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3 years ago

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Please help me thanks

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Answer:

2:1____________________________________

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A closed cylindrical vessel contains a fluid at a 5MPa pressure. The cylinder, which has an outside diameter of 2500mm and a wal

Julli [10]

Answer:

1) Increase in the diameter equals 3.5 mm

2) Increase in the length equals $0.0003724L_{i}$ where $L_{i}$ is the initial length of the vessel.

Step-by-step explanation:

The diametric strain in the vessel is given by

$\epsilon_{D} =\epsilon_{diam}-\nu \epsilon _{axial}$

We have

$\epsilon _{diam}=\frac{\sigma _{hoop}}{E}\\\\\sigma _{hoop}=\frac{\Delta P\times D}{2t}\\\\\therefore \epsilon _{diam}=\frac{\Delta P\times D}{2t\times E}$

Applying values we get

$\therefore \epsilon _{diam}=\frac{5\times 10^{6}\times 2.5}{2\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{5}{3088}$

Similarly axial strain is given by

$\epsilon _{diam}=\frac{\sigma _{axial}}{E}$

$\sigma _{axial}=\frac{\Delta P\times D}{4t}\\\\\therefore \epsilon _{axial}=\frac{\Delta P\times D}{4t\times E}$

Applying values we get

$\therefore \epsilon _{axial}=\frac{5\times 10^{6}\times 2.5}{4\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{2.5}{3088}$

Hence The effect of axial strain along the diameter is given by

$-\nu \epsilon _{axial}$

Applying values we get

$-\nu \epsilon _{axial}=-0.27\times \frac{2.5}{3088}=-0.0002185$

hence

$\epsilon _{D} =\frac{5}{3088}-0.0002185\\\\\epsilon =0.00140$

Now by definition of strain we have

$\epsilon _{D} =\frac{D_{f}-D_{i}}{D_{i}}\\\\\therefore D_{f}=D_{i}+\epsilon D_{i}\\\\D_{f}=2.5+0.0014\times 2.5\\\\\therefore D_{f}=2503.5mm$

Increase in the diameter is thus 3.5 mm

Using the same procedure for axial strain we have

$\epsilon_{axial} =\epsilon_{axial}-\nu \epsilon _{diam}$

Applying values we get

$\epsilon_{axial} =\frac{2.5}{3088}-0.27\times \frac{5}{3088}$

$\epsilon_{axial} =0.0003724$

Now by definition of strain we have

$\epsilon _{axial} =\frac{L_{f}-L_{i}}{L_{i}}\\\\\therefore \Delta L=0.0003724L_{i}$

where $L_{i}$ is the initial length of the cylinder.

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