Axial Deformation & Thermal Stress

Worked: a tie rod

A 1 in diameter steel rod (A = 0.785 in², E = 29,000 ksi) 10 ft long carries 15 kips. σ = 15/0.785 = 19.1 ksi and δ = 15 × 120/(0.785 × 29,000) = 0.079 in. L goes in inches because E is in ksi.

Now the same rod sits between rigid walls with a 0.02 in gap, unloaded, and warms 50°F (α = 6.5 × 10⁻⁶ /°F). Free expansion would be 6.5 × 10⁻⁶ × 120 × 50 = 0.039 in. The first 0.02 in closes the gap; the remaining 0.019 in is prevented, so σ = E·δ/L = 29,000 × 0.019/120 = 4.6 ksi compression.