Click here to load reader
Upload
christian2424
View
33
Download
5
Embed Size (px)
Citation preview
Linear responseLinear
responsePurdue University
CE571 - Earthquake EngineeringSpring 2003
Mete A. Sozen and Luis E. García
Purdue UniversityCE571 - Earthquake Engineering
Spring 2003
Mete A. Sozen and Luis E. García
Newton (1642-1727) 2nd LawNewton (1642-1727) 2nd Law
"The force acting on a body and causing its movement,
is equal to the rate of change of momentum in the body."
Momentum Q, is equal to the product of the body mass by its velocity:
"The force acting on a body and causing its movement,
is equal to the rate of change of momentum in the body."
Momentum Q, is equal to the product of the body mass by its velocity:
Q m v mdxdt
mx== == == &Q m v mdxdt
mx== == == &
Newton 2nd Law (cont.)Newton 2nd Law (cont.)
Under the premise that the mass remains constant, the forces that act on the body are equal to the rate of change in momentum:
Under the premise that the mass remains constant, the forces that act on the body are equal to the rate of change in momentum:
FdQ
dt
d
dtmv m
dv
dtm
dx
dtmx ma== == == == == ==( )
& &&FdQ
dt
d
dtmv m
dv
dtm
dx
dtmx ma== == == == == ==( )
& &&
F ma==F ma==
D’Alambert PrincipleD’Alambert Principle
D’Alambert (1717-1783) suggested that Newton's 2nd Law should be written in a similar way that the principle of equilibrium in statics (F=0):
D’Alambert (1717-1783) suggested that Newton's 2nd Law should be written in a similar way that the principle of equilibrium in statics (F=0):
F ma−− == 0F ma−− == 0inertial force
“A simple device can yield perhaps 80% of the truth whereas the next 10% would be difficult to obtain and the last 10% impossible …”
“A simple device can yield perhaps 80% of the truth whereas the next 10% would be difficult to obtain and the last 10% impossible …”
H. M. WestergaardH. M. Westergaard
Response to base excitation of a single degree of freedom (SDOF) oscillator
Response to base excitation of a single degree of freedom (SDOF) oscillator
l Some analytical devices are efficient as design tools and some are good to organize experience, but very few can serve both functions.
l The SDOF oscillator is better at organizing and understanding experience.
l Some analytical devices are efficient as design tools and some are good to organize experience, but very few can serve both functions.
l The SDOF oscillator is better at organizing and understanding experience.
Base excitationBase excitation
0 z u
(b)
0 0 z u
mass
structural element
(a)
damper
0 0 z u
(b)
0 0 z u
mass
structural element
(a)
damper
0
00 zz uu00
Absolute position of the mass with respect to a distant frame of reference
Absolute position of the mass with respect to a distant frame of reference
Position of the base of
the structure with respect
to the ground
Position of the base of
the structure with respect
to the ground
massmass
structural elementstructural element
damperdamper
using equilibrium and D’Alambert principleusing equilibrium and D’Alambert principle
00 zz uu00
inertial forceinertial force
damping forcedamping force
Structural element forceStructural element force
mu&&&&mu&&&&
(( ))c u z−−& && &(( ))c u z−−& && &
(( ))k u z−−(( ))k u z−−
(( )) (( ))mu c u z k u z 0+ − + − =+ − + − =&& & &&& & &(( )) (( ))mu c u z k u z 0+ − + − =+ − + − =&& & &&& & &
We now do a simple transformationWe now do a simple transformation
(( ))x u z= −= −(( ))x u z= −= −
We use the relative displacementWe use the relative displacement
Deriving against time:Deriving against time:
And deriving again:And deriving again:
(( ))x u z= −= −& & && & &(( ))x u z= −= −& & && & &
(( ))x u z= −= −&& && &&&& && &&(( ))x u z= −= −&& && &&&& && &&
Now the equilibrium equation:Now the equilibrium equation:
m = mass concentrated at its center of gravity
= acceleration of the mass relative to the base
= base acceleration
c = damping coefficient
= velocity of the mass relative to the base
k = stiffness coefficient of the structural element
x = displacement of the mass relative to the base
m = mass concentrated at its center of gravity
= acceleration of the mass relative to the base
= base acceleration
c = damping coefficient
= velocity of the mass relative to the base
k = stiffness coefficient of the structural element
x = displacement of the mass relative to the base
(( ))m x z cx kx 0+ + + =+ + + =&& && &&& && &(( ))m x z cx kx 0+ + + =+ + + =&& && &&& && &
x&&&&x&&&&z&&&&z&&&&
x&&x&&
By assuming that we have an idea of the variation of the mass displacement with time, we can now attempt to find a description of the mass movement in terms of displacement, velocity and acceleration by finding them at particular times (dates) using a time interval (∆∆t).
This leads us to the central-difference equation for acceleration:
By assuming that we have an idea of the variation of the mass displacement with time, we can now attempt to find a description of the mass movement in terms of displacement, velocity and acceleration by finding them at particular times (dates) using a time interval (∆∆t).
This leads us to the central-difference equation for acceleration:
Central-Difference equation for accelerationCentral-Difference equation for accelerationD
ispl
acem
ent
Dis
plac
emen
tA
ccel
erat
ion
Acc
eler
atio
nV
eloc
ityV
eloc
ity
TimeTime
TimeTime
TimeTime
∆∆t∆∆t ∆∆t∆∆t
1x1x 2x2x 3x3x
11 3322
∆∆t 2 11.5
x xx
t
−−==
∆∆&& 2 11.5
x xx
t
−−==
∆∆&&
3 22.5
x xx
t
−−==
∆∆&& 3 22.5
x xx
t
−−==
∆∆&&
(( ))2.5 1.5 1 2 3
2 2
x x x 2x xx
t t
− − +− − += == =
∆∆ ∆∆
& && &&&&&(( ))
2.5 1.5 1 2 32 2
x x x 2x xx
t t
− − +− − += == =
∆∆ ∆∆
& && &&&&&
By solving the central-difference equation
for the displacement at date 3, we obtain:
This means that if we know the displacement at date 1, and the displacement and acceleration at date 2, we can obtain the displacement at date 3.
By solving the central-difference equation
for the displacement at date 3, we obtain:
This means that if we know the displacement at date 1, and the displacement and acceleration at date 2, we can obtain the displacement at date 3.
(( ))1 2 3
2 2
x 2x xx
t
− +− +==
∆∆&&&&
(( ))1 2 3
2 2
x 2x xx
t
− +− +==
∆∆&&&&
(( ))23 1 2 2x x 2x x t= − + + ∆= − + + ∆&&&& (( ))2
3 1 2 2x x 2x x t= − + + ∆= − + + ∆&&&&
We know that in order to meet equilibrium, at any given date the relationship between displacement, velocity and acceleration is fixed in terms of the mass, damping and stiffness. For example for date No. 2:
We now extrapolate the velocity from date 1.5 to date 2:
This means that we have assumed the acceleration to be constant and equal to that at date 2 in the half interval between dates 1.5 and 2.
We know that in order to meet equilibrium, at any given date the relationship between displacement, velocity and acceleration is fixed in terms of the mass, damping and stiffness. For example for date No. 2:
We now extrapolate the velocity from date 1.5 to date 2:
This means that we have assumed the acceleration to be constant and equal to that at date 2 in the half interval between dates 1.5 and 2.
(( ))2 2 2 2m x z cx kx 0+ + + =+ + + =&& && &&& && &(( ))2 2 2 2m x z cx kx 0+ + + =+ + + =&& && &&& && &
2 1.5 2t
x x x2
∆∆= += +& & &&& & &&2 1.5 2
tx x x
2
∆∆= += +& & &&& & &&
We know use these values at date No. 2:
And rearranging:
We have obtained a way to find the displacement, velocity and acceleration at any date (time).
We know use these values at date No. 2:
And rearranging:
We have obtained a way to find the displacement, velocity and acceleration at any date (time).
2 12 2 2 2
x x tmx mz c cx kx 0
t 2
−− ∆∆+ + + + =+ + + + =
∆∆&& && &&&& && &&2 1
2 2 2 2x x t
mx mz c cx kx 0t 2
−− ∆∆+ + + + =+ + + + =
∆∆&& && &&&& && &&
2 12
2
x xmz c kx
txt
m c2
−−− − −− − −
∆∆==∆∆
++
&&&&&&&&
2 12
2
x xmz c kx
txt
m c2
−−− − −− − −
∆∆==∆∆++
&&&&&&&&