four bar mechanism 1

[elogesh]

Dear members
we are executing a simple mechanism for one of our applications.It is a four bar link mechanism, in the input link we have given a constant torque of 100 lb-in.we expect that the input link will have constant angular acceleration as per equation T= I*alpha. but we found that the angular acceleration is not constant with in a cycle and also that angular acceleration is increasing from one cyle to next cycle.we could not understand the logic.we have given an end time of one second.we dont know how to give the end time.can you kindly help us in understanding the above mentioned details.

regards
logesh

 

[elogesh]

we forgot to indicate in the above thread that the gravity effects are deactivated...

 

[GregLocock]

The driven bars in the mechanism are being accelerated, so they alter the acceleration of the input link.

I assume you are using ADAMS/View?

Try setting the mass and inertia properties of the other two links to zero, you should then see your expected result.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[GregLocock]

Also remember to switch gravity off

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[elogesh]

Dear Greg,

Thanks for your valuable comments.

Yes. The input acceleration is constant now.

But in the out put link, the angular accleration gradually increases from one cycle to the subsequent cycles.

The ouput link is connected to the ground by revolute joint....

I am stiil not clear about end time relevance to the dynamic analysis using torque....?

Looking forward for your suggestions.

Regards,
Logesh.E

 

[GregLocock]

Here's my guess:

As the speed of a mechanism like this increases the bars will accelerate more. That is, you have a steady increase in acceleration due to the input torque, and a second, speed dependent term

Incidentally I don't really like running models with zero mass/inertia parts.

Here's the solver file (/file/export/solver )

UNITS/FORCE = NEWTON, MASS = KILOGRAM, LENGTH = MILLIMETER, TIME = SECOND
PART/1, GROUND
MARKER/7, PART = 1, QP = -450, 200, 0
MARKER/14, PART = 1, QP = -50, 0, 0
MARKER/21, PART = 1, QP = -50, 0, 0
MARKER/22, PART = 1, FLOATING
PART/2, CM = 17, IP = 0, 0, 0
MARKER/1, PART = 2, QP = -300, 350, 0, REULER = 315D, 0D, 0D
MARKER/2, PART = 2, QP = -50, 100, 0, REULER = 315D, 0D, 0D
MARKER/10, PART = 2, QP = -300, 350, 0
MARKER/11, PART = 2, QP = -50, 100, 0
MARKER/17, PART = 2, QP = -175, 225, 0, REULER = 225D, 90D, 90D
PART/3, CM = 18, IP = 0, 0, 0
MARKER/3, PART = 3, QP = -300, 350, 0, REULER = 225D, 0D, 0D
MARKER/4, PART = 3, QP = -450, 200, 0, REULER = 225D, 0D, 0D
MARKER/8, PART = 3, QP = -450, 200, 0
MARKER/9, PART = 3, QP = -300, 350, 0
MARKER/18, PART = 3, QP = -375, 275, 0, REULER = 315D, 90D, 90D
PART/4, MASS = 0.04202, CM = 19, IP = 41.03, 40.78, 0.4311
MARKER/5, PART = 4, QP = -50, 100, 0, REULER = 270D, 0D, 0D
MARKER/6, PART = 4, QP = -50, 0, 0, REULER = 270D, 0D, 0D
MARKER/12, PART = 4, QP = -50, 100, 0
MARKER/13, PART = 4, QP = -50, 0, 0
MARKER/19, PART = 4, QP = -50, 50, 0, REULER = 180D, 90D, 90D
MARKER/20, PART = 4, QP = -50, 0, 0
GRAPHICS/1, FORCE, ETYPE = VTORQUE, EID = 1, EMARKER = 20
JOINT/1, REVOLUTE, I = 7, J = 8
JOINT/2, REVOLUTE, I = 9, J = 10
JOINT/3, REVOLUTE, I = 11, J = 12
JOINT/4, REVOLUTE, I = 13, J = 14
VTORQUE/1, I = 20, JFLOAT = 22, RM = 21, TX = 0\ TY = 0\ TZ = 1000
VARIABLE/1, FUNCTION = RTOD*WDTZ(18,0,0)
ACCGRAV/
OUTPUT/REQSAVE, GRSAVE
RESULTS/XRF
END





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[GregLocock]

Also, this mechanism has no damping, friction, or compliance. Therefore it will 'rattle', in both mathematical and real world senses. These spikes will cause large peaks in acceleration.

I very rarely run such naked systems, eg I would replace at least one revolute by a 6 dof bush, or add some friction across one revolute, or add a discrete spring and damper somewhere. This will calm the thing down analytically and reduce the number of simulation steps needed.

The trick with ADAMS is that it behaves more like the real world than a back of the envelope calculation - so the same problems you see in physical testing will also occur in ADAMS. One problem that then occurs is that the results become as hard to interpret as real world test results.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[elogesh]

Greg,

Sorry for the belated reply.

I am on travel and couldn't respond you.

Thanks for your invaluable timely help.

Regards,
E.Logesh