Question 3.17: A stepped shaft consisting of solid circular segments (D1 = ...
A stepped shaft consisting of solid circular segments (D_{1} = 44 mm and D_{2} = 53 mm, see Fig. 3-60) has a fillet of radius R = 5 mm.
(a) Find the maximum permissible torque T_{\max }, assuming that the allowable shear stress at the stress concentration is 63 MPa.
(b) If the shaft is to be replaced with a shaft with allowable shear stress of 86 MPa, D_{2} = 53 mm with a full quarter-circular fillet, carrying a torque of T = 960 N m, find the smallest acceptable value of diameter D_{1}.
(a) Maximum permissible torque. If we compute the ratio of the shaft diameters (D_{2}/D_{1} = 1.2) and the ratio of the fillet radius R to diameter D_{1} (R/D_{1} = 0.114), we can find the stress concentration factor K to be approximately 1.3 from Fig. 3-59 (repeated here). Then, equating the maxi-mum shear stress in the smaller shaft to the allowable shear stress \tau_{a}, we ge
\tau_{\max }=K\left(\frac{16 T}{\pi D_{1}^{3}}\right)=\tau_{a} (a)
Solving Eq. (a) for T_{\max }, we get
T_{\max }=\tau_{a}\left(\frac{\pi D_{1}^{3}}{16 K}\right) (b)
Substituting numerical values gives
T_{\max }=(63 \mathrm{MPa})\left[\frac{\pi(44 \mathrm{~mm})^{3}}{16(1.3)}\right]=811 \mathrm{~N} \cdot \mathrm{m}(b) Smallest acceptable value of diameter D_{1}. In the shaft redesign, a full quarter-circular fillet is being used so
D_{2}=D_{1}+2 R \quad \text { or } \quad R=\frac{D_{2}-D_{1}}{2}=\frac{53 \mathrm{~mm}-D_{1}}{2}=26.5 \mathrm{~mm}-\frac{D_{1}}{2} (c)
Next, solve Eq. (a) for diameter D_{1} in terms of the unknown stress con-centration factor K
D_{1}=\left[K\left(\frac{16 T}{\pi \tau_{a}}\right)\right]^{\frac{1}{3}}=\left[K\left[\frac{16(960 \mathrm{~N} \cdot \mathrm{m})}{\pi(86 \mathrm{MPa})}\right]\right]^{\frac{1}{3}}=\left(\frac{7680 K \mathrm{~N} \cdot \mathrm{m}}{43 \pi \mathrm{MPa}}\right)^{\frac{1}{3}} (d)
Solving Eqs. (c) and (d) using trial and error and using Fig. 3-59 to obtain K, we have the following results:
Trial #1.
From Fig. 3-59: K = 1.24:
D_{1 b}=\left(\frac{7680 K \mathrm{~N} \cdot \mathrm{m}}{43 \pi \mathrm{MPa}}\right)^{\frac{1}{3}}=41.31 \mathrm{~mm}Trial #2.
D_{1 a}=41.3 \mathrm{~mm} \quad R=26.5 \mathrm{~mm}-\frac{D_{1 a}}{2}=5.85 \mathrm{~mm} \quad \frac{R}{D_{1 a}}=0.142From Fig. 3-59: K = 1.26:
D_{1 b}=\left(\frac{7680 K \mathrm{~N} \cdot \mathrm{m}}{43 \pi \mathrm{MPa}}\right)^{\frac{1}{3}}=41.53 \mathrm{~mm}Trial #3.
D_{1 a}=41.6 \mathrm{~mm} \quad R=26.5 \mathrm{~mm}-\frac{D_{1 a}}{2}=5.7 \mathrm{~mm} \quad \frac{R}{D_{1 a}}=0.137From Fig. 3-59: K = 1.265:
D_{1 b}=\left(\frac{7680 K \mathrm{~N} \cdot \mathrm{m}}{43 \pi \mathrm{MPa}}\right)^{\frac{1}{3}}=41.59 \mathrm{~mm}Use D_{1} = 41.6 mm. Check the maximum shear stress:
\tau_{\max }=K\left(\frac{16 T}{\pi D_{1}^{3}}\right)=(1.265)\left[\frac{16(960 \mathrm{~N} \cdot \mathrm{m})}{\pi(41.6 \mathrm{~mm})^{3}}\right]=86 \mathrm{MPa}A stepped shaft with D_{2} = 53 mm, D_{1} = 41.6 mm, and a full quarter-cir-cular fillet of radius R = 5.7 mm will carry the required torque T without exceeding the allowable shear stress in the fillet region.
