CE 581

Reinforced Concrete Structures

Lecture 11: Performance Based
Assessment and Strengthening

1
 Current Status
• Equivalent Lateral Force Procedure
- Assume global ductility (Ra)
- Detail accordingly
• Modal Superposition Procedure
- Include higher mode effects
• Time History Analysis
- Rarely used
W A(T1 )
- Tedious and requires hysteretic models Vt 
Ra (T1 )
2
 Target Performance
Dictated by codes (DBYBHY 2007, Section 1.2.1):
“....The objective of seismic resistant design is to
have no structural/nonstructural damage in low
magnitude earthquakes, limited and repairable
damage in moderate earthquakes and life safety
for extreme earthquakes...”

3
 Critique of Current Practice
Advantages :
- Simple to use
- Have proven to work
- Became a tradition all over the world
- Uncertainty is lumped and easier to deal with
Disadvantages :
- No clear connection between capacity and demand
- No option for interfering with the target performance
- No possibility of having the owner involved in the decision
process
- Not easily applicable to seismic assesment of existing structures

4
 Vision 2000
• First complete document published by SEAOC in 1995 addressing
performance based design principles.
• Then ATC (1996) and NEHRP (1997) guidelines follow.

5
6
7
 Performance Based Engineering
Methodology:
- Define Performance Levels (operational, immediate occupancy etc.)
- Define Response Targets (Damage, drift ratios, rotations etc.)
- Define Design Earthquakes (Response Spectrum, Synthetic G.M.)
- Explicit Check of Structural Performance against Design Earthquakes Using
Linear and Nonlinear Analysis
Concern:
- Uncertainties due to stiffness, strength, ductility, demand (earthquakes)
- Complications introduced (more loop holes)

CAN WE ACTUALLY ESTIMATE PERFORMANCE ?

8
 Turkish Earthquake Code-Chapter 7
- For Evaluation and Strengthening of Existing Buildings
(Is it easier to establish PBD guidelines for existing structures ?)
(Economical and academical reasons !!)
- Steps:
• Collect information from an existing structure
• Assess whether info is dependable and penalize accordingly
• Conduct structural analysis
- Linear static analysis
- Nonlinear static analysis (Pushover analysis)
- Incremental pushover analysis
- Time history analysis
• Identify for each member the damage level
• Decision based on number of elements at certain damage levels

9
 Linear Elastic Assesment

- Obtain base shear from design response spectrum

- Assume R = 1 and apply full earthquake force
- Compute member forces (Demand)
- Compute DCR = Demand / Capacity
10
 DCR Limits
If member satisfies
transverse steel
requirements of
Chapter 2, consider
as confined.

Based on these
tables, determine
performance level
of each member.

11
 Inelastic Assessment

F D V

Vy

V
dy d
Spectral
Acceleration
1.4
Spectral Acceleration,g

Capacity Capacity Curve

1.2
10% Response Spectra
1 5% elastic
7.5% spectra
0.8 Demand
0.6 Say
0.4

0.2

0
0 10 20 30 40 50

Spectral Displacement, in dy Spectral

Displacement
12
 Obtain a Capacity Curve
Base Shear

Bilinear approximation

Vy
Capacity curve from analysis

Roof Displacement
dy dt
Target displacement

A simple visual representation of strength, stiffness, ductility.

How about demand?

13
 How much to push?
dt = Co CR Sde
Co = First mode participation
CR = Ratio of inelastic to elastic displacement
Sa

TB
Te
CR = 1 if Te > TB (equal displacement)
CR = f (RY) if Te < TB (CR ≥ 1)
Sae
RY = Sae / Say
1  ( Ry  1)TB / Te
Say CR 
Ry

Sd
Sde
14
 Can we do better?
- Actual earthquake response is hard to predict anyways.
- Closest estimate can be found using inelastic time-history analysis.
- Difficulties with inelastic time history analysis:
- Suitable set of ground motion (Description of demand)
- hysteretic behavior models (Description of capacity)
- Computation time (Time)
- Post processing (Time and understanding)
Alternative approach is pushover analysis.

Düzce Ground Motion

0.6

0.4
Acceleration (g)

0.2

0 Sec.
0 5 10 15 20 25 30
-0.2

-0.4

-0.6

15
 Pushover Analysis
• Definition: Inelastic static analysis of a structure
using a specified (constant or variable) force
pattern from zero load to a prescribed ultimate
displacement.
• Use of it dates back to 1960s to1970s to
investigate stability of steel frames.
• Many computer programs were developed since
then with many features and limitations.

16
 Available Computer Programs
• Design Oriented:
SAP 2000, GTSTRUDL, RAM etc.
• Research Oriented:
Opensees, IDARC, SeismoStrut etc.

What is different?
• User interface capabilities
• Analysis options
• Member behavior options

17
 Section Damage Levels
Internal Force

Minimum Significant Severe Collapse

Damage Damage Damage Damage
Zone Zone Zone Zone

Strain

Damage levels are established based on concrete outermost

compressive fiber strain and steel strain (for nonlinear analysis
procedure).

18
 Section Damage Levels

How should these values be decided?

- Construction practice
- Experience of engineers
- Input of academicians
19
Curvature demand at target displacement (d=dt)

600

500

AK
400
GV GÇ

Moment
(kN.m)
Φp = θp / Lp 300

Φt = Φy + Φp 200

100

0
0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200

(Φy) Curvature
Eğrilik (Φt)
(rad/m)

20
 How do we estimate strains from a
structural analysis?
Moment Moment
My

Curvature Strain
øy øu

Moment
θpu =(øu – øy) Lp OR
My Utilize this idealized
θp =(ø – øy) Lp moment-rotation
Where Lp = 0.5h response during
inelastic structural
analysis
Plastic
θpu Rotations
21
 ATC-40/SAP 2000
AB : Linear elastic
b BC : Yielding
a
CD : Strength decay
C DE : Residual strength
Force
Yük
a, b and c values: depend
B
on type of inelasticity
D E
c
A
Deformation
Deformasyon

22
 Definition of Potential Plastic Hinges
• End regions of columns and beams (center for gravity loads)
are the potential plastic hinges
• Plastic hinges are hinges capable of resisting My (not
significantly more, hardening allowed) undergoing plastic
rotations
Rigid End
zones

Elastic
h Beam- Plastic
Column Hinges
Element

Lp
23
 Elastic Parts
For regions other than plastic hinging occurs, cracking is expected therefore
use of cracked stiffness is customary (0.4-0.8) EIo

0.4-0.8EIo
Moment

EIo

Eğrilik
Curvature 24
Pushover Analysis

25
 Steps of Pushover Analysis:
A Simple Incremental Procedure
1. Build a computational model of the structure

26
 Steps of Pushover Analysis

2. Define member behavior

– Beams: Moment-rotation relations
– Columns: Moment-rotation and Interaction Diagrams
– Beam-column joints: Assume rigid (DBYBHY 2007 )
– Walls: Model as beam columns but introduce a shear
spring to model shear deformations
– Use cracked rigidities for elastic portions

27
 Steps of Pushover Analysis

3. Apply gravity loads

1.0 G + n Q n=0.3 (live load reduction factor)
(if the interaction diagrams will not be used a good
estimate of the moment capacity of column hinges
needs to be made)
Possibilities:
- Based on initial gravity load analysis
- Based on a beam hinging mechanism
- Based on elastic lateral force analysis with an assumed
reasonable Ra value.

28
 Steps of Pushover Analysis

4. Specify a Lateral Load Profile:

(Inverted triangular, constant, first mode shape are some of the possibilities)
It is a good idea to have a spreadsheet page ready indicating all
members, current load increment
5. Lateral Load Incrementation:
Step 1:
Elastic analysis is valid up to the formation of the first hinge, i.e.
when the first critical location reaches its moment capacity.
• Find the lateral loads that cause first hinge formation (V1).
• Record all member forces and deformations (F1, d1).

29
 Steps of Pushover Analysis
Step 2:
Beyond Step 1, yielded element’s critical location can not take
any further moment. Therefore place an actual hinge at that
location. Conduct an analysis increment for this modified
structure. This load increment should be selected such that
upon summing the force resultant from this incremental step
and previous step, second hinge formation is reached.

V2 = V1 + ΔV
Results from Step 1 + Results from an
F2 = F1 + ΔF incremental analysis with a hinge placed at first
yield location = Second Hinge formation
d2 = d1 + Δd

30
 Steps of Pushover Analysis
.
.
Step i:
Similar to step 2 but additional hinges form and incremental
analysis steps are conducted for systems with more hinges.
Results are added to those from the previous step

Vi = Vi-1 + ΔV
Results from Step 1 + Results from an
Fi = Fi-1 + ΔF incremental analysis with a hinge placed at first
yield location = Second Hinge formation
di = di-1 + Δd

31
 Steps of Pushover Analysis
Step n:
Sufficient number of plastic hinges have formed and system
has reached a plastic mechanism. Note that this could be a
partial collapse mechanism as well. Beyond this point, system
rotated as a rigid body.

ANALYSIS DONE
- Plot Base Shear- Roof Displacement
- Check member rotations and identify performance levels

32
Example Application: 3 Story- 2 Bay RC Frame

MODEL

12 J8 15
J4 J12

3 6 9
3m

11 J7 14
J3 J11

2 5 8 3m

10 J6 13
J2 J10

4 3m
1 7

J1 J5 J9

6m 6m

33
 Assumptions
Assume
• Constant Axial Load on Columns for Analysis Steps
• Rigid-plastic with no hardening or softening moment-rotation behavior for columns
and beams
• Plastic hinging occurs when moment capacity is within 5% tolerance
• Load combinations 1.0 DL + 0.3 LL and 1.0 DL + 0.3 LL+1.0EQ to compute axial load
levels

SABİT YÜK HAREKETLİ YÜK YATAY YÜK

DL=10kN/m LL=2kN/m
EQ=60kN

DL=15kN/m LL=2kN/m
EQ=40kN

DL=15kN/m LL=2kN/m
EQ=20kN

34
 DATA
Columns Beams
10-f10
3-f10
60cm 50cm

3-f10

60cm 25cm

Steel (fyd=495 Mpa) M+ is the same as M-

Concrete (fcd=25 Mpa)
Clear cover=5 cm
E=2.779E+4 MPa

Note that if this is a seismic evaluation problem strength values obtained at

site should be used!

35
 Section Capacities
Elemnaların Moment-eğrilik ilişkileri
Idealized member moment curvature
elasto-plastik, pekleşmesiz
relations for estimated axial load level Φ

Φy
u l t

Eleman
Member N My
kN kNm rad/m rad/m
1 -83,786 124 0,0055 0,111
2 -51,347 115,5 0,0056 0,115
3 -19,872 107,5 0,0056 0,119 My
4 -253,392 166 0,0059 0,085
5 -158,905 143 0,0060 0,099
6 -64,797 119 0,0060 0,113

Moment
7 -124,104 133,5 0,0056 0,105
8 -77,747 122 0,0057 0,112
9 -31,201 110 0,0054 0,118
10 5,606 49 0,0073 0,103
11 1,421 50 0,0069 0,102
12 -17,233 53 0,0069 0,099 fy Eğrilik f ult
13 5,606 49 0,0073 0,103
14 1,421 50 0,0069 0,102
15 -17,233 53 0,0069 0,099

To be conservative smaller axial load from two load

combinations can be selected (as long as N<Nb)
36
 Step 1 First yielding stage
Total Base Shear (kN)=
Lateral Disp. at J4 (mm)=
6
0.2947
Frame Joint Myield M
Condition
J4 (monitored node ) Element Label kNm kNm
J1 124.0 -4.33
1 J2 124.0 20.60
J2 115.5 -22.14
COMBO2: 1.0 DL + 0.3 LL + 1.0 EQ 2 J3 115.5 21.00
J3 107.5 -22.23
DL=10kN/m 3 J4 107.5 27.35
EQ=3kN
LL=2kN/m J5 166.0 6.23
4 J6 166.0 -0.60
J6 143.0 3.50
DL=15kN/m 5 J7 143.0 -2.94
EQ=2kN J7 119.0 1.52
LL=2kN/m
6 J8 119.0 -3.29
J9 133.5 16.03
7 J10 133.5 -20.07
DL=15kN/m
EQ=1kN J10 122.0 26.88
LL=2kN/m 8 J11 122.0 -24.83
J11 110.0 22.95
9 J12 110.0 -30.82
J2 49.0 -42.74
10 J6 49.0 -49.58 YIELDED
J3 50.0 -43.24
11 J7 50.0 -49.28
J4 53.0 -27.35
12
Detection of first yield (moment J8 53.0 -34.34
J6 49.0 -45.48
reaches My±5%My ) 13 J10 49.0 -46.95
J7 50.0 -44.83
14 J11 50.0 -47.79
J8 53.0 -31.05
15 J12 53.0 -30.82 37
 Total Incremental Load (kN)= 6
Total Base Shear (kN) = 12

Step 2 (Incremental) Frame

Inc. Lateral Disp. at J4 (mm)=
Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
0.2865
0.5812
Condition
Element kNm kNm (kNm)
-4.33 6.39 2.06
1 20.60 0.76 21.36
-22.14 2.05 -20.10
2 21.00 -2.18 18.82
-22.23 0.24 -21.99
3 27.35 -1.82 25.53
ΔEQ=3kN
6.23 6.47 12.71
4 -0.60 0.39 -0.21
3.50 2.79 6.29
5 -2.94 -3.15 -6.09
ΔEQ=2kN
1.52 1.56 3.08
6 -3.29 -3.43 -6.72
New locations 16.03 6.48 22.51
7 -20.07 0.20 -19.87
ΔEQ=1kN at which yield 26.88 2.57 29.45
moments 8 -24.83 -2.26 -27.09
within 9
22.95 0.15 23.10
tolerance are -30.82 -1.80 -32.62
-42.74 1.29 -41.46
reached 10 -49.58 0.00 -49.58 YIELDED
-43.24 2.42 -40.82
11 -49.28 -2.36 -51.64 YIELDED
-27.35 1.82 -25.53
12 -34.34 -1.73 -36.07
Actual hinge at previously yielded 13
-45.48 2.40 -43.08
-46.95 -2.38 -49.33 YIELDED
location for the incremental analysis -44.83 2.35 -42.48
14 -47.79 -2.41 -50.19 YIELDED
-31.05 1.71 -29.34
15 -30.82 -1.80 -32.62
38
 Total Incremental Load (kN)= 42
Total Base Shear (kN) = 54
Inc. Lateral Disp. at J4 (mm)= 2.94

Step 3 (Incremental) Frame

Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
3.5212
Condition
Element kNm kNm (kNm)
2.06 57.79 59.85
1 21.36 12.12 33.48
-20.10 24.68 4.58
2 18.82 -16.19 2.64
-21.99 -2.12 -24.11
3 25.53 -18.94 6.58
ΔEQ=21kN 12.71 56.85 69.56
4 -0.21 12.18 11.97
6.29 24.58 30.87
5 -6.09 -13.41 -19.49
ΔEQ=14kN 3.08 0.99 4.07
6 -6.72 -34.94 -41.67
New location at 22.51 53.65 76.16
7 -19.87 18.00 -1.88
ΔEQ=7kN which yield 29.45 18.00 47.45
moment within 8 -27.09 -8.15 -35.24
tolerance are 23.10 -8.15 14.95
9 -32.62 -18.38 -51.00
reached
-41.46 12.56 -28.90
10 -49.58 0.00 -49.58 YIELDED
-40.82 14.07 -26.75
11 -51.64 0.00 -51.64 YIELDED
-25.53 18.94 -6.58
12 -36.07 -17.61 -53.68 YIELDED
Actual hinges at previously yielded -43.08 12.40 -30.68
13 -49.33 0.00 -49.33 YIELDED
location for the incremental analysis -42.48 14.40 -28.08
14 -50.19 0.00 -50.19 YIELDED
-29.34 17.33 -12.01
15 -32.62 -18.38 -51.00
39
 Total Incremental Load (kN)= 6
Total Base Shear (kN) = 60

Step 4 (Incremental) Frame

Inc. Lateral Disp. at J4 (mm)=
Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
0.4692
3.9904
Condition
Element kNm kNm (kNm)
59.85 8.59 68.44
1 33.48 2.00 35.48
4.58 3.91 8.49
2 2.64 -1.96 0.67
-24.11 0.29 -23.82
3 6.58 -1.96 4.63
ΔEQ=3kN 69.56 8.43 77.99
4 11.97 2.07 14.04
30.87 3.95 34.82
5 -19.49 -1.77 -21.26
ΔEQ=2kN 4.07 0.50 4.57
6 -41.67 -3.40 -45.07
New location at 7
76.16 7.95 84.12
which yield -1.88 2.90 1.02
ΔEQ=1kN 47.45 2.90 50.35
moment within 8 -35.24 -0.50 -35.74
tolerance are 14.95 -0.50 14.45
9
reached -51.00 -3.35 -54.36
-28.90 1.91 -26.99
10 -49.58 0.00 -49.58 YIELDED
-26.75 2.26 -24.49
11 -51.64 0.00 -51.64 YIELDED
-6.58 1.96 -4.63
12 -53.68 0.00 -53.68 YIELDED
Actual hinges at previously yielded -30.68 1.88 -28.79
13 -49.33 0.00 -49.33 YIELDED
location for the incremental analysis -28.08 2.27 -25.81
14 -50.19 0.00 -50.19 YIELDED
-12.01 3.40 -8.61
15 -51.00 -3.35 -54.36 YIELDED
40
 Total Incremental Load (kN)= 36
Total Base Shear (kN) = 96

Step 5 (Incremental) Frame

Inc. Lateral Disp. at J4 (mm)=
Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
3.41
7.4004
Condition
Element kNm kNm (kNm)
68.44 55.34 123.78
1 35.48 15.86 51.34
8.49 28.66 37.15
2 0.67 -6.38 -5.71
-23.82 10.42 -13.40
3
ΔEQ=18kN 4.63 -15.82 -11.19
77.99 54.50 132.49
4 14.04 16.03 30.06
34.82 28.70 63.52
5 -21.26 -6.00 -27.26
ΔEQ=12kN 4.57 10.75 15.33
6 -45.07 -15.83 -60.90
84.12 51.48 135.60 YIELDED
7 1.02 21.43 22.45
ΔEQ=6kN 8
50.35 21.43 71.78
-35.74 1.18 -34.57
14.45 1.18 15.62
9 -54.36 0.00 -54.36
-26.99 12.80 -14.19
10 -49.58 0.00 -49.58 YIELDED
-24.49 16.80 -7.69
11 -51.64 0.00 -51.64 YIELDED
-4.63 15.82 11.19
12 -53.68 0.00 -53.68 YIELDED
-28.79 12.68 -16.12
13 -49.33 0.00 -49.33 YIELDED
-25.81 16.75 -9.05
14 -50.19 0.00 -50.19 YIELDED
-8.61 15.83 7.22
15 -54.36 0.00 -54.36 YIELDED
41
 Total Incremental Load (kN)= 0.12
Total Base Shear (kN) = 96.12

Step 6 (Incremental) Frame

Inc. Lateral Disp. at J4 (mm)=
Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
0.01277
7.41317
Condition
Element kNm kNm (kNm)
123.78 0.25 124.03 YIELDED
1 51.34 0.03 51.38
37.15 0.08 37.23
2 -5.71 -0.03 -5.74
-13.40 0.03 -13.37
3 -11.19 -0.06 -11.25
ΔEQ=0.06kN 132.49 0.26 132.75
4 30.06 0.02 30.09
63.52 0.07 63.60
5 -27.26 -0.02 -27.29
ΔEQ=0.04kN 6
15.33 0.04 15.36
-60.90 -0.06 -60.96
135.60 0.00 135.60 YIELDED
7 22.45 0.09 22.54
71.78 0.09 71.87
ΔEQ=0.02kN 8 -34.57 0.00 -34.57
15.62 0.00 15.63
9 -54.36 0.00 -54.36
-14.19 0.05 -14.14
10 -49.58 0.00 -49.58 YIELDED
-7.69 0.06 -7.63
11 -51.64 0.00 -51.64 YIELDED
11.19 0.06 11.25
12 -53.68 0.00 -53.68 YIELDED
-16.12 0.05 -16.07
13 -49.33 0.00 -49.33 YIELDED
-9.05 0.06 -8.99
14 -50.19 0.00 -50.19 YIELDED
7.22 0.06 7.28
15 -54.36 0.00 -54.36 42
YIELDED
 Total Incremental Load (kN)= 9.6
Total Base Shear (kN) = 105.72

Step 7 (Incremental) Inc. Lateral Disp. at J4 (mm)=

Total Lateral Disp. at J4 (mm)=
1.3
8.71317
Frame M ΔM M + ∆M
Condition
Element kNm kNm (kNm)
124.03 0.00 124.03 YIELDED
1 51.38 4.04 55.42
37.23 8.81 46.05
2 -5.74 -3.63 -9.37
-13.37 2.07 -11.30
3 -11.25 -5.15 -16.40
ΔEQ=4.8kN 4
132.75 35.16 167.90 YIELDED
30.09 -3.63 26.45
63.60 2.03 65.63
5 -27.29 -2.56 -29.84
15.36 3.01 18.38
ΔEQ=3.2kN 6 -60.96 -5.18 -66.14
135.60 0.00 135.60 YIELDED
7 22.54 5.95 28.49
71.87 5.95 77.82
8
ΔEQ=1.6kN -34.57 -1.02 -35.58
15.63 -1.02 14.61
9 -54.36 0.00 -54.36
-14.14 4.77 -9.37
10 -49.58 0.00 -49.58 YIELDED
-7.63 5.70 -1.93
11 -51.64 0.00 -51.64 YIELDED
11.25 5.15 16.40
12 -53.68 0.00 -53.68 YIELDED
-16.07 5.67 -10.40
13 -49.33 0.00 -49.33 YIELDED
-8.99 5.57 -3.42
14 -50.19 0.00 -50.19 YIELDED
7.28 5.18 12.46
15 -54.36 0.00 -54.36 YIELDED

43
 Total Incremental Load (kN)= 39
Total Base Shear (kN) = 144.72

Step 9 (Incremental) Inc. Lateral Disp. at J4 (mm)=

Total Lateral Disp. at J4 (mm)=
M ΔM M + ∆M
12.69
21.40317
Condition
kNm kNm (kNm)
124.03 0.00 124.03 YIELDED
55.42 -46.64 8.78
46.05 5.74 51.79
-9.37 -44.15 -53.51
-11.30 1.29 -10.01
-16.40 -38.69 -55.09
ΔEQ=19.5kN 167.90 0.00 167.90 YIELDED
26.45 -46.22 -19.76
65.63 6.05 71.68
-29.84 -43.74 -73.58
ΔEQ=13kN 18.38 1.72 20.10
-66.14 -38.78 -104.91
135.60 0.00 135.60 YIELDED
28.49 -24.15 4.35
77.82 -24.15 53.68
ΔEQ=6.5kN -35.58 -21.98 -57.57
14.61 -21.98 -7.37
-54.36 0.00 -54.36
-9.37 52.37 43.00
-49.58 0.00 -49.58 YIELDED
-1.93 45.43 43.51
-51.64 0.00 -51.64 YIELDED
16.40 38.69 55.09 YIELDED
-53.68 0.00 -53.68 YIELDED
-10.40 52.27 41.87
-49.33 0.00 -49.33 YIELDED
-3.42 45.46 42.03
-50.19 0.00 -50.19 YIELDED
12.46 38.78 51.24
-54.36 0.00 -54.36 44
YIELDED
 Frame M ΔM M + ∆M
Step 10 (Incremental) Element
1
kNm
124.03
kNm
0.00
(kNm)
124.03
Condition
YIELDED
8.78 -1.83 6.95
51.79 0.44 52.22
2 -53.51 -1.74 -55.25
-10.01 0.30 -9.71
3 -55.09 0.00 -55.09
167.90 0.00 167.90 YIELDED
4 -19.76 -1.82 -21.59
ΔEQ=0.75kN 5
71.68 0.44 72.12
-73.58 -1.44 -75.02
20.10 0.64 20.74
6 -104.91 -1.86 -106.77
ΔEQ=0.50kN 135.60 0.00 135.60 YIELDED
7 4.35 -0.84 3.50
53.68 -0.84 52.83
8 -57.57 -0.54 -58.11
ΔEQ=0.25kN -7.37 -0.54 -7.91
9 -54.36 0.00 -54.36
43.00 2.27 45.27
10 -49.58 0.00 -49.58 YIELDED
43.51 2.03 45.54
11 -51.64 0.00 -51.64 YIELDED
55.09 0.00 55.09 YIELDED
12 -53.68 0.00 -53.68 YIELDED
41.87 2.26 44.13
13 -49.33 0.00 -49.33 YIELDED
42.03 2.08 44.11
14 -50.19 0.00 -50.19 YIELDED
51.24 1.86 53.10 YIELDED
15 -54.36 0.00 -54.36 YIELDED

45
 Total Incremental Load (kN)= 4.2
Total Base Shear (kN) = 150.42

Step 11 (Incremental)
Inc. Lateral Disp. at J4 (mm)= 1.94
Total Lateral Disp. at J4 (mm)= 23.90917
Frame M ΔM M + ∆M
Condition
Element kNm kNm (kNm)
124.03 0.00 124.03 YIELDED
1 6.95 -5.34 1.61
52.22 2.18 54.40
2 -55.25 -4.04 -59.29
-9.71 3.14 -6.57
3 -55.09 0.00 -55.09
ΔEQ=2.1kN 4
167.90 0.00 167.90 YIELDED
-21.59 -5.17 -26.76
72.12 2.35 74.47
5 -75.02 -4.19 -79.21
20.74 3.00 23.73
ΔEQ=1.4kN 6 -106.77 0.00 -106.77
135.60 0.00 135.60 YIELDED
7 3.50 -2.09 1.41
52.83 -2.09 50.74
ΔEQ=0.7kN 8 -58.11 0.16 -57.95
-7.91 0.16 -7.75
9 -54.36 0.00 -54.36
45.27 7.52 52.79 YIELDED
10 -49.58 0.00 -49.58 YIELDED
45.54 7.18 52.72 YIELDED
11 -51.64 0.00 -51.64 YIELDED
55.09 0.00 55.09 YIELDED
12 -53.68 0.00 -53.68 YIELDED
44.13 7.52 51.65 YIELDED
13 -49.33 0.00 -49.33 YIELDED
44.11 7.18 51.30 YIELDED
14 -50.19 0.00 -50.19 YIELDED
53.10 0.00 53.10 YIELDED
15 -54.36 0.00 -54.36 YIELDED

46
 Collapse Mechanism

160
SYSTEM IS UNSTABLE
140

120

Base Shear (kN)

100

80

60

40

20

0
0 5 10 15 20 25 30
Roof Displacement (mm)

Beam sway mechanism is observed

No lateral load incrementation possible (only
rigid body motion)

47
 What did we obtain?
• A simple representation of the capacity curve
• Plastic mechanism and sequence of hinge formation
• Lateral load and displacement capacity
• Ductility and plastic rotation demand
160

140

120 SAP 2000 built in pushover

Incremental analysis options include:
Total Base Shear(kN)

100
SAP2000
80 • hardening/loss of strength
• P-M interaction
60

40

20
• Systematic stiffness approach
0
0 5 10 15 20 25 30
Top Displacement (mm)

48
Lumped Inelasticity Approach: A detailed
model (Filippou et.al.1998)

49
Elastic Element
[d ]  [ f ] [ F ]

50
Plastic Hinges
[d ]  [ f ] [ F ]

51
Effect of Axial Force

• Compute moment rotation

relationships accounting for axial
force variation
• Always remain on the yield
surface

52
 Element Stiffness Matrix
Element Tangent
Flexibility Matrix
Invert
Element Stiffness
Matrix

Transformation 0
crd
Matrix 0

53
 Solution Algorithm
Assemble structural stiffness matrix with
conventional assembly procedures

Minimize the residual until a

specified convergence
criterion is satisfied.
Newton-Raphson Solution Procedure

54
Distributed Inelasticity (T.H.Lee 2005)

55
Section Deformations
İth fiber

εo

Strain
Displacement
Relations

56
 Interpolation Functions
u( x)  N1u1  N 4u4 Relates nodal forces to internal
v( x)  N 2u2  N3u3  N5u5  N6u6 displacements
N4 N2 N5
N1
1 1
1 N3
N6
Linear Interpolation for
Cubic Interpolation for
axial displacements
transverse
displacements

Strain and curvature at any point on the element as a

function of nodal displacements
57
Virtual Work

Weak Form of
Equilibrium

58
 Section and Member Stiffness

Element Stiffness
Matrix

Section Stiffness and

Resisting Force

Assemble structural stiffness matrix and proceed with the nonlinear

solution 59
 Advantages
• Monitor section deformations and strains during
nonlinear analysis
• Account of axial force-moment interaction automatically
• Ductility is incorporated through material stress-strain
behavior
• Cracking is automatically accounted
• Advancements in element formulation such as flexibility
that works based on moment distribution rather than
assumed curvature profiles

60
 Why Strengthening?
• Insufficient capacity (deformation or strength)
• Increased seismicity
• Updated codes
• Changes in target performance

61
 Definitions
• Strengthening: Upgrades on undamaged
members or structures to restore member
capacities
• Repair: Rehabilitation of damaged members
or structures to restore member capacities

62
Strengthening Methods

Member Strengthening
(Beam-Column Strengthening) 63
Repair / Strengthening Methods

Global Strengthening
(Shear Wall Addition, Bracing)
64
 Member Strengthening
Members
• Column jacketing
• Beam strengthening
Materials
- Reinforced concrete
- Structural steel
- Fiber reinforced polymers

65
 Column Strengthening
• Objectives
– Strength (axial, bending, shear)
– Deformation capacity
• Stiffness enhancement is usually a by-product.

66
66
 Steel Jacketing
• Corner elements
• Welded steel plates

67
67
Steel Jacketing
• Corner L plates
• Welded trans. Plates

Confinement
Shear Strength

68
68
 Steel Jacket Shear Contribution

• Provide good contact area for shear strength

enhancement.
tj b d
Vj  f yw
s
– tj plate thickness
– b plate width
– s spacing
– d effective depth

69
69
 Kesit büyütülerek
eğilme kapasitesi
artırılabilir.

Steel Jacketing
70
 Steel Jacketing

Çelik Plaka L- D Kutu

A Manto B Berkitme C Berkitme Berkitme

Tamir Harcı
Dolgusu

t jb d
Vm  f yw Çelik Plaka Mantolama
s 71
Steel Jacketing

D
C

B
A

72
 Steel Jacketing

• Fire Proofing
• For moment capacity enhancement longitudinal
continuity is needed
• Otherwise used only for shear and axial capacity

73
73
 RC Jacketing

• Continuous reinforcement for moment capacity

• Stirups for confinement and shear
• New concrete layer
• Engage old and new sections with dowels embedded
with epoxy

74
74
75
75
COLUMNS – RC JACKETING
Jacketing Concrete
Available Concrete

Embeded
Dowel

Long.
Reinfor. 76
for Jacket 76
 Use of Fiber reinforced Polymers
Axial Strength Enhacement
Confinement
Shear Strength Enhancement
Insuff. Lap Splice Strengthening for Deformed Bars

77
77
炭素繊維ストランド巻付け耐震補強

Fiber Reinforced Polymer Strengthening

78
テープ状アラミド繊維, 居着き工事, 足場, 重機, 溶接作業不可, 狭い空間, 作業性, 接着性

ADVANTAGES
•Lightweight
• No welding
• Short construction period
• No heavy equipment
• Less workmanship
• Easy handling
DISADVANTAGE
•Cost
• Total cost? 79
 Continuous Fiber Sheets
Cloth Type CF Sheet Prepreg Type CF Sheet

AF Tape
80
 Stress-Strain Relationship of Fibers
Stress Carbon
(GPa)
3.0 Aramid

2.0
Glass

1.0 Mild Steel

Strain
0.0 (%)
1.0 2.0 3.0 4.0
81
82
 Drift angle

Shear force (kN)

Horizontal displacement (mm)

83
 COLUMN TESTS by Matsuzaki et al.

Pw=0.13% Pw=0.62% Pw=1.43% 84

 Design of FRPs
Shear strength Enhancement

85
 Design of FRPs
Axial Strength and Ductility Enhancement

86
 Design of FRPs
Axial Strength and Ductility Enhancement

FRP Confined Concrete Ultimate Strain

fcc
fc’
Axial stress-strain model for
FRP confined concrete

0.002 εcc 87
 Design Example
P Long. Reinforcements : 12ø26
Stirrups : ø10 @ 400 mm
fc = 15 MPa
fy = 420 MPa
Clear Cover = 50 mm
L D P =1800 kN (%27Po)
D = 750 mm, L = 5 m
Properties of FRP :
Ef = 230000 MPa
ffu = 3450 MPa
Thickness of 1 layer = 0.165 mm

Design the above cantilever column to enhance its displacement ductility to 1.5
times the available one by utilizing FRP.

88
 Design Example
For the available column: 1200

My = 970 kNm 1000

øy = 4.8 rad/km 800

Moment (kNm)
Øu = 15 rad/km 600

Displacement Capacity : 400

d = d e + dp 200

de = øy L2/3 = 40 mm 0
0 2 4 6 8 10 12 14 16 18
Curvature
Eğrilik (rad/km)
dp = (øu - øy) L Lp = 20 mm
d = 60 mm d εc = 0.0035
Target d = 90 mm
(unconfined)

89
 Design Example
For column confined by 3 layers of
1200
FRP:
1000
fl = ½ κa ρf εf Ef
800

Moment (kNm)
ρf = (4x3x0.165/750)=0.0026
600

εf = 0.004 (including shear) 400

fl/fc = 0.081 200

fcc/fc = (1+2.4x0.081) =1.2 (min) 0

0 5 10 15 20 25 30
Eğrilik (rad/km)
εcc = 0.002(1+15x0.0810.75)
σ
εcc = 0.00655 Displacement Capacity :
18 d = de + dp
My = 775 kNm
15
de = øy L2/3 = 66.6 mm
øy = 8 rad/km dp = (øu - øy) L Lp = 32 mm
d = 99 mm > 90 mm
Øu = 25 rad/km
0.002 0.0066 ε 90
 Design Example

Lateral Shear Force:

V = Mu/L = 1035/5 = 207kN
Shear Capacity:
Vc = 0.65 x fctd D d
0.5
Vc = 0.65 x 0.53 15 0.9 x 7502
0.8 x Vc = 540 kN 1 cm
boşluk
No need for shear strengthening.

D (veya h) kadar sargı

uzatılmalı Enine Karbon Lifli
Sargı
91
 CF-Anchor

Column CF sheet split by

walls

CF sheet

Penetrating hole
Bundle of CF
strands
CF-anchor Wing wall

92
 When there are spandrel walls

Surface preparation
-Remove finishing
- Grind the damaged part
of the concrete
93
Drilling

Penetrating holes through spandrel wall

close to the column 94
Restoration of defoliated part after drilling

Restore the defoliated parts with

polymer cement mortar 95
 Application of epoxy primer

Coat the epoxy resin primer evenly with a roller

96
40
Correction of irregular surface

Filling epoxy resin

putty in concave
parts to make the
concrete surface
smooth

Epoxy resin putty

wall column

Penetrated97holes
Cutting of Fiber Sheets

98
Application of Carbon Fiber Sheet （CFS)

Installation of CFS Installation of CFS

with epoxy resin in with epoxy resin in
axial direction hoop direction
99
 Preparation of CF anchor

Winding CF strands
around the reel
and bundling them up

Wire Wall thickness

100
 Application of CF anchor
Resin impregnation
before installation

Wall thickness
20～30mm

Resin impregnation into

CF strands

Setting CF-anchor with epoxy

resin through the holes
101
 Application of CF anchor

Spread and set the both

ends of CF strands onto
CFS surface
Impregnate the epoxy
resin into CF strands of
spread portion
102
Curing epoxy resin

103
 Columns with
window frame

CF-anchor
Closed wrapping
without removing
window frame

CF-anchor
104
Application to beams with slab

105
System Retrofit

106
 System Retrofit
• Insufficient Lateral Building Stiffness
• Numerous Members require retrofit
• Impractical or ineconomical member
rehabilitation/retrofit

System Enchancement: a new lateral loading system

capable of resisting whole lateral load demands

107
107
 System Retrofit
• Limits lateral displacement demands,
• Avoids the uneconomical strengthening of
circumferential members having insufficient
capacity and/or ductility,
• Eliminates system related weaknesses like soft
story, short column, strong beam – weak
column, etc.

108
108
Strengthening with Steel Braces Strengthening with RC Shearwalls
109
SUGGESTION FOR INSUFFICENT STRUCTURES
IN OUR COUNTRY

A w  0.0025 Api
 0.01Apt
SAw: total shear wall area for each direction
SApi: total floor area of entire building
Apt : building floor area
Aw : shearwall thickness × distance between outer
faces of end columns

110
110
 RC Shearwall Layout
• RC shearwalls should be placed to satisfy the symmetry
along each direction.
• If the above limitation is not fulfilled, the extra
torsional moment demands would occur due to the
shift of center of rigidity.
• Newly formed RC frames decrease the fundamental
period of structure in concern due to the stiffness
enchancement.
• Therefore, it should be proved that the capacity
enchancement is larger than the increase in the
demands.
111
111
 Strengthening with RC Infill Walls
• Do not seperate the RC
infill walls from the frame.
• Dowels
– Deformed bars
– For frames 10-15f
– Inside shearwall 35f
– Hole diameter > Dowel
diameter + 5 mm
– Minimum f16

112
112
 Strengthening with RC Infill Walls
• Total dowel area embedded inside both columns and
beams should be larger than total reinforcement
area placed inside RC wall along each direction.
• The spacing of anchorages should be calculated to
account for the story shear.
• Anchorage rods should be mounted to the drilled
holes by utilizing a suitable epoxy based adhesive.
• Before applying adhesive material inside the
anchorage zone, the dust inside the anchorage holes
should be cleared with the help of air compressors
after removing the damaged concrete portions inside
the holes by using steel brush.
113
113
 Strengthening with RC Infill Walls
• To satisfy the continuity between stories, the
whole height of the beams should be drilled and
the vertical anchorage rods should be placed.
• If the concrete quality of columns are not
adequate, column jacketing might be a good
solution.
• Mesh reinforcements should be placed at the
both sides of shearwall and the reinforcement
ratio should be compatible with the limits of
TEC2007 for ductile RC shearwalls.
114
114
 Strengthening with RC Infill Walls

• The grout should be placed so that there will be 10-15

cm void at the top of RC shearwall. This void must be
filled by a special grout that has no shrinkage tendency.
• Generally, the lap splices in the old columns are not
adequate. So, at least four large-diameter
reinforcements should be placed at the ends of the RC
shearwall to form two hidden end columns.
• These large-diameter reinforcements must be confined
by using seismically detailed strirrups.
• The reinforcement amount inside these hidden
columns must be larger than 50% of the old columns.
115
115
 Strengthening with RC Infill Walls

• The voids inside the grout should be prevented as

much as possible.
• If voids are mandatory due to some architectural
reasons, some specially designed reinforcements
may be placed diagonally.
• The capacity and stiffness of the RC shearwalls with
voids should be reduced.
– If the void length to RC shearwall length ratio is less than
or equal to 25%, the reduction factors for capacity and
stiffness could be taken as approximately %20 and %50,
respectively.

116
116
117
117
 Strengthening with RC Infill Walls
• RC shearwall has to be connected to a suitable
foundation.
• The dowel length for a continuous foundation
should be at least 20f.
• If the foundation underneath RC shearwall is
inadequate or there is no foundation, a new
foundation should be built.
• This new foundation should be connected to the
available foundation system with a proper
anchorage mechanism.
118
118
119
119
120
Strengthening with Steel Braces

121
Strengthening with interior steel frames
Column

IPE200

122
 Strengthening of Infill Walls
Requirements:
• In-plane capacity and ductility enhancement
• Out-of-plane stability
• Limited displacement capacity (<%0.5)

Light Damage Moderate Damage Heavy Damage

123
STRENGTHENING OF INFILL WALLS:
Concrete Grout with Wire Mesh

124
STRENGTHENING OF INFILL WALLS:
Concrete Grout with Wire Mesh

125
 DOLGU DUVARLARIN
GÜÇLENDİRİLMESİ: Hasır Donatılı Sıva

126
STRENGTHENING OF INFILL WALLS:
Fiber Reinforced Polymer

127
 FRP Diagonals
Sıvalı Duvar

LP şerit

LP ankraj

LP dübel
LP çekme çubuğu
Dolgu duvar basınç çubuğu

128
Strengthening with Lifli Polimer ile
Güçlendirme

129
STRENGTHENING OF INFILL WALLS:
Precast Concrete Panels

130
Strengthening with Precast Concrete
Panels

131