Polar Moment / Moment of Inertia for Punching Shear Design in ACI 318 and CSA A23.3

The following section explains the calculation method according to ACI 421.1R-20 [1] using a rectangular interior column as an example. The results of analytical formulas are shown and, finally, compared with the RFEM 6 results.

1. Formula According to ACI 421.1R-20 – Example of Inner Column

\(
J_x = d \sum\left[ \dfrac{l}{3} \cdot \left( y_i^2 + y_i y_{i+1} + y_{i+1}^2 \right) \right]
\)

\(
J_y = d \sum\left[ \dfrac{l}{3} \cdot \left( x_i^2 + x_i x_{i+1} + x_{i+1}^2 \right) \right]
\)

Structural system
The column cross-section and the effective depth of the slab are given for the calculation of the moments of inertia.

Ground Plan of Inner Column with Dimensioning

• Column cross-section: b = 12 in
• Column cross-section: h = 16 in
• Effective depth of the slab: d = 8.5 in

The control perimeter is located at a distance of 0.5d from the edges of the perimeter section. To calculate the moment of inertia, the distances from the corner points on the control perimeter to the center of mass of the control perimeter are required.

This results in the following coordinates for the corner nodes.

• Point 1 (-10.25 / 12.25)
• Point 2 (10.25 / 12.25)
• Point 3 (10.25 / -12.25)
• Point 4 (-10.25 / -12.25)

In the following example, the moments of inertia are calculated for illustrative purposes. The formula values y_i and x_i are the respective coordinates of Points 1 through 4.

The distances of each side are determined from the previously calculated distance coordinates.

• l_1-2 = 20.5 in
• l_2-3 = 24.5 in
• l_3-4 = 20.5 in
• l_4-1 = 24.5 in

Calculation of Moment of Inertia I_x:

I_x1 = d * [l_1-2/3 * (y₁² + y₁ * y₂ + y₂²)]
I_x1 = 8.5 * [20.5/3 * (12.25² + 12.25 * 12.25 + 12.25²)]
I_x1 = 26148 in⁴

I_x2 = d * [l_2-3/3 * (y₂² + y₂ * y₃ + y₃²)]
I_x2 = 8.5 * [24.5/3 * (12.25² + 12.25 * -12.25 + (-12.25))²)]
I_x2 = 10417 in⁴

I_x3 = d * [l_3-4/3 * (y₃² + y₃ * y₄ + y₄²)]
I_x3 = 8.5 * [20.5/3 * ((-12.25)² - 12.25 * -12.25 + (-12.25)²)]
I_x3 = 26148 in⁴

I_x4 = d * [l_4-1/3 * (y₄² + y₄ * y₁ + y₁²)]
I_x4 = 8.5 * [24.5/3 * ((-12.25)² - 12.25 * 12.25 + 12.25)²)]
I_x4 = 10417 in⁴

∑ I_x = 73130 in⁴

Calculation of Moment of Inertia I_y:

I_y1 = d * [l_1-2/3 * (x₁² + x₁ * x₂ + x₂²)]
I_y1 = 8.5 * [20.5/3 * ((-10.25)² - 10.25 * 10.25 + 10.25)²)]
I_y1 = 6102 in⁴

I_y2 = d * [l_2-3/3 * (x₂² + x₂ * x₃ + x₃²)]
I_y2 = 8.5 * [24.5/3 * (10.25² + 10.25 * 10.25 + 10.25²)]
I_y2 = 21879 in⁴

I_y3 = d * [l_3-4/3 * (x₃² + x₃ * x₄ + x₄²)]
I_y3 = 8.5 * [20.5/3 * (10.25² + 10.25 * -10.25 + (-10.25)²)]
I_y3 = 6102 in⁴

I_y4 = d * [l_4-1/3 * (x₄² + x₄ * x₁ + x₁²)]
I_y4 = 8.5 * [24.5/3 * ((-10.25)² - 10.25 * -10.25 + (-10.25)²)]
I_y4 = 21879 in⁴

∑ I_y = 55962 in⁴

2. Comparison with Analytical Formulas – Inner Column

• 2.1 – Formula from ACI 318-19 [2], page 108

\(
J_x = \dfrac{d(h + d)^3}{6} + \dfrac{d^3(h + d)}{6} + \dfrac{d(b + d)(h + d)^2}{2}
\)

\(
J_y = \dfrac{d(b + d)^3}{6} + \dfrac{d^3(b + d)}{6} + \dfrac{d(h + d)(b + d)^2}{2}
\)

b = 12 in
h = 16 in
d = 8.5 in

I_x = 75638 in⁴
I_y = 58061 in⁴

• 2.2 – Formula from Literature Jordahl [3] USA, Inc. | Studrails | 03-2023

\(
J_x = d \left( \dfrac{L_y^3 }{6} + \dfrac{L_x L_y^2}{2} \right)
\)

\(
J_y = d \left( \dfrac{L_x^3 }{6} + \dfrac{L_y L_x^2}{2} \right)
\)

b = 12 in
h = 16 in
d = 8.5 in
L_x = 20.5 in
L_y = 24.5 in

I_x = 73130 in⁴
I_y = 55963 in⁴

3. Comparison with Analytical Formulas – Edge Column

Due to their location at the edge of the slab, edge columns exhibit an asymmetry in their basic control perimeter. This shifts the center of gravity, and it is necessary to take it into account in subsequent calculations.

• 3.1 – Formula from CAC Concrete Design Handbook – 4th ^th Edition [4]

Ground Plan of Edge Column with Dimensioning

\(
J_x = \dfrac{(h + d)d^3 + (h + d)^3 d}{12} + 2\left(b + \dfrac{d}{2}\right)d(e_2)^2
\)

\(
J_y = \dfrac{(b + \dfrac{d}{2})d^3 + (b + \dfrac{d}{2})^3 d}{6} + \left(h + d\right) d(e_1)^2 + 2\left(b + \dfrac{d}{2}\right)d\left(\dfrac{b + \dfrac{d}{2}}{2 - e_1}\right)^2
\)

b = 12 in
h = 16 in
d = 8.5 in

e₁ = (h + d/2)² / (2h + b + 2d) = 4.63 in
e₂ = (b + d)/ 2 = 12.25 in

I_x = 53125 in⁴
I_y = 15581 in⁴

Result Comparison

As already mentioned at the beginning of this technical article, the polar moment of inertia in RFEM 6 is calculated according to the approach specified in ACI 421.1R-20 [1]. Compared to manual calculations using the analytical formulas, the results are slightly lower; this is due to the neglect of a component of the moment of inertia (for rectangular geometries where b*d < sup > 3 < /sup >/6). Because of the lower moments of inertia, the results of the punching shear design are on the safe side. However, the real advantage of the implemented formula is its universal applicability; all cross-section shapes and geometric conditions can be taken into account.