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Advice on drawing the FBD (Free Body Diagram)

Advice on drawing the FBD (Free Body Diagram)

Advice on drawing the FBD (Free Body Diagram)

(OP)
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.

RE: Advice on drawing the FBD (Free Body Diagram)

To me it there are two moments working on the boomerang. One is the hydraulic cylinder force times the perpendicular distance between cylinder centerline and point A. This moment is working clockwise.
The other moment is the vertical force in point D times the perperdicular distance between D and A. This one is working counter-clockwise. For every position during rotation of the boomerang around 'pivot' point A above moments are equal, so the system is in balance. During rotation both the perpendicular distances and the required cylinder force are changing.

RE: Advice on drawing the FBD (Free Body Diagram)

thx for the very detailled description ... it is something we don't often get here.

However ...
if the link is pinned at A (please don't use the word "fixed" 'cause that means something else) then the link will rotate about A (and D won't move horizontal), and
"point D we have the actual load" is followed by "A is attached to a 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)?" i think the answer is yes ... this is a three force body, so the forces intersect at a point. From the text description, if A is in a slot (allowed to move vertically but restrained horizontally) and the load at D is vertical (ground contact without drag/friction) then the force through E would be through the intersection of these two forces.

However, the end of the shock absorber (at E) is going to be constrained somewhere, presumably pinned, and i think this point needs to be considered. Since point E is moving it's hard to see how the line of action of the shock absorber can be described that way (in reality).

Also "I am trying to figure out if my hydraulic cylinder can lift the load" ... I imagine this link is like a trailing link for a vehicle, the ground is pushing up (against the weight of the vehicle) so the shock absorber is pushing down against the link (not up, trying to lift it).

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

maybe a way to look at it is the line of action of the shock absorber is through the point (xD, yA), yes?

at a prescribed angle (looks artifical to me).

so now you've got a locus of points for E. you also know the distance between DE (and AE) ... ie an arc from D will intersect this line at E. and another arc from E, together with the angles between DE, DA, and AE, will define A.

now you've determined the link's position you can sum moments about A (remember E does not need to be aligned with A) to determine the required shock absorber force.

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

I believe you have a very simple problem equivalent to a bulging ladder against a wall. An oblique force on the ladder ( the cylinder) causes the ladder to slide along the base and up the wall.And there is a force downward along the wall (the load).The force from the base (at D) is vertical since you have a wheel rolling horizontally.
You now have 3 unknown forces , namely the cylinder,the vertical force at D and the horizontal force at A.
You should be able to write the 3 equilibrium equations for this.
sum of horz forces=0
sum of vert forces =0
Moment about any point = 0




RE: Advice on drawing the FBD (Free Body Diagram)

i think the biggest part of the problem is to find the distance between the points (since the lines of actions of the forces are defined).

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

LOKI:
I agree with Rb, that you did a good job of defining you problem, and that’s really good to see, and half the battle. The sketch is particularly good because as Rb suggested, you have to be very careful with the exact meaning of some of the words you used in your word picture. This is also sort of a kinematics problem, a study of motion problem, isn’t it?

I think you need a few more pieces of info. on you sketch. You need a point ‘G’ which is the pinned end point on the back end of the hydraulic cylinder. This will help you define the line of action of the cylinder. I also assume that points A & G are fixed to the same machine frame so they can’t translate in the X or Y directions w.r.t. each other, but they are pinned so they can rotate. Then you need the X & Y coordinates of points E, D & G, assuming point A is the origin; also, Y is positive upward from point A and X is positive to the right from point A; Z is positive out of the page toward you. We also know that points E & D rotate about point A, at their respective radii from point A, and we can now calc. all of the needed dimensions. The angles btwn. the various lines of action now fall out.

Now, you can rotate point E by 5̊, and it will define the location of point D on its circle. Alternatively, you can move point D at increments and define the location of E. As Zekeman suggested sum the H & V forces, and calc. the moments about point A. Note that the lever arm for the vert. force at D is the horiz. distance btwn. A & D; and the lever arm for the cylinder is the perpendicular distance from its line of action to point A. Both of these lever arm lengths are always perpendicular to their lines of action, and measured to A.

Finally, you have a somewhat suspect mechanism here, in that, as the cylinder lever arm decreases much more this will become unstable, and snap through, at the extreme when the line of action of the cylinder gets near the line btwn. A & G. You probably want to put a stop block on the machine frame, so that the cylinder lever arm never gets too small.

RE: Advice on drawing the FBD (Free Body Diagram)

1) draw a FBD of the whole assembly ... loaded by the red weight, reacted at two points ... i think Fx = 0

2) it not clear how point A is constrained ... you said it was constrained to move only vertically ... doesn't look like that ...

3) ah, the problem with 1) is that you don't know the precise geometry, where is W ? it now looks like points A and C are fixed (onto the yellow piece), yes?

the problem is you need the spring curve of the actuator. it changes length with load, this changes the geometry of the ground contact point, ...

4) then you want to lift the weight, pic2. i think you need to consider the work done by the actuator against the weight. actuator work done comes from it's spring curve.

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
Hello rb1957...
indeed A and C are pined to the yellow structure.
I know the center of gravity of the entire structure and from this I know which is the vertical reaction in B ( lets say Wy=1000N) and F (Fy=1000N). From this I can make the analysis of the cylinder separately by drawing the FBD of the lifting arm.I just confused where the force of the cylinder is acting: is it at the intersection of the line connecting the A and B? look at those 2 drawings sketched on the first picture.
If i have the answer for this , the rest is just easy: sum M A -FEcos alfa * c -Fsin alfa * a + Wy(d+c)=0 ???corect???
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
sorry for the mistake: it is not sin or cos alfa but the sin or cos of the angle between the direction of the cylinder force and a horizontal line.

RE: Advice on drawing the FBD (Free Body Diagram)

i guess it depends on how precise you want to be.

you might know where the load is acting (the cg), but you don't know "accurately" when the ground contact point is, 'cause it depends on the actuator force (which you haven't determined yet) so you can't "accurately" calculate the ground contact force. maybe you know the angle the arm makes with the ground ?

there's going to be some load on the actuator in the down position (unless the load is taken by a stop ...
and then there's additional load to raise the yellow beam
and that determines if the actuator is strong enough

i think you're missing a ground friction force ... in pic2, the actuator is pushing the tire into the ground, creating a Fx. without this you get to the odd situation when D is below A that the actuator force goes to zero !?

what defines the raised position ? the top of the yellow beam being horizontal ??

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

Hi there,

Please see the attached solution to your link problem. I assumed your were looking for the Force (Fe) from your cylinder to lift (or balance) a load at 'A' in the vertical direciton. I called this load (Ray) in the attached analysis. You will need to provide the values Xae, Yae, Xad from your geometry.

Also, just some advice on your diagrams, please draw them to correctly represent the contraints.

Please see the attached and let me know if you have further questions.

Best,

OQ172

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
rb1957 I have the 3D model of the analyzed structure at any given angle I can find the required measures.I also I can find the angles which the arm makes with the ground at minimum stroke (closed cylinder) and maximum stroke (maximum opening of the cylinder)?

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
oq172 thanks for explaining ... I am just getting confused, I need to drop the idea of intersecting points of action? It is enough to know the distances between the points and angles?
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

@oq172 ... loki has changed the problem (i believe) so that A is now a pinned joint (not in a slot). loki can you confirm ... is A a pin or a roller in a slot (in the yellow beam) ?

@loki ... the link is a three force member; the forces intersect at a point. as for the ground reaction, consider the forces when D is directly under A ... the actuator force goes to zero ('cause there's no other Fx to balance it). consider the situation just before D is under A ... the reaction at A has a small Fx component balancing the large actuator force, so the reaction at A is very large. this ground reaction, FxD is balanced by the difference between FxA and FxC, and this force also shows up at F ... keeping everything in balance.

now, if you're not being "accurate" then Wy is almost constant (at W) unless the yellow beam's weight is significant. not sure if this gets you very far ??

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
rb1957..indeed A is a pin connection.
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

@ Loki1983 - Yes, I think it's a pretty straight forward problem and you should be able to sum forces and moments to arrive at the correct answer. I am not sure about trying to find out lines of action and intersecting points. For now, try shifting gears a bit and take a closer look at what I attached in my earlier response. I am fairly certain that will get you most of the way there. Even if you don't explicitly solve for the reactions the way I did, at least consider setting up the FBD and the Force and moment eqations in that fashion.

@rb1957 - thanks for taking a look. With regard to 'A' now pinned, I still think you can represrnt it the way I have shown as long as the 'yellow' link in Loki's diagram is sufficiently long compared to the smaller link. In this fashion, 'A' will move mostly vertical with very little horizontal movement. I will take closer look tonight and confirm.

Best regards,

Oq172

RE: Advice on drawing the FBD (Free Body Diagram)

i don't think it's that easy to solve.

the actuator extends under hydraulic pressure.

in the initial position, the ground force may be vertical, in which case it's easy to solve the FB, knowing the CG of the yellow beam and weight (and FxF = 0)
If you know the geometry well (and don't rely on the actuator load/extension to balance things).
knowing the ground force FyD you should be able to solve the link (three unknowns, FxA, FyA, FA, three equations of equilibrium)

mind you, i think there's FxD (acting to the right, opposing the actuator) and that makes this indeterminate.

in the raised position, you can do the FB again but now i really think there's an FxD (i see the actuator pushing the link into the ground).

remember FBD are in static equilibrium, and we're talking about a moving link so there's a dynmaic thing to be considered. granted the effect is probably small.

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

@rb1957

As with most things you have to make the best reasonable assumptions and shoot for a good approximation in your results. I would think Loki1983 would want to get close in his cylinder sizing and then take the rest emprically once he builds his prototype

You say the link is being driven into the ground, I agree, but the roller should reduce alot of the horizontal force. You could add in rolling friction, but that would be a function of Rdy, in which case you would add another equation to solve for the additional unknown.

I also would not worry about the dynamic thing right now and until there is clarity on the staic condition. I would be inclined to get all the geomtery from the CAD model for Xae, Yae,Xad & the angle of the piston for 5 or so positions spanning the complete range of motion for the mehcanism. Then, in a spreadsheet calculate Fe, Rax, RDx(if you include rolling friction)and finally Rdy for all 5 positions. This will give a nice representation of how the forces are changing as funciton of the different positions.

I'll chew on this more tonight.

OQ172

RE: Advice on drawing the FBD (Free Body Diagram)

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.

RE: Advice on drawing the FBD (Free Body Diagram)

@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

RE: Advice on drawing the FBD (Free Body Diagram)

"@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.


RE: Advice on drawing the FBD (Free Body Diagram)

"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

RE: Advice on drawing the FBD (Free Body Diagram)

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=8...

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.

RE: Advice on drawing the FBD (Free Body Diagram)

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.






RE: Advice on drawing the FBD (Free Body Diagram)

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

RE: Advice on drawing the FBD (Free Body Diagram)

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

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.

RE: Advice on drawing the FBD (Free Body Diagram)

"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.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
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????

RE: Advice on drawing the FBD (Free Body Diagram)

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.

RE: Advice on drawing the FBD (Free Body Diagram)

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

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
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? :(((

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
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 !!!

RE: Advice on drawing the FBD (Free Body Diagram)

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

RE: Advice on drawing the FBD (Free Body Diagram)

"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

RE: Advice on drawing the FBD (Free Body Diagram)

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

RE: Advice on drawing the FBD (Free Body Diagram)

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.

RE: Advice on drawing the FBD (Free Body Diagram)

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.

RE: Advice on drawing the FBD (Free Body Diagram)

LOKI
"What do you mean Fcyl about 2.5W."

2.5W=2.5*W

RE: Advice on drawing the FBD (Free Body Diagram)

some practical questions ...
do the actuators have enough stroke ?
can you fit two of them into your structure ?
can the actuators react tension ? what happens if D moves to the other side of A ? it can't if there is
1) a stroke limit in the actuaor (it has "bottomed out")
2) mechanical stops preventing the link rotating further.
i'd suggest that the link contacts stops in the down position, so that there is no load in the SA and so that the yellow beam would be rigid.

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
rb1957 there are only 2 positions of the hydraulic cylinder: closed cylinder and maximum stroke.
This means that the arm must remain always in quadrant 4 which by design it is.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
As I had promised this is the the real sketch of the problem.
Node E can rotate,but is constrained along X and Y
Node B can rotate and move along X, but in Y direction can't because of the load.
Nodes A,E can rotate but are fixed to the structure (black line) and will move in X and Y direction only accordingly is the black line changes its angle.
E and D is the connection of both ends of the hydraulic cylinder.One end of the hydraulic cylinder is connected to the structure (black line) and the other end to the lifting arm.
A,D,B is the lifting arm.
Hope this is explanatory enough.
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

The problem is more in the realm of dynamics than static analysis. You should consider the inertias of the boomerang element, rollers, load and the frame supporting the load as well as the frictional forces from bearings and the rolling resistance. You will need to estimate an acceptable either linear or angular accelerations in order to carry out this dynamic problem.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
By the calculation which where made from the file delivered by oq172 I had obtained that I need a cylinder force of aprox 300kN. :((((
Can someone else do the calculations and tell me if I my calculations are correct?
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

by that geometry i get actuator load = 1.3 FyD
(simple enough three force body)
FyD intersects the actuator line of action at (566,428.3)
line of action of reaction at A is through this point.

once you know the directions of the three forces, you can use a force polygon to solve magnitudes (the OP seems to like graphical methods)

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
chicopee indeed I agree with you.
But for now I need my statical analysis in order to see if from this very basic point of view I am wrong. Tomorrow I need to do something, I hope I shall not do something foolish in panic.

RE: Advice on drawing the FBD (Free Body Diagram)

LOKI1983, you should add the location of the center-of-mass of your weight to your diagram. The picture from 26 Feb 13 13:49 shows the weight as a box... the CM would be in the center of the box (which is proud of the platform by a bit). Also, you probably want to consider the mass of the L-shaped piece... it looks like it might weigh a lot.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
flash3780 the mass of the lifting arm is 80 kg.

RE: Advice on drawing the FBD (Free Body Diagram)

and the CG would be where ?
of the truck ?
of the load?

i assume you have two links, one on each side of the truck, both driven (since you need two actuators). i guess some care needs to taken to synchronise these two.

how are you lifting the back of the truck ? (or someone else's job ?)

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
rb1957...both hydraulic cylinders are synchronized by a mechanical linkage.
The back of the truck is lifted by another 2 hydraulic cylinders.
The CG i can deliver only tomorrow, because the person who has the data gave me only the load repartition for the front bogie (22 tons) and rear bogie (12 tons).
I know that there are a lot of unknowns,but for my boss this does not matter.I am also the only engineer and the project manager for the this project which is a total aberration, but can I say something?
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

we can help with the analysis without knowing the ground load ... i got actuator load = 1.3*FyD ... anyone else ?

what's the link geometry with the truck lifted ?

my concern is that i see the actuator load, from static analysis, decreasing which sounds wrong. the way i think it's working is with constant hydraulic pressure (which would mean constant load in a simple actuator, no?). but maybe you can look at the work done comparison ... the initial work done by the actuator is Pact*(dy/sin(23.5deg)), and the work done on the load is W*dy ... yes?

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
what's the link geometry with the truck lifted ?
I don't understand the question...
The hydraulic system is a pressure sensing load - the variable displacement pump shall deliver the flow in accordingly with load.

RE: Advice on drawing the FBD (Free Body Diagram)

Hey Rb1957 - How can you possibly do any meaningful analysis without knowing the center of gravity of the most significant load??

Guys - I really want to help here, and believe me, this IS a simple analysis. What is making this so atrocious is the lack of geometric detail. It seems as though the details of the loads and constraints are just dribbling out with each posting. This is not the wayt to do it.

Loki - Please look at the anlysis I submitted this morning and put REAL numbers in for all the lengths and angles that I assumed. I got these values from scaling your original picture; you know, the one where you show a red block for the load. Now you're saying that the C.G of that load isn't known????

Guys please!!

RE: Advice on drawing the FBD (Free Body Diagram)

"what's the link geometry with the truck lifted ?" ... you've given us the geometry of the link with the truck in the low position.
as the actuator extends the link rotates about A, the co-ords of the points D and E change, the angle of the actuator changes ... if you could draw this in CAD then it'll help.

@oq172 ... i have calc'd the actuator load in terms of the ground load (using today's pic, sent at 12:03), it is (as you say) a simple analysis ... there is a geometrical relationship between the loads. we know the line of action of two of the loads, so we can calc the loads in terms of one another. for whatever the ground load, PyD, is the actuator load is 1.3*PyD.

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
oq172...tomorrow first morning I shall post the CG location.

RE: Advice on drawing the FBD (Free Body Diagram)

Loki - Please see the attached.

This is a plot showing the ratio of Fdc (cylinder load) to the c.g. location of the weight (W). This will give you an idea of what the cylinder force will be as the c.g is moved from 0% (right over the pivot F) to 100% (at the location shown by your RED block).

To get the cylinder load, muliply the Fdc/W ratio by the actual weight W at the corresponding c.g. -

For example.....

if your c.g. is 100% (128mm in my scaling image) away from the pivot F, Fdc/W = 3.07

If you're at 50% (64mm in my scaling image) away from the pivot F, Fdc/W = 1.50

See things are looking alot better, right........ah, but you need the c.g.!!!!!!!!!!!

please also refer to my attachment posted at 7:06 today.........

RE: Advice on drawing the FBD (Free Body Diagram)

in the raised position i get Pact = 0.09FyB (i hadn't noticed the OP changed the contact point label)

i'm surprised the sketches aren't to scale ... the first one (for the lower position) looked pretty reasonable, the one for the raised position is clearly not to scale ("50" vs "250") ... i imagine you've got the data from CAD, and are using sketches you made earlier.

i'd expect FyB to be slightly lower for the raised position (the two supports are further apart)

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

LOKI,
Calm down,

First of all, the worst case for load on the cylinder is in the low position and the "answers" you got so far are probably way off the mark.
For example we ( those who worked the problem) scaled your original drawing which turns out to be fiction,since we really didn't know where the the CM was located. If we were off by a little bit, the answer could easily have been

Fcl=W

which is the case if the CM is in a vertical line with the wheel bearing.
If this isn't the case then relocating the load or the cylindrical arm may mitigate the problem.

In any event, we have all been through similar circumstances and have survived and you will too; if necessary,we can certainly help you get through this.



RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
zekeman I hope from all my heart that you are right and I am wrong.

RE: Advice on drawing the FBD (Free Body Diagram)

Hello Loki, I have good news for you! Your cylinders are OK !
Based on your two sketches 13959877 and 68396129 and the known reaction force in B 120kN per boomerang I modelled the system into the SAM software.
The maximum required cylinder force is 168 kN (so app 140 bar).
See the attached results in pdf form.
So you can continu production of your cylinders.

BTW: Don't know why I am doing this, since it costed me 3,5 hours...

RE: Advice on drawing the FBD (Free Body Diagram)

jlnsol - Sorry, seems like you wasted 3.5hrs. - Look, I don't know why people insist on providing solutions when the position of the C.G is not know. This is the largest load in the system and it is NOT a known location. You say you calculated the cylinder force from the ground force Rby = 120kN - is this true? Then where did the 120kN come from? Who provided that?

I am truly sorry if I missed something, but Rby = 120kN is pure rubbish until someone can tell me where it came from, and they won't be able to until LOKI publishes the C.G. of the 1000N load.

Is the load really even 1000N (220 lbs) this seems kind of light for what I see in the picture http://img811.imageshack.us/img811/356/railroadtru... that LOKI posted.

RE: Advice on drawing the FBD (Free Body Diagram)

Don't think so oq172. Forget about the 1000N !

Yesterday 14.08 Loki provided the front bogie reaction load of 22 tons, so 215 kN.
In his two sketches he calculated with 120kN reaction force in B.
Thus two times 120kN = 240kN. That is even 2,5 ton more so a good and safe value to calaculate with.
So with this known reaction force B the CG is not nescessary any more for the calculations.
The mass of the truck and the load of the truck can very good be 22 ton + 12 ton = 34 ton in total.
In my calculations I entered a mass at an arbitrary postion that caused the reaction of 120kN. Peroid.

The load of 1000N mentioned in Lokis post at 26-02 15:11 hrs is per example ('lets say')
And also post 28-02 4:14hr is 'Example'

But...let Loki himself confirm.

RE: Advice on drawing the FBD (Free Body Diagram)

the solution for the link depends on FyB (the ground contact load). this can be determined from the external loads.
that's why i solve the actuator load in terms of the ground load, using simple geometry.

it looks like jlnsol came up with a different solution ... 168 = 1.4*120 ?

E' 566 428.2843054 this is the point of intersection of the two forces B and SA (i used the angle 23.5deg ... the dimensions give 23.46deg)
B 566 0
A 0 656 610.0905159 the length AE'

x y
SA -0.917060074 -0.398749069
B 0 1
A 0.927731189 -0.373249032 0.927731189 A = 0.917060074 SA ... sumFx
direction cosines of the three forces 0.373249032 A + 0.398749069 SA = 1 B ... sumFy
0.373249032 A = 0.368955781 SA ... (1)*0.373249032/927731189
0.76770485 SA = 1 B ... subs (3) into (2)
SA = 1.30258393 B
this was maybe 15min work

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

the difference is the angle of the actuator ...
i used the 23.5deg given (from dimensions its 23.46deg)
i think your upper actuator point is a little off ...
your (1500,776) should be (1495,831.5) from LOKI's post 12:03 28 Feb.

sure the difference is small, but the problem (and the math) is simple

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
Hello.
Unfortunately the person with CG data is on a vacation, and I can get the data only Monday.
Hope that you will have the patience.
I wish you a good weekend.
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
@jlnsol...the truck with cargo has maximum 34 tons.Truck (10 tons) + cargo (16 tons) + some auxiliaries (8 tons) = total weight 34 tons.
They had calculated that with regard to all CG (truck,cargo,auxiliaries) when the truck is lifted on the 2 bogies the weight distribution is : front bogie 22 tons, rear bogie 12 tons.
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

LOKI - Thanks for helping to clarify the load distribution. Sorry I missed it earlier.

I have a question, if the left side is 22tons (215kN) - how did you arrive at Rby =120 kN. I am sorry if I am getting confused. Should Rby = 22 tons (215kN) ? - Just checking.

RE: Advice on drawing the FBD (Free Body Diagram)

two front bogies ... 215kN/2 = 107.5kN ... 120kN is a little conservative, maybe assuming not 50/50 distribution, 55/45 would be 118.25/96.75; so 120/95 sounds reasonable ?

Quando Omni Flunkus Moritati

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
Hello Indeed:
I just wanted to do the calculation for one side.
Meaning: 1 cylinder = 245 kN, for the left side I have 22 tons - calculating for 1 cylinder will be 22/2=11 tons.I made some rough assumptions because I don't have all the data so I choose Rby=120kN.If Rby=22 tons than take in account the force developed by 2 cylinders which is 490kN.
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
Hello.
I found out where is the center pf gravity. It is 1815 mm away to the right from A.The total mass of the truck is 33 tons = 323619 N. The force developed by hydraulic cylinders is 490625 N.
From the first look I don't think the cylinders can lift the load.
Please advise.
Thank You.

P.S.: I shall work right now to make a better sketch of my system in order to avoid any confusion.

RE: Advice on drawing the FBD (Free Body Diagram)

Hello Loki,

This doesn't change much. The cylinders are strong enough. See also my previous post at 1 March 5:23 . I updated the calculations with your latest c.g. info and the largest cylinder force is now 172 kN per cylinder (per side).
The cylinder stroke is app 316 mm. Horizontal displacement of B = 497 mm.
Since a cylinder can deliver 245 kN @ 200 bars, they are OK.

RE: Advice on drawing the FBD (Free Body Diagram)

Loki,
Why the pessimism,
I basically agree that you have ample cylinder.
From your new CM , I got
Fby=239Kn at the wheel bearing
and since Fcyl=1.3Fby, I got
Fcy=310 Kn total for both cylinders and
155Kn for each.
In practice you should add the effect of bearing friction but in this case your cylinders are so ample that you should be good to go.

RE: Advice on drawing the FBD (Free Body Diagram)

Loki - Your c.g. location along with the known weight of 33 Tons (323.6 kN) yield the following in my calcs:

Rby - Total ground force at B = 243 kN

Fcycl (tot) - Total force at the cylinders = 316.5 kN

Fcyl (ea.) - The force required by each cylinder =158.3 kN

So, as others are also saying, you're in good shape. What are you concerned about. Is there more you're not telling us??? winky smile




RE: Advice on drawing the FBD (Free Body Diagram)

(OP)
Guys what I am doing wrong in my calculations?
The summation of moments of Fx component and the Fy component of the hydraulic cylinders around pint A (clockwise) is lower than the moment of By at A (counterclockwise).
Is there a problem that I tried to solve the linkage putting the conditions that A can move vertically and B only horizontally?
Can someone show me the FBD they developed after we knew the CG?

oq172 - I have nothing to hide, just the shame that I can't solve a simple FBD (some years ago I was good at it) :((
Thank You.

RE: Advice on drawing the FBD (Free Body Diagram)

"A can move vertically" ? ... this was discussed (at length) previously and determined that A is pinned and not in a vertical slot.

the CG is used only to determine the ground reaction, at ptB.

the link is a "three force body", all the forces pass through a single intersection point. If we know two directions (actuator laod and ground load) then we can figure out the third (reaction at ptA), and draw a force polygon. once we know one of the forces, the triangle shows us the other two. the math way is solve sumFx and sumFy (since you know the directions of the forces, the only unknowns are the magnitude of the two other forces ... you know one force, the ground contact force at ptB; and you can use components to sum in x- and y-.

now if pt A is in a slot, then your geometry won't solve. But "all" that'll happen is the link will rotate a more to allow the actuator force to pass throu the intersection of the vertical ground force and the horizontal reaction at ptA.

but, aft 80+ posts, we should be able to state with some assurance whether point A is able to translate or not ... sigh

Quando Omni Flunkus Moritati

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