📏 Beam Deflection

Written and maintained by the PhDino author · Last reviewed 21 September 2026 · Checked against 4 independent reference calculations · how PhDino checks its numbers

Related standards: ACI 318, AISC 360

How much a beam sags under a uniform load, and the bending moment that causes it.

When a beam carries a load, it bends. Deflection is how far it moves from its unloaded position; the bending moment is the internal force couple that causes that curvature at any point along the span. Both depend on how the beam is supported — simply supported, cantilevered, or fixed at both ends — because the support conditions change how the load is carried to the ground.

For a uniformly distributed load (the same load per foot along the whole span), each support case has a closed-form solution from classical beam theory (Euler–Bernoulli). PhDino covers the three most common cases.

Key formula

Simply supported: δ_max = 5wL⁴/(384EI), M_max = wL²/8
Cantilever: δ_max = wL⁴/(8EI), M_max = wL²/2 (at the fixed end)
Fixed-fixed: δ_max = wL⁴/(384EI), M_max = wL²/12 (at the supports)
Deflection ratio = L / δ_max

Variables

w

uniform load per unit length

L

span length

E

Young's modulus of the beam material

I

moment of inertia of the cross-section about the bending axis

How to use the Beam Deflection calculator

Use this calculator to check how far a beam sags under a load spread evenly along its length, and how large the bending moment is that the beam must resist. It answers the question a floor joist, a header, a shelf or a machine frame usually raises first: will it feel bouncy, or crack the ceiling below, before it ever gets close to breaking?

It checks stiffness, not strength. Deflection and bending stress are separate checks and a beam can pass one and fail the other, so treat the deflection ratio here as one half of the answer. Choose the support condition first, because it changes the result by a factor of five or more, then enter the span, the load per foot, and the stiffness of the material and section.

Beam Type

Most joists, rafters and headers rest on two supports and are treated as simply supported. Choose cantilever only for a member genuinely fixed at one end and free at the other, such as a balcony slab or a shelf bracket. Fixed-fixed assumes both ends are held perfectly against rotation; real framing connections are far more flexible than that, so it flatters ordinary timber and light-steel construction.

Span Length (ft)

The clear distance between the supports, not the overall length of the piece. For a cantilever it is the distance from the fixed end to the free tip.

Uniform Load (lbs/ft)

Total load per foot of beam. For a joist, multiply the area load in lb/ft² by the joist spacing in feet (16 in is 1.33 ft). Add the beam's own weight when it is not small next to the load it carries.

Youngs Modulus (E) (ksi)

Young's modulus, the material's stiffness: about 29,000 ksi for steel and 10,000 ksi for aluminum. Sawn softwood lumber runs from roughly 1,200 to 1,900 ksi depending on species and grade, so use the published design value for your grade, and use the modulus, not a strength.

Moment of Inertia (I) (in⁴)

How stiff the cross-section is in bending, about the axis the beam bends around (the strong axis for a joist on edge). A rectangle has b × d³ ÷ 12 using the actual dimensions; a rolled steel shape is listed in the manufacturer's section tables. This input carries the largest leverage of all, because depth is cubed.

Worked example: a 2×10 floor joist spanning 14 feet

A floor is framed with nominal 2×10 joists 16 in apart, spanning 14 ft between bearing walls. The floor carries 40 lb/ft² of live load and 10 lb/ft² of dead load. The lumber's modulus is taken as 1,600 ksi. Is one joist stiff enough?

Worked by hand:

1. Load per joist. The floor carries 40 + 10 = 50 lb/ft². Each joist collects a strip 16 in (1.33 ft) wide, so w = 50 × 1.333 = 66.7 lb/ft.
2. Stiffness of the section. A nominal 2×10 measures 1.5 in × 9.25 in, so I = 1.5 × 9.25³ ÷ 12 = 98.9 in⁴. (Using the nominal 2 × 10 would give 166.7 in⁴ and overstate the stiffness by two thirds.)
3. Consistent units. The span is 14 ft × 12 = 168 in, the load is 66.7 ÷ 12 = 5.556 lb/in, and E = 1,600 ksi × 1,000 = 1,600,000 psi.
4. Deflection. δ = 5wL⁴ ÷ (384EI) = 5 × 5.556 × 168⁴ ÷ (384 × 1,600,000 × 98.9) = 0.364 in, which is span ÷ 461 under the full 50 lb/ft².
5. Compare with the limits. Floors are usually held to L/360 under live load only: 168 ÷ 360 = 0.467 in. Live load is 40 of the 50 lb/ft², so the live-only deflection is 0.364 × 40 ÷ 50 = 0.291 in (span ÷ 577). For total load the common limit is L/240 = 0.700 in.
6. Bending moment. M = wL² ÷ 8 = 1,633 ft-lb. To carry that, the section modulus S = 1.5 × 9.25² ÷ 6 = 21.4 in³ gives a bending stress of M × 12 ÷ S = 916 psi.

Stiffness passes comfortably: the live-load sag of 0.29 in is well inside the 0.47 in limit, and the total-load sag is inside L/240 as well. The bending stress is a different question. About 916 psi is in the range that common softwood grades are rated for, so strength, not deflection, is the check to make next, against the published bending design value for the actual species and grade.

Reading the result: how much sag is too much

Sag is judged as a fraction of the span, because the same half inch is invisible on a 30 ft beam and alarming on a 4 ft one. The deflection ratio on the results panel is the span divided by the deflection, so a bigger number means a stiffer beam: a ratio of 360 means the beam sags one 360th of its span.

Building codes and product standards set the limits, and finishes tighten them. The figures below are typical starting points, not a substitute for the limit that applies to your project.

• About L/360 for floors under live load, which keeps plaster ceilings from cracking and floors from feeling springy.
• About L/240 for total load (dead plus live) on floors, with looser limits such as L/180 or L/240 for roofs that carry no plaster ceiling.
• L/480 and stiffer where brittle finishes sit on the beam: tile, stone, glass or a long run of cabinetry.
• Sensitivity is steep. Deflection grows with the fourth power of the span, so doubling the span multiplies it by 16, while it falls in direct proportion to E and I. A 2×12 is about 1.8 times as stiff as a 2×10, which is why going up one depth is often the cheapest fix.
• Vibration is separate. A floor can meet L/360 and still feel lively underfoot; shorter spans and deeper joists help more than a slightly stiffer species.

Notes & limitations

• Keep the units consistent: with the load in lb/in, the span in inches, E in psi and I in in⁴, the deflection comes out in inches. PhDino converts the feet, lb/ft and ksi you type, so you never do that by hand — but it is the most common way a hand calculation goes wrong (an error of 12× or 1,000× is easy to make).
• Deflection is usually judged as a fraction of the span, not as an absolute number: for example L/360 is a common limit for floors under live load. The calculator reports the ratio (span ÷ deflection) so you can compare directly; larger is stiffer.
• Deflection scales with L⁴ — doubling the span multiplies deflection by 16, which is why long spans need much deeper (higher-I) sections.
• These formulas assume linear-elastic material behavior and a load that stays uniform along the whole span; point loads or partial loads need different formulas.
• Serviceability limits (how much deflection is acceptable) are set by the governing code, not by this formula — a beam can be "strong enough" and still deflect more than a floor or ceiling can tolerate.

Common mistakes

• Using nominal lumber sizes for the moment of inertia. Depth is cubed, so treating a 2×10 as 2 in × 10 in overstates its stiffness by about two thirds.
• Checking total load against a limit that applies to live load only, or the reverse. Look up which load case each limit governs before comparing.
• Assuming fixed ends for a beam that sits in a pocket, on a hanger or on a wall plate. Those supports rotate, so the simply supported result is the honest one.
• Forgetting long-term creep in timber. Wood under sustained load keeps sagging over the years, and design standards multiply the dead-load part of the deflection to allow for it; a joist that passes on the day it is built can look different a decade later.
• Entering a strength where the modulus goes. Bending strength in psi and Young's modulus in ksi are different properties, and swapping them gives a result that looks plausible and is wrong.
• Applying a uniform-load formula to a concentrated load. A post or heavy appliance at midspan needs its own expression (δ = PL³ ÷ 48EI for a simply supported beam), and mixing the two loads means adding their separate deflections.

Frequently asked questions

Papers worth reading

On the correction for shear of the differential equation for transverse vibrations of prismatic bars Timoshenko, S. P. (1921), Philosophical Magazine. The standard deflection formulas ignore shear deformation; this is the paper that introduced the correction, which matters for short, deep beams.