Structural Dynamics - Theory and Computation 4th Ed (1997)

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~~~'~;~'~~;~~~~;.~ Dyn;t.mics, 4th programs In 5trUcturai edition features a ti7:~~:f~::~;,::,,~:~~'~r~~~1!11 '~;

~ an 'introduction to the dynamtc analysis of s'U'Uctu;es

Method

• a new addition to the chaptet on Random Vibration' ~ "~i~nse of strtu:tuni~.c·.... •. ,', modeled as 11 multi degrea-of.freedom 5},st.em,5ubje

PLANE

0, 1

(3) Establish a grid with two diVIsions in the X and Y directions, then use the scale command;

1000

Llk= 5753 Ib/in

vn:w_PAR 1,

(2) Define the XY plane at Z ~ 0: PL>.NE, Z,

co =2..".

21

(Ans.)

UNDAMPED SINGLE-DEGREE-OF-FR!::EDOM SYSTEMS USING COSMOS

> C,

0, 1,

2, 2, 2

DISPLAY SCALE, 0

>

DISP_PAR

>

Example 1.6. An instrument of mass m = 0.026 (lb . secL/in) is mOl;.nted on isolation springs of total spring constant k = 29.30 (lb/in). Model the system as an undamped single-degree-of-freedom system (Fig. 1.14) and de;tennine its natural frequer:cy.

SCALE

(4) Generate a curve from 1, 0, O. to I, 1, 0: GEOMETRY

>

CRPCORD,

The foi;owing example is presented to illustrate the use of the program COSMOS in the analysis of structures modeled as single degree-of-freedom systems. Detailed explanations for the llse of COSMOS including numerous examples with data preparation and results are presented in the supplement STRUCTURAL DYNAMICS USING COSMOS .•

GlUDON 1.,

GRIDON,

CiJRVES > CRPCORD , \.I, 0, :, ~, C-

(5) Define element group two nodes:

using the SPRING element formulation with

PROPSETS > EGROUP EGROUP, 1, SPRING, 0, 2, 1, 0, 0, 0, 0

(6) Define real constant for spling element: k PROPSETS

ReOM'ST,

>

29.3 (ib/in):

=

RCONST

1, I, }, 1,

29.3

(7) Generate one spring element along curve 1: MESHING ~CR,

L

> 1f

>

PAR..b.Y"...J,!:ESH

L

2,

1,.

}CCR

1

(8) Define element group 2 using the MASS element fonnulation: PROPSETS

Fig. 1.14 Mathematical model for Example 1.6,

>

EGROUP

EGHOUP, 2. MASS, 0,

0, 0, 0, 0,

C,

°

(9) Define real constant for mass element: m = 0.026 (lb . sec2 /in): PROPSETS • A tonVenlelll foffi'l ror or>Jering ~his supplement is provided :in the lacs! page of this volume,

~CON$'l',

>

RCONS1'

2, 2, 1, 7,

0,

0.026,

0,

0,

0,

0, Q

22

Structures Modeled as a Single-Deg.ee-ol-Free:dom Syslem

Undamped Single-Oegree-of-Freedom System

(10) Gene,rate one mass element at point 2: MESHING

>

>

PARhlCMESH

l"CFT .. 2, 2,

(5)

!'CPT

The differential equalion of the undamped simpie oscHlator in free motion is

~

(11) Merge nodes; MESH ING

and its general solution is

>

NODES.

NMERG£

)' = A cos

NMEHGE, 1, 3. 1, O.aOOL O. 0, 0

(12) App!y constraints in an degrees of freedom at node 1. and an degrees of freedom exce.pt UY at node 2: LOADS-Be > DP1', 1, l\L,

STRUCTURAL 0, 1, 1

>

;)ISPLHKT'S

>

DPT

FREQ!BlJCK

>

>

FREQ/BUCK

>

w=Jkim is the r.atural frequency in rad/sec w

f = ...-...~ is the nalural frequency in cps 2.".

A_FREQUENCY ~

T=

(6)

The equation of motion may be written in the alternate fonns:

(14) List the natural frequency of the system:

>

LIST

>

y

1

1.11

= C sin (UJI + a)

FREQLIST

or

F:~E:QLIST

FREQUENCY#

71 is the natural period in seconds

R_FREQUENCY

~FREQUENCY

RESULTS

FREQU£K'CY (2.AD / SEC)

FREQUENCY (CYCLES/SEC)

{SECONDS)

3,35597E+Ol

5.34278£:+:):)

:.87:68£-:)1

PERIOD

)' """ C eos((ul - f3) where

c

SUMMARY

Several '::>as1c concepts were inLroduced in this chapter.

and

(1)

The mathematical model of a structure is an idealized representation for its analysis.

tan a

{2)

The number of degrees of freedom of a system is equal to the number of independent coordinates necessary to describe its ,Position. The free body diagram (FBD) for dynamic equilibrium (to allow application of D' A~embert's Principle) is a diagram of the system isolated from all orher bodies. showing the external forces on the system. including the inertial force.

llnfJ

(3)

an

(4)

mt

B =v(jiw

A-FP£QUCNCY, 1, 5, 16, 0, 0, 0, 0, IE-OS, 0, 1£-06, 0, 0, 0, ANALYSIS

+ B sin

A; Yo

(13) Set the options for the frequency analysis to extract one frequency using the Subspace Iteration Method with a maximum of 16 iterations, and run the frequency analysis:

>

fJJJ

where A and Bare conslants of integration determined from initial condiiions:

VPT, 2, UX, 0, 2, 1, UZ, RX, RY. RZ

A,NAI.''iSIS

23

The stiffness or spring constant of a iinear system is the force necessary to produce a unir displacement

PROBLEMS L 1 Determine the natural period [or the system in Fig. P U. Assume Lhllt the, bellm and springs supporting the weight Ware massless. 1.2 The foHowing numerical values are given in Problem 1.1; L = 100 in, E! lOlI(lb_inl), W=3000 Ib, and k 2000!biin. If the weight W has an initial

24

Struclures Modeled as a Single-Degrce-Ql-Freedom SysLer

Undampeij

1.S

EI

T

,

System

25

Detennine the natural frequency of the fixed beam in Fig. PLS carrying a concentrated weight W at its center. Neglect the mass of the beam.

~.

..

~

fl

I

ml../2

Y

Sjngle~Degree-Qj-Freedom

····-1.12

=I ~

Y

Fig. Pl.1.

Fig. P1.5.

OJ'=' 20 in/sec, cetermlne the displacement and the veloctry 1 sec later. 1.3 Determjne the natural frequency for norlzontaf motion of the sreel frame in Fjg.

displacement of Yo'= 1.0 in and an initial velocity

1.6

The numerical values for Problem 1,5 are given as; L = 120 in, El = 10 9(Ib . in2), and W 5000 lb. If [he initial displacement and (he initial velociry of the weight are, respectiveiy, Yo = 0.5 in and Vo = 15 in/sec, determine the displacement, vel~ odty, and acceleration of W when time ( = 2 sec,

1.1

Consider the simple pendulum of weight W illustrated in Fig. Pl.?, A simple pendulum is a particle or concentrated weight that oscillates in a vertical arc and is supported by a weightless cord. The only forces acting are those of gravity and cord tension (I.e., frictional resistance is neglected). If the cord length is L, determine the motion ir the maximum 9sdllation angle B is small and the lnllial displacement and' velocity are 80 and 80 • respectively.

P1.3. Assume the horizontal girder to be Infinitely rigid and neglect the mass of the columns.

~ I\~

Fig, Pl.3.

1.4 Calculate the natural frequency in the horizontal mode of the steel frame in Fig. PL4 for the following cases: (a) me horiz,ontal member is assumed:o be infmitely rigid; (v) the horizontal member is flexible and made of steel~WJ:O X 33.

I I

\

I

:-' ;

W"" 25 Kips

;

\

~

W

Fig. Pl.7. 15'

W10

;:

II

il 'I

11

x 33

"

I 1.8 A diver standing at the end of a. diving board that cantilevers 2 ft osciIlates at a frequency 2 cps. Determine the flexural rigidity EI of the diving board. The

_L

wejght of the dIver is 180 lb. (Neglect the mass of the diving board).

1---'5'

1.9

Fig. PlA.

,,j; ..

A bullet weigh1ng 0.2 Ib is fired at a. speed of 100 ft/sec into a wooden block weighing 50!b and supported by a spring of stiffness 300 IbJin (Fig, PL9). Determine the displacement y{t) and velocity v(t) of the block after r sec,

26

Structures Modeled as a Single-Degrae-o!·Freedom System

Undamped S:ngle'Degree-of-Freedom System

1.12

An c;evator weighing 500 Ib is suspended from" spring having a stiffness of 600 IbJin. A weight of 300 Ib is suspended through a cable to the elevator as shown schematically in Fig. PLIO, Determine the equation of motion of the elevator if the ca:::'Je of the suspended weight suddenly breaks.

where w mg, ¥.It, is the critical budding weigbt, and 10 is the natlJral frequency neglecting the effect of gravity. 1.13

=p,

A vertical pole of length L and DexuaJ rigidity EI carries a mass m at ~ts top. as shown in Fig. P LJ 3. Neglecting the weight of the pole, derive the differential equation for small horizontal vibrations of the maSs, and find the natural frequency, Assume tba( Lhe effect of grav!:y is small and nonlinear effects may be neglected,

t--r

CPT~

Fig. F1.l0.

1.11

Sbow thilt lhc nalural frequency for the system of Problem L 11 may be expressed as

f= ie

Fig. P1.9.

1.10

27

I

I

Write the differential equation of motion fo~ lhe inverted pendulum shown in Fig, P Lll and determine its natur and opposite to the direction of motion. This type of damping is known as viscous damping; it is the type of damping force that could be developed in a body restramed in its motion by a surrounding viscous fluid. There are situations in which the assumption of viscous damping is realistic and in which the dissipative mechanism is approximately viscous. Nevertheless, the assumption of viscous damping is often made regardless of the actual dissipative characteristics of the system, The primary reason for such wide use of this method is that it leads to a relatively simple mathematical analysis.

2.2

which, after cance1!aiion of the common factors, reduces to an equation called the characteristic equation: for the system, namely

mp2 + cp + k == 0

Let us assume that we have modeled a str"JclUral system as a simple oscillator with viscous damping, as shown in Fig. 2.1(a). In this figure, In and k are, respectively. the mass and spring constant of the osci!!ator and c is the viscous damping coefficIent We proceed. as in the case of the undamped oscillator, to draw the free body diagram (FED) and apply Newton's Law to obtain the differential equation of motion, Figure 2.1(b) shows the FED of the camped oscillator in which the inertial force my is also shown, so that we can use D' Alernbert's Principle. The summation of forces in the y direction gives the differential equation of motion,

my+cy+ky

(2,1)

0

The reader may verify that a trial solution y ;;;; A sin wt or y = B cos t:J! will not satisfy eg. (2.l). However, the exponential function yo:: Cef'i coes satisfy this equation. Substitution of this function into eg. (2.1) results in the equation

The roots of this ;:;uadratic equation are

P,= P2

~--~-... :t It~r-~ 2m'~

+ kCe"

=0

y(l)

= C,e'"

l2m)

m

(2,3)

+ C,e"

(2,4)

where C\ and C1 are constants of integration to be determined from the initial conditions, The final fonn of eq. (2.4} depends on the sign of the expression under the radical in eq. (2.3). Three distinct cases may occur: the quantity under the racical may either be zero, positive, or negative. The limiting case in which the quantity under the radical is zero is treated first The damping present in this case is called crilical damping,

2,3 CRITICALLY DAMPED SYSTEM For a system osciUating with critical camping, as defined above, the expression under the radical jn eq" (2.3) is e:.-jl

(A cos woe

+ n sin

(/Jot)

where A and B are redefined conStants of integration and frequency of the system, is given by

(2.15) WD,

the damped

(2.16)

or (uv=

(2.17)

W

This lost result is obtained after substituting, in eq, (2.16), the expression for tbe undamped natural frequency

(2.18)

and defining the damping ratio of the system as

c

Fig. 2.2 Free-vibraLion response with critical damping,

where the critical damping coefficient c'" JS given by eq. (2.6).

(2.19)

36

structures Modeled as a

Singje~Degree·ol-Freedom

System

Damped

Finally, when the initial conditions of dispLacement and veiodty, Yo and VOl are introduced, the constants of integration can be evaluated and substituted ilito eq. (2,15), giving y(t):;:; e

'~I ' Yo cos

by the viscous force in eq. (3JO), we need to substitute the sine of the phase angle 8 intO eq. (3.31). Thus, from eq. (3.2i), we have sin 8 2fT cos8 =1=7 sin'

e

··si·~rfT+ cos 2 0 = (1 -

3.6

EQUIVALENT VISCOUS DAMPING

As mentioned in the introductory sectio:1s of Chapter 2, the mechanism by which structures dissjpr.:e energy during vibratory motion is usually assumed to be viscous_ This assumption provides the enormous advantage that the differentiai equation of motion remains linear for damped dynamiC syslems vibrating in the elastic range, Only for some exceptional situations such as the use of frictional devices installed in buildings to ameliorate damage rest.;,lting from strong motion earthquakes, viscous damping is usually used [0 account for frictional or damping forces in structurr.l dynamics. TIle numerical v PD·-CURVES

;. PD-CURDEF

1000, 11, 1000

PD_CURDEF, 1, 1, 2,

(17) Apply dynamic force function at node 2 in the Y direction: (3.64)

The steady-state response of this system expressed as the dynamic magnification factor is then given from eg. (3.23) by

CONTROL )< ACTIVATE )- AC'I'SET ACTSET, TC, 1 L.()p.D- EC

STRUCTUR.L.~

>'

;. FORCE ;. ?ND

FN'b, 2, FY, 1, 2, I, 2, 1, 0

(3.65)

This equation is represented graphically in Fig. 3.3. In this case, it can be seen 0.7, the value of the response is from this figure that for a damping ratio nearly constant in the frequency range 0 < r< OA Thus it is clear from eq. (~.65) that the response indicated by this instrument wi!: be directly proportlOnal to the base-acceleration ampIitude for frequencies up to about six .tenths of the natural frequency. Its range of appHcabiHty wUl be :ncreased by increas~ ing the natural frequency, that is, by increasing the stiffness of the spring or by decreasing the mass of the oscillator. Such an instrument 1S an accelerometer.

(18) Request output for the amplitude of the displacement and of the acceleration in function of the frequency: ANALYS:S

)<

PD~PR::;:N7,

POS_DYN ;. PD_OUTPUT 1, Q, 1, 0, 0, 10, L

>'

PD_PRINT 1000, 1

(19) Request plots for displacement and for accelerarion at node 2 in function of the frequency: P.NA!,YSIS

PD_NRESP, fu~ALY$!$

PD_PLO~,

:>

POS_DYN

:>

PD.. _OUTPUT

:>

P::LNRESP

1, 2, 0 :> P05_DYN > PD 1, 1000, 10, 0

OC~PU~

) ?D_PL07

:

82

Structures Modeled as

a.

Single-Degree-of-Freedom SYSlem

.' ..""

Response of One-Degree-oj·Freedom System 10 Harmonic Loading

83

(20) Execute harmonic analysis: RESULTS

;I>

POS_DYN

R.....!)YNAMIC

:>

R_DYNAMIC

,--1--11

(21) Activate XY plot infonnarion for Y displacement at node 2 as a func~ tion of frequency, and pJot the displacement YS, frequency for node 2: orSPI,AY

XV_PLOTS

:>

>

XYPLOT,

v y

In

the range set

to

, ,,

,I

1

(22) Replot the displacements repajnt the SCreen:

- '-+----\~U '" 1 I I I -~ i '"

ACTXYPLOT

." ~'"'-

1

--"

,

1

I,

I

,I

f--! -

1

,'II

'-i'\

ACTXYPLOT, l, FREQ, OY, 2, C, 12, 1, 0, 2K DISPLAY > XY_PLOTS ? XYPLOT

H

,--

,i

i

\+1 b-I:::t:, f-~I 1 J==t=bk-i

1000 maximum and

'"

i

,

I

,

I,

7 : \

DISPLAY ~ XY_PLOTS :> XYRANGE XYRAAGE, 1, L 2, 10, 0, 1000

l---J,

( L

-'-~,--

:1

DISPLAY REPAINT

LI1

L~

J.'6

'.l.l

\i,\!:

""I.~

8.4

9.2

j$

F!l(C Ifll)

(23) Activate XY plot !nformation for Yacceleration at node 2 as a function

(a)

,of frequenc)', and generate plot of the acceleration vs. frequency for node 2, Then replot with acceleration range from 6000 to lE6: lE'S£t-,,~-~-···~

DISP:0AY ;> XY_PLOTS ACTXYPOST, 1, FREQ,

::>

AC?XYPOS'I' AY, 2, 0, 12,

1,

0,

s.!'i&S€ XY_PLOTS > XYPLOT XYPLOT, 1 DISPLAY XYRANGE.

- - '_ _

4~

I,

,!.

II'

1

'



,

,

!+---'--\ j - I L., : ~

, _ I __

~'

..

1

'~

,. I, . ,"=,.~,

:

DEVICE > LASERJET 150,0

FRSI)!'U)

Figure 3.18 shows the frequency response in terms of the displacement and of the acceleration,

Example 3.11. Solve Example 3,10 assuming that the damping in the system is equal to 10% of the critical damping.

(b)

, \ I

i

i

t. ';',

" !;

~'

~

Fig. 3,18 Frequency response for Exnnlp1e 3.10. (il) Amplllude of the (b) Amplitude of tbe acceleratiol:.

d~splacemem;

i' u

Ii

84

Response of One-Degree-of-Freedom System to Harmonic Loading

Struc1ures Modeled as a Single-Degree-of-Freedom System

Solution: The analysis is a continuation of Examples 1.6 and 3.10: A harmonic analysis is performed over the frequency range of 2 to to Hz. 10% of critical damping is assumed. The following commands in COSMOS follow those of Examples 1.6 and 3.10:

'"

_~ALYSIS

) POST_DYN

PD_MDAMP I L L

1,

>

PD_DN1P!GAP

>

i

I \ I \

'" , '"

.1

ANALYSIS > POST_DYN ) R_DYNAMIC R_DYNAMIC

, ,

(27) Activate XY plot information for X displacement at node 2 as a function of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 200 maximum:

'\

i

,"

PD_MDAMP

(26) Execute hannonic analysis:

\

!I II

V

I

1\

.\

IL

,

'------

,

DISPLAY ) XY_PLOTS ) ACTXYPLOT ACTXYPOST, 1, FREQ, UY, 2, 0, 12, 1, 0, 2N DISPLAY XYPLOT XYPLOT, 1 DISPLAY ) XV_PLOT XYRANGE XYRk~GE, I, 1, 2, 10, 0, 200 DISPLAY ) VIEvCPAR ) REPAINT REPAINT

6.6

-

1$

9.2

FRE01H,1

(a)

2E·('l~

m

I

1.8E+G5

(\

I

1.6E+GS

I --\

! .4E+.:IlS

0, 2N

I I II

1.2E+GS

,,

,

lE+~S

1 8G.:Il.:ll!l

---- .--.. ---- --- ---

1\

\

."

6SGGS

DISPLAY REPAINT

6.4

7.6

(28) Activate XY plot infonnation for X acceleration at node 2 as a function of frequency, and plot the acceleration vs. frequency for node 2. Then replot with acceleration range set to 2E5 maximum: DISPLAY ) XV_PLOTS ) ACTXYPLOT ACTXYPLOT, 1, FREQ, AY, 2, 0, 12, I, DISPLAY XYPLOT XYPLOT, 1 DISPLAY ) XY PLOT XYRANGE XYRANGE, 1, 1, 2, 10, 0, 2E5

'"

'" '"

(25) Define modal damping for mode 1 as 10% of critical damping (0.10):

85

I'---

/

4GGGS

-=-

2GSn

Figure 3.19 shows the frequency response plot for Example 3.11 in terms of the amplitudes of the displacement and of the acceleration.

Example 3.12. Detennine the steady-state response of the system of Example 1.6 when a vertical harmonic acceleration of magnitude 0.3 g and frequency UJ in the 2 to 10 cps is applied at the support. Neglect damping. Solution: The analysis is a continuation of Example 1.6. A harmonic analysis is perfonned over the frequency of 2 to 10Hz. The following commands in COSMOS follow those of Example 1.6:

,~ 2

2.8

3.6 ~

.4

6.8

5.2

7.6

8.4

9.2

lG

FREOIH,)

(b)

Fig. 3.19 Frequency response for Example 3.11. (a) Amplitude of the displacement; (b) Amplitude of the acceleration.

86

Structures Modeled as a Slngle·Oeg:ee-of-Freedom Syslem

Response of One-Degree-of· Freedom System to HarmoniC Loading

.,

(15) Define analysis type as hannonic using one natural frequency, units of ex.citing frequency as Hz, frequency range from 2 to [0 Hz, 1000 frequency points in frequency range, linear inrerpolalion, and reque..')t relative displacement and relative velocity response printout: ANALYS:::S

J

POST_DYN )-

:

(16) Define dynamic forcing function as frequency dependent base acceleration in the Y direction: ANALYSIS

,

,

,

POST_DYN PD_CURVES L 1, 1 POST_uYN PD_CURVES PD ... CUR::>EF, 1. 1, 2, .3, 10, O,}

PD_CURTYP, M1"ALYSIS

,

ANALYSIS ) POST_JYN

,

?D_CURTY? PD_CURDEF

PD_BEXC'!) PD_BASE

PD_BASE, 1, 0, 386.4, 0,

, ,, "

0

(17) Request response at node 2: k~ALYSI$

PO_NRESP, 1, 2, 0 l-.NALYS'1S > POS~DYN

~

PD~PRIN'I',

0,

1, 0, L

:>

!

,.

./

,.,

'-

I

..

J.

,

•••

"

PD._OUTPUT » PD_PRINT 0, 10, 1, 1000, 1

ANALYSIS ) POS_DYM > PD OUTPUT PD_PI,OT, 1, 1COe, 10. 0 (18) Execute harmonic analysis; AJ,"VALYSIS

I



,

)- POST_DYN )- Pu_OUTPUT ) PD_NRESP

fJ~1

i

", ,

;

=-t-=

~

J.

, •

"



, ,

?D_ATYPE

PDyTYPE, 5, 1, 1, :;L 10, 1000, 1, 0, 0

87

>

(a)

PO_PLOT

I

POST_OYN' ) R_DYNAHIC

FLJYNAMIC

(19) Activate XY plot information for Y displacement at node 2 as a fune·· tion of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 10 maximum: DISPLAY ) XY~PLOTS ) ACTXYPOST ACTXYPOST, 1, FREQ, UY, 2, 0, 12, 1, 0, 2N DISPLAY ) XY_PLOTS ) XY~~GE XYRANGE, 1, 1, 2, 10, 0, 10

, ,,

REPAINT

(20) Activate XY plot information for Yacceleration at node 2 as a function of frequency. and piot the acceleration vs. frequency for node 2. Then replot with acceleration range set LO 10,000 maximum: DISPLAY ) XY_?LOTS ) ACTXYPOST ACTXYPOST, 1, FREQ, AY, 2, 0, 12. 1, 0, 2N DISP~AY , XV_PLOTS > XY~~GE XYRANGE, 1, 1, 2, 10, 0, 10000 REPAINT

Fig:lre 3,20 shows the frequency response for Example 3.12 in terms of the amplitudes of the cisplacement and of the acceleration,

). I !

(b)

Fig. 3.20 Frequency response. fer Example 3,12, (a) Amplilude of the displacement;

(b) Amplitude of the acceleration,

Structures Modeled as a Single-Degree-of-Freedom SySlem

88

Response of

One~Degree-of·Freedom

System 10 Harmonic Loading

89

Example 3.13. Solve Example 3.12 assuming that the damping in the system is equal to 15% of the critical damping.

Solution: The analysis IS a continuation of ExampJe 3.12. Damping is increased to 15% of critical damping. (21) Define modal damping for mode I as 15% of critical damping (0.15): O.J:O;

]\l\AL'fSIS > POST_DYN > P;:}_DAMP iGAP ?D_HDA.,'\J?, 1, 1, 1, 0.15

"y, ,

{22) Execute hannonic analysis: m,ALYSIS

>

POST_DYN

>

'.

R_D~NAMIC

R~DYNA1HC

0,

L

/1

:

,

I

\

I

.\

!

,,1--

2N

I

'" I---I

III.SS

, .!"

,

I

, ., 9.1S'

(23) Activate XY P!ot information for Y displacement at node 2 as a function of frequency, and plot the displacement vs. frequency for node 2. Then replot with displacement range set to 0.5 maximum: DISPLAY> XY_?[,OTS ) AC'rXYP['OT ACTXYPIJOT, _, FREQ, UY, 2, 0, 12, DISPLAY > XV_PLOT ) XYRA~GE XY?.ANGE, 1., 1, 2, 10, 0, 0.:) DISPLAY > VIEW~PAR > REPp_IN·1' REPAINT

II

1.. 25

,

'\

/

,.,

., , ,.,

,

! I

i

'.'

'iL7

"

FfI€QHhl

(a) .......~.~.----.:~--------,

(24) Activate XY plot information for Yacceleration at node 2 as a function of frequency, and plot the acceleration vs. frequency for node 2. Then repiot with acceleration range set to 500 maximum: DISPLAY:> XY_PLOTS > ACTXY?LO'f ACTXYPLOL l, FREQ, AY, 2, 0, 12, 1, 0, 2N DISPLAY :> XY_P:01' :> XY?JL~GE XYRANGS, L L 2, 10, 0, 500 DIS?LAY :. VIE~CPAR > REPA:)lT REPAINT

"y

,,

,

Figure 3.21 shows the frequency response for Example 3.13 in terms of (he amplitudes of the displacement and of the acceleration.

3.11

SUMMARY

In this chapter, we have determined the response of a single degree-of~freedom system SUbjected to harmonic loading. This type of loading is expressed as a sine, cosme, or exponential function and can be handled mathematically with minimum difficulty for the undamped Or damped structure. The dIfferential equation of motion fo:- a linear single degree'of-freedom system is the secoodo:-der diffe:-e:1tJai equation

my + cy + ky = Fo sin w!

eq. (3.10)

(b)

Fig. 3.21 Frequency response for Example 3,13. (a) Amplitude of the displacement; (b) Amplitude of the acceleration.

90

Structures Model€ld as a Sing!e-Degree-of"Freedom Syt')tem

Response 01 One"Degree-of·Freedom System 10 Harmonic Loading

or

with respect to the support, In this latter case, tbe equation assumes a much simpler and more convenient form, namely F m

._

y + 2~wY + w Y = -~a Slfi 2

in which

91

w to

wt

+ cu + ku.

mil

Fef! (l)

eq. (3.50)

in whkh

is the forced frequency,

F.;.:r(t)

=

m)'J(t)

is the effective force

is the damping ratio

" u

and and

it =}' _.}',-

r;;W

is the natural frequency

= /~ It!

The ueneral solution of eq, (3.10) is obtained as the summalion of the com-

plem~n\ary

Ii

(transient) and y

= e -,~ (A ,~

cos

~he

particular

WDI

_____

~

+B

. Slfl

(steady~slate)

soh1l10ns. namely

. Folk sin (WI - 0) wvt) ~ r-:--:.I' " j'

J (1

____

~J

transient solution

, r

r

+ (lrt;)

~~--'

steady-state solulion

in which A and B are constaf'lt.5 of integration,

r

w"

€v

=-

w

is the relative displacement

[i,

For harmonic excitation of tbe foundation, the solution of eq, (3.s0) in terms of the relative motion is of the same form as the solution of eq, (3.10) in which the force lS acting On the mass, In this chapter, we have also ShOWli that the equivalent damping in the system may be evaluated experimentally either from the peak amplitude or from the bandwidth obtained from a plot of the ,1mpJitude-frequency curve when the system is forced to harmonk vibnuion. Most commonly, equivaient viscous damping is evaluated by equating the experimentally measured energy dissipated in the system during a vibratory cycle at the resonant fret:;uency to the theoretically calculated energy that the system, assumed viscously damped, would dissipate in a cycle_ This approach leads to the following expression for the equivalent viscous damping:

is the frequency ratio

wJ 1 -

47Tr

/

is the damped natural frequency

and

,I 2rt \ 0= tan - '--,1 \ 1 - r'l

is the phase angle

The transient part of the solution vanishes rapIdly to zer~ because ~f lhe negati ve exponential factor leaving only the steady-state SOIU1l0D: Of partlcui~r significance is the condition of resonance (r = iiJlw =:. 1) for whlch the ampltttldes of motion become very large for the damped system and lend 10 become .. infinity for the undamped system. The response of the structure 10 support or foundaHo.n motl~n can, be obtained in terms of the absolute motion of the mass or of tis relatlve motJon

E~

in which

It = energy dissipated in the system during a cycle of harmonic vibration at reSOnance E~ strain energy stored at max;mum displacement jf the system were elastic r = ralio of forced vibration frequency to the natural frequency of the system.

'i

I I

Two related problems of vibrating isolation were discussed in this chapter: (1) the motion transmissibility, that is, the relative motion transmitted from the foundation to the structure; and (2) the force transmissibility which is the relative magnitude of the force transmitted from the structure to the foundation" For both of these prob!ems, tbe transmissibility js given by

T< -

1 (1

~1+(2rt;)'-­

\

r'l' + (:irt;)'

I

III

,I

I

i

92

Response of One·Degree~or-Freedom System to Harmonic Loading

Str.;clures Modefed as a Singie-Degree-o:-Freedom System

the stiffness of the isolation springs required to reduce the vertical motion amplitude of the instrument to 0,01 in. Neglect damping,

PROBLEMS 3.1

93

An elec(;-ic motor of total weight W = 1000 Ib is mounted at (he cemer of a slm;!ly supported beam as shown in rig. P3.L The unbalance in the rolOr is W' e "" I lb· in, Determine the stef = 6El

(4.14)

The calculation of Duhamel's integral thus requires the evaluation of the integrals A (0 and B (I) numerically. Several numerical integration techniques have been used for this evaluation. In these techniques the integrals are replaced by a suitable summation of the function under the integral and evaluated for convenience at n equal time increments, liT. The most popular of these methods are the trapezoidal rule and the SImpson's rule. Consider the integation of a general function J(T)

J\zfaximum displacement:

:i. T

B (I) cos WI} Imw

where

m= 5000 = 12.95341b.sec'lin T= 211' '

wi -

.

15,083 pSI

(Ans.)

+ 21, + ... + 4/"_, + I")

(4.17)

where n ~ l/1J'T must be an even number for Simpson's rule. The implementation of these rules is straightforward. The response obtajned will be approxi-

'. ':.;

106

Structures Modeled as a Sing!e-Deg;ee,o(-Freedom System

Response 10

F(r)

General Dynamic Loading

107

where

I

t1F, = F(I,) - F(lI_ ,) F(r'>1 1

and

--------"---~---~--~----

L1I'=(;-[;_1

; I I I

F(t,l -.----"'~ ;

1

I

I

"

I

F!r)

L

The substitution of eg. (4.20) jnlo eq_ {4.18) and integration yield

I I

£IF;

I I I

_-:-'-~_._L~_"--~ _~

r,

f,_ 1

t"

________

t1F· +---2j' {cos wt,-COS (ot;_I+w(t;s;nW(,-I;_1 sin wt i - r»)

T

Cd Ll.t,

!

i (4.2])

-:-£l(,~-----l

I

Analogously from eq. (4.19).

Fig. 4.7 Segmentally linenr loading function.

II

B(lI)

mate since these rules arc based on the substitution of the function 1 (r) for a piecewise linear function for the trapezoidal rule, or piecew~se parabolic function for Simpson's rule. An alternative approach to the evalualion of Duhamel's integral is based on obtaining the exact analytical solution of the integral for the loading function assumed to be given by a succession of linear segments. This method does n?t introduce numerical approximations for the integration olher than lhose inherent in the round off error, so in this sense it is an exact method. In using this method. it is assumed that F (r), the forcing function. may be approx1mated by a segmentaHy linear function as shown in Fig. 4,7. To provide a complete response history, i~ is more convenient to express the integrations in eg. (4.15) in incremental form, namely A(r,)=A(I,_,)+

r

F(r) cos ""dr

f.,

+ ~ {sin we; - sio wrj_! U?

w(!, cos Wi j

-

(i_I

cos

Wi_I)}

Equations (4.21) and (4.22) are recurrent formulas for the evaluatlOO of the integrals in eq, (4,15) at any lime!:.:-.;; I;.

Exnmple 4,2. Determine the dynamic response of a tower subjected to a blaslloading. The idealization of lhe stucture and (he blaSlloadJog are shown

in Fig. 4.8. Neglecl damping.

F(r)

F(r) sin wrdr

(4.19) K'" 100 kim.

where A (l,J and B(t,) represent the values of the integals in eq. (4,15) at time tl_ Assuming that the forcing function F(T) is approxin;ated by a piecewise linear function as shown in Fig. 4.6, we may write

(4.22)

JJ.t;

(4.18)

li.1

B(t,)=B(IH)+

i

L1F;

-Y, {"

(4.W) Fig. 4.8

'"

Idealiz.ing srructure and Joadmg for Exan'lple 4. L

!

108

Structures Modeled as a Single-Degree-ci-Freedom SYSlem

Response to General Dynamic Loading

For this system, the natural frequency is

Solurion:

4.3

Since the loading is given as a segmented linear function, the response obtained using Duhamel's integral, eq, (i!.14), with the coefficients A (/) and B (t) detennined from eqs. (4.2]) and (.:!.22), will be exact The necessary calculations are presented in a convenient tabular fonnar in Table 4.f for a few time steps. The integrals in eqs. (4. J 8) and (4.19) are labeled L1A (t) and LlB (I) in this table, since

109

NUMERICAL EVALUATION OF DUHAMEL'S INTEGRAL-DAMPED SYSTEM

The response of a damped system expressed by the Duhamel's integral is obtained in a manner entirely equivalent to the undamped analysis except that the impulse F( r) dr producing an intral velocity dv = F(r) drlm is substituted into the corresponding damped free-vibration equation. Setting Yo = 0, Vo = F(T) dTlm, and substituting I for t - r in eq. (2.20), we obtain the differ-ential displacement at time t as -,f",(I-T)

dy (/) = e '

F(r) dr

.

. ---'-- sm 'Yo (t - r mwo

)

(4.23)

,.[;

L1A(ti)=A(ti)

A(ti~:)=) ~

1,-_,

Surruning these differential response terms over the entire roading interval results in

F(r) cos urrdT

and

y(i) =_1_

LlB(li)

B(I,)

"

B(li~l)

{'., F(r) sin wrdr

Since the blast terminates at t = 0,060 seCt the values of A and B remain constant after this time. Consequently) the free vibration that foHoVots is obtained by substituting these values of A and B evaluated at t = 0.060 sec into eq. (4.14), that is,

mwo

r

Jo

F(T)e--! or in terms of the relative motion u =)' -)'1> for the strucrure excited at its support.

4.4.1.

113

Solution of the Equation of Motion

The method of solution for the differential equation of motion implemented in the computer program presented in this chapter is exact for an excitation function described by linear segments, The process of solution requires for convenience that the excitation function be calculated at equal time intervals LIt. This result is accomplished by a lineer interpolation between points defining the excitation, Thus, the time duration of the excitation, including a suitable extension of time after cessation of the excitation, is divided into N equal time intervals of duration For each interval .1t, the response is caiculated by considering the initial conditions at the beginning of that time interval and the linear excita~ion during the intervaL The initial conditions are, in thjs case, the displacement and velocity at the end of the preceding time interval. Assuming that the excitation function F(r) is approximated by a piecewise linear function as shown in Fig, 4.7, we may express this function by

(4.45)

which upon its substitution into eg, (4.42) gjves I-I)

11-1 '\

-~' F+l~-' F I Lli ' Lli i ,+ where A, and BI are constants for the interval t,.:s t S ti + L1t and where we use the notation F; F (t/) and FJ + I = F(t, + Llt). Establishing the identity of terms between the left- and right-hand sides, that is, between the constant terms and the tenns with a factor (t - t;) and solving tbe resulting equations, we obtain

(4.46)

Fi-cA; ··~C-

.at.

The substitution into eq. (4.43) of li}e complementary solution Yc from eq. (4.44) and of the particular solution YP from eq. (4.45) gives the total solution as (4.47)

The velocity is then given by the derivative of eq, (4.47) as (4.41) in which ti = i· t3t for equal intervals of duration .11 and i 1,2,3" .. , N. The differential equation of motion, eq. (4.35), is then given by

my+ :OEX (;>.CCELEPJI!tlOO C1" G"G>Vl1'Y C'R ZSROj

C " (;

0.150

'tItE

Excr'!',",?lON

TIME

;,;xcrT,~':'XC"

O.l)1C

120000.0:)

() OISC

O.OC

:).00

OL"!'l'vT RBsm.:rs:

4.5

117

PROGRAM 2-RESPONSE BY DIRECT INTEGRATION

The computer program described in this section calculates the response of the simple oscillator excited by a time~dependent external force acting on the mass or by an acceleration applied to the support. The excitation is assumed to be piecewise linear between defining points. The response consislS of a table giving at equal increments of time the displacement. the velocity, and the acceleration of the mass for the case of the oscillator excited by a force applied to the mass. For the case of the osciHator excited at its support, the response is given in terms of the displacement and the velocity of the mass relative respectively to the djsp]ac,ement and velocity of the support ar:.d of the absolute acceieratJon of the mass. The program implements the direct integration method to calculate the response. The program has been written in an interactive mode with the user. It starts by requesting information about the data needed in the solution of the

(1.000

0.000

0.0:'0

0.074

to.692

0.0;\0

0.451 O. 92~

2'),,155

43(;. 7~()

17.096

-t1~:z.SU

(L

aco

9!n.:;:n

L044

-4.821

~1Jl2.581

O. ,-CJ

J.ns

'2

C .072 0.242

:;'1.10$

-l1LJU

5.6n

"3D.4?7

.

D=SP"~ACl':.'{£N"l'

)'AX. VE;t.OCl'J'Y

Ace.

DIS?!...

C .Ji:O C ,060

w.x.

VC".JX:

7::ft:

o.c,jO

>i.63j

(9) Graph eQ. f4.S2i

2 For excitation at the supper:: Displacement :tnd velocity ue n.~~ative 10 the sup]Xlrt. whereas

the acceleration

)$

absoiute vall;.e.

Fig.4_12

Graphs for excitation

func~ions

grven by eqs. ('L56) through (4.63).

II

120

StructureS Modeled as a Sing'e-Degree-oi·Frcedom System

E,(T)

= P(l}*T

Os T,;; P (2)

=P(l)*P(2)

T>P(2)

E,(T)=P(1)*(I

TIP (2»)

=0

(4.59)

=2*P(I),[l

TIP (2)],

o

OST5P(2)12

Natural frequency: w=~ =3l.623 rad/sec Natural pedod; T

(460)

T> P (2)

E,(T) = P (I) * [2 *T IP(2)J

o

= 0.20 sec

Imput Datn and Output Results (4.6,)

P(2)12 1'(2)

E, (T) = P(l)IEXP [I' (2) * 71

2'lf/w

Select time step: ..1T = 0.01 sec

T> 1'(2)

E,(T)=P(l)*sin [P(3) *71

E, (T)

Os T,;; P(2)

Response 10 Genera! Dynamic Loading

Tn!e S"!'£P Of W'l'EGRA'l""l:)M

(4.63)

= !'lEW FU!'ICTION

R£S?OMS£ K1.X. T!M£

(a) Determine the dynamic response of [he tower shown

in Fig. 4.13 subjected to the sinusoidal force F(l) Fa sin WI applied at its top for 030 sec, (b) Check results using the exact solution which in this case is C

)1) ~

0

O;nPJT RElhIL1'S:

DIS7: ••

VH.rA:.

Ace

o.oue

0,(::)0

0.010

0.0:10 C .00':

o .O?t

0.016

L1SS 5_ ,,$5

52S.-lSa

0.030

(Lt:n

11. $70

';S(L

O.iHO

c,zn

211.229HZ

6~0.10.j

T;"'£

0.000

:l.()SO

O.

4S~

J.060

o.

]~,

0.0;0

1.c:n

o.oan

l.J96

0-.090

1.199

"" '"

" "

"

'~().H9

~8J

$1:1:.513 ~16.Z&~

16$

-).67.611

:89-

~S20. 916 -len] .9$0 -145·1 .He

6 S~

O,Hl::>

1. SS5

O.H{>

1.5.4;

-13.C1<

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0.120

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o .DC

0.9J7

-17.;;-.'5

-1£:1S ~H

0.1'"

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-61.205

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O.J.!)O

-0.29: -l m 171

-n.2?9 -10.S.n

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~SjI.H5

(1.1,70

-,

::. !}IE

-~2

.(0)

'"

-6B6. Hi )1.0112

BH_732.

::L 1\;0

-, -, nc -,

~)i!.0:;'$

:lS~L~O,

(I.NC

:it}

-lL60,

2SU.269

O.

Tower for Example 4.5.

~

1\l001l0

PARNren:R ... (2) i'AiVt.M£T£R PO)

0.1£0

Fig.4.13

0

THAX



HS.EOl

lOS: .496

121

122

Stn;clures Modeled as a

Singla-Degree-of~F(aadom

System

O.21S0

LL?S

l21,lS1l

~126.4116

0,,;10

:L ~:>o 1. )98

US.574

-1:'09.642 -Z5B,161

(). ?:SC 0,300

4.213 4.674

,,$.996 $5.:n5 2$.&1::;

C.310

4.711

-18.7;;5

o.n::;

Response to General Dynamic Loading

Results shown in the above table are sufficiently close to corresponding values given by the computer in part (a) of this problem. .

-354,LSS4 -O.6~,)76

Example 4.7. Solve Example 4.6 selecting the time step 111 equal to and 0.005. Then compare the displacements at time 1= 0.1, 0.2, and 0.3 sec with the response obtained in Example 4.5 using the exact solution of the differential equation,

~45$tI.3;;)

0.02,

.. n

},.AX . or SPLACEMElIlT'

K'X. V£LCCITY ACC;;;LtAA'l'XON

">-X.

1;)(

f:"O(;()

C,OO"

J.n"

6:. ;$7

0.040

(l,015

~. 22~

lH.2SS

~

.691

124..894

'Z .110 '0.5-60 usn t?lI -1:'''1 ~H

O. J6C

C,1)S

!).:lee

O,27S

7.922

o .lDO

{j, 4~6

0.120

O,57{i

8.646 L39"l

0.140

0.51'

_8,O!

c· .1&0

0,26$

PROPSE7S )- EGROUP EGROtJP, 2, Ml\SS, 0,

us

(l.2:(0

.iL t&Z

0.:'10

-;; .4~()

HH"*'1 410_ :iliS

0,

0,

0,

(7) Define the real constant for mass element: PROPSETS > RCONST RCONST, 2, 2, L 7,

lOa, 0,

0, m;;;:

0

100 lb· sec 1 /in:

0, 0, 0, '0, 0

(8) Activate element group I (spring) and real constant 1 (k ~ 29,30 Ib/in), and generate one element along curve 1:

"173 ?EO

-c.on

0,

10

CONTRa!... > ,'1CTIVE )- ACTSET l"..':,;>.CEY;;NT ,. P.AX. V£LOC:Ti '"

ACTSET, EG, 1 ACT SET RC 1 M_CR, 1, 1, 1, 2, 1. 1

1,,';>10

1S .le

I

S9l_2C

4,7

(9) Activate e~ement group 2 (mass) and real constant 2 (m = 100 lb' sec!./

RESPONSE TO GENERAL DYNAMIC LOADING USING COSMOS Example 4.9,

I

in), and generate one element at point 2: CONTROL ~ ACTIVE ACTSE'J', EG, 2

Solve Example 4.2 using program COS}'10S.

Solution: The analysis is pre:onned using a single spring element wIth one concentrated mass element The following commands are implemented:

ACTSET, He,

Ii

2

>

ACTSET

126

Structures Mode!ed as a. Singla-Oegree-of·Freedom Syslem

Response to General DynamIc Loading

(10) Merge nodes; MESHING

07) Define modal damping for mode 1

NY-ERG?:, 1, 3,

1, 0,0001, 0,

ANALYS:S :;. POS1'_D":'~J > PD~DAMP/GAP PD-HD.r..Mp, L 1, 1, .2

0, 0

(11) Apply constraints in all degrees of freedom

1,

A_FREQUENCY,

0,

G,

D,

0

ANALYS:S

>

S,

0,

0,

0,

0,

lE-OS,

0,

18··06,

(13) List the natural frequency of the system:

LIST

1-

:R£

>

R_DYNAHIC

R_DYNAr~rC

(20) Activate XY piot information for X displacement at node 2 as a function of time, and plot lhe displacement VS. lime for node 2:

D::SPLAY ) XY_PLO?S >ACTXYPOS'T DISPLAY

:>

XYPLOT,

1

1, TIME, OX,

2,

12, 1, 0,

2N

XY_?LOTS ) XYPLCT

CISPLAY ~ AC?XYPOST,

XY~PLOTS

D!SPLAY

:>

XV_PLOTS ) XYPLOT

XYPLO'!',

1

FORCES, FP'I'

> ACTXYPOST

1. TIME, AX,

2,

12, 1, 0, 2N

i. !

,.I'!

Figure 4.15 shows the displacement and acceleralion response for the first 1 sec

,

,i·

(21) Activate XY plot informalion for X acceleration at node 2 as a functio:1 of time, and plot the acceleration VS. lime for node 2:

FREQr..IST

:>

A;-.JALYSIS

ACTXYPOS?,

PREQ/8UCK ) R_FREQUENCY

R_?REQUENCY

HESUL':'S

·

POST_DYN ~ PC_OUTPUT) PD_KRESF 1, 2, 0

:>

(19) Execcte modai time his:ory analysis

RZ

A_FREQUEKCY

16,

·· ,,

(18) Request response at node 2: ANALYSIS PD__NRESP.

DPT

(12) Set the options for analysis to exlract one frequency using the Subspace Iteration Method with a maximum of 16 iterations, and run the frequency analysis: ANALYS1S

127

VIb"'f,' PAR> v!EW 1. 0

>

0,

0,

(2) Define XY plane at Z = 0: GEOMETRY PLANE,

Z,

>

GRIJ 0,

1

:>

PLANE

II

Structures Modeled as a

128

1.1)2

& .6-1

&.65 J

x

t:l,! :2

)

.~2

-s .S

I

i , ....

•I

,

,

0.0

\.

• .,' '"

s.<

'-'

(al

"

I

i

?ROPSETS EGROUP,

I /'""'

¥2~tl

PROPSETS :> HGROUP RCONST, 1, 1, 1, L

'--'

3.1

2, L

0,

0,

0,

0

lES

(6) Define element group 2 using the ~1ASS element fomulation:

i

V

0,

(5) Define a real constant for spring element: k = lOO,OOOlblin:

.,

-nIJ I--H>'

,

EGROUP

1,. 'SPRING,

'-...--

-~8&'

-12SG .

>

using the SPRING element formulation with

I"-'~

\

-SSi'il

Water tower and excitation ror Example 4,10.

(4) Define element grQup two nodes:

/\

'"

(b)

Fig.4.16

!

no

,,s

--"~'l

aId

i

G.5

I

'50

,

(t2

--+ 0,

(a)

A

(in/aec')

'i"!\----";f;;1

'-

I



"0

a(t)"'300 ain/3t

i,

/" ..........

129

2M1

,!

(

=-1\\

-tl,24

Response to General 0ynamic Loao:ng

I ,

I

System

i

,

-S.96

-&

- ....

i

•• 3

L

, ...

I

0.48

R E

,

I

-

t-I

Sjngle·Degree~of~Freedom

,

"

.:,

0.'

.:,

, .1

3.a

PROPSETS > EGROUP EGROUP, 2. ¥~SS, 0, 0, 0, 0, 0,

•. g

TIME

(7) Define a real constant for mass element: m = 15.5 lb· sec1.lin:

(bl Fig. 4.15 Computer plot for rhe response of Exarr:p!e 4.9. (a) Displaceme':'lt. (b) AcceleratioL

PROPSETS > RCONST RCONST, 2, 2, 1, 7, 15.5, 0, 0, 0, 0, 0, 0

(8) Activate element group I (spring) and real constant I (k ~ 100,000 Ibl in)~

(3) Generate a curve from 0, 0, 0 to I, O. 0, GEOMETKY > CURVES > CRPCORD CRPCORD, 1, 0, 0 1 0, 1, 0, C

0, 0

and generate one element along cunte 1:

CONTROL :> ACTIVE > ACTSET AC'ISE'L EG, 1 ACTSE'!, RC, 1 MESHING " PAR!>'LMESH ) M_CR M_CR, L L 1, 2, .3., 1

130

Structures Modeled as a Smgle-[)egree-of-Freedom System

Respon..l;c 10 General Dynamic Loading

(9) Activate element group 2 (mass) and real constant 2 (n:::::: 155 Jb . sec 1 f in), and generate one element at point 2: CONTROL AC7SET. ACTSE'!', MESHING M~PT,

;.. ACTIVE ;.. ACTSE':' EG r 2 Ret 2 ). PARt.l·CH2:SH

131

(16) Define base excitation as acceleration in the X direction: ANl\LYSlS ~ POS':'_DYN } PD_£5EXCIT > PC_BASE PD_BASE, 1, 1, G, 0, 0

(I7) Req"Jest response at node 2: ANALYSIS ) POST_CYN PJ_N RES?, 1, 2, 0

2, 2,

(10) Merge nodes:

~

PD_OU'l'PU'T ,

PD_NRESP

(18) Execute modal time history analysis:

MESHING > NCDED ;.. t4}1ERGE

NMERGE,

1, ), 1, 0,0001,

0, C,

0

ANALYSIS ) POS'l'_DYN > R_DVNAM!C

(11) Apply constraints in aH degrees of freedom at node 1, and all degrees of freedom except UX at node 2: LOADS-Be ;.. STRUCTURAL DPT. 1, ALL, 0, L 1

~

DISPLMNTS ) CP'r

DPT, 2, OY, 0, 2, 1, UZ, RX,

PREQ !BUCK > A .. FREQUENCY A_FREQUENCY, ~, S, 15, 0, 0, O. 0,

Or 0,

>

lE-C5,

0,

>

DISPLAY ;. XY_PLOTS ) ACTXYPQS,=, AC'l'XYPOST, 1, ':'J:ME, UX, 2, 12, 1, DJ:SPLAY ) XY_PLOTS > XYPLO'l' X,{PLOT, 1

(Figure 4.17 shows the plot of the displacement a: node 2 as a function of time.)

R.]REQUENCY

A.:,"JALYS:r:S > POST_DYN ) PD_OUTPUT :> P:U1AXMIN PD_M..Ji.X:-Hl';, L 1. 10, 2, 2. 0, 2 ANALYSIS } POS'T'~ .. DYN > PD_PREP.lJ..RB

(13) List the natural frequency of the system~ RESUL7$ ~ LIST > FREQLIST F'HEQLrS?

PD_PREPARE,

Frequency#-

~requency

Frequency

1

(rad/sec} 8.032:ge+C1

{cycles/sec) L27B36e..-01

7.82250e-02

A,,),

0,25,

1, 2,

1

ANA:'YSIS ;,

POST~DYN

PD__CURD£F,

1,0,

>-

PD_CURVES

PD~Ct:RDEF

.2,:3CO, 100, 0, 0, 0, 0, C

MaxilTlIJm displacement:

Type

Positive Negative

Slep. no.

Time (sec)

Value (in~

70 87

0.l40 0.174

~0.18413

0.18323

{)

(15) Define dynamic forcing function as time-depe!1dent harmonic base excitation: ANALYSIS > POST_DYN

1

ANALYSIS > POST_DYN ;. PC OU'J'PtJ?

Period (seconds)

(14) De5.ne analysis type as modal time history llsing one natural frequency, 1000 time steps starting at t = 0 with a time increment of 0,002 sec; use default values for all integration paramelers: and request pri:1tom of relative displacements and relative velocities:

PD_CURTYP,

2N

(20) Request SCJn for maximum displacement in the X direction at node 2 from t = 0 to 0.2 sec.

R_FREQUENCY

ANALYSIS ) POST_:)YI", :. PD_ATYl?E PD_ATYPE, 2, 1, 1000, 0, .002, 0,

0,

1E-06,

0, C

ANALYSIS :. FREQ !BtXK

(19) Activate Xl' plot infomuHion for X displacement at node 2 as a function of lime, and plot the displacement VS. time for node 2:

RY, RZ

(12) Set the options for the frequency analysis to exJract one frequency using the S'Jbspace Iteration Method witb a maximum of 16 iterations, and n.l!1 the frequency analysis: A..~ALYSIS

R_DYNAMIC

4.8

SUMMARY

In this chapter, we have shown that the differential equation of motioc for a linear system can be solved for any forcing function in terms of Duhamel's integral. The numerical evaluation of this integra! can be accomplished by any standard method sHch as the trapezoidal or the Simpson's rule. We have

Sir,gle·Deg(ee~or~Freedom

S:ructu:es Modeled as a

'32

System

Response :0 Generar Dynamic Loading

,., ~

Q .l2

,

R E l

/\

a.ea

u

i I





I II

!

0 04

• 'f.lJI4

-e.l.iB

,i

I

/

[sV ·····r

I\ j

i

-$.12

,

I

-iil.2 Q

0.$2

,

I

I

rL~4

~.~&





1!l.1i8

r

\

\/

.\

!/ J!m~

11

!, _.. ,I

G,!



f;

+-V mJ ;

I 3.12

:

Ii I Ii !-I! i ; ,

\

\

\,

m

-oS .IS

!

1\

i



\

@.l4

W8X 24

r

,v, :lLlG

~.18

.., 'J

Fig. P4.l

TIME

i N 12, the corresponding frequencies are equal to the negative of frequencies of order N n. This fact restricts the harmonic components that may be represe:1ted in the series to a maximum of N 12. The frequency correspondIng to this maximum order WI-In. = (N fl)liJ is known as the Nyquisl frequency or sometimes as [he joldlng frequency. :\1oreover, if [here are harmonjc components above W,v/2 in the original functior.. these higher compor.ents wilt introduce dis.tortions in the lower harmonic corr.ponents of the series, This phenomenon is called aliasing [Newland., D.E., 1984, p. 118). In view of this fact, it is recommended that the number of intervals or sampJed pOints N should be at least twice the highest harmonic compone:1t present in the function, The Nyquist :requency &J y is given in radians per second by

(5.33)

H(w,,)

,,=0

UpO:l ir.troducing t!le frequency :-atio

and the damping ratio

c

['=-= -

C

Cet

eq. (5,33) becomes

H(w,,)="--

k(l

I 1.

+ ro + 2/f,/;)

Therefore, the response)'4 (;j) al tirt.e {j = jiJr to a harmoniC force cOr:1ponent of arr.pliwde en indicated :n eq. (5.28) is given by (5.34)

(5.29) ar.d the total response due to (he N hannonic force components by

and in cycles per second by

f ; 2." = .y

(cps)

(5.30)

(5.35)

148

Structures Modeled as a

Single~Cegree~ct-Freedom

Fourier Ar_alysis ard Response in the Frequency Domain

System

where ell is expressed in discrete form by eg. (5.27). In the determination of the response y (t) using eg. (5.35), it is necessary CO bear in mind that in eg. (5.28) the force component of [he frequency of order n is equal co [he cegative of [he component of (he frequency of orde-r N - n. ThIs fact may be verified by substitcting - (N n) for n i!1 the exponemial factor of eq. (5.28), In this case we obtain (5.36)

since e -2""/:= COS i sin = l for all integer values of j. Equation (536) together wiili eq. (5.28) shows that harmonic components of the force corres~ ponding to frequencies of orders Ii and - (N - n) have the same value. As a consequence of this fact, r'l w,,/w, where (seJecting N as an even number) as bJ,,=niiJ

for

w=.i KIm

(539) where lY! is an imeger. In this case, :he integers j and n can be expressed in binary form. For the pUf'I,?ose of iHustrution, we wi!: consider a very simple case where :he load period is divided ~nto only eight time increments, that 1s. N = 8, M = 3. In this case, rhe icdices in eqs, (5,27) and (5.35) will have the bicary ::eprese:1tatlon

j

=

jo + 2j, + 4[, ~5.40}

and eq, (5.37) may be written as

n::S;Nt2

(SAl)

A(j)

(N-n)w

for

n>NI2

The exponential factor can be written as

where the frequency corresponding 1:0 fi = tV /2, as already mentioned. is the highest frequency [bat can be considered in the discrete Fourier series. The evaluation of [he sums necessary to determine the respoGO

coooo

-0. aDeD

-5. SC'OC

-0. )OCO

wO.CG}S

-B.?SOO -i$.0000

0.0000 ·(l.OGeo

-LSO\lO GOOl'

(I.oooc

. 0 ,0S-;1

~O.

0000

,,).O;;S{

(1.0(100

-0.0091 ·0.C11 = 3 for the exponent in N 214 compare results with those in the solution of Example 5.3.

the force at intervals of D. iO

0: Fourier series;

input

,~ec.

5.14

Repeat Probiem 5. J 3 as:;uming: that (he systel7'l has 20% of [he critical darr.ping.

5.15

Ob:ro;n the close solution for the system in Proble:n 5.12 by considering (he half-circle sinusoidal excitation as the s"Jperposition of two sinusoidal functions: PI = Pa sin Tt1 smrring ,n'" 0 and P! and Pc sin 17(1 - I) stanir:g at r = 1 sec as shown in Fig. PS.iS.

p~

P 1

1r\;stnw: L:::, _ ,

Fitl

t ft; ""

200 K/in.

Fig, PS.15.

0.1 0.2

I O.S

• r sec

,

1.0

r--~''''''- 10"'~

5.16

i,

(,j

(b)

Fig. PS.IO.

5.11

Solve Problem 5.10 in the frequer:cy domain l'sing Program 4. Take M 5 for the expOJ1eIU in N 2 M.ColT.pare resules with those in the solutIon of Example 5.3.

A single~degree.of-f:eedom system having a natural period of ;).8 sec ar.d stiff" ness of 5000 Iblin subjected to an impulse of dUr" (x)Y (I))' dx

or K·= fEJ(X){4>" (x)}'dx

E1

(6.1l)

since dy /dx e, where B, being assumed smaH, is :aken as the slope of the elastic curve, Consequently, the strain energy is given by

v~ f1M(X)d8

(6.14)

where

The generali7.ed force F"{t) may be found from the virtual displacement OI(t) of the generalized cnordinate Y(l) upon equating the work performed by

the external forces in the structure to the work done by the generalized force in the equivalent single-degree~of~freedom system. The work of the distributed external force p (x, t) due to this virtual displacement is given by

W= Substituting

oy

LLp (X,t)8ydx

4>(x)8Y from eg. (6.7) gives W

f

p (x, 1)4> (x)OI dx

(6.11)

(6.15)

The work of tlle generaUzed force F·(t) in tbe equivalent system corresponding to the virtual displacement OY of tbe generalized coordinate is

W· M(x)

(6.13)

)0

or d8=--dx

169

(6.8)

Equating this expression for the kjne~ic energy of the continuous system to the kinetic energy of the equivalent single~deg:ee-of~freedom system t.M"Y(r)l and solving the resulting equation for the generalized mass, we obtain M·

and Rayleigh's Method

= ret) IlY

(6.16)

Equating eq. (6.15) with eq. (6.16) and canceling the factor oY, which is taken to be different from zero; we obtain tbe generalized force as r(t) ~

1 rL

p(x,t)4>(x)dx

(6.17)

Similarly, to detennine the generalized damping coefficient) assume a virtual d~spiacement and equate the work of the damping forces in the physical

170

Generalized Coor;j:na\es and Rayleigh's Method

Structures MOdeled as a Slnglc-Oegree-of-Freedom System

sysrem with the work of the dampmg force in the eqUivalent of-freedom system. Hence

. J'

C'Y6Y =

0

sing~e-degree

c (x)y8y dx

where c (x) is the distributed damping coefficient per unit length along the beam. Substituting 6y = ¢(x)oY and ]I: ¢(x)'i' from eq. (6.7) and canceling the COITLt'J:lon factors. we obtain

C': fC(X)[¢(X))'dx

expression waS conveniently set equa! to the initiaL length of the beam L instead of to its borizontal component L I. Now we define a new stiffness coeftlcient to be called the generalized geometric stiffness K~ as the stiffness of the equivalent system requlfe.d to store the same potentia! energy as lhe potentia! energy stored by the normal force N, that is, IK~Y(!)' = No(!)

SUbstituting 8(1) fro:n eg. (6.21) and [he derivative dyldx from eq. (6.7),

(6.18)

we have

which is the expression for the generalized damping coefficient To calculate tne porentlal energy of the axial force N which is unchanged during the vibration of the beam and consequenlly is a conservative force., it is necessary to evaluate the horizontal component of the mOlion 8(f) of the free end of the beam as indicated in FIg. 6.6. For tnis purpose, we consider a differential element of length dL along the beam as Sh?wn in Fig. 6.6(.a} ,\he length of this element may be expressed as

or (6,l9) Now, integrating over the horizontal projection of the lenglh of beam (L f) and expanding in series the binomIal expre.sslon, we obtain

r'l,Y(t)d]' dx

I, 2 I -KeY(,) : -N)

2'

2,[

dx

or (6.22) Equations (6.9), (6,14), (6.17), (6.18), and (6.22) give, respectively, the generalized expression for the mass, stiffness, force, damping,_ and geometric stiffness for 3 beam with distribuled properties and load, modellng it 3S a , simple oscillator. For the case of an axial compressive force, the potentiai energy in the beam decreases with a loss of stiffness in tbe beam. The opposite is true for a tensile axial force, which resuhs in an increase of the flexural stiffness of the beam. Customarily, the geometric stiffness IS determined for a compressive axial force. Consequently, the combined generalized stiffness K; is then given by (6.23)

Finally> the differential equation for the equivalent system may be written as (6,24)

Retaining only the first two terms of the series results in

dy " , Irq- f-) dx ~O 2 \d:x,

L=L +

(6.20)

or

The critical buckling load N n is defined as the axial compressive load that reduces the combined stiffness to zero, that is,

K; = g' 0(1)

=L

Y )'x '_J'lld L--d (} 2 \ dx

(6.21)

The reader should realize that eqs. (6.20) and (6.21) ll\voive approximations since lhe series was truncated and the upper limit of the integral in the final

K~ ~O

The substitutian of K' and Ko from eqs. (6.14) and (6.22) gives

172

General!zed'Coordinales and Rayleigh's Melhod

Structures ,\1odeled as a Single-Degree-or·Freedom System

and solving for the critical buckling load, we obtain E1(d'¢ldx')'dx

(6.25)

173

less accurate than the displacemeflls, because the derivatives of approximate shape functions tends to increase errors in the shape function. A better approach to estimate the equivalent forces is to compute the inertial forces that resulted in the calculated displacements. These inertial forces are

given by

(d¢ldx)' dx

Once the generalized stiffness K~ and generalized mass M'" have been determined, the system can be analyzed by any of the methods presented in the

preceding chapters for single-deg~ee-of-freedom systems. In particular, the square of .he natural frequency, 0'-, is given from eqs. (6.9) and (6.14) by

,

p (x, /)

111 (x)ji(x,

I)

(6.30)

or substituting Y(x, t) from eq. (6.7) by p (x, I)

m (x). and T mm we obtain (v

Fig. 6.17 Mathematical mode! for structure of Example 6,'),

= 12,57 rad/sec

Of

Equating maximum pOlential energy with max.imum kinetic energy and solving for the naturai frequency gives w~

13.69 fadlsec

Of

w 2".

~.• ~

2.1& cps

The natural frequency calculated as f;;;:; 2,18 cps is only an approximation to the exact value, since the deformed shape was assumed for the purpose of apply:ng Rayleigh's Method. To improve this calculated value the ~atu~al frequency. :et us load the mathematical model in Fig. 6. 17(a) with the mertlal load calculated as

.of

= (136)( 13.69), (l) = 25,489 F, = m"iy, = (66) (13.69)' (2) = 24.739 F, ~

Ill,

w'y,

The equilibrium equations obtained by equatjng to zero tbe sum of the forces in the free body diagram shown in Fig. 6,l7(b) gives 30,700y,

44,300 (y, - y,)

= 25,489

44,300(y, - y,)

24,739

and solving

f= 2.00 cps This last calculated value for the natural frequency f= 2.00 cps could be further improved by applying a new inertial load in the system based on the last value of the natural frequency and repealing a new cycle of calculations. Table 6,3 shows results obi.ained for four cycles. The exact natural frequency and deformed shape, which are calculated for this system in Chapter 10, Example 10.1, as a twowdegrees~of-freedorn system, checks with the values obtained in tl1e last cycle of the calculations shown in Tab!e 6.3.

6.9

Horizontal forces in buildings, such as those produced by eanhquake molion or wind action, are often resisted by structural elements called shear walls. TABLE 6,3

y,

= 1.64 =

2.19

Improved Rayleigh's Method Applied to Example 6.6

Deformed Cycle

Shape

Inertial Load

Fl

'-~-'~~~-~-~

2 YI

SHEAR WALLS

1:2.00 1 :1.34

J

I: 1.32

4

1.1.27

25,489 21,489 19,091

Natura]

F2

Frequency

"-~-'--

24,739 ;8,725 12,230

2.18 cps 2.00 1.88 1.88

i

. j

:1 !

I

I

196

Structures MOdeied as a

Slng\e~Degree-of·Freedom

Sysler-,

Generalized Coordinates and Raylefgh's MeL'lod

197

x

---+. x Floor mass

L y

r

----L

Fig. 6.19 Assumed deflection curve for Example 6.7,

The de.formed shape equation is assumed as the deflection curve produced on a cantdever beam supporting three concentrated weights lV, as shown in Fig. 6.19. The static deflections Yit Y:h and Y3 cafculated hy using basic knowledge of strength of materials are

----/

i-··o----j Fig. 6.18 Mathematical model for shear wall and rigid floors

These structural elements are generally designed as reinforced concrete walls fixed at the foundation. A single cantilever shea! wall can be expected to behave as an ord;nary flexural member if its length-to-depth ratio (LlD) is greater than about 2. For short shear walls (LID < 2), the shear strength assumes preeminence and both flexural and shear deformations should be considered in the analysis. When the floor syscem of a multistory buildIng is rigid, the structure's weights or masses at each floor may be treated as concentrated loads, as shown in Fig. 6.18 for a three-story building. The response of the structure is then a function of these masses and of the stiffness of the shear walL In practice a mathematical model is developed in which the mass as weI! as the stiffness of the structure are combined at each fioor ieveL The fU.I1damental frequency (lowest natural freqoency) for such a structure can then be obtained using Rayleigh's Method. as shown in the following illustrative example.

y,

IS WL' WL' -::ccc -:::- ~ 0.0926 El

49 V(L J

WL 3

El

0.3025 EI

92 WL'

WL'

y, ~ 162 y, ~ W

El ~ 0.5679

EI

(b)

The natural frequency hy eq. (a) js then cak'uiated as

_ {386 (0.0926 + 0.3025 + 05679)----;;.r· w - nO.092-6-'-+··O;;-.;;-30;;;2"'5""-:+"'O"'.s::-6C:7;;;9-l:i):..·· wL'

I

El

29.66~ WL' radlsec

or Example 6.8. Determine> using Rayleigh's Method, the natural period of the three-story building shown in Fig. 6.18. All the floors have equal weight W. Assume the mass of the wall negligible compared to the floor masses and consider only flexural deformations (1... fD > 2). Solution: The naturai frequency can be calculated using eq. (6,83), which is repeated here fOf convenience:

,---,,-,

i= 1

Y ~WDI

lEI

Example 6.9. For the mathematical model of a three-storv buHdincr shown in Fig. 6.18 determine the total deflections at the floor levels "considering both fleXUf2.1 and shear deformations. j

Solution: The lateral deflection Lly,,> conSidering only shear deformation for a beam segment of length L1x, is given by

iIg"Wv. L.. UI

w= 1-,,-··-

w

f= 21T =4.72 y We cps

(a)

VLIx .dy,~ aAG

(a)

i ~

,

'",."". ,

198

Generalized Coordinates arid

Structures Modeled as a Single·Degree-of-Freedom System

where

TABLE 6.4

v=

shear force

A

cross~sectional

Yl" (in)

y, " (in)

w·" (radisec)

0,09259 0,)3600

0.30247 0.37483

0~56790

0.65465

29,66 27.67

O~OO

G = shear modulus of elasticity

O~50

. At the first story, V

3W. Therefore, by eq. (a) the shear deflection at lhe

first story is 3W(Ll3) WL y" = ~~ aAG -- - aAG

(b)

1.00 1.50 2.00 2.50 3~00



O~26620

O~59193

O~91489

23.30

0.48322

O~95376

19~O5

O~78704

1.46032 2.11161

\.34862 1.95585 2~736S8

13~21

3.69079

11.33

Ll77GS 1.65509

2~90764

~-----~

At the second floor the shear deflection is equal to the first floor deflection plus the relative deflection between floors, that is Y~I

YJl

2W(U3)

+

aAG

6WL 3aAG

£i '

49 WL'

Y,

~

5WL

92 wL' 6WL y, = -[:6=2=~~~~ +-3aA~G-

(e)

We can see better the relative importance of the shear contribution to the total deflection by factoring the first terms in eqs. (e). Considering a rectangular waH for which A = D X 1, E 10 2~S, 1= ID'1i2, a = 1.2 (I = thickness of the wall), we obtain IS WL ' [ 1+1.8751-~1 ID\'] y,=-~-

y,

49

'D)"

WL) [

~--I+O~9S7l· 162 El L i 92 WL' r1 +O~611\{D)'1,

y,~~-

162 El ,

L

Example 6.10. For lhe structure modeled as shown in Fig. 6.18, study the relative imporlance of shear deformation in calculatin.g the natural frequency. SolWton: In {his study we will consider, for the wall, a range of values from 0 to 3.0 for the ratio D IL (deplh-to~length ratio). The deflections Yh Yl, y} al the floor levels are given by eqs. (f) of Example 6.9 and the natural frequency by eq. (6.83). The necessary calculations are conveniently shown in. Table 6.4. It may be seen from the last column of Table 6.4, that for this example the natural frequency neglecting shear deformation (D /L = 0) . is

= 162 Ei~ + 3~G

;L j

fE/iwI.',

(d)

WL aAG

162 EI

~~~--~-~~-~~~--~~--~~--

The next iHUSlrative example presents a table showing the relative import~ ance that shear deformation has in calculating the natural frequency for a series of values of the ratio D i L

The total deflection is then obtained by adding the flexural deflection determined in Example 6.8 to the above shear deflections. Hence,

IS WL)

L50 2.10 180

(c)

since the shear force of the second story is V= 2W, and at the third floor

Y ,162

15.71

4.72 4.40 3.71 3.03

'" DIL-O is equivalent. to neglecL shear defoml O~S)

the effecl of shear defor-

SUMMARY

The concept of generalized coordJnate presented in (his Chapter permits the analysis of multiple interconnected rigid or elastic bodies with distributed ProPM erties as single·degrce-of-freedom systems. The analysis as one.,-degree~of freedom systems can be made provided that by the specification of a single coordi~ nate (the generalized coordinate) the configuration of the whole system is determined. Such a system may then be modeled as the simple osciilntor with its various parameters of mass, stiffness, damping, and load, calculaLed to be

200

Structures MOdeled as a

Single-Degree~of·Freedom

System

Generalized Coordinates and Rayleigh's Method

dynamica!ly equivalent to the actual system to be analyzed. The sOlution of thIs model provides the response in terms of the generalized coordinate. The principle of virtual work which is applicable to systems in static or dynamic equHibdum is a powerful method for obtaining the equations of motIon as an alternative to the direct appiication of Newton's law. The principle of virtual work states that for a system In equil!brium the summation of the work done by ail its forces during any displacement compatible with the constraints of the system is equal to zero. Rayleigh's Method for detennining the natural frequency of a vibrating system is based on the principle of conservation of energy_ In practice, it is applied by equating the maximum potential energy with the maximum kinetic energy of the system. To use Ray[eigh's Method for the detennining of the natural frequency of a discrete or a continuous system, it is necessary to assume a defonned shape. Often, this shape is selected as the one produced by gravitational loads acting in the direction of the expected dispJaceme,lts. This approach leads to the following formula for calculating the nmuml freqc:ency:

,Y. Un!fOf/l'l disk

TOldl mass'" m

: !

V

Rigid beaM Total mn

&:. r--

~1T'l>no.riV-)"'-;""';

u

:.

'D

.,..,

0 oc 0 ,... '" '"~ '"""""oZ N =r--0'" '" '" "" NI '"'"I "'"I '" i '-"I r-'"

-31 . ?S5e

$:;10

-0.,,1;64

.{1 )1$90

GEOME~RY

BC52

_:4 936:

?:"?-..ol\lE,

lCO

n258 1122

oS ]77';

:leo

120Z

400 l. ,c:;

PH

:!ss:.

;

6~ZS

n57

5:1,12$5

50C

6911

\,. 2lsa

-i.3aM

7-:;0

59/1

.'1471

-s~. 5~6~

,.

,

,sec: ooc-

7.9

~9';;()

630'<

.,

st:i

Fig. 7.7 Comparison of elastoplastic behavoir with elasTic response for Example 7.2.

s. dSJ

~Ja,

"as

on]

0'

,

~;56

B:.2?

(2) Define the XY plane ar Z = 0:

~

. :C2S

il"-3e

~7?_

, c

liiO 3

$753

-,

~;;J.S

-59 441{

Ol';' .,

-'t ga97

·s . .ns~

DISPL.?,';( VI~\\!,

0,

the XY plane; :>

V:::E.:'i1 J?AR )

Dt

1,

C

PLk~B

CEOME-~RY

). C0RVE$

CRPCORD.

1. C, 0, 0, 1, G, 0

:>

capCORD

(4) Define element group 1 using the TRL'SS2D element formulation with etastoplastic behavior PROPS 2'rS ). EGROH? GROUP, 1, 'fRUSS2C, 0,

0, 0,

C, 1, 0, 0

(5) Define real consta!1ts for truss element: A = 1.0 in::!; PRO?SE':'S ). ?CC:TST RCONS?, 1, k, I, ,;. 1, 0, ::::, 0

(6) Define elastic modulus for truss elernem base OIl 1 in length and area I in' (k = AEIL = 12,346 Iblin). Also define yield 'tress as 15.000 Ibl mZ and tangem modulus as tE~7 Ib/tn 2: ?ROPSETS )

\rIE'd

~

1

(3) Generate a curve from 0, 0, 0, to 1, 0, 0

5051

Solve Example 7.1 uSi3g the program COSMOS

to

G.

56.7'~S

Solution: The analysis is perfonned using a single truss element with one cO:1cemrated mass element. The following commands are implemented: (I) Set view

) GRID

Z,

-9.4756

..

NONLINEAR STRUCTURAL RESPONSE USING COSMOS

Example 7.3

, Yielding/""

.72Y1

,

oe·::;

E1utic (~cnse

1/

01--~+---4---~--+---'.;:+--~---+--~,.c.+- .. ~ Time 1.8 (~cJ 0.4 e.8 0.6

::J~s"'t.

,





+

1':;-;£

t 20D

CU':?'..:'..' ,,;;:5(,.':"75.

1.CGD

~;//.-

30~

,

.

MPROP

t1J?ROP, 1. EX, ~2346 MP?.CP, 1, SIGYLC, 15000, ETP,_N, 1E-7

226

,\onlinear Structural Response

Structures Modeled as a Single-Oegree-of-Freedcm System

227

(15) Request response at node 2:

(7) Generate one truss element along curve 1:

:CNRES?

(8) Define element group 2 using the MASS element formulation:

(16) Set analysis to start at

I

= 0 and

end at l -= '1 sec, with n time increment

of 0.05 sec: PROPSETS EG~OU?,

j

2,

EGROUP K~SS,

0, 0, O. O. 0, 0, 0

LOADS-BC > LOAD_OPS > l'IHES TIMES, 0, 2, .OS

(9) Define real constant for mass element: m ;:;;; 200 !b· sec:? lin:

PROP$ETS ReOl'1ST,

2.

~

RCONST

I

2,

7, 200, 0,

1,

0, 0,

0,

0I

(7) Set the options for [he nonlinear dynamic analysis using default values for toierance parameters and run analysls:

0

t>.NALYSIS ~ NOKL::NE:A::\ > A~K0NLINS:l-_)=-

~O:.qCES

)

FPT

(14) Define Rayleigb damping coefficients [C = ak + pm]. C" = 2(km),n = 3143 10' sec lin; C 0.087 (C,,) = 273.4 lb· sec lin: Using a= 0.01, fJ [273.4-(0.01)(12346)11200 0.75 ANALiSIS-NONSINEA:R NL-ROf..Hi? NL_RDAMP, 1, 0.01, 0.75

C:S?LAY ~ XY_?LOTS ~ ACTXYPOST ACTXYPOST, 1, TH!E, UX. 2, 12 • .\., DISPLAY XY~?=-O'T'S:> XYPLO'!' XYPLO?, 1

v,

2N

(19) Activate XY plot information for X velocity at node 2 as a functio;i of lirr:e, and plot the veiociry vs. time for ;iode 2:

HZ

d~rection:

CuRC2F,TIME.

0,

R_NOtt,;L'::~SAR

(11) Merge nodes:

X

Y,

,;';"lALYS!S > NOl'lLINEAR

2, 2,. 1

LOADS-Be )

N,

0.001, 0.01. 0

DISPwAY > XY P~OTS } l-CTXiPOST ACTXYPOST, 1, TIME, VX, 2, 12, .\., DIS!?LAY > XY ... PLor{~S } xY2LOT XY?LCYr, 1

0,

2N

(20) Activare XY plo~ informatio:l for X acceleration at node 2 as a function of time, and plOl the acceleration vs. ~~me for :lode 2: DISPL.!;'Y ) XV_PLOTS } ACTXYf'O$T ,l:!,C'rXYPOS'L L 'rIME, AX, 2, 12, 1, DISP.!:.AY XY ...,?LCTS > XYP:"'OI' x YPL01' , 1

Figure 7.8 shows the response for Example 7.3 in (b) velocity, and (c) aeeelerarion.

0,

[elmS

2N

of (:1)

djspJaceJ:len~,

Nonlinear Structural Response

Structures Modeled as a Single-Degree-ol-Freedom System

228

~

i

!

[

,I

I

u

I

I

x

I

I

2

II

,

1.5 I

....r,

,,

I

I '\ I I \!

!

;

I/bl II

'~ ___ ~ '\

/ -f~ I

~

!

I

i-~t

I

I

, .4

.

I

,

I

I I~', .8

-

=~

I

!

I

~

[

'ti

I I

,,

I

I

!

:r , /

'.5

lit

i

r-

J.5

,i

,

.

~

1.4

I

48

I

!

12 16

/

,

A X

-! 5

·31

-qs

-5, -go

:?

I

I .8

1.2

I 64 I

i ,

,

EJ

,

OJ.2

1.8

i I

,, ,

,i ~ I, ~ i

-~r-

1

,,

~"KI ~

I

. I

I I

I

,

, ,

I

.4 !;!

i

! I I

.5

I

\

\

!

II I , ,

r-

1\ ,i

I

I .2

2N

\

II

i

,I

? '\

,7

i

I~

I

7

! i

i I

)

I

I

I

[

,I

229

I ,I I.,

I, ! I .6

\ II ,

1-1, 1.8

T!HE

,,' i

i

j

I

./

/ i

v

~

I

I

" '\ ,

I

I

-4

I

-8

,

L-

Y

I \ I \

II

I

,

, .8

I

,

,

, .4

'.6

/ , .2

I

l

\

I

IT

!

"-...

I '"I

I

I ! \ I \ I/ \

,

-, -co

i

i\

/

x

Fig. 7 ..Be

,

,

iol

\

I

I ! 1.\

i l.6

I I

I I

\

I

1.8

T!HE Ib)

F;g.7.8

7.10

Response for Example 7.3. (a) Displacement; (b) Velocity; (c) Accelenuion.

motion of an eanhquake or the effects of nuclear explosion. In these cases, it is not realistic [0 assume that the structure will remain linearly elastic and it is then necessary [0 design the Structure to withstand deformation beyond the elastic limiL The simplest and most accepted assumption for the design beyond the elastic limit is [0 assume an elaswpi (d is of the fo~ given by eG:. (8,5) wIth the constants of integration determined from known values of displacement and velocity calculated from eg. (8.lOa) at time 1 ld_ The final expression for the response is then given by ttl _ ~~.;:..- COS 1}-- Sin

)'

)'.s:

2 7T

I'

-I

{i (

~ -

1.1

\1'

.for

{:.?:: t.1

(S.IOb)

It may be seen from eq, (S.lO) that the response in terms of yly" is a function of the ratio of the pulse duration 10 the natural perIod of the system (talT) and of time expressed as liT. Hence for any fixed value of the parameter t,Jr, we can obtain the maximum response from eq. (8.1 0). The plot in Fig. 8.3 of these maximum values as a function of I"IT is the response spectrum for the half~siflusoidal force duration considered in thjs case. It can be seen from the response s;>ectrum in Fig. 8.3 that lhe maximum vaiue of the response (amplification factor) }'IY~I = 1.76 occurs for this particular pulse when (uIT~ 0.8. Owing to the simplicity of the ip.put force. it was possible in this case to obtain a closed soiution and to plot the response spectrum in teons of dimen~ ;:;ion!ess ratios, thus making this plot valid for any impulsive force described by one-half of the sine cycle. However, in general, for an arbitrary input load, we cannot expect to obtain such a general plot of the response spectrum and we normally have to be satisfied with the response spectrum plotted for a completely specified input excitation,

fj--klY -y,~ t}-+ C{y~ Yl1~,-_ _ __ Fig. 8A (a} Damped si'TIple oscillator subjected to support excit

, 1

0,02

O.OS 0.1

tal

\8 \J]

V

!

i

,

·5

2O

i

___ M

o. 1

1.S ~

!

o.2 -

0.05

I

\

i i

!

i

! '~I i '

I

0.2

Fig.8.8 Response spectra for elastic system for :he t940 EI Centro earthquake {f:oom Blume et al. 1961.)

Natural frequency, I:pS

Fig. 8.6 Displacement response spectrum for elastk system, subjected to the ground motion of 1940 El Cemro earthquake, (from Design of Multisfory Reinforced Building for Earthquake MOlions by J. A. Blum, N. M. Newmark, and L R Corning, Portland

For constant values of So, eq, (8.20) is the equation of a straight !jne of iog versus log/with a slope of 45", Analogously, from eq. (8,19)

To demonstrate the construction of a tripartite diagram such as the one of Fig, 8.8, we write eq. (8.19) in terms of the natural frequency / jn cycles per second (cps) and take the logarithm of the tenus, so that

S, log S,

coSo = 2'iT/SD log! ~ Jog (2OTSD)

(8.20)

Y/9 (

c

_:~h~I\~httl~tJ.i~i~/v.,~I/Wii~~f,\JlrM~'N*/~"~ ~q o

11

:

5

"I,

10

'~'~-!'~'~~"-';;:'~'~'.L-J....-.i.~~'~-'-Tj(m:.~i: IS

M

25

Fig. 8.7 Ground acceleration record for El Centro, California earthquake of May 1g, 1940 north-south compOnent.

S~

5,,= S" ~~_ -

Cement Association 1961.)

"i

241

~-'

I

i

~

'"~

I

i

,

I

i .~

I

Response Spectra

Jog S"

w

2r.f

So

- log! ~ Jog27r

(8.21)

For a constant value of SI>' eq. (S.2i) is the equation of a straight Hne of log 5 . ."" '~ -. .... '

STl FF 50 J LS «200 FT) - 31

~ECOROS:

PERI DO, sec

Fig. 8.11 Average acce),i:f0

" "" ""

20 .31

25. SS

1.41 e.54 jJ.

$:

H,

"

n n.n

"

26. $I' lc. ;)6

';;!i .02

""

6'1.25

1S.17

LaO

,

1S

127.4()

2.41

n.n

lU_S7

.t~

25.H

)16.111

".H

"

"

~:).

H

&1.

$~

]. CO

2

L.50

1.17

H, U

H9.~()

) .oc

O. 7~

1 J. 96

261.1.\

J. 'jU

O.

$)

1l.H

~ ...

coo

\lA, 0.31

lo,a

15'1.

9.1>&

lSS.

$.0

to.oo

".JO I). OS

294.

15.00

0.02

L,--'

" '" "

20.CO

1).l)l

". SOl

120.01'

4. ~o

1."0

,HI>

(,lOsee/SECl

1. 50

S _flO

"

Sf>E:(:T

1)'1!)

O.

somewhat different than those obtair:ed from the respo:lse spectrum of the EI Centro earthouake of J940, Also, if we compare these rest.;lls ror the elastoplaSlic benavi~r Wilh the results in Example 8.1 for the elastic slructure, we observe that the maximum relative. disp!aceme:lt has essentially the same magdtude whereas the acceleration and the reli'ltive pseuGovelocity are appreciably Jess. This obSe;vtllion is in general true for «roy structure when inelas~k response is compared with the response based On elastic behavior.

G

lJ,05

0.20

As can be seen, these spectfJl values based On the design spectrl·.m are

,

·

, · ," ", · "

nlEQU!':1ICY INCREN;;;l4i IC.f.".l

\. )!i

""

liS .0.4

I

III

I

N

~

~.

TABLE 8.3

Digitized Values of the Accel~ration Recorded 10r the First Ten Seconds for the EI Centro Earthquake 011940 ~~"

.~~-~~

Ace,

Time (sec)

(ace. g ')

0.0000

O.OlOS

0.2210 0.3740 06230 0.7890 0.9410 10760 1.3840 1.5090 1.8550 2.2150 2.4500 2.7080 3.0680 3.3860 3.6680 4.0140 4.3140 4.6650 5.0390 5.3020 5.5100 5.8000

0.0189 0.0200 0.0094 - 0.0387 - 0.0402 .• 0.0381 - 0.0828 - 0.1080 0,1428 0.2952 0.2865 0.1087 0.0520 0.1927 0.0365 0.0227 0.1762 - 0.2045 0.0301 0.1290 -0.1021 - 0.0050

TABLE 8.3

Time (sec)

Ace.

Time

(ace. g ')

(se.c)

0.0420 0.2630 0.4290 0.6650 0.8290 0.9610 10940 1.4120 1.5370 1.8800 2.2700 2.5190 2.7690 3.1290 34190 3.7380 4.0560 44160 4.7560 5.1080 5.3300 5.6060 5.8090

0.0020 0.0001 .. 0.0237 0.0138 - 0.0568 0.0603 ~. 0.0429 - 0.0828 - 0.1280 0.l777 0.2634 - 0.0469 - 0.0325 0.1547 - 0.0937 -0.0736 - 0.0435 0.1460 0.060S 0.2183 0.1089 0.0141 - 0.0275

0.0970 0.2910 0.4710 0.7200 0.8720 09970 1.1680 14400 1.6280 1.9240 2.3200 2.5750 2.R930 3.2120 3.5300 3.8350 4.1060 4.4710 4.S31O 5.1990 5.3430 5.6900 5.S690

Ace. (ace. g ')

Time (sec)

-

- ------

Time (sec)

Ace. (ace. g ')

0.0159 0.0059 00076 0.0088 0.0232 - 0.0789 0.OR97 - 0.0945 0.1144 0.2610 - 0.2984 0.1516 0.1033 0.0065 0.1708 0.031 I il.0216 - 0.0047 - 0.2733 0.0267 0.0239 -0.)949 - 0.0573

0.1610 0.3320 0.5810 0.7400 0.9020

Ace, (ace, 8 ')

Time (sec)

]0660 1.3150 1.4810 1.7030 2.0070 2.3950 2.6520 2.9760 3.2530 3.5990 3.904il 42:>20 4.6180 4.9700 5.2330 5.4540 5.7730 5.8830

Ace. (ace. g '; ~

00001 0.0012 0.0425 - 0.0256 - 0.0343 - 0.0666 0.1696 0.0885 0.2355 - 0.3194 0.0054 0.2077 - 0.0803 0.2060 0.0359 0.1833 - 0,1972 0.2572 0.1779 0.1252 0.1723 0.2420 .- 0.0327

(continuation} ~~~-.---

~

v

~

Time

Ace.

Time

(sec)

(ace. g ')

(sec)

5.9250 61320 62290 6.3820 6.5200 6.6030 67280 0.8520 71210 72260 'i4250 76000 7.7520 7.9600 8.1260 8.2780 85330 8.8180 8.9560 9.1500 9.4410 9.8150 )0.0200 )0.1500

0.0216 0.0014 - 0.0381 - 0.0162 0.0043 - 0.0170 0.0009 0.0022 0.0078 0.0576 0.0186 - 0.0628 - 0.0054 - 0.0140 0.0260 0.Q305 - 0.0344 .. 0.0028 OJ 849 0.1246 - 0.1657 - 0.0881 - 00713 0.0024

5.9800 6.1740 6.2790 6.4090 65340 6.6450 6.7490 6.9080 7.1430 7,2950 74610 76;10 7.7940 7.98'10 8.1660 8.3340 8.5960 8.8600 9.0530

92530 9.5100 9.8980 10.0500 10.1900

0.0108 0.0493 0.0207 0.0200 00040 00373 0.0288 0.0092 - 0.0277 - 0.0492 - 0.2530 0.0280 0.0603 - 0.0056 - 0.0335 0.D2.46 -0.0104 0.0233 0.1260 - 00328 0.0419 0.0064 .. 0.0448 0.0510

6.0130 6.1880

6.31.60 6.4590 6.5620 6.6860 6.7690 6.9910 7.1490 7.3700 7.5250 7.6690 7.8350 8.0010 8.1950 8.4030 8.6380 8.8820

9,0950 9.2890 9.6350 9.9390 10.0800

0.0235 0.0149 - 0.0058 - 0.1760 .~ 0.0099 0.0457 0.0016 - 00996 0'{1026 0.0297 0.0347 0.0196 - 0.0357 0.0222 - 0.0128 0,0347 - 0.0260 0.0261 0.0320 - 0.0451 - 0.0936 - 0.0006 .. 0.0221

6.0850 6.1980 6.3680 6.4780 6.5750 6.7140 6.8110 7.0740 7.1710 7.4060 7.5720 7.6910 7.8770 8JI700 8.2230 8.4580 8.7350 8.9150 9.1230 9.4270 9,7040 9.9950 10.1000

Ace.

(ace. g ') -0.0665 - 0 0200 - 0.0603 0.0033 0.0017 0.0385 0.0113 0.0360 0.0272 00109 0.0036 0.0068 0.0716 00468 0.066) - 0.0369 0.1534 - 0.0022 0.0955 0.)301 0.0816 0.0586 0.0093

------_...

~-----

---

---.-.,.=.~~.

Structures Mode!ed as a Single-Oegree-o1-Frcedom System

260

8.9

Response Speclra

RESPONSE SPECTRA USING COSMOS

"'f-:'- r

JIS

Example 8.4 Use the Program COSMOS to genenHe response spectra for :.elastic system subjected to the 1940 EI Centro earhqunke. Assume damping equal (Q 10% of the crillcsl dampir.g.

I

I

I

..

,,-1 \ I (\ , i \/j

I

'\

i

I

....

......

Ii .)2

,1·

.$ .IS

&.tB

, -&

i

'

i

,

I

-9.15

,

<

, I,

,..

,I

I I

:,VY

...............

,

,

i ,

SL:i

I

15 87.:;

I 12 • ~

1?S

[0

the XY plane: >

VIEW

(2) Define the XY plane at Z ~ 0: G~OHE~RY

~ GRID PLANE, Z, 0, 1

>

PLANE

(3) Generate, a curve from 0, 0, 0, to 1. 0, 0,". i

GEOMETRY ) CURVES ) CRPCORD CRPCORD, 1, 0, 0, 0, 1, 0,

~

.......

•------J

i

II

"

I

1'1

Solution: The analysis is performed ~!iing n single spri:Jg element wHh one concentrated mass element The following commands are implemented in COSMOS:

I

• E! '- vr~'._",

I

..

:

Fig. S.22 Ac::elel:likm response spcctwrn ;){ Ihe base of an e!;istic syste:n subjected 10 ground nlOllOO of 1940 E! Cenlro carthquilke,

,

!

i ~

:

DISP::.. AY j VIEW_PAR VIEi' l-CCR 1, 1. 2, 1, 1

PROPSS'I'S '

EGROUP

MASS,

0,

J,

(8) Define renl constant for mass elemenr: m

=l

2,

?ROPSETS

0,

0,

0,

C,

)

POST~DYN

ANALYSIS ) PD~ATYPE,

(7) Define element group 2 using the MASS elemenr formulation: EGROUP,

263

Response Speclra

2.

1,

2000,

?D_A?YPE

Ot

.01,

0,

0.5,

C.25,

0

(15) Prepare file ELCENTRO,GEO in a formal suilable for input in COSMOS, cont

.J10S,

.o,n .002, .0)7, .015'S,

.H1.

-.0001

.221,

.018S . . 263,

.0001, .291, .0059, .3J2,

S,

. }?4,

.02,

0::37,

PD_CUKDEF,

(9) Generate one mass element at point 2:

G,

1,

PD_C::K::n::? . 1. S, PD_CCRDEF, PD_CURDEfl,

"

.6:n,

13,

"P • .

.42S,

0;)94,

769, -,0331,

,665,

,471, .0076, .Sill,

.013S,

.72. -.00S6,

-.O~12

.0425

.71, -.0256

,S29. -.0568 . . 312. -.0232 • . 902,

-.tlJ'D

MESHING } PAR.l..,"LMESH

:1_?T, 2, 2,

1

(16) Define dynamic forcing function as time-dependent hase acceleration in the X direction, Read forCing function curve from file "ELCENTRO.GEO":

(10) Merge nodes: MESiUNG ,. NODES, 1"mERGE NMERGE. 1, 3, L 0.0001,

o.

C,

(l:) Apply constraints in all degrees of freedom at node I, and aU degrees

of freedom except UX at node 2:

(l2) Set the options for the frequency analysis to extract ~ frcquency using the Subspace Iteration Method with .t: maximum of 16 jterations. and [un tte freqJency analysis:

0,

0,

U'l'ILI'1'Y ) "!?ILS

l',

386.4, 0,

0,

0

DISPLAY )- XY._PL07S :'> ACTXYPRE l, ;., TIt-!£, L 12, L DISPLAY X':CPLO?S) XYPLOT ACTXYPRE,

XYP:"OT,

p~ALysrs

C,

~

PD~C()R?YP

(17) Activate XY plot information for acceleration input at node 1 as a function of time and generate plot (see Fig. 8.20:

DPT, 2, UY, 0, 2, 1, UZ, RX, RY, HZ

1E--05,

0,

1£-06,

0

ANALYSIS > FREQ/BUCK ) R_FREQUSKCY

CONTRCL

>

l, 0, 1

1?l.LE, ELeEN'PRO.GEO, L 1, 1, ::. ANALYSIS > POST_DYN > PD_BECIT

PO_BASE,

:OADS-BC ) S'I'RUCTURAL ;. DISP::"'Yl.JTS ,. DP}' DPT, 1, AL, 0, 1, .:

> FREQ/B~CK > A_FREQUENCY A_FREQUENCY, L S, 15. 0, 0, 0, 0,

POST_D'iN ,. ::URVF::S

ANALYSIS> PD_CURTYP,

0

R_FREQUENCY

0

1

(18) Define modal damping for mode 1 as. 10% of critical damping (0.10): ANA!. 'ISIS ~ ?OST. __ DYN > Pi:LDhliP;'GAP ) PD_MDAN?, 1, 1, 1, .1

PD_HDA.."1P

(3) LiSl the natural frequency of the system:

(l9) Request response at nodes 1 and 2: RES0L'l'S )

LIST) FREQLIST

FREQUENCY

FREQUSNCY

FREQUENCY

PERIOD

imMBER

(RJ:i.D/SEC)

{CYC!~ES!SEC)

(SECONDS)

C 1000000E+01

0.1591549E+00

0.6283185£+01

(14) Define analysis type as modal time history using one natural frequency, 2000 time steps starting at 1=0 with a time increment of 0.01 sec;

ANA~YSIS

>

PD...:.Nf:£sP,

POS'I'_:nfN ) OUTPUT) PD_!'lRESP 1,

1,

2,

0

(20) Execute modal lime history analysis: ANALYSIS

R_DYNM{IC

;>

POST_DYN > R_DYN1\MIC

264

Response Spectra

Stn.lclures Modeled as a Single-Degree-o!-Freedorn System

(21) Define analysis type as response spectrum generation using one natural frequency, starting at 0.314 fad/sec (0.05 Hz). ending at :25.7 TJdJsec (20 Hz), linear scale, with 200 points in frequency range, 10% of critical damping, for node 1 in the X direction:

Use an editor to display the output file (EX8w4.0UT) and print the firsl nine lines of the response spt"~[ra: RISSfOl'lS~

SIT.c7I\lJIf

~N

ANALYSIS ,

?OST_DYN )

3,

PD.J,':''::.'PE. Q,

Q,

1,

PD~ATYPE

.314,

125.7,

1,

1,

200 ..

0.1,

I,

0,

0, 0

'OR

!)~l\gCT;OI

I"REQ\jH1CY fREQUENCY

SEC!

>

NOD:;

,

POST~CYN

) R_DYNAM:C

R_DYN~J"GC

(23) Activate XY plot information for X acceleration at ~?de 1 as a fUfictlor: of frequency (response spectrum), and generate plo~ (see Fig. 8.20):

PElno!)

(S!tCl

D::;:SPLAY

1, FREQ, AX,

1, 12, 1,

0,

The following table reproduces the first len lines of the list;ng shown in the screen:

, 3

5

6 7

8

9 10

'tRUE

AU$

Rtt.

m

Dr$p

\itt.

;£c

VEL

Ace

'tRUE

".:n10E.n~

Q.H11t,01 1.20HE.C)

'.1J7S)SE.ry,

C.';6!HE.Oj!

.Q .2H2~JE.'.t

O.1"f)f..:j1

~.,>~;.;:.c'

J. ,tf"lUE.t2

Q_::tlln;,~:

-0. BO~S;;t.1)2 -$.191SllalS.O'

:';.1Il00E.01 1).1592t,00 C.i.10jE.Ol

:.lun~E.;JZ

C.~~6?Ht.Ol

-O.tUi1:;E.Ol .Q.\%Oilf.llt ·0.))1»9E.01

c. ts.,£.(l. 0.1"(>$E.1)0 t.)SSlt.nl

b.lsn48E4': .n.1)a%tE,ol ·jj_il~J1n:.;n

C.~2SH(f.'n

~_:17H-SE'01

C.no{,>ol

O.)S081'.\j0 O.zaOSE.:l -O.! i'l41St.(>?

-O.2S~SHE.(,:

O.nH~.Oi

8.4";\

;E'O~

O.HnB'Cl

o.a~5L9£.Cl

O_l)~S?lE.Ol

O.29snBl:'Q -0,]2505£.02 -0. );]&3):E,02 O.1?SSOGE.C2 ·O.67501/E.()2 Q.)HUle,l}"l -~ _6n,es!':.o)

1),}HH~(t\

O.5~Ht.CI}

O.HU!:.O:

". ,'64,1,.0:

O.'n~9J!t.h

-().S;;$MS!:.Ol

-q.lO~lH!t.Q:;

OAC94C.Ol ",o6I1t.un (l.lS)$E.¢!

o.n~9'n>~,

J.;1S2{$ XY_PLOTS > ,t..CTXYPOST

ACTXYPOST,

,

01.AC/SEC) iCYCt.Il:/

(22) Execute response spectrum generation: ANALySIS

265

FR8Qt:ENCY (BAD/SEC)

SPECTRAL

NO::;:E 1 U.!1/ sec')

3 .140e-Ol 9 441e-01 1. aOOe-Ol 1.S74e-Ol 2.204e-01 2.834e-Ol 3.464e--01 4.09Se-01 4.72Se-Ol S_3SSe·-Ol

(25) LIst the response spectrum from the output file: CONTROL ) UTILITY :- SySTEM

ACe.

2 .390e+00 1 .353e+01 1 .313e+01 3 .282e+OJ 5, 9S5e+01

6, 925 •.,:+01 6 ,085e+Ol

8 .17ge+01 9 .2S6e401 9.413e+01

SUMMARY

Response spectra are pl0iS tha~ give the maximum response for a single~ degree-of-freedom system subjecied LO a specified excitation, The construction of these ploes requires tbe solution of single-degree-of~f:-eedom systems for a sequence of values of the natural frequency and of the damping ratio in the range of interest. Every so;ut:on provides only one point (the maximum value) of the response spectrum. In solving the single~degree-of-freedom systems, use is made of Duhamel's integrak or of t!:e direct method (Chapter 4) for elastic systems and of the step~by~step linear accelerJ.£ion method for ir,elastk behavior (Chapter 7). Since a la:-ge number of systems must be analyzed to fully plot each response spectmm, the task is Jengthy and time consuming eveu with the use of a computer. However, once these curves 3re constructed and are availabJe for Lhe excitation' of interest, the analysis for the design of sl.rlJctures subjected to dynamic loading is reduced to a simpJe calculation of the natural frequency of the system and the use of the response spectrum. In the following chapters dealing (hat structures which are modeled as systems with many degrees of freedom, it will be shown that the dynamic analysis of 3 system with Jt degrees of freedom can be transformed to the problem of solving n systems in which ead1 one 1S a single·degree~of·freedom system. ConsequentlY, this transformation ex.tends the usefulness of response spectra for single-degree-of-freedom systems to the solution of systems of any number of degrees of freedom. The reader should thus reaiize the full importance of a thorough understanding and mastery of the cor.cepts and methods of solutions for single~degree-of~

:~

i

l

I

I

1

Response SpeClr3

Sln.:ctures Moceled a,s a Singha·Oegree·o(·Freedom System 8.4

freecom SVSlems, since tbese S'lme. methods "fE also llpplicabk to systems of many deg;ees of freedom afler {he prob~eflj has been transformed to if\depcn~ dent s;nglt>degree·of-rreedom systems,

8.5

267

Derennine the lll~iiCirnum S((CSS in (!.e COhOllllS or the frame of Proble:':) g.] The frame sbowll ;;~ Fig. Pil.t i;; sub;ee\O!d to Ihe excll:!.liol\ prowLced by the El Centro eanhqu~ke of i940. A~$ume 10% camping ano from. lhe appropnllle chnrt de:ermine lilt specllal v;)ttJe~ fOf di~;):aCtmeIH, velocity. and :,cceleftIlJon . ..,\s· $.urne e)-..Slic b.;ll,,"VlOr.

8.G

Repeal Pmbkm 8 .5 using thl: b;\sic Gcsign spCriofl. Neglect daruping.

Ro::pe(l.l Problem 8"7

8.9

D::!ermin;: the force lr;,Hlsmiul.'c Pwblem 8.8,

8.10

dlt: system lW.$ 10% Df !he ClltiC:!\ tbmping,.

(0

the rOundaliOl1 (or /he sySleti, .,::,,,,1 y-z.eJ In

C(}ns~der Ihe '1jJring-m>lss .'YSIClll of Problem 8.7 ,\I\J tiS$UI11¢ lila! the spring dement follOws, :;,;1.1 tlun c!aslOp!:I~!ic b;:hnvior with a l\l.J.:;.iwutll valu~ fOr (he f~SIOfL!g fo(c~ ill tc.cSIOIl '1

", i

l

(b)

Fig. 9.1 Single-bay model representation of a shear building.

.

,~

I

i

,

rYz~ Fl

:()--+! *!Y;

.J I")

Fig. 9.3 Multimass spring model

repreSenla~fOn

of a shCO

0,OGDO;;:~0()

c. "c:::n~JC

.

1)

C JOCJS~OO 0. Jih;)OS. 30

,

c.:sn~>o~

1.C~O

0;;)

o .1"':lH:~C$

:.;;00

!SCl.GO

(].:S9~r->-O$

l.CO::

:$:). all

O.l':l~:£~j~

l.OCO

,C0

~eQ,

M'j~S·CO

£.Sq59l.

)

15CX) Jb/h

F1irl

Y

)2'

2000 IbJh Y

30'

"""w

.a &

.a;,&

J

Fig. P9.3.

9.4

Write the differential equation of motion using the stiffness formulation for (he shear building in Fig. P9.2. Model the structure by a

multimass~spring

system.

9.5

Write the differential equation for the motion of the shear building In Fig. P9,],

9.6

The three-story shear building in Fig. P9.5 IS subjected t;::; a foundation motion which is given as an accelennion function Y.(r). Obtain the stiffness differemjllJ: equation of motion. Ex?ress the disj:llacement of the floors relative to the foun~ dllticn displacement (i.e .• u, "" )" Yf)

9.9

Model \he stmclure as a shear column with lumped masses as the floor levels.

9.7

Generalize the results of Problem 9.6 and obtain the equations of motion for a shear building of n. slories.

Use Program 7 Ie de,ermine the stiffness and mass matrices for (he she;:;r buliding shown in Fig. P9.9. The modulus of elasticity is f. "" 30 X !OI> pst. Neglec( the p. Ll dfecL

9.10

Solve Problem 9.9 inclUding lhe P-::l

9.11

Use Program 7 to de!er:nlne the stiffness i:.nd mass matrices for a uni form (en-srory shear bui!ding: m which the mass :u each level it: 20 {lb· sec'llin}; ~he imerstory heigh, is 12 ft; rhe mcdulus of elasticity is 3.0 X 106 psi; the ~()tol 9 flexur

~~

Or

l/w'{a} = [DJ {a}

(107)

"t

I.)

Fig. 10.1 Two~story shear building for Example 10.1.

290

Free Vibration of a Shear SUi!ding

S1Jucture5 Modeled as Shear Buildings

291

In the usual manner, these equatjons of motion are solved for free vibration by substituting

YI = Cl; sin (WI

a)

Y1 =

a)

a2

sin (M -

(l0.1l)

for the displacements and - a,w'Y. sin (wt - a)

YI (h;

h Fig. 10,2 Mult!mass~spring moce! for a two-story sheor building. (u) Model (b) frcc

=

- G.7.W2

sin (

o.

= 147.9

Q:;.

= 1,24

{4>} T = {LOO

1.24} yields

Free Vibratior. of a Shear BUi!dir.[,;

Structures Modeled as Shear Buildings

300

This value of fJ} is virtuaily equal to the solution w~ = i40 calculated in. Example 10.1. Anothe:- popular iterative method to solve an eigenproblern, that is, for structural dynamics, to calculate nmura! frequencies and modal shapes, is. [he 'subspace ireration method. Computer programs for the solution of an eigenproblem using the Jacobi method or the s:.lbspace iteration method are included as alternative methods in [he set of educat.:onal programs accompanYlng thi.s textbook. Commercial computer program for the solurion of an eigenproblem problem such as COSMOS usually provide severa! other methods in addition to the Jacobi and subspace iteration method. The computer program COSMOS gives the user the option of solving for natural frequencies and modal shapes by any of the following methods·, Jacobi, subspace iteratio:l, Lanczos, inverse iteration power, and ::omptex eigenvalue analysis.

Input Data and Output Results

O. GOCoo,:_co

O~OC');:.'N;

;},' S;)'J'Jt>C;)

C.

:'.10000a.01

:;,MtlST t 1, 3, 1, 1, 30QO

LCADS~BC

RCONS'1'

j

tCC~,

2,

2,

CONTROL

?

.f..CTSE?,

Re,

MESHING

:>

::-CCFL

3,

1, 2,

3

M_CR

?Af ?AR3M.MSSH fo M_CR l, 2, 1,

0

) STRCC":'URAL ) D!SPLr-0-!T$ > DP'I'

DP7, 1, AL,

ANALYSIS

{7) Generate spring and mass elements with appropriate e!ement gro~p and real constants actlv

XE':SHI:JG ) NO::JES ) NCOM?RSSS NCOV.PRESS, 1, 0:

?RO?S£'1'S

,

1

(8) Merge find compress nodes:

0

(5) Defi:lt real conSi.ftnts for spring eiemems: k = 50,000 Ib lin. 40,000 Iblil1, and 30,000 Ib/in:

?RO?SE'1'S

P.\RP."M_~lESn

::;,

> ACTLVE: ACTSE:7 Po.C'!'SE'I', Re. 5 !1ESHING > PAR.A'!'LM£SH M_I?'T·, 4., 4, 1

PROPS2TS ' EGR00? SCRO~P,

~

CONTR.OL

(4) Define element group I using the SPRING element formulmion with two- nodes and element group 2 using (he MASS etemem fomlUlation:

EGROUf:'. 2, HASS,

3,

I

(12) List (he natural frequencies of the system: RESULTS > LIST ) FREQLXS':'

Frequency l 2 J

Frequen.cy (Raciisec)

Ft'eC;:".lency

Period

(cycle/sec)

(seconds)

2.S1B60e+Ol

4.00848ef-OO

::.49471e-Ol

S.€9871e .... Ol 8.J6760e,..Cl

9:.Q697ge+OO 1.J3174e+Ol

1.10256e-01 7,50595e-02

304

Free Vl!Jratlon of a Shear Building

Structures Modeled as Shear Buildings

constit~tes the ~odal matrix [

LIST

>

DISLIS'r

, 1, 1, 1, , , 1, 0 2, c, 1, 4, ", 0 , DISL!ST, 3, 1, l, " 1. 0

DIS1,:31', D:;::S;:'IS'l',

)~o,-;:Q

,

1 . SBe-:>"

?;':

',>?

~.(j0e~()t'

'J. :;:;"" Oi}

,

;) .::::;e~v7

:}~"'~oc

S.5s",~i)~

:>;e'(;'

0. JGe~";; ,;;;;",~:)C

:).00>[;),,, Co:.

,.

OCe~C,;

0.00'11-00 ,'!o:le sic"p'"

,w

0%

;;":'

:';1.

rtx

0.309-00

C _0Cq~CS

O. CDe-JC

(; . (;C",,,,C;)

j.GJe w ;)0

,

-3.66.;\-:)2

C. G()e .. Ct

·?0]f~C()

c.rte-C]

;;. DOe.O:::

C .C~",·CO

.,

0.:;':2-00

:.OiJe~OC

G. aCe'CU

C.CC?-Co

C,;)0e~:·!)

Ooe~DO

G.00".]C

C. ;;C",.JJ

c. :)O~~~~

.). :;Q!!-_()J

L}i?e~:::

}%,n

,

L

;=1,2, .. ,n

where the normalized modal vec~ors {1->00

().

305

{¢};(M]{¢h=O

for i

{¢h;M]{¢},=!

fori=J

{d>};[X]{¢};=O

for iFj

ar.d

l\Z C~",-Oj

{¢};[X] {¢}j

wi

for i=j

The above relations are equivalent to

10.6

rn = - 0.0924

1281/(1) in which /(1) ~ 1 - III" for IS; 0.1 "nd fU) ~ 0 for I> 0.1. The maximum values for Zt and z~ are (hen obtained from available spec~ral charts such .as the one shown in 4.5. For [his exam?le, I,

= lO,OOO (l

F, (I) = 20,000 (l - I lid) Ib,

Id

0.1

TJ

O.l91

- = - - .. =0524

1>"

- Ill d ) Ib

0.188

and

in which the modal natural periods are calculated as

The forces acting on the frame which are shown in Fig, 1 LI(b) may be expressed by

F, (I)

0.1

T,

211' T\ = - ..- = 0.532 sec w, From

and

2IT T, =-- = 0.191 sec

w,

4.5, we obtain

for I s; 0.1 sec 0.590

in which

fd

= 0. 1 sec and F,(t) = F,(I) =0,

fOfi>O,1 sec

122

316

Structures Modeled as Shear Buildings

Forced Motion cf Shear Buildings

where in this cas.e the static deflections are calculated as

Fo.. z!q

=

?'770

The max.:mum shear force

Vi!

at story i corresponding to mode j [s given by

- 1281

(l U9)

_ . _ = -1.18

16.3 wf =----= 140 '

1082.41

where Z:;,.,..u is tl':e m~mum modal response, (,; - OF'

U • . OJ,

r~:'l'Ii:::;flA1'!Otj

C~C

GRAV!1'h1'IOtl';L HiQ£;{

force F F'J sla Wr which is i.:pplied at ~he level of the second floor. In this = Fo s:n Wf become case eqs. (l L I) With FI (I) 0 and

r;

(11.43) Fo: the steady-sli).te response we seek a solution of the form I)

OQOO"v(lO~

.OOM

I). tO~o

c.oooo

J.4000

{l.jCCOl

3.0000

2.000C.:HiCO

J.!OOO

v.coco

C ••WOO

:l.onno

Y::

= y.~ sin WI

(11.44)

A:rer substitution of eqs. (l1.44) inro eqs_ (11,43) and cancellation of [he commOn factor si;1 wr. we then obttin

!;:XC!1'>.rtON

(k, + k, -

m,w') Y,

k,Y,

=0 (11.45)

tv.X. V£i.OC.

H.:'>.x. Ace,

c.s-nn

?64.33

l1>'7.{10C

o. i5t\$~

~J. ~851

:'!AX.

11.4

!HS?t...

HARMONIC FORCED EXCITATION

When :he excitation, that is, (he external forces or base motion, is ha:r.uook (sine or cosine function), the analysis is quite simple and [he response can readHy be found without the use of modd analysis. Let us consider [he two'-s£Ory shear building as shown in 11.4 subjected £0 a single harmonic

which is a sysrem of two equations in two unknowns, Y I > and Yl - This system always has a un:que SOlutiOn except in [he case when the determinant formed by the coe:ficiems of [he unknowns is equal :0 zerO, The reader should remember [hat in this case the forced frequency w would equal one of the natu!'al frequencfes, since this de(erminam when equated co zero is precisely [he condition used for determining t:-te natural frequencies. In other words, unless the structure is forced to vibra(e at one of the resonam frequencies, the algebraic system 0: eqs. (l L43) has ~ unique sol:Jtion for YJ, and Y::. Example 11.5. De(ermine the steady-state respor:se of the two-Story shear LO,OOO sin 201 is appHed to [:te buildir:g of Example 10.1 when a force F?(r) second story as shown in Fig_ [lA,

328

Fo:"ced Motion of Shear Buildings

Structures Modeled as Shear Buildings

Solution: The natural frequencies for this frame were dete:mined in Example 10, I to be fiJi

= II,83

fJ)2

= 32.89 rad/sec

~(tl

~

_____

m-,'~_-;>--+

329

V,

rad/sec F;{tl

Since the forcing frequency is 20 rad/sec, the system is not at reSonance. The steady-state response is then given by solving eqs. (11 A5) for Y! and Y::. Substituting numerical values in this system of equations, we have (75,000 - 136 X 20') Y, - 44,300 Y, = 0 - 44,300 Y

+ (44,300 - 66 X 20') Y,

10,000

So[v:r.g these equations simultaneously resr.;.lts in 0.13 in

Therefore, according to eqs, 0

~

44), the

steady~state

response is (b)

y,

=

0,28 sin 20l in

Y2 = ~ 0,]3 sin 20t in

(Ans,)

Damping may be considered in the analysis by simply including dampjr,g elements in the ~nodel as is shO\~ln in Fig, 11.5 for a two-story shear building. The equations of motion which a:-e obtained by equatir.g to zero the sum of the forces in the free body diagram sbown in Fig. j L.5(c) are

:t

m,Y, + (0,

+ (k, + k,)y,

-

= F; (t) + klY2 = F, (t) - k,y2

(ll ,46)

Ie)

Fig, 11.5 (a) Damped shear Ol!ilding with harmonic load. (b) Multidegree model, eel Free body diagram,

Now, considering [be general case of applied forces of the fo!T.l given by

Real {(Fe

iF;)e'Wt}

(l1.48)

with the tacit understanding tbat oniy the real part of the eq. (11.48) is the applied force. Vile show that [he real part of the complex force in eq. (11.48)

=

Real {(F"

= F"

(11.47) we, conveniently, express such force in complex fonn as

= cos W[

is precisely the force in (11.47). Using Euler's formuia we obtain

cos

!F;)(cos

wt + Fs sin

[;J{ -7

mass~$pring

+ i sin

rut,

i sin Wi)}

(11.49)

lu(

wbet, is the expression in eq. (l1.47). Assuming that the forces FI (I) and F~(t) in eg. (J 1.46) are in the fonn giver. by eg, (11.47), we substitute eg, (l L48) into eq, (l 1.46) :0 obtain mlil

+ (c: + c:) Yl + (k + k:) Y: j

mlh - c:!J'; - k:YI

~ cJ: - k;:)'-;

(Fd - iF,I) e'i;,!

+ c:!.-y: + ki:ll = (Fd

i

iFde ''''

(11.50)

330

f"orced Motion of Shear Buildings

Slfuctures Modeled as Shear BUi!dings

The solution o~ the complex system of eqs. (11.50) will, in general, be of the form Y{i) = (Yo: +- iY,;)e''&':. Since on!y the re.al parts cf::he fcrces in eqs. (1 LSO) are applied, the sOlution is also only the real part of y (i). Then, analogously to ego (1 1.49). (1151) For the stea.dy~s~ate response of eqs. (11.50), we seek so;utlor!s 1!1 the form of

Solwiofl:

The su::,stitution of egs. (11 Yz, namely

The damping constantS are calculated as c, = a01< ,

3071b·seclin (a)

The substitution of numerical values for this example into eqs. (i J .53) gives the following system of equations: (20,600 + 15.000i) Y, - (44,300 + 8860,) Y, = 0

Y\ = Yle'~ Y2::::

331

yl/,j.;

- (44,300

(11.52)

aed the first and second derivatives of YI and

~

8860i) Y,

+ ( 17,900 + 8860£) Y, = - lQ,OOOi

(b)

The solution of this sys[em of equations is

y, = 00006814 + 0.26865; Y,

into eqs. (l: .50) and leteing Yl Yd + i Y,I and Yz ;;;;: Yc:; + i Yib reselts 1:1 the foEowing syS(em of complex algeb:-aic equfI[ions:

=

0.06309

+ O. 137775i

Therefo::e, by (he re:2.tio:1 established in eq. (j [.54), [he steady-stare response is given by

= 0.0006814 cos 201 y, (f) = - 0.06309 cos 201

y, (r)

(l1.53)

Ytt cos fur - Yt • sin

Y: (fJ

Yd cos

wr -

wt

Ys1 sin Wi

(11.54)

in wpjch YI = Yt ' l + iYs !, Y2 Yel + iYs:; is the solution of the complex equations. (11.53), The necessary calculations are better explained lhrough the use of a !1ur:1ericai example.

Example 11.6. Determine [he steady-scale response for the tWQ-story shear building of Example 11.5 in which damping is considered in me analysis (Fig. 11.5). Assume for this example that the damping constants Cl and C2 are, respectively, proportional to the magnitude of spring constants k; and k'l in which the f.acmr of proportionality, ao = 0.01.

0.137775 sin 201

which also may be wrtnen as

y,

AS aiready stated, the response is (hen ~ound 'JY solving !he complex system of equations (11,53) and considering only the real part of the solution. Hence, from eq. (11.51), we have

YI (t)

0.26865 si, 201

0.2686 sin (20t

+ 3.144) in

y,=O.1516 sin (20t+3.57])in

(Ans.)

Wher: these rest.:lrs a;e compa;ed with those obmined fo~ the undamped Struc~ lure in Exampie 11.5, we nOie only a smail change in the amplitL.de of motion. This is at ways the case for systems ligh(1y damped and subjected to harmonic excitation of a frequency [hal is not close to one of the natt.:ftll frequencies of the system. For (his example. the forced frequency Wj = 20 rod/sec is relativelY far from the naU!;al frequenctes WI ! l "83 rad lsec Or UJz = 32.89 rad Isec which were calculated in Example : 0.1.

11.5

PROGRAM 10-HARMONIC RESPONSE

Prog:am 10 calculates the ;esponse 10 hannonic excitations of a strucrurru sys:em for wh:ch the s:if::1ess and mass matrices have been determined by one

332

Forced Motion :::f Shear BuHdings

Structures Modeled as Shear BuHdings

y, (I)

of the programs modeling the structure (Program 7 to model the structure as a shear building). DampIng in the system is assumed to be proportional to the stiffness and lor mass coefficients, that is, the damping matrix is calculated as (11.55)

[C]=ao[M]+a,[K]

b which ao and at are constants specified in the input data. The program calculates the steady-state response for structures subjected ro harmonic forces applied at the nodal coordinates or a harmonic acceleration applied at the base of the structure. Example 11.7, Obtain tbe response of tbe damped two-degree-of-freedom shear building of Example 1 L6 using computer Program 10. Solution:

333

0.00068 cos 20, - 0.2686 sin 20t

y, (I)

0.0631 eos 20r - 0.1378 sin 20,

or y:(t) =0.2686

y,(t)

s;n (20,+3.'44)

0,1516 s'n (20,.,-3.571)

The results given by the computer, as expected. are the same as the values calcdated in Example I 1.6.

Example 11.8. For the structure modeled as a four~story shear bullding shown in Fig. 1L6, determine the steady-state response when subjected to the force FCt) = 10,000 sin 20r (1b) appl:cd at the top floor of the building. Assume dampIng proportional to stiffness (factor of proportionality al = 0.01. Modulus of elasticity E 30 X 10'" pSI.

Input D.t. and Output Results

m?""

OA'!.''; OAMr:NG S1"U'?Ne:S$ :t.;C':'OZ:

K?AC '"

.c:.

O;;''1?!NO ¥Jl.$S FACTO"

Solution.." The solution of this prOblem is initialed by executing Program 7 to model the structure as a shear building followed by the executior:, of Program 10 to obtain the response to hannonic forces. Modeling of this strucmre has been undertnken in the soiution of Example 9.1. '"

rCRC:;:p tlREQ!;;;:OIC:" l:AAP / $"C::

-~ .4300S~G{

Fltl

~ .4~CO.>o~

V

-1 I '" 7955 in.4 .......... 180 in. m"'1

V

,

,

I "" 79.55 in.

4

180 in.

-........

180 in.

m- Zl"_r~rtEQU2NCY A_FREQ:.J3NCY, 3, s, :6, 0, 0, 0, ;),

0,

C.

(}, 0

ANALYSIS > r:'?\EQ /8UCK

= 0.3886 Ib

R_FREQUSNCY

sec: lin: (13) List the natuml frequenCies of the system:

RCONST,

;>

RCONST

2 , 2 , 1 , 7,

3886,

(L

0,

°

0

?E\O?SETS ;. RCONS'fRCONST, L L 1, 1, 1500

(6) Define the reat constant for mass eiements: m

0,

(12) Set the options for the frequency analysis to extract three frequencies using the Subspace Iteration Method with a maximum of 16 iter2.tions, and run the frequency analysis:

(5) Define the real COnstant for spring elements: k = 1500 lb 1m:

pRQ?SSTS

!

LIST

:>

?:KEQLIS':'

1S-05,

0,

12-06,

338

f

Structures Modeled as Shear 8uildings

Frequency#

1 2 3

Frequency \ Rad Isec; 2,76500t'HOl

fr.s-quency {cycles /.secl

7.7473Ge+C1

1.23303e+Cl L78178e1'Ol

R_OYNAMIC

DISPLAY

R ;)YNAtlIC

~

2, TINE, ), '":'XME,

XY PLOTS

>

UX,

),

0, 10. 1, 0,

UX,

4,

8,

L

0,

2N 3N 4~

XYPLO?S

XYPLOT, 1, 1, 1

(19) Request Scan for m PD ...OUTPUT 1, 1, 10, 2, 4, 0, .5

ANALYSIS > ?05'T'_DY!\AM:CC

PO_PREPARE, ANALYS~S

~5GI

fPT

>

1 1 1

POS':'_DYN

1,

&

Fig. 11.8 Latera! disploce.ment funcdons at the three-story building of Example J 1.9.

response at nodes 2, 3, and 4:

ANALYSIS )

~.5

iIl.q01J;2"

JGG:'

"

~.es.s'.;!

1 ;. POST_DYNAMIC

PD_MAXLlS1'

~

PD_HAXMIN

PD~PREPARE

(2!) Activule XY plot information for X velocity at nodes 2, 3, and 4 as a function of time, and viOl the velocity vs. time fOf these modes (see Fig. 11.9); DISPLAY

~

XY_?~CTS

~

ACTXYPCST

L TI1·1E, VX, 2, 12, L 0, 2N ACTXYPOS?, 2, ':'!ME, VX, 3. 10, 1, 0, IN ACTXYPOST, ), l'It1E, VX. 4, 8, I, 0, 4N ACTXYPOST,

DISPLAY ) XV_PLOTS ) XY2LOT, l, 1,

XYPLO~

34Q

t

Structures Modeled as Shear Bulldings

Forced Motion of Shear 8 .•dldings 829.2-2 667'1.3 SaSE.

4

3Q36.S

, ", ,

•••

RRR

A AA

Bas

2S2.S1

L L L

" s (inJsecl )

-14 ! S • 2

~

E"

(i'1isec)

X X

162i'1.Ei

-3@33.' -«65l

.2"

."

-3¢5,31

+---l. -+-+f'-'+,~--+ -+-+-1--+,1-[..... ; L \.

$1 llGl!

!:' , I ~1'WS $. 2&ISJS (il Gl.QSI!I9 (i,!SS? (il,2S!]:;

,3~i94

(\) , (il,J5QJ

-62$9

4~G,

e

ACTXY?OST, ACTXYi?OST, ACTXYPOST,

XY_PLOTS > ACTXYPCST 1, TIME, AX, 2, 12, 1, Of 2N 2, TIME, AX, 3, 10, L 0, 3N 3, TIME, AX, 4, 8, L 0, 4N

DISPLAY ) XY_PLOTS > XYPLO'::' XYPLOl', 1, L 1

Example 11.10. For the structure modeled as a four-story shear building shown in Fig. 11.11 a. determine the response when it is subjected to the force shown in Fig. ll.llb applied at the top level of the building. The modulus of elnSlicity is E::;.; 2.0 X 10 6 psi. Assume dampi!1g in the system proportionaJ ro the stiffness coeffiCIent (Co 0.01). The cross-sectional area of the columns is 0.5 in X 5.0 in.

DISPLAY ,. VIE_PAR ) VIE VIE..), 0, 0, 1, 0

(2} Define [he XY plane Z = 0:

,,

GECMETRY PL,b_I\lE,

0,

) PLANE

1

(3) Gene;:ate four curve segments from 0, 0, 0 to 4. 0, 0:

GEOMETRY ) CURVES

~

CRPCORD

CRPCORD, 1, 360, 0, 0, 360, lac, 0, 360, 180, 0 CRPCORD, 2, 360, 18C, 0, 360, 360, 0, 360, 360, C, CRPCORD, 3, 360, 36G, 0, 360, 540. 0, 360, 540, 0, CRPCORD. 4, 360, 540, 0, 360, 720, 0, 360, 728,

°

(4) Define element group i using the BEAM2D elemerHs, tT.2.;:enal properties, and cross~sectiDnal consrants and genera[e mesh with two nodes and one element along each Cl!rve: ?ROPSE?S EGROIJP,

Solution: The analysis is perfonned using four beam eiemenrs and four concer.trated mass elemems.

> G~ID

Z,

>

L

EGROUP BEAM2D,

PROPSETS :;. HPROP 11?ROP,

1,

EX,

2E6,

0, 0, 0,

0, 0,

0,

0,

t! ,

Struclures Modeled as Sheaf Buildings

342

Piti

m" ,

I

! .. 79551f'!'

ANALYSIS

F(t)(:b}

180'm,

T

m" 1

--+

ANALYSIS)

y,

0,

Ib]

PROPS E'I'S

H_CR,

L

1,

1,

lS9,~,

5,

}

L

5,

0,

0,

G,

0,

0,

M~CR

1

(5) Define group 2 using mass eiemems, real constants, and assign a mass PROPSE':'S

>

EGROUP

e,

2, HASS,

0

v,

0, 8,

0,

0,

L

12,

0,

0,

Q,

0,

0,

0,

MESHING

>

NODES

>

1,

0,

OPT, 2, Uy, 0,

~

A~fREQUENCY

16,

0,

0,

0,

0,

iE-OS,

:),

1E-C6,

R... FREQUENCY

}

FREQLIST

>

Period ( seconds}

1 2

0, 8886530£:+01 C " 2S5877f~£~02 0, 3920269£+02 0 .4808922£:+02

0.1414335E+01 0.4072416£+01 0.6239303£+-01 0 7653638£+0::'

a .707046CE+OO

4

0,

0,

~

POST_DYN )

2,

~

DISPL}-l)l'TS

RX,

RY,

JOG,-

PD_ATYPE

O. . Cl,

0,

,05,

0.25,

0

:n

the X direction: ~

)

POST_CVN "-.

~

PD_CURVES

0, 0

POST_DYN > PD_GJRVES

1,

L

0,

1000,

.25,

PO COROE?

1000,

.5,

0,

5,

0

CONTRO:" ) ACTIVE ,. ACTSST

NCOV.PRESS

L 1 5, 1, UZ,

4,

(12; Define the dynamic forcing function as a time-depender.t force, and

PO_CURDEF,

0

O.245S5~5£ .. OO 0.:602743£:+00 0.:'306568£+00

(IJ) Define analYSIS type as modal time history using four natural frequen~ cies, 300 time steps st2.ning at l 0 wirh a rime increment of 0.01 sec; use default values for all inl:egration parameters; and request printout of relative displacements and reliHive velocities;

ANA;"·YSrS

0,0:)01.

STRUCTUR..A.L

ALL,

0,

Frequency (cyc:::'es Isecl

PD~CURTYP,

(7) Apply constraints in all degrees of freedom at node 1, and in a!i degrees of freedom except UX at nodes, 2, 3, 4, and S: DP':',

G,

{radl5
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