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Mechanics of Materials

Topic: Area Moment of Inertia

Here, you can learn what an area moment of inertia (also known as second moment of area, or second-order area moment) is, where it finds application in engineering mechanics, and how to calculate it.

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2.1.2 Vertical Infinitesimal Area Strip

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

Vertical strip of area within an arbitrary surface. Coordinates arranged in a freely positioned x, y, z coordinate system, with y being positive to the left and z being positive downward. The strip is located at position y, with height h(y), and has a width of dy.
Fig. 6.2.2: Calculation Method Vertical Infinitesimal Area Strip in Cartesian Coordinates

The height \(h(y)\) of the strip depends on the \(y\)-coordinate. Therefore, the calculation for the area element \(\mathrm{d}A\) in the formulas (6.2), (6.3) and (6.4) is given by

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

In order to perform the calculation of area moments of inertia using this method, it is necessary to determine the functional relationship \(h(y)\). This allows us to obtain the upper and lower bounding functions, which, for each \(y\), provide the upper \(\left(z^+(y)\right)\) and lower \(\left(z^-(y)\right)\) bounds of the area element.

Thus, for any arbitrary \(y\):

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

For any arbitrary \(y\), we can determine the infinitesimal area moments of inertia using the formulas (6.2) and (6.4) by utilizing equations (1) and (2) while keeping the independent variable \(y\) constant:

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

We obtain the sought area moments of inertia for the entire area by summing, or integrating, the infinitesimal quantities \(\mathrm{d}I_y(y)\), \(\mathrm{d}I_z(y)\), and \(\mathrm{d}I_{yz}(y)\) over the independent variable \(y\):

$$ \begin{alignat}{3} \tag{6} I_y&= \int\limits_{y_1}^{y_2} \mathrm{d}I_y(y) &&= \dfrac{1}{3} \int\limits_{y_1}^{y_2}\left[ \left(z^+(y)\right)^3 - \left(z^-(y)\right)^3 \right]\ \mathrm{d}y \\[7pt] \tag{7} I_z&= \int\limits_{y_1}^{y_2} \mathrm{d}I_z(y) &&= \int\limits_{y_1}^{y_2} y^2 \left[ z^+(y) - z^-(y)\right]\ \mathrm{d}y \\[7pt] \tag{8} I_{yz} &= \int\limits_{y_1}^{y_2} \mathrm{d}I_{yz}(y) &&= -\dfrac{1}{2}\int\limits_{y_1}^{y_2} y \left[ \left(z^+(y)\right)^2 - \left(z^-(y)\right)^2 \right]\ \mathrm{d}y \end{alignat} $$

Compared to Formula (6.5), this calculation method essentially simplifies only Equation (7) for \(I_z\), since for the vertical strip, the following holds:

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

Thus, our result is

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

(6.6)

Example 6.2: Determining Area Moment of Inertia using the Calculation Method for Arbitrary Shapes Vertical 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 vertical infinitesimal area strip in Cartesian coordinates:

A vertical strip of area dA(y) with side lengths dy and z(y) is exemplified within a quarter-circle region. Yellow coordinate system, with 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.
Fig. B6.2.1: Quarter-Circle Area
  1. Axial Area Moment of Inertia \(I_y\)
  2. Axial Area Moment of Inertia \(I_z\)
  3. Biaxial Area Moment of Inertia \(I_{yz}\)
Solution

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

Figure
Fig. B6.2.2: Functional Relationship, Circular Equation

As shown in Fig. B6.2.2, the lower bounding function is

$$ \begin{align} \tag{B6.2-1} z^-(y) &= 0 \end{align} $$

, which corresponds to the \(y\)-axis.

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