Increasing Material Strength w/ Cold Work/Plastic Deformation; True vs. Engineering Stress & Strain
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- Опубликовано: 2 авг 2024
- LECTURE 02a
Playlist for MEEN361 (Advanced Mechanics of Materials):
• MEEN 361: Advanced Mec...
Playlist for MEEN462 (Machine Element Design):
• MEEN 462: Machine Elem...
This lecture was recorded on December 9, 2019. All retainable rights are claimed by Michael Swanbom.
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Dear Sir,
How many times can the strengthening mechanism be performed? Is there a limit?
Realistically, if one was to only use the engineering stress and strain values you would get a more conservative results in your design. This is because the values of stress and strain are calculated lower which is beneficial in cases where your loading conditions aren't well known.
Thank you so much :) hugs from Portugal
We'll have to do a fist-bump instead... COVID and all :)
perfect sir saved my degree,from south africa
I'm glad I could help! Thanks for watching!
Thank you for such a great series of lectures.
Question: The "Ultimate Tensile Strength" of the material is increased when we compare the graphs of the original specimen and the elongated specimen. How does it happen within the material?
Because, in the original case, we assumed constant area (even after deformation)by which we calculated the stress. Now when we plot the sceond graph, we correct that area value and thereby observe the increase in ultimate tensile strength. So, technically, there is no increase in the strength. All we have done is make the values more accurate by considering the true area at that point.
Correct?
by strain hardening
In 56:50, Is the green line’s slope same as the 0.2% offset line?
thank you very much
You're welcome, I'm glad it was helpful! Thanks for watching!
can't the revised yield stress be calculated the same way as the ultimate stress is computed, using the Cold work factor?
Which Textbook you have referred to?
I use Shigley's Mechanical Engineering Design. I teach two courses out of this book, and I have a lot of videos from those courses collected in these playlists:
MEEN361: ruclips.net/p/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS and
MEEN462: ruclips.net/p/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB
Check them out if you're interested! Thanks for watching!
Which book you are referring.where can i get the formula someone help me
In this video, I am referring to Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
dear sir,when we reach UTS why engineering stress began to decrease?I mean as the specimen is held in UTM,how stress is decreased bcz area is constant,It means load applied by UTM, is decreased after uts,If it is the case,how UTM managed to increase or decrease load?
My second question is that why some materials like yield mild steel have two yield points?
The yield point phenomenon occurs due to the segregation of impurity solute atoms(C and/ or N in Fe) around dislocations so as to reduce the strain energy associated with the distorted atomic arrangement (maximum in bcc metals, less in hcp metals and least in fcc metals).
This additional stress required to free the dislocations and set them in motion needed for plastic deformation is called the upper yield point. Once dislocations have been freed, the stress needed for their motion drops abruptly and is called the lower yield point.
There comes slight variation in its value due to the interaction of moving dislocations with the impurity solute atoms obstructing their paths
The hexagonal closest packed (hcp) has a coordination number of 12 and contains 6 atoms per unit cell.
The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell.
The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.
Aluminum FCC
Nickel FCC
Cadmium HCP
NiobiumBCC
ChromiumBCC
PlatinumFCC
CobaltHCP
SilverFCC
CopperFCC
TitaniumHCP
GoldFCC
VanadiumBCC
IronBCC
ZincHCP
LeadFCC
ZirconiumHCP
MagnesiumHCP
after necking, the area becomes soo small that it requires less and less stress to strain ..... as we talk about engineering stress strain diagram ...
stress = Force / Area