Stress Distribution in Compound Cylinders

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Homework Statement


A long, hollow, thick elastic cylinder (E , v) is surrounded by a thin elastic band of thickness t made of a different material ([itex]E_2[/itex] , [itex]v_2[/itex]). The fit is ideal, such that neither a gap nor a pressure exists. An internal pressure p is then applied at the inside radius (r=a) of the thick cylinder. Assuming that the radial displacements is continuous at the interface (r=b), prove that the stress in the thick cylinder are given by
[itex]\sigma_{rr} = -p\frac{\beta(1+(b/r)^2)+(1-(1-((b/r)^2)}{\beta(1+(b/a)^2)+(1-((b/a)^2)}[/itex]

where
[itex]\beta= \frac{(1-v^2)}{(1+v) v-(bE)(1-v_b^2)/(E_b h)}[/itex]

Homework Equations


Lame's Equations. If a long cylinder is subject to internal pressure [itex]p_i[/itex] at [itex]r_i[/itex] and external pressure [itex]p_o[/itex] at [itex]r_o[/itex] then the radial stress distribution is

[itex]\sigma_{rr}=\frac{p_i r_i^2-p_o r_o^2}{r_o^2-r_i^2}+\frac{r_i^2 r_0^2(p_o-p_i)}{r^2(r_o^2-r_i^2)}[/itex]

and the tangential stress distribution
[itex]\sigma_{tt}=\frac{p_i r_i^2-p_o r_o^2}{r_o^2-r_i^2}-\frac{r_i^2 r_0^2(p_o-p_i)}{r^2(r_o^2-r_i^2)}[/itex].

and finally the displacement is

[itex]u_r=\frac{1-v}{E}\frac{(p_i r_i^2 - p_o r_o^2)}{r_o^2-r_i^2} r + \frac{1+v}{E}\frac{r_i^2 r_o^2 (p_i-p_o)}{r_o^2-r_i^2}\frac{1}{r}[/itex].

The Attempt at a Solution



I attempted to match the displacements at the interface and solve for the interface pressure [itex]P[/itex]. However, I can not get the correct solution with this approach.
The displacement of the inner cylinder at the interface is given by
[itex]u_r(r=b)=\frac{1-v}{E}\frac{(p_i a^2 - P b^2)}{b^2-a^2} b + \frac{1+v}{E}\frac{a^2 b^2 (p_i-P)}{b^2-a^2}\frac{1}{b}[/itex].

Canceling terms yields
[itex]u_r(r=b)=\frac{1}{E(b^2-a^2)}(2p_i a^2 b - (1-v)P b^3 - (1+v)P a^2 b)[/itex]
The displacement at the interface of the outer cylinder is given by:

[itex]\sigma_\theta ≈ E \epsilon_\theta→ u(r=b) = \frac{Pb^2}{E_2 t}[/itex]

because the outer elastic band is thin. Equating the displacements I get the interface pressure to

[itex]P = \frac{2 p_i a^2}{\frac{Eb(b^2-a^2)}{E_2 t}+(1-v)b^2+(1+b)a^2}[/itex].

I know this is the wrong interface pressure by check the solution given in the problem statement and through FEA analysis.
 

Answers and Replies

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Answer

There is a typo in my post above because [itex](1+b)[/itex] should be [itex](1+v)[/itex] in the denominator of the expression for [itex]P[/itex]. The mistake I made was using the lame's equations for plain stress, while the problem wanted to plane strain solution. The expression for [itex]P[/itex] matches the given result by making the plane stress to plane strain substitutions [itex]E\rightarrow E/(1-v^2)[/itex] and [itex]v\rightarrow v/(1-v)[/itex].
 
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