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1-Visitor
August 9, 2013
Question

How to measure torque, when applying force on a leaver arm?

  • August 9, 2013
  • 1 reply
  • 6235 views

Hello,

May be it seem very easy task to do but beginner like me in Creo 2.0 need your help.
Please if anyone knows how I can measure torque, when apply force on leaver arm.

I am sure there will be a very straight forward way to do it so please help me.

Actually I want to simulate for a 4 links robotic arm with 4 rotational joints and I want to know if robotic arm lift 2 kg load from robotic arm end-effector then how much torque I need to has for my joint actuators. Will Static and dynamic analysis help me?

Please answer me or refer me to any document or tutorial.

Looking forward for your help

1 reply

1-Visitor
August 9, 2013

It might be easier/faster to just do this by hand. Find the max distance from the pivit joint to the load and use the standard equation torque = force x distance. No need to get too complicated. Do not forget to include the requirements to accelerate the load (how quickly you want to get it moving), the mass/inertia of the arms, and any friction from the pivot joints/bearings.

For instance, if each link is .5 meters, and the arm can stretch completely horizontal (worse case), the torques needed would be:

(1kgm @ gravity = 9.8N)

Joint 1: 19.6N X .5m = 9.8 Nm torque

Joint 2: 19.6N X 1m = 19.6 Nm torque

Joint 3: 19.6N x 1.5m = 29.4 Nm torque

Joint 4: 19.6N x 2 m = 39.2 Nm torque

Keep in mind that would be a static torque to hold the load with a zero mass arm. Since each arm segment will have a mass associated with it, and this needs to be included as well. Assuming each arm has a mass of 5 kg, and is evenly distributed, the worse case equation would look like this:

Joint 1: (19.6N x .5m) + (49N X .25m) = 22.05 Nm torque

Joint 2: (19.6N x 1m) + (98N X .5m ) = 68.6 Nm torque

Joint 3: (19.6N x 1.5m) + (147N x .75m) = 139.65 Nm torque

Again, that is static torque to hold the load against gravity, not to accelerate the load. If you need to accelerate the load at 2 m/sec, then the worse case Newton force would be 11.8N (9.8 N to overcome gravity, plus 1 N per kg mass per meter/sec acceleration required).

Also, don't forget braking, or else the load will just keep moving past your required position. Either the motor or brake system will need to apply a certain braking torque to slow down the arm and stop the movement.

1-Visitor
August 9, 2013

By the way, to get this information from Creo 2, you will need to have access to the Mechanism Dynamics Option (MDO). MDO provides more options than just MDX (Mechanism Design Extension). I have never used it, but if you own it I would image the help section would get you pointed in right direction. I am not sure if it would get you a torque measure, or just reaction forces.

https://www.ptc.com/appserver/cs/view/solution.jsp?n=CS42033