
In this lecture we'll bring together and link much of what we've talked about in the previous few lectures. After completing this lecture you'll have a clear understanding of what orthogonality actually means and exactly how it decouples our equations of motion.

Recall that we said previously that the undamped free vibration of a MDoF system is in fact a specific type of problem in mathematics known as an eigenproblem. We can rearrange the equations of motion into the standard eigenproblem structure as follows,

$$
\begin{bmatrix}
K
\end{bmatrix}
\begin{Bmatrix}
U
\end{Bmatrix} =
\omega^2
\begin{bmatrix}
M
\end{bmatrix}
\begin{Bmatrix}
U
\end{Bmatrix}
\tag{1}
$$

where $\begin{Bmatrix}U\end{Bmatrix}$ represents the eigenvectors and $\omega^2$ represents the eigenvalues of the system. These eigenvectors posses a special property called **orthogonality**. It is this orthogonality that allows us to uncouple the equations of motion by performing a **transformation of the problem into modal coordinates** (also commonly referred to as generalised coordinates).

### Orthogonality

At this point let's take a step back, and discuss orthogonality in a more familiar context. Then we'll map this understanding onto the dynamic behaviour of our system. We can start by considering a simple 3-dimensional axis system, Fig 1.

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/orthogonality-and-decoupling-equations-of-motion/img1.jpg"
	alt="3-dimensional axis system. | EngineeringSkills.com"
	width="300"
	height="254"
	caption="Fig 1. 3-dimensional axis system."
/>

We can represent this axis system with three unit vectors, one defining each coordinate axis,

$$
\begin{Bmatrix}
x
\end{Bmatrix} =
\begin{Bmatrix}
1\\0\\ 0
\end{Bmatrix},
\:
\begin{Bmatrix}
y
\end{Bmatrix} =
\begin{Bmatrix}
0\\1\\ 0
\end{Bmatrix},
\:
\begin{Bmatrix}
z
\end{Bmatrix} =
\begin{Bmatrix}
0\\0\\ 1
\end{Bmatrix}
\tag{2}
$$

We usually bundle these vectors into a single matrix that conveniently packages up the _vector basis_ for a 3-dimensional space,

$$
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{bmatrix}
\tag{3}
$$

We can say that these three vectors (represented as the columns of the vector basis matrix above), **span the space**, a normal 3D Euclidean space in this case. Note that the vectors are all orthogonal (mutually perpendicular) to each other and are therefore linearly independent from one another.

**Here's the key point;** because the three vectors are orthogonal (they are an orthogonal vector basis for the space) any other vector in the space can be obtained from a linear combination of the basis vectors. We can say this set of **3** orthogonal vectors are a valid basis set to represent any other vector in the **3** dimensional space.

Now recall that the eigenvectors representing the mode shapes from the free vibration problem are also an orthogonal set of vectors,

$$
\begin{bmatrix}
\Phi
\end{bmatrix} =
\begin{bmatrix}
\phi_{1,1} & \phi_{1,2} & \phi_{1,3}\\
\phi_{2,1} & \phi_{2,2} & \phi_{2,3}\\
\phi_{3,1} & \phi_{3,2} & \phi_{3,3}\\
\end{bmatrix}
\tag{4}
$$

So, we can also say that they are a valid set of basis vectors for an $N$ dimensional space where $N=3$ in our example. Think about what this means; we can represent any $N^{\text{th}}$ order vector in the space with a linear combination of the basis vectors (mode shapes).

So, let's consider the parallels between the vector basis of a 3D Euclidean space and a $N^{\text{th}}$ order space spanning the space of possible dynamic behaviour of our structure. Consider the vector $\alpha$, Fig 2.

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/orthogonality-and-decoupling-equations-of-motion/img2.jpg"
	alt="3-dimensional vector. | EngineeringSkills.com"
	width="250"
	height="267"
	caption="Fig 2. 3-dimensional vector"
/>

We know that $\alpha$ can be obtained as the linear combination of the basis vectors,

$$
\alpha = \gamma_x
\begin{Bmatrix}
1\\0\\ 0
\end{Bmatrix} +
\gamma_y
\begin{Bmatrix}
0\\1\\ 0
\end{Bmatrix} +
\gamma_z
\begin{Bmatrix}
0\\0\\ 1
\end{Bmatrix}
\tag{5}
$$

where $\gamma_x$, $\gamma_y$ and $\gamma_z$ are the scale factors applied to each (unit) basis vector. The scale factors can be thought of as the length of the projection of $\alpha$ onto each basis vector.

Applying this same theory to an NDoF system, the total response $u(t)$ will be obtained by superimposing the vibration responses _'along each axis'_ or put another way, the responses in each mode,

$$
u(t) =
\underbrace{
Y_1(t)
\begin{Bmatrix}
\phi_{1,1} \\ \phi_{2,1} \\ \phi_{3,1}
\end{Bmatrix}}_{\text{mode 1 response}} +
\underbrace{
Y_2(t)
\begin{Bmatrix}
\phi_{1,2} \\ \phi_{2,2} \\ \phi_{3,2}
\end{Bmatrix}}_{\text{mode 2 response}} +
\underbrace{
Y_3(t)
\begin{Bmatrix}
\phi_{1,3} \\ \phi_{2,3} \\ \phi_{3,3}
\end{Bmatrix}}_{\text{mode 3 response}}
\tag{6}
$$

where $Y_1(t)$, $Y_2(t)$ and $Y_3(t)$ are the responses in each mode, modelled using a SDoF system with the modal dynamic characteristics (mass and stiffness). Remember the axes (mode shapes in this case) are linearly independent, so the vibration in each mode is also independent. This is what allows us to model the NDoF system with $N$ independent SDoF systems.

The previous equation is a statement of modal superposition and can be condensed into the form commonly found in textbooks,

$$
\begin{Bmatrix}
u
\end{Bmatrix}  = \sum_{k=1}^N
\begin{Bmatrix}
\phi
\end{Bmatrix}_k \:Y_k
\tag{7}
$$

where $Y_k$ is the response in mode $k$ and $\begin{Bmatrix}\phi_k\end{Bmatrix}$ is the mass normalised eigenvector for mode $k$.

### Decoupling the equations

Now that we understand at least conceptually what orthogonality and decoupling mean, we can start to discuss the details of how to implement the decoupling. Due to the orthogonality property of the mode shapes we get the mass and stiffness orthogonality relationships,

$$
\begin{bmatrix}
\Phi
\end{bmatrix}^T
\begin{bmatrix}
M
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}  = \text{diagonal}
\tag{8}
$$

$$
\tag{9} \begin{bmatrix}
\Phi
\end{bmatrix}^T
\begin{bmatrix}
K
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}  = \text{diagonal}
$$

If we consider a single mode shape vector, as opposed to the matrix of all vectors, $\begin{bmatrix}\Phi\end{bmatrix}$, we could say,

$$
\begin{Bmatrix}
\phi
\end{Bmatrix}_i^T
\begin{bmatrix}
M
\end{bmatrix}
\begin{Bmatrix}
\phi
\end{Bmatrix}_j = 0 \:\: i\neq j
\tag{10}
$$

$$
\begin{Bmatrix}
\phi
\end{Bmatrix}_i^T
\begin{bmatrix}
K
\end{bmatrix}
\begin{Bmatrix}
\phi
\end{Bmatrix}_j = 0 \:\: i\neq j
\tag{11}
$$

In other words, for all non-diagonal values in the resulting matrix, the multiplication yields a zero (i.e. a diagonal matrix). Transforming $\begin{bmatrix}M\end{bmatrix}$ and $\begin{bmatrix}K\end{bmatrix}$ into diagonal matrices essentially has the effect of decoupling our equations of motion (note that for our relatively simple structures $\begin{bmatrix}M\end{bmatrix}$ is already diagonal).

By pre-multiplying $\begin{bmatrix}M\end{bmatrix}$ and $\begin{bmatrix}K\end{bmatrix}$ by $\begin{bmatrix}\Phi\end{bmatrix}^T$ and post-multiplying by $\begin{bmatrix}\Phi\end{bmatrix}$, we perform the coordinate transformation that uncouples the mass and stiffness matrices. The diagonal matrices $\begin{bmatrix}\Phi\end{bmatrix}^T\begin{bmatrix}M\end{bmatrix}\begin{bmatrix}\Phi\end{bmatrix}$ and $\begin{bmatrix}\Phi\end{bmatrix}^T\begin{bmatrix}K\end{bmatrix}\begin{bmatrix}\Phi\end{bmatrix}$ can now be used to build an equivalent uncoupled system of equations.

To build the uncoupled system of equations that we need to solve, consider the statement of modal superposition from above, but represented in matrix form,

$$
\begin{Bmatrix}
u
\end{Bmatrix} =
\begin{bmatrix}
\Phi
\end{bmatrix}
\begin{Bmatrix}
Y
\end{Bmatrix}
\tag{12}
$$

The system acceleration is obtained as the second derivative in the usual way,

$$
\begin{Bmatrix}
\ddot{u}
\end{Bmatrix} =
\begin{bmatrix}
\Phi
\end{bmatrix}
\begin{Bmatrix}
\ddot{Y}
\end{Bmatrix}
\tag{13}
$$

Recall now the coupled system of equations,

$$
\begin{bmatrix}
M
\end{bmatrix}
\begin{Bmatrix}
\ddot{u}
\end{Bmatrix} +
\begin{bmatrix}
K
\end{bmatrix}
\begin{Bmatrix}
u
\end{Bmatrix} =
\begin{Bmatrix}
P
\end{Bmatrix}
\tag{14}
$$

Substituting in for $\begin{Bmatrix}\ddot{u}\end{Bmatrix}$ and $\begin{Bmatrix}u\end{Bmatrix}$ yields,

$$
\begin{bmatrix}
M
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}
\begin{Bmatrix}
\ddot{Y}
\end{Bmatrix} +
\begin{bmatrix}
K
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}
\begin{Bmatrix}
Y
\end{Bmatrix} =
\begin{Bmatrix}
P
\end{Bmatrix}
\tag{15}
$$

If we now pre-multiply both sides by the transpose of the modal matrix, we note that the equations become decoupled by virtue of the orthogonality relations. This pre-multiplication is the final operation required to decouple the equations of motion, allowing them to be solved independently.

$$
\underbrace{
\begin{bmatrix}
\Phi
\end{bmatrix}^T
\begin{bmatrix}
M
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}}_{\text{diagonal modal mass}}
\begin{Bmatrix}
\ddot{Y}
\end{Bmatrix} +
\underbrace{
\begin{bmatrix}
\Phi
\end{bmatrix}^T
\begin{bmatrix}
K
\end{bmatrix}
\begin{bmatrix}
\Phi
\end{bmatrix}}_{\text{diagonal modal stiffness}}
\begin{Bmatrix}
Y
\end{Bmatrix} =
\underbrace{
\begin{bmatrix}
\Phi
\end{bmatrix}^T
\begin{Bmatrix}
P
\end{Bmatrix}}_{\text{modal force vector}}
\tag{16}
$$

We note that the mass and stiffness terms are both diagonal and therefore the total response of the undamped NDOF system has been transformed into the summation of responses for a series of SDOF systems.

Recall that a few lectures ago, we mass normalised the modal matrix and cryptically said that this was a convenient form of normalisation. Well now we see that mass normalisation has given us a unit mass in each of our SDoF systems. This is the convenience we spoke of.

The system described by the preceding equation is said to have been transferred into modal coordinates. This is by virtue of the fact that an orthogonal vector set (matrix of eigenvectors) has been used to decouple the dynamic equations.

We can now evaluate the SDoF response in each mode and recombine these using the mode shapes, in much the same way that we reconstruct a 3D vector as the linear combination of the unit basis vectors, discussed above.

In the next lecture we'll put this into practice and determine the response of our 3DoF shear building using a modal superposition approach.


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