Bending Stress Calculator

Compute bending stress from moment, geometry, and section properties.

Home σ = Mc/I 3 shapes MPa + psi
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Bending Stress

σ = M·c / I · rectangle, circle, custom · bidirectional

Instructions — Bending Stress Calculator

1

Pick section and mode

Rectangle, circle, or custom (I and c entered directly). Then choose solve mode: stress from moment, or moment from allowable stress.

2

Enter the values

For rectangle, b × h in mm. For circle, diameter in mm. Bending moment in kN·m, stress in MPa.

3

Read results

Get stress in MPa and psi, section modulus W, moment of inertia I, and extreme-fiber distance c.

Formulas

Bending stress (general)
$$ \sigma = \frac{M \cdot c}{I} = \frac{M}{W} $$
M is bending moment, c is distance from neutral axis to extreme fiber, I is moment of inertia. W = I/c is section modulus.
Rectangle section properties
$$ I = \frac{b h^3}{12},\quad W = \frac{b h^2}{6} $$
For a solid rectangular cross-section. W scales as h² — doubling height gives 4× the section modulus.
Solid circle properties
$$ I = \frac{\pi d^4}{64},\quad W = \frac{\pi d^3}{32} $$
For a solid circular cross-section like a round bar or pipe (with d as outer diameter).
Stress varies linearly through depth
$$ \sigma(y) = \frac{M \cdot y}{I} $$
σ is zero at the neutral axis (y = 0), maximum at y = c. Top fiber is in compression, bottom in tension for a sagging beam.
Allowable moment from σ_allow
$$ M_{allow} = \sigma_{allow} \cdot W $$
Back-solving for the maximum moment that keeps stress below the material limit. Useful for sizing.

Reference

Common shapes
SectionIWc
Rectangle b × hbh³/12bh²/6h/2
Solid circle dπd⁴/64πd³/32d/2
Hollow tube D, t≈πD³t/8≈πD²t/4D/2
Square s × ss⁴/12s³/6s/2
I-beam (tabulated)see AISCsee AISCd/2
Triangle b × hbh³/36bh²/242h/3

Article — Bending Stress Calculator

Bending Stress Calculator: σ = M·c / I, Explained

Bending stress equals M·c / I, the cornerstone formula of beam flexure. A 10 kN·m moment on a 100 × 200 mm rectangular section produces 15 MPa of stress at the extreme fiber — well below the 160 MPa allowable for structural steel.

The bending stress formula appears in every mechanics of materials textbook, every structural engineering exam, and every plan-review checklist. This page walks through what each variable means, how to compute the section properties (W, I, c) for common shapes, where the formula breaks, and how engineers convert between stress in MPa and psi when crossing the Atlantic.

What is bending stress?

Bending stress is the internal normal stress that develops in a beam under flexure. When a load bends a beam, the top fiber compresses while the bottom fiber stretches (or vice versa for an inverted beam). The stress varies linearly from compressive at one extreme fiber, through zero at the neutral axis, to tensile at the other extreme fiber.

The magnitude depends on three quantities: the bending moment M (how hard you're bending), the section modulus W (how well the cross-section resists bending), and the location y within the section (because stress isn't uniform). At the extreme fiber where y = c, the stress is maximum and equal to M/W.

Did you know

The bending stress formula is sometimes called the Euler-Bernoulli equation after the 18th-century mathematicians who derived it. Their assumption — that plane sections remain plane after bending — holds well for slender beams but breaks down for deep beams where shear deformation matters.

The bending stress formula

Two equivalent forms appear in textbooks. They describe the same physics.

Bending stress, two forms
σ = M · c / I using moment of inertia
σ = M / W using section modulus
where W = I / c relationship
σ(y) = M · y / I at any height y

The first form is cleaner mathematically but requires computing both I and c. The second form is more practical: engineering tables list W directly for standard sections. AISC tables for I-beams, IPE tables for European sections, and lumber span tables all give W (sometimes called S in US references).

Bending stress and section modulus

Section modulus W is the geometric quantity that links bending moment to stress. It captures both how much material is in the cross-section and how far that material sits from the neutral axis. For a rectangular cross-section b × h, W = bh²/6. For a solid circle of diameter d, W = πd³/32.

Section modulus scales aggressively with depth. Doubling the height of a rectangular section gives 4× the section modulus (since W = bh²/6 contains h²). Doubling the width only doubles W. This asymmetry drives every I-beam design — push the material as far from the neutral axis as possible to maximize W per unit weight.

Section moduli for common shapes
Rectangle b × h W = bh²/6
Square s W = s³/6
Solid circle d W = π d³/32
Hollow tube D, t W ≈ π D² t / 4
Triangle b × h (base) W = bh²/24

Bending stress distribution through depth

Bending stress isn't uniform across the section. At the neutral axis (y = 0) it's zero. At any height y above or below the neutral axis it equals M·y/I. At the extreme fibers (y = c = h/2 for a symmetric rectangle) it reaches the maximum value σ_max = M·c/I.

This linear distribution has practical consequences. Reinforced concrete beams put steel rebar near the tension face — where stress is highest — because concrete can't carry tension. Pre-stressed concrete reverses the situation: jacking force creates compression in the bottom fiber, which gravity loads then reduce rather than reverse. Wooden beams sometimes use a stronger species or denser grain on the bottom face for the same reason.

Bending stress by section shape

Different shapes give different W per unit weight. The efficient ones cluster material far from the neutral axis.

Solid rectangle
W = bh²/6
baseline efficiency
I-beam (same area)
W ≈ 2.5× rectangle
flange-dominant

An I-beam achieves 2-3× the section modulus of a solid rectangle with the same cross-sectional area. The flanges (top and bottom) sit at y = ±c, contributing the most to W. The web (vertical middle) contributes little to bending but resists shear and keeps the flanges aligned.

Tip

If you only know the moment of inertia I and not the distance c, compute W = I / c and use σ = M/W. If you only have W tabulated and need to estimate deflection, deflection depends on I not W, so you must recover I = W·c from the table's c value.

Bending stress and material limits

The stress σ from M·c/I is meaningful only against a material limit. Three thresholds matter.

  • Yield strength σ_y: Where the material starts to deform permanently. Steel S235: 235 MPa.
  • Ultimate strength σ_u: Where the material breaks. Steel S235: 360 MPa.
  • Allowable stress σ_allow: Code-mandated working stress, well below yield. Steel S235: 160 MPa.
  • Tension vs. compression: Most metals are symmetric; concrete is not (compression strong, tension weak).
  • Fatigue limit: For cyclic loading, σ must stay below the endurance limit — typically half of σ_y.
  • Brittle vs. ductile: Brittle materials (cast iron, concrete) fail at yield; ductile materials (steel) yield and redistribute.

Worked bending stress example

A wood floor joist measures 50 × 200 mm (nominal 2×8). A 1.2 kN point load at the midspan of a 4 m simply-supported span produces M_max = PL/4 = 1.2 kN·m.

Section modulus W = b·h²/6 = 50 · 40,000 / 6 = 333,333 mm³ = 3.33 × 10⁵ mm³.

Bending stress σ = M / W = 1.2 × 10⁶ N·mm / 3.33 × 10⁵ mm³ = 3.6 MPa.

For grade-stamped lumber with σ_allow = 10 MPa, the joist passes with a safety factor of 2.8. If the joist were turned 90° (oriented with the 50 mm dimension vertical), W would shrink to 50²·200/6 = 83,333 mm³ — one-quarter of the strong-axis value, and stress would jump to 14.4 MPa, exceeding the allowable. Orientation matters.

Common bending stress mistakes

Five recurring errors trip up new engineers:

Strong axis vs. weak axis

A 50×200 mm joist with the long side vertical has W = 333,333 mm³. Flip it horizontal and W drops to 83,333 mm³. Same beam, four times less moment capacity. Always specify orientation on drawings and check it on inspection.

  • Strong vs. weak axis confusion: A rectangular section is far stronger about one axis than the other.
  • Wrong c distance: For asymmetric sections like a T-shape, c differs above and below the neutral axis.
  • Unit mismatch: M in kN·m and I in mm⁴ requires conversion factor 10⁶ to get MPa.
  • Forgetting shear stress: Short beams may fail by shear before bending. Both must check.
  • Using nominal lumber dimensions: A 2x10 is actually 1.5x9.25 inches. Use actual values in W.
  • Ignoring stress concentrations: Holes, notches, and welds amplify local stress by 2-3×. Smooth design only.

FAQ

σ = M · c / I, where M is bending moment, c is distance from neutral axis to extreme fiber, and I is moment of inertia. Often written as σ = M / W where W = I/c is the section modulus. This is the fundamental equation of beam flexure.
The imaginary line through the beam's cross-section where bending stress is zero. Above the neutral axis the material compresses; below it stretches (for a sagging beam). For symmetric sections the neutral axis passes through the centroid.
I (moment of inertia) is a purely geometric property: ∫y²dA. W (section modulus) is I divided by the distance c to the extreme fiber: W = I/c. For computing stress, W is more direct: σ = M/W. For computing deflection, I appears: δ ∝ 1/EI.
For a rectangular section, W = bh²/6 — section modulus scales as h². Doubling h gives 4× W and 4× moment capacity at the same stress. The moment of inertia I = bh³/12 scales as h³, so doubling h gives 8× stiffness against deflection.
For safe design, σ ≤ σ_allow. σ_allow embeds a safety factor below the material's yield strength. Steel S235 has yield 235 MPa and σ_allow ≈ 160 MPa (factor 1.47). Wood typically uses σ_allow around 10-12 MPa for joists and 15-20 MPa for glulam.
1 MPa = 145 psi. A 100 MPa stress equals 14,500 psi. US codes typically use psi (allowable stress 24,000 psi for A36 steel). European codes use MPa (160 MPa allowable for S235 steel). Same beams, different units.
For ductile materials yes — they yield and redistribute. Steel beams in plastic design routinely operate above yield in extreme fibers. Brittle materials (concrete in tension, glass) fracture at first yield. Allowable-stress design keeps σ well below yield to avoid relying on plastic behavior.
Use the tabulated section modulus W from AISC or Eurocode manuals. For a W12×26 (US) or IPE 300 (Europe), W is listed in steel tables. Apply σ = M / W. No need to recompute I from geometry — manufacturers test and tabulate it.