This problem can also be solved using three-moment equation. First by identifying the reactions which is Ra = Re = 75 kN, Then followed by getting the moment equation of the entire beam starting at point A which is M = 75x - 150, therefore, other parameters will be: Ma = 0 Mb = 75(3) - 0 = 225 kNm Md = 75(9) - 150 = 225 kNm L1 = 3 m L2 = 6 m I_2 = 2(I_1) EI = 200(10^6)*(300*10^-6) = 60000 kNm^2 6Aa/L = 0 6Ab/L = (Pb/L)(L^2 - b^2) = [150(3)/6](6^2 - 3^2) = 2025 Since the deflection at B is equal to the deflection at D, therefore, we use the reference line at deflection at B which means that h1 = ?? and h2 = 0 Substitute all the values to three-moment equation 0 + 2(225)(3 + 6/2) + 225(6/2) + 0 + 2025/2 = 6(60000)(h/3 + 0/6) Therefore, h = 0.0365625 m = 36.5625 mm
@maditmichaelmalangatem63662 жыл бұрын
Thanks I understand your lectures very well. 💯
@samuelgosaye507111 ай бұрын
well explained, we love you man
@esthermuthomi6003 жыл бұрын
Finally,I understood virtual method...thumbs up😘
@nyakunimartin223016 күн бұрын
Is this the same method used when solving engineering Mechanics principles with the principle of virtual work
@florenceting9824 Жыл бұрын
May i know how to determine the boundary of equation? i am confused when you always start x from A but set your boundary within the segment itself only... Does the x always start from same point? and the boundary always follow back the segment itself?
@mihrabanjabar160710 ай бұрын
Thank you ❤️
@vicentssekyanzi95672 ай бұрын
Moment equation for section DE on the real part is75x-150(x-9)
@RakiaKun2 ай бұрын
no, only x-6
@natanimkore Жыл бұрын
thank you sir
@ryansingculan4533 жыл бұрын
Hi Sir! Very helpful tutorial. I have a question about whether to consider the segments CD and DE when computing for the deflection at B?
@viralcat-w4x2 жыл бұрын
By symmetry, we can see that the deflection at B is just equal to deflection at D
@Msaigon13 Жыл бұрын
12:29 sir hindi po ba kau nagkamali ng input sa point load na 1? sa D po ba talaga siya or sa C?