Question 20.P.11: The truss shown in Fig. P.20.11 carries a train of loads con...
The truss shown in Fig. P.20.11 carries a train of loads consisting of, left to right, 40, 70, 70 and 60 kN spaced at 2, 3 and 3 m, respectively. If the self-weight of the truss is 15 kN/m, calculate the maximum force in each of the members CG, HD and FE.
From the method of sections F_{\mathrm{CG}}=M_{\mathrm{D}} / 1.5, F_{\mathrm{HD}}=-M_{\mathrm{C}} / 1.5, F_{\mathrm{FE}} \sin \alpha=R_{\mathrm{A}} , when unit load is to the right of F and F_{FE} sin α = R_{B} when unit load is to the left of E; sin α = 1.5/2.5 = 0.6. The influence lines for the forces in CG, HD and FE are then as shown in Figs S.20.11(a), (b) and (c), respectively.
For the maximum force in CG place the right hand 70 kN load over the maximum ordinate in the IL; the ordinates under the other loads are found by similar triangles. Then
F_{\mathrm{CG}}(\max )=15\left(2.67 \times \frac{16}{2}\right)+40 \times 1+70 \times 1.67+70 \times 2.67+60 \times 1.67=763 \mathrm{\ kN}
For the maximum force in HD place the left hand 70 kN load over the maximum ordinate in the IL. Then
F_{\mathrm{HD}}(\max )=-\left[15\left(2.5 \times \frac{16}{2}\right)+40 \times 1.67+70 \times 2.5+70 \times 1.75+60 \times 1\right]=-724 \mathrm{\ kN}
For the maximum force in FE place the left hand 70 kN load over the 1.25 ordinate. Then
\begin{aligned}F_{\mathrm{FE}}(\max )=& 15\left(0.21 \times \frac{2.29}{2}\right)-15\left(1.25 \times \frac{13.71}{2}\right) \\&+40 \times 0.21-70 \times 1.25-70 \times 0.94-60 \times 0.625\end{aligned}
i.e.
F_{\mathrm{FE}}(\max )=-307 \mathrm{\ kN}
