for the bending moment diagram, if the moment is negative (clockwise as shown) how would the moment diagram close out with a positive 180 momnet? (2:00) Wouldn't it be further negative?
appreciate the help but youre not really providing any work for your answer. You're just assuming that we already understand how to do this, in which we dont because thats why we're here
I think this would be simpler to solve by creating a function for the Shear and then integrating it, which would give you a function of for Moment. Especially since this only has 1 discontinuity which is easy to determine logically, so you can ignore that half of the piecewise function. If memory serves, shear is the integral of the distributed load. And since the distributed load is linear as a function of of X, that means that the Shear will take the form V=1/2 a*x^2 + c. Where 'a' is a constant you determine from the slope of the distributed load. Which is a=w_max/L 'C' can be found by solving the Shear function at point B, which you can determine from the moment equilibrium. This way you can solve the moment at any point, rather than just the peaks and intersections. Which will be helpful when you want to calculate the displacement. Or if you want to engineer a constant strain beam (2nd moment varies to keep strain constant at every length).
If someone is wondering, how the value of 135lb was found. q(x) or distributed load is 30/9 lb/ft^2 x (Since, the maximum distributed load is 30lb/ft, it is not the slope of the load and distance is 9 ft. So, slope of the distributed load is 30/9 lb/ft^2). Integrating q(x) with limits 0 and 9, we get 135 lbs which is the total force exerted by the distirbuted load.
everything is clear up to Mx= Ay*x+25*1.29 assume x=1.29 but unclear to me why x is that value geometrically, This is super easy method than using integration. Please send or reply with the logic I don't understand Ay*x+25*1.29. Thanks
What I couldn’t understand by my professor’s notes in the last 3 hours, I understood now in 3 minutes..
Thank you for your greatness!!
for the bending moment diagram, if the moment is negative (clockwise as shown) how would the moment diagram close out with a positive 180 momnet? (2:00) Wouldn't it be further negative?
what i couldn't understood in my whole semester, understood in 3 min, thanks man, u r gem
I love your videos! Easy to understand and less boring. This helps me to pass my Civil Engineering Board exam here in the Philippines :)
Great video, great help with understanding the subject with very easy examples
For those who are wondering where he got the 135lb from, you can get it by multiplying 30lb/ft with the length, 9ft and then divided by 2.
appreciate the help but youre not really providing any work for your answer. You're just assuming that we already understand how to do this, in which we dont because thats why we're here
Ppp😅po
Thank you so much!
Your playlist is a great refresher, thanks
I think this would be simpler to solve by creating a function for the Shear and then integrating it, which would give you a function of for Moment.
Especially since this only has 1 discontinuity which is easy to determine logically, so you can ignore that half of the piecewise function.
If memory serves, shear is the integral of the distributed load. And since the distributed load is linear as a function of of X, that means that the Shear will take the form V=1/2 a*x^2 + c. Where 'a' is a constant you determine from the slope of the distributed load. Which is a=w_max/L
'C' can be found by solving the Shear function at point B, which you can determine from the moment equilibrium.
This way you can solve the moment at any point, rather than just the peaks and intersections. Which will be helpful when you want to calculate the displacement. Or if you want to engineer a constant strain beam (2nd moment varies to keep strain constant at every length).
How did you get 1.29ft though? You didnt really explain the calculation behind that.
If someone is wondering, how the value of 135lb was found.
q(x) or distributed load is 30/9 lb/ft^2 x (Since, the maximum distributed load is 30lb/ft, it is not the slope of the load and distance is 9 ft. So, slope of the distributed load is 30/9 lb/ft^2).
Integrating q(x) with limits 0 and 9, we get 135 lbs which is the total force exerted by the distirbuted load.
What do you use to animate these?
how to calculate to 3 ft disctance for the load, I dont understand
why the moment arm of Ay in finding mmax is 1.29?
Please make a video of conjugated beam
what app do you use to draw like this kind sir? TIA
Finding the X on those curves really throws me. I can't get it. Ty
THANK YOUUUUUUUUU
bless your soul
everything is clear up to Mx= Ay*x+25*1.29 assume x=1.29 but unclear to me why x is that value geometrically, This is super easy method than using integration. Please send or reply with the logic I don't understand Ay*x+25*1.29. Thanks
way easier to understand in this form, I can kinda get lost in the moment and lose sight of what I'm doing when we spend 20 mins on a problem
is this correct?
🔥
👍👍
Shouldn’t the slope of the shear diagram be linear? Yours appears to be exponential
Bro is too fast wtf is this
r u joking this is a mess, too fast. way way too fast- no explanation just stating numbers
im so tred
im ded
help me :P
i cant go on
You mean this is less boring?
It sure is compared to my professor
This is trash. Why so fast? Couldn't grab anything
No way your living in 2023 and don't know about the playback setting to slow the video down. lmao
hehe