Mechanics of Materials: Lesson 17 - Axial Elongation Due to Axial Load Example

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[@YDAB123]

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.

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[@leafrancis400]

so do the two 2kN forces on section AB not matter? I'm confused why we didn't account for them anywhere

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Any chance we can get a video dedicated to Bearing Stress?😁

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[@The.Ananse.Effect]

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[@aminathakmal7097]

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]

I was wondering this

[@michell9328]

That confused me too lol

[@ChuckYuuu]

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

[@aislincerron287]

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?

[@T181T]

Wouldn't E=73100 Pa where did the mega come from. Shouldn't we divide by 1000^2

[@koaw4792]

How do u know which sections to cut

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Do you mind making a video explaining the St Venats Principle

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Will do!

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Great!

[@erayerkul9185]

Why we did not take these 2kN's for calculation? I am confused. If anyone knows the answer, pls reply it.Thank you!!

[@ahmaranees]

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]

@@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 [e-mail] Kind regards :)

[@nickborghesan2296]

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]

@@ahmaranees I didnt understand, would you be happy to help me?

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I love you

[@A.Hisham86]

What's the name of the book you're using? :)

[@minhtienle1124]

Dear professor, why you know which force is in tension and which force is in compression ? Thank you

[@kyleboyles9563]

if its going away from the cut, its in tension. if its going towards, then compression

[@ahmaranees]

Assume loads moving left as positive (tension) and loads moving right as negative ( compression )

[@mathewpalacios7605]

Why use 25 when the diameter of that section is 50mm?

[@fuadislamovic8688]

radius is 25

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