## Answers

Let's divide the cross - section into (3) areas Area of region 1(A_1) = 625 mm^2 (50 times 12.5) Area of region 2(A_2) = 75 times 75 = 5625 mm^2 Area of region 3(A_3) = = 100 times 100 = 10,000mm^2 Therefore Required Area of cross-section (A) = 10,000- 5625 - 625 A = 3750 mm^2 Now, calculating the centroid distance to cross section NKt y^- = Sigma y_1 A_i/Sigma A_i A_i = Individual segment area y_i = Individual segment's centroid distance from a datum y^-_1 = 6.25 mm, y^-_2 = 50 mm y^-_3 = 50 mm y^- = 10,000 (50) - 5625 (50) - 625 (6.25)/3750 y^- = 57.29 mm c_1 = 100 - 57.29 = 42.71 mm Area moment of inertia for rectangular section I = bh_3/12 I_1 = 50 times (12.5)^3/12 = 8133 mm^4 I_2 = 75 times 75^3/12 = 2.63 times 10^6 mm^4 I_3 = 100 times 100^3/12 = 8.33 times 10^6 mm^4

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