Sign in to follow this  

Spring constant calculation (help needed)

This topic is 4490 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

i got stucked on a question...hope someone here can help me out.. a 254mm long, 3kg/mm compression spring is now cut till 178mm long, wat will be the new spring constant/stiffness in (kg/mm)? is this calculation cotrrect?: K1 X length 1 = K2 X length 2 hence, K2 = (length 1/length 2) X K1 = (254mm/178mm) X 3kg/mm = (4.28kg/mm) thanks in advance

Share this post


Link to post
Share on other sites
Quote:
Original post by Eelco
kg/mm,wht kind of unit is that, btw?


kilograms per millimetre...

which seem strange for a spring constant. I would expected Newtons per unit length.


-Josh

Share this post


Link to post
Share on other sites
kg/mm is not an SI unit....just a given unit.

actually shud change to Newton/meter


btw....if the calculation is correct...

like

K1 X length 1 = K2 X length 2

let's say u compress spring 1 for a length of "length 1" (it's own length, tht means u compress it fully, though it's not legical to do so, just assume it can be done).

u will need F1 = K1 X length 1 = 4.4kg/mm X 254mm =1117.6 Kg.

since K1 X length1 = K2 X length 2
which gives us F1 = F2

which also mean, we need F2 = F1 = 1117.6kg aswell, to compress the short spring fully?

so no matter how we cut a spring, we still need the same force to compress it fully???
since F1 = F2 = F3 = F4.........


is tht true?


and i would like to know how to calculate the spring constant using

given spring diameter
given spring coil diameter
given turns/wounds per unit length
given length
given modulus of elasticity of material used

thx again

[Edited by - TauhuTauhu on August 30, 2005 5:32:06 AM]

Share this post


Link to post
Share on other sites
Quote:
Original post by jjd
which seem strange for a spring constant. I would expected Newtons per unit length.


Maybe not so odd, in a sense. I can imagine it is convenient to factor out graviational acceleration, however, I think the final units should be equivalent to Newtons per unit length.


-Josh

Share this post


Link to post
Share on other sites
Quote:
Original post by jjd
Quote:
Original post by jjd
which seem strange for a spring constant. I would expected Newtons per unit length.


Maybe not so odd, in a sense. I can imagine it is convenient to factor out graviational acceleration, however, I think the final units should be equivalent to Newtons per unit length.


-Josh


meh, what about springs on the moon? springs loaded by other factors than gravity? kg is not a measure of force. period.

Share this post


Link to post
Share on other sites
Quote:
Original post by TauhuTauhu
kg/mm is not an SI unit....just a given unit.

actually shud change to Newton/meter


btw....if the calculation is correct...

like

K1 X length 1 = K2 X length 2

let's say u compress spring 1 for a length of "length 1" (it's own length, tht means u compress is fully, though it's not legical to do so, just assume it can be done).

u will need F1 = K1 X length 1 = 4.4kg/mm X 254mm =1117.6 Kg.

since K1 X length1 = K2 X length 2
which gives us F1 = F2

which also mean, we need F2 = F1 = 1117.6kg aswell, to compress the short spring fully?

so no matter how we cut a spring, we still need the same force to compress it fully???
since F1 = F2 = F3 = F4.........


is tht true?

yes. the force it exerts depends on the deformation. just picture a spring, exerting a load F, and then picture the two virtual halves of it: they both also exert this same force F, and the same goes for any subsection of that same spring.

Quote:

and i would like to know how to calculate the spring constant using

given spring diameter
given spring coil diameter
given turns/wounds per unit length
given length
given modulus of elasticity of material used

thx again


ill think about it.

Share this post


Link to post
Share on other sites
ignoring any moments induced in the spring (only sheer):

K = G * crossection / (2Pi*r * (turns/m) * L)

G being the sheer modulus, which you can calculate given E and nu.
r is the radius of the whole spring, crosssection the crossectional area of one wind of the spring as cut though from the direction of loading.

Share this post


Link to post
Share on other sites

This topic is 4490 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this