Question

A 4-in. wide by 1.125-in. a rectangular steel bar supports a load in tension.

The stress in the bar if = 32,000 lb (precision of 100)
The load that can be supported by the bar if the axial stress must not exceed 25,000 psi (precision of 10,000)

Answer 1

Step-by-step explanation:

First, let's calculate the cross-sectional area of the bar:

A = width x height

A = 4 in x 1.125 in

A = 4.5 in^2

The stress in the bar is given as 32,000 lb, and we know that stress = force / area. Solving for force:

force = stress x area

force = 32,000 lb x 4.5 in^2

force = 144,000 lb

So the bar can support a load of 144,000 lb.

To find the load that can be supported if the axial stress must not exceed 25,000 psi, we need to rearrange the stress equation:

stress = force / area

force = stress x area

We know that the area is 4.5 in^2 and the stress cannot exceed 25,000 psi, so we can solve for the maximum load:

force = 25,000 psi x 4.5 in^2

force = 112,500 lb

Therefore, the maximum load that can be supported by the bar if the axial stress must not exceed 25,000 psi is 112,500 lb.