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CONICAL SPRING 2
- Thread starter pardal
- Start date
Hi Pardal
The formula for conical springs is based on ordinary compression springs ie for each turn on the conical spring
here are some formula :-
the rate for each individual turn or
fraction of a turn = k=Gd^4/(8*D^3*Na)
total rate k for a complete spring
of different diameter turns = k= 1/((1/k1)+(1/k2)etc)
To calculate stress in the wire the mean diameter of the
largest active coil is used at the required load ie:-
S= 8*P*D*Kw/(3.142*d^3)
The solid height of a conical spring of a uniform taper but not telescoping, with squared and ground ends can estimated
from :-
La=Na*(d^2-u^2)^0.5 + 2d
where G= modulus of rigidity of material
d= wire dia
D= mean coil diameter
Na = number of active turns
k = spring stiffness
S = stress
P= spring load
La= solid height of spring
u = o.d of large coil - o.d. of small coil/(2*Na)
Kw1= (4*C-1/(4*C+1))+ 0.615/C
C = mean dia of largest coil/ wire diameter
The rate formula is valid only if the pitch angle is less than 15 degrees or the deflection per turn is less than D/4
hope this helps
regards desertfox
The formula for conical springs is based on ordinary compression springs ie for each turn on the conical spring
here are some formula :-
the rate for each individual turn or
fraction of a turn = k=Gd^4/(8*D^3*Na)
total rate k for a complete spring
of different diameter turns = k= 1/((1/k1)+(1/k2)etc)
To calculate stress in the wire the mean diameter of the
largest active coil is used at the required load ie:-
S= 8*P*D*Kw/(3.142*d^3)
The solid height of a conical spring of a uniform taper but not telescoping, with squared and ground ends can estimated
from :-
La=Na*(d^2-u^2)^0.5 + 2d
where G= modulus of rigidity of material
d= wire dia
D= mean coil diameter
Na = number of active turns
k = spring stiffness
S = stress
P= spring load
La= solid height of spring
u = o.d of large coil - o.d. of small coil/(2*Na)
Kw1= (4*C-1/(4*C+1))+ 0.615/C
C = mean dia of largest coil/ wire diameter
The rate formula is valid only if the pitch angle is less than 15 degrees or the deflection per turn is less than D/4
hope this helps
regards desertfox
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2
- #3
Conical helical spring can be designed in two ways
1. As a linear spring where the pitch is variable so that all coils will come to touch each other at the same time when pressed to solid.
2. As a non-linear spring where the pitch is constant. Therefore the largest coil will touch the coil above it first and then the second and so on thereby, creating a non-linear rate spring where each time a coil reaches the coil above it the number of active coils decrease and the rate of the spring increase.
The rate of the first type of spring can be calculated/estimated as a standard helical coil using the average diameter of the conical spring. The stresses should be calculated with respect to the largest diameter because the stresses are the highest at the largest coil.
Calculating the variable rate of the second type is more complicated and done in steps. First you treat the spring as a normal spring until the largest coil touches the coil above it. Then you deduct one coil and do the calculations until the next coil touches the coil above it and so on. Since the smallest coil will be the last to work, the stresses are calculated with respect to the smallest coil.
1. As a linear spring where the pitch is variable so that all coils will come to touch each other at the same time when pressed to solid.
2. As a non-linear spring where the pitch is constant. Therefore the largest coil will touch the coil above it first and then the second and so on thereby, creating a non-linear rate spring where each time a coil reaches the coil above it the number of active coils decrease and the rate of the spring increase.
The rate of the first type of spring can be calculated/estimated as a standard helical coil using the average diameter of the conical spring. The stresses should be calculated with respect to the largest diameter because the stresses are the highest at the largest coil.
Calculating the variable rate of the second type is more complicated and done in steps. First you treat the spring as a normal spring until the largest coil touches the coil above it. Then you deduct one coil and do the calculations until the next coil touches the coil above it and so on. Since the smallest coil will be the last to work, the stresses are calculated with respect to the smallest coil.
- Thread starter
- #4
What about a spring that shall close to the wire diameter?
for the deflection I had got this formula from a old book
deflect= 16 P /(d^4 G) * ( R1^2 +R2^2) (R1+R2)
given
P= load
d= wire diameter
G= shear modules
R1 R2 = mean radii of two end coils
It is aplicable yet??
On the same book i read
m= 2 * R2 / d
What does it mean??
Thanks in advance (TIA)
Pardal
for the deflection I had got this formula from a old book
deflect= 16 P /(d^4 G) * ( R1^2 +R2^2) (R1+R2)
given
P= load
d= wire diameter
G= shear modules
R1 R2 = mean radii of two end coils
It is aplicable yet??
On the same book i read
m= 2 * R2 / d
What does it mean??
Thanks in advance (TIA)
Pardal
pardal
Something in the formula for the deflection does not seams right. The number of active coils "Na" is missing. Which book are you refering to? Anyway, for a conical spring to have a solid height of one wire diameter, the coils should be designed such that each coil outside diameter is smaller than the coil underneath inside diamter. The formula that desertfox gave for the spring constant is valid and the spring will have a linear rate characteristics (you can use the (R1+R2) instead of the D in the formula
k=Gd^4/(8*D^3*Na)
Something in the formula for the deflection does not seams right. The number of active coils "Na" is missing. Which book are you refering to? Anyway, for a conical spring to have a solid height of one wire diameter, the coils should be designed such that each coil outside diameter is smaller than the coil underneath inside diamter. The formula that desertfox gave for the spring constant is valid and the spring will have a linear rate characteristics (you can use the (R1+R2) instead of the D in the formula
k=Gd^4/(8*D^3*Na)
- Thread starter
- #6
Pardal
I still think the formula is not correct it should be 8 instead of the 16 and it should be R1^3+3*R1^2*R2+3*R2^2*R1+R2^3 which is (R1+R2)^3=D^3 instead of the ( R1^2 +R2^2)*(R1+R2) which is R1^3+R1^2*R2+R2^2*R1+R2^3.
You didn't answer which book you are refering to.
I still think the formula is not correct it should be 8 instead of the 16 and it should be R1^3+3*R1^2*R2+3*R2^2*R1+R2^3 which is (R1+R2)^3=D^3 instead of the ( R1^2 +R2^2)*(R1+R2) which is R1^3+R1^2*R2+R2^2*R1+R2^3.
You didn't answer which book you are refering to.
- Thread starter
- #8
Israelkk , thanks for your help .
About the book , it is a old one, author Celso Maximo del Cosso , edited in 1952 at Madrid Spain .
but this formula is equal to one I got from other old book
Machines from Germany and I found it too here
I had change J [Polar area moment] = Ix + Iy =
(pi d^4 / 64 )* 2 = pi d^4/32
thus the term
pi/(2*j) = pi/2 * (32 /(pi d^4))= 16/d^4
about the term
((r1^2+r2^2)* (r1+r2))
it is the same at all books
I'm new on this topic , and I want to learn.
I'm a 56 years old student since the last 39 years when I begun my technician carrer, of course I got my degree by 1969, but allways I kept learning a bit every day.
In this case I need a conical spring 100 mm free length , to be closed at the wire diameter, the outside big diameter is 70 mm , the inside small diameter 38 mm.
the whole stroke [100 mm] is done in about 10 seconds , and the run to the whole length by the spring own speed .
It guess the load would be about 5 to 7 Kg.
It will be used on a two diameter hole saw to extract the piece of material cuted.
So the compression is done while cuting the 100 mm and the expansion is done when the cut is end through the material tickness.
The piece to extrac is like a washer 76 mm od 34 id the 100 mm is because the material is diplayed in 20 each 5 mm layers.
Hope it make clear.
I had done a calc , I will post it here later , I had to copy and paste.
Thanks again
Pardal
About the book , it is a old one, author Celso Maximo del Cosso , edited in 1952 at Madrid Spain .
but this formula is equal to one I got from other old book
Machines from Germany and I found it too here
I had change J [Polar area moment] = Ix + Iy =
(pi d^4 / 64 )* 2 = pi d^4/32
thus the term
pi/(2*j) = pi/2 * (32 /(pi d^4))= 16/d^4
about the term
((r1^2+r2^2)* (r1+r2))
it is the same at all books
I'm new on this topic , and I want to learn.
I'm a 56 years old student since the last 39 years when I begun my technician carrer, of course I got my degree by 1969, but allways I kept learning a bit every day.
In this case I need a conical spring 100 mm free length , to be closed at the wire diameter, the outside big diameter is 70 mm , the inside small diameter 38 mm.
the whole stroke [100 mm] is done in about 10 seconds , and the run to the whole length by the spring own speed .
It guess the load would be about 5 to 7 Kg.
It will be used on a two diameter hole saw to extract the piece of material cuted.
So the compression is done while cuting the 100 mm and the expansion is done when the cut is end through the material tickness.
The piece to extrac is like a washer 76 mm od 34 id the 100 mm is because the material is diplayed in 20 each 5 mm layers.
Hope it make clear.
I had done a calc , I will post it here later , I had to copy and paste.
Thanks again
Pardal
pardal
I looked up my literature and found your formula. It is originated from Timoshenko Strength of materials Part 1 page 276. I assume it is the exact solution for the deflection. Notice that if you use R1=R2=R=D/2 the formula reduces to the one gave by Desertfox.
I am no longer uses the basic formulas I use a computer program I wrote in 1986 thats why I didn't recognise the formula in this form.
I looked up my literature and found your formula. It is originated from Timoshenko Strength of materials Part 1 page 276. I assume it is the exact solution for the deflection. Notice that if you use R1=R2=R=D/2 the formula reduces to the one gave by Desertfox.
I am no longer uses the basic formulas I use a computer program I wrote in 1986 thats why I didn't recognise the formula in this form.
- Thread starter
- #10
I have a design situation where I need to fit a spring into a space of 0.088" to 0.097" to provide a force around 70 to 80lbs. The max dia. of spring is only 0.8". Can someone please let me know if this is a possibility and if there is such a spring. Thank you in advance.
Ency
Ency
This is not a simple case. There may be a way to do it but some assumptions are needed and some special processes too.
However there is more info needed.
For how long the spring will sit under this load? If this is for a long time then relaxation should be considered too.
Is it a real life design and if so how many springs are needed?
What is the price target for this spring?
However there is more info needed.
For how long the spring will sit under this load? If this is for a long time then relaxation should be considered too.
Is it a real life design and if so how many springs are needed?
What is the price target for this spring?
Israelkk,
Thanks much for this response. This is a real life situation and is a working pump design. The force of 70 - 80lbs is required to seal off (2) ceramic plates, one sliding on top of the other (which is stationary) ceramic plate. The spring is located behind the stationary ceramic plate. Ceramic surfaces are ground and lapped to 2 light bands. Available space to fit the spring behind this ceramic plate is only 0.085" to 0.108". The max. dia, of spring (if circular) is 0.800". However, someone in my company before my time has selected a smalley spring which is now destroying the ceramics and causing havoc. I understand from Smalley engineers that this spring cannot be compressed more than 0.024 to 0.028" and we are apparently compressing the spring beyond this limit to seal off leaks between the plates.
The current spring rate is around 6000lbs/in. This is about 6lbs per every (1)thou compressed and if we compress, say, (28) thous, then the force is 168 lbs and is double of what is required. Is there a better type and design of spring that I can use in this application?
Thank you in advance.
Thanks much for this response. This is a real life situation and is a working pump design. The force of 70 - 80lbs is required to seal off (2) ceramic plates, one sliding on top of the other (which is stationary) ceramic plate. The spring is located behind the stationary ceramic plate. Ceramic surfaces are ground and lapped to 2 light bands. Available space to fit the spring behind this ceramic plate is only 0.085" to 0.108". The max. dia, of spring (if circular) is 0.800". However, someone in my company before my time has selected a smalley spring which is now destroying the ceramics and causing havoc. I understand from Smalley engineers that this spring cannot be compressed more than 0.024 to 0.028" and we are apparently compressing the spring beyond this limit to seal off leaks between the plates.
The current spring rate is around 6000lbs/in. This is about 6lbs per every (1)thou compressed and if we compress, say, (28) thous, then the force is 168 lbs and is double of what is required. Is there a better type and design of spring that I can use in this application?
Thank you in advance.
How do you compensate for the reduction of the force due to the wear of the ceramic plates especially with this high rate? The "This is about 6lbs per every (1)thou compressed and if we compress, say, (28) thous, then the force is 168 lbs and is double of what is required" is nor clear. If you use a stack 28 springs or even a stack of 14 springs (to get 84 lbf) the thickness of the stack of spring is larger than 0.108"?
Israelkk,
Thanks again for the feedback. When I assemble the parts and fit them inside the pump 'housing', the clearance available behind the stationary ceramic plate and the inside of the pump housing, to fit the spring, ranges from 0.085" to 0.108".
To explain this pumpo design further, I have (4) cylinder/piston assemblies in this pump, located say, 12 O'clock position, 6 O'clock, 9 O'clock and 3 O'clock positions, named as cylinders 1,2,3 & 4 respectively. It operates on a crankshaft which allows each cylinder to operate at different times. The ceramic plates serves as porting valves to open and close in-take and discharge of fluid. The crankshaft assembly design has a sliding ceramic plate on a stationary ceramic plate. This spring in question is located behind the stationary ceramic plate.
The current smalley spring we use has a free height of around 0.120". When this spring is fitted behind the stationary ceramic plate and the inside edge of the housing, it can be precompressed by as much as 0.028".
Hope this information helps.
Thanks again for the feedback. When I assemble the parts and fit them inside the pump 'housing', the clearance available behind the stationary ceramic plate and the inside of the pump housing, to fit the spring, ranges from 0.085" to 0.108".
To explain this pumpo design further, I have (4) cylinder/piston assemblies in this pump, located say, 12 O'clock position, 6 O'clock, 9 O'clock and 3 O'clock positions, named as cylinders 1,2,3 & 4 respectively. It operates on a crankshaft which allows each cylinder to operate at different times. The ceramic plates serves as porting valves to open and close in-take and discharge of fluid. The crankshaft assembly design has a sliding ceramic plate on a stationary ceramic plate. This spring in question is located behind the stationary ceramic plate.
The current smalley spring we use has a free height of around 0.120". When this spring is fitted behind the stationary ceramic plate and the inside edge of the housing, it can be precompressed by as much as 0.028".
Hope this information helps.
If the 0.085 to 0.108 range can not be adjusted or can not be changed then you have to assume 0.085" when you design/select a spring.
Getting the 80 lb using a helical conical spring in the 0.085 clearance is probably not possible while if the clearance can be closer to 0.108" it will be possible to design one.
I am not sure about of the shelf wave spring (Smally springs) but after a quick check I am sure a custom one can be designed.
Bellevile springs can do the job but again I am not sure about of the shelf springs but again a custom one can be desigened for sure. A belleveile spring can even be designed to give a fairely constant force over a specified range of deflection which makes it insensitive to the high rate of the spring you had.
Getting the 80 lb using a helical conical spring in the 0.085 clearance is probably not possible while if the clearance can be closer to 0.108" it will be possible to design one.
I am not sure about of the shelf wave spring (Smally springs) but after a quick check I am sure a custom one can be designed.
Bellevile springs can do the job but again I am not sure about of the shelf springs but again a custom one can be desigened for sure. A belleveile spring can even be designed to give a fairely constant force over a specified range of deflection which makes it insensitive to the high rate of the spring you had.
Here is a website that you might want to take a look at.
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