Circular Drive Shaft Torsional Shear Stress (Mechanical Engineering) Solution

A mechanical design engineer at the UNLV Howard R. Hughes College of Engineering is analyzing the torsional load capacity of a solid alloy steel drive shaft powering an industrial centrifugal water chiller.

The drive shaft has a circular cross-section of radius $R = 3\text{ cm}$. When transmitted engine torque twists the shaft, internal elastic torsional shear stress $\tau$ develops. According to the mechanics of materials elastic torsion formula, shear stress varies directly with radial distance from the central rotational axis, modeled in Cartesian coordinates as:

where $x^2 + y^2 \le 9$.

Step-by-Step Solution:

1. Define Polar Domain of Integration:
   State the Circle Polar Radius Identity:

   $$x^2 + y^2 \le R^2 \implies r^2 \le 3^2 \implies 0 \le r \le 3\text{ cm}$$
   Because the shaft is a complete solid cylinder, the angular sweep traverses the full circle:
   $$0 \le \theta \le 2\pi\text{ rad}$$
   Thus, the integration domain is $D = \{(r, \theta) \mid 0 \le r \le 3, \, 0 \le \theta \le 2\pi\}$.

2. Transform Integrand and Area Differential:
   State the Polar Coordinate Radial Distance Identity:

   $$\sqrt{x^2 + y^2} = r$$
   Substitute into the shear stress function:
   $$\tau(r) = 12r\text{ N/cm}^2$$
   The internal moment arm multiplying the shear stress is $r$. Incorporating the Polar Area Element Formula with Jacobian ($dA = r \, dr \, d\theta$):
   $$dT = (r) \cdot \tau(r) \, dA = (r)(12r)(r \, dr \, d\theta) = 12r^3 \, dr \, d\theta$$
   Formulate the double integral for total torque $T$:
   $$T = \iint_D 12r^3 \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{3} 12r^3 \, dr \, d\theta$$

3. Evaluate the Double Integral:
   Apply Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):

   $$T = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{3} 12r^3 \, dr \right)$$

   Evaluate the radial integral:

   $$\int_{0}^{3} 12r^3 \, dr = \left[ 3r^4 \right]_{0}^{3} = 3(3)^4 - 0 = 3(81) = 243\text{ N}\cdot\text{cm}$$

   Evaluate the angular integral:

   $$\int_{0}^{2\pi} d\theta = [\theta]_0^{2\pi} = 2\pi\text{ rad}$$

   Multiply and convert units:

   $$T = 2\pi \times 243 = 486\pi\text{ N}\cdot\text{cm} \approx 1526.814\text{ N}\cdot\text{cm}$$
   Convert to standard SI units ($1\text{ m} = 100\text{ cm}$):
   $$T = \frac{486\pi}{100}\text{ N}\cdot\text{m} = 4.86\pi\text{ N}\cdot\text{m} \approx 15.268\text{ N}\cdot\text{m}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total internal elastic resisting torque generated across the cross-section is $486\pi\text{ N}\cdot\text{cm} \approx 15.27\text{ N}\cdot\text{m}$.
• Engineering Application: In mechanical driveline design, integrating shear stress over polar coordinates confirms the classical torsion formula $T = \frac{\tau_{\max} J}{R}$. For a solid circular shaft of radius $R = 3\text{ cm}$, the polar moment of inertia is $J = \frac{\pi}{2} R^4 = \frac{\pi}{2}(81) = 40.5\pi\text{ cm}^4$. At the outer boundary ($r = 3\text{ cm}$), $\tau_{\max} = 12(3) = 36\text{ N/cm}^2$, giving $T = \frac{(36)(40.5\pi)}{3} = 486\pi\text{ N}\cdot\text{cm}$. Knowing the exact resisting torque enables engineers to size shaft diameters to avoid plastic deformation under dynamic starting torque surges.