problem 6-010

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R Software Verification

PROGRAM NAME: SAP2000

REVISION NO.: 0

EXAMPLE 6-010 - 1

EXAMPLE 6-010

LINK SUNY BUFFALO EIGHT-STORY BUILDING WITH RUBBER ISOLATORS

PROBLEM DESCRIPTION

This example is presented in Section 2, pages 5 through 23, of Scheller and

Constantinou 1999 (the SUNY Buffalo report). It is an eight-story building that

is 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 displacements

relative 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. The

superstructure stick connects joints 23 and 55 through 62. The floor masses are

concentrated at the eccentric joints, joints 46 through 54. Diaphragm constraintsare used at each floor level above the isolation system to connect the mass to the

superstructure. Only the Ux, Uy and Rz degrees of freedom are active for the

analysis. 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 145

are in the same location as joints 1 through 45, respectively, and are fully

restrained (fixed to the ground). Zero-length link elements with rubber isolatorproperties connect joints 1 through 45 to joints 101 through 145. The properties

for all of the link elements in the model are presented in the section titled Link

Element Properties later in this example.

The SAP2000 model used in this verification example differs from that used in

the Scheller and Constantinou 1999 report as follows. First, this verification

example uses a linear link element for the stick superstructure rather than thedamper element used in the report. The linear link is a simpler and more

appropriate 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 in

the report. This item is explained in more detail in the section titled Linear

Effective Stiffness of Rubber Isolator Elements later in this example.

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REVISION NO.: 0

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'=

160'

CG8'

46

Two joints at the same

location with a rubberisolator 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

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REVISION NO.: 0

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 asdiaphragm, typical atlevels 1 through 8

Linear link element typicalat each level for the stickrepresenting the buildingsuperstructure

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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, typicalfor joints 1 through 45 and

101 through 145.

3, 103

43, 143

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REVISION NO.: 0

EXAMPLE 6-010 - 5

ANALYSIS CASES USED

Three different analysis cases are run for this example. They are described in the

following table. It is important to note that the 3D-BASIS-ME model uses 3%modal damping for all modes associated with the superstructure. No modal

damping is associated with the isolation system in 3D-BASIS-ME.

Analysis Case Description

RITZ Modal analysis case for Ritz vectors. Ninety-nine modes

are requested. The program will automatically determinethat a maximum of twenty-seven modes are possible andthus reduce the number of modes to twenty-seven. The

starting vectors are Ux acceleration, Uy acceleration, and all

link 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 Elementslater in this example for more information.

NLDHIST1 Nonlinear direct integration time history analysis case. This

case includes proportional damping, which is defined to

provide damping similar to, but not exactly the same as, thedamping for the modal time history. See the section titled

Proportional Damping for Direct Integration Time

History 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, the

nonlinear modal time history in SAP2000, and the nonlinear direct integration

time 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) are

expected.

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REVISION NO.: 0

EXAMPLE 6-010 - 6

EARTHQUAKE RECORD

The following figures show the earthquake records used in this example. They

are the recorded pair of horizontal ground acceleration time histories from the

1971 San Fernando earthquake at station number 211. The earthquake recordsare provided in files named EQ6-010-trans.txt and EQ6-010-long.txt. Those files

have one acceleration value per line, in g. The acceleration values are provided at

an 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 in

the 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)

GroundAcceleration(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)

GroundAc

celeration(g)

Ground Acceleration for Longitudinal (Y) Direction

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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 is

parallel 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/in

DJ U3 = 72 in

ke 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 in

ke U3 = 2551.3 k/in

DJ U3 = 72 in

ke 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 in

ke U3 = 1700.9 k/in

DJ 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, and

U3 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 k

k = 25.6k/in r = 0.1887

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REVISION NO.: 0

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 small

effective 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 SAP2000

results underestimated the 3D-BASIS-ME results.

The calculated isolator effective stiffness of 6.55 kips/in is the appropriate

value to use and, as shown in this verification example, leads to results thatmatch the 3D-BASIS-ME results when the SAP2000 model is made

essentially equivalent to the 3D-BASIS-ME model.

It is important to recognize that the 3D-BASIS-ME model has 3% modal

damping 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 to

the 3D-BASIS-ME model, modes 1, 2 and 3 in the SAP2000 model must be

assigned 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 isolators

because the SAP2000 model had 3% damping in all modes, including modes 1, 2

and 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 the

isolated 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 making

the 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 an

artificially small effective stiffness.

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REVISION NO.: 0

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 time

history analysis case. The linear effective stiffness of the isolators is only used in

linear analysis cases. In this verification example, the only linear analysis case isthe modal analysis case, RITZ. Both the nonlinear modal time history and the

nonlinear direct integration time history analysis cases use the nonlinear

properties of the isolator, not the linear properties. However, the nonlinear modaltime history uses modes from the modal analysis case RITZ that are based on the

linear effective stiffness of the isolators. Thus, the nonlinear modal time history

analysis case is indirectly affected by the linear effective stiffness of the isolators,

whereas the nonlinear direct integration time history analysis case is unaffected

by 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 to

designate appropriate proportional damping that approximates 3% damping in all

modes, 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 the

superstructure periods range from approximately 0.06 to 0.60 second.

For this example, theproportional damping is

specified as stiffnessproportional damping

only. The mass coefficient

for the damping is setequal to zero and the

stiffness coefficient is set

equal to 0.0040. The solidline in the chart to the

right plots the resulting

proportional 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)

DampingRatio

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REVISION NO.: 0

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 modal

time history). The results for SAP2000 analysis case NLDHIST1 (nonlinear

direct 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 isolation

system are determined using generalized displacements. The generalized

displacements are defined to subtract the displacement or rotation at joint 23

from that at joint 62.

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REVISION NO.: 0

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)

Level8Ux

Displacem

entRelativetoIsolationSystem(

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)

Level8Ux

Displacem

entRelativetoIsolationSystem(

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)

Level8Uy

Displa

cementRelativetoIsolationSystem(

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)

yDispla

cementRelativetoIsolationSystem(

in)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 8 Uy Displacement Relative to Isolation System

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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)

Level8Rz

RotationRelativetoIsolationSystem(radians)

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)

Level8Rz

RotationRelativetoIsolationSystem(radians)

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)

BaseFx

(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)

BaseFx

(kip)

3D-BASIS-ME

SAP2000 NLMHIST1

Base Shear Fx

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EXAMPLE 6-010 - 13

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30

Time (sec)

Level3UxA

bsoluteAcceleration(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)

Level3UxA

bsoluteAcceleration(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)

Level3U

yAbsoluteAcceleration(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)

Level3U

yAbsoluteAcceleration(in/sec

2)

3D-BASIS-ME

SAP2000 NLMHIST1

Level 3 Uy Absolute Acceleration

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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)

Isolator23Fx

Force(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)

Isolator23Fx

Force(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)

Isolator23Fy

Force(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)

yForce(kip)

3D-BASIS-ME

SAP2000 NLMHIST1

Isolator 23 Force-Deformation in the Y Direction

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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 for

both the modal time history analysis case, NLMHIST1, and the direct integration

time history analysis case, NLDHIST1.

OutputParameter

Dir andMax/Min

AnalysisCase SAP2000

Independent3D-BASIS-ME

PercentDifference

NLMHIST1 3.521 +1%UxMax

NLDHIST1 3.504

3.494

0%NLMHIST1 -2.875 +3%Ux

Min NLDHIST1 -2.911-2.804

+4%

NLMHIST1 2.642 +4%UyMax NLDHIST1 2.729

2.538+8%

NLMHIST1 -2.167 +7%

Level 8

displacement

relative toisolation system

(in)

UyMin NLDHIST1 -2.211

-2.029+9%

NLMHIST1 0.00080 +7%Rz

Max NLDHIST1 0.00077 0.00075 +3%

NLMHIST1 -0.00077 +1%

Level 8 rotationrelative to

isolation system

(rad)Rz

Min NLDHIST1 -0.00076-0.00076

0%

NLMHIST1 1854 +2%FxMax NLDHIST1 1873

1818+3%

NLMHIST1 -2109 +1%

Base shear in the

X direction

(kip) FxMin NLDHIST1 -2091

-20890%

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EXAMPLE 6-010 - 16

OutputParameter

Dir andMax/Min

AnalysisCase SAP2000

Independent3D-BASIS-ME

PercentDifference

NLMHIST1 65.76 -3%UxMax NLDHIST1 70.03

67.73+3%

NLMHIST1 -72.71 +2%UxMin NLDHIST1 -72.85

-71.47+2%


64.96-5%

NLMHIST1 -58.35 +1%

Level 3 (jt 57)absolute

acceleration

(in/sec2)


-57.780%

NLMHIST1 48.66 +1%FxMax NLDHIST1 48.18

48.170%

NLMHIST1 -43.55 +2%FxMin NLDHIST1 -43.54

-42.68+2%

NLMHIST1 36.65 +1%FyMax NLDHIST1 36.53

36.380%

NLMHIST1 -30.91 +1%

Isolator 23

shear force

(kip)

FyMin NLDHIST1 -30.78

-30.54+1%

NLMHIST1 7.935 +1%UxMax NLDHIST1 7.828

7.8450%

NLMHIST1 -6.916 +3%UxMin NLDHIST1 -6.905

-6.746+2%


6.150+1%

NLMHIST1 -4.419 +3%

Isolator 23

deformation(in)


-4.304+2%

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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 and

solution techniques for the isolated structure. The clearest comparison of results

is evident in the graphical comparisons.

problem 6-010

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