Linkage Torque/calculation

[gmcgrory]

Hi guys, im just new to this forum as a member but have been reading it for advice for a few months now, some expert advice indeed.

I am basically looking for some guidance on how to calculate the forces and torques involved in a basic linkage, which is the linkage used to curl the bucket on a back actor of an excavator. I am using it to rotate a conveyor through 180degrees. As the linear actuator (in my case an hydraulic ram) retracts it produces a moment on the arced swinging link which causes the straight link to provide a moment arm on the central large pin providing me with the rotation i need.

i would assume my ram creates a moment on the arced swinging link of 1400mm x max ram force( 53KN). At this point i dont know how to continue with the calculation to determine the maximum torque available at my main pin.

could any one tell me what i need to do to solve this problem? i have attached a jpeg of the linkage with a few dimensions i think i need

many thanks for reading and i look forward to some discussion on this

 

[Occupant]

From your Pdf I've come up with AD->F=28.62kN and M=3.816N-m, CD->F=62.26kN in the shown endpoint position. Just for comparison.

 

[gmcgrory]

bradley rather than using the linkage above to curl a bucket, i am turning it on its flat in order to rotate a mass of 4000kg using the main pin B. is there not a simple method or calculation to determine what torque is needed to rotate a certain mass about a point? is it not an inertial calculation maybe?

 

[gmcgrory]

occupant, is M that you stated meant to be 3.816KNm? or Nm

 

[hydtools]

For this linkage see also the analysis of the 6-bar linkage, pp. 16-17, in this attachment:

Ted

 

[gmcgrory]

Nice artcicle ted... just shows, linkages are a whole science in their own respect. smart minds initially thought of such stuff.

So now that we have more or less sorted the problem of calculating the force in the link AD, therefore multiplied by the perpindicular distnce to the pin AB gives us the max torque availale with our chosen cylinder. how can we determine if this will allow us to easily rotate our 4000kg turntable load which acts at the main pin B?

 

[desertfox]

Hi gmcgrory

To find out whether you can rotate the 4000kg mass, we need to first fully understand how it attached to pin B, is it sat a distance from pivot B thereby creating torque? is it in fact attached to the pin A? thats what we need to know to help you further, also we need more dimensions of the pivot centres, link lengths etc, we can't analyse the mechanism as the information is incomplete.
Also to determine the maximum torque on pivot B the mechanism needs to be stepped through a number of positions and a force analysis done in each position.
Example:- rotate the link A-B through a angle of say 2 degrees,draw all the positions of the connecting links and cylinder for this new position, do a force analysis for all the links including the force in link A-D and determine the torque on pivot B, then move the link A-B another 2 degrees and repeat the force analysis, keep doing this till you reach the final position for the mechanism, this will give you the minimum and maximum torques on pivot B throughout the mechanism movement.
I'll try and post an example later.

desertfox

 

[Occupant]

occupant, is M that you stated meant to be 3.816KNm? or Nm

Yes, should read kN-m. By the way, the equivalent numbers for the 0.0 degree position are: A-D->F=77.85kN, M=12.12kN-m and C-D->22.29kN

 

[bradleyelwood]

gmc,

Thanks for the clarification about how you intend to use the mechanism for slewing rather than lifting....

You can use the virtual work method to equate cylinder displacement times force to torque times angle. Use angles in radians, then be sure your units are consistant for work.

By one of the methods presented in all the responces you should be able to find the available torque from the cylinder and mechanism.


Several bigger problems:

1) How much torque is necessary to slew the conveyor? Add up unbalanced wind load, vertical slewing axis not true vertical, friction in the pivots, and a bit to accelerate the conveyor. If you have trouble arriving at a reasonable value, take the worst case unbalanced vertical moment and multiply by a reasonable factor, say 20 or 30 per cent. Keep in mind that a conveyor will at one time be completely full of material, and the drive will quit, then people will shovel the product off the side to get it going again... (from many years of experience.)

2) Will the backhoe arm and mechanism be strong enough to carry the conveyor? For a start, find the worst case of unbalanced vertical plane moment. Then compare it to the section modulus of the backhoe boom times its strength. If you dont know the materials of the boom, assume mild steel.

3) Backhoes are designed to curl the bucket in a few seconds. You will need to slow the slewing motion by providing a very small hydraulic flow.

Again, I am making an assumption....that the backhoe mechanism is "carrying" the conveyor. If the conveyor is suspended by a separate king post and bearing, you can set aside the comment number (2) about backhoe arm strength.

 

[gmcgrory]

the turn table will in fact be attached to the pin at B through a keyed arrangement to transmit the torque. it is on this turntable that the 4000kg conveyor will be mounted.

I dont know if you have seen the DWG file i attached desertfox. i will attach a better PDF tomorrow with more measurements.

i thought it would be easier to determine what torque we would need to get the mass to start rotating........

 

[desertfox]

hi gmcgrory

I can't see your dwg file and thats why I requested the pdf.
I have uploaded a file to try to explain what I wrote i my last post, hope it helps.

desertfox

 

[gmcgrory]

Thank you desertfox. I have now attached a PDF showing the linkage fully open and also fully closed. I have added as many dimensions as i think you might need to give it a bash. if you need any more let me know.

thanks

 

[gmcgrory]

looking at rb1957 equations for point D, they seem logical and correct to me. working them out for when our cylinder is at full stroke of 1400mm:

approximately, there is a force of 26KN in AD in tension and a force of 58KN in CD in compression.

as a few have said i would have to carry out this analysis for a number of positions of the linkage to determine max and min forces in AD there fore max and min torques available at the pin B.

 

[gmcgrory]

when the linkage is closed and the ram is now pushing out, i calculate that AD is nw in compression at 80KN and CD is in compression at 36KN.

interested to see what others get.

 

[gmcgrory]

CD in tension sorry, when ram is pushing out from fully closed position

 

[desertfox]

hi gmcgrory

I get 37.47kN and 80.91kN in the respective links where you get 36 and 80 so were close, however thats based on 53KN from the cylinder, i haven't seen yet how 53KN will be sufficient to cope with the 4000kg mass, what is the torque created on the links from the actual load your trying to move?

desertfox

 

[gmcgrory]

"what is the torque created on the links from the actual load your trying to move?"

Desertfox, so we now have determined how to find out the forces in the links AD and cd, there fore multiplying the AD link foce by the perpindicular distance to the link B will give us the torque at various angles.

the question nw is, hw do i simply determine what is the resisting torque we must over come to allow us to rotate the turntable attached at pin B. bear in mind the 4000kg conveyor is nt acting directly through the pin, its centre of gravity will be acting at a distance from the pin. due to it acting at a distance, i belive the inertial force we need to overcome is greater.

 

[hydtools]

gmcgory,
Did you calculate the forces or derive them graphically?

Will the axis of pinB be vertical?
Is the rotation axis of the turntable the same as the axis of pinB?
How far is the 4000kg from the rotation axis? Do you lift the 4000kg against gravity or move it in a plane perpendicular to gravity?
How fast do you want to accelerate?

Sketch?



Ted

 

[gmcgrory]

i calculated the forces using the geometry hydtools as advised by rb1957. i made sure i fully understood the mechanics again before just simply applying the equations.

i have attached a pdf of the proposed idea. i will be accelerating at a slow rate, im not sure exactly what. it wont be 1mm/sec nor it wont be 10m/s..... just a reasonable rate at which a radial conveyor would normally be turned.

 

[potteryshard]

In looking at your drawing, you seem to have a tensile support cable or chain attached to the boom inboard of the center of gravity of the boom. At the end of a run (when load material is on the outboard end of the conveyor, but not on the inboard end) it seems like this arrangement could create considerable uplift on the turntable bearing, possibly requiring additional bearings.

 

[gmcgrory]

this is really only a provisional drawing. just illustrating the idea. the chain would just take the load of the conveyor when it is in its working position rather than letting the rams hold the load.

Could you maybe explain what you mean further by it creating uplift?

Potteryshard, do you have any idea how i easily calculate what torque is required to turn this arrangement with my linkage?