CE573 – Structural Dynamics
SDOF – HARMONIC EXCITATION
UNDAMPED SDOF:
mx(t )  kx(t )  F0 sin(wt )
x (t )  xh (t )  x p (t )
Solution:
 xh (t )  A sin(wnt )  B cos(wnt )
 x p (t )  ?
Let’s guess a particular solution be in the form of x p (t )  D sin( wt )  E cos( wt )
x p (t )  w  D cos( wt )  E sin( wt )

x p (t )   w2  D sin( wt )  E cos( wt )   w2 x p (t )
Substitute into the dynamic equilbrium eqn.
mxp  kx p  F0 sin( wt )
 mw
2
x p (t ) 
 k  x p (t )  F0 sin( wt )
F0 sin( wt ) F0 sin( wt )

w2
k  mw2
k
1 2
wn
F0
Note that the maximum amplitude of x p (t ) is
F
k
where 0
is the static deflection.
2
k
 w
1  
 wn 
Sketching
1/8
Solution: x (t )  xh (t )  x p (t )
x(t )  A sin( wnt )  B cos( wn t ) 
Initial conditions:
F0 sin( wt )
2
k
 w
1  
 wn 
x(0)  x0 and x (0)  x0
x(0)  B  x0
x (0)  Awn 
A
F0
k
F
1
x0  0
wn
k
w
 w
1  
 wn 
 w
 
 wn 
 w
1  
 wn 
2
 x0
2
Solution in full form:
 w
 
x0
F0  wn 
F0
x(t )  x0 cos( wnt )  sin( wnt ) 
sin( wnt ) 
sin( wt )
2
wn
k
  w 2 
 w
1  
k 1    
  wn  
 wn 


If
x0  0, x0  0
 w
 
F0  wn 
F0
x(t )  
sin( wnt ) 
sin( wt )
2
k
  w 2 
 w
1  
k 1    

w 
 wn 
  n 
x(t ) 
F0
k


w
 sin( wt )  sin( wn t ) 
wn
 w 

1  
 wn 
1
2
2/8
Case 1:
w  wn

w
1
wn
Case 2:
w  wn

w
1
wn
Case 3:
w  wn
Using L’Hospital’s rule, we can find the form of the solution to be x (t ) 
F0
F
sin( wn t )  0 wn t cos( wn t )
2k
2k
Sketching
3/8
w  wn
Case 4:
say,
w  wn  w with w  0 but not zero.
F0
F0
k
k
x(t ) 
sin( wt )  1sin( wn t )  
2 
2
 w
 w
1  
1  
 wn 
 wn 

  w  wn  
  w  wn   
 2sin  
 t   cos  
 t 
 2  
  2  

Sketching,
4/8
DAMPED SDOF:
A SDOF linear system subject to harmonic excitation with forcing frequency w .
mx(t )  cx (t )  kx(t )  F0 sin(wt )
In mass-normalized form, the differential equation of motion is

x (t )  2 wn x (t )  wn2 x (t ) 
F0
sin( wt )
m
The solution is in the form of x (t )  xh (t )  x p (t )
We know the solution to the homogeneous (unforced) problem 
x (t )  2 wn x (t )  wn2 x (t )  0 with I.C.s:
xh (t )  e  wnt 1 cos  wd t    2 sin  wd t  
For the particular solution, assume the solution is in the form of
x p (t )  1 sin( wt )   2 cos( wt )
Substitute into the differential equation of motion and equate like terms. The resulting system of linear
equations are:
w
2
n
 w2  1  2 wn w 2 
F0
2 wn w1   w  w   2  0
2
n
Solve for
m
2
1 , 2
 w2 
1
k  w2 
n 

1 
2
2
2
 w  
w
1  2    2

 wn   wn 
F0
2 

F0
k
2 w
2
wn
 w2  
w
 1  2    2

 wn   wn 
2
Complete solution,
x(t )  e  wnt 1 cos  wd t    2 sin  wd t    1 sin( wt )   2 cos( wt )
Note that the part from homogeneous problem diminishes with time; it is “transient”. The particular solution,
however, keeps going and repeating itself; it makes the system reach a “steady-state” response.
5/8
Case 1:
w  wn
Case 2:
w  wn
The steady-state solution x p (t ) 
F0
x p (t ) 
1 sin( wt )   2 cos( wt )
k
2
 w2  
w
 1  2    2

 wn   wn 
2
sin( wt   )
can be rewritten as follows.
where tan( ) 
2
1
We see that the steady-state response has a pure harmonic oscillation at a frequency w . The amplitude of
the steady-state response is
F0
Amplitude  x p (t ) 
Note that
F0
k
k
2
 w2  
w
1  2    2

 wn   wn 
2
is the static deflection, i.e. the amount of deflection the SDOF system would have assumed
if the load with amplitude F0 were applied statically.
6/8
AF 
Definition: Amplification Factor (AF)
x p (t )

F0
 w2
k
1  2
 wn
1
2
 
w
   2

  wn 
2
Plotting AF
2


2 1   w    2 w   4 2  w  2
wn 
 wn 
d ( AF )
1   wn   
0  
0
2 32
2
d ( w wn )
 w2  2 

w
1  2    2
 
 wn   wn  
Maximum values of AF:
2

 w   w 
2
Setting numerator equal to zero, i.e. 
 1     2   0 , gives two roots:

 wn    wn 
(  1
1.
w0
2.
w
 1  2 2
wn
2
 70%) for which AFmax  1
(  1
Note that for lightly damped systems, i.e.
2
 70%) for which AFmax 
  1 , AFmax 
1
2 1   2
1
2
These results indicate that
x max

 static
2 1   2
if   1
 static
2
if   1
  static
if   1

2
2
7/8
Response for
w  wn
x(t )  e  wnt 1 cos  wd t    2 sin  wd t    1 sin( wt )   2 cos( wt )
1  0
2 

F0
k    static
2
2
Let’s consider a system that is at rest initially: x (0)  0, x (0)  0 .
1 
 static
2 1 
x(t )   static
2
2 
 static
2


1   wnt 

e
sin  wd t    cos  wnt  
 cos  wd t  


2 

1  2



Note that as time increases,
x(t )  
 static
cos  wn t 
2
[steady-state response]
Sketching,
8/8