Spring force after fixating with limited stiffness

In summary, the springs on a press jig prevent it from rotating, and the jig is considered to be very stiff. The jig and any deformations in it can be considered part of spring k2. The press action removes only a small amount of the tension/compression force on the springs.
  • #1
rmwanders
4
0
Hi folks,

I have an interesting problem here from the real world, it's a design i am working on.

So I have an object that is pressed by an hydraulic press with 50kN, let's call it F_before. Then I drive in a jig to fixate it. But the part that holds the jig has a limited stiffness. Hence if I release the hydraulic press the system will bounce up and the force in the system will be less than 50 kN, but how much?

The way I see it the springs are in series and I can use the conservation of spring energy to find the ratio between F_after and F_before.
F_after/F_before = sqrt(k2/(k1+k2))

Is this correct? The square root in the solution looks kinda odd to me.
 

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  • #2
Does the jig prevent rotation? Is the jig considered very stiff or do we need to consider deformations of the jig itself?
 
  • #3
Hi, yes the jig prevents rotation and is considered very stiff.

In reality the spring k2 is a bus with limited stiffness that goes around k1. The jig goes through the bus supporting against the inner spring on the left and the right sides. see picture
I think the jig and any deformations in it can be considered part of spring k2 once I transfer my conceptual design to a finite element program.

But for now I am trying to get a more basic understanding of my design.
 

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  • #4
rmwanders said:
Hi, yes the jig prevents rotation and is considered very stiff.
So then you can consider the springs to act as one joint spring. The stiffness of the joint spring is just ##k_{joint}=k_1+k_2##
 
  • #5
As I see it, after the press action is removed, spring 1 will be under compression load, which will be equal to the tension load of spring 2.
Is that statement correct?
 
  • #6
Yes indeed. Because the springs are in series. But the question is how much remains of the tension/compression force after the jig is placed and the press is released. It should be less that the 50 kN the press applied
 
  • #7
rmwanders said:
F_after/F_before = sqrt(k2/(k1+k2))

Is this correct? The square root in the solution looks kinda odd to me.
I also don't think the root should be there.

Consider the force over jig displacement (relative to the release position) graphs of the two springs:

spring_1: intercept = F_before, slope = -k1
spring_2: intercept = 0, slope = k2

F_after
is where they intersect after some jig displacement d:

(F_before - F_after) / d = k1
F_after / d = k2


Solving for F_after leads to:

F_after = F_before * k2/(k1+k2)
 
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  • #8
I agree with @A.T.

And to be pedantic (it is what we do! ) the energy in the springs is not conserved
rmwanders said:
Summary:: spring force after fixating with limited stiffness

The way I see it the springs are in series and I can use the conservation of spring energy to find the ratio between F_after and F_before.
This is because you do negative work on them during the relaxation. If you didn't, they would oscillate (until "friction" dissipated the energy).
 
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  • #9
hutchphd said:
And to be pedantic (it is what we do! ) the energy in the springs is not conserved
Yes, the springs change because they tend towards a lower potential energy state.

In particular, the force on the jig by spring_2 is not equal to the force by spring_1, while the displacement is the same for both. So the positive work by the jig on spring_2 is smaller than the negative work on spring_1. The jig would gain kinetic energy, which you either remove during the process or it dissipates in osculations.
 
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  • #10
Thanks a lot guys! It all makes sense now :)
 

What is spring force?

Spring force is the force exerted by a spring when it is compressed or stretched. It is proportional to the displacement from its equilibrium position and follows Hooke's law.

How does fixation affect spring force?

Fixation refers to the process of securing a spring to a fixed point. When a spring is fixated, its stiffness increases and therefore the spring force also increases. This is because the spring is less able to stretch or compress, resulting in a stronger force being required to displace it from its equilibrium position.

What is limited stiffness?

Limited stiffness refers to the maximum amount of force that a spring can exert before it reaches its elastic limit. This means that beyond a certain point, the spring will no longer follow Hooke's law and will not return to its original length when the force is removed.

How does limited stiffness affect spring force?

When a spring is fixated with limited stiffness, it means that the maximum force it can exert is limited. This can affect the spring force by either reducing or increasing it, depending on the amount of force applied. If the force applied is within the limited stiffness, the spring will behave according to Hooke's law. However, if the force exceeds the limited stiffness, the spring will become permanently deformed and the spring force will decrease.

Why is understanding spring force after fixating with limited stiffness important?

Understanding spring force after fixating with limited stiffness is important in various fields such as engineering, physics, and biomechanics. It allows us to accurately predict the behavior of springs in different situations and design systems that rely on springs with the appropriate stiffness for their intended use. It also helps in understanding the limitations and potential failures of spring-based systems.

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