Shaft overhang: maximum deflection formula

Given data

Shaft ABC, simply supported at A and B ($AB = l$), overhanging to C ($BC = a$). A transverse force $P$ acts at C. $EI$ constant. To show: maximum deflection in AB is $\dfrac{Pal^2}{9\sqrt3\,EI}$.

 A ---------------- B ------- C
 ^        l         ^    a     | P

Reactions

Moments about B: $R_A\,l = -Pa$, so $R_A = -\dfrac{Pa}{l}$ (downward). $R_B = P\left(1 + \dfrac{a}{l}\right)$.

Bending moment in AB

Taking $x$ from A:

$$M = -\frac{Pa}{l}\,x\quad(\text{hogging}),\qquad 0 \le x \le l$$

Double integration

$$EI\frac{d^2y}{dx^2} = M = -\frac{Pa}{l}x$$ $$EI\frac{dy}{dx} = -\frac{Pa}{2l}x^2 + C_1$$ $$EIy = -\frac{Pa}{6l}x^3 + C_1x + C_2$$

Boundary conditions: $y = 0$ at $x = 0$ gives $C_2 = 0$; $y = 0$ at $x = l$ gives

$$-\frac{Pal^2}{6} + C_1l = 0 \;\Rightarrow\; C_1 = \frac{Pal}{6}$$

Therefore (positive y = upward)

$$y = \frac{Pa}{6EIl}\left(l^2x - x^3\right)$$

Position of maximum deflection

$$\frac{dy}{dx} = 0 \;\Rightarrow\; l^2 - 3x^2 = 0 \;\Rightarrow\; x = \frac{l}{\sqrt3}$$

Maximum deflection

$$y_{max} = \frac{Pa}{6EIl}\left(\frac{l^3}{\sqrt3} - \frac{l^3}{3\sqrt3}\right) = \frac{Pa}{6EIl}\cdot\frac{2l^3}{3\sqrt3}$$ $$\boxed{y_{max} = \frac{Pal^2}{9\sqrt3\,EI}}$$

The shaft bows upward in AB (opposite to the load at C), at a distance $l/\sqrt3 = 0.577\,l$ from A. This proves the required result.