Transcript Document

ISIS Educational Module 4:
An Introduction to FRPStrengthening of Concrete Structures
Updated October 2010 for ISIS Canada
ISIS EC Module 4
Module Objectives
The use of FRPs in civil infrastructure is steadily increasing in Canada and
around the world. This module is directed mainly to students to:
• Provide a background and general awareness of FRP materials, their
properties , their behaviour and their potential uses
• Introduce the philosophies and procedures for strengthening structures
with FRPs
• Familiarize the students with designing using the Canadian Highway
Bridge Design Code (CHBDC)
ISIS EC Module 4
Overview
1. Introduction
2. FRP Materials
3. Evaluation of Existing Structures
4. Flexural Strengthening
5. Shear Strengthening
6. Column Strengthening
7. Installation of FRP strengthening systems
8. Quality control and quality assurance
9. Additional applications
10. Field applications
ISIS EC Module 4
1 - Introduction
• The world’s population depends on an extensive
infrastructure system
• Roads, sewers, highways, buildings
• The system has suffered in past years
• Neglect, deterioration, lack of funding
Global Infrastructure Crisis
ISIS EC Module 4
1 - Introduction
Why is strengthening needed?
• Many structures, including bridges and parking garages, have become
structurally deficient due to deterioration.
• In Canada, more than 40% of the bridges currently in use were built
more than 40 years ago.
• Many structures are also becoming functionally obsolete due to
increased loading.
ISIS EC Module 4
2 - FRP Materials
• Why repair with the same materials?
• Why repeat the cycle?
Light weight
High Strength
Easy to install
5x steel
FRP Materials
Corrosion resistant
Highly versatile
Durable structures
Suits many
projects
ISIS EC Module 4
2 - FRP Materials
FRP is a composite consisting of fibres and matrix.
2400-4300
Fibres
Stress [MPa]
Fibres:
– Provide strength and
stiffness
–Their quality, orientation
and shape affect the
final product
50-90
Matrix (resin):
–Coats the fibres
–Protects the fibres from abrasion
–Transfers stresses between the fibres
ISIS EC Module 4
FRP
Matrix
0.5-4.8
2-8
Strain [%]
2 - FRP Materials
FRP material properties are a function of:
•Fibre quality, orientation and shape
•Fibre volumetric ratio
•Adhesion to the matrix
•Manufacturing process (additives and fillers)
ISIS EC Module 4
2 - FRP Materials
Wide range of FRP products are available:
• Plates (Rigid strips)
• Sheets (Flexible fabric)
• Rods
The fibres could be:
• Carbon
• Glass
FRP sheet
• Aramid
ISIS EC Module 4
2 - FRP Materials
FRP advantages:
• Does not corrode
• High strength to weight ratio
• Reduced installation time and cost
• Non-conductive and non-metallic
• Low maintenance requirements
Disadvantage:
High temperature is a serious concern
ISIS EC Module 4
3 - FRP Materials
FRP properties versus steel:
2500
Stress [MPa]
• Linear elastic behaviour
to failure
• No yielding
• Higher ultimate
strength
• Lower strain at
failure
CFRP
2000
1500
GFRP
1000
Steel
500
1
2
Strain [%]
ISIS EC Module 4
3
2 - FRP Materials
FRP System
Fiber Type
Weight
[g/m2]
Thickness
[mm]
Tensile
Strength [MPa]
Fyfe Co. LLC [www.fyfeco.com]
915
1.3
460
505
0.5
417
644
1.0
834
Hughes Brothers Inc. [www.hughesbros.com]
-6.4-12.7
2068-1724
Tensile Elastic
Modulus [GPa]
Strain at
Failure [%]
20.9
20.9
82
1.8
1.8
0.9
124
0.017-0.015
Tyfo SEH51A
Tyfo SEH25A
Tyfo SCH41
Glass
Glass
Carbon
Aslan 200
Carbon
Aslan 500 #2
Carbon
--
2.0
2068
124
1.7
Aslan 500 #3
Carbon
--
4.5
1965
124
1.5
Aslan 400 CFRP Laminates
Carbon
2400
131
1.9
SikaWrap 430G
SikaWrap 100G
SikaWrap 230C
SikaWrap 103C
CarboDur S
Glass
Glass
Carbon
Carbon
Carbon
504
558
715
717
2800
24.6
24.4
61.0
65.1
165
1.9
2.2
1.1
1.0
1.7
CarboDur M
Carbon
2400
210
1.2
CarboDur H
Carbon
-1.4
1300
BASF Building Systems Inc. [www.BASFBuildingSystems.com]
Carbon
-1.4
2700
300
0.5
159
1.7
MBrace S&P Laminate
-1.4
Sika Canada Inc. [www.sika.ca]
430
0.5
915
1.0
230
0.4
610
1.0
-1.2-1.4
--
1.4
MBrace EG 900
Glass
900
0.37
1517
72.4
2.1
MBrace CF 130
Carbon
300
0.17
3800
227
1.7
MBrace CF 160
Carbon
600
0.33
3800
227
1.7
MBrace CF 530
Carbon
300
0.17
3500
373
0.94
MBrace AK 60
Aramid
600
0.28
2000
120
1.6
ISIS EC Module 4
2 - FRP Materials
Installation:
1) Wet lay-up system:
- The system is installed on the surface of the concrete
element while the resin matrix is still “wet”, and the
polymerization occurs on site
- Saturate sheets with epoxy adhesive, then place on
the concrete surface and press with a roller
- Used with flexible sheets
- Multiple layers can be used
Epoxy
Roller
ISIS EC Module 4
2 - FRP Materials
2) Pre-cured system:
- Used with FRP plates or laminates
- Used for surface bonded plates or near
surface mounted reinforcements
- Not as flexible for variable structural
shapes
- The pre-cured laminates should be
placed on or into the wet adhesive
- Place on the concrete surface
- Multiple layers can not be used
ISIS EC Module 4
2 - FRP Materials
3- Near Surface Mounted (NSM)
•
It is a newer class of FRP strengthening technique.
Un-strengthened Longitudinal grooves
concrete T-beam
cut into soffit
FRP strips placed Grooves filled
in grooves
with epoxy grout
• Research indicates that NSM reinforcement is effective and efficient
for strengthening.
ISIS EC Module 4
2 - FRP Materials
FRP-Strengthening Applications :
Type
Flexural
Application
Confinement
Schematic
Along long.
axis of the
beam
Section
Side face of the beam Perpendicular
(U or closed wrap) to long. axis
of the beam
Section
Tension and/or side
face of the beam
Shear
Fibre Dir.
Around the column
Circumferential
Section
ISIS EC Module 4
3 - Evaluation of Existing Structures
Repair process includes:
1. Evaluation of the existing structure
 Understanding the cause and the effect of the deterioration
2. Determining if repair is required and its extent (quantify)
3. Analysis and design
4. Introducing a repair strategy
ISIS EC Module 4
3 - Evaluation of Existing Structures
Fixing the effect without understanding the cause is likely to result
in premature failure of repair. Proper repair requires an
understanding of the cause to eliminate the effect.
• Evaluation is important to:
Determine concrete condition
Identify the cause of the deficiency
Establish the current load capacity
Evaluate the feasibility of FRP strengthening
ISIS EC Module 4
3 - Evaluation of Existing Structures
Evaluation should include:
All past modifications
Actual size of elements
Actual material properties
Location, size and cause of cracks and spalling
Location and extent of corrosion
Quantity and location of rebar
ISIS EC Module 4
3 - Evaluation of Existing Structures
Problems in a structure could be due to:
Defects:
In design or material or during construction
Damage:
Due to overloading, earthquake or fire
Deterioration:
Due to corrosion or sulphate attack
ISIS EC Module 4
3 - Evaluation of Existing Structures
• Examples of some deficiencies:
1. Environmental effects
Chloride Ingress
Wet-Dry
Freeze-Thaw
ISIS EC Module 4
3 - Evaluation of Existing Structures
• A primary factor leading to extensive degradation…..
2. Corrosion
Concrete
Reinforcing
Steel
Moisture, oxygen and
chlorides penetrate
Corrosion products form
Volume expansion occurs
Through concrete
More cracking
Through cracks
Corrosion propagation
ISIS EC Module 4
End result
3 - Evaluation of Existing Structures
• Deficiencies could be due to:
Then
3. Updated design loads
4. Updated design code
procedures
5. Increase in traffic loads
ISIS EC Module 4
Now
3 - Evaluation of Existing Structures
• Deficiencies could be due to:
6. Fire damage
7. Earthquakes
ISIS EC Module 4
4 - Flexural Strengthening
Material resistance factors (CHBDC):
- Concrete c = 0.75
- Steel reinforcement:
- Reinforcing bars s = 0.90
- Prestressing strands p = 0.95
- Base FRP for pultruded FRP:
- AFRP FRP = 0.55 (for externally bonded applications)
- AFRP FRP = 0.65 (for NSMR)
- CFRP FRP = 0.80 (for externally bonded applications and NSMR)
- GFRP FRP = 0.70 (for externally bonded applications and NSMR)
- Non-pultruded FRP made by wet lay-up: 0.75 times base FRP
ISIS EC Module 4
4 - Flexural Strengthening
Modes
The analysis of the flexural strength of FRP strengthened elementsFailure
is based
on the following assumptions:
1) The internal stresses are in equilibrium with the applied loads.
2) Plane sections remain plane.
3) Strain compatibility exists between adjacent materials.
(ie. Perfect bond between: concrete and steel, concrete and FRP)
4) The maximum tensile strain of the FRP (eFRPt ) is 0.006.
5) The maximum compressive strain in the concrete (ecu ) is 0.0035.
6) The contributions of FRPs in compression and of the concrete in
tension are neglected.
ISIS EC Module 4
4 - Flexural Strengthening
Failure Modes
The potential modes of failure are:
1) Concrete crushing before steel yielding or rupture of the FRP.
2) Steel yielding followed by concrete crushing before rupture of
the FRP.
3) Steel yielding followed by rupture of the FRP.
4) Peeling, debonding, delamination or anchorage failure of the
FRP (considered premature tension failures to avoid).
ISIS EC Module 4
4 - Flexural Strengthening
Rectangular section without compression steel:
b
d
h
As
bFRP
Cross Section
Ts = sAsfs
a1Φcf’c
ec
a = b1c
c
es
fs
fFRP
eFRP
Strain Distribution
Stress Distribution
TFRP = FRPAFRPEFRPeFRP
ISIS EC Module 4
Cc
Ts
TFRP
Equiv. Stress
Distribution
Cc = ca1f’cb1bc
4 - Flexural Strengthening
The equilibrium equations are:
1) Force equilibrium in the section:
Compression forces =Tension forces
Cc = Ts + TFRP
2) Moment equilibrium in the section:
External applied moment= Internal moment
a
a
+ TFRP h Mapplied = Ts d 2
2
ISIS EC Module 4
4 - Flexural Strengthening
Rectangular section with compression steel :
b
ecu
d’
h
A’s
e’s
c
d
As
a1Φcf’c
f’s
es
fs
fFRP
eFRP
bFRP
Cross Section
Strain Distribution
Equiv. Stress
Distribution
Stress Distribution
Cs = sf’sA’s
Mr = Ts
d-
+TFRP
a
h2
ISIS EC Module 4
+Cs
a
2
Cc s
Ts
TFRP
Add a compressive stress resultant
a
2
C
a = b1c
- d′
4 - Flexural Strengthening
Iterative design procedure:
1. Assume initial strains
2. Select FRP material and AFRP
3. Assume a failure mode
If concrete crushing before steel yields, then:
ecu = 0.0035
and esteel < eyield
If concrete crushing after steel yields, then:
ecu = 0.0035
and
esteel > eyield
If FRP ruptures after steel yields, then:
eFRP = eFRPt
and esteel > eyield
ISIS EC Module 4
b
ec
d
h
As
bFRP
c
es
eFRP
4 - Flexural Strengthening
4. Determine the compressive stress block factors (a1, b1)
b
h
a1Φcf’c
d
a = b1c
Cc
As
Ts
TFRP
bFRP
b1 = 0.97 – 0.0025 f’c > 0.67
a1 = 0.85 – 0.0015 f’c > 0.67
ISIS EC Module 4
4 - Flexural Strengthening
5. Calculate c (neutral axis position)
Using equilibrium equation the following equations can be derived and used:
- Concrete crushing before steel yields (es and e’s < ey )
a1c f’cb1bc2 +
sEs ecu (As’+As)+ FRPEFRP (ecu+efi)AFRP c -
sEs ecu (As'd'+Asd)+ FRPEFRP ecuAFRP h = 0
-Steel yielding followed by concrete crushing (es and e’s > ey )
a1c f’cb1bc2 + sfy(As’-As)+ FRPEFRP (ecu+efi)AFRP cFRPEFRP ecuAFRP h = 0
- Steel yielding followed by FRP rupture (es and e’s > ey )
c=
sfy(As-As’)+ FRPEFRP eFRPtAFRP
a1c f’cb1b
ISIS EC Module 4
4 - Flexural Strengthening
6. Check failure mode assumption with the material strains
If failure is initiated by:
Concrete crushing:
FRP rupture:
ecu= 0.0035
eFRP = eFRPu ≤ eFRPt
d-c
es= ecu
c
eFRP= ecu h - c - efi
c
'
'
c
d
e =e
'
c-d
(efi+eFRPu)
≤ es'= (efi+eFRPt)
h-c
(efi+eFRPu) d-c ≤ es= (efi+eFRPt)
h-c
c
(efi+eFRPu)
≤ ec= (efi+eFRPt)
h-c
s
cu
c
c-d'
h-c
d-c
h-c
c
h-c
- If the assumption is proven to be false, go back to step 3 and make another
assumption
- If the assumption is correct, proceed to the next step
ISIS EC Module 4
4 - Flexural Strengthening
7. Compute internal forces
Cs = sfsA's
Ts = sAsfs
TFRP = FRPAFRPEFRPeFRP
8. Calculate the section moment resistance (Mr)
a
a
a
hMr = Ts d+TFRP
+Cs
2
2
2
9. Compare Mr to the applied moment (Mapplied)
– If Mr < Mapplied, go to step 2 and change AFRP
– If Mr > Mapplied, then the design is safe
ISIS EC Module 4
- d′
4 - Flexural Strengthening
Optimized determination of AFRP
Assuming:
All the steel has yielded and combining the equilibrium equations:
1. Determine c using the following equation
a1c f’cbwb12c2 -a 1 c f’cbwhb1c- Cs(h-d′)+Ts(ds-h) - Mf = 0
2
2. Determine strain in FRP
eFRP = ecu h - c - efi ≤ 0.006 ≤ eFRPu
c
ISIS EC Module 4
4 - Flexural Strengthening
3. Determine successively:
Cc = ca1f’cb1bc
TFRP = Cc + Cs - Ts
4. Optimize value of AFRP
AFRP =
TFRP
FRPEFRP eFRP
Use this AFRP as an input for the iterative design method
ISIS EC Module 4
4 - Flexural Strengthening
T-section:
be
c
hf
h
• If the neutral axis lies in the web (c < hf),
then treat it as a rectangular section with a
compression zone width = be.
t FRP
b FRP
Geometric parameters
• If the neutral axis lies outside the web
(c > hf), then treat it as a T-section.
ISIS EC Module 4
4 - Flexural Strengthening
be
c
hf
=
h
t FRP
b FRP
+
As
A FRP
Mr
Geometric parameters
A sw
A FRP
A sf
=
+
Mrf
Mrw
Factored moment subdivision
The section is treated as the summation of: flange (Mrf) and web (Mrw).
- Flange (Mrf)
a1c f’c(be-bw)hf
Asf =
sfy
- Web (Mrw)
Asw = As –Asf
Mrf = s fy Asf d -
Mrw = s fy Asw d ISIS EC Module 4
a
2
hf
2
+TFRP
h-
a
2
4 - Flexural Strengthening
Design procedure:
1. Select FRP material and AFRP
2. Determine behaviour of the section ( Rect. or T)
sfyAs+FRP EFRP eFRP A FRP
If
hf ≥
a1c f’c b1be
Then, it is rectangular behaviour
Else, it is a T-section
3. Determine Asf and Mrf
4. Determine AFRP to obtain required Mrw
ISIS EC Module 4
4 - Flexural Strengthening
Example:
Calculate the moment resistance (Mr) for an FRP-strengthened
rectangular concrete section
h = 360 mm
d = 320 mm
Section information:
f’c = 40 MPa
AFRP = 110 mm2
fy = 400 MPa
eFRPu = 1.26 %
Es = 200 GPa
EFRP = 210 GPa
2-15M bars
CFRP
b = 125 mm
ISIS EC Module 4
4 - Flexural Strengthening
Solution:
Step 1: Assume failure mode
Assume failure of beam due to crushing of concrete in
compression after yielding of internal steel reinforcement
Thus,
ecu= 0.0035
and
esteel > eyield
ISIS EC Module 4
4 - Flexural Strengthening
Step 2: Calculate concrete stress block factors
a1 = 0.85 – 0.0015 f’c > 0.67
a a1 = 0.85 – 0.0015 (40) = 0.79
b1 = 0.97 – 0.0025 f’c > 0.67
a b1 = 0.97 – 0.0025 (40) = 0.87
ISIS EC Module 4
4 - Flexural Strengthening
Step 3: Find depth of neutral axis, c
Steel yielding followed by concrete crushing
a1c f’cb1bc2 + sfy(As’-As) + FRPEFRP (ecu+efi)AFRP c - FRPEFRP ecuAFRP h = 0
0.79(0.75)40(0.87) +
125(c2)
0.9(400) (0-400)+
0.80(210000)(0.0035+0)110
c - 0.80(210000)
(0.0035)(110)360
2577.375(c2)-79320(c)-23284800 = 0
c = 111.7 mm or c = -80.9 mm (rejected)
ISIS EC Module 4
=0
4 - Flexural Strengthening
Step 4: Check mode of failure
Steel yielding followed by concrete crushing
ecu= 0.0035
es= ecu
eFRP= ecu
ds-c
= 0.0035
c
h-c = 0.0035
c
320-111.7
111.7
360-111.7
111.7
= 0.0065 > 0.002(ey)
= 0.0078 > 0.006
not O.K.
Trial 2, assume the steel yields and the strain in the FRP is 0.006
ISIS EC Module 4
4 - Flexural Strengthening
Step 5: Trial 2
Compression in concrete = Tension in ( Steel + FRP)
a1c f’cbb1c = sfy As + FRPEFRP eFRPAFRP
0.79(0.75)40(0.87)125(c) = 0.9(400)400 + 0.80(210000)0.006(110)
c = 98.9 mm
ISIS EC Module 4
4 - Flexural Strengthening
Step 6: Check mode of failure
eFRP = 0.006
es= eFRP
ecu= eFRP
ds-c
= 0.006
h-c
c = 0.006
h-c
320-98.9
360-98.9
98.9
360-98.9
= 0.0051 > 0.002(ey)
= 0.0023 < 0.0035
O.K.
The assumed mode of failure is correct
ISIS EC Module 4
4 - Flexural Strengthening
Step 7: Moment Resistance
Mr = Ts
ds-
b1 c
2
= sfy As ds-
+TFRP
b1c
h-
b1 c
2
b1 c
+ FRPEFRP eFRPAFRP h2
2
0.87 (98.9)
= 0.9(400)400 3202
0.87 (98.9)
+0.80(210000)0.006(110)
3602
Mr = 75  106 N.mm = 75 kN.m
ISIS EC Module 4
4 - Flexural Strengthening
Example:
The T-beam requires strengthening to upgrade its moment
capacity to 600 kN-m. Calculate the required area of FRP (AFRP).
650
Section information:
70
f’c = 30 MPa
510 600
fy = 400 MPa
As = 8 x 300 mm2
eFRPu = 1.26 %
Es = 200 GPa
EFRP = 155 GPa
250
ISIS EC Module 4
4 - Flexural Strengthening
Step 1: Calculate concrete stress block factors
a1 = 0.85 – 0.0015 f’c > 0.67
a a1 = 0.85 – 0.0015 (30) = 0.805
b1 = 0.97 – 0.0025 f’c > 0.67
a b1 = 0.97 – 0.0025 (30) = 0.895
ISIS EC Module 4
4 - Flexural Strengthening
Step 2: Evaluating the moment capacity of the existing section
 Assume neutral axis is inside the flange (c < hf)
Compression in concrete = Tension in steel
a1c f’cbb1c = sfy As
0.805(0.75)30(0.895)650(c) = 0.9(400)(300× 8)
c = 82mm > hf
The assumption (c < hf) is wrong.
ISIS EC Module 4
4 - Flexural Strengthening
Step 2: Evaluating the moment capacity of the existing section
 Assume neutral axis is outside the flange (c > hf)
Asf =
Asf =
a1c f’c(be-bw)hf
sfy
0.805(0.75)30(650-250)70
=1409 mm2
0.9(400)
Asw = As –Asf = 2400 -1409 = 991 mm2
Mrf = 0.9 (400)1409 (510 -
70
) = 240.939 ×106 N.mm
2 = 240.939 kN.m
ISIS EC Module 4
4 - Flexural Strengthening
Step 2: Evaluating the moment capacity of the existing section
Compression in concrete = Tension in steel (Asw)
a1c f’cbb1c = sfy Asw
0.805(0.75)30(0.895)250(c) = 0.9(400)(991)
c = 88.03mm > hf
Mrw= 0.9 (400)991(510 -
0.895 (88.03)
2
) = 167.89 ×106 N.mm
= 167.89 kN.m
Mr= 167.89 + 240.939 =408.8 kN.m
ISIS EC Module 4
Moment resistance of the
section
4 - Flexural Strengthening
Step 3: Optimized determination of AFRP
1) Determine c using the following equation:
a1c f’cbwb12c2 - a1c f’cbwhb1c - Cs(h-d′)+sfy Asw(ds-h)-(Mf -Mrf ) = 0
2
0.805(0.75)30(250)(0.895)2c2 - 0.805(0.75)30(250)600(0.895)c
2
- 0.9(400)991(510-600) + (600-240.9)x106 = 0
1813.57c2 - 2431603.1c + 391208400 = 0
c= 1153 (rejected) or 187mm (accepted)
ISIS EC Module 4
4 - Flexural Strengthening
Step 3: Optimized determination of AFRP
2) Strain in FRP
eFRP = ecu h-c = 0.0035 600-187 = 0.0077 > 0.006
c
187
eFRP = 0.006
TFRP = a1c f’cbwb1c - sfy Asw
=0.805(0.75)30(250)0.895(187) - 400(0.9)991=401 089.6 N
AFRP =
TFRP
FRPEFRP eFRP
=
401 089.6
0.75x0.8×155×103×0.006
= 718.8 mm2
Select: 2 layers b = 250mm and tFRP = 1.5mm, AFRP = 750mm2
ISIS EC Module 4
4 - Flexural Strengthening
Step 4: Check the design
Assume tension failure of the FRP and yielding of steel
1) Neutral axis location
c = TFRP + sfy Asw = 0.75(0.8)(155000)0.006(750)+ 0.9(400)991
0.805(250)0.75(30)0.895
a1c f’cbb1
= 191.3 mm >hf …………O.K
2) Check strains
ec= eFRP c = 0.006 191.3
h-c
600-191.3
es= eFRP ds-c = 0.006 510-191.3
h-c
600-191.3
= 0.0028 < 0.0035
= 0.00467 > ey ………….O.K
ISIS EC Module 4
4 - Flexural Strengthening
Step 4: Check the design
3) Moment resistance of the section
b1 c
b1 c
+TFRP h Mrw = Ts ds 2
2
=
Mrw = 0.9(400)991 510 - 0.895(191.3) × 10-6+
2
0.895(191.3)
0.75(155000)0.006(600) 600 2
= 366.68 kN.m
× 10-6
Mrf = 0.9(400)1409(510 - 70 ) = 240.939 ×106 N.mm = 240.939 kN.m
2
Mrt = Mrw + Mrf = 366.68 + 240.94 = 607.62 kN.m
ISIS EC Module 4
5 - Shear Strengthening
• The shear resistance of the concrete element depends on
the interaction between the concrete and the reinforcement.
• FRP sheets can be applied to increase shear resistance.
• The sheets are placed perpendicular or at an angle to the
beam’s longitudinal axis. The shear capacity from the FRP
stirrups is related to the angle of the cracks in the concrete,
the direction and the effective strain of the FRP.
ISIS EC Module 4
5 - Shear Strengthening
• dFRP is the effective shear
depth for FRP
b
• sFRP is the spacing of the FRP
stirrups
d FRP

sFRP
w FRP
• wFRP is the width of the FRP
stirrup
•  is the angle of inclination of diagonal cracks in the
concrete.
• b is the angle of the FRP stirrups
ISIS EC Module 4
5 - Shear Strengthening
• Many different possible configurations:
1) Continuous wraps or finite width sheets (width and spacing)
Continuous
Finite
2) Angle between the sheet and the beam’s axis
b  90
b = 90
3) Wrap configuration with respect to the cross section
Fully Wrapped
U-Wrap
ISIS EC Module 4
5 - Shear Strengthening
Shear resistance of a beam (Vr ):
1) Existing capacity
- Resistance from concrete (Vc)
- Resistance from steel (Vs)
2) Additional capacity
- Resistance from FRP wraps (VFRP)
Vr = Vc + Vs + VFRP
ISIS EC Module 4
5 - Shear Strengthening
Shear resistance of a beam (Vr ):
1) Resistance provided by concrete (Vc)
Vc = 2.5bvcfcr bvdv
dv ≥ (0.72h, 0.9d)
2) Resistance provided by steel (Vs)
s fyAv dv(cot+cota)sina
Vs =
s
ISIS EC Module 4
5 - Shear Strengthening
3) Resistance provided by FRP:
VFRP = FRP AFRP EFRP eFRPe dFRP (cot + cotb) sinb
sFRP
• AFRP = 2 tFRP wFRP
• eFRPe is the effective strain in the FRP stirrups
• dFRP is the effective depth
• sFRP is the spacing of the FRP stirrups
ISIS EC Module 4
5 - Shear Strengthening
a Effective depth of FRP, dFRP:
d
Closed wrap shear FRP
No flexural FRP
Closed wrap shear FRP
Tension FRP for flexure
dFRP ≥ (0.9d, 0.72h)
dFRP ≥ 0.9h
ISIS EC Module 4
5-Shear Strengthening
a Effective depth of FRP, dFRP:
hfrp
U-Shaped FRP stirrup
No flexural FRP
U-Shaped FRP stirrup
Tension FRP for flexure
dFRP ≥ (0.9hFRP, 0.72h)
dFRP ≥ (0.72h,hFRP)
ISIS EC Module 4
5 - Shear Strengthening
a Effective strain in FRP, eFRPe:
• efrpe = 0.004 ≤ 0.75 efrpu (For completely wrapped sections)
• efrpe = Kvefrpu ≤ 0.004
(For other configurations)
where:
K v=
K1K2Le
11900 eFRPu
≤ 0.75
K1=
fc
’
27
2/3
K2 =
dFRP-Le
dFRP
ISIS EC Module 4
Le =
23300
(tFRPEFRP)0.58
5 - Shear Strengthening
Checks:
- Spacing of strips, sFRP:
d FRP
sFRP ≤ wFRP +
4
- Maximum allowable shear strengthening, VFRP :
Vc+ Vs+ VFRP ≤ 0.25cf’c bvdv
ISIS EC Module 4
5 - Shear Strengthening
Shear Strengthening
Example
Example:
Calculate the shear capacity (Vr) for an FRP-strengthened
concrete section
b = 150 mm
Section
Section
f’c = 45 MPa
h=600 mm
eFRPu = 1.5%
ds =550mm
hFRP = 450 mm
150mm
Section information
tFRP = 1.02 mm
wFRP = 100 mm
sFRP = 200 mm
EFRP = 230GPa
s = 225 mm c/c
fy = 400 MPa (re-bar & stirrup)
CFRP wrap
Elevation
ISIS EC Module 4
Steel used is 10M
5 - Shear Strengthening
Solution:
1) Resistance provided by concrete (Vc)
Vc = 2.5bvcfcr bvdv
fcr = 0.4* √ f’c = 0.4* √45=2.68
dv ≥ (0.72h and 0.9d)
≥ (0.72*600 and 0.9*550)
≥ (432 and 495) = 495mm
Vc = 2.5*0.18*0.75*2.68*150*495*10-3 = 67.24 kN
ISIS EC Module 4
5 - Shear Strengthening
2) Resistance provided by steel (Vs)
Vs =
Vs =
s fyAv dv(cot + cota)sina
s
(0.9)400(200)495(cot42 + cot90)sin90
225
Vs = 175,921 N = 175.9 kN
ISIS EC Module 4
5 - Shear Strengthening
3) Resistance provided by GFRP (VFRP)
VFRP= FRP AFRP EFRP eFRPe dFRP (cot + cotb) sinb
sFRP
dFRP ≥ (0.9 hFRP,0.72h) ≥ (0.9 × 450, 0.72 × 600)
≥ (405,432) = 432mm
AFRP = 2 tFRP wFRP = 2(1.02)(100) = 204 mm2
ISIS EC Module 4
5 - Shear Strengthening
3) Resistance provided by FRP:
K1 =
Le =
fc’
2/3
27
23300
45
=
2/3
27
= 1.406
23300
=
=
17.888mm
(tFRPEFRP)0.58 (1.02 x 230 000)0.58
K2=
Kv=
dFRP-Le
dFRP
432 - 17.88
=
K1K2Le
11900 eFRPu
432
=
= 0.959
(1.406)(0.959)(17.888)
11900 (1.5)(10-2)
ISIS EC Module 4
= 0.135 < 0.75
=0.135
5 - Shear Strengthening
a Effective strain in FRP, efrpe:
•eFRPe ≤ 0.004
•eFRPe ≤ KveFRPu = 0.135 (1.5)(10-2)= 0.002025
eFRPe= 0.002025
FRP AFRP EFRPeFRPe dFRP (cot + cotb) sinb
sFRP
0.6(204)(230000)(0.002025)(432)(cot42)
=136.8 kN
VFRP =
200(1000)
VFRP =
ISIS EC Module 4
5 - Shear Strengthening
Total resistance of the section (Vr):
Vr = Vc + Vs + VFRP
Vr = 67.24 + 175.9 + 136.8 = 379.9 kN
ISIS EC Module 4
5 - Shear Strengthening
Checks:
1) Maximum allowable shear strengthening, VFRP :
Vc + Vs + VFRP ≤ 0.25cf’c bvdv
379.9 ≤ 0.25(0.75)(45)(150)(495)(10-3)
379.9 ≤ 626.48 kN…………………………O.K.
ISIS EC Module 4
5 - Shear Strengthening
Checks:
2) Spacing of strips, sFRP:
d FRP
sFRP ≤ wFRP +
4
432
200 ≤ 100 +
4
200 ≤ 208 mm…………………….O.K
ISIS EC Module 4
6 - Column Strengthening
• FRP sheets can be wrapped around
concrete columns to increase strength
• How it works:
Internal reinforcing steel
Concrete
FRP wrap
Concrete
shortens…
…and dilates…
…FRP confines
the concrete…
ISIS EC Module 4
flFRP
…and places it in
triaxial stress…
6 - Column Strengthening
• The result:
Increased load capacity
Increased deformation capability
ISIS EC Module 4
6 - Column Strengthening
• Confinement efficiency
– Best: circular cross-section
– Worst: rectangular section
• Areas of concrete unconfined by the small bending stiffness of FRP
system
• Stress concentration at corners
confined
f FRP
f FRP
f FRP
f FRP
unconfined
f FRP
f FRP
Uniform stress distribution
in circular section
ISIS EC Module 4
Stress distribution in
rectangular section
6 - Column Strengthening
Slenderness of the column
 If the column is not slender, then the column is designed and analyzed
for axial load only (short column).
 If the column is slender, then the column is designed and analyzed for
combined axial load and bending moment.
ISIS EC Module 4
6 - Column Strengthening
Slenderness of the column
Slenderness could be ignored if:
klu
r
klu
r
< 34 - 12 M1
M2
Braced columns
< 22
Un-braced columns
Where:
k is the effective length factor for the column
lu is the unsupported length of the column
r is the radius of gyration of the section
M1 is the smaller end moment at ULS due to factored loads
M2 is the larger end moment at ULS due to factored loads
ISIS EC Module 4
6 - Column Strengthening
1 - Short column (axial load only)
ISIS EC Module 4
6 - Column Strengthening
1) Confinement Pressure (flFRP):
flFRP=
2tFRPFRPfFRPu
…………………………Eq 6-2
Dg
Where:
flFRP is the confinement pressure
tFRP is the thickness of the FRP confining system
Dg is the external diameter of the circular section or the diagonal
of the rectangular section
ISIS EC Module 4
6 - Column Strengthening
Confinement Limits:
Minimum
confinement
pressure
To ensure
Why? adequate
ductility of
column
Maximum
confinement
pressure
To prevent
Why? excessive
Limit
deformations
of column
Limit
ISIS EC Module 4
flFRP ≥ 0.1fc′
flfrp ≤ 0.33 fc′
6 - Column Strengthening
2) Confined concrete strength (fcc′ ):
The benefit of the confining pressure is to increase the confined
compressive concrete strength, fcc′
fcc′ = fc′+ 2 flFRP
…………………………Eq 6-3
Where:
fc′ is the unconfined specified concrete strength
ISIS EC Module 4
6 - Column Strengthening
3) Axial Load capacity (Pr):
The factored axial load resistance for an FRP-confined reinforced
concrete column, Pr is given by:
Pr= 0.8 a1c fcc′ (Ag-As)+ s fyAs …………………………Eq 6-5
Where:
Ag is the gross area of the cross section
As is the total cross- sectional area of the longitudinal steel
reinforcing bars
ISIS EC Module 4
6 - Column Strengthening
Design steps for short column (axial load only):
′ ) strength according to
1) Determine the required confined concrete (fcc
Equation 6-5.
2) Determine the required confinement pressure (flFRP) from Equation
6-3.
3) Using the properties of the selected FRP system, determine a
minimal thickness for the FRP (tFRP) from Equation 6-2.
4) Check for the confinement limits.
ISIS EC Module 4
6 - Column Strengthening
2 - Slender Column (axial load + moment)
ISIS EC Module 4
6 - Column Strengthening
Section analysis is based on stress and strain compatibility
e c  e cc'
d'
e s'
FRP
e
c
'
c
c'
Ccc
Cs
Cc
confined concrete
d sj
unconfined concrete
h
e sj
Fsj
d
Steel bars
side FRP
es
tension face FRP
e FRP  0.006
Cross
section
Axial strain
distribution
Equivalent stress
distribution
TFRP,side
Ts
TFRP, face
Internal
forces
ec
e′c
e′s
esj
es
eFRP
…………………Eq 6-6
=
=
=
=
=
c
c′
c-d′
dsj-c d-c
h-c
′
′cc
fcc
e
Ccc+ Cc+ Cs – Fsj – Ts - TFRP,side - TFRP,face = Pr ≥ Pf
=5
-1 +1
′
fc
e′c
ISIS EC Module 4
6 - Column Strengthening
1) Assuming concrete crushing
Internal force
′
fc′ +fcc
c e′
c
b cc
e
2
cc
c
ca1 f′c b b 1 e′ ec′
Ccc
Cc
cc
Cs
Fsj
Ts
TFRP,side
TFRP,face
s fy A′ s
e′cc (d -c)
Es Asj ≤ s fy A sj
sj
c
s fy A s or s es Es A s if es < ey
s
FRP eccc′ (h-c) EFRP (h-c) tFRP
FRP eccc′ (h-c) EFRP btFRP
ISIS EC Module 4
Distance from the centre of
the section
h - c - c e′ 2fc′ +fcc′
c
′
3fc′ +3fcc′
e
cc
2
h
b1 c e ′
- c + 1c
′
e
2
2
cc
h /2 -d′
dsj - h
2
d – h/2
h (h-c)
2
3
h
2
6 - Column Strengthening
2) Assuming maximum FRP tension (eFRP = eFRPt ): Dfc= (f′cc-f′c) ec-e′c
e′cc - ec′
Internal force
Ccc
Cc
Cs
Fsj
Ts
TFRP,side
TFRP,face
Distance from the centre of the section
′
c fc+Dfc b c-eh-c e′c
FRPt
2
h-c e′
′
c
ca1 fc b b1 e
FRPt
s fy A ′s or fs es′ Es A′s if es′ < ey
e
s FRPt (dsj-c) Es Asj ≤ s fy A sj
h-c
s fy A s
FRP fFRPu (h-c) tFRP
≤ FRP EFRPeFRPt (h-c) tFRP
FRP fFRPu b tFRP ≤FRP EFRPeFRPt b tFRP
ISIS EC Module 4
h c - h-c e′ 3fc′ +Dfc
eFRPt c 6fc′ +3Dfc
2
h
b1 h-c e′
- c + 1c
e
2
2
FRPt
h - d′
2
dsj - h
2
d-h/2
h (h-c)
2
3
h
2
6 - Column Strengthening
Design steps for slender column:
1) Assuming a linear distribution of strain, identify the relationship of
strain in the various materials as a function of the assumed failure
strain.
2) Determine the resultant force for each material.
3) Calculate the position of the neutral axis using equilibrium of forces.
4) Check the validity of the assumptions of strains and stresses for all
materials.
ISIS EC Module 4
6 - Column Strengthening
Design steps for slender column:
5) Determine Pr as the sum of the resultant force from each material.
6) Determine Mr as the sum of the internal resultant forces multiplied
by their respective distances to the centroid of the section.
ISIS EC Module 4
6 - Column Strengthening
Rectangular Columns
• External FRP wrapping may be used with rectangular columns.
However, strengthening is not as effective and is more complex.
Confinement all around
Confinement only in some areas
ISIS EC Module 4
6 - Column Strengthening
Some geometrical limitations are imposed:
• Sharp edge concrete should be rounded to promote an intimate
and continuous contact of the FRP with the concrete.
- minimum radius is 35 mm
• The aspect ratio of the section (h/b) ≤ 1.5
• The smaller cross section dimension (b) ≤ 600 mm
The equations used are the same. Dg is taken as the diagonal of
the cross section.
flFRP=
2tFRPFRPFFRPu
Dg
=√ h2+b2
ISIS EC Module 4
6 - Column Strengthening
Additional Considerations
• External FRP wrapping may also be used with circular and
rectangular RC columns to strengthen for shear.
• Particularly useful in seismic upgrade situations where
increased lateral loads are a concern.
ISIS EC Module 4
6 - Column Strengthening
• The confining effects of FRP wraps are not activated until
significant radial expansion of concrete occurs.
• Therefore, ensure service loads are kept low enough to
prevent failure by creep and fatigue
• To avoid creep failure:
PD ≤ 0.85 0.8a1c f`c (Ag-As)+ fsAs
fs ≤ 0.0015 Es ≤ 0.8fy
Where:
PD is the dead load
fs is the stress in the axial steel reinforcement
ISIS EC Module 4
6 - Column Strengthening
Example
Example:
Determine the number of layers of GFRP wrap that are required to
increase the factored axial load capacity of the column to 3450 kN.
Information
RC column factored axial resistance
(after strengthening) = 3450 kN
lu = 2500 mm
f’c = 30 MPa
Dg = 450 mm
fFRPu = 600 MPa
Ag = 159000 mm2
tFRP = 1 mm
As = 2500 mm2
FRP = 0.70*0.75
fy = 400 MPa
ISIS EC Module 4
6 - Column Strengthening
Solution:
Step 1: Check for the slenderness effect
klu
r
< 34 - 12 M1
M2
k =1.0, M1=0 and M2=0
2500 = 22.2 < 34
112.5
Thus, the slenderness effect can be ignored
ISIS EC Module 4
6 - Column Strengthening
Step 2: Determine the required confined concrete strength, fcc′
Pr = 0.8 a1c fcc′ (Ag-As)+ s fyAs
a1 = 0.85 – 0.0015 f’c > 0.67
a1 = 0.85 – 0.0015 (30) = 0.81
3450 000 = 0.8 0.81(0.75) fcc′ (159000-2500)+ 0.9(400)2500
fcc′ = 35.9 MPa
ISIS EC Module 4
6 - Column Strengthening
Step 3: Determine the required confinement pressure (flFRP)
fcc′ = fc′+ 2 flFRP
35.9 = 30+ 2 flFRP
flFRP = 2.95 MPa
Step 4: Check for the confinement limits
flFRP ≥ 0.1fc′ =0.1(30) = 3 MPa
flFRP ≤ 0.33 fc′ =0.33(30) = 9.9 MPa
flFRP = 3 MPa
ISIS EC Module 4
6 - Column Strengthening
Step 5: Determine the minimal thickness for the FRP (tFRP) and
number of layers
flFRP =
3=
2tFRPFRPFFRPu
Dg
2tFRP(0.70×0.75)600
450
tFRP = 2.14 mm
Since tGFRP = 1.0 mm, 3 layers of GFRP are required.
ISIS EC Module 4
6 - Column Strengthening
Example
Example:
Check the design of the following column. It is required to resist a
factored axial load of 6000 kN and a factored moment of 1600
kN.m.
Information
75
Axial FRP
2 layers
fy = 400 MPa
fFRPu = 3450MPa
Ast = 4000 mm2
eFRPu = 0.015
Hoop FRP
6 layers
tFRP = 0.167 mm
Steel bars
f’c = 30 MPa
325
800
325
75
600
ISIS EC Module 4
6 - Column Strengthening
Step 1: Determine the properties of the confined concrete
Equivalent diameter:
h
b < 800
= 1.25 ≤ 1.5
b
Dg=√ b2+h2 =√ 6002+8002 = 1000 mm
Confining pressure:
2(6 × 0.167)(0.75 ×0.70)3450
2tFRPFRPfFRPu
flFRP =
=
= 3.63 MPa
Dg
1000
Confinement limits:
0.33 fc′ ≥ flFRP ≥ 0.1fc′
10 ≥ flFRP ≥ 3………………….O.K
ISIS EC Module 4
6 - Column Strengthening
Step 1: Determine the properties of the confined concrete
Confined concrete strength:
fcc′ = fc′ + 2 flFRP
fcc′ = 30+2×3.63 = 37.26 MPa
Concrete strain:
′
e′cc = 5 fcc
-1 +1
ec′
fc′
ecc′ = 0.0035( 5 37.26 -1 +1) = 0.0077
30
ISIS EC Module 4
6 - Column Strengthening
Step 2: Determine the equations of the resultant forces
The following assumptions were made:
- Compression failure (concrete crushing)
′
- fc varies linearly from f’c to fcc
- Yielding of both tension and compression steel
- Intermediate steel in elastic domain
a1 = 0.85 – 0.0015 f’c > 0.67
a1 = 0.85 – 0.0015 (30) = 0.805
b1 = 0.97 – 0.0025 f’c > 0.67
b1 = 0.97 – 0.0025 (30) = 0.895
ISIS EC Module 4
6 - Column Strengthening
Assuming concrete crushing:
Ccc= c
′
fc′ +fcc
2
30+37.26
c × 0.0035
c
′
ec = 0.75
b c600 c e′cc
2
0.0077
c × 0.0035
c
=15133.5
0.0077
c e′
′
Cc= ca1 fc b b 1 e′ c = 0.75(0.805)30(600)0.895
cc
c × 0.0035
= 9726.4
0.0077
Cs=s fy A ′s = 0.9(400)1500 = 540 000 N
ISIS EC Module 4
c × 0.0035
0.0077
6 - Column Strengthening
′
200 000×1000
Fsj = s eccc (dsj-c) Es Asj = 0.9 0.0077
c (400-c)
=180 × 106
0.0077 (400-c)
c
Ts= s fy A s = 0.9 (400) 1500 = 540 000 N
′
TFRP,side= FRP eccc (h-c) EFRP (h-c) tFRP = 0.75×0.70 0.0077(800-c) 2 230 000 ×
c
0.334
= 40330.5 0.0077 (800-c) 2
c
TFRP,face= FRP ec′cc (h-c) EFRP btFRP= 0.75×0.70 0.0077 (800-c) 230 000 ×600
c
×0.334
= 24198300 0.0077 (800-c)
c
ISIS EC Module 4
6 - Column Strengthening
Step 3: Determine the position of the neutral axis, c:
Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face= Pr
c × 0.0035
c
×
0.0035
+ 9726.4
+ 540 000
15133.5 c 0.0077
0.0077
0.0077 (400-c)
0.0077 (800-c) 2
-180 × 106
-540
000
40330.5
c
c
- 24198300 0.0077 (800-c) = 6000 × 103
c
c = 472 mm
ISIS EC Module 4
6 - Column Strengthening
Step 4: Check the assumptions for strains:
e′s = e′cc (c-d′ ) = 0.0077 × 472-75 = 0.0065 > 0.002  OK
c
472
esj = e′cc (dsj-c) = 0.0077 × 400-472 = -0.0012 < ± 0.002  OK
c
472
es = e′cc (d-c ) = 0.0077 × 725-472 = 0.0041 > 0.002  OK
472
c
eFRP= e′cc (h-c ) = 0.0077 × 800-472 = 0.0054 < 0.006  OK
c
472
c′ = c e′c = 0.0035 × 472 = 214.5 mm
e′cc
0.0077
ISIS EC Module 4
6 - Column Strengthening
Step 5: Determine Pr:
Pr = Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face
=15133.5 472-214.5 + 9726.4 214.5 + 540 000 -180 × 106 -0.0012
- 540 000 - 40330.5
- 24198300
0.0077(800-472) 2
472
0.0077 (800-472) = 5997 × 103 N
472
ISIS EC Module 4
6 - Column Strengthening
Step 6: Determine Mr:
Ccc
h - c - c e′ 2fc′ +fcc′
c
′
e
3fc′ +3fcc′
cc
2
2X30+37.3
800 - 472-214.5
= 3897000
3X30+3X37.3
2
Cc h - c + 1- b1
2
2
= 1075 X 106 N.mm
c e′
c
e′cc
800
0.895
=2086000
- 472 + 12
2
X 214.5
= 97 X 106 N.mm
Cs h - d ′ = 540 000 800 -75 = 176 X 106 N.mm
2
2
ISIS EC Module 4
6 - Column Strengthening
Step 6: Determine Mr:
Fsj dsj - h = 0 N.mm
2
Ts d - h
2
= 540 000 725 - 800
2
= 176 X 106 N.mm
TFRP,side h - (h-c) = 71400 800 - (800-472) = 21 X 106 N.mm
2
2
3
3
TFRP,face h
2
800
=131 000
= 52 X 106 N.mm
2
Total = 1597 X 106 N.mm
The flexural resistance is adequate Mr = 1597 kN.m ≈1600 kN.m
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
Includes:
1) Approval of FRP materials
2) Handling and storage of FRP materials
3) Staff and contractor qualifications
4) Concrete surface preparation
5) Installation of FRP systems
6) Curing the FRP system
7) Protection and finishing for FRP system
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
1) Approval of FRP materials:
The use of certified FRP materials is recommended.
Qualification testing can be used for the approval of the FRP materials.
2) Handling and storage of FRP materials:
- Must be carried out in accordance with manufacturer specifications.
- Contractor and supplier must ensure that FRP materials are shipped in adequate
conditions. Do not use opened or damaged containers.
- FRP components must be stored in clean & dry area, sheltered from sun rays.
- Do not use material that has exceeded its shelf life.
- Material safety data sheet for all FRP materials and components should be obtained from
the manufacturer and should be accessible at the job site.
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
3) Staff and contractor qualifications:
The workers must have a basic knowledge of all stages of the installation
of the FRP systems. The minimum required knowledge includes:
- An understanding of the security instructions
- Mixing proportions of resins
- Application rates
- Pot life and curing times
- Installation techniques
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
4) Concrete surface preparation:
-Repair of existing substrate:
- The concrete surfaces must be free of particles and pieces that no
longer adhere to the structure.
- The surface must be cleaned from oil residuals or contaminants.
- Rough surface should be smoothed.
- Sections with sharp edges must be rounded.
- Surface preparation for contact critical applications
- A continuous contact between the concrete and the FRP confinement
system should be guaranteed.
- Rounding of corners, filling holes and elimination of depression are of
prime importance.
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
5) Installation of FRP systems:
- Primer, putty, saturating resin and fibres should be a part of the same system.
- All equipment should be clean and in good operating condition
- Ambient air and concrete surface temperature should be 10°C or more
- The mixing of resins should be done in accordance with the FRP system
manufacturer recommended procedure. All components should be mixed at a
proper temperature and in the correct ratio until there is a uniform mix, free from
trapped air.
- The installation of FRP is either hand wet applied system or precured system.
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
6) Curing the FRP system
- FRP materials should be cured according to the recommendations of the
manufacturer unless the curing process is accelerated by heating,
chemical reactant or other external supply.
- The curing time should not be less than 24 hours before further work is
done on the repaired surface.
- Chemical contamination from gases, dust or liquid must be prevented
during the cure of all materials.
ISIS EC Module 4
7 - Installation of FRP Strengthening Systems
7) Protection and finishing for FRP system
- When the surface of the FRP materials is sufficiently dry or hard, a
protection system and/or paint compatible with the installed
reinforcement can be added.
- The coating must dry for a minimum of 24 hours .
- A certificate of compatibility of the protection system with the selected
type of FRP reinforcement must be obtained from the manufacturer of
the FRP materials.
ISIS EC Module 4
8 - Quality Control and Quality Assurance
The FRP material suppliers, the FRP installation contractors and all others
associated with the FRP strengthening project should maintain a
comprehensive quality assurance and quality control program.
1) Material qualification and acceptance:
The FRP manufacturer, distributor or their agent should provide
information demonstrating that the proposed FRP meets all mechanical,
physical and chemical design requirements.
 Tensile strength, type of fibres, resins, durability, etc.
2) Qualification of contractor personnel:
The selection of contractors should be based on evidence regarding
their qualifications and experience for FRP strengthening projects.
ISIS EC Module 4
8 - Quality Control and Quality Assurance
3) Inspection of concrete substrate:
- The concrete surface should be inspected and tested before
application of FRP. The inspection should include:
- Smoothness or roughness of the surface
- Holes and cracks
- Corners radius
- Cleanliness
- Pull-off tests should be performed to determine the tensile strength of
the concrete for bond-critical applications.
ISIS EC Module 4
8 - Quality Control and Quality Assurance
4) FRP material inspection:
Inspection of the FRP materials shall be conducted before, during and after
their installation.
- Before Construction
The FRP supplier should submit certification & identification of all the FRP
materials to be used. The installation procedure should be submitted as well.
- During Construction
Keep records for:
- Quantity and mixture proportions of resin
- The date and time of mixing
- Ambient temperature & humidity
- All other useful information
Visual inspection of fibres orientation and waviness should be carried out.
ISIS EC Module 4
8 - Quality Control and Quality Assurance
4) FRP material inspection:
- At completion of the project:
A record of all final inspection and test results related to the FRP material
should be retained.
Samples of the cured FRP materials should be retained as well.
5)Testing:
- Qualification testing:
It is a specification for the product certification of FRPs used for rehabilitation.
It includes some guidelines as:
- FRP systems whose properties have not been fully established should
not be considered
- Constituent materials, fibres, matrices and adhesives, should be
acceptable by the applicable code and known for their good performance.
ISIS EC Module 4
8 - Quality Control and Quality Assurance
5) Testing:
- Field testing:
Confirmatory test samples of the FRP material systems should be prepared at
the construction site and tested at an approved laboratory.
In-place load testing can be used to confirm the behaviour of the FRP
strengthened member.
ISIS EC Module 4
9 - Additional Applications
Prestressed FRP Sheets
• One way to improve FRP effectiveness is to apply prestress to the
sheet prior to bonding
• This allows the FRP to contribute to both service and ultimate loadbearing situations
• It can also help close existing cracks, and delay the formation of
new cracks
• Prestressing FRP sheets is a promising technique, but is still under
development
ISIS EC Module 4
10 - Field Applications
Maryland Bridge
- Winnipeg, Manitoba
- Constructed in 1969
- Twin five-span
continuous precast
prestressed girders
- CFRP sheets to
upgrade shear capacity
ISIS EC Module 4
10 - Field Applications
John Hart Bridge
- Prince George, BC
- 84 girder ends were
shear strengthened with
CFRP
- Increase in shear
capacity of 15-20%
- Upgrade completed in
6 weeks
Locations for FRP shear reinforcement
ISIS EC Module 4
10 - Field Applications
Country Hills Boulevard Bridge
- Calgary, AB
- Deck strengthened in
negative bending with
CFRP strips
- New wearing surface
placed on top of FRP
strips
ISIS EC Module 4
Design Guidance
CAN/CSA-S806-02:
Construction
of Building
A CanadianDesign
code and
exists
for the design
of
Components with Fibre Reinforced Polymers
FRP-strengthened concrete members
(Currently under revision)
CAN/CSA-S6-10: Canadian Highway Bridge Design Code
ISIS EC Module 4
Additional Information
Available from www.isiscanada.com
ISIS EC Module 1: Mechanics Examples Incorporating FRP Materials
ISIS EC Module 2: An Introduction to FRP Composites for Construction
ISIS EC Module 3: An Introduction to FRP-Reinforced Concrete
ISIS EC Module 5: Introduction to Structural Health Monitoring
ISIS EC Module 6: Application & Handling of FRP Reinforcements for
Concrete
ISIS EC Module 7: Introduction to Life Cycle Engineering & Costing for
Innovative Infrastructure
ISIS EC Module 8: Durability of FRP Composites for Construction
ISIS EC Module 9: Prestressing Concrete Structures with Fibre Reinforced
Polymers
ISIS EC Module 4