3D Force Analysis of a Hydraulic Crawler

[ChrisDanger]

I'm busy with a force analysis of a crawler, pictured below, using Excel. I've colour-coded the main parts into (static) body, yoke, [COLOR= #204A87]boom[/color] and nozzle. I've quickly drawn in the cylinders, but there are 3 pairs: the body/yoke pair swivels the yoke, the yoke/boom pair lifts and lowers the boom, and the final pair lifts and lowers the nozzle.

Here's a simplified diagram of the forces, where (generally) the cylinder forces are labeled F, and the reaction forces are labeled R. (I'm neglecting the weights and the nozzle suction force.)

All I need to calculate is the resultant force on the nozzle for activation of the cylinders for any arbitrary position. I have the cylinder forces, and all the displacements calculated for whatever angles the respective hinges may be in. So now I just need the final piece of the puzzle: the nozzle force!

I'm running into several issues. I thought I could do this like a mechanism, using vectors. But with so many unknowns and the complexity of vector algebra, solving for multiple unknown vectors is impossible (it seems). I'm just really unsure how to proceed. Can I disregard the reaction forces and consider the cylinder forces acting in isolation, and simply sum their effect on the nozzle? Or do I need to write out the full equations for each "link", summing forces and moments to yield a full set of simultaneous equations, then solve that system?

Everything I've seen online just has simplified 2D analyses, like for backhoe excavators. I assume this is a fairly trivial problem for something like an FEA solver, but even that methodology is not the right tool I don't think.

 

[desertfox]

Hi

I might of misunderstood but isn't the force to lift the nozzle simply the forces in Fkl and Fkr balanced by the mass of the nozzle acting at its own centre of gravity?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[GregLocock]

Desertfox is taking moments about Rkr_Rkl, that seems fine in the absence of dynamics. I haven't seen a simple tool for 3d force diagrams, the old Working Model 3d is overkill, as are MBD programs like MSC Adams, however

http://www.freebyte.com/cad/dynamic.htm

throws up some interesting possibilities.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[ChrisDanger]

Sorry, maybe I wasn't clear enough. All 6 cylinders could be activated together, so it's unlike a backhoe excavator, which is first positioned then has a single activation point, which is probably what you were thinking.

 

[ChrisDanger]

Thanks Greg, I'll look into that. I see some are open source, which might be useful.

 

[ChrisDanger]

Just some further information, I just need to provide a quasi-static analysis, and I'm hoping to integrate the solution into an existing spreadsheet. So far I have the below, which I've been generating as I go along (mostly to check that it's all working). I'll probably add a sheet with an array of plots showing general behaviour of the nozzle force as the position of the structure changes.

 

[rb1957]

the nozzle force in two (or three) components; one component loads up FBU and FBL, the other FYL and FYR. The loads on FKL and FKR can be determined from moments of the nozzle force about RKL and RKR and then the loads on RKR and RKL can be determined from a free body.

another day in paradise, or is paradise one day closer ?

 

[BUGGAR]

You can do this with vectors in 3d cad. Isolate each cylinder (force vector) in combination with the nozzle and find the resultant nozzle vector for that cylinder. You can write a script to draw a collection of the resultant nozzle vectors vs. the various cylinder positions.

 

[ChrisDanger]

Okay, so that means the superposition of the reaction force from each cylinder would work. I considered this early on but dismissed the idea after a simple thought experiment, although I can't remember exactly why now. Why then do analyses separate the links and find the reaction forces? Is this because they specifically need those reaction forces, or is this not really a mechanism (or one simple enough with which to do a straightforward analysis)? I read through this paper on a backhoe excavator: 689 kB PDF

 

[rb1957]

i think it just depends on how you define the device's functioning envelope.

if you define it as a force at the nozzle, then you work backwards through the structure, and size the actuators accordingly; if you define it as the capacity of the actuators, then you work forwards to find out what force can be applied at the nozzle and ask "is this enough ?"

another day in paradise, or is paradise one day closer ?

 

[pylfrm]

ChrisDanger,

Is the following a fair assessment of the problem?

There are nine independent input variables (rotational position at each of three pivots and force at each of six cylinders) and three output variables (three mutually orthogonal forces applied to some known point on the nozzle).


If so, the following seems like it should work:

Set up an equation for each pivot where the moments of the five forces (two cylinder forces and three nozzle forces) about the axis of rotation sum to zero. There will be three equations and three unknowns, so I'd expect the system to be solvable.


pylfrm

 

[ChrisDanger]

Yeah, I was recently getting stuck on solving the fairly trivial problem, M = r x F, with M and r known, and solving for F. I found a solution where
[tab]F = M x r / r • r + k * r, with k = any scalar.
Hence there are an infinite number of solutions, I imagine in a half-plane, since any force can give an equivalent moment depending on the angle. But after much reading and playing around it finally dawned on me that the force generated by the moment will always be orthogonal to r, which is when k = 0. So finally I can solve this thing with simple vector algebra!

Wow. That took a while. I think I'm a little rusty.

Thanks for everyone's help though. I think it edged me closer to the solution and helped define the strategy more clearly.

 

[pylfrm]

ChrisDanger,

So are you independently determining a force vector for each pivot axis that is applied to the point near the end of the nozzle, is orthogonal to the plane containing the pivot axis and the force application point, and balances the moment of the two cylinder forces? Are you then summing these three vectors to get a total resultant force vector?

If so, I don't think that will work. Have I misunderstood?


pylfrm

 

[ChrisDanger]

Hi. Yes, that would be what I'll be doing. What is wrong with that approach, and do you have any suggestions? It's effectively like fixing everything and activating each cylinder pair separately (which is practically possible), then saying "okay, now turn them all on!" Surely since the system is linear superposition applies?

I found your previous statement (quoted below) a bit confusing. You mentioned 5 forces (2 cylinder and 3 nozzle). If you consider the nozzle as 3 forces (really 1 force with 3 components) then surely you have to consider the cylinder-pair as 6 forces (2 forces with 3 components each)? Or what did you mean?

Previously I calculated the moment equations from M = r x F but this system is unsolvable (3 simultaneous equations with 3 unknowns, yet they yield a zero-determinant matrix) until you introduce a "reality-check" condition. The way I see it a simple moment will generate an orthogonal force, unless you mean that the force will be dependent on the relative angle of the contact surface and maybe the friction forces, but then we'd introduce all kinds of complications and for this analysis I just need to get an idea of the relative maximal forces depending on system pressure and cylinder sizing.

Please set me straight if anything I say doesn't make sense.

Set up an equation for each pivot where the moments of the five forces (two cylinder forces and three nozzle forces) about the axis of rotation sum to zero. There will be three equations and three unknowns, so I'd expect the system to be solvable.

 

[jlnsol]

Hi.
To answer your question about whether it is a mechanism or not: Yes it is a mechanism because it has freedom to move. It is only holded in position by hydraulic pressure in the cilinders.
Talking about hydraulics: I suggest looking into the working principle of the cilinders first before solving the equations in excel.
as I see it: the two cilinders FKL and FKR work as a team in lifting or lowering the nozzle. If these cilinders are not properly controlled by hydraulic valves they can easily work against each other (action in one gives reaction in the other) and giving additional load on the pivot points and torsion in the nozzle structure.
So I recommend to use one valve and a T splitted piping that feeds these two cilinders simultaneously.
That way FKL will be equal to FKR all the time and in your calculations these two cilinders can be treated as just one cilinder.
This principle is also valid for the cilinder pair FBU and FBL: they can also work against each other easily if not properly laid out hydraulically.
I think you do not want to put unneeded load on yoke/boom pivot points and in the yoke/boom structure.
For the pair FYL and FYR applies the same in my view.
How about this?

 

[ChrisDanger]

Hi jlnsol,

I think the way I'm looking at it, where the nozzle is constrained by contact with the earth to yield the applied force, this stucture can be viewed as not-a-mechanism. At least this is the view from a source I encountered during my recent research. In particular:

Indeterminate structures are mechanisms

An indeterminate structure cannot carry all loads and, if not also redundant, has more
equilibrium equations than unknown reaction or interaction force components. Such
a structure is also called a mechanism. The stamp machine below is a mechanism if
there is assumed to be no contact at D.

You make some fair points about the hydraulic system, and thank you for that, but that's unfortunately outside my scope, as I'm just providing a tool for resizing the cylinders. However, his vehicle has been in service for some time, and as it's been modified with extra weight and capacity, the system pressure has slowly been increased, with the result that cylinder seals are now regularly being blown. This has been passed onto me, just to find a way for someone to see the effects of making changes to the pressure and cylinder sizes to get whatever force they think they need at the nozzle. But it's safe to say that other than the above problem the system is perfectly operational.

 

[jlnsol]

Hi Chris,
Impressive picture. I think were on the same page here regarding the mechanism/structure once fixated by hydraulics and acting on the ground. Shall I try to develop a formula for this case in Excel tomorrow? It would then start with the given cilinder forces and the angles variation and resulting in nozzle force achievable.

 

[pylfrm]

ChrisDanger,

I suppose the "five forces" bit wasn't as clear as it could have been. I think of each piston as providing a single force of known magnitude and known direction. For the reaction force at the nozzle, I think of is as being broken up into the three forces of unknown magnitudes and arbitrarily assigned directions. I said mutually orthogonal because that is often the most convenient form for the result, but it probably isn't actually necessary. Perhaps I am making things more complicated than necessary here, but that's what came to mind first.


To illustrate the problem I see with your latest approach, consider the following simplified example:
[ul]
[li]body/yoke pivot will be ignored, and yoke assumed static.[/li]
[li]yoke/boom pivot axis is parallel to the z-axis and located at x = 0, y = 0.[/li]
[li]boom/nozzle pivot axis is parallel to the z-axis and located at x = 1, y = 0.[/li]
[li]nozzle reaction force is applied at x = 2, y = 0, z = 0.[/li]
[li]yoke/boom cylinders apply forces resulting in a moment of (0, 0, 2) about the yoke/boom pivot axis.[/li]
[li]boom/nozzle cylinders apply forces resulting in a moment of (0, 0, -1) about the boom/nozzle pivot axis.[/li]
[/ul]
As I understand it, your method would calculate a reaction force vector of (0, -1, 0) to balance moments at the yoke/boom pivot axis, and (0, 1, 0) to balance moments at the boom/nozzle pivot axis. Add these up and you get a total of (0, 0, 0). However, the x-axis component of the reaction force is actually undefined (positive or negative infinity in the limit depending on direction of approach).

I picked a special case for this example just so I wouldn't have to do any significant calculations. For a better example not involving infinity, perhaps move the boom/nozzle pivot to x = 1, y = 0.1 or similar. There should be a very large component of the reaction force in the positive x-axis direction, which I believe your method will greatly underestimate.


As for the unsolvable system of equations, unfortunately I have no great insights at the moment. Can you get the method to work for a simplified planar version of the problem with two pivots?


pylfrm

 

[ChrisDanger]

Hi. Thanks for the detailed reply. I just woke up (don't worry, it's only 5 a.m. here) and will work through this soon and post my findings/results.

 

[ChrisDanger]

Yes, I see what you did there... Effectively the you're applying moments that result in equal and opposite reaction forces. This would lead to the statically indeterminant case like in my reply to jlnsol above, where there is no contact force at D in the example mechanism pictured. And the way I'm calculating things the resultant force simply becomes zero. But around this point it has values that increase in magnitude at you move away. I don't see any infinities.

Here's what I did (similar to your example but reoriented so as not to confuse myself: I used x longitudinal and z vertical).

Example case:

Changing the Boom moment:

The question is, would this be a problem in reality? I assume if moments did cancel then it would be a transitory case, although if the operator simultaneously lowered the boom and raised the knuckle* he would hardly be trying to maximise any force and would likley be repositioning the nozzle, where there would be no force on the nozzle in any case..

* - Sorry to have introduced the knuckle terminology so late. I was starting to confuse myself with the ambiguity of referring to both the final structure/axis and the resultant force as the nozzle

I had a day off yesterday but will be working on this throughout the day and am quite confident I can get a meaningful answer now. I think the most complicated thing will be doing the coordinate transformations between hinge axes to get the nozzle forces, but I'll probably either post back with more problems or hopefully just some results.

Of course if anything I posted here looks odd/wrong please let me know.