Punching In the example calculation, a punching calculation will be made for a column with whose dimensions of 100x50 in the sample calculation for V10.19are 100x50. Other parameters in the account are listed as follows.flooring
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fctd = 1380.42 kN/m2 (Concrete design tensile strength)
The slab utility useful height, d , calculationis calculated;
d = (slab Slab thickness - slab spacers Slab cover ) = 27 - 3 = 24cm24 cm
Column dimensions
c1 = 100 cm c2 = 50 cm
The punch punching circumference (up) is calculated at a distance d/2 from the column surface and is shown in the image picture below. In this case, we can add distance d to the column dimensions the following operations are performed to find the edges of the punch perimeter (b1, b2) of the punching.
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| body | $$ \normalsize b_1=c_1 |
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+ d = 100 cmb| Mathinline |
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| body | $$ \normalsize b_2=c_2 |
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+ d = 100 cm...
In this case, the punching circumference, (up ), and the punching area(Az) obtained by multiplying the punch punching perimeter by the slab flooring useful height, d , is calculated as shown below.
**| \times(124+74) = 396 \; cm = 3.96 \; m $$ |
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| body | --uriencoded--$$ \normalsize A_z=u_p |
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* * 095040 m 2The shear stresses plotted below are the punch punching stresses perpendicular to the floor slab plane.
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The J values are the sum of the polar inertia and second moments of inertia of the surfaces forming the punching area (Az). Also According to TBDY Equation 7.28 according γ f ,this value 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.considering the γf coefficient .
Calculation of J and γf Coefficients for Strong Axis (Major Aspects) to J and γ f Coefficient AccountThe following steps are followed for the calculation of the coefficient j Direction)
The coefficients γf(maj) and γv(maj) are calculated as follows.
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| body | --uriencoded--$$ \normalsize \gamma_%7Bf(maj)%7D = \frac %7B1%7D %7B1+\frac 2 3 \sqrt%7Bb_1/b_2%7D%7D = \frac %7B1%7D %7B1+\frac 2 3 \sqrt%7B1.24/0.74%7D%7D = 0.537 \; $$ |
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| body | --uriencoded--$$ \normalsize \gamma_%7Bv(maj)%7D =1- \gamma_%7Bf(maj)%7D = 1-0.537 = 0.463 $$ |
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To find the value of J (maj) value, the following way is followedoperations are performed.
c (maj) is the center of gravity distance perpendicular to the moment vector, which is considered whenfinding the J value ( J (maj) ) on the strong axis of the section. Since the punching circumference is rectangular;c, the c (maj) value is calculated as follows.
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| body | --uriencoded--$$ \normalsize c_%7B(maj)%7D = b_1/2 = 124/2=62 \; cm = 0.62 \; m $$ |
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| body | --uriencoded--$$ \normalsize c'_%7B(maj)%7D = b_1-c_%7B(maj)%7D= 124-62=62 \; cm = 0.62 \; m $$ |
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In this case, the The polar inertia and the two second moments of inertia are respectively calculated as follows, respectively.
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| body | --uriencoded--$$ \normalsize J_%7B1(maj)%7D = \left( \frac%7Bb_1d%5e3%7D%7B12%7D + \frac%7Bb_1%5e3d%7D%7B12%7D \right) \times 2 = \left( \frac%7B1.24 \times 0.24%5e3%7D%7B12%7D + \frac%7B1.24%5e3 \times 0.24%7D%7B12%7D \right) \times 2 = 0.079122 \; m%5e4 $$ |
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| body | --uriencoded--$$ \normalsize J_%7B2(maj)%7D = b_2 d \times c%5e2_%7B(maj)%7D \times 2 = 0.74 \times 0.24 \times 0.62%5e2 \times 2= 0.1365388 \; m%5e4 $$ |
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In this case, the sum of the polar inertia and the second moments of inertia J about the strong (majmajor) with respect to the major axis of the section, J (maj) , is found as follows.
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| body | --uriencoded--$$ \normalsize J_%7B(maj)%7D = J |
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1 2 | _%7B2(maj)%7D = 0.079122 + 0.1365388 = 0. |
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21566080 m 4 | 2156608 \; m%5e4 = 21566080 |
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cm 4Calculation of J and γf Coefficients for Weak Axis (Minor Direction) to J and γ f Coefficient AccountFor
The following steps are followed for the calculation of the γ f(min) and γ v (min) coefficient, the following steps are followed.
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The following method is used to find the (min) coefficients.
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| body | --uriencoded--$$ \normalsize \gamma_%7Bf(min)%7D = \frac %7B1%7D %7B1+\frac 2 3 \sqrt%7Bb_2/b_1%7D%7D = \frac %7B1%7D %7B1+\frac 2 3 \sqrt%7B0.74/1.24%7D%7D = 0.660 \; $$ |
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| body | --uriencoded--$$ \normalsize \gamma_%7Bv(min)%7D =1- \gamma_%7Bf(maj)%7D = 1-0.66 = 0.340 $$ |
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The J(min) value is found as follows.
The c(min) value is the center of gravity distance perpendicular to the moment vector, which is considered whenfinding the J value (J (min)) on the strong weak axis of the section. Since the punching circumference is rectangular;
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| body | --uriencoded--$$ \normalsize c_%7B(min)%7D = b |
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1 | _2/2 = 74/2=37 \; cm = 0.37 \; m $$ |
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| body | --uriencoded--$$ \normalsize c'_%7B(min)%7D = b |
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1 | _2-c_%7B(min)%7D= 74-37=37 \; cm = 0.37 \; m $$ |
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In this case, the polar moments of inertia and the two second moments of inertia are respectively found as follows, respectively.
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| body | --uriencoded--$$ \normalsize J_%7B1(min)%7D = \left( \frac%7Bb_2d%5e3%7D%7B12%7D + \frac%7Bb_2%5e3d%7D%7B12%7D \right) \times 2 = \left( \frac%7B0.74 \times 0.24%5e3%7D%7B12%7D + \frac%7B0.74%5e3 \times 0.24%7D%7B12%7D \right) \times 2 = 0.01791392 \; m%5e4 $$ |
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| body | --uriencoded--$$ \normalsize J_%7B2(min)%7D = b_1 d \times c%5e2_%7B(min)%7D \times 2 = 1.24 \times 0.24 \times 0.37%5e2 \times 2= 0.08148288 \; m%5e4 $$ |
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In this case, the sum of the polar inertia and the second moments of inertia J about the weak (minminor) with respect to the minor axis of the section, J (min) , is found as follows.
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| body | --uriencoded--$$ \normalsize J_%7B(min)%7D = J |
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1 2 017914 081483 m 4 cm 4Finding Punching Stresses
The forces to be considered for punching stresses and the values obtained obtained from hand the geometry are given below.
Vd = 679.31 kN
DMd(maj) = 394.624 kNm
DMd(min) = 65.3914 kNm
Az = 0.
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9504 m2
γf(maj) = 0.537
γv(maj) = 0.463
γf(min) = 0.660
γv(min) = 0.340
J(maj)
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J(min) = = 0.0993968 m4 ,
c(maj) = 0.62 m
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If we substitute The punching stresses are found by substituting all the above values we found above for in the stress formulas together with the internal forces based on the punching design;
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Among the values of essential to the punching design.
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| body | --uriencoded--$$ \normalsize \tau_%7Bpd,1%7D = \frac %7BV_d%7D%7BA_z%7D + \frac %7B \gamma_%7Bv(maj)%7DDM_%7Bd(maj)%7Dc_%7B(maj)%7D %7D %7BJ_%7B(maj)%7D%7D + \frac %7B \gamma_%7Bv(min)%7DDM_%7Bd(min)%7Dc_%7B(min)%7D %7D %7BJ_%7B(min)%7D%7D $$ |
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| body | --uriencoded--$$ \normalsize \tau_%7Bpd,2%7D = \frac %7BV_d%7D%7BA_z%7D - \frac %7B \gamma_%7Bv(maj)%7DDM_%7Bd(maj)%7Dc_%7B(maj)%7D %7D %7BJ_%7B(maj)%7D%7D - \frac %7B \gamma_%7Bv(min)%7DDM_%7Bd(min)%7Dc_%7B(min)%7D %7D %7BJ_%7B(min)%7D%7D $$ |
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| body | --uriencoded--$$ \normalsize \tau_%7Bpd,1%7D = \frac %7B679.31%7D%7B0.9504%7D + \frac %7B 0.463 \times 394.624 \times 0.62 %7D %7B0.2156608%7D + \frac %7B 0.34 \times 65.39 \times 0.37 %7D %7B0.0993968%7D = 1322.795 \; kN/m%5e2 $$ |
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| body | --uriencoded--$$ \normalsize \tau_%7Bpd,2%7D = \frac %7B679.31%7D%7B0.9504%7D - \frac %7B 0.463 \times 394.624 \times 0.62 %7D %7B0.2156608%7D - \frac %7B 0.34 \times 65.39 \times 0.37 %7D %7B0.0993968%7D = 106.73 \; kN/m%5e2 $$ |
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The absolute value of τ pd,1, τ pd,2, the absolute value is τ pd,1 = 13231322.0 795 kN/m 2 . If we compare this value with the This obtained value is compared with the fctd value, which is the concrete designtensile strength f ctd value;.
τpd,1 = 13231322.0 795 kN/m2 <f < fctd = 1380.42 kN/m2
We reach the result. From here, we can conclude that the Upholstery punching strength of this column is sufficient. These values can also be compared with the report results.
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Related Topics
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Punching Shear Check
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Punching Shear Stress at Critical Section (7.11.8)
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(8.3.1) TS500 Puching Shear Design Conditions
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Punching Shear Stress Check Report
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