The bending stress formula, two ways
Bending stress is highest at the outer fibre of the section, distance c from the neutral axis (c = h/2 for a symmetric section). Divide the bending moment M by the moment of inertia I and multiply by c. Since the section modulus S = I / c bundles the geometry into one number, engineers usually just write σ = M / S.
Section properties for common shapes
| Shape | Moment of inertia I | Section modulus S |
|---|---|---|
| Rectangle (b × h) | b·h³ / 12 | b·h² / 6 |
| Solid circle (dia. d) | π·d⁴ / 64 | π·d³ / 32 |
| Hollow round (D, d) | π(D⁴ − d⁴) / 64 | I / (D/2) |
The h³ in the rectangle's I is the whole reason depth matters so much more than width. An I-beam of the same mass as a solid rectangle can have 3 to 5 times the I, and so 3 to 5 times less stress.
Max moment depends on how the beam is held
Before you can find stress you need Mmax, and that changes with the support and load. Same load, same span, very different moment.
| Case | Point load P | Uniform load w |
|---|---|---|
| Simply supported | P·L / 4 (at mid-span) | w·L² / 8 |
| Cantilever | P·L (at the fixed end) | w·L² / 2 |
| Fixed both ends | P·L / 8 | w·L² / 12 |
A cantilever carries eight times the mid-span moment of the same simply supported beam under a central point load. Confusing the two support conditions is a classic sizing error. Watch units too: kN·m with I in mm⁴ needs the 10⁶ conversion.
Allowable stress by material
| Material | Allowable σ (MPa) |
|---|---|
| Structural steel S235 | 160 |
| Higher-grade steel S355 | 245 |
| Aluminium 6061-T6 | 160 |
| Oak | 30 |
| Pine | 25 |
| Concrete C25/30 | 13 |
Allowable stress is roughly the yield stress divided by a safety factor of about 1.5. Concrete is weak in tension and cracks, so reinforced-concrete beams are analysed differently, with steel carrying the tension. Don't forget the beam's own weight, which often adds 10 to 20% to the load.
Common questions
What is the formula for bending stress?
Bending stress equals the bending moment times the distance to the outer fibre, divided by the moment of inertia: sigma = M times c divided by I. Because the section modulus S is I divided by c, the same thing is often written sigma = M divided by S.
Is it better to make a beam deeper or wider to carry more load?
Deeper. The moment of inertia grows with the cube of depth but only linearly with width, so a 10% increase in depth cuts stress about three times more than a 10% increase in width. This is why I-beams put most of the material in top and bottom flanges, far from the neutral axis.
Does low stress mean the beam is fine?
Not on its own. A beam can pass the stress check and still sag too much. Deflection has its own limit, usually span over 360 for a plastered floor, and it depends on span to the third or fourth power. Check both.
