 Hello students. I am Dr. Bhagesh Deshmukh from Mechanical Engineering Department, Vulture Institute of Technology, Solapur. This session is on machine design one, design of spring. The learning outcome of this session is, at the end of this session, the student will be able to design a spring for a given load. We know that the spring can be designed for various types of loading. One is direct load, second a given range of load and third is the energy type of the problem. In this session, we'll be only concentrating on design of a spring for given load. Let's see the problem. It is required to design a helical compression spring subjected to a maximum force of 1250 Newton. Here the maximum force value is given. The deflection of the spring corresponding to the maximum force should be approximately 30 millimeter. The spring index is given as C, spring index is C6. The spring is made of patented cold-drawn steel wire for which the ultimate tensile strength and the modulus of rigidity of the spring material are 1090 and 81370 Newton per mm square respectively. The permissible shear stress for the spring wire should be taken as 50% of the SUT. Design the spring and calculate what we are asked to calculate. First is the wire diameter, second is the mean coil diameter, third the number of active coils, then the total number of coils, free length of the spring and the pitch of the coil. Let us begin with the problem. At the end we need to draw a neat sketch of the spring showing various dimensions. If we check the problem statement the data we have is the direct force is P 1250 Newton, the maximum deflection is 30 millimeter, spring index is given as 6, the SUT value is 1090 Newton per mm square, the G value is 81370 Newton per mm square and the relation between tau and SUT is given as tau equals 0.5 SUT. With the help of this data we need to design the spring. Now the first part, the wire diameter. Let us begin with the wire diameter. How to calculate this wire diameter? The first step should be calculation of permissible shear stress. As we have the relation the given data is tau equals 0.5 SUT and SUT is 1090. Get the value of the working shear stress with the help of this relation. It comes out to be 545 Newton per mm square. Now with the help of this shear stress see what we can do further. In order to calculate the diameter the further part required is the valves factor. We know that valves factor is given as K equals 4C minus 1 upon 4C minus 4 plus 0.615 divided by C where in this equation C corresponds to the spring index. If I put the value of C equals 6 I can get the valves factor equal to 0.1 0.52 from the resultant shear stress equation that we have studied. We know that tau which is the combination of this tau is the combination of tau 1 and tau 2 direct shear stress and the torsional shear stress. But this equation is least useful for us. We need to get the other form of the equation. Let us simplify the equation. What we did? Instead of D we need to replace it by C. The reason is in the first equation D and capital D both are unknown and hence if we know the value of C we can rewrite the equation in the terms of tau C and small d. In this equation if I put the value tau we are calculated as 545 then valves factor we have already calculated as 1.252. The other values load maximum force is given as 1250 Newton. The spring index is 6 and we need to know what is the value of D. Solving this equation I can get small d which is the diameter of the spring wire 6.63. We can round it up to 7 millimeter. If you have the SWG standard wire gauge you can calculate this diameter on the basis of SWG. Right now for this instance let us follow the diameter D 6.63 is approximately equal to 7 millimeter. With this data we will proceed now. Now we know that wire diameter is 7. Let us proceed. The mean coil diameter. Mean coil diameter is capital D. We know the relation of mean coil diameter and the wire diameter capital D equals C spring index multiplied by the wire diameter. We can get this value. Now the next part I need to get the number of active coils. How to get the number of active coils? The relation is of load deflection. The load and the deflection we need to put the corresponding values. The deflection given for load 1250 Newton is 30 millimeter. Other values are used. 42 is the capital D then G is mentioned as 81370 and small d we have already calculated it as 47. Solving this equation I can get the number of coils. Remember these are active coils n equals 7.91. We cannot use 7.91 as number of coils. We need to make it round. Round figure it is 8. We cannot make it 7. Now with the help of this eight coil we need to proceed. Remember these are active coils which do take part in the deflection. You can please recall the concept of total number of coils. Total number of coils we are given by some different equations. Whether active number of coils and total number of coils are equal? You can think upon this. With this concept you need to proceed. Now the total number of coils. How to proceed with total number of coils? We know that if the assumption is squared and ground end of the spring what we need to do is number of inactive coils are 2 for such kind of a spring. Hence to get total number of turns what I need to do is I need to add these two turns in the active turns. Active turns you have already calculated as 8. If I add 2 to it I will get 10 as a total number of coils. Remember this total number of coils includes active and inactive number of coils. Now I need to get the free length of the spring. How I can get the free length of spring? It is a combination of different lengths. Let us calculate these lengths. Deflection delta is given by the classical equation we have solved it. We need to put the actual value of number of turns active turns which was 7 point the fraction but we need to take it as 8. The dimension is obtained. Then solid length of the spring it is given as Nt into D. The gap between adjacent coil is 1 millimeter. We assume it as 1 millimeter. Therefore total axial gap is 10 minus 1 into 1 equals 9. If I take the free length of the spring it is equal to the deflection plus solid length plus the axial gap total axial gap. If I do it the total length of spring comes out to be 109 we approximate it to 110 millimeter. Then pitch of the coil it is free length upon Nt minus 1 which is given as 12.15 millimeter. Then the dimensions of the spring as we are asked to draw the spring and calculate the dimension let us draw the spring. All the dimensions are mentioned the wire diameter then the mean coil diameter then the inside diameter 35 outside diameter 49 then total free length and the pitch of the coil. Thank you.