# Punching Shear Check Example 1

**Punching Shear Design Example**

In the example calculation, a punching calculation will be made for a column whose dimensions are 100x80. Other parameters in the calculation are listed as follows.

: 27 cm**Slab thickness**: 3 cm**Slab cover**: C35/S420**Material**= 1380.42 kN/m**f**_{ctd}^{2}(Concrete design tensile strength)

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The slab useful height, * d* , is calculated;

* d* = (

*Slab thickness*-

*Slab cover*) = 27 - 3 = 24 cm

Column dimensions

* c_{1} *= 100 cm

*= 80 cm*

**c**_{2}The punching circumference **(*** u_{p}) is calculated at a distance d/2* from the column surface and is shown in the picture below. In this case, punching is done by adding

*to the column dimensions to find the perimeter's edges (b1, b2).*

**d**

In this case, the punching circumference** (*** u_{p })* and the punching area

**(A**_{z}**)**obtained by multiplying the punching perimeter by the flooring useful height,

*, is calculated as shown below.*

**d**

The shear stresses plotted below are the punching stresses perpendicular to the slab plane.

The * J* values are the sum of the polar inertia and second moments of inertia of the surfaces forming the punching area

**(A**_{z}**)**

*according to the loading direction considering the γ*

**. TBDY Equation 7.28 calculates this value**_{f}coefficient

**.****Calculation of J and γ _{f} **

**Coefficients for Strong Axis (Major Direction)**The coefficients * γ_{f(maj)}* and

*are calculated as follows.*

**γ**_{v(maj)}The following operations are performed to find the J (maj) value.

* c _{(maj)}* is the center of gravity distance perpendicular to the moment vector, which is considered when

*finding the*

*value (*

**J***) on the strong axis of the section. Since the punching circumference is rectangular,*

**J**_{(maj)}The polar inertia and second moments of inertia are calculated as follows, respectively.

In this case, the sum of the polar inertia and second moments of inertia about the section's strong (major) axis, * J _{(maj)}*, is found as follows.

**Calculation of J and γ _{f} Coefficients for Weak Axis ( Minor Direction )**

The following steps are followed to calculate the* γ _{f(min)}* and

*coefficients.*

**γ**_{v(min)}

The * J_{(min)}* value is found as follows.

The * c_{(min)}* value is the center of gravity distance perpendicular to the moment vector, which is considered when

*finding the*

*value*

**J****(**

**J**_{(min)}**)**on the weak axis of the section. Since the punching circumference is rectangular,

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

In this case, the sum of the polar inertia and second moments of inertia about the section's weak (minor) axis, J _{(min)}, is found as follows.

**Finding Punching Stresses**

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

kN**V**_{d}= 683.13= 143.53 kNm**DM**_{d(maj)}= 546.65 kNm**DM**_{d(min)}= 1.0944 m**A**_{z}^{ 2}= 0.579**γ**_{f(maj)}= 0.421**γ**_{v(maj)}= 0.621**γ**_{f(min)}= 0.379**γ**_{v(min)}= 0.27101448 m**J**_{(maj)}^{ 4}= 0.2083328 m**J**_{(min)}^{ 4}= 0.62 m**c**_{(maj)}= 0.52m**c**_{(min)}= 0.62 m**c'**_{(maj)}= 0.52 m**c'**_{(min)}

The punching stresses are found by substituting all the above values in the stress formulas and the internal forces essential to the punching design.

The absolute value of** τ _{pd,1}**

_{ },

**τ**_{pd,2}_{ is τ pd,1}= 1279.56 kN/m

^{ 2}. This obtained value is compared with the

**f**value, which is the concrete design

_{ctd}*tensile strength.*

* τ_{pd,1}* = 1279.56 kN/m

^{2}<

*= 1380.42 kN/m*

**f**_{ctd}^{2}

Upholstery punching strength is sufficient. These values can also be compared with the report results.

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