Stresses with thickwalled tube theory

  • Thread starter Thread starter ladil123
  • Start date Start date
  • Tags Tags
    Theory Tube
Click For Summary
SUMMARY

The discussion focuses on calculating the stress in a ring mounted on a solid shaft using thick-walled tube theory under plane strain conditions. The inner diameter of the ring is 90mm - 50μm, while the shaft's diameter is 90mm + 20μm, resulting in a radial mismatch of 110μm. The radial stress equations are established with constants A and B derived from boundary conditions, leading to the formulation of stress in both the ring and the shaft. The user seeks guidance on incorporating the radial mismatch into their calculations and determining the resultant stress based on given material properties.

PREREQUISITES
  • Understanding of thick-walled tube theory
  • Knowledge of plane strain conditions
  • Familiarity with stress-strain relationships in materials
  • Basic concepts of boundary conditions in mechanics
NEXT STEPS
  • Research methods for incorporating radial mismatches in thick-walled tube calculations
  • Learn about stress distribution in composite materials
  • Explore finite element analysis (FEA) for stress evaluation
  • Study the effects of Poisson's ratio on stress calculations
USEFUL FOR

Mechanical engineers, materials scientists, and anyone involved in stress analysis of cylindrical structures or components requiring precise calculations of stress due to geometric mismatches.

ladil123
Messages
42
Reaction score
0
Hello


A ring should be mounted on a solid shaft. The rings inner diameter is smaller than than the shaft diamater so it will be forced on.
for my problem the inner diamater for the ring is 90mm -50 \mum
the shaft is 90mm + 20\mum

so there can be a difference of 110 \mum.
This should give a certain stress after mounting the ring.

That stress is what I need to calculate

Im using thickwalled tube theory. plane strain.
I have attached a figure to make it easier...

If the radial stress in the ring starts with
S_radial=A - B/r^2
A, B are constants as we know.

with BC:
S_radial(r=OR)=0
S_radial(r=IR)= -Pi

so A=B/(OR^2)

and at the inner radius the stress is equal to the inner pressure: -Pi
So A=-Pi/(1/OR^2 -1/IR^2)/)

So the radial stress for the ring is :
SigmaRing=[-Pi/(1/OR^2 -1/IR^2)]/OR^2 + [Pi/(1/OR^2 -1/IR^2)]*1/r^2

If I do equlibrium for the solid shaft:
Witch BC:
Sigmaradial(r=0)=0
and Sigmaradial(r=IR)= -Pi

The constants A=0 and B=Pi*(IR^2)
So that stress is Sigmar=-Pi(IR^2)/r^2

What should I do to introduce the radial mismatch of 110 micro meters?
And after that how do I get the stress in the ring with my material data below ?

E-modulus ring =540 GPa
Poissons ring = 0.24
E-modulus shaft=205 GPa
Poissons shaft = 0.3


Thanks for any help
 

Attachments

  • mounting.jpg
    mounting.jpg
    24 KB · Views: 514
Last edited:
Engineering news on Phys.org
ladil123: For an answer to this question, see my post in thread 289475[/color].
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 104 ·
4
Replies
104
Views
12K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
8K
Replies
19
Views
9K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
1K
Replies
4
Views
2K
Replies
4
Views
3K