Problem on Design of Helical Compression Spring – Springs – Design of Machine


Hello friends here in this video we will see a problem on Design of Helical Compression spring and for that here we have a question I’ll read what is given here design a helical compression spring to carry a load of 500 Newton with a deflection of 25 mm the spring index may be taken as 8 assume following values for spring material tau that is shear stress is equal to 350 Newton per mm square capital G that is modulus of rigidity 84 kilo Newton per mm square now for these data which is given we have to design a legal compression spring from the data here I will write whatever is given here we have to design a helical compression spring to carry a load of 500 Newton so W is equal to 500 Newton with a deflection of 25 mm deflection that is 25 mm denoted by Delta the spring index may be taken as 8 so capital C is equal to 8 assume the following values for spring material tau and G values are given tau is 350 Newton per mm square and G which is modulus of rigidity 84 kilo Newton per mm square I will convert this so this becomes 84 into 10 raised to 3 Newton per mm square now for these data available let us try to design the spring so in the solution for this problem since the value of load and deflection they are given the first thing which I can calculate here is the stiffness of the spring so here I will write down stiffness of the spring is given by stiffness is denoted by capital K and that is equal to load upon deflection here I will put the values K is equal to load it is 500 deflection is 25 so here I get the stiffness as 20 Newton per mm so the spring will deflect by 1 mm when the load on the spring is 20 Newton now after getting the stiffness I’ll write down since spring index is given by spring index is capital C and that is equal to capital D upon small D where capital D it is called as mean coil diameter and small D is called as vial diameter this is the formula now from this I can say that therefore the value of capital D or small D if I am able to get any one of them I can get the other value because C spring index is given in the problem as eight so if you know any one of the values you can calculate the other values so here I’ll write down therefore capital D will be C into small D I’ll keep this as my equation number one once I know the value of small D I will put it in equation one to get the answer of capital D next now in this problem as shear stress is given tau is equal to 350 Newton per mm square so here I will calculate the shear stress factor since shear stress factor is given by case of X s is equal to 1 plus 1 upon 2 see this is the formula of shear stress factor and in this problem as the value of C is given I can find out K suffix s that is 1 plus 1 upon 2 into C that is 8 so if I calculate from this the answer of shear stress factor it comes out to be 1 point 0 6 to 5 now after getting the shear stress factor I can use the formula of resultant shear stress because this tau value which they have given in the problem it is the resultant shear stress so I will write down since resultant shear stress is given by the formula s tau is equal to 8 w d upon PI D cube multiplied by KS this is the formula of resultant shear stress now here as I can see in this formula there is one relation of capital D upon small D means capital D upon small D that is mean coil diameter upon the diameter of wire that is equal to C which is spring index so here in this formula I can simplify it in one more way that shear stress will be equal to 8w capital D upon small D is C so in the denominator there is PI and D Square will be left into KS I will write down into the brackets since C is equal to capital D upon small D now if I look into this equation shear stress is given in the problem as 350 load is their spring index and shear stress factor we have calculated just now so only unknown which is left is d square which is the mean coil which is the major diameter of Y of small D so here I will write it down as therefore shifting D square on one side I’ll shift this term because I want to calculate D so D square will be equal to 8 WC upon PI into tau multiplied by KS here I have shifted D square on left hand side tau will be here in the denominator on right hand side so now I will just go on putting the values that is d square will be equal to 8 into w is 500 see is 8 upon pi into tau it is 350 x KS shear stress factor was one point zero six to five therefore now from this if I calculate here I’ll get an answer and if I take the square root of that answer I will get the value for D and small D value it comes out to be five point five six I can round off this value I say that or this is 6mm so now you’re after getting the value of small D this will be our answer number one that is here we have designed how much should be the wire diameter and its cross section will be circular and here this is the wire diameter denoted by small D its value is 6 mm now after getting the value of small D here I’ll put this value in equation number one and I will get the value of capital D therefore put small D is equal to 6 mm in equation number one so therefore we have capital D is equal to C into D C is 8 and small D is 6 so here I get the value of capital D as 48 mm this is our second answer so here I have got the mean coil diameter next after this I use the formula deflection to get the number of turns which are there in a spring I write down since deflection is given by 8w D cube n upon GD raised to 4 this is the formula of deflection now again in this formula what we can do is that either we can put the values of capital D or small D or we can write the formula in the form of small C which is the spring index so even if I simplify this I will get it as deflection is equal to 8w capital D upon small D is spring index so spring index C cube into n upon year I weigh I am left with only G into D into bracket since capital C is equal to capital D by small D so therefore here I want to calculate n that is the number of active coils so n will be equal to deflection this GD term will go on to the left hand side and in the denominator it will be 8 WC cube so putting the values here deflection it was given in the problem as 25 G it was 84 into 10 raised to 3 small D value we have calculated it is 6 upon 8 into w 500 see it was 8 cube so from this if I calculate I will get the number of turns as 6.15 which I can round it off to 8:00 means 8:00 number of active coils are required so I will say that therefore number of active coils is equal to 8 so here after getting the values of Y diameter and milk mean coil diameter the next value is number of active coils this is our third answer now number of active coils means those coils which take part in the deflection apart from this there would be one coil at the top and another at the bottom which do not take part in the deflection so here I will find out total number of coils therefore I’ll write down assuming squared and ground ends total number of coils will be equal to number of active coils which is n plus 2 when we are using squared and ground ends then the total number of coils would be n plus 2 so therefore total number of coils I’ll denote it by n dash is equal to n it is 8 plus 2 so the total number of coils required are 10 next your eyes write down after getting the total number of coils this will be the fourth answer for us next after getting the total number of coils now I can say that since solid length of spring solid length is denoted by L suffix s and that is n dash multiplied by the diameter of wire so here we have n dash as 10 diameter of is 6 so therefore solid length becomes 60 millimeters solid length I can explain it with the help of a diagram Here I am drawing the diagram for solid lengths now this solid length indicates this is the length of the spring in completely closed condition that is when we are applying the load this spring would be compressed under the action of load since we are using or we are designing the helical compression spring so when the load is applied and then there should not be any gap between the terms of the spring or coils of the spring so when it is completely closed each of the wild diameters they are in contact then this complete length would be called as a solid length of the spring so while designing we should even know how much should be the solid length next I’ll calculate free length of spring free length of spring that is L F is equal to solid length plus maximum deflection plus 15 percent if I write it in the form of number it is 0.15 of maximum deflection so this is the formula to calculate the free length of the spring I’ll go on putting the values L F is equal to solid length is 60 maximum deflection is given in the question it is 25 mm yes it is 25 plus 15 percent of maximum deflection which is 25 and so my answer of free length it comes out to be eighty eight point seven five mm this will be our sixth answer after getting the free length of the string next I can calculate the pitch of spring so after this I’ll say that therefore now before going on to pitch of the spring here I will explain what is free length I will explain what is the meaning of free length before going on to the pitch of the spring here I am drawing the diagram for the free length of spring explaining what it is suppose we have a spring now the length of spring in the unloaded condition that is when it is not loaded then that indicates the free length of the spring so here this complete length is the free length of the spring and maximum deflection it would be we can measure it from the total this we are getting the deflection at each of the ends so that would be the maximum deflection and here is the pitch so free length indicates the length of the spring in the unloaded condition now after getting the free length of the spring the next thing which we have to calculate it is the pitch so it is called as since pitch of coil which is p is equal to the formula is free length upon total number of coils which is n – -1 from this formula we can calculate how much is the pitch so therefore pitch will be equal to free length is eighty eight point seven five upon n – -1 n – was the total number of coils and the value was 10 so 10 minus 1 so here pitch comes out to be 9 point 8 6 mm this is the seventh answer for us now after we have reached up till pitch next I will say that whatever dimensions we have found out I will denote it on a diagram here I am again drawing a helical compression spring this is the diagram for the helical compression spring here I have drawn the diagram now this indicates a miracle compression spring when we are applying load over this load is given in the problem it is 500 Newton so under the action of 500 Newton load this spring will compress and here the first thing which we have found out was the wire diameter wire diameter is a small diameter here with the help of which the spring is made it is this diameter these are the wire diameters so here in this problem we have got the wire diameter as small D is equal to 6 mm then the mean coil diameter that is the line which is passing through the centre of this wire on both the sides this was the mean coil diameter and it was 48 mm next after this we have calculated the solid length solid length was 60 mm free length of the spring that was eighty eight point seven five mm then after getting solid length and free length we had calculated the number of active coils which was in equal to eight active coils are these coils which are subjected to compression or which take part in the deflection apart from p.m. one and two these two tons of coil which do not take part in the compression because they are fixed so total number of coils in – it was ten because we were assuming that these coils they were ground they were squared and ground grinding was performed so that they can they can be seated here so in – is equal to ten so in total here we have calculated one parameter which is the wire diameter second parameter which was the mean coil diameter third parameter it was the number of turns fourth total number of turns then we have calculated solid length and at last free length and finally even pitch now what was pitch if we take any point any arbitrary point like if I take this Center then on the next wire I have to take the centre that is the same point and if I measure the distance between these two points it would be called as one pitch and the pitch value is nine point eight six mm so here we have completely designed this helical compression spring helical compression spring with all the dimensions on it and in this video we have seen how to design a helical compression spring subjected to load and there was some parameters given with the help of that we have completely designed the spring

76 thoughts on “Problem on Design of Helical Compression Spring – Springs – Design of Machine

  1. i have another question about converting pressure to load. Can you inform me about converting Bar to newtons. i am designing a compression spring but the fiven load is in Bar…Thank you in advace!

  2. Hey! I really like your video. We are a 3D CAD program company seeking for youtubers which could created content for us! If you want to know more please take a look at:

  3. Its really super , great explanations . can you please privide vedio for DESIGN OF SPRIAL SPRINGS

  4. Can you tell me something about the material and their related properties that can be used in design the helical spring?

  5. easy and simple explanation thats like a great influencer thank you sir for your eas for explaining design of helical spring

    because of you i will pass my exam easily

  6. Sir I want to known about spring test after form coil finished and we should be check chemical composition before or after.

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  8. Sir I have only 2tyoes of value that is max force ,min force and material which is to be used so is it possible to design helical compression spring pls reply me as soon as possible

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