Strain and Stress in Thin-Walled Hoops Under Pressure

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Discussion Overview

The discussion revolves around the mechanics of a thin-walled ring or hoop subjected to internal pressure, focusing on the relationship between stress, strain, and material properties. Participants explore the application of Hooke's law and the implications of Poisson's ratio in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the formula for hoop stress as σh = p d / (2 t) and questions the application of Hooke's law to relate stress and strain.
  • Another participant introduces the 3D version of Hooke's law, suggesting the inclusion of Poisson's ratio in the analysis of stress and strain.
  • A participant expresses unfamiliarity with the strain-displacement equation for hoop strain, prompting further clarification from others.
  • There is a discussion about the relationship between strain and displacement, with references to specific equations that incorporate material properties like Young's modulus and Poisson's ratio.
  • Participants confirm the correctness of the derived equations and clarify the conditions under which they apply.
  • One participant poses a question about the stress conditions when pressure is applied externally, leading to a confirmation that the stress is indeed related to the pressure difference between inside and outside.

Areas of Agreement / Disagreement

Participants generally agree on the application of the stress and strain relationships, but there are nuances regarding the inclusion of different stress components and the implications of external versus internal pressure that remain open for further exploration.

Contextual Notes

Some assumptions regarding material properties and the specific conditions of the hoop are not fully detailed, and the discussion does not resolve all mathematical steps or definitions related to the equations presented.

Jony S
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Homework Statement



Hi all, this isn't exactly homework, but it is nonetheless a problem I would like to solve, so here goes.

Consider a thin-walled ring/hoop with diameter "d" and thickness "t". I want to know the change of diameter "Δd" due to the stress caused by pressure "p" being applied in this hoop (from inside to outside).

Homework Equations



So, EngineeringToolBox tells me that the stress in the hoop is σh = p d / (2 t)

The Attempt at a Solution



Can I apply Hooke's law directly here ? knowing the young's modulus of elasticity "E" for my material, can I say that σh = E * strain, where strain = Δd/d ? if not, what is the relation between the stress calculated from the first formula, the strain, and "E" ?
 
Last edited:
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Jony S said:

Homework Statement



Hi all, this isn't exactly homework, but it is nonetheless a problem I would like to solve, so here goes.

Consider a thin-walled ring/hoop with diameter "d" and thickness "t". I want to know the change of diameter "Δd" due to the stress caused by pressure "p" being applied in this hoop (from inside to outside).

Homework Equations



So, EngineeringToolBox tells me that the stress in the hoop is σh = p d / (2 t)

The Attempt at a Solution



Can I apply Hooke's law directly here ? knowing the young's modulus of elasticity "E" for my material, can I say that σh = E * strain, where strain = Δd/d ? if not, what is the relation between the stress calculated from the first formula, the strain, and "E" ?
Are you familiar with the 3D version of Hooke's law, which includes all the stress and strain components.
 
Chestermiller said:
Are you familiar with the 3D version of Hooke's law, which includes all the stress and strain components.

These concepts are fairly new to me so I'm not "familiar" with much. Are you talking about Poisson's ratio, i.e., the expansion/compression of the material in perpendicular directions ? so I would add this:

σh - v(σr+σz)= E * strain

?
 
Jony S said:
These concepts are fairly new to me so I'm not "familiar" with much. Are you talking about Poisson's ratio, i.e., the expansion/compression of the material in perpendicular directions ? so I would add this:

σh - v(σr+σz)= E * strain

?
Yes. This is one of the equations you would use. The other relationship is the strain-displacement equation for the hoop strain. Do you know that equation?
 
Chestermiller said:
Yes. This is one of the equations you would use. The other relationship is the strain-displacement equation for the hoop strain. Do you know that equation?

No.

Searched a bit more and found this:

https://nptel.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength of materials/lects & picts/image/lect16/lecture16.htm

In the end of the first part there is a formula for displacement. This appears to be the combination of the two equations that we already had, i.e., strain = deltaD/D, and
strain = PD/4tE * (2-v) (σr is zero and σz is σh/2 ?)

v and E are characteristics of the material so we have everything ?

edit: corrected some mistakes
 
Last edited:
Jony S said:


No.

Searched a bit more and found this:

https://nptel.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength of materials/lects & picts/image/lect16/lecture16.htm

In the end of the first part there is a formula for displacement. This appears to be the combination of the two equations that we already had, i.e., strain = deltaD/D, and
strain = PD/4tE * (2-v) (σr is zero and σz is σh/2 ?)

v and E are characteristics of the material so we have everything ?

edit: corrected some mistakes
Yes, that's what I get (assuming that the ##(2-\nu)## is in the numerator
 
Chestermiller said:
Yes, that's what I get (assuming that the ##(2-\nu)## is in the numerator
Yes it is :)

Thanks for the help!
 
Jony S said:
Yes it is :)

Thanks for the help!
No problem. Nicely done.
 
Chestermiller said:
No problem. Nicely done.
Extra question for clarification, the stress is the same (opposite signal) if the pressure is applied externally, right ? If the pressure is the same inside and outside the total stress has to be zero.
 
  • #10
Jony S said:
Extra question for clarification, the stress is the same (opposite signal) if the pressure is applied externally, right ? If the pressure is the same inside and outside the total stress has to be zero
Yes. p is really the pressure difference between inside and outside. For a cylinder in atmospheric air, it's the gauge pressure inside.
 

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