For the rectangle, compute the following: (1) the moment of inertia about the x-axis by integration; (2) the moment of inertia about the centroidal axis that is parallel to the x-axis; and (3) the polar moment of inertia about the centroid.

Question

For the rectangle, compute the following: (1) the moment of inertia about the [latex]x[/latex]-axis by integration; (2) the moment of inertia about the centroidal axis that is parallel to the [latex]x[/latex]-axis; and (3) the polar moment of inertia about the centroid.

Accepted Answer

Part 1

The area of the differential element shown in the figure is dA=b dy. Because all parts of the element are a distance y from the x-axis, we can use Eq. (9.2):

[latex]I_{x}=\int_{\mathscr{A}}{y^2\:dA}=b\int_{0}^{h}{y^2\:dy}=\frac{by^3}{3}\big]_{0}^{h}=\frac{bh^3}{3}[/latex]

This result agrees with the information listed for a rectangle in Table 9.1. If we had chosen to use double integration with dA=dx dy, the analysis would yield

[latex]\bar{I}_{x}=\frac{bh^3}{12}[/latex] [latex]\bar{I}_{y}=\frac{b^3h}{12}[/latex] [latex]\bar{I}_{xy}=0[/latex]

[latex]I _{x}=\frac{bh^3}{3}[/latex] [latex]I_{y}=\frac{b^3h}{3}[/latex] [latex]I_{xy}=\frac{b^2h^2}{4}[/latex]

[latex]I_{x}=I_{y}=\frac{πR^4}{4}[/latex] [latex]I_{xy}=0[/latex]

[latex]\bar{I}_{x}=\frac{bh^3}{36}[/latex] [latex]\bar{I}_{y}=\frac{b^3h}{36}[/latex] [latex]\bar{I}_{xy}=\frac{b^2h^2}{72}[/latex]

[latex]I _{x}=\frac{bh^3}{12}[/latex] [latex]I_{y}=\frac{b^3h}{12}[/latex] [latex]I_{xy}=\frac{b^2h^2}{24}[/latex]

[latex]\bar{I}_{x}=0.1098R^4[/latex] [latex]\bar{I}_{xy}=0[/latex]

[latex]I_{x}=I_{y}=\frac{πR^4}{8}[/latex] [latex]I_{xy}=0[/latex]

[latex]\bar{I}_{x}=\frac{bh^3}{36}[/latex] [latex]\bar{I}_{y}=\frac{b^3h}{48}[/latex] [latex]\bar{I}_{xy}=0[/latex]

[latex]I _{x}=\frac{bh^3}{12}[/latex] [latex]I_{xy}=0[/latex]

Table 9.1 Inertial Properties of Plane Areas: Part 1

[latex]I_{x}=\int_{\mathscr{A}}{y^2\:dA}=\int_{0}^{h}\int_{0}^{b}{y^2\:dx\:dy}=\frac{bh^3}{3}[/latex]

which is identical to the previous result.

Part 2

We can calculate [latex]\bar{I}_{x}[/latex] from the parallel-axis theorem and the result of Part 1. Substituting [latex]I_{x}=bh^3/3[/latex] into the parallel-axis theorem, and recognizing that the transfer distance [latex]d[/latex] (the distance between the [latex]x[/latex]-axis and the centroidal [latex]x[/latex]-axis) is [latex]h/2[/latex], we find that

[latex]\bar{I}_{x}=I_{x}−Ad^2=\frac{bh^3}{3}−bh\left(\frac{h}{2}\right)^2=\frac{bh^3}{12}[/latex]

This answer also agrees with the results in Table 9.1.

Part 3

One method of computing [latex]\bar{J}_{C}[/latex] is to use [latex]\bar{J}_{C}=\bar{I}_{x}+\bar{I}_{y}[/latex]. From the results of Part 2, or Table 9.1, we have

[latex]\bar{J}_{C}=\bar{I}_{x}+\bar{I}_{y}=\frac{bh^3}{12}+\frac{hb^3}{12}=\frac{bh}{12}(h^2+b^2)[/latex]

Another method of computing [latex]\bar{J}_{C}[/latex] is to first compute [latex]J_{O}=I_{x}+I_{y}[/latex] and then transfer this result to the centroid. From the results of Part 1, we have

[latex]J_{O}=I_{x}+I_{y}=\frac{bh^3}{3}+\frac{hb^3}{3}=\frac{bh}{3}(h^2+b^2)[/latex]

The transfer distance is the distance between point [latex]O[/latex] and the centroid of the rectangle; that is, [latex]d=\sqrt{(b/2)^2+(h/2)^2}[/latex]. From the parallel-axis theorem, we obtain

[latex]\bar{J}_{C}=J_{O}−Ad^2 =\frac{bh}{3}(h^{2}+b^{2})−bh\left(\frac{b^2}{4}+\frac{h^2}{4}\right)[/latex]

or

[latex]\bar{J}_{C}=\frac{bh}{12}(h^2+b^2)[/latex]

which agrees with the previous result.

Question 9.2: For the rectangle, compute the following: (1) the moment of ...

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