Bending

Bending creates tension on one side of a beam and compression on the other. Learn the flexure formula σ = My/I, neutral axis, section modulus, beam selection and combined bending and axial load.

When a beam bends, its top fibres shorten and its bottom fibres lengthen (for a sagging beam). Somewhere between is a neutral axis with no stress. Stress grows linearly with distance from it.

The flexure formula connects the bending moment from equilibrium to the stress in the material — it sizes floor joists, crane girders, axles and aircraft wings.

An analogy

Bend a thick foam rubber block: the top surface wrinkles (compression) and the bottom stretches (tension). A line through the middle doesn't change length.

Tension, compression and the neutral axis

In a simply supported beam under load the bottom is in tension and the top in compression. The neutral axis passes through the centroid. Fibres farthest from it carry the highest stress.

Section modulus and beam selection

σ_max = M/Z, so the required Z = M/σ_allow. Steel tables list Z for standard sections — pick the lightest one that meets it. Deeper sections give much more Z per kg.

Combined loading

When a member carries axial force and bending (a column with an eccentric load, a crane hook), stresses add: σ = P/A ± M/Z. The side where both are tensile usually governs.

Plastic bending and unsymmetric sections

Beyond yield, the stress block flattens; the plastic modulus Z_p (1.5 Z for rectangles, ≈ 1.12 Z for I-beams) gives the moment at which a plastic hinge forms — the basis of plastic design. Unsymmetric sections bend about principal axes, and brittle materials (cast iron) need separate tension and compression checks.

Formulas

• Flexure formula: σ = M · y / I
• Maximum stress: σ_max = M / Z
• Bending equation: M/I = σ/y = E/R
• Rectangle / solid circle Z: Z = bh²/6, Z = πd³/32
• Combined: σ = P/A ± M/Z

Worked example: Stress in a timber beam

Given: 100 × 200 mm (on edge); M = 20 kN·m

Find: Maximum bending stress.

• Z = 100 × 200²/6 = 666 667 mm³
• σ = 20 × 10⁶/666 667 = 30 MPa

Answer: 30 MPa (top compression / bottom tension)

Worked example: Selecting a steel beam

Given: M_max = 40 kN·m; σ_allow = 165 MPa

Find: Required section modulus.

• Z = 40 × 10⁶/165 = 242 400 mm³ = 242 cm³
• From tables pick e.g. IPE 220 (Z ≈ 252 cm³)

Answer: Z ≥ 242 cm³ → IPE 220

Common mistakes

• Wrong: Using M in kN·m with dimensions in mm. — Right: Convert M to N·mm (× 10⁶) so σ comes out in MPa.
• Wrong: Measuring y from the bottom of the section. — Right: y is measured from the neutral axis (centroid).

Applications

• Floor & roof beams: Joists and girders sized from M_max/Z (and deflection).
• Axles & shafts: Rotating axles see fully reversed bending stress every revolution — a fatigue case.
• Crane jibs & wings: Tapered sections match Z to the bending-moment diagram to save weight.

Key takeaways

• σ = My/I, max = M/Z.
• Neutral axis passes through the centroid.
• Required Z = M/σ_allow → pick section from tables.
• Combined: σ = P/A ± M/Z.

Going further

• Lateral-torsional buckling: Slender beams without lateral restraint can twist and buckle sideways before reaching M = σZ — codes reduce capacity accordingly.
• Curved beams: Crane hooks and C-frames have a shifted neutral axis and higher inner-fibre stress than straight-beam theory predicts.
• Composite beams: Transformed-section method converts steel into equivalent concrete (or vice versa) using the modular ratio.