**1. Required reinforcement**

Punching shear reinforcement according to ACI is calculated as;

Now punching ratio in SAFE is given by;

Rearranging this, we get;

**where;**

vu = ultimate max applied shear stress including effects of moments; Mpa

Vu = ultimate max applied shear force including effects of moments; N

Ac = area of concrete bounded by critical parameter; Ac = bo.d in mm²

fc = shear stress capacity of unreinforced concrete and is given by following formula in Mpa;

Phi Vc = shear capacity of reinforced concrete and is given by following formula in N;

The term Vu-Phi Vc in ACI formula then becomes;

Left side term Av/s can also be simplified as;

where db is the diameter of shear bar and x is total number of legs in one peripheral line around the column.

Rearranging the main equation we get;

Going further and putting values of fc and Ac, we get;

or

where bo is the critical parameter around column in mm and is given by;

For a given problem or project, values of f’c, fy, bo.d and s/db² are constant and x can be easily calculated as a function of punching ratio in SAFE.

Alternatively, if x is kept constant, maximum punching ratio in SAFE can be calculated for the specified punching reinforcement. Having determined that ratio, a uniform detail can be applied to columns having punching ratio less than this value. Above equation can be rearranged for that purpose and is given below;

**Example 1**

600mm circular columns with 780mm thick drop panels are used in a flat slab system with effective depth equal to 735mm. It is proposed to use 41 dia 10 bars at 150mm spacing around the column for punching reinforcement. Concrete strength is 45Mpa. Calculate the max SAFE punching ratio up to which this reinforcement can be applied safely. fys is 420 N/mm².

For 600mm circular column, bo is: pi(600+735) = 4194 mm

Putting this value in Eq: 2, we get;

See these calculations from SAFE with similar parameters;

**2. Extent of punching reinforcement**

Punching reinforcement is required up to the point beyond which unreinforced concrete alone can carry shear stress. Shear stress capacity of unreinforced concrete in MPa is given by;

and that for reinforced concrete is half of above value. To calculate shear force capacity we need to multiply this equation by concrete area Ac equal to bo’.d. From this we can calculate bo’, the critical parameter around column outside which concrete can carry entire shear. Value of this critical parameter is simply;

Next we need to calculate the distance from edges of column to this parameter to place punching shear reinforcement. Following are the examples of interior columns;

i. Square columns

From geometry this bo’ for a square column is equal to

B is the column dimension in x and y directions; cx and cy. Putting above equation equal to bo’ and rearranging, we get A in mm as;

ii- Rectangular columns

For rectangular columns critical parameter is equal to

and in terms of punching ratio we can write that;

iii- Circular columns

Critical parameter in terms of punching ratio will be;

Also we can relate A of circle and A of square column as;

In Example 1 above, A is calculated as;

**3. Detailing of punching reinforcement**

The most important part of a punching problem is detailing.

The first line of shear reinforcement should be put from face of the column a distance not less than 0.35d and not more than 0.4d. These lines of shear reinforcement should be put at spacing s that is equal or less than 0.5d up to a distance that is 0.5d away from critical parameter boundary.

Lets apply this to example 1; A is calculated as [1.47*2(600+735)-600]/2 = 1662mm

l = A-0.85d = 1662-0.85*735 = 1038mm

No. of single leg stirrups required are: l/s + 1 = 1038/150 + 1 = 8.

This is different from SAFE results of 3 legs. First, because we used a ratio of 1.47 and not 1.44 as SAFE calculated, second, in SAFE vu is integrated from stresses along punching parameter. It does not take max values. Maximum limit on shear stress vu as per ACI is 0.5(f’c^0.5). This corresponds to SAFE punching ratio of 0.5/0.33 = 1.52 or simply 1.5. But SAFE often calculates punching reinforcement up to ratios as high as 1.75 for the same reason mentioned above.

You can use the approach defined in this article by using equation 1 above to arrive at specified punching details and calculate its corresponding punching ratios. Then just simply look at SAFE model and specify the appropriate detail you have developed.

It can also be calculated conservatively that zone outside the drop panel will not fail in punching if;

is there feature in SAFE to actually detail the punching shear reinforcement ?

Dear John, I am not aware of any such feature.

Hi Rana the last image is still not updated, “It should be Ratio > bo.dcol / bo.dslab” as this was your comment last time. thanks just wanted clarification.

Hii Rana, Thank you very much for your great effort, nice example and perfect illustration this really helped me alot.. please forgive my weak english..

(ACI-318M-11/ eq 11.2.1.1) says that [ fc=0.17 sqr(f’c) ] is the basic expression for shear strength of members without shear reinforcement. I Iooked in ACI and did’t find [fc=0.33 sqr(f’c)] .

ACI says that both [Unreinforced concrete (concrete alone)] and [Reinforced concret] can carry shear stress by [ fc=0.17 sqr(f’c)] but the shear carrying mecanism varies between Unreinforced and Reinforced concrete section but both have the ” 0.17 ” .

I don’t know … I’m really confused ..

fc = 0.17 sq (45) =1.14 N/mm^2

fc = 0.33 sq (45) =2.21 N/mm^2

both are pretty far from the SAFE12 value =1.67 N/mm^2 … why !! please help !!

Egyptain COD XD says that [fc=0.24 sq(fc)] for Unreinforced and reduced to half for reinforced concrete…

0.24 sq (45) = 1.61 N/mm^2 XD XD XD very close !! ..

the eq 11.2.1.1 in ACI dose not give different reinforcement fom

[fc=0.33 sq(f’c)] the 0.17 in ACI only helps to put the shear reinforcement a little earlier than eq of the 0.33 but in case of Reinforced concrete section both are 0.17 because 0.33*0.5 = 0.17 so same reinforcement amount…

Please the program required 3 leg per set. Your drawing 2 set of leg .what the spacing between leg

I mean you have 41 set . The No of set divided by 4 (four face) this mean we need 11 set for each face. For each set three legs mean we need three vertical bars. Your drawing is 2 bars this mean you used two leg for set.

Waiting your reply if there is any mistake