Question 1.12: Three steel rods, as shown in Figure 1.32(a) are pin-connect...

The downward force, 15 kN will be balanced by the forces developed in the three rods. If the forces developed in individual rods are F_1, F_2 \text { and } F_3 as shown in Figure 1.32(b), we can write F_1+F_2 + F_3 = 15 kN. This is an equilibrium equation along the vertical direction. Another equilibrium equation can be the moment about B (or any other suitable point).

\sum M_{ B }=0=-F_1 \times 0.8+15 \times 0.6-F_2 \times 0.4

We have two equations but three unknowns. So, the problem is statically indeterminate. We need compatibility equation, in this case, a load–displacement equation. How can we get it? As the bar AB is rigid, it does not bend and remains straight. So, deformations of the vertical rods \delta_1, \delta_2 \text { and } \delta_3 are such that

\frac{\delta_1-\delta_3}{0.8}=\frac{\delta_2-\delta_3}{0.4}

as evident from Figure 1.32(c) which shows the new position of AB. So,

\delta_2=\frac{\delta_1+\delta_3}{2}           (1)

Now, δ = PL/AE , therefore

\delta_1=\frac{F_1 \times 0.5}{A_1 E}, \quad \delta_2=\frac{F_2 \times 0.5}{A_2 E}, \quad \delta_3=\frac{F_3 \times 0.5}{A_3 E}

Thus, Eq. (1) can be rewritten as

\frac{F_2 \times 0.5}{15 \times E}=\left\lgroup \frac{F_1 \times 0.5}{25 \times E}+\frac{F_3 \times 0.5}{25 \times E} \right\rgroup \times \frac{1}{2}

or        F_2=0.3 F_1+0.3 F_3              (2)

Solving equilibrium force equation, moment equation and Eq. (2), we get

F_1=9.52  kN , \quad F_2=3.46  kN \text { and } \quad F_3=2.02  kN