Find the distance to the centroid of the channel section shown in figure, from the left edge \({(\bar{X})}\).

Explanation:

Concept:

Location of N/A of complete section from \(\bar{x} \)

\(\bar{x} \) = \(\frac{(A_{1} \times x_{1})+ (A_{2} \times x_{2})+ (A_{3} \times x_{3})}{(A_{1}+A_{2}+A_{3})}\)

\(A_{1}\) = Area of 1st section

\(A_{2}\) = area of 2nd section

\(A_{3}\) = area of third section

\(x_{1}\) = location of neutral axis of 1st section

\(x_{2}\) = location of neutral axis of 2nd section

\(x_{3}\) = location of neutral axis of 3rd section

Calculation:

\(\bar{x} \) = \(\frac{(10 \times 100 \times 50)+ (10 \times 100 \times 50)+ (180 \times 10 \times 5)}{(10 \times 100)+(10 \times 100)+(180 \times 10)}\)

\(\bar{x} \) = 28.68 mm