| Problem | Hints |
Answers |
| 1.1 | Make sure you do the review problem. This is a fill in the blank problem. Please write the complete sentence with the correct answer. You can also make a photocopy of the problem statement and tape it to green engineering paper. | |
| 1.2 | Make sure you do the review problem. This is a multiple choice problem. Please write the complete question and then indicate the correct answer. You can also make a photocopy of the problem statement and tape it to green engineering paper and circle the correct answer. | |
| 2.1 |
b)
36.9 rad/s c) 15.5 rad/s |
|
| 2.2 | Be careful with units since f1 is in Hz and wn is in rad/s. Also be careful with the units of density. The units lbf/in3 is a weight density, not mass. You will need to divide by 386.4 to have the mass density in units of lbf-s2/in4. | b) 12.2 in. |
| 3.1 | 22.1 rad/s (or 3.52 Hz) | |
| 3.2 | The object rolls without slipping, so don't forget the friction force. |
 |
| 3.3 | a)
208 rad/s (33.1 Hz) b) 108 rad/s (17.2 Hz) |
|
| 4.1 | 2.71 N-m-s | |
| 4.2 | Remember this problem is modified from the problem statement in the book. c = 12 N-s/mm so the system is underdamped. | xmax
=
1.05 m t = 0.252s |
| 4.3 | For c = 905.1 n-s/m, k 88,830 N/m | |
| 5.1 | When using Simulink with Coulomb damping, use a fixed time step or it will sometimes lock up. | ii)
a) k =
40,740 N/m, C =
283 N-s/m b) k = 40,240 N/m, m = 0.050 |
| 5.2 | The data can be downloaded by clicking here | |
| 6.1 | Remember this is a review problem. | |
| 6.2 | Don't forget to include the equivalent mass of the spring and be careful with units. | 2.526 sin8t in |
| 6.3 | w < 61.6 rad/s or w > 64.8 rad/s | |
| 7.1 | There is no need to combine the sin and cos term in the force transmitted. | Answer not provided. |
| 7.2 |
This answer assumed the maximum occurs at r= sqrt(1-2zeta^2). |
k = 1.007 x 105 N/m, C = 633.4 N-s/m |
| 8.1 | 1.696x 10-4 m | |
| 8.2 | Don't forget to lump the mass of the long rod to the mass of the oil. |
21.7 bbl/h (15.0 bbl/h if rod was rigid) |
| 9.1 | xss
= 0.397 mm FT = 32.9 N |
|
| 9.2 | The answer shown to the right are using r2 = 1. If you use r2 = frequency for peak response you get different answers (which the problem statement also asks you to find). |
me = 9.91 kg-m z = 0.396 |
| 10.1 | 1.52 x 105 N/m | |
| 10.2 | Be sure to check the entire frequency range. The initial forces given are simply the force transmitted if the engine was rigidly attached to the ground that is mew2. | 6.44 x 105 N/m |
| 14.1 |
Here is a snapshot of my time
response. |
|
| 14.1 | Just solve this problem in Maple. The input will not look exactly like the one in the problem statement. The maple worksheet tends to work better if you use fractions in your piecewise definition (3/2 instead of 1.5, etc.) Also, when you plot you may need to increase the number of points (plot(y(70),t=0..2*T,numpoints=500) | a)
40.82+60.81cos1.047t+19.68cos(2.0944t) b) 0.0408+0608cos(1.047t-.0003) + 0.197cos(2.0944t-.00063) |
| 15.1, 15.2 | Answer not provided | |
| 16.1 | zmax = 0.708 ft | |
| 17.1 | w1 = 0.518sqrt(k/m) w2 = 1.932sqrt(k/m) mode1 = {1 1.366}T mode 2 = {1 -0.366}T x1 = -0.366cosw1t + 1.366cosw2t x2 = -0.5cosw1t -0.5cosw2t |
|
| 17.2 | Don't forget to put mg on your FBD. Assume small angles. | w1 = sqrt(g/L) w2 = sqrt(g/L+2ka2/(mL2)) mode1 = {1 1}T mode 2 = {1 -1}T |
| 18.1 | After finding the natural frequencies, determine what velocities would result in those frequencies. | 49.69 km/hr and 194.2 km/hr |
| 18.2 | I found the center of gravity to be 3 feet from the left side. Use the parallel axis theorem to find the mass moment of inertia about G. | w1 = 3.83 rad/s w2 = 8.91 rad/s mode1 = {1 -.049}T (if x = 1, q = -0.049) mode 2 = {1 8.57}T |
| 19.1 | This is not a forced response problem! You will have to go back to ES204 material and remember how to do an impact problem and how to use the coefficient of restitution. You need to do this to find an initial velocity for the anvil. You should find omega1 = 66.3 rad/s and omega2 = 208.9 rad/s. The frequencies do not include the mass of the tup. I defined x1 to be the foundation and x2 to be the upper mass. | x1(t)=0.0818sinw1t
- 0.026sinw2t
in. x2(t)=0.101sinw1t + 0.0295sinw2t in. |
| 19.2 | For your Simulink solution be sure your input is [50 0] | x1(t) =
0.00977sin4pt x2(t) = 0.0161sin4pt |
| 20.1 | You must derive this. After this problem you may just state this as one of the conditions for a vibration absorber. | k2/m2 = w2 |
| 20.2 | Assume the mass m2' was chosen arbitrarily and that the original system was being forced at resonance. There is not a unique solution to this problem. One answer that works is shown to the right. The equivalent mass of the pipe is 20.16 kg (be sure you show this!) and k1 = 141,500 N/m | m2 =
5.68 kg k2 = 39,860 N/m (these give natural frequencies of 615 rpm and 1040 rpm) |
| 21.1 | Be sure to use Lagrange's Equations. |
 (click on image) |
| 21.2 | Assume small angles. |
 (click
on image) |
| 21.2 | For the modes the first coordinate is x and the second q. The hardest part of this problem is the kinetic energy for Lagrange's equations. Also, be careful with your derivatives. |
 (click
on image)w1 = 0.68 rad/s w2 = 2.29 rad/s mode1 = {0.1756 0.0092}T mode2 = {-0.050 0.365}T |
| 25.1 | Remember that Ig for a slender bar is mL^2/12 |
w1
= 39.7 rad/s w2 = 87.2 rad/s w3 = 122.5 rad/s
|
| 26.1 | The matrix shows up better using Internet Explorer rather than Netscape. |
w1
= 0.735 sqrt(k/m) rad/s w2 = 1.66 sqrt(k/m) rad/s w3 = 2.17 sqrt(k/m) rad/s  |
| 26.2 |
a) mode 1 is not mass
normalized b) -0.562 |
|
| 27.1 | Don't forget to compare your theoretical answer to a Simulink model. | (click here answer) |
| 28.1 |
Since we are just interested in
the steady state solution, assume the homogeneous solution is transient
(even though in this case it is not since there is zero damping). Your answers may be slightly different depending on round-off error. |
b) w1
= 19.23 rad/s w2 = 44.72 rad/s w3 = 58.14 rad/s c) mode1 = {0.00117 0.00186 0.00137}T mode2 = {-0.001054 0.001054 -0.00527}T mode 3 = {-0.00159 0.000667 0.00451}T d) x1(t) = 0.154x10-5sin40t x2(t) = -0.257x10-5sin40t x3(t) = 0.428x10-5sin40t |
