0
16kviews
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.

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

2 Answers
0
2.1kviews

1 2

0
992views

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=σnσl2=Pd8t

qmax=50252=3.125 N/mm2

D] Longitudinal strain =

El=δlL=σlμ 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

Please log in to add an answer.