Find the MOI about x:x & y:y axes

We set the origin at the bottom-left corner.

• $A_1 = 70 \times 10 = 700 \text{ mm}^2$, $x_1 = 35 \text{ mm}$, $y_1 = 5 \text{ mm}$
• $A_2 = 15 \times 50 = 750 \text{ mm}^2$, $x_2 = 7.5 \text{ mm}$, $y_2 = 10 + 25 = 35 \text{ mm}$
• $A_3 = 15 \times 50 = 750 \text{ mm}^2$, $x_3 = 70 - 7.5 = 62.5 \text{ mm}$, $y_3 = 10 + 25 = 35 \text{ mm}$
• Total Area $A = 700 + 750 + 750 = 2200 \text{ mm}^2$

The centroid coordinates are:

$$\bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A} = \frac{700(35) + 750(7.5) + 750(62.5)}{2200} = \frac{24500 + 5625 + 46875}{2200} = \frac{77000}{2200} = 35 \text{ mm}$$ $$\bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A} = \frac{700(5) + 750(35) + 750(35)}{2200} = \frac{3500 + 26250 + 26250}{2200} = \frac{56000}{2200} = \frac{280}{11} \approx 25.45 \text{ mm}$$

We use the parallel axis theorem: $I_{xx} = \sum (I_{xi} + A_i d_{yi}^2)$, where $I_{xi} = \frac{b_i h_i^3}{12}$.

• For $R_1$: $I_{x1} = \frac{70 \times 10^3}{12} = \frac{17500}{3} \text{ mm}^4$. $d_{y1} = \bar{y} - y_1 = \frac{280}{11} - 5 = \frac{225}{11} \text{ mm}$. $A_1 d_{y1}^2 = 700 \left(\frac{225}{11}\right)^2 = \frac{35437500}{121} \text{ mm}^4$.
• For $R_2$: $I_{x2} = \frac{15 \times 50^3}{12} = 156250 \text{ mm}^4$. $d_{y2} = y_2 - \bar{y} = 35 - \frac{280}{11} = \frac{105}{11} \text{ mm}$. $A_2 d_{y2}^2 = 750 \left(\frac{105}{11}\right)^2 = \frac{8268750}{121} \text{ mm}^4$.
• For $R_3$: $I_{x3} = \frac{15 \times 50^3}{12} = 156250 \text{ mm}^4$. $d_{y3} = y_3 - \bar{y} = \frac{105}{11} \text{ mm}$. $A_3 d_{y3}^2 = 750 \left(\frac{105}{11}\right)^2 = \frac{8268750}{121} \text{ mm}^4$.

$$I_{xx} = \left(\frac{17500}{3} + \frac{35437500}{121}\right) + \left(156250 + \frac{8268750}{121}\right) + \left(156250 + \frac{8268750}{121}\right)$$ $$I_{xx} = \frac{271480000}{363} \approx 747878.8 \text{ mm}^4$$

The Moment of Inertia about the x-x axis is $\boxed{\text{747878.8 mm}^4}$.

We use the parallel axis theorem: $I_{yy} = \sum (I_{yi} + A_i d_{xi}^2)$, where $I_{yi} = \frac{h_i b_i^3}{12}$.

• For $R_1$: $I_{y1} = \frac{10 \times 70^3}{12} = \frac{857500}{3} \text{ mm}^4$. $d_{x1} = \bar{x} - x_1 = 35 - 35 = 0 \text{ mm}$. $A_1 d_{x1}^2 = 0$.
• For $R_2$: $I_{y2} = \frac{50 \times 15^3}{12} = 14062.5 \text{ mm}^4$. $d_{x2} = \bar{x} - x_2 = 35 - 7.5 = 27.5 \text{ mm}$. $A_2 d_{x2}^2 = 750 (27.5)^2 = 567187.5 \text{ mm}^4$.
• For $R_3$: $I_{y3} = \frac{50 \times 15^3}{12} = 14062.5 \text{ mm}^4$. $d_{x3} = x_3 - \bar{x} = 62.5 - 35 = 27.5 \text{ mm}$. $A_3 d_{x3}^2 = 750 (27.5)^2 = 567187.5 \text{ mm}^4$.

$$I_{yy} = \left(\frac{857500}{3} + 0\right) + (14062.5 + 567187.5) + (14062.5 + 567187.5)$$ $$I_{yy} = \frac{857500}{3} + 581250 + 581250 = \frac{857500}{3} + 1162500 = \frac{4345000}{3} \approx 1448333.3 \text{ mm}^4$$

The Moment of Inertia about the y-y axis is $\boxed{\text{1448333.3 mm}^4}$.

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Shape 2: I-Beam

First, we divide the I-beam into three rectangles:

• $R_1$: Top flange (width $b_1 = 40$ mm, height $h_1 = 10$ mm)
• $R_2$: Web (width $b_2 = 10$ mm, height $h_2 = 30$ mm)
• $R_3$: Bottom flange (width $b_3 = 80$ mm, height $h_3 = 10$ mm)

We set the origin at the bottom-left corner of the bottom flange.

• $A_1 = 40 \times 10 = 400 \text{ mm}^2$, $x_1 = 40 \text{ mm}$, $y_1 = 50 - 5 = 45 \text{ mm}$
• $A_2 = 10 \times 30 = 300 \text{ mm}^2$, $x_2 = 40 \text{ mm}$, $y_2 = 10 + 15 = 25 \text{ mm}$
• $A_3 = 80 \times 10 = 800 \text{ mm}^2$, $x_3 = 40 \text{ mm}$, $y_3 = 5 \text{ mm}$
• Total Area $A = 400 + 300 + 800 = 1500 \text{ mm}^2$

The centroid coordinates are:

$$\bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A} = \frac{400(40) + 300(40) + 800(40)}{1500} = \frac{16000 + 12000 + 32000}{1500} = \frac{60000}{1500} = 40 \text{ mm}$$ $$\bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A} = \frac{400(45) + 300(25) + 800(5)}{1500} = \frac{18000 + 7500 + 4000}{1500} = \frac{29500}{1500} = \frac{59}{3} \approx 19.67 \text{ mm}$$

We use the parallel axis theorem: $I_{xx} = \sum (I_{xi} + A_i d_{yi}^2)$, where $I_{xi} = \frac{b_i h_i^3}{12}$.

• For $R_1$: $I_{x1} = \frac{40 \times 10^3}{12} = \frac{10000}{3} \text{ mm}^4$. $d_{y1} = y_1 - \bar{y} = 45 - \frac{59}{3} = \frac{76}{3} \text{ mm}$. $A_1 d_{y1}^2 = 400 \left(\frac{76}{3}\right)^2 = \frac{2310400}{9} \text{ mm}^4$.
• For $R_2$: $I_{x2} = \frac{10 \times 30^3}{12} = 22500 \text{ mm}^4$. $d_{y2} = y_2 - \bar{y} = 25 - \frac{59}{3} = \frac{16}{3} \text{ mm}$. $A_2 d_{y2}^2 = 300 \left(\frac{16}{3}\right)^2 = \frac{25600}{3} \text{ mm}^4$.
• For $R_3$: $I_{x3} = \frac{80 \times 10^3}{12} = \frac{20000}{3} \text{ mm}^4$. $d_{y3} = \bar{y} - y_3 = \frac{59}{3} - 5 = \frac{44}{3} \text{ mm}$. $A_3 d_{y3}^2 = 800 \left(\frac{44}{3}\right)^2 = \frac{1548800}{9} \text{ mm}^4$.

$$I_{xx} = \left(\frac{10000}{3} + \frac{2310400}{9}\right) + \left(22500 + \frac{25600}{3}\right) + \left(\frac{20000}{3} + \frac{1548800}{9}\right)$$ $$I_{xx} = \frac{4228500}{9} = 469833.33 \text{ mm}^4$$

The Moment of Inertia about the x-x axis is $\boxed{\text{469833.3 mm}^4}$.

We use the parallel axis theorem: $I_{yy} = \sum (I_{yi} + A_i d_{xi}^2)$, where $I_{yi} = \frac{h_i b_i^3}{12}$.
Since the y-y axis is the axis of symmetry and all component centroids lie on this axis, $d_{xi} = 0$ for all components.

• For $R_1$: $I_{y1} = \frac{10 \times 40^3}{12} = \frac{160000}{3} \text{ mm}^4$. $d_{x1} = 0$.
• For $R_2$: $I_{y2} = \frac{30 \times 10^3}{12} = 2500 \text{ mm}^4$. $d_{x2} = 0$.
• For $R_3$: $I_{y3} = \frac{10 \times 80^3}{12} = \frac{1280000}{3} \text{ mm}^4$. $d_{x3} = 0$.

$$I_{yy} = \frac{160000}{3} + 2500 + \frac{1280000}{3}$$ $$I_{yy} = \frac{1440000}{3} + 2500 = 480000 + 2500 = 482500 \text{ mm}^4$$

The Moment of Inertia about the y-y axis is $\boxed{\text{482500 mm}^4}$.