Shear and Torsion Worked Examples

All examples use the NSCP 2015 / adopted ACI 318-14 basis stated in the lesson. Unless noted otherwise, concrete is normal weight (λ=1.0\lambda=1.0) and ϕ=0.75\phi=0.75 for shear and torsion.

Example 1 — A beam has bw = 300 mm, d = 500 mm, f'c = 28 MPa, and factored shear Vu = 250 kN. Determine Vc, the concrete design strength, and the nominal shear contribution required from stirrups.

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Example 2 — Continue Example 1 using 10 mm two-leg U-stirrups, Av = 157.1 mm², and fyt = 275 MPa. Complete the stirrup design including minimum reinforcement and spacing.

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Example 3 — For the same 300 mm web and 500 mm effective depth with f'c = 28 MPa, take Vu = 80 kN and fyt = 420 MPa. Determine the governing requirement for 10 mm two-leg stirrups.

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Example 4 — A normal-weight beam has bw = 300 mm, d = 500 mm, f'c = 28 MPa, and Vu = 500 kN. Check whether adding stirrups can make this section adequate under the adopted shear-strength ceiling.

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Example 5 — A solid 400 mm by 600 mm beam has f'c = 28 MPa, no axial force, and factored torque Tu = 5.0 kN·m. Determine whether torsion may be neglected.

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Example 6 — A solid 400 mm by 600 mm beam has Vu = 200 kN, Tu = 40 kN·m, f'c = 30 MPa, d = 540 mm, and 40 mm clear cover to a 10 mm closed stirrup. Check the combined shear-torsion compression-strut section-size limit.

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Example 7 — For the beam in Example 6, use Aoh = 158,100 mm², Ao = 0.85Aoh, ph = 1640 mm, Tu = 40 kN·m, fyt = fy = 420 MPa, and theta = 45°. Determine strength-required At/s and associated longitudinal torsion steel.

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Example 8 — A beam has bw = 300 mm, d = 450 mm, f'c = 28 MPa, and requires Vs = 150 kN. Using 10 mm two-leg stirrups with Av = 157.1 mm² and fyt = 420 MPa, determine a valid spacing.

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