To determine the moment of inertia ($I_x$) of the given composite shape about the x-axis, I'll break it into simpler geometric shapes, calculate their respective moments of inertia, and then use the parallel axis theorem as needed.

Here's the updated problem where $d = 11 \, \text{m}$. Let me walk through the process.

Step 1: Decompose the shape

The shape consists of:

1. A rectangle of height $a = 7 \, \text{m}$ and width $b = 7 \, \text{m}$,
2. A triangle with a base $c = 5 \, \text{m}$ and height $d = 11 \, \text{m}$.

Step 2: Calculate individual contributions

(a) Rectangle:

The moment of inertia about its centroidal axis is: $I_{\text{rect}, x} = \frac{b \cdot a^3}{12} = \frac{7 \cdot 7^3}{12}.$

Shift to the x-axis using the parallel axis theorem: $I_{\text{rect, x-total}} = I_{\text{rect}, x} + A_{\text{rect}} \cdot (y_{\text{centroid}})^2,$ where $y_{\text{centroid}} = \frac{a}{2}$.

(b) Triangle:

The centroid of the triangle is located at $\frac{d}{3}$ from the base (measured vertically). The moment of inertia about its centroidal axis is: $I_{\text{tri}, x} = \frac{b \cdot d^3}{36}.$

Shift to the x-axis using the parallel axis theorem: $I_{\text{tri, x-total}} = I_{\text{tri}, x} + A_{\text{tri}} \cdot (y_{\text{shift}})^2,$ where $y_{\text{shift}} = \frac{2d}{3}$.

Step 3: Sum up total moment of inertia

The total moment of inertia about the x-axis is: $I_x = I_{\text{rect, x-total}} + I_{\text{tri, x-total}}.$

Let me compute these values for $a = 7 \, \text{m}$, $b = 7 \, \text{m}$, $c = 5 \, \text{m}$, and $d = 11 \, \text{m}$. Given values a = 7 # height of rectangle (m) b = 7 # width of rectangle (m) c = 5 # base of triangle (m) d = 11 # height of triangle (m)

Rectangle calculations

I_rect_centroid = (b * a3) / 12 # Rectangle's I_x about centroidal axis A_rect = b * a # Area of rectangle y_rect_shift = a / 2 # Distance from rectangle centroid to x-axis I_rect_total = I_rect_centroid + A_rect * (y_rect_shift2) # Parallel axis theorem

Triangle calculations

I_tri_centroid = (c * d3) / 36 # Triangle's I_x about centroidal axis A_tri = 0.5 * c * d # Area of triangle y_tri_shift = (2 * d) / 3 # Distance from triangle centroid to x-axis I_tri_total = I_tri_centroid + A_tri * (y_tri_shift2) # Parallel axis theorem

Total moment of inertia about x-axis

I_x_total = I_rect_total + I_tri_total

I_rect_total, I_tri_total, I_x_totalThe total moment of inertia of the shape about the x-axis is calculated as follows:

1. Moment of inertia of the rectangle about the x-axis:
   $I_{\text{rect, x-total}} = 800.33 \, \text{m}^4$

2. Moment of inertia of the triangle about the x-axis:
   $I_{\text{tri, x-total}} = 1663.75 \, \text{m}^4$

3. Total moment of inertia:
   $I_x = 2464.08 \, \text{m}^4$

Let me know if you'd like a detailed explanation or have any questions!

Here are five follow-up questions to deepen understanding:

1. How does changing $d$ affect the centroid location and the moment of inertia?
2. Why is the parallel axis theorem necessary for composite shapes?
3. How do we calculate the centroid for more complex composite shapes?
4. What are the implications of using higher moments of inertia in structural engineering?
5. How would this calculation change if the triangle were inverted?

Tip: When applying the parallel axis theorem, always ensure that the correct distance to the axis of interest is used!