2023-03-23 14:33:46 +00:00
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---
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author: Akbar Rahman
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date: \today
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title: MMME2053 // Thick Walled Cylinders
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tags: [ thick_walled_cylinders ]
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uuid: b53973dc-2c57-4e37-8409-96875125f4de
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lecture_slides: [ ./lecture_slides/MMME2053_TC1_Intro.pdf, ./lecture_slides/MMME2053_TC2.pdf, ./lecture_slides/MMME2053_TC3.pdf ]
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lecture_notes: [ ./lecture_notes/MMME2053_TC_Notes.pdf ]
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exercise_sheets: [ ./exercise_sheets/Thick Cylinders Exercise Sheet.pdf, ./exercise_sheets/Thick Walled Cylinders Exercise Sheet Solutions.pdf ]
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worked_examples: [ ./worked_examples/MMME2053_TC_WE1.pdf, ./worked_examples/MMME2053_TC_WE2.pdf, ./worked_examples/MMME2053_TC_WE3.pdf ]
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---
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2023-03-23 15:49:05 +00:00
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# Lame's Equations
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Derivation in lecture slides 2 (pp. 3-11)
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$$\sigma_h = A + \frac{B}{r^2}$$
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$$\sigma_r = A - \frac{B}{r^2}$$
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where $A$ and $B$ are *Lame's constants* (constants of integration).
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Note that $\sigma_r$ does not vary with radius, $r$.
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## Obtaining Lame's Constants
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The constants can be obtained by using the boundary conditions of the problem:
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At the inner radius ($r = R_i$) the pressure is only opposing the fluid inside:
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$$\sigma_r= -p_i$$
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At the outer radius ($r = R_o$) the pressure is only opposing the fluid outside (e.g. atmospheric
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pressure):
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$$\sigma_r = -p_o$$
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Therefore:
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\begin{align*}
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-p_i &= C - \frac{D}{R_i^2}
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-p_o &= C - \frac{D}{R_o^2}
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\end{align*}
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where $C$ and $D$ are constants which can be determined.
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## Cylinder with Closed Ends
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$$\sigma_z = \frac{R_i^2p_i - R_o^2p_o}{R_o^2-R_i^2}$$
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## Cylinder with Pistons
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No axial load is transferred to the cylinder.
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$$\sigma_z = 0$$
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## Solid Cylinder
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$$\sigma_r = \sigma_\theta = A$$
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