Question

Please show all work and formulas used to solve. Thank you for your help.

Calculate the moment of inertia of the composite area below about: a) The x- and y-axes shown (Ix and ly). b) Thex- and y-axes through the centroid of the area (Ix and Iy). 160 40 mm 40 EMm 40 mm 120

Answer 1

The moment of area about x axis for area 1 I_x1 = (160 - 40) times 40^3/12 + 120 times 40 times (200 - 20)^2 = 1.5616 times 10^8 m m^4 the moment of area about x axis for area 2 I_x2 = 40 times 200^3/12 + 200 times 40 times 100^2 = 1.067 times 10^8 m m^4 The moment of area about x axis I_x = I_x1 + I_x2 + I_x3 = 2.735 times 10^8 m m^4 The moment of area about y axis for area 1 I_y1 = (160 - 40)^3 times 40/12 + 120 times 40 times (40 + 120/2)^2 = 5.376 times 10^7 m m^4 The moment of area about x axis I_y = I_y1 + I_y2 + I_y 3 = 8.0214 times 10^7 m m^4 B) centroid of area 1 (x_1, y_1) (120, 180) centroid of area 2

a) The Moment of Area about x axis for Area 1.

$I_{x1} = \frac{(160-40)*40^3}{12} +120*40*(200-20)^2 = 1.5616*10^8 mm^4$

The Moment of Area about x axis for Area 2.

$I_{x2} = \frac{40*200^3}{12} +200*40*100^2 = 1.067*10^8 mm^4$

The Moment of Area about x axis for Area 3.

$I_{x3} = \frac{(120-40)*40^3}{12} +80*40*20^2 = 1.067*10^7 mm^4$

Total Moment of Area about x Axis

I_x= I_x1+I_x2+I_x3 = 2.735*10⁸ mm⁴

The Moment of Area about y axis for Area 1.

$I_{y1} = \frac{(160-40)^3*40}{12} +120*40*(40+120/2)^2 = 5.376*10^7 mm^4$

The Moment of Area about x axis for Area 2.

$I_{y2} = \frac{200*40^3}{12} +200*40*20^2 = 4.267*10^6 mm^4$

The Moment of Area about x axis for Area 3.

$I_1 = \frac{(120-40)^3*40}{12} +80*40*(40+80/2)^2 = 2.219*10^7 mm^4$

Total Moment of Area about x Axis

I_y= I_y1+I_y2+I_y3 = 8.0214*10⁷ mm⁴

b) Centroid of Area 1 (x₁,y₁) (120,180)

Centroid of Area 2 (x₂,y₂) (20,100)

Centroid of Area 3 (x₃,y₃) (80,20)

$\bar{x} = \frac{x_1A_1+x_2A_2+x_3A_3}{A_1+A_2+A_3} = \frac{120*120*40+20*200*40+80*80*40}{40*120+200*40+80*40}$

$\bar{x} = 62$

$\bar{y} = \frac{y_1A_1+y_2A_2+y_3A_3}{A_1+A_2+A_3} = \frac{180*120*40+100*200*40+20*80*40}{40*120+200*40+80*40}$

$\bar{y} = 108$

$I_{\bar{x}} = I_x +(A_1+A_2+A_3)\bar{y}^2=2.735*10^8 +(40*120+200*40+80*40)*108^2$

$I_{\bar{x}} = 4.60124*10^8 mm^4$

$I_{\bar{y}} = I_y +(A_1+A_2+A_3)\bar{x}^2=8.0214*10^7 +(40*120+200*40+80*40)*62^2$

$I_{\bar{y}} = 1.4172*10^8 mm^4$