Question 16.9: Design of a Parabolic Pressure Vessel A parabolic shell is c...
Design of a Parabolic Pressure Vessel
A parabolic shell is closed at the top by a thick plate and subjected to internal pressure p (Figure 16.21a). At level A–A, calculate the minimum permissible thickness t_{all} of the shell.
Given: p = 200 psi
Design Decision: The parabola is y = x^{2} /4, in which x and y are in inches.
Assumptions: At level A–A, the allowable membrane stress is σ_{all} = 16.5 ksi. Section A–A is away from the top.
Let the load resultant for the portion of the shell below plane A–A be F (Figure 16.21b). At level A–A, y = 25 in.:
x=r_0=\sqrt{4(25)}=10 in.
and hence, dy/dx = x/2 = 5. From geometry,
r_\theta=10\left(\frac{\sqrt{26}}{5}\right)=10.2 in.
The familiar expression for the curvature then gives
r_\phi=\frac{\left[1+( d y / d x)^2\right]^{1.5}}{ d ^2 y / d x^2}=\frac{\left(1+5^2\right)^{1.5}}{1 / 2}=265.15 in.
The membrane forces per unit length at A–A can now be obtained by applying Equation 16.64:
\begin{aligned} N_\phi &=-\frac{F}{2 \pi r_0 \sin \phi}=-\frac{-p \pi r_0^2}{2 \pi r_0(5 / \sqrt{26})}=\frac{\sqrt{26} p r_0}{10} \\ &=\frac{\sqrt{26}(200)(10)}{10}=1.02 kips \end{aligned}
Solving, N_{\theta} = 2.0 kips. Inasmuch as N_\theta>N_\phi , we have
t_{\text {all }}=\frac{N_\theta}{\sigma_{\text {all }}}=\frac{2.0}{16.5}=0.121 in.
Comment: The required shell thickness should be ⅛ in. at level A–A.