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Akbar Rahman 2023-02-05 16:33:11 +00:00
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uni/mmme/2xxx/2053_mechanics_of_solids/asymmetrical_bending.md
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@ -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)