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Monday, June 18, 2018

Design Example of Punching Shear in Slabs (Eurocode 2)





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

Thank you for visiting Structville today and God bless you.


5 comments:

  1. Keep up this great work Bro. Thanks alot

    ReplyDelete
  2. Hi.. it would be nice sir to do a sketch of the punching shear rebar arrangement. So as aid me in understanding the example well.. Cheers

    ReplyDelete
  3. Thank you for this detailed example. Keep going.

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