Mechanics of Materials: Lesson 4 - Shear Stress, Single and Double Shear Example

[@jerrysmith9780]

As a retired PE that still enjoys learning, this is where I come to get refreshed. Thank you Dr..

[@kevcopo]

you retired potential energy? I guess you're always on the move huh

[@PolyStylized]

@@kevcopo bruvv this killed me 🤣🤣

[@kevcopo]

@@PolyStylized this was the lamest comment I’ve ever made but I appreciate the love 😭😂

[@josephr1727]

Graduated from Tech 7 years ago and needed a refresher on double and single shear. Glad to see you are still an awesome instructor!!

[@robpratt811]

This man is the GOAT for mechanics courses... thank you for your service sir.

[@venusjajie2853]

That was really great. My prof explained it in an hour and I didnt understand a single piece of it. Not until this man came up. Thumbs up for you sir👍

[@logan9699]

Good sir, I am currently attending NDSU and have an absolutely horrible professor for mechanics of materials (the name of this course at NDSU). He doesn't explain everything, simply writes down equations and says (this is verbatim) 'this is so easy, how can you guys not get it just plug numbers into the equations', all while not explaining any of the theory behind anything of why we use each equation or anything. I've a test tomorrow, and I think you're going to be the man who saves me.

[@ihassan32755]

@11:23 The fly also loves your lectures!

[@mahdi7d1rostami]

I think after reading other comments now I undrestand why we are using resultant force instead of the Ax. The pin itself is perpendicular to the whole plane (you can imagine it sticking out of board) So because this pin is perpendicular to all other elements and the board (the plane which all elements are on it) the both Ay and Ax are perpendicular to pin and considered shear force.

[@mitchell6973]

It looked to me that Ay should have been the one used. I am not sure why he is using the resultant either as the components seemingly act on different areas.

[@640everett]

@@mitchell6973 The shear force would be the resultant because the pin is oriented with the end facing you. This means the normal force would be directed either into or out of the board.

[@raymondtrann]

Adding some further clarification. Background: our previous understanding of shear, is based on a *2D* diagram of a beam, with the shear force perpendicular to the longitudinal axis of the beam. For example we have the beam extending along the x axis, and thus the shear force extends along the y axis. In this problem: - Support A has support forces along x and y - Pin at A is oriented along z, with its faces/cross sections in the xy plane - The diagram in the example is a 2D depiction of an 3D structure (recall the pin has a longitudinal axis in the z-direction) Now, recall the shear force is perpendicular to the longitudinal axis. Since we have a 3D structure here, and the longitudinal axis of the pin is along the z axis, the shear forces on the pin will occur in both the x and y axes (both x and y axes are perpendicular to the z axis). Thus we use the resultant vector, or we can use the individual x and y components, and find the resultant.

[@shanthanramsamy7421]

Thank you so much, I was confused on this same thing

[@sydneykv]

Finished Statics and am so excited, and relieved, to see that you have a whole Solids and thermodynamics video set! You are a wonderful teacher. Thank you so much for your work in helping us all!

[@danielpaul460]

What major are u taking?

[@rutaremwa69]

This channel carried me through the statics part of the class now I'm trusting he'll get me through mechanics, so far so good

[@1234jhanson]

Best of luck to you this semester!

[@rutaremwa69]

@@1234jhanson thank you, I really appreciate everything you do!

[@CluckYou21]

Professor Hanson, keep up the amazing work! Your sense of humour is what keeps us engaged! Awesome!

[@ShelleySkyDoe]

This was the video that finally made me understood how simple shear stress is

[@wikiwaka7862]

Awesome video, love the fly who’s there for the lesson

[@moonchild5412]

Sir I cant thank you enough seriously Wishing you all the happiness and good health out there ♡

[@leafrancis400]

Jeff you're saving my grades right now i love you

[@leenardlunas2942]

Thank god I found this channel, It is a lot better than my professors.

[@ksfishchannel]

How can you teach me something perfectly in 15 minutes that my professor can't explain thoroughly in 55.... why am I paying tuition

[@laughtherapey]

Touche dawg. Some guys are just dumb. they can do things but can;t explain shisse.

[@thehoiya5280]

i was about to comment the same thing lol

[@camerongrabowski7734]

Funny how the internet as a free educational resource that anyone can use is commonly better than 40k a year in tuition...... Makes me wonder how much more educated the world could be if money was never a factor in education.

[@jakob.conrad]

@@camerongrabowski7734 Dr. Hansen’s salary is paid by Texas Tech tuition, soooooo

[@camerongrabowski7734]

@@jakob.conrad yeah but Texas tech tuition is not paying him to post this video. Edit: I’m guessing here that it’s not specifically for his students could be wrong though.

[@CaptainMoroni9]

11:20 that bug SHEAR seems STRESSed out !

[@chriscopeland1318]

All my profs at UT Austin had terrible accents. This is amazing…so clear. Freaking hilarious!

[@bnbmjtc]

i'm seeing this show from Brazil. Amazing! Fantastic. Congrat, professor!

[@diegoguatemala1520]

that " shingle" part kill me hahhahaha, best professor ever

[@giftonjalloh9644]

😂

[@mohanbabupm5778]

Gem of a professpr jef hanson amazing expln 0f single double shears high regards for you sir

[@PeterParker-sz2vm]

Legend! Best explanation I've heard

[@melikehabeeby]

You saved my life! I was gonna quit

[@venusjajie2853]

I'll share your vids prof Hanson, you're our hope👍👍

[@ledylan]

Finally, I can understand shear stress

[@TanzinaAfrin-z2x]

Hello, Thank you for sharing nice videos. I had a question, for calculating shear stress at A & B why are we taking resultant force? Will not be the force working on perpendicular to the cross-section, that mean Ay and Fbc?

[@דניאלאורהטל]

first of all shear strees creat by parallel force to the cross section. if you understand this then Ax and Ay are both parallel forces to the cross section.so you most calculate the resultant RA .

[@ChuckYuuu]

11:29 bug

[@TheRandyWCW]

Agree! Best professor ever!

[@alalebrahim1156]

you are the best man

[@harirusamiru2836]

bu videolar olmasa asla anlayamazdım bu konuyu. gerçekten iyi bir öğretmensiniz emekleriniz için teşekkür ederim. çok sağolun

[@yigitcan824]

Knk bu adamı dinlemek yeterli oluyor mu geçmede

[@harirusamiru2836]

@@yigitcan824 dersi geçemedim.

[@yigitcan824]

@@harirusamiru2836 Umarım bir dahakine geçersin,geçmek için ne yspılmasını önerirsin bu arada

[@harirusamiru2836]

@@yigitcan824 soruları önce kendim çözmeye çalışarak sonra da izleyerek konuyu net bir şekilde anladım. ama sorun anlamak değil. Hocaların sizi geçirmeyi isteyip istemediğine bağlı.

[@biridun9999]

ben önce bu hocadan izleyip sonra hibblerin kitabındaki soruları çözüyorum ve içinden geçiyorum sınavların. çok havalıyım.

[@sawenmahmud1998]

I'm in Kurdistan in Iraq. You're really good. I didn't understand this subject. You taught me very easily. Thank you very much I used to have a lot of fear of taking the Mechanics of Material exam but you made me do it but now I am looking forward to it thanks

[@greenphantom1955]

Thank you for this i am 1 step closed to passing my exams

[@mahdi7d1rostami]

10:45 We want to calculate the shear stress on pin so shouldn't we just use Ax because it's the shear force apllied on pin.

[@samvalmassoi4237]

Think of shear force as a “tearing” force. Both Ax and Ay are trying to tear the pin apart.

[@mitchell6973]

@@samvalmassoi4237 Yes, but they act on different cross sectional areas so how can you just get a resultant like that?

[@bahramaghakhan1024]

That’s my question too. How can we add shear to normal the way he did?

[@dannydang1507]

@@mitchell6973 they both acting on the same cross sectional area of the pin which is circular

[@abbastufan7508]

@@samvalmassoi4237 i got your point, thanks. samely, why dont we take into consider Ax and use resultant force when we consider normal stress?

[@lemmereact7589]

THANK YOU SO MUCH PROFESSOR .LOVE ALL THE WAY FROM PAKISTAN

[@mitchell6973]

Why are you taking the resultant on A for the shear stress? The x component acts on a different cross sectional area on the internals of the pin vs the y-component.

[@howardbaxter2514]

Both are acting across the same cross sectional area. Both are acting across the xz-face.

[@yakupcaferdemir3116]

SUCH AFUNNY MAN THANKS MR. JEFF

[@MixerRWB]

I jumped back from that interrupting fly

[@e6294]

thanks from Turkey❤️❤️❤️

[@panchita0476]

Thank you very much, very helpful and funny to watch !

[@meshalalkorbi7220]

can you explain more on why you used the resultant force at point A and not Ay to get the shear stress

[@mitchell6973]

Same problem here.

[@fadelity4444]

because the support at A has two components or reactions, Ay and Ax. So, using the resultant force of the two components gives you the more accurate reaction that is applied at support A, therefore, using resultant force to get the shear stress.

[@AlbertVathapally]

@@fadelity4444 I am still not following. If I was to obtain normal stress, id only use the x component of the force, but for shear why arent we using just y component of the force?

[@doublemg23]

love the sound effect at the end lol

[@manuboker1]

Wonderful Lectures ! Thanks.

[@oz6763]

God bless you Jeff

[@abdullahal-asmari5847]

Why did we take the resultant force since for shear stress the force needed that is tangential to the area correct? Im pretty sure the resultant of the forces isn't tangential correct?

[@panirys1226]

Because we are calculating shear stress acting on a pin

[@bermudezavilasandramariana6466]

Greta clases 🎉 Enjoyed yogur videos

[@Smileyyyy1806]

Thanks Prof. I thought at pin A we only include V as it is the perpendicular force.

[@x2iseynedir]

Did you understand why is it like that? If you do, can you explain?

[@marcosdyonel]

@@x2iseynedir I think, you always need to use the resultant force

[@hancar8142]

@@x2iseynedir pin bağlantısında pin, her iki yönde de kesmeye maruz kalıyor. Bu nedenle bileşke kuvveti aldı.

[@lucastadesse2678]

V is the force parallel to the cross-sectional area of the object. The pin is oriented so that it faces the whiteboard.

[@arpansingh2116]

why is the resultant force being considered for calculating shear stress.. shouldn't it be the tangential one...

[@surferrosa7285]

That's what I was thinking. For the single shear case wouldn't it be the 4/5 Fbc used as that is acting parallel to the plain in question?

[@hancar8142]

if it was a beam yes you are right. But it is a pin connection so pin is under shear stress either x and y direction. because of this we take the resultant force at the pin.

[@bardia8225]

@@hancar8142 Man I hate statics and solids. It makes no sense. It is never clear. Are we trying to calculate stresses on the pin or on the member. Last lesson he mentioned that axial forces acting on a member are considered normal stress. Then Ax has to be normal stress of the AB member. But if we talking about the pin, then Ax is obviously perpendicular to pin and is a shear stress. So which one is it.

[@hancar8142]

​@@bardia8225 if you look at the question its asking to find shear stress at pins. So if you analyse pin connection and if there are force acting tangential direction of the pin, there occurs shear stress on the pin. But it depends on what you are analyse on which part of the system. in order to be a normal stress on the pin, there must be a force acting z direction I mean the direction must be normal/perpendicular to the pin/body. English is not my native language I hope I helped you understand man.

[@bardia8225]

@@hancar8142 yeah kind of. thank you. One more question, why is the Ay considered in shear stress too? The Ay force only applies tension force on the pin. is that not true?

[@gabrielfonseca4014]

why do you set V to be the resultant of Ay and Ax when you said at the beginning V is the vertical component of the reaction force

[@aliarshad9739]

Thanks so much, professor!!!!!!!!!!!!!

[@MrLegendkiller230]

You are truly wonderful!

[@abduahmad4167]

I like your humor mixed lecture

[@johnrayapale2845]

Sir hanson, about the area of a circle. Isn't it pi*d^2 over 4?

[@edwardli9856]

pi*r^2 is the same

[@arsenioferrao4553]

Professor shouldn’t be used the force of 12.5 kN to calculate the shear stress in the pin A, being the force that will shear the pin because is actuating on the horizontal direction!?

[@howardbaxter2514]

Both Ay and Ax are acting parallel across the face of the pins. Therefore, both forces are shear forces. If there was a Az force, that would be the normal force.

[@howardbaxter2514]

Think about how the pin is along the z-plane

[@kingnegative44]

you blow my boring ass teacher outta the water sir. thanks professor hanson.

[@chriscopeland1318]

Loved it!

[@hunnybun4267]

"shingle" what a king

[@Sultanalrashidi]

Doctor, thank you for your lectures. I have learned a lot from them. I have one question. Why do we neglect point C I'm not sure if it is pin?

[@ShAlAmAnAyA3]

C isn't neglected. It's one end of the two force member F_BC and its value is already calculated.

[@monaji222]

10:51 please HELP !!! How do he got 35.6 ? Can someone explain how to find the 'V'.

[@monaji222]

Never mine, just got it. V^2 = 12.52^2 + 33.342^2 V = 35.6

[@josehernandeznegrete637]

Damn I found these videos after rating 1/5 to my Mechanics of Materials professor

[@tomacap7772]

Pull up to UT fr, we need you here

[@savinduperera6454]

great help sir

[@melanieabreu9268]

The fly is a paid actor lolll

[@AmmarAlGhafri-h8y]

shouldn't we say the v for the single shear is 20.8kN -12.5kN instead of 20.8kN?? thancks for the great lecture

[@giftonjalloh9644]

Thanks prof.

[@ryanisaacson118]

can you do a theory of structure class

[@danmarkcacatian5874]

Area is given right 20mm? is a radius or diameter? because the formula of diameter is π/4 (d)^2?

[@AB-gu9ui]

The diametre of the pin was given as 20mm so we can half it to find the radius and use π*r^2 or use π/4 (d)^2 with the diametre

[@DrJuiceMan]

I’m not sure about the area of the circle because we were given the diameter, can’t it be rewritten pi*d^2/2 or something else

[@bag-aogianresterdaquis2296]

its okay men cause d/2 = r so the given diameter is 20 and you divide it by two so that you can get the radius and use the πr² as a formula. anyways you can use also the πd²/4

[@Abdulwahab-bl2vd]

Thank you Prof for this video. Although I hae got a question. Shouldn't the Force used in calculating the Shear stress be the shear force and not the resultant force as used in the calculation. Thank you as you dispel my misconception and calrifying imthe resultant force should be used. Thank you sir

[@multipleorganfailure3449]

Professor, YOU ARE MY FUCKING GOD.

[@bardia8225]

I don't get it. Why do you consider Ax as part of the shear stress. On lesson 2 you said that the forces acting axially on a member are considered normal stress. So Ax must be a normal stress

[@AB-gu9ui]

Its because we are calculating the shear stress on the pin, not the beam. Think about the pin and you will realise that both Ax and Ay is perpendicular to the pin (as the pin is going into the page)

[@brandonf3865]

@@AB-gu9ui thank you this was the best explanation. The top down view I think confused me some

[@abbastufan7508]

@@AB-gu9ui so do we have to use Ay with Ax when calculating the normal stress on the pin?

[@bahramaghakhan1024]

Thanks Professor The only thing i'm trying to understand is how you did a sum between normal & shear stress while they are not on the same surface

[@PRAKETAKSHANTALA]

When drawing the FBD of bar AB, why do we neglect drawing the reaction forces on Pin B? Does that mean that we can eliminate drawing reaction forces at a joint if there is another two force member which emerges from that same joint?

[@insmnac6526]

thank you

[@insmnac6526]

so much

[@insmnac6526]

because i have no idea what im doing here

[@reubendsouza6465]

thanks

[@hilarymartins-udeze4315]

How can you tell that it is a double shear stress or a single stress? Is it given in the question?

[@paezmayor]

¿Dónde estabas cuando te necesité en el 2016? :(

[@dikansijus]

at 7:24 i believe its 4.8 Fbc, none of it makes a lot of difference

[@michaelmurakami3487]

examples for inclined beam sir pls. im confused in that thanks

[@thegreatestotaku9567]

This online lectures is gonna carry me, my prof. is terrible lecturer

[@rcp_clsr999]

why did we use the resultant force while calculating the shear stress? i thought we were supposed to use the force component which is parallel to the surface area

[@shadkamal1669]

Ok but we just need vertical force for shear or both normal and vertical????

[@tolgaaka2110]

Bu nabarlaaaa

[@moanakeni2734]

why you don't use only y axis forces

[@merttrenova8700]

Allah izin verirse sensin abimmm

[@ashikrahman5370]

0:30 I hold my breath if you didn't do miki mouse voice.

[@smoothsmith123]

these ads have to stop

[@FHale-fw5ny]

2x3 6 alright calculator :D

[@worldnews5790]

Can you have your Facebook account?

[@thatamazingguy6926]

ucalgary enme317 gang?

[@jorgetrevino325]

11:30 bug

[@jorgetrevino325]

Haha