 Hello students, I am Bhargesh Deshmukh from Valchandri Institute of Technology in Solapur. This course is Machine Design 1 and the topic is Desire of Knuckle Joint. At the end of this session, you will be able to learn to design a knuckle joint. This knuckle joint is used to connect two rods whose axes either coincide or intersect or lie in one plane. These are used to transmit axial force. About the pin axis, there is a limited angular movement between the rods, which was not possible in the cotter joint. It is not used for connecting shafts that rotate and transmit the torque. Typical applications are tie bars or tie rods in roof trusses, links for a suspension bridge, fulcrum of the levers, and the links of a bicycle chain. There is an eye component, there is a fork and these two are connected by means of a pin and there is a locking arrangement for the pin so that it cannot move out. These knuckle pins are used to connect the two rods. This is the knuckle joint, fork, this is the eye end and the pin is inserted from the top, a collar and a split pin we use. The notations are these are the diameter of the rod, this is the diameter of the rod d, there is enlarged diameter of this one, on both the ends we can see it is d1, small d is the diameter of this pin, small d, d o is the outside diameter of the eye, this point to this point, outside diameter, a is the thickness of each eye of fork, this is a, this is also a, which is the thickness of eye of the fork, b is the thickness of eye end, eye end is in between the fork, x, you can find this distance x, this is how the arrangement is fork is on the left end, pin is at the central portion and eye end is from the right end, force p is exerted over here, reaction on the pin, p by 2, p by 2 we can divide the force which is acting on the each end of the fork, p by 2, p by 2 on the pin, then first the rod in the tensile failure, this is the rod with this cross section, circular cross section, that capital D is the diameter of the rod which is subjected to tensile force p, let us write the equation, tensile force p is given by area multiplied by the tensile stress, area is pi by 4 D square and stress is sigma t, you can think upon why the failure will not happen at the area indicated by this vertical line or it is the enlarged diameter D 1, you can think upon it, the enlarged diameter is taken as D 1 equals 1.1 times the D rod diameter, let us design the pin now, the pin is considered in shear, the half part, middle part remains in the eye end and two parts, one is the upper part and another is the lower part, these two, this one and this one, these two will remain in the fork, eye will carry away this on the right end, that means there will be two shear areas, we need to write the equation, shear force p is equal to the area under shear multiplied by the stress, force p equals pi by 4 D square is the shear area, but as there are 1, 2, two such areas we need to use two times, then stress is tau, shear stress, the standard proportion is to be used as small d equals capital D, the pin in crushing, this is the pin which is in the eye, L by D is the projected area of this pin which is subjected to this crushing force, this L by D is subjected to force p, we need to write the equation for force, projected area multiplied by the stress, it is given as L into D which is the projected area multiplied by the corresponding stress which is sigma c, L by D we need to change as B by D because the dimensions are given as B into D, stress is sigma c, this projected area of the pin in the eye is B into D, we are considering this section, this B multiplied by the diameter, hence we have changed L by D, this L and D into B into D, then pin can fill the crushing failure in the fork also, let us write the equation, the area is same for pin in the upper end of the fork and the lower end of the fork, that means there are two such areas, L into D is the area which is subjected to the compressive or the crushing stress, this P equals the crushing area L into D, it is the projected area into sigma c, let us rewrite the equation L by D, this L into D is written as two times A into D for the fork end, as there are two such areas, this area is two times A into D, first area is over here, this half part and the second one this half, therefore it is two times A into D, then the pin in bending, the pin, this pin part we have seen that P by 2 force is acting on the fork end and P is acting on the eye, this P by 2 is acting at a distance equal to z from the center, second P by 2 we need to check what is the distance, it is this distance half of the B plus some distance over here, that distance is given as x, therefore this x we can calculate as x equals one third of A, one third of A because it is a triangular distribution, for triangular distribution this x equals one third of A, then z, this z can be calculated as z equals half of half of B, how this is half of the B, this half of that is z, therefore half of B half of that is z, therefore it is one fourth of B, the bending moment M B we can calculate force multiplied by distance minus force multiplied by the distance we are calculating the net bending moment, the net bending moment is obtained by this equation and then it is simplified force multiplied by B by 4, this distance B by 4 plus A by 3 is the net bending moment, then M I about the bending axis is given as I equals pi by 64 d raised to 4, the distance from the neutral axis is given as y equals d by 2, we can write the equation for bending this is sigma B equals M B y by I, from the previous equation we can write sigma B equals this is the bending moment divided by the I value pi by 64 d raised to 4, simplifying we can get that sigma B in a simplified equation, that two such ends representing 2 I n, it is also called as 2 I n, 4 also fails in shear we need to establish the relation for it, it is the similar failure as that of I n, but only there are two such I s therefore we need to multiply the equation by 2, d o and d i are the diameter and height is A, shear force P is given as again area multiplied by the stress force P and the area this area but two times this area because there are two such I s, two times A into d o minus d multiplied by the shear stress, standard proportions are thickness A equals 0.75 times d, thickness B equals 1.25 times d and the diameter of the pin head d 1 equals 1.5 times d, thank you.