Shear stress

[@oecekoc]

i hope that this playlist could be helpful to understand the logic and to pass my strength of materials course that i'm taking once again...

[@Engineer4Free]

Yes hopefully! If you have Time it's worth watching them all. Good luck!!

[@mangoharun]

did you pass it

[@oecekoc]

I even graduated last year! 😂

[@mangoharun]

@@oecekoc ahaha Congratulations, if you dont mind i would like to have your notes of this class thanks

[@farahalyahya1026]

@@mangoharun me too! if you still have them i'd really appreciate sharing ur notes/cheat sheet or any advice you could give!

[@tcutoast1]

These vids are so CLUTCH

[@Engineer4Free]

tks bruh

[@donyah7471]

what is the difference between shear stress and average shear stress, and normal stress and average normal stress?

[@jrohit1110]

I'm not entirely sure, but I think the term 'Shear Stress' is a general term to denote a typically observed value whereas the term 'Average Shear Stress' is introduced during cyclic loading, especially(but not exclusively) for fatigue stress analysis. Let's say that we have maximum stress of +15Mpa and a minimum stress of -5Mpa(opposite direction loading) acting on a member, we can then the average stress is ((+15)+(-5))/2=(+5)Mpa. The next term 'Normal Stress' would be the stress acting perpendicular to the plane of cross section of the material. Now if this Normal stress has a minimum and a maximum limit, then the arithmetic(similar to (+15)+(-5)/2(+5)Mpa) average of that would be 'Average Normal Stress'

[@dannydang1507]

Hi why do you decided to choose the cross section area by slicing the object vertically but not horizontally? That way you would get a rectangular as the cross section area and the force is acting on the sides too.

[@Engineer4Free]

The blocks that P and P' are acting on apply a force to the rod that would make the rod break clean in half. The broken surfaces would expose the circular cross section of the rod. The stress that is caused by this type of force and the resulting failure are called "shear stress" and "shear failure." The circular cross section is what we are mathematically interested in and doing the vertical cut gives us that area. If we were to cut horizontally, that would give us a rectangular cross section that is as wide as the diameter and as long as the rod which is a meaningless area. Notice I didn't even specify the length of the rod because it doesn't matter.Anything beyond that flat 2D circular cross section that forms the breaking surface is irrelevant to the problem. Hope that clears it up.

[@rakeshk.r.3238]

Thanks for your videos. They are my life savior. Presentation is neat and to the point. I have question regarding the magnitude of shear used in this example. Since 20 KN load is applied on both the sides, shouldn't that be additive at the cross section considered ? Implying shear force considered for stress calculations to be 40 KN instead of 20 KN which would have been the case where load is applied only on one side. Thoughts ?

[@Engineer4Free]

You would think so but no. If you draw a shear force diagram for this member, it would look weird, basically being zero everywhere except a positive and negative value of 20KN infinitesimally close to each side of the point of contact. The jump from positive 20 to negative 20 is 40, but the value of shear experienced by the member is never greater than magnitude 20.

[@a.j.shileikis4456]

Great Video, Thank You! Would you mind sharing what software you're using for this Demonstration?

[@Engineer4Free]

Thanks for watching! I have a full list of all the hardware and software used in this video at engineer4free.com/tools :) cheers

[@a.j.shileikis4456]

Wonderful!! Thank You for responding!!

[@Engineer4Free]

No worries, sorry that it took so long!

[@carlosgarcia-ortiz1879]

Which program are you using as the notebook? I need to do explanation videos for my work and I love how you can color areas and do straight lines and other geometry. I tried looking for the answer in other areas but could not find it.

[@Engineer4Free]

This video was done with Sketchbook, but my more recent ones are done in Krita. Both are good programs. You can find all the hardware and software that I use here: engineer4free.com/tools

[@zhirnawzad3413]

There are two forces acting on the area how come when we a do a FBD we only use one to determine the stress!

[@Engineer4Free]

It’s a good question. The two reactions make a plane of contact basically, but we take the virtual cut basically right on the plane, which we can consider to be between the forces. So the FBD of either side will then only show one force. Best is to not overthink it at this stage when you are learning, and with practice it will start to become more intuitive.

[@jacobhouston1655]

Watched previous videos and came back. You never explained what "tau" symbol is.

[@Engineer4Free]

Tau is the symbol that is used to denote shear stress.

[@zhieima]

your area must have over 4 right?, A= pie d^2 / 4

[@Engineer4Free]

Hey thanks for the question, pi*(r^2) is the same as pi*(d^2)/4. The work is as follows: Area = pi*(r^2) = pi*((d/2)^2) = pi*(d^2)/(2^2) = pi*(d^2)/4. Does that clear it up?

[@rebkh7354]

its the same, because in your formula you are using diameter instead of radius and radius equals diameter/2, when you square that you have r^2= (d^2)/4

[@Engineer4Free]

yup! thanks for contributing :)

[@temi6034]

@@Engineer4Free OMG. i hate that people actually use the (d^2)/4. it was used in my textbook and i was so confused. when it's literally pi*(r^2). that's the thing with math stuff. some things are just unnecessary and confusing . thank for clearing that up

[@Engineer4Free]

Happy that you found this old comment reply and it helped!!

[@Attalla-t9b]

Thanks!

[@aaravsingh3027]

gotchu

[@Engineer4Free]

🤙

[@jacobhouston1655]

I didn't understand.

[@Engineer4Free]

I recommend starting at the beginning of the playlist: kzbin.info/aero/PLOAuB8dR35oft2ZLc1sHseypNMAiG_TeJ

[@mitchellvandoorn2625]

ez

[@Engineer4Free]

Good! =) ready to move through the whole playlist then: engineer4free.com/mechanics-of-materials