Beam deflection

Engineering · First-year engineering · about 10 min

Deflection is the movement of a loaded beam away from its original position. Even when a beam is strong enough not to fail, excessive deflection can prevent it from working properly. The simplest exam cases use standard elastic beam formulas and careful SI unit conversions.

The key idea

For a constant-section cantilever fixed at one end and carrying a downward point load $P$ at its free end, the free-end deflection magnitude is

\delta_{\mathrm{tip}}=\frac{PL^3}{3EI}

Here $L$ is span, $E$ is Young's modulus and $I$ is the second moment of area about the bending axis. The tip moves downwards for a downward load.

Side view of a horizontal slender cantilever with its left end embedded in a fixed vertical wall (short diagonal hatch marks), and its right end free. At the free end draw a vertical downward applied-force arrow P. Draw the original centreline as a straight dashed horizontal line and a separate smooth deformed centreline beginning tangent and coincident at the left fixed end, curving gradually down to a maximum downward displacement at the right. Draw a short vertical dimension arrow from the original tip level to the deformed tip labelled delta. Add a horizontal dimension line from wall to unloaded tip labelled L. Do not exaggerate the curvature so far that the tip displacement is comparable with L. — A cantilever with a downward end load has its greatest displacement at the free end.

Try it yourself

Change the span, load and stiffness to see which adjustment reduces tip deflection most.

Match the support and loading

A cantilever is fixed against translation and rotation at one end, with its other end free. Its maximum deflection under a single downward end force is at the free end. A different support arrangement requires a different formula: for a simply supported beam under a central point load $P$ , the midspan deflection magnitude is $PL^3/(48EI)$ , not $PL^3/(3EI)$ . Distributed loading also changes the expression. Before choosing an equation, sketch the supports and check where and how the load acts.

Understand flexural rigidity

The product $EI$ is the beam's flexural rigidity. A larger Young's modulus $E$ means a less deformable material, while a larger second moment $I$ means material is distributed more effectively away from the neutral axis. Both reduce deflection. Length has a particularly strong effect: for an end-loaded cantilever, doubling $L$ while keeping $P$ , $E$ and $I$ fixed increases the tip deflection by a factor of $8$ . This formula assumes small deflections, linear elasticity and a slender beam with constant $E$ and $I$ .

Keep units and direction straight

In $\delta=PL^3/(3EI)$ , use $P$ in $\mathrm{N}$ , $L$ in $\mathrm{m}$ , $E$ in $\mathrm{Pa}=\mathrm{N/m^2}$ and $I$ in $\mathrm{m^4}$ . The result is in metres. Convert to millimetres only at the end if the answer is small. If upward displacement is defined positive, a downward-load deflection is negative; most standard formula tables instead report its magnitude as a positive number. Deflection and slope are different: deflection has length units, while slope is an angle.

Worked example: Free-end deflection of a steel cantilever

A straight cantilever beam is $1.50\,\mathrm{m}$ long and carries a downward force of $500\,\mathrm{N}$ at its free end. Its Young's modulus is $E=200\,\mathrm{GPa}$ and its constant section has $I=2.00\times10^{-6}\,\mathrm{m^4}$ . Assuming small elastic deflections, find the magnitude and direction of the free-end deflection.

Draw a horizontal straight cantilever fixed into a hatched wall on the left and free on the right. From the free tip draw a vertical downward arrow labelled 500 N. Under the beam add a horizontal dimension line running from the left wall to the right tip, labelled 1.50 m. Use a faint dashed undeformed line and a gently downward-curving elastic deflected outline toward the right. Do not put a measured deflection, numerical slope or worked answer on the figure. — The fixed left end restrains rotation, while the downward load moves the right tip.

Show the solution

This is a fixed-free beam with a point load at the free end, so use the cantilever formula.

Write the quantities in SI base units:

P=500\,\mathrm{N},\quad L=1.50\,\mathrm{m},\quad E=200\times10^9\,\mathrm{Pa},\quad I=2.00\times10^{-6}\,\mathrm{m^4}.

Calculate flexural rigidity:

EI=(200\times10^9)(2.00\times10^{-6})=4.00\times10^5\,\mathrm{N\,m^2}.

Substitute without rounding intermediate values:

\delta_{\mathrm{tip}}=\frac{PL^3}{3EI}=\frac{500(1.50)^3}{3(4.00\times10^5)}=1.40625\times10^{-3}\,\mathrm{m}.

Convert metres to millimetres: $1.40625\times10^{-3}\,\mathrm{m}=1.40625\,\mathrm{mm}$ .

Answer: the tip deflects $1.41\,\mathrm{mm}$ downwards.

Common mistakes

Using a simply supported formula for a cantilever

Identify the fixed and free ends before selecting the coefficient.

Substituting $E=200$ for $200\,\mathrm{GPa}$

Convert gigapascals into pascals by multiplying by $10^9$ .

Forgetting that length is cubed

Calculate $L^3$ carefully and check that the final units reduce to length.

Confusing deflection with slope

Report displacement in $\mathrm{m}$ or $\mathrm{mm}$ ; slope is an angle, not a length.

Check your understanding

1. If a cantilever's length doubles with $P$ , $E$ and $I$ unchanged, by what factor does its end-load tip deflection increase?

2. A cantilever has $P=400\,\mathrm{N}$ , $L=2.00\,\mathrm{m}$ , $E=100\,\mathrm{GPa}$ and $I=4.00\times10^{-6}\,\mathrm{m^4}$ . What is its tip deflection magnitude?

3. Keeping cantilever load, length and material unchanged, what happens to tip deflection when $I$ doubles?

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