In the link at 12:35 , the author show the equation of the displacement of the beam without explaining , can someone try to explain why we need to use 3 Φ , 2EI / L , 2θa and 2θB ? I have no idea at all .
I would guess this was covered in an earlier lecture.
In the left hand picture, we have a beam subject to a moment M at x=0.
For equilibrium, there would also be an upward force F at x=0 and downward at x=L. To balance the moments we have M=FL.
Consider a point at x. The moments from the left of that are Fx-M, so EI y" = Fx-M. Integrating, EI y' = Fx2/2-Mx+constant.
For small angles, tan θ≈θ, so from the boundary condition at x=0, the constant is EI θ11.
Integrating again, EI y = Fx3/6-Mx2/2+ EI θ11x+constant.
The boundary condition (L, 0) gives θ11=ML/(3EI).