How Do I Calculate the Shearing Force for Cutting Plastic?

[raniero]

I have been assigned by my tutor to design a plastic granulator. After designing its features of the shaft, rotor etc I tried to calculate the force needed of cutting a piece of plastic (POM/Polycarbonate) having an ultimate tensile strength of 70 Mpa (shear strength assumed to be equal to tensile). The maximum thickness of the plastic slab was fixed to 0.01m and assumed to be the length of the blades, 0.6m. This set up would cater for the maximum cutting action since plastic material is spread all over the blades.

The equation I used was: F = A x Ss

where F is the cutting force, A is the shear area (0.01 x 0.6) , and Ss is the shear strength.

Using this equation the cutting force would be ridiculously large, 420 kN.

Can someone please point me in the right direction ? Books, examples and explanations are appreciated

Thanks

 

[Bystander]

What is your objection to 420 kN?

 

[billy_joule]

420 kN sounds right for shear...

Can someone please point me in the right direction? Books, examples and explanations are appreciated

Ever tried to cut a tomato with a spoon? Why does a sharp knife require less force than spoon? How do you think you would calculate the cutting force required of a spoon vs a knife?

As far as cutting sheet goes, look at a pair of scissors, do they cut the entire sheet at once? Have you used a guillotine to cut sheet stock? they cut along the sheet like scissors.

 

[raniero]

What is your objection to 420 kN?

The work was calculated using W=Fd, the rotor cuts 6 times per rotation and when calculating for 200 rpm the energy requirement is huge, in the tune of 7MW. Similar granulators have extremely lower power ratings.

 

[Bystander]

And, if you examine them closely, you will notice that the blades are helices about the axis of the cutter --- they don't sever the entire 60 cm² at once. Shears are built as guillotines for the same reason.

 

[raniero]

420 kN sounds right for shear...
Ever tried to cut a tomato with a spoon? Why does a sharp knife require less force than spoon? How do you think you would calculate the cutting force required of a spoon vs a knife?

As far as cutting sheet goes, look at a pair of scissors, do they cut the entire sheet at once? Have you used a guillotine to cut sheet stock? they cut along the sheet like scissors.

Thanks for your insight. A sharp knife has a very low area increasing pressure whick makes cutting 'easier'. Analysing the different areas of a knife and a spoon would show the force needed to cut.

Scissors do not cut at once no, but the design was made to do so (similar to marketed granulators)

I was suspecting that the cutting force I obtained was not realistic

 

[raniero]

And, if you examine them closely, you will notice that the blades are helices about the axis of the cutter --- they don't sever the entire 60 cm² at once. Shears are built as guillotines for the same reason.

Apart from the helical blades, if you consider the height, the blade does not penetrate the whole height (in my case 0.01m) in an instant but rather depend on the cutting velocity (in my case on rotational velocity since its a rotor). Does this make sens ? If so, how can I represent it mathematically to obtain maximum force needed ?

 

[Bystander]

If so, how can I represent it mathematically to obtain maximum force needed ?

You have just exceeded my ability to assist --- I understand the question, but I can't tell you what to do to set it up, or how to solve it. Tool shape is going to sneak in on it as well --- hollow ground, wedge, relief, etc..

 

[billy_joule]

Merchant's circle is the most basic model:

http://www.learnengineering.org/2013/01/cutting-force-analysis-merchant-circle.html

A textbook on manufacturing technology is a good place to start.

 

[raniero]

Merchant's circle is the most basic model:

http://www.learnengineering.org/2013/01/cutting-force-analysis-merchant-circle.html

A textbook on manufacturing technology is a good place to start.

Isn't merchant's circle based on the orthogonal cutting method ? If I recall correctly, the orthogonal cutting method can be used only when you can physically measure certain cutting forces, using these forces to 'derive' other forces to be used theoretically.