Mechanics of Materials: Lesson 17 - Axial Elongation Due to Axial Load Example
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Пікірлер: 47
Your dedication to help students is priceless. I wish that every lecturer could be like you.For the solids series, lesson 22 is not available. Please resolve this issue.
@robinjaikar6636
2 жыл бұрын
watch more kzread.info/dash/bejne/oJZmqbCCaavXh8Y.html
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Best teacher i have ever seen
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Wonderful Lectures ! Thanks.
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Incredibly helpful
so do the two 2kN forces on section AB not matter? I'm confused why we didn't account for them anywhere
Thanks a lot sir🙌
thank you Jeff
This is Tutor Jeff who went to private school. Hanson waku lekafye 🙏🏼
Galing mo talaga tanda
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Professor, when you take 8kNs for the calculation of elongation in A isn't there a (4+4) compression also acting there? total of forces going to the right (tension) is 10kNs and the total of forces going to the left (compression) is 10kNs. Is this how to calculate the total forces P in the points or am I wrong somewhere?
@parthpatel4125
Жыл бұрын
I was wondering this
Any chance we can get a video dedicated to Bearing Stress?😁
Great!
Hey Jeff, Ive been watching your solids lactures in preparation of the fall semester but I'm not able to view lectures 9, 10, or 12. I'm hoping you can sort that out so I don't miss anything, your videos have been very helpful
For sections AC and AD why didnt we add up the lengths of each section? Shouldnt length for section AC 6m and length for AD 8m?
Wouldn't E=73100 Pa where did the mega come from. Shouldn't we divide by 1000^2
Do you mind making a video explaining the St Venats Principle
@1234jhanson
2 жыл бұрын
Will do!
How do u know which sections to cut
Dear professor, why you know which force is in tension and which force is in compression ? Thank you
@kyleboyles9563
3 жыл бұрын
if its going away from the cut, its in tension. if its going towards, then compression
@ahmaranees
3 жыл бұрын
Assume loads moving left as positive (tension) and loads moving right as negative ( compression )
Why we did not take these 2kN's for calculation? I am confused. If anyone knows the answer, pls reply it.Thank you!!
@ahmaranees
3 жыл бұрын
Hi, you can use axial load diagram to solve this even faster. You can take load at A as 8kn, load at B as 4kn, load at C as 6 kn and load at D as 2kn. Assume load going left as positive and loads moving right as negative. Axial load diagram would be as follows: +8kn. 8-4=4kn. 4-6=-2kn A. B. C. D When the -2kn reaches point D, it meets the left side moving load of 2kn. -2+2=0 Since there is no load as I explained above after point D, you do not need to calculate anything. If you still did not understand, let me know you email and o will send you a solution of this problem.
@erayerkul9185
3 жыл бұрын
@@ahmaranees hey firstly thank you for responding my question. I still don’t get it. If it is no problem to send solution of question, you could send to erayerkul7@gmail.com Kind regards :)
@nickborghesan2296
3 жыл бұрын
Hi. He looks at one side of each section as we know it is in equilibrium. Forces will be equal on both sides of the imaginary midpoint squiggle he drew. On his first section he looked at (8kN) he looked at the left side as it was more simple. He then showed adding the right side did also equal 8kN to show this. I was confused at first also.
@majidalbusaidi7319
Жыл бұрын
@@ahmaranees I didnt understand, would you be happy to help me?
I love you
What's the name of the book you're using? :)
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Why use 25 when the diameter of that section is 50mm?
@fuadislamovic8688
2 жыл бұрын
radius is 25
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