Axial Stress & Deformation

σ = \frac{P}{A}, δ = \frac{P L}{A E}, ε = \frac{δ}{L} = \frac{σ}{E}

Uniform axial stress is load over area. Elastic deformation uses Hooke’s law with modulus E from the Handbook materials tables. Thermal strain is αΔT, independent of stress if free to expand.

$P$ (N): Axial force
$A$ (m²): Cross-sectional area
$E$ (Pa): Modulus of elasticity

Step-by-step solved example

A 2 m steel rod, A = 400 mm², E = 200 GPa, carries 80 kN. Find δ and σ.

Axial bar: σ = P/A, uniform on the cut.

1. Stress
σ = 80e3 / 400e−6 = 200 MPa.
2. Elongation
δ = PL/AE = 80e3 × 2 / (400e−6 × 200e9) = 2.00 mm.
$δ = 2.00 mm$

Answer: σ = 200 MPa, δ = 2.00 mm

10 practice questions

0/10 correct

1.P = 50 kN, A = 250 mm². σ =

2.Strain if δ = 1.5 mm over L = 3 m is

3.For steel, E is typically

4.A stiffer member (larger AE) deforms

5.Thermal strain αΔT with α = 12e−6 /°C, ΔT = 50°C is

6.Poisson’s ratio ν is

7.Shear modulus G relates to E, ν by

8.Statically indeterminate two-bar system needs

9.A 10 mm diameter rod, P = 15.7 kN. σ ≈

10.Factor of safety on yield is