The assembly in Fig. (a) consists of two identical, thin plates, each of mass M and thickness t. For point O, determine (1) the inertia tensor with respect to the axes shown; and (2) the principal moments of inertia and the principal axes.

Question

Accepted Answer

Part 1

The inertia tensor at point O for the assembly is found by summing the moments and products of inertia of the two plates about the xyz-axes. We use the thin-plate approximations described in Arts. F.3 and F.6
Plate 1 Referring to Fig. (a) and the table on page 652, the moments and products of inertia for the area of Plate 1 are

[latex](I_x)_{area} = \frac{a(2a)^3}{12} = \frac{2a^4}{3}[/latex]

[latex](I_y)_{area} = \frac{2a(a)^3}{3} = \frac{2a^4}{3}[/latex]

[latex](J_O)_{area} = (I_x)_{area} + (I_y)_{area} = \frac{2a^4}{3} +\frac{2a^4}{3} = \frac{4a^4}{3}[/latex]

[latex](I_{xy})_{area} = 0[/latex] (x-axis is an axis of symmetry)

[latex](I_{yz})_{area} = (I_{zx})_{area} = 0[/latex] (area lies in x y-plane)

Using the thin plate equations [Eqs. (F.2) and (F.7)] and M = ρ At = ρ2a²t, where A = 2a² is the plate area and ρ the mass density, the mass properties become

[latex]I_x = ρt (I_x)_{area} I_y = ρt (I_y)_{area}[/latex]
[latex]I_z = ρt (J_O)_{area}[/latex] (F.2)

[latex]I_{xy} = ρt (I_{xy})_{area} I_{yz} = ρt (I_{yz})_{area}[/latex]
[latex]I_{zx} = ρt (I_{zx})_{area}[/latex] (F.7)

[latex]I_x = ρt (I_x)_{area} = ρt\frac{2a^4}{3} = \frac{1}{3}Ma^2[/latex]
[latex]I_y = ρt (I_y)_{area} = ρt\frac{2a^4}{3} = \frac{1}{3}Ma^2[/latex]
[latex]I_z = ρt (J_O)_{area} = ρt\frac{4a^4}{3} = \frac{2}{3}Ma^2[/latex]
[latex]I_{xy} = I_{yz} = I_{zx} = 0[/latex]

Plate 2 The area moments and products of inertia for Plate 2 are

[latex](I_y)_{area} = \frac{a(2a)^3}{3} = \frac{8a^4}{3}[/latex]

[latex](I_z)_{area} = \frac{2a(a)^3}{3} = \frac{2a^4}{3}[/latex]

[latex](J_O)_{area} = (I_y)_{area} + (I_z)_{area} = \frac{8a^4}{3} +\frac{2a^4}{3} = \frac{10a^4}{3}[/latex]

[latex](I_{xy})_{area} = (I_{zx})_{area} = 0[/latex] (the area lies in the yz-plane)

[latex]I_{yz} = \frac{a^2(2a)^2}{4} = a^4[/latex]

Substituting these results and M = ρ At = ρ2a²t into Eqs. (F.2) and (F.7), the mass properties of Plate 2 become

[latex]I_x = ρt (J_O)_{area} = ρt\frac{10a^4}{3} = \frac{5}{3}Ma^2[/latex]
[latex]I_y = ρt (I_y)_{area} = ρt\frac{8a^4}{3} = \frac{4}{3}Ma^2[/latex]
[latex]I_z = ρt (I_z)_{area} = ρt\frac{2a^4}{3} = \frac{1}{3}Ma^2[/latex]
[latex]I_{xy} = I_{zx} = 0[/latex]
[latex]I_{yz} = ρt (I_{yz})_{area} = ρta^4 = \frac{1}{2}Ma^2[/latex]

Assembly The mass properties of the assembly are found by summing the properties of the two plates:

[latex]I_x = Ma^2 (\frac{1}{3} +\frac{5}{3})= 2Ma^2[/latex]
[latex]I_y = Ma^2 (\frac{1}{3} +\frac{4}{3})= \frac{5}{3}Ma^2[/latex]
[latex]I_z = Ma^2 (\frac{2}{3} +\frac{1}{3})= Ma^2[/latex]
[latex]I_{xy} = 0 + 0 = 0[/latex]
[latex]I_{yz} = 0 +\frac{1}{2}Ma^2 = \frac{1}{2}Ma^2[/latex]
[latex]I_{zx} = 0 + 0 = 0[/latex]

Substituting these values into Eq. (19.11), the inertia tensor at point O for the assembly becomes

I [latex]= \begin{bmatrix} I_{x} & -I_{xy} & -I_{xz} \ -I_{yx} & -I_{y}& -I_{yz}\ -I_{zx} & --I_{zy}& -I_{z} \end{bmatrix} [/latex] (19.11)

I [latex]= \begin{bmatrix} 2 & 0 & 0 \ 0 & \frac{5}{3}& -\frac{1}{2}\ 0 & -\frac{1}{2}& 1 \end{bmatrix} Ma^2[/latex]

Part 2

Substituting the inertial properties of the assembly obtained in Part 1 into Eqs. (F.11) yields the following equations that must be solved for the principal moments of inertia and the direction cosines of the principal axes:

[latex](I_x − I)λ_x           −I_{xy}λ_y           −I_{xz}λ_z           = 0[/latex]
[latex]−I_{xy}λ_x           (I_y − I)λ_y           −I_{yz}λ_z           = 0[/latex]           (F.11)
[latex]−I_{zx}λ_x           −I_{zy}λ_y           (I_z − I)λ_z           = 0[/latex]
[latex]λ_x^2 + λ_y^2 + λ_z^2 = 1[/latex]

[latex](2Ma^2 − I)λ_x + 0λ_y    + 0λ_z = 0[/latex]           (a)
[latex]0λ_x +(\frac{5}{3}Ma^2 − I)λ_y    − \frac{Ma^2}{2}λ_z = 0[/latex]           (b)
[latex]0λ_x    − \frac{Ma^2}{2}λ_y + (Ma^2 − I)λ_z = 0[/latex]           (c)
[latex]λ_x^2 + λ_y^2 + λ_z^2 = 1[/latex]           (d)

The characteristic equation is obtained by setting the determinant of the coefficients in Eqs. (a) through (c) equal to zero:

[latex]\begin{vmatrix} 2Ma^2 − I & 0 & 0 \ 0 & \frac{5}{3}Ma^2 − I& -\frac{Ma^2}{2}\ 0 & -\frac{Ma^2}{2}& Ma^2 − I \end{vmatrix} = 0[/latex] (e)

Expanding the determinant and simplifying, we get

[latex](2Ma^2 − I) (I^2 − \frac{8}{3}Ma^2 I + \frac{17}{12}Ma^2 a^4)[/latex] (f)

Solving Eq. (f) and labeling the roots as [latex]I_1[/latex], [latex]I_2[/latex], and [latex]I_3[/latex] we obtain

[latex]I_1[/latex] = 0.7324Ma² [latex]I_2[/latex] = 1.9343Ma² [latex]I_3[/latex] = 2.0000Ma²

as the principal moments of inertia (eigenvalues).
The direction cosines of a principal axis can be found by substituting the corresponding eigenvalue into Eqs. (a) through (d) and then solving these equations for the direction cosines [latex]λ_x[/latex] , [latex]λ_y[/latex] , and [latex]λ_z[/latex]. Substituting [latex]I = I_3 = 2.0Ma^2[/latex], Eqs. (a) through (c) become

0 = 0
[latex]−\frac{1}{3}λ_y − \frac{1}{2}λ_z = 0[/latex]
[latex]−\frac{1}{2}λ_y − λ_z = 0[/latex]

Because the second and third equations are linearly independent, their solution is [latex]λ_y[/latex] = [latex]λ_z[/latex] = 0. Combining these results with Eq. (d), [latex]λ_x^2 +λ_y^2+λ_z^2 = 1[/latex], we conclude that [latex]λ_x[/latex] = ±1. Hence the unit vector in the direction of the axis associated with [latex]I_3[/latex]—that is, the third eigenvector—is

λ[latex]_3[/latex] = ±i

The plus-or-minus sign in this answer reflects the fact that an eigenvector defines only the direction of the principal axis; the sense of the unit vector along that axis is arbitrary. In other words, if λ is an eigenvector, then −λ is also an eigenvector.

With [latex]I = I_2 = 1.9343Ma^2[/latex], Eqs. (a) through (c) are

[latex]0.0657λ_x = 0[/latex]
[latex]−0.2676λ_y − 0.5000λ_z = 0[/latex]
[latex]−0.5000λ_y − 0.9343λ_z = 0[/latex]

From the first equation we get [latex]λ_x = 0[/latex]. The second and third equations are not independent, because both yield [latex]λ_z = −0.5352λ_y[/latex] . Substituting these results into Eq. (d), we obtain

[latex]0^2 + λ_y^2 + (−0.5352λ_y)^2 = 1[/latex]

which yields[latex]λ_y[/latex] = ±0.8817. Consequently, [latex]λ_z[/latex] = −0.5352(±0.8817) = ∓0.4719, from which the second eigenvector is

λ[latex]_2[/latex] = ±0.882j ∓ 0.472k

The direction of the first eigenvalue, corresponding to [latex]I = I_1[/latex] = 0.7324 Ma², can be obtained in an analogous manner, with the result being

λ[latex]_1[/latex] = ±0.472j ± 0.882k

The three principal axes are shown in Fig. (b). Note that these axes are mutually orthogonal, as they should be, because the principal moments of inertia are distinct. Furthermore, as shown in Fig. (b), the signs of the eigenvectors are usually chosen so that λ[latex]_1[/latex], λ[latex]_2[/latex], and λ[latex]_3[/latex] form a right-handed triad.

Question F.7: The assembly in Fig. (a) consists of two identical, thin pla...

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