Question

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.

Answer 1

Answer:

The pressure that the container will start to leak is 154.87 psi

Explanation:

Force on end plates F = p(πR²) ⇒ pπR²

There is an equal reaction from four bolts ∴ F = pπR² / 4 .............. (1)

due to force on bolt, each bolt stretch by $\sigma_b$

Therefore; $\sigma_b = \frac{F_bL}{A_bE_b}$ .........................(2)

Leaking will start when tightening is outstretched by half turn per 16 turns per inch.

Cylinder experience a radial expansion as p increases. it is accompanied by poisson contraction $\sigma_c$ in axial direction. That is, the bolt do not have to experience a long stretch before the pressure breaks off the plate.

$\sigma_b + \sigma_c = \frac{1}{2} *\frac{1}{16} = \frac{1}{32}$ ...................... (3)

Axial deformation $\sigma_c$ of cylinder = L times axial ∈₀

see continuation in attached picture in writing to get the value of Pressure: since we have the value of P

R=8", L=16", t=0.125"

Area of bolts $A_b = \frac{\pi}{4}(\frac{7}{16})^2 = 0.150 inch^2$

$E_c = E_{cylinder} = 125$

$E_b = E_{bolt} = 200$

V for brass = 0.3

$P = \frac{1}{8*8*16} [\frac{0.150*12.5*200*0.125*10^9}{\pi*8*125*0.125+4*0.3*0.150*200}]$

p = 1.67797.183 P ⇒ 154.87 psi