# Homework Help: Levers and UTS

1. Apr 18, 2006

### tompotts

Hi everyone,

Im designing a new device for openeing bottle caps. But Im stuck,

basically I have a second class lever (pivot at one end, effort at the other and weight in the middle). The weight is 72.21N the overall length from the effort to the fulcrum is 64mm and the length from the weight to the fulcrum is 15mm.

I need to find out how I can convert the effort that needs to be exerted into units suitable for choosing materials (ie. ultimate tensile strength). How do I convert the (effort) force from Newtons to UTS or yeild strength. I have tried and tried to get my head around this one but i just cant figure it out.

Do I need to find the beding moment within the lever? or can i convert the effort force? This has never been covered in my course, but I need to find out how I can get a value that relates to a UTS so I can choose appropriate materials.

I can imagine that this is really pretty simple, but I cant get my brain into gear. If anyone can help it would be greatly appreciated.

Thanks guys, have fun

Last edited: Apr 18, 2006
2. Apr 18, 2006

### Astronuc

Staff Emeritus
Given the reactions (forces) on the lever, one must determine the stress distribution in the lever, looking at both shear and normal stresses. Normally, one would pick yield strength, and perhaps some margin, as the limiting stress (von Mises).

The stress is related to the force applied to the cross-sectional area, and bending moments are involved.

This is a basically a loaded beam problem.

3. Apr 18, 2006

### FredGarvin

I would do a quick and dirty look at it by taking the load point as the fixed end of a cantilever beam and the force applied at the free end. That will give you the max bending moment and the resultant max bending stress. With only 72 N as your force, you will have plenty of saftey factor if you go with aluminum or steel with a decent cross sectional area. It shouldn't have to be very big. I would assume that dynamic forces will not be an issue here, i.e the speeds are low and the impact forces are negligible...

4. Apr 18, 2006

### tompotts

Thanks Astronuc,

Ive got the equations for calculating the sheer stress etc in a cantilever beam or a simply supported beam, but this being a lever is confusing me.

Can I convert each side of the lever into one cantilever beam to work out the maths that way, or is this totally the wrong idea.

Thanks Tom

5. Apr 18, 2006

### tompotts

Thanks Garvin, your post pretty much answered my last question (should refresh my browser more often).

Like with the last question, should I combine the forces etc on both sides of the lever to create on simple cantilever beam?

6. Apr 18, 2006

### tompotts

OK, I think ive got it.

I simply convert the lever into a cantilever beam and then calculate the sheer and bendiong forces on that beam, the only load being on the end.

Have I got the right idea, or am I way off target here?

Thanks a lot guys.

Tom

7. Apr 18, 2006

### FredGarvin

That's what I would do as a first pass. Then I would model it and put it into our highly sophisticated FEA analysis package and create 3D rendition and animations showing the flexures and stresses. No. Not really.

8. Apr 18, 2006

### haynewp

Get the section modulus (S) of the lever then solve for the bending stress, fb=M/S, this is the stress to compare to the yield stress of the material (for practical purposes). Most likely, yielding of a metal bottle opening lever from bending would be the failure mode and not shear. You would want the lever to return to its original undeformed shape after every use, so look at yield strength with a safety factor and not plastic (ultimate) strength. Max moment is (effort*distance to resisting force).

9. Apr 19, 2006

### tompotts

Thanks haynewp, ive tried to figure out the section modulus. The equation for it is (moment of inertia)/(distance to neutral axis). The 'bar' in question of rectangular uniform cross-section. I have found hundreds of solutions for moments of inertia, but none that I can make sense of. Is there a simple equation for moment of inertia for a rectangular section?

10. Apr 19, 2006

### tompotts

Thanks haynewp, ive tried to figure out the section modulus. The equation for it is (moment of inertia)/(distance to neutral axis). The 'bar' in question of rectangular uniform cross-section. I have found hundreds of solutions for moments of inertia, but none that I can make sense of. Is there a simple equation for moment of inertia for a rectangular section?

11. Apr 19, 2006

### tompotts

Ok sussed it, almost.

Ive calculated the moment of imertia as 142.92mm^2. The cross section is 5mm wide, 7mm thick, I used the equation frm here http://darkwing.uoregon.edu/~struct/courseware/461/461_example_problems/ex_prob_lecture_28/461_example28-2.html [Broken] to find the moment across the x axis. Im assuming the neutral axis runs through the centre of the bar intersecting the centroid perpendicular to the x and y axis.

This gives the distance from the neutral axis as 3.5mm, when dividing the moment of inertia by this i get a section modulus of 4.08333E-8m^3.

Putting this into the equation given by haynewp (fb=M/S) I am stuck, I just want to make sure that I am correct in thinking that 'M' is the max moment of the lever.

Thankyou guys

Last edited by a moderator: May 2, 2017
12. Apr 19, 2006

### tompotts

THE LEVER
Effort= 85N
Length to fulcrum from effort = 26mm
Resitant force = 310N
length to fulcrum from resistance = 7mm

Converted to a beam I get:
Length = 0.02m
Force = 433.706N

Giving a moment of 8.674Nm

Dividing this by the section modulus I get
212424490Nm^2 (i think thats the correct units)

This looks ridiculously too big, any advice?

13. Apr 19, 2006

### haynewp

The required applied force is about 83.46N. The max moment is this force times the distance to the resisting force.

14. Apr 19, 2006

### tompotts

Sorry to keep bothering you guys, but Ive calculated the max moment. To be 2.752 (by doing - 83.405N * 0.033m).

Ive calculated the section modulus to be 0.00007m^3 (thats - 7E-5m^3).

Dividing the moment by the section modulus I get 39314.29, This seems far to big to be the bending force.

am I missing something really obvious here?

15. Apr 19, 2006

### tompotts

Ive tried it keeping the units as millimetres, which gives fb = 67.4104 which appears a lot better (although I doubt it is), but I still dont know what the units are for this, am I right in assuming its Nmm?

Its been a few years since I have had to any of this, and my brain is totally fried on it.

Thanks for all the help so far

16. Apr 19, 2006

### tompotts

Sorry haynewp, its my silly fault, for not explaining things better. I was working with a second class lever on one part of the opener, but now im looking at another part which is a first class lever. I thought I had mentioned it before, but I havnt, so I am terribly sorry for making this very confusing (havnt had a lot sleep lately - 8 days till final handin).

The lever that I am looking at is:
26mm to the fulcrum from the effort
7mm from the fulcrum to the load

Sorry for the confusion

Last edited: Apr 19, 2006
17. Apr 19, 2006

### haynewp

I understood your opener to be a total of 26mm long. I don't understand how you are adding the 26 and 7 to get 33. What you should have is this:

http://en.wikipedia.org/wiki/Image:C...ualization.png [Broken]

Where the distance from the force to the first support (or resistance in your case) is 26-7=19mm. So the max moment is 83.4*19. Am I misunderstanding the set up?

Bending stress is in units of force/area.

Last edited by a moderator: May 2, 2017
18. Apr 19, 2006

### tompotts

Sorry haynewp, its my silly fault, for not explaining things better. I was working with a second class lever on one part of the opener, but now im looking at another part which is a first class lever. I thought I had mentioned it before, but I havnt, so I am terribly sorry for making this very confusing (havnt had a lot sleep lately - 8 days till final handin).

The lever that I am looking at is:
26mm to the fulcrum from the effort
7mm from the fulcrum to the load

Sorry for the confusion

19. Apr 19, 2006

### haynewp

Sorry, I tried to rearrange my last post to the end before you posted again.

For a first class lever;
http://en.wikipedia.org/wiki/Lever
The max moment will occur at the fulcrum (not like in second class), so now Mmax=effort*26mm which is the same as the resistance*7mm. Now that you have rearranged it from a second class to a first class lever, I have a feeling you did not recalculate the required effort correctly. What is the resistance force?

Last edited: Apr 19, 2006
20. Apr 19, 2006

### tompotts

The original lever I calculated as a resistance of (about) 310N a distance 7mm from the fulcrum with an applied force of 85N a distance of 26mm from the fulcrum.

After this I have multiplied each lengh of the side of the lever by the force on the same side and have come up with a balanced lever 10mm on both sides with a weight of 216.9N on each.

Which I believe can be converted into a cantilever beam 20mm long with a weight of 433.7N at the end. If I calculate the total moment of this beam I get 8674.12.

Divide this by the section modulus gives 212.5MPa, which seems a reasonable answer.

Does this look right to you?

21. Apr 19, 2006

### haynewp

"The original lever I calculated as a resistance of (about) 310N a distance 7mm from the fulcrum with an applied force of 85N a distance of 26mm from the fulcrum."

For a second class lever, this is close enough to right.

"After this I have multiplied each lengh of the side of the lever by the force on the same side and have come up with a balanced lever 10mm on both sides with a weight of 216.9N on each."

Here, I think you are trying to convert the second class lever to a first class lever. There is a force of 216.9N on each end of the first class apparatus with the fulcrum in the middle, equal distance from the fulcrum at each side, therefore obviously balanced.

"Which I believe can be converted into a cantilever beam 20mm long with a weight of 433.7N at the end. If I calculate the total moment of this beam I get 8674.12."

This is where you lost me. The max moment from your second statement is 216.9*10=2169Nmm, that is the force at either end time the distance to the fulcrum. You don't need to convert it to a cantilever beam twice as long. There is a serious misunderstanding here.

22. Apr 20, 2006

### tompotts

Ive got you now, its earlier in the thread there was talk about converting it to cantilever beam, so figured that was what I needed to do.

If I just convert one side of the 1st class lever I get an answer like yours (2168.53), Divided by the section modulus gives 53.11MPa.

23. Apr 20, 2006

### FredGarvin

There's no reason why you can't look at it like a cantilever beam. The thought was that it would make it a bit easier to examine. I guess the note should have been made that you have to make sure you have equivilent loading between cases. Either way, a while back you calculated the moment of inertia in units of mm^2. The moment of inertia should be in units of mm^4 or m^4 since it is $$I = \frac{1}{12}bh^3$$

53.1 MPa seems a bit high since that equates to about 7.7 ksi. I have to do the numbers myself, but you look like you've got it down so I'll just do it to prove it to myself.

24. Apr 20, 2006

### haynewp

It is a cantilever beam, cantilevered on each side of the fulcrum. But don't try to equate the first class lever to a single cantilever of the entire lever length with the combined load on the end. That is why you got 4 times the moment the first time (8674).