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

Question #201702

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)



1
Expert's answer
2021-06-03T16:06:01-0400

"\\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.


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