Question
Download Solution PDFA seamless pipe having a diameter of 600 mm and thickness of 9 mm, contain the fluid under a pressure of 4 MPa, find the longitudinal stress developed in the pipe.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
By thin-walled cylinder, we mean that the thickness ‘t' is very much smaller than the radius Ri and we may quantify this by stating that the ratio t / Ri of the thickness of radius should be less than 0.1.
In a thin shell circumferential stress is \({\sigma _c} = \frac{{Pd}}{{2t}}\;\) and
Longitudinal stress will be half of the circumferential stress i.e. \({\sigma _l} = \frac{{Pd}}{{4t}}\).
where p is the pressure, t is the thickness of cylinder, d is the diameter of the cylinder
Calculation:
Given:
p = 4 MPa, t = 9 mm and d = 600 mm.
Longitudinal stress
\({\sigma _l} = \frac{{Pd}}{{4t}}\)
\({\sigma _l} = \frac{{4\times 600}}{{4\times 9}} =66.66~MPa\)
|
Hoop stress \({\sigma _h} = \frac{{Pd}}{{2t}} = {\sigma _1}\) |
Hoop strain \({\epsilon_h} = \frac{{Pd}}{{4tE}}\left( {2 - \mu } \right)\) |
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Longitudinal stress \({\sigma _L} = \frac{{Pd}}{{4t}} = {\sigma _2} = \frac{{{\sigma _1}}}{2}\) |
Longitudinal strain \({\epsilon_L} = \frac{{Pd}}{{4tE}}\left( {1 - 2\mu } \right) = \frac{{{\sigma _2}}}{E}\left( {1 - 2\mu } \right)\) |
Last updated on Jul 1, 2025
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