Answer
given,
l₀ = 50 mm
d₀ = 13.1 mm
l₁ = 53.7 mm
d₁ = 12.6 mm
Area of cross section
[tex]A = \dfrac{\pi}{4}d_0^2[/tex]
[tex]A = \dfrac{\pi}{4} \times 13.1^2[/tex]
A = 134. 782 mm²
The strain is
[tex]\epsilon = \dfrac{l_f-l_0}{l_0}[/tex]
[tex]= \dfrac{53.7-50}{50}[/tex]
= 0.074
the tensile modulus of A356 aluminium is
E = 72.4 GPa
The stress is
σ = εE
= 72.4 × 10⁹ × 0.074
= 5357.6 × 10⁶ Pa
σ = 5357.6 MPa
[tex]UTS = \dfrac{ultimate\ load}{cross\ sectional\ area}[/tex]
[tex]UTS = \dfrac{ultimate\ load}{134.782}[/tex]
