Engineer's Notebook §3.2

Bending stress

A beam bending under a moment M is squeezed on one face and stretched on the other. The stress grows linearly with distance y from the neutral axis:

σ =M·yI
σ
bending stress
Pa
M
bending moment
N·m
y
distance from the neutral axis
m
I
second moment of area
m⁴
Z
section modulus, I/ymax
m³
b
section width
m
h
section depth
m
R
radius of curvature
m
E
Young's modulus
Pa
Left, a rectangular cross-section with its neutral axis and a marked fibre. Right, the bending stress diagram: compression at the top, tension at the bottom, zero at the neutral axis. NEUTRAL AXIS σ = 3.75 MPa C −7.5 T +7.5
Fig. 3.2-1Under a sagging moment the top is in compression (C) and the bottom in tension (T).
I = bh³/126.67 × 10⁷ mm⁴
σmax = M/Z±7.5 MPa
σ = My/I3.75 MPa

Worked example

Uses the values in Fig. 3.2-1. Change them and the working updates.

Given
  • Msagging bending moment5 kN·m
  • b×hrectangular section100 × 200 mm
  • yfibre below the neutral axis50 mm
Find

The bending stress at the fibre y from the neutral axis.

Solution
I = bh³/12 = 100 mm × (200 mm)³ / 12 = 6.667 × 10⁷ mm⁴ σ = My/I = 5 × 10⁶ N·mm × 50 mm / 6.667 × 10⁷ mm⁴ = 3.75 N/mm² = 3.75 MPa
Answer
σ = 3.75 MPa

The extreme fibres, at y = ±100 mm, carry ±7.5 MPa.

Related cases

  • Extreme fibre, rectangle

    σmax = MZZ = bh²6

  • Solid round bar

    σmax = 32Mπd³

  • Axial force plus bending

    σ = FA ± MyI

Assumptions
  • Plane sections stay plane and perpendicular to the neutral axis as the beam bends.
  • The material is linear-elastic, with the same modulus in tension and compression.
  • The section is symmetric about the plane of bending, and the moment acts about a principal axis.
  • Here y is positive below the neutral axis, so a sagging moment gives tension there.

Derivation
  1. A fibre at distance y from the neutral axis, in a beam bent to radius R, has strain ε = y/R.
  2. Hooke's law gives σ = Ey/R, linear in y.
  3. There is no net axial force: ∫σ dA = (E/R)∫y dA = 0, so the neutral axis passes through the centroid.
  4. The stresses add up to the moment: M = ∫σy dA = (E/R)∫y² dA = EI/R.

Eliminate E/R between the last two steps: σ/y = M/I = E/R. This is the engineer's bending equation.