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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--$$ \normalsize J_%7B(maj)%7D = J |
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1 2 | _%7B2(maj)%7D = 0.079122 + 0.1365388 = 0. |
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21566080 m 4 | 2156608 \; m%5e4 = 21566080 |
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cm 4Calculation of J and γf Coefficients for Weak Axis (Minor Direction) to J and γ f Coefficient AccountFor
The following steps are followed for the calculation of the γ f(min) and γ v(min) coefficient, the following steps are followed.
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The following method is used to find the coefficients.
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| body | --uriencoded--$$ \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 all the values we found above for the stress formulas together with the internal forces based on the punching design;
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