Step-by-step solved example
P = 120 kN, L = 1.5 m, A = 600 mm², E = 200 GPa. Find δ.
1. SI units
A = 600×10⁻⁶ m² = 6.00×10⁻⁴ m². P = 1.20×10⁵ N.
2. Elongation
δ = PL/AE = (1.20×10⁵)(1.5)/[(6.00×10⁻⁴)(2.00×10¹¹)] = 1.50 mm.
Answer: 1.50 mm
P = 120 kN, L = 1.5 m, A = 600 mm², E = 200 GPa. Find δ.
1. SI units
A = 600×10⁻⁶ m² = 6.00×10⁻⁴ m². P = 1.20×10⁵ N.
2. Elongation
δ = PL/AE = (1.20×10⁵)(1.5)/[(6.00×10⁻⁴)(2.00×10¹¹)] = 1.50 mm.
Answer: 1.50 mm
1.P = 50 kN, L = 2 m, A = 500 mm², E = 200 GPa. δ =
2.Doubling A (same P, L, E) will
3.Same P, L, A: aluminum (E ≈ 70 GPa) vs steel (200 GPa)
4.δ = 3.0 mm over L = 2.0 m. Axial strain ε =
5.The product AE has units of
6.A statically determinate bar uses P from
7.P = 40 kN, L = 4 m, A = 200 mm², E = 200 GPa. δ =
8.If P = 0, the mechanical δ is
9.A stiffer member means
10.L doubled, P, A, E unchanged: δ