This article presents a worked example on the design of punching shear for flat slabs to BS EN 1992-1-1-2004 (Eurocode 2).
The flat slab to be designed for punching has already been analyzed using equivalent frame method here, and then designed for flexural strength here. In this article we shall design the same slab for punching shear.
The shear force diagram of the flat slab from the analyses is reproduced below. The maximum shear force 326.93KN is the most critical of the shear forces. A punching shear check shall be performed on this and the result will be generalized on all other points for practicality.
Some of the parameters necessary for the design of punching shear are listed as follows:
Maximum axial force on column VEd = 326.93kN
Slab thickness = 250 mm
Dimension of column = 300 x 300 mm
Reinforcement of slab in the longer direction = T16@250mm (As,prov = 718.18 mm2)
Reinforcement of slab in the shorter direction = T16@250mm (As,prov = 718.18mm2)
Grade of concrete = C25/20
Yield strength of reinforcement = 500 Mpa
Concrete cover to slab = 25mm
Check Punching Shear at column Perimeter
Calculate the punching shear stress at column perimeter
(vEd) = 𝛽 VEd/u0x d
𝛽 = 1.15 (internal columns)
Column perimeter (u0) = 4 x 300 = 12000
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
d = (217 + 209)/2 = 213mm
(vEd) = (1.15 x 326.93 x / (12000 x 213) = 1.47N/mm2
Calculate the punching shear capacity of concrete
vRDmax = 0.3 (1 -fck/250)fcd
fcd = fck/1.5 = 25/1.5 = 16.67 N/mm2
vRDmax = 0.3 (1 -25/250)16.67
=4.5 N/mm²
vRDmax > vEd, the depth of slab is sufficient.
Check Punching Shear at Basic Control Perimeter
Calculate the punching shear stress at basic control perimeter
(vED) = 𝛽VED /u1 x d
𝛽 = 1.15 (internal columns)
d = 213mm
Column Perimeter (u1) = 2(C1 + C2) + 4πd
Column Perimeter (u1) = 2(300 + 300) + 4 x 3.142 x 213
Column Perimeter (u1) = 3876.98mm
(vED) = (1.15 x 326.93)/ (3876.98 x 213)
vED = 0.46 N/mm²
Calculate the shear capacity without shear reinforcement
vRdc = {0.12K(100ρLfck)1/3} ≥ vmin
K = (1 +√200/d) ≤ 2.0
K = (1 +√200/213) = 1.97
ρL = √ρLx. ρLy ≤ 0.02
ρL = Asl/bwd ; ρL = Asl/bwd
ρly = (718) / (1000 × 217) = 0.0034
ρlx = (718) / (1000 × 209) = 0.0034
ρL = √0.0034 . 0.0034
ρL = 0.0034
vRdc = {0.12 x 1.97 (100 x 0.0034 x 25)1/3} ≥ vmin
vRdc = 0.4 N/mm2
vmin = {0.035K3/2fck1/2}
vmin = 0.035 x 1.97³/2 x 251/2
vmin = 0.48 N/mm2
Since vRdc (0.4N/mm2) < vmin (0.48 N/mm2); hence vRdc = 0.48 N/mm2
Since vED < vRdc then no punching shear reinforcement is required.
At this point since no shear reinforcement is required at the first column perimeter, then no further design is required, and the section is sufficient. However, for the sake of learning we shall reduce the thickness and effective depth of slab to 200 and 163 respectively so that the section becomes insufficient, and we can then proceed to design for shear reinforcement.
If we go through the calculation all over again with d = 113, then vED and vRdc becomes:
vED = 0.7N/mm²; vRdc = 0.49N/mm²
Punching Shear Reinforcement Design
Calculate the perimeter (Uout) beyond which shear reinforcement is not required
Uout = 𝛽 vED/(VRdc x d)
Uout = (1.15 x 326.93 x 10³)/(0.49 x 163)
Uout = 4659.9
Calculate the distance of ( Uout) from the column face
aout=( Uout – 2(2C1 + 2C2))/2π
aout = (4659.9 – 2(300 + 300))/2 x 3.142
aout = 550.598mm
Detailing consideration for the shear reinforcement
The farthest perimeter of shear reinforcement to the column should not be greater than ( aout – 1.5 d) from the column face.
(550.598 – 1.5 d) = 306mm
The distance between the first shear reinforcement perimeter and the loaded area should be between 0.3d and 0.5d. So, let’s take the position of the first shear reinforcement perimeter to be 0.4d
0.4d = 0.4 x 163 = 65.2
Let’s take it as 65
This means at least 2 perimeters of shear reinforcement with a spacing not exceeding 0.75d (122mm) should be placed between the distances of 65mm and 306mm from the column face.
Let’s position the outermost perimeter of shear reinforcement be 300mm
The next perimeter should be at most 0.75d (122mm) from the last so let us place it at 120mm from the outermost which corresponds to 180mm from the column face.
There needs to be another perimeter between 0.3d and 0.5d from the column face. So, let’s place another shear reinforcement at 65mm from the column face. This corresponds with a distance of 115 from the second perimeter.
In summary, we have three shear perimeters, and they are 65mm, 180mm, and 300mm from the face of the column. They are detailed as shown in the end of this article.
Calculate the Area of Shear Reinforcement
The area of shear reinforcement required for resisting punching shear is given by the expression:
Asw ≥ (VEd – 0.75VRd,c)sru1/(1.5fywd,ef)
fywd,ef = 250 + 0.25d ≤ fywd
fywd,ef = 250 + 0.25 x 163 ≤
fywd,ef = 290.75MPa
Asw ≥ (0.7 – 0.75 x 0.49)122 x 3876.98/(1.5 x 290.75)
Asw = 309mm2
Calculate the number of reinforcements in each perimeter
Tangential spacing between links is 1.5d = 1.5 x 163 = 244.5mm
For first perimeter
Distance to column face = 65mm
Perimeter of first ring = column perimeter + 8 x 65 = 1200 + 8 x 65 = 1720mm
No of shear links in perimeter = perimeter/tangential spacing
Take the tangential spacing to be 240mm rather than 244.5mm
No of links = 1720/240 = 7
For second perimeter
Distance to column face = 180mm
Perimeter of second ring = column perimeter + 8 x 180 = 1200 + 8 x 180 = 2640mm
No of shear links in perimeter = perimeter/tangential spacing
No of links = 2640/240 = 11
For third perimeter.
Radius = 300
Perimeter of third ring = column perimeter + 8 x 300 = 1200 + 8 x 300 = 3600mm
No of shear links in perimeter = perimeter/tangential spacing
No of links = 3600/240 = 15
The area of single 10mm rebar is 78.55mm2. If you multiply this area by the number of rebars required in each perimeter then you will have 562mm2, 864mm2, 1178mm2. This area is greater than the required area of 309mm2 so 10mm rebar should be used as shear links
Detailing
