diff --git a/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md b/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md
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+---
+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:
+
+
+
+
+\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.
+
+
+
+$$C = \frac{I_{xx} + I_{yy}}{2}$$
+
+$$R = \sqrt{\left(\frac{I_{xx}-I{yy}}{2}\right)^2 + I_{xy}^2}$$
+
+$$I_p = C + R$$
+
+$$I_q = C - R$$
+
+$$\sin{2\theta} = \frac{I_{xy}}{R}$$
+
+# Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Princial Axes
+
+
+
+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
+
+
+
+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:
+
+
+
+\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}
+
+
+
+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)
+
+
+
+$$\sigma = \frac{M_PQ}{I_P} - \frac{M_QP}{I_Q}$$
+
+Note the -ve sign, as a positive stress results in a -ve moment about the y-axis.
+
+## Position of the Neutral Axis
+
+
+
+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)
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