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Q. 6.9

Calculate the maximum longitudinal and circumferential stresses in the cylinder shown in Fig. 6.16 due to internal pressure P.

Verified Solution

From Fig. 6.16

f + F = H                                     (1)

where from Section 6.4.1,

$H=\frac{P r \tan \alpha^{\prime}}{2}$.

The deflection compatibility between the cylinder and cone is given by

deflection of cylinder at junction due to M and f

= deflection of cone at junction due to M and F                                  (2)

Similarly,

rotation of cylinder at junction due to M and f

= rotation of cone at junction due to M and F                                     (3)

Using Tables 5.2 and 6.4 and solving Eqs. 1,2 , and 3 result in the following expressions:

 Table 5.2 Various Discontinuity Functions Functions* Edge Functions W $\frac{-M_o}{2 \beta^{2} \cdot D}$ $\frac{Q_o}{2 \beta^{3} D}$ Δo 0 θ $\frac{M_{o}}{\beta . D}$ $\frac{-Q_o}{2 \beta^{2} \cdot D}$ 0 $θ_o$ $M_θ$ $M_o$ 0 $2 \beta^{2} \cdot D \cdot \Delta o$ $2 \beta \cdot D \cdot \theta_{o}$ $N_θ$ $-2 M_{o} \beta^{2} \cdot r$ $2 \beta \cdot r \cdot Q_o$ $\frac{E \cdot t \cdot \Delta o}{r}$ 0 $Q_o$ 0 $Q_o$ $4 \beta^{3} \cdot 0 . \Delta o$ $2 \beta^{2} \cdot D \cdot \theta_{o}$ General Functions W $\frac{-M_{o}}{2 \beta^{2} \cdot D} B_{\beta x}$ $\frac{Q_o}{2 \beta^{3} D} C_{\beta X}$ $\Delta_{o}\left(2 C_{\beta x}-B_{\beta x}\right)$ $\frac{\theta_o}{\beta}\left(C_{\beta x} -B_{\beta x}\right)$ θ $\frac{M_{o}}{\beta . D} C_{\beta x}$ $\frac{-Q_o}{2 \beta^{2} D} \cdot A_{\beta x}$ $2 \beta \Delta_{o}\left(A_{ \beta x}-C_{ \beta x}\right)$ $\theta_{0}\left(A_{\beta x}-2 C_{\beta x}\right)$ $M_X$ $M_{o} \cdot A_{\beta x}$ $\frac{Q_o}{\beta} \cdot D_{\beta X}$ $2 \beta^{2} \cdot D \cdot \Delta_o\left(A_{ \beta x}-C_{ \beta x}\right)$ $2\beta \cdot D \cdot \theta_{0} \left(D_{\beta x}- A_{\beta x}\right)$ $N_θ$ $-2 M_{o} \beta^{2} \cdot r \cdot B_{\beta x}$ $2 \beta \cdot r \cdot Q_{o} \cdot C_{\beta x}$ $\frac{E \cdot t}{r} \cdot \Delta_o\left(2 C_{\beta x}-B_{\beta x}\right)$ $\frac{\text { E.t. } \theta_{o}}{r \beta}\left(C_{\beta x}-B_{\beta x}\right)$ $Q_X$ $-2 \beta . M_o . D_{\beta X}$ $Q_{o} \cdot B_{\beta X}$ $4 \beta^{3} D \cdot \Delta_{o}\left(B_{\beta X}-D_{\beta X}\right)$ $2 \beta^{2} \cdot D \cdot \theta_{o}\left(2 D_{\beta x}+B_{\beta x})\right.$ $^*$Clockwise moments and rotation are positive at point 0 . Outward forces and deflections are positive at point 0 . $M_{\theta}=\mu M_{x}$.

 Table 6.4 $M_{\theta}=\mu M \phi$. Q $-\sqrt{2} e^{-\beta x} \sin \phi \cos (\beta x+\pi / 4) H_o$ $\frac{-2 \beta e^{-\beta x} \sin \phi}{l} \sin (\beta x) M_o$ $N_θ$ $\frac{-2 r_{2} \cdot \beta e^{-\beta x}}{l} \sin { }^{2} \phi \cos (\beta x) H_o$ $\frac{2 \sqrt{2} r_{2} \beta^{2}e^{-\beta x} }{l^{2}} \sin ^{2} \phi \cos (\beta x+\pi / 4) M_o$ $N_\phi$ $-\sqrt{2} e^{-\beta x} \cos \phi \cos \left(\beta x+\frac{\pi}{4}\right) H_o$ $\frac{-2 \beta}{\ell} e^{-\beta x} \cos \phi \sin (\beta x) M_o$ $M_\phi$ $\frac{\ell}{\beta} e^{-\beta x} \sin \beta x H_o$ $-\sqrt{2} e^{-\beta x} \sin (\beta x+\pi / 4) M_o$ δ $\frac{ H_ol ^{3} e^{-\beta x}}{2 D \beta^{3} \sin \phi}\left[\cos \beta x-\frac{\mu \ell}{\sqrt{2}r_{2} \beta}\frac{\cot \phi}{\sin \phi} \cos (\beta x+\pi /4)\right]$ \begin{aligned}&\frac{-\ell^{2} e^{-\beta x}}{2 D \beta^{2} \sin \phi}[\sqrt{2} \cos (\beta x+\pi / 2) \\&\left.+\mu \frac{1}{r_{2}} \frac{\cos \phi \sin \beta x}{\beta \sin ^{2} \phi}\right] M_o \end{aligned} θ $\frac{-l^{2} e^{-\beta x} H_o}{\sqrt{2} D \beta^{2} \sin \phi} \text { sin }(\beta x+\pi / 4)$ $\frac{le^{-\beta x}}{D \beta s i n \phi} \cos \beta x \quad M_o$
\begin{aligned}f &=H V_{1} \\F &=H\left(1-V_{1}\right) \\M &=H\left(\frac{2}{\beta}\right) V_{2}\end{aligned}

where

\begin{aligned}H &=\frac{P r \tan \alpha^{\prime}}{2} \\V_{1} &=\frac{\cos ^{2} \alpha^{\prime}\left(3+\cos ^{2} \alpha^{\prime}\right)}{1+\cos ^{2} \alpha^{\prime}\left(6+\cos ^{2} \alpha^{\prime}\right)} \\V_{2} &=\frac{V_{1}}{\left(3+\cos ^{2} \alpha^{\prime}\right)} \\\beta &=\sqrt[4]{\frac{3\left(1-\mu^{2}\right)}{r^{2} t^{2}}}\end{aligned}

The maximum longitudinal stress due to M and pressure is expressed as

$\sigma_{x}=\frac{P r}{t}\left(0.5+4.559 V_{2} \sqrt{\frac{r}{t}} \tan \alpha^{\prime}\right)$

whereas the maximum circumferential stress due to M and pressure is given by

$\sigma_{\theta}=\frac{P r}{t}\left[1.0-1.316 \sqrt{\frac{r}{t}}\left(V_{1}-2 V_{2}\right) \tan \alpha^{\prime}\right]$