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

4:01 "E for Eluminium!"

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

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

This channel is a treasure trove

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

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

Thanks Jeff. It is very easy to understand

BLESS YOU FOR MAKING THIS VIDEOS

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

Dr. Hanson I appreciate this sooooooooo much.

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

Thanks Jeff!

i wish i had a teacher like you

Jeff, you're a legend

Galing mo talaga tatang

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

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

Thank you.

thank you

you are the best

thank you, u just saved me

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

Thank you sir

Thank u Dr

Saved my grades.

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

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

4 жыл бұрын

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

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

That helped

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

How about change in the y detection?

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 :)

Better than my lecturer

If temperature decrease whether we take ^T negative

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

Is this course the same as Mechanics of Materials?

@1234jhanson

7 жыл бұрын

Peter Zablocki definitely!

@coolonthelowe

6 жыл бұрын

Pete yessir

@Skystrike70

4 жыл бұрын

That's what it's called at my school

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

5 жыл бұрын

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

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!!

E is for Modulus of Elasticity.

@njabulobuthelezi4036

4 жыл бұрын

Yes

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

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

@mariagranda9258

4 жыл бұрын

Finally someone else who noticed the mistake. Thanks

Sir you know Indian languages

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

3 жыл бұрын

I have the same question.

@MultiJames236

3 жыл бұрын

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.

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

@berkantelgin

5 жыл бұрын

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

5 жыл бұрын

@@berkantelgin It's 208 kip

u picked the wrong aluminum its 2014

Türkçe add subtitle to know you

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.

208N IS WRONG , ACTUALLY THAT COMES .000208N

@Rycamcam

5 жыл бұрын

208 kip