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7.0 Punching Check
7.1 Punching of Column in Column Capital
Data Input
Size of Column :
Diameter of Column = 1900 mm (D)
Depth of Column Capital = 1900 mm
Axial load in column above base ( = 114185 kN (Member 31032)
Diameter of punching area = 1900 + 1760 =
Relief from Punching Area (ULS) = 12000 x 10.52 =
(Assuming Rock End bearing of 8000 kN/m2 on conservative side)
Axial Load on Column = Pu = -12066 kN (Maximum of all cases)
Transverse Moments
LHS Moment M1 = 20398 kN-m (Member 10427)
RHS Moment M2 = 19951 kN-m (Member 10424)
Shear Resistance of Section = ts
ts = ks*tc
Where,
ks = ( 0.5+βc )
βc = Ratio of Short side to long side of column capital
tc = 0.25*(fck)0.5
For Our Case
fck = 40 MPa
fys = 415 MPa (For Stirrups)
tc = 1.58 MPa
βc = 1.00
ks = 1.00
Shear Resistance = ts = 1.58 Mpa
Total Load above Roof Slab = 14.185 x10
141.85 kN/m2 (SLS)
As per IS-456, Critical Section for Punching in Flat Slabs are at distance d/2 from the face of column
Effective depth = d = 1760 mm
d/2 = 880 mm
Critical Section Perimeter b01 = 11498 mm
τv = τv1 + τv2
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τv1 = Shear Stresses due to vertical Load
τv2 = Additional Shear Stresses due to Unbalanced moments at both end of
τv1 = Vu
b01*d
τv2 = Muv * c
Jc
Where,
Vu = Shear force due to vertical loads, for our case considered to be equal to Pu
b01 = Perimeter of critical section i.e. perimeter of the section at distance d/2 from column capital
Mu = Unbalanced moment at critical section from in both sides of section
Muv = Moment which is transferred through shear
Where, Muv = (1-α ) * Mu
c= Distance of point under consideration on face of critical section to centroidal axis of critical
Jc = Property of critical section analogous to the polar moment of Intertia
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As per IS-456 (2000) clause 31.6.2.2,
For unbalanced gravity load, wind, eqrthquake or other forces causes tranfer of bending moment betwee
fraction (1-α) of moment shall be considered transferred by eccenticity of shear about the centeroid of cr
This shear stresses is considered to be varying linearly about the centroid of critical section
1
α=
(1+2/3*(a1/a2)0.5)
a1 = Overall dimension of critical section for shear in the direction in which moment acts
a2 = Overall dimension of critical section for shear transverse to the direction in which moment
α= 0.600
(1-α )= 0.400
Mu = M1-M2 = 447 kN-m
Muv = (1-α ) * Mu = 179 kN-m
For interior Column,
c = (b+d)/2 = 950 mm
Jc =[(b+d)*(d3)+(b+d)3*d] / 6 + (b+d)2*b*d /2 = 9.774E+12 mm4
Vu Muv * c
τv = b01*d + Jc
τv = -0.5962 + 0.0174 = -0.579 Mpa
Vu1 = τv*b01*d = -11714 kN
Where,Vu1 = Shear Force Due to vertical load and unbalanced moments
7.0.1 Check for Section Adequacy :
As per IS 456-2000 clause 31.6.3.2., if shear stresses at critical section (ts) exceeds (1.5 x tv ), section shou
1.5* ts = 2.37 Mpa
τv = -0.579 < 2.37
Mpa Mpa
Designed Section is O.K.
7.0.2 Check for Requirement of Shear R/f :
As per IS 456-2000 clause 31.6.3.2., if shear stresses at critical section (ts) exceeds (tv )and less than (1.5
Shear reinforement to be provided
τv = -0.579 < 1.58
Mpa Mpa
Shear R/f Not Required
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3660
126251 kN
ll cases)
e of column
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s at both end of section
column capital face
axis of critical section
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moment between slab and column,
centeroid of critical section.
ent acts
which moment acts
tv ), section should be revised
d less than (1.5 x tv )
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