The Design of Fiber Reinforced Polymers for Struc Strengthening ACI 440.pdf

The Design of Fiber Reinforced Polymers for Structural Strengthening An Overview of ACI 440 Guidelines Sarah Witt Fyfe Company November 7, 2008 1

GUIDE FOR THE DESIGN AND CONSTRUCTION OF EXTERNALLY BONDED FRP SYSTEMS FOR STRENGTHENING CONCRETE STRUCTURES

ACI Document 440.2R-08 Printed July 2008 2

Outline  Strengthening Concrete Structures Reasons for strengthening Types of FRP strengthening systems Materials and properties of FRP strengthening systems



Substrate Preparation/FRP Application Repair Proper detailing and installation methods Quality control



Design Principles Strengthening limits Flexural strengthening Shear strengthening Axial strengthening



Reinforcement Details Bond and delamination Detailing of laps and splices

3

 Design Examples and Case Studies

Reasons for Strengthening

Change in use  Construction or design defects  Code changes  Seismic retrofit Deterioration 

4

Excessive Loading

5

Flexural Cracking

6

Overloading

7

Seismic Loads

8

Improper Steel Placement

9

Impact Damage

10

Typical FRP Systems for Strengthening Structures Section 3.2, Guide:

 Wet lay-up systems

Unidirectional fiber sheets Multidirectional fiber sheets Mechanically applied fiber tows  Prepreg systems

Unidirectional fiber sheets Multidirectional fiber sheets Mechanically applied fiber tows  Precured systems

Unidirectional laminates Multidirectional grids Shell elements  Other forms not covered

11

Typical FRP Systems for Strengthening Structures

12

Typical FRP Systems for Strengthening Structures

13

Typical Fiber Properties Carbon

Aramid

E-Glass

14

Substrate Preparation / Repair Bond vs. Contact Critical  Contact Critical Requires intimate contact between the FRP System and the concrete c Confinement of columns

 Bond Critical Requires an adhesive bond between the FRP system and the concrete c Beam, slab and wall strengthening

15

Substrate Preparation / Repair Section 6.4, Guide:  Substrate issues: c ACI 503 c ICRI 03730

 200 psi (1.4 MPa)

minimum tensile strength  2500 psi minimum compressive strength of concrete

16

Removal / replacement of unsound concrete

Substrate Preparation Section 6.4, Guide:

Minimum ICRI CSP 3

Preparation of concrete surface

17

Epoxy Injection Section 6.4, Guide:  Cracks wider than

0.010 in (0.3 mm) should be injected prior to application of the FRP system. ACI 224.1

 Smaller cracks in

aggressive environments may require sealing

18

Quality Control & Assurance During-construction:  Bond testing

ACI 503R ASTM D4541 Tension adhesion strengths should exceed 200 psi (1.4 MPa), exhibit failure of the concrete substrate.  Cured thickness

Extract small core samples less than 0.5 in (13 mm) diameter Avoid sampling in high stress areas if possible Repair using overlapping sheets on filled core. 19

Quality Control & Assurance Post-construction:  General Acceptance Criteria for Delaminations

Wet Layup c Delaminations less than 2 in2 (1300 mm2) each are permissible: ¡ No more than 10 delaminations per 10 ft2 of laminate area ¡ Total delamination area less than 5% of total laminate area c Delaminations less than 25 in2 (16,000 mm2) may be repaired by resin

injection or ply replacement, depending upon the size, number and location of delaminations. c Delaminations greater than 25 in2 (16,000 mm2) should be repaired by selectively cutting away the affected sheet and applying an overlapping sheet patch of equivalent plies.

Precured systems c Each delamination must be inspected and repaired in accordance with

the engineer’s direction

20

Design Guidelines

21

FRP Strengthening Applications Flexural Strengthening Beams, Slabs, Walls, etc. Shear Strengthening Beams, Columns, Walls, etc. Axial Enhancement Column Wrapping, Pressure Vessels 22

Strengthening Limits Section 9.2, Guide:  Limited by strength of other structural components Columns, footings, etc.  Limited by other failure mechanisms Punching shear  Loss of FRP should not result in immediate

collapse

(φRn )existing ≥ (1.1SDL + 0.75 SLL )new 23

(9-1)

Structural Fire Endurance  Glass Transition Temperatures of most FRP

systems is typically in the range of 140 - 180oF (60 - 80oC)  Use of an insulation system can improve the overall fire rating of the strengthened reinforced concrete member  Insulation system can delay strength degradation of concrete and steel, increasing the fire rating of the member  The contribution of the FRP system can be considered if it is demonstrated that the FRP temperature remains below a critical temperature 24

Rational Fire Endurance Check ACI 216R:  Given cover and fire endurance requirement  Find the temperature of the steel & concrete  Find a reduced steel & concrete material

strength  Find the associated reduced section strength  Reduced strength > Unfactored demand  No phi factors or load factors

25

Rational Fire Endurance Check Section 9.2.1, Guide:  From ACI 216R - Reduce material strengths at elevated

temperature:

Steel: f y → f yθ Concrete: f 'c → f 'cθ FRP: f fu → 0 *

(Rn )existing ≥ (S DL + S LL ) 26

(9-2)

Maximum Service Temperature Section 1.3.3, Guide:  Typical glass transition temperature (Tg) for epoxy 140 -

180oF (60 - 80oC)  Above Tg mechanical properties start to degrade  Service temperature should not exceed Tg - 27°F (Tg – 15°C)

27

Flexural Strengthening Chapter 10, Guide  Typical flexural strength increases up to 40%    

This limit is based on the Guide’s requirements Positive and negative moment strengthening Add strength to RC and PC members Reduce crack widths Seismic loadings not covered

φM n > M u 28

(10-1)

Assumptions Section 10.2.1, Guide:  Design calculations are based on actual      

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dimensions and material properties. Plane sections remain plane (including FRP). Maximum compressive strain ε cu = 0.003 Tensile strength of concrete is ignored. FRP has linear-elastic relation to failure. Perfect bond between FRP and concrete (no slip). The shear deformation within the adhesive layer is neglected.

Verification of Shear Capacity Section 10.2.1, Guide:

Section shear capacity must be

sufficient to handle shear forces associated with increased flexural capacity.

30

Failure Modes Section 10.1.1, Guide: 1. 2. 3. 4. 5.

crushing of concrete prior to steel yield yield of steel followed by concrete crushing yield of steel followed by FRP failure shear / tension delamination in concrete cover FRP debonding from substrate

The desired mode of failure is usually mode 2 or 3.

31

Effective Strain in FRP 600

Rupture Strain

Stress (ksi)

500 400

Effective Strain

300 200 100 0 0

32

0.005

0.01

Strain (in/in)

0.015

0.02

Limitation on Strain in FRP To prevent debonding in regions away from FRP Termination

ε fd

f c' = 0.083 ≤ 0.9 ε fu nE f t f

ε fd

f c' = 0.41 ≤ 0.9 ε fu nE f t f

⎛h−c⎞ ε fe = ε cu ⎜ ⎟ − ε bi ≤ ε fd ⎝ c ⎠ 33

(10-2) US

(10-2) SI

(10-3)

Calculation Procedure Determine initial strain in substrate

Estimate neutral axis, c Compute Moment Capacity

Determine failure mode

Calculate material strain

Check service conditions

Calculate stresses and forces Check Equilibrium (Calculate c)

No

34

Estimated c = c for Equilibrium?

Yes

Estimate the Neutral Axis Depth  No closed form

solution exists  Must find depth to the neutral axis by trial and error  As a starting point, a good rule of thumb is 20% of the effective section depth

c ≈ 0.20 d 35

εc c

εs εfe εbi εb

Determine Mode of Failure ⎛h−c⎞ ε fe = ε cu ⎜ ⎟ − ε bi ≤ ε fd ⎝ c ⎠ ⎛h−c⎞ ε fe = ε cu ⎜ ⎟ − ε bi ≤ ε fd ⎝ c ⎠

Concrete Crushing Controls

⎛h−c⎞ ε fe = ε cu ⎜ ⎟ − ε bi ≥ ε fd FRP Rupture Controls ⎝ c ⎠

36

(10-3)

Concrete Stress Block  Whitney stress block is

γf 'c

valid only when concrete crushing β1c c governs failure (i.e., εc=0.003)  If FRP rupture controls, a stress block Actual Equivalent appropriate for the Stress Stress concrete strain level Distribution Distribution should be used

37

Concrete Stress Block

[

εc < 0.003

(

(

γf'c β1c

38

]

4 (ε c ε c′ ) − tan −1 (ε c ε c′ ) β1 = 2 − (ε c ε c′ ) ln 1 + ε c2 ε c′2

0.90ln 1 + ε c2 ε′c2 γ= β1ε c ε′c

)

)

1.71f c′ ε′c = Ec

Calculation of Flexural Strain  Assume strain

compatibility  Based on failure mode  Calculate the strain in each material by similar triangles

εc c

εs

⎛ d −c ⎞ ⎟ ε s = (ε fe + ε bi ) ⎜⎜ ⎟ d − c ⎝ f ⎠ 39

(10-10)

εfe εbi εb

Steel – Elastic / Plastic:

f s = Esε s ≤ f y

(10-11)

Stress

Calculation Of Stress FRP

FRP – Elastic:

f fe = Esε fe

(10-9)

Steel

Strain

40

Check Force Equilibrium  Sum forces in the horizontal

direction  If forces do not equilibrate, revise “c”  Repeat previous steps

cest

41

As f s + A f f fe = β1α 1 f c′ b

β1c

α1 f 'c

Asfs Afff

Ultimate Strength Model εc c

εs Af = n tf wf

εfe

εbi

fs ffe = Ef εfe

β1c ⎞ β1c ⎞ ⎛ ⎛ M n = As f s ⎜ d − ⎟ ⎟ + ψ f A f f fe ⎜ h − 2 ⎠ 2 ⎠ ⎝ ⎝ 42

(10-13)

Loss in Ductility ≈0.75ρb

φ

ACI 318 : A section with lower 0.90 ductility should compensate with a 0.65 higher reserve of strength

ρb

εsy

⎧ 0.90 ⎪ 0.25(ε t − ε sy ) ⎪ φ = ⎨0.65 + 0.005 − ε sy ⎪ 0.65 ⎪⎩ 43

for for for

0.005 Steel Strain at ε t ≥ 0.005 Ultimate ε sy < ε t < 0.005 (10-5)

ε t ≤ ε sy

Design Flexural Strength

⎡ β1c ⎞ β1c ⎞⎤ ⎛ ⎛ φM n = φ ⎢ As f s ⎜ d − ⎟ +ψA f f fe ⎜ h − ⎟⎥ 2 ⎠ 2 ⎠⎦ ⎝ ⎝ ⎣ Reduction factor for FRP contribution:

ψ = 0.85

φM n > M u 44

(10-13)

(10-1)

Serviceability At service, stress in steel should be limited to 80% of yield strength:

Moment

Mu My Ms

FRP Strengthened

Unstrengthened

f s , s ≤ .80 f y Curvature

45

(10-6)

Prestressed Concrete Members Assumptions  Assumptions for concrete members apply  strain compatibility for strain or change in strain

in the prestressing steel  prestressing steel rupture mode should be investigated  where prestressing steel is draped several sections should be evaluated

46

Prestressed Concrete Members Failure Modes 1. Strain level in FRP governed by strain limitations due: 1. 2. 3. 4.

47

concrete crushing FRP rupture FRP debonding Prestressing steel failure

Prestressed Concrete Members Strength Reduction Factor To maintain a sufficient ductility the nominal strain in the prestressing steel should be higher than 0.013. If this strain is not achieved a lower strength factor should be used

0.90 ⎧ ⎪⎪ 0.25 (ε ps − 0.010) φ = ⎨0.65 + (0.013 − 0.010) ⎪ 0.65 ⎪⎩

48

for

ε ps ≥ 0.013

for 0.010 < ε ps < 0.013 for

ε ps ≤ 0.010

(10-19)

Prestressed Concrete Members Serviceability In service stress in the prestressing steel should be prevented from yielding:

f ps , s ≤ 0.82 f py

f ps , s ≤ 0.74 f pu

49

(10-20a)

(10-20b)

Prestressed Concrete Members Nominal Strength •The calculation procedure for nominal strength: • satisfy should strain compatibility •satisfy force equilibrium •consider mode of failure •similar to method for reinforced members

50

Prestressed Concrete Members Nominal Strength For a given value of the neutral axis, c: Stress level in the FRP

Strain in the tendon

51

ε ps

f fe = E f ε fe P = ε pe + e Ac Ec

(10-21)

⎛ e2 ⎞ ⎜⎜1 + 2 ⎟⎟ + ε pnet ≤ 0.035 (10-22) ⎝ r ⎠

Prestressed Concrete Members Nominal Strength The value of enet depends on the mode of failure

concrete crushing

⎛ dp −c ⎞ ⎟ ε pnet ≤ 0.003 ⎜⎜ ⎟ c ⎠ ⎝

FRP rupture or debonding

⎛ dP − c ⎞ ⎟ ε pnet ≤ (ε fe + ε bi )⎜⎜ ⎟ d − c ⎝ f ⎠

52

(10-23a)

(10-23b)

Prestressed Concrete Members Nominal Strength Force equilibrium can be checked by satisfying:

c=

53

Ap f ps + A f f fe

α f β1b ' 1 c

(10-25)

Case Study – Slab Upgrade P/T flat slab live load increase: 50 – 100 psf

54

Case Study – Slab Upgrade Positive moment upgrade to column strip

55

Shear Strengthening Chapter 11, Guide  Increase shear capacity of beams or columns Amount of increase depends on section geometry, existing reinforcement, and a variety of additional factors.  Change failure mode to flexural Typically results in a more ductile failure

φVn > Vu 56

(11-1)

Wrapping Schemes

Overlap

Fully Wrapped

57

“U-wrap” Two sides bonded

Effective Strain in FRP  Maximum strain that can be achieved in the FRP

system at the ultimate load stage  Governed by the failure mode of the FRP system and the strengthened member.

ε fe = 0.004 ≤ 0.75ε fu for completely wrapped members (11-6a) ε fe = κ vε fu ≤ 0.004 for bonded U − wraps or face plies (11-6b)

58

Effective Strain Limitations for FRP Determination of bond-reduction coefficient κv:

k 1 k 2 Le κv = ≤ 0.75 468 ε fu

(11-7) US

k1k 2 Le ≤ 0.75 κv = 11,900ε fu

(11-7) SI

⎧ d f − Le ⎪ ⎪ df k2 = ⎨ d − 2 Le ⎪ f ⎪⎩ d f 59

⎛ f ⎞ ⎟ k 1 = ⎜⎜ ⎟ 4000 ⎝ ⎠ ' c

⎛ ⎞ k 1 = ⎜⎜ ⎟⎟ ⎝ 27 ⎠ f c'

for U wraps (11-10)

for two − sides bonded

2/3

(11-9) US

2/3

(11-9) SI

Effective Strain Limitations for FRP Determination of active bond length Le:

Le =

2500

(n t

f

Ef

)

0.58

Le =

(11-8) US

Le 60

23 ,300

(n t f E f )

0.58

(11-8) SI

Effective Strain Limitations for FRP Determination of bond-reduction coefficient κv:

k 1 k 2 Le κv = ≤ 0.75 468 ε fu

(11-7) US

k1k 2 Le ≤ 0.75 κv = 11,900ε fu

(11-7) SI

⎧ d f − Le ⎪ ⎪ df k2 = ⎨ d − 2 Le ⎪ f ⎪⎩ d f 61

for U wraps

(11-10)

for two − sides bonded

⎛ f ⎞ ⎟ k 1 = ⎜⎜ ⎟ 4000 ⎝ ⎠ ' c

⎛ ⎞ k 1 = ⎜⎜ ⎟⎟ ⎝ 27 ⎠ f c'

2/3

(11-9) US

2/3

(11-9) SI

ε fe = κ vε fu ≤ 0.004

Pertinent Shear Dimensions sf

sf

df

βα wf

Vf =

A fv f fe (sin α + cos α )d f

A fv = 2 nt f w f f fe = ε fe E f 62

wf (11-3)

sf (11-4) (11-5)

Design Shear Capacity

φVn = φ (Vc + Vs +ψ f V f )

(11-2)

φVn = φ (Vc + Vs +ψV f ) φ = 0.85 ( ACI 318) ψ f = 0.95 ( fully wrapped ) ψ f = 0.85 (bonded U − wraps or face plies )

63

Spacing, Reinforcing Limits s f ,max

d = wf + 4

Section 11.1, Guide:

Based on ACI 318-05, Section 11.5.6.9:

Vs + V f ≤ 8 f 'c bw d Vs + V f ≤ 0.66 f c′bd 64

(11-11) US

(11-11) SI

Case Study – Precast Garage Installed FRP “U” Wraps

65

Confinement Chapter 12, Guide  Increase in member axial compressive strength  Enhance the ductility of members subjected to

combined axial and bending forces  Increase the strength of members subjected to combined axial and bending forces

66

Axial Compression  Fibers oriented transverse to the longitudinal

axis of the member Contribution of any longitudinal fibers to axial strength is negligible  Results in an increase in the apparent strength

of the concrete and in the maximum usable compressive strain in the concrete  Passive confinement Intimate contact between FRP system and member is critical

67

Confinement

Confining Pressure

Stress

FRP Confined Concrete Behavior FRP Confined Concrete

f cc′ εL εT f c′ 0.85 f c′

Transverse Strain

Transverse Strain (Dilation)

69

ε fu

ε fe

Unconfined Concrete

ε c′

ε cu = 0.003

ε ccu

Longitudinal Strain

Longitudinal Strain

FRP Confined Concrete

 Strain Limitation For pure axial loading:

ε fe = κ ε ε fu

(12-5)

κ ε = 0.55

Recommended value (accounts for premature failure strain of FRP)

For combined axial + bending:

ε fe = 0.004 ≤ κ ε ε fu Transverse Strain (Dilation)

70

(12-12)

Limit to maintain shear integrity of concrete

Longitudinal Strain

Stress

FRP Confinement Model

f cc′ E2

FRP Confined Concrete

2 ⎧ ( Ec − E 2 ) 2 ⎪E ε − εc fc = ⎨ c c 4 f c′ ⎪ f ′+ E ε 2 c ⎩ c

f c′ Unconfined Concrete

Ec

Transverse Strain (Dilation)

71

(12-2a)

for ε t′ ≤ ε c ≤ ε ccu

Where,

E2 =

ε c′ ε t′

for 0 ≤ ε c ≤ ε t′

ε ccu

Strain

ε t′ =

f cc′ − f c′

ε ccu

2 f c′ Ec − E 2

(12-2b)

(12-2c) Longitudinal Strain

Stress

FRP Confinement Model

f cc′ E2

FRP Confined Concrete

f c′

f cc′ = f c′ + ψ f 3.3κ a f l ε ccu

Unconfined Concrete

0.45 ⎛ ⎞ ε ⎞ ⎛ f fe = ε c′ ⎜1.50 + 12κ b l ⎜⎜ ⎟⎟ ⎟ (12-6) ⎜ f c′ ⎝ ε c′ ⎠ ⎟ ⎝ ⎠

Where,

Ec

ε c′ ε t′ Transverse Strain (Dilation)

72

(12-3)

ε ccu

Strain

fl is the confining pressure exerted by the FRP jacket

κa and κb are shape factors Longitudinal Strain

Circular Sections FRP Jacket Confining pressure:

fl

fl = Ef εfe

fl

Ef εfe

2 E f n t f ε fe D

Shape factors:

κ a = κ b = 1.0 Concrete

fl 73

(12-4)

Rectangular Sections Equivalent circular column

Confining pressure:

fl = D2 D = b + h2

h

74

b

2 E f n t f ε fe D

(12-4)

Rectangular Sections Shape factors: Ae ⎛ b ⎞ κa = ⎜ ⎟ Ac ⎝ h ⎠

Effective confinement area, Ae

b

Ae ⎛ h ⎞ κb = ⎜ ⎟ Ac ⎝ b ⎠

2

0.5

h Confining stress concentrated at corners

75

(12-9)

(12-10)

Rectangular Sections

 Ratio of effective confinement area to total area

of concrete

Ae = Ac

76

⎡⎛ b ⎞ ⎛h⎞ 2 2⎤ ⎢⎜ h ⎟(h − 2rc ) + ⎜ b ⎟(b − 2rc ) ⎥ ⎝ ⎠ ⎝ ⎠ ⎦ −ρ 1− ⎣ g 3 Ag 1− ρg

(12-11)

Using the Confinement Model Compressive Strength: with existing steel spiral reinforcing

φPn = 0.85φ [0.85 f cc' (Ag − Ast ) + f y Ast ]

(12-1a)

with existing steel-tie reinforcing:

φPn = 0.80φ [0.85 f cc' (Ag − Ast ) + f y Ast ]

(12-1b)

Use the confined concrete compressive strength in ACI 318 equations

77

Serviceability ConsiderationsAxial Compression Section 12.1.3, Guide:  To ensure radial cracking will not occur under

service loads,

f c ≤ 0.65 f c '  To avoid plastic deformation under sustained or

cyclic loads,

f s ≤ 0.60 f y 78

Reinforcement Details Chapter 13, Guide  General Guidelines:

Do not turn inside corners; Provide a minimum 1/2 in. (13 mm) radius when the sheet is wrapped around outside corners Provide adequate development length Provide sufficient overlap when splicing FRP plies.

79

Allowable Termination Points – Simply Supported Beams Section 13.1.2, Guide

 Plies should extend a

distance equal at least to ldf past the point along the span corresponding to the cracking moment, Mcr,  If Vu > 0.67Vc at the termination point the FRP laminate should be anchored with transverse (“clamping”) reinforcement

80

Bond and Delamination Transverse (“clamping”) reinfocement  Area of transverse (“clamping”) FRP U-Wrap

reinforcement to prevent concrete cover layer from splitting:

A f anchor

81

(A = (E

f f

f fu )longitudinal

κ v ε fu )anchor

(13-1)

Development Length  The bond capacity of FRP is developed over a critical

length:

nE f t f

ldf =0.057

ldf = 82

f

nE f t f f

' c

' c

in in.-lb units (13-2)

in SI units

Detailing of NSM bars  groove dimensions shall be at least 1.5 times the

diameter of the bar  For a rectangular bar the minimum groove size shall be 3ab x 1.5bb

83

Development Length of NSM bars  Development length of NSM bar:

84

ldb =

db f fd 4(0.5τ max )

ldb =

ab d b f fd 2(a b +b b )(0.5τ max )

for circular bars

for rectangular bars

(13-3)

(13-4)

QUESTIONS?

Thank You

85

Design Example Flexural Strengthening of Interior Beam

86

Design Example: Flexural Strengthening of an Interior Beam 12” 24’-0” DL,wLL

21.5”

f’c=5000 psi

24”

3-#9 bars fy=60 ksi

2-12”x 23’-0” FRP plies

φMn=266 k-ft w

FRP

(w/o FRP)

ELEVATION

SECTION

Manufacturer’s reported FRP-system properties

87

Thickness per ply,

0.040 in.

1.016 mm

Ultimate tensile strength

90 ksi

0.62 kN/mm2

Rupture strain,

0.015

0.015

Modulus of elasticity of FRP laminates,

5360 ksi

37 kN/mm2

Design Example: Flexural Strengthening of an Interior Beam

Loadings and corresponding moments Loading/Moment Dead loads,wDL Live load,wLL Unfactored loads, (wDL + wLL) Unstrengthened load limit (1.1wDL +0.75wLL) Factored loads, (1.2wDL +1.6wLL)

Existing loads 1.00 k/ft 14 N/mm 1.20 k/ft 17 N/mm

Anticipated loads 1.00 k/ft 14 N/mm 1.80 k/ft 26 N/mm

2.20 k/ft

32.1 N/mm

2.80 k/ft

40.9 N/mm

n/a

n/a

2.45 k/ft

34.9 N/mm

3.12 k/ft

50.2 N/mm

4.46 k/ft

65.1 N/mm

Dead-load moment,MDL Live-load moment,MLL Service-load moment,Ms Unstrengthened moment limit (1.1MDL +0.75MLL) Factored moment,Mu

72 k-ft 86 k-ft 158 k-ft

96.2 kN-m 114.9 kN-m 211.1 kN-m

72 k-ft 130 k-ft 202 k-ft

96.2 kN-m 173.6 kN-m 269.8 kN-m

n/a

n/a

177 k-ft

240 kN-m

224 k-ft

303.6 kN-m

294.4 k-ft

399.2 kN-m

Two, 12 in. wide by 23 ft. long plies are to be bonded to the soffit of the beam using the wet-lay-up technique.

88

Design Example: Flexural Strengthening of an Interior Beam • Step 1 - Compute the FRP-system design material

properties

For an interior beam, an environmental-reduction factor (CE ) of 0.95 is suggested.

89

f fu = C E f fu*

f fu = (0.95)(90 ksi) = 85ksi

ε fu = CE ε *fu

ε fu = (0.95)(0.0 15in./in.) = 0.0142in./ in.

Design Example: Flexural Strengthening of an Interior Beam • Step 2 - Preliminary calculations Properties of the concrete: β1 from ACI 318-05, Section 10.2.7.3. β1 = 1.05 − 0.05

f 'c = 0.80 1000

Ec = 57,000 5000 psi = 4,030,000 psi

90

Design Example: Flexural Strengthening of an Interior Beam

• Step 2 - Preliminary calculations Properties of existing reinforcing steel: As = 3(1.00 in. 2 ) = 3.00 in. 2 As ρs ≡ bd

91

3.00 in. 2 ρs = = 0.0116 (12 in. )(21.5 in. )

Design Example: Flexural Strengthening of an Interior Beam • Step 2 - Preliminary calculations Properties of the externally bonded FRP reinforcement:

A f = nt f w f ρf ≡

92

Af bd

A f = (2 plies)(0.040 in. ply )(12 in.) = 0.96in.2 0.96 in.2 ρf = = 0.00372 (12 in.)(21.5 in.)

Design Example: Flexural Strengthening of an Interior Beam • Step 3 - Determine the existing state of the strain on the soffit The existing state of strain is calculated assuming the beam is cracked and the only loads acting on the beam at the time of the FRP installation are dead loads. A cracked section analysis of the existing beam gives k=0.334 and Icr=5937 in.4 M DL (h − kd ) ε bi = I cr Ec

93

ε bi =

(864 k ⋅ in.)[24 in. − (0.334 )(21.5 in.)] (5,937 in.4 )(4,030 ksi )

ε bi = 0.00061

Design Example: Flexural Strengthening of an Interior Beam

• Step 4 – Determine the design strain of the FRP System

ε fd

5000 psi = 0.083 ≤ 0.9 ε fu 2 (5360000 psi ) (0.04in )

ε fd = 0.0113 ≤ 0.9(0.0142) = 0.0128

94

Design Example: Flexural Strengthening of an Interior Beam

• Step 5 - Estimate c, the depth to the neutral axis

A reasonable initial estimate of c is 0.20d. The value of c is adjusted after checking equilibrium. c = 0.20d

95

c = (0.20)(21.5 in.) = 4.30 in.

Design Example: Flexural Strengthening of an Interior Beam • Step 6 - Determine the effective level of strain in the FRP reinforcement ⎛df −c⎞ ⎟⎟ − ε bi ≤ ε fd ⎝ c ⎠

ε fe = 0.003 ⎜⎜

⎛ 24 − 4.3 ⎞ ⎟ − 0.00061 ≤ 0.009 4 . 3 ⎝ ⎠

ε fe = 0.003 ⎜

ε fe = 0.0131 ≤ 0.009 ε fe = 0.009

96

Design Example: Flexural Strengthening of an Interior Beam

Since FRP controls the section failure, the concrete strain is less than 0.003:

ε c = (ε f e + ε bi )

⎛ c ⎞ ⎜ ⎟ ⎜d −c⎟ ⎝ f ⎠

⎛ 4 .3 ⎞ ⎟ = 0.0021 ⎝ 24 − 4.3 ⎠

ε c = (0.009 + 0.00061) ⎜

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Design Example: Flexural Strengthening of an Interior Beam

• Step 7 - Calculate the strain in the existing reinforcing steel

⎛ d −c ⎞ ⎟ ε s = (ε fe + ε bi ) ⎜⎜ ⎟ d − c f ⎝ ⎠ ⎛ 21.5 − 4.30 ⎞ ε s = (0.009 + 0.00061) ⎜ ⎟ = 0.0084 ⎝ 24 − 4.30 ⎠

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Design Example: Flexural Strengthening of an Interior Beam

• Step 8 - Calculate the stress level in the reinforcing steel and FRP

f s = Es ε s ≤ f y

f fe = E f ε fe

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f s = (29,000 ksi)(0.0084) ≤ 60 ksi f s = 348 ksi ≤ 60 ksi

f fe = (5,360 ksi )(0.009 ) = 48.2 ksi

Design Example: Flexural Strengthening of an Interior Beam • Step 9a - Calculate the internal force resultants Approximate stress block factors may be calculated using the parabolic stress-strain relationship of concrete as follows:

1.7 f c' 1.7 (5,000) ε = = = 0.0021 6 Ec 4,030 ×10 ' c

4ε c' − ε c 4(0.0021) − 0.0021 β1 = ' = = 0.749 6ε c − 2ε c 6(0.0021) − 2(0.0021) 3ε c' ε c − ε c2 3(0.0021) (0.0021) α1 = = = 0.886 '2 2 3γε c 3(0.749) (0.0021) 100

Design Example: Flexural Strengthening of an Interior Beam • Step 9b – Check equilibrium Force equilibrium is verified by checking the initial estimate of the neutral axis, c

c=

As f s + A f A fe

α f c'cβ1 b 1

=

(3)(60) + (0.96)(48.2) = 5.87in (0.886)(5)(0.749)(12)

c = 5.87in ≠ 4.30in

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NG

Design Example: Flexural Strengthening of an Interior Beam • Step 10 – Iterate on c until force equilibrium is satisfied c = 5.17 in. ε s = 0.0083 f s = f y = 60 ksi

(3.00 in. )(60 ksi) + (0.96 in. )(48.2 ksi) c= 2

2

(0.928)(5 ksi)(0.786)(12 in.)

ε fd = 0.009 f fe = 49.8 ksi

c = 5.17 in. = OK .

ε c = 0.0027 β1 = 0.786 α = 0.928 1

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The value of c selected for the final iteration is correct.

Design Example: Flexural Strengthening of an Interior Beam • Step 11 – Calculate reinforcement and FRP contribution to strength

0.786(5.17) ⎞ β1c ⎞ ⎛ ⎛ M ns = As f s ⎜ d − ⎟ = (3.00)(60)⎜ 21.5 − ⎟ = 3,504k − in = 292k − ft 2 2 ⎝ ⎠ ⎝ ⎠

β c⎞ 0.786(5.17) ⎞ ⎛ ⎛ M sf = A f f fe ⎜ d f − 1 ⎟ = (0.96)(48.2)⎜ 24 − ⎟ = 1,017 k − in = 85 k − ft 2 ⎠ 2 ⎠ ⎝ ⎝

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Design Example: Flexural Strengthening of an Interior Beam • Step 11 – Calculate design flexural strength of the section The flexural strength is calculated using the reduction factor. Since εs =0.0083>0.005, the value of Ф is 0.9

φ M n = φ [M ns + ψ M nf ]= 0.9(292 + 0.85(85) ) = 327 k − ft φ M n = 327 k − ft ≥ M u = 294 k − ft The strengthened section is capable of sustaining the new required moment strength 104

Design Example: Flexural Strengthening of an Interior Beam • Step 12 – Check service stresses in the reinforcing steel and the FRP

Calculate the elastic depth to the cracked neutral axis by summing the first moment of the areas of the transformed section.

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Design Example: Flexural Strengthening of an Interior Beam • Step 13 – Check service stresses in the reinforcing steel and the FRP

2

Ef ⎞ E f ⎛ h ⎞ ⎞ ⎛ Es Ef ⎞ ⎛ Es ⎛ Es ⎟⎟ ⎟⎟ + 2⎜⎜ ρ s + ρf + ρf + ρf k = ⎜⎜ ρ s ⎜ ⎟ ⎟⎟ − ⎜⎜ ρ s Ec ⎠ Ec ⎝ d ⎠ ⎠ ⎝ Ec Ec ⎠ ⎝ Ec ⎝ Ec

k = 0.343 kd = (0.343)(21.5 in.) = 7.37 in.

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Design Example: Flexural Strengthening of an Interior Beam • Step 13 – Check service stresses in the reinforcing steel and the FRP Calculate the stress level in the reinforcing steel:

f s,s

⎡ kd ⎞⎤ ⎛ M A E h ε + − ⎟⎥ (d − kd )Es bi f f ⎜ ⎢ s 3 ⎠⎦ ⎝ ⎣ ≤ 0. 8 f y = kd ⎞ kd ⎞ ⎛ ⎛ As Es ⎜ d − ⎟(d − kd ) + A f E f ⎜ d f − ⎟(d f − kd ) 3 ⎠ 3 ⎠ ⎝ ⎝

f s , s = 40.4 ksi ≤ (0.80)(60 ksi ) = 48 ksi OK The stress level in the reinforcing steel is within the recommended limit 107

Design Example: Flexural Strengthening of an Interior Beam • Step 14 – Check creep rupture limit t service for the FRP Calculate the stress level in the FRP:

f f ,s

⎛ Ef = f s , s ⎜⎜ ⎝ Es

⎞⎛ d f − kd ⎞ ⎟⎟⎜⎜ ⎟⎟ − ε bi E f ≤ 0.55 f fu ⎠⎝ d − kd ⎠

f f , s = 5.60 ksi ≤ (0.55)(85) = 50 ksi OK

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Design Example: Flexural Strengthening of an Interior Beam • Step 14 – Detailing Requirements Detail the FRP reinforcement as follow: 1. Check that shear force at termination is less than shear force that causes end-peeling (estimate as 2/3 of concrete shear strength). 2. Terminate FRP at ldf (per Eq. 12.2) past cracking moment. a) If shear force is higher extend FRP beyond and/or use FRP U-wraps.

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