 Hello students, I am Bhargesh Deshmukh from Mechanical Engineering Department, Valchin Institute of Technology, Sulapur. This session is on design of bevel gear part 2. At the end of this session, the learner will be able to derive the equations of forces on the bevel gear. Then what are the different forces on the bevel gear? Let us consider this as schematic representation of a bevel gear. It is a cone, the center line, the line of rotation or the axis of rotation. It is mentioned, it is counterclockwise looking from this end. This is a pitch plane or the plane of rotation. At this point I have drawn a tangent and particular pitch point is considered over here. You can think upon why we have considered the pitch point. This is the plane in which we will find the resultant force which is indicated as Pn or P. Pn is always inclined at an angle alpha which is equal to the pressure angle. This is the box where we need to analyze the forces. All the lines, this line must be parallel to the axis of rotation. One line must be perpendicular to the axis of rotation. And third line, it must be also perpendicular to the axis at the pitch point. I can resolve this force and get the first force which is the tangential force Pt. Pt must be opposite to the direction of rotation as per the convention. Then the other force, if we resolve this force into two, one is horizontal, another is vertical component. This component, vertical obviously in the blue plane. This separating force Ps, I need to check this force. This is neither parallel to vertical, this horizontal or the third component and hence I need to resolve this force also into possible two components. This is inclined at an angle equal to gamma. I need to consider this frame of reference green. First force is the axial force as it is parallel to the axis of the shaft. Second force is the radial force which is always directed towards the center of the gear. At the pitch point, we can show all the forces. Assumption in this analysis is, the force between the meshing tooth is concentrated at the midpoint of the paste width of the tooth. Pt is the tangential or useful component, Pr is the radial component and Pa is the axial or the thrust component. This is the analysis gear rotation, point O is the apex, point D is the pitch point line OD. Direction of rotation is counter clockwise. There is a triangle on which we are assuming that all the forces are acting at this particular point which is a pitch point. AD is the force which is perpendicular to this line OD. A red zone this, it is the plane where force P or the resultant force is acting. This is the resultant force. We may call it as Pn, same view. Now this is the plane under consideration. I need to resolve the forces, same plane. But I need to show it in the 2D view. Let us show it in the 2D view. BD represents the force P. I need to resolve this force. The pressure angle is alpha, PS is the separating force and Pt is the tangential force. In this triangle BCd, BCd, then alpha is BC by CD or it is PS upon Pt and therefore I can get the equation as PS equals Pt tan alpha. PS is perpendicular to the pitch line OD. PS, this is the separating force. PS acting along AD, it is perpendicular to OD. Now let us consider the plane which is shown by red color. I can see that this is the plane. Let us consider its 2D view. Here I can say that force PS is acting along this AD. This angle is gamma. This force I need to resolve into two components. One is radial which is directed towards the center and PA is the axial force which is along the axis of the shaft of the gear. The angle between lines OD and OX, line OD and line OX is gamma, this angle. Therefore the angle between lines AD and FD, AD and FD is equal to the pitch angle gamma. You can think upon why this angle equals gamma. I can establish the relations. This is PS. Its adjacent component is cos component which is radial force. Therefore PR equals Pt tan alpha cos gamma. It is the cos component of the force. The sine component will be equal to PA, this force. And hence PA equals Pt tan alpha sine gamma. We know that Pt equals MT upon RM, where RM is the mean radius. RT is the torque transmitted by the gears. We need to now check what is the mean radius. RM equals DP by 2. What is DP by 2? It is the pitch diameter. This DP by 2 is the radius minus B sine gamma as this is inclined. I need to find out this particular point and hence I need to reduce the distance, the radial distance. I need to find out this difference. Therefore RM equals DP by 2 minus B sine gamma by 2, where B is the face width of the tooth in millimeter. Now the Bevel gear and pinion forces. The radial component of the gear is equal to the axial component on the pinion. This is very important. Radial component of the gear is equal to the axial component on the pinion. The axial component on the gear is also equal to the radial component PR on the pinion. Let us see the 2D views, pinion and the gear. It is the mean radius. Mean radius is also over here. Angle gamma corresponding angle will be over here. This is the pitch point shown on the pinion. This is the pitch point shown on the gear. There is no issue about the tangential force, but the force exists PS, it is a separating force. Why it is called as separating? Here we can understand that this force tries to separate the pinion and the gear and we need to prevent it. Therefore the mounting methods of the Bevel gears have come in picture. However right now we are going to consider only the resolution of forces. Now this PS we have considered it is a separating force and hence we need to analyze this force separately PS. If I check over here the forces how these are shown PR and PA are the outcomes of PS, which are the forces first one is PN or P which is the normal force or the resultant force acting on the tooth. It has to be resolved into 2 components which we have resolved as PS and PT. But for our understanding we need to find out axial force and radial force and hence we have established PA and PR. Axial force on the pinion is parallel to the axis of the pinion. Radial force on the pinion is perpendicular to the axis of the pinion. Correspondingly if I consider this gear the radial force on the gear is acting perpendicular to the axis of the gear and axial force on the gear is acting parallel to the axis of the gear. As per these 2 statements first one and second one PR on pinion the radial force on the pinion I am talking about this force is equal to the axial force on the gear these need to keep each other in equilibrium. The other force axial force on the pinion I am talking about this force and radial force on the gear this force these 2 are equal and keeps each other in equilibrium these are mutually opposite in direction PT of pinion is equal to PT of gear.