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Old 01-14-2011, 03:42 PM
t walgamuth's Avatar
t walgamuth t walgamuth is online now
dieselarchitect
 
Join Date: Apr 2005
Location: Lafayette Indiana
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Quote:
Originally Posted by moon161 View Post
Wrong.

Shear does not accumulate like displacement through the length of the spring. If it did, there would be no reaction at one of the supports- either the spring perch or the LCA.

Viewing each coil as a free body is useful, and spring body connects the equal and opposite (in a static case) forces at the spring perch and LCA via shear.

Consider the spring again 10" 10 coils, 100lb/inch. For a 100lb load, each coil sees 100lb, and deflects .1". The spring constant of each coil is the force over the deflection, or 100/.1 = 1000lb/in.

Cut a coil off, 100lb load, deflection is 9x.1" or .9". To achieve 1" deflection, each coil deflects 1/9" = .111", the reaction is 1000*.111=111 lb. The new spring constant is 111 lb/in.

Another way to look at it is each coil being in series w/ the next.

http://en.wikipedia.org/wiki/Hooke%27s_law#Multiple_springs

Springs in series add like resistors in parallel:
If added in series, 1/k = Sum(1/k1 + 1/k2 ....)

10 coils 1/k = 10 * 1/1000 =.01
k = 1/.01 = 100

9 coils 1/k = 9 * 1/1000 = .009
k = 1/.009 = 111
So you are saying cutting a coil off increases rate by 10%, right?
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