Response to Step & Pulse force

In regard to this topic, we are interested in the reaction solution and maximum reaction like other studies. In addition, we still use the zero initial condition for convenience.

1.Step force

1.1Response Solutions

EOM:
\[
m\ddot{u}+c\dot{u}+ku=p(t)=p_0
\]

1.1.1 Classic Method

General solution EOV:
\[
u(t) = e^{-\zeta w_nt}(Acosw_Dt+Bsinw_Dt)+\frac{p_0}{k}
\]
Substitute \(u(0)=\dot{u}(0)=0\) to simplify the EOV:
\[
u(t) = (u_{st})_0[1-e^{-\zeta w_nt}(cosw_Dt+\frac{\zeta}{\sqrt{1-\zeta ^2}}sinw_Dt)]
\]
Where \( (u_{st})_0 = \frac{p_0}{k} \) is the static deformation caused by \(p_0\)

1.1.2Duhamel Integral Method

Since we have already derived the Duhamel integral which could be used to express the any arbitrary forces, we can also use it to derive the EOV of step excitation.

The Response(EOV) expresses in Duhamel integral:
\[
u(t) = \frac{1}{mw_0}\int_0^t p_0 e^{-\zeta w_n(t-\tau)}sin[w_D(t-\tau)]d\tau
\]
Integration table:
\[
\int e^{\alpha x}sin(\beta x)dx = \frac{e^{\alpha x}[\alpha sin(\beta x) - \beta cos(\beta x)]}{\alpha^2 +\beta ^2}+C
\]
Now:
\[
\alpha = -\zeta w_n; \beta = w_D = W_n\sqrt{1-\zeta ^2}
\\ \alpha ^2 +\beta ^2 = w_n^2
\]
Hence:
\[
u(t) = (u_{st})_0[1-e^{-\zeta w_nt}(cosw_Dt+\frac{\zeta}{\sqrt{1-\zeta ^2}}sinw_Dt)]
\]

1.2Observations

The max. response occurs when \(\dot{u}(t) = 0\).

1.2.1For Undamped System

\[
\dot{u}(0) = 0
\\ \Rightarrow w_nt_0 = j\pi
\\ \Rightarrow t_0 = \frac{j}{2}T_n…j=1,3,5,…,2n+1
\]

Substitute it into the EOV:
\[
u_0 = 2(u_{st})_0
\]
So, we find an interesting conclusion:

A sudden application of a force causes twice as much deformation as a slow application of a force. 所以“趁它不注意，你会引起两倍于原来的位移。”

1.2.2For Damped System

\[
u_0 < 2(u_{st})_0
\\ u_0 \rightarrow (u_{st})_0 \Rightarrow \frac{u_0}{(u_{st})_0} \rightarrow 1
\]

As for the response to damped system, the conclusion is similar to the response of harmonic vibration with damping ratio. The response is tending to move around the steady state. And the damping ratio determines the overshoot magnitude and decay rate beyond the equilibrium position.

2.Response to Step Force (with finite rise time)

2.1Response Solutions

EOM:

When \( t<t_r\):
$$
p(t) = p_0(\frac{t}{t_r})
\\ u(t) = (u_{st})_0(\frac{t}{t_r}-\frac{sinw_nt}{w_nt_r})
$$

When \(t≥t_r\):

\[p(t) = p_0 \]

\[\begin{align}
u(t)& = u(t_r)cosw_n(t-t_r) + \frac{\dot{u}(t_r)}{w_n}sinw_n(t-t_r)…①
\\ & + (u_{st})_0[1-cosw_n(t-t_r)]…②
\\ &= (u_{st})_0 { 1 + \frac{1}{w_nt_r}[(1-cosw_nt_r)sinw_n(t-t_r)-sinw_nt_rcosw_n(t-t_r)}
\\ &= (u_{st})_0 {1-\frac{1}{w_nt_r}[sinw_nt-sinw_n(t-t_r)]}
\end{align}\]

2.2Observations

The response spectrum shows maximum deformation at different \(t_d/T_n\).

□System oscillates at the natural period \(T_n\) during both phases;

□If \(t_r/T_n = 1,2,3…,\) no vibration during the constant force phase (the velocity is zero at the end of the ramp);

□For short time rise (\(t_r/T_n<<1\)), the response is similar to a sudden step force;

□If \(t_r/T_n>>1\), the dynamic effect is small;

□If \(t_r<T_n/4\), \(R_d≈2\), similar to suddenly applied load;

□If \(t_r>3T_n\), \(R_d≈1\), similar to static load;

□If \(t_r/T_n=1,2,3…\), velocity is zero at the end of the force-rise phase, no oscillation and \(R_d=1\).

3.Response to Rectangular Pulse Force

3.1Response Solutions

$$
u(t)=\frac{p_0}{k}\left[1-\cos \omega_n t\right], t<t_d\
u(t)=\frac{p_0}{k}\left[\cos \omega_n\left(t-t_d\right)-\cos \omega_n t\right], t \geq t_d
$$

3.2Observations

□Response depends only on \(t_d/T_n\);

□Nature of response can vary greatly with \(t_d\);

□Oscillates freely about the two equilibrium positions;

□No decay of motion (undamped);

□If \(t_d/T_n = 1,2,3…\), the system stops moving during the free vibration phase due to zero initial displacement and velocity at the beginning of the free vibration phase.

□If \(t_d≥T_n/2\), a least on peak during the forced vibration phase, i.e. \(R_d=2\);

□If \(t_d<T_n/2\), the peak occurs at the end of the forced vibration phase, \(R_d = 1-cos(2\pi t_d/T_n)\).

4.Response to Half-Cycle Sine Pulse Force

4.1 Response solutions

$$
p(t) = \begin{cases}
p_osin(\frac{\pi t}{t_d}) &t<t_d
\\ 0 & t≥t_d
\end{cases}
$$

When \(t_d/T_n≠1/2\):
$$
u(t)=\frac{p_0}{k} \frac{1}{1-\left(T_n / 2 t_d\right)^2}\left[\sin \left(\frac{\pi t}{t_d}\right)-\frac{T_n}{2 t_d} \sin \left(\frac{2 \pi t}{T_n}\right)\right],\ t<t_d(Forced\ phase)\
u(t)=\frac{p_0}{k} \frac{\left(T_n / t_d\right) \cos \left(\pi t_d / T_n\right)}{\left(T_n / 2 t_d\right)^2-1} \sin \left[2 \pi\left(\frac{t}{T_n}-\frac{t_d}{2 T_n}\right)\right],\ t≥t_d(Free\ phase)
$$
When \(t_d/T_n=1/2\):
$$
u(t)=\frac{p_0}{k} \frac{1}{2}\left[\sin \left(\frac{2 \pi t}{T_n}\right)-\frac{2 \pi t}{T_n} \cos \left(\frac{2 \pi t}{T_n}\right)\right],\ t<t_d(Forced\ phase)\
u(t)=u\left(t_d\right) \cos \omega_n\left(t-t_d\right)+\frac{\dot{u}\left(t_d\right)}{\omega_n} \sin \omega_n\left(t-t_d\right),\ t≥t_d(Free\ phase)
$$

4.2Observations

□Response depends only on \(t_d/T_n\);

□Nature of response can vary greatly with \(t_d\);

□No decay of motion (undamped);

□Dynamic effects are small when \(t_d/T_n>3\) (slow varying force);

□If \(t_d/T_n = 1.5, 2.5…,\) the system stops moving during the free vibration phase due to zero initial disp. and velocity at the beginning of the free vibration phase.

5.Response to Symm. Triangular Pulse Force

5.1Response Solutions

\[
u(t)=\frac{p_0}{k} 2(\frac{t}{t_d}-\frac{T_n}{2 \pi t_d} \sin (2 \pi \frac{t}{T_n})) , 0 \leq t \leq \frac{t_d}{2}
\]

\[
u(t)=\frac{p_0}{k} 2\{1-\frac{t}{t_d}+\frac{T_n}{2 \pi t_d}(2 \sin (\frac{2 \pi}{T_n}(t-\frac{t_d}{2}))-\sin (2 \pi \frac{t}{T_n}))\} ,\frac{t_d}{2} \leq t \leq t_d
\]

\[
u(t)=\frac{p_0}{k}2\{\frac{T_n}{2\pi t_d}(2sin(\frac{2\pi}{T_n}(t-\frac{t_d}{2}))-sin(\frac{2\pi}{T_n}(t-t_d))-sin(\frac{2\pi}{T_n}) )\} ,t>t_d
\]

5.2Observations

The spectrum is quite similar to that of half-cycle sine pulse force. The only difference is that if \(t_d/T_n = 2,3…,\) the system stops moving during the free vibration phase due to zero initial disp. and velocity at the beginning of the free vibration phase.