Finding area of cubic curve isn't difficult. You can finding area of any degree curve using the genral formula... Area of nth curve: An= b*h/(n+1) So cubic means n=3 so....A= bh/4...where b would be the base(x) and h be the height(y). The CG in x direction in figure....from B X'=b (n+1)/(n+2) In y direction from the base... Y'= h (n+1)/(2n + 1) So for cubic...X= 4/5 b Y=4/7 h Also for parabola n=2 obviously Also if it's outward parabola or any curve...just minus the area from bh So outward cubic is bh-(bh/4) is 3bh/4... i mean the outward and inward together makes a rectangle anyway.
Thank you very much , you made it very simple finding deflection by this method, I was trying to find deflection between A and D but not able to solve by integration method in finding moment area between point say A and E , as limits are not known of unknown points. Kindly look into this problem if possible.
Sir, In the BMD for portion B to C, shouldn't the point go upto -120 not -90 at B from point C ,considering the moment of the UDL is 120kN.m? Can you describe please?
The region between A and B is hogging as per the moment which means the slope you are calculating for B is opposite in direction. I think your elastic curve is wrong
The 30kN force is attempting to make the beam “hog”, however the UDL of 20kN/m * 3m overpowers that load and creates the deflected shape as shown. He is right. :)
Finding area of cubic curve isn't difficult. You can finding area of any degree curve using the genral formula...
Area of nth curve: An= b*h/(n+1)
So cubic means n=3 so....A= bh/4...where b would be the base(x) and h be the height(y).
The CG in x direction in figure....from B
X'=b (n+1)/(n+2)
In y direction from the base...
Y'= h (n+1)/(2n + 1)
So for cubic...X= 4/5 b
Y=4/7 h
Also for parabola n=2 obviously
Also if it's outward parabola or any curve...just minus the area from bh
So outward cubic is bh-(bh/4) is 3bh/4... i mean the outward and inward together makes a rectangle anyway.
A wonderful explanation 🎉
Thank you very much , you made it very simple finding deflection by this method, I was trying to find deflection between A and D but not able to solve by integration method in finding moment area between point say A and E , as limits are not known of unknown points. Kindly look into this problem if possible.
Nice video sir..
Thank you 😘
Wonderful sir you made the concept so easy
Thanks for clarification
Well done man... You describe all the thing 😊😊👍👍
where can i learn more on the table you used for this method. youre the first video ive come across using that
Which table? At what position in the video can you mention
Sir, In the BMD for portion B to C, shouldn't the point go upto -120 not -90 at B from point C ,considering the moment of the UDL is 120kN.m? Can you describe please?
such a good video. thanks!
What is difference between slope ab and slope ba
Thanks
Bro you are too good ❤💕
lovely video
😊
But when i took the t(c/a) the thita a is coming out as (52.5)/EI anticlock wise is it correct plz...clarify
sir i have solved it myself. but not getting your answer. i have checked my paper again and again.
The region between A and B is hogging as per the moment which means the slope you are calculating for B is opposite in direction. I think your elastic curve is wrong
Isnt that the hogging elastic curve at 24:41 ? what is wrong in that ?
The 30kN force is attempting to make the beam “hog”, however the UDL of 20kN/m * 3m overpowers that load and creates the deflected shape as shown. He is right. :)
Bro point B bm is wrong on left
18:50 to 19:60
14:44