Car Modeling and Simulation
July 30, 2017 | Author: shrikantbagal | Category: N/A
Short Description
Half Car Model Analysis of Suspensions...
Description
Table of contents Introduction………………………………………………….….2 Model of the system…………………………………………..3 Model of the bump……………………………………………..4 Differential Equation of Motion (2DOF)…………………….6 Natural Frequency and Mode Shapes……………………...8 General Solution……………………………………………...19 D.E of Motion using Lagrange Equation…………………20 Differential Equation of Motion (4DOF)…………………..22 Nonlinear Differential Equation of Motion……………….24 Solution with Laplace Method……………………………..25 Numerical Solution…………………………………………..26 Sample of Excel worksheet………………………………...31 VBA code for numerical solutions………………………..32
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Introduction In this project, I’m trying to model a car which could be a Sedan car, a truck, or an SUV, when passing through a Bump. This bump is acting like a base excitation for our system. First of all we will assume a 2DOF linear system; however, I will derive the differential equation for 4DOF and Non linear system.
The different parts of this project are: 1. Modeling of the system 2. Modeling of the bump 3. Differential Equation of Motion 2DOF (linear) 4. Calculation of Natural Frequency and Mode shapes 5. Differential Equation of Motion Using Lagrange Eq. 6. Differential Equation of Motion 4DOF (linear) 7. Differential Equation of Motion 2DOF(Non linear) 8. Calculating X(t),θ(t) with Laplace Method 9. Calculating X(t),θ(t) with Numerical method Runge Kutta(4) 10. State variables for 2DOF (linear) 11. State variables for 2DOF (non linear) 12. VBA codes of 4-th order Runge Kutta for this problem 13. Graphs of X(t),θ(t) with some inputs (with Excel)
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Modeling of the System
Model of a car: • •
First assumption: The effects of the tires are negligible. Second assumption: The left side and the right side of the car are symmetric.
Dynamic model:
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Modeling of the Bump • First assumption: We assume that the bump is like a half sine wave. So; the effect of bump is like base excitation on the system: z1, z2.
Top graph: Front wheel
h: height of the bump d: horizontal coordinate D: length of the bump L: distance between the front and back tires
Bottom graph: rear wheel
•
Second assumption: We assume that the car’s speed is constant (acceleration is zero) at this period.
z1 = h0 . sin
z1 =
π .d
h0 . sin
D
=V .t d → t p1 =
π .d
V : Car Speed
if → t < tp1
D if → t ≥ tp1
0, 0 z 2 = h0 . sin
π .V .t D
0
Note: t l =
D V
if → t < t l if → t l < t < tp 2 if → t > tp 2
L L+D → tp 2 = V V
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Other form of z1, z2 function: Introduce U c (t ) =
(
0 t≤C 1 t >C
)
z1 = 1 − U tp1 (t ) .h0 . sin
[
π .V0 .t D
]
z 2 = U tl (t ) − U tp2 (t ) .h0 . sin
π .V0 .t D
t < tp1 → z1 = (1 − 0).h0 . sin t < t L → z 2 = (0 − 0).h0 . sin
π .V0 .t D
π .V0 .t D
Check, t L ≤ t ≤ tp 2 → z 2 (1 − 0).h0 . sin t > tp 2 → z 2 = (1 − 1).h0 . sin
!
=0
π .V0 .t D
π .V0 .t D
= h0 . sin
π .V0 .t D
=0
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Differential Equation of Motion 2DOF (linear)
Assume:
z 2 < x − l 2θ z1 < x + l 2θ ..
Fx = m. x ..
− Fs 2 − Fc 2 − Fc1 − Fs1 = m x Fs 2 = k 2 ( x − l1θ − z 2 ) Fs1 = k1 ( x − l1θ − z 2 )
.
.
.
Fc 2 = c 2 ( x − l 2 θ − z 2 ) .
.
.
Fc1 = c1 ( x − l1 θ − z 2 ) .
.
.
.
..
..
..
− k 2 x + k 2 l 2θ + k 2 z 2 − c 2 x + c 2 l 2 θ + c 2 z 2 − k1 x − k1l1θ + k1 z1 − c1 x − c1l1 θ + c1 z 1 = m x ..
.
.
.
.
m x + (c1 + c 2 ) x + (c1l1 − c 2 l 2 ) θ + (k1 + k 2 ) x + (k1l1 − k 2 l 2 )θ = c1 z 1 + c 2 z 2 + k1 z1 + k 2 z 2
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) ..
McG = I θ ..
Fs 2 .l 2 + Fc 2 l 2 − Fc1 .l1 − Fs1l1 = I θ .
.
.
.
.
.
..
k 2 l 2 ( x − l 2θ − z 2 ) + c 2 l 2 ( x − l 2 θ − z 2 ) − c1l1 ( x + l1 θ − z 1 ) − k1l1 ( x + l1θ − z1 ) = I θ ..
.
.
.
.
I θ + (c 2 l 22 + c1l12 ) θ + (c1l1 − c 2 l 2 ) x + (k 2 l 22 + k1l12 )θ + (k1l1 − k 2 l 2 ) x = c1l1 z 1 − c 2 l 2 z 2 + k1l1 z1 − k 2 l 2 z
If we put 1, 2 in matrix format;
m 0 0
I
..
. x.. +
θ .
=
c1
c1l1 − c 2 l 2
c1l1 − c 2 l 2
c 2 l 22 + c1l12
.
. x. +
θ
k1 + k 2
k1l1 − k 2 l 2
k1l1 − k 2 l 2
k 2 l 22 + k1l12
.
c1 z 1 + c 2 z 2 + k1 z1 + k 2 z 2 .
.
c1l1 z 1 − c 2 l 2 z 2 + k1l1 z1 − k 2 l 2 z 2
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x
θ
*
Calculation of Natural Frequency and Mode shapes Assuming Reasonable values for all parameter, we have:
I = m.r 2
r : Radius gyration of the car
Reasonable approximate values for a Sedan car; r 2 = 0.6m 2
m = 1000kg
k1 = k 2 = 5000 N m
c1 = c 2 = 500 N .s m l1 = 0.8m
l 2 = 1 .2 m
Reasonable approximate values for a SUV;
r 2 = 0.65m 2
m = 3000kg
k1 = k 2 = 1500 N m
l1 = 1m
c1 = c 2 = 1500 N .s m l 2 = 1 .5 m
Solved for a bump with h = 0.1m , D = 1.5m
Eigen Value Problem and Mode Shapes For our model we assume c1 = c2 = c and k1 = k2 = k eq .3
→
m 0 0
I
..
.
. x.. +
θ
2c c(l1 − l 2 ) x 2k k (l1 − l 2 ) x g (t ) . . + . = 1 2 2 2 2 c(l1 − l 2 ) c(l1 + l1 ) θ k (l1 − l 2 ) k (l1 + l1 ) θ g 2 (t )
Where:
[
]
[
]
g1 (t ) = h0 1 − u tp1 (t ) .(k . sin ω f t + cω , f . cos ω f t ) + h0 u tL (t ) − u tp2 (t ) .(k . sin ω f t + cω , f . cos ω f t )
[
]
[
]
g 2 (t ) = h0 .l1 1 − u tp1 (t ) .(k . sin ω f t + cω , f . cos ω f t ) + h0 .l 2 u tL (t ) − u tp2 (t ) .(k . sin ω f t + cω , f . cos ω f t )
Note: ω f =
π .V D !
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Eigen Value Problem:
([k ] − ω .[m]) = 0 2
det
2k − ω 2
k (l1 − l 2 )
k (l1 − l 2 ) k (l12 + l 22 )ω 2 .I
ω4 −
=0
k 2 (l1 + l 2 ) 2 2k k 2 2 + (l1 + l 2 ) .ω 2 + =0 m I M .I
2k k 2 2 ω = + (l1 + l 2 ) ± m I 2
2k k 2 2 + (l1 + l 2 ) m I
2
k 2 (l1 + l 2 ) 2 −4 m.I
For a sedan car we have: m = 1000kg , I = m.r 2
and
l1 = 0.8
k1 = k 2 = 5000 N m
l 2 = 1 .2 ,
r 2 = 0.6
I = 600 Nm 2 c1 = c 2 = 500 N .s m
If we plug in eq.(5)
ϖ4 −
10000 5000 (5000) 2 .(0.8 + 1.2) 2 + ( 0 .8 2 + 1 .2 2 ) ω 2 + =0 1000 600 1000 ⋅ 600
ω 4 − 27.334ϖ 2 + 166.677 = 0 \ ω12 = 9.182 ω1 = 3.03(rad sec)
Back sub
1
ω 22 = 18.151 ω 2 = 4.26(rad sec)
in (4)
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10000 − (9.182).(1000) 5000.(0.8 − 1.2) 2k − ω12 m k (l1 − l 2 ) = 2 2 2 5000.(0.8 − 1.2) 5000.(0.8 2 + 1.2 2 ) − (9.182).(600) k (l1 − l 2 ) k (l1 + l 2 ) − ω1 .I =
818 − 2000 − 2000 4890.8
Sub
2
ϕ1 =
1 → ( First.Mode) 0.409
in (4)
10000 − (18.151).(1000)
2
5000.(0.8 − 1.2)
ϕ2
1 − 4.075
5000.(0.8 − 1.2) 2
5000.(0.8 + 1.2 ) − (18.151).(600)
=
− 8151
− 2000
− 2000 − 490.6
→ ( Second .Mode)
∴ Approximate Mode Shapes and Natural Frequencies For Sedan Cars are,
ω1 = 3.03 rad sec
ϕ1 =
ω 2 = 4.26 rad sec
ϕ2 =
1 0.409 1 − 4.075
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For an SUV we have m = 3000kg I = m / r 2 k1 = k 2 = 15000 N m c1 = c 2 = 1500 N m
and
r = 0.64 → I = 1920 N / m 2 l1 = 1 l 2 = 1 .5
Plug the numbers in (5),
ω4 −
30000 15000 2 (15000) 2 .(1 + 1.5) 2 + (1 + 1.5 2 .ω 2 + =0 3000 1920 (3000).(1920)
ω 4 − 35.39ω 2 + 244.14 = 0 → ω12 = 9.39 , ω 22 = 26
ω1 = 3.064 rad sec ,
ω 2 = 5.1 rad sec
Back sub ω 1 in (4),
30000 − (9.39).(3000) 15000.(1 − 1.5) 1830 − 7500 = 2 15000.(1 − 1.5) 15000.(1 + 1.5 ) − (9.39).(1920) − 7500 30721.2
→ ϕ1 =
1 0.244
Back sub
2
in (4),
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30000 − (26).(3000) 15000.(1 − 1.5) − 48000 − 7500 = 2 15000.(1 − 1.5) 15000.(1 + 1.5 ) − (26).(1920) − 7500 − 1170 → ϕ2 =
1 − 6.4
∴ Approximate Mode Shapes and Natural Frequencies For an SUV is:
ω1 = 3.06 rad sec
ϕ1 =
ω 2 = 5.1 rad sec
ϕ2 =
1 0.244 1 − 6 .4
Conclude:
If we vibrate a sedan and a SUV at their second Natural Frequency, the SUV is more capable of rotating than a sedan. However, at their first Natural Frequency, the situation is vice versa.
Important Note:
[k ] =
2.k
k .(l1 − l 2 ) 2 1
2 1
k .(l1 − l 2 ) k .(l + l )
=k
2
l1 − l 2
l1 − l 2
l12 + l12
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[C ] =
C.(l1 − l 2 )
2.C
2 1
= C.
2 1
C.(l1 − l 2 ) C.(l + l )
2
l1 − l 2
l1 − l 2
l12 + l12
As we can see above, the [C] matrix is a factor of [k]; so we will be able to calculate modal C.
Modal Mass: (for sedan car)
ϕ=
1000 0 1 1 , M = 0 600 0.409 − 4.075
m1 0
0 1100 0 T = [ϕ ] .[m][ . ϕ] = m2 0 10963
m1 = 1100kg , m2 = 10963kg
Modal Damping: (sedan) C1
0
0
C2
. ϕ ], [C ] = = [ϕ ] .[C ][
1000
− 200
C1
0
− 200
1090
0
C2
T
=
1010
0
0
19900
C1 = 1010 N .s m , C 2 = 19900 N .s m
Modal Stiffness: (sedan) k1
0
0
k2
= [ϕ ] .[m][ . ϕ ], [k ] = T
k1 = 10100 N .m
10000
− 2000
k1
0
− 2000
10400
0
k2
=
10100
0
0
199000
k 2 = 199000 N .m
∴ Our uncoupled System is, x
θ
= [ϕ ].
η1 η2
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( ..
.
m1 .η1 + C1 .η1 + k1η1 = F1 (mod al _ force) ..
.
m2 .η 2 + C 2 .η 2 + k 2η 2 = F2 (mod al _ force) m1, c1, k1, m2, k2, c2, are already calculated. We should find only F1 and F2 which are model forces.
Modal Damping: (For SUV Car)
C1' 0
C.(l1 − l 2 )
2.C
[C '] =
C.(l1 − l 2 ) C.(l12 + l12 )
=
3000
− 750
− 750
4875
2920 0 0 T = [ϕ '] .[C '][ . ϕ '] = ' 0 21280 C2
C1' = 2920 N .s m , C 2' = 212280 N .s m
Modal Stiffness: (For SUV Car) k .(l1 − l 2 )
2.k
[k '] =
2 1
2 1
k .(l1 − l 2 ) k .(l + l )
k1'
0
0
' 2
k
=
29200
0
0
212800
=
30000
− 7500
− 7500
48750
k1' = 29200 N .m , k 2' = 2122800 N .m
∴ Our uncoupled system is: x
ϕ
= [ϕ ].
η1 η2
..
.
..
.
m1' .η + C1' .η + k1'η = F1' (mod al _ force) m2' .η + C 2' .η + k 2' η = F2' (mod al _ force)
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Calculation of Modal forces:
F1 F2
=[
g1(t)
]T
(6)
g2 (t)
Finding g1 and g2
Calculate g 1 (t ) and g 2 (t ) g 1 (t ) = c1 z1 + c 2 z 2 + k1 z1 + k 2 z 2 z1 = [1 − U tA (t )] h o sin
πvt D
z 2 = [U tL (t ) − U tP (t )] h o sin 2
πv D
→ z1 =
πvt D
πv D
[1 − U tP (t )] h o cos
πvt
1
→ z2 =
πv D
D
[U tL (t ) − U tP (t )] h o cos 2
πvt D
= ωf
g1(t) = c1wf ho[1−UtP (t)]coswf t + c2wf ho[UtL(t) −UtP (t)]coswf t + kiho[1−UtP (t)]sinwf t + k2ho[UtL(t) −UtP (t)]sinsinwf t 1
2
1
2
g1(t) = ho[1−UtP (t)](k1 sinwf t + c1wf coswf t) + ho[UtL(t) −UtP (t)](k2 sinwf t + c2wf coswf t) 1
2
The same method for g2(t) = c1l1z1 + k1l1z1 − c2l2z2 − k2l2z2 g2 (t) = hol1[1−UtP (t)](k1 sinwf t + c1wf coswf t) − hol2[UtL(t) −UtP (t)](k2 sinwf t + c2wf coswf t) 1
2
Byusing VBA Codesin RungeKuttaMethod,we willhave a numericalsolutionfor x,x, θ, θ D L L+ D Note thatin codes: tP = , tL = , tP = V V V 1
2
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So we should calculateg1(t) andg2(t) and put themin(6) withho = 0.1m D =1.5 (heightand lengthof thebumb) andL = L1 + L2 = 0.8+1.2 = 2 m, K = 5000,c = 500
1) V = 15 mph = 6.7 m/sec → ω f =
πv D
=
π × 6.7 1.5
= 14.03 rad/sec
D 1.5 = = 0.224 sec ≈ t P = 0.22 sec v 6.7 v 2 tL = = = 0.298 sec ≈ t L = 0.3 sec L 6.7 L+D tP = = 0.522 sec ≈ t P = 0.52 sec v tP = 1
1
2
2
z1, z2 vs. time 0.12
z1, z2 (height, meters)
0.1 0.08 0.06
z1 z2
0.04 0.02 0 0
0.1
0.2
0.3
0.4
0.5
0.6
-0.02 time
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Insert the numbers into formula, g1 (t ) = [1 − U 0.22 (t )][500sin(14.03t ) + 701.5 cos(14.03t )] + [U 0.3 (t ) − U 0.52 (t )][500sin(14.03t ) + 701.5 cos(14.03t )] g 2 (t ) = [1 − U 0.22 (t )][400sin(14.03t ) + 562.2 cos(14.03t )] + [U 0.3 (t ) − U 0.52 (t )][600sin(14.03t ) + 841.8 cos(14.03t )]
2) v = 27 mph = 12 m/sec→ w f =
πv π × 12 D
=
1.5
= 25.13 rad/sec
D 1.5 = = 0.125sec ≈ t P = 0.12 sec v 12 v 2 t L = = = 0.167 sec ≈ t L = 0.17 sec L 12 L + D 3.5 tP = = = 0.292 sec ≈ t P = 0.29sec v 12 tP = 1
1
2
2
With this velocity the graphs for g1 (t ) & g 2 (t ) are: g1, g2 vs. time 1500 1000
g1, g2
500 0 0
0.1
0.2
0.3
0.4
0.5
0.6
g1 g2
-500 -1000 -1500 -2000 time
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2) v = 27 mph = 12 m/sec→ wf =
πv D
=
π ×12 1.5
= 25.13 rad/sec
D 1.5 = = 0.125sec ≈ t P 1 = 0.12 sec v 12 v 2 t L = = = 0.167 sec ≈ t L = 0.17 sec L 12 L + D 3.5 = = 0.292 sec ≈ t P2 = 0.29sec t P2 = v 12 tP1 =
Insert he t number into formula, g1(t) = [1 −U0.12(t)][500sin(25.13t) +1256.5cos(25.13t)] + [U0.17(t) −U0.29(t)][500sin(25.13t) +1256.5cos(25.13t)] g2 (t) = [1 −U0.12(t)][400sin(25.13t) +1005.2cos(25.13t)] + [U0.17(t) −U0.29(t)][600sin(25.13t) +1507.8cos(25.15t)]
3) v = 47 mph = 21m/sec→ wf =
πv
= 44rad/sec D t P 1 = 0.07sec, tL = 0.09sec, tP2 = 0.16sec g1(t) = [1 −U0.07(t)][500sin(44t) + 2200cos(44t)] + [U0.09(t) −U0.16(t)][500sin(44t) + 2200cos(44t)] g2 (t) = [1 −U0.07(t)][400sin(44t) +1760cos(44t)] + [U0.09(t) −U0.16(t)][600sin(44t) + 2640cos(44t)] F1
′ =
F2
1 1
0.409 - 4.075
F1 = g1 (t ) + 0.409 g 2 (t )
g1 (t ) g 2 (t ) and F2 = g1 (t ) − 4.075 g 2 (t )
(8)
For all cases , we have " Uncoupled Equation" 1100η1 + 1010η1 + 10100η1 = g1 (t ) + 0.409 g 2 (t ) 10963η2 + 19900 η 2 + 199000 η 2 = g1 (t ) − 4.075 g 2 (t )
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General solution
η k = e −ξ
k wk t
[ Ak cos wdk t + ( Bk +
w dk = wk 1 −ξ k
Where
wdk
ϑ0
mk
t
) sin wdk ] +
1 Fk′(τ )eξ mk wdk 0
k w w Ak
sin wdk (t − τ ) dτ
(7)
2
x0
[Φ ]T × [m] × Ak =
ξ k ww Ak
T
Φ k × [m] × Bk =
,
x0
ϑ0
wdk × mk
By assuming that initial condition is zero, we have A k = B k = 0 t
1 Fk′(τ )eξ wdk × mk 0
ηk =
k w w Ak
sin wdk (t − τ ) dτ
k = 1 .2
w1 = 3.03, m1 = 1100 ,
ξ1 =
w2 = 4.26, m 2 = 10963, ξ 2 =
c1 ≈ 0.151 2m1w1 c2 ≈ 0.213 2 m2 w2 2
and wd 1 = w1 1 -ξ 1 = 2.99 rad/sec 2
wd 2 = w2 1 -ξ 2 = 4.16 rad/sec t
η1 =
1 F1 ( z )e −0.46 z sin 2.99(t − z )dz 328 0
η2 =
1 F2 ( z )e −0.9 z sin 4.16(t − z )dz 45606 0
(9)
t
Substitude from (8) in (9) with different speed which caused if different F1 (t), F2 (t) we can calculated those integral but even in MATLAB it is so complecated to be evluated, so the numerical solutions are advised x
θ
= [Φ ]
η1 1 = η2 0.409
1 - 4.075
η1 η2
x = η1 + η 2
θ = 0.409η1 − 4.075η 2
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Differential Equation of Motion Using Lagrange Eq
1 m x + I 0θ 2 2 1 1 V = k 2 ( x − l 2θ ) 2 + k 1 ( x + l1θ ) 2 2 2 T=
R=
1 1 c 2 ( x − l 2θ ) 2 + c1 ( x + l1θ ) 2 2 2
∂w = Q1 ∂x + Q 2 ∂θ
1)
∂ ∂T ∂T ∂R ∂V − + + = Q1 ∂t ∂x ∂x ∂x ∂x
2)
∂ ∂T ∂T ∂R ∂V − + + = Q2 ∂t ∂θ ∂ θ ∂θ ∂θ
x.equation :
∂T =0 ∂x
∂ ∂T ∂T = mx → = mx ∂t ∂x ∂x
∂R = c 2 ( x − l 2θ ) + c1 ( x + l1θ ) ∂x
∂V = k 2 ( x − l 2θ ) + k 1 ( x + l1θ ) ∂x ∂w = ( c 1 z 1 + k 1 z 1 + c 2 z 2 + k 2 z 2 ) ∂x + ( c 1 l 1 z 1 − c 2 l 2 z 2 + k 1 l 1 z 1 − k 2 l 2 z 2 ) Q1 = c1 z 1 + k 1 z 1 + c 2 z 2 + k 2 z 2 mx + c 2 ( x − l 2θ ) + c1 ( x + l1θ ) + k 2 ( x − l 2θ ) + k 1 ( x + l1θ ) = Q1 mx + (c1 + c 2 ) x + (c1 l1 − c 2 l 2 )θ + ( k 1 + k 2 ) x + ( k 1 l1 − k 2 l 2 )θ = c1 z 1 + c 2 z 2 + k 1 z 1 + k 2 z 2
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mx + c2 ( x − l2θ ) + c1 ( x + l1θ ) + k2 ( x − l2θ ) + k1 ( x + l1θ ) = Q1 mx + (c1 + c2 ) x + (c1l1 − c2l2 )θ + ( k1 + k2 ) x + (k1l1 − k2l2 )θ = c1 z1 + c2 z2 + k1 z1 + k2 z2
θ .equation :
∂T ∂ ∂T = Iθ → = Iθ ∂t ∂θ ∂θ
∂T =0 ∂θ
∂R = - c2l2 ( x − l2θ ) + c1l1 ( x + l1θ ) ∂θ
∂V = −k2l2 ( x − l2θ ) + k1l1 ( x + l1θ ) ∂θ and Q1 = c1l1 z1 − c2l2 z2 + k1l1 z1 − k2l2 z2 2
2
2
2
Iθ + +(c1l1 − c2l2 ) x + (c2l2 + c1l1 )θ + (k1l1 − k2l2 ) x + ck1l1 + k2l2 )θ = c1l1 z1 − c2l2 z2 + k1l1 z1 − k2l2 z2
Put them in matrix format: m 0 0
I
..
. x.. +
θ .
=
c1 c1l1 − c 2 l 2
.
c1l1 − c 2 l 2 x k1 + k 2 . . + 2 2 c 2 l 2 + c1l1 θ k1l1 − k 2 l 2
k1l1 − k 2 l 2 x . k 2 l 22 + k1l12 θ
.
c1 z 1 + c 2 z 2 + k1 z1 + k 2 z 2 .
.
c1l1 z 1 − c 2 l 2 z 2 + k1l1 z1 − k 2 l 2 z 2
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Differential Equation of Motion 4DOF (linear)
T=
1 1 1 2 2 2 mx1 + I 0θ 2 + m3 x 3 + m 2 x 2 2 2 2
V=
1 1 1 1 k 2 ( x1 − l 2θ − x 3 ) 2 + k 1 ( x1 + l1θ − x 2 ) 2 + k 3 ( x 3 − z 2 ) 2 + k 4 ( x 2 − z1 ) 2 2 2 2 2
R=
1 1 c 2 ( x1 − l 2θ − x 3 ) 2 + c1 ( x1 + l1θ − x 2 ) 2 2 2
In this case we bring
z1 , z2 in the potential energy, so we don' t have any Q1 , Q2
1)
∂ ∂T ∂T ∂R ∂V − + + =0 ∂t ∂x1 ∂x1 ∂x1 ∂x1
2)
∂ ∂T ∂T ∂R ∂V − + + =0 ∂t ∂θ ∂θ ∂θ ∂θ
3)
∂ ∂T ∂T ∂R ∂V − + + =0 ∂t ∂x2 ∂x2 ∂x2 ∂x2
4)
∂ ∂T ∂T ∂R ∂V − + + =0 ∂t ∂x3 ∂x3 ∂x3 ∂x3
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1) m1 x1 + c2 ( x1 − l2θ − x3 ) + c1 ( x1 + l1θ − x2 ) + k2 ( x1 − l2θ − x3 ) + k1 ( x1 + l1θ − x2 ) = 0 m1 x1 + (c1 + c2 ) x1 + (c1l1 − c2l2 )θ − c1 x2 − c2 x3 + (k1 + k2 ) x1 + (k1l1 − k2l2 )θ − k1 x2 − k2 x3 = 0
2) Iθ + (−c2l2 )( x1 − l2θ − x3 ) + c1l1 ( x1 + l1θ − x2 ) + (−k2l2 )( x1 − l2θ − x3 ) + k1l1 ( x1 + l1θ − x2 ) = 0 2
2
2
2
Iθ + (c2l2 + c1l1 )θ + (c1l1 − c2l2 ) x1 − c1l1 x2 + c2l2 x3 + (k1l1 + k2l2 )θ + (k1l1 − k2l2 ) x1 − k1l1 x2 + k2l2 x3 = 0
3) m2 x2 + (−c2 )( x1 + l1θ − x2 ) + (−k1 )( x1 + l1θ − x2 ) + k4 ( x2 − z1 ) = 0 m2 x2 + c1 x2 − c1 x1 − c1l1θ + (k1 + k4 ) x2 − k1 x1 − k1l1θ − k4 z1 = 0
4) m3 x3 + (−c2 )( x1 − l2θ − x3 ) + (−k2 )( x1 − l2θ − x3 ) + k3 ( x3 − z2 ) = 0 m3 x3 + c2 x3 − c2 x1 + c2l2θ + (k2 + k3 ) x3 − k2 x1 + k2l2θ − k3 z2 = 0
Put them in matrix format:
m1
0
0
0 0
I 0
0 m2
0 0
θ
0
0
0
m3
x3
k1 + k 2 +
c1 + c 2
x1
0
x2
+
k 1 l1 − k 2 l 2 2
2
c1l1 − c 2 l 2 2
2
− c1
− c2
x1
- c1l1
c2l2
θ
c1l1 − c 2 l 2
c1 l 1 + c 2 l 2
− c1
- c1l1
c1
0
x2
-c2
c2l2
0
c2
x3
− k1
− k2
x1
- k 1 l1
k 2l2
θ
k 1 l1 − k 2 l 2
k 1 l1 + k 2 l 2
- k1
- k 1 l1
k1
0
x2
-k2
k2l2
0
k2
x3
=
0 0 k 4 z1 k3 z2
Note : In this case we have eight state variables.
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Differential Equation of Motion 2DOF (Non linear) 1 mx + I 0θ 2 2 1 1 V = k 2 ( x − l 2 sin θ ) 2 + k1 ( x + l1 sin θ ) 2 2 2 T=
R=
1 1 c 2 ( x − l 2 cos θ . θ ) 2 + c1 ( x + l1 cos θ . θ ) 2 2 2
1)
∂ ∂T ∂T ∂R ∂V − + + = Q1 ∂t ∂x ∂x ∂x ∂x
2)
∂ ∂T ∂T ∂R ∂V − + + = Q2 ∂t ∂θ ∂θ ∂θ ∂θ
1) mx + c2 ( x − l2 cos θ . θ ) + c1 ( x + l1 cos θ . θ ) + k 2 ( x − l2 sin θ ) + k1 ( x + l1 sin θ ) = c1 z1 + c2 z 2 + k1 z1 + k 2 z 2 mx + (c1 + c2 ) x + (c1l1 cos θ − c2l2 cos θ )θ + (k1 + k 2 ) x + (k1l1 − k 2l2 ) sin θ = c1 z1 + c2 z 2 + k1 z1 + k 2 z2
2) Iθ − c2l2 cos θ ( x − l2 cos θ . θ ) + c1l1 cos θ ( x + l1cosθ ) − k 2l2 cos θ ( x − l2 sin θ ) + k1l1 cos θ ( x + l1 sin θ ) = Q2 where Q2 = c1l1 cos θ z1 − c2l2 cos θ z 2 + k1l1 cos θ z1 − k 2l2 cos θ z 2
2
2
2
2
Iθ + (c2l2 cos 2 θ + c1l1 cos 2 θ )θ + (c1l1 cos θ − c2l2 cos θ ) x + (k1l1 cos θ + k 2l2 cos θ ) sin θ + (k1l1 cos θ − k 2l2 cos θ ) x = (c1l1 z1 − c2l2 z 2 + k1l1 z1 − k 2l2 z 2 ) cos θ
Note: The damping constant is also Non-linear, but we assume that it is linear
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Calculating X(t), θ(t) θ( with Laplace Method x1
m1 0 0
+
θ
I
c1 + c2 c1l1 − c2l2
c1l1 − c2l2 2
c1l1 + c2l2
assume : x(0) = x(0) = 0,
2
x1
θ
+
k1 + k 2
k1l1 − k 2l2 2 11
k1l1 − k2l2
k l + k 2 l2
x 2
θ
=
g1 (t ) g 2 (t )
θ (0) = θ (0) = 0
[ms 2 + (c1 + c 2 ) s + (k1 + k 2 )] Χ + [(c1l1 − c 2 l 2 ) s + k1l1 − k 2 l 2 ]θ = L [g 1 (t )] 2
2
2
2
[(c1l1 − c 2 l 2 ) s + k1l1 − k 2 l 2 ] Χ + [ Is 2 + (c1l1 + c 2 l 2 )s + (k1l1 + k 2 l 2 )]θ = L [ g 2 (t )]
Det Χ= Det
θ=
L [g 1 (t )]
(c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 )
L [ g 2 (t )]
Is 2 + (c1l1 + c 2 l 2 )s + (k1l1 + k 2 l 2 )
2
2
2
2
ms 2 + (c1 + c 2 ) s + (k1 + k 2 ) (c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 ) 2
2
2
2
(c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 ) Is 2 + (c1l1 + c 2 l 2 )s + (k1l1 + k 2 l 2 )
ms 2 + (c1 + c 2 ) s + (k1 + k 2 )
L [g1 (t )]
(c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 )
L [ g 2 (t )]
2
Det
ms + (c1 + c 2 ) s + (k1 + k 2 ) (c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 ) 2
2
2
2
(c1l1 − c 2 l 2 ) s + (k1l1 − k 2 l 2 ) Is 2 + (c1l1 + c 2 l 2 )s + (k1l1 + k 2 l 2 )
Note: It is obvious that for solution we must have ay least one of the “Nonzero initial condition Or Force (displacement) excitation By taking an inverse, we will find the solution; however this method usually is not easy.
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Calculating X(t), θ(t) θ( with Numerical method Runge Kutta (4) The general format of Runge Kutta method is based on state variables; however the formulation of this method is provided: ∂y1 = f1 (t , y1 , y 2 ,... y m ) ∂t ∂y 2 = f1 (t , y1 , y 2 ,.. y m ) ∂t
∂y m = f 2 (t , y1 , y 2 ,... y m ) ∂t y m,(i +1) = y m,(i ) +
k1, ( m ) + 2k 2, ( m ) + 2k 3,( m ) + k 4,( m )
6 “m” represent the number of state variables For example: k1,1 : k 1 → state var iable 1
k1, 2 : k1 → state var iable2
k 2 ,1 : k 2 → state var iable 1
k 2, 2 : k 2 → state var iable2
k 3,1 : k 3 → state var iable 1
k 3, 2 : k 3 → state var iable2
k 4 ,1 : k 4 → state var iable 1
k 4, 2 : k 4 → state var iable2
k1,1 = h * f1 (t i , y1( i ) , y 2 ( i ) ) k1,1 k1, 2 h k 2,1 = h * f1 (t i + , y1( i ) + , y 2(i ) + 2 2 2 k h 3,1 k 3,1 = h * f 1 (t i + , y1,( i ) + , y 2 ( i ) + k 3, 2 ) 2 2 k 4,1 = h * f1 (t i + h, y1,( i ) + k 3,1 , y 2 ( i ) + k 3, 2 )
Note: The main point is to finding the State variables
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State variables for 2DOF (linear) mx + (c1 + c2 ) x + (c1l1 − c2l2 )θ + (k1 + k 2 ) x + (k1l1 − k 2l2 )θ = c1 z1 + c2 z2 + k1 z1 + k2 z2 2
2
2
2
Iθ + (c1l1 − c2l2 ) x + (c2l2 + c1l1 )θ + (k1l1 − k 2l2 ) x + ck1l1 + k 2l2 )θ = c1l1 z1 − c2l2 z2 + k1l1 z1 − k 2l2 z2 Assume : c1 z1 + c2 z2 + k1 z1 + k2 z2 = g1 (t ) c1l1 z1 − c2l2 z2 + k1l1 z1 − k2l2 z2 = g 2 (t ) y1 = x
y2 = x
y3 = θ
y4 = θ
my 2 + (c1 + c2 ) y 2 + (c1l1 − c2l2 ) y 4 + (k1 + k2 ) y1 + (k1l1 − k 2l2 ) y 2 = g1 (t ) 2
2
2
2
I y 4 + (c1l1 − c2l2 ) y 2 + (c2l2 + c1l1 ) y 4 + (k1l1 − k2l2 ) y1 + (k1l1 + k 2l2 ) y3 = g 3 (t ) y1 = y 2
, y3 = y 4
In this case we have four state variables
y1 y2 y3 y4
0 (k + k 2 ) - 1 m = 0 -
( k1l1 − k 2 l 2 ) I
1 (c + c 2 ) - 1 m 0 -
(c1l1 − c 2 l 2 ) I
0
0 (c l − c 2 l 2 ) - 11 m 1
(k l − k 2 l 2 ) - 11 m 0 2
-
2
( k1l1 + k 2 l 2 ) I
2
-
y2 y3
2
y4 c1l1 + c 2 l 2 I Matrix A
Each line gives us one gradient function
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0 g 1 (t )
y1
&
+
0 g 2 (t )
m
I
*
With previous state variable and using runge kutta method the solution is: (Note: the related code is available in appendix) linear silution v=6.7 m/sec runge kutta method 0.15
0.1
x
0.05 Series1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
-0.05
-0.1
-0.15 time
linear solution with runge kutta v=6.7 m/sec 0.15
0.1
Tetha
0.05
0
Series1 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
-0.05
-0.1
-0.15 time
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State variables for 2DOF (non linear) 1) mx + (c1 + c 2 ) x + (c1l1 cos θ − c 2 l 2 cos θ )θ + ( k1 + k 2 ) x + (k1l1 − k 2 l 2 ) sin θ = c1 z1 + c 2 z 2 + k1 z1 + k 2 z 2
2
2
2
2
2) Iθ + (c 2 l 2 cos 2 θ + c1l1 cos 2 θ )θ + (c1l1 cos θ − c 2 l 2 cos θ ) x + (k1l1 cos θ + k 2 l 2 cos θ ) sin θ + (k1l1 cos θ − k 2 l 2 cos θ ) x = (c1l1 z1 − c 2 l 2 z 2 + k1l1 z1 − k 2 l 2 z 2 ) cos θ
We have four state variables in this case. State Variables y1 = x 1) y 1 = y 2 y2 = x
3) y 3 = y 4
y3 = θ y4 = θ 2) my 2 + (c1 + c 2 ) y 2 + (c1 l1 cos y 3 − c 2 l 2 cos y 3 ) y 4 + ( k 1 + k 2 ) y 1 + ( k 1 l1 − k 2 l 2 ) cos y 3 = g 1 (t )
y2 =
− (c1 + c 2 ) y 2 − (c1 l1 cos y 3 − c 2 l 2 cos y 3 ) y 4 − ( k 1 + k 2 ) y 1 − ( k 1 l1 − k 2 l 2 ) cos y 3 + g 1 (t ) m
where : g 1 (t ) = c1 z 1 + c 2 z 2 + k 1 z 1 + k 2 z 2 2
y4 =
2
2
2
− (c1 l1 cos 2 y 3 + c 2 l 2 cos 2 y 3 ) y 4 + (c1 l1 cos y 3 − c 2 l 2 cos y 3 ) y 2 + ( k 1 l1 cos y 3 + k 2 l 2 cos y 3 ) sin y 3 I
where : g 3 (t ) = (c1 l1 z 1 − c 2 l 2 z 2 + k 1 l1 z 1 − k 2 l 2 z 2 ) cos y 3
The graphs are in the next page
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nonlinear v=6.7 runge kutta 0.15
0.1
X
0.05 Series1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
-0.05
-0.1
-0.15 time
nonlinear 0.15
0.1
tetha
0.05
0 0
0.1
0.2
0.3
0.4
0.5
0.6
Series1
0.7
-0.05
-0.1
-0.15 time
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Sample of Excel worksheet:
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( * ( * ) ( ) )( " " )))*) ( * " )(( ( ( (" ) "" "
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VBA codes: Option Base 1 Function u_tp1(t_in, t_p1) If t_in >= t_p1 Then u_tp1 = 1 ElseIf t_in < t_p1 Then u_tp1 = 0 End If End Function Function u_tp2(t_in, t_p2) If t_in >= t_p2 Then u_tp2 = 1 ElseIf t_in < t_p2 Then u_tp2 = 0 End If End Function Function u_tL(t_in, t_L) If t_in > t_L Then u_tL = 1 ElseIf t_in t_p1 Then u_tp = 1 ElseIf t_in
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