Discussion
A channel has flanges \( 50 \) mm wide, a web \( 100 \) mm deep, uniform thickness \( 2 \) mm, and \( I_{xx} = 6.67 \times 10^5 \) mm\( ^4 \). Its shear centre lies from the web centreline at
Answer: D
Using \( e = b^2 h^2 t/(4I_{xx}) \), the numerator is \( 2500 \times 10\,000 \times 2 = 5 \times 10^7 \) and the denominator is \( 4 \times 6.67 \times 10^5 = 2.67 \times 10^6 \), giving \( 18.8 \) mm. The point lies outside the section, on the far side of the web from the flanges.
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