diff --git a/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md b/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md index e51fcd2..af28121 100755 --- a/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md +++ b/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) 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