Question

A thin cylindrical shell 3 m long has 1m internal diameter and 15 mm metal thickness. Calculate the circumferential and longitudinal stresses induced and also the change in the dimensions of the shell, if it is subjected to an internal pressure of1.5 N/mm2 Take E = 2x105 N/mm2 and poison’s ratio =0.3. Also calculate change in volume.

Answer 1

Answer:

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

∴δl=(σl−μ σnE) l

∴δl=(25–0.3×502×105)×3000

δl=0.15 mm and El=5×10−5

E] Hoop strain:

En=δdd=σn−μ σlE

i.e. δd=(σn−μ σlE) d

∴ δd=(50–0.3×252×105)×1000

δd=0.212 mm and En=2.125×10−4

F] Volumetric strain Ev=δvV=2 En+El

i.e. δv=(2×En+El)π4×d2×l

∴ δv=(2×2.125×10−4+5×10−5)×π4×10002×3000 (V=π4×d2×l)

δV=1.119×106 mm3