Question 8.3: Ball Bearing Load Capacity A single-row ball bearing support......
Ball Bearing Load Capacity
A single-row ball bearing supports a radial load F as shown in Figure 8.11a. Calculate:
a. The maximum pressure at the contact point between the outer race and a ball.
b. The factor of safety, if the ultimate strength is the maximum usable stress.
Given: F=1.2 kN, E=200 GPa, \nu =0.3, and S_u=1900 MPa. Ball diameter is 12 mm; the radius of the groove, 6.2 mm; and the diameter of the outer race, 80 mm.
Assumptions: The basic assumptions listed in Section 8.6 apply. The loading is static.
See Figure 8.11a and Table 8.5.
For the situation described, r_1=r_1^{\prime}=0.006 m , r_2=-0.0062 m \text {, and } r_2^{\prime}=-0.04 m .
a. Substituting the given data into Equations (8.8) and (8.10), we have
m=\frac{4}{\frac{1}{r_1}+\frac{1}{r_1^{\prime}}+\frac{1}{r_2}+\frac{1}{r_2^{\prime}}} \quad n=\frac{4 E}{3\left(1-\nu^2\right)} (8.8)
A=\frac{2}{m}, \quad B=\pm \frac{1}{2}\left[\left(\frac{1}{r_1}-\frac{1}{r_1^{\prime}}\right)^2+\left(\frac{1}{r_2}-\frac{1}{r_2^{\prime}}\right)^2+2\left(\frac{1}{r_1}-\frac{1}{r_1^{\prime}}\right)\left(\frac{1}{r_2}-\frac{1}{r_2^{\prime}}\right) \cos 2 \theta\right]^{1 / 2} (8.10)
m=\frac{4}{\frac{2}{0.006}-\frac{1}{0.0062}-\frac{1}{0.04}}=0.0272, \quad n=\frac{4\left(200 \times 10^9\right)}{3(0.91)}=293.0403 \times 10^9
A=\frac{2}{0.0272}=73.5294, \quad B=\frac{1}{2}\left[(0)^2+(-136.2903)^2+2(0)^2\right]^{1 / 2}=68.1452
From Equation (8.9),
\cos \alpha=\frac{B}{A} (8.9)
\cos \alpha=\frac{68.1452}{73.5294}=0.9268, \quad \alpha=22.06^{\circ}
Corresponding to this value of a, interpolating in Table 8.5, we obtain c_a =3.5623 and c_b =0.4255. The semiaxes of the ellipsoidal contact area are found by using Equation (8.7):
a=c_a \sqrt[3]{\frac{F m}{n}} \quad b=c_b \sqrt[3]{\frac{F m}{n}} (8.7)
a=3.5623\left[\frac{1200 \times 0.0272}{293.0403 \times 10^9}\right]^{1 / 3}=1.7140 mm
b=0.4255\left[\frac{1200 \times 0.0272}{293.0403 \times 10^9}\right]^{1 / 3}=0.2047 mm
The maximum contact pressure is then
p_o=1.5 \frac{1200}{\pi(1.7140 \times 0.2047)}=1633 MPa
b. Since contact stresses are not linearly related to load F, the safety factor is defined by Equation (1.1):
n=\frac{\text { Failure load }}{\text { Allowable load }} (1.1)
n=\frac{F_u}{F} (a)
in which F_u is the ultimate loading. The maximum principal stress theory of failure gives
S_u=\frac{1.5 F_u}{\pi a b}=\frac{1.5 F_u}{\pi c_a c_b \sqrt[3]{\left(F_u m / n\right)}}
This may be written as
S_u=\frac{1.5 \sqrt[3]{F_u}}{\pi c_a c_b(m / n)^{2 / 3}} (8.11)
Introducing the numerical values into the preceding expression, we have
1900\left(10^6\right)=\frac{1.5 \sqrt[3]{F_u}}{\pi(3.5623 \times 0.4255)\left(\frac{0.0272}{293.0403 \times 10^9}\right)^{2 / 3}}
Solving, F_u =1891 N. Equation (a) gives then
n=\frac{1891}{1200}=1.58
Comments: In this example, the magnitude of the contact stress obtained is quite large in comparison with the values of the stress usually found in direct tension, bending, and torsion. In all contact problems, 3D compressive stresses occur at the point, and hence a material is capable of resisting higher stress levels.
| TABLE 8.5 Factors for Use in Equation (8.7) |
|||||
| \alpha\left({ }^{\circ}\right) | c_a | c_b | \alpha\left({ }^{\circ}\right) | c_a | c_b |
| 20 | 3.778 | 0.408 | 60 | 1.486 | 0.717 |
| 30 | 2.731 | 0.493 | 65 | 1.378 | 0.759 |
| 35 | 2.397 | 0.530 | 70 | 1.284 | 0.802 |
| 40 | 2.136 | 0.567 | 75 | 1.202 | 0.846 |
| 45 | 1.926 | 0.604 | 80 | 1.128 | 0.893 |
| 50 | 1.754 | 0.641 | 85 | 1.061 | 0.944 |
| 55 | 1.611 | 0.678 | 90 | 1.000 | 1.000 |