Data: Cylindrical Shell:
L = 3m = 3000 mm
D = 1m = 1000 mm, 1/m=μ=0.3
P=1.5 Mpa=1.5 N/mm2 t=15 mm,E=2×105 N/mm2
A] σn
B] σl
C] Maximum shear stress.
D] δd
E] δl
F] δv
A] Circumferential stress σn
σn=Pd2t=1.5×10002×15=50
B] Longitudinal stress σl
σl=Pd4t=σn2=502=25 N/mm2
C] Maximum shear stress qmax.
qmax=σn−σl2=Pd8t
qmax=50–252=3.125 N/mm2
D] Longitudinal strain =
El=δlL=σl−μ σnE
∴
\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