 Hello students, I am Bhangesh Deshmukh from Valchin Institute of Technology, Department of Mechanical Engineering. This session is on design of a cotter joint, which is from the course of machine design 1. At the end of this session, the learner will be able to design a cotter joint. More specifically, will be able to derive the equations to design a cotter joint. Cotter joint is used to connect two coaxial rods, which are subjected to either axial tensile force or axial compressive force. It is not used to connect shafts or the components that rotate and transmit torque. It is only to transmit axial forces. Typical applications are piston rod crosshead of a steam engine, spindle and fork of the wall mechanism, piston rod and the tail or pump rod, foundation bolts. It has one socket in and another is the spigot in. We can see that there is a slot on the socket in and the spigot in. Cotter is used, spigot is put in the socket and from the top the cotter is inserted and hammered. This cotter is waist shaped piece made of a steel plate. It provides a wedge action, cotter is tapered, its one side is tapered, another side is straight. It causes wedge action in the slot and provides a fitting force. We can see that this red line corresponds to the cotter. It is also represented in the top view and similarly it is represented in the side view. Second component we can see that this is the spigot in the front view. That is also represented in the top view. It is hidden as it is inside the socket. The slot which accommodates the cotter, here we can see that slot. Next line in the side view we can see where is the spigot and one line for cotter, another line for cotter. When this starts the socket we can see that the socket is represented in the front view by the green line. Similarly its inner portion it has a slot. Socket is represented in the top view, it has a slot for cotter, its slot and the slot of the spigot shall match. Socket is represented in the side view, slot for the cotter. This is the notation P is the force, D is the diameter of rod, D1, D2, D3, D4 are the dimensions for the socket and spigot. We can see what is ABC, then T is the thickness of the cotter and L is the cotter length. Let us see the first failure, rod in tension, this is the area under tension. For this rod diameter is D and force is P, we can very easily write the equation tensile force P equals area multiplied by the stress P equals area pi by 4D square into sigma T. Next is spigot end under tension, this spigot is under tension you can see that section xx is the weak section. The rod of diameter D2 is subjected to tensile force P, this rod is of diameter D2. Let us write the equation for the failure by the classical equation force equals area pi by 4D2 square is the total area minus I need to do this vertical zone because there is no material D2T multiplied by the stress. This I need to do minus second part, I can get the total area. We can use T equals 0.31 times D which is the empirical relation. Think upon why the failure will not happen at this section, then design of socket, socket is under tension, this zone, this is section yy, we can see that outer diameter, inner diameter, outer dia, inner dia and I need to reduce this, this from the total ring area. I can write this equation outer diameter is D4, this diameter is D2 D1 force P is given as pi by 4D1 square minus D2 square D1 is outside diameter D2 is inside diameter minus D1 minus D2. I can get this equation further the quarter under shear, you can see that this upper zone, lower zone, spigot will pull it out and there will be a failure of quarter, shear failure, area multiplied by stress, force is P, this is the zone of shear. I need to use 2 times Bt, B is the width, T is the thickness of the quarter, multiplied by tau is the shear stress, with B we can obtain from this equation, spigot end in shear, for the spigot this slice will come out, I need to establish the area for this slice and need to equate it to the force, length is A, this height is D2, I need to find out the equation for the shear force P, area multiplied by stress, in this case area is given as A into D2, but there is one more area behind this zone, it is double shear to AD2, we can calculate the length A, socket we can have the same failure for the socket, however the area is different D4 multiplied by C minus this D2C, only that bracket of shear failure area will change and this red zone we need to reduce from the total area, 2 times D4 minus D2 into C, we can calculate C from this equation, spigot under crushing, there is a compressive force at this zone of width T, because quarter is continuously having compressive force on that zone, I need to establish the crushing force equation, crushing force P equals the area resisting the crushing failure, which is T into D2 multiplied by the corresponding shear crushing stress, induced compressive stress can be checked from this equation, socket under crushing, I can establish similar equation, this zone is under compression, I need to establish the area, the only change is in the area, outer diameter is D4, inner diameter is D2, D4 minus D2 into T is the area that is resisting the crushing failure multiplied by sigma C, we can here check the compressive stress which is induced in the socket, final failure is quarter in bending, where this quarter is considered as a beam, this is the central line, vertical line, force P by 2 is acting at a distance at, resisting force P by 2 is acting at a distance, this is D2, therefore from center it is D2 by 2 plus this x distance, we need to calculate these two distances, for this distribution, triangular distribution the force is acting at one-third of the distance y, therefore it is one-third of D4 minus D2 by 2, which is D4 minus D2 by 6, the bending moment is given by bending moment equals P by 2, the force P by 2, which is acting at distance x plus D2 by 2, this minus P by 2 multiplied by z, effectively we can calculate P by 2 into D2 by 4 plus D4 minus D2 by 6, this is the final equation for bending of the quarter, we need to find out how the quarter behaves in bending, the Mi about the axis of bending I is given as Tb cube by 12, Tb cube by 12, the bending stress sigma B is classically given as Mby by I or we can write the equation sigma B equals this bending moment divided by the value of I Tb cube by 12, thank you.