504 kN · 113,300 lbf
Shear rupture (0.6·Fu·Anv) governs the shear term — the usual case.
Rₙ = min(0.6·Fu·Anv, 0.6·Fy·Agv) + Ubs·Fu·Ant — the shear term is capped by gross-section yield.
Also computed
378 kN · 84,980 lbf
φRₙ with φ = 0.75 — must be ≥ the required strength from LRFD (factored) load combinations.
252 kN · 56,650 lbf
Rₙ/Ω with Ω = 2.00 — must be ≥ the required strength from ASD (service) load combinations.
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