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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 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 # Product Moments of Area
To analyse asymmetrically loaded sections, we need the second moments of area $I_{yy}$, and $I_{xx}$ 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
$$C = \frac{I_{xx} + I_{yy}}{2}$$ $$C = \frac{I_{xx} + I_{yy}}{2}$$
$$R = \sqrt{\left(\frac{I_{xx}-I{yy}}{2}\right)^2 + I_{xy}^2}$$ $$R = \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$
$$I_p = C + R$$ $$I_p = C + R$$
@ -82,22 +94,22 @@ resolved onto the principal axes, P and Q:
\begin{align*} \begin{align*}
\cos\theta &= \frac{M_{P_x}}{M_x} \rightarrow M_{P_x} = M_x\cos\theta\\ \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\\ \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) ![](./images/vimscrot-2023-02-02T14:54:32,264829706+00:00.png)
Similarly we get: Similarly we get:
\begin{align*} \begin{align*}
M_{P_y} = M_y\sin\theta\\ M_{P_y} = M_y\sin\theta\\
M_{Q_y} = M_y\cos\theta\\ M_{Q_y} = M_y\cos\theta
\end{align} \end{align*}
Therefore: Therefore:
\begin{align*} \begin{align*}
M_P = M_{P_x} + M_{P_y} = M_x\cos\theta + M_y\sin\theta \\ 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*} \end{align*}
## Bending Stress at Position (P, Q) ## Bending Stress at Position (P, Q)
@ -122,15 +134,3 @@ The neutral axis is where $\sigma = 0$:
\end{align*} \end{align*}
The maximum stress is located in cross section point which is furthest from the neutral axis. 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)