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 …

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asked Dec 5, 2020 215k views

1 Answer

Dertoni · Answer 1 · 2020-12-09T03:59:40+0000

Answer:

Final diameter = 1.0032 m

Final length =3.0027 m

Step-by-step 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

$\sigma _h=(Pd)/(2t)$

$\sigma _h=(15* 1)/(2* 0.01)$

$\sigma _h=750 \ MPa$

Longitudinal stress

$\sigma _l=(Pd)/(4t)$

$\sigma _l=(15* 1)/(4* 0.01)$

$\sigma _l=375 \ MPa$

Hoop strain

$\varepsilon _h=(\sigma _h)/(E)-\mu (\sigma _l)/(E)$

Take μ=0.26

$\varepsilon _h=(750)/(200* 1000)-0.26* (375)/(200* 1000)$

$\varepsilon _h=0.0032$

$\varepsilon _h=(\Delta d)/(d)$

$0.0032=(\Delta d)/(1)$

$\Delta d=0.0032$

$d_f=1+0.0032\ m$

Final diameter = 1.0032 m

Longitudinal strain

$\varepsilon _l=(\sigma _l)/(E)-\mu (\sigma _h)/(E)$

$\varepsilon _l=(375)/(200* 1000)-0.26* (750)/(200* 1000)$

$\varepsilon _l=0.0009$

$\varepsilon _l=(\Delta L)/(L)$

$0.0009=(\Delta L)/(3)$

So the final length = 3.0027 m