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Chapter 5

Q. 5.8

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

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
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


or from Tables 5.2 and 5.3 with x = 0,

Table 5.2 Various Discontinuity 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}


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)


\theta_{p}=\frac{429.33  P}{E}

The rotation compatibility equation at point A is



\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}}


\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}