Beam Deflection Formula

A beam deflection formula tells you how far a beam bends under load. If you searched for the beam deflection formula, the key point is this: there is no single formula for every beam. The correct expression depends on the support condition, the load pattern, and the bending stiffness $E I$.

One of the most common cases is a cantilever beam with a point load at the free end:

$$\delta_{max} = \frac{P L^3}{3 E I}$$

This formula is useful only when that exact beam case and the usual small-deflection, linear-elastic assumptions are reasonable.

What beam deflection depends on

The physical idea is simple. Loads create bending moments, and the beam resists that bending through $E I$.

• $E$ is Young's modulus, so it tells you how stiff the material itself is.
• $I$ is the second moment of area, so it tells you how the cross section resists bending about a chosen axis.
• $E I$ is called the flexural rigidity.

For an Euler-Bernoulli beam with small slopes, the curvature is related to bending moment by

$$\kappa(x) = \frac{M(x)}{E I}$$

up to sign convention. That is why beam deflection formulas all follow the same pattern: larger moments produce more bending, while larger $E I$ reduces it.

A common beam deflection formula: cantilever with end load

For a cantilever beam of length $L$ with a point load $P$ at the free end, the maximum deflection occurs at the tip and has magnitude

$$\delta_{max} = \frac{P L^3}{3 E I}$$

Here,

• $P$ is the applied load
• $L$ is the beam length
• $E$ is Young's modulus
• $I$ is the second moment of area

Use this formula only if these conditions match the problem:

• the beam is fixed at one end and free at the other
• the load acts at the free end
• the material stays in the linear elastic range
• deflections and slopes are small enough for beam theory to be a good approximation
• the beam is slender enough for Euler-Bernoulli theory to be reasonable
• shear deformation is neglected

The $L^3$ term is the part worth noticing. If all else stays the same and the length doubles, the tip deflection becomes $2^3 = 8$ times larger.

Worked example with numbers

Suppose a cantilever beam has

• $P = 120\ \mathrm{N}$
• $L = 1.5\ \mathrm{m}$
• $E = 200 \times 10^9\ \mathrm{Pa}$
• $I = 4.0 \times 10^{-6}\ \mathrm{m^4}$

Use the cantilever tip-load formula:

$$\delta_{max} = \frac{P L^3}{3 E I}$$

Substitute the values and keep SI units throughout:

$$\delta_{max} = \frac{120(1.5)^3}{3(200 \times 10^9)(4.0 \times 10^{-6})}$$

Since $(1.5)^3 = 3.375$, this becomes

$$\delta_{max} = \frac{405}{2.4 \times 10^6}\ \mathrm{m}$$ $$\delta_{max} \approx 1.69 \times 10^{-4}\ \mathrm{m}$$

So the tip deflection is

$$0.000169\ \mathrm{m} = 0.169\ \mathrm{mm}$$

That is very small compared with a $1.5\ \mathrm{m}$ span, so the small-deflection assumption is at least plausible in this example.

Common mistakes with beam deflection formulas

Treating one formula as universal

The cantilever tip-load formula does not apply to a simply supported beam, a uniformly distributed load, or a beam with a different support condition. The correct expression changes with the setup.

Mixing up $I$

Here, $I$ means the second moment of area of the cross section. It is not electric current, and it is not the mass moment of inertia.

Ignoring units

Beam formulas are very sensitive to units because $I$ often has units of $\mathrm{m^4}$ or $\mathrm{mm^4}$. A unit mismatch can change the answer by a factor of millions.

Using the formula outside its assumptions

If deflections are large, the material yields, the beam is not slender, or $E I$ changes along the span, a simple textbook formula may no longer be reliable.

When the beam deflection formula is used

Beam deflection formulas are used when you need stiffness, not just strength. A beam can be strong enough to avoid breaking and still deflect too much for the job.

That matters in structures, machine components, lab fixtures, shelves, and long members where alignment or serviceability is important. In practice, engineers often check both stress and deflection because those are different design limits.

Try a similar problem

Keep the same cantilever example and double only the length $L$. Predict the new tip deflection from the $L^3$ term before you calculate it. Then try a different support condition or load type and compare which parts of the formula change.