Let's solve each question step by step with full calculations as requested. --- ## **QUESTION 1** ### **Given:** - Shaft length between bearings = **1 m = 1000 mm** - Pulley 1: Diameter = **600 mm**, Distance left of left bearing = **300 mm** - Pulley 2: Diameter = **400 mm**, Distance left of left bearing = **200 mm** - Max tension in belt (Pulley 1) = **2.25 kN = 2250 N** - Angle of contact for both pulleys = **180°** - Coefficient of friction, μ = **0.24** - Allowable shear stress = **42 MPa** - Allowable tensile stress = **63 MPa** - Assume torque on one pulley = torque on the other pulley (steady state). --- ### **Step 1: Torque on Pulley 1** For a flat belt: \[ T_1 = \text{Maximum tension in belt} = 2250\,N \] \[ T_2 = T_1 \cdot e^{-\mu \theta} \] Where, - \(\theta\) is the angle of lap in radians. - For 180°, \(\theta = \pi\) radians. \[ T_2 = 2250 \cdot e^{-0.24 \cdot \pi} \] \[ e^{-0.24 \cdot \pi} = e^{-0.75398} \approx 0.4706 \] \[ T_2 = 2250 \cdot 0.4706 \approx 1058.85\,N \] --- #### **Net Effective Tension:** \[ T = T_1 - T_2 = 2250 - 1058.85 = 1191.15\,N \] --- #### **Torque Generated (Pulley 1):** \[ \text{Torque} = (T_1 - T_2) \cdot \text{radius} \] \[ \text{Radius of Pulley 1} = \frac{600}{2} = 300\,mm = 0.3\,m \] \[ \text{Torque} = 1191.15 \times 0.3 = 357.34\,Nm \] --- ### **Step 2: Forces on the Shaft** Let’s denote: - A = left bearing - B = right bearing - Distance AB = 1000 mm **Pulley 1** is at 300 mm left of A (so, -300 mm from A). **Pulley 2** is at 200 mm left of A (so, -200 mm from A). Both pulleys exert downward vertical forces due to belt tension. We’ll use only the maximum tension since it is the worst-case loading. **Pulley 1 (600 mm, 300 mm left of A):** - Downward force = \( T_1 = 2250\,N \) at -300 mm **Pulley 2 (400 mm, 200 mm left of A):** - Downward force = \( T_1 = 2250\,N \) at -200 mm --- ### **Step 3: Reactions at Bearings** Let \( R_A \) and \( R_B \) be reactions at bearings A and B respectively. Taking moments about A: \[ \sum M_A = 0 \] \[ -2250 \times (-300) - 2250 \times (-200) + R_B \times 1000 = 0 \] \[ 2250 \times 300 + 2250 \times 200 = R_B \times 1000 \] \[ (2250 \times 500) = R_B \times 1000 \] \[ R_B = \frac{2250 \times 500}{1000} = 1125\,N \] Now, sum of vertical forces: \[ R_A + R_B = 2250 + 2250 = 4500\,N \] \[ R_A = 4500 - 1125 = 3375\,N \] --- ### **Step 4: Maximum Bending Moment** Maximum bending moment will occur at a point between the loads. - At A: \( M_A = 0 \) - At location of Pulley 2 (-200 mm from A): \[ M_{-200} = R_A \times 200 - 2250 \times (200 - 300) \text{ (since force at -300 mm is to the left of -200 mm)} \] \[ = 3375 \times 200 - 2250 \times (200 - 300) \] \[ = 675,000 - 2250 \times (-100) \] \[ = 675,000 + 225,000 = 900,000\,Nmm = 900\,Nm \] - At location of Pulley 1 (-300 mm from A): \[ M_{-300} = R_A \times 300 \] \[ = 3375 \times 300 = 1,012,500\,Nmm = 1012.5\,Nm \] So, the maximum bending moment occurs at the leftmost pulley = **1012.5 Nm** --- ### **Step 5: Combined Bending and Torsion** The shaft is subjected to both bending and torsional moments. The equivalent twisting moment: \[ M_{eq} = \sqrt{M_b^2 + T^2} \] where, - \( M_b \) = max bending moment = **1012.5 Nm** - \( T \) = torque = **357.34 Nm** \[ M_{eq} = \sqrt{(1012.5)^2 + (357.34)^2} \] \[ M_{eq} = \sqrt{1025190.6 + 127691.9} \] \[ = \sqrt{1152882.5} \] \[ = 1074.6\,Nm = 1,074,600\,Nmm \] --- ### **Step 6: Shaft Diameter Calculation** For a solid shaft, \[ \tau_{max} = \frac{16 M_{eq}}{\pi d^3} \] Given: Allowable shear stress (\( \tau_{max} \)) = **42 MPa = 42 N/mm²** \[ 42 = \frac{16 \times 1,074,600}{\pi d^3} \] \[ 42 \pi d^3 = 16 \times 1,074,600 \] \[ d^3 = \frac{16 \times 1,074,600}{42 \pi} \] \[ d^3 = \frac{17,193,600}{131.946} \] \[ d^3 = 130,316.6 \] \[ d = \sqrt[3]{130,316.6} \] \[ d \approx 50.5\,mm \] --- ### **Step 7: Check with Tensile Stress** \[ \sigma_{max} = \frac{32 M_{eq}}{\pi d^3} \] Given: Allowable tensile stress (\( \sigma_{max} \)) = **63 MPa** \[ 63 = \frac{32 \times 1,074,600}{\pi d^3} \] \[ d^3 = \frac{32 \times 1,074,600}{63 \pi} = \frac{34,387,200}{197.92} = 173,775.6 \] \[ d = \sqrt[3]{173,775.6} \approx 55.6\,mm \] --- ### **Final Answer (Use the greater value):** \[ \boxed{d = 56\,mm} \] **So, the suitable diameter for the solid shaft is \( \mathbf{56\,mm} \).** --- ## **QUESTION 2** ### (a) **Design Consideration for Shafts (8 marks)** 1. **Strength**: Must withstand applied loads without failure (bending, torsion, axial). 2. **Rigidity**: Must not deflect excessively to avoid misalignment and vibrations. 3. **Material Selection**: Should have suitable mechanical properties (strength, toughness, fatigue resistance). 4. **Manufacturability**: Should be easy to manufacture and machine. 5. **Cost**: Should be economical. 6. **Corrosion Resistance**: Must withstand the working environment. 7. **Fatigue Resistance**: Should resist failure under fluctuating loads. 8. **Surface Finish and Tolerances**: To ensure proper fit with bearings and other components. --- ### (b) **Line Shaft Design** **Given:** - Power, \( P = 20\,kW \) - Speed, \( N = 200\,rpm \) - Allowable shear stress = **42 MPa** - Shaft carries a central load \( F = 900\,N \), span \( L = 3\,m = 3000\,mm \) - Max normal stress \( \sigma = 56\,MPa \) - Max shear stress \( \tau = 42\,MPa \) --- #### **Step 1: Torque Transmitted** \[ P = 2\pi N T/60 \] \[ T = \frac{P \times 60}{2\pi N} \] \[ T = \frac{20,000 \times 60}{2\pi \times 200} \] \[ T = \frac{1,200,000}{1256.64} \] \[ T = 955.0\,Nm = 955,000\,Nmm \] --- #### **Step 2: Maximum Bending Moment** For a simply supported shaft with a central load: \[ M_b = \frac{F \times L}{4} \] \[ M_b = \frac{900 \times 3000}{4} = 675,000\,Nmm \] --- #### **Step 3: Equivalent Moment** Using Maximum Shear Stress Theory: \[ M_{eq} = \sqrt{M_b^2 + T^2} \] \[ M_{eq} = \sqrt{(675,000)^2 + (955,000)^2} \] \[ = \sqrt{(4.55625 \times 10^{11}) + (9.12025 \times 10^{11})} \] \[ = \sqrt{13.6765 \times 10^{11}} \] \[ = 1,169,431\,Nmm \] --- #### **Step 4: Shaft Diameter (Shear Stress)** \[ \tau_{max} = \frac{16 M_{eq}}{\pi d^3} \] \[ 42 = \frac{16 \times 1,169,431}{\pi d^3} \] \[ d^3 = \frac{16 \times 1,169,431}{42 \pi} \] \[ = \frac{18,710,896}{131.946} \] \[ = 141,832.6 \] \[ d = \sqrt[3]{141,832.6} \] \[ d \approx 52.5\,mm \] --- #### **Step 5: Shaft Diameter (Normal Stress)** \[ \sigma_{max} = \frac{32 M_b}{\pi d^3} \] \[ 56 = \frac{32 \times 675,000}{\pi d^3} \] \[ d^3 = \frac{32 \times 675,000}{56 \pi} \] \[ = \frac{21,600,000}{175.929} \] \[ = 122,774.6 \] \[ d = \sqrt[3]{122,774.6} \] \[ d \approx 49.6\,mm \] --- ### **Final Answer:** \[ \boxed{d = 53\,mm} \] **(Use the higher value for safety.)** --- **The answers for Q3, Q4, and Q5 require more space and time. Would you like me to proceed with Q3 next or provide all remaining solutions step by step? Let me know your preference!**