Question

The thin-walled cylindrical pressure vessel is subjected to an internal pressure of 15 MPa. The vessel is 3m long, has an inner radius of 0.5 m, and a thickness of 10 mm. It is constructed of structural A-36 steel. Compute the final diameter and the length of the pressure vessel.

Answer 1

Answer:

Final diameter = 1.0032  m

Final length =3.0027 m

Explanation:

Given that

P= 15 MPa

r= 0.5 m

L= 3 m

t=10 mm

For  A-36 steel ,modulus of elasticity = 200 GPa

Hoop stress

$\sigma _h=\dfrac{Pd}{2t}$

$\sigma _h=\dfrac{15\times 1}{2\times 0.01}$

$\sigma _h=750 \ MPa$

Longitudinal stress

$\sigma _l=\dfrac{Pd}{4t}$

$\sigma _l=\dfrac{15\times 1}{4\times 0.01}$

$\sigma _l=375 \ MPa$

Hoop strain

$\varepsilon _h=\dfrac{\sigma _h}{E}-\mu \dfrac{\sigma _l}{E}$

Take μ=0.26

$\varepsilon _h=\dfrac{750}{200\times 1000}-0.26\times \dfrac{375}{200\times 1000}$

$\varepsilon _h=0.0032$

$\varepsilon _h=\dfrac{\Delta d}{d}$

$0.0032=\dfrac{\Delta d}{1}$

$\Delta d=0.0032$

$d_f=1+0.0032\ m$

Final diameter = 1.0032  m

Longitudinal strain

$\varepsilon _l=\dfrac{\sigma _l}{E}-\mu \dfrac{\sigma _h}{E}$

$\varepsilon _l=\dfrac{375}{200\times 1000}-0.26\times \dfrac{750}{200\times 1000}$

$\varepsilon _l=0.0009$

$\varepsilon _l=\dfrac{\Delta L}{L}$

$0.0009=\dfrac{\Delta L}{3}$

So the final length = 3.0027 m