--- author: Akbar Rahman date: \today title: MMME2053 // Asymmetrical Beam Bending tags: [ mmme2053, beam_bending, asymmetrical_beam_bending ] uuid: 7afb5f13-4d55-4e00-927a-5d622520d844 --- # Product Moments of Area To analyse asymmetrically loaded sections, we need the second moments of area $I_{yy}$, and $I_{xx}$ but we also need $I_{xy}$, the product moment of area: $$I_{xy} = \int_A xy \mathrm{d}A$$ # Parallel Axis Theorem The parallel axis theorem allows calculation of 2nd moments of area and product moments of area with respect to $x'$ and $y'$ axes: ![](./images/vimscrot-2023-02-02T14:40:20,244467338+00:00.png) \begin{align*} I_{x'x'} &= \int_A y'^2 \mathrm{d}A \\ &= \int_A (y+b)^2 \mathrm{d}A \\ &= \int_A (y^2 + b^2 + 2by) \mathrm{d}A \\ \\ I_{x'x'} &= I_{xx} + Ab^2 \end{align*} Similarly you can get \begin{align*} I_{y'y'} &= I_{yy} + Aa^2 \\ I_{x'y'} &= I_{xy} + abA \end{align*} # Principal Axes and Principal 2nd Moments of Area Once the second moments of area and product moments are found, they can be used to plot a Mohr's circle where: - Point A is plotted at $(I_{xx}, I_{xy})$ - Point B is plotted at $(I_{yy}, -I_{xy})$ - Points P and Q show the positions of the principal 2nd moments of area, $I_p$, and $I_q$. - $\theta$ is the angular position $e$ of the principal axes with respect to the $x$-$y$ axes The principal axes are the axes where the product moment of area is 0. ![](./images/vimscrot-2023-02-02T14:44:15,574579179+00:00.png) $$C = \frac{I_{xx} + I_{yy}}{2}$$ $$R = \sqrt{\left(\frac{I_{xx}-I{yy}}{2}\right)^2 + I_{xy}^2}$$ $$I_p = C + R$$ $$I_q = C - R$$ $$\sin{2\theta} = \frac{I_{xy}}{R}$$ # Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Princial Axes ![](./images/vimscrot-2023-02-02T14:53:05,040998048+00:00.png) If a bending moment, $M_x$ is applied about the x-axis only, then the stress in the flanges will cause bending to takeplace about both x and y axes. This is a consequence of $I_{xy} \neq 0$. To avoid this moment coupling effect, it is usually convenient to solve bending problems by considering bending about the principal axes, for which $I_{xy} = 0$. ## Resolving onto Principal Axes ![](./images/vimscrot-2023-02-02T14:54:13,309546586+00:00.png) If bending moments $M_x$ and $M_y$ are applied about the x and y axes respectively, these can be resolved onto the principal axes, P and Q: ![](./images/vimscrot-2023-02-02T14:54:26,513070327+00:00.png) \begin{align*} \cos\theta &= \frac{M_{P_x}}{M_x} \rightarrow M_{P_x} = M_x\cos\theta\\ \sin\theta &= \frac{-M_{Q_x}}{M_x} \rightarrow M_{Q_x} = -M_x\sin\theta\\ \end{align} ![](./images/vimscrot-2023-02-02T14:54:32,264829706+00:00.png) Similarly we get: \begin{align*} M_{P_y} = M_y\sin\theta\\ M_{Q_y} = M_y\cos\theta\\ \end{align} Therefore: \begin{align*} M_P = M_{P_x} + M_{P_y} = M_x\cos\theta + M_y\sin\theta \\ M_Q = M_{Q_x} + M_{Q_y} = -M_x\sin\theta + M_y\cos\theta \\ \end{align*} ## Bending Stress at Position (P, Q) ![](./images/vimscrot-2023-02-02T15:02:36,806587572+00:00.png) $$\sigma = \frac{M_PQ}{I_P} - \frac{M_QP}{I_Q}$$ Note the -ve sign, as a positive stress results in a -ve moment about the y-axis. ## Position of the Neutral Axis ![](./images/vimscrot-2023-02-02T15:09:13,105535645+00:00.png) The neutral axis is where $\sigma = 0$: \begin{align*} \frac{M_PQ}{I_P} &= \frac{M_QP}{I_Q} \\ \frac Q P &= \frac{M_QI_P}{M_PI_Q} \\ \\ \alpha &= \arctan\frac{Q}{P} \end{align*} The maximum stress is located in cross section point which is furthest from the neutral axis. ## Procedure for Solving for Bending Stress and Neutral Axis Position in Asymmetrical Bending Problems 1. [Determine the principal axes](#Principal-Axes-and-Principal-2nd-Moments-of-Area) of the section (about which $I_{xy}= 0$) 2. [Resolve bending moments onto these axes](#resolving-onto-principal-axes) 3. [Determine angle of neutral axis](#position-of-neutral-axis) 4. Evaluate bending stress at any poisition in the section, such as extremes away from neutral axis, which give maximum bending stress # Worked Example - [Worked Example 1 (PDF)](./worked_examples/MMME2053 AB WE1 Slides.pdf) - [Worked Example 2 (PDF)](./worked_examples/MMME2053 AB WE2 Slides.pdf)