Compact Steel Flexural Strength

ϕ M_{n} = ϕ F_{y} Z_{x} ϕ = 0.90

For compact I-shaped members laterally supported, the LRFD design moment is 0.90 Fy Zx. If the section is noncompact, Mn is reduced toward Fy Sx. Always compare with Mu from load combinations.

$ϕ$: Resistance factor
$F_{y}$ (ksi or MPa): Yield stress
$Z_{x}$ (in³): Plastic modulus

Step-by-step solved example

W16×26, Zx = 44.2 in³, Fy = 50 ksi. φMn?

Compact steel beam, fully braced: φMn = 0.90 Fy Zx.

1. Mn
Mn = Fy Zx = 50×44.2 = 2210 kip·in = 184.2 kip·ft.
2. Design
φMn = 0.90×184.2 = 166 kip·ft.
$ϕ M_{n} = 166 kip\cdotft$

Answer: 166 kip·ft

10 practice questions

0/10 correct

1.φ for steel flexure (LRFD) is

2.Compactness depends on

3.If Mu > φMn the member is

4.Zx for a W-shape is listed in

5.Fy = 36 ksi, Z = 30 in³. Mn =

6.Lateral-torsional buckling reduces Mn when

7.ASD steel flexure uses

8.Units: Z in in³, Fy in ksi → Mn in

9.Mp = Fy Z is the

10.My = Fy S is reached