Q. Find out M.I. of T section as shown in figure given below about X-X and Y-Y axis through center of gravity of the section.
Figure
Sol.: As the diagram is symmetrical about y axis that is X = 0
A1 = 150 × 50 = 7500 mm^{2}
A2 = 50 × 150 = 7500 mm^{2}
y_{1} = (150 + 50/2) = 175 mm
y_{2} = 150/2 = 75 mm
Y = (A_{1}y_{1} + A_{2}y_{2})/(A_{1} + A_{2})
= (7500 × 175 + 7500 × 75)/(7500 + 7500) = 125 mm
C.G. = (0,125)
Moment of inertia about x-x axis = I_{X}_{X} = I_{XX1} + I_{XX2}
Moment of inertia (M.I.) about y-y axis = I_{yy} = I_{yy1} + I_{yy2 }Since