The document analyzes the shock response of a multi degree-of-freedom spring-mass system to a general pulse load. It derives the equations of motion using a modal transformation, solves for the natural frequencies and mode shapes, and uses this to determine the force transmitted through each spring over time. The results show the immediate response of the leftmost spring compared to the delayed response of the rightmost spring, indicating it takes time for the wave to travel through the system.

1. 1
Shock Response to General Pulse Loads - Multi DOF Systems
x1 x2 x3 x4 x5
f0!(t) k
m m m m
m
k k k k
EOM for System
M
[ ]˙˙
x + K
[ ]x = f0 ! t
( ) (1)
where
M
[ ]= m I
[ ]
K
[ ] =
k !k
!k 2k !k
!k 2k !k
!k 2k !k
!k 2k
"
#
$
$
$
$
%
&
'
'
'
'
f0 = f0 0 0 0 0
{ }T
Modal EOM’s
Using the modal transformation:
x t
( ) = ˜
!
j
( )
qj t
( )
j=1
5
" (2)
produces the following (uncoupled) modal EOM’s:
˙˙
qj + !j
2
qj = ˆ
Fj t
( ) (3)
where
!j = natural frequencies (4)
˜
!
j
( )
= mass normalized modal vectors (5)
2
ˆ
Fj t
( ) = ˜
!
j
( )T
f0 " t
( ) = ˜
!1
j
( )
f0 " t
( ) (6)
Solution of Modal EOM’s
For zero initial conditions, the response of each modal equations is given by:
qj t
( ) =
˜
!1
j
( )
f0
"j
# $
( )sin "j t %$
( )d$
0
t
& =
˜
!1
j
( )
f0
"j
sin "jt (7)
Total Solution of System
Using (2), (4), (5) and (7) gives:
x t
( ) = ˜
!
j
( )
qj t
( )
j=1
5
"
= f0
˜
!
j
( ) ˜
!1
j
( )
#j
sin #jt
j=1
5
"
(8)
or
xp t
( ) = ˜
!p
j
( )
qj t
( )
j=1
5
"
= f0
˜
!p
j
( ) ˜
!1
j
( )
#j
sin #jt
j=1
5
"
(9)
The force transmitted by the spring between the pth and (p+1)th springs for p =
1, 2, 3, 4 is given by:
Fp t
( ) = k xp+1 t
( ) ! xp t
( )
[ ]
= k f0
˜
"p+1
j
( )
! ˜
"p
j
( )
#
$
%
&
˜
"1
j
( )
'j
sin 'jt
j=1
5
(
)
*
+
,
-
.
(10)
Numerical Results
Solving for the five natural frequencies and mass-normalized modal vectors
gives:

2. 1
Shock Response to General Pulse Loads - Multi DOF Systems
x1 x2 x3 x4 x5
f0!(t) k
m m m m
m
k k k k
EOM for System
M
[ ]˙˙
x + K
[ ]x = f0 ! t
( ) (1)
where
M
[ ]= m I
[ ]
K
[ ] =
k !k
!k 2k !k
!k 2k !k
!k 2k !k
!k 2k
"
#
$
$
$
$
%
&
'
'
'
'
f0 = f0 0 0 0 0
{ }T
Modal EOM’s
Using the modal transformation:
x t
( ) = ˜
!
j
( )
qj t
( )
j=1
5
" (2)
produces the following (uncoupled) modal EOM’s:
˙˙
qj + !j
2
qj = ˆ
Fj t
( ) (3)
where
!j = natural frequencies (4)
˜
!
j
( )
= mass normalized modal vectors (5)
2
ˆ
Fj t
( ) = ˜
!
j
( )T
f0 " t
( ) = ˜
!1
j
( )
f0 " t
( ) (6)
Solution of Modal EOM’s
For zero initial conditions, the response of each modal equations is given by:
qj t
( ) =
˜
!1
j
( )
f0
"j
# $
( )sin "j t %$
( )d$
0
t
& =
˜
!1
j
( )
f0
"j
sin "jt (7)
Total Solution of System
Using (2), (4), (5) and (7) gives:
x t
( ) = ˜
!
j
( )
qj t
( )
j=1
5
"
= f0
˜
!
j
( ) ˜
!1
j
( )
#j
sin #jt
j=1
5
"
(8)
or
xp t
( ) = ˜
!p
j
( )
qj t
( )
j=1
5
"
= f0
˜
!p
j
( ) ˜
!1
j
( )
#j
sin #jt
j=1
5
"
(9)
The force transmitted by the spring between the pth and (p+1)th springs for p =
1, 2, 3, 4 is given by:
Fp t
( ) = k xp+1 t
( ) ! xp t
( )
[ ]
= k f0
˜
"p+1
j
( )
! ˜
"p
j
( )
#
$
%
&
˜
"1
j
( )
'j
sin 'jt
j=1
5
(
)
*
+
,
-
.
(10)
Numerical Results
Solving for the five natural frequencies and mass-normalized modal vectors
gives:

3. 3
!1 = 0.2846 k /m
!2 = 0.8308 k /m
!3 = 1.3097 k/m
!4 = 1.6825 k /m
!5 = 1.9190 k /m
(11)
˜
P
[ ]= ˜
!
1
( ) ˜
!
2
( ) ˜
!
3
( ) ˜
!
4
( ) ˜
!
5
( )
"
#
$
%
=
0.5969 &0. 5485 0.4557 &0.3260 &0.1699
0.5485 &0.1699 &0.3260 0.5969 0.4557
0.4557 0.3260 &0.5485 &0.1699 &0.5969
0.3260 0.5969 0.1699 &0.4557 0.5485
0.1699 0.4557 0.5969 0.5485 &0.3260
"
#
'
'
'
'
$
%
(
(
(
(
1
m
(12)
With these results, we can write:
Fp t
( ) = k f0
˜
!p+1
j
( )
" ˜
!p
j
( )
#
$
%
&
˜
!1
j
( )
'j
sin 'jt
j=1
5
(
)
*
+
,
-
.
=
k
m
f0
"0.1014
"0.1946
"0.2720
"0.3274
/
0
1
2
1
3
4
1
5
1
sin '1t +
"0.2500
"0. 3274
"0.1788
0.0932
/
0
1
2
1
3
4
1
5
1
sin '2t +
"0.2720
"0.0774
0.2500
0.1486
/
0
1
2
1
3
4
1
5
1
sin '3t
)
*
+
+
+
+
"0.1788
0.1486
0.0554
"0.1946
/
0
1
2
1
3
4
1
5
1
sin '4t +
"0.0554
0.0932
"0.1014
0.0774
/
0
1
2
1
3
4
1
5
1
sin '5t
,
-
.
.
.
Shown below is the force in two springs for an equivalent nine-DOF model. Here
we see that the deformation in the spring to the left responds immediately to the
impact load. There is a time delay before the spring on the far right responds
indicating that it takes a finite time for the wave to travel along the system. Note
that the shape of the response is not preserved (the motion is NOT periodic).
This is in contrast with the shock response of the rod (also presented as an
example in this lecture) for which the shape of the traveling wave is preserved in
time.
4

4. 3
!1 = 0.2846 k /m
!2 = 0.8308 k /m
!3 = 1.3097 k/m
!4 = 1.6825 k /m
!5 = 1.9190 k /m
(11)
˜
P
[ ]= ˜
!
1
( ) ˜
!
2
( ) ˜
!
3
( ) ˜
!
4
( ) ˜
!
5
( )
"
#
$
%
=
0.5969 &0. 5485 0.4557 &0.3260 &0.1699
0.5485 &0.1699 &0.3260 0.5969 0.4557
0.4557 0.3260 &0.5485 &0.1699 &0.5969
0.3260 0.5969 0.1699 &0.4557 0.5485
0.1699 0.4557 0.5969 0.5485 &0.3260
"
#
'
'
'
'
$
%
(
(
(
(
1
m
(12)
With these results, we can write:
Fp t
( ) = k f0
˜
!p+1
j
( )
" ˜
!p
j
( )
#
$
%
&
˜
!1
j
( )
'j
sin 'jt
j=1
5
(
)
*
+
,
-
.
=
k
m
f0
"0.1014
"0.1946
"0.2720
"0.3274
/
0
1
2
1
3
4
1
5
1
sin '1t +
"0.2500
"0. 3274
"0.1788
0.0932
/
0
1
2
1
3
4
1
5
1
sin '2t +
"0.2720
"0.0774
0.2500
0.1486
/
0
1
2
1
3
4
1
5
1
sin '3t
)
*
+
+
+
+
"0.1788
0.1486
0.0554
"0.1946
/
0
1
2
1
3
4
1
5
1
sin '4t +
"0.0554
0.0932
"0.1014
0.0774
/
0
1
2
1
3
4
1
5
1
sin '5t
,
-
.
.
.
Shown below is the force in two springs for an equivalent nine-DOF model. Here
we see that the deformation in the spring to the left responds immediately to the
impact load. There is a time delay before the spring on the far right responds
indicating that it takes a finite time for the wave to travel along the system. Note
that the shape of the response is not preserved (the motion is NOT periodic).
This is in contrast with the shock response of the rod (also presented as an
example in this lecture) for which the shape of the traveling wave is preserved in
time.
4