We shall isolate the segments AB, BC and CD of the bar and compute the deformations in them. Then we use principle of superposition and algebraically add the deformations.

Portion DC
In the portion DC, we consider 32 kN force only as shown in Figure 2.32(b). This portion is under tension.

\begin{aligned} & \text { Change in length of } DC =\frac{P_{ DC } L_{ DC }}{A_{ DC } E} \\ & =\frac{32 \times 10^3 \times 500}{225 \times 200 \times 10^3} \\ & =0.355  mm \end{aligned}

Portion CB
The portion CB has to sustain 40 kN load due to addition of 8 kN. This portion is also under tension.
Refer Figure 2.32(c)

Elongation of CB =\frac{40 \times 10^3 \times 500}{225 \times 200 \times 10^3}=0.444  mm

Portion BA
This portion receives a net load of 32 kN as shown in Figure 2.32(d), this portion is also in tension.

\text { Elongation of } BA =\frac{32 \times 10^3 \times 500}{225 \times 200 \times 10^3}=0.355  mm

Now as per principle of superposition, net elongation of the bar or deflection of point D is

= 0.355 + 0.444 + 0.355 = 1.154 mm

The bar elongates.