Question 7.8: Figure 7–23 shows a portion of a round shaft where a gear is...
Figure 7–23 shows a portion of a round shaft where a gear is mounted. A bending moment of 30 N · m is applied at this location. Compute the stress due to bending at sections 1, 2, 3, 4, and 5.
Objective Compute the stress due to bending at sections 1, 2, 3, 4, and 5.
Given Beam geometry in Figure 7–23; M = 30 N · m
Analysis Stress concentrations must be considered because of the several changes in geometry in the area of interest. Equation (7–6) will be used to compute the maximum stress at each section. Appendix A–18 is the source of data for stress concentration factors, Kt .
Results Section 1: This is the part of the shaft where the diameter is 40 mm and no changes in geometry occur. Therefore, the stress concentration factor is 1.0 and the stress is the same
as that which would be computed from the flexure formula alone. That is, σ = M/S.
S_{1} = \frac{ \pi (D)^{3}}{32} = \frac{ \pi (40 mm)^{3}}{32} = 6283 mm³ (Appendix A-1)
\sigma_{1} = \frac{MK_{t1}}{S_{1}} = \frac{30 N·m}{6283 mm^{3}} \times \frac{10^{3} mm}{m} = 4.77 N/mm² = 4.77 MPa
Section 2: The step in the shaft causes a stress concentration to occur.
Then, the stress is
\sigma_{2} = \frac{MK_{t2}}{S_{2}}
The smaller of the diameters at section 2 is used to compute S_{2} . From Appendix A–1,
S_{2} = \frac{ \pi (D)^{3}}{32} = \frac{ \pi (25 mm)^{3}}{32} = 1534 mm³
The value of K_{t2} depends on the ratios r/d and D/d. (See Appendix A–18–9.)
\frac{r}{d} = \frac{2 mm}{25 mm} = 0.08
\frac{D}{d} = \frac{40 mm}{25 mm} = 1.60
From Appendix A–18–9, K_{t2} = 1.87. Then,
\sigma_{2} = \frac{MK_{t}}{S_{2}} = \frac{(30 N·m)(1.87)}{1534 mm^{3}} \times \frac{10^{3} mm}{m} = 36.6 N/mm²
\sigma_{2} = 36.6 MPa
Section 3: The key seat causes a stress concentration factor of 2.0 as listed in Appendix A–18–11. S_{3} is based on the full 25 mm diameter of the shaft. Therefore, S_{3} = S_{2} = 1534 mm³. Then,
| A–18–11 Shafts with keyseats: Bending and torsion. | |
| Type of keyseat | K_{t}^{a} |
| Sled runner | 1.6 |
| Profile | 2.0 |
| ^{a} K_{t} is to be applied to the stress computed for the full nominal diameter of the shaft where the keyseat is located. | |
\sigma_{3} = \frac{MK_{t3}}{S_{3}} = \frac{(30 N·m)(2.0)}{1534 mm^{3}} \times \frac{10^{3} mm}{m} = 39.1 N/mm²
\sigma_{3} = 39.1 MPa
Section 4: The groove requires the use of Appendix A–18–9 again to find K_{t4} nominal stress is based on the root diameter of the groove, d_{g} . Note that the. For the groove,
\frac{r}{d_{g}} = \frac{1.2 mm}{20 mm} = 0.06
\frac{d}{d_{g}} = \frac{25 mm}{20 mm} = 1.25
Then, K_{t4} = 1.93. The section modulus at the root of the groove is
S_{4} = \frac{ \pi (d)_{g}^{3}}{32} = \frac{ \pi (20 mm)^{3}}{32} = 785 mm³
Now, the stress at section 4 is
\sigma_{4} = \frac{MK_{t4}}{S_{4}} = \frac{(30 N·m)(1.93)}{785 mm^{3}} \times \frac{10^{3} mm}{m} = 73.8 N/mm²
\sigma_{4} = 73.8 MPa
Section 5: This is the part of the shaft where the diameter is 25 mm and no changes in geometry occur. Therefore, the stress concentration factor is 1.0 and S_{5} = 1534 mm³ as computed for section 2.
\sigma_{5} = \frac{MK_{t5}}{S_{5}} = \frac{(30 N·m)(1.0)}{1534 mm^{3}} \times \frac{10^{3} mm}{m} = 19.6 N/mm² = 19.6 MPa
Comment Observe the large variation in stress levels that exist over this relatively small portion of the shaft. Figure 7–24 is a graph showing that variation. The stress at section 4 is by far the largest because of the small diameter at the base of the groove and the rather high stress concentration factor, K_{t4} . The design of the shaft to determine a suitable material must use this level of stress, 73.8 MPa, as the actual maximum stress.
