sISIS EC Module 4 - Handout

Repair with

FRP reinforcement

ISIS Educat Educat ional Modul e 4:

An Int rodu ro duct ctio ionn to t o FRPStrengthening Strengthening of Concrete Structures

Module Objectives

• To provide provide studen students ts with with a general general awareness of FRP materials and their potential uses • To introdu introduce ce students students to the general general philosophies and procedures for strengthening structures with FRPs

Produc ed by ISIS Canada Canada

ISIS EC Module 4

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FRP

FRP

Overview

reinforcement

reinforcement

Introduction

Additional Info Field Applications

FRP Materials

Advanced Applications Specifications & Quality Control

Evaluation of Existing Structures

Column Strengthening ISIS EC Module 4

Introduction

Section: 1

• The world’s world’s populati population on depends depends on an extens extensive ive infrastructure system • Roads, Roads, sewers, sewers, highways, highways, buildings buildings • The system system has has suffered suffered in past past years years • Neglect, Neglect, deteriora deterioration, tion, lack of funding funding

Beam & One-Way Slab Strengthening Global Infrastructure Crisis ISIS EC Module 4

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FRP

Introduction

reinforcement

Section: 1

• A primary factor factor leading leading to extensive extensive degrada degradation tion… … Corrosion

FRP

Introduction

reinforcement

Section: 1

• Why repair repair with the the same same materia materials? ls? • Why repeat repeat the cyc cycle? le? Lightweight

Concrete

High Strength

Easy to install

Reinforcing Steel

5x steel

FRP Materials Moisture, oxygen and chlorides penetrate

Corrosion products form Volume expansion occurs More cracking Corrosion propagation

Through concrete Through cracks

End result

Corrosion resistant

ISIS EC Module 4

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Suit any project

Durable structures

ISIS EC Module 4

FRP

Highly versatile

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FRP Materials

Section: 1

FRP reinforcement

FRP Materials

General

FRP-Strengthening FRP-Strengthening Applic ations

Type Flexural

Shear

Confinement

Application

Fibre Dir.

Schematic

Tension and/or Along long. side face of axis of beam beam

Section

Side face of Perpendicular axis beam (u-wrap) tooflong. beam

Section

Around column

Section: 2

• Longstan Longstanding ding reputat reputation ion in automotiv automotivee and aerospace industries • Over the the past 15 15 years years have FRP FRP materials materials been been increasingly considered for civil infrastructure applications FRP costs have decreased New, innovative solutions needed!

Circumferential Section ISIS EC Module 4

ISIS EC Module 4

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FRP

FRP Materials

reinforcement

Section: 2

FRP

FRP Materials

reinforcement

Section: 2

General

Constituents

• What is FRP?

• Wide range of FRP products available: • Plates

Fibres

Matrix

Provide strength and stiffness

• Rigid strips • Formed through pultrusion

Protects and transfers load between fibres

Carbon, glass, aramid

Epoxy, polyester, vinyl ester

• Sheets • Flexible fabric

Carbon FRP sheet

Fibre

Composite

Matrix

Creates a material with attributes superior to either component alone! ISIS EC Module 4

ISIS EC Module 4

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FRP

FRP Materials

reinforcement

Section: 2

FRP

FRP Materials

reinforcement

Properties

• Typical FRP stress-strain behaviour 1800-4900

Wet lay-up

Used with flexible sheets Saturate sheets with epoxy adhesive Place on concrete surface

Fibres

] a P M [ s s e r t S

Section: 2

Installation Techniques

FRP

Epoxy Matrix

34-130 0.4-4.8

>10

Roller Resin acts as adhesive AND matrix

Strain [%] ISIS EC Module 4

ISIS EC Module 4

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FRP

FRP Materials

reinforcement

Section: 2

FRP reinforcement

FRP Materials

Section: 2

Installation Techniques

Pre-cured

Used with rigid, pre-cured strips Apply adhesive to strip backing Place on concrete surface Not as flexible for variable structural shapes Resin acts as adhesive AND matrix

Properties

• FRP properties (versus steel): • Linear elastic behaviour to failure • No yielding • Higher ultimate strength • Lower strain at failure

ISIS EC Module 4

reinforcement

] 2000 a P M [ 1500

CFRP GFRP

s s e 1000 r t S

Steel

500 1

2 Strain [%]

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FRP Materials

Section: 2

FRP reinforcement

FRP Materials

Properties

FRP material properties are a function of:

3

ISIS EC Module 4

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FRP

2500

Type of fibre and matrix Fibre volume content Orientation of fibres

Section: 2

Pro/Con

FRP advantages Will not corrode High strength-to-weight ratio Electromagnetically inert FRP disadvantages High initial material cost But not when life-cycle costs are considered

ISIS EC Module 4

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FRP reinforcement


Section: 3

FRP reinforcement


Deficiencies

• Deficiencies due to:

Section: 3

Deficiencies

• Deficiencies due to: Then

Now

Chloride Ingress

Freeze-Thaw

Wet-Dry

Environmental



Effects



ISIS EC Module 4

ISIS EC Module 4

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FRP reinforcement

Updated Design Loads Updated design code procedures

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Section: 3

FRP reinforcement


Deficiencies

• Deficiencies due to:

Section: 3

Evaluation

• Evaluation is important to: Then

Now

Determine concrete condition Identify the cause of the deficiency Establish the current load capacity Evaluate the feasibility of FRP strengthening Increase

in Traffic Loads

ISIS EC Module 4

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Section: 3

FRP


reinforcement

Evaluation

• Evaluation should include:

Section: 3

Concrete Surface

• One of the key aspects of strengthening: State of concrete substrate

All past modifications Actual size of elements

• Concrete must transfer load from the elements to the FRPs through shear in the adhesive

Actual material properties Location, size and cause of cracks, spalling Location, extent of corrosion

• Surface modification required where surface flaws exist

Quantity, location of rebar ISIS EC Module 4

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FRP reinforcement

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Beam/One-Way Slab Strengthenin g FRP rupture

Section: 4

FRP

Beam/One-Way Slab Strengtheni ng

reinforcement

Flexural Strengthening Ass ump tio ns

Failure caused by:

Section: 4

Resistance Factors

Material

Bridge

Building

Steel

φS =0.90

φS =0.85

Concrete

φC =0.75

φC =0.6

Concrete crushing Plane sections remain plane Perfect bond between steel/concrete, FRP/concrete Ad equat e anch or age & d evel opm ent l engt h pr ov id ed fo r FRPs FRPs are linear elastic to failure Concrete compressive stress-strain curve is parabolic, no strength in tension Initial strains in FRPs can be ignored ISIS EC Module 4

FRP

Carbon

φfrp = 0.75

Glass

φfrp = 0. 50

ISIS EC Module 4

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FRP

Beam/One-Way Slab Strengthenin g

reinforcement

Section: 4


Failure Modes

• Four potential failure modes:

Perform analysis

d

a = β1c

c

As

εs

f s f frp

εfrp

bfrp Cross Section

Check failure mode

α1Φcf’c

εc

h

Debonding is prevented through special end anchorages

Section: 4

General Design

b

Concrete crushing before steel yields Steel yielding followed by concrete crushing Steel yielding followed by FRP rupture Debonding of FRP reinforcement Assume failure mode

FRP reinforcement

Strain Distribution

Stress Distribution

• Force equilibrium in section: Ts + Tfrp = Cc

Cc Ts Tfrp

Equiv. Stress Distribution

Eq. 4-1

*** Assume initi al strains at th e time of strengthening are zero ***

Ts = φsAsf s

*** Refer to EC Module 4 Notes ***

Tfrp = φfrpAfrpEfrpεfrp

ISIS EC Module 4

Cc = φcα1f’cβ1bc

ISIS EC Module 4

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FRP


reinforcement

Section: 4

FRP


reinforcement

General Design

b d

h As

bfrp Cross Section

b

α1Φcf’c

εc

a = β1c

c εs

f s f frp

εfrp

Strain Distribution

Stress Distribution

Cc Ts Tfrp


• Apply strain compatibility and use these equations to solve for neutral axis depth, c • Section capacity: Mr = Ts d − a + Tfrp h − a Eq. 4-5

2

ISIS EC Module 4

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Section: 4

Anal ysi s Pro cedu re εcu

d

h As

bfrp Cross Section

c εs εfrp

Strain Distribution

Step1: Assume failure mode

Assume that section fails by concrete crushing after steel yields εc = εcu = 0.0035 Thus: εfrp = εcu (h-c)/c Eq. 4-6 εs = εcu (d-c)/c Eq. 4-7 ISIS EC Module 4

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FRP


reinforcement

Section: 4

FRP


reinforcement

Anal ysi s Pro cedu re

b d

h

a = β1c

c

As

εs

bfrp Cross Section

b

α1Φcf’c

εc

f s f frpu

εfrpu

Strain Distribution

Cc

c

d

As


Step 9: Calculate factored moment resistance

εcu

A’s

h

Ts Tfrp

Stress Distribution

Section: 4

With Compression Steel α1Φcf’c

ε’s

f’s

εs

f s f frp

εfrp

bfrp Cross Section

Strain Distribution

Stress Distribution

a = β1c

Cc

Cs

Ts Tfrp


• Similar analysis procedure

Add a compressive stress resultant

Mr = φsAsf y d − a + φfrpAfrpEfrpεfrpu h − a 2 2

Eq. 4-17

ISIS EC Module 4

ISIS EC Module 4

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FRP reinforcement

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Section: 4

FRP reinforcement


Tee Beams

bf

Problem statement

hf c h

= bfrp

Section: 4

Flexural Example

Afrp

Mr

• Similar analysis procedure

Neutral axis in flange: treat as rectangular section Neutral axis in web: treat as tee section

Calculate the moment resistance (M r ) for an FRPstrengthened rectangular concrete section

+ Mrf

Mrw

Section information m m 5 2 3 = d

m m 0 5 3 = h

3-10M bars CFRP

b = 105 mm ISIS EC Module 4

Afrp = 60 mm2 f’c = 45 MPa

εfrpu = 1.55 %

f y = 400 MPa

Efrp = 155 GPa

Es = 200 GPa ISIS EC Module 4

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Section: 4

FRP reinforcement


Flexural Example

Section: 4

Flexural Example

Solution

Solution

Step 1: Assumed failure mode

Step 2: Calculate concrete stress block factors α1 = 0.85 – 0.0015 f’c > 0.67

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

 α1

= 0.85 – 0.0015 (45) = 0.78

β1 = 0.85 – 0.0025 f’c > 0.67  β1

= 0.85 – 0.0025 (45) = 0.86

ISIS EC Module 4

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FRP reinforcement

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Section: 4

FRP reinforcement


Flexural Example

Solution

Solution

Step 3: Find depth of neutral axis, c

Step 4: Check failure mode

Use Equation 4-10:

εfrp = εcu (h-c)/c

φcα1f’cβ1bc = φsAsf s + φfrpAfrpEfrpεfrp

0.6 (0.78) (45) (0.86) (105) c

0.75 (60) (155000) 0.0035

c = 90.5 mm ISIS EC Module 4

vs.

εfrp = 0.0035

0.85 (300) (400) 350 - c c

Section: 4

Flexural Example

εfrp = 0.01

εfrpu = 0.0155

Eq. 4-11

350 - 90.5 90.5

< ε frpu = 0.0155

Therefore, FRP rupture does NOT occur and assumed failure mode is correct ISIS EC Module 4

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Section: 4

FRP


reinforcement

Flexural Example

Solution

Solution

Step 4: Check failure mode

Step 5: Calculate moment resistance

To promote ductility, check that steel has yielded: εs = εcu εs = 0.0035

Mr = φsAsf y d − a 2

d-c c

325 - 90.5 = 0.009 > 0.002 = εy 90.5

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

a 2

Eq. 4-12

Mr = 50.9 × 106 N· mm = 50.9 kN· m

65% increase over unstrengthened beam!

ISIS EC Module 4

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+ φfrpAfrpEfrpεfrp h −

0.85 (300) (400) 325 - 0.86 x 90.5 2 0.75 (60) (155000) (0.01) 350 - 0.86 x 90.5 2

ISIS EC Module 4

FRP

Section: 4

Flexural Example

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Section: 4

FRP reinforcement


Shear Strengthening Ass ump tio ns

• FRP sheets can be applied to provide shear resistance • Many different possible configurations May be aligned at any angle to the longitudinal axis

• FRP sheets can be applied to provide shear resistance • Many different possible configurations ne = 2

May be applied on sides only or as U-wraps

May be applied in continuous sheets or in finite widths

Section: 4


Section

ne = 1 Section

*U-wraps also improve the anchorage of flexural FRP external reinforcement ISIS EC Module 4

ISIS EC Module 4

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reinforcement

Section: 4

FRP reinforcement



Section: 4

Shear Strengthening Design Principles

External strengthening with FRPs: wfrp

To avoid stress concentrations, allow for a minimum radius of 15 mm

β

sfrp

Flexural failure

Generally fairly ductile

Shear failure

Sudden and brittle Undesirable failure mode

Section

Control shear deformation

to avoid sudden failure ISIS EC Module 4

ISIS EC Module 4

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FRP reinforcement

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Section: 4

FRP reinforcement



Shear resistance of a beam: Vr = Vc + Vs + Vfrp

Shear resistance of a beam: Eq. 4-18

Vc = 0.2 φc√f’c bwd Vs =

ISIS EC Module 4

Section: 4


φs f y Av d

s

Eq. 4-19

Eq. 4-20

ISIS EC Module 4

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FRP


reinforcement

Section: 4

FRP reinforcement



Shear resistance of a beam: Vfrp =



φfrp Afrp Efrp εfrpe dfrp (sinβ + cosβ)

sfrp

Effective strain in FRP, εfrpe: εfrpe = R εfrpu ≤ 0.004

Eq. 4-21

Afrp = 2 tfrp wfrp



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



Ensures aggregate interlock!

Reduction factor, R: f’c2/3 R = αλ1 ρ E frp frp

λ2 Eq. 4-24

Carbon: λ1 = 1.35, λ2 = 0.30 Glass: λ1 = 1.23, λ2 = 0.47

0.8 ISIS EC Module 4

ISIS EC Module 4

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Eq. 4-23

Prevents shear cracks from widening beyond acceptable limits



FRP

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Section: 4

FRP reinforcement


Shear Strengthening Design Principles 

Section: 4


FRP shear reinforcement ratio, ρfrp: 2 tfrp wfrp ρfrp = bw sfrp

Another limit on effective strain in FRP, εfrpe: αk1k2Le

εfrpe ≤

9525

Eq. 4-25

Eq. 4-26

0.8 

Parameters, k1 and k2: f’ k1 = c 27.65

ISIS EC Module 4

Section: 4


2/3 Eq. 4-27

k2 =

dfrp- ne Le dfrp

Eq. 4-28

ISIS EC Module 4

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Section: 4

FRP reinforcement


Shear Strengthening Design Principles 

Effective anchorage length, L e: Le = 25350 0.58 tfrpEfrp

Limit on spacing of strips, s frp: sfrp ≤ wfrp +

Eq. 4-29

ISIS EC Module 4

reinforcement

d 4

Eq. 4-30

ISIS EC Module 4

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FRP

Section: 4


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Section: 4

FRP reinforcement



Limit on maximum allowable shear strengthening, V frp: Vr ≤ Vc + 0.8λφc√f’c bwd

Problem statement Calculate the shear capacity (V r ) for an FRPstrengthened concrete section

Eq. 4-31

Shear contribution due to steel stirrups and FRP strengthening must be less than this term

Section information 4.76 mm Ø

λ = 1.0

m m 5 2 3 = d

m m 0 5 3 = h

f’c = 45 MPa εfrpu = 2.0 %

3-10M bars GFRP wrap

b = 105 mm Section ISIS EC Module 4

Section: 4

Shear Strengthening Example

Elevation

tfrp = 1.3 mm wfrp = 100 mm sfrp = 200 mm

Efrp = 22.7 GPa ss = 225 mm c/c f y = 400 MPa (rebar) f y = 400 MPa (stirrup)

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Section: 4


Solution Step 1: Calculate concrete and steel contributions

φs f y Av d

0.85 (400) (36) (325) Vs = = s 225 Vs = 17680 N = 17.68 kN

Steel:


Solution Afrp:

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

ρfrp:

ρfrp =

2 tfrp bw

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Section: 4


Solution

Step 2: Determine Afrp, ρfrp, Le for effective strain calculation

Le:

2 (1.3) 100 105 200

=

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reinforcement

wfrp sfrp

ρfrp = 0.0124

ISIS EC Module 4

FRP

Section: 4


Step 2: Determine Afrp, ρfrp, Le for effective strain calculation

Vc = 0.2 φc√f’c bwd Vc = 0.2 (0.6) √45 (105) (325) Vc = 27470 N = 27.47 kN

Concrete:

FRP reinforcement

Le =

25350 tfrpEfrp

0.58

=

25350 1.3 x 22700

0.58

FRP reinforcement


Section: 4


Solution

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

k1:

f’ k1 = c 27.65

2/3

2/3

45 = = 1.38 27.65

Because of u-wrap

Le = 64.8 mm k2: ISIS EC Module 4

k2 =

dfrp- ne Le 325 – 1 (64.8) = = 0.80 dfrp 325 ISIS EC Module 4

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Section: 4


Solution

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

FRP


reinforcement

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

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

εfrpe:

εfrpe ≤

εfrpe =

R:

αk1k2Le

R = αλ1

Eq. 4-26

9525

0.8 (1.38) (0.80) (64.8) 9525

45 2/3 0.0124 (22700)

0.47

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Section: 4


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

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

εfrpe:

λ2

R = 0.229

ISIS EC Module 4

reinforcement

f’c2/3 ρfrp Efrp

R = 0.8 (1.23)

εfrpe = 0.0060

FRP

Section: 4


εfrpe = R εfrpu ≤ 0.004 εfrpe = 0.229 (0.02) εfrpe = 0.0046

ISIS EC Module 4

Eq. 4-23

FRP


reinforcement

Section: 4


Solution Step 5: Determine governing effective strain, εfrpe

For design purposes, use the smallest limiting value of: εfrpe = 0.0046

Eq. 4-23

εfrpe = 0.0040

Eq. 4-23

εfrpe = 0.0060

Eq. 4-26

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reinforcement

Section: 4


Solution Step 6: Calculate contribution of FRP to shear capacity

Vfrp:

Vfrp = Vfrp =

FRP reinforcement


Section: 4


Solution Step 7: Compute total shear resistance of beam

φfrp Afrp Efrp εfrpe dfrp (sinβ + cosβ)

sfrp

Eq. 4-21

Vr :

Vr = Vc + Vs + Vfrp

Eq. 4-21

Vr = 27.5 + 17.7 + 19.2

0.5 (260) (22700) (0.004) (325) (sin90 + cos90) 200

Vr = 64.4 kN

Vfrp = 19200 N = 19.2 kN

ISIS EC Module 4

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FRP


reinforcement

Section: 4


Solution Step 8: Check maximum shear strengthening limits

Eq. 4-31 Vr ≤ Vc + 0.8λφcf’cbwd 64400 ≤ 27500 + 0.8 (1) (0.6) (45) (105) (325)

64400 ≤ 137400 

OK

FRP reinforcement


Solution Step 9: Check maximum band spacing

d 4 325 200 ≤ 100 + 4 sfrp ≤ wfrp +

Eq. 4-30

200 ≤ 181 

ISIS EC Module 4

Section: 4


Not true, therefore use 180 mm spacing ISIS EC Module 4

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Section: 4

FRP

Column Strengthening

reinforcement

Section: 5

Add ’l Co nsi derat ion s

• FRP sheets can be wrapped around concrete columns to increase strength • How it works:

Additional factors to consider: FRP anchorage and development length Deflections Crack widths

Creep-rupture stress limits sometimes govern FRPstrengthened design

External strengthening with FRPs may reduce flexural deformability

Vibrations

3-layers FRP 1-layer FRP

d a o L

Creep

Overview

Internal reinforcing steel Concrete FRP wrap

No FRP

Fatigue Ductility

Deflection

…and dilates…

ISIS EC Module 4

reinforcement

…FRP confines the concrete…

…and places it in triaxial stress…

ISIS EC Module 4

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FRP

f lfrp

Concrete shortens…

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Section: 5

FRP reinforcement


Overview

• The result: Increased load capacity Increased deformation capability

Section: 5

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

ISIS EC Module 4

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Section: 5

Circular Columns Slenderness Limits

• Strengthening equations only valid for nonslender columns. Thus, from CSA A23.3: lu

Dg

6.25 Pf / f’cAg

FRP reinforcement

• Strengthening equations only valid for nonslender columns. Thus, from CSA A23.3: lu

Dg

Ag = gross cross-sectional area of column f’c = concrete strength Pf = factored axial load lu = unsupported length Dg = column diameter

Eq. 5-1

0.5

Column may become slender! Ensure that column remains short

ISIS EC Module 4

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6.25 Pf / f’cAg

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

ISIS EC Module 4

FRP

Section: 5

Circular Columns Slenderness Limit s

Eq. 5-1

0.5


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Section: 5

FRP reinforcement


Circular Columns Confinement

• Based on equilibrium, the lateral confinement pressure exerted by the FRP, f lfrp: f lfrp =

2 Nb φfrp f frpu tfrp Dg

Eq. 5-2

Nb = number of FRP layers φfrp = material resistance factor for FRP f frpu = ultimate FRP strength tfrp = FRP thickness ISIS EC Module 4

Section: 5


• The benefit of a confining pressure is to increase the confined compressive concrete strength , f’cc f’cc = f’c + k1 f lfrp

Eq. 5-3

f’c = ultimate strength of unconfined concrete k1 = empirical coefficient from tests

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f’cc = f’c + k1 f lfrp = f’c (1 + αpcωw) αpc = performance coefficient depending on: (currently taken as 1.0)

2 f lfrp φc f’c

FRP


reinforcement

Circular Columns Confinement Limits

Minimum confinement pressure

Eq. 5-4

FRP type f’c member size

Maximum confinement pressure

Why?

Why?

To ensure adequate ductility of column To prevent excessive deformations of column

Eq. 5-5

Limit

f l frp

f l f rp

4 MPa

f’ c 2

1 pc

ke

-

c

ISIS EC Module 4

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reinforcement

Limit

= 0.85 (Strength reduction factor to account for unexpected eccentricities)

ISIS EC Module 4

FRP

Section: 5


• ISIS design guidelines suggest a modification to f’cc:

ωw =

Section: 5

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Section: 5

Circular Columns Axi al Lo ad Resi stan ce

• Factored axial load resistance for an FRP-confined reinforced concrete column, P rmax: Prmax = ke [α1φcf’cc (Ag-As) + φs f y As]

Eq. 5-9

FRP reinforcement


Section: 5

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

Same equation as for conventionally RC column, except includes confined concrete strength, f’cc Confinement all around ISIS EC Module 4

Confinement only in some areas ISIS EC Module 4

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Section: 5

FRP reinforcement


Add ’l Co nsi derat ion s Shear

Section: 5

Add ’l Co nsi derat ion s Strengthening Limits

• External FRP wrapping may be used with circular and rectangular RC columns to strengthen also for shear

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

• Particularly useful in seismic upgrade situations where increased lateral loads are a concern

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

ISIS EC Module 4

ISIS EC Module 4

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Section: 5

FRP reinforcement


Section: 5

Example

Example

Solution

Problem statement

Step 1: Check if column remains short after strengthening

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

6.25 Pf / f’cAg

lu

Dg

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

= 3000 mm Dg = 500 mm Ag = 196350 mm2 Ast = 2500 mm2

lu

New axial live load requirement P L = 1550 kN New axial dead load requirement PD = 1200 kN New factored axial load, Pf = 4200 kN

f’c = 30 MPa f frpu = 1200 MPa tfrp = 0.3 mm f frp = 0.75

f y = 400 MPa ISIS EC Module 4

3000 500

0.5

Eq. 5-1

6.25 4200000/(30 x 196350) 0.5 6 

7.4 OK ISIS EC Module 4

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FRP


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Section: 5

FRP reinforcement


Section: 5

Example

Example

Solution

Solution

Step 2: Compute required confined concrete strength, f’ cc Take equation 5-9 and rearrange for f’ cc:

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

Prmax = ke [α1φcf’cc (Ag-As) + φs f y As]



f’cc =

Pf ke

α1:

α1 = 0.85 – 0.0015f’c = 0.85 – 0.0015 (30) = 0.81

Eq. 5-9

f’cc:

− φs f y As

α1φc (Ag-As)

f’cc =

4200000 − 0.85 (400) (2500) 0.85 0.81 (0.6) (196350-2500)

f’cc = 43.4 MPa

ISIS EC Module 4

ISIS EC Module 4

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Repair wit h

FRP


reinforcement

Section: 5

FRP reinforcement


Section: 5

Example

Example

Solution

Solution

Step 3: Compute volumetric strength ratio, ωw Take equation 5-4 and rearrange for ωw:

Step 4: Compute required confinement pressure, f lfrp Take equation 5-5 and rearrange for f lfrp:

f’cc = f’c + k1 f lfrp = f’c (1 + αpcωw) ωw:

ωw = ωw =

f’cc -1 f’c αpc

=

43.4 -1 30

1

Eq. 5-4

ωw =

f lfrp:

f lfrp =

ρfrp φfrp f frpu φc f’c ωw φc f’c

2

=

=

2 f lfrp φc f’c

Eq. 5-5

0.447 (0.6) (30) 2

f lfrp = 4.02 MPa

0.447 ISIS EC Module 4

ISIS EC Module 4

23

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FRP


reinforcement

Section: 5

FRP reinforcement


Section: 5

Example

Example

Solution

Solution

Step 4: Compute required confinement pressure, f lfrp Check f lfrp again confinement limits:

Step 5: Compute required number of FRP layers Take Equation 5-2 and rearrange for N b:

f lfrp = 4.02 > 4.0 f’c 1 - φ  Maximum: f lfrp = 4.02 < 2 αpc ke c



Minimum:

Nb:

30 1 - 0.6 = 8.65 f lfrp = 4.02 < 2 (1) 0.85 

Nb =

f lfrp Dg 2 φfrp f frpu tfrp  Use

Eq. 5-2

=

4.02 (500) 2 (0.75) (1200) (0.3)

4 layers

ISIS EC Module 4

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reinforcement

2 Nb φfrp f frpu tfrp Dg

Nb = 3.72

OK, limits met ISIS EC Module 4

FRP

f lfrp =

Repair wit h


Section: 5

FRP reinforcement


Example

Example

Solution

Solution

Step 6: Compute factored axial strength of FRP-wrapped column Use Equations 5-2, 5-5, 5-4 and 5-9: 2 Nb φfrp f frpu tfrp f lfrp: f lfrp = = 4.32 MPa Dg

Step 6: Compute factored axial strength of FRP-wrapped column Use Equations 5-2, 5-5, 5-4 and 5-9:

ωw :

ωw =

2 f lfrp = 0.48 φc f’c

Section: 5

f’cc: Prmax:

f’cc = f’c (1 + αpcωw) = 44.4 MPa Prmax = ke [α1φcf’cc (Ag-As) + φs f y As] Prmax = 4230 kN > Pf = 4200 kN Note: Additional checks should be performed for creep and fatigue

ISIS EC Module 4

ISIS EC Module 4

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Specifications & Quality Control

Section: 6

FRP reinforcement


Section: 6

Specifications Approval of FRP materials

• Strengthening of structures with FRP is a relatively simple technique • However, it is essential to performance to install the FRP system properly

Handling and storage of FRP materials Staff and contractor qualifications Concrete surface preparation

Specifications

Installation of FRP systems Adequate conditions for FRP cure

Quality Control / Quality Assurance

Protection and finishing for FRP system ISIS EC Module 4

ISIS EC Module 4

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FRP reinforcement

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Section: 6

FRP reinforcement

Addi tion al App licatio ns

Section: 7

Prestressed FRP Sheets

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

ISIS EC Module 4

• 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 ISIS EC Module 4

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Ad ditional A ppli cations

Section: 7

FRP reinforcement

Field Applications

Section: 8

NSM Techniq ues

Maryland Bridge

• Newer class of FRP strengthening techniques: near surface mounti ng reinforcement (NSMR)

Winnipeg, Manitoba Constructed in 1969 Unstrengthened Longitudinal grooves concrete T-beam cut into soffit

FRP strips placed Grooves filled in grooves with epoxy grout

• Research indicates NSMR is effective and efficient for strengthening

Twin five-span continuous precast prestressed girders CFRP sheets to upgrade shear capacity

ISIS EC Module 4

ISIS EC Module 4

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FRP reinforcement

Repair wit h

Field Applications

Section: 8

John Hart Bridge

FRP reinforcement

Field Applications

Section: 8

Country Hills Boulevard Bridge

Prince George, BC

Calgary, AB

64 girder ends were shear strengthened with CFRP Increase in shear capacity of 15-20% Upgrade completed in 6 weeks

Deck strengthened in negative bending with CFRP strips New wearing surface placed on top of FRP strips

Locations for FRP shear reinforcement ISIS EC Module 4

ISIS EC Module 4

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Field Applications

Section: 8

FRP reinforcement

Design Guidance

Section: 9

Canadian codes exist for the design of FRPreinforced concrete members

St. Émélie Bridge Sainte-Émélie-de-l'Énergie, Quebec Single-span, simply supported tee-section bridge Strengthened for flexure and

CAN/CSA-S6-00: The Canadian Highway Bridge Design Code (CHBDC) CAN/CSA-S806-02: Design and Construction of Building Components with Fibre Reinforced Polymers

shear

Site preparation: 3 weeks, FRP installation: 5 days

ISIS EC Module 4

ISIS EC Module 4

Repair with

FRP reinforcement

Ad di tion al Information

Section: 9

Available from www.isiscanada.com ISIS Design Manual No. 3: Reinforcing Concrete Structures with Fiber Reinforced Polymers ISIS Design Manual No. 4: Strengthening Reinforced Concrete Structures with Externally-Bonded Fiber Reinforced Polymer ISIS EC Module 1: An Introduction to FRP Composites for Construction ISIS EC Module 3: An introduction to FRP-Reinforced Concrete Structures ISIS EC Module 4: An Introduction to FRP-Strengthening of Reinforced Concrete Structures ISIS EC Module 4

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sISIS EC Module 4 - Handout

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