Beam deflection | Engimations

By the end of this topic you will be able to

Relate curvature to bending moment through EI.
Use standard deflection formulae for common loads and supports.
Combine cases by superposition and compare the result with a serviceability limit.

Why deflection matters

A beam that is strong enough may still be unusable if it sags too much: floors feel bouncy, plaster cracks, doors jam. Designers therefore check deflection as well as stress. A common limit for floor beams is span/360.

The governing equation

For small deflections the curvature of a beam is proportional to the bending moment:

EI\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = M(x)

$EI$ is the flexural rigidity, the product of the material's stiffness and the section's shape. Integrate once for the slope and twice for the deflection, using the boundary conditions at the supports to find the constants:

simple (pin or roller) support: $y = 0$ ;
fixed support: $y = 0$ and slope $= 0$ .

Standard cases

Integration has been done for the common cases, and the results are tabulated:

Beam and load	Max deflection
Simply supported, central point load $P$	$PL^3/48EI$ at mid-span
Simply supported, UDL $w$	$5wL^4/384EI$ at mid-span
Cantilever, end load $P$	$PL^3/3EI$ at the free end
Cantilever, UDL $w$	$wL^4/8EI$ at the free end

Note the high powers of span: doubling the span multiplies deflection by 8 (point load) or 16 (UDL).

Superposition

For a linear-elastic beam, the deflection under several loads is the sum of the deflections under each load alone. Use the model to confirm: set $P$ and $w$ separately, note both deflections, then use both together.

Interactive modelChange the load, span and stiffness. The beam is drawn exaggerated.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

A 6 m simply supported steel beam ( $E = 200$ GPa, $I = 100\times10^6$ mm⁴) carries a 20 kN load at mid-span. Find the maximum deflection and compare it with the limit span/360.

Flexural rigidity. $EI = 200\,000\times100\times10^6 = 2.0\times10^{13}$ N·mm² $= 20\,000$ kN·m².

Deflection.

\delta = \frac{PL^3}{48EI} = \frac{20\times6^3}{48\times20\,000} = 0.0045 \text{ m} = 4.5 \text{ mm}

Check. Limit $= 6000/360 = 16.7$ mm. Since $4.5 < 16.7$ , the beam is acceptable.

Practice problems

A cantilever 3 m long, $EI = 20\,000$ kN·m², carries 10 kN/m over its whole length. Find the tip deflection.

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$\delta = wL^4/8EI = 10\times81/(8\times20\,000) = 5.06\times10^{-3}$ m $= 5.06$ mm.
A beam's span is doubled with the same load and section. By what factor does mid-span deflection change for a central point load?

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$\delta \propto L^3$ , so it increases 8 times.
Show that making a beam twice as deep (same width) reduces deflection by a factor of 8.

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$I = bh^3/12$ increases by $2^3 = 8$ and $\delta \propto 1/I$ .
In the model, a beam deflects 4.5 mm under a 20 kN central load. What will it be for 20 kN at mid-span plus 10 kN/m over the span?

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UDL part: $5\times10\times6^4/(384\times20\,000) = 8.44$ mm. Total by superposition: $4.5 + 8.44 = 12.9$ mm.

Key formulae

Moment–curvature: $EI\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = M$
Simply supported, central point load: $\delta_{\text{max}} = \frac{PL^3}{48EI}$
Simply supported, full-span UDL: $\delta_{\text{max}} = \frac{5wL^4}{384EI}$
Cantilever, end load: $\delta_{\text{max}} = \frac{PL^3}{3EI}$
Cantilever, full-span UDL: $\delta_{\text{max}} = \frac{wL^4}{8EI}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (50 min). 10 min: demonstrate with a ruler (span, width and depth effects); 15 min: concept of curvature and EI; 10 min: standard cases; 15 min: superposition with the model.

Activity. Ask: "Which is more effective for reducing sag: a stiffer material or a deeper section?" (E: steel vs aluminium ×3; depth ×2 gives ×8.)

Discussion. What deflection limits are used in building codes you know?

Assessment idea. Combined loading by superposition with a limit check.