problem 6 003

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R Software Verification

PROGRAM NAME: SAP2000

REVISION NO.: 0

EXAMPLE 6-003 - 1

EXAMPLE 6-003

LINK GAP ELEMENT

PROBLEM DESCRIPTION

This example uses a single-bay, single-story rigid frame to test the gap link

element. This link element carries compression loads only; it has zero stiffness

when subjected to tension. The gap element is paced at the bottom of the right-hand column in the frame. The frame is then loaded with a gravity load P (10kips) at the center of the beam. Once the full load P is applied, a lateral load V

(20 kips) is applied, pushing the frame from right to left. The compression loadin the gap element after the full load P has been applied and the uplift at the gapafter the full load V has been applied are compared with independent hand

calculated results.

The model is created in the XZ plane. Only the Ux, Uzand Rydegrees of freedom

are active for the analysis. The gap element is modeled as a single-joint link

element at joint 2. This means that one end of the gap element is connected to theground and the other end is connected to joint 2. The gap element is oriented

such that its positive local 1 axis is parallel to the positive global Z axis. This is

the default orientation of single joint link elements. Only U1degree of freedomproperties are defined for the gap element.

All three frame elements have identical properties. The compression stiffness ofthe gap element is chosen to be approximately 100 times the axial stiffness of the

frame element directly above it, frame element 2. The initial gap element

opening is zero inches.

The loading is applied as follows. First the full load P is applied. Then, while theload P is maintained, the full load V is applied. The analysis is run several times

using different types of analysis cases. Nonlinear static, nonlinear modal time

history and nonlinear direct time history analysis cases are used. See the

subsequent section titled Summary of Analysis Cases for more information.

Two different models are used in this example. In Model A the linear effectivestiffness of the gap element is set equal to zero. In Model B the linear effective

stiffness of the gap element is set equal to the nonlinear stiffness of the gap

element.

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EXAMPLE 6-003 - 2

The gap linear effective stiffness is only used for the linear analysis cases, whichin this example are analysis cases P, V and MODAL. The gap linear effective

stiffness is not used in the other analysis cases and thus has no direct effect on

them; however, it does indirectly affect thenonlinear modal time history cases named

NLMHIST1 and NLMHIST2 because those

cases are solved using the modes from theanalysis case named MODAL.

The lumped joint masses shown in the tableto the right are used for the modal and

modal time history analyses.

GEOMETRY, PROPERTIES AND LOADING

1

P = 10 k

Z

Y

X

Single-joint link element (gap)with compression stiffnessdefined for the U1 degree of

freedom

2

3 4

1 2

3

V = 20 k

144"

144"

Loading

First apply full load P; thenapply full load V

Frame Material Properties

E = 29,900 k/in2

= 0.3G= 11,500 k/in2

Frame Section Properties

A = 10 in2

I = 100 in4

Av = 2 in2 (shear area)

Link Properties (Gap U1 DOF)

Linear Ke (Model A) = 0 k/inLinear Ke (Model B) = 200,000 k/inLinear Ce = 0 k-sec/inNonlinear K = 200,000 k/in

Nonlinear Open = 0 in

Active Degrees of Freedom

Ux, Uz, R

Joint Mass kip-s2/in

Joint DOF Ux DOF Uz

2 0 0.001

3 0.3 0.1

4 0.3 0.1

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EXAMPLE 6-003 - 3

SUMMARY OF ANALYSIS CASES

The following table summarizes the analysis cases that are used in this example.

Analysis Case Description

PA linear static analysis case with a 10 kip gravity load applied in the

negative global Z direction at the center of frame element 3.

VA linear static analysis case with a 20 kip lateral load applied in thenegative global X direction at joint 2.

MODALA ritz-type modal analysis case with starting loads defined for load P, load

V and the link element.

NLMHIST1

A nonlinear modal time history starting from zero initial conditions and

using a ramp load to apply the load P.

20 second ramp load rise time.

99.9% modal damping for all modes.

20 output steps and a 2 second output time step (40 seconds ofoutput total).

Default values for all nonlinear parameters (tolerances).

NLMHIST2

A nonlinear modal time history starting from the conditions at the end ofNLMHIST1 and using a ramp load to apply the load V.


20 output steps and a 4 second output time step (80 seconds of

output total).

99.9% modal damping for all modes.

Default values for all nonlinear parameters (tolerances), except inModel B the Force Convergence Tolerance is reduced from the

default 1E-05 to 1E-11.

NLSTAT1A nonlinear static case starting from zero initial conditions and applyingthe load P. This analysis case uses default nonlinear parameters

(tolerances).

NLSTAT2A nonlinear static case starting from the conditions at the end of NLSTAT1and applying the load V. This analysis case uses default nonlinear

parameters (tolerances).

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EXAMPLE 6-003 - 4

Analysis Case Description

NLSTAT3A nonlinear static case starting from the conditions at the end of the directintegration time history NLDHIST1 and applying the load V. This analysis

case uses default nonlinear parameters (tolerances).

NLDHIST1

A nonlinear direct integration time history starting from zero initial

conditions and using a ramp load to apply the load P.


400 output steps and a 0.1 second output time step (40 seconds of

output total). Mass and stiffness proportional damping is used with damping set

to 99.9% at periods of 1 and 3.4 seconds.

Default values for all nonlinear parameters (tolerances) except forthe Maximum Iterations per Substep and the Iteration Convergence

Tolerance. The Maximum Iterations per Substep item is increased

from the default 10 to 100. The Iteration Convergence Tolerance isreduced from the default 1E-04 to 1E-06.

NLDHIST2

A nonlinear direct integration time history starting from the conditions at

the end of NLDHIST1 and using a ramp load to apply the load V.

40 second ramp load rise time. 800 output steps and a 0.1 second output time step (80 seconds of

output total).

Mass and stiffness proportional damping is used with damping set

to 99.9% at periods of 1 and 3.4 seconds.

Default values for all nonlinear parameters (tolerances), except forthe Maximum Iterations per Substep and the Iteration Convergence

Tolerance. The Maximum Iterations per Substep item is increased

from the default 10 to 100. The Iteration Convergence Tolerance isreduced from the default 1E-04 to 1E-06.

NLDHIST3 A nonlinear direct integration time history that is the same as NLDHIST2,except it starts from the conditions at the end of nonlinear static caseNLSTAT1 instead of nonlinear direct integration time history NLDHIST1.

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EXAMPLE 6-003 - 5

The ramp times for the analysis cases are selected using the rule of thumb thatthe ramp loading time should be approximately 10 times the period of interest.

The period of interest is assumed to be the period of the first mode. The

following table shows the first mode period obtained for the two models.

Model

Gap Effective Stiffness for

Modal (Linear) Analysis

k/inMode 1 Period

sec

A 0 3.42

B 200,000 1.90

The load P is applied first and it causes the gap element to always be in

compression. Thus the 1.90 sec period, which is the period when the gap stiffness

is 200,000 k/in, is taken as the period of interest for application of the load P. The

ramp rise time for this load is chosen as 20 seconds, which is approximately 10times the period of interest.

The load V is applied after load P and it causes the gap element to eventually

uplift. Thus the 3.42 sec period, which is the period when the gap stiffness is

0 k/in, is taken as the period of interest for application of the load V. The ramprise time for this load is chosen as 40 seconds, which is approximately 10 times

the period of interest.

Analysis case NLMHIST2 in Model B requires 200 output steps (800 seconds

total) instead of the 20 output steps (80 seconds total) required in Model A. This

occurs because in Model B the modes are calculated using the gap effectivestiffness of 200,000 k/in and the 99.9% damping is applied to those modes.

When the gap opens and the gap effective stiffness becomes zero, the period

lengthens. Because the modal damping coefficient is held constant, the percent ofcritical damping increases in proportion to the period lengthening, thus over-

damping the system. This over-damping that initiates when the gap opens is thereason that it takes so long for analysis to reach its final value in case

NLMHIST2 for Model B.

Analysis case NLMHIST2 in Model B also requires a force convergencetolerance of 1E-11 rather than the default 1E-05. The following equation, which

is equation of motion used to solve nonlinear modal time history analyses, helps

explain the rationale for the change in the convergence tolerance.

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EXAMPLE 6-003 - 6

KLu(t) + KNu(t) + Cu(t) +Mu(t) + rN(t) = r(t) + KNu(t). ..

KLu(t) + KNu(t) + Cu(t) +Mu(t) + rN(t) = r(t) + KNu(t). ..

In the preceding equation KLis the stiffness of all the linear elements, including

the linear degrees of freedom of the link elements; KN is the linear effective

stiffness matrix for all link element nonlinear degrees of freedom; C is theproportional damping matrix; M is the diagonal mass matrix; r is the vector of

applied loads; and rN is the vector of forces from the nonlinear degrees of

freedom of the link elements that is computed by iteration at time t.

In the preceding equation, if the KN term is very large compared to the other

terms, the relative force convergence tolerance needs to be very small to capturean accurate representation of rN. This is why the 1E-11 force convergence

tolerance is required for analysis case NLMHIST2 in Model B.

The nonlinear direct integration analysis cases need the maximum number of

iterations per substep set to 100 rather than the default 10 and the iterationconvergence tolerance set to 1E-06 rather than the default 1E-04. Reducing the

iteration convergence tolerance to 1E-06 eliminates the slight fluctuations (high

frequency chatter) in the results.

For this particular example, increasing the maximum allowed number of

iterations per substep to 100 decreases the running time of the problem by severalorders of magnitude. Using the default 10 maximum iterations per substep causes

the program to have to cut the time steps from the initially specified 0.1 second

to as low as approximately 1E-07 second to solve the problem. When 100

maximum iterations per substep are allowed, the program does not have toreduce the time step size below 0.1 second to solve the problem.

TECHNICAL FEATURES OF SAP2000 TESTED

Gap element links Force-controlled nonlinear static analysis

Nonlinear modal time history analysis Nonlinear direct time history analysis

Frame point loads

Joint force loads

Joint mass assignments Ramp loading for time histories

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EXAMPLE 6-003 - 7

RESULTS COMPARISON

Independent results are hand calculated using the unit load method described on

page 244 in Cook and Young 1985. The results are presented separately forModel A and Model B

Results for Model A (Gap Ke= 0 k/in)

OutputParameter

AnalysisCase SAP2000 Independent

PercentDifference

NLMHIST1 -4.534 0%

NLSTAT1 -4.534 0%

Link force after

full load P is

applied

kip NLDHIST1 -4.534

-4.534

0%

NLMHIST2 3.917 0%

NLSTAT2 3.917 0%

NLSTAT3 3.917 0%

NLDHIST2 3.917 0%

Link

deformation

after full loadV is applied

inch

NLDHIST3 3.917

3.917

0%

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EXAMPLE 6-003 - 8

Results for Model B (Gap Ke= 200,000 k/in)

OutputParameter

AnalysisCase SAP2000 Independent

PercentDifference

NLMHIST1 -4.534 0%

NLSTAT1 -4.534 0%

Link force afterfull load P is

applied

kip NLDHIST1 -4.534

-4.534

0%

NLMHIST2 3.917 0%

NLSTAT2 3.917 0%

NLSTAT3 3.917 0%

NLDHIST2 3.917 0%

Link

deformationafter full load

V is applied

inch

NLDHIST3 3.917

3.917

0%

COMPUTER FILE: Example 6-003a, Example 6-003b

CONCLUSION

The SAP2000 results show an acceptable comparison with the independentresults both for Model A where the gap linear effective stiffness, Ke, is 0 k/in and

for Model B where the gap linear effective stiffness, Ke, is 200,000 k/in. The

comparison is exact when sufficiently small convergence tolerances are used.

This example illustrates that solution of nonlinear problems using gap elements

can be sensitive to the convergence and iteration tolerances that are used. For thisverification example, it is easy to determine if the tolerances used are sufficient

because the hand calculated results were available. In other situations it is helpful

to run the analysis two or more times using tolerances that are different by anorder of magnitude or more and verify that the results are the same. This is a

good check that the tolerances used are sufficient.

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EXAMPLE 6-003 - 9

HAND CALCULATION

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EXAMPLE 6-003 - 10

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EXAMPLE 6-003 - 11

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EXAMPLE 6-003 - 12

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EXAMPLE 6-003 - 13

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EXAMPLE 6-003 - 14

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EXAMPLE 6-003 - 15

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EXAMPLE 6-003 - 16

problem 6 003

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