Design Moment of a Compact Steel Beam

Easyflexureplastic moment

A compact, doubly symmetric steel beam section is sufficiently laterally braced to develop its full plastic moment. Fy = 250 MPa, plastic section modulus Zx = 800,000 mm³, elastic section modulus Sx = 700,000 mm³, and φb = 0.90. Use Mn = Fy·Zx and ignore other limit states. Find the design flexural strength φbMn.

A200 kN·m
B180 kN·m
C157.5 kN·m
D162 kN·m

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Answer: B. 180 kN·m

Equation: φbMn = φb Fy Zx. Substitution: Mn = 250(800,000) = 200×10⁶ N·mm = 200 kN·m; φbMn = 0.90(200) = 180 kN·m. Wrong: 200 kN·m is the nominal Mn without φb; 157.5 kN·m uses elastic Sx (0.90·250·700,000), which is not the plastic limit; 162 kN·m applies φb twice (0.90²·200).

ϕ_{b} M_{n}

= 0.90 (250) (800, 000)

= 180 \times 1 0^{6}

N·mm

= 180

kN·m

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Design Moment of a Compact Steel Beam

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