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A 4.0 in. wide by 1.125 in. thick rectangular steel bar supports a load of P in tension. Determine:
• The stress in the bar for F = 32,000 lb.
• The load P that can be supported by the bar if the axial stress must not exceed 25,000 psi.

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;

[tex]Stress, \sigma = \dfrac{Force, F}{Area, A}[/tex]

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;

[tex]The \ stress \ in \ the \ bar , \sigma = \dfrac{ F}{A} = \dfrac{32,000 \ lb}{4.5 \ in.^2} = 7,111.\overline 1[/tex]

The stress in the bar when F is 32,000 is σ = 7,111.[tex]\overline 1[/tex] 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;

[tex]\sigma = \dfrac{ P}{A}[/tex]

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.