Question

1. problem 9.43 determine the moments of inertia ( overline{i}_{x} ) and ( overline{i}_{y} ) of the area shown with respect to centroidal axes respectively parallel and perpendicular to side ( a b ). hint: first, locate the centroid of the area.

Explanation:

Step1: Calculate the area and centroid of each part

• Part 1 (Large rectangle):
• \(A_1=(0.9 + 2.0+2.1)\times(1.2 + 5.0+1.8)=5\times8 = 40\space in^2\)
• \(\overline{x}_1=\frac{0.9+2.0 + 2.1}{2}=2.5\space in\), \(\overline{y}_1=\frac{1.2 + 5.0+1.8}{2}=4\space in\)
• Part 2 (Small rectangle - to be subtracted):
• \(A_2 = 2.0\times5.0=10\space in^2\)
• \(\overline{x}_2=0.9+\frac{2.0}{2}=1.9\space in\), \(\overline{y}_2=1.8+\frac{5.0}{2}=4.3\space in\)
• Net area \(A=A_1 - A_2=40 - 10=30\space in^2\)
• Centroid \(\overline{x}=\frac{A_1\overline{x}_1 - A_2\overline{x}_2}{A}=\frac{40\times2.5-10\times1.9}{30}=\frac{100 - 19}{30}=\frac{81}{30}=2.7\space in\)
• \(\overline{y}=\frac{A_1\overline{y}_1 - A_2\overline{y}_2}{A}=\frac{40\times4-10\times4.3}{30}=\frac{160 - 43}{30}=\frac{117}{30}=3.9\space in\)

Step2: Calculate moments of inertia using the parallel - axis theorem \(I = I_{cm}+Ad^2\)

• For \(I_x\):
• \(I_{x1}=\frac{1}{12}(5)(8)^3+40\times(4 - 3.9)^2=\frac{1}{12}\times5\times512+40\times0.01=\frac{2560}{12}+0.4\approx213.33+0.4 = 213.73\space in^4\)
• \(I_{x2}=\frac{1}{12}(2)(5)^3+10\times(4.3 - 3.9)^2=\frac{1}{12}\times2\times125+10\times0.16=\frac{125}{6}+1.6\approx20.83+1.6 = 22.43\space in^4\)
• \(\overline{I}_x=I_{x1}-I_{x2}=213.73 - 22.43 = 191.3\space in^4\)
• For \(I_y\):
• \(I_{y1}=\frac{1}{12}(8)(5)^3+40\times(2.7 - 2.5)^2=\frac{1}{12}\times8\times125+40\times0.04=\frac{1000}{12}+1.6\approx83.33+1.6 = 84.93\space in^4\)
• \(I_{y2}=\frac{1}{12}(5)(2)^3+10\times(2.7 - 1.9)^2=\frac{1}{12}\times5\times8+10\times0.64=\frac{10}{3}+6.4\approx3.33+6.4 = 9.73\space in^4\)
• \(\overline{I}_y=I_{y1}-I_{y2}=84.93 - 9.73 = 75.2\space in^4\)

Answer:

The moments of inertia are \(\overline{I}_x = 191.3\space in^4\) and \(\overline{I}_y = 75.2\space in^4\)