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

 

[desertfox]

Hi gmcgrory

I'm having difficultly relating the orientation of your mechanism with the device your trying to move.
The best way to solve the problem is increment your load in stages ie step through the load positions and find the greatest torque or force position it exerts on your mechanism, then draw your linkages for that point and resolve the forces in them, it seems to me your doing the calculations the wrong way round.


desertfox

 

[gmcgrory]

the main pin going thorugh the turn table, this is the pin that linkage will be operating on. i.e Pin B in the linkage we have been looking at is the the pin that the turntable and conveyor rotate around.if you can imagine the boom/conveyor rotating +-90deg about the Z axis.

i am nw having difficulty determining what torque is needed to turn pin B. and yes, i have worked backwards really to understand how i work out what torque such a linkage was capable of with the ram i chose as a starting point.

So im at a point now where i really need to know how to determine what torque is required to turn a a load which overhangs its axis of rotation

 

[Occupant]

The way I understand this is you want to rotate this conveyor hanging on pin B through 180 degrees. That requires to know the inertia of the conveyor. Then you can calculate the angular acceleration anlogous to F=m*a and develop a velocity profile for the move. However, the term "velocity profile" assumes that you have some kind of servo control on your hydraulic system. I don't think a bang-bang approach would work here.

 

[hydtools]

gmcgrory, I'll take a shot at this.
You must first overcome friction of the pin bearing/bushing and turntable bearing.
The pinB reacts to the 4000kg, W, located distance R, 3.65m, from the pinB axis. The pinB reaction is W*R/d, d being the assumed distance <= length of the pin or length of the pin bushing over which the pinB reactions act. This pin reaction, fb, is the normal force used to calculate the frictional force and the rotation resistance of the pin/bushing. The friction force is fb*mu. Its torque of resistance is fb*mu*(bushing diameter).
If the turntable is supported by a bearing, then figure the friction force by W*mu and estimate the diameter at which this force acts to find the torque resistance, W*mu*D.
The sum of these two friction torques must be overcome by the linkage torque to get the rotation started.

Once started the friction torque may be neglected. The weight, W, is then accelerated at some rotational acceleration, alpha. (W/g)*alpha*R = M, the torque required to drive the swing movement. By kinematic equations, alpha = 2*theta/t^2, where theta is the angle in radians of the swing and t is the time in seconds to move through that angle and initial velocity is 0.

Then, the torque to swing is M = (W/g)*R*alpha = (W/g)*R*(2*theta/t^2)

Certainly open to critique.

Ted

 

[desertfox]

Hi gmcgrory

Okay I think I get the picture, now I agree with others about friction and inertia etc, however if your moving this thing really slow you can usually ignore the dynamic aspects,then your static analysis might just be enough of course you need to build in margin on your cylinder to make sure it does move.
So maximum torque on pin B would be 4000kg * max lever arm length, then draw your load in various positions throughout the mechanism movement and again draw link positions and calculate forces.

desertfox

 

[williedawg]

as a sidenote...
this very mechanism was used in some early washing machines

The main shaft being the agitator in the tub (can't say for sure if there was more than 180deg rotation in the one I tore apart as a kid)

The long link (shown as the actuator in your example) was an arm attached to a crank driven by a reduction gear from the elect motor.

All of the mecahnism was inside a cast alum gearbox.
It was intriguing to see how it worked.

 

[gmcgrory]

so say the max lever arm is as i said it was 3.65m and the mass is 4000kg, then the torque on pin is 4000 x 9.81 x 3.65 Nm.

is this the max torque i need to provide using the cylinder to rotate the conveyor? i should add 20-30% for friction plus margin for error i guess?

 

[desertfox]

Yes, that sort of approach I would agree with. And yes, your lever arm would be at maximum torque at 3.65m.
desertfox

 

[hydtools]

No. 4000*9.81*3.65 would be the moment applied to tilting/prying/bending the pinB in its bushing longitudinally to the pinB axis, not twisting the pinB around its axis. The pinB and its bushing must resist being bent by this moment. This is not the moment against which you must apply torque through the pinB to move the conveyor.
As I understand your conveyor arrangement, it swings in a horizontal plane, not a vertical plane. The pinB axis is vertical, not horizontal. Or is my understanding of your conveyor arrangement incorrect?

Ted

 

[gmcgrory]

hydtools, ur understanding is correct, the conveyor does indeed swing in the horizontal plane. what do you suggest knowing this?

 

[desertfox]

I thought it was in the vertical plane, so hydtools is correct the moment you need to overcome is that due to fricton from the 3.65m * 4000*9.81 and because the pin axis is at 90 degrees to the movement your pin will have to resist bending and shear.

desertfox

 

[SnTMan]

A problem well stated is a problem half solved

 

[hydtools]

gmcgrory,
Estimates and guesses:
PinB diameter = 100mm
PinB bush length = 400mm
PinB reaction forces separated by 340mm
friction coefficient = .16
PinB friction torque = 4000*9.81*3.65*(1000/340)*.16*(100/1000) = 6740Nm
Turntable bearing is ring shaped:
150mm id, 300mm od, average radius 112.5mm
friction coefficient .16
Turntable friction torque = 4000*9.81*.16*(112.5/1000) = 706Nm
Total static friction torque = 6740 + 706 = 7446Nm

Inertia torque:
guess 45deg rotation in 3 secs.; theta = .785rad
4000*3.65*2*.785/9 = 2547Nm

For order of magnitude estimate.

Ted