 Hello student, I am Professor Vargage Deshmukh from Mechanical Engineering Department, Valshan Institute of Technology. In the previous lecture, we have studied the introduction of spring and the terminology related to the springs. In this session, what are the learning outcomes we need to derive the equation to calculate the stresses in the spring and to derive the load life relationship on the spring. Let us see what are the end styles of the spring for typical helical compression spring. The first one is the plain end, second is the plain and ground end, third is the square or squared end and the last one is the squared and ground end. We can see that for the first one, the ends are sharp, those are ground and it is now flat, third is the squared end. This section of the wire is perpendicular to the spring axis and third is the squared and ground. For these typical springs, first one plain, the total number of turns are given as NT, these are active turns. For plain and ground end, we need to reduce half turn from the total number of turns to get the number of active turns. For the squared ends, we need to reduce two turns from the total number of turns to get the active turns. For the squared and ground ends, we need to again reduce two number of turns from the total number of turns. The turns at the end do not affect the deflection or do not take part in the deflection. For helical extension spring, we have different types of hook ends. One is VU, second the rectangular one, third is the full hook and fourth is the extended hook. The effect of stress concentration at these ends is so severe that the spring body, this body becomes weak than the total spring. Number of active coils N is same as that of number of total coils. Let us consider spring as a state bar, this is a spring or helical compression spring. We are going to consider it as a state bar. For the helical compression spring, capital D and small D are the mean coil diameter and the wire diameter respectively. We can say that this is wire diameter and the mean coil diameter. N are the total number of turns and P is the axial force applied on the spring. It can be assumed as a state bar, a straighten bar of length pi dn. If I extend this spring, if I make it straight, it will be a bar of length equal to pi dn. And further we are going to assume that this bracket is attached at the distance d by 2 to the end of the spring and a torsional moment is applied at the end of the spring or at the end of a bar formed from the spring. You can recall and write the design equation for the shaft subjected to torsion only. Further please recall and write the equation for direct shear of a rod. You need to use the pure torsion equation first and second a pure direct shear. Now let us see what are the resultant stresses in the spring. Torsional moment due to the force d by 2, torque is equal to P into d by 2. The shear stress in the bar is given as tau 1, it is a stress due to torsional shear stress or torsional moment. We can simplify the equation and ultimately we can get tau 1 equals 8 P d upon pi d cube. Let us see in the figure how tau 1 can be represented. This is the cross section of the spring, axis of the spring and on this end we can see that how the shear stress is distributed. This is a component of torsional shear stress. Then second part is the direct shear stress. In direct shear stress tau 2 is given as P upon pi by 4 d square. We can rewrite it but the distribution for the shear stress is shown as it is same throughout as compared with the distribution maximum at the inner end and minimum at the outer end 0 at the center. If I superimpose these two stresses I can get the resultant stresses wherein at the inner position the stress is highest and at the outer end the stress is minimum. Direct stress as we have established the relation tau 2 equals 4 upon the cross section of the spring as the spring we have taken is a circular one, it is pi by 4 d square. We can simplify and see what we did. We have rewritten it in the form of some readjustments, 8 if I multiply with 0.5 it is nothing but 4. That means the effect is same. If I multiply numerator and denominator by d I can cancel that effect. Third this d and d cube in the denominator is equivalent to the d square in the denominator. That means 4 P upon pi d square is equivalent to 8 P d upon pi d cube multiplied by 0.5 d divided by capital D. The resultant shear stress if I add these two stresses torsional shear stress and direct shear stress first term is torsional shear stress second is the direct shear stress I can get the common term 8 P d upon pi d cube and in the bracket what remains is 1 plus 0.5 d by d. But we know I am going to mention it as shear stress concentration factor. Then whether you know this d by d yes we have used it as 1 by c hence I have replaced this term I have rewritten it as ks a factor what is that factor shear stress concentration factor 1 plus 0.5 divided by c. Hence the equation changes to tau equals ks into bracket 8 P d upon pi d cube. Then these two factors ks and kc ks accounts for the direct shear stress and kc it is a stress concentration due to curvature effect when we coil a rod into a spring. But these two are acting simultaneously let us introduce a factor k ks into kc. This factor is 4 c minus 1 upon 4 c minus 4 plus 0.615 by c where c is the spring index hence the resultant shear stress is rewritten as tau equals k into 8 P d upon pi d cube. This k is the stress factor or walls factor. Next equation is load deflection equation of the spring. Angle of twist theta at the end is the effect caused by the force P but our interest is delta. We know this angle theta by the torsion equation we can rewrite tau is P d by 2 L is pi dn then j and g. The equation is simplified axial deflection which is equal to theta into d by 2. We can replace the value of theta and then delta changes to delta equals 8 P d cube n upon gd raise to 4. This is what is the load deflection equation. The reference is design of machine element by VB Bhandari thank you.