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 ).
b1 = c1 + d = 100 + d= 100 + 24 = 124 cm
b2 = c2 + 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.
up = 2*(b1 + b2) = 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) = b1 / 2 = 124/2 = 62 cm = 0.62 m c'(maj) = b1 - 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) = 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.