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CHAPTER
11
THE FLEXURE AND SHEAR
DESIGN
OF
CORBEL
(BRACKET)
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 in corbel.
11.2 BEHAVIOR 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 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 of corbel are : yielding of the tension tie, failure of the end anchorage of the
tension tie, failure of concrete by compression or shearinga and bearing failure.
The followings figure shows the mode of failure of corbel.
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Vu
DIAGONAL SHEAR
Vu
SHEAR FRICTION
Vu
Vu
Nu
ANCHORAGE SPLITING
FIGURE 11.1
VERTICAL SPLITING
MODES OF FAILURE 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 :
Vn Vu
where :
11 - 2
Vn
= nominal shear strength of shear friction reinforcement
Vu
= ultimate shear force
= strength reduction factor ( = 0.85)
[11.1]
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Vu
SHEAR FRICTION REINFORCEMENT
Avf f y
Avf f y
ASSUMED CRACK
FIGURE 11.1
SHEAR FRICTION THEORY
The nominal shear strength of shear friction reinforcement is :
TABLE 11.1
SHEAR FRICTION REINFORCEMENT STRENGTH
VERTICAL
INCLINED
SHEAR FRICTION
SHEAR FRICTION
REINFORCEMENT
REINFORCEMENT
Vn
Avf
A vf =
Vn = A vf fy
Vn
Vn
fy
Vu
A vf =
Avf
A vf =
Vn = A vf fy ( sin f + cos f )
fy
Vn
fy ( sin f + cos f )
Vu
A vf =
fy ( sin f + cos f )
where :
Vn
= nominal shear strength of shear friction reinforcement
Avf
= area of shear friction reinforcement
Fy
= yield strength of shear friction reinforcement
= coefficient of friction
TABLE 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
COEFFICIENT OF FRICTION
1.4
1.0
0.6
0.7
The value of is :
= 1.0
normal weight concrete
= 0.85
sand light weight concrete
= 0.75
all light weight concrete
11 - 3
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The ultimate shear force must follows the following condiitons :
Vu (0.2f 'c )b w d
[11.1]
Vu (5.50 )b w d
where :
Vu
= ultimate shear force
(N)
fc
= concrete cylinder strength
(MPa)
bw
= width of corbel section
(mm)
= effective depth of corbel
(mm)
11.3.3
STEP BY 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.
Vu (0.2f 'c )b w d
Vu (5.50 )b w d
Calculate the area of shear friction reinforcement Avf.
VERTICAL
INCLINED
SHEAR FRICTION
SHEAR FRICTION
REINFORCEMENT
REINFORCEMENT
Vn
Avf
A vf =
Vn = A vf fy
Vn
fy
Vu
A vf
Vn
Avf
A vf =
Vn = A vf fy ( sin f + cos f )
=
fy
Vn
fy ( sin f + cos f )
Vu
A vf =
fy ( sin f + cos f )
The design must be follows the basic design equation as follows :
Vn 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.
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Vu
a
d
h
min d/2
Nuc
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 :
[11.2]
Nuc
fy
An =
where :
An
= area of tension reinforcement
Nuc
= ultimate horizontal force at corbel
fy
= yield strength of the tension reinforcement
= strength reduction factor ( = 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 shear.
FLEXURAL REINFORCEMENT
Vu
a
Ts
Nuc
d
h
jd
11.4.3
Cc
FIGURE 11.3
ULTIMATE FLEXURE MOMENT AT CORBEL
11 - 5
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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
= distance of Vu from face of column
Nuc
= ultimate horizontal force at corbel
= height of corbel
= effective depth of corbel
The resultant of tensile force of tension reinforcement is :
Tf = A f fy
[11.4]
where :
Tf
= tensile force resultant of flexure reinforcement
Af
= area of flexure reinforcement
fy
= yield strength of the flexure reinforcement
The resultant of compressive force of the concrete is :
Cc = 0.85 f 'c ba(cos )
[11.5]
where :
Cc
= compressive force resultant of concrete
fc
= concrete cylinder strength
= width of corbel
= depth of concrete compression zone
The horizontal equilibrium of corbel internal force is :
H = 0 Cc =Ts
[11.6]
0.85 f 'c ba(cos ) = A f fy
a=
A f fy
0.85f 'c b(cos )
The flexure reinforcement area is :
Af =
11 - 6
Mu
a
fy d
2
[11.7]
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Af =
Mu
A f fy
0.85 f ' b(cos )
fy d
Cos value can be calculated based on the Tan value as follows :
Tan =
[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 :
Af =
Af =
[11.9]
Mu
fy (jd)
Mu
fy (0.85d)
where :
Af
= area of flexural reinforcement
Mu
= ultimate flexure moment at corbel
fy
= yield strength of the flexural reinforcement
= strength reduction factor ( = 0.9)
= effective depth of corbel
11.4.4
DISTRIBUTION OF CORBEL REINFORCEMENTS
Vu
Avf +An
Nuc
As= Af +An
Nuc
Ah=
1
3
Avf
d
h
(2/3)d
2
3
(2/3)d
As=
Vu
Ah=
1
2
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.
TABLE 11.3
DISTRIBUTION OF CORBEL REINFORCEMENTS
CASE
CLOSED
PRIMARY
As
STIRRUP
REINFORCEMENT
Ah
As
2
A vf + A n
3
A s A f + An
As =
2
A vf + A n
3
A s = A f + An
LOCATION
Ah =
1
A vf
3
2
d
3
Ah =
1
Af
2
2
d
3
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
= 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 :
As
f'
0.04 c
bd
fy
[11.10]
where :
As
= area of primary tension reinforcement
= width of corbel
= effective depth of corbel
The limits of horizontal closed stirrup reinforcement at corbel design is :
A h 0.5(A s A n )
[11.11]
where :
As
= area of primary tension reinforcement
An
= area of tension reinforcement
11.4.6
STEP BY 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 An.
An =
Calculate the area of flexural reinforcement Af.
Af =
Nuc
fy
Mu
fy (0.85d)
Calculate the area of primary tension reinforcement As.
CASE
As
CLOSED
PRIMARY
STIRRUP
REINFORCEMENT
Ah
As
2
A vf + A n
3
A s A f + An
As =
2
A vf + A n
3
A s = A f + An
LOCATION
Ah =
1
A vf
3
2
d
3
Ah =
1
Af
2
2
d
3
Check the reinforcement for minimum reinforcement.
As
f'
0.04 c
bd
fy
A h 0.5(A s A n )
11.5 APPLICATIONS
APPLICATION 01 DESIGN OF CORBEL
Vu=150000 N
100
Nuc
200
400
11.5.1
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PROBLEM
Design the flexural and shear friction reinforcement of corbel structure above.
MATERIAL
Concrete strength
= K 300
Steel grade
= Grade 400
Concrete cylinder strength
= f 'c = 0.83 30 = 24.9 MPa
1 = 0.85
DIMENSION
= 200
mm
= 400
mm
Concrete cover
= 30
mm
= 370
mm
DESIGN FORCE
Vu = 150000 N
Nuc = 0.2Vu = 0.2 150000 = 30000 N
Mu = Vu (a ) + Nuc (h d) = 150000 (100 ) + 30000 (400 370 ) = 15900000 Nmm
LIMITATION CHECKING
(0.2f 'c )b w d = 0.85(0.2 24.9 )200 370 = 313242 N
(5.5 )b w d = 0.85 5.5 200 370 = 345950 N
Vu = 150000 < (0.2f 'c )b w d = 313242 < (5.5 )b w d = 345950
SHEAR FRICTION REINFORCEMENT
= 1 .4 = 1 .4 1 .0 = 1 .4
Vu
A vf =
=
fy
150000
0.85 = 315 mm2
400 1.4
TENSION REINFORCEMENT
An =
Nuc
30000
=
= 88 mm2
fy
0.85 400
FLEXURAL REINFORCEMENT
Af =
11 - 10
Mu
15900000
2
=
= 140 mm
fy (0.85d) 0.9 400(0.85 370 )
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PRIMARY TENSION REINFORCEMENT
As
CASE
CLOSED
PRIMARY
2
(mm )
STIRRUP
REINFORCEMENT
Ah
(mm )
LOCATION
2
(mm )
As
2
A vf + A n
3
As
2
(315 ) + 88 298
3
A s A f + An
2
A s 140 + 88 228
Ah =
A s = 298
Ah =
1
A vf
3
1
(315 ) = 105
3
A s = 228
(mm)
2
d
3
247
The reinforcement of the corbel are :
A s = 298 mm2
A h = 105 mm2
CHECK FOR AS MINIMUM AND AS MAXIMUM
min = 0.04
=
f 'c
24.9
= 0.04
= 0.00249
fy
400
As
298
=
= 0.00402 > min = 0.00249
bd 200 370
OK
A hmin = 0.5(A s A n ) = 0.5(298 88 ) = 210 mm2
A h = 105 < A hmin = 210 A h = 210 mm2
The final reinforcement of the corbel are :
A s = 298 mm2
A h = 210 mm2
CORBEL REINFORCEMENT
As
Ah
2
(mm )
(mm )
As=3D16
Ah=3(2 Legs D10)
A s = 3 D2 = 3 162 = 603
4
4
1
1
A s = 3 2 D2 = 3 2 102 = 471
4
4
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SKETCH OF REINFORCEMENT
247
3D16
2 LEGS 10
11 - 12
