A cylindrical shell 1m in diameter and 3m in length has a metal thickness of 10 mm. If it is subjected to an internal pressure of 3 N/mm^2, determine change in length, change in diameter and change in volume. Take E=2x10^5 N/mm^2 and Poisson’s ratio=0.3

Data:

Dia. (d) = 1m = 1000 mm

Length (l) = 3m = 3000 mm

Thickness (t) = 10 mm

Internal pressure (p) = $3 N/mm^2$

$E = 2 \times 10^5 \ N/mm^2 $ & $\mu = 0.3$

A] $\delta d = ?$

B] $\delta l = ?$

C] $\delta v = ?$

For thin cylindrical shell,

$Hoop \ stress = \ \sigma_n = \frac{Pd}{2t} = \frac{3 \times 1000}{2 \times 10} = 150 \ N/mm^2$

$Longitudnal \ stress = \sigma_l = \frac{Pd}{1t} = \frac{\sigma n}{2} = \frac{150}{2} = 75 \ N/mm^2 $

Now, Hoop strain due to internal pressure:

$\epsilon n = \frac{\sigma n}{E} - \mu \frac{\sigma l}{E} = \frac{1}{E} (\sigma n - \mu \sigma l)$

$En = \frac{1}{2 \times 10^5} (150 – 0.3 \times 75) = 6.375 \times 10^{-4}$

Now, $E n = \frac{\delta d}{d} $ i.e. $\delta d = E n \times \ d$

A] $\therefore \ \delta d = 6.37 \times 10^{-4} \times 1000 = 0.637 \ mm$

Longitudinal strain due to internal pressure:

$El = \frac{\sigma L}{E} - \frac{ \mu \ \sigma n}{E} = \frac{1}{E} (\sigma l - \mu \ \sigma n)$

$El = \frac{1}{2 \times 10^5} (75 – 0.3 \times 150) = 1.5 \times 10^{-4}$

Now, $El = \frac{\delta l}{l} $

i.e. $\delta l = E l \times l$

B] $\therefore \ \delta l = 1.5 \times 10^{-4} \times 3000 = 0.45 \ mm$

Now, volumetric strain = 2 x hoop strain + Longitudinal strain.

$Ev = \frac{\delta v}{v} = 2 \times 6.375 \times 10^{-4} + 1.5 \times 10^{-4}$

$Ev = \frac{\delta v}{v} = 1.425 \times 10^{-3}$

i.e. $\delta v = 1.425 \times 10^{-3} \times V$

$= 1.425 \times 10^{-3} \times \frac{\pi}{4} \times d^2 \times l$

$= 1.425 \times 10^{-3} \times \frac{\pi}{4} \times 1000^2 \times 3000$

C] $\delta V = 3357577.1 \ mm^3$