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11.1 INTRODUCTION Corbel or bracket is a reinforced concrete member is a short-haunched cantilever used to support the reinforced concrete beam element. Corbel is structural element to support the pre-cast structural system such as pre-cast beam and pre-stressed beam. The corbel is cast monolithic with the column
element or wall element.
This chapter is describes the design procedure of corbel or bracket structure. Since the load from precast structural element is large then it is very important to make a good detailing i n corbel.
11.2 BEHA VIOR OF CORBEL The followings are the major items show the behavior of the reinforced concrete corbel, as follows :
The shear span/depth ratio is less than 1.0 , it makes the corbel behave in two-dimensional manner.
Shear Shear deformation is significant is the corbel.
There is large horizontal force transmitted from the supported beam result from long-term shrinkage and creep deformation .
Bearing failure due to large concentrated load.
The cracks are usually vertical or inclined pure shear cracks .
The mode of failure failure of corbel are : yielding of the tension tie, failure of the end anchorage of of the tension tie, failure of concrete by compression or shearinga and bearing failure.
The followings figure shows the mode of failure of corbel.
11 - 1
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Vu
DIAGONAL SHEAR
Vu
SHEAR FRICTION
Vu
Vu Nu
ANCHORAGE SPLITING
FIGURE 11.1
VERTICAL SPLITING
MODES OF F AILURE OF CORBEL
11.3 SHEAR DESIGN OF CORBEL 11.3.1
GENERAL
Since the corbel is cast at different time with the column element then the cracks occurs in the interface of the corbel and the column. To avoid the cracks we must provide the shear friction reinforcement perpendicular with the cracks direction.
ACI code uses the shear friction theory to design the interface area.
11.3.2
SHEAR FRICTION THEORY
In shear friction theory we use coefficient of friction
to transform the horizontal resisting force
into vertical resisting force .
The basic design equation for shear reinforcement design is :
φVn ≥ Vu where :
11 - 2
Vn
= nominal shear strength of shear friction reinforcement
Vu
= ultimate shear force
φ
= strength reduction factor (φ = 0.85)
[11.1]
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Vu SHEAR FRICTION REINFORCEMENT
Avf f y μ Avf f y
α
f
ASSUMED CRACK
FIGURE 11.1
SHEAR FRICTION THEORY
The nominal shear strength of shear friction reinforcement is :
T ABL E 11.1
SHEAR FRICTION REINFORCEMENT STRENGTH
VERTICAL
INCLINED
SHEAR FRICTION
SHEAR FRICTION
REINFORCEMENT
REINFORCEMENT
Vn
A vf
A vf = Vn = A vf f yμ
Vn
A vf
Vn
A vf =
f yμ Vu
Vn
Vn
f y (μ sin α f + cos α f )
= A vf f y (μ sin α f + cos α f )
φ A vf = f yμ
Vu A vf =
φ
f y (μ sin α f + cos α f )
where : Vn
= nominal shear strength of shear friction reinforcement
Avf
= area of shear friction reinforcement
Fy
= yield strength of shear friction reinforcement
μ
= coefficient of friction
T ABL E 11.2
COEFFICIENT OF FRICTION METHOD
Concrete Cast Monolithic Concrete Placed Against Roughened Hardened Concrete Concrete Placed Against unroughened Hardened Concrete Concrete Anchored to Structural Steel
The value of
COEFFICIENT OF FRICTION
1.4λ 1.0λ 0.6λ 0.7λ
is :
λ = 1.0
normal weight concrete
λ = 0.85
sand light weight concrete
λ = 0.75
all light weight concrete
11 - 3
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The ultimate shear force must follows the following condiitons :
≤ φ(0.2f 'c )b w d Vu ≤ φ(5.50 )bw d
[11.1]
Vu
where : Vu
= ultimate shear force
(N)
f’c
= concrete cylinder strength
(MPa)
bw
= width of corbel section
(mm)
d
= effective depth of corbel
(mm)
11.3.3
STEP – B Y – STEP PROCEDURE
The followings are the step – by – step procedure used in the shear design for corbel (bracket), as follows :
Calculate the ultimate shear force Vu .
Check the ultimate shear force for the following condition, if the following condition is not achieved then enlarge the section .
≤ φ(0.2f 'c )b w d Vu ≤ φ(5.50 )b w d
Vu
Calculate the area of shear friction reinforcement A vf . VERTICAL
INCLINED
SHEAR FRICTION
SHEAR FRICTION
REINFORCEMENT
REINFORCEMENT
Vn
A vf
A vf = Vn = A vf f yμ
Vn
A vf
Vn f yμ Vu
A vf = Vn
f y (μ sin α f + cos α f )
= A vf f y (μ sin α f + cos α f )
φ A vf = f yμ
Vn
Vu
A vf =
φ f y (μ sin α f + cos α f )
The design must be follows the basic design equation as follows :
φVn ≥ Vu
11.4 FLEXURAL DESIGN OF CORBEL 11.4.1
GENERAL
The corbel is design due to ultimate flexure moment result from the supported beam reaction Vu and horizontal force from creep and shrinkage effect Nu.
11 - 4
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Vu a
Nuc
2 / d n i m
d h
FIGURE 11.2
11.4.2
DESIGN FORCE OF CORBEL
TENSION REINFORCEMENT
The ultimate horizontal force acts in the corbel Nuc is result from the creep and shrinkage effect of the pre-cast or pre-stressed beam supported by the corbel. This ultimate horizontal force must be resisted by the tension reinforcement as follows :
An
=
[11.2]
Nuc
φf y
where : An
= area of tension reinforcement
Nuc
= ultimate horizontal force at corbel
f y
= yield strength of the tension reinforcement
φ
= strength reduction factor (φ = 0.85)
Minimum value of Nuc is
0.2Vuc .
The strength reduction factor is taken 0.85 because the major action in corbel is dominated by sh ear .
11.4.3
FLEXURAL REINFORCEMENT
Vu a
Ts
Nuc
β
d j a
d h
Cc FIGURE 11.3
ULTIMATE FLEXURE MOMENT AT CORBEL
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The ultimate flexure moment Mu result from the support reactions is :
Mu
= Vu (a) + Nuc (h − d)
[11.3]
where : Mu
= ultimate flexure moment
Vu
= ultimate shear force
a
= distance of Vu from face of column
Nuc
= ultimate horizontal force at corbel
h
= height of corbel
d
= effective depth of corbel
The resultant of tensile force of tension reinforcement is :
Tf = A f f y
[11.4]
where : Tf
= tensile force resultant of flexure reinforcement
Af
= area of flexure reinforcement
f y
= yield strength of the flexure reinforcement
The resultant of compressive force of the concrete is :
Cc
= 0.85f 'c ba(cos β )
[11.5]
where : Cc
= compressive force resultant of concrete
f’c
= concrete cylinder strength
b
= width of corbel
a
= depth of concrete compression zone
The horizontal equilibrium of corbel internal force is :
∑ H = 0 ⇒ Cc =Ts 0.85 f 'c ba(cos β) = A f f y a=
[11.6]
A f f y 0.85f 'c b(cos β )
The flexure reinforcement area is :
A f =
11 - 6
Mu
⎛ a ⎞ φf y ⎜ d − ⎟ ⎝ 2 ⎠
[11.7]
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A f =
Mu
⎛ ⎛ A f f y ⎞ ⎞⎟ ⎜ ⎜ ⎟ ⎜ ⎜⎝ 0.85 f 'c b(cos β) ⎠⎟ ⎟ φf y ⎜ d − ⎟ 2 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠
Cos β value can be calculated based on the Tan β value as follows :
Tanβ =
[11.8]
jd a
where : a
= distance of Vu from face of column
jd
= lever arm
Based on the equation above we must trial and error to find the reinforcement area Af . For practical reason the equation below can be used for preliminary :
A f = A f =
[11.9]
Mu
φf y ( jd) Mu
φf y (0.85d)
where : Af
= area of flexural reinforcement
Mu
= ultimate flexure moment at corbel
f y
= yield strength of the flexural reinforcement
φ
= strength reduction factor (φ = 0.9)
d
= effective depth of corbel
11.4.4
DISTRIBUTION OF CORBEL REINFORCEMENTS Vu
a
As= 23 Avf+An
Vu Nuc
d ) 3 / 2 (
d h
Ah= 13 Avf
a
As= Af +An
Nuc
d ) 3 / 2 (
d
Ah= 12 Af
FRAMING REBAR
CASE 1 FIGURE 11.4
FRAMING REBAR
CASE 2 DISTRIBUTION OF CORBEL REINFORCEMENTS
11 - 7
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From the last calculation we already find the shear friction reinforcement Avf , tension reinforcement An and flexural reinforcement Af . We must calculate the primary tension reinforcement As based on the above reinforcements.
T ABL E 11.3
DISTRIBUTION OF CORBEL REINFORCEMENTS
CASE
CLOSED
PRIMARY
As
STIRRUP
REINFORCEMENT A h
1
2
A s
2
≥ A vf + An 3
A s ≥ A f + A n
A s
2
= A vf + An 3
A s = A f + An
LOCATION
1
2
3
3
1
2
2
3
Ah
= A vf
Ah
= A f
d d
where : As
= area of primary tension reinforcement
Avf
= area of shear friction reinforcement
An
= area of tension reinforcement
Af
= area of flexure reinforcement
Ah
= horizontal closed stirrup
d
= effective depth of corbel
The reinforcements is taken which is larger, case 1 or case 2, the distribution of the reinforcements is shown in the figure above.
11.4.5
LIMITS OF REINFORCEMENTS
The limits of primary steel reinforcement at corbel design is :
ρ=
A s bd
≥ 0.04
f 'c f y
[11.10]
where : As
= area of primary tension reinforcement
b
= width of corbel
d
= effective depth of corbel
The limits of horizontal closed stirrup reinforcement at corbel design is :
A h
≥ 0.5( A s − An )
[11.11]
where : As
= area of primary tension reinforcement
An
= area of tension reinforcement
11.4.6
STEP – B Y – STEP PROCEDURE
The followings are the step – by – step procedure used in the flexural design for corbel (bracket) , as follows :
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Calculate ultimate flexure moment Mu.
Mu
= Vu (a ) + Nuc (h − d)
Calculate the area of tension reinforcement A n.
An
Nuc
φf y
Calculate the area of flexural reinforcement Af .
A f =
=
Mu
φf y (0.85d)
Calculate the area of primary tension reinforcement As .
CASE
As
CLOSED
PRIMARY
STIRRUP
REINFORCEMENT A h
1
2
A s
2
≥ A vf + An 3
A s ≥ A f + A n
A s
2
= A vf + An 3
A s = A f + An
1
2
3
3
1 2
2 d 3
Ah
= A vf
Ah
= A f
d
Check the reinforcement for minimum reinforcement .
ρ= Ah
11.5
LOCATION
A s bd
≥ 0.04
f 'c f y
≥ 0.5( A s − An )
APPLICATIONS 11.5.1
APPLICATION 01 – DESIGN OF CORBEL Vu=150000 N 100
Nuc
200
0 0 4
11 - 9
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PROBLEM Design the flexural and shear friction reinforcement of corbel structure above.
M ATERIA L Concrete strength
= K – 300
Steel grade
= Grade 400
Concrete cylinder strength
= f 'c
= 0.83 × 30 = 24. 9 MPa
β1 = 0.85 DIMENSION b
= 200
mm
h
= 400
mm
Concrete cover
= 30
mm
d
= 370
mm
DESIGN FORCE Vu = 150000 N Nuc Mu
= 0.2Vu = 0.2 × 150000 = 30000 N = Vu (a ) + Nuc (h − d) = 150000(100) + 30000(400 − 370) = 15900000 Nmm
L IMITATION CHECKING
φ(0.2f 'c )b w d = 0.85(0.2 × 24.9)200 × 370 = 313242 N φ(5.5 )b w d = 0.85 × 5.5 × 200 × 370 = 345950 N Vu
= 150000 < φ(0.2f 'c )bw d = 313242 < φ(5.5)b w d = 345950
SHEAR FRICTION REINFORCEMENT
μ = 1.4λ = 1.4 × 1.0 = 1.4 Vu A vf =
150000
φ = 0.85 = 315 mm2 f yμ 400 × 1.4
TENSION REINFORCEMENT An
=
Nuc φf y
=
30000 0.85 × 400
= 88 mm2
FLEXURAL REINFORCEMENT A f =
11 - 10
Mu
φf y (0.85d)
=
15900000 0.9 × 400(0.85 × 370 )
= 140 mm2
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PRIMARY TENSION REINFORCEMENT
A s
CASE
CLOSED
PRIMARY
STIRRUP
REINFORCEMENT
2
(mm )
A h
2
(mm )
LOCATION 2
(mm )
A s 1
A s
2
≥ A vf + An 3
2 (315) + 88 ≥ 298 3
≥
A s ≥ A f + An 2
A s
≥ 140 + 88 ≥ 228
1
= A vf
Ah A s = 298 A h
=
3
1 (315) = 105 3
A s = 228
(mm)
2 d 3 247
–
–
The reinforcement of the corbel are : 2
A s = 298 mm
2
Ah = 105 mm
CHECK FOR AS MINIMUM AND AS M AXIMUM
ρmin = 0.04 ρ=
A s bd
Ah−min
Ah
=
f 'c
= 0.04
f y
298 200 × 370
24.9 400
= 0.00249
= 0.00402 > ρmin = 0.00249
OK
= 0.5( A s − An ) = 0.5(298 − 88) = 210 mm2
= 105 < Ah−min = 210 ⇒ Ah = 210 mm2
The final reinforcement of the corbel are : 2
A s = 298 mm
2
Ah = 210 mm
CORBEL REINFORCEMENT A s
A h 2
A s
2
(mm )
(mm )
As=3D16
Ah=3(2 Legs D10)
⎛ 1 ⎞ ⎛ 1 ⎞ = 3⎜ πD2 ⎟ = 3⎜ π × 162 ⎟ = 603 ⎝ 4 ⎠ ⎝ 4 ⎠
A s
⎛ 1 ⎞ ⎛ 1 ⎞ = 3⎜ 2 × πD2 ⎟ = 3⎜ 2 × π × 102 ⎟ = 471 ⎝ 4 ⎠ ⎝ 4 ⎠
11 - 11
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SKETCH OF REINFORCEMENT
3D16
7 4 2
2 LEGS Ø10
11 - 12
