RANDOM DYNAMICS
Dynamic response to a mono-variate seismic acceleration
Expressing the equations of motion using relative coordinates (with respect to soil) (Fig. 1a):
Mx  Cx  Kx  Mru
(1)
where r is the vector of the influence coefficients.
The dependence on time is neglected for sake of simplicity.
Fig. 1
Expressing the equations of motion using absolute coordinates (Fig. 1b):
Mq  Cq  Kq  0
(2)
or, in partitioned form:
Mff
M
 sf
Mfs  qf  Cff
 
Mss   qs  Csf
Cfs  qf  K ff
 
Css   qs  K sf
K fs  qf  0f 
  
K ss   qs   0 
(3)
where q f is the vector of the absolute displacements of the free D.O.F.s, qs is the absolute
displacement of the support. Moreover, let us assume:
q  qp  x
(4)
where q p is the vector of the pseudo-static displacements, i.e. of the absolute displacements due to
the quasi-static application of the motion u(t), x  q - q p is the vector of the relative or vibratory
displacements. In partitioned form:
qf  qfp  xf 
     
 qs  qsp   x s 
(5)
where q sp  u is the imposed displacement at the support, x s  0 is the relative displacement of the
support,
1
q fp  ru
(6)
being r the vector of the absolute displacements of the free nodes, due to the static application of
u  1 ( r is also known as the vector of the influence coefficients), x f is the unknown vector.
Applying the above definitions:
q  ru  x  r 
x 
q =  f      f    u   f 
0
 qs   u   0  1
(7)
Substituting Eq. (7) into Eq. (3):
 M ff M fs  r 
 M ff M fs  xf 
 u  
M

 
 sf M ss  1
M sf Mss   0 
Cff Cfs  r 
Cff Cfs  xf 
 u  
C

 
 sf Css  1
Csf Css   0 
 K ff K fs  r 
 K ff K fs  xf  0f 
 u  
K

   
 sf K ss  1
K sf K ss   0   0 
(8)
Expanding the first row of the above equation:
M ff ru  M fs u Mff xf  Cff ru  Cfs u  Cff xf  K ff xf  0f 
Mff xf  Cff xf  K ff xf    Mff r  Mfs  u   Cff r  Cfs  u
(9)
The problem can be simplified introducing the following hypotheses:
1)
2)
the structural system has lumped masses  M fs  0
C = K  Cff r  Cfs   K ff r  K fs   0
C  K  Cff r  Cfs
0
In this case:
Mff xf  Cff xf  K ff xf  Mff ru
(10)
Finally, the vector r shall be determined. Assuming u = 1 in Eq. (8):
K ff
K
 sf
K fs  r  0f 
  
Kss  1  0 
Expanding the first row:
K ff r  K fs  0 
r  K -1ff K fs
(11)
2
Dynamic response to a multi-variate seismic acceleration
Le us consider a structural system subjected to a multi-variate seismic acceleration (Fig. 2).
Fig. 2
3
Expressing the equations of motion using absolute coordinates, analogously to Eqs. (2) and (3):
Mq  Cq  Kq  0
Mff
M
 sf
(12)
Mfs  qf  Cff
 
Mss  qs  Csf
Cfs  qf   K ff
 
Css  qs  K sf
K fs  qf  0
  
K ss  qs  0
(13)
where q f is the vector of the absolute displacements of the free D.O.F.s (Nf), qs is the vector of the
absolute displacements of the supported D.O.F.s (Ns). Moreover, let us assume:
q  qp  x
(14)
where q p is the vector of the pseudo-static displacements, i.e. of the absolute displacements due to
the quasi-static application of the seismic motion at the supports; x  q - q p is the vector of the
relative or vibratory displacements. Using the partitioned form:
q  qfp  x 
q =  f     f 
qs  qsp   xs 
(15)
T
q sp  u   u1u 2 ...u Ns  being the vector of the seismic motion applied at the supports, xs  0s ,
Ns
qfp   iri u i  Ru
(16)
1
where ri = q fp (for u i  1, u j  0 with j  i) is the i-th vector of the influence coefficients.
R  r1r2 ...rNs  is the matrix of the influence coefficients. Thus, Eq. (15) becomes:
q  Ru  x  R 
x 
q =  f       f    u   f 
qs   u   0s  Is 
 0s 
(17)
Substituting Eq. (17) into Eq. (13):
 M ff
M
 sf
C
  ff
Csf
K
  ff
K sf
M fs  R 
 M ff
 u  

M ss  I s 
M sf
M fs  x f 
 
M ss   0s 
Cfs  R 
Cff Cfs  x f 
 u  

 
Css  I s 
Csf Css   0s 
K fs  R 
 K ff K fs  x f  0f 
 u  

   
K ss  I s 
K sf K ss   0s   0s 
(18)
Expanding the first row:
Mff xf  Cff xf  K ff xf    Mff R  MfsIs  u   Cff R  CfsIs  u
(19)
4
The problem can be simplified introducing the following hypotheses:
the structural system has lumped masses  M fs  0
1)
C = K  Cff R  CfsIs    K ff R  K fsIs   0f
2)
C  K  Cff R  Cfs Is
0f
In this case it results:
Mff xf  Cff xf  K ff xf  Mff Ru
(20)
where:
Ns
Ru   iri u i
1
Thus, Eq. (20) may be re-written as:
Ns
xf   i xfi
1
(21)
M ff xf  Cff xf  K ff xf   M ff ri u i
Finally, the matrix R shall be determined.
The i-th column ri may be evaluated assuming u i  1, u j  0 with j  i . From Eq. (18):
K ff
K
 sf
K fs   ri  0f 
  
K ss  Is  0s 
where I si is a nil vector except for the i-th unit component.
Expanding the first row:
K ff ri  K fs Isi  0f 
ri  K ff-1 K fs I si
(22)
5
Example. Consider a single-storey shear-type r.c. building subjected to different seismic
accelerations at the base of the two columns (Fig. 3).
Fig. 3
The equilibrium equations are:
EJ
EJ
q  12 3 q 3
3 1
h
h
EJ
EJ
0  12 3 q 2  12 3 q 3
h
h
EJ
EJ
EJ
0  24 3 q 3  12 3 q1  12 3 q 2
h
h
h
0  12
Thus:
 1 0 1
 q1 
m 0 0 
12EJ 
 

K  3  0 1 1 q  q 2  M   0 m 0 
h
 
 1 1 2 
q 3 
 0 0 M 
Moreover, let us assume:
C  K
Rewriting the equations of motion in partitioned form:
 M 0 0  q 3 
q 3 
 2 1 1 q 3 
0 
 0 m 0   q  + C  q  + 12EJ  1 1 0   q  q  0 
 1
 
3 

 1
 1
q  h  1 0 1  q 
 
 0 0 m  q 2 
 2

 2
0 
where :
q 3   q 3P   x 3  q 3P   x 3 
         
 q1    q1P    x1    u1    0 
q  q   x   u   0 
 2   2P   2   2   
q 3p being a linear combination of u1 and u2:
6
2
q 3P   i ri u i  r1u1  r2 u 2
1
Fig. 4
It follows:
1  1
h 3 12EJ
r1  

1

1


 
24EJ h 3
0  2
3
0  1
h 12EJ
r2  

1

1


 
24EJ h 3
1  2
1
R   r1 r2   1 1
2
The equation providing the vibratory motion of the free D.O.F. is:
Mx 3 
u 
24EJ
24EJ
1
M
x 3  3 x 3  M 11  1     u1  u 2 
3
h
h
2
2
u 2 
Finally:
1
 u  u 2   x 
q3   2 1
3
  
  
u1
 q1   
 0 
q  
 0
u2
 2 
  


7
Modal analysis
Consider Eq. (20) and evaluate the eigenvalue problem:
K
ff
 2k M ff   k  0f
  1  2 ...  Nf 
Let us apply the principal transformation law:
xf  p 
 T Mff p +  TCff p +  TK ff p   TMff Ru
and let us introduce the definitions:
 T Mff   
 T M ff R  G
In the hypothesis of classical damping:
Ns
pk  2k k pk  k2 pk   G k u
k  1,...Nf
1
where G k is the generic term of the matrix G.
Frequency domain analysis
Applying the Fourier transform of both sides of Eq. (20):
2Mff xf  iCff xf  K ff xf  Mff Ru
where x f , u are the Fourier transforms of x f , u .
H is the complex frequency response function:
H     2Mff  iCff  K ff 
1
In the case of classically damped systems:
H    HP    T
HP   is a diagonal matrix whose i-th term is HPi   .
xf    H   Mff Ru  
8
Random dynamic analysis
Let us consider Eq. (20) and let us assume:
F t   Mff Ru(t)
It follows that:
Sxf    H*   SF   HT  
SF    Mff RSU   RTMffT 
Sxf    H*   Mff RSU   RTMff HT  
where SU   is the power spectral density matrix of the seismic acceleration.
9