# Torsion of a Solid Circular Shaft: Theory and Derivation Let's break down the derivation and meaning behind the relationship between **shear stress** and **angle of twist** in a solid circular shaft under pure torsion. --- ## **Step 1: Assumptions in the Derivation** 1. **Material is homogeneous and isotropic:** Properties are the same at every point and in every direction. 2. **Shaft is solid and circular in cross-section.** 3. **Pure torsion:** Only torque is applied; no bending or axial forces. 4. **Deformations are small:** Linear elastic behavior (Hooke’s law applies). 5. **Plane sections remain plane:** Cross sections remain perpendicular to the shaft axis during twist. 6. **Uniform shear strain across the shaft's length.** --- ## **Step 2: Derivation Using Uniform Shear Strain** ### **A. Shear Strain and Angle of Twist** - **Let:** \( L \) = Length of shaft \( r \) = Radius at which we measure strain \( \theta \) = Total angle of twist (radians) \( \phi = \frac{d\theta}{dx} \) = Rate of twist per unit length - **Shear Strain (\( \gamma \)):** \[ \gamma = r \frac{d\theta}{dx} = r \phi \] ### **B. Shear Stress** - **From Hooke’s Law for shear:** \( \tau = G \gamma \) Where \( G \) = Shear modulus of the material \[ \tau = G r \frac{d\theta}{dx} \] ### **C. Torque and Polar Moment of Inertia** - **At radius \( r \), elemental torque:** \[ dT = \tau \cdot r \cdot dA \] Where \( dA \) = Elemental area at radius \( r \) - **Total torque \( T \):** \[ T = \int_A \tau r\, dA = \int_A G r \frac{d\theta}{dx} r\, dA \] \[ T = G \frac{d\theta}{dx} \int_A r^2\, dA = G \frac{d\theta}{dx} J \] Where \( J = \int_A r^2 dA \) is the **polar moment of inertia** for the shaft. --- ## **Step 3: The Torsion Equation** \[ \boxed{ \frac{T}{J} = \frac{\tau}{r} = G \frac{d\theta}{dx} } \] - \( T \) = Applied torque - \( J \) = Polar moment of inertia (\( \frac{\pi d^4}{32} \) for a solid shaft of diameter \( d \)) - \( \tau \) = Shear stress at radius \( r \) - \( r \) = Radius where stress is calculated (maximum at surface, \( r = R \)) - \( G \) = Shear modulus - \( \frac{d\theta}{dx} \) = Twist per unit length --- ## **Step 4: Relationship Between Variables** - **Polar moment of inertia (\( J \))**: Measures resistance to twisting, depends on cross-sectional shape and size. - **Shear modulus (\( G \))**: Material property, relates shear stress and strain. - **Twist per unit length (\( \frac{d\theta}{dx} \))**: How much the shaft twists per unit length. ### **Solving for Angle of Twist** For total angle of twist over length \( L \): \[ \theta = \frac{T L}{G J} \] --- ## **Step 5: Physical Significance** - **Torsion equation** tells us how much a shaft will twist and how much shear stress develops for a given torque. - **Engineers use this to ensure**: - Shaft does not fail in shear (check \( \tau_{max} \)) - Shaft does not twist excessively (\( \theta \) is within acceptable limits) - Shaft’s dimensions are suitable for the torque and material used. --- ### **Summary Table** | Term | Meaning | |---------------|------------------------------------------------------------------| | \( T \) | Applied torque (Nm) | | \( J \) | Polar moment of inertia (\( m^4 \)), \( = \frac{\pi d^4}{32} \) | | \( \tau \) | Shear stress (Pa or N/m²) | | \( r \) | Radial distance from shaft center (m) | | \( G \) | Shear modulus (Pa) | | \( \theta \) | Total angle of twist (radians) | | \( L \) | Length of shaft (m) | --- ## **Final Summary Equation** \[ \boxed{ \frac{T}{J} = \frac{\tau}{r} = G \frac{d\theta}{dx} } \] This relates torque, shear stress, shaft geometry, material properties, and angle of twist for a circular shaft under torsion.