Exercise F-6.2.2 TechMech 2 - Mechanics of Materials

Practice Exercise F-6.2.2

Area Moment of Inertia: Determine Area Moment of Inertia for a Coordinate System Parallel to the Principal Axis System

Problem Statement

The depicted rectangular cross-section has an area of $A=72~\mathrm{cm}^2$.

The illustration shows a rectangular cross-sectional area with a width of b and a height of h. A yellow y, z-coordinate system has its origin at the centroid of the area. A green y-bar-subscript-1 axis is located at a-subscript-1 distance above the yellow y-axis. A red y-bar-subscript-2 axis is located at a-subscript-2 distance below the yellow y-axis. The y-axes are positive to the left, and the z-axis is positive downward. — Fig. 1: Rectangular Surface

The axial area moment of inertia with respect to the $\overline{y}_1$-axis ($a_1 = 5~\mathrm{cm}$) is known and is $I_{\overline{y}_1}=2664~\mathrm{cm}^4$.

Calculate the axial area moment of inertia $I_{\overline{y}_2}$ with respect to the $\overline{y}_2$-axis ($a_2 = 2~\mathrm{cm}$).

Short Solution

$$ \begin{aligned} I_{\overline{y}_2} &= 1152~\mathrm{cm}^4 \end{aligned} $$

Comprehensive Solution

The reference axis of the sought area moment of inertia, $\overline{y}_2$, is parallel to the reference axis of the given area moment of inertia, $\overline{y}_1$.

However, a direct conversion of the given area moment of inertia $I_{\overline{y}_1}$ to the $\overline{y}_2$ axis using the Steiner's theorem according to Formula (6.10) is not permissible, as a fundamental prerequisite for the application of this formula is missing: Neither of the two axes, $\overline{y}_1$ and $\overline{y}_2$, is the centroid axis of the given rectangular area.

When the area moment of inertia, as in this case, needs to be converted from any axis to another parallel axis, i.e., both axes are not centroid axes, then two steps are necessary:

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