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 5/8in., D = 1 7/8 in.,

Question

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 [latex]d[/latex] to [latex]D[/latex] with a full fillet where there is reversed bending and no torsion, as shown in Figure 7.10.

Given: [latex]d = 1 5/8 in., D = 1 7/8 in., S_{u} =100 ks.[/latex]

Accepted Answer

We now have, by Equation 7.1, [latex] S_e^{\prime}=0.5(100)=50 ksi .[/latex] From the given dimensions, the full fillet radius is [latex] r=(1 7 / 8-1 5 / 8) / 2=0.125 ext { in. } [/latex] Therefore,

[latex] \frac{r}{d}=\frac{0.125}{1.625}=0.08, \quad \frac{D}{d}=\frac{1.875}{1.625}=1.15 [/latex]

Referring to Figure C.9, [latex]K_{t} [/latex]= 1.7. For [latex]r[/latex] = 0.125 in. and [latex]S_{u}[/latex] = 100 ksi, by Figure 7.9a, [latex]q[/latex] = 0.82. Hence, through the use of Equation 7.13b,

[latex] \begin{aligned} K_f &=1+q\left(K_t-1 ight) \ &=1+0.82(1.7-1)=1.57 \end{aligned} [/latex]

The endurance limit, given by Equation (b) of Example 7.1, becomes

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

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

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