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8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.

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rafaleo84

Answer:

The stress S = 1935 [Psi]

Explanation:

This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.

The initial data:

P = internal pressure [Psi] = 90 [Psi]

Di= internal diameter [in] = 22 [in]

t = wall thickness [in] = 0.25 [in]

S = stress = [Psi]

Therefore

ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]

And using the expression to find the stress:

[tex]S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi][/tex]

In the attached image we can see the stress σ1 & σ2 = S acting over the point A.

Ver imagen rafaleo84

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