Question 7.2: Endurance Limit for a Stepped Shaft in Reversed Bending Rewo...
Endurance Limit for a Stepped Shaft in Reversed Bending
Rework Example 7.1 for the condition that the critical point on the shaft is at a diameter change from d to D with a full fillet where there is reversed bending and no torsion, as shown in Figure 7.10.
Given: d = 1 5/8 in., D = 1 7/8 in., S_{u} =100 ks.
We now have, by Equation 7.1, S_e^{\prime}=0.5(100)=50 ksi . From the given dimensions, the full fillet radius is r=(1 7 / 8-1 5 / 8) / 2=0.125 \text { in. } Therefore,
\frac{r}{d}=\frac{0.125}{1.625}=0.08, \quad \frac{D}{d}=\frac{1.875}{1.625}=1.15
Referring to Figure C.9, K_{t} = 1.7. For r = 0.125 in. and S_{u} = 100 ksi, by Figure 7.9a, q = 0.82. Hence, through the use of Equation 7.13b,
\begin{aligned} K_f &=1+q\left(K_t-1\right) \\ &=1+0.82(1.7-1)=1.57 \end{aligned}
The endurance limit, given by Equation (b) of Example 7.1, becomes
\begin{aligned} S_e &=C_f C_r C_s C_t\left(1 / K_f\right) S_e^{\prime} \\ &=(0.80)(0.84)(0.85)(0.71)(1 / 1.57)(50) \\ &=12.92 ksi \end{aligned}
