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

See Figures 7.16 and 7.17.

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

\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}

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

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

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

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

Equation 7.14 results in

\begin{aligned} \sigma_m &=\frac{1}{2}\left(\sigma_{\max }+\sigma_{\min }\right) \\ \sigma_a &=\frac{1}{2}\left(\sigma_{\max } \sigma_{\min }\right) \end{aligned}       (7.14)

\sigma_a=\sigma_m=\frac{26.71}{2}=13.36 ksi

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

At point E,

\frac{r}{d}=\frac{0.1}{1.0}=0.1, \quad \frac{D}{d}=\frac{3.0}{1.0}=3.0

K_{t} = 1.8 (from Figure C.9)

At point F,

\frac{r}{d}=\frac{0.1}{1.0}=0.1, \quad \frac{D}{d}=\frac{1.6}{1.0}=1.6,

K_{t} = 1.7 (from Figure C.9)

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

K_f=1+q\left(K_t-1\right)       (7.13b)

\begin{array}{l} \left(K_f\right)_E=1+0.86(1.8-1)=1.69 \\ \left(K_f\right)_F=1+0.86(1.7-1)=1.60 \end{array}

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

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

S_e=C_f C_r C_s C_t\left(\frac{1}{K_f}\right) S_e^{\prime}

where

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

Hence,

\begin{aligned} S_e &=(0.886)(0.75)(0.85)(1.0)\left(\frac{1}{1.6}\right)(65) \\ &=22.95  ksi \end{aligned}

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

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

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.