Punching is a failure mechanism when structural elements are subjected to a concentrated load. Failure occurs within a defined radius away from the concentrated load. Slabs, pile tops, foundations, slab foundations, etc. These are elements subject to puncturing and must be sized accordingly.
Punch cutting is considered a critical section where shear failure can occur. The following figure shows typical punch cutting. The selection of the critical cut may vary depending on the relevant standard.
As shown in the figure above, different standards use different values to define the punching circumference. Scope selection may also vary with other factors.
Punching stress = V/bd
Where'
V – applied force
b – circumference length
d – effective depth.
The punching stress must be lower than the allowable shear stress; v
To prevent punching shear failure, punching shear checks are generally carried out for the following elements.
- Normal dish
- Flat roof
- pile blocks
- Foundations
- Slab foundations
As explained above, the amount of punching may vary from pattern to pattern.
Let's discuss the variation of punching circumference according to different standards
Punching range according to Eurocode 2
The consideration of different perimeters is the main difference between most methods, in addition to the different equations used to calculate the shear capacity.
The distance from the concentrated load to be considered for the critical span also depends on the arrangement of the concentrated load. Furthermore, when defining the critical shear extension, the location where the concentrated load is applied must also be taken into account.
The following figure shows the method for defining the punching perimeter of a plate.
The shear perimeter is considered to be 2d from the front of the column.
The type of support or concentrated load must be taken into account when defining the shear extension. The following figure shows the method for determining the extent of shear.
Furthermore, the position of the support must be taken into account to define the extent of the shear. The length of the shear circumference is one of the factors that affect the shear stress.
In general we need to respect the 2D distance. However, it should be based on the other boundary conditions. The total length must be calculated as indicated above if we design according to Eurocode 2.
Punching range to BS 5400 Part 4
BS 5400 Part 4 also provides clear guidance on how to calculate the punching shear circumference depending on the different arrangement of boundary conditions and the support/load area.
The information provided in BS 5400 is very useful. And they cover almost all the different cases that may arise in the design of slabs, pile caps, flat slabs, footings and slab foundations.
The distance to the critical shear band is given in BS 5400 Part 4 as 1.5 d. The same amount of shear is assumed in BS 8110.
Furthermore, these definitions can also be used when designing according to other standards. The distance to the thrust circumference can be found in the relevant standard.
Punching circumference according to ACI
ACI defines a much smaller drilling range compared to other codes. It is 0.5d The following figure shows the amount of shear to be considered according to the ACI standard.
Although the scope is smaller, we cannot assume that it is always a conservative design without considering the other factors. However, it appears that ACI has major concerns about punching punch design.
Example of punching shear design
Consider foundation sizing for punching punching in accordance with BS 8110.
According to BS 8110 Part 1 the shear range should be 1.5d.
Consider the following data
- Effective depth 350 mm
- Force applied in the range of 1100kN
- Column size 300mm
- Characteristic strength of concrete 25N/mm 2
- For this example, we will assume an allowable shear stress of Vc = 0.6 N/mm. two
Check the drilling in front of the column
Shear stress = 1100×1000 / (4x300x350) = 2.62N/mm 2
Allowable voltage = 0.8 (fcu) 0.5 = 0.8 (25) 0.5 = 4N/mm 2 <5N/mm 2
So, okay.
Consider the shear magnitude to be 1.5d
Action circumference length = 4 x (1.5 d + 300 + 1.5 d) = 4 x (1.5 × 350 + 300 + 1.5 × 350) = 5400 mm
Shear stress, v = 1100×1000 / (5400×350) = 0.582 N/mm 2
v
In a similar way, punching shear reinforcement can also be designed for other elements.
The requirement V < Vc cannot always be met. If the shear stress exceeds the shear capacity, reinforcements must be provided.
Typically, the shear reinforcement is omitted and the thicknesses and reinforcements are adapted to the transverse connections. For some elements, such as B. slab foundations, however, it is not advisable to significantly increase the cross-section and area of the reinforcement.
This leads to higher construction costs. If we provide shear joints, the cost can be comparatively low. If we increase the thickness, the entire area may need to be increased. However, there are methods to increase section depth in a critical section.
Both options are applicable, especially for slab foundations. Option 01 is more difficult to construct compared to Option 02. When using option 01, continued waterproofing, fixing reinforcement, etc. must be carried out with additional care.
Option 02 is much easier than option 01. The lead could be created in the second phase. However, if the slab is used as a basement or ground floor, the overhang will create an obstacle, therefore, option 01 is preferable in these situations.
Reinforcement detailing for punching punching
Let us discuss this with an example where we need to provide shear joints to accommodate shear stress.
Design of a slab foundation against punching shear
Data taken into account e.g.
- Designed to BS 8110 Part 1
- Support axial load 3000 kN
- Column size 450×450 mm
- Punching force 1.5d of the pillar face 500kN/m
- Effective depth of the raft 550mm
- Tensile reinforcement provided T25@150 (3272mm 2 /M)
Drilling in front of the column
Shear stress = 3000×1000 / (4x450x550) = 3.03N/mm 2
Allowable voltage = 0.8 (fcu) 0.5 = 0.8 (25) 0.5 = 4N/mm 2 <5N/mm 2
So, okay.
Consider the shear extension to be 1.5d
Shear stress = 500×10 3 /(1000×550) =0.91 N/mm 2
According to the intended reinforcement area and useful depth
Shear capacity, Vc = 0.532 N/mm 2
Spreadsheet could be used for simple calculation of reinforcement and shear connections.
In accordance with BS 8110 Part 1 Table 3.16,
Vc + 0.4 = 0.532 + 0.4 0.932 N/mm 2
Vc < V < Vc+0.4
Therefore, ensure a minimum reinforcement area.
Note: Depending on the applied cutting force, the appropriate formula must be selected from Table 3.16.
A SV ≥ 0.4bs against /0.95f jv
We make the shear connections in both directions, unlike other elements. In beams there are usually only two legs. Likewise, we must take into account the number of legs per meter of length to calculate the distance of the connections in the other direction.
We typically consider main reinforcement spacing when selecting shear connection spacing. As the main reinforcement is spaced 150 mm apart, the connections must be spaced 300 mm apart.
Then,
A SV = 377mm 2 /M
377≥ 0.4x1000xs versus /(0.95×500)
SV ≤ 478mm
So look at T12@300 in the other direction.
In the range up to V
The following types of connections are provided in the reinforcement design.
The method described above can be used to construct slabs, foundations and punched slab foundations. However, a slightly different approach is used in the design of pile caps.
A gain factor is also taken into consideration in the design. A similar method is used in the construction of flat ceilings.
The article Pile Head Construction For more information on these types of designs, you can refer to this page.
