1. FRPs are being planned in the potential rehabilitation of a slab-on-girder bridge. The bridge is a one-way slab supported by prestressed concrete girders. Make a recommendation for what type of FRP (e.g., fibre and resin) would be appropriate. 2. Eight different unidirectional FRPs (4 with a fibre volume fraction of 70% and 4 with a fibre volume fraction of 35%) are going to be fabricated. For each FRP, find the modulus in both the fibre direction and the direction perpendicular to the fibres and sketch the stress-strain curve for each FRP. 3. Using the S806 design standard for a building application, calculate the factored moment resistance, Mr, in positive bending, for the precast (φc = 0.70) FRP-reinforced concrete

1. CIVL 838 - Design of Concrete Structures with FRP
Department of Civil Engineering
Date of Submission
10 Feb 2018 (Saturday)
Submitted by
Seyed Mohammad Sadegh (Arshia) Mousavi
20079128
Submitted to
Prof. Mark F. Green
Assignment 1: FRP Properties & R/C Design with FRP
Civil (Structural) Engineering
Queen’s University
Page of1 15

2. Question 1 3
(a) Replacement of the temperature and shrinkage reinforcement in the deck slab 3
(b) Flexural strengthening of the bridge girders 4
(c) Shear strengthening of the bridge girders 6
(d) External post-tensioning of the bridge girders to account for damage to pre-
stressing tendons 7
Question 2 8
a) For each FRP, find the modulus in both the fibre direction and the direction
perpendicular to the fibers. 8
b) Sketch the stress-strain curve for each FRP. 9
c) Comment on the properties of the different FRPs. 9
Question 3 10
Solution 11
Page of2 15

3. Question 1
FRPs are being planned in the potential rehabilitation of a slab-on-girder bridge. The
bridge is a one-way slab supported by pre-stressed concrete girders. Make a
recommendation for what type of FRP (e.g., fiber and resin) would be appropriate for
the following applications:
(a) Replacement of the temperature and shrinkage reinforcement in the deck slab
Shrinkage and thermal stresses in concrete have been known to be principal factors for the
incipient cracking in concrete bridge decks.
FRP: One of the most important points for using the FRP in the structure is the cost
efficiency. Therefore GFRP (Glass FRP) is the best choice for this issue. For instance, the R-
Glass fibers grade has temperature resistance and high tensile strength & modulus or AR-
Glass is alkali resistance.
CFRP composites are corrosion resistant, but CFRP can contribute to increased corrosion of
steel components through galvanic corrosion because it should not be permitted to come in to
direct contact with steel or aluminum in structures. So, we have to use some plastic spacers
for the internal reinforcement or applying a thin layer of epoxy or GFRP sheet between steel
and CFRP bars for the external strengthening.
Also, considering Table 1.1, The thermal expansion coefficient of the GFRP is the most
compatible fibre among the others with concrete.
Table 1.1 - Thermal expansion coefficient of the bars and concrete 1
Bar Type
Thermal expansion
coefficient (1/°C)
Values in the literature
(1/°C)
Concrete 8 x 10-6 7 - 14 x 10-6
Steel 11.3 x 10-6 11.7 x 10-6
GFRP 4.43 x 10-6 6 - 10 x 10-6
CFRP 1.05 x 10-6 (-9) - (0) x 10-6
AFRP -5.2 x 10-6 (-6) - (-2) x 10-6
BFRP 1.92 x 10-6 4 x 10-6
Construction and Building Materials (Experimental investigation of thermal expansion and concrete strength effects on1
FRP bars behavior embedded in concrete), Ferhat Aydin, Sakarya University, Sakarya, Turkey - Elsevier Journal
Page of3 15

4. Resins: Among two different polymers (Thermoplastics & Thermosets), the thermosets
polymers have widely used in the structures and more interesting because it is easier to use in
actual application and manufacturing process. Among the different thermosetting resins,
which are good in different aspects but basically they do not satisfy the requested conditions
which are the temperature and shrinkage. Therefore, Phenolic that it has a good reputation at
high-temperature resistance, low/zero shrinkage and its thermal expansion coefficient (30-45
[10
-6
/°C]) is very close to the concrete (10 [10
-6
/°C]) but it is a very expensive material.
Eventually, we can use Epoxy which has a superior high-temp resistance for this case which
is not expensive and has a good material properties.
(b) Flexural strengthening of the bridge girders
In order to satisfy the flexural strengthening of the bridge girders, several materials have been
used to retrofit the bridges with the wrapping method, but the most common types of fibers
used are Glass and Carbon. Glass fibers have been used with a polyester, vinyl ester, or epoxy
matrix, and carbon fibers are used with epoxy resins. There are many techniques such as wet2
lay-up and CFRP preperg strips and if it is possible filament winding, but the most common
technique is CFRP prepreg strips and easy to handle in strengthening period.
However, according to the following figure (Figure 1.1), the high tensile strength and elastic
modulus of CFRP composites make them ideal candidates for upgrading structures, although
it is necessary to appreciate the possible limitations of their mechanical properties, their
interaction with the substrate and their long-term behavior in harsh environments.
Figure 1.1. Stress - Strain diagram
The use of CFRP for girder strengthening is similar to the attachment of steel plates onto
concrete girders. However, CFRP presents the advantages of easier handling, higher
corrosion resistance, and since fatigue is among the most critical forms of damage potentially
Bridge Engineering Handbook, edited by Wai-Fah Chen, Lian Duan2
Page of4 15

5. occurring in bridges, CFRP has the excellent fatigue behavior - even better than steel - and
GFRPs have satisfactory fatigue resistance. An important factor to be considered when using
this technique is the adhesion between the girder and the CFRP strip. The contact surfaces
must be adequately prepared, and an effective bond must be developed.
Figure 1.2. Strengthening of reinforced concrete girder using CFRP laminate
In principle, reinforcing with CFRP laminates consists of a CFRP strip bonded onto the
tension surface of the girder to increase or restore its original flexural capacity. The CFRP
strip can be applied either non-tensioned or tensioned. The high strength and stiffness of the
CFRP allow the use of very thin layers to achieve the desired capacity.
Incompatible thermal expansion coefficient should be considered in consistent thermal
expansion rate between fibre and resin; and between FRP and concrete.
For instance, the Ste. Emilie de L’Energie Bridge, Quebec, 1999 is an example of this
method.
• Externally-bonded CFRP strips for flexural strengthening of a reinforced concrete
bridge girder
• Light bands are GFRP U-wraps for shear reinforcement and anchorage of the flexural
sheets
Figure 1.3. Ste. Emilie de L’Energie Bridge, Quebec, 1999
Page of5 15

6. (c) Shear strengthening of the bridge girders
According to the different properties of FRPs which are described in the following table, we
can conclude that which FRPs would be the best choice among the others.
Considering table 1.2, carbon fibers have a very high tensile strength and elastic modulus.
The elastic modulus of CFRP is higher than steel but in contrast, it is very low for GFRP.
Table 1.2 - FRPs Properties
Considering the shear strengthening of the bridge girders and several tests have been
performed in this field, CFRP is the best choice for this issue. Since it is a bridge girder there
are many harsh scenarios such as long-term deflection, fatigue failure, or creep-rupture in the
structural component and as discussed before, the durability and strength of CFRP is much
greater than the other materials like GFRPs. Also, among the other shapes of CFRP like T-
shaped or I-shaped, L-shaped is very popular in the shear strengthening of the bridge girders
and indicated an excellent behavior in many experimental tests.
Figure 1.4. Externally-bonded CFRP sheets for shear strengthening of an R/C bridge girder
Reinforcing
Materials
Tensile Strength
(Mpa)
Elastic Modulus
(Gpa)
Steel 400-600 200
CFRP 3500-5200 250-400
GFRP 2400-3100 72.4
Page of6 15

7. (d) External post-tensioning of the bridge girders to account for damage to pre-
stressing tendons
According to the properties of different FRPs such as elastic modulus and strength that
discussed before, CFRP (strips, sheets, L-shaped elements and so on) is the best and only
choice for the strengthening of the bridge girders. GFRP is not proper because based on
research started in Germany in the 70’s, to investigate the use of prestressed GFRP tendons
but it showed to be less appropriate due to its low elastic modulus. In contrast, CFRP has a
high strength to weight ratio, its light weight, high modulus of elasticity, outstanding fatigue
strength, is resistant to chemicals - high durability - but proper spacing should be considered
between steel and CFRP in order to prevent the galvanic corrosion, and the repair methods
are usually inexpensively and rapidly applicable in the field with little to no disturbance to
traffic; the repairs also maintain the over-height clearance and original configuration of the
structure .3
There are different available methods of externally strengthening of the bridge girders by
post-tensioning technique such as CFRP bonded, and un-bonded sheets and, tendons.
Shin and Lee, 20033
Page of7 15

8. Question 2
8 different unidirectional FRPs (4 with a fibre volume fraction of 70% and 4 with a fibre
volume fraction of 30%) are going to be fabricated with the two following fibres and
resins:
a) For each FRP, find the modulus in both the fibre direction and the direction
perpendicular to the fibers.
Properties of the composite depend on the component volume fractions and mechanical
properties.
• Ef - Fiber modulus
• Vf - Fiber volume fraction
• Em - Matrix modulus
• Vm - Matrix volume fraction
Unidirectional Elastic Properties:
- Modulus in the fiber direction - E11
E11= Em Vm + Ef Vf
- Modulus ⊥ to fiber direction - E22
E22= Ef Em / (Em Vf + Ef Vm)
Table 2.1 - Volume Fraction (Vf) Fibre (70%) & Resins (30%)
Page of8 15
Volume
Fraction
Ultimate
Stress
Ultimate
Strain
Elastic
Modulus
Longitudinal
Modulus
Transveral
Modulus
Strain Stress
[%] [Mpa] [%] [Gpa] [Gpa] [Gpa] [%] [Mpa]
Fiber 1 0.7 2200 3.4% 65
Resin 1 0.3 120 3.0% 4.0
Fiber 1 0.7 2200 3.4% 65
Resin 2 0.3 60 2.4% 2.5
Fiber 2 0.7 3300 1.5% 220
Resin 1 0.3 120 3.0% 4.0
Fiber 2 0.7 3300 1.5% 220
Resin 2 0.3 60 2.4% 2.5
FRP 1-1 46.7 1587.83.4%11.7
FRP 1-2 46.3 7.6 3.4% 1572.5
FRP 2-2 154.8 8.1 1.5% 2321.3
FRP 2-1 155.2 12.8 1.5% 2328.0

9. Table 2.2 - Volume Fraction (Vf) Fibre (30%) & Resins (70%)
b) Sketch the stress-strain curve for each FRP.
Figure 2.1. σ-ε diagrams of different Vf of fibers & resins
c) Comment on the properties of the different FRPs.
Considering the results that provided in the above tables and diagram, it is obvious that the
fibre playing an important role and by decreasing the volume fraction from 70% to 30%, the
stiffness of FRP tends to increase more than double. On the other hand, by increasing the
volume fraction, the strength of FRP will be increased.
Page of9 15
Volume
Fraction
Ultimate
Stress
Ultimate
Strain
Elastic
Modulus
Longitudinal
Modulus
Transveral
Modulus
Strain Stress
[%] [Mpa] [%] [Gpa] [Gpa] [Gpa] [%] [Mpa]
Fiber 1 0.3 2200 3.4% 65
Resin 1 0.7 120 3.0% 4.0
Fiber 1 0.3 2200 3.4% 65
Resin 2 0.7 60 2.4% 2.5
Fiber 2 0.3 3300 1.5% 220
Resin 1 0.7 120 3.0% 4.0
Fiber 2 0.3 3300 1.5% 220
Resin 2 0.7 60 2.4% 2.5
FRP 2-1 [R] 68.8 5.7 1.5% 1032.0
FRP 2-2 [R] 67.8 3.6 1.5% 1016.3
FRP 1-1 [R] 22.3 5.6 3.4% 758.2
FRP 1-2 [R] 21.3 3.5 3.4% 722.5
0
1587.8
0
758.2
0
1572.5
0
722.5
0
2328
0
1032.0
0
2321.25
0
1016.3
0
500
1000
1500
2000
2500
0.0% 0.5% 1.0% 1.5% 2.0% 2.5% 3.0% 3.5% 4.0%
σ-Stress
Ɛ - Strain
σ -Ɛ diagram
FRP 1-1
FRP 1-1 [R]
FRP 1-2
FRP 1-2 [R]
FRP 2-1
FRP 2-1 [R]
FRP 2-2
FRP 2-2 [R]
F2(70%)-R1(30%)
F2(70%)-R2(30%)
F2(30%)-R1(70%)
F2(30%)-R2(70%)
F1(30%)-R1(70%)
F1(30%)-R2(70%)
F1(70%)-R1(30%)
F1(70%)-R2(30%)

10. Question 3
Using the S806 design standard for a building application, calculate the factored
moment resistance, Mr, in positive bending, for the precast (φc = 0.70) FRP-reinforced
concrete section shown below. Assume that the beam has an interior exposure condition.
Comment on your results.
Material Properties:
Concrete Compressive Strength, f’c = 50 Mpa
FRP Ultimate Strength, ffrp,u = 1430 MPa
FRP Elastic Modulus, Efrp = 120 GPa
Area of FRP Bars, Abar = 70 mm2
Figure 3.1. Stress block
Page of10 15

11. Solution
Solve for balanced reinforcement ratio (ρfrp, bal)
According to CSA Standards S806-12, we have:
Strains
The ultimate strain at the extreme concrete compression fibre shall be assumed to be 0.0035.
Concrete:
FRP:
Minimum clear concrete cover in R/C members shall be 2dFRP or 30 mm, whichever is
greater.
Effective depth:
Calculate reinforcement Ratio:
Stress Distribution
The real stress block is non-linear. Instead, replace with equivalent rectangular stress block.
The factors α1 and β1 shall be taken as: (S806 Code - 8.4.1.5)
(8-5)
(8-6)
Although, we can assume α1=0.8 and β1=0.9, but in this case those values are obtained from
the aforementioned formulas in order to accuracy.
Balanced reinforcement ratio (ρfrp,bal)
Checking the reinforcement ratios
• When —> Tension Failure (under-reinforced) - FRP Rupture.
The least desirable situation is sudden and violent rapture.
• When —> Balanced - FRP Rupture and Concrete Crushing
• When —> Compression Failure (over-reinforced) - Concrete Crushing.
The most desirable situation is warning of failure.
εc
= εcu
= 0.0035
ε frp
= ε fu
=
f fu
Efrp
d ' = 30 +
dFRP
2
deff
= h − d '
ρfrp
=
Afrp
b.deff
α1
= 0.85− 0.0015 fc
'
≥ 0.67
β1
= 0.97 − 0.0025 fc
'
≥ 0.67
ρfrp,bal
=
Afrp,bal
bd
= α1
β1
φc
φfrp
⎛
⎝
⎜
⎞
⎠
⎟
fc
'
f frp,u
⎛
⎝
⎜
⎞
⎠
⎟
εcu
εcu
+ ε frp,u
⎛
⎝
⎜
⎞
⎠
⎟
ρfrp
< ρfrp,bal
ρfrp
= ρfrp,bal
ρfrp
> ρfrp,bal
Page of11 15

12. Table 3.1 - Type of failure
As it is mentioned in the above table, FRP Rupture has happened. In this case, the least
desirable situation is the sudden and violent rupture and the strain in the FRP is equal to:
εfrp = εfrp,u
The strain in the extreme compression fibre in the concrete:
εc < εcu
Page of12 15
Balanced
b 300 mm
h 600 mm
Фcarbon v-rod 10 mm
f'c 50 Mpa
ffrp,u 1430 Mpa
Efrp 120000 Mpa
Abar 70 mm2
A'frp 140 mm2
Afrp 210 mm2
ФC 0.7 Code
Фfrp 0.75 Code
Ɛc=Ɛcu 0.0035 Assumed
Ɛfrp=Ɛfrp,u 0.01192
d' 35 mm
deff 565 mm
α1 0.78 ≥ 0.67
β1 0.85 ≥ 0.67
ρfrp 0.0012
ρfrp, bal 0.0049
ρfrp/ρfrp, bal 0.26
Reinforcement
Ratio
FRP Rupture
Under Refinforced C.S
S806-12 Code

13. Position of neutral axis:
First suggestion of position of neutral axis is obtained from the following formula:
Calculate Ɛc:
By knowing Ɛc, we can find α and β from Table B.2 or Figures 6.4 and 6.5 of M3-Design
Manual:
New α and β
Computing the resultant compressive and tensile forces:
Bottom Reinforcement:
Top Reinforcement:
Ɛfrp,st: Top FRP bars strain
Concrete:
Equilibrium Check:
Tfrp,sb + Tfrp,st - Cc = 0
If the equilibrium condition is not satisfied, the new position of neutral axis should be
selected and the values of α and β can be adopted from Table B.2 or Figures 6.4 and 6.5 of
M3-Design Manual.
cb
=
εcu
εcu
+ ε frp,u
× d
C =
ρfrp
ρfrp,bal
× cb
C
deff
=
εc
εc
+ ε frp,u
→ εc
=
C
deff
− C
× ε frp,u
Tfrp,sb
=φfrpε frp,u
Efrp
Afrp
Tfrp,st
=φfrpε frp,st
Efrp
A'
frp
ε frp,st
=
C − d '
C
⎛
⎝⎜
⎞
⎠⎟ × εc
Cc
= −φcα1 fc
'
β1C
a
! "#
b
Page of13 15

14. Table 3.2 - 1st iteration (Assumed C)
* minus means, the resultant force is compression force.
Based on the first assumption of position of neutral axis, which is 25.65 mm, the top FRP bars
are located in the tension zone. but since the equilibrium is not satisfied, the assumed value of
neutral axis should be changed and iterate the procedures until the equilibrium condition is
satisfied. Therefore, all these equations and computations are defined in an excel spreadsheet
in order to save time and accuracy.
Table 3.3 - 2nd iteration
The new value of C has been considered, hence the top FRP bars are positioned in the
compression zone. Even in this case, the equilibrium is not satisfied and the next iteration
should be calculated.
Page of14 15
2nd Iteration
C 50.00 mm
Ɛc 0.00116
Ɛst 0.00035
α1 0.67 ≥ 0.67
β1 0.67 ≥ 0.67
Tfrp, sb 225.23 kN
Tfrp, st -4.373 kN
Cc -235.67 kN
EQ -14.82 ≠ 0
EQ Check EQ Not Satsfied
S806-12 Code
1st Iteration
C 25.65 mm
Ɛc 0.00054
Ɛst 0.00149
α1 0.68 ≥ 0.67
β1 0.67 ≥ 0.67
Tfrp, sb 225.23 kN
Tfrp, st 18.71 kN
Cc -122.72 kN
EQ 121.22 ≠ 0
EQ Check EQ Not Satsfied
S806-12 Code

15. Eventually, finding the moment resistance.
Table 3.4 - 3rd iteration & resistance bending moment of cross section
After many iterations, the exact position of neutral axis is about 47.04 mm and the moments
resistance of cross section, which is obtained from the above equation, is equal to 123.64
kN.m.
Mr
= Tfrp,sb
d −
a
2
⎛
⎝⎜
⎞
⎠⎟ +Tfrp,st
d '−
a
2
⎛
⎝⎜
⎞
⎠⎟
Page of15 15
3rd Iteration
C 47.04 mm
Ɛc 0.00108
Ɛst 0.00028
α1 0.67 ≥ 0.67
β1 0.67 ≥ 0.67
Tfrp, sb 225.23 kN
Tfrp, st -3.491 kN
Cc -221.73 kN
EQ 0.00 = 0
EQ Check EQ Satisfied
Moment
Resistance (Mr)
123.64 kN.m
S806-12 Code