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

F = 59250 pounds

Step-by-step 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* 10^(-6) \F`

We know that

`\sigma  =  Ee`

`\sigma = E ∝ \Delta T`

`(F)/(A) = E ∝ \Delta T`

`F = A* E ∝ \Delta T`

`F =(\pi)/(4) D^2* E* ∝* \Delta T`

`F =(\pi)/(4)* 2.3^2* 29* 10^6* 4 * 10^(-6)* 123`

F = 59250 pounds