# Spring constant calculation (help needed)

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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

yep

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kg/mm,wht kind of unit is that, btw?

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Quote:
 Original post by Eelcokg/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

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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]

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Quote:
 Original post by jjdwhich 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

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Quote:
Original post by jjd
Quote:
 Original post by jjdwhich 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.

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Quote:
 Original post by TauhuTauhukg/mm is not an SI unit....just a given unit.actually shud change to Newton/meterbtw....if the calculation is correct...like K1 X length 1 = K2 X length 2let'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 = F2which 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 usinggiven spring diametergiven spring coil diametergiven turns/wounds per unit lengthgiven lengthgiven modulus of elasticity of material usedthx again

ok. thx again.

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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.

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