1. **Problem statement:** A horizontal shaft is supported at two bearings A and B. Forces of 25 kN and 35 kN act at points C and D, located 150 mm from A and 200 mm from B respectively, with 600 mm between C and D. We need to find the diameter of the central portion of the shaft so that the stress does not exceed 100 MPa. 2. **Given data:** - Force at C, $F_C = 25$ kN = 25000 N - Force at D, $F_D = 35$ kN = 35000 N - Distance $AC = 150$ mm = 0.15 m - Distance $DB = 200$ mm = 0.20 m - Distance $CD = 600$ mm = 0.60 m - Maximum allowable stress, $\sigma_{max} = 100$ MPa = $100 \times 10^6$ Pa 3. **Step 1: Calculate reactions at bearings A and B** Using equilibrium of moments about A: $$ \sum M_A = 0 = R_B \times (0.15 + 0.60 + 0.20) - 25,000 \times 0.15 - 35,000 \times (0.15 + 0.60) $$ $$ R_B \times 0.95 = 25,000 \times 0.15 + 35,000 \times 0.75 $$ $$ R_B = \frac{(25,000 \times 0.15) + (35,000 \times 0.75)}{0.95} $$ $$ R_B = \frac{3,750 + 26,250}{0.95} = \frac{30,000}{0.95} = 31,578.95 \text{ N} $$ 4. Using vertical force equilibrium: $$ \sum F_y = 0 = R_A + R_B - 25,000 - 35,000 $$ $$ R_A = 60,000 - 31,578.95 = 28,421.05 \text{ N} $$ 5. **Step 2: Calculate the bending moment at the central portion (between C and D)** Maximum bending moment occurs between the forces. Calculate moment at point between C and D (e.g., at C): $$ M_C = R_A \times 0.15 - 25,000 \times 0 = 28,421.05 \times 0.15 = 4,263.16 \text{ Nm} $$ 6. **Step 3: Calculate the required diameter using bending stress formula:** The bending stress formula is: $$ \sigma = \frac{32 M}{\pi d^3} $$ Rearranged to solve for diameter $d$: $$ d = \sqrt[3]{\frac{32 M}{\pi \sigma}} $$ Substitute values: $$ d = \sqrt[3]{\frac{32 \times 4,263.16}{\pi \times 100 \times 10^6}} $$ $$ d = \sqrt[3]{\frac{136,421.12}{314,159,265}} = \sqrt[3]{0.0004345} $$ $$ d \approx 0.076 \text{ m} = 76 \text{ mm} $$ **Final answer:** The diameter of the central portion of the shaft should be approximately **76 mm** to ensure the stress does not exceed 100 MPa.