An Introduction to FRP-Strengthening of Concrete Structures
ISIS Educational Module 4:
Updated October 2010 for ISIS Canada
ISIS EC Module 4
Module Objectives
ISIS EC Module 4
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)
Overview
ISIS EC Module 4
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
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
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.
1 - Introduction
ISIS EC Module 4
2 - FRP Materials
• Why repair with the same materials?• Why repeat the cycle?
Light weightEasy to install
High Strength5x steel
Corrosion resistant
Durable structures
Highly versatileSuits many
projects
ISIS EC Module 4
ISIS EC Module 4
FRP is a composite consisting of fibres and matrix.
Fibres:– Provide strength and stiffness–Their quality, orientation and shape affect the final product
Matrix (resin):–Coats the fibres–Protects the fibres from abrasion–Transfers stresses between the fibres
Strain [%]0.5-4.8 2-8
50-90
2400-4300
Stre
ss [M
Pa]
Fibres
Matrix
FRP
2 - FRP Materials
ISIS EC Module 4
FRP material properties are a function of:
•Fibre quality, orientation and shape
•Fibre volumetric ratio
•Adhesion to the matrix
•Manufacturing process (additives and fillers)
2 - FRP Materials
2 - FRP Materials
Wide range of FRP products are available:
• Plates (Rigid strips)
• Sheets (Flexible fabric)
• Rods
The fibres could be:
• Carbon
• Glass
• Aramid
ISIS EC Module 4
FRP sheet
2 - FRP Materials
ISIS EC Module 4
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
3 - FRP Materials
ISIS EC Module 4
FRP properties versus steel:
• Linear elastic behaviour
to failure• No yielding• Higher ultimate
strength• Lower strain at
failure
Steel
CFRP
GFRP
ISIS EC Module 4
2 - FRP Materials
FRP System Fiber TypeWeight[g/m2]
Thickness[mm]
Tensile Strength [MPa]
Tensile Elastic Modulus [GPa]
Strain at Failure [%]
Fyfe Co. LLC [www.fyfeco.com]Tyfo SEH51A Glass 915 1.3 460 20.9 1.8Tyfo SEH25A Glass 505 0.5 417 20.9 1.8Tyfo SCH41 Carbon 644 1.0 834 82 0.9
Hughes Brothers Inc. [www.hughesbros.com]Aslan 200 Carbon -- 6.4-12.7 2068-1724 124 0.017-0.015
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 -- 1.4 2400 131 1.9Sika Canada Inc. [www.sika.ca]
SikaWrap 430G Glass 430 0.5 504 24.6 1.9SikaWrap 100G Glass 915 1.0 558 24.4 2.2SikaWrap 230C Carbon 230 0.4 715 61.0 1.1SikaWrap 103C Carbon 610 1.0 717 65.1 1.0CarboDur S Carbon -- 1.2-1.4 2800 165 1.7
CarboDur M Carbon -- 1.4 2400 210 1.2
CarboDur H Carbon -- 1.4 1300 300 0.5BASF Building Systems Inc. [www.BASFBuildingSystems.com]
MBrace S&P Laminate Carbon -- 1.4 2700 159 1.7
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
1) Wet lay-up system:
- Used with flexible sheets
- Multiple layers can be used
- Saturate sheets with epoxy adhesive, then place on the concrete surface and press with a roller
Epoxy
Roller
Installation:
- The system is installed on the surface of the concrete element while the resin matrix is still “wet”, and the polymerization occurs on site
2 - FRP Materials
2 - FRP Materials
- Used with FRP plates or laminates
- Used for surface bonded plates or near surface mounted reinforcements
- The pre-cured laminates should be placed on or into the wet adhesive
- Place on the concrete surface
- Multiple layers can not be used
- Not as flexible for variable structural shapes
2) Pre-cured system:
ISIS EC Module 4
ISIS EC Module 4
3- Near Surface Mounted (NSM)• It is a newer class of FRP strengthening technique.
Un-strengthened concrete T-beam
Longitudinal grooves cut into soffit
FRP strips placed in grooves
Grooves filled with epoxy grout
• Research indicates that NSM reinforcement is effective and efficient for strengthening.
2 - FRP Materials
2 - FRP Materials
ISIS EC Module 4
Type Application
Schematic
FRP-Strengthening Applications :Fibre Dir.
Confinement Around the column Circumferential
Section
ShearSide face of the beam
(U or closed wrap)
Perpendicular to long. axisof the beam Section
Flexuralface of the beam
Tension and/or sideaxis of the
beam
Along long.
Section
3 - Evaluation of Existing Structures
ISIS EC Module 4
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
3 - Evaluation of Existing Structures
ISIS EC Module 4
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
3 - Evaluation of Existing Structures
ISIS EC Module 4
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
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
3 - Evaluation of Existing Structures
ISIS EC Module 4
3 - Evaluation of Existing Structures
ISIS EC Module 4
• Examples of some deficiencies:
1. Environmental effects
Freeze-Thaw
Chloride Ingress
Wet-Dry
3 - Evaluation of Existing Structures
• A primary factor leading to extensive degradation…..
2. Corrosion
Moisture, oxygen and chlorides penetrate
ConcreteReinforcing
Steel
Through concrete
Through cracks
Corrosion products formVolume expansion occurs
More cracking
Corrosion propagation
End result
ISIS EC Module 4
3 - Evaluation of Existing Structures
ISIS EC Module 4
• Deficiencies could be due to:
3. Updated design loads
4. Updated design code procedures
5. Increase in traffic loads
ThenNow
3 - Evaluation of Existing Structures
ISIS EC Module 4
• Deficiencies could be due to:
6. Fire damage
7. Earthquakes
ISIS EC Module 4
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
4 - Flexural Strengthening
ISIS EC Module 4
Failure ModesThe analysis of the flexural strength of FRP strengthened elements 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 (FRPt ) is 0.006.
5) The maximum compressive strain in the concrete (cu ) is 0.0035.
6) The contributions of FRPs in compression and of the concrete in
tension are neglected.
4 - Flexural Strengthening
ISIS EC Module 4
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).
4 - Flexural Strengthening
ISIS EC Module 4
Rectangular section without compression steel:
b
d
Cross Section
As
Strain Distribution
FRP
c
h
bFRP
s
c
Stress Distribution
fs
fFRP
Equiv. Stress Distribution
a =1c
1Φcf’c
Ts
TFRP
Cc
Cc = c1f’c1bcTFRP = FRPAFRPEFRPFRPTs = sAsfs
4 - Flexural Strengthening
The equilibrium equations are:
1) Force equilibrium in the section:
Compression forces =Tension forces
2) Moment equilibrium in the section:
External applied moment= Internal moment
ISIS EC Module 4
Cc = Ts + TFRP
Mapplied = Ts d -a
2+ TFRP
h -a
2
4 - Flexural Strengthening
ISIS EC Module 4
b
d
Cross Section
As
Rectangular section with compression steel :
h
bFRP
Strain Distribution
FRP
s
c
Stress Distribution
fs
fFRP
Equiv. Stress Distribution
a =1c
1Φcf’c
Ts
TFRP
CcA’s
cu
’s f’s Cs
Add a compressive stress resultant
Cs = sf’sA’s
d’
4 - Flexural Strengthening
Mr = Ts d-a2
+TFRPh-
a2
+Cs
a2
- d′
b
d
As
FRP
c
h
bFRP
s
c
ISIS EC Module 4
4 - Flexural Strengthening
4. Determine the compressive stress block factors (1, 1)
1 = 0.85 – 0.0015 f’c > 0.671 = 0.97 – 0.0025 f’c > 0.67
d
As
b
a =1c
1Φcf’c
Ts
Cc
h
bFRP
TFRP
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 (s and ’s < y )
-Steel yielding followed by concrete crushing (s and ’s > y )
- Steel yielding followed by FRP rupture (s and ’s > y )
1c f’c1bc2 + sEs cu (As’+As)+ FRPEFRP (cu+fi)AFRP
sEs cu (As'd'+Asd)+ FRPEFRP cuAFRP h = 0
c -
sfy(As’-As)+ FRPEFRP (cu+fi)AFRP c-
FRPEFRP cuAFRP h = 0
ISIS EC Module 4
c =sfy(As-As
’)+ FRPEFRP FRPtAFRP
1c f’c1b
4 - Flexural Strengthening
1c f’c1bc2 +
6. Check failure mode assumption with the material strains
If failure is initiated by: Concrete crushing: FRP rupture:
- 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
cu= 0.0035
s= cu
d - cc
FRP= cuh - c
cfi
s'= cu
c - d'
c
FRP = FRPu ≤ FRPt
fi+FRPu) ≤ s'= fi+FRPt)
fi+FRPu) ≤ s= fi+FRPt)
fi+FRPu) ≤ c= fi+FRPt)
c-d'
h-cc-d'
h-cd-ch-c
d-ch-c
ch-c
ch-c
4 - Flexural Strengthening
7. Compute internal forces
8. Calculate the section moment resistance (Mr)
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
Cs = sfsA's TFRP = FRPAFRPEFRPFRPTs = sAsfs
ISIS EC Module 4
Mr = Ts d-a2
+TFRPh-
a2
4 - Flexural Strengthening
+Cs
a2
- d′
Optimized determination of AFRP
Assuming: All the steel has yielded and combining the equilibrium equations:
1. Determine c using the following equation
2. Determine strain in FRP
ISIS EC Module 4
1c f’cbw12c2
2 1 c f’cbwh1c Cs(h-d′)+Ts(ds-h) - Mf
FRP = cuh - c
c- fi ≤ 0.006 ≤ FRPu
4 - Flexural Strengthening
3. Determine successively:
4. Optimize value of AFRP
Use this AFRP as an input for the iterative design method
ISIS EC Module 4
AFRP =TFRP
FRPEFRP FRP
Cc = c1f’c1bc
TFRP = Cc + Cs - Ts
4 - Flexural Strengthening
Geometric parameters
h
FRPt
FRPb
eb
c fh • If the neutral axis lies in the web (c < hf), then treat it as a rectangular section with a compression zone width = be.
• If the neutral axis lies outside the web (c > hf), then treat it as a T-section.
ISIS EC Module 4
T-section:
4 - Flexural Strengthening
Factored moment subdivisionGeometric parameters
h
FRPt
FRPb
eb
c fh
FRPA FRPAsA sfA swA
rfM rwMrM = +
= +
The section is treated as the summation of: flange (Mrf) and web (Mrw).
- Flange (Mrf)
- Web (Mrw)
ISIS EC Module 4
sfy
1c f’c(be-bw)hf Asf =
Asw = As –Asf Mrw = s fy Asw d -a
2+TFRP
h -a
2
Mrf = s fy Asf d -hf
2
4 - Flexural Strengthening
Design procedure:
1. Select FRP material and AFRP
2. Determine behaviour of the section ( Rect. or T)
If
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
hf ≥sfyAs+FRP EFRP FRP A FRP
1c f’c 1be
4 - Flexural Strengthening
ISIS EC Module 4
Example:
Calculate the moment resistance (Mr) for an FRP-strengthened rectangular concrete section
Section information:
f’c = 40 MPaFRPu = 1.26 %
AFRP = 110 mm2
fy = 400 MPa
Es = 200 GPa EFRP = 210 GPa
b = 125 mm
h =
360
mm
2-15M bars
d =
320
mm
CFRP
4 - Flexural Strengthening
ISIS EC Module 4
Solution:Step 1: Assume failure mode
Assume failure of beam due to crushing of concrete in compression after yielding of internal steel reinforcement
4 - Flexural Strengthening
ISIS EC Module 4
Step 2: Calculate concrete stress block factors
1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (40) = 0.79
1 = 0.97 – 0.0025 f’c > 0.67
1 = 0.97 – 0.0025 (40) = 0.87
4 - Flexural Strengthening
ISIS EC Module 4
Step 3: Find depth of neutral axis, cSteel yielding followed by concrete crushing
1c f’cbc2 + sfy(As’-As) + FRPEFRP (cu+fi)AFRP c - FRPEFRP cuAFRP 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
= 0
2577.375(c2)-79320(c)-23284800 = 0
c = 111.7 mm or c = -80.9 mm (rejected)
4 - Flexural Strengthening
ISIS EC Module 4
Step 4: Check mode of failureSteel yielding followed by concrete crushing
cu= 0.0035
s= cu= 0.0035
Trial 2, assume the steel yields and the strain in the FRP is 0.006
ds-cc
320-111.7
111.7= 0.0065 > 0.002(y)
FRP= cuh-cc
= 0.0035 360-111.7111.7
= 0.0078 > 0.006 not O.K.
4 - Flexural Strengthening
ISIS EC Module 4
Step 5: Trial 2
sfy As + FRPEFRP FRPAFRP
Compression in concrete = Tension in ( Steel + FRP)
1c f’cbc =
0.79(0.75)40(0.87)125(c) 0.9(400)400 + 0.80(210000)0.006(110)=
c = 98.9 mm
4 - Flexural Strengthening
ISIS EC Module 4
Step 6: Check mode of failure
FRP = 0.006
s= FRP = 0.006
The assumed mode of failure is correct
ds-c 320-98.9= 0.0051 > 0.002(y)
cu= FRPc
h-c= 0.006 98.9
360-98.9= 0.0023 < 0.0035
O.K.
h-c 360-98.9
4 - Flexural Strengthening
ISIS EC Module 4
Step 7: Moment Resistance
sfy As + FRPEFRP FRPAFRP=
0.9(400)400
+0.80(210000)0.006(110)
=
Mr = Ts ds-1c
2+TFRP
h-1c
2
ds-1c
2h-
1c
2320-
0.87 (98.9)
2
360-2
Mr = 75 106 N.mm = 75 kN.m
4 - Flexural Strengthening
ISIS EC Module 4
Example:The T-beam requires strengthening to upgrade its moment capacity to 600 kN-m. Calculate the required area of FRP (AFRP).
Section information:f’c = 30 MPa
FRPu = 1.26 %As = 8 x 300 mm2
fy = 400 MPa
Es = 200 GPa EFRP = 155 GPa600
250
650
70
510
4 - Flexural Strengthening
ISIS EC Module 4
Step 1: Calculate concrete stress block factors
1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (30) = 0.805
1 = 0.97 – 0.0025 f’c > 0.67
1 = 0.97 – 0.0025 (30) = 0.895
4 - Flexural Strengthening
ISIS EC Module 4
Step 2: Evaluating the moment capacity of the existing section
sfy As
Compression in concrete = Tension in steel
1c f’cb1c =
0.805(0.75)30(0.895)650(c) 0.9(400)(300× 8)=
c = 82mm > hf
Assume neutral axis is inside the flange (c < hf)
The assumption (c < hf) is wrong.
4 - Flexural Strengthening
ISIS EC Module 4
Step 2: Evaluating the moment capacity of the existing section
sfy
1c f’c(be-bw)hf
=1409 mm2
Assume neutral axis is outside the flange (c > hf)
Asf =
0.9(400)
0.805(0.75)30(650-250)70 Asf =
Asw = As –Asf = 2400 -1409 = 991 mm2
Mrf = 0.9 (400)1409 (510 -70
2) = 240.939 ×106 N.mm = 240.939 kN.m
4 - Flexural Strengthening
ISIS EC Module 4
Step 2: Evaluating the moment capacity of the existing section
sfy Asw
Compression in concrete = Tension in steel (Asw)
1c f’cb1c =
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 Moment resistance of the section
4 - Flexural Strengthening
ISIS EC Module 4
Step 3: Optimized determination of AFRP
1) Determine c using the following equation:
1c f’cbw12c2
c= 1153 (rejected) or 187mm (accepted)
21c
f’cbwh1cCs(h-d′)+sfy Asw(ds-h)-(Mf -Mrf ) = 0
0.805(0.75)30(250)(0.895)2c2
2- 0.805(0.75)30(250)600(0.895)c
- 0.9(400)991(510-600) + (600-240.9)x106 = 0
1813.57c2 - 2431603.1c + 391208400 = 0
4 - Flexural Strengthening
ISIS EC Module 4
Step 3: Optimized determination of AFRP
2) Strain in FRP
Select: 2 layers b = 250mm and tFRP = 1.5mm, AFRP = 750mm2
FRP = cuh-c
c= 0.0035 600-187
187= 0.0077 > 0.006
FRP = 0.006
TFRP =1c f’cbw1c -sfy Asw
=0.805(0.75)30(250)0.895(187) - 400(0.9)991=401 089.6 N
AFRP =TFRP
FRPEFRP FRP
=401 089.6
0.75x0.8×155×103×0.006= 718.8 mm2
4 - Flexural Strengthening
ISIS EC Module 4
Step 4: Check the designAssume tension failure of the FRP and yielding of steel
= 191.3 mm >hf …………O.K
c = TFRP +sfy Asw
1c f’cb1
= 0.75(0.8)(155000)0.006(750)+ 0.9(400)9910.805(250)0.75(30)0.895
1) Neutral axis location
2) Check strains
c= FRPc
h-c= 0.006 191.3
600-191.3= 0.0028 < 0.0035
s= FRPds-ch-c
= 0.006 510-191.3600-191.3
= 0.00467 > y ………….O.K
4 - Flexural Strengthening
ISIS EC Module 4
Step 4: Check the design
Mrw =
=
0.75(155000)0.006(600)
3) Moment resistance of the section
Mrw = Ts ds -1 c
2+TFRP
h -1 c
2
0.9(400)991
600 - 0.895(191.3)
2
510 - 0.895(191.3)2
× 10-6+
= 366.68 kN.m × 10-6
Mrf = 0.9(400)1409(510 - 702
) = 240.939 ×106 N.mm = 240.939 kN.m
Mrt = Mrw + Mrf = 366.68 + 240.94 = 607.62 kN.m
4 - Flexural Strengthening
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
• dFRP is the effective shear depth for FRP
• sFRP is the spacing of the FRP stirrups
• wFRP is the width of the FRP stirrup
5 - Shear Strengthening
FRPd
FRPwFRPs
• is the angle of inclination of diagonal cracks in the concrete.
• is the angle of the FRP stirrups
ISIS EC Module 4
• Many different possible configurations: 1) Continuous wraps or finite width sheets (width and spacing)
2) Angle between the sheet and the beam’s axis
3) Wrap configuration with respect to the cross section
U-Wrap
Continuous Finite
= 90
5 - Shear Strengthening
Fully Wrapped
90
ISIS EC Module 4
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+ +
5 - Shear Strengthening
ISIS EC Module 4
Shear resistance of a beam (Vr ):
1) Resistance provided by concrete (Vc)
dv ≥ (0.72h, 0.9d)
2) Resistance provided by steel (Vs)
Vc = 2.5vcfcr bvdv
5 - Shear Strengthening
ISIS EC Module 4
3) Resistance provided by FRP:
VFRP = FRP AFRP EFRP FRPe dFRP (cot + cot) sinsFRP
• AFRP = 2 tFRP wFRP
•FRPe is the effective strain in the FRP stirrups
• dFRP is the effective depth
• sFRP is the spacing of the FRP stirrups
5 - Shear Strengthening
ISIS EC Module 4
Effective depth of FRP, dFRP:
Closed wrap shear FRPNo flexural FRP
Closed wrap shear FRPTension FRP for flexure
dFRP ≥ (0.9d, 0.72h) dFRP ≥ 0.9h
d
5 - Shear Strengthening
ISIS EC Module 4
Effective depth of FRP, dFRP:
U-Shaped FRP stirrupNo flexural FRP
U-Shaped FRP stirrupTension FRP for flexure
dFRP ≥ (0.9hFRP, 0.72h) dFRP ≥ (0.72h,hFRP)
hfrp
5-Shear Strengthening
ISIS EC Module 4
•frpe = 0.004 ≤ 0.75 frpu (For completely wrapped sections)
•frpe = Kvfrpu ≤ 0.004 (For other configurations)
where:
Kv=
K1=
Effective strain in FRP, FRPe:
K1K2Le
11900 FRPu
≤ 0.75
fc’
27
2/3
K2 =dFRP-Le
dFRP
Le=23300
(tFRPEFRP)0.58
5 - Shear Strengthening
ISIS EC Module 4
Checks:
- Spacing of strips, sFRP:
sFRP ≤ wFRP + d FRP
4
- Maximum allowable shear strengthening, VFRP :
Vc+ Vs+ VFRP ≤ 0.25cf’c bvdv
5 - Shear Strengthening
Shear Strengthening
ISIS EC Module 4
ExampleExample:Calculate the shear capacity (Vr) for an FRP-strengthened concrete section
Section information
Sectionb = 150 mm
h FRP =
450
mm
d s =55
0mm
CFRP wrap
Section Elevation
f’c = 45 MPa
FRPu = 1.5%
fy = 400 MPa (re-bar & stirrup)
Steel used is 10M
EFRP = 230GPa
s = 225 mm c/c
tFRP = 1.02 mm
wFRP = 100 mmsFRP = 200 mm15
0mm
h=600 mm
5 - Shear Strengthening
ISIS EC Module 4
Solution:
1) Resistance provided by concrete (Vc)
Vc = 2.5vcfcr 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
5 - Shear Strengthening
ISIS EC Module 4
2) Resistance provided by steel (Vs)
Vs = 175,921 N = 175.9 kN
5 - Shear Strengthening
ISIS EC Module 4
3) Resistance provided by GFRP (VFRP)
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
VFRP= FRP AFRP EFRP FRPe dFRP (cot + cot) sinsFRP
5 - Shear Strengthening
ISIS EC Module 4
3) Resistance provided by FRP:
27
fc’
2/3 =27
45 2/3 = 1.406
K2=dFRP-Le
dFRP
Le=23300
(tFRPEFRP)0.58=
23300
(1.02 x 230 000)0.58
= 17.888mm
=432 - 17.88
432= 0.959
Kv=
K1=
K1K2Le
11900 FRPu
=(1.406)(0.959)(17.888)
11900 (1.5)(10-2)= 0.135 < 0.75
=0.135
5 - Shear Strengthening
ISIS EC Module 4
•FRPe ≤ 0.004
•FRPe ≤ KvFRPu = 0.135 (1.5)(10-2)= 0.002025
FRPe= 0.002025
Effective strain in FRP, frpe:
VFRP = FRP AFRP EFRPFRPe dFRP (cot + cot) sinsFRP
VFRP =0.6(204)(230000)(0.002025)(432)(cot42)
200(1000)=136.8 kN
5 - Shear Strengthening
ISIS EC Module 4
Total resistance of the section (Vr):
Vr = Vc Vs VFRP+ +
Vr = 67.24 + 175.9 + 136.8 = 379.9 kN
5 - Shear Strengthening
ISIS EC Module 4
Checks:
1) Maximum allowable shear strengthening, VFRP :
Vc + Vs + VFRP ≤ 0.25cf’c bvdv
379.9 ≤ 0.25(0.75)(45)(150)(495)(10-3)
379.9 ≤ 626.48 kN…………………………O.K.
5 - Shear Strengthening
ISIS EC Module 4
Checks:
2) Spacing of strips, sFRP:
sFRP ≤ wFRP + d FRP
4
200 ≤ 100 + 4324
200 ≤ 208 mm…………………….O.K
5 - Shear Strengthening
ISIS EC Module 4
6 - Column Strengthening
• FRP sheets can be wrapped around concrete columns to increase strength
• How it works:
Concrete shortens…
…and dilates……FRP confines the concrete…
flFRP
…and places it in triaxial stress…
Internal reinforcing steelConcrete
FRP wrap
ISIS EC Module 4
• The result:
Increased load capacity
Increased deformation capability
6 - Column Strengthening
ISIS EC Module 4
• 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
FRPf FRPf
FRPf
confined
unconfined
FRPfFRPf
FRPf
Stress distribution in rectangular section
Uniform stress distribution in circular section
6 - Column Strengthening
ISIS EC Module 4
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.
6 - Column Strengthening
ISIS EC Module 4
Slenderness of the column
Slenderness could be ignored if:
klur
< 34 - 12 M1
M2
Where:k is the effective length factor for the columnlu is the unsupported length of the column
r is the radius of gyration of the sectionM1 is the smaller end moment at ULS due to factored loads
M2 is the larger end moment at ULS due to factored loads
Braced columns
klur
< 22 Un-braced columns
6 - Column Strengthening
ISIS EC Module 4
1 - Short column (axial load only)
6 - Column Strengthening
ISIS EC Module 4
1) Confinement Pressure (flFRP):
flFRP=
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
2tFRPFRPfFRPu
Dg
…………………………Eq 6-2
6 - Column Strengthening
ISIS EC Module 4
Minimum confinement pressure
Maximum confinement pressure
Why?To ensure adequate ductility of column
LimitflFRP ≥ 0.1fc
To prevent excessive deformations of column
LimitWhy?flfrp ≤ 0.33 fc
′
′
Confinement Limits:
6 - Column Strengthening
ISIS EC Module 4
2) Confined concrete strength (fcc):
Where:
fc is the unconfined specified concrete strength
fcc= fc+ 2 flFRP′ ′
′
′
The benefit of the confining pressure is to increase the confined compressive concrete strength, fcc
′
…………………………Eq 6-3
6 - Column Strengthening
ISIS EC Module 4
3) Axial Load capacity (Pr):
Where:
Ag is the gross area of the cross section
As is the total cross- sectional area of the longitudinal steel reinforcing
bars
Pr= 0.8 c fcc(Ag-As)+ s fyAs′
The factored axial load resistance for an FRP-confined reinforced concrete column, Pr is given by:
…………………………Eq 6-5
6 - Column Strengthening
ISIS EC Module 4
Design steps for short column (axial load only):
1) Determine the required confined concrete (fcc) strength according to
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.
′
6 - Column Strengthening
ISIS EC Module 4
2 - Slender Column (axial load + moment)
6 - Column Strengthening
ISIS EC Module 4
Section analysis is based on stress and strain compatibility
Equivalent stress distribution
confined concrete
unconfined concrete
Internal forces
side FRP
tension face FRP
sC
sT
sjF
cC
ccC
faceFRPT ,
sideFRPT ,
Axial strain distribution
's
s
006.0FRP
sj
'ccc
'c 'c
'd
h
d
sjdc
Cross section
FRP
Steel bars
c
c
=c
c
=s
c-d=
sj
dsj-c=
s
d-c
=FRP
h-c
′′
′′
…………………Eq 6-6
Ccc+ Cc+ Cs – Fsj – Ts - TFRP,side - TFRP,face = Pr ≥ Pf
6 - Column Strengthening
′
′= 5
cc
′
c
fcc
fc
-1′ +1
ISIS EC Module 4
1) Assuming concrete crushing Internal force
fc+fcc
cc
Ccc
′′2
b c- c c
′
cc′Cc ′ c
c ′
Cs ′cc′Fsj c
Ts s fy A s or s s Es A s if s <y
′TFRP,side FRP c EFRP (h-c) tFRP (h-c)cc
TFRP,face FRP c EFRP btFRP(h-c)cc′
6 - Column Strengthening
c
c1 fc b 1
s fy A s
s ≤ s fy A sjEs Asj (dsj-c)
cc- c - c
c ′
′′
′′2
h 2fc+fcc3fc+3fcc′
Distance from the centre of the section
c
cc
c2h
- c + 1-21
′′
′h /2 -d
dsj - 2h
2h
(h-c)2h
3-
d – h/2
ISIS EC Module 4
2) Assuming maximum FRP tension (FRP =FRPt ):
fc+fc FRPtCcc c
′2
b c-h-c c
′
FRPtCc c1 fc b 1
′ h-c c
′
Cs s fy A s or fs s Es A s if s < y′FRPtFsj s ≤ s fy A sjh-c
Es Asj (dsj-c)
Ts s fy A s
TFRP,side FRP fFRPu (h-c) tFRP
≤ FRP EFRPFRPt (h-c) tFRP
fc= (fcc-fc)′′cc - c
c-c′′′
′ ′′
TFRP,face FRP fFRPu b tFRP ≤FRP EFRPFRPt b tFRP
6 - Column Strengthening
Internal force Distance from the centre of the section
′2h - d
d-h/2
2h
dsj - 2h
(h-c)2h
3-
c FRPt
h-c2h
- c + 1-21 ′
FRPtc - h-c c
′ ′2
h 3fc+fc6fc+3fc
′
ISIS EC Module 4
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.
6 - Column Strengthening
ISIS EC Module 4
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.
6 - Column Strengthening
ISIS EC Module 4
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
6 - Column Strengthening
ISIS EC Module 4
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 mmThe equations used are the same. Dg is taken as the diagonal of
the cross section.
6 - Column Strengthening
flFRP=2tFRPFRPFFRPu
Dg =√ h2+b2
ISIS EC Module 4
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.
6 - Column Strengthening
ISIS EC Module 4
• 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:
6 - Column Strengthening
PD ≤ 0.85 0.81c 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
Example
ISIS EC Module 4
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.
InformationRC column factored axial resistance (after strengthening) = 3450 kN
lu = 2500 mm
Dg = 450 mm
Ag = 159000 mm2
As = 2500 mm2
fy = 400 MPa
f’c = 30 MPa
fFRPu = 600 MPa
tFRP = 1 mm
FRP = 0.70*0.75
6 - Column Strengthening
ISIS EC Module 4
Solution:Step 1: Check for the slenderness effect
klur
< 34 - 12 M1
M2
k =1.0, M1=0 and M2=0
2500112.5
= 22.2 < 34
Thus, the slenderness effect can be ignored
6 - Column Strengthening
ISIS EC Module 4
Step 2: Determine the required confined concrete strength, fcc
Pr = 0.8 1c fcc(Ag-As)+ s fyAs′
1 = 0.85 – 0.0015 f’c > 0.67
1 = 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
6 - Column Strengthening
′
′
′
ISIS EC Module 4
Step 3: Determine the required confinement pressure (flFRP)
35.9 = 30+ 2 flFRP
fcc= fc+ 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
6 - Column Strengthening
ISIS EC Module 4
Step 5: Determine the minimal thickness for the FRP (tFRP) and number of layers
flFRP =2tFRPFRPFFRPu
Dg
3 =2tFRP(0.70×0.75)600
450
tFRP = 2.14 mm
Since tGFRP = 1.0 mm, 3 layers of GFRP are required.
6 - Column Strengthening
Example
ISIS EC Module 4
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
Ast = 4000 mm2
fy = 400 MPa
f’c = 30 MPa
fFRPu = 3450MPa
tFRP = 0.167 mm
FRPu = 0.015
75
325
325
75
800
600
Axial FRP2 layers
Hoop FRP6 layers
Steel bars
6 - Column Strengthening
ISIS EC Module 4
Step 1: Determine the properties of the confined concrete
flFRP =
bDg=√ b2+h2 =√ 6002+8002 = 1000 mm
= 1.25 ≤ 1.5
Confinement limits:
h b < 800
Equivalent diameter:
Confining pressure: 2tFRPFRPfFRPu
Dg
2(6 × 0.167)(0.75 ×0.70)3450
1000= = 3.63 MPa
0.33 fc ≥ flFRP ≥ 0.1fc
10 ≥ flFRP ≥ 3………………….O.K
′ ′
6 - Column Strengthening
ISIS EC Module 4
Step 1: Determine the properties of the confined concrete
Confined concrete strength:
Concrete strain:
fcc = fc+ 2 flFRP′ ′
fcc= 30+2×3.63 = 37.26 MPa′
= 5cc
′c
fcc -1 +1′
′fc′
cc
′ = 0.0035( 5 -1 +1) = 0.0077 37.26
30
6 - Column Strengthening
ISIS EC Module 4
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
′
1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (30) = 0.805
1 = 0.97 – 0.0025 f’c > 0.67
1 = 0.97 – 0.0025 (30) = 0.895
6 - Column Strengthening
ISIS EC Module 4
Assuming concrete crushing:
fc+fcc
cc
Ccc= c
′′
2b c- c
c ′
cc′Cc= c1 fc b 1
′ c c
′
Cs=s fy A s ′
′= 0.75
30+37.262
600 c - c × 0.0035 0.0077
= 0.75(0.805)30(600)0.895 c × 0.0035 0.0077
=15133.5 c - c × 0.0035 0.0077
= 9726.4c × 0.0035 0.0077
= 0.9(400)1500 = 540 000 N
6 - Column Strengthening
ISIS EC Module 4
cc′Fsj = s c Es Asj (dsj-c)
Ts= s fy A s = 0.9 (400) 1500 = 540 000 N
′TFRP,side= FRP c EFRP (h-c) tFRP = 0.75×0.70 (h-c)cc
TFRP,face= FRP c EFRP btFRP= 0.75×0.70 (h-c)cc′
= 0.9 c (400-c)0.0077 200 000×1000
c (400-c)0.0077=180 × 106
230 000 × 0.334c
(800-c) 20.0077
= 40330.5c
(800-c) 20.0077
c(800-c)0.0077
230 000 ×600 ×0.334
= 24198300c
(800-c)0.0077
6 - Column Strengthening
ISIS EC Module 4
Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face= Pr
15133.5 c - c × 0.0035 0.0077
+ 9726.4c × 0.0035 0.0077
+ 540 000
c (400-c)0.0077-180 × 106 -540 000 - 40330.5c
(800-c) 20.0077
- 24198300c
(800-c)0.0077 = 6000 × 103
c = 472 mm
Step 3: Determine the position of the neutral axis, c:
6 - Column Strengthening
ISIS EC Module 4
Step 4: Check the assumptions for strains:
=cc
c
s = 0.0077 × = 0.0065 > 0.002 OK′′
sj
(c-d )′ 472-75472
cc
c
′= (dsj-c) = 0.0077 × 400-472
472= -0.0012 < ± 0.002 OK
=cc
c
s = 0.0077 × = 0.0041 > 0.002 OK′ (d-c ) 725-472472
=cc
c
FRP
= 0.0077 × = 0.0054 < 0.006 OK′ (h-c ) 800-472472
=cc
cc
c = 0.0035 × = 214.5 mm′′ 4720.0077′
6 - Column Strengthening
ISIS EC Module 4
Pr = Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face
=15133.5 472-214.5 + 9726.4 214.5 + 540 000 -0.0012-180 × 106
- 540 000 - 40330.5472
(800-472) 20.0077
- 24198300472
(800-472)0.0077 = 5997 × 103 N
Step 5: Determine Pr:
6 - Column Strengthening
ISIS EC Module 4
Step 6: Determine Mr:
cc Ccc - c - c
c ′
′
Cc
′
c
′
Cs
′
cc
′
c
2h 2fc+fcc
3fc+3fcc′
2h - d
2h
- c + 1-21
′′
- 472-214.52
800 2X30+37.33X30+3X37.3= 3897000 = 1075 X 106 N.mm
=20860002
800- 472 + 1-
20.895
X 214.5 = 97 X 106 N.mm
= 540 000 2
800 -75 = 176 X 106 N.mm
6 - Column Strengthening
ISIS EC Module 4
Step 6: Determine Mr:
Fsj dsj -
TFRP,side(h-c)
TFRP,face2h
2h
3-
2h
= 176 X 106 N.mm
= 0 N.mm
Ts d -2h = 540 000 725 -
2800
= 71400 (800-472)2800
3- = 21 X 106 N.mm
=131 0002
800= 52 X 106 N.mm
Total = 1597 X 106 N.mm
The flexural resistance is adequate Mr = 1597 kN.m ≈1600 kN.m
6 - Column Strengthening
ISIS EC Module 4
Includes:
2) Handling and storage of FRP materials
3) Staff and contractor qualifications
4) Concrete surface preparation
5) Installation of FRP systems
7) Protection and finishing for FRP system
7 - Installation of FRP Strengthening Systems
6) Curing the FRP system
1) Approval of FRP materials
ISIS EC Module 4
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.
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.
7 - Installation of FRP Strengthening Systems
ISIS EC Module 4
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
7 - Installation of FRP Strengthening Systems
ISIS EC Module 4
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.
7 - Installation of FRP Strengthening Systems
ISIS EC Module 4
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.
7 - Installation of FRP Strengthening Systems
ISIS EC Module 4
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.
7 - Installation of FRP Strengthening Systems
ISIS EC Module 4
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.
7 - Installation of FRP Strengthening Systems
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 load-bearing 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
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
10 - Field Applications
ISIS EC Module 4
Country Hills Boulevard Bridge
- Calgary, AB- Deck strengthened in negative bending with CFRP strips- New wearing surface placed on top of FRP strips
10 - Field Applications
ISIS EC Module 4
A Canadian code exists for the design of FRP-strengthened concrete members
CAN/CSA-S806-02: Design and Construction of Building Components with Fibre Reinforced Polymers
(Currently under revision)
Design Guidance
CAN/CSA-S6-10: Canadian Highway Bridge Design Code
Additional Information
ISIS EC Module 4
Available from www.isiscanada.comISIS EC Module 1: Mechanics Examples Incorporating FRP MaterialsISIS EC Module 2: An Introduction to FRP Composites for ConstructionISIS EC Module 3: An Introduction to FRP-Reinforced ConcreteISIS EC Module 5: Introduction to Structural Health MonitoringISIS EC Module 6: Application & Handling of FRP Reinforcements for ConcreteISIS EC Module 7: Introduction to Life Cycle Engineering & Costing for Innovative InfrastructureISIS EC Module 8: Durability of FRP Composites for ConstructionISIS EC Module 9: Prestressing Concrete Structures with Fibre Reinforced Polymers
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