Do subscribe to Ekeeda Channel and press bell icon to get updates about latest engineering HSC and IIT JEE MAIN and advanced videos Today we are starting a new topic this name as Deflection of beam using using energy methods in energy method we are seen here as first we are seeing over here the virtual work method or unit load method or castigliano’s theorem all the three names are same in exam you have written the virtual work method or unit load method or castigliano’s theorem all three names are same now in in this we have to find the slope and deflection for deflection we have to apply for deflection we have to apply the unit load and for slope we have to apply the unit movement let’s see in details now let’s start with it say this is our question given over here and I’ll start the solution now first what we have to draw first draw the same diagram as it is and here load a and B and this is ready you okay now see this is this is what support this is rollers of hinge support so they’re 22 reactions this is your he this is VA and this is a chain and this is see this is roller support roller support that’s why it will be this okay now first what you have to find this name as M diagram M this is big am not small and this is Big M diagram okay now what you have to do first first find the reactions first herpes fine find support reaction okay now in support reaction what we are taking taking moment at a so you are finding over here taking moment at T equal to zero clockwise positive okay clockwise positive means now see we are taking moment here okay so this step will become clockwise see put a rod and put a rod over a and apply include alike this will rotate it clockwise direction so then in two threes fours in two perpendicular distance then Plus this UDL plus 10 into 4 into 4 by 2 this is nothing but force into perpendicular distance of UDL then minus VD into 4 equal to 0 this is your VD okay 2 BD into 4 equal to 0 while solving this you get the value of VD will be twenty seven point five kilo Newton okay for direction now after that summation summation FY equal to zero upward positive C we got the value of VD will be twenty seven point five kilo Newton then this VA will be minus 40 plus VA this will be minus minus 10 in to 4 then plus twenty seven point five is equal to zero VA will be get the value of V is twelve points you all the external loads now remove the all external nodes which is over here like see this is external node means this thank you out of this UDL this only okay you have to remove remove all the external nodes loads and only draw the diagram okay now we are the moving on external Lord this will be a this is B this is C and this one is D and this part is e he is nothing like this one this is your a part okay now in this what we have to do see you know not read the question carefully first C find the horizontal displacement of roller support means we have to apply the word we have to apply the one can outer horizontal and for rural support means roller is wave is roller roller is here so we have to apply here horizontally one kilo Newton okay now instead find using this one kilonewton we have to balance the see there is have here I have hinge support so this will come this is also coming and hisses rule of support is also coming okay so what we have to do see this 1 kiloton is going in this direction when we are taking X then this will go this direction this is also have 1 kilo Newton now see then vertical they have no reactions over here that’s why it will come 0 but we also get 0 this is your small M diagram so this is a small M diagram okay now after that what we have to do after draw the table first and take the different different sections let’s see how it gets lawrence draw the cable region origin limit M is small M then e aí and science follow this angle of a so this means we are taking we are taking section at certain sections and we have to find that section we have to find the moment moment depending we went over there okay now see what have to come for C region what is our first reason will be see we are taking first reason is a and E so write it over here reason to tradition over here heyy okay now what is origin we have to take origin now take origin as a this is your your limit will be in this year limit will be 0 to 3 0 to 3 now now what you have to do you have to take a section take a section K and E like this and this is as origin ok that this sensor is nothing but X and this is your origin now you have to take point here we have to take depending moment and see here this side not this side because we are taking origin here that sir we have to make this side only okay when we were taking bending moment over here see what will come see this will come plus sign this will come then into this distance is nothing but as 10 into X see like this we are taking section over here this 10 will go like this is your 10 kilo Newton okay this 10 will be go like this now like this so we are taking positive so write it in M and into 10 in to X okay this is your M now let’s move to M now let’s move to small M for small M we are taking section over here and this is your origin we are taking bending moment here what will come this one into X positive so 1 into X positive now whatever ei is constant so ei as it is now we have done the first part okay now let’s see the next one now see now what is the region over here this we are taking Region II and V we are taking region P and B E and B and what is the origin the origin will be e so region will be e okay what is a limit let’s say matter will be 0 to 2 0 to 2 now see we are taking section over here we are taking section over here and this is the distance X now I have to take the bending moment here when we are taking bending moment over here we have to see all this vanna not dude don’t see only this 10 kilonewton because in as you all know bending movement either you better take left order you can take right but hole not there in some parts or clothing so please listen first see we have to taking a bending moment over here so we have to take this all but this 10 will come in the picture and this 10 also come in the picture okay now let’s see how its come see just simply like root over like put here like this now see we are taking moment over here so this distance will be 10 into this distance if this is X and what is this distance whole distance this is the thing but 10 into 3 plus X or X plus 3 we are taking bending moment over here this point so we are taking this this into 2 X this is directly this come – okay see this what we come plus n into this 3 plus X so X plus 3 distance will be 3 plus X so right over here n into X plus 3 okay X plus 3 now we have done with me this far see now we have done with thee this pertaining to X sorry 10 into 3 plus X means X plus 3 and same now if this deserves about this say direction so minus minus 10 into what religious systems this is nothing but X so minus minus 10 into X okay now let’s move with the small M diagram a smaller diagram we are taking section over here this is your this is your existence so what has come 1 into X 1 into 3 plus X as this is this is 3 meter this is 2 meter this is 5 meter I think for a 4 meter and this is 5 meter okay now see we are taking section over here so what we come to distance 1 into 3 plus X this distance will be 3 this one has 3 now this will come plus X so directly we have 2 right over here sorry 1 in 2 X plus 3 ok 1 2 X + 3 1 into this is so distant taking section away at this ability 1 into this is X this is X plus is this whole 3 ok now Y I will be constrained because is the same okay now let’s move to the another part the other part will be me and see so I did over here reject will be B and C and what about the origin now now you can take the origin B also you can take see also but see when we are taking move agent at B this will be taking origin at B we are taking section over here this will see this part is so little bit complicated for us for time saving you have to take the origin at C and I have to see this side because this set is not only one load is here okay now take origin at C the limit will be 0 2 4 0 2 4 now take a section over here take a section over here and this distance will be X we are taking bending moment over here and what will become C if as this value is coming this value is coming 27 point 5 into X positive twenty seven point five in to X positive now now after that this UDL will come in the picture so minus minus 10 into X into X by 2 3 arrange it once times for this equation see twenty seven point five X minus five X square this will come finally okay we have to this rearrange over here okay 5 X square now now after that let’s move to the M 1 sorry M small M for smile small M the cross section over here this will be X take a bending moment and see this side what will come see this one is only coming the picture so force in two perpendicular distance to please the Sun students don’t need to write one e to this existence we have to OB one in a moment force it to that perpendicular distance that’s why we have to take one in to five that’s it one in to five that’s it this is your E I you have done with this part also okay I hope you understanding this videos now now after that after that let’s see with the another part this region will remaining is C and B you know we are solving C and D origin will be C origin so you don’t take the region C because then we are taking origin and said you have to see all this part only take originality and take this one also this is also the region at T limit will be 0 to 5 now move to this one we are taking section over here this will be for eg this is your existence ok now you are taking bending moment over here to see in bending when we’re taking bending moment over here there is no horizontal load C there is no horizontal load satisfied this value will become 0 okay now in small small I am take a bending moment over here this distance will be XC in here this one is there so 1 into this x+ C will put a like over here and this will come positive so 1 okay now we are competing with the table now let’s see how to put in the equations first now see we have to find the deflection at roller support in horizontal direction that’s why we have to write Y at D we have to fight deflection Y stands for deflection these transfer we have to find it joint D or bracket bracket occur at yd at horizontal okay what is the formula or formula will be 0 to L upon C I D X okay now put all the values notice in the table take take 1 upon e I constant take 1 upon e I constant now C right no see this is the 0 to 3 so limit will be 0 to 3 after that n into X then in to X ok now this will turn into small this M into small M so 10 into X into this X so this will come 1 X DX okay now see students we have done with the first part into X into 1 now now see this will have that we have the now after that we have to do plus plus now integration for this exact value will be 0 to 2 0 to 2 now what is this any 10 X plus 3 minus 10 X okay so right like this 10 into X plus 3 minus 10 X this will come in the first bracket and this will come in the another bracket 1 into X plus 3 into DX okay now 1 X plus 3 after that plus integration 0 to 4 now this value 0 to 4 this better will come in the 1 bracket 7.5 X minus 5 X square and in two small M will be 1 into 5 so 1 into 5 after this one after that plus integration 0 to 5 0 to 5 0 to 5 this will get 0 and this will get 1 X DX this is also with DX ok DX now close the bracket now you have done with the Alton Americal parts and lost solve it solve it one by one sulfur’s this part in this part in this part in this part 90 plus 240 plus 566 point double 6 plus 0 this is where we are getting Christmas now we got the value of yd while solving this we get the yd will be the 896 point double 6 upon e I now see we have to put the simply value put the value of e and I the e value will be given 210 into 10 raised to 3 Newton per mm square convert this to connect per meter square okay now ILO will be 2 into 10 raise to 8 mm raised to 4 simply put the head over here convert and then put the value okay now see I am converting directly as you all know he will be 210 into 10 raised to 3 still be Newton to Fonda kilo Newton 10 raise to minus 3 10 raise to minus 3 upon 10 raise to 10 raise to 6 upon C upon 10 raise to minus 6 okay because this is per meter so that’s why it will come up it will go plus into into 10 raised to plus 6 into this is what with the evil you know I will do will be 2 into 10 raised to hit into 10 raise to minus 2 this will come in the meter the circum 21 point 3 4 into 10 raise to minus 3 meter know why he will be 21 point 3 4 mm in this direction this is your final answer I hope you understand five videos and please subscribe the Ikeda channel thank you

Sir please upload one more example.

Check for the Integration limits you have applied to the different parts of the frame.

sir structural analysis-2 ki vedio send please reply

I think the conversion of I from mm^4 to m^4 is wrong it should be (2×10^8)×(10^-12)

Check the limit of integration

U made mistake do the correction

Worst explanation and poor English.

How many times you say …."see" "see" horrible video🤮

last tk M n m kya h ye ni bataya..

sir aap apni shakal board k samne mat laya karo apki awaaz kaafi hai samajne k liye.

Good lecture

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Bhai unit load or castigiliaan alag alag hai

Bhai unit load or castigiliaan alag alag hai

bhai…bohot ganda padhate ho..papa ko bhejo apne..mera 20 mint wapas do 🙁

Please come up with a good teacher for structural analysis