
In this lecture we want to introduce the influence of support motion on a dynamic system. For us, support motion can be thought of as earthquake-induced ground motion. We're going to keep things relatively simple initially. Let's start by evaluating a single degree of freedom system subject to a harmonic ground motion.

Now, we're specifying a harmonic ground motion. So this means we're going to be able to establish a closed-form analytical solution, an equation in other words. And in particular, our aim is to derive an equation for the steady-state response of our system. We're not interested in the transient response. This will set us up nicely to discuss the concept of <b>transmissibility</b> in the next lecture.

So, our prescribed ground motion is given by,

$$
u_s(t) = U_0\sin(\bar{\omega}t)
\tag{1}
$$

We will let $u(t)$ represent the motion of the mass in our SDoF system. It's important to note at this stage that all motions are absolute motions measured with reference to a fixed global reference frame. In other words, $u(t)$ is not measured relative to the support motion (this will come later). Fig 1. below represents the system graphically.

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/capturing-the-influence-of-ground-motion/img1.jpg"
	alt="Dynamic system indicating support motion u_s(t) and (absolute) motion of the mass, u(t). | EngineeringSkills.com"
	width="450"
	height="362"
/>
<Caption>
	Fig 1. Dynamic system indicating support motion $u_s(t)$ and (absolute) motion
	of the mass, $u(t)$.
</Caption>

Next, we can evaluate a free-body diagram of the mass, Fig 2 below. Note that the stiffness and damping forces are proportional to the motion of the mass relative to the support, while the inertia force is proportional to the absolute motion of the mass.

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/capturing-the-influence-of-ground-motion/img2.jpg"
	alt="Free-body diagram of the mass. | EngineeringSkills.com"
	width="450"
	height="227"
	caption="Fig 2. Free-body diagram of the mass."
/>

The free-body diagram leads to the following equation of motion,

$$
m\ddot{u} + c(\dot{u}-\dot{u}_s) + k(u-u_s) = 0
\tag{2}
$$

Now we can substitute in expressions for the support displacement and velocity,

$$
u_s(t) = U_0\sin(\bar{\omega}t)
\tag{3}
$$

$$
\dot{u}_s(t) = U_0\bar{\omega}\cos(\bar{\omega}t)
\tag{4}
$$

which yields,

$$
m\ddot{u} + c\dot{u} - cU_0\bar{\omega}\cos(\bar{\omega}t) + ku - kU_0\sin(\omega t) = 0
\tag{5}
$$

Rearranging this gives,

$$
m\ddot{u} + c\dot{u} + ku = cU_0\bar{\omega}\cos(\bar{\omega}t) + kU_0\sin(\omega t)
\tag{6}
$$

The right hand side of this equation can be replaced by the following expression,

$$
D\sin(\bar{\omega}t + \gamma)
\tag{7}
$$

where,

$$
D = U_0 k \sqrt{1+(2\xi\beta)^2}
\tag{8}
$$

in which $\beta$ is the ratio of support motion frequency to system natural frequency and $\xi$ is the damping ratio. Note also that,

$$
\tan \gamma = \frac{c\bar{\omega}}{k} = 2\xi\beta
\tag{9}
$$

We'll prove these relationships at the end of this lecture, but for now let's keep moving forward. Assuming for now that these statements are true, we can rewrite our equation of motion in a more familiar form,

$$
m\ddot{u} + c\dot{u} + ku = D\sin(\bar{\omega}t + \gamma)
\tag{10}
$$

The forcing term on the right hand side is now just a simple harmonic forcing function. This equation of motion has a standard closed-form solution that we saw while studying the [Fundamentals of Engineering Structural Dynamics](/course/fundamentals-of-structural-dynamics-with-python/characterising-the-steady-state-response). The steady-state solution is given by,

$$
\boxed{
u(t) = \frac{D}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2} \right]\left[(1-\beta^2)\sin(\bar{\omega}t+\gamma) - 2\xi\beta\cos(\bar{\omega}t+\gamma) \right]
}
\tag{11}
$$

Now we can plot both the support motion and steady-state mass motion.

### Plotting the response

We can start by importing the relevant libraries. I’ll omit this step in future text lectures.

```python
import math #Basic math functionality
import numpy as np #Numpy for working with arrays
import matplotlib.pyplot as plt #Plotting functionality
```

Then we can establish the system parameters, construct a time vector and process the response equation.

```python
# System parameters
m = 1000 #(kg) Mass
xi = 0.05 #Damping ratio
f = 1 #(Hz) Natural frequency
w = 2*math.pi*f #(rad/s) Angular natural frequency
k = m*w**2 #(N/m) Stiffness

# Time vector
tmax = 10 #(sec) The max time
delta_t = 0.01 #(sec) The timestep
nPoints = tmax/delta_t #Number of equally spaced data points
t = np.linspace(0,tmax, int(nPoints)) # Time vector

# Support motion parameters
U0 = 0.1 #(m) Base motion amplitude
fb = 1 #(Hz) Base motion frequency
wb = 2*math.pi*fb #(rad/s) Angular base motion frequency
us = U0*np.sin(wb*t) #(m) Base motion

# Response
beta = wb/w
D = U0*k*math.sqrt(1+(2*xi*beta)**2)
gamma = math.atan(2*xi*beta)

# Calculate the response (break the equation up)
uConst = (D/k)*(1/((1-beta**2)**2+(2*xi*beta)**2))
uVar = (1-beta**2)*np.sin(wb*t+gamma) - 2*xi*beta*np.cos(wb*t+gamma)
u = uConst * uVar
```

And finally, plot the response.

```python
#Plotting
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,2])
axes.plot(t,us,'-',label='Support motion')
axes.plot(t,u, '-', label='Absolute steady-state response')
axes.set_xlim([0,tmax])
axes.set_xlabel('Time (sec)')
axes.set_ylabel('Displacement (m)')
axes.set_title('Support and mass motion')
axes.grid()
axes.legend(loc='lower right')
plt.show()
```

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/capturing-the-influence-of-ground-motion/img3.jpg"
	alt="Plot of absolute support and mass motion. | EngineeringSkills.com"
	width="1000"
	height="527"
	caption="Fig 3. Plot of absolute support and mass motion."
/>

By changing the frequency ratio we observe that the motion induced in the mass varies significantly. This leads us nicely onto the topic of **transmissibility**. Transmissibility describes the degree of relative isolation between the base and mass motion. We will discuss this in the next lecture and get more of an understanding of the relationship between base and mass motion.

### Derivation of $D$ and $\gamma$

The last thing we need to do in this lecture is prove the statement made earlier, that,

$$
cU_0\bar{\omega}\cos(\bar{\omega}t) + kU_0\sin(\omega t) = D\sin(\bar{\omega}t + \gamma)
\tag{12}
$$

where,

$$
D = U_0 k \sqrt{1+(2\xi\beta)^2}
\tag{13}
$$

and,

$$
\tan \gamma = \frac{c\bar{\omega}}{k} = 2\xi\beta
\tag{14}
$$

We will start by making the statements,

$$
D = \sqrt{(kU_0)^2+(c\bar{\omega}U_0)^2}
\tag{15}
$$

which can be easily shown to be equivalent to $D$ from equation (13), and

$$
\tan \gamma = \frac{c\bar{\omega}U_0}{kU_0}
\tag{16}
$$

We can visualise both of these statements in the context of a right-angle triangle, Fig. 4 below.

<Image
	src="/images/courses/multi-degree-of-freedom-dynamics-modal-analysis-and-seismic-response-simulation-in-python/capturing-the-influence-of-ground-motion/img4.jpg"
	alt="Right-angle triangle showing geometric relationship between D and gamma. | EngineeringSkills.com"
	width="400"
	height="253"
/>
<Caption>
	Fig 4. Right-angle triangle showing geometric relationship between $D$ and
	$\gamma$.
</Caption>

From this visualisation we see that,

$$
\sin \gamma = \frac{c\bar{\omega}U_0}{D}
\tag{17}
$$

and

$$
\cos \gamma = \frac{kU_0}{D}
\tag{18}
$$

Remember, our task is to show that the left side of equation (12) equals the right; to do this, multiple the left side of equation (12) by $D/D$, giving us

$$
D\left[\frac{cU_0\bar{\omega}}{D}\cos(\bar{\omega}t) + \frac{kU_0}{D}\sin(\bar{\omega_D}t)\right]
\tag{19}
$$

Now we can substitute in our expressions for $\sin \gamma$ and $\cos \gamma$ to give us,

$$
D\underbrace{\left[\sin \gamma \:\cos(\bar{\omega}t) + \cos \gamma\:\sin(\bar{\omega}t)\right]}_{=\:\sin(\bar{\omega}t + \gamma)}
\tag{20}
$$

Using a standard trigonometric identity, we replace the quantity in square brackets with $\sin(\bar{\omega}t+\gamma)$, leaving us with the left side of equation (12) as

$$
D \sin(\bar{\omega}t + \gamma)
\tag{21}
$$

Which matches the right side shown above.


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