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 2
Materials: M-35 grade concrete (f ck ck =35 N/mm ), 2
Fe-415 HYSD bars (f y = 415 N/mm ) Dimension of corbel
Bearing Length
= Width of Column = 300 mm
Using a bearing plate plate of length = 300 mm , the bearing bearing pressure is calculated calculated 2
Bearing pressure = 0.8f ck ck = 0.8x 35 = 28 N/mm
Width of plate
=
= 59.52 mm
Therefore provide a minimum width of 100 mm and adopt a be aring plate 100 x 300 mm Estimation of depth:
τc,max =
2 3.7 N/mm ……………………………………. (From table -20, IS 456 for M-35 grade
concrete)
d= = = 450.45 mm Adopt effective depth,
d
= 475 mm
Total depth at support, D s = (d + cover + ½ 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
= 0.213
Substituting, we have
Solving we get,
0.74
Therefore
z = 0.74 x d = 0.74 x 475 = 351.5mm
d – z = 0.4x 475 – 351.5 = 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 =
=
Ft not less than 0.5F v = 0.5 x 500
= 284.5 = 250 kN = 285.5 kN
Area of Tension reinforcement:
= 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 ξs = 0.00188 , fs = 322 N/mm
2
But Fh = 0
2
= 882.71 mm
2
Use 5 bars of 16mm diameter (Ast = 1005.31 mm ) Check for minimum and maximum reinforcement:
= 0.6 > 0.4 but < 1.3 percent Hence satisfactory. Area of shear reinforcement:
Asv(min) =
=
2
= 502.66mm
Provide 4 numbers of 10 mm diameter 2 legged horizontal links in the upper two 2
third (Asv = 628 mm ).
Spacing of links = S v =
= 75mm
Shear capacity of section :
Using table -19 of IS 456-2000, M-35 grade concrete and 0.6 percent steel. 2
) = 0.42 =
τc = 0.536 N/mm and (
Enhanced Shear Strength
= 2.55 N/mm2
Shear Capacity of Concrete, V c =
= 363.71 kN Shear Capacity 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 face of 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 2
Materials: M-25 grade concrete (f ck =25 N/mm ), 2
Fe-415 HYSD bars (f y = 415 N/mm ) Dimension of corbel
Bearing Length
= Width of Column = 300 mm
Using a bearing plate of length = 300 mm , the bearing pressure is calculated 2
Bearing pressure = 0.8f ck = 0.8x 25 = 20 N/mm
Width of plate
=
= 41.67 mm
Therefore provide a minimum width of 100 mm and adopt a be aring plate 100 x 300 mm Estimation of depth:
τc,max =
2 3.1 N/mm ……………………………………. (From table -20, IS 456 for M-35 grade
concrete) d=
= = 268.82 mm Adopt effective depth,
d
= 350 mm
Total depth at support, D s = (d + cover + ½ 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
=
= 0.57
= 0.84
= 0.16
Substituting, we have
Solving we get,
Therefore
0.77 z = 0.83x d = 0.77 x 350 = 269.5mm
d – z = 0.4x 350 – 269.5 = 0.4x x = 201.25 mm
Therefore
= = 0.575 < 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 =
=
Ft not less than 0.5F v = 0.5 x 250
= 185.53 = 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 ξs = 0.00259 , fs = 347.43 N/mm
2
But Fh = 0
= 524 mm2 2
Use 3 bars of 16mm diameter (Ast = 603.19 mm ) Check for minimum and maximum reinforcement:
= 0.57 > 0.4 but < 1.3 percent Hence satisfactory. Area of shear reinforcement:
Asv(min) =
=
= 301.6 mm
2
Provide 2 numbers of 10 mm diameter 2 legged horizontal links in the upper two 2
third (Asv = 314.16 mm ).
Spacing of links = S v =
= 58.33mm = 50 mm
Shear capacity of section :
Using table -19 of IS 456-2000, M-25 grade concrete and 0.6 percent steel. 2
) = 0.57
τc = 0.5124 N/mm and (
Enhanced Shear Strength
=
= 1.8 N/mm2 Shear Capacity of Concrete, V c =
= 188.79 kN Shear Capacity of Steel
=
=
= 396.8 kN Total Shear Capacity
= 188.79 + 396.8 = 585.59 kN > 500 kN
