Let's solve the problem step by step. --- ## **Step 1: Understanding the Problem** - We have a shaft carrying pulleys A and B, transmitting torque via a belt drive. - Pulley A receives torque from a motor, and pulley B delivers it to the shaft. - Tensions at A: \( T_1(A) = 750 \) N, \( T_2(A) = 313 \) N (tight/loose sides). - At B: \( T_1(B) = 180 \) N, \( T_2(B) = 549 \) N. - At B, the loose side is 85% of the tight side. - We are to **determine principal stresses and maximum shear stress** at the location of **maximum bending moment**. --- ## **Step 2: Calculate Net Belt Forces** ### **Pulley A:** - Net force, \( F_A = T_1(A) - T_2(A) = 750 - 313 = 437 \) N (downward). ### **Pulley B:** - Net force, \( F_B = T_2(B) - T_1(B) = 549 - 180 = 369 \) N (upward). --- ## **Step 3: Find Maximum Bending Moment Location** Let’s recall the setup: - Pulley A is at \( x = a \) - Pulley B is at \( x = a + b \) - Supports at O and C Let’s call the distances: - \( OA = a \) - \( AB = b \) - \( BC = c \) Let’s use the standard approach for a simply supported beam with point loads. --- ### **A. Reaction Forces at O and C** Let’s call: - \( R_O \): Reaction at O (upward) - \( R_C \): Reaction at C (upward) Sum of forces (vertical equilibrium): \[ R_O + R_C = F_A + F_B \] \[ R_O + R_C = 437 + 369 = 806\text{ N} \] Sum of moments about O: \[ R_C \times (a+b+c) = F_A \times a + F_B \times (a + b) \] Plug in location values from the figure (assume \( a = .15 \, \text{m} \), \( b = .55 \, \text{m} \), \( c = .25 \, \text{m} \)): \[ R_C \times (.15+.55+.25) = 437 \times .15 + 369 \times (.15 + .55) \] \[ R_C \times .95 = 437 \times .15 + 369 \times .7 \] \[ 437 \times .15 = 65.55 \] \[ 369 \times .7 = 258.3 \] \[ R_C \times .95 = 65.55 + 258.3 = 323.85 \] \[ R_C = \frac{323.85}{.95} = 341 \text{ N} \] \[ R_O = 806 - 341 = 465 \text{ N} \] --- ### **B. Shear and Bending Moment at Each Section** At **A** (\( x = .15 \, \text{m} \)): - Shear just left: \( V = R_O = 465 \) N - Shear just right: \( V = 465 - 437 = 28 \) N At **B** (\( x = .15 + .55 = .70 \, \text{m} \)), just left: - Shear: \( V = 465 - 437 = 28 \) N - Shear just right: \( 28 + 369 = 397 \) N At **C** (\( x = .95 \, \text{m} \)), shear back to zero. #### **Maximum Bending Moment: Under Pulley A or B** Let's check at A and B: **At A:** \[ M_A = R_O \times a = 465 \times .15 = 69.75 \text{ Nm} \] **At B:** \[ M_B = R_O \times (a+b) - F_A \times b \] \[ = 465 \times .70 - 437 \times .55 \] \[ = 325.5 - 240.35 = 85.15 \text{ Nm} \] So, **maximum bending moment occurs at B** and is **85.15 Nm**. --- ## **Step 4: Calculate Principal Stresses and Max Shear at Maximum Bending Moment** ### **A. Bending Stress** \[ \sigma_{\text{bending}} = \frac{M \cdot c}{I} \] - \( M = 85.15 \) Nm \( = 85150 \) Nmm - \( c = \frac{d}{2} \) (shaft radius) - \( I = \frac{\pi}{64} d^4 \) Assume shaft diameter \( d \) is **not given**, so keep in terms of \( d \): \[ \sigma_{\text{bending}} = \frac{M \cdot (d/2)}{(\pi/64) d^4} = \frac{M \cdot 32}{\pi d^3} \] ### **B. Torsional Shear Stress** \[ \tau_{\text{torsion}} = \frac{T \cdot c}{J} \] - \( J = \frac{\pi}{32} d^4 \) - \( c = d/2 \) Net torque at B: \[ T_B = r_B \times (T_1(B) - T_2(B)) \] Assume pulley radius \( r_B = .09 \, \text{m} \) (from figure): \[ T_B = .09 \times (549 - 180) = .09 \times 369 = 33.21 \text{ Nm} = 33210 \text{ Nmm} \] \[ \tau_{\text{torsion}} = \frac{T \cdot (d/2)}{(\pi/32) d^4} = \frac{T \cdot 16}{\pi d^3} \] --- ## **Step 5: Principal Stresses and Maximum Shear Stress** ### **A. Principal Stresses** \[ \sigma_1, \sigma_2 = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau^2} \] Since only bending (fiber at surface) and torsion: \[ \sigma_x = \sigma_{\text{bending}} \] \[ \tau_{xy} = \tau_{\text{torsion}} \] So, \[ \sigma_{1,2} = \frac{\sigma_{\text{bending}}}{2} \pm \sqrt{\left(\frac{\sigma_{\text{bending}}}{2}\right)^2 + \tau_{\text{torsion}}^2} \] ### **B. Maximum Shear Stress** \[ \tau_{\text{max}} = \sqrt{\left(\frac{\sigma_{\text{bending}}}{2}\right)^2 + \tau_{\text{torsion}}^2} \] --- ## **Step 6: Final Answers (in terms of d)** ### **A. Values** \[ \sigma_{\text{bending}} = \frac{85150 \times 32}{\pi d^3} = \frac{2,724,800}{\pi d^3} \text{ N/mm}^2 \] \[ \tau_{\text{torsion}} = \frac{33210 \times 16}{\pi d^3} = \frac{531,360}{\pi d^3} \text{ N/mm}^2 \] ### **B. Principal Stresses** \[ \sigma_{1,2} = \frac{1}{2} \left( \frac{2,724,800}{\pi d^3}\right) \pm \sqrt{ \left( \frac{1}{2} \frac{2,724,800}{\pi d^3} \right)^2 + \left( \frac{531,360}{\pi d^3} \right)^2 } \] ### **C. Maximum Shear Stress** \[ \tau_{\text{max}} = \sqrt{ \left( \frac{1}{2} \frac{2,724,800}{\pi d^3} \right)^2 + \left( \frac{531,360}{\pi d^3} \right)^2 } \] --- ## **Summary Table** | Quantity | Expression (in terms of \( d \)) | |----------------------|-------------------------------------------------| | \(\sigma_{\text{bending}}\) | \(\frac{2,724,800}{\pi d^3}\) N/mm\(^2\) | | \(\tau_{\text{torsion}}\) | \(\frac{531,360}{\pi d^3}\) N/mm\(^2\) | | Principal stresses \(\sigma_{1,2}\) | See above formula | | Max shear stress \(\tau_{\text{max}}\) | See above formula | **To get numeric values, plug in the actual shaft diameter.** --- ### **Final Answers (in terms of d):** - **First principal stress:** \[ \sigma_1 = \frac{1}{2} \left( \frac{2,724,800}{\pi d^3}\right) + \sqrt{ \left( \frac{1}{2} \frac{2,724,800}{\pi d^3} \right)^2 + \left( \frac{531,360}{\pi d^3} \right)^2 } \] - **Second principal stress:** \[ \sigma_2 = \frac{1}{2} \left( \frac{2,724,800}{\pi d^3}\right) - \sqrt{ \left( \frac{1}{2} \frac{2,724,800}{\pi d^3} \right)^2 + \left( \frac{531,360}{\pi d^3} \right)^2 } \] - **Maximum shear stress:** \[ \tau_{\text{max}} = \sqrt{ \left( \frac{1}{2} \frac{2,724,800}{\pi d^3} \right)^2 + \left( \frac{531,360}{\pi d^3} \right)^2 } \] --- If you provide the shaft diameter, I can give the final numeric answers.