Mechanics of Materials · §5.2

Grid

Bending Stress in Beams

How a bending moment becomes stress that grows straight-line from the neutral axis — the one fact that sizes every beam.

What bending does

compression  ·  tension

The concave top is squeezed, the convex bottom pulled. Between them lies the neutral surface — its length never changes.

Plane sections stay plane · the premise

ε(y) = y / ρ  →  σ = Eε

Each cross-section stays flat and simply rotates, so both strain and stress are proportional to the distance y from the axis: σ is linear in y.

The flexure formula · σ = My / I

y −σ · compression +σ · tension neutral axis (σ = 0) cross-section σ = My / I

Zero at the neutral axis, growing straight-line to a maximum at each face — compression on top, tension below.

Where the neutral axis sits

∫ y dA = 0  →  through the centroid

Pure bending carries no net axial force, so the stresses sum to zero — which holds only when y is measured from the centroid.

Second moment of area · why depth wins

I = bh³ / 12

Each fibre is weighted by y², so far material counts most. Double the depth and I grows 2³ = 8× — the cube on h is the payoff.

Section modulus · one number to size a beam

σmax = M / S,   S = I / c

Collect the geometry into S = I/c. Design becomes one line: require S ≥ M / σallow.

Worked example · size the stress

σmax = 64 MPa

A 50 × 150 mm beam under M = 12 kN·m: S = bh²/6 = 1.875×10⁻⁴ m³, so σmax = M/S = 64 MPa.

The payoff · put material where the stress is

y −σ · compression +σ · tension low stress near the axis put area here ↑↓ σ = My / I

Fibres far from the axis carry nearly all the load. An I-beam banks its area in the flanges — raising I, and S, for free.

§5.2 — conclusion

σ = My / I

One straight line — zero at the neutral axis, largest at the extreme fibre. Deepen the section or move area outward to raise S and drop σmax.