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MECH315:Mechanics3 

SOLUTIONS 1 Question 1C.  1 4 2  GIVEN: matrix   4 0 8   2 8 2    GOAL: Find the eigenvalues and eigenvectors of the matrix using MATLAB. SOLUTION: Enter the following commands: >> matrix = [1 4 -2; 4 0 8; -2 8 2]; >> [eigenvector, eigenvalue] = eig(matrix) The output is obtained as shown below: eigenvector = -0.4007 -0.9022 -0.1594 0.7035 -0.1915 -0.6844 -0.5870 0.3864 -0.7115

eigenvalue = -8.9532 0 0 0 2.7054 0 0 0 9.2478 Finally, analyze the output to obtain the results shown below.







  0.4007 The first eigenvalue is -8.9532 and the corresponding eigenvector is  0.7035  .     0.5870   0.9022   The second eigenvalue is 2.7054 and the corresponding eigenvector is  0.1915 .    0.3864    0.1594   The third eigenvalue is 9.2478 and the corresponding eigenvector is  0.6844 .     0.7115

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MECH315:Mechanics3 

Question 2.

GOAL: Solve the second-order, ordinary differential equation: 󰇘  5.6󰇗  18.5  6.5 exp󰇛2.3󰇜

with initial conditions 󰇛0󰇜  0.5󰇟󰇠;󰇗 󰇛0󰇜  1.5󰇟/󰇠

and plot the results using MATLAB. APPROACH: Reformulate problem as a system of coupled first-order ODEs, and solve the system of equations using the ode23 function in MATLAB. IMPLEMENTATION OF APPROACH: Use the substitution   ;  󰇛0󰇜  0.5󰇟󰇠    󰇗 ; 󰇛0󰇜  1.5󰇟/󰇠

to rewrite the problem as 󰇗     󰇗    5.6  18.5  6.5exp󰇛2.3󰇜

The corresponding MATLAB function can then be constructed as shown below. function f = x_dot(t,x) % x = [x1; x2] f = [x(2); -5.6*x(2) - 18.5*x(1) + 6.5*exp(-2.3*t)];

This system can be solved using ode23, and the results plotted on a graph, using the code shown below. x0 = [0.5;1.5]; % Initial conditions [x1(0); x2(0)] t = [0 2]; % Range of time values [t,x] = ode23(@x_dot, t, x0); % x = [x1, x2] plot(t,x(:,1),'b-',t,x(:,2),'r-') title('Dynamic response x_1 and x_2 with ODE23'); xlabel('Time t'); ylabel('Solution'); legend('x_1','x_2')

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MECH315:Mechanics3 

Question 3.

TO FIND: Energy dissipated per cycle of vibration by a viscous dashpot, c, with one end fixed and the other end undergoing simple harmonic motion given by y  A sin  t . APPROACH: Use the defining properties of a viscous dashpot to set up the equation for energy dissipation. SOLUTION: By definition, the energy dissipated per unit time by the dashpot is given by dEd  c  x  y 2 dt

(1)

where x and y are the velocities of the two terminals of the dashpot. In this problem, one end is fixed and the other end undergoes simple harmonic motion. Hence, x  0

y 

and

d ( A sin t )  A cos t . dt

(2)

Combining Eq. (1) and Eq. (2), we obtain dEd  cA2 2 cos 2 t dt

(3)

Eq. (3) gives the energy dissipated per unit time. To get the energy dissipated over one cycle of vibration, we integrate over time from t  0 to t   , where  is the time period of oscillation. For the simple harmonic motion under consideration,   2 /  . Hence, 





0

0

0

dE E d   d dt   cA2  2 cos2  t dt  cA2 2  cos2  t dt dt 2 / 

Using the standard integral (



0

cos2  t dt 

(4)

 ) and substituting this result into Eq. (4), we obtain 

Ed   cA2

(5)

Question 4.

STEP 1: Kinematics For oscillation with small amplitude, we have x1  L2  and x 2  L1 . STEP 2: Kinetic Energy Using the defining properties of particles and rigid bodies, the total kinetic energy of the system is given by T 

1 1 2 2 m1x1  J 0 . 2 2 3

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MECH315:Mechanics3 

Finally, substituting x1  L2 in the expression for the kinetic energy, we obtain





1 1 1 T  m1L222  J 0 2  m1 L22  J 0 2 . 2 2 2

Question 5. The total kinetic energy of the system is given by 1 1 T  (m1  m2 ) x2  m2 L2 2  m2 L x cos . 2 2

Question 6.

The total elastic potential energy of the system is given by













2 1 2 1 2 1 1 5 V  k x12  k x1  x 2  k x2  x3  k x32  k x1  x3 2 2 2 2 2

BONUS QUESTION Please consult Section 1.6 of the textbook for a description of modeling of mechanical systems using discrete elements (also known as lumped-parameter elements). Thus, inertia is represented using point masses and rigid bodies; elasticity is represented using springs and torsional springs; and dissipation is represented using viscous dampers (also known as dampers). Figure 1.16 shows the iconic models for a forging hammer, and Figure 1.17 illustrates four different iconic models for the vibrations of a motorcycle. In addition, consult the Case Study on the vibrations of the Millennium Bridge (available on myCourses). This article describes the primary steps of vibration analysis consisting of: (i) a clear and succinct statement of the vibration problem; (ii) modeling the Millennium Bridge as a system with a single degree of freedom; (iii) deriving the equation of motion for the iconic model; and (iv) solving the equation of motion to obtain insights into the vibrations of the bridge.

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