Statically Indeterminate Axially Loaded Rod Example 2 - Mechanics of Materials

[@jacquesnicolay9221]

Thanks man. Last minute studying for a Solids Exam. Nice to be able to listen to someone do a problem while I'm eating.

[@jeniferjanet4809]

I. LOVE. YOU. Seriously, I'm a UofT student and you teach this better than everyone. Also, you are very cool to listen to. Keep up the great work!!

[@binladin01]

I have a exam coming up soon and this video helped me a lot! Thanks.

[@pegatrisedmice]

i will become homeless if i don't find a video on axial deformation of slanted rods

[@structurefree]

since the beam is assumed rigid, it's linear. If it were not rigid, then the deformation would be cubic. thanks for your question.

[@MrPattara23]

Excellent! Once you get a hang of some of the basic concepts your videos are great since you explain it very straight forward without wasting time on the trivial things one is assumed to know by now :P. Keep up the good work!

[@popedope4842]

What an awesome vid man! I love the way how you solve the problem with a cool style. Really really helped me a lot on my exam. Wish me luck

[@structurefree]

+Vic Leus Thanks, I hope you crushed the exam.

[@nishellavey8509]

Dude I love you Thanks so much for this. I have this same exact problem in my text book and I was so confused with the solution they'd given. This really helped Thanks so much!!

[@structurefree]

@TheGsheth no problem. glad i could help.

[@structurefree]

@MrsNaina I'm glad the video is helpful. maybe you and axial deformation were meant to be.

[@duz2270]

thanks a lot!! my instructor have'nt explained about this that much! :D

[@1StraightPath2Islam]

I really like that you explain everything and do not take it upon the student to read your mind of some of the parts. For example I was lost at the last part where you started using deltaA-deltaE. What was the purpose of the red line?

[@henrybartholomew3314]

I know this might be 8 years too late, but what if the P force isn't equally in-between 2 of the supports? Would similar triangle still work? The problem im working on has d=72in. And P's distance from point A is 60in. So a bit closer to C than A

[@structurefree]

Yes.

[@jacobwenzel8722]

Great Video. So is this problem completely solvable since we need to use the equation sigma=N/A and we don't have A when trying to solve for sigma

[@chael88]

These videos are great! I have a way better understanding through the videos than my professor cuz i cant understand his thick asian accent! No offense to anyone.

[@justsomeoneonline437]

will the deformation always form a nice trapezoidal shape where I can easily use similar triangles to get a nice simple expression for the tensions in the rods. Or are there instances where the deformation will form other shapes. If so what are methods I can use to get the expression for the tensions in the rods?

[@structurefree]

Hi JsO....as long as the horizontal beam ACE can be considered rigid and the deformations are small, you can assume the nice trapezoidal shape.

[@theotema2053]

You are the best❤

[@gojisandhu]

cheers buddy, plain and simple

[@無名-t4c]

what if the load is directly in the middle, i.e. right under cable DC? does the compatibility diagram end up looking like two identical right triangles where the maximum delta is right under point B?

[@andrewglaros3]

beautiful!

[@ravikohli4397]

Thank you very much for this video- it was very clearly explained and the working was clear. Why can you not make more equations by taking moments about other points (e.g.B and C) to form more equations, perhaps allowing you to solve the system of equations? I understand that for a given system of n variables, you must have n equations, but I just fail to see why other moment equations are not counted. Any help is greatly appreciated.

[@Yo-ru8ch]

I was first feeling kind of weird why it is a linear related deformation between point a c and e , but I realized the bar is a “rigid body”, Since it is a rigid body , point c won’t have the most deformation or else it will be bend from the middle and look like a v shape. And since the loading is closer to left so point a will maybe the most deformation, and because the bar is a rigid body, the deformation will be a linear equation from left to right ,( since the bar won’t be bent , there will not be situation that some point didn’t follow linear relation or else the bar must be bent and means it is not rigid)

[@carlosperaltacrisanto5703]

Wich book?

[@ethanboba5262]

Bless your soul

[@chris42119]

im not sure about this, wouldnt the greatest deflection actually be directly under P, making the deflection parabolic, not allowing for a similar triangle analysis?

[@katherinepadden8626]

Chris Baker haha only 6 years later, but we “ignore” deformation and assume that it remains uniform

[@pooguhaneswari]

@@katherinepadden8626 What do you mean? Why is it that we ignore that the greatest deflection should be directly under P? 🤔

[@aerostructures814]

nice analysis

[@deucedroppar]

You are the best! Thank you!!!!!!

[@MrsNaina]

Thanks alot, btw this vid was uploaded on my bday haha :p

[@azizyes]

What is the area!

[@structurefree]

the A = area. It's a symbolic problem.

[@dannyhagos6716]

thank you man

[@anuradhajayasuriya2221]

how is deta = NL/EA ???

[@valentinfinozzi9358]

Using N = (sigma)*A and sigma = E * (epsilon) where epsilon is the engineering strain: epsilon = delta/L Then: N = E * (delta / L) * A solving for delta: delta = N*L / A *E Good luck to anybody reading this

[@Buildings1772]

Couldnt you just do 3 different moment equations at the 3 different points for the extra equations?

[@onelove637]

that was epic

[@noobie379]

"don't be homeless", now that is how you get people to subscribe

[@TheGsheth]

thanks a lot

[@Ahmad-Haydar]

Thx ...very helpful :)

[@muhammadmazhari6900]

Thanks sir

[@chael88]

I love you man

[@krishkrishna7481]

you ddint consider the weight of rod

[@FilthySoapCore]

Thank you so much,! You're an amusing and interesting teacher :)