Cam and Follower Stress Analysis of an Intermittent-Motion Mechanism Figure 7.16 shows a camshaft and follower of an intermittent-motion mechanism. For the position indicated, the cam exerts a force Fmax on the follower. What are the maximum stress at the contact line between the cam and follower,

Question

Cam and Follower Stress Analysis of an Intermittent-Motion Mechanism

Figure 7.16 shows a camshaft and follower of an intermittent-motion mechanism. For the position indicated, the cam exerts a force [latex] F_{max} [/latex] on the follower. What are the maximum stress at the contact line between the cam and follower, and the deflection?

Given: The shapes of the contacting surfaces are known. The material of all parts is AISI 1095 steel carburized on the surfaces, oil quenched, and tempered (Q&T) at 650°C.

Data:

[latex] F_{\max }=P_{\max }=1.6 kips , \quad r_c=1.5 ext { in., } \quad D_f=L_4=1.5 in [/latex]

[latex] E=30 imes 10^6 psi , \quad S_y=80 ksi [/latex]

Assumptions: Frictional forces can be neglected. The rotational speed is slow so that the loading is considered static.

Accepted Answer

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 [latex] a [/latex] of the contact patch. Since [latex] E_1=E_2=E ext { and } \Delta=2 / E, [/latex] we have

[latex] a=1.076 \sqrt{\frac{F_{\max }}{L_4} r_c \Delta} [/latex]

Substitution of the given data yields

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

The rectangular patch area is

[latex] 2 a L_4=2\left(11.113 imes 10^{-3} ight)(1.5)=33.34\left(10^{-3} ight) ext { in. }^2 [/latex]

Maximum contact pressure is then

[latex] \begin{aligned} p_o & =\frac{2}{\pi} \frac{F_{\max }}{a L_4} \ & =\frac{2}{\pi} \frac{1600}{\left(11.113 imes 10^{-3} ight)(1.5)}=61.11 ksi \end{aligned} [/latex]

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

[latex] \delta=\frac{0.579 F_{\max }}{E L_4}\left(\frac{1}{3}+\ln \frac{2 r_c}{a} ight) [/latex]

Introducing the numerical values,

[latex] \begin{aligned} \delta & =\frac{0.579(1600)}{30 imes 10^6(1.5)}\left(\frac{1}{3}+\ln \frac{2 imes 1.5}{11.113 imes 10^{-3}} ight) \ & =0.122\left(10^{-3} ight) in \end{aligned} [/latex]

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.

TABLE 8.4 Maximum Pressure $p_o$ , and Deflection $\delta$ of Two Bodies at the Point of Contact
Configuration	$\text { Spheres: } p_o=1.5 \frac{F}{\pi a^2}$	$\text { Cylinders: } p_o=\frac{2}{\pi} \frac{F}{a L}$
$\quad\quad\quad\quad\quad\quad\quad\quad$	$\text { Sphere on a flat surface } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F r_1 \Delta} \\ & \delta=0.775 \sqrt[3]{F^2 \frac{\Delta^2}{r_1}} \end{aligned}$	$\text { Cylinder on a flat surface } \\ \\ \begin{aligned} & a=1.076 \sqrt{\frac{F}{L} r_1 \Delta} \\ & \text { For } E _1= E _2= E \\ & \delta=\frac{0.579 F}{E L}\left(\frac{1}{3}+\ln \frac{2 r_1}{a}\right) \end{aligned}$
$\quad\quad\quad\quad\quad\quad\quad\quad$	$\text { Two spherical balls } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F \frac{\Delta}{m}} \\ & \delta=0.775 \sqrt[3]{F^2 \Delta^2 m} \end{aligned}$	$\text { Two cylindrical rollers } \\ \\ a=1.076 \sqrt{\frac{F \Delta}{L m}}$
$\quad\quad\quad\quad\quad\quad\quad\quad$	$\text { Sphere on a spherical seat } \\ \\ \begin{aligned} & a=0.880 \sqrt[3]{F \frac{\Delta}{n}} \\ & \delta=0.775 \sqrt[3]{F^2 \Delta^2 n} \end{aligned}$	$\text { Cylinder on a cylindrical seat } \\ \\ a=1.076 \sqrt{\frac{F \Delta}{L n}}$
Source: [9]. Notes: $\Delta=\frac{1}{E_1}+\frac{1}{E_2}, m=\frac{1}{r_1}+\frac{1}{r_2}, n=\frac{1}{r_1}-\frac{1}{r_2}$ , where the modulus of elasticity ( $E$ ) and radius ( $r$ ) are for the contacting members, 1 and 2. The $L$ represents the length of the cylinder (Figure 8.7). The total force pressing two spheres or cylinder is $F$ . Poisson’s ratio $\nu$ in the formulas is taken as 0.3.

TABLE B.4 Mechanical Properties of Selected Heat-Treated Steels
AISI Number	Treatment	Temperature (°C)	Ultimate Strength $S_u$ (MPa)	Yield Strength $S_y$ (MPa)	Elongation in 50 mm (%)	Reduction in Area (%)	Brinell Hardness (HB)
1030	WQ&T	205	848	648	17	47	495
	WQ&T	425	731	579	23	60	302
	WQ&T	650	586	441	32	70	207
	Normalized	925	521	345	32	61	149
	Annealed	870	430	317	35	64	137
1040	OQ&T	205	779	593	19	48	262
	OQ&T	425	758	552	21	54	241
	OQ&T	650	634	434	29	65	192
	Normalized	900	590	374	28	55	170
	Annealed	790	519	353	30	57	149
1050	WQ&T	205	1120	807	9	27	514
	WQ&T	425	1090	793	13	36	444
	WQ&T	650	717	538	28	65	235
	Normalized	900	748	427	20	39	217
	Annealed	790	636	365	24	40	187
1060	OQ&T	425	1080	765	14	41	311
	OQ&T	540	965	669	17	45	277
	OQ&T	650	800	524	23	54	229
	Normalized	900	776	421	18	37	229
	Annealed	790	626	372	11	38	179
1095	OQ&T	315	1260	813	10	30	375
	OQ&T	425	1210	772	12	32	363
	OQ&T	650	896	552	21	47	269
	Normalized	900	1010	500	9	13	293
	Annealed	790	658	380	13	21	192
4130	WQ&T	205	1630	1460	10	41	467
	WQ&T	425	1280	1190	13	49	380
	WQ&T	650	814	703	22	64	245
	Normalized	870	670	436	25	59	197
	Annealed	865	560	361	28	56	156
4140	OQ&T	205	1770	1640	8	38	510
	OQ&T	425	1250	1140	13	49	370
	OQ&T	650	758	655	22	63	230
	Normalized	870	870	1020	18	47	302
	Annealed	815	655	417	26	57	197
Source: ASM Metals Reference Book, 3rd ed. Materials Park, OH, American Society for Metals, 1993.
Notes: To convert from MPa to ksi, divide given values by 6.895. Values tabulated for 25 mm round sections and of gage length 50 mm. The properties for quenched and tempered steel are from a single heat: OQ&T, oil-quenched and tempered; WQ&T, water-quenched and tempered.

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

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