Moment Distribution Method Type 2 (With concepts)

[@sailendranathpal8]

The word" teacher" meaning define by ur teaching style and explanation & thanks fram JADAVPUR UNIVERSITY.....😍

[@priyharshgangwar]

"Notice that you should not add these distribution factors" this was funny. But seriously man you just saved my grades. Great Work.

[@divyakola117]

Great lecture..👍 Thank you so much sir😊

[@vatsaakhil]

Thanks man, content was crystal clear!

[@ShahidIqbal-wp6kz]

Thumbs up for the lecturer...

[@DCBAonline]

Shahid Iqbal thank you so much, appreciated.

[@prajvaldudhbale5418]

Awesome Lecture....Thank You For This Video

[@vlogs4youlike]

thanks sir go forward you are my professor.

[@alialhameedi990]

Thank so much for that

[@kanusharma1577]

Very well explained , sir

[@mytahoorq6662]

Sir in practical life how does we apply the load on hinged As u said we are applying -ve 10 Newton and which is further transmitted to the support c (i.e, -5N) And at support D load zero . Actually in practical point of view what we are doing sir?? Please do clear my question..

[@souvik.fingtore]

😄 thanks sir 😍

[@mytahoorq6662]

sir why we are multiplying by half before transferring -10 N from D to C? In iteration of moment distribution ,when the carryover is done at that time also we are multiplying by half(0.5)...why sir ? Reason behind multiplying...

[@naveensingh9237]

Nice headings..

[@yuvrajpanchal4432]

Nice

[@migmartsering1010]

Column analogy method should be ur next....

[@UpperXEgo]

thx

[@mukeshjatav4287]

thankyou so much jiii

[@allhailvegeta-theprinceofs5485]

i was watching this before my semester exams...now that i am done with engg....i am so fucking glad that i wont have to deal with this shit anymore...fuck it

[@DCBAonline]

Hahaha ,sadly you will have to deal with this for your life if you plan to do your career in structural engineering.

[@allhailvegeta-theprinceofs5485]

DCBA online u right...but only if u make a career as a st. engg..

[@indiramishra7013]

Can't believe someone can hate it so much! I love this subject

[@junaidscreation3044]

Upload video about moment distribution method for sway type which is inclined from both sides plzzz............ I have papers very soon

[@sahaldhapa3139]

Sir in this beam for BC and CB, the far end is Rolling support but why the relative stiffness is EI/L.

[@DCBAonline]

Sahal Dhapa The joints B & C are intermediate joints and our first primary assumptions is that all joints except end joints are considered to be fixed. Hence the relative stiffness is EI/L

[@gauravthakur1655]

For point load at an equal distance, we take moment wL/8

[@DCBAonline]

+gaurav thakur are you talking about fixed end moment? For point load at equal distance or at center of the span its not wL*L/8, its wL/8. However you can also use the formula for an eccentric point load that is a point load at a distance of 'a', in this case a=L/2

[@gauravthakur1655]

thnxx...ya it was the same i was supposed to write

[@askatv1015]

I have a question after the moment distribution calculations how to get the net span moment?

[@positivevibes4265]

By simple method, using 3 degree of freedom

[@kpuniyal3594]

Sir thr is one doubt .. cb and cd are not also fixed then why EI/l for cb

[@civilguy7818]

Intermediate joints are treated as fixed joints in moment distribution method ...so

[@reymartcabajes3805]

What if you already have a balance moment on FEM when plotting it in table

[@DCBAonline]

I didnt get your exact question.

[@pradeeppatil8338]

Sir put the problem on uvl load sfd &bmd

[@sushiljalwal8914]

Sir please add a video with sinking of beam

[@DCBAonline]

sushil jalwal its already there on channel, it has beem solved by slope deflection method.

[@sushiljalwal8914]

DCBA online sir i need by this method

[@sauravbhagat2467]

can i get tutorial of portal method

[@DCBAonline]

you can find it here Sourab : kzbin.info/www/bejne/qqHSh3SGZ7aphZY

[@kanusharma1577]

Sir agr humare pass A support pe hinged support hai toh B support carry over kregi A Support ko

[@DCBAonline]

nahi,agar support A hinged hai to uspe carry over nahi hoga.

[@kanusharma1577]

DCBA online ok sir thanku

[@maheshmalletula6872]

sir plz add cantilever and portal method

[@shantanujaiswal3851]

halo sir apne 4th lecture me re stiffness nikalne ke lye 4EI/L ka use kya h or es lecture me EI/L yes Kyo

[@DCBAonline]

shantanu jaiswal aap 4EI/L ya EI/L koi bhi formula use kar sakte ho dono ek hi hai agar detail me batana ho to Humein humare clg me sikhaya gaya tha ki jab far end fixed hai tab rel stiff EI/L hai aur jab far end hinged hai to rel stiff 3EI/4L hai. Par kahi saare books me jab fix hota hai to 4EI/L liya jata hai aur hinge hota hai to 3EI/L liya jata hai jo mein bhi ab use karne laga hun agar aap inmein se 4 common nikaloge to aapko jo meine is video me use kiya hai wo mil jaega rel stiff. To agar short me batana ho to aap koi bhi use kar sakte hai jo aapko yaad rakhne me aasan ho Fixed ke liye 4EI/L ya EI/L Hinge ke liye 3EI/L ya 3EI/4L

[@shantanujaiswal3851]

tnks sir

[@dattatraydurgoli5768]

How to solve for Frame structure?? please upload the video for frame

[@DCBAonline]

+Dattatray Durgoli inclined frame with sway is there kn channel

[@azizelbussaidy5893]

fixed end moment for roller/pined support must be zero

[@diresquid]

How are you determining the value for I please?

[@DCBAonline]

'I' is the Moment of Inertia and it is the property of the section that is used for analysis. For eg a Rectangular Beam will have I= (BD^3)/12 A Circular c/s beam will have I= 𝜋/64 𝐷^4 Hence it varies as per the section considered. You will learn more about it when you will be designing the sections.

[@diresquid]

To be clear, as in the 1.5I or 2I selections for the stiffnes calculations

[@DCBAonline]

The Moment of Inertias are already given in the question. Further to describe in easier words you can relate Moment of Inertia directly to the strength of the members used in this particular problem. How to decide the value of I depends upon which cross section has been used while designing the beam, this is not significant for us at this stage as it is directly provided to us in the form of 'I". So you can say in simple language that strength of member CD is 1.5 times the strength of member AB provided they both are of same material. Further this value of 'I' has to be put into the equations for calculating stiffness.

[@nyabutonelson2128]

moment BC is wl/4 or wab/l

[@DCBAonline]

Its the same WL/4 is special case of Wab/L when a=b= L/2

[@nyabutonelson2128]

exactly confirm in that question span BC is 40KN under 2m /2m span @@DCBAonline

[@QUANT.369]

Bro some author fixed support relative stiffness is 4EI/L someone EI/L

[@DCBAonline]

Both are correct; When you use 4EI/L formula for far end fixed ,you need to use 3EI/L for far end hinged When you use EI/L formula for far end fixed ,you need to use 3EI/4L for far end hinged You can use whichever set of formulaes is easier for you to remember ,just dont jumble up between them.

[@AnkitPatelAP]

Which software ?

[@DCBAonline]

Ankit Patel MS Office PowerPoint 2016

[@AnkitPatelAP]

thank you very much

[@zzero8815]

why does some text shows 3EI/L and 4EI/L as for finding stiffness???

[@DCBAonline]

Relative stiffness when the far off end is hinged is given as 3EI/L and when far off end is fixed it is given as 4EI/L i have also derived these formulaes you can find them here: kzbin.info/www/bejne/m3WYlaaGn9-fhZo

[@zzero8815]

but you used here EI/L and 3EI/4L...whats the difference in using this and the one I mentioned above??

[@DCBAonline]

there is absolutely no difference actually we were taught that way to use as EI/L and 3EI/4L it is nothing but if you take 4 common from 4EI/L and 3EI/L you get those values. However now even i have started using 4EI/L and 3EI/L as they are whole numbers and easier to use and remember.

[@zzero8815]

thnx for the reply ;) Keep bringing up the content..its really helpful.

[@DCBAonline]

sure, i'm glad it helped you.

[@diflection1]

I sent you a compact matrix solution....

[@diflection1]

this method fail with mine......

[@amirb948]

I think M (FDC) should be 0, not 10 since support D is a hinge.

[@DCBAonline]

amir b so we did make it zero, we have released that moment on that joint so at the end there is a zero moment on D

[@amirb948]

thanks for your reply, but you did some calculations for it and I think it should be considered zero at the beginning.

[@amirb948]

also based on fixed end moment table FEM(CD) formula is (p/l^2)(b^2a+0.5 a^2b)=25 kN.m

[@DCBAonline]

amir b no, the formula i have used is correct, and the fixed end moment should not be considered as 0 in begening because of a simple assumption that is prerequisite that all the joints are considered to be rotationally fixed. Therefore every joint before doing MDM iterations will have some fixed end moment.

[@amirb948]

engineering.purdue.edu/~ce474/Docs/Fixed%20End%20Moments.pdf please see second-row right side.

[@diflection1]

I got a surprise for you

[@DCBAonline]

+THE STRUCTURAL ANALYSIS MATRIX FORMULATION , Thank you I have recieved the mail, I will check the solution in some time.