🏗️ Beam Stress & Deflection Calculator

Last updated: June 21, 2026

Beam Stress & Deflection Calculator

Exact closed-form solutions · Euler–Bernoulli beam theory

metres (m)
GPa (steel ≈ 200, alum ≈ 70)
mm⁴
mm (from N.A. to outer fibre)
N (Newtons)
Results
Maximum Bending Moment
Maximum Bending Stress (σmax)
σ = M · c / I
Maximum Deflection (δmax)
Section Modulus (S)
S = I / c
Active Formula Set
Select a configuration above to see formulas.

Beam Stress and Deflection: The Physics Behind Every Floor Joist, Bridge Girder, and Shaft

Every structural element that carries a transverse load — a floor beam, a crane girder, a helicopter rotor blade, a silicon wafer on probe pins — obeys the same underlying mechanics. Euler–Bernoulli beam theory, first codified in the eighteenth century, gives exact closed-form solutions for the four or five loading configurations that appear in nine out of ten real design problems. This article goes deep on those solutions: where they come from, what they assume, and where they break down.

The Flexure Formula and What It Really Means

The central result of linear beam theory is the flexure formula:

σ = M · c / I

Here, M is the bending moment at the cross-section of interest (N·m), c is the perpendicular distance from the neutral axis to the fibre you are checking (m), and I is the second moment of area of the cross-section about the bending axis (m⁴). The stress σ is normal stress — tension on one side of the neutral axis, compression on the other.

The ratio I/c is called the section modulus, S. Larger S means lower stress for the same moment. This is why I-beams are efficient: material is concentrated far from the neutral axis, maximising I without maximising area (and therefore weight). A hollow rectangular tube does the same trick. A solid circular bar is the worst shape for bending — most of the area is near the neutral axis, contributing little to I while still carrying weight.

Simply Supported Beam, Centre Point Load

This is the textbook configuration, but it is also genuinely common: a beam resting on two knife-edge supports (or wide flanges approximated as such) with a single concentrated load at mid-span. The bending moment diagram is triangular, peaking at the centre:

Mmax = P·L / 4

The deflection at mid-span follows from two integrations of the moment-curvature relationship M = EI·d²y/dx²:

δmax = P·L³ / (48·E·I)

Notice the cubic dependence on length. Doubling the span while keeping everything else the same multiplies deflection by eight. This is why long beams almost always fail on deflection criteria (serviceability) long before they fail on stress (strength). Building codes typically limit beam deflections to L/360 for floor beams under live load — not because a beam at L/200 collapses, but because floor bounce is perceptible and plaster cracks.

Uniform Distributed Load: The More Realistic Case

Self-weight, stored goods, and pressure loads are all distributed. For a simply supported beam carrying w N/m over its full length:

Mmax = w·L² / 8

δmax = 5·w·L⁴ / (384·E·I)

The moment is again maximum at mid-span. A useful check: if the total load W = wL, then M = WL/8, which is exactly half what an equivalent centre point load would produce. This is intuitive — a distributed load spreads its moment more gently.

The deflection coefficient 5/384 (≈ 0.01302) versus 1/48 (≈ 0.02083) for the point-load case confirms this: the UDL beam deflects about 62% as much as the same total load applied at one point.

Cantilever Beams: The Severity of Fixed-End Bending

A cantilever fixed at one end and free at the other is structurally very different from a simply supported beam. The bending moment is maximum not at mid-span but right at the fixed support — and it is larger for the same load and span. For an end point load:

Mmax = P·L

δmax = P·L³ / (3·E·I)

Compare the deflection coefficient: 1/3 for a cantilever versus 1/48 for a simply supported beam under the same point load. The cantilever deflects sixteen times more. This is why cantilever balconies and overhanging eaves require far heavier sections than equivalent interior spans, and why the depth-to-span ratio for cantilevers is typically half that of simply supported spans in rule-of-thumb design.

For a cantilever with UDL:

Mmax = w·L² / 2

δmax = w·L⁴ / (8·E·I)

The free-end deflection coefficient 1/8 versus 1/3 for the point-load cantilever again reflects the more gradual moment build-up under distributed load.

The Modulus of Elasticity: Material's Role

Stress depends only on geometry (moment, section) — not on the material. A steel beam and an aluminium beam of identical cross-section under the same load will experience identical stress. What the material controls is deflection. Steel (E ≈ 200 GPa) deflects about three times less than aluminium (E ≈ 70 GPa) for the same geometry and load, because deflection is inversely proportional to E.

This has a practical consequence in aerospace and automotive design. You cannot reduce mass by switching from steel to aluminium while maintaining the same beam geometry — the aluminium beam will meet the stress criterion (aluminium alloy yield strength ≈ 270–500 MPa, comparable to structural steel) but fail on stiffness. You must increase the section depth, partially recovering the mass savings by using material more efficiently.

Second Moment of Area: Why Section Shape Dominates

The second moment of area I scales as the fourth power of a linear dimension. A rectangular section of width b and depth d gives I = b·d³/12. Doubling the depth multiplies I — and therefore reduces stress and deflection — by a factor of eight, at only double the material cost. This is the core reason structural beams are deep, not wide.

Typical I values for reference: a 100×50 mm solid rectangular steel bar has I ≈ 1.04×10⁶ mm⁴; a W200×100 I-beam has I ≈ 113×10⁶ mm⁴ about the strong axis while using far less material than the equivalent solid section. The difference is a factor of roughly 100 in bending stiffness for maybe 30% more steel — this is why structural engineers reach for I-sections almost reflexively.

Assumptions and Failure Modes Not Captured Here

Euler–Bernoulli beam theory assumes: plane sections remain plane after bending, the beam is straight and prismatic, deflections are small relative to span, and the material is linearly elastic. These assumptions break down in several important cases.

Shear lag in wide flanges means the outer portions of the flange are not fully effective in bending — the simple flexure formula is non-conservative for very wide box girders. Lateral-torsional buckling can cause a beam to fail at a fraction of the calculated bending moment if the compression flange is unrestrained — an I-beam can suddenly twist and buckle sideways. Web buckling occurs in thin-webbed sections under high shear near supports. Plastic hinge formation in ductile steel allows redistribution of moments beyond the elastic limit, which the elastic formulae ignore.

For preliminary design and most standard structural checks, the closed-form solutions here are the right starting point. For unusual geometry, high spans, or safety-critical structures, finite element analysis combined with code-specific checks (Eurocode 3, AISC LRFD, IS 800) is essential.

Reading the Results: Practical Benchmarks

Structural steel (S275/S355): yield strength 275–355 MPa. Aluminium alloy 6061-T6: 276 MPa. Douglas fir (timber): 40–55 MPa in bending. If your calculated maximum bending stress exceeds 60–70% of the material yield strength, your design factor of safety is marginal for most building applications. For deflection, L/360 (≈ 0.28% of span) is the typical floor live-load limit; L/240 for roofs with no attached finishes.

These benchmarks, combined with the exact formulas embedded in the calculator above, give you a complete rapid-design workflow for the most common beam loading scenarios encountered in structural, mechanical, and civil engineering.

FAQ

What is the difference between bending stress and shear stress in a beam?
Bending stress (also called flexural or normal stress) acts longitudinally along the beam fibres and is highest at the top and bottom surfaces, zero at the neutral axis. It is computed as σ = Mc/I. Shear stress acts transversely and is highest at the neutral axis, zero at the outer fibres. For most standard beams where span-to-depth ratio exceeds 10, bending stress governs design. This calculator computes bending stress only; for short deep beams, shear stress must be checked separately.
What units should I enter for the second moment of area (I)?
Enter I in mm⁴ (millimetres to the fourth power), which is the standard unit in structural section tables. The calculator automatically converts this to m⁴ for the deflection formula. For reference: a 100×100 mm solid square section has I = 100⁴/12 ≈ 8.33×10⁶ mm⁴. Standard steel sections (UB, UC, IPE) list I values in cm⁴ — multiply by 10,000 to convert to mm⁴.
Why does my cantilever beam deflect so much more than a simply supported beam of the same span?
A simply supported beam has two reaction points and the moment builds from zero at each support to a maximum at or near mid-span. A cantilever has only one fixed support and the full moment concentration happens there, while the free end has no restraint at all. The end deflection coefficient for a cantilever under point load is PL³/(3EI) versus PL³/(48EI) for simply supported — a factor of 16 difference. Cantilevers also have all their stiffness coming from one connection, making that joint design critical.
What does the neutral axis distance (c) input represent?
The neutral axis is the horizontal plane through the cross-section where bending stress is exactly zero. For symmetric sections (I-beams, rectangles, circles), it coincides with the geometric centroid. The value c is the perpendicular distance from this neutral axis to the outermost fibre — the point where stress is maximum. For a rectangle of depth d, c = d/2. For an I-beam with depth D, c = D/2. If your section is asymmetric (T-beam, angle), the top and bottom c values differ and you should check the larger one.
Can I use this calculator for timber or concrete beams?
Yes for timber — enter the modulus of elasticity for the timber species (typically 8–14 GPa for structural grades) and use the bending stress result against the characteristic bending strength from your design code. For concrete, use it only as a rough preliminary check: reinforced concrete beams are not linearly elastic, the neutral axis shifts after cracking, and design is dominated by reinforcement layout, not E and I of the gross section. For concrete, use dedicated RC design tools.
What is the L/360 deflection rule and how do I check it?
Most building codes require that floor beams under live load deflect no more than the span divided by 360 (L/360) to prevent perceptible bounce and damage to brittle finishes like ceramic tiles. To check: divide your beam span in mm by 360, then compare against the deflection result in mm from this calculator. For example, a 5 m beam has a limit of 5000/360 ≈ 13.9 mm. If your calculated deflection exceeds this, you need to increase I (deeper section or wider flange), reduce the span with an intermediate support, or use a stiffer material.