fix typos

uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md

@@ -6,6 +6,18 @@ tags: [ mmme2053, beam_bending, asymmetrical_beam_bending ]
 uuid: 7afb5f13-4d55-4e00-927a-5d622520d844
 ---
 
+## 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)
+
 # Product Moments of Area
 
 To analyse asymmetrically loaded sections, we need the second moments of area $I_{yy}$, and $I_{xx}$
@@ -51,7 +63,7 @@ Once the second moments of area and product moments are found, they can be used
 
 $$C = \frac{I_{xx} + I_{yy}}{2}$$
 
-$$R = \sqrt{\left(\frac{I_{xx}-I{yy}}{2}\right)^2 + I_{xy}^2}$$
+$$R = \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$
 
 $$I_p = C + R$$
 
@@ -82,7 +94,7 @@ 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}
+\end{align*}
 
 ![](./images/vimscrot-2023-02-02T14:54:32,264829706+00:00.png)
 
@@ -90,14 +102,14 @@ Similarly we get:
 
 \begin{align*}
 M_{P_y} = M_y\sin\theta\\
-M_{Q_y} = M_y\cos\theta\\
-\end{align}
+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 \\
+M_Q = M_{Q_x} + M_{Q_y} = -M_x\sin\theta + M_y\cos\theta
 \end{align*}
 
 ## Bending Stress at Position (P, Q)
@@ -122,15 +134,3 @@ The neutral axis is where $\sigma = 0$:
 \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)