Note: At 3:52, I said the slope at the free end is equal to zero. I meant to say shear. Link to an additional, more difficult practice problem: ua-cam.com/video/OSU0ZnJyqtg/v-deo.html
Thanks as always guys! Love your videos :) How much better would we have done in university if channels like there were around then? I was a freshman in 2005 and am pretty sure there were noooooo engineering example problem channelsssss. Later!
Good question! To the left or right of B will give you the same result, but if you're looking at the right side of B, you'll notice that you have reactions that must be considered. The shear at B on the conjugate beam is equal to the slope, the moment is equal to the deflection. Thus, if you look at the right, you'll have reaction Cy as well as the area under the curve, as well as a moment Mc which will also have to be considered. Because of this, you'll always want to cut a cantilever beam on the side that doesn't have the reactions in order to skip this messy and time consuming step. For simply supported beams there's usually no way to avoid this, but you can skip finding the reactions in cantilevers. It's a good trick for exams :)
Note: At 3:52, I said the slope at the free end is equal to zero. I meant to say shear.
Link to an additional, more difficult practice problem: ua-cam.com/video/OSU0ZnJyqtg/v-deo.html
love this example. Simple, but has its trick to it drawing the conjugate beam diagram. Thank you so much!
That's exactly what we were going for, glad.you enjoyed it!
Thanks as always guys! Love your videos :) How much better would we have done in university if channels like there were around then? I was a freshman in 2005 and am pretty sure there were noooooo engineering example problem channelsssss. Later!
Much love bud! We try to put out content that we wished we had in these courses when we took them!
Exactly!! That's the way to add value to the world :)
And your point load produces a positive moment (i.e. counter-clockwise) at support A so your moment diagram is flipped (should be positive)
Well, thank you so much!!
Very helpful!!
Sir please why can't you treat the first as a trapeziod instead considering the rectangular and triangle
How did you get 0.45in?
if you used the right side of 'B' to solve for the slope at 'B', how would that look like?
Good question! To the left or right of B will give you the same result, but if you're looking at the right side of B, you'll notice that you have reactions that must be considered. The shear at B on the conjugate beam is equal to the slope, the moment is equal to the deflection. Thus, if you look at the right, you'll have reaction Cy as well as the area under the curve, as well as a moment Mc which will also have to be considered. Because of this, you'll always want to cut a cantilever beam on the side that doesn't have the reactions in order to skip this messy and time consuming step. For simply supported beams there's usually no way to avoid this, but you can skip finding the reactions in cantilevers. It's a good trick for exams :)
Hey, really great video, how would you solve the deflection at C?
Same way, form conjugate beam model, take moment and shear at c instead.
Don't you have to convert ft to inches to have consistent units ?
Consistant with what, E? As long as all of your units are the same it doesnt matter if you work with feet or inches.
how did u come up with 0.0043 rad?
Convert ft to in
It is determinate beam Or indeterminate beam
Refer to our determinate frame video and we'd suggest calculating for yourself for practice!
Why is I.=6000,I=..?
FUCK ME, YOUR VOICE IS MAKING ME DRIP.
Also, thanks for the very helpful video.