A thin cylindrical shell, 3.25 m long and 1 m internal diameter, is subjected to an internal fluid pressure of 1.2 MPa. The shell thickness is 10 mm. Find the circumferential and longitudinal stresses. Find also the maximum shear stress and change in dimensions. Also, determine the change in volume.Take E=200 GN/m^2 and Poisson’s ratio=0.3

Data:

A thin cylindrical shell.

L = 3.25 m = 3250 mm

D = 1m = 1000 mm

P = 1.2 Mpa = 1.2 $N/mm^2$

T = 10 mm

E = 200 $GN/m^2$

$E = 200 \times 10^3 \ N/mm^2$

$\mu = 0.3$

To Find:

1] Maximum shear stress.

2] $\sigma n$

3] $\sigma l$

4] $\delta l , \delta d , \delta v$

A] Longitudinal stress $\sigma l = \frac{Pd}{4 t} = \frac{1.2 \times 1000}{4 \times 10} = 30 \ N/mm^2$

B] Hoop stress $\sigma n = \frac{Pd}{2 t} = \frac{1.2 \times 1000}{2 \times 10} = 60 \ N/mm^2$

C] $Maximum \ shear \ stress = q_{max} = \frac{\sigma n - \sigma l}{2}$

i.e. $q_{max} = \frac{\frac{pd}{2t} - \frac{pd}{4t}}{2} = \frac{pd}{8t}$

$q_{max} = \frac{1.2 \times 1000}{8 \times 10} = 15 \ N/mm^2$

($q_{max} = \frac{pd}{8t}$)

OR

$q_{max} = \frac{60 – 30}{2} = 15 \ N/mm^2$

($q_{max} = \frac{\sigma n - \sigma l}{2}$)

D] Now, Longitudnal strain $El = \frac{\sigma l}{E} - \frac{\mu \ \sigma n}{E}$

$= \frac{1}{E} (\sigma l - \mu \ \sigma n) = \frac{1}{200 \times 10^3} \times (30 – 0.3 \times 60)$

$El = 6 \times 10^{-5}$

But $El = \frac{ \delta l}{l}$ i.e. $\delta l = E l \times l$

$\therefore \ \delta l = 6 \times 10^{-5} \times 3250 = 0.195 \ mm$

E] Hoop strain $En = \frac{\sigma n}{E} - \frac{ \mu \ \sigma l}{E} = \frac{1}{E} \ (\sigma n - \mu \ \sigma l)$

$En = \frac{1}{2 \times 10^{-5}} \times (60 – 0.3 \times 30) = 2.55 \times 10^{-4}$

Now,

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

$\therefore \ \delta d = 2.55 \times 10^{-4} \times 1000 = 0.255 \ mm$

F] Volumetric strain $Ev = 2 \sigma n + \sigma l$

$= 2 \times 2.55 \times 10^{-4} + 6 \times 10^{-5}$

$Ev = 5.7 \times 10^{-4}$

Now, $Ev = \frac{\delta v}{v}$

i.e. $\delta v = Ev \times V = E_v \times \frac{\pi}{4} \ \times d^2 \times l$

$\therefore \ \delta V = 5.7 \times 10^{-4} \times \frac{\pi}{4} \times 1000^2 \times 3250$

$\delta V = 1.4549 \times 10^6 \ mm^3$