Let's break down the problem step by step: --- ## **Given Data** - Pulleys A and B transmit forces at different points. - Tension in the belts: - \( T_1(A) = 560 \) lb - \( T_2(A) = 93.3 \) lb - \( T_1(B) = 700 \) lb - \( T_2(B) = 60 \) lb - Distances: - \( a = 9 \) in - \( b = 12 \) in - \( c = 6 \) in - \( d = 18 \) in - Shaft diameter \( d_{shaft} = 2 \) in --- ## **Step 1: Find Forces at Each Pulley** ### Pulley A - Net force at A: \[ F_A = T_1(A) - T_2(A) = 560 - 93.3 = 466.7 \text{ lb (downward)} \] (Assume direction per figure.) ### Pulley B - Net force at B: \[ F_B = T_1(B) - T_2(B) = 700 - 60 = 640 \text{ lb (upward)} \] --- ## **Step 2: Find Reactions at Bearings** Let’s denote: - Left bearing = O - Right bearing = C Sum of vertical forces: \[ R_O + R_C - F_A + F_B = \] \[ R_O + R_C - 466.7 + 640 = \] \[ R_O + R_C = -173.3 \text{ lb} \] Take moments about O: - \( F_A \) acts at \( a = 9 \) in from O - \( F_B \) acts at \( a + b = 9 + 12 = 21 \) in from O - \( R_C \) acts at \( a + b + c + d = 9 + 12 + 6 + 18 = 45 \) in from O \[ \text{Sum of moments about O:} \] \[ - F_A \cdot 9 + F_B \cdot 21 - R_C \cdot 45 = \] \[ -466.7 \cdot 9 + 640 \cdot 21 - R_C \cdot 45 = \] \[ -4,200.3 + 13,440 - 45R_C = \] \[ 9,239.7 = 45R_C \] \[ R_C = \frac{9,239.7}{45} = 205.33 \text{ lb} \] Now back to force equilibrium: \[ R_O + 205.33 = -173.3 \] \[ R_O = -173.3 - 205.33 = -378.63 \text{ lb} \] --- ## **Step 3: Shear and Bending Moment Diagrams** Key points: - O (x=) - A (x=9 in) - B (x=21 in) - C (x=45 in) ### Shear at each point: - \( V_O = R_O = -378.63 \) lb - At A: \( V_A = V_O - F_A = -378.63 - 466.7 = -845.33 \) lb - At B: \( V_B = V_A + F_B = -845.33 + 640 = -205.33 \) lb - At C: \( V_C = V_B + R_C = -205.33 + 205.33 = \) lb ### Bending Moment at each section: - At O: \( M_O = \) - At A: \( M_A = R_O \cdot a = -378.63 \cdot 9 = -3,407.67 \) in-lb - Between A and B (at B, x=21 in): \[ M_B = R_O \cdot 21 - F_A \cdot (21-9) \] \[ = -378.63 \cdot 21 - 466.7 \cdot 12 \] \[ = -7,951.23 - 5,600.4 = -13,551.63 \text{ in-lb} \] - At C: \( M_C = \) **Maximum moment is at B: \( M_{max} = |M_B| = 13,551.63 \) in-lb** --- ## **Step 4: Find Torsional Moment at B** Torsional moment at B is due to the net belt tensions at B: \[ T_B = (T_1(B) - T_2(B)) \cdot r_B \] But since the shaft transmits the torque, the **torque at B** is: \[ T = (T_1(B) - T_2(B)) \cdot r_B = 640 \cdot 9 = 5,760 \text{ in-lb} \] (Assuming pulley B radius is 9 in per diagram.) --- ## **Step 5: Calculate Bending and Shear Stresses** ### **Bending Stress \(\sigma_b\):** \[ \sigma_b = \frac{M_{max} \cdot c}{I} \] Where: - \( M_{max} = 13,551.63 \) in-lb \( = 162,619.56 \) lb-in (\(1 \text{ ft-lb} = 12 \text{ in-lb}\)) - \( c = \frac{d}{2} = 1 \) in - \( I = \frac{\pi d^4}{64} \) for a circular shaft, \( d = 2 \) in Calculate \( I \): \[ I = \frac{\pi (2)^4}{64} = \frac{\pi \cdot 16}{64} = \frac{\pi}{4} = .785 \text{ in}^4 \] \[ \sigma_b = \frac{13,551.63 \cdot 1}{.785} = 17,273 \text{ psi} \] --- ### **Torsional Shear Stress \(\tau_{max}\):** \[ \tau_{max} = \frac{T \cdot c}{J} \] Where: - \( T = 5,760 \) in-lb - \( c = 1 \) in - \( J = \frac{\pi d^4}{32} = \frac{\pi (2)^4}{32} = \frac{\pi \cdot 16}{32} = \frac{\pi}{2} = 1.57 \) in\(^4\) \[ \tau_{max} = \frac{5,760 \cdot 1}{1.57} = 3,668 \text{ psi} \] --- ## **Summary Table** | Quantity | Value | Unit | |-------------------------------|-----------|------| | Max bending moment (\(M_{max}\)) | 13,551.63 | in-lb | | Max bending stress (\(\sigma_b\)) | 17,273 | psi | | Max torsional shear (\(\tau_{max}\))| 3,668 | psi | --- ## **Final Answers** \[ \boxed{ \begin{aligned} \text{The bending stress at the point of maximum bending moment is:} \quad & 17,273~\text{psi} \\ \text{The torsional shear stress at the point of maximum bending moment is:} \quad & 3,668~\text{psi} \end{aligned} } \] --- **Let me know if you need a combined stress analysis or have further questions!**