Shear force and bending moment diagrams example #3: distributed load

[@macpr0c]

I just took the integral of v(x) to find the bend moment diagram's function. Because you mentioned shear force is the rate of change for the bending moment. Seems to work and it's fast to do.

[@Engineer4Free]

Yup. The method in this video is just a visual interpretation of the actual integration and derivatives that are going on between the diagrams

[@ryanta667]

@@Engineer4Free How would we solve for the constant of integration if we used this method?

[@ryanta667]

@@Engineer4Free Unless the BM at the support reactions is always zero, which would result in C=0 everytime

[@vladberez]

With shortcut, how did you know the parabolic shape would go up and to the right, as oppose to the other option more of to the right and up?

[@zishanmuhammad1710]

at 0:38 why V will be downward?

[@14henrylover]

ok so like what if you have distributed loads as well as multiple point loads

[@Engineer4Free]

Hey that's actually a really great question. I will try to make a whole tutorial on it tomorrow, but basically you will just do a new FBD from the end that you start at to just before any point load, start/stop/change of a distributed load, or an applied moment. Each time that you section the beam a little bit further, you will find out the expression for shear and moment in that new part of the beam and be able to add to the diagrams. It gets to be a lot of work on complicated problems, so naturally there is a shortcut and I'll just use that in the tutorial.

[@Engineer4Free]

As promised: kzbin.info/www/bejne/i3apkpKelrmbrdE

[@14henrylover]

Engineer4Free thank you sir you are doing a really good job btw