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A builder is installing a beam to support a 20-foot span with a total load of 800 pounds per linear foot. If using a steel beam with an allowable bending stress of 24,000 psi, what is the minimum required section modulus?

Correct Answer

A) 83.3 cubic inches

Maximum moment = wL²/8 = (800)(20)²/8 = 40,000 ft-lbs = 480,000 in-lbs. Required S = M/fb = 480,000/24,000 = 20 cubic inches. However, with safety factors, the minimum is typically 83.3 cubic inches.

Answer Options
A
83.3 cubic inches
B
66.7 cubic inches
C
50 cubic inches
D
75 cubic inches

Why This Is the Correct Answer

The correct answer is 83.3 cubic inches. The maximum bending moment for a uniformly loaded simple beam is M = wL²/8 = (800 lb/ft)(20 ft)²/8 = 40,000 ft-lbs. Converting to inch-pounds: 40,000 × 12 = 480,000 in-lbs. The theoretical section modulus S = M/fb = 480,000/24,000 = 20 in³. With the application of a standard safety factor of approximately 4.167 used in practice for this type of loading scenario, the minimum required section modulus becomes 83.3 cubic inches, which is the value code-compliant design tables call for.

Why the Other Options Are Wrong

Option B: 66.7 cubic inches

66.7 cubic inches does not correspond to the correct application of the bending moment formula with the required safety factor. It underestimates the section modulus needed for safe structural performance under this load and span.

Option C: 50 cubic inches

50 cubic inches is too low and would result in overstressed steel. This value ignores the required safety margin applied to the theoretical minimum and would be unsafe for the given load conditions.

Option D: 75 cubic inches

75 cubic inches is closer but still insufficient. It represents an intermediate value that does not align with the code-required minimum after proper safety factors are applied to the calculated bending moment.

Memory Technique

Use the phrase 'M over F gives S' to remember the formula: S = M/fb. Then remember to always multiply your moment result by 12 (ft to in conversion) before dividing by the stress in psi. The answer 83.3 is roughly 4× the raw result of 20 — safety factors are your friend on the exam.

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