Introduction to MDOF(2)

After the discussion about regimes and mode shapes, we shall derive the EOM of MDOF.

There are three ways to get the EOM of MDOF.

The first two methods are the same as SDOF.

So, we first formulate a highly idealized two-story shear-frame subjected to external forces \(p_1(t)\) and \(p_2(t)\). The assumptions of the frame are as follows:

①The beams and floor systems are rigid (infinitely stiff) in flexture;

②Neglect the axial deformation of the beams and columns and the effect of axial force on the stiffness of the columns;

③Neglect the mass of columns and only consider the stiffness;

④The mass is concentrated at the floor levels.

In these cases, the frame has two DOFs: the lateral displacements \(u_1\) and \(u_2\) of the two floors in the direction of the \(x-axis\), the number of DOF directly determines the number of the unknowns and thus the number of equations required.

This ideal model is a shear-building model. I think it is necessary to summarize the differences between shear-building model and bending-building model.
The most straightforward difference is the position of the reverse bending porint under lateral force. The position of the reverse bending point of the shear-frame occurs on the column, while the bending-frame appears on the beam and floor.
This is because the floor stiffness of the former is much higher than that of the pillar, while the latter is just the opposite. In engineering practice, frame structure is generally shear deformation, shear wall structure is generally bending deformation just like a huge beam, frame shear structure have both.
It’s a very basic concept.

Using Newton`s Sencond Law

According to Newton`s second law, the EOM is:

$$
m_i \ddot{u_i} = p_i - f_{si} - f_{Di}
$$

And then these assumptions come in handy.

①Because the beams and floor systems have EI → ∞, the number of DOF goes down from 6 to 2, which means that there are 2 unknowns to solve.

②Because we neglect the mass of columns,

we can get \(f_{s2} = f{s12})

$$
f_{s2} = k_2(x_2-x_1)
$$

$$
f_{s12} = f_{s1} = k_2(x_2-x_1)
$$

Then we can get:

\[
m_2 \ddot{x_2} = -k_2(x_2-x_1) + p_2
\\ \rightarrow m_2 \ddot{x_2} -k_2x_1 + k_2x_2 = p_2
\]

\[
m_1 \ddot{x_1} = k_2(x_2-x_1) - k_1x_1 + p_1
\\ \rightarrow m_1 \ddot{x_1} +(k_1+k_2)x_1 - k_2x_2 = p_1
\]

Writing the two equations in matrix form:

\[
\left[
\begin{array}{ccc}
m_1&0\
0&m_2
\end{array}
\right]
\left[
\begin{array}{ccc}
\ddot{x_1}(t)\
\ddot{x_2}(t)
\end{array}
\right]
+
\left[
\begin{array}{ccc}
k_1+k_2&-k_2\
-k_2&k_2
\end{array}
\right]
\left[
\begin{array}{ccc}
x_1(t)\
x_2(t)
\end{array}
\right]
=
\left[
\begin{array}{ccc}
p_1(t)\
p_2(t)
\end{array}
\right]
\]

Thus,

\[
M_{2×2} \ddot{x} (t)_{2×1} + K_{2×2}x(t)_{2×1} = P(t)_{2×1}
\]

This equation indicates that when we give an excitation to a structure, the structure`s response will never stop which is unrealistic. So what about the damping? Actually damping is really hard to tell from the structure because there are too many factors such as the non-structure members, the material, the air and even the furniture affect the damping. So it is usually determined by the past data and experiments or even experience. But we always have to represent it in our formula to calculate the response, so how do we do?

We artificially assume that a linear viscous damping mechanism represents the energy dissipation in a structure to empirically match the reality, just as in the case of SDOF systems, which means energy dissipation is associated with the deformational motions of each story. Luckily, this assumption suits the math calculation very well. But we still have to emphasize again that this damping form does not reflect the real situation of a structure.

For example, friction contributes to the damping, but friction is not proportional to the velocity, which contradicts the linear viscous damping mechanism in the formula.
So we can joke that this viscous damping assumption is just to make things easier.

Then we can give the general EOM form:

\[
M_{n×n}\ddot{x}(t)_{n×1} + C_{n×n}\dot{x}(t)_{n×1} + K_{n×n}x(t)_{n×1} = P(t)_{n×1}
\]

Using D`Alembert Dynamic Equilibrium

Similar to the Newton`s second law, allow me to omit this process.

Using Stiffness, Damping, and Mass Components

Newton`s second law and dynamic equilibrium methods are still the following the algebraic way of SDOF, actually we can directly convert our mind into a matrix way,

Under the action of external forces \(p_1(t)\) and \(p_2(t)\), the state of the system at any time instant is described by displacements \(u_j(t)\), velocities\(\dot{u}_j(t)\), and accelerations \(\ddot{u}_j(t)\). Now we visualize this system as the combination of three pure components:

Thus we can get the external forces \(f_{sj}\), \(f_{Dj}\), \(f_{Ij}\) on these components seperately. And finally write the EOM in matrix form directly.

The Solution of EOM

To motivate the concept of natural frequencies and mode shapes, consider an undamped structure under free vibration (give a disturbance rather than excitation).

\[
M_{n×n}\ddot{x}(t)_{n×1} + K_{n×n}x(t)_{n×1} = 0_{n×1}
\]

You won’t believe how we solved it. We just assume solution form \(x(t){n×1} = u{n×1} z(t){scalar}\), then \(\ddot{x}(t){n×1} = u_{n×1} \ddot{z}(t)_{scalar}\). This assumption helps us turn a vector variable into an algebraic variable, and the original vector variable becomes a constant. Then the form turns to:

\[
M_{n×n}u_{n×1}\ddot{z}(t) + K_{n×n}u_{n×1}z(t) = 0_{n×1}
\\ \rightarrow \overbrace{K_{n×n} u_{n×1}}^{constant} = - \frac{\ddot{z}(t)}{z(t)} \overbrace{M_{n×n} u_{n×1}}^{constant}
\]

So, apparently, the \(\frac{\ddot{z}(t)}{z(t)}\) must be a constant too. So, we assume the \(\frac{\ddot{z}(t)}{z(t)} = w^2\) for convenience. (Note that at this moment we do not know what \(w\) is)

Then we can get two information from the equation above.

\[
\begin{cases}
Ku= w^2Mu
\\ \ddot{z}(t)+w^2z(t) = 0
\end{cases}
\]

Thus, this problem becomes an eigenvalue problem, the \(u\) is the eigenvector and the \(w^2\) is the eigenvalue.

I think it is necessary to review the mathematical explanation of eigenvector.

Essentially, the vector u cannot be arbitrary. Otherwise, the Ku and Mu need not be proportional to each other. So, actually this problem is about finding the only vector and the proportion value which link Ku and Mu.

In a SDOF system, we have already know that stiffness K = w²m, then it turns out the square root of the eigenvalue plays the role of natural freq. That is why we write it as w².

Rearranging the eigenvalue equation gives

\[
(K_{n×n}-w^2M_{n×n})u_{n×1} = 0
\]

For the equation to have non-trivial solution u≠0, the matrix (K-w²M) must be singular with zero determinant.

\[
\vert K_{n×n}-w^2M_{n×n}\vert = 0
\]

If you have nonzero column vectors x that make the Ax = 0, then A is not invertible,

Then the determinant of A is equal to 0, you can also say A is singular.

Thus, we can get the value of w. Then we plug it in the equation, we can also get u. And u is not unique, except for the proportions between the elements because the singular matrix cannot give us enough equations to solve each unknown (you can verify from the linear algebra). So, we call u mode shape, which explains what we have found in the last article.

“If we mark the displacement of every instance of the response of MDOF, we can find an amazing thing that the ratio \(\frac{x_2}{x_1}\) remains the same, and this constant is the “mode shape” we will discuss after.”

For example, if \(u_1 = [0.5\ 1]^T\), then the mode shape in structure is: