## Chapter 2

## Q. 2.23

A horizontal spacer bar of uniform cross section is fastened at its two ends to unyielding walls as shown in Figure 2.36. Determine the reaction exerted by the wall at end B, if loads are applied at intermediate points as shown.

## Step-by-Step

## Verified Solution

Had there been no wall at B, there wouldn’t be a reaction at B. We release the bar from wall at B, then we write an expression for elongation of the entire bar assuming a free movement of end B.

Elongation of bar AB =\frac{450 \times 10^3 \times 100}{A E}+\frac{300 \times 10^3 \times 200}{A E}+0(\text { no load in the portion DB })

=\frac{1050 \times 10^3}{A E}

Since the wall is unyielding, it will not let the movement of B to happen. This is because of compressive effect of reaction R_{ B } .

\text { Contraction of bar } AB \text { due to } R_{ B }=\frac{R_{ B } \times 1000(100+200+300)}{A E}

As there is no change in the length of spacer bar,

Elongation caused by loads = Contraction caused by the reaction

∴ \frac{1050 \times 10^3}{A E}=\frac{R_{ B } \times 1000 \times 600}{A E}

R_{ B }=\frac{1050 \times 10^5}{6 \times 10^5}=175 kN

Reaction at A can also be found out by applying equation of equilibrium:

\begin{gathered} \sum F_X=0 \\ R_{ A }+175-300-150=0, \quad \therefore \quad R_{ A }=275 kN \end{gathered}