Stirrup Spacing for Required Shear
Hardshearstirrup-spacingdetailing
At a reinforced concrete beam section, factored shear Vu = 150 kN, nominal concrete shear Vc = 100 kN, φ = 0.75, total stirrup area Av = 200 mm² (all legs of one stirrup), stirrup fy = 420 MPa, and d = 500 mm. Use Vs = Avfyd/s. The maximum permitted spacing is supplied as 250 mm, and all other requirements are satisfied. What is the largest usable stirrup spacing?
- A420 mm
- B840 mm
- C250 mm
- D3,360 mm
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Answer: C. 250 mm
Equation: strength requires φ(Vc + Vs) ≥ Vu, with Vs = Avfyd/s. Inputs: Vs,req = Vu/φ − Vc = 150/0.75 − 100 = 100 kN. Substitution: s ≤ (200)(420)(500)/100,000 N = 420 mm. The supplied cap is 250 mm, so the smaller value, 250 mm, controls (at 250 mm, φVn = 0.75(100 + 168) = 201 kN > 150 kN). Wrong: 420 mm satisfies strength but ignores the 250 mm detailing cap; 840 mm forgets φ (Vs,req = 50 kN); 3,360 mm multiplies Vu by φ instead of dividing (Vs,req = 12.5 kN).
kN mm mm
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