Download - 10 CE225 MDOF Seismic Inertial Forces
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MDOF Seismic Inertial Forces
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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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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
(Loading Applied on Mass)
p1(t)
p2(t)
m1
m2
u1
u2
c1
c2
k1
k2
u2
u1
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Equation of Motion of Two-Story Shear
Building
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)
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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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Equation of Motion of a Two-
Degree-of-Freedom System
(Base Excitation, Undamped)
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2-DOF System Two-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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fixedr
ef
erencea
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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vgdisplacement of support/ground from a fixed reference axis
v1displacement of mass1 relative to the base
v2displacement 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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If damping is considered:
+ = -
m1
0
0
m2
1
1
vg
= - 1 vg
Equation of Motion (Ground Acceleration)
m1 0
0 m2
u1
u2
c1+c2 -c2
- c2 c2
+
k1+k2 -k2
- k2 k2
u1
u2
u1
u2
m u + c u +k u m
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Equation of Motion (Loading on Mass)
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)
+ = -
m1
0
0
m2
1
1
vg
= - 1 vg
Equation of Motion (Ground Acceleration)
m1 0
0 m2
u1
u2
c1+c2 -c2
- c2 c2
+
k1+k2 -k2
- k2 k2
u1
u2
u1
u2
m u + c u +k u m
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Equivalence:
m1
m2
vg
m1
m2
==== -m1vg
-m2vg
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Disregard damping; solve eigenvalues and
eigenvectors. Uncouple the equation of motion:
+ = -
m1
0
0
m2
1
1
vg
= - 1 vg
Equation of Motion (Ground Acceleration)
m1 0
0 m2
u1
u2
c1+c2 -c2
- c2 c2
+
k1+k2 -k2
- k2 k2
u1
u2
u1
u2
m u + c u +k u m
1
T m 1 +
z1
1
T c 1
+z1 1
T k 1
z1
= 1
T m- 1 vg
2
T m 2
+z2
2
T c 2
+z2
2
T k 2
z2
= 2
T m- 1 vg
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1
T m 1
+z1
1
T c 1
+z1
1
T k 1
z1
= 1
T m- 1 vg
2
T m 2
+z2
2
T c 2
+z2
2
T k 2
z2
= 2
T m- 1 vg
The above equations can be combined as:
T
m +z T
c +z T
k z = T
m- 1 vg
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1
T m 1
+z1
1
T c 1
+z1
1
T k 1
z1
= 1
T m- 1 vg
2
T m 2
+z2
2
T c 2
+z2
2
T k 2
z2
= 2
T m- 1 vg
m1* c 1* k 1* p 1*
m1*
+z1
c 1*
z1
+ k 1* z
1= - p 1
*vg
m2* c 2*
k 2*
p 2*
m2*
+z2
c 2*
z2
+ k 2* z
2= - p 2
*vg
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m1*
+z1
c 1*
z1
+ k 1* z
1= - p 1
*vg
m2
*
+z2 c 2*
z2 + k 2* z
2 = - p 2*
vg
The general form of the above equations is:
mi*
+zi
c i*
zi
+ k i* z
i= - p i
*vg
These are equations of motion of single-
degree-of-freedom systems, vibrating under
each mode.
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Divide each term by :
mi*
+zi
c i*
zi
+ k i* z
i= - p i
*vg
mi*
+zi i
vg2 i zi + i z
2
i= -
p i*
mi*
=p
i
*
mi*i =
iT m 1
iT m i
i is called mode participation factor (for mode i)
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Since vgis a function of time, we can express it as
maxgv
+zi i
2 i zi + i z
2i
=
=
gv f(t)
gv= kcg
maxgv
t
gv = f(t)kcg
f(t)kcgi-
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The general form for the acceleration response is:
+zi i
2 i zi + i z
2i = f(t)kcgi-
+
zi
i2 i zi + i z
2
i
= f(t)kcgi- + i (t, , )
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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 - k2 + k2v2
2 x 1
=
- m1 vg
- m2 vg
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
=
- m1vg
- m2vg
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
=
- m1vg
- m2vg
2 x 1
m1v1+ 0 v2( k1+k2 )v1 - k2v2+ = - m1vg
0 v1 m2+ v2 - k2 + k2v2 - m2vg=
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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION
m1v1( k1+k2 )v1 - k2v2+ = - m1vg
m2v2 - k2 + k2v2 - m2vg=
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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION
m1v1( k1+k2 )v1 - k2v2+ = - m1vg
m2v2 - k2 + k2v2 - m2vg=
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 kg x1
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/mm4= 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