Leverage problem -- Need to calculate why my mechanism is breaking

  • Thread starter Ted Farkas
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  • #1
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Hi,

I'm new to the forum and I am after some help to blow the cobwebs regarding a leverage problem I have, I am a draftsman and did structural engineering diploma many years ago that had limited statics and dynamics, I have forgotten all of it....

In a nut shell we have a working model of both pdf's below.

Model in problem 2 works and we are trying to replicate the action/leverage force in problem 1:

The end state is that the digging arm can resist up to 3 ton of force before the ram (nitrogen between 100 to 150 bar) compresses, then when the obstacle is passed will then push the digging arm back into the ground.

Refering to problem 2: (note: the machine works in practice)

I make it to be a class 1 lever wit the fulcrum (a) and the ration of Lin and Lout to be 583/530=1.1 hence 1.1 x 3 ton means the ram would need to hold 3.3 ton before moving.?????

Refering to problem 1: (this machine breaks out of the ground too easily)

This can work in 2 ways:

Both actions probably work in tandem however:

On striking a force digging arm pivots at point C for 25 degrees (pin at point B stops extension at 25 degrees): I make that to be 938/131= 7 approx meaning I would need 3 x 7 or approx 21 tonne to hold it (hence the easy breakout).

Once the 25 degree angle is reached it will start to rotate at point B (dimension not shown on drawing) but it would be 1100/516= 2 hence 2 x 3 meaning 6 ton at ram should hold digger from rotating (this should be the other way around e.g hard force to overcome at first then lower force after more rotation).

The implement can be 'locked out at point F with a pin through slave arm hence rotation at Point A only being 803/516 = 1.5 hence 1.5 x 3 = 4.6 ton. Meaning ram would need to hold 4.6 tone before breakout.
However machine is not holding in ground. ( have a video which I can try an upload to show mechanism).

My calcs maybe all wrong however the main thing I am trying to understand is that I have the Mechanical advantage distances and fulcrum positions right especially since it is not a standard 'seesaw' drawing.

Our end state is to be hold the digger in the ground until 3 ton is reached - this means if it hits a bolder or tree root - digger or other immovable force digger bends out of the way.

Thanks
Brett
 

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Answers and Replies

  • #2
Tom.G
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For Problem 1, I have trouble understanding the details of how much force there is where under exactly what conditions. Do you know the force at the tip when it starts to move, when pin B is hit, at least two points of tip travel after pin B hits, the operating curve of the ram (force versus movement)? Or three points of force versus tip travel with the F pin in place.

Once the 25 degree angle is reached it will start to rotate at point B (dimension not shown on drawing) but it would be 1100/516= 2 hence 2 x 3 meaning 6 ton at ram should hold digger
When the digger hits something it rotates around C. When it rotates to 25 degrees it hits the pin at B, forcing the rotation to point A. The same as if F was pinned but at a different location.

The problem then becomes solving the triangle AED for the ram. When the AED triangle collapses, points A, E, D are in line and infinite ram force will not move the digger. Seems like ram force required would be related to TAN(90 - ∠AED).

Mechanics is not my specialty, so the above may be a little off. Hopefully one of the mechanical specialists will jump in and carry on.

Cheers,
Tom
 
  • #3
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Hi,

Thanks for having a look. You have got the concept right regarding the movements. The only thing I failed too mention is that there is stopper at point A preventing the ram from rotating in line preventing AED aligning.

I am trying to see if I have my leverage ratios right .

thanks
Brett
 
  • #4
JBA
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Ted,
After further review of your design vs the Problem 2 design, difference, a leverage issue, standout clearly by simply looking at your drawing and the Problem 2 design and that is, at 0 cylinder stroke, the distance from your A pivot to the plow tip of 803 mm vs his distance from his A pivot to the plow tip of 583 mm which alone gives his design a 38% greater resistance to an obstruction force than your design . Also, at 25° rotation (38.5 mm stroke), this differential is even greater for a 25° rotation by your tip about the C pivot because that increases both your cutting depth and your A pivot to plow tip distance to 994 mm, while his distance remains at 583 mm which results in a 70% greater plow tip force resistance than your design.

There is an additional advantage to the Problem 2 design because it has a greater cylinder stroke at a 25° rotation and, therefore greater cylinder force, than your design. I have done a full analysis of your Problem 1 design vs. the Problem 2 design and have the details for all of the above for both your Problem 1 and the other problem 2 design. The results are very close but not necessarily exact because I had to manually determine the required angles required for the calculations from screen prints of your drawings.

Let me know if you are interested in seeing resulting values and/or my Excel calculations for the above.
 
  • #5
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HI JBA,
Thankyou, I did some rough calcs after dragging out my old engineering mechanics book. I will do a sketch to see if my calcs agree, I would be happy to see yours. I think I got 750 kg for the first action hinge point, I'm dubios if was even that as it would not break hard ground at all. I'm wondering if the x and y forces of the ram lessen force as well???

My main aim is to ultimately calculate myself what is happening as I need to get the new machinery piece to reist 2.5 tonne min at the tip, via moving pivot points.

thanks
Ted
 
  • #6
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IN caLc 1 I worked out it would take 2.4 tonne at tip to start to compress cylinder.

Calc 2 I worked out 612 kg to start to compress cylinder

Calc 3 - it would be 1.5 tonne min to compress past the 25 degree

just using distances from fulcrum and class 1 and 2 levers
 

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  • #7
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JBA did a stick diagram in another thread that may be easy to understand.

thanks
 
  • #9
JBA
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I will post more information later, but for now, here are my results for a 100 bar pressure charge in the cylinder. I started at the cylinder rod force and calculated the resulting horizontal resisting force of the tip of the blade for each case.

Problem 1 Design
Case 1 - D to C to Tip at 0 stroke: F = .57 Tonnes
Case 2 - D to C to Tip @ 25° Rotation (No Rotation about A), (38.5 mm stroke): F = .74 Tonnes
Case 3 - D to A to Tip @ 0°Rotation w/ Tip Fully Rotated 25° (38.5 mm stroke): F = .71 Tonnes
Case 4 - D to A to Tip @ 25° Rotation w/ no D-C-Tip Rotation (38.5 mm stroke): F = .71 Tonnes
Case 5 - D to A to Tip @ 0° Rotation w/ no D-C-Tip Rotation (0 mm stroke): F = .73 Tonnes

Problem 2 Design
0° Rotation (0 mm Stroke): F = 1.53 Tonnes
25° Rotation (63 mm Stroke): F = 2.47 Tonnes
 
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  • #10
JBA
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In recalculating the Case 1 - D to C to Tip at 0 stroke case today using the force vectoring and moment method I gave in my response in your other thread, I discovered that I misread the angle on my protractor between the cylinder C/L and D-C arm as 50° which gave me the: F = .57 Tonnes; The correct angle, as I can best determine it is 40° with a corrected plow tip horizontal F = .43 Tonnes.

For the other cases, I scaled that effective lever length from your drawing so they are not affected by my above angle error but are subject to potential scaling errors, so all of my results should considered approximate.

Edit:
I was concerned about the earlier Case 5 - D to A to Tip @ 0° Rotation w/ no D-C-Tip Rotation (0 mm stroke) result; so I checked it using the lever and moment procedure and the result F = .76 Tonnes is essentially the same effective lever procedure used for the previous post F = .73 Tonnes

That is, even without misread values, one of the problems working by scaling for lengths and measuring angles from prints of your drawings; my calculated value are always going to have a variance from the exactly correct result.
 
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