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1
An Introduction to FRP-Strengthening of Concrete Structures
ISIS Educational Module 4:
Produced by ISIS Canada
FRPRepair with
reinforcementModule Objectives
• To provide students with a general awareness of FRP materials and their potential uses
• To introduce students to the general philosophies and procedures for strengthening structures with FRPs
ISIS EC Module 4
FRPRepair with
reinforcementOverview
Introduction
FRP Materials
Evaluation of Existing Structures
Beam & One-Way Slab Strengthening
Column Strengthening
Specifications & Quality Control
Advanced Applications
Field Applications
Additional Info
ISIS EC Module 4
FRPRepair with
reinforcement Section: 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
2
FRPRepair with
reinforcement Section: 1 Introduction
• A primary factor leading to extensive degradation…
ISIS EC Module 4
Corrosion
Moisture, oxygen and chlorides penetrate
Concrete
Reinforcing Steel
Through concreteThrough cracks
Corrosion products formVolume expansion occursMore crackingCorrosion propagation
End result
FRPRepair with
reinforcement Section: 1 Introduction
• Why repair with the same materials?• Why repeat the cycle?
ISIS EC Module 4
FRP Materials
LightweightEasy to install
High Strength5x steel
Corrosion resistantDurable structures
Highly versatileSuit any project
FRPRepair with
reinforcement Section: 1FRP Materials
ISIS EC Module 4
Type Application SchematicFRP-Strengthening Applications
Fibre Dir.
Confinement Aroundcolumn Circumferential
Section
Shear Side face of beam (u-wrap)
Perpendicular to long. axis
of beam Section
Flexural side face of Tension and/or
beamaxis of beamAlong long.
Section
FRPRepair with
reinforcement Section: 2 FRP Materials
• Longstanding reputation in automotive and aerospace industries
• Over the past 15 years have FRP materials been increasingly considered for civil infrastructure applications
ISIS EC Module 4
FRP costs have decreasedNew, innovative solutions needed!
General
3
FRPRepair with
reinforcement Section: 2 FRP Materials
• Wide range of FRP products available:• Plates
• Rigid strips• Formed through pultrusion
• Sheets• Flexible fabric
ISIS EC Module 4
General
Carbon FRP sheet
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Constituents
• What is FRP?
FibresProvide strength and stiffness
Carbon, glass, aramid
MatrixProtects and transfers load
between fibres
Epoxy, polyester, vinyl ester
Fibre MatrixCompositeCreates a material with attributes superior to either component alone!
Strain [%]0.4-4.8 >10
34-130
1800-4900
Stre
ss [M
Pa]
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Properties
• Typical FRP stress-strain behaviour
FRP
Fibres
Matrix
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Installation Techniques
Wet lay-upUsed with flexible sheetsSaturate sheets with epoxy adhesivePlace on concrete surface
Epoxy
RollerResin acts as adhesive
AND matrix
4
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Installation Techniques
Pre-curedUsed with rigid, pre-cured stripsApply adhesive to strip backingPlace on concrete surfaceNot as flexible for variable structural shapes
Resin acts as adhesive AND matrix
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Properties
• FRP properties (versus steel):• Linear elastic behaviour
to failure• No yielding• Higher ultimate strength• Lower strain at failure Strain [%]
1 2 3
500
1000
1500
2000
2500
Stre
ss [M
Pa]
Steel
CFRPGFRP
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Properties
FRP material properties are a function of:
Type of fibre and matrix
Fibre volume content
Orientation of fibres
FRPRepair with
reinforcement Section: 2 FRP Materials
ISIS EC Module 4
Pro/Con
FRP advantages
FRP disadvantages
Will not corrodeHigh strength-to-weight ratioElectromagnetically inert
High initial material cost
But not when life-cycle costs are considered
5
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Deficiencies
• Deficiencies due to:
Environmental Effects
Freeze-Thaw
Chloride Ingress
Wet-Dry
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Deficiencies
• Deficiencies due to:
Updated Design Loads Updated design code procedures
Then Now
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Deficiencies
• Deficiencies due to:
Increase in Traffic Loads
Then Now
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Evaluation
• Evaluation is important to:
Determine concrete condition
Identify the cause of the deficiency
Establish the current load capacity
Evaluate the feasibility of FRP strengthening
6
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Evaluation
• Evaluation should include:
All past modifications
Actual size of elements
Actual material properties
Location, size and cause of cracks, spalling
Location, extent of corrosion
Quantity, location of rebar
FRPRepair with
reinforcement Section: 3 Evaluation of Existing Structures
ISIS EC Module 4
Concrete Surface
• One of the key aspects of strengthening: State of concrete substrate
• Concrete must transfer load from the elements to the FRPs through shear in the adhesive
• Surface modification required where surface flaws exist
FRPRepair with
reinforcement Section: 4 Beam/One-Way Slab Strengthening
ISIS EC Module 4
FRP ruptureFailure caused by:
Flexural StrengtheningAssumptions
Concrete crushingPlane sections remain planePerfect bond between steel/concrete, FRP/concrete
Adequate anchorage & development length provided for FRPsFRPs are linear elastic to failureConcrete compressive stress-strain curve is parabolic, no
strength in tensionInitial strains in FRPs can be ignored
FRPRepair with
reinforcement Section: 4 Beam/One-Way Slab Strengthening
ISIS EC Module 4
Resistance Factors
Material Bridge Building
Steel φS =0.90 φS =0.85
Concrete φC =0.75 φC =0.6
FRPφfrp = 0.75φfrp = 0. 50
CarbonGlass
7
FRPRepair with
reinforcement Section: 4 Beam/One-Way Slab Strengthening
ISIS EC Module 4
Failure Modes
Concrete crushing before steel yields
• Four potential failure modes:
Steel yielding followed by concrete crushingSteel yielding followed by FRP rupture
Debonding of FRP reinforcement
Assume failure mode Perform analysis Check failure modeDebonding is prevented through special end anchorages
*** Assume initial strains at the time of strengthening are zero ***
*** Refer to EC Module 4 Notes ***
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
• Force equilibrium in section:
Beam/One-Way Slab StrengtheningGeneral Design
b
d
Cross Section
As
Strain Distribution
εfrp
εc
h
bfrp
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Ts + Tfrp = Cc Eq. 4-1
Cc = φcα1f’cβ1bcTfrp = φfrpAfrpEfrpεfrpTs = φsAsfs
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
• Apply strain compatibility and use these equations to solve for neutral axis depth, c
Strain Distribution
εfrp
εc
Beam/One-Way Slab StrengtheningGeneral Design
h
bfrp
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
• Section capacity:Mr = Ts d − a
2Eq. 4-5+ Tfrp h − a
2
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step1: Assume failure modeAssume that section fails by concrete crushing after steel yields
Strain Distribution
εfrp
εcu
εs
c
Thus: εfrp = εcu Eq. 4-6
εc = εcu = 0.0035(h-c)/c
εs = εcu (d-c)/c Eq. 4-7
8
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 2: Determine compressive stress block factorsStrain Distribution
εfrp
εcu
εs
c
Eq. 4-8α1 = 0.85-0.0015f’c > 0.67Eq. 4-9β1 = 0.97-0.0025f’c > 0.67
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 3: Determine neutral axis depth, cStrain Distribution
εfrp
εcu
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Eq. 4-10φcα1f’cβ1bcφfrpAfrpEfrpεfrp =φsAsfs +
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 4: Check if assumed failure mode is correctStrain Distribution
εfrp
εcu
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
?>εfrp = εcu (h-c)/c εfrpu Eq. 4-11
If true, go to Step 6 If false, go to Step 5
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 5: Calculate factored moment resistanceStrain Distribution
εfrp
εcu
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Mr =φsAsfy d − a2
Eq. 4-12+ h − a2
φfrpAfrpEfrpεfrp
9
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 5: Calculate factored moment resistanceStrain Distribution
εfrp
εcu
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Check if internal steel yields to ensure adequate deformability
εs = εcu (d-c)/c > εy?If yes, OK
If no, reduce FRP amount & recalculate
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrp
Step 6: Assume different failure modeStrain Distribution
εfrpu
εc
εs
c
Stress Distribution
fsffrpu
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Assume failure occurs by tensile failure of FRP
Thus:εfrp = εfrpu
εc < εcu
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrpStrain Distribution
εfrpu
εc
εs
c
Stress Distribution
fsffrpu
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Eq. 4-15φcα1f’cβ1bcφfrpAfrpEfrpεfrpu =φsAsfy +
Step 7: Determine depth of neutral axis
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrpStrain Distribution
εfrpu
εc
εs
c
Stress Distribution
fsffrpu
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Step 8: Check if assumed failure mode is correct
εc < εcu
εfrpu c / (h-c) < εcu
10
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningAnalysis Procedure
h
bfrpStrain Distribution
εfrpu
εc
εs
c
Stress Distribution
fsffrpu
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
Step 9: Calculate factored moment resistance
Mr =φsAsfy d − a2
Eq. 4-17+ h − a2
φfrpAfrpEfrpεfrpu
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
b
d
Cross Section
As
Beam/One-Way Slab StrengtheningWith Compression Steel
h
bfrpStrain Distribution
εfrp
εs
c
Stress Distribution
fsffrp
Equiv. Stress Distribution
a = β1c
α1Φcf’c
TsTfrp
Cc
• Similar analysis procedure
A’s
εcu
ε’s f’s Cs
Add a compressive stress resultant
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Beam/One-Way Slab StrengtheningTee Beams
• Similar analysis procedureNeutral axis in flange: treat as rectangular sectionNeutral axis in web: treat as tee section
bf
hf
h
bfrp
Afrp
c
Mr Mrw
= +
Mrf
FRPRepair with
reinforcement Section: 4
Flexural Example
ISIS EC Module 4
Problem statementCalculate the moment resistance (Mr) for an FRP-strengthened rectangular concrete section
Section information
Beam/One-Way Slab Strengthening
f’c = 45 MPa εfrpu = 1.55 %
Afrp = 60 mm2
fy = 400 MPaEs = 200 GPa
Efrp = 155 GPa
b = 105 mm
h = 35
0 mm
3-10M bars
d = 32
5 mm
CFRP
11
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 1: Assumed failure mode
Flexural Example
Beam/One-Way Slab Strengthening
Assume failure of beam due to crushing of concrete in compression after yielding of internal steel reinforcement
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 2: Calculate concrete stress block factors
Flexural Example
Beam/One-Way Slab Strengthening
α1 = 0.85 – 0.0015 f’c > 0.67α1 = 0.85 – 0.0015 (45) = 0.78
β1 = 0.85 – 0.0025 f’c > 0.67β1 = 0.85 – 0.0025 (45) = 0.86
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 3: Find depth of neutral axis, c
Flexural Example
Beam/One-Way Slab Strengthening
Use Equation 4-10:
φcα1f’cβ1bc = φfrpAfrpEfrpεfrpφsAsfs +
0.6 (0.78) (45) (0.86) (105) c 0.85 (300) (400)350 - c
0.75 (60) (155000) 0.0035c
c = 90.5 mm
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 4: Check failure mode
Flexural Example
Beam/One-Way Slab Strengthening
Therefore, FRP rupture does NOT occur and assumed failure mode is correct
εfrp = 0.0035 350 - 90.590.5
εfrp = 0.01 < εfrpu = 0.0155
εfrp = εcu (h-c)/c εfrpu = 0.0155 Eq. 4-11vs.
12
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 4: Check failure mode
Flexural Example
Beam/One-Way Slab Strengthening
εs = εcud - c
c
To promote ductility, check that steel has yielded:
εs = 0.0035 325 - 90.590.5 > 0.002 = εy
If the steel had NOT yielded, the beam failure could be expected to be less ductile, and we would need to carefully check the
deformability of the member
= 0.009
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 5: Calculate moment resistance
Flexural Example
Beam/One-Way Slab Strengthening
Mr =φsAsfy d − a2
Eq. 4-12+ h − a2
φfrpAfrpEfrpεfrp
0.85 (300) (400) 325 - 0.86 x 90.5 2
0.75 (60) (155000) (0.01) 350 - 0.86 x 90.5 2
Mr = 50.9 × 106 N· mm = 50.9 kN· m 65% increase over unstrengthened beam!
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
May be aligned at any angle to the longitudinal axis
Assumptions• FRP sheets can be applied to provide shear resistance• Many different possible configurations
May be applied in continuous sheets or in finite widths
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Assumptions
• FRP sheets can be applied to provide shear resistance• Many different possible configurations
May be applied on sides only or as U-wraps
Section
Section
*U-wraps also improve the anchorage of flexural FRP external reinforcement
ne = 1
ne = 2
13
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Assumptions
Section
To avoid stress concentrations,
allow for a minimum radius
of 15 mm
wfrp
sfrp
β
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
External strengthening with FRPs:
Flexural failure Generally fairly ductile
Shear failure Sudden and brittle
Undesirable failure modeControl shear deformation
to avoid sudden failure
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Shear resistance of a beam:
Vr = Vc Vs Vfrp+ + Eq. 4-18
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Shear resistance of a beam:
Vs =φs fy Av d
sEq. 4-20
Vc = 0.2 φc√f’c bwd Eq. 4-19
14
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Shear resistance of a beam:
Vfrp =φfrp Afrp Efrp εfrpe dfrp (sinβ + cosβ)
sfrpEq. 4-21
Afrp = 2 tfrp wfrp
dfrp: distance from free end of FRP to bottom of internal steel stirrups
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Eq. 4-23εfrpe = R εfrpu ≤ 0.004
Prevents shear cracks from widening beyond acceptable limitsEnsures aggregate interlock!
Effective strain in FRP, εfrpe:
Reduction factor, R:
0.8 Carbon: λ1 = 1.35, λ2 = 0.30Glass: λ1 = 1.23, λ2 = 0.47
Eq. 4-24R = αλ1f’c2/3
ρfrp Efrp
λ2
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
FRP shear reinforcement ratio, ρfrp:
Eq. 4-25ρfrp = 2 tfrpbw
wfrpsfrp
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Another limit on effective strain in FRP, εfrpe:
Eq. 4-26εfrpe ≤αk1k2Le
95250.8
Parameters, k1 and k2:
Eq. 4-27k1 = f’c
27.65
2/3
Eq. 4-28k2 = dfrp- ne Le
dfrp
15
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Effective anchorage length, Le:
Eq. 4-29Le = 25350tfrpEfrp
0.58
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Limit on spacing of strips, sfrp:
Eq. 4-30sfrp ≤ wfrp + d4
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Design Principles
Limit on maximum allowable shear strengthening, Vfrp:
Shear contribution due to steel stirrups and FRP
strengthening must be less than this term
Eq. 4-31Vr ≤ Vc + 0.8λφc√f’c bwd
FRPRepair with
reinforcement Section: 4
Shear Strengthening
ISIS EC Module 4
Beam/One-Way Slab Strengthening
Example
Problem statementCalculate the shear capacity (Vr) for an FRP-strengthened concrete section
Section information
Sectionb = 105 mm
h = 35
0 mm
3-10M bars
d = 32
5 mm
4.76 mm Ø
GFRP wrap
Section Elevation
λ = 1.0f’c = 45 MPaεfrpu = 2.0 %
fy = 400 MPa (rebar)
Efrp = 22.7 GPa
fy = 400 MPa (stirrup)
ss = 225 mm c/c
tfrp = 1.3 mmwfrp = 100 mmsfrp = 200 mm
16
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 1: Calculate concrete and steel contributions
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Concrete:
Steel: Vs =φs fy Av d
s = 0.85 (400) (36) (325)225
Vs = 17680 N = 17.68 kN
Vc = 0.2 φc√f’c bwdVc = 0.2 (0.6) √45 (105) (325)Vc = 27470 N = 27.47 kN
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 2: Determine Afrp, ρfrp, Le for effective strain calculation
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Afrp: Afrp = 2 tfrp wfrp = 2 (1.3) (100)Afrp = 260 mm2
ρfrp: ρfrp = 2 tfrp
bw
wfrp
sfrp=
2 (1.3)105
100200
ρfrp = 0.0124
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 2: Determine Afrp, ρfrp, Le for effective strain calculation
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Le: Le = 25350tfrpEfrp
0.58 = 253501.3 x 22700 0.58
Le = 64.8 mm
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
SolutionStep 3: Determine k1, k2 and effective strain, εfrpe [Limit 2]
Beam/One-Way Slab StrengtheningShear Strengthening
Example
k1: k1 = f’c
27.65
2/3
= 45
27.65
2/3
= 1.38
k2: k2 = dfrp- ne Le
dfrp=
325 – 1 (64.8)325 = 0.80
Because of u-wrap
17
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
εfrpe:
εfrpe =0.8 (1.38) (0.80) (64.8)
9525εfrpe = 0.0060
εfrpe ≤αk1k2Le
9525Eq. 4-26
Note: This strain is one of three limits placed on the FRPStep 3: Determine k1, k2 and effective strain, εfrpe [Limit 2]
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
R:
Step 4: Determine R and effective strain, εfrpe [Limit 1]
R = 0.229
R = αλ1f’c2/3
ρfrp Efrp
λ2
R = 0.8 (1.23)45 2/3
0.0124 (22700)
0.47
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
εfrpe:
Step 4: Determine R and effective strain, εfrpe [Limit 1]Note: This strain is one of three limits placed on the FRP
Eq. 4-23εfrpe = R εfrpu ≤ 0.004
εfrpe = 0.229 (0.02)
εfrpe = 0.0046
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Step 5: Determine governing effective strain, εfrpe
For design purposes, use the smallest limiting value of:
εfrpe = 0.0060 Eq. 4-26
εfrpe = 0.0040 Eq. 4-23
εfrpe = 0.0046 Eq. 4-23
18
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Step 6: Calculate contribution of FRP to shear capacity
Vfrp: Vfrp =φfrp Afrp Efrp εfrpe dfrp (sinβ + cosβ)
sfrpEq. 4-21
Vfrp = 0.5 (260) (22700) (0.004) (325) (sin90 + cos90)200
Vfrp = 19200 N = 19.2 kN
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Step 7: Compute total shear resistance of beam
Vr: Vr = Vc Vs Vfrp+ + Eq. 4-21
Vr = 27.5 + 17.7 + 19.2
Vr = 64.4 kN
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Step 8: Check maximum shear strengthening limits
Eq. 4-31Vr ≤ Vc + 0.8λφcf’cbwd64400 ≤ 27500 + 0.8 (1) (0.6) (45) (105) (325)
64400 ≤ 137400
OK
FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Solution
Beam/One-Way Slab StrengtheningShear Strengthening
Example
Step 9: Check maximum band spacing
dEq. 4-30sfrp ≤ wfrp + 4
200 ≤ 100 + 3254
200 ≤ 181
Not true, therefore use 180 mm spacing
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FRPRepair with
reinforcement Section: 4
ISIS EC Module 4
Add’l Considerations
FRP anchorage and development length
Beam/One-Way Slab Strengthening
Additional factors to consider:
Deflections
Crack widths
VibrationsCreep
FatigueDuctility
Creep-rupture stress limits sometimes govern FRP-strengthened design
External strengthening with FRPs may reduce flexural deformability
Deflection
Load
No FRP
1-layer FRP
3-layers FRP
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Overview
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
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Overview
Column Strengthening
• The result:
Increased load capacity
Increased deformation capability
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Overview
Column Strengthening
• Design equations are largely empirical (from tests)• ISIS equations are applicable for the following cases:
Undamaged concrete column
Short column subjected to concentric axial load
Fibres oriented circumferentially
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FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Slenderness Limits• Strengthening equations only valid for non-
slender columns. Thus, from CSA A23.3:
Ag = gross cross-sectional area of columnf’c = concrete strengthPf = factored axial loadlu = unsupported lengthDg = column diameter
lu
Dg≤ Eq. 5-1
6.25Pf / f’cAg
0.5
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Slenderness Limits• Strengthening equations only valid for non-
slender columns. Thus, from CSA A23.3:
lu
Dg≤ Eq. 5-1
6.25Pf / f’cAg
0.5
The axial load capacity is increased by the confining effect of the wrap
Ensure that column remains short
Column may become slender!
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Confinement• Based on equilibrium, the lateral confinement
pressure exerted by the FRP, flfrp:
flfrp = Eq. 5-22 Nb φfrp ffrpu tfrp
Dg
Nb = number of FRP layersφfrp = material resistance factor for FRPffrpu = ultimate FRP strengthtfrp = FRP thickness
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Confinement
• The benefit of a confining pressure is to increase the confined compressive concrete strength, f’cc
f’cc = f’c + k1 flfrp Eq. 5-3
f’c = ultimate strength of unconfined concretek1 = empirical coefficient from tests
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FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Confinement• ISIS design guidelines suggest a
modification to f’cc:
f’cc = f’c + k1 flfrp = f’c (1 + αpcωw) Eq. 5-4
αpc = performance coefficient depending on:FRP type
f’cmember size(currently taken as 1.0)
=ωw =2 flfrpφc f’c
Eq. 5-5
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Confinement Limits
Minimum confinement pressure
Maximum confinement pressure
Why?To ensure adequate ductility of column
Limitflfrp ≥ 4 MPa
To prevent excessive deformations of column
LimitWhy?
= 0.85 (Strength reduction factor to account for unexpected eccentricities)
flfrp ≤f’c
2 αpc
1ke
- φc
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Circular Columns
Column Strengthening
Axial Load Resistance
• Factored axial load resistance for an FRP-confined reinforced concrete column, Prmax:
Prmax = ke [α1φcf’cc (Ag-As) + φs fy As] Eq. 5-9
Same equation as for conventionally RC column, except includes confinedconcrete strength, f’cc
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Rectangular Columns
Column Strengthening
• External FRP wrapping may be used with rectangular columns• There is far less experimental data available for
rectangular columns• Strengthening is not nearly as effective
Confinement all around Confinement only in some areas
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FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Add’l Considerations
Column Strengthening
• External FRP wrapping may be used with circular and rectangular RC columns to strengthen also for shear
• Particularly useful in seismic upgrade situations where increased lateral loads are a concern
Shear
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
Add’l Considerations
Column Strengthening
• The confining effects of FRP wraps are not activateduntil significant radial expansion of concrete occurs
• Therefore, ensure service loads kept low enough to prevent failure by creep and fatigue
Strengthening Limits
FRPRepair with
reinforcement Section: 5
Example
ISIS EC Module 4
Problem statementDetermine the FRP wrap details for an RC column as described below
InformationRC column factored axial resistance (pre-strengthening) = 3110 kN
Column Strengthening
New axial live load requirement PL = 1550 kNNew axial dead load requirement PD = 1200 kN
New factored axial load, Pf = 4200 kN
lu = 3000 mmDg = 500 mmAg = 196350 mm2
Ast = 2500 mm2
fy = 400 MPa
f’c = 30 MPaffrpu = 1200 MPatfrp = 0.3 mmffrp = 0.75
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 1: Check if column remains short after strengthening
Column StrengtheningExample
Eq. 5-1lu
Dg≤ 6.25
Pf / f’cAg0.5
OK
3000500
≤ 6.254200000/(30 x 196350) 0.5
6 ≤ 7.4
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FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 2: Compute required confined concrete strength, f’cc
Column StrengtheningExample
Prmax = ke [α1φcf’cc (Ag-As) + φs fy As] Eq. 5-9
Take equation 5-9 and rearrange for f’cc:
f’cc =
Pf
ke− φs fy As
α1φc (Ag-As)
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 2: Compute required confined concrete strength, f’cc
Column StrengtheningExample
f’cc =
4200000 0.85 − 0.85 (400) (2500)
0.81 (0.6) (196350-2500)
α1 = 0.85 – 0.0015f’c = 0.85 – 0.0015 (30) = 0.81α1:
f’cc:
f’cc = 43.4 MPa
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 3: Compute volumetric strength ratio, ωw
Column StrengtheningExample
ωw:
f’cc = f’c + k1 flfrp = f’c (1 + αpcωw) Eq. 5-4
Take equation 5-4 and rearrange for ωw:
ωw =
f’ccf’c
- 1
αpc=
43.430
- 1
1
ωw = 0.447
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 4: Compute required confinement pressure, flfrp
Column StrengtheningExample
flfrp:
Take equation 5-5 and rearrange for flfrp:
ωw =ρfrp φfrp ffrpu
φc f’c=
2 flfrpφc f’c
Eq. 5-5
flfrp = 2ωw φc f’c = 2
0.447 (0.6) (30)
flfrp = 4.02 MPa
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FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 4: Compute required confinement pressure, flfrp
Column StrengtheningExample
Check flfrp again confinement limits:
flfrp = 4.02 > 4.0 Minimum:
flfrp = 4.02 < Maximum:f’c
2 αpc
1ke
- φc
flfrp = 4.02 <302 (1)
10.85
- 0.6 = 8.65
OK, limits met
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 5: Compute required number of FRP layers
Column StrengtheningExample
Take Equation 5-2 and rearrange for Nb:
flfrp = Eq. 5-22 Nb φfrp ffrpu tfrp
Dg
Nb: Nb =flfrp Dg
2 φfrp ffrpu tfrp=
4.02 (500)2 (0.75) (1200) (0.3)
Nb = 3.72 Use 4 layers
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 6: Compute factored axial strength of FRP-wrapped column
Column StrengtheningExample
Use Equations 5-2, 5-5, 5-4 and 5-9:
flfrp: flfrp =2 Nb φfrp ffrpu tfrp
Dg= 4.32 MPa
ωw : ωw = =2 flfrpφc f’c
0.48
FRPRepair with
reinforcement Section: 5
ISIS EC Module 4
SolutionStep 6: Compute factored axial strength of FRP-wrapped column
Column StrengtheningExample
Use Equations 5-2, 5-5, 5-4 and 5-9:
f’cc:
Prmax:
f’cc = f’c (1 + αpcωw) = 44.4 MPa
Prmax = ke [α1φcf’cc (Ag-As) + φs fy As]Prmax = 4230 kN > Pf = 4200 kN
Note: Additional checks should be performed for creep and fatigue
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FRPRepair with
reinforcement Section: 6
ISIS EC Module 4
Specifications & Quality Control
• Strengthening of structures with FRP is a relatively simple technique
• However, it is essential to performance to installthe FRP system properly
Specifications
Quality Control / Quality Assurance
FRPRepair with
reinforcement Section: 6
ISIS EC Module 4
Specifications & Quality ControlSpecifications
Approval of FRP materials
Handling and storage of FRP materials
Staff and contractor qualifications
Concrete surface preparation
Installation of FRP systems
Adequate conditions for FRP cure
Protection and finishing for FRP system
FRPRepair with
reinforcement Section: 6
ISIS EC Module 4
Specifications & Quality Control
Quality Control and Quality Assurance
Material qualification and acceptance
Qualification of contractor personnel
Inspection of concrete substrate
FRP material inspection
Testing to ensure as-built condition
FRPRepair with
reinforcement Section: 7
ISIS EC Module 4
Additional ApplicationsPrestressed FRP Sheets
• One way to improve FRP effectiveness is to applyprestress 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 delaythe formation of new cracks
• Prestressing FRP sheets is a promising technique, but is still in initial stages of development
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FRPRepair with
reinforcement Section: 7
ISIS EC Module 4
Additional ApplicationsNSM Techniques
• Newer class of FRP strengthening techniques: near surface mounting reinforcement (NSMR)
Unstrengthened concrete T-beam
Longitudinal grooves cut into soffit
FRP strips placed in grooves
Grooves filled with epoxy grout
• Research indicates NSMR is effective and efficient for strengthening
FRPRepair with
reinforcement Section: 8
ISIS EC Module 4
Field Applications
Maryland Bridge
Winnipeg, Manitoba
Constructed in 1969
Twin five-span continuous precast prestressed girders
CFRP sheets to upgrade shear capacity
FRPRepair with
reinforcement Section: 8
ISIS EC Module 4
Field Applications
John Hart Bridge
Prince George, BC
64 girder ends were shearstrengthened with CFRP
Increase in shear capacity of 15-20%
Upgrade completed in 6 weeks
Locations for FRP shear reinforcement
FRPRepair with
reinforcement Section: 8
ISIS EC Module 4
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
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FRPRepair with
reinforcement Section: 8
ISIS EC Module 4
Field Applications
St. Émélie Bridge
Sainte-Émélie-de-l'Énergie, Quebec
Single-span, simply supported tee-section bridgeStrengthened for flexure and
shearSite preparation: 3 weeks,
FRP installation: 5 days
FRPRepair with
reinforcement Section: 9 Design Guidance
ISIS EC Module 4
Canadian codes exist for the design of FRP-reinforced concrete members
CAN/CSA-S806-02: Design and Construction of Building Components with Fibre Reinforced Polymers
CAN/CSA-S6-00: The Canadian Highway Bridge Design Code (CHBDC)
Additional Information
ISIS EC Module 4
ISIS Design Manual No. 3: Reinforcing Concrete Structures with Fiber Reinforced Polymers
ISIS EC Module 1: An Introduction to FRP Composites for Construction
ISIS Design Manual No. 4: Strengthening Reinforced Concrete Structures with Externally-Bonded Fiber Reinforced Polymer
ISIS EC Module 4: An Introduction to FRP-Strengthening of Reinforced Concrete Structures
ISIS EC Module 3: An introduction to FRP-Reinforced Concrete Structures
FRPRepair with
reinforcement Section: 9
Available from www.isiscanada.com