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Problem on Design of Helical Compression Spring Video Lecture | Design of Machine Elements - Mechanical Engineering
FAQs on Problem on Design of Helical Compression Spring Video Lecture - Design of Machine Elements - Mechanical Engineering
| 1. How do I calculate the spring constant of a helical compression spring? |  |
Ans. The spring constant of a helical compression spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula to calculate the spring constant is given by k = (Gd^4)/(8nD^3), where k is the spring constant, G is the shear modulus of the material, d is the wire diameter, n is the number of active coils, and D is the mean coil diameter.
| 2. What factors affect the maximum load a helical compression spring can withstand? | |
Ans. Several factors can affect the maximum load a helical compression spring can withstand. These include the wire diameter, coil diameter, number of active coils, material properties (such as tensile strength and yield strength), and the design of the spring ends. Additionally, the stress-relieving process during manufacturing can also influence the maximum load capacity of the spring.
| 3. How do I determine the required number of active coils for a helical compression spring? | |
Ans. The number of active coils in a helical compression spring can be determined based on the available space, the desired spring rate, and the required overall length of the spring. A general guideline is to have at least two-and-a-half active coils in compression to ensure stability and prevent buckling. However, it is recommended to consult spring design tables or software to accurately determine the required number of active coils.
| 4. What are the common materials used for helical compression springs? | |
Ans. The common materials used for helical compression springs include high-carbon steel, stainless steel, phosphor bronze, and various alloys. High-carbon steel is widely used due to its excellent strength and durability. Stainless steel is preferred for applications requiring corrosion resistance. Phosphor bronze is commonly used for applications requiring good electrical conductivity. The selection of material depends on factors such as environmental conditions, load requirements, and operating temperatures.
| 5. How do I ensure the optimal performance and lifespan of a helical compression spring? | |
Ans. To ensure the optimal performance and lifespan of a helical compression spring, it is important to consider proper design, material selection, and regular maintenance. The design should take into account factors such as load requirements, stress levels, and space limitations. Proper material selection based on the application's needs and environmental conditions is crucial. Regular maintenance, including periodic inspection for signs of wear, lubrication if necessary, and replacing worn-out springs, can help extend the lifespan of the spring.
Text Transcript from Video
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
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
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Oct 07, 2025
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