Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
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)

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)

Refrigeration and Air Conditioning Technology (MindTap Course List)
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter42: Heat Gains And Heat Losses In Structures
Section: Chapter Questions
Problem 12RQ
See similar textbooks
icon
Related questions
Question

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
Transcribed Image Text:+++ ++++ Figure 1: Thick walled cylinder subject to internal and external pressures
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

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
Transcribed Image Text: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
Solution
Bartleby Expert
SEE SOLUTION
Knowledge Booster
Pressure Vessels
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Refrigeration and Air Conditioning Technology (Mi…
Refrigeration and Air Conditioning Technology (Mi…
Mechanical Engineering
ISBN:
9781305578296
Author:
John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS