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

where

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

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