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 V

_{Ed}= 400 kN

Thickness of slab = 250 mm

Dimension of column = 450 x 230 mm

Reinforcement of slab in the longer direction = H16@150mm (A

_{s,prov}= 1340 mm

^{2})

Reinforcement of slab in the shorter direction = H16@175mm (A

_{s,prov}= 1149 mm

^{2})

Grade of concrete = C30

Yield strength of reinforcement = 500 Mpa

Concrete cover to slab = 25mm

**Solution**

_{y}= 250 - 25 - (16/2) = 217 mm

Effective depth of slab in x-direction d

_{x}= 250 - 25 - 16 = 209 mm

ρ

_{ly}= (1340) / (1000 × 217) = 0.00617 (reinforcement ratio)

ρ

_{lx}= (1149) / (1000 × 209) = 0.00549 (reinforcement ratio)

**(a) Check shear at the perimeter of the column**

**V**

_{Ed}= β V

_{Ed}/(u

_{0}d) < V

_{Rd,max}

_{}

From figure 6.21N of EN 1992-1-1;

β = 1.40

d = (217 + 209)/2 = 213 mm

u

_{0}= c

_{2}+ 3d < c

_{2}+ 2c

_{1}For edge columns (clause 6.4.5(3))

u

_{0}= 230 + (3 × 213) < (230 + 2 × 450)

u

_{0}= 869 mm

V

_{Ed}= 1.40 × 400 × 1000/(869 × 213) = 3.025 MPa

V

_{Rd,max}= 0.5 ν f

_{cd}

= 0.5 × 0.6(1 - f

_{ck}/250) × α

_{cc}f

_{ck}/γ

_{m}

= 0.5 × 0.6(1 - 30/250) × 1.0 × (30 /1.5) = 5.28 MPa

V

_{Ed}< V

_{Rd,max}...OK

**(b) Check shear at u**

_{1}, the basic control perimeterV

_{Ed}= β V

_{Ed}/(u

_{1}d) < V

_{Rd,c}

β,V

_{Ed}as before

u

_{1}= c

_{2}+ 2c

_{1}+ π × 2d

u

_{1 }= 230 + (2 × 450) + (π × 2 × 213) = 2468 mm

V

_{Ed}= 1.4 × 400 × 1000/(2468 × 213) = 1.065 MPa

V

_{Rd,c}= 0.12 k(100 ρ

_{l}f

_{ck})

^{1/3}

k = 1 + (200/d)

^{1/2}= 1 + (200/213)

^{1/2}= 1.969

ρ

_{l}= (ρ

_{ly}ρ

_{lx})

^{1/2}= (0.00617 × 0.00549)

^{1/2}= 0.00582

V

_{Rd,c}= 0.12 × 1.969(100 × 0.00582 × 30)

^{1/3}= 0.613 MPa

V

_{Ed}> V

_{Rd,c}?

1.065 MPa > 0.613 MPa ... Therefore punching shear reinforcement required

NA check:

V

_{Ed}≤ 2.0V

_{Rd,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**

u

_{out}= β V

_{Ed}/(dV

_{Rd,c})

= 1.4 × 400 × 1000/(213 × 0.613) = 4289 mm

Rearrange: u

_{out}= c

_{2}+ 2c

_{1}+ π r

_{out}

r

_{out}= (u

_{out}– (c

_{2}+ 2c

_{1}))/π

r

_{out }= (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:

s

_{r,max}= 0.75 × 213 = 159.75 mm, say 150 mm

**(d) Area of reinforcement**

A

_{sw}≥ (V

_{Ed}– 0.75V

_{Rd,c})s

_{r}u

_{1}/(1.5f

_{ywd,ef})

f

_{ywd,ef}= (250 + 0.25d) = 303 MPa

A

_{sw}≥ (1.065 – 0.75 × 0.613) × 150 × 2468/(1.5 × 303)

≥ 492 mm

^{2}per perimeter

Provide 7H10 (A

_{sprov}= 549 mm

^{2}per perimeter)

Within the u

_{1}perimeter the link spacing around a perimeter,

s

_{t}≤ 1.5d = 1.5 × 213 = 319.5 mm

Outside the u

_{1}perimeter the link spacing around a perimeter,

s

_{t}≤ 2d = 426 mm

Use say s

_{t,max}= 300 mm

Minimum area of a link leg:

A

_{sw,min}≥ [0.053 s

_{r}s

_{t}sqrt(f

_{ck})] /f

_{yk}= (0.053 × 150 × 300 × √30) / 500

≥ 26 mm

^{2}

Use H10s (78.5 mm

^{2}) and 7 per perimeter.

@ 300 mm tangential spacing and @150 mm radial spacing

Thank you for visiting Structville today and God bless you.

Keep up this great work Bro. Thanks alot

ReplyDeleteHi.. 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

ReplyDeleteThank you for this detailed example. Keep going.

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