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]

gmcgrory, I think if you review your statics, you will find that the moment on the large pin at any position is the product of the force applied by the actuator and the perpendicular distance pin-to-actuator. Neglecting losses.

I agree with SnTMan, but I don't think you can neglect the losses. The tricky part is, of course, nailing down the geometry and I don't know if you can get away with just a static analysis.

 

[gmcgrory]

Occupant, u think that it is just as simple as multiplying the cylinder force by the perpindicular distance from the central pivot point at different positions of the linkage?

im now in a situation where i dont know what road to go down with this tricky problem.

i hope some one can shed some light on this and help me nail it down

 

[rb1957]

i think you're on the right track.

let's start with your FBD. look at the linear actuator, and the two links that attach to it. set up a co-ordinate at end D, x-axial and y-transverse. the two other links, AD and CD, balance this load. as a simple example, assume AD is perpendicular to DE (the rod). then CDcos(theta) = 53kN and AD = CDsin(theta). in reality you'll have components from both CD and AD in both of the axes at D ... two unknowns, two equations ... GTG.

now i think 53kN is a guess (it's your maximum load, right?, but you're asking yourself waht load is it ? damnit, i'll put in the maximum and see what happens). a good enough way to start, but at the end of the day you'll need the right answer. the good thing is everything is linear (if the rod load 1/2s, the loads in CD and AD 1/2 as well). the rod is reacting torque about B (yes?). the applied load (4000kg) is through A ?, is also creating torque about B (weight?). so you can see how to figure out the rod load to react the applied load, yes?

 

[gmcgrory]

i basically chose this ram as i had a determined space to fit my mechanism within. this hydraulic ram with a 6oomm stroke allowed me to rotate this linkage through my required 180degrees.

the maximum retraction force this ram can provide is the 53KN i have illustrated (just a starting point i guess).

the whole linkage in essence is trying to rotate the pin B which is attached to a turntable with 4000kg on top.

i am almost understanding this now rb1957, however, when u say "set up a co-ordinate at end D, x-axial and y-transverse", what exactly do you mean?

sorry for so many questions, but im keen to get to the bottom of this challenging problem. once i solve this, ill be alot more confident doing something similar in the future

 

[rb1957]

you need to determine the forces acting at D. you know the rod force (even if it's an assumption for now and you'll calculate the correct value later), so you have 2 unknowns, the load in rods AD and CD. you can sum forces in two directions (x- and y-) to put the point in equilibrium. you can use any set of axes, aligning them to the rod direction makes things a little easier; using you geometry axes would work just as well

 

[Occupant]

Can you put a sheet here with the linkages fully defined?

 

[rb1957]

show the load (ie the force due to 4000kg) on your FBD

 

[gmcgrory]

its hard to show the load of 4000kg rb, as its basically sitting on top of the link B. the actuator is trying to rotate the link B which goes through a turntable with a mass on top of it.

my linkage is in effect being opposed by a dead load sittin on link B

i have attached a DWG file for reference guys

many thanks

 

[hydtools]

This may help. Find shaft torque in terms of the link force. Find the link force in terms of the cylinder force and link geometry. The two links and hydraulic cylinder are all two-force links. By the link geometry the force directions are known. The cylinder force is known. Solve the force triangle at the pin common to the three links. Law of Sins relates forces in the triangle to the angles between the forces/links. Use geometry to solve for various angles.
See my attached sketch.

Ted

 

[desertfox]

hi gmcgrory

Can't see attachment, however your FBD in the other one looks okay, except it looks like you have some very acute angles in that mechanism, does it actually work? its looks like the links might bind up to me.
The torque on pin marked B will be the force will be the force in the link A-D multiplied by the 132.9?? dimension.

desertfox

 

[Occupant]

Can't open your drawing, have only 2006.

 

[gmcgrory]

desert fox, the straight link is actually in reality a curved link, but a few on here said i should make it a straight link for calculation purposes. so it doesnt actually bind up.

i have been trying to calculate this with all the help, but i still cannot come up with a resonable force in the link AD.

has anyone had a go at actually calculating this to see what force is present in link AD? i am coming up with a wild force of 865KN.

this cannot be correct i dont think

 

[gmcgrory]

occupant, ive reattached as an earlier cad version (2004). you should now be able to open it

 

[rb1957]

assume for simplicity that AD is perpendicular to the actuator rod. then the load in CD = 53kN/cos(26.73deg) = 59.34kN and the load in AD is 59.34kN*sin(26.73deg) = 26.7kN. check 26.7^2+53^2 = 59.34^2.

 

[desertfox]

Hi

please reattach as pdf as well I'm happy to have a go calculating it.

desertfox

 

[rb1957]

full equations for ptD ...
theta = angle between CD and AD
53kn = CD*cos(26.73)+DA*cos(26.73+theta)
0 = CD*sin(26.73)+DA*sin(26.73+theta)

simplified example, theta = 90-26.73 = 63.27deg

53kN = CD*cos(26.73)+DA*cos(90)
0 = CD*sin(26.73)*DA*sin(90)

note CD is in compression, and DA is in tension

 

[hydtools]

As an observation, you will not get 180 degrees rotation of the rotation shaft and still have torque to drive it. When the line of action of the driving link force passes through the shaft rotation center, input torque is zero. It dead-centers. Top or bottom.
Inertia may carry the drive through. Or an over-running load may push it through.

Ted

 

[hydtools]

Please disregard my observation comment. It is incorrect for the excavator linkage. You can get 180 degree rotation.
My error.

Ted

 

[gmcgrory]

guys, i will be back at work on monday and post a pdf as requested FAO desert fox.

all help appreciated thus far. great forum where minds can be brought together to solve engineering problems.

 

[bradleyelwood]

Here is suggestion for a much simpler way to determine whether the mechanism will do what you want. It's called the method of virtual work.

In your drawing, attach the object and arm in place of the backhoe bucket. Then move the cylinder a SMALL amount, say 10 mm. Find the amount the center of the weight moves lets say the weight moves 52 mm. For that small increment, 52 X weight = 10 X cylinder force, pretty simple. (Obviously you will need the cylinder area and available pressure to get the maximum cylinder force available.) Then continue the increments thru the full range of desired motion.

Another method which is even simpler, although not very elegant, is to fire up the backhoe machine, clamp a piece of wood to the bucket, with a target in place of the intended 4000 Kg. With a measuring tape, measure the vertical location of the target as a function of the cylinder displacement. Again, use the information in increments.

If you want to get fancy, put in two targets, one for the 4000 Kg weight, and another for the center of gravity of the arm you are going to add...then sum both the arm work and the weight work.

With either of the above methods, you need to deduct a reasonable amount for pin friction, say 10 to 20 per cent.

The above method works for backhoe mechanisms, scissor lifts, dump trucks, articulated cranes, etc.