 Hello, I am Sanju B. Knight, Assistant Professor in Mechanical Engineering Department at Vulture Institute of Technology, Swalapur. In this video, I am explaining static load capacity of rolling contact bearings. At the end of the session, learners will be able to derive the equation for static load carrying capacity of ball bearings. Static load is the load acting on stationary shaft. So whenever the shaft is stationary, whatever load is acting on the shaft is defined as static load. So if you see that this is a static load, when it is acting on the shaft, it is acting on the bearing on the balls and the races. So these are the balls which support this static load like P1, P2, P3, etc. So this load for the ball is acting on the races and as the load is acting between the contact point, we find there is a deformation of the ball and the races. So as this load increases, the values of P1, P2 also increase and that is why the deformation at particular point of contact will also increase. So if this deformation goes beyond certain allowable limit, then the performance of the bearing is lost. And then bearing may start giving vibrations and noise. And that is why for all practical purposes, this deformation should have some allowable limit beyond which it should not be allowed. And by past experience, it has been found that the maximum allowed limit for this deformation is 0.0001 of the ball diameter. That means whatever the maximum load is acting at particular ball and race point of contact, there the maximum deformation is expected and that maximum deformation should have our limited value equal to 0.001 of the ball diameter. And that is why the static load carrying capacity of the bearing is defined as the static load which corresponds to a total permanent deformation of balls and races at the most heavily stressed point of contact equal to 0.0001 of the ball diameter. Strybek has provided analysis for estimation of the static load. He has derived an equation by which we can calculate the static load acting on the bearing or static load which can be supported by the bearing. He has made certain assumptions in his analysis that the races are rigid and retain their circular shape under applied load. So it has been sure that whenever the load is acting between point of contact, deformation is there. However, he made the assumption that the races will maintain their circular shape and they do not consider the change in the radius of curvature. So that is a rigid races will remain as it is. The second assumption he has made is the balls are equally spaced. That these balls are arranged in a bearing having the same angle between the adjacent balls and that's why they are positioned at equal angle and that's why they are equally spaced. The third assumption he has made is the balls in the upper half do not support the load. As the load is acting downward on the bearing, the balls in the upper half will not contribute this load. Only the balls which are arranged in the lower half will contribute the load and that's why they have been subjected to forces P1, P2, P3, etc. and all these balls will be only considered on the bottom. Now look to this figure that as the load is acting downward, the load acting on the balls will be shown in this direction P1, P2, P3, etc. So just let us recall the resolution of forces because P1 is acting vertically up whereas P2, P3 they are acting as the inclined forces. So this is a vertical force so P1 is acting exactly as a vertical force. This is the inclined force that's why we know that we have to resort into two components. So P2, P3, etc. are there which are acting inclined so vertical components must be vertical and horizontal components we have to calculate for inclined forces. Look this figure P2 is acting along this line even P3 is also acting along this line. Now pause and think that can we take full force P2 to support CO that is if I want to support CO can I consider P2 as a force to support entire value. Let us consider the equilibrium of forces in vertical direction because this vertical force has to be supported by all these balls in opposite direction that is the vertical direction. So equilibrium equation will be C0 equal to P1 because already it is in vertical direction whereas P2 is acting along inclined line it is to be resolved into component vertical and horizontal and here vertical component of P2 will support C0. So that vertical component of P2 is P2 cos of beta so P2 cos of beta because beta is angle between this vertical plane and line so cos beta component so P2 cos beta P2 cos beta of this P2 and P2 cos beta of this ball it is 2 P2 cos of beta. Similarly P3 is making angle 2 beta with vertical line and that is why its component is cos 2 beta. So P3 cos 2 beta is vertical force here also vertical force that is why it is 2 P3 cos of 2 beta. So how many balls are arranged at the lower half they will contribute the components. Now because of these loads P1, P2 etc there is a deformation at the point of contact between the ball this this etc so it is once again delta 1 is the proportionate deformation according to P1 delta 2 is proportionate deformation according to P2 and once again being the same angle between them we get delta 2 by delta 1 is equal to cos of beta. Now it has been found that the deformation because of the load at the point of contact is proportionate as delta is proportionate to load raise to 2 by 3 this has been found by Hertz context so according to Hertz he found that the deformation is proportional to load into 2 by raise to 2 by 3 so therefore delta 1 is C1 P1 raise to 2 by 3 and delta 2 is C1 P2 raise to 2 by 3 and delta 2 by delta 1 will give this P2 upon P1 raise to 2 by 3. So we know that delta 2 by delta 1 is also cos of beta and therefore we get P2 by P1 raise to 2 by 3 is cos of beta and hence P2 is equal to P1 cos of beta raise to 3 by 2. So if I use this equation I can found that I can find that P2 is equal to P1 cos of beta 3 by 2 similarly P3 is equal to P1 cos of 2 beta raise to 3 by 2 so we get component P2 P3 in terms of P1 and angle so substituting in equation of equilibrium we find that C0 is P1 it was 2 P2 cos beta so 2 P2 is this value P1 cos beta raise to 3 by 2 into cos beta it was 2 P2 cos 2 beta so it is 2 P1 cos 2 beta raise to 3 by 2 into cos of 2 beta and so on. So we just replace P2 P3 value in terms of P1 and we get this equation so we get here P1 common 1 plus 2 cos of beta raise to 5 by 2 because it is 3 by 2 and this cos beta it makes 5 by 2 plus 5 by 2 2 cos 2 beta 5 by 2 and so on. So being a beta constant for given number of balls arranged in 360 degree the angle between 2 balls will be constant and that is why this bracket will be constant substituted as m so m is substituted as this bracket and now if I consider z is the number of the balls then beta is found out by 365 number of balls so normally in a bearing we find that the balls numbers are 8, 10, 12, 15 something for which we get particular value of m and if I get z by m ratio we find it is almost constant 4.35, 4.36 etc so it is almost constant. So this is considered by Strybek as 5 so m is z by 5 so we know that C0 is P1 m so m is replaced by z by 5 so I get static load capacity C0 is equal to P1 z by 5. So we know that by experiential evidence that whatever maximum deformation is limited to 0.0001 times the diameter of the ball we find that the maximum load corresponding to that deformation is Kd square P1 is equal to Kd square d is a ball diameter K is a factor depending upon the material and that is why if I substitute C0 is equal to Kd square by z so finally you found that the static load capacity C0 can be calculated by bearing construction having the material constant and d is the diameter of ball and z is the number of ball. So this is known as Strybek equation to estimate the static load capacity of bearing. So these are my reference thank you.