Effect of the load sequence on the deformation of a spring

[Nayef]

Consider a spring balance with no initial deflection. Let an object of mass 'm' be attached to it. We allow the spring to come into equlibrium, and 'd' is the deflection at this eqb position. We add another object of mass 'M', while m is also present, so that the final position is x, and hence deflection between the two equilibrium stages is x-d. Now, let 'm' be attached on its own from the zero position of spring and let 'l' be the deflection produced. Will 'l' be less than x-d?

 

[jrmichler]

The deflection of a spring is always proportional to the force on the spring, regardless of load sequence. The proportionality constant is the spring rate, normally measured in N/meter or lbs/inch.

At steady state (nothing is moving), the force is due to the mass times gravity. If the mass is dropped onto the spring, an additional force due to the mass times the acceleration is added to the force due to gravity.

 

[Nayef]

For the purpose of clarity, does your answer imply that l = x-d?

 

[jrmichler]

The definition of a spring is Force = Spring Rate X Distance (F = K*x). The distance is the amount of spring compression/tension from its free length. Therefore, if you compress a spring a little bit, you will get a force. Compress it a little farther, and the force will increase. The difference between the two forces is equal to the amount of additional compression times the spring rate. The total force is equal to the the total amount of compression from the free length times the spring rate.

You can calculate two different forces, and subtract. Or you can take the difference between the two compressed lengths and multiply by the spring rate. The force difference will be same either way. You can prove it with some high school algebra.