Pushover Example Large

8/13/2019 Pushover Example Large

1/39

CE 58A - PUSHOVER ANALYSIS HANDOUT #11

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

P1

P2

7 m

3.5 m

B101 B102

B201 B202

C 1 0 1

C 1 0 2

C 1 0 3

C 2 0 1

C 2 0 2

C 2 0 3


2/39

2

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

φ8 STIRRUPS AT A SPACING OF 200 mm PROVIDED ALONGTHE ENTIRE LENGTH OF THE COLUMNS.

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

3-φ22 AT THE BOTTOM

φ8 STIRRUPS AT A SPACING OF 150 mm PROVIDED ALONG

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

TRANSVERSE REINFORCEMENT)


3/39

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


4/39

4

• SEISMIC ZONE 1 : A0 = 0.40

LOCAL SOIL TYPE Z2 : T A = 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, wi = (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 : wi = 322 kN (ON SINGLE FRAME)

STORY MASSES : mi = wi / g = (322 x 103 N) / (9.81 m/s2)

mi = 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

Residental buildings, office buildings, hotels, hospitals, etc.


5/39

5

• SECTION STIFFNESS (TDY-2007):

FOR C20 : Ec 28 = 28,000 MPA

FOR ALL BEAMS : I 0 = (1/12)(0.450 m)(0.600 mm)3 = 0.0081 m4

EI 0 = 226800 kN.m2

EI = 0.40( EI 0)

FOR COLUMNS : Acf cm = (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)

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

Interpolation between 0.40EI 0 – 0.80EI 0 for intermediate valuesof axial load

f cm : existing concrete compressive strength (no material

factor)

(Beams)

(Columns,

Shear Walls)


6/39

6

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

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

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

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

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

SECTION STIFFNESS COEFFICIENTS:

USE EI = 0.40( EI 0) FOR ALL COLUMNS

Ec 28 = 28,000 MPA

I 0 = (1/12)(0.450 m)(0.450 mm)3 = 0.00342 m4

EI 0 = 95800 kN.m2

7 m

3.5 m

P1

P2

7 m

3.5 m

B101 B102

B201 B202

C 1 0 1

C 1 0 2

C

1 0 3

C 2 0 1

C 2 0 2

C 2 0 3


7/39


8/39

8

• 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

φ8 STIRRUPS AT A SPACING OF 150 mm PROVIDED ALONGTHE ENTIRE LENGTH OF THE BEAMS.

COVER TO THE CENTER OF LONGITUDINAL

REINFORCEMENT = 40 mm

Eksenel b h d C c

Yük f ck γmc f yk f u γms E s 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. φ Adet Alan

(mm) (mm2)

1 22 3 1140

2 22 3 1140

3

4

5

6

7

8

9

10

f yw (MPa) 420

φe (mm) 8

s (mm) 150

bk (mm) 4 02

hk (mm) 5 52

b'k (mm) 402

h'k (mm) 552

ess 1

ETR İ YE Bİ LG İ LER İ

M om

ent E ğ r i l i k P r

og r amı

(mm)

MALZEME ÖZELLİ KLER İ Beton

-260

260

Uzaklık

BETON KES İ T

BOYUNA DONATI DÜZENLEMES İ 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

E ğ rilik (rad/m)

M o m e n t

( k N . m

)

HESAPLA

YIELD MOMENT My = 250 kN.m

YIELD CURVATURE y = 0.008 rad/m


9/39

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

f ck γ mc f yk γ ms E s

(MPa) (MPa) (MPa)

20 1,00 420 1,00 200.000

Geni şlik

(b)

Yükseklik

(h)

(mm) (mm)

450 450

No. Donat ı Alanı

Kesit

Merkezinden

Uzakl ık (x i )

(mm2 ) (mm)

1 1140 -185

2 760 0 N (kN): 0,03 1140 185 M (kN.m): 240,34

5

6

BETON ve ÇEL İ K MODELLER İ

DONATI DÜZENLEMES İ

Dİ K DÖ R T G E N K E S İ T

AN

A

Lİ Z İ

KES İ T GEOMETR İ S İ

-2000

-1000

0

1000

2000

3000

4000

5000

0 50 100 150 200 250 300 350 400

Moment, M (kN.m)

E k s e n e l Y ü k ,

P ( k N )

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)


10/39

10

FOR EXAMPLE, FOR COLUMN C102, THE AXIAL LOAD DUE TO

VERTICAL LOADS ONLY IS ND = [(0.50)(2x322 kN)] = 322 kN.

f ck γ mc f yk γ ms E s

(MPa) (MPa) (MPa)

20 1,00 420 1,00 200.000

Geni şlik

(b)

Yükseklik

(h)

(mm) (mm)

450 450

No. Donat ı Alanı

Kesit

Merkezinden

Uzakl ık (x i ) (mm

2 ) (mm)

1 1140 -185

2 760 0 N (kN): 322,03 1140 185 M (kN.m): 292,14

5

6

BETON ve ÇEL İ K MODELLER

İ

DONATI DÜZENLEMES İ

Dİ

K DÖ R T G E N K E S İ

T

AN

ALİ

Z İ

KES İ T GEOMETR İ S İ

-2000

-1000

0

1000

2000

3000

4000

5000

0 50 100 150 200 250 300 350 400

Moment, M (kN.m)

E k s e

n e l Y ü k ,

P ( k N )

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.

Eksenel b h d C c


(Basınç +) 450 450 410 20


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

No. φ Adet Alan

(mm) (mm2)

1 22 3 1140

2 22 2 760

3 22 3 1140

4

5

6

7

8

9

10

f yw (MPa) 420

φe (mm) 8

s (mm) 200

bk (mm) 402

hk (mm) 402

b'k (mm) 402

h'k (mm) 402

ess 1


185

M o

ment E ğ r i l i k P

r og r amı

(mm)


-185

0

Uzaklık

BETON KES İ T


Boyuna Donatı

0

50

100

150

200

250

300

350

0,0000 0,0500 0,1000 0,1500

E ğ rilik (rad/m)

M o m e n t

( k N . m

)

HESAPLA

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

(CHECK OK)


11/39

11

NOTE: DO NOT USE MATERIAL FACTORS IN THE MOMENT-

CURVATURE OF P-M INTERACTION ANALYSES FOR

ASSESSMENT OF EXISTING BUILDINGS.

(ALWAYS TAKE γmc =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)

E k s e n e l Y ü k ,

P ( k N )


12/39

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:

0.40 FOR MOMENT OF INERTIA ABOUT 2 AND 3 AXES


13/39

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


o ALL BEAMS:

END I : 0.225 m

END J : 0.225 m


• 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 MASSIN 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)

= 33 kN.sec2/m (33 tons)


14/39

14

• 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 ANALYSISo 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

mm

m

m

φ φ −⎡ ⎤

Φ = =⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

⎡ ⎤Φ Φ = ⎢ ⎥

⎣ ⎦


15/39

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: Zo 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.

During the pushover analysis, the distribution (pattern) of

lateral story forces acting on the building can be assumed to

be constant

The distribution of the lateral story forces shall be proportional

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


16/39

16

THEREFORE, CAN ASSIGN THE LATERAL LOADS AT THE

EXTERIOR BEAM-COLUMN JOINTS AT EACH STORY.

o ASSIGN JOINT LOADS: FORCESo 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 = My = 250 kN.m


17/39

17

o CURVATURE = φy = 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

o MOMENT-CURVATURE DEPENDENCE IS SYMMETRIC

o

MODIFY/SHOW P-M3 INTERACTION SURFACE DATA


18/39

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 01 50 (ANY LARGE VALUE)

0.2 50 (SUDDEN DROP)

0.2 60 (RESIDUAL)


19/39

19

• ASSIGN PLASTIC HINGES:

FOR BEAMS:

o SELECT ALL BEAMS

o ASSIGN/FRAME/HINGES

o HINGE PROPERTY: BEAMHINGE

o RELATIVE DISTANCE = 0 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




RIGID END OFFSETS

• DEFINE ANALYSIS CASES:

NONLINEAR ANALYSIS UNDER GRAVITY LOADS:

o ADD NEW CASEo ANALYSIS CASE NAME: PUSHOVER-GRAVITY

o ANALYSIS CASE TYPE: STATIC

o ANALYSIS TYPE: NONLINEAR

o ZERO INITIAL CONDITIONS

o MODAL ANALYSIS CASE: MODAL (IRRELEVANT)

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


20/39

20

o LOAD APPLICATION: FULL LOAD

o RESULTS SAVED: FINAL STATE ONLY

o NONLINEAR PARAMETERS: DEFAULTNONLINEAR 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


RIGID END OFFSETS

o NONLINEAR PARAMETERS: DEFAULT


21/39

21

• AT THIS POINT, CAN RUN THE PUSHOVER ANALYSIS

• ANALYZE: RUN ANALYSIS

o RUN THE PUSHOVER-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 (LATERALDISPLACEMENT AT THE TOP – DOF 1 AT JOINT 3)


22/39

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


23/39

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


24/39

24

• CONVERSION OF THE PUSHOVER CURVE INTO THE MODAL CAPACITYCURVE (TDY-2007):

Pushover Curve

Modal CapacityCurve

Modal Hysteresis

Skeleton

curve

Modal spectral displacement M o d a l s p e c t r a l a c c .

(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

21

1

1

1 11

21 1

1

11

1

x x

N

x i xii

N

i xii

x x

L M

M

L m

M m

L

M

=

=

=

= Φ

= Φ

Γ =

∑

∑


25/39

25

11

21

1

2

0.0841

0.1524

33

33

x

x

m

m

Φ =

Φ =

==

2

1 11

(33)(0.0841) (33)(0.1524) 7.8045 x i xii

L m=

= Φ = + =∑

22 2 2

1 11

(33)(0.0841) (33)(0.1524) 1.0i xii

M m=

= Φ = + =∑

11

1

7.8045 x x L

M Γ = =

21

1

1

60.91 x x L

M M

= =

• CAN GENERATE THE FOLLOWING TABLE:

uxN1 (m) Vx1 (kN) Mx1 x21 x1 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


26/39

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 t a l B a s e S h e a r , V x 1 ( 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 t r a l A c c . , a 1

( m / s 2 )


27/39

27

• RECALL (TDY-2007):

(OK – CHECKS)

Pushover analysis method can be used only if:

Number of stories (not including basement) < 8

Torsional irregularity constant for the building < 1.4

The ratio of the effective mass corresponding to the first modeof vibration to the total mass of the building > 0.70

1

1

0.70 x N

ii

M

m=

>∑

1

2

1

12

1

60.91

33 33 66

0.92 0.70

x

ii

x

ii

M tons

m tons

M

m

=

=

=

= + =

= >

∑

∑


28/39

28

• DETERMINATION OF THE MODAL DISPLACEMENT DEMAND:

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

(TB)(TA) (TB)(TA)

(Equal Disp. Rule)(Equal Disp. Rule)

SSdi1di1 = S= Sde1 (linear elastic)de1 (linear elastic)

SSdi1di1 > S> Sde1 (linear elastic)de1 (linear elastic)

(TA) (TB)

For rigid structures (low period of vibration) (TFor rigid structures (low period of vibration) (T11


29/39

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: a y1 = 7.62 m/s2

Spectral Acceleration Coefficient

Spectral Acceleration

1 1

1

1

11

1

1 ( 1) /1

y B

R

y

ae y

y

R T T C

R

S R

a

+ −= ≥

=

[ ] 21 0 ( ) (0.40)(1.0) 2.5 (9.81) 9.81 /aeS A IS T g m s= = =

11

1

9.811.2874

7.62ae

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

+ − + −= = =


30/39

30

S di1 = C R1S de1

S di1 = C R1S de1 = (1.016)(0.035) = 0.036 m

( IS THE MODAL DISPLACEMENT DEMAND)

• CONVERSION OF THE MODAL DISPLACEMENT DEMAND INTODISPLACEMENT DEMAND (TARGET DISPLACEMENT):

IS THE DISPLACEMENT DEMAND (TARGET DISPLACEMENT

AT THE TOP OF THE BUILDING)

11 2 2

1

9.810.035

(2 / ) (2 / 0.373)

aede

S S m

T π π = = =

( )1 1 0.036 p

did S m= =

( )1 pd

(Ötelenme İstemi)

Modal Capacity Curve Pushover Curve

Back-conversion:

Modal displacement demand for the first modeDisplacement 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 xu d m= Φ Γ = =

( )21 p

xu


31/39

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


32/39

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

C 1 0 1

C 1 0 2

C 1 0 3

C 2

0 1

C 2 0 2

C 2 0 3


33/39

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 PLASTICCURVATURE DEMAND :

φP = θP / LP = (6.20x10-3 rad)/(0.6m / 2) = 0.021 rad/m

o CONVERT THE PLASTIC CURVATURE DEMAND TO TOTAL

CURVATURE DEMAND:

φT = φY + φP


34/39

34

o OBTAIN THE YIELD CURVATURE (φY) FROM THE RESULTS OF

THE MOMENT-CURVATURE ANALYSIS (ALREADY PERFORMED)

RECALL FOR ALL BEAMS: YIELD MOMENT MY = 250 kN.m

YIELD CURVATURE φY = 0.008 rad/m

Eksenel b h d C c


(Basınç +) 450 600 560 20


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

No. φ Adet Alan

(mm) (mm2)

1 22 3 1140

2 22 3 1140

3

4

5

6

7

8

9

10

f yw (MPa) 420

φe (mm) 8

s (mm) 150

bk (mm) 4 02

hk (mm) 5 52

b'k (mm) 402

h'k (mm) 552

ess 1


M o

ment E ğ r i l i k P

r og r amı

(mm)


-260

260

Uzaklık

BETON KES İ T


Boyuna Donatı

0

50

100

150

200

250

300

350

400

0,0000 0,0500 0,1000 0,1500 0,2000 0,2500

E ğ rilik (rad/m)

M o m e n t

( k N . m

)

HESAPLA

o THEREFORE, THE TOTAL CURVATURE DEMAND ON BEAM 102 IS

CALCULATED AS:

φT = φY + φP = 0.008 + 0.021 = 0.029 rad/m


35/39

35

o FROM THE RESULTS OF THE MOMENT-CURVATURE ANALYSIS,

FOR φ = 0.029 rad/m,

ci = 0.00183 (AT THE EXTREME FIBER IN COMPRESSION)

c = 6.33 cm,

εs = (εc)[(d-c)/c]

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

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


36/39

36

o CONVERT THE PLASTIC ROTATION DEMAND TO PLASTIC

CURVATURE DEMAND :

φP = θP / LP = (3.25x10-3 rad)/(0.45m / 2) = 0.014 rad/mo CONVERT THE PLASTIC CURVATURE DEMAND TO TOTAL

CURVATURE DEMAND:

φT = φY + φP

o OBTAIN THE YIELD CURVATURE (φY) 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)

Eksenel b h d C c


(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. φ Adet Alan

(mm) (mm2)

1 22 3 1140

2 22 2 760

3 22 3 1140

4

5

6

7

8

9

10

f yw (MPa) 420

φe (mm) 8

s (mm) 200

bk (mm) 4 02

hk (mm) 4 02

b'k (mm) 402

h'k (mm) 402

ess 1


185

M oment E ğ r i l i k P

r og r amı

(mm)


-185

0

Uzaklık

BETON KES İ T


Boyuna Donatı

0

50

100

150

200

250

300

350

0,0000 0,0500 0,1000 0,1500

E ğ rilik (rad/m)

M o m e n t

( k N . m

)

HESAPLA

o APPROXIMATELY: φY = 0.01 rad/m


37/39

37

o THEREFORE, THE TOTAL CURVATURE DEMAND ON COUMN

C102 IS CALCULATED AS:

φT = φY + φP = 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,

ci = 0.0027 (AT THE EXTREME FIBER IN COMPRESSION)

c = 11.34 cm,

εs = (εc)[(d-c)/c]

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

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 εs)


38/39

38

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

o THE CALCULATED STRAIN DEMANDS ON CONCRETE ANDSTEEL (εci AND εs) 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:

εci = 0.00183 < (εcg)MN = 0.004

εs = 0.0143 < (εsg)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)

(Unconfinedconcrete)

(Confinedconcrete)


39/39

o FOR COLUMN C102:

εci = 0.0027 < (εcg)MN = 0.004

εs = 0.0071 < (εsg)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).

Documents

Pushover Example Large