Bending Stress and Torsion

Bending (Flexure) Stress

In pure bending, plane sections stay plane and the normal stress varies linearly from the neutral axis, which passes through the centroid:

σ = M y / I

where M is the bending moment, y is the distance from the neutral axis, and I is the centroidal moment of inertia. Stress is zero at the neutral axis and largest at the extreme fiber, where y = c:

σ_max = M c / I = M / S

The section modulus S = I/c packages the geometry into one number, so σ_max = M/S — a larger S means lower stress for the same moment. For a sagging (positive M) beam, the top fiber is in compression and the bottom in tension.

Section properties of common shapes

Worked example — maximum bending stress

A rectangular beam 50 mm wide × 100 mm tall carries M = 10 kN·m. Find σ_max.

1. S = bh²/6 = (50)(100²)/6 = 500,000/6 = 83,333 mm³.
2. Convert moment to N·mm: M = 10 kN·m = 10×10⁶ N·mm.
3. σ_max = M/S = 10×10⁶ / 83,333 = 120 MPa.

The trap is the moment unit: 10 kN·m = 10,000 N·m = 10,000,000 N·mm. Forgetting the ×1000 (kN→N) and the ×1000 (m→mm) is the classic error.

Transverse Shear Stress

Where the bending moment changes (dM/dx = V ≠ 0), a transverse shear stress acts on the cross-section:

τ = V Q / (I b)

• V = shear force at the section
• Q = first moment of the area above (or below) the level of interest, taken about the neutral axis: Q = A′ ȳ′
• I = moment of inertia of the whole section
• b = section width at the level of interest

Because Q is largest at the neutral axis, maximum transverse shear occurs at the neutral axis — exactly where bending stress is zero. This complementary pattern is a favorite FE distinction.

Torsion of Circular Shafts

For a circular shaft carrying torque T, the shear stress grows linearly from the center:

τ = T r / J

with the maximum at the outer surface (r = c = d/2), τ_max = Tc/J. The polar moment of inertia J is:

• Solid shaft: J = πd⁴/32
• Hollow shaft: J = π(d_o⁴ − d_i⁴)/32

Worked example — polar moment and shaft stress

For a solid shaft d = 50 mm carrying T = 2 kN·m:

1. J = πd⁴/32 = π(50⁴)/32 = π(6,250,000)/32 = 613,592 mm⁴.
2. c = d/2 = 25 mm; T = 2×10⁶ N·mm.
3. τ_max = Tc/J = (2×10⁶)(25)/613,592 = 50×10⁶/613,592 = 81.5 MPa.

Angle of Twist

φ = T L / (G J) (radians)

The twist is proportional to torque and length and inversely proportional to the shaft's stiffness GJ. For shafts in series, sum φ segment by segment, just as with axial δ.

Power Transmission

A rotating shaft transmits power:

P = T ω, where ω = 2πn/60 for n in rpm

P is in watts when T is in N·m and ω in rad/s. Rearranged to size a shaft from a motor rating: T = P/ω.

Worked example — torque from motor power

A motor delivers 50 kW at 1,500 rpm. Find the shaft torque.

1. ω = 2π(1,500)/60 = 157.08 rad/s.
2. T = P/ω = 50,000 / 157.08 = 318.3 N·m.

Note the inverse relationship: at fixed power, higher rpm means lower torque, so high-speed shafts can be slimmer.

Putting It Together — Common Traps

• Mc/I vs. Tr/J. Bending uses the area moment of inertia I = πd⁴/64 for a circle; torsion uses the polar moment J = πd⁴/32. Note J = 2I for a circle. Mixing them halves or doubles the stress.
• Shear location. Bending stress is max at the extreme fiber; transverse shear is max at the neutral axis. They never peak at the same place.
• Units. Convert kN·m to N·mm (×10⁶) before dividing by mm³ or mm⁴ section properties.
• Hollow shafts are efficient. Removing the low-stress core near the center barely reduces J but saves weight — why drive shafts are tubes.

Quick reference