FORCE IN TRUSS DIAGONAL CAUSED
BYA MOVING UNIFORM LOADThe bridge floor in Fig. 48« carries a moving uniformly distributed load. The portion of
the load transmitted to the given truss is 2.3 kips/lin ft (33.57 kN/m). Determine the limit-
ing values of the force induced in member Cd by this load.
Calculation Procedure:- Locate the neutral point, and compute dh
The force in Cd is a function of Vcd. Locate the neutral pointy in Fig. 48c and compute dh.
From Eq. c of the previous calculation procedure, Vcd = -jg/24 + S = 0;yg = 72 ft (21.9
m). From Eq. a of the previous procedure, dh = 60/120 = 0.5. - Determine the maximum shear
To secure the maximum value of Vcd, apply uniform load continuously in the interval jg.
Compute Vcd by multiplying the area under the influence line by the intensity of the ap-
plied load. Thus, Vcd =^1 / 2 (72)(0.5)(2.3) = 41.4 kips (184.15 kN). - Determine the maximum force in the member
Use the relation Cdmax = Vcd(csc 6), where esc 6 = [(2O^2 + 252 )/25^2 ]^0 -^5 = 1.28. Then
Ctfmax = 41.4(1.28) = 53.0-kip (235.74-kN) tension. - Determine the minimum force in the member
To secure the minimum value of Vcd, apply uniform load continuously in the interval of.
Perform the final calculation by proportion. Thus, Cdmin/Cdmax = area azy/area jhg =
-(2/3)^2 = 9. Then O/min = -<4/9)(53.0) = 23.6-kip (104.97-kN) compression.
FORCE IN TRUSS DIAGONAL CAUSED BY
MOVING CONCENTRATED LOADSThe truss in Fig. 49a supports a bridge that transmits the moving-load system shown in
Fig. 496 to its bottom chord. Determine the maximum tensile force in De.
Calculation Procedure:- Locate the resultant of the load system
The force in De (Fig. 49) is a function of the shear in panel de. This shear is calculated
without recourse to a set rule in order to show the principles involved in designing for
moving loads.
To locate the resultant of the load system, take moments with respect to load 1.
Thus, R = SQ kips (222.4 kN). Then SM 1 = 12(6) + 18(16) + 15(22) = 50*; x = 13.8 ft
(4.21 m). - Construct the influence line for Vde
In Fig. 49c, draw the influence line for Vde. Assume right-to-left locomotion, and express
the slope of each segment of the influence line. Thus slope of ik = slope of ma = 1/200;
slope of km = -7/200.