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
+ C_{1}
The 4th boundary condition can be used to determine the integration constant, C_{1},
V(x=L) = -wL
+ C_{1} = 0
C_{1} = wL
Integrating again gives the moment equation,
M(x) = EIv'' = -wx^{2}/2 + xwL +
C_{2}
The 3rd boundary conditions gives,
0 = -w 0^{2}/2 + x 0 L +
C_{2}
C_{2} = 0
Thus, the final moment equation is
M(x) = -wx^{2}/2 + xwL
Next, this equation can be integrated to give the rotation equation,
θ(x) = EIv' = -wx^{3}/6 + wLx^{2}/2 + C_{3}
Using boundary condition 1 gives,
0 = -wL^{3}/6 + wL^{3}/2 + C_{3}
C_{3} = -wL^{3}/3
The final rotation equation is,
θ(x) = EIv' = -wx^{3}/6 + wLx^{2}/2 - wL^{3}/3
The deflection equation is just the integral of the rotation equation,
EIv = -wx^{4}/24 + wLx^{3}/6 - wL^{3}x/3 + C_{4}
Applying the last unused boundary conditions, number 2, gives,
0 = - 0 + 0 - 0 + C_{4} ==> C_{4} = 0
The final deflection equation is
v = -w (x ^{4}/8 - Lx^{3}/2 + xL^{3}) / (3EI) |