link – suny buffalo eight-story building with...

C OMP UTE R S &S TRU CTU R ES

IN C.

R Software VerificationPROGRAM NAME: SAP2000REVISION NO.: 0

EXAMPLE 6-010 - 1

EXAMPLE 6-010LINK – SUNY BUFFALO EIGHT-STORY BUILDING WITH RUBBER ISOLATORS

PROBLEM DESCRIPTION

This example is presented in Section 2, pages 5 through 23, of Scheller andConstantinou 1999 (“the SUNY Buffalo report”). It is an eight-story building thatis seismically isolated using rubber bearings. The model is subjected to arecorded pair of scaled horizontal ground acceleration histories from the 1971San Fernando earthquake. SAP2000 results for superstructure displacementsrelative to the isolation system, superstructure accelerations, and isolator forcesand deformations are compared with results obtained using the computerprogram 3D-BASIS-ME (see Tsopelas, Constantinou and Reinhorn 1994).

The SAP2000 model is shown in the figures on the following three pages. Thesuperstructure is modeled as a stick using linear link elements. Thesuperstructure stick connects joints 23 and 55 through 62. The floor masses areconcentrated at the eccentric joints, joints 46 through 54. Diaphragm constraintsare used at each floor level above the isolation system to connect the mass to thesuperstructure. Only the Ux, Uy and Rz degrees of freedom are active for theanalysis. The superstructure is assumed to have 3% modal damping and theisolation system to have 0% modal damping.

Joints 1 through 45 define the location of the 45 rubber isolators in the model.Those joints are constrained to joint 46 using a body constraint. Joints 101 to 145are in the same location as joints 1 through 45, respectively, and are fullyrestrained (fixed to the ground). Zero-length link elements with rubber isolatorproperties connect joints 1 through 45 to joints 101 through 145. The propertiesfor all of the link elements in the model are presented in the section titled “LinkElement Properties” later in this example.

The SAP2000 model used in this verification example differs from that used inthe Scheller and Constantinou 1999 report as follows. First, this verificationexample uses a linear link element for the stick superstructure rather than thedamper element used in the report. The linear link is a simpler and moreappropriate element to use, but it was not available when the report was written.

Second, this verification example uses the actual linear effective stiffness for theisolators, 6.55 kip/in, rather than the artificially small effective stiffness used inthe report. This item is explained in more detail in the section titled “LinearEffective Stiffness of Rubber Isolator Elements” later in this example.


IN C.


EXAMPLE 6-010 - 2

GEOMETRY AND PROPERTIES

1, 101

Y

X

5, 1052 3 4

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23, 123 24 25, 125

26 27 28 29 30

31 32 33 34 35

36 37 38 39 40

41, 141 42 43 44 45, 145

4 @ 20' = 80'

8 @

20'

= 1

60'

CG

8'

46

Two joints at the same location with a rubber isolator link element connecting them, typical for joints 1 through 45 and 101 through 145

Note: Joints 1 through 46 are constrained using a body constraint

Plan View at Isolator Level


IN C.


EXAMPLE 6-010 - 3

43,143 38,138 33,133 28,128 23,123 18,118 13,113 8,108 3,10346

47 55

48 56

49 57

50 58

51 59

52 60

53 61

54 62

Isolator Level

Level 1

Level 2

Level 3

Level 4

Level 5

Level 6

Level 7

Level 8

Y

Z

8'

8 @

12'

= 9

6'

8 @ 20' = 160'

Longitudinal Section

Joints constrained as diaphragm, typical at levels 1 through 8

Linear link element typical at each level for the stick representing the building superstructure


IN C.


EXAMPLE 6-010 - 4

1, 101

23, 123

XY

Z

46

54

5, 105

41, 141

45, 145

62

2, 102

21, 121

25, 125

Active degrees of freedom are Ux, Uyand Rz

Linear link element typical at each level for the stick representing the building superstructure

Two joints at the same location with a rubber isolator link element connecting them, typical for joints 1 through 45 and 101 through 145.

3, 103

43, 143


IN C.


EXAMPLE 6-010 - 5

ANALYSIS CASES USED

Three different analysis cases are run for this example. They are described in thefollowing table. It is important to note that the 3D-BASIS-ME model uses 3%modal damping for all modes associated with the superstructure. No modaldamping is associated with the isolation system in 3D-BASIS-ME.

Analysis Case Description

RITZ Modal analysis case for Ritz vectors. Ninety-nine modesare requested. The program will automatically determinethat a maximum of twenty-seven modes are possible andthus reduce the number of modes to twenty-seven. Thestarting vectors are Ux acceleration, Uy acceleration, and alllink element nonlinear degrees of freedom.

NLMHIST1 Nonlinear modal time history analysis case that uses themodes in the RITZ analysis case. This case includes 3%modal damping in all modes, except that modes 1, 2 and 3,which are the modes associated with the isolation system,are assigned 0% modal damping. See the section titled“Linear Effective Stiffness of Rubber Isolator Elements”later in this example for more information.

NLDHIST1 Nonlinear direct integration time history analysis case. Thiscase includes proportional damping, which is defined toprovide damping similar to, but not exactly the same as, thedamping for the modal time history. See the section titled“Proportional Damping for Direct Integration TimeHistory” later in this example for more information.

Note that the inherent viscous damping in the superstructure, which is specifiedto be 3% of critical damping, is accounted for differently in 3D-BASIS-ME, thenonlinear modal time history in SAP2000, and the nonlinear direct integrationtime history in SAP2000. Thus, slight differences in the results for each of thethree time history analyses (one in 3D-BASIS-ME and two in SAP2000) areexpected.


IN C.


EXAMPLE 6-010 - 6

EARTHQUAKE RECORD

The following figures show the earthquake records used in this example. Theyare the recorded pair of horizontal ground acceleration time histories from the1971 San Fernando earthquake at station number 211. The earthquake recordsare provided in files named EQ6-010-trans.txt and EQ6-010-long.txt. Those fileshave one acceleration value per line, in g. The acceleration values are provided atan equal spacing of 0.02 second.

Inside SAP2000 each of the two components is multiplied by a factor of 2.345,as described in the SUNY Buffalo report, and also by a factor of 386.22 toconvert from g to in/sec2. The recorded north and west components are applied inthe transverse and longitudinal directions of the model, respectively.

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0 5 10 15 20 25 30 35 40 45

Time (sec)

Gro

un

d A

ccel

erat

ion

(g

)

Ground Acceleration for Transverse (X) Direction

-0.10

-0.05

0.00

0.05

0.10

0.15

0 5 10 15 20 25 30 35 40 45

Time (sec)

Gro

un

d A

ccel

erat

ion

(g

)

Ground Acceleration for Longitudinal (Y) Direction


IN C.


EXAMPLE 6-010 - 7

LINK ELEMENT PROPERTIES

This section presents the properties used for all of the link elements in the model.All link elements in the model are oriented such that the positive local 1 axis isparallel to the positive global Z axis, the positive local 2 axis is parallel to thepositive global X axis and the local 3 axis is parallel to the positive global Y axis.

The superstructure linear link elements have an effective stiffness, ke, and for theshear degrees of freedom, a distance from the J-end to the shear spring, DJ.Properties are specified for the U2, U3 and R1 degrees of freedom.

Between the Isolator Level and Level 3 (Property Name LINST123)ke U2 = 3401.8 k/inDJ U2 = 72 inke U3 = 3401.8 k/inDJ U3 = 72 inke R1 = 3.996 E+09 k-in/radian

Between the Level 3 and Level 6 (Property Name LINST456)ke U2 = 2551.3 k/inDJ U2 = 72 inke U3 = 2551.3 k/inDJ U3 = 72 inke R1 = 2.997 E+09 k-in/radian

Between the Level 7 and Level 8 (Property Name LINST78)ke U2 = 1700.9 k/inDJ U2 = 72 inke U3 = 1700.9 k/inDJ U3 = 72 inke R1 = 1.998 E+09 k-in/radian

The rubber isolator link elements have a linear effective stiffness, ke, a nonlinearinitial stiffness, k, a nonlinear yield strength, Fy, and a post yield stiffness ratio, r.See the following section titled “Linear Effective Stiffness of Rubber IsolatorElements” for more information about ke. Properties are specified for the U2, andU3 degrees of freedom and the properties are the same for the two degrees offreedom. The rubber isolator property name is BILIN and its properties are:

ke = 6.55 k/in Fy = 12.8 kk = 25.6k/in r = 0.1887


IN C.


EXAMPLE 6-010 - 8

LINEAR EFFECTIVE STIFFNESS OF RUBBER ISOLATOR ELEMENTS

This verification example uses the calculated linear effective stiffness, ke, of 6.55kip/in for the isolators. The SUNY Buffalo report used an artificially smalleffective stiffness of 0.0001 kip/in in their SAP2000 model to match the 3D-BASIS-ME results. The report further shows that when they used the actualisolator effective stiffness of 6.55 kip/in in their SAP2000 model, their SAP2000results underestimated the 3D-BASIS-ME results.

The calculated isolator effective stiffness of 6.55 kips/in is the appropriatevalue to use and, as shown in this verification example, leads to results thatmatch the 3D-BASIS-ME results when the SAP2000 model is madeessentially equivalent to the 3D-BASIS-ME model.

It is important to recognize that the 3D-BASIS-ME model has 3% modaldamping in the superstructure and no modal damping in the isolation system.Thus, for the SAP2000 model to be essentially equivalent to the 3D-BASIS-MEmodel, it must have 3% damping in all modes, except for modes 1, 2 and 3,which are dominated by the isolation system behavior. Thus, to be equivalent tothe 3D-BASIS-ME model, modes 1, 2 and 3 in the SAP2000 model must beassigned 0% damping with all other modes assigned 3% damping.

The SUNY Buffalo report SAP2000 model underestimated the 3D-BASIS-MEresults when using a linear effective stiffness of 6.55 kip/in for the isolatorsbecause the SAP2000 model had 3% damping in all modes, including modes 1, 2and 3. Thus, the SAP2000 model was not equivalent to the 3D-BASIS-MEmodel.

When the SUNY Buffalo report SAP2000 model used a linear effective stiffnessof 0.0001 kip/in for the isolators, the results matched the 3D-BASIS-ME results.Using 0.0001 kip/in effective stiffness for the isolators made the periods of theisolated modes (modes 1, 2 and 3) 528, 512 and 435 seconds, respectively.Noting that the entire earthquake duration is approximately 44 seconds, it isapparent that very little energy can be absorbed by modes 1, 2 and 3, thus makingthe damping associated with them approximately 0%, which is consistent withthe 3D-BASIS-ME model.

Again, we recommend using the actual effective stiffness and adjusting themodal damping associated with the isolated modes, rather than using anartificially small effective stiffness.


IN C.


EXAMPLE 6-010 - 9

Note that the preceding explanation is only relevant for the nonlinear modal timehistory analysis case. It is not relevant for the nonlinear direct integration timehistory analysis case. The linear effective stiffness of the isolators is only used inlinear analysis cases. In this verification example, the only linear analysis case isthe modal analysis case, RITZ. Both the nonlinear modal time history and thenonlinear direct integration time history analysis cases use the nonlinearproperties of the isolator, not the linear properties. However, the nonlinear modaltime history uses modes from the modal analysis case RITZ that are based on thelinear effective stiffness of the isolators. Thus, the nonlinear modal time historyanalysis case is indirectly affected by the linear effective stiffness of the isolators,whereas the nonlinear direct integration time history analysis case is unaffectedby the linear isolator properties.

PROPORTIONAL DAMPING FOR DIRECT INTEGRATION TIME HISTORY

The nonlinear direct integration time history analysis case NLDHIST1 uses massand stiffness proportional damping. For this analysis case the challenge is todesignate appropriate proportional damping that approximates 3% damping in allmodes, except the isolator modes (modes 1, 2 and 3), which are to have 0%damping. The isolator modes have periods of approximately 2 seconds, and thesuperstructure periods range from approximately 0.06 to 0.60 second.

For this example, theproportional damping isspecified as stiffnessproportional dampingonly. The mass coefficientfor the damping is setequal to zero and thestiffness coefficient is setequal to 0.0040. The solidline in the chart to theright plots the resultingproportional dampingused. The dashed lineshows a constant 3%damping.

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Period (sec)

Dam

pin

g R

atio


IN C.


EXAMPLE 6-010 - 10

TECHNICAL FEATURES OF SAP2000 TESTED

Rubber isolator links Linear links Zero-length, two-joint link elements Diaphragm constraints Modal analysis for ritz vectors Nonlinear modal time history analysis Nonlinear direct integration time history analysis Generalized displacements

RESULTS COMPARISON

Independent results are obtained using the computer program 3D-BASIS-ME(see Tsopelas, Constantinou and Reinhorn 1994).

The eight figures shown on the following four pages plot results from 3D-BASIS-ME and from the SAP2000 analysis case NLMHIST1 (nonlinear modaltime history). The results for SAP2000 analysis case NLDHIST1 (nonlineardirect integration time history) are similar. The following plots are shown:

Level 8 X direction displacement relative to the isolation system Level 8 Y direction displacement relative to the isolation system Level 8 rotation about Z relative to the isolation system Base shear in the X direction Level 3 absolute acceleration in the X direction Level 3 absolute acceleration in the Y direction Link 23 force-deformation in the X direction Link 23 force-deformation in the Y direction

In SAP2000, the level 8 displacements and rotations relative to the isolationsystem are determined using generalized displacements. The generalizeddisplacements are defined to subtract the displacement or rotation at joint 23from that at joint 62.


IN C.


EXAMPLE 6-010 - 11

-3.00

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Ux

Dis

pla

cem

ent

Rel

ativ

e to

Iso

lati

on

Sys

tem

(in

)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Ux Displacement Relative to Isolation System

-3.00

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Ux

Dis

pla

cem

ent

Rel

ativ

e to

Iso

lati

on

Sys

tem

(in

)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Ux Displacement Relative to Isolation System

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Uy

Dis

pla

cem

ent

Rel

ativ

e to

Iso

lati

on

Sys

tem

(in

)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Uy Displacement Relative to Isolation System

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Uy

Dis

pla

cem

ent

Rel

ativ

e to

Iso

lati

on

Sys

tem

(in

)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Uy Displacement Relative to Isolation System


IN C.


EXAMPLE 6-010 - 12

-0.0008

-0.0006

-0.0004

-0.0002

0.0000

0.0002

0.0004

0.0006

0.0008

0.0010

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Rz

Ro

tati

on

Rel

ativ

e to

Iso

lati

on

Sys

tem

(ra

dia

ns)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Rz Rotation Relative to Isolation System

-0.0008

-0.0006

-0.0004

-0.0002

0.0000

0.0002

0.0004

0.0006

0.0008

0.0010

0 5 10 15 20 25 30

Time (sec)

Lev

el 8

Rz

Ro

tati

on

Rel

ativ

e to

Iso

lati

on

Sys

tem

(ra

dia

ns)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Rz Rotation Relative to Isolation System

-2500

-2000

-1500

-1000

-500

0

500

1000

1500

2000

0 5 10 15 20 25 30

Time (sec)

Bas

e F

x (

kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Base Shear Fx

-2500

-2000

-1500

-1000

-500

0

500

1000

1500

2000

0 5 10 15 20 25 30

Time (sec)

Bas

e F

x (

kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Base Shear Fx


IN C.


EXAMPLE 6-010 - 13

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30

Time (sec)

Lev

el 3

Ux

Ab

solu

te A

ccel

erat

ion

(in

/sec

2 )

3D-BASIS-ME

SAP2000 NLMHIST1

Level 3 Ux Absolute Acceleration

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30

Time (sec)

Lev

el 3

Ux

Ab

solu

te A

ccel

erat

ion

(in

/sec

2 )

3D-BASIS-ME

SAP2000 NLMHIST1

Level 3 Ux Absolute Acceleration

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30

Time (sec)

Lev

el 3

Uy

Ab

solu

te A

ccel

erat

ion

(in

/sec

2 )

3D-BASIS-ME

SAP2000 NLMHIST1

Level 3 Uy Absolute Acceleration

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30

Time (sec)

Lev

el 3

Uy

Ab

solu

te A

ccel

erat

ion

(in

/sec

2 )

3D-BASIS-ME

SAP2000 NLMHIST1

Level 3 Uy Absolute Acceleration


IN C.


EXAMPLE 6-010 - 14

-50

-40

-30

-20

-10

0

10

20

30

40

50

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Isolator 23 Ux Deformation (in)

Iso

lato

r 23

Fx

Fo

rce

(kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Isolator 23 Force-Deformation in the X Direction

-50

-40

-30

-20

-10

0

10

20

30

40

50

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Isolator 23 Ux Deformation (in)

Iso

lato

r 23

Fx

Fo

rce

(kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Isolator 23 Force-Deformation in the X Direction

-40

-30

-20

-10

0

10

20

30

40

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Isolator 23 Uy Deformation (in)

Iso

lato

r 23

Fy

Fo

rce

(kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Isolator 23 Force-Deformation in the Y Direction

-40

-30

-20

-10

0

10

20

30

40

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Isolator 23 Uy Deformation (in)

Iso

lato

r 23

Fy

Fo

rce

(kip

)

3D-BASIS-ME

SAP2000 NLMHIST1

Isolator 23 Force-Deformation in the Y Direction


IN C.


EXAMPLE 6-010 - 15

The following table compares the maximum and minimum values of the outputitems shown in the charts on the previous four pages. Results are compared forboth the modal time history analysis case, NLMHIST1, and the direct integrationtime history analysis case, NLDHIST1.

OutputParameter

Dir andMax/Min

AnalysisCase SAP2000

Independent3D-BASIS-ME

PercentDifference

NLMHIST1 3.521 +1%Ux

Max NLDHIST1 3.5043.494

0%

NLMHIST1 -2.875 +3%Ux

Min NLDHIST1 -2.911-2.804

+4%

NLMHIST1 2.642 +4%Uy


+8%

NLMHIST1 -2.167 +7%

Level 8displacement

relative toisolation system

(in)

Uy


+9%

NLMHIST1 0.00080 +7%Rz

Max NLDHIST1 0.000770.00075

+3%

NLMHIST1 -0.00077 +1%

Level 8 rotationrelative to

isolation system(rad)

Rz

Min NLDHIST1 -0.00076-0.00076

0%

NLMHIST1 1854 +2%Fx

Max NLDHIST1 18731818

+3%

NLMHIST1 -2109 +1%

Base shear in theX direction

(kip) Fx

Min NLDHIST1 -2091-2089

0%


IN C.


EXAMPLE 6-010 - 16

OutputParameter

Dir andMax/Min

AnalysisCase SAP2000

Independent3D-BASIS-ME

PercentDifference

NLMHIST1 65.76 -3%Ux


+3%



+2%



-5%

NLMHIST1 -58.35 +1%

Level 3 (jt 57)absolute

acceleration(in/sec2)

Uy


0%

NLMHIST1 48.66 +1%Fx


0%

NLMHIST1 -43.55 +2%Fx


+2%

NLMHIST1 36.65 +1%Fy


0%

NLMHIST1 -30.91 +1%

Isolator 23shear force

(kip)

Fy


+1%

NLMHIST1 7.935 +1%Ux


0%



+2%



+1%

NLMHIST1 -4.419 +3%

Isolator 23deformation

(in)

Uy


+2%


IN C.


EXAMPLE 6-010 - 17

COMPUTER FILE: Example 6-010

CONCLUSION

The SAP2000 results show an acceptable comparison with the independentresults considering that SAP2000 and 3D-BASIS-ME use different modeling andsolution techniques for the isolated structure. The clearest comparison of resultsis evident in the graphical comparisons.

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