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vazorg [7]
3 years ago
6

Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is

inflated to a gage pressure of 78 kPa.
Engineering
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

<em>The normal stress is 0.7821 MPa</em>

<em></em>

Explanation:

The external diameter D = 160 mm

The thickness t = 3.8 mm = 3.8 x 10^-3 m

gauge pressure P = 78 kPa = 78 x 10^3 Pa

The maximum shear stress τmax = ?

The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm

The internal radius of the shell r = R - t

==> 80 - 3.8 = 76.2 mm

Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm

==> d = 152.4 x 10^-3 m

The normal stress σ = \frac{Pd}{4t} = \frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} } = 782052.63 Pa

==>  σ = <em>0.7821 MPa</em>

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