Punching Shear Check Example 2

In the example calculation, a punching calculation will be made for a column whose dimensions are 100x50. Other parameters are listed as follows.

• Slab thickness : 27 cm

• Slab cover: 3 cm

• Material : C35/S420

• f_ctd = 1380.42 kN/m² (Concrete design tensile strength)

The slab useful height, d , is calculated;

d = ( Slab thickness - Slab cover ) = 27 - 3 = 24 cm

Column dimensions

c₁ = 100 cm c₂ = 50 cm

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, the following operations are performed to find the edges (b₁, b₂) of the punching.

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, d , is calculated as shown below.

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 TBDY Equation 7.28, this value is calculated according to the loading direction considering the γ_f coefficient .

Calculation of J and γ_f Coefficients for Strong Axis (Major Direction)

The coefficients γ_f(maj) and γ_v(maj) are calculated as follows.

To find the J _(maj) value, the following operations are performed.

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

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) = b₁ / 2 = 74/2 = 37 cm = 0.37 m c'_(min) = b₁ - 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.