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Derive the expression for the volumetric strain for thin cylindrical shells.

Derive the expression for the volumetric strain for thin cylindrical shells.

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p is the internal pressure, t is the thickness and d is diameter. 1 2 3 4 5

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Let $E_l$ and $E_h$ be the longitudinal strain and hoop strain.

And $\mu$ is the poison ration

And E is the modulus of elasticity.

Now, Longitudinal strain $E_L = \frac{P_d}{2 \ t E} - \frac{p_d}{4 t E} \times \mu$

$E_L = \frac{pd}{2 t E} ( \frac{1}{2} - \mu)$

$Circumferential / hoop \ strain = \frac{d}{2 t E} - \frac{2d \mu}{4 t E}$

$E_n = \frac{pd}{2 t E} ( 1 - \frac{1}{2} - \mu)$

$E_n = \frac{pd}{4 t \ E} (2 - \mu)$

$Hoop \ strain \ or \ circumferential \ strain = \frac{change \ in \ circumference}{original \ circumference}$

$En = \frac{\pi \delta d}{\pi d} = \frac{\delta d}{d}$ ($\delta d$ is change in diameter.)

But $\frac{\delta d}{d} = strain \ of \ diameter.$

$\therefore \ Change \ in \ diameter = E_h \times Original \ diameter$

And also, $Longitudinal \ strain = \frac{\delta l}{l} = E_L$

$\therefore \ Change \ in \ length = E_l \times Original \ length$

Now, $capacity \ of \ shell = \ V \ = \frac{\pi}{4} \times d^2 \ l$

$\therefore$ $Change \ in \ capacity \ of \ shell = \delta V = \frac{\pi}{4} d^2 \times \delta l + \frac{\pi}{4} \times 2d \ \delta d \ l$

(i.e. taking differential.)

Now,

$\frac{\delta V}{V} = \frac{\delta l}{L} + \frac{2 \ \delta d}{d} $ $\rightarrow$ [1]

$\frac{\delta V}{V} = El + En = \frac{Pd}{2 t E} (\frac{1}{2} - \mu) + \frac{ 2 pd}{2 t E} ( 1 - \frac{1}{2} \ \mu)$

$\frac{\delta V}{V} = \frac{Pd}{2 t E} [\frac{1}{2} - \mu + 2 - \mu] = \frac{pd}{2 t E} ( \frac{5}{2} - \ 2 \mu)$

$\therefore$ $\frac{\delta V}{V} = \frac{pd}{4 t E} [5 – 4 \mu]$ $\rightarrow$ [2]

Equation [1] and [2] are the volumetric strain for thin cylindrical shell.

Note:

If the compressibility of the fluid in cylinder is also considered then the change in capacity of shell.

$\delta V = \frac{pd}{4tE} (5 – 4 \mu) V + \frac{p}{k} \ V$

i.e.

$ Volumetric \ Strain = \frac{\delta V}{V} = \frac{pd}{4tE} (5 – 4 \mu) V + \frac{p}{k} $

( Where K is bulk modulus of fluid).

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