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from Ekeeda Hello friends welcome back to the subject of machine design 1 we

are right now learning about design of various Springs in the last lecture we

have seen the introduction regarding helical Springs and their lip Springs

today we are going to look at the design aspects of helical Springs and then we

will continue with the numerical based on the design of helical Springs so let

us begin so students we need to understand the

design aspects of helical spinning because it is a complex structure like

we have mentioned this in the previous lecture also now why it is compressed or

why it is the structure is complex if you see this properly a spring if you

can imagine right now is a structure made up of different wires or one wire

which has got a coil structure but that’s not a simple coil again it got

certain angle also and that’s called helix angle so the complexity are mainly

in nature if you see this properly not only a bar or we can say if this is

the bar which is the wire we have to paint it in coil manner so and that’s

called spiral of course the second thing is it will have certain angle at which

it is a spiral so that’s why it is a helix angle so the basic structure which

was a simple bar is now converted into a complex structure where the nature of

forces acting will vary and since the structure is complex we have to analyze

it using maximum possible aspects so let us look at them basically there are two

aspects which are used and which have proven very strong for the design of

Springs the first aspect is called load stress equation and second is called

load deflection equation so based on these two aspects we can design this

spring called the helical spring you see this load stress equation is based on

the modified bar like I said if there is a spring given of certain number of coils let me unwind

these coils let me make them straight so this straight things will take a form of

bar of certain length now this bar and this spring will have only one thing

that is in common is the cross-section small diameter D whereas the total

length of this modified bar will be equal to the circumference of one coil

into number of coils so in that case this is how the simplified bar is or the

modified bar is now this particular spring is now converted into bar and

then it is analyzed for various stresses let us look at them the possible

stresses that are possible or the possible stresses that are causing the

failure that may cause the failure in the spring are pure torsional stresses

then direct shear stresses then comes the combined torsional stresses where

the torsion and other direct stresses will be combined and the last one is the

curvature shear stress curvature is nothing but this curved portion which

may under water which may experience a shear so these are the possible four

types of stresses that will induce in a spring which is actually a simple member

and that’s why we need to analyze using all of them

now since the derivation is not a part of our syllabus we are going to look at

it very quickly the stresses are combined in the form of this particular

formula where shear stress is prominent one if you see torsion shear and torsion

and that’s why everywhere this stress that is occurring is the shear stress

and that’s why the shear strength of this particular member will be given by

this particular formula where there are certain terminologies we need to

understand this is the constant of equality let’s call it a four-time being

into eight P is the maximum force acting on the spring it can be a compressive

force or it can be tensile force D is the main diameter of the

spring and small D is the diameter of wire so using this particular formula

where all these forms of stresses are combined we can do the design of the

spring where K is a constant of proportionality and it is called Vols

factor the vols factor is given by this empirical relation you can Bahat it or

you will find this formula in the spring section of your PhD data book so this is

the formula now if you retail it from the previous chapter see is nothing but

this pain index and that is nothing but the ratio of the mean diameter capital D

to the wire diameter small D so using this empirical relation and this

particular empirical relation we can always find out the maximum stress that

is induced in the body or if the material is given based on the maximum

allowable stress we can always find out the wire diameter let’s move ahead with

the second approach yeah basically this equation is called load stress equation

where load and the possible stress are equated in the form of or they are put

into one format which is called an empirical relation let us go to the load

deflection equation now if you see properly again since it’s a spring it

will have certain angular deflection now angular deflection is given by this

particular formula which is called a torsional formula but now in that case

the maximum deflection which is going to be the linear one if we have to relate

them they found out force very small angle theta the formula is this of

course is not a part of the syllabus we are not going into the derivation of

this particular relation but we can call it an empirical relation where the

linear deflection is connected with the angular deflection theta or twist angle

you can say by means of mean diameter divided by two now in such case

when this particular formula was modified in terms of the torsion it was

found that Delta is equal to this particular relation where P is again the

force acting is the nominal diameter or the mean diameter and this n or we can

use small in in this case which is nothing but the number of coils of this

thing given G is the torsional rigidity or constant of rigidity and D is again

the wire diameter now there was another relation found was a small key factor

now this K and this K are different if you see it’s a capital K it’s a small K

so small K is given by the ratio of force to the deflection which is nothing

but the spring constant it gives the unit Newton per millimeter so it is a

spring constant so spring constant can be modified after using these two terms

P and Delta we will get the formula for spring factor is this one so this is the

this is called district corrections are called load deflection equation whereas

this particular equation is for load stress equation now further analysis of

this particular spring will be based on these two equations the students before

we move to the actual design procedure we need to understanding material that

can be used for Springs now there are certain properties which are expected

from the materials or the springs that are going to undergo certain environment

for example the metals that are chosen for the springs based on the loads

acting the lower can be compressive it can be tensile or it can be torsion in

nature and accordingly spring metal will be defined and design for example for

the compressive strength or the compressive kind of spring the material

use will be different than the material used for the tensile kind of spring or

the application so load can be compressive or can be tensile or can be

torsional all so the next thing is the range of

stresses no students like me have seen in the load stress activation we have

considered four different kind of stresses which are possible which may

feel this spring so we always have to consider the range of or the ranges of

this Springs or the stresses which may occur if there are only two stressors

possible we have to design and select the material only for those two stresses

if you are doing it for all of the stresses in that case the design will be

over designed so according to the stresses or the range of stress we need

to consider the materials then size constraints you know students spring is

a small member that is generally fitted in multiple sections of a machine so in

that case it always have a constraint related with it dimension may be length

may be diameter may be the wire diameter available so with respect to all these

constraints we have to select a proper material for example if there is a heavy

application where the space for spring is very small in that case we have to

use a strength material or very strong material for the small application also

the next is fighting like expected the springs are mainly use or many times

they use in certain applications where cyclic loads come into picture also

their dynamic in nature due to this variability the spring member or other

members but we are concerned about the spring spring may undergo the fatigue

failure so there is some life expectancy and or the life expected by the

manufacturer for that particular machine in such case depends depending upon the

life expected we should use a proper material which will have a more or which

will have a better endurance life or endurance limit the next important thing

is the possibilities of temperature and corrosion no springs are use in certain

environment where corrosion or the liquid may come into contact with this

particular spring in such cases we must consider the

of temperature as well as the corrosion also because accordingly we will have to

select the material so these are some of the important aspects that we must

consider there comes the different types of the materials I am NOT talking about

the names of materials different types of the materials where the metal can be

a cold drawn Springs the cold drawn Springs are generally used where the

orientation of crystals are changed so as to suit the requirement of the given

conditions then oil hardened steel wires now we know that the springs are made up

of wires the steel wires which may undergo oil hardening so as to suit the

given environment there come these stainless steel springs many a kind

applications where corrosion is possible we may have to use stainless steel

springs and there comes another format where alloy steels or allowed wires are

used for making this fins for example if there is an application where the

conventional material like spring is like steel is not possible in that is we

may have to go for a higher end where the combined properties of different

metals can be used which are further allow it vias

allured wires can also be used for making the springs now let us look at

some of the most famous metals which are used now the names of materials which I

use for making the Springs the very first one and which is very prominently

used is called high speed or high carbon steel where the carbon percentage is

quite high now we know that Springs are used for absorbing shocks so castings

are the material where the carbon content is more they have got good

capacity or capability of absorbing the shocks and that’s why we may go for the

high carbon steel the next one is the chromium vanadium steel chromium

vanadium steel is a kind of a Lloyds wire which can be used for making the

springs for special applications then comes the chromium silicon steel it also

is an alloy steel where or which can be used for specific

applications then stainless steel for corrosion material or corrosion

environment we can use and there are few more materials like Monell then

phosphorus bronze then bronze silicon bronze etc etc which are again alloys or

non ferrous materials which can be used for making Springs for given certain

applications specified applications so these are what the important matters is

that we can use for the springs now let us move at D main power part that is

called a design aspect like if you have discussed this that the load stress

equation is a prominent one where is the design of spring is concerned let us

look at it torsion is main form of the load that act on Springs and that’s why

shear stress is the prominent one where the spring safety is concerned in that

case we already know about the formula that is given by the induced stress is

given by walls factor which is key in to this particular empirical relation

where applied load the nominal or the mean diameter of the spring coil and the

diameter of wire coil is used using this empirical relation let miracle like that

either we can find out the diameter of the spring which is going to be saved

based on the metal selected or we have to use different trial methods where the

diameter of this spring wire will be considered based on which we will reach

a safe value of the induced stress if we consider this spring constant or spring

index which is called C this particular D will be eliminated with one degree of

small D because C is nothing but the ratio of capital D by small D this is

another aspect that is called material property and which is the shear stress

since we are considering the shear stress shear stress is having somewhat

this particular value for this preen design that’s a standard

value given by the Association which deals with this spring design aspects

earlier we used to have this value is equal to exactly half but this value can

be used for the yield stress which is approximately equal to 0.3 times or 30%

the yield strength of the given material as far as the indian standards are

considered for the springs they are suggested not to take point 3

they are suggested to take point five times the ultimate strain so whenever

the criterias are specified we are going to take the value of shear stress equal

to either 0.5 times this or 0.3 times the ultimate strength and accordingly we

are going to design the further part of our design aspect now before we conclude

this topic let us look at the important steps which I use for the designing of a

helical spring the very first thing is the estimation of Delta based on the P

max in majority of the cases we may have to use a formula or the empirical

relation which is used to connect this to if the value of P max is given we can

definitely estimate this value of Delta or if both the values are given we can

use them for the further steps second step which is very common in almost all

design is the selection of material now the material will be the either provided

or we have to select a proper material from the design data book based on the

given application now in the second part of the metal selection is nothing but

determining its shear strength now like we have discussed in the previous slide

we can design or we can select that based on Indian Standard or based on the

rough value of 0.3 or 30 percent of the ultimate strength the next one is the

assumption of finding of C now C can be found out using applications or based on

the applications from design data book or in most of the cases they are either

assumed or they are generally provide it now the standard value for

the spring index is greater than three it should never be smaller than three

that is what the suggested value is the value may lie between five to ten and

generally values they are taken are between eight to ten so the values I

hated of fire are generally considered safe for index or spring index the next

is the determination of walls factor wall factor k is already given in the

formula which is equal to using this formula we have to determine

the walls factor the next thing is to find out the diameter or the wire

diameter which is small d we already know this empirical relation once we

have known about the safe value was factor and all other indices we can

definitely find out the wire diameter now this is going to be the same

diameter because you have already considered the safe value of the shear

strength then comes the determination of different specifications of a spring the

very first specification that comes afterwards is the mean coil diameter D

which is generally given by this particular relation where this is the

value of C which is spring index into diameter small D then comes the number

of active coils number of active coils are the coils which actually undergo the

deformation because the end points are generally not movable they are attached

to certain surfaces or some application so the active coils are the coils which

lie between the ends of the coil and that’s why the active coils will be

given by this particular empirical relation where generally in case of the

spring materials that we are going to use the constant of rigidity will be

somewhere required to 81.4 gigapascal the next thing is selection of instyle

like we have discussed for different aides there are flat heads flat plus

ground ends then there are square ends and square plus ground ends so

accordingly to total number of coils will change generally in that case the

total number of coils will be equal to the small n which are active coils plus

one or two number of coils which are the end of the coils then comes this solid

length now solid length also is given by the diameter into the actual or the

total number of coils this is the wire diameter the next thing is the actual

deflection Delta now using the same formula of the Delta which we have seen

in the previous slide the same formula we can use and based on the new values

of found out things we can find out the actual deflection that the spring is

going to undergo there comes the total gap

between the springs and that is nothing but the total number of coils minus one

into the gap between one of them so gap is generally taken 0.5 to 2 millimeters

the standard value if it is not provided we’ll take the average of that the next

thing comes out to it is the free length now the free length as we know is the

summation of this particular terms where this is the solid length the actual

deflection and the total gap then comes the pitch which is a very important

parameter and pitch will be given on or the based on free lengths and the total

number of coils minus 1 because we take the pitch between the two coils so the

end curl will not be considered and the last thing is the spring rate or spring

constant which is also called constant of

stiffness or stiffness value that will be given by this particular empirical

relation which we have seen in the derivation part of this particular

design aspect so these are what the steps that should be used for the

procedure to design a helical spring and certain formula which are associated now

students you need not buy Hardy’s formula because you can find or you can

always find dis formally in this segment of Springs in the book of designed it

about PhD data so you need not buy hot D values rather you look at the steps

which are carried out or the procedure which is carried out so this was from my

end about it is an aspect in the next section we are going to solve few

numerical based on that is enough helical Springs thank you so much for

watching this video if you like this video please subscribe to Akira thank

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18:14 I think that the Wahl factor formula dinominator should be 4c+4