Bending stress in beams

Engineering · First-year engineering · about 10 min

A beam carrying transverse loads bends, so fibres on one side shorten while fibres on the other side lengthen. Bending stress quantifies the normal stress caused by that deformation. Exam questions typically ask for the largest stress and whether it is tensile or compressive.

The key idea

For a straight, homogeneous, linearly elastic beam in small-deflection bending,

\sigma_x(y)=-\frac{My}{I}

Here $M$ is the signed bending moment at the section, $I$ is the second moment of area about its centroidal neutral axis, and $y$ is measured upwards from that axis. Positive sagging $M$ gives compression above the axis and tension below it.

Create a section-view schematic with one tall upright rectangular beam cross-section at the left, width b and depth h. Draw its horizontal neutral axis across mid-depth as a dashed line labelled NA. On the right, draw a normal-stress diagram sharing the same vertical positions: a left-pointing horizontal triangular wedge above the neutral axis whose width grows steadily from zero at mid-depth to maximum at the top (compression), and a right-pointing triangular wedge below the axis growing to maximum at the bottom (tension). Include a thin vertical reference line at zero stress and label the two wedges. Add a vertical arrow y pointing upward from the neutral axis. The triangles represent signed bending normal stress, not shear force. — Positive sagging bending gives zero stress at the neutral axis and opposite signs at the faces.

Read the stress distribution

Bending normal stress varies linearly with distance from the neutral axis. At $y=0$ , bending stress is zero; at the extreme fibres, its magnitude is largest. For a symmetric rectangular section of height $h$ , the extreme-fibre distance is $c=h/2$ , so $|\sigma_{\max}|=|M|c/I$ . Under a positive sagging moment, the top fibres are in compression and the bottom fibres are in tension. A negative hogging moment reverses those signs. Keep this bending normal stress separate from transverse shear stress, which has a different distribution.

Use the correct section geometry

For a rectangle of width $b$ and vertical depth $h$ , the centroidal second moment is $I=bh^3/12$ . Substituting $c=h/2$ gives $|\sigma_{\max}|=6|M|/(bh^2)$ . The cubed depth in $I$ makes increasing beam depth an effective way to reduce bending stress. For a built-up or asymmetric section, calculate the centroid and $I$ about the relevant neutral axis. The distances to its top and bottom extreme fibres may differ, so evaluate both stresses before deciding which is largest.

Find the critical section

First obtain support reactions from equilibrium, then find the bending moment along the beam using cuts or a bending-moment diagram. The location of greatest $|M|$ normally controls the largest bending normal stress when the cross-section is constant. Use consistent units: $M$ in $\mathrm{N\,m}$ , $I$ in $\mathrm{m^4}$ and $y$ in $\mathrm{m}$ produce stress in $\mathrm{Pa}$ . Alternatively, $\mathrm{N\,mm}$ with $\mathrm{mm^4}$ and $\mathrm{mm}$ produces $\mathrm{N/mm^2}$ , numerically equal to $\mathrm{MPa}$ .

Worked example: Maximum stress in a simply supported beam

A simply supported beam has a span of $4.0\,\mathrm{m}$ and a downward point load of $6.0\,\mathrm{kN}$ at midspan. Its rectangular section is $80\,\mathrm{mm}$ wide and $160\,\mathrm{mm}$ deep. Assume elastic bending of a prismatic beam. Determine the maximum bending normal stress and identify which face is in tension.

Side view of a straight horizontal simply supported beam of total span 4.0 m. At left end A, draw a pinned support shown as a triangle; at right end B, draw a roller support with two small rollers. A single downward force arrow of 6.0 kN acts at the exact midpoint, not at a support. Draw a span dimension line beneath from A to B showing 4.0 m. On the right, show a separate small rectangular cross-section sketch with horizontal width dimension 80 mm and vertical depth dimension 160 mm. Do not include reactions, moment values, stress values or answers. — The central point load creates a maximum sagging moment at midspan.

Show the solution

By symmetry, each vertical support reaction is $6.0/2=3.0\,\mathrm{kN}$ .

The largest positive bending moment occurs at midspan:

M_{\max}=\frac{PL}{4}=\frac{(6000)(4.0)}{4}=6000\,\mathrm{N\,m}.

Convert the cross-section dimensions to metres and calculate its second moment:

I=\frac{bh^3}{12}=\frac{(0.080)(0.160)^3}{12}=2.73067\times10^{-5}\,\mathrm{m^4}.

The extreme-fibre distance is $c=0.160/2=0.080\,\mathrm{m}$ . Use the unrounded $I$ value:

|\sigma_{\max}|=\frac{M_{\max}c}{I}=\frac{(6000)(0.080)}{2.730666\ldots\times10^{-5}}=17.578125\times10^6\,\mathrm{Pa}.

The moment is sagging, so the top face is compressed and the bottom face is stretched.

Answer: maximum bending stress magnitude $17.6\,\mathrm{MPa}$ , with tension at the bottom and compression at the top.

Common mistakes

Putting the applied force into $\sigma=My/I$ instead of moment

Find the internal bending moment $M$ at the critical section first.

Using the full beam depth as $y$ for a symmetric rectangle

Measure from the neutral axis: the extreme distance is $h/2$ .

Mixing $\mathrm{kN\,m}$ with $\mathrm{mm^4}$ without conversion

Choose a consistent length and force system before substituting.

Calling both faces tensile

For positive sagging bending, the top is compressive and the bottom tensile.

Check your understanding

1. A rectangular beam is $50\,\mathrm{mm}$ wide and $100\,\mathrm{mm}$ deep. Under $|M|=2.0\,\mathrm{kN\,m}$ , what is the extreme-fibre bending stress magnitude?

2. For a homogeneous beam under positive sagging bending, which statement is correct?

3. At the same $|M|$ , what happens to the maximum bending stress of a rectangular beam if its depth doubles but its width stays fixed?

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