Answer to Question #201702 in Civil and Environmental Engineering for Akshat Garg

Question

Two vertical rods, one of steel and the other of bronze, each rigidly fastened at the

upper end, are at a horizontal distance of 1.05m apart. Each rod is 2.5m long and

20mm in diameter. A horizontal plate connects the lower end of the rods. Where

should a load of 30kN be placed on the horizontal plate. so that it remains horizontal

after being loaded. Es-200 GN/m² and Ebronze = 100 GN/m² (refer Fig-4)

Answer 1

$\frac{P_s}{E_s}=\frac{P_b}{E_b}$

$\frac{P_s}{200*10^5}=\frac{P_b}{100*10^5}=P_s=P_s*\frac{200*10^5}{100*10^5}=P_s=2P_b$

From equilibrium

$P_s+P_b=30*10^3N$

$2P_b+P_b=30*10^3N$

$3P_b=30*10^3$

$P_b=10*10^3N$

Taking moments

$(P_b)1.05=30*10^5*x$

$x=\frac{10*10^3*1.05}{30*10^5}$

$x=0.0035=3.5mm$

The load should be placed at a distance $x=3.5$ from steel bar.