C5.Numerical Methods Compatibility Mode

8/10/2019 C5.Numerical Methods Compatibility Mode

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DYNAMIC OF STRUCTURES

CHAPTER 5

NUMERICAL EVALUATION OF

DYNAMIC RESPONSE

Department of civil engineering, University of North Sumatera

Ir. DANIEL RUMBI TERUNA, MT;IP-U

1


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NUMERICAL METHODS

A. Time Stepping Method

For inelastic system with a linier viscous damping, the equation ofmotion can be written as

( ) ( )1)(, tpuufucum s =++ &&&&

The applied force is given by a set of discrete values)(tp

.0),()( Ntoitptp i == The time interval

)2(1 iii ttt = +

2


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TIME STEPPING METHODS

The equation (1) can be given at time

3iisii pfucum =++ &&&

where iisis kufitimeatforceresistingtheisf =;

At the time interval eq.(1) lead to1+i

:i

for linier elastic system

( ) ( )41111 ++++ =++ iisii pfucum &&&

3


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Fig.1 Notation for time stepping methods

ipp

t

op 1p 2p

1+ip

ot 1t 2t it 1+it

it

u

t1u

2u

1+iu

1t 2t it 1+it

it

iu

4


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The three(3) important requirement for a numerical procedure are:

CONVERGENCE

as the time step decreases, the numerical solution should approachthe exact solution

STABILITYthe numerical solution should be stable in the presence ofnumerical round-off errors

ACCURACYthe numerical solution should provide result that are close enoughto the exact solution

5


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They are several methods of time stepping procedures :

Methods Based on Interpolation of the Excitation Function

6

Finite Diffrence Methods

Newmarks Methods

Houmbolt Methods

Wilson Methods


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Methods Based on Interpolation of Excitation

( ) )5(ii

i pt

pp

+=

)6(1 iii ppp = +

In the time intervals , the excitation function is given by

where

1+ ii ttt

7

1+ip

ipp

t

op

ot it 1+it

it

( )p ip


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8


The diffrential equation of the system without damping at the time canbe given as

)7(ii

i pt

pkuum

+=+

&&

The response over the time interval is the sum of threeparts

( )uit0

1. Free vibration due to initial conditions (displacement and velocityat )

iu iu&

0=

2. Response to step force with zero initial conditions, andip

3. Response to ramp force with zero initial conditionsii tp /


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( ) ( ) )8(sin

cos1sincos

+++=

in

n

i

i

n

i

n

n

i

ni

ttk

p

k

puuu

&


9

The solution of this diffrential equation is

At gives the displacement and velocity at timeit= 1+iu 1+iu& 1+i

( ) ( )

( )[ ] ( )[ ]ininin

iin

i

in

n

i

inii

tttk

ptkp

tu

tuu

+

++=+

sin1cos1

sincos1&

(9)


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10


( ) ( )

( ) ( )[ ]in

in

i

in

i

in

n

i

ini

n

i

ttk

p

tk

p

tu

tuu

+

++=+

cos1

1

sin

cossin1 &&

(10)

These equations can be rewritten after substituting Eq.(2) as recurrenceformula:

)11(11 ++ +++= iiiii DpCpuBAuu &

)12(11 ++ +++= iiiii pDpCuBuAu &&


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11


These formula also apply for damped system with the following coeficients:

+

=

tteA DD

tn

cossin1 2

=

teB D

D

tn

sin1

+

+

=

tt

te

kC D

tn

D

D

t

tn

n

cos2

1sin1

2121

2

2

+

+

=

tt

te

kD D

tn

D

D

t

tn

n

cos2

sin122

11 2


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12


=

teA Dntn

sin1 2

+

+

+

=

t

tt

te

tkC DD

ntn

cos1

sin11

11

22

+

=

tte

tkD DD

tn

cossin1

11

2

= tteB DDtn

sin

1cos

2


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13


( )tp

mmkgm /det2533.0 2=

mmkgk /10=

05.0=

kgp,

.sec,t

6.0/sin10 ttp =

1.0 6.0

5

66.810

8095.0cos5871.0sin275.619691.0 2 ===== tte DDnDtn

006352.001236.009067.08129.0 ==== DCBA

1871.01709.07559.05795.3 ==== DCBA


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14


iiiiiiii uAuuuBDpCppt &&1+

0.0 0.000 0.0000 0.0318 0.0000 0.0000 0.0000 0.0000

0.1 5.000 0.0618 0.0550 0.0848 0.9354 0.0258 0.03180.2 8.660 0.1070 0.0635 0.2782 3.0679 0.1849 0.22740.3 10.00 0.1236 0.0550 0.4403 4.8558 0.5150 0.63360.4 8.660 0.1070 0.0318 0.4290 4.7318 0.9218 1.13390.5 5.0000.6 0.0000.7 0.0000.8 0.000

0.9 0.000 0.0000 0.0000 -0.6272 -6.9177 -0.6160 -0.75771.0 0.000 0.0000 0.0000 -2.5171 -1.2432


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15

iiiiiiii uuBuuApDpCpt && +1

0.0 0.000 0.0000 0.9354 0.0000 0.0000 0.0000 0.0000

0.1 5.000 0.8544 1.6201 -0.1137 0.0318 0.7071 0.93540.2 8.660 1.4799 1.8707 -0.8140 0.2274 2.3192 3.06790.3 10.00 1.7088 1.6201 -2.2679 0.6336 3.6708 4.85880.4 8.660 1.4799 0.9354 -4.0588 1.1339 3.5771 4.73180.5 5.0000.6 0.0000.7 0.0000.8 0.000

0.9 0.000 0.0000 0.0000 2.7124 -0.7577 -5.2295 -6.91771.0 0.000 0.0000 0.0000 -1.2432 -2.5171



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Central Difference Method

The central difference expression for velocity and acceleration at time arei

16

( )21111 2

2 t

uuuu

t

uuu iiii

ii

i

+=

= ++ &&& (13)

Substituting Eq.(13) into Eq.(3), gives

( ) ( ) ( ) iiii u

t

mku

t

c

t

mpu

t

c

t

m

+

=

+

+ 21212

2

22

(14)

In this equation and are assumed known (from implementation ofthe procedure for the preceding time step).

( ) ii

iiiii pkut

uuc

t

uuum =+

+

+ ++

2

2 112

11

iu

1i

u

(15)


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or

17

(16)

where

ii puk

1 =+

(17)

(19)

( )

+

=t

c

t

mk

2

2

( ) ( ) iiii u

t

mku

t

c

t

mpp

+

= 212

2

2 (18)

The unknown is then given by1+iu

k

pu ii

1 =+

or iiii buaupp = 1

(20)


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The values and are required to determine

18

For , from Eq. (13) we obtain

(21)

(20)

0u

Solving for from the first equation and substituting in the second gives

( )0

2

001 2 u

t

tuuu &&

+=

1u 1u

0=i

( )2101

011

0

2

2 t

uuuu

t

uuu

+=

= &&&

1u

The initial displacement and initial velocity are given, and the

equation of motion at time 0

0u 0u&

00 =t


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19

(24)

(23)

The specific requirement for stability

m

kuucpu 0000

=

&&&

1

C5.Numerical Methods Compatibility Mode

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