Question 12.5: Determine the moment of inertia Ic with respect to the horiz......

We will determine the moment of inertia I_c with respect to axis C-C by applying the parallel-axis theorem to each individual part of the composite area. The area divides naturally into three parts: (1) the cover plate. (2) the wide-flange section, and (3) the channel section. The following areas and centroidal distances were obtained previously in Example 12-2:

A_1 = 3.0 in.²        A_2 = 20.8 in.²        A_3 = 8.82 in.²

\bar{y}_1 = 9.485 in.          \bar{y}_2 = 0        \bar{y}_3 = 9.884 in.        \bar{c} = 1.80 in.

The moment of inertia of the three parts with respect to horizontal axes through their own centroids C_1,C_2, and C_3 are as follows:

I_1=\frac{bh^3}{12}=\frac{1}{12}(6.0\ in.)(0.5\ in.)^3=0.063\ in.^4 \\ I_2=1170\ in.^4 \quad \quad I_3=3.94\ in.^4

The moments of inertia I_2 and I_3 are obtained from Tables E-1 and E-3. respectively, of Appendix E.

Now we can use the parallel-axis theorem to calculate the moments of inertia about axis C-C for each of the three parts of the composite area:

I_{c1}=I_1+A_1(\bar{y}_1+\bar{c})^2=0.063\ in.^4+(3.0\ in.^2)(11.28\ in.)^2=382\ in.^4 \\ I_{c2}=I_2+A_2\bar{c}^2=1170\ in.^4+(20.8\ in.^2)(1.80\ in.)^2=1240\ in.^4\\ I_{c3}=I_3+A_3(\bar{y}_3\ -\ \bar{c})^2=3.94\ in.^4+(8.82\ in.^2)(8.084\ in.)^2=580\ in.^4

The sum of these individual moments of inertia gives the moment of inertia of the entire cross-sectional area about ifs centroidal axis C-C:

I_c=I_{c1}+I_{c2}+I_{c3}=2200\ in^4

This example shows how to calculate moments of inertia of composite area by using the parallel-axis theorem.