Inclined load on a rectangular cantilever

Section properties (principal axes)

$x$ horizontal (width 80 mm), $y$ vertical (depth 120 mm):

$$I_x = \frac{80\times120^3}{12} = 1.152\times10^{7}\ \text{mm}^4,\qquad I_y = \frac{120\times80^3}{12} = 5.12\times10^{6}\ \text{mm}^4$$

The section is symmetric, so $x, y$ are principal axes ($I_{xy}=0$).

Load components

$$\begin{aligned} P_y &= 2000\cos30^\circ = 1732.1\ \text{N} \\ P_x &= 2000\sin30^\circ = 1000.0\ \text{N} \end{aligned}$$

Moments at the fixed end ($L = 1500$ mm):

$$\begin{aligned} M_x &= P_yL = 1732.05\times1500 = 2.598\times10^6\ \text{N mm} \\ M_y &= P_xL = 1000\times1500 = 1.500\times10^6\ \text{N mm} \end{aligned}$$

(a) Maximum stress

It occurs at a corner farthest from the neutral axis, at $x = \pm40$, $y = \pm60$ mm:

$$\begin{aligned} \sigma_{max} &= \frac{M_x\,y}{I_x} + \frac{M_y\,x}{I_y} = \frac{2.598\times10^6\times60}{11.52\times10^6} + \frac{1.5\times10^6\times40}{5.12\times10^6} \\ &= 13.53 + 11.72 = 25.25\ \text{MPa} \end{aligned}$$

The top fibre is in tension from $P_y$. The horizontal component bends the beam sideways, so the side away from it is in tension at the support. The corner where both tensions add carries $+25.3$ MPa and the diagonally opposite corner carries $-25.3$ MPa.

(b) Neutral axis

$$\tan\alpha = \frac{M_yI_x}{M_xI_y} = \tan\phi\,\frac{I_x}{I_y} = \tan30^\circ\times\frac{11.52}{5.12} = 0.5774\times2.25 = 1.299$$ $$\alpha = 52.41^\circ$$

The neutral axis passes through the centroid at $52.4^\circ$ to the x-axis. It is turned towards the weaker axis (the y-axis, which has the smaller moment of inertia), compared with the position $30^\circ$ that would be perpendicular to the load plane.

      y
      |  load plane (30 deg from y)
      | /     NA (52.4 deg from x)
      |/    /
  ----+----/----- x
     /|   /

(c) Deflection of the free end

$$\begin{aligned} \delta_y &= \frac{P_yL^3}{3EI_x} = \frac{1732.05\times1500^3}{3\times200\times10^3\times11.52\times10^6} = 0.846\ \text{mm} \\ \delta_x &= \frac{P_xL^3}{3EI_y} = \frac{1000\times1500^3}{3\times200\times10^3\times5.12\times10^6} = 1.099\ \text{mm} \end{aligned}$$ $$\delta = \sqrt{\delta_x^2 + \delta_y^2} = 1.386\ \text{mm}$$

The direction is at $\tan^{-1}(\delta_x/\delta_y) = 52.4^\circ$ from the vertical: this is not the $30^\circ$ of the load, and it is perpendicular to the neutral axis.

Answer: (a) $\sigma_{max} = 25.3$ MPa at a corner; (b) neutral axis at $52.4^\circ$ to the horizontal axis through the centroid; (c) resultant deflection $1.39$ mm at $52.4^\circ$ to the vertical.