At 41:40, why do you get to assume the shear stress is uniformly distributed in the bolt's cross section? You are only dividing by area, when maximum shear stress is 1.33 times as high as average shear stress in a circular section.
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The "Coulomb Stress" in the plane of a fault between slabs of Earth's crust is essentially the same formula as these from Mechanical Engineering. And it's already known that Deep Injection Wells can infuse faults with fluid, pressurizing & lubricating them until they slip. So, it has already occurred, many thousands of times, that Engineers have "enabled" earthquakes, having allowed faults to "vent" the unstoppable Tectonic Forces which inexorably build over time. (They did not cause earth-quakes, only Earth causes earth-quakes, but Engineers did "vent" the faults of their already-long-pent-up power.) For thousands of years, stone age man knapped flint. All of our ancestors knew how to separate flakes of flint from cores in numerous ways to make numerous stone tools. I predict that, for an Engineer well versed in programs like NASGRO or ABAQUS, it would take far longer to kinesthetically relearn how to knapp flint, than it would for them to write a basic program, modelling a section of a (generally vertical) fault plane, and including the effects of Deep Injections, which would have the effect of pressurizing, prying apart, and lubricating the fault (reducing the Coulomb Stress), so that it began to slip, under human-controlled conditions (zero net force). In practice, injection wells would have to be slant drilled down into the fault plane, at all locations, and at all depths, creating a vast 2D array of injection sites, each one of which could be independently pressurized (at up to a few thousand atmospheres or more) & depressurized, so as to maintain zero net force equilibrium at all times, so that the slabs would ease past each other, slowly, without slipping in a jolt (Fnet > 0) and triggering a major quake. At $5-20M per injection well, "venting" (say) the San Andreas fault would cost billions of dollars. But so will the inevitable inescapable "Big One". Earth wants the west coast of CA to be several meters north of its current location. And Earth will accomplish that feat, one way or another. The options are "Big One" tomorrow destroying billions of dollars... or something else. But Earth demands about 3m of displacement, and CA is going to "eat it", The End. The "catapult" is already "cocked". The only question, is how (and when)? In principle, I predict that it would be more difficult & time consuming to relearn knapping flint than to figure out how to completely control earthquakes (with already existing drilling & fluid injection technologies -- the San Andreas is all near the ocean, use brine. CA does not need to keep allowing the "Big One" to keep looming over its population. Earthquakes can be controlled, probably pretty easily. It's just rock. Not conceptually too different from figuring out how to pressurize the gap between blocks in a Giza Pyramid, allowing one to be slid over the other. You could think of it as "chiropracty" for Earth's crust :)
I'm glad it helped! Check out my other Shigley-based playlists too: MEEN361: th-cam.com/play/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS.html and MEEN462: th-cam.com/play/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB.html Thanks for watching!
55 mins in when working out the y component of force in bolt 3, why have you only used the unit vector? Should it not be 3.971+ 27.869(4/sqrt(4^2+2.8235^2))?
Hi Mr @TheBom_PE, thanks for your videos How can analyze a bolted joint in a high temperature application? Ambient to 200C and so on. Could you please give us some references?
The force that generates primary shear is applied at the centroid of the group. With a force applied at this location, every bolt experiences equal shearing strain because the force will create no tendency to rotate. Because they have equal shearing strain, they will also have equal shearing stress because they are assumed to be the same kind of material.
At 41:40, why do you get to assume the shear stress is uniformly distributed in the bolt's cross section? You are only dividing by area, when maximum shear stress is 1.33 times as high as average shear stress in a circular section.
If you found this video useful, consider helping me upgrade the old tablet PC I use to create these videos! Thanks!
www.gofundme.com/help-replace-my-2011-tablet-pc
The "Coulomb Stress" in the plane of a fault between slabs of Earth's crust is essentially the same formula as these from Mechanical Engineering. And it's already known that Deep Injection Wells can infuse faults with fluid, pressurizing & lubricating them until they slip. So, it has already occurred, many thousands of times, that Engineers have "enabled" earthquakes, having allowed faults to "vent" the unstoppable Tectonic Forces which inexorably build over time. (They did not cause earth-quakes, only Earth causes earth-quakes, but Engineers did "vent" the faults of their already-long-pent-up power.)
For thousands of years, stone age man knapped flint. All of our ancestors knew how to separate flakes of flint from cores in numerous ways to make numerous stone tools. I predict that, for an Engineer well versed in programs like NASGRO or ABAQUS, it would take far longer to kinesthetically relearn how to knapp flint, than it would for them to write a basic program, modelling a section of a (generally vertical) fault plane, and including the effects of Deep Injections, which would have the effect of pressurizing, prying apart, and lubricating the fault (reducing the Coulomb Stress), so that it began to slip, under human-controlled conditions (zero net force).
In practice, injection wells would have to be slant drilled down into the fault plane, at all locations, and at all depths, creating a vast 2D array of injection sites, each one of which could be independently pressurized (at up to a few thousand atmospheres or more) & depressurized, so as to maintain zero net force equilibrium at all times, so that the slabs would ease past each other, slowly, without slipping in a jolt (Fnet > 0) and triggering a major quake. At $5-20M per injection well, "venting" (say) the San Andreas fault would cost billions of dollars. But so will the inevitable inescapable "Big One".
Earth wants the west coast of CA to be several meters north of its current location. And Earth will accomplish that feat, one way or another. The options are "Big One" tomorrow destroying billions of dollars... or something else. But Earth demands about 3m of displacement, and CA is going to "eat it", The End. The "catapult" is already "cocked". The only question, is how (and when)?
In principle, I predict that it would be more difficult & time consuming to relearn knapping flint than to figure out how to completely control earthquakes (with already existing drilling & fluid injection technologies -- the San Andreas is all near the ocean, use brine. CA does not need to keep allowing the "Big One" to keep looming over its population. Earthquakes can be controlled, probably pretty easily. It's just rock. Not conceptually too different from figuring out how to pressurize the gap between blocks in a Giza Pyramid, allowing one to be slid over the other. You could think of it as "chiropracty" for Earth's crust :)
Really helpful Video, good explanation! Thank you!
I'm glad it helped!
Check out my other Shigley-based playlists too:
MEEN361: th-cam.com/play/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS.html and MEEN462: th-cam.com/play/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB.html
Thanks for watching!
Thank's alot sir 💙💐
I'm glad it helped, thanks for watching!
@@TheBomPE 2:36
55 mins in when working out the y component of force in bolt 3, why have you only used the unit vector? Should it not be 3.971+ 27.869(4/sqrt(4^2+2.8235^2))?
Hi Mr @TheBom_PE, thanks for your videos How can analyze a bolted joint in a high temperature application? Ambient to 200C and so on. Could you please give us some references?
Thanks alot for your job. could you please tell me what is your text book ?
I teach this course out of Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!
@@TheBomPE in 28:40 did you mention to shigley? could you please tell me what page was that?
I suppose the direction of secondary shear forces should be opposite direction??
Anyone can tell me why primary tou (shears) should be equal?
The force that generates primary shear is applied at the centroid of the group. With a force applied at this location, every bolt experiences equal shearing strain because the force will create no tendency to rotate. Because they have equal shearing strain, they will also have equal shearing stress because they are assumed to be the same kind of material.
What book are you referring to?
Shigley's Mechanical Engineering Design
You forgot to multiply 27.869 times 4/root(4^2+2.8235^2)... Minor mistake there... Good video tho... The Boom_PE
This happens at min 55!!!
Sure enough, I neglected to write it down, didn't I! Good thing I had the calculation correct in the notes I was referring to! Thanks for the note!
@@TheBomPE Yes sir!!! You had the right answer!!! Not a big deal tho... Most important thing is how the lecture was developed.