Bending stresses in beams

Why it matters

Chassis rails, axle beams, leaf springs, crane jibs and lifting arms fail most often by bending. The bending (flexure) formula turns the bending moment from the BMD into the stress at the top and bottom fibres, which is then compared with the allowable stress. It also explains why I-sections, box sections and hollow tubes are so efficient: they put material far from the neutral axis, where it does the most work.

Key ideas

Theory of simple (pure) bending: assumptions.

• The beam is initially straight and has a constant cross-section with a plane of symmetry; loads act in that plane.
• The material is homogeneous, isotropic and linear elastic, with the same E in tension and compression.
• Plane sections remain plane after bending (Bernoulli–Euler).
• Stresses stay below the proportional limit, and the radius of curvature is large compared with the depth.
• Shear deformation is negligible (true for slender beams, span much larger than depth).

Strain and stress distribution. When a beam bends, fibres on one side shorten and on the other lengthen. Somewhere between, a layer (the neutral surface) has no strain. Its intersection with a cross-section is the neutral axis (NA). Strain varies linearly with distance y from the NA: ε = y/R. With Hooke's law, stress also varies linearly: σ = E·y/R. It is zero at the NA and maximum at the extreme fibres.

Location of the NA. For pure bending with no net axial force, ∫σ dA = 0, which puts the NA through the centroid of the section (for a homogeneous elastic beam). Moment equilibrium then gives M = (E/R)·I, where I is the second moment of area about the NA.

Flexure formula. Combining the two gives M/I = σ/y = E/R. Under sagging moment, fibres above the NA are in compression and those below in tension; under hogging, the reverse.

Section modulus. Z = I/y_max, so σ_max = M/Z. For a given area, a larger Z means a stronger section. Values (about the centroidal axis that bending happens about):

• Rectangle b × d: I = b·d³/12, Z = b·d²/6. Standing a plank on edge greatly increases strength.
• Solid circle d: I = π·d⁴/64, Z = π·d³/32.
• Hollow circle D, d: Z = π·(D⁴ − d⁴)/(32·D).
• I-section: compute by parts (flanges contribute most of I).

Unsymmetrical sections. For T, channel or angle sections bent about the axis through the centroid, the distances to the top and bottom fibres differ, so the extreme stresses differ: σ_top = M·y_top/I, σ_bottom = M·y_bottom/I. For materials weaker in tension (cast iron), put the larger part of the section on the tension side so the tension fibre is closer to the NA.

Moment of resistance. The largest moment a section can carry at allowable stress: M_r = σ_allow·Z (for unequal allowables, the smaller of σ_t·I/y_t and σ_c·I/y_c).

Combined axial load and bending. Superpose: σ = P/A ± M·y/I. The NA then shifts away from the centroid.

Beyond the elastic limit. At first yield the outer fibres reach σ_y; full plastic moment M_p = σ_y·Z_p, where the shape factor Z_p/Z is 1.5 for a rectangle and about 1.12–1.15 for I-sections.

Formulas

M / I = σ / y = E / R

• M: bending moment at the section (N·m); I: second moment of area about the neutral axis (m⁴); σ: bending stress at distance y from the NA (Pa); 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).

I = b·d³/12, Z = b·d²/6 (rectangle); I = π·d⁴/64, Z = π·d³/32 (solid circle); Z = π·(D⁴ − d⁴)/(32·D) (hollow circle)

I_NA = Σ (I_c + A·h²) (parallel-axis theorem)

• I_c: second moment of area of each part about its own centroidal axis; A: its area; h: distance from its centroid to the NA.

σ = P/A ± M·y/I

• Combined axial force P and bending.

Worked examples

Example 1 (standard): rectangular beam. Given: a simply supported steel beam, span 4 m, UDL 10 kN/m, rectangular section 100 mm wide × 200 mm deep, E = 200 GPa. Find the maximum bending stress and the radius of curvature at mid-span.

1. M_max = w·L²/8 = 10 × 4²/8 = 20 kN·m = 20 × 10⁶ N·mm.
2. I = b·d³/12 = 100 × 200³/12 = 66.67 × 10⁶ mm⁴; Z = b·d²/6 = 100 × 200²/6 = 666.7 × 10³ mm³.
3. σ_max = M/Z = 20 × 10⁶ / 666.7 × 10³ = 30 MPa (compression at top, tension at bottom).
4. R = E·I/M = 200 000 × 66.67 × 10⁶ / 20 × 10⁶ = 666.7 × 10³ mm. Answer: σ_max = 30 MPa; R ≈ 667 m.

Example 2 (GATE level): T-section with unequal allowable stresses. Given: a T-section with flange 100 mm × 20 mm on top and web 20 mm × 80 mm below (overall depth 100 mm) is used as a simply supported beam of span 2 m with a central point load W. Allowable stresses: 40 MPa in tension, 80 MPa in compression. Find the maximum W.

1. Centroid from the top: flange area 2000 mm² at 10 mm, web area 1600 mm² at 60 mm. ȳ = (2000 × 10 + 1600 × 60)/3600 = 32.22 mm. So y_top = 32.22 mm and y_bottom = 67.78 mm.
2. I_NA = [100 × 20³/12 + 2000 × 22.22²] + [20 × 80³/12 + 1600 × 27.78²] = (66 667 + 987 654) + (853 333 + 1 234 568) = 3.142 × 10⁶ mm⁴.
3. Sagging moment: top in compression, bottom in tension.
4. Tension limit: M = 40 × 3.142 × 10⁶ / 67.78 = 1.854 × 10⁶ N·mm.
5. Compression limit: M = 80 × 3.142 × 10⁶ / 32.22 = 7.80 × 10⁶ N·mm.
6. The smaller governs: M_r = 1.854 kN·m. With M_max = W·L/4, W = 4 × 1.854 / 2 = 3.71 kN. Answer: W ≈ 3.71 kN, governed by tension in the bottom of the web. Turning the T upside down would put the tension fibre only 32.22 mm from the NA and allow a much larger load, which is why cast-iron brackets are shaped with the flange on the tension side.

Common mistakes

• Using the full depth instead of the distance from the NA to the extreme fibre.
• Taking the NA at mid-depth for an unsymmetrical section.
• Forgetting the parallel-axis term A·h² when building up I.
• Unit errors: kN·m with mm⁴ (convert M to N·mm to get MPa directly).
• Confusing I about the wrong axis, for example b·d³/12 when the beam bends about the other axis (d·b³/12).
• Checking only the larger stress when tension and compression allowables differ.

For GATE ME

Expect maximum bending stress for given loads and sections, ratio of strengths of rectangular, square, circular and hollow sections of equal area or weight, extreme stresses in T and I sections, radius of curvature from M/I = E/R, and combined axial and bending stress. Practise computing I and the centroid quickly for composite sections.

Quick check

1. A square and a circle have the same area. Which has the larger section modulus?
2. A rectangular beam is turned from flat (b > d) to on edge. By what factor does Z change?
3. What is the bending stress at the neutral axis?
4. For a solid circular section of diameter 0.3 m and allowable stress 150 MPa, what moment can it carry?

Answers: 1. The square (about 18 % larger). 2. By the ratio of the longer to the shorter side. 3. Zero. 4. M = σ·π·d³/32 = 150 × 10⁶ × 2.651 × 10⁻³ ≈ 398 kN·m.