Question

A thin cylindrical shell 1.00 m diameter and 3 m in length has metal thickness of 8 mm. if it is subjected to internal pressure 2.5 MPa, determine change in length. Take E = 200 GPa and Poisson's ratio 0.3​

Answer 1

Step-by-step explanation:

Dia. (d) = 1m = 1000 mm

Length (l) = 3m = 3000 mm

Thickness (t) = 10 mm

Internal pressure (p) = 3N/mm2

E=2×105 N/mm2 & μ=0.3

A] δd=?

B] δl=?

C] δv=?

For thin cylindrical shell,

Hoop stress= σn=Pd2t=3×10002×10=150 N/mm2

Longitudnal stress=σl=Pd1t=σn2=1502=75 N/mm2

Now, Hoop strain due to internal pressure:

ϵn=σnE−μσlE=1E(σn−μσl)

En=12×105(150–0.3×75)=6.375×10−4

Now, En=δdd i.e. δd=En× d

A] ∴ δd=6.37×10−4×1000=0.637 mm

Longitudinal strain due to internal pressure:

El=σLE−μ σnE=1E(σl−μ σn)

El=12×105(75–0.3×150)=1.5×10−4

Now, El=δll

i.e. δl=El×l

B] ∴ δl=1.5×10−4×3000=0.45 mm

Now, volumetric strain = 2 x hoop strain + Longitudinal strain.

Ev=δvv=2×6.375×10−4+1.5×10−4

Ev=δvv=1.425×10−3

i.e. δv=1.425×10−3×V

=1.425×10−3×π4×d2×l

=1.425×10−3×π4×10002×3000

C] δV=3357577.1 mm3