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

Click For Summary
SUMMARY

The discussion centers on calculating the maximum diameter of a hole that can be punched into a 6 mm thick sheet of polystyrene using a punch exerting a force of 45 kN, with a required average shearing stress of 55 MPa. The initial calculation incorrectly used the cross-sectional area as A = (0.006 m)(D), leading to a diameter of 68.3 mm. The correct approach involves using the circumference of the hole (πD) to determine the shear area, resulting in the accurate diameter of 43.4 mm as stated in the textbook.

PREREQUISITES
  • Understanding of shear stress calculations (τ = P/A)
  • Familiarity with the concept of cross-sectional area in mechanical engineering
  • Knowledge of material properties, specifically for polystyrene
  • Basic geometry, particularly relating to circles and circumferences
NEXT STEPS
  • Study the mechanics of materials, focusing on shear stress and failure criteria
  • Learn about the properties of polystyrene and its applications in engineering
  • Explore advanced calculations involving shear areas in circular sections
  • Investigate the effects of varying thickness and force on material failure
USEFUL FOR

Engineering students, mechanical engineers, and anyone involved in materials science or structural analysis will benefit from this discussion, particularly those focused on stress analysis and material failure mechanisms.

tadisc
Messages
3
Reaction score
0

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??
 
Physics news on Phys.org
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:
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
9K
  • · Replies 13 ·
Replies
13
Views
9K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 5 ·
Replies
5
Views
13K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K