74. Modern Bridge Theory.—The evolution of bridge design having reached that point where necessity of accurate analysis began to make itself felt, it is necessary to recognize some of the fundamental theoretical considerations which lie at the base of modern bridge theory, and which involve to a considerable extent that branch of engineering science known as the elasticity or strength of the materials used in engineering construction.

The entire group of modern bridge structures may be divided into simple beams or girders, trusses, arches, suspension bridges, and arched ribs, each class being adapted to carry either highway or railway traffic. That class of structure known as beams or girders is characterized by very few features. There are solid beams like those of timber, with square or rectangular cross-sections, and the so-called flanged girders which are constituted of two horizontal pieces, one at the top and the other at the bottom, connected by a vertical plate running the entire length of the beam. The fundamental theory is identically the same for both and is known as the “common theory of flexure,” i.e., the theory of beams carrying loads.

If an ordinary scantling or piece of timber of square or rectangular cross-section, like a plank or a timber joist, so commonly used for floors, be supported at each end, it is a matter of common observation that it will sustain an amount of load depending upon the dimensions of the stick and length of span. When such a bar or piece is loaded certain forces or stresses, as they are called, are brought into action in its interior. The word “stress” is used simply to indicate a force that exists in the interior of any piece of material. It is a force and nothing else. It is treated and analyzed in every way precisely as a force. If the stresses or forces set up by the loading in the interior of the bar become greater than the material can resist, it begins to break, and the breaking of that portion of the timber in which the stresses or forces are greatest constitutes its failure. The load which produces this failure in a beam is called the breaking load of the beam. In engineering practice all beams are so designed or proportioned that the greatest load placed on them shall be only a safe percentage of the breaking load; the safe load usually being found between ⅓ and ⅙ of the breaking load. In most buildings the safe or working load, as it is called, is probably about ¼ of the breaking load.

75. The Stresses in Beams.—The proper design of beams or girders to carry prescribed loads is based upon the stresses which are developed or brought into action by them. It can easily be observed that if a beam supported at each end be composed of a number of thin planks or boards placed one upon the other, it will carry very little load. Each plank or board acts independently of the others and a very small load will cause a sag, as shown in Fig. 6. If there be taken, on the other hand, a beam made of a single stick of timber of the same width and depth as the number of planks shown in Fig. 6, so as to secure the solid beam shown in Fig. 7, it is a further common observation that this latter beam may carry many times the load which the laminated beam, shown in Fig. 6, sustains. The thin planks or boards readily slide over each other, so that the ends present the serrated form shown in Fig. 6. The preventing of this sliding is the sole cause of the greatly increased stiffness of the solid beam shown in Fig. 7, for there is thus developed along the imaginary horizontal sections in the solid beam of Fig. 7 what are called shearing forces or stresses; and since they exist on horizontal sections or planes running throughout the entire length of the beam, they are called horizontal shears.

At each end of the beam shown in Fig. 7 there will be an upward or supporting force exerted by the abutments on which the ends of the beam rest. Those upward or supporting forces are shown at R and R′ and are called reactions, because the abutments, so to speak, react against the ends of the beam when the latter is loaded. These reactions depend for their value on the amount and the location of the loading which the beam carries. Obviously these upward forces or reactions tend to cut or shear off the ends of the beam immediately above them, and if the loads were sufficiently large and the beam kept from bending, the reactions would actually shear off those ends, just as punches or shears in a machine-shop actually shear off the metal when the rivet-hole is punched, or when a plate is cut by shearing into two parts. The beam, however, bends or sags before shearing apart actually takes place.

76. Vertical and Horizontal Shearing Stresses.—If it be supposed that the length of the beam is divided into a great number of parts by imaginary vertical lines, like those shown in Fig. 8, then vertical shearing forces will be developed in those vertical planes and sometimes, though not often, they are enough to cause failure. It is not an uncommon thing, on the other hand, in timber to have actual shearing failure take place along a horizontal plane through the centre of the beam. Indeed this is recognized frequently as the principal method of failure in very short spans. When this horizontal shearing failure takes place, the upper and lower parts of the beam slide over each other and act precisely like the group of planks shown in Fig. 6.

If, then, the loaded beam be divided by vertical and horizontal planes into the small rectangular portions shown in Figs. 8 and 9, on each such vertical and horizontal imaginary plane there will be respectively vertical and horizontal shearing forces, which are shown by arrows in Fig. 9. It will be noticed in that figure that in each corner of the rectangle the two shearing forces act either toward or from each other; in no case do the two adjacent shearing forces act around the rectangle in the same direction. This is a condition of shearing stresses peculiar to the bent beam. It can be demonstrated by theory and is confirmed by experiment. There is a further peculiarity about these shearing forces which act in pairs either toward or from the same angle in any rectangle, and it is that the two stresses adjacent to each other have precisely the same value per square inch (or any square unit that may be used) of the surface on which they act. These stresses per square inch vary, however, either along the length of the beam or as the centre line of any normal cross-section is departed from. They are greatest along the centre line or central horizontal plane represented by AB, and they are zero at the top and bottom surfaces of the beam.

Inasmuch as the horizontal shear along the plane AʹBʹ is less than that along AB in Fig. 9, a part of the latter has been taken up by the horizontal fibres of the beam lying between the two planes. In other words, the horizontal layer of fibres at AʹBʹ is subjected to a greater stress or force along its length than at AB. The same general observation can be made in reference to any horizontal layer of fibres that is farther away from the centre than another. Hence the farther any fibre is from the centre the greater will be the stress or force to which it is subjected in the direction of its length. It results, then, that the horizontal layers of fibres which are farthest from the centre line of the beam, i.e., those at the exterior surfaces, will be subjected to the greatest force or stress, and that is precisely what exists in a loaded beam whatever the material may be.

77. Law of Variation of Stresses of Tension and Compression.—Since a horizontal beam supported at each end is deflected or bent downward when loaded, it will take a curved form like that shown in either Fig. 7 or Fig. 10; but this deflection can only take place by the shortening of the top of the beam and the lengthening of its bottom. This shows that the upper part of the beam is compressed throughout its entire length, while the lower part is stretched. In engineering language, it is stated that the upper part of the beam is thus subjected to compression and the lower part to tension. The horizontal layers or fibres receive their tension and compression from the vertical and horizontal shearing forces in the manner already explained. If the conditions of loading of the bent beam should be subjected to mathematical analysis, it would be found that throughout the originally horizontal plane AB, Fig. 7, passing through the centre of each section there would be no stress of either tension or compression, although the horizontal shearing stress there would be a maximum. Further, as this central plane is departed from the stress of tension or compression per square inch in any vertical section would be found to increase directly as the distance from it. This is a very simple law, but one of the greatest importance in the design of all beams and girders, whatever may be the form or size of cross-section. It is a law, which applies equally to the solid timber beam and to the flanged steel girder, whether that girder be rolled in the mill or built up of plates and angles or other sections in the shop. It is a fundamental law of what is called the common theory of flexure, and is the very foundation of all beam and girder design. The horizontal plane represented by the line AB in Fig. 8, along which there is neither tension nor compression, is called the “neutral plane,” and its intersection with any normal cross-section of the beam is called the “neutral axis” of that section. Mathematical analysis shows that the neutral plane passes through the centres of gravity of all the normal sections of the beam and, hence, that the neutral axis passes through the passes through the centre of gravity of the section to which it belongs.

78. Fundamental Formulæ of Theory of Beams.—The fundamental formulæ of the theory of loaded beams may be quite simply written. Fig. 10 exhibits in a much exaggerated manner a bent beam supporting any system of loads W₁, W₂, W₃, etc., while Fig. 11 shows a normal cross-section of the same beam. In Fig. 10 AB is the neutral line, and in Fig. 11 CD is the neutral axis passing through the centre of gravity, c.g., of the section.

If a is the amount of force or stress on a square inch (or other square unit), i.e., the intensity of stress, at the distance of unity from the neutral axis CD of the section, then, by the fundamental law already stated, the amount acting on another square inch at any other distance z from the neutral axis will be az. This quantity is called the “intensity of stress” (tension or compression) at the distance z from the neutral axis. Evidently it has its greatest values in the extreme fibres of the section, i.e., ad and ad₁. At the neutral axis az becomes equal to zero. FG in Fig. 11 represents the same line as FG in Fig. 10. If the line FH in Fig. 11 be laid down equal to ad and at right angles to FG, and if O represent the centre of gravity, c.g., of the section, then let the straight line LH be drawn. Any line drawn parallel to FH from FG to LH will represent the intensity of stress in the corresponding part of the beam’s cross-section. Obviously, as these lines are drawn in opposite directions from FG, those above O will indicate stress of one kind, and those below that point stress of another kind, i.e., if that above be tension, that below will be compression. It can be demonstrated by a simple process that the total tension on one side of the neutral axis is just equal to the total compression on the other side, and from that condition it follows that the neutral axis must pass through the centre of gravity or centroid of the section.

Returning to the left-hand portion of Fig. 11, let dA represent a very small portion of the cross-section; then will az. dA be the amount of stress acting on it. The moment of this stress or force about the neutral axis will be

azdA·z = az²· dA.

If this expression be applied to every small portion of the entire section, the aggregate or total sum of the small moments so found will be the moment of all the stresses in the section about the neutral axis. That moment will have the value

In equation (1) the symbol ∫ means that the sum of all the small quantities to the right of it is taken, and I stands for that sum which, in the science of mechanics, is called the moment of inertia of the cross-section about its neutral axis. The value of the quantity I may easily be computed for all forms of section. Numerical values belonging to all the usual forms employed in engineering practice are found in extended tables in the handbooks of the large iron and steel companies of the country, so that its use ordinarily involves no computations of its value.

Equation (1) may readily be changed into two other forms for convenient practical use. In Fig. 10 mn is supposed to be a very short portion of the centre line of the beam represented by dl. Before the beam is bent the section FG is supposed to have the position MN parallel to PQ. Also let u be the small amount of stretching or compression (shortening) of a unit’s length of fibre at unit’s distance from the centre line AB of the beam, then will udl and uzdl be the short lines parallel to GN in the triangle GmN shown in the figure. The point C is the centre of curvature of the line mn, and Cn = Cm is the radius. The two triangles Cnm and mNG are therefore similar, hence

If the quantity called the coefficient or modulus of elasticity be represented by E, then, by the fundamental law of the theory of elasticity in solid bodies,

a = Eu.      (3)

As has already been shown, the greatest stresses (intensities) in the section are +ad (tension) and -ad₁ (compression). If K represent that greatest intensity of stress, then

If the value of a from equation (4) be substituted in equation (1),

79. Practical Applications.—Equation (5) is a formula constantly used in engineering practice. All quantities in the second member are known in any given case. K is prescribed in the specifications, and is known as the “working resistance” in the design of beams and girders. For rolled steel beams in buildings it is frequently taken at 16,000 pounds, i.e., 16,000 pounds per square inch, about one fourth the breaking strength of the steel. In railroad-bridge work it may be found between 10,000 and 12,000 pounds, or approximately one fifth of the breaking strength of the steel. The quantities I and d depend upon the form and dimensions of the cross-section, and are either known or may be determined. The quotient I ÷ d is now known as the “section modulus,” and its numerical values for all forms of rolled beams can be found in published tables. The use of equation (5) is therefore in the highest degree convenient and practicable.

80. Deflection.—It is frequently necessary, both in the design of beams and framed bridges, to ascertain how much the given loading will cause the beam or truss to sag, or, in engineering language, to deflect below the position occupied when unloaded. The deflection is determined by the sagging in the vertical plane of the neutral line below its position when the structure carries no load. In Fig. 10 the curved line AB is the neutral line of the beam when supporting loads. If the loads should be removed, the line AB would return to a horizontal position. The line drawn horizontally through A and indicated by x is the position of the centre line of the beam before being bent. The vertical distance w below this horizontal line shows the amount by which the point at the end of the line x is dropped in consequence of the flexure of the beam. The vertical distance w is therefore called the deflection. Evidently the deflection varies with the amount of loading and with the distance from the end of the beam. The curved line AB in one special case only is a circle. The general character of that curve is determined by the loading and the length of span.

In order that the deflection may be properly considered it is necessary that the relation between x and w shall be established for all conditions of loading and length of span. If the value of u from equation (2) be placed in equation (3), there will result

If the value of a from equation (6) be substituted in the last member of equation (1), there will at once result

It is established by a very simple process in differential calculus that

Hence, substituting from equation (8) in equation (7),

Equation (9) may be used by means of some very simple operations in integral calculus to determine the value of w in terms of x and the loads on the beam when the value of the bending moment M is known, and the procedures for determining that quantity will presently be given.

Using the processes of the calculus, the two following equations will immediately be found:

As already explained, numerical values for both E and I may be taken at once from tables already prepared for all materials and for all shapes of beams ordinarily employed in structural work, so that equation (11) enables the deflection or sag of the bent beam to be computed in any case. The expression dw/dx is the tangent of the angle made by the neutral line of a bent beam with a horizontal line at any given point, and it is a quantity that it is sometimes necessary to determine. dw and dx are indefinitely short vertical and horizontal lines respectively, as shown immediately to the left of B in Fig. 10.

Equation (11) is not used in structural work nearly as much as equation (5), but both of them are of practical value and involve only simple operations in their use.

81. Bending Moments and Shears with Single Load.—The second members of equations (5) and (9) exhibit values of the moments of the internal forces or stresses in any normal cross-section of a bent beam about the neutral axis of the section, while the values of Mmust be expressed in terms of the external forces or loading. Inasmuch as the latter moment develops just the internal moment, it is obvious that the two must be equal. In order to write the value of the external moment in terms of any loading, it is probably the simplest procedure to consider a beam carrying a single load. In Fig. 12, AB is such a beam, and W is a load which may be placed anywhere in the span, whose length is l. The distances of the load from the abutments are represented by x₁ and x₂. The reactions or supporting forces exerted under the ends of the beam at the abutments are shown by R and R′. The reactions, determined by the simple law of the lever, are

The greatest bending moment in the beam will occur at the point of application of the load, and its value will be

The bending moments at the end of the beam are obviously zero, and the second and fourth members of equation (13) show that the moment increases directly as the distance from either end. Hence in the lower portion of Fig. 12, at D, immediately under the load W, the line DC is laid off at any convenient scale to represent the moment M₁. The straight lines AC and CB are then drawn. Any vertical intercept, as FH or KL, between AB and either AC or CB will represent the bending moment at the corresponding point in the beam. The simple triangular diagram ACB therefore represents the complete condition of bending of the beam under the single load W placed at any point in the span.

The beam AB is supposed for the moment to have no weight. Consequently the only force acting upon the portion of the beam AO is the reaction R, and, similarly, Rʹ is the only force acting upon the portion OB. Obviously so far as the simple action of these two forces or reactions is concerned, the tendency of each is to cause vertical slices of the beam, so to speak, to slide over each other. In other words, in engineering language, the portion AO of the beam is subjected to the shear S = R, while OB is subjected to the shear Sʹ = -Rʹ. The cross-sectional area of the beam must be sufficient to resist the shear S or Sʹ. The upper part of Fig. 13 shaded with broken vertical lines indicates this condition of shear. It is evident from this simple case that the total vertical shears at the ends of any beam will be the reactions or supporting forces exerted at those ends, and that each will remain constant for the adjoining portion of the beam.

The third member of equation (13) shows that the greatest bending moment M₁ in the beam varies as the product x₁x₂ of the segments of the span. That product will have its greatest value when x₁ = x₂. Hence a simple beam loaded by a single weight will be subjected to the greatest possible bending moment when the weight is placed at the middle of the span, at which point also that moment will be found.

82. Bending Moments and Shears with any System of Loads.—The general case of a simple beam loaded with any system of weights whatever may be represented in Fig. 13, in which the beam of Fig. 12 is supposed to carry three loads, w₁, w₂, w₃. The spacing of the loads is as shown. The reactions or supporting forces Rʹ are determined precisely as in Fig. 12, each reaction in this case being the resultant of three loads instead of one. Applying the law of the lever as before, the reaction R will have the value

A similar value may be written for Rʹ, but it is probably simpler, after having found one reaction, to write

R′ = W₁ + W₂ + W₃ - R.(15)

As the beam is supposed to have no weight, no load will act upon the beam between the given weights. The bending moments at the points of application of the three weights or loads will be

After substituting the value of R from equation (14) in equations (16) the values of the latter are at once known.

The bending produced by each weight will also be represented precisely like that in Fig. 12. The triangle ANB represents the bending produced by W₁; AOB the bending produced by W₂; and APB the bending produced by W₃. The resultant bending effect produced by the three loads or weights acting simultaneously is simply the summation of the three effects each due to a single load. Hence DC is erected vertically through the point of application of W₁, so as to equal DN added to the two vertical intercepts between AB and AP, and AB and AO. Similarly, HF is equal to HO added to the intercepts between AB and AP, and AB and BN. Finally, KL is equal to PL added to the other two intercepts, one between AB and BN, and the other between AB and BO. Straight lines then are drawn through A, C, F, K, and B. Any vertical intercept between AB and ACFKB will represent the bending moment in the beam at the corresponding point. Obviously any number of loads of any magnitude, or a uniform load, may be treated in precisely the same way.

An important practical rule can readily be deduced from the equations (16), each one of which may be regarded as a general equation of moments. If the system of three, or any other number of loads, be moved a small distance Δx, while they all remain separated by the same distances as before, the bending moment M will be changed by the amount shown in equation (16a):

ΔM = RΔx - W₁Δx - W₂Δx - etc.(16a)

If the notation of the differential calculus be used by writing the letter d instead of Δ, and if both members of equation (16a) be then divided by dx, equation (16b) will result:

The second member of this equation shows the sum of all the external forces acting on one portion of the beam, that portion being limited by the section about which the moment M acts. That sum of all the external forces, as given by the second member of equation (16b), is evidently the total transverse shear at the section considered. Equation (16b) then shows, in the language of the differential calculus, that the first derivative of M in respect to x is equal to the total transverse shear. It is further established in the differential calculus that whenever a function, such as M, the bending moment, is a maximum or a minimum, the first derivative is equal to zero. The application of this principle to equation (16b) shows that the bending moment in any beam or truss has its greatest value wherever the shear is zero. Hence, in order to determine at what section the bending moment has its greatest value in any loaded beam carrying a given system of loads, it is only necessary to sum up all the forces or loads, including the reaction R, on that beam from one end to the point where that sum or shear is zero; at this latter point the greatest moment sought will be found. This is a very simple method of determining the section at which the greatest moment in the beam exists.

The preceding formulæ and diagrams may be extended to include any number of loads, and they are constantly used in engineering practice, not only for beams and girders in buildings, but also for bridges carrying railroad trains. Whatever may be the number of loads, the expressions for the bending moments at the various points of application of those loads are to be written precisely as indicated in equations (16). When the number of loads becomes great the number of terms in the equations correspondingly increase, but in reality they are just as simple as those for a smaller number of loads.

The diagram for the vertical shear in this beam is the lower part of Fig. 13. As in the case of Fig. 12 the shear at A is the reaction R, as it is Rʹ at the other end of the beam. The shear in the portion AD of the beam has the value R, but in passing the point D to the right the weight W₁ represented by OT must be subtracted from R, so that the shear over the section b of the span is R - W₁ or QV in the diagram. Similarly, in passing the point H toward the right, both W₂ and W₁ must be subtracted from R, giving the negative shear (the previous shear being taken positive) VW. The negative shear VW remains constant throughout the distance c, but is increased by W₃ at the point L, so that throughout the distance d the shear Sʹ = -Rʹ. These shear values are all shown in the lower portion of Fig. 13 by the vertical shaded lines. Obviously it is a matter of indifference whether the shear above the straight line GJ is made positive or negative; it is only necessary to recognize that the signs are different.

In the case of heavy beams, either built or rolled, as in railroad structures, it is of the greatest importance to determine both the bending moments and the shears, as represented in the preceding equations and diagrams, and to provide sufficient metal to resist them.

The case of Fig. 13 is perfectly general for moments and shears, and the methods developed are applicable to any amount or any system of loading whatever.

83. Bending Moments and Shears with Uniform Loads.—Fig. 14 represents what is really a special case of Fig. 13, in which the loading is uniform for each unit of length of the beam throughout the whole span l. Inasmuch as the load is uniformly distributed, it is evident that the reaction at each end of the beam will be one half the total load, or

The general expression for the bending moment at any point G in the span, and located at the distance x from the end A, will take the form

This equation, giving the value of M, is the equation of a parabola with the vertex over the middle of the span. The bending moment at the latter point will be found by placing x = l/2 in equation (18), which will give

Hence, in Fig. 14, if the vertical line DC be erected at D, so as to represent the value of M in equation (19) to a convenient scale, the parabola ACB may be at once drawn. Any vertical intercept, as GF between AB and the curve AFCB, will represent by the same scale the bending moment in the beam at the point indicated by the intercept. Equation (19), giving the greatest external bending moment in a simple beam due to a uniform load, is constantly employed in structural work, and shows that that moment is equal to the total load multiplied by one eighth of the span.

It has already been shown, in connection with Fig. 12, that when a single centre weight rests on a beam the centre bending moment is equal to that weight multiplied by one fourth the span. If the total uniform load in the one case is equal to the single load in the other, these equations show that the single centre load will produce just double the bending moment due to the same load uniformly distributed over the span. Wherever it is feasible, therefore, the load should be distributed rather than concentrated at the centre of the span.

That portion of Fig. 14 shaded with vertical lines shows the shear existing in the beam. Evidently the shear at each end is equal to the reaction, or one half the total load on the span. The expression for the shear at any point, as G, distant x from A will be

If x = l/2 in equation (20), S becomes equal to zero. In other words, there is no shear at the centre of the span of a beam uniformly loaded. Hence, if at each end of the span a vertical line AK or BL be laid off downward, and if straight lines KD and DL be drawn, any vertical intercept, as GH, between these lines and AB will represent the shear at the corresponding point. Equation (20) also shows that the shear S at any point is equal to the load resting on the beam between the centre D and that point. Although this case of uniform loading is a special one it finds wide application in practical operations.

84. Greatest Shear for Uniform Moving Load.—The preceding loads have been treated as if they were occupying fixed positions on the beams considered. This is not always the case. Many of the most important problems in connection with the loading of beams and bridges arise under the supposition that the load is movable, like that of a passing railroad train. One of the simplest of these problems, although of much importance, consists in finding the location of a uniform moving load, like that of a train of cars, which will produce the greatest shear at a given point of a simple beam, such as that represented in Fig. 15, in which a moving load is supposed to pass continuously over the span from the left-hand end A. It is required to determine what position of this uniform load will produce the greatest shear at the section C.