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 mm²

A2 = 50 × 150 = 7500 mm²

y₁ = (150 + 50/2) = 175 mm

y₂ = 150/2 = 75 mm

Y = (A₁y₁  + A₂y₂)/(A₁  + A₂)

= (7500 × 175 + 7500 × 75)/(7500 + 7500) = 125 mm

C.G. = (0,125)

Moment of inertia about x-x axis = I_XX  = I_XX1  + I_XX2

Moment of inertia (M.I.) about y-y axis =  I_yy =  I_yy1 +  I_yy2
Since