Minimum Flexural Steel in a Beam
Mediummin-steelunitsflexure
A normal-weight rectangular concrete beam has b = 12 in, d = 20 in, f′c = 4,000 psi, and fy = 60,000 psi. Use the supplied rule As,min = max(3√f′c/fy, 200/fy)·b·d with stresses in psi and dimensions in inches; no exception applies. What is the required minimum tension-steel area?
- A0.759 in²
- B0.800 in²
- C1.559 in²
- D0.024 in²
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Answer: B. 0.800 in²
Equation: As,min = max(3√f′c/fy, 200/fy)bd. Inputs: bd = 12(20) = 240 in². Substitution: first expression 3√4,000/60,000 × 240 = 0.759 in²; second expression (200/60,000) × 240 = 0.800 in². The larger controls: 0.800 in². Wrong: 0.759 in² takes only the square-root expression (the smaller branch); 1.559 in² adds the two branches instead of taking the maximum; 0.024 in² enters f′c in ksi (4) and so uses the wrong units for the psi-based constant.
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