Dynamics of Stuctures

7/31/2019 Dynamics of Stuctures

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Dynamics of Structures

Yvona Koleková

Slovak University of Technology

Günther SchmidRuhr University Bochum

Ss Kyril and Methodius University Skopje

Andrej Tosecký

Ruhr University Bochum

October 2005



Dynamics of structures _____________________________________________________________________________________________________________________

2

Quantities And Units

The following basic quantities and units will be used:

Basic Quantity Unit

Length L, [L], m;

Mass M, [M], kg;

Time t, [Z], s.

The mechanical quantities refer to the basic units as follows:

Quantity UnitDisplacement u, v, w, ieIδA=δ-δ- [L] , m;

Velocity u& , v& , w& , [LZ-1], m/s;

Acceleration u&& , v&& , w&& , [LZ-2], m/s2;

Force = mass * acceleration [MLZ-2], 1kg . m/s2 = 1 N

or 103 kg . m/s2 = 1 kN ;

Work = force * displacement [ML2Z-2], N . m ;

Density = mass /volume [ML

-3

], kg/m

3

or 103 kg/m3 = t0 /m3 ;

Mass moment of inertia [ML2] , kg . m2.

Impulse [MLZ-1] kg . m/s = N . s




3

1 Single-Degree-of-Freedom System

The system consists of a (rigid) mass, a spring and a damper.

2.1 Formulation of The Equations of Motion

The equation of motion is based on the dynamic equilibrium. The inertia force is opposite to the

acceleration (d’Alembert’s principle). Three different methods can be employed for the

formulation of the equation of motion:

2.1.1 Direct Equilibrium of Forces

Note: This is in general a vector equation

I D Sf (t)+f (t)+f (t)=P(t)

where in case of a linear system

f I (t) = mv&& (t) inertial force;

f D (t)= cv& (t) damping force;

f s (t)= kv (t) spring force.

The positive sense of inertia, spring and

damper force is opposite to the motion and the load.

Figure 1 – equilibrium of forces

2.1.2 Principle of Virtual Displacement

Note: This is a scalar equation

i eδA=δA - δA =0

where

Ai work of internal forces;

Ae = AP + AI work of external forces P and inertial forces f I .

Note: The virtual displacement is an arbitrary small displacement (compatible with the

kinematics of the system).




4

2.1.3. The Hamilton’s Principle

The principle is in is original form only valid for conservative systems.

Define the Lagrange’s function as F = T – U where T is the kinetic energy and U = A i – Ae the

total potential energy.

If F = F(v,v& ) than the functional

( )2

1

t

t

I(v,v)= F v,v dt⎡ ⎤⎣ ⎦∫& &

is minimum for that motion which satisfies the equation of motion under the constrain that the

variation of the motion at t = t1 and t = t2 is zero.

The functional I is minimum if

2

1

t

tδI = δFdt = 0∫ .

This leads, considering the constraints at t1 and t2, to the equation of motion

d F F0

dt v v

∂ ∂− =

∂ ∂&

Note:

In the case of a n-degree of freedom system with linearly independent coordinates (DOF’s)

qi, i = 1,2,…,n one obtains

( )2

1

t

i i i

t

I(q ,q )= F q ,q dt⎡ ⎤⎣ ⎦∫& &

and the equations of motion become

i i

d F F0

dt q q

∂ ∂− =

∂ ∂&i = 1,2,…,n

In the case of a n-degree of freedom system with m linearly independent coordinates qi, i=1,2,

…,m and r constraint equations f j(qi)=0, j=1,2,…,r one obtains

( ) ( )2

1

t

j j ji i i i

t

I(q ,q ,λ )= F q ,q +λ f q dt⎡ ⎤⎣ ⎦∫& & , i = 1,2,….,m; j = 1,2,…,r; m + r = n

whereby

qi linearly independent degrees of freedom;

λ r Lagrange’s factors.

The stationary condition δI = 0 of the functional leads to the equation of motion.

i i

d F F0

dt q q

∂ ∂− =

∂ ∂&; i = 1, 2, ,m;




5

and to the constraint equations

f j(qi) = 0, i = 1,2,….,m; j = 1,2,…,r .

For systems with non-conservative forces Hamilton’s principle can be extended as follows:

2 2

1 1

t t

nc

t tδI = δFdt+ δA dt=0∫ ∫

where Anc is the work of non-conservative forces.

2.2 Equation of Motion of a Single-Degree-of-Freedom System Subjected to a

Dynamic Load

Each of the above mentioned methods leads for a linear system to the equation of motionmv(t) + cv(t) + kv(t) = P(t)&& &

Note:

a) m, c, k, v and P can be, in a more general case, be understood as generalized quantities

related to a deformation state ( mechanism) with one degree of freedom.

b) If a rotational DOF instead of a translational DOF is considered, than v corresponds to the

angel of rotation and m to the moment of inertia; c and k than correspond to rotational

damping and stiffness, respectively.

2.2.1 Equation of Motion under Dead Weight

The dead weight W does not dependent on time.

For static condition we have: k vst = W

Total displacement: stv(t) = v + v(t)

The equation of motion with the effect of gravity is as……… .follows:

Figure 2 – Single-degree- mv(t) + cv(t) + kv(t) = W + P(t)&& &

of-freedom system

Whence we obtain the motion about the static equilibrium position vst :

mv(t) + cv(t) + kv(t) = P(t)&& &




6

2.2.2 Equation of Motion due to Support Excitation

Definition:

vtot = vg(t) + v(t)

where

v(t) : elastic displacement

gv (t) : ground displacement

totv (t) : total displacement

3

12EIk=2 H

Figure 3 - Single-degree-of –freedom system subjected to support excitation

The equation of motion has to be written for the total deformation, considering that damping and

spring forces depend only on the relative (elastic; internal) deformation. Whence:

totmv (t) + cv(t) + kv(t) = 0&& & .

We can now express the motion in total deformation or in elastic deformation.

Total deformation:

Withtot gv(t)= v (t)-v (t)

one obtains

tot tot tot g g eff mv (t) + cv (t) + kv (t) = cv (t) + kv (t)=P (t)&& & &

Elastic deformation:

With

vtot = vg(t) + v(t)

one obtainseff gmv(t) + cv(t) + kv(t) = - mv ( ) P (t)t =&& & &&

EI

EI




7

2.2.3 Equation of Motion of Rigid Beam Element Under Influence of Axial Force

Figure 4 – System under influence of axial force

The equation of motion, derived with the Principle of Virtual Work (PVW) and considering

terms of 2nd

order from axial force, becomes:

V2

k 1mv(t) + - P v(t) = P(t)

H H

⎛ ⎞⎜ ⎟⎝ ⎠

&&

or

( )E Gmv(t) + k - k v(t) = P(t)&&

where Ek and Gk are the elastic and geometric stiffness, respectively.

The total system stiffness reduces with growing axial force VP . The system fails at zero stiffness

under the critical load

v,cr

k P =

H

2.3 Harmonic Vibration

Harmonic vibration; Representation in the complex plane.

Euler’s pair of equations serve to transform from trigonometric to exponential functions:

iωte = cos(ωt) + i sin(ωt)

-iωte = cos(ωt) - i sin(ωt)

2 2sin cos 1i t t t e

ω ω ω + = =

Figure 5 – Representation of the function eiωt

in the complex plane




8

Using the exponential function we can write for the displacements as :

iωtv(t) = ve% %

whereR I

v= v + iv% is a complex quantity.

Or using the modulus (amplitude) and the argument (phase angle) of the motion

i(ωt+θ)v(t) = v e% %

with the amplitude: R 2 I 2v = (v ) + (v )%

and the phase angle θ:I

R

vtan θ =

v

Figure 6 – Complex plane representation of the vector v(t)% .

Using the Figure 6, one can easily imagine the vibration as a projection of the rotating vector v%

onto the reel axis.

{ } ( )1v(t) Re v(t) v(t) v(t)2

= = +% % % , where v% is the complex conjugate of v% .

Re

Im

|v|

v

vI

vR

α

v

v I

~

~

~

ω

ωv(t)

Figure 7: Free-vibration in complex plane




9

2.3.1 Undamped Free-Vibration

The motion taking place without applied force is called free vibration.

The equation of motion is than: mv(t)+cv(t)+kv(t)=0&& &

If the damping coefficient c is set to zero, the only possibility for motion is a harmonic motion

with natural circular frequencyk

mω ≡ [rad/s]

Whence the displacement (velocity, acceleration) has the form of sinus, or cosine function,

respectively:

The general solution to the equation of motion is:{

1

2

αv

v

v(t)= Acosωt + Bsin ωt1424314243

The constants A and B can be determined from the initial conditions of the displacement v0=v(0)

and velocity 0=v v(0)& & at time t = 0. One obtains

A =v(0),v(0)

B=ω

&

Using this, the solution is of the form:

0

v(0)v(t)=v cos sin

ωt t ω ω +

&

or:

v( ) cos( )t t ω ϕ = − ,

with amplitude of the vibration

2

2 00

vv +

ω ρ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

&and the phase angleϕ from 0

0

vtan

ωvϕ =

&

ωv1

ρv2

v0ω

ϕ

Re

Im

v0

Time t = 0

v1

ρ

v2

v0ω

ϕ

Re

Im

v0

Time t > 0

v cos ωτ0

ρ ω ϕcos( t- )sin tωv0ω

ωt

. ( t- )ω ϕ

.

.

.

ωt

.

Figure 8 – Representation of free vibration with initial conditions v0 and 0v& in complex plane

...

( ) sin

c. o. s.

v t t

t

ω

ω




10

Definitions:

Natural period of the vibration:2

T π

ω = [T] = s

Natural frequency of the vibration:1

f

T

= [f] = s-1 = Hz

Relationship between the circular frequency ω and the frequency f: 2 f ω π =

k

0 0.5 1 1.5 2 2.5 3 3.5 42

1

0

1

2

k

0 0.5 1 1.5 2 2.5 3 3.5 42

1

0

1

2

0 0.5 1 1.5 2 2.5 3 3.5 42

1

0

1

2

k

T

v cos tω

v0

0

1

v0. v0

.

ωπ

2π π

23 π2

cos t- 2)ω( π/v

0ω

t(s)

t(s)

t(s)

v

1π

2 π π23

π2

ϕ/ω

ρcos( t- )ρ ω ϕ

t’= t-ϕ/ω

v0

a)

b)

c)

v

v

Figure 9: Representation of the free-vibration with initial conditions

a) with initial displacement v0 and initial velocity 0v =0& ;

b) with initial displacement v0 = 0 and initial velocity 0v& ;

c) with initial displacement and velocity v0 and 0v& .




11

2.3.2 Damped Free Vibration

The equation of motion is:c

v+ v+kv=0m

&& & Initial conditions:0

v ,0

v&

The vibration is of the form: v st Ce Where C is a constant.

Using this form the equation of motion becomes: 2 2 0i t cs s Ce

m

ω ω ⎛ ⎞+ + =⎜ ⎟⎝ ⎠

With 2 k

mω = , where ω is the eigenfrequency of the undamped system.

Solution of the equation of motion:

2

2

1,22 2

c cs

m mω

Δ

⎛ ⎞= − ± −⎜ ⎟⎝ ⎠14243

Definition:

ccr = 2 m ω − critical damping

cr

cξ =

c − damping ratio

Three types of motion are represented by this expression, according to whether the quantity

under the square-root sign is positive, negative or zero.

a) Over-critically Damped System

c>ccrit, ξ > 1

2

1,2ˆξ ξ 1 ξs ω ω ω ω = − ± − = − ±

in which 2ˆ ξ 1ω ω ≡ − .

Solution:

( ) ( )

( ) ( )

( ) ( )( ) ( )

2 2

1 2ξ + ξ 1 ξ ξ 1

ˆ ˆξ ξ

1 2 1 2 1 2

ˆ ˆ ξ ξ

1 2 1 1 2 2

ξ t

1 2 1 2

v( )

ˆ ˆ ˆ ˆsinh cosh sinh cosh

ˆ ˆ ˆ ˆsinh cosh sinh

......

..... cosh.

t t s t s t t t t t

t t t t

t G e G e G e G e G e e G e e

G e G e e G t G t G t G t e

G G t G G t e A t B t e

ω ω ω ω ω ω ω ω

ω ω ω ω

ω

ω ω ω ω

ω ω ω ω

− − − − − − − −

− − −

−

= + = + = + =

= + = + − + =

= − + + = + ξ tω −

The constants A and B depend on the initial conditions.

v(t)

t

v(0)

v(0).

α

tgα=

Figure 10: Over-critically damped free motion




12

b) Under-critically Damped System

c<ccrit, ξ < 1

2

1,2 D1 ξ i2 2

c c

s im mω ω ω ω

⎛ ⎞

= − ± − = − ±⎜ ⎟⎝ ⎠ ;

where ( )2

D 1ω = ω − ξ is the damped natural frequency.

Solution:

D Dξ t + i ξ t - i ξ tv( ) ( ) D Dt t i t i t t Ae Be e Ae Ae

ω ω ω ω ω ω ω − − −−= + = +% % %%

where the constants A% and B% are complex conjugated pairs: R I

A A iA= +%

,

R I

B A A iA= = −%%

The solution in the equivalent trigonometric form:ξ t

1 2v( ) [ cos sin ] D Dt C t C t eω ω ω −= +

where1 2

2 , 2 R I C A C A= = −

or

ξ t0 00

D

v v ξv( ) [v cos sin ] D Dt t t e

ω ω ω ω

ω

−+= +

&

Introducing the amplitude and the phase angle leads to the form:

( )ξv( ) cost

Dt e t ω ω ϕ −= −

where

12 2

2 0 00

D

v v ξv

ω ρ

ω

⎡ ⎤⎛ ⎞+⎢ ⎥= + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

&and 0 0

D 0

v v ξtan

v

ω ϕ

ω

+=

&

Figure 11 – Under-critically-damped free-vibration;0

v 0=&




13

c) Critically Damped System

c = ccrit, ξ = 1

1 22

krit cs sm

ω = = − = −

Solution:

1 2v( ) t t t G e G te

ω ω − −= +

Because the function t e

ω − is real, the both constants G1 and G2 must be real too. Through

introducing the initial conditions v(0), v(0)& the constants can be evaluated and the solution

becomes:

( )0 0v( ) v 1 vt

t t t eω

ω −

= − +⎡ ⎤⎣ ⎦&

Note that this free response of a critically-damped system does not include oscillation about the

zero-deflection position; instead it simply returns to zero asymptotically in accordance with the

exponential term. However, a single zero-displacement crossing could occur if the signs of the

initial velocity and displacement were different from each other.

v(t)

t

v(0).

v(0)

t

v(t)

v(0).

a) b)

Figure 12 – Critically-damped free-vibration:

a) with positive initial displacement and velocity

b) with positive initial displacement and negative initial velocity

2.3.3 Logarithmic decrement of damping δ

Consider any two successive positive peaks such as vn and vn+1 which occur at times ( )2 / D

n π ω

and ( )( 1) 2 / D

n π ω + , respectively. Taking use of the damped free-vibration response function ,

the ratio of these successive values is given by:




14

D D

D

D

ωξ 2π

ξωi(ω t - θ) ω-ξωt 2πωn

i(ω (t + T) - θ)-ξω(t + T) i2π

n+1

v v(t) ρe e e= = = = e

v v(t + T) ρe e e;

Taking the natural logarithm of both sides of this equation and substituting 2ξ 1≡ − Dω ω , one

obtains the so-called logarithmic decrement of damping δ, defined by:

n

2n+1

v 2πξδ = ln =

v 1 - ξ.

For small values of damping (ξ << 1) one has approximately δ 2πξ .

One can expand for small damping values:

2 3δ 2πξn

n+1

v (2πξ) (2πξ)=e e = 1 + 2πξ + + + ...

v 2! 3!

The damping ratio is obtained with sufficient accuracy by retaining only the first two terms in the

Taylor’s series expansion on the right hand side, in which case:

n n + 1

n + 1

v - vξ =

2πv

or in a more general form

n n + k

n + k

v - vξ =2πkv

.




15

3 Harmonic Loading

The equation of motion of system subjected to a harmonically varying load, e.g. 0P(t) = P cos ωt ,

is:

0mv(t) + cv(t) + kv(t) = P cos ωt&& &

with the initial conditions:

0

0

v(0) = v

v(0) = v& &

where

ω is the excitation circular frequency.

The general solution includes two parts:

c pv(t) = v (t) + v (t)

where cv (t) is the complementary solution or the transient response.

and pv (t) is the particular solution or the steady-state response.

3.1 Undamped Vibration

Before considering the damped case, it is instructive to examine the behavior of an undamped

system as described by:

0v( ) v( ) sinm t k t P t ω + =&&

Complementary Solution

v( ) v( ) 0m t k t + =&&

The complementary solution is the same as that one of undamped free-vibration:

Cv (t) = A sin ωt + B cos ωt .

The real constants A and B can be evaluated using the initial conditions in general solution.

Particular Solution

The particular solution depends upon the type of dynamic loading. If the loading is a sine

(cosine) function, it is reasonable to assume that the corresponding motion is also a sine (cosine)

function and in phase with the loading; thus, the particular solution is:

pv (t) = C sinωt




16

Substituting this into the equation of motion and taking the second derivation, leads to:

2

0sin sin sinm C t kC t P t ω ω ω ω − + =

Dividing this by sin t ω (which is nonzero

in general) and by k noting that

2/k m ω = one obtains:

0

2

1

1

PC

k β

⎡ ⎤= ⎢ ⎥−⎣ ⎦

whence 0

2 2

1 1v( ) v

1 1st

pt

k β β

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

Figure 13 – response ratio vs. β

β is the so-called frequency ratio and is defined as: / β ω ω ≡ .

General solution

The general solution is obtained by combining the complementary and particular solutions:

0

2

1v( ) v ( ) v ( ) cos sin sin

1c p

Pt t t A t B t t

k ω ω ω

β

⎡ ⎤= + = + + ⎢ ⎥−⎣ ⎦

The constants A and B are depending upon initial conditions. For the system starting from rest,

i.e., v(0) = 0, v(0) 0=& , it is easily shown that:

A = 0 and 0

2

1

1

P B

k

β

β

⎡ ⎤= − ⎢ ⎥−⎣ ⎦

Through introducing these constants becomes the response function form:

( )0

2

1v( ) sin sin

1

Pt t t

k ω β ω

β

⎡ ⎤= −⎢ ⎥−⎣ ⎦

where 0 / vst P k = is the static displacement and 21/(1 ) β − is the dynamic magnification factor

R(t).

3.2 Damped Vibration

The equation of motion including viscous damping and noting that c/m = 2ξω is as follows:

2 ( )0v( ) 2 v( ) v( ) i t Pt t t e

m

ω φ ξω ω −+ + =&& &% % %

Definition of the phase angle φ is important when an arbitrary periodic loading applied. Such a

loading is usually given as a series of harmonic, in general, each having different phase angle.




17

When just only harmonic loading applied it is advantageous set the phase angle to be zero. So it

is done in the next discussion.

General solution is obtained by combination of complementary and particular solution.

Complementary solution is the already known solution to the equation of motion of a damped

free-vibration, i.e.:

ξ tv ( ) [ cos sin ]c D Dt A t B t e ω ω ω −= +

Particular Solution

The particular solution to the equation of motion and its first and second time derivatives are:

2

v ( )

v ( )

v ( )

i t

p

i t

p

i t

p

t Ge

t i Ge

t Ge

ω

ω

ω

ω

ω

=

=

= −

%

%&

%&&

Where G is a complex constant. After substituting these terms into the equation of motion,

performing some rearrangements and canceling out expressions common for all terms, for G

yields:

2

0 0

2 2 2 2

1 (1 ) (2 )

(1 ) (2 ) (1 ) (2 )

P P iG

k i k

β ξβ

β ξβ β ξβ

⎡ ⎤⎡ ⎤ − −= = ⎢ ⎥⎢ ⎥− + − +⎣ ⎦ ⎣ ⎦

%

General solutionIs obtained as a summation of the complementary and the particular solution, also:

ξ t 20

2 2 2

1v( ) [ cos sin ] (1 ) cos 2 sin

(1 ) (2 ) D D

Pt A t B t e t i t

k

ω ω ω β ω ξβ ω β ξβ

− ⎡ ⎤⎡ ⎤= + + − −⎢ ⎥ ⎣ ⎦− +⎣ ⎦

The first part of the expression represents the transient response (the damped free-vibration)

which damps out according to the exponential function. This part is usually of little interest.

When necessary, the constants A and B can be evaluated using the given initial conditions.

The second part represents the steady-state vibration which will continue indefinitely.

Steady-state response to the time t = 0 is represented in the following figure. Real part of the

applied loading 0 0( ) (cos sin )i t p t p e p t i t ω ω ω = = + at this time is identical with its amplitude,

since the imaginary part vanishes.

The real and the imaginary part of the response are:

( ) ( )

( ) ( )

2Re

2 22

Im2 22

1v

1 2

2v1 2

i

β

β ξβ

ξβ β ξβ

−=

− +

−=− +




18

Figure 14: Steady-state displacement response in complex plane.

The steady-state displacement response is in complex plane represented by two rotating vectors

vRe and vIm. The projection of their resultant into the real axis gives the steady-state response in

the form:

v( ) cos( )t t ρ ω ϕ = −

having an amplitude1/ 2

2 2 20 (1 ) (2 )P

k ρ β ξβ

−⎡ ⎤= − +⎣ ⎦ and a phase angle ϕ by which the response

lags behind the applied loading 1

2

2tan

1

ξβ ϕ

β

− ⎡ ⎤= ⎢ ⎥−⎣ ⎦

.

In case without damping the phase angle θ is equal to zero or π. There is a tendency for the two

components (loading and displacement) to get in phase and then out of phase again.

During the motion, of course, all forces taking part on the vibration must be in equilibrium.

Knowing the displacement response function v(t) and the applied loading, are the participating

forces as follows:

- the applied loading: 0( )i t

p t P eω =

- inertial force: 2 ( )( ) v ( )P

i t

I pF t m t m e ω ϕ ω ρ −= = −&&

- damping force: ( )( ) v ( )P

i t

D pF t c t ic e ω ϕ ωρ −= =&

- spring force: ( )( ) v ( )P

i t

S pF t k t k e ω ϕ ρ −= =

p(t)

F (t)Sp

F (t)Ip

F (t)Dp

Figure 15 – Steady-state harmonic vibration – equilibrium of forces.

Re

Im

p0

ϕ

ρ

vRe

vIm




19

3.3 Dynamic Magnification Factor

The ratio of the resultant harmonic response amplitude to the static displacement which would be

produced by the force P0 is called the dynamic magnification factor D, thus:

1/ 22 2 2

0 (1 ) (2 )/ D P k β ξβ

−

⎡ ⎤≡ = − +⎣ ⎦ .

Dynamic magnification factor D and the phase angle θ depend on the frequency ratio β and the

damping ratio ξ. See the sketches below.

Figure 16 – Dynamic magnification factor – dependency on frequency ratio β (the horizontal

.axis) for discrete values of damping ratio ξ.

Figure 17 – Phase angle - dependency on frequency ratio β (the horizontal axis) for discrete

values of damping ratio ξ.




20

3.4 Resonant Response

Going out from the expression for the dynamic magnification factor in case of undamped

harmonic vibration2

1( )

1 R t

β =

−, it is apparent, that the steady-state response amplitude of an

undamped system tends towards infinity as the frequency ratio β approaches unity (the excitation

frequency approaches the natural frequency of the system). This fact can be seen as well from the

Figure 14 for ξ = 0. For low values of damping the maximum amplitude of a steady-state

response occurs at frequency ratio slightly less then unity.

To find the maximum (peak) value of dynamic magnification factor, one must differentiate the

expression for dynamic magnification factor including damping with respect to β and set the

resulting expression equal to zero obtaining the desired value of β at which the peak value of D

occurs:

1/ 22 2 2 2(1 ) (2 ) 0 1 2 peak

D β ξβ β ξ

β β

−∂ ∂⎡ ⎤≡ − + = ⇒ = −⎣ ⎦∂ ∂

Harmonic excitation at which occur the maximal amplitudes is called resonance and the

frequency of the loading causing this condition is called the resonance frequency. The resonance

frequency is identical with the natural frequency of the system in case of undamped system.However, entirely undamped system is purely hypothetical, but for low values of damping ratio

(say ξ < 0.1, which is the usual amount of damping of structures) the difference can be neglected.

On the other hand, the resonance frequency becomes significantly lower in comparison with the

natural frequency as the amount of damping increases (see Figure 16).

Substituting the obtained value of β back into the expression for the dynamic magnification

factor one obtains the maximal value of D:

2

1 1

22 1 peak

D

Dω

ξ ω ξ ξ = =

−.




21

4 Impulsive Load

4.1 Unit Impulse Response Function

Assumptions:

- Duration τ of the load is much shorter than the period of vibration T of the system: τ << T

- Initial conditions: v(0) = 0 and v(0) 0=&

-

t

P(t)

0 t'τ

k C

m

v(t)

P(t)

Figure 18 – Impulsive loading. Loading duration τ .

The response due to impulsive loading consist of two phases: The loading phase (0<t<τ) and the

free vibration phase (t<τ).

We define t’ = t - τ .

For t’>0 we have a free vibration with initial condition:

=0v v(t'=0) = v( )τ and =0v v(t'=0) = v( )τ & & & .

The free response is

( ) -ξ t'0 00 D

D

v + v ξv(t') = v cos(ω t') + sin ' e Dt

ω ω ω

ω

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

&

The equation of motion during the loading phase is:

mv(t) + cv(t) + kv(t) = P(t)&& &

From this equation the acceleration is obtained as:




22

P(t) c k v(t) = - v(t) - v(t)

m m m&& &

Through integration one obtains at time τ with zero initial conditions the velocity at v(0) = 0 and

v(0) 0=& :

τ

0

1 P(t) c k v( ) = { - v(t) - v(t)}dt

m m m mτ ∫& &

With the assumption τ << T the 2nd

and 3rd

term can be neglected as τ tends to zero. Whence

τ τ

0 0

1 P(t) c k 1 P(t)v( ) = { - v(t) - v(t)}dt dt

m m m m m mτ ≈∫ ∫& &

In the limit the free vibration response becomes for τ = 0 and t = t’ the vibration due to the initial

conditions v(0)=0 andτ

0

1 P(t)v(0) = { dt

m m∫&

With the definition of the impulse0

I = P(t) dt = mv(0)τ

∫ &

One obtains

Iv(0) =

m& .

Note the dimension of I: [I] = N s = kg m/s

The Impulse with intensity I can be understood as the impulse produced by a force acting in an

infinitesimal small time step. The time history of such a force is represented by the Dirac- Delta

function δ(t).

The response to an impulse of infinitesimal length is therefore

1v(t) = sin ωt h(t)

mω

≡ [1] = Ns ⇒ [h] = 1/Ns

The exact solution for a damped system is:

-ξωt

D

D

Iv(t) = e sin ω t

mω

The special case of the excitation through a unit impulse is then:

-ξωt

D

D

1v(t) = e sin ω t h(t)

mω≡

The function h(t) is called unit impulse response function, where [h] = m / N s .




23

4.2 General Loading P(t)

The total response of an arbitrary loading can be obtained as summation of all the differential

responses dv due to impulses dI, which represent the loading (see Fig 19). Due to the

superposition involved the procedure can only be applied for linear behavior of the system.

Let dI(τ) = P(τ)dτ

The response due to dI is than

v = dI h(t - τ) = P(τ) h(t - τ)dτd

and the response due to all infinitesimal Impulses up to the time t

t t

0 0

v(t) = dv = P(τ) h(t - τ)dτ∫ ∫

with-ξω(t - τ)

D

D

1

h(t - τ) = e sin ω (t - τ)mω .

The Integral

t t

0 0

P(τ) h(t - τ)dτ P(t - τ) h(τ)dτ=∫ ∫

is called Duhamel integral

Figure 19 – Derivation of the Duhamel integral




24

The integral in its general application

t t

1 2 1 2

0 0

f (τ) f (t - τ) dτ f (t - τ) f (τ) dτ=∫ ∫ is the so-called

convolution integral.

4.3 Application of Duhamel Integral to Base Excitation

m

k,c

v..

g (t)

or

ω, ξ

Peff =mv..

g

Peff

Figure 20 – Base excitation

The elastic displacement due base acceleration is given by

D

t

-ξω(t - τ)1eff Dmω

0

v(t) = P (τ) e sin (ω (t-τ)) dτ ∫

or

t t

-ξω(t - τ) -ξω(t - τ)

g D g D

D D D0 0

m 1 1v(t) = - v (τ)e sin (ω (t-τ)) dτ = - v (τ)e sin (ω (t-τ)) dτ = V(t)

mω ω ω∫ ∫&& &&

The maximum displacement vmax due to earthquake loading (base excitation) is of significance

since it is direct proportional to internal stresses in the system.

For sufficiently small damping we set Dω ω

=

Thus define:

max d1

v = max V(t) =S ( , ) ω

ξ ω

Since Sd is a function of the frequency ω of the system it is called displacement response

spectrum or spectral displacement response of the considered earthquake.




25

5 Direct Integration of The Equation of Motion

Direct integration allows to consider non-linear behavior

5.1 Linear Acceleration Method (Wilson θ- method)

In the Wilson θ method a linear acceleration is assumed in the time increment from time t to

time τ = t + θΔt. In case of θ = 1 this method is reduced to the linear acceleration scheme

(Wilson Method).

Note: All equations are written in matrix notation for a multi-degree of freedom system. These

equation can be understood as a scalar equation for a single degree of freedom system.

A. Initial calculations:

1. Establish K, M and C

2. Introduce the initial conditions0 0 0

v , v and the load P& .

3. Calculate the initial acceleration 0v&& from

0 0 0 0M v = P - C v - K v&& &

4. Choose the time-step Δt such that the load is well approximated and that the relevant

eigenvibrations are well represented (about 10 time steps per period).

Note: For value θ ≥ 1.37 becomes this method unconditionally stable. Usually θ = 1.4 .

and1

Δt T10

≤ , where T is the highest considered eigenperiode of the system

5. Calculate the integration constants

1

3a =

τ; 2

6a =

τ; 3

τa =

2; 4 2

6a =

τ

6. Calculating the effective stiffness

4 1K = K + a M + a C

B. For ever time-step i calculate:

1. Loading increment




26

[ ]i i i+1 iˆ P = θ ΔP = θ P - PΔ

Effective loading increment for time increment τ

i i 2 i 3 iˆ ˆΔP = ΔP + (a M + 3 C) v + (3 M + a C) v& &&

2. Displacement increment for time increment τ:

-1

i i i iˆ ˆ ˆ ˆˆ ˆK Δv = ΔP Δv = K ΔP→

3. Acceleration increment for time increment τ:

i 4 i 2 i iˆ ˆΔ v = a Δv - a v - 3 v&& & &&

4. Reduction to the time increment Δt =θ

τ

ii

ΔvΔv =

θ

&&&& (linear interpolation)

5. i i i

ΔtΔ v = v Δt + Δv

2& && &&

2 2

i i i i

Δt ΔtΔ v = v Δt + v v

2 6+ Δ& && &&

6. State at the timei+1 1t = t + Δt

i + 1 i iv = v + Δv

i+1 i iv = v + Δv& & &

-1

i + 1 i + 1 i + 1 i + 1v = M (P - C v - K v )&& &

7. Go to next load increment (step1)

Remark: K, M and C can be functions of time (e.g. K ( v )= K (v(t)) ).




27

6 Harmonic Motion; Periodic Motion; Fourier Transform

We consider a linear system:mv(t) + cv(t) + kv(t) = P(t)

6.1 Harmonic Loading

0 0P(t) = P cos ωt or P(t) = P sin ωt

In complex form

iωt

0P(t) = P e

Solution:

iωt0

2

Pv(t) = e

k-ω m + iωc

iωt

0v(t) = H(iω) P e

where

2

1 1H(iω) =

k (1 - β ) + i2ξβ

H(iω) is the transfer function. It is the dynamic flexibility or the inverse dynamic stiffness due

to a harmonic unit load.

6.2 Periodic Loading

We define:

1

2πω =

T

2 1ω = 2 ω

.

.

.

n 1

2πω = n ω = n

T

The periodic load is given as Fourier series:




28

0 n n n n

n=1 n=1

P(t) = a + a cos ω t + b sin ω t∞ ∞

∑ ∑

where

T

00

1

a = P(t)dtT ∫ ;

T

n n

0

2= P(t) cos ω t dt

Ta ∫

T

n n

0

2 b = P(t) sin ω t dt

T ∫

In complex form:

n 1iω t inω t

n n

n= - n= -

P(t) = c e = c e∞ ∞

∞ ∞∑ ∑

with n

T

- iω t

n n 1

0

1 2πc = P(t) e dt and ω = nω = n

T T∫

Solution:

For the nth component:

niω t

n n nv = H(ω ) c e

where

( )( ) ( )n 2 2 2

n n n 1 1 n

1 1 1 1H ω = =

k k 1 - β + i2β ξ 1 - n β + i2nβ ξ

with nn 1

ωβ = = nβ

ω

( ) ni nω t

1 nv(t) = H nω c e

For all the components:

( ) ni ω t

n n

n = -

v(t) = H ω c e∞

∞∑

or

( ) 1i nω t

1 n

n = -

v(t) = H nω c e∞

∞∑




29

6.3 Arbitrary Loading; Fourier Transform

It is possible to consider a general loading as a periodic loading with the period T → ∞.

In this case one obtains

1 dω

T 2π→

1

2πω = = Δω => dω

T

nω ω→ , nc c(ω)→

and correspondingly

iωt iωt

- -

1 1P(t) = c(ω) e dω= P(ω) e dω

2π 2π

∞ ∞

∞ ∞∫ ∫ Inverse Fourier Transform ofP(ω)

( ) -iωt

-

P( ) c P(t) e dtω ω∞

∞

≡ = ∫ Fourier Transform of P(t)

The condition for the existence of the Fourier Transform is:

P(t) dt

∞

−∞

<∞∫

Solution for the frequency component ω with amplitude ( )

2π

c P(ω) dωTω =

is than

iωt1dv(t) = H(iω) P(ω) e dω

2π

and for all frequency components

iωt

-

1v(t) = H(iω) P(ω) e dω

2π

∞

∞∫ .

The Fourier Transforms pairs for the load are given as:

iωt1P(t) = P(ω) e dω

2π

∞

−∞∫

- iωtP(ω) = P(t) e dt

∞

−∞∫

And correspondingly for the displacements:

iωt

-

1v(t) = v( ) e dω

2πω

∞

∞∫




30

- iωt

-

v( ) = v(t) e dtω

∞

∞∫ .

We consider as loading the Dirac impulse:

P(t) = δ(t)

and obtain the Fourier transformed load

- iωtP(ω) = (t) e dt = 1δ

∞

−∞∫

The response is on the one hand the impulse response h(t) and on the other hand the inverse

Fourier response as given from the formula above. .

Whence

iωt

1v(t) h(t) = H( ) 1e d2 ω ωπ

∞

−∞≡ ∫

This means that h(t) (impulse response) and H( )ω (transfer function) are Fourier Transforms

pairs.

6.4 Discrete Fourier Transforms (DFT)

Given the time series Pm= P(tm), m = 0,1,2, …, N – 1 which is assumed to be periodic with period

T. The period is divided into N equal time increments. Whence: T = NΔt and tm = mΔt. One

obtains with these assumptions:

nm N - 1- i2π

Nn m

m = 0

1P(ω ) = P(t ) e

N∑ Inverse Discrete Fourier Transforms (I.D.F.T)

and correspondingly:

nm N - 1i2π

Nm n

n = 0

P(t ) = P( ) eω∑ Discrete Fourier Transforms (D.F.T)

where the N circular frequencies n

2πω = nΔω with Δω =

Tand n = 0,1,2, …, N – 1 will be

considered in the series.

The highest circular frequency which can be considered is:

max

N N 2π πω = Δω = =

2 2 T Δt.

For numerical analysis the calculations are performed with the Fast Fourier Transforms (FFT)

(see e.g. Clough and Penzien Dynamics of Structures)




31

7 Multi-Degree-of-Freedom Systems

rigid, m

EI

v1(t)

v2(t)

vi(t)

vn(t)

Figure 21 – Computational model of multi story building

Nodal forces and displacements of a n-degree-of-freedom system:

1

2

(n×1)

n

P

P=

P

⎡ ⎤⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

P

M

1

2

(n×1)

n

v

v=

v

⎡ ⎤⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

v

M

;

The equation of motion in matrix form is as follows:

(t) + (t) + (t) = (t)M v C v K v P&& &

The system matrices M(n×n), C(n×n) and K (n×n) are symmetric, MT

= M; CT

= C; K T

= K and are

positive definite.vTMv > 0, if all degrees of freedoms have inertia (masses)

vTCv > 0, if all degrees of freedoms have damping

vTKv > 0, if no rigid body motions are possible

If all masses (and mass moments) are concentrated at the nodes the mass matrix is diagonal.

7.1 Homogenous Equation Without Damping; C=0, P=0

The equation of motion is(t) + (t) =M v K v 0&&




32

We assume a solution of the form

1

2iωt iωt

n

v

v(t) = e = e

v

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

v v

%

%%

M

%

, where the components iv% are complex.

which leads to the eigenvalue problem

( )2- ω =K M v 0% .

Its n eigenvalues, ωi2 , i=1,2,…,n can be determined from

2- ω 0=K M

and are usually different and, in the case M and K being positive definite, non-zero. For each

eigenvalue ωi2follows from equation ( )2- ω =K M v 0% an eigenvector φi .

Note that iαφ , with α a non-zero constant, is again an eigenvector belonging to the ith

eigenvalue. This fact can be used to scale the eigenvectors appropriately.

Note: In dynamics ωi are the eigenfrequencies and φi are the eigenvibration mode shapes.

7.1.1 Properties of the eigenvectors

Let φi and φ j be two different eigenvectors. It then holds:

TiMi i = *

M ; T 0, j i i j= ≠M

TK i i i= *

K ;T

0, j i i j= ≠ .

Mi*

and K i*

are the generalized mass and stiffness, respectively corresponding to the ith

eigenmodeshape for which holds

2

i

K ω M

i

i=

*

*

If we collect the n eigenvectors in the modal matrix [ ]1 2 ... nφ φ φ =Φ these orthogonality conditions

can be written as

T = *Φ MΦ M ; T = *

Φ K Φ K ; =* *M K .

with 2 2 2

1 2, , , nω ω ω ⎡ ⎤Λ = ⎣ ⎦L

The matrixes M*, K * and are diagonal matrixes.

Note: It is always possible to write these conditions in ortho-normalized form.




33

Normalizing of eigenvectors with respect to mass matrix:*

1ˆi i

i M

φ φ =

Where T *

i i i M φ φ =M

Then it holds:

Tˆ ˆ 1i i =M ; T 2ˆ ˆi i

iω =K or Tˆ ˆ =Φ MΦ I ; Tˆ ˆ =Φ K Φ

7.2 Modal Analysis

For the purpose of dynamic response analysis, it is often an advantage to express the displaced

position in terms of the mode shapes weighted by the functions ( )i

t η called mode coordinates.

Displacement vector written in this way is:

1 1 2 2 n(t) = (t)= ( ) ( ) ( )nt t t φ η φ η φ η + + +v Φη L

The new formulation using the mode coordinates will be introduced into the equation of motion

in order to obtain:

T T T(t) + (t) = (t)Φ M Φ η Φ K Φ η Φ P&&

Orthogonality conditions:

T *

i i i=Φ M Φ M (modal mass)

T T

i j i j 0= =Φ M Φ Φ K Φ for i ≠ j

*T 2

ii i iK ω Mi= =Φ K Φ (modal stiffness)

Orthonormality:

T

T 2

i

ˆ ˆ 1

ˆ ˆ ω

i i

i i

φ φ

φ φ

=

=

M

K

where iφ is the ith mode shape vector normalized with respect to mass matrix.

The initial conditions for the mode coordinates result from v(0) = Φη(0) and (0) = (0)v Φη&& with

the condition of orthonormality conditionTˆ ˆ =Φ M Φ I :

T

i i

T

i i

ˆη (0)

ˆη (0)

=

=

Φ M v(0)

Φ M v(0)& &

The equation of motion using the mode coordinates is uncoupled:

* * * *i i i i i i iM η (t) + C η (t) + K η (t) = P (t)&& &

or




34

*2 i

i i i i i i *

i

P (t)η (t) + 2ξ ω η (t) + ω η (t) =

M&& &

with * *

i i i iC =2M ω ξ and * 2 *

i i iK = ω M .

For ortho-normalized eigenmodes:

*

iM =1, *C 2i i iξ ω = and * 2K i iω =

The equation of motion corresponds to a single-degree-of-freedom system.

8 Seismic Excitation (Earthquake)

An earthquake releases deformation energy from the earth crust. This deformation energy

originates from movement and friction of the earth crusts. The energy will be transported in the

form of seismic waves. Depending on the intensity and duration of the earthquake this can cause

severe damage.

The focus of the earthquake, also called hypocenter , is the place where the rupture of an earth

crust begins. Usually the focus is assumed in the center of the rupture. The focus is also

designated as the seismic source.

The epicenter is the place on the surface above the focus.

8.1 Waves

Waves are place and time dependent processes. They transport earthquake energy through the

transmitting medium, the soil.

To model wave propagation in the soil we assume that the soil is a homogeneous, isotropic

linear-elastic medium. Two types of independent spherical waves exist in an (infinitely extended)

3-dimensional medium: the compression waves (dilatation wave, P- wave, primary wave) and the

shear waves (rotational wave, S- wave, secondary wave). The square of the propagation velocity

of waves is proportional to the stiffness and inversely proportional to the density of the

transmitting medium. Therefore the propagation velocity CS of the shear waves is lower than the

velocity CP of the compression waves; S-waves arrive later. The relationship CP/CS between the

velocities depends on Poisson’s ratio ν. For ν = 0.5 (incompressible) come the pressure wave

velocity CP → ∞, i.e.: there exist no pressure waves. When we consider as region a half-space as

a model for the earth than P-Waves and S-waves are coupled to fulfill the traction free boundary

condition at the surface of the half-space. This creates new type of cylindrical waves which

propagate along the surface of the half-space and decay rapidly with depth. Therefore thesewaves are called surface waves or also Rayleigh waves. In compression waves the material




35

particles move in the direction of the wave propagation, inducing an alternation tension and

compression. In shear waves the material particles move in a plane perpendicular to the direction

of wave propagation.

The waves transport deformation energy. As the spherical or cylindrical surface of the wave front

increases with distance from the source, and as the transported energy is conserved, the energy

density of the wave-front (energy per unit surface) diminishes. This results in a decrease of the

wave amplitudes with increasing distance from the source.

Figure 22: Soil waves




36

The vibration of the particles creates an elliptical trace. The propagation velocity, CR, of

Rayleigh waves is about 90% of the shear wave velocity, depending on ν.

The propagation velocities for some well-known materials can be seen in the following table:

Fluidsc = K/ρ Air 340 m/s

Water 1450 m/s

Elastic mediums

CP [m/s] CS [m/s]

P Sc = E /ρ sand 450 230

granite 4000 2400

Sc = G/ρ concrete 3800 2200

steel 5200 2500

Table 1 – propagation velocities

K = modulus of compression2(1 ) (1 )

1 2 (1 )(1 2 )S

G E E

ν ν

ν ν ν

− −= =

− + −

8.2 Strength of The Earthquake

In order to characterize an earthquake its magnitude and intensity will be introduced.

8.2.1 Magnitude

The most important measure of the size of the earthquake is the amount of strain energy E

released at its source. It is any function of rupture length, soil characteristics and duration of the

earthquake. The released energy is related to the so called magnitude M (Richter’s scale)

by the formula

Log E [erg] = 11.8 + 1.5 M

By this formula, the released energy increases by a factor of 32 if the magnitude increases by 1.




37

M

Grand global earthquake 8 ~ 8.9

Global earthquake 7 ~ 7.9

Destroying earthquake 6 ~ 6.9

Damage earthquake 5 ~ 5.9

Controllable earthquakes 4 ~ 4.9

Very weak earthquake 3 ~ 3.9

Table 2 – Magnitude M

8.2.2 Intensity

The severity of the ground motion observed at any location of the earth surface is called the

earthquake intensity, denoted by I. It expresses through observation local destructive power of

the earthquake. The intensity is a subjective measure. It depends e.g. on state of buildings, on soil

condition and distance from the earthquake focus. To measure the intensity the MSK scale (scale

by Medwedew – Sponheuer – Karnik) is mostly used in Europe. This is a 12-point scale ranging

from I (not felt by anyone) to XII (total destruction).

The MS scale (Mercalli – Sieberg) is also a 12-point scale. It is also used in Germany.

8.3 Response Spectrum Analysis

An other objective way to characterize the strength of a considered earthquake at a selected

location is to measure the soil acceleration at this location and to calculate the response of a

single degree of freedom system with eigenfrequency ω to this acceleration. The plot of the

maximal displacement of this response as a function of the eigenfrequency is called the spectral

displacement response.

The spectral displacement, pseudo-spectral velocity and pseudo-spectral acceleration will be used

to describe an earthquake with ground acceleration gv (t)&& . The spectral values are the maximum

values of a single-degree-of-freedom system with natural frequency ω and damping ratio ξ due to

the considered earthquake.

A linear single-degree-of-freedom system is characterized by ξ and ω.




38

The equation of motion for elastic deformation v(t) due to the base acceleration is:

g

2

g

mv(t) + cv(t) + kv(t) = - mv (t)

or

v(t) + 2ξωv(t) + ω v(t) = - v (t)

&& & &&

&& & &&

The minus sign on the right side of the equation express the direction of the inertial force relative

to the ground motion. By earthquake has the sign no great relevance because mostly just the

maximal amplitude of vibration is important.

v(t)

ω, ξ

v (t)g..

Figure 23 – Computational model for spectral values

The displacement can be obtained for the linear system through the Duhamel integral. In case of

Dω ω≈ for ξ << 1 one obtainst

-ξω(t - τ)

g

0

1 1v(t) = - v (τ) sinω(t - τ) e dτ = - V(t)

ω ω∫ &&

[ ] dmaxmax

1v( ) V( ) = S

ωt t = .

Sd(ω, ξ) is the spectral displacement response.

One defines also:

Sv(ω, ξ) = ω Sd as the pseudo-spectral velocity response;

Sa(ω, ξ) = ωSv = ω2Sd as the pseudo-spectral acceleration response.

Alternatively the spectral responses are given as a function of the natural period of the SDOF

system:

Sa(Τ, ξ) = ωSv(Τ, ξ) = ω2Sd (Τ, ξ)




39

Sv(ξ,T) cm/s

undamped natural period T(s)

Figure 24 – Response spectra for El Centro earthquake,.1940.

8.3.1 Generalized Single-Degree-of-Freedom System

v (t)g..

H

x

m(x), EI(x)

v(x, t) = x) (t)φ( η

Figure 25 – Generalized single-degree-of-freedom system

8.3.1.1 Elastic Deformation

With the assumption of

v(x,t) = φ(x) η(t)

the equation of motion for a generalized single-degree-of-freedom system becomes:

* * * *

m η(t) + c η(t) + k η(t) = P(t)&& &




40

with

[ ]H*

2

0

m = m(x) φ(x) dx∫ ;

[ ]H*

2

0

k = EI(x) φ''(x) dx∫ ;

H*

g g

0

P = m(x)φ(x)v ( )dx = v (t)t ∫ L

and

H

0

= m(x)φ(x)dx∫ L .

With the Duhamel integral one obtains the response in generalized coordinates:

-ξ (t - )

*

0

1(t) = - ( )e sin ( )

t

g Dv t d

m

ω τ η τ ω τ τ ω

−∫ L

With Dω ω≈ one obtains in case of an earthquake where just the maximum value is of

relevance:

[ ] [ ]* * *max max1 1( ) V(t)

d vt S S

m m m

ηω ω

= = =L L L

and finally

max max * *

1( , ) φ(x) ( ) φ(x) φ(x)

d V v x t t S S

m m

ηω

= = =L L

8.3.1.2 Effective Earthquake Force

The effective earthquake force producing the elastic deformation is given as* * *

2EP (t) = k η(t) = ω m η(t)

or with the definition of the generalized quantities

H H

2

E

0 0

φ(x)q (x,t)dx = ω φ(x)m(x)φ(x)η(t)dx∫ ∫

Consequently the effective distributed loading is (in a weak or average sense) given as:

2

Eq (x,t)=ω m(x)φ(x)η(t)




41

If one is only interested in the maximal stresses which produces the earthquake one uses the

maximal displacement in the above equation which is equal to the spectral-displacement

response of this earthquake. The maximal effective distributed loading is than:

2

E E* *

[q (x,t)]max = q (x)=ω m(x) φ(x) m(x)φ(x)

m md a

S S=L L

With this loading the stresses are obtained with the usual procedures of static analysis.

8.3.2 Multi-Degree-of-Freedom Systems

Displacement v(n × 1)

Mass matrix M(n × n)

Stiffness matrix K (n × n)

Loading P(n × 1)

Equation of motion for elastic deformation due to rigid base excitation

g g(t) + (t) + (t) = (t) = v (t)Mv Cv Kv Mv Mr

where r follows of the transformation:( ,1) ( ,1) (1,1)

v r vg g

n n

g= .

Modal coordinates will be introduced:

m

1 1 2 2 m m i

i = 1

= = η + η + ... + η = ; m nv Φη Vφ φ φ ≤

∑

Equation of motion in modal coordinates ηi:

2 ii i i i i i g*

i

η (t) + 2ξ ω η (t) + ω η (t) = v (t)

M

L

where

*

M T i i i

Mφ φ= ;

iT i Mrφ=L .

Maximal response of the generalized coordinate iη (t) :

[ ] i ii d i i v i i* *max

ii i

1η (t) = S (ξ ,T ) = S (ξ ,T )

ωM M

L L

The part of the displacements belonging to the ith eigenmode are

[ ] [ ]

[ ]

i max max

i

i d i i*max

i

S (ξ ,T )M

i i

i

v

v

φ η

φ

=

=

L




42

Because the maximal values of the individual modes do not occur at the same time, one chooses

as the expected value of the displacement component v j the mean square of all components of the

considered modes:

1/2m

2

j j,imaxi = 1

v = v

⎡ ⎤

⎡ ⎤ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦∑

The stresses (e.g. shear forces Q j) are obtained through the effective forces of the considered

modes

2 2 2 i iE,i i i,max i i i,max i i d i a* *

i i

= ω = ω = ω S = S

M M

P Mv Mφ η Mφ MφL L

The usual methods of structural analysis gives than the corresponding stresses (e.g. Q j ) from

where the mean value is obtained

1/2m

2

j j,imaxi = 1

Q = Q⎡ ⎤

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦∑

8.4 Structural Model According to DIN 4149

The effective earthquake forces for ith mode for the given earthquake are

2 2 2 i iE,i i i,max i i i,max i i d i a* *

i i

= ω = ω = ω S = S

M M

P Mv M η M Mφ φ φL L

where

i

T

iMrφ=L ;

*T

i i i M Mφ φ= .

E,i E,i≡H P

where HE,i is the loading in horizontal direction from the i th modal shape for the jth degree-of-

freedom.

iE,j,i j j,i a i i*

i

H = m S (T ,ξ )

M

φL

where

n

i j,i j

j = 1

= mφ∑L ;

n*2

i j,i j

j = 1

M = mφ∑ .




43

In accordance with DIN 4149:

E,j,i j j,iH = m r β cal a

whereby

β = β(Ti): a factor of the normalized response spectra according to DIN 4149

i j,i j,i *

i

r = φ

M

‹;

0cal a = a κα ;

aS = β cal a

a0 earthquake strength, earthquake zone

κ soil, underground

α diminution factor depending on construction class

β(T)

Figure 26 – Design response spectra

The course of curve β(Ti) is introduced in DIN 4149.




44

9 Soil – Structure Interaction

9.1 Flexible Structures on Rigid Foundation

9.1.1 Sub-Structure methodWe assume the system consist of sub-structures. And it is also further assumed that all sub-

structures and then also the complete system behaves linearly.

9.1.1.1 Structure We assume that the structure is discretized by finite elements leading to the equation of motion:

( )t =Mu +Cu + Ku P&& &

For harmonic excitationi t

Peω % with frequency ω we obtain the steady-state response:

( ) i t t e ω =u u%

where u% is determined from:

( ){ }2 iω ω − + =K M C u P%%

or K u = P% %%

K % is the impedance or dynamic stiffness of the system, u% and P% are the displacement amplitudes

vector and force amplitudes vector, respectively.

Real part of the impedance: 2 R ω = −K K M

Imaginary part: I ω =K C

For each degree of freedom:

( ) ( )

( ) ( )

2 2

2 2

, t. an

, t

.

. n. a

i

i

i R I

i i i

i R I

i i i i

I R I i

i i i i R

i

I R I i

i i i i R

i

P P iP P e

u u iu u e

PP P P

P

uu u u

u

ψ

ϕ

ψ

ϕ

= + =

= + =

= + =

= + =

% %

% %

%

%




45

9.1.1.2 Soil

The soil is represented by the impedance matrix of the mass-less rigid foundation described with

respect to the center of the lower soil-structure interface.In general case we have an embedded 3-D rectangular foundation.

rigid

0

Foundation

Soil

X1

X3

X2

Figure 27 – rigid foundation, degrees of freedom

The motion of the rigid basement may be described by the displacement 0u (6,1)at the point 0

(center of the base interface). The corresponding forces acting at the point 0 are 0P :

1

2

3

1

2

3

u

u

u

θ

θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

0u

1

2

3

1

2

3

P

P

P

M

M

M

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

0P

Forces and displacements are related through the dynamic matrix of the foundation

0 0 0( 6,6) (6,1) ( 6,1)

=K u P

0K is complex and frequency dependent.

9.1.1.3 Coupling of Structure And Soil

Step 1: Kinematics constraint of basement

Structural

model

rigid

rigid

interactionnodes

Figure 28 – Kinematics constraint through rigid basement




46

We define all nodes of the structure, the motion of which is restricted through the rigid

foundation at interaction node (index I), the remaining nodes are the structure nodes (index S)

and partition the equation of motion of the structure correspondingly:

S S

S SSS SI

S S

I I IS II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦

u PK K

u PK K

The kinematical connection between I

u and0

u is expressed by the transformation:

0 I =u au

One obtains a by applying the principle of virtual works.

0 0

S SS SSS SI

T S T S

IS II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦

u PK K a

u Pa K a K a

Step 2: CouplingThe soil-structure system is coupled by considering the dynamic stiffness of the rigid foundationas a hyper-element stiffness matrix. The direct stiffness method yielding immediately:

n-6 6

n-6

m

0 00

S S

S SSS SI

T S T S

IS II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

+ ⎣ ⎦ ⎣ ⎦⎣ ⎦

u PK K a

u Pa K a K a K




47

9.2 Flexible Structure And Flexible, Mass-less Foundation9.2.1 Soil

The impedance matrix (dynamic stiffness matrix) of the soil is defined by the dynamic stiffness

matrix of the nodes on the interface between structure and soil.

We assume theory of elasticity and assume a discretization of the (excavated) soil with elements

and (if necessary) condense the number of degrees of freedom to those of the interface.

Boundary elementdiscretization(Integral equationmethod)

FEM+Thin Layer Method

brick elements

Thin Layer Method/Flexible VolumeMethod

Figure 29 – Various methods for solving soil-structure interaction problems

Through semi-analytical or numerical methods the impedance matrix of the soil-structure

interface F K (index F for foundation) interface can be obtained as the relation between

displacement and forces related to the m degrees of freedom at the interface nodes:

( , ) ( ,1) ( ,1)F F F

m m m m

=K u P

F K can be understood as hyper-element matrix. The direct stiffness matrix couples the two sub-structures.

n-m m

n-m

m

S S S

SSS SI S

S S F

I IS II II I

⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥

+ ⎣ ⎦ ⎣ ⎦⎣ ⎦

uK K P

uK K K P

with I

I II i=P K u

or ( , ) ( ,1)( ,1)n n nn

=K u P% %%




48

All matrices involved are in general complex and frequency dependent. For a selected frequency

ω for a specified harmonic loads and/or displacements the solution can be obtained from the

complex linear system of equations.

Note:

1)

For each degree of freedom either iP or iu i = 1, 2, 3...m has to be given. Their corresponding unknown values are calculated.

2) SinceF K is regular for unbounded soil (no rigid body motion),

( , )n mK is also regular.

In the TLM/FVM, the m interaction nodes are defined as the nodes of the intersection of thehorizontal layers and the volume elements representing the volume to be excavated.

The impedance matrix of the interaction node is calculated as the inverse of the dynamic

flexibility matrix of the m degrees of freedom of the interaction nodes.

( , ) ( ,1) ( ,1)F F F

m m m m=K u P% %%

where 1

( , ) ( , )F F

m m m m

−=K F% %

The elements Fij

% of the matrix F% or obtained as a displacementsiu% due to a harmonic unit load

1 j

P =% .,i j

F % corresponds to the numerical Green’s function of the interaction region.

To couple the soil with the structure, the foundation volume has to be excavated. This is done by

subtracting from the dynamic stiffness matrix of the interaction node the dynamic stiffnessmatrix of the foundation volume (to be excavated) discretized by volume brick elements.

In practical calculations, this matrix will be subtracted from the structural matrix (therefore the

basement nodes of the structure have to be identical with the finite volume model).Soil-structure coupling results finally in:

S S

S SSS SI

S S E F

I I IS II II II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + ⎣ ⎦ ⎣ ⎦⎣ ⎦

u PK K

u PK K K K

with F F

I II =P K u

Load cases:

a) Specified loads- Wind

- Traffic- Explosion

- Machines

Wind

Traffic

Machine

Traffic




49

Loads may be specified at any nodal point of the structure, above the soil surface or below the

soil surface. If, for example, wind loads are specified I P would be zero. If traffic load are

Figure 30 – cases of dynamic loading

considered SP would be zero and I P would be

specified at “interaction nodes” on the soil surface or below it. (Note that we define here as

interaction nodes these points, where loads are specified). The governing equation is:

S SS SSS SI

S S F I I IS II II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ⎣ ⎦ ⎣ ⎦⎣ ⎦

u PK K

u PK K K

The impedance matrix of the soil, F

II K , may be obtained by any method described above, i.e. with

rigid or flexible foundation; or using Boundary Element or Thin Layer Method.

b) Seismic loads

Most simple case

Figure 31 – Various cases of dynamic loading

Definitions:

Free-field 'u : wave field at the site without structure.

Scattered field ′′u : wave field at the side with excavation but without structure.

At the interaction nodes the forces stemming from the structure and the forces stemming from the

soil have to add up to zero. In case the scattered field is known the forces stemming from the soilare proportional to the difference of the total displacement field minus the scattered field:

' '( )S S F

IS S II I II I I + + − =K u K u K u u 0

Whence the equation of motion becomes

S S

SSS SI

S S F I I IS II II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ⎣ ⎦ ⎣ ⎦⎣ ⎦

u 0K K

u PK K K

withF

I II I ′′=P K u

=>Free field problem




50

In the case, when the free-field is known, the equation of motion is:

S SSSS SI

S S E F

I I IS II II II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′−− + ⎣ ⎦⎣ ⎦⎣ ⎦

u 0K K

u u 0K K K K or

S SSSS SI

S S E F

I I IS II II II

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + ⎣ ⎦ ⎣ ⎦⎣ ⎦

u 0K K

u PK K K K

where 'F

I II I =P K u and E

II K is the stiffness of the excavated soil.

Documents

Dynamics of Stuctures