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TITLE
DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

1.0 General
Corbels are short cantilevers having shear span-to-depth(av/d) ratio
not greater than unity and whose depth Df at the end face is not less
than half the depth Ds at the support.when this criterion is
satisfied,the design of the corbel is governed mostly by shear.
However,if av>d,it is termed as a bracket which is designed for
flexure and shear as cantilever beams as per usual procedures.
The corbel may fail by shearing along the interface between the
column and the corbel,by yielding of the tension tie ,by crushing or
splitting of the compression strut,or by localized bearing or shearing
failure under the loading plate.
When av/d ratio is low,the load transfer to the support can be
assumed to take place more by 'strut & tie' action than by simple
bending.Under these circumstances,'Enhanced shear strength' may
be taken into account by increasing the design shear strength(tc) of
concrete to (2d/av)xtc provided that the design stress at the face of
the support remains less than t c,max.
Thus,Enhanced shear strength tce = t c x (2d/av) ≤ t c,max.
The depth of corbel at the face of the support is determined on the
basis of this enhanced shear strength.

2.0 Analysis of forces in corbels

av = distance of vertical load from the face of the support

b = breadth of the corbel
d = effective depth of the corbel at the face of the support
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DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

z = lever arm distance (distance of the center of compression from the
centre of the tension steel)
Pu = Design vertical load on the corbel
Hu = Design horizontal load on the corbel
Ft = tension in the horizontal direction
Fc = compression developed in concrete as strut action to support Pu

Using triangle law of forces, we get

Fc = Pu
(a +z )
v
2 2 1/2
z

Fc = Pu (a 2+z2)1/2 ----- 11
v

z 1
1
where cos b = av & sin b = z
(a +z )
v
2 2 1/2
(av2+z2)1/2

z = d - 0.45xu or xu = (d- z)/0.45

From the Strut & Tie Diagram,

Ft = H u + T u
= Hu + Fc cos b

Substituting the values of Fc and cos b,we get

Ft = Hu + Pu(av/z)

The value of compressive force can be found from the strength of

concrete,nothing but the area perpendicular to the force Fc over which
the compression is builtup is 0.9b xu cos b.

Fc =[(0.67/1.5)fck b]*(0.9xu cos b )

(where b is the width of corbel)

Fc =0.402 fck b xu cos b ----- 22

1
1 b
Pu =Fc sin b = 0.402 fck b xu cos b sin
From equation 1 and 2,

Equating the values of Fc , we get

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TITLE
DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

0.402 fck b xu cos b = Pu (av2+z2)1/2
z

Substituting the values of cos b & z in the above equation, we get

0.402 fck b (d- z) av = Pu (a 2+z2)1/2

v
0.45 (a +z )
v
2 2 1/2
z

0.402 fck b d (1 - (z/d)) av z = Pu (a 2+z2)

v

0.45

(1 - (z/d)) av z = 0.45 Pu (av2+z2)

0.402 fck b d

(1 - z/d) (av/d) ( z/d ) = 0.45 Pu ((av/d)2+(z/d)2)

0.402 fck b d

Putting K = 0.45 Pu & r = av , we get

0.402 fck b d d

(1 - (z/d)) ( r ) (z/d) = k [ r2 + (z/d)2 ]

r (z/d) - r (z/d)2 = k r2 + k (z/d)2

(z/d) 2 ( k+ r )- r (z/d) +k r2 = 0

(z/d) 2 - r (z/d) + k r2 = 0
(k + r ) (k + r)

Solving the above quadratic equation & taking the +ve root,we get

z = 0.5 r {1+ √ [1 - 4k(k + r)]}

d (k + r)

Once the value of 'z/d' is known,Ft & xu values can be computed as below
es
xu = d [ 1 - (z/d) ] & Ft = Hu + Pu(av/z)
d-xu

0.45

The Area of tension steel Ast = Ft

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DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

fs

xu
where fs is the stress in steel corresponding to the strain es in steel
ec = 0.0035
es = 0.0035 (d - xu )
xu

3.0 Design procedure:

The design of corbels is done in the following steps

3.1 Fixation of dimensions of bearing plate :-

In order that ultimate bearing pressure on concrete is not exceeded,fix
the dimensions of the bearing plate or bearing area taking the following
values of bearing pressure :
a) Bearing with no padding material : 0.4 fck
b) Bearing in cement mortar : 0.6 fck Cl 5.2.3.4 c ,
c) bearing on steel plate cast into member : 0.8 fck BS 8110 -1 : 1997

Area of bearing plate = Pu

sbr

3.2 Fixation of dimensions of corbel :-

Fix the enhanced value of shear (tce) as given tby
ce
= t c * (2d/av) .
To start with,assume tce nearer to tc,max . Compute the effective depth of
corbel at the root on the basis of vertical shear (d) : ( Pu / ( b *tce )) ,
where tce ≤ tc,max . Hence, Ds = ( d + cc + dia /2) Round off this
Df ≥ 0.5 Ds
value of Ds.Compute the total depth at the face of the corbel:
Check the dimension for 'corbel action' wherein the value of av/d <1.0 . Cl 28.1 a) , IS 456
However,it is preferable to keep av/d ratio less than 0.6 .

3.3 Determination of Lever arm and N.A.Depth :-

Knowing b,d & av ,determine the value of z/d using the following

z = 0.5 r {1+ √ [1 - 4k(k + r)]}

d (k + r)

where , K = 0.45 Pu & r = av

0.402 fck b d d

knowing Z,compute xu= d [ 1 - (z/d) ] & check that xu < xu,max

0.45
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DESIGN OF CORBEL

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Note : if xu > xu,max, adequate steel should be provided in

compression also. However, the support steel for main reinforcement
and horizontal shear reinforcement can satisfy this condition.

3.4 Computation of tensile force Tu :-

Compute T u= Pu ( av ) ['Tu' should be taken at least = 0.5Pu ]

z

3.5 Computation of tensile steel (Ast) :-

Compute es = 0.0035 (d - xu ) = 0.0035*(d/xu - 1)

xu

Find stress fs in tensile steel from the stress - strain curve relationship. Table A , SP 16
where fs is the stress in steel corresponding to the strain es in steel
Compute Ast = Ft = (Tu + Hu)
fs fs

The value of Ast found above should neither be less than 0.004 bd(i.e. Cl 7.7.2 , SP 34
0.4% bd) nor more than 0.013 bd (i.e.1.3% bd).If the value of Ast comes
out to not more than 1.3% bd,increase the depth (d) and redesign.

Determination of fs corresponding to strain εs Table A

SP16:1980
Fe 250 Fe 415 Fe 500
εs fs (N/mm2) εs fs (N/mm2) εs fs (N/mm2)
<0.00109 ε x Es < 0.00144 ε x Es < 0.00174 ε x Es
≥ 0.00109 217.39
Note 0.00144 288.7 0.00174 347.8
0.00163 306.7 0.00195 369.6
0.87 x 250 = 0.00109 0.00192 324.8 0.00226 391.3
200000 0.00241 342.8 0.00277 413
0.00276 351.8 0.00312 423.9
0.8x0.87x415 = 0.00144 ≥0.0038 360.9 ≥0.00417 434.8
2 x 10 ^5

0.8x0.87x500 = 0.00174
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DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

2 x 10 ^5

Note
Linear Interpolation shall be done for intermediate values.
Modulus of Elasticity of steel(for all grades) = 2 x 10^5 N/mm2

3.6 Computation of horizontal shear steel (Ash) :-

Cl 28.2.3 , IS 456
Ash ≥ 0.5 Ast
Provide these in the form of horizontal hoops in the upper two-third part
of the total depth of the corbel at the support.

3.7 Check for vertical shear :-

Since the exact value of Ast is known,find tc from Table 19 of IS 456:200

& Compute enhanced shear stress tce fromtce = tc (2d)
av
Compute the actual shear stress (tv) at the section :

tv = P u check that tv ≤ tc,max

bd

Design Shear Strength of Concrete tc N/mm2 Table 19 , IS 456

Cl 40.5.1 , IS 456
Grade of concrete
pt M15 M20 M25 M30 M35 ≥ M40
≤ 0.15 0.28 0.28 0.29 0.29 0.29 0.30
0.25 0.35 0.36 0.36 0.37 0.37 0.38
0.50 0.46 0.48 0.49 0.50 0.50 0.51
0.75 0.54 0.56 0.57 0.59 0.59 0.60
1.00 0.60 0.62 0.64 0.66 0.67 0.68
1.25 0.64 0.67 0.70 0.71 0.73 0.74
1.50 0.68 0.72 0.74 0.76 0.78 0.79
1.75 0.71 0.75 0.78 0.80 0.82 0.84
2.00 0.71 0.79 0.82 0.84 0.86 0.88
2.25 0.71 0.81 0.85 0.88 0.90 0.92
2.50 0.71 0.82 0.88 0.91 0.93 0.95
2.75 0.71 0.82 0.90 0.94 0.96 0.98
≥ 3.00 0.71 0.82 0.92 0.96 0.99 1.01
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DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

Maximum Shear Strength of Concrete tc,max N/mm 2

Grade of concrete M15 M20 M25 M30 M35 ≥M40 Table 20 , IS 456
(tcmax) Cl 40.5.1 , IS 456
N/mm 2
2.5 2.8 3.1 3.5 3.7 4.0

If tv comes out to be greater than t c,max ,redesign the section by taking higher
value of d.

Corbels should also be provided with Vertical stirrups that satisfies the
Minimum Shear Reinforcement requirements of cl.26.5.1.6 of IS 456:2000

Sv = 0.87 * fy *Asv
0.4 b
where,
Asv = total cross-sectional area of stirrup legs effective in shear.
Sv = stirrup spacing along the length of the member.
b = breadth of the beam or breadth of the web of flanged beam.
fy = characteristic strength of the stirrup reinforcement in N/mm2,
which shall not be taken greater than 415 N/mm2.

3.8 Provisions of compression steel :

Compression steel is not theoretically required. However, the practice

illustrated in SP 34 is as follows: ( Refer Fig.7.17 to 7.19 of SP 34)
a ) For Main bars > 16mm dia , the main tension bars at the front face of the
corbel are welded to transverse bars, thus providing nominal compression steel
to anchor the horizontal stirrups.
b) For Main bars ≤ 16mm dia , the main tension bars are bent back to form
hoops to act as compression steel.

3.9 Detailing of reinforcement :-

As per the standard practice given in SP34.

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DESIGN OF CORBEL

CL. NO. DESIGN CALCULATIONS REFERENCES

 PAGE 9 .

CL. NO. CORBEL DESIGN CALCULATIONS REFERENCES

1.0 Basic Input Data

Grade of Concrete = M 30
Grade of Steel = Fe 500
Design Vertical Load on the corbel Pu = 75 kN
Dist. of Pu from the face of the column av = 500 mm
Design Horizontal Load on the corbel Hu = 18.75 kN
Nominal cover cc = 40 mm
Diameter of bar in Tension dia = 12 mm
Width of the Corbel b = 450 mm
Over all Depth of corbel @ the face of the Column Ds = 600 mm
Depth of corbel @ the end face ( ≥ Ds / 2 ) Df = 600 mm Cl 28.1 b) , IS 456

2.0 Design of Bearing plate :

Length of bearing plate(taken as width of corbel) = 450 mm

Bearing Strength ( 0.8 fck) sbr = 24 N/mm2 Cl 5.2.3.4 c) ,
Width of the bearing plate( Pu / ( b *sbr )) = 7 mm BS 8110 -1 : 1997
Provide the width of the Bearing plate = 450 mm

3.0 Check for Strut Action :

Effective Depth from bottom of corbel to centre of

Tensile Reinforcement ( Ds - cc - dia / 2 ) d = 554 mm
Shear span to depth Ratio av / d = 0.9025 <### Cl 28.1 a) , IS 456
Hence designed as a corbel and strut & tie action will be available

4.0 Determination of N.A Depth & Lever Arm :

Lever Arm to depth Ratio Z/d = 0.978

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CL. NO. CORBEL DESIGN CALCULATIONS REFERENCES

Depth to Neutral Axis from compression Edge xu = 27.08 mm
Maximum Depth of Neutral Axis xu,max = 252.62 mm > xu Table B , SP 16

Note : if xu > xu,max, adequate steel should be provided in

compression also. However, the support steel for main reinforcement
and horizontal shear reinforcement can satisfy this condition.

5.0 Computation of Area of Tension Reinforcement :( Ast )

Tensile Force in Reinforcement ( Pu x av / z ) Tu = 69.21 kN

Min. Tensile Force in Reinforcement ( Pu / 2) Tu,min = 37.50 kN
Strain in the Tensile Reinforcement εs = 0.0681 0.0035*(d/xu - 1)
Stress in the Tensile Reinforcement fs = 434.8 N/mm2 Table A , SP 16
Ast,reqd = (Tu + Hu)/ fs = 202.3 mm 2

5.0.1 check for Minimum and Maximum steel

Ast,min = 0.4 % b d = 997.20 mm2 Cl 7.7.2 , SP 34

Ast,max = 1.3 % b d = 3240.90 mm2 Cl 7.7.2 , SP 34
Provide 4 Nos of Y 20 bars Ast,provided = 1256.64 mm 2

997.20 < 1256.64 < 3240.90 mm 2

Area of Steel is with in the permissible limit ; Hence O.K

5.0 Computation of Area of Horizontal Shear Reinfocement ( Ash ) Cl 28.2.3 , IS 456

Minimum Area of Horizontal Reinforcement Ash,min = 628.32 mm2 ( ≥ Ast / 2 )

This Area of steel has to be provided in the form of horizontal Loops in the
Upper 2/3 of the effective depth of the corbel at the support . i.e., over a
distance equals to(2/3) * ( ### ) =
369 mm
Provide 5 Nos of Y 10 bars 2 leg. sh,provided =
A 785.40 mm2
Spacing of Y 10 stirrups 2 legged in the 2/3 d = 70 mm

6.0 Check for Vertical Shear capacity :

% of Tensile Reinforcement provided pt = 0.504 %

Design shear strength of concrete tc = 0.5 N/mm2 Table 19 , IS 456
Enhanced Shear Strength t ce = t c * (2d/av) tce = 1.11 N/mm 2
Cl 40.5.1 , IS 456
Actual shear stress tv = 0.3 N/mm2
Maximum shear stress tc,max = 3.5 N/mm2 Table 20 , IS 456
tce subjected to a Maximum of tc,max ; Hence tce = 1.11 N/mm 2 Cl 40.5.1 , IS 456
Shear capacity of concrete = tce x bd = 276.72 kN
Shear capacity of steel = (0.87 fy Ash d / Sv ) = 448.85 kN
Total Shear capacity of the section = 725.57 kN
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CL. NO. CORBEL DESIGN CALCULATIONS REFERENCES

Design Vertical Load on the corbel Pu = 75 kN
725.57 kN > 75.00 kN Hence Safe
Also tv < tc,max Hence Safe

Corbels should be provided with Vertical stirrups that satisfies the Minimum
Shear Reinforcement requirements of cl.26.5.1.6 of IS 456:2000
Spacing of Y 10 stirrups 2 legged = 289.72 mm
Maximum Spacing of stirrups = 289.7 mm
Provide Y 10 stirrups 2 legged of spacing = 280 mm

7.0 Provisions of compression steel :

Compression steel is not theoretically required. However, the practice

illustrated in SP 34 is as follows: ( Refer Fig.7.17 to 7.19 of SP 34)
a ) For Main bars > 16mm dia , the main tension bars at the front face of the
corbel are welded to transverse bars, thus providing nominal compression steel
to anchor the horizontal stirrups.
b) For Main bars ≤ 16mm dia , the main tension bars are bent back to form
hoops to act as compression steel.

8.0 Dimensions of Corbel:

Minimum Length = 880 mm Width of corbel b = 450 mm

Depth @ support = 600 mm Depth @ free end Df = 600 mm
 Contents:

1 IS 456:2000 Codal Provisions for Corbel.

2 SP 34:1987 Specification for corbel.

3 Technical Notes

4 Corbel Design Template.

Horizontal
Node L/C Fx kN
Max Fx 230 1 WL+X 0
Min Fx 230 1 WL+X 0
Max Fy 269 16 1.5(DL+LL) 0
Min Fy 269 23 1.5(DL+WL+ 0
Max Fz 262 16 1.5(DL+LL) 0
Min Fz 269 16 1.5(DL+LL) 0
Max Mx 230 1 WL+X 0
Min Mx 230 1 WL+X 0
Max My 230 1 WL+X 0
Min My 230 1 WL+X 0
Max Mz 230 1 WL+X 0
Min Mz 230 1 WL+X 0

Horizontal Vertical
Node L/C Fx kN Fy kN
1 10 1.0(DL+WL+X) 1.768 -38.63
11 1.0(DL+WL-X) 17.731 -101.42
12 1.0(DL+WL+Z) 12.485 -88.582
13 1.5(DL+LL) -25.831 183.358
14 1.2(DL+LL+WL+X -13.633 60.402
15 1.2(DL+LL+WL-X 5.522 -14.947
16 1.2(DL+LL+WL+Z -0.773 0.459
17 1.5(DL+WL+X) 2.652 -57.944
18 1.5(DL+WL-X) 26.596 -152.13
19 1.5(DL+WL+Z) 18.728 -132.872

11 10 1.0(DL+WL+X) 33.059 -155.268

11 1.0(DL+WL-X) 18.208 -65.804
12 1.0(DL+WL+Z) 33.672 -137.384
13 1.5(DL+LL) -61.699 277.569
 14 1.2(DL+LL+WL+X 3.161 -22.972
15 1.2(DL+LL+WL-X -14.661 84.385
16 1.2(DL+LL+WL+Z 3.897 -1.511
17 1.5(DL+WL+X) 49.589 -232.901
18 1.5(DL+WL-X) 27.311 -98.705
19 1.5(DL+WL+Z) 50.509 -206.075

50.509 277.569

12 10 1.0(DL+WL+X) -7.227 -54.841

11 1.0(DL+WL-X) 12.428 -19.914
12 1.0(DL+WL+Z) 2.523 -47.258
13 1.5(DL+LL) 7.604 96.562
14 1.2(DL+LL+WL+X -4.258 -8.859
15 1.2(DL+LL+WL-X 19.328 33.054
16 1.2(DL+LL+WL+Z 7.442 0.24
17 1.5(DL+WL+X) -10.84 -82.262
18 1.5(DL+WL-X) 18.642 -29.871
19 1.5(DL+WL+Z) 3.784 -70.887

19.328 96.562

59 10 1.0(DL+WL+X) -6.34 -80.371

11 1.0(DL+WL-X) -41.812 -132.052
12 1.0(DL+WL+Z) -31.039 -130.265
13 1.5(DL+LL) 61.764 263.663
14 1.2(DL+LL+WL+X 28.838 58.809
15 1.2(DL+LL+WL-X -13.729 -3.208
16 1.2(DL+LL+WL+Z -0.802 -1.063
17 1.5(DL+WL+X) -9.509 -120.557
18 1.5(DL+WL-X) -62.718 -198.078
19 1.5(DL+WL+Z) -46.559 -195.397

61.764 263.663

85 10 1.0(DL+WL+X) -7.736 -5.543

11 1.0(DL+WL-X) -12.029 -17.799
12 1.0(DL+WL+Z) -12.001 -15.153
13 1.5(DL+LL) 18.162 37.896
14 1.2(DL+LL+WL+X 2.123 14.11
15 1.2(DL+LL+WL-X -3.029 -0.598
16 1.2(DL+LL+WL+Z -2.996 2.578
17 1.5(DL+WL+X) -11.604 -8.314
18 1.5(DL+WL-X) -18.044 -26.699
19 1.5(DL+WL+Z) -18.002 -22.729

18.162 37.896
Vertical HorizontalMoment
Fy kN Fz kN Mx kNm My kNm Mz kNm
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0
269.104 -56.843 0 0 0
-223.208 47.476 0 0 0
251.397 50.07 0 0 0
269.104 -56.843 0 0 0
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0
-22.825 -11.061 0 0 0

269.104 50.07

HorizontalMoment
Fz kN Mx kNm My kNm Mz kNm Mz kNm
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0