Bending stresses in beams

Assumptions of simple bending, the flexure formula M/I = σ/y = E/R, section modulus, neutral axis of unsymmetric sections and the governing fibre.

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Why it matters

Most aircraft structural members, from wing spars and stringers to floor beams and landing-gear legs, are sized by bending. The flexure formula tells you the stress at any fibre of a section for a given bending moment, which is why spar caps are placed as far from the neutral axis as possible and why I- and box-sections dominate aerospace structures.

Key ideas

Assumptions of simple (Euler–Bernoulli) bending theory:

  • The beam is initially straight, and the material is homogeneous, isotropic and linear elastic with the same E in tension and compression.
  • Plane cross-sections remain plane and perpendicular to the deformed axis.
  • The section has an axis of symmetry in the plane of loading, so bending occurs in that plane (no twisting).
  • The radius of curvature is large compared with the depth, and the beam is long compared with its depth.
  • Strictly the theory is for pure bending (constant M, no shear); it is used for ordinary transversely loaded beams because the shear correction is small for slender beams.

Strain and stress distribution. A fibre at distance y from the neutral surface, bent to radius R, has strain ε = y/R. With Hooke's law, σ = E·y/R: stress varies linearly with y, zero at the neutral axis and largest at the extreme fibres. For a sagging moment, fibres above the neutral axis are in compression and those below in tension.

Neutral axis. Since there is no net axial force, ∫σ dA = 0, so ∫y dA = 0: for a homogeneous beam the neutral axis passes through the centroid of the section. The moment balance ∫σ·y dA = M then gives M = E·I/R, where I is the second moment of area about the neutral axis.

The flexure formula. Combining the two results gives M/I = σ/y = E/R. The extreme-fibre stress is σ_max = M/Z, where the section modulus Z = I/y_max measures bending strength. For a symmetric section the top and bottom stresses are equal in magnitude; for an unsymmetric section (T, channel) they are not, and the side farther from the neutral axis has the larger stress.

Efficient sections. Material near the neutral axis is lightly stressed. Moving area outward (I-sections, box beams, sandwich panels) raises I and Z for the same weight. For a rectangle, Z = b·h²/6, so doubling the depth quadruples the strength while doubling the width only doubles it.

Beams of uniform strength vary the section so that M/Z, and hence σ_max, is constant along the span.

Composite sections (for example a timber beam with steel plates) are handled by the transformed-section method with modular ratio m = E₂/E₁. This is beyond the basic formula above.

Formulas

M / I = σ / y = E / R M = bending moment (N·m), I = second moment of area about the neutral axis (m⁴), σ = bending stress (Pa) at distance y (m) from the neutral axis, E = Young's modulus (Pa), R = radius of curvature of the neutral surface (m).

σ_max = M / Z, Z = I / y_max Z = section modulus (m³), y_max = distance to the extreme fibre (m).

I = b·h³ / 12, Z = b·h² / 6 Rectangle of width b and depth h (m), about the centroidal axis parallel to b.

I = π·d⁴ / 64, Z = π·d³ / 32 Solid circle of diameter d (m).

I = π·(D⁴ − d⁴) / 64, Z = π·(D⁴ − d⁴) / (32·D) Hollow circle, outer D, inner d (m).

I = Σ (I_c + A·h²) Parallel-axis theorem for built-up sections; I_c = each part's own centroidal I, A = its area, h = distance from its centroid to the neutral axis.

ȳ = Σ A·y / Σ A Centroid location of a composite section (m).

Worked examples

Example 1 (standard). A simply supported beam of span 4 m carries a UDL of 6 kN/m. Its section is a rectangle 120 mm wide and 250 mm deep. Find the maximum bending stress.

  1. M_max = w·L²/8 = 6 × 4²/8 = 12 kN·m = 12 × 10⁶ N·mm.
  2. I = b·h³/12 = 120 × 250³/12 = 1.5625 × 10⁸ mm⁴.
  3. y_max = 250/2 = 125 mm.
  4. σ = M·y/I = 12 × 10⁶ × 125 / 1.5625 × 10⁸ = 9.6 MPa.
  5. Check with Z = b·h²/6 = 120 × 250²/6 = 1.25 × 10⁶ mm³; 12 × 10⁶ / 1.25 × 10⁶ = 9.6 MPa.

Answer: σ_max = 9.6 MPa (compression at the top, tension at the bottom).

Example 2 (GATE level). A T-section has a flange 100 mm × 20 mm on top and a web 20 mm thick × 80 mm deep below it (overall depth 100 mm). The beam carries a sagging moment. Allowable stresses are 60 MPa in tension and 90 MPa in compression. Find the neutral axis, I and the largest allowable moment.

  1. Areas: flange A₁ = 2000 mm² with centroid 10 mm from the top; web A₂ = 1600 mm² with centroid 20 + 40 = 60 mm from the top.
  2. ȳ = ΣA·y/ΣA = (2000 × 10 + 1600 × 60)/3600 = 116 000/3600 = 32.22 mm from the top.
  3. Flange: 100 × 20³/12 + 2000 × (32.22 − 10)² = 66 667 + 987 654 = 1 054 321 mm⁴.
  4. Web: 20 × 80³/12 + 1600 × (60 − 32.22)² = 853 333 + 1 234 568 = 2 087 901 mm⁴.
  5. I = 1 054 321 + 2 087 901 = 3.142 × 10⁶ mm⁴.
  6. Under sagging, the bottom fibre (y = 100 − 32.22 = 67.78 mm) is in tension: M ≤ 60 × 3.142 × 10⁶ / 67.78 = 2.78 × 10⁶ N·mm.
  7. The top fibre (y = 32.22 mm) is in compression: M ≤ 90 × 3.142 × 10⁶ / 32.22 = 8.78 × 10⁶ N·mm.
  8. The smaller value governs.

Answer: ȳ = 32.2 mm from the top, I ≈ 3.14 × 10⁶ mm⁴, M_allow ≈ 2.78 kN·m (tension at the bottom governs). Turning the T upside down would be far better for this material.

Common mistakes

  • Taking I about the base of the section instead of about the neutral axis through the centroid.
  • Using y = h/2 for an unsymmetric section; measure to each extreme fibre separately.
  • Forgetting the A·h² term of the parallel-axis theorem for flanges.
  • Unit slips: N·m with mm⁴ gives nonsense. Use N·mm and mm (result in MPa) or N·m and m (result in Pa).
  • Believing a stiffer material lowers the bending stress. For a given M and section, σ = M·y/I does not depend on E; E affects only curvature and deflection.

For GATE AE

Expect numericals on maximum bending stress for rectangular, circular, hollow and I- or T-sections, locating the neutral axis of unsymmetric sections, choosing the governing fibre when tension and compression limits differ, and comparing section moduli of different shapes of equal area. Conceptual questions test the assumptions and the linear stress distribution. Practise the parallel-axis theorem until it is quick and error-free.

Quick check

  1. Where is bending stress zero in a homogeneous beam?
  2. A rectangle is turned so its depth doubles and width halves. By what factor does Z change?
  3. Write Z for a solid circular section of diameter d.
  4. Does the maximum bending stress for a given M depend on E?

Answers: 1. At the neutral axis, through the centroid. 2. It doubles. 3. Z = π·d³/32. 4. No.

Interview questions

All Mechanics of Solids interview questions

Try answering each one aloud before you open it.

  1. 1.What is bending stress in beams?Concept

    Bending stress in beams is the internal stress induced in a beam when an external bending moment is applied. It is a measure of the distribution of internal forces within the beam that resist the bending. The bending stress varies linearly from the neutral axis, being maximum at the outermost fibers.

  2. 2.Explain the bending equation σ = M·y / I.Concept

    The bending equation σ = M·y / I relates the bending stress (σ) in a beam to the bending moment (M), the distance from the neutral axis (y), and the moment of inertia (I) of the beam's cross-section. It shows that the bending stress is directly proportional to the bending moment and the distance from the neutral axis, and inversely proportional to the moment of inertia.

  3. 3.What is the neutral axis in a beam?Concept

    The neutral axis is the line in the cross-section where bending strain and stress are zero; it is the trace of the neutral surface, whose fibres do not change length. Because the net axial force in pure bending is zero, for a homogeneous linear-elastic beam it passes through the centroid of the section. It separates the compression side from the tension side, and stress grows linearly with distance from it.

  4. 4.Why is the moment of inertia important in analyzing bending stresses?Application

    The moment of inertia is a measure of a beam's resistance to bending and is crucial in analyzing bending stresses. It depends on the shape and size of the beam's cross-section. A higher moment of inertia indicates that the beam can resist more bending, leading to lower bending stresses for the same applied moment.

  5. 5.What happens to the bending stress if the beam's cross-section is changed from rectangular to circular of the same area?Application

    For the same area, a circle is a poor bending section: its I is A²/(4π) ≈ 0.080A², slightly less than a square's A²/12 ≈ 0.083A², and far less than a deep rectangle's. Its section modulus is also lower, so for the same moment the maximum bending stress rises. Circles are chosen for torsion or for loads in any direction, not for bending efficiency.

  6. 6.Explain why I-beams are commonly used in construction.Application

    I-beams are commonly used in construction because their shape provides a high moment of inertia with less material, making them efficient in resisting bending stresses. The flanges resist bending, while the web resists shear forces, providing a strong and lightweight structural element.

  7. 7.What is the effect of increasing the beam length on bending stress?Application

    Increasing the beam length generally increases the bending moment for a given load, which can lead to higher bending stresses. This is because the bending moment is a function of the distance from the load to the point of interest, and a longer beam increases this distance.

  8. 8.Calculate the maximum bending stress in a simply supported beam with a span of 4 meters, subjected to a central point load of 10 kN. The beam has a rectangular cross-section with a width of 100 mm and a height of 200 mm.Numerical

    M_max = P·L/4 = 10 × 4/4 = 10 kN·m = 10 × 10⁶ N·mm. I = b·h³/12 = 100 × 200³/12 = 6.667 × 10⁷ mm⁴ and y_max = 100 mm. σ_max = M·y/I = 10 × 10⁶ × 100 / 6.667 × 10⁷ = 15 MPa, compressive at the top and tensile at the bottom.

  9. 9.A cantilever beam of length 3 meters is subjected to a uniform distributed load of 5 kN/m. Calculate the maximum bending stress if the beam has a circular cross-section with a diameter of 150 mm.Numerical

    M_max = w·L²/2 = 5 × 3²/2 = 22.5 kN·m at the fixed end. Z = π·d³/32 = π × 150³/32 = 3.313 × 10⁵ mm³. σ_max = M/Z = 22.5 × 10⁶ / 3.313 × 10⁵ = 67.9 MPa, tensile at the top and compressive at the bottom of the fixed end.

  10. 10.How does material selection affect the bending stress in beams?Application

    For a statically determinate beam the bending stress σ = M·y/I depends only on the moment and the section, not on the material. Material choice decides whether that stress is acceptable (yield or ultimate strength, fatigue strength) and how much the beam deflects, through E. For weight-critical aerospace parts the useful comparison is strength-to-density and stiffness-to-density.

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