# Design of Helical Compression Springs Design Aspects – Design of Springs – Machine Design I

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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
you you

## 3 thoughts on “Design of Helical Compression Springs Design Aspects – Design of Springs – Machine Design I”

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2. Sohit singh kshatriye says:
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