Locate the centroid x y of the composite area shown and then

Locate the centroid x^-, y^- of the composite area shown, and then calculate the moment of inertia about the X\' axis. (Show all your work)

Solution

>> Taking Ground as the reference line.

Considering Semicircle as Part 1 & Rectangular Portion as Part 2

>> Considering Part 1

y1 = Centroid from reference line = r - 4r/3*pi

r = 200 mm

=> y1 = 0.085 m = 84.85 mm

A1 = Area = pi*r² /2 = (22/7)*200² /2 = 6.3*10⁴ mm²

>> Considering part 2

y2 = Centroid = r + 300/2 = 350 mm

A2 = Area = 300*400 = 12*10⁴ mm²

>> Total Area, A = A1 + A2 = 18.3*10⁴ mm²

>> Let \'y\' be the centroid of this figure

=> y*A = y1*A1 + y2*A2

=> y*18.3*10⁴ = 84.85*6.3*10⁴ + 350*12*10⁴

Solving,

=> y = 258.72 mm ......ANSWER......

I1\" = Inertia of Part 1 about axis X\' = I1 + A*(y - y1)² (Using Parallel Axis Theorem)

>> As, I1 = 0.110*r⁴ = 0.110*200⁴ = 1.76*10¹⁶ mm⁴

=> I1\" = 1.76*10⁸ + 6.3*10⁴ *(258.72 - 84.85)² = 20.81*10⁸ mm⁴

I2\" = Inertia of Part 2 about axis X\' = I2 + A2*(y2 - y)² (Using Parallel Axis Theorem)

>> As, I2 = ab³/3 (base of rectangle) = 400*300³/3 = 36*10⁸ mm⁴

=> I2\" = 36*10⁸ + (12*10⁴)*(350 - 258.72)² = 46*10⁸ mm⁴

>> As, Ix*A = I1*A1 + I2*A2

=> Ix*18.3*10⁴ = 20.81*10⁸*6.3*10⁴ + 46*10⁸*12*10⁴

=> Solving

=> Ix = 37.32*10⁸ mm⁴ ......ANSWER........