The Flexure Formula — Bending Stress Distribution

Why a bending moment creates tension on one side of a beam, compression on the other, and zero stress exactly at the neutral axis.

Mechanics of MaterialsMechanical Engineering Year 1Free preview
⏱️ About 18 min

Place a wooden ruler flat on two supports and press down in the middle. The top surface of the ruler gets shorter — it is in compression. The bottom surface gets longer — it is in tension. Somewhere in between, a layer stays exactly the same length. That invisible layer is the neutral axis, and the stress there is precisely zero. Engineers use the flexure formula to predict how large those tensile and compressive stresses are before the beam is ever built.

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The big idea: For a straight beam in elastic bending about a principal centroidal axis, the normal stress at any distance y from the neutral axis is σ = −M·y/I, where M is the bending moment, I is the second moment of area, and the negative sign ensures compression at positive y (above the neutral axis for a sagging moment). The stress varies linearly from zero at the neutral axis to maximum values at the outer fibres. The section modulus S = I/c collapses the geometry into a single number used for design.
🎯 By the end, you'll be able to
  • State the flexure formula and identify each symbol
  • Compute the maximum bending stress for symmetric cross-sections
  • Define section modulus S and use it for quick hand calculations
  • Sketch the linear bending-stress distribution across a beam section
📎 Helpful to know first

From beam curvature to the flexure formula

When a beam bends, longitudinal fibres on the concave side shorten and those on the convex side lengthen. The plane where length does not change is called the neutral surface; its intersection with the cross-section is the neutral axis (NA). For elastic, symmetric bending, the NA passes through the centroid of the cross-section.

Strain is proportional to the distance y from the NA: the farther a fibre is from the neutral axis, the more it stretches or compresses. Invoking Hooke's law (σ = Eε) gives a linear stress distribution. Integrating the moment produced by these stresses over the entire cross-section must equal the applied bending moment M, which yields the flexure formula:

\[ \sigma = -\frac{M\,y}{I} \]
Flexure formula: normal stress at distance y from the neutral axis. Sign convention: negative (compression) above the NA for a positive (sagging) moment.
\[ \sigma_{\max} = \frac{M\,c}{I} = \frac{M}{S} \qquad S = \frac{I}{c} \]
Maximum bending stress occurs at the outer fibre distance c. S is the section modulus, a convenient geometric property for design.
🔑 The neutral axis sits at the centroid — but only for elastic symmetric bending

If the material is homogeneous and the bending moment acts about a principal centroidal axis, the neutral axis coincides with the centroidal axis. This is the case for most introductory problems. When the material is non-homogeneous (a steel plate bonded to timber) or the moment is not aligned with a principal axis, the NA shifts or tilts — topics covered later in this module.

Side-by-side rectangular beam cross-sections showing linear bending stress (left) and parabolic transverse shear stress (right). Bending is maximum at the outer fibres and zero at the neutral axis; shear is maximum at the neutral axis and zero at the outer fibres. NA Comp Tens Bending stress σ Linear: max at outer fibres NA τ_max Shear stress τ Parabolic: max at NA Same beam, different distributions
Bending stress and transverse shear stress have fundamentally different distributions. Never confuse them — a common source of exam and design errors.
🎮 Interactive: cross-section stress heatmap LIVE
Predict first: Set width to 150 mm, height to 300 mm, and moment to 50 kN·m. Switch to bending-stress mode. Where is the stress zero, and where is it maximum? Now double the height — how does the maximum stress change?

An interactive rectangular beam cross-section simulator showing a colour heatmap of bending stress with sliders for width, height, moment, and shear force.

Explore how beam dimensions and applied moment affect the bending-stress distribution. Toggle between bending stress and transverse shear stress to see the contrasting patterns.
📝 Worked example: A rectangular timber beam is 100 mm wide × 200 mm deep. The maximum bending moment is 30 kN·m. Find the maximum bending stress and state whether it occurs at the top or bottom fibre.
  1. I = b·h³/12 = 100 × 200³ / 12 = 6.667 × 10⁷ mm⁴ = 6.667 × 10⁻⁵ m⁴.
  2. c = h/2 = 100 mm = 0.100 m.
  3. σ_max = M·c / I = 30 000 × 0.100 / 6.667×10⁻⁵ = 45.0 MPa.
  4. For a sagging moment, the top fibre is in compression and the bottom fibre is in tension; both have magnitude 45.0 MPa.
✓ σ_max = 45.0 MPa, occurring at both the top (compression) and bottom (tension) fibres.
✏️ Practice: A rectangular steel beam 80 mm wide × 160 mm deep carries a bending moment of 12 kN·m. What is the maximum bending stress, in MPa?
MPa
Solution
  1. I = 80 × 160³ / 12 = 2.731 × 10⁷ mm⁴ = 2.731 × 10⁻⁵ m⁴.
  2. c = 80 mm = 0.080 m.
  3. σ_max = 12 000 × 0.080 / 2.731×10⁻⁵ = 35.2 MPa.
✏️ Practice: A beam must carry a bending moment of 24 kN·m with an allowable bending stress of 150 MPa. What is the minimum required section modulus S, in cm³?
cm³
Solution
  1. S_req = M / σ_allow = 24 000 / (150 × 10⁶) = 1.60 × 10⁻⁴ m³.
  2. Convert to cm³: 1.60 × 10⁻⁴ m³ × 10⁶ = 160 cm³.

Check your understanding

1. In elastic bending of a symmetric homogeneous beam, the bending stress at the neutral axis is:
σ = −M·y/I. At y = 0 (the neutral axis), the bending stress is exactly zero.
2. The section modulus S is defined as:
S = I/c. It condenses the cross-section geometry into a single number for quick design checks: σ_max = M/S.
3. If the depth of a rectangular beam is doubled while the width and load stay the same, the maximum bending stress:
For a rectangle, I ∝ h³ and c ∝ h, so S = I/c ∝ h². Doubling h quadruples S, reducing σ_max to one-quarter.
🔎 Related lessons
Transverse Shear Stress in Beams · soon Selecting Beam Sections — an Intro to Design · soon
✅ Key takeaways
  • Flexure formula: σ = −M·y/I (linear distribution, zero at the neutral axis).
  • Maximum stress: σ_max = M·c/I = M/S, where S = I/c is the section modulus.
  • For elastic symmetric bending of a homogeneous beam, the neutral axis passes through the centroid.
  • Bending stress is maximum at the outer fibres and zero at the neutral axis.
➡️ The flexure formula assumes a smooth, undisturbed cross-section. But real beams have holes, fillets, and notches that locally amplify stress. The next lesson quantifies that amplification with stress concentration factors.
Want to test yourself on this? Try the Mechanical Aptitude test →
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