punching shear failure

Punching Shear Design to Eurocode 2 – an overview

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Punching shear occurs as a result of concentrated load or reaction acting on localized area of a slab. This mostly occurs in slabs which are directly supporting columns such as pad foundation, or slabs that are directly being supported by columns such as flat slabs. Flat slabs, however, are much more vulnerable to punching failure than pad foundation as there is always greater allowance to increase the slab thickness against punching shear for the case of pad foundations.

Punching failure starts from the joint between a column and a slab and then penetrates into the slab at an angle of 45degrees to the horizontal. The failure is brittle as it is always sudden, leaving little or no prior warning.  And often times, punching failure leads to progressive collapse in building/vertical structures as the floor at the level of failure collapses on the lower floor which in turn triggers progressive collapse of successive floors below. This underscores that the devastating effect of punching failure cannot be overstated.

punching shear failure
punching shear failure

The stress due to punching within a perimeter u1 of a slab of effective depth d, which is subjected to a concentrated load can be calculated using the expression:

Punching shear stress (VED) = VED/u1 x d

Where;

VED is the concentrated load

d is the effective depth of reinforcement

u1 is the perimeter under consideration (whether control perimeter or otherwise)

The Basic Control Perimeter

As can be deduced from the formular above, punching shear is always meant to be checked within a localized perimeter around a concentrated load. So, Eurocode 2 stipulates that the punching shear is to be checked at a distance of 2d from the loaded area. The perimeter of the slab at a distance of 2d from the column is known as the basic control perimeter. The shape of the control perimeter varies according to the shape of the column as shown below:

control perimeter for different shapes
control perimeter for different shapes (figure 6.13 of BS EN 1992-1-1-2004)

A rectangular column has a basic control perimeter of: u1 = 2(C1 + C2) + 4πd

Where: C1 and C2 are the cross-sectional length of the column (C2 is the longer length).

Likewise, a circular column as a basic control perimeter of: u1 = π (D + 4d)

Where, D is the diameter of the column

The effective depth d can be taken as the average of the effective depths in two orthogonal directions. This is given as: deff = dy + dx/2

 

Control Perimeter near openings/Holes

If openings are present at a distance not greater than 6d from the perimeter of the loaded area, allowance shall be made to accommodate the effect of the opening by reducing the length of the critical perimeter. How much length to be reduced is that part of the control perimeter contained between two tangents drawn to the outline of the opening from the center of the loaded area as shown below.

control perimeter near opening
control perimeter near opening (fig 6.14 of BS EN 1992-1-1-2004)

Two instances of control perimeter near openings

From the above figure, the two different openings marked A shows two different situations:

  1. The first to the left shows when the longer sider of the hole (l2) is parallel to the longer side of the column (C2),

 the critical perimeter length is reduced by u = ℓ2 × [(0.5C1 + 2d)/ (a + 0.5C1)

 

  1. The second one shows when the shorter side of the hole (l1) is parallel to the longer side of the column (C2)

the critical perimeter length is reduced by u = √(l1l2) × [(0.5C1 + 2d)/ (a + 0.5C1)]

 

Control perimeter when Loaded area situated near edge/corner

When the loaded area is situated near a corner or an edge, the control perimeter should be such that the length of the free edge(s) is excluded as demonstrated in the figure below

Excluded free edges
Excluded free edges

 

Punching Shear stress when acting simultaneously with Moment

When an unbalanced moment is transferred between slab and columns, EC2 introduces a magnification factor to increase the punching shear force thereby increasing the shear stress.

The shear stress then becomes:

(vED) = 𝛽VED /u1 x d

𝛽 = 1 + k (MED/VED ) (u1/W1)

Where:

u1 is the length of the basic control perimeter

k (6.39) is a coefficient dependent on the ratio between the column dimensions C1 and C2. its value is a function of the proportions of the unbalanced moment transmitted by uneven shear and by bending and torsion (see Table 6.1 of EN 1992-1-1-2004).

W1 corresponds to a distribution of shear, and it is a function of the basic control perimeter:

Although clause 6.4.3 (3), (4), and (5) gives different equations for 𝛽 for different column shapes and positions, for structures in which the lateral stability does not depend on frame action between the slabs and the columns, and where the adjacent spans do not differ in length by more than 25%, approximate values for 𝛽 may be used for simplicity. The approximate value of 𝛽 increases VEd by 1.15 for internal columns, 1.4 for edge columns, and 1.5 for corner columns as shown in the figure below:

value of B for different column positions
value of B for different column positions (fig 6.21N of BS EN 1992-1-1-2004)

 

Compressive Struct Capacity Check

It is important to check whether the design shear stress at column face is not too large to cause the crushing of the inclined compressive strut of the concrete. To ensure this, the maximum capacity of the section must be greater than the design shear stress at column face. The maximum capacity of the section is given as:

vRDmax = 0.3 (1 -fck/250)fcd

And for the stress at the column face given as (vED) = 𝛽/u0 x d

The ‘column perimeter’ u0 is given by

Interior column: u0 = 2(C1 + C2)

Edge column: u0 = C2 + 3d ≤ (2C1 + C2)

C2 = column dimension parallel to the free edge

Corner column: u0 = 3d ≤ (C1 + C2)

 

Punching Shear capacity without Shear reinforcement

The punching shear capacity without shear reinforcement is to be checked at the basic control perimeter u1.

The shear resistance of reinforced concrete without reinforcement is given in EN 1992-1-1-2004 as:

VRdc = (CRdcK(100ρLfck)1/3 + K1σcp) bwd ≥ (vmin + K1σcp) bwd

For punching resistance without reinforcement in terms of stress (capacity) along the control section considered it becomes:

vRdc = (CRdcK(100ρLfck)1/3 + K1σcp) ≥ (vmin + K1σcp)

σcp = (σcy + σcz)/2

σcy = NEdy/Acy;    σcz = NEdz/Acz

NEdy and NEdz 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.

Ac is the area of concrete

When there is no longitudinal (axial) force in the member, then the punching shear strength becomes:

vRdc = (CRdcK(100ρLfck)1/3)  ≥ vmin 

Where:

CRdc  =  0.18/ϒ

ϒ = 1.5 (partial factor of safety for concrete)

K =    (1 +√200/d)    ≤   2.0

ρL   =  √ρLx. ρLy    ≤        0.02

ρL = Asl/bwd    ; ρL = Asl/bwd

vmin = {0.035K3/2fck1/2}

When the value of CRdc and  ϒ is substituted into the equation it becomes:

vRdc  = {0.12K(100ρLfck)1/3}

Punching shear capacity with Shear Reinforcement

When shear reinforcement is required, the shear resistance (strength) should be calculated in accordance with the Expression:

vRdcs = 0.75vRdc + 1.5(d/Sr)Aswfywd.ef(1/(u1d)) sinα

 

vRdcs  is the design value of the punching shear resistance of a slab with punching shear reinforcement along the control section considered.

Asw is the area of one perimeter of shear reinforcement around the column [mm)

Sr is radial spacing of perimeters of shear reinforcement in mm

fywd.ef is the effective design strength of the punching shear reinforcement = 250 + 0.25d ≤ fywd

d is the mean of the effective depths in the orthogonal directions (mm)

α is the angle between the shear reinforcement and the plane of the slab (sin α is 1.0 for vertical shear reinforcement)

When shear reinforcement is required, it is important to calculate further another perimeter beyond which reinforcement is not required (Uout). This is to enable the code provision that stipulate that the shear reinforcement should be placed at a maximum distance of 1.5d from

Uout  =  𝛽 vED/(VRdc x d)

 

Summary of Steps in Designing for Punching Shear

STEP 1; Check whether the punching shear stress (vED) at the column face (loaded area) exceeds the maximum capacity of the compressive struct (VRDmax)

If vED ≥ VRDmax, resize the section, if otherwise then proceed to Step 2.

 

STEP 2; Check whether the shear stress (vED) at the basic control perimeter exceeds the shear strength without reinforcement (vRdc)

If vED ≤ vRdc, then the section is adequate and no further design is required. If not proceed to step 3

STEP 3; Check whether punching shear stress (vED) exceeds twice the shear strength without reinforcement (vRdc) at basic control perimeter. (it should be noted that this is exclusive provision from UK National Annex base on Uk experience)

 

If vED ≤ 2 x vRdc, then we return to step 1 to resize the section. If not, we proceed to step 4

 

STEP 4; Design for shear reinforcement

Click here to study a worked example where this steps are applied in designing for punching shear in a flat slab

Detailing consideration of Punching Shear Reinforcement

After designing the shear reinforcement, these are some of the detailing considerations when placing the reinforcements:

  1. Shear reinforcements should be placed between the loaded area and within 1.5d from the perimeter at which reinforcement is no longer required (Uout).
  2. The distance between the first reinforcement and the loaded area must not be more than 0.5d
  3. Shear reinforcements should be placed in at least two perimeters of link legs and the spacing (radial spacing) should not be more than 0.75d
  4. The spacing of the link legs around (tangential spacing) the perimeter within basic control perimeter should not exceed 1.5d
  5. The spacing of the link legs around perimeter (tangential spacing) outside the basic control perimeter should not exceed 0.2d
  6. When the radial and tangential spacing are known then the Area of links can be derived from the expression: ((1.5Aswmin) / (sr .st) ) ≥ 0.08fck/fyk

 

Other solutions when punching shear stress is greater than shear resistance

When punching shear stress exceeds the punching shear capacity without shear reinforcement, other than designing shear reinforcement, other probable solutions include:

  • Introducing stud rails
  • Introducing column capital (increasing the head of the column)
  • Introducing drop panels

 

Control Radius of Columns with Capitals or Slab with drops

Introducing column capitals or slab drops are effective methods of reducing punching shear stress. Column capitals reduce shear stress by increasing the perimeter of the loaded area, while slab drops reduced the punching stress by increasing the effective depth of the slab. Both methods also lead to increasing the length of basic control perimeter which is also a huge factor in reducing punching stress. The increment in length of basic control perimeter for columns with capitals and slab with drops is further discussed below:

  1. Slabs with circular column heads: For slabs with circular column heads for which lH < 2hH (see Figure below) a check of the punching shear stresses is only required on the control section outside the column head. The distance of this section from the centroid of the column should be taken as:

rcont= 2d + lH + O.5c

 

  1. Rectangular column with rectangular column heads: If lH < 2hH (see Fig. below for notation) of overall dimension ℓ1 × ℓ2, then the shear stress needs to be checked at a section outside column head at a radius of

rcont = 2d + 0.56ℓ1√ (ℓ2/ℓ1), for (ℓ2/ℓ1) ≤ 1.5

rcont = 2d + 0.69ℓ1, for (ℓ2/ℓ1) > 1.5

 

Column with enlarge head when LH < 2Hh
Column with enlarge head when LH < 2hH (fig 6.17 of BS EN 1992-1-1-2004)

 

        3. for slabs with enlarge column head If ℓH > 2then the shear stress needs to be checked at sections both inside as well as outside the column head at a radius of (see Fig. below):

rcont, ext = 0.5c + 2d + ℓH,:  effective depth = d

rcont, int = 0.5c + 2(d + hH), effective depth = d + hH – hH× 2(d + hH)/ ℓH

column with enlarge head with lh > 2hH
column with enlarge head with lh > 2hH

 

 

Author: Amuletola Rasheed

You can reach Amuletola Rasheed via amuletola@fppengineering.com

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