Torsion moment, inertia moment 1

[VT600]

Small problem, maybe one of you can help?

I have a machine with an arm and a weight, this rotates horizontal over 90° and hits a stop plate. The weight gives a force vertical of 1750N
The distance from the center of the weight to the center of the machine where it rotates is 286mm.
The machine is rotated by hand, begin speed is zero and end speed is also zero because of the stop plate. Lets say that the rotate speed is 1 sec for 90°.

I want to know, what moment I get when the machine hits the stop plate. I need this for my torsion moment om my rectangular profile.

Tanks,
Johnny (Belgium)

 

[Cockroach]

Multiply your force, 1750N, by the length of the arm from center of rotation to point of load. Force times distance or WORK, measured in Newton-meters is the moment.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[sreid]

Are you looking for the momentum when the weight crashes into the stop plate? If so it is mass x velocity.

 

[desertfox]

Hi VT600

When the arm and weight are vertical the moment about the arm is zero because the force acting by the weight is pulling vertically through the pivot.
When the arm is lifted to 90 degrees before release the maximum torque will depend on the combined weight in Newtons
of the (arm) and (mass) multipled by the distance of the centre of gravity of the combined (arm)and (mass)from the arm pivot point. The first post by Cockroach shows the equation for maximum torque as being the weight x the distance from the pivot, this is only true for a simple pendulum ie:- where the mass of the arm is negligible in comparison to the weight on the end. You give no details regarding the size of the arm other than its length to the centre of the weight. However when the arm with its weight strikes the plate it has momentum and impact ensues now it would be virtually impossible to work out the striking force
on the plate because of the small interval of time the force lasts, the best I can help you with is to work out the energy the plate absorbs. This will only be approximate as I am sure that the weight after striking the plate will bounce back off the plate and therefore will not impart all its kinetic energy.

so to find the striking energy of the pendulum with the assumption that all the kinetic energy is absorbed by the plate.

1.Find the potential energy in the pendulum arm

potential energy = mass*g*h

where g=gravity constant

h= the vertical distance through which
(centre of gravity )of the combined arm and weight
fall travelling through 90 degrees.

now assuming no friction losses :-

m*g*h = 0.5*m*V^2

where 0.5*m*V^2 = kinetic energy

where m=mass

and V= striking velocity

so the kinetic energy the weight strikes the late is equal to the potential energy


regards desertfox.

 

[VT600]

Desertfox and the rest,

First of all thanks for the replie(s).
I'll try to give some more info.
The arm is vertical, the force (1750N) of the weight also. The force acts on a distance of 286mm from center arm. The weight rotates round the arm (horizontal) over 90°.
If it was fix, then I could use Force times Arm, but unfortunately it isn't. And there is the fact that it doesn't rotate by gravitie but by a man who operates the machine.

I want to know the impulsmoment (force) when it hits the stop plate, to see what torque I get om my arm.

It's a shame there isn't an option to add a picture to this thread to make things more clear.

Greetz,
Johnny

 

[notnats]

Unless you know something about the deflection of the stop plate you can not work out the force.

If you can make an estimate of how far the stop will move when struck by the arm, and assume constant deceleration (unlikely)then you can estimate the force on the stop.

If the kinetic energy (KE)is 1/2 mv^2, the energy dissipated at the stop = KE = Fs where F is the force on the stop and s is the deflection of the stop.

If the stop behaves like a spring, the deceleration is proportional to the deflection and the maximum force is twice the above. The truth is somewhere between.

It sounds like your application is some kind of screw press, so the characteristics of the work piece when acted on by the mass will effect the deceleration of the arm.

Jeff

 

[VT600]

I think I found a formula,

Impulsforce equals the integral of F dt
Impulsforce = A force F working on a body (i.e. a collsion)can be seen as a derivative of the time. (hope I said it correct)

Impulsmoment equals the integral of M dt
Impulsmoment = On a body with a mass m works a force F, this gives a momentum around the point O.

I'll try to find out if I can use these ones.

 

[sreid]

Also, in a simplified form

Impulse = Momentum

or

Force x Time = Mass x Velocity

For two solid objects hitting each other it's difficult to calculate the time or force of the impact. But the time will be short and the force high.

If you want to reduce the force an oil or pneumatic snubber can be used.

 

[desertfox]

VT600

For your situation the formula you need is:-

Ft= mv-mu

where Ft = impulse

mv = final momentum

mu = initial momentum

and v= e*u where e= coefficient of restitution.

So in order to get any answers from this formula you need to know the following:-

1/ initial velocity that mass strikes the plate ie:- u

2/ a final velocity that the arm with mass attached
leaves the plate after impact (Newtons laws of
restitution).(ie:- the arm and mass will
bounce back off the plate
just after impact)

If the pendulum speed is determined by someone operating it
then the impulse will vary each time and calculating the initial velocity of the striking arm will be impossible.
I assumed the pendulum machine was like that of an impact tester where the pendulum was realised manually ie:- by operator but was allowed to fall under gravity.
The final velocity will depend on the "coefficient of restitution" you choose ie:- 0 is for particles that fully coalesce and 1 is for perfectly elastic impacts.
Your case will fall somewhere between the two values.
If you assume there is no bounce back (very unlikly) of the pendulum then
the impulse is as follows:-

Ft = mu

u can be calculated from the kinetic energy ie:- 1/2 mV^2 by
equating the latter to the potential energy of the arm and mass (assuming that it falls under gravity only) see my earlier post.


best of luck


desertfox

 

[flamby]

If the stop behaves like a spring, the deceleration is proportional to the deflection and the maximum force is twice the above.

Nonsensical.

 

[GregLocock]

The sooner that particular concept gets buried the better, I have rarely seen a simple observation misused and misunderstood as much as that one. It's a shame that it is in Shigley, or at least his explanation of why you shouldn't use it is not written in bold capitals.

Here's where it comes from:

Apply a force to an undamped single degree of freedom spring mass system. The deflection will be... F/k

Now remove the same force 'suddenly' from the system. The subsequent amplitude of the deflection in the spring is 2F/k

That's it. That is the mysterious factor of 2. Banal heh?

(OK I'm typing this from memory, but that is the gist of it)



Cheers

Greg Locock

 

[CostasV]

VT600

If I understand correctly, you want to calculate the force ON THE ARM, when this rotating arm has a weight at the free end and goes from vertical to horizontal position.

In my opinion, the weight does not play an important role because the weight hits the horizontal plane. Only the length and the rate of decreasing velocity (decellaration?) are to be considered.

I hope I understand your question right.

Regards
Costas

 

[VT600]

Varsamidis,

The arm rotates in the horizontal plane, not vertical.
Then it would be easy because you can use the gravitational acceleration.

Greetz,
Johnny (Belgium)

 

[notnats]

Flame,
I will stand by what I said:

"If the stop behaves like a spring, the deceleration is proportional to the deflection and the maximum force is twice the above."

In trying to be brief I might not have been clear.

I was referring to the mass stopping in the same distance against an elastic stop as it would stop against a constant force. Against the elastic stop the force Fe= k*s (Where Fe is the elastic force at deflection s)and the work done against the stop = 1/2* k*s^2

If the work done against a constant force Fc is Fc*s, then

Fc*s=1/2*k*s^2
Fc=1/2*k*s = 1/2*Fe or Fe=2*Fc
So the maximum force against an elastic stop is twice the constant force which will stop the mass in the same distance. Such a constant force would be a stop of zero mass held by friction.

I am ignoring coefficients of restitution and have not tried to estimate forces based on contact time, because basically some guesswork will be required for all the methods of calculation suggested here,

Also Greg, I wasn't saying that "a suddenly applied force will cause twice the load". I agree that is a much misunderstood concept. We had this explained and demonstrated in materials 1.01. It really refers to a suddenly applied MASS applied to an undamped spring. That is why my bathroom scales overshoot when I step on them.

Jeff

 

[flamby]

Jeff,

Now I see what you were driving at.I am sorry for my failure on getting your point. I have no problem with the scales shooting up but when someone tells me that it will shoot twice the load, I get agitated.
Regards.

 

[notnats]

Thanks Flame,
Now if you can suggest a way that my scales will stabilise about 10kg less I would appreciate it.

Jeff

 

[Tmoose]

"someone tells me that it will shoot twice the load, I get agitated."

For a rigid or point mass in a gravitational system (that's all I can afford) is lowered to just touch a beam, then suddenly released, I would expect the displacement to look like this over time.

http://mail.bris.ac.uk/~aemtak/damp/Image112.gif

As the applied force (ma)is gravitational and constant (neglecting the small variations in altitude during oscillation) the frequency of the applied force is zero, so I do not think we need to worry about amplification factors and phase lags of force and response.

The scaling on the x or time axis is related to a particular system which would have a particular resonant frequency, so should be considered somewhat arbitrary.
The 1/0/-1 Y-axis scale could be labeled something like no-load/stabilized-load/over-shoot-load. For the first few cycles the value of overshoot-load looks to be darned close to 2X the stabilized-load.

As far as the beam is concerned I think the instantaneous stresses are pretty much related to the instantaneous deflection. If the completely linear beam is instrumented to be part of an electronic scale it will read 2X stabilized load at the instant of full overshoot-load deflection. If I took off the mass and put a load cell on top center of the beam, and used a jack screw to slowly crank in a downward deflection that was equal to the over-shoot-load deflection I would expect the load cell force reading to be 2X the stabilized load. I believe this means that at the brief moment when the dropped mass is in the max downward position the beam is exerting an upward force = 2X. If it only exerted 1X force the mass would have not incentive to move upward from the lowest position.

 

[MintJulep]

You need to know the angular acceleration vs. rotation characteristic in order to solve this.

Torque = angular acceleration x polar moment of inertia.

Of course, a hard stop does not apply a torque to a pivoting arm, it applies a force. You can convert via the distance between the stop and the pivot point.

If the operator is attempting to slow down the arm before it hits the stop, then he is applying some of the total force calculated by the above equation.