Physical characteristics of good bearing material- compressive strength, fatigue strength, embeddability, bondability, corrosion resistant, thermal conductivity, thermal expansion, conformability
(i) definition of (1) Basic static load rating-static radial load or axial load which corresponds to a total permanent deformation of the ball and race,at the most heavily stressed contact,equal to 0.001times the ball diameter. (2) basic dynamic load rating- the constant stationary radial load or a constant axial load which a group of of apparently bearings with stationary outer ring can endure for a rating life of one million revolutions with only 10% failure.
The following steps must be adopted in selecting the bearing from the manufacturer’s catalogue: 1. Calculate the radial and axial load reaction (Fa and Fr) acting on the bearing. 2. Decide the diameter of the shaft on which the bearing is to be mounted. 3. Select the proper size of bearing suitable for given application, specified with speed and available space. 4. Find the basic static rating Co of the selected bearing from the catalogue. 5. Calculate the ratio (Fa / VFr) and (Fa / Co). 6. Find the value of x and y i. e. radial and thrust factor from the catalogue.
(i) Industrial and automotive gear boxes. (ii) Electric motors and machine tool spindles. (iii) Small size centrifugal pumps. (iv) Automobile front and rear axles
The life of an individual bearing is defined as the total number of revolutions (or the number of hours at a given constant speed) which the bearing can complete before the evidence of fatigue failure develops on the balls or races. The bearing life can be defined by rating life. The rating life of a group of apparently identical bearing is defined as the number of revolutions (or the number of hours at a given constant speed) that 90 percent of a group of bearings will complete or exceed before the first evidence of fatigue failure develops. It is also known as L10 life.
Problem on Foot step bearing D = 225 mm = 0.225 m W = 7500 N µ = 0.09 N = 600 rpm ω = 2 π N / 60 =62.83 rad/sec Uniform pressure condition Frictional torque T = 2/3 µ W R = 50.625 Nm Power lost in friction = T x ω = 50.625 x 62.83 = 3180.8 W --------Ans Uniform wear condition Frictional torque T = 1/2 µ W R = 37.98 Nm Power lost in friction = T x ω = 37.98 x 62.83 = 2385.57 W -------- Ans
b) Function of the Clutch 1. Function of transmitting the torque from the engine to the drive train. 2. Smoothly deliver the power from the engine to enable smooth vehicle movement. 3. Perform quietly and to reduce drive-related vibration. WORKING PRINCIPLE OF CLUTCH It operates on the principle of friction. When two surfaces are brought in contact and are held against each other due to friction between them, they can be used to transmit power. If one is rotated, then other also rotates. One surface is connected to engine and other to the transmission system of automobile.
