Question

Calculate the hoop stress in an aluminum can containing a carbonated beverage at 4

atmospheres of gauge pressure (4 atm of differential pressure between inside and outside of

can) if the can is 6.2 inches tall, 3.1 inches wide, and its wall thickness is 0.0076 inches.

Answer 1

From the definition of stress, we have

F_H = t L _H                                                                                     { eq.1 }

The wall must support the difference in (vertical) forces on the inner and outer sides of the wall due to a pressure difference.

Then, we have

2 F_H = (P L R Sin ) d

2 F_H = 2 P L R

F_H = P L R                                                                              { eq.2 }

Equating eq.1 & 2, we get

t L _H = P L R

_H = P (R / t)

where, t = thickness of the wall = 0.0076 inches = 0.00019304 m

R = radius of the can = 1.55 inches = 0.03937 m

P = differential pressure b/w inside & outside of can = 4 atm = 405300 Pa

then, we get

_H = (405300 Pa) [(0.03937 m) / (0.00019304 m)]

_H = [(405300 Pa) (203.9)]

_H = 82640670 Pa

converting Pa into MPa :

_H = 82.6 MPa