Answer:
See explaination
Step-by-step 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