Question

The cylindrical aluminum air tank below is to be rated for 300 psi and it must comply with the ASME Boiler Code which requires a factor of safety of 4.0. If the inside diameter is 12 inches and the material is 6061-T6 aluminum with yield stress equal to 32.5 ksi, determine the minimum allowable wall thickness of the tank using: a) the Tresca criterion and b) the von Mises criterion. (neglect any effects of the welds)

Answer 1

Answer:

Detailed solution is given below:

Answer 2

Answer:

• 0.2215 inch
• 0.1918 inch

Explanation:

Make assumptions like considering the cylinder to be thin

first we calculate the hoop stress

∝1 = $\frac{pd}{2t}$  

p = 300 psi

d = 12 inches

   = $\frac{300 * 12 }{2t}$  = 1800 / t

secondly we calculate the longitudinal stress

∝2 = $\frac{pd}{4t}$ = $\frac{300 *12}{4t}$  = 900 / t

∝3 = 0

These stress are the principal stresses found in thin cylinders

to determine the allowable wall thickness

A) using the Tresca criterion

( т max) absolute = $\frac{Syt}{2n}$

∝1 - ∝3 / 2 = $\frac{Syt}{2n}$

= 1800 / t = 32.5 * 10^3 / 4

t = 7200 / 32500 = 0.2215

B) using the Von Mises criterion  

∝1² + ∝2² - ∝1∝2 = $(\frac{Syt}{n})^{2}$

( 1800 / t )^2 + ( 900 / t ) ^2 - $\frac{1800}{t} * \frac{900}{t}= (\frac{32.5 *10^{3} }{4} )^{2}$

= (243 *10^4 ) /  t^2  = ( 32500 / 4 )^2

= (243 * 10^4) / t^2  = 66*10^6

t = $\sqrt{0.0368}$ = 0.1918 inches