How Do You Calculate Pocket Depth for Different Compression Springs?

[helpinghand]

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

 

[hutchphd]

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?

 

[helpinghand]

The original spring compresses by 9mm

 

[hutchphd]

So the new stiffer spring will compress by (6.75/7.5)x9mm=8.1mm to produce the same force. Good?

 

[Lnewqban]

Are you sure that the spring rates are in Newton / meter?

 

[helpinghand]

Are you sure that the spring rates are in Newton / meter?

Yes, based on the data provided - N/m

 

[helpinghand]

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?

 

[hutchphd]

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.

 

[JBA]

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.

 

[Lnewqban]

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).

 

[helpinghand]

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?

 

[JBA]

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.

 

[hutchphd]

What does "compression amount" mean ?

 

[JBA]

@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

 

[hutchphd]

And this is less than the solid height of the new spring so it can get mushed if force is large enough!

 

[JBA]

@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.

 

[helpinghand]

What does "compression amount" mean ?

It is the amount the springs are going to be compressed by. I.e the spring displacement.

 

[helpinghand]

@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.