Find out change in diameter:

A cylindrical shell, 0.8 m in a diameter and 3 m long is along with 10 mm wall thickness. If the shell is subjected into an internal pressure of 2.5 N/mm², find out

(a) change in diameter,

(b) change in length, and

(c) change in volume.

Take E = 200 GPa and Poisson's ratio = 0.25.

Solution

Diameter of the shell, d = 0.8 m = 800 mm.

Thickness of the shell, t = 10 mm.

Internal pressure, p = 2.5 N/mm².

Hoop stress,

σ_h  = pd/ 2t

= 2.5 × 800/(2 × 10 ) = 100 N/mm²

Longitudinal stress,

σ_l  =  pd /4t

 = 2.5 × 800/(4 × 10) = 50 N/mm²

Hoop strain, ε_n = E (σ_h  - ε σ_l ) =( 1/(2 × 10⁵))  (100 - 0.25 × 50)

= 4.375 × 10^{- 4}

Longitudinal strain, ε_l  = E (σ_l  - ν σ_h ) = 2 × 10⁵  (50 - 0.25 × 100)

= 1.25 × 10^{- 4}

Change in volume /Original volume = 2εh  + εl  = 2 × 4.375 × 10^{- 4}  + 1.25 × 10^{- 4}

= 10 × 10^{- 4}  = 10^{- 3}

Increase in diameter = Hoop strain × Original diameter

4.375 × 10- 4 × 800 = 0.35 mm

Enhance in length = Longitudinal strain × Original length

1.25 × 10^{- 4} × 3000 = 0.375 mm

Increase in internal volume = =( δv/ v )× Original length

 Original volume =( π d ² /4 )× l = (π /4)× 800² × 3000 = 1507 × 10⁶  mm³

Increase in volume = 10^{- 3} × 1507 × 10⁶ = 1507 × 10³ mm³