Answer:
a) The stress in the bar when F is 32,000 is approximately 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed is approximately 110,000 lb
Explanation:
The question topic relates to stresses in structures;
The given parameters of the steel bar are;
The width of the steel bar, W = 4.0 in.
The thickness of the steel bar, t = 1.125 in.
The formula for stress in a bar is given as follows;

The cross sectional area of bar, A = W × t = 4.0 in. × 1.125 in. = 4.5 in.²
∴ The cross sectional area of bar, A = 4.5 in.²
a) The stress in the bar for F = 32,000 lb, is given as follows;

The stress in the bar when F is 32,000 is σ = 7,111.
psi ≈ 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed, σ = 25,000 psi is given as follows;

Therefore;
P = σ × A = 25,000 psi × 4.5 in² = 112,500 lb
For the axial stress of 25,000 psi not to be exceeded, the maximum load that can be supported by the bar, P = 112,500 lb ≈ 110,000 lb.