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.

 

[rb1957]

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

 

[zekeman]

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.

 

[LOKI1983]

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

 

[jlnsol]

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...

 

[oq172]

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/railroadtruck.jpg

that LOKI posted.

 

[jlnsol]

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.

 

[rb1957]

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

 

[rb1957]

well that didn't post well ... try this

Quando Omni Flunkus Moritati

 

[jlnsol]

rb 1957,
it would be fine if you enter the dimensions like I used in this sketch into your calculations, just to check
on the other hand 1,3 or 1,4 times RB is fair enough for Loki to give him a go ahead isn't it?

 

[rb1957]

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

 

[LOKI1983]

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.

 

[LOKI1983]

@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.

 

[oq172]

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.

 

[rb1957]

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

 

[LOKI1983]

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.

 

[LOKI1983]

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.

 

[jlnsol]

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.

 

[zekeman]

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.

 

[oq172]

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??? ;-)

 

[LOKI1983]

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.