2 6-DOF sensors attached to the same plate

[Sandstone86]

For a project I'm using 2 6 DOF-sensors attached to the same plate. I'm using 2 sensors because the sensors can be loaded to a maximum torque of 20 Nm and my test need sensors that withstand 50 Nm.

I've attached a sketch of the setup. (the circles are the sensors, the plate will be loaded by a force and a moment)

My problem is that 1 sensor will fully determine the plate, adding the second sensor will make it statically indeterminate. First problem I need to determine how far the sensors should be placed apart to withstand the loading situation. Second problem how can I attach the two sensors (statically indeterminate, so temperature changes will cause large forces..)

 

[MintJulep]

Uhm.

If you need something to withstand 50 using two of something good for 20 doesn't get you there.

And not because 20+20=40 < 50

 

[rb1957]

but wouldn't some of the applied torque be reacted as a couple ? wouldn't most of the torque be reacted as a couple ?

 

[MintJulep]

I'll admit to not looking at the attachment before my first reply.

Now that I have looked at the attachment...

 

[IRstuff]

I don't recall much of my high school physics, but I know that you can't divide acceleration measurements between two sensors, and since torque is a function of angular acceleration, I'd expect the same. Acceleration is the derivative of the velocity, so if both sensors have the same initial and final velocities, then they must both have had the same accelerations.

TTFN
faq731-376

 

[rb1957]

yeah, that's something i didn't like with the initial post. it's written as though these sensors are "constraints". won't the torque applied rotate the plate these sensors are mounted on ? and probably only a tiny amount, assuming the edge of the plate is properly restrained ?

don't sensors respond to displacement (linear and rotational) ? a torque applied to the plate will rotate the plate as a whole ?

 

[BrianE22]

I assumed they were force sensors. One side of the sensor attached to the plate and the other side to ground.

 

[Sandstone86]

Ah I could have been more clear I see.. BrianE22 was right, 1 side of the force sensor is fixed to a rigid frame, the other side fixed to the plate on which the moment and force work.

The sensors are Force-Torque sensors.

 

[rb1957]

ok, so they a essentially a spring to ground.

the applied torque will try to rotate the plate.

if the springs were Rigid, then the plate won't rotate, the two points are fixed, and the torque will be reacted by a couple.
if the springs are very flexible, then the plate will rotate a lot (unlikely, i think); say 50% of the applied torque will be reacted at each point.

if the springs are reasonably stiff, then the plate will rotate a little; maybe 10% of the applied torque will be reacted as moment at the constraints (5% per) and 90% will be reacted as a couple between the points.

 

[zekeman]

I would go with the assumption that the you could frame the problem by assuming a linear and torsional motion of the sensors each with their "spring constants" in the sensors assuming plate infinitely stiff.
You then can write the 3 equations either static or dynamic,using the CM as the reference coordinate for dynamic case.
BUT, the problem is you have 9 variables with 3 equations .
Statically indeterminate, and BTW, the single sensor problem is equally indeterminate.
I'll pass on this one.
You need a canned program.

 

[zekeman]

"BUT, the problem is you have 9 variables with 3 equations .
Statically indeterminate, and BTW, the single sensor problem is equally indeterminate."
Correction
should be
6 variables,3 equations, single sensor determinate

 

[zekeman]

One more crack at this

You translate and rotate the plate at the origin, x,y,@.
This motion by the transformation matrix results in new positions at the sensors at the plate to be
x1,y1, and x2,y2 each a linear function of x,y,@.
Now you can write the 3 force equations
Fx=k(x1-x10)+k(x2-y20)
Fy similar
T= kt@1+kt@2+L1k(x1-x10)+L2k(x2-y20)+L3k(y1-y10)+L4k(y2-y20)
where
L's are arms to origin
kt torsion spring constant
k=linear spring constant

Now there are 5 variables,namely
x,y,@,@1,@2
and we so far have 3 equations
You next write the strain energy equation ( I'll leave that to you)
To get the final 2 equations, invoke the minimum energy theorem by minimizing with respect to @1 and @2.

 

[zekeman]

Oh, I forgot to mention
x10,y10,x20,y20 are original ground positions of sensor

 

[rb1957]

six equations of equilibrium, no?

more than six unknowns (so statically indeterminate)

what happened to Fz ? Mx, My (assuming T = Mz) ?

 

[zekeman]

"..what happened to Fz ? Mx, My (assuming T = Mz) ?+


I believe the OP shows motion only in x-y plane

 

[rb1957]

if there are out-of-plane forces then there'll be out-of-plane rotations, no?

 

[GregLocock]

I suggest you mount one force transducer rigidly, and the other compliantly. This will resolve your temperature problem.

I'm not convinced there is a big issue with the outputs

Simply, if we mount one on the x axis of the other at distance d and make their axes parallel and z vertical

FX=fx1+fx2
similar FY and FZ

RX=rx1+rx2
RY=ry1+ry2+d(fz1-fz2)
RZ=rz1+rz2+d(fy1-fy2)

or something like that

Cheers

Greg Locock


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