---
author: Akbar Rahman
date: \today
title: MMME2053 // Asymmetrical Beam Bending
tags: [ mmme2053, beam_bending, asymmetrical_beam_bending ]
uuid: 7afb5f13-4d55-4e00-927a-5d622520d844
lecture_notes: [ ./lecture_notes/Asymmetrical Bending Notes.pdf ]
exercise_sheets: [ ./exercise_sheets/Asymmetrical Bending Exercise Sheet.pdf, ./exercise_sheets/Asymmetrical Bending Exercise Sheet Solutions.pdf ]
lecture_slides: [ ./lecture_slides/MMME2053 AB L1 Slides.pdf, ./lecture_slides/MMME2053 AB L2 Slides.pdf, ./lecture_slides/MMME2053 AB WE1 Slides.pdf ]
---

## 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-the-neutral-axis)
4. Evaluate bending stress at any position 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 $I_{xx}$
but we also need $I_{xy}$, the product moment of area:

$$I_{xy} = \int_A xy \mathrm{d}A$$

# Parallel Axis Theorem

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-2023-02-05T16:58:38,302909671+00:00.png)

\begin{align*}
I_{x'x'} &= \int_A y'^2 \mathrm{d}A \\
&= \int_A (y+b)^2 \mathrm{d}A \\
&= \int_A (y^2 + b^2 + 2by) \mathrm{d}A \\
\\
I_{x'x'} &= I_{xx} + Ab^2
\end{align*}

Similarly you can get

\begin{align*}
I_{y'y'} &= I_{yy} + Aa^2 \\
I_{x'y'} &= I_{xy} + abA
\end{align*}

# Principal Axes and Principal 2nd Moments of Area

Once the second moments of area and product moments are found, they can be used to plot a Mohr's circle where:

- Point A is plotted at $(I_{xx},  I_{xy})$
- Point B is plotted at $(I_{yy}, -I_{xy})$
- Points P and Q show the positions of the principal 2nd moments of area, $I_p$, and $I_q$.
- $\theta$ is the angular position $e$ of the principal axes with respect to the $x$-$y$ axes

  The principal axes are the axes where the product moment of area is 0.

![](./images/vimscrot-2023-02-05T16:59:07,932521138+00:00.png)

$$C = \frac{I_{xx} + I_{yy}}{2}$$

$$R = \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$

$$I_p = C + R$$

$$I_q = C - R$$

$$\sin{2\theta} = \frac{I_{xy}}{R}$$

# Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Principal Axes

![](./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.
This is a consequence of $I_{xy} \neq 0$.

To avoid this moment coupling effect, it is usually convenient to solve bending problems by
considering bending about the principal axes, for which $I_{xy} = 0$.

## Resolving onto Principal Axes

![](./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-2023-02-05T17:02:36,461283527+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-2023-02-05T17:02:45,578419970+00:00.png)

Similarly we get:

\begin{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
\end{align*}

## Bending Stress at Position (P, Q)

![](./images/vimscrot-2023-02-05T17:03:08,322998726+00:00.png)

$$\sigma = \frac{M_PQ}{I_P} - \frac{M_QP}{I_Q}$$

Note the -ve sign, as a positive stress results in a -ve moment about the y-axis.

## Position of the Neutral Axis

![](./images/vimscrot-2023-02-05T17:03:20,351506450+00:00.png)

The neutral axis is where $\sigma = 0$:

\begin{align*}
\frac{M_PQ}{I_P} &= \frac{M_QP}{I_Q} \\
\frac Q P &= \frac{M_QI_P}{M_PI_Q} \\
\\
\alpha &= \arctan\frac{Q}{P}
\end{align*}

The maximum stress is located in cross section point which is furthest from the neutral axis.