A symmetrical T-section has its flange horizontal on top. Its dimensions are: Flange: Width = 100 mm, thickness = 24 mm; Web: Height = 84 mm, thickness = 20 mm. Its moment of inertia about a vertical axis through its centroid parallel to the web is (in mm⁴):

Concept:

The moment of inertia of a symmetrical T-section about its vertical centroidal axis is calculated by summing the contributions of the flange and the web, using the parallel axis theorem if necessary.

Given:

• Flange: Width = 100 mm, Thickness = 24 mm
• Web: Height = 84 mm, Thickness = 20 mm
• The T-section is symmetrical, with the flange horizontal on top.

Step 1: Break the T-section into two rectangles

The section is divided into:

1. Flange (top part): Width = 100 mm, Thickness = 24 mm
2. Web (vertical part): Height = 84 mm, Thickness = 20 mm

Step 2: Calculate moment of inertia for each part

The moment of inertia for a rectangle about its centroidal axis is:

\[ I = \frac{b h^3}{12} \]

Flange:

\[ I_{\text{flange}} = \frac{24 \times 100^3}{12} = 2,000,000 \, \text{mm}^4 \]

Web:

\[ I_{\text{web}} = \frac{84 \times 20^3}{12} = 56,000 \, \text{mm}^4 \]

Step 3: Sum the contributions

Since the centroids of both parts align with the global centroid (due to symmetry), their individual moments of inertia can be directly added:

\[ I_{yy} = I_{\text{flange}} + I_{\text{web}} = 2,000,000 + 56,000 = 2,056,000 \, \text{mm}^4 \]

Moment of inertia about the vertical centroidal axis = \({2056 \times 10^3 \, \text{mm}^4}\)