Deflection at A: beam with load and moment

Assumptions: cantilever of length $L$, fixed at B, free end A carries a downward load $P$ and a couple $M$ whose sense also tends to bend the beam downward (same direction as $P$). $EI$ constant.

Take $x$ measured from the free end A and $y$ positive upward. Both $P$ and $M$ cause hogging, so the bending moment at $x$ gives:

$$EI\frac{d^2y}{dx^2} = -(Px + M)$$

Integrate:

$$EI\frac{dy}{dx} = -\left(\frac{Px^2}{2} + Mx\right) + C_1$$

At the fixed end B ($x = L$) the slope is zero:

$$C_1 = \frac{PL^2}{2} + ML$$

Integrate again:

$$EI\,y = -\left(\frac{Px^3}{6} + \frac{Mx^2}{2}\right) + C_1 x + C_2$$

At B ($x=L$), $y = 0$:

$$C_2 = \frac{PL^3}{6} + \frac{ML^2}{2} - \frac{PL^3}{2} - ML^2 = -\frac{PL^3}{3} - \frac{ML^2}{2}$$

At A ($x=0$), $EI\,y_A = C_2$:

$$y_A = -\left(\frac{PL^3}{3EI} + \frac{ML^2}{2EI}\right)$$

The negative sign means the deflection is downward.

(Superposition gives the same: $PL^3/3EI$ from the point load plus $ML^2/2EI$ from the couple.)

Answer: deflection at A $= \dfrac{PL^3}{3EI} + \dfrac{ML^2}{2EI}$ downward. If the couple acts in the opposite sense, subtract the second term.