An aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in.) on an edge is pulled in tension with a load of 66,700 n (15,000 lbf) and experiences an elongation of 0.43 mm (1.7 * 10-2 assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

7.12x10¹⁰ N/m²

Explanation:

The modulus of elasticity (E) is the pressure (P) divided by the deformation (ε), where:

P = F/A₀, where F is the tension and A₀ the initial square cross-section

ε = Δl/l₀, where Δl is the length variation (elongation), and l₀ is the initial length.

A₀ = (0.0165m)² =2.7225x10⁻⁴m²

E = P/ε = F/A₀/Δl/l₀

E = (F*l₀)/(A₀*Δl)

E = (66,700 N * 0.125m)/(2.7225x10⁻⁴m² * 4.3x10⁻⁴m)

E = (8,337.5 N.m)/(1.171x10⁻⁷ m³)

E = 7.12x10¹⁰ N/m²

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