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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.

Respuesta :

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

[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]

[tex]\sigma _h=\dfrac{15\times 1}{2\times 0.01}[/tex]

[tex]\sigma _h=750 \ MPa[/tex]

Longitudinal stress

[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]

[tex]\sigma _l=\dfrac{15\times 1}{4\times 0.01}[/tex]

[tex]\sigma _l=375 \ MPa[/tex]

Hoop strain

[tex]\varepsilon _h=\dfrac{\sigma _h}{E}-\mu \dfrac{\sigma _l}{E}[/tex]

Take μ=0.26

[tex]\varepsilon _h=\dfrac{750}{200\times 1000}-0.26\times \dfrac{375}{200\times 1000}[/tex]

[tex]\varepsilon _h=0.0032[/tex]

[tex]\varepsilon _h=\dfrac{\Delta d}{d}[/tex]

[tex]0.0032=\dfrac{\Delta d}{1}[/tex]

[tex]\Delta d=0.0032[/tex]

[tex]d_f=1+0.0032\ m[/tex]

Final diameter = 1.0032  m

Longitudinal strain

[tex]\varepsilon _l=\dfrac{\sigma _l}{E}-\mu \dfrac{\sigma _h}{E}[/tex]

[tex]\varepsilon _l=\dfrac{375}{200\times 1000}-0.26\times \dfrac{750}{200\times 1000}[/tex]

[tex]\varepsilon _l=0.0009[/tex]

[tex]\varepsilon _l=\dfrac{\Delta L}{L}[/tex]

[tex]0.0009=\dfrac{\Delta L}{3}[/tex]

So the final length = 3.0027 m

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