Beam Deflection Basics: Spans, Stiffness, and the Standard Formulas

Deflection Is a Separate Check From Stress

Bending stress and bending deflection come from the same load but are governed by different properties, so they can fail independently. Stress depends on the section modulus, S, and on the material's strength. Deflection depends on the product of elastic modulus and second moment of area, E·I, a quantity called flexural rigidity. Nothing about a high-strength alloy makes it stiffer.

That has a consequence people find surprising: swapping mild steel for a high-strength alloy raises the allowable stress substantially but changes deflection almost not at all, because nearly all steels share an elastic modulus near 200 GPa. Aluminum is the opposite case. Typical alloys sit near 69 GPa, roughly a third of steel, so an identical section under an identical load deflects about three times as much even at comparable yield strength.

Deflection limits come from the application, not the material. Building floors are commonly held to a fraction of the span, such as span over 360 under live load, where occupants start to notice movement and brittle finishes crack. Machine structures are often far tighter, since an alignment tolerance rather than human perception sets the budget.

• Strength check: compares peak bending stress to an allowable stress derived from yield and a safety factor.
• Stiffness check: compares predicted deflection to a limit set by the application, with no reference to strength at all.
• E is the property that matters for deflection, and within a material family it barely varies with grade or heat treatment.

A stronger alloy does not make a stiffer beam. Only a larger E·I, a shorter span, or different end restraint will reduce deflection.

The Four Standard Cases

Most everyday beam problems reduce to one of four textbook cases, and each formula is valid only for the boundary conditions it was derived under. All four assume a prismatic beam of constant cross-section, linear elastic material well below yield, small deflections relative to the span, and loads applied in the plane of bending.

For a simply supported beam, one held at both ends by supports that carry vertical load but apply no restraining moment, a single point load P at midspan gives a maximum midspan deflection δ = P·L³ / (48·E·I). Under a uniformly distributed load w, meaning force per unit length spread over the whole span, the midspan deflection is δ = 5·w·L⁴ / (384·E·I).

For a cantilever, rigidly fixed at one end and completely free at the other, a point load P at the free end gives a tip deflection δ = P·L³ / (3·E·I), and a uniformly distributed load w over its full length gives δ = w·L⁴ / (8·E·I). In all four expressions L is the full span or cantilever length, E is the elastic modulus, and I is the second moment of area about the bending axis.

• Simply supported, central point load: δ = P·L³ / (48·E·I), maximum at midspan.
• Simply supported, uniform load: δ = 5·w·L⁴ / (384·E·I), maximum at midspan.
• Cantilever, point load at free end: δ = P·L³ / (3·E·I), maximum at the tip.
• Cantilever, uniform load: δ = w·L⁴ / (8·E·I), maximum at the tip.

Boundary conditions are part of the formula. A beam whose ends are bolted into rigid connections is no longer simply supported, and using the simply supported result will overpredict its deflection.

A Worked Example, Units Included

Take a steel beam spanning 3.0 m between two simple supports, carrying a 5 kN point load at midspan. Use E = 200 GPa as a typical structural steel value and a section with I = 5.0 × 10⁶ mm⁴. Convert to consistent SI base units first, because this is where most deflection calculations go wrong: E = 200 × 10⁹ Pa and I = 5.0 × 10⁻⁶ m⁴, since 1 mm⁴ equals 10⁻¹² m⁴.

Substituting gives δ = (5000 N × 3.0³ m³) / (48 × 200 × 10⁹ Pa × 5.0 × 10⁻⁶ m⁴) = 135000 / (4.8 × 10⁷) = 2.81 × 10⁻³ m, or about 2.8 mm. The units resolve cleanly: N·m³ divided by (N/m²)·m⁴ leaves metres, which is the check worth doing every time before trusting a result.

Now spread the same 5 kN uniformly across the span instead, so w = 1667 N/m. Then δ = 5 × 1667 × 3.0⁴ / (384 × 200 × 10⁹ × 5.0 × 10⁻⁶) = 1.76 mm, exactly five-eighths of the point-load result. That ratio is a useful sanity check: distributing a load always deflects a span less than concentrating it at the centre.

If the same 3 m span were limited to span over 360, the allowance would be 8.3 mm, so this beam passes comfortably on stiffness. Rebuild it in 6061-T6 aluminum with the identical cross-section and E drops to roughly 69 GPa, pushing the point-load deflection to about 8.2 mm, right at the limit rather than a third of it.

Why Span Dominates Everything

Look at where L appears in all four formulas: cubed under a point load, and to the fourth power under a distributed load. No other variable enters with an exponent anywhere near that. Doubling the span of a simply supported beam under a central point load multiplies deflection by eight, and under a uniform load by sixteen.

In the worked example, stretching the span from 3.0 m to 6.0 m takes the deflection from 2.8 mm to 22.5 mm, a factor of exactly eight, with nothing else changed. That is why an intermediate support, or any change that shortens the effective span, is the most powerful intervention available, and why problems that resist material and section changes often vanish once the span is broken up.

The second-strongest lever is section depth, since I for a rectangle is proportional to the cube of its depth. Going from 100 mm deep to 150 mm at the same width multiplies I by 3.375 and cuts deflection by the same factor, for fifty percent more material. Standing a rectangular section on edge instead of flat is the free version of that trick.

Boundary conditions are the third lever. A cantilever under an end point load deflects sixteen times as much as a simply supported beam of the same span, section, and load, purely because of how it is restrained. Adding a support, or fixing an end that was free to rotate, changes the denominator without adding a gram of material.

Deflection scales with L³ or L⁴. Halving the span beats almost any change of material or section you could make.

Fixing a Beam That Deflects Too Much

Work the levers in order of effectiveness, not convenience. Shorten the span or add an intermediate support first, since the exponent on L makes that the highest-leverage change available. Next deepen or reorient the section to raise I. Then reconsider the end restraints, since fixing rotation at the supports cuts midspan deflection substantially. Only then reach for a stiffer material, and within one material family that option barely exists.

Two cautions apply to any deflection result. These formulas describe elastic behaviour only, so a beam that has yielded anywhere will deflect more than predicted and will not fully recover. And real structures rarely match one textbook case, so combined loads are handled by superposition, adding the deflections computed separately for each load case.

Deflection and stress are both required checks and neither substitutes for the other. Size the section for bending stress using the section modulus, verify deflection against the application's limit using E·I, and check shear near the supports separately. The companion guide on bending stress and section modulus covers the strength side of the same problem.