
In this lecture, we'll work through the calculation of the complete (transient and steady-state) response for the first $100\:s$. Recall from an earlier lecture that we said the complete response of the system to harmonic loading consisted of a transient component and a steady-state component. The complete equation was,

$$
u(t) = \underbrace{e^{-\xi\omega_nt}[A\sin(\omega_dt) + B\cos(\omega_dt)]}_{\text{Transient}} + \underbrace{\frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right][(1-\beta^2)\sin(\omega t) - 2\xi\beta\cos(\omega t)]}_{\text{Steady-state}}
\tag{1}
$$

In order to use this equation we must first determine the constants $A$ and $B$ within the transient component. We do this by applying the initial conditions, in this case at $t=0$, the mass is at rest and therefore the position, $u(t=0)$ and velocity, $\dot{u}(t=0)$ are zero.

To apply the second initial condition on velocity we must differentiate our expression for $u(t)$. First we will make some substitutions to make the equation easier to digest, let,

$$
\Omega = \frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right]
\tag{2}
$$

$$
C_1 = \Omega\:(1-\beta^2)
\tag{2}
$$

$$
C_2 = -\Omega\:2\:\xi\:\beta
\tag{3}
$$

This allows us to rewrite the equation for $u(t)$ as,

$$
u(t) = A\underbrace{\left[e^{-\xi\omega_nt}\:\sin(\omega_d t) \right]}_{f_1(t)} + B\underbrace{\left[e^{-\xi\omega_nt}\:\cos(\omega_d t) \right]}_{f_2(t)} + C_1\:\underbrace{\sin(\omega t)}_{f_3(t)} + C_2\:\underbrace{\cos(\omega t)}_{f_4(t)}
\tag{4}
$$

Now each of the four functions of $t$ can be differentiated in turn. Both $f_1(t)$ and $f_2(t)$ can be processed using the product rule while $f_3(t)$ and $f_4(t)$ and be processed using the chain rule. Evaluating each in turn,

$$
\frac{\mathrm{d}}{\mathrm{d}t}f_1(t) = e^{-\xi\omega_nt}\omega_d\cos(\omega_dt) - \sin(\omega_dt)\: \xi\omega_n \: e^{-\xi\omega_nt}
\tag{5}
$$

$$
\frac{\mathrm{d}}{\mathrm{d}t}f_2(t) = -e^{-\xi\omega_nt}\omega_d\sin(\omega_dt) - \cos(\omega_dt)\:\xi\omega_n\:e^{-\xi\omega_nt}
\tag{6}
$$

$$
\frac{\mathrm{d}}{\mathrm{d}t}f_3(t) = \omega\cos(\omega t)
\tag{7}
$$

$$
\frac{\mathrm{d}}{\mathrm{d}t}f_4(t) = -\omega\sin(\omega t)
\tag{8}
$$

Combining these four expressions we obtain our expression for the velocity, $\dot{u}(t)$,

$$
\begin{split}
\dot{u}(t) =  & \:A\left[e^{-\xi\omega_nt}\omega_d\cos(\omega_dt) - \sin(\omega_dt)\: \xi\omega_n \: e^{-\xi\omega_nt} \right] + \\  & \:B\left[ -e^{-\xi\omega_nt}\omega_d\sin(\omega_dt) - \cos(\omega_dt)\:\xi\omega_n\:e^{-\xi\omega_nt}\right] + \\ &\: C_1\:\omega\cos(\omega t) - C_2\: \:\omega\sin(\omega t)
\end{split}
\tag{9}
$$

Expressions for $u(t)$ and $\dot{u}(t)$ can now be evaluated at $t=0$. Helpfully many of the terms disappear due to the fact $t=0$. Evaluating the two equations for the two unknowns, $A$ and $B$ yields,

$$
A = -\frac{1}{\omega_d}(C_1\:\omega + C_2\:\xi\:\omega_n)
\tag{10}
$$

$$
B=-C_2
\tag{11}
$$

Now that we know the constants $A$ and $B$, we're in a position to plot the response. We start by defining the constants derived above.

```python
#New constants from this lecture
O = (P/k)*(1/((1-beta**2)**2 + (2*xi*beta)**2)) #Capital Omega defined above
C1 = O*(1-beta**2) #Defined above
C2 = -O*2*xi*beta #Defined above
omega_d = omega_n*math.sqrt(1-xi**2) #Damped natural frequency
A = (-1/omega_d)*(C2*xi*omega_n + C1*omega) #Constant of integration
B = -C2 #Constant of integration
```

Next we define a time vector for the first $60$ seconds.

```python
tmax = 60 #(sec) The max time
delta_t = 0.001 #(sec) The time step
nPoints = tmax/delta_t #Number of equally spaced data points
t = np.linspace(0,tmax, int(nPoints)) # Time vector
```

Now we can generate the response vector. We're dividing up the calculation and computing the transient and steady-state responses separately, partly because we want to visualise them separately and partly because it's a very long equation to code out without mistakes.

```python
ut = math.e**(-xi*omega_n*t)*(A*np.sin(omega_d*t) + B*np.cos(omega_d*t)) #Transient response
us = O*((1-beta**2)*np.sin(omega*t) - 2*xi*beta*np.cos(omega*t)) #Steady-state response
```

Next we can plot the steady-state and transient response, separately at first.

```python
#Plotting (0-60)
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,2])
axes.plot(t,ut,'-',label='Transient response')
axes.plot(t,us, '-', label='Steady-state response')
axes.set_xlim([0,tmax])
axes.set_xlabel('Time (sec)')
axes.set_ylabel('Displacement (m)')
axes.set_title('Components of displacement-time history')
axes.grid()
axes.legend(loc='lower right')
plt.show()
```

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-3-part-b/img1.jpg"
	alt="Transient and steady-state components of displacement-time history. | EngineeringSkills.com"
	width="1000"
	height="676"
	caption="Fig 1. Transient and steady-state components of displacement-time history."
/>

We can clearly see that after about $50$ seconds, the transient response has decayed to a negligible amplitude due to damping and we're left with the steady-state response alone. Of course the actual response of the system is the combination of both of these waveforms. We can visualise that next.

```python
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,2])
axes.plot(t,(ut+us), label='Combined response')
axes.set_xlim([0,tmax])
axes.set_xlabel('Time (sec)')
axes.set_ylabel('Displacement (m)')
axes.set_title('Combined displacement-time history')
axes.grid()
axes.legend(loc='lower right')
plt.show()
```

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-3-part-b/img2.jpg"
	alt="Combined displacement-time history. | EngineeringSkills.com"
	width="1000"
	height="676"
	caption="Fig 2. Combined displacement-time history."
/>

Again, the decaying influence of the transient response is clearly visible. We can see that after the transient response decays away, the system behaviour is well described by the steady-state response. It is for this reason that we often ignore the transient response.

Before we conclude this part of the solution, let's plot the static deflection and the steady-state amplitude predicted by the dynamic magnification factor. First we need to restate the DMF from the previous lecture.

```python
DMF = 1/math.sqrt((1-beta**2)**2 + (2*xi*beta)**2) #Dynamic Magnification Factor
```

```python
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,2])
axes.plot([0,tmax],[P/k,P/k],'g', label='Static displacement')
axes.plot([0,tmax],[DMF*P/k,DMF*P/k],'r', label='Steady-state amplitude')
axes.plot(t,us, '-', label='Steady-state response')
axes.set_xlim([0,tmax])
axes.set_xlabel('Time (sec)')
axes.set_ylabel('Displacement (m)')
axes.set_title('Steady-state displacement-time history')
axes.grid()
axes.legend(loc='lower right')
plt.show()
```

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-3-part-b/img3.jpg"
	alt="Steady-state displacement-time history. | EngineeringSkills.com"
	width="1000"
	height="676"
	caption="Fig 3. Steady-state displacement-time history."
/>

We can clearly see that the DMF correctly predicts the amplitude of the steady-state response. There is no real surprise here, but this is a nice confirmation and underlines the utility of the DMF as a quick way of assessing the response to harmonic loading. In the next lecture we'll finish out the solution by considering the free vibration response.


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