# Punching Shear Check Example 2

Punching calculation will be made for a column with dimensions of 100x50 in the sample calculation for V10.19. Other parameters in the account are listed as follows.

flooring thickness: 27 cm, floor spacers: 3 cm,

Material: C35 / S420, Concrete design tensile strength f _{ctd} = 1380.42 kN / m ^{2}

slab utility height, d, calculation;

d = (slab thickness - slab spacers) = 27 - 3 = 24cm

Column dimensions c _{1} = 100 cm c _{2} = 50 cm

The punch circumference (u _{p} ) is calculated at a distance d / 2 from the column surface and is shown in the image below. In this case , we can add distance d to the column dimensions to find the edges of the punch perimeter (b _{1} , b _{2} ).

b_{1} = c_{1} + d = 100 + d= 100 + 24 = 124 cm

b_{2} = c_{2} + d = 100 + d= 50 + 24 = 74 cm

In this case, the punching circumference (u _{p} ), and the punching area (A _{z} ) obtained by multiplying the punch perimeter by the slab useful height, d, is calculated as shown below.

u_{p} = 2*(b_{1} + b_{2}) = 2*(124+74) = 396 cm = 3.96 m

A _{z} = u _{p} * d = 3.96 * 0.24 = 0.95040 m ^{2}

The shear stresses plotted below are the punch stresses perpendicular to the floor plane.

J values are the sum of the polar inertia and second moments of inertia of the surfaces forming the punching area (A _{z} ). Also **TBDY Equation 7.28** according γ _{f} is calculated according to the loading direction coefficient taken into account. Here, separate calculations are made in the X and Y directions of the Column for both values.

**Strong Axis (Major Aspects) to J and γ _{f} Coefficient Account**

The following steps are followed for the calculation of the coefficient _{j f (maj)} and γ _{v (maj)} .

To find the value of J _{(maj)} , the following way is followed.

c _{(maj)} is the center of gravity distance perpendicular to the moment vector, considered when _{finding the} J value (J _{(maj)} ) on the strong axis of the section . Since the punching circumference is rectangular;

c_{(maj)} = b_{1} / 2 = 124/2 = 62 cm = 0.62 m c'_{(maj)} = b_{1} - c_{(maj)} = 124 - 62 = 62 cm = 0.62 m

In this case, the polar inertia and the two moments of inertia are respectively as follows.

In this case, the sum of the polar inertia and the second moments of inertia J _{(maj)} with respect to the major axis of the section is found as follows.

J _{(maj)} = J _{1 (maj)} + J _{2 (maj)} = 0.079122 + 0.1365388 = 0.21566080 m ^{4} = 21566080 cm ^{4}

**Weak Axis (Minor Direction) to J and γ _{f} Coefficient Account**

For the calculation of γ _{f (min)} and γ _{v (min)} coefficient, the following steps are followed.

The following method is used to find the J _{(min)} value.

c _{(min)} is the center of gravity distance perpendicular to the moment vector, which is considered when finding the J value (J _{(min)} ) on the strong axis of the section . Since the punching circumference is rectangular;

c_{(min)} = b_{1} / 2 = 74/2 = 37 cm = 0.37 m c'_{(min)} = b_{1} - c_{(min)} = 74 - 37 = 37 cm = 0.37 m

In this case, the polar inertia and the two moments of inertia are respectively as follows.

In this case, the sum of the polar inertia and the second moments of inertia J _{(min)} with respect to the minor axis of the section is found as follows.

J _{(min)} = J _{1 (min)} + J _{2 (min)} = 0.017914 + 0.081483 = 0.0993968 m ^{4} = 9939680 cm ^{4}

**Finding Punching Stresses**

The forces to be considered for punching stresses and the values obtained from hand geometry are given below.

V _{d} = 679.31 kN DM _{d (maj)} = 394.624 kNm DM _{d (min)} = 65.3914 kNm

A _{z} = 0.95040 m ^{2}

γ _{f (maj)} = 0.537 γ _{v (maj)} = 0.463 γ _{f (min)} = 0.660 γ _{v (min)} = 0.340

J _{(maj)} = = 0.21566080 m ^{4} , J _{(min)} = = 0.0993968 m ^{4} ,

c _{(maj)} = 0.62 m, c _{(min)} = 0.37 m, c ' _{(maj)} = 0.62 m, c _{(min)} = 0.37 m

If we substitute all the values we found above for the stress formulas together with the internal forces based on the punching design;

Among the values of τ _{pd, 1} , τ _{pd, 2, the} absolute value is τ _{pd, 1} = 1323.0 kN / m ^{2} . If we compare this value with the concrete design tensile strength f _{ctd} value;

τ _{pd, 1} = 1323.0 kN / m ^{2} <f _{ctd} = 1380.42 kN / m ^{2}

We reach the result. From here, we can conclude that the punching strength of this column is sufficient. These values can also be compared with the report results.