Bending and shear stresses in beams

Why it matters

Once the bending moment and shear force diagrams are known, the beam's cross-section must be chosen so that the stresses they cause are safe. Bending stresses usually decide the size of beams, axles and machine frames; shear stresses decide web thickness, short heavily loaded beams, and the welds, bolts or glue that join built-up sections.

Key ideas

Assumptions of simple bending theory (Euler–Bernoulli).

1. The beam is initially straight, with a constant cross-section symmetric about the plane of loading.
2. The material is homogeneous, isotropic and linearly elastic, with the same E in tension and compression.
3. Plane sections remain plane and perpendicular to the neutral axis after bending.
4. Deflections are small and the radius of curvature is large compared with the depth. Strictly these hold for pure bending (constant M, no shear); the error with shear present is small for slender beams.

Neutral axis. Under bending, fibres on one side stretch and on the other side shorten; between them lies the neutral surface, which has no strain. For pure bending its trace in the section (the neutral axis) passes through the centroid. Strain, and hence stress, varies linearly with the distance y from it. In a sagging region the top is in compression and the bottom in tension; in a hogging region the opposite.

Flexure formula. M/I = σ/y = E/R. The largest stress is at the extreme fibre: σmax = M/Z with section modulus Z = I/ymax. For unsymmetric sections (T, channel) the two extreme fibres have different distances, so compressive and tensile maxima differ; this matters for cast iron, which is weak in tension.

Efficient sections. For a given area, putting material far from the neutral axis raises I and Z. That is why I-sections and box sections are used, and why a rectangular beam is much stronger on edge (Z = bh²/6 grows with depth squared).

Shear stress in beams. A varying moment (V ≠ 0) produces horizontal shear between layers, and equal complementary vertical shear on the section: τ = VQ/(Ib). Q is the first moment about the neutral axis of the area beyond the level considered, and b is the width at that level. τ is zero at the extreme fibres and usually maximum at the neutral axis (not for every shape: for a triangle it is at mid-depth).

• Rectangle: parabolic distribution, τmax = 1.5 τavg at the neutral axis.
• Solid circle: τmax = (4/3) τavg.
• I-section: the web carries almost all the shear and τ is nearly uniform in it; τ jumps at the flange–web junction because b changes.

Combined check. Bending stress is maximum at the extreme fibre of the maximum-moment section; shear stress is maximum at the neutral axis of the maximum-shear section. Check both. Where both are significant (e.g. at the flange–web junction near a support), combine them with a failure theory.

Formulas

M / I = σ / y = E / R

• M = bending moment (N·mm), I = second moment of area about the neutral axis (mm⁴), σ = bending stress (MPa) at distance y (mm), E = Young's modulus (MPa), R = radius of curvature (mm).

σmax = M / Z, Z = I / ymax

I = b h³ / 12, Z = b h² / 6 (rectangle b × h) I = π d⁴ / 64, Z = π d³ / 32 (solid circle) I = I_G + A d² (parallel-axis theorem for built-up sections)

τ = V Q / (I b)

• V = shear force (N), Q = first moment of area beyond the level (mm³), b = width at that level (mm).

τmax = 1.5 V / (b h) (rectangle), τmax = (4/3) · V / (π d²/4) (solid circle)

Worked examples

Example 1 (standard). A simply supported beam of span 4 m carries a UDL of 10 kN/m. Its section is rectangular, 100 mm wide and 200 mm deep. Find the maximum bending and shear stresses.

1. M_max = wL²/8 = 10 × 16 / 8 = 20 kN·m = 20 × 10⁶ N·mm; V_max = wL/2 = 20 kN.
2. I = 100 × 200³ / 12 = 6.667 × 10⁷ mm⁴, ymax = 100 mm.
3. σmax = 20 × 10⁶ × 100 / 6.667 × 10⁷ = 30 MPa (compression at top, tension at bottom, mid-span).
4. τmax = 1.5 × 20 000 / (100 × 200) = 1.5 MPa (neutral axis, at the supports).

Answer: σmax = 30 MPa, τmax = 1.5 MPa.

Example 2 (GATE level, T-section). A T-beam has a flange 100 mm × 20 mm and a web 20 mm × 100 mm (overall depth 120 mm). At one section it carries a sagging moment of 5 kN·m and a shear force of 10 kN. Find the extreme bending stresses and the shear stress at the neutral axis and at the flange–web junction.

1. Centroid from the top: ȳ = (2000 × 10 + 2000 × 70) / 4000 = 40 mm.
2. I = 100 × 20³/12 + 2000 × 30² + 20 × 100³/12 + 2000 × 30² = 66 667 + 1 800 000 + 1 666 667 + 1 800 000 = 5.333 × 10⁶ mm⁴.
3. Top (40 mm above NA, compression): σ = 5 × 10⁶ × 40 / 5.333 × 10⁶ = 37.5 MPa.
4. Bottom (80 mm below NA, tension): σ = 5 × 10⁶ × 80 / 5.333 × 10⁶ = 75.0 MPa.
5. Shear at NA: area below NA is the web, 20 × 80 mm, centroid 40 mm below NA: Q = 1600 × 40 = 64 000 mm³; τ = 10 000 × 64 000 / (5.333 × 10⁶ × 20) = 6.0 MPa.
6. At the junction: Q = 2000 × 30 = 60 000 mm³. In the web (b = 20 mm): τ = 5.63 MPa; in the flange (b = 100 mm): τ = 1.13 MPa.

Answer: 37.5 MPa compression at the top, 75.0 MPa tension at the bottom; τ = 6.0 MPa at the NA, 5.63 MPa (web) and 1.13 MPa (flange) at the junction.

Common mistakes

• Using the distance from the top of the section instead of from the neutral axis for y.
• Using I about the base instead of about the centroidal axis; forgetting the A d² terms.
• Taking the same ymax for both faces of an unsymmetric section.
• Using the full flange width b in τ = VQ/(Ib) for a point in the web.
• Unit slips: N·m with mm⁴. Keep N and mm throughout so stresses come out in MPa.
• Assuming shear stress is always largest at the neutral axis for every shape.

For GATE PI

Expect section modulus and maximum stress for rectangles, circles, hollow and I/T sections, the ratio of maximum to average shear stress (1.5 for rectangle, 4/3 for circle), choice between orientations of a rectangular beam, and beams of uniform strength. Practise finding the centroid and I of built-up sections quickly.

Quick check

1. A rectangular beam is turned from flat (b > h) to on edge. How does Z change?
2. What is τmax/τavg for a rectangular section?
3. Bending stress at the neutral axis?
4. Z of a 100 mm × 200 mm rectangle (bending about the stronger axis)?

Answers: 1. It increases by the ratio of the depths (Z = bh²/6). 2. 1.5. 3. Zero. 4. 6.67 × 10⁵ mm³.