Advice on drawing the FBD (Free Body Diagram)

[LOKI1983]

Hello.
Sorry for the foolish question but I am somehow got lost and feel like cornered in solving a simple problem.
I had attached a picture: the "boomerang" look like thing is a arm which is fixed at A (pin type connection) (allows the arm to rotate around A),in point E we have a hydraulic cylinder, and in point D we have the actual load.Actually when the hydraulic cylinder is applying force to point E, point D is moved horizontally left (in point D we have a wheel which is moved on a flat surface) , while the A is moving the Up (A is attached to a load - in order to avoid further questions point A can move up and down and rotate,right and left is constrained). I am trying to figure out if my hydraulic cylinder can lift the load.My question is: when the hydraulic cylinder is applying the force the point of action where it is: at the intersection of the direction of force (B) on the line AD or at the intersection of the perpendicular on line AD ( point C)?
Thank You.

 

[zekeman]

You can't get a FBD immediately on the arm A-B-D since you have no vector direction at A .
You must first get a FBD on the total structure where the external forces are the weight (vertical force), the vertical force at the wheel bearing,Wy and the the force at F which must be vertical since the 2 other forces are vertical. taking moments about F and setting to 0 will give Wy from which you get Fy=F.
Next you do a FBD on the yellow link taking moments about A . Setting =0 you now get the force in the cylinder.
Now you can do A FBD on the A-B-D link, since you have 2 of the 3 force vectors, the cylinder and the Wy.

 

[rb1957]

@zeke, i agree about the overall FB, but we don't know (as far as i can tell) where the ground contact D is and we don't know Fx ... it might be zero, or it migt be FxD =-FxF

we don't know the line of action of the shock absorber (it's pivoting about E), but we do know the link is pivoting about A.

i suspect in the lowered position the shock absorber is unloaded (that the link is resting on a stop on the yellow beam).

a simple approximation may be to say work done by the shock absorber (in lifting the weight) = work done on the weight.
by geometry you should be able to figure the extension of the SA.

Quando Omni Flunkus Moritati

 

[zekeman]

"@zeke, i agree about the overall FB, but we don't know (as far as i can tell) where the ground contact D is and we don't know Fx ... it might be zero, or it migt be FxD =-FxF"

Fx must be zero, since the 2 remaining forces, W, the weight and Wy on the overall system are clearly vertical. statics 101

What do you mean that you can't find the ground contact. Obviously we are talking for a given extension of the cylinder and the OP has to look at all the positions of the cylinder.

 

[rb1957]

"Obviously we are talking for a given extension of the cylinder" ... i guess it wasn't obvious to me, particularly as there isn't a dimension for xAD ... i expect the OP knows this given his beam model.

as for Fx ... of course, you could say the wheel releases Fx, but when D is below A, then the SA force is zero if FxD is zero and i'm uncomfortable with that ... but maybe that's what happens (as the SA load changes from compression to tension)

certainly if FxD is zero, it's a much easier problem to solve ... FyD from overall free body (getting D from the CAD model), the line of action of the SA from the CAD model, three forces intersect. i suspect that as D gets closer to A, the load in the SA will decrease, as the reaction at A approaches the load at D.

Quando Omni Flunkus Moritati

 

[flash3780]

When it comes to free body diagrams, I find it easiest to keep my signs straight if I draw all of my arrows in the positive x and y directions. This one's pretty straightforward, but in more complicated analyses things can get confusing if you try and decide what's positive and negative in advance.

http://files.engineering.com/getfile.aspx?folder=8e2d9f95-59f6-4f12-b289-54212e738f77&file=FBD.png

I think you could make a similar FBD for the yellow L-shaped part and you'll have enough equations to solve for the actuator forces as a function of the actuator length.

 

[zekeman]

Rb,
Why is it so difficult for you to understand that Fx=0 under ANY scenario?
3 force vectors on the body, 2 of which are vertical so the 3rd force vector MUST be vertical.

LOKI,
BTW, I did the problem using the first drawing for the first case and got
Fcyl about 2.5W. Also did it by kinematic analysis and energy conservation and got 2.8W.

 

[oq172]

Zekeman - where did you get the dimensions of the assmebly and link to come up with 2.5W? Sorry, maybe Loki posted them earlier and I just missed it.

Thanks,

OQ172

 

[dhengr]

I’m sure glad I didn’t butt in on this one on 26FEB13 @ 13:43.

Sounds like LOKI should put down his CAD and 3-D modeling programs for a few hours and dig out his Engineering Mechanics, Strength of Materials and Machine Design text books and study them for a while.

 

[zekeman]

"Zekeman - where did you get the dimensions of the assmebly and link to come up with 2.5W? Sorry, maybe Loki posted them earlier and I just missed it."

I assumed the drawing was to scale. Sorry, should have said so.

And I second Dehener' comment. You don't need a steamroller to kill a fly.

 

[LOKI1983]

zekeman...my heart just stopped ((.
I don't have now the time for the calculation,but i shall do it for tonight.
What do you mean Fcyl about 2.5W.
Ex: W=1000N
Fcyl=1000/2.5=400N
Fcyl=2.5 *1000=2500N????

 

[oq172]

Loki - Scaling your image for the dimensions in the same fashion that Zekeman did, I came up with 3.07*W. Please see the attached analysis.

Why did you say "zekeman...my heart just stopped ((." ???

 

[jlnsol]

Loki: if you ever want to check whether your cacluations are correct, I can do that easily for you. I use a piece of mechanism design software for that, see attached pdf as example of your mechanism. Dimensions and masses are estimates for now, but yet already it can be seen that the hydraulic cylinder force varies by roughly a factor 6. See the red curve with values (N) along right vertical axis and the blue handwritten values for the two end positions: cyl in and cyl out.

 

[oq172]

jlnsol - Looks pretty cool - what's the name of this s/w package?

 

[LOKI1983]

dhengr indeed you are right...I need to read some books to remember what I had learned in faculty.
oq172 Because my hydraulic cylinder is rated to work at 200 bar, i have a diameter of the piston of 125 mm this means that one hydraulic cylinder is capable of 245 kN.I have 2 hydraulic cylinders and I have to lift about 240 kN also....so 240 kN I have to lift with 2 hydraulic each rated at 245 kN (total 490 kN). By your results this means I can't lift the load and my hydraulic cylinders are already in progress.I am so deep in s**t.
What can I do? ((

 

[LOKI1983]

jlnsol what's the name of the software?
Tonight I shall post the real world problem with all dimensions.I am really in need for help.
Thank You All for your assistance !!!

 

[oq172]

Loki - don't panic .

-You need to 1st agree on the analysis the Zekeman, jlnsol and myself are suggesting. Seems as though we're roughly in the ball park with only approximtely dimesnions and rough scaling to work with.

-Later today, please submit the actual dimensions of your system. We can then recalculate and see what the actual situation is.

-Finally, once we are all in agreement on the math, put the equations in a spreadhseet and begin to vary some of the link parameters. You might be surpirsed to find that a small change in length or angle could significantly alter the cylinder force.

BTW - I was noticing some of the semi rude and pompus comments left by folks on here. I would just ignore it. This formum, in so far as I can tell is set up to help folks out, not to put them down. Seems like you're on critical path with your design, the last thing you need is to deal with arrogant people.

good luck - I'll check in later.

OQ172

 

[rb1957]

"What do you mean Fcyl about 2.5W.
Ex: W=1000N
Fcyl=1000/2.5=400N
Fcyl=2.5 *1000=2500N????"

oh dear, oh dear, oh dear ... Fcyl = 2.5W is pretty clear, and cannot be construed as W/2.5 ???


Quando Omni Flunkus Moritati

 

[LOKI1983]

rb1957 sorry but for now everything I see is only negative feedback.I am down with the morale.

 

[jlnsol]

Loki, it is SAM mechanism software from ARtas see here:

http://www.artas.nl/en

I use it professionally in my engineering tasks.
If you post the real dimensions and masses I return the exact cylinder force curve. No problem.

 

[flash3780]

LOKI1983, did you see my example FBD? Since the sum of the forces and moments for a static free body must be zero, you should be able to come up with some equations relating the forces (e.g. sum of forces in the x-direction must equal zero, sum of moments about any point must equal zero).

I think that if you make a similar free body diagram for the L-shaped part, you'll have enough equations to solve for the forces at the various points and in the actuator.

Once you have a system of equations, you may need to use matrices to solve them.

When drawing a FBD, I always start by drawing all of the arrows in the positive x and y direction and let the signs fall where they may. Once you've solved the problem, you can express the arrows in their proper directions.

I think that your goal, at the end of the day, is to relate the force on the actuator to the length of the actuator, determine the maximum, and size your actuator accordingly (with a factor of safety, of course). There's some trigonometry required to relate the length of the actuator to the angle at which the force is acting, but that's doable.