Section Modulus

S = \frac{I}{c} σ = \frac{M}{S}

Elastic section modulus S = I/c converts moment to extreme-fiber elastic stress. Plastic modulus Z is used for compact steel φMn = φ Fy Z. Larger S means more elastic bending strength.

$S$ (m³): Elastic section modulus
$I$ (m⁴): Moment of inertia
$c$ (m): Distance to extreme fiber
$M$ (N·m): Bending moment

Step-by-step solved example

I = 1.2×10⁸ mm⁴, h = 400 mm (c = 200 mm). S? Stress if M = 80 kN·m?

S = I/c. Elastic flexure σ = M/S.

1. S
S = 1.2e8 / 200 = 6.0×10⁵ mm³ = 6.0×10⁻⁴ m³.
$S = 6.0 \times 1 0^{5} mm^{3}$
2. Stress
σ = M/S = 80e6 N·mm / 6.0e5 mm³ = 133 MPa.
$σ = 133 MPa$

Answer: S = 6.0×10⁵ mm³, σ = 133 MPa

10 practice questions

0/10 correct

1.S = I/c. If c doubles and I is fixed, S

2.Rectangle b×h about strong axis: S =

3.Allowable moment for elastic design is

4.Plastic modulus Z compared with S is

5.I = 300 in⁴, c = 6 in. S =

6.Selecting a W-shape from tables often uses required

7.σ = M c / I is identical to

8.Circle radius r: S =

9.Increasing depth is more efficient than width because S scales as

10.Section modulus is NOT used for