For the flat slab with the general arrangement as shown below, let us design the punching shear for column B1 given the following design information;


Ultimate axial force on column VEd = 400 kN
Thickness of slab = 250 mm
Dimension of column = 450 x 230 mm
Reinforcement of slab in the longer direction = H16@150mm (As,prov = 1340 mm2)
Reinforcement of slab in the shorter direction = H16@175mm (As,prov = 1149 mm2)
Grade of concrete = C30
Yield strength of reinforcement = 500 Mpa
Concrete cover to slab = 25mm


Solution

Effective depth of slab in y-direction dy = 250 – 25 – (16/2) = 217 mm
Effective depth of slab in x-direction dx = 250 – 25 – 16 = 209 mm

ρly = (1340) / (1000 × 217) = 0.00617 (reinforcement ratio)
ρlx = (1149) / (1000 × 209) = 0.00549 (reinforcement ratio)

(a) Check shear at the perimeter of the column

VEd = β VEd/(u0d) < VRd,max

From figure 6.21N of EN 1992-1-1;
β = 1.40
d = (217 + 209)/2 = 213 mm


u0 = c2 + 3d < c2 + 2c1 For edge columns (clause 6.4.5(3))

u0 = 230 + (3 × 213) <  (230 + 2 × 450)
u0 = 869 mm
VEd = 1.40 × 400 × 1000/(869 × 213) = 3.025 MPa
VRd,max = 0.5 ν fcd
= 0.5 × 0.6(1 – fck/250) × αcc fckm
= 0.5 × 0.6(1 – 30/250) × 1.0 × (30 /1.5) = 5.28 MPa
VEd < VRd,max …OK


(b) Check shear at u1, the basic control perimeter
VEd = β VEd/(u1d) < VRd,c

β,VEd as before
u1 = c2 + 2c1 + π × 2d
u= 230 + (2 × 450) + (π × 2 × 213) = 2468 mm

VEd = 1.4 × 400 × 1000/(2468 × 213) = 1.065 MPa
VRd,c = 0.12 k(100 ρl fck)1/3

k = 1 + (200/d)1/2 = 1 + (200/213)1/2 = 1.969


ρl = (ρlyρlx)1/2 = (0.00617 × 0.00549)1/2 = 0.00582

VRd,c = 0.12 × 1.969(100 × 0.00582 × 30)1/3 = 0.613 MPa

VEd > VRd,c ?
1.065 MPa > 0.613 MPa … Therefore punching shear reinforcement required

NA check:
VEd ≤ 2.0VRd,c at basic control perimeter
1.06 MPa ≤ 2 × 0.613 MPa = 1.226 MPa – OK

(c) Perimeter at which punching shear no longer required
uout = β VEd/(dVRd,c)
= 1.4 × 400 × 1000/(213 × 0.613) = 4289 mm

Rearrange: uout = c2 + 2c1 + π rout
rout = (uout – (c2 + 2c1))/π
rout = (4289 – 1130)/π = 1005 mm

Position of outer perimeter of reinforcement from column face:
r = 1005 – 1.5 × 213 = 686 mm

Maximum radial spacing of reinforcement:
sr,max = 0.75 × 213 = 159.75 mm, say 150 mm

(d) Area of reinforcement
Asw ≥ (VEd – 0.75VRd,c)sru1/(1.5fywd,ef)
fywd,ef = (250 + 0.25d) = 303 MPa

Asw ≥ (1.065  – 0.75 × 0.613) × 150 × 2468/(1.5 × 303)
≥ 492 mm2 per perimeter

Provide 7H10 (Asprov = 549 mm2 per perimeter)

Within the u1 perimeter the link spacing around a perimeter,
st ≤ 1.5d = 1.5 × 213 = 319.5 mm

Outside the u1 perimeter the link spacing around a perimeter,
st ≤ 2d = 426 mm
Use say st,max = 300 mm

Minimum area of a link leg:
Asw,min ≥ [0.053 sr st sqrt(fck)] /fyk = (0.053 ×  150 ×  300 ×  √30) / 500
≥ 26 mm2

Use H10s (78.5 mm2) and 7 per perimeter.
@ 300 mm tangential spacing and @150 mm radial spacing

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