Combined Axial and Bending Stress

σ = \frac{P}{A} \pm \frac{M c}{I}

Elastic superposition of uniform axial stress and linearly varying flexure. The ± selects the fiber: compression adds on the concave side. An eccentric axial load P at eccentricity e is equivalent to P plus M = Pe. Apply only when the member is short enough that buckling is checked separately.

$σ$ (Pa): Normal stress at a fiber
$P$ (N): Axial force
$A$ (m²): Area
$M$ (N·m): Bending moment
$c, I$ (m, m⁴): Extreme-fiber distance and I

Step-by-step solved example

Rectangle 100×200 mm, strong-axis bending. P = 400 kN compression, M = 20 kN·m. Extreme-fiber stresses?

σ = P/A ± Mc/I. Superpose axial and flexure on the same section.

1. Axial and section
A = 0.020 m² → P/A = 20.0 MPa C. I = bh³/12 = 6.667×10⁻⁵ m⁴, c = 0.10 m.
2. Bending and combine
Mc/I = 20e3×0.10 / 6.667e−5 = 30.0 MPa. σ = 20±30 → 50.0 MPa C and 10.0 MPa T.
$σ = 50.0 MPa C, 10.0 MPa T$

Answer: 50.0 MPa C and 10.0 MPa T

10 practice questions

0/10 correct

1.Elastic combined stress is superposition of

2.For compression P, bending adds extra compression on

3.The kern of a rectangle (no tension under P+M) is

4.Section modulus S is I/c so Mc/I equals

5.P = 120 kN, A = 3000 mm². P/A =

6.The ± sign selects

7.Eccentric axial load P at eccentricity e is equivalent to

8.At the centroidal fiber y = 0, bending stress is

9.P = 0 reduces the formula to

10.This combined-stress formula is for