Bending Stress in Beams — Flexure Formula Derivation, Section Modulus, and Examples

When a beam bends, its fibres on one side stretch and on the other side compress, producing bending (flexural) stresses. The relationship that governs them — the flexure formula M/I = σ/y = E/R — is the single most used equation in structural design. This article derives it cleanly, explains every term, and applies it to worked examples.

Assumptions of the Theory of Simple Bending

• The beam material is homogeneous, isotropic and obeys Hooke's law (stress ∝ strain).
• The beam is initially straight and has a constant cross-section.
• Plane sections before bending remain plane after bending (Bernoulli's assumption).
• The radius of curvature is large compared with the cross-sectional dimensions.
• The beam bends about a principal axis, and stress stays within the elastic limit.

Derivation

Consider a small length of beam bent into an arc of radius R subtending an angle dθ at the centre of curvature. The neutral axis (NA) is the layer whose length does not change.

A fibre at distance y from the NA originally had length R·dθ. After bending it becomes (R + y)·dθ. Its strain is:

ε = change in length / original length = (y·dθ)/(R·dθ) = y / R

By Hooke's law, σ = E·ε, so:

σ = E · y / R   →   σ / y = E / R     ...(1)

This shows bending stress varies linearly with distance from the neutral axis.

Now relate this to the applied moment. The force on an elemental area dA is σ·dA, and its moment about the NA is σ·dA·y. Summing over the whole section gives the resisting moment M:

M = ∫ σ · y dA = ∫ (E·y/R) · y dA = (E/R) ∫ y² dA = (E/R) · I

where I = ∫y²dA is the second moment of area (moment of inertia) about the NA. Therefore:

M / I = E / R     ...(2)

Combining (1) and (2) gives the complete flexure formula:

M / I = σ / y = E / R

Where Is the Neutral Axis?

For pure bending, the net axial force is zero: ∫σ dA = ∫(E·y/R) dA = 0, which requires ∫y dA = 0. That condition is satisfied only when the neutral axis passes through the centroid of the cross-section. So for elastic bending, the neutral axis is always the centroidal axis.

Section Modulus

The maximum bending stress occurs at the extreme fibre, y = y_max:

σ_max = M · y_max / I = M / Z,   where Z = I / y_max

Z is the section modulus. Designers compare beams by Z because a larger Z gives a lower stress for the same moment.

Worked Example 1 — Rectangular Beam

A timber beam 200 mm wide × 300 mm deep carries a maximum bending moment of 30 kN·m. Find the maximum bending stress.

1. I = bd³/12 = 200 × 300³ / 12 = 4.5 × 10⁸ mm⁴
2. y_max = 300/2 = 150 mm
3. Z = I / y_max = 4.5×10⁸ / 150 = 3.0 × 10⁶ mm³
4. σ_max = M / Z = 30 × 10⁶ / 3.0 × 10⁶ = 10 N/mm² (MPa)

The top fibre is in compression at 10 MPa, the bottom in tension at 10 MPa, and the stress is zero at mid-depth.

Worked Example 2 — Choosing a Steel Section

A simply supported steel beam spans 5 m with a UDL of 24 kN/m. The permissible bending stress is 165 MPa. Find the required section modulus.

1. M_max = wL²/8 = 24 × 5² / 8 = 75 kN·m
2. Z_req = M / σ_perm = 75 × 10⁶ / 165 = 4.55 × 10⁵ mm³ = 455 cm³
3. Select a rolled section (e.g. ISMB 250, Z ≈ 410 cm³ — slightly low; ISMB 300, Z ≈ 573 cm³ — adequate). Adopt ISMB 300.

Why I-Sections Dominate Beam Design

Bending stress grows with distance from the neutral axis, so material near the NA does little work. An I-section concentrates material in the flanges (far from the NA), maximising I and Z for a given weight. That is the entire reason steel beams look like an "I".

Common Mistakes

• Using I about the wrong axis (it must be about the centroidal axis perpendicular to the bending plane).
• Forgetting to convert units — mix of kN·m and mm gives wrong stresses by factors of 1000.
• Confusing section modulus Z (elastic) with plastic modulus Z_p used in limit-state steel design.