Mechanical engineering - Stress concentration

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SUMMARY

The discussion focuses on calculating the maximum force (F) that can cause failure in a block with two drilled holes, utilizing stress concentration factors (K) from a Stress Concentration Chart. The user initially calculated the D/W ratios for both holes and derived the stress formula σ = F/S, where S is defined as S = C * (B-d-D). The challenge arises in determining how to combine the K factors for multiple holes, with suggestions to treat the block as two semi-infinite plates if the holes are sufficiently close. The conversation emphasizes the need for specific dimensions to refine the analysis.

PREREQUISITES
  • Understanding of stress concentration factors and their application in mechanical engineering.
  • Familiarity with the equations σ = F/S and S = C * (B-d-D).
  • Knowledge of Stress Concentration Charts for holes in plates.
  • Basic principles of materials science, particularly regarding brittle materials.
NEXT STEPS
  • Research methods for combining stress concentration factors for multiple holes in a material.
  • Learn about the analysis of semi-infinite plates in mechanical engineering.
  • Explore advanced topics in materials science related to brittle failure and stress distribution.
  • Investigate online engineering forums for practical examples and community insights on similar problems.
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Mechanical engineers, materials scientists, and students tackling complex stress analysis problems involving multiple stress concentration points in structural components.

Mechstudent
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1. In my materials science's exam, I had the following question: What would be the maximum force F to cause failure in a block that has been drilled at two place?
See following drawing:

drawing.png

(Lame paint skills, I know..)
2. Homework Equations

1) D/W
2) σ = F/S
chart.png

The Attempt at a Solution



1)First, I calculated both D/W ratios. In this case, D/B and d/B
2)Once I had this ratio, I found the corresponding K factors by using the Stress Concentration Chart for hole in a plate.
3) This is where I get blocked. I am unsure what to do with the factors. What I did was pick the hole with the highest corresponding K factor, and ignore the other. I do not know if I have to interpolate between or add them somehow.
4) Using σ = F/S, I know that σ*S = F and that:
S = C * (B-d-D) which is the area of the top rectangle
Since the material is brittle, The ultimate strength is the maximum
So:
x * C * (B-d-D) = F
Because there is stress concentration, The result has to be divided by the factor K.
So my answer was
(x * C * (B-d-D))/K = F max
I wasn't able to find an example where there is more than a single stress concentration area(in this case, two holes instead of one), So I do not know how it adds up, and whether all the necessary information is in the image or not.
 

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This is not an area I know anything about, and all I could find on the net that might handle this level of complexity is behind pay walls.
That said, a possible approach is notionally to split the block with a vertical partition between the holes. The idea is that the split should be where the "flow line" is straight. The external load would be apportioned in proportion to the width of the block piece.
If you have a formula/chart, whatever, that handles a single hole placed asymmetrically in a block then you could define this split such that each hole would fail at the same external load.

Edit2:
Do you have the actual dimensions? If the gap between the holes is no greater than the distance from each hole horizontally to the edge of the block then you can probably treat the plate as infinite and just concentrate on failure of the section between the holes. So each side of the partition is a semi-infinite plate. I do at least see links to such analysis, but, again, behind pay walls.

Edit: this is probably the wrong forum for such a question. Have you tried posting it on an engineering forum here? https://www.physicsforums.com/forums/engineering-and-computer-science-homework.158/
 
Last edited:

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