diff --git a/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md b/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md new file mode 100755 index 0000000..d81ed3f --- /dev/null +++ b/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md @@ -0,0 +1,136 @@ +--- +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) diff --git a/uni/mmme/2xxx/2053_mechanics_of_solids/exercise_sheets/Asymmetrical Bending Exercise Sheet Solutions.pdf b/uni/mmme/2xxx/2053_mechanics_of_solids/exercise_sheets/Asymmetrical Bending Exercise Sheet Solutions.pdf new file mode 100644 index 0000000..d309882 Binary files /dev/null and b/uni/mmme/2xxx/2053_mechanics_of_solids/exercise_sheets/Asymmetrical Bending Exercise Sheet Solutions.pdf differ diff --git a/uni/mmme/2xxx/2053_mechanics_of_solids/exercise_sheets/Asymmetrical Bending Exercise Sheet.pdf b/uni/mmme/2xxx/2053_mechanics_of_solids/exercise_sheets/Asymmetrical Bending Exercise Sheet.pdf new file mode 100644 index 0000000..2f30fb7 Binary files /dev/null and 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