Design of Corbel

8/10/2019 Design of Corbel

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Design a corbel to support a factored load of 500KN at a distance of 200mm from the face

of a column 300mm x 300mm. Adopt M-35 grade concrete and Fe-415 HYSD bars. Sketch

the details of reinforcement details.

DATA:

Assume the load as factored load

Factored load = Fv = 500KN

Width of column = Length of Corbel = 300 mm

Shear span = av = 200 mm

Materials: M-35 grade concrete (fck=35 N/mm2),

Fe-415 HYSD bars (fy= 415 N/mm

2

)

Dimension of corbel

Bearing Length = Width of Column = 300 mm

Using a bearing plate of length = 300 mm , the bearing pressure is calculated

Bearing pressure = 0.8fck = 0.8x 35 = 28 N/mm2

Width of plate =

= 59.52 mm

Therefore provide a minimum width of 100 mm and adopt a bearing plate 100 x 300

mm

Estimation of depth:

c,max = 3.7 N/mm2.(From table -20, IS 456 for M-35 grade

concrete)

d =

=

= 450.45 mm

Adopt effective depth, d = 475 mm


2/9

Total depth at support, Ds = (d + cover + diameter of bar)

= (475 + 40 +10)

= 525 mm

Depth at face, Df = (0.5 Ds)

= (0.5 x 525)

= 263 mm

Check for strut action:

Ratio (av / d) =

= 0.42 < 0.6 , Hence acts as a corbel

Determination of Lever arm (z):

Using the Eq.

()

Where, Kv=

=

= 0.114

=

= 0.42

= 0.79


3/9

= 0.213

Substituting, we have

Solving we get, 0.74

Therefore z = 0.74 x d

= 0.74 x 475

= 351.5mm

dz = 0.4x

475351.5 = 0.4x

x = 308.75 mm

Therefore=

= 0.65 > the limiting value of 0.48 for Fe-415 bars (pg 70 IS 456)

Hence adequate steel should be used in compression also.

Resolution of forces:

Ft=

=

= 284.5

Ftnot less than 0.5Fv= 0.5 x 500 = 250 kN

= 285.5 kN

Area of Tension reinforcement:


4/9

= 0.0035 x

= 0.00188

From fig 3 of SP-16 or T.A pg 6 of SP-16 read out stress corresponding to strain s =

0.00188 , fs = 322 N/mm2

But Fh= 0

= 882.71 mm2

Use 5 bars of 16mm diameter (Ast = 1005.31 mm2)

Check for minimum and maximum reinforcement:

= 0.6 > 0.4 but < 1.3 percent

Hence satisfactory.

Area of shear reinforcement:

Asv(min) =

=

= 502.66mm2

Provide 4 numbers of 10 mm diameter 2 legged horizontal links in the upper two

third (Asv = 628 mm2).

Spacing of links = Sv=

= 75mm

Shear capacity of section :

Using table -19 of IS 456-2000, M-35 grade concrete and 0.6 percent steel.

c= 0.536 N/mm2 and (

) = 0.42

Enhanced Shear Strength =

= 2.55 N/mm2


5/9

Shear Capacity of Concrete, Vc=

= 363.71 kN

ShearCapacity of Steel =

=

= 359 kN

Total Shear Capacity = 363.71 + 359

= 722.71 kN > 500 kN

Design a corbel to support a factored load of 250KN at a distance of 200mm from the faceof a column 300mm x 300mm. Adopt M-25 grade concrete and Fe-415 HYSD bars. Sketch

the details of reinforcement details.

DATA:

Assume the load as factored load

Factored load = Fv = 250KN

Width of column = Length of Corbel = 300 mm

Shear span = av = 200 mm

Materials: M-25 grade concrete (fck=25 N/mm2),

Fe-415 HYSD bars (fy= 415 N/mm2)

Dimension of corbel

Bearing Length = Width of Column = 300 mm

Using a bearing plate of length = 300 mm , the bearing pressure is calculated

Bearing pressure = 0.8fck = 0.8x 25 = 20 N/mm2

Width of plate =

= 41.67 mm

Therefore provide a minimum width of 100 mm and adopt a bearing plate 100 x 300

mm

Estimation of depth:


6/9

c,max = 3.1 N/mm2.(From table -20, IS 456 for M-35 grade

concrete)

d =

=

= 268.82 mm

Adopt effective depth, d = 350 mm

Total depth at support, Ds = (d + cover + diameter of bar)

= (350 + 40 +10)

= 400 mm

Depth at face, Df = (0.5 Ds)

= (0.5 x 300)

= 200 mm

Check for strut action:

Ratio (av / d) =

= 0.57 < 0.6 , Hence acts as a corbel

Determination of Lever arm (z):

Using the Eq.

()

Where, Kv=

=

= 0.108


7/9

=

= 0.57

= 0.84

= 0.16

Substituting, we have

Solving we get, 0.77

Therefore z = 0.83x d

= 0.77 x 350

= 269.5mm

dz = 0.4x

350269.5 = 0.4x

x = 201.25 mm

Therefore=

= 0.575 < the limiting value of 0.48 for Fe-415 bars (pg 70 IS 456)


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Hence adequate steel should be used in compression also.

Resolution of forces:

Ft=

=

= 185.53

Ftnot less than 0.5Fv= 0.5 x 250 = 125 kN

= 185.53 kN

Area of Tension reinforcement:

= 0.0035 x

= 0.00259

From fig 3 of SP-16 or T.A pg 6 of SP-16 read out stress corresponding to strain s =

0.00259 , fs = 347.43 N/mm2

But Fh= 0

= 524 mm2

Use 3 bars of 16mm diameter (Ast = 603.19 mm2)

Check for minimum and maximum reinforcement:

= 0.57 > 0.4 but < 1.3 percent

Hence satisfactory.

Area of shear reinforcement:

Asv(min) = =

= 301.6 mm2

Provide 2 numbers of 10 mm diameter 2 legged horizontal links in the upper two

third (Asv = 314.16 mm2).


9/9

Spacing of links = Sv=

= 58.33mm = 50 mm

Shear capacity of section :

Using table -19 of IS 456-2000, M-25 grade concrete and 0.6 percent steel.

c= 0.5124 N/mm2 and (

) = 0.57

Enhanced Shear Strength =

= 1.8 N/mm2

Shear Capacity of Concrete, Vc=

= 188.79 kN

ShearCapacity of Steel =

=

= 396.8 kN

Total Shear Capacity = 188.79 + 396.8

= 585.59 kN > 500 kN

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Design of Corbel