**ABC—XYZ**

320 B B I D G E S [FRAMES. Appropriate signs must be given to the arithmetical values of the forae and the alteration of length; thus thrust and compression may Fig. 78. be called negative, pull and extension positive. If every other nieiu- i L4-AL * ber of the frame were absolutely rigid, the span S of the bridge would xmdergo an alteration AS. The ratio T- will depend merely on the geometrical form of the frame ; let q be the value of this constant ratio, so that L Let /be the force produced on the member A by a horizontal force /;. acting between the springing!* ; then by the principle of virtual velocities we have / : h = AS : AL, for we may consider the structure merely as a kind, of lever, by which the force h exerts a force on A, the fulcrum being the joint opposite A ; then the above proportion expresses the fact that the forces k and /are inversely proportional to the spaces which the ends of the lever would move through when a small displacement occurs. Thus we may write f=qh. Similarly let /j be the stress which would be produced on A by a vertical force v, applied at one springing, while the other springing was held rigidly so that the whole frame could not turn ; the ratio between /, and v is constant, depending merely on the form of the frame, so that we may write where p is a constant, to be found in the same manner as q ; p may be defined as equal to the whole stress produced on the given mem ber by a unit vertical force at tiie springing, and q as the whole stress produced on the same member by a unit horizontal force at the springing, the frame being held rigidly at the opposite abutment. Then any thrust t at the springing having the vertical and hori- zontal components v and h will produce a stress F on A, equal to the sum of the stresses of/ and /j, or F qh+pv . Substituting this value of F in equation 1 we have where e = =n-, a constant for each member. Now, let the values of q, p, and c be calculated for every member of the frame ; then, calling the whole elongation 2AS = k, we have If the abutments do not yield k = 0, and for this case 2,cpqv 3 h ~= 2 v is to be found as for a girder similarly loaded, and t, the required thrust or tension, is the resultant of h and v. The calculation is best made as follows : Construct tables of the values of d and q for each member of the frame ; the method of sections or moments will answer best for the top and bottom members, and that of reciprocal figures for the diagonals ; assume a cross section for each member, based on a probable assumed value of t ; for the required load make tables of epq and c<f, or what is equivalent, when E is constant, make tables of the values of - and - . The (Jfj (M o Li sum of cq 2 or can then be made. If there be a load on one joint ct only, find the values of v and v lt for the right and left abutments, then find "Sepqv, using the value v for all members to the right of the load, and % for all members to the left of the load ; equation 3 will now give the value of h for this single load. The process of finding 2epqv must be repeated for each joint which is loaded, and the whole horizontal thrust due to the load on any number of joints will be the sum of the separate values of h. When the horizontal thrust is known, the thrusts t and ^ are obtained at the two abutments by compounding the horizontal thrust with the vertical weight borne at that abutment. When t and ^ are known, the stresses on each member are to be computed by recipro cal figures or any other convenient method. The process must be repeated for each combination of passing and permanent load, so as to find the maximum stress to which any member can be subjected. If the assumed cross sections are not suitable for these stresses, fresh cross sections must be assumed and the whole calculation repeated. The change in cross section will cause some change in the values of h, but this tentative Drocess need seldom be gone through more than once. 4-20 +49 Fig. S0. Fig. 80 shows half the frame of a bridge for which the calcula tions have been made. The span is 120 feet, the rise 12 feet, and the truss 5 feet deep at the crOwn. The load assumed is 10 tons led p F c? __?__ c- V Fig. 81. permanently on each top joint, and 10 tons of passing load. On fig. 80 are marked the stresses when the bridge is wholly covered with the passing load. On fig. 80 are marked the maximum

stresses with any distribution of load. Figs. 81 and 82 show graphi-