Question 9.SP.7: For the section shown, the moments of inertia with respect t......

MODELING and ANALYSIS: Divide the area into three rectangles as shown (Fig. 1). Note that the product of inertia I_{x^{\prime} y^{\prime}} with respect to centroidal axes parallel to the x and y axes is zero for each rectangle. Thus, using the parallel-axis theorem

I_{x y}=I_{x^{\prime} y^{\prime}}+\bar{x} \bar{y} A

you find that I_{x y} \text { reduces to } \bar{x} \bar{y} A for each rectangle.

I_{x y}=\Sigma \bar{x} \bar{y} A=-6.56 ~\mathrm{in}^4

a. Principal Axes. Since you know the magnitudes of I_x, I_y, \text { and } I_{x y}, you can use Eq. (9.25) to determine the values of \theta_m (Fig. 2):

\tan 2 \theta_m=-\frac{2 I_{x y}}{I_x-I_y}=-\frac{2(-6.56)}{10.38-6.97}=+3.85

2 \theta_m=75.4^{\circ} \text { and } 255.4^{\circ}

\theta_m=37.7^{\circ} \text { and } \theta_m=127.7^{\circ}

b. Principal Moments of Inertia. Using Eq. (9.27), you have

I_{\max }=15.45 ~\mathrm{in}^4 \quad I_{\min }=1.897 ~\mathrm{in}^4

REFLECT and THINK: Note that the elements of the area of the section are more closely distributed about the b axis than about the a axis. Therefore, you can conclude that I_a=I_{\max }=15.45 ~\mathrm{in}^4 \text { and } I_b=I_{\min }= 1.897 ~\mathrm{in}^4. You can verify this conclusion by substituting θ = 37.7° into Eqs. (9.18) and (9.19).

I_{x^{\prime}}=\frac{I_x+I_y}{2}+\frac{I_x-I_y}{2} \cos 2 \theta-I_{x y} \sin 2 \theta (9.18)

I_{y^{\prime}}=\frac{I_x+I_y}{2}-\frac{I_x-I_y}{2} \cos 2 \theta+I_{x y} \sin 2 \theta (9.19)