Flexure formula

Bend a ruler between your fingers and one face stretches while the other squeezes — somewhere in between, a line of fibres does neither. That silent line is the neutral axis, and the stress that builds up on either side of it is what stops beams from snapping under everyday loads: floor joists, bridge girders, aircraft wing spars, the shaft of a wrench. The flexure formula is the one equation that turns an internal bending moment $M$ , found from a bending-moment diagram, into an actual stress number at any point in the cross-section. It is the single most-used design equation in beam theory, and every later idea in this chapter — section modulus, allowable-stress design, safety checks — is built directly on top of it.

The core insight

Bending stress in a beam varies linearly across the cross-section, from zero at the neutral axis to a maximum at the extreme fibre, following $\sigma = My/I$ .

The core idea

A beam in pure bending carries a constant internal moment $M$ with no shear. Under the Euler–Bernoulli assumptions — linearly elastic, homogeneous material, and plane cross-sections that stay plane and perpendicular to the beam axis after bending — the longitudinal strain grows linearly with distance $y$ from the neutral axis. For symmetric, linear-elastic bending the neutral axis passes through the centroid of the cross-section, so it is a purely geometric line, not something you have to guess or measure. Combining that linear strain with Hooke's law, and requiring the internal stress distribution to balance the applied moment $M$ , gives the flexure formula:

\sigma = \frac{My}{I}

where: $\sigma$ is the bending (normal) stress at a point, $M$ is the internal bending moment at that section (from the bending-moment diagram), $y$ is the signed distance from the neutral axis to the point, and $I$ is the second moment of area of the cross-section about the neutral (centroidal) axis. Stress is zero right on the neutral axis and grows linearly out to the extreme fibre, $y = c$ , where it peaks:

\sigma_{max} = \frac{Mc}{I}

For design work it is convenient to fold $I$ and $c$ into one property, the section modulus $S = I/c$ , so that $\sigma_{max} = M/S$ and a required section can be sized directly as $S_{req} = M_{max}/\sigma_{allow}$ . For a rectangle of width $b$ and depth $h$ : $I = bh^3/12$ , $c = h/2$ , giving $S = bh^2/6$ . For a solid circular section of radius $c$ : $I = \pi c^4/4$ , giving $S = \pi c^3/4$ .

Figure

Cross section of a beam showing tensile stress above the neutral axis and compressive stress below it varying linearly with distance from the axis

Derivation of the flexure formula

Derivation ADeriving sigma = My/I from plane sections remaining plane

The goal is to show, starting only from the assumption that cross-sections stay plane and perpendicular to the beam axis during bending, that bending stress must be proportional to distance from the neutral axis and that the constant of proportionality is exactly $M/I$ , which is the working formula every later beam-design calculation in this chapter depends on.

Step 1. Fix the geometry. Take a small beam segment bent into a circular arc of radius $R$ about the neutral axis. A fibre at distance $y$ from the neutral axis has length that changes by an amount proportional to $y$ , because the fibre at $y = 0$ (the neutral axis) does not change length at all while fibres farther away are stretched or compressed more. This gives the longitudinal strain $\epsilon = y/R$ : strain is linear in $y$ .

Step 2. Apply Hooke's law. For a linear-elastic material, stress is proportional to strain, $\sigma = E\epsilon$ , so substituting Step 1 gives $\sigma = Ey/R$ . Since $E$ and $R$ are constants across the section at a given point along the beam, this already shows $\sigma$ is linear in $y$ — zero at $y = 0$ , growing toward the outer fibres.

Step 3. Enforce equilibrium with the applied moment. The internal stress on every small area element $dA$ contributes a moment $\sigma\, y\, dA$ about the neutral axis, and the sum of all these must equal the applied bending moment $M$ : $M = \int \sigma y\, dA = \int (Ey/R) y\, dA = (E/R)\int y^2\, dA$ . The integral $\int y^2\, dA$ is by definition the second moment of area $I$ about the neutral axis, so $M = EI/R$ , which gives $E/R = M/I$ .

Step 4. Substitute back into Step 2. Replacing $E/R$ with $M/I$ in $\sigma = Ey/R$ gives $\sigma = My/I$ , the flexure formula. Setting $y = c$ , the distance to the outermost fibre, gives the extreme value $\sigma_{max} = Mc/I$ . Both forms use the same $I$ about the neutral (centroidal) axis, which is why finding $I$ correctly is the first job in any flexure problem.

Worked example

A simply supported beam has a rectangular cross-section $b = 100\ \text{mm}$ wide and $h = 200\ \text{mm}$ deep. At the section of maximum bending moment, $M = 24\ \text{kN}\,\cdot\,\text{m}$ . Find the maximum bending stress and the stress at a point $50\ \text{mm}$ above the neutral axis.

Step 1. Moment of inertia about the centroidal (neutral) axis: $I = bh^3/12 = (0.1)(0.2)^3/12 = 6.667\times10^{-5}\ \text{m}^4$ .

Step 2. Extreme fibre distance $c = h/2 = 0.1\ \text{m}$ , so $\sigma_{max} = Mc/I = (24{,}000)(0.1)/(6.667\times10^{-5}) = 36.0\times10^{6}\ \text{Pa}$ .

Step 3. Stress at $y = 0.05\ \text{m}$ : $\sigma = My/I = (24{,}000)(0.05)/(6.667\times10^{-5}) = 18.0\ \text{MPa}$ — exactly half of $\sigma_{max}$ , since $0.05\ \text{m}$ is half of $c = 0.1\ \text{m}$ and stress is linear in $y$ .

Final answer: $\sigma_{max} = 36.0\ \text{MPa}$ at the extreme fibres, and $\sigma = 18.0\ \text{MPa}$ at $y = 50\ \text{mm}$ .

Competitive stretch

A cantilever of circular cross-section, diameter $d = 80\ \text{mm}$ , carries $P = 3\ \text{kN}$ at its free end, $L = 1.5\ \text{m}$ from the fixed support. The maximum moment occurs at the support: $M_{max} = PL = (3000)(1.5) = 4500\ \text{N}\,\cdot\,\text{m}$ . With $c = d/2 = 0.04\ \text{m}$ , $I = \pi c^4/4 = 2.011\times10^{-6}\ \text{m}^4$ , so $\sigma_{max} = Mc/I = (4500)(0.04)/(2.011\times10^{-6}) = 89.5\ \text{MPa}$ . Against an allowable stress of $150\ \text{MPa}$ , the beam is safe with a factor of safety of $150/89.5 \approx 1.68$ .

Common traps

Using the full depth $h$ instead of the neutral-axis distance $c = h/2$ in $\sigma_{max} = Mc/I$ for a symmetric section.
Using the moment of inertia about the wrong axis (for example a vertical axis instead of the horizontal centroidal axis the beam actually bends about).
Forgetting that stress is zero at the neutral axis, not at the top or bottom — the linear variation is often mis-sketched as constant.
Plugging in a moment other than $M_{max}$ from the bending-moment diagram when checking whether a design is safe.

Ask the tutor: If you are unsure how to find the neutral axis, $c$ , or $I$ for a cross-section that is not a plain rectangle or circle, ask the AI tutor to walk through the centroid and second-moment-of-area steps with you before applying the flexure formula.

Why does the flexure formula only apply to pure bending, and does that limit its use in real problems

The derivation assumes a constant bending moment and no shear force, so cross-sections stay perfectly plane. Real beams almost always carry shear too, which technically warps the cross-section slightly. In practice this warping barely changes the bending stress, so engineers apply $\sigma = My/I$ at every section using the local $M(x)$ from the bending-moment diagram, even where shear is present — this is standard engineering-beam theory, not an approximation you need to worry about at this level.

Why is stress zero at the neutral axis instead of at the edges where the beam looks most stressed visually

Stress comes from strain, and strain comes from how much a fibre's length changes. The neutral axis is defined as the one layer of fibres whose length does not change at all during bending, so by Hooke's law its stress must be zero. The edges look "most stressed" to the eye because they deflect the most, but deflection and local fibre stress are different things — the edges have the largest y, not the largest curvature.

Do I need calculus to find I for every cross-section, or can I always use a formula

For the standard shapes in this course — rectangles and solid circles — you use the ready-made formulas $I = bh^3/12$ and $I = \pi c^4/4$ directly; the calculus that produces them is done once and then reused. For composite shapes built from these standard pieces, you combine them with the parallel-axis theorem from the earlier moment-of-inertia chapter rather than integrating from scratch each time.