Determine the moment of inertia Ix of the area shown with respect to the centroidal x axis

Question

Determine the moment of inertia [latex]\bar{I}_{x}[/latex] of the area shown with respect to the centroidal [latex]x[/latex] axis (Fig. A.21).

Accepted Answer

Location of Centroid. The centroid [latex]C[/latex] of the area must first be located. However, this has already been done in Example A.02 for the given area. We recall from that example that [latex]C[/latex] is located [latex]46 \mathrm{~mm}[/latex] above the lower edge of the area [latex]A[/latex].

Computation of Moment of Inertia. We divide the area [latex]A[/latex] into the two rectangular areas [latex]A_{1}[/latex] and [latex]A_{2}[/latex] (Fig. A.22), and compute the moment of inertia of each area with respect to the [latex]x[/latex] axis.

Rectangular Area [latex]A_{1}[/latex]. To obtain the moment of inertia [latex]\left(I_{x} ight)_{1}[/latex] of [latex]A_{1}[/latex] with respect to the [latex]x[/latex] axis, we first compute the moment of inertia of [latex]A_{1}[/latex] with respect to its own centroidal axis [latex]x^{\prime}[/latex]. Recalling the formula derived in part [latex]a[/latex] of Example A.04 for the centroidal moment of inertia of a rectangular area, we have

[latex]\left(\bar{I}_{x^{\prime}} ight)_{1}=\frac{1}{12} b h^{3}=\frac{1}{12}(80 \mathrm{~mm})(20 \mathrm{~mm})^{3}=53.3 imes 10^{3} \mathrm{~mm}^{4}[/latex]

Using the parallel-axis theorem, we transfer the moment of inertia of [latex]A_{1}[/latex] from its centroidal axis [latex]x^{\prime}[/latex] to the parallel axis [latex]x[/latex] :

[latex] \begin{aligned} \left(I_{x} ight)_{1}=\left(\bar{I}_{x^{\prime}} ight)_{1}+A_{1} d_{1}^{2}= & 53.3 imes 10^{3}+(80 imes 20)(24)^{2} \ = 975 × 10³ ext{ mm}^4 \end{aligned} [/latex]

Rectangular Area [latex]A_{2}[/latex]. Computing the moment of inertia of [latex]A_{2}[/latex] with respect to its centroidal axis [latex]x^{\prime \prime}[/latex], and using the parallel-axis theorem to transfer it to the [latex]x[/latex] axis, we have

[latex] \begin{aligned} \left(\bar{I}_{x^{\prime \prime}} ight)_{2} & =\frac{1}{12} b h^{3}=\frac{1}{12}(40)(60)^{3}=720 imes 10^{3} \mathrm{~mm}^{4} \ \left(I_{x} ight)_{2} & =\left(\bar{X}_{x^{\prime \prime}} ight)_{2}+A_{2} d_{2}^{2}=720 imes 10^{3}+(40 imes 60)(16)^{2} \ & =1334 imes 10^{3} \mathrm{~mm}^{4} \end{aligned} [/latex]

Entire Area A. Adding the values computed for the moments of inertia of [latex]A_{1}[/latex] and [latex]A_{2}[/latex] with respect to the [latex]x[/latex] axis, we obtain the moment of inertia [latex]\bar{I}_{x}[/latex] of the entire area:

[latex] \begin{aligned} & \bar{I}_{x}=\left(I_{x} ight)_{1}+\left(I_{x} ight)_{2}=975 imes 10^{3}+1334 imes 10^{3} \ & \bar{I}_{x}=2.31 imes 10^{6} \mathrm{~mm}^{4} \end{aligned} [/latex]

Question A.06: Determine the moment of inertia Ix of the area shown with re...

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