13
Reinforced Concrete Design
Bond, Anchorage, and Development Length
Bond Stress in Beam
Mechanism of Bond Transfer
Development of Bars in Tension
Development of Bars in Compression
Development of Bundled Bars
Angchorage of Tension Bars by Hooks
Mongkol JIRAVACHARADET
SURANAREE
INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY
SCHOOL OF CIVIL ENGINEERING
Bond Stresses in Beam
Concrete
Reinforcing bar
End slip
P
Greased or lubricated
Free slip εs ≠ εc
Bond forces acting on concrete
Bond forces acting on steel
Tied-arch action in a beam with little or no bond
(Round bar reinforcing)
Mmax
jd
Little or no bond
Bond Stress Based on Simple Cracked Analysis
[ ΣMO = 0 ]
C + dC
C
j d (dT ) = V (dx )
jd
V
dT V
=
dx jd
V
[ ΣFx = 0 ]
T + dT
T
O
dx
u Σo dx = dT
u
1
T + dT
T
ก
= As f s =
= π d Lu
4
fs
= ก
ก
2
2
u=
V
Σo j d
ก
π d2
1
Elastic Cracked
Section Equation
Critical section
L
π d Lu =
π d2
L =
4
fs
d fs
4u
(WSD)
ก
:
ก
ua =
ก
ua =
ก
:
2.29 f c′
db
≤ 25 kg/cm2
3.23 f c′
≤ 35 kg/cm 2
db
ua = 1.72 f c′ ≤ 28 kg/cm 2
* กกก
!"!ก 11 กก./&.2
* ก(ก)
ก *! ก+ก 30 &.
Nominal Bond Strength
ก
:
ก
un =
ก
un =
ก
:
4.51 f c′
db
≤ 39.4 kg/cm 2
6.39 f c′
≤ 56.2 kg/cm 2
db
un =
3.44 f c′
db
≤ 56.2 kg/cm2
* กกก
!"!ก 17.6 กก./&.2
Steel Force and Bond Stress
Cracked concrete segment
M
M
T
T
u stresses on bar
steel tension T
T=
M
jd
u=
1 dT
Σ o dx
bond stress u
Actual Distribution of Flexural Bond Stress
CL
T=
M
jd
Actual T
u=
V
Σ o jd
Actual u
Mechanism of Bond Transfer
R
u
∆T
∆T
Friction & chemical adhesion
between concrete and steel
R
R
Reaction of concrete on ribs
R
R
Forces exerted by ribs on concrete
Failure modes
R
R
Side-split failure
V-notch failure
R
Vertical cracking of bottom cover
Ultimate Bond Strength
Cylindrical zones of
circumferential tension
Reinforcement
Radial component
of bearing pressure
Circumferential
tensile stresses
Minimum Bar Covering and Spacing
Minimum bar covering ( cb )
cb
2cs
cs
Minimum bar spacing ( 2cs )
Splitting of concrete along reinforcement
DEVELOPMENT LENGTH
Embedment length to develop full tensile strength of bar
ld
T=0
a
T = Ab fs
Requirement: l ≥ ld : development length
Factors influencing ld
- Tensile strength of concrete fct
- Steel covering
- Bar spacing
- Stirrup
ACI 12.2 – Development of deformed bars in tension
Development length Ld shall be determined from either 12.2.2 or 12.2.3, but shall
not be less than 12 in.
0.28 fy α β γ λ
12.2.3 – For deformed bars Ld shall be:
Ld =
db
fc′  c + K tr 


 db 
Atr fyt
in which the term (c + Ktr)/db shall not be taken greater than 2.5, and K tr =
105 s n
c = the smaller of
(a)
distance from center of bar to the nearest concrete surface
(b)
half the center-to-center spacing of the bars
Atr = total cross-sectional area of all transverse reinforcement within spacing s
fyt = yield strength of transverse reinforcement
s = maximum spacing of transverse reinforcement within Ld
n = number of bars being developed along the plane of splitting.
It shall be permitted to use Ktr = 0 as a design simplification even if transverse
reinforcement is present.
Atr
Definition of Atr
Potential plane
of splitting
ACI 12.2.5 – Excess reinforcement
Reduction in Ld shall be permitted where reinforcement in a flexural member is in
excess of that required by analysis
(As required)/(As provided) Ld
Reduction for large cover and wide bar spacing:
0.8
Heavily confined reinforcement:
0.75
Modifying Multipliers of Development Length in Tension
α Bar location:
β Coating factor:
(a) Top reinforcement
α = 1.3
(b) Bottom bars
α = 1.0
(a) Epoxy-coated cover < 3db or
clear spacing < 6db
β = 1.5
(b) All other epoxy-coated
β = 1.2
(c) Uncoated reinforcement
β = 1.0
Note: α × β need not be taken > 1.7
γ Bar-size factor:
λ Lightweight
aggregate factor:
(a) DB20 and smaller
γ = 0.8
(b) DB25 and larger
γ = 1.0
(a) Lightweight aggregate
concrete
λ = 1.3
(b) Normal weight concrete
λ = 1.0
ACI 12.2.2 – Simplified Equation for Development Length
For deformed bars Ld shall be as follows:
c + K tr
= 1.5
db
DB20 and smaller
(γγ = 0.8)
DB25 and larger
(γ = 1.0)
Case A:
(1) covering = db
clear c-c = db
min. stirrup
ld 0.15 f y
=
αβλ
db
f c′
ld 0.19 f y
=
αβλ
db
f c′
(2) covering = db
clear c-c = 2db
(A - 1)
(A - 2)
ld 0.23 f y
=
αβλ
db
f c′
ld 0.28 f y
=
αβλ
db
′
fc
(B - 1)
(B - 2)
set
Case B: others
Development Length of Bars in Compression
ldb =
0.075 db f y
f c′
≥ 0.0043 db f y
Modification factors:
- Excess reinforcement
As req’d / As prv’d
- Spiral stirrup
dia. ≥ 6 mm, pitch ≤ 10 cm
0.75
- Tied stirrup
DB12, spacing ≤ 10 cm
0.75
Development Length for Bundled Bars
when space for proper clearance is restricted
- No more than 4 bars bundled in contact
- Bar larger than DB36 shall not be bundled in beams
- Termination of each bar must be different at least 40db
- Development length based on single bar in the bundle
- Increase 20% for 3 bars
- Increase 33% for 4 bars
Anchorage of Tension Bars by Hooks
When straight embedment is not enough
a) Standard Bar Hook
D = diam of bend
db
T
180o Hook
4 db ≥ 6 cm
D
db
Min. Dia. of Bend (D)
T
90o Hook
DB10 - DB25
DB28 - DB36
DB40 - DB60
12 db
For stirrup and tie anchorage only:
db
(a) For DB16 or smaller
D
T
6db
db
(b) For DB20 or DB25
D
T
12db
db
(c) For DB25 or smaller
T
135o
6db
6db
8db
10db
b) Development length and Hook
Combined actions:
Ldh
Ld
- Bond along straight length
- Anchorage provided by hook
T
db
12db
Critical section
full bar tension
T
db
4db ≥ 6 cm
4db
Ldh
Basic development length
with fy = 4,000 kg/cm2
Lhb =
318d b
f c'
Ldh = Modification factor × Lhb
Modification factors:
- Bar with fy other than 4,000 ksc
- DB25mm and smaller
side cover not less than 4 cm
90o hook cover hook not less than 5 cm
fy / 4,000
0.7
- DB25mm and smaller
hook enclosed within stirrup along ldh
stirrup spacing not greater than 3db
0.8
- Excess reinforcement
As reqd / As prvd
- Lightweight concrete
1.3
- Epoxy coating
1.2
EXAMPLE 11-1 Anchorage of a Straight Bar
A 40-cm-wide cantilever beam frames into the edge of a 40-cm-thick wall, as shown.
At ultimate, the DB25 bars at the top of the cantilever are stressed to their yield
strength at point A at the face of the wall. Compute the minimum embedment of the
bars into the wall. The concrete is normal-weight concrete with a strength of 240 ksc.
The yield strength of the flexural reinforcement is 4,000 ksc.
DB16 at 30 cm O.C.
B
Ld
Construction joint
45 cm
3DB25
A
39 cm
Construction joint
40 cm
Wall
1.5 m
1. Find the spacing and confinement case.
There are no stirrups or ties in the wall, but there are DB16 at 30 cm o.c. vertical
bars outside of the 3DB25
Clear side cover = 4 + 1.6 = 5.6 cm (2.24db)
Clear spacing = (40 – 2(4+1.6) – 3x2.5) / 2 = 10.65 cm (4.26db)
ก
α = 1.3
ก"!01ก&
β = 1.0
ก +ก0ก λ = 1.0
Since the clear spacing between the bars is not less than 2db and the clear cover
exceeds db, case (A-2) applies.
2. Compute the development length. From (A-2) :
Ld =
0.19 fy α β λ
fc′
db =
0.19(4,000)(1.3)(1.0)(1.0)
× 2.5 = 159.4 cm
240
Use development length Ld = 160 cm
Ld =
3. Compute the development length. Detailed equation :
0.28 fy α β γ λ
c = the smaller of
(a)
 c + K tr 
fc′ 

 db 
db
distance from center of bar to the nearest concrete surface:
the side cover is 4 + 1.6 + 2.5/2 = 6.85 cm;
 40 − 2 × 6.85 
half the center-to-center spacing of the bars, or 0.5 
 = 6.58 cm
2


Therefore, c = 6.58 cm.
K tr =
A tr fyt
105 sn
=
4.02 × 4,000
= 1.7 cm
105 × 30 × 3
s = spacing of the transverse reinforcement within Ld = 30 cm
Atr = transverse steel area crossing splitting plane within spacing s
= DB16 at 30 cm each face = 2 x 2.01 = 4.02 cm2
fyt = 4,000 kg/cm2 for the wall steel
n = number of bars being anchored = 3
6.58 + 1.7
c + K tr
=
= 3.31 > 2.5
2.5
db
0.28 fy α β γ λ
 c + K tr 
fc′ 

 db 
= 122.2 cm
db =
0.28 × 4,000 × 1.3 × 1.0 × 1.3 × 1.0
× 2.5
240 × 2.5
Use development length Ld = 125 cm
Construction joint
A
3DB25
B
Ld = 125 cm
Construction joint
Wall
1.5 m
45 cm
Ld =
USE 2.5
39 cm
(b)
Bar Cutoff and Bend Points in Beams
Moment capacity of beam:
2 bars
a

M = As f y  d − 
2

4 bars
C
L
5 bars
Ld
Ld
Ld
5M
Required moment Mu / φ
Moment capacity Mn
Face of support
Bar Cutoff requirements of the ACI Code
Moment capacity
Inflection point
for +As
Inflection point for -As
-M
of bars M
Ld
Bars M
+M
Moment capacity
of bars O
Greatest of d , 12db or Ln/16
CL of span
for at least 1/3 of -As
d or 12db
Ld
Bars L
Bars N
15 cm for at least
1/4 of +As
Ld
d or 12db
Ld
Bars O
Development of Reinforcement at Simple Supports
La
At point A, Mu = φMn of bars continuing
into the support
Mn /Vu
ACI: Distance from A to bar end ≥ Ld
≥ Ld
Approximate
A
Mu
Vavg (say, 0.9Vu )x = φ M n
φMn
→
x=
Mn
Vu
La = Straight length beyond c.g. support
Vu
ACI: Embedment length
Vavg
La + 1.3
x
Mn
≥ Ld
Vu
1.3 for less tendency to split due to
confinement at support
Development of Reinforcement at Inflection Points
Locate away from support → no confinement → 1.3 factor not apply
Actual La
Usable La
ACI: Embedment length
Mn /Vu
Mu
A
φMn
Inflection point
Vu
Usable La +
Mn
≥ Ld
Vu
Usable La = Actual La but
not exceed the larger of
12db or d
Standard Cutoff and Bend Points for Bars
For approximately equal spans with uniformly distributed loads
L1/4
L1/3
L2/3
L2/3
0 cm
15 cm
15 cm
L2/8
L1/8
L1
L2
L1/3
L1/7
L2/3
L1/4
L2/4
L2/4
6.3 4546ก7ก! !
!ก8999
+ก!07)+9 wu = 8.0 ก9 f’c = 280 กก./&.2, fy = 4,000 กก./&.2,
b = 40 &., h = 60 &. 7ก 8 4 &.
Exterior column
Interior column
wu
Ln = 7.6 m
1. กก!"##$
a. ก7?076!@ ?7A
Interior face of
exterior support
-Mu = wuLn2/16 = 8(7.6)2/16 = -28.88 t-m
Mid span positive
+Mu = wuLn2/14 = 8(7.6)2/14 = 33.01 t-m
Exterior face of first
interior support
-Mu = wuLn2/10 = 8(7.6)2/10 = -46.21 t-m
Exterior face of first
interior support
Vu = 1.15wuLn/2 = 1.15(8)(7.6)/2 = 34.96 t-m
b. 546ก@ ?99 @9ก 8 4 &. ก0ก DB10 7ก
@ ?99 DB25 DB28 ! d ≈ 60-4-1.0-1.4 ≈ 53.6 &.
Mu
As required
Bars
As provided
- 28.88 t-m
15.97 cm2
4DB25
19.63 cm2
+ 33.01 t-m
18.44 cm2
4DB25
19.63 cm2
- 46.21 t-m
26.76 cm2
2DB25+3DB28
28.29 cm2
A
4DB25
B
2DB25+3DB28
4DB25
B
A
40 cm
40 cm
4DB25
DB10@0.20m
60 cm
2DB25
3DB28
60 cm
DB10@0.20m
4DB25
Section A-A
4DB25
Section B-B