Question 8.CS.1: Cam and Follower Stress Analysis of an Intermittent-Motion M......

See Figure 7.16, Table 8.4, and Tables B.1, and B.4 in Appendix B.

Equations on the second column of case A of Table 8.4 apply. We first determine the half width a of the contact patch. Since E_1=E_2=E \text { and } \Delta=2 / E, we have

a=1.076 \sqrt{\frac{F_{\max }}{L_4} r_c \Delta}

Substitution of the given data yields

\begin{aligned} a & =1.076\left[\frac{1600}{1.5}(1.5)\left(\frac{2}{30 \times 10^6}\right)\right]^{1 / 2} \\ & =11.113\left(10^{-3}\right) in . \end{aligned}

The rectangular patch area is

2 a L_4=2\left(11.113 \times 10^{-3}\right)(1.5)=33.34\left(10^{-3}\right) \text { in. }^2

Maximum contact pressure is then

\begin{aligned} p_o & =\frac{2}{\pi} \frac{F_{\max }}{a L_4} \\ & =\frac{2}{\pi} \frac{1600}{\left(11.113 \times 10^{-3}\right)(1.5)}=61.11  ksi \end{aligned}

The deflection \delta of the cam and follower at the line of contact is obtained as follows:

\delta=\frac{0.579 F_{\max }}{E L_4}\left(\frac{1}{3}+\ln \frac{2 r_c}{a}\right)

Introducing the numerical values,

\begin{aligned} \delta & =\frac{0.579(1600)}{30 \times 10^6(1.5)}\left(\frac{1}{3}+\ln \frac{2 \times 1.5}{11.113 \times 10^{-3}}\right) \\ & =0.122\left(10^{-3}\right)  in \end{aligned}

Comment: The contact stress is determined to be less than the yield strength and the design is satisfactory. The calculated deflection between the cam and the follower is very small and does not affect the system performance.