01 ce225 multi-degree of freedom systems
TRANSCRIPT
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Multi-Degree-of-Freedom Systems
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Multi-Degree-of-Freedom Systems
Degree-of-Freedom is two or more Degree-of-Freedom is the number of independent
displacement coordinates necessary to describe the
motion of the system.
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k
u
m
Examples of Single-Degree-of-Freedom Systems
u
k
m
u
k
m
m
k
u
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c1
u1
c2
k1 k2
m1 m2
p1(t) p2(t)
u2
Examples of Two-Degree-of-Freedom Systems
u1
k1
m1
k2
m2
u2
2-DOF
2-DOF
u1
u2
2-DOF
Rigid bar
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Example of Two-Degree-of-Freedom System
k1
k2
m1
m2
u1
u2
2-DOF
Shearbuilding
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Example of Three-Degree-of-Freedom System
c1
u1
c2 c3
k1 k2 k3
m1 m2 m3
p1(t) p2(t) p3(t)
u2 u3
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Example of Three-Degree-of-Freedom System
p1(t)
p2(t)
m1u1
u2
c1
c2
k1
k2
p3(t) u3
k3
m2
m3
c3
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c1
u1
k1
m1 m2
p1(t) p2(t)
u2
Degree of Freedom does not necessarily mean the number
of lumped masses
u1
k1
m1
m2
u2
1-DOF
1-DOF
Rigid bar
u1 =u2
Rigid bar
u1 =u2
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Equation of Motion of a Two-
Degree-of-Freedom System
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Two-Degree-of-Freedom Spring System (Classical)
c1
u1
c2
k1 k2
m1 m2
p1(t) p2(t)
u2
c1
u1
c2
k1 k2
m1 m2
p1(t) p2(t)
u2
u1
p1(t) p2(t)
u2
c1u1c2(u2-u1)
k1u1 k2(u2-u1)
c2(u2-u1)
k2(u2-u1)
m1u1 m1u1
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Spring System
u1
p1(t)
p2(t)
u2
c1u1 c2(u2-u1)
k1u1 k2(u2-u1)
c2(u2-u1)
k2(u2-u1)
m1u1
m2u2
m1u1+ + = p1(t)c1u1 k1u1c2(u2-u1)- k2(u2-u1)-
m2u2+ + = p2(t)k2(u2-u1)c2(u2-u1)
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Spring System
c1
u1
m u(t) + c u(t) + k u(t) = p(t)
c2
k1 k2
m1 m2
p1(t) p2(t)
u2
Equation of Motion
m1 0
0 m2
u1
u2
+
c1+c2 -c2
- c2 c2
u1
u2
+
k1+k2 -k2
- k2 k2
u1
u2
=
p1(t)
p2(t )
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Spring System
c1
u1
m v(t) + c v(t) + k v(t) = p(t)
c2 c3
k1 k2 k3
m1 m2 m3
p1(t) p2(t) p3(t)
u2 u3
Equation of Motion
m1 0 0
0 m2 0
0 0 m3
+
c1+c2 -c2 0
- c2 c2+c3 -c3
0 -c3 c3
+
k1+k2 -k2 0
- k2 k2+k3 -k3
0 -k3 k3
=
p1(t)
p2(t)
p3(t )
u1
u2
u3
u1
u2
u3
u1
u2
u3
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Two-Story Shear Building
(Assumptions)
Beams and floor systems are rigid (infinitely stiff)
in flexure
Axial deformations of beams and columns are
neglected Effect of axial force on stiffness of the columns
are neglected
Mass is concentrated at the floor levels Linear viscous damping is associated with
deformational motions of each story
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Two-Story Shear Building
p1(t)
p2(t)
m1
m2
u1
u2
c1
c2
k1
k2
u2
u1
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Using Newtons Second Law of Motion
p2(t)
p1(t)
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Using Newtons Second Law of Motion
p2(t)
p1(t)
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Using Newtons Second Law of Motion
p2(t)
p1(t)
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Using Newtons Second Law of Motion
p2(t)
p1(t)
m2u2
k2(u2-u1)
k1u1 c1u1
m2u2+ += p2(t)
m1u1+ + = p1(t)
m1u1
c1u1 k1u1
c2(u2-u1)
k2(u2-u1) c2(u2-u1)
k2(u2-u1)c2(u2-u1)
c2(u2-u1)- k2(u2-u1)-
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Using Newtons Second Law of Motion
m2u2+ + =p2
(t)
m1u1+ + = p1(t)c1u1 k1u1
k2(u2-u1)c2(u2-u1)
c2(u2-u1)- k2(u2-u1)-
m2u2- + =p2(t)k2u2c2u1 c2u2+ k2u1-
m1u1+ - = p1(t)k2u2(c1+c2)u1 c2u2- + (k1+k2)u1
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Using Newtons Second Law of Motion
m2u2- + =p2(t)k2u2c2u1 c2u2+ k2u1-
m1u1+ - = p1(t)k2u2(c1+c2)u1 c2u2- + (k1+k2)u1
m1u1+ 0 u2 + (k1+k2)u1k2u2-(c1+c2)u1 c2u2-
= p1(t)
0 u1+ m2u2 - c2u1 + c2u2
+
k2u1- + k2u2= p2(t)
m1 0
0 m2
u1
u2
+c1+c2 -c2
- c2 c2
u1
u2
+
k1+k
2-k2
- k2 k2
u1
u2
=
p1(t)
p2(t )
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Using Newtons Second Law of Motion
m1 0
0 m2
u1
u2
+
c1+c2 -c2
- c2 c2
u1
u2
+
k1+k2 -k2
- k2 k2
u1
u2
=
p1(t)
p2(t )
m u+
c u+
k u = p(t)
mass
matr ix
acc.
vector
damping
matr ix
vel.
vector
st i f fness
matr ix
disp.
vector
loading
vector
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Using Newtons Second Law of Motion
m1 0
0 m2
u1
u2
+
c1+c2 -c2
- c2 c2
u1
u2
+
k1+k2 -k2
- k2 k2
u1
u2
=
p1(t)
p2(t )
m u+
c u+
k u = p(t)
This is the equation of motion of the two-story shear
building
The matrix equation represents two ordinary differential
equations
Each equation contains both u1 and u2, and, therefore,
coupled.
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General Form of the Equation of Motion
m11 m12
m21 m22
u1
u2
+
u1
u2
+
u1
u2
=
p1(t)
p2(t )
Physical meaning of each element of the matrices mij, cij, kijis the force at the i
thmass due to a unit
acceleration, velocity or displacement at the jthmass,
respectively, with all other accelerations, velocities
and displacements equal to zero.
c11 c12
c21 c22
k11 k12
k21 k22
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Using Newtons Second Law of Motion
p2(t)
p1(t)
m2u2
k2(u2-u1)
k1u1 c1u1
m2u2+ += p2(t)
m1u1+ + = p1(t)
m1u1
c1u1 k1u1
c2(u2-u1)
k2(u2-u1) c2(u2-u1)
k2(u2-u1)c2(u2-u1)
c2(u2-u1)- k2(u2-u1)-
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Equation of Motion of a Three-
Degree-of-Freedom System
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Equation of Motion of a 3-DOF System
m1 0 0
0 m2 0
0 0 m3
u1
u2
u3
+
c1+c2 -c2 0
- c2 c2+c3 -c3
0 -c3 c3
u1
u2
u3
+
k1+k2 -k2 0
- k2 k2+k3 -k3
0 -k3 k3
u1
u2
u3
=
p1(t)
p2(t)
p3(t )
p1(t)
p2(t)
m1u1
u2
c1
c2
k1
k2
p3(t) u3
k3
m2
m3
c3
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Problem: Formulate the equation of motion of the given system.
Level Weight (ton) Stiffness
(ton/cm)
5 100 53
4 100 61.5
3 100 96
2 100 115
1 100 114
5
4
3
2
1
m = W/g = W ton/980 cm/sec2
100 0 0 0 0 y1 -115 -115 0 0 0 y1 0
0 100 0 0 0 y2 -115 211 -96 0 0 y2 0
0 0 100 0 0 y3 + 0 -96 157.5 -61.5 0 y3 = 0
0 0 0 100 0 y40 0 -61.5 114.5 -53
y4 00 0 0 0 100 y5 0 0 -53 53 y5 0
1/980
ton-sec2/cm cm ton/cmcm ton
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Equation of Motion of a Two-
Degree-of-Freedom System
(Base Excitation, Undamped)
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SDOF SystemOne-storey building with rigid girder
(Influence of Support Excitation)
fixedr
eferen
cea
xis
vg(t)
vT(t)
v(t)
fs(t)
2
fs(t)
2
fD(t)
fI(t)
m
k
2
k
2fD(t)= c v(t)
fs(t)= k v(t)
fI(t)= m vT(t)
F = 0 inertia force
- fS(t) - fD(t) - fI(t) = 0
- k v(t) - c v(t) - m (vg(t) + v(t) ) = 0
m v(t)+ + k v(t) =c v(t) - m vg (t)
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2-DOF SystemTwo-storey building with
rigid girder (Influence of Support Excitation)
fixedr
eferencea
xis
vg
v1T
v1
m1
k
2
k
2
m2
k
2
k
2
v2T
v2
Applications:
1. Motion of building caused byearthquake
2. Motion of equipment due tomotion of building where it ishoused
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2-DOF SystemTwo-storey building with
rigid girder (Influence of Support Excitation)
= + v1
disp of mass1 rel to
moving support/base
total disp of mass1 rel
to fixed reference axis
vgv1T
disp of frame support
rel to fixed reference
axis
fixedr
eferencea
xis
vg
v1T
v1
m1
k
2
k
2
m2
k
2
k
2
v2T
v2
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2-DOF SystemTwo-storey building with
rigid girder (Influence of Support Excitation)
= +
disp of mass2 rel to
moving support/base
total disp of mass2 rel
to fixed reference axis
vgv2T
disp of frame support
rel to fixed reference
axis
fixedr
eferencea
xis
vg
v1T
v1
m1
k
2
k
2
m2
k
2
k
2
v2T
v2
v2
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fixedr
eferencea
xis
v
m1 v1T
k1v1
v1
v2
v1)-k2(v2
m2 v2T v1)-k2(v2+ = 0
v1)-k2(v2
m1 v1T+ k1v1 - v1)-k2(v2 = 0
m2 v2T
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vg displacement of support/ground from a fixed reference axis
v1 displacement of mass1 relative to the base
v2 displacement of mass2 relative to the base
m2v2T v1)-k2(v2+ = 0
m1v1T + k1v1 - v1)-k2 (v2 = 0
= + v1vgv1T
= +vgv2T v2
v1 = + v1vgv1T
= +vgv2T v2
v1
m2v2 k 2v2+k2v1- =
m1v1 + (k1+ k2) v1 - k2 v2 = - m1vg
- m2vg
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m2v2 k 2v2+k2v1- =
m1v1 + (k1+ k2) v1 - k2 v2 = -m1vg
- m2vg
m1
0
0
m2
+
v1
v2
k1+k2
- k2
- k2
k2
v1
v2= -
m1
0
0
m2
1
1
vg
M v + K v = - M 1 vg
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Quiz
m1
0
0
m2
+
v1
v2
k1+k2
- k2
- k2
k2
v1
v2= -
m1
0
0
m2
1
1
vg
MULTIPLY THE MATRICES
2 x 2 2 x 1 2 x 2 2 x 1 2 x 2 2 x 1
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Quiz
m1
0
0
m2
+
v1
v2
k1+k2
- k2
- k2
k2
v1
v2= -
m1
0
0
m2
1
1
vg
MULTIPLY THE MATRICES
ANSWER
m1v1
0 v1
+ 0 v2
m2+ v2
+( k1+k2 )v1 - k2v2
- k2v1 + k2v2
=
- m1 vg
- m2 vg
2 x 1 2 x 1 2 x 1
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QuizADD THE MATRICES
m1v1
0 v1
+ 0 v2
m2+ v2
+( k1+k2 )v1 - k2v2
- k2v1 + k2v2
=
- m1 vg
- m2 vg
2 x 1 2 x 1 2 x 1
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QuizADD THE MATRICES
ANSWER
m1v1
0 v1
+ 0 v2
m2+ v2
+( k1+k2 )v1 - k2v2
- k2v1 + k2v2
=
- m1 vg
- m2 vg
2 x 1 2 x 1 2 x 1
m1v1+ 0 v2( k1+k2 )v1 - k2v2+
0 v1 m2+ v2- k
2 + k2v2
2 x 1
=
- m1 vg
- m2vg
2 x 1
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QuizWHAT ARE THE RESULTING TWO EQUATIONS?
m1v1+ 0 v2( k1+k2 )v1 - k2v2+
0 v1m2+ v2 - k2 + k2v2
2 x 1
=
- m1 vg
- m2 vg
2 x 1
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QuizWHAT ARE THE RESULTING TWO EQUATIONS?
ANSWER
m1v1+ 0 v2( k1+k2 )v1 - k2v2+
0 v1m2+ v2 - k2 + k2v2
2 x 1
=
- m1 vg
- m2 vg
2 x 1
m1v1+ 0 v2( k1+k2 )v1 - k2v2+ = - m1 vg
0 v1 m2+ v2 - k2 + k2v2 - m2 vg=
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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION
m1v1( k1+k2 )v1 - k2v2+ = - m1 vg
m2v2 - k2 + k2v2 - m2 vg=
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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION
m1v1( k1+k2 )v1 - k2v2+ = - m1 vg
m2v2 - k2 + k2v2 - m2 vg=
ANSWER
m1
0
0
m2+
v1
v2
k1+k2
- k2
- k2
k2
v1
v2= -
m1
0
0
m2
1
1vg
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k4= 1600 kN/mk4= 1600 kN/m
k3= 1200 kN/mk3= 1200 kN/m
m4= 4500 kg
k2= 800 kN/mk2= 800 kN/m
m3= 3000 kg
m2= 3000 kg
k1= 400 kN/mk1= 400 kN/m
m1= 1500 kg
x1
x2
x3
x4
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k4= 1600 kN/mk4= 1600 kN/m
k3= 1200 kN/mk3= 1200 kN/m
m4= 4500 kg
k2= 800 kN/mk2= 800 kN/m
m3= 3000 kg
m2= 3000 kg
k1= 400 kN/mk1= 400 kN/m
m1= 1500 kgx1
x2
x3
x4
x1=1
x2=0
x3=0
x4=0
800 kN
800 kN
800 kN
- 800 kN
0
0
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k4= 1600 kN/mk4= 1600 kN/m
k3= 1200 kN/mk3= 1200 kN/m
m4= 4500 kg
k2= 800 kN/mk2= 800 kN/m
m3= 3000 kg
m2= 3000 kg
k1= 400 kN/mk1= 400 kN/m
m1= 1500 kgx1
x2
x3
x4
x1=0
x2=1
x3=0
x4=0
800 kN
800 kN
1600 kN
1600 kN
- 800 kN
2400 kN
- 1600
0
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k4= 1600 kN/mk4= 1600 kN/m
k3= 1200 kN/mk3= 1200 kN/m
m4= 4500 kg
k2= 800 kN/mk2= 800 kN/m
m3= 3000 kg
m2= 3000 kg
k1= 400 kN/mk1= 400 kN/m
m1= 1500 kg
x1
x2
x3
x4
x1=0
x3=1
x2=0
x4=0
1600 kN
1600 kN
2400 kN
2400 kN
0
- 1600 kN
4000 kN
-2400 kN
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k4= 1600 kN/mk4= 1600 kN/m
k3= 1200 kN/mk3= 1200 kN/m
m4= 4500 kg
k2= 800 kN/mk2= 800 kN/m
m3= 3000 kg
m2= 3000 kg
k1= 400 kN/mk1= 400 kN/m
m1= 1500 kg
x1
x2
x3
x4
x1=0
x4=1
x2=0
x3=0
2400 kN
2400 kN
3200 kN
3200 kN
0
0
-2400 kN
5600 kN
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800
- 800
0
0
- 800
2400
- 1600
0
0
- 1600
4000
-2400
0
0
-2400
5600
F1
F2
F3
F4
=
x1
x2
x3
x4
1 2 3 4
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1500
0
0
0
0
3000
0
0
0
0
3000
0
0
0
0
4500
F1
F2
F3
F4
=
x1
x2
x3
x4
1 2 3 4
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0
0
x4
1500
0
0
0
0
3000
0
0
0
0
3000
0
0
0
0
4500
x1
x2
x3
x4
1 2 3 4
+
800
- 800
0
0
- 800
2400
- 1600
0
0
- 1600
4000
-2400
0
0
-2400
5600
x1
x2
x3
1 2 3 4
=0
0
M x + K x = 0
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M x + K x = 0
Equation of motion:
Let x = a sin ( t )
x = a sin ( t )- 2
(a)
(b)
Substitute (b) to (a) results in
K - 2 Ma = 0
=12
.
.
n
a1 =a11a21
.
.
an1
a2 =a12a22
.
.
an2
etc.
Mode1 Mode2
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END