HW help Finding the diameter of a hole punched in a wall.

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Homework Help Overview

The problem involves determining the maximum diameter of a hole that can be punched into a sheet of polystyrene, given its thickness, the force exerted by the punch, and the required shearing stress for material failure.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the diameter using the cross-sectional area and shearing stress formula but arrives at a different answer than the textbook. Some participants suggest that the relevant area for shear stress should consider the circumference of the hole rather than just its diameter.

Discussion Status

Participants are actively engaging in clarifying the assumptions regarding the area used in the shear stress calculation. Guidance has been provided regarding the correct interpretation of the area involved in the shearing process.

Contextual Notes

The original poster expresses confusion over the discrepancy between their calculated diameter and the answer provided in the textbook, indicating a potential misunderstanding of the problem setup.

tadisc
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Homework Statement


Like the title says, I have a homework problem that asks for the maximum diameter of a hole that can be punched into a sheet of polystryrene 6 mm thick, knowing that the force exerted by the punch is 45 kN and that a 55 MPa average shearing stress is required to cause the material to fail.


Homework Equations



τ = P/A
τ = ((1/2)F)/A

The Attempt at a Solution



I simply made the cross sectional area A = (0.006 m)(D) then plugged that in and solved for D.

55 x 10^6 Pa = 45000 N/(0.006 m)(D)

and I keep getting D = 68.3 mm. BUT the back of the book claims it to be 43.4 mm. What am I doing wrong??
 
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Welcome to PF tadisc! :smile:
tadisc said:

Homework Statement


Like the title says, I have a homework problem that asks for the maximum diameter of a hole that can be punched into a sheet of polystryrene 6 mm thick, knowing that the force exerted by the punch is 45 kN and that a 55 MPa average shearing stress is required to cause the material to fail.


Homework Equations



τ = P/A
τ = ((1/2)F)/A

The Attempt at a Solution



I simply made the cross sectional area A = (0.006 m)(D) then plugged that in and solved for D.

55 x 10^6 Pa = 45000 N/(0.006 m)(D)

and I keep getting D = 68.3 mm. BUT the back of the book claims it to be 43.4 mm. What am I doing wrong??

It seems like the area that is relevant for shear stress is the area of the "inner lining" of the hole. If you imagine "unwrapping" this circular strip, you end up with a rectangle whose width is the hole thickness, and whose length is the circumference of the hole. But the circumference of the hole is πD, not just D. :wink:
 
The shearing stress is developed around the circumference of the hole. Try using the circumference of the hole and the thickness of the styrene to calculate shear area.
 
wow, thank you so much. that does make sense now that you put it that way. I plugged it an got the right answer. This is my first time using this site, and i was beyond impressed with the time of the response, and the quality of the responses, especially at 3:00 am. Thank you so much! :D
 
tadisc said:
wow, thank you so much. that does make sense now that you put it that way. I plugged it an got the right answer. This is my first time using this site, and i was beyond impressed with the time of the response, and the quality of the responses, especially at 3:00 am. Thank you so much! :D

Yeah, and the sad part is, I actually AM in the Eastern time zone, so it really is 3:30 am for me. Time to go to sleep...:zzz:
 

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