CivilSteel DesignmediumFE ExamCourseworkPE / Advanced

Design Flexural Strength of a Compact Steel Beam

A compact, laterally braced steel beam has plastic section modulus $Z_x = 1.0 \times 10^{6}\ \text{mm}^3$ and yield strength $F_y = 250\ \text{MPa}$ .

Using LRFD with $\phi_b = 0.90$ , find the design flexural strength $\phi_b M_n$ .

Given

Zx	1000000 mm^3	Plastic section modulus
Fy	250 MPa	Yield strength
φb	0.9 -	Flexural resistance factor

Hint 1

Compact and laterally braced means the section can reach its full plastic moment.

Hint 2

M_p = F_y·Z, using the plastic section modulus.

Hint 3

Then multiply by φ_b = 0.90 for the LRFD design strength.

Worked solution — try the problem first

For a compact, laterally braced section, the nominal flexural strength is the full plastic moment — no reduction for local or lateral-torsional buckling:

$M_n = M_p = F_y Z_x$

$M_p = 250\ \text{N/mm}^2 \times 1.0\times10^{6}\ \text{mm}^3 = 250\times10^{6}\ \text{N·mm} = 250\ \text{kN·m}$

Note $1\ \text{MPa} = 1\ \text{N/mm}^2$ , so working in N and mm needs no conversion at all.

Apply the resistance factor:

$\phi_b M_n = 0.90 \times 250 = 225\ \text{kN·m}$

$\boxed{\phi_b M_n = 225\ \text{kN·m}}$

Why $Z$ and not $S$ ? The elastic section modulus $S$ marks first yield at the extreme fibre; the plastic modulus $Z$ marks full plastification of the section. For a typical W-section $Z \approx 1.1S$ , so using $S$ here would understate capacity by roughly 10%.

Concepts:LRFDPlastic momentSteel beam designSection modulus

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