Beam Bending Stress⚠ unverified

Understanding the Parameters

• Bending moment $M$ — the internal moment the section must resist, found from the loads and supports (e.g. from Beam Moment At X or Continuous Beam Moment). Stress is directly proportional to $M$: double the moment, double the stress.

• Distance from the neutral axis $y$ — the through-depth coordinate. At the neutral axis ($y=0$) the bending stress is exactly zero; it grows linearly to a maximum at the outermost fibre. This linear distribution is the geometric heart of beam theory. Using the extreme fibre $y = c$ gives the peak stress that governs design.

• Second moment of area $I$ — the section's resistance to bending, measuring how far the material is spread from the neutral axis (see Rectangular Moment of Inertia, Moment Of Inertia I Beam). Because it appears in the denominator, a larger $I$ lowers stress; because $I$ grows with the cube of depth, making a beam deeper is dramatically more effective than making it wider.

The grouping $I/y_{\max} = S$ is the section modulus (Section Modulus), so the peak stress is often written $\sigma_{\max} = M/S$ — a compact form for design tables.

Derivation (Approaching a Proof)

The flexure formula follows from Euler–Bernoulli beam kinematics under the assumptions that plane sections remain plane and normal to the neutral axis, the material is linear-elastic, and deflections are small.

1. Kinematics. A beam bent to radius of curvature $\rho$ stretches a fibre at height $y$ by a strain proportional to its distance from the neutral axis (see Beam Strain): $$\varepsilon = \frac{y}{\rho}.$$

2. Constitutive law. For linear-elastic material, Hooke's law gives the stress: $$\sigma = E\varepsilon = \frac{E y}{\rho}.$$

3. Equilibrium of the internal moment. The stresses over the cross-section must sum to the applied moment $M$. Taking moments of the fibre forces $\sigma\,\mathrm{d}A$ about the neutral axis: $$M = \int_A \sigma\, y\, \mathrm{d}A = \frac{E}{\rho}\int_A y^2\, \mathrm{d}A = \frac{E I}{\rho},$$ where $I \equiv \int_A y^2\,\mathrm{d}A$ is the second moment of area by definition. This is the moment–curvature relation $M = EI/\rho$.

4. Eliminate $\rho$. From step 2, $E/\rho = \sigma/y$; substituting into step 3 gives $M = (\sigma/y) I$, i.e. $$\sigma = \frac{M y}{I}.$$

The zero-net-axial-force condition $\int_A \sigma\,\mathrm{d}A = 0$ is what forces the neutral axis to pass through the centroid of the section — the reason $I$ is taken about the centroidal axis.

Dimensional check. $[\sigma] = \dfrac{\text{N}\cdot\text{m} \cdot \text{m}}{\text{m}^4} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$. ✓

History and Development

The problem of the strength of beams occupied Galileo (1638), who got the depth-dependence qualitatively but the stress distribution wrong. Euler and Daniel Bernoulli established the moment–curvature relation and the elastica in the 18th century. The correct linear stress distribution and the flexure formula in its modern form are credited to Claude-Louis Navier (1820s), whose work founded the engineering theory of elasticity. It remains the backbone of strength-of-materials courses and every structural design code (AISC, Eurocode).

Related Concepts: Rectangular Moment of Inertia, Section Modulus, Beam Strain, Moment Of Inertia I Beam, Continuous Beam Moment, Shear Stress, Von Mises Stress

Notes: Valid for linear-elastic, symmetric bending about a principal axis. For combined loading use Von Mises Stress; for unsymmetric or composite sections use the generalised flexure formula or a transformed section. Beyond yield, the plastic section modulus Plastic Section Modulus governs.