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MAE 340 – Vibrations
Numerical Simulation and DesignSection 2.8
Non-linear Response PropertiesSection 2.9
Forced Vibration with Coulomb Damping
• A single degree-of-freedom system with mass 10 kg, spring stiffness of 1000 N/m and a Coulomb damping coefficient of 0.3 is excited by a harmonic force of 100 N amplitude at 100 rad/s.
• Plot the EXACT response x(t). Assume x0 = -0.8683 mm and v0 = 35 mm/s.
MAE 340 – Vibrations 2
k
m
x
μF0cosωt
MAE 340 – Vibrations 3
taking derivative wrt time
Section 2.8
Numerical Simulation & Design
is: )()(
)()(
2
1
txtz
txtz
&=
=
212
21
zm
cz
m
kxz
zxz
−−==
==
&&&
&&
−−=
)(
)(10
)(
)(
2
1
2
1
tz
tz
m
c
m
ktz
tz
&
&
Azz =&
)(∆)()( 1 iii tttt Azzz +=+
Putting into matrix form
Approximating
t
ttt ii
i∆
)()()( 1 zz
z−
≈ +&
0=++ kxxcxm &&&
• Recall that the “state-space” formulation for free vibration
MAE 340 – Vibrations 4
taking derivative wrt time
Numerical Simulation & Design
• If we include the harmonic force term
)()(
)()(
2
1
txtz
txtz
&=
=
tm
Fz
m
cz
m
kxz
zxz
ωcos0212
21
+−−==
==
&&&
&&
+
−−=
t
m
Ftz
tz
m
c
m
ktz
tz
ωcos
0
)(
)(10
)(
)(0
2
1
2
1
&
&
)(tfAzz +=&
)(∆)(∆)()( 1 iiii tttttt fAzzz ++=+
Putting into matrix form
Approximating
t
ttt ii
i∆
)()()( 1 zz
z−
≈ +&
tFkxxcxm ωcos0=++ &&&
it becomes:
MAE 340 – Vibrations 5
taking derivative wrt time
Section 2.9
Non-linear Response Properties
• General formulation for any spring and damper forces:
)()(
)()(
2
1
txtz
txtz
&=
=
tfzzfxz
zxz
ωcos],[ 0212
21
+−==
==
&&&
&&
)()( tfzFz +=&
tttttt iiii ∆)(∆)]([)()( 1 fzFzz ++=+
Putting into matrix form
Approximating
t
ttt ii
i∆
)()()( 1 zz
z−
≈ +&
tftxtxftx ωcos)](),([)( 0=+ &&&
• Setting up “state space” numerical time-integration:
+
−=
tfzzf
tz
tz
tz
ωcos
0
),(
)(
)(
)(
011
2
2
1
&
&
MAE 340 – Vibrations 6
taking derivative wrt time
Non-linear Response Properties
• For Coulomb damping:
)()(
)()(
2
1
txtz
txtz
&=
=
tm
Fz
m
Nz
m
kxz
zxz
ωµ
cos)sgn( 0212
21
+−−==
==
&&&
&&
tFkxxNtxm ωµ cos)sgn()( 0=++ &&&
• Setting up “state space” numerical time-integration solution:
tt
m
Fttz
m
Ntz
m
k
tz
tz
tz
tz
tz
iii
i
i
i
i
i∆
+∆
−−+
=
+
+
)cos(
0
))(sgn()(
)(
)(
)(
)(
)(0
21
2
2
1
12
11
ωµ
MAE 340 – Vibrations 7
Numerical (Actual) Nonlinear Response compared with Linearized Response
• Actual problem using solved numerically
• Linearized problem using solved algebraically.
)sgn(xmgFdamping&µ=
X
mgceq
πω
µ4=
MAE 340 – Vibrations 8
Numerical (Actual) Nonlinear Response compared with Linearized Response
• Same as above with higher friction coefficient.
MAE 340 – Vibrations 9
Comparing Actual NonlinearResponse with Linearized Response• Non-linear spring (Fspring = kx – k1x
3) versus linear spring (Fspring = kx)
ω=ωn/2.964
MAE 340 – Vibrations 10
Comparing Actual NonlinearResponse with Linearized Response• Non-linear spring (Fspring = kx – k1x
3) versus linear spring (Fspring = kx)
ω=ωn/1.09
MAE 340 – Vibrations 11
Comparing Actual NonlinearResponse with Linearized Response
• Displacement-squared damping ( )
versus use of ceq ( )
2)sgn( xxFdamping&&α=
πω3
4dXceq =
α=0.005
MAE 340 – Vibrations 12
Comparing Actual NonlinearResponse with Linearized Response
• Displacement-squared damping ( )
versus use of ceq ( )
2)sgn( xxFdamping&&α=
πω3
4dXceq =
α=0.5