Q1. Find the centroidal axes, Moment of Inertia, and Polar moment of inertia of the section shown below. Q2a. Write briefly on the following: Proof Resilience, Strain Energy, Impact Loading, Modulus of Resilience, Obliquity and how it is determined, Principal plane, Mohr's Circle

We use the parallel axis theorem: $I_x = \sum (I_{xi} + A_i d_{yi}^2)$, where $I_{xi} = \frac{b_i h_i^3}{12}$ and $d_{yi}$ is the distance from the component's centroid to the overall centroidal x-axis.

• For Rectangle 1 (Flange):

  • $I_{x1} = \frac{160 \text{ mm} \times (40 \text{ mm})^3}{12} = \frac{160 \times 64000}{12} = 853333.33 \text{ mm}^4$
  • $d_{y1} = y_1 - \bar{y} = 180 \text{ mm} - 130 \text{ mm} = 50 \text{ mm}$
  • $A_1 d_{y1}^2 = 6400 \text{ mm}^2 \times (50 \text{ mm})^2 = 16000000 \text{ mm}^4$
  • $(I_{x1} + A_1 d_{y1}^2) = 853333.33 + 16000000 = 16853333.33 \text{ mm}^4$

• For Rectangle 2 (Web):

  • $I_{x2} = \frac{40 \text{ mm} \times (160 \text{ mm})^3}{12} = \frac{40 \times 4096000}{12} = 13653333.33 \text{ mm}^4$
  • $d_{y2} = \bar{y} - y_2 = 130 \text{ mm} - 80 \text{ mm} = 50 \text{ mm}$
  • $A_2 d_{y2}^2 = 6400 \text{ mm}^2 \times (50 \text{ mm})^2 = 16000000 \text{ mm}^4$
  • $(I_{x2} + A_2 d_{y2}^2) = 13653333.33 + 16000000 = 29653333.33 \text{ mm}^4$

Total $I_x$:

Since the section is symmetrical about the y-axis, the centroidal y-axis passes through the centroid of each component, so $d_{xi} = 0$. We sum the individual moments of inertia $I_{yi} = \frac{h_i b_i^3}{12}$.

• For Rectangle 1 (Flange):

  • $I_{y1} = \frac{40 \text{ mm} \times (160 \text{ mm})^3}{12} = \frac{40 \times 4096000}{12} = 13653333.33 \text{ mm}^4$

• For Rectangle 2 (Web):

  • $I_{y2} = \frac{160 \text{ mm} \times (40 \text{ mm})^3}{12} = \frac{160 \times 64000}{12} = 853333.33 \text{ mm}^4$

Total $I_y$:

The polar moment of inertia about the centroid is the sum of the moments of inertia about the centroidal x and y axes: