Answered: A thick walled cylinder, of internal…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A thick walled cylinder, of internal diameter 160 mm and external diameter 320 mm, is subject to an internal and external pressure as shown in Figure 1. If the external pressure is 10 MPa and the maximum circumferential (hoop) stress permitted on the inside wall of the cylinder is limited to 30 N.mm^-2, what is the maximum internal pressure that can be applied?

A) What is the maximum internal pressure that can be applied?

B) What will be the change in the outside diameter when this pressure is applied? (E = 207 GPa, ν = 0.29)

+++
++++
Figure 1: Thick walled cylinder subject to internal and external pressures

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Hi,

Thank you for the solution on this, we have been taught in university an alternative way of tackling this problem and it results in different final answers, please see in image attached. Now im wondering which one is correct? this is a new subject to us and im hoping to better understand the principles and workings of this, I come to bartleby to offer alternate ways of solving problems to better my understanding of given concepts. Does this example read true to yourself or is your way the correct way according to an external verified reference? Thanks again!

From the known boundary conditions at the outside diameter, where the external pressure is applied:
B
σ₁ = A - 2
B
160²
r = 160 mm and or=-Po = -10 N. mm2 -10 = A -
B = 25600A + 256000 (1)
From the known boundary conditions at the inside diameter, where the maximum circumferential (hoop) stress occurs:
B
OH = A + 2
Combining (1) and (2) gives:
The permissible internal pressure is then:
r = 80 mm and = 30 N. mm² 30 = A +
B
80²
OL =
Ori
Answer
B 192000 6400A (2)
25600A + 256000= 192000 6400A
= A -
B
32000A - 64000
A = 2 and B= 204800
204800
80²
P₁ = -0,₁ = 34 N.mm-2
=-2-
B
To calculate the change in the outside diameter, the stress state at this location needs to be known:
204800
-= 6 N.mm-²
160²
JH₂ = A + ====2+.
B
OTO
= A== -2-
Ar =
-
204800
160²
The change in the outside diameter of the cylinder is then given by:
Pir-Por (34 x (80²)) (10 x (160²))
r-r
160²80²
Ad₁ = 24r₁ = 2 (0-v(or + 0₂)) =
2%
E
H₂
= -34 N.mm-²
2 x 160
207000
= -10 N.mm-²
r
(H
- (OH - V(O₂ + O₂))
= -2 N.mm-²
(6 -0.29(-10-2)) = 0.0147 mm

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