Sheer stress due to a hole puncher

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The discussion revolves around calculating the shear stress exerted by a hole punch on multiple sheets of paper. The hole punch has a diameter of 8mm and applies a force of 6.7kN on 10 sheets, each 0.2mm thick. Participants emphasize the importance of using the correct area for shear stress calculations, specifically the shear area that is parallel to the applied force. One contributor calculates the shear stress but finds their result significantly lower than the expected value, indicating a misunderstanding of the relevant area. The conversation highlights the need to consider both the circumference of the punch and the thickness of the paper in the calculations.
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A hole punch has a diameter of 8mm and presses onto 10 sheets of paper with force of 6.7kN. IF each sheet of paper is of thickness .2mm, find the shear stress. [HINT: Be careful in decing what area to use. Remember that a shear force acts parallel to the surface whose area is relavant.]

shear stress= F/A

can anybody help me with this question from Physics by Giambattista.
The answer is supposed to be 1.3x10^8 Pa but i can't go about figuring out how they reach the answer.

tia
 
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maxxed said:
A hole punch has a diameter of 8mm and presses onto 10 sheets of paper with force of 6.7kN. IF each sheet of paper is of thickness .2mm, find the shear stress. [HINT: Be careful in decing what area to use. Remember that a shear force acts parallel to the surface whose area is relavant.]

shear stress= F/A

can anybody help me with this question from Physics by Giambattista.
The answer is supposed to be 1.3x10^8 Pa but i can't go about figuring out how they reach the answer.

tia
you were given the hint on which area to use. The paper must punch through by forces acting parallel to the direction of the applied force. Consider the circumference of the punch and the depth of the punctured paper.
 
hmm

I get an answer an order of magnitude off by just taking 6.7E3 / (10*(pi*(0.004m)^2)) = 1.33E7.

That's just using 10*[area one side of circle formed by holepunch]. Doesn't take into account the paper thickness, which may be why it's wrong. Maybe your book has a relevant example in the section?
 
sippyCUP said:
I get an answer an order of magnitude off by just taking 6.7E3 / (10*(pi*(0.004m)^2)) = 1.33E7.

That's just using 10*[area one side of circle formed by holepunch]. Doesn't take into account the paper thickness, which may be why it's wrong. Maybe your book has a relevant example in the section?
You're using the wrong values. The shear area which is parallel to the applied force has nothing to do with the surface area of the punch. And the shear stress depends on the shear area which must depend on the thickness of the paper sheets and the circumference of the 8mm circle.
 
Diagram attached.
 

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thanks for the input and the diagram everybody.
 
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