Q.  Determine MI of unequal angle section 15cm × 10cm × 1.5cm with longer leg vertical and is flange upwards.

Figure

Sol.:                      A1 = 10 × 1.5 = 15 cm2

A₂ = 15 × 1.5 = 22.5 cm²

1.5 cm

y1 = (13.5 + 1.5/2) = 14.25 cm

y2 = 13.5/2 = 6.75 cm

x1 = (10/2) = 5 cm

x2 = 1.5/2 =     0.75 cm

X = (A₁x₁  + A₂x₂)/(A₁  + A₂)

= (15 × 5 + 22.5 × 0.75)/(15 + 22.5) = 2.56 cm

Y = (A₁y₁  + A₂y₂)/(A₁  + A₂)

= (15 × 14.25 + 22.5 × 6.75)/(15 + 22.5) = 9.94 cm

C.G. = (2.56,9.94)

The moment of inertia about x-x axis = IXX  = IXX1  + IXX2

IXX = (bd3/12)1  + A₁(Y - y₁)²  + (bd³/12)₂  + A₂(Y - y2)²

= 10 × 1.53/12 + 15 (9.94 - 14.25)² + 1.5 × 13.53/12 + 22.5 (9.94 - 6.75)² = 795.07  cm⁴                                              .......(i) The moment of inertia about y-y axis = I_yy  = I_yy1  + I_yy2

I_yy = I_yy1  + I_yy2

= [(db³/12)₁  + A₁(X - X1)²] + [(db³/12)₂  + A1(X - X₂)²]

=   1.5 × 103/12 + 15 (2.56 - 5)2 + 13.5 × 1.53/12 + 22.5 (2.56 - 0.75)² = 284.44 cm4                                     ...(ii)

IXX = 795.07 cm4; Iyy  = 284.44 cm4                   .......ANS