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## Q. 5.8

Determine the maximum stress at point A of the thin cylinder in Fig. 5.14a. Let μ = 0.3.

## Verified Solution

Solution. A free-body diagram of junction a is shown in Fig. 5.14 b. The deflection at point A in the thick cylinder due to P is obtained from Table 5.3 by letting βx equal to βI. Hence,

 Table 5.3 Various Functions of Short Cylinders Function w $\frac{M_{0}}{2 \beta^{2} D}\left[\frac{-C_{2}}{C_{1}} V_{7}+\frac{C_{3}}{C_{1}} V_{2}-V_{8}\right]$ $\frac{Q_{0}}{2 \beta^{3} D}\left[\frac{C_{4}}{C_{1}} V_{7}-\frac{C_{5}}{C_{1}} V_{5}-\frac{C_{6}}{C_{1}} V_{6}\right]$ $\frac{\theta_{0}}{\beta}\left[\frac{C_{6}}{C_{1}} V_{5}-\frac{C_{5}}{C_{1}} V_{6}-\frac{C_{4}}{C_{1}} V_{8}\right]$ $\Delta_{0}\left[V_{7}-\frac{C_{3}}{C_{1}} V_{1}-\frac{C_{2}}{C_{1}} V_{8}\right]$ θ $\frac{M_{0}}{2 \beta D}\left[\frac{C_{2}}{C_{1}} V_{1}+\frac{2 C_{3}}{C_{1}} V_{7}-V_{2}\right]$ $\frac{-Q_{0}}{2 \beta^{2} D}\left[\frac{C_{4}}{C_{1}} V_{1}+\frac{C_{5}}{C_{1}} V_{4}+\frac{C_{6}}{C_{1}} V_{3}\right]$ $\theta_{0}\left[\frac{C_{6}}{C_{1}} V_{4}-\frac{C_{5}}{C_{1}} V_{3}-\frac{C_{4}}{C_{1}} V_{2}\right]$ $\beta \Delta_{0}\left[-V_{1}+2 \frac{C_{3}}{C_{1}} V_{8}-\frac{C_{2}}{C_{1}} V_{2}\right]$ $M_{x}$ $M_{0}\left[\frac{C_{2}}{C_{1}} V_{8}-\frac{C_{3}}{C_{1}} V_{1}-V_{7}\right]$ $\frac{Q_{0}}{\beta}\left[-\frac{C_{4}}{C_{1}} V_{8}-\frac{C_{5}}{C_{1}} V_{6}+\frac{C_{6}}{C_{1}} V_{5}\right]$ $2 \beta D \theta_{0}\left[\frac{C_{6}}{C_{1}} V_{6}+\frac{C_{5}}{C_{1}} V_{5}-\frac{C_{4}}{C_{1}} V_{7}\right]$ $2 \beta^{2} D \Delta_{0}\left[-V_{8}+\frac{C_{3}}{C_{1}} V_{2}-\frac{C_{2}}{C_{1}} V_{7}\right]$ $N_{\theta}$ $2 B^{2} r M_{0}\left[-\frac{C_{2}}{C_{1}} V_{7}+\frac{C_{3}}{C_{1}} V_{2}-V_{8}\right]$ $2 \beta r Q_{0}\left[\frac{C_{4}}{C_{1}} V_{7}-\frac{C_{5}}{C_{1}} V_{5}-\frac{C_{6}}{C_{1}} V_{6}\right]$ $\frac{E t \theta_{0}}{\beta}\left[\frac{C_{6}}{C_{1}} V_{6}+\frac{C_{5}}{C_{1}} V_{5}-\frac{C_{4}}{C_{1}} V_{7}\right]$ $\frac{E t \Delta_{0}}{r}\left[V_{7}+\frac{C_{3}}{C_{1}} V_{1}-\frac{C_{2}}{C_{1}} V_{8}\right]$ $Q_{x}$ $-\beta M_{0}\left[\frac{C_{2}}{C_{1}} V_{2}-\left(\frac{2 C_{3}}{C_{1}}+1\right) V_{8}\right]$ $Q_{0}\left[\frac{C_{4}}{C_{1}} V_{2}+\frac{C_{5}}{C_{1}} V_{3}-\frac{C_{6}}{C_{1}} V_{4}\right]$ $2 \beta^{2} D \theta_{0}\left[\frac{C_{6}}{C_{1}} V_{3}+\frac{C_{5}}{C_{1}} V_{4}+\frac{C_{4}}{C_{1}} V_{1}\right]$ $-2 \beta^{3} D \Delta_{0}\left[-V_{2}+2 \frac{C_{3}}{C_{1}} V_{7}+\frac{C_{2}}{C_{1}} V_{1}\right]$ Constants Variables $C_{1}=\sinh ^{2} \beta l-\sin ^{2} \beta l$ $V_{1}=\cosh \beta x \sin \beta x-\sinh \beta x \cos \beta x$ $C_{2}=\sinh ^{2} \beta l+\sin ^{2} \beta l$ $V_{2}=\cosh \beta x \sin \beta x+\sinh \beta x \cos \beta x$ $C_{3}=\sinh \beta l \cosh \beta l+\sin \beta l \cos \beta l$ $V_{3}=\cosh \beta x \cos \beta x-\sinh \beta x \sin \beta x$ $C_{4}=\sinh \beta l \cosh \beta l-\sin \beta l \cos \beta l$ $V_{4}=\cosh \beta x \cos \beta x+\sinh \beta x \sin \beta x$ $C_{5}=\sin ^{2} \beta l$ $V_{5}=\cosh \beta x \sin \beta x$ $C_{6}=\sinh ^{2} \beta l$ $V_{6}=\sinh \beta x \cos \beta x$ $V_{7}=\cosh \beta x \cos \beta x$ $V_{8}=\sinh \beta x \sin \beta x$
$w_{p}=\frac{-P}{2 \beta_{1}^{3} D_{1}}\left(\frac{C_{4}}{C_{1}} V_{7}-\frac{C_{5}}{C_{1}} V_{5}-\frac{C_{6}}{C_{1}} V_{6}\right)$

For $\beta_{1}=0.7421, D_{1}=0.01145 E$, and $\beta x=\beta l$, the following terms are obtained:

$C_{1}$ = 0.2028               $C_{2}$ = 1.1164

$C_{3}$ = 1.5444               $C_{4}$ = 0.5481

$C_{5}$ = 0.4568               $C_{6}$ = 0.6596

$V_{1}$ = 0.2721                $V_{3}$ = 0.4006

$V_{4}$ = 1.4984              $V_{5}$ = 0.8707

$V_{6}$ = 0.5986              $V_{7}$ = 0.9495

Thus the expression for $w_{p}$ due to P is given by

$W_{p}=\frac{144.02 P}{E}$

The deflection compatibility equation at point A is

$\left.\delta_{1}\right|_{x=0}=\left.\delta_{2}\right|_{x=0}$

or from Tables 5.2 and 5.3 with x = 0,

 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 . $M_{\theta}=\mu M_{x}$.
$\frac{144.02 P}{E}-\frac{Q_{0}}{2 \beta_{1}^{3} D_{1}}\left(\frac{C_{4}}{C_{1}}\right)-\frac{M_{0}}{2 \beta_{1}^{2} D_{1}}\left(\frac{-C_{2}}{C_{1}}\right)=\frac{Q_{0}}{2 \beta_{2}^{3} D_{2}}-\frac{M_{0}}{2 \beta_{2}^{2} D_{2}}$

and with $\beta_{2}=1.0495$ and $D_{2}=0.00145 E$, the equation becomes

$144.02 P-288.82 Q_{0}+436.59 M_{0}=298.30 Q_{0}-313.07 M_{0}$

or

$5.206 M_{0}-4.077 Q_{0}=-P$                          (1)

The rotation at point A due to P is obtained from Table 5.3 as

$\theta_{p}=\frac{P}{2 \beta_{1}^{2} D_{1}}\left(\frac{C_{4}}{C_{1}} V_{1}+\frac{C_{5}}{C_{1}} V_{4}+\frac{C_{6}}{C_{L}} V_{3}\right)$

or

$\theta_{p}=\frac{429.33 P}{E}$

The rotation compatibility equation at point A is

$\left.\theta_{1}\right|_{x=0}=\left.\theta_{2}\right|_{x=0}$

Hence,

$\frac{429.33 P}{E}+\frac{M_{0}}{2 \beta_{1} D_{1}}\left(\frac{2 C_{3}}{C_{1}}\right)+\frac{Q_{0}}{2 \beta_{1}^{2} D_{1}}\left(\frac{C_{5}}{C_{1}}+\frac{C_{6}}{C_{1}}\right)=\frac{-M_{0}}{\beta_{2} D_{2}}+\frac{Q_{0}}{2 \beta_{2}^{2} D_{2}}$

or

$\frac{429.33 P}{E}+896.45 M_{0}+436.59 Q_{0}=-657.13 M_{0}+313.07 Q_{0}$

which reduces to

$12.58 M_{0}+Q_{0}=-3.48 P$                          (2)

Solving Eqs. 1 and 2 yields

\begin{aligned}&M_{0}=-0.27 P \\&Q_{0}=-0.08 P\end{aligned}

Hence, maximum axial stress is

$\sigma=\frac{6 M_{0}}{t^{2}}=\frac{6(0.27 P)}{0.25^{2}}=25.9 P \text { psi }$

The circumferential bending moment is given by

$M_{\theta}=\mu M_{x}=0.08 P$

The circumferential force $N_{\theta}$ is given by Eq. 5.19 as

$N_{\theta}=\frac{-E t w}{r}$                              (5.19)

\begin{aligned}N_{\theta} &=\frac{E t w}{r} \\&=\frac{E t_{2}}{r}\left(W_{Q_{0}}-W_{M_{0}}\right) \\&=\frac{E t_{2}}{R}\left(\frac{-0.08 P}{2 \beta_{2}^{3} D_{2}}-\frac{-0.27 P}{2 \beta_{2}^{2} D_{2}}\right) \\&=\frac{P}{24}(-23.86+84.53)=2.53 P\end{aligned}

\begin{aligned}\sigma_{\theta} &=\frac{N_{\theta}}{t}+\frac{6 M_{\theta}}{t^{2}} \\&=\frac{2.53 P}{0.25}+\frac{6(0.08 P)}{0.25^{2}} \\&=17.80 P\end{aligned}