# Problem Summary - **:** Solid titanium alloy - **Given Data:** - Shear modulus, \( G = 114\, \text{GPa} \) - Length of shaft, \( L = 900\, \text{nm} \) - Torque, \( T = 160\, \text{N·m} \) - Maximum shear stress, \( \tau_{max} = 165\, \text{MPa} \) - Twist angle, \( \phi \leq 5^\circ \) # Objective Determine the **minimum diameter** \( d \) of the shaft such that: - Shear stress does not exceed 165 MPa - Twist angle does not exceed 5° --- # Step 1: Convert units to SI | Quantity | Value | Units | |------------|---------|---------| | Shear modulus, \( G \) | 114 | GPa = \( 114 \times 10^9 \) Pa | | Length, \( L \) | 900 | nm = \( 900 \times 10^{-9} \) m | | Torque, \( T \) | 160 | N·m | | Max shear stress, \( \tau_{max} \) | 165 | MPa = \( 165 \times 10^6 \) Pa | | Twist angle, \( \phi \) | 5 | degrees = \( \frac{5\pi}{180} \) radians ≈ .0873 rad | --- # Step 2: Express shear stress Shear stress in a circular shaft: \[ \tau = \frac{T r}{J} \] Where: - \( r = \frac{d}{2} \) - \( J = \frac{\pi d^4}{32} \) (polar moment of inertia) Rearranged: \[ \tau = \frac{16 T}{\pi d^3} \] To ensure \( \tau \leq 165\, \text{MPa} \): \[ \frac{16 T}{\pi d^3} \leq 165 \times 10^6 \] Solve for \( d \): \[ d \geq \left( \frac{16 T}{\pi \times 165 \times 10^6} \right)^{1/3} \] Plug in values: \[ d \geq \left( \frac{16 \times 160}{\pi \times 165 \times 10^6} \right)^{1/3} \] Calculate numerator and denominator: \[ \text{Numerator} = 256 \] \[ \text{Denominator} = \pi \times 165 \times 10^6 \approx 3.1416 \times 165 \times 10^6 \approx 518.4 \times 10^6 \] Thus: \[ d \geq \left( \frac{256}{518.4 \times 10^6} \right)^{1/3} \approx \left( 4.94 \times 10^{-7} \right)^{1/3} \] Cube root: \[ d \geq (4.94 \times 10^{-7})^{1/3} \approx 7.7 \times 10^{-3} \text{ m} = 7.7\, \text{mm} \] --- # Step 3: Check twist constraint Twist angle: \[ \phi = \frac{T L}{G J} \] Where: \[ J = \frac{\pi d^4}{32} \] Rearranged for \( d \): \[ \phi = \frac{16 T L}{\pi G d^4} \] Solve for \( d \): \[ d^4 = \frac{16 T L}{\pi G \phi} \] \[ d = \left( \frac{16 T L}{\pi G \phi} \right)^{1/4} \] Plug in values: \[ T = 160\, \text{N·m} \] \[ L = 900 \times 10^{-9}\, \text{m} \] \[ G = 114 \times 10^9\, \text{Pa} \] \[ \phi = .0873\, \text{rad} \] Calculate numerator: \[ 16 \times 160 \times 900 \times 10^{-9} = 16 \times 160 \times 900 \times 10^{-9} \] \[ = 16 \times 160 \times 900 \times 10^{-9} = 16 \times 144000 \times 10^{-9} = 2,304,000 \times 10^{-9} = 2.304 \times 10^{-3} \] Calculate denominator: \[ \pi \times 114 \times 10^9 \times .0873 \approx 3.1416 \times 114 \times 10^9 \times .0873 \] \[ = 3.1416 \times 114 \times .0873 \times 10^9 \approx (3.1416 \times 114 \times .0873) \times 10^9 \] \[ = (3.1416 \times 114) \times .0873 \times 10^9 \approx 357.1 \times .0873 \times 10^9 \approx 31.2 \times 10^9 \] Thus: \[ d^4 = \frac{2.304 \times 10^{-3}}{31.2 \times 10^9} \approx 7.38 \times 10^{-14} \] \[ d = (7.38 \times 10^{-14})^{1/4} \] Fourth root: \[ d \approx (7.38 \times 10^{-14})^{.25} \approx 1.65 \times 10^{-3}\, \text{m} = 1.65\, \text{mm} \] --- # **Final Minimum Diameter:** - **Shear stress constraint:** \( d \geq 7.7\, \text{mm} \) - **Twist constraint:** \( d \geq 1.65\, \text{mm} \) **Therefore,** the **minimum diameter** to satisfy both constraints is: \[ \boxed{ d \approx \mathbf{7.7\, \text{mm}} } \] This ensures the shear stress does not exceed 165 MPa and the twist angle remains within 5°.