Slope and Deflection (Double Integration Method) - Problem 1 - Slope and Deflection of Beams

Subject - Strength of Materials
Video Name - Slope and Deflection (Double Integration Method) - Problem 1
Chapter - Slope and Deflection of Beams
Faculty - Prof. Zafar Shaikh
Watch the video lecture on Topic Slope and Deflection (Double Integration Method) - Problem 1 of Subject Strength of Materials by Professor Zafar Shaikh. Watch the Complete lecture on Engineering Academics with Ekeeda.
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[@boomdaus]

one of my ears enjoyed the vid

[@THE_BLACKMAN]

😂😂😂😂😂😂

[@nishantgawai4547]

😂😂

[@adityasharma11057]

The way You Taught Me In Macaulay Method Solving technique Was Very Good. Day After Tomorrow is my exam And I did not prepare this topic numericals. But Now After watching This Video I understood How to do this Numerical Thank you Sir.

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[@sankojukarthik1621]

A Great THANKS to you sir with your explanation i had pass this SOM with 20marks. I am commenting on your video after seeing my results immediately. Saw multi times until you understand practice well, practice to do it fast because it takes much time. A Good thanks to you sir. GOD BLESS YOU SIR

[@mightmech292]

I literally listen this topic in online class no. Of times but no use when I watch ur video the way u teach very simple tq sir ❤

[@yadikishameer9587]

When my friends have doubts ,I just help them by refering this channel 🤗🤗🤗

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[@tutingrubena1584]

ekeeda why are you not replying to some concerns regarding the video itself?

[@rajatchoudhary7772]

bro i read it one day befor exam and i got it completly clear my rest of my life thankyu bro :)

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[@X-MAX11]

MESMERISING TEACHING SKILL..... JUST AMAZING..... MORE POWER TO U SIR.... LOVE FROM A DUFFER FROM KASHMIR

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[@lightningmcdweeb]

khud ko duffer volta hai xD

[@vuyisananisipango3796]

Hello sir thank you so much for your tutorial, but I'm a bit confused. According to intergration method ain't [x-a]

[@sandeepsingh4042]

if moments is a acting at the end b of the beam then is not considered because x-L = L-L=0 m i right?

[@fariakhan9829]

when my explained this method i didn't understood. .but now i can do it....bcz u explained superb

[@bomzan_22]

its so great to learn from u sir.

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[@yashsingh917]

As per macaulay's method, you have to take the expressions zero whenever the negative or zero occurs in any expression. but here, while calculating deflection you didn't take into this considerations which gives wrong answer. please clarify this. Thanks.

[@vuyisananisipango3796]

Exactly where my confusion is, this guy here doesn't neglect simplifications that gives out -ve

[@marykathleensatuita129]

so it should be that the value of x which is equal to zero, must be substituted to all terms? Im confused too, sorry :((

[@tutingrubena1584]

@@marykathleensatuita129 that's what also I was thinking

[@yash_bot]

So the ans is, the when the value inside the brackets become negative, then the term is neglected

[@micky6045]

Fantastic Video Sir....Wish to know more but it is quite unfortunate the Ekeeda app is not in my dear country, Ghana..I know it is going to be a great app when I start using it and which of course I will gladly recommend to my friends.....Counting on your cooperation.

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[@suchithraperera]

sir double integration method not equal to Macaulay's Method. PLEASE CORRECT ME IF IM WRONG

[@manishmessi6423]

The method of solving this question using the double integration method involves finding moment equations for 3 cross sections. Find the displacement and slope equations using double integration and integration of momentum equations. Using the principle of continuity and boundary conditions, find the constants, which in this case would be 6 constants. Substitute the respective constants in the slope and displacement equations of the middle section and then substitute 4.5 in the place of x to find the displacement and slope at the middle of the beam. I don't understand in the video how only the momentum equation of the third cross section is used to represent the momentum equation for the whole beam as done by you. Can you please clarify?

[@GULAA]

i am a little bit confuse , is it allright to only cut one time ? since my lecterur taught me to cut into a few section .she said that cut only one section is for Macaulay only, cant u please explain to me ?

[@sulthanahaneefa4543]

Thankz a lot sir...

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[@sasisavocations5990]

Brilliant teaching

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[@Manjunath_R]

Sir while finding (dy/dx)max (0-4)^2 it will be assumed as o right in our class sir did like that

[@aliafarhanna6826]

Thanks for sharing such a good video for my studies 😄

[@patohnaxh8194]

i just learnt a lot. thanks so much am actually preparing for my structures paper....would you share the link for downloading the app?

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[@factfenz7032]

Thanks u sir your way of teaching is very good But sir i have little confusion my teacher teaches that when we put x=0 then we have to consider only first section not the whole section but in this while you finding slope you put value of x=0 in all the sections Plz plz plz reply me sir my semester is coming near plz sir

[@swapnilk.b1396]

still waiting for reply?

[@venapusasravan3969]

Thank you, sir, for such a nice explaination

[@frogswell8964]

Sir udl Ko point load mei convert krne k liye , purey span ka square krke usko divided by 2 krna hota hai , aur usko multiplied by self weight. Fir yaha self weight ko sidha multiplied kyu Kia span length se?

[@pavansankar2901]

Sir you are little bit slow at explanation but very clear....... THANK YOU SIR

[@phanimaxx4196]

Thanks for your videos !

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[@amutong8856]

Sir u made the problm very easy thank you.

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[@syedubaidkashif2940]

Sir u r doing the exact questions with the exact readings as mentioned in my IA question papers Love u 3000❤️

[@ratanmee4581]

why did it take so longg " We recorded your response. We will get back to you shortly"

[@kariukinyaga9531]

Thanks very much

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[@PankajMadwan]

Thank you so much!

[@chakshumotta225]

You are THE BEST!

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[@aniljoshi1001]

sir this video is awesone but i have one question and i.e. Why 1st boundary condition is applied only upto c2 at 18:03?

[@Ekeeda]

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[@Sumitsingh-th3wl]

my question is that how can we take at x=9 y=0?

[@shehbaz149]

Sumit singh because in simply supported beam 'y' or deflection is zero at end supports

[@bheeshmad6562]

@@Sumitsingh-th3wl for simply supported beam, deflection is zero at supports

[@bheeshmad6562]

according to macaulay method ...if the value in brackets(even if it is square) is negative then we should not consider those values

[@laksita6445]

Thank you very much sir 🥰

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[@laksita6445]

@@Ekeeda thank you very much sir, you are a good teacher 😇

[@nadineblackrose]

when you were substituting with x=0 at(18:00), why did you substitute only for the first section of integration. Thank you

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[@muzamelpamir4336]

Nadine Hossam Because in simply supported beam slope and deflection is zero at the ends . And also we are neglecting the parts that is (-ve) so if u apply the Conditions for all the equation again u will get the same value.

[@nomipakistanli]

you mean if we put x = 0 to all the equation C2 will turn out to be zero again?. İ TRİED BUT its not zero

[@afijithsp5413]

same dought ?

[@amoorhjeer380]

exact same question for my assignment dang!!! all i have to do is copy and past !!!! (wonder how lucky i got)

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[@vinodvinnu1121]

Sir excellent teaching

[@pjanardhanreddy92]

Sir please add conjugate beam concepts and solved problems

[@kieranbutroid4631]

Hello there, very informative video. I have one question, at 10.45, the UDL is 26x^2/2. Why is x squared? Thanks

[@nainam007don]

Kieran Butroid 26x is the total weight due to UDL upto distance x.. Also centriod of UDL is at half of the length which means total load due to UDL acts at x/2 Thus Moment due to UDL = (Total load due to UDL) x (Distance of centroidal point of UDL from the point about which moment is to be found) = (26x . x/2) = 26x² /2

[@Babe_Chinwendum]

It does not have sound?

[@budahtsiki4723]

Thank you✨🌠

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[@cricketnews_guru]

What are the units of deflection (y) ..m or mm

[@matovuhuzaifah1764]

Thanx alot

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[@rajata2819]

Simple explained good video sir

[@malisettisrikrishnasanthos3308]

why you didn't substitute x=0 in the whole equation for 1st boundary equation

[@rakshitsharma5177]

Yeah Bro same problem

[@tutingrubena1584]

until now he didn't reply

[@tutingrubena1584]

i badly need his answers rn 😢

[@shreyashavaldar2076]

+1, same doubt

[@DebasmitaSahoo-ve3uy]

What is the difference between double integration method and Macaulay method??

[@muhammadsheharyar2001]

Sir for maximum slope at first support we need to put the value of x is zero in first equation??it's just a point of view

[@kishanpatel7090]

please make more video about other beams and suppors and really you are best sir for us thanks sir and please make more video

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[@user-mp5dg9tb4j]

If the value in the bracket is negative, it supposed should be ignored, right?

[@gopikanchapogu5536]

Excellent explanation

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[@priyadharshinipriya5703]

Is double integration and Macaulay method are same?

[@__saqulainsayeed_SVNIT]

u r putting 4.5 in the bracket (x-6)^3 which is not valid , we need to put according to domain defined earlier . put only x>= 6 in bracket (x-6)^3

[@SamoJamo]

love you man

[@civilexplorer7038]

Very well explained.

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[@vudumalachowdaiah841]

Exallent sir

[@ethandavies3016]

Thank you sir!

[@itsabhi4ru]

Thank you sir. Your teaching style is awesome.

[@srujithappu5318]

Our semester god😍

[@karanbanga8790]

Hello sir, Well explained! i have a doubt see the equation at 11:29 ... instead of 125{x-6) it should be 125(x-8) ? or either DB Should be 3.. THANKS

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[@micky6045]

With reference back to the main question...It said the beam carried 125KN at 6m from the left support.😉

[@mandlasiddhartha6531]

Nice sir 👍❤️

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[@sumitkale]

nice sir very good explain thanky you

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[@MukeshMaurya-kc2cx]

Thank you sir 🙏

[@phillipzikhali3593]

Well explained..thank you!

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[@sajad7935]

Thanks very clear and easy 😍😍😍😍😍

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[@atsm141]

sir when we calculate slope at zero according to Macaulay's brackets we not consider negitive sign.but here ( x-2)² negitive distance because we consider (x-2) whole plzz explain ..sir

[@mgudisilibazi9607]

i'm also suprised with the way he did it

[@shivambarbate7842]

Thankyou very much sir !

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[@sidheshwaran9205]

Its very usefull...tq

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[@keshavrajjoshi9336]

thanks dear

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[@princerajput7506]

sir in this i have confusion that while finding slope i.e max slope at x=0 i.e at point A...u have taken the whole slope=q for finding slope at point A.......but in RK Rajput book they are taking the section up to the distance........i.e if same question i would solve according to d book it whould be......at point A x=0......they take d slope =q till d section A -C i.e (322÷2)x^2 - 2160.86 and that gives d complete differt answer

[@princerajput7506]

sir plzz do comment and another question from my side is......if in d same question if they asked to find d max . deflection i.e( at what section and at what distance from A)how do we solve

[@ManishKumar-ci2tu]

Also guide us on calculating slopes at supports of SSB carrying moments at supports only.

[@speakingbot608]

thank you sir

[@anilchandgarabarapu2113]

EI(d^2y/dx^2)= -Mx(negative of M) i think so..what do you say sir

[@Kuldeeprao447]

sir 6:00 to 6:15 the distance is taken 3m but distance is 1m.???????????????

[@rashidulislam1156]

Sir the beam is symmetric the slopes at both the ends should be equal na

[@sandeepsingh4042]

hi sir if there is moment at B then how we will solve it

[@EngineeringPlanets]

No need to waste time in colleges u teach better than them.

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[@rakeshmahato3347]

I am little bit confuse a another video of Ekeeda i saw 125(X-2)/6 Taken be zero when x =0

[@freeforall3160]

Sir, Please make video on moment area method....

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[@tutingrubena1584]

why you only put x = 0 on the first term of equation 2 sir? how about the other terms? isnt c2 = 390??

[@vax8790]

Thanx sir.....

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[@akhlaqurrahman1722]

Appreciable

[@ShahbazAlam-gi2fs]

When put the first boundary condition x=0 only first section why put in overall equation plz solve my problem sir

[@reshmithramesh7110]

well explained. thanks.

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[@GUGULOTHU_RAJKUMAR]

Tq... Sir...... 👍👍👍👍👍👍👍

[@kajalmane6400]

Sir can you please make videos on moment area theorem and conjugate beam method...

[@ashokgaikwad9866]

why Rb is not taken

[@kashibhojia3797]

Sir how we find x for maximam drflection if we don't know where maximam deflection occure

[@Chandu_devavarapu]

kashi Bhojia for a simply supported beam maximum deflection will always occur at the center

[@mubasshirhussain1622]

Simply put slope equation equal to zero and find x Because at the point where deflection is maximum slope is zero

[@princerajput7506]

mubasshir hussain .....yha but d calculations part will be difficult......

[@princerajput7506]

CHANDU DEVAVARAPU yha but only for symmetrical loading.....what about unsymmetrical loading

[@jaisingal1998]

Hlo Sir In this question you have taken boundary condition as the initial point x=0 and x=9.So i want to ask, it is neceassary to take the initial and final values of x.Or any mid point can be taken??.Plz ans. Sir

[@kos6779]

Usually its initial and final, every question i solved i used initial and final

[@suhailrather985]

Sir max deflection galt nekalrehey hai ap quu kee 4.5 lies in 2nd part of the beam

[@imbernardbayla]

nice tutorial! may I ask if why wont you double integrate the 4th equation? EIy?

[@walswals175]

Sir, thanks.

[@xavialonso7539]

why is c2 always 0???

[@princerajput7506]

xavi alonso ....bcz it depend on d end condition or boundary condition to find C1nd C2.......nd end conditions r only used to find d value of C1 nd C2.....nd this end condition says that we have to put deflection or slope =0.....since the slope at d end is max....nd deflection at support is zero...thats y we put y=o in deflection =q but if same question whould be asked for cantilever beam than d end conditions will be changed. ....nd that is......at fixed slope there is no slope nd no deflection.....so from here u put slope = 0 in slope =q to find C1 nd than y=0 in deflection=q to find C2

[@harshadkamble8214]

Respect from Mumbai sir thanks u so muchh

[@TejSwaroop]

12:45 : How is integral of (x-2) = (x-2)^2/2 ? Should it be x^2/2+2X .

[@darsafwan6951]

Tej Swaroop here integration is taken as a whole... i.e (x-2) as a whole is taken as variable

[@nimalsenna]

for this method, integration should taken as whole to get correct answer...just follow the steps

[@uniconhub]

Thank u

[@engineersplay8536]

Great Work Done Sir!

[@najamshah5186]

Sir while cutting section we ignored the moment caused by reaction at support B. Can you please tell me the reason why can't we consider that moment for this question? Because for deflection we have to consider all moments caused by all foces and reactions

[@Ekeeda]

Kindly download the Ekeeda Learning App ekeeda.com/download.html & ask your doubts. Our experts help you to resolve queries on chat.

[@PawanKumar-di9mn]

Sir this is not double integration method... This is Macaulay's method.. Please reply..?

[@kaparthijanibasha42]

Sir at 10.52 we take x but may doubt taking (x-9)

[@djoshi1565]

After obtaining eqn @ 12:14 can we not solve the eqn a little further & simplify it? I did but the answer is different. Can someone plz xplain it?