Strength of Materials | How to draw Mohr's circle? | Determination of Principal stresses and Plane

Dr. Michael Thomas Rex, National Engineering College, Kovilpatti, Tamil Nadu, INDIA
This video lecture explains
1. How to draw Mohr's Circe?
2. How to determine Principal stresses
3. How to determine the orientation of Principal Plane
4. How to determine the maximum shear stress
Answers for Exercise Problem:
The principal stresses: 88.8 Mpa & 78.8 Mpa
Position of Principal Plane :36.3 degree (CCW)

[@DadaAdebayo-nf9sg]

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!❤❤❤

[@michaelthomasrex]

You're very welcome! Kindly subscribe to the channel and share it with your friends.

[@mosesthantzin]

For me too.🎉🎉🎉

[@ketankarkare6180]

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.

[@michaelthomasrex]

We can take that. However, the sign convention must be same throughout the problem.

[@muhdlufy6185]

VERYVERY GOODSIR,GOOD TO KNOW SOMEONE LIKE U EXISTED ❤

[@michaelthomasrex]

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[@NTMEPavanpatil]

Supeeeerrrrrr sarrrrr, nala iruku teaching 🔥

[@michaelthomasrex]

Thanks for the comment. kindly subscribe the channel and keep watching.

[@santhoshramv9036]

Very useful and understandable sir

[@michaelthomasrex]

Thanks for your comments .

[@mdistaz2002]

It is so simply way to explain a a lot of thanks sir

[@michaelthomasrex]

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[@sundaramsingh752]

Amazing explanation sir , thank you so much Sir for a good video.

[@michaelthomasrex]

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[@tehoais1790]

Excellent!!!Understandable!!👍🏻

[@michaelthomasrex]

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[@nsulwachugambuli7001]

What about to draw a failure envelope and estimate the coefficient of friction for the failure

[@shukurullomeliboyev2004]

thank you a lot, you made it very easy and understandable, thanks again😊👌🇺🇿

[@michaelthomasrex]

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[@Rbsk25]

Thanks Sir for this video and way of explanation

[@michaelthomasrex]

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[@nafisa_ashimi]

You have my heart ❤️ Thank you so so much

[@michaelthomasrex]

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[@talha9276]

Nice video to watch before exam

[@michaelthomasrex]

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[@ameerqamar3605]

Thank you so much you helped me so much may allah bless you

[@michaelthomasrex]

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[@md.alimalrabby2713]

mind blowing teaching

[@michaelthomasrex]

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[@Epic_multiverse]

Nice explanation. While orientation, how did you recognised max and min force will be along y and x axis

[@michaelthomasrex]

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.

[@LeMishecki]

superb🏁

[@michaelthomasrex]

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[@tagorebhukya9667]

pls try to make a video with using angles like 30 with counter clock with respect to X face

[@michaelthomasrex]

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[@shubhammathur4054]

Thanx sir☺️☺️

[@michaelthomasrex]

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[@nsulwachugambuli7001]

Your the great thank you

[@michaelthomasrex]

Thanks for the compliment

[@ernestmaisela1211]

I like the video and I I want to know where to find more videos for different exercises

[@michaelthomasrex]

I will upload it soon

[@zohainapasandalan4324]

Thank you so much for the precise and clear explanation sir❤

[@michaelthomasrex]

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[@godwinayamga9156]

Please how can I find the magnitude and direction of principal stresses?

[@michaelthomasrex]

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.

[@sundaresanmurali7949]

Thanks to my guru.

[@michaelthomasrex]

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[@raoneialves1138]

Excellent..

[@michaelthomasrex]

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[@zirotube123]

isnt it counterclockwise positive and clockwise negetive???

[@michaelthomasrex]

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.

[@edn8624]

Thank you so much

[@michaelthomasrex]

You're most welcome

[@nanivignesh5194]

Can u share answers for exercise problem

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@venuveeramalla7966]

@@michaelthomasrex keep the answer bro ones

[@user-fz4cf1vv1o]

Hlo sir can u keep the answer for practice problem

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@user-fz4cf1vv1o]

Thank you sir

[@bhanuprasaddharavath7]

Tq sir

[@michaelthomasrex]

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[@DerrickTenkorang]

Can i compare answers for the 2nd question

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@osba9306]

I want the answer for the exercise please

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@vivanmeena7651]

correctly done @@michaelthomasrex

[@abdulsamad-tr1jw]

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

[@michaelthomasrex]

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

[@AliHamza._]

Can centre be negative?

[@michaelthomasrex]

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)).

[@manikandaprabhug4471]

Thankyou Sir

[@michaelthomasrex]

You are welcome

[@venuveeramalla7966]

Exercise problem where is answer bro

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@reinahadassa]

Sir, I think the sign convention for Normal Stress is wrong. The tension should be negative and compression is positive.

[@michaelthomasrex]

Any Sign convention is correct. However, you have to maintain it throughout the problem. The answer will not change.

[@richbob5]

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?

[@michaelthomasrex]

The answer is available in the description of the video

[@tgshamimamelin1621]

Where can i check my ans

[@michaelthomasrex]

Thanks for the comment. It is now available in the description.

[@michaelthomasrex]

The principal stresses: 88.8 Mpa & 78.8 Mpa Position of Principal Plane :36.3 degree (CCW)

[@user-tl2vd4rm5p]

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

[@michaelthomasrex]

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

[@TroyAlfred-li6rf]

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?.

[@michaelthomasrex]

Thanks for watching. The answers are correct.

[@user-uq5yy6nh1u]

Help I'm getting 87,32 as my R 😭

[@farismashaqbah5791]

👍🇯🇴

[@michaelthomasrex]

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