capacity-based equivalent linear modeling of the spear test frame

CAPACITY-BASEDEQUIVALENT LINEAR MODELING OF THESPEAR TEST FRAMEHaluk SUCUOGLU and M. Selim GUNAYMiddle East Technical UniversityAnkara, Turkey

SPEAR TEST FRAMEC51.03.0 5.0 1.7B1 C1 B2 C2Frame X35.5B11B9B76.0B3C9C3B4C4Frame X25.0B12B10B84.0C8B5C6B6C7Frame X1Frame Y1 Frame Y2 Frame Y3

SIMULATION OFLINEAR ELASTIC RESPONSEPGA = 0.02 g

MODAL PROPERTIES (0.02 g)Period (s) Damping (%)Mode Test Result Model Result Test Result Analytical Model1 0.85 0.87 3.7 3.72 0.78 0.76 4.3 4.33 0.66 0.63 5.0 5.04 0.35 0.30 3.4 3.45 0.31 0.25 3.9 3.96 0.24 0.20 2.4 2.47 0.22 0.20 2.2 2.28 0.20 0.16 3.2 3.29 0.15 0.12 1.2 1.2

RESULTS (0.02g RUN)1st Level X Direction0.0030.002TestModelDisplacement (m)0.0010-0.001-0.002-0.0030 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)2nd Level X Direction0.0060.004TestModelDisplacement (m)0.0020-0.002-0.004-0.0060 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)3rd Level X Direction0.0080.006TestModel0.004Displacement (m)0.0020-0.002-0.004-0.006-0.0080 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)1st Level Y Direction0.0030.002TestModelDisplacement (m)0.0010-0.001-0.002-0.0030 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)2nd Level Y Direction0.0080.006TestModel0.004Displacement (m)0.0020-0.002-0.004-0.006-0.0080 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)3rd Level Y Direction0.0100.008TestModel0.006Displacement (m)0.0040.0020.000-0.002-0.004-0.006-0.008-0.0100 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)1st Level θ Direction0.00060.0004TestModelRotation (rad)0.00020-0.0002-0.0004-0.00060 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)2nd Level θ Direction0.00150.0010TestModelRotation (rad)0.00050.0000-0.0005-0.0010-0.00150 2 4 6 8 10 12 14 16 18Time (sec)

RESULTS (0.02g RUN)3rd Level θ Direction0.00150.0010TestModelRotation (rad)0.00050.0000-0.0005-0.0010-0.00150 2 4 6 8 10 12 14 16 18Time (sec)

DISPLACEMENT PROFILES AT MAXIMUM DISPLACEMENTX - DISPLACEMENTY - DISPLACEMENTROTATION333222sto rysto rysto ry1MODELTEST00 1 2 3 4 5 6 7 8displacement (mm)1MODELTEST00 1 2 3 4 5 6 7displacement (mm)1TESTMODEL00 2 4 6 8 10 12 14 16rotation (radX10E- 4)X Y θ

SIMULATION OF INELASTIC RESPONSEBYEQUIVALENT LINEARIZATIONPGA = 0.20 g

A) LINEAR ELASTIC DYNAMIC ANALYSIS• Cracked concrete sections• Internal forces at maximum displacementB) CAPACITY ANALYSISStep 1 : Calculation of Moment Capacities of BeamsStep 2 : Calculation of Axial Forces in ColumnsN = N G + N EQN GN EQLinear Elastic Static AnalysisLimit Analysis(All beam ends are assumed to yield)

Step 3 : Calculation of Moment Capacities of ColumnsNSection at N =N iM cYN iM cXM cYM cXMcX=f⎛ M⎜N,⎝ McXcY⎞⎟⎠& McY=⎛ Mg ⎜N,⎝ McXcY⎞⎟⎠M cX =McYMMaXaY

Step 4 : Calculation of Beam/Column Capacity Ratios(BCCR’s) at all JointsM CBNM CBEM CCYU+ M CCYBM CCXU+M CCXBM CBWBCCRBCCRXYM=MM=MCBNCCXUCBECCYU+ M+ M+ M+ MCBSCCXBCBWCCYBYXM CBS

Step 5 : Correction of Axial Forces in ColumnsIF AND (BCCR X < 1; BCCR Y < 1) No modification for beam-end momentsIF AND (BCCR X > 1; BCCR Y < 1)IF AND (BCCR X < 1; BCCR Y > 1)No modification of beam-end momentsfor NS beams, M CCXU +M CCXB is distributedto EW Beam endsNo modification of beam-end momentsfor EW beams, M CCYU +M CCYB is distributedto NS Beam endsIF AND (BCCR X < 1; BCCR Y < 1) M CCXU+M CCXB is distributed to EW Beam ends,M CCYU +M CCYB is distributed to NS Beam endsAxial forces in columns are recalculated

Step 6 : Recalculation of Column Moment CapacitiesStep 7 : Recalculation of BCCR’s at all JointsStep 8 : Decision on the Potential Yielding Member EndsIF AND (BCCR X < 1; BCCR Y < 1)IF AND (BCCR X > 1; BCCR Y < 1)ALL BEAM ENDS MAY YIELDNS BEAM ENDS AND COLUMN ENDS MAY YIELDIF AND (BCCR X < 1; BCCR Y > 1) EW BEAM ENDS AND COLUMN ENDS MAY YIELDIF AND (BCCR X < 1; BCCR Y < 1)BOTH COLUMN ENDS MAY YIELD

Step 9 : Determination of DCR’s for Potential Yielding Member EndsDCR =Demand (Internal force from LE analysis)CapacityStep 10 : Determination of Final Yielding DistributionAt a potential yielding member end;IF DCR > 1IF DCR < 1Member end yieldsMember end does not yield

Step 11 : Equivalent Stiffness (and Damping) ValuesCaseNo yieldingYielding at one endII crIcr⎛ 1 ⋅ ⎜1+2 ⎝ DCRY⎞⎟⎠Yielding at both endsMAXIcr( DCR ,DCR )IJDCR Y: DCR at the yielding end DCR I: DCR at end I DCR J: DCR at end J

Step 12 : Linear time history analysis with equivalent stiffnessand damping, and calculation of chord rotationsStep 13 : Controlling Component AcceptabilityStep 14 : Decision on the Seismic Performance of Structure

MODAL PROPERTIES (0.20 g)Period (s) Damping (%)Mode Test Result Model Result Test Result Analytical Model1 1.77 1.78 2.7 12.92 1.43 1.44 13 12.93 1.17 1.18 8 12.94 0.61 0.59 13 12.95 0.50 0.45 10 12.96 0.43 0.43 4.3 12.97 0.37 0.35 8 12.98 0.31 0.31 6.4 12.99 0.24 0.19 1.9 12.9

Predicted Yielding DistributionsFrame Y2Frame X2

RESULTS (0.2g RUN)1st Level X0.030.02TestModelDisplacement (m)0.010-0.01-0.02-0.030 3 6 9 12 15Time (sec)

RESULTS (0.2g RUN)3rd Level X0.150.10TestModelDisplacement (m)0.050.00-0.05-0.10-0.150 3 6 9 12 15Time (sec)

RESULTS (0.2g RUN)1st Level Y0.030.02TestModelDisplacement (m)0.010-0.01-0.02-0.03-0.040 3 6 9 12 15Time (sec)

RESULTS (0.2g RUN)2nd Level Y0.080.06TestModel0.04Displacement (m)0.020-0.02-0.04-0.06-0.080 3 6 9 12 15Time (sec)

RESULTS (0.2g RUN)3rd Level Y0.150.10TestModelDisplacement (m)0.050.00-0.05-0.10-0.150 3 6 9 12 15Time (sec)

RESULTS (0.2g RUN)1st Level θ0.0050.0040.0030.002Rotation (rad)0.0010-0.001-0.002-0.003-0.004-0.0050 3 6 9 12 15Time (sec)TestModel

RESULTS (0.2g RUN)2nd Level θ0.0150.0100.005Rotation (rad)0.000-0.005-0.010-0.015-0.0200 3 6 9 12 15Time (sec)TestModel

RESULTS (0.2g RUN)3rd Level θ0.0200.0150.010Rotation (rad)0.0050.000-0.005-0.010-0.015-0.020-0.0250 3 6 9 12 15Time (sec)TestModel

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