VE 9962 | Punching Shear Design of Edge Columns in Flat Slab According to CSA A23.3

Description

This example investigates the punching shear design of an edge column according to CSA A23.3-19 [1]. The geometry and loading were taken from the edge column D2 in Example 1 of the “CAC Concrete Design Handbook – 4th Edition,” page 5-19 [2].

55x

7x

Verification Example | Punching Shear Design of Edge Column According to CSA A23.3

Verification Example 9962 | Column Dimensions

Analytical Solution:

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\)

1. Determination of Geometric Quantities

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\)
Dimension of the critical perimeter parallel to the eccentricity:
\(
\begin{aligned}
\mathsf{b_{1}}
&= \mathsf{ b + \dfrac{d}{2} + a } \\
&= \mathsf{ 600 + 105 + 100 } \\
&= \mathsf{ 805\,mm }
\end{aligned}
\)

\(
\begin{aligned}
\mathsf{b_{2}}
&= \mathsf{ h + d } \\
&= \mathsf{ 400 + 210 } \\
&= \mathsf{ 610\,mm }
\end{aligned}
\)
\(
\)
Length of the critical perimeter:
\(
\begin{aligned}
\mathsf{b_{o}}
&= \mathsf{ 2 \cdot \left( b + a + \dfrac{d}{2} \right) + h + d } \\
&= \mathsf{ 2 \cdot (600 + 100 + 105) + 400 + 210 } \\
&= \mathsf{ 2220\,mm }
\end{aligned}
\)
\(
\)
Eccentricities of the critical perimeter
\(
\begin{aligned}
\mathsf{e_{1}}
&= \mathsf{ \dfrac{ b_{1}^{2} }{ 2\,b_{1} + b_{2} } } \\
&= \mathsf{ \dfrac{ 805^{2} }{ 2 \cdot 805 + 610 } } \\
&= \mathsf{ 292\,mm }
\end{aligned}
\)

\(
\begin{aligned}
\mathsf{e_{2}}
&= \mathsf{ \dfrac{ b_{2} }{ 2 } } \\
&= \mathsf{ \dfrac{ 610 }{ 2 } } \\
&= \mathsf{ 305\,mm }
\end{aligned}
\)

2. Calculation of Actions:

The indices "1" and "2" refer to the main axes of the structural system.

• The index "1" refers to the global X-axis.

• The index "2" refers to the global Y-axis. It will not be considered further below, as there is only a moment about the X-axis.

\(
\)
The reduction factors are calculated as follows:
\(
\begin{aligned}
\mathsf{\gamma_{v,1}}
&= \mathsf{
1
-
\frac{1}{
1
+
\frac{2}{3}
\sqrt{
\frac{
b + \frac{d}{2} + a
}{
h + d
}
}
}
} \\
&= \mathsf{
1
-
\frac{1}{
1
+
\frac{2}{3}
\sqrt{
\frac{
600 + \frac{210}{2} + 100
}{
400 + 210
}
}
}
} \\
&= \mathsf{ 0.434 }
\end{aligned}
\)
\(
\)

Thus, the following “polar” moments of inertia are obtained:

\(
\begin{aligned}
\mathsf{J_{1}}
&= \mathsf{
2 \left(
\frac{b_{1}^{3} \cdot d}{3}
+
\frac{d^{3} \cdot b_{1}}{12}
\right)
-
b_{o} d e^{2}
} \\
&= \mathsf{
2 \left(
\frac{805^{3} \cdot 210}{3}
+
\frac{210^{3} \cdot 805}{12}
\right)
-
2220 \cdot 210 \cdot 292^{2}
} \\
&= \mathsf{3.453 \cdot 10^{10}\,\mathrm{mm^{4}}}
\end{aligned}
\)

Shear force to be reduced due to load redistribution within the critical perimeter:

\(
\begin{aligned}
\mathsf{\Delta V_{f}}
&= \mathsf{ p \cdot b_{1} \cdot b_{2} } \\
&= \mathsf{ 11.6\,\mathrm{kN/m^{2}} \cdot 0.805\,\mathrm{m} \cdot 0.610\,\mathrm{m} } \\
&= \mathsf{ 5.70\,\mathrm{kN} }
\end{aligned}
\)
\(
\)
The reduced shear force is calculated as follows:
\(
\begin{aligned}
\mathsf{V_{f,\mathrm{res}}}
&= \mathsf{ V_f - \Delta V_f } \\
&= \mathsf{ 339.26\,\mathrm{kN} - 5.70\,\mathrm{kN} } \\
&= \mathsf{ 333.56\,\mathrm{kN} }
\end{aligned}
\)
\(
\)
The component of the moment transferred in the first principal direction:

\(
\begin{aligned}
\mathsf{M_{f,1,\mathrm{sl}}}
&= \mathsf{ M_{f,1} - V_{f,\mathrm{res}} \cdot e_{1,\mathrm{sl}} } \\
&= \mathsf{ 167.62\,\mathrm{kNm} - 333.56\,\mathrm{kN} \cdot 0.1131\,\mathrm{m} } \\
&= \mathsf{ 129.89\,\mathrm{kNm} }
\end{aligned}
\)
\(
\)

Shear stress from the reduced shear force:
\(
\begin{aligned}
\mathsf{\nu_{fv}}
&= \mathsf{ \frac{ V_{f,\mathrm{res}} }{ b_{o} \cdot d } } \\
&= \mathsf{ \frac{ 333.56\,\mathrm{kN} }{ 2.220\,\mathrm{m} \cdot 0.210\,\mathrm{m} } } \\
&= \mathsf{ 0.715\,\mathrm{MPa} }
\end{aligned}
\)
\(
\)

The maximum acting shear stress is obtained as follows:
\(
\begin{aligned}
\mathsf{\nu_{f}}
&= \mathsf{\nu_{fv} + \frac{\gamma_{v,1} \cdot M_{f,1,sl} \cdot e_{1}}{J_{1}}} \\
&= \mathsf{0.715\,\mathrm{MPa} + \frac{0.434 \cdot 129.89\,\mathrm{kNm} \cdot 292\,\mathrm{mm}}{3.453 \times 10^{10}\,\mathrm{mm^{4}}}} \\
&= \mathsf{0.715 + 0.477} \\
&= \mathsf{1.192\,\mathrm{MPa}}
\end{aligned}
\)
\(
\)

3. Calculation of Resistance:

Ratio of the long to the short side of the column:

\(
\begin{aligned}
\mathsf{\beta_{c}}
&= \mathsf{ \dfrac{ \max\!\left( b,\, h \right) }{ \min\!\left( b,\, h \right) } } \\
&= \mathsf{ \dfrac{ \max\!\left( 0.600\,\mathrm{m},\, 0.400\,\mathrm{m} \right) }{ \min\!\left( 0.600\,\mathrm{m},\, 0.400\,\mathrm{m} \right) } } \\
&= \mathsf{ 1.500 }
\end{aligned}
\)

Punching shear resistance of the slab without shear reinforcement according to 13.3.4.1 (a):

\(
\begin{aligned}
\mathsf{\nu_{c(a)}}
&= \mathsf{
\left( 1 + \dfrac{2}{\beta_{c}} \right)
\cdot 0.19
\cdot \lambda
\cdot \Phi_{c}
\cdot \min\!\left( \sqrt{f'_{c}},\, f'_{c,\max} \right)
} \\
&= \mathsf{
\left( 1 + \dfrac{2}{1.50} \right)
\cdot 0.19
\cdot 1.00
\cdot 0.65
\cdot \min\!\left( \sqrt{25.00\,\mathrm{MPa}},\, 8.00\,\mathrm{MPa} \right)
} \\
&= \mathsf{ 1.441\,\mathrm{MPa} }
\end{aligned}
\)

Punching shear resistance of the slab without shear reinforcement according to 13.3.4.1 (b):
\(
\begin{aligned}
\mathsf{\nu_{c(b)}}
&= \mathsf{
\left(
\dfrac{ \alpha_{s} \cdot d }{ b_{o} }
+ 0.19
\right)
\cdot \lambda
\cdot \Phi_{c}
\cdot \min\!\left( \sqrt{f'_{c}},\, f'_{c,\max} \right)
} \\
&= \mathsf{
\left(
\dfrac{ 3.00 \cdot 0.210\,\mathrm{m} }{ 2.220\,\mathrm{m} }
+ 0.19
\right)
\cdot 1.00
\cdot 0.65
\cdot \min\!\left( \sqrt{25.00\,\mathrm{MPa}},\, 8.00\,\mathrm{MPa} \right)
} \\
&= \mathsf{ 1.540\,\mathrm{MPa} }
\end{aligned}
\)

Punching shear resistance of the slab without shear reinforcement according to 13.3.4.1 (c):
\(
\begin{aligned}
\mathsf{\nu_{c(c)}}
&= \mathsf{
0.38
\cdot \lambda
\cdot \Phi_{c}
\cdot \min\!\left( \sqrt{f'_{c}},\, f'_{c,\max} \right)
} \\
&= \mathsf{
0.38
\cdot 1.00
\cdot 0.65
\cdot \min\!\left( \sqrt{25.00\,\mathrm{MPa}},\, 8.00\,\mathrm{MPa} \right)
} \\
&= \mathsf{ 1.235\,\mathrm{MPa} }
\end{aligned}
\)

Minimum punching shear resistance of the slab without shear reinforcement:
\(
\begin{aligned}
\mathsf{\nu_{c}}
&= \mathsf{ \min\!\left( \nu_{c(a)},\, \nu_{c(b)},\, \nu_{c(c)} \right) } \\
&= \mathsf{ \min\!\left( 1.441\,\mathrm{MPa},\, 1.540\,\mathrm{MPa},\, 1.235\,\mathrm{MPa} \right) } \\
&= \mathsf{ 1.235\,\mathrm{MPa} }
\end{aligned}
\)

4. Comparison of Actions and Resistance:

\(
\mathsf{\eta_{13.3.4} = \dfrac{\nu_{f}}{\nu_{r}}}
\)

\(
\mathsf{\eta = \dfrac{1.192\,MPa}{1.235\,MPa}}
\)

\(
\mathsf{\eta \approx 0,97}
\)

\(
\mathsf{\eta = 0,97 \;\leq\; 1 \quad \Rightarrow \quad \text{Verification satisfied}}
\)

The arrangement of punching shear reinforcement is not required.
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\)

Results

The results from RFEM 6 are shown below.

VE 9962 – Design Check Details

The results from RFEM 6 are compared with the reference solution below.

VE 9962 – Design Ratio

Evaluation

The results from RFEM 6 agree very well with the reference solution.

RFEM 6 calculates slightly lower (approx. 3.5%) values for the polar moment of inertia than the manual calculation. RFEM 6 calculates the polar moment of inertia according to ACI 421.1R.

The approach in ACI 421.1R is applicable to all cross-section geometries and is well-suited for a software solution.

In contrast to the analytical formula, which is applicable only to rectangular cross-sections, ACI 421.1R neglects the term \(\mathsf{\dfrac{b\,d^{3}}{6}}\).

The smaller polar moment of inertia according to ACI 421.1R results in a conservative approach to the punching shear design.

55x

7x

Verification Example | Punching Shear Design of Edge Column According to CSA A23.3