MULTICALCI
Euler–Bernoulli beam theory · L/360 serviceability check

Beam Deflection Calculator

Maximum deflection, bending moment, shear and bending stress for simply supported, cantilever and fixed-end beams under point or uniformly distributed load. Includes ten rolled sections plus rectangular, circular, I-beam and hollow custom shapes.

7 load cases IPE · HEA · UB sections Steel · Alu · Timber · Concrete L/360 limit

Calculate beam deflection and stress

Span, Section & Loading

Enter your values and select Calculate.

Beam deflection formulas

All cases use Euler–Bernoulli theory, in which deflection is governed by the flexural rigidity EI and bending stress by the section modulus Z. Span enters deflection to the third or fourth power, which is why long beams are almost always governed by stiffness rather than strength.

— Simply supported — UDL δ = 5wL⁴ / 384EI M = wL²/8 V = wL/2 Point, mid δ = PL³ / 48EI M = PL/4 V = P/2 Point, at a δ = Pab(a+2b)√(3a(a+2b)) / 27EIL M = RA·a — Cantilever — UDL δ = wL⁴ / 8EI M = wL²/2 V = wL Point, tip δ = PL³ / 3EI M = PL V = P — Fixed both ends — UDL δ = wL⁴ / 384EI M = wL²/12 V = wL/2 Point, mid δ = PL³ / 192EI M = PL/8 V = P/2 σ = M / Z — bending stress, MPa SF = Fy / σ — safety factor against yield δlimit = L / 360 — serviceability check

δ maximum deflection · w distributed load (N/m) · P point load (N) · L span (m) · E Young's modulus (GPa) · I moment of inertia (mm⁴) · Z section modulus (mm³) · Fy yield strength (MPa)

Depth beats width. Moment of inertia grows with the cube of depth but only linearly with width. Doubling the depth of a rectangular beam makes it eight times stiffer; doubling the width makes it twice as stiff for the same increase in material.

Worked example

An IPE300 steel beam spans 5 m, simply supported, carrying a uniformly distributed load of 20 kN/m.

Given
Section
IPE300 — I = 8.356 × 10⁷ mm⁴, Z = 5.574 × 10⁵ mm³
Span L
5 m, simply supported
Load w
20 kN/m = 20 000 N/m
Material
Steel — E = 200 GPa, Fy = 250 MPa
Step 1 — moment and shear
M = wL²/8 = 20 000 × 25 / 8
62 500 N·m = 62.5 kN·m
V = wL/2 = 20 000 × 5 / 2
50 kN (each reaction)
Step 2 — bending stress
σ = M/Z = 62.5 × 10⁶ / 557 400
112.13 MPa
SF = 250 / 112.13
2.23
Step 3 — deflection
EI = 200 × 10⁹ × 83.56 × 10⁻⁶
16 712 kN·m²
δ = 5wL⁴/384EI
9.74 mm
Limit = L/360 = 5000/360
13.89 mm
σ 112.13 MPa (SF 2.23) · δ 9.74 mm vs 13.89 mm limit → beam adequate

Both checks pass with reasonable margin. Deflection uses 70 % of its allowance while stress uses only 45 % of yield — the usual pattern, and a reminder that stiffness normally governs before strength on spans of this order.

Units and input ranges

QuantitySymbolUnitAccepted range
SpanLm, mm, ft or in> 0
Distributed loadwkN/m or N/m> 0
Point loadPkN, N, kip or lbf> 0
Young's modulusEGPa> 0
Yield strengthFyMPa> 0
Maximum deflectionδmmoutput
Bending momentMkN·moutput
Bending stressσMPaoutput
Moment of inertiaImm⁴output

Imperial inputs are converted before calculation: 1 ft = 304.8 mm, 1 in = 25.4 mm, 1 kip = 4 448.22 N and 1 lbf = 4.44822 N.

Rolled section properties

SectionI (mm⁴)Z (mm³)A (mm²)
IPE1608.693 × 10⁶123 0002 009
IPE2001.943 × 10⁷194 2002 848
IPE2403.892 × 10⁷324 3003 912
IPE3008.356 × 10⁷557 4005 381
HEA1003.490 × 10⁶72 7602 124
HEA1401.033 × 10⁷173 5003 142
HEA1802.790 × 10⁷324 0004 525
HEA2003.692 × 10⁷388 8005 383
UB 203×133×302.850 × 10⁷279 0003 820
UB 305×165×541.170 × 10⁸765 0006 860

Material properties

MaterialE (GPa)Fy (MPa)Density (kg/m³)
Structural steel2002507 850
Aluminium692762 700
Timber1230500
Concrete30252 400

Frequently asked questions

How is beam deflection calculated?

Deflection depends on the load, the span raised to the third or fourth power, and the flexural rigidity EI. For a simply supported beam under uniform load the maximum deflection is 5wL⁴/384EI. For a central point load it is PL³/48EI. Because span appears to the fourth power for distributed load, doubling the span increases deflection sixteen-fold at the same load intensity.

What deflection limit should a beam meet?

This calculator checks against L/360, the common limit for floors supporting brittle finishes such as plaster or tile. Other limits are used in practice: L/250 for general floor beams, L/200 for roof members, and L/180 for purlins and secondary members. Cantilevers are usually checked on twice the projection. Confirm the limit required by your governing code.

What is the difference between moment of inertia and section modulus?

Moment of inertia I governs stiffness and therefore deflection, with units of mm⁴. Section modulus Z governs strength and therefore bending stress, with units of mm³. Z equals I divided by the distance from the neutral axis to the extreme fibre. A beam can be strong but flexible, or stiff but weak, which is why both checks are needed.

Why does my beam pass on stress but fail on deflection?

Strength scales with section modulus while stiffness scales with moment of inertia, and deflection grows much faster with span than stress does. On long spans deflection almost always governs before yield. Increasing the depth of the section is far more effective than increasing width, because moment of inertia grows with the cube of depth while area grows only linearly.

Which support condition should I choose?

Simply supported assumes both ends are free to rotate, the safe default for beams on bearings or simple connections. Fixed-fixed assumes both ends fully restrained against rotation, which reduces mid-span moment and deflection substantially but requires genuinely rigid connections. Cantilever assumes one end fully fixed and the other free. Real connections usually fall between simply supported and fixed, so simply supported is the conservative choice unless the restraint is verified.

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