Question

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1. problem 9.53 a channel and a plate are welded together as shown to form a section that is symmetrical with respect to the y axis. determine the moments of inertia of the combined section with respect to its centroidal x and y axes. hint: evaluate the centroid of the composite section, c, with respect to the centroidal x axis of the channel:

Explanation:

Step1: Calculate the centroid \(y\)

For the plate: \(A_1 = 12\times0.5=6\ in^2\), \(y_1 = 0.25\ in\)
For the channel: \(A_2\) (from standard properties, for \(C8\times11.5\), \(A_2 = 3.38\ in^2\)), \(y_2=2.26 + 0.5/2=2.51\ in\)
Using the formula \(y=\frac{A_1y_1 + A_2y_2}{A_1+A_2}\)
\(y=\frac{6\times0.25+3.38\times2.51}{6 + 3.38}=\frac{1.5+8.4838}{9.38}=\frac{9.9838}{9.38}\approx1.064\ in\)

Step2: Calculate \(I_x\)

For the plate: \(I_{x1}=\frac{1}{12}\times12\times0.5^3+6\times(1.064 - 0.25)^2=\frac{1}{12}\times12\times0.125+6\times(0.814)^2=0.125 + 6\times0.6626=0.125+3.9756 = 4.1\ in^4\)
For the channel: \(I_{x2}=I_{x'}+A_2\times(2.51 - 1.064)^2\) (from standard \(I_{x'}=9.3\ in^4\) for \(C8\times11.5\))
\(I_{x2}=9.3+3.38\times(1.446)^2=9.3+3.38\times2.09=9.3 + 7.0642=16.3642\ in^4\)
\(I_x=I_{x1}+I_{x2}=4.1+16.3642 = 20.4642\ in^4\)

Step3: Calculate \(I_y\)

For the plate: \(I_{y1}=\frac{1}{12}\times0.5\times12^3=\frac{1}{12}\times0.5\times1728 = 72\ in^4\)
For the channel: \(I_{y2}=I_{y'}\) (from standard \(I_{y'}=0.336\ in^4\) for \(C8\times11.5\))
\(I_y=I_{y1}+I_{y2}=72+0.336=72.336\ in^4\)

Answer:

\(I_x\approx20.5\ in^4\), \(I_y = 72.3\ in^4\)