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Chapter 6

Q. 6.3

Calculate the head discontinuity forces of the head-to-shell junction shown in Fig. 6.8a. Let p = 300 psi and μ = 0.3.

Calculate the head discontinuity forces of the head-to-shell junction shown in Fig. 6.8a. Let p = 300 psi and μ = 0.3.


Verified Solution

From Fig. 6.8 b and Table 6.1 the deflection in the head due to pressure is

Table 6.1 Membrane Forces & Deflections In Spherical Shells
P_r -P\cos \phi -P \cos ^{2} \phi P
P_\phi -P \sin\phi P \cos \phi \sin \phi 0
N_θ -P \cdot r\left(\cos \phi-\frac{1}{1+\cos \phi}\right) \frac{-P \cdot r \cdot \cos 2 \phi}{2} \frac{P \cdot r}{2}
N_\phi \frac{-P \cdot r}{1+\cos \phi} \frac{-P \cdot r}{2} \frac{P \cdot r}{2}
δ \frac{P \cdot r^{2}}{E t} \sin \phi\left[-\cos \phi+\frac{1+\mu}{\sin ^{2} \phi}(1-\cos \phi)\right] \frac{P \cdot r^{2}}{E t} \sin \phi\left[-\cos ^{2} \phi+\frac{1+\mu}{2}\right] \frac{P \cdot r^{2}}{2 E t}(1-\mu) s i n \phi
θ \frac{-P \cdot r}{E t}(2+\mu) \sin \phi \frac{-P \cdot r}{E t}(3+\mu) \sin \phi \cos \phi 0
P_r -\gamma[h+r(1-\cos \phi)] 0
P_\phi 0 0
N_θ \frac{-\gamma \cdot r^{2}}{6}\left(-1+3 \frac{h}{r}-\frac{4 \cos ^{2} \phi-6}{1+\cos \phi}\right) \frac{+F\sin \phi_{0}}{\sin ^{2} \phi}
N_\phi \frac{-\gamma \cdot r^{2}}{6}\left(-1+3 \frac{h}{r}-\frac{2 \cos ^{2} \phi}{1+\cos \phi}\right) \frac{-F \sin \phi_{0}}{\sin ^{2} \phi}
δ \begin{aligned}&\frac{-\gamma \cdot r^{3}}{6 E t} \sin \phi\left[3\left(1+\frac{h}{r}\right)(1-\mu)\right. \\&\left.-6 \cos \phi-\frac{2(1+\mu)}{\sin^2 \phi}(\cos \phi-1)\right]\end{aligned} \frac{-F_{\cdot} r}{E t}(1+\mu) \frac{\sin \phi_{0}}{\sin \phi}
θ \frac{\gamma \cdot r^{2}}{E t} \sin \phi 0
\begin{aligned}\delta_{p} &=\frac{P r^{2}}{E t}(1-\mu) \sin \phi \\&=\frac{(300)(50)^{2}}{E(0.50)}(1-0.3) \\&=\frac{1,050,000}{E}\end{aligned}


\begin{aligned}\lambda &=\sqrt[4]{3\left(1-0.3^{2}\right)\left(\frac{50}{0.5}\right)^{2}} \\&=12.038\end{aligned}

From Table 6.2,

Table 6.2 Approximate Force and Deflection Functions for Spherical Segments
Q -\sqrt{2} e^{-\lambda \gamma} \sin \phi_{0} \cos (\lambda \gamma+\pi / 4) Ho \frac{2 \lambda}{r} e^{-\lambda \gamma} \sin (\lambda \gamma) M 0
N_{\theta} 2 \lambda e^{-\lambda \gamma} \sin \phi_{0}(\cos \lambda \gamma) \text { Ho } 2 \sqrt{2} \frac{\lambda^{2}}{r} e^{-\lambda \gamma} \cos (\lambda \gamma+\pi / 4) Mo
N _{\phi} \sqrt{2} e ^{-\lambda \gamma} \sin \phi_o \cot \phi \cos (\lambda \gamma+\pi / 4) Ho -\frac{2 \lambda}{r} e^{-\lambda \gamma} \cot \phi \sin (\lambda \gamma) M_{0}
M _{\theta} \mu . M_{\phi} \mu M_{\phi}
M _{\phi} \frac{r}{\lambda} e^{-\lambda \gamma}\sin  \phi_{0} \sin (\lambda \gamma) H_{0} \sqrt{2} e^{-\lambda \gamma}\sin (\lambda \gamma+\pi / 4) M_{0}
\delta \begin{aligned}\frac{H_{0}}{E t} &\left\{r e^{-\lambda \gamma}_{s i n} \phi_{0}[2 \lambda \sin \phi \cos \lambda \gamma\right.\\&-\sqrt{2} \mu \cos \phi \cos (\lambda \gamma+\pi / 4)]\}\end{aligned} \frac{M_{0}}{E t}\left\{2\lambda e^{-\lambda \gamma }\left[\sqrt{2}\lambda \sin \phi \cos (\lambda \gamma +\pi /4)+\mu \cos \phi \sin \phi \lambda \right] \right\}
\alpha \frac{H o}{E t}\left[-2 \sqrt{2} \lambda^{2} e^{-\lambda \gamma}\sin \phi_{0} \sin (\lambda \gamma+\pi / 4)\right] \frac{M_{0}}{E t}\left(-\frac{4 \lambda^{3}}{r} e^{-\lambda \gamma} \cos \lambda \gamma\right)
\delta_{H_{0}}=\frac{2407.6}{E} H_{0}


\alpha_{H_{0}}=\frac{-579.65}{E} H_{0}


\delta_{M_{0}}=\frac{579.65}{E} M_{0}


\alpha_{M_{0}}=\frac{-279.11}{E} M_{0}

Similarly, the deflection in the shell due to pressure is obtained from Eq. 1 of Example 5.5 as

\begin{aligned}\delta &=\frac{\operatorname{Pr}^{2}}{E t}\left(1-\frac{\mu}{2}\right) \\&=\frac{(300)(50)^{2}}{E(1.0)}(1-0.15) \\&=\frac{637,500}{E}\end{aligned}


\begin{aligned}\beta &=\sqrt[4]{\frac{3\left(1-0.3^{2}\right)}{(50)^{2}(1.0)^{2}}} \\&=0.1818\end{aligned}


D=\frac{E(1.0)^{3}}{12\left(1-0.3^{2}\right)}=0.0916 E

and from Table 6.2,

\delta_{H_{0}}=\frac{-908.43}{E} H_{0}


\theta_{H_{0}}=\frac{-165.15}{E} H_{0}


\delta_{M_{0}}=\frac{165.15}{E} M_{0}


\theta_{M_{0}}=\frac{60.05}{E} M_{0}

total deflection of head = total deflection of shell

\frac{1,050,000}{E}+\frac{2407.6}{E} H_{0}+\frac{579.65}{E} M_{0}=\frac{637,500}{E}-\frac{908.43}{E} H_{0}+\frac{165.15}{E} M_{0}


8 H_{0}+M_{0}=-995.17                                (1)


rotation of head = rotation of shell

\frac{-579.65}{E} H_{0}-\frac{279.11}{E} M_{0}=-\frac{165.15}{E} H_{0}+\frac{60.05}{E} M_{0}


H_{0}=-0.818 M_{0}

and from Eq. 1,

M_{0}=179.4  lb – in . / in.


H_{0}=-146.8  lb / in .

From Table 6.2, N_{\theta} at discontinuity is

\begin{aligned}N_{\theta} &=2 \lambda H_{0}+\frac{2 \lambda^{2}}{r} M_{0}+\frac{P r}{2} \\&=2(12.038)(-146.8)+\frac{2(12.038)^{2}(179.4)}{50}+\frac{300(50)}{2}\end{aligned}




N_{\theta}=5006  \text { lb/in. }


N_{\phi}=\frac{p r}{2}=7500  lb / in.


M_{\phi}=179.4  lb – in . / in.


M_{\theta}=53.8  lb / \text { in. }