Answer:
Tensile load = 2261.94 lb
Explanation:
Given data:
Tensile strength of aluminium = 18,000 lbs/in^2
yield strength = 8000 lb/in^2
diameter of rod = 0.4 inches
ultimate tensile strength is given as [tex]S_{ut} = \frac{Permissible/ load}{Area}[/tex]
[tex]18,000 = \frac{P}{\frac{pi}{4} D^2}[/tex]
solving for tensile load P
[tex]18000 = \frac{P}{\frac{\pi}{4} \times 0.4^2}[/tex]
Tensile load = 2261.94 lb
