A walkway suspended across a hotel lobby is supported at numerous points along i

Question

A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above each point and a vertical column underneath. The steel cable is 1.27 cm in diameter and is 5.75 m long before loading. The aluminum column is a hollow cylinder with an inside diameter of 16.14 cm, an outside diameter of 16.24 cm, and an unloaded length of 3.25 m. When the walkway exerts a load force of 9900 N on one of the support points, through what distance does the point move down?

Explanation / Answer

The stress on the steel is F/A = 8500/(pi*r^2) = 8500/(pi*(1.27x10^-2/2)^2) = 6.71x10^7Pa Strain = Stress/Y = 6.71x10^7/2x10^11 = 3.36x10^-4 So the elongation = L*strain = 5.75*3.36x10^-4 = 1.93x10^-3m Stress on the AL = 8500/(pi*(.0812^2-.0807^2) ) = 3.34x10^7Pa Strain = 3.34x10^7/7x10^10 = 4.77x10^-4 elongation = 4.77x10^-4*3.25m = -1.55x10^-3m However the change in length is the same So solve each for this change Fs = Y*A/5.75*deltaL = 2x10^11*pi*(1.27x10^-2/2)^2/5.75 deltaL = 4.406x10^6*deltaL and FA = Y*A/3.25*deltaL = 7x10^10*pi*(.0812^2 - .0807^2)/3.25*deltaL = 5.477x10^6*deltaL Adding these Fs + Fa = 8500 + (4.406x10^6+5.477x10^6)*deltaL So deltaL = 8500/(4.406x10^6+5.477x10^6) = 8.61x10^-4m

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