Punching shear failure occurs in a slab when the magnitude of a concentrated load (such as that from a column) exceeds the shear strength or resistance of the slab or the column punches through the slab. The failure plane is located at a certain distance from the perimeter of the column and has a funnel-shaped failure pattern. The design for punching shear resistance in flat slabs normally involves controlling the thickness of the slab or providing punching shear reinforcement.

Generically, punching is a three-dimensional, brittle failure mechanism leading to a truncated cone that separates from the slab. The shear crack develops from tangential flexural cracks and propagates into the direction of the compression zone near the column edge constricting the circumferential compression ring with increasing loads. Once the punching shear resistance is reached the shear crack intersects the uncracked compression ring leading to a sudden penetration of the column into the slab.

The recommendations found in BS EN 1992 (Eurocode 2) are usually followed when designing punched shear reinforcement. To assess whether punched shear reinforcement is necessary, the shear stress in the concrete is computed at the column face and at the fundamental control perimeter u_{1} (2d from the column face).

The position of the outside control perimeter where shear reinforcement is no longer needed (u_{out}) is then determined if reinforcement is necessary. Shear studs are placed starting at 0.3d or 0.5d from the column face to within 1.5d of the outer control perimeter (u_{out}), with intermediate studs spaced at 0.75d intervals.

The most cost-effective approach will often be a radial arrangement, with rails spaced either 30° or 45° apart. To meet this requirement, extra secondary rails are installed as necessary. The tangential spacing between studs is kept to within 1.5d for studs inside the basic control perimeter (u_{1}) and 2d for studs outside the basic control perimeter.

The suggested values from BS EN 1992 can be used to design shear load V_{Ed} for internal, edge, and corner columns where lateral stability does not depend on frame action between slabs and columns and where neighbouring spans differ by less than 25%.

Design for punching shear should take moment transfer into account at the intersection of the column and slab. The design punching shear can be obtained for structures whose lateral stability is not dependent on the frame action between the slab and columns and where adjacent spans do not differ in length by more than 25% by increasing V_{Ed} by 1.15 for internal columns, 1.4 for edge columns, and 1.5 for corner columns.

Generally, the following checks should be carried out:

- Ensure that maximum punching shear stress is not exceeded, i.e.
*v*_{Ed}<*v*_{Rd,max}at the column perimeter - Determine whether punching shear reinforcement is required, i.e. whether
*v*_{Ed}>*v*_{Rd,c}at the basic perimeter, u_{1} - Determine whether punching shear reinforcement is required, i.e. whether
*v*_{Ed}>*v*_{Rd,c}at at successive perimeters to establish*u*= the length of the perimeter where_{out}*v*_{Ed}=*v*_{Rd,c}. Perimeters within 1.5 d from u_{out}need to be reinforced.

Where required provide reinforcement such that *v*_{Ed }≤ *v*_{Rd,cs}.

where

*v*_{Ed} = applied shear stress. The shear force used in the verification should be the effective force taking into account any bending moment transferred into the slab (see above)

*v*_{Rd,max }= design value of the maximum punching shear resistance, expressed as a stress *v*_{Rd,c} = design value of punching shear resistance of a slab *without *punching shear reinforcement, expressed as a stress *v*_{Rd,cs }= design value of punching shear resistance of a slab *with *punching shear reinforcement, expressed as a stress.

*v*_{Rd,cs} = 0.75 *v*_{Rd,c} + 1.5 (*d*/*s*_{r})*A*_{sw} *f*_{ywd,ef }(1/*u*_{1}*d*)sin a

where:

*A*_{sw} = area of shear reinforcement in one perimeter around the column (subject to *A*_{sw,min})*s*_{r} = radial spacing of perimeters of shear reinforcement*f*_{ywd,ef }= effective design strength of reinforcement (250 + 0.25*d*) ≤ *f*_{ywd}*d *= mean effective depth in the two orthogonal directions (in mm)*u*_{1} = basic control perimeter at 2*d *from the loaded area

sin a = 1.0 for vertical shear reinforcement

Where required each perimeter should have*A*_{sw} = (*v*_{Ed} – 0.75 *v*_{Rd,c})*s*_{r} *u*_{1}/(1.5 *f*_{ywd,ef})

The punching shear resistance of a slab should be assessed for the basic control section (see Figure 6.12). The design punching shear resistance [MPa] may be calculated as follows:

`v _{Rd,c}` =

`C`

_{Rd,c}`k`(100

`ρ`

_{l}`f`)

_{ck}^{1/3}+

`k`

_{1}`σ`≥ (

_{cp}`v`+

_{min}`k`

_{1}`σ`)

_{cp}where:`f _{ck}` is the characteristic compressive strength of concrete, see Table 3.1

`k`= 1 + √200/

`d`≤ 2.0

where

`d`is the effective depth, in [mm]

`ρ`= (

_{l}`ρ`⋅

_{ly}`ρ`)

_{lz}^{1/2}≤ 2%

`ρ`,

_{ly}`ρ`are longitudinal reinforcement ratios in y- and z- directions respectively. Their values should be calculated as mean values taking into account a slab width equal to column width plus 3

_{lz}`d`each side

`σ`= (

_{cp}`σ`+

_{cy}`σ`)/2,

_{cz}where

`σ`,

_{cy}`σ`are the normal concrete stresses in the critical section in y- and z- directions (in [MPa], positive if compression):

_{cz}`σ`=

_{cy}`N`/

_{Ed,y}`A`and

_{cy}`σ`=

_{cy}`N`/

_{Ed,z}`A`

_{cz}`N`,

_{Ed,y}`N`are the longitudinal forces across the full bay for internal columns and the longitudinal force across the control section for edge columns. The force may be from a load or prestressing action.

_{Ed,y}`A`,

_{cy}`A`are the areas of concrete according to the definition of

_{cz}`N`,

_{Ed,y}`N`respectively

_{Ed,y}`C`is a Nationally Determined Parameter, see § 6.4.4 (1)

_{Rd,c}`v`is a Nationally Determined Parameter, see § 6.4.4 (1), or (6.3N) for the calculation of

_{min}`v`following the Eurocode recommendation

_{min}`k`is a Nationally Determined Parameter, see § 6.4.4 (1).

_{1}**Punching Shear Design Example**

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**

Effective depth of slab in y-direction d_{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 perimeter**

V

_{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

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

Thank you for this detailed example. Keep going.

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It’s perfectly crystal clear! thanks a lot

In example, are you using slab top or bottom reinforcement (B16-150 & B16-175) ?

Hi

I think ”dx” must be replaced.

The correct calculation is:

dx=250-25-16-(16/2)=201 mm

So there is some variation in the rest of the calculations.