CHAPTER 5 RESPONSE OF SDOF SYSTEMS TO NONPERIODIC EXCITATION (General Load) p( t T 1 )
p( t )
periodic load
Harmonic loads Forms of Nonperiodic Excitation p( t )
p( t )
p0
p0 t
t
t r
Step Load
Ramp Load t r : Rise Time
p( t )
p( t )
p0
p0 t
t d
t
Rectangular Pulse Load
Triangular Pulse
t d : Duration
p( t )
p( t )
p0 t
t
Half-sine pulse load
Explosion load 1
Ref: Introduc Introduction tion to Structura Structurall Dynamics, Dynamics, J. M. Biggs, Biggs, McGraw-Hill, McGraw-Hill, 1982 §5.1 Response of a SDOF System to an Ideal Step Load Undamped System
p( t ) && mu
ku
u(0)
t 0
p0 ,
u& (0)
p0
0
t Figure 5.1. Step Load. u( t )
u p ( t ) p0 k
Apply u(0) A1
u( t )
uh ( t )
A1 cos
u& (0)
k p 0 k R( t )
t A2 sin
n
t
0 , then
p0 k p0
n
0
A2 (1 cos
n
t )
R( t ) 1
cos
n
t
Load Factor Factor (DLF) (DLF) R( t ) : Response Ratio or D D ynamic Load
2
Ref: Introduc Introduction tion to Structura Structurall Dynamics, Dynamics, J. M. Biggs, Biggs, McGraw-Hill, McGraw-Hill, 1982 §5.1 Response of a SDOF System to an Ideal Step Load Undamped System
p( t ) && mu
ku
u(0)
t 0
p0 ,
u& (0)
p0
0
t Figure 5.1. Step Load. u( t )
u p ( t ) p0 k
Apply u(0) A1
u( t )
uh ( t )
A1 cos
u& (0)
k p 0 k R( t )
t A2 sin
n
t
0 , then
p0 k p0
n
0
A2 (1 cos
n
t )
R( t ) 1
cos
n
t
Load Factor Factor (DLF) (DLF) R( t ) : Response Ratio or D D ynamic Load
2
Damped System
&& mu
ku
cu&
p0 ,
t 0
u(0)
u& (0)
0
u( t )
u p ( t )
uh ( t )
po
e
k
Apply u( 0)
nt
( A1 cos
0
A1
0
k
u& ( t )
n
e
e u& (0)
(5.2)
u& (0)
po
u(0)
nt
nt
d
po k
{1
( A1 cos
[cos
(5.4)
po k
t A2 sin
t ) +
d
t A2 cos
t )
d
0
n t
t )
d
d
( A1 sin A2 d
e
t A2 sin
d
A1
A1 n
u( t )
(5.1)
d
po
A2
n
t
d
n
k
d
sin
d
t ] }
(5.5)
d
po
u( t )
R( t ) k
R( t ) : Response Response Ratio Ratio , or D ynamic Load Factor Factor (DLF) (DLF) (5.6)
R( t )
1 e
nt
[cos
t
n
d
sin
d
3
t ]
d
Rmax
2
(5.7)
Figure 5.2. Response ratio plot for a step input. p0 u( t ) 0, (1 cos n t ) If k
(5.8)
§5.2 Response of an Undamped SDOF System to Rectangular Pulse and Ramp Loadings ● Rectangular Pulse Load: Undamped System ku p( t )
&& mu
p( t ) p0 0 u(0)
0
t t d
t d
t
u& (0)
p( t )
p0 0
t
t d : Duration
t d
Rectangular Pulse Load (a) For 0 ≤ t ≤ t d (Forced Vibration Era)
&& mu u(0) u( t )
k u p0 u& (0) u p ( t )
0 uh ( t )
4
p 0
u1 ( t )
A1 cos
k
Apply u(0)
u& (0)
A2
0
cos
n
k
p0
u1 ( t )
1
k p0 k
R1 ( t )
1
t A2 sin
n
t
0
p0
A1
n
t
R1 ( t )
cos
n
t
(5.10) (b) For td ≤ t (Free Vibration Era) Method 1 (Conventional Method) && ku o mu p0 1 cos n t d u1 ( t d ) k u2 ( t )
u1 ( t d ) cos
u& 1 ( t d )
( t t d ) n
p0
u& 1 ( t d ) sin
n
n
k
sin
t
n d
( t t d )
n
u2 ( t )
p0 k
[ R1 ( t d ) cos
k
cos R2 ( t )
(1
n
( t t d )
sin
n
( t t d )]
(5.11)
n
p0 R2 ( t )
n
R& 1 ( t d )
R2 ( t ) cos
t ) cos
n d
( t t d ) cos
n
cos
( t t d )
n
( t t d )
t cos
n d
cos
n
t
5
sin
n
t sin
n d
( t t d )
sin
n
( t t d )
t sin
n d
n
( t t d )
Method 2 (Superposition Method)
p0
p0
+
= t d
u2 ( t )
p( t ) u1 ( t )
p0
p0
u1 ( t t d )
k
( t 0)
p( t ) p0
R2 ( t )
R2 ( t ) R1 ( t ) R1 ( t t d ) 1 R2 ( t )
cos
cos
n
n
t
( t t d )
1
cos cos
n
n
( t t d )
t
● Plot R1 ( t )
1
cos
R2 ( t )
cos cos 2
n
t 1 n ( t t d ) t
t d
T n
T n
t
cos 2 cos
0≤
T n n
t T n
≤
t d T n
t t
cos 2
T n
6
t
t d
T n
T n
p0
( t t d )
●Ramp Load: Undamped System
p(t )
t r :
p0
rise time
t r
Figure 5.5 Ramp Load. t
&& mu
u(0)
ku
t r p0
u& (0)
0
p0
t t r
t r
(5.14)
t
0
(5.15)
(a) For 0 ≤ t ≤ t r u p ( t )
u( t )
Apply u(0)
u1 ( t )
t
p0
t r
k
t t r
p0
A1 cos
k u& (0)
(5.16)
n
t A2 sin
n
t
(5.17)
0 , then
p0
t
1
k
t r
t n r
sin
n
7
t
(5.18)
3T n (
If t r
u pseudostat ic
6 ) , the second term is small
t
n r
p0
t
k
t r
0
t t r
(5.20)
(b) For t r ≤ t u2
p0 k
1
1 t
[sin
n
( t t r )
n r
● Plot
8
sin
n
t ]
(5.19)
§5.3 Response of Undamped SDOF System to a Short-Duration Impulse: Unit Impulse Response t d
I
0
impulse
p( t )dt
(5.21)
u( t )
k
p(t )
m
Figure 5.7. Undamped SDOF system subjected to short-duration impulse.
For an undamped system
&& mu
p( t )
ku
0
0
t t d
t d
t
(5.22)
● Method 1 t
&& ( mu 0 u(0)
ku)dt u& (0)
m u& ( t d ) t d 0
t 0
p( t )dt
0
(5.23)
kuavg t d I
(5.24)
udt uavg t d uavg : average displacement in 0
For t d
0( t d
t t d
T n ), ignore the second term in (5.24)
9
m u& (0 ) I u& ( 0 ) u(0 )
I m
(5.26a) 0
0 since t d
&& Recall, for m u u( t )
(5.25)
ku
u0 cos
t n
(5.26b)
0
u& 0
sin
n
t
n
Therefore u( t )
h( t )
I m
n
t
sin
n
t
impulse response
(5.27)
n
1 m
sin
unit impulse response function
(5.28)
n
For a viscous-damped system u( t )
h( t )
I m
n t
e
n t
sin
n
t
(5.29)
sin
n
t
(5.30)
d
1 m
e
d
10
● Method 2 Using the response of a system under rectangular pulse load
&& mu I
0
0
ku( t ) p( t )
cu&
p( t ) dt :
impulse load
p( t )
t
I
Assume u(0)
u& (0)
p( t ) dt t
0 T n
Then
&&(0) mu u(0)
0
cu& (0)
I
ku( 0)
0 Thus
u& (0)
I
&&(0) u
m
at t u& ( )
I
&&(0) u 1
u( )
2
&& u
m 1 I
u& ( t )
0
2 m
t t
free vibration The time is measured from && c u& k u 0 mu
u& 0
0
u0 u( t )
e
n t
t
I m
u0 cos
t t d
u0
n d
u( t )
I m
&
u
e
n t
sin
d
d
11
u0
sin
t
d
u( t ) I h( t ) Impulse Response h( t )
I m
e
n t
sin
t
d
d
Unit Impulse Response Function
● Method 3 Using the response of a system under rectangular pulse load Ref: Example 5.1
&& mu
ku p( t )
p( t ) p0 0 u(0)
u& (0)
p( t )
0
t t d
t d
t p0
0
t d << T n
t d
(a) For 0 ≤ t ≤ t d (Forced Vibration Era) p0 u1 ( t ) (1 cos n t d ) k u& 1 ( t )
n
T n 2 and t d
t
n d
lim (1 t d
0
lim sin t d
0
Rectangular Pulse Load
p0
sin n t k p0 (1 cos n t d ) u1 ( t d ) k p0 u& 1 ( t d ) sin n t d n k
Since
t
2 and cos
t &
n d
t ) & n d
(1) (2)
T n t
n d
1 2
0 t n d
2
t
n d
12
u1 ( t d ) &
1 p0 2
u& 1 ( t d ) &
t ) 2 n d
(
k p0 n k
(
I
t d
2
m
I
t )
n d
(3)
m
(b) For td ≤ t (Free Vibration Era)
&& mu
ku
u2 ( t )
0
u1 ( t d ) cos
( t t d ) n
u& 1 ( t d )
sin
n
( t t d )
n
lim u2 ( t ) t d
0
u( t )
lim t d
0
I m
It d 2m sin
cos
n
( t t d ) n
I m
sin
n
( t t d )
(4)
n
(5)
t
n
13
5.4 Response of a SDOF System to General Dynamic Excitation: Duhamel Integral (Convolution Integral) Method
Figure 5.8 Incremental response of an undamped SDOF system.
dI
du(t )
m
sin
m
) undamped system
( t
(5.31)
n
1
u( t )
n
t
p( ) sin
0
n
)d Duhamel Integral
( t
(5.32)
n
or u( t )
t 0
p( )h( t
h( t
)
)d
1 m
sin
n
( t
)
(5.33)
n
It can be shown that t
p( ) h( t 0
) d
t 0
p( t
) h( ) d
14
convolution integral
u( t )
1
t
m
0
n ( t
p( )e
)
sin
d
( t
)d
damped system
(5.34)
d
For integration, use the MATLAB program. ● General solution
u( t )
1
t 0
m
p( ) sin
n
( t
)d
n
u0 cos
undamped system (5.36)
u& 0
t n
sin
n
t
n
u( t )
1
t
m
0
n ( t
p( )e
)
sin
d
( t
)d
d n t
u0 e 1
cos
damped system
t
d
u& 0
n
u0 e
n t
sin
(5.37)
t
d
d
● Comments (1) : time of loading
t : time of response
(2) The Duhamel integral was derived assuming u(0)
u& (0)
0
(3) The Duhamel integral is a convolution integral. (4) From Eqs. (5.32) and (5.34), it can be seen that as increases, u( t ) becomes smaller. (5) The integration may not be easy.
&& mu u( t )
ku p0 cos t
cu&
1 m
t 0
cos
sin
n
( t
)d
n
Assume u( t )
Ue i
t
or u( t )
U cos t
15
complicate
n
or
d
● Numerical Computation of Duhamel Integration 1
u( t )
t
m
0
p( ) sin
n
( t
)d
(5.32)
n
Use Simpson’s Rule or Trapezoidal Rule
● Step Load: Undamped System
&& mu
ku p0 ,
p( t ) p0 u(0)
p( t )
t 0
t 0
u& (0)
p0
0
t
Step Load
General solution of undamped system u( t )
1 m
t 0
p( ) sin
n
( t
)d
n
u0 cos
u& 0
t n
sin
n
t
n
u(0) u( t )
u& (0)
0
1 m
t 0
n
( t
)d
n
p0 m u( t )
p0 sin
p0 k
(1
2
cos
( t n
)
t 0
n
cos
n
t )
16
(5.36)
● Rectangular Pulse Load: Undamped System p( t )
&& ku p(t ) mu u(0) u& (0) 0 p( t ) p0 0 0
p0
t t d
t d
t
t d
Rectangular Pulse Load
t t d u( t )
t d
0
p0 k
(1
cos
n
t )
t
u( t )
1
t
m
0
t d 0
u( t )
p0 k
( t
)d
p0 sin
n
( t
)d
n
p0 m
n
n
1 m
p( t ) sin
2
cos
n
( t
)
t d 0
n
cos
n
( t t d )
cos
17
n
t
t
Example 5.2 Triangular Pulse Load: Undamped System p( t ) p0 t t d
p( t )
p( t ) 0
t
p0 1
0
t d
0
t t d
(1)
t
(2)
t d
t t d u( t )
1 m
t 0
p0 (1
t d
n
p0 k p0 k
sin
t n
cos
t n
t
) sin
(1
0
t 0
t d
(1
n
( t
) cos
n
t d
) sin
)d
n
d (
n
d (
n
)
(3)
)
Using integration by parts, we get cos
n
d (
n
)
sin
1
sin
n
n
d (
n
sin
n
)
1
cos
n
(4)
n
n
Similarly, sin
n
d (
n
)
cos
1 n n
Hence,
18
sin
n
(5)
p0
u( t )
sin
k cos
t sin n
cos
t n
t
t n
sin
t d t
t 1 n
t d
1
t n
cos
t
1
t n
t
n d
cos
1
t n
t
n d
sin
n
(6)
t
n d
Simplifying this expression, we get u( t )
R1 ( t ) t d
p0
R1 ( t )
k
t
1
cos
t d
1
t n
t
sin
n
t
(7)
n d
t
u( t )
1 m
t d 0
p0 (1
n
t d
) sin
n
( t
)d
(8)
Note that this is the same as Eq. 3 except that it is evaluated at since p( )
t d ,
0 for t t d . Equation 6 can be used by setting t t d
within the square brackets. Thus, u( t )
p0
sin
k
1
t n
t
cos
1
t n d
t
n d
- cos
n
t 1
1
t
n d
sin
t
n d
n d
Thus R2 ( t )
1 t n d
sin
n
t (1 cos
19
t )
n d
cos
n
t (
t
n d
sin
t ) (9)
n d
● Plot
20
5.5 Response Spectra A response spectrum is a plot of maximum “response” (e.g., displacement, stress, acceleration, etc.) of SDOF systems to a given input versus some system parameter, generally the undamped natural frequency. A set of such curves, for example, curves plotted for various levels of system damping, may be referred to as response spectra. umax
umax ( ,
n
)
Figure 5.9 Example of a design problem amenable to solution using a response spectrum. (a) Vehicle moving over specified bump. (b) Relative displacement response spectra.
21
● Rectangular Pulse Load: Undamped System ku p( t )
&& mu
p( t ) p0 0 u(0)
p( t )
0
t t d
t d
t
u& (0)
p0
0 t
t d
R1 ( t )
1
R2 ( t )
cos
cos
n
n
t
( t t d )
cos
n
Rectangular Pulse Load
t
(5.11)
● ( R1 ) max dR1 dt
0
n
T n
( t 1 ) max
( R1 )max
R1 (
T n 2
n
)
t 0 ( t 1 ) max
or
2
n
sin
T n
1
)
2
2
(5.10)
● ( R2 )max dR2 dt 2 sin ( t 2 )max
R2
max
0
t
n d
2
cos
1 2
n
( 2t t d ) 2
t d n
2 sin
n
0
( 2t t d ) 2
2
T n
t d
( t 2 ) max
1
4
2
T n
4
t d T n
22
1 t d ( ) 2 T n
Figure 5.3. Response to a rectangular pulse input. (a) Rectangular Pulse. (b) Response Ratios. T n T n (b ) t d (a ) t d 2 2
Figure 5.4. Maxi
( R1 )max
mum
R2
T n
Free
--
T n 2
2 sin
max
( t 1 ) max
Figure 5.4. Maximum
R1 (
)
)
2
t d T n
1 2
( t 2 ) max
1
T n
4
1 t d ( ) 2 T n
Forced Era
Fig. 5.4 Maximum response ratio for rectangular pulse excitation (a) maximum response amplitude (b) time at which maximum response occurs
23
Note (1) If t d
T n / 2 , the maximum occurs in the forced-vibration era,
that is, prior to t d (2) If t d
T n / 2 , the maximum occurs in the residual vibration era,
that is, after t d , ● Ramp Load-Undamped system
p(t )
p0
t
t r
Figure 5.5 Ramp Load. t
&& mu u(0)
ku
u& (0) R1 ( t )
R2
1
R& 1 ( t )
0
p0
t r p0
t t r
t r
(5.14)
t
0 t
1
t r
t n r 1 t n r
[sin
1
1
t r
t r
sin
n
n
( t t r )
cos
(5.18)
t
t
n m
sin
0
24
n
t ]
(5.19)
cos
t =1
n m
( R1 ) max R& 2
1 n
t r
[sin
n
( t t r )
sin
n
t ]
0
( R2 ) max
Figure 5.6. Response of an undamped SDOF system to ramp inputs. (a) Response to ramp inputs. (b) Maximum response to ramp inputs.
25
● Plot: Time of maximum
26
● Example 5.2: Triangular Load-Undamped system p( t ) p0
t t d p( t )
t
p0 1
t d
0
0
t t d
(1)
t d
t
(2)
Solution u( t )
p0 k
R1 ( t )
For t d
t n d
t
1
t d
cos
1
t n
sin
t
n
t
0
t t d
t (
n d
(7)
n d
t 1
R2 ( t )
t n
R( t )
t n d 2 T n 2
t 2
sin
n
t ( 1
t d
t
T n
t d
cos
cos
t )
n d
since
t d T n
27
2 n
T n
n
t
sin
t )
n d
(9)
Example 5.3 Plots of R(t) and R max Using the response ratio expressions determined in Example 5.2, a. Plot the response ratio versus t / t d for t d / T n and 1.5 for 0 b. Determine
t / t d
0.25, 0.5, 1.0
2.0 .
expressions
for
the
maximum
response
for
t t d and t t d . Using this information, plot a response
spectrum in the form of Rmax versus f n . (a) Plots R1 ( t )
R2 ( t )
t
1
t d 1 t n d
cos
( 1 cos
1
t n
sin
t n d t ) sin
n d
n
n
t (
Fig. 1 Response Ratio Plots
28
t
t
n d
(1)
sin
t ) cos
n d
n
t (8)
● Rmax R1 ( t )
t
1
t d 1
R& 1 ( t )
cos
t sin
n d
1 cos
t
n d
n m
sin
t n d
t sin
1
t d
1
t n
n
t cos
n
t
(1)
t =0
(2)
n
t
(3)
n m
Note ( 1
2 sin 2
cos 2 )
sin 2
2 sin cos
cos 2
2 cos 2 t
1
2 tan (
t
R1
t
max
(5)
n d
2
n m
(4)
1
n m
tan
t m
1
(6)
cos
t d
1
R2 ( t )
t )
n d
( 1 cos
t
1
t n m
t n d
t ) sin
n d
n
sin
t (
(7)
t
n m
t
n d
sin
t ) cos
n d
n
t (8)
n d
R2
1 max
t n d
( 1 cos
t )2 n d
t
2 f n t d
Note n
2 f n
n d
29
(
t n d
sin
t )2 n d
1/ 2
(9)
● Plots of Rmax
Figure 2 Response Spectrum for a Triangular Pulse Load Note f n t d 0.371
( R2 )max
( R1 ) max
f n t d 0.371
( R1 )max
( R2 ) max
30
31
Example 5.4: Application of Response Spectrum
u(t)
p(t)
p(t) m
t d
k /2
k /2 t
t
d
Blast Force Given: k = 9.0 GN/m m = 10 Mg Case 1 t d
0.4 sec
umax
5 mm
Case 2 t d
0.04 sec
umax
5 mm
For: Compare ( p0 )max Solution Find the maximum response ratio from the figure 2 shown in Example 5.2 a. Determine the natural frequency k n
f n
m n
2
1/ 2
9 10 9 10 10
6
1/ 2
30 rad/s
(1)
4.77 Hz
b. Determine the maximum response ratio from the figure 2 shown in Example 5.2:
32
Figure 2 Response Spectrum for a Triangular Pulse Load For Case 1, 4.77 0.4
f n t d
1.91
(2a)
From the response spectrum of Example 5.2 1.75
Rmax
(2b)
For Case 2, 4.77 0.04
f n t d
0.191
(3a)
and from the response spectrum of Example 5.2, 0.58
Rmax
(3b)
c. Determine the static displacement for each case and the p0 : umax
5 mm
umax
Rmax
(4)
p0
p0
k
kumax Rmax
(5)
Thus, for Case 1, 9( 10 9 )( 5 )( 10 3 )
p0
1
p0
1
1.75
( t d
25.7 MN
33
0.4 sec)
(6)
9( 10 9 )( 5 )( 10 3 )
p0
2
p0
2
0.58
( t d
77.6 MN
0.04 sec)
Comment ( t d )1
0.4
( t d ) 2
0.04
10
( p0 )1
25.7
1
( p0 ) 2
77.6
3
p( t ) 77.6 MN
Case 2 Case 1
25.7 MN t 0.04 sec
0.4 sec
34
(7)
● Support Motion z ( t )
) u(t k m
p(t)
c
Figure 5.10 Prototype Relative Motion SDOF system .
&& mu
k ( u z )
w
u z
&& mw
cw&
0
(5.39)
kw
&& m z
(5.40)
&&( t ) z &&max f a ( t ) z p( t )
&&( t ) m z &&max f a ( t ) m z
p( t ) p0 cos t p0 w ( t )
&&max m z 1
t
m
n
1
t
0
0
&&( ) sin ( )m z
n
( t
)d
undamped system
&&( ) sin ( ) z
n
( t
)d
n
1
t 0
&&max ) f a ( ) sin ( z
n
( t
n
35
)d
p 0
w ( t )
k
k
t n 0
R( t )
&&max z
w max
&&max m z
R( t )
&&max z
R( t )
2
R( t )
n
f a ( ) sin
n
)d
( t
(5.43)
Rmax
2
(5.44)
n
1
w ( t )
t
&&( )e z 0
n ( t
)
sin
d
( t
damped system
)d
d
2 t 0
n
R( t )
f a ( )e
n ( t
)
sin
d
( t
)d
(5.42)
d
Undamped system under support motion
&& mu
k ( u z )
&& mu
kw
0
k
&&( t ) u
m 2
&&max u
0
n
(5.45)
w
w max
(5.46)
Using (5.44) w max
1 2
(5.44)
&&max Rmax z
n
&&max u
(5.47)
&max Rmax &z
36
Example 5.5 Support Motion
&& z
&&max z 2t d
t
&&max z
&& mw
kw
&& m z
&& z &&max f a ( t ) z 1
f a ( t )
t
0
t d
0
t 2t d
2t d
t
Solution From Eq. 7 of Example (5.2) R1 ( t )
1
t t d
cos
1
t n
t n d
sin
n
t
(1)
From Eq. 8 of Example (5.2) R2 ( t ) R1 ( 2t d ) cos
n
( t 2t d )
R& 1 ( 2t d ) n
37
sin
n
( t 2t d )
(2)
1
R& 1 ( t )
1
t d
(
1
t
2 tan (
t
2 2 tan (
n m
max
t cos
n
t ) p
n d
t
t
R1 ( 2t d )
max
(3)
(4)
cos
(6) 1
t n m
t
n d
R2
t
t )
n m
1
n
n d
1
n m
R1
t ) sin
n d
2
sin
t
n m
(7)
n d
R& 1 ( 2t d )
2
1/ 2
(8)
n
1
wmax 2
2 2
t
&&max t d z
Rmax
(9)
n d
w max
1 2
&&max z
Rmax
n
Recall
&&max u
2
n
w max
(5.46)
Therefore
&&max u &&max z
Rmax
(10)
&&max ? Why should we consider w max and u w max :
related to spring force
&&max : u
related to the safety of human body
38
39
