Bending stresses in beams

Why it matters

Most structural members in a machine are loaded in bending: robot arm links, aluminium extrusion profiles on linear axes, gantry beams, sensor brackets and PCB stiffeners. The flexure formula tells you the peak stress for a given bending moment and, just as important, how cross-section shape changes it. Choosing a tube instead of a solid bar, or an I-profile instead of a flat bar, can cut mass several-fold for the same stress, which matters for moving axes where mass costs motor torque.

Key ideas

Assumptions of simple (Euler-Bernoulli) bending theory.

• The beam is initially straight, with a constant cross-section symmetric about the plane of loading.
• Material is homogeneous, isotropic and linear-elastic, with the same E in tension and compression.
• Plane sections remain plane after bending and perpendicular to the deformed axis.
• Stresses stay below the proportional limit; deflections are small compared with the depth; shear deformation is neglected.

Pure bending and the neutral axis. Under a bending moment the beam bends into an arc of radius R. Fibres on the concave side shorten, those on the convex side stretch, and one surface in between, the neutral surface, keeps its length. Its intersection with the 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.

Locating the NA. With no axial force, the net normal force on the section must be zero, so ∫σ dA = 0, which puts the NA through the centroid of the section (for a homogeneous elastic beam). For an unsymmetric section such as a T or channel, find the centroid first.

The flexure formula. Moment equilibrium of the stresses gives M = (E/R)·∫y² dA = E·I/R. Combining: M/I = σ/y = E/R. Peak stress occurs at the outermost fibre, y_max: σ_max = M/Z, with section modulus Z = I/y_max. For an unsymmetric section the tension and compression faces are at different distances, so the two peak stresses differ.

Second moment of area (I). I = ∫y² dA about the neutral axis. Material far from the NA counts most, which is why I-sections and tubes are efficient. Use the parallel-axis theorem, I = I_c + A·d², to build composite sections.

Sagging and hogging. In a sagging (positive) moment the top fibres are in compression and the bottom in tension; in a hogging moment the reverse. For materials weaker in tension (cast iron, concrete), place more material on the tension side, for example a T-beam with the flange in compression.

Beam of uniform strength. If Z is varied along the length so that M/Z is constant, every section reaches the allowable stress together, saving material, as in leaf springs and tapered cantilever arms.

Section shapes compared. For the same area, a hollow tube or I-section has a much larger Z than a solid round or square. A rectangle is twice as strong laid on edge as flat if h = 2b, because Z = b·h²/6.

Formulas

M / I = σ / y = E / R

• M: bending moment at the section (N·m); I: second moment of area about the NA (m⁴); σ: bending stress at distance y from the NA (Pa); y: distance from the NA (m); 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 of outer diameter D and inner diameter d (m).

I = I_c + A·d²

• Parallel-axis theorem; I_c: about the part's own centroid; A: area of the part (m²); d: distance between the axes (m).

Worked examples

Example 1 (standard): aluminium robot arm tube. Given: an aluminium tube, outer diameter 60 mm, inner diameter 50 mm, acts as a cantilever arm 0.8 m long carrying a 200 N load at the tip. E = 70 GPa. Find the maximum bending stress and the radius of curvature at the root.

1. Root moment: M = 200 × 0.8 = 160 N·m = 160 000 N·mm.
2. I = π·(D⁴ − d⁴)/64 = π × (60⁴ − 50⁴)/64 = π × 6 710 000/64 = 329 400 mm⁴.
3. Z = I/y_max = 329 400/30 = 10 980 mm³.
4. σ_max = M/Z = 160 000/10 980 = 14.6 MPa (tension on top, compression on bottom, since a cantilever with a downward load hogs).
5. R = E·I/M = 70 000 × 329 400/160 000 = 144 100 mm ≈ 144 m. Answer: σ_max ≈ 14.6 MPa, R ≈ 144 m at the root. The stress is low; for a robot arm the stiffness (deflection) usually governs, not strength.

Example 2 (GATE level): T-section beam. Given: a T-section with flange 100 mm × 20 mm and web 20 mm × 80 mm (total depth 100 mm), flange on top, used as a simply supported beam with a sagging moment of 5 kN·m at the critical section. Find the NA position, I and the extreme fibre stresses.

1. Centroid from the bottom: flange area 2000 mm² at 90 mm; web area 1600 mm² at 40 mm. ȳ = (2000 × 90 + 1600 × 40)/3600 = 67.78 mm.
2. Flange: I = 100 × 20³/12 + 2000 × (90 − 67.78)² = 66 667 + 987 654 = 1 054 321 mm⁴.
3. Web: I = 20 × 80³/12 + 1600 × (67.78 − 40)² = 853 333 + 1 234 568 = 2 087 901 mm⁴.
4. Total I = 3 142 222 mm⁴ ≈ 3.142 × 10⁶ mm⁴.
5. Top fibre (y = 100 − 67.78 = 32.22 mm, compression): σ = 5 × 10⁶ × 32.22/3.142 × 10⁶ = 51.3 MPa.
6. Bottom fibre (y = 67.78 mm, tension): σ = 5 × 10⁶ × 67.78/3.142 × 10⁶ = 107.9 MPa. Answer: NA 67.8 mm above the base; I ≈ 3.14 × 10⁶ mm⁴; 51.3 MPa compression at the top, 107.9 MPa tension at the bottom. The tension face is about twice as far from the NA, so it is the critical face for a sagging moment.

Common mistakes

• Taking y from the top or bottom edge instead of from the neutral axis.
• Assuming the NA is at mid-depth for unsymmetric sections.
• Forgetting the A·d² term in the parallel-axis theorem, or applying it about the wrong axis.
• Mixing units: with N·mm and mm⁴, stress is in N/mm² (MPa); with N·m and m⁴, it is in Pa.
• Using b·h³/12 with b and h swapped, or about the weak axis by mistake.
• Thinking material (E) affects bending stress in a homogeneous beam; it affects curvature and deflection, not σ = M·y/I.

For GATE ME

Expect maximum bending stress for standard sections, ratio of strengths of different shapes (square versus circle of equal area, rectangle on edge versus flat), stresses in T and I sections, radius of curvature from M/I = E/R, and beams of uniform strength. Practise centroid and parallel-axis calculations quickly; they are the usual source of lost marks.

Quick check

1. If I is doubled with y_max unchanged, what happens to σ_max?
2. Find Z for a 0.1 m wide, 0.2 m deep rectangle.
3. What is the bending stress at the neutral axis?
4. A 100 mm diameter solid shaft carries a bending moment of 1 kN·m. What is σ_max?
5. Which face of a cantilever with a downward tip load is in tension?

Answers: 1. It halves. 2. Z = 0.1 × 0.2²/6 = 6.67 × 10⁻⁴ m³. 3. Zero. 4. 32 × 10⁶/(π × 10⁶) = 10.2 MPa. 5. The top face.