Answer:
The resulting strain is 4.05 x 10⁻³
Explanation:
Given;
dimension of the specimen = 10 mm x 12.7 mm
Cross sectional area of the aluminum specimen = 10 mm x 12.7 mm = 127 mm² = 1.27 x 10⁻⁴ m²
applied force, F = 35,500 N
Young's modulus is given by;
[tex]E = \frac{Stress}{Strain}\\\\E = \frac{F}{A(strain)}\\\\E = \frac{F}{A(\epsilon)}\\\\\epsilon = \frac{F}{A E}\\\\[/tex]
Where;
ε is the resulting strain
E is Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
[tex]\epsilon = \frac{35500}{(1.27*10^{-4}) (69*10^{9})}\\\\ \epsilon = 4.05*10^{-3}[/tex]
Therefore, the resulting strain is 4.05 x 10⁻³
