Area Moment of Inertia - Horizontal Strip of Area

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2.1.3 Horizontal Infinitesimal Strip of Area

In Figure 6.2.3, instead of using a tiny area element $\mathrm{d}A$ with side lengths $\mathrm{d}y$ and $\mathrm{d}z$ (Figure 6.2.1) or a vertically arranged infinitesimal strip of thickness $\mathrm{d}y$ parallel to the $z$-axis (Figure 6.2.2), we use a horizontally arranged infinitesimal strip of thickness $\mathrm{d}z$ parallel to the $y$-axis. This ensures that all points on the strip have the same $z$-coordinate with respect to the $y$-axis.

Horizontal strip of area within an arbitrary surface. Arbitrarily arranged x, y, z coordinate system, y positive to the left, z positive downward. The strip is located at position z, with a height of dz and a width of b(z). — Fig. 6.2.3: Calculation method of horizontal infinitesimal strip in Cartesian coordinates

The base $b(z)$ of the strip depends on the $z$-coordinate. Therefore, the calculation for the area element $\mathrm{d}A$ in formulas (6.2) - (6.4) is given by

$$ \begin{align} \tag{1} \mathrm{d}A &= b(z) \cdot \mathrm{d}z \end{align} $$

In order to perform the calculation of area moments of inertia using this method, it is necessary to determine the functional relationship $b(z)$. This provides the upper and lower boundary functions that, for each $z$, give the upper $\left(y^+(z)\right)$ and lower $\left(y^-(z)\right)$ boundaries of the area differential.

For any arbitrary $z$, the following holds:

$$ \begin{align} \tag{2} b(z) &= \int\limits_{y^-(z)}^{y^+(z)} \mathrm{d}y \end{align} $$

Thus, for any arbitrary $z$, we can determine the infinitesimal area moments of inertia using formulas (6.2) and (6.4) by using equations (10) and (11) and keeping the independent variable $z$ constant:

$$ \begin{alignat}{9} \tag{3} I_y&= \int\limits_{(A)} z^2 \ \mathrm{d}A \quad &&\Rightarrow \quad &&\mathrm{d}I_y(z) &&= z^2 \left( \int\limits_{y^-(z)}^{y^+(z)} \ \mathrm{d}y\right)\ \mathrm{d}z &&= z^2 \left[ y^+(z) - y^-(z)\right]\ \mathrm{d}z \\[7pt] \tag{4} I_z&= \int\limits_{(A)} y^2 \ \mathrm{d}A \quad &&\Rightarrow \quad &&\mathrm{d}I_z(z) &&= \left( \int\limits_{y^-(z)}^{y^+(z)} y^2 \ \mathrm{d}y\right)\ \mathrm{d}z &&= \dfrac{1}{3} \left[ \left(y^+(z)\right)^3 - \left(y^-(z)\right)^3 \right]\ \mathrm{d}z \\[7pt] \tag{5} I_{yz} &= -\int\limits_{(A)} y \cdot z \ \mathrm{d}A \quad &&\Rightarrow \quad &&\mathrm{d}I_{yz}(z) &&= -z \left( \int\limits_{y^-(z)}^{y^+(z)} y \ \mathrm{d}y\right)\ \mathrm{d}z &&= -\dfrac{1}{2}z \left[ \left(y^+(z)\right)^2 - \left(y^-(z)\right)^2 \right]\ \mathrm{d}z \end{alignat} $$

We obtain the desired area moments for the entire area by summing, i.e., integrating, the infinitesimal quantities $\mathrm{d}I_y(z)$, $\mathrm{d}I_z(z)$, and $\mathrm{d}I_{yz}(z)$ over the independent variable $z$.

$$ \begin{alignat}{3} \tag{6} I_y&= \int\limits_{z_1}^{z_2} \mathrm{d}I_y(z) &&= \int\limits_{z_1}^{z_2} z^2 \left[ y^+(z) - y^-(z)\right]\ \mathrm{d}z \\[7pt] \tag{7} I_z&= \int\limits_{z_1}^{z_2} \mathrm{d}I_z(z) &&= \dfrac{1}{3} \int\limits_{z_1}^{z_2}\left[ \left(y^+(z)\right)^3 - \left(y^-(z)\right)^3 \right]\ \mathrm{d}z \\[7pt] \tag{8} I_{yz} &= \int\limits_{z_1}^{z_2} \mathrm{d}I_{yz}(z) &&= -\dfrac{1}{2}\int\limits_{z_1}^{z_2} z \left[ \left(y^+(z)\right)^2 - \left(y^-(z)\right)^2 \right]\ \mathrm{d}z \end{alignat} $$

In comparison to formula (6.5), this calculation method essentially simplifies equation (15) for $I_y$, as for the horizontal strip, the following holds:

$$ \begin{align} \tag{9} y^+(z) - y^-(z) &= b(z) \end{align} $$

Therefore, the result obtained is

$$ \begin{align} \tag{1} I_y&= \int\limits_{(A)} z^2 \ \mathrm{d}A = \int\limits_{z_1}^{z_2} z^2 \cdot b(z)\ \mathrm{d}z \\[7pt] \tag{2} I_z&= \int\limits_{(A)} y^2 \ \mathrm{d}A = \dfrac{1}{3} \int\limits_{z_1}^{z_2}\left[ \left(y^+(z)\right)^3 - \left(y^-(z)\right)^3 \right]\ \mathrm{d}z \\[7pt] \tag{3}I_{yz} &= -\int\limits_{(A)} y \cdot z \ \mathrm{d}A = -\dfrac{1}{2}\int\limits_{z_1}^{z_2} z \left[ \left(y^+(z)\right)^2 - \left(y^-(z)\right)^2 \right]\ \mathrm{d}z \qquad \end{align} $$

(6.7)

Example 6.3: Determining Area Moment of Inertia using the Calculation Method for Arbitrary Shapes Horizontal Infinitesimal Area Strip in Cartesian Coordinates

For the quarter-circle area depicted, the following area moments of inertia with respect to the illustrated $y$, $z$-coordinate system are to be determined using the calculation method horizontal infinitesimal area strip in Cartesian coordinates:

A horizontal strip of area dA(z) with side lengths dz and y(z) is exemplified within a quarter-circle region. Yellow coordinate system, y being positive to the left, and z being positive downward. The position of the area element is dimensioned from the origin in terms of z. — Fig. B6.3.1: Quarter-Circle Area

Axial Area Moment of Inertia $I_y$
Axial Area Moment of Inertia $I_z$
Biaxial Area Moment of Inertia $I_{yz}$

Solution

In order to apply the required calculation method, it is necessary to determine the functional relationship $b(z) = y(z)$ and thus also the upper and lower bounding functions $y^+(z)$ and $y^-(z)$.

Fig. B6.3.2: Functional Relationship, Circular Equation

As shown in Fig. B6.3.2, the lower bounding function is

$$ \begin{align} \tag{B6.3-1} y^-(z) &= 0 \end{align} $$

, which corresponds to the $z$-axis.

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2.1.4 Summary for Practical Application

When calculating area moments of inertia for arbitrary surfaces, we need to do so using equations (6.2) - (6.4) via the area differential $\mathrm{d}A$. This means that we typically need to perform double integration, as we need to determine the infinitesimal area element $\mathrm{d}A$ for each component in the $y$- and $z$-direction.

This applies to all formulas after (6.5), (6.6), and (6.7). Simplifications occur when calculating $I_y$ according to Formula (6.7) and $I_z$ according to Formula (6.6), as the initial integrations based on equations (4) and (12) result in $h(y)$ and $b(z)$. Thus, in these cases, we only need to deal with one integration.

The prerequisite for the applicability of these formulas is that the functions $b(z)$ and $h(y)$ can be established. If this is the case, the limiting functions $y^+(z)$, $y^-(z)$, $z^+(y)$, and $z^-(y)$ can also be determined. Therefore, for the calculation of the biaxial area moment of inertia $I_{yz}$, one can choose to proceed either according to Formula (6.6) or (6.7).

Therefore, it is advisable to use the following formulas when calculating sarea moments of inertia in Cartesian coordinates, based on the calculation methods of horizontal and vertical infinitesimal strips of area:

$$ \begin{align} \tag{1} I_y&= \int\limits_{(A)} z^2 \ \mathrm{d}A = \int\limits_{z_1}^{z_2} z^2 \cdot b(z)\ \mathrm{d}z \\[7pt] \tag{2} I_z&= \int\limits_{(A)} y^2 \ \mathrm{d}A = \int\limits_{y_1}^{y_2} y^2 \cdot h(y)\ \mathrm{d}y \\[7pt] \tag{3a} I_{yz} &= -\int\limits_{(A)} y \cdot z \ \mathrm{d}A = -\dfrac{1}{2}\int\limits_{z_1}^{z_2} z \left[ \left(y^+(z)\right)^2 - \left(y^-(z)\right)^2 \right]\ \mathrm{d}z \\[7pt] \tag{3b} I_{yz} &= -\int\limits_{(A)} y \cdot z \ \mathrm{d}A = -\dfrac{1}{2}\int\limits_{y_1}^{y_2} y \left[ \left(z^+(y)\right)^2 - \left(z^-(y)\right)^2 \right]\ \mathrm{d}y \qquad \end{align} $$

(6.8)

If $b(z)$ and/or $h(y)$ cannot be found, alternative methods for calculation must be found according to Formula (6.5), (6.6), or (6.7). If integration is not possible here either, it is possible to switch from Cartesian coordinates to polar coordinates.

2.2 Use of Polar Coordinates

2.2.1 Infinitesimal Area Element

We are working with a tiny area element $\mathrm{dA}$. The infinitesimal area element $\mathrm{dA}$ is defined as the product of the width and height of the infinitesimal rectangle that represents this area element.

In polar coordinates, the width of this area element corresponds to the arc length of a circular segment bounded by two adjacent radii that enclose the infinitesimal angle $\mathrm{d}\varphi$.

A rectangular area element dA with side lengths dr and rdPhi is exemplified within an arbitrary cross-sectional area. Its position is dimensioned based on an arbitrarily arranged yellow x, y, z coordinate system with variables y = r cos(Phi) and z = r sin(Phi). — Fig. 6.2.4: Calculation method infinitesimal area element in polar coordinates

This width or arc length changes with the distance of the arc from the origin of the coordinate system, so it depends on $r$. Specifically, the arc length $b$ of our infinitesimal circular segment is calculated as:

$$ \begin{align} \tag{19} b &= r \mathrm{d}\varphi \end{align} $$

The height of the area element is given by the distance between two concentric circles, $\mathrm{d}r$. Thus, the area of the rectangle is given by

$$ \begin{align} \tag{20} \mathrm{d}A &= r \mathrm{d}\varphi \mathrm{d}r \end{align} $$

Since this area element is infinitesimal in both coordinate directions, we need a double integral to perform the calculation. If we replace the coordinates $y$ and $z$ of this infinitesimal area element with

$$ \begin{align} \tag{21} y &= r \cos(\varphi) \\[7pt] \tag{22} z &= r \sin(\varphi) \end{align} $$

we obtain

$$ \begin{align} I_y&= \int\limits_{(A)} z^2 \ \mathrm{d}A = \int\limits_{(\varphi)} \int\limits_{(r(\varphi))} \left(r \sin(\varphi)\right)^2 r \ \mathrm{d}\varphi \mathrm{d}r = \int\limits_{(\varphi)} \int\limits_{(r(\varphi))} r^3 \sin^2(\varphi) \ \mathrm{d}\varphi \mathrm{d}r \\[7pt] I_z&= \int\limits_{(A)} y^2 \ \mathrm{d}A = \int\limits_{(\varphi)} \int\limits_{(r(\varphi))} \left(r \cos(\varphi)\right)^2 r \ \mathrm{d}\varphi \mathrm{d}r = \int\limits_{(\varphi)} \int\limits_{(r(\varphi))} r^3 \cos^2(\varphi)\ \mathrm{d}\varphi \mathrm{d}r\\[7pt] I_{yz} &= -\int\limits_{(A)} y \cdot z \ \mathrm{d}A = -\int\limits_{(\varphi)} \int\limits_{(r(\varphi))} r^3 \cos(\varphi)\sin(\varphi) \ \mathrm{d}\varphi \mathrm{d}r \end{align} $$

(6.9)

To apply this formula, it is necessary to determine the functional relationship $r(\varphi)$. If this is not possible, an alternative option is to use the integration limits $r$ and $\varphi(r)$.

Example 6.4: Determining Area Moment of Inertia using the Calculation Method for Arbitrary Shapes Infinitesimal Area Element in Polar Coordinates

For the quarter-circle area depicted, the following area moments of inertia with respect to the illustrated $y$, $z$-coordinate system are to be determined using the calculation method infinitesimal area element in polar coordinates:

A rectangular area element dA with side lengths dr and rdPhi is exemplified within a quarter-circle region. Yellow coordinate system, y being positive to the left, and z being positive downward. The position of the area element is dimensioned from the origin in terms of y and z. — Fig. B6.4.1: Quarter-Circle Area

Axial Area Moment of Inertia $I_y$
Axial Area Moment of Inertia $I_z$
Biaxial Area Moment of Inertia $I_{yz}$

Solution

To apply the required calculation method, it is necessary to determine the functional relationship $r(\varphi)$.

We find this by illustrating the relationship between any angle $\varphi$ and the corresponding radius $r(\varphi)$ of the (quarter) circle: The radius is constant. This property of the circle simplifies the calculation using this method. Therefore, it is preferable to Cartesian coordinate methods when calculating circular areas.

a) Axial Area Moment of Inertia $I_y$