Bearing Load Analysis for Combined Loads

We defined the following equations in Bearing Load Analysis and Size Selection…

• $P_0\ge X_0{F}_r+Y_0{F}_a$
• $P=X{F}_r+Y{F}_a$ These factors are obtained from empirical tables…
• For dynamic equivalent loads…
  • Identify ${\frac{F_a}{F_r}}_{}$
  • Identify $C_{0r}$ and $f_0$ and compute $\frac{f_0{F}_a}{C_{0r}}$
  • Interpolate from table closest to e (or just select the next highest e value and ratio)
  • Compare $\frac{F_a}{F_r}$ and e to find X and Y
• For static equivalent loads…
  • Identify ${\frac{F_a}{F_r}}_{}$
  • Compare $\frac{F_a}{F_r}$ and 0.8 to find $X_0$ and $Y_0$

Tables

Bearings with Angles in Pars

• Angular contact ball bearings and taper roller bearings are widely used for (large) combined radial and axial loads, they are used in pairs to support both axial directions
• Due to the “effective load center“ and the “induced axial load”, a different load analysis is required. We want to use the effective load center for moment equations instead of where the center actually is (due to the angle)

Important

At shaft $F_{ae}=0$ then bearings still have an induced axial load

Angular Contact Ball Bearings B2B

$F_{ae}$ is the external axial load and $F_{rn}$ is the external radial load at bearing n…
$F_r$ is calculated from $\Sigma F$ and $\Sigma M$

• Equivalent loads are computed based on the effective axial loads
  • Find the applied forces from load analysis using the effective load center
    • Here, we must take into account the center offset in our moment calculations to find $F_r$ (effective load center)
  • At each angular bearing location assigned induced axial forces $F_{an}$ for up to bearing n
    • These are from $F_r$ as if we don’t have axial loads, this only applies to angular ball bearings or taper ball bearings
    • $F_a=\frac{0.6F_r}{Y}$ to compute the induced axial load where $Y$ is the axial bearing factor
  • Find the sum of forces in the axial direction where $F_{ae}+F_{a2}=F_{a1}$ and assuming equality…
    • Where you can sub in the definition of $F_{a1}$ or $F_{a2}$ based on the $F_a$ equation
  • If the net force to the right is larger, bearing 1 (the left bearing) takes all the axial load, otherwise, bearing 2 (the right bearing) takes all the axial load… so…
    • if $F_{ae}+\frac{0.6}{Y_2}{F}_{r2}\ge \frac{0.6}{Y_1}{F}_{r1}$ then
      • $\{\begin{array}{l}{P}_I=X{F}_{rI}+Y_I\left(\underset{=F_{aI}}{\underbrace{F_{ae}+\frac{0.6}{Y_{II}}{F}_{rII}}}\right),\\ P_{II}=F_{rII}(\text{i.e. }{F}_{aII}=0).\end{array}$
      • Otherwise: $\{\begin{array}{l}{P}_I=F_{rI}(\text{i.e. }{F}_{aI}=0),\\ P_{II}=X{F}_{rII}+Y_{II}\left(\underset{=F_{aII}}{\underbrace{\frac{0.6}{Y_I}{F}_{rI}-F_{ae}}}\right).\end{array}$