Mechanics of Materials · Chapter 5 · Bending

Bending Stress in Beams

How a bending moment becomes stress that grows straight-line with distance from the neutral axis — and how that one fact sizes every beam.

bookSHelf · deck style sample · zhangzara-grid · cobalt grid palette

Titlepage in the Zhangzara grid grammar: italic-serif title on the graph-paper canvas, a cobalt double rule, a one-sentence lede, hairline rule, mono byline; the pixel-glitch scanline stair carries the decorative signature.

Explain why bending strain — and so stress — varies linearly across the depth premise
Apply the flexure formula $\sigma = My/I$ to find the stress at any fibre definition
Locate the neutral axis and tell tension from compression figure 1
Use the section modulus $S = I/c$ to size a beam against an allowable stress table 1
Predict where and when a beam first yields under bending failure case

A ledger-style outline: mono leading-zero numerals, italic-serif entries on hairline rules, a right-aligned tag column. Each objective reveals on click.

A transverse load makes a beam bend. On the concave (top) side the fibres are squeezed shorter — compression; on the convex (bottom) 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.

Figure 1: A simply-supported beam under a growing load $P$ . Top fibres shorten, bottom fibres stretch; the neutral axis (dashed) keeps its original length.

Figure-right convention with a live manim animation: prose LEFT, the deflecting-beam video RIGHT. The moment-as-resultant idea reveals on click.

Premise 5.2.1 — Euler–Bernoulli kinematics

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}.

Now 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.

The geometric premise: plane sections rotate, strain is proportional to y, Hooke's law makes stress proportional to y. Two clicks build the chain, then the takeaway.

Definition 5.2.2 — Bending (flexure) stress

Requiring the linear stress field to sum to the applied moment $M$ — 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.

Stress is zero at the neutral axis and grows straight-line toward each face — watch it sweep the depth.

Figure 2: Linear stress distribution over the section depth — zero at the neutral axis, equal-and-opposite at the extreme fibres.

Definition box LEFT, the live stress-distribution animation RIGHT. The fibre marker sweeps top to bottom so students see the linear profile and the sign change at the neutral axis.

Result 5.2.3 — The neutral axis is centroidal

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.

The 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.

For a symmetric section the centroid is at mid-depth, so $c = h/2$ top and bottom.

Figure 3: The neutral axis passes through the centroid; for a symmetric shape it lies at mid-depth.

Static inline SVG (figure-right): the zero-net-force argument places the neutral axis at the centroid; mid-depth for a symmetric section.

Worked Example 5.2.1 — Centroid of a T-section

A T-section has a $120\times30\ \text{mm}$ top flange sitting on a $30\times120\ \text{mm}$ 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}.

So the neutral axis lies $97.5\ \text{mm}$ above the base — pulled toward the flange, where the area is banked.

Figure 3a: The centroid of a T-section lies toward the flange; the neutral axis passes through it.

Commit-first worked example: take moments of area about the base to locate the centroid ȳ = 97.5 mm. SVG of the T-section RIGHT with the neutral axis marked.

Try it now 5.2.1 — your turn

Find the neutral axis

An inverted-T has a $100\times20\ \text{mm}$ flange at the bottom and a $20\times100\ \text{mm}$ web standing on it. How far above the base is the neutral axis?

Hint: both parts have area $2000\ \text{mm}^2$ ; the flange centroid sits $10\ \text{mm}$ up, the web centroid $70\ \text{mm}$ up. Take the area-weighted average.

Answer

\bar{y} = \frac{2000(10)+2000(70)}{4000} = 40\ \text{mm above the base.}

Learner locates the centroid of an inverted-T; answer ȳ = 40 mm revealed behind a click after an area-weighted-average hint.

Definition 5.2.4 — Second moment of area $I$

$I$ 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.

Read it: 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.

I = ∫ y² dA; the y² weighting is why deep sections dominate — bh³/12 for a rectangle grows with the cube of depth.

Worked Example 5.2.2 — Second moment of area of a rectangle

Find the second moment of area $I$ 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, $I = \dfrac{b h^3}{12}$ . 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.

Read it: the depth enters as $h^3$ — the same width at $h = 240\ \text{mm}$ would give eight times this $I$ .

Direct application of I = bh³/12 for a 50×120 mm rectangle → I = 7.2×10⁶ mm⁴. The h³ dependence is underscored in the takeaway.

Worked Example 5.2.3 — Second moment by subtraction

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 $100\times200$ box and remove the two side voids (each $40\times160$ ):

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.

Figure 3b: Flanges banked far from the axis do almost all the work; the box-minus-voids trick gives $I$ quickly.

Composite I by subtraction: full 100×200 box minus two 40×160 voids → I ≈ 3.94×10⁷ mm⁴. SVG of the I-section RIGHT.

Try it now 5.2.2 — your turn

Compute the second moment $I$

A rectangular beam is $b = 40\ \text{mm}$ wide and $h = 120\ \text{mm}$ deep. Find its second moment of area $I$ about the neutral axis.

Hint: use $I = \dfrac{b h^3}{12}$ with $(120)^3 = 1\,728\,000\ \text{mm}^3$ .

Answer

I = \frac{(40)(120)^3}{12} = 5.76\times10^{6}\ \text{mm}^4.

Learner applies I = bh³/12 to a 40×120 mm rectangle → I = 5.76×10⁶ mm⁴, revealed on click.

Definition 5.2.5 — Section modulus $S$

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.

S = I/c folds the whole cross-section into one number; sizing is S ≥ M/σ_allow. Handbooks list S for rolled shapes.

Worked Example 5.2.4 — Maximum stress in a rectangular beam

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 $S = \dfrac{b h^2}{6}$ , so

S = \frac{(0.05)(0.15)^2}{6} = 1.875\times10^{-4}\,\text{m}^3, \qquad \sigma_{\max} = \frac{M}{S} = \frac{12\times10^{3}}{1.875\times10^{-4}} = 64\,\text{MPa}.

Commit-first: the prompt shows, then one click reveals the full solution. S = bh²/6 gives σ_max = M/S = 64 MPa.

Try it now 5.2.3 — your turn

Find the maximum stress

A rectangular beam $75\ \text{mm}\times200\ \text{mm}$ (width $\times$ depth) carries a bending moment $M = 20\ \text{kN}\cdot\text{m}$ . What is the maximum bending stress?

Hint: the section modulus of a rectangle is $S = \dfrac{b h^2}{6}$ ; then $\sigma_{\max} = M/S$ .

Answer

\begin{aligned} S &= \frac{(0.075)(0.200)^2}{6} = 5\times10^{-4}\ \text{m}^3, \\ \sigma_{\max} &= \frac{M}{S} = \frac{20\times10^{3}}{5\times10^{-4}} = 40\ \text{MPa}. \end{aligned}

Learner computes S = bh²/6 then σ_max = M/S = 40 MPa for a 75×200 mm beam under 20 kN·m.

Worked Example 5.2.5 — The flexure formula away from the face

The $50\times150\ \text{mm}$ 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}.

Read it: 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.

σ = My/I at an interior fibre y = 25 mm gives 21.3 MPa = one-third of the 64 MPa extreme-fibre stress, reinforcing linearity.

Try it now 5.2.4 — your turn

Stress at a fibre

For the same $50\times150\ \text{mm}$ 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?

Hint: stress is linear in $y$ , so $\sigma = \sigma_{\max}\,(y/c)$ with $\sigma_{\max} = 64\ \text{MPa}$ and $c = 75\ \text{mm}$ .

Answer

\sigma = \frac{(12\times10^{3})(0.050)}{1.406\times10^{-5}} = 64\cdot\frac{50}{75} \approx 42.7\ \text{MPa}.

Learner finds σ at y = 50 mm = two-thirds of c, giving 42.7 MPa via linearity.

Give three cross-sections the same area, $A = 7500\,\text{mm}^2$ , and compare their section moduli $S = I/c$ . Only the distribution of that area differs.

Cross-section (equal area $A = 7500\,\text{mm}^2$ )	$S = I/c$	$S\ (\text{mm}^3)$
Rectangle, $50 \times 150$ — deep	$b h^2/6$	187,500
Solid circle, $d \approx 97.7$	$\pi d^3/32$	91,600
Rectangle, $150 \times 50$ — flat	$b h^2/6$	62,500

Table 1: Section modulus of three equal-area cross-sections. Turning the same rectangle on edge (deep, not flat) roughly triples its bending resistance.

Read it: 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.

Booktabs discipline: cobalt top/bottom rules, one hairline under the header, no vertical rules; the winning cell is bold cobalt. Concrete equal-area numbers make the depth argument land.

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.

At constant area the section modulus climbs in direct proportion to depth. Every millimetre of extra height buys bending resistance for free — the geometry, not the material, is doing the work.

Figure 4: Constant area, growing depth. As $h$ rises the section modulus $S = Ah/6$ rises with it.

Live section-modulus animation RIGHT: constant area, deeper section, S ∝ h. The takeaway reveals on click.

Worked Example 5.2.6 — Choosing the depth of a timber beam

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 $S = b h^2/6$ , so solve for $h$ :

\begin{aligned} 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}. \end{aligned}

Read it: round up to a stock size — a $100\times275\ \text{mm}$ beam clears the allowable stress; anything shallower does not.

Design in reverse: required S = M/σ_allow = 1.25×10⁻³ m³ → h = √(6S/b) ≈ 274 mm, round up to 275 mm.

Try it now 5.2.5 — your turn

What section modulus is needed?

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?

Hint: keep $\sigma_{\max} = M/S \le \sigma_{\text{allow}}$ , so $S \ge M/\sigma_{\text{allow}}$ .

Answer

S \ge \frac{25\times10^{3}}{150\times10^{6}} = 1.67\times10^{-4}\ \text{m}^3 = 167\times10^{3}\ \text{mm}^3.

Learner finds the required S ≥ M/σ_allow = 1.67×10⁻⁴ m³ (167 cm³); pick any rolled shape at or above it.

The headline result

Put material where the stress is

Because $\sigma = My/I$ grows linearly with $y$ , the fibres far from the neutral axis carry almost all the load — and the ones near it barely work. An I-beam answers this literally: it banks its area in top and bottom flanges and leaves a thin web between.

For the $50 \times 150\,\text{mm}$ beam the extreme fibre reaches σ_max = 64 MPa — the number you compare against the material's strength.

Move that same area outward into flanges and $I$ — hence $S$ — climbs sharply, dropping $\sigma_{\max}$ for the very same moment. That is the whole engineering argument for the I-section: maximum $S$ per kilogram of steel.

The key result in a ruled, cobalt-topped box and a boxed value — stated inline, never a lone 180px numeral. The I-beam footnote reveals last.

Failure Case 5.2.1 — First yield in the $50 \times 150$ beam

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 $\sigma_Y \approx 250\,\text{MPa}$ . Our beam 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.

Failure case with real steel numbers: factor of safety ≈ 3.9; first-yield moment M_Y = σ_Y·S ≈ 46.9 kN·m; yielding starts at the extreme fibre and spreads inward.

Try it now 5.2.6 — your turn

Factor of safety

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 $n$ .

Hint: $M_Y = \sigma_Y S$ , and $n = M_Y/M$ (equivalently $\sigma_Y/\sigma_{\max}$ ).

Answer

\begin{aligned} M_Y &= \sigma_Y S = (250\times10^{6})(5\times10^{-4}) = 125\ \text{kN}\cdot\text{m}, \\ n &= \frac{M_Y}{M} = \frac{125}{40} \approx 3.1. \end{aligned}

Learner computes M_Y = σ_Y·S = 125 kN·m and n = M_Y/M ≈ 3.1 for a beam with S = 5×10⁻⁴ m³.

Sign — which side pulls

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 — symmetric means equal

Because $\sigma$ is odd in $y$ , a symmetric section feels the same magnitude at equal $|y|$ : the top-fibre compression equals the bottom-fibre tension. Asymmetric sections (a T, a channel) do not — the fibre with the larger $c$ governs.

Design habit: always check the fibre with the largest $|y|$ ; for an unsymmetric section that may be the tension face, not the compression face.

Two cards: sign convention (compression above, tension below) and equal magnitude at equal |y| for symmetric sections; asymmetric shapes are governed by the larger c.

Worked Example 5.2.7 — Unequal stresses in a T-section

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:

\begin{aligned} \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)}. \end{aligned}

Read it: 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.

Ties the T-section back: σ = Mc/I gives 66 MPa tension at the bottom (larger c) vs 36 MPa compression at the top — the larger-c fibre governs.

5.2

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. Back to start.

Closing argument in two ruled cards — the core idea under a cobalt top rule, the sizing recipe beside it — over a ghost section numeral. The next-step line reveals last.

Attribution — openly licensed content

This section adapts foundational material on stress, strain, and elastic moduli from OpenStax, University Physics Volume 1 (Chapter 12, Static Equilibrium and Elasticity), together with the standard linear-elastic (Euler–Bernoulli) beam theory that gives the flexure formula $\sigma = My/I$ .

OpenStax content is published by Rice University under the Creative Commons Attribution 4.0 International (CC BY 4.0) license. Access the original for free at openstax.org.

Figures 1, 2 and 4 were redrawn for bookSHelf as dual-theme manim animations. Deck styled in the zhangzara-grid visual language (adapted from Open Design's Zhangzara html-ppt templates, MIT).

Colophon: cites OpenStax University Physics Vol 1 Ch 12 (stress/strain/elastic moduli) under CC BY 4.0 plus standard Euler–Bernoulli beam theory; notes the manim figures and the zhangzara-grid style provenance.

Bending Stress in Beams

Objectives

What bending does

Plane sections stay plane

The flexure formula

Where the neutral axis sits

Locate the neutral axis

Find the neutral axis

Second moment of area

Compute I for a rectangle

Compute I for an I-section

Compute the second moment $I$

Section modulus

Worked example

Find the maximum stress

Stress at an interior fibre

Stress at a fibre

Shape at equal area

Deeper is stronger

Size a beam to an allowable stress

What section modulus is needed?

When a beam fails

Factor of safety

Reading the stress

Sign — which side pulls

Magnitude — symmetric means equal

Which fibre governs?

What to carry forward

The one idea

The sizing recipe

Sources & license

Bending Stress in Beams

Objectives

What bending does

Plane sections stay plane

The flexure formula

Where the neutral axis sits

Locate the neutral axis

Find the neutral axis

Second moment of area

Compute I for a rectangle

Compute I for an I-section

Compute the second moment III

Section modulus

Worked example

Find the maximum stress

Stress at an interior fibre

Stress at a fibre

Shape at equal area

Deeper is stronger

Size a beam to an allowable stress

What section modulus is needed?

When a beam fails

Factor of safety

Reading the stress

Sign — which side pulls

Magnitude — symmetric means equal

Which fibre governs?

What to carry forward

The one idea

The sizing recipe

Sources & license

Compute the second moment $I$