An Introduction to FRP- Strengthening of Concrete Structures ISIS Educational Module 4: Produced by ISIS Canada.

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

FRPRepair with


• 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


• Why repair with the same materials?• Why repeat the cycle?

ISIS EC Module 4

FRP Materials

Lightweight

Easy to install

High Strength

5x steel

Corrosion resistant

Durable 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

FRPRepair with


• 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


ISIS EC Module 4

Constituents

• What is FRP?

Fibres

Provide strength and stiffness

Carbon, glass, aramid

Matrix

Protects and transfers load between fibres

Epoxy, polyester, vinyl ester

Fibre MatrixComposite

Creates 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


ISIS EC Module 4

Properties

• Typical FRP stress-strain behaviour

FRP

Fibres

Matrix

FRPRepair with


ISIS EC Module 4

Installation Techniques

Wet lay-up

Used with flexible sheetsSaturate sheets with epoxy adhesivePlace on concrete surface

Epoxy

Roller Resin acts as adhesive

AND matrix

FRPRepair with


ISIS EC Module 4

Installation Techniques

Pre-cured

Used 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


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

CFRP

GFRP

FRPRepair with


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


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

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


ISIS EC Module 4

Deficiencies


Updated Design LoadsUpdated design code procedures

Then Now

FRPRepair with


ISIS EC Module 4

Deficiencies


Increase in Traffic Loads

Then Now

FRPRepair with


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

FRPRepair with


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


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 plane

Perfect bond between steel/concrete, FRP/concrete

Adequate anchorage & development length provided for FRPs

FRPs are linear elastic to failure

Concrete compressive stress-strain curve is parabolic, no strength in tensionInitial strains in FRPs can be ignored

FRPRepair with


ISIS EC Module 4

Resistance Factors

Material Bridge Building

Steel S =0.90 S =0.85

Concrete C =0.75 C =0.6

FRPfrp = 0.75

frp = 0. 50

Carbon

Glass

FRPRepair with


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 mode

Debonding 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

frp

c

h

bfrp

s

c

Stress Distribution

fs

ffrp

Equiv. Stress Distribution

a =1c

1Φcf’c

Ts

Tfrp

Cc

Ts + Tfrp = Cc Eq. 4-1

Cc = c1f’c1bcTfrp = frpAfrpEfrpfrpTs = sAsfs

FRPRepair with


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

frp

c

Beam/One-Way Slab StrengtheningGeneral Design

h

bfrp

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

• Section capacity:Mr = Ts d a

2Eq. 4-5

+ Tfrp

h a2

FRPRepair with


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

frp

cu

s

c

Thus: frp = cu Eq. 4-6

c = cu = 0.0035(h-c)/c

s = cu (d-c)/c Eq. 4-7

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 2: Determine compressive stress block factorsStrain Distribution

frp

cu

s

c

Eq. 4-81 = 0.85-0.0015f’c > 0.67Eq. 4-91 = 0.97-0.0025f’c > 0.67

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 3: Determine neutral axis depth, cStrain Distribution

frp

cu

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

Eq. 4-10c1f’c1bcfrpAfrpEfrpfrp = sAsfs +

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 4: Check if assumed failure mode is correctStrain Distribution

frp

cu

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

?frp = cu (h-c)/c frpu Eq. 4-11

If true, go to Step 6 If false, go to Step 5

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 5: Calculate factored moment resistanceStrain Distribution

frp

cu

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

Mr =sAsfy d a2

Eq. 4-12+ h a2

frpAfrpEfrpfrp

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 5: Calculate factored moment resistanceStrain Distribution

frp

cu

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

Check if internal steel yields to ensure adequate deformability

s = cu (d-c)/c > εy?If yes, OK

If no, reduce FRP amount & recalculate

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Step 6: Assume different failure modeStrain Distribution

frpu

c

s

c

Stress Distribution

fs

ffrpu


a =1c

1Φcf’c

Ts

Tfrp

Cc

Assume failure occurs by tensile failure of FRP

Thus:frp = frpu

c < cu

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Strain Distribution

frpu

c

s

c

Stress Distribution

fs

ffrpu


a =1c

1Φcf’c

Ts

Tfrp

Cc

Eq. 4-15c1f’c1bcfrpAfrpEfrpfrpu = sAsfy +

Step 7: Determine depth of neutral axis

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Strain Distribution

frpu

c

s

c

Stress Distribution

fs

ffrpu


a =1c

1Φcf’c

Ts

Tfrp

Cc

Step 8: Check if assumed failure mode is correct

c < cu

frpu c / (h-c) < cu

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As


h

bfrp

Strain Distribution

frpu

c

s

c

Stress Distribution

fs

ffrpu


a =1c

1Φcf’c

Ts

Tfrp

Cc

Step 9: Calculate factored moment resistance

Mr =sAsfy d a2

Eq. 4-17+ h a2

frpAfrpEfrpfrpu

FRPRepair with


ISIS EC Module 4

b

d

Cross Section

As

Beam/One-Way Slab StrengtheningWith Compression Steel

h

bfrp

Strain Distribution

frp

s

c

Stress Distribution

fs

ffrp


a =1c

1Φcf’c

Ts

Tfrp

Cc

• Similar analysis procedure

A’s

cu

’s f’s Cs

Add a compressive stress resultant

FRPRepair with


ISIS EC Module 4

Beam/One-Way Slab StrengtheningTee Beams

• Similar analysis procedureNeutral axis in flange: treat as rectangular section

Neutral axis in web: treat as tee section

bf

hf

h

bfrp

Afrp

c

Mr Mrw

= +

Mrf

FRPRepair with


Flexural Example

ISIS EC Module 4

Problem statement

Calculate the moment resistance (Mr) for an FRP-strengthened rectangular concrete section

Section information

Beam/One-Way Slab Strengthening

f’c = 45 MPa frpu = 1.55 %

Afrp = 60 mm2

fy = 400 MPa

Es = 200 GPa

Efrp = 155 GPa

b = 105 mm

h =

350

mm

3-10M bars

d =

325

mm

CFRP

FRPRepair with


ISIS EC Module 4

SolutionStep 1: Assumed failure mode

Flexural Example


Assume failure of beam due to crushing of concrete in compression after yielding of internal steel reinforcement

FRPRepair with


ISIS EC Module 4

SolutionStep 2: Calculate concrete stress block factors

Flexural Example


1 = 0.85 – 0.0015 f’c > 0.67

1 = 0.85 – 0.0015 (45) = 0.78

1 = 0.85 – 0.0025 f’c > 0.67

1 = 0.85 – 0.0025 (45) = 0.86

FRPRepair with


ISIS EC Module 4

SolutionStep 3: Find depth of neutral axis, c

Flexural Example


Use Equation 4-10:

c1f’c1bc = frpAfrpEfrpfrpsAsfs +

0.6 (0.78) (45) (0.86) (105) c 0.85 (300) (400)

350 - c0.75 (60) (155000) 0.0035

c

c = 90.5 mm

FRPRepair with


ISIS EC Module 4

SolutionStep 4: Check failure mode

Flexural Example


Therefore, FRP rupture does NOT occur and assumed failure mode is correct

frp = 0.0035 350 - 90.590.5

frp = 0.01 < frpu = 0.0155

frp = cu (h-c)/c frpu = 0.0155 Eq. 4-11vs.

FRPRepair with


ISIS EC Module 4

SolutionStep 4: Check failure mode

Flexural Example


s = cud - c

c

To promote ductility, check that steel has yielded:

s = 0.0035 325 - 90.590.5

> 0.002 =y

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


ISIS EC Module 4

SolutionStep 5: Calculate moment resistance

Flexural Example


Mr =sAsfy d a2

Eq. 4-12+ h a2

frpAfrpEfrpfrp

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


Shear Strengthening

ISIS EC Module 4


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


Shear Strengthening

ISIS EC Module 4


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

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Assumptions

Section

To avoid stress concentrations,

allow for a minimum radius

of 15 mm

wfrp

sfrp

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

External strengthening with FRPs:

Flexural failure Generally fairly ductile

Shear failure Sudden and brittle

Undesirable failure mode

Control shear deformation to avoid sudden failure

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Shear resistance of a beam:

Vr = Vc Vs Vfrp+ + Eq. 4-18

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles


Vs =s fy Av d

sEq. 4-20

Vc = 0.2 c√f’c bwd Eq. 4-19

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles


Vfrp =frp Afrp Efrp frpe dfrp (sin + cos)

sfrp

Eq. 4-21

Afrp = 2 tfrp wfrp

dfrp: distance from free end of FRP to bottom of internal steel stirrups

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Eq. 4-23frpe = R frpu ≤ 0.004

Prevents shear cracks from widening beyond acceptable limitsEnsures aggregate interlock!

Effective strain in FRP, frpe:

Reduction factor, R:

0.8 Carbon: 1 = 1.35, 2 = 0.30Glass: 1 = 1.23, 2 = 0.47

Eq. 4-24R1

f’c2/3

frp Efrp

2

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

FRP shear reinforcement ratio, frp:

Eq. 4-25frp = 2 tfrp

bw

wfrp

sfrp

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Another limit on effective strain in FRP, frpe:

Eq. 4-26frpe ≤k1k2L

e 9525

0.8

Parameters, k1 and k2:

Eq. 4-27k1 = f’c

27.65

2/3

Eq. 4-28k2 = dfrp- ne Le

dfrp

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Effective anchorage length, Le:

Eq. 4-29Le = 25350tfrpEfrp

0.58

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Limit on spacing of strips, sfrp:

Eq. 4-30sfrp ≤ wfrp + d4

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Design Principles

Limit on maximum allowable shear strengthening, V frp:

Shear contribution due to steel stirrups and FRP strengthening

must be less than this term

Eq. 4-31Vr ≤ Vc + 0.8λc√f’c bwd

FRPRepair with


Shear Strengthening

ISIS EC Module 4


Example

Problem statement

Calculate the shear capacity (Vr) for an FRP-strengthened concrete section

Section information

Sectionb = 105 mm

h =

350

mm

3-10M bars

d =

325

mm

4.76 mm Ø

GFRP wrap

Section Elevation

λ = 1.0f’c = 45 MPafrpu = 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

FRPRepair with


ISIS EC Module 4

SolutionStep 1: Calculate concrete and steel contributions

Beam/One-Way Slab StrengtheningShear Strengthening

Example

Concrete:

Steel: Vs =s fy Av d

s=

0.85 (400) (36) (325)225

Vs = 17680 N = 17.68 kN

Vc = 0.2 c√f’c bwd

Vc = 0.2 (0.6) √45 (105) (325)

Vc = 27470 N = 27.47 kN

FRPRepair with


ISIS EC Module 4

SolutionStep 2: Determine Afrp, frp, Le for effective strain calculation


Example

Afrp: Afrp = 2 tfrp wfrp = 2 (1.3) (100)Afrp = 260 mm2

frp: frp = 2 tfrp

bw

wfrp

sfrp

= 2 (1.3)

105

100

200

frp = 0.0124

FRPRepair with


ISIS EC Module 4

SolutionStep 2: Determine Afrp, frp, Le for effective strain calculation


Example

Le: Le = 25350tfrpEfrp

0.58=

253501.3 x 22700 0.58

Le = 64.8 mm

FRPRepair with


ISIS EC Module 4

SolutionStep 3: Determine k1, k2 and effective strain, frpe [Limit 2]


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

FRPRepair with


ISIS EC Module 4

Solution


Example

frpe:

frpe = 0.8 (1.38) (0.80) (64.8)

9525

frpe = 0.0060

frpe ≤k1k2

Le 9525Eq. 4-26

Note: This strain is one of three limits placed on the FRP

Step 3: Determine k1, k2 and effective strain, frpe [Limit 2]

FRPRepair with


ISIS EC Module 4

Solution


Example

R:

Step 4: Determine R and effective strain, frpe [Limit 1]

R = 0.229

R1

f’c2/3

frp Efrp

2

R = 0.8 (1.23)45 2/3

0.0124 (22700)

0.47

FRPRepair with


ISIS EC Module 4

Solution


Example

frpe:

Step 4: Determine R and effective strain, frpe [Limit 1]

Note: This strain is one of three limits placed on the FRP

Eq. 4-23frpe = R frpu ≤ 0.004

frpe = 0.229 (0.02)

frpe = 0.0046

FRPRepair with


ISIS EC Module 4

Solution


Example

Step 5: Determine governing effective strain, frpe

For design purposes, use the smallest limiting value of:

frpe = 0.0060 Eq. 4-26

frpe = 0.0040 Eq. 4-23

frpe = 0.0046 Eq. 4-23

FRPRepair with


ISIS EC Module 4

Solution


Example

Step 6: Calculate contribution of FRP to shear capacity

Vfrp: Vfrp =frp Afrp Efrp frpe dfrp (sin + cos)

sfrp

Eq. 4-21

Vfrp =0.5 (260) (22700) (0.004) (325) (sin90 + cos90)

200

Vfrp = 19200 N = 19.2 kN

FRPRepair with


ISIS EC Module 4

Solution


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


ISIS EC Module 4

Solution


Example

Step 8: Check maximum shear strengthening limits

Eq. 4-31Vr ≤ Vc + 0.8λcf’cbwd

64400 ≤ 27500 + 0.8 (1) (0.6) (45) (105) (325)

64400 ≤ 137400

OK

FRPRepair with


ISIS EC Module 4

Solution


Example

Step 9: Check maximum band spacing

dEq. 4-30sfrp ≤ wfrp +

4

200 ≤ 100 + 3254

200 ≤ 181

Not true, therefore use 180 mm spacing

FRPRepair with


ISIS EC Module 4

Add’l Considerations

FRP anchorage and development length


Additional factors to consider:

Deflections

Crack widths

Vibrations

Creep

Fatigue

Ductility

Creep-rupture stress limits sometimes govern FRP-strengthened design

External strengthening with FRPs may reduce flexural deformability

DeflectionLo

ad

No FRP

1-layer FRP

3-layers FRP

FRPRepair with


ISIS EC Module 4

Overview


• 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


ISIS EC Module 4

Overview


• The result:

Increased load capacity

Increased deformation capability

FRPRepair with


ISIS EC Module 4

Overview


• 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

FRPRepair with


ISIS EC Module 4

Circular Columns


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. 5-16.25

Pf / f’cAg0.5

FRPRepair with


ISIS EC Module 4

Circular Columns


Slenderness Limits• Strengthening equations only valid for non-

slender columns. Thus, from CSA A23.3:

lu

Dg

≤ Eq. 5-16.25

Pf / f’cAg0.5

The axial load capacity is increased by the confining effect of the wrap

Ensure that column remains short

Column may become slender!

FRPRepair with


ISIS EC Module 4

Circular Columns


Confinement• Based on equilibrium, the lateral confinement

pressure exerted by the FRP, flfrp:

flfrp = Eq. 5-22 Nb frp ffrpu tfrp

Dg

Nb = number of FRP layersfrp = material resistance factor for FRPffrpu = ultimate FRP strengthtfrp = FRP thickness

FRPRepair with


ISIS EC Module 4

Circular Columns


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

FRPRepair with


ISIS EC Module 4

Circular Columns


Confinement• ISIS design guidelines suggest a

modification to f’cc:

f’cc = f’c + k1 flfrp = f’c (1 + pcw) Eq. 5-4

pc = performance coefficient depending on:

FRP type f’c

member size(currently taken as 1.0)

=w =2 flfrpcf’c

Eq. 5-5

FRPRepair with


ISIS EC Module 4

Circular Columns


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 pc

1

ke

- c

FRPRepair with


ISIS EC Module 4

Circular Columns


Axial Load Resistance

• Factored axial load resistance for an FRP-confined reinforced concrete column, Prmax:

Prmax = ke [1cf’cc (Ag-As) + s fy As] Eq. 5-9

Same equation as for conventionally RC column, except includes confined concrete strength, f’cc

FRPRepair with


ISIS EC Module 4

Rectangular Columns


• 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

FRPRepair with


ISIS EC Module 4



• 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


ISIS EC Module 4



• The confining effects of FRP wraps are not activated until significant radial expansion of concrete occurs

• Therefore, ensure service loads kept low enough to prevent failure by creep and fatigue

Strengthening Limits

FRPRepair with


Example

ISIS EC Module 4

Problem statement

Determine the FRP wrap details for an RC column as described below

InformationRC column factored axial resistance (pre-strengthening) = 3110 kN


New axial live load requirement PL = 1550 kN

New axial dead load requirement PD = 1200 kN

New factored axial load, Pf = 4200 kN

lu = 3000 mm

Dg = 500 mm

Ag = 196350 mm2

Ast = 2500 mm2

fy = 400 MPa

f’c = 30 MPa

ffrpu = 1200 MPa

tfrp = 0.3 mm

ffrp = 0.75

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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

3000

500≤

6.25

4200000/(30 x 196350) 0.5

6 ≤ 7.4

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ISIS EC Module 4

SolutionStep 2: Compute required confined concrete strength, f’cc


Prmax = ke [1cf’cc (Ag-As) + s fy As] Eq. 5-9

Take equation 5-9 and rearrange for f’cc:

f’cc =

Pf ke

s fy As

1c (Ag-As)

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ISIS EC Module 4

SolutionStep 2: Compute required confined concrete strength, f’cc


f’cc =

4200000 0.85

0.85 (400) (2500)

0.81 (0.6) (196350-2500)

1 = 0.85 – 0.0015f’c = 0.85 – 0.0015 (30) = 0.811:

f’cc:

f’cc = 43.4 MPa

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ISIS EC Module 4

SolutionStep 3: Compute volumetric strength ratio, w


w:

f’cc = f’c + k1 flfrp = f’c (1 + pcw) Eq. 5-4

Take equation 5-4 and rearrange for w:

w =

f’cc

f’c - 1

pc=

43.4

30 - 1

1

w = 0.447

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ISIS EC Module 4

SolutionStep 4: Compute required confinement pressure, flfrp


flfrp:

Take equation 5-5 and rearrange for flfrp:

w =frpfrpffrpu

cf’c=

2 flfrpcf’c

Eq. 5-5

flfrp = 2

w c f’c = 2

0.447 (0.6) (30)

flfrp = 4.02 MPa

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ISIS EC Module 4

SolutionStep 4: Compute required confinement pressure, flfrp


Check flfrp again confinement limits:

flfrp = 4.02 > 4.0 Minimum:

flfrp = 4.02 < Maximum:

f’c2 pc

1

ke

- c

flfrp = 4.02 <30

2 (1)

1

0.85- 0.6 = 8.65

OK, limits met

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ISIS EC Module 4

SolutionStep 5: Compute required number of FRP layers


Take Equation 5-2 and rearrange for Nb:

flfrp = Eq. 5-22 Nb frp ffrpu tfrp

Dg

Nb: Nb =flfrp Dg

2 frp ffrpu tfrp

=4.02 (500)

2 (0.75) (1200) (0.3)

Nb = 3.72 Use 4 layers

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ISIS EC Module 4

SolutionStep 6: Compute factored axial strength of FRP-wrapped column


Use Equations 5-2, 5-5, 5-4 and 5-9:

flfrp: flfrp =2 Nb frp ffrpu tfrp

Dg

= 4.32 MPa

w : w = =2 flfrpcf’c

0.48

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ISIS EC Module 4

SolutionStep 6: Compute factored axial strength of FRP-wrapped column


Use Equations 5-2, 5-5, 5-4 and 5-9:

f’cc:

Prmax:

f’cc = f’c (1 + pcw) = 44.4 MPa

Prmax = ke [1cf’cc (Ag-As) + s fy As]

Prmax = 4230 kN > Pf = 4200 kN

Note: Additional checks should be performed for creep and fatigue

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ISIS EC Module 4


• Strengthening of structures with FRP is a relatively simple technique

• However, it is essential to performance to install the FRP system properly

Specifications

Quality Control / Quality Assurance

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ISIS EC Module 4


Specifications

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

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ISIS EC Module 4


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

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ISIS EC Module 4

Additional ApplicationsPrestressed 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 in initial stages of development

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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

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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

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ISIS EC Module 4

Field Applications

John Hart Bridge

Prince George, BC

64 girder ends were shear strengthened with CFRP

Increase in shear capacity of 15-20%

Upgrade completed in 6 weeks

Locations for FRP shear reinforcement

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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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ISIS EC Module 4

Field Applications

St. Émélie Bridge

Sainte-Émélie-de-l'Énergie, Quebec

Single-span, simply supported tee-section bridge

Strengthened for flexure and shear

Site preparation: 3 weeks, FRP installation: 5 days

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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

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Available from www.isiscanada.com

An Introduction to FRP- Strengthening of Concrete Structures ISIS Educational Module 4: Produced by ISIS Canada.

Documents

frp materials isis ec

frp costs

typeapplicationschematic

feasibility of frp

reinforcement section

properties frp properties

stress mpa frp repair

frp materials lightweight