love the vid but how did you get 250MPa did you use the yield strength and divided it by the safety factor of 2? please let me know I just need a confirmation.
Hi, I have a slightly strange question if you’re still around on here. I have been speaking to a flat earther who seems to think he’s found ‘proof’ that the earth should rip itself apart due to hoop stress caused by the centrifugal force due to rotation (as in a flywheel). I’ve tried to get through to him that due to the far greater opposing force of gravity, the net acceleration experienced by any mass within the earth is towards the axis of rotation and therefore no hoop tension could be present in the crust of the earth - instead it would be under compression, not tension. My analogy for this would be a pressure vessel with an internal pressure of 1psi situated within a pressurised environment at 300psi, resulting in a net external pressure of 299psi which will place the vessel walls under compression, not tension. Could you confirm that this analogy is correct, with respect to the forces experienced by a theoretical ‘hoop’ around the earths equator? Thanks in advance
Centrifugal force of the Earth can be easily calculated and is much less than force of gravity, you are correct there. That would be a decent analogy to use in my opinion but I'm certainly no expert.
Could this calculation be used for a piece of pipe laid down in a trench and up the other end...simulating the pipe in 'compression', and not 'tension', and the compression load is the weight of the tube?
Hoop stress should be thought of as applying (on a cylinder) to thin strips of the cylinder. if you have a varying radius then you have varying hoop stress as you go along - if you think about the direction of the forces associated with the stress they only act along a circle of a fixed radius for each of these different strips, hence if you wanted to find the average hoop stress, you would just find the average radius and perform the calculation with that. In a real world scenario this wouldn't be too useful - the only hoop stress you would be interested in would be the maximum hoop stress, and you only need one radius value (the largest) to perform this calculation. If you are talking about axial stress, then since this acts along the direction of the cylinder where the radius is changing, it would certainly affect your calculations. I'm entirely sure how you would go about finding the axial stress from this, but I have a feeling you find the average out the axial stress by integrating along the length of the cylinder? I'm kind of new to this topic so if anyone would like to add/correct anything I've said it would really help me out
@@Tanoforfucksake you're completely correct, this equation simply decides to use radius instead of diameter. If you sub r=d/2 into the equation in the video then you get your equation of pd/2t
Excellent explanation. Right to the point. Nice job Dr Megan
Thank you. Finally somebody explains how it correlates to if you have a material yield stress value.
Could use your clear explanations in a tutoring session right about now for a solids class Im taking online. Great job!
love the vid but how did you get 250MPa did you use the yield strength and divided it by the safety factor of 2? please let me know I just need a confirmation.
Yeah
Mam can we say that Radial & Hoop Stress is one kind of NORMAL Stress..??
Safety factor is 2 so it should be 25Mpa not 12.5.. right??
Hi, I have a slightly strange question if you’re still around on here. I have been speaking to a flat earther who seems to think he’s found ‘proof’ that the earth should rip itself apart due to hoop stress caused by the centrifugal force due to rotation (as in a flywheel). I’ve tried to get through to him that due to the far greater opposing force of gravity, the net acceleration experienced by any mass within the earth is towards the axis of rotation and therefore no hoop tension could be present in the crust of the earth - instead it would be under compression, not tension. My analogy for this would be a pressure vessel with an internal pressure of 1psi situated within a pressurised environment at 300psi, resulting in a net external pressure of 299psi which will place the vessel walls under compression, not tension.
Could you confirm that this analogy is correct, with respect to the forces experienced by a theoretical ‘hoop’ around the earths equator?
Thanks in advance
Centrifugal force of the Earth can be easily calculated and is much less than force of gravity, you are correct there. That would be a decent analogy to use in my opinion but I'm certainly no expert.
great video
Could this calculation be used for a piece of pipe laid down in a trench and up the other end...simulating the pipe in 'compression', and not 'tension', and the compression load is the weight of the tube?
the little example at the end was nice, but wouldn't it be better to actually show the derivation of where the formulae for hoop stress comes from?
I Finally get it thanks, understanding should help my rocket motor from exploding.🙂👍
Thank you ,very helpful this video for me
th-cam.com/video/i7PH9_KnlP0/w-d-xo.html
this is why i use wraps during squats
Is hoop stress the same as circumferential stress?
Yes
can you help me on Explain the derivation of hoop stress and radial stress for sphere shell and its application in modern technology.
wow, that was great!
Thank you and appreciated! How do calculate hoop stress when the radius vary along the pipe. Rahman
Hoop stress should be thought of as applying (on a cylinder) to thin strips of the cylinder. if you have a varying radius then you have varying hoop stress as you go along - if you think about the direction of the forces associated with the stress they only act along a circle of a fixed radius for each of these different strips, hence if you wanted to find the average hoop stress, you would just find the average radius and perform the calculation with that. In a real world scenario this wouldn't be too useful - the only hoop stress you would be interested in would be the maximum hoop stress, and you only need one radius value (the largest) to perform this calculation. If you are talking about axial stress, then since this acts along the direction of the cylinder where the radius is changing, it would certainly affect your calculations. I'm entirely sure how you would go about finding the axial stress from this, but I have a feeling you find the average out the axial stress by integrating along the length of the cylinder? I'm kind of new to this topic so if anyone would like to add/correct anything I've said it would really help me out
Integration
Great explanation. Thanks!
Short and nice video for quick understanding
hold on, why is axial stress divided by 2t? can anyone explain a little bit?
same question here
Also, why is it p•r & not p•d?
I thought hoop stress was p•d / 2t?
@@Tanoforfucksake you're completely correct, this equation simply decides to use radius instead of diameter. If you sub r=d/2 into the equation in the video then you get your equation of pd/2t
@@thomaswilliams9320 search in thin walled formula derivation.you will understand.
KLASS ...!!! 👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻
Thank you so much!!
Thank you, very helpful!
Thanks!
Thank you!!!!
Absolutely! BTW your voice is so good👌
Mam make video on hoop tension
Good explanation 👍
love ur voice
Pdhai pr dhyaan dena chahiye bro😅😅
thank you sir
That's a woman, but okay ;)