Pushover Example Large

(1)

INCREMENTAL PUSHOVER ANALYSIS FOR SEISMIC

PERFORMANCE ASSESSMENT

EXAMPLE OF INCREMENTAL PUSHOVER ANALYSIS TO ASSESS

THE PERFORMANCE OF AN EXISTING REINFORCED CONCRETE

FRAME: (TDY–2007 APPROACH)

7 m 3.5 m

P

1

P

2 7 m 3.5 m B101 B102 B201 B202 C101 C102 C103 C201 C202 C203

(2)

• COLUMNS : 450 mm x 450 mm WITH 8-φ22

φ8 STIRRUPS AT A SPACING OF 200 mm PROVIDED ALONG

THE ENTIRE LENGTH OF THE COLUMNS. • BEAMS : 450 mm x 600 mm WITH 3-φ22 AT THE TOP

3-φ22 AT THE BOTTOM

THE ENTIRE LENGTH OF THE BEAMS. • CONCRETE COVER : CLEAR COVER = 20 MM

COVER TO THE CENTER OF LONGITUDINAL

REINFORCEMENT = 20 MM + φ8 + (φ22)/2 = 39 MM

TAKE COVER TO THE CENTER OF

LONGITUDINAL REINFORCEMENT AS 40 mm FOR

BOTH THE BEAMS AND THE COLUMNS

(NOTE: TOP AND BOTTOM REINFORCEMENT IN THE BEAMS IS ASSUMED EQUAL IN ORDER TO SIMPLIFY THE ANALYSIS (Mult+ = Mult-). TYPICALLY,

THE AMOUNT OF TOP REINFORCEMENT WOULD BE LARGER SINCE NEGATIVE MOMENTS RESULTING FROM BOTH GRAVITY AND

EARTHQUAKE LOADS NEED TO BE RESISTED AT THE BEAM/COLUMN JOINTS)

• MATERIALS : C20 CONCRETE

S420 STEEL (FOR BOTH LONGITUDINAL AND

(3)

• BUILDING : OFFICE BUILDING LOCATED IN SEISMIC ZONE 1

LOCAL SOIL CLASS: Z2

• LOADS : DEAD LOAD (SELF WEIGHT) : G = 20 kN/m LIVE LOAD : Q = 10 kN/m

(DISTRIBUTED LOAD ON THE BEAMS)

(NOTE: IF GIVEN THE PLAN VIEW OF THE BUILDING, NEED TO CALCULATE THE DISTRIBUTED DEAD AND LIVE LOAD PER UNIT LENGTH ON THE

BEAMS, BASED ON THE TRIBUTARY WIDTH OF EACH BEAM AND THE DEAD AND LIVE LOAD VALUES DISTRIBUTED OVER THE FLOOR AREA)

• RESPONSE SPECTRUM VARIABLES (TDY-2007):

EQ Zone

Soil Type

Spectral Acceleration Coefficient Spectral Acceleration

(4)

• SEISMIC ZONE 1 : A0 = 0.40

LOCAL SOIL TYPE Z2 : TA = 0.15 sec TB = 0.40 sec

BUILDING IMPORTANCE COEFFICIENT:

ALWAYS TAKE BUILDING IMPORTANCE COEFFICIENT AS

I = 1.0 FOR ASSESSMENT OF EXISTING BUILDINGS USING

PUSHOVER ANALYSIS METHOD (SECTION 7.4.2 IN TDY-2007)

• STORY MASSES (TDY-2007):

SHALL BE DEFINED IN ACCORDANCE WITH STORY WEIGHTS DEFINED IN SECTION 2.7.1.2:

THEREFORE,

w

i = (20 kN/m)(14 m) + (0.30)(10 kN/m)(14 m)

= (20 kN/m)(14 m) + (3 kN/m)(14 m)

= (23 kN/m)(14 m)

= 322 kN

STORY WEIGHTS :

w

i = 322 kN (ON SINGLE FRAME)

STORY MASSES :

m

i =

w

i

/ g

= (322 x 103 N) / (9.81 m/s2)

m

i = 32,823 kg = 33 tons (ON SINGLE FRAME)

Live load participation factors (n)

Storage facilities

Schools, dorms, sports facilities, theatres, concert halls, garages, restaurants, stores, etc.

Building Type and Function

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• SECTION STIFFNESS (TDY-2007):

FOR C20 : Ec 28 = 28,000 MPA

FOR ALL BEAMS : I0 = (1/12)(0.450 m)(0.600 mm)3 = 0.0081 m4

EI0 = 226800 kN.m2

EI = 0.40(EI0)

FOR COLUMNS : Acfcm = (0.45 m)(0.45 m)(20x103 kN/m2)

= 4050 kN

ASSUME INTERIOR COLUMNS RESIST 50% OF THE

TOTAL VERTICAL LOAD, WHEREAS EACH EXTERIOR COLUMN RESISTS 25% OF THE TOTAL VERTICAL

LOAD AT EACH STORY.

(SINCE THE TRIBUTARY AREA OF THE INTERIOR COLUMNS ARE TWICE OF THE EXTERIOR COLUMNS)

¾ Use cracked section stiffness when modeling the structure ¾ N_D : axial load under vertical loads only (service loads)

¾ Interpolation between 0.40EI₀ – 0.80EI₀for intermediate values

of axial load

¾ f_cm : existing concrete compressive strength (no material

factor)

(Beams)

(Columns, Shear Walls)

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STORY WEIGHT = 322 kN ( FROM g + 0.30q ) THEREFORE:

C201 AND C203 : (ND/ Acfcm) = [(0.25)(322 kN)]/(4050 kN)] = 0.02

C202 : (ND/ Acfcm) = [(0.50)(322 kN)]/(4050 kN)] = 0.04

C101 AND C103 : (ND/ Acfcm) = [(0.25)(2 x 322 kN)]/(4050 kN)] = 0.04

C102 : (ND/ Acfcm) = [(0.50)(2 x 322 kN)]/(4050 kN)] = 0.08

SECTION STIFFNESS COEFFICIENTS:

USE EI = 0.40(EI0) FOR ALL COLUMNS

Ec 28 = 28,000 MPA I0 = (1/12)(0.450 m)(0.450 mm)3 = 0.00342 m4 EI0 = 95800 kN.m2 7 m 3.5 m

P

1

P

2 7 m 3.5 m B101 B102 B201 B202 C 101 C 102 _C10 3 C 201 C 202 _C203

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• ADDITIONAL CHECKS (TDY-2007):

• NUMBER OF STORIES = 2

• ASSUMING THE FRAMES IN THE BUILDING ARE SYMMETRICCALLY PLACED, THE TORSIONAL IRREGULARITY CONSTANT = 1.0

• NEED TO CALCULATE THE MODE SHAPES IN ORDER TO DETERMINE THE EFFECTIVE MASS RATIO CORRESPONDING TO THE FIRST MODE OF VIBRATION (COMPUTER MODEL)

Pushover analysis method can be used only if:

¾ Number of stories (not including basement) < 8 ¾ Torsional irregularity constant for the building < 1.4

¾ The ratio of the effective mass corresponding to the first mode

of vibration to the total mass of the building > 0.70

1 1

0.70

x N i i

M

m

=

>

∑

¾ During the pushover analysis, the distribution (pattern) of lateral story forces acting on the building can be assumed to be constant

¾ The distribution of the lateral story forces shall be proportional to the product of the mass of each story and the amplitude of

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•

PLASTIC MOMENT CAPACITIES FOR COMPUTER MODELING: CAN USE THE PROGRAM BETONARME TO DERIVE THE MOMENT- CURVATURE RELATIONSHIPS FOR THE BEAMS AND THE P-M INTERACTION DIAGRAMS FOR THE COLUMNS:

(http://www.ce.metu.edu.tr/betonarme)

• BEAMS : 450 mm x 600 mm WITH 3-φ22 AT THE TOP

3-φ22 AT THE BOTTOM

THE ENTIRE LENGTH OF THE BEAMS.

COVER TO THE CENTER OF LONGITUDINAL REINFORCEMENT = 40 mm

Eksenel b h d Cc

Yük fck γmc fyk fu γms Es εsh εsu (mm) (mm) (mm) (mm)

(Basınç +) 450 600 560 20

(kN) (MPa) (MPa) (MPa) (MPa)

0,00 20,00 1,00 420 600 1 200.000 0,01 0,1

No. φA Adet Alan

(mm) (mm2) 1 22 3 1140 2 22 3 1140 3 4 5 6 7 8 9 10 fyw (MPa) 420 φe (mm) 8 s (mm) 150 bk (mm) 402 hk (mm) 552 b'k (mm) 402 h'k (mm) 552 ess 1 ETRİYE BİLGİLERİ M o m e n t E ğ r i l i k P r o g r a m ı (mm) MALZEME ÖZELLİKLERİ Beton -260 260 Uzaklık BETON KESİT

BOYUNA DONATI DÜZENLEMESİ

Kesit Merkezine Boyuna Donatı 0 50 100 150 200 250 300 350 400 0,0000 0,0500 0,1000 0,1500 0,2000 0,2500 Eğrilik (rad/m) Moment (k N .m) HESAPLA YIELD MOMENT My = 250 kN.m

(9)

• COLUMNS : 450 mm x 450 mm WITH 8-φ22

φ8 STIRRUPS AT A SPACING OF 200 mm PROVIDED

ALONG THE ENTIRE LENGTH OF THE COLUMNS.

COVER TO THE CENTER OF LONGITUDINAL REINFORCEMENT = 40 mm

fck γmc fyk γms Es

(MPa) (MPa) (MPa)

20 1,00 420 1,00 200.000 Genişlik (b) Yükseklik (h) (mm) (mm) 450 450

No. Donatı Alanı

Kesit Merkezinden Uzaklık (xi) (mm2₎ (mm) 1 1140 -185 2 760 0 N (kN): 0,0 3 1140 185 M (kN.m): 240,3 4 5 6

BETON ve ÇELİK MODELLERİ

DONATI DÜZENLEMESİ D İ K D Ö R T G E N K E S İ T A N A L İ Z İ KESİT GEOMETRİSİ -2000 -1000 0 1000 2000 3000 4000 5000 0 50 100 150 200 250 300 350 400 Moment, M (kN.m) Eks en el Yü k, P ( kN ) BU PROGRAMDA:

1) Betonun çekme dayanı ihmal edilmektedir. 2) Beton basınç dağılımı dikdörtgen alınmaktadır. 3) Çelik modelinde pekleşme ihmal edilmektedir. 4) Sargı etkisi göz önüne alınmamaktadır.

(+) xi

(-) xi

NOTE THAT WE NEED THE P-M INTERACTION CURVES FOR THE COLUMNS SINCE THE AXIAL LOAD ON EACH COLUMN IS NOT THE SAME AND ALSO SINCE THE AXIAL LOAD WILL CHANGE WHEN LATERAL LOADS ARE APPLIED.

CAN ALSO RUN MOMENT-CURVATURE ANALYSES FOR THE COLUMNS TO CHECK (TO COMPARE THE RESULTS OF THE P-M INTERACTION ANALYSES WITH THE RESULTS OF THE MOMENT-CURVATURE ANALYSES)

(10)

FOR EXAMPLE, FOR COLUMN C102, THE AXIAL LOAD DUE TO VERTICAL LOADS ONLY IS ND = [(0.50)(2x322 kN)] = 322 kN.

fck γmc fyk γms Es

(MPa) (MPa) (MPa)

20 1,00 420 1,00 200.000 Genişlik (b) Yükseklik (h) (mm) (mm) 450 450

No. Donatı Alanı

Kesit Merkezinden Uzaklık (xi) (mm2₎ (mm) 1 1140 -185 2 760 0 N (kN): 322,0 3 1140 185 M (kN.m): 292,1 4 5 6

BETON ve ÇELİK MODELLERİ

DONATI DÜZENLEMESİ D İ K D Ö R T G E N K E S İ T A N A L İ Z İ KESİT GEOMETRİSİ -2000 -1000 0 1000 2000 3000 4000 5000 0 50 100 150 200 250 300 350 400 Moment, M (kN.m) Ek se nel Yük , P ( kN ) BU PROGRAMDA:

1) Betonun çekme dayanı ihmal edilmektedir. 2) Beton basınç dağılımı dikdörtgen alınmaktadır. 3) Çelik modelinde pekleşme ihmal edilmektedir. 4) Sargı etkisi göz önüne alınmamaktadır.

(+) xi

(-) xi

FROM P-M INTERACTION ANALYSIS: Mult = 292 kN.m FOR N = 322 kN.

(Basınç +) 450 450 410 20

(kN) (MPa) (MPa) (MPa) (MPa)

322,00 20,00 1,00 420 420 1 200.000 0,01 0,1

No. φA Adet Alan (mm) (mm2₎ 1 22 3 1140 2 22 2 760 3 22 3 1140 4 5 6 7 8 9 10 fyw (MPa) 420 φe (mm) 8 s (mm) 200 bk (mm) 402 hk (mm) 402 b'k (mm) 402 h'k (mm) 402 ess 1 ETRİYE BİLGİLERİ 185 M o m e n t E ğ r i l i k P r o g r a m ı (mm) MALZEME ÖZELLİKLERİ Beton -185 0 Uzaklık BETON KESİT

Kesit Merkezine Boyuna Donatı 0 50 100 150 200 250 300 350 0,0000 0,0500 0,1000 0,1500 Eğrilik (rad/m) Mo m en t (kN.m ) HESAPLA

FROM MOMENT-CURVATURE ANALYSIS: My = 291 Kn.m for N=322 kN

(11)

NOTE: DO NOT USE MATERIAL FACTORS IN THE MOMENT- CURVATURE OF P-M INTERACTION ANALYSES FOR

ASSESSMENT OF EXISTING BUILDINGS.

(ALWAYS TAKE

γ

mc =1 AND γms = 1 IN THE PROGRAMS)

CAN LINEARIZE THE P-M INTERACTION DIAGRAM FOR THE COLUMNS (FOR EASY INPUT INTO THE STRUCTURAL ANALYSIS PROGRAM):

N = 4650 kN (COMPRESSION): M = 0 kN.m (PURE COMPRESSION) N = 1500 kN (COMPRESSION): M = 370 kN.m (BALANCED POINT) N = 0 kN : M = 240 kN.m (PURE BENDING) N = 1275 kN (TENSION) : M = 0 kN.m (PURE TENSION)

-2000 -1000 0 1000 2000 3000 4000 5000 0 50 100 150 200 250 300 350 400 Moment, M (kN.m) Ek sen el Yü k, P (kN )

(12)

•

COMPUTER MODELING USING SAP2000:

• NEED TO DETERMINE THE PERIODS OF VIBRATION, THE MODE

SHAPES OF VIBRATION, AND THE PUSHOVER CURVE TO PROCEED WITH THE PUSHOVER ANALYSIS.

• SET UP THE MODEL FOR THE FRAME USING CENTER-TO-CENTER DIMENSIONS BETWEEN THE BEAMS AND COLUMNS.

• ASSIGN FIXED ENDS AT THE BOTTOM (RESTRAINTS) • DEFINE MATERIALS:

o CONC:

MASS AND WEIGHT PER UNIT VOLUME = 0

MODULUS OF ELASTICITY = 28000000 kN/m2

POISSON’S RATION = 0.2 • DEFINE FRAME SECTIONS:

o ADD RECTANGULAR: BEAM (FOR ALL BEAMS)

0.45 m x 0.60 m

MATERIAL: CONC

SET MODIFIERS:

0.40 FOR MOMENT OF INERTIA ABOUT 2 AND 3 AXES 0 FOR MASS AND WEIGHT

LARGE VALUE (E.G., 1000000) FOR CROSS-SECTIONAL AREA, SHEAR AREAS, AND TORSIONAL CONSTANT

o ADD RECTANGULAR: COLUMN (FOR ALL COLUMNS) 0.45 m x 0.45 m

MATERIAL: CONC

SET MODIFIERS:

(13)

0 FOR MASS AND WEIGHT

LARGE VALUE (E.G., 1000000) FOR CROSS-SECTIONAL AREA, SHEAR AREAS, AND TORSIONAL CONSTANT

• ASSIGN FRAME SECTIONS

• ASSIGN END (LENGTH) OFFSETS TO THE BEAMS AND COLUMNS: (FOR RIGID BEAM-COLUMN JOINTS)

o FIRST STORY COLUMNS:

END I : 0 END J : 0.3 m

RIGID ZONE FACTOR : 1

o SECOND STORY COLUMNS: END I : 0.3 m END J : 0.3 m

RIGID ZONE FACTOR : 1

o ALL BEAMS:

END I : 0.225 m END J : 0.225 m

RIGID ZONE FACTOR : 1

• ASSIGN JOINT MASSES:

MASSES WILL BE DEFINED ONLY IN LATERAL DIRECTIONS.

SINCE THE BEAMS ARE AXIALLY RIGID (INFINITE CROSS-SECTIONAL AREA, IT DOES NOT MATTER WHICH JOINT YOU ASSIGN THE MASS IN A PARTICULAR STORY.

THEREFORE, CAN ASSIGN THE MASSES AT THE INTERIOR BEAM- COLUMN JOINTS AT EACH STORY.

o ASSIGN MASSES IN DIRECTION 2: (322 kN) / (9.81 m/sec2)

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• AT THIS POINT, CAN RUN A MODAL ANALYSIS TO DETERMINE THE NATURAL PERIODS OF VIBRATION AND MODE SHAPES OF

VIBRATION

o DEFINE ANALYSIS CASE: MODAL

o ANALYSIS CASE TYPE: MODAL

o TYPES OF MODES: EIGEN VECTORS

o START FROM UNSTRESSED STATE • ANALYZE: RUN ANALYSIS

o CASE NAME: MODAL

o RESULTS: MODE 1: T1 = 0.373 sec

MODE SHAPE VECTOR: φ1 = {0.0841, 0.1524}T

(CAN OBTAIN FROM DEFORMED SHAPE) MODE 2:

T2 = 0.113 sec

MODE SHAPE VECTOR: φ2 = {-0.1524, 0.0841}T

o NOTE THAT THE MODE SHAPES GIVEN ARE MASS NORMAL:

[

1 2

]

1 2 0.0841 0.1524 0.1524 0.0841 0 33 0 0 0 33 1 0 0 1 T m m m m φ φ ⎡ − ⎤ Φ = _{= ⎢} _⎥ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ =_⎢ _{⎥ ⎢}= _⎥ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ Φ _{Φ = ⎢} _⎥ ⎣ ⎦

(15)

• RECALL (TDY-2007):

BACK TO THE SAP2000 MODEL • DEFINE LOAD CASES:

o LOAD NAME: GRAVITY TYPE: DEAD, SELF WEIGHT MULTIPLIER = 1

o LOAD NAME: LATERAL TYPE: QUAKE, SELF WEIGHT

MULTIPLIER = 1, AUTO LATERAL LOAD: NONE • ASSIGN FRAME LOADS ON ALL BEAMS:

o DISTRIBUTED

o LOAD CASE NAME: GRAVITY

o COORD SYS: GLOBAL o DIRECTION: Z

o UNIFORM LOAD = g + nq = -[(20 kN/m)+(0.3)(10 kN/m)]= -23 kN/m • ASSIGN LATERAL LOADS AT STORY LEVELS

o SINCE THE BEAMS ARE AXIALLY RIGID (INFINITE CROSS-SECTIONALAREA, IT DOES NOT MATTER WHICH JOINT YOU ASSIGN THE LATERAL LOADS AT A PARTICULAR STORY. ¾ During the pushover analysis, the distribution (pattern) of

lateral story forces acting on the building can be assumed to be constant

¾ The distribution of the lateral story forces shall be proportional

to the product of the mass of each story and the amplitude of the first mode shape of vibration of that story.

(16)

THEREFORE, CAN ASSIGN THE LATERAL LOADS AT THE EXTERIOR BEAM-COLUMN JOINTS AT EACH STORY. o ASSIGN JOINT LOADS: FORCES

o LOAD CASE NAME: LATERAL o COORDINATE SYSTEM: GLOBAL

o LATERAL LOADS SHOULD BE PROPORTIONAL TO THE PRODUCT OF THE STORY MASS AND THE AMPLITUDE OF THE FIRST MODE SHAPE AT THAT STORY (TDY-2007).

o STORY MASSES ARE EQUAL IN THE PRESENT EXAMPLE,

THEREFORE, THE LATERAL LOADS WILL BE PROPORTIONAL TO THE FIRST MODE SHAPE.

o FIRST MODE SHAPE: φ1 = {0.0841, 0.1524}T

o 0.1524 / 0.0841 = 1.81

o THEREFORE, ASSIGN A FORCE GLOBAL Z OF 1 kN AT THE

FIRST STORY EXTERIOR JOINT AND A FORCE FLOBAL Z OF 1.81 kN AT THE SECOND STORY EXTERIOR JOINT.

• DEFINE PLASTIC HINGES:

o DEFINE: HINGE PROPERTIES

FOR BEAMS:

o ADD NEW PROPERTY

o HINGE PROPERTY NAME: BEAMHINGE o DEFORMATION CONTROLLED

o MOMENT M3

o MODIFY/SHOW PROPERTY o MOMENT SF = M_y = 250 kN.m

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o CURVATURE =

φ

y = 0.008 rad/m

o TYPE: MOMENT-CURVATURE

o HINGE LENGTH = 0.3 m (h/2 FOR THE BEAM) o DO NOT CHECK RELATIVE LENGTH

o DISPLACEMENT CONTROL PARAMETERS: CHECK SYMMETRIC

MOMENT/SF CURVATURE/SF 1 0

1 50 (ANY LARGE VALUE) 0.2 50 (SUDDEN DROP) 0.2 60 (RESIDUAL)

LOAD CARRYING CAPACITY BEYOND POINT E DROPS TO ZERO

FOR COLUMNS:

o ADD NEW PROPERTY

o HINGE PROPERTY NAME: COLUMNHINGE o DEFORMATION CONTROLLED

o INTERACTING P-M3

o MODIFY/SHOW HINGE PROPERTY

o MOMENT-CURVATURE TYPE

o HINGE LENGTH = 0.225 m (h/2 FOR THE COLUMN)

o DO NOT CHECK RELATIVE LENGTH

o SCALE FACTOR FOR CURVATURE: USER SF=1

o LOAD CARRYING CAPACITY BEYOND POINT E DROPS TO ZERO

(18)

o INTERACTION SURFACE: USER DEFINITION

o AXIAL LOAD – DISPLACEMENT: ELASTIC-PERFECTLY PLASTIC o DEFINE/SHOW USER INTERACTION SURFACE

INTERACTION CURVE IS SYMMETRIC NUMBER OF POINTS ON EACH CURVE = 4 SCALE FACTORS: kN.m C

P = 1500 kN (Pbalanced) (REFERENCE POINT)

M3 = 370 kN (Mbalanced) (REFERENCE POINT)

FIRST AND LAST POINTS: (FACTORS TO BE MULTIPLIED BY THE REFERENCE POINT)

POINT 1: P = -3.1 M3 = 0 (PURE COMPRESSION) POINT 4: P = 0.85 M3 = 0 (PURE TENSION)

INTERACTION CURVE DATA: (FACTORS TO BE MULTIPLIED BY THE REFERENCE POINT)

POINT 2: P = -1 M3 = 1 (BALANCED POINT) POINT 3: P= 0 M3 = 0.65 (PURE BENDING) NOTE THAT IN SAP, TENSION IS POSITIVE WHEN

DEFINING THE P-M INTERACTION CURVE.

o MODIFY/SHOW MOMENT-CURVATURE CURVE DATA

MOMENT/YIELD MOM CURVATURE/SF 0 0

1 0

1 50 (ANY LARGE VALUE) 0.2 50 (SUDDEN DROP) 0.2 60 (RESIDUAL)

(19)

• ASSIGN PLASTIC HINGES: FOR BEAMS:

o SELECT ALL BEAMS

o ASSIGN/FRAME/HINGES

o HINGE PROPERTY: BEAMHINGE o RELATIVE DISTANCE = 0 ADD o RELATIVE DISTANCE = 1 ADD

o NOTE THAT SAP WILL ASSIGN THE HINGES OUTSIDE THE RIGID END OFFSETS

FOR COLUMNS:

o SELECT ALL COLUMNS

o ASSIGN/FRAME/HINGES

o HINGE PROPERTY: COLUMNHINGE

o RELATIVE DISTANCE = 0 ADD

o RELATIVE DISTANCE = 1 ADD

• DEFINE ANALYSIS CASES:

NONLINEAR ANALYSIS UNDER GRAVITY LOADS: o ADD NEW CASE

o ANALYSIS CASE NAME: PUSHOVER-GRAVITY o ANALYSIS CASE TYPE: STATIC

o ANALYSIS TYPE: NONLINEAR o ZERO INITIAL CONDITIONS

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o LOAD APPLICATION: FULL LOAD

o RESULTS SAVED: FINAL STATE ONLY o NONLINEAR PARAMETERS: DEFAULT

NONLINEAR ANALYSIS UNDER LATERAL LOADS: o ADD NEW CASE

o ANALYSIS CASE NAME: PUSHOVER-LATERAL o ANALYSIS CASE TYPE: STATIC

o ANALYSIS TYPE: NONLINEAR

o CONTINUE FROM STATE AT END OF NONLINEAR CASE: PUSHOVER GRAVITY (REQUIRED BY TDY-2007)

o MODAL ANALYSIS CASE: MODAL (IRRELEVANT)

o LOAD TYPE: LOAD, LOAD NAME: LATERAL, SCALE FACTOR: 1

o LOAD APPLICATION: DISPLACEMENT CONTROL

USE MONITORED DISPLACEMENT

LOAD TO A MONITORED DISPLACEMENT MAGNITUDE OF 0.25 m (ARBITRARY – TAKE REASONABLY LARGE) o MONITORED DISPLACEMENT:

DOF1 (HORIZONTAL) AT JOINT 3 (SECOND STORY EXTERIOR COLUMN JOINT)

o RESULTS SAVED: MULTIPLE STATES

MINIMUM NUMBER OF SAVED STEPS = 50 MAXIMUM NUMBER OF SAVED STEPS = 50

CHECK SAVE POSITIVE DISPLACEMENT INCREMENTS ONLY

(21)

• AT THIS POINT, CAN RUN THE PUSHOVER ANALYSIS • ANALYZE: RUN ANALYSIS

o RUN THE GRAVITY AND THE

PUSHOVER-LATERAL CASES TOGETHER (PUSHOVER-GRAVITY SHOULD COME FIRST

• RESULTS:

o DISPLAY: SHOW STATIC PUSHOVER CURVE: TOTAL BASE

SHEAR VERSUS MONITORED DISPLACEMENT (LATERAL DISPLACEMENT AT THE TOP – DOF 1 AT JOINT 3)

(22)

o DISPLAY: SHOWHINGE RESULTS

A BEAM HINGE IS SHOWN BELOW

NOTE THAT THE PLASTIC ROTATION OF THE HINGE AT ANY POINT DURING THE ANALYSIS IS PROVIDED BY THE PROGRAM

(23)

A COLUMN HINGE IS SHOWN BELOW

NOTE THAT THE PLASTIC ROTATION OF THE HINGE AT ANY POINT DURING THE ANALYSIS IS PROVIDED BY THE PROGRAM

(24)

•

CONVERSION OF THE PUSHOVER CURVE INTO THE MODAL CAPACITY CURVE (TDY-2007): Pushover Curve Modal Capacity Curve Modal Hysteresis Skeleton curve

Modal spectral displacement

M o da l sp ec tr al acc .

(Modal mass defined for the first mode of vibration)

Effective mass defined for the first mode of vibration in the x-direction

Amplitude of the first mode shape at the top of the building (N’th story) defined for the first mode of vibration in the x-direction

Participation factor defined for the first mode of vibration in the x-direction

2 1 1 1 1 1 1 2 1 1 1 1 1 1 x x N x i xi i N i xi i x x L M M L m M m L M = = = = Φ = Φ Γ =

∑

(25)

11 21 1 2

0.0841

0.1524

33

x x

m

Φ

=

Φ

=

2 1 1 1

(33)(0.0841)

(33)(0.1524)

7.8045

x i xi i

L

m

=

_∑

Φ =

+

=

2 2 2 2 1 1 1

(33)(0.0841)

(33)(0.1524)

1.0

i xi i

M

m

=

_∑

Φ =

+

=

1 1 1

7.8045

x x

L

M

Γ =

=

2 1 1 1

60.91

x x

L

M

=

• CAN GENERATE THE FOLLOWING TABLE:

uxN1 (m) Vx1 (kN) Mx1 φx21 Γx1 d1 (m) a1 (m/s2) 0,000 0 60,91 0,1524 7,805 0,000 0,00 0,005 73 60,91 0,1524 7,805 0,004 1,20 0,010 146 60,91 0,1524 7,805 0,008 2,40 0,015 219 60,91 0,1524 7,805 0,013 3,60 0,025 328 60,91 0,1524 7,805 0,021 5,38 0,030 371 60,91 0,1524 7,805 0,025 6,09 0,038 409 60,91 0,1524 7,805 0,032 6,71 0,045 426 60,91 0,1524 7,805 0,038 6,99 0,050 434 60,91 0,1524 7,805 0,042 7,13 0,080 454 60,91 0,1524 7,805 0,067 7,45 0,100 462 60,91 0,1524 7,805 0,084 7,58 0,170 464 60,91 0,1524 7,805 0,143 7,62 0,210 464 60,91 0,1524 7,805 0,177 7,62 0,250 464 60,91 0,1524 7,805 0,210 7,62

(26)

PUSHOVER CURVE 0 50 100 150 200 250 300 350 400 450 500 0,000 0,050 0,100 0,150 0,200 0,250 0,300 Top Displacement, uxN1 (m) T o ta l B ase S h e a r, V x1 (k N )

MODAL CAPACITY CURVE

0,00 1,00 2,00 3,00 4,00 5,00 6,00 7,00 8,00 0,000 0,050 0,100 0,150 0,200 0,250

Modal Spectral Displacement, d1 (m)

M o d a l S p e c tr a l A c c ., a 1 ( m /s 2 )

(27)

•

RECALL (TDY-2007):

(OK – CHECKS)

Pushover analysis method can be used only if:

¾ Number of stories (not including basement) < 8 ¾ Torsional irregularity constant for the building < 1.4

¾ The ratio of the effective mass corresponding to the first mode

of vibration to the total mass of the building > 0.70

1 1

0.70

x N i i

M

m

=

>

∑

1 2 1 1 2 1

60.91 33 33

66

0.92

0.70

x i i x i i

M

tons

m

tons

M

m

= =

=

+

=

>

∑

(28)

•

DETERMINATION OF THE MODAL DISPLACEMENT DEMAND:

For flexible structures (high period of vibration) (T

For flexible structures (high period of vibration) (T₁₁>T>T_B_B):):

(T_B) (T_A) (T_B) (T_A)

(Equal Disp. Rule)

S

S_di1_di1 = S= S_{de1 (linear elastic)}_{de1 (linear elastic)}

S

_di1_di1

> S

_{de1 (linear elastic)}_{de1 (linear elastic)}

(T_A) (T_B)

For rigid structures (low period of vibration) (T

(29)

• FOR THE FRAME IN THIS EXAMPLE:

T1 = 0.373 sec.

LOCAL SOIL TYPE Z2: TB = 0.40 sec. T1 < TB

FROM THE MODAL CAPACITY CURVE: ay1 = 7.62 m/s2

Spectral Acceleration Coefficient Spectral Acceleration 1 1 1 1 1 1 1

1 (

1)

/

1

y B R y ae y y

R

T T

C

R

S

R

a

+

−

=

≥

=

[ ]

2 1 0

( )

(0.40)(1.0) 2.5 (9.81)

9.81 /

ae

S

=

A IS T g

=

m s

1 1 1

9.81 1.2874

7.62

ae y y

S

R

a

=

1 1 1 1

1 (

1)

/

_{1 (1.2874 1)(0.4 / 0.373)}

1.016 1.2874

y B R y

R

T T

C

R

+

−

₊

₋

=

(30)

Sdi1 = CR1Sde1

Sdi1 = CR1Sde1 = (1.016)(0.035) = 0.036 m

( IS THE MODAL DISPLACEMENT DEMAND)

•

CONVERSION OF THE MODAL DISPLACEMENT DEMAND INTO

DISPLACEMENT DEMAND (TARGET DISPLACEMENT):

IS THE DISPLACEMENT DEMAND (TARGET DISPLACEMENT

AT THE TOP OF THE BUILDING)

1 1 2 2 1

9.81

0.035 (2 / )

(2 / 0.373)

ae de

S

m

T

π

=

( ) 1 1

0.036

p di

d

=

S

=

m

( ) 1 p

d

(Ötelenme İstemi)

Modal Capacity Curve Pushover Curve

Back-conversion:

Modal displacement demand for the first mode

Displacement demand(target displacement)at the top of the building (N’th story) in the x-direction

( ) ( ) 21 21 1 1

(0.1524)(7.8045)(0.036)

0.043

p p x x x

u

= Φ Γ

d

=

m

( ) 21 p x

u

(31)

•

THE COMPUTER MODEL OF THE FRAME (ALREADY DEVELOPED) NEEDS TO BE PUSHED UP TO THE DISPLACEMENT DEMAND:

o DEFINE: ANALYSIS CASES

o MODIFY/SHOW CASE: PUSHOVER LATERAL o LOAD APPLICATION: MODIFY/SHOW

o LOAD TO A MONITORED DISPLACEMENT MAGNITUDE OF 0.043 m

(32)

•

DETERMINATION OF TOTAL CURVATURE AND STRAIN DEMANDS:

o AT THE TARGET DISPLACEMENT (DISPLACEMENT DEMAND), PLASTIC HINGES HAVE FORMED ON: C101, C102, C103, B101, B102

o NEED TO CALCULATE THE TOTAL CURVATURE DEMANDS AT THE SECTIONS WHERE THE PLASTIC HINGES HAVE

DEVELOPED. B101 B102 B201 B202 C101 C102 C103 C201 C202 _C203

(33)

• FOR EXAMPLE, FOR THE PLASTIC HINGE ON BEAM B102, THE PLASTIC ROTATION DEMAND (AT THE TARGET DISPLACEMENT) IS θP = 6.20x10-3 rad

o CONVERT THE PLASTIC ROTATION DEMAND TO PLASTIC CURVATURE DEMAND :

φ

P

= θ

P

/ L

P = (6.20x10-3 rad)/(0.6m / 2) = 0.021 rad/m

o CONVERT THE PLASTIC CURVATURE DEMAND TO TOTAL CURVATURE DEMAND:

(34)

o OBTAIN THE YIELD CURVATURE (

φ

_Y) FROM THE RESULTS OF THE MOMENT-CURVATURE ANALYSIS (ALREADY PERFORMED) RECALL FOR ALL BEAMS: YIELD MOMENT MY = 250 kN.m

YIELD CURVATURE

φ

Y = 0.008 rad/m

(Basınç +) 450 600 560 20

(kN) (MPa) (MPa) (MPa) (MPa)

0,00 20,00 1,00 420 600 1 200.000 0,01 0,1

No. φA Adet Alan (mm) (mm2) 1 22 3 1140 2 22 3 1140 3 4 5 6 7 8 9 10 fyw (MPa) 420 φe (mm) 8 s (mm) 150 bk (mm) 402 hk (mm) 552 b'k (mm) 402 h'k (mm) 552 ess 1 ETRİYE BİLGİLERİ M o m e n t E ğ r i l i k P r o g r a m ı (mm) MALZEME ÖZELLİKLERİ Beton -260 260 Uzaklık BETON KESİT

Kesit Merkezine Boyuna Donatı 0 50 100 150 200 250 300 350 400 0,0000 0,0500 0,1000 0,1500 0,2000 0,2500 Eğrilik (rad/m) Mome nt (k N.m) HESAPLA

o THEREFORE, THE TOTAL CURVATURE DEMAND ON BEAM 102 IS CALCULATED AS:

(35)

o FROM THE RESULTS OF THE MOMENT-CURVATURE ANALYSIS, FOR

φ =

0.029 rad/m,

ε

ci = 0.00183 (AT THE EXTREME FIBER IN COMPRESSION)

c = 6.33 cm,

ε

s = (εc)[(d-c)/c]

= (0.00183)[(56 cm - 6.33 cm)/ 6.33 cm] = 0.0143

ε

_s = 0.0143 (AT THE OUTER LAYER OF TENSION STEEL)

• SIMILARLY, FOR THE PLASTIC HINGE ON COLUMN C102, THE

PLASTIC ROTATION DEMAND (AT THE TARGET DISPLACEMENT) IS θP = 3.25x10-3 rad

(36)

o CONVERT THE PLASTIC ROTATION DEMAND TO PLASTIC CURVATURE DEMAND :

φ

P

= θ

P

/ L

P = (3.25x10-3 rad)/(0.45m / 2) = 0.014 rad/m

o CONVERT THE PLASTIC CURVATURE DEMAND TO TOTAL CURVATURE DEMAND:

φ

T

= φ

Y

+ φ

P

o OBTAIN THE YIELD CURVATURE (

φ

_Y) FROM THE RESULTS OF A

MOMENT-CURVATURE ANALYSIS ON COLUMN C102 UNDER THE AXIAL LOAD THAT THE COLUMN EXPERIENCES AT THE

TARGET DISPLACEMENT.

o AT THE TARGET DISPLACEMENT, THE AXIAL LOAD ON COLUMN C102 IS 351 kN (FROM SAP2000 – AXIAL FORCE DIAGRAM)

(Basınç +) 450 450 410 20

(kN) (MPa) (MPa) (MPa) (MPa)

351,00 20,00 1,00 420 420 1 200.000 0,01 0,1

No. φA Adet Alan (mm) (mm2₎ 1 22 3 1140 2 22 2 760 3 22 3 1140 4 5 6 7 8 9 10 fyw (MPa) 420 φe (mm) 8 s (mm) 200 bk (mm) 402 hk (mm) 402 b'k (mm) 402 h'k (mm) 402 ess 1 ETRİYE BİLGİLERİ 185 M o m e n t E ğ r i l i k P r o g r a m ı (mm) MALZEME ÖZELLİKLERİ Beton -185 0 Uzaklık BETON KESİT

Kesit Merkezine Boyuna Donatı 0 50 100 150 200 250 300 350 0,0000 0,0500 0,1000 0,1500 Eğrilik (rad/m) Moment (k N .m) HESAPLA o APPROXIMATELY:

φ

Y = 0.01 rad/m

(37)

o THEREFORE, THE TOTAL CURVATURE DEMAND ON COUMN C102 IS CALCULATED AS:

φ

T

= φ

Y

+ φ

P = 0.01 + 0.014 = 0.024 rad/m

o FROM THE RESULTS OF THE MOMENT-CURVATURE ANALYSIS FOR COLUMN C102 ABOVE, FOR

φ =

0.024 rad/m,

ε

ci = 0.0027 (AT THE EXTREME FIBER IN COMPRESSION)

c = 11.34 cm,

ε

s = (εc)[(d-c)/c]

= (0.0027)[(41 cm - 11.34 cm) / 11.34 cm] = 0.0071

ε

s = 0.0071 (AT THE OUTER LAYER OF TENSION STEEL)

•

(NEED TO REPEAT THESE CALCULATIONS FOR COLUMNS C101 AND

C103, AND FOR BEAM B101 (AT EVERY SECTION WHERE A PLASTIC HINGE HAS FORMED)

•

AT EACH SECTION WHERE A PLASTIC HINGE HAS FORMED, NEED TO

CALCULATE THE STRAIN DEMANDS ON CONCRETE IN COMPRESSION AND STEEL IN TENSION (εci

AND ε

s

)

(38)

•

STRAIN CAPACITIES GIVEN IN TDY-2007 FOR CONCRETE AND STEEL:

o

THE CALCULATED STRAIN DEMANDS ON CONCRETE AND

STEEL (εci

AND ε

s) NEED TO BE COMPARED WITH THE CODE

DAMAGE LIMITS TO ASSESS THE LEVEL OF DAMAGE IN EACH MEMBER WHERE A PLASTIC HINGE HAS FORMED.

o

FOR EXAMPLE, FOR BEAM B102:

ε

_ci = 0.00183 < (ε_cg)MN = 0.004

ε

_s = 0.0143 < (ε_sg)MN = 0.010

THEREFORE, BEAM B102 IS IN THE “BELİRGİN HASAR

BÖLGESİ” (VISIBLE DAMAGE ZONE)

For confined concrete:

(Volumetric ratio of existing confinement steel) (Volumetric ratio of confinement steel

that is required for design of a new building)

(Unconfined concrete)

(Confined concrete)

(39)

o

FOR COLUMN C102:

ε

_ci = 0.0027 < (ε_cg)MN = 0.004

ε

_s = 0.0071 < (ε_sg)MN = 0.010

THEREFORE, BEAM B102 IS IN THE “MINIMUM HASAR

BÖLGESİ” (MINIMUM DAMAGE ZONE)

• BASED ON THE DISTRIBUTION OF DAMAGE IN THE BEAMS

AND COLUMNS AT EACH STORY, DETERMINE THE

PERFORMANCE LEVEL OF THE STRUCTURE AND MAKE

APPROPRIATE DECISIONS TOWARD REHABILITATION.

(PER SPECIFICATIONS OF TDY-2007).