torsion

Torsion

Torsion occurs when a member, typically a circular shaft, is subjected to a twisting moment or torque about its longitudinal axis. This creates shearing stresses that vary linearly from zero at the center to a maximum at the outer surface.

Relevance to Philippine Standards: The National Structural Code of the Philippines (NSCP) 2015, Volume 1, addresses torsion in structural members, particularly in reinforced concrete beams and steel sections, ensuring they can withstand twisting forces without failure.

Torsional Stress ($\tau$)

For a circular shaft (solid or hollow) subjected to a torque $T$, the shear stress $\tau$ at any radial distance $\rho$ from the center is given by: $\tau = \frac{T\rho}{J}$ where:

• $T$ is the applied torque (N$\cdot$m or lb$\cdot$in).
• $\rho$ is the radial distance from the center to the point of interest.
• $J$ is the polar moment of inertia of the cross-sectional area.

Maximum Shear Stress ($\tau_{max}$)

Occurs at the outer surface where $\rho = c$ (radius of the shaft): $\tau_{max} = \frac{Tc}{J}$

Angle of Twist ($\theta$)

The angle of twist measures the rotation of one end of the shaft relative to the other. $\theta = \frac{TL}{JG}$ where:

• $L$ is the length of the shaft.
• $G$ is the shear modulus of elasticity (Modulus of Rigidity).
• $\theta$ is in radians.

Polar Moment of Inertia ($J$)

• Solid Shaft: $J = \frac{\pi c^4}{2} = \frac{\pi d^4}{32}$
• Hollow Shaft: $J = \frac{\pi (c_o^4 - c_i^4)}{2} = \frac{\pi (d_o^4 - d_i^4)}{32}$

Power Transmission

Shafts are often used to transmit power. The relationship between power ($P$), torque ($T$), and angular velocity ($\omega$) is: $P = T\omega = 2\pi f T$ where:

• $P$ is power in Watts (W).
• $T$ is torque in N$\cdot$m.
• $\omega$ is angular velocity in rad/s.
• $f$ is frequency in Hz (revolutions per second).

Examples

Example 1: Solid Shaft Stress

Problem: A solid steel shaft with a diameter of 50 mm is subjected to a torque of 2 kN$\cdot$m. Determine the maximum shear stress in the shaft.

Solution:

1. Given:

   • $d = 50 \text{ mm} = 0.05$ m, so $c = 0.025$ m.
   • $T = 2000 \text{ N}\cdot\text{m}$.

2. Calculate $J$: $J = \frac{\pi (0.05)^4}{32} \approx 6.136 \times 10^{-7} \text{ m}^4$

3. Calculate Maximum Shear Stress: $\tau_{max} = \frac{Tc}{J}$ $\tau_{max} = \frac{(2000)(0.025)}{6.136 \times 10^{-7}}$ $\tau_{max} \approx 81.49 \times 10^6 \text{ Pa} = 81.49 \text{ MPa}$

Result: The maximum shear stress is 81.49 MPa.

Example 2: Angle of Twist

Problem: A hollow aluminum shaft ($G = 26$ GPa) has an outer diameter of 80 mm and an inner diameter of 60 mm. If the shaft is 2 m long and subjected to a torque of 1.5 kN$\cdot$m, determine the angle of twist.

Solution:

1. Given:

   • $d_o = 0.08$ m, $d_i = 0.06$ m.
   • $L = 2$ m.
   • $T = 1500 \text{ N}\cdot\text{m}$.
   • $G = 26 \times 10^9$ Pa.

2. Calculate $J$: $J = \frac{\pi (0.08^4 - 0.06^4)}{32} = \frac{\pi (4.096 \times 10^{-5} - 1.296 \times 10^{-5})}{32}$ $J \approx 2.749 \times 10^{-6} \text{ m}^4$

3. Calculate Angle of Twist: $\theta = \frac{TL}{JG}$ $\theta = \frac{(1500)(2)}{(2.749 \times 10^{-6})(26 \times 10^9)}$ $\theta = \frac{3000}{71474} \approx 0.042 \text{ rad}$

Result: The angle of twist is 0.042 radians (approx. $2.4^\circ$).