reaction), and then constructing the stress diagram up to the member whose stress is required. In a truss with parallel chords it is only necessary to calculate the stress in the first web member for any given reaction, since the shear is constant between the left reaction and the panel in question.

The live load web stresses may all be obtained from a single diagram as follows: With an assumed left reaction of, say, 100,000 lb. construct a stress diagram on the assumption that the truss is a cantilever fixed at the right abutment, and that there are no loads on the truss. Then the maximum stress in any web member will be equal to the stress scaled from the diagram, divided by 100,000, multiplied by the left reaction that produces the maximum stress. This method is a very convenient one for finding the stresses in a truss with inclined chords. For examples, see Problem 19, and Problem 20, Chapter VII.

In calculating the maximum and minimum stresses in a bridge truss by graphic resolution the labor in constructing the stress diagram may be reduced by replacing the truss to the left of the panel by a triangle as in (a) or (b) in Fig. 6. In (a) the correct stresses will be given in UsLs' or Us'L3, but the correct stress will not be given in U3L3.

ALGEBRAIC MOMENTS.-The dead and live load stresses in a truss with inclined chords are calculated by algebraic moments in Fig. 7. The conditions for maximum loading are the

same in this truss as in a truss with parallel chords, and are as follows: Maximum chord stresses occur when all loads are on; minimum chord stresses occur when no live load is on; maximum web stresses in main members occur when the longer segment of the truss is loaded; and minimum stresses in main members and maximum stresses in counters occur when the shorter segment of the truss is loaded. For a proof of this criterion, see Fig. 6, Chapter IV. An apparent exception to the latter rule occurs in post U2L2, which has a maximum tensile stress when the truss is fully loaded with dead and live loads.

To calculate the stress in member UL2, take moments about point A, the intersection of the upper and lower chords produced and pass a section cutting U,U2, UL2 and L1L2, and cutting away the truss to the right. The dead load stress is then given by the equation

The maximum live load stress occurs when all loads are on except L1, and

The maximum live load stress in counter UL1 occurs with a load at L1, and is given by the equation

The dead load stress in counter U1⁄2L1 when main member U1L2 is not acting will be

9.84 tons, and the minimum
=
4.10 2.18 tons, and the

stress is zero. The maximum stress in counter U2L1 is + 1.92

minimum stress is zero.

To calculate the stress in member U1U2, take the center of moments at L2, and pass a section cutting U1U2, U¿L2 and L¿L2', and cutting away the truss to the right. The dead load stress is

The stresses in the remaining members may be found in the same manner.

To obtain stress

in upper chord U2U2, take moments about L2 as a center; to obtain stress in lower chord LoLi take moments about U1 as a center. The dead load and maximum live load tensile stress in post U¿L is equal to the vertical component of the dead and live loads, respectively, in upper chord U1U2. The stresses in LoU1, LcL1, L2L2', U¿U1⁄2′ and U1⁄2L1⁄2' are most easily found by algebraic resolution.

For additional problems, see Chapter VII.

GRAPHIC MOMENTS.-The dead load stresses in the chords of a Warren truss are calculated by graphic moments in Fig. 8.

Bending Moment Polygon.-The upper chord stresses are given by the ordinates to the bending moment parabola direct, while the lower chord stresses are arithmetical means of the upper chord stresses on each side, and are given by the ordinates to the chords of the parabola as shown in Fig. 8.

The parabola is constructed as follows: The mid-ordinate, 4-j, is made equal to the bending moment at the center of the truss divided by the depth; in this case the mid-ordinate is the stress in 6-x; if the number of panels in the truss were odd, the mid-ordinate would not be equal to any chord stress. The parabola is then constructed as shown in Fig. 8. The live load chord stresses may be found from Fig. 8 by changing the scale, or by multiplying the dead load chord stresses by a constant.

Shear Polygon.-In Chapter II it was shown that the maximum shear in a beam at any point could be represented by the ordinate to a parabola at the point. The same principle holds for a symmetrical bridge truss with equal panels and loaded with equal joint loads, as will now be proved.

In Fig. 9 assume that the simple Warren truss is fixed at the left end as shown, and that the right reaction R2 is not acting. Then with all joints fully loaded with a live load P, construct a

force polygon as shown, with pole O and pole distance, H = span L, and beginning at point 4 in the load line of the force polygon, construct the equilibrium polygon a-g-h for the cantilever truss.

Now the bending moment at the left support will be equal to ordinate y, multiplied by the pole distance H. But the truss is a simple truss and the moment of the right reaction will be equal to the moment at the left abutment, and

=3

=

Now, with the loads remaining stationary, move the truss one panel to the right as shown by the dotted truss. With the same force polygon draw a new equilibrium polygon as above. This equilibrium polygon will be identical with a part of the first equilibrium polygon as shown. As above, the bending moment at left reaction is yз H y3 LR3 L, and yз R3. In like manner ys can be shown to be the right reaction with three loads on, etc. Since the bridge is symmetrical with reference to the center line, the ordinates to the shear polygon in Fig. 9 are equal to the maximum shears in the panel to the right of the ordinate as the load moves off the bridge to the right.

For a method of drawing the shear parabola direct, without the use of the force and equilibrium polygons, see Problem 18, Chapter VII.

CHAPTER IV.

STRESSES IN RAILWAY BRIDGE TRUSSES.

LOADS.-The dead load of a railway bridge is assumed to act at the joints the same as in a highway bridge. The dead joint loads are commonly assumed to act on the loaded chord, but may be assumed as divided between the panel points of the two chords, one-third and two-thirds of the dead loads usually being assumed as acting at the panel points of the unloaded and the loaded chords, respectively.

The live load on a railway bridge consists of wheel loads, the weights and spacing of the wheels depending upon the type of the rolling stock used. The locomotives and cars differ so much that it would be difficult if not impossible to design bridges on a railway system for the actual conditions, and conventional systems of loading, which approximate the actual conditions are assumed. The conventional systems for calculating the live load stresses in railway bridges that have been most favorably received are: (1) Cooper's Conventional System of Wheel Concentrations; (2) the use of an Equivalent Uniform Load; and (3) the use of a uniform load and one or two wheel concentrations. In addition to these some railroads specify special engine loadings. The first and second methods will be discussed in this chapter.

Cooper's Conventional System of Wheel Concentrations.-In Cooper's loadings two consolidation locomotives are followed by a uniformly distributed train load. The typical loading for Cooper's Class E 40 is shown in Fig. 3, Chapter IX. The loads on the drivers in thousands of pounds and the uniform train load in hundreds of pounds are the same as the class number. The wheel spacings are the same for all classes. The stresses for Cooper's loadings calculated for one class may be used to obtain the stresses due to any other class loading. For example, the stresses in any truss due to Cooper's Class E 50 are equal to of the stresses in the same truss due to Class E 40 loading. The E 55 and the E 60 loadings are those most used for steam railways in the United States. In bridges designed for Class E 40 loading and under the floor system must in addition be designed for two moving loads of 50,000 lb. each, spaced 6′ o" apart on each track. The corresponding loads for Class E 50 are 60,000 lb. with the same spacing. The American Railway Engineering Association has adopted Cooper's loadings, except that the special loads are spaced 7' o". The values for moment, M, shear, S, and floorbeam reaction, R, for Class E 60 are given in Table I.

Equivalent Uniform Load System.-The equivalent uniform load for calculating the stresses in trusses and the bending moments in beams, is the uniform load that will produce the same bending moment at the quarter points of the truss or beam as the maximum bending moment produced by the wheel concentrations. The equivalent uniform loadings for different spans for Cooper's E 40 loading are given in Fig. 4, Chapter IX. In calculating the stresses in the truss members select the equivalent load for the given span, and calculate the chord and web stresses by the use of equal joint loads, as for highway bridges. In designing the stringers for bending moment take a loading for a span equal to one panel length, and for the maximum floorbeam reaction take a loading for a span equal to two panel lengths. It is necessary to calculate the maximum end shears and the shears at intermediate points by wheel concentrations, or to use equivalent uniform loads calculated for wheel concentrations.

Live load stresses calculated by the method of equivalent uniform loads are too small for the chords and webs between the ends of the truss and the quarter points, and are too large between the quarter points. The stresses obtained for the counters are too large. The live load

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