Mechanics of Materials: Lesson 12 - Strain Energy; Example Problems From Stress Strain Diagram

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Dr. Hanson using his powers of Statics, Dynamics and Strengths to levitate that broom with his mind, Truly a legendary man.

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loll

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hey so I just wanted to say thank you becuse you helped me so much in this midterm season. I was working so I couldnt go to the courses and I found you in youtube and you made my life so much easier. 2 days after I have the mechanics of materials midterm and I was never confident about myself in this course before you. thank you very much

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The man the myth the legend is back

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İ love you Hanson sensei, Even the way you talk puts me at ease. I'm getting away from my worries. Thank you I'm glad I met you, I'm glad you exist. (from a civil engineering student, Turkey)

[@Fahimsy3d]

First!!! I have never been happier to see you back!!!!!! So much love from Canada

[@sgpark5056]

My solution for Number 9 is as follows. The total deformation when the rod is loaded: 0.003 + (701.7-600)/(780-600) * (0.005 - 0.003) = 4.13e-3 mm/mm. The deformation that is removed when the rod is unloaded: (701.7 MPa)/(2e5 MPa) = 3.508e-3 mm/mm. The permanent deformation is 4.13e-3 - 3.508e-3 = 6.22e-4 mm/mm or delta = 6.22e-4 * 600 mm = 0.3732 mm.

[@maize3724]

i got the same thing

[@rogerdelbarco]

Same for me

[@KIMIRAIKKONE198N4]

U (resilience) = Area untill proportional limit (MPa) U (toughness) = Area untill fracture point (MPa) Energy absorbed = U x Volume (J)

[@RK-tx5xb]

Welcome back professor Hanson !! Pleasure to see you live again. Lots of best wishes from Alberta, Canada !!!!

[@reyjr4]

Yes! I’m currently taking it. Thanks for this.

[@scottybarra5075]

Shouldn't number 8's answer be 241.75 N.m?

[@yk6474]

Shake hands,we got the same number.

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The boss is backk

[@tblol2039]

I have an exam in 2 days for solids any tips? Edit: I GOT 95 FOR THE UNIT Thanks Jeff Hanson

[@KIMIRAIKKONE198N4]

what kind of tips or questions do you have ?

[@tblol2039]

@@KIMIRAIKKONE198N4 I got 95 Let's go!!!!

[@usandmexico]

25:50 Poisson's ratio is not always positive. There are negative cases, which exhibit the behavior expected from a negative value.

[@aslsahdemirtas8454]

i wish you have other engineering classes on youtube....... thank you jeff hanson

[@epicfailled]

In the 4th section, isnt the max tensile stress before fracture supposed to be 780MPa, as it is the UTS

[@amgil475]

i was searching for this comment to see if someone noticed , but is it correct?

[@enggurux3986]

i got confused at that point

[@TruePreView]

Hi, thank you for the excellent video. I am trying to calculate the compaction energy when compressing soil. I have a stress / strain curve for this and have followed the same approach detailed in the video. So that I can compare compaction energy with samples of different size, how do I calculate energy density per m3 of material?

[@patricklubisi]

The Professor is back

[@XinLiu-t8g]

I am sorry but seems that there is a mistake about the strain engery density, after some unit deduction, you will find that (stress * strain)/2 is the density and the strain engery absorbed should be the density multiple the volume.

[@TruePreView]

I agree. The method shown would work if you had ((load x deformation)/2) / volume.

[@A.Hisham86]

Prof. why you didn't choose the sigma ultimate as max load that the rod can handle before failure? Well, after the ultimate load, the specimen will start to let go itself before the force on it, then practically, the max load that the rod sustain before fracture is the max load sigma ultimate after that, the material won't sustain any load anymore, it just suspended like that. Isn't it?

[@NexiOHome]

i was waiting for someone to say this, because this is the MAX load it can sustain. Not the load right before it fractures...

[@aramhadizadeh7753]

doctor hanson you are freaking awesome

[@anasabdelfattah5285]

hello Doctor, i did not understand why we choose 450 MPa to calculate the maximum Tensile Load, according to the diagram if the material subjected to 500 MPa it will not fail, also it will back it it's regular shape. so how come 450MPa will make a fracture

[@danbo528]

I have the same question.

[@faiblesse9739]

excellent video Dr. Hanson. thank you

[@rogerdelbarco]

Thank you Dr. Hanson. I have one little obsrvation: I think for question 4 we should have taken the 780 MPa value. Am I right? Because 450 MPa it is still under the yield strenth.

[@NexiOHome]

I hope that if Jeff has kids, his dad is the coolest person ever

[@manuboker1]

Wonderful Lectures ! Thanks.

[@christopherabdelhay959]

thank you Dr

[@mohammadali0019]

Damet garm dr 😂 👌🏻👌🏻 Its a persian term means Well Done 👍🏻 ✌️

[@saddlepiggy]

On number 9 I’m very confused about how this method works. Don’t you need to account for the different rate stress/strain rate/slope in the yielding region of the graph? If you do that, you get a final answer of .3729mm.

[@kyleshum5941]

I think he might be wrong does anyone know

[@markangelodavid8531]

Thank you sir Jeff!!

[@Jtruongg]

Love you jeff

[@suvinduamarasekara7550]

strain energy density is strain energy / volume right not multiplied by volume?

[@subramanivasu3458]

First comment. Amazing

[@usandmexico]

Until he explains otherwise, viewers should regard his answer to question 4 as incorrect. He answers what the load is immediately before fracture, assuming no necking. The way his question is phrased requires you to use the max stress on the graph.

[@mdlass4124]

You can check your answer for Q5 by PL/AE

[@leowjiahao3299]

@changjunyang any mistake here?

[@A.Hisham86]

Why not Joules per mmcc (volume)?

[@JuxSwampy]

Very effective

[@Adam_mohammed_]

Thanks ! great example. But I think this video should be the number 14 lesson. Because in my book the lesson 14 is actually the beginning of ch 4 While this video belongs to ch3.

[@sajeevantony9098]

The addition for Q8 in 34-35 min is 2.28. Kindly make correction.

[@XerathMTA]

dont understand last question at all can someone explain it to me

[@bArda26]

hahaha great intro

[@danielniels22]

16:01

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tamamdır efe altun - jorge jesus

[@buhyeah69]

Shouldn’t the area under the curve for number 8 be 2.28N/mm^2 not 2.43N/mm^2? I ran this calculation through a few times and I kept getting 2.28. Is anyone else getting 2.28?

[@antoups]

yes you're absolutely right and the energy is 241.74J