Aircraft Wing Torsional Polar Moment of Inertia (Aerospace & Mechanical Engineering) Solution

An aerospace structural dynamics engineer at the UNLV Howard R. Hughes College of Engineering is designing a lightweight carbon-fiber composite internal spar bracket for an unmanned aerial vehicle (UAV).

The cross-section of the structural bracket occupies a planar wedge region $R$ in the $xy$-plane bounded below by the horizontal axis $y = 0$, on the right by the vertical root rib $x = 4$, and above by the aerodynamic camber curve $y = \sqrt{x}$ (with dimensions in millimeters).

Step-by-Step Solution:

1. Extract Original Boundaries and Invert Boundary Curve: State the Type I Region Definition:

   $$0 \le x \le 4, \quad 0 \le y \le \sqrt{x}$$
   Invert the upper boundary function solving for $x$ in terms of $y$:
   $$y = \sqrt{x} \implies x = y^2 \quad (\text{for } y \ge 0)$$

2. Re-slice as a Type II General Region: Apply the 4-Step Re-slicing Protocol (defined in Switching the Order of Integration):

   • Determine the vertical span of the region: at $x = 0 \implies y = 0$; at $x = 4 \implies y = \sqrt{4} = 2$.
   • The outer constant interval for $y$ is $0 \le y \le 2$.
   • A horizontal ray entering the region from the left hits the curve $x = y^2$ and exits on the right at the vertical wall line $x = 4$. Thus, the Type II formulation is:
     $$R = \{(x, y) \in \mathbb{R}^2 \mid 0 \le y \le 2, \, y^2 \le x \le 4\}$$

3. Formulate and Evaluate Switched Double Integral: State the Type II Second Moment of Area Integral:

   $$J_y = \int_{0}^{2} \int_{y^2}^{4} x^2 \, dx \, dy$$

   Evaluate Inner Integral ($dx$):

   $$\int_{y^2}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{x=y^2}^{x=4} = \frac{(4)^3}{3} - \frac{(y^2)^3}{3} = \frac{64}{3} - \frac{y^6}{3}$$

   Evaluate Outer Integral ($dy$):

   $$J_y = \int_{0}^{2} \left( \frac{64}{3} - \frac{y^6}{3} \right) \, dy = \left[ \frac{64}{3}y - \frac{y^7}{21} \right]_{0}^{2}$$
   Evaluate at the upper limit $y = 2$:
   $$J_y = \left( \frac{64}{3}(2) - \frac{(2)^7}{21} \right) - 0 = \frac{128}{3} - \frac{128}{21}$$
   Express with common denominator ($21$):
   $$J_y = \frac{128 \times 7}{21} - \frac{128}{21} = \frac{896 - 128}{21} = \frac{768}{21}$$
   Divide numerator and denominator by $3$:
   $$J_y = \frac{256}{7}\text{ mm}^4 \approx 36.571\text{ mm}^4$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The second moment of area of the bracket section with respect to the root axis is $J_y = \frac{256}{7}\text{ mm}^4 \approx 36.57\text{ mm}^4$. This parameter captures how structural mass is concentrated away from the fuselage root chord.
• Engineering Application: In aeronautical wing design, aerodynamic lift forces generate torsional twisting moments around the elastic axis. Calculating moments of inertia over reversed Type II horizontal strips eliminates fractional exponent radicals in finite element analysis (FEA) verification routines, ensuring rapid calculation of critical flutter speeds to prevent catastrophic aeroelastic wing failure in flight.