A cylindrical cells 3 m long closed at the ends having 1m of internal diameter is subjected to an internal pressure of 1.5 MPa. If the thickness of the shell wall is 15mm. Find the circumferential, longitudinal stress and maximum shear stress. Find the change in diameter, length and volume of the shell. E=2x10^5 N/mm^2 and 1/m=0.3

Data: Cylindrical Shell:

L = 3m = 3000 mm

D = 1m = 1000 mm, $1/m = \mu = 0.3$

$ P = 1.5 \ Mpa = 1.5 \ N/mm^2$ $t = 15 \ mm, E = 2 \times 10^5 \ N/mm^2$

A] $\sigma n$

B] $\sigma l$

C] Maximum shear stress.

D] $\delta d$

E] $\delta l$

F] $\delta v$

A] Circumferential stress $\sigma n$

$\sigma n = \frac{Pd}{2 t} = \frac{1.5 \times 1000}{2 \times 15} = 50$

B] Longitudinal stress $\sigma l$

$\sigma l = \frac{Pd}{4 t} = \frac{\sigma n}{2} = \frac{50}{2} = 25 \ N/mm^2$

C] Maximum shear stress $q_{max}$.

$q_{max} = \frac{\sigma n - \sigma l}{2} = \frac{Pd}{8 t}$

$q_{max} = \frac{50 – 25}{2} = 3.125 \ N/mm^2$

D] Longitudinal strain =

$E l = \frac{\delta l}{L} = \frac{\sigma l - \mu \ \sigma n}{E}$

$\therefore \delta l = (\frac{\sigma l - \mu \ \sigma n}{E}) \ l$

$\therefore \delta l = (\frac{25 – 0.3 \times 50}{2 \times 10^5}) \times 3000$

$\delta l = 0.15 \ mm$ and $El = 5 \times 10^{-5}$

E] Hoop strain:

$E n = \frac{ \delta d}{d} = \frac{\sigma n - \mu \ \sigma l}{E}$

i.e. $\delta d = (\frac{ \sigma n - \mu \ \sigma l}{E}) \ d$

$\therefore$ $\delta d = (\frac{50 – 0.3 \times 25}{2 \times 10^5}) \times 1000$

$\delta d = 0.212 \ mm$ and $En = 2.125 \times 10^{-4}$

F] $Volumetric \ strain \ Ev = \frac{\delta v}{V} = 2 \ En + El$

i.e. $\delta v = (2 \times En + El) \frac{\pi}{4} \times d^2 \times l$

$\therefore \ \delta v = (2 \times 2.125 \times 10^{-4} + 5 \times 10^{-5}) \times \frac{\pi} {4} \times 1000^2 \times 3000$ $(V = \frac{\pi}{4} \times d^2 \times l)$

$\delta V = 1.119 \times 10^6 \ mm^3$