Practice Question: Section Properties of Steel Members

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Solution

The position of the elastic neutral axis (centroid) using the bottom edge of the section as the reference point is given by:

\[\breve{y}=\frac{\sum{y_iA_i}}{\sum{A_i}}=\ \frac{0.5\times 8+7\times 12+13.5\times 8}{8+12+8}=7\]

Using the Parallel Axis Theorem, the moment of inertia I_xc is given by:

\[I_{xc}=\frac{1}{12}\times 8\times 1^3+8\times {\left(13.5-7\right)}^2+\ \frac{1}{12}\times 1\times {12}^3+12\times {\left(7-7\right)}^2+\ \frac{1}{12}\times 8\times 1^3+8\times {\left(1-7\right)}^2\]

\[=821.33\ {in}^4\]

 

The farthest fibre from the neutral axis is at a distance \(c=14-7 =7\ in\)

 

Elastic section modulus: 

\[S_x=\frac{I_{xc}}{c}=\ \frac{821.33}{7}=117.33\ {in}^3\]

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