ABC Tools

Beam Deflection, Explained

After reading this you can predict how far a loaded beam sags, pick the right formula for your support and load, and check the numbers by hand before you trust any calculator.

Load a beam and it bends. A shelf full of books dips in the middle. A diving board droops under the diver. That vertical drop at any point is the deflection, and the largest value along the span is what usually matters for design. Too much sag looks alarming, cracks plaster, and jams doors long before the beam actually breaks.

This tool computes the maximum deflection for the four textbook cases: a cantilever or a simply supported beam, carrying either a single point load or a load spread evenly along its length. Those four cover most quick checks you will ever do.

What deflection is and why it grows so fast

A steel shelf bracket 0.3 m long barely moves under a 200 N load. Stretch that same bracket to 0.6 m and it sags eight times as far, even though you only doubled the length. That cube relationship is the single most important fact about beams, and it comes straight from the physics.

Bending stiffness depends on two things about the beam itself: how hard the material resists stretching, and how the cross-section is shaped. The material property is Young's modulus E. The shape property is the second moment of area I. Their product EI is the flexural rigidity, and every deflection formula divides by it.

Deeper beams win by a lot. For a rectangle, I = \frac{b h^3}{12}. Doubling the depth h raises I by a factor of 8. Turning a plank on edge instead of laying it flat is the cheapest stiffness you can buy.

When these four formulas apply, and when they do not

The formulas here assume a prismatic beam (constant cross-section), a linear elastic material (stress proportional to strain), small deflections, and one of the four idealized support and load arrangements. Inside those limits they are exact.

They break down when reality drifts away from the ideal. If the beam sags more than about 1/20 of its span, small-deflection theory starts to under-predict. If the material yields, the elastic formula no longer holds. Real supports are never perfectly pinned or perfectly fixed: a bolted connection sits somewhere between the two, so the true deflection lands between the simply supported and cantilever answers for the same load.

These formulas ignore the beam's own weight unless you add it as a uniform load. A long steel joist can sag noticeably under nothing but itself. To include self-weight, add its weight per metre to your uniform load w.

The four formulas and the intuition behind them

Each maximum deflection has the same skeleton: a load times a power of the span, divided by EI and a numeric constant. Only the constant and the span power change.

\delta_{\text{cant, point}} = \frac{P L^3}{3 E I}

A cantilever is fixed at one end and free at the other. A point load P at the free end gives the largest deflection of the four, because nothing supports the far end. Here L is the span in metres and the result is in metres.

\delta_{\text{cant, udl}} = \frac{w L^4}{8 E I}

Spread the same total load evenly and the cantilever sags less, because much of the load sits near the fixed end where it barely bends anything. Note the fourth power of L: a uniform load w (in newtons per metre) is even more sensitive to span than a point load.

\delta_{\text{ss, point}} = \frac{P L^3}{48 E I}

Simply supported means pinned at both ends, free to rotate but not to move down. A point load at the centre deflects 16 times less than the same load on a cantilever of equal span, because both ends now push back.

\delta_{\text{ss, udl}} = \frac{5 w L^4}{384 E I}

The uniform case for a simply supported beam carries the awkward-looking factor \frac{5}{384}. That is the exact result of integrating the bending curve twice; there is nothing to memorize beyond the shape of the fraction.

A worked example you can reproduce

Simply supported steel beam, point load at centre

Use the demo values: a simply supported beam, span L = 2 m, point load P = 1000 N, steel with E = 200 GPa, and I = 800 cm⁴. First convert every quantity to base SI units.

  1. Young's modulus: E = 200 \text{ GPa} = 200 \times 10^9 Pa.
  2. Second moment of area: I = 800 \text{ cm}^4 = 800 \times 10^{-8} \text{ m}^4 = 8 \times 10^{-6} m⁴. (One cm⁴ is 10^{-8} m⁴.)
  3. Flexural rigidity: E I = 200 \times 10^9 \times 8 \times 10^{-6} = 1.6 \times 10^6 N·m².
  4. Numerator: P L^3 = 1000 \times 2^3 = 8000 N·m³.
  5. Divide: \delta = \frac{8000}{48 \times 1.6 \times 10^6} = \frac{8000}{7.68 \times 10^7}.
  6. Result: \delta \approx 1.042 \times 10^{-4} m, or about 0.104 mm.

That is a tenth of a millimetre, invisible to the eye. Now swap the same beam to a cantilever with the load at the free end. The constant drops from 48 to 3, so the deflection climbs by a factor of 16 to about 1.67 mm. Same beam, same load, sixteen times the sag.

All four cases with L = 2 m, E = 200 GPa, I = 800 cm⁴, using P = 1000 N or w = 500 N/m. The cantilever point load sags most; the simply supported uniform load sags least.

How span dominates everything

Because deflection scales with L^3 or L^4, span is the parameter to watch. Halve the span and the point-load sag drops to one eighth; double it and the sag grows eightfold. The chart below plots the simply supported point-load case as you stretch the span, holding the load and beam fixed.

The curve is a cubic. Going from 2 m to 4 m span multiplies the sag by 8, from 0.104 mm to 0.833 mm.

A simply supported steel beam with P = 1000 N and I = 800 cm⁴ deflects 0.104 mm at 2 m span, 0.833 mm at 4 m, and 2.813 mm at 6 m. Doubling the span multiplies deflection by 8 for a point load and by 16 for a uniform load.

Reading and interpreting the result

The number the tool returns is the maximum vertical drop, in millimetres, of the most-deflected point (the free end for a cantilever, the mid-span for a simply supported beam). Two questions decide whether it is acceptable.

Is it serviceable? Engineers cap deflection with a span ratio, commonly L/360 for floors carrying plaster or L/250 for general use. A 2 m span at L/360 allows 2000 / 360 \approx 5.6 mm. Our 0.104 mm result is far inside that, so stiffness is not the concern here.

Is it still elastic? Deflection tells you nothing about whether the beam has yielded. A beam can pass the deflection check and still be overstressed, or vice versa. Always check bending stress separately.

Turn a deflection into a span ratio to judge it fast. Divide the span by the deflection: 2000 \text{ mm} / 0.104 \text{ mm} \approx L/19000. Anything stiffer than about L/250 feels solid underfoot.

Common mistakes

Nearly every wrong answer comes from units, not physics. The formulas need base SI throughout.

Mixing GPa and Pa
Young's modulus is entered in GPa but the formula wants Pa. Multiply by 10^9. Forgetting this makes the beam look a billion times floppier.
Leaving I in cm⁴
One cm⁴ equals 10^{-8} m⁴, not 10^{-2}. This is the most frequent slip. An IPE 160 section is 869 cm⁴, which is 8.69 \times 10^{-6} m⁴.
Confusing point load and uniform load
A point load P is a force in newtons. A uniform load w is a force per metre. If you have a total distributed weight W over span L, then w = W / L.
Using the wrong support case
A bracket welded to a wall is a cantilever, not simply supported. Choosing the wrong case here changes the answer by a factor of 16 for point loads.

Related tools

Once the beam is standing, other quick calculators help with the rest of the physics around it. The Projectile Motion Calculator handles range and trajectory when something leaves that beam in flight, and the Free Fall Calculator gives impact speed from a drop height. For rotating machinery mounted on your structure, the Gear Ratio Calculator converts speed and torque through a gear train. If you are analysing vibration or wave behaviour, the Wavelength to Frequency Converter relates the two, and the Wind Chill, Heat Index and Dew Point tool covers outdoor exposure conditions.

Frequently asked questions

Why does a simply supported beam deflect 16 times less than a cantilever?

Compare the constants: the cantilever formula divides P L^3 by 3, the simply supported formula divides by 48. Since 48 / 3 = 16, the same load on the same beam sags 16 times more as a cantilever. Both ends of the supported beam carry half the load and push back.

How do I find I for my beam?

For a solid rectangle of width b and depth h, I = \frac{b h^3}{12}. A 50 mm by 150 mm timber joist, loaded on edge, gives I = \frac{0.05 \times 0.15^3}{12} = 1.406 \times 10^{-5} m⁴, or 1406 cm⁴. For standard steel sections, read I from a section table.

Does the calculator include the beam's own weight?

No. Add self-weight as part of the uniform load. A steel section weighing 15.8 kg/m exerts about 155 N/m of downward load (15.8 \times 9.81), which you would add to w.

What deflection is too much?

It depends on use, not on a single number. Floors that carry brittle finishes are often limited to L/360, general beams to L/250. Convert your result to a ratio by dividing span by deflection and compare against the limit for your case.

Can I use these formulas for a beam with the load off-centre?

Not directly. These four cases assume the point load sits at the free end (cantilever) or at mid-span (simply supported). An off-centre load needs a different formula and shifts the location of maximum deflection away from the centre.

Try it yourself in the tool

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