Designing Machine for Chew Toy Testing

[dkelly]

I am designing a machine using a cam to generate a down force on a customers product (chew toy). The customer is wanting to try out different materials against his current production model. I was wondering if anyone could help me calculate the cam force in the veticle direction using a 110 in.lb motor. Attached is concept design of the machine (jpeg).

 

[Dr.D]

Assuming that the motor torque is sufficient to drive the system, there is no answer to your question with the information given. It depends on the amount of interference that exists between the mechanism and the frame, that is, how much deformation will have to occur for the cycle to complete. The force will develop in response to the required deformation.

You need to get some data about the closure of the gap on the product, and you also need to specify the lift of the cam before you can talk about a force analysis.

 

[dkelly]

Assuming that the motor torque is sufficient to drive the system, there is no answer to your question with the information given. It depends on the amount of interference that exists between the mechanism and the frame, that is, how much deformation will have to occur for the cycle to complete. The force will develop in response to the required deformation.

You need to get some data about the closure of the gap on the product, and you also need to specify the lift of the cam before you can talk about a force analysis.

Thank you for your reply Dr. D.
I am mainly looking to see if I have at least 220 lbs. in the vertical direction over the part.

I redesigned the cam to only rise once and also decresed the sharp rise in the cam. This should help lessen the force of the pressure angles on the cam. Please see new JPEG

I researched the force an average male can bite with his back molars and found (2) sources on the web that confirmed this number.

The machine is mainly going to repeatedly go to a set distance or compression of the tip being squeezed. This force must be at least greater than 112 lbs.

The cam rise is 3/4". I read the CAM section in the Macininery's hand book, but I keep getting confused on how to calculate the force the cam will deliver. I am pretty sure I have to calculate the pressure forces that act horizontal to the cam follower, but I am not sure.

Thanks again,

Dan

 

[Dr.D]

Dan, your mechanism is just a simple lever system. The fact that it is a cam doing the lifting is of very little consequence.

You say that you are willing to specify the force at the product end, F = 220 lb.

Just as a rough analysis, assuming that the follower beam is horizontal (which is not quite correct, but is approximately true), then if Fc is the contact force, and considering only static loading (very slow operations) a moment balance gives
Fc * 5.47 = F * 4.44
if I got those dimensions correct off of your drawing. From here, the max contact force is
Fc = (4.44/5.47)*220
and you can finish the arithmetic.

In your original geometry, the analysis would be more complicated simply because of the geometry.

 

[dkelly]

Dan, your mechanism is just a simple lever system. The fact that it is a cam doing the lifting is of very little consequence.

You say that you are willing to specify the force at the product end, F = 220 lb.

Just as a rough analysis, assuming that the follower beam is horizontal (which is not quite correct, but is approximately true), then if Fc is the contact force, and considering only static loading (very slow operations) a moment balance gives
Fc * 5.47 = F * 4.44
if I got those dimensions correct off of your drawing. From here, the max contact force is
Fc = (4.44/5.47)*220
and you can finish the arithmetic.

In your original geometry, the analysis would be more complicated simply because of the geometry.

Dr D.

I calculated Fc. through the simple lever equation. However is'nt this force perpendicular to the center of rotation of the motor shaft. How do I know that the torque of the motor will be enough to exert Fc on the cam follower. Motor torque is 110 in.lbs. Here's where I get confused. How do you convert Motor Torgue to Cam force perpendicular to rotation.?

Thanks again for your help Dr. D.

 

[Dr.D]

You do have a legitimate problem here. I sent you a PM on this.

 

[dkelly]

You do have a legitimate problem here. I sent you a PM on this.

Dr. D

I am proceding with the design. If I use 112lb (bite force at end of lever), Fc comes out to 90 lbs. The 110 in.lbs of motor is sufficient to give enough downward force.
I now understand that Fc is the upward force of the cam.


Thanks again for the help Dr. D.

 

[Dr.D]

Good luck, Dan.

 

[irian]

Good luck, Dan.

good luck dund

oh ... have tutorial CAM, because i have class CAM for mechanical for this years..
Help me please...

irian

 

[Dr.D]

irian, what are you saying?

 

[xxChrisxx]

Dont you wish you had a babel fish.

 

[irian]

Heloo


I need information about CAM

can I get a tutorial

 

[xxChrisxx]

www.google.com

 

[Dr.D]

irian, are you talking about Computer Aided Manufacturering, or do you mean the physical device called a cam?

 

[dkelly]

Irian, I have been offline for a while. Are you looking for info on mechanical info or Computor Aided Manufacturing like Dr.D stated.

 

[dkelly]

Good luck, Dan.

Dr. D,

Thanks for the help on this particular project.

I tried to attach a picture of the finished design but when I finally got the picture to under 300KB (jpg) the file would not upload.

The customer accepted the machine and is running the parts unattended.

I am currently helping write test procedures for baselining his product.

Dan

 

[Dr.D]

Congratulations, Dan. I'm glad to hear that it was a success. Can you send your jpg to me at drd_dbq@yahoo. com? I 'd like to see it. Thanks.