Question

A rigid bar is suspended by three rods made of the same material as shown in the figure. The area and length of the central rod are $3A$ and $L$ respectively, while that of the two outer rods are $2A$ and $2L$ respectively. If a downward force of $50kN$ is applied to the rigid bar, the forces in the central and each of the outer rods will be

• A. $16.67kN$ each
• B. $30kNand15kN$
• C. $30kNand10kN$
• D. $21.4kNand14.3kN$

Answer 1

The correct option is C $30kNand10kN$

Outer rods are of same length and same cross section,

Let force resisted by each outer rod is $P_1$

Let force resisted by central rod is $P_2$

$2P_1+P_2=50...\left(i\right)$



All rods are connected by rigid bar so deflection in each rod will be same
${\delta }_1={\delta }_2$

$\frac{P_{1}2L}{2AE}=\frac{P_{2}L}{3AE}$

$P_1=\frac{P_2}{3}...\left(ii\right)$

Solving eq. (i) & (ii)
$2P_1+P_2=50$

$\frac{2P_2}{3}+P_2=50$

$\Rightarrow P_2=30kN$

$P_1=10kN$