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
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?
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
Thanks
I understand your lectures very well. 💯
well explained, we love you man
Finally,I understood virtual method...thumbs up😘
Thank you ❤️
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?
well explained sir❤
thank you sir
12:29 sir hindi po ba kau nagkamali ng input sa point load na 1? sa D po ba talaga siya or sa C?
Is this the same method used when solving engineering Mechanics principles with the principle of virtual work
Moment equation for section DE on the real part is75x-150(x-9)
no, only x-6
Thanks
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?
By symmetry, we can see that the deflection at B is just equal to deflection at D
Section DE.its x-9