3. Bond , Anchorage and Shear • This chapter will discuss the

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3. Bond , Anchorage and Shear
• This chapter will discuss the following topics:
• Outline the theory of calculating the anchorage bond
length.
• Determination of anchorage bond length,
length tension lap
length and compression lap length by using Table 8.4
and Table 8.5 of the code.
• Outline the design formulae for the design of shear
reinforcement and the use of Table 6.3 of the Code.
RC Design and Construction – HKC 2004 (2nd Edition)
3-1
3. Bond, Anchorage and Shear
• Minimum Distance Between Bars
– The min. distance between bars are given in terms of
hagg (the maximum size of aggregate).
aggregate)
– Clear Horizontal and Vertical Distance between bars
≥ bar diameter or
≥ ( hagg + 5mm) or
≥ 20 mm
whichever is greater.
greater
RC Design and Construction – HKC 2004 (2nd Edition)
3-2
3.1 Minimum and Maximum Percentages of
Reinforcement in Beams, Slabs and Columns
•
•
Minimum % : Table 9.1 of the code
Maximum %
(a) Beam (9.2.1.3)
Neither the area of tension reinforcement nor the area of compression
p
reinforcement should exceed 4% of the gross cross-sectional area of concrete.
((b)) Column ((9.5.1))
The longitudinal reinforcement should not exceed the following
amounts, calculated as percentages of the gross cross-sectional area of the
concrete :
~ vertically cast columns
: 6%
~ horizontally cast columns
: 8%
p in the verticallyy or horizontallyy cast column
: 10%
~ laps
(c) Wall (9.6.2)
The area of vertical reinforcement should not exceed 4% of the concrete
cross-sectional area of the wall.
3-3
RC Design and Construction – HKC 2004 (2nd Edition)
3.2 Anchorage Bond
• The steel bar subjected to tension shown in Fig.
3.1 shall be firmly anchored in concrete. The
anchorage length (embedded length) depends on
the bond between the bar and the concrete, and the
contact area.
L
Φ
F
Fig.
g 3.1 Anchorage
g Bond
RC Design and Construction – HKC 2004 (2nd Edition)
3-4
3.2 Anchorage Bond
L = Min. anchorage length to prevent pull out.
φ = Bar size or nominal diameter.
diameter
fbu= Ultimate anchorage bond stress.
fs = Direct tensile or compressive stress in the bar.
Consider the forces on the bar,
Tensile pull-out force = Area of bar * Stress of bar
πφφ 2
⋅ fs
=
4
RC Design and Construction – HKC 2004 (2nd Edition)
3-5
3.2 Anchorage Bond
Anchorage force
= Contact Area * Bond Stress
= (Lπφ)⋅fbu
2
πφ
∴
φ) bu
⋅ f s = ((Lπφ)⋅f
4
L=
0.87f y
fs
⋅φ =
⋅φ
4 f bu
4 f bu
The ultimate anchorage bond stress, fbu = β fcu .
Values of β are given in table 8.3.
∴
anchorage length L can be written as
L = KAφ
(
(see
t bl 8.4
table
8 4 off the
th Code)
C d )
RC Design and Construction – HKC 2004 (2nd Edition)
3-6
3.2 Anchorage Bond
Table 8.3 - Values of the bond coefficient β
β
B type
Bar
t
Plain bars
Type 2 : deformed bars
Fabric
Bars in tension
0.28
0.50
0.65
Bars in compression
0.35
0.63
0.81
• Normally deformed type 2 bars are used in
construction works.
• Furthermore anchorage may be provided by hooks
and bends in the reinforcement.
reinforcement (see Fig.8.2).
Fig 8 2)
RC Design and Construction – HKC 2004 (2nd Edition)
3-7
Fig. 8.2 Anchorage of Links
Reproduce
d
ffrom HK Code
d
RC Design and Construction – HKC 2004 (2nd Edition)
3-8
3.3 Lapping of Reinforcement
• Lapping of reinf. may be required if the reinf. is of
insufficient length or there is curtailment (cutting
off of reinf. when it is no longer required in order
to save money) of reinf.
• Rules
R l for
f lapping
l i are:• The laps should be staggered and be away from sections
with high stresses.
stresses (i.e.
(i e avoid lapping of reinf.
reinf at sections
with large bending moment as it induces large tensile stress
in reinforcement)
• Min. lap length should be not less than the greater of 15Φ
or 300mm.
RC Design and Construction – HKC 2004 (2nd Edition)
3-9
3.3 Lapping of Reinforcement
The tension lap length should be at least equal to the design tension
anchorage length. Lap lengths for unequal size bars may be based
upon the
h diameter
di
off the
h smaller
ll bar.
b The
h following
f ll i provisions
ii
also
l
apply:
– where a lap occurs at the top of a section as cast and the
minimum cover is less than twice the diameter of the lapped
reinforcement, the lap length should be increased by a factor
of 1.4
– where a lap occurs at a corner of a section and the minimum
cover to either face is less than twice the diameter of the
lapped reinforcement, the lap length should be increased by a
factor of 1.4
– where both conditions above apply, the lap length should be
increased by a factor of 2.0
Figure
i
8.5 andd table
bl 8.5 are referred.
f
d
RC Design and Construction – HKC 2004 (2nd Edition)
3-10
3.3 Lapping of Reinforcement
• Rules for lapping are:• Compression laps should be at least 25% greater than the
compression anchorage length.
• Lap lengths for unequal size bars may be based on the
smaller bar.
x 1.4
x 2.0
If Cover < 2φ, multiple lap length by
1.4 or 2.0
x 1.4
14
x 1.0
(Lap length = anchorage length x 1.0)
Fig. 3.3 Lapping of tension reinforcement
RC Design and Construction – HKC 2004 (2nd Edition)
3.4
3-11
Shear
• Shear reinforcement can be in the form of shear
links and inclined bars (bent
(bent-up
up bars). However,
inclined bars are less frequently used in
construction today.
• Sh
Shear Links
Li k (Stirrups)
(Sti
)
• Links can be broadly classified into shear link and
nominal link (minimum link) and the design
formulae for these types of links are as follows:
RC Design and Construction – HKC 2004 (2nd Edition)
3-12
Design Formulae for Shear Reinforcement
f fcu ≤ 40 MPa
for
MP
• Shear Links:Links:
Asv b(v − vc )
≥
sv
0.87 f yv
• Minimum (nominal) Links:Asv
0.4b
≥
sv 0.87 f yv
RC Design and Construction – HKC 2004 (2nd Edition)
3-13
Design Formulae for Shear Reinforcement
Where:Asv = Total cross-sectional
cross sectional area of the vertical legs of the
link.
b = Breadth of beam
fyv = Characteristic strength of the shear reinf. = fy
sv = Longitudinal
L it di l spacing
i off shear
h li
links
k
v = Design shear stress
= Design
D i shear
h fforce/(bd)
/(bd) = V/(bd)
vc = The ultimate shear stress that can be resisted by the
concrete.
RC Design and Construction – HKC 2004 (2nd Edition)
3-14
Design Formulae for Shear Reinforcement
The values of vc is given in table 6.3 of the Code.
• It shall be noted that the values given in table 6.3 is for
grade 25 conc. For other grades of conc., adjustment shall
be made by multiplying ⎛
1
fcu ⎞ 3
⎜
⎝ 25 ⎠
• The values of vc increases for shallow members and those
with larger percentages of tensile reinforcement.
RC Design and Construction – HKC 2004 (2nd Edition)
3-15
3.5 Enhanced Shear Resistance
• Within a distance of 2d from a support or a
concentrated load, the design concrete shear stress
vc may be increased to
vc⋅(2d / av)
• The distance av is measured from the support or
concentrated load to the section being designed.
– Average shear stress should never exceed the
lesser of 0.8 fcu or 7 N/mm2.
RC Design and Construction – HKC 2004 (2nd Edition)
3-16
Figure 6.3 – Shear failure near supports
R
Reproduce
d
from
f
HK C
Code
d
3-17
RC Design and Construction – HKC 2004 (2nd Edition)
Table 6.3 - Values of vc Design Concrete Shear Stress
R
Reproduce
d
from
f
HK C
Code
d
RC Design and Construction – HKC 2004 (2nd Edition)
3-18
Table 9.1 - Minimum Percentage of Reinforcement
R
Reproduce
d
from
f
HK C
Code
d
RC Design and Construction – HKC 2004 (2nd Edition)
3-19
Table 8.4 – Ultimate Anchorage Bond Lengths (lb)
Reproduce from HK Code
RC Design and Construction – HKC 2004 (2nd Edition)
3-20
Table 8.5 Ultimate Lap Lengths
Reproduce from HK Code
RC Design and Construction – HKC 2004 (2nd Edition)
3-21
Figure 8.5 - Factors for lapping bars
Reproduce from HK Code
RC Design and Construction – HKC 2004 (2nd Edition)
3-22
Self-Assessment Questions
Q1.
Determine the anchorage bond length from the first
principle for a T32 bar in tension (deformed type 2)
and fcu = 35 N/mm2. the bars as listed below:
Choices:
(a)
1000 mm
(b)
1082 mm
(c)
1360 mm
RC Design and Construction – HKC 2004 (2nd Edition)
3-23
Self-Assessment Questions
Q2. Determine the anchorage bond length by using table 8.4
of HKC 2004 for a R16 bar in compression and fcu = 30
N/mm2 .
Choices:
(a)
464 mm
(b)
528 mm
(c)
540 mm
RC Design and Construction – HKC 2004 (2nd Edition)
3-24
Self-Assessment Questions
Q3. Determine the ultimate shear resistance of concrete vc
for the singly reinforced concrete section shown in Fig.
Q3, without shear enhancement. Given that fcu = 35
N/mm2 and fy = 460 N/mm2.
400
350
250
2T32
Choices:
(a)
(b)
(c)
Fig. Q3
0.70 N/mm2
0.90 N/mm2
1.10 N/mm2
3-25
RC Design and Construction – HKC 2004 (2nd Edition)
Assignment No. 3
AQ1
Fig. AQ1 shows the lapping of steel reinforcement of a beam.
Determine the lap length for the bar groups 1, 2, 3 and 4. Given
that the nominal cover is less than 2 times the diameter of the bar
and fcu = 35 N/mm2. The spacing between group 1 and 2 bars is
greater than s and the spacing between group 3 & 4 bars is less
than
h s.
(a) The lapping of steel reinforcement is at tension zone.
(b) The lapping of steel reinforcement is at compression zone
(1)
(2)
(3)
(4)
3T25
3T16
3T32
3T32
Fi AQ1
Fig.
RC Design and Construction – HKC 2004 (2nd Edition)
3-26
Assignment No. 3
AQ2 Determine the nominal cover and breadth of the web for the
simply supported beams whose sections are shown in Fig.
AQ2a, AQ2b & AQ2c. The maximum size of the aggregate
in each case is 20 mm, fcu = 40 N/mm2 and the link size is
10mm (min.
(min cement content = 300 kg/m3 and max waterwater
cement ratio = 0.55).
( )
(a)
Exposure condition:
E
diti
severe
Fire resistance: 2 hours
(b)
Exposure condition: moderate
Fire resistance: 1 hours
(c)
Exposure condition: severe
Fi resistance:
Fire
it
4 hours
h
3-27
RC Design and Construction – HKC 2004 (2nd Edition)
Assignment No. 3
3T40
g AQ1a
Q
Fig.Fig.AQ2a
Fig
4T20
g AQ1b
Q
Fig.Fig.AQ2b
Fig
2T32
Fig.Fig.
AQ2c
AQ1c
RC Design and Construction – HKC 2004 (2nd Edition)
3-28
Assignment No. 3
AQ3. Determine the average shear stress for the section
shown in Fig. AQ3a, AQ3b & AQ3c, and state whether
shear reinforcement is required. Given that fcu = 30
N/mm2 and fy = 460 N/mm2.
200
300
450
400
350
300
260
300
225
2T16
2T20
3T25
Ultimate Shear = 250 kN
Ultimate Shear = 150 kN
Fig. AQ2a
Fig
Fig.
AQ3a
Fig. AQ2c
Fig.
AQ3c
Ultimate Shear = 60 kN
AQ2b
Fig.Fig.AQ3b
3-29
RC Design and Construction – HKC 2004 (2nd Edition)
Assignment No. 3
AQ4 The sections in Fig. AQ4a & AQ4b are subject to shear
forces at the ultimate limit state as shown. Determine
the spacing of R10 links. Given that fcu = 35 N/mm2, fy
= 460 N/mm2 and fyv = 250 N/mm2.
300
450
400
350
310
225
2T20
3T25
Ultimate Shear = 40 kN
Fig. AQ3a
Fig.
Fi
AQ4a
AQ4
Ultimate Shear = 160 kN
Fig. AQ3b
Fig. AQ4b
RC Design and Construction – HKC 2004 (2nd Edition)
3-30
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