The centroid of structural sections like T-section, I-section, L-section etc. are obtained by splitting them into rectangular components. Then equations (4.1) and (4.2) are used.

**Problem 4.7. ***Find the centre of gravity of the T-section shown in Fig. 4.10.*

**Sol. **The given T-section is split up into two-rectangles ABCD and EFGH as shown in Fig. 4.10 (a). The given T-section is symmetrical about Y-Y axis. Hence the C.G. of the section will lie on the axis. The lowest line of the figure is line GF. Hence the moments of the areas are taken about this line GF, which is the axis of reference in this case.

Fig. 4.10

Fig. 4.10(a)

**Sol. **The I-section is split up into three rectangles ABCD, EFGH and JKLM as shown in Fig. 4.11 (b). The given I-section is symmetrical about Y-Y axis. Hence the C.G. of the section will lie on this axis. The lowest line of the figure line is ML. Hence the moment of areas are taken about this line, which is the axis of reference.

**Problem 4.9. ***Find the centre of gravity of the L-section shown in Fig. 4.12.*

Fig. 4.12

**Sol. **The given L-section is not symmetrical about any section. Hence in this case, there will be two axis of references. The lowest line of the figure (i.e., line GF) will be taken as axis of reference for calculating . And the left line of the L-section (i.e., line AG) will be taken as axis of reference for calculating .

The given L-section is split up into two rectangles ABCD and DEFG, as shown in Fig. 4.12

Hence the C.G. of the L-section is at a distance of 4.33 cm from the bottom line GF and 2.33 cm from the left line AG. **Ans.**

**Problem. 4.10. ***Using the analytical method, determine the centre of gravity of the plane uniform lamina shown in Fig. 4.13.*

Fig. 4.13

**Sol. **Let be the distance between C.G. of the lamina and the bottom line AB.

**Problem 4.11. ***From a rectangular lamina ABCD 10 cm × 12 cm a rectangular hole of 3 cm × 4 cm is cut as shown in Fig. 4.14.*

*Find the C.G. of the remainder lamina.*

Fig. 4.14

**Sol. **The section shown in Fig. 4.14, is having a cut hole. The centre of gravity of a section with a cut hole is determined by considering the main section first as a complete one, and then subtracting the area of the cut-out hole, i.e., by taking the area of the cut-out hole as negative.

Let is the distance between the C.G. of the section with a cut hole from the bottom line DC.

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