygon affords a complete solution, but for a general determination of displacements in the plane of the truss, a second deflection polygon would be required, giving deflections perpendicular to the direction chosen for the first polygon. In the above illustration this would be parallel to the x axis and the resultant displacement of any point would be the resultant of the horizontal and vertical displacements of that point. The displacements in the x direction could be computed from the same Eq. (37isr) by substituting cot X for +tan A in each case, but a better solution is given below. The method fails when any angle ^=90, hence no member of the kinematic chain should be parallel to the direction of the deflections. If this occurs then a pair of substitute members must be employed as was done in Fig. 36B. For this reason also, the w loads from Eq. (37x) cannot usually be employed for finding horizontal displace- ments as per Art. 38. For a straight chord, the Eqs. (37F) and (37*) give the following simple formula for w n ......... (37o) Example. The above method will now be illustrated by the following example of a two-hinged framed arch as shown in Fig. 37c. The same problem is also solved in Art. 50, using the method given in Art. 36. The unit stresses and angle functions appear in the truss diagram of Fig. 37c arid the nomenclature used in Eqs. (37o) as indicated on a small triangle, is successively applied to all the triangles of the frame. The deflection polygons are drawn first for the bottom chord and then for the top chord, both for the conventional loading X a =l, and the computations of the Ew loads are given in detail for the bottom chord only, because these represent the general case necessitating the use of Eqs. (37D), Eq. (37r) and Eq. (3?N). For the top chord, the computations do not require the use of Eq. (3?N) and the resulting values of the loads Ew only are given without the details of the figuring. In the present problem, the method would not apply to the kinematic chain f/0^1^1^2^2, etc., because the verticals are parallel to the direction of the vertical deflections. Hence, the chords are treated separately, thus avoiding all vertical members in the chains. 112 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VI] 15* 15' U f g -O.OI07 3 The ^ ig" r "bo'v* +he members reprlcserrl unit SfreSSesfl Kip. BOTTOM CHORD DEFLECTION POLYGON All deflections af*- times actual 'in f>et, when meaeUred + scale of lengths . The figures in ft. 'are E time* actual I I I I I I I I I LENGTHS' DEFLECTIONS. FIG. 37c. AKT. 157 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 113 The elastic load Ew Q for the bottom chord has no effect on the deflection of this chord and is, therefore, neglected. Computation of the elastic loads Ew for the bottom chord panel points. For panel point L i : By Eqs. (3?D), #Ji =(-0.0257 -0.0503)3.059 - (0.0255+0.0257)0.454 = -0.2557 =0.0 (0.0503+0.0107)0.890 = -0.0543 = (0.0625-0.0347)0.1125 -(-0.0342-0.0625)1.394 ==+0.1379 ByEq. (37F), -EAfa =EI>Aa = -0.1721 By Eq. (37N), Ewi =0.1721 -0.0255x0.660+0.0347X0.5596= +0.175 For panel point L 2 : By Eqs. (37D), EActi =(-0.0342-0.0625)1.394 - (0.0347+0.0342)0.5596= -0.1733 EAa% =0.0 (0.0625 +0.0352)0.7173 = -0.0701 EJci 3 = (0.0791-0.0493)0.4192 -(-0.0498-0.0791)0.963 ==+0.1366 By Eq. (37r), -EJfa =EI>Aa = -0.1068 By Eq. (37N), Ew 2 =0.1068 -0.0347X0.5596 +0.0493x0.4312= +0.109 For panel point L :; : By Eqs. (37D>, EActi= (-0.0498 -0.0791)0.963 - (0.0493+0.0498)0.431 =-0.1668 #J 2 =0.0 (0.0791+0.0736)1.038 =-0.1585 = (0.1065-0.0687)0.821 -(-0.0657-0.1065)0.649 ==+0.1428 By Eq. (37r), -EAfa =EI>Aa = -0.1825 By Eq. (37N), Ew 3 =0.1825-0.0493 X0.4312 +0.0687 X0.3147 = +0.183 For panel point L 4 : By Eqs. (37o), EAon =(-0.0657-0.1065)0.6487- (0.0687+0.0657)0.3147 = -0.1540 =0.0 (0.1065+0.1000)1.542 =-0.3184 = (0.0889-0.0990)1.409 -(-0.0579-0.0889)0.462 ==+0.0536 By Eq. (37r), -EA$*=ET>Aa = -0.4188 ByEq. (37N), Ew =0.4188 -0.0687X0.3147 +0.0990X0.1865= +0.416 For panel point L- : By Eqs. (37D), EAa^ =(-0.0579-0.0889)0.462 - (0.0990+0.0579)0.1865= -0.0971 EJa 2 =0.0 (0.0889+0.1081)2.164 = -0.4263 = (0.0541-0.1207)2.111 -(-0.0270-0.0541)0.400 -0.1082 By Eq. (37r), -EA^=E^Aa = -0.6316 ByEq. (37ar), Ew 5 =0.6316 -0.0990X0. 1865 +0.1207 X0.0621 = +0.621 For panel point L, ; : By Eqs. (37D),jBJi= (-0.0270 -0.0541)0.400 - (0.1207+0.0270)0.062 =-0.0416 JJa 2 ==0.0 (0.0541+0.1290)2.500 = -0.4578 By Eq. (37p), -i^J0 6 =i^SJa= -0.4994 ByEq. (37u), ^Ewe =0.4994 -0.1207 X0.0621 = +0.492 114 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII The minus sign in Eq. (37r) applies here because the angles a are on the side of the chord opposite to the angle -dL^ FIG. 38A. by making the sides perpendicular to the respective rays of the force polygon and maintaining the same order of succession in the pin points formerly used in constructing the deflection polygon for vertical deflections. The horizontal abscissa B Q B" is then the horizontal displacement of the point B. Similarly the abscissa cfod" is the horizontal displacement of the point d, etc. For any horizontal member, as ef, the horizontal displacement of the point / with respect to e must naturally be the M for that member. Hence since_the scale of deflec- tions is m times actual to the scale of lengths the displacement e"f" =mAl to the scale of lengths used. If the point B is made to roll on some inclined plane ik instead of the horizontal, then a new closing line A7% must be determined for the two deflection polygons, as follows: draw a line 'B^' \\~ik and erect a perpendicular to AB prolonged, and passing through B". The intersection of these two lines gives k" and the vertical ordinate 3 B represents the vertical displacement of the support B. 116 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII Hence, making B'k' =d B the closing line A'k' may be drawn. Also a line A"k J_ A'k' becomes the required closing line for the horizontal deflection polygon. The same solution will apply to deflections in any direction other than 90 with the vertical. Also, when the deflection polygon for vertical deflections of all points of a truss is given, as in Fig. 36A, then the horizontal deflections of all pin points are found precisely as above by adhering strictly to the sequence in which these points occur on the given deflection polygon. The change d AB in the length of the chord AB, Fig. 38A, may easily be found by taking the loads w parallel to AB. In this case the displacement d^B becomes the intercept on AB produced and included between the extreme rays of the equilibrium polygon. Hence this displacement is equal to ,ww (38A) wherein y is the ordinate of any pin point measured normally to A B. This same result may also be found by computation from (38u) which follows for a pole distance of unity. For the case given in Fig. 37fi, where the angle changes A$ and the changes in the lengths of the members are given, the total effect on the length of span AB then becomes for n members, ^ B = Si"~V0 + SrJZ cos (;-), .--.. ,>.;-. . (38c) where a is the angle which the span AB makes with the horizontal. Example. The lengthening d^B for the span AB due to the loading X a = l in Fig. 37o is computed from Eq. (38c) , using the values E Al as given in Fig. 50A. The value \Ed AB was found to check the value determined graphically in Fig. 50s. The values Ed<{> are those above computed for the bottom chord panel points and a =0 because the span is symmetric. Point. E4$ Feet. EyJ E4l Fig. 50A. Feet. cos >l KJl co* X A 0.1721 6.188 1.0649 . 286 0.834 0.2385 L, 0.1068 14.58 1.5571 0.597 0.873 0.5212 L 3 0.1825 21.05 3.8416 0.805 0.918 0.7390 L< 0.4188 25.77 10.7925 1.080 0.954 1.0303 L, 0.6316 28.568 18.0435 1.511 0.983 1.4853 L, = 0.4994 29.50 14 . 7323 1.814 0.998 1.8104 = 50.0319 5.8247ft. These values substituted into Eq. (38c) give for a lengthening E times actual, 4B =iE27/^ + A#2J/cos >l= 50.03 19 +5.8247 =55.8566 ft., where -29000 kips per sq. inch. r. 39 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 117 ART. 39. DEFLECTION OF SOLID WEB BEAMS (a) Deflection due to moments. The differential equation of the elastic curve is dx2 = -E~r ' ' ' ( 39A > | Eq. (36s) gives the elastic loads w, for only one chord, in terms of moment and truss dimensions as Mp Ml 7/1 = L = r 2 EFr 2 ' Considering both chords, each of area F, and neglecting the web effect, which is usually quite small, then for a unit length of girder EFr 2 El ' Hence Eq. (39A) gives directly the elastic loads w per unit length of girder as d 2 v M (39s) An equilibrium polygon drawn for these loads, with pole distance equal to unity, gives deflections to the scale of lengths chosen for the drawing. By treating the moments M per unit length of girder, as loads which now become El times too large, and constructing an equilibrium polygon for these loads M with a pole distance H=El, the same deflection polygon is again obtained, giving deflections to the scale of lengths of the drawing. In other words the moment polygon for any case of loading becomes the load line for the deflection polygon corresponding to such case of loading. Deflections are usually drawn several times actual size, in which case the pole must be divided by as many times the scale of the drawing. Thus, if the scale of lengths is 1 :n and the deflections should appear twice actual, then the pole H =EI/2n. When the value / is variable, the pole distance is made to vary as a function of 7, as illustrated in the example below. See also Art. 47. (b) Deflection due to shear. From Eq. (15M) the deflection of a straight beam when acted on by shearing forces only, is sn ^f) (39c) Disregarding all unnecessary refinements, since shear deflections are small compared to the total deflection, tire value /? may be taken as constant and Q will be assumed uniformly distributed over the web plate of sectional area FI =^ 118 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII For a single concentrated load 'dQ/'dP m = 1 and Q =R =the end reaction, hence 3 / 30 , 8 / Taking (? =0.3332? and Fi=F/3 and assuming an average value for ,3=2.5 for| ordinary girder sections, then approximately (c) Deflection due to combined shear and moments. Ordinarily it will suffice to figure the deflection due to shear for a point at or near the point of maximum moments and to determine the percentage which this d m is of the moment deflection ordinate at the same point. All other moment deflection ordinates may then be increased in the same proportion to obtain the deflection polygon for combined shear and moment effects. For a uniformly loaded beam or constant / and neglecting shear effect, the deflection becomes 1 r = EIJ When the depth of a girder is constant, but the moment of inertia varies so as to maintain a constant stress on the extreme fiber at all sections, then the ratio M/EI becomes constant and the deflection for such a case would be -if: where I m is the moment of inertia at the section of maximum stress. When the moment of inertia varies abruptly the deflection may be expected to fall between the two above values, making the numerical coefficient close to 1/9. Taking in the shear effect, the approximate formula for maximum deflection, of a beam of variable /, becomes % '"max^ | ""*max fon^\ "-*"' ' ' However, for general cases of loading and variable 7 the graphic solution, above given under Art. 39A, should be employed. (d) Example. Given a plate girder of variable section and uniformly varying water load as shown in Fig. 39A, required to find its deflection polygon. The girder is designed as a double web beam and normally occupies a vertical position so that there are no dead load stresses. These girders are spaced 9.175 ft. center to center so that each will carry a full water load P=31.25X9.175(?-a) 2 giving rise to the end reactions A and B, and moments M, all as indicated in the diagrams, Fig. 39A. DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 119 FIG. 39A. 120 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VII The moments of inertia were computed for the various sections on gross areas, and are plotted in the upper diagram. The bending moments were also computed, using the formula (39H) and the results were plotted in millions of foot-pounds in the second diagram. The w loads were then taken as the moments M over certain convenient lengths Ax, instead of unit lengths according to the formula (39s) , whence The lengths Ax are usually chosen with respect to the girder sections, so that / is constant over each length Ax. Where the depth of section is variable the mean value of / is taken for each length Ax. Since M and Ax are both expressed in feet, while E and 7 are both for inches, the factor 12X12 = 144 is introduced into Eq. (39 j) to establish a true ratio between like units. Also, M and E are both expressed in millions of pounds, making the unit weight one million pounds. This furnishes the true values for w in terms of the given data as 144MAx MAx MAx ,_. . --' (39K) 144 wherein E =28 is chosen low rather than high. Now the scale of lengths was taken as 1 : 120, and if the deflections are to appear twice actual size then the pole must be made equal to (* OO T The pole distances are thus a constant function of 7 and may be found for all values of 7. The force polygon is now easily drawn to any convenient scale of forces, using the same scale for the loads w and the pole distances 77. This scale has no influence on the deflection polygon, but merely affects the size of the force polygon. The deflection polygon is then constructed as the equilibrium polygon of the w loads with the various pole distances, and the ordi nates included between this polygon and the closing line A'B' will represent the deflections to twice natural scale. Had the pole distances been taken twice as long, then the deflections would have been actual. The deflection due to shear, at the point of maximum moments, is found from 5x537X12 Eq. (39n) as o m = = ; ^-p-=0.09S inch which is 10.5 per cent of the maxi- mum deflection due to bending alone. ART. 39 DEFLECTION POLYGONS BY ANALYTICS AND GRAPHICS 121 The total deflection of the girder is therefore 10.5 per cent greater than the amounts given on the deflection polygon of Fig. 39A. The maximum deflection due to combined shear and bending effect is thus 0.93 +0.098 = 1 .028 inches, for the point 7. The approximate formula (39o) gives for this same point A 5.37X12X56.5 2 X12 2 . 07 = 9X28X134000 ' + See Art. 47 for another method of dealing with variable moment of inertia, by treating the quantities MAx/I as elastic loads and making the pole distance equal to E. CHAPTER VIII DISPLACEMENT INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURE! ART. 40. INFLUENCE LINES FOR ELASTIC DISPLACEMENTS Deflection influence lines could not receive adequate treatment in Chapter T\ because these depended on a knowledge of deflection polygons. The, latter were taker up in Chapters VI and VII. The subject is now treated from the most general aspec covering the influence line for any elastic displacement. Proceeding from Maxwell's law, Professor Mohr, in 1875, first developed displacemen' influence lines. The application results from a consideration of two conventiona loadings, of unit work each, applied to any frame so that one of the loadings represent; the desired influence at some point n, and the other loading is a vertical moving loac P = l applied at any load point m of the loaded chord. This includes all cases of con- ventional loadings given in Art. 9, besides the simpler problems pertaining to displace ments of points. As applied to any simple beam or truss, the following theorem is now established A deflection polygon, drawn for a load unity acting in a fixed direction on some detiniti point n of any frame, is the deflection influence line for the deflection of the point n, in thi flxed direction for any system of parallel moving loads applied to the loaded chord. Wher the system of moving loads does not consist of parallel loads, then, of course, no influence line is possible. To prove this theorem, the simple beam, Fig. 40A, is loaded with a vertical loac P n = l applied at the point n. A deflection polygon, drawn for this case of loading will then be the deflection influence line for the point n, according to the following demonstration. The method of drawing the deflection polygon is given in Art. 39. The moment diagram is first drawn for the conventional loading P n = l kip, appliec at n and thus making the maximum ordinate under n equal to M n = l-aa'/l=4.8 kip ft. The symbol M is used to indicate a moment due to a conventional unit load. This diagram is divided into suitable sections (2 ft. apart) and the moment areas MAx are computed and treated as elastic loads w, for which the deflection poly- gon A "B" is finally drawn, all as shown in Fig. 40A. Assume the modulus '=28000 kips per sq. inch, 7=2087, in inches, M in kip feet, and Jo? in feet. Then the pole distance for the force polygon becomes H =7/144 when the w loads are made equal to MJx, by Eq. (39j), for deflections of actual size, However, the scale of lengths of the drawing was made 1 :36 and the deflections should appear 200 times actual, hence H must be made 7/200X36X144=56.35 w units. 122 40 DISPLACEMENT INFLUENCE LINES 123 With a pole H= 56.35 and the w loads MAx, construct the force and equilibrium polygons as shown, and the deflection polygon so found represents the deflection influence ine for the point n according to the following application of Maxwell's law. The vertical deflection of any point m, for the given load P n = l, is represented by :he vertical ordinate rj m of the deflection polygon A"B", vertically under the point m. For, by Maxwell's law, rj m =d mn =d nm , or in words, T? OT equals the elastic displacement :>f the point n for a unit load P m acting at the point m. Hence m being any position |g.iKip. yn 24-" SOIb. I Beam , I =2O87.z in* ; I _ % , i- fcw 7 rl M DIAGRAM FOR P = I KIP. B Scale for len^thsland momenta. DEFJ.ECTIO|^ INFLL|ENCE L|NE FOR; POINT n. B Deflections zoo times actual. FIG. 40A. Df the moving load point, it follows that all ordinates rj m represent deflections for the point n due to a unit load at the variable point m. This makes the polygon A"B" the iesired deflection influence line for the point n. Therefore, for any train of moving loads, the deflection at the point n becomes (40A) Since Maxwell's law is generally applicable to any case of conventional loading, whether for a single point, a pair of points or a pair of lines, as illustrated in Art. 9, 124 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VIII therefore, the above application affords a solution for any displacement influence line other than those for simple vertical deflections. However, the train of loads must consist of parallel forces P, as otherwise no influence line is possible. ART. 41. SPECIAL APPLICATIONS OF DISPLACEMENT INFLUENCE LINES (a) Required the deflection influence line for the point n of the simple cantilever beam, Fig. 4lA. The deflection polygon is drawn according to the method of Art. 31) and illustrated in Art. 40, for the conventional loading P n = l at the point n. This then becomes the deflection influence line for the point n. FIG. 4lA. The beam is anchored at A and supported by a hinged bearing at R and a roller bearing at C. The point n is the hinged connection at the end of the cantilever arm Bn. The following moments M and reactions R, resulting from the conventional loading P n = l, are then found: A = --^, B=- L\ M A =0, MK-- ^ ^ V M n =0. The M diagram is constructed by making the ordinate at B' equal to 1 2 , and after dividing this diagram into suitable sections and computing the loads w\ to w 8 , the deflection polygon is easily drawn, observing the method followed in Art. 40. The closing line of the deflection polygon is then found to be A"B"C", which completes the influence line. The point B" is the intersection of the deflection line ART. 41 DISPLACEMENT INFLUENCE LINES with the vertical through B and n"C" is a straight line. Upward deflections are negative. The support at A is elastic and its displacement o Aa , due to the conventional loading P n =-l, should be computed from Eq. (4A) including temperature effect if desirable. This displacement is then applied at the point A" and the final closing line Ai"B"C\" i-s thus made to include this effect. By applying any train of loads the resulting actual deflection of the point n, according to Eq. (40A) becomes d n = HPi). For any case of variable moment of inertia of the given beam the pole distance H is made variable, as was done in Fig. 39A. (b) Given a simple truss, Fig. 41 B, on two supports, to find the displacement influ- ence line for any panel point n for displacements d n in the fixed direction rm 7 . The loading is to consist of a system of vertical concentrated loads P acting on the bottom chord. FIG. 41s. Apply the conventional loading P n = l in the given direction nn' and compute the reactions H, A and B for this load, then determine t-he stress S\ and resulting change Jl in length, for each member. Finally compute the w loads for the several panel points and draw the deflection polygon for the loaded chord (here the bottom chord) using any of the methods of Chapter VI, but preferably the one given in Art. 36 (c) . The direction of the w loads is always taken parallel to the system of loads P, hence the displacement influence ordinates jj will be vertical ordinates, measured vertically under the respective loads P. This is necessary because the direction of the deflections is determined by the direction assigned to the w loads. The proof that the polygon A 'B' is the desired influence line again follows from Maxwell's law, viz., for any ordinate rj m =d mn = d nm . The displacement d n of the point n in the direction nn', caused by the system of vertical loads P is then (c) Given the three-hinged arch, Fig. 41c, to find the influence line ior the angular rotation 3 n between the two lines An and Bn. The train loads are to be carried bv the top chord. 126 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. VIII The conventional loading now becomes one of loading a pair of lines with a positive moment equal to unity applied to each line in the direction in which the angle J would increase for vertical loads acting on the top chord. The horizontal thrust for this conventional loading is obtained by taking moments about the crown n and gives H = (2-{-AV)/ < 2f and the reactions A and :i are found by taking the moments of the external forces about B and A respectively. Then for this case of conventional loading determine the stresses S\ and changes Al in the lengths of the members and compute the w loads for the half span, which is all that is required for a systematic structure. FIG. 41c. The moment diagram, drawn for the w loads, will then be the displacement influence line for the change d n in the angle J. The d n for any particular set of loads is then d n =2Prj, measured in arc. The above problems will suffice to show the perfectly general solution of displace- ment problems afforded by the application of Maxwell's law. These problems can also be solved analytically with the aid of Mohr's work equation as indicated in Art. G. CHAPTER IX INFLUENCE LINES FOR STATICALLY INDETERMINATE STRUCTURES ART. 42. INFLUENCE LINES FOR ONE REDUNDANT CONDITION The principle, illustrated in Chapter VIII, for drawing deflection influence lines, is easily applied to the construction of the influence line for any external or internal redundant condition. Eqs. (7n) and (So) for one redundant condition and a single external force P = 1 become ' a a0 aa f- and < 42i > "a 1 ' "ma ~ X a O aa = X a f) a after substituting the value d a as obtained from Eq. (4A) and neglecting temperature and abutment displacements. This gives X a in terms of displacements or stresses, whichever is preferred, as Y - ' where the redundant may be external or internal. If the single load P = l is vertical and applied to some panel point of the loaded chord then Eq. (42s) represents the influence line for X a for a unit load applied at any panel point m. A deflection polygon, drawn for the conventional loading or condition X a = l, will give the ordinates r} m =d ma =d am = 'ZS S a p for the displacement- of the point a, for a unit load at any point m of the loaded chord. Hence the deflection polygon drawn for condition X a = 1 is the X a influence line provided all the ordinates are divided by the constant denominator d aa +p a = 'Sc?o + p a , The constant displacement d a a=2>S a 2 o is found by computation or otherwise, and p a is a given constant depending on the length and section of the redundant member. When X a is a reaction then ,o a =0 for the case of immovable supports. Accordingly the following theorem may be stated: The ordinates to the influence line of any redundant X a are some constant factor p. 1 ~ (d^ +p a ) times the ordinates to a deflection polygon drawn for the loaded chord, for the conventional loading X a = l (P=0), applied to the principal system. 127 128 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IX (a) When the redundant X=0', etc. ; and the following simple equations are obtained 1 . 9* X^ 7~> S ~V & \^ Z? Af* 1 J} 1 O b = Ztr'mOmb AfcOfcfc Ltti b ar + 0bt (44A) ART. 44 INFLUENCE LINES FOR INDETERMINATE STRUCTURES 133 involving only one redundant X in each equation. The same assumptions may be applied to Eqs. (7n). This method of analysis was first introduced by Professor Mueller-Breslau, and serves a most valuable purpose, especially when applied to fixed arches. As will be shown presently, this disposition of the redundants is easily made when their number does not exceed three. Beyond this number the analysis leads to many complications. Fortunately, the important cases of redundancy are nearly always limited to from one to three redundants, and then there is no difficulty in following the scheme here proposed. The solution of Eqs. (So) for simultaneous values of the X's is thus avoided and the other chances for serious errors are greatly lessened. It is usually expedient to treat temperature stresses and abutment displacements separately from the primary stresses, and then Eqs. (44A) become for load effects only and immovable abutments, Z2->imma -\r ^*m mb /it \ a = ^ ; x b = 5 ; etc (44s) *!=!--; *6i=l-rJ etc (44c) Oaa Obb By placing 2.R a Jr=0, d a also becomes zero, likewise for S-fi^Jr and d b , a circum- stance which should be noted in writing Eqs. (44s) and Eqs. (44c). In all of the following investigations, the X's will represent either a single force or a couple applied to a principal system. The d's in the first case will then represent linear displacements and in the second case they will be angular distortions expressed in arc. Thus, if a redundant X a is applied to the point a then ^ is the displacement of the point a in the direction of X a for a force X a = 1. When X a is a moment or couple, then daa becomes the angular rotation of a rigid principal system as produced by a couple X a --=l. The displacements d a , d b , etc., and the conventional loadings X a = l', X b = l, etc., are always positive in the opposite sense in which the redundants X a , X b , etc., are supposed to act. It should also be noted that in all cases here considered, the points a and b are coincident and these designations are retained merely to distinguish the particular forces from each other. The Eqs. (44A) to (44c) are equally applicable to graphic and analytic solutions, but the latter method is useful only when there are not more than three redundant conditions, and great accuracy is desired. The graphic method is less laborious and leads to a more comprehensive presentation. The above will now be applied to general cases of two and three redundants. (a) Structures having two redundant conditions. Here the two redundant forces can always be applied at the same point and the case can be solved by applying Eqs. (44s) and (44c), provided the directions of X a and X b are so chosen that #&, =&=(). 134 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. IX To accomplish this, assume the direction of X a as may seem most convenient, and find the displacement a\a of the point a for a load X a = l, Fig. 44A. Now take X b acting at the same point a and at right angles with aid. Then ^=0, because it represents the displacement of the point 6 in the direction of Xj, when only the force X a = l is active. Therefore, dab also becomes zero by Maxwell's law. If then the displacement 6j6 of X b is found for the load Xb = l, this in turn must be perpendicular to X a because d ab is zero. This always fur- nishes a valuable check on the solu- tion when the graphic method is employed. The example of a two-hinged arch with a column support at the center, Figs. 44B to 44D, is used to illustrate the application of the method to two redundants. Fig. 44B shows the given struc- ture with hinged supports at A, B and C, thus involving two external redundants. By removing the sup- port at A the structure becomes determinate, involving a simple truss on supports B and C, and an over- hanging cantilever AC'. The redun- FIG. 44A. dant forces X a and X b are then applied to the principal system at A and are treated as external forces. The first redundant X a is assumed to act horizontally and the second redundant X b is applied at the same point and making the angle 6 with X a , such that d a b=dba=0. Fig. 44c represents the conventional loading X a = l and a Williot-Mohr displace- ment diagram, drawn for this loading, will furnish the displacement aa t for the point A and the displacement mm a for the point m. Hence d ao becomes the pro- jection of act! on the direction of X a and a which makes it possible to represent the S influence area as the area inclosed between the X a influence line and the S /S a influence line, which latter is drawn with as much ease as the simple S line. The resulting S influence area will then require a factor S a applied to all its ordinates. These influence areas may be drawn in either of two ways: First, by plotting both the X a and the S /S a lines from a common straight base, observing the rule that areas above this base are negative, see Fig. 45u. The S line ordinates TJ will then_ be measured between the X a line and the S /S a line and these ordinates will be positive when measured down from the X a line. Second, by first plotting the X a line from a straight base and then constructing the S area by applying the +S /S a ordinates y' down from the X a line, see Fig. 45c. In this case the +X a ordinates are best plotted above the base so that the 4- TJ ordinates of the S area will appear below the base in the customary way. The first method is more generally used, though the second leads to a very interesting property by which the S line may be derived from the X a line when one point of the S line and its zero points are known. This property is shown in Fig. 45c, where it is seen that over any distance, as m'B', for which the ~S~ /S a line is straight, the corresponding elements of the S line and the X a line will intersect in points b', b" , &'"_which are in a straight line parallel to P and passing through the end zero point of the S /S a line. Further details and applications are not given here, as these will be illustrated in connection with specific problems in Chapter X. CHAPTER X SPECIAL APPLICATIONS OF INFLUENCE LINES TO STATICALLY INDETERMINATE STRUCTURES ART. 46. SIMPLE BEAM WITH ONE END FIXED AND OTHER END SUPPORTED The solution by influence lines is illustrated by Figs. 46A, using the general method developed in Art. 45B. Applying the criterion of Eq. (3c) to this structure, where m = l, 2r=4 and 2p=4, gives ra + 2r-2p = l. Hence there is one redundant condition and by Eq. (SB) this condition must be external. Accordingly, the reaction X a , acting at the expansion end B, is chosen as the redundant support, reducing the beam to a simple cantilever arm as the principal determinate system. The problem is considered solved when the shear and moment influence lines are found for any point n of the beam. The case of direct loading is assumed. The X a influence line is represented by Eq. (42o) as i ^ 1 . Ont, , . where dma is the variable influence line ordinate at any point m with [1 = 1/8^ is the constant factor, when d ma and daa may be acutal or measured to the same scale. The equations for moment and shear at any point n and for the vertical reaction A , are given by Eqs. (?A) written in the form of Eq. (45E) as ~J A Y -- 4 Y \ ** A a-A-a ~ -^rtl ~r~ -A-a \Aa (46B) wherein M is the moment about the point n produced in the principal system by a load unity acting at any point m of the structure. M a is the moment at the point n due to the conventional loading X a = l. Similar definitions follow for Q~ , Q*, also A^ and A a . 140 ART. 46 SPECIAL APPLICATIONS OF INFLUENCE LINES 141 The following solution by influence lines is based on these Eqs. (46B), when the influence line for X a is known or found from Eq. (46A) . The ordinates d ma are ordinates of a deflection polygon drawn for the conventional loading X a = l kip, being a load unity acting downward at the point B when the beam is treated as a cantilever. This deflection polygon is drawn as described in Art. 40, and as shown in Fig. 46Aa. The M moment diagram for the conventional loading X a = l, is drawn by making the end ordinate A^C' =1 and drawing C^B'. This end ordinate may be measured to any convenient scale and was drawn to half the scale of lengths, making all the moment ordinates to half scale of lengths. The figured ordinates represent moments, in kip feet, and are scaled at points Jz=2 ft. apart. The areas w =Jfdx are now computed and treated as elastic loads for the construc- tion of the force polygon (a) and the deflection polygon (b) when the pole distance H=EI Takino- E =28000 kips and 7=2087 in. 4 and reducing Ax to inches and M to kip inches then the pole would be . When the scale of lengths is 1:60 then, , ,, . ,, TT 28000X2087 O7n ^ nnita for deflections 25 times actual, this pole becomes #=744x60x25 =270 - When / is variable the pole is made variable as was done in Fig. 39A, or the w loads may be divided by / as is done in Art. 47. The force and equilibrium polygons are now drawn, using any convement_scale for the w forces and the same scale for H . This gives the deflection polygon A'C', Fig. 46Ab with ordinates 25 times actual. Thus the actual end ordinate ^ = 1.96/25 inches. This 'same polygon is also the X a influence line with a factor p = l/i) B irrespective of the scale of ordinates, since all ordinates rj m and TJ B are measured with the same scale. To obtain the X a influence line with a factory 1, divide all the ordinates of Fig. 46Ab by the end ordinate , Bf whence the curve A'C', in Fig. 46AC, is obtained with the end ordinate WC' = l, drawn to any convenient scale of ordinates. The redundant react X can thus be found for any case of concentrated loads P and is X a = ^Pr l /t) B . To obtain the M n influence area for moments about the point n, the X a influence line is now combined with the M /M a influence line, with a factor p = M a according t q ' In Fio- b the line ^C 7 represents the W influence line provided the end ordinate WC' is made equal to z, but this end ordinate is actually , fl> hence the linem question must have a factor p=z/y B . But since M a =z, therefore, the same line n C becomes the W /M a influence line with a factor p-lfa which is the same as the factor for the X a line. Hence the area IVC 7 represents the M n influence area with a fad / ' = ^irnifarly / ?n 5 'Fi g . c, the area A^V represents the M n influence area with the factor a=M a =z because ,' fl was here made equal to unity. This was done to show the two solutions and to illustrate the fact that the second step of drawing the Z influence line with a = 1 is really unnecessary and adds nothing except to simplify the final factor. The algebraic sign of the M n area is derived from_Eq. (46B) since the larger _ X area (which is positive) is subtracted from the positive M /M a area, leaving a negat area as the remainder. 142 KINETIC THEORY OF ENGINEERING STRUCTURES Oi- o/Q, LINE. C' FIG. 46A. ART. 47 SPECIAL APPLICATIONS OF INFLUENCE LINES 143 The moment area M m , for moments about a point m, is indicated to show a case where a portion of the area below the X a line is positive, creating a load divide at the point i. The moment area MA, for moments about the abutment A, is also shown. It is the area between the line A'C' and the X a line, which is entirely a positive area. In all of these moment influence areas, the moments obtained will be expressed in the same units as the applied loads P and the ordinate 2. Thus for P in Ibs. and z in feet the moments will be ft. Ibs. The scale of the ordinates y is immaterial and is eliminated in the factor p, Thus if Fig. b is used, the moment about n for any train of loads would be ii;-- S and using Fig. c it would be and the units would depend solely on those chosen for z and the loads P. To obtain the Q n influence area for shear at the point n, the X a influence line is now combined with the Q /Q a line, with a factor /*=Q a = l, according to Eq. (46e), see Fig. 46Ad. For a load unity to the right of n, the shear in the principal system would be Q) = l and for X a = l the shear Q a = l, hence Q /Q a = ^ so that the broken line A'n'DC' represents the Q /Q a influence line with a factor unity, and the Q n influence area is the shaded area with a factor /*=Q a = l, with the portion below the X a line positive, as before. Were Fig. b used, the factor would be / w=Q a /'?B == l/'?s- The point n is always a load divide for Q H . The end reaction A for any system of loads is simply HP X a , from Eq. (46B), because A = DP and A a = l. A uniform live load covering the whole span would give X a =3pl/8, as may be found from Eq. (15j). The graphic diagram, Fig. 46Ab, gave X a =7Al2p=0.372pl ART. 47. PLATE GIRDER ON THREE SUPPORTS The example chosen is a general case of unequal spans and variable moment of inertia as illustrated in Figs. 4?A. All dimensions in inches, and the loads are directly applied to the girder. According to the criterion of Eq. (3c) this structure involves one external redundant condition. Any one of the three supports may be taken as the redundant condition, but it is generally most convenient to assume the middle support at C and thus obtain the principal system as a simple beam on two supports A and B. The Xc influence line for the redundant support is again given by Eq. (42n) as (47A) where S mc is the variable influence line ordinate at any point m with the constant factor fi = \/d ect and d mc and d cc must be measured to the same scale. The ordinates d mc are 144 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x the ordinates of a deflection polygon drawn for the conventional loading X c = \ kip, acting downward at C on the principal system or simple beam AB. This deflection polygon is the equilibrium polygon drawn for a series of elastic loads* w = MJx/I when the pole H=E. The variable moment of inertia is not taken care of] in the manner previously given, by varying the pole distance, but the alternative method of involving / in the w loads is here employed. The moments of inertia are given for the various sections of the beam in Fig. 4?Aa. The M diagram for X c = 1000 Ibs. is drawn in Fig. 47Ab, with an ordinate under the point C equal to 1000 l^lo/l = 175,000 in. Ibs. The several M ordinates are scaled from the diagram or computed, and these values divided by the corresponding 7's furnish the figures for plotting the ~kZ/7 diagram^ Thus the ordinate at C is 1 75000 H- 34982 =5. 00. The w loads are then computed aa the areas of the M/l diagram using the horizontal distances between the ordinates in inches. These loads are applied at the centers of gravity of the respective areas. The w loads are now combined into a force polygon, Fig. 47AC, with pola H =El 100 XI 20 -=2333 w units, all drawn to the same scale, but this scale may be taken as any convenient one without regard to the deflections. The particular pole will then] give deflections one hundred times actual, because the scale of lengths for the girder span was chosen 1:120 and the pole was made 100X120 = 12000 times too small. I/ =28,000,000 Ibs. per sq. inch. The resulting equilibrium polygon 'A'C'B', Fig. 47ACJ is the deflection polygon for the load X C = IQOO Ibs. with ordinates 100 times actual and it is also the X c influence line with a factor 1/c where c is the ordinate under C. measured to the same scale as the other influence ordinates. It is clear from Eq. (4?A) that the scale of the influence ordinates is immaterial so; long as the same scale is used for the ordinate c, but when actual deflections are sought then a natural scale, as inches, must be employed. The reaction X c for any case of concentrated loads may then be found from Fig. 47AC, by multiplying the loads by the d ordinates and dividing the sum of these products by c, thus X c = ^Pd/c, where c and the d's are measured to the same convenient scale. The A and B influence areas are easily found when the X c line is given. Thus in Fig. 47 AC, the line B'C'A" represents the A /A a influence line with some factor, and the shaded area is the A influence area with a factor I/TJA to be proven. For this proof the end ordinate A' A" is evaluated according to Eq. (46s), and the factor required for this ordinate will be the factor for all ordinates TJ of the A influence area. For the point A, A = l and A c = l-l 2 /l and X c has the factor 1/c. Hence ~A f A c = 1/1 2 and the end ordinate should be i 2 \i x c -\ i 2 \d I = ~ = '' ...... (7B) where the Tactor is l 2 /cl. But = - = , hence the factor is simplv \fr\A. d C T) A ' ART. 47 SPECIAL APPLICATIONS OF INFLUENCE LINES 145 k 9i rlW P n t 5 1= fZ.U 6 7 a 3 10 u i i 1 1 3 S.94 *u= FIG. 47x. 146 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP.X . Similarly \/r IB is the factor for the B influence area indicated in Fig. 47 AC as the area included between the X c line and the line A'C'B". Hence the reactions A and B, for any train of concentrated loads, become A= 2Pr y and B= 2Pr i '. (47 c) f)A "T iB .-. The algebraic signs of the areas are determined as before. All areas below the X c line are positive and those above this line are negative. The ordinates may be measured with any scale so long as the same scale is used for all. The M influence area is a portion of the A area for all points between A and (7, while for points between C and B the M area is a portion of the B area. See Fig. 4?Ad. Thus for the point 4 the M 4 influence area is shown as the shaded area between the X c line and the broken line A'-i'B'. The factor is M C /C=X/TJ A , where x is the ordinate of the point 4. When x is expressed in feet the resulting moment is in ft. Ibs. Similarly the M-j influence area is indicated by dotted lines and the factor is M c /c=x/r) B . The usual directions as to algebraic signs and scales apply as before. The Q influence area for shear is shown, in Fig. 4?Ae, for the point 4 as the shaded area with a factor Q C /C = \/T)A the same as for the A influence are.a. The line A'4 7 is parallel to B'C'4". Similarly the Q 7 influence area is indicated by dotted lines and has a factor Q c /c = I/T)B' B'7' is parallel to A'C'7". The signs are again determined with reference to the areas above and below the X c line. All of these influence areas have a load divide at the point C' and the Q areas have two load divides each. This determines the positions of train loads for positive and negative effects, and to obtain maximum effects, the heaviest loads should be placed over the maximum ordinates. Temperature effects. A uniform change in temperature will expand the girder equally in all directions and will produce a slight lifting of the ends A and B equal to d(72- 12) =0.0105 inch, for =25 F. and =0.000007. When the three supports are on the same level then no stress will be produced by uniform temperature changes. However, when the sun shines down on such a structure experience teaches that the top flange is heated much more than the bottom flange and this difference in tem- perature may become quite considerable and will cause the girder to assume a curved position, convex upward, when the top flange is warmer than the bottom. The maximum difference in temperature between the two flanges is bound to remain somewhat problematic, but observations indicate that differences of 30 F. are quite common. In the present example Jf=+20 F. will be assumed and on this basis and that the temperature varies uniformly between the two flanges, a set of w t elastic loads is now computed from the formula w t = -sJUx/h = -0.00014 dx/h, where Ax is the horizontal distance between sections and h is the depth of girder, both in inches. AKT. 47 SPECIAL APPLICATIONS OF INFLUENCE LINES The following table gives the w t loads: 147 Point. dx Inches. h Inches. l 10,000u> ( Point. dx Inches. h Inches. w t 10,000u>, 1 60 19.5 0.00043 4.3 7 60 72 0.000117 1.17 ') 60 34.5 0.00024 2.4 8 60 72 0.000117 1.17 3 60 49.5 0.00017 1.7 9 45 64.5 0.000098 0.98 4 60 64.5 0.000115 1.15 10 45 49.5 0.000127 1.27 5 90 72 0.000175 1.75 11 45 34.5 0.000183 1.83 6 90 72 0.000175 1.75 12 45 19.5 0.000323 3.23 An equilibrium polygon drawn for these w t loads with a pole distance unity, would represent the deflection curve of the beam due to M. However, for convenience it is better to multiply these small loads w t by some factor as 10,000 and then make the pole distance 10,000 instead of unity. The scale of lengths for the girder was made 1 : 120 and if the ordinates are to appear say 5 times actual, the pole distance should be made equal to H = 10,000/5X120 = 16.67, using the scale chosen for the w t loads. Fig. 47Af shows the force and equilibrium polygons for the w t loads, with ordinates five times actual. Since these w t loads are all negative for +At, the 3 ct deflection curve was drawn above the closing line A 'B'. The ordinate under the point C represents 5d ct and this must be increased by the small ordinate 5x0.0105, previously found for the uniform rise in temperature to obtain total effect. According to Eq. (44c), the redundant reaction produced by this temperature effect is X ct =d ct /d ee . The deflection polygon for X e = l kip, gives 100 cc =0".93, whence + cc =0".0093, and the d ct deflection polygon gives -5d c< =0".74+0.05=0".79, making 3 ct = 0'M6. Hence X ct =d et /d ce = -0.16/0.0093 = -17.2 kips, which indicates a downward reaction at C to maintain the girder on the middle support. As the entire girder has an approxi- mate weight of 7440 Ibs. the above temperature effect would lift the girder off the center support and cause the beam to carry itself on two supports, producing a compression of about 3100 Ibs. per sq.in. on the extreme fiber. This temperature stress would not, however, be fully developed unless the span is loaded down in contact with the center support. The dead and live load stresses must finally be combined with the temperature effect to obtain the real stress at any point. The application of these influence areas is illustrated by placing a single load P n at any point n of the span and showing the values of the several functions. The same process is followed for each load of a train of loads and the total effect of the train is the sum of the individual load effects. The ordinates r? under the point n, in each of the influence areas, are all scaled to the same scale to which 7j_ 4 and rj B are measured. This may be any convenient scale, which is universally used for all ordinates of the drawing. 148 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X The values of the various functions for the load P n are Y fnOn _ ' -p _ A c = -- /" C 57. B = _M = _-M- PB = - .207P n ; ijjs 15.94 +-B=P, which checks to 1 percent. n =49.4P n in.lbs. for P n Ibs. X ct l 2 a 17200X300X194 '. M 4t J 2 - = - 72Q =1,390,330 in.lbs. ART. 48. TRUSS ON THREE SUPPORTS Figs. 48A illustrate the analysis by influence lines of a truss on three supports. The bottom chord is the loaded chord and the lengths and cross-sections of the members are written over the members in Fig. a, while the stress SA, produced by a unit reaction at A on the simple span AB, are written below the members. The stress diagram, from which the SA stresses are found, is shown in Fig. b. The solution is similar to that given in Art. 47. As in the previous problem, Art. 47, the present structure involves one external redundant condition, which is again taken as the middle support C, reducing the truss to a principal system on two supports, A and B. The X c influence line is derived from Eq. (42o) as in the previous problem and becomes (48A) where the variable influence line ordinate d mc is the ordinate for a deflection polygon, drawn for the loaded chord and for the conventional loading X c = 1 kip acting downward at C on the prinicpal system or simple truss AB. The influence line factor ju = l/ cc . The elastic loads for this deflection poylgon are found from Eqs. (36s) as : M c I - (48B) for each pin point of the top and bottom chords. The effect of the web members is usually neglected as being insignificant, but when it is desired to include their effect, Eqs. (36c) will give the w loads due to the web members and these are added to the ART. 48 SPECIAL APPLICATIONS OF INFLUENCE LINES 149 chord loads found from Eq. (48R). See Art. 36, for a complete discussion of this method, and Art. 50 for a complete problem. A Williot-Mohr diagram might also be employed to obtain this deflection polygon. See also the example in Art. 50, Fig. 50B. In the present example the web system will be_ neglected and the w loads arc found for the chords, using Eq. (48 B). The moments M are those produced by the conven- tional loading X c = 1000 Ibs. and are represented in the moment diagram, Fig. b, expressed in inch-lbs. The ordinate at the point C will be P//4 =528,000 in. Ibs. Using these moments and the values for /, F and r given, in Fig. a, for each member, the following computation of the w forces is made: w Member. A/c Inch-lbs. I Inch. F Square Inches. r Inch. Md Fr* 1 L L 2 127,125 508.5 22.7 384 14.8 2 t/!^ 254,250 509.0 42.0 384 20.8 3 L 2 L 4 381,375 508.5 22.7 384 57.7 4 C7 3 f7 5 528,000 299.3 44.7 434.9 2X18.7 5 L 4 L 6 381,375 508.5 22.7 384 57.7 6 7 5 t/ 7 254,250 509.0 42.0 384 20.8 7 L 6 L 8 127,125 508.5 22.7 384 14.8 * z. 224.0 The modulus #=29,000,000 /6s. per xq. inch, was not embodied in the tabulated values, hence the w loads are E times too large and the deflections would be natural size for a pole H=E. But the scale of lengths of the drawing was 1:300, and wishing to make the deflections 400 times actual the pole must be made equal to E/3QQ X400 =241. 1 'w units. The force and equilibrium polygons, Fig. c, are then drawn, using any convenient scale for w forces. The X c influence line is thus found as the polygon A CB, with a factor ft = l/c.' The actual deflection of the point C is d ee = l". 07/400=0". 0027, obtained by measuring the ordinate c in inches. All influence ordinates are measured with the same scale, which may, however, be any convenient scale, because the factor JJL makes the quotients d mc /d cc independent of an absolute scale. For this same reason the M diagram might have been drawn with any middle ordinate as unity, though when the actual deflection d cc is wanted for temperature investigations, the method here given is less confusing. The A influence area is now found by combining the X c influence line with the ordinary A line for the span AB, in such manner as to make the ordinate at C equal to zero and then finding the factor u which reduces the ordinate at A to unity. The line A'CB, Fig. c, is the ordinary A influence line with a factor I/IJA, and when this is combined with the X c influence line, the areas included between the two lines will represent the A influence area with a factor ^ = !/TJA =0.0187. Since it is more difficult to redraw the X c line so as to make the ordinate c = l, than it is to apply the factor p, the latter method is decidedly preferable to the one frequently given. 150 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X t i ? 1 I A INFLUENCE AREA , |fX/rk=+0.|Ql87 INFLUECE AREA , Upper chord I Lower chord Tension is + , compreision is ART. 48 SPECIAL APPLICATIONS OF INFLUENCE LINES 151 The point C is a load divide for both reactions A and B, and in the present case the same influence area will serve for both these reactions, because l\ =1 2 . When the spans are unequal, then the B influence area is included between the X c line and the line AC prolonged to the vertical through B and the factor then becomes ,B = I/>?,B. The influence area for a chord member L 2 L 4 is obtained by combining the ordinary S stress influence line of the member for the principal system with the A" c influence line, Fig. d. A load at C can produce no stress in any member of the indeterminate truss, hence the ordinate under C, for every stress influence line, must be zero. Therefore, the line BCA' must be common to all stress influence lines of the members between A and C, web members included. The point C must be a load divide for all members. The center of moments for the chord L 2 L 4 is at C7 3 , hence the line A3, Fig. d, completes the S stress influence line with a factor SA/T)A, where SA is the stress in the chord L 2 L 4 for a unit reaction at A. The area included between the X c polygon and the S influence line is thus the required influence area for the chord L 2 L 4 with the factor JJ.=SA/^A- The area below the X c polygon is positive as usual. The stress *SU = +1.99, hence the factor /i =+1.99/53.6 =+0.0371. If the vertical L 3 t/ 3 were absent, then the S line would have to be straight over the panel L 2 Z/ 4 and a line 2C would be necessary to complete the S influence line. With the member L 3 t/ 3 acting, the chord stress L 2 L 4 is, therefore, greater than when the vertical is omitted. The S influence lines and their factors for all chord members are shown by dotted lines in Fig. d, and the final influence areas are always positive below the X c line. The factors ft are negative for the top chords because SA is then negative. The stress in the member L 2 L 4 , for any train of moving loads, is expressed by (48c) where each ij is the ordinate of the shaded influence area vertically under its lespective load P and S A is given in the same units as the loads. This influence area is also the M 3 influence area with a factor fi=x 3 /i)A and gives moments in foot units when x 3 is measured in feet as indicated. The scale of the influence ordinates is immaterial so long as all ordinates are measured with the same scale. This is apparent from Eq. (48A). The influence area for a web member L 2 U 3 is found from exactly similar considerations as those shown to exist for the chords. The shaded area in Fig. e is the influence area for the web member L 2 U 3 . The S lines for the other web members are indicated by dotted 'lines. The \me-BCA', Fig. e, is again one of the limiting lines of the S influence line for each web member, and when the chords are parallel in the panel containing the particular web member, then the other -limiting line A2' will be parallel to BC and the line 2'3 completes the S line for the member L 2 U S . The load divide i for this member, considering the principal system, must fall 152 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x vertically over the point i' where the line 2'3 intersects the base line AH. This serves as a check or when the chords are not parallel, as in the panel L 3 L 4 , then it may be used to complete the S line as indicated for the member t/ 3 Z/ 4 with the load divide t. In case the chords are not parallel and the center of moments for the diagonal falls outside the drawing, then the ordinate 753 as for the member U S L 4 , may be com- puted from the formula r/ 3 x 3 +a 3 Jx 3 +a 3 \ /63.6+91\. Q u-ir or ^ 3= t^> 4 = r^r-r 3 - 6 < ^ where x 3 is the distance L L 3 from A to the panel containing the web member, and a s is the distance from A to the center of moments of the member, both in the same units, as feet. The factor {J.=SA/T)A again applies and the sign of SA determines the sign for //. Stresses due to temperature. When the three supports A, B and C are on the same level then a uniform change in temperature produces no stresses in the structure. However, when the sun illuminates the bridge from above, there is usually a difference in the temperature of the two chords. Assuming this difference as Jf=20 F. as in Art. 47, then the bottom chord would be cooler than the top chord by this amount and the changes in the lengths of the bottom chord members would be eMl and the w t elastic loads would be (48 E ) These loads being extremely small, it is well to multiply them by say 10,000 and then using a pole distance of 10,000, the d cf deflections would be natural size. However. the scale of lengths on the drawing was 1 : 300 and for deflections five times natural the pole becomes 10000 rr _ = 5X300 * units. The loads w t are figured for the panel points 1, 3, 5 and 7, taking =0.000007, J*=20 F. and r=384 inches. The d ct deflection polygon is shown in Fig. d, and has negative deflections, five times actual. Hence o ct =0".95/5 =0".19 and d cc was previously found from Fig. e, as 0".0027, hence from Eq. (44c) This value is in kips because d cc is the actual deflection for one kip, and X ct being negative produces the same effect on the principal system as a load X ct hung at the point C. The reaction at A, due to a load of 70.4 kips at C, would then be A t =70.4Z 2 /(Zi +Z 2 ) = +35.2 kips. The stresses in the members are then S t =S A A t =35.2 S A kips. The stresses S A , due to a unit reaction at A, are already found in Fig. b, hence the tem- perature stresses are readily determined. Thus for the member U 3 U 4 , S A = -2A2, hence S t = -2.42 X35.2 = -85.2 kips. The negative sign indicates compression. ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 153 ART. 49. TWO-HINGED SOLID WEB ARCH OR ARCHED RIB This and the two-hinged framed arch are perhaps the most common structures involving external redundancy which are met with in practice. They will receive special attention here as deserving a prominent place among commendable structures. The external redundancy may be said to offer less objection here than in any other class of structures. The present theory will be developed in its most general application to unsym- metric arches and will be equally applicable to masonry, concrete or steel arched ribs. Fig. 49A represents an unsymmetric two-hinged arched rib of any cross-section and the lettered dimensions are in general the same as those previously employed for three- hinged arches in Art. 28. The arch thrust along AB is treated as the external redundant condition. When it is removed, by replacing the hinged support at A by a roller bearing, the simple beam AB (though curved) on two supports, then becomes the principal system. FIG. 49A. According to Eqs. (7 A), the following values for the reactions, thrusts and moment for any point ra may be written: A =A A a X a , where B=B -B a X a , where and A a =-l-sina and = l- since M m =M -M a X a , where M =A x-P(a' -x') and M a = l-ycosa N m = N - N a X a , where N = Q sin ; N a = - 1 cos ( -a) T m = T -T a X a , where T =Q cos = (A - X p) cos<; 7 7 = l-sin(0-a) (49A) 154 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X in which A and B are the vertical end reactions; M m is the moment of the external forces about any axial point m; R m is the resultant thrust, on the normal section mm', produced by all the external forces acting on one side of this section; N m is the com- ponent of R m , normal to the section mm'; and T m is the tangential component of R m in the section mm'. Q is the vertical shear at any point m produced by all the loads P acting on a simple beam AB. The conventional loading X a = \ is applied as indicated in the opposite direction of X a . Eqs. (49A), with the special values inserted, thus become SPa' . A = , -- 1- X a sm a v . X a sm a I M m =M X a y cos a N m =Q sin +X a cos (< -a) T m =Q cos < X a sin ( -a) J / SPa A,, sm a (49s) where M and Q have the values in Eqs. (49A). From Fig. 49A, M m =N m v; R m =VN m 2 + T n ?; H=X a cosa. ^ . . . In the above equations, all terms except X a are derived from the static conditions and the solution becomes possible when X a is determined. The X a influence line for the redundant haunch thrust. When the haunches are on the same level then this thrust X a becomes the horizontal thrust H. The equation for X a is given by Eq. (42o) as where d ma is the vertical deflection ordinate of any point m of a deflection polygon, drawn for the loaded chord and for the conventional loading X a = l applied to the principal system, and o^ is the change in the length AB, due to X a = \ acting on the principal system. Since dma and d^ are not parallel displacements, they cannot be obtained from the same deflection polygon as in the previous examples, Arts. 46^8. This circum- stance necessitates the construction of an extra displacement polygon for daa, if the graphic solution is strictly followed, or of finding d^a by computation, which is usually advisable. The X a influence line thus becomes the d TOa deflection polygon with a factor p. = l/aa- Should this deflection polygon be constructed for a pole distance H =daa then the factor p-1. According to Art. 40, the dma deflection polygon is the equilibrium polygon drawn for a set of elastic loads w with a pole distance H = 1. These elastic loads represent partial areas of a moment diagram drawn for the conventional loading X a = l. ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 155 The moment produced by X a = 1 about any axial point m is M, n = 1 y cos a .......... (49E) and the elastic loads, by Eq. (39i) become M Au Axy cos a ^-JT--J -- -r, El El cos (j> ' (49F) where Au is the width of a partial moment area measured along the axial line, and Ax is the horizontal distance between the vertical moment ordinates, making Au=4x/cos. The deflection d aa may be obtained from a deflection polygon drawn for the same set of w loads by allowing these loads to act in a direction parallel to X a , as per Art. 38. According to Eq. (38B) the deflection of any point in any direction, may be expressed as the sum of the moments of all w loads on one side of the point, when the direction of the w loads is taken parallel to the required direction of the deflection. Hence a daa = ^T ^(J/ COS ) (49o) which affords a purely analytic solution for computing the X a influence line ordinates and also the d m displacement. Since the w loads are best found by computation, it is a comparatively easy task also to compute the values wy cos a and thus obtain d aa = I>wy cos a, which is the required pole distance for the d ma deflection polygon. Hence according to Eq. (49n) i and the X a influence line is an equilibrium polygon drawn for the elastic loads w with a pole distance H = Swn/ cos a. Since E enters into w, and hence into each term of Eqs. (49o), as a constant factor, which cancels in the numerator and denominator of Eq. (49n), it would be proper to multiply Eq. (49r) by E and thus make the elastic loads equal to Ew without affecting the ordinates r? a from Eq. (49n). Should it be desirable to obtain values of d ma from the X a influence line, then the jj a ordinates must be divided by E and multiplied by cos a. . For very flat arches, the effect of the axial thrust should be considered in the determination of X a , by allowing for the quantity d f m produced by N a , Eq. (49A), on the displacement daa- Thus for N a = -I cos (< -a) and Au = Jar/cos , then from Eq. (15.v), 2 /j - 156 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X which is the effect due to N a only, and this added to the displacement previously found for moments only, gives the total displacement 2 (49HH) Hence, when axial thrust is to be considered in Eq. (49n), then the value 3 aa should be computed from Eq. (49HH) and used as the pole distance H in drawing the equilibrium polygon for the elastic loads w. When the loads w are taken E times actual, then the E in the second term of Eq. (49HH) is omitted. Temperature stress. For any change in temperature above or below the normal, X at is expressed by Eq. (44c) as X at =m^, ..... V . '.I/ .. . (49i) Oaa wherein d al is the change in the distance AB due to any change in temperature of acting on the principal system. When the structure retains a uniform temperature then the change t will be uniform in all directions so that the shape of the structure will always be similar to its normal geometric shape. Hence, the distance AB will change as for any case of linear expan- sion and dl 0.000488/ for e =0.0000065 per 1F., and Z = 75. When the change in temperature is not uniform, as when the upper flange is +Jl c warmer than the lower flange, then d at must be found from a deflection polygon drawn for a set of w t elastic loads computed from the formula w t = edtJu/D, where D is the depth of section and Au is the length between sections measured along the axis. The pole distance, if made equal to unity, will give the actual o at , and if the pole be made equal to d aa = 'Zwy cos a as for X a , then the displacement found will be X nt measured parallel to AB. The w t loads must likewise be taken parallel to AB in drawing the force and equilibrium polygons since d at is a displacement along AB. X at will be positive for +t, while for + At the w t loads are negative and X at would also be negative. In any case, when X at is determined, then the quantities M mt , N mt , and T mt are found from Eqs. (49fi) by omitting all the terms involving the effects of the loads P, thus: M mt = ~X at y cos a =N mt v N m t= X at cos((f>-a) T mt = -X at sin (< -a) Abutment displacements. When the abutments undergo displacements Ar in the directions of the reaction forces R, then, for the case of external redundancy where d a =0, Eq. (8D) gives for P=0, ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 157 where R a has the several values A a =-l-sina, B a = \- sin a, and X a = l, while Jr represents the vertical displacements A A, AB and a change M in the length of span I. Hence the redundant X ar , due only to abutment displacements, is S/g Jr _ A a 4 Xar ix or JB sin a A A sin a + M sec a. X-ar = 5~~' ' " " When AA=AB, which is usually the case, then . (49L) showing the effect due to an elongation M, in the span I, to be precisely the same in character as that due to a uniform fall in temperaturegiven by Eq. (49i). For a two-hinged arch with a tension member AB, and one expansion bearing, the problem becomes one involving an internal redundant member X a and Eq. (42s) is then used vvith the term p a in the denominator. The solution would otherwise remain unchanged. . Stresses on any arch section. Neglecting curvature, which is always permissible, the stress on the extreme fiber of any arch section mm' is given by Navier's law as N My (4j)M) J~F I ' where N is the normal thrust on the section, M=Nv is the bending moment about the gravity axis F is the area of the section, y is the distance from the gravity axis to the extreme fiber and I=Fr* is the moment of inertia of the section about the gravity axis. Fig 49B gives a graphic representation of the distribution of stress on any arch section mm' in accordance with Eq. (49M). The center of gravity is at c and the uniform axial stress f m =N/F is the stress ordinate at c, while the stresses on the extreme fibers /. and /,-, are applied as ordinates giving the line tt' as representing the uniform , tribution of stress over the section. . The point e is the kernel point for the extrados while * is the kernel point for the mtrados. These points are determined by the distances k.=i*/e and k^fi/i, where r is the radius of gyration of the section and e and i are the respective distances to the extreme f the section measured from the gravity axis c. When N and * are given, the line tt' is easily constructed as indicated by prolong- ing i?_to 6 to find t, and by prolonging ?ito a to find f. The stress /. is then the inter- cept bn while the stress f t is the intercept an. 158 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X Using e and i as subscripts, referring respectively to extrados and intrados, then for M=Nv and I^Fr 2 , Eq. (49M) gives and (4*0 The moments about the kernel points e and i are evaluated from the figure as N(v+k e )=M e and N(v-kd=Mi ........ (49o) FIG. 49B. For r 2 /e=k e and r' 2 /i=k{, Eqs. (49N) become - _i 4. = _ - _j? F\ + k e ) F\ k e I Fk e F \ ki/ ''F \ ki i Fki Also solving Eqs. (49o) for N and v, then M e Mi N=, -r and v = (49?) Me -Mi (49 Q ) Hence Eqs. (49p) and (49cj) furnish a complete solution for the stresses f e and /, on the extreme fibers, the normal thrust N and its distance v from the gravity axis, for any unsymmetric section tt' of an arch ring, in terms of M et Mi, k e and &,. ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 159 When the section is symmetric about a horizontal gravity axis, which is usually the case, then these equations give, for e=i=D/2, or the half depth between extreme fibers, _- __ _ = ,-i _______ M e Mi =s ~ and i== Me-Mi 1. T "~ , Ctllvl i) M e +Mi (49R) 2k M e -Mi 2N For a rectangular section of unit thickness and of depth D, making e=i=D/2. fc _* x> F==1 . D} /=i^- 3 M e +Mi N=- and ?'= : 2N (49s) In all the above formulae M e and Mi have opposite signs when N acts between the two kernel points e and i. When v>ki, both moments are negative and when v is negative and larger than k e , both moments are positive. The figure shows v to be positive when measured from the gravity axis at c toward the extrados. The stresses f e and fi take their signs from M e and Mi respectively, and compression is regarded as a negative stress. The stresses on any arch section mm' may, therefore, be found for any simultaneous position of a moving train of loads when the influence areas for M et M t and T m have been drawn. The resultant polygon for any simultaneous case of loading may also be drawn by plotting the offsets v from the gravity axis of each section examined, observing that +v is toward the extrados and v is toward the intrados. The method of combining the X a influence line with ordinary moment and shear influence lines to obtain the M e , Mi and T m influence areas for any section mm' will now be illustrated. Kernel moment influence areas. The moment equations for the kernel points accord- ing to Eq. (49B) are ,_ \M oe ! [ M oe 1 , =M ae \ -a-, X n \=y e cos a \- X a \ lM ae J [j/ecos a ' (49r) . The intercept of the M oe influence line on the vertical through A is simply the ordinate x e of the point e. Similarly for the M oi influence line this intercept is the ordinate 160 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. 3 Xi of the point i. Hence the positive intercepts on the vertical through A, of th M oe /y e cos a and of the M^/yi cos a influence lines, are easily computed when the coordi- nates of the kernel points e and i are given, provided the X a influence line was drawn for a factor /* = !. The coordinates of the kernel points are derived from the coordinates (x m y m ) of the axial point for the same section, when the kernel distances k and the angle are known for that section. From Fig. 49s, , x e = x m +k e sm, y e = y m -k e cos(j> } I, V 49u) X{ =x m -ki sin 0, yi =y m +k { cos ^ J which apply to all points from A to the crown, and from there to the abutment B the signs of the last terms are reversed. By using the I x ordinates the intercepts on the vertical through B are found. One intercept only need be computed as the two limiting rays of the M /y cos a. line must intersect on the vertical through the center of moments. See Fig. 49c. By constructing the X a influence area for // = !, which may always be done by making the pole distance H = 2wy cos a, and since both the X a and the M /y cos a lines are pos- itive, then by applying them below the closing line A'W, the area included between the two lines will represent the M m influence area. The portion below the X a line will be the positive area because X a is subtractive in Eqs. (49-r). The entire M m influence area has a factor p. =y cos a =M a . The T m influence area for tangential force on any section mm'. From Eqs. (49u) the following equation for T m is obtained. From this the end ordinate at A, for the T /T a Yme, becomes 1- cos <^ sin (0 a) because the end ordinate for the Q line is unity. In this case the end ordinate at B is numerically the same but negative. For axial points to the right of the crown the end ordinate at B is positive and the one at A is negative. Other details are illustrated in connection with the example. The final T m influence area has a factor /i=sin (<-a). Example. A two-hinged arched rib, modeled after the Chagrin River Bridge near Bentleyville, O., was selected to illustrate the application of the previous theory. The structure was made unsymmetric by shortening the span at the right-hand end as shown in Figs. 49c. The clear span thus became 164 ft. and the rise 27.89 ft. This bridge was designed to carry a live load of 24 kips per truss per panel. The arch section is composed of a f-inch web plate and 4-6" X6" X 11/16" angles, with 2-14" X7/16" flange plates on each flange. Besides these there is one 14" X|" plate on each flange extend- ing from sections 2 to 5 and 7 to 10. The sections are all symmetric about the gravity axis and all general dimensions are given on the drawing and in Table 49A. The bridge consists of two steel arched ribs 27 ft. between centers and carrying a total dead load of 1,058,000 pounds. The following values are assumed as a basis ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 161 X ordinores are ?.a times I toicole of Ittjfeth ,. M e INFLUENCE AREA. >\=y g CoiqC-18.-l B' FIG. 49c. 162 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X for the analysis: #=29,000 kips per sq.in. =4,176,000 kips per sq.ft.; e -0.0000065 per 1F.; andZ = 75F. The first step in the analysis is to compute the Ew loads from Eq. (49r) using the tabulated dimensions in Table 49A. In the same table the values Ewy cos a are computed, giving the required pole distance H=E2wycosa for constructing the X a influence line in accordance with Eq. (49n). TABLE 49A Xa INFLUENCE LINE Sec- tion. Coordinates of Gravity Axis. <*> l / in. 4 / ft. 6 = 1 Jx ft. 0J.r = Ax Ew = Axy cos a Ewy cos a X ft. y ft. cos cos ft, 1 COS (j> ft. I COS (j> ft. 36 15' 1.240 74,180 3.578 0.3466 9.4 15 15 15 15 15 15 15 15 15 15 4.6 0.652 1 9.4 6.01 32 15 1.182 74,180 3.578 0.3260 4.954 29.8 179 2 24.4 14.12 26 00 1.113 64,710 76,530 3.121 3.691 0.3567 0.3017 4.939 69.7 984 3 39.4 20.30 20 00 1.064 66,170 3.191 0.3332 5.014 101.8 2065 4 54.4 24.73 14 15 1.032 56,750 2.737 0.3770 5.673 140.3 3468 5 69.4 27.24 7 30 1.009 48,160 40,420 2.323 1.949 0.4352 0.5176 7.198 196.0 5339 6 84.4 27.89 00 1.000 33,590 1.620 0.6173 9.011 251.3 7007 7 99.4 26.77 7 30 1.009 40,420 48,160 1.949 2.323 0.5176 0.4352 7.198 192.6 5156 8 114.4 23.58 14 15 1.032 56,750 2.737 0.3770 5.673 133.7 3153 9 129.4 18.58 20 00 1.064 66,170 3.191 0.3332 5.014 93.1 1730 10 144.4 11.82 26 00 1.113 76.530 64,710 3.691 3.121 0.3017 0.3567 4.939 58.4 690 11 159.4 3.14 32 15 1.182 74,180 3.578 0.3260 3.983 12.5 39 12 164.0 34 20 1.211 0.3260 0.0 EZwycos a = 29,810 NOTE. The x abscissae are measured horizontally from A. The y ordinates are measured vertically from AB. The Jj are measured horizontally between sections. a=l 06', cos a = 0.9998, sin a = 0.0192. The products QAx in Table 49A are summed by the prism oidal formula treating Jx as the length of a prismoid whose middle area is 6 and whose end areas are the means of the successive tabulated values of 6. This is necessary because 6 is a variable quantity. AHT. 19 SPECIAL APPLICATIONS OF INFLUENCE LINES 163 Calling 0o-i the mean between and 0i, and 0]- 2 the mean between 6\ and 2 etc., then the values 6 Ax become: Q j x =^[200 +0 _!] =^[0.6932 +0.3363] =0.652; o o 0! Jz = ^[0o-i +40i +0i_ 2 ] =[0.3363 + 1.3040 +0.3413] =4.954. 6 2 Ax =^[0i -2 +20 2 +20 2 ' +0 2 - 3 ] =^?[0.3413 +0.7134 +0.6034 +0.3174] =4.939. 6 z Ax = ^[0 2 -3 +403+03-4] =y [0.3 174 + 1.3328 +0.3551] =5.014, etc., etc. After collecting the tabulated data from the drawing, Fig. 49c, the remaining computations are quite simple and no further comment is necessary here, as the several operations are indicated in Table 49A. The X n influence line is now drawn by combining the Ew loads into a force polygon and making the pole distance equal to E2wy cos a. The resulting equilibrium polygon represents the X a influence line with ordinates to the scale of lengths and /* = !, in accordance with Eq. (49n). In the drawing, the pole distance was made 1/20 of the actual length so that the ordinates ?? are twenty times too large. For this reason a special scale of ordinates, twenty times the scale of lengths, was constructed and used for all influence ordinates. The scale of forces for the force polygon may be any convenient scale, so long as the pole distance is measured with the same scale as the forces. The influence areas for the kernel moments and tangential force are now drawn by com- bining the M /M a lines and the T /T a line each with the X a line. According to Eqs. (49x), (49u), (49v), this will require computing the kernel distances k, and the coordi- nates of the kernel points and finally the end ordinates at A of all influence lines. The end ordinates at B may be obtained by substituting for the abscissa? x the values I x. These computations are given in Table 49s, which is self-explanatory. In Fig. 49c the influence areas M e and Mi are constructed for section 3. The X a polygon is copied from the original one by transferring down the several j) a ordinates from a horizontal base' ~A r W. The M /M a lines are constructed from the end ordinates A'A"=x c /y e cosa and A f A 7r '=Xi/y i cosa, using the coordinates for the two kernel, points as given in Table 49u. The lines WA" and B'A'" are thus determined. The kernel points e and i are the centers of moments and have the abscissae x e and x i} as given in the table, and from these the points e' and i' are located. The two moment influence lines are completed by drawing the lines A'e' and A'i'. The M oe /M ae influence line is thus found to be the broken line A'e'B' and the shaded area included between it and the X a line is the M e influence area for section 3. The fac- tor for the ordinates i) e , when measured to the same scale as the rj a ordinates, is i = cos a =18.41. 164 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X TABLE 49n KERNEL POINTS AND END ORDINATES FOR Al me , M mi , AND T m AREAS Sec- tion. F Sq. in. D 2 in. i 2I * k sin < ft. k cos ft. Coordinates of kernel points, Eqs. (49u). End Ordinatea. \-2FD Eq. (49R) ft. x e ft, Ve ft. x i ft. * ft. x e. x i COS a) 1 2 3 4 5 6 7 8 9 10 11 12 81.6 79.0 89.5 87.8 86.3 85.7 75.2 73.6 75.2 85.7 86.3 87.8 89.5 79.0 81.6 36.0 35.5 33.4 33.8 31.6 29.5 27.4 27.0 24.9 27.0 27.4 29.5 31.6 33.8 33.4 35.5 35.7 2.13 2.04 2.11 1.99 1.86 1.71 1.64 1.53 1.64 1.71 1.86 1.99 2.11 2.04 2.13 36 15 32 15 26 00 20 00 14 15 7 30 00 7 30 14 15 20 00 26 00 32 15 34 20 1.14 0.91 0.68 0.46 0.23 0.00 0.23 0.46 0.68 0.91 1.14 1.80 1.86 1.87 1.79 1.66 1.53 1.66 1.79 1.87 1.86 1.80 10.54 25.31 40.08 54.86 69.63 84.40 99.17 113.94 128 . 72 143.49 158.26 4.21 12.26 18.43 22.94 25.58 26.36 25.11 21.79 16.71 9.96 1.34 8.26 23.49 38.72 53.94 69.17 84.40 99.63 114.86 130.08 145.31 160.54 7.81 15.98 22.17 26.52 28.90 29.42 28.43 25.37 20.45 13.68 4.94 2.504 2.064 2 . 174f 1.058 1.470 1.747t 1.633 2.135 2 . 895f 4.307 8.884 55.555 8.884* 4.307 2.895 2.135 1.633 3.202 2 . 582* 2 . 869 2.265* 4.284 0.700 NOTE. D is measured between extreme fibers. For all dimensions in inches, k is in inches. The coordinates of the axial points are given in Table 49A. * For sections 7 to 12 the ordinates are for the B end. t Values used in Fig. 49c. Similarly the dotted line A'i'B' represents the M oi /M ai influence line and the area included between it and the X a line represents the Mi influence area with factor Pi=yi cos a =22.13 for section 3. Hence for any simultaneous position of moving loads the two kernel moments for section 3 are represented by the expressions and (49w) where the subscripts refer to the kernel points. The T m influence area for section 3, is found by laying off the end ordinate A' A" = cos a) =2.895 and drawing the line A"B'. The negative end ray A' 3' is parallel to A"B', and the T /T a line is thus the broken line A'3'3"B' and the shaded area is the T 3 influence area with a factor /*=sin (< a). It will be seen that for sections near the crown the end ordinate at A becomes very large and when <=, this ordinate approaches infinity, while becomes zero. Hence there will always be several sections near the crown for which the T m influence area must be found by making n = l and reducing the X a line accordingly. This is done by using the original form of Eq. (49s) which is T m =Q cos < X a sin ( a). ART. 49 SPECIAL APPLICATIONS OF INFLUENCE LINES 165 For the A ordinate Q = l, hence this ordinate is simply cos < and by reducing all the T) U ordinates by multiplying them by sin (<-), the .T 5 area is found with a factor 1=1. The rj a ordinates are easily reduced by graphics, laying off the line B'z such that its deviation from the vertical is sin (0 -a) in a distance unity. The rj a ordinates pro- jected over horizontally are then reduced to the small horizontal intercepts between Wz and the vertical. The T 5 influence area is thus constructed and is represented by the shaded area included between the broken line A'5'5"B' and the reduced X a line. The factor p = l. The tangential force for section 3, for any train of moving loads is expressed by (49x) and the tangential stress by T Z /F 3 , where rj t is any ordinate to the T 3 area under some particular load P. The stress due to a unifo (49 J) whence for t = 75 F., particular load P. The stress due to a uniform change in temperature is found from Eqs. (49i) and l-.a*_ etf 0.000488 1 Xat 5 it ?, d aa COS Oi d aa COS a Eq (49o) gives d aa = Hwycosa and Table 49A furnishes EZwij cos < =29810, hence L=29810/JS ft. for X a = l kip. Making =4,176,000 kips per sq.ft., Z = 164 ft. and cos a a =0.999S, then 0^-0.00714 ft. and X at = ^mfxo.9998 = n - 21 ki P s - * Eqs. (49K) , will furnish the means of finding the values M mt , N mt , T mt , and v=M mt /N mt and with these^and Eqs. (49?) the temperature stresses f e and f t may be computed for any section mm'. Stress due to abutment displacements. Eq. (49L) furnishes the haunch thrust X ar for any change Al in the length of span. The quantity Al must be estimated from the elastic properties of the abutments and is always more or less problematic, though it is well to investigate the probable stress which might be created by such a displacement. In the present example it is assumed that for the maximum case of loading the haunches will spread an amount JZ=0.03 ft, then for d aa =0.0071< ft, X a - kip, and cos a =0.9998, Eq. (49L) gives y & - 03 = -4.20 kips. ar = ~ according to Eq. (45E) for one external redundant condition. This equation is later employed to construct the stress influence lines for the members. The X u influence line for the haunch thrust is represented by Eq. (49n) as Z* ' Oma _ ! ' O-ma f ~p. \ a = K """v* 'Ja? ........ (OUB) daa 2jWy COS a where d^a is the vertical deflection ordinate of any point m of a deflection polygon, drawn for the loaded chord and for the conventional loading X a = l applied to the principal system; and d aa = 'wy cos a is the change in the span AB, due to X a = l acting on the principal system. According to Art. 40, the d ma deflection polygon is the equilibrium polygon drawn for a set of elastic loads w, with a pole distance H = l, according to the method given in Art. 36c. ART. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 167 The elastic loads w are functions of the changes in the lengths of all the members of the frame as given by Eqs. (36fi) and (36c). These are algebraically summed for all the panel points to obtain the total loads w. Thus the elastic loads for the chords alone are w c = Al/r and each web member contributes two elastic loads w u = M/r u and w n = M/r n , acting at the two adjacent panel points u and n of the loaded chord. Hence when the loads P are to be applied to the top chord, then the w u and w n elastic loads are com- puted for the top chord panel points. The lever arms r for the chords are measured as shown in Fig. 36A and the lever arms r u and r n for the web members are measured as explained in Figs. 36s and 36r. The details of the computation of the w loads are illus- trated in Table oOu, and Fig. 50A, in connection with a complete example. For any braced arch with parallel chords, the w loads for the web members may always be neglected. The displacement d aa = 'Swy cos a is computed for the same w loads by taking the sum of their moments about the line A B joining the haunches. The lever arms for an unsymmetric span thus become the vertical ordinates of the respective pin points times cos a. See also Table 50A. The X a influence line thus becomes the equilibrium polygon drawn for the elastic loads w with a pole distance o aa = ^wy cos a. Still another method of finding this influence line consists in drawing the deflection polygon for the loaded chord by means of a Williot-Mohr displacement diagram, which also furnishes the value d^ from which the influence line ordinates rj a may be computed and plotted. This solution is illustrated in Fig. 50B. Since the modulus E does not affect the X a influence line it is more convenient to comput-e all displacements and the w loads E times too large, thus avoiding the small quantities resulting in many decimals. Temperature Stress- Calling Al t the change in the length of any member due to any temperature effect, then from Mohr's work Eq. (5n) the change d at in the length of the span AB becomes l.d at =ZS a 4l t =2S a dl .......... (50c) The redundant thrust X at , due to any temperature effect, is found from Eq. (44c) as . at _ a _ g at ~~d^~ daa 'EZwycosa For a uniform change in temperature of t, above or below a certain normal, the change o^ in the length of the span 'AB, is found from Eq. (49j) as dAB cos a In any case the stresses in the members are best found from S t = ^SaXat for X at , or from a Maxwell diagram drawn for the external loading X at , acting on the principal system. For a uniform change of t from the normal temperature, the resulting stresses become S t = ^S a X at . The effect on the final stresses S for full loading, will then be additive or S max = (S +S t .) 168 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X Abutment displacements will affect the stresses in the members by producing a redundant thrust X ar as found from Eq. (49L) when JZ is an increase in the length of span AB. The stresses S r would be found from a Maxwell diagram drawn for X ar acting on the principal system. They could also be found as S r = =F*S a X ar , for X ar , since *S a is the stress produced in any member S by X a = 1 acting outward. The stress influence area for any member $ is derived from the previous Eq. (50 A) from which any stress influence ordinate TJ becomes s=s a ^-x a ^=y). .:.; . . (SOP) These influence areas are alike in principle for all members, whether chords or web members, and represent the area inclosed between the X a influence line and the S /S a influence line, times a factor S a . The lSo/S a influence line is the ordinary stress influence line S for any determinate frame, nultiplied by the factor l/S a . Hence the end ordinates for the S line may be found in the usual way. Thus the stress SA in the member S due to the upward reaction .4=1 is the end ordinate of the S line at the point A, and this ordinate divided by S a becomes the required end ordinate rj A =S A /S a of the S /S a line at A. Similarly the end ordinate of the S /S a line at B is -fjB=SB/S a , where SB is the stress in the member due to the upward reaction B = \ acting on the principal system. Hence if the stresses SA, SB and S a , for all the members of the principal system, are found either by computation or from three Maxwell diagrams, then all the data for the several stress influence lines are at hand provided the X a influence line is known. In drawing the S /S a influence lines the following points should be observed: 1, That the end bounding lines must always intersect in a point i' on the vertical through the center of moments i of the particular member treated. 2, that this S /S a line must be a straight line between adjacent panel points of the loaded chord. 3, that when one of the end ordinates is too large to be conveniently measured, then half this ordinate may be laid off at the center of the span. This is frequent!}' done as in the example which follows, see Figs. oOc. The signs of the end ordinates and of the factors fJi=S a all follow from the signs of the stresses SA, SB and S a . Deflection of any point m. Applying Mohr's work equation (Gfi) the deflection d, n of any point m, becomes (600) SI wherein Si is the stress in any member due to a load P = l, applied at the point m in the direction of the desired deflection ; and S is the corresponding stress due to any cause or actual condition of simultaneous loading for which the deflection is sought, includ- ing load effects as well as temperature changes and abutment displacements. The sum- mation covers all the members of the principal system. ART. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 169 For any redundant condition X a the stress S by Eq. (7 A) becomes S=S -S a X a +S t , . (50H) where S t = S a X at is the stress due to temperature effect. Example. A two-hinged, riveted, spandrel-braced arch, taken from a thesis by Mr. A. V. Saph, 1901, Cornell University, is used to illustrate the above method. The arch has a span of 168.75 ft. between pin supports, a rise of 29.5 ft., and weighs 1,060 ,320 Ibs. without abutment shoes, making a uniform dead load of 43.2 kips per truss per panel. The top chord is the loaded chord. The abutments are symmetric and, therefore, a =0. Fig. 50A shows the half span with the lengths of members in feet below the lines, and the values EM, in feet, as found from Table 50A, above the lines. The various lever arms used in the computation of the stresses S A , S B and S a , and of the w loads, are also shown. The actual values Al in feet =EM + 29000, because the areas F were not reduced to square feet. See also the example in Art. 52, where M is actual. TABLE 50A COMPUTATION OF EAl, j) A , TJ B AND E2S a 4l t . Stresses in Kips. Unit End Ordinates, Temperature Effect. Area, ..enscth Stress, S a l Oo *ES a 4lt Member S A SB S a F i Oa F *EM t for for for * SA SB -stlh A = l 5=1 a Sq.in. ft. Kips. sq. in. ft. / Sa T>B Sa ft. Kip.ft. Kip. ft. uu - 0.320 - 5.437 -0.211 19.80 15.000 -0.0107 -0.1605 1.517 25.770 197.9 41.8 u c/ - 1.165 - 6.902 -0.697 19.80 15.000 -0.0352 -0.5280 1.671 9.917 197.9 137.9 u.u, - 2 725 - 8.953 -1.457 19.80 15.000 -0.0736 - 1 . 1040 1.870 6.145 197.9 288.3 - 5 588 -11.755 -2.649 26.48 15.000 -0.1000 - 1 . 5000 2.109 4.438 197.9 524.2 ifu* -10.008 -14.336 -4.121 38.11 15.000 -0.1081 -1.6215 2.429 3.479 97 9 815.5 u*u* -14.062 -14.062 -4.917 38.11 15.000 -0.1290 -1.9350 2.859 2.859 197.9 973.1 T V " 01 79 + 1 085 42 50 11 228 + 0.0255 0.2863 -0.159 88.9 96.5 LI 366 6.231 1.388 40.00 17.190 0.0347 0.5965 0.263 4.481 136.1 188.9 LI* 1 269 7.516 1.848 37.50 16.335 0.0493 0.8053 0.687 4.067 129.4 239.1 L L 2 857 9 . 386 2.575 37 . 50 15.725 0.0687 1.0803 1.109 3.645 124.5 320.6 5 685 11.958 3.712 37.50 15.258 0.0990 1.5105 1.532 3.221 120.8 448.4 J* L 10 027 14 . 363 5.131 42.50 15.029 0.1207 1.8140 1.954 2.799 119.0 610.6 T? L 917 -0.605 23 . 52 35.942 -0.0257 -0.9237 1.516 379 . 5 229.6 UT 1 179 - 2.043 -0.678 19.80 29.312 -0.0342 -1.0025 1.739 3.014 309.5 209.8 r/ 1 /" 1 1 502 - 1.976 -0.732 14.70 20.918 -0.0498 -1.0417 2.052 2.700 220.9 161.7 U>L 3 - 1.857 2 042 - 1.817 - 1 193 -0.773 -0.681 11.76 11.76 14.450 -0.0657 9.730-0.0579 -0.9494 -0.5634 2.402 2.999 2.351 1.752 152.6 102.7 118.0 70.0 U L - 1.622 + 0.109 -0.318 11.76 6.932 -0.0270 -0.1872 5.101 -0.343 73.2 23.3 77 V 5 0.0 11.76 6.000 0.0 0.0 0.0 0.0 63.4 0.0 rr r If) 71 13 34 32 928 . 0503 1.6563 1.515 347.7 233.3 T7T .UI/ 1 451 2 514 0.834 13.34 25.741 0.0625 1.6088 1.740 3.015 271.8 226.7 1 2.166 3.413 4.869 4.367 2.848 3.339 2.843 - 0.294 1.055 1.421 1.622 0.857 13.34 13.34 18.25 15.84 20.828 17.871 16 . 524 16.156 0.0791 0.1065 0.0889 0.0541 1.6475 1.9033 1.4690 0.8640 2.053 2.402 3.002 5.096 2.700 2.350 1.753 -0.343 220.0 188.7 174.4 170.6 Tot'k 232.1 268.1 282.9 146.2 3293.4 3593 . 2 * Where =29,000 kips instead of 29,000X144. lE^S a M t - 299.8 kip. ft. 170 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X U ' The f i^ure] obove-the members| represent EA] other dijnen&ions are lengths in ft. DEF-LEcflONS ARE \E TIlllES ACTUAL IN F*ET. E= 29OOto KIPS PER SQ.IN. S \ .XI " " : r " AKT. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 171 Table 50A gives these stresses, in kips, as found by computation, also the areas F of the members, being gross for compression and net for tension members. The lengths I of the members, unit stresses/ and quantities EAl (for shortening) are also included in this table, and finally the end ordinates of the S /S a influence lines are obtained. The X a influence line is now found by two different methods to illustrate the applica- tions frequently referred to elsewhere. The first method is by constructing a deflection polygon of the top chord by means of a Williot-Mohr displacement diagram, Fig. 50B, using the quantities EM, in feet, as the changes in the lengths of the members for the conventional loading X a = l, producing stresses S a . The second method is by finding this deflection polygon from the w loads. The first method requires little description other than to say that the displacement diagram Fig. 50B is drawn for displacements EAl, Table 50A, on the assumption that the point LQ and the direction of the member L 6 U 6 remain fixed and since the span is symme- tric about this member and the point L 6 , therefore no rotation diagram is necessary. The deflection polygon of the top chord is then found by projecting the vertical deflec- tions of the top chord panel points onto the verticals through these points, furnishing the polygon ~A rT lT y with a closing line A'A", horizontally through A' '. These deflec- tions are E times actual, in feet, measured to the scale of displacements. The horizontal displacement \Ed aa between A and L 6 is also obtained from the same diagram as the horizontal distance between A' and L' 6 . The vertical ordinates of the deflection polygon are values of dma, and hence the X a influence line ordinates y a are found by dividing the several deflection ordinates Eo ma by the constant Ed m according to Eq. (50s), giving This shows that the factor E and the scale of the ordinates do not affect X a . The ordinates r] f! , for all panel points, are plotted to any convenient scale to obtain the X a influence line. Second method. The same displacements EM, in feet, are here employed to com- pute the elastic loads Ew, using the lever arms, also in feet, as given on Fig. 50A. See Table 50s. In the present example all the members are included and the table indicates exactly how much each member contributes to the several total Ew loads. The method of Art. 35 is rigidly followed. By using displacements EM which are E times too large, the w loads are also multiplied by E. Each chord member furnishes one w load which acts at the center of moments for that chord while each web member contributes two loads w u and w n acting at the two adjacent panel points u and n of the loaded chord or the chord for which the deflection polygon is to be drawn. " The w c loads resulting from the chord members are always positive as found by Eq. (36s), thus: M EM ro , uv = or Ew c = ......... (DUK; 172 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, x The negative and positive loads for a web member are Al Al EM EAl w u =- and w n = or Ew u = and Ew n = -- . . (50L) TU r n TU T n as given by Eq. (36o). The lever arm r for a chord, is the distance from the center of the moments for such chord and perpendicular to the chord. The lever arms r u and r n are perpendicular distances onto a web member as described in Figs. 36s, and 36F. They may be identified by comparing the values in Table 50s with the dimensioned lever arms on Fig. 50A. The rule for the signs of the two loads for any web member was stated in Art. 36, and is as follows: Calling all top chord members negative and all bottom chord members positive, and giving the proper sign to Al for the web member in question, then the positive w load is found on that side of the panel where the sign of Al coincides mth the sign of the adjacent chord. For the panel point f/ the total load Ew (Table O()B) is made up of the positive load produced by the chord L Li and one of the loads from each of the web members U Li and U\L\, the signs of which are negative according to the above rule. The w load for point A produces zero effect on the deflection and need not be considered. For the panel point U l the total load Ew\ becomes ALL JUpLi AVjLz . JE/nZq J7 2 L 2 " ""' ~~ =auby 25.7 13.4 17.15 "' 11.7 and similarly for the other panel points of the loaded (top) chord. The bottom chord points receive only the loads produced by the top chord members except at LI where the member U L , being the end post, contributes a load Jt/ L /10'.2. The total Ew loads acting in the same vertical are then summed for the seven panel points of the half span and used in constructing the deflection polygon with a pole dis- tance H =E2yw. The y ordinates are measured vertically from the line A~B to the points of application of the respective Ew loads, distinguishing between the loads acting at the top and bottom panel points. In order that the X a influence line ordinates may appear to a scale twenty times as large as the scale of lengths, the pole distance was made equal to E2yw/20, see Fig. 50c. Note the agreement of the ordinates here found with those obtained in Fig. 50fi. Stress influence areas. The end ordinates i) A and r) B of the S^/Sa lines are com- puted in Table 50A and these serve to construct all stress influence areas for the several members. Fig. 50c shows stress influence areas for six typical members, and each one has a factor S a as per Eq. (50A). The end ordinates are applied down from the closing line when positive and up when negative. The signs of the influence areas are uniformly + for areas below the .Y a influence line and the factor p takes the sign of S a . SPECIAL APPLICATIONS OF INFLUENCE LINES 173 P^ o CO co co 00 f^. ,_! a; ' ~ "3 *** s & II - -i <*- *? 1C cb T 1 o i*"! "S-2" ^ 'M O 1 *x ei *& O O S | _j_ ii II 2 II UJ C ^^ 5 CD 03 c In sC *3 ii II 11 || || ii to S S~i .5 aj S > Q) tit; 1 ii ii S N OJ Sl*^ " -i ai "Sto >s * * w Tl ,- g sT fiq I 1 S, BJJ SjJl o3 ^ = 2^^ o^ .2 ~ o Sji 4 J- K-f 4 4 2ls sl o ItB^l g g 2 ; b" b ^ b b b b ^ CC C H-. g5s c ^ - * r+K 1 T-l CD CO O CO 1 C5 M CD rf S i-C us CO CO to q oo CD Ss R3 ic IN CO rf 06 co T * 1C d d ^ co CO O5 w * 1 + t-_ a i-C co us 1C CO c co CO CO 1C co 1C i-C CO CD oo iC rf i I q (M (N 1C 00 IN 1C O5 (N 1 -3- n i-H a CD O i i o o T I oo CO . "^ C^ cO 2 ^ eo ^** 00 C^ ^ . q T-H O i-i O i 1C I I-H CO O5 O ?i T-* d ^ d ^ d ^ d d ^ g iC C '^i. ** J2 " J^ II ^ " > II b |l b || ^ " 1 O r ^> OOICD rf ^o " ""3 Oi CO O- co "" . d CO I> CO jX Ci C- ^ oc <*>. CC c-i "s ] 1 1 I-H T 1 14 ^ *~ o | 3 BB i a CO CO co co o CO CO O5 (N CO >o '" (N "i <=> S. d ^ ii S, d r^\ ii S, d ^ 11 b (| *~i d b 11 * ""p KJ !l KJ ii ""rf V j ^8 ^ | CD rl H CO CC CD ff = c3 CO C 3 2 " i 3"= oo co 8 C3 i co (N CD *4 ^5 iC rf CO (N - 2 ^ d rf ^ d 1C (N co b d u? II "2 4* d >-3 d *3 d *s. ^ S, ^(* II ^J il b* II b" II b ii b il b o .a ^ " ^ d? 4 *"1 00 IN 5 (N O- i ^ C 1 O C^ "^ rf u: O rf '"J rf 1 t T N S CO CT 1-H tf S c i-! tf 5 1 5 d ^ 3 d c O i- t b b b" b" b ^ b"* 1 4* 4 4" 4 4" 4- i 174 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X FIG. 50c. ART. 50 SPECIAL APPLICATIONS OF INFLUENCE LINES 175 The method of computing stresses for any train of loads need not be repeated here except to call attention to the load divides, which must be carefully observed for each member in placing the loads on the span. Temperature effects. For a general case of unequal temperatures in the several mem- bers the following assumptions are made: Let the normal . temperature be 65 F., for which the structure has no temperature stresses. Then assume a case where the top chord is heated to 130 F.; the bottom chord to 104 F., and the web system to 117 F. For s=0.000007, and #=29000 kips per sq.in. the values of Ed become Top chord Z = 130-65 =65, Eet = 13.20; WebSystem t = 117 -65 -52, #rf = 10.56; Bottom chord = 104 -65 -39, Eet= 7.92. The values E Al t = dlE and ES a Al t are computed in Table 50A, using I in feet and E in kip feet. Finally the half sum E2S a 4l t is found to be 3293.4 -3593.2 = -299.8 kip feet. Table 50fi gives ^E^lyw=54.7Q and from Eq. (50o) for cosa = l, - *. *. * . i * ' == ~ = ~ 5 ' 7 PS ' which is a thrust acting outward the same as the conventional loading X a = Hence the temperature stress in any member becomes wherein S t has the same sign as S a . A more severe stress would be produced when the top chord is colder than the bottom chord, or when the top chord has a temperature of +10 F. at the same time that the bottom chord has a temperature of -16 F, a case which might occur on a clear, cold day. The stresses S t produced by such a condition would have the opposite sign of S a when found from +X at . For a uniform rise of 65 above the normal, the elongation for the whole span becomes, by Eq. (50E), _.! A D 771 = 2226.6 ft., cos a and from Eq. (SOo), Edat 2226.6 giving stresses S t = S a X ai . For a uniform fall of 65 below the normal ^ == -2226.6 ft. and X at = -20.33 kips, giving stresses S t =S a X at , Compare these results with those of the example in Art. 59, for a solid web arch o about the same dimensions. 176 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X Abutment displacements. Assuming that as a result of yielding abutments the length of span is increased an amount J/=0.03 ft. then from Eq. (49L) 0.03 X 29 ,000 _ _ . e ~ 109.53 '' lpS ' Comparing this result with the one obtained for the solid web arch in Art. 49, it is clear that the framed arch is almost twice as stiff. Deflection of the crown due to the temperature effect producing X at = 5.48 kips, Table 50A, and Eq. (50M). In this case the abutments are assumed rigid making Jr=0 ? and the stresses S are not included, since temperature effects alone are desired. This reduces Eq. (50c) to i /St i , where S t = -S a X at and = - = -EAlX at ; also S l = for a load unit Y at the r r Zi center of a symmetric span, and dlE =EAlt. Hence which can easily be computed for any X at using the values S A) EM and EAl t given in Table 50A. ART. 51. TWO-HINGED ARCH WITH CANTILEVER SIDE SPANS Occasionally a structure of this type is peculiarly adapted to certain sites as the one at High Bridge, Ky. Its application has, however, received adverse criticism because the analysis of stresses was considered too complicated. This objection might apply to any of the ordinary methods of analysis, but not when the problem is solved by influence areas. As a type of bridge it is commendable and deserves careful consideration whenever a particular site offers a suitable profile and good foundations for the center span. Fig. 5lA represents one of several forms which might be employed. The center span is the same as the arch in the previous article with the exception of the end posts which are here made vertical, thus increasing the span to 180 ft. Owing to this dif- ference, the computations in Tables 50A and 50s no longer apply because the stresses SA, SB and S a are now slightly changed. Otherwise the preliminary computations would be precisely the same in both structures. The difference between the two-hinged arch and the present structure with cantilever side spans is entirely in the principal system which results when the redundant thrust X a is removed. In the first case the principal system is a simple truss on two supports A and B, while in Fig. 5lA the principal system becomes a cantilever when the hinged support at A is made movable for the purpose of analysis. The supports at D and F are roller bearings and the simple trusses DC and EF are hinged at C and E respectively. ART. 51 SPECIAL APPLICATIONS OF INFLUENCE LINES 177 178 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X The two-hinged central span, as in any ordinary two-hinged arch, is made deter- minate by substituting a roller bearing for the hinged bearing at A. The XQ influence line is found as in the previous problem and the computation of the Ew loads is not repeated here. However, the loads Ew and_EVt- 2 , must_now be included because these loads affect the directions of the extreme rays C'A' and B'E' of the equilib- rium polygon outside of the span AB while they have no bearing on 'the A" a influence line within this span. The complete X a influence line for the whole span DF is found by prolonging the extreme rays of the equilibrium polygon A'B' to the points C" and E' and finally draw- ing the lines D'C' and E'F'. The middle ordinate A may be computed from the force polygon as follows: X= " measured to scale of lengths, 4/2 or measured to scale of ordinates. The lines A'C' and B'E', found by laying off A, are respectively parallel to the extreme rays of the force polygon. Stress influence areas- These are found precisely as in Art.' 50, so far as the span AB is concerned, by laying off end ordinates T,A =&A/Sa and fjB = ^B/Sa and the factor becomes a=S a . In each case the two end rays so determined will intersect in a point i r , which is vertically under the center of moments i of the member in q^stion. Also the S /S a influence lines must be straight over the panel containing the member. Outside of the span A'B the S /S a lines- are drawn as for any cantilever system as per Art. 26. ART. 52. FIXED FRAMED ARCHES A framed arch with fixed ends has three external redundant conditions according to Eq. (3c) and Fig. 3j, and hence requires for its analysis three elasticity equations either of the form of Eqs. (7n) or Eqs. (8D) . Temperature changes and abutment displacements come into prominence here. These effects are bound to remain more or less problematic because the actual circum- stances attending the construction and later life of the structure cannot be foretold with any high degree of certainty. Therefore, it would seem unnecessary to attempt the analysis of the load effects with any extraordinary refinement and some assumptions may be made to simplify the work, provided they are on the side of safety. It is nearly always permissible to neglect the effect of the web system in comput- ing the elastic loads w. In preliminary work it is also admissible to choose a constant cross-section for the chord members. As a general criticism it might be added that fixed framed arches are not commendable for the flat type, since the temperature and reaction effects produce very considerable ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 179 stresses which increase as the rise diminishes. Good rock foundations must be available under all circumstances, otherwise the fixed arch is prohibitive. It is always advisable to separate the computation of the load stresses from the temperature and reaction displacement stresses so as to determine the relative importance of the latter. General relations between the external forces and the principal system. There are several wavs in which the three external redundant conditions may be applied, depending on the choice of the principal system. One of these was illustrated in Figs. 44E to J, where the principal system was a cantilever fixed at one end. Another assumption might be made by cutting the arch at the crown and creating two cantilevers fixed at the respective abutments of the arch. However, the simplest determinate structure is always a truss on two supports, and in the present case that disposition will be made, as it affords the most compre- hensive solution. FIG. 52A. All three methods have been used by Professor Mueller-Breslau and others, and the final results, are identical and involve about the same amount of labor. In each case the external redundant conditions may be so chosen that the application of the plified Eqs. (44A) becomes possible in accordance with the discussion in Art 44 Fi 52A illustrates the relation of the external forces for a single applied load I , producing reactions R, and R z , intersecting in the point C on this load. This fixes the points ttl and b, on the verticals through the outer supports a and ftjnth span I. ine triangle ^Cb[ thus becomes a resultant polygon with the closing line a^. R l may be resolved into the vertical component A and the haunch thrust H along afr. The reaction R 2 may similarly be resolved into the vertical reaction B and the H' which latter is equal and opposite to the H' acting at ai. ' The vertical reactions A and B are the same as for a simple beam o* with deter- minate supports A and B . Hence A =P(l-e)+l and B = Pe/l. Also the moment, f, P intTof the simple beam, equals M^-KH, wh_ere_H is the honzonta componen ' and K is the vertical ordinate of the triangle a.Cb, through the point m. Hence 180 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X K = M om /H=M om /H' cos a. Therefore, the resultant polygon a l Cb l is determined when H' and the closing line a^bi are found. The external redundant conditions are now removed by relieving the fixed ends and resting the curved structure on two determinate supports at the extreme outer points a and b, Fig. 52B. This simple truss of span /, provided with a hinged bearing at a and a roller bearing at b, constitutes the principal system. To each end of the principal system an imaginary rigid disk is attached as shown in the figure by the two shaded triangles. The redundant conditions are now applied to these disks as external forces or moments, thereby re-establishing conditions of stress in the principal system which are identical with those produced in the original structure while the redundant conditions were active. The two disks are not connected at 0, but are free to transmit a set of forces independently to each abutment. The equal and opposite forces Xb are chosen vertically, and the one acting upward is supposed to act on the disk OA. The forces X c are equal and opposite, but of unknown direction /? with the horizontal, the one acting to the right is applied to the disk OA. The FIG. 52s. two moments X a are also equal and opposite and act separately, one on each disk. The pole is not yet determined, but will be fixed by certain geometric conditions to be established later. To determine the exact relations between these redundant forces and the principal system it is preferable to discuss only the forces acting on one end of the span. In Fig. 52c, the left-hand abutment is shown with the redundant forces which are active at that end only. A similar set, not shown, would be active at the right-hand abutment. The structure is now referred to coordinate axes (x, y) with origin at 0. The y axis is made vertical, and the x axis is coincident with the redundant X c making the angle /? with the horizontal. The location of and the angle /? are still unknown. The ordi- nate y m of any point m is measured vertically from the x axis, while the abscissa x m of this point is measured horizontally from the y axis instead of parallel to the x axis. This is more convenient Jnjihe considerations which follow. The rigid disk aa'O connects the origin with the arch along aa' ', and to this origin are applied two equal and opposite forces H' , which are equal and parallel to the original haunch thrust acting at a^ The equilibrium of the principal system and of the fixed ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 181 arch thus remains undisturbed. In the original fixed condition the force P produced the reactions RI and R2 and these were resolved into components A and H' acting at ai and B and H' acting at 61. Suppose now that all the external forces to the left of a section tt act on the principal system only, and that the three forces H' and the vertical reaction A are applied to the rigid disk aa'O and are thence transmitted to the principal system. Then the force H' at a\ and the force H' ', in opposite direction at 0, form a couple with lever arm z cos a producing a moment X a =H'z cos a. The other force H', acting at and to the right, may be resolved into two components X b and X c , where X b is vertical along the y axis, and X c acts along the x axis. The external forces to the left of the section and acting on the principal system, are then P, A , X b , X c and a moment X a =H'z cos a =Hz . Of these the two forces Xb and X c and the moment X a constitute the redundant condi- tions while the forces P and A are known and all are applied to the principal system FIG. 52c. to the left of the section tt. A similar set of external forces acts on the principal system to the right of the section, but these are not shown in Fig. 52c. The moment of all external forces about any point m of any frame, involving thre redundant conditions, is expressed by Eq. (7 A) as M m =M om -M a X a -M b X b -M c X c (52A) wherein M om =A (l i -x m ) -Pd-^ the moment about m due to the load P acting on the principal system. This is condition X=0. . . M a = l=the moment about m due to the moment X a = l applied to the principal system. Condition X a =1. . . , M 6 = l.* m =the moment about m due to the force X b = l acting on the prmcipa system. Condition X b = l. M c = l. &b , ana o c = , . . . (o2c) I'm I'm ?"m ^m I'm where the lever arm r m is constant for the same member. Therefore, Eq. (52s) will furnish the stress in any member as M 1 S = - = [M om -l-X a -x m X b -y m cos px e ] =S -S a X a -S b X b -S C X C , (o2c) ' m 'm M a I M b x m M c y m cos 8 where S a = - = ; S b = ?= ; and S c = = if - E ....... (52E) * * - ' 4* 4" V / ' m 'm 'm 'm 'm 'm Location of the coordinate axes. According to the conditions imposed by the simplify- ing process of Eqs. (8n), discussed in Art. 44, the coordinate axes must be so located that the displacements d, bearing different double subscripts, must be made zero. These displacements d as given by Eqs. (SB), then become =0', and substituting the values for the stresses from Eqs. (52E) then ^ xl v 1 ^ 2j XW = (J . ^xycosfll v =04, = 1 -^f = cos P2xyw a =0 (52r) ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 183 where w a = l-l/EFr 2 for a moment unity, according to Eqs. (36B) and (36o), giving w a =4l/r=Sl/EFr=Ml/EFr 2 and representing a certain geometric function called an elastic load for some particular pin point with coordinates (x, y). Eqs. (52r) then represent the conditions which determine the location of the coordinate axes such that the simple work Eqs. (44A) become applicable. The first two conditions (52r) imply that the origin of the (ar, ?/) coordinate axes is the center of gravity of the several elastic loads of all panel points of the principal sys- tem. This must be so because the moments ^lxw a and %yw a could not be zero unless the resultant Hw a passes through the origin. According to the third condition, the angle between the axes must be such that the centrifugal moment 2>xyw a =Q, which is true when the axes are related as conjugate axes. The origin may then be located with respect to any assumed pair of axes as the (z, v) axes in Fig. 52c, where the z axis is taken vertically through the point a and the v axis is any horizontal axis conveniently located say through a'. The coordinates z, i\ of all the panel points, are then determined from the arch diagram and tabulated. The w a elastic loads are computed from Eq. (36fl) (and Eqs. (36D) if the web members are included) for each pin point. See the problem in Art. 51. Finally the moments zw a and vw a are found and from these the coordinates li and z ' for the origin are obtained from , ^vw a , 2zw a ._ , ''- and * ~ ..... This fixes the y axis, which is parallel to the vertical z axis through the center of gravity 0. The x axis, while passing through 0, makes some angle ft with the horizontal such that I,xyw a =0, according to the last of Eqs. (52r). The (x, y) coordinates are derived from the (z, v) coordinates when the angle /? is determined. Taking /? positive when measured to the left of the origin and below the horizontal, or to the right and above the horizontal as shown in Fig. 52c, then x=li v and y=z z ' +x tan /? ...... (52H) The angle is found by substituting the value for y from Eqs. (52n) into the condition equation, giving z f +x tan fi]w a =0; or I>xzw a -z f Zxw a + tan p2x 2 w a =0; and noting that 2xw a =Q, then (52j) The abscissa? x being known from Eqs. (52n) the values Xxzw a and Xx 2 w a are readily found and tan /? is then obtained from Eq. (52j). Finally, the y ordmates are computed by the second Eq. (52n) and the new axes and (x, y) coordinates are thus determined. For symmetric arches the x axis is horizontal and tan =0, thus greatly lessening the foregoing computations. 184 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X Influence lines for X a , X&, X c and M m . The coordinate axes (x, y) having been located to fulfill the requirements making ^=^=0, d ac =d ca =0 and d bc =o cb =0, the simplified Eqs. (44A) become applicable to the present problem. Since it is desirable to investigate separately the effects due to loads, temperature and abutment displacements, the load effect will be considered first for immovable abut- ments. Eqs. (44A) then take the simple form of Eqs. (44B), as YP ft *^*- m.u tnn. -*T -. jji^ jnu i -\r *-* * m^ jnr. / ~c\ \ (,)2K) and Y p ^ *->* m u mc r s ' Eqs. (SB) give the displacements dm, w> and cc in terms of the stresses in the members. Noting that p =l/EF, and substituting the values for S a , S b and S c from Eqs. (52E) , then Eqs. (SB) give EFr 2 ' d bb = 'EFr 2 ' and calling l/EFr 2 =w a ; x 2 l/EFr 2 =x 2 w a =xw b and y 2 l/EFr 2 =ij 2 w a =yw c then Eqs. (52x) become for a single load unity at any point ra : 1 $mb 1 ' ' * ' " ' " (52L) Eqs. (52L) furnish the values X a , X b and X c for any position m of a single moving load P TO = 1, as functions of the deflections d ma , d^ and d mc of the point m resulting from the conventional loadings X a = ], X b = l, and X c = l. The exact significance of the deflections 3 ina , dmb and d mc may be determined from their values as given by Eqs. (8 A), noting that S = M om /r m for a load P^-l applied at m from Eqs. (52c), also using the values given by Eqs. (52E). Hence is> v^oo ' y "* om' EF J_ b lEF I M om xl =cos (52M) Now ^ ma is the deflection of a point m due to X a = l and P TO =0, and by Eqs. (52M) it is equal to the sum of the moments M om w a of the loads w a about m. Hence d ma must represent the ordinate to a moment digram drawn for the loads w a with a pole unity. Also, if this moment diagram be drawn with a pole H a = 2w a , then the resulting ordinates i) a will represent ordinates of the X a influence line according to Eqs. (52L). ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 185 Similarly the moment diagram drawn for the loads w b =xw a with a pole H b = will furnish ordinates jj b for the X b influence line and a third moment diagram drawn 'or the loads w c =yw a with a pole H c = cos {32>yw c will give ordinates )j c for the X c influence ine, noting that one of the cos /? factors cancels. Hence the equations for the influence lines of the three redundant conditions from Eqs. (52L), become 1 /^ <> (52N) dr, me TTvi TT cos Lw c H *-< These influence lines remain the same for the same structure and hence are drawn only once. The moment influence line M m , for any point TO with coordinates x m and y m , is derived from Eq. (52s) , giving any moment ordinate as r, m =M m =l\l f > m -\X a +x m X b +y m cos8X c ]=rj om -[-n a +x m -f) b +y m cos fa], . (52o) where T lom is any ordinate of the ordinary moment influence line M om , drawn for the point TO of a simple beam on two determinate supports A and B , which is a different line for each point TO. The ordinates r) a , y b and TJ C are those respectively of the X a , X b and X c influence lines all under the same panel point of the truss. Hence, the moment influence line M m for any point TO is drawn by computing the ordinates -[rj a +x m rj b +y m cos fa] for _all panel points of the loaded cord and plotting these ordinates negatively from the M om influence line. Positive areas thus correspond to positive moments. This M m influence line will serve to find the maximum and minimum bending moments for the point TO due to any system of concentrated loads. The influence line gives the load divides. The stress influence line for any member may be derived from the moment influence line drawn for the center of moments TO for that particular member. Thus if r m is the lever arm for a certain member S with center of moments m, then S=M m /r m , and the ordinates T?, for this stress influence line by Eq. (52o) become (52P) where the several 9 ordinates are as above defined and all measured under the same panel point. Positive areas then give positive or tensile stress. The moment influence line for the center of moments for any member may be directly used by applying the factor l/r w to obtain the stress. The rather lengthy operation of computing all these ordinates for the several stress influence lines for all the members may be considerably shortened by employing the 1S6 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X following suggestions: Thus the multiplications x m -fj b and y m cos /?j? c , may be performed graphically by laying off angles whose tangents are respectively x m and y m cos /?, as illus- trated in Fig. 52o, using the scale of the TJ ordinates. Two such diagrams are required for each M m or S line. If a proportional divider is at hand, then it may be used to perform these multiplications by setting the divi- ders first for the ratio 1 : x m and then for 1 : y m cos /?. Aside from the above methods, the stress influence line for any web member may be derived from the influence lines of two adjacent chords by using the following method given by Professor Mueller-Breslau. When the top chord is the loaded chord, then the FIG. 52o. adjacent members of the bottom chord are used, and vice versa. In Fig. 52E, assume a unit positive stress in L\ and resolve the same into com- ponents 1 parallel to D\ and +2 parallel to D 2 . FIG. 52E. FIG. 52r. Then assume a unit positive stress in L 2 and resolve this into components + n\ parallel to D\ and /i 2 parallel to D 2 . Then for the top chord loaded and no load at tr>, the stresses D\ and D 2 become D 2 =+e 2 L l - +L 2 (52Q) By substituting influence ordinates for the stresses D and L this furnishes a ready means of deducing a D influence line from the two influence lines of the adjacent chords. Thus the DI influence area is the area between the L 2 line and the z\L\/ // t line and the ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 187 final Di area thus obtained will have a factor m. The ordinates to be used in the equa- tions are always under the same load point. When the bottom chord is the loaded chord and there is no load at the point n, then using the stresses as indicated in Fig. 52F, the diagonals D 2 and D 3 become (52 R) The value of D 2 will be the same in both equations Q and R and the signs will come alike by noting that the influence ordinates for the L and U lines have opposite signs and hence the coefficients s and have opposite signs from those used in Eqs. (52 Q ). The resultant polygon for any case of applied loads P may be located by finding the corresponding values of X a , X b , and X c from the three influence lines for thes having previously located the (a;, y) axes. The redundants X e and X b , Fig. 52c ? were taken as the components of H , respec- tively coincident with the x and y axes. The angle /?, which the x axis makes with t horizontal is given by Eq. (52j) and the y axis was taken vertically, izontal component H of X c is H =X C cos /?, and from Fig. 52c, tan a.=t&n (3+-jj cos a z ~ cos a X a = Xg H'cosa H H . . (52s) X a = COS = These dimensions fix the location of the closing line a of the abutments also the haunch thrust H' all in terms of on the two end verticals X and X r, 188 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X A force polygon is drawn by laying off all the loads P in proper succession, dividing this load line into the parts A and B and at the dividing point draw a line parallel to a[bi of length equal to H' '. This determines the pole of the force polygon from whicli the reactions RI and R 2 and the resultant polygon through ai and 61 are easily drawn. See also the force polygon in Fig. 52A, showing how the pole is located when H' , A and B are given. Temperature stresses. For the general case of temperature effects, each member may be supposed to undergo some change 4l t =etl in its length. These changes will pro- duce a deformation in the frame, giving rise to external redundant forces X at , X bt and X ct , which may be found from Eqs. (44c), where abutment displacements and load effects are excluded. Using the values of d a t, $bt and d ct as given by Eqs. (Sc) and then noting the sub- stitutions for Saa = w a , dbb = 2>xw b and d cc = *>yw c made in Eqs. (52N) , the Eqs. (44c) become Y _dat_'S a Jlt -A at * vi Oaa _d bt _ J\.bt ^ COS (52T) where d at represents a rotation measured in arc, while d bt and d ct are the displacements of the origin 0, due to the assumed temperature changes measured in the directions of the y and x axes respectively. These redundant conditions would then produce a moment M m t about any point m as given from Eq. (52s), where M om =Q, thus -M mt =X at +x m X b t+y m cos^X ct . ..... (52r) Since the changes Al t in the lengths of the members represent a simultaneous condition and the stresses S a , S b and S c were previously found in computing the w loads, it is best to solve Eqs. (52T) analytically, using the denominators previously found for the pole distances of the X influence lines. The moment Eq. (52u) can then be solved for any moment point to obtain the stress in .a corresponding member. For a uniform change in temperature of t from the normal, an approximate deter- mination of temperature stresses may be obtained for arches of high rise. For flat arches this assumption would not be permissible. When the effect of the redundants X a and X b is neglected then the stresses may be found from Eqs. (52T) and (52u) as v -- ........ (52v) \ The effect of abutment displacements may be investigated in a similar manner, using Eqs. (44A) and omitting the terms representing the loads P and the temperature. The \BT. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 189 iisplacements Ar must then be assumed or estimated and o a , d b and d c become zero. Hence the redundants and moments may be found in precisely the same manner above illustrated for temperature effects, thus Xar **. (52w) where the d's in the denominators are those used in Eqs. (52T). Example. Owing to the comparatively few fixed arches of the framed type in existence, it was difficult to find a suitable example to use in illustrating the above method of analysis. For this reason the present example, Figs. 52a, was taken with slight modi- fication, from the one given by Professor Mehrtens in his " Statik der Baukonstruk- tionen," Vol. Ill, p. 343. The bridge has a clear span of 162 ft. and rise of 23.69 ft. The roadway is 30 ft. wide and the estimated weight is about 400,000 Ibs., making 20 kips per truss per panel. The uniform live load is taken at 110 Ibs. per sq.ft. of roadway or 30 kips per truss per panel, for medium highway loading. The top chord is the loaded chord and the arch is symmetric, making /? 0. " The top chord panel points are on the arc of a circle whose radius is R u = 163.7 ft. The chords are parallel and the radius for the bottom chord points is R t = 148.7 ft. Ordinarily the computation might be carried out by neglecting the effect of the web members, but for the sake of completeness all members will be included. The cross-sections F, lengths I, and lever arms r of all the members, are tabulated in Tables 52A and 52n, and the Ew a loads are computed by successive steps indicated by the headings of the columns. The lever arms r u and r n are found as described in Figs. 36B and 36F In the present solution the areas were converted into sq.ft. in computing Al so that the modulus E enters with its real value of 4,176,000 kips per sq.ft. instead of 29 000 kips per sq.in. as used in the problem of Art. 51. The stresses S a due to a moment fc-1 kip ft., are easily found, as the reciprocals of the lever arms r, being careful 1 observe the signs of the stresses. The arch being symmetric, the y axis is known to be the vertical axis of symmetry, and the angle ft, which the x axis makes with the horizontal, becomes zero. Hence, the (x y) axes are located by computing the ordinate z ' =E2zw a /EXw a -.21.927 *; thus determining the distance from the v axis to the center of gravity of the w a loads, bee Table 52c. The y ordinates then become y=z-z '. The (x, z) coordinates must of course be determined from the arch dimensions or the equations of the chord curves. Table 52c then gives the functions Ew b =Exw a , Ew c =Eyw a , Exw b and Eyw c all in terms of the w a loads. The sums EXw a , EXxw b and E2yw e represent the pole distances for the X a , X b and X e influence lines such that the influence ordinates will when measured to the scale of lengths. 190 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. X X c INFLUENCE LIKI^-ORDINATES IOO TIMtSfO SCALE OF LENGTHS NCE LINE-ORDIMATCSACTUALTOSCALEOFLfNGTHS. ! I FIG. 52c. SPECIAL APPLICATIONS OF INFLUENCE LINES 191 TABLE 52A w a LOADS FOR CHORD MEMBERS l .S',, S a l Chord. F I r *EJl = jr *Ew a = ^ Panel Point. Sq.in. ft. ft. Kips. Kips sq.ft. ft. 2 100 20.73 14.92 -0.0670 -0.0965 -2.002 0.1341 1 2 4 110 19.51 14.84 -0.0674 -0.0883 -1.728 0.1165 3 4 6 120 18.72 14.79 -0.0676 -0.0811 -1.512 . 1022 5 fi 8 124 18.25 14.77 -0.0677 -0.0786 -1.440 0.0975 7 8 10 130 18.03 14.79 -0.0676 -0.0737 -1.325 0.0896 9 82 20.59 15.14 0.0661 0.1161 2.390 0.1580 2 Q K 94 19.33 15.21 0.0657 0.1008 1.944 0.1277 4 ^ 7 100 18.55 15.25 0.0656 0.0945 1.757 0.1152 6 . 7 9 104 18 . 13 15.28 0.0654 0.0906 1.642 0.1074 8 1(9-9') 128 9.00 15.31 0.0653 0.0734 0.662 0.0432 10 * Above w a loads are for ' = 29,000X144=4,176,000 kipa per sq.ft. TABLE 52s w a LOADS FOR WEB MEMBERS Member. F Sq.in. I ft. r ft. *- Kips. S a f-p ips. sq.ft. .? c 9* 2 8.2 0.078 0.100 14.4 4.21 0.79 1.20 13.5 3.51 0.49 2. 28 4 14.6 0.104 0.330 28.8 5.62 2.61 5.97 27.0 4.68 1.62 9.34 6 18.2 0.088 0.535 25.2 4.75 4.23 -1.98 31.5 3.96 2.62 11.96 8 20.0 0.048 0.647 21.6 2.59 5.11 -6.10 27.0 2.16 3.17 8.01 10 20.5 0.000 0.657 18.0 0.00 5.19 -7.69 22.5 0.00 3.22 5.22 8' 20.0 -0.048 0.647 14.4 -2.59 5.11 -8.12 18.0 -2.16 3.17 3.33 6' 18.2 -0.088 0.535 10.8 -4.75 4.23 -6.88 13.5 -3.96 2.62 1.8S 4' 14.6 -0.104 330 7.2 -5.62 2.61 -4.39 9.0 -4.68 1.62 0.70 2' 8.2 -0.078 0.100 3.6 -4.21 0.79 -1.18 4.5 -3.51 0.49 0.30 All ordinates are in feet. ART. 52 SPECIAL APPLICATIONS OF INFLUENCE LINES 193 The three influence lines for the redundants are now drawn by constructing the force polygons, using the w a , w b and w c forces in the order of the loaded panel points and drawing the corresponding equilibrium polygons. The scales used for the w forces arc any convenient scales, noting that the pole distance must of course be laid off to the same scale as the forces. The poles H b and H c were divided by one hundred, thus making the tjb and rj c ordinatcs hundred times actual when measured to the scale of lengths. The w c forces being of both signs it is best to plat their algebraic sums from some initial point of the load line and number the points in the order of the loads, thus 0, 1,2, 3, etc., to 10. The pole rays are drawn in the same order. Note the check by which the end rays of the X a line intersect in a point on the y axis. If the w a forces were made to act horizontally, then by the method given in Art. 38, the horizontal resultant of these forces would be obtained acting at the intersection of the outer rays. This resultant EHw a would intersect the y axis in the center of gravity 0. However, the method of finding zj by computation is preferable, as the point must be accurately located. The moment influence line for any point as panel point 4, is now constructed by computing the i? w ordinates from Eq. (52o) as illustrated in Table 52o. The coordi- nates of point 4 are x =54 ft. and y=7.9 ft. and the rj a , T) b and rj c ordinates are scaled from the three X influence lines. The rj ordinates are obtained from the M o4 influence line which is the ordinary moment influence line for point 4 of a simple beam of span Z = 180 ft. The table indicates the details of combining these several ordinates to obtain the ordinates which are finally plotted (Figs. 52c) , giving the M 4 influence line. The MS line is similarly found. The M 4 influence line may be used to obtain the stress S 3 _ 5 in the chord 3-5 with lever r = 15.21 ft. Since S=M/r, this same influence line becomes a stress influence line with a factor l/r. The M 5 influence line may thus be regarded as the stress influence line for the chord 4-6 with a factor l/r = 1 -*- 14.79 =0.068. Stress influence lines might also be developed from the formula S =S S a X a S b X b S C X C , but this would require computing the stresses S b and S c in addition to the S a stresses already found, and the stresses S A and S B required to draw the ordinary S line for any member in question. The influence lines for the web members can best be found from the chord influence lines as described in the theoretical portion of this article. It is not considered necessary to go further into this problem, as the theory is fully treated and the general method is illustrated by finding a single stress influence line. The temperature stresses are determined precisely as previously outlined. CHAPTER XI DESIGN OF STATICALLY INDETERMINATE STRUCTURES ART. 53. METHODS FOR PRELIMINARY DESIGNING The term indeterminate, according to previous definitions given in Art. 2 and the tests for identification discussed in Art. 3, is always used in the sense that the laws ot pure statics fail to give a solution when applied to structures involving statically inde- terminate or redundant conditions. In all structures of this class the analysis of stresses can be accomplished only by resorting to the theory of elasticity. Hence, indeterminate is not synonymous with impossible only in so far as the application of the laws ot statics is concerned. The complete analysis of a statically indeterminate structure, according to Eqs (7 A) and (7n), involves the solution of three separate problems as follows: 1. The determination of all internal stresses So, resulting in the principal system from the externally applied loads P. 2. The determination of all internal stresses St and Sr produced in the principal system by the temperature changes and abutment displacements. 3. The determination of elastic deformations in the principal system produced by the several stresses S a , Sb, S c , etc., S t and S r from which the redundant conditions X a , Xb, X c , etc., X t and X r may be evaluated. Eqs. (7n) and (SD) , furnish the solution of these three problems for any indeterminate structure by means of considerations which are of similar character in each problem. A statically determinate structure can always be analyzed by the methods known in statics, when the live loads and the dead loads of the structure are known. Experience in shop practice supplies approximate data for the weights of structures in common use, but new types of structures, departing from the ordinary forms, necessitate repeated approximations to determine the dead loads before a final analysis for the given live loads becomes possible. A statically indeterminate structure is likewise incapable of analyis until its dead weight is known with some degree of certainty and will require a preliminary design which is much more difficult to approximate than in the case of determinate structures, because this involves a knowledge of the magnitude of the redundant conditions. These redundant conditions in turn require that the cross-sections of the members be also known. Fortunately, the influence of a system of loads can be made to depend on relative cross-sections, thus rendering an approximate solution of the redundants X possible, 194 ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 195 before the actual cross-sections are definitely evaluated. This is done by making cer- tain assumptions like adopting a uniform cross-section for all chord members and neglecting the web members entirely, whence the variable F, involved in p=l/EF or Eq. (42u) , is treated as a constant. The w loads required for constructing the deflection polygon for the loaded chord for any condition X = l may then be made EF times the values given by Eq. (36s) , using a pole distance of EF. The details of this process will be discussed later; suffice it to say now that it becomes possible to construct an approx- imate influence line for any redundant condition without knowing the final cross-sections of the members. With the aid of these approximate X influence lines the redundants may be evaluated for any given live loads and some assumed dead loads. Then applying all these loads simultaneously to the principal system, the first approximate values for the stresses S may be determined from a Maxwell .diagram. The redundants X ar*e applied along with the external forces. These stresses S will now serve as a basis for a close approximation of the cross- sections F from which our new influence lines for the redundants X may be found in the usual manner as illustrated by the problems in Chapter X. A final analysis of the stresses is now possible and from this the final sections are found. If the first assumption for dead load was grossly in error then the first X influence lines might be used again, employ- ing revised dead loads before the final analysis is made. This process is on the whole similar to the one above cited for new types of deter- minate structures, the dead loads for which are not accurately known. The additional difficulty here encountered consists in evaluating the redundants, which depend both on the sections and loads at the same time. Therefore, all such work is likely to be some- what tedious. The method of determining approximate values for the redundants in terms of the mutual relations of the sections of the members among themselves, instead of the actual sections, will now be described. In some cases as for arches with parallel chords it is always permissible to assign a constant section to all chord members and neglect the web system entirely. This is then a very simple case and affords a ready solution for the X influence lines by com- puting all w loads EF times too large. When there is only one redundant condition or when there are two or three of these so chosen as to comply with the conditions discussed in Art. 44 then for the above- mentioned case of constant chord sections and disregarding the web system, the redundants may be found analytically from Eq. (42o) whence S , 53 , - . . where the stresses S must be obtained from a Maxwell diagram drawn for a maximum case of combined live and assumed dead loads for the principal system. The stresses 196 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI S a are found for the conventional loading X a = 1 from a similar stress diagram, and these are independent of any actual loads. Eq. (53 A) shows that whenever any of the variables common to the numerator and denominator become constant, they cease to affect the value of X a . The stress in each member of the principal system may now be found from S=S S a X a , ... ........ (53s) and the first approximate sections F may be evaluated from these stresses. The approximate effect due to changes in temperature may also be included b} finding X at from Eq. (44c) and Eqs. (SB) and (Sc) as o at ZS a tl A a< =^ = -- . , ......... (o3c ; Oaa * using some assumed constant value for F. If for any reason the lengths of the chord members are all equal, then I, in Eqs, (53A) and (53c) , would likewise be eliminated. A somewhat more comprehensive method, especially when the chord sections are not assumed equal, is obtained by drawing the influence lines for the redundants. This method will now be outlined. When the redundants are so chosen that the simplified Eqs. (44s) become applicable then the X influence lines may be approximated by computing w loads for relative sections of the members. Thus for any redundant where F c is some constant chosen about equal to a typical chord section as described below. The w a loads from Eqs. (36s) and (36c), for condition X a = l, then become and multiplying each by EF C , then, ^. . (53 E ) ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 197 M a is the moment, with lever arm r, produced by the conventional loading Z a = l. If a deflection polygon be drawn for these EF c w loads with pole distance unity, the ordinates r? will represent values EF c d ma measured to the scale of lengths. Hence the numerator of Eq. (53n) represents the sum of the products Prj, and the denominator is 2>S%pr = ^p/--jr=EF c 'ZM a w, according to Eq. (53E), thus making it possible to construct any X influence line provided suitable assumptions for the values F C /F can be made. The equation of the X influence line for a load P = 1 then becomes 77T TJT ^ 77T TJ1 JJt Zj2jr c O ma Hir c u ma - = . f^F^ 7^ FW^ 1\/T iii * ^uor^ 2>S~l-pT in which F c need not be numerically known, but the ratios F C /F, for different members of the frame, must be approximated in order to compute the EF c w loads from Eqs. (53E). The value of F c should be chosen equal to that of a chord section of most frequent occurrence, which would usually be nearest the crown of an arch or where the chord approaches the horizontal. This will make F c /F = l for most chord members and usually that assumption may be made for both top and bottom chord members. The web system may be entirely neglected in the first approximation or if it is thought desirable to include the web members then F C /F may be given some constant value ranging from 2 to 10 for these members, depending on the character of the structure. The F c should not be regarded as an average value of the chord sections because the deflection polygon of a frame is governed largely by the chords near the center of the span. Thus, if the depth of an arch is much greater at the springing than at the crown, then the crown sections will govern. From the approximate X influence lines the redundant conditions may be evaluated for any case of total loading and the stresses S , S a , etc., can be found by Maxwell dia- grams or otherwise. Then Eq. (53B) will furnish the stresses S from which the sections of the several members are derived. The stresses S may also be obtained from a single stress diagram by applying all redundants and external loads simultaneously to the prin- cipal system. Temperature effect may be included in the redundants at the same time. Another method of making a preliminary design would result from the use of a Williot-Mohr displacement diagram as used in Art. 50. However, the deflection poly- gons as found from the w loads are more easily obtained except for oblique loads P. To illustrate the entire procedure of making a preliminary design for a structure involving redundant conditions, the fixed arch in Art. 52 will now be investigated. Example. The data for this arch are given in the example of Art. 52 and need not be repeated here. The EF c w a loads will be computed from Eq. (53E) making F C /F = 1 and neglecting the web system. The moment M a = I and the lengths I, lever arms r and ordinates z are taken from Tables 52A and 52c. 198 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI TABLE 53A COMPUTATION OF APPROXIMATE LOADS EF c w c Pin Point. Chord. I Feet. r Feet. 1 ~2 Feet. z Feet, EF c w a ^~ zEF c w n 1 0-2 20.73 14.92 . 00449 0.00 0.0931 0.000 3 2-4 19.51 14.84 0.00454 9.99 0.0886 0.885 5 4-6 18.72 14.79 0.00457 17.03 0.0856 1.458 7 6-8 18.25 14.77 0.00458 21.53 0.0836 1.800 9 8-10 18.03 14.79 0.00457 23 . 69 0.0824 1.952 2 1-3 20.59 14.14 0.00500 22.31 0.1030 2.300 4 3-5 19.33 15.21 0.00432 29 . 83 0.0835 2.491 6 5-7 18.55 15.25 0.00430 34 . 99 . 0798 2.792 8 7-9 18.33 15.28 0.00428 38.01 . 0784 2.980 10 4(9-9') 9.00 15.31 0.00427 39.00 0.0384 1.498 0.8164 18.156 This makes z ' = 2zEF c w a 18.156 = 22.24 ft. from which the origin can be located ZEF c w a 0.8164 and the ordinates y=zz ' be computed. The remaining w loads and pole distances are then easily found. TABLE 53 B APPROXIMATE w LOADS AND POLE DISTANCES Coordinates. Pin Point. EF c Wa X y EF c w b = xEF c W a EF c w c = yEF c w a xEF c Wb yEF c Wc Feet. Feet. 1 0.0931 81 -22.24 7.541 -2.070 610.82 46: 04 2 0.1030 72 + 0.07 7.416 + 0.007 533.95 0.00 3 0.0886 63 -12.25 5.582 -1.085 351.67 13.29 4 0.0835 54 + 7.59 4.509 + 0.634 243.49 4.81 5 0.0856 45 - 5.21 3.852 -0.446 173.34 2.32 6 0.0798 36 + 12.75 2.873 + 1.017 103.43 12.97 7 0.0836 27 - 0.71 2.257 -0.059 60.94 0.04 8 0.0784 18 + 15.77 1.411 + 1.236 25.40 19.49 9 0.0824 9 + 1.45 0.742 + 0.120 6.68 0.17 4(10) 0.0384 + 16.76 0.000 + 0.644 0.00 10.79 Totals.. 0.8164 36 183 3 660 2109 72 109 92 H a = 1.633 + 3.658 H b = 4219. 4 tf c = 219.8 The three influence lines for X a , X b , and X c may now be drawn precisely as was done in Figs. 52c, by using the values in Table 53B. ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 19i) Comparing the ordinates found from these influence lines, Fig. 53A, with those pre- viously obtained in Art. 52, it will be seen that the former are fairly close to the correct values, but inclined to be a little large. X,, INFLUENCE LINE - ORDtNATES ACTUAtTOSCALEOF LENGTHS- ISO' FlG. 53A. The live load per truss per panel was stipulated at 30 kips and for an assumed dead load of 20 kips the total load would be 50 kips. . Using the ordinates from the influence lines in Fig. 53A and the total load of 50 kips per truss per panel, the redundants X a , X b and X c are computed in the following Table 53c. For a symmetric loading X b =0 and need not be considered. 200 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI TABLE 53c REDUNDANTS A' a AND X c FOR TOTAL LOADING Ordinates. Redundants. Panel Panel Point. f]a TjC Xa Xc Feet. Feet. Kips. Kip- feet. Kips. 2 8.4 0.086 50 420 4.3 X a = 2X3730 = 7460 k. ft. 4 14.5 0.298 50 725 14.9, X c = 2X97.9 = 195.8 kips 6 19.0 0.520 50 950 26.0 A = iSF = 225 kips 8 21.5 0.684 50 1075 34.2 i(10) 22.4 0.740 25 560 18.5 Totals 225 3730 97.9 The redundants might be obtained for any position of the live load, but since the chords are stressed to their maximum for total loading over the span, and since the web system plays a rather unimportant part in structures of the class here considered, no further investigation of stresses is warranted at this point. The stresses for the total loading, including the redundants X a and X c applied as external forces, are now found from a Maxwell stress diagram in Fig. 53 B, and from these the preliminary cross-sections of the members are deduced. From the sums in Table 53A, the value of z=Z#/4X C = 103.4 ft. The stresses S are now tabulated and cross-sections F are determined in Table 53D, using a rather low unit stress of say 9200 Ibs. per sq.in., instead of an allowable 15,000 ART. 53 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 201 Ibs. This then covers extra metal for joints, splices, latticing, etc., and for reduction to net sections for tension members, and for l/r for columns. o BOO KIM. FIG. 53s. TABLE 53D PRELIMINARY STRESSES AND SECTIONS Top Chord. Bottom Chord. Diagonals. Diagonals. Mem. 8 Kips. F Sq.in. Mem. S Kips. F Sq.in. Mem. S Kips. F Sq.in. Mem. S Kips. F Sq.in. 0-2 2-4 4-6 6-8 8-10 - 920 - 1028 -1114 -1178 -1210 100 110 120 124 130 1-3 3-5 5-7 7-9 9-9' 748 846 930 970 988 82 94 100 104 128 0-1 2-3 4-5 6-7 8-9 - 70 + 166 + 160 + 117 + 95 8 18 18 14 10 1-2 3-1 5-6 7-8 9-10 140 - 90 - 55 + 10 + 48 14 10 6 6 6 202 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI These cross-sections were used in the example of Art. o2, in making the complete analysis. Naturally a two-hinged arch or other structure involving one redundant only, would be susceptible to the same method of design above outlined, though the labor would be greatly lessened. Foi an unsymmetric arch the angle /? would have to be determined- and the closing line aibi would not be horizontal. ART. 54. ON THE CHOICE OF THE REDUNDANT CONDITIONS In the analysis of a structure involving redundancy, it becomes necessary to remove the redundant conditions, whether external or internal, and thereby reduce the indeter- minate system to a determinte principal system to which the redundant conditions are applied along with the external loading. The particular reactions or members best suited to represent the redundant con- ditions designated by X a , Xj,, X c , etc., are those which reduce the given structure to the simplest possible principal system. This will usually offer no difficulty. However, many cases present themselves wherein it would be difficult to decide which of several possible assumptions would represent the most judicious selection of the redundants in the direction of simplifying the analysis. Therefore, a few suggestions along these lines will not be out of place. According to the general method just referred to, the elastic deformation of an indeterminate structure subjected to loads P, will be exactly the same as that of its prin- cipal system loaded with loads P, X a , X b , X c , etc. In each case the deformation is entirely derived from the elastic changes Al in the lengths of the members of the principal system. The structural deformation is thus affected by the redundant conditions merely to the extent of altering the stresses in the necessary members of the principal system. Hence, for the same case of loading, the stresses S in the principal system are always diminished by the redundant conditions provided temperature stresses and abutment displacements are excluded. Generally speaking, any member or reaction of an indeterminate structure may be removed to produce the principal system so long as the latter still remains a stable, determinate structure and does not become subject to infinitesimally small rotation, a condition described in Art. 3, Fig. 3c. The rule should be to select a principal system of the simplest possible form, always avoiding composite structures, such as the three-hinged arch or a cantilever, whenever a simple beam or truss could as well be used. Solid web structures should always be transformed into externally determinate beams by assigning the redundant conditions to the supports. Framed structures, if externally indeterminate, should always be so transformed as to remove the external conditions. The only exception to this rule might be a con- tinuous girder wherein a top chord member over each intermediate pier might be treated as a redundant member. ART. 55 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 203 In the case of the fixed arch, always involving three redimdants, several assumptions may be made. First removing one fixed end and thus reducing the arch to a simple cantilever arm acted upon by a moment X a , a vertical force X b and a horizontal force X c . Second, by removing two members adjacent to one end support and one member adjacent to the other end support, thus forming a girder on two determinate supports acted upon by three external forces replacing the three members thus removed. Also three members may be removed at the crown converting the arch into two cantilever arms with three external forces X a , X b , and X c applied to the end of each arm to replace the redundant members. Frequently all the redundant conditions (not exceeding three) may be applied at one point as described in Art. 44. Then for certain assumed directions of the X's, the work equations become exceedingly simple and afford excellent graphic solutions. The best of these possible solutions was chosen in Art. 52, where a complete fixed arch problem is solved. Indeterminate structures, having a vertical axis of symmetry, will always have two d's bearing double subscripts of like letters, which become equal. Thus d aa =d bb , which condition greatly simplifies the determination of X a and X b , as illustrated in the case of a girder over four supports with the two outer spans equal. See Fig. 43A. In this and similar problems, when the above mentioned symmetry exists, the two influence areas for X a and X b will also be equivalent but symmetrically placed. When the redundant conditions are internal then the only way of deriving the princi- pal system is to remove such redundant members and replace them by external forces A'. Composite structures, or those composed of several determinate frames combined into an indeterminate system, are best treated by assigning the redundant conditions to the reactions and treat as for external redundancy. ART. 55. SOLUTION OF THE GENERAL CASE OF REDUNDANCY In the following it will be assumed that the preliminary design is completed and it is now desired to make the final analysis for stresses due to any causes such as loads, temperature and abutment displacements. Problems of the general type are usually quite complicated, and since it is important to know the stresses due to loads, to temperature changes and to yielding supports sepa- rately, these should always be dealt with in this manner. This is not done merely as a matter of convenience, but it is necessary to know the relation between the load and temperature stresses as a criterion in judging the merits of any particular structure under consideration. In the general discussion which follows here, only the load effects will be included, while the other matters will be taken up later. Hence the quantities 2#Jr and d t , S t , etc., will all be neglected, assuming for the present that they are all zero. To make the case perfectly general, no distinction will be made between external and internal redundancy, hence any of the redundants may be assumed to belong to either class. For n redundant conditions, Eqs. (?A), offer a general solution for the stress S in any 204 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI member, the moment M about any point, the shear Q on any section or any of the reactions R, all for the principal system only. The redundant conditions X, involved in these equations, are treated as external forces applied to the principal system along with the loads P and their reactions. These redundants may be found from Mohr's work equations (?H) or (80) of which there will always be as many equations as there are redundants, rendering the solution of the X's possible by solving for simultaneous values of n unknowns X in n equations. It is usually perferable to employ Eqs. (80) in terms of deflections which are obtained from deflection polygons of the loaded chord, though in principle the solution remains the same whether dealing with conventional stresses or conventional deflections. In general then for n redundants, including effects due to temperature changes and yielding supports, Eqs. (?A) give (55A) S=S S a X a SfrXb etc. S n X n M=M -M a X a -M b X b - etc. -M n X n M t +M r R=R - R a X a - R b X b - etc. -R n X n R t +R Q=*Qo - Q a X a - Q b X b - etc. - Q n X n Q t +Q wherein tne quantities S , M , R and Q are all linear functions of the externally applied loads P acting on the principal system as a result of what is known as condition .X"=-0. The quantities bearing subscripts a, b, c, etc., to n, are constants due to conventional loadings X a = l, X b = l, X c = l, etc., to X n = l, while the subscript prefers to temperature effects and the subscript r to effects produced by yielding supports. The n redundants may be obtained from Eqs. (7n) in terms of the constant stresses due to the conventional loadings or from Eqs. (So) in terms of certain deflection constants obtained from the same conventional loadings. The latter equations are d n = B X a d na etc. X^d an e tc. X n d bn etc. X n d cn etc. X n d nn +d c (55B) wherein the o's have the definitions given in Art. 8, and all those bearing double sub- scripts of like letters are equal by Maxwell's law. Thus =d c =d etc. d = etc. meaning that the order of the subscripts is immaterial and that only half of the d's need be determined, or if they are all determined, that they must check. Hence, either one of the subscripts may be made to refer to the point of application of a load, while the other subscript deals with the conventional loading or condition X \. The terms RAr express the effect due to yielding supports and the d t quantities ART. 55 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 205 express the effect of temperature changes, both of which will be placed equal to zero for the present, though they will be considered in following articles. The redundants X may represent indeterminate reaction conditions or they may be stresses in redundant members, and in either case the displacements d a , d b , S c , etc., o n are the displacements of the points of application of the respective redundants in the directions of their lines of action. These values o may thus be expressed in terms of the redundants themselves in accordance with Eqs. (7j). When the point of application of a certain redundant is rigid, then its d =0. For each of the n redundant conditions there will be one work equation of the form of Eqs. (55B) involving deflections due to conventional loadings. These together with the case of actual loading due to loads P, will constitute in all n + l cases of loading on the principal system to solve for one position of a moving train of loads. The aim, in all practical problems of this nature, consists in representing all the required stresses or other functions by influence lines, thus requiring n -f 1 influence line-* to determine each such stress or function. However, the n influence lines for the n redun- dants will be the same for all cases and all stresses or functions of the same structure, while the influence line for the load effect must be separately found for each stress, moment, shear or reaction. For influence lines the applied load becomes unity and the functions ^P m d m = 1 d m . Each of the n redundants will furnish a deflection polygon for condition X = l, drawn for the loaded chord of the principal system. The n deflection polygons will then furnish all the conventional deflections in Eqs. (55s) for a load P = l. Also, since the subscripts may be interchanged, one such deflection polgyon drawn for X A = l will fur- nish all the double subscript bearing o's of the first equation. Thus as a matter of convenience, all of the d coefficients in a single equation can be determined from one deflection polygon drawn for the loaded chord of the prin- cipal system. The same might also be accomplished by drawing a Williot-Mohr dis- placement diagram for the principal system, though this would usually be more laborious. Therefore, the n Eqs. (55s) can be solved successively with the aid of n deflection polygons, in accordance with the following form with transposed subscripts and for a single load P = l. X I y- 7 X2 X b O hc -X c d cc . . . - X ^nc = , . . . (5oc) where the values of d a , 3 b , S c , etc., were obtained from Eqs. (7j) and may be chosen to represent anv of the possible cases of redundancy, as changes in the lengths of redundant members elastic displacements of redundant supports or angular changes between pairs 206 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI of lines according to Art. 9. When the redundants are immovable reactions, then these S's become zero. The d coefficients in Eqs. (55c) can thus be obtained from n deflection polygons drawn as above described and the n equations solved for simultaneous values of the X's thus furnish the values for a complete solution of Eqs. (OOA). Eqs. (55c) are written for a single load P = l intended specifically for use with influence lines, which are most valuable when dealing with concentrated load systems. Examples involving dead loads only, or when the live load cannot occupy more than one or two positions, could be solved advantageously without resorting to influence lines according to the very interesting example of a lock gate in Chapter XIV. For solid web structures and masonry arches, all the above is applicable, remember- ing that in these cases the redundants must all be external. As soon as the X's are found the stresses on any section are readily ascertained. The solution of the n simultaneous equations is a matter of considerable labor and may offer some difficulty. The method best suited to problems of the kind here con- sidered is given in connection with the lock gate problem referred to above. See Chap- ter XIV, Art. 68. ART. 66. EFFECT OF TEMPERATURE ON INDETERMINATE STRUCTURES Determinate systems are not materially affected by temperature changes. But it is not correct to say that the temperature stresses are zero, because any structure which, in consequence of changes in temperature, undergoes such deformations as to change the positions of the points of application of the external forces must thereby create some stresses. These will usually be small and are entirely negligible except possibly in very flat suspension systems or very flat three-hinged arches where the horizontal pull or thrust is a function of the middle ordinate. For the case of indeterminate structures, wherein the redundant conditions ares direct functions of elastic deformations, the general rule implies that stresses always accompany changes in the lengths of structural dimensions, no matter what external cause may have produced such changes. All temperature changes in externally indeterminate structures will affect the reac- tions and these in turn set up temperature stresses in the members. The only exception to this case is a continuous girder with supports on the same level and subjected to abso- lutely uniform temperature throughout. When such a structure is unequally heated then excessive temperature stresses may result, as shown in the examples of Arts. 47; and 48. On the other hand where the redundancy is entirely internal, no temperature stresses are produced so long as the whole structure retains a uniform temperature throughout. When this uniformity does not exist, then temperature stresses are created, though the reactions may or may not be materially affected. Hence, in all cases of redundancy, it will be necessary to investigate the question of temperature stresses in a very thorough manner, as in many instances these may assume dangerous proportions. ART. so DESIGN OF STATICALLY INDETERMINATE STRUCTURES 207 For this reason also, statically determinate structures should always receive the preference, other considerations being nearly equal. Internal redundancy is less objectionable than external, and according to the best modern engineering experience, no design should be made to include more than one redundant condition of any kind. This is especially true of steel structures, though to a lesser degree applicable to masonry arches of short spans where the poor conductivity of masonry acts as a protective agency against excessive temperature deformations. For long-span masonry arches this consideration assumes greater importance. Even though monumental structures of this class have been built and are regarded with pride and admiration from an esthetic standpoint, they cannot be accepted as representative of the best practice when viewed from the point of modern and progressive engineering. The general effect of temperature changes on indeterminate systems is thus to pro- duce deformations and resultant stresses of the same kind as those created by externally applied loads in accordance with Eqs. (55A). Since the temperature effects may be plus or minus in character, depending on the existing conditions, it is always possible to increase the stresses due to the loading by choosing appropriate changes in temperature. For this reason the temperature effect is applied with a plus sign in Eqs. (55A), meaning thereby that the function, whether posi- tive or negative, suffers a numerical increase. Of course some assumptions will produce a decrease, but these are not vital to the problem and require no consideration. It is always desirable to determine the temperature stresses apart from all other influences so as to furnish a clear conception of their relative importance to the load stresses. This then offers a criterion by which to judge the feasibility of a structure involving redundancy. The temperature stresses are determined from Eqs. (7n) and (80) by dropping out all terms depending on P m , S and R, thus reducing these equations to the following forms, covering all members of the principal system: d a =ea a = ^S a stt-X at ^S^p -X bt 2S a S b p . . . etc. 1 k ... (56A) d b = eil b = '2S b etl-X at 2SJS b p-X bt Z,Szp . . . etc. J or from Eqs. (80) 8a=X a tpa=8 at -X a td aa -X b t8 ab etc. ] \ (o6B) 8 b =X bt p b =d bt -X a t8 btt -X bt 8 bb ... etc. J In both Eqs. (56A) and (56s) the displacements d a and d b become zero for external redundancy with immovable supports. Accordingly the redundant conditions X at , X bt , etc., may be found from either set of the above equations, depending on the method used in finding the load effects from Eqs. (7n), or (80) since both these and Eqs. (56A) and (56u) involve the same constants. The stresses S t are then easily found from a Maxwell stress diagram drawn for the principal system with the forces X at , X bt , etc., applied as external forces. See also examples in Arts. 47, 48, 49 and 50. 208 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI Regarding the assumptions which may be made as a basis for computing temperature stresses, the following conclusions are taken from some experiments on steel arches at Lyons, France, and given in An. d. ponts et chaus., 1893, II, p. 438-444. For an air temperature in the sun of 5 F. higher than the shade temperature, when the latter was 90 F., the steel had an average temperature of 115 F., while the parts exposed directly to the sun were heated to about 130 F. and the shaded portions indicated about 104 F. At the same place the coldest winter temperature was about 15 F. The steel was thus subjected to a mean range of 15 to +115 = 130 F. giving an average of +65 F. Therefore, such a structure should be designed for a mean temperature of 65 F., allowing for uniform changes of 65 F. from this mean. The difference between maximum and minimum simultaneous temperatures in the steel, amounting to 26 F. in these experiments, may cause very serious stresses in certain structures like fixed arches and continuous girders over several supports. In the latter case the entire top chord would elongate relatively to the bottom chord and thus set up an arching effect, relieving the intermediate supports and increasing the two end supports, while a uniform change in temperature would produce no stresses. Structures over a single span would not be so seriously stressed for unequal temper- atures in the two chords. In fact this assumption might work to advantage in the case of arches, though similar conditions for a clear cold day might prove more severe. The painting of steel structures of the indeterminate class thus assumes considerable importance, since light colors will obviously keep down the temperature while black paint will absorb heat. Masonry structures are not so severely distorted by temperature changes, but the elasticity of masonry being proportionately lower, the temperature stresses may prove equally dangerous. ART. 57. EFFECT OF SHOP LENGTHS AND ABUTMENT DISPLACEMENTS Every structure is designed for a certain geometric figure for a condition of no stress. This is the figure which the structure should present when at a mean temperature and when carrying no loads of any kind. Naturally the unavoidable errors in the shop lengths of the members as built and the small inaccuracies in the joints and in the loca- tion of the supports during erection, preclude the practical possibility of accomplishing this end. For structures of the determinate class this will have no significance, but for indeter- minate types, these errors introduce initial stresses which may assume momentous proportions. To avoid such stresses requires the utmost care during construction and erection, and even then only a partially satisfactory outcome can be expected. The difficulties attending the erection of a structure so that no member will be stressed prior to the introduction of the closing link, at the proper temperature, are known to every practical bridge man. In any event the final member should be inserted at the calculated mean tem- perature. If this cannot be accomplished, then the exact length which this member ART. 57 DESIGN OF STATICALLY INDETERMINATE STRUCTURES 209 should have must be ascertained by measuring the space which is actually available when the structure is all in place and under no stress. Then applying proper tem- perature corrections to the entire structure, the required length of the final member must be ascertained before attempting to force the latter into its place. When the end supports are not correctly placed or the entire structure has an eror in length, then all the values S , S a , etc., and the redundant conditions will differ from their computed values. If such displacements are in the same direction with a redundant X, then the stress X is affected thereby, otherwise the principal system only receives the additional stress. Should a redundant member possess an erroneous length, then all the members will receive certain initial stresses, while errors in the lengths of principal members would change somewhat the geometric figure, but could not effect the stresses S , S a , etc., of the principal system. In dealing with abutment displacements it is necessary to distinguish between elastic changes which the- supports may undergo as a result of loading, and permanent settlement. An example of elastic displacement is given in Fig. 45A, where the supporting column CD will be shortened by an amount d b , which can be determined and then be included in the computations as for any other member of the structure. However, should the pier at C undergo permanent settlement, or should one abut- ment, as B, be pushed out horizontally or settle vertically, then serious difficulties would arise unless these displacements could be determined beforehand, which is usually quite impossible. In either case the stresses occasioned by such displacements only, may be found as follows, provided the displacements are known or can be estimated with reasonable accuracy. If, in Eqs. (55s), the terms involving the temperature effects be dropped, then for internal redundancy, the displacements d a , d b , etc., are evaluated from Eqs. (7j), and the work of the reactions is found as shown in the derivation of Eqs. (?E). When the redundancy is purely external then the Eqs. (55s) again apply by treat- ing the reactions R as the reactions of the principal system and evaluating the elastic displacements d a , d b , etc., and the Ar for each reaction R, using such considerations as given in Art. 12, or estimating these displacements as elastic yielding in the masonry supports, etc. See also examples in Articles 49 and 50. In any case the combined stresses from all causes are given by Eqs. (55A) and no further comment is necessary here except to emphasize the inadvisability of adopting externally indeterminate structures whenever immovable supports are not available. Even for bed rock foundations, this will depend largely on the depth to which the masonry must be carried and also on the quality of masonry used. When steel towers or pendulum piers are employed to support an indeterminate structure, then the elastic deformation of such supports must be considered in comput- ing the structural stresses. CHAPTER XII STRESSES IN STATICALLY DETERMINATE STRUCTURES ART. 58. DEAD LOAD STRESSES (a) General considerations. The purpose of taking up the question of stresses ir statically determinate structures is not with any intention of covering this subject exhaust- ively, but merely to present in the briefest possible space the methods which are besl suited to the analysis of all ordinary structures. It was not deemed advisable to take up the present chapter before having treated influence lines and developing the general criteria for the position of loads for maximum and minimum stresses in Chapter IV. The fundamental conceptions there presented are very material to a broader understanding of determinate structures and hence the present order of subjects was considered justifiable. Since it is desired to treat methods of analysis and not types of structures, only the general case of non-parallel chords will be considered. Whenever a structure is sim- plified by introducing parallel chords and otherwise simple relations between its dimen- sions, then naturally the analysis becomes less complicated, until it might be said that the problem is reduced to a mere application of arithmetic. The so-called algebraic methods are, therefore, passed over without further consideration. Since the dead loads are fixed in position and magnitude, the stresses produced by them in any structure must be absolutely invariable. The live loads, however, owing to their shifting position, may produce a variety of conditions and stresses which may tend to increase or to diminish dead load stresses. Therefore, every member may be said to be subjected to a maximum and a minimum stress corresponding to the peculiar or critical positions of the live loads. These critical positions are always determined from certain tests or criteria, differing for different members of the same structure according to the principles discussed in Articles 20, 23 and 24. It was found there that for the chord members the maximum stresses are produced for the case of maximum loading over the entire span, while the minimum stresses in these members result from the minimum total load, which is the dead load. This condition is very different for the web members, where the dead load rarely produces minimum stresses because different positions of partial live load may pro- duce live load stresses of opposite signs. When these are combined with the dead load 210 ART. 58 STRESSES IX STATICALLY DETERMINATE STRUCTURES 211 stresses to obtain total stresses, then the minimum total stresses are less than the dead load stresses, while the maximum total stresses exceed the dead load stresses. When a web member is designed to take only one kind of stress like tension, and this stress is reversed by the live loads, then the panel containing such member must be supplied with a counter web member or the original web member must be differently designed so as to resist both tension and compression. This latter is considered the best modern practice. Hence, in order that the relation between the dead and live load stresses may be clearly brought out in the analysis, it is always necessary to determine these stresses separately. Also, the methods which apply to the solution of dead load stresses are usually not the best adapted to finding the live load stresses, which presents another reason for dealing with the general subject under separate headings. However, each method given will be applied directly to all the members of a struc- ture, showing the application both to chords and web members before proceeding further. (b) Aug. Hitter's methods of moments (1860). This method is based on the gen- eral theorem of moments. Accordingly for all external forces in the same plane and constituting a system in equilibrium, the sum of the static moments, about any point in this plane, must equal zero. If now such a set of forces or loads be applied to a frame, Fig. 58A, and these loads, together with the two end reactions A and B, form a system in equilibrium, then if the FIG. 58A. frame be cut by a section tt, the stresses in the members cut, if acting as external forces, must maintain equilibrium of all the external forces on either side of the section. If, further, such planes can be passed which do not cut more than three members of the frame, then the stresses in the members cut may be determined one at a time, by taking as the center of moments the intersection of two of the members. The moments of the two intersecting members thus become zero. The following designations are used: For member U the center of moments is n and the ever arm is r n ; For member L the center of moments is m and the lever arm is r m ; For member D the center of moments is i and the lever arm is r ; 41so calling M the sum of the moments of the external forces A, P, and P 2 to the 212 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII left of the section it and using subscripts n, m and i to designate the centers about which the moments are taken, then for positive moments in a clockwise direction Ur n +M n =Q or U = ~ " -Lr m +M m =Q or M. -Dri-Mi =0 or Z>=- -1 (5SA) where a positive stress indicates tension and a negative stress indicates compression. Thus a top chord member is always in compression and a bottom chord member is always in tension, while a web member may have either stress depending on its inclination with respect to its center of moments. Hence, the stress S, in any member of a determinate frame, may always be expressed in terms of a moment M and its related lever arm r, by (58B) which is Ritter's fundamental moment equation so extensively used in the previous chapters. The centers of moments for the chord members are thus seen to be located opposite the chord in question, while for a web member the center of moments may be anywhere, depending upon the relative inclinations of the two chords composing the panel. The lever arms should be scaled from a large scale drawing or be computed. When the three members cut by a section intersect in the same point, the method fails. Also, when the chords are parallel the lever arm for the diagonal becomes infinite. Hence the stress in such a diagonal cannot be found by a direct application of Eq. 1.58B) without some modification. For the case of chords which are parallel or nearly so, the stress in a, web member may be found by first computing the stresses in the chords and then finding the web stresses. This is done by choosing for the center of moments for a web member, any convenient point of one chord and then writing a moment equation, including the other chord of the panel cut among the external forces. However, there is a simpler method of finding the web stresses in structures with parallel chords and that is from the shear in the panel cut. Thus in the center panel mq the chords are parallel and the diagonal pq is cut by a section ft'. The lever arm for pq would be infinite. Since the sum of the moments of the external forces to the left of the section and of the three members cut must be zero, therefore the sum of the vertical components of these forces and stresses to the left of the section must likewise be equal to zero. The sum of the vertical components to the left of the section Ft' is called the shear Q. ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES 213 It is equal to the end reaction A minus the sum of the panel loads between A and the section. Thus Q=A-*lP, .......... (58c) where the summation covers only the panel loads to the left of the section and the shear is taken positive upward on the left of the section. The chords, being horizontal, cannot have a vertical component, hence the only stress in any of the three members cut, which can have a vertical component, is the stress in the diagonal pq=D. The vertical component of D l is then DI cos 0, and the sum of the vertical components of all forces and members cut on the left of the section W becomes or When DI is vertical, then 0=0 and the stress becomes equal to the shear. The above moment equation also admits of a graphic solution which will not be considered here. (c) The method of stress diagrams. The first description of these diagrams seems to have come from Bow, and in 1864 Clerk-Maxwell published a paper " On Reciprocal Figures and Diagrams of Forces " in which he presents a scientific treatment of the sub- ject. Cremona, in 1872, discussed the geometric properties of stress diagrams, showing their general usefulness in connection with graphostatics. English and American writers, therefore, call such diagrams "Maxwell stress diagrams," while in Germany and France they bear Cremona's name. The Maxwell stress diagram, so called in the present work, serves a most valuable purpose in the graphic analysis of all determinate frames, and is generally applicable to all cases which are susceptible to treatment by Ritter's moment method. A peculiar relationship exists between a frame and its stress diagram by which each member of the frame is parallel to a line of the diagram and each pin point of the frame is represented by a force polygon in the diagram. It is thus equally possible to construct a stress diagram for a given frame or to construct a frame from a given stress diagram. Hence, the term " reciprocal figures " used by Maxwell. The closed force or funicular polygon, which constitutes the basis for the Maxwell diagram, was known to Stevin, 1608, and Varignon, 1725, and marks the beginning of graphics. Such a polygon may be drawn for any set of forces in equilibrium. Since all the forces meeting in a pin point must constitute a system in equilibrium, a closed force polygon may be drawn for each such point. Hence a Maxwell diagram is merely a succession of closed force polygons drawn for all the pin points of a frame. For any given frame, the directions of all forces and members are given or must be assumed before proceeding to an analysis of stresses. Also, if the frame constitutes a system in equilibrium, then the externally applied loads must be in static equilibrium with the supports or reactions, and all these in turn must be in equilibrium with the internal stresses in the members. KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII Any pin point in equilibrium and acted upon by any number of forces of known direc- tions, may then be represented in the force polygon by a succession of forces and stresses respectively equal and parallel to the forces and stresses meeting in that pin point. Since all directions are known, two magnitudes may be found by inserting the unknown members in such a way as to close the force polygon. This is the fundamental principle of the Maxwell diagram. It is illustrated in Fig. 58s, where the force P and stresses S\ and 82 are known and the stresses S 3 and S 4 are of unknown magnitudes, but of given positions. All the forces meet in the point A and are supposed to be in equilibrium. The method of drawing the stress diagram and nomenclature used in the figure is precisely the same in all cases and affords an easy way of deciding the direction of action of each unknown force. The clockwise arrow indicates the order in which the forces are assembled in the stress diagram. Passing around the point A in this direction the first known force reached is P. Hence the letters a, b, c, d, e are supplied as shown in the angles between the Forces Acting on A. FlG. 58B. Stress Diagram. forces, such that the force P is included between a and 6. In speaking of the force ab in the stress diagram, we mean a force equal and parallel to the force P and acting in the given direction from a to 6 in the stress diagram. Observing this designation, the other known forces are added in their proper order in the stress diagram and made to act in the given directions a-b-c-d as found by going around the point A in a clockwise direction. The stress diagram from a to d is thus obtained and may now be closed by drawing a line de' \ S% through d and another line ae" || 84 through a. The intersection e thus found completes the force polygon and determines the stresses 3 and S both in magnitude and direction of action. Thus the arrows around the force polygon must be in the same direction as indicated by the initial given force P. Accordingly the force 83 acts in the direction from d to e and the force S^ acts in the direction from e to a. All forces in the force polygon are laid off to a certain scale by which the unknown forces are finally determined. A force or stress acting away from the pin point exerts a positive or tensile stress. It is thus seen that the unknown stresses in two of the members meeting in any pin point may be determined by means of a closed force polygon. However, if a given frame presents no pin point involving only two unknown forces, then the method can- ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES 215 not be applied except by first finding the forces in excess of two, by some other means. When there are three unknown forces acting on the point, and there are no redundant conditions involved, then Ritter's moment method will always furnish the necessary solution for one of the unknowns, prior to drawing the Maxwell diagram. The forces in Fig. 58B were arbitrarily assembled in a clockwise direction. A counter clockwise direction might have been chosen with equal right, though the stress diagram would then have occupied a symmetric position with respect to the present case. To illustrate the method, a simple truss, Fig. 58c, is used, are each 10 kips and the reactions A =B =40 kips. k -43.0 B-40K. The panel loads P FIG. 5Sc. The reactions must always be computed in order to supply the forces which are necessary to establish a system of external forces in absolute equihbrium. The diagram is then lettered and the direction of assembling the forces is chosen so as to locate the stress diagram properly on the paper. A pin point embracing only two unknown members must be fleeted for a point beginning and by inspection it is seen that either abutmentsaUsfies this condition theTress diagram iscommenced by drawingjhe triangle * inwhich the vertica upward reaction is known and the two forces cb and tajire found, lollo^ -ound triangle in the direction a-c-b, it is seen that the force cb acts toward the 216 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII The stress in ba is found produces negative or compressive stress in the member cb. to act away from A , and is thus positive or tensile. The construction of the diagram is now continued by assembling the forces around pin point 1, beginning with the first known member be, then adding the known force cg = Pi and finally closing the figure by drawing gf and/6, which are the two unknowns for point 1 . The members ub and bf being_known_the closed force polygon for pin point 2 may next be drawn to find the stresses fe and ed. -*TTZ\__ \_LI_U. _v- ' - FIG. 58o. The process is continued, taking the pin points in the following order: A-l-2-3-4-5-6- 7-8. As a final check rp= B. AH the stresses are then scaled from the stress diagram and written on the truss diagram with proper signs, for compression and + for tension. The present example shows a truss in which the chord stresses are nearly all equal and the web stresses are small. It is similar to the Pauli truss and is economical in design. A very interesting Maxwell diagram is presented in Fig. 53e, where some of the external loading consists of a moment. It frequently happens that loads are not applied directly to the pin points, in which ART. 58 STRESSES IN STATICALLY DETERMINATE STRUCTURES 217 case certain load concentrations must be effected before proceeding to draw a stress- diagram. The members so loaded will usually receive a combination of direct stress and bending as illustrated by the member AC, Fig. 58D, representing a portion of a roof truss. The parallel loads PI to P 5 , acting on the member AC, may be combined (graphically or analytically) into a resultant R of kncfwn position and magnitude, and R may then be resolved into the components R \ and R%, constituting the pin point concentrations at A and C respectively. FIG. 58e. After all loads have been thus concentrated on the several pin points of the struc- ture the total reactions may be computed in the usual manner, taking into account only the loads R,, R 2 , etc., which now replace the loads P, and then the stress diagram is drawn in the usual way and the direct stresses in the members are obtained. The reactions R, and R 2 are the same as for a span d, taken perpendicular to the direction of the forces, and the bending moment M, for any point k distant x from A , may be found from the equilibrium polygon or by computation. For a direct stress - S, as obtained from the stress diagram, the total thrust m the member becomes N=-S-(Ri -ft -ft -ft) cos a. and the moment M-- 218 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP, xil r]H, where H is the pole distance measured to the scale of forces and T? is the ordinate of the equilibrium polygon measured to the scale of lengths. The unit stresses on the extreme fibers of the column A C then become by Eq. (49ivi) f _, J p j , where y is the distance of the extreme fiber from the gravity axis of the section. The dead weight of inclined or horizontal members would be considered in this manner, only that for uniform loads the treatment is simplified. The example given in Fig. 58E belongs to the class previously pointed out, in which not less than three unknown members meet in every pin point, and hence a Maxwell diagram cannot be constructed without first computing one member by Ritter's method of moments. The resultant R is found as in the previous figure, and supposing the support at .4 to be a roller bearing, the reaction at R\ must be vertical and RZ must pass through B and d. Hence the reactions of known directions may be found from the force polygon by resolving R into R \ and R%. The stress in the member AB is then found by passing a section tt', cutting only three members. Then with C as center of moments Eq. (58s) gives AB-H-(Rr-R, l Y-(Rr-R r V-- R (r- r ' C" Kl 2)Ji ( K 2]h J\ r 2 RI and H being now known the Maxwell diagram can be commenced for the pin point A and continued throughout the truss. For vertical loads the solution may be conducted analytically. ART. 69. LIVE LOAD STRESSES (a) The critical positions of a train of moving loads to produce maximum and minimum stresses in any member of a given determinate structure must be known prior to applying any method for finding the stresses themselves. This question was fully discussed in Arts. 20, 23 and 24, and in the General Considerations of Art. 58. All the methods for the analysis of stresses which follow here presuppose a knowl- edge of the criteria for the positions of loads. The reader is referred to the articles just mentioned without repeating this discussion here. (b) The method of influence lines, fully treated in Chapters IV and V, may be men- tioned as the most universal, answering as it does all questions relating to criteria for positions of loads and magnitudes of stresses for any determinate structure. The method requires no further explanation here except to point out the types of structures which warrant its application. In general, the more complicated the geometric figure of a structure, the greater the advisability of employing the method of influence lines, since the geometric relations of the truss dimensions are thereby expressed^, independently of the loads. ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 219 Hence it would scarcely be advisable to apply influence lines to any truss with parallel chords except in cases of complicated cantilever systems. On the other hand a saving of labor might be expected in analyzing structures in which either or both chords are curved. The method certainly offers obvious advantages in all cases illustrated in Chapter V, especially when concentrated loads are employed. It may be added that any structure which does not warrant the accurate analysis by concentrated load systems should not be dignified to the extent of being called a modern bridge. Nor should anyone not familiar with these more exact methods be intrusted with the design of bridges. It should also be emphasized that all computations of stresses should err decidedly on the side of safety, since the secondary stresses produced by the friction on pin-con- nected joints are frequently a very considerable quantity, to say nothing of the failure to analyze properly the effect of the riveted connections. (c) Discussion of methods in common use. In general, all methods involve the solution of the following distinct problems: To find certain moments, shears and end reactions resulting from certain critical positions of the train of loads and then to deter- mine the maximum and mimimum stresses in a particular member in terms of these moments or shears. Both analytic and graphic solutions have been proposed to solve all of these prob- lems in their various phases. Some are applicable only to trusses with parallel chords while others are more general. Still others are used when the position of the train is chosen so as to avoid the complications arising from the exact loadings required by the criteria for maximum and minimum stresses in a certain member. It would be outside the province of the present chapter to present all of these methods in detail, though many of them are exceedingly interesting and ingenious. The author would offer the general criticism that the methods in common use are too special in their application, and when a more general problem is encountered no one method will suffice, but several methods must be combined, thus necessitating an intimate knowl- edge of many limited methods to accomplish a complete analysis. The following method is an attempt to determine the live load stresses in each of the members of any simple truss by applying the same process throughout. (d) The author's method of determining live load stresses is based on Ritter's com- prehensive and universal method of moments. The moment of the live loads on one side of the section taken about the center of moments for any particular member is derived from the principles of the sum A line, see Fig. 22fi. Thus the stress in any member of any determinate truss is given by Eq. (58s) as except for the web members of a truss with parallel chords where the stress is derived from the shear according to Eq. (58o) as ( 59B ) 220 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII The values of M and Q for any position of a live load will now be derived and evaluated first from the sum A line, and then from a single tabulation of moments. The critical position of the loads for any particular effect on a certain member is supposed to have been determined in accordance with Arts. 20, 23 and 24. M 1 P il 11 1 1 *B , *m - 1 lam - ., ... | - _, FIG. 59A. Referring to Fig. 59A and calling MB the sum of the moments of all loads on the span about the reaction B and M m the sum of the moments about any point m; also calling Q m the vertical shear at the point m; then for loads 1 to q } covering a distance xq=ljr (l !*P; (59c) (59D) (59E) (59r) or M m =l am (A- lam (59c) wherein A am = 3. Pe/l am may be defined as the end reaction at A of the loads 1 to m covering the distance x m of a simple span Z am . The values A and ^4 aw may be obtained from a reaction summation influence line, or sum A line, drawn for the reaction A and span I, according to Art. 22, Figs. 22A and 22s. The ordinates of the sum A line may be computed from Eq. (59o), or constructed graphically by the method given in Art. 22. Such a sum A line is shown in Fig. 59s, drawn for a span of 200 feet and using a train of Cooper's EGO loading, consisting of two locomotives followed by a uniform load of 3000 Ibs. per linear foot per rail. The ordinate y x under the first load PI at x, represents the end reaction A for the position of loads indicated, when the span A B is loaded for a distance 61 from B. The reaction A am , for the same train covering the distance x m from m on the span l am> may be obtained from, the same sum A line as follows: Had the sum A line ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 221 been drawn for a span l am then the ordinate at x m from B' would be the required reaction A am . However, this polygon was actually drawn for a span I, and hence the ordinate i) xm distant x m from B' is l/l am times too small, therefore, L (59n) This illustrates how a sum A line drawn for any span may be used to obtain end reactions for any other span. However, in order to avoid the multiplication of all ordi- o . SCALE OF LOADS. 300 KIPS. 222 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XII nates by I, it is best to draw the sum A line for a span of 1000 ft., and then derive all ordinates according to Eq. (o9n) ? by which the ordinates of one A line are to those of another A line in the inverse proportion of the spans for which the lines were drawn. A portion of such a 1000 ft. sum A line is shown in Fig. 59B, from which the actual reactions A and A am , for a span /=200 ft., are obtained as and ^= ....... (59.) I where JJA is the ordinate distant 61 from B' and f) xm is the ordinate distant x m from B'. Any moment M TO or chord stress is now readily found from Eq. (59o) , using the values from Eq. (59i) as follows: l a , ...... (59j) and the stress in any chord member having the point m for its center of moments with a lever arm r m , becomes by Eqs. (59A) and (59j) (59K) The web stresses for trusses with non-parallel chords are derived from moments Mi of the external forces about the center of moments i for any particular web member. Calling the lever arm r t - for a web member with moment center i distant Z at - to the left of A then for x m d, a case which happens very rarely, indicating that some loads extend to the left of n, then *" -"-w -Anmv/ai -rl an ) ^ ^ r(l n i **^ x wherein the two last terms must be computed. It should be remembered that this last case does not occur more than once or twice, if at all, for a whole analysis and hence involves only a slight amount of work. In any case Eq. (59A) gives the stress in a web member for non-parallel chords, as (59x) The web stresses for trusses with parallel chords are derived from the shear accord- ing to Eq. (59E) ART. 59 STRESSES IN STATICALLY DETERMINATE STRUCTURES 223 Thus the vertical shear to the left of a section through any panel when there are no loads to the left of n, making x m M 1000^ m y m ^A-A nm = ^ ~^~, (59o) wherein A nm is the reaction at n for the loads in the panel ^m extending over a distance x m on a span d. When x m >d, which would be the rare case for one or two loads to the left of n, then r -2, R -3~2,, P > . . .. (Glx) from Eqs. (6 IK) and ( +K l>a -K ea ^ -44,350 + 16.2^ At panel point C, by Eq. (3), K ee =O.Q9K A -K M = -45,816+26.4^+2.38^. I By Eq. (6lN), K ec =EJa 2 -2K ee -2K A -# 6c = 147,2 17 -76.6^-6.76^ II \ By Eq. (61o), K bc =E4p 2 +K cc -K hc +K ec -=71,3Ql -41.6^-2.38^ By Eq. (61p), K eb =EA r2 +K cb +K bc -K ec = - 157,577 +84.2# a6 +6.767^. ART -6l SECONDARY STRESSES 241 At panel point B, by Eq. (2) : ^ M = l-23X te -0.12X 6c +0.27X 6e = 15,780-10.2X o6 +0.83X 6([ . [ ByEq. (61 N ), K db =EJa 3 -2K bd -2K be -K eb = -31,502 + 19.4X a6 -3.66X 6o AIII j ByEq. (61o), K ed ~EJ i 8 3 +K M -K eb +K db = l 27,037 -75.0X a6 -9.59Xj >a I By Eq. (61p), K de =E4 r3 +K be +K eb -K db = -25,129 +23.2X a6 +8.04X 6a . At panel point E, by Eq. (5) : K eg =0.493X ec -0.72X e6 +0.82X ed =290,203 - 159.9X n6 - 16.07X 6a . I" By Eq. (6lN), K ge =EJa^-2K eg -2K (id -K de = -811,157+446.6X a6 +43.28X 6a AIV I ByEq. (61o), K dg =EJ l 3 4 +K eg -K de +K ge = -501,161 +263.5X^ + 19.17X6,, I By Eq. (61p), K gd =E4 r4 +K ed +K de -K ge =920,207 -498.4X tt6 -44.83X 6a . An panel point D, by Eq. (4) : K d1 =0.932X d6 -0.288X de +0.252X dff = -148,414 +77.80 K ab -O.gOX^. ByEq. (6lN), K fd =EAa5-2K df -2K dg -K gd =374,963 -184.2X a6 +8.29X 6a AV ByEq. (61o), K g f=E^s + K df -K ad + K fd = -703,644 +392.0X ab +52.22X fca By Eq. (61?) , K fa =EJ r5 +K dg +K gd -K fd =58,049 -SOJX^ -33.95^. The moment Eqs. (6) and (7) for panel points F and G now furnish independent values for X/ ff and K g f and by combining these with the last two equations of AV, the values K ab and X^ may be found as follows: By Eq. (6), X /ff =6.942X /d =2,602,993 -1278.7X a6 +57.55X 6a . By Eq. (7), ^--2.433^+1.746^-3,580,226-1056.81^-183.57^. Substituting these values of Kf g and K g j into the last two equations of AV, then, 2, 544,954 -1228.0X aft + 91.50X^=0, 4,283,870 -2348.8X a6 -235.79X fea =0, from which K ab = 1947.5 and K ba = -1676.8. These values of K ab and X^ are now substituted into the above twenty equations, furnishing all the values X as follows: (1) K ab = 1947.5 (2) K ba =- 1676.8 (3) X ac =6.6X ab = 12,854 (4) # = 15,215- 29,602+ 1,677= -12,710 242 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII (5) K bc = 29,135- 16,748+ 3,354= 15,741 (6) K cb =- 44,350+ 31,549- 3,354 = -16,155 (7) K cc =- 45,816+ 51,414- 3,991= 1,607 (8) K ec = 147,217-149,178 + 11,335= 9,374 (9) K be = 71,391- 81,016+ 3,991= - 5,634 (10) K eb =- 157, 577 + 163 ,979 -11, 335=- 4,933 (11) K bd = 15,780-19,864- 1,392- - 5,476 (12) Kdb=-- 31,502+ 37,781+ 6,137= 12,416 (13) K ed = 127,037-146,062 + 16,081 = - 2,944 (14) K de =- 25,129+ 45,182-13,481= 6,572 (15) K eg = 290,203-311,405+26,946= 5,744 (16) K ge = -811,157 +869,753 -72,572 = -13,976 (17) X d ,= -501,161 +513,166 -32,144 = -20,139 (18) K gd = 920,207-970,634+75,171= 24,744 (19) K df =- 148,414 + 151,515+ 1,509= 4,610 (20) K id = 374,963-358,730-13,901= 2,332 (21) K gf = -703,644 +763,420 -87,562 = -27,786 (22) K ta = 58,049- 98,738+56,927= 16,238 Eq. (6lM) now furnishes a convenient check on the above numerical results, by tak- ing the sum of all the six values K for each triangle. Thus for AI, the values (1) to (6) give S#i =30,542 -30,542 = and for All, the values (5) to (10) give 2K 2 = 26,722 -26,722= and for AIII, the values (9) to (14) give 2# 3 = 18,988 -18,987= ?. and for AIV, the values (13) to (18) give 2 A' 4 =37,060 -37,059 = 1 and for AV, the values (17) to (22) give 2^5=47,924-47,925= -1 The secondary stresses due to the bending at each end of each member are now found from Eq. (61v) as given in the following Table 6lB. ART. (51 SECONDARY STRESSES 243 TABLE 6lB SECONDARY STRESSES DUE TO BENDING Mem. End. 6y I in. K /.- 5 ? Eq. (61v). Lbs. sq.in. '- Lbs. sq.in. 67 K " SI in. Mem. End. E D E G D G D F G F 6.V I in. K '.= 6J P Eq. (61y), Lbs. sq.in. /-I Lbs. sq. in. 61 K ~ SI in. AB AC EC CE BE BD A B A C B C C E B E B D 0.103 + 1,948 - 1,677 + 12,854 -12,710 + 15,741 -16,155 + 1,607 + 9,374 - 5,634 - 4,933 - 5.476 + 12,416 200 172 1645 1627 1621 1664 205 1200 434 380 849 1924 -7310 -7310 + 9960 + 9960 + 4960 + 4960 + 9960 + 9960 + 8875 + 8875 -9510 -9510 0.15 0.13 0.23 0.23 0.46 0.47 0.03 0.17 0.15 0.13 0.34 0.72 ED EG DG DF GF 0.100 - 2,944 + 6,572 + 5,744 -13,976 -20,139 + 24,744 + 4,610 + 2,332 -27,786 + 16,238 294 657 741 1803 1551 1905 715 361 2779 1624 -3010 -3010 + 9370 + 9370 + 2750 + 2750 -9640 -9640 - 460 - 460 0.32 0.72 0.13 0.32 1.76 2.16 0.28 0.14 19.62 11.46 0.128 0.129 0.103 0.077 0.128 0. 155 0.077 0.100 0.155 The total stress on the extreme fiber is f s +/, noting that no increase was made in / for buckling effect in compression members. The actual signs of the values K, and hence also of the moments M^&KI/l, now being determined, the real character of the distortions may be represented diagrammatically in the following Fig. 6 IN. For compression members the most severely stressed fibers will occur on the side where /. is negative and for tension members on the side where / is positive. Thus in the compression member AB, the critical stressesoccur on the upper side at A and on the lower side at B, while for the tension member AC, these occur on the upper and on the lower side at C. 244 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII ART. 62. SECONDARY STRESSES IN RIVETED CROSS FRAMES OF TRUSSES The analysis of cross frames, so far as this is possible, presents many obstacles, some of which cannot be overcome owing to the variable character of the external loads. The bending effect due to all forces acting on a frame, even when direct stresses are neglected, leads to very complicated formulae of more or less questionable value, and no attempt is made here to discuss the general problem. A cross frame is usually subjected to the dead load of the bridge floor concentrated on the floor beam; the live load, impact and centrifugal force applied to the floor; wind pressure against the moving load and the vertical trusses; and unequal deflection of the main trusses due to a variety of causes, but particularly to unequal temperature. The magnitudes of the secondary stresses depend of course on the details of construc- tion and bracing employed in any special frame, so that many different forms would require investigation. However, only two principal types will be discussed and the same formulae are applicable to both through and deck bridges. In the following, the unbraced and one type of braced cross frames are considered first for direct loading and then for wind effect. (a) Dead and live load effects. Unbraced cross frame with rigid post connections. The construction, shown in Fig. 62A with lettered dimensions, is analyzed by assuming equality between the unknown bending moments induced in the posts by the symmetric loads P, as follows: M = M 2 and M] =M S , positive as indicated when they produce compression on the outer fibers of the posts. Call M the bending moment on the floor beam due to symmetrically placed loads P acting on a simple beam which rests on two supports C and D. Also call mi the moment at any point of the post AC distant x from A ; m?. the moment at any point of the strut AB; and m^ the moment at any point of the beam CD. Then from Fig. 62A. m l =M 1 +( M ~ Ml \x, m 2 =M 1 and m 3 =M +M. . . . (62A) \ h / Considering only the effect due to bending, by neglecting shear and direct thrust, the virtual work of deformation for the entire frame would be by Eq. (15n) ~ Cmfdz. .- . : . . . . . . . . (62n) -&/3./0 ~ -&/3 Substituting the values from Eqs. (62A) into Eq. (62p,) then which integrated by considering everything constant except x, gives ?^.+ * \M *b+2M f b Mdx+ P^WI, (62c) /2 bl3\_ J Jo ART. 62 SECONDARY STRESSES 245 which is the general expression for any case of loading where M must be evaluated for each case. Now since the unknown moments M and MI must have such values as will make the first differential derivative of W with respect to each, equal to zero, then after some reduction 'dW and 3M 3/i _ ^ =f) - " " (62n) ?*" m * II -j ,-"" "* .-I, I,- T 50 ITIj Ij - D FIG. 62A. FIG. 62B. Solving Eqs. (62n) gives ,. 6l\o . \ oil = -Mn(-r^-+2 -TJ- and from which (62s) Mi = ^i- 1 Mdx 3/i6 (62p) -1 1/2 '-/\ W 3 For the symmetric loading shown in Fig. 62A, the integral Mdx becomes and for a uniform load p per foot of floor beam, fb f)b 3 JQ (62G) 246 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII MQ and MI may thus be found from Eqs. (62F) and (62r), and from these the moments TOI, m-2 and ra 3 , for any point of the frame, are given by Eqs. (62 A). The maximum fiber stress in the posts must be combined with the direct stress sus- tained as a truss member. The upper strut receives a compression of (MoMJ/h and a bending moment MI over its entire length. The floor beam will receive a bending moment of Mo+M and a direct stress of (Mi-Mo) /h. Example. Given the cross frame at DE, Fig. 61i, with the following dimensions: h =28 ft., b -=17 ft., a = 7 ft., P -59000 Ibs., h =226 in. 4 = 0.0109 ft. 4 , 7 2 = 1134 in. 4 =0.0547 ft., 4 and 7 3 = 15262 in. 4 =0.736 ft. 4 The style of the frame is as shown in Fig. 62A. Using the value from Eq. (62c) in Eq. (6.2r), and substituting the above data, then 3X00109 59000 __ 28X0.736* 4 (U n __ /3X0.0109X17 W3X0.0109X17 \ - > V 28X0.0549 )\ 28X0.736 / and from Eq. (62s) For t/=5.5 inches, Eq. (61v) then gives the secondary stress in the post DE. t ,, ,Miy 1499X12X5.5 for the upper end D, f d = -^~ = =438 Ibs. per sq.in.; /I t ., . ,M v 3541X12X5.5 for the lower end E, f e = p- = = 1034 Ibs. per sq.m. 1\ 22u The actual deformation of this frame is indicated in Fig. 62s. A braced cross frame with rigid post connections Fig. 62c, will now be treated as in the previous case. Neglecting shear and direct stress as before, and dealing only with the effect due to bending, then Eq. (62B) becomes Integrating this expression as above, and then differentiating first with respect to MQ and then with respect to MI and placing these derivatives equal to zero, the follow- ing equations are obtained: ~~ = A! M + 2hi M l + 2eMi - 0. ART. 62 SECONDARY STRESSES Solving these two zero equations and noting that h=h\+e, then 247 i M and f Jo Mdx (62n) /*6 wherein I Mcfo is given by one of Eqs. (62a) and the moments MQ and M\ may Jo thus be found. The stresses in the upper struts then become M I M, = -- and /S e / = -- 1 e e t. *, I. B 11 / H 1L. -m, -I. K > .. _ FIG. 62c. FIG. 620. and the stresses in the post will be and FIG. 62E. Example. Taking dimensions as in the previous example and making hi =19.5 ft. then M =4770 ft.-lbs., and ^=1661 ft.-lbs. giving /, = 1393 Ibs. per sq.in., and f c = Ibs. per sq.in. . (b) Wind effects. Unbraced cross frame with rigid post connections. wind loads on the trusses of a bridge may be carried to the abutments by means of com- plete horizontal trusses in the planes of the top and bottom chords respectively, pro vided the end reactions of the top chord wind system can be carried to the abutments by suitable end postal bracing. In this case the intermediate cross frames suff< little or no distortion and hence carry no bending effects. However when the wind pressure along the top chord is carried down to the I torn chord locally at each cross frame, then the latter must be distorted and thus resi bending In this case the total wind pressure is carried to the abutments through the bottom chord lateral system. The external forces on a cross frame are then as repn 248 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII sented in Fig. 62D. The wind loads w are transmitted through the posts to the lower lateral system, producing bending effects as shown in Fig. 62E. Assuming that the wind loads w are equal and calling T the tangential stress at the point of counterflexure and S the normal stress at the same point, then H=w = (62i) The general work equation for the frame, considering bending effects only, is the same as Eq. (62fi) in which the moments mi, m 2 and ra 3 must be evaluated for the wind forces now acting. The moments of the external forces about A, B, C, D and 0, from Fig. 62o (count- ing clockwise moments positive) are respectively M 3 =- T(h -ho) -Sb =H(h ~ho) = - (62J) M 2 = Th Q Sb 2wh = Hh because -Sb-2w(h-ho)=0 The moments mi, m 2 and m^ thus become mi=Mi + ( ^- - J x=H(x +hoh) with origin at A \ h I ( ^-r -\x=H(hho)(-rl} with origin at A ~with origin at C (62K) Substituting these values into Eq. (62s) then 2# 2 C h , W =ETJ O (x+h ~v which gives by integration, - dx (62L) Since &o must be so chosen that the internal work will be a minimum, this particular value may be obtained by equating to zero the differential derivative of W with respect to ho, thus obtaining ^ W h SRI QM _i_ b (i> i \ j_^ & n 5r- =7-(6rto 3ft) +- r (h -hi) +-j- =0, from which b_ Oh b' /3 /I /2 (62M) ART. 62 249 Substituting this value of ho into Eqs. (62X), will give the moments at all points of the frame and from Eqs. (62j) the stress S and the moments at the ends of the post may be found. Example. For the cross frame of the first example and a horizontal wind load of Ibs. the following values are obtained: oo 2S 3X28 (XOl09 0.0547 / 17 6X28 17_ 0.736 0.0109 0.0547 - 14.25 ft. Substituting values into Eqs. (62j), then, Mi = -H(h-ho) = -3100(28-14.25) = -42,625 ft.-lbs. MO =Hho =3100 X 14.25 -44,175 ft.-lbs. _ 2w(h -ho) _ 6200(28 - 14.25) b 17 This gives for the stress at the bottom of the post AC, = -5197 Ibs. M y_S ' /i F 44175X12X5.5 5197 , 100ftn . ~T4~7 = 12,899 3o3 Ibs. per sq.m. 226 For a braced cross frame with rigid post connections Fig. 62r, the analysis is con- ducted in precisely the same manner as in the previous case, making as before, H =w = T. /t Counting clockwise moments positive, then the moments of the external forces about the points E, C and are respectively M l = -T(h l -ho) = -H(h l -ho) ' M = Tho=Hho .... , -. .. . . (62N) -Sb-2w(h-ho)=0 250 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII The moments at any point x from C then become: for Cl, Mo + ( Ml ^ M -}x = H &o -*) , i,\ (62o) for EA, Mi (x -h,) =H(h l -h ) (-}, e \ e / for CD, Mo ~x =HhJl ~ . These values inserted into the general work Eq. (62B) give for the frame i . --- dx+-- e from which is obtained after integrating i *e^^H *o) q- + Q o J o Taking the first differential derivative with respect to /? and equating the same to zero then 3fto ^i from which MM 0= - .............. ^ ^ Example. Given a cross frame as per Fig. 62r, for which ft =28 ft., 7^ =19. 5 ft., 6 = 17 ft., e=8.5 ft., /i =226 in. 4 , 7 3 = 15262 in. 4 and w>=3100 Ibs. to find /% M and M l . From Eq. (62p) 19.5(56 + 19.5) ho= - iT><226 = 10 - 96ft -' 2(28+39)+^ and from Eqs. (62N) MI = -3100(19.5-10.96) = -26,474 ft.-lbs. M =3100X19.96 =33,976 ft.-lbs. ^=-^^0) = -6215 Ibs. 6 The maximum stress in the post ~AC is then My _S^ _ 88976X12 XBJ _^18 _ 9928 _ 423 Ibs. per sq.in. /i /'ac 226 14.7 The stresses in the top struts and diagonal are not so easily determined when the double latticed type of bracing is used. In the present case with one diagonal AF, the ART. 03 SECONDARY STRESSES 251 stresses are determined by passing a section tt through the three members and the point of counterflexure, and taking moments about A, then, P-^T 3100(28-10.96) _T(hhv) -eEF=Q, or EF=- ^ - = -62i5 Ibs. o.o The stress in AJ3 is found by taking moments about F, obtaining, - T(hi -h ) -Sb -we +eAB =0, whence T ( hl _^) + Sb +we 3100(19.5 -10.95) -6215X17 +3100X8.5 = AB= - T 8.5 Also from the vertical shear on the section ji a =0, or TF = --^- = +^a = !3,842 Ibs. In concluding the subject of cross frames it might be added that good designs should aim at deep floor beams and slender vertical posts, as is clearly indicated by Eq. (62n) , which shows that M is diminished when 7 3 is increased and increased when I r is increased. It has been stated by Mr. Grimm, " Secondary Stresses in Bridge Trusses," p. 80, that the assumption of fixed connections between floor beams and posts is not verified by investigations. The author suggests an explanation for this by calling attention to the fact that recent tests indicate that compression members as formerly built are not nearly as stiff against buckling as was supposed and furthermore, a slight initial deformation approaching the elastic curve which the stressed member might attain, would almost obviate secondary stresses in the cross frames. That such deformations actually exist, or might be produced in overloaded posts by a permanent set, there can be little doubt, and hence it is quite easy to understand why some of the high theoretical stresses do not appear to exist when the actual stresses are measured with mstrum ART. 63. SECONDARY STRESSES DUE TO VARIOUS CAUSES (a) Bending stresses in the members due to their own weight. The maximum bending moment, in any member, resulting from its own weight, when supported on fnctionless pin bearings at the ends, will occur at the center of the member and i (63A) where I is the length of the member, p its weight per unit of length and 6 is the angle which the member makes with the horizontal. WhenThe member is fixed at the ends and there is no direct stress the bendmg moments M at the ends of the member, and M c at the center, become respe, and 252 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII In any case these moments produce a unit stress / on the extreme fibers of the mem- bers which is given by Navier's law, Eq. (49M) as (630 where y is the normal distance from the neutral axis to the extreme fiber. . It is thus seen that M increases as the square of I while / is inversely proportional to 7. Hence, long members of small moment of inertia may receive severe secondary stress due to their own weight. When the member is in compression or tension, the combined bending stress on the extreme fiber, due to the direct stress and the uniform load, must be investigated. This cannot be accomplished by algebraic summation of the separate bending effects, because axial compression increases the deflection due to cross binding, while axial tension dimin- ishes this deflection. The bending stress, resulting from the simultaneous loading must, therefore, be found. The analysis for riveted end connections, while very complicated, is not usually necessary. For pin-connected members where the cross bending effect is always more severe, the following approximate solution is given. A horizontal compression member, with centric pin connection at each end, is shown in Fig. 63A. Let M c designate the maximum bending moment which, in this case, occurs at the center of the member. Calling p the weight of the member per unit of length, S the axial stress, I the length of the member and 3 the deflection at the center, then I ,'.'.-.. . . (63D) _ = _ 8Sd ~48ET~3MEI P from which Assuming the effect of the direct stress on the deflection d to be the same in character as that of the uniform load p, which is not quite true, then (63E) (63F) 48EI -5S1 2 ' This value of d substituted into Eq. (63D) gives 6pm ART. 63 SECONDARY STRESSES 253 The combined direct and bending unit stress on the extreme fiber of the member then becomes by Eq. (49M) (030) where F is the cross-section and y is the normal distance from the neutral axis to the extreme fiber of the member. Horizontal tension members with pin-connected ends, when similarly treated, lead to the following formulae: , 6pPE7 " "* M <=48W+ (63H) The combined direct and bending stress on the extreme fiber is again found by Eq. (63o). Example- Given an eye bar 2 X 15 in., making F =30 sq.in.; 5=600,000 Ibs.; Z=660 in.; 7=562.5 in. 4 ; p=8.5 Ibs. per in.; and E= 29, 000 ,000 Ibs. per sq.in. Then from Eqs. (63n), 3 =0.4824 in.; and M e --= 173 ,450 in. Ibs. The stress on the extreme fiber then becomes by Eq. (63c) ,600,000 173,450X7.5 =2313 lbg . J 30 562.5 showing that the bending stress alone increases the tension in the lower fiber by 11.5 per cent. (b) Secondary stresses in pin-connected structures. In all usual stress computations the question of how equilibrium is established between the stresses in the several mem- bers meeting at a pin or panel point is not considered. It is thus assumed that the individual members extend to the pin centers with full effective sections where the stres are suddenly balanced in a point. In reality the case is very different and no such balance in the stresses can be accom- plished The nearest approach to a realization of the ideal condition is in the case of two abutting compression members and then only when there are no bending stres be transmitted. ,, , , When all the members are connected by pins according to the usual methods of construction, the stresses are transmitted past the panel points in very much the same manner as for riveted connections on account of the factional resistance on the pins created when the structure is distorted by superimposed loads^ The advantages uu^ claimed for pin connections are not all realized in pract.ce and wh, e cent connections are best made by this stvle of panel joint there may be suffic.ent fnctional resistance on the pin to produce secondary bending stresses qnite equal to those occurring m rive. is thus an erroneous idea to regard a pin-connected structure as free from secondary 254 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII stresses. This is never so and the rnalysis may be even more complicated, though the results are probably more reliable than for riveted structures. The exact behavior of a member with pin bearings depends on many circumstances, especially the diameters of the pins. When these are sufficiently large the member can- not turn under load, and secondary stresses must result. These may be partially relieved by the vibratory action of the live loads which would probably allow the structure to adjust itself for some mean condition of loading and reduce the secondary stresses to a minimum. However, temperature changes would enter as a disturbing element to pre- vent such favorable action from establishing permanent relief. The frictional resistance offered by a pin connection is now analyzed. Let S =the stress in any member. R =the total frictional resistance on the surface of contact between the pin and the member. r=the radius of the pin. ^o the angle of repose, making tan ^o = ( a=the coefficient of friction between the two metals. c =the eccentricity of S required to resist the moment of the frictional force R. e=the actual eccentricity of S required to resist the bending moment M a due to secondary stress. The line of stress S, Fig. 63s, is supposed to suffer a displacement c from the pin center A of an amount which will make the moment Sc exactly equal to the opposing moment of the frictional resistance R. FIG. 63B. For a frictionless pin the eccentricity c would be zero and the stress would pass through the pin center. Hence from the figure Sc=Rr and c=r sw.p, ......... (63i) showing that c is independent of the stress S. When c>e no rotation will take place as a result of deformation of the truss due to load effect, and when c < e the member will turn on the pin and the full amount of bend- ing cannot be developed by the frictional resistance. The secondary stresses in a pint-connected structure are found in precisely the same manner as shown in Art. 61, for riveted connections, except that for all members where ART. 63 SECONDARY STRESSES 255 c members subject to a rapid change of stress. The causes enumerated under (2) and (3) are really of the most serious character, and while much may be accomplished by a careful maintenance of the track and rolling stock, a certain average condition will usually prevail beyond which no remedial measures will be possible or feasible. In point of design and construction of a railroad bridge the following suggestions may be offered: The bridge and its approaches back for some distance should be straight and when curves are absolutely unavoidable speed restrictions would seem proper. The rails should be long and the joints and tie connections of the best and most rigid construction, carefully maintained. The connections at the abutments require the most attention, avoiding uneven settlements and loose rails. The intro- duction of continuous roadbed over bridges is very desirable as affording some elasticity to absorb the impact rather than to transmit the same to the structural portions of the floor and trusses. A similar effect is produced by long ties over widely spaced stringers. Very rigid floors and rails directly on stringers may be classed as objectionable. The greater the proportion of dead to live load and the longer the span, the less will be the effect due to impact. The avoidance of synchronous vibrations between the moving load and the structure is of the utmost importance and may be practically accomplished by properly stiffening all the members and by providing thorough bracing in the lateral and sway systems. The effect of vertical centrifugal force might be separately estimated, but the amount is small and the attending conditions are too uncertain to warrant this. As early as 1859, Gerber proposed a general formula providing a certain reduction in the allowable unit stresses to cover these several sources of dynamic impact. This formula is wherein /' is about the elastic limit of the material and S t and S d are the respective live and dead load stresses in any member. The following formula are at present in use and give the factor by whic live load stress should be multiplied, so as to include the dynamic effects considered under the present, heading. Calling S t the live load stress, S d the dead load stress and I the strength of span for chord members, or the loaded distance producing maximum stress in a web member, then

33, = 1.50 (10)* American Railway Eng'r and Maintenance of Way Ass'n, 1910, , _l 2 +40,000 * 1 2 + 20,000 Table 64A gives values of as obtained from each of the above ten formula for a few even lengths I. Only those values of are comparable which are intended for the .ame cla of structures as (1), (3), (5), (6), (8) and (10). It is clearly seen that formula (lO),ives the highest values for short spans I and the lowest values for long spans, noting that in '. . . (64a) * Where I is the span length and all members receive the same impact. ART. 64 SECONDARY STRESSES 265 this formula I represents the length of span and not the loaded distance as in formula? (1) to (9). VALUES OF TABLE 64 A. AS FOUND FROM THE VARIOUS EQS. (64c) I Feet. (l) (2) (3) (4) (5) (6) (7) (8) (9) (10) 10 1.97 1.48 1.78 1.63 Cannot 1.73 1 .55 1.97 1.66 1.99 50 1.86 1.43 1.73 1.50 be 1.44 1.33 1.69 1.50 1.89 100 1.75 1.38 1.67 1.40 solved 1.32 1.24 1.67 1.50 1.67 200 1.60 1.30 1.57 1.28 without 1.24 1.18 1.67 1.50 1.33 300 1.50 1.25 1.50 1.22 knowing 1.21 1.16 1.67 1.50 1.18 400 1.43 1.21 1.44 1.18 the 1.20 1.14 1.67 1.50 1.11 500 1.38 1.19 1.40 1.15 stresses 1.18 1.13 1.67 1.50 1.07 600 1.33 1.17 1.36 1.13 1.18 1.13 1.67 1.50 1.05 l The American Railway Engineer and Maintenance of Way Association formulae is undoubtedly entitled to the greatest confidence, being based on very extensive experiments which were carried out by the committee on impact tests under actual conditions of traffic. The formula is not as yet officially adopted by the association. However, no allowance was made for secondary stresses in the members and hence the formula (10) may be said to include more than actual impact effect which is probably true of all the Eqs. (64o). (d) Fatigue of the material. Based on the classic experiments of Wohler (1859-1870) which were continued by Professor J. Bauschinger, a formula was proposed by Launhardt (1873) and later modified by Weyrauch, aiming to apply the principles of the fatigue of material to the design of bridge members. Wohler's law states that for a large number of repeated load applications, rupture of a material is produced under stress which is below the ordinary (static) breaking strength of that material. The conditions under which Wohler's experiments were made differ so radically from those attending the actual operating conditions of bridges, that it is questionable whether anyone is justified in applying the Launhardt- Weyrauch formula? to bridge designs. Wohler's load repetitions followed in quick succession and were continued without interruption (several million times) until failure was produced. Bridge members are subjected to a repetition of stress which is always followed by a rather long period of rest, and few structures, even under the heaviest traffic, are retained in service long enough to receive several million applications of the moving load. Also, a well-designed bridge is never calculated for stresses approaching e ven the elastic limit, while Wohler bases all his observations on stresses exceeding this limit. In addition to these facts, modern experiments made under conditions which respond quite closely with bridge practice, though limited in extent, point uniformly con against the existence of fatigue in bridges. The above mentioned Launhardt-Weyrauch formulae have been extensively used 266 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XI I] in designing bridge members and are still retained in many specifications. However, this practice does not seem justifiable, especially when allowance is made for impact and secondary stresses generally. The fatigue formula undoubtedly served a good purpose in the days when so many important factors were neglected, but at the present time the aim should be to make proper allowance for all the known stress elements and thereby reduce the " factor of ignorance " to a minimum. (e> Unusual load effects may be produced by loads of unusual magnitude applied at rare intervals or by loads which occur under unusual circumstances. It is possible that the maximum combination of moving loads, wind pressure and temperature effects may take place, under which the structure might be stressed to its utmost capacity, while under ordinary conditions of traffic the stresses would be far below those for which the design was made. The maximum load basis for a design is then an unusual load effect for the average use of a structure. At times it may be necessary to transport some abnormal piece of freight or a train of locomotives, which would exceed the loads assumed in the computations. This may be done without danger even if the members are stressed quite near the elastic limit, provided the design is made with some forethought. For most ordinary structures such a practice might prove very disastrous. It should be observed that, while the stress in any member for a particular position of a moving load is exactly proportional to the intensity of this load, the combination of maximum and minimum stresses for such member, which fixes the required cross- section, might not be tlms proportional. In the case of chord members, all governing stresses are proportional to the loads because the critical positions of the loads are alike for all chords, that is, for maximum chord stress the whole span is fully loaded while the minimum chord stress occurs for dead load only. Hence, the chord sections are directly proportional to the total moving load, while most of the web sections are not so proportional and increase more rapidly than the loads. This is particularly the case with all counter web members wherein the stress is the difference between that produced by the moving load and that due to the dead load. It also applies to the sections of such web members as are subject to stress reversal. This is an important item in making provision for a future increase in train loads, where 25 to 50 per cent additional carrying capacity might easily be secured by providing a slight increase in the counters and web members having stress reversals. Many old structures would still be usable for considerable overload had such provision been made in their design. The best way to provide for such overload in view of future increase in train loads, is to design the structure for the given case of loading and allowable unit stresses and then to increase the sections of the counters, and web members with stress reversals, for the desired overload, which of course could not exceed a reasonable amount according to the limitations imposed by the chord sections. A structure so designed might then carry say 30 per cent overload without imposing unduly on certain few web members, while the ordinary structure would fail by the buckling of these members long before the chords had received any serious stresses. iART. 05 SECONDARY STRESSES 267 In the same way a highway bridge may be designed to carry an occasional overload W a certain heavy road roller, etc., without increasing the sections of the chords and [main web members, but by providing for the extra counter stresses due to the excess load. To the second class of unusual load effects belong derailments, collisions with drift- wood during times of high water and possible settlements of piers or abutments. The absolute prevention of such occurrences may be classified with the impossible. However, the ultimate destruction of a structure when they do occur may be safeguarded by applying certain precautionary measures which should never be overlooked. Thus the bridge floor should be made sufficiently strong to allow a derailed train to pass over without breaking through, and the wheels should be guided between guard rails or by other means to prevent collision with the main truss members, necessitating certain clearances for through bridges to accomplish this. In the case of deck bridges t may be advisable to utilize the top chord as a barrier or protection to prevent the train 'rom leaving the structure. The design of the floor should, therefore, aim at the use of solid web, riveted girders and braces in preference to built-up frames, and the soft, ductile varieties of steel are more desirable than the harder, brittle varieties. In cases where high water may reach the bottom chord, slender eye-bar members ire decidedly objectionable. ART. 65. CONCLUDING REMARKS (a) Features in design intended to diminish secondary and additional stresses. The 'ollowing suggestions should be given careful consideration: Skew structures and bridges on curves should never be built except in very rare cases where no other disposition is possible. These types should be regarded as measures )f last resort. The axes of all members of a truss should be in the same plane and should intersect .n a point at all connections. Special attention should be directed toward a careful design of all connections with i view of reducing the secondary stresses. Thick gusset plates and large diameter rivets materially reduce the sizes of these plates and are thus desirable from this aspect. The widths of members, in the plane of the truss, should not be chosen larger than s absolutely required to secure proper connections and stiffness against buckling -in compression members. It may thus be desirable to taper compression members from ihe center toward both ends. The cross-sections of members should be so chosen that the material is concentrated as far from the neutral axis as possible, thus securing the largest moments of inertia for the smallest over all dimensions. The cross form is thus the least advantageous, while a square box form is most desirable. In rare instances the cross form may be acceptable, when the width of a member is determined by other conditions. Secondary stresses occurring simultaneously in the plane of the truss and in a cross frame are additive in members of the box type while in the cross form they are not additive, a consideratK which may become important at times. 268 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII The secondary stresses may be reduced to a certain degree by shortening all tension members and lengthening all compression members by amounts Al, determined for the case of maximum total load. This practice is generally followed and provides a camber in the unloaded structure which exactly disappears under the full load when the structure assumes its true geometric shape. The plane of the lateral system should coincide with that of the chords and the plane of the floor should be as close as possible to that of the lateral system. The web members of the lateral systems should present considerable stiffness against buckling, making l/r not greater than 140 for all these members, whether in compression or tension. The end portals, or cross frames, should be heavy to carry the wind reactions and dynamic effects to the supports. Floor beams should be made deep to reduce secondary stresses in the cross frames- When this cannot be done then flexible connections should be provided between the truss members and the beams, a type which is best suited to large bridges, while riveted connections are desirable for small spans. The stringers should be made heavy and continuous and should be designed tot transmit the tractive forces to the panel points of the loaded chord instead of to the floor beams by inserting proper tie members between the stringers. Long ties over widely spaced stringers tend to relieve impact vibrations. The use of pin connections is not admissible for bridges of less than 200 ft. span, and even in larger structures the advantages to be gained may not be so easily demon- strated. Some benefits may be derived from pins for the web system but certainly not for the compression chord. According to prevalent practice in designing pins, the diameters are usually so large that the friction produces secondary stresses quite equal to those resulting from riveted; connections. Perhaps this condition might be remedied. Also pin-connected columns are not as stiff as those with riveted ends, though the great convenience offered by pin connections during erection and the saving of material and other advantages in point of design, speak greatly in their favor. The advisability of using the higher classes of steel for the main truss members of large structures and employing soft steel for the floor system was previously mentioned. ' In choosing between different styles of trusses, those of the statically determinate class should always receive preference, other things being equal. The primary stresses will usually be less than in similar indeterminate systems, especially when temperature stresses are included. Yet the deformations and secondary stresses may be less and frequently the connections may be simpler for the indeterminate types. The use of light-colored paints is advisable to reduce temperature effects, especially in structures involving redundant conditions. (b) Final combination of stresses as a basis for designing. If the several stresses resulting in a bridge member from all causes could be absolutely known, then there is nqj good reason why a design should not be based on a unit stress of say four-fifths the elastic limit of the material and still allow ample leeway for some deterioration and unusual effects. The low unit stresses of one-third to one-half the elastic limit, generally employed ART. 65 SECONDARY STRESSES 269 are intended to cover, more or less blindly, the unknown stresses, on the presumption that these are in a way proportional to the known primary stresses. In reality, few, if any, of the secondary and additional stresses are directly related to the primary stresses, but are produced by totally different causes. Our knowledge of the properties of material, while of a purely empiric nature, undoubtedly approaches the truth as closely as the knowable accuracy of the moving loads would require. The behavior of full-size bridge members has also been investigated to an extent which should make it possible to design such members with a reasonable expectancy of developing a strength commensurate with that observed on test specimens. This was perhaps not possible in the past, but should be expected when the results of the elaborate tests of the American Society for Testing Materials become available. If the stresses in the members of a structure are determined with similar exactness and combined into totals representing the maximum and minimum stresses for each member as a result of all known material causes, and on a basis of equivalent quiescent loads, then there is no valid argument why the safe unit working stress / should not be taken at least equal to two-thirds the elastic limit of the material. If also the main members and counters, subjected to reversals of stress, are designed for a 30 per cent overload, then such a structure should still be usable even for a 30 per cent increase in the assumed moving loads. Furthermore, since the combined effect of the maximum values of all the known primary, secondary and additional stresses is one of very rare occurrence, this provides still greater safety and subjects the structure to a less severe average usage than that intended by the maximum combination of all loads. Assuming then that the following stresses have been computed, or approximately estimated when no other means is afforded, then the required area for any member may be determined from the formulae given below. Let Sd =the stress in any member due to dead load. Sz=max. stress in this member due to moving load. S'i=mm. stress in this member due to moving load. 5,0= the max. wind stress due to Wi and w 2 , Art. 64b. S r =the stress due to tractive forces, Art. 64a. S t =the temperature stress when redundant conditions exist. S c =the stress due to centrifugal force from curved track. Art. 64b. ^=the impact coefficient or moving load factor Eq. (64c). Z=the length of a member in inches. r=the least radius of gyration in inches. /=the allowable unit stress = two-thirds the elastic limit of the material. M 8 =the bending moment produced by rigidity of joints, weight of the member, eccentricity, etc., representing the total secondary stress, Arts. 61, 62, 63. Then the required area F for any tension member becomes: t-8m4-8*+&l+-rar. ( 65A ^ 270 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIII Whenever the area My/fr 2 can be added to the area given by the first term of Eq. (65A) to obtain F without materially altering y and r, then it may be done, otherwise a new section must be chosen to satisfy the equation. Also, that combination of the stresses in the parenthesis which gives maximum must be used. Eq. (65A) will also apply to any compression member when / is reduced for buckling effect according to a modified form of Mr. T. H. Johnson's formula, for values of Z/r< 120: For hinged ends and soft steel of 30,000 Ibs. per sq.in. elastic limit, /= 20,000 -88^- 1 T I ..... ..... (65B) for flat ends, / = 20,000 -7 1 1 - \ For hinged ends and medium steel of 35,000 Ibs. per sq.in. elastic limit, /=24,000-105- T (65c) for flat ends, /=24,000- 85- When S'i is opposite in sign from Sd and Si, then the member must be capable of carrying an additional stress l.3'S'i in the place of Si, using the maximum combina- tion of the same sign as S'i in Eq. (65A). In this case, which is one of stress reversal, the factor 1.3 should also be applied to the stress Si. The impact factors and TTD* LO.CTH '- 17 FT. "T ' ~ ' --- j DEFLECTION ^OLYGON FOR GIRDER I. M/I POLYGON FOR GIRDERS II TO V. DEFLECTION "POLYftOIH FOR GIRDERS II TOY h^a**C' All deflections are 20,000 times actual measured to the scale of lengths. FIGS. 68F. ART. 68 MITERING LOCK GATES and 20,000 times actual in feet becomes ^=0.00499(^5^) =8.32 feet. \ iz / This crown deflection d c =8.32 ft. is now incorporated in the deflection polygon for Girder I,J^ig. 68r, by laying off the ordniate C 7 C 77 =d c and then drawing the "final closing line D'C" to complete the deflection polygon. The ordinates written in this diagram represent deflections 20,000 times actual in feet. In similar manner the deflection polygon for girders II to V is drawn. 9/ &) v 1 2 The actual moment deflection at point 6 is now "' =0.00171 inch, and the 20,000 actual deflection due to shear at the same point is &MR 5x2436 d * =2EF\ = 2 X29000 X 30 X0.375 =0 - 00019 inch > being about 11 per cent of that due to moments. Hence, the deflection ordinates are increased 11 per cent to obtain the combined moment and shear deflection polygon. The crown deflection d c is found as before, and all numerical values are the same except the cross-section F of the girder, which is now 31.3 sq.in. Hence 1.4X292 1.4X24.8 . ' c 29000 X 31.3 X 0.367 + 30X864X0.367 or 20,000 times actual in feet gives d c =S.l5 feet. This value of d c is now used to complete the deflection polygon for girders II to V. The average deflection d av for each girder is now computed from the ordinates taken from the deflection polygons Figs. 68F, and the distances Az between these ordinates. The values Az were taken in inches and the areas finally divided by the length Z in inches, which eliminates the unit of length. The computations are given in the following table : TABLE 68c DEFLECTIONS d av FOR THE HORIZONTAL GIRDERS. 4z Horizontal Girder I. Horizontal Girders II to V. d* Areas d* Areas Inches. Feet. dJz. Feet. dJj. 26.5 + 1 . 58 20.94 + 1 . 50 19.88 27.5 1.58+3.12 64.63 1.50+2.96 61.33 27.5 3.12+4.64 106.70 2.96+4.38 100.93 31.2 4.64 + 6.10 167.54 4.38+5.80 158.81 32.5 6.10+7.30 217.75 5.80+6.92 206.70 32.4 7.30+8.14 250.13 6.92+7.76 237.82 32.3 8.14 + 8.52 269.06 7.76+8.20 257.75 31.8 8.52 + 8.68 273.48 3.20+8.35 263.15 27.0 8.68+8.62 233.55 8.35+8.30 224.78 26.3 8.62 + 8.48 224.86 8.30+8.22 217.24 21.8 8.48+8.32 183.12 8.22 + 8.15 178.43 316.8 Tntn.1 2011.76 Total 1926.82 The deflection ordinates d are 20.000 times actual in feet. 288 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV This gives for horizontal girder I, 2011.76 .. 6.35X12 d av = 3168 =6.35 feet, or actually 2000Q -- 0.00381 inch, and for girders II to V, 89 fi - =6-082 feet, or actually - =0.00365 inch, which values will be used in Eqs. (67o) to obtain the -deflections d^ to 5 , observing that d'a v for arches II, III, IV and V is the same quantity. Attention is called to a slight oversight on Figs. 68F, where the deflection ordinates should have been measured at the numbered points 2 to 10, instead of on the lines of the w loads. The effect on the results is scarcely appreciable and therefore, the error was not corrected. The conventional deflections d of the principal system are determined graphically in Figs. 68c and 68n, for the case of full contact at the miter sill. This requires drawing a deflection polygon for the principal system for each of the five cases of conventional loading. The conventional loading for the horizontal arches was made equal to 1 kip, uni- formly distributed over the wetted length of a gate leaf, being 1/1 = 1/27 =0.037 kip per foot of arch. Hence the same load \/l =0.037 kip must be employed as the con- ventional load for the principal system. The deflection of the principal system is made up of the deflection of the vertical girder and the elastic displacements of the two determinate supports A and B. For full contact at the miter sill the B support is assumed as being immovable and the A support is subject to the elastic displacement of the top horizontal arch caused by a con- ventional load X = 1/1 acting on the principal system. The elastic displacement of the point A is determined from the deflection d av just found for the top horizontal arch. For the condition X\ l/l, there are no moments produced in any part of the vertical girder and hence the deflection polygon for this case of loading would show no bending effect of the vertical girder, but merely an elastic displacement of the A support which was previously found to be d\-\=d av for the top horizontal girder. Hence, this con- ventional deflection polygon is easily drawn and becomes a triangle aa'b, Fig. 680, wherein the ordinate aa'=d av -= di_i for the top horizontal girder. This deflection line also furnishes the values d l _ 2 , ^1-3, ^1-4 and != 655.2 M 2 = 0.2341X157.0 =36.75 M 2 = 0.5695 X 48.00= 27.3 40.25 w 2 = 1289.6 w 2 = 1561.7 M 3 = 0.2341 X 1 16.75 = 27.33 M 3 = 0.5695 X 88.25= 50.3 40.25 w> 3 = 910.5 w 3 = 1674.4 MI = 0.2341 X 76.5 =17.91 vl/ 4 = 0.4305= 76.50= 32.9 40.25 w 4 = 539.4 w t = 976.1 M 5 = 0.2341 X 36.25= 8.49 M, = 0.4305 X 36.25= 15.6 36.25 u> 5 = 153.9 w> s = 282.8 M 6 = 0.2341X 0.0 = 0.0 SM> =3775.4 M 8 = 0.0 ~Lw = 5150.2 w loads for X 4 = 1 Kip. w loads for X b = 1 Kip. Ax inches. M for X 4 = l Kip. Kips and inches. w = MAx MforX 5 =l Kip. Kips and inches. -** A 4 = 0.627 4 = 0.373 /1 5 = 0.823 B & = 0.177 48 00 M, = 0.0 Wl = 429.6 1 u>,= 204.0 M 2 = 0.373 X 48.0 =17.9 w 2 = 1022.4 M 2 = 0.177 X 48.00= 8.5 w; 2 = 485.0 M 3 = 0.373 X 88.25X32.9 w 3 = 1626.1 M 3 = 0.177 X 88.25 = 15.6 M> 3 = 770.8 40.25 36.25 M 4 = 0.373 X 128.5 =47.9 M 5 = 0.627 X 36.25 = 22.7 w> 4 = 1420.8 M> 5 = 411.4 M 4 = 0.177 X 128.5 =22.7 M 5 = 0.177 X 168.75 = 29.9 w t = 1058.6 tw s = 541.9 Su> = 3060.3 290 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV The several conventional deflection polygons, for loadings X^l/l =0.037 kip, are now drawn as illustrated in Figs. 68c and 68n. The factor 30,000 was introduced to obtain dia- grams of convenient size to the scale of the drawing, and it is worthy of comment that the highest desirable accuracy may be obtained from small drawings by an appropriate choice of the scale and this factor. The scale employed for the force polygon is entirely immate- rial so long as it is convenient, noting that the w loads and the pole must be measured to the same scale. The pole distance remains the same for all the diagrams and is 25,360 w units. The moment diagrams were drawn for the conventional loadings X = 1 to correspond with the moments in Table 68D, and serve merely to locate the centers of gravity gi, <7 2 , etc., which are the points of application for the w loads. The value d av =0.00381 inch, as previously found for the top horizontal arch, now furnishes <5i_ 1 =30,000 d av = 114.3 inches, from which the deflection line for X^=l/l is drawn. This diagram then furnishes the deflections #1 -2=^2-1, di_ 3 -=d s _i, ^ 1 _ 4 =o 4 _ 1 and ^1-5=^5-1, which are necessary in fixing the closing lines of the several other deflec- tion polygons. With this explanation the drawings should be readily understandable and will furnish all the deflections d required in solving Eqs. (67c) . The problem is solved completely for the case of full contact at the miter sill, and the method of deriving the deflections for the case of no sill contact is shown in Figs. 68H. When there is no sill contact then the bottom support of the vertical girder becomes the bottom horizontal arch in the same manner as the principal system was previously f ormed for the top support at A , requiring now the determination of d' av for the bottom arch (for a loading q =0.037 k. per foot of arch) as was done for the other arches. This d'^ then gives <5e -6=30,000 d' av for the principal system and furnishes the means of draw- ing the deflection polygon for condition X 6 = l/l from which ^0-5=^5-6, ^6-4=^4-6, e~tc., are obtained. These deflections then serve to complete the conventional deflection polygons for the case of no contact at the miter sill in the manner shown for the two conditions X 4 = l/l and X 5 = l/l in Figs. 68n. Hence, all the work previously accom- plished in solving the case for full contact is also usable for the case of no contact, though the final deflections d are different in the two problems. The formation of the final equations for the solution of the redundants X is now pos- sible by introducing the numerical values for all the deflections d into Eqs. (67c) , remem- bering that these were all taken 30,000 times actual, which in no wise interferes with their use in the equations. The numerical terms 27> TO l7n , Zp m d 2m , etc., must first be computed for the actual water loads p per foot of the horizontal arches. These loads are given on the diagram Fig. 68c and represent the actual water pressures on the girders for vertical depths extending from center to center of the respective spaces between arches. Thus the load pi is the water pressure per horizontal foot of arch and over a depth extending from the water surface down to a line 24 inches below the top arch. The load p% is the pressure per horizontal foot of the arch included between the depths 38 inches and 82.12 inches below the surface, and so on for each girder. The double-subscript-bearing deflections are taken from the deflection polygons in ART. 08 MITERING LOCK GATES DEFLECTION POLYGONS FOR THE PRINCIPAL SYSTEM FOR FULL CONTACT AT MITER . THE CONVENTIONAL LOAD *1/1= 1/27O.O37 KIP CON. 291 * 2 MOMENT DjIAGRAM FORjX^=IKIR All deflections are 30.000 times actual measured to the scale of lengths. FlGS. 68G. 292 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV DEFLECTION POLYGONS FOR THE PRINCIPAL SYSTEM FOR FULL CONTACT AT MITER. THE CONVENTIONAL LOAD = 1/1= 1/27= O.037 KIP CONCENTRATED. . . ._!. .._, v upstream chord. n Gravity axis. t tn~ 4>" 1 i 2 3 4 r 5 L AX=4ft^ I W-'iS J 4o'.'z5 > ' 4o'/25 > < 36.25 MOMENT DIAGRAM FOR|X = I I 21 __jPi oi'o, nit r i i MOMENT DIAGRAM ^0^X5= I KIP I DEFLECTION POLYGON FOR X ft 1/l- 0.037 K- NO MITER SILL CONTACT. SCALES BO LOADS. 1000 O 5000 10,000 \ All deflections ire 30,000 times actual measured to the scale of lengths. FlGS. 68H. ART. 68 MITERING LOCK GATES 293 Figs. 68a to H, using the values 30,000 times actual, in inches. The deflections di to S 5 are obtained according to Eqs. (67o), using 30,000 times the deflection d av , found for the horizontal arches. All these numerical values are now entered in Table 68E, and the sum products SpwAm are computed, giving all required data for writing out the general equations for the redundants. TABLE 68E COMPUTATION OF THE TERMS K = Zp m dam FOR EQS. (67c) 3 = davX Pt. Pm s im Pm "im *m Pm 3 tm S 3m Pm 3 vn 3 4 m Pm 3 -38.65 -38.02 -37.79 -37.64 37.58 x t c t - 7.16 - 4.70 - 3.72 -37.64 x 3 c s -13.58 - 8.50 -41.51 3.17 = 0.3358 x 3 ' 17 .AjCj 18.73 51.22 .64 4 ~4 87 6 64 -78.12 10.85 = 0.6509 .A 7.53 10 85 20.95 = 0.8818 2 10.66 20.95 = 1.0178 1 20.2 = 1.0371ki ps per horizontal "oot of gate. C = the coefficients from the last of Tables 68F. The redundants X are thus expressed in kips per horizontal foot of gate, representing the average intensity of these forces for the average vertical stiffness of a gate leaf. The reactions A and B and the moments M for this average vertical girder, may now be found from Eqs. (67e) employing the values in the following table. 296 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XIV TABLE 68n REACTIONS AND MOMENTS, VERTICAL GIRDER Point. p Kips per ft. X Kips per ft. p-X Kips per ft. h Inches. hi -h Inches. (p-X)h Kip-inches. (p-X)(hi-h) Kip-inches. 1 0.314 1.0371 -0.7231 205.0 0.0 -148.236 0.0 2 1.150 1.0178 + 0.1322 157.0 48.0 + 20.724 6.336 3 1.783 0.8818 0.9012 116.75 88.25 105.192 79.513 4 2.475 0.6509 1.8241 76.5 128.5 139.536 234 . 380 5 3.015 0.3358 2.6792 36.25 168.75 97.120 452.115 Totals 8.737 3 . 9234 + 214.336 772.344 Eqs. (67fi) then give for A, B and M 3 779 -h) =i_' =3.767 kips. \ 341.848 = 196.057 kip-inches, t , My 196.057X23.4 __,_ . giving for f=j= -- =- ~ - = 4.727 kips per sq.m. on the most extreme fiber of J- \j i J. .O the vertical girder at the panel 3. In like manner the stress may be found for any other point of the vertical girder, using the results given in Table 68n. B+PQ represents the pressure which the miter sill must sustain on the assumption of full contact. Finally the static condition ././ -. . . . . . . (680) must be fulfilled by the above numerical results. The deflections of the vertical girder are now readily found from Eqs. (67c) and the redundants X which are now known. The deflections d av of the horizontal girders were found for a load of 1 kip uniformly distributed over the wetted length of 27 ft. Hence, for a load of 27.X" uniformly dis- tributed over the same length, the deflection would be 27d av X. Accordingly the deflections d to d$ become .r- ^, \ 1 _,-" ~~*T 1 FIG. 7lA. The thickness of the arch ring at the crown is approximated from the formula? of A. Tolkmitt,* which are rational but approximate, and, therefore, furnish much better results than purely empiric formula?.. Thus for a uniform live load p/2 over the whole span, and an allowable unit stress 0.4/, which is supposed to be 0.4 as great as for the unsymmetric critical loading, the crown thickness should be 4+T5)' t for feet or meters, (71 A) where the allowable unit stress / is likewise expressed in cu. ft. of masonry, as was done for the loads. * Leitfaden fuer das Entwerfen Gewoelbter Bruecken, 1895. ART - 71 FIXED MASONRY ARCHES 307 For the unsymmetric live load, extending over half the span to the center, the following value of Z) is obtained : where G = $(q + ?-+*} f for feet or meters, (71s) \ 2i Au/ j When the value from Eq. (7 IB) exceeds that given by Eq. (71 A) then the larger dimension should be adopted. It is clearly seen that the full allowable unit stress / could not be taken in the above formulae, because this average case of loading will stress the arch ring only about half as much as when the resultant polygon passes through the middle third points. In all arch designs where tensile stresses are prohibitive, this condition will govern. The next step is to choose a preliminary shape for the intrados such that, for the case of dead load plus half live load over the whole span, the resultant polygon will exactly coincide with the arch center line. At this point all theory fails and the judgment and experience of the engineer must guide in making a suitable first approximation. However, the following practical suggestions will serve a valuable purpose. Theoretically the intrados can never be a true circle nor a true parabola for any arch which is made to follow the resultant polygon, as above required, for economic reasons. The resultant polygon becomes a parabola when the total load is uniform- per foot of arch. This can never happen in a real arch, even if the spandrel filling were neglected, because the stresses in the arch ring increase from the crown toward the abutments, thus necessitating a variable thickness of ring, increasing with the stress. On the other hand the resultant polygon becomes a circle when the superimposed load at the springing points becomes infinite, a condition which is never attainable. Hence an arch of economic design and shape to fit the resultant polygon for average loading will have an intrados which ^neither a circle nor a parabola, but a curve lying between these two. Having thus established the limits between which the true intrados must be situated and knowing from experience that the true line falls nearer to the mean of the two curves than to either one of them, for all open spandrel arches, it becomes an easy matter to approximate the intrados. Hence, for open spandrels with roadway supported on piers or columns, the intrados may safely be taken half way between the circle and the parabola, both drawn through the three given points of the intrados. The radius of the circle is given by the formula and the parabola passing through the same three points is which gives the coordinates y =ho/l and x =/ /4 for the quarter points, see Fig. 7U. 308 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV From this a mean point at each quarter span is readily determined and a three- center curve is then made to pass between these two and the given crown and abutment points. When the spandrel filling is solid, then the intrados approaches nearer to the circle and for a plain arch ring without any spandrel loading, the true intrados approaches the parabola. Between these limits the designer will soon be able to make very close approximations even for the first trial. To draw the extrados, no good rules can be given, though a fair approximation can be made by applying the formula for the thickness of any point of an arch ring in terms of the thickness DO at the crown, as follows: = V / .- . . . /;.;. (71B) cos 9 For flat arches this gives good approximations, but for semicircular arches the formula should not be used for angles greater than 30 from the crown. Even for <=30 the values are a little large and for (/> =90, D = GO , which is clearly impossible. However, the 30 from the crown include the most important portion of the arch ring, so that the remainder of the extrados can be approximated with sufficient accuracy for the first trial. Another method is to sweep an arc with radius r'=r+Z)o through the center of the crown joint, which will be a good approximation for a center line between crown and quarter points. It is not prudent to spend more time in preliminary guessing, but with the dimensions above given proceed at once to the graphic analysis. For this purpose the arch should be drawn on a scale of at least 1 : 100, using a scale 1 ft. =3000 cu. ft. of masonry for the scale of forces. For small structures larger scales may be employed. The arch ring is now divided up into suitable sections corresponding with the panel lengths chosen for the spandrel piers and the areas of these sections and of the piers are then computed from the drawing. Having thus found all the areas, including the uniform floor load q per ft. of bridge, we may proceed to draw the resultant polygon for a live load p/2 extending over the whole span. This polygon should be passed through the center points of the crown and abutment joints and if it coincides quite closely with the assumed arch center line, then the investigation may be continued for the unsymmetric loading, otherwise the shape of the arch ring must be corrected and the first resultant polygon must be recon- structed. A second trial nearly always gives acceptable results, and sometimes the first trial is sufficiently close. For symmetric loading the resultant polygon is drawn for the half span only. With the dead loads previously found and a live load covering half the span and extending past the crown to the load divide i for the critical point m, see Fig. 7lA, now pass a resultant polygon through the points a, b, and n. The point a is the outer third point of the arch section at A, the point b is the inner third point at B and the point n is about Do/8 above the center line at the crown. This is the first approximation. ART. 72 FIXED MASONRY ARCHES 309 If this resultant polygon remains within the middle third at every point and crosses the axial line in at least three points, then the design is acceptable. If the polygon does not touch the middle third point at the two critical sections m and s, Fig. 7lA, then the ring is too thick. If the polygon goes outside the middle third at one of these sections and remains inside at the other, then the point n should be shifted and a new polygon should be passed through a, b, and n. However, if the resultant polygon passes outside the middle third at both critical sections, then the arch ring must be made thicker and the investigation is then repeated. Lastly the stresses at the four sections a, m, n, and b must not exceed the allowable unit stress as given by Eq. (69A). When all these requirements are fulfilled, the crown section should be tested for full live load over the entire span, which is easily done from the loads already found for the half span, since this is a case of symmetric loading. This then constitutes a complete design according to the ordinary graphic method, and the arch ring so found should now be subjected to a rigid analysis according to the theory of elasticity which follows. The above outline will be exemplified more fully when presenting a problem at the close of this chapter. The criterion for position of moving loads to produce maximum and minimum stress is discussed in another of this point, is measured horizontally from the y axis instead of parallel to the x axis. This is a mere matter of convenience. The span AB=l is taken as the line joining the intersections between the axial line ART. 72 FIXED MASONRY ARCHES 311 and the verticals through a'\ and b'\. The support at a\ is assumed as hinged while the one at b\ is made movable for the principal system. The single load P then produces reactions RI and RZ, intersecting in the point C and passing respectively through the' unknown points a^ and 61 on the verticals through the supports a\ and b\. The triangle aiC&i thus becomes a resultant polygon with the closing line a\b\. The reaction RI may be resolved into the vertical component A and the haunch thrust H' along a\b\. The reaction R% may be similarly resolved into the vertical reaction B and the thrust H', which latter is equal and opposite to H' acting at a\. The vertical reactions A and B are the same as for a simple beam of span I on determinate supports. Hence, A =j(l-e) and ' B =*. . ....... (72 A ) Also, the moment for any point m of the simple beam, equals -\f i\f osa or tf--i--, .... (72 B ) where K is the ordinate ig through m, and H is the horizontal component of H' . Therefore, the resultant polygon a\Cb\ becomes fixed whenever H' and the closing line a\b\ are found. The origin is connected to the arch along aa' by the rigid disc aa'O and to this origin are applied two equal and opposite forces H f , which are equal and parallel to the original hanuch thrust acting at a\. The equilibrium of the principal system and of the fixed arch thus remains undisturbed. Suppose now that all the external forces to the left of a section tt' act on the principal system only, and that the three forces H' and the vertical reaction A are applied to the rigid disc and are thence transmitted to the principal system. Then the force H' at a] and the opposing force H' acting at 0, form a couple with lever arm z cosa, producing a moment X a =H'z cosa. The other force H', acting at and to the right, may be resolved into two components X b and X c , where X b is vertical along the y axis, and X c acts along the x axis. The external forces to the left of the section and applied to the principal system are then P, A , X b , X c , and a moment Xa=H'z cos a = Hz . Of these the two forces X b and X c and the moment Xa constitute the redundant conditions while the forces P and A are known, and all are applied to the principal system to the left of the section tt'. A similar set of external forces (not shown) acts on the principal system to the right of this section. The moment of all the external forces abou'c any point m of any structure, involving three external redundant conditions, is given by Eq\ (7 A) as M m ^M om -M a X a -M b X b -M c X c , ....... (72c) wherein M om =-A (li x m )Pd=the moment about m due to the load P acting on the principal system, This is condition X=0. 312 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV M a = l is the moment about m due to the moment X u = l applied to the principal system. Condition X a 1 . Mb = 1 x m is the moment about m due to the force X b = 1 acting on the principal system. Condition X b = 1 . M c = l-y m cos /? is the moment about m due to the force X c = l acting on the principal system. Condition 'X c = 1 . Substituting these values into Eq. (72c), gives the following fundamental moment equation for fixed arches: M m =M om -l-X a -x m X b -y m cospX e , . (72D) Before M m can be determined for any point m of the arch, the three redundants X a , Xb, and X c must be evaluated from three simultaneous work equations of the form of Eqs. (44s) which may be made to apply to the present problem by so locating the (x, y) axes that d ab =d ba =Q, ^=^=0 and d bc =d cb =0. The Redundant conditions for a single load P 1 then become ^w>f " "wZ> " "me for fixed abutments and omitting temperature effects. Neglecting the term involving the axial thrust N in Eq. (15N) the deflections in Eqs. (72E) may be expressed as virtual work in terms of the moments M om , M a = l, M b =x and M e =y cos/?, as follows: ~M om du C Mldu Cdu _ fM om M a du _ ( dma ~J~~~ET ~J rM om M b du fM om xdu C 8* =) - - Er -J -^7; fa - J El j~. r ~\/r1j,. r-.vj,. (72r) El CM om M c du CM om y cos 8du C M 2 c du A/ 2 cos 2 3du O mc =J ^J E j -I ^ = J E J~ = J gf- Introducing the elastic loads du du du v7 =w a, x~T^T =w b, and y-j^f =w c (72c) Hit til ' hi into Eqs. (72r), then substituting these values into Eqs. (72E) and replacing the inte- grations by summations, then V _m _ - om b -^ _ mc _ j omc . . ~~~ ; Ac ~~ The modulus E, being involved in both numerator and denominator, of Eqs. (72n), has no effect on the redundants except when temperature and reaction displacements are included. If the axial thrust is to be considered, then the deflections due to this effect alone ART. 72 FIXED MASONRY ARCHES 313 and for N a =Q, A\ = 1 sin (f> and A' c = 1 cos , according to Eq. (15N), become for du=dx/cos (f): /A T ~ / fl ' 2 rk A -fij- = J E p CO^A fN c 2 du _ fcos 2 fi du _ C l cos < Ax CC= J ~Ej r= J EY Jo ~EF~ (72j) These values added for total effect includin axial thrust in the determination of the redundants will change Eqs. (72n) to the following: X a =- I,M om w c sn (72K) EF cos (f> Eqs. (72n) and (72x) being written for a single load P = l represent the equations of the influence lines of the redundants. If axial thrust is to be considered in Eqs. (72K) it will be well to compute the two functions in the denominators and thus include them in the pole distances when drawing the Xb and the X c influence lines. It should be pointed out that in general, axial thrust becomes important for very flat arches, in which case the function S sin 2 Jx/EF becomes large. Hence it is never necessary to include the thrust function in determining X^, while for X c it may take on considerable proportions. Regarding the curvature of an arch, which is involved in the integral J du and which could be fully considered should the necessity arise, it must be remembered that the depth of the arch ring is. almost without exception, very small compared with the radius of curvature of the axial line, and no appreciable error is committed when this effect is entirely neglected. Some comparative analyses have been worked out by various investigators, which show quite conclusively that this error is smaller than the knowable accuracy of the strength of materials, to say nothing about temperature stresses and abutment displacements, which may be estimated but never become known. For these reasons, which are considered ample, the question of axial curvature will not be treated here. The location of the coordinate axes was to be so chosen that the displacements o, where < is the angle which the normal section makes with the vertical. The Eqs. (72c) then become Au Ax xAx , yAx after com- the values Ix'^ix* T I X , ^ T N * M i ' i - I't-M 'I i If B +\ *** i x^x^x-*.x4 *-" i - FIG. 73A. Wa El El w b =xw a = El cos and El cos > [ (73 A) Each half of the arch is now spaced off into an even number of spaces Ax, beginning at the crown. When the spaces Au appear to get too large they may be shortened toward the haunches of the arch, but it is best to keep as many as possible of the Ax alike. Now tabulate the depths D of the arch ring and angles < for the axial points from A to B and figure the reciprocal values of El cos for each. If the section is one of a steel girder the real moment of inertia is to be used while for a masonry arch it is best to take a section of width unity and depth D making I=--D 3 /12. Simpson's rule is now employed to sum the values w a in order to obtain finite r/* I x / w=== ~~~~ for the ART. 73 FIXED MASONRY ARCHES 315 The quantities 1/EIcoscf) are treated as parallel ordinates to some irregular curve and the volume of length I, which is divided into an even number n of equal spaces Ax-, is then iven from W a =-^-[w +4wi+2w 2 +4 : W3+2w 4 to 4u' n _i o (73s) One such compression can be written for any even number of equal spaces Ax, Ax', or Ax". To find W a =\w a for each of the sections, Simpson's rule for three quantities is then applied. This is the well-known prismoidal formula for volume and is used here to figure volumes between successive mean areas taking the originally computed values w for the middle terms. Since the loads W a are wanted for the points m themselves and not between these, we interpolate mean values between consecutive pairs of computed values and then figure the sums W a , using the computed W a as the middle term in the formula. Thus the formula, Ax w a3 =-^-[w 2 +4:W 3 +w 4 ], (73c) as here applied, becomes Ax \w 2 +w s , w 3 +WA~\ w ^=-^- -V--+4w 3 +- M, (730) 6 [ 2 2 J when the interpolated means are used. This gives a much better check on Eq. (73s) than could be obtained by the use of Eq. (73c). Beginning at A, and making - I .=w _i, - -=wi_ 2 , etc., the several values of W a are as follows: Ax' i'Wo = ^-\2wo +WQ _il = half load for an end section D W - 6 W a2 = - _ 2 Ax a3 = [MJ 2 _ 3 Ax etc., w ^x W a s=- Q - etc., etc. Ax" _ 8 +2lV 8 ] +[2Wg +W> 8 _9 etc. =- -Wn -12 (73E) 316 KINETIC THEORY OF ENGINEERING STRUCTURES and as a final check the sum of all W a loads must be by Eq. (73s) CHAP. XV Xi A X ' j x w a =-Hi0 +wi +w 2 ] + -x- o o 4 +4w 5 +2w 6 Ax" +4w 7 +w 8 ] + [w s +4w g +2w w +4wn o /dx C Ax w a . these are now used El cos Jo to determine the center of gravity and the angle /? which the x axis makes with the horizontal. The y axis was taken vertically and is thus fixed as soon as the origin is located. +v FIG. 73s. The center of gravity is located with respect to any assumed rectangular axes (z, v), Fig. 73B, with origin at the axial point A the same as shown in Fig. 73A. It is most convenient to assume the z axis vertical and the v axis horizontal. The coordinates (z, v) are found for each of the several axial points m previously used for the computation of the W a loads, and all are tabulated together. Finally the moments zW a and vW a are found and from these the coordinates l\ and z, 7 of the origin are obtained from / v rr a , . *-*a Z I=^TI^; and -^ (73F) This also fixes the y axis which is parallel to the vertical z axis through the center of gravity 0. The x axis, while passing through 0, makes some angle /? with the horizontal such that I,xyW a =Q, according to the last of Eqs. (72L). The (x, y) coordinates may be derived from the (z, v) coordinates when the angle ,5 is determined. Taking /? positive when measured to the right of the origin and above the horizontal, as shown in Fig. 73s, then x =li v and y =z z ' +x tan /? (73o) ART. 74 FIXED MASONRY ARCHES 317 The angle /? is found by substituting the value of y from Eq. (73c) into the condition equation %xyW a =Q, giving, as in Eq. (52j), (73H) The abscissae x being known from Eqs. (73c) the values HxzW a and 2x 2 W a are readily found and tan /? is then obtained from Eq. (73n) and the new axes and (x, y) coordinates are thus determined. For symmetric arches the above calculations are greatly simplified and the coordinate axes are readily located as follows: The y axis, being the axis of symmetry, is practically given when the shape of the arch is given, and the angle 0=0 because for symmetric forces the centrifugal moment ( xyW a =0 for any pair of rectangular gravity axes. Hence the y axis and the direction of the x axis are given and the origin is located by computing the ordinate z ' from any assumed horizontal v axis, giving for symmetric arches, SzW ' ; 0=0; and y=z-z r ....... (73j) In any case the W a , W b , and W c loads are now readily figured and from these the equations for the influence lines for the redundant conditions X a , Xb, and X c are evaulated. Solid web circular arches which are only very slightly unsymmetric may be approx- imately analyzed by extending the short half to a point of symmetry and treating the structure as a symmetric arch. However, when the difference in the elevations of the springing points is appreciable, such approximations should not be made. The center of gravity may be found graphically by combining the W a loads into two equilibrium polygons, for the loads acting first vertically and then horizontally through the centers of gravity of the arch sections. The two resultants thus obtained will intersect in the required point 0. The moments of inertia 5ix 2 W a and Hiy 2 W a may be obtained from the same polygons by Professor Mohr's inethod. ART. 74. INFLUENCE LINES FOR X a , X 6 , X c , AND M m Having located the coordinate axes (x, y) to fulfill the requirements d a b^ae ^bc = the Eqs. (72E) and (72H) now become applicable to the present problem, giving X a = ~"2W ~JT~ riar j vv & J-^a 1<\ du/EF =cos ^yW c + 2 cos Ax/EF from Eq. (72K) . The M m influen.ce line for moments, about any point m- of the axial line and a moving load P m =l, is now derived from Eq. (72D). Accordingly the ordinates jj m , of the M m influen.ce line, must represent the algebraic summation of the influences due to Mom, X a , Xb, and X c expressed by the equation y m =M m =M om -X a -x m X b -y m cos t 8X c ...... (74s) The influence line for M om is the ordinary moment influence line drawn for the point m and for a simple beam on two determinate supports at a'\ and b'i, Fig. 72A. It is a different line for each axial point m. The influence lines for X n , X b , and X c remain the same for the same structure, but the ordinates of the Xb and X c lines, according to Eq. (74s), require multiplication by the variable coordinates x and y cos {3, which differ for each axial point m. Hence the M m influence line is best constructed by computing the ordinates (X a +x m X b +y m cos 3X C ) for the- particular values of x m and y m cos/? and plotting these ordinates negatively from the M om influence line. The M m line, so found, will serve to determine the bending moment for the axial point m due to any position of a system of concentrated loads. The normal thrust N m , and stresses on any normal arch section through m, are then computed as described in Art. 49 and further discussed below by involving the moments about the kernel points e and i. ART. 75. TEMPERATURE STRESSES The effect ,of a uniform rise in temperature is to expand the principal system uniformly in all directions. But since a uniform vertical expansion is not resisted by the fixed abutments, it may be assumed that the whole temperature stress is produced by horizontal expansion. This manifests itself by a bending moment extending over the entire arch ring, and produced by resisting abutments. Thus for a rise in temperature, the extreme elements of the extrados at the crown are in tension while at the haunches they are in compression, when the arch is not otherwise loaded. The simultaneous condition of the intrados elements is exactly the reverse. ART. 75 FIXED MASONRY ARCHES 319 In Eqs. (72E) the temperature term was neglected. It is now separately considered without regard to the effect of the loads P. According to Eqs. (44c), which apply to the present problem, the redundant temperature stresses, for rigid abutments, become: X at =^=0; A" M =A'=0; and X ct = d ^ ....... (75A) Oaa Obb O cc The first two are placed equal to zero b.ecause uniform temperature changes cannot produce rotation nor -vertical deflection of the principal system, which latter is here regarded as a simple beam on two supports. The distance over which the change d ct is cumulative must be the projection of the span / on the x axis, hence d ct =etl/cos^ and from Eqs. (72F) d cc = 2?/ cos 2 (3W C , giving for the last of Eqs. (75A) : ' where w --* w '- (75B) The moment about any axial point m due to temperature alone, may now be found from Eq. (72o), wherein M ow =0, X at ^0, X bt =Q, and X^ is given from Eq. (75s). Then -M mt =y m cos ?X et =^ 2 *. . . . , .. . (75c) According t'o Eq. (75c) the only variable affecting the moment M mt is the ordinate y m . Hence for y m =0, M mt =0 and for y m maximum, M mt becomes maximum. The points where the x axis intersects the axial line are thus points of zero moment for temperature effects, while the crown and haunch points receive maximum effects. The temperature effects must finally be combined with the dead and live load stresses to obtain absolute maxima and minima. It is a well-established fact that when masonry arches open up cracks this generally occurs during the coldest winter weather. This is because masonry or concrete is scarcely ever placed during freezing temperatures, but generally in the warmest summer months. Hence, the variations in temperature from that existing at the time of closing the arch ring are nearly always greater in the negative direction. Thus, if the ring was closed at 70 F. the completed structure might at some future period attain a temperature of perhaps 90, but it is quite certain to fall to about 10 if the climate is severe. This then would subject the arch to temperature stresses resulting from +20 and 80 from the closing temperature. When it is considered that a lowering in temperature will induce tensile stresses in the spandrels and thus cause rupture at the points of least strength, it is clear why this phenomenon is of such frequent occurrence. Rising temperature would tend to increase the compressive stresses, but probably not to any alarming extent. 320 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV ART. 76. STRESSES ON ANY NORMAL ARCH SECTION This subject is fully discussed in Art. 49 and illustrated by Fig. 49s and will not be repeated here except in so far as to show the application of Eqs. (49s) to the fixed masonry arch. For any rectangular arch section the kernel points e and i are located on opposite sides of the axial line at equal distances k=D/Q from the axis. The kernel point i for the intrados is situated above the gravity axis, while the kernel point c for the extrados is below this axis as shown in Fig. 49s. The moments M e and M t - are found from Eq. (74s) in which the coordinates of the axial point m are replaced by the coordinates (x e , y e ) for an extrados kernel point and by the coordinates (xi, yj) for an intrados kernel point. This is permissible because this equation is perfectly general and is true for any point in the plane of the structure. The moment equations for the kernel points e and i, of any normal section, are thus M e =M oe -[X a +x e X b +y e c.oa t 3X e ] j ( Mi = M oi -[X a + XiX b + yi cos pX e ] I Designating all influence line ordinates by y and using proper subscripts as a distinctive feature, then the ordinates of the kernel point moment influence lines may be expressed as wherein )j , x, and y are special for each normal arch section it' while the ordinates jj a , jjfe, and rj c are the same for all influence lines but differ for each point of the span. When the kernel moments are known for any rectangular section and due to any position of a moving train of loads, the stresses on the extreme fibers of the arch section and the normal thrust N and its distance v from the axial line, may be determined from Eqs. (49s) thus: J)2 ' (76c) i=5 ; F^l-D; and /=~ where D is the depth and F is the area of a normal arch section of unit thickness. ART. 76 FIXED MASONRY ARCHES 321 In Eqs. (76c) M and Mi have opposite signs when TV acts between the two kernel points e and i. The stresses f e and /; take their signs from M e and Mi respectively, and compression is regarded as a negative stress. The offset v is positive when measured from the gravity axis toward the extrados and when so applied determines a point on the resultant polygon for that section. It is not necessary to draw the M, n influence lines because the M- e and Mi Tines cannot be derived from the M m line, nor is it possible to determine N and v when M m alone is known except by constructing a resultant polygon. Hence the stresses on any arch section are best found from the kernel point moments M e and Mi of that section. If the redundants X a , X^, and X c are evaluated from their respective influence lines for a certain position of a train of loads, then the kernel moments, for any section, may be computed from Eqs. (76A). However, it is preferable to conduct the entire solution by means of influence lines, as will be illustrated later. Since different portions of the arch ring are situated in the four quadrants of the coordinate axes, it may be well to show how to pass from the coordinates of any axial point m to those of the kernel points of the normal section through ra. For a rectangular arch section k e =ki D/Q, so that the following relations may be written out, from Fig. 76A, in terms of the lettered variable dimensions D and for each point, thus: For point A, x e = x-s' For point m, For point n j x e = x -~- sin Xi = x+-^-sm 6 x e =0 and and and and and and D sr COS For point m', x e = x -f-^- sin and x -^ sin 6 and For point B, -x e = -x + sin (j> and yi = y ~i~ ~TT cos <^> D 7/ e = y COS COS (f> D D , = y cos (p ;= 2/ + " COS^> D x ~-si and iji = y + cos 96 . (76D) 322 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV The general equations for the moment influence lines of the two kernel points may then be expressed in terms of the ordinates rj and coordinates (x, y) of the axial point. Thus D , A J D r / -Tioe |^a ^ COS 01 , (76E) where the coordinates with their proper signs are to be taken from Eqs. (76D). ART. 77. MAXIMUM STRESSES AND CRITICAL SECTIONS The maximum stress for any given section is determined from the moment influence line for that section by placing the maximum train of live loads over the positive portion of the influence line. Similarly the minimum stress in found by loading the negative portion of the moment influence line. Any point for which the TJ ordinate of a moment influence line- is zero, must be a load divide, and there is no satisfactory way of locating the load divides other than by drawing the influence lines. This will be illustrated in connection with the problem which follows. By the method of influence lines we are thus enabled to find the maximum and minimum stresses for any particular section. However, this does not afford sufficient knowledge regarding the safety of a structure unless those sections are examined which receive the greatest maximum stresses for the most unfavorable positions, of the live load considering the structure as a whole. These sections of greatest maxima are called critical sections, and for any given structure all of ART. 77 FIXED MASONRY ARCHES 323 the critical sections must be investigated while all other sections are of minor importance and usually receive no consideration. Unfortunately the theory of elasticity does not afford a direct solution for finding the critical sections. The critical sections might be located after a number of sections all along the arch ring had been examined, and the points of greatest maxima would correspond to the critical sections. However, this would involve much labor and from the nature of the case is not warranted, since the stresses in an arch ring always change gradually and not abruptly. An approximate knowledge of the location of the critical sections is, therefore, all that is required and this may be had from a discussion of the moment equation. Thus the general equation for moments about any point m by Eq. (74s) is M m =Mom -[X a +x m X b +y m cos pX c ] wherein the parenthetic quantity may be positive or negative, depending on the location of the point m. Since the stresses are direct functions of M e and M{ for a given section, then these moments and stresses attain maximum values simultaneously: 1. For X a negative and Xj, and X c both positive, when y m =0 or when x m =0. This would locate three critical sections, one at the crown for x m =0, and two on the x axis for y m =Q. 2. For X a , X b , and X c all positive, M m may become maximum when y m has its; greatest negative value or when both x m and y m ha-ve their greatest negative values. This locates two other critical points, one in each abutment. Hence, there are five critical sections in every arch which must be carefully examined for maximum and minimum stresses, and the positions of the moving loads for these limiting stresses are given from the influences lines drawn for the five critical sections. In the case of symmetric arches, only three critical sections require investigation. The two critical sections for ?/=0 fall very close to the quarter points of the arch, a fact which is important in making preliminary designs by the method of resultant polygons. The load divide may then be approximately located in the manner given for three-hinged arches by treating the crown as a hinged point. The maximum stresses are not appreciably altered by taking the quarter point critical sections a little to one side or the other of the real theoretical point, while the stresses are greatly affected by the shape of the axial line of the arch ring. Thus for maximum economy, the axial line should be a line of zero moments for the case of average loading. That is M m should be zero for every point of the axis of the ring when the arch is carrying a dead load and half the maximum live load over the whole svan. 324 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV ART. 78. RESULTANT POLYGONS The resultant polygon for any case of applied loads P may be located by finding the corresponding values of X a , X^, and X c from the three influence lines for these redundants, having previously located the (x, y) axes. The redundants X c and X b , Fig. 78A, were taken as the components of H' , respectively coincident with the x and y axes. The angle /?, which the x axis makes with the hori- zontal, is given by Eq. (73n) and the y axis was taken vertically. Hence the horizontal component H , of X c , is H=X c cos{3 and, from Fig. 72A, tan a= H' = H H X c cos /? cos a cos a X a _X a H' cos a ~ H H Co =^! ^ (tan a -tan /?) ti X a =X c z cos fi=X b z - c (78A) These dimensions fix the closing line a\b\ on the two end verticals of the abutments, also the haunch thrust H' , all in terms of X a , Xb, and X c . The reactions R\ and #2 may be found from the vertical reactions A and B and the thrust H'. The vertical reactions are those due to loads P acting on a simple beam = ^(le) P/l and B =^Pe/l. The and may be computed from Eqs. (72A) as reactions R\ and R% must then intersect on the line of the resultant R of all the applied loads P. A force polygon, Fig. 78A, is drawn by laying off all the loads P in proper succession, dividing this load line into the parts A and B and at the dividing point drawing a line parallel to aib\ of length equal to H'. This determines the pole of the force polygon from which the reactions RI and R 2 and the resultant polygon through i and bi are easily drawn. See also the force polygon in Fig. 52A, showing how the pole is located when H', A , and B are given. The resultant polygon thus found must intersect the arch center line at least three times for unsymmetric loading on an unsymmetric arch, and four times for a symmetric arch, provided the arch center line was so chosen as to be coincident with the resultant polygon drawn for the case of average loading. This average loading consists of the total dead load, plus half the live load uniformly distributed over the entire span. It ART. 78 FIXED MASONRY ARCHES 325 will be so understood whenever average loading is referred to in connection with arch designs. The most favorable resultant polygon is the one which coincides with the axial line of the arch ring. Hence the most economic shape for an arch is the one for which the axial line is chosen to be the resultant polygon drawn for the case of average loading as just defined. The live load, in its critical positions, will then produce minimum values for the moment M m =N m v by making the offsets v, between the axial line and the resultant polygon, all minimum. The absolute minimum for M m =N m v can occur FIG. 78A. omy for v=Q at all arch sections, and this condition can be produced only for one case of loading, which is taken as the average loading, when maximum economy is to be achieved in the design. For all cases of live loads other than the average, the moments M m --=N m v could never become zero for all points of the axial line, though these moments would be minimum for an arch properly designed for the average loading. 326 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV ART. 79. EXAMPLE 150 FT. CONCRETE ARCH (a) Given data and preliminary design. The data. Required to design a single track railway arch of 150 ft. clear span and 50 ft. clear rise with profile of surface and foundations as shown in Fig. 79fi. The loading to be Cooper's EGO, allowing full impact according to formula (1), Eqs. (64c), 300 -r-[l +300], where I is the portion of span covered by loads, for any case of loading. The material shall be concrete, mixed in the proportion of one part American Portland cement to two parts sand to four parts crushed granite or limestone. Cubes of 12 inches on each side, at the age of 6 months, shall sustain a compressive stress of not less than 3600 Ibs. per sq. in. and briquettes 90 days old, mixed one part cement to two parts sand, shall not break below 300 Ibs. per sq. in. in tension. Accordingly the weight of one cu. ft. of concrete is assumed at 140 Ibs., and for stresses between 100 and 600 Ibs. per sq. in. "=3,670,000 Ibs. per sq. in., and 13' FIG. 79A Pier of Roadway 3'X 13'. e =0.0000054 per 1 F. The allowable unit compressive stress will be 500 Ibs. per sq. in. or 515 cu. ft. of concrete per sq. ft. of surface. All pressures will be expressed in cu. ft. per sq. ft., a unit which is very convenient and nearly the same as Ibs. per sq. in. Thus if 7--=144 Ibs., then Ibs. per sq. in. =cu. ft. per sq. ft. Design of roadway It is .proposed to place the track on a floor built of concrete and reinforced by rolled I-beams of sufficient strength to carry the whole wheel loads over spans 10 ft. in length. This floor system is supported by concrete piers 3 ft. thick and 13 ft. wide and these in turn transmit the loads to the arch ring. This floor is shown in section Fig. 79A. Each I-beam must carry two 30,000 Ib. wheel loads spaced five feet apart and producing a maximum bending moment of M =30,000X60-30,000X30=900,000 in. Ibs. Allowing 97 per cent impact for a 10 ft. span, this moment becomes 1,733,000 in. Ibs., and for an allowable unit stress of 16,000 Ibs., the required section modulus is ART. 79 FIXED MASONRY ARCHES 327 A// 16,000 = 111 in inches. One 20-inch I, 65 Ibs., has a section modulus of 117, which is ample. Weight of floor. 2 ft. of ballast at 120 Ibs. per cu. ft. . .3850 Ibs. Concrete '. .5000 " 2 I-beams and cor. bars 150 " 2 rails 200 " Total per foot, of roadway 9200 Ibs. Total per sq. ft. of arch ring q=7lO Ibs. =5.07 cu. ft. Live loads. E60 loading for 150 ft. span, all locomotives, gives 8400 Ibs. per ft. of one track, and this, distributed over a 13 ft. wide arch, gives 640 Ibs. per sq. ft. Adding 67 per cent, impact for 150 ft. span, then the total live load p becomes 1.67x640 = 1070 Ibs. per sq. ft. -=7.64 cu. ft. concrete per sq. ft. Preliminary design by resultant polygons. The methods outlined in Art. 71 are applied here and the design is shown in Fig. 79s. Taking dimensions above given and assuming that the stress for dead load plus half uniform live load will be about 0.4 of the allowable, then Eq. (71 A) gives . ho and Eqs. (71s) give D =6.00 ft. which value is accepted for the crown thickness of the arch. Eq. (71c) gives 81.25 ft. for the radius of the intrados at the crown, and the quarter point for the parabola passing through the same three points at crown and springing, is from Eq. (7 ID) y J* = 12.5 feet and x -? =37.5 feet. 4 4 The point on the parabola is plotted at s, Fig. 79s, and the point s' is on the circle drawn with radius 81.25 ft. The point s", midway between s and s', is chosen and a three-center curve is drawn through s" and the given crown and springing points. The thickness of the arch at the point s" is found to be 6.9 ft., using Eq. (7lE). The remainder of the ring is drawn according to good judgment and the thickness at the springing was taken at 10 ft. The arch ring is now divided up into an even number of sections as described in Art. 73, so that the same subdivisions may be used throughout the final analysis. The dead loads are then computed and tabulated together with the live loads as required for the construction of the resultant polygons. All this is given in Table 79*. 328 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV m Oi t> d h-l fe ART. 79 FIXED MASONRY ARCHES 329 TABLE 79A DEAD LOADS AND EQUIVALENT UNIFORM LIVE LOADS Ordinates V from .4 ft. Length on Axis, Jw ft. D ft. DEAD LOADS IN CUBIC FEET. Live Load, P cu.ft. tf cu.ft. Q+P cu.ft. Arch Sections, sq.ft. Piers. sq.ft. Road- way. sq.ft. Total Q. cu.ft. 0- 0.0 1-5.9 2-12 A 3-18.9 4-25.4 5-31.9 6-38.4 7^4.9 8-51.4 9-57.9 10-64.4 11-71.9 12 79 4 10.00 9.30 8.60 8.00 7.60 7.20 6.90 6.64 6.46 6.30 6.20 6.10 6.00 193.44 145.40 115.47 95.82 87.33 91.62 108.00 64.80 42.00 21.00 14.80 6.00 65.91 65.91 65.91 65.91 65.91 76.05 Totals <3 1 = 367.35 #3 = 276.11 Q 5 = 223.38 Q 7 = 182.73 #, = 168.04 Q u = 173.67 99.32 99.32 99.32 99.32 99.32 114.60 ^ = 417.01 P 3 = 325.77 P 5 = 273.04 P 7 =232.39 P 9 = 217.70 P u = 230.97 P' 23 = 466.67 P' n = 375.43 P' 1B = 322.70 P' 17 = 282.05 P'is = 267.36 P' 13 = 288.27 20.80 18.10 16.00 14.40 13.84 15.02 1391.28 611.20 1696.88 2002.48 Above loads are all for an arch ring one foot thick. The resultant polygon for the symmetric loading Q+^P. given in Table 79A as loads PI to PI i, may now be drawn so as to pass through the center of the arch ring at the joint a and at the crown n. This polygon is shown in Fig. 79B, as a fine line almost coincident with the center line, and the shape of the arch ring is thus found to be acceptable. Had this coincidence not been so close then the shape of the arch would require modification. For the above case of symmetric shape and loading, the resultant polygon for the half span only is required, which is best constructed by computing the horizontal thrust instead of finding it graphically. This is done by computing the moments of the forces PI to PII about n and dividing the sum of these moments by # = 2P to obtain the lever arm x of R. The vertical component of the reaction at A is equal to R and hence H is found by taking moments about n of all the external forces to the left of n. This gives H=Q h 1696.88(79.4-45.83) 50.4 = 1130.2 cu.ft. The force polygon is thus easily constructed and the resultant polygon is then drawn through the two section centers and the crown. The resultant polygon for maximum one-sided loading, is now constructed. To do this the load divide for the quarter point m is found as indicated in Fig. 79s by drawing lines am and bn which intersect in the required point i. The full load p =7.64 cu.ft. per foot is allowed to cover the span from the right up to the point i and the left portion is acted on by dead loads only. The loads Q\ to QQ and P'n to P'zz, from Table 79A, are now united into a force polygon with assumed pole 330 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV distance E'\ and an equilibrium polygon a'cb' is drawn for the purpose of finding the reactions R\ and R'-z and the correct pole 3 . It is then required to pass a resultant polygon through the three points a, n and 6, where a, and b are middle third points and n is D/8 above the axial line. This is done as follows: project the points a, n and b down vertically on the equilibrium polygon, giving the points a', c and b' respectively. Draw a'c and b'c. Then in the force polygon make 0> 2 c' \\ a'c and O^cf' \\cb' thus establishing the points c' and c". Now draw the lines c'0 3 || an, and c"0 s \\bn giving the intersection 3 as the required pole and the line U^T [| ~ah will be the true pole distance. An equilibrium polygon drawn through the point a, using the force polygon OsMN, will then pass through the other two points n and b. The loads are divided at T into the reactions .4 = If rand B =NT&s per Eqs. (72A) &ndO s M=R'i and 3 N*R>3. By the construction given in Fig. 79s, the correct pole H' =1220 cu.ft., and the required resultant polygon anb are found. The polygon intersects the arch center line in three points and remains just inside the middle third of the arch ring at all critical sections. It is seen by inspection that a more favorable resultant polygon could not be drawn, hence the solution is considered completed provided the stresses resulting from this load- ing do not exceed the allowable limits. The stresses are now found from Eq. (69A) which, for a rectangular section of depth D and width unity, becomes B D (79B) where v is the lever arm of the normal N about the axial point of the section. TABLE 79B STRESSES FOUND FROM PRELIMINARY DESIGN Section. V ft. D ft. UNSYMMETKIC LOAD Q + P. LOADS Q + P OVER WHOLE SPAN. N cu.ft. V ft. fe cu.ft. ft cu.ft. N cu.ft. V ft. fe cu.ft. ft cu..t. a 0.0 10.00 1980 1.67 396.0 0.0 2372 237.2 237.2 m 41.3 6.77 1390 1.13 0.0 410.7 n 79.4 6.00 1220 0.75 406.6 0.0 1266 422. 422.0 - 94.4 6.20 1234 1.03 398.0 0.0 b 158.8 10.00 2290 1.67 0.0 458. 2372 237.2 237.2 It is thus seen that the stresses are all well on the safe side and the preliminary design is, therefore, acceptable. Had it been impossible to construct a resultant polygon for the unsymmetric loading, which would remain within the middle third of the arch ring, then the thickness of the ring would have to be altered accordingly. Also, if the unit stresses had exceeded the allowable limits the thickness of the arch would have to be increased. ART. 79 FIXED MASONRY ARCHES 331 The structure thus designed and found adequate to carry the required loads safely is now subjected to a rigid analysis by applying the foregoing formulae, derived from the theory of elasticity. (b) Analysis of the preliminary design by the theory of elasticity treating the structure as symmetric. The computations are all given in tabular form, thus illustrating more clearly the method of conducting such analyses and furnishing a record of the numerical work which is well suited for reference and checking. The computations for the elastic loads W, and the location of the (x, y) axes are carried out in Table 79c, following closely the method described in Art. 73. The original subdivisions of the arch ring are still retained and each space is again divided into two, making 24 sections in the whole ring between a and b. The programme carried out in the table is as follows: Compute the loads W a from Eq. (73A) then determine z ' and transform the original (v, z) axes to the (x, y) axes according to Eqs. (73j). Finally compute the Wb and W c loads by Eqs. (73A), and the pole distances H a , H b and H c from Eqs. (74A). This then furnishes all the data for con- structing the X a , X b and X c influence lines. The y axis is known to be the axis of symmetry and must, therefore, pass through the crown joint n or 12. The location of the x axis is found by computing the ordinate 2,,' of the origin from some assumed v axis which was taken 26.4 ft. below the springing points and 24, Fig. 79u. In Table 79c, the dimensions D, z and sec are tabulated from the drawing and from these the values I/I, w a and zw a are derived. The abscissa; x and increments Ax are also taken from the drawing, and the other values result from performing the operations indicated in the headings. It should be noted that the constant modulus E was not considered in computing the elastic loads W so that all these loads are E times actual. The secants are computed from R 115.8 , sec = 7^-^ = ^r^~ f or P om ts between a and ra, 2+OO.O 2 +00.;) sec & 7= '^-^ for points between m and n. 2+0.9 The sums 'w a = W a , 2zw a =zTF , etc., are computed by Simpson's rule for distance increments Ax, according to Eqs. (73 E). The ordinate z ' , of the center of gravity or origin of coordinates, is then found to be 22TF a /2PF a =64.72 ft. Also, the sum 2zzTF a =0 when extended over the whole arch, hence tan /3=0, making y=zz '. The functions xw a , x 2 w a , yw a and y 2 w a are now figured and from these the sums 2.ru? =Wb, H>x 2 w n =xWb, Hyw n = W c and ^y 2 w a =yW c are again derived by applying Simpsons's rule for the distance increments Ax. It should be noted that for the whole arch ring 2 JF& =0 and 2JF =0, serving as a check on the computations. If these sums d6 not become zero, then errors have occurred which must be rectified. Small discrepancies will always be found owing to insufficient accuracy in the data and not carrying sufficient decimals. These may be adjusted by proportionate distribution. However, this must 332 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV GO w o c PH C o b 05 *< *~ w o s tf 3 H ^ H N| ^ 02 w i EH fe O K o I I I ? o s H 10 CH O , s > "si >i 6- C5COOOr(H>-H IOOC5OOO1 I i-l i l + ooooooooooooo I I I I I I + 1 + ?O(MOC5t-HCDGOTtl'Ttl' IT-HT^T- 1 i I O5 iO CC CO LO Cl i i CO (N(N(N(MIM(MGOGOOOOOOOXC U5 t>- O i i Oi-Hi-HlMIMfNINIMCOCOCOeOIN OOOOOOOOOOOOO co co co co co co pppooooooopop ooooooooooooo -- ooppppppppppp odo'do'o'ooooo'oo I * N O II -S 55 -I M was omil o H ART. 79 FIXED MASONRY ARCHES 333 be confined to small errors not exceeding ^ per cent. Mistakes must be corrected and not distributed. Lisa's FIG. 79c. X Influence Lines for "Symmetric Arch. The influence lines for X a , X 6 , and X c were drawn in Fig. 79c, using the results from Table 79c, and neglecting the effect Scos (j>Jx/EF due to the axial thrust N on the pole distance H c of the X c influence line. If this effect is to be considered then the value * cos (f>4x/EF would be determined in the manner later indicated in Table 79M, omitting 334 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV the constant E, since it was also omitted in computing the values w a . This is permissible because E being involved as a reciprocal factor in all elastic loads, as per Eqs. (72o), may be considered as canceled in Eqs. (72ii). It must be remembered, however, that the elastic loads are then actually E times too large, a point which must be considered in computing temperature effects by Eqs. (7oB). The X influence lines are the equilibrium polygons for the W loads when certain pole distances are employed in the respective force polygons. It is generally convenient to draw these diagrams so that the influence line ordinates may appear a certain number of times actual when scaled to the scale of lengths of the drawing. The X a influence line is the equilibrium polygon for the W a loads when the pole distance is made equal to H a = HW a =6.564, as given by Table 79c. In the present case the pole was made ^TF a so that the ordinates >j a of the X a influence FOR INTRADOS KERNEL POINTS. 37/0 m n B FIG. 79o. Mo Influence Lines for Symmetric Arch. line become twice actual to the scale of lengths. The scale of the force polygon is chosen so as to give a reasonably large figure and to permit of accurate construction of the equilibrium polygon, otherwise the scale of loads need not bear any specified relation to the scale of ordinates. As a check on the drawing, the outer rays of this influence line must intersect on the y axis, in fact this is the graphic method of locating this axis. The Xb influence line is the equilibrium polygon for the Wb loads when the pole dis- tance is made equal to H b = 2xW b =2X5482.9 = 10965.8, from Table 79c. As this pole is very large in comparison with the 2^ = 107.4, it is advisable to take Hb=Q.QlHxWb = 109.66, making the ordinates r jb one hundred times actual to the scale of lengths. The scale of the Wb loads was again chosen as a matter of convenience. The closing line AB divides the influence line into a positive and a negative area. The X c influence line is the equilibrium polygon for the W c loads when the pole is made equal to # c =cos 8^yW c =2X560.9 = 1121.8. Here also, the pole is too large ART. 79 FIXED MASONRY ARCHES 335 to furnish a well shaped force polygon and hence the pole was made # C =0.02?/TF C , giving influence line ordinates >j c fifty times actual to the scale of lengths. The scale of W c loads is also chosen as a matter of convenience. The M influence lines are ordinary moment influence lines drawn for the simple span AB and the several kernel points of. the critical sections. Fig. 79o. The ordinates may be made actual to the scale of lengths or, as in the case of the M ot line, the ordinates were made five times actual. The M; and M e influence lines for the kernel points of the critical sections are drawn by plotting the coordinates computed according to Eqs. (76E). See Fig. 79E and Tables 79E and 79F. The coordinates of the kernel points i and e of the three critical sections at t, m and n of the arch ring, are computed from the coordinates (x, y) of the axial points, using Eqs. (76o) as given in Table 79D. TABLE 79D DATA RELATING TO THE CRITICAL SECTIONS AXIAL POIXT. KERNEL, POINTS. D . n Point. D * sin c COS < X V x i Vi x e Ve ft. ft. ft. ft ft. ft, ft, ft. ft. t 9.75 56 30' 78.05 -36.19 1.36 0.89 79.4 -35.3 76.7 -37.1 m 6.90 30 30' 41.8 0.0 0.62 0.99 42.4 + 0.99 41.2 - 0.99 n 6.00 00' 0.0 + 12.08 0.00 1.00 0.0 + 13.08 0.0 + 11.08 From the coordinates of the kernel points Table 79o, and the ordinates in Tables 79E and 79r, taken from the X a , X b , X c and M influence lines, t) e and TH are computed for each section of the span and for the critical sections t, m and n. Thus, for the section at m, Eqs. (76B) may be written (79c) The computations are all indicated in the headings of the tables and no further comment is necessary here. With the use of M f and M e influence lines for the three critical sections of a symmetric arch, the stresses in these sections may be found for any possible case of loading. The resultant polygon for any particular case of simultaneous loading, is located when the redundants A" a , X b and X c are known. Eqs. (78A) then give the necessary dimensions from which the resultant polygon may be drawn. Table 79o gives the computations for the resultant polygon for the case of average loading Q +%P, over the entire span. This is the case for which the resultant polygon should coincide with the axis and the results are comparable with those on Fig. 79s. 336 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV The lettered dimensions in the table are all shown in Fig. 7SA, but the resultant polygon is not drawn for this case. The influence line ordinates y a , TJ& and TJ C are taken from Fig. 79c and the sums of the products of the loads and ordinates give the respective redundants X from which z ,c and c 0} also A and B are readily found by Eqs. (78A). TABLE 79s MOMENT INFLUENCE LINES FOR EXTRADOS KERNEL POINTS. SYMMETRIC ARCH Pt. r]a ft. r/b ft. f)c ft. SECTION t J2=>?o [i}a+76.7i}6 37.l7j c ] SECTION m. T)m = ijo - [>?a+ 4 1 .2-fjb - 0.99r; c ] SECTION n. T)n=T)o [ija+ll.l'jc] 90 76.7 T) b 37.lTjc fit ? 41.27JC 0.99i? c Tim Vo H.lTJc 9 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1 2.9 0.031 0.014 2.61 2.38 - 0.52 -2.15 4.5 1.28 -0.01 0.33 2.9 0.14 -0.14 2 6.0 0.062 0.056 9.4 2.55 -0.05 0.90 6.3 0.62 -0.32 3 8.8 0.086 0.120 2.39 6.59 - 4.45 -8.55 14.2 3.54 -0.12 1.98 9.5 1.33 -0.63 4 11.4 0.102 0.200 19.1 4.20 -0.20 3.70 12.8 2.22 -0.82 5 13.9 0.110 0.296 2 17 8.44 -10.98 -9.19 24.0 4.53 -0.29 5.86 16.0 3.29 -1.19 6 16.0 0.111 0.396 29.0 4.57 -0.39 8.82 19.3 4.40 -1.10 7 17.8 0.105 0.496 1.95 8.05 -18.40 -5.50 27.5 4.33 -0.49 5.86 22.6 5.51 -0.71 8 19.4 0.095 0.588 25.9 3.91 -0.58 3.17 25.9 6.53 -0.03 9 20.7 0.079 0.672 1.73 6.06 -24.93 -0.10 24.2 3.25 -0.67 0.92 29.1 7.46 0.94 10 21.7 0.058 0.734 22.7 2.39 -0.73 -0.66 32.3 8.15 2.45 11 22.4 0.030 0.782 1.48 2.30 -29.01 + 5.69 21.0 1.24 -0.77 -1.87 36.0 8.68 4.92 12 22.6 0.000 0.800 19.2 0.00 -0.79 -2.70 40.0 8.88 8.52 13 22.4 -0.030 782 1 23 -2.30 -29.01 10.14 17.4 -1.24 -0.77 -2.99 36.0 8.68 4.92 14 21.7 -0.058 734 15.6 -2.39 -0.73 -2.98 32.3 8.15 2.45 15 20.7 -0.079 0.672 0.98 -6.06 -24.93 11.27 14.0 -3.25 -0.67 -2.78 29.1 7.46 0.94 16 19.4 -0.095 0.588 12.4 -3.91 -0.58 -2.51 25.9 6.53 -0.03 17 17.8 -0.105 0.496 0.77 -8.05 -18.40 9.42 10.8 -4.33 -0.49 -2.18 22.6 5.51 -0.71 18 16.0 -0.111 0.396 9.3 -4.57 -0.39 -1.74 19.3 4.40 -1.10 19 13.9 -0.110 0.296 0.55 -8.44 -10.98 6.07 7.7 -4.53 -0.29 -1.38 16.0 3.29 -1.19 20 11.4 -0.102 0.200 6.1 -4.20 -0.20 -0.90 12.8 2.22 -0.82 21 8.8 -0.086 0.120 0.32 -6.59 - 4.45 2.56 4.6 -3.54 -0.12 -0.54 9.5 1.33 -0.63 22 6.0 -0.062 0.056 3.1 -2.55 -0.05 -0.30 6.3 0.62 -0.32 23 2.9 -0.031 0.014 0.10 -2.38 - 0.52 0.10 1.5 -1.28 -0.01 -0.11 2.9 0.14 -0.14 24 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 All ordinates ij in above table are expressed in feet. Table 79n presents the case of unsymmetric loading shown in Fig. 79s. The results ere plotted in Fig. 79r, and the resultant polygon is finally drawn in accordance with the method given in Art. 78. The arch being symmetric /3=0 and Zi =2 hence with the values of X a , Xb and X c from Table 79n, the resulting data in that table were found. In Fig. 79r the arch ring and the (x, y] axes are given to locate the points a and b on the verticals through the end supports and then to draw the resultant polygon. ART. 79 FIXED MASONRY ARCHES 337 /\ ' I 2 \ \ \ \ AT SECTION t. 1-158'.8 8 \> 10 II U l 14 IS 16 17 18 >? 30 21 22 B LENGTHS. Ijllll.M.I I I __l___L_ 1| 10 IO 20 30 40 SOFT. OROI NATES. I// itifrados kernel points. extrados kernel points. FIG. 79E. M Influence Lines for the Kernel Points. Symmetric Arch. 338 KINETIC THEORY OF ENGINEERING STRUCTURES. CHAP. XV TABLE 79F MOMENT INFLUENCE LINES FOR INTRADOS KERNEL POINTS. SYMMETRIC ARCH Pt. Tja ft. T,b ft, 1)c ft. SECTION t. rj = ^,0 [rja + 79.4,6 35. 3, c ] SECTION m. ,m = ,o- [,3+42.4,6+0.99,0] 1 SECTION n T)n=T)o [,-} ! rl tf H g O O OQ 3 3 fe O HH H I RESULTING DATA. d II H SB +3 II ^2. W O 4* o -> . 3 ^ *- O5 ^ ic 2 4 II ^ c I- II - 1 ^ % ~ " Jfcj **? w II 06 133 * 2 a II ^ - H u O h o II 05 -1&5 -S 11 P C * co 1C M 2 II 14 ft.| CO /? + 1 * co o r^ t~- TjH Tfl Ttl * t^ 1 ^ O CO O ^ O5 t- Tfl O CD OS O CO . i 1 1C i 1 rH 1C r- 1 1 rfi TJI tc >C -<^ * C * r-- o S CO O ^ GO (N O (N rH f:^ e ^ 2g 3 ""On IN*; O 00 O * O O O l> * < D 00 O X ^ I * H| ^ 3 I> i CO 1-1 (N t^ TJH CO C^ (N t^ I-H T-H t^ (N C CO 1 CO CO -4 CO t (M (N (N (N (N (N C O >C t^ >. C^ rH ^ CO -* 1 co >^o 1C iO 1C CO cO CO 1C 1C O O >C 1C ' co co co co co o c C 1C 1C O CO CD a 1 |CI O C5 O5 ^* ^^ ^ C5 C5 t^ t^ O O5 C TjH TfH >C 1C Tt< Tjl 35 C5 * Tj( TfH 73 -S" * i ' TjH 1 Tt< t> O t^ 1^ O t^ - * rH T(H go> "3 0,3 3 i^ co eo CO t^ - CO R- * O CO r-H C^ 05 O ^H (M CO (M (N (M C^J CO C o> t^ oo oo t^ O5 c rfn CO t^ t-- CO Th C o o ** R (M rH ^ rH O o o o o o o o o o c D -f^> rH CO O O O i i >C C5 O O C5 1C C O t^ CO CO t^ O i- T-I O O O O I-H ' D CO rH H x co H O O o o o o o o o o o c: 1 1 I D 1 1 1 O 00 O5 00 t- rH TfH t>- 00 C s oo as C- (M QO CO t^ O (M iM O b- C r-4 . os i i co >c t>- c i 1 I 1 rH T 1 T- 5 rH CO H IM N 340 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV B C5 - II 14 gO 1 O ^J ti -^ co CO 2 8 co cs 00 ^H + o n . C> !> O CO Oi O^ Oi 0i O5 O^ O5 O^ Ol O5 O^ 03 iOGO^-*Tfl^-^-tCDOCOcDO5C.iCiCCOi IOOO1C i-H C^ CO CO CO CO CD 1C 1C "^ CO i-H 1C CD CO C^l 00 OO OO I s * C3 C^l 1C CD 00 ^^ cD(^-CMOOCOOOOOCDOO CO 00 CN t^ CO CO- 00 OS OO OS OS i I CO 1C f>- OS T-l CO ART. 79 FIXED MASONRY ARCHES 341 342 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV The moments M e and M, and the resulting stresses on any critical section, and for any case of loading, may be found from the moment influence lines Fig. 79E, drawn for the kernel points of the critical sections. The stresses for the two cases of loading, Q+^P and the one-sided load Q+P, are computed in Tables 79i and 79J, using the influence line ordinates from Tables 79E and 79r, or from the diagrams Fig. 79E. At the bottom of the tables the values R, N, v, i g and fi are computed for each section, using Eqs. (76c) . The sections t and t', at the abutments, are so located that the intrados kernel points i fall on the verticals through the end axial points, thus avoiding negative TJ O values in Eqs. (79c), and other difficulties affecting the X influence lines. It is seen from Table 79i, that the actual resultant polygon for the- case of average loading really coincides very closely with the arch center line as designed. The offsets v being + and show that the polygon intersects the center line four times, as it should. MOMENTS M e AND Mi FOR THE CRITICAL SECTIONS, LOADS Q+- SYMMETRIC ARCH Loads SECTION I. SECTION m. SECTION" n. Pt. >S 1e ,. M e Mi 9 ti Me Mi 1* T) Me Mi cu.ft. ft. ft. ft. ft. ft. ft. 1 417.0 -2.15 - 4.87 - 896.6 -2030.8 0.33 0.28 137.6 116.8 -0.14 '-0.18 - 58.4 75.1 3 325.8 -8.55 -11.39 -2785.6 .-3710.9 1.98 1.73 645.1 563.6 -0.63 -0.87 - 205.3 - 283.4 5 273.0 -9.19 -12.18 -2508.9 -3325.1 5.86 5.35 1599.8 1460.6 -1.19 -1.78 - 324.9 - 485.9 7 232.4 -5.50 - .63 -1278.2 -2005.6 5.86 3.86 1366.9 897.1 -0.71 -1.70 - 165.0 - 395.1 9 217.7 -0.10 - 3.25 21.8 - 707.5 0.92 -1.22 "200.3 - 265.6 4-0.94 -0.21 4- 204.6 - 45.7 11 231.0 4-5.69 + 2.82 4-1314.4 4- 651.4 -1.87 -4.14 - 432.0 - 956.3 4.92 4-3.36 1136.5 + 776.2 13 231.0 10.14 7.58 2342.3 1751.0 -2.99 -5.10 -690.7 -1178.1 4.92 4-3.36 1136.5 4- 776.2 15 217.7 11.27 9.29 2453.5 2022.4 -2.78 -4.42 - 605.2 - 962.2 4-0.94 ^0.21 + 204.6 45.7 17 232.4 9.42 8.05 2189.2 1870.8 -2.18 -3.24 - 506.6 - 753.0 -0.71 -1.70 - 165.0 - 395.1 19 273.0 6.07 5.28 1657 . 1 1441.4 -1.38 -2.03 - 376.7 - 554.2 -1.19 -1.78 - 324.9 - 485.9 21 325.8 2.56 2.27 834.0 739.6 -0.54 -0.87 - 175.9 - 283.4 -0.63 -0.87 - 205.3 - 283.4 23 417.0 0.10 0.05 41.7 20.9 -0.11 -0.20 - 45.9 83.4 -0.14 -0.18 - 58.4 75.1 R = 3393.5 D= 9'. 75 4-3341.1 -3282.4 D= 6'. 90 4-1116.7 -1998.1 Z> = 6'. 00 4-1175.0 -1018.0 N= 2037.9 N 1354.3 JV = 1096.5 v= 4-0'. 014 v= -0'.325 v = 4-0'. 071 /= - 210.9 - 207.2 f= - 140.8 - 251.8 J~ - 195.9 - 169.7 Eqs. (76c), f O'M e /e= -^ = , * v 1S positive when measured from the axis 2N D* ' * W ' toward the extrados. All loads in above table are expressed in cubic feet of masonry and the stresses are cubic feet per square foot. Cubic feet per square foot X 0.972 = pounds per square inch. The stresses in Table 79j are those due to the case of loading shown in Fig. 79r, and hence the offsets v are those which the resultant polygon in that figure actually present. The smallness of the scale, however, prevents a very close agreement. The resultant polygon for the unsymmetric loading is seen to intersect the center line three times as it should, since two values v are negative and three positive. ART. 79 FIXED MASONRY ARCHES 343 TABLE 79j MOMENTS Me AND M FOR CRITICAL SECTIONS. TOTAL DEAD AND ONE-SIDED LIVE SYMMETRIC ARCH SECTION I. SECTION m. SECTION n. Ft. Loads Q and P', i rje fji M e Mi Tje T)i M e M{ fje rji M e \ Mi cu.ft. ft. ft. ft. ft. ft. ft. 1 Qi = 367.4 - 2.15 - 4.87 - 789.9 -1789.2 0.33 0.28 121.2 102.9 -0.14 -0.18 51.4 - 66.1 3 Q 3 = 276.1 - 8.55 -11.39 -2360.7 -3144.8 1.98 1.73 546.7 477.6 -0.63 -0.87 - 173.9 - 240 . 2 5 Ob = 223.4 - 9.19 -12.18 -2053.0 -2721.0 5.86 5.35 1309 . 1 1195.2 -1.19 -1.78 - 265.8 -397.6 7 Qi= 182.7 - 5.50 - 8.63 -1004.9 -1576.7 5.86 3.86 1070.6 705.2 -0.71 -1.70 - 129.7 -310.6 9 Qo= 168.0 - 0.10 - 3.25 - 16.8 - 546.0 0.92 -1.22 154.6 - 205.0 + 0.94 -0.21 + 157.9 - 35.3 11 Pn' = 288.3 + 5.69 + 2.82 + 1640.4 + 813.0 -1.87 -4.14 -539.1 -1193.6 4.92 + 3.36 1418.4 + 968.7 13 Pis' = 288.3 10.14 7.58 2923.4 2185.3 -2.99 -5.10 -862.0 -1470.3 4.92 + 3.36 1418.4 + 968.7 15 Pis' = 267.3 11.27 9.29 3012.5 2483 . 2 -2.78 -4.42 - 743 . 1 -1181.5 + 0.94 -0.21 + 252.3 - 56.1 17 PIT' = 282.0 9.42 8.05 2656.4 2270.1 -2.18 3.24 -614.8 - 913.7 -0.71 -1.70 - 200.2 -479.4 19 P 19 ' = 322.7 6.07 5.28 1958.8 1703.8 -1.38 -2.03 -445.3 - 653.0 -1.19 -1.78 - 384.0 -574.4 21 P 2 i'= 375.4 2.56 2.27 961.0 852.2 -0.54 -0.87 -202.7 - 326.6 -0.63 -0.87 - 236.5 -326.6 23 Pi/ = 466.7 0.10 0.05 46.0 23.3 -0.11 -0.20 - 51.3 - 93.3 -0.14 -0.18 - 65.3 - 84.0 *- 3508.3 D= 9'.75 + 6973.9 + 453.2 D= 6'.90 -256.1 -3556.1 D = e'.oo + 1740.2 -632.9 N= 2006.4 N= 1434.8 N= 1186.6 v= + 1'.851 v = -1'.328 v = +0'.467 /= - 440.2 t28.5 /= + 32.3 - 448.2 != - 290.0 -105.5 Ft. Loads Q and P', cu.ft. SECTION m'. SECTION V. ft. ft. M e Mi ft. ft. M e Mi 1 Qi = 367.4 -0.11 -0.20 40.4 - 73.5 0.10 0.05 36.7 18.3 3 Os = 276.1 -0.54 -0.87 - 149.1 - 240.2 2.56 2.27 706.8 626.7 5 #6 = 223.4 -1.38 -2.03 - 308.3 - 453.5 6.07 5.28 1356.0 1179.6 7 Q 7 = 182.7 -2.18 -3.24 - 398.3 - 591.9 9.42 8.05 1721.0 1470.7 9 Q 9 = 168.0 -2.78 -4.42 - 467.0 - 742.6 11.27 9.29 1893.4 1560.7 11 Pii' = 288.3 -2.99 -5.10 - 862.0 -1470.3 10.14 7.58 2923.4 2185.3 13 Pis' -288.3 -1.87 -4.14 - 539.1 -1193.6 5.69 2.82 1640.4 813.0 15 Pis' = 267. 3 +0.92 -1.22 + 245.9 - 326.0 -0.10 - 3.25 26.7 - 868.7 17 Pn' = 282.0 5.86 + 3.86 1652.5 + 1088.5 -5.50 - 8.63 -1551.0 -2433.7 19 Pi 9 ' = 322.7 5.86 5.35 1891.0 1726.4 -9.19 -12.18 -2965.6 -3930.5 21 P 2 i' = 375.4 1.98 1.73 743.3 649.4 -8.55 -11.39 -3210.5 -4275.8 23 P 24 ' = 466.7 0.33 0.28 154.0 130.7 -2.15 - 4.87 -1003.4 -2272.8 7- 3508.3 D- 6'. 90 + 1922.5 -1496.6 Z>= 9'.75 + 1520.5 -5927.2 if- 1485 . 3 N= 2291.4 v = + OM44 v= -0'.962 /= - 242.3 - 188.7 /= - 96.0 - 374.1 The influence line ordinates ij for points TO' and tf are symmetric with those of points m and t. See note, foot of Table 79i. 344 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV By referring to Fig. 79E, it is seen that this position of loading gives maximum stresses for the section I and almost maximum for section m, but not for the other sections. In the present instance the effects at the five critical sections were found for one position of the moving load, giving a simultaneous condition of stress, but not maximum stresses at all sections. The maximum stresses for each of the critical sections must be computed for special positions of the moving train as illustrated in Table 79K, for the crown section. In the general investigation of stresses it is best to obtain the dead load moments separately for each section and combine these with the live load moments which are found without regard to impact and then increased for impact prior to adding the dead load moments, as shown in Table 79K. TABLE 79K INVESTIGATION FOR MAXIMUM STRESSES AT THE CROWN SECTION "n." SYMMETRIC ARCH Dead Total Dead Load Moments. No. l-I'L 1 Max. + Live Load Moments. Point. Loads vv neel Live from Loads, Q ije 1,1 M e Mi Head of Train. f)e 'Ji M e Mi cu.ft. ft. ft. cu.ft. ft. ft. 1 367.4 -0.14 -0.18 - 51.4 - 66.1 1 16.5 0.00 -1.19 0.0 - 19.6 3 276.1 -0.63 -6.87 -173.9 -240.2 2 33.0 1.30 + 0.10 42.9 + 3.3 5 223.4 -1.19 -1.78 -265.8 -397.6 3 33.0 2.55 1.10 84.2 36.3 7 182.7 -0.71 -1.70 -129.7 -310.6 4 33.0 4.10 2.55 135.3 84.2 9 168.0 + 0.94 -0.21 + 157.9 - .35.3 5 33.0 6.10 4.45 201.3 146.9 11 173.7 4.92 + 3.36 854.6 + 583.6 6 21.4 6.40 4.80 137.0 102.7 13 173.7 4.92 + 3.36 854.6 + 583.6 7 21.4 4.40 2.80 94.2 59.9 15 168.0 0.94 -0.21 157.9 - 35.3 8 21.4 2.55 1.00 54.6 21.4 17 182.7 -0.71 -1.70 -129.7 -310.6 9 21.4 1.30 0.00 27.8 0.0 19 223.4 -1.19 -1.78 -265.8 -397.6 10 0.0 0.00 -1.19 0.0 0.0 21 27fi 1 fi^ 087 mo 74O 9 ft A 23 fff \f A 367.4 \J . UO -0.14 . 01 -0.18 . U - 51.4 HI . , - 66.1 Max.+ L.L.= + 777.3 + 435.1 84% 652.9 + 365.5 R = 2782.6 Z> = 6'. 00 + 783.4 -932.4 N = 857.9 TotaFD.L.= 783.4 -932.4 v = -0'.087 Total max. + A/ = 2213.6 -131.8 f= -130.6 -155.4 max.A' r = 1190.8 =-=6'.0, y e = ll'.OS and i/; = 13'.08, then for the maximum moments from Table 79K M et = 10.08 X 230.9 = + 2558.4 M e from Table 79K - + 2213.6 Total M e = +4772.0 or/ e = -795.3 cu.ft., and M it = 13.08 x 230.9 = + 3020.2 Mi from Table 79K = - 131.8 Total M;= +2888.4 or /= +481.4 cu.ft. It is readily seen that these stresses are excessive and would undoubtedly cause cracks on the tension elements occurring in the intrados at the crown and in the extrados at the haunches. (c) Analysis as under (b), treating the structure as unsymmetric. It is presumed that for good and sufficient reasons the B abutment had to be founded on a lower stratum of rock, a condition which developed after the original design for the symmetric arch was accepted. The question to be answered here is whether such a change in the location of a foundation would seriously alter the resulting stresses and necessitate a change in the design. The location of the (x, y) axes and computation of the elastic loads is given in Tables 79L and 79.M, following the method outlined in Art. 73, and using the same data as in the previous case of the symmetric arch with the addition of two sections at the right-hand end, producing the unsymmetric condition illustrated in Fig. 79u. . The (z, v) axes are located through the axial points of the haunches, which is the same position previously chosen for these axes in dealing with the symmetric condition. The dimensions D, v and z are tabulated in Table 79L, and from these the values 1/7, sec (j>, w a , vw a and zw a are computed, using the formulae for sec < above given, p. 331. ART. 79 FIXED MASONRY ARCHES 347 O O O C5 co TP co K + I I I I I I I I I I I I I B tf H i + r I I I I J M I I I I J5 N 'T* * .p "5 h-*. W I! *-H iO 00 C^ CM ^O 00 C^ CM CO t^ ^2 *"i CO I^ CO CM CO 00 *O C^ ^D GO ^O 01 i~H Cw CO **"* PH *" f-H T i I-H I-H C^4 CM C^ C^ CO CO CO CM CM C^J C^l i~~i *~H I-H i^ CM M t>. .^ !>. CO OXOOCOCOCOQOOOOO' i^QOO' iLOOOOOCOT-HOOr^OOO * "5 CC - C^^CCOCN' iiOiOC v J^ ( O l OQO'^t | OOC^lT-HCM <^ fOOC^QO'^O'~'Ot^- CO O5 Ti* S r 1 P K fe^ OOOOOOOOOOOOOOOOOOOOOOOOOOO CD i "- a ^ ^ PN fa ^ o ^ ~ T ^ ^ ^ ^ ^ ~ ~ il, " ~ ~ ^-, ^. ^- .- ^ ^ ~ ^ N a ?) OOi-Hi-H'-t(N(MCOCOCOCO^Tt < TtiCOCOCOCOCMCM'-li-H' lOOOO H O g Or-HCOCOGO(M_COO^OOCOOOTt p-HOOt--^H M C5 H 2. __^ SS S S PI 1 c^^i^^^^ooot-Cicoooococnt-ooo^^LCic^^as-H T T ^ -j i"^ OOOOOOOOOOOOOOOOOOOOOOOOOOO do'dddooodoooooooooooooooooo fa CO I - "- coooooOOTc^t^^^^c^doocM^^^t^cM'osboQGOcocD'-o o H^ H f^ II *j p p p p p c5 p p p p p p p p p p p p p p p p p p p ~ ^ r-i i >~H o' o' o o o' o d d o d o o' o' o o' o o' d d o o o' d o o o o o p 2 CMCO^f^O^COCOcOt-t^-t^-t^-t t t- t^-t 5>cOCD'O l O^COCM> I p ^ t-H 1 1 ^~* OOOOOOO-^cO. OOOOOOOcOrPOOpOOOOOO ___^ co^f O I-H (N CO -^ 1C CO t^'OO PH 348 KINETIC THEORY OF ENGINEERING STRUCTURES TABLE COMPUTATION OF W FORCES AND POLE Pt. W a XZW a .V. = xz W a ^ Eq. (73c), ?/ 2 2 ' ~\~ X T ' ill 3 XW a x tan 3 2 2 ' y ft. 0.0232 49.8 153.2 155.1 446.6 -4.22 -37.17 -41.39 1.885 1 0.0251 66.6 142.6 414.5 885.0 -3.91 -28.37 -32.28 1.892 2 0.0285 85.0 135.2 551.2 877.8 -3.57 -20.27 -23.84 1.963 3 0.0325 101.4 126.4 656.0 817.5 -3.24 -13.57 -16.81 2.027 4 0.0354 110.4 110.5 716.8 718.8 -2.90 - 7.77 -10.67 1.978 5 0.0394 118.3 96.0 765.2 622.3 -2.56 - 2.77 - 5.33 1.945 6 0.0429 119.7 78.8 774.2 511.5 -2.22 + 1.53 - 0.69 1.839 7 0.0458 114.4 60.6 739.5 393 . 9 -1.89 5.13 + 3.24 1.666 8 0.0477 102.0 42.6 661.0 278.3 -1.55 8.03 6.48 1.425 9 0.0499 86.1 27.3 556.8 178.8 -1.21 10.23 9.02 1.166 10 0.0513 65.2 14.6 450.6 102.8 -0.88 11.73 10.85 0.865 11 0.0530 38.0 4.6 283.2 38.0 -0.49 12.93 12.44 0.497 12 0.0555 8.0 0.2 51.2 5.2 -0.10 13.23 13.13 0.104 13 0.0530 - 22.8 1.7 -168.6 16.3 + 0.29 12.93 13 . 22 -0.298 14 0.0513 - 50.7 8.8 -349.1 62.8 0.68 11.73 12.41 -0.674 15 0.0499 - 72.3 19.2 -467.3 126.4 1.02 10.23 11.25 -0.980 16 0.0477 - 89.2 32.6 -577.8 213.4 1.36 8.03 9.39 - 1 . 246 17 0.0458 -102.7 48.8 -663.6 317.5 1.69 5.13 6.82 -1.494 18 0.0429 -109.3 65.7 -706.6 426.6 2.03 1.53 3.56 - 1 . 679 19 0.0394 -109.3 82.0 -706.8 531.7 2.37 - 2.77 - 0.40 - 1 . 798 20 0.0354 -103.0 96.2 -668.5 625.8 2.71 - 7.77 - 5.06 -1.845 21 0.0325 - 95.3 111.7 -615.4 722.4 3.04 -13.57 -10.53 -1.906 22 0.0285 - 80.4 120.9 -521.2 784.9 3.38 -20.27 -16.89 - 1 . 856 23 0.0251 - 63.3 128.8 -388.0 782.1 3.72 -28.37 -24.65 - 1 . 798 24 0.0174 - 35.6 104.6 -214.3 621.6 4.02 -37.17 -33.15 -1.349 25 0.0129 - 16.7 89.8 - 99.6 544.1 4.33 -48.07 -44.74 -1.076 26 0.0131 0.0 104.5 - 8.2 301.0 4.64 -63.57 -58.93 -1.170 + 620.3 11953.1 All elastic loads W The sums ^w a = W a , ^vw a =vW a and Hzw a =zW a are now computed by Simpson's rule according to Eqs. (73E), using the lengths Jv for the horizontal distances between the arch sections. Eqs. (73r) then give the coordinates l\, z ' of the new origin with respect to the (v, z) axes. Thus 2vW a 545.7264 6.7149 =81.272 feet, =63.574 feet, and x=li v. ART. 79 FIXED MASONRY ARCHES 349 79M DISTANCES. UNSYMMETRTC ARCH ft. z^-r, yw a 2 yWa We y 2 * cos 2 cos Jx Point. D D 5.56 -0.960 - 2.76 1713.1 39.74 110.5 052 0.158 5.9 11.77 -0.810 - 5.04 1042.0 26.15 163 . 4 0.064 0.394 1 6.5 12.77 -0.679 - 4.43 568.3 16.20 106 8 0.077 0.500 2 6.5 13.13 -0.546 - 3.54 282 6 9.18 60.6 090 0.583 3 6.5 12.88 -0.378 - 2.47 113.8 4.03 27.4 0.101 0.657 4 6.5 12.61 -0.210 - 1.36 28.4 1.12 8.3 0.113 0.732 5 6.5 11.92 -0.030 - 0.13 0.5 0.02 1.0 0.123 0.799 6 6.5 10.80 + 0.148 + 0.95 10.5 0.48 3.7 0.135 0.872 7 6.5 9.26 0.309 1.99 42.0 2 00 13.3 0.144 0.930 8 6.5 7.56 0.450 2.90 81.4 4.06 26.3 0.153 0.987 9 6.5 5.99 0.557 3.89 117.7 6.04 42.6 0.158 1.110 10 7.5 3.71 0.659 4.91 154.8 8.20 61.0 0.163 1.225 11 7 .5 0.73 0.729 5.39 172.4 9.57 70.7 0.167 1.250 12 7.5 - 2.22 0.701 5.22 174.8 9.26 68.8 0.163 1.225 13 7.5 4.65 0.637 4.45 154.0 7.90 55.3 0.158 1.110 14 6.5 6.35 0.561 3.62 126.6 6.32 40.8 0.153 0.987 15 6.5 - 8.08 0.448 2.89 88.2 4.21 27.4 0.144 0.930 16 6.5 - 9.66 0.312 2.01 46.5 2.13 - 14.1 0.135 0.872 17 6.5 - 10.86 0.153 0.99 12.7 0.54 4.1 0.123 0.799 18 6.5 - 11.63 -0.016 - 0.08 0.2 0.01 0.8 0.113 0.732 19 6.5 - 11.98 -0.179 - 1.16 25.6 0.91 6.9 0.101 0.657 20 6.5 - 12.31 -0.342 - 2.21 110.9 3.60 24.4 0.090 0.583 21 6.5 - 12.05 -0.481 - 3.14 285.3 8.13 54.2 0.077 0.500 22 6.5 - 10.94 -0.619 - 3.75 607.6 15.25 92.6 0.064 0.393 23 5.9 8.03 -0.577 - 3.44 1098.9 19.12 114.2 0.047 0.275 24 5.9 6.53 -0.577 -3.52 2001 . 7 25.82 158.7 0.032 0.187 25 5.9 3.40 -0.772 - 2.18 3472.7 45.49 124.5 0.020 0.064 26 + 118.69 + 39.21 1482.4 19.511 -118.69 -39.21 .#:,*. = 1480.6 are E times actual. When the ordinates x are found then the quantities xzw a and x 2 w a and their sums xzW a and xWi, are determined in Table 79M, using Simpson's rule in summing for the horizontal increments Ax. Eq. (73n) then gives 620.3 tan 8 = 11953.1 = -0.0519. Eq. (73o) now furnishes the ordinates y and from these the quantities yw a and y 2 w a are computed. As a check STF&=0 and SW C =0, and any small discrepancies must be adjusted. 350 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV ART. 79 FIXED MASONRY ARCHES 351 The W loads were again derived without including E so that these loads are all E times actual, a fact which should not be overlooked if temperature stresses are to be found from Eq. (75c). In Table 79M the quantities cos (f>dx/D are given, showing that the pole distance H c =cos 52?/TF c + 2cos (f>Ax/D from Eq. (72ic), would be increased by 1.3 per cent if axial thrust is included in the determination of X c . This would tend to diminish X c , and the ordinates of the X c influence line. The influence lines for X a , X b and X c are shown in Fig. 79i, noting that the length of span /, is now taken as the horizontal distance between the intrados kernel points of the haunch sections. This was done to obviate the negative M moments otherwise encoun- tered here and leading to complications which are rather far reaching. In the previous case of symmetric arch, the real haunch sections could not be examined becuase the span was chosen as the distance between the verticals through the axial points and the sec- tion t was accordingly taken as the nearest approach to the haunch section. It is thus interesting to note the two methods of treating this matter. The length does not enter into the computations, but must be considered in drawing all the influence lines. The detailed description of the method of drawing these influence lines need riot be repeated here, and the reader is referred back to the treatment of the symmetric arch in the first part of this article. Axial thrust was not included in Fig. 79i, thus making all the results of both investiga- tions comparable. However, the effect is so small that it may be regarded as a negligible quantity in the present case. Having located the coordinate axes, this determines the critical sections, and the data required for the computation of the ordinates for the M e and Mi influence lines are given in Table 79N. TABLE 79N. DATA RELATING TO THE CRITICAL SECTIONS UNSYMMETRIC ARCH Coordinates, Coordinates Extrados Kernel Coordinates Intrados Kernel Axial Point. Points. Points. T> ' D . D , X y g sm g COS $ x e Ve cos @y e *i Vi COS &/j ft. ft. ft. ft. ft. ft. ft. ft. ft. ft. ft. A 10.0 58 51' + 81.27 -41.39 1.43 0.86 + 79 . 85 -42.25 -42.20 + 82.70 -40.53 -40.48 m 6.9 31 04 + 41.20 0.00 0.59 1.00 + 40.61 - 1.00 - 1.00 + 41.79 + 1.00 + 1.00 n 6.0 00 0.00 + 13.25 0.00 1.00 0.00 + 12.25 + 12.23 0.00 + 14.25 + 14.23 m" 7.2 34 58 -43.90 0.00 0.69 0.98 -43.21 - 0.98 - 0.98 -44.59 + 0.98 + 0.98 B 14.7 73 11 -89.33 -58.93 2.35 0.71 -86.98 -59.64 -59.56 -91.68 -58.22 -58.14 /?=2 58' and cos = 0.9987 from Table 79L The M influence lines for the kernel points of the several critical sections are shown in Fig. 79J, the data for the construction of these being taken from Table 79x. It should be noted that for the haunch sections A and B, M ot -=0. 352 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV 1=174138 FIG. 79i. X Influence Lines for Unsymmetric Arch ART. 79 FIXED MASONRY ARCHES 353 From the coordinates of the kernel points in Table 79N, and the ordinates in Tables 79o and 79p, taken from the X a , X b , X c and M influence lines, Figs. 79i and 79J, the T) e and j)i are computed for the odd points and for the five critical sections according to Eqs. (76B). The operations are all indicated in the headings of Tables 79o and 79p, thus requiring no further description here. These ordinates, plotted in Fig. 79K, furnish the M e and M{ influence lines from which the moments and stresses for any position of the live loads may be found for each of the five critical sections. In practice it is well to compute all the ordinates TJ instead of merely the odd ones as given here. FOR INTRADOS KERNEL POINTS. FORCXTRAOOS KERNEL POINTS, FIG. 79j. M Influence Lines for Unsymmetric Arch. Table 79o, gives the computations for the resultant polygon for the same case of loading previously used in Table 79n, adding one additional load P^ 5 at a point on the right abutment. The results of this computation are shown in Fig. 79n, where the resultant polygon is drawn through the points a and b which are located on the verticals through the intrados kernel points of the A and B sections. It should be noted that c and c are both negative, indicating that they must be meas- ured up from the line ah when found. Hence z is measured down on the y axis to locate the point s and c =ss' is then laid off upward from s. The line as'||z axis, locates the point a at the left abutment and the line as prolonged fixes the point 6 at the right abutment. The force polygon is therl easily constructed by laying off the reactions A and B and finding the pole 0\ by drawing the line 0\T \\ ab and equal in length to H', 354 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV TABLE 79o. MOMENT INFLUENCE LINES FOR Point. 90 ft. fib ft. rye ft. SECTION A. r> A = r> - [ij a + 79.857/6 - 42.27j c ] SECTION TO. rjm = ijo [i)a + 40.6j?& l.Oqc] r lo 79.85 t)b -42.2r,c IA T>0 + 40.67,6 T)m 1 3.5 0.036 0.012 2.74 2.87 - 0.51 - 3.12 5.2 1.46 0.2t> 3 9.8 0.082 0.100 2.52 6.55 - 4.22 - 9.61 15.2 3.33 2.17 5 15.2 0.102 0.252 2.31 8.14 -10.63 -10.40 25.1 4.14 6.01 7 19.7 0.092 0.424 2.10 7.35 -17.89 - 7.06 31.0 3.74 7.98 9 22.9 0.064 0.580 1.88 5.11 -24.48 - 1.65 28.0 2.60 3.08 11 25.0 0.014 0.692 1.64 1.12 -29.20 + 4.72 24.5 0.57 -0.38 13 25.5 -0.044 0.720 1.40 - 3.51 -30.38 9.79 21.0 -1.79 -1.99 15 24.5 -0.100 0.652 1.18 - 7.99 -27.51 12.18 17.4 -4.06 -2.39 17 22.1 -0.132 0.520 0.97 -10.54 -21.94 11.35 14.2 -5.36 -2.02 19 18.7 -0.148 0.360 0.75 -11.62 -15.19 8.86 11.1 -6.01 -1.23 21 14.3 -0.136 0.196 0.53 -10.86 - 8.27 5.36 8.0 -5.52 -0.58 23 9.1 -0.100 0.070 0.32 - 7.99 - 2.95 2.16 4.8 -4.06 -0.17 25 3.9 -0.046 0.016 0.13 - 3.67 - 0.68 0.58 2.0 -1.87 -0.01 All ordinates in above TABLE 79p. MOMENT INFLUENCE LINES FOR Point. T/a ft. >?& ft. T)c ft. SECTION A. ^ = 0-[7j a +82.77j6-40.5rjc] SECTION m. fjm = rjo [f)a + 41 .87J6 + 1.0)jf] S2.7rj b -40.57y c r >A 1)0 41.87/6 l.OlJc TJm 1 3.5 0.036 0.012 2.98 - 0.49 - 5.99 5.4 1.50 0.01 0.39 3 9.8 0.082 0.100 6.78 - 4.05 -12.53 15.4 3.43 0.10 2.07 5 15.2 0.102 0.252 8.44 -10.21 -13.43 25.2 4.26 0.25 5.49 7 19.7 0.092 0.424 7.61 -17.17 -10.14 30.0 3.85 0.42 6.03 9 22.9 0.064 0.580 5.29 -23.49 - 4.70 27.0 2.68 0.58 0.84 11 25.0 0.014 0.692 1.16 -28.03 + 1.87 23.8 0.59 0.69 -2.48 13 25.5 -0.044 0.720 - 3.64 -29.16 7.30 20.2 -1.84 0.72 -4.18 15 24.5 -0.100 0.652 - 8.27 -26.41 10.18 17.0 -4.18 0.65 -3.97 17 22.1 -0.132 0.520 -10.92 -21.06 9.88 14.0 -5.52 0.52 -3.10 19 18.7 -0.148 0.360 -12.24 -14.58 8.12 11.0 -6.18 0.36 -1.88 21 14.3 -0.136 0.196 -11.25 - 7.94 4.89 7.8 -5.68 0.20 -1.02 23 9.1 -0.100 0.070 - 8.27 - 2.84 2.01 4.8 -4.18 0.07 -0.19 25 3.9 -0.046 0.016 - 3.80 - 0.65 0.55 2.0 -1.92 0.02 0.00 All ordinates in above It is seen that the resultant polygon intersects the axial line four times, showing con- sistency with Professor Winkler's theorem. As a final check, the normal thrusts TV, offsets v and stresses /are computed in Table 79n for the same case of loading; used in Table 79Q, and the close agreement of the results shows both solutions to be satisfactory. The only stresses, however, which are maximum, are those on the section A, the others are simply for the simultaneous case of loading and have no special interest. The stresses on the haunch section A are thus found to be/ e = 549 and/ t -= +152.9 ART. 79 FIXED MASONRY ARCHES 355 EXTRADOS KERNEL POINTS UNSYMMETRIC ARCH SECTION n. , = ,- [, a +12.23,c] SECTION m'. SECTION B. Point. r/o 12.23 7j c r/n VO -43.2,6 -.098, c T)m fjo -87.0,6 -59.56, c r ,B 3.7 0.14 + 0.05 1.9 -1.56 -0.01 -0.03 0.18 - 3.13 - 0.71 0.52 1 10.6 1.22 -0.42 5.5 -3.54 -0.10 -0.66 0.54 - 7.13 - 5.96 3.83 3 17.2 3.08 -1.08 9.1 -4.41 -0.25 -1.44 0.89 - 8.88 -15.01 9.58 5 24.2 5.19 -0.69 13.0 -3.97 -0.41 -2.32 1.24 - 8.00 -25.25 14.79 7 31.1 7.10 + 1.10 16.4 -2.76 -0.57 -3.17 1.59 - 5.57 -34.54 18.80 9 38.5 8.46 5.04 20.3 -0.60 -0.68 -3.42 1.99 - 1.22 -41.22 19.43 11 40.8 8.80 6.50 24.6 + 1.90 -0.71 -2.09 2.38 + 3.83 -42.88 15.93 13 34.2 7.97 + 1.73 28.4 4.32 -0.64 + 0.22 2.76 8.70 -38.83 8.39 15 28.1 6.36 -0.36 32.0 5.70 -0.51 4.71 3.10 11.48 -30.97 0.49 17 22.0 4.40 -1.10 33.4 6.39 -0.35 8.66 3.47 12.88 -21.44 - 6.67 19 15.6 2.40 -1.10 23.7 5.88 -0.20 3.72 3.82 11.83 -11.67 -10.64 21 9.5 0.85 -0.45 14.4 4.32 -0.07 + 1.05 4.17 8.70 - 4.17 - 9.46 23 4.0 0.20 -0.10 5.8 1.99 -0.02 -0.07 4.50 4.00 - 0.95 - 2.45 25 table are expressed in feet. INTRADOS KERNEL POINTS UNSYMMETRIC ARCH SECTION n. 'Jn= >?o [)Ja+ 14.23^ c ] SECTION m'. 5}m /==) ?o -[Jjo 44.6i;6 + 0.987j c ] SECTION B. ?B = 0-[?-9l.7ij6-58.14ij c ] Point. rio 14.23ij c Tjn 90 -44.67?6 0.98)? c Tim -91.7>j6 -58.14/jc IB 3.7 0.17 + 0.03 1.8 -1.61 0.01 -0.09 -3.30 - 0.70 + 0.50 1 10.6 1.42 -0.62 5.4 -3.66 0.10 -0.84 -7.52 - 5.81 3.53 3 17.2 3.58 -1.58 8.9 -4.55 0.25 -2.00 -9.35 -14.65 8.80 5 24.2 6.03 - 1 . 53 12.4 -4.10 0.41 -3.61 -8.44 -24.65 13.39 7 31.1 8.25 -0.05 16.0 -2.85 0.57 -4.62 -5.87 -33.72 16.69 9 38.5 9.85 + 3.65 19.6 -0.62 0.68 -5.46 -1.28 -40.23 16.51 11 40.8 10.24 5.06 23.8 + 1.96 0.71 -4.37 + 4.03 -41.86 12.33 13 34.2 9.28 0.42 27.8 4.46 0.64 -1.80 9.17 -37.91 4.24 15 28.1 7.40 -1.40 31.2 5.88 0.51 + 2.71 12.10 -30.23 - 3.97 17 22.0 5.12 -1.82 33.4 6.60 0.35 7.75 13.57 -20.93 -11.34 19 15.6 2.84 -1.54 24.2 6.06 0.20 3.64 12.47 -11.40 -15.37 21 9.5 0.99 -0.59 14.6 4.46 0.07 0.97 9.17 - 4.07 -14.20 23 4.0 0.23 -0.13 6.0 2.05 0.02 0.03 4.22 - 0.93 - 7.19 25 table are expressed in feet. cu.ft., which are quite excessive as compared to those previously obtained for section t of the symmetric arch, where f e = 440 and J\ = +28.5 cu.ft. If the investigation of stresses were extended to all the critical sections it would be found that the dimensions found safe for the symmetric arch are now quite insufficient when the arch is treated as an unsymmetric structure using the same axial line as before. The shape of the ring might be so adjusted as to reduce the stresses within the required limits by computing a resultant polygon for the average loading Q+%P, and shifting the axial line by the amounts v obtained for the several critical sections. 356 KINETIC THEORY OF ENGINEERING STRUCTURES CHAP. XV A ^^rrr^r^ SECTION n. ^r^rrri' rf rl'"-T T^jNa ip n la is K i5/fcJ"'T T m 5*7 8S X ] T / T>I7 18 19 ; V intrados kernel points. \ extrados ke rnel points. \ //uml \ / '~~^ / LENGTHS. / l...iliml I I I I l| / 10 O |O ZO 3O 4O SO PT. OROINATCS. III! SFT. v' FIG. 79K. M Influence Lines for the Kernel Points. Unsymmetric Arch. ART. 79 FIXED MASONRY ARCHES 357 TABLE 79q. THE RESULTANT POLYGON FOR LOADS Q + P UNSYMMETRIC ARCH Dead Live Total r,(Q + P) for Moments about a Point. r ia 16 r, c Loads. Loads. Loads. Q P x a x b X c r r(Q+P)' ft. ft. ft. cu.ft. cu.ft. cu.ft. ft. 1 3.5 0.036 0.012 367.4 0.0 367.4 1285.9 13.23 4.40 5.9 2167.7 3 9.8 0.082 0.100 276.1 0.0 276.1 2705.8 22.64 27.61 18.9 5218.3 5 15.2 0.102 0.252 223 .4 0.0 223.4 3395.7 22.82 56.30 31.9 7126.5 7 19.7 0.092 0.424 182.7 0.0 182.7 3599.2 16.81 77.46 44.9 8203.2 9 22.9 0.064 0.580 168.0 0.0 168.0 3847.2 10.75 97.44 57.9 9727.2 11 25.0 0.014 0.692 173.7 114.6 288.3 7207 . 5 4.03 199 . 50 71.9 20728.8 13 25.5 -0.044 0.720 173.7 114.6 288.3 7344.8 -12.68 207.59 86.9 25053.3 15 24.5 -0.100 0.652 168.0 99.3 267.3 6548.9 -26.73 174.28 100.9 26970.6 17 22.1 -0.132 0.520 182.7 99.3 282.0 6232.2 -37.22 146.64 113.9 32119.8 19 18.7 -0.148 0.360 223.4 99.3 322.7 6034.5 -47.76 116.17 126.9 40950.6 21 14.3 -0.136 0.196 276.1 99.3 375.4 5368.2 -51.05 73.58 139.9 52518.5 23 9.1 -0.100 0.070 367.4 99.3 466.7 4247.0 -46.67 32.67 152.9 71358.4 25 3.5 -0.040 0.012 680.0 84.0 764.0 2674.0 -30.56 9.17 165.9 126747.6 4272.3 60490.9 -162.39 1222.81 428890.5 = # = X a =x b =X C = M Z = 174'.38, r, = 82'.7, / Then H = X c cos /?= 1221.22 cu.ft, TT H' = = 1242 cu.ft. cos a I c = -X&=-23.19ft. n M r = 7r= 100.388 ft. ti tan 13= -0.0519, cos/3 = 0.9987 V tan a = tan -77 1 = -0.1849 = = 49.53 ft. -^Xi-- 11.00 ft. H B = =2494.6 cu.ft. A = R B = 1777 7 cu.ft. The investigation for temperature is not repeated here except to evaluate the moment Eq. (75c) for this case using the same data just given for the symmetric arch. For cos,/? =0.9987 span 1 = 174.4 ft., and 2t/ W c = 1482.6 from Table 79M, then etl = -0.0752 ft. and _0.0752X3,775,720 from which the temperature stresses for any section are readily obtained by computing the moments about the two kernel points of the section. A complete investigation of the stresses in aiiy arch would then consist in finding the kernel moments due to the dead loads, the kernel moments due to the train of con- centrated live loads, Fig. 79o, making proper allowance for impact; and finally the kernel moments due to temperature effects; all for each of the five critical sections. These moments for the same section are algebraically combined into the total M e and M t - moments and from them and Eqs. (76c) the maximum stresses are found. 358 KINETIC THEORY OF ENGINEERING STRUCTURES CHAT. XV TABLE 79E COMPUTATION OF MOMENTS AND STRESSES ON THE CRITICAL SECTIONS FOR LOADS Q+P UNSYMMETRIC ARCH SECTION A. SECTION m. SECTION n. Point. Loads Q + P fie m Me Mi i?e ft Me Mi to w M e *< cu.ft. ft. ft. ft. ft. ft. ft. 1 367.4 - 3.12 - 5.99 -1146.3 -2200.7 0.25 0.39 91.9 143.3 + 05 + 0.03 + 18.4 + 11.0 3 276.1 - 9.61 -12.53 -2653.3 -3459.5 2.17 2.07 599.1 671.5 -0.42 -0.62 - 116.0 - 171.2 5 223.4 -10.40 -13.43 -2323.4 -3000.3 6.01 5.49 1342.6 1226.5 -1.08 -1.52 - 241.3 - 339.6 7 182.7 - 7.06 -10.14 -1289.9 -1852.6 7.98 6.03 1457.9 1101.7 ^0.69 -1.53 - 125.1 - 279.5 9 168.0 - 1.65 -4.70 - 277.2 - 789.6 3.08 0.84 517.4 141.1 4-1.10 -0.05 + 184.8 - 8.4 11 288.3 + 4.72 + 1.87 + 1360.8 + 539.1 -0.38 -2.48 - 109.5 - 715.0 5.04 + 3.65 1453.0 + 1052.3 13 288.3 9.79 7.30 2822 . 4 2104.6 -1.99 -4.18 - 573.7 -1205.1 6.50 5.06 1874 . 1 1458. S 15 267.3 12.18 10.18 3255.7 2721 . 1 -2.39 -3.97 - 638.8 -1051.2 -1-1.73 -1-0.42 -1- 462.4 + 112.3 17 282.0 11.35 9.88 3200 . 7 2786.2 -2.02 3.10 - 569.6 - 874.2 -0.36 -1.40 - 101.5 394.8 19 322.7 8.86 8.12 2859 . 1 2620 . 3 -1.23 -1.88 - 396.9 - 606.7 -1.10 -1.82 355.0 587.3 21 375.4 5.36 4.89 2012.1 1835.7 -0.58 -1.02 - 217.7 - 382.9 -1.10 -1.54 - 412.9 - 578.1 23 466.7 2.16 2.01 1008 . 1 938.1 -0.17 -0.19 - 79.4 - 88.7 -0.45 -0.59 - 210.0 - 275.4 25* 764.0 0.42 0.40 320.9 305.6 -0.01 0.00 7.6 - 0.0 -0.08 -0.11 - 61.1 - 84.0 Totals 4272.3 D= 10'. + 9149.7 + 2548.0 D= 6'.90 + 1415.7 -1739.7 D= 6'. 00 + 2369.8 - S3.S N= 1980.5 N= 1372.0 A" = 1226.9 v = + 2'.953 v = -OM19 v = -f 0'.932 f - 549.0 + 152.9 /= - 178.4 - 219.7 f - 395.0 - 14.0 SECTION in'. SECTION B. Point Loads. Q + P T ie 9 M e Mi 1e fc M e tfi cu ft. ft. ft. ft. ft. 1 367.4 -0.03 -0.09 11.0 - 33.0 0.52 0.50 191.0 183.7 3 276.1 -0.66 -0.84 - 182.2 - 231.9 3.83 3.53 1057 . 5 974.6 5 223.4 -1.44 -2.00 - 321.7 - 446.8 9.58 8.80 2140.2 1965.9 7 182.7 -2.32 -3.61 - 323.9 - 659.5 14.79 13.39 2702.0 2446.4 9 168.0 -3.17 -4.62 - 532.6 - 776.2 18.80 16.69 3158.4 2803.9 11 288.3 -3.42 -5.46 - 986.0 - 1574 . 1 19.43 16.51 5601.7 4759.8 13 288.3 -2.09 -4.37 - 602.5 -1259.9 15.93 12.33 4592.6 3554.7 15 267.3 + 0.22 -1.80 + 58.7 - 481.1 8:39 4.24 2242 . 6 1133.4 17 282.0 4.71 + 2.71 1328.2 + 764.2 0.49 - 3.97 138.2 -1119.5 19 322.7 8.66 7.75 2794.6 2500.9 - 6.67 -11.34 -2152.4 -3659.4 21 375.4 3.72 3.64 1396.5 1365.4 -10.64 -15.37 -3994.3 -5769.9 23 466.7 + 1.05 0.97 + 490.1 452.7 - 9.46 -14.20 -4415.0 - 6627 . 1 25i 764.0 -0.05 0.02 - 38. 2 15.3 - 1.76 - 6.80 -1344.6 -5195.2 Totals 4272.3 D = 7'. 20 + 3070.0 - 374.0 D = 14'. 7 + 9917.9 -4548.7 N= 1435.0 ;V = 2952.4 V + 940 V + 0' . 909 /= - 355.3 43.3 /- - 274.7 - 126. C Eqs. (76c, 3 -Jj { D 2 v is positive when measured from the axis toward the extrados. All loads in above table are expressed in cubic feet of masonry and the stresses are cubic feet per square foot. Cubic feet per square foot X 0.972 = pounds per square inch. BIBLIOGRAPHY THEORY OF ELASTICITY HOOKE. Philosophical tracts and collections, 1678. Hooke's law of proportionality between stress and strain. BERNOULLI, 1667-1748, was the first to give an analytic solution for the elastic curve. EULER. Methodus invemendi hneas curvas, 1741-4. COULOMB. Memoires de Pacademie des Sciences, 1784. LAGRANGE. Mecamque analytique, Paris, 1788. YOUNG. A course of lectures on natural philosophy and mechanical arts, 1807 Gives Young 3 modulus of elasticity E. NAVIER. Memoire sur la flexion des verges elastiques courbes, 1827. Law of distribution of stress on any section subjected to bending. POISSON. Memoire de 1'Academie des Sciences, VIII, 1829. CLAPEYRON. Sur 1 equilibre interieur des corps solides homogcnes. Memoires des savants etrangers, IV, 1833. LAME. Lemons sur la theorie mathematique de 1'elasticite, 1852. DE SAINT-VENANT. Memoires des savants etrangers, XIV, 1856. MENABREA. Nouveau principe sur la distribution des tensions dans les systemes elastiques Comptes rendus, 1858. CASTIGLIANO. Theorie de 1'equilibre des systemes elastiques, 1879. FRAENKEL. Das Prinzip der kleinsten Arbeit, etc., Zeitsch. des Arch.- und Ing.-Vereins, Hannover, 1882. MOHR. Ueber die Darstellung des Spannungszustandes und des Deformationszustandes eines Koerperelementes. Civilingenieur, 1882. MOHR. Ueber die Elastizitaet der Deformationsarbeit. Zeitschr. des Arch.- und Ing.-Vereins, Hannover, 1886. MOHR. Welche Umstaende bedingen die Elastizitaetsgrenze und den Bruch eines Materials. Zeitshcr. des Verems deutscher Ing., 1900. GRAPHOSTATICS PONCELET. Cours de Mecanique industrielle, 1826-9. CULMANN. Graphische Statik, 1866. CLERK MAXWELL. On reciprocal figures and diagrams of forces. Phil. Mag., 1864. WINKLER. Vortrag ueber die Berechnung von Bogenbruecken, Mitth. des Arch.- und Ing.-Vereins, Boehmen, 1868. Introduces influence lines. 360 BIBLIOGRAPHY MOHR. Beitrag zur Theorie der Holz und Eisenkonstruktionen. Zeitschr. des Arch.- und Ing. Vereins, Hannover, 1868. Introduces influence lines. CREMONA. Le figure reciproche nella Statica Grafica, 1872. MOHR. Die graphische Statik und das graphische Rechnen. Civiling., 1875. WILLIOT. Notions pratiques sur la statique graphique. Genie Civil, Oct., 1877. MUELLER-BRESLAU. Die graphische Statik der Baukonstruktionen, 1881-1905. MOHR. Ueber Geschwindigkeitsplane und Beschleunigungsplane. Civiling., 1887. W. HITTER. Anwendungen der graphischen Statik, 1890-1906. GENERAL TREATISES SCHWEDLER. Theorie der Brueckenbalkensysteme. Zeitschr. fuer Bauwesen, 1851. A. RITTER. Elementare Theorie und Berechnung eiserner Dach- und Brueckenkonstruktionen, 1863. CLERK MAXWELL. On the calculation of the equilibrium and stiffness of frames. Phil. Mag., 1864. MOHR. Beitrag zur Theorie des Fachwerks. Zeitschr. des Arch.- und Ing.-Vereins, Hannover, 1870, 1874, 1875, 1877. WEYRAUCH. Theorie und Berechnung der Kontinuierlichen und einfachen Traeger, 1873. WINKLER. Vortraege ueber Brueckenbau, 1873-1884. FRAENKEL. Anwendung der Theorie des augenblicklichen Drehpunktes auf die Bestimmung der Formaenderung von Fachwerken. Civiling., 1875. FRAENKEL. Ueber die unguenstigste Einstellung eines Systemes von Einzellasten auf Fachwerk- traeger mit Hilfe von Influenzkurven. Civiling., 1876. FOEPPL. Theorie des Fachwerks, 1880. MUELLER-BRESLAU. Graphische Statik der Baukonstruktionen. 1881-1908. SWAIN. On the application of the principle of virtual velocities to the determination of the deflec- tion and stresses of frames. Jour. Franklin Inst., 1883. MELAN. Beitrag zur Berechung statischunbestimmter Stabsysteme. Zeitschr., des Oestr. Arch. und Ing. Vereins, 1884. LAND. Die Gegenseitigkeit elastischer Formaenderungen, etc. Wochenblatt fuer Baukunde, Jan., 1887. LAND. Beitrag zur Ermittelung der Biegungslinien ebener und elastischer Stabwerke. Civiling., 1889. L. v. TETMAJER. Angewandte Elastizitaets und Festigkeitslehre. 1888, 1902. MUELLER-BRESLAU. Festigkeitslehre, 1893. LANDSBERG. Beitrag zur Theorie des raemulichen Fachwerks. Zentr. der Bauverwaltung. 1898 and 1903. MEHRTENS. Statik der Baukonstruktionen. 3 vol., 1903-5. MOHR. Abhandlungen aus dem Gebiete der technischen Mechanik. 1905. KECK-HOTOPP. Elastizitaetslehre, 1905. Handbuch der Ingenieurwissenschaften. Der Brueckenbau. SECONDARY AND ADDITIONAL STRESSES ENGESSER. Ueber die Druchbiegung von Fachwerktraegern und die hierbei auftretenden zusaetz- lichen Spannungen. Zeitschr. fuer Baukunde, 1879. ASIMONT. Hauptspannung und Sekundaerspannung. Zeitschr. fuer Baukunde, 1880. MANDERLA. Die Berechnung der Sekundaerspannungen, etc., Allgemeine Bauzeitung, 1880. MUELLER-BRESLAU. Ueber Biegungsspannungen in Fachwerken. Allgemeine Bauzeitung, 1885. BIBLIOGRAPHY 361 LANDSBERG. Ebene Fachwerksysteme mit festen Knotenpunkten, etc. Centralb. der' Bauver- waltung, 1885. LANUSBERG. Beitrag zur Theorie des Fachwerks. Zeitschr. des Arch.- und Ing.-Vereins, Han- nover, 1885 and 1886. MUELLER-BRESLAU. Zur Theorie der Biegungsspannungen in Fachwerktraegern. Zeitschr. des Arch.- und Ing.-Vereins, Hannover, 1886. WINKLER. Querkonstruktionen. Zeitschr. des Arch.- und Ing.-V., Hannover, 1886. Handbuch der Ingenieurwissenschaften, Vol. II, 1890. MOHR. Die Berechnung der Fachwerke mit starren Knotenverbindungen. Civiling., 1891 and 1892. ENGESSER. Die Zusatzkraefte und Nebenspannungen eiserner Fachwerke, 1892-3. HIROI. Statically Indeterminate Bridge Stresses, 1905. MEHRTENS. Statik der Baukonstruktionen, Vol. Ill, 1905. MOHR. Abhandlungen aus dem Gebiete der technischen Mechanik, 1906. GRIMM. Secondary Stresses in Bridge Trusses, 1908. SPECIAL TREATISES ON ARCHES LAME and CLAPEYRON. Journal des Voies de Communication, 1826. This contains the first applica- tion of force and equilibrium polygons to fixed arches. PONCELET. Memorial de 1'officier du genie, 1835. GERSTNER. Handbuch der Mechanik, 1831. Line of thrust method. MOSELEY. Phil. Mag., 1833. Discussion of most probable position of line of thrust. HAGEN. Ueber Form und Staerke gewoelbter Bogen, 1844. SCHWEDLER. Theorie der Stuetzlinie. Zeitschr. fuer Bauwesen, 1859. CULMANN. Die graphische Statik, 1866. FRAENKEL. Berechnung eiserner Bogenbruecken. Civiling., 1867, 1875. MOHR. Beitrag zur Theorie der elastischen Bogentraeger. Zeitschr. des Arch.- und Ing.-Ver.. Hannover, 1870, 1874, 1881. WINKLER. Beitrag zur Theorie der elastischen Bogentraeger. Zeitschr. des Arch.- und Ing.-Ver., Hannover, 1879. WEYRAUCH. Theorie der elastischen Bogentraeger. Zeitschr. fuer Baukunde, 1878. ENGESSER. Theorie und Berechnung der Bogenfachwerktraeger, 1880. MUELLER-BRESLAU. Theorie und Berechnung der eiserner Bogenbruecken, 1880. W. RITTER. Der elastische Bogen berechnet mit Hilfe der graphischen Statik, 1886. MUELLER-BRESLAU. Graphische Statik der Baukonstruktionen, 1892, 1907. WEYRAUCH. Die elastischen Bogentraeger, 1897. TOLKMITT. Leitfaden fuer das Entwerfen gewoelbter Bruecken, 1895. MEHRTENS. Statik der Baukonstruktionen, 3 vols., 1903-5. Handbuch der Ingenieurwissenschaften, Vol. I, 1904. INDEX Abutment displacements, fixed framed arches . 188 for general case of re- dundancy 208 in two-hinged arches, 156, 165, 168 stresses due to 32 Additional stresses due to dynamic influences . 256 Arches, bibliography of special treatises on 361 fixed framed 178, 197 fixed solid web or masonry 298 three-hinged 72,. 78 two-hinged 153, 166, 176 without hinges 178, 298 B Beams, deflection of 117, 118, 123, 125, 126 work of deformation for 41 Betti's law 28 Bibliography of treatises, etc 359 Cantilever bridges by influence lines 67 Castigliano's law, derivative of work equation . 30, 31 Centrifugal force due to curved track 260 Changes in the angles of a triangle due to stress 107 Choice of the redundant conditions 202 Clapeyron's law 2, 14 Column formulae 270 Combination of stresses as a basis for design- ing 268 Continuous girder over four supports 130 three supports. 129, 143, 148 Critical loading for max. moments, direct loading 59 Critical loading for max. moments, indirect loading 60 Critical loading for max. shear, direct loading. . 62 , indirect loading 62 D PAGE Dead-load stresses in determinate structures . . 210 Definitions of terms used throughout this work . vii Deflection influence lines 49 of a solid w r eb beam 117 polygons according to Prof. Laud . . 107 polygons according to Prof. Mohr, 100, 104 polygons for determinate structures. 99 polygons for the loaded chord 104 Deformations 11 Deformation, work due to 1, 29 Derivative of the work equation; Castigliano's law... .\ 30, 31 Designing, combination of stresses as a basis for 268 Determinate structures, dead load stresses in .. 210 displacements of points 18 displacement influence lines for 122 influence lines for. 46, 65 stresses in 210, 218 Direct and indirect loading 47, 48, 50 Displacement influence lines, determinate structures 122 for a cantilever. . 124 for a simple beam ortruss.123,125,126 Displacements, horizontal 114 of points for determinate struc- tures 18 Distortion of a statically determinate frame by graphics. 87 Distortions due to changes in the lengths of members 87 Double intersection trusses by influence lines . . 65 Dynamic impact 262 work of deformation due to. ... 43 influences ' 256 363 364 INDEX E PAGE Eccentric connections, stresses produced by. . . 255 Effects due to unusual loads 266 Effect of shop lengths on determinate structures 256 indeterminate struc- tures 208 Elastic deformations 11 Elastic loads w in terms of angle changes. 109, 111 Equation of an influence line 47 Equations, solution of simultaneous 295 Externally indeterminate, definition of 6 External redundant, influence line for one . 128, 140 F Fatigue of material 265 Features in design intended to diminish secon- dary stresses 267 Fixed framed arches 178 effect of abutment dis- placements 188 example 189 influence lines for the re- dundants 184 location of the coordinate axes 182 resultant polygon for .... 187 stress influence lines for.. 185 temperature stresses in . . 188 Fixed masonry arches 298 coordinate axes and elastic loads, 314, 331, 346 critical sections 322 determination of the redundants 309 example of symmetric. . 326 example of unsymmet- ric 346 influence lines for the redundants. 317, 333, 351 maximum stresses, 322, 344, 357 modern methods of con- struction 301 preliminary design for, 304, 326 resultant polygons, 324, 335, 357 stresses on any normal section 320, 344 temperature stresses in, 318, 345, 357 Framed structures, theorems, laws and formulae for 11 G General case of redundancy 203 Girder continuous over four supports 130 three supports. 129, 143, 148 fixed at one end, supported at the other . 140 Glossary of terms vii Graphostatics, bibliography 359 H Horizontal displacements 114 Impact due to dynamic effects ............... formulae ........................... Indeterminate, externally ................... internally .................... Indeterminate straight beams, work of defor- mation .................... structures by Maxwell's law. ... by Mohr's work equa- tion .............. effect of abutment displacements. 32, effect of shop lengths effect of temperature influence lines for, 127, influence line for one redundant ....... stress influence lines for ...... 137, 139, Influence line, equation for .................. defined ....................... for cantilever bridges .......... deflections ................ determinate structures, 46, 65, direct loading .............. displacements of determinate structures ............... double intersection trusses ... indeterminate structures. 127, indirect loading ............ moments ................ 49 one external redundant. 128, one internal redundant ..... one redundant condition. . . . shear ................... 43 262 264 6 41 24 19 208 208 206 140 127 140 47 46 67 49 218 48 122 65 140 50 , 50 140 128 127 , 50 INDEX 365 PAGE I Influence line for skew plate girder on curved double track 80 stresses in truss members ... 52 reactions 48, 50 redundants in fixed masonry arches 317, 333 three-hinged arches 72 three-hinged masonry arches 78 three-hinged solid web arches 78 trusses with subdivided panels 83 two redundant conditions. . . 130 Internal redundant, influence lines for one .... 128 Internally indeterminate 6, 7 Introduction 1 Isotropic bodies, theorems, laws and formulae for 34 K Kernel moment influence areas for three- hinged solid web arch 79 Kernel moment influence areas for two-hinged solid web arches 159 Lateral trusses, stresses due to wind, etc . . 258, 261 vibrations 260 Least work, theorem of 29 Live load stresses 218, 219 author's method 219 Load divide for a truss 51 Loading, direct and indirect . 47, 48, 50 M Masonry arches, see under Arches. Maxwell's theorem of reciprocal displacements . 27 Menabrea's law of least work 29, 31 Mitering lock gates 271 example 279 theory of the analysis .... 273 Mohr's rotation diagram 89 work equations 16, 17 Moment influence lines 49, 50 Moving load stresses 218, 219 Multiple redundancy 132, 137 simplification of influence lines for 132, 135 N Nature of secondary stresses 226 Plate girder continuous over four supports .... 130 over three supports, 129, 143, 148 deflection of 117 fixed at one end and supported at the other 140 Positions of a moving load for max. moments. . . 57 for max. shears 62 Preface v Principle of virtual velocities 2, 16 work 1, 15 R Reaction conditions 5 influence lines 48, 50- summation influence lines 54 Redundancy, simplification of influence lines for 132, 135 solution of the general case of ... 203- stress influence lines for struc- tures involving same . 137,139,140 Redundant conditions. 5 influence lines for one ... 128 influence lines for two . . . 130 on the choice of 202 Ritter's method of moments 211 Rotation diagram, Mohr's 89 of a rigid frame about a fixed point. . 89 S Secondary and additional stresses, bibliog- raphy 360 Secondary stresses 226 as effected by certain de- signs 267 concluding remarks on 267 due to riveted connections, 227, 235 due to various causes 251 in members due to eccentric connections 255 in members due to fheir own weight 251 in pin-connected structures. 253 in riveted cross-frames, load effects 244 in riveted cross-frames, wind effects 247 the nature of 226 Shear influence lines 49, 50 Shearing stress, work of deformation due to. . 38 366 INDEX Simplification of influence lines for multiple redundancy 132, 135 Skew plate girder on curved double track by influence lines 80 Solid web beam, deflection of 117 Solution of simultaneous equations 295 Special applications of influence lines to inde- terminate structures 140 Statically determinate frame, distortion of by graphics 87 structures, deflection polygons for 99 indeterminate structures 140, 194 influence lines for. 140 methods of pre- liminary designing 194 Stresses by Ritter's method of moments 211 the method of influence lines 218 stress diagrams 213 Stresses, combination of as a basis for designing 268 due to abutment displacements 32 in statically determinate structures, 210, 218 on any arch section 157, 320 Stress influence lines for indeterminate struc- tures 137, 139, 140 for truss members 52 Structural redundancy, tests for 6,7 Summation influence lines for reactions . . 54 Temperature effects on determinate structures . 256 indeterminate struc- tures 206 stresses follow Menebrea's and Castigliano's laws 31 fixed framed arches 188 fixed masonry arches, 318, 345 girder on three supports . 146 truss on three supports .. 152 two-hinged arches, 156, 165, 167 Tests for structural redundancy 7 Theorem of least work 29 Theorems relating to work of deformation .... 29 Theory of elasticity, bibliography 359 Three-hinged framed arches by influence lines . 72 solid web and masonry arches by influence lines. . 78 Tractive forces, effect of j Trusses with subdivided panels ; Truss on three supports, problem 1 Two-hinged arch with cantilever side spans. . . 170 tension member 1 framed arch, abutment displace- ments 168, 1 deflection for any point 168, 1 example 166, 1 9 stress* influence areas for 168, 1 temperature stresses in 167, 17" with column supports I.'/ X a influence line for, 166, 1 solid web 'arch, abutment displace- ments 156, ] example 153, 1 kernel moment in- fluence areas . . . . ] temperature stresses, 156, ] X a influence line for, 154, U Unusual load effects. V Virtual velocities , 2, work, principle of 1 , W Williot displacement diagrams Williot-Mohr diagrams Wind pressure 2 stresses in cross-frames in main and lateral trusses 2 Work equations for any frame 14 for isotropic solids of deformation 1 due to dynamic impact.. . due to shearing stress .... due to moments for any indeterminat i straight beam theorems relating to THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO S1.OO ON THE SEVENTH DAY OVERDUE. SEP. DEC 10 1932 MAR IB 1934 MAR 23 1S23 J940 APR 28 1947 14 r LD 21-50?re-8,'32 YD 02782 UNIVERSITY OF CALIFORNIA LIBRARY