The assembly shown is formed by joining two pieces of 1.5 mm steel plate. Determine its products of inertia with respect to the axes shown. The weight density of steel is γ = 7900 kg/m³.

Question

Accepted Answer

In this problem, we are justified in using the thin plate approximation of Eqs. (F.2) or (F.3), because the thickness of each component plate is much smaller than its in-plane dimensions.

[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_x = \frac{m}{A} (I_x)_{area} I_y = \frac{m}{A} (I_y)_{area}[/latex]
[latex]I_z = \frac{m}{A} (J_O)_{area}[/latex] (F.3)

Triangular Plate

Because the triangular area lies in the x y-plane, [latex](I_{yz})_{area} = (I_{zx})_{area} = 0[/latex]. From the table on page 652, we obtain

[latex](I_{xy})_{area} = \frac{(0.3)^2(0.23)^2}{24} = 1.98 × 10^{−4} m^4[/latex]

Using Eqs. (F.2), the mass products of inertia of the triangular plate become

[latex]I_{xy} = ρt (I_{xy})_{area} = 7900[0.0015] (1.98 × 10^{−4})= 2.346 × 10^{−3}[/latex] kg · m²
[latex]I_{yz} = I_{zx} = 0[/latex] (a)

Semicircular Plate

Noting that the semicircular area lies in the yz-plane, we have [latex](I_{xy})_{area} = (I_{zx})_{area} = 0[/latex]. We use the parallel-axis theorem to compute [latex](I_{yz})_{area}[/latex]. Referring to the table on page 652, the centroid of the semicircular area is located at [latex]\bar x[/latex] = 0, [latex]\bar y[/latex] = 0.15 m, [latex]\bar z[/latex] = 4(0.15)/3π m. Also observe that [latex](\bar I_{yz})_{area}[/latex] = 0, because the centroidal z-axis is an axis of symmetry. The parallel-axis theorem thus yields

[latex](I_{yz})_{area} = (\bar I_{yz})_{area} + A\bar y\bar z[/latex]
[latex]= 0 +\frac{π}{2} (0.15)^2(0.15)\frac{4}{3π}(0.15) = 3.375 × 10^{−4} m^4[/latex]

The mass products of inertia of the semicircular plate can now be computed from Eqs. (F.2)—remembering to convert inches to feet and weight density to mass density:

[latex]I_{xy} = I_{zx} = 0[/latex] (b)
[latex]I_{yz} = ρt (I_{yz})_{area} = 7900[0.0015] (3.375 × 10^{−4})= 4 × 10^{−3}[/latex] kg · m²

Assembly

The mass products of inertia of the assembly are found by adding the results for the triangular and semicircular plates in Eqs. (a) and (b), which gives

[latex]I_{xy} = 23.46 × 10^{−4}[/latex] kg · m²
[latex]I_{yz} = 40 × 10^{−4}[/latex] kg · m²
[latex]I_{zx}[/latex] = 0

Question F.5: The assembly shown is formed by joining two pieces of 1.5 mm...

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