Exercise F-6.1.3 TechMech 2 - Mechanics of Materials

Practice Exercise F-6.1.3

Second Moment of Inertia: Determine the Axial and Biaxial Area Moment of Inertia

Problem Statement

Triangular area for which the moment of inertia (axial, biaxial) is to be calculated. Right-angled triangle, right angle in the top-right corner. Base: b, Height: h, Coordinate system y, z with center at the centroid (b/3 from the right side, h/3 from the top), y positive to the left, z positive downward. — Fig. 1: Triangle Area with given Principal Axes

For the triangle area depicted, the following area moments of inertia are to be determined with respect to the illustrated $y$, $z$-coordinate system:

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

Short Solution

a) Determine the Axial Area Moment of Inertia $I_y$

$$ \begin{aligned} I_y&= \dfrac{b \cdot h^3}{36} \end{aligned} $$

b) Determine the Axial Area Moment of Inertia $I_z$

$$ \begin{aligned} I_z &= \dfrac{h \cdot b^3}{36} \end{aligned} $$

c) Determine the Biaxial Area Moment of Inertia $I_{yz}$

$$ \begin{aligned} I_{yz} &= \dfrac{b^2 \cdot h^2}{72} \end{aligned} $$

Comprehensive Solution

One approach to determine the area moments of inertia is to perform the calculation in Cartesian coordinates.

To minimize computational effort, we can avoid two double integrals. Therefore, it is advisable to use the following formulas in the calculation, based on the methods of horizontal and vertical infinitesimal area strips:

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