Problem:
From a rectangular lamina ABCD 10 cm * 14 cm a rectangular hole of 3 cm * 5 cm is cut as shown in Fig. 5.42. Find the center of gravity of the remainder lamina.
Solution:
Let the big rectangle be BD and smaller(hollow) one be bd,
and let C1(X1, Y1) and C2(X2,Y2) be their centroids w.r.t. the lower left corner of the total section.
Given values:
B = 10 cm, D = 14 cm
b = 3 cm, d = 5 cm
X2 = 7.5 cm Y2 = 4.5 cm
Area A1 = BD = 10.014.0= 140.0 cm^2, Centroid C1(X1, Y1) = B/2, Y/2= (5.0, 7.0) cm
Area A2 = bd = 3.05.0= 15.0 cm^2, Centroid C2 (X2, Y2) = ((7.5, 4.5)) cm
Total Area, A = A1 – A2 (subtracting the hollow part) = 140.0- 15.0= 125.0 cm^2
let C(Xc,Yc) be the centroid of the total area.
Equating the Moment of area, AXc = ∑AiXi,
or 125.0*Xc = 140.0*5.0 – 15.0*7.5 solving gives, Xc = 4.7 cm
Similarly, AYc = ∑AiYi, or 125.0Yc = 140.07.0 – 15.04.5
solving gives, Yc = 7.3 cm
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Using the parallel axis theorem, Moment of Inertia about the neutral axes is given as
Ixx = ∑(Ic.g.x + Aidyi^2), Iyy = ∑(Ic.g.y + Aidxi^2)
hence, Ixx = [(B*D^3)/12 + (B*D)*(Yc-Y1)^2] – [(b*d^3)/12 + b*d*(Yc-Y2)^2]
or Ixx = 2150.417 cm^4
Similarly, Iyy = (D*B^3)/12 + (B*D)(Xc-X1)2) – (d*b^3)/12 + b*d(Xc-X2)^2)
or Iyy = 3085.083 cm^4
