Question 9.SP.6: Determine the product of inertia of the right triangle shown......
Determine the product of inertia of the right triangle shown (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.
STRATEGY: You can approach this problem by using a vertical differential strip element. Because each point of the strip is at a different distance from the x axis, it is necessary to describe this strip mathematically using the parallel-axis theorem. Once you have completed the solution for the product of inertia with respect to the x and y axes, a second application of the parallel-axis theorem yields the product of inertia with respect to the centroidal axes.
MODELING and ANALYSIS:
a. Product of Inertia I_{x y}. Choose a vertical rectangular strip as the differential element of area (Fig. 1). Using a differential version of the parallel-axis theorem, you have
d I_{x y}=d I_{x^{\prime} y^{\prime}}+\bar{x}_{e l} \bar{y}_{e l} d A
The element is symmetrical with respect to the x^{\prime} \text { and } y^{\prime} \text { axes, so } d I_{x^{\prime} y^{\prime}}=0. From the geometry of the triangle, you can express the variables in terms of x and y.
y=h\left(1-\frac{x}{b}\right) \quad d A=y d x=h\left(1-\frac{x}{b}\right) d x
\bar{x}_{e l}=x \quad \bar{y}_{e l}=\frac{1}{2} y=\frac{1}{2} h\left(1-\frac{x}{b}\right)
Integrating d I_{x y} \text { from } x=0 \text { to } x=b \text { gives you } I_{x y}:
I_{x y}=\int d I_{x y}=\int \bar{x}_{e l} \bar{y}_{e l} d A=\int_0^b x\left(\frac{1}{2}\right) h^2\left(1-\frac{x}{b}\right)^2 d x
=h^2 \int_0^b\left(\frac{x}{2}-\frac{x^2}{b}+\frac{x^3}{2 b^2}\right) d x=h^2\left[\frac{x^2}{4}-\frac{x^3}{3 b}+\frac{x^4}{8 b^2}\right]_0^b
I_{x y}=\frac{1}{24} b^2 h^2
b. Product of Inertia \bar{I}_{x^{\prime \prime} y^{\prime \prime}}. The coordinates of the centroid of the triangle relative to the x and y axes are (Fig. 2 and Fig. 5.8A)
\bar{x}=\frac{1}{3} b \quad \bar{y}=\frac{1}{3} h
Using the expression for I_{x y} obtained in part a, apply the parallel-axis theorem again:
\begin{aligned}I_{x y} & =\bar{I}_{x^{\prime \prime} y^{\prime \prime}}+\bar{x} \bar{y} A \\\frac{1}{24} b^2 h^2 & =\bar{I}_{x^{\prime \prime} y^{\prime \prime}}+\left(\frac{1}{3} b\right)\left(\frac{1}{3} h\right)\left(\frac{1}{2} b h\right) \\\bar{I}_{x^{\prime \prime} y^{\prime \prime}} & =\frac{1}{24} b^2 h^2-\frac{1}{18} b^2 h^2\end{aligned}
\bar{I}_{x^{\prime \prime} y^{\prime \prime}}=-\frac{1}{72} b^2 h^2
REFLECT and THINK: An equally effective alternative strategy would be to use a horizontal strip element. Again, you would need to use the parallel-axis theorem to describe this strip, since each point in the strip would be a different distance from the y axis.
