Beam Deflections with Singularity Functions (Example 2) - Mechanics of Materials

[@MrSmileyFella2]

Thank you for this. The whole social distancing thing forced us to figure out these topics ourselves cause the lecturers can't do much now. The material they gave was so insufficient. I was about to give up until I found this video. p/s: 6 years later and this guy's still hearting comments. What a legend.

[@chloeheartsdaisies]

Same here! A literal lifesaver thank you

[@alexm9141]

same🙏🏼

[@ewan12kd]

James Franco teaching me maths. Thank you so much you have saved me for this exam.

[@Anonymous-ur3te]

you are a life saver saw 100 vids yours was best explained and simplified

[@Euphoricbryanna]

i love you. cramming for my final tomorrow with a newborn asleep on my chest! these videos are so amazing! thank you so much !!!

[@shwater5898]

you are the best teacher. you earned a sub please never stop teaching.

[@Mtmtmtmtmtmtmtmtmtmtmt]

You're an absolute life saver!! I could not, for the life of me, grasp this method so tq so much

[@SIMPLYxISxPRO]

I love these videos. For starters, you sound like James Franco and that just makes learning so much more enjoyable. Secondly, your teaching style just flows really well and you always use great examples. Quality videos every time.

[@structurefree]

SIMPLYxISxPRO Thank you for the very nice and detailed comment about my teaching style! Although, I suspect the only thing James Franco and I have in common is a love-hate relationship with spiderman.

[@SIMPLYxISxPRO]

structurefree Your next video should be on calculating the stress Harry Osborn went through when spiderman killed his dad

[@5stepshred300]

Great example!! I learned so much from this video. I did not have things in my notes such as if you have a distributed load in the center, you have to grab both bounds. Very helpful!

[@baohandsome5706]

I love you, my prof was not doing great with his lecturing XD This helps a lot, thank you

[@ashkan76]

Thanks so much for posting these videos.. they seriously saved me.

[@cherrytree5026]

such an amazing channel man you just saved me from the quiz i have tomorrow

[@luisramrod9121]

Explained better than my professor which I am paying hundreds of dollars to teach me.

[@gillian5917]

Great!! Finally understood how to make a moment equation using singularity functions. Thank you so much for this!!!!

[@strugglers34]

Its too good and easy to understand,, Thank you very much..

[@nittikorn]

You make Mechanical Eng fun! lol Been watching since I was in Foundation, currently in Year 2~

[@structurefree]

🙏

[@interestingcan]

Thanks for the videos, helps a lot. Really appreciate them!

[@SunderlandMorrow]

love this shit, thanks buddy hope you're out there in the world crushin it

[@BboyDaquack]

solid explanation! thanks for your vids. you're saving my butt here :D

[@onedepressedengineer_2003]

What will happen if we place the concentrated moment somewhere along the length and not on the edge??? Will we count only 5KNm?.. 4:40

[@perpetututuator]

sory why when 2kn/m uniformal distrubuted loads moment is divided to 2?

[@moderatex6049]

if you look at the formula sheet you'll see for a dispersed load M = (pressure/2)^2

[@AshanPeirisNCVU]

Thanks man,You really saved my day :)

[@christinaadjiman5695]

Hello, thank you so much for the video I would like to know what would have happened for the very first singularity function if there was a roller at C? Thank you

[@vandalucas5999]

Your videos are amazing and SUPER helpful! Why go to lectures when you have structurefree?

[@crechecreepy4190]

ur right, but am kinda regretting not going to lectures given that i gotta learn the whole curriculum by watching this guy the day before the test hahaha

[@brunoprado8192]

Bro, you are insane. Thanks !!

[@structurefree]

insane in the membrane. :)

[@leunggordon766]

Hi thx for the videos, they helped me a lot. However, not sure if you will still answer my question now... or anyone can help... So, in 3:18, the first value of the moment function is in positive instead of negative? It is in clockwise direction so I think it should be in negative... this confusing me, and thx

[@NathanaelTak]

It comes from integrating the (x-2)^1 from the shear equation... it becomes more easily apparent when you start with the loading function W(x) and integrate to get V(x) and again to M(x)->slope->deflection

[@gryonr96]

Great video, thank you very much! But I have a question. If EI not constant, i.e. if the some section of the beam at the middle has different EI, can I still use the singularity functions to find deflection equationof the beam?

[@lauragiselle1429]

how do we know what power to use

[@TylerTimeee]

yooo thanks so much man. I love your videos

[@JaxRayne]

Would you be able to find the moment diagram through this method? If so, is it possible if you could explain how? Thanks :)

[@rohitk8690]

Hi. Thanks for the video. I just tried example my own. I took different Boundary condition. I consider at roller support slope is 0. It gave me different value for C1. So it means that final answer depends on boundary condition which you apply.

[@nathanielpeza8504]

When we start integrating from the loading function, why isn't there any constants of integration? Because I've noticed that the constants only show up for the slope and deflection functions.

[@CK-zj4ts]

Thank you. You are awesome!

[@Xari0n92]

Cool vid man grtz from holland!

[@structurefree]

+XariOn vondutch Thank you from calif., usa! #placesthatarefarapartinreallifebutcloseontheinternet

[@yusufmoola9937]

Hi. How would you go about solving a problem where there is a triangular distributed load starting at 0 and increasing to half or quarter of the beam? with the linearly distributed load you mentioned that according to the equation, the load continues to the end and has to have the non existent part cancelled off. how would you cancel off a triangular distributed load?

[@elta8064]

does this method work with any type of support ?

[@evanburgs5104]

So, anti clockwise and clockwise moment are both positive on the singularity function? Great video,anyways!!

[@subratkumar7534]

hi! why dont we directly integrate the load function. I guess we would get 4 boundary conditions instead of 2.Thank you! Anyways great video for learners! Keep it up!

[@md.hossainnadim1243]

As per as I know, singularity method can't be directly used in beams with internal hinge (slope discontinuity). But I have also seen only one lecture note where a slope discontinuity function was added into the equation similar to load and moment. Do you know something about this or can you suggest some literature regarding this. Thanks

[@GC25-s6f]

@structurefree, I am just wondering ,how do you determine the powers of the singularity function when you start with the moment equation? I am a little confused on that

[@xAzx2]

How do you go about finding the max deflections using these equations, i try making the slope equal to 0 and solving for x on a beam with symmetrical loading, but the x value doesn't come out to be the halfway point as expected? What am i doing wrong?

[@structurefree]

Az your method is good. for this example the beam geometry and loading are not symmetrical, therefore the max deflection does not occur at midspan.

[@WonderingNobody]

What if the loading is not UDL.If triangular starting from different point of the Beam?How to write loading function using heavy side notation then?

[@structurefree]

I Wonder i think i have an explanation video on singularity functions that might be helpful.

[@jfluckey]

This was kind of asked below, but if you have a plywood beam that slopes from its lowest points from its ends to its maximum point at the center how do you handle the varying moment of inertia? In the standard example, when I is a constant, EIy''=M, EIy`=EI*slope, and EIy=EI*deflection. Is it just a straight replacement of I with I(x) [EI(x)y] or does I being a function of location complicate the integration?

[@structurefree]

Yes, for a tapered beam the MOI would be included with the integration. Ooooh, that would be a fun and challenging.

[@jfluckey]

Not too challenging when you use math software. I created a Mathcad calculation that will accept 7 supports (fixed or pinned), 10 point loads, 10 distributed loads, and 10 ramp loads. Every thing gets thrown in a matrix that solves for all the reactions and c1/c2. Plot out the shear, moment, and deflection diagrams and you're the talk of the town (at least I imagine that is what is town worthy talk). Very handy for this structural engineer.

[@structurefree]

Juston Fluckey Sounds cool. I'm a big fan of Mathcad too. Definitely, "talk of the town" for a city full of engineers.

[@kaankorkmaz93]

How would you set up the load equation if there is a triangular distributed load? Thanks for the videos!

[@structurefree]

+Kaan Korkmaz I think this introductory video on using singularity functions might help you figure out how to handle a triangular (or linearly distributed) load. kzbin.info/www/bejne/fJrCZaGPbZVrh5o

[@kaankorkmaz93]

I appreciate it!

[@floormatt3]

+Kaan Korkmaz in case you did not want to follow the video: The singularity function for a triangularly distributed load is M = (m/6)^3. m is the slop, which is rise/run.

[@zalianurshahirah6473]

How about if udl. hinge and roller at the back. inverted version? should i include the Rb?

[@byronlarsen1936]

I understand you sample but and I have seen another exaple with a l uniform type wedge load that starts at point away from the left reaction and increases short of the right reaction. I now that the original load has to continue to the right reaction and a counter wedge shown from the end of the wedge load to the right reation. but I can't come up with a formula that gives me the proper deflections and curvature. I have it working fine when the wedge goes all the way to the right reaction

[@engniazi8581]

Hi, How will you define/describe the equation if the distributed load is all over the beam?

[@conorlee5812]

Does anyone know how you are meant to know what powers to use for the moment function?

[@structurefree]

you may want to watch this one...kzbin.info/www/bejne/fJrCZaGPbZVrh5o

[@ginnys9937]

Why is it that you can just jump to the moment equations with no constants of integration from going to shear and then to the moment equation? You only use constants for the last two equations. Do they just always cancel out to zero?

[@structurefree]

Ginny Schilling you could start from a loading equation and integrate 4 times to get displacements. There will be constants with each integration and for shear and moment functions those constants would be the support reactions of the beam.

[@jacobbelmontes4899]

I know your videos are already quite long but I wish you took it just 1 step further to find slope and deflection at some point, any point for that matter. My confusion is what do we do when a term inside the parenthesis is negative? For example, lets say we wish to know deflection at 6m. Your 4th term would look like this: (4.625/6)KN(6m-8m)^3 Does this mean that this term is zero since according to step functions x is less than or equal to 8m ? Or does step function rules not apply at this point in the method and the actual value for that term is -6.1666667KN*m^3 Please let me know at your earliest convenience and thank you in advance

[@brandonmaddox5565]

how could we find the max deflection?

[@structurefree]

One way would be to identify where the slope is zero.

[@mehedimasudrafi9145]

How can i understand the powers will be either 0/1/2 over singularly brackets?

[@Minecraftmayhem92]

Is it fair to say that the slope of the beam is 0 at x = 0 and then you plug that into the slope equation getting C1 = 0

[@owenbarnes4066]

So is double integration method with a singularity function just the same as Macaulay's method? Also man your videos are so helpful it's unreal.

[@structurefree]

Owen Barnes Yes. Thank you for the kind words.

[@alabaska182]

what if it's a triangle distribution I'm a bit confused from the solution in my text book. Can you still "turn it on and off" but instead of 2 / 2 it would be the slope / 6 ? but wouldn't the "on" position be 0 ?

[@thatcooldude8422]

hi , can u explain how can we choose which power of n we shud use , like in previous video you used for distributed load the power of 0 but here u are choosing it to be 1, why is that? P.S. u and ur videos are awesome

[@steaminggreentea]

+Surayans Tiwari in this video he's using the singularity function for M(x), so for UDL it would be -w/2^2, while the previous video was describing the w(x) function which is why n = 0

[@Captain14414]

Wouldn't the integral of the moment function be the shear function? How does the integral of (d^2v/dx^2) get you the slope? shouldn't the integral of d^2v/dx^2 = d^3v/dx^3 ?

[@derekhanson9899]

If you use Slope at x=0 is 0, then you get C1=0. Is this correct, because I am a little confused on why you chose to use y(8)=0 instead.

[@zackarystewart9214]

what if you only have 1 boundary condition. lets say its positioned in the center of the beam and you want to know the deflection on both sides. Do you just plug in your solo boundary condition and hope that both c1 and c2 are zero

[@kindingiliray2169]

great video. but would be better if you started the video from the 4th derivative equation( load equation) EIv= -q(x)

[@hweinheimer18]

What program do you use to draw this all out?

[@hanzalakhan1401]

using singularity function and double integration method both are same or difference between this two method

[@sabaalkhulaifi8272]

thank you so much for this information

[@dvdcar90]

Great explanation, but I have one question why is the last part of the equation M(x)/EI or w(x) always zero in this case 5(x-11)^0? I noticed you did this in example 1, but I'm not quite understand why. Thanks

[@ogunstega7348]

Because X will never ge greater than 11 (which is the length of the beam) so that singularity function will not be active

[@azrarajwani8023]

For the distributive loading, why is it 2kNm/2? thanks!

[@hanzalakhan1401]

when we finding reation at A and B, during summation of moment at A. the concentrated load at C 5 k.N. you only subtract it. are we can multiply with the span upto A means -5*11

[@edwardRsa]

Great stuff,thank you

[@dan0themano575]

I can I use singularity equations to solve ANY deflection slope problem? I'm wondering can I use singularity on a beam that has different moments of inertia (fat part and then a small part), do I Just use 4 equations, 2 for each slope/deflection with each corresponding moment of inertia? Thanks =d

[@dustinbrennan5271]

Ya. you can do that. You just need to do like in the double integration method where you had different equations for different sections of the beam except in this method you have different equations for different material properties. it would be a huge pain to do with more complicated beams

[@structuralanalysis2790]

info:I designed a 5 by 5 matrix by using An Integral method to solve for the unknown variables :R(a)=27/8,R(b)=37/8,Slope(a)=68/3,Slope(b)=1 and Y(c)=-51/2 end of the beam.

[@bossla331]

Can we find Max Deflection using singularity functions?

[@ameyerstv]

Why do you find v''(x) at first rather than v''''(x) and integrating 4 times?

[@chillouttheworld6516]

How to find C1 if we have only one boundary condition only one end fixed ?

[@structurefree]

° Chill out the World ° on a fixed end there are two boundary conditions. displacement = 0 and slope = 0.

[@ahmed.alshawi]

structurefree Hello sir, How to find the C1 if we have beam with three hinges supported, it is symmetric with three unknowns. The slope for hinge is not = 0, right? Thanks in advance.

[@jfluckey]

Ahmed Alshawi You are correct. One of your boundary conditions would be that the moment at the hinge location is zero.

[@sunmichoi6888]

Thank you so so so so so so much

[@puihangyan5416]

i have one question sir,y c2 is0 when u do x=10?

[@haidilhelqi5519]

why you put +ve sign in calculation for moment 5kN.m anticlockwise ? thank you..

[@lloyddull6170]

haidil helqi He does assume that anti-clockwise is positive, but as you know that in the bending moment diagram when there is an external moment acting clockwise on the beam the graph shoots upwards in the positive direction. Thus making it positive in his singularity equation. That is what I understand it to be. Hope that helps.

[@Mamoonaria]

which software can we use for lectures, I need to have the name of software that u already have it tnx :)

[@comeatmebrahhhhhhhh]

YOur a hero!

[@marcuss5330]

What shall I do if my EI is not constant?

[@carultch]

Then EI will be a function of x, when you integrate curvature to find slope and deflection. You'll have a product of two functions, M(x) and EI(x) as your initial integrand.

[@muhammadrifqifauziwijaya8230]

how to get the deflection of beam on 8m is zero?

[@giftd4002]

Anyone know why the distributed load was divided by 2?

[@structurefree]

Watch this introductory video and you will know....kzbin.info/www/bejne/fJrCZaGPbZVrh5o

[@celine5461]

Thanks for the video!

[@raj76471]

Thank you! Sir.

[@prasaddeshmukh2605]

c1 should be unitless , how u can write it . it is scalar as i know

[@siddiqisah1452]

That was good but you sped up a little at the end...thanks a lot though!

[@dustinbrennan5271]

khan ain't got noth on this =)

[@josueestradamartinez1601]

I love you

[@fangli2867]

C2 was 181 when i put it in my calculator

[@othescientist]

YEEEEEEEEEEEE

[@structurefree]

+Fricks and Fracks YEEEEEEEEEEE...S?

[@puihangyan5416]

x=8 sorry

[@batuhan7796]

please check gmail please

[@jacobbelmontes4899]

I know your videos are already quite long but I wish you took it just 1 step further to find slope and deflection at some point, any point for that matter. My confusion is what do we do when a term inside the parenthesis is negative? For example, lets say we wish to know deflection at 6m. Your 4th term would look like this: (4.625/6)KN(6m-8m)^3 Does this mean that this term is zero since according to step functions x is less than or equal to 8m ? Or does step function rules not apply at this point in the method and the actual value for that term is -6.1666667KN*m^3 Please let me know at your earliest convenience and thank you in advance