Non Symmetrical Bending of Straight Beams

Meaning

Bending is unsymmetrical when (a) the load plane does not coincide with a principal axis of the cross-section (e.g. an inclined load on a rectangular beam), or (b) the section has no axis of symmetry (angle, Z-section). The beam then deflects in a plane different from the plane of loading.

Derivation

Let $x, y$ be centroidal axes, $z$ the beam axis, $M_x$ the moment about $x$ (positive when it causes tension at $+y$) and $M_y$ the moment about $y$ (positive moment vector along $+y$, causing compression at $+x$). Plane sections remain plane, so the longitudinal strain varies linearly and

$$\sigma_z = a + bx + cy$$

Conditions of equilibrium:

$$\begin{aligned} \int \sigma_z\,dA &= 0 \ \Rightarrow\ a = 0 \quad (x, y \text{ centroidal}) \\ \int \sigma_z\,y\,dA &= M_x \ \Rightarrow\ b\,I_{xy} + c\,I_x = M_x \\ -\int \sigma_z\,x\,dA &= M_y \ \Rightarrow\ b\,I_y + c\,I_{xy} = -M_y \end{aligned}$$

Solving for $b$ and $c$ with $D = I_xI_y - I_{xy}^2$:

$$\boxed{\sigma_z = \frac{(M_xI_y + M_yI_{xy})\,y - (M_yI_x + M_xI_{xy})\,x}{I_xI_y - I_{xy}^2}}$$

Special cases:

• Principal axes ($I_{xy} = 0$): $\sigma_z = \dfrac{M_x y}{I_x} - \dfrac{M_y x}{I_y}$ (the magnitudes simply superpose).
• Only one moment, symmetric section: the ordinary $\sigma = My/I$.

Neutral axis

Put $\sigma_z = 0$:

$$\frac{y}{x} = \tan\alpha = \frac{M_yI_x + M_xI_{xy}}{M_xI_y + M_yI_{xy}}$$

This is a straight line through the centroid. In the principal-axes form ($I_{xy}=0$): $\tan\alpha = \dfrac{M_y I_x}{M_x I_y}$. If the load makes angle $\phi$ with the y-axis (so $M_y/M_x = \tan\phi$), then

$$\tan\alpha = \tan\phi\,\frac{I_x}{I_y}$$

Why the neutral axis is not perpendicular to the load plane

The neutral axis would be perpendicular to the load plane only if $\alpha = \phi$ (or $90^\circ$ apart), which needs $I_x = I_y$ (a circle, square) or load along a principal axis. Otherwise $I_x \ne I_y$ rotates the axis towards the axis of least moment of inertia. The deflection is perpendicular to the neutral axis, so the beam deflects in a plane different from the loading plane.

     y
     |   load plane
     |  /
     | / phi
  ---+/------- x
    /|
   / | neutral axis (alpha > phi if Ix > Iy)

The maximum stress occurs at the point of the section farthest from the neutral axis.

Section properties (principal axes)

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

$$I_x = \frac{80\times120^3}{12} = 1.152e+07\ \text{mm}^4,\qquad I_y = \frac{120\times80^3}{12} = 5.120e+06\ \text{mm}^4$$

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, i.e. it is rotated away from the load plane towards the weaker axis.

      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.

(a) Principal axes

$$\begin{aligned} I_{avg} &= \frac{I_x + I_y}{2} = 4.250e+06\ \text{mm}^4 \\ R &= \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2} = \sqrt{(1.75)^2 + (1.6)^2}\times10^6 = 2.371e+06\ \text{mm}^4 \\ I_{max,min} &= I_{avg} \pm R = 6.621e+06,\ 1.879e+06\ \text{mm}^4 \end{aligned}$$ $$\tan 2\theta_p = \frac{-2I_{xy}}{I_x - I_y} = \frac{3.2}{3.5} \Rightarrow \theta_p = 21.22^\circ$$

The axis of $I_{max}$ is at $21.2^\circ$ anticlockwise from the x-axis. Check: $I_{max}I_{min} = 12.44\times10^{12} = I_xI_y - I_{xy}^2$. $\checkmark$

(b) Neutral axis

With $M_y = 0$, the stress equation reduces to

$$\sigma_z = \frac{M_x\,(I_y\,y - I_{xy}\,x)}{I_xI_y - I_{xy}^2}$$

$\sigma_z = 0$ gives $y/x = I_{xy}/I_y$:

$$\tan\alpha = \frac{-1.6\times10^6}{2.5\times10^6} = -0.64 \quad\Rightarrow\quad \alpha = -32.62^\circ$$

The neutral axis passes through the centroid at $-32.6^\circ$ to the x-axis (not along the x-axis, although $M_y = 0$).

(c) Stress at $(x, y) = (25, 40)$ mm

$$D = I_xI_y - I_{xy}^2 = 15\times10^{12} - 2.56\times10^{12} = 12.44\times10^{12}\ \text{mm}^8$$ $$\begin{aligned} I_y\,y - I_{xy}\,x &= (2.5\times10^6)(40) - (-1.6\times10^6)(25) = 100\times10^6 + 40\times10^6 = 140\times10^6 \\ \sigma_z &= \frac{2\times10^6\times140\times10^6}{12.44\times10^{12}} = 22.51\ \text{MPa (tension)} \end{aligned}$$

If $I_{xy}$ had been ignored, $M_xy/I_x = 13.3$ MPa would have been obtained, i.e. an error of about 41 percent.

Answer: $I_{max} = 6.62\times10^6$ mm$^4$, $I_{min} = 1.88\times10^6$ mm$^4$, $\theta_p = 21.2^\circ$; neutral axis at $-32.6^\circ$; $\sigma = 22.5$ MPa (tension) at $(25, 40)$ mm.