Production of pressure vessels is fastening an open-ended cylinder and two rigid plates with bolts. The cylinder made of brass. The plates are attached with four W 7/16" steel bolts and nuts. These bolts have 16 threads per inch. An additional half turn is given to each of the nuts after they snugged. Calculate the pressure that the container will start to leak. Does the cylinder or bolt fail under this pressure? Take D= 16", L=16" and t= 0.125" Warning: Take into account the Poisson effect.

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Explanation:

As pressure 'p' inside cylinder increases.

Force excuted on end plates F=p(\pi R^{2})

This force is equally reacted by four bolts

F - PAR +1 4

due to this forcce on bolt each bolt stretch by \delta _{b}

hence,

\delta _{b}=\frac{F_{b}L}{A_{b}E_{b}}\, \, \, \, \, \, \, \, \, \, \rightarrow 2

vessel will start to leak when the bolts have stretched by an amount equal to the original tightening=1/2 turn/ 16 turns per inch.

But as p increases,cylinder experience a radial expansion.The radial expansion by itself does not cause leakage, but it is accompanied by a poisson contraction \delta _{c} in the axial direction.This means the bolts don't have to stretch as far before the restraining plates are lifted clear.

Hence,

дь +5 x - 3 2 16 32

Axial deformation (\delta _{c}) of cylinder = L times axial strain \epsilon _{z}

\delta _{c}=\epsilon _{z}L=\frac{L}{E_{c}}[\sigma _{z}-V \sigma _{\theta }]

Since \sigma _{z} becomes zero as plate lifts off

\sigma _{\theta }=\frac{pR}{t_{c}}

\Rightarrow \delta _{c}=\frac{LV pR}{E_{c}t_{c}}\, \, \, \, \, \, \, \, \, \, \, \rightarrow 4

Put values of \delta _{c} and \delta _{b} in equation 3

\frac{F_{b}L}{A_{b}E_{b}}+\frac{LV pR}{E_{c}t_{c}}=\frac{1}{32}

\frac{p \pi R^{2} L}{4A_{b}E_{b} }+\frac{LV pR}{E_{c}t_{c}}=\frac{1}{32}

pRL=\left [ \frac{\pi RE_{c}t_{c}+4V A_{b}E_{b}}{4A_{b}E_{b}E_{c}t_{c}} \right ]=\frac{1}{32}

p=\frac{1}{8RL}=\left [ \frac{A_{b}E_{b}E_{c}t_{c}}{\pi R E_{c}t_{c}+4V A_{b}E_{b}} \right ]

subsititute all values we get

R=8'',\, L=16'',\, t_{c}=0.125''

\textrm{Area of bolts}\, \, A_{b}=\frac{\pi}{4}\left ( \frac{7}{16} \right )^{2}=0.150\,\, inch^{2}E_{c}=E_{cylinder}=125\, \, \, \, E_{bolt}=200=E_{b}\, \, \textrm{taking V for brass = 0.3}

p=\frac{1}{8\times 8\times 16}\left [ \frac{0.150\times12.5 \times 200\times 0.125\times 10^{9}}{\pi \times 8 \times125 \times 0.125 +4\times 0.3 \times 0.150\times 200}\right ]

p=1.67797.183\,Pascle=154.87\,psi

explanation:

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explanation: