A bending moment stretches one side of a beam and compresses the other. The normal stress varies linearly from zero at the neutral axis to a maximum at the outer fibre, given by the flexure formula σ = My/I. The largest stress is σ_max = Mc/I = M/S, where c is the distance to the farthest fibre and S = I/c is the section modulus.
Last updated
Where the flexure formula comes from
In pure bending, plane cross-sections stay plane and rotate about a line called the neutral axis. Fibres on one side lengthen and those on the other shorten, in proportion to their distance y from that axis. In the linear-elastic range Hooke’s law turns that linear strain into linear stress.
Two equilibrium conditions finish the derivation. The net axial force on the section must be zero, which puts the neutral axis through the centroid of the cross-section. The moment of the stresses about that axis must equal M, which brings in the second moment of area I.
Using it
Get M from the shear and moment diagram: the largest |M| is usually the critical section. Get I about the centroidal axis that bending happens about. Then check both extreme fibres. For a symmetric section they carry equal and opposite stress; for a T or a channel, the fibre farther from the neutral axis carries more.
Positive moment (concave up, as in a simply supported beam under downward loads) puts the top in compression and the bottom in tension.
Negative moment (over supports, or along a cantilever) does the reverse.
Orientation matters: a rectangle loaded on edge has I = bh³/12 with h the larger dimension, so it is much stronger and stiffer than the same bar laid flat.
Limits
The formula assumes a straight, prismatic, homogeneous beam in the elastic range, with loads in a plane of symmetry. It ignores stress concentrations at holes and notches, and it is only approximate for deep beams, curved beams and very short spans where shear deformation matters. Combined with the shear stress τ = VQ/(It), the result feeds a Mohr’s circle for principal stresses.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Flexure formula
σ = −M·y / I
σ = normal stress (MPa), M = internal bending moment (N·mm), y = distance from the neutral axis, positive upward (mm), I = second moment of area about the neutral axis (mm⁴). The minus sign gives compression above the axis under positive M.
Gere & Goodno, Mechanics of Materials
Maximum bending stress
σ_max = M·c / I = M / S
c = distance from the neutral axis to the extreme fibre (mm), S = I/c = elastic section modulus (mm³)
Hibbeler, Mechanics of Materials
Rectangular section
I = b·h³ / 12, S = b·h² / 6
b = width (mm), h = depth in the plane of bending (mm)
Beer & Johnston, Mechanics of Materials
Midspan moment, centre point load on a simple span
M_max = P·L / 4
P = point load (N), L = span (mm)
Hibbeler, Mechanics of Materials
Worked examples
Example: A rectangular aluminium bar on edge and laid flat
Given
A simply supported aluminium bar spans 2.4 m with a 3 kN point load at midspan. Self-weight is neglected.
The bar is 40 mm × 80 mm in cross-section.
Find
The maximum bending stress with the 80 mm side vertical (on edge), the stress 20 mm above the neutral axis, and the maximum stress with the bar laid flat.
Solution
Maximum moment at midspan, in N·mm:M = P·L / 4 = 3000 × 2400 / 4 = 1.8 × 10⁶ N·mm
On edge: b = 40 mm, h = 80 mm, c = 40 mm.I = 40(80)³ / 12 = 1.707 × 10⁶ mm⁴
Maximum stress (compression at the top, tension at the bottom):σ_max = M·c / I = 1.8 × 10⁶ × 40 / 1.707 × 10⁶ = 42.2 MPa
Stress varies linearly with y, so 20 mm above the neutral axis it is half the peak, in compression:σ = M·y / I = 1.8 × 10⁶ × 20 / 1.707 × 10⁶ = 21.1 MPa (C)
Laid flat: b = 80 mm, h = 40 mm, c = 20 mm.I = 80(40)³ / 12 = 0.427 × 10⁶ mm⁴; σ_max = 1.8 × 10⁶ × 20 / 0.427 × 10⁶ = 84.4 MPa
Answer: On edge, σ_max = 42.2 MPa (compression on top, tension on the bottom) and 21.1 MPa at 20 mm above the neutral axis. Laid flat, the same bar sees 84.4 MPa: twice the stress, because S depends on h².
Example: Tension and compression in a T-section cantilever
Given
A cantilever 1.5 m long carries a 2.5 kN point load at its free end, flange up. Self-weight is neglected.
The cross-section is the T-section from the centroids lesson: centroid 82.73 mm above the bottom of the web, total depth 120 mm, I = 5.674 × 10⁶ mm⁴.
Find
The bending stress at the top of the flange and at the bottom of the web at the fixed end.
Solution
The moment is largest at the fixed end. It is negative (hogging), so the top is in tension.M = −P·L = −2500 × 1500 = −3.75 × 10⁶ N·mm
Distances from the neutral axis to each extreme fibre:c_top = 120 − 82.73 = 37.27 mm, c_bottom = 82.73 mm
Top of the flange:σ_top = |M|·c_top / I = 3.75 × 10⁶ × 37.27 / 5.674 × 10⁶ = 24.6 MPa (T)
Bottom of the web:σ_bottom = |M|·c_bottom / I = 3.75 × 10⁶ × 82.73 / 5.674 × 10⁶ = 54.7 MPa (C)
Ratio check: the stresses are in the ratio of the distances.54.7 / 24.6 = 2.22 = 82.73 / 37.27 ✓
Answer: At the fixed end the flange top carries 24.6 MPa tension and the web bottom 54.7 MPa compression. The fibre farther from the neutral axis governs, so for a T both extremes must be checked, especially in materials weaker in one sign of stress.
Common mistakes
Measuring y from the bottom of the section. y and c are measured from the neutral axis, which passes through the centroid. Find the centroid first for any unsymmetric section.
Using the wrong I for the bending axis. Use I about the axis the beam bends around. For a rectangle on edge h is the vertical dimension; laid flat, it is the thickness.
Mixing kN·m with mm⁴. Convert M to N·mm (1 kN·m = 10⁶ N·mm) so that σ comes out in MPa.
Checking only one extreme fibre of an unsymmetric section. Compute both c_top and c_bottom. They differ for T, L and channel sections.
Using the flexure formula above yield. It assumes linear-elastic behaviour. Compare σ_max with the yield strength; plastic bending needs a different analysis.
In industry
Frequently asked questions
Why is bending stress zero at the neutral axis?
Fibres on the neutral axis do not change length as the beam bends, so their strain is zero, and in the elastic range so is their stress.
What is the section modulus used for?
S = I/c packages the geometry into one number, so σ_max = M/S. Steel and aluminium tables list S for standard shapes, which makes sizing a beam a one-line check.
Does the material affect bending stress?
Not for a statically determinate beam: σ = Mc/I uses only the moment and the geometry. The material decides whether that stress is acceptable and how much the beam deflects.
How is bending stress different from shear stress in a beam?
Bending stress is a normal stress, largest at the outer fibres. Transverse shear stress τ = VQ/(It) acts along the section and is largest at the neutral axis. Both act together and can be combined with stress transformation.
Can I use the flexure formula for a beam with a hole?
For the nominal stress, yes, using the net section’s I. The actual peak near the hole is higher, by a stress concentration factor.