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]

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

2.5W=2.5*W

 

[rb1957]

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

 

[LOKI1983]

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.

 

[LOKI1983]

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.

 

[chicopee]

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.

 

[LOKI1983]

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.

 

[rb1957]

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

 

[LOKI1983]

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.

 

[flash3780]

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.

 

[LOKI1983]

flash3780 the mass of the lifting arm is 80 kg.

 

[LOKI1983]

In order to avoid any confusion anymore I had attached the picture of the equipment which was designed.
Thank You.

 

[rb1957]

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

 

[LOKI1983]

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.

 

[rb1957]

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

 

[LOKI1983]

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.

 

[oq172]

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!!

 

[rb1957]

"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

 

[LOKI1983]

oq172...tomorrow first morning I shall post the CG location.

 

[LOKI1983]

@rb1957 ...the picture with the truck lifted.

 

[oq172]

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.........