A thin-walled pressure vessel is constructed by rolling a 6 mm thick steel sheet into a cylindrical shape, welding the seam along line A-B, and capping the ends. The vessel is subjected to an internal pressure of 1.25 MPa. What is the normal stress normal to line A-B?

41.015

Explanation:

Solution

Given that:

Th first step to take is to find  the longitudinal stress in the cylinder

σl = PD/4t

P = the pressure

D = the diameter

t = the thickness

Thus,

σl = 1.25 * ^ 6 * 0.45 / 4 * 6 * 10 ^ ⁻3

=23.475Mpa

Now. we find the hoop stress in the cylinder∠

σh = PD/2t

σh = 1.25 * ^ 6 * 0.45/ 2 * 6 * 10 ^ ⁻3

σh  =46.875 Mpa

Then

we find the normal stress in the line of the 30° angle with the longitudinal axis stated below:

σab = σh + σl/2 + ( σh - σl/2) cos 2θ + t sin 2θ

So,

σab  =46.875 + 23.4375/2 + ( 46.875 - 23.4375/2) cos 2(30°) + 0

σab= 41.015

Therefore the normal stress to line A-B is 41.015

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explanation: