Bearing stress

[@DJLOG74]

How is it possible that someone is that good of a person, share their knowledge and explain by going straight to the point? Thank you so much for your work!😊

[@Engineer4Free]

Thanks!!! Happy to help =)

[@alfonsomaurice7421]

i dont mean to be off topic but does anybody know a method to get back into an instagram account? I stupidly forgot my account password. I would love any tricks you can give me

[@Hud_Adnan]

I really don't know how to thank you , I would really appreciate your videos ... I finished my first year studies I'm a second year student now and here I am ... learning from your videos ... your just such a great man ! NOT ALL HEROS WEAR CAPES ^-^

[@Engineer4Free]

Thanks Hud, I really appreciate hearing from you =)

[@justinrodsicat7132]

Thank God I found your channel. It's really helpful. I hope to see more from you. Thank you so much for your hard work!

[@Engineer4Free]

You're welcome Justin, thanks for watching!! =)

[@rogerarbizo1242]

Ah! Okay thank you very much for this visual explanation. My professor just could not make it that clear. I was very confused on how the projection was being taken. Thanks again

[@Engineer4Free]

Really glad that it helps! Feel free to share engineer4free.com/mechanics-of-materials with the rest of your class :)

[@ScholarStream_25]

Man appreciate your work , straight to the point with neat explanation and overall presentation really made the work look simple . Simply cheers mate 👍💪💪

[@mitsuya6625]

This vid cleared my confusion. Thanks a lot

[@Engineer4Free]

Glad it helped!! More at engineer4free.com/mechanics-of-materials =)

[@Maybe.019]

I am very thankful for you and I really do appreciate your work . I always watch your videos because you are explaining the things in the exact way I am looking for . I would like to say thank you so much 🌹

[@Engineer4Free]

Thanks so much for the nice feedback, I’m really happy to hear it! I hope you continue to get lots more use out of the videos as you continue your studies 🙂

[@Maybe.019]

@@Engineer4Free You are making engineers understand how the things work perfectly not memorizing it. I really do appreciate your work Thank you 🌹

[@mac-donaldduru6687]

Wow just wow, I believe this is my 4th video into your mechanics of material contents and I'm simply loving it. Quick Question: What software do you use in taking your notes?

[@Engineer4Free]

Thanks!! I hope you get a lot of use from the rest of the course =). And I have a full list of the hardware and software that I use to make the videos here: engineer4free.com/tools

[@anotherwasiangirl]

you just saved my life thank u

[@Engineer4Free]

I gochu 🤜🤛

[@antoniodbz5850]

So what if we have pressure only from one side of the rod? will the calculation be different I need this kind of example please. Thank you

[@hyoukagaming_n_graphics2150]

Tq for the great video.. Helps a lot and easy to understand.. Now I can start making my thesis👍

[@Engineer4Free]

Glad I can help, good luck!

[@hz-vb6hn]

Thanks .... the way u put it was simple and helpful thanks

[@SK-cs5pk]

As clear as crystal.🙏🏽👍🏽

[@Engineer4Free]

Thanks Sarthak 💎

[@ciiil8802]

Finnaly i can understand bearing stress

[@guguya3675]

Wondering if there is a video about punching shear? I’d like to learn something about it. Thank u

[@Sana-zm3tn]

If I understand this right, bearing stress is no internal stress? It is just the stress the other body exerts on another one, right?

[@Engineer4Free]

Yeah

[@Jason-o5s]

Cheer~~~pressure or tension exerted on a material object.😊

[@laurasy5416]

I understand your videos more than I understand my professor lol

[@Engineer4Free]

Mission accomplished!! =)

[@عاصمالفايز-ث4ض]

the area is not the half of circumference of a circle multiplied by t so A=((pi*d)/2) * t A=(pi*20)/2 *15 A= 471.23mm^2 isn't it ? .

[@Engineer4Free]

Yes and no. The actual contact area between the rod and the plate is 471.23mm^2 like you have calculated, but because of the complex nature of the stress distributions in the curved surface, it suffices to take the projection of the rod onto the connection (so diameter of the rod times the width of the plate), and consider that projected area to be subjected to a normal stress (equal to the applied load divided by the rectangular projected area). It is a simplification, but in practice is good enough!

[@sohaibalam4168]

Can someone guide me on which software is used to make such tutorials? Need to make some for my students at school.

[@Engineer4Free]

Hey Sohaib I made a full list of all the hardware and software that I use at engineer4free.com/tools you should check it out

[@jessevicente1361]

Nice video! But I would like to ask if what is the main difference between shear stress and bearing stress? since I'm still a little confused.

[@Engineer4Free]

Take a look at this video: www.engineer4free.com/4/shearing-stress Shearing stress is more like trying to force a clean break, whereas bearing stress is more trying to crush the rod. The blocks in the other video would technically be exerting a bearing stress too, there is shearing stress in the rod occurring in the plane the lays between them.

[@yasirqazi2853]

can we have a video about stress concentration?

[@Cantthinkofaname40]

Thanks this was really useful!

[@Engineer4Free]

Awesome thanks for leaving the comment!!

[@Getcho33]

im still confused between block shear and bearing stress I know they use different equations. i cant find images of bearing stress failures. They just come out a little nudged near the bolt.

[@Engineer4Free]

Yeah I can see how they appear similar, watch videos 3 and 4 here: engineer4free.com/mechanics-of-materials

[@Getcho33]

Oh wow!! Thanks for this!!

[@Dulycup]

what is the app that this guy using to draw ?

[@Engineer4Free]

I've got a full list of the hardware and software that I use for the videos here: engineer4free.com/tools =)

[@reyortuoste2896]

Well explained! May I know the name of the program/software that you used in this video presentation? Thank you very much

[@Engineer4Free]

Thanks! Yep the full list of hardware and software that I use to make the videos is here: engineer4free.com/tools

[@reyortuoste2896]

@@Engineer4Free thank you once again. :)

[@georgepp98]

great videos!

[@Engineer4Free]

Thanks!!! 😁

[@vy4109]

I wish my lecturer can simplify the learning material like you do :) Btw, what app did you use for these presentations?

[@Engineer4Free]

Thanks! You can find all of the hardware and software that I use here: engineer4free.com/tools

[@sumiyaafrose]

The visual explanation is much helpful. Can I know which software you used for making these tutorials?

[@Engineer4Free]

Thanks for the comment Sumiya! I have a list of all the hardware and software that I use to make the videos at engineer4free.com/tools

[@MahKuanMing]

I've been wondering about your bearing stress...isn't supposed to be: pi x d x t ?

[@Engineer4Free]

See Mechanics of Materials by Beer and Johnston. It suffices at this level for bearing stress to be P/dt

[@parusharora9499]

Hey can you explain how is bearing stress different from compressive stress ?

[@Aki-wd7zx]

Thank you!

[@Engineer4Free]

You're welcome!

[@shwetagiram7273]

Very helpful

[@Engineer4Free]

Glad it helped, more at engineer4free.com/mechanics-of-material =)

[@miravchauhan8076]

What is the permissible/allowable bearing stress of plate?

[@TheHamzaalikhan]

It is a yield strength of a material multiply by 0.6. this will give you the allowable stress of a material.

[@mdwaqar7401]

thanks for the video helped alotttt....

[@Engineer4Free]

Awesome thanks for the comment! :)

[@a93397]

what is the area if the rod is not circular?

[@Engineer4Free]

If you are studying at the undergraduate level, just take the projection of the rod onto the other member such that direction of the force is normal to the projected area, like I did in this example.

[@a93397]

how to determine the projection area? lets say a square which it slightly rotated.

[@Engineer4Free]

Use basic geometry. If you know the angle it's been rotated at, draw it out. Use SOH CAH TOA to determine the new distance from axis of rotation to the top most fibres and bottom most fibres of the rod.

[@FGspa-eb9ti]

thank you sir

[@Engineer4Free]

Most welcome, thanks for watching! =)

[@Dulycup]

Thanks alot

[@HashemAljifri515]

This area is only for bearing stress which diameter*thickness

[@laweenkakil9681]

Where does 20 KN load come from ?

[@Engineer4Free]

It's the sum of the two 10kN point loads acting on the cylindrical member, but it transfers to the block on the single contact surface as 20kN.

[@zai6nab]

Thanks 🌹🌹

[@Engineer4Free]

You/re welcome, thanks for the roses!

[@johnwesleyballestar6174]

is it not the area πDT ?

[@Engineer4Free]

Hey John thanks for the question. You're right that the actual contact surface is πdt but because of the complicated nature of that surface being curved, it suffices to take the rectangular projection of the rod on to block, effectively making A=td. You will see this simplification made in mechanics of materials textbooks too.

[@cristianconstantin6496]

Also, what we are calculating is average bearing stress, so it probably is fair to just use the projected area. Keep in mind the stress will be different in every part because of the shape (more stress in center rather than top and bottom) - I think

[@williamhensley8322]

@@cristianconstantin6496 Yes, maximum stress is located in the same direction as the direction the force is being applied in and the minimum stress is going to be at +- pi/2 radians. The pressure increases and decreases in a sinusoidal manner, where at theta = 0, pressure equals maximum and theta = 90, pressure equals 0. In other words, cosine function would be used.

[@rabeulalambijoy]

Thanks

[@عاصمالفايز-ث4ض]

Now its clear thank you so much

[@Engineer4Free]

Ok glad to hear it. Thanks for commenting!!

[@md.isharulhaquenuhas4508]

❤

[@Engineer4Free]

💕