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Chapter 6

Q. 6.9

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

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

Step-by-Step

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]