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

−−− − −− − −

∆∆==∆∆++

&&&&&&&&