Beam Bending Stress Calculator
Maximum bending stress in a beam from the flexure formula σ = M·c ÷ I.
Tension on one face, equal compression on the other. Section modulus S = 83330 mm³, so σ = M ÷ S. 1 MPa = 1 N/mm².
Maximum bending stress is the flexure formula σ = M·c ÷ I = M ÷ S, where M is the bending moment, c the distance to the extreme fibre, I the area moment of inertia, and S = I ÷ c the section modulus. A 5 kN·m moment on a 50 × 100 mm rectangle (I = 4.167×10⁶ mm⁴, c = 50 mm) gives σ = 60 MPa.
What bending stress is
When a beam bends, the material on one face is stretched while the opposite face is squeezed. Somewhere between them lies the neutral axis — a layer that is neither in tension nor compression and carries no bending stress at all. The stress grows linearly with distance from that axis, so the largest values sit at the extreme fibres, the surfaces farthest from the neutral axis. The flexure formula turns the internal bending moment M into that peak stress using the section’s geometry.
σ = bending stress, M = bending moment, c = distance from the neutral axis to the extreme fibre, I = area moment of inertia, S = I ÷ c = section modulus. Use N·mm and mm so σ comes out in MPa (N/mm²).
Worked example
A rectangular beam 50 mm wide and 100 mm deep carries a maximum bending moment of 5 kN·m. For a rectangle, I = b·h³ ÷ 12 = 50 × 100³ ÷ 12 = 4.167×10⁶ mm⁴, and c is half the depth, 50 mm.
- 1 Find the maximum bending moment. Take M at the section of interest — usually where the moment peaks. Here M = 5 kN·m = 5×10⁶ N·mm. Working in N·mm and mm makes σ come out in MPa.
- 2 Locate the neutral axis and the extreme fibre. For a symmetric section the neutral axis passes through the centroid, so c is half the depth: c = 100 ÷ 2 = 50 mm.
- 3 Get the area moment of inertia I. For a rectangle, I = b·h³ ÷ 12 = 50 × 100³ ÷ 12 = 4.167×10⁶ mm⁴. Use the Area Moment of Inertia tool for other shapes.
- 4 Compute the section modulus (optional). S = I ÷ c = 4.167×10⁶ ÷ 50 = 83,333 mm³. Bending tables often list S directly, letting you skip straight to σ = M ÷ S.
- 5 Apply the flexure formula. σ = M·c ÷ I = 5×10⁶ × 50 ÷ 4.167×10⁶ = 60 MPa. Equivalently σ = M ÷ S = 5×10⁶ ÷ 83,333 = 60 MPa.
Unit combinations that give σ in MPa
Bending stress in MPa (N/mm²) needs consistent units. Mixing N·m with mm inflates σ by 1000, so convert the moment first.
| M (moment) | c and I (or S) | Result σ |
|---|---|---|
| N·mm | c in mm, I in mm⁴ | MPa (N/mm²) directly |
| kN·m | multiply by 1×10⁶ → N·mm | MPa after converting M |
| N·mm | S in mm³ (S = I ÷ c) | MPa via σ = M ÷ S |
| N·m | c in m, I in m⁴ | Pa — divide by 1×10⁶ for MPa |
Reading and using the result
The flexure formula assumes pure bending of a straight, prismatic beam made of a linear-elastic material, with plane sections staying plane. Under those conditions the top and bottom fibres see equal and opposite stresses on a symmetric section — one in tension, one in compression. Compare the peak σ against the material’s allowable stress (yield strength divided by a factor of safety) to judge whether the section is adequate.
Because σ depends on c ÷ I, pushing material away from the neutral axis is the most effective way to cut stress: it raises I faster than c, so the section modulus S climbs and σ falls. That is exactly why I-beams concentrate area in the flanges. This calculator handles the bending stress once M is known; finding M itself, and the resulting deflection, are separate steps.