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) Replace the shaft with a shaft with allowable shear stress of 86 MPa, and 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.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Maximum permissible torque.
1,2. Conceptualize, Categorize: 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) to find the stress concentration factor K to be approximately 1.3 from Fig. 3-59 (repeated on next page). Then, equate the maximum shear stress in the smaller shaft to the allowable shear stress \tau _a to get
\tau _{max}=K\left\lgroup\frac{16T}{\pi D^3_1} \right\rgroup =\tau _a (a)
3,4. Analyze, Finalize: Solve Eq. (a) for T_{max} to get
T _{max}=\tau _a\left\lgroup\frac{\pi D^3_1}{16K} \right\rgroup (b)
Substituting numerical values gives
T _{max}=(63 \ MPa)\left[\frac{\pi (44 \ mm)^3}{16(1.3)}\right] =811 \ N\cdot m \hookleftarrow
Part (b): Smallest acceptable value of diameter D_1.
1,2. Conceptualize, Categorize: In the shaft redesign, a full quarter-circular fillet is being used, so
D_2=D_1+2R or R=\frac{D_2-D_1}{2}=\frac{53 \ mm-D_1}{2}=26.5 \ mm-\frac{D_1}{2} (c)
Next, solve Eq. (a) for diameter D_1 in terms of the unknown stress concentration factor K:
D_1=\left[K\left\lgroup\frac{16T}{\pi \tau _a} \right\rgroup\right]^\frac{1}{3} = \left[K\left[\frac{16(960 \ N\cdot m)}{\pi (86 \ MPa)}\right] \right]^\frac{1}{3}=\left\lgroup\frac{7680K \ N\cdot m}{43 \ MPa} \right\rgroup^\frac{1}{3} (d)
3. Analyze: Solving Eqs. (c) and (d) using trial and error and using Fig. 3-59 to obtain K produces the following results:
Trial 1:
D_{1a}=38 \ mm R=26.5 \ mm -\frac{D_{1a}}{2} =7.5 \ mm \frac{R}{D_{1a}} =0.197
From Fig. 3-59, K = 1.24, so
D_{1b} =\left\lgroup\frac{7680K \ N\cdot m}{43 \ MPa} \right\rgroup^\frac{1}{3} =41.31\ mmTrial 2:
D_{1a}=41.3 \ mm R=26.5 \ mm -\frac{D_{1a}}{2} =5.85 \ mm \frac{R}{D_{1a}} =0.142
From Fig. 3-59, K = 1.26, so
D_{1b} =\left\lgroup\frac{7680K \ N\cdot m}{43 \ MPa} \right\rgroup^\frac{1}{3} =41.53\ mmTrial 3:
D_{1a}=41.6 \ mm R=26.5 \ mm -\frac{D_{1a}}{2} =5.7 \ mm \frac{R}{D_{1a}} =0.137
From Fig. 3-59, K = 1.265, so
D_{1b} =\left\lgroup\frac{7680K \ N\cdot m}{43 \ MPa} \right\rgroup^\frac{1}{3} =41.59\ mmUse D_1 = 41.6 mm. Check the maximum shear stress:
\tau _{max}=K\left\lgroup\frac{16T}{\pi D^3_1} \right\rgroup =(1.265)\left[\frac{16(960 \ N\cdot m)}{\pi (41.6 \ mm)^3} \right]=86 \ MPa4. Finalize: A stepped shaft with D_2= 53 mm, D_1= 41.6 mm, and a full quarter-circular fillet of radius R = 5.7 mm will carry the required torque T without exceeding the allowable shear stress in the fillet region. The shaft is approximately the same size as that in part (a), but the higher allowable shear stress leads to greater permissible torque.
