Question

A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball is inflated to a 108-kPa gage pressure. (Round the final answer to two decimal places.)

Answer 1

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

$\sigma = \frac{p*r}{2t}$

Where:

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm  

Hence, we need to find r, as follows:

$r_{inner} = r_{outer} - t$    

$r_{inner} = \frac{d}{2} - t$

Where:

d: is the outer diameter = 300 mm

$r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm$

Now, we can find the normal stress (σ) in the wall of the basketball:

$\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa$

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!