Camshaft Fatigue Design of Intermittent-Motion Mechanism Figure 7.16 illustrates a rotating camshaft of an intermittent-motion mechanism in its peak lift position. The cam exerts a force P on the follower, because of a stop mechanism (not shown), only during less than half a shaft revolution.

Question

Camshaft Fatigue Design of Intermittent-Motion Mechanism

Figure 7.16 illustrates a rotating camshaft of an intermittent-motion mechanism in its peak lift position. The cam exerts a force [latex]P[/latex] on the follower, because of a stop mechanism (not shown), only during less than half a shaft revolution. Calculate the factor of safety for the camshaft according to the Goodman criterion.

Given: The geometry is known and the shaft supports a pulsating force with [latex]P_{max}[/latex] and [latex]P_{min}[/latex]. The material of all parts is AISI 1095 steel, carburized on the cam surface and oil quenched and tempered (OQ&T) at 650°C. The fillet and adjacent surfaces are fine ground.

Data:

[latex] \begin{array}{l} P_{\max }=1.6 kips , \quad P_{\min }=0 \ S_u=130 ksi , \quad S_y=80 ksi \quad ( ext { from Table B.4) } \ L_1=2.8 ext { in., } \quad L_2=3.2 ext { in., } \ L_3=0.5 ext { in., } \quad L_4=1.5 ext { in., } \ L_5=L_1+\frac{1}{2}\left(L_3+L_4 ight)=3.8 ext { in., } \ L_6=L_2+\frac{1}{2}\left(L_3+L_4 ight)=4.2 ext { in., } \ D_s=1 ext { in., } \quad D_c=1.6 ext { in., } \ r_c=1.5 ext { in., } \quad r=0.1 ext { in., } \ I / c=\pi D_s^3 / 32=98.175\left(10^{-3} ight) ext { in. }^3 \end{array} [/latex]

Assumptions:

1. Bearings act as simple supports.

2. The operating temperature is normal.

3. The torque can be regarded negligible.

4. A material reliability of 99.99% is required.

Accepted Answer

See Figures 7.16 and 7.17.

Alternating and mean stresses. The reactions at the supports [latex]A[/latex] and [latex]B[/latex] are determined by the conditions of equilibrium as

[latex] \begin{aligned} R_A &=\frac{L_6}{L_5+L_6} P_{\max } \ &=\frac{4.2}{8}(1600)=840 lb \ R_B &=P_{\max }-R_A=760 lb \end{aligned} [/latex]

and noted in Figure 7.17a.

The plot of the moment diagram, from a maximum moment of 760 × 4.2 = 3192 lb·in., is shown in Figure 7.17b. We observe that the moment on the right side

[latex] \begin{aligned} M &=R_B\left(L_6-\frac{1}{2} L_4 ight) \ &=760(3.45)=2622 lb \cdot in . \end{aligned} [/latex]

is larger than (2562 lb·in.) at the left side. We have

[latex] \begin{aligned} \sigma_{\max } &=\frac{M}{I / c}=\frac{2622}{98.175\left(10^{-3} ight)}=26.71 ksi \ \sigma_{\min } &=0 \end{aligned} [/latex]

Equation 7.14 results in

[latex] \begin{aligned} \sigma_m &=\frac{1}{2}\left(\sigma_{\max }+\sigma_{\min } ight) \ \sigma_a &=\frac{1}{2}\left(\sigma_{\max } \sigma_{\min } ight) \end{aligned} [/latex] (7.14)

[latex] \sigma_a=\sigma_m=\frac{26.71}{2}=13.36 ksi [/latex]

Stress-concentration factors. The step in the shaft is asymmetrical. Stress at point [latex]E[/latex] is influenced by the radius [latex]r[/latex] = 1.5 in. (equivalent to a diameter of 3.0 in.) and at point [latex]F[/latex] by the 0.8 in. cam radius (equivalent to [latex]D_{c}[/latex] = 1.6 in. diameter). Hence, we obtain the following values:

At point [latex]E,[/latex]

[latex] \frac{r}{d}=\frac{0.1}{1.0}=0.1, \quad \frac{D}{d}=\frac{3.0}{1.0}=3.0 [/latex]

[latex]K_{t}[/latex] = 1.8 (from Figure C.9)

At point [latex]F[/latex],

[latex] \frac{r}{d}=\frac{0.1}{1.0}=0.1, \quad \frac{D}{d}=\frac{1.6}{1.0}=1.6, [/latex]

[latex]K_{t}[/latex] = 1.7 (from Figure C.9)

For [latex]r[/latex] = 0.10 in. and [latex]S_{u}[/latex] = 130 ksi, by Figure 7.9a, [latex]q[/latex] = 0.86. It follows, from Equation 7.13b, that

[latex] K_f=1+q\left(K_t-1 ight) [/latex] (7.13b)

[latex] \begin{array}{l} \left(K_f ight)_E=1+0.86(1.8-1)=1.69 \ \left(K_f ight)_F=1+0.86(1.7-1)=1.60 \end{array} [/latex]

Comments: Note that the maximum stress in the shaft is well under the material yield strength. The stress concentration at [latex]E[/latex] is only 5% larger than that at [latex]F[/latex]. Therefore, fatigue failure is expected to begin at point [latex]F[/latex], where the stress pulses are tensile and compressive at [latex]E[/latex].

Modified endurance limit. Through the use of Equation 7.6, we have

[latex] S_e=C_f C_r C_s C_t\left(\frac{1}{K_f} ight) S_e^{\prime} [/latex]

where

[latex] \begin{array}{l} C_f=1.34(130)^{-0.085}=0.886 \quad( ext { from Table 7.2) } \ C_r=0.75 \quad( ext { by Table 7.3) } \ C_{ s }=0.85 \quad( ext { from Equation 7.9) } \ C_t=1.0 \quad( ext { room temperature) } \ K_f=1.6 \ S_e^{\prime}=0.5(130)=65 ksi \quad ext { (by Equation 7.1) } \end{array} [/latex]

Hence,

[latex] \begin{aligned} S_e &=(0.886)(0.75)(0.85)(1.0)\left(\frac{1}{1.6} ight)(65) \ &=22.95 ksi \end{aligned} [/latex]

Factor of safety. The safety factor guarding against fatigue failure at point [latex]F[/latex] is determined using Equation 7.22:

[latex] n=\frac{S_u}{\sigma_m+\frac{S_u}{S_e} \sigma_a}=\frac{130}{13.36+\frac{130}{22.95}(13.36)}=1.46 [/latex]

Comments: If the load is properly controlled so that there is no impact, the foregoing factor seems well sufficient. Inasmuch as lift motion is involved, the deflection needs to be checked accurately by FEA. Case Study 8.1 analyses contact stresses between cam and follower.

Table 7.2 Surface Finish Factors [latex]C_{f}[/latex]
	[latex]A[/latex]
Surface Finish	MPa	ksi	[latex]b[/latex]
Ground	1.58	1.34	−0.085
Machined or cold drawn	4.51	2.7	−0.265
Hot rolled	57.7	14.4	−0.718
Forged	272.0	39.9	−0.995

Table 7.3 Reliability Factors
Survival Rate (%)	[latex]C_{r}[/latex]
50	1.00
90	0.89
95	0.87
98	0.84
99	0.81
99.9	0.75
99.99	0.70

Question 7.CS.1: Camshaft Fatigue Design of Intermittent-Motion Mechanism Fig...

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