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Michael Thomas Rex F
India
เข้าร่วมเมื่อ 26 มี.ค. 2020
This channel facilitates Mechanical Engineering Students to learn concepts in the courses Engineering Mechanics, Mechanics of Materials, Dynamics of Machinery and Project Management in an effective way.
Shear Force and Bending Moment Diagram of Cantilever Beam with Point load and UDL |Example Problem
Dr. Michael Thomas Rex, National Engineering College, Kovilpatti, Tamil Nadu, INDIA
This video lecture explain
1. What is shear force?
2. What is bending moment?
3. How to draw shear force and bending moment diagrams for cantilever beam subjected to point loads and UDL?
4. How to determine sign convention of shear force and bending moment?
Problem description:
A cantilever of 14 m span carries loads of 6 kN, 4 kN, 6kN and 4 kN at 2 m, 4 m, 7 m, and 14 m respectively from the fixed end. It also has a uniformly distributed load of 2 kN/m run for the length between 4 m and 10 m from the fixed end. Draw the shear force and bending moment diagrams.
This video lecture explain
1. What is shear force?
2. What is bending moment?
3. How to draw shear force and bending moment diagrams for cantilever beam subjected to point loads and UDL?
4. How to determine sign convention of shear force and bending moment?
Problem description:
A cantilever of 14 m span carries loads of 6 kN, 4 kN, 6kN and 4 kN at 2 m, 4 m, 7 m, and 14 m respectively from the fixed end. It also has a uniformly distributed load of 2 kN/m run for the length between 4 m and 10 m from the fixed end. Draw the shear force and bending moment diagrams.
มุมมอง: 841
วีดีโอ
Shear Force and Bending Moment Diagram of Cantilever Beam with Point loads | Example Problem
มุมมอง 6565 หลายเดือนก่อน
Dr. Michael Thomas Rex, National Engineering College, Kovilpatti, Tamil Nadu, INDIA This video lecture explains: 1. What is shear force? 2. What is bending moment? 3. How to draw shear force and bending moment diagrams for cantilever beam subjected to point loads? 4. How to determine sign convention of shear force and bending moment? Problem Description: A cantilever of 10 m span carries loads ...
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Sir You take minimum to maximum..... maximum to minimum 🤔🤔🤔
Sir minimum energy is 270 . but you take 295 ? how sir posible
thank you❤
You're welcome 😊
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
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
Thankyou Sir
You are welcome
Thank you so much sir 🙏 Nice explanation 👍
Most welcome
Thank you so much
You're most welcome
Sir Ra+Rb=48+45=93 But u consider 48+40=88
"There's a typo in the question; please consider replacing '45' with '40'."
How can ı reach you?
Mail me at michealrex20@gmail.com
Solution : drive.google.com/file/d/1MO7O5Bd06reEJLVu0cGAnfS6ku3_5qqA/view?usp=drivesdk
I like the video and I I want to know where to find more videos for different exercises
I will upload it soon
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
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
You have my heart ❤️ Thank you so so much
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Perfect ❤
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@@michaelthomasrex ok
👍🇯🇴
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Subscribing because of your this wonderful explanation!!! Thanks sir🫂
Thanks and welcome
Tutor?
I will update it soon. Kindly subscribe to the channel.
❤
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Thanks to my guru.
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Sir kindly help..how to give proper constraints in case of both end fixed??
I will post the video soon. Kindly subscribe to the channel.
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.
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 😭
Excellent..
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Thank you sir
Keep watching. Kindly subscribe to the channel 🙂
Nice explanation sir
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mind blowing teaching
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thank you a lot, you made it very easy and understandable, thanks again😊👌🇺🇿
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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.
Perfect sir😊❤
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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)
Nice video to watch before exam
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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.
Nice.Informative
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omega r(sintheta+ cos2theta/n)=0 why remain ncostheta+cos2theta =0, where omega r ??
Thank you so much for the precise and clear explanation sir❤
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VERYVERY GOODSIR,GOOD TO KNOW SOMEONE LIKE U EXISTED ❤
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Thanks Sir for this video and way of explanation
Most welcome.kindly subscribe to the channel.
Sir I didn't understand why theeta is only pi/2 in tmax
The maximum torque is obtained at exactly middle of the power stroke. Hence the angle is pi/2. The total angle covered during the power stroke is pi.
Nicely explained professor!
Many thanks!
It is so simply way to explain a a lot of thanks sir
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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.🎉🎉🎉
I think I found one mistake, in 6(circle) the x distance you write 1 , while in drawing it's 10 cm , please ans me ,may be i am wrong !
It is mentioned as 1.0 cm in the drawing. I think, the dot may be not visible. Thanks for the comments. kindly subscribe to the channel and share the video lecture with your friends.
You have not converted kg/mm2 into N/mm2 by multiplying by 9.81 Except that everything is perfect
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Good expalination brother, easy to understand❤
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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.
Amazing explanation sir , thank you so much Sir for a good video.
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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)