Question

Question 16 1 points Save Answer A thin-wall aluminum pop can is subjected to an internal pressure of 100 psi. The diameter and the wall thickness of the can are 2.5 inch and 0.004 inch respectively. The Young's modulus and Poisson's ratio for aluminum are E = 10 x 10ps and v- 0.33 , respectively. The axial stress (the stress along the longitudinal direction) is closest to 32000 psi 40000 psi 16000 psi 8000 psi

Answer 1

As per HOMEWORKLIB POLICY, only one question can be answered. But, I am answering all questions for your reference.

P = 100psi

d = 2.5 inch

t = 0.004 inch

E = 10E7 psi

Axial stress: $\sigma_a = \frac{Pd}{4t} = \frac{100*2.5}{4*0.004}= 15625 psi$

Circumferential stress: $\sigma_h = \frac{Pd}{2t} = \frac{100*2.5}{2*0.004}= 31250 psi$

Axial strain: $\epsilon_a = \frac{\sigma_a}{E} = \frac{15625}{10^7} = 1.5675*10^{-3} in/in$