Calculating Force for Punching a Hole in a Plate

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Discussion Overview

The discussion revolves around calculating the force required to punch a hole in a plate, specifically focusing on the correct area to use in the shear strength formula. The problem involves a 20-mm-diameter hole in a 25-mm-thick plate with a given shear strength.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant initially calculates the force using the area of the hole but arrives at a different answer than expected, indicating a potential misunderstanding of the area calculation.
  • Another participant emphasizes the importance of using the correct area for shear force, noting that the shear force is not applied perpendicular to the plate.
  • A participant questions how to derive the correct area, referencing a solution that involves multiplying π by the diameter and the thickness of the plate.
  • It is clarified that the area used should be the circumference of the punched hole multiplied by the thickness of the plate.
  • Further discussion suggests that multiplying the circumference by the thickness indeed gives the shear area relevant for the calculation.
  • One participant expresses confusion about the formulas being discussed, indicating a need for further clarification.

Areas of Agreement / Disagreement

Participants generally agree on the need to use the correct area for the shear calculation, but there is some confusion regarding the derivation of the formulas and the application of the concepts involved.

Contextual Notes

The discussion highlights potential misunderstandings regarding the application of shear strength calculations and the importance of considering the geometry of the problem, particularly the thickness of the plate and the nature of the shear area.

Who May Find This Useful

Students and individuals interested in mechanics, engineering principles related to material strength, and those working on similar homework problems may find this discussion beneficial.

tsukuba
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Homework Statement


What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2.

Homework Equations


σ = V / A

The Attempt at a Solution


The first thing I did was rearrange the formula to isolate for V.
I got the area which (A=πr2) gave me 314.15mm2. I change that to m and then I wne ahead and calculate the force which gave me 110.11 MN

But the answer is 549.8 kN ( they multiplied 20x25xπ to get the area)

Not sure where I am going wrong.
 
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You have to be careful which area you use for A. When the punch punches a hole in a plate, the shear force is not applied perpendicular to the plate. It's the circular punching which shears from the rest of the plate.

If you make a sketch looking at the edge of the plate, you can see quite clearly what the shear area must be.
 
The problem shows me a picture and yes I do see that I use include the thickness of the plate in there but how can I get the value?
The solutions says to multiply pi x 20 x25 but I don't understand where that formula comes from or I get the same value by multiplying 2 x pi x 10 x 25 but again not sure what formula that is
 
It's the circumference of the punched hole multiplied by the thickness of the plate. ;)
 
so multiplying circumference by height gives area as well :S ?
 
Obviously. Look, shearing takes place when two surfaces want to slide over one another. The circumference of the piece being punched out slides over the circumference of the hole left in the plate. Both the plate and the punch out obviously have the same thickness, so the only other property you need to calculate the shear area is the circumference of the hole left in the plate.
 
thank you for the explanation :)
 

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