Resistenza a taglio di travi in calcestruzzo
fibrorinforzato
Giovanni Plizzari
Università degli Studi di Brescia
giovanni.plizzari@unibs.it
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Giovanni Plizzari
Outlines
 Shear Action
 Factor affecting the shear strength
 Experimental tests on PC and FRC beams
 Wide-shallow beams in FRC
 Shear Design of FRC beams
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Giovanni Plizzari
Optimized reinforcement: definition
Place the best performing reinforcement
(fibers and/or rebars) where required by
tensile stresses in the structural elements
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Reinforcement use in structural elements
• In structural elements both distributed and localized
stresses are generally present
• Conventional rebars represent the best
reinforcement for localized stresses
• Fibers represent the best reinforcement for diffused
stresses
• Structural optimization generally requires the use of
a combination of rebars and fibers
• Structural ductility is generally enhanced
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Mechanisms of Shear Transfer
Uncracked concrete zone
Unterface shear transfer
Dowel action
Residual tensile stress across
cracks
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Giovanni Plizzari
Shear in beams without stirrups
In FRC elements there is an additional contribution to shear
resistance provided by fiber reinforcement:
V = Vc + Vf
Vc represents the concrete contribution.
Vf represents the fiber contribution (post cracking strength).
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Shear in beams with stirrups
V = Vc + Vw + Vf
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Factor affecting the shear strength
Giovanni Plizzari
Tensile strength of concrete
Longitudinal reinforcement ratio
Shear span-to-Depth ratio (a/d)
Axial forces
Coarse aggregate size
Size effect
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Giovanni Plizzari
Factors affecting the Shear Strength
Distribution of longitudinal reinforcement along beam
height (fiber?)
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Size effect
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Experimental campaign on PC and FRC beams
V
V
Campaign I
2Ø24 Bars, L=4550 mm
480 mm
d
a
4350 mm
200
8 beams NSC
3 beams HSC
45
480
Steel Plate
200x90x30 mm
2Ø24 Deformed
Bars
45
a/d=2.5
Longitudinal reinforcement
ratio 1%
No stirrups
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Experimental campaign on PC and FRC beams
400
Normal Strength Concrete, f'c = 24.8 MPa
30 kg/m3,
30/0.6
30 kg/m3, 30/0.6 + 15 kg/m3, 12/0.18
30
200
V
100
NSC1-PC
NSC1-FRC1
NSC1-FRC2
V

30
Ø 0.38
Load [kN]
300
0
60
f’c = 24.8 MPa30
3
30 kg/m3,
30/0.6
+
15 kg/m3,
12/0.18
30 kg/m3,
30/0.6
Fibers:
30
0.38% macro-fibres, 30 mm long
with an aspect ratio of
50
50
V
1
TPT
V
V
NSC1-PC
NSC1-FRC1
NSC1-FRC2
0.19% micro-fibres, 12 mm long
12
with an aspect
ratio of 66.7
0
0
100
200

Load
[kN]
300
12
Ø 1.0
V
Average First
CPT
Cracking
30
Ø 0.18
Crack Width [mm]
Normal Strength Concrete, f'c = 24.8 MPa
2
30
80
Ø 0.18
40
Displacement [mm]
Ø 0.62
20
Ø 0.6
0
400
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Experimental campaign on PC and FRC beams
500
50 kg/m3, 80/30
Fibers:
50 kg/m3, 45/30
300
30
Ø 0.62
0.6% macro-fibers,
200
Normal strength fibers
(45/30)
HSC-PC
HSC-FRC1
HSC-FRC2
100

0
0
20
40
Displacement [mm]
60
80
V
CPT
HSC-PC
HSC-FRC1
HSC-FRC2
1
0.5
V
50 kg/m3,
45/30
A relative small amount of
fibers signifiantly enhances
both the shear capacity and the
ductility, both in case of normal
strength concretes and high
strength concretes
V
V
12
TPT
1.5
50
50
3

30
Ø 0.18
Crack Width [mm]
Average First
Cracking
fibers
30
High Strength Concrete
2
30
Ø 0.18
2.5
High strength
(80/30)
30
ØØ0.38
0.6
V
Ø 0.62
V
50 kg/m ,
80/30
12
0
0
100
200
300
Load [kN]
400
500
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Ø 0.6
Load [kN]
400
Ø 0.38
f’c = 60 MPa.
High Strength Concrete
Giovanni Plizzari
Experimental campaign on PC and FRC beams
Fibre 50/1.0
No fibre
PC beams: very brittle and sudden failure
FRC: prior warning of impending collapse,
more distributed crack pattern
Fibre 80/30
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Final crack patterns
PC
Giovanni Plizzari
Fibres 45/30
90 kN
150 kN
Failure
Fibres control and stabilize
the shear crack propagation
The combination of macrofibres with micro-fibres lead
to a better crack control at
small and high values of
crack opening.
Fibres 80/30
90 kN
150 kN
Failure
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Size effect
Beams without web reinforcement fail
when inclined cracking occurs or shortly
afterwards.
The inclined cracking load of a beam is
affected by many principal variables. One
of these is the size of beam (size effect).
An increase in the effective depth of a
beam results in a decrease of the shear
bearing capacity.
d
vu =
Vu
bw  d
Ohio shear collapse (1955)
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Experimental campaign II
a
a
d
d
2Ø24 Bars, L=4550 mm
4350 mm
200
L=4600
mm
6Ø206Ø20
Bars, Bars,
L=4600
mm
4550 mm
4550 mm
4600 mm
4600 mm
Steel Plate
200x90x30 mm
200
200
2Ø24 Deformed
Bars
Steel Plate
200x90x30 mm
1000
45
1000
45
480
P
P
1000 mm
V
1000 mm
V
480 mm
d
a
45
11 beams
Steel Plate
6Ø20
Deformedmm
Bars
200x90x30
45
6Ø20 Deformed Bars
a/d=2.5
Longitudinal reinforcement
ratio 1%
NSC e HSC (f’c =25.7 and 55
MPa)
H=500 mm and H=1000 mm
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Giovanni Plizzari
Experimental campaign II
500
800
Load Displacement Curve
Load-Displacement Curve
Small Size Specimens H = 500 mm; fc = 25.7 MPa
Large Size Specimens H = 1000 mm; fc = 25.7 MPa
400
300
P
200
Load [kN]
Load [kN]
600
d
PC-100
MSR-100
PC-50
FRC-50 20 kg/cm Test 1
FRC-50 20 kg/cm Test 2
MSR-50 Test 1
MSR-50 Test 2
100
400
200
P
FRC-100 20 kg/cm
d
0
0
0
3
6
9
12
15
0
3
6
9
12
15
Displacement [mm]
Displacement [mm]
Small and Large Size Specimens, f’c=25.7 MPa.
FRC beams: Steel fibres 0.25% l=50 mm Aspect ratio= 50
Minimum shear reinforcement: MSR-50 with 2f8@300mm,
2f8@650mm
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MSR-100 with
18
Giovanni Plizzari
Experimental campaign III
1000
5
Crack
Width
[mm]
4
Large Size Specimens
Load
[kN]
H = 1000 mm; fc = 55 MPa
800
P
Large Size Specimens
H = 1000 mm; fc = 55 MPa
P
TPT5
TPT2
d
TPT4
TPT1
P
600
3
TPT6
400
TPT3
2
H-PC-100
H-PC-100
H-MSR-100
H-MSR-100
H-FRC-100
200
1
H-FRC-100
Displacement [mm]
Load [kN]
0
0
0
5
10
15
20
0
200
400
600
800
1000
Large Size Specimens, f’c=55.0 MPa.
The same amount of fibers
Stable propagation of cracks, higher ductility and shear capacity, prior
warning of impeding collapse.
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Experimental campaign IV
SPECIMEN
PC-50
MSR-50 1
MSR-50 2
FRC-50 1
FRC-50 2
PC-100
MSR-100 NSC
FRC-100
PC-100
MSR-100 HSC
FRC-100
Pu
vu
[kN] [MPa]
216
1.22
346
1.93
302
1.69
388
2.16
308
1.72
365
1.07
635
1.81
494
1.42
393
1.14
880
2.48
656
1.86
vu/(fc)1/2
u
Vu/Vu,FL
[-]
0.24
0.38
0.33
0.43
0.34
0.21
0.36
0.28
0.15
0.33
0.25
[mm]
2.74
9.33
7.03
10.95
4.77
7.60
12.60
11.05
9.79
18.62
12.01
[-]
0.52
0.82
0.72
0.92
0.73
0.43
0.73
0.57
0.45
0.99
0.74
Fibres can completely substitute the
minimum amount of shear
reinforcement
20-30 kg/m3 is the amount required
in many structural applications
Fibres resulted less effective than the
minimum amount of shear reinforcement
on the size effect.
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PC H100
MSR H100
FRC-20 H100
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Giovanni Plizzari
Numerical Analyses NFEA
1.4
Size Effect on PC and FRC Members
Vu,num
v=
bw  d
gc= 1, rs= 1%, NSC f'c= 22 MPa, HSC f'c= 58.3 MPa
1.2
Skin Reinforcement Recommended
v/vmax,flex [-]
1
M u,flex
vmax,flex =
2.5  b w  d 2
0.8
feq(0.6-3)
0.6
HSC-FRC3
NSC-FRC2
NSC-FRC1
NSC-PC
exp PC
exp FRC
0.4
0.2
0
0
500
1000
1500
2000
2500
Effective Depth [mm]
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d from 250 to 2000
mm.
Increasing the fiber
toughness, the size
effect influence
decreases
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Giovanni Plizzari
Experimental Program on DEEP BEAMS
H=500 mm
9 full scale beams
a/d=3.0
ρ= 1%
NSC
(fcm =35 MPa)
H=1000 mm
H=500 mm
H=1000 mm
H=1500 mm
H=1500 mm
0, 50 or 75 kg/m3 of steel
fibers (SFRC)
No stirrups anywhere
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Deep beams: Experimental results
Fibers enhance the ductility and the shear bearing capacity (50 kg/m3
of fibers doubled the shear strength).
ULS
Fibers increase the load at which the shear crack becomes unstable.
PC - max shear crack = 0.2-0.5 mm; FRC - max shear crack = 3-4 mm
Beams H1000
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Deep beams: Experimental results
Fibers enhance the behavior at Serviceability Limit State as well.
A significant enhanced post-cracking stiffness is observed in FRC beams,
which is mainly due to the bridging effect of fibers.
SLS
Fibers improve the durability of structural elements
Beams H1500
Beams
H1500
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Deep beams: Crack patterns
In FRC elements:
MULTIPLE SHEAR CRACKS
STABLE SHEAR CRACKS
Progressive formation of multiple cracks, with a
stable propagation.
Fibers are highly effective in controlling
development and propagation of cracking
Fibers
more distributed crack pattern, with
more closely spaced and smaller cracks
H1500 FRC50
H=1500 mm
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Deep beams in PFRC: Experimental results
Unlike MSR samples (rmin=0.17%),
shear failure was observed in
PFRC beams.
PP fibres enhanced the ultimate
shear strength and ductility of deep
beams, which are much more
critical in shear than WSBs.
13 kg/m3 of PP fibres was able to
double both the shear strength and
the ductility, as compared to control
samples
PP fibres were found to be effective
also in presence of prestressing
Vu
vu
vu/(fcm)1/2
[kN]
[MPa]
[-]
Shear
183
0.80
0.15
MSR 300x800-1
Flexure
424
1.86
0.34
MSR 300x800-2
Flexure
436
1.91
0.35
PFRC 300x800-1
Shear
381
1.67
0.29
PFRC 300x800-2
Shear
405
1.77
0.30
PC 150x800-1
Shear
91
0.80
0.15
PC 150x800-2
Shear
101
0.88
0.16
MSR 150x800-1
Flexure
244
2.13
0.39
MSR 150x800-2
Flexure
250
2.18
0.40
PFRC 150x800-1
Shear
205
1.79
0.31
PFRC 150x800-2
Shear
247
2.16
0.37
PC 150x800 PT-1
Shear
199
1.73
0.31
PFRC 150x800 PT-1
Shear*
284
2.49
0.42
PC 150x600-1
Shear
89
1.05
0.19
PC 150x600-2
Shear
64
0.76
0.14
MSR 150x600-1
Flexure
185
2.19
0.40
MSR 150x600-2
Flexure
204
2.42
0.44
PFRC 150x600-1
Shear
166
1.96
0.34
PFRC 150x600-2
Shear
198
2.35
0.40
Specimen
PC 300x800-1
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Failure mode
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Giovanni Plizzari
Deep beams in PFRC: Experimental results
PP fibres and MSR increase the load at which shear crack becomes
instable
PP fibres have been able to slightly increase the post-cracking stiffness
Flexural crack spacing was 20% smaller in PFRC deep beams
1000
8
Deep Beams 300x800
Load
900
[kN]
d=761 mm
800
6
5
PC 300x800-1
MSR 300x800-1
500
MSR 300x800-2
400
PFRC 300x800-1
d=761 mm
PC 300x800-1
MSR 300x800-1
MSR 300x800-2
PFRC 300x800-1
PFRC 300x800-2
700
600
Deep Beams 300x800
Crack
Width
7
[mm]
4
P/2
V
3
PFRC 300x800-2
300
P/2
2
P/2
200
1
100
Mid-span deflection [mm]
Load [kN]
0
0
0
5
10
15
20
25
30
35
40
45
0
200
400
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600
800
1000
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Prestressed Double tees
6 full-scale double tees
H=500mm; d=390 mm;
L=6000mm
bw=120 mm
3 tendons 0.6”
ρl=0.89%
2 double tees WR1 DT
1 double tees WR2+SCPFRC DT
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Prestressed Double Tees
Giovanni Plizzari
3 double tees SCPFRC-DT
10kg/m3 of PP fibres (Vf=1.10%)
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Tests in zone with uniform prestressing
3 experimental tests
a/d ≈ 3
WR1 DT-1 (RC)
SCPFRC DT-1 (only PP fibres)
SCPFRC DT-2 (only PP fibres)
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Tests in zone with uniform prestressing
All specimens showed a shear
failure
+10-15%
+70-240%
Both PP fibres and MSR led
to:
- An increment of the shear
capacity of about 10-15%
- An increment of the ductility
at least of 70%.
Shear crack
PP fibres can completely
substitute
the
minimum
amount of shear reinforcement
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Tests in the end zone
Giovanni Plizzari
6 experimental tests
a/d ≈ 3
WR1 DT-2 a/b (RC)
SCPFRC DT-3 a/b (only PP fibres)
WR2+SCPFRC DT a/b (RC + PP fibers)
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Tests in the end zone
+15%
The more effective solution in
the end zones seems to be the
combination of PP fibres and
conventional
reinforcement
(steel wire mesh Ø5 200x300),
in which after shear cracking a
stable behaviour with an
increment (+20%) of the
ultimate bearing capacity was
observed.
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Wide-Shallow Beams (WSBs)
Typical structure typology of residential building in: Southern Europe,
Australia, Middle-East, Central and South-America.
Beam
Topping concrete layer
RC spandrel
wide-shallow
beam
Lightweight ribbed
one-way reinforced
concrete slab
RC central
wide-shallow
d
beam
b
d≥b
d
b
Wide-shallow
beam
d
2d ≤ b
b
Benefits: floors completely flat, more architectural flexibility, easier
formwork
Disadvantages: deformability, small net span, transfer of load to column.
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Experimental campaigns on WSBs
FIRST EXP. CAMPAIGN
SECOND EXP. CAMPAIGN
16 full scale WSBs
14 full scale WSBs
Main purposes:
1) Study both shear and flexure
behavior of WSBs
2) Evaluate the possibility of
completely substitute the
minimum amount of shear
reinforcement by steel fibers
(SFRC)
Main purposes:
1) Study the shear behavior of
PC wide-shallow beams
2) Evaluate structural polymer
fibers in full scale tests
3) Replace the minimum shear
reinforcement by polymer fibers
(PFRC)
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Giovanni Plizzari
SFRC and PFRC WSBs: Experimental results
Either minimum shear reinforcement or fibers greatly influences the shear
behavior of wide-shallow beams (by altering the collapse from shear to
flexure, with enhanced bearing capacity and ductility).
Both STEEL FIBERS and POLYPROPYLENE FIBERS have been able to
completely substitute the minimum amount of transverse reinforcement.
13 kg/m3
SFRC
PFRC
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Giovanni Plizzari
FRC WSBs: Flexure behavior
Steel fibers even in small amount, provides a greater flexure bearing
capacity and increases the overall ductility under flexure, due to the
positive effect in increasing the compression softening.
In the case of PFRC WSBs this behavior is less pronounced.
μδ=4.3
SFRC
(+6%)
(+10%)
+40%
μδ=6.1
(+7%)
(+10%)
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Giovanni Plizzari
Shear design of FRC elements
Shear strength without shear reinforcement
 0.18

1
3
VRd  
 k  (100  r1  f ck )  0.15   CP   bW  d
 gc

Modified in MC2010 in order to consider fibres contribution
- Fibres contribution included in the shear concrete strength
- Fibres contribution is taken into account considering the residual strengths
according to EN14651 (3PBT)
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Shear design of FRC elements
Giovanni Plizzari
3
 0.18 



f Ftuk 



 k  100  r1  1  7.5 

f

0
.
15


 ck 
cp   bW  d
f ctk 



 g c

1
VRd,F
[tensio
fFtuk
characteristic value of the residual
strength (SLU), considering wu = 1.5 mm, in
MPa;
f Ftu
f R3

3
This model has been adopted by Model Code 2010 (MC2010)
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Shear Strength vs. Toughness
2.40
Range used
in structural
application
New model for FRC w/o stirrups
2.20
Vu,FRC/Vu,EC2 [-]
2.00
1.80
1.60
fctm = 2.0 MPa
fctm = 2.5 MPa
1.40
fctm = 3.0 MPa
fctm = 3.5 MPa
1.20
[MPa]
eq(0.6-3)
MPa

ffFtu
fctm = 4.0 MPa
fctm = 5.0 MPa
1.00
0
1
2
3
4
5
6
7
8
9
The shear strength significantly increases using fibres
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MC2010 model vs experimental results
2.50
fib vs. RILEM
2.25
2.00
RILEM DATA
Conservative
BRESCIA DATA
Vu,exp/Vu,model
1.75
1.50
1.25
1.00
0.75
0.50
fib
Unconservative
RILEM
0.25
0.00
0
5
10
15
20
25
30
35
40
45
50
55
60
Test #
MC2010 model seems to be consistent
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Minimum shear reinforcement
1.75
• Crack control
• Ductility
• Warning of impending
collapse
Minimum FRC Transverse Reinforcement
1.50
f eq , min  0 . 7 
feq(0.6-3) [MPa]
1.25
f ck
4 .5
Calculation
Design Formulation
Best Fitting Formulation
1.00
0.75
0.50
r w,min
0.08  f ck

f yk
f eq ,min
0.25
3



0.008  f ck
f ct 



 1  1

2.5  0.18  k  100  r  f ck 1 / 3




0.00
20
V Rd ,ct , FIBERSVRd ,ct  0.08 
f eq,min
f ck  bw  d
3



0
.
008

f
f ct 
ck



 1  1
1/ 3


2.5  0.18  k  100  r  f ck 




40
60
100
Compressive Strength [MPa]
f Ftuk 
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f ck
20
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Giovanni Plizzari
MC2010 – Shear Workshop Salò Ottobre 2010
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Example of Application for Shear: I
2Ø24 Bars
500
200
500 mm
d
p = 35 kN/m
2Ø24 Deformed
Bars
6m
1
1
2
M max   p  l   35  62  157.5 kN  m
8
8
h  500 mm; d  460 mm
1
1
Vmax   p  l   35  6  105 kN
f ck  30 MPa; f yk  500 MPa
2
2
g c  1.5; g s  1.15
As
904 mm 2
rl 

 0.98%
30
500
bw  d 200 mm  460 mm
f cd 
 20 MPa; f yk 
 435 MPa
1.5
1.5
M u  161 kN  m
f ctk  2 MPa ( EC 2)
pu  35 kN / m (ULS )
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Example of Application for Shear: I
VRd ,ct
 0.18

1
3

 k  (100  r1  f ck )  0.15   CP   bW  d  49 kN
 gc

Minimum Shear Reinforcement
1.6
1.4
3.2 meters requiring design shear reinforcement; 2.8 meters requiring
minimum shear reinforcement.
Design Shear Reinforcement:
Minimum Shear Reinforcement:
s  0.75  d  345 mm
Asw
r w,min  0.08
f ck
f yk
 2 6 @ 300 mm
VR ,ds 
 0.0009
 z  f yd  VRd  VRd ,ct  56 kN
s
s  321 mm
 28@ 300 mm
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Example of Application for Shear: I
Giovanni Plizzari
Assume 30 kg/m3 of steel fibers having l/f =67 and fFtk,u=0.90 MPa (tested at the
University of Brescia)
f Ftk ,u
 0.18

1
VRd , F  
 k  (100  r1  (1  7.5 
)  f ck ) 3  0.15   CP   bW  d
f ctk
 gc

VRd , F
 0.18 

200 
0.90
1
3

 1 
)  20)   200  460  81 kN
  (100  0.0098  (1  7.5 
460 
2
 1.5 

Minimum shear reinforcement
f Ftuk 
f ck
20

Minimum Shear Reinforcement
30
 0.27 MPa
OK
20
Design Shear Reinforcement
0.7
2.3
A

VR ,ds  sw  z  f yd  VRd  VRd ,ct  24 kN 
s
  26@ 300 mm

s  420 mm
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Example of Application for Shear: I
2Ø8@300mm
2Ø6@300mm
Giovanni Plizzari
Plain concrete
2Ø6@300mm
FRC
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Giovanni Plizzari
Example of Application for Shear: WSBs
500 cm
500 cm
g = 40 kN/m
WSBs
q = 10 kN/m
H = 30 cm
Concrete
B450C
h  500 mm; C30/37
d  460 mm
d = 26 cm
ϒfck  30 MPa; f yk  500 MPa
b = 80 cm
pu  35 kN / m (ULS )
g c  1.5; g s  1.15
30
500
f cd 
 20 MPa; f yk 
 435 MPa
1.5
1.5
f ctk  2 MPa ( EC 2)
80
ϒ
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d=26cm
40+10 kN/m
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Giovanni Plizzari
Example of Application for Shear: WSBs
Shear design according to Level I of Model Code 2010
(inclination of the compressive stress field of 30°).
MSR
MSR
Steel incidence on concrete volume: 145 kg/m3
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Giovanni Plizzari
Example of Application for Shear: WSBs
FRC classified as 3c (fR1k=3MPa; fR3k=3MPa, i.e. 35 kg/m3
of steel fibres 60 mm long with diameter of 0.75 mm)
MSR by fibres
MSR by fibres
Steel incidence on concrete volume: 115 kg/m3
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Giovanni Plizzari
Thank you for your kind attention
University of Brescia
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Giovanni Plizzari
Examples of Shear Design: II
500
160
L=10 m
160
120
680
40
1000
p (UDL)
1- Given material properties (NSC) and reinf. details, calculation of Mrd;
2- Corresponding ultimate load (UDL) pu=8Mrd/l2;
3- Design ultimate shear Vrd=1.1 pu l (-pu 1m) where 1.1 is a
incremental safety factor to guard against brittle collapse phenomena;
4- Concrete contribution to shear without fibers, fiber contribution,
concrete+fiber (Vcd,FIBERS) contribution;
5- Asw/s ratio necessary to fully reach the required shear capacity Vrd,
being Vwd=Asw/s 0.9 d fywd = Vrd – Vcd,FIBERS;
6- Transverse reinforcement saving ratio due to fibers as:
 Asw / s FIBRES
S .R.  1 
 Asw / s NO FIBRES
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Examples of Shear Design: II
90
Fibre Contribution to Shear
Prestressed and not-Prestressed Member
80
CNR rho = 0.8%
70
Fibre Contribution [kN]
Parameters:
-prestressing;
-Fiber toughness;
-Reinforcement ratio.
RILE
M
Model
CNR rho = 1%
CNR rho = 1.2%
60
r
CNR rho = 1.4%
CNR rho = 1.6%
50
CNR model is
equal to
MC2010 one
Model
RILEM all rho
40
30
RILEM does not
depends on the
reinforcement ratio.
20
fR3=0.90feq(0.6-3)
10

1
1.5
2
2.5
3
3.5
4
4.5
5
feq(0.6-3) [MPa]
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
V f  0.7  k f  kl  fd  bw  d
0
Linear trend vs. feq
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Giovanni Plizzari
Examples of Shear Design: II
180
Concrete + Fibre Shear Contribution
Prestressed Member
300
Concrete + Fibre Contribution [kN]
Concrete + Fibre Contribution [kN]
350
r
250
200
150
100
CNR
RILEM
Concrete + Fibre Shear Contribution
Not-Prestressed Member
160
140
r
120
100
80
CNR
RILEM
60
50
40
0
1
1.5
2
2.5
3
3.5
4
4.5
5
1
1.5
feq(0.6-3) [MPa]
2
2.5
3
3.5
4
4.5
feq(0.6-3) [MPa]
Good agreement up to feq(0.6-3)<3 MPa, which covers most of practical
applications (for strain-softening materials).
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Giovanni Plizzari
Examples of Shear Design: II
Stirrup Saving ratio
S .R.  1 
 Asw / s FIBRES
 Asw / s NO FIBRES
1.20
0.40
1-(Asw/s,FIBRES)/(Asw/s,NO FIBRES) [-]
0.35
1-(Asw/s,FIBRES)/(Asw/s,NO FIBERS) [-]
Transverse Shear Reinforcement Saving Ratio
Not-Prestressed Member
Transverse Shear Reinforcement Saving Ratio
Prestressed Member
0.30
CNR
0.25
RILEM
0.20
0.15
0.10
r
1.00
CNR
0.80
RILEM
0.60
0.40
r
0.20
0.05
0.00
0.00
1
1.5
2
2.5
3
3.5
4
4.5
5
1
1.5
2
2.5
3.5
3
4
4.5
feq(0.6-3) [MPa]
feq(0.6-3) [MPa]
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Pictures at failure
W770 PC AND W770 FRC-2
PC 15x60 A and FRC 15x80 B
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Examples of Shear Design: II
1.00
0.90
0.80
0.80
0.70
0.70
0.60
0.60
0.50
CNR rho = 0.8%
0.40
L'/L RILEM
Not-Prestressed Member
0.90
L'/L [-]
L'/L [-]
1.00
L'/L CNR
Not-Prestressed Member
0.50
RILEM rho = 0.8%
0.40
RILEM rho = 1%
CNR rho = 1%
0.30
0.30
RILEM rho = 1.2%
CNR rho = 1.2%
0.20
0.10
CNR rho = 1.4%
0.20
CNR rho = 1.6%
0.10
RILEM rho = 1.4%
RILEM rho = 1.6%
0.00
0.00
1
1.5
2
2.5
3
3.5
4
4.5
5
1
1.5
2
2.5
3
3.5
4
4.5
5
feq(0.6-3) [MPa]
feq(0.6-3) [MPa]
L’: Length of the central portion of a beam where no traditional
reinforcement is necessary.
L: Total length.
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Examples of Shear Design: II
2Ø8@300mm
2Ø6@300mm
Plain concrete
2Ø6@300mm
FRC
The introduction of fibers is industrially convenient, cost-effective
and time-saving when the design process can be suitably optimized.
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