Consider a mass-spring-dashpot system, described by the equation: mu** + cu* + k

Question

Consider a mass-spring-dashpot system, described by the equation:

mu** + cu* + ku = f where m denotes mass, c is the damping coefficient, k is the stiffness of the spring, u = u(t) is the displacement of the mass, f = f (t) is the external force applied to the mass, and an asterik (*) denotes differentiation with respect to time. We are interested in computing the displacement u(t) as a function of time t . Assume the initial displacement of the mass is u(0) = 0.5m, and its initial velocity is v(0) =1.0 m/s . Moreover, assume:

f(t) = 0 N

k = 4 N/m

c = 0.05 N.s/m

m = 1.0 kg

1. Write the second-order differential equation as a system of first-order differential equations.

2. Write a MATLAB script that implements the forward Euler (FE), backward Euler (BE), and Crank-Nicolson (CN) method for the computation of the displacement, for a period of T = 50s. Your program should accept the time step (delta t) as input. You should run 3 simulations for each method:

a. For delta t = 0.1s

b. For delta t = 0.01s

c. For delta t = 0.001s

For each time step delta t , plot the displacement as a function of time for the three methods in a single figure, and compare them with the exact solution. For instance, the first figure compares FE, BE, and CN, for delta t =0.1s, against the exact solution (total 3 figures). Use MATLAB for generating the plots, and use appropriate labels and legends.

3. Compute the maximum absolute error for each method, for each time step. For instance, maximum absolute error of BE when delta t = 0.1s ; maximum absolute error of BE when delta t = 0.01s , and so on (total 9).

Explanation / Answer

Consider the single-degree-of-freedom (SDOF) system shown at the right that has both a spring and dashpot. If we examine a free-body diagram of the mass we see that an additional force is provided by the dashpot. The force is proportional to the velocity of the mass

. Fdamping = cu&

where c is the viscous dashpot coefficient. Summing forces in the vertical direction yields:

W k(u ) cu mu static - + d - & = &&

After simplifying using Eq. 2, the equation of motion of the system is: m&u& + cu& + ku = 0 (

or &u& & c m u k m + + u = 0 (16b)

To solve this differential equation, assume a solution of the form:

u t Aert ( ) = (17) Take the first and second derivative with respect to time of Eq. 17

yielding: u&(t) rAert = (18a) &u&(t) r Aert = 2 (18b) and substitute Eqs. 18a and b into giving:

r Ae c m rAe k m Ae 2 rt rt rt + + = 0

which simplifies to: m k c u(t) m W k(u + dstatic ) cu& r c m r k m 2 + + = 0

Solving for r produces two possible roots: r c m c m k m = - ± æ è ç ö ø ÷ - 2 2 2

Thus the solution is given by: u t Ae Be r t r t ( ) = + 1 2

We define a critical value of c such that the term inside the radical equals 0: ccrit = 2 km

and a fraction of critical damping: b w = = c c c crit m n 2

Rearranging yields: c m n 2 = w b

which can be substituted into to yield: r i = -wnb ± wn 1- b 2

After substituting, becomes: u t e [Ae Be ] n n n t i t i t ( ) = + -w b - w 1-b w 1-b 2 2

To evaluate the coefficients A and B in

, we need two initial conditions. Usually, these are the initial displacement and velocity of the mass:

u(t = 0) = u0 (27a) u&(t = 0) = u& 0 (27b)

After substituting these expressions into Eq. 26 and its first derivative, we obtain: ( ) ( ) u t e u i u u e u i u u e n n n t n n i t n n i t

( ) & & = + - - æ è ç ç ö ø ÷ ÷ + - - - æ è ç ç ö ø ÷ ÷ é ë ê ê ù û ú ú -w b - w -b w -b w b w b w b w b 0 0 0 2 1 0 0 0 2 1 2 2 1 2 2 1 2 2 (

Consider the solution for three different values of b: 1. b = 1 (critically damped) For b = 1 reduces to: r = -wn

and the partial differential equation has repeated roots. As a result, the solution takes the form: u t [u (u u )t]e n t n ( ) = + & + - 0 0 w 0 w

shows the response of a critically damped SDOF for three initial conditions. The initial displacement is equal to 1.0, but the initial velocity of the mass is 1.0, 0.0, and –1.0 for the three different cases. Notice that the motion quickly diminishes to zero because of the large damping in the system

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