I would like to know what the dynamic force load is....

[Divya Shyam Singh]

You have not shown your working. Have you accounted for the To fact that the piston force has to overcome gravity as well as produce an acceleration?
Also, as the angle increases, the angle of the piston must change. This may reduce its moment.

Yes, you are right but in the light of the complexity of the problem wouldn't it be a safe assumption to ignore the mass of the piston with respect to the system?
Similarly, the change of the angle of the force provided by the piston will change but it will be better to assume to be same.

 

[Divya Shyam Singh]

Also, what is the approximate mass of the actuator of that particular model?

 

[Divya Shyam Singh]

I haven't checked the actual numbers but that seems like a realistic figure .

You should ideally repeat the calculation with the mechanism in different positions to see how much the force varies . You can do that later when you do the tidied up calculations for your project submission .

Anyway - we now have a known value for B_y . That's the hard work done .

From that you can work out the motion profile for the platen using standard angular motion calculations .

Need any more ?

What standard angular motion calculations are we exactly talking about here?
Please elaborate

 

[omarmorocci]

Hello guys,

So I took the part out with the forces in order to find the force B
Using moments about A I get
754 * 0.520 - By *0.12
By= 3267 NThen I used the diagram bellow

And I used the formula
Sum of moment = mr2 * angular accelration
Hence I get

3267*0.086 - 640cos 20 * 0.44 = mr^2*a
754 * 0.520 - 3267 *0.086 = 64 *0.44^2 * a

a= 1.27 rad/ sec and my tested value is approximately 0.7 rad/s2 so I think that this is a good estimate ?Since the angle of the pneumatic cylinder changes as the arm goes up then I assume the force supplied by cylinder is constant
But since the angle of the plate changes , the moment due to the weight will degrees hence results in a higher angular acceleration across time

 

[Divya Shyam Singh]

Yes, i think that's correct.
At the same time, the value of angular acceleration will change with θ(increase, since the cos component of the weight will decrease)

 

[haruspex]

wouldn't it be a safe assumption to ignore the mass of the piston

I was referring to the lack of mention of the weight of the load in this equation:

I can use Sum of moment = Inertia * angular accelration right ?

By- Ay = I * a

But maybe omarmorocci was taking moments about the mass centre.
Also, it seems to me that Ay and By are forces, not moments.
And I cannot find what omar is using for I.

 

[omarmorocci]

I was referring to the lack of mention of the weight of the load in this equation:

But maybe omarmorocci was taking moments about the mass centre.
Also, it seems to me that Ay and By are forces, not moments.
And I cannot find what omar is using for I.

I believe that you are right.
Btw please just look at my latest answer and ignore my previous workings.
For mass moment of inertia I took the moments about point A and used I = mr^2 ( hence I believe I might have made a mistake there)
But i'

 

[haruspex]

I believe that you are right.
Btw please just look at my latest answer and ignore my previous workings.
For mass moment of inertia I took the moments about point A and used I = mr^2 ( hence I believe I might have made a mistake there)
But i'

3267*0.086 - 640cos 20 * 0.44 = mr^2*a
754 * 0.520 - 3267 *0.086 = 64 *0.44^2 * a

As I have pointed out a couple of times, you are using the wrong expression for moment of inertia about A.
Do you know the formula for moment of inertia of a uniform rod (or rectangular plate) about one end? Or, if you don't know that, the formula for the moment of inertia about the centre of the rod, and the parallel axis theorem?

 

[omarmorocci]

As I have pointed out a couple of times, you are using the wrong expression for moment of inertia about A.
Do you know the formula for moment of inertia of a uniform rod (or rectangular plate) about one end? Or, if you don't know that, the formula for the moment of inertia about the centre of the rod, and the parallel axis theorem?

Using the parallel axis therom
To find the mass moment of inertia about point A , I assume the total weight acts as a point mass at 440mm from A ( for simplicity reasons)

Hence I = mass (64 kg) * distance from mass to axis which I am taking ( 0.44^2)

Is there something that I am not seeing?

 

[haruspex]

assume the total weight acts as a point mass at 440mm from A

But that is not correct. That gives you mL²/4 for a rod of length L about one end. It should be mL²/3. This will get you closer to your measured result.

 

[omarmorocci]

But that is not correct. That gives you mL²/4 for a rod of length L about one end. It should be mL²/3. This will get you closer to your measured result.

But is actually a rectangle , is possible to just take consider it a rod ?

And yes that does give me a value closer to the actual one

 

[haruspex]

But is actually a rectangle , is possible to just take consider it a rod

Yes, it's the same formula. Think of it as a lot of rods side by side.

 

[Delta2]

The mass of 64kg is that of the rotating rod? or we have a rotating rod (or plating) that supports a mass of 64kg at distance 440mm from the point of rotation? Cause some of the figures at the first posts indicate the latter...

Anyway, if the mass of 64kg has dimensions comparable to the moment arm of 440mm then you just can not approximate sufficiently enough its moment of inertia by just 64*(0.440)^2, that's a very raw approximation..

 

[Divya Shyam Singh]

But that is not correct. That gives you mL²/4 for a rod of length L about one end. It should be mL²/3. This will get you closer to your measured result.

Actually, he is assuming the whole system as a point mass denoted by a square mass of 64kg. Hence its moment of inertia will be Mr2.
Although you are right in saying that the wooden plate has to be treated correctly by using correct formulae and parallel axis theorem for correct moment of inertia for more precise value. However he is assuming the system to be a point mass.

 

[omarmorocci]

Yes, it's the same formula. Think of it as a lot of rods side by side.

Alright then, I am surprised I got such a close value even though I estimated and assummed many things for simplicity ( for example that they are all a comined mass )

The mass of 64kg is that of the rotating rod? or we have a rotating rod (or plating) that supports a mass of 64kg at distance 440mm from the point of rotation? Cause some of the figures at the first posts indicate the latter...

Anyway, if the mass of 64kg has dimensions comparable to the moment arm of 440mm then you just can not approximate sufficiently enough its moment of inertia by just 64*(0.440)^2, that's a very raw approximation..

The 64 Kgs is the weight of the plate + the sheet metal + the weight on the rotating plate

Actually, he is assuming the whole system as a point mass denoted by a square mass of 64kg. Hence its moment of inertia will be Mr2.
Although you are right in saying that the wooden plate has to be treated correctly by using correct formulae and parallel axis theorem for correct moment of inertia for more precise value. However he is assuming the system to be a point mass.

Yes, you are right , that was the assumption that I have made, but treating it like a rod does infact give a closer number to the actual value

 

[Delta2]

Can you tell us each weight separately? What is the weight of the plate, what is the weight of the sheet metal and what is the weight of the weight on the rotating plate?

 

[omarmorocci]

Can you tell us each weight separately? What is the weight of the plate, what is the weight of the sheet metal and what is the weight of the weight on the rotating plate?

Ws = weight of sheet metal plate = 6 kg
Ww= weight of wood = 8 kg
Wp= weight on the wood is 50 kg

 

[Delta2]

So does the "main weight" Wp of 50Kg extends over the wood like the figure shows, or is it concentrated at a small region.

From the figure (I suppose Wp is the blue body) it looks like its moment of inertia is that of a rod with length 86+115+230+444
And ofcourse different moment of inertia for the wood (slightly bigger than the blue body) and another for the metal plate, cause each has different length as I see from this figure.

(Sorry if I made you repost the figure)..

The best approximation would be to take the moment of inertia as 64*(0.875^2)/3 I believe that would take you much closer to the result.

 

[omarmorocci]

So does the "main weight" Wp of 50Kg extends over the wood like the figure shows, or is it concentrated at a small region.

From the figure (I suppose Wp is the blue body) it looks like its moment of inertia is that of a rod with length 86+115+230+444
And ofcourse different moment of inertia for the wood (slightly bigger than the blue body) and another for the metal plate, cause each has different length as I see from this figure.

(Sorry if I made you repost the figure)..

The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3

(No need to appologize , thank you guys for everything so far)

 

[Delta2]

The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3

(No need to appologize , thank you guys for everything so far)

yes but I edited my post, I believe the best approximation would be to take the moment of inertia as 64*(0.875^2)/3 I believe that would take you much closer to the result.

 

[haruspex]

The weight is assumed to be equally distributed over the whole plate.
So you are suggesting to treat the 3 bodies as 3 different rods with different lengths using the formula mL2/3

you could do that, but I don't think it will make enough difference to matter. There are probably other aproximations being made that are more significant, like the way the torque from the load decreases as the angle increases. (How did you measure the angular acceleration?)

 

[omarmorocci]

you could do that, but I don't think it will make enough difference to matter. There are probably other aproximations being made that are more significant, like the way the torque from the load decreases as the angle increases. (How did you measure the angular acceleration?)

For that I will make an excel document and formula that connects the angle to the torque of the load hence giving me different acceleration.
But for now it's just the initial position,

 

[omarmorocci]

What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?

 

[haruspex]

What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?

The problem with that formula is that not all the forces and accelerations are parallel. So it works as a vector equation (if you drop the cos theta) but not as a scalar one.
Choose two perpendicular directions to resolve the forces in (either vertical and horizontal, or parallel to the plate and perpendicular to the plate) and write one equation for each.

 

[omarmorocci]

The problem with that formula is that not all the forces and accelerations are parallel. So it works as a vector equation (if you drop the cos theta) but not as a scalar one.
Choose two perpendicular directions to resolve the forces in (either vertical and horizontal, or parallel to the plate and perpendicular to the plate) and write one equation for each.

I am talking about the tangential acceleration tangential acceleration = angular acceleration * radius
and I am resolving the forces in the direction of the plate and perpendicular to it ( The same as the Ax and Ay)

 

[Delta2]

What I am actually also confused about is how to calculate the dynamic load on point A ( pivoting point )
now that I know the acceleration should I just use the following formula
By - Ay - Wt cos tetha = mass * acceleration ?

given eveything is know , I should be able to calculate Ay , which is the dynamic load right ?

What you calculated is the angular acceleration $a$. The acceleration in your equation above should be $a_y$ which is equal $a_y=0.440a$

But there is also the $A_x$ component of the load which relates to the centripetal force and centripetal acceleration $a_x$ which doesn't relate to the angular acceleration, but it relates to the angular velocity $\omega$ it is $a_x=\omega^2(0.440)$ and it should be
$A_x+W_tsin(\theta)=M_ta_x$ , where $M_t$ total mass of 64kg.

The total load will be $A=\sqrt{A_x^2+A_y^2}$

 

[haruspex]

View attachment 103561
I am talking about the tangential acceleration tangential acceleration = angular acceleration * radius
and I am resolving the forces in the direction of the plate and perpendicular to it ( The same as the Ax and Ay)

Sorry, yes, I see what you are doing. As Delta² says, you need to consider Ax as well. However, I don't think you need to worry about centripetal acceleration. It will be insignificant.

 

[omarmorocci]

Excellent work guys. I will tidy up all the calculations and make less assumptions since now I know how to work it out.

One last question I have is how to I get the compressive pressure ?
http://www.igus.com/wpck/3726/iglidur_clipslager?C=US
Im looking to use these bearings, and I need to know the compressive pressure.
the static load on each bearing is approx 500 N
and dynamic load is approx 1100 N
the thickness of the sheetmetal is 2 mm

Thanks in advance

 

[Divya Shyam Singh]

Come in contact with someone from the company and talk to them. They usually have a software where you 'd have to enter your value. Trust me, they are better at selecting the right bearings for you. I tried doing the same calculations, got some values, fixated on some model of bearings, talked to their technical guy and it turns out that whatever i had done was all in vain.
You could also ask them to suggest you the right bearning model for your use.
Please do tell them the correct conditions in which the bearing is to be used. (like environmental conditions, loads, Life of the component that you want)
One important aspect is serviceability which basically means how often can you get the bearing serviced.

Try to get as much information regarding bearing as possible from the person you talk to. He will be more than happy to help.

Hope this helps.

 

[Nidum]

Before doing anything else work out whether your calculated acceleration figure gives you total motion time for the platen consistent with your practical observations .