|
| |
HOMEWORK SET #1
| Problem |
Hints |
Answers |
| 1.22 |
|
|
| 2.1 |
|
|
| 2.2 |
The answer to the right assumes
6061-T6 Aluminum. |
12.03 in |
| 2.3 |
The easiest way to find the
static equilibrium point is to define a coordinate system where the origin
is at the location where the nonlinear spring is undeflected so the
nonlinear spring force is f(x)=24,000x2. If this is the
case, what is the force in the linear spring in term of x? |
b) 50 rad/s
when L = 0.5m
c) 22.36 rad/s
when L = 0.75 |
| 3.1 |
Note that the answer given on
the handout is missing a term. The equation of motion should have an
"x" term on the right hand side of the equation. This will
come from the spring force which should have an (x-y) or (y-x) depending on
what direction you draw the force in your FBD. |
y(t) = -0.18 cos8t + 0.18 |
| 3.2 |
|
18.54 rad/s |
| 3.3 |
|
a) 208
rad/s (33.1 Hz)
b) 108 rad/s (17.2 Hz) |
| 4.1 |
|
Decay
rate is 1/2 of the original. |
| 4.2 |
|
n = 2 cycles for
z
= 0.04
n = 115 cycles for z
= 0.0005 |
| 4.3 |
|
a) 3.65 N-m-s
b) 2e-9.13t(cos18.85t+0.484sin18.85t) (deg)
c) 0.095 degrees |
HOMEWORK SET #2
| Problem |
Hints |
Answers |
| 5.1 |
|
ii) a) k = 50,916 N/m, C = 353 N-s/m
b) k = 50,304 N/m, m = 0.0503 |
| 5.2 |
Click here to download data.mat. If just
clicking on this link doesn't give you the option of saving the file, then
right click on it.
There is a typo in the
handout. The last two plots I want are semilog plots of abs(h1) vs.
time and abs(h2) vs. time. |
|
| 7.1 |
|
2 rad/s |
| 7.2 |
|
w
< 61.6 rad/s or w
> 64.8 rad/s |
| 7.3 |
|
-8.57 x 10-4
sinwt rad |
| 8.1 |
|
5.2 x 10-4
m |
| 8.2 |
|
0.118 |
HOMEWORK SET #3
| Problem |
Hints |
Answers |
| 9.1 |
You only want to determine the stress in the spring, so you want to use the
force in the spring, not the total force transmitted. |
The stress is about 1.31 x 106 N/m2
so the maximum stress is exceeded. |
| 9.2 |
|
1.70 x 10-4 m |
| 10.1 |
Use e from problem 2.72, that is, e = 0.3 m. |
FT = 99.2 N |
| 10.2 |
|
me = 9.91 kg-m
z = 0.396 |
HOMEWORK SET #4
| 13.1 |
|
4.55% (so it went down), but the phase error will be about 5.2degrees. |
| 13.2 |
|
|
| 14 |
|
|
|
15.1 |
|
1.52 x 105
N/m |
|
15.2 |
|
6.44 x 105
N/m |
HOMEWORK SET #5
| Problem |
Hints |
Answers |
| 17.1 |
|
|
| 17.2 |
|
|
| 18.1 |
The figure in the book is not
correct. At t = 5 the curve should be going down. |
0.611*sin(0.785*t)+ 0.
257*sin(2.356*t)-0.0215*sin(3.927*t) m (no damping) |
| 18.2 |
|
a)
4.082+6.081cos1.047t+1.968cos(2.0944t)
(mm)
b) 4.082+6.098cos(1.047t-0.0000086) + 1.990cos(2.0944t-.000070)
(mm) |
| 19.1 |
|
|
| 19.2 |
|
|
| 20.1 |
Note that the modes may be
multiplied by any constant and it is still a mode. |
w1 =
521 rad/s
w2
= 2575 rad/s
mode1 = {1 0.955}T
mode 2 = {1 -0.105}T |
| 20.2 |
|
w1 =
0.874sqrt(k/m)
w2 =
2.288sqrt(k/m)
mode1 = {0.618 1}T
mode 2 = {-1.62 1}T |
|
20.3 |
Don't forget the weight on your
FBD's. |
w1 =
2.29 rad/s
w2
= 2.40 rad/s
mode1 = {1 1}T
mode 2 = {1 -1}T |
HOMEWORK SET #6
| Problem |
Hints |
Answers |
|
21.1 |
|
x1 =
0.212cosw1t
+ 0.384sinw1t
+ 0.789cosw2t
- 0.240sinw2t
x2 =0.289cosw1t
+ 0.524sinw1t
- 0.289cosw2t
+ 0.088sinw2t |
|
21.2 |
|
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 |
| |
|
|
| |
|
|
| |
|
|
HOMEWORK SET #7
| Problem |
Hints |
Answers |
| 25.1 |
|
k2/m2
= w2
= 100 |
| 25.2 |
For your Simulink solution be sure your input is [50 0] |
x1(t) =
0.00977sin4pt
x2(t) = 0.0161sin4pt
Click here for no damping
Click here for with damping |
| 26.1 |
|
m2
=
0.194 kg
k2 = 7
N/m |
| 26.2 |
|
a) Wa = 489.4 lbf
b) w1
49.3 rad/s
w2
80.7 rad/s
c) m2
21.13 lbf-s2/ft
k2
83439 lbf/ft |
| |
|
|
HOMEWORK SET #8
| Problem |
Hints |
Answers |
| 28.1 |
|
 (click on image) |
| 28.2 |
|
 (click
on image) |
| 29.1 |
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 |
HOMEWORK SET #9
| Problem |
Hints |
Answers |
| 31.1 |
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
|
| 31.2 |
|
a) mode 1 is not mass
normalized
b) -0.562 |
| 32.1 |
The modes are mass normalized
so you know what [F]T[K][F]
equals. |
x1(t)
= 0.5 cos(t)-0.5 cos(sqrt(3)t)
x2(t) = cos(t)
x3(t) = 0.5 cos(t)+0.5 cos(sqrt(3)t)
|
| 33.1 |
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).
These answers have not been checked so if you get something different please
let me know. |
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
|
|
