an introduction to frp- strengthening of concrete structures isis educational module 4: updated...

An Introduction to FRP-Strengthening of Concrete Structures

ISIS Educational Module 4:

Updated October 2010 for ISIS Canada

ISIS EC Module 4

Module Objectives

ISIS EC Module 4

The use of FRPs in civil infrastructure is steadily increasing in Canada and around the world. This module is directed mainly to students to:

•Provide a background and general awareness of FRP materials, their properties , their behaviour and their potential uses

•Introduce the philosophies and procedures for strengthening structures with FRPs

•Familiarize the students with designing using the Canadian Highway Bridge Design Code (CHBDC)

Overview

ISIS EC Module 4

1. Introduction

2. FRP Materials

3. Evaluation of Existing Structures

4. Flexural Strengthening

5. Shear Strengthening

6. Column Strengthening

7. Installation of FRP strengthening systems

8. Quality control and quality assurance

9. Additional applications

10. Field applications

1 - Introduction

• The world’s population depends on an extensive infrastructure system• Roads, sewers, highways, buildings

• The system has suffered in past years• Neglect, deterioration, lack of funding

Global Infrastructure Crisis

ISIS EC Module 4

Why is strengthening needed?

•Many structures, including bridges and parking garages, have become structurally deficient due to deterioration.

•In Canada, more than 40% of the bridges currently in use were built more than 40 years ago.

•Many structures are also becoming functionally obsolete due to increased loading.

1 - Introduction

ISIS EC Module 4

2 - FRP Materials

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

Light weightEasy to install

High Strength5x steel

Corrosion resistant

Durable structures

Highly versatileSuits many

projects

ISIS EC Module 4

ISIS EC Module 4

FRP is a composite consisting of fibres and matrix.

Fibres:– Provide strength and stiffness–Their quality, orientation and shape affect the final product

Matrix (resin):–Coats the fibres–Protects the fibres from abrasion–Transfers stresses between the fibres

Strain [%]0.5-4.8 2-8

50-90

2400-4300

Stre

ss [M

Pa]

Fibres

Matrix

FRP

2 - FRP Materials

ISIS EC Module 4

FRP material properties are a function of:

•Fibre quality, orientation and shape

•Fibre volumetric ratio

•Adhesion to the matrix

•Manufacturing process (additives and fillers)

2 - FRP Materials

2 - FRP Materials

Wide range of FRP products are available:

• Plates (Rigid strips)

• Sheets (Flexible fabric)

• Rods

The fibres could be:

• Carbon

• Glass

• Aramid

ISIS EC Module 4

FRP sheet

2 - FRP Materials

ISIS EC Module 4

FRP advantages:

• Does not corrode

• High strength to weight ratio

• Reduced installation time and cost

• Non-conductive and non-metallic

• Low maintenance requirements

Disadvantage: High temperature is a serious concern

3 - FRP Materials

ISIS EC Module 4

FRP properties versus steel:

• Linear elastic behaviour

to failure• No yielding• Higher ultimate

strength• Lower strain at

failure

Steel

CFRP

GFRP

ISIS EC Module 4

2 - FRP Materials

FRP System Fiber TypeWeight[g/m2]

Thickness[mm]

Tensile Strength [MPa]

Tensile Elastic Modulus [GPa]

Strain at Failure [%]

Fyfe Co. LLC [www.fyfeco.com]Tyfo SEH51A Glass 915 1.3 460 20.9 1.8Tyfo SEH25A Glass 505 0.5 417 20.9 1.8Tyfo SCH41 Carbon 644 1.0 834 82 0.9

Hughes Brothers Inc. [www.hughesbros.com]Aslan 200 Carbon -- 6.4-12.7 2068-1724 124 0.017-0.015

Aslan 500 #2 Carbon -- 2.0 2068 124 1.7

Aslan 500 #3 Carbon -- 4.5 1965 124 1.5

Aslan 400 CFRP Laminates Carbon -- 1.4 2400 131 1.9Sika Canada Inc. [www.sika.ca]

SikaWrap 430G Glass 430 0.5 504 24.6 1.9SikaWrap 100G Glass 915 1.0 558 24.4 2.2SikaWrap 230C Carbon 230 0.4 715 61.0 1.1SikaWrap 103C Carbon 610 1.0 717 65.1 1.0CarboDur S Carbon -- 1.2-1.4 2800 165 1.7

CarboDur M Carbon -- 1.4 2400 210 1.2

CarboDur H Carbon -- 1.4 1300 300 0.5BASF Building Systems Inc. [www.BASFBuildingSystems.com]

MBrace S&P Laminate Carbon -- 1.4 2700 159 1.7

MBrace EG 900 Glass 900 0.37 1517 72.4 2.1

MBrace CF 130 Carbon 300 0.17 3800 227 1.7

MBrace CF 160 Carbon 600 0.33 3800 227 1.7

MBrace CF 530 Carbon 300 0.17 3500 373 0.94

MBrace AK 60 Aramid 600 0.28 2000 120 1.6

ISIS EC Module 4

1) Wet lay-up system:

- Used with flexible sheets

- Multiple layers can be used

- Saturate sheets with epoxy adhesive, then place on the concrete surface and press with a roller

Epoxy

Roller

Installation:

- The system is installed on the surface of the concrete element while the resin matrix is still “wet”, and the polymerization occurs on site

2 - FRP Materials

2 - FRP Materials

- Used with FRP plates or laminates

- Used for surface bonded plates or near surface mounted reinforcements

- The pre-cured laminates should be placed on or into the wet adhesive

- Place on the concrete surface

- Multiple layers can not be used

- Not as flexible for variable structural shapes

2) Pre-cured system:

ISIS EC Module 4

ISIS EC Module 4

3- Near Surface Mounted (NSM)• It is a newer class of FRP strengthening technique.

Un-strengthened concrete T-beam

Longitudinal grooves cut into soffit

FRP strips placed in grooves

Grooves filled with epoxy grout

• Research indicates that NSM reinforcement is effective and efficient for strengthening.

2 - FRP Materials

2 - FRP Materials

ISIS EC Module 4

Type Application

Schematic

FRP-Strengthening Applications :Fibre Dir.

Confinement Around the column Circumferential

Section

ShearSide face of the beam

(U or closed wrap)

Perpendicular to long. axisof the beam Section

Flexuralface of the beam

Tension and/or sideaxis of the

beam

Along long.

Section

3 - Evaluation of Existing Structures

ISIS EC Module 4

Repair process includes:

1.Evaluation of the existing structure

Understanding the cause and the effect of the

deterioration

2.Determining if repair is required and its extent (quantify)

3.Analysis and design

4.Introducing a repair strategy


ISIS EC Module 4

Fixing the effect without understanding the cause is likely to result in premature failure of repair. Proper repair requires an understanding of the cause to eliminate the effect.

• Evaluation is important to:

Determine concrete condition

Identify the cause of the deficiency

Establish the current load capacity

Evaluate the feasibility of FRP strengthening


ISIS EC Module 4

Evaluation should include:

All past modifications

Actual size of elements

Actual material properties

Location, size and cause of cracks and spalling

Location and extent of corrosion

Quantity and location of rebar

Problems in a structure could be due to:

Defects:

In design or material or during construction

Damage:

Due to overloading, earthquake or fire

Deterioration:

Due to corrosion or sulphate attack


ISIS EC Module 4


ISIS EC Module 4

• Examples of some deficiencies:

1. Environmental effects

Freeze-Thaw

Chloride Ingress

Wet-Dry


• A primary factor leading to extensive degradation…..

2. Corrosion

Moisture, oxygen and chlorides penetrate

ConcreteReinforcing

Steel

Through concrete

Through cracks

Corrosion products formVolume expansion occurs

More cracking

Corrosion propagation

End result

ISIS EC Module 4


ISIS EC Module 4

• Deficiencies could be due to:

3. Updated design loads

4. Updated design code procedures

5. Increase in traffic loads

ThenNow


ISIS EC Module 4

• Deficiencies could be due to:

6. Fire damage

7. Earthquakes

ISIS EC Module 4

Material resistance factors (CHBDC):

- Concrete c = 0.75

- Steel reinforcement:

- Reinforcing bars s = 0.90

- Prestressing strands p = 0.95

- Base FRP for pultruded FRP:

- AFRP FRP = 0.55 (for externally bonded applications)

- AFRP FRP = 0.65 (for NSMR)

- CFRP FRP = 0.80 (for externally bonded applications and NSMR)

- GFRP FRP = 0.70 (for externally bonded applications and NSMR)

- Non-pultruded FRP made by wet lay-up: 0.75 times base FRP

4 - Flexural Strengthening

ISIS EC Module 4

Failure ModesThe analysis of the flexural strength of FRP strengthened elements is based on the following assumptions:

1) The internal stresses are in equilibrium with the applied loads.

2) Plane sections remain plane.

3) Strain compatibility exists between adjacent materials.

(ie. Perfect bond between: concrete and steel, concrete and FRP)

4) The maximum tensile strain of the FRP (FRPt ) is 0.006.

5) The maximum compressive strain in the concrete (cu ) is 0.0035.

6) The contributions of FRPs in compression and of the concrete in

tension are neglected.


ISIS EC Module 4

Failure Modes

The potential modes of failure are:

1) Concrete crushing before steel yielding or rupture of the FRP.

2) Steel yielding followed by concrete crushing before rupture of the

FRP.

3) Steel yielding followed by rupture of the FRP.

4) Peeling, debonding, delamination or anchorage failure of the

FRP (considered premature tension failures to avoid).


ISIS EC Module 4

Rectangular section without compression steel:

b

d

Cross Section

As

Strain Distribution

FRP

c

h

bFRP

s

c

Stress Distribution

fs

fFRP

Equiv. Stress Distribution

a =1c

1Φcf’c

Ts

TFRP

Cc

Cc = c1f’c1bcTFRP = FRPAFRPEFRPFRPTs = sAsfs


The equilibrium equations are:

1) Force equilibrium in the section:

Compression forces =Tension forces

2) Moment equilibrium in the section:

External applied moment= Internal moment

ISIS EC Module 4

Cc = Ts + TFRP

Mapplied = Ts d -a

2+ TFRP

h -a

2


ISIS EC Module 4

b

d

Cross Section

As

Rectangular section with compression steel :

h

bFRP

Strain Distribution

FRP

s

c

Stress Distribution

fs

fFRP

Equiv. Stress Distribution

a =1c

1Φcf’c

Ts

TFRP

CcA’s

cu

’s f’s Cs

Add a compressive stress resultant

Cs = sf’sA’s

d’


Mr = Ts d-a2

+TFRPh-

a2

+Cs

a2

- d′

b

d

As

FRP

c

h

bFRP

s

c

ISIS EC Module 4


4. Determine the compressive stress block factors (1, 1)

1 = 0.85 – 0.0015 f’c > 0.671 = 0.97 – 0.0025 f’c > 0.67

d

As

b

a =1c

1Φcf’c

Ts

Cc

h

bFRP

TFRP

ISIS EC Module 4


5. Calculate c (neutral axis position)

Using equilibrium equation the following equations can be derived and used:

- Concrete crushing before steel yields (s and ’s < y )

-Steel yielding followed by concrete crushing (s and ’s > y )

- Steel yielding followed by FRP rupture (s and ’s > y )

1c f’c1bc2 + sEs cu (As’+As)+ FRPEFRP (cu+fi)AFRP

sEs cu (As'd'+Asd)+ FRPEFRP cuAFRP h = 0

c -

sfy(As’-As)+ FRPEFRP (cu+fi)AFRP c-

FRPEFRP cuAFRP h = 0

ISIS EC Module 4

c =sfy(As-As

’)+ FRPEFRP FRPtAFRP

1c f’c1b


1c f’c1bc2 +

6. Check failure mode assumption with the material strains

If failure is initiated by: Concrete crushing: FRP rupture:

- If the assumption is proven to be false, go back to step 3 and make another assumption

- If the assumption is correct, proceed to the next step

ISIS EC Module 4

cu= 0.0035

s= cu

d - cc

FRP= cuh - c

cfi

s'= cu

c - d'

c

FRP = FRPu ≤ FRPt

fi+FRPu) ≤ s'= fi+FRPt)

fi+FRPu) ≤ s= fi+FRPt)

fi+FRPu) ≤ c= fi+FRPt)

c-d'

h-cc-d'

h-cd-ch-c

d-ch-c

ch-c

ch-c


7. Compute internal forces

8. Calculate the section moment resistance (Mr)

9. Compare Mr to the applied moment (Mapplied)

– If Mr < Mapplied, go to step 2 and change AFRP

– If Mr > Mapplied, then the design is safe

Cs = sfsA's TFRP = FRPAFRPEFRPFRPTs = sAsfs

ISIS EC Module 4

Mr = Ts d-a2

+TFRPh-

a2


+Cs

a2

- d′

Optimized determination of AFRP

Assuming: All the steel has yielded and combining the equilibrium equations:

1. Determine c using the following equation

2. Determine strain in FRP

ISIS EC Module 4

1c f’cbw12c2

2 1 c f’cbwh1c Cs(h-d′)+Ts(ds-h) - Mf

FRP = cuh - c

c- fi ≤ 0.006 ≤ FRPu


3. Determine successively:

4. Optimize value of AFRP

Use this AFRP as an input for the iterative design method

ISIS EC Module 4

AFRP =TFRP

FRPEFRP FRP

Cc = c1f’c1bc

TFRP = Cc + Cs - Ts


Geometric parameters

h

FRPt

FRPb

eb

c fh • If the neutral axis lies in the web (c < hf), then treat it as a rectangular section with a compression zone width = be.

• If the neutral axis lies outside the web (c > hf), then treat it as a T-section.

ISIS EC Module 4

T-section:


Factored moment subdivisionGeometric parameters

h

FRPt

FRPb

eb

c fh

FRPA FRPAsA sfA swA

rfM rwMrM = +

= +

The section is treated as the summation of: flange (Mrf) and web (Mrw).

- Flange (Mrf)

- Web (Mrw)

ISIS EC Module 4

sfy

1c f’c(be-bw)hf Asf =

Asw = As –Asf Mrw = s fy Asw d -a

2+TFRP

h -a

2

Mrf = s fy Asf d -hf

2


Design procedure:

1. Select FRP material and AFRP

2. Determine behaviour of the section ( Rect. or T)

If

Then, it is rectangular behaviour Else, it is a T-section

3. Determine Asf and Mrf

4. Determine AFRP to obtain required Mrw

ISIS EC Module 4

hf ≥sfyAs+FRP EFRP FRP A FRP

1c f’c 1be


ISIS EC Module 4

Example:

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

Section information:

f’c = 40 MPaFRPu = 1.26 %

AFRP = 110 mm2

fy = 400 MPa

Es = 200 GPa EFRP = 210 GPa

b = 125 mm

h =

360

mm

2-15M bars

d =

320

mm

CFRP


ISIS EC Module 4

Solution:Step 1: Assume failure mode

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


ISIS EC Module 4

Step 2: Calculate concrete stress block factors

1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (40) = 0.79

1 = 0.97 – 0.0025 f’c > 0.67

1 = 0.97 – 0.0025 (40) = 0.87


ISIS EC Module 4

Step 3: Find depth of neutral axis, cSteel yielding followed by concrete crushing

1c f’cbc2 + sfy(As’-As) + FRPEFRP (cu+fi)AFRP c - FRPEFRP cuAFRP h = 0

0.79(0.75)40(0.87)125(c2)

0.9(400) (0-400)+ 0.80(210000)(0.0035+0)110

c - 0.80(210000)(0.0035)(110)360

= 0

2577.375(c2)-79320(c)-23284800 = 0

c = 111.7 mm or c = -80.9 mm (rejected)


ISIS EC Module 4

Step 4: Check mode of failureSteel yielding followed by concrete crushing

cu= 0.0035

s= cu= 0.0035

Trial 2, assume the steel yields and the strain in the FRP is 0.006

ds-cc

320-111.7

111.7= 0.0065 > 0.002(y)

FRP= cuh-cc

= 0.0035 360-111.7111.7

= 0.0078 > 0.006 not O.K.


ISIS EC Module 4

Step 5: Trial 2

sfy As + FRPEFRP FRPAFRP

Compression in concrete = Tension in ( Steel + FRP)

1c f’cbc =

0.79(0.75)40(0.87)125(c) 0.9(400)400 + 0.80(210000)0.006(110)=

c = 98.9 mm


ISIS EC Module 4

Step 6: Check mode of failure

FRP = 0.006

s= FRP = 0.006

The assumed mode of failure is correct

ds-c 320-98.9= 0.0051 > 0.002(y)

cu= FRPc

h-c= 0.006 98.9

360-98.9= 0.0023 < 0.0035

O.K.

h-c 360-98.9


ISIS EC Module 4

Step 7: Moment Resistance

sfy As + FRPEFRP FRPAFRP=

0.9(400)400

+0.80(210000)0.006(110)

=

Mr = Ts ds-1c

2+TFRP

h-1c

2

ds-1c

2h-

1c

2320-

0.87 (98.9)

2

360-2

Mr = 75 106 N.mm = 75 kN.m


ISIS EC Module 4

Example:The T-beam requires strengthening to upgrade its moment capacity to 600 kN-m. Calculate the required area of FRP (AFRP).

Section information:f’c = 30 MPa

FRPu = 1.26 %As = 8 x 300 mm2

fy = 400 MPa

Es = 200 GPa EFRP = 155 GPa600

250

650

70

510


ISIS EC Module 4

Step 1: Calculate concrete stress block factors

1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (30) = 0.805

1 = 0.97 – 0.0025 f’c > 0.67

1 = 0.97 – 0.0025 (30) = 0.895


ISIS EC Module 4

Step 2: Evaluating the moment capacity of the existing section

sfy As

Compression in concrete = Tension in steel

1c f’cb1c =

0.805(0.75)30(0.895)650(c) 0.9(400)(300× 8)=

c = 82mm > hf

Assume neutral axis is inside the flange (c < hf)

The assumption (c < hf) is wrong.


ISIS EC Module 4


sfy

1c f’c(be-bw)hf

=1409 mm2

Assume neutral axis is outside the flange (c > hf)

Asf =

0.9(400)

0.805(0.75)30(650-250)70 Asf =

Asw = As –Asf = 2400 -1409 = 991 mm2

Mrf = 0.9 (400)1409 (510 -70

2) = 240.939 ×106 N.mm = 240.939 kN.m


ISIS EC Module 4


sfy Asw

Compression in concrete = Tension in steel (Asw)

1c f’cb1c =

0.805(0.75)30(0.895)250(c) 0.9(400)(991)=

c = 88.03mm > hf

Mrw= 0.9 (400)991(510 -0.895 (88.03)

2) = 167.89 ×106 N.mm = 167.89 kN.m

Mr= 167.89 + 240.939 =408.8 kN.m Moment resistance of the section


ISIS EC Module 4

Step 3: Optimized determination of AFRP

1) Determine c using the following equation:

1c f’cbw12c2

c= 1153 (rejected) or 187mm (accepted)

21c

f’cbwh1cCs(h-d′)+sfy Asw(ds-h)-(Mf -Mrf ) = 0

0.805(0.75)30(250)(0.895)2c2

2- 0.805(0.75)30(250)600(0.895)c

- 0.9(400)991(510-600) + (600-240.9)x106 = 0

1813.57c2 - 2431603.1c + 391208400 = 0


ISIS EC Module 4

Step 3: Optimized determination of AFRP

2) Strain in FRP

Select: 2 layers b = 250mm and tFRP = 1.5mm, AFRP = 750mm2

FRP = cuh-c

c= 0.0035 600-187

187= 0.0077 > 0.006

FRP = 0.006

TFRP =1c f’cbw1c -sfy Asw

=0.805(0.75)30(250)0.895(187) - 400(0.9)991=401 089.6 N

AFRP =TFRP

FRPEFRP FRP

=401 089.6

0.75x0.8×155×103×0.006= 718.8 mm2


ISIS EC Module 4

Step 4: Check the designAssume tension failure of the FRP and yielding of steel

= 191.3 mm >hf …………O.K

c = TFRP +sfy Asw

1c f’cb1

= 0.75(0.8)(155000)0.006(750)+ 0.9(400)9910.805(250)0.75(30)0.895

1) Neutral axis location

2) Check strains

c= FRPc

h-c= 0.006 191.3

600-191.3= 0.0028 < 0.0035

s= FRPds-ch-c

= 0.006 510-191.3600-191.3

= 0.00467 > y ………….O.K


ISIS EC Module 4

Step 4: Check the design

Mrw =

=

0.75(155000)0.006(600)

3) Moment resistance of the section

Mrw = Ts ds -1 c

2+TFRP

h -1 c

2

0.9(400)991

600 - 0.895(191.3)

2

510 - 0.895(191.3)2

× 10-6+

= 366.68 kN.m × 10-6

Mrf = 0.9(400)1409(510 - 702

) = 240.939 ×106 N.mm = 240.939 kN.m

Mrt = Mrw + Mrf = 366.68 + 240.94 = 607.62 kN.m


ISIS EC Module 4

5 - Shear Strengthening

• The shear resistance of the concrete element depends on the interaction between the concrete and the reinforcement.

• FRP sheets can be applied to increase shear resistance.

• The sheets are placed perpendicular or at an angle to the beam’s longitudinal axis. The shear capacity from the FRP stirrups is related to the angle of the cracks in the concrete, the direction and the effective strain of the FRP.

ISIS EC Module 4

• dFRP is the effective shear depth for FRP

• sFRP is the spacing of the FRP stirrups

• wFRP is the width of the FRP stirrup


FRPd

FRPwFRPs

• is the angle of inclination of diagonal cracks in the concrete.

• is the angle of the FRP stirrups

ISIS EC Module 4

• Many different possible configurations: 1) Continuous wraps or finite width sheets (width and spacing)

2) Angle between the sheet and the beam’s axis

3) Wrap configuration with respect to the cross section

U-Wrap

Continuous Finite

= 90


Fully Wrapped

90

ISIS EC Module 4

Shear resistance of a beam (Vr ):

1) Existing capacity

- Resistance from concrete (Vc)

- Resistance from steel (Vs)

2) Additional capacity

- Resistance from FRP wraps (VFRP)Vr = Vc Vs VFRP+ +


ISIS EC Module 4

Shear resistance of a beam (Vr ):

1) Resistance provided by concrete (Vc)

dv ≥ (0.72h, 0.9d)

2) Resistance provided by steel (Vs)

Vc = 2.5vcfcr bvdv


ISIS EC Module 4

3) Resistance provided by FRP:

VFRP = FRP AFRP EFRP FRPe dFRP (cot + cot) sinsFRP

• AFRP = 2 tFRP wFRP

•FRPe is the effective strain in the FRP stirrups

• dFRP is the effective depth

• sFRP is the spacing of the FRP stirrups


ISIS EC Module 4

Effective depth of FRP, dFRP:

Closed wrap shear FRPNo flexural FRP

Closed wrap shear FRPTension FRP for flexure

dFRP ≥ (0.9d, 0.72h) dFRP ≥ 0.9h

d


ISIS EC Module 4

Effective depth of FRP, dFRP:

U-Shaped FRP stirrupNo flexural FRP

U-Shaped FRP stirrupTension FRP for flexure

dFRP ≥ (0.9hFRP, 0.72h) dFRP ≥ (0.72h,hFRP)

hfrp

5-Shear Strengthening

ISIS EC Module 4

•frpe = 0.004 ≤ 0.75 frpu (For completely wrapped sections)

•frpe = Kvfrpu ≤ 0.004 (For other configurations)

where:

Kv=

K1=

Effective strain in FRP, FRPe:

K1K2Le

11900 FRPu

≤ 0.75

fc’

27

2/3

K2 =dFRP-Le

dFRP

Le=23300

(tFRPEFRP)0.58


ISIS EC Module 4

Checks:

- Spacing of strips, sFRP:

sFRP ≤ wFRP + d FRP

4

- Maximum allowable shear strengthening, VFRP :

Vc+ Vs+ VFRP ≤ 0.25cf’c bvdv


Shear Strengthening

ISIS EC Module 4

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

Section information

Sectionb = 150 mm

h FRP =

450

mm

d s =55

0mm

CFRP wrap

Section Elevation

f’c = 45 MPa

FRPu = 1.5%

fy = 400 MPa (re-bar & stirrup)

Steel used is 10M

EFRP = 230GPa

s = 225 mm c/c

tFRP = 1.02 mm

wFRP = 100 mmsFRP = 200 mm15

0mm

h=600 mm


ISIS EC Module 4

Solution:

1) Resistance provided by concrete (Vc)

Vc = 2.5vcfcr bvdv

fcr = 0.4* √ f’c = 0.4* √45=2.68

dv ≥ (0.72h and 0.9d)

≥ (0.72*600 and 0.9*550)

≥ (432 and 495) = 495mm

Vc = 2.5*0.18*0.75*2.68*150*495*10-3 = 67.24 kN


ISIS EC Module 4

2) Resistance provided by steel (Vs)

Vs = 175,921 N = 175.9 kN


ISIS EC Module 4

3) Resistance provided by GFRP (VFRP)

dFRP ≥ (0.9 hFRP,0.72h) ≥ (0.9 × 450, 0.72 × 600)

≥ (405,432) = 432mm

AFRP = 2 tFRP wFRP = 2(1.02)(100) = 204 mm2

VFRP= FRP AFRP EFRP FRPe dFRP (cot + cot) sinsFRP


ISIS EC Module 4

3) Resistance provided by FRP:

27

fc’

2/3 =27

45 2/3 = 1.406

K2=dFRP-Le

dFRP

Le=23300

(tFRPEFRP)0.58=

23300

(1.02 x 230 000)0.58

= 17.888mm

=432 - 17.88

432= 0.959

Kv=

K1=

K1K2Le

11900 FRPu

=(1.406)(0.959)(17.888)

11900 (1.5)(10-2)= 0.135 < 0.75

=0.135


ISIS EC Module 4

•FRPe ≤ 0.004

•FRPe ≤ KvFRPu = 0.135 (1.5)(10-2)= 0.002025

FRPe= 0.002025

Effective strain in FRP, frpe:

VFRP = FRP AFRP EFRPFRPe dFRP (cot + cot) sinsFRP

VFRP =0.6(204)(230000)(0.002025)(432)(cot42)

200(1000)=136.8 kN


ISIS EC Module 4

Total resistance of the section (Vr):

Vr = Vc Vs VFRP+ +

Vr = 67.24 + 175.9 + 136.8 = 379.9 kN


ISIS EC Module 4

Checks:

1) Maximum allowable shear strengthening, VFRP :

Vc + Vs + VFRP ≤ 0.25cf’c bvdv

379.9 ≤ 0.25(0.75)(45)(150)(495)(10-3)

379.9 ≤ 626.48 kN…………………………O.K.


ISIS EC Module 4

Checks:

2) Spacing of strips, sFRP:

sFRP ≤ wFRP + d FRP

4

200 ≤ 100 + 4324

200 ≤ 208 mm…………………….O.K


ISIS EC Module 4

6 - Column Strengthening

• FRP sheets can be wrapped around concrete columns to increase strength

• How it works:

Concrete shortens…

…and dilates……FRP confines the concrete…

flFRP

…and places it in triaxial stress…

Internal reinforcing steelConcrete

FRP wrap

ISIS EC Module 4

• The result:

Increased load capacity

Increased deformation capability


ISIS EC Module 4

• Confinement efficiency– Best: circular cross-section – Worst: rectangular section

• Areas of concrete unconfined by the small bending stiffness of FRP system

• Stress concentration at corners

FRPf FRPf

FRPf

confined

unconfined

FRPfFRPf

FRPf

Stress distribution in rectangular section

Uniform stress distribution in circular section


ISIS EC Module 4

Slenderness of the column

If the column is not slender, then the column is designed and analyzed

for axial load only (short column).

If the column is slender, then the column is designed and analyzed for combined axial load and bending moment.


ISIS EC Module 4

Slenderness of the column

Slenderness could be ignored if:

klur

< 34 - 12 M1

M2

Where:k is the effective length factor for the columnlu is the unsupported length of the column

r is the radius of gyration of the sectionM1 is the smaller end moment at ULS due to factored loads

M2 is the larger end moment at ULS due to factored loads

Braced columns

klur

< 22 Un-braced columns


ISIS EC Module 4

1 - Short column (axial load only)


ISIS EC Module 4

1) Confinement Pressure (flFRP):

flFRP=

Where:

flFRP is the confinement pressure

tFRP is the thickness of the FRP confining system

Dg is the external diameter of the circular section or the diagonal of the

rectangular section

2tFRPFRPfFRPu

Dg

…………………………Eq 6-2


ISIS EC Module 4

Minimum confinement pressure

Maximum confinement pressure

Why?To ensure adequate ductility of column

LimitflFRP ≥ 0.1fc

To prevent excessive deformations of column

LimitWhy?flfrp ≤ 0.33 fc

′

′

Confinement Limits:


ISIS EC Module 4

2) Confined concrete strength (fcc):

Where:

fc is the unconfined specified concrete strength

fcc= fc+ 2 flFRP′ ′

′

′

The benefit of the confining pressure is to increase the confined compressive concrete strength, fcc

′

…………………………Eq 6-3


ISIS EC Module 4

3) Axial Load capacity (Pr):

Where:

Ag is the gross area of the cross section

As is the total cross- sectional area of the longitudinal steel reinforcing

bars

Pr= 0.8 c fcc(Ag-As)+ s fyAs′

The factored axial load resistance for an FRP-confined reinforced concrete column, Pr is given by:

…………………………Eq 6-5


ISIS EC Module 4

Design steps for short column (axial load only):

1) Determine the required confined concrete (fcc) strength according to

Equation 6-5.

2) Determine the required confinement pressure (flFRP) from Equation

6-3.

3) Using the properties of the selected FRP system, determine a

minimal thickness for the FRP (tFRP) from Equation 6-2.

4) Check for the confinement limits.

′


ISIS EC Module 4

2 - Slender Column (axial load + moment)


ISIS EC Module 4

Section analysis is based on stress and strain compatibility

Equivalent stress distribution

confined concrete

unconfined concrete

Internal forces

side FRP

tension face FRP

sC

sT

sjF

cC

ccC

faceFRPT ,

sideFRPT ,

Axial strain distribution

's

s

006.0FRP

sj

'ccc

'c 'c

'd

h

d

sjdc

Cross section

FRP

Steel bars

c

c

=c

c

=s

c-d=

sj

dsj-c=

s

d-c

=FRP

h-c

′′

′′

…………………Eq 6-6

Ccc+ Cc+ Cs – Fsj – Ts - TFRP,side - TFRP,face = Pr ≥ Pf


′

′= 5

cc

′

c

fcc

fc

-1′ +1

ISIS EC Module 4

1) Assuming concrete crushing Internal force

fc+fcc

cc

Ccc

′′2

b c- c c

′

cc′Cc ′ c

c ′

Cs ′cc′Fsj c

Ts s fy A s or s s Es A s if s <y

′TFRP,side FRP c EFRP (h-c) tFRP (h-c)cc

TFRP,face FRP c EFRP btFRP(h-c)cc′


c

c1 fc b 1

s fy A s

s ≤ s fy A sjEs Asj (dsj-c)

cc- c - c

c ′

′′

′′2

h 2fc+fcc3fc+3fcc′

Distance from the centre of the section

c

cc

c2h

- c + 1-21

′′

′h /2 -d

dsj - 2h

2h

(h-c)2h

3-

d – h/2

ISIS EC Module 4

2) Assuming maximum FRP tension (FRP =FRPt ):

fc+fc FRPtCcc c

′2

b c-h-c c

′

FRPtCc c1 fc b 1

′ h-c c

′

Cs s fy A s or fs s Es A s if s < y′FRPtFsj s ≤ s fy A sjh-c

Es Asj (dsj-c)

Ts s fy A s

TFRP,side FRP fFRPu (h-c) tFRP

≤ FRP EFRPFRPt (h-c) tFRP

fc= (fcc-fc)′′cc - c

c-c′′′

′ ′′

TFRP,face FRP fFRPu b tFRP ≤FRP EFRPFRPt b tFRP


Internal force Distance from the centre of the section

′2h - d

d-h/2

2h

dsj - 2h

(h-c)2h

3-

c FRPt

h-c2h

- c + 1-21 ′

FRPtc - h-c c

′ ′2

h 3fc+fc6fc+3fc

′

ISIS EC Module 4

Design steps for slender column:

1) Assuming a linear distribution of strain, identify the relationship of

strain in the various materials as a function of the assumed failure

strain.

2) Determine the resultant force for each material.

3) Calculate the position of the neutral axis using equilibrium of forces.

4) Check the validity of the assumptions of strains and stresses for all

materials.


ISIS EC Module 4

Design steps for slender column:

5) Determine Pr as the sum of the resultant force from each material.

6) Determine Mr as the sum of the internal resultant forces multiplied

by their respective distances to the centroid of the section.


ISIS EC Module 4

Rectangular Columns

• External FRP wrapping may be used with rectangular columns. However, strengthening is not as effective and is more complex.

Confinement all around Confinement only in some areas


ISIS EC Module 4

Some geometrical limitations are imposed:• Sharp edge concrete should be rounded to promote an intimate

and continuous contact of the FRP with the concrete. - minimum radius is 35 mm

• The aspect ratio of the section (h/b) ≤ 1.5

• The smaller cross section dimension (b) ≤ 600 mmThe equations used are the same. Dg is taken as the diagonal of

the cross section.


flFRP=2tFRPFRPFFRPu

Dg =√ h2+b2

ISIS EC Module 4

Additional Considerations

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

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


ISIS EC Module 4

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

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

• To avoid creep failure:


PD ≤ 0.85 0.81c f`c (Ag-As)+ fsAs

fs ≤ 0.0015 Es ≤ 0.8fy

Where:PD is the dead load

fs is the stress in the axial steel reinforcement

Example

ISIS EC Module 4

Example:Determine the number of layers of GFRP wrap that are required to increase the factored axial load capacity of the column to 3450 kN.

InformationRC column factored axial resistance (after strengthening) = 3450 kN

lu = 2500 mm

Dg = 450 mm

Ag = 159000 mm2

As = 2500 mm2

fy = 400 MPa

f’c = 30 MPa

fFRPu = 600 MPa

tFRP = 1 mm

FRP = 0.70*0.75


ISIS EC Module 4

Solution:Step 1: Check for the slenderness effect

klur

< 34 - 12 M1

M2

k =1.0, M1=0 and M2=0

2500112.5

= 22.2 < 34

Thus, the slenderness effect can be ignored


ISIS EC Module 4

Step 2: Determine the required confined concrete strength, fcc

Pr = 0.8 1c fcc(Ag-As)+ s fyAs′

1 = 0.85 – 0.0015 f’c > 0.67

1 = 0.85 – 0.0015 (30) = 0.81

3450 000 = 0.8 0.81(0.75) fcc(159000-2500)+ 0.9(400)2500

fcc= 35.9 MPa


′

′

′

ISIS EC Module 4

Step 3: Determine the required confinement pressure (flFRP)

35.9 = 30+ 2 flFRP

fcc= fc+ 2 flFRP′ ′

flFRP = 2.95 MPa

Step 4: Check for the confinement limits

flFRP ≥ 0.1fc =0.1(30) = 3 MPa′

flFRP ≤ 0.33 fc =0.33(30) = 9.9 MPa ′

flFRP = 3 MPa


ISIS EC Module 4

Step 5: Determine the minimal thickness for the FRP (tFRP) and number of layers

flFRP =2tFRPFRPFFRPu

Dg

3 =2tFRP(0.70×0.75)600

450

tFRP = 2.14 mm

Since tGFRP = 1.0 mm, 3 layers of GFRP are required.


Example

ISIS EC Module 4

Example:Check the design of the following column. It is required to resist a factored axial load of 6000 kN and a factored moment of 1600 kN.m.

Information

Ast = 4000 mm2

fy = 400 MPa

f’c = 30 MPa

fFRPu = 3450MPa

tFRP = 0.167 mm

FRPu = 0.015

75

325

325

75

800

600

Axial FRP2 layers

Hoop FRP6 layers

Steel bars


ISIS EC Module 4

Step 1: Determine the properties of the confined concrete

flFRP =

bDg=√ b2+h2 =√ 6002+8002 = 1000 mm

= 1.25 ≤ 1.5

Confinement limits:

h b < 800

Equivalent diameter:

Confining pressure: 2tFRPFRPfFRPu

Dg

2(6 × 0.167)(0.75 ×0.70)3450

1000= = 3.63 MPa

0.33 fc ≥ flFRP ≥ 0.1fc

10 ≥ flFRP ≥ 3………………….O.K

′ ′


ISIS EC Module 4

Step 1: Determine the properties of the confined concrete

Confined concrete strength:

Concrete strain:

fcc = fc+ 2 flFRP′ ′

fcc= 30+2×3.63 = 37.26 MPa′

= 5cc

′c

fcc -1 +1′

′fc′

cc

′ = 0.0035( 5 -1 +1) = 0.0077 37.26

30


ISIS EC Module 4

Step 2: Determine the equations of the resultant forces

The following assumptions were made:

- Compression failure (concrete crushing)

- fc varies linearly from f’c to fcc

- Yielding of both tension and compression steel

- Intermediate steel in elastic domain

′

1 = 0.85 – 0.0015 f’c > 0.671 = 0.85 – 0.0015 (30) = 0.805

1 = 0.97 – 0.0025 f’c > 0.67

1 = 0.97 – 0.0025 (30) = 0.895


ISIS EC Module 4

Assuming concrete crushing:

fc+fcc

cc

Ccc= c

′′

2b c- c

c ′

cc′Cc= c1 fc b 1

′ c c

′

Cs=s fy A s ′

′= 0.75

30+37.262

600 c - c × 0.0035 0.0077

= 0.75(0.805)30(600)0.895 c × 0.0035 0.0077

=15133.5 c - c × 0.0035 0.0077

= 9726.4c × 0.0035 0.0077

= 0.9(400)1500 = 540 000 N


ISIS EC Module 4

cc′Fsj = s c Es Asj (dsj-c)

Ts= s fy A s = 0.9 (400) 1500 = 540 000 N

′TFRP,side= FRP c EFRP (h-c) tFRP = 0.75×0.70 (h-c)cc

TFRP,face= FRP c EFRP btFRP= 0.75×0.70 (h-c)cc′

= 0.9 c (400-c)0.0077 200 000×1000

c (400-c)0.0077=180 × 106

230 000 × 0.334c

(800-c) 20.0077

= 40330.5c

(800-c) 20.0077

c(800-c)0.0077

230 000 ×600 ×0.334

= 24198300c

(800-c)0.0077


ISIS EC Module 4

Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face= Pr

15133.5 c - c × 0.0035 0.0077

+ 9726.4c × 0.0035 0.0077

+ 540 000

c (400-c)0.0077-180 × 106 -540 000 - 40330.5c

(800-c) 20.0077

- 24198300c

(800-c)0.0077 = 6000 × 103

c = 472 mm

Step 3: Determine the position of the neutral axis, c:


ISIS EC Module 4

Step 4: Check the assumptions for strains:

=cc

c

s = 0.0077 × = 0.0065 > 0.002 OK′′

sj

(c-d )′ 472-75472

cc

c

′= (dsj-c) = 0.0077 × 400-472

472= -0.0012 < ± 0.002 OK

=cc

c

s = 0.0077 × = 0.0041 > 0.002 OK′ (d-c ) 725-472472

=cc

c

FRP

= 0.0077 × = 0.0054 < 0.006 OK′ (h-c ) 800-472472

=cc

cc

c = 0.0035 × = 214.5 mm′′ 4720.0077′


ISIS EC Module 4

Pr = Ccc+Cc+Cs-Fsj-Ts-TFRP,side-TFRP,face

=15133.5 472-214.5 + 9726.4 214.5 + 540 000 -0.0012-180 × 106

- 540 000 - 40330.5472

(800-472) 20.0077

- 24198300472

(800-472)0.0077 = 5997 × 103 N

Step 5: Determine Pr:


ISIS EC Module 4

Step 6: Determine Mr:

cc Ccc - c - c

c ′

′

Cc

′

c

′

Cs

′

cc

′

c

2h 2fc+fcc

3fc+3fcc′

2h - d

2h

- c + 1-21

′′

- 472-214.52

800 2X30+37.33X30+3X37.3= 3897000 = 1075 X 106 N.mm

=20860002

800- 472 + 1-

20.895

X 214.5 = 97 X 106 N.mm

= 540 000 2

800 -75 = 176 X 106 N.mm


ISIS EC Module 4

Step 6: Determine Mr:

Fsj dsj -

TFRP,side(h-c)

TFRP,face2h

2h

3-

2h

= 176 X 106 N.mm

= 0 N.mm

Ts d -2h = 540 000 725 -

2800

= 71400 (800-472)2800

3- = 21 X 106 N.mm

=131 0002

800= 52 X 106 N.mm

Total = 1597 X 106 N.mm

The flexural resistance is adequate Mr = 1597 kN.m ≈1600 kN.m


ISIS EC Module 4

Includes:

2) Handling and storage of FRP materials

3) Staff and contractor qualifications

4) Concrete surface preparation

5) Installation of FRP systems

7) Protection and finishing for FRP system

7 - Installation of FRP Strengthening Systems

6) Curing the FRP system

1) Approval of FRP materials

ISIS EC Module 4

2) Handling and storage of FRP materials:

- Must be carried out in accordance with manufacturer specifications.

- Contractor and supplier must ensure that FRP materials are shipped in adequate

conditions. Do not use opened or damaged containers.

- FRP components must be stored in clean & dry area, sheltered from sun rays.

- Do not use material that has exceeded its shelf life.

- Material safety data sheet for all FRP materials and components should be obtained from

the manufacturer and should be accessible at the job site.

1) Approval of FRP materials:

The use of certified FRP materials is recommended.

Qualification testing can be used for the approval of the FRP materials.


ISIS EC Module 4

3) Staff and contractor qualifications:

The workers must have a basic knowledge of all stages of the installation of the FRP systems. The minimum required knowledge includes:

- An understanding of the security instructions

- Mixing proportions of resins

- Application rates

- Pot life and curing times

- Installation techniques


ISIS EC Module 4

4) Concrete surface preparation:-Repair of existing substrate:- The concrete surfaces must be free of particles and pieces that no longer

adhere to the structure.- The surface must be cleaned from oil residuals or contaminants.- Rough surface should be smoothed.- Sections with sharp edges must be rounded.

- Surface preparation for contact critical applications- A continuous contact between the concrete and the FRP confinement system should be guaranteed.- Rounding of corners, filling holes and elimination of depression are of prime importance.


ISIS EC Module 4

5) Installation of FRP systems:

- Primer, putty, saturating resin and fibres should be a part of the same system.

- All equipment should be clean and in good operating condition

- Ambient air and concrete surface temperature should be 10°C or more

- The mixing of resins should be done in accordance with the FRP system

manufacturer recommended procedure. All components should be mixed at a

proper temperature and in the correct ratio until there is a uniform mix, free from

trapped air.

- The installation of FRP is either hand wet applied system or precured system.


ISIS EC Module 4

6) Curing the FRP system

- FRP materials should be cured according to the recommendations of the

manufacturer unless the curing process is accelerated by heating, chemical

reactant or other external supply.

- The curing time should not be less than 24 hours before further work is

done on the repaired surface.

- Chemical contamination from gases, dust or liquid must be prevented

during the cure of all materials.


ISIS EC Module 4

7) Protection and finishing for FRP system

- When the surface of the FRP materials is sufficiently dry or hard, a

protection system and/or paint compatible with the installed

reinforcement can be added.

- The coating must dry for a minimum of 24 hours .

- A certificate of compatibility of the protection system with the selected

type of FRP reinforcement must be obtained from the manufacturer of

the FRP materials.


ISIS EC Module 4

8 - Quality Control and Quality Assurance

The FRP material suppliers, the FRP installation contractors and all others associated with the FRP strengthening project should maintain a comprehensive quality assurance and quality control program.

1) Material qualification and acceptance:The FRP manufacturer, distributor or their agent should provide information demonstrating that the proposed FRP meets all mechanical, physical and chemical design requirements.

Tensile strength, type of fibres, resins, durability, etc.

2) Qualification of contractor personnel:The selection of contractors should be based on evidence regarding their qualifications and experience for FRP strengthening projects.

ISIS EC Module 4


3) Inspection of concrete substrate:

- The concrete surface should be inspected and tested before application of FRP. The inspection should include:

- Smoothness or roughness of the surface- Holes and cracks- Corners radius- Cleanliness

- Pull-off tests should be performed to determine the tensile strength of the concrete for bond-critical applications.

ISIS EC Module 4


4) FRP material inspection:

Inspection of the FRP materials shall be conducted before, during and after their installation.

- Before Construction The FRP supplier should submit certification & identification of all the FRP materials to be used. The installation procedure should be submitted as well .

- During Construction Keep records for:

- Quantity and mixture proportions of resin- The date and time of mixing- Ambient temperature & humidity- All other useful information

Visual inspection of fibres orientation and waviness should be carried out.

ISIS EC Module 4


4) FRP material inspection:- At completion of the project:A record of all final inspection and test results related to the FRP material should be retained. Samples of the cured FRP materials should be retained as well.

5)Testing:- Qualification testing:It is a specification for the product certification of FRPs used for rehabilitation. It includes some guidelines as:

- FRP systems whose properties have not been fully established should not be considered- Constituent materials, fibres, matrices and adhesives, should be acceptable by the applicable code and known for their good performance.

ISIS EC Module 4


5) Testing:- Field testing:

Confirmatory test samples of the FRP material systems should be prepared at the

construction site and tested at an approved laboratory.

In-place load testing can be used to confirm the behaviour of the FRP

strengthened member.

ISIS EC Module 4

9 - Additional Applications

Prestressed FRP Sheets• One way to improve FRP effectiveness is to apply prestress to the

sheet prior to bonding

• This allows the FRP to contribute to both service and ultimate load-bearing situations

• It can also help close existing cracks, and delay the formation of new cracks

• Prestressing FRP sheets is a promising technique, but is still under development

ISIS EC Module 4

10 - Field Applications

Maryland Bridge

- Winnipeg, Manitoba

- Constructed in 1969

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

ISIS EC Module 4

John Hart Bridge

- Prince George, BC

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

Locations for FRP shear reinforcement


ISIS EC Module 4

Country Hills Boulevard Bridge

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


ISIS EC Module 4

A Canadian code exists for the design of FRP-strengthened concrete members

CAN/CSA-S806-02: Design and Construction of Building Components with Fibre Reinforced Polymers

(Currently under revision)

Design Guidance

CAN/CSA-S6-10: Canadian Highway Bridge Design Code

Additional Information

ISIS EC Module 4

Available from www.isiscanada.comISIS EC Module 1: Mechanics Examples Incorporating FRP MaterialsISIS EC Module 2: An Introduction to FRP Composites for ConstructionISIS EC Module 3: An Introduction to FRP-Reinforced ConcreteISIS EC Module 5: Introduction to Structural Health MonitoringISIS EC Module 6: Application & Handling of FRP Reinforcements for ConcreteISIS EC Module 7: Introduction to Life Cycle Engineering & Costing for Innovative InfrastructureISIS EC Module 8: Durability of FRP Composites for ConstructionISIS EC Module 9: Prestressing Concrete Structures with Fibre Reinforced Polymers

an introduction to frp- strengthening of concrete structures isis educational module 4: updated...

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