Question

A 2.3-in-diameter steel rod 3 ft long is used between two immovable (rigid) walls. The rod is welded to the two walls at each of its two ends. The ambient temperature increases by 123 °F. The modulus of elasticity for the steel is 29 x 10^6 psi and the CTF is 4.0 x 10^-6/F. Calculate the force exerted by the rod on the walls

Answer 1

Answer:

F = 59250 pounds

Explanation:

given data;

Diameter = 2.3 inch

length = 3 ft = 36 inch

ambient temp = 73.4 degree F

$\Delta T = 123 degree\ F$

cofficient of thermal expansion$∝ = 4\times 10^{-6} \F$

We know that

$\sigma  =  Ee$

$\sigma = E ∝ \Delta T$

$\frac{F}{A} = E ∝ \Delta T$

$F = A\times E ∝ \Delta T$

$F =\frac{\pi}{4} D^2\times E\times ∝\times \Delta T$

$F =\frac{\pi}{4}\times 2.3^2\times 29\times 10^6\times 4 \times 10^{-6}\times 123$

F = 59250 pounds