
<Callout type="success" heading="💬 Mid-course check-in - how's it going??">

I hope you’re enjoying the course so far.

<Image
	src="/images/mid-course-check-in.jpg"
	alt="How's the course going? | EngineeringSkills.com"
	width="400"
	height="438"
/>

I’d appreciate it if you would take 5 minutes to let me know how it’s going by leaving a quick mid-course review. You can update your review again at the end of the course.

[Complete your review here](/leave-a-review) by selecting your course from the dropdown and telling me how it's going.

Thanks, and enjoy the rest of the course.

Seán

</Callout>

In this lecture we'll work through another example question that should stretch you a little more than the last one. Again, make sure to try solving this on your own first.

Consider the lightweight steel frame structure shown below. It supports heavy machinery with a total mass of $m = 10,000\: kg$. We can neglect the weight of the steel frame. Based on an experimental impact test, the inherent structural damping has been estimated to be $2\%$ $(\xi=0.02)$.

A load test has shown that a lateral force of $P=1500\: N$ induces a lateral displacement of $\Delta = 7 \:mm$.

The steel frame is constrained such that it can only move laterally (i.e. there is no twisting or significant vertical movement). As such it can be modelled as a SDoF system as shown on the right below.

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-2/img1.jpg"
	alt="Single-storey lightweight steel frame supporting heavy machinery (left), idealised SDoF dynamic model (right). | EngineeringSkills.com"
	width="600"
	height="312"
	caption="Fig 1. Single-storey lightweight steel frame supporting heavy machinery (left), idealised SDoF dynamic model (right)."
/>

With this information, determine the following:

1. The undamped natural frequency

2. The damping coefficient

3. The logarithmic decrement

4. The damped angular natural frequency

5. The number of cycles for the amplitude to reduce to $0.001 \:m$

Plot the free vibration response, assuming the structure was released from its initial displacement of $7 \:mm$

Again, we start by defining the system constants before working our way down through the questions.

```python
# Constants
m = 10000 # (kg) Mass
xi = 0.02 # Damping ratio
P = 1500 # (N) Static force magnitude
Delta = 7 # (mm) Static displacement
```

### 1. Undamped natural frequency, $\omega_n$ & $f_n$

In order to determine the natural frequency, we need to know the stiffness of the SDoF system. We can obtain this directly from the static load test,

$$
k = \frac{P}{\Delta}
\tag{1}
$$

Then the undamped natural frequency can be determined in the usual way.

```python
# Undamped natural frequency
k = P/(Delta/1000) #(N/m) Stiffness
omega_n = round(math.sqrt(k/m),3) #(rads/sec) Angular natural frequency
fn = round(omega_n/(2*math.pi),3) # (Hz) Natural frequency
T = round(1/fn,2) # (sec) Period of oscillation
print(f"1.0 The angular natural frequency is {omega_n} radians per second or
{fn} Hz with a period of {T} seconds")
```

- `1.0 The angular natural frequency is 4.629 radians per second or 0.737 Hz with a period of 1.36 seconds`

### 2. Damping coefficient, $c$

```python
c = round(xi*2*m*omega_n,2) #(Ns/m) Damping coefficient
print(f"2.0 The damping coefficient is {c} Ns/m")
```

- `2.0 The damping coefficient is 1851.6 Ns/m`

### 3. Logarithmic decrement, $\delta$

```python
delta = round(2*math.pi*xi,3) # Logarithmic decrement
print(f"3.0 The logarithmic decrement is {delta}")
```

- `3.0 The logarithmic decrement is 0.126`

### 4. Damped angular natural frequency, $\omega_d$

```python
omega_d = round(omega_n*math.sqrt(1-(xi**2)),3) #(rads/sec)
print(f"4.0 The damped angular natural frequency is {omega_d} radians per second")
```

- `4.0 The damped angular natural frequency is 4.628 radians per second`

### 5. Number of oscillations before the amplitude reduces to $0.001 \:m$

We know the height of the first peak $(7\:mm)$ and the height of the last peak $(1\:mm)$ and that this occurs over $n$ successive cycles. So, we can make the following statement,

$$
\frac{u_1}{u_n} =  \frac{7}{1}
\tag{1}
$$

This can also be represented as a succession of oscillation cycles,

$$
\frac{u_1}{u_n} = \frac{u_1}{u_2} \times \frac{u_2}{u_3}\times \frac{u_{n-1}}{u_n}
\tag{2}
$$

Taking the natural log of both sides, remember, taking the log maps multiplication to addition,

$$
\ln\left(\frac{u_1}{u_n}\right) = \delta + \delta+...\delta = n\delta
\tag{3}
$$

So, the number of cycles required to reduce the amplitude is given by,

$$
n = \frac{\ln\left(\frac{u_1}{u_n}\right)}{\delta}
\tag{4}
$$

```python
u_start = 0.007
u_finish = 0.001
n = math.log(u_start/u_finish)/delta
n_full = math.ceil(n) t_full = T*n_full
print(f"5.0 It takes {n_full} complete cycles and {t_full} seconds for the oscillation amplitude to reduce to {u_finish} mm")
```

- `5.0 It takes 16 complete cycles and 21.76 seconds for the oscillation amplitude to reduce to 0.001 mm`

### 6. Free vibration response

To plot the free vibration response, we could repeat exactly what we did at the end of the last lecture, i.e. use the following equation,

$$
u(t) = e^{-\xi\omega_n t}\left[\frac{v_0 + u_0\xi \omega_n}{\omega_d}\sin(\omega_d t) + u_0\cos(\omega_d t) \right]
\tag{5}
$$

However, let's demonstrate the alternative approach of solving for numerical values of the constants of integration, using the initial conditions.

We know the general solution to the free vibration response is,

$$
u(t) = e^{-\xi\omega_n t}[A \sin(\omega_d t) + B\cos(\omega_d t) ]
\tag{6}
$$

We have the following initial conditions,

$$
\begin{align}
u(t=0) &= 0.007 \:\text{m}\\
\dot{u}(t=0) &= 0 \:\text{m/s}
\end{align}
\tag{7}
$$

In order to use the initial velocity condition, we need to differentiate our expression for displacement,

$$
\dot{u}(t) = A\left[e^{-\xi\omega_nt}\omega_d\cos(\omega_dt) - \sin(\omega_dt)\: \xi\omega_n \: e^{-\xi\omega_nt}\right] + B\left[-e^{-\xi\omega_nt}\omega_d\sin(\omega_dt) - \cos(\omega_dt)\:\xi\omega_n\:e^{-\xi\omega_nt} \right]
\tag{8}
$$

Now we can substitute our initial conditions into the equations for displacement and velocity and solve the two simultaneous equations to yield,

$$
\begin{align}
A &= 0.00014\\
B &= 0.007
\end{align}
\tag{9}
$$

Therefore the free vibration response is given by,

$$
u(t) = e^{-\xi\omega_n t}[0.00014\: \sin(\omega_d t) + 0.007\:\cos(\omega_d t)]
\tag{10}
$$

We can now plot this response in the usual way. We also plot on a vertical and horizontal line to indicate the point at which we predicted the amplitude would fall below $0.001\:m$.

```python
tmax = 60 #(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

#The damped oscillation
u = math.e**(-xi*omega_n*t)*(0.00014*np.sin(omega_d*t) + 0.007 *np.cos(omega_d*t))

#Plotting
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,1.5])
axes.plot(t,u,label='Damped oscillation')
axes.plot([0, tmax],[u_finish,u_finish],'r--',lw=0.75)
axes.plot([t_full, t_full],[u_start,-u_start],'r--',lw=0.75)
axes.set_xlim([0,tmax])
axes.set_xlabel('Time (sec)')
axes.set_ylabel('Displacement (m)')
axes.set_title('Displacement-time history')
axes.grid()
axes.legend()
plt.show()
```

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-2/img2.jpg"
	alt="Damped free vibration response indicating the time at which the response amplitude reduces to 0.001 m | EngineeringSkills.com"
	width="1000"
	height="523"
/>

<Caption>
	Fig 2. Damped free vibration response indicating the time at which the response
	amplitude reduces to $0.001\:m$.
</Caption>

This concludes our analysis of undamped free vibration. In the next section we'll start to examine the response of SDoF systems to harmonic excitation.


Please log in or enroll to continue

Please log in or enroll to continue