solhw05 (1) 3.8-3.14
TRANSCRIPT
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8/3/2019 SolHW05 (1) 3.8-3.14
1/3
German University in Cairo
Modeling and Simulation in
Mechanics Department
Introduction to Vibration
ENME 502 (Mechanics II)
1
Problem 3.8
Solve the differential equation:
( )tkAkxxcxm sin=++ &&& Assuming )sin()()( = tXtx
Solution
)sin()()( = tXtxp
)sin()cos()( tBtAtx ppp +=
Where:
=+=
p
p
ppB
ABAX 122 tan,
By differentiation:)cos()sin()( tBtAtx ppp +=&
)sin()cos()( 22 tBtAtx ppp =&&
Substitute in the original differential equation:
( ) ( )tAxxxtkAkxxcxm nnn sin2sin22 =++=++ &&&&&&
)sin()sin()cos()cos(2)sin(2)sin()cos( 22222 tAtBtAtBtAtBtA npnpnpnpnpp =+++
( ) ( ) )sin()sin(2)cos(2 22222 tAstABBtABA npnppnpnpnp =+++ Equate the coefficients the sine and cosine in both sides:
02 22 =++ pnpnp ABA
AABB npnppn 222 2 = Put the above two equation in matrix form:
=
AB
A
np
p
nn
nn
222
22 0
2
2
From the above matrix we get:
( ) ( )( )
( ) ( )( )ABAA n
nn
npn
nn
np
2
2222
222
2222 2
)(,
2
2
+
=
+
=
( )
( ) ( ) ( ) ( )2222
2
2222
22222
22
22
)2(
nn
n
nn
nnn
pp
AABAX
+
=
+
+=+=
( ) ( ))sin(
2
)sin()(2222
2
+
++= tA
tCetx
nn
nd
tn
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8/3/2019 SolHW05 (1) 3.8-3.14
2/3
German University in Cairo
Modeling and Simulation in
Mechanics Department
Introduction to Vibration
ENME 502 (Mechanics II)
2
Problem 3.14
Required:
Calculate the response of the vehicle z(t). Force transmitted to the vehicle.
Solution
Since the vehicle is moving with constant velocity, we can find relation between x
and t:
vdt
dx
=
vtxvdtdx by = = nintegratio Substitute this relation into the y(x) to get:
=
= t
L
vAtyx
LAxy
2sin)(
2sin)(
( )L
vtAty
2:where,sin)( ==
Free-Body Diagram for the mass:
The force transmitted to the vehicle is: )()( yzcyzkF && +=
The equation of motion of the mass:
=
L
xAxy
2sin)(
m
)( yzk )( yzc &&
z
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8/3/2019 SolHW05 (1) 3.8-3.14
3/3
German University in Cairo
Modeling and Simulation in
Mechanics Department
Introduction to Vibration
ENME 502 (Mechanics II)
3
= zmF && zmyzcyzk &&&& = )()(
Rearrange the equation: kyyckzzczm+=++ &&&&
Solve the above equation assuming particular solution
)sin()( tYty =
Following the same procedure used in the lecture, the general solution will be:
( ) = tZtz cos)(Where:
=
22
1 2tan
n
n
( )
( ) ( )2222
21
12
rr
rYZ
+
+=
n
r
=
L
v
2=