Hello friends here in this video we will see a problem on Helical compression spring for given load range and for that purpose here we have a question I will read what is given here a helical spring is to be designed for an operating load range of approximately 90 Newton to 135 Newton the deflection of the spring for the load range is 7.5 mm assume our spring index of 10 permissible shear stress for spring material is 480 mega Pascal modulus of rigidity 80 kilo Newton per mm square here we have to design the spring so this is the question what we have now whatever is given in this question I will write that in the form of data now here as we look into this problem it has given that a helical spring is to be designed for an operating load range here some load range is given means not a single value of load but there would be one is n1 lowest value so approximately 90 Newton to 135 Newton this is the load range so I will write down 90 Newton as w1 and 135 Newton is w2 now after reading w1 and w2 what is the meaning of this load range is that here it means that whenever there is a load of 90 Newton then above 90 Newton the spring will start deflecting before 90 you turn it won’t deflect because this is the minimum load which is required on the spring and maximum load is 135 Newton so the spring will be deflecting only between 90 and 135 Newton so here we can find out the total load therefore W will be the net load which is W 2 minus W 1 so it is 135 minus 90 so this w comes out to be 45 Newton so now this is very very important this 45 Newton is the value we have to use for designing such a spring this 45 Newton is above 90 means once we reach the value of 90 then from that we have to start and we have to reach up to 135 Newton so the difference between them is 45 Newton and hence this spring would be designed for 45 Newton next the deflection of the spring for the load range is 7.5 mm deflection is denoted by Delta assume a spring index of 10 spring index is denoted by C also called as spring capacity and that is equal to capital D by small D then permissible shear stress for spring material that is tau it is 480 tau is 480 mega Pascal mega Pascal means it is in terms of Newton per mm square modulus of rigidity is given 80 kilo Newton per mm square so G is equal to 80 kilo Newton per mm square I will convert this it becomes 80 into 10 raised to 3 Newton per mm square and here the question is design the spring so for the delta given here we have to design the spring this is the question for us now let us try to get the solution that is the design for this problem solution since load and deflection values are known so I can get the stiffness of the spring so stiffness of spring k is equal to load upon deflection so therefore k will be equal to load is 45 deflection is given in the problem as 7.5 therefore K value it comes out to be 6 Newton per mm this is the stiffness of the spring it means to deflect the spring by 1 mm we require a minimum load of 6 Newton next after getting the stiffness since in this problem shear stress tau is given this tau is called as a resultant shear stress and I can say that since shear stress Factor case of X is that will be equal to 1 plus 1 upon 2 C so therefore case of X s that will be equal to 1 plus 1 upon 2 C and C is given in the problem as 10 so from this I will get the value of shear stress factor and the answer is one point zero five after getting the shear stress factor I can use the formula of resultant shear stress here I will write down since resultant shear stress is given by tau is equal to it was 8 WD upon PI D cube multiplied by KS that is the shear stress factor now if we look into this shear stress factor then shear stress is known the value of load is known KS we have calculated so capital D and small D both are unknown but here we have only one equation and there are two unknowns so we cannot get the values so what we can do we can use a relation that therefore tau will be equal to 8w capital D upon small D that is called as spring index denoted by C so I will write down C upon pi into small d square because one small D I have used so n into K s this that is the shear stress factor I will write down since spring index is equal to capital D upon small D so therefore here I’ll shift d square on one side so D Square will be equal to 8 WC upon PI into tau will be in the denominator multiplied by K suffix s here I will go on putting the values 8 into load is 45 C value it was 10 PI into shear stress is given in the problem tau it is 480 Newton per mm square 480 x KS KS value it was one point zero five so from this first I will get the value of D square and then when I take the square root my answer of small D it comes out to be one point five eight mm so here the wire diameter is one point five eight but we need to increase the strength so what I will do I will increase this value and suppose I take the value as 4mm because if I increase the diameter area increases and stress value will go on decreasing so here I take the wire diameter is 4 mm and this will be the first answer for us now after getting small D that is the wire diameter is 4 mm wire diameter means it time a circular wire of 4 mm daya will be used to make this helical spring now after getting small D I’ll see that since spring index C is equal to capital D upon small D so therefore capital D will be equal to C into small D C is given as 10 small D we have calculated it is 4 so therefore capital D value comes out to be 40mm this will be the second answer now as we are designing the spring we have calculated the wire diameter and we have calculated the main coil diameter after getting this the next step is we will design the number of coils in the spring so I will say that number of coils we can get it from the formula of deflection so I will say that there for deflection of spring is given by deflection is equal to 8w C cube n upon Gd this is the formula from which we can get the number of turns in this formula small n indicates the number of active coins so that we can found it out here I will write down therefore small n will be equal to this is deflection GD will get multiplied in the denominator here I have 8w into C cube so next I will go on putting the values deflection it is given in the question deflection was 7.5 mm so here 7.5 next capital G its value was also given capital G was 80 into 10 raised to 3 Newton per mm square so 80 into 10 raise to 3 small D is 4 we have calculated in the denominator we have 8 into 45 because that is the load for which we are designing the spring into C cube C is 10 so 10 cube so if I calculate this from this I will get the answer of n and it comes out to be six point six seven or we can round it up to seven so therefore I see that number of active coils is equal to seven now these are the active coils they are they take part in the deflection they take part when the load is applied now from this I can even calculate the two number of coils so I will say that assuming square and ground ends squared and ground n means since the spring is in the form of a wire it is circular so if we keep that spring on any surface there are chances of the spring to slip because the surface is circular so we are grinding that ends to make it in the form of a square so that it can be seated flat so therefore assuming number of square and ground ends total number of coils will be equal to number of active coils which is n plus 2 so therefore total number of coils becomes n – L denoted by n – n is 7 plus 2 so here I am getting the total number of coils as 9 and this will be the third answer now after getting the total number of coils next we can find the solid length of the spring so I will say that therefore solid length of spring else of X s that will be equal to n – into D which is the wire diameter so therefore solid length is equal to n – it is 9 D is 4 so therefore solid length comes out to be 36 mm after getting solid lengths I will explain what is the meaning of this suppose we have two supports Here I am explaining what is the meaning of solid length now here I have drawn the diagram and when the spring is closed that is after the application of load when this spring it compresses completely that is each of the wire diameter will coincide here I can denote the wire diameters so these are the wire diameters if I can show the cut section of the spring and Here I am drawing the section so now when we are compressing the spring in such a way that there is no gap in between and each of the wire diameters are in contact so that would be called as solid length and this solid length value we have calculated next after finding the solid length I will say that there for free length of spring this solid length we have found out solid length value was 36 mm that is in the closed condition when the spring is completely closed its length will be 36 mm then for free lengths the formula is LF is equal to solid length plus maximum deflection plus 15% if I write it in the number format is 0.15 of maximum deflection so this is the formula of free length here therefore L F will be equal to solid length is 36 maximum deflection is given it is seven point five fifteen percent 0.15 into seven point five of maximum deflection so there for free length value it comes out to be forty four point six three mm this will be the fifth answer for us now once we have calculated free length I will explain what is the meaning of relent for that suppose we have a diagram here Here I am explaining what is the meaning of free length of spring now to explain free length here we have a diagram this is called as a free length of the spring it is the length of the spring in the unloaded condition this is the free length of the spring when it is not loaded in the unloaded condition and that is solid length plus maximum deflection plus 15 percent of maximum deflection the value we have got it as forty four point six three mm now after getting this free length of the spring free length as I have told LF indicates the length of spring in unloaded condition so here we have got free length now after grading free length next I can calculate the pitch and what is pitch if I take a point on a spring that is on a single turn if I take a point then on the next turn I have to take the point at the same location and if I measure this linear distance this linear distance would be called as the pitch so here we even want to know how much should be the pitch for this helical spring so here I will say that therefore pitch of spring is given by P is equal to LF upon n – -1 so therefore which will be equal to free length we have got just now it was forty four point six three upon n – -1 the total number of turns in this problem we have calculated it was nine so nine minus one from this I will get the answer of pitch the value is five point five eight mmm so after the free length the next thing which we have calculated it is the pitch free length it was our answer number fifth which will be the sixth answer now once we have found out the pitch here I can denote all the values of the dimensions on the spring so here again I will be drawing a spring I will explain this diagram here again I am drawing a helical spring here I have drawn the diagram of a helical compression spring now in this helical compression spring this section is the wild diameter I will mark on the dimensions first of all the wire diameter we have calculated this diameter of wire it was 4mm then we had even calculated that the mean coil diameter it is this distance you can say the mean diameter outer diameter and then there would be coal diameter if we take the mean this becomes the mean coil diameter and the value in this problem it was 40 mm then after that we have calculated the solid length solid length value was 36 mm then we had calculated how much was the stiffness of the spring and that stiffness value was stiffness was 6 Newton per mm then free length of the spring three length was forty four point six three mm after free length we had even calculated the total number of coils and total number of active coils so I will write down therefore total number of coils and active coils active coils were denoted by n value was seven and total number of coils were denoted by n – value was nine so these parameters we have calculated in this problem and even pitch which distance this is 5.5 8mm this which we have calculated and for a helical compression spring this pitch value will remain same throughout so your the diagram is of a helical compression spring as we look into the problem there was load range given because this problem was based on the load range it was 90 Newton to 135 Newton and as I have explained in this problem when the spring would be loaded once the load is up to 90 Newton it is 90 then only this spring will deflect otherwise spring won’t deflect so the minimum load is 90 under that it will deflect and this value of load will go on increasing up to 135 so the load range we have taken from 90 to 135 the difference was 45 Newton and we have designed this spring considering for divided Newton above 90 Newton so for that purpose whatever the unknown values were there which are necessary for designing the spring that we have found out starting with the diameter of wire mean coil diameter then we have calculated stiffness after that we have calculated the solid length next there was number of active coils number total number of coils because this upper and below two coils do not take part in the deflection then we have calculated free lengths and finally we have calculated the pitch and with this we complete this design

sir plz release a video on lewis equation for static beam strength for spur gear teeth

sir

at 10.05

you got "d" value as 1.58 but to make it as a whole no. you should approximate it as 2 mm but you gave the reason for increasing the strength we will take 4mm, can you justify your answer in detail?

bcoz if to increase strength i can also take "d" as 8mm or 16 etc.

45 wala logic kuch smgh ni aaya bdhia se

Why every time u take shear stress factor not wahl correction factor

I think u r absolutely wrong. ….in many books like bhandari they have taken W =(W1+W2)/2………..not ….W1-W2? U CAN check

Give my answers. …

Sir why are we taking the dia of the wire 4mm which is more than squared value of the term.

Sir,can you explain Semi elliptical leaf spring? will be really helpful ,thankyou

Sir please upload jntuk kakinada syllabus on syllabus book please sir I would like to prefer all my friends of this ekaada adress please up load sir please

Sir open coil helical spring… Ki video uplaod nhi hai kya???

Sir plzz tell me your Facebook profile name actually m b. Tech krta hu but ek ree appear aagyi thi or vo abhi tk clear nhi hoparhi plzz btado kse tyari krni h ek pattern btaado padhne ka m krunga plzz sir

Sir please tell why we have assume the spring index between 6-12

Show using data book

Why do we increase wire diameter for solving our numerical

In resultant shear stress calculation part you should take 135N instead of 45N !

K Mahadevan ,K Balveer Reddy Data handbook

Sir which data hand book you have ise

Is there any spring designer, i have some project for design with payable , contact ASAP

09646549568, [email protected]

Aoa sir plz can you explain how you rounnd off 1.58 to 4 incase of spring dia

when ever load range is given, the it is wrong to use the method which is used in this video, the correct method is to calculate Twisting moment and find the dia of spring wire.

i.e we have to use these formulae.

T=W2 x (D/2)

T=(pie/16) x d-cube x toe

Sir for maximum deflection you can't take 7.5 . Because in question it have been mention that 7. 5 is for load range so for maximum deflection you have to re- calculate by using same formulae and then n= 7 for finding free length and load will be 135

You should actually justify for considering wire Dia as 4 instead of 2mm.

Kw={(4c-1)/(4c-4)+(0.651/c)}

Where the use in this formula

In helical comp.spring of circular wire

Sir 1.58 aaya h but 4 kaise kr diya

Sir why u take d =4mm I'm not understand this logic

sir is it necessary to denote all the value with diagram ..

can we list them ?

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1.75×

Sir in shear stress , Wmax is to be taken for calculation . Load range is used for finding deflection.