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V
V
CASE 1: av/d 6
av
d
V CASE 2: 2 av/d 6
av
d
V
av
V
CASE 3: av/d 2
d
Types of Shear Failure
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Shear force is transmitted through the crack member by a combination of the uncracked concrete in compression
zone, Vcz, the dowelling action of the flexural reinforcement, Vd and aggregate interlocking across
tension cracks, Va
V
Vcz = Shear force in the compression zone
(20 – 40%)
Va = Interlocking between aggregates (35 – 50%)
Vd = Dowel action (35 – 50%)
Concrete compression
Steel tension
d
Shear Resistance
EC2 Shear Design
EC 2 introduces the strut inclination method
for shear capacity checks. In this method the
shear is resisted by concrete struts acting in
compression and shear reinforcement acting in
tension.
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EC2: Cl. 6.2.3
EC2: Cl. 6.2.3
z cot
VEd
Longitudinal steel in tension
Concrete strut in compression Vertical shear steel in tension
z =
0.9
d
d
bw
z cos
z
fcdbwz cos VRd, max
VEd
sin
VEdcot
VEd
Assumed truss model for the strut inclination method
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(1) Diagonal Compressive Strut
VRd, max = [fcd (bw z cos θ)] sin θ
= fcd bw z cos θ sin θ
In EC2 this equation is modified by the inclusion of a strength
reduction factor for concrete cracked in shear v1 and the introduction
of coefficient taking account of the state of the stress in compression
chord αcw thus,
VRd, max = αcwv1fcd bw z / (cot θ + tan θ)
= αcwv1(fck/1.5) bw (0.9d) / (cot θ + tan θ)
= αcwv10.6fck bwd / (cot θ + tan θ)
)tan(cot
)250/1(36.0 ckwckmaxRd,
fdbfV
It is set by the EC2 to limit the θ value between 22 to 45.
The recommended value for αcw and v1 are given in Cl. 6.2.3
EC2. For the purpose of this module the following values are
used: αcw = 1.0 and v1 = 0.6 (1 – fck/250), hence
(1) Diagonal Compressive Strut
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(2) Vertical Shear Reinforcement
The shear resistance of the link is given by
VEd = VRd, s = fywdAsw
= (fyk / 1.15) Asw
= 0.87fykAsw
If the links are spaced at a distance s apart, then
the shear resistance of the link is increased
proportionately and is given by;
VEd = VRd, s = 0.87fykAsw (z cot θ / s)
= 0.87fykAsw (0.9d cot θ / s)
= 0.78fykAswd (cot θ / s)
(2) Vertical Shear Reinforcement
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Thus, rearranging them gives;
cot78.0 yk
Edsw
df
V
s
A
(2) Vertical Shear Reinforcement
EC2 (Cl. 9.2.2) specifies a minimum value for Asw/s
such that;
yk
ckwsw08.0
f
fb
s
A
EC2 (Cl. 9.2.2) also specifies that the maximum
spacing of vertical link, s should not exceed 0.75d.
(2) Vertical Shear Reinforcement
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(3) Additional Longitudinal Force
The longitudinal tensile force, Ftd is caused by the
horizontal component required to balance the
compressive force in the inclined concrete strut;
Longitudinal force = (VEd/sin ) cos
= VEd cot
Assumed half of this force is carried by the reinforcement
in the tension zone of the beam, then the additional tensile
force provided in the tensile zone is given by;
∆Ftd = 0.5VEd cot θ
Shear Design Procedure in EC2
Start
Determine shear
force, VEd
Determine VRd, max for
cot = 1.0 ( = 45) and
cot = 2.5 ( = 22)
If VEd VRd, max
cot = 1.0
If VRd, max cot θ = 2.5
VEd VRd, max cot θ = 1.0
Red
esig
n s
ection
If VEd VRd, max
cot = 2.5
A B
)tan(cot
)250/1(36.0 ckwckmaxRd,
fdbfV
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Shear Design Procedure in EC2
Calculate shear
reinforcement
using cot = 2.5
A
df
V
df
V
s
A
yk
Ed
yk
Edsw 513.0
cot78.0
C
Shear Design Procedure in EC2
Calculate
B
)250/1(18.0sin5.0
ckck
Ed1
fdfb
V
w
Calculate shear
links cot78.0 yk
Edsw
df
V
s
A
C
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Shear Design Procedure in EC2
Calculate additional
longitudinal tensile
force caused by shear
C
End
Calculate minimum links
required by EC2: Cl.
9.2.2(5) and s 0.75d
Flanged
beam?
Yes
No
D
Example 1
Design the required shear reinforcement from the beam section shown
below. Take fyk = 500 N/mm2 and fck = 30 N/mm2.
100 kN/m 225 mm
d = 500 mm
2H16
3H25
(1473 mm2)
8 m
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Example 1: Solution
Maximum shear force;
VEd = wL/2 = 100 8.0 /2 = 400 kN
Concrete strut capacity;
VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )
= 0.36 225 500 30 (1 – 30/250)
(cot + tan )
for = 22, cot = 2.5 VRd, max = 371 kN
= 45, cot = 1.0 VRd, max = 535 kN
VRd, max cot = 2.5 (371 kN) VEd (400 kN) VRd, max cot = 1.0 (535 kN)
Therefore, 22
Example 1: Solution
= 0.5sin-1 [VEd / 0.18bwdfck(1 – fck/250)]
= 0.5sin-1 400 103
0.18 225 500 30 (1 – 30/250)
= 24.2
tan = 0.45 ; cot = 2.22
Shear links;
Asw/s = VEd / 0.78fykdcot
= 400 103 / (0.78 500 500 2.22)
= 0.923
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Example 1: Solution
Try link: H10 Asw = 157 mm2
Spacing, s = 157/0.923
= 170 mm smax = 0.75d (375 mm)
Provide H10-150 mm
Minimum links;
Asw/s = 0.08fck1/2bw / fyk
= 0.08 (30)1/2 225 / 500
= 0.197
Try link: H10 Asw = 157 mm2
Spacing, s = 157/0.197
= 797 mm smax = 0.75d (375 mm)
Provide H10-350 mm
Example 1: Solution
400
400
217
217
H10-350 H10-150 H10-150
1.83 m 1.83 m 4.33 m
Vmin = (Asw/s)(0.78d fykcot )
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Example 1: Solution
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot
= 0.5 400 2.22
= 445 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 445 103 / 0.87 500
= 1022 mm2
Provide 3H25 ( 1473 mm2)
Flanged Beam
Shear between the web and flanged of a flanged section
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Flanged Beam
).(
dEd
xh
Fv
f
f
wf
f
d
2/)(x
)2/( b
bb
hd
MF
The longitudinal shear stress, vEd at the web-flange interface is determine
according to;
M = the change in moment over the distance ∆x
where:
x = half the distance between the sections with zero moment and that
where maximum moment occurs. Where point loads occur, x should
not exceed the distance between the loads.
Flanged Beam
The permitted range of the values of cot f are recommended as follows:
1.0 cot f 2.0 for compression flanges (45 f 26.5)
1.0 cot f 1.25 for tension flanges (45 f 38.6)
The concrete strut capacity of the flange is given by;
vRd = v fcd sin f cos f
= 0.6 (1 – fck/250) (fck/1.5) sin f cos f
= 0.4fck (1 – fck/250)
(cot f + tan f)
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Flanged Beam: Design Procedure
D
Calculate longitudinal
design shear stress, vEd
vEd 0.27fctk 0.4fctd = 0.4(fctk/1.5) = 0.27fctk
No
Yes
Check the stresses in the
incline strut. Compare vEd
with vRd, max
).(
dEd
xh
Fv
f
vEd vRd, max Use min
Yes No E F
F
Flanged Beam: Design Procedure
E
Calculate
and cot f
Calculate transverse
shear reinforcement
)250/1(2.0sin5.0
ckck
Ed1
ff
v
fyk
Ed
f
sf
cot87.0 f
hv
s
A f
Calculate minimum
transverse steel area
Calculate additional
longitudinal tensile
force caused by shear
End
F
where b = 1000 mm
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Example 2
Design the required shear reinforcement from the beam section shown
below. Take fyk = 500 N/mm2 and fck = 25 N/mm2.
90 kN/m
9 m
250 mm
d =
53
0 m
m
3H20
d’ =
45
mm
2H12
600 mm
110 mm
Example 2: Solution
Maximum shear force;
VEd = wL/2 = 90 9.0 /2 = 405 kN
Concrete strut capacity;
VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )
= 0.36 250 500 25 (1 – 25/250)
(cot + tan )
for = 22, cot = 2.5 VRd, max = 373 kN
= 45, cot = 1.0 VRd, max = 537 kN
VRd, max cot = 2.5 (373 kN) VEd (405 kN) VRd, max cot = 1.0 (537 kN)
Therefore, 22
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Example 2: Solution
= 0.5sin-1 [VEd / 0.18bwdfck(1 – fck/250)]
= 0.5sin-1 400 103
0.18 250 500 25 (1 – 25/250)
= 24.5
tan = 0.46 ; cot = 2.19
Shear links;
Asw/s = VEd / 0.78fykdcot
= 400 103 / (0.78 500 530 2.19)
= 0.893
Example 2: Solution
Try link: H10 Asw = 157 mm2
Spacing, s = 157/0.893
= 176 mm smax = 0.75d (398 mm)
Provide H10-150 mm
Minimum links;
Asw/s = 0.08fck1/2bw / fyk
= 0.08 (25)1/2 250 / 500
= 0.200
Try link: H10 Asw = 157 mm2
Spacing, s = 157/0.200
= 786 mm smax = 0.75d (398 mm)
Provide H10-375 mm
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Example 2: Solution
405
405
214
214
H10-375 H10-150 H10-150
2.12 m 2.12 m 4.76 m
Vmin = (Asw/s)(0.78d fykcot )
Example 2: Solution
Transverse steel in the flange;
x = 0.5(L/2) = 9000/4 = 2250 mm
Change of moment over distance x from zero moment:
M = (wL/2)(L/4) – (wL/4)(L/8) = 683.44 kNm
Change in longitudinal force;
= 683.44 103 (600 – 250)
(530 – 55) (2 600)
= 420 kN
f
wf
f
d
2/)(x
)2/( b
bb
hd
MF
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Example 2: Solution
Longitudinal shear stress;
vEd = Ftd / (hf x)
= 420 103 / (110 2250)
= 1.70 N/mm2
Since vEd (1.70 N/mm2) 0.27fctk = 0.27 1.80 = 0.49 N/mm2
Transverse steel reinforcement is required
Example 2: Solution
Concrete strut capacity in the flange;
vRd, max = 0.4fck (1 – fck/250) / (cot + tan )
= 0.4 25 (1 – 25/250)
(cot + tan )
for = 27, cot = 2.0 vRd, min = 3.59 N/mm2
= 45, cot = 1.0 vRd, max = 4.50 N/mm2
vEd (1.70 N/mm2) vRd, min cot = 2.0 (3.59 N/mm2)
and
vEd (1.70 N/mm2) VRd, max cot = 1.0 (4.50 N/mm2)
Therefore, use = 27 ; tan = 0.50 ; cot = 2.0
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Example 2: Solution
Transverse shear reinforcement;
Asf / sf = vEdhf / 0.87fykcot
= 1.70 110 / (0.87 500 2.0)
= 2.0
Try H10: Asf = 79 mm2
Spacing, sf = 79/0.21 = 367 mm
Minimum transverse steel area;
As, min = 0.26 (fctm/fyk) bhf
= 0.26 (2.60/500) bhf
= 0.0013bhf 0.0013bhf
= 0.0013 1000 110 = 147 mm2
Provide H10 – 300 (As = 262 mm2/m)
Example 2: Solution
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot
= 0.5 405 2.19
= 444 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 444 103 / 0.87 500
= 1021 mm2
Provide 3H25 ( 1473 mm2)
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