Mechanics of Materials: Stress, Strain, and Torsion

Axial stress, strain, and Hooke's law

Normal (axial) stress is force per unit area over the resisting cross-section:

$\sigma = \frac{P}{A}$

Normal strain is the fractional change in length, ε = δ/L (dimensionless). In the linear-elastic range these obey Hooke's law:

$\sigma = E\,\varepsilon$

where E is the modulus of elasticity (Young's modulus). For structural steel E ≈ 200 GPa (29×10⁶ psi); aluminum ≈ 70 GPa. Worked example: a 12 mm diameter steel rod (A = π(0.012)²/4 = 1.131×10⁻⁴ m²) carries 20 kN. σ = 20 000/1.131×10⁻⁴ = 177 MPa. Strain ε = σ/E = 177×10⁶/200×10⁹ = 8.85×10⁻⁴. Over a 2 m length, δ = εL = 1.77 mm, which also follows from δ = PL/(AE).

The stress-strain diagram

A tensile test traces the material's signature curve, and FE questions probe its named points:

The slope of the initial straight line is E. The area under the whole curve up to fracture is the modulus of toughness (energy absorbed); the area under the elastic portion is the modulus of resilience. Ductile materials (mild steel) show large plastic strain and necking; brittle materials (cast iron, ceramics) fracture with little warning near the proportional limit.

Axial deformation and thermal effects

For a prismatic bar, total elongation is

$\delta = \frac{PL}{AE}$

Stiffness AE/L behaves like a spring constant. Thermal strain from a temperature change is ε_T = αΔT, where α is the coefficient of thermal expansion (steel ≈ 12×10⁻⁶ /°C). A bar free to expand develops strain but no stress. If both ends are fixed so it cannot lengthen, the prevented strain produces thermal stress σ = EαΔT. A fully restrained steel bar heated 50°C carries σ = (200×10⁹)(12×10⁻⁶)(50) = 120 MPa compression — large enough to matter even without any mechanical load.

Poisson's ratio and elastic-constant relations

When a bar stretches axially it contracts laterally. Poisson's ratio is the magnitude of that coupling:

$\nu = -\frac{\varepsilon_{lateral}}{\varepsilon_{axial}}$

Most metals have ν ≈ 0.3; the thermodynamic upper bound is 0.5 (incompressible). The three isotropic elastic constants are not independent — the Handbook gives

$G = \frac{E}{2(1+\nu)}$

so for steel G = 200/(2·1.3) = 76.9 GPa. The bulk modulus follows from K = E/[3(1 − 2ν)].

Torsion of circular shafts

A torque T twisting a circular shaft produces shear stress that varies linearly from zero at the center to a maximum at the surface:

$\tau = \frac{T r}{J}, \qquad \tau_{max} = \frac{T R}{J}$

where J is the polar moment of inertia: J = πd⁴/32 for a solid shaft, J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist over length L is

$\phi = \frac{TL}{JG}$

Worked example: a solid 40 mm diameter steel shaft (J = π(0.040)⁴/32 = 2.51×10⁻⁷ m⁴, G = 80 GPa) carries 500 N·m. Surface stress τ = TR/J = 500(0.020)/2.51×10⁻⁷ = 39.8 MPa. Over 1.5 m, twist φ = TL/(JG) = 500(1.5)/(2.51×10⁻⁷·80×10⁹) = 0.0373 rad = 2.14°. Power transmitted by a rotating shaft is P = Tω, a common combined torsion question: a shaft at 1800 rpm (ω = 188.5 rad/s) delivering 50 kW carries T = P/ω = 50 000/188.5 = 265 N·m. Hollow shafts give nearly the same strength at less weight because material near the center carries little shear stress.

Factor of safety and allowable stress

Design rarely loads a part to failure. The factor of safety is FS = S_y/σ_allow (or S_u/σ_allow for brittle materials), so a steel part (S_y = 250 MPa) designed to FS = 2 may see at most 125 MPa working stress. FE problems run this both ways: given a load and required FS, size the area (A ≥ FS·P/S_y); or given dimensions, report the resulting FS. Always match the strength basis (yield for ductile, ultimate for brittle) to the material.

Thin-walled pressure vessels

A cylindrical thin-walled pressure vessel (wall thickness t much less than radius r) carries two membrane stresses from internal pressure p. The hoop (circumferential) stress is σ_h = pr/t and the longitudinal (axial) stress is σ_l = pr/(2t) — hoop is exactly twice longitudinal, which is why cylindrical tanks split along a longitudinal seam. A spherical vessel carries σ = pr/(2t) in every direction. Worked example: a pipe with r = 0.3 m, t = 6 mm, p = 2 MPa gives σ_h = 2(0.3)/0.006 = 100 MPa and σ_l = 50 MPa. These are the σ_x and σ_y that feed directly into a Mohr's-circle or failure-theory check.

Shear and bearing connections

Bolted and riveted joints introduce two extra stress checks. Direct (average) shear on a pin or bolt is τ = V/A, where A is the bolt cross-section (πd²/4); double shear doubles the area. Bearing stress is the contact pressure σ_b = P/(t·d), using the projected area of the hole (plate thickness × bolt diameter). For example, a 20 mm bolt bearing P = 30 kN on an 8 mm plate gives σ_b = 30000/(0.008 × 0.020) = 188 MPa. FE connection problems ask you to find which mode — bolt shear, bearing, or plate tension on the net section — governs by comparing each against its allowable, the lowest capacity controlling the joint.