Open Educational Resources

Review of single and multi-degree of freedom (mdof) systems: Transient Vibrations

For dynamical systems excited by non-periodic forces, displacements, accelerations, etc., the response over the initial few periods of the system is termed transient response. This is important for many diverse situations including packaging of equipment, protecting people from automobile impacts, and loading of structures due to blasts and earthquakes. The modeling of these situations is often difficult and may require specific detail of the excitation as the response can vary dramatically depending on the detail of the input.

For a SDOF system the solution requires the general solution of m\ddot{x} + c\dot{x} + kx = F(t) and becomes difficult for complicated F(t)

Example

Consider first the undamped case for a step input:

Solution

    \[ m\ddot{x} +kx = F_0, t>0\]

    \[ \implies x_H = A\sin{pt} + B\cos{pt} \]

Assume x_p = C

    \[ \implies kC = F_0 \]

    \[C = \frac{F_0}{k} \]

and the total solution:

    \[\begin{split} x(t) &= x_H +x_p \\&= \frac{F_0}{k} + A\sin{pt} + B \cos{pt} \end{split} \]

If t=0, x(0) = 0, \dot{x}(0) = 0, then:

    \[  B = -\frac{F_0}{k} \]

    \[A = 0 \]

    \[x(t) = \frac{F_0}{k}(1-\cos{pt}) \]

While this case is straightforward, the solution is more difficult with damping or for different forcing functions.

Example

The response can be determined in a general form by considering F(t) as being made up from a series of short impulses. A general short impulse at some time t' is shown. Consider what happens(response) if only this impulse is applied to the system:

If \Delta t is vanishingly small the impulse can be considered as changing the velocity of m using impulses/momentum relations so that from initial (i) (just before) to final(F) condition:

    \[(m\dot{x})_i + F(t')\Delta t = (m\dot{x})_F \]

Therefore:

    \[ \dot{x}_F - \dot{x}_i = \frac{F_0(t')}{m}\Delta t \]

This impulses will result in a free vibration with initial conditions:

    \[ x(0) = 0 \]

    \[\dot{x}(0) = \frac{F(t')}{m} \Delta t\]

If this is the only impulse then the response is:

    \[ x(t) = e\strut^{-\zeta pt}\big[A\cos{\sqrt{1-\zeta^2}pt} + B\sin{\sqrt{1-\zeta^2}pt}\big] \]

with x(0) = 0, \dot{x}()0) = \frac{F(t')}{m}\Delta t

    \[ \implies A=0 \]

    \[ \begin{split}\dot{x}(0) &= B\sqrt{1-\zeta^2}p \\&= \frac{F(t')}{m}\Delta t \end{split}\]

    \[ \implies B = \frac{F(t')\Delta t}{mp\sqrt{1=\zeta^2}} \]

Therefore:

    \[x(t) = \frac{F(t') \Delta t}{mp\sqrt{1-\zeta^2}}e^{-\zeta pt}\sin{\big(\sqrt{1-\zeta^2}pt\big)} \]

Where t is the time since the impulse.

Therefore the response at t due to the impulse at t' is x'(t):

    \[x'(t) = \frac{F(t')\Delta t'e^{-\zeta p(t-t')}}{mp\sqrt{1-\zeta^2}}\sin{\big[\sqrt{1-\zeta^2}p(t-t')\big]}\]

The response due to all the impulses from t = 0 to t = t is the superposition of all the impulses’ response.

Therefore:

    \[x(t) = \frac{1}{mp\sqrt{1-\zeta^2}} \int _0^\tau F(t')e^{-\zeta p(t-t')}\sin{\sqrt{1-\zeta^2}p(t-t')}\,dt'\]

Assuming x(0) = \dot{x}(0) = 0 i.e. the mass was stationary prior to F(t) being applied.

For the step input F_0

    \[x(t) = \frac{F_0}{k}\biggr[1-\frac{e^{-\zeta pt}}{\sqrt{1-\zeta^2}}\cos{\big(\sqrt{1-\zeta^2}pt - \Psi\big)}\biggr]\]

    \[ \tan{\psi} = \frac{\zeta}{\sqrt{1=\zeta^2}}\]

If the initial conditions are x(0) = x_0, \dot{x}(0) = v_0, then:

    \[\begin{split}x(t) &= e\strut^{-\zeta pt}\biggr[x_0\cos{\sqrt{1-\zeta^2}}pt+\frac{V_0 + \zeta px_0}{\sqrt{1-\zeta^2}p}\sin{\sqrt{1-\zeta^2}pt}\biggr] \\&+\frac{1}{mp\sqrt{1-\zeta^2}} \int _0^\tau F(t')e^{\zeta p(t-t')}\sin{\sqrt{1-\zeta^2}p(t-t')}\,dt'\end{split} \]

Example

For a package being dropped or for an airplane landing, a SDOF model can be utilized.

For the undamped case, as it strikes, x(0) = 0 and \dot{x}(0) = \sqrt{2gH}, and the forcing function is mg (constant). NOTE: x is not measured from the static equilibrium configuration. Using the general result:

    \[ \begin{split} x(t) &= \frac{\sqrt{2gH}}{p}\sin pt +\frac{1}{mp} \int_{0}^{t} mg\sin (t-t') dt' \\&= \frac{\sqrt{2gH}}{p}\sin pt + \frac{g}{p} \Big[\frac{\cos p(t-t')}{p}\Big]_{o}^{t} \\&= \frac{\sqrt{2gH}}{p}\sin pt + \frac{g}{p^2}\Big[1 - \cos pt \Big] \end{split} \]

    \[x(t) = \sqrt{\frac{2gH}{p^2} + \Big(\frac{g}{p^2}\Big)^2} \sin(pt + \phi) + \frac{g}{p^2}\]

    \[\ddot{x}(t) = -p^2\sqrt{\frac{2gH}{p^2} + \Big(\frac{g}{p^2}\Big)^2} \sin(pt + \phi) \]

Therefore, the maximum acceleration is:

    \[ \frac{\ddot{x}_\text{MAX}}{g} = -\frac{1}{\delta_\text{ST}}\sqrt{2H\delta_\text{ST} + \delta_\text{ST}^2} \]

    \[ \frac{\ddot{x}_\text{MAX}}{g} = -\sqrt{\frac{2H}{\delta_\text{ST}} + 1} \]

With:

    \[\begin{split} \delta_\text{ST} &= \frac{mg}{k} \\ &= \frac{p}{g^2} \end{split}\]

In forensic studies of automobile collisions, the concern is the acceleration of a passenger’s head and the potential for internal damage. Assume the body/skull is modelled as a SDOF system.

Biomechanical studies gives the spinal stiffness k = 81000 \ \frac{\text{N}}{\text{m}} \ \text{or} \ 458 \ \frac{\text{lb}}{\text{in}}. Assume the person weighs 160 lbs and falls 6 inches because of being unrestrained. Thus:

    \[\begin{split} \delta_\text{ST} &= \frac{160}{458} \\&= 0.35 \ \text{inches} \end{split} \]

Therefore:

    \[ \begin{split} \frac{\ddot{x}_\text{MAX}}{g} &= -\sqrt{\frac{12}{0.35} + 1} \\&= -5.94 \end{split} \]

If we add a seat cushion to the seat (cushion stiffness is k_c = 51 \frac{\text{lb}}{\text{in}}). What is the change in maximum acceleration?

The effective stiffness is now:

    \[\frac{1}{k_\text{eff}} = \frac{1}{458} + \frac{1}{51}\]

    \[\begin{split} k_\text{eff} &= \frac{(51)(458)}{458 + 51} \\& = \ 45.9 \text{ lbs/in} \end{split}\]

Thus:

    \[\begin{split} \delta_\text{ST} &= \frac{160}{45.9} \\&= 3.49 \ \text{in} \end{split}\]

Therefore:

    \[ \begin{split} \frac{\ddot{x}_\text{MAX}}{g} &= -\sqrt{\frac{12}{3.49} + 1} \\&= -2.1 \end{split} \]

The cushion reduces the acceleration drastically.

For many situations, the support of the system is subjected to the motion specified by its displacement, velocity, or acceleration.

Therefore, for base acceleration:

    \[ m\ddot{x}+ k(x-y) + c(\dot{x} - \dot{y}) = 0 \]

    \[m(\ddot{x} - \ddot{y}) + k(x-y) + c(\dot{x} - \dot{y}) = -m\ddot{y}\]

Using the relative displacement z = x - y:

    \[m\ddot{z} + c\dot{z} + kz = -m\ddot{y} \]

The convolution integral for this case is:

    \[ z(t) = -\frac{1}{\sqrt{1-\zeta^2}p}\int_0^t \ddot{y}(t') e^{\zeta(t-t')}\sin\sqrt{1-\zeta^2}p(t-t') dt' \]

The same approach can be used for a base excitation with velocity.

Examples of Dynamic Responses of Undamped SDF Systems to Different Pulse Forces

Note that the static solution is shown in the dashed lines.

Rectangular Pulse Force
Half Cycle Sine Pulse Force
Triangular Pulse Force

Shock Response Spectrum

Previously we developed a technique to find the response of a damped spring mass system to an arbitrary excitation. When the duration of the pulse is small compared to the natural period (\tau=\frac{2\pi}{P}), the excitation is called a shock. It has been found that the concept of the shock response spectrum is useful for these situations.

The Shock Response Spectrum (SRS) is a plot of the maximum peak response of a SDOF oscillator as a function of the natural period of the oscillator. The maximum of the peaks (called maxima) represents only a single point on the time response curve.

To illustrate the SRS concept consider a SDOF undamped system with no initial motion (\dot{x}(0)=x(0)=0). The SRS is

    \[ x(t)\bigg|_\text{MAX}=\bigg|\frac{1}{mP}\int_0^t F(t')\sin p(t-t')dt'\bigg|_\text{MAX} \]

If F(t) is a rectangular pulse of length t^*

then during the pulse

    \[ x(t)=\frac{F_0}{k}(1-\cos pt) \]

where t<t^* and after

    \[ x(t) = \frac{F_0}{k}\big[\cos p(t-t^*)-\cos pt\big] \]

where t>t^*. Often the results are determined for the two sections (during & after) individually then combined as the primary response (during) and the residual (after pulse) response.

During the pulse

    \[ x(t)=\frac{F_0}{k}(1-\cos pt) \]

Where \tau = \frac{2\pi}{p} for a maximum:

    \[ \frac{d}{dt}\Big(\frac{xk}{F_0}\Big)=0=p\sin pt' \]

therefore:

    \[ pt'=\pi, 2\pi, … \]

    \[ t'=\frac{\pi}{p}, \frac{\pi}{p}… \]

therefore up to t=\frac{\pi}{p}, the maximum is the value at the time considered as it is still increasing therefore:

    \[ x_\text{MAX} = \frac{F_0}{k}(1-\cos pt) \]

where t^*<\frac{\pi}{p} after t^*=\frac{\pi}{p} the maximum is:

    \[ \begin {split} x_\text{MAX} &=\frac{F_0}{k}\Big(1-\cos\Big(\frac{p\pi}{p}\Big)\Big) \\&=2\frac{F_0}{k} \end{split} \]

These results are called the primary values as they correspond to the result during the pulse.

after t^*

    \[ \frac{kx(t)}{F_0}=\big[\cos p(t-t^*)-\cos pt\big] \]

where t>t^*

and we find the maxima by differentiating

    \[ \frac{d}{dt}\Big(\frac{x(t)k}{F_0}\Big)=p\big[\sin p(t_1^*-t^*)-\sin pt_1^*\big]=0 \]

at t_1

    \[ 0 = p\big[\sin pt_1 \cos pt^* - \cos pt_1 \sin pt^* -\sin pt_1\big] \]

therefore:

    \[ \begin{split} \sin pt_1(1-\cos pt^*) &= -\sin pt^* \cos pt_1 \\ \tan pt_1 &= \frac{-\sin pt^*}{(1-\cos pt^*)} \end{split} \]

This value for t_1 must be inserted into the original expression to find the maximum displacement. This will require \cos pt_1 and \sin pt_1 in terms of \sin pt^* and \cos pt^*.

therefore consider the triangle

[pt_1 > \pi as tan is negative], sin is negative

therefore:

    \[ \sin pt_1 = \frac{\sin pt^*}{\sqrt{2(1-\cos pt^*)}} \]

    \[\begin{split} \cos pt_1 &= \frac{-(1-\cos pt^*)}{\sqrt{2(1-\cos pt^*)}} \\ &= -\frac{1}{\sqrt{2}} \sqrt{1-\cos pt^*} \end{split} \]

now

    \[ \begin{split} \frac{x(t_1)k}{F_0}\bigg|_{\text{MAX}} &= \cos pt^* \cos pt_1 + \sin pt^* \sin pt_1 - \cos pt_1 \\ &= -\frac{1}{\sqrt{2}}\sqrt{1-\cos pt^*}[-1+ \cos pt^*] + \frac{\sin^*2pt^*}{\sqrt{2(1-\cos pt^*}} \\ &=\frac{\frac{\sqrt{2}}{\sqrt{2}}(1-\cos pt^*)^2+\sin^*2pt^*}{\sqrt{2(1-\cos pt^*)}} \\ &= \frac{2(1-\cos pt^*)}{\sqrt{2(1-\cos pt^*)}} \\ &=\sqrt{2(1-\cos pt^*)} \\ &= 2\sin p\frac{t^*}{2} \end{split} \]

but \tau = \frac{2\pi}{p} therefore:

    \[ \frac{x(t)k}{F_0}\bigg|_{\text{MAX}} = 2\sin \frac{\pi t^*}{\tau} \]

Residual Response

The total SRS can now be determined

If damping is added then the magnitude can be reduced considerably. However, the calculation of the SRS must be done numerically

For cases which the pulses variation is less than one-half the period (< \frac{\tau}{2}), the overall maximum occurs after the pulse in the “free” vibration phase. As the pulse duration compared to the period becomes smaller we can approximate the response where it is the magnitude of the impulse(not the details of the shape) that matters. In the development of the theory, the response for a damped system was:

    \[x(t) = \frac{F(t^{'}) \Delta t^{'}}{mp \sqrt{1- \zeta ^{2}}} e^{-\zeta pt} \sin\big(\sqrt{1-\zeta ^{2}}pt\big)\]

Where t is the time since the impulse and for for the undamped situation

    \[x(t) = \frac{F \Delta t}{mp} \sin pt\]

Where F \Delta t is the impulse, I , and the maximum displacement is:

    \[\begin{split} x_\text{MAX} &= \frac{I}{mp} \\&= \frac{I}{k} \frac{k}{mp} \\&= \frac{Ip}{k} \\&= \frac{I}{k}\frac{2 \pi}{\tau}  \end{split}\]

Consider three different impulses:

1) rectangular 2) half sine 3) triangular

1)

    \[I = F_o t^{\star}\]

Therefore, x_\text{MAX} = \frac{F_o}{k} 2\pi \frac{t^{\star}}{\tau} compared to the static response due to F_o (x_\text{STATIC} = \frac{F_o}{k})

    \[\frac{x_\text{MAX}}{x_\text{STATIC}} = \frac{2\pi t^{\star}}{\tau}\]

2)

    \[\begin{split} I &= F_o \int_{0}^{t^{\star}} \sin \frac{\pi t}{t^{\star}} \,dt \\&= -F_o \frac{t^{\star}}{\pi} \cos\frac{\pi t}{t^{\star}}  \Big|_ 0^ {t^{\star}} \\&= \frac{2F_o t^{\star}}{\pi} \end{split}\]

Therefore,

    \[\begin{split} x_\text{MAX} &= \frac{2\pi}{\tau} \frac{2F_o t^{\star}}{\pi} \\&= \frac{4 t^{\star}}{\tau} (\frac{F_o}{k}) \end{split}\]

3)

    \[I = \frac{F_o t^{\star}}{2}\]

    \[x_\text{MAX} = \frac{\pi t^{\star}}{\tau} (\frac{F_o}{k})\]

If the effect of damping is included then:

    \[x_\text{MAX}(t) = \Big[\frac{I}{mp \sqrt{1-\zeta ^{2}}} e\strut^{-\zeta pt} \sin\big(\sqrt{1-\zeta ^{2}}pt\big)\Big]_\text{MAX} \]

and for \zeta ^{2} << 1 we assume the maximum occurs at (or near) where \sin(\sqrt{1-\zeta ^{2}}pt) is 1.

Therefore \hat{t} (time at t_\text{MAX}) is \frac{1}{4} of a so-called damped period:

    \[\begin{split} \hat{t} &= \frac{\pi}{2p \sqrt{1-\zeta ^{2}}} \\& \approx \frac{\pi}{2p} \end{split}\]

Therefore:

    \[\begin{split} x_\text{MAX} &= \frac{I}{mp} e\strut^{-\zeta p \hat{t}} \\&= \frac{I}{mp} e\strut^{-\zeta \frac{\pi}{2}} \end{split}\]

and for \zeta ^{2} << 1,  e^{-\zeta \frac{\pi}{2}} \approx (1- \frac{\zeta \pi}{2}):

    \[x_\text{MAX} = \frac{I}{k} \frac{2\pi}{\tau} (1- \frac{\zeta \pi}{2})\]

and we can multiply the undamped values by (1- \frac{\zeta \pi}{2}):

\zeta0.050.100.150.200.250.3
1- \frac{\zeta \pi}{2}0.9210.8430.7640.6860.6070.529

Leave a Reply

Your email address will not be published. Required fields are marked *