## Q. 2.22

A member ABCD is subjected to point loads $P_1$ = 45 kN, $P_2$ = 100 kN, and $P_4$ = 130 kN. Determine the net elongation of the member. Assume E = 210 GPa. Refer Figure 2.35(a).

## Verified Solution

For the external equilibrium, algebraic sum of horizontal forces on the bar must be
zero.

$\begin{array}{r} -P_1+P_2-P_3+P_4=0 \\ -45+100-P_3+130=0 \end{array}$

∴        $P_3=185 kN$

Now, we draw FBD (Figures 2.35b, c and d) of three portions of the member and superpose their deformations

Therefore, Net deformation = Algebraic sum of deformations in three portions

\begin{aligned} & =\frac{10^3}{210 \times 10^3}\left[\frac{130 \times 1000}{1250}-\frac{55 \times 600}{2500}+\frac{45 \times 1200}{625}\right] \\ & =\frac{1}{210}(177.2)=0.844 mm \end{aligned}