How Do You Calculate Pocket Depth for Different Compression Springs?

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Discussion Overview

The discussion revolves around calculating the necessary pocket depth for different compression springs to achieve equivalent force under constant conditions. Participants explore the implications of spring dimensions and rates on pocket depth, considering both theoretical and practical aspects of spring mechanics.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a scenario involving two compression springs with different dimensions and rates, seeking to determine the appropriate pocket depth for the new spring.
  • Another participant notes that to answer the question, either the applied force or the amount of compression in the original spring must be known.
  • It is stated that the original spring compresses by 9mm.
  • A calculation is proposed that suggests the new stiffer spring will compress by 8.1mm to produce the same force.
  • Concerns are raised about the correctness of the spring rate units, which are confirmed to be in Newton/meter.
  • Discussion includes the need to maintain the same exposed spring length when switching to the new spring.
  • A participant suggests that the required depth of the pocket for the new spring can be calculated by subtracting the desired exposed height from the required compressed height.
  • Another participant points out that the solid height of the new spring must be considered in the new geometry.
  • One participant calculates the necessary pocket depth based on the difference in compressed lengths of the two springs.
  • Concerns are raised about the potential for the new spring to become "mushed" if the force is large enough, given the calculated pocket depth.
  • Clarification is sought regarding the term "compression amount," which refers to the displacement of the springs when compressed.
  • Ultimately, one participant concludes that the new spring cannot be used based on the solid height considerations.

Areas of Agreement / Disagreement

Participants express differing views on the implications of spring dimensions and the necessary calculations for pocket depth. There is no consensus on the final determination of pocket depth, and the discussion remains unresolved regarding the feasibility of using the new spring.

Contextual Notes

Limitations include assumptions about constant applied force and compression, as well as the dependence on specific spring dimensions and rates. The discussion does not resolve the implications of these factors on the overall design.

helpinghand
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TL;DR
Determining a pocket depth for a compression spring
Hi Guys,

Forgive me, as it has been quite sometime since I have done my spring theory.

The problem I am having is the following:

This is the current situation:
- In a steel block I have a pocket depth of 15mm
- I have a compression spring with the following information:
  • Outside diameter is 34mm
  • Free length 31.75mm
  • Solid height 14mm
  • Spring rate 6.75 N/m
- I have an unknown force applied (assumed x)

Given that the applied force never changes, I need the put in a different compression spring the with following information:
  • Outside diameter is 34mm
  • Free length 25.4mm
  • Solid height 11mm
  • Spring rate 7.5 N/m

How do I go about determining what the pocket depth should be, to get an equivalent force?

I know hooks law, but I am not too sure exactly how to apply it to this situation, or am having an absolute brain fart.

Cheers,
Ed
 
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You need to know either the applied force or the amount of compression (height above "flush"
caused by it) in the original geometry to answer this question. Am I missing something?
 
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The original spring compresses by 9mm
 
So the new stiffer spring will compress by (6.75/7.5)x9mm=8.1mm to produce the same force. Good?
 
Are you sure that the spring rates are in Newton / meter?
 
Lnewqban said:
Are you sure that the spring rates are in Newton / meter?

Yes, based on the data provided - N/m
 
hutchphd said:
So the new stiffer spring will compress by (6.75/7.5)x9mm=8.1mm to produce the same force. Good?

This part I understand. But how does it affect the depth of the pocket retaining the new spring, would it mean I would need a new pocket depth of 14mm to get the same effect?
 
I assume you want the same "exposed" spring length for the new one at load ?
So the original spring has a compressed length of (31.75 - 9)=22.75mm. This leaves (22.75 - 15)=7.75mm exposed.

The new spring has a compressed length (25.4 - 8.1)=17.3mm. So how deep is the necessary hole? (14 mm is incorrect)

I believe you may need to worry about the solid height of the new spring for the new geometry.

Hope this helps.
 
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Take the required compressed height of the new spring for the same load as the current spring and subtract the desired exposed height and the result is the required depth of the hole for the new spring.
 
  • #10
helpinghand said:
Yes, based on the data provided - N/m
These are really weak springs.
Wire should have a diameter of around 0.8 mm.

As the original spring compresses by 9 mm, the reactive force is 9 mm x 6.75 N/m x 0.001 m/mm = 0.06075 Newtons (6.2 gram-force).
 
  • #11
Hi guys,

Cheers for all the help. The data provided was just put together. The main information is trying to determine the new pocket depth required for the shorter spring.
1. I have a long spring in a pocket with a fixed depth.
2. I have a shorter spring, but need to determine the new pocket depth.

Both applied force and compression amount are assumed constant. Thus, with the shorter spring, what is the pocket depth required to exert the same spring force?
 
  • #12
helpinghand said:
Both applied force and compression amount are assumed constant. Thus, with the shorter spring, what is the pocket depth required to exert the same spring force?

If the compression and force for the the new spring are the same as the old one then the pocket depth does not change.
 
Last edited:
  • #13
What does "compression amount" mean ?
 
  • #14
@helpinghand :
Ignoring all posts after @hutchphd post #8, if the two compressed lengths hutchpad gives in his post are correct and your original hole depth is 15 mm then: the difference in the installed height is: 22.75 mm - 17.3 mm = 5.45 mm and the hole depth for the new spring = 15 mm - 5.45 mm = 9.55 mm
 
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  • #15
And this is less than the solid height of the new spring so it can get mushed if force is large enough!
 
  • #16
@hutchphd
Thanks for that. I was only dealing with the required hole depth and failed to look at the solid height of the spring.

Edit: So bottom line, he can't use the new spring.
 
Last edited:
  • #17
hutchphd said:
What does "compression amount" mean ?
It is the amount the springs are going to be compressed by. I.e the spring displacement.
 
  • #18
JBA said:
@hutchphd
Thanks for that. I was only dealing with the required hole depth and failed to look at the solid height of the spring.

Edit: So bottom line, he can't use the new spring.
All good. I will need to apply this to what I am currently doing and double check if it will be a problem.

really appreciate all the help and assistance.
 

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