Net Section Rupture of a Tension Member
Mediumtensionnet sectionrupture
A steel tension member has net area An = 1,500 mm² (hole deductions already included), shear-lag factor U = 0.80 and Fu = 450 MPa. For net-section rupture use Ae = U·An and φ = 0.75. Gross-section yielding has a higher strength. Find the design rupture strength.
- A506.25 kN
- B324 kN
- C405 kN
- D540 kN
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Answer: C. 405 kN
Equation: φPn = φ Fu U An. Substitution: Ae = 0.80(1,500) = 1,200 mm²; Pn = 450(1,200) = 540 kN; φPn = 0.75(540) = 405 kN. Wrong: 506.25 kN ignores shear lag (0.75·450·1,500); 324 kN applies U twice (0.75·450·0.8²·1,500); 540 kN is the nominal strength without φ.
N kN
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