28 April 202028 April 2020 by Omotoriogun Victor

Designing for Punching Shear in Concrete Slabs

Punching shear is a failure mechanism of shallow concrete elements when subjected to concentrated forces. It arises when high concentrated loads are applied to smaller portions of a slab, or as a result of high column reactions in flat slabs, pad footings or very shallow beams. The resulting failure mode is a punch through the concrete element

Figure 1: Punching Failure in a Flat Slab

In order to avoid this kind of failure in slabs, there are two options available to a structural engineer: (1) is to size the concrete element such that the punching shear stress is less than the concrete shear strength of the concrete element. (2) Design shear reinforcement to resist punching shear.

In this post, we are more concerned about the second option i.e. the provision of shear reinforcement to resist punching shear, particularly in flat slabs.

The Eurocode requires that the punching shear verification is carried out at the column face and at the basic control perimeter 2d from the face of the column. Figures 2 & 3 shows the basic control perimeter for different types of columns in a flat slab arrangement.

Figure 2: Basic control perimeter of internal rectangular and circular columns.

Figure 3: Basic control perimeter of columns at an edge or corner of a slab

Procedure for Design

The procedure for carrying out punching design can be divided into four basic steps. This steps are explained below

Step 1: Punching Shear Verification at the Column Face

The first step towards designing for punching shear is to verify its effect at the column face. At the column face, the punching shear stress must be less than the maximum shear strength of the concrete element, else the section will have to be resized or the strength of the concrete element increased. Mathematically, this can be expressed as

\left( { v }_{ Ed }=\frac { \beta { V }_{ Ed } }{ { u }_{ o }{ d }_{ min } } \right) \le \left( { v }_{ Rd,max }=0.2\left( 1-\frac { { f }_{ ck } }{ 250 } \right) { f }_{ ck } \right)

Where:

v_Ed= applied punching shear stress
v_Rd,max = maximum concrete shear strength

f_ck = is the characteristic concrete strength.
V_Ed= applied punching force

β is a factor that deals with the eccentricity of the applied loads. Its value is defined in clause 6.4.3 of BS EN 1992-1-1. However, for structures where the lateral stability does not rely on the frame action between the slab and columns and in which adjacent spans do not differ in length by more than 25% figure 4 may be used.

u_o= column perimeter. Its values are defined as:

For interior column u_o= 2(h+b)
For edge column u_o= b + 3d ≤ h + 2b

For corner column u_o=3d ≤ h + 2b

d_min = is the mean effective depth taken as (d_x+ d_y)/2: where d_y& d_zare the effective depth in the y and z-axis respectively.

Haven verified that the punching shear stress is less than the maximum concrete shear strength, proceed to punch verification at the basic control perimeter.

Step 2: Punching Shear Verification at the Basic Control Perimeter

At the basic control perimeter, if the punching shear stress is less than the concrete shear strength, shear reinforcement is not required and need not be designed for. However, if the shear stress exceeds the concrete shear strength punching shear reinforcement must be provided in the slabs. In addition to this, the U.K National Annex requires that the punching shear stress must be less than twice the concrete shear strength, else the element must be resized.

{ v }_{ Ed }\le { v }_{ Rd,c } \quad No\quad punching\quad reinforcement\quad required

{ 2{ v }_{ Rd,c }\le v }_{ Ed }\le { v }_{ Rd,c }\quad punching\quad reinforcement\quad required

{ v }_{ Ed }>2{ v }_{ Rd,c }\quad resize\quad element

Where:

v_Ed= applied punching shear stress

v_Rd, = concrete shear strength

{ v }_{ Ed }=\frac { \beta { V }_{ Ed } }{ { u }_{ 1 }{ d }_{ min } }

β & d_min, V_Ed are as defined in the preceding section.

u₁= basic control. Its values are defined in figures 2 and 3

{ v }_{ Rd,c }=0.12k(100{ \rho }_{ l }{ f }_{ ck })^{ 1/3 }\le 0.035{ k }^{ 3/2 }\sqrt { { f }_{ ck } }

k=1+\sqrt { \frac { 200 }{ { d }_{ min } } } \le 2

{ \rho }_{ l }=\sqrt { { \rho }_{ y }\cdot { \rho }_{ z } } \le 0.02\quad \quad \rho =\frac { { A }_{ s } }{ bd }

Where p_ly and p_lz = the mean ratios of reinforcement in each direction over a width equal to the column dimension plus 3d on each side.

Step 3: Punching Shear Reinforcement

If punching shear exceeds the concrete shear strength, shear reinforcement is provided accordingly.

{ A }_{ sw }\ge\frac { \left( { v }_{ Ed }-0.75{ v }_{ Rd,c } \right) { s }_{ r }{ u }_{ 1 } }{ \left( 1.5{ f }_{ ywd,ef } \right) }

A_sw= area of vertical links required in slab

v_Ed=punching shear stress
v_Rd,c=concrete shear strength
u₁=basic control perimeter.

s_r=radial spacing of vertical links
f_{_ywd=}effective design strength of the punching shear reinforcement.

{ f }_{ wyd,eff }=250+0.25{ d }_{ min }\le { f }_{ ywd }

After obtaining the area of punching shear reinforcement, we must determine at what point, punching reinforcement is no longer required. This is defined as the outer perimeter u_out

{ u }_{ out }=\frac { \beta { V }_{ Ed } }{ { v }_{ Rd,c }{ d }_{ min } }

Step 4: Detailing Requirements

Haven provided the punching reinforcement, the detailing checks involves verifying the minimum area of vertical links and checking the spacing of bars. The minimum area of a vertical link is defined as

{ A }_{ sw,mim }\ge 0.053{ s }_{ r }{ s }_{ t }\left( \frac { \sqrt { { f }_{ ck } } }{ { f }_{ yk } } \right)

Finally, the following rules must be kept in mind as regards to spacing when detailing the slab for punching reinforcement

There should be at least two perimeters of shear links

The radial spacing of the links should not exceed 0.75d:
The tangential spacing of the links should not exceed 1.5d within 2d of the column face
The tangential spacing of the links should not exceed 2d for any other perimeter

The distance between the face of the column and the nearest shear reinforcement should be less than 0.5d.

Worked Example

Design the shear reinforcement around the interior column of a flat slab 275mm thick and reinforced with T16 bars @ 125 mm centres in both directions. The interior columns are 400mm x 400mm thick and spaced at 7.5m centres in both directions. The design load on the slab is 20.5kN/m². Use C25/30mpa concrete and steel bars with 460mpa yield strength.

Step 1: Punching Verification at Column Face

{ V }_{ Ed }=20.5\times 7.5\times 7.5=1153.125kN

\beta =1.1(interior\quad column)

{ u }_{ o }=2\left( b+h \right) =2\left( 400+400 \right) =1600mm

{ d }_{ y }=h-\left( { c }_{ nom }+{ \phi }_{ y }/2 \right) =275-(25+16/2)=242mm

{ d }_{ z }=h-({ c }_{ nom }+{ \phi }_{ y }/2+{ \phi }_{ z })=275-(25+16/2+16)=226mm

{ d }_{ min }=\frac { { d }_{ y }+{ d }_{ z } }{ 2 } =\frac { 242+226 }{ 2 } =234mm

{ v }_{ Ed }=\frac { 1.15\times 1153.125\times { 10 }^{ 3 } }{ 1600\times 234 } =3.54Mpa

{ v }_{ Rd,max }=0.2\left( 1-\frac { { f }_{ ck } }{ 250 } \right) { f }_{ ck }=0.2\left( 1-\frac { 30 }{ 250 } \right) \cdot 30=5.28Mpa

{ (v }_{ Ed }=3.54Mpa)\le { (v }_{ Rd,max }=5.28Mpa)\quad o.k

Step 2: Punching Verification at the Basic Control Perimeter

{ v }_{ Ed }=\frac { { \beta V }_{ Ed } }{ { u }_{ 1 }d }

{ u }_{ 1 }=2\left( b+h \right) +4\pi d

=2\left( 400+400 \right) +4\pi (234)=4541.71mm

=\frac { 1.15\times { 1153.125\times 10 }^{ 3 } }{ 4541.71\times 234 } =1.25Mpa

{ v }_{ Rd,c }=0.12k\left( 100{ \rho }_{ l }{ f }_{ ck } \right) ^{ \frac { 1 }{ 3 } }\ge 0.035{ k }^{ \frac { 3 }{ 2 } }\sqrt { { f }_{ ck } }

k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 234 } } =1.93<2

{ \rho }_{ y }={ \rho }_{ z }=\frac { { A }_{ s } }{ bd } =\frac { 1608 }{ { 10 }^{ 3 }\times 234 } =0.0068

{ \rho }_{ l }=\sqrt { { \rho }_{ y }{ \rho }_{ z } } =0.0068<0.02

{ v }_{ Rd,c }=0.12\times 1.93\left( 100\times 0.0068\times 25 \right) ^{ \frac { 1 }{ 3 } }\ge 0.035\times { 1.93 }^{ \frac { 3 }{ 2 } }\sqrt { 25 }

{ v }_{ Rd,c }=0.63Mpa

\left( 2{ v }_{ Rd,c }=1.26Mpa \right) <\left( { v }_{ Ed }=1.25Mpa \right) <\left( { v }_{ rd,c }=0.63Mpa \right)

Since the punching shear stress is greater than concrete shear strength, punching reinforcement is required in the slab.

Step 3: Punching Shear Reinforcement.

{ A }_{ sw }=\frac { \left( { v }_{ Ed }-0.75{ v }_{ Rd,c } \right) { s }_{ r }{ s }_{ t } }{ 1.5{ f }_{ ywd,eff } }

{ s }_{ r }=0.75d=0.75(234)\approx 175mm

{ f }_{ ywd,eff }=250+0.25d=250+0.25(234)=308.5mpa

{ A }_{ sv }=\frac { \left( 1.25-0.75\times 0.63 \right) \cdot 175\times 4541.71 }{ 1.5\times 308.5 } =1335.40{ mm }^{ 2 }/prm

{ u }_{ out }=\frac { \beta { V }_{ Ed } }{ { v }_{ Ed }{ d }_{ min } } =\frac { 1.15\times 1153.125\times { 10 }^{ 3 } }{ 4541.71\times 234 } =8995mm

The radius of the of the point at which punching shear reinforcement is no longer required can be evaluated using the same expression for the control perimeter as

{ r }_{ out }=\frac { { u }_{ out }-2\left( h+b \right) }{ 2\pi } =\frac { 8995-2(400+400) }{ 2\pi } =1176.5mm

Therefore, the position of outer perimeter of reinforcement from column face is

r={ r }_{ out }-1.5d=1176.5-1.5(234)=825.5mm

Step 4: Detailing requirements

minimum\quad area\quad of\quad vertical\quad link

{ s }_{ t }=1.5d=1.5(234)\approx 350mm

{ A }_{ sw,min }=0.053{ s }_{ r }{ s }_{ t }\left( \frac { \sqrt { { f }_{ ck } } }{ { f }_{ yk } } \right)

0.053\times 175\times 350\times \frac { \sqrt { 30 } }{ 460 } =38.66{ mm }^{ 2 }

THEREFORE\quad USE\quad 18T10\quad ({ A }_{ sprov }=1413{ mm }^{ 2/ }perimeter)

The Layout of the punching reinforcement is shown as figure 5

Figure 5: Layout of Punching Reinforcement in Slab

THANK YOU FOR READING!!! KINDLY LEAVE A COMMENT.

Published by Omotoriogun Victor

A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices View all posts by Omotoriogun Victor