Essentials in the theory of framed structures . nate, cannot be accurately computed until the truss has beendesigned. In order that a preliminary design may be made;the reactions will be tentatively determined by assuming thatthe truss functions as a beam of uniform cross-section, con-tinuous over three level supports. Only the end reactions Roare necessary. They are given in Table II as determined fromEqs. (9) and (11) page 257. By this process it is possible tocompute the stresses and make a preliminary design, after whichthe true reactions may be determined. Talbe II r lb. at i?o 1,000 lb.
Essentials in the theory of framed structures . nate, cannot be accurately computed until the truss has beendesigned. In order that a preliminary design may be made;the reactions will be tentatively determined by assuming thatthe truss functions as a beam of uniform cross-section, con-tinuous over three level supports. Only the end reactions Roare necessary. They are given in Table II as determined fromEqs. (9) and (11) page 257. By this process it is possible tocompute the stresses and make a preliminary design, after whichthe true reactions may be determined. Talbe II r lb. at i?o 1,000 lb. at ^0 Xi +0-793 L-, — U +OS93 U — L, U U L, U + ill — 188. Positive Shear in Panel o-i. Case III.—The influenceline abc for shear in the panel is shown in Fig. 197, from whichthe following criterion for maximum positive shear may bedeveloped. Sec. I SWING BRIDGES 307 If Pi = the load in panel o-i and P = the total load on the arm 0-6 the criterion is Ui Ug Ua U4 Us U6 Ut Ur U9 Uio Un. CaseH FIG. 198 A When the train is moving to the left, wheel 4 passing Zi satisfiesthis criterion and the maximum shear in the panel is +162,000 308 THEORY OF FRAMED STRUCTURES Chap. VIII lb. The area abc is and the equivalent uniform load perlinear foot is 162,0001 = . = 2, Case IV.—The influence line for positive shear is defghij (Fig. 197), from which the following criterion may be developed. If Pi = the load in panel o-i Pi = the load in panel 1-2 P3 = the load in panel 2-3, etc. the criterion is 793^1 $ 200P2 + 187P3 + 165^4 + 138^6 + 103P6Try wheel 4 at Li, train moving to the left. Wheel 4 approaching Li793P1 = 793 X 50 = 39,650 200P2 = 200 X 66 = 13,200 187P3 = 187 X 56 = 10,472165P4 = 165 X 73 = 12,045138P6 = 138 X 57 = 7,866103P6 = 103 X 50 = 5,150 48,73339,650 <48,733 therefore the shear is increasing. Wheel 4 having passed Li793-Pi = 793 X 70 = 55,510 200P2 = 200 X 59 = 11,800 187P3 = 187 X 43 = 8,041165P4 =
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