From 1f8a5382d8baae104f8753a793de1b08b133b1ca Mon Sep 17 00:00:00 2001 From: Alvie Rahman Date: Sun, 5 Feb 2023 16:33:11 +0000 Subject: [PATCH] fix typos --- .../asymmetrical_bending.md | 36 +++++++++---------- 1 file changed, 18 insertions(+), 18 deletions(-) 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 d81ed3f..e51fcd2 100755 --- a/uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md +++ b/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 $$C = \frac{I_{xx} + I_{yy}}{2}$$ -$$R = \sqrt{\left(\frac{I_{xx}-I{yy}}{2}\right)^2 + I_{xy}^2}$$ +$$R = \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$ $$I_p = C + R$$ @@ -82,22 +94,22 @@ 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) Similarly we get: \begin{align*} -M_{P_y} = M_y\sin\theta\\ -M_{Q_y} = M_y\cos\theta\\ -\end{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 \\ +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)