Identify given parameters and relationships.

Step 1: Identify given parameters and relationships. The given parameters are: • Inner diameter of bronze sleeve (and outer diameter of steel shaft), d_i = d_s = 80 \, mm = 0.08 \, m. • Torque carried by bronze sleeve T_b = 2 T_s. • Shear modulus of steel G_s = 2.5 G_b. • Maximum allowable shear stress in steel _s, max = 60 \, MPa = 60 × 10^6 \, Pa. • Maximum allowable shear stress in bronze _b, max = 40 \, MPa = 40 × 10^6 \, Pa. • Rotation speed N = 200 \, r/min. Step 2: Calculate the outside diameter of the sleeve (d_o). For a compound shaft, the angle of twist per unit length () is the same for both components: (T_s)/(G_s J_s) = (T_b)/(G_b J_b) Substitute the given relationships T_b = 2 T_s and G_s = 2.5 G_b: (T_s)/(2.5 G_b J_s) = (2 T_s)/(G_b J_b) (1)/(2.5 J_s) = (2)/(J_b) J_b = 5 J_s The polar moment of inertia for a solid shaft is J = ()/(32) d^4. For a hollow shaft, J = ()/(32) (d_o^4 - d_i^4). For the steel shaft (solid): J_s = ()/(32) d_s^4. For the bronze sleeve (hollow): J_b = ()/(32) (d_o^4 - d_i^4). Substitute these into J_b = 5 J_s: ()/(32) (d_o^4 - d_i^4) = 5 ()/(32) d_s^4 d_o^4 - d_i^4 = 5 d_s^4 Since d_i = d_s: d_o^4 - d_s^4 = 5 d_s^4 d_o^4 = 6 d_s^4 d_o = (6)^1/4 d_s Substitute d_s = 80 \, mm: d_o = (6)^1/4 × 80 \, mm ≈ 1.56508 × 80 \, mm d_o ≈ 125.206 \, mm The outside diameter of the sleeve is 125.21 mm. Step 3: Determine the limiting shear stress and calculate individual torques. The shear stress in a shaft is given by = (Tr)/(J). For the maximum stress, r is the outermost radius. For the steel shaft: _s = (T_s (d_s/2))/(J_s) = (T_s (d_s/2))/()32 d_s^4 = (16 T_s)/( d_s^3). For the bronze sleeve: _b = (T_b (d_o/2))/(J_b) = (T_b (d_o/2))/()32 (d_o^4 - d_i^4) = (16 T_b d_o)/( (d_o^4 - d_i^4)). From Step 2, we know T_b = 2 T_s, d_o = (6)^1/4 d_s, and d_o^4 - d_i^4 = 5 d_s^4. Let's find the ratio of the stresses: (_b)/(_s) = (16 T_b d_o)/( (d_o^4 - d_i^4))(16 T_s)/( d_s^3) = (T_b d_o d_s^3)/(T_s (d_o^4 - d_i^4)) Substitute the relationships: (_b)/(_s) = (2 T_s) ((6)^1/4 d_s) d_s^3T_s (5 d_s^4) = 2 (6)^1/45 (_b)/(_s) ≈ (2 × 1.56508)/(5) ≈ 0.626032 So, _b ≈ 0.626032 _s. Now, check which material limits the torque: • If _s = _s, max = 60 \, MPa, then _b = 0.626032 × 60 \, MPa ≈ 37.56 \, MPa. This is less than _b, max = 40 \, MPa, so it is acceptable. • If _b = _b, max = 40 \, MPa, then _s = 40 \, MPa0.626032 ≈ 63.89 \, MPa. This is greater than _s, max = 60 \, MPa, so it is not acceptable. Therefore, the maximum torque is limited by the steel shaft reaching its maximum allowable shear stress, _s = 60 \, MPa. Calculate the torque carried by the steel shaft (T_s): T_s = ( d_s^3 _s)/(16) = (0.08 \, m)^3 (60 × 10^6 \, Pa)16 T_s = × 0.000512 \, m^3 × 60 × 10^6 \, Pa16 = 1920 \, N · m T_s ≈ 6031.858 \, N · m Calculate the torque carried by the bronze sleeve (T_b): T_b = 2 T_s = 2 × 1920 \, N · m = 3840 \, N · m T_b ≈ 12063.716 \, N · m Step 4: Calculate the total torque that may be transmitted (T_total). The total torque is the sum of the torques carried by each component: T_total = T_s + T_b = 1920 \, N · m + 3840 \, N · m = 5760 \, N · m T_total ≈ 18095.574 \, N · m The torque that may be transmitted is 18.10 kN·m. Step 5: Calculate the power transmitted. First, convert the rotation speed from r/min to rad/s: N = 200 \, r/min = (200)/(60) \, r/s = (10)/(3) \, r/s The angular velocity is: = 2 N = 2 × (10)/(3) \, rad/s = (20 )/(3) \, rad/s ≈ 20.94395 \, rad/s The power transmitted P is given by P = T_total : P = (5760 \, N · m) × ((20 )/(3) \, rad/s) P = (5760 × 20 ^2)/(3) \, W = 1920 × 20 ^2 \, W = 38400 ^2 \, W P ≈ 38400 × (3.14159)^2 \, W ≈ 379800.36 \, W P ≈ 379.80 \, kW The power transmitted is 379.80 kW. Send me the next one 📸