Question 7.2: Endurance Limit for a Stepped Shaft in Reversed Bending Rewo...

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

$\begin{aligned} S_e &=C_f C_r C_s C_t\left(1 / K_f\right) S_e^{\prime} \\ &=(0.80)(0.84)(0.85)(0.71)(1 / 1.57)(50) \\ &=12.92  ksi \end{aligned}$