How to calculate the stiffness of a spring wire cantilever beam?

  • #1
argyrg
3
1
TL;DR Summary
A length of spring wire, fixed along part of its length and free at the other end. For a given diameter of wire, fixed length and free length how can the torque or lifting power be calculated
I am designing a small box with a lid. The box and its lid measure 100mm x 70mm x 25mm tall..

The lid is spring loaded by a suitably designed torsion spring. When the lid is closed it is latched in place horizontally. The latch mechanism consists of a fixed item in the lid and a moveable latch in the base. When the lid closes, its lever pushes the base lever out of the way until the base lever snaps in to place. The latching faces overlap by 1.5mm.

I have a horizontal lever which is 15mm long and 2mm square. The lever operates the base latch. When the lever is pressed it releases the latch and allows the lid to raise under the tension of the torsion spring.

At one end of the horizontal lever, inside the box, there is a hinge / pivot point. At 90 degrees vertical to the horizontal lever is the latch. The lever passes through an aperture in the base of the box which constrains its moveable range from horizontal to -18 degrees. This translates to a movement of more than 1.5mm at the latch interface, therefore sufficient to release the latch

The aperture through which the klever passes in the base of the box is 5mm away from the centre of the hinge. The aperture is 3.3mm tall which means that the 2mm tall lever can be depressed by 1.3mm.

When the lever is depressed (by pressing with a finger or thumb) the spring will act to oppose the movement. When the lever is released the spring will force the lever back to horizontal.

I need to specify a simple spring which will provide sufficient force and a reasonable amount of feedback to the user as the lever moves over this 1.3mm range. I want to avoid creating a spring which is too strong such that it becomes uncomfortable to operate or one that is too weak to return and hold the latch in place.

My expectation is that a short length of music wire or stainless wire can be set into a horizontal well in the base of the box on its inside face. The diameter of the well will be slightly larger than that of the music wire spring. The position of the well will allow a length of spring wire to be inserted in to it and for the free end to extend beyond the well and then sit just above the underside of the lever. In this way one end of the wire will be fixed and the free end can be pushed down to put the lever under the desired tension. When the lever is fully depressed it will deflect the spring by a further 1.3mm.

I would like to know how to specify the force that the free end of the wire will create. I think there are three variables.

1. The diameter of the wire to be used
2. The length of the leg in the well
3. The distance between the exit of the well and the far side of the horizontal latch lever (or just beyond it)

Thanks in anticipation

Russell
 
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  • #2
Welcome to PF. :smile:

Can you post a sketch of the mechanism? That would help a lot. Use the "Attach files" link below the Edit window to upload a PDF or JPEG image. Also, are you familiar with using Free Body Diagrams (FBDs) to analyze forces on mechanisms?
 
  • #3
Thanks @berkeman for your input. I've attached a couple of sketches of the system.

I'll look into the Free Body Diagrams

The component parts are
Cyan colour is the base of the box, cut away for clarity
Magenta is the lever with the pivot shown in red
Green is the lid's latch
White is the well which I will use to hold the spring wire
Black is the spring wire, shown to be supporting the magenta lever.
When the lever is pressed it will rotate around the red pivot. This will withdraw the movable latch from the fixed latch of the lid and therefore allow the lid to rise

Hopefully with the original description this will illustrate what I'm trying to accomplish.

Thanks again in anticipation of any further input

Russell
 

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Likes berkeman
  • #4
Looking around I think I'm going to need a course on cantilever beam deflection

What I know is the modulus of elasticity and the amount of deflection, I can vary the diameter and free length of the spring as well as the point at which the load is applied so I should be able to solve the equation for the force. I've looked at a couple of calculators on-line but my physics is letting me down, for example the 'moment of inertia' in this calculator (https://calcresource.com/statics-cantilever-beam.html)

Without the underlying formula it's a bit tricky to play with the variables...

Anyone know of a calculator that will allow me to play with a set of numbers?
 

1. What is the formula to calculate the stiffness of a spring wire cantilever beam?

The stiffness (k) of a spring wire cantilever beam can be calculated using the formula: \( k = \frac{E I}{L^3} \), where \( E \) is the Young's modulus of the material, \( I \) is the second moment of area (also known as the area moment of inertia), and \( L \) is the length of the beam.

2. How do you determine the Young's modulus (E) for the material of the beam?

The Young's modulus (E) of a material, which measures the stiffness of the material itself, is typically found in tables of material properties or can be determined experimentally. It is a constant that depends on the material of the beam, such as steel, aluminum, or plastic.

3. What is the second moment of area (I), and how is it calculated?

The second moment of area (I), also known as the area moment of inertia, is a geometrical property that reflects how the cross-sectional area is distributed about an axis. It is calculated based on the shape of the cross-section of the beam. For a rectangular cross-section, \( I = \frac{b h^3}{12} \), where \( b \) is the base width and \( h \) is the height. For a circular cross-section, \( I = \frac{\pi d^4}{64} \), where \( d \) is the diameter of the circle.

4. Does the orientation of the beam’s cross-sectional area affect its stiffness?

Yes, the orientation of the beam's cross-sectional area significantly affects its stiffness. A beam with a larger moment of inertia (I) about the axis of bending will be stiffer. This means that changing the orientation of a rectangular cross-section (flipping it on its side, for example) can increase or decrease the stiffness depending on how the height and width are aligned relative to the bending axis.

5. How does the length of the cantilever beam affect its stiffness?

The length of the cantilever beam has a cubic relationship with its stiffness: as the length increases, the stiffness decreases proportionally to the cube of the length. This means that even a small increase in length can lead to a significant decrease in stiffness, making the beam more flexible.

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