Mechanics of Materials: Lesson 16 - Thermal Coefficient of Expansion Problem

[@Eightarmedscales]

Love you Jeff. My whole engineering class owes you their grades.

[@JonathanBlackwood1]

The best explanation I’ve seen so far !!! Literally learned the method by watching the video once !!!

[@pradanator]

"Low stress man" in a hippie voice, man I love Dr. Hanson😂

[@McBobtheruggaman]

4:01 "E for Eluminium!"

[@Skystrike70]

This channel is a treasure trove

[@imGool]

THANK YOU JEFF HANSON!!! You are the savior of my solids grade

[@DavidDiaz-sq5ws]

Dr. Hanson I appreciate this sooooooooo much.

[@whatruwaiting4859]

i wish i had a teacher like you

[@rahmanrahmani2896]

Thanks millions for all your tutorial and very much appreciated. They are very well presented and easy to follow. Could you do a few steel portal frames with different base connections. You are a living legend and I am sure your outstanding work get appreciated by your followers !!! Kind regards, Rahman

[@Come2Daclub]

Hello Dr. Hanson, I would like to let you know, that you have been mistaken while you looked up the value of the coefficient of expansion, as the material that the question has is 2014-T6 Aluminum not 6061-T6 Aluminum. thanks, best Regards

[@anthonysawaya6142]

you are amazing you solve totally different from the book and from my doctor

[@tammyhuynh2060]

BLESS YOU FOR MAKING THIS VIDEOS

[@jordicoenders2588]

Jeff, you're a legend

[@fahriali3041]

Also, KSI means Kips per square inches, so it's better to use inches for this types of question.

[@BritPopLivesOn72]

I think that the main take away from this video is that since these materials are fixed in place, the total deformation of all three rods together is zero. All rods expand due to temperature rise, so the walls respond with a compressive force, which decreases the length of each member. I think this was a good video but writing out the entire equation of total deformation would help clarify the process.

[@njabulobuthelezi4036]

So it the temperature was increasing, the force would have been the tensile force?

[@felipegarcia2269]

I believe that at @ the co. of thermal expansion for 2014-T6 is 12.8 not 13.1 (6061-T6). and E is 10.6 not 10.0

[@danielbadel1226]

A quick comment Normal stress it's not force over area. Dimensionally speaking indeed it's. How ever it's more than that. Sigma=F/A is only valid for pure traction force, however with bending we can obtain a contribution for normal stress. Recalling the DSV solution (Valid for slender, isotropic, homogeneous and compact beam -shear contribution can be neglected) Normal stress=T/A+Mx/Ixx (y) -My/Iyy (x). However ti ringrazio pero il video. Great explanation

[@ellijahabarientos6244]

It helped a looot thaaaanks!!!! be to God ❤

[@ronniebarba8477]

Galing mo talaga tatang

[@Js_ping]

Better than my lecturer

[@MattyP62]

Oh how I wish you taught at Penn State. Great videos! Super helpful

[@josemariafernandez6450]

Thank you so much!! I´m enjoying this stuff

[@thutieu3258]

Thanks Jeff. It is very easy to understand

[@LongTran-nf1hr]

Saved my grades.

[@pete6461]

Is this course the same as Mechanics of Materials?

[@1234jhanson]

Peter Zablocki definitely!

[@coolonthelowe]

Pete yessir

[@Skystrike70]

That's what it's called at my school

[@Jaycdiamond]

Thanks Jeff!

[@Google245gubd]

thank you, u just saved me

[@anthonysawaya6142]

I have a question doctor jeff is there a systematic way you follow to come up with the compatibility equation or is it just logical depending on the case of each exercise

[@puteridania00]

i have question regarding this one. what if there is a gap between aluminium and stainless steel ? not the bronze. hope you can answer me asap thank you :)

[@Vishnudhumale]

If temperature decrease whether we take ^T negative

[@phophiligidima7889]

Thank u Dr

[@sanaullahkhan6911]

E is for Modulus of Elasticity.

[@njabulobuthelezi4036]

Yes

[@njabulobuthelezi4036]

Thank you.

[@tanujgupta8381]

After I have calculated the delta for all the material, I know that force in the sections will be same. So let's say we have force P, based on delta values if we calculate the force: All the sections will have different forces. Why does this logic does not work I am confused!!

[@jggarcia6537]

you are the best

[@puravgandhi3843]

Thank you sir

[@nathanielllanos8577]

That helped

[@evancruz4058]

Dr. Hanson, for the force F, I got 1172.64 kips using the same numbers that you used in the deformation equation equaling .05 inches. How did you get 208 kips? That is not matching up with my calculator's answer?

[@Hmansqd]

he incorrectly wrote the coefficient of heat expansion for aluminum. maybe that should help although this message might be late :D

[@siten1]

Coef. of Therm. for Aluminum should be 12.8E-6 NOT 13.1E-6

[@mariagranda9258]

Finally someone else who noticed the mistake. Thanks

[@miseantropatagonianbaggins1057]

How about change in the y detection?

[@spikeoac4248]

Why couldnt u do for each material: strain = L - L0 / L0 where the numerator is equal to the axial elongation, subbing in u get strain = thermal coeff * delta T then stress = E * thermal coeff * delta T I got the wrong answer doing this and im not sure why.

[@MultiJames236]

I have the same question.

[@MultiJames236]

Wait I figured it out, you cant do that because the bars never actually stretch because of the wall. So you need to find out the force the wall is putting in order to keep the bars from elongating.

[@stanleylebrun6160]

I am not sure how the are is (pi)*(r^2) using diameter, I thought it would have been (pi/4)*(d^2)

[@berkantelgin]

The unit of the answer needs to be in the unit of Ibf I guess, not the Newton.

[@berkantelgin]

He tried to make it correvt at the end of the video but still it's wrong. Cause, it is not 208 kips it is 0.208 kips.

[@Rycamcam]

@@berkantelgin It's 208 kip

[@karanchowdhary5969]

Sir you know Indian languages

[@PCgamerChannel]

u picked the wrong aluminum its 2014

[@alisaleem5451]

you're doing this problem wrong,, after you get the elongation for each bar you can apply Hook law stress = E*strain----> for Al ; stress = (10,000 ksi)*(0.0189''/48'')=3.9375 ksi.

[@umutkececi7241]

Türkçe add subtitle to know you

[@garcha5725]

208N IS WRONG , ACTUALLY THAT COMES .000208N

[@Rycamcam]

208 kip

[@FahadUS1]

thank you