Stresses with thickwalled tube theory

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The discussion focuses on calculating the stress induced in a ring mounted on a solid shaft, where the ring's inner diameter is smaller than the shaft's outer diameter, creating a radial mismatch of 110 micrometers. The user employs thick-walled tube theory and plane strain conditions to derive the radial stress equations for both the ring and the shaft. Key constants A and B are determined from boundary conditions, leading to expressions for radial stress in both components. The user seeks guidance on incorporating the radial mismatch into their calculations and how to derive the resulting stress in the ring using specified material properties. The inquiry emphasizes the need for clarity on the impact of the dimensional difference on stress calculations.
ladil123
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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
 

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