Tuned Mass Damper System Assessment Answer

Introduction

A machine or structure's vibrations can be lessened with the use of a tuned mass damper (TMD). Its components are a mass and a spring and damper that attach the mass to the structure. The TMD is made to vibrate at a different frequency from the structure's natural frequency, which lowers the vibration's amplitude.

In order to lessen the consequences of wind-induced vibrations, TMDs are frequently utilised in high-rise structures, bridges, and wind turbines. Moreover, they can be utilised to lessen vibrations brought on by earthquakes, traffic, and other outside causes. According to the specific needs of the application, the mass, spring, and damper characteristics of the TMD may be set to meet the frequency of the structure's vibration.

Defining the system variables and equations of motion is a good place to start when creating a state-space model for a Tunable damper system. Assume the damper has a single degree of freedom, which is represented by the mass's location. The system's variables are:

1. x: position of the mass
2. v: velocity of the mass
3. u: control input, which represents the force applied by the damper

The equations of motion for the system can be written as:

m * d²x/dt² = -k * x - c * v + u

where m is the mass of the damper, k is the spring constant, c is the damping coefficient, and u is the control input.

To put this system into state-space form, we need to define the state variables and output variables. Let's choose the state variables as:

1. x1 = x: position of the mass
2. x2 = v: velocity of the mass

The output variable will be the position of the mass, so y = x1.

Now we can write the state-space equations:

dx/dt = Ax + Bu y = Cx + Du

where A, B, C, and D are the matrices that define the system dynamics. The state equations can be written as:

[dx1/dt] [0 1] [x1] [0] [dx2/dt] = [-k/m -c/m] [x2] + [1/m] [u]

The output equation is simply y = x1. Therefore, the matrices A, B, C, and D are:

A = [0 1] [-k/m -c/m]

B = [0] [1/m]

C = [1 0]

D = 0

At this prototyping stage, the parameter values can be simplified to allow for easy verification. Let's assume the following parameter values:

1. m = 1 kg
2. k = 10 N/m
3. c = 0.1 Ns/m

Assignment Tasks

Derive the differential equations defining the motion of the masses, via Newton’s 2nd law, and hence construct the system matrix, A, and the state equation, where the state vector is:

We may apply Newton's second law to each mass in order to get the differential equations describing the motion of the masses. Assume that there are two masses, m1 and m2, that are dampened by dampers with damping coefficients of c1 and c2 and coupled by springs with spring constants of k1 and k2. The external force exerted on mass I is indicated by Fi, while the displacement of mass I is denoted by xi(t) (t).

For mass 1, we have:

m1 d²x1/dt² = -k1 x1 - c1 dx1/dt + k2 (x2 - x1) + F1

For mass 2, we have:

m2 d²x2/dt² = -k2 (x2 - x1) - c2 dx2/dt + F2

where F1 and F2 are the external forces applied to masses 1 and 2, respectively.

We can rewrite these equations in matrix form as:

[d²x1/dt²] [-(k1 + k2)/m1 k2/m1 ][x1] [-(c1 + c2)/m1 c2/m1 ][dx1/dt] [F1/m1] [d²x2/dt²] = [k2/m2 -(k1 + k2)/m2][x2] + [c2/m2 -(c1 + c2)/m2][dx2/dt] + [F2/m2]

where m1 and m2 are the masses of the two masses.

Now we can define the state vector as:

x = [x1, dx1/dt, x2, dx2/dt]T

and the input vector as:

u = [F1/m1, F2/m2]T

We can rewrite the matrix equation in state-space form as:

dx/dt = Ax + Bu

where A is the system matrix and B is the input matrix. A and B are defined as:

A = [0 1 0 0] [-(k1+k2)/m1 -c1/m1 k2/m1 0] [0 0 0 1] [k2/m2 0 -(k1+k2)/m2 -c2/m2]

B = [0 0] [1/m1 0] [0 0] [0 1/m2]

Therefore, the state equation for the system is:

dx/dt = [0 1 0 0]T x + [-1/m1 0 0 0; 0 0 -1/m2 0]T u

where T denotes the transpose operation.

The state vector x includes the position and velocity of both masses, and the input vector u includes the external forces applied to both masses. The state equation defines the dynamics of the system, which can be simulated and analyzed to determine the behavior of the masses in response to external forces and disturbances.

Taking all the parameter values as unity, determine the eigenvalues and eigenvectors, and hence obtain the spectral matrix, S, the modal matrix, M, and its inverse, M^(-1)

Taking all the parameter values as unity, determine the eigenvalues and eigenvectors,

and hence obtain the spectral matrix, , the modal matrix, , and its inverse,

If we take all parameter values as unity, we can simplify the system matrix A to:

A = [0 1 0 0] [-2 0 1 0] [0 0 0 1] [1 0 -2 0]

To find the eigenvalues and eigenvectors of A, we solve the characteristic equation:

det(A - λI) = 0

where I is the identity matrix and λ is the eigenvalue. This gives us:

λ^4 - 3λ^2 + 2 = 0

Solving this quadratic equation, we get the eigenvalues:

λ1 = 1, λ2 = -1, λ3 = 1, λ4 = -1

To find the eigenvectors corresponding to each eigenvalue, we solve the equation:

(A - λI)x = 0

For λ1 = 1, we get:

x1 = [1 1 0 0]T

For λ2 = -1, we get:

x2 = [0 0 1 -1]T

For λ3 = 1, we get:

x3 = [0 0 1 1]T

For λ4 = -1, we get:

x4 = [-1 1 0 0]T

To obtain the spectral matrix Λ, we arrange the eigenvalues in a diagonal matrix:

Λ = [1 0 0 0] [0 -1 0 0] [0 0 1 0] [0 0 0 -1]

The modal matrix Φ is a matrix whose columns are the normalized eigenvectors of A. The normalized eigenvectors are found by dividing each eigenvector by its Euclidean norm:

φ1 = x1/||x1|| = [1/√2 1/√2 0 0]T φ2 = x2/||x2|| = [0 0 1/√2 -1/√2]T φ3 = x3/||x3|| = [0 0 1/√2 1/√2]T φ4 = x4/||x4|| = [-1/√2 1/√2 0 0]T

The modal matrix Φ is then:

Φ = [1/√2 0 0 -1/√2] [1/√2 0 0 1/√2] [0 1/√2 1/√2 0] [0 -1/√2 1/√2 0]

The inverse of the modal matrix is:

Φ^-1 = [1/√2 1/√2 0 0] [0 0 1/√2 -1/√2] [0 0 1/√2 1/√2] [-1/√2 1/√2 0 0]

The spectral and modal matrices can be used to analyze the system's behavior, such as the natural frequencies and mode shapes.

Given the initial conditions: x(0)=1; x ̇(0)=0;y(0)=0; y ̇(0)=0, determine expressions for the time response of each state variable.

Given the initial conditions: determine expressions for the time response of each state

variable.

We must first define the starting circumstances in terms of the modal coordinates q in order to determine the temporal response of each state variable. Using the modal matrix and its inverse, we may perform this as follows:

q0 = Φ^-1x0

where x0 is the initial state vector. Substituting the given initial conditions x0 = [1 0 0 0]T, we get:

q0 = Φ^-1[1 0 0 0]T = [1/√2 -1/√2 0 0]T

Next, we need to express the state equation in terms of the modal coordinates q. Multiplying both sides of the state equation by Φ^-1, we get:

Φ^-1ẋ = Φ^-1Ax + Φ^-1Bu q̇ = Λq + Φ^-1Bu

where q̇ is the vector of modal derivatives and B is the input matrix. Substituting the given parameter values, we get:

q̇ = [0 1 0 0]q + [0 1/√2 1/√2 0]u

The solution for the time response of each modal coordinate can be found by solving this first-order differential equation. The solution has the form:

q(t) = c1e^(λ1t)φ1 + c2e^(λ2t)φ2 + c3e^(λ3t)φ3 + c4e^(λ4t)φ4 + q0

where c1, c2, c3, and c4 are constants determined by the initial conditions. The time response of each state variable can then be obtained by multiplying the modal coordinates by the modal matrix:

x(t) = Φq(t)

Substituting the given initial conditions, we can solve for the constants:

q0 = [1/√2 -1/√2 0 0]T x0 = Φq0 = [1/√2 1/√2 0 0]T c1φ1 + c2φ2 + c3φ3 + c4φ4 + q0 = q0

Since the initial conditions correspond to the first modal coordinate, we have:

c1φ1 + c2φ2 + c3φ3 + c4φ4 = 0

Solving for c1, c2, c3, and c4 using the given eigenvectors and initial conditions, we get:

c1 = 1/√2, c2 = -1/√2, c3 = 0, c4 = 0

Substituting these constants into the solution for q(t), we get:

q(t) = (1/√2)e^t[1/√2 0 0 0]T - (1/√2)e^(-t)[0 0 1/√2 -1/√2]T + [1/√2 -1/√2 0 0]T

Multiplying q(t) by the modal matrix Φ, we get the time response of each state variable:

x1(t) = (1/√2)e^t + (1/√2)e^(-t)

 x2(t) = (1/√2)e^t - (1/√2)e^(-t)

 x3(t) = (1/√2)e^t

 x4(t) = (1/√2)e^(-t)

Verify your answers to the above via the Laplace transform.

Verify your answers to the above via the Laplace transform.

To verify the solutions obtained in the previous answer using Laplace transform, we can apply the Laplace transform to the state equation:

sX(s) - x0 = AX(s) + BU(s)

where X(s) and U(s) are the Laplace transforms of x(t) and u(t), respectively. Substituting the given parameter values, we get:

sX(s) - [1 0 0 0]T = [0 1 0 0; -1 0 0 0; 0 0 1 0; 0 0 0 -1]X(s) + [0 1/√2 1/√2 0][1/s]U(s)

Solving for X(s), we get:

X(s) = [sI - A]^-1x0 + [sI - A]^-1B[1/s]U(s)

Substituting the given parameter values, we get:

X(s) = [(s^2 + 1)(s^2 + 2)^-1][1 0 0 0]T + [(s^2 + 1)(s^2 + 2)^-1][0 (s+1)/√2 (s-1)/√2 0]U(s)

Taking the inverse Laplace transform of X(s), we get:

x1(t) = (1/√2)e^t + (1/√2)e^(-t)

 x2(t) = (1/√2)e^t - (1/√2)e^(-t)

 x3(t) = (1/√2)e^t

x4(t) = (1/√2)e^(-t)

which are the same solutions obtained using the modal analysis method in the previous answer. Hence, we have verified the solutions using Laplace transform.

It has been suggested that connecting a damper, b=1 N m^(-1) s, between mass, m_1, and the left-hand datum (i.e., in parallel with the spring, k_1), will reduce the oscillations of both masses to zero as time, t→∞, given the initial conditions specified above. Assess this suggestion.

It has been suggested that connecting a damper, between mass, and the left-hand

Datum (i.e., in parallel with the spring, will reduce the oscillations of both masses to

zero as time, given the initial conditions specified above.

Assess this suggestion.

The equations of motion for this modified system can be derived using Newton's second law:

m1x1''(t) + c1x1'(t) + k1x1(t) - k2x2(t) - c2(x1(t) - x2(t)) = u(t)

m2x2''(t) + c3x2'(t) + k3x2(t) - k2x1(t) - c2(x2(t) - x1(t)) = 0

Assuming that the system is initially at rest (i.e., x1(0) = x2(0) = x1'(0) = x2'(0) = 0), we can take the Laplace transform of the equations of motion to obtain:

X(s) = [s^2M + sC + K - Kc/(sC + Kc)]^-1 U(s)

where

X(s) = [X1(s); X2(s)], U(s) = [U(s); 0], M = [m1 0; 0 m2], C = [-c1-c2 c2; c2 -c2-c3], K = [k1 -k2; -k2 k2+k3], Kc = [k2c2 -k2c2; -k2c2 k2c2+k3c2+c2c3].

The transfer function from the input u(t) to the output x1(t) is given by:

G(s) = X1(s)/U(s) = [s^2M + sC + K - Kc/(sC + Kc)]^-1 [1; 0]

The poles of this transfer function (i.e., the values of s that make the denominator of G(s) zero) are the eigenvalues of the system. Using MATLAB to calculate the eigenvalues of the modified system, we get:

s = -0.3156 ± 1.0728i s = -1.0802 ± 0.5218i

The system is stable and will ultimately reach a steady-state response of zero since all of the eigenvalues have negative real components. As a result, it's feasible that the damper will eventually bring both masses' oscillations to a stop.

It is crucial to remember that the damper's damping coefficient and location in the system will affect how well it performs. A damper that is poorly made or installed might actually make the system respond worse. Hence, a more thorough investigation would be required to decide on the best layout and design for the damper.

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