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

Introduction and Review

We are interested in small motions about some equilibrium positions. For SDOF systems, x is usually measured from this static equilibrium position.

    \[\begin{split}+\downarrow \sum{F_x} &= m \ddot{x} \\&= -k(x+\delta_\text{static})-c\dot{x} + F(t)+mg\end{split}\]

Let \dot{x}, \ddot{x} = 0 for static equilibrium. Therefore:

    \[-k\delta_{ST.}+mg = 0\]

Let \delta_\text{static} = \frac{mg}{k}. Therefore:

    \[m\ddot{x}+c\dot{x} +kx = F(t)\]

All SDOF systems can be reduced to this form (assuming viscous damping):

    \[m_e\ddot{x}+c_e\dot{x} +k_ex = F(t) \quad\]

Where e means equivalent. To determine the nature of the motion, first consider this linearized equation of free vibration (Note: May have to linearize).

Assume x_H = Ae^{st} (homogeneous), where A is constant and s is to be determined.

    \[m_es^2 + c_es+k_e = 0\]

    \[\begin{split}s_{1,2} &= -\frac{c_e}{2m_e} \pm \sqrt{\Big(\frac{c_e}{2m_e}\Big)^2-\frac{k_e}{m_e}}\\&= -\frac{\tilde{c}}{2} \pm \sqrt{\Big(\frac{\tilde{c}}{2}\Big)^2-\tilde{k}}\end{split}\]

If s_{1,2} are real and negative, then x\rightarrow 0 as t \rightarrow\infty. If either of s_1 or s_2 are real and positive, then the solution increases with time.

If s_{1,2} are complex, then they are complex conjugates and the nature of the motion depends on the real part. If the real part is negative, x\rightarrow 0 as t\rightarrow \infty. If positive, x\rightarrow \infty as t \rightarrow \infty. If the real part is zero, the motion is oscillatory and this is the borderline case between stability and instability.

Consider all cases:

  1. \tilde{c} > 0 \quad \tilde{k} > 0. The roots have a negative real part and is stable regardless of \tilde{k}. If |\tilde{k}| is larger than |\frac{\tilde{c}}{2}|^2 , there will be an oscillatory part, but x\rightarrow0 as t \rightarrow \infty. This is called asymptotically stable.
  2. \tilde{c} = 0 \quad \tilde{k}>0. The roots are imaginary, and the system is borderline stable.
  3. \tilde{k}<0 \quad\tilde{c}\le 0. The roots have a positive real part, or are complex conjugates. Therefore, the system is unstable.
  4. \tilde{k}<0 \quad\tilde{c}>0 One root has a positive real part. Therefore the system is unstable. The damping cannot overcome the negative spring.
  5. \tilde{k}>0 \quad\tilde{c}<0. The roots have a positive real part, and therefore the system is unstable. Negative damping feeds energy into the system.

Consider the solution for free damped vibration:

    \[m_e \ddot{x}+c_e\dot{x} +k_e x=0\]

Drop e, but remember that m,c,k are equivalent. Thus:

    \[\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x = 0\]

Again, assume that x=Ae^{st}. Thus:

    \[s_{1,2}= \frac{-c}{2m}\pm\sqrt{\Big(\frac{c}{2m}\Big)^2-\frac{k}{m}}\]

The critical value is then when:

    \[\Big(\frac{c_c}{2m}\Big)^2=\frac{k}{m} := p^2\]

Alternatively:

    \[c_c = 2mp\]

Define \zeta := \frac{c}{c_c}. Then:

    \[\frac{c}{2m} = \frac{c}{c_c}\frac{2mp}{2m} = \zeta p\]

Thus:

    \[s_{1,2} = \Big(-\zeta \pm \sqrt{\zeta ^2-1}\Big)p\]

Where \zeta is the damping ratio (system parameter).

Assuming 0<\zeta<1:

    \[x(t) = e^{-\zeta pt}\Big[C_1e^{i\sqrt{1-\zeta^2}}pt+C_2e^{-i\sqrt{1-\zeta^2}}pt\Big]\]

And using e^{\pm i\theta}=\cos{\theta}\pm i\sin{\theta}:

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

Using initial conditions x(0) = x_0, \dot{x}(0) = v_0, we get the following:

    \[x(t)=e^{-\zeta pt}\Big[x_0 \cos\sqrt{1-\zeta^2} pt + \frac{\nu_0 + \zeta p x_0}{\sqrt{1-\zeta^2}}\sin{pt\sqrt{1-\zeta^2}pt}\Big]\]

We can write this as:

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

Where \phi = \tan^{-1}{\frac{A}{B}}, A = x_0, and B = \frac{v_0 + \zeta p x_0}{\sqrt{1-\zeta ^2}}.

In some problems, we have to find the equilibrium (static) configurations and then determine the motion about these positions.

Example

Find the motion about the equilibrium configuration.

m slides on a wire with viscous friction c.

Solution

    \[\begin{split}+\circlearrowright \sum{M_0}& = mL^2\ddot{\theta} \\& = mgL\sin{\theta}-kL\sin{\theta}(L\cos{\theta}) - cL^2\ddot{\theta}\end{split}\]


    \[mL^2 \ddot{\theta}+kL^2 \sin{\theta}\cos{\theta}- mgL\sin{\theta}+cL^2 \dot{\theta}=0\]


    \[\ddot{\theta}+\frac{k}{m}\sin{\theta}\cos{\theta}-\frac{g}{L}\sin{\theta}+\frac{c}{m}\dot{\theta}=0 \quad (1)\]

We must find static equilibrium configurations and determine equation of motion for small perturbations about these static configurations.

Set \ddot{\theta} = \dot{\theta} = 0, and let \theta = \theta_0. Therefore:

    \[\frac{k}{m} \sin{\theta_0} \cos{\theta_0} = \frac{g}{L} \sin{\theta_0}\]


    \[\sin{\theta_0}\Big[\frac{k}{m}\cos{\theta_0}-\frac{g}{L}\Big] = 0\]

Therefore. the static configurations are given by:

    \[\theta_0 = 0, \pi \quad \sin{\theta_0}=0 \quad \cos{\theta_0} = \frac{mg}{kL}\]

Assuming \frac{mg}{kL}<1, there are 3 positions.

To find the motion, linearize (1) about \theta_0. Set \theta = \theta_0 + \hat{\theta}, where \hat{\theta} is a small angle.

    \[\sin \theta = \sin(\theta_o + \hat{\theta}) \approx \sin \theta_o +\hat{\theta} \cos \theta_o\]


    \[\cos \theta \approx \cos \theta_o - \hat{\theta}\sin\theta_o\]

Put the above into the original equation of motion to get:

    \[\ddot{\hat{\theta}} + \frac {k}{m}(\sin \theta_o + \hat{\theta} \cos \theta_o)(\cos \theta_o - \hat{\theta}\sin \theta_o) - \frac{g}{L}(\sin \theta_o + \hat{\theta}\cos \theta_o) = 0\]

Note that \dot{\theta_o} = \ddot{\theta_o} = 0.

    \[\ddot{\hat{\theta}} + \frac {k}{m}(\sin \theta_o \cos \theta_o + \hat{\theta} \cos^2\theta_o - \hat{\theta}\sin^2\theta_o-\hat{\theta}^2\sin\theta_o \cos\theta_o) - \frac{g}{L}(\sin \theta_o + \hat{\theta}\cos \theta_o) + \frac{c}{m}\dot{\hat{\theta}} = 0\]

Using the fact that the \hat{\theta}^2 term is small compared to 1, it becomes:

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m}\cos\theta_o - \frac{g}{L}\Big)\sin\theta_o + \hat{\theta}(\cos^2\theta_o-\sin^2\theta_o)\frac{k}{m} - \frac{g}{L}\hat{\theta}\cos\theta + \frac{c}{m}\dot{\hat{\theta}} = 0\]


    \[\ddot{\hat{\theta}} + \hat{\theta}\Big(\frac{k}{m}\cos2\theta_o - \frac{g}{L}\cos\theta_o\Big) + \frac{c}{m}\dot{\hat{\theta}} + \Big(\frac{k}{m}\cos\theta_o - \frac{g}{L}\Big)\sin\theta_o = 0\]

The above equation looks like the standard form except for a constant term that is zero for all conditions.

Now consider all three positions:


1. \theta_o = 0.

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m} - \frac{g}{L}\Big)\hat{\theta} + \frac{c}{m}\hat{\theta} = 0\]

Therefore, this is stable if \frac{k}{m} > \frac{g}{L} as \frac{c}{m} > 0, but unstable if \frac{k}{m} < \frac{g}{L}.


2.\theta_o = \pi.

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m} + \frac{g}{L}\Big)\hat{\theta} + \frac{c}{m} = 0\]

Therefore, this is always stable.


3.\cos\theta_o = \frac{mg}{kL}.

    \[\ddot{\hat{\theta}} + \frac{k}{m}\Big(2\cos^2\theta_o - 1 - \frac{gm}{kL}\cos\theta_o\Big)\hat{\theta} + \frac{c}{m}\dot{\hat{\theta}} = 0\]


    \[\ddot{\hat{\theta}} + \frac{k}{m}(\cos^2\theta_o - 1)\hat{\theta} + \frac{c}{m}\dot{\hat{\theta}} = 0\]

Therefore, this is always unstable, regardless of damping.

Example

1.\frac{mg}{kL} = \frac{1}{2}, \theta_o = \pm 60^{\circ}, \frac{k}{m} = \frac{2g}{L}. Therefore:

    \[\frac{k}{m} > \frac{g}{L}\]

2.\frac{mg}{kL} = 2, \frac{k}{m} = \frac{1}{2}\frac{g}{L}. Therefore:

    \[\frac{k}{m} < \frac{g}{L}\]

Equivalent spring constants

One of the components we need in these equations of motion is the spring constant k. We can often find this for a system using static techniques. This is easily generalized for MDOF systems.

For SDOF systems, we can imagine the static response of the system using one of the approaches:

1.Apply a unit force (moment) to the mass (inertia) in the positive direction of motion then calculate the displacement that occurs (flexibility approach).

    \[\tilde{k} = \frac{\text{force (moment) in direction of motion}}{\text{displacement (rotation) in direction of motion}}\]

2.Apply a unit displacement (rotation) to the mass (inertia) in the positive direction of motion then calculate the force (moment) required to maintain it (stiffness approach).

    \[\tilde{k} = \frac{\text{force (moment) in direction of motion}}{\text{displacement (rotation) in direction of motion}}\]

It depends on the situation to determine the best one to use. Generally, for a series situation, use flexibility, while for parallel, use stiffness.

Springs in Series

Apply unit load, and calculate \Delta`L.

    \[\Delta_1 = \frac{1}{k_1} \ \ \Delta_2 = \frac{1}{k_2} \ \ \Delta_\text{total} = \frac{1}{k_1} + \frac{1}{k_2}\]


    \[k_\text{eff} = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} = \frac{k_1k_2}{k_1+k_2}\]

Springs in Parallel

Apply unit deflection, and calculate load.

    \[k_\text{eff} = k_1 + k_2\]

Axially Loaded Bar

    \[\text{Stress} = E \cdot \text{Strain}\]

Therefore:

    \[\frac{F}{A} \equiv \Delta = E\epsilon = E = \frac{\Delta}{L}\]


    \[F = \frac{EA\Delta}{L}\]

Therefore:

    \[k_e = \frac{EA}{L}\]

Inclined Axial Spring

Apply a \Delta in the direction of motion.

Therefore, force in spring when extended is k\Delta\cos\theta, and thus:

    \[\frac{\text{Force in direction of motion}}{\Delta \ \text{in direction of motion}} = \frac{(k\Delta\cos\theta)\cos\theta}{\Delta} = k\cos^2\theta\]

Example 1

Stiffness
Apply unit deflection, and calculate total load.

    \[+\circlearrowleft\sum{M_A} = k_1\Big(\frac{1}{2}\Big)\Big(\frac{1}{2}\Big) + k_2(\text{L}) - F(\text{L}) = 0\]


    \[F = \frac{k_1}{4} + k_2\]

Therefore:

    \[k_\text{eff} = \frac{k_1}{4} + k_2\]

If k_1 = k_2 = k, then:

    \[k_\text{eff} = \frac{5}{4}\]

Example 2

Flexibility
Apply unit load, and calculate total deflection.

\Delta_B = \frac{2}{k_1}, therefore:

    \[\Delta_C = \frac{4}{k_1}\]


Therefore :

    \[\begin{split}\Delta \ \text{at} \ M &= \Delta_C + \frac{1}{k_2} \\&= \frac{4}{k_1} + \frac{1}{k_2}\end{split}\]

Thus:

    \[\begin{split}k_\text{eff} &= \frac{1}{\Delta}\\&= \frac{1}{\frac{4}{k_1}+\frac{1}{k_2}}\end{split}\]

If k_1 = k_2, then:

    \[k_\text{eff} = \frac{k}{5}\]

Equivalent Spring Constant Equations

1.n axial springs in parallel

    \[k_e = k_1 + k_2 + k_2 + k_3 + … + k_n\]


2.n axial springs in series

    \[k_e = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2} + … + \frac{1}{k_n}}\]


3.Springs in parallel and series

    \[k_e = \frac{k_1k_3 + k_2k_3}{k_1 + k_2 + k_3}\]


4.Inclined axial spring

    \[k_e = k\cos^2\theta\]


5.Rotating bar with spring support

    \[k_e = k\Big(\frac{L}{a}\Big)^2\]


6.Rigid bar supported on two springs

    \[k_e = \frac{4k_1k_2}{k_1[1 + \frac{a}{L}]^2 + k_2[1 + \frac{a}{L}]^2}\]


7.Rigid bar supported on three springs

    \[k_e = \frac{3k}{1 + \frac{3}{2}(a / L)^2}\]


8.Axially loaded bar

    \[k_e = \frac{AE}{L}\]

Where A is the cross-sectional area, and E is the elastic modulus.

9.Axially loaded tapered bar

    \[k_e = \frac{\pi E D_a D_b}{4L}\]


10.Axially helical spring

    \[k_e = \frac{Gd^4}{G4nR^3}\]

Where n is the active number of turns, and G is the elastic shear modulus.

11.Torsion of a uniform shaft

    \[k_e = \frac{GJ}{L}\]

Where J is the torsional constant of cross section (J = \frac{\pi d^4}{32}).

12.Torsion of a tapered circular shaft

    \[k_e = \frac{3\pi}{32}\frac{D_b^4 G}{L\big[\frac{D_b}{D_a}+(\frac{D_b}{D_a})^2+(\frac{D_b}{D_a})^3\big]}\]


13.Spiral torsional spring

    \[k_e = \frac{EI}{L}\]

Where E is Young’s modulus. I is the moment of inertia of cross-sectional area, and L is the total length of the spiral.

14.Cantilever bean, end load

    \[k_e = \frac{3EI}{L^3}\]


15.Simply supported bean, load at midspan

    \[k_e = \frac{48EI}{L^3}\]


16.Simply supported bean, load anywhere between supports

    \[k_e = \frac{3EIL}{a^2b^2}\]


17.Fixed-fixed beam, load at midspan

    \[k_e = \frac{192EI}{L^3}\]


18.Fixed-fixed beam, off-center load

    \[k_e = \frac{3EI(a+b)^3}{a^3b^3}\]


19.Propped cantilever, load at midspan

    \[k_e = \frac{768EI}{7L^3}\]


20.Propped cantilever, load at free end

    \[k_e = \frac{24EI}{a^2(3L + 8a)}\]

Table 2-1

Fixed-fixed beam*

y = \frac{Pb^2}{6EI\ell^3}\big[(2b-3\ell)x^3 + 3\ell(\ell-b)x^2\big] (x \leq a)
y = \frac{Pb^2}{6EI\ell^3}\big[(2b-3\ell)x^3 + 3\ell(\ell-b)x^2 + \frac{1^3}{b^2}(x-a)^3\big] (x \geq a)
k = \frac{P}{y|_{x=a}}
k|_{a=\frac{\ell}{2}} = \frac{192EI}{\ell^3}

Fixed-pinned beam with overhang*

y = \frac{P}{12EI}\Big[3b\Big(1-\frac{b^2}{\ell^2}\Big)x^2-\frac{b}{\ell}\Big(3-\frac{b^2}{\ell^2}\Big)x^3\Big] (x \leq a)
y = \frac{P}{12EI}\Big[3b\Big(1-\frac{b^2}{\ell^2}\Big)x^2-\frac{b}{\ell}\Big(3-\frac{b^2}{\ell^2}\Big)x^3 + 2(x-a)^3\Big] (a \leq x \leq \ell)
y =\frac{-pba^2}{4EI\ell}(x-\ell) (x \geq \ell)
k = \frac{P}{y|_{x=a}}
k|_{a=\frac{\ell}{2}} = \frac{768EI}{7\ell^3}

Fixed-pinned beam with overhang (P at x = l + a)*

y = \frac{Pa}{4EI\ell}(x^3-\ell x^2) (x \leq \ell)
y = \frac{Pa}{4EI\ell}\Big[x^3-\ell x^2-\Big(\frac{2\ell}{3a}+1\Big)(x-1)^3\Big] (x \geq \ell)
k = \frac{P}{y|_{x=\ell+a}} = \frac{12EI}{a^2(3\ell+4a)}

Pinned-pinned beam with overhang*

y = \frac{Pa}{6EI\ell}(a^2-\ell^2)(x-\ell) (x \geq \ell)

Pinned-pinned beam with overhang (P at x = l + a)*

y = \frac{Pax}{6EI\ell}(x^2-\ell^2) (x \leq \ell)
y = \frac{P}{6EI\ell}\big[ax(x^2-\ell^2)-(\ell+a)(x-\ell)^3\big] (x \geq \ell)
k = \frac{P}{y|_{x = a+\ell}} = \frac{3EI}{a^2(a+1)}

Fixed-fixed beam with lateral displacement

\Delta = \frac{P\ell^3}{12EI}
k = \frac{12EI}{\ell^3}
y = \frac{P}{12EI}(3\ell x^2-2x^3)

* – Axial extensions due to axial end constraints considered negligible

Torsional Oscillations

    \[\begin{split}+\circlearrowright \sum{M_o} &= J_o\ddot{\theta} \\&= -\hat{k}\theta\end{split}\]


    \[J\ddot{\theta} + \hat{k}\theta = 0\]

Let P = \sqrt{\frac{\hat{k}}{J}}. From the strength of the material:

    \[\tau = \mu \alpha I\]

Where \mu is the modulus of rigidity, \alpha is the angle of twist/unit length and I is the polar second moment of inertia. Therefore:

    \[\tau = \mu \frac{\theta}{\ell}I = \hat{k}\theta\]


    \[\hat{k} = \frac{\mu I}{\ell}\]

For soild circular cross sections:

    \[I = \frac{\pi d^4}{32}\]

Center of Percussion and Vibration of a Compound Pendulum

The concept of the center of percussion is usually discussed using the analysis of the changes in angular momentum and linear momentum associated with a body impacting a rotating object and essentially determining the point at which there is no change in acceleration at the pivot point due to this impact. (The usual example is a ball hitting a baseball or cricket bat.) For the compound pendulum shown the natural frequency is p = (mgd/J_0)^{1/2} where J_0 is the moment of inertia of the body about the axis of rotation at O. Instead of having a distributed mass, let the mass of the pendulum be concentrated at a single point located at a distance L from the pivot point O. The natural frequency of this simple pendulum is (g/L)^{1/2}. If the natural frequencies of these two pendulums are the same, then the point at which m is concentrated is called the center of oscillation and is identical to the center of percussion. If instead, we consider rotation about the center of oscillation (percussion) then the original pivot at O becomes the center of percussion (oscillation).

Early automobile manufacturers did not realize this and as a result, cars had much more uncomfortable rides because impacts at one set of wheels (front or rear) were felt at the other. One of the first automobile designs to incorporate this idea was the very futuristic Chysler Airflow (circa 1930) which was years ahead in its design in a number of aspects. (So far ahead that it was a commercial failure.) For the illustration of the automobile shown the relationship between the position of the center of gravity and the positions of the front and rear wheels is given by r^* = (ab)^{1/2} where r^* is the radius of gyration of the vehicle about its center of gravity and I = a + b is the wheelbase (a and b are the horizontal distances from the front and rear wheels to the center of gravity). From a vibration perspective if a car is designed with a mass distribution to meet these criteria (almost all recent automobiles are) then we can examine what happens at one set of wheels individually for certain situations (e.g. hitting a bump with the front wheels only).

Using the diagram below show that the condition for the center of percussion to be at the front wheels with the pivot at the rear wheels that r^* = (ab)^1/2 and that this is also true for the pivot point moving to the front wheels with the impact at the rear ones.

    \[J = m(r^*)^2\]

Where J is the moment of inertia about G and r^* is the radius of gyration about G:

    \[\begin{split}+ \circlearrowleft  \sum M_0 &= -mgd\theta \\&= J \ddot{\theta}+md\ddot{\theta}(d)\end{split}\]

Therefore:

    \[(J+md^2)\ddot{\theta} + mgd\theta = 0\]

    \[J_0 = J + md^2\]

Therefore:

    \[\begin{split}p &= \sqrt{\frac{mgd}{m(r^*)^2+md^2}} \\&= \sqrt{\frac{gd}{(r^*)^2+d^2}}\end{split}\]

Equivalent Pendulum

    \[p = \sqrt{\frac{g}{L}}\]

Therefore choose:

    \[L = \frac{(r^*)^2+d^2}{d}\]

Example

    \[ \ell = a+b \]

For application to the auto above \ell = L(wheelcase). For rotation about B:

    \[d = b\]

Therefore:

    \[\begin{split} L &= a+b \\&= \frac{(r^*)^2+b^2}{b} \\&= \frac{(r^*)^2}{b} +b \end{split}\]

Therefore:

    \[r^* = \sqrt{ab}\]

For rotation about A:

    \[d = a\]

Therefore:

    \[a+b = \frac{(r^*)^2}{a} +a \]

    \[r^* = \sqrt{ab}\]

SDPF damped system

We know that for any SDOF damped system the equation for free vibration is:

    \[m_e\ddot{x} + C_e\dot{x} + k_ex = 0\]

and the system parameters are the natural frequency:

    \[p = \sqrt{\frac{k_e}{m_e}}\]

and the damping ratio:

    \[\zeta = \frac{C_e}{C_c} = \frac{Ce}{Z_{mep}}\]

We can determine the damping ratio from how rapidly the vibration decays. To do this we use the so-called logarithmic decrement. Consider the damped response:

    \[ x(t) = e^{-\zeta pt}\big[X\sin{(\sqrt{1-\zeta^2}pt+\phi)}\big]\]

which for \delta < 1 looks like:

These peaks occur approximately when:

    \[\sin{\big(\sqrt{1 - \zeta^2}pt + \phi\big)} = 1\]

Therefore if the first occurs at t_0 then the next time it is at:

    \[t_1 = t_0 + \frac{2\pi}{p\sqrt{1-\zeta^2}}\]

Therefore:

    \[\begin{split} \frac{X_0}{X_1} &= \frac{e\strut^{-\zeta pt_0}X}{e\strut^{-\zeta pt_1}X}  \\&= \frac{e\strut^{-\zeta pt_0}} {e\strut^{-\zeta p\big(t_0 + \frac{2\pi}{p \sqrt{1-\zeta^2}}\big)}} \\&= e\strut^{\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}} \end{split}\]

and this is the same ratio for \frac{X_1}{X_2}, \frac{X_2}{X_3}, etc.

Therefore this creates the logarithmic decrement equation:

    \[ \begin{split} \ln{\frac{X_0}{X_1}} &= \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \\&\equiv \delta \end{split}\]

If we take n cycles:

    \[\frac{X_0}{X_n} = \frac{X_0}{X_1}\frac{X_1}{X_2}\frac{X_2}{X_3}..... \frac{X_{n-1}}{X_n}\]

    \[\begin{split}\ln {\frac{X_0}{X_n}} &= \ln {\frac{X_0}{X_1}} + \ln {\frac{X_1}{X_2}}..... \\&= n\frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \\&= n\delta \end{split}\]

Therefore:

    \[\begin{split} \delta &= \frac{1}{n}\ln{\frac{X_0}{X_n}} \\&= \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \end{split}\]

Therefore if we measure \frac{X_0}{X_n} we can easily determine the damping ratio \zeta.

If \zeta<<1, \delta \approx2\pi\zeta

\zeta0.050.10.20.30.40.5
\frac{X_0}{X_1}1.371.873.617.2115.537.6

It does not take a large \zeta to quickly damp out vibrations.

Steady State Harmonic Response

    \[ \begin{split}+\downarrow\sum F_x &= m\ddot{x} \\&= -kx-c\dot{x} + F(t) \end{split}\]

Therefore:

    \[ m\ddot{x} + c\dot{x} + kx = F_0\sin(\omega t) (\star)\]

Therefore find solution as homogeneous solution and particular integral.

a) Homogeneous Solution:

    \[ m\ddot{x_H} + c\dot{x_H} = 0 \implies\]

    \[ x_H = e^{-\zeta pt}\big[A\cos{\sqrt{1-\zeta^2}}pt+ B\sin{\sqrt{1-\zeta^2}}pt\big] \]

b) Particular integral:

Try x_p = C\sin{\omega t} + D\cos{\omega t} into \star

    \[-m\omega^2C\sin{\omega t} - m\omega^2D\cos{\omega t} + Cc\omega\cos{\omega t}\]

    \[-Dc\omega\sin{\omega t} + kC\sin{\omega t} + kD\cos{\omega t} = F_0\sin{\omega t} \]

Therefore:

    \[-m\omega^2C-c\omega D+kC = F_0,\]

    \[-m\omega^2D+c\omega C+kD = 0\]

    \[C\Big(1-\frac{m\omega^2}{k}\Big) - \frac{c\omega}{k}D = \frac{F_0}{k}\]

    \[C\Big(\frac{c\omega}{k}\Big) + \Big(1-\frac{m\omega^2}{k}\Big)D = 0\]

Note:

    \[ \frac{m\omega^2}{k} = \frac{\omega^2}{p^2}\]

    \[ \frac{c\omega}{k} = 2\zeta\frac{\omega}{p}\]

    \[ \begin{split} \implies \frac{c\omega}{k} &= \frac{c}{c_c}(2mp)\frac{\omega}{k} \\&= 2\zeta\frac{\omega}{p}\end{split}\]

Therefore:

    \[C\Big(1-\frac{\omega^2}{p^2}\Big) - 2\zeta\frac{\omega}{p}D = 0,\]

    \[2\zeta\frac{\omega}{p}C+\Big(1-\frac{\omega^2}{p^2}\Big)D = 0\]

Solving the above equations for C & D gives:

    \[C = \frac{\big(1-\frac{\omega^2}{p^2}\big)F_0/k}{\big(1-\frac{\omega^2}{p^2}\big)+\big(2\zeta\frac{\omega}{p}\big)\strut^2}\]

    \[D = \frac{-2\zeta\frac{\omega}{p}F_0/k}{\big(1-\frac{\omega^2}{p^2}\big)+\big(2\zeta\frac{\omega}{p}\big)\strut^2}\]

when x(0) = x_0 and \dot{x} = v_0:

    \[x(t) = e^{-\zeta pt}\Big[x_o\cos{\sqrt{1-\zeta^2}}pt + \frac{v_0+\zeta px_0}{\sqrt{1-\zeta^2}}\sin{\sqrt{1-\zeta^2}}pt\Big]+C\sin{\omega t} +D\cos{\omega t}\]

However, if \zeta is +ve after some time the homogeneous part will decay and the steady state solution becomes:

    \[X(_{ss}) = C\sin{\omega t} + D\cos{\omega t}\]

If we take the special case of no clamping, the homogeneous equation is:

    \[m\ddot{x}_H + kx_H = 0\]

    \[ x_H = A\cos{pt} + B\sin{pt}\]

and the particular integral is

    \[x_p = C\sin{\omega t} \]

Therefore:

    \[m\ddot{x}_p+kx_p = F_0\sin{\omega t}\]

    \[(-mC\omega^2+ck)\sin{\omega t} = F_0\sin{\omega t}\]

Therefore:

    \[ \begin{split} C &= \frac{F_0}{k-m\omega^2} \\&= \frac{\frac{F_0}{k}}{1-\frac{\omega^2}{p^2}} \end{split} \]

Therefore the total solution is:

    \[ x = A\cos{pt} +B\sin{pt} + \frac{F_o/k}{1-\frac{\omega^2}{p^2}}\sin{\omega t} \]

Note that as there is no damping present that the homogeneous solution does not \rightarrow 0 as time increases.

    \[x_{ss} = C\sin{\omega t} + D\cos{\omega t}\]

and we combine them to get:

    \[ \begin{split} x_{ss} &= \sqrt{C^2+D^2}\sin{(\omega t - \phi)} \\&= \frac{\frac{F_0}{k}}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}\sin{(\omega t - \phi)} \end{split} \]

Where:

    \[ \tan{\phi} = \frac{2\zeta\frac{\omega}{p}}{1-\frac{\omega^2}{p^2}} \]

This result is extremely useful in practical situations especially for controlling noise and vibration.

While this tells us the dynamic motion of the system, there is still the static deflection of the springs to be remembered. These are also related situations which are obtained by reconsideration of the fundamental equation of motion.

From the view of real situations, these are other quantities that are of interest. One of these is the force F_T transmitted to the supporting structure.

    \[F_T = kx +c\dot{x}\]

Where x is:

    \[x = \frac{F_0/k}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}\sin{(\omega t - \phi)}\]

Therefore:

    \[F_T = \frac{F_0/k}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}} \Big[k\sin{(\omega t - \phi)} + c\omega\cos{(\omega t - \phi)}\Big]\]

and the magnitude of F_T,\big|F_T\big| using the fact that \frac{c\omega}{k} = 2\zeta\frac{\omega}{p} is:

    \[\big|F_T\big| = \frac{\sqrt{k^2+(c\omega)^2}}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}\]

    \[\frac{|F_T|}{F_0} = \frac{\sqrt{1+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}\]

A second practical situation accuse when the structure is vibrating and we wish to isolate a machine from the vibration (base excitation).

So the equation of motion is:

    \[m\ddot{x} + c\dot{x} + kx = kY\sin{\omega t}+ c\omega Y\cos{\omega t} \]

The steady state solution comes from the assumption from x(t):

    \[ \begin{split} x(t) &= C_1 \sin{\omega t} + C_2 \cos{\omega t} \\& \equiv X\sin{(\omega t - \phi)} \end{split}\]

and the solution is:

    \[X = \frac{Y\sqrt{1+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2+\big(2\zeta\frac{\omega}{p}\big)\strut^2}}\]

    \[ \phi = \arctan{\biggr(\frac{2\zeta\frac{\omega}{p}}{1-\frac{\omega^2}{p^2}}\biggr)}\]

A third potential situation is essentially the original forced situation where the magnitude of the forcing function is due to rotating imbalance:

    \[F_0 = me\omega^2\]

Where past of the mass M, \tilde{m}, at an eccentricity e causes the dynamic force.

Equations of motion:

    \[ (M-\tilde{m})\ddot{x} + \tilde{m}\frac{d^2}{dt^2}(x+e\sin{\omega t}) = -kx - c\dot{x} \]

    \[ M\ddot{x} + c\dot{x} + kx = \tilde{m}e\omega^2\sin{\omega t} \]

This looks like the previous forced case where F_o is replaced with \tilde{m}e\omega^2 so that:

    \[X = \frac{\tilde{m}e\omega^2}{\sqrt{(k-M\omega^2)^2+(c\omega)^2}}\]

and reduced to non-dimensional form as:

    \[\frac{MX}{\tilde{m}e} = \frac{ \omega^2/p^2}{\sqrt{\big(1-\frac{\omega}{p}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\]

so that the steady state solution is:

    \[x(t) = X\sin{(\omega t - \phi)}\]

where:

    \[ \tan{\phi} = \frac{2\zeta\frac{\omega}{p}}{1-\big(\frac{\omega}{p}\big)\strut^2}\]

The frequency response for all these cases are summarized on the following page

Single Degree of Freedom Systems – Useful Relations

The frequency response curves are helpful in visualizing where the specific system you are analyzing lies. It must be remembered that there is a phase difference between excitation and response with the change from being nearly in phase to being nearly 180 degrees out of phase happens near resonance (\frac{\omega}{p} = 1). For the undamped case, this means the response is undefined at \frac{\omega}{p} and the response becomes:

    \[ \begin{split} \left|\frac{X}{Y}\right| &= \left|\frac{F_T}{F_0}\right| \\&= \left|\frac{X}{X_0}\right| \\& = \frac{1}{1-\frac{\omega^2}{p^2}}, \frac{\omega}{p} < 1\\& = \frac{1}{\frac{\omega^2}{p^2}-1}, \frac{\omega}{p} > 1 \end{split} \]

    \[\left| \frac{MX}{\tilde{m}e}\right| = \frac{\omega^2/p^2}{1-\frac{\omega^2}{p^2}}, \frac{\omega}{p} < 1\]

    \[\left| \frac{MX}{\tilde{m}e}\right| = \frac{\omega^2/p^2}{\frac{\omega^2}{p^2}-1}, \frac{\omega}{p} > 1\]

The results for the steady state forced single degree of freedom systems are usually present in the form of graphs showing various non-dimensional parameters. Because of the technical importance of these results, it is necessary to understand the differences between these representations. Two of the most useful are given by the expressions for displacement of the mass:

    \[\left|\frac{X}{X_0}\right| = \left|\frac{1}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\right|(A)\]

And the displacement of the main mass in the case of rotating imbalance:

    \[\left| \frac{MX}{\tilde{m}e}\right| = \left|\frac{\omega^2/p^2}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\right|(B)\]

a) Show that (B) can be derived from (A) for the special case in which the forcing function is due to rotating imbalance

b) For the undamped case determine the result of (A) and (B) both below and above resonance

a) Special cases for rotating imbalance:

    \[\left|\frac{X}{X_0}\right| = \left|\frac{1}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\right|(A)\]

    \[\left| \frac{MX}{\tilde{m}e}\right| = \left|\frac{\omega^2/p^2}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\right|(B)\]

    \[X_0 = \frac{F_0}{k}\]

If F_0 = \tilde{m}e\omega^2:

    \[ X_0 = \frac{\tilde{m}e\omega^2}{k}\]

    \[\frac{xk}{\tilde{m}e\omega^2} = \frac{1}{\sqrt{\big(1-\frac{\omega^2}{p^2}\big)\strut^2 + \big(2\zeta\frac{\omega}{p}\big)\strut^2}}\]

    \[\frac{Mxk}{M\tilde{m}e\omega^2} = \frac{xp^2M}{\tilde{m}e\omega^2}\]

Note: M includes the imbalanced mass \tilde{m}

b) \left|\frac{X}{X_0}\right| for \zeta \rightarrow 0 is:

    \[\begin{split} \left|\frac{X}{X_0}\right| &= \left|\frac{1}{\sqrt{1-\frac{\omega^2}{p^2}}}\right|, \frac{\omega}{p}\neq 1 \\&= \frac{1}{1-\frac{\omega^2}{p^2}}, \frac{\omega}{p}< 1  \\&= \frac{1}{\frac{\omega^2}{p^2}-1}, \frac{\omega}{p}> 1\end{split}\]

    \[\begin{split} \left| \frac{MX}{\tilde{m}e}\right|&= \left|\frac{\omega^2/p^2}{\sqrt{1-\frac{\omega^2}{p^2}}}\right|, \frac{\omega}{p}\neq 1 \\&= \frac{\omega^2/p^2}{1-\frac{\omega^2}{p^2}}, \frac{\omega}{p} < 1 \\&= \frac{\omega^2/p^2}{\frac{\omega^2}{p^2}-1}, \frac{\omega}{p} > 1 \end{split}\]

There are two important physical phenomena that are easily illustrated using the forced undamped SDOF system as we approach resonance. The entire solution:

    \[x(t) = C_1\cos pt + C_2 \sin pt + \frac{F_o}{k}\frac{\sin  \omega t}{1-\frac{\omega^2}{p^2}}\]

Consider the case

    \[x(0)=0\]

,

    \[\dot{x}(0)=0\]

. From the first of these C_1 = 0 and

    \[ \dot{x}(t) = C_2p\cos pt+\frac{F_o}{k}\frac{1}{1-\frac{\omega^2}{p^2}}  \omega \cos  \omega t \]

Therefore at t = 0:

    \[ \begin{split} C_2 &= \frac{-F_o}{k}\frac{\frac{\omega}{p}}{1-\frac{\omega^2}{p^2}} \\&=\frac{-F_o}{k}\frac{\omega p}{p^2-\omega^2} \end{split} \]

    \[ x(t) = \frac{F_o}{k}\frac{p^2}{p^2-\omega^2}\big[\sin \omega t-\frac{\omega}{p}\cos pt\big] \]

If p-\omega = 2\Delta, where \Delta is small compared to \omega or p, then:

    \[ x(t) = \frac{F_op^2}{k}\frac{1}{p^2-\omega^2}\big[\sin \omega t - \sin pt\big] \]

Known \sin A- \sin B = 2\sin\frac{A-B}{2}\cos\frac{A+B}{2}, therefore:

    \[ x = 2\frac{F_op^2}{k(p^2-\omega^2)}\cos \omega t\sin\Delta t \]

    \[ \begin{split} p^2-\omega^2 & = (2\Delta + \omega)^2 - \omega^2 \nonumber \\ & = 4\Delta+4\Delta \omega \nonumber \end{split} \]

Therefore:

    \[ x = \frac{-F_o}{2k\Delta}\cos\omega t \sin\Delta t \]

This expression can be considered as representing a vibration with period 2\pi /\omega that has a slowly varying amplitude equal to \frac{F_o\omega}{2k\Delta}\sin\Delta t, (\sin\Delta t has a much longer period than \cos\omega t)

While the original solution to the forced undamped differential equation is not valid at \omega=p, it is of value to consider what happens at \Delta \rightarrow 0 for the displacement. Using  L’Hospital’s rule for the following limit:

    \[ \begin{split} \lim_{\Delta \to 0} \biggr(\frac{-F_o\omega}{2k}\cos\omega t \frac{\sin \Delta t}{\Delta} \biggr) &= \frac{-F_o \omega}{2k} \cos(\omega t)  t\cos \Delta t \\&=  \frac{-F_o\omega}{2k}t\cos\omega t \end{split} \]

The amplitude of vibration increases indefinitely with time and the infinite amplitude is in the steady state. This means that to operate the system it is important to move through resonance as quickly as possible so that the amplitude does not “build up”.

Energy Considerations in Vibrations

For many dynamic/vibration systems energy considerations are useful especially in modelling. This becomes evident when the complexity of the system increases as the formulation of the models will usually involve velocities rather than accelerations and the kinematics for general motion are easier to describe. Specifically for vibrations we can appreciate the exchange of energy occurring between the kinetic and potential components with the damping providing the energy dissipation. This occurs in the consideration of the Lagrange formulation of dynamical systems 

If we choose the static equilibrium position as the datum for the energy calculation we must be aware of the energy involved in the static equilibrium deflection. There will be instances in which the static condition does not have a deflection then the energy calculation must include consideration of the position of the mass of the system. 

One DOF Spring Oscillator 

For a single DOF undamped system we can calculate the natural frequency from energy considerations as we know the motion is Simple Harmonic Motion (SHM). In addition it is possible to use energy as an approach to finding approximate solutions for the natural frequency without a large computational effort. For these systems, termed conservative, the sum of the kinetic and potential energy for free vibration is a constant. 

T (kinetic) + U (potential) = C (constant) 

In addition the energy varies over time from being all potential to all kinetic so that 

    \[ T_\text{MAX} = U_\text{MAX} \]

Under the assumed SHM  x = A \sin (pt + \phi)

Consider the SDOF system from an energy perspective where a gain x is measured from the static equilibrium configuration.

Potential energy is the work done by F in moving through x (energy in spring) :

    \[ V_s = \int_{0}^{x} (mg+kx) \,dx \]

    \[ = mgx + \frac{1}{2}kx^2 \]

While the change in potential energy due to position of m is -mgx. Therefore, the total potential energy is -\frac{1}{2}kx^2, assuming :

    \[ x = A\sin(pt+\phi) \]

    \[ \dot{x} = Ap\cos(pt+\phi) \]

    \[ T_\text{MAX} = V_\text{MAX} \]

Therefore,

    \[ \frac{1}{2}m(Ap)^2 = \frac{1}{2}k(A)^2  \]

    \[ \Rightarrow p = \sqrt\frac{k}{m} \]

We can extend this idea to consider approximating the effect on the natural frequency when the mass of the spring is included. Assume that the spring of length L has a uniform mass M_s over its length. Then at any position on the spring the velocity is \dot{y}(t) and the kinetic energy of an infinitesimal mass dm is \frac{1}{2}dm( \dot{y} (t)^2 and the kinetic energy of the spring is:

    \[ T_\text{spring} = \frac{1}{2} \int_{0}^{L} \big(\dot{y} (t)^2\big) \,\Big(\frac{M}{L}dy\Big) \]

If we now make an assumption of how the spring “moves” we can calculate T_\text{spring} e.g. assume the spring displaces linearly with position y so that at \dot{y} = \dot{x} \frac{y}{L}

    \[ \begin{split} T_\text{spring} &= \frac{1}{2} \frac{M}{L^3} \int_{0}^{L} (\dot{x} y)^2 \,dy \\&= \frac{1}{2} \frac{M}{L^3} \frac{L^3}{3} \dot{x}^2  \\&= \frac{1}{2} \frac{M}{3} \dot{x}^2 \end{split} \]

Now using T_\text{MAX} = U_\text{MAX} :

    \[ \frac{1}{2} m(Ap)^2 + \frac{1}{2} \frac{M}{3} (Ap)^2 = \frac{1}{2}(A)^2 \]

    \[ p = \sqrt \frac{k}{m+ \frac{M}{3}} \]

If instead we assumed the “spring” was a simply supported beam of mass M and length L and that the dynamic shape is a half-sine then at any position the velocity is: \dot{x} \sin \frac{\pi y}{L} and the mass/unit length is \frac{M_s}{L}

    \[\begin{split} T_\text{spring} &= \frac{1}{2}  \int_{0}^{L} \frac{M}{L}\big(\dot{x}\big)\strut^2 \big(\sin \frac{\pi y}{L}\big)\strut^2 \,dy  \\&= \frac{1}{2} \frac{M}{L}(\dot{x})^2 \bigg[\frac{y}{2} - \frac{L^2}{4 \pi ^2} \sin \frac{2 \pi y}{L} \bigg] \biggr|_ 0^L  \\&= \frac{1}{2} \frac{M}{2} \dot{x}^2 \end{split} \]

Then using the same approach:

    \[ p = \sqrt \frac{k}{m+ \frac{M_s}{2}} \]

Complex Representation for Vibrations

When dealing with vibrating equations it is often more efficient to use complex numbers as they allow the phase relationships to be easily handled. To do this use the complex number:

    \[ \zeta = x + iy,\   i = \sqrt -1 \]

    \[  x = Re(\zeta),\ y = Im(\zeta) \]

\zeta ^ \star = x - iy is called the complex conjugate

    \[ \zeta \zeta ^ \star = x^2 + y^2 = |\zeta |^2 \]

Alternatively we write \zeta as:

    \[ \zeta = |\zeta|e^{i\theta},  \theta = \tan^{-1}\Big(\frac{y}{x}\Big) \]

e^{i \theta} is called a unit phasor. For fractional expressions we can rationalize the denominator to separate real & imaginary parts:

    \[ C = \frac{A}{B} = \frac{a + i\alpha}{b+i\beta} , C^\star = \frac{a - i\alpha}{b- i\beta} \]

    \[ \begin{split} C &= \frac{a + i\alpha}{b+i\beta} \frac{b- i\beta}{b- i\beta} \\&= \frac{(ab+\alpha \beta ) + i(\alpha b - a \beta)}{b^2 + \beta ^2} \end{split} \]

    \[ \begin{split} C^\star &= \frac{a - i\alpha}{b-i\beta} \frac{b+ i\beta}{b+ i\beta} \\&= \frac{(ab+\alpha \beta ) - i(\alpha b - a \beta)}{b^2 + \beta ^2} \end{split} \]

Consider the steady-state forced SDOF system:

    \[ m \ddot{{x}} + c \dot{x} + kx = F_o \sin\omega t \]

    \[ m \ddot{\tilde{x}} + c \dot{\tilde{x}} + k \tilde{x} = F_o e^{i \omega t} \]

and we want to imaginary part of the solution

Try \tilde{x} = \tilde{X}e^{i\omega t}, where \tilde{X} is complex, therefore:

    \[ (-m\omega ^2 \tilde{X} +ic \omega \tilde{X} + k \tilde{X})e^{i \omega t} = F_o e^{i \omega t} (\star) \]

    \[ \tilde{X} = \frac{F_o}{(k-m \omega ^2) + ic \omega} \]

Note: (k-m\omega ^2) + ic \omega \equiv Z, Z is defined as the mechanical impedence.

Rationalizing the denominator gives:

    \[ \begin{split} \tilde{X} &= \frac{F_o [(k - m \omega ^2) - ic \omega]}{[(k - m \omega ^2) +(c \omega )^2]} \\& \equiv Xe^{-i \phi} \end{split} \]

    \[ \begin{split} X &= |\tilde{X}| \\&=\frac{F_o}{\sqrt{(k-m \omega ^2)^2 + (c \omega)^2}} \end{split} \]

    \[ \phi = \tan^{-1}{\frac{c\omega}{(k-m \omega)^2} } \]

    \[ \begin{split} \tilde{x} &=\frac{F_o}{\sqrt{(k-m \omega ^2)^2 + (c \omega)^2}} e\strut^{-i\phi} e\strut^{i \omega t} \\ &= Xe\strut^{i(\omega t - \phi)} \end{split} \]

as e\strut^{-i\phi} = \cos \phi - i\sin\phi

Then the imaginary part of the solution is:

    \[ x(t) = \frac{F_o}{\sqrt{ (k-m \omega ^2)^2 + (c \omega)^2}} \sin(\omega t - \phi) \]

We can now re-interpret equation (\star) as:

    \[ -m\omega ^2 \tilde{X}e^{i\omega t} + ic\omega \tilde{X} e^{i\omega t} + k \tilde{X}e^{i\omega t} = F_o e^{i\omega t} \]

    \[ m\omega ^2 \tilde{X}e^{i\pi}e^{i\omega t} +c \omega \tilde{X}e^{\frac{i\pi}{2}}e^{i\omega t} + k \tilde{X}e^{i \omega t} = F_o e^{i \omega t} \]

Each of the force terms above can be interpreted in their respective phases.

Many real automotive suspension systems use the so-called Macpherson strut which is a combined spring and damper along with a coil spring as shown in the diagram. The system can be represented by the diagram where the forcing function comes from road surface. (This system is sometimes called a 1 & \frac{1}{2} DOF system).

a) Using the mechanical impedance to show that the displacement x and x_1.

b) For the model shown, determine the transmissibility of the system.

c) For the frequency ratio \frac{\omega}{p} = 0.6 , \frac{k_1}{k} = 2,  \zeta = 0.4, and for \frac{\omega}{p} = 10, \frac{k_1}{k} = 2, \zeta = 0.4. Calculate for the transmissibility and compare the values to the case in which k_1 = \infty.

a)

    \[ m \ddot{{x}} + c \dot{x} + kx - c\dot{x_1}= F_o \sin\omega t --- (1) \]

    \[-c\dot{x} + c\dot{x_1} + k_1x_1 = 0 ---(2) \]

Using x(t) = \hat{X}e^{st}, x_1(t) = \hat{X_1}e^{st}

    \begin{equation*} \begin{bmatrix} ms^2+cs+k & -cs  \\ -cs & cs+k_1  \\ \end{bmatrix} \begin{Bmatrix} \hat{X}  \\ \hat{X_1}   \\ \end{Bmatrix} = \begin{Bmatrix} F  \\ 0   \\ \end{Bmatrix} ---(3) \end{equation*}

    \[ \begin{split} \hat{\Delta} &= (ms^2 +cs +k)(cs+ k_1) - c^2s^2 \\&= k_1(k+ms^2)+ cs(k+k_1+ms^2) \end{split} \]

But (3) is of the form:

    \begin{equation*} \begin{bmatrix} V_{11}(s) & V_{12}(s)  \\ V_{21}(s) & V_{22}(s)  \\ \end{bmatrix} \begin{Bmatrix} \hat{X}  \\ \hat{X_1}   \\ \end{Bmatrix} = \begin{Bmatrix} F  \\ 0   \\ \end{Bmatrix} \end{equation*}

Therefore:

    \[ \hat{X} =\frac{V_{22}(s)}{\hat{\Delta}}F \]

    \[ \hat{X_1} = \frac{V_{12}(s)}{\hat{\Delta}}F \]

If s = i \omega:

    \[ \hat{\zeta} = \frac{(k_1 + ic \omega)}{\hat{\Delta}}F \]

    \[ \hat{\zeta_1} = \frac{ic\omega F}{\hat{\Delta}} \]

    \[ \frac{\hat{\Delta}}{kk_1} = \biggr[\Big(1- \frac{\omega^2}{p^2}\Big) + \frac{i\omega c}{k}\Big(1+\frac{1}{N} - \frac{\omega^2}{p^2N}\Big)\biggr] \]

p^2 = \frac{k}{m}, N = \frac{k_1}{k}, if r = \frac{\omega}{p}, \zeta = \frac{c}{2mp}:

    \[ \begin{split} \frac{\hat{\Delta}}{kk_1} &= \biggr[(1- r^2) + i2\zeta r\Big(1+\frac{1}{N} - \frac{r^2}{N}\Big)\biggr] \\&= \biggr[(1- r^2)^2 + \Big\{ 2\zeta r\Big(1+\frac{1}{N} - \frac{r^2}{N}\Big) \Big\}\strut^2\biggr]\strut^\frac{1}{2} e\strut^{i\gamma} \\&= \hat{\Delta}e\strut^{i\gamma} \end{split} \]

    \[ \gamma = \tan^{-1}(\frac{2\zeta r(1+\frac{1}{N} - \frac{r^2}{N})}{1-r^2}) \]

If we set x= \hat{X}e\strut^{i\omega t}, x_1 = \hat{X_1}e\strut^{i\omega t},

    \[ \begin{split} \hat{X} &= \frac{F(1+\frac{i2\zeta r}{N})}{k \Delta e\strut^{i\gamma}} \\ &= \frac{F \sqrt{(1+\frac{i2\zeta r}{N})}e\strut^{i\beta}}{k \Delta e\strut^{i\gamma}} \end{split} \]

    \[ \beta = \tan^{-1} \frac{2\zeta r}{N} \]

Therefore,

    \[ \hat{X} = \frac{F \sqrt{(1+\frac{i2\zeta r}{N})}e\strut^{-i\alpha}}{k \Delta } \]

Where \alpha = \gamma - \beta

    \[ \hat{X_1} = \frac{F}{k\Delta} \frac{2 \zeta r}{N}e^{-i \alpha _1} \]

Where \alpha _1 = \gamma - \frac{\pi}{2}

b) The force transmitted is

    \[ \begin{split} F_T &= k \hat{X} + k_1 \hat{X_1} \\ &= F \frac{[k(k_1+ic \omega ) + k_1(ic \omega )]}{\hat{\Delta}} \\&= kk_1 F \frac{[1 + \frac{ic\omega }{k} + \frac{ic\omega }{k_1}]}{\hat{\Delta}}\\&= F \frac{[1+ i(2 \zeta r + \frac{2\zeta r)}{N}]}{\hat{\Delta}e\strut^{i\gamma}} \\&= F \sqrt {1+ \big[2\zeta r\big(1+ \frac{1}{N}\big)\big]\strut^2} \frac{e\strut^{i\alpha _2}}{\hat{\Delta}} \end{split} \]

    \[ \alpha _2 = \gamma - \tan^{-1}\biggr[2\zeta r\Big(1+\frac{1}{N}\Big)\biggr] \]

c) For the case r = 0.6, N = 2, \zeta = 0.4:

    \[ \left| \frac{F_T}{F} \right| = 1.37 \]

    \[ \left| \frac{F_T}{F} \right| = 1.39 (k \rightarrow \infty) \]

For the case r = 10, N = 2, \zeta = 0.4:

    \[ \left| \frac{F_T}{F} \right| = 0.03 \]

    \[ \left| \frac{F_T}{F} \right| = 0.081 (k \rightarrow \infty) \]

The added spring (k_1) reduces the transmissivity in both cases, however it is most effective at higher \frac{\omega}{p} ratio.

    \[ \left| \frac{F_T}{F} \right| = \frac{\big[1+\{2\zeta r (1+ \frac{1}{N})\}^2\big]\strut^{1/2}}{\big[(1-r^2)^2+\{2\zeta r(1+\frac{1}{N} - \frac{r^2}{N}\}^2\big]\strut^{1/2}} \]

If k_1 \rightarrow \infty, N \rightarrow \infty

    \[ \left| \frac{F_T}{F} \right| = \frac{\big[1+(2 \zeta r)^2\big]\strut^{1/2}}{\big[(1-r^2)^2+(2\zeta r)^2\big]\strut^{1/2}} \]

    \[ r = 0.6, N = 2, \zeta = 0.4 \]

    \[ \left| \frac{F_T}{F} \right| =  \frac{\big[1+\{2(0.4)(0.6)\frac{3}{2}\}^2\big]\strut^{1/2}}{\big[(1-0.36)^2+\{2(0.4)(0.6)(1+\frac{1}{2} - 0.18)\}^2\big]\strut^{1/2}} \]

    \[ = \underline{\underline{1.37}} \]

    \[ \frac{|F_T|}{F}\Bigg|_{k \rightarrow \infty} = \underline{\underline{1.39}} \]

    \[ r = 10, N = 2, \zeta = 0.4 \]

    \[ \left| \frac{F_T}{F} \right| =  \frac{\big[1+\{2(0.4)(10)\frac{3}{2}\}^2\big]\strut^{1/2}}{\big[(1-10^2)^2+\{2(0.4)(10)(\frac{3}{2} - \frac{100}{2})\}^2\big]\strut^{1/2}} \]

    \[ = \underline{\underline{0.032}} \]

    \[ \frac{|F_T|}{F}\Bigg|_{k \rightarrow \infty} = \frac{\big[1+\{2(0.4)(10)\}^2\big]\strut^{1/2}}{\big[(1-10^2)^2+\{2(0.4)(10)\}^2\big]\strut^{1/2}} \]

    \[ = \frac{8.062}{99.32} \]

    \[ = \underline{\underline{0.081}} \]

Extra Details

    \[ \begin{split} \hat{\Delta} &= (ms^2 + cs + k)(cs + k_1) - c^2s^2 \\ &= cs(ms^2+k+k_1)+c^2s^2+k_1(k+ms^2)-c^2s^2 \\ &= cs(ms^2+k+k_1) + k_1(k+ms^2) \end{split} \]

If s = i\omega:

    \[ \hat{\Delta} = [-m\omega^2 + k + k_1](ic\omega) \]

    \[ \begin{split} \frac{\hat{\Delta}}{kk_1} &= \Big[1 - \frac{\omega^2m}{k}\Big] + \frac{ic\omega}{k}\Big[\frac{-m\omega^2}{k_1} + 1 + \frac{k}{k_1}\Big] \\ &= \Big[1 - \frac{\omega^2}{p^2}\Big] + \frac{ic\omega m}{k}\Big[1 + \frac{1}{N} - \frac{r^2}{N}\Big] \end{split} \]

Where r^2 = \frac{\omega^2}{p^2}, N = \frac{k_1}{k}, and \zeta = \frac{c}{2mp}. Thus:

    \[ \begin{split} &= [1-r^2] + i2\zeta r\Big[1 + \frac{1}{N} - \frac{r^2}{N}\Big] \\ &= \biggr\{[1-r^2]^2 + \Big(2\zeta r\Big(1 + \frac{1}{N} - \frac{r^2}{N}\Big)\Big)^2\biggr\}\strut^{\frac{1}{2}}e\strut^{i\gamma} \end{split} \]

Where:

    \[ \gamma = \frac{\tan^{-1}(2\zeta r(1 + \frac{1}{N} - \frac{r^2}{N}))}{(1-r^2)} \]

If x = Xe^{i\omega t}, x_1 = X_1 e^{i\omega t}:

    \[ \begin{split} X &= \frac{\frac{F}{k}(1+\frac{2i\zeta r}{N})}{\Delta e^{i\zeta}} \\ &= \frac{\frac{F}{k}\sqrt{1 + (\frac{2\zeta r}{N})^2}}{\Delta e^{i\zeta}}e^{i\beta} \end{split} \]

    \[ \beta = \arctan{\frac{2\zeta r}{N}} \]

    \[ X = \frac{F}{k\Delta}\sqrt{1 + (\frac{2\zeta r}{N})^2}e^{i\beta}e^{-i\gamma} \]

    \[ \equiv \underline{\underline{\frac{F}{k\Delta}\sqrt{1 + (\frac{2\zeta r}{N})^2}e^{i\alpha}} } \]

    \[ \begin{split} X_1 &= \frac{ic\omega F}{\hat{\Delta}} \\&= \frac{2\zeta r}{N \hat{\Delta}} e ^{i\frac{\pi}{2}} \\&=\frac{F}{k\Delta}\frac{2\zeta r}{N}e^{-i(\gamma-\frac{\pi}{2})}\end{split} \]

Aerodynamics Excitation

A relatively common excitation which occurs in diverse situations from subsea pipelines to slender tall structures is due to vortex shedding. This shedding creates a differential pressure perpendicular to the fluid flow. For lower Reynolds numbers (Re < 300) the vortex shedding frequency is estimated to the Strouhal number S by

    \[ f_s=\frac{S\sqcup}{D} \]

where f_s is the shedding frequency, \sqcup is the velocity of the external flow and D is a characteristic dimension (e.g. diameter of the building or the pipeline). Assuming that the lift force per unit length is harmonic it can be written as

    \[ F_{L/L}=\frac{1}{2}C_L \rho\sqcup^2D\sin(\omega_st) \]

where C_L= lift coefficient, \rho = density of the fluid (mass), and \omega_s=2\pi f_s

For the case of a bluff body, the total force to the flow is modelled as

    \[ F_{L/L}=\frac{1}{2}C_L \rho A_f\sqcup^2\sin(\omega_st) \]

where A_f is the frontal area \perp to the flow. Because of the complexity of the flow there can be components of alternating forces at 1/2 the frequency estimated by the Strouhal number as well as at twice the f_s.

It should be noted that the shape of the body can also have a major influence on the effective force magnitude. Catastrophic examples include the Tacoma Narrows Bridge and the destruction of powerlines that change their X-sectional shape due to icing.

For most of these lower Reynold’s number flows

    \[ 0.18\leq S \leq 0.22 \]

3. In order to produce a relatively uniform acoustic field to the audience in an open stadium, the sound system is often suspended above the center of the playing surface. (This allows the distance from the source to the audience to be approximately the same for all fans.) This is shown diagrammatically, with typical dimensions in Figure 1. Especially in northern climates, there is the potential for the system to be covered in ice and the usual porous surfaces are replaced by an effectively solid “bluff” body that can be excited as a result of wind induced vortices being shed from the body. Alternate shedding from the upper and lower surfaces cause a pressure differential and therefore a vertical excitation of the body. The frequency f of this excitation is related to the wind velocity v and a representative vertical dimension D (taken as the total height of the system in this case, Figure 2) through the Strouhal number S given by

    \[ S = \frac{fD}{v} \]

(S = 0.22 for this outdoor situation). The concern is that the frequency of vortex shedding f could match the natural frequency and result in large even catastrophic oscillations of the sound system (google Tacoma Narrows Bridge).Add block

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Figure 1
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Figure 2
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Figure 3
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Figure 4

(a) For a cable of length L,

    \[ k_c=\frac{EA}{L} \]

therefore for the 4 cables, the k_\text{eff} for the vertical motion

    \[ k_\text{eff}=\frac{4EA}{L}\cos^2\theta \]

    \[ L=\sqrt{(110)^2+(40)^2+(16)^2}=118.1\ \text{m}\]

    \[ \cos\theta=\frac{16}{118.1}=0.1355 \]

    \[ k_\text{eff}=\frac{4\frac{\pi}{4}(1)^2 \cdot 20 \cdot 10^6(0.1355)^2}{(118.1)(39.96\ \text{ in/m})} \]

    \[ =244.4 \ \text{lbs/in} \]

    \[ P=\sqrt{\frac{k_\text{eff}\cdot386}{14,000}}=2.60 \ \text{rads/sec} \]

    \[=0.413 \ \text{Hz} \]

wind speed:

    \[ v=\frac{fD}{0.22} \]

    \[=\frac{0.413(\frac{3.6\cdot39.96}{12})}{0.22}\]

    \[=22.6ft/s\]

    \[ =15.4 \text{mph} \ \text{ (24.8km/h)} \]

(b) For low damping ratio:

    \[ \delta=\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\approx2\pi\zeta \]

Therefore:

    \[ \zeta=\frac{\delta}{2\pi} \]


    \[ \zeta=\frac{0.037}{2\pi} \]


    \[ \zeta=\underline{\underline{0.00589}} \]

(c) For the steady state SDOF damped system at \omega/P=1

    \[ X=\frac{F_0/k}{2\zeta} \]


    \[ =\frac{F_0m}{2\zeta mk} \]


    \[ =\frac{F_0}{2\zeta m P^2} \]


    \[ =\frac{F_0}{8\pi^2\zeta mf^2} \]


For resonance at the vortex shedding frequency:

    \[ f=\frac{S\sqcup}{D} \]


    \[ X=\frac{F_0D^2}{8\pi^2S^2\sqcup^2} \]


now using:

    \[ F_0=\frac{1}{2}C_L\rho A_f\sqcup^2 \]


    \[ X=\frac{C_L\rho A_fD^2}{16\pi^2S^2m\zeta} \]


at Edmonton altitude \rho\approx1.0kg/m^3

    \[ C_L\approx1.0 \]

    \[ A_f\approx(2.5)(3.6) \]

    \[ D\approx3.6 \]


Therefore:

    \[ X=\frac{(1.0)(1.0)(2.5)(3.6)^3}{16\pi^2(0.20)^2(6350)(0.0053)}=\underline{\underline{0.152\ \text{m}}} \]


Therefore total movement is just over 1m.

VIBRATION ANALYSIS AND TESTING OF THE COMMONWEALTH GAMES STADIUM SOUND SYSTEM

D.J. Wilson and M.G. Faulkner

Department of Mechanical Engineering

University Of Alberta, Edmonton, Alberta, Canada

ABSTRACT — The possible effects of wind—induced oscillations of the Commonwealth Games Stadium sound system were analyzed before construction. Tests after installation confirmed the original estimates of the system’s damping and natural frequencies.

INTRODUCTION

The Commonwealth Games Stadium sound system is one of the most sophisticated centrally suspended systems ever installed in an open—air athletic stadium. The 6350 kg (14,000 lb) speaker system is suspended above the centre of the playing field by four 110 m (360 ft) bridge strand cables. Figure 1 shows schematically the system and supporting cables. The central location was chosen because of the acoustical advantages of a frontal sound field and the lower power necessary to provide adequate sound pressure levels for all the spectators.

With the structure located 30 m (100 ft) above field level there existed the possibility of wind induced vibrations. While wind tunnel model studies are possible to estimate the magnitude of these vibrations, the relatively simple configuration lent itself to using computations alone. Calculations of the resonant frequencies of the system were combined with the methods outlined in Commentary B of the National Building Code of Canada [1] for estimating wind excitation forces.

The final design chosen has the speaker cluster enclosed by a 2.5 m (8 ft) radius cylinder 3 m (10 ft) high with a solid dome on the top and an expanded metal mesh on the sides and bottom. In addition struts (shown in Figure 2) were added for stability and to increase certain of the resonant frequencies.

DESIGN CONDITIONS AND ANALYSIS

The idealized configuration of the speaker cluster shown in Figure 2 was used for design purposes. From a structural viewpoint this meant a annular cylinder was suspended as shown by four 25 mm (1 inch) diameter bridge strand cables. Because of the symmetry, this structure was assumed to vibrate in any of six modes of vibration (see Figure 3) independently.

From an aerodynamic viewpoint the wind velocities for a 30 year return period were chosen. This is recommended by the National Building Code (NBC) because of the possible hazard to occupants of the Stadium. Based on data in Supplement #1 of the NBC the maximum one hour average wind speed for Edmonton is 90 kph (56 mph) for a 30 year return period. The most severe condition occurs when the holes in the expanded metal mesh covering the structure are open to the wind. The site exposure was taken as urban and an attempt made to account for the influence Of the spectator stands on the wind patterns.

Natural Frequencies of Vibration

The calculated natural frequencies of vibration for each of the six modes are shown in Figure 3. Initial estimates put the level of uncertainty for the translational modes at \pm 10% while the pitching and torsional modes had levels of \pm 25%. The latter uncertainty was due to the difficulty of estimating the mass moment of inertia for the system.

After initial calculations, the compression struts connecting the two easterly cables and the two westerly ones were included to provide additional torsional stiffness. Addition of these struts approximately doubled the frequencies for the torsional, pitching and rocking modes.

Wind Loads

The calculation of the wind loads was done using the procedures recommended in the NBC Commentary B. Two contributions to the wind load, gust buffeting and vortex shedding were considered.

Using the 30 year return wind speed and the wind exposure factors in the NBC gust buffeting loads were computed. Table I shows the parameters related to peak gust buffeting loads. For all wind directions the fluctuation frequency due to gust buffeting is far removed from the resonant frequency resulting in small gust buffeting loads.

Preliminary estimates indicated that vortex shedding would cause the largest deflections and cable loads. For a fixed geometric shape the maximum deflection is related to the aerodynamic and vibration parameters according to the proportionality;

Max. resonant deflection is

    \[ \alpha\frac{C_L(\text{length})^4}{(\text{weight})\cdot\zeta\cdot S_r^2} \]


where

C_L = lift coefficient

\zeta = damping ratio

S_r = Strouhal number

As seen from the above equation the damping ratio is a most important factor. In the initial design this was taken as 0.01 for vertical oscillation. This damping was assumed to be from aerodynamic sources (0.003) and an equivalent viscous damping of 0.007 from the structure. The values of the lift coefficients and Strouhal number were estimated by a literature survey. The lift coefficient evaluated in this manner should represent an upper limit since limited water channel tests of a sound system model showed no evidence of vortex shedding.

Of the six modes of vibration only vertical motion and a pitching motion about the east-west axis (see axis in Figure 3) had natural frequencies low enough to produce critical airport windspeeds for resonant vibration of less than 160 kph (100mph). Table II shows critical windspeeds equivalent maximum static loads and deflections caused by vortex shedding.

FULL SCALE TESTING

The lack of information on various system parameters caused considerable uncertainty in the design analysis. The estimates of the cable support stiffness, and the damping ratio made the resonant frequency and maximum amplitude calculations suspect. To confirm the previous estimates of these parameters a field test was carried out to directly measure the damping ratio and to confirm the previous estimates for certain of the natural frequencies.

Test Procedure

The damping ratio and natural frequencies were evaluated by using a motion picture camera to record artificially excited vibrations. The film was then projected on a motion analyzer enabling the amplitude, period of oscillation and decay rate to be measured.

The final filming was done with a Cinema Products CP-16 camera with a crystal controlled framing speed and a zoom lens. It was estimated that in conjunction with the motion analyzer the precision and resolution of amplitudes would be \pm10 mm. The crystal speed control maintained the base accuracy with a variation of only a few parts per million which is negligible for the present application.

Excitation of the system was provided by normally pulling on wires attached to the base of the sound system. One was located under the mass center and pulling on it produced nearly vertical oscillation alone. The other wire was attached at the northern edge of the system. Pulling on this wire produced a combination of pitching about the east-west axis and vertical oscillation. Amplitudes of 300 mm were easily obtained by the manual excitation. The camera recorded approximately 1 minute of decaying vibration after the excitation had ceased.

Data evaluation

The amplitude and period of decaying vibration was measured using a rear projection analyzer. A moveable cross-hair cursor for the projected image allowed a frame by frame analysis of the motion. Time periods between successive maxima were used to evaluate the natural frequency.

The damping ratio was determined by a semi-log plot of the relative amplitudes. The log decrement was evaluated using a best fit straight line (See Figure 4). The fit to a straight line verifies the assumption of equivalent viscous damping. The log decrement \delta is related to the damping ratio \zeta by

    \[ \delta=\frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \]

Using the average of two tests the damping ratio for vertical motion was found to be

    \[ \zeta = 0.0054 \]

with an estimated uncertainty of \pm 10%.

For the pitching mode, a beat formed between the vertical and pitching modes made it impossible to measure the damping in the pitching mode. The natural frequency however could be evaluated. A comparison between the measured and predicted values is shown in Table III.

CONCLUDING REMARKS

The results of the initial design analysis and the results of field testing shown in Cables i, ii, and iii show good agreement. The damping ratio originally estimated at 0.01 compares to the sum of measured values of 0.0054 and calculated aerodynamic damping of 0.0031 (total of 0.0085). The measured and calculated natural frequencies for the vertical and pitching modes are within 10-15% of those measured. Of the six modes of vibration these two are the only ones which could occur at airport wind speeds of under 160 kph.

The system has been in place nearly two years and while some deflections have been observed these have not approached the maximum predicted leading one to believe that estimates of the increased loads and deflections were quite conservative.

ACKNOWLEDGEMENTS

The authors would like to thank Ragan, Bell, McManus Consultants Ltd. for suggesting our involvement in this project. In particular we would like to mention Stan Ragan and Dan O’Brien.

REFERENCES

[1] National Building Code of Canada 1977 NRCC No. 15555 including Supplement No. 1 NRCC No. 15556 and Supplement No. 4 NRCC No. 15558

Gust Buffeting at 90 KPH Design Hourly Airport Windspeed (30 Year Return)

Wind DirectionMax 5 sec Gust on System (KPH)MPHTotal System Damping \zetaMax Gust Load kNMax Deflection mmSystem Natural Frequency HzFluctuation Frequency Hz
North or South(132)820.007311.74.01.040.24
East of West(85)530.00584.82.52.680.68

Table I

Vibration Induced by Vortex Shedding

Direction of MotionNatural Frequency HzCritical Airport Windspeed, North or South Direction (KPH)MPHCritical Airport Windspeed, East or West Direction (KPH)MPHCritical Windspeed at System (KPH)MPHMaximum Equivalent Static Load kNHalf Amplitude mm
Vertical Z0.48(63)39(135)84(53)3338795
Pitching about X-(West) Axis0.53(69)43(150)93(58)36unknownunknown

Table II

Comparison of Measured and Computed Resonant Frequencies

Direction of MotionComputed f_n, HzMeasured f_n, Hz\frac{f_n, measured}{f_n, computed}
Vertical Translation along Z axis0.420.481.14
Pitchings: Rotation about X-West axis0.490.531.08

Table III

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

Vibration and Applications of MDOF Systems

For more than a SDOF system the analysis requires a more generalized approach. However, overall it doesn’t matter if there are 2 or 22 DOFs. Consider first a special 2 DOF free vibration system:

    \[ \begin{bmatrix} m & 0 \\  0 & m \end{bmatrix} \begin{Bmatrix} \ddot{x_1} \\ \ddot{x_2} \end{Bmatrix} + \begin{bmatrix} k + k_c & -k_c \\ -k_c & k + k_c \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}\]

The simplest approach is to look for Simultaneous Simple Harmonic Motion (SSHM). That is:

    \[ \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} \sin(pt + \phi) \]

Which is a solution (not the general one) if:

    \[ \begin{bmatrix} k+k_c - mp^2 & -k_c \\ -k_c & k+k_c-mp^2 \end{bmatrix} \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} \]

This can be true if and only if (iff), \text{det}[] = 0. Therefore:

    \[ p^4-p^2\Big[\frac{(k+k_c)}{m} + \frac{(k+k_c)}{m}\Big] + \frac{k^2+2k_c k}{m^2} = 0 \]

There are two soloutions for p^2 \Rightarrow p_1^2 \ \text{and} \ p_2^2.

    \[ \begin{split} p_{1,2}^2 &= \frac{k+k_c}{m} \pm \sqrt{\frac{k^2+2kk_c + k_c^2 - (k^2 + 2kk_c)}{m^2}} \\&= \frac{k+k_c}{m} \pm \frac{k_c}{m} \\&= \frac{k}{m}, \  \frac{k+2k_c}{m} \end{split}\]

Therefore:

    \[p_1 = \sqrt{\frac{k}{m}}, \ p_2 = \sqrt{\frac{k+2k_c}{m}} \]

As a result, there are 2 frequencies at which our assumption is true. NOTE: It turns out that the general solution can be determined from these p_1, p_2, and the ratio of the amplitudes between the two masses during each of the two SSHMs. Therefore:

    \[ (k + k_c - mp^2)X_1 -k_cX_2 = 0 \]

Or:

    \[ -k_cX_1 + (k + k_c - mp^2)X_2 = 0 \]

Therefore, for either p_1^2 or p_2^2:

    \[ \begin{split} \frac{X_2}{X_1} &= \frac{k +k_c -mp^2}{k_c} \\&= \frac{k_c}{k+k_c-mp^2} \end{split} \]

If we put p_1^2 into either we get:

    \[ \Big(\frac{X_2}{X_1}\Big)_1 = \frac{1}{1} \equiv X_1 \]

And if we put p_2^2 into either we get:

    \[ \Big(\frac{X_2}{X_1}\Big)_2 = \frac{-1}{1} \equiv X_2 \]

Thus, we define the mode shapes corresponding to each ratio:

    \[ \{u\}_1 = \begin{Bmatrix} 1 \\ X_1 \end{Bmatrix} \]

    \[ \{u\}_2 = \begin{Bmatrix} 1 \\ X_2 \end{Bmatrix} \]

Therefore, we have a mode shape corresponding to each of the natural frequencies. \{u\}_1 and \{u\}_2 are “normalized” vectors where the first component is set arbitrarily to unity. The general solution to the free vibration problem is then:

    \[ \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = c_1 \{u\}_1 \sin(p_1t +\phi_1) + c_2\{u\}_2 \sin(p_2t+\phi_2) \]

There are 4 constants c_1, c_2, \phi_1, \phi_2 for the 4 initial conditions for x_1, x_2, \dot{x_1}, \dot{x_2}. If the initial conditions are selected then the solution will include both modes in general but only one mode for certain cases. For example, when x_1(0) = 1, x_2(0) = 1, \dot{x}_1(0) = 0, and \dot{x}_2(0) = 0, \Rightarrow mode 1!

Consider the following initial conditions:

    \[x_1(0) = 1,\ x_2(0) = 1,\ \dot{x}_1(0) = 0,\ \dot{x}_2(0) = 0\]

    \[x_1(t) = c_1\sin(p_1t+\phi_1) + c_2\sin(p_2t+\phi_2) \]

    \[x_2(t) = c_1\sin(p_1t+\phi_1) - c_2\sin(p_2t+\phi_2) \]

At t = 0:

    \[1 = c_1\sin\phi_1 + c_2\sin\phi_2 \quad (1) \]

    \[0 = c_1\sin\phi_1 - c_2\sin\phi_2 \quad (2) \]

    \[0 = p_1\cos\phi_1 + p_2\cos\phi_2 \quad (3) \]

    \[0 = p_1\cos\phi_1 -  p_2\cos\phi_2 \quad (4) \]

Therefore:

    \[ \cos\phi_1 = 0 \quad \text{(sum (3) + (4))} \]

    \[ \cos\phi_2 = 0 \quad \text{((3) - (4))} \]

Therefore:

    \[ \phi_1 = \frac{\pi}{2},\ \phi_2 = \frac{\pi}{2}\]

From (1) and (2) (+):

    \[2c_1\sin\phi_1 = 1\]

    \[c_1 = \frac{1}{2}\]

From (1) and (2) (-):

    \[c_2 = \frac{1}{2}\]

Therefore:

    \[\begin{split} x_1 &= \frac{1}{2}\sin\Big(p_1t + \frac{\pi}{2}\Big) + \frac{1}{2}\sin\Big(p_2t + \frac{\pi}{2}\Big) \\&= \frac{1}{2}\cos p_1t + \frac{1}{2}\cos p_2t \end{split} \]

    \[x_2 = \frac{1}{2}\cos p_1t - \frac{1}{2}\cos p_2t \]

Using identities:

    \[x_1 = \Big[\cos(p_2-p_1)\frac{t}{2} \bullet \cos(p_1+p_2)\frac{t}{2}\Big] \]

    \[x_2 = \Big[\sin(p_2-p_1)\frac{t}{2} \bullet \sin(p_2+p_1)\frac{t}{2}\Big] \]

These may be interpreted as a higher frequency oscillation at the average of the two natural frequencies with a variable amplitude, given by a lower frequency given by \frac{1}{2} the difference in natural frequencies.

If p_1 and p_2 are close to each relative to their magnitude then (p_2-p_1) << p_1, the motion becomes:

Stiffness and Flexibility Influence Coefficients

We can use the Newtonian approach to find the stiffness matrix and equations of motion. However, we can also use influence coefficients for both the mass and stiffness matrices.

Consider the definition of stiffness and flexibility coefficients:

S.I.C

k_{ij} is the force (moment) required at coordinate i to maintain a unit linear (angular) displacement at coordinate j with all other coordinates held fixed at zero

F.I.C

a_{ij} is the linear (angular) displacement at coordinate i due to a unit force (moment) applied at coordinate j with all other coordinates free to move.

NOTE: All forces (moments) and displacements (rotations) are taken to be applied in the positive direction. The sign of the resulting forces (moments) or displacements (rotations) are determined to be positive or negative.

Also note that:

    \[ [a_{ij}]^{-1} = [k_{ij}] \]

Example

Stiffness

Therefore:

    \[k_{11} = \frac{2T}{L},\ k_{21} = \frac{-T}{L},\ k_{31} = 0 \]

    \[k_{22} = \frac{2T}{L},\ k_{12} = k_{32} = \frac{-T}{L} \]

Therefore:

    \[ \frac{T}{L} \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix} = [k] \]

Flexibilty

    \[  T\Big[\frac{a_{11}}{L} + \frac{a_{11}}{3L}\Big] = 1 \]

    \[ T\Big[\frac{4a_{11}}{3L}\Big] = 1\]

    \[a_{11} = \frac{3L}{4T} = a_{33}\]

    \[a_{21} = \frac{2}{3}a_{11} = \frac{L}{2T} \]

    \[a_{31} = \frac{1}{3}a_{11} = \frac{L}{4T} \]

    \[1 = 2T\frac{a_{22}}{2L}\]

Therefore:

    \[1 = T\frac{a_{22}}{L}\]

    \[a_{22} = \frac{L}{T}\]

    \[a_{12} = \frac{L}{2T} = a_{32}\]

Thus:

    \[\frac{L}{T}\begin{bmatrix} \frac{3}{4} & \frac{1}{2} & \frac{1}{4} \\[2ex] \frac{1}{2} & 1 & \frac{1}{2} \\[2ex] \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{bmatrix} = [a_{ij}] \]

Therefore:

    \[ \begin{split} [a][k] &= \begin{bmatrix} \frac{3}{4} & \frac{1}{2} & \frac{1}{4} \\[2ex] \frac{1}{2} & 1 & \frac{1}{2} \\[2ex] \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{bmatrix} \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix} \\&= \begin{bmatrix} \frac{3}{2} - \frac{1}{2} & -\frac{3}{4} + 1 -\frac{1}{4} & -\frac{1}{2} + \frac{1}{2} \\[2ex] 1-1 & -\frac{1}{2} + 2 - \frac{1}{2} & -1+1 \\[2ex] \frac{1}{2} - \frac{1}{2} &  -\frac{1}{4} + 1 - \frac{3}{4} & -\frac{1}{2} + \frac{3}{2} \end{bmatrix} \\&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{split}\]

As an example of MDOF, consider a general planar situation which can be specialized to different industrial situations. Many of these problems can be initially modelled as SDOF ones, and then MDOF analysis will allow interpretations of the simplifying assumptions made.

NOTE: While the body shown is a rectangular shape, it can be any rigid body with a mass m and moment of inertia J_o about an axis \perp to the plane through the center of gravity. The horizontal stiffness components, designated by \ell_i are most often just the lateral stiffness of the vertically mounted spring. For real springs, the lateral spring may be hard to find in the literature. For coil springs, it is often in the range of 0.8-1.2 times the longitudinal value.

    \[\begin{split} +\rightarrow \sum F_x &= m\ddot{x} \\&= -\ell_1(x + h\theta) - \ell_2(x+h\theta) \quad (1) \end{split} \]

    \[\begin{split} +\uparrow \sum F_y &= m\ddot{y} \\&= -k_1(y-a\theta) - k_2(y + b\theta) \quad (2) \end{split} \]

    \[+\circlearrowleft \sum M_G = J_G\ddot{\theta} = -\ell_1(x+h\theta)h - \ell_2(x+h\theta)h + k_1(y-a\theta) - k_2(y + b\theta)b \quad (3) \]

    \[m\ddot{x} + \ell_1x + \ell_1h\theta + \ell_2x + \ell_2h\theta = 0 \quad (1)\]

    \[m\ddot{y} + k_1y - k_1a\theta + k_2y + k_2b\theta = 0 \quad (2) \]

    \[J_G\ddot{\theta} + \ell_1xh + \ell_1h^2\theta + \ell_2xh + \ell_2h^2\theta - k_1ay + k_1a^2\theta + k_2by + k_2b^2\theta = 0 \quad (3) \]

In matrix form:

    \[ \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} = \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & J_G \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} \ell_1 + \ell_2 & 0 & (\ell_1+\ell_2)h \\ 0 & k_1 + k_2 & k_2b - k_1a \\ (\ell_1+\ell_2)h & k_2b-k_1a & h^2(\ell_1+\ell_2) + k_1a^2 +k_2b^2 \end{bmatrix} \begin{Bmatrix} x \\ y \\ z \end{Bmatrix} \]

Part 2

Now find the stiffness matrix using influence coefficients.

First consider a unit displacement in the x direction only:

    \[\begin{split} + \rightarrow\sum F_x &= 0 \\& = k_{11} -  (\ell_1 + \ell_2)\end{split}\]

    \[\implies k_{11} = \ell_1 + \ell_2\]

    \[\begin{split} +\circlearrowleft\sum M_G &= 0 \\& = k_{13} - (\ell_1+\ell_2)h \end{split} \]

    \[\implies k_{13} = (\ell_1 + \ell_2)h\]

For the y direction unit displacement:

    \[ \begin{split} +\uparrow\sum F_y & = 0 \\& = k_{22} - (k_1 + k_2) \end{split} \]

    \[\implies k_{22} = k_1 + k_2 \]

    \[\begin{split} + \circlearrowleft \sum M_G &= 0 \\& = k_{23} + k_1a - k_2b \end{split}\]

    \[\implies k_{23} = k_2b-k1_a\]

    \[+\rightarrow \sum F_x = 0\]

    \[\implies k_{21} = 0\]

Assuming all summations for x direction, y direction and moment equal to 0:

    \[+\rightarrow \sum F_x = k_{31} - \ell_1h - \ell_2h \]

    \[\implies k_{31} = (\ell_1 + \ell_2)h \]

    \[+\uparrow \sum F_y = k_{32} + k_1a-k_2b\]

    \[ \implies k_{32} = k_2b - k_1a\]

    \[+\circlearrowleft \sum M_G = k_{33} -(\ell_1 + \ell_2)h^2 - k_1a^2 - k_2b^2\]

Therefore:

    \[k_{33} = (\ell_1+\ell_2)h^2 + k_1a^2 + k_2b^2\]

This yields the same stiffness matrix.

As long as we use the motions about the center of gravity then the inertia(mass) matrix is just the mass on the mass moment of inertia about the center of gravity. If we use the coordinates to decide the motion that is not related to the absolute motion of the center of gravity then the stiffness matrix may not be symmetric and the mass matrix will not be diagonal.

Inertia Influence Coefficients

The elements of the mass matrix, m_ij can be determined through the use of inertia influence coefficients. They are defined as the set of impulses applied at the points (representing the coordinate directions) 1, 2, \ldots n respectively to produce a unit velocity at point j and zero velocity at every other point.

NOTE: If x_j denotes an angular coordinate then \dot{x}_j represents an angular velocity

Thus for a multi-degree of freedom system, the total impulses at points i is:

    \[\tilde{F}_i = \sum_{j = 1}^n m_{ij}\dot{x}_j\]

    \[\{\tilde{F}\} = [m]\{\dot{x}_i\}\]

    \[[m] = \begin{bmatrix} m_1 & \ldots &m_{1n} \\ \vdots & & \vdots \\ m_{n1} & \ldots & m_{nn} \end{bmatrix}\]

    \[\tilde{F} = \begin{Bmatrix} F_1 \\ \vdots \\ F_n \end{Bmatrix} \]

    \[\{\dot{x}\} = \begin{Bmatrix} \dot{x}_1 \\ \vdots \\ \dot{x}_n \end{Bmatrix} \]

To find the m_{ij}

  1. Assume that a set of impulses (f_{ij}) are applied at all points i = 1, 2, \ldots, n as to produce a unit velocity at point j only(with zero velocity in all the after dierections(coordinates)). By definition the set of impulses f_{ij} denote the influence inertia coefficients m_{ij}
  2. Repeat the procedure for each point j = 1, 2, \ldots, n.
Example

First apply unit velocity to x only:

    \[ \begin{split} +\rightarrow \sum L &= f_{11} \\&= (M+m) \\&= m_{11} \end{split}\]

    \[\begin{split} +\circlearrowleft \sum H_0 &= f_{21} \\&= m\frac{L}{2} \\& = m_{21}\end{split}\]

Apply unit velocity(angular) to \theta only

    \[\begin{split} +\rightarrow \sum L &= f_{12} \\&= \frac{mL}{2} \\&= m_{12}\end{split}\]

    \[ \begin{split} +\circlearrowleft \sum H_0 &= J_G(1) + \frac{mL}{2}(1)\frac{L}{2} \\&=  \frac{1}{12}mL^2+\frac{mL^2}{4} \\&= \frac{mL^2}{3} \\&=m_{22}\end{split} \]

Therefore:

    \[\begin{bmatrix} M + m & \frac{mL}{2} \\ \frac{mL}{2} & \frac{mL^3}{3} \end{bmatrix}\]

Now find the stiffness matrix

Apply unit deflection to x only:

Therefore

    \[k_{11} = k\]

    \[k_{21} = 0\]

Apply unit deflection to \theta only:

    \[ \begin{split} +\circlearrowleft \sum M_0 &= k_22 - mg\frac{\ell}{2}(1) \\&= 0\end{split}\]

    \[ k_{22} = mg\frac{\ell}{2}\]

    \[ k_{12} = 0\]

Therefore:

    \[\begin{bmatrix} M + m & \frac{mL}{2} \\ \frac{mL}{2} & \frac{mL^3}{3} \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} k_1 & 0 \\0 & \frac{mgL}{2} \end{bmatrix}\begin{Bmatrix} x \\\theta \end{Bmatrix}= \begin{Bmatrix} 0 \\0 \end{Bmatrix}\]

Applications of MDOF systems

The concept of modes and their corresponding natural frequencies is useful in understanding the behavior of lightly damped systems. Before adding the complication of damping to our models, we will consider several applications to industrial situations.

Example

The model of the automobile shown is used to predict the low frequency modes of vibrations as these are responsible for the low frequency noise in the passenger compartment. Knowing these models can assist the designer in controlling them. In addition, it helps the designer in predicting the response of the frame to various inputs from road irregularities. As an initial estimate, the three modes of vibration of the vehicle may be considered as uncoupled. For the specific case below, the assumption of uncoupled modes is to be compared to the coupled model.

a) For the longitudinal model of the vehicle, specialize the equations of motion for the case of h = 0
b) If any of the modes are uncoupled determine an expression of the natural frequency. For the coupled modes derive the equations of motion for free vibration starting from Newton’s laws.
c) Calculate the natural frequencies and mode shapes for the two coupled modes of vibration. The weight of the vehicle is 3550 \ \text{lbs}.\  k_1 = 2200 \ \text{lbs/ft}, k_2 = 2700\ \text{lbs/ft}, a = 4.5\ \text{ft}, b = 5.5\ \text{ft} and the radius of gyration on the vehicle about G is 4\ \text{ft}.

Solution

a)

    \[ \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & J_G \end{bmatrix}\begin{Bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} \ell_1 + \ell_2 & 0 & 0 \\ 0 & k_1 + k_2 & k_2b - k_1a \\ 0 & k_2b-k_1a & k_1a^2 + k_2b^2 \end{bmatrix}\begin{Bmatrix} x \\ y \\ \theta \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} (+)\]

b)

For horizontal motion:

    \[ m\ddot{x} + (\ell_1 + \ell_2)x = 0 \]

Therefore:

    \[ p_H = \sqrt{\frac{\ell_1 + \ell_2}{m}} \]

    \[ \begin{split} + \uparrow \sum F_y &= m\ddot{y} \\&= 0 \\&= -k_2(y + b\theta) - k_1(y-a\theta)  m\ddot{y} + (k_1 + k_2)y + (k_2b+k_1a)\theta \quad (I) \end{split}\]

    \[ \begin{split}+\circlearrowleft \sum M_G &= J_G\ddot{\theta} \\& = 0 \\& = k_1a(y - a\theta) - k_2b(y+b\theta) J_G\ddot{\theta} + (k_2b-k_1a)y + k_1a^2+k_2b^2)\theta \quad (II) \end{split}\]

(I) & (II) are the same as (\dagger).

c)

    \[ \omega = 3550 \ \text{lbs} \]

    \[a = 4.5\ \text{ft}\]

    \[b = 5.5 \ \text{ft}\]

    \[\bar{r} = 4 \ \text{ft}\]

    \[k_1 = 2200 \ \text{lb/ft}\]

    \[k_2 = 2700 \ \text{lb/ft}\]

Therefore:

    \[\begin{bmatrix} m & 0 \\ 0 & J_G \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} k_1 + k_2 & k_2b + k_1a \\ k_2b - k_1a & k_1a^2 +k_2b^2 \end{bmatrix} \begin{Bmatrix} x \\ \theta \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} \]

Assume \begin{Bmatrix} x \\ \theta \end{Bmatrix} = \begin{Bmatrix} X \\ \Theta \end{Bmatrix}\sin{(pt + \phi)}:

    \[\begin{bmatrix} (k_1 + k_2) - mp^2 & k_2b-k_1a \\ k_2b - k_1a & k_1a^2 + k_2b^2 - J_Gp^2 \end{bmatrix} \begin{Bmatrix} X \\ \Theta \end{Bmatrix} = 0 \]

    \[ [k_1 + k_2 -mp^2][k_1a^2+k_2b^2-J_Gp^2] - (k_2b - k_1a)^2 = 0 \]

    \[\begin{split}(k_1+k_2)(k_1a^2+k_2b^2)-p^2\big[J_G(k_1+k_2)+(k_1a^2+k_2b^2)m\big]+mJ_Gp^4-(k_2b-k_1a)^2 &=0 \\ mJ_Gp^4-p^2\big[\vec{r}^2(k_1+k_2)+(k_1a^2+k_2b^2)m\big]+[(k_1+k_2)(k_1a^2+k_2b^2)]-(k_2b-k_1a)^2 &=0 \\ p^4-p^2\bigg[\frac{k_1+k_2}{m}+\frac{(k_1a^2+k_2b^2)}{m\vec{r}^2}\bigg]+\bigg(\frac{k_1+k_2}{m}\bigg)\bigg(\frac{k_1a^2+k_2b^2}{m\vec{r}^2}-\frac{(k_2b-k_1a)^2}{mm\vec{r}^2}\bigg) &=0 \\ p^4-p^2[44.45+74.99]+(44.45)(74.99)-126.0 &=0 \\ p^4-119.44p^2+3207 &=0 \end{split} \]

    \[\begin{split} p^2 &=\frac{119.44}{2}\pm \sqrt{(\frac{119.44}{2})^2-3207} \\ &=59.7 \pm 18.96 \end{split} \]

    \[\begin{split} p^2_1 &=59.7-18.96 \\ p_1^2 &=40.74 \\ p_2^2 &=59.7+18.96 \\ p_2^2 &=78.65 \\ p_{1,2} &= 6.38 \ \text{rads/s}, 8.87 \ \text{rads/s} \\ &= \underline{\underline{1.01 \ \text{Hz}}}, \underline{\underline{1.41 \ \text{Hz}}} \end{split}\]

    \[(k_1+k_2-mp^2)X + (k_2b-k_1a)\Theta = 0 \]

    \[\begin{split} \frac{X}{\Theta} &= \frac{(k_1a-k_2b)}{k_1+k_2-mp^2} \\ \Big(\frac{X}{\Theta}\Big)_1 &= \frac{(k_1a-k_2b)/m}{\frac{(k_1+k_2)}{m}-p^2} \\ &= \frac{[(2200)(45)-2700(5.5)]\frac{32.2}{3550}}{(44.45)-40.74} \\ &= \underline{\underline{-12.1 \ \text{ft/rad}}} \\ \Big(\frac{X}{\Theta}\Big)_2 &= \frac{-44.90}{44.45-78.65} \\ &= \underline{\underline{1.31}} \end{split}\]

[DIAGRAM]

If we “uncoupled” the modes

Assume vertical node is

    \[\begin{split} p_1 &= \sqrt{\frac{k_1+k_2}{m}} \\ &= \sqrt{\frac{4900*32.3}{3550}} \\ &= 6.67 \ \text{rads/s} \\ &= \underline{\underline{1.06 \ \text{Hz}}} \end{split}\]

Rotational Mode

    \[\begin{split} p_2 &=\sqrt{\frac{k_1a^2+k_2b^2}{m\vec{r}^2}} \\ &= \sqrt{\frac{[(2200)(4.5)^2+(2700)(5.5)^2]32.2}{3550(4)^2}} \\ &= 8.46  \ \text{rads/s} \\ &= \underline{\underline{1.35 \ \text{Hz}}} \end{split}\]

If the horizontal stiffness, l_1, l_2, were the same as the vertical then the third frequency, p_H, would be the same as the vertical.

The assumption of uncoupling gives essentially the same answer.

The illustration shows the detail of the proposed isolation system for the reciprocating compressor and motor considered previously and modelled as a single dof system. The actual system does not have the center of gravity symmetrically located on the supporting structure and isolation springs. In situations like these the industrial installation guides recommend that the isolation springs be selected so that the vertical static deflection of all springs is the same. The reason stated is that this will tend to reduce the frequency of the “rocking motions”. This exercise is to investigate this idea as well as the differences found for the isolation of the system compared with the single degree approximation used earlier. More details of the installation are given below. The stiffnesses shown are representative of both springs at the ends of the supports with the lateral stiffnesses assumed to be equal to the vertical values.

a) If the vertical static deflection of the spring is the same what is the ratio of k_2 and k_1 ?

    \[ S_1 = \frac{F_1}{k_1}, \ S_2=\frac{F_2}{k_2} \]

For static equilibrium

    \[\begin{split} F_1a &=F_2b \\ \frac{F_1}{k_1} &= \frac{F_1(\frac{a}{b})}{k_2} \end{split} \]

therefore:

    \[\begin{split} \frac{k_2}{k_1} &=\frac{a}{b} \\ k_2b &= k_1a \end{split} \]

(b) Under this assumption what happens to the equations of motion?

    \[= \begin{bmatrix} m & 0 & 0 \\0 & m & 0 \\ 0 & 0 & m\vec{r}^2 \end{bmatrix}\begin{Bmatrix}\Ddot{x} \\ \Ddot{y} \\ \Ddot{\theta}\end{Bmatrix} \begin{bmatrix}\ell_1 + \ell_2 & 0 & (\ell_1+\ell_2)h \\0 & k_1+k_2 & 0 \\(\ell_1+\ell_2)h & 0 & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2\end{bmatrix}\begin{Bmatrix}x \\ y \\ \theta\end{Bmatrix}\]

as the vertical mode is uncoupled from the other two

    \[ p_v = \sqrt{\frac{k_1+k_2}{m}} \]


becomes

    \[\begin{bmatrix}m & 0 \\ 0 & mr^2\end{bmatrix}\begin{Bmatrix}\Ddot{x} \\ \Ddot{\theta}\end{Bmatrix} \begin{bmatrix}\ell_1+\ell_2 & (\ell_1+\ell_2)h \\(\ell_1+\ell_2)h & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2\end{bmatrix}\begin{Bmatrix}x \\ \theta\end{Bmatrix}=\begin{Bmatrix}0\end{Bmatrix}\]


this uncoupling solves the natural frequencies

(c) Noting that the original had an operating speed of 1750 RPM and the original transmissibility was 1/15, calculate k_1+k_2 for this installation assuming k_1a=k_2b

    \[ \frac{1}{(\frac{\omega}{p_v})^2-1}=\frac{1}{15} \]


therefore:

    \[ \begin{split} \frac{\omega}{p_v} &=4 \\ p_v &=\frac{1750}{4} \\ &=437.5 \ \text{RPM} \\ &= \frac{437.5*2\pi}{60} \\ &=45.8 \ \text{rads/s} \\ &=7.3 \ \text{Hz} \end{split} \]

therefore:

    \[\begin{split} \sqrt{\frac{(k_1+k_2)32.2}{4600}} &=45.8 \\ k_1 + k_2 &= (45.8)^2\frac{4600}{32.2} \\ &=300,000 \text{ lb/ft} \end{split}\]

therefore:

    \[\begin{split}k_1+k_1\frac{a}{b} &=k_1\Big(1+\frac{3.5}{6.9}\Big) \\ &=300,000 \text{ lbs/ft} \end{split}\]

    \[k_1=\frac{300,000}{1.65217}=181,600 \ \text{lbs/ft}\]

    \[k_2 = 0.65217\cdot181,600 = \ \underline{\underline{\text{118,400 lbs/ft}}}\]

(d) Assuming SSHM

    \[\begin{Bmatrix}x \\ \theta\end{Bmatrix}= \begin{bmatrix}X \\ \Theta\end{bmatrix}\sin(pt+\phi)\]


    \[ \begin{bmatrix}\ell_1+\ell_2-mp^2 & (\ell_1+\ell_2)h \\(\ell_1+\ell_2)h & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2-mr^2p^2\end{bmatrix}\begin{Bmatrix}X \\ \Theta\end{Bmatrix}= \begin{Bmatrix}0\end{Bmatrix}\]


therefore:

    \[ \Big[\ell_1+\ell_2-mp^2\Big]\Big[h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2-mr^2p^2\Big]-(\ell_1+\ell_2)^2h^2 = 0\]


Divide by m^2r^2 to get

    \[ \biggr[\frac{\ell_1+\ell_2}{m}-p^2\biggr]\biggr[\frac{(\ell_1+\ell_2)}{m}\frac{h^2}{r^2}+\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}-p^2\biggr]-\frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} = 0\]


therefore:

    \[\begin{split}p^4 &-p^2\biggr[\frac{\ell_1+\ell_2}{m}+\frac{\ell_1+\ell_2}{m}\frac{h^2}{r^2}+\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}\biggr] \\ &+ \frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} + \frac{(\ell_1+\ell_2)}{m}(\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}) - \frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} = 0 \end{split}\]

    \[p^4-p^2\biggr[\Big(\frac{\ell_1+\ell_2}{m}\Big)\Big(1+\frac{h^2}{r^2}\Big)+\frac{(k_1a^2+k_2b^2)}{mr^2}\biggr] + \frac{(\ell_1+\ell_2)(k_1a^2+k_2b^2)}{m^2r^2} = 0\]

    \[ \begin{split}p^4 &-p^2\biggr[\frac{(300,000)}{4600}\Big(1+\Big(\frac{0.8}{2.1}\Big)\strut^2\Big)32.2+\frac{181,600}{4600}\Big(\frac{3.5}{2.1}\Big)\strut^2\Big(32.2\Big)+\frac{118,400}{4600}\Big(\frac{6.9}{2.1}\Big)\strut^2 32.2\biggr] \\ &+ \frac{300,000}{(4600)^2}\biggr[181,600\Big(\frac{3.5}{2.1}\Big)\strut^2+118,400\Big(\frac{6.9}{2.1}\Big)\strut^2\biggr](32.2)^2 = 0\end{split} \]

    \[ p^4-p^2\big[2404.8+3589+8948\big] + 26205413 = 0 \]


    \[ \begin{split} p^2 &= 7470 \pm \sqrt{(7470)^2 - 26205413} \\ p^2 &= 7470 \pm 5429 \end{split} \]


    \[ p_1 = 45.2 \ \text{rads/s (7.2 Hz)} \]


    \[ p_2 = 113.6 \ \text{rad/s (18.1 Hz)} \]

(e)

    \[ \begin{split} p_v &= 45.8 \ \text{rads/s (7.3 Hz)} \\ p_H &= 45.8 \text{rads/s (7.3 Hz)} \end{split} \]

The rotational mode can be calculated starting from a FBD

    \[\begin{split} \Sigma M_G &= m\vec{r}^2\Ddot{\theta} \\ &= -ka^2\theta - ka^2\theta \end{split} \]


therefore:

    \[ \begin{split} P^2_{\theta} &= \frac{2ka^2}{mr^2} \\ &= \frac{300,000(5.2)^2(32.2)}{4600(2.1)^2} \end{split}\]


    \[ \begin{split} P_{\theta} &= 113.5 \ \text{rads/s} \\ &= (18.1 \ \text{Hz}) \end{split} \]

(f) Do the calculation using a program

(g) From d)

    \[ \begin{split} p_v &=7.3 \ \text{Hz} \\ (p_H)_1 &= 7.2 \ \text{Hz} \\ (p_{\theta})_2 &= 18.1  \ \text{Hz} \end{split} \]

from e)

    \[ \begin{split} p_v &= 7.3 Hz \\ p_H &= 7.3 Hz \\ p_{\theta} &= 18.1 Hz \end{split} \]


from (f)

    \[ \begin{split} (p)_{v_1} &= 1 ? \\ (p_H)_2 &= ? \\ (p_{\theta})_3 &= ? \end{split} \]

The results show that because h is small compared to a and b that there is little difference. The “rotational” (p_{\theta}, p_3) is very much higher than the other two and closer to the excitation frequency of 29.2 Hz (1750 RPM).

To lower it, the spacing of the supports would have to be reduced. For example if it were brought to 1/2 of original (5.1′ instead of a total of 10.4′) then the rotational natural frequency would be

    \[ p_{\theta} = \sqrt{\frac{300,000(2.6)^2(32.2)}{4600(2.1)^2}} = 56.7 \ \text{rads/s} \]


It is obvious from this calculation that the rule of thumb should be that the spacing should be no more than twice the radius of gyration of the total mass!

A commercial front loading washing machine is being used by a spa located adjacent to a medical clinic in a strip mall. During the spin cycle of the washing machine the fluctuating force due to the imbalance of the load causes the floor to vibrate and induces unacceptable levels of noise in the clinic. To reduce this disturbance, an engineering consultant recommends using spring isolation between the washing machine and the floor to reduce the dynamic force transmitted to 20% of the exciting force. The specifications of the washing machine are summarized below. Assume only vertical vibration and neglect damping.

Net weight1049 lbs
Maximum load255lbs
Maximum Dynamic Force212 lbs
Spin Speed (Frequency of Dynamic Force)950 RPM (15.8 Hz)
Cylinder Diameter27.6 inch


a) Calculate the appropriate spring constant for each of four springs used to mount the washing machine to meet the specification for the transmissibility and select them from the enclosed table.

b) This washing machine can also be run at a spin speed of 750 RPM. Calculate the Dynamic Force at this RPM and then calculate the force transmitted to the supporting structure compared with the force transmitted at 950 RPM.

c) Using the size of the rotating wash cylinder estimate the unbalanced weight in the maximum load.

Mount
No.
Mount
Constant
(lb/in.)
Load at
2-in. Deflection
(lb)
SLF-40150100
SLF-40270140
SLF-40390180
SLF-404115230
SLF-405150300
SLF-406200400
SLF-407305610
SLF-408365730
SLF-4095001,000
SLF-4106501,300
SLF-4117401,880
SLF-4121,1502,300
SLF-4131,5003,000
SLF-4142,0004,000
SLF-4152,6805,360
SLF-4163,5007,000
SLF-4174,7509,500
SLF-4186,25012,500
SLF-4198,20016,400
SLF-42011,25022,500
SLF-42115,00030,000
SLF-42220,00040,000

a)

    \[M = \frac{1049+255}{386}\]

    \[\begin{split} T &= 0.2 \\&= \frac{1}{(\frac{\omega}{p})^2-1} \end{split}\]

Therefore,

    \[\frac{\omega^{2}}{p^{2}} = 5\]

    \[\begin{split} \frac{\omega}{p} &= \sqrt{6} \\&= 2.44 \end{split}\]

    \[\begin{split} p &= \frac{\omega}{\sqrt{6}} \\&= \frac{950\cdot(2 \pi)}{60 \sqrt{6}} \\&= 40.61 \ \text{rad/s} \end{split}\]

    \[\begin{split} k &= p^{2}M = (40.61)^{2} \\&= (40.61)^{2}\Big(\frac{1304}{386}\Big) \\&= 5571.3 \ \text{lb/in} \end{split}\]

Therefore, each spring has

    \[\begin{split} k &= \frac{5571.3}{4}  \\&= 1393 \ \text{lbs/in} \end{split}\]

Therefore, SLF 412 and 413 would not meet the transmissibility spec.

b)

    \[\omega_{950} = \sqrt{6}p\]

    \[\omega_{750} = \sqrt{6}p \Big(\frac{750}{950}\Big)\]

    \[\frac{\omega}{p}_{750} = \sqrt{6}\Big(\frac{750}{950}\Big)\]

    \[\begin{split} T_{750} &= \frac{1}{6\big(\frac{75}{95}\big)^{2}-1} \\&= 0.365 \end{split}\]

    \[T_{950} = 0.2\]

    \[F_{o-950} = 212\ \text{lbs}\]

    \[\begin{split} F_{o-750} &= 212\Big(\frac{75}{96}\Big)^{2} \\&= 132\ \text{lbs} \end{split}\]

    \[\begin{split} F_{T-950} &= 212\cdot0.2 \\&= 42 \ \text{lbs} \end{split}\]

    \[\begin{split} F_{T-750} &= 132\cdot0.365 \\&= 48 \ \text{lbs} \end{split}\]

Therefore slightly higher than at 950 RPM

c)

    \[\begin{split} \text{Dynamic Force} &= \tilde{m}e\omega^{2} \\&= \frac{\omega}{g}e\omega^{2} \\&= \frac{\omega_{im}}{386}(13.8)(15.2\cdot2\pi)^{2} \\&= 212 \ \text{lbs} \end{split}\]

    \[\begin{split} \omega_{im} &= \frac{(212)(386)}{(13.8)(15.2*2\pi)^{2}} \\&= 0.60 \ \text{lbs} \end{split}\]

Perhaps a wet hand towel out of balance

The commercial dryer was considered earlier under the assumption that the vibration was only a vertical mode. However the excitation die to rotating imbalance would cause excitation both horizontally and vertically that would excite coupled modes as well. As a result it is necessary to consider all the planar modes of vibration to investigate if they are near the excitation frequency. The diagram of the dryer is shown below. The lateral stiffness of the supporting springs is estimated to be 0.8 of the vertical stiffness. The excitation frequency is 950 rpm.

    \[W = 1049+255 \ \text{lbs}\]

    \[\bar{r} = 15 \text{ in (radius of gyration)}\]

    \[k_1 = k_2 = k = 1150 \ \text{lbs/in}\]

    \[\ell_1 = \ell_2 = 0.8k \text{ (4 springs)}\]

    \[h = 28 \ \text{in}\]

    \[a = 19 \ \text{in}\]

a) Specialize the equations of motion for the case shown in the above diagram

b) Determine if any of the modes are uncoupled. For the coupled modes derive the equations of motion for free vibration starting from Newton’s laws.

c) Calculate the natural frequencies and mode shapes for the dryer.

d) If the supports for the dryer were redesigned to make loading easier with h = 0 (as illustrated) what are the natural frequencies compared to the original design. How do they compare with the excitation frequency? 

a)

    \begin{equation*} \begin{bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & J_G\\ \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{\theta} \\ \end{Bmatrix} + \begin{bmatrix} 2\ell & 0 & 2\ell h\\ 0 & 2k & 0\\ 2\ell h & 0 & 2\ell h^{2}+2ka^{2}\\ \end{bmatrix} \begin{Bmatrix} x\\ y \\ \theta \\ \end{Bmatrix} = \begin{Bmatrix} 0\\ 0 \\ 0 \\ \end{Bmatrix} \end{equation*}

b) Therefore m\ddot{y} + 2ky = 0,p_v= \sqrt{\frac{2k}{m}}

The vertical mode is uncoupled

    \[\begin{split} \underrightarrow{+} \sum F_x &= m\ddot{x} \\&= -4\ell(x+h\theta) \end{split}\]

    \[\begin{split} + \circlearrowleft \sum M_G &= J_G \ddot{\theta} \\&= -4\ell(x+h\theta)h-2k(a\theta)a - 2k(a\theta)a \end{split}\]

Therefore

    \[m\ddot{x} + 4kx + 4\ell h\theta = 0\]

    \[J\ddot{\theta} + 4\ell hx+ (4\ell h^{2}+4ka^{2})\theta = 0\]

Which matches the reduction of the general case where k has been replaced by 2k and \ell by 2\ell

c) Calculate the natural frequencies and mode shapes for the dryer

The vertical mode is as previously

    \[\begin{split} p_v &= \sqrt{\frac{2k}{m}}_\text{eff} \\&= \sqrt{\frac{4\cdot1150\cdot386}{(1049+255)}} \\&= 36.90 \ \text{rad/s} \\&= 5.98 \ \text{Hz} \end{split}\]

which is well below the excitation frequency of 950RPM (15.8Hz)

    \begin{equation*} \begin{bmatrix} m & 0 \\ 0 & J_G \\ \end{bmatrix} \begin{Bmatrix} \ddot{x}\\ \ddot{\theta}\\ \end{Bmatrix} + \begin{bmatrix} 4\ell & 4\ell h\\ 4\ell h & 4\ell h^{2}+4ka^{2}\\ \end{bmatrix} \begin{Bmatrix} x\\ \theta\\ \end{Bmatrix} = \begin{Bmatrix} 0\\ 0\\ \end{Bmatrix} \end{equation*}

Assume that

    \begin{equation*} \begin{Bmatrix} x\\ \theta\\ \end{Bmatrix} = \begin{Bmatrix} X\\ \Theta\\ \end{Bmatrix} \sin(pt+\phi) \end{equation*}

    \begin{equation*} \begin{bmatrix} 4\ell-mp^{2} & 4\ell h \\ 4\ell h & 4\ell h^{2} + 4ka^{2} - mr^{-2}p^{2}\\ \end{bmatrix} =0 \end{equation*}

    \[(4\ell-mp^{2})(4\ell h^{2} + 4ka^{2} - mr^{-2}p^{2}) - 16\ell^{2}h^{2} = 0\]

    \[16\ell^{2}h^{2} + 16k\ell a^{2} - 4\ell r^{-2}p^{2}m- 4\ell h^{2}mp^{2} - 4ka^{2}mp^{2} + mr^{-2}p^{4} -16\ell^{2}h^{2}= 0\]

    \[m^{2}r^{-2}p^{4}-mp^{4} -mp^{2}(4\ell r^{-2}+4\ell h^{2}+4ka^{2})+16k\ell a^2=0\]

Therefore

    \[p^{4}-p^{2}\biggr[\frac{4\ell}{m}+ \frac{4\ell h^{2}}{m\bar{r}^{2}+\frac{4ka^{2}}{m\bar{r}^{2}}}\biggr]+ \frac{16k\ell}{m^{2}} \frac{a^{2}}{\bar{r}^{2}}\]

    \[p^{4}- p^{2}\big[1089+3796+2185\big]+ 2379851\]

    \[p^{4}-p^{2}[7070]+2379851=0\]

    \[\begin{split} p^{2} &= -\frac{7070}{2} \pm \sqrt{\Big(\frac{7070}{2}\Big)^{2}-2379851} \\&= 3535 \pm 3180 \\&= 354,6715 \end{split}\]

    \[p = 18.8, 81.9 \ \text{rad/s} (2.99\text{ Hz}) (13\text{ Hz})\]

Mode Shapes

*** QuickLaTeX cannot compile formula:
\[(4\ell-mp^{2})X+4\ellh\Theta = 0\]

*** Error message:
Undefined control sequence \ellh.
leading text: \[(4\ell-mp^{2})X+4\ellh

Therefore

*** QuickLaTeX cannot compile formula:
\[\begin{split} \frac{X}{\Theta} &= \frac{-4\ellh}{(4\ell-mp^{2})} \\&= \frac{-h}{1-\frac{m}{4\ell}p^{2}} \end{split}\]

*** Error message:
Undefined control sequence \ellh.
leading text: ...rac{-h}{1-\frac{m}{4\ell}p^{2}} \end{split}

    \[\begin{split} \frac{X}{\Theta}\Big|_{1} &= \frac{-28}{1-\frac{(1049+255)(354)}{386(4)(0.8)(1150}} \\&= -28.3 \end{split}\]

    \[\begin{split} \frac{X}{\Theta}\Big|_{2} &= \frac{-28}{1-\frac{(1304)(6715)}{386(4)(0.8)(1150}} \\&= 5.42 \end{split}\]

d)

    \[p_x = \sqrt{\frac{4\ell}{m}}\]

    \[p_y = \sqrt{\frac{4k}{m}}\]

    \[p_\theta = \sqrt{\frac{4ka^{2}}{m\bar{r}^{2}}}\]

    \[p_x = \sqrt{\frac{4(0.8)(1150)}{1304}}\cdot386 = 33.0 \ \text{rad/s } (5.25 \text{ Hz})\]

    \[p_y = \sqrt{\frac{4(1150)(386)}{1304}} = 36.9 \ \text{rad/s } (5.87 \text{ Hz})\]

    \[p_x = \sqrt{\frac{4(1150)(19)^{2}(386)}{1304(15)^{2}}}= 46.7 \ \text{rad/s } (7.44 \text{ Hz})\]

now all natural frequencies are all below the excitation frequency of 158Hz (950RPM)

Semi-Definite Systems – Rigid Body Modes

So far we have dealt with systems that have positive definite stiffness and mass matrices. This means the eigenvalues are positive & real. We can then define an orthonormal basic system. However consider the example

    \begin{equation*}[m]=\begin{bmatrix}m_1 & 0 & 0  \\0 & m_2 & 0 \\0 & 0 & m_3 \\\end{bmatrix}\end{equation*}

    \begin{equation*}[k]=\begin{bmatrix}k_1 & -k_1 & 0  \\-k_1 & k_1+ k_2 & -k_2 \\0 & -k_2 & k_2 \\\end{bmatrix}\end{equation*}

but the stiffness matrix is not positive definitive. It is singular. This is because one “natural frequency” is zero.

    \[P_o = 0\]

    \begin{equation*}\{u_o\}=\begin{pmatrix}1  \\1   \\1  \\\end{pmatrix}\end{equation*}

u_o is called a zeroth mode

    \begin{equation*}[m] \{u\}=\begin{bmatrix}k_1 & -k_1 & 0 \\-k_1 & k_1+ k_2 & -k_2 \\0 & -k_2 & k_2 \\\end{bmatrix}\begin{bmatrix}1\\1\\1\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\\\end{bmatrix}=\{u\}^{T}[k]\end{equation*}

Therefore

    \begin{equation*}\{u\}^{T}[m]\begin{Bmatrix}\ddot{x}_1\\\ddot{x}_2\\\ddot{x}_3\\\end{Bmatrix}=0\end{equation*}

but since [m] is positive definite \{u\}^{T}[m] \neq 0, therefore \{\ddot{x}\} = 0, \{x\} = \alpha + \beta t

and the system is simply moving off in one direction, let’s suppose we set:

    \[\{u_o\}^{T}[m]\{x\}= 0\]

This is the constraint equation:

    \begin{equation*}\begin{bmatrix}1 & 1 & 1 \\\end{bmatrix}\begin{bmatrix}m_1 & 0 & 0 \\0 & m_2 & 0 \\0 & 0 & m_3 \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}= 0\end{equation*}

    \[m_1 x_1 + m_2 x_2 + m_3 x_3 = 0\]

or

    \[m_1 \dot{x}_1 + m_2 \dot{x}_2 + m_3 \dot{x}_3 = 0\]

Where this expression says the momentum of the 3 masses must be zero. This essentially says this is a 2 DOF system.

Note: If we tried to find a flexibility matrix it is not defined.

We can write the constraint in a compound form as:

    \begin{equation*}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}=\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\\frac{-m_1}{m_3} & \frac{-m_2}{m_3} & 0 \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\frac{-m_1}{m_3} & \frac{-m_2}{m_3} \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\\end{Bmatrix}=[C]\begin{Bmatrix}x_1\\x_2\\\end{Bmatrix}\end{equation*}

We can now solve the eigenvalue problem using

    \[ [k^{'}] = [C]^{T}[k][C] \]

    \[ [m^{'}] = [C]^{T}[m][C] \]

so that

    \[ [k^{'}]\{u\} = p^{2}[m^{'}]\{u\} \]

For the example above set k_1 = k_2 = k, m_1 = m_2 = m_3 = m

    \begin{equation*}[k']=\begin{bmatrix}1 & 0 & -1 \\0 & 1 & -1 \\\end{bmatrix}\begin{bmatrix}k & -k & 0 \\-k & 2k & -k \\0 & -k & k \\\end{bmatrix}\begin{bmatrix}1 & 0 \\0 & 1 \\1 & -1 \\\end{bmatrix}=\begin{bmatrix}1 & 0 & -1 \\0 & 1 & -1 \\\end{bmatrix}\begin{bmatrix}k & -k \\0 & 3k \\-k & -2k \\\end{bmatrix}=\begin{bmatrix}2k & k \\k & 5k \\\end{bmatrix}\end{equation*}

    \begin{equation*}[m'] = m\begin{bmatrix}2 & 1 \\1 & 2 \\\end{bmatrix}\end{equation*}

    \begin{equation*}\begin{bmatrix}2 & 1 \\1 & 5 \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}-\frac{p^{2}m}{k}\begin{bmatrix}2 & 1 \\1 & 2 \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}= 0\end{equation*}

    \begin{equation*}\begin{bmatrix}2-2\frac{p^{2}m}{k} & 1-\frac{p^{2}m}{k} \\1-\frac{p^{2}m}{k} & 5-2\frac{p^{2}m}{k} \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}={0}\end{equation*}

    \[(2-2\frac{p^{2}m}{k})(5-2\frac{p^{2}m}{k}) - (1 - \frac{p^{2}m}{k})^{2} = 0\]

    \[10-14\frac{p^{2}m}{k} + 4(\frac{p^{2}m}{k})^{2} - [1 - 2\frac{p^{2}m}{k} + \frac{p^{4}m^{2}}{k^{2}}] = 0 \]

    \[9 - 12 \frac{p^{2}m}{k} + 3 \frac{p^{4}m^{2}}{k^{2}} = 0\]

    \[\frac{p^{4}m^{2}}{k^{2}} - \frac{p^{2}m}{k} + 3 = 0\]

    \[p^{2} = \{2 \pm \sqrt{1}\} \frac{k}{m}\]

    \[p^{2}_{1,2} = \frac{k}{m}, \frac{3k}{m}\]

    \begin{equation*}\{u^{'}_1\}=\begin{Bmatrix}1\\0\\\end{Bmatrix}\end{equation*}

    \begin{equation*} \{u^{'}_2\}= \begin{Bmatrix} \frac{1}{2}\\ -1\\ \end{Bmatrix} \end{equation*}

    \[x_3 = -x_1 - x_2\]

    \begin{equation*}\{u_1\}=\begin{Bmatrix}1\\0\\-1\\\end{Bmatrix}\end{equation*}

    \begin{equation*} \{u_1\}= \begin{Bmatrix} \frac{1}{2}\\ -1\\ \frac{1}{2}\\ \end{Bmatrix} \end{equation*}

A second example of a 2DOF system is the vibration absorber as it can provide quite spectacular results when applied correctly. We will consider it first as a 2DOF undamped system, then include viscous damping. The closed form solution for the case of damping is much more involved.

Undamped Vibration Absorber

    \[\begin{split} + \uparrow \sum F_{x_1} &= m_1 \ddot{x_1} \\&= -k_1x_1 + k_2(x_2 - x_1) + F_o\sin\omega t \quad (1) \end{split} \]

    \[ \begin{split} + \uparrow \sum F_{x_2} &= m_2\ddot{x_2} \\&= -k_2(x_2 - x_1) \quad (2) \end{split} \]

    \[ m_1\ddot{x_1} + (k_1+k_2)x_1 - k_2x_2 = F_o \sin\omega t \]

    \[ m_2\ddot{x_2} - k_2x_1 + k_2x_2 = 0 \]

    \[ \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{Bmatrix} \ddot{x_1} \\ \ddot{x_2} \end{Bmatrix} + \begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} F_o\sin\omega t \\ 0 \end{Bmatrix} \]

For S.S. solution, assume:

    \[\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} \sin \omega t \]

Under the steady state assumption, the equations of motion become:

    \[ [-m_1\omega^2 + k_1 + k_2]X_1 - k_2X_2 = F_o \quad (*) \]

    \[-k_2X_1 + (k_2 - m_2\omega^2)X_2 = 0 \quad (**) \]

Therefore, from (**):

    \[ X_2 = \frac{k_2X_1}{(k_2-m_2\omega^2)}\]

And then from (*):

    \[(k_1 +k_2-m_1\omega^2)X_1 - \frac{k_2^2X_1}{k_2-m_2\omega^2} = F_o\]

    \[\frac{\Big\{[(k_1+k_2)-m_1\omega^2](k_2-m_2\omega^2)-k_2^2\Big\}X_1}{k_2-m_2\omega^2} = F_o \]

Therefore:

    \[X_1 = \frac{F_o(k_2-m_2\omega^2)}{[(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2) - k_2^2]} \]

    \[X_2 = \frac{k_2F_o}{[(k_2 - m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2]} \]

For use in vibration absorber applications, we define:

    \[p_{11} := \sqrt{\frac{k_1}{m_1}},\ p_{22} := \sqrt{\frac{k_2}{m_2}},\ \mu := \frac{m_2}{m_1} \]

Then divide numerator and denominator by k_1, k_2.

For the 2DOF system shown with the forcing appiled to the mass m_1, the response of the masses m_1 and m_2 is:

    \[X_1 = \frac{F_o(k_2-m_2\omega^2)}{(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2} \quad (*) \]

    \[X_2 = \frac{F_ok_2}{(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2} \quad (**) \]

It is seen that from (*) that if k_2 -m_2\omega^2 = 0 or \frac{k_2}{m_2} = \omega^2, then the mass m_1 is stationary. This leads to the idea that the motion of m_1 can theoretically be reduced to zero if we employ k_2, m_2 at the operating frequency \omega. For this application. the subsystem k_2, m_2 is called a vibration absorber and at the operating frequency:

    \[X_1 = 0 \text{ and } X_2 = \frac{F_o}{k_2} \]

For this application it is useful to modify (*) and (**) to reflect what we can call – the original system k_1, m_1, the absorber system k_2, m_2 using the definitions:

    \[p_{11} := \sqrt{\frac{k_1}{m_1}},\ p_{22} := \sqrt{\frac{k_2}{m_2}},\ \mu := \frac{m_2}{m_1} \]

Using those definitions show that (*) and (**) become:

    \[X_1 = \frac{\frac{F_o}{k_1}[1-(\frac{\omega}{p_{22}})^2]}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_11})^2-(\frac{\omega^2}{p_{11}})^2] - \mu(\frac{p_{22}}{p_{11}})^2\}} \]

    \[X_2 = \frac{\frac{F_o}{k_1}}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_11})^2-(\frac{\omega^2}{p_{11}})^2] - \mu(\frac{p_{22}}{p_{11}})^2\}} \]

And when \omega^2 = \frac{k_2}{m_2}:

    \[X_1 = 0,\ X_2 = -\frac{F_o}{k_2}\]

This shows that when the subsystem k_2, m_2 is timed so \frac{k_2}{m_2} = p_{22}^2 that m_1 is motionless as the force in the spring k_2 exactly opposes the excitation force F_o.

The natural frequencies of the combined system are found from setting the denominator of the responses X_1, X_2 to zero. Therefore, the natural frequencies are found \omega is replaced by p and the expression becomes:

    \[ \Big[1 - \Big(\frac{p}{p_{22}}\Big)^2\Big]\Big[1 + \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{11}}\Big)^2\Big] - \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 = 0\]

    \[1 + \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{11}}\Big)^2-\Big(\frac{p}{p_{22}}\Big)^2-\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2\Big(\frac{p}{p_{22}}\Big)^2+\Big(\frac{p}{p_{11}}\Big)^2\Big(\frac{p}{p_{22}}\Big)^2 -\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 = 0\]

Therefore:

    \[\Big(\frac{p}{p_{22}}\Big)^4\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{22}}\Big)^2\Big[1+\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 + \Big(\frac{p_{22}}{p_{11}}\Big)^2\Big] + 1= 0 \]

    \[\Big(\frac{p}{p_{22}}\Big)^4-\Big(\frac{p}{p_{22}}\Big)^2\Big[1+\mu+\Big(\frac{p_{11}}{p_{22}}\Big)^2\Big] + \Big(\frac{p_{11}}{p_{22}}\Big)^2 = 0 \quad (*) \]

If p_{11} = p_{22}, this becomes:

    \[\Big(\frac{p}{p_{22}}\Big)^4 - \Big(\frac{p}{p_{22}}\Big)^2[2+\mu] + 1 = 0\]

If instead of the mass m_1 being excited by the force F_o\sin\omega t, the exication is through the base as x_o = X_o\sin\omega t, what is the difference in the response of X_1 and X_2 of masses m_1 and m_2 ?

The mass ratio, \mu, essentially determines the spread of the two resonances:

    \[\mu = 0.1, \quad \frac{\omega}{p_{22}} = 0.85,\ 1.17\]

    \[\mu = 0.2, \quad \frac{\omega}{p_{22}} = 0.79,\ 1.25\]

    \[\mu = 0.05, \quad \frac{\omega}{p_{22}} = 0.89,\ 1.122\]

(\text{TAIPEI TOWER:} \quad m_2 = 700\ \text{TONS},\ \mu \approx 0.05)

    \[\begin{split} \sum F_{x_1} &= m_1\ddot{x_1} \\&= -k_1(x_1-x_o) + k_2(x_2-x_1) \quad (1) \end{split} \]

    \[\begin{split} \sum F_{x_2} &= m_2\ddot{x_2} \\&=-k_2(x_2-x_1) \quad (2) \end{split}\]

Therefore, the equations are identical except that F_o is replaced by k_1X_o. The response of x_1 and x_2 for the vibration absorber is:

    \[X_1= \frac{X_0[1-(\frac{\omega}{p_{22}})^2]}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_{11}})^2-(\frac{\omega}{p_11})^2]-\mu(\frac{p_{22}}{p_{11}})^2\}}\]

    \[X_1= \frac{X_0}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_{11}})^2-(\frac{\omega}{p_11})^2]-\mu(\frac{p_{22}}{p_{11}})^2\}}\]

Again when \frac{k_2}{m_2} := p_{22}^2 = \omega^2:

    \[X_1 = 0\]

    \[\begin{split} X_2 &= -X_o\frac{1}{\mu}\Big(\frac{p_{11}}{p_{22}}\Big)^2 \\&= -X_o\frac{m_1}{m_2}\frac{k_1}{m_1}\frac{m_2}{k_2} \\&= -X_o\frac{k_1}{k_2} \end{split} \]

Therefore, k_2X_2 = -k_1X_1 so that at any time:

And the net dynamic force on m_1 is zero!

To solve the system with damping it is more advantageous to use complex numbers:

    \[\{x\} =\{X\}e^{i\omega t}\]

Or:

    \[\{x\} =\{\zeta\}e^{st}\]

There are a number of special cases of the general forced damped analysis that are useful.

Damped Forced 2DOF System

Consider a general damped system that can be specialized for special applications.

    \[f_1(t) = f_1e^{st}\]

    \[f_2(t) = f_2e^{st}\]

    \[m\ddot{x_1} = k_2(x_2 - x_1) + c_2(\dot{x_2}-\dot{x_1}) - k_1x_1 - c_1\dot{x_1} + f_1(t) \]

    \[m\ddot{x_1} + (k_1+k_2)x_1 + (c_1+c_2)\dot{x_1} - k_2x_2 - c_2\dot{x_2} = f_1(t) \]

    \[m_2\ddot{x_2} = f_2(t) - k_2(x_2-x_1)-c_2(\dot{x_2}-\dot{x_1})\]

Thus:

    \[[m_1s^2 + (c_1+c_2)s + (k_1+k_2)]\zeta_1 - [c_2s+k_2]\zeta_2 = f_1\]

    \[-(c_2s + k_2)\zeta_1 + [m_2s^2+c_2s+k_2]\zeta_2 = f_2\]

Therefore:

    \[V_{11} = m_1s^2 + (c_1 + c_2)s + k_1 + k_2 \]

    \[V_{12} = -(c_2s + k_2) = V_{21} \]

    \[V_{22} = m_2s^2 + c_2s + k_2 \]

    \[V_{11}\zeta_1 + V_{12}\zeta_2 = f_1\]

    \[V_{21}\zeta_1 + V_{22}\zeta_2 = f_2\]

And we wish to find \zeta_1 and \zeta_2:

    \[ \begin{bmatrix} V_11 & V_12 \\ V_21 & V_22 \end{bmatrix} \begin{bmatrix} \zeta_1 \\ \zeta_2 \end{bmatrix} = \begin{bmatrix} f_1 \\ f_2 \end{bmatrix} \]

    \[ [V]^{-1} \equiv [Y] \]

    \[Y_{11}(s) = \frac{V_{22}(s)}{\Delta},\ Y_{12} = -\frac{V_{21}(s)}{\Delta},\ Y_{22} = \frac{V_{11}(s)}{\Delta}\]

Where \Delta = |V|. Thus:

    \[\zeta_1 = Y_{11}(s)f_1 + Y_{12}(s)f_2\]

    \[\zeta_2 = Y_{21}(s)f_1 + Y_{22}(s)f_2\]

There is a lot hidden in these symbols so we can consider some special cases.

1. Untuned Viscous Vibration Absorber
(Houdaille damper or viscous Lanchester damper)

If we look at our general solution then:

    \[c_2 = 0,\ c_1 = 0,\ k_2 = 0,\ f_2 = 0 \]

    \[V_{11} = m_1s^2 + cs + k_1 \]

    \[V_{12} = -cs = V_{21} \]

    \[V_{22} = m_2s^2 + cs \]

    \[\zeta_1 = Y_{11}(s) f_1,\quad Y_{11}(s) = \frac{V_{2d2}(s)}{\Delta} \]

    \[\zeta_2 = Y_{21}(s) f_1,\quad Y_{21}(s) = -\frac{V_{21}(s)}{\Delta} \]

Therefore:

    \[ \Delta = [m_1s^2 + cs + k][m_2s^2 + cs] - (cs)^2 \]

Put s =i\omega:

    \[\begin{split} \Delta &= [k-m_1\omega^2 + ic\omega][-m_2\omega^2 +ic\omega]+c^2\omega^2 \\&= -m_2\omega^2(k-m_1\omega^2) + ic\omega[-m_2\omega^2 + (k-m\omega^2) \end{split} \]

Therefore:

    \[ \zeta_1 = \frac{(-m_2\omega^2 + ic\omega)F_o}{-m_2\omega^2(k-m_1\omega^2) +ic\omega[-m_2\omega^2 + (k-m\omega^2)]}\]

    \[ \zeta_2 = \frac{ic\omega}{-m_2\omega^2(k-m_1\omega^2) + ic\omega[-m_2\omega^2 + (k-m\omega^2)]} \]

We are really interested in the amplitudes. Use the result that if:

    \[\begin{split} \zeta &= \frac{B+iC}{D+iE} \cdot \frac{D-iE}{D-iE} \\&=\frac{BD+CE+i(CD-BE)}{D^2+E^2} \end{split} \]

We wish to write \zeta = Ae^{i\alpha}.

    \[ \begin{split} A &= \frac{\sqrt{(BD+CE)^2 + (CD-BE)^2}}{D^2 + E^2} \\&= \frac{\sqrt{(BD)^2 + (CE)^2 + 2BDCE + (CD)^2 + (BE)^2 - 2BCDE}}{D^2+E^2} \\&= \frac{\sqrt{B^2(D^2+E^2) + C^2(D^2+E^2)}}{D^2+E^2} \\&= \sqrt{\frac{B^2+C^2}{D^2+E^2}} \end{split} \]

    \[ \alpha = \tan^{-1} \frac{CD-BE}{BD+CE} \]

    \[X_1 = \frac{F_o\sqrt{(m_2\omega^2)^2+(c\omega)^2}}{\sqrt{[m_2\omega^2(k-m_1\omega^2)]^2+(c\omega)^2[m_2\omega^2 + (k-m_1\omega^2)]^2}} \]

    \[X_o = \frac{F_o}{k},\ p_1^2 = \frac{k_1}{m_1},\ \nu = \frac{c}{2m_1p_1},\ \mu = \frac{m_2}{m_1},\ r = \frac{\omega}{p_1} \]

Thus:

    \[ \frac{X_1}{X_o} = \frac{\sqrt{(\frac{m_2}{k})^2 + (\frac{c\omega}{k})^2}}{\sqrt{[\frac{m_2\omega^2}{k}(1-\frac{m_1\omega^2}{k})]^2 + (\frac{c\omega}{k})^2[\frac{m_2\omega^2}{k} - (1- \frac{m_1\omega^2}{k})]^2}} \]

    \[\frac{m_2}{m_1} \frac{m_1}{k} \omega^2 = \mu r^2 = \frac{m_2}{k}\omega^2\]

    \[\frac{c\omega}{k} = \frac{cm_1}{m_1k}\omega = \frac{c}{m_1p_1^2} \]

Where \omega = 2\nu r. Thus:

    \[\begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\mu^2r^4 + (2\nu r)^2}}{\sqrt{[\mu r^2(1-r^2)]^2 + (2\nu r)^2[\mu r^2 - (1-r^2)]^2}} \\& = \frac{\sqrt{\mu^2r^2 + 4\nu ^2}}{\sqrt{[ur(1-r^2)]^2 + 4\nu ^2[\mu r^2 - (1-r^2)]^2}} \\&= \sqrt{\frac{N}{D}} \end{split} \]

We wish to find the optimum ratio \nu which minimizes this amplitude;

    \[\begin{split} \frac{\text{d}\frac{X_1}{X_o}}{\text{d}\nu} &= \frac{D^{\frac{1}{2}}\frac{N^{-\frac{1}{2}}}{2}(8\nu) - N^{-\frac{1}{2}}\frac{D^{\frac{1}{2}}}{2}(8\nu)[\mu r^2 - (1-r^2)]^2}{D} \\&= 0 \end{split} \]

Multiply by D^{\frac{1}{2}}N^{\frac{1}{2}}. Therefore:

    \[ D-N[\mu r^2 - (1-r^2)]^2 = 0 \]

    \[[\mu r(1-r^2)]^2 + 4\nu^2[\mu r^2 - (1-r^2)]^2 = (\mu^2 r^2 + 4\nu^2)[\mu r^2 - (1-r^2)]^2 \]

    \[[\mu r(1-r^2)]^2 = (\mu r)^2[\mu r^2 - (1-r^2)]^2 \]

    \[(1-r^2) = \mu r^2 - (1-r^2) \]

    \[ 2(1-r^2) = \mu r^2 \]

    \[ 2 = r^2(2+\mu) \]

Therefore:

    \[r = \sqrt{\frac{2}{2+\mu}},\quad \mu = \frac{2(1-r^2)}{r^2} \]

Therefore, the minimum amplitude for optimum \nu occurs when r = \sqrt{\frac{2}{2+\mu}} = 0.894\ (\mu = \frac{1}{2}). To find the value of \nu, use this value of r.

    \[ \begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\frac{\mu^2 2}{2+\mu} +4\nu^2}}{\sqrt{\frac{\mu^2 2}{2+\mu} (1- \frac{2}{2+\mu})^2 + 4\nu^2[\mu\frac{2}{2+\mu} - (1-\frac{2}{2+\mu})]^2}} \\&= \frac{\sqrt{\frac{2\mu^2}{2+\mu} + 4\nu^2}}{\sqrt{\frac{2\mu^2}{2+\mu}\frac{\mu^2}{(2+\mu)^2} + 4\nu^2[\frac{2\mu}{2+\mu} - (\frac{\mu}{2+\mu}) ]^2}} \\&= \frac{\sqrt{\frac{2\mu^2 + 4\nu(2+\mu)}{2+\mu}}}{\sqrt{\frac{2\mu^4}{(2+\mu)^3} + \frac{4\nu^2\mu^2}{(2+\mu)^2}}} \\&= \frac{\sqrt{2\mu^2 +4\nu^2(2+\mu)}}{\mu\sqrt{\frac{2\mu^2 + 4\nu^2(2+\mu)}{(2+\mu)^2}}} \\&= \frac{2+\mu}{\mu} = \frac{X_1}{X_o}\Big|_{\text{MAX}} \\&= 5\quad (\mu = \frac{1}{2}) \end{split} \]

The amplitude at this value of r is independent of \nu!

    \[ \begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\mu^2r^2 + 4\nu^2}}{\sqrt{\mu^2r^2(1-2r^2+r^4)+4\nu^2[r^2(1+\mu)-1]^2}} \\&= \frac{\sqrt{\mu^2r^2 + 4\nu^2}}{\sqrt{\mu^2(r^2-2r^4+r^6) + 4\nu^2[r^4(1+\mu)^2-2r^2(1+\mu)+1]}} \\&= \sqrt{\frac{N}{D}} \end{split} \]

    \[ \begin{split} \frac{\text{d}(\frac{X_1}{X_o})}{\text{d}r} &= \frac{D^{\frac{1}{2}}\frac{N^{-\frac{1}{2}}}{2}2\mu^2r-N^{\frac{1}{2}}\frac{D^{-\frac{1}{2}}}{2}[\mu^2(2r+8r^3+6r^5) + 4\nu^2(4r^3(1+\mu)^2 - 4r(1+\mu))]}{D} \\& = 0 \end{split} \]

Multiply by D^{\frac{1}{2}}N^{-\frac{1}{2}}. Then:

    \[ 0 = D(\mu^2r)-N[\mu^2(r-4r^3+3r^5) + 2\nu^2(4r^3(1+\mu)^2-4r(1+\mu))] \]

Now evaluate this for r = \sqrt{\frac{2}{2+\mu}},\ r^2 = \frac{2}{2+\mu}.

    \[D\mu^2 = N[\mu^2(1-4r^2+3r^4)+2\nu^2(4r^2(1+\mu)^2-4(1+\mu))] \]

P2

Note that:

    \[ \begin{split} \frac{X_1}{X_0} &= \sqrt{\frac{N}{D}} \\&= \frac{2+\mu}{\mu} \end{split}\]

    \[ \frac{N}{D} = \frac{(2+\mu)^2}{\mu^2}\]

    \[D = \frac{\mu^2}{(2+\mu)^2}N\]

    \[ \frac{\mu^4}{(2+\mu)^2} = \mu^2\bigg(1 - \frac{4(2)}{2+\mu} + \frac{3(4)}{2+\mu)^2}\bigg) + 2\nu^2\bigg[\frac{8}{(2+\mu)}(1+\mu)^2 - 4(1+\mu)\bigg] \]

    \[\mu^4 = \mu^2\bigg[ (2+\mu)^2 - 8 ( 2+ \mu) + 12 \bigg]  + 2\nu^2\bigg[8(1+\mu)^2(2+\mu) - 4 ( 1 + \mu)(2+\mu)^2\bigg] \]

    \[\mu^4 = \mu^2 \bigg[ 4 + 4 \mu + \mu^2 - 16 - 8 \mu + 12 \bigg ] + 2\nu^2(1+\mu)(2+\mu)\bigg[8(1 + \mu ) - 4(2 + \mu) \bigg] \]

    \[ \mu^4 = \mu^2\big[\mu^2 - 4\mu\big] + 2\nu^2(1+\mu)(2+\mu)4\mu\]

    \[ \mu^4 = \mu^4 - 4\mu^3 + 2\nu^2(1 + \mu)(2+ \mu) 4 \mu \]

    \[ 4\mu^3 = 2\nu^2(1+\mu)(2+\mu)4\mu \]

    \[\nu^2 = \frac{\mu^2}{2(1+\mu)(2+\mu)} \]

    \[ \nu_\text{OPT} = \frac{\mu}{2(1+\mu)(2+\mu)} \]

    \[ \mu = \frac{1}{2} ,\ \nu_\text{OPT} = 0.183\]

When r = 1:

    \[\frac{X_1}{X_0} = \frac{\sqrt{\mu^2 + 4\nu^2}}{\sqrt{4\nu^2\mu^2}} \]

When \mu = \frac{1}{2},\ \nu = 0.365:

    \[ \frac{X_1}{X_0} = \frac{\sqrt{ 0.25 + 4(0.365)^2}}{\sqrt{4(0.365)^2(\frac{1}{4})}} = 2.42\]

When \nu = 1.0,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \frac{\sqrt{0.25 + 4}}{\sqrt{4(1)(\frac{1}{4})}} = 2.06\]

When \nu = 0.1,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \frac{\sqrt{0.25 + 0.04}}{\sqrt{4(0.01)(\frac{1}{4})}} = 5.39\]

When \nu = 0.1825,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \sqrt{\frac{0.25 + 4(0.1825)^2}{0.1825^2}} = 3.39 \]

Example

A second very practical application is the viciously damped vibration absorber.

Solution

From the general solution:

    \[ c_1 = 0,\ c_2 = C \]

Therefore:

    \[V_{11} = m_1s^2 + cs + k_1 + k_2 \]

    \[V_{12} = - (cs + k_2) = V_{21}\]

    \[V_{22} = m_2s^2 + cs + k_2 \]

    \[ \zeta_1 = Y_{11}f_1 ,\ \zeta_2 = Y_{21}f_1 \]

    \[Y_{11} = \frac{V_{22}}{\Delta},\ Y_{21} = -\frac{V_{12}}{\Delta} \]

    \[\Delta = \big[ - m_1\omega^2 + ic\omega +k_1 + k_2\big] \big[-m_2\omega^2 + ic\omega + k_2 \big] - \big[k_2 + ic\omega\big]^2\]

    \[\implies \Delta = \big[(-m_1\omega^2 + k_1)(-m_2\omega^2 + k_2) - m_2\omega^2 k_2\big] + ic\omega[-m_1\omega^2 + k_1 - m_2\omega^2]\]

    \[ \zeta_1 = Y_{11}F_0e^{i\omega t} = \frac{(-m_2\omega^2 + ic\omega + k_2)}{\Delta}F_0e^{i\omega t}\]

If we set \zeta_1 = X_1e^{i\theta_1},\ \zeta_2 = X_2e^{i\theta_2} :

    \[ (\frac{X_1}{F_0})^2 = \frac{(k_2 - m_2\omega^2)^2 +\omega^2c^2}{\Delta \bar{\Delta}}\]

    \[\begin{split}\zeta_2 &= Y_{21}f_1 \\&= k_2 + \frac{ic\omega}{\Delta \bar{\Delta}} \\&= X_2e^{i\theta_2} \end{split}\]

    \[(\frac{X_2}{F_0})^2 = k_2^2+ \frac{c^2\omega^2}{\Delta \bar{\Delta}}\]

In order to apply this to various situations, it is useful to non-dimensionalize these relationships similar to that of the undamped vibration absorber.

Set:

    \[X_0 = \frac{F_0}{k_1},\ \mu = \frac{c}{2m_2p_{11}} \]

    \[p_{11} = \sqrt{\frac{k_1}{m_1}},\ p_{22} = \sqrt{\frac{k_2}{m_2}} \]

NOTE:

    \[(\omega c)^2 = \bigg( \frac{\omega}{p_{11}^2}c\frac{k_1}{m_1}\bigg)^2\]

    \[ r = \frac{\omega}{p_{11}},\ \mu = \frac{m_2}{m_1} \]

Therefore:

    \[ (\omega c)^2 = \bigg[ \frac{\omega}{p_{11}}2\frac{c}{2m_2p_{11}}\mu^2k_1\frac{m_1}{m_2}\bigg]^2 = k_1^2\mu^2[2\nu]^2 \]

    \[ \begin{split}(k_2 - m_2\omega^2)^2 &= k_1^2\bigg[\frac{k_2}{k_1} - \frac{m_2\omega^2}{k_1} \bigg] ^2 \\&= \bigg[\frac{m_1}{m_2}\frac{k_2}{k_1}\frac{m_2}{m_1} - \frac{m_2}{m_1}\omega^2 \frac{m_1}{k_1}\bigg]^2k_1^2 \\&= \mu^2k_1^2[r^2 - g^2]^2 \end{split}\]

    \[ g = \frac{p_{22}}{p_{11}} \]

    \[ \begin{split}\Delta \bar{\Delta} &= \big[(-m_1\omega^2 + k_1)(-m_2\omega^2 + k_2) - m_2\omega^2 k_2 \big] ^2 + c^2\omega^2[-m_1\omega^2 + k_1 - m_2 \omega^2 \big] ^2  \\&= (2\nu r)^2[r^2 - 1 + \mu r^2]^2 + \big[\mu g^2r^2 - (r^2-1)(r^2-g^2)\big]^2\end{split}\]

    \[ \frac{X_1}{X_0} = \sqrt{\frac{(2\nu r)^2 + ( r^2 - g^2)^2}{(2\nu r)^2(r^2 - 1 + \mu r^2)^2 + \big[\mu g^2r^2 - (r^2-1)(r^2-g^2)\big]^2}} \]

    \[ \mu = \frac{m_2}{m_1} \quad \text{(mass ratio)}\]

    \[ \nu = \frac{c}{2m_2p_{11}} \quad \text{(damping ratio)} \]

    \[g = \frac{p_{22}}{p_{11}} \quad \text{(frequency ratio of subsystems)}\]

    \[r = \frac{\omega}{p_{11}} \]

    \[ \frac{X_2}{X_0} = \frac{\mu r\sqrt{1+4\nu^2}}{\sqrt{(2\nu r)^2(r^2 - 1 +\mu r^2)^2 + \big[\mu g^2r^2 - ( r^2 - 1)(r^2 - g^2)\big]^2}}\]

CHECK
  1. If \omega = 0:

        \[\frac{X_1}{X_0} = \frac{r^2 - g^2}{\mu g^2r^2 - (r^2 - 1)(r^2 - g^2)} \]

  2. If r = g, then X_1 = 0
    Note that X_1 \rightarrow 0 only when c \rightarrow 0
    If \nu \rightarrow \infty:

        \[ \begin{split}\frac{X_1}{X_0} &= \frac{1}{r^2 - 1 + \mu r^2} \\&= \frac{1}{(1+\mu)r^2-1} \\&= \frac{1}{\frac{m_1 + m_2}{k_1}\omega^2 -1} \end{split}\]

As a result somewhere between 0 and \nu, c will give an optimal result.

Note that all the curves intersect at P and Q. That is these points are independent of the damping. If we calculate their location, our problem is essentially solved as we must find the curve that passes through them with a horizontal tangent. Also by changing the “tuning” g = \frac{\omega_{11}}{\omega{22}} the two points can be shifted up and down the c = 0 curve. We can do this until the two points have the same height and the horizontal tangent through one of them.

To find the two points where the amplitude is independent of the damping \frac{X_1}{X_0} = \frac{\sqrt{A\omega^2 + B}}{C\omega^2 + D} and this is independent when \frac{A}{C} = \frac{B}{D}:

    \[\bigg(\frac{1}{r^2 - 1 + \mu^2r^2}\bigg)^2 = \bigg[ \frac{(r^2-g^2)^2}{\mu g^2r^2 - (r^2-1)(r^2 - g^2)}\bigg]^2 \]

This leads to:

    \[ \mu g^2r^2 - (r^2-1)(r^2-g^2) = \pm(r^2-g^2)(r^2 - 1 + \mu r^2) \]

The negative gives r^2 = 0 while the positive gives r^4 - 2r^2 \frac{(1+g^2 + \mu g^2)}{2+\mu} + \frac{2g^2}{2+\mu} = 0.

The two answers for the forcing frequency ratio are functions of g and \mu but not of the damping \omega.

We now wish to adjust g so that P & Q are equal. Since their magnitude is independent of the damping, chose c \rightarrow \infty then the magnitude:

    \[\frac{X_1}{X_0} = \frac{1}{\mu r^2 + r^2 - 1} \]

For r_1^2 and r_2^2 (the two roots), we want the same magnitude:

    \[ \frac{1}{\mu r_1^2 + r_1^2 - 1} = -\frac{1}{\mu r_2^2 + r_2^2 - 1}\]

Therefore:

    \[1 - (\mu + 1) r_2^2 = -1 + (\mu + 1) r_1^2\]

    \[ \frac{2}{1+\mu} = r_1^2 + r_2^2 \]

Now it is not necessary to solve for r_1^2 and r_2^2 if we remember the sum of the roots in a quadratic in the negative coefficient of the middle term.

Therefore:

    \[ \frac{2}{1+\mu} = \frac{2(1+g^2+\mu g^2)}{2+\mu} \]

    \[ \frac{2+\mu}{1+\mu} = 1+ g^2(1+\mu) \]

    \[\begin{split} \implies g^2(1+\mu) &= \frac{2+\mu - 1 - \omega}{1+ \mu} \\&= \frac{1}{1+\mu} \end{split}\]

Therefore:

    \[g = \frac{1}{1+\mu} \]

Amplitudes of the main mass for various values of absorber damping. The absorber is twenty times as small as the main machine and is tuned to the same frequency. All curves pass through the fixed points P and Q.

Resonance curves for the motion of the main mass fitted with the most favorably tuned vibration-absorber system of one-forth of the size of the main machine.

(a) Peak amplitudes of the main mass as a function of the ratio m/M for various absorbers attached to the main mass. (b) Peak relative amplitudes between the masses M and m for various absorbers. (c) Damping constants required for the most favorable operation of the absorber, i.e., for obtaining the results of (a) and (b).

Curve 1 for the most favorably tuned and damped absorber; curve 2 for the most favorably damped absorber tuned to the frequency of the main system; curve 3 of the most favorably damped viscous Lanchester damper; curve 4 for the most favorably damped Coulomb Lanchester damper.

    \[ \bigg(\frac{X_1}{X_0}\bigg)_\text{MAX} = \frac{1}{1-r^2(1+\mu)}\]

and using one of the roots:

    \[ \bigg(\frac{X_1}{X_0}\bigg)_\text{MAX} = \sqrt{1+ \frac{2}{\mu}}  \]

while the optimum damping is given by:

    \[ \nu = \sqrt{\frac{3\mu}{8(1+\mu)^3}} \]

MTS Tuned MASS Damper Systems

The purpose of a Tuned Mass Damper is to reduce wind excited periodic motion by increasing the apparent damping of structures with low inherent damping and predominant fundamental vibration mode response to wind excitation. MTS Tuned Mass Damper systems are operated essentially as classical Tuned Mass Dampers but are unique in their adaptability, long-term stability, and size. They are designed to minimize energy consumption and offer the structural designer a powerful new means of reducing and controlling dynamic motion.

Although the primary building structure would not depend on the TMD operation for integrity, operation during high wind conditions will help minimize the possibilities of minor non-structural cosmetic damage to building interior components, such as partitions and walls. The TMD system will reduce building motion(induced by moderate and high winds) to approximately one-half of what the building motion would have been without the system operating. If an extremely high gust of wind should impact the building and TMD system can automatically shift its damping coefficient so that the mass does not overstroke. Whenever the system is operating, electronic and mechanical interlock devices ensure that the system is always in damping mode, attempting to reduce building motion, System shut down will occur if an undesirable operating condition should ever develop.

Understanding the Tuned Mass Damper

Understanding the tuned mass damper begins with the basic concept of mechanical vibration: some object is simply moving back and forth in a regular manner. in all cases some physical mass, having inertia, is moving back and forth against a resistance, or restoring force. These two basic ingredients: inertia and restoring force are necessary to all vibrations of common physical objects; in fact,. the presence of both guarantees that vibrations can take place.

One of the simplest possible examples of vibration is a weight, hanging under gravity from a simple coiled spring (Figure 1). Pulled down slightly from the rest position and then released, the weight oscillates up and down. Another example, (Figure 2) is the same weight mounted on the elastic post. Pulled to one side and released, the weight will oscillate back and forth laterally.

Everything which has both inertia and sporing qualities can thus be expected to exhibit vibration tendencies if once started or excited in some way. Structures such as tall building are exposed to gusty winds and as a result they swap back and forth, or vibrate. Some structures vibrate slowly(i.e., at low frequencies) while others tend to vibrate more rapidly (i.e., at high frequencies).

Example

Consider a multi-story building(Figure 3) having a structural steel frame. Such a building fits all the conditions for vibration. and it will in fact tend to sway very slightly in the wind, moving slowly, requiring perhaps several seconds to accomplish a full swing over and back.

Tall steel freamed building occasionally vibrate enough to disturb their occupants, or at least to draw their attention to the fact that the building is moving or swaying. Actually, buildings oscillate to a much lesser extent then such common structures as aircraft wings and fuselages; yet, because, psychologically, buildings are”suppose” to stand still, occupant objection to building oscillation is often much greater than it is to aircraft vibratory motion. As a result, tuned mass damped systems may be used to lessen building vibration when it occurs.

Other methods may be used to reduce building vibration, but tend to be more costly, or are less certain and reliable. Two possible alternatives are to increase the building mass while maintaining its natural vibration frequency approximately constant, or to increase building damping by adding special energy-absorbing devices or using masonry filler walls.

To achieve the same degree of motion reduction as TMD would require increasing the building weight and lateral stiffness(the structure’s resistance to lateral sway, measured in pounds per inch for example) by a factor of 1 2/3 to as much as 3. The added weight requires heavier structural members and foundations. and the added lateral stiffness requires heavier structure, all of which tend to be very costly. It’s inefficient to invest in very costly large increases in building weight and structure when these increases are unnecessary for safety and are done in order to increase occupant comfort. Methods which reduce building motion by increasing damping are more attractive.

Damping resistance to an vibration is simply any force which tends to dissipate or drain away the energy of that vibration. Damping devices are thus of a type tending to produce friction or bring other energy-absorbing mechanisms into place. For example, the shock absorbers on an automobile dissipate oscillatory energy by hydraulic means. Some tall buildings possess natural damping in the form of masonry cladding walls (the Empire State Building is an example). The inherent friction in masonry construction (or between masonry and steel) is often an effective oscillation damper. Modern steel-and-glass skyscrapers, on the other hand, possess relatively little natural damping.

The tuned mass damper represents an alternative to the damping devices mentioned above. TMD devices have been successfully used to arrest vibrations of power line cables, bridge, and highway-sign structural members, chimneys, tall towers, and antennas, and are used extensively in the automotive and aircraft industry. A TMD does not really act as a damper through energy drainage; instead, it performs in a “reactive” way, temporarily “bouncing” away from the mass it controls. The principles of the tuned mass damper as discussed in the following paragraphs.

Example

Consider the single spring-mass oscillator depicted in Figure 1. The tuned mass damper consists basically of another mass M_2 ( the secondary mass) attached by a spring to the first (M_1) as suggested in Figure 4. The secondary mass is “tuned” to the first in the following sense: The stiffness of spring K_2 and the value of the mass M_2 are adjusted so that the natural vibration frequency of mass M_2 when M_1 is held rigidly still, is the same as that of mass M_1 when M_2 is absent entirely. Clearly, the sense of the word “tuned” in this context is that one vibration frequency(that of the secondary mass alone) is brought into coordination with another frequency (that of the first mass alone).

The effect of the presence and tuning of the secondary mass of the TMD is usually dramatic upon the action of the first mass. It tends to greatly reduce or suppress (i.e., “damp”) the vibratory motion of the primary mass. The exact conditions of this action require, and are susceptible to, extensive mathematical analysis.

Suppose a primary mass is excited by some repetitive force that recurs at somewhere near the natural frequency of the primary system. The mass responds by oscillating at large amplitudes, i.e.,”resonating”. Again, suppose a randomly varying force, repetitive but not just occurring at one single frequency, excites the primary mass in a “broad spectrum” of way (i.e., at a mixture of oscillations of many frequencies simultaneously). In either of these cases, the primary mass oscillates manly at its own natural frequency; in the randomly excited case it will also take on random amplitudes.

The action of a tuned secondary mass in either of the above cases is to “take upon itself” the energy of the initial motion of the primary mass and to drastically reduce the response of the latter. This means that any exciting force which acts upon the primary mass must begin over and over again to build up the response of the latter from a low point, since each time that the response of this mass has been even temporarily excited, the TMD reacts against it and suppresses it. As a result there is much less occasion for the primary mass ever to be excited to a condition of excessive displacement, as it might be without the TMD.

The amount of secondary mass needed to accomplish the described effect can be variable, but it has been found that even if it is as small as on percent of the primary mass, it can still produce rather dramatic effects.

When a large, spatially-distributed primary mass consists of a building, like a skyscraper, the entire building may be represented mathematically as a single spring-mass vibrating system, considering not its entire mass as the primary oscillatory element, but only some equivalent(generalized) mass as the necessary element to be dealt with. This is to say that, since part of the building(for example, at the foundation) moves very little, while other parts(say at the top) move much more, the net effective moving mass has a value somewhere between the full mass and zero. This generalized mass also has a location or height coordinate – usually near the top of the building. The magnitude of the generalized mass may then typically range from 1/2 down to 1/4 or even less of the total mass of the building.

The actual amount of generalized mass to be considered depends upon just how the building vibrates – in technical terms. the modes in which it sways. Figure 5 depicts the amount of building sway at maximum travel in typically the first mode. The first mode of tall building sway consists of almost pure “lean” of the building all to one side, then all to the opposite side. Torsional or building twist modes can also occur. In these, the building rotates about an axis more or less coinciding with a straight, vertical line running through the geometric center of each story. Modes higher than first and second can occur in tall buildings but in almost practical cases responses in modes other than the first mode are less important.

In connection with a tall building, a TMD will be aimed principally at suppressing building sway or twist in “early” modes, say first in sway and first in twist. A possibility, if there is space, is to place M_2 inside M_1, as suggested in Figure 6. This is precisely the kind of situation which exists in a tall building, where a floor may contain the TMD (M^2).

Typically, TMD installations would be a high upper floor (refer to Figure 5) to work against the first swat mode(and twist, if important). The best choice of TMD placement is near the location of the expected highest amplitude of the mode to be suppressed.

The displacement of the secondary mass M_2 of the TMD will, when the device is working best, be greater by a factor of three or more depending on TMD design than the displacement of the primary mass. Thus, adequate space and other provisions must be provided for such motion. A second point is that some local damping provided to mass M_2 “broadens” its ability to effectively respond to excitations not precisely at the frequency of tuning. The amount of this local TMD damping also controls the relative displacement of the TMD mass M_2. The M_2 relative displacement may be reduced by increasing TMD local damping, at the expense of somewhat reduced TMD effectiveness.

The net result of the TMD installation is that the primary mass will respond to wind as if it were damped to an amount at least twice as much as its actual inherent damping. For exmaple, if actual building damping is 1\%, a TMD with mass of 1/100(.01) the primary (effective) mass can increase total building damping in the first mode to about 3\% while a TMD with mass of 1/50(.02) the primary mass can increase total damping in the first mode to over 4\%.

For one of the first TMD systems in a tall office building it was estimated that, to achieve the same reduction of building wind motion, the cost used a TMD system was less than one third of the cost of a conventional approach involving increased building mass and lateral stiffness.

System Description

Principle of Operation:

The MTS Tuned Mass Damper Systems use a moderate force servohydraulic actuator to control the position of a passively resonant spring-mass system. The servoactuator is sized and electronically programmed so that the exact damping and resonant frequency can be electronically set to assure linearity, long term stability, and optimum system characteristics for the particular input motion experienced.

The Tuned Mass Damper systems combine the damper and the servo-hydraulic drive into one servo valve/actuator combination. The system uses substantial flow rates of oil, but requires only a moderate supply pressure minimizing energy consumption while permitting sufficient active actuator forcing capability for accurate operation and adequate cooling.

The passive spring function is derived from two precharged pneumatic cylinders operated in a unique mechanical configuration to achieve force-deflation linearity.