5.2 Bending Stress in Beams
- Explain why bending strain — and so stress — varies linearly across a beam's depth.
- Apply the flexure formula \(\sigma = My/I\) to find the stress at any fibre.
- Locate the neutral axis and tell tension from compression.
- Use the section modulus \(S = I/c\) to size a beam against an allowable stress.
- Predict where and when a beam first yields under bending.
A transverse load makes a beam bend. On the concave side the fibres are squeezed shorter — compression; on the convex side they are pulled longer — tension. Between them lies one surface whose length never changes: the neutral surface. Its trace on any cross-section is the neutral axis. The bending moment \(M\) at a section is exactly what those internal normal stresses add up to — find how \(\sigma\) is distributed and you can find \(M\), and vice versa. The rest of this section is that distribution, worked out and put to use.
Figure — A simply-supported beam under a growing load \(P\). Top fibres shorten, bottom fibres stretch; the neutral axis (dashed) keeps its original length.
5.2.1 Plane Sections Stay Plane
As the beam bends, each flat cross-section stays flat and simply rotates about the neutral axis. A fibre a distance \(y\) from that axis therefore stretches in direct proportion to \(y\): if \(\rho\) is the local radius of curvature, the normal strain is
$$ \varepsilon(y) = \frac{y}{\rho}. $$Apply Hooke's law \(\sigma = E\varepsilon\) for a linear-elastic material:
$$ \sigma(y) = E\,\varepsilon(y) = \frac{E}{\rho}\,y \quad\Longrightarrow\quad \sigma \text{ is linear in } y. $$That single line — linear strain times a constant modulus — is the whole reason bending stress grows in a straight line from zero at the neutral axis to a maximum at the outer fibre.
5.2.2 The Flexure Formula
Requiring the linear stress field to sum to the applied moment — that is, \(M = \int_A \sigma\,y\,dA\) — pins the constant \(E/\rho\) and gives
$$ \sigma = \frac{M\,y}{I}, $$where \(I = \int_A y^2\,dA\) is the second moment of area about the neutral axis. The extreme fibre \(y = c\) carries the largest stress.
Figure — Linear stress distribution over the section depth — zero at the neutral axis, equal-and-opposite at the extreme fibres.
5.2.3 Where the Neutral Axis Sits
In pure bending there is no net axial force, so the normal stresses must sum to zero over the area:
$$ \int_A \sigma\,dA = \frac{M}{I}\int_A y\,dA = 0 \;\Longrightarrow\; \int_A y\,dA = 0. $$That last integral is zero only when \(y\) is measured from the centroid. So the neutral axis always passes through the centroid of the cross-section, and for a symmetric section the centroid sits at mid-depth, so \(c = h/2\) top and bottom.
A T-section has a \(120 \times 30\,\text{mm}\) top flange sitting on a \(30 \times 120\,\text{mm}\) web. Locate the neutral axis — the centroid — measured from the base.
Solution
Two rectangles of equal area \(A_1 = A_2 = 3600\,\text{mm}^2\); flange centroid \(y_1 = 135\,\text{mm}\), web centroid \(y_2 = 60\,\text{mm}\). Take moments of area about the base:
$$ \bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{3600(135) + 3600(60)}{7200} = 97.5\,\text{mm}. $$The neutral axis lies \(97.5\,\text{mm}\) above the base — pulled toward the flange, where the area is banked.
An inverted-T has a \(100 \times 20\,\text{mm}\) flange at the bottom and a \(20 \times 100\,\text{mm}\) web standing on it. How far above the base is the neutral axis? Both parts have area \(2000\,\text{mm}^2\); the flange centroid sits \(10\,\text{mm}\) up, the web centroid \(70\,\text{mm}\) up.
Answer
5.2.4 Second Moment of Area
\(I\) measures how the area is spread about the neutral axis. Because each element is weighted by \(y^2\), material far from the axis counts far more:
$$ I = \int_A y^2\,dA, \qquad I_{\text{rectangle}} = \frac{b\,h^3}{12}. $$The cube on \(h\) is the punchline: double a rectangle's depth and \(I\) grows \(2^3 = 8\times\) for the same width — the fibres you added are the ones farthest from the axis. A large \(I\) means a stiff, strong section not because there is more material, but because the material is placed where the lever arm \(y\) is longest.
Because \(I\) grows with the cube of distance from the axis, area parked next to the neutral axis barely moves \(I\) at all, while the same area moved out to the flanges does most of the work. An I-beam is the second-moment argument built in steel: thin flanges carry the bending, and a thin web mostly just holds them the right distance apart.
Find \(I\) about the neutral axis for a rectangular section \(b = 50\,\text{mm}\) wide and \(h = 120\,\text{mm}\) deep.
Solution
For a rectangle bent about its centroidal axis, \(I = bh^3/12\). Substitute:
$$ I = \frac{(50)(120)^3}{12} = \frac{(50)(1{,}728{,}000)}{12} = 7.2\times10^{6}\,\text{mm}^4 = 7.2\times10^{-6}\,\text{m}^4. $$The depth enters as \(h^3\) — the same width at \(h = 240\,\text{mm}\) would give eight times this \(I\).
A symmetric I-section is \(100\,\text{mm}\) wide and \(200\,\text{mm}\) deep, with \(20\,\text{mm}\) flanges and a \(20\,\text{mm}\) web. Find \(I\) about the neutral axis.
Solution
Take the full \(100 \times 200\) box and remove the two side voids (each \(40 \times 160\,\text{mm}\)):
$$ I = \frac{100(200)^3}{12} - \frac{80(160)^3}{12} \approx 3.94\times10^{7}\,\text{mm}^4. $$Nearly all of that \(I\) comes from the two flanges — the material farthest from the axis.
A rectangular beam is \(b = 40\,\text{mm}\) wide and \(h = 120\,\text{mm}\) deep. Find \(I\) about the neutral axis.
Answer
5.2.5 Section Modulus
Evaluate the flexure formula at the extreme fibre \(y = c\) and collect the geometry into a single number, the section modulus \(S = I/c\):
$$ \sigma_{\max} = \frac{M\,c}{I} = \frac{M}{S}, \qquad S = \frac{I}{c}. $$Design turns into one line: to keep \(\sigma_{\max}\) below an allowable stress \(\sigma_{\text{allow}}\), require \(S \ge M/\sigma_{\text{allow}}\). Handbooks tabulate \(S\) for every rolled shape for exactly this reason.
| Section | \(I\) | \(S = I/c\) |
|---|---|---|
| Rectangle | \(bh^3/12\) | \(bh^2/6\) |
| Solid circle (dia. \(d\)) | \(\pi d^4/64\) | \(\pi d^3/32\) |
A rectangular beam \(50\,\text{mm} \times 150\,\text{mm}\) (width \(b\), depth \(h\)) carries a bending moment \(M = 12\,\text{kN}\cdot\text{m}\). Find the maximum bending stress.
Solution
The section modulus of a rectangle is \(S = bh^2/6\), so
$$ S = \frac{(0.05)(0.15)^2}{6} = 1.875\times10^{-4}\,\text{m}^3. $$ $$ \sigma_{\max} = \frac{M}{S} = \frac{12\times10^{3}}{1.875\times10^{-4}} = 64\,\text{MPa}. $$The extreme fibre — the top and bottom faces, \(75\,\text{mm}\) from the neutral axis — sees this full \(64\,\text{MPa}\); every fibre closer in sees proportionally less.
A rectangular beam \(75\,\text{mm} \times 200\,\text{mm}\) (width \(\times\) depth) carries a bending moment \(M = 20\,\text{kN}\cdot\text{m}\). What is the maximum bending stress?
Answer
The \(50 \times 150\,\text{mm}\) beam of Worked Example 5.2.4 carries \(M = 12\,\text{kN}\cdot\text{m}\), for which \(I = 1.406\times10^{-5}\,\text{m}^4\). Find the stress at a fibre \(y = 25\,\text{mm}\) from the neutral axis.
Solution
Apply the flexure formula at \(y = 0.025\,\text{m}\):
$$ \sigma = \frac{M\,y}{I} = \frac{(12\times10^{3})(0.025)}{1.406\times10^{-5}} \approx 21.3\,\text{MPa}. $$At one-third of the way to the face (\(y = c/3\)) the stress is one-third of \(\sigma_{\max} = 64\,\text{MPa}\) — the straight-line law in action.
For the same \(50 \times 150\,\text{mm}\) beam (\(M = 12\,\text{kN}\cdot\text{m}\), \(I = 1.406\times10^{-5}\,\text{m}^4\)), what is the bending stress at \(y = 50\,\text{mm}\) from the neutral axis? Stress is linear in \(y\), so \(\sigma = \sigma_{\max}(y/c)\) with \(\sigma_{\max} = 64\,\text{MPa}\) and \(c = 75\,\text{mm}\).
Answer
5.2.6 Shape at Equal Area
Give three cross-sections the same area, \(A = 7500\,\text{mm}^2\), and compare their section moduli. Only the distribution of that area differs.
| Cross-section | \(S = I/c\) | \(S\ (\text{mm}^3)\) |
|---|---|---|
| Rectangle, \(50 \times 150\) — deep | \(bh^2/6\) | 187,500 |
| Solid circle, \(d \approx 97.7\,\text{mm}\) | \(\pi d^3/32\) | 91,600 |
| Rectangle, \(150 \times 50\) — flat | \(bh^2/6\) | 62,500 |
Turning the same rectangle on edge — deep, not flat — roughly triples its bending resistance. For a rectangle \(S = bh^2/6\) scales with \(h^2\), so orientation alone changes \(S\) threefold: the deep rectangle beats the flat one and the circle for the very same amount of material.
5.2.7 Deeper Is Stronger
Hold the area fixed and let the section grow taller — the width \(b = A/h\) must shrink to compensate. Then
$$ S = \frac{b\,h^2}{6} = \frac{A\,h}{6} \;\propto\; h. $$Figure — Constant area, growing depth. As \(h\) rises the section modulus \(S = Ah/6\) rises with it.
A rectangular timber beam of fixed width \(b = 100\,\text{mm}\) must carry \(M = 15\,\text{kN}\cdot\text{m}\) without exceeding \(\sigma_{\text{allow}} = 12\,\text{MPa}\). Find the minimum depth \(h\).
Solution
Sizing requires \(S \ge M/\sigma_{\text{allow}}\); for a rectangle \(S = bh^2/6\), so solve for \(h\):
$$ S_{\text{req}} = \frac{15\times10^{3}}{12\times10^{6}} = 1.25\times10^{-3}\,\text{m}^3 = 1.25\times10^{6}\,\text{mm}^3. $$ $$ h = \sqrt{\frac{6\,S_{\text{req}}}{b}} = \sqrt{\frac{6(1.25\times10^{6})}{100}} \approx 274\,\text{mm}. $$Round up to a stock size: a \(100 \times 275\,\text{mm}\) beam clears the allowable stress; anything shallower does not.
A steel beam must carry \(M = 25\,\text{kN}\cdot\text{m}\) with an allowable stress \(\sigma_{\text{allow}} = 150\,\text{MPa}\). What minimum section modulus \(S\) must the chosen shape provide?
Answer
5.2.8 When a Beam Fails
A beam fails in bending when the extreme-fibre stress reaches the material's strength — exactly where \(\sigma\) is largest. Take structural steel, yield strength \(\sigma_Y \approx 250\,\text{MPa}\). The beam of Worked Example 5.2.4 runs at \(\sigma_{\max} = 64\,\text{MPa}\), so the factor of safety is
$$ n = \frac{\sigma_Y}{\sigma_{\max}} = \frac{250}{64} \approx 3.9. $$Raise the load until \(\sigma_{\max} = \sigma_Y\) and yielding begins. The moment at first yield is
$$ M_Y = \sigma_Y\,S = (250\times10^{6})(1.875\times10^{-4}) \approx 46.9\,\text{kN}\cdot\text{m}. $$Beyond \(M_Y\) the outer fibres yield first and the plastic zone spreads inward toward the neutral axis — failure starts at the top and bottom faces, never at the centre.
A beam with section modulus \(S = 5\times10^{-4}\,\text{m}^3\) is made of steel with yield strength \(\sigma_Y = 250\,\text{MPa}\). It currently carries \(M = 40\,\text{kN}\cdot\text{m}\). Find (a) the first-yield moment \(M_Y\) and (b) the factor of safety \(n\). \(M_Y = \sigma_Y S\), and \(n = M_Y/M\) (equivalently \(\sigma_Y/\sigma_{\max}\)).
Answer
5.2.9 Reading the Stress: Sign and Magnitude
sign — for a sagging (positive) moment, fibres above the neutral axis shorten (compression, \(\sigma \lt 0\)) and fibres below stretch (tension, \(\sigma \gt 0\)); flip the moment and the two zones swap.
magnitude — because \(\sigma\) is odd in \(y\), a symmetric section feels the same magnitude at equal \(|y|\): top-fibre compression equals bottom-fibre tension. Asymmetric sections (a T, a channel) do not — the fibre with the larger \(c\) governs.
Always check the fibre with the largest \(|y|\); for an unsymmetric section that may be the tension face, not the compression face. The T-section worked out earlier makes the point concrete.
The T-section of Worked Example 5.2.1 has its neutral axis \(97.5\,\text{mm}\) from the base (\(c_{\text{bot}} = 97.5\,\text{mm}\), \(c_{\text{top}} = 52.5\,\text{mm}\)) and \(I = 1.47\times10^{-5}\,\text{m}^4\). For a sagging moment \(M = 10\,\text{kN}\cdot\text{m}\), find the stress at the top and bottom fibres.
Solution
Apply \(\sigma = Mc/I\) at each face:
$$ \sigma_{\text{bot}} = \frac{(10\times10^{3})(0.0975)}{1.47\times10^{-5}} \approx 66\,\text{MPa (tension)}. $$ $$ \sigma_{\text{top}} = \frac{(10\times10^{3})(0.0525)}{1.47\times10^{-5}} \approx 36\,\text{MPa (compression)}. $$The fibre with the larger \(c\) — here the bottom — carries the larger stress and governs the design. On an unsymmetric section, the tension face can be the critical one.
5.2.10 What to Carry Forward
The one idea: \(\sigma = My/I\) is linear in \(y\) — zero at the neutral (centroidal) axis, largest at the extreme fibre. Everything else follows from that straight line.
The sizing recipe: compute \(\sigma_{\max} = M/S\) with \(S = I/c\); require \(S \ge M/\sigma_{\text{allow}}\). Deep sections and I-beams win because they raise \(S\) by banking area far from the axis.
Next: §5.3 Shear Stress in Beams — the companion stress that acts along the section.