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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
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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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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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?
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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
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.
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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 ?
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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
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?
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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.
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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
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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
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
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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
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What will be the effect of C1&C2 if we will not write the moment eq. in the same manner. As C1 & C2 can be written any where in that case the same value of C1 & C2 is not same as u have written
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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
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