fix images

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

@@ -30,8 +30,7 @@ $$I_{xy} = \int_A xy \mathrm{d}A$$
 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)
+![](./images/vimscrot-2023-02-05T16:58:38,302909671+00:00.png)
-
 
 \begin{align*}
 I_{x'x'} &= \int_A y'^2 \mathrm{d}A \\
@@ -59,7 +58,7 @@ Once the second moments of area and product moments are found, they can be used
 
   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)
+![](./images/vimscrot-2023-02-05T16:59:07,932521138+00:00.png)
 
 $$C = \frac{I_{xx} + I_{yy}}{2}$$
 
@@ -71,9 +70,9 @@ $$I_q = C - R$$
 
 $$\sin{2\theta} = \frac{I_{xy}}{R}$$
 
-# Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Princial Axes
+# Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Principal Axes
 
-![](./images/vimscrot-2023-02-02T14:53:05,040998048+00:00.png)
+![](./images/vimscrot-2023-02-05T16:59:59,303885015+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.
@@ -84,19 +83,19 @@ 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)
+![](./images/vimscrot-2023-02-05T17:00:19,100442577+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)
+![](./images/vimscrot-2023-02-05T17:02:36,461283527+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)
+![](./images/vimscrot-2023-02-05T17:02:45,578419970+00:00.png)
 
 Similarly we get:
 
@@ -114,7 +113,7 @@ M_Q = M_{Q_x} + M_{Q_y} = -M_x\sin\theta + M_y\cos\theta
 
 ## Bending Stress at Position (P, Q)
 
-![](./images/vimscrot-2023-02-02T15:02:36,806587572+00:00.png)
+![](./images/vimscrot-2023-02-05T17:03:08,322998726+00:00.png)
 
 $$\sigma = \frac{M_PQ}{I_P} - \frac{M_QP}{I_Q}$$
 
@@ -122,7 +121,7 @@ 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)
+![](./images/vimscrot-2023-02-05T17:03:20,351506450+00:00.png)
 
 The neutral axis is where $\sigma = 0$: