Civil Engineering Tools
BeamTrussMohr's CircleUnit ConversionsCentroidDeflectionBucklingConcreteRebarFE Prep

Design Tensile Strength by Gross Yielding

EasyLRFDtension member
A steel tension member has a gross area of 2,000 mm² and Fy = 250 MPa. Using LRFD with φ = 0.90 for yielding on the gross section, its design tensile strength is most nearly:
  1. A400 kN
  2. B450 kN
  3. C500 kN
  4. D556 kN
Show solution

Answer: B. 450 kN

Equation: LRFD tensile yielding on the gross section, φPn = φFyAg. Values: φ = 0.90, Fy = 250 MPa = 250 N/mm², Ag = 2,000 mm². Step 1: Pn = 250 × 2,000 = 500,000 N = 500 kN. Step 2: φPn = 0.90 × 500 = 450 kN. 500 kN forgets φ; 556 kN divides by φ; 400 kN uses φ = 0.80.
Practice this deckUntimed, one problem at a time, with your progress saved.

Related problems

Keep studying