## Q. 2.20

A steel bar is supported and loaded as shown in Figure 2.33(a). The cross sectional area of the bar is 250 mm². Determine the force P so that the lower end of the bar does not move vertically when the loads are applied. Take E = 200 GPa for the material of the bar.

## Verified Solution

Since the point D has no vertical movement, the net deformation due to forces acting on the bar must be zero.
Deformation of DC [Figure 2.33(b)]

$d l_{ CD }=\frac{9 \times 1000 \times 300}{250 \times 200 \times 10^3}=0.054 mm$

Deformation of CB (Figure 2.33(c))

$d l_{ CB }=\frac{24 \times 1000 \times 200}{250 \times 200 \times 10^3}=0.096 mm$

Deformation of BA (Figure 2.33(d))
In order to nullify the elongations of the portions DC and CB, the portion BA should be in compression, therefore, it is subjected to compressive force given by (24 – P) kN.

Therefore,  $d l_{ BA }=\frac{(24-P) 10^3 \times 500}{250 \times 200 \times 10^3}=0.01(24-P) mm$

In order that the point D should not move

0.054 + 0.096 + 0.01(24 – P) = 0

P = 39 kN