Understanding the Moment of Inertia of a Triangle

The moment of inertia of a triangle is an important concept in mechanics, especially in analyzing the rotational properties of triangular laminae. It plays a significant role in determining how the mass of the triangle is distributed with respect to an axis and influences calculations in engineering and physics problems.

Definition of Moment of Inertia for a Triangle

The moment of inertia of a triangle refers to its second moment of area about a given axis. For a triangle with base $b$ and height $h$, the value varies depending on the chosen axis. The axis may pass through the base, the centroid, or be perpendicular to the plane.

Moment of Inertia about the Base of the Triangle

For a triangle of base $b$ and height $h$, the moment of inertia about its base (the axis along the base of the triangle) is given by the formula:

$I_{\text{base}} = \dfrac{b h^3}{12}$

This expression is derived by integrating over the area of the triangle. It is important in structural analysis and mechanical engineering applications, as triangles are common in truss and framework designs.

For further understanding of this property, refer to the Moment Of Inertia page.

Moment of Inertia about the Centroidal Axis (Parallel to Base)

The centroidal axis of a triangle is the axis passing through its centroid and parallel to the base. The moment of inertia about this axis can be found using the parallel axis theorem, since the centroid is located at a distance of $\dfrac{h}{3}$ from the base.

The moment of inertia about the centroidal axis parallel to the base is:

$I_{\text{centroid}} = \dfrac{b h^3}{36}$

This formula is essential for calculating stresses and deformations in beams and other structures made of triangular elements.

Derivation: Moment of Inertia of a Triangle about its Base

To derive the moment of inertia of a triangle about its base, consider a horizontal strip at a distance $y$ from the base with thickness $\mathrm{d}y$. The width of the strip is $x = \dfrac{b}{h}(h - y)$ by the property of similar triangles.

The elemental area is $\mathrm{d}A = x \, \mathrm{d}y$. Its moment of inertia about the base is $y^2 \mathrm{d}A$. Integrating over the height gives:

$ I_{\text{base}} = \int_0^h y^2 \left[ \dfrac{b}{h}(h-y) \right] \mathrm{d}y $

Evaluating the integral yields the result $I_{\text{base}} = \dfrac{b h^3}{12}$.

For the moment of inertia of shapes such as discs and spheres, visit Moment Of Inertia Of A Disc and Moment Of Inertia Of A Hollow Sphere.

Parallel Axis Theorem Application to a Triangle

The parallel axis theorem allows calculation of the moment of inertia about any axis parallel to the centroidal axis. It states:

$I = I_{\text{centroid}} + A d^2$

Here, $A$ is the area of the triangle and $d$ is the distance between the centroidal axis and the new axis. For a triangle, $A = \dfrac{1}{2} b h$ and $d = \dfrac{h}{3}$ between centroidal axis and base.

This theorem is fundamental for engineering applications and analysis involving complex or composite bodies.

Moment of Inertia about the Vertex and Perpendicular Axes

The moment of inertia of a triangle about an axis passing through its vertex and parallel to the base can be derived by shifting the axis using the parallel axis theorem. For axes perpendicular to the base or in the plane, the calculation involves considering subdivision of the triangle into right triangles and summing their moments of inertia.

Refer to Moment Of Inertia Of A Circle for comparative analysis with other geometric figures.

Table: Common Moments of Inertia for a Triangle

Solved Example: Calculating Moment of Inertia

Consider a triangle where $b = 6$ m and $h = 4$ m. The moment of inertia about the base is:

$ I_{\text{base}} = \dfrac{6 \times (4)^3}{12} = \dfrac{6 \times 64}{12} = \dfrac{384}{12} = 32 \, \text{m}^4 $

The moment of inertia about the centroidal axis is:

$ I_{\text{centroid}} = \dfrac{6 \times (4)^3}{36} = \dfrac{384}{36} = 10.67 \, \text{m}^4 $

Application to Equilateral and Laminar Triangles

For an equilateral triangle or a lamina of uniform mass, the same formula applies with the appropriate values for base and height. The center of mass is considered for axis placement. These principles extend to composite bodies comprising triangular sections.

For advanced applications on other geometries, see Moment Of Inertia Of A Cube and Moment Of Inertia Of An Ellipse.

Summary of Key Formulae

• About the base: $I = \dfrac{b h^3}{12}$
• About centroidal axis: $I = \dfrac{b h^3}{36}$
• Area of triangle: $A = \dfrac{1}{2} b h$
• Distance from base to centroid: $d = \dfrac{h}{3}$