That is an amazing video, sir! Thank you !! In some cases, I wondered if anticlockwise shear is taken as positive and sometimes negative, will that make a difference.
This video was so helpful. However i just wanna confirm my answers. For the radius: R=83.82 For the Max shear stress: τmax=83.82MPa For principal stresses: σmin=78.82MPa and σmax=88.82MPa For the angle at which we have to rotate to get to get the point A&B at in the horizontal plane is: 2θp=72.64° And the orientation of the principal plane is: θp=36.32° Can someone confirm these answers for me?.
In this problem, the point A denotes the X face. When is rotated to the horizontal in CCW direction, we have got principal stresses. So the intersection point is 9.66 Mpa. Hence the 9.66 Mpa is considered along X axis while orienting principal stresses. Similarly, the point B denotes the Y face. When is rotated to the horizontal in CCW direction, we have got principal stresses. So the intersection point is 13.66 Mpa. Hence the 13.66 Mpa is considered along Y axis while orienting principal stresses. Hence, the maximum and minimum just represent the numerically maximum and minimum value among principal stresses.
For the radius: R=87.32 For the Max shear stress: τmax=87.32MPa For principal stresses: σmin=82.32MPa and σmax=92.32MPa For the angle at which we have to rotate to get to get the point A&B at in the horizontal plane is: 2θp=66.37° And the orientation of the principal plane is: θp=33.19° Can someone confirm these answers for me?
The clockwise rotation created by the shear forces at any face is always positive. Similarly, the counter clockwise rotation created by the shear forces at any face is always negative.
When considering the Mohr circle for stress analysis, it's essential to understand the coordinate system used. Typically, the x-axis represents the normal stress, and the y-axis represents the shear stress. In the center of mohr's circle, shear stress is zero because it represents the stress state where the maximum and minimum principal stresses act on the same plane
The magnitude of the principal stresses are the distance of extreme stress values from the centre to the points where the circle intersects the horizontal axis. Please refer the video in time line 17.52 for orientation of principal plane.
There is no thumb rule in selecting positive and negative axes, so it is not a mistake. You can choose as you wish. However, the x-axis represents normal stress, and the y-axis represents shear stress
The centre value may be negative, based on the given problem. For example, if no value is provided for the normal stress along y face, we would have got a negative centre value in the same problem (-8/2=-4; centre(-4,0)).
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I've been watching series of videos on this particular Mohr's circle and I'd say this is the best so far!!! You're a great teacher sir!❤❤❤
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For me too.🎉🎉🎉
Nice video to watch before exam
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VERYVERY GOODSIR,GOOD TO KNOW SOMEONE LIKE U EXISTED ❤
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It is so simply way to explain a a lot of thanks sir
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thank you a lot, you made it very easy and understandable, thanks again😊👌🇺🇿
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That is an amazing video, sir! Thank you !! In some cases, I wondered if anticlockwise shear is taken as positive and sometimes negative, will that make a difference.
We can take that. However, the sign convention must be same throughout the problem.
Excellent!!!Understandable!!👍🏻
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This video was so helpful. However i just wanna confirm my answers.
For the radius:
R=83.82
For the Max shear stress:
τmax=83.82MPa
For principal stresses:
σmin=78.82MPa and σmax=88.82MPa
For the angle at which we have to rotate to get to get the point A&B at in the horizontal plane is:
2θp=72.64°
And the orientation of the principal plane is:
θp=36.32°
Can someone confirm these answers for me?.
Thanks for watching. The answers are correct.
Help I'm getting 87,32 as my R 😭
Thanks Sir for this video and way of explanation
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Supeeeerrrrrr sarrrrr, nala iruku teaching 🔥
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Very useful and understandable sir
Thanks for your comments .
Amazing explanation sir , thank you so much Sir for a good video.
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You have my heart ❤️
Thank you so so much
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mind blowing teaching
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Thank you so much you helped me so much may allah bless you
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Excellent..
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I like the video and I I want to know where to find more videos for different exercises
I will upload it soon
Thank you so much for the precise and clear explanation sir❤
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Thanks to my guru.
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superb🏁
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What about to draw a failure envelope and estimate the coefficient of friction for the failure
Your the great thank you
Thanks for the compliment
Thanx sir☺️☺️
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Nice explanation. While orientation, how did you recognised max and min force will be along y and x axis
In this problem, the point A denotes the X face. When is rotated to the horizontal in CCW direction, we have got principal stresses. So the intersection point is 9.66 Mpa. Hence the 9.66 Mpa is considered along X axis while orienting principal stresses.
Similarly, the point B denotes the Y face. When is rotated to the horizontal in CCW direction, we have got principal stresses. So the intersection point is 13.66 Mpa. Hence the 13.66 Mpa is considered along Y axis while orienting principal stresses.
Hence, the maximum and minimum just represent the numerically maximum and minimum value among principal stresses.
Thank you so much
You're most welcome
For the radius:
R=87.32
For the Max shear stress:
τmax=87.32MPa
For principal stresses:
σmin=82.32MPa and σmax=92.32MPa
For the angle at which we have to rotate to get to get the point A&B at in the horizontal plane is:
2θp=66.37°
And the orientation of the principal plane is:
θp=33.19°
Can someone confirm these answers for me?
The answer is available in the description of the video
Sir, I think the sign convention for Normal Stress is wrong. The tension should be negative and compression is positive.
Any Sign convention is correct. However, you have to maintain it throughout the problem. The answer will not change.
pls try to make a video with using angles like 30 with counter clock with respect to X face
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isnt it counterclockwise positive and clockwise negetive???
The clockwise rotation created by the shear forces at any face is always positive. Similarly, the counter clockwise rotation created by the shear forces at any face is always negative.
Thankyou Sir
You are welcome
Why the Mohor circle the y components of central of circle is taken is zero
And why the x component is not taken is zero
When considering the Mohr circle for stress analysis, it's essential to understand the coordinate system used. Typically, the x-axis represents the normal stress, and the y-axis represents the shear stress. In the center of mohr's circle, shear stress is zero because it represents the stress state where the maximum and minimum principal stresses act on the same plane
Please how can I find the magnitude and direction of principal stresses?
The magnitude of the principal stresses are the distance of extreme stress values from the centre to the points where the circle intersects the horizontal axis. Please refer the video in time line 17.52 for orientation of principal plane.
Man you make a mistake as in mohr circle y axis in 3 or 4 block must be +ve while -ve in 1 and 2 block
There is no thumb rule in selecting positive and negative axes, so it is not a mistake. You can choose as you wish. However, the x-axis represents normal stress, and the y-axis represents shear stress
Tq sir
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Can u share answers for exercise problem
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
@@michaelthomasrex keep the answer bro ones
Can i compare answers for the 2nd question
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
Hlo sir can u keep the answer for practice problem
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
Thank you sir
I want the answer for the exercise please
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
correctly done
@@michaelthomasrex
Can centre be negative?
The centre value may be negative, based on the given problem. For example, if no value is provided for the normal stress along y face, we would have got a negative centre value in the same problem (-8/2=-4; centre(-4,0)).
Where can i check my ans
Thanks for the comment. It is now available in the description.
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
Exercise problem where is answer bro
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)
👍🇯🇴
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