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.

 

[jlnsol]

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.

 

[rb1957]

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

 

[rb1957]

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

 

[zekeman]

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

 

[rb1957]

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

 

[dhengr]

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.

 

[LOKI1983]

Hello.
I had attached bellow a detailed sketch of my problem.
Remember: I just need to find the reaction in the cylinder, so I'll separate the arm and do the FBD like in picture1, i just need to know which FBD I draw is correct?
Thank You.

 

[rb1957]

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

 

[LOKI1983]

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 [sub]A[/sub] -F[sub]E[/sub]cos alfa * c -F[sub]sin[/sub] alfa * a + W[sub]y[/sub](d+c)=0 ???corect???
Thank You.

 

[LOKI1983]

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.

 

[rb1957]

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

 

[oq172]

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

 

[LOKI1983]

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)?

 

[LOKI1983]

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.

 

[rb1957]

@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

 

[LOKI1983]

rb1957..indeed A is a pin connection.
Thank You.

 

[oq172]

@ 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

 

[rb1957]

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

 

[oq172]

@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