If the load on a ball bearing is halved, its life
Increases eight times
This question explores the relationship between the load applied to a ball bearing and its operational lifespan. Understanding this relationship is crucial in mechanical design and maintenance.
The life of a ball bearing is not linear with respect to the load it carries. It is typically estimated using the following formula, known as the basic rating life ($L_{10}$):
$$ L_{10} = \left( \frac{C}{P} \right)^p $$
Where:
$L_{10}$ represents the basic rating life, often expressed in millions of revolutions.$C$ is the basic dynamic load rating of the bearing, a constant value provided by the manufacturer.$P$ is the equivalent dynamic bearing load, the actual load experienced by the bearing.$p$ is the life exponent. For ball bearings, the value of $p$ is 3.Let's analyze how the bearing's life changes when the load is altered. Assume the initial conditions are:
$P_1$$L_1$Using the formula:
$$ L_1 = k \cdot \left( \frac{C}{P_1} \right)^3 $$
(Note: We include a constant factor $k$, which represents other factors influencing life, to make the comparison clearer.)
Now, consider the scenario where the load is halved. The new load, $P_2$, becomes:
$$ P_2 = \frac{P_1}{2} $$
Let the new life be $L_2$. Applying the formula again with the new load:
$$ L_2 = k \cdot \left( \frac{C}{P_2} \right)^3 $$
Substitute $P_2 = P_1 / 2$ into the equation:
$$ L_2 = k \cdot \left( \frac{C}{P_1 / 2} \right)^3 $$
Simplify the expression:
$$ L_2 = k \cdot \left( \frac{2C}{P_1} \right)^3 $$
$$ L_2 = k \cdot \frac{2^3 C^3}{P_1^3} $$
$$ L_2 = k \cdot 8 \cdot \left( \frac{C}{P_1} \right)^3 $$
Now, compare $L_2$ with $L_1$:
Since $L_1 = k \cdot \left( \frac{C}{P_1} \right)^3$, we can substitute this back:
$$ L_2 = 8 \cdot L_1 $$
The calculation shows that when the load on a ball bearing is halved ($P$ becomes $P/2$), its life increases by a factor of $2^3 = 8$ times, assuming the basic dynamic load rating $C$ and the exponent $p=3$ remain constant.
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