Stress Distribution in Compound Cylinders

AI Thread Summary
The discussion focuses on the stress distribution in a long, hollow, thick elastic cylinder surrounded by a thin elastic band under internal pressure. The primary equation for radial stress in the thick cylinder is derived, emphasizing the importance of continuous radial displacements at the interface. The attempt to solve for interface pressure initially used incorrect assumptions, leading to discrepancies with the expected results. A correction was made regarding the application of Lame's equations, clarifying that a plane strain solution was required instead of a plane stress solution. The final expression for interface pressure aligns with the problem statement after appropriate substitutions are applied.
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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 (E_2 , v_2). 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
\sigma_{rr} = -p\frac{\beta(1+(b/r)^2)+(1-(1-((b/r)^2)}{\beta(1+(b/a)^2)+(1-((b/a)^2)}

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

Homework Equations


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

\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)}

and the tangential stress distribution
\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)}.

and finally the displacement is

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}.

The Attempt at a Solution



I attempted to match the displacements at the interface and solve for the interface pressure P. However, I can not get the correct solution with this approach.
The displacement of the inner cylinder at the interface is given by
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}.

Canceling terms yields
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)
The displacement at the interface of the outer cylinder is given by:

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

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

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

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

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