The loading is a constant distributed load, or
EI v'''' = -w
where 'w' is acting downward. Integration this gives the shear function,
V(x) = EIv''' = -wx
+ C1
The 4th boundary condition can be used to determine the integration constant, C1,
V(x=L) = -wL
+ C1 = 0
C1 = wL
Integrating again gives the moment equation,
M(x) = EIv'' = -wx2/2 + xwL +
C2
The 3rd boundary conditions gives,
0 = -w 02/2 + x 0 L +
C2
C2 = 0
Thus, the final moment equation is
M(x) = -wx2/2 + xwL
Next, this equation can be integrated to give the rotation equation,
θ(x) = EIv' = -wx3/6 + wLx2/2 + C3
Using boundary condition 1 gives,
0 = -wL3/6 + wL3/2 + C3
C3 = -wL3/3
The final rotation equation is,
θ(x) = EIv' = -wx3/6 + wLx2/2 - wL3/3
The deflection equation is just the integral of the rotation equation,
EIv = -wx4/24 + wLx3/6 - wL3x/3 + C4
Applying the last unused boundary conditions, number 2, gives,
0 = - 0 + 0 - 0 + C4 ==> C4 = 0
The final deflection equation is
v = -w (x 4/8 - Lx3/2 + xL3) / (3EI) |