Question:

Give step-by-step solution with explanation and final answer: Given (from the tables in your image — I transcribed the important values) Engine power ? = 120  kW = 120000  W P=120 kW=120000 W Engine speed ? = 5000  rpm N=5000 rpm Output gearbox ratio ? ? = 1.8 i g ​ =1.8 Final drive ratio ? ? = 3.2 i f ​ =3.2 Transmission efficiency ? = 0.92 η=0.92 Wheel radius ? ? = 0.32  m r w ​ =0.32 m Flywheel moment of inertia ? ? = 0.6   kg \cdotp m 2 I f ​ =0.6 kg\cdotpm 2 (table) Bending moment at point G: ? ? = 300   N \cdotp m M b ​ =300 N\cdotpm (table) Stress concentration factor (given): ? ? = 2.5 K t ​ =2.5 (table) Modified endurance stress ? ? = 300  MPa S e ​ =300 MPa and design factor ? = 3 n=3 → allowable fatigue/shear stress ? ? ? ? ? ? = ? ? / ? = 100  MPa τ allow ​ =S e ​ /n=100 MPa (table) Reaction forces at bearing (for bearing sizing): ? ? = 1150  N ,    ? ? = 1300  N F x ​ =1150 N,F y ​ =1300 N (table) I use standard engineering formulae below and state any extra assumptions explicitly. 1) Engine maximum torque (straight numerical) Use ? = ? ? T= ω P ​ with ? = 2 ? ? / 60 ω=2πN/60. ? = 2 ? 5000 60 = 523.5988  rad/s ω=2π 60 5000 ​ =523.5988 rad/s ? ? ? ? ? ? ? = 120000 523.5988 = 229.18   N \cdotp m T engine ​ = 523.5988 120000 ​ =229.18 N\cdotpm Answer (1): ? ? ? ? ? ? ? ≈ 229.2   N \cdotp m T engine ​ ≈229.2 N\cdotpm. 2) Tractive force at the wheel Wheel torque = engine torque × (gearbox ratio × final drive ratio) × efficiency. Total ratio ? ? ? ? = 1.8 × 3.2 = 5.76 i tot ​ =1.8×3.2=5.76. ? ? ℎ ? ? ? = ? ? ? ? ? ? ? ⋅ ? ? ? ? ⋅ ? = 229.18 × 5.76 × 0.92 ≈ 1214.49   N \cdotp m T wheel ​ =T engine ​ ⋅i tot ​ ⋅η=229.18×5.76×0.92≈1214.49 N\cdotpm Tractive force ? ? ? ? ? ? ? ? ? = ? ? ℎ ? ? ? ? ? F tractive ​ = r w ​ T wheel ​ ​ : ? ? ? ? ? ? ? ? ? = 1214.49 0.32 ≈ 3795.27  N F tractive ​ = 0.32 1214.49 ​ ≈3795.27 N Answer (2): ? ? ? ? ? ? ? ? ? ≈ 3.80  kN F tractive ​ ≈3.80 kN (≈3795 N). 3) Clutch slipping work and slipping time — method + example estimate Method (exact): If the flywheel (inertia ? ? I f ​ ) at angular speed ? ? ω i ​ is brought to a new angular speed ? ? ω f ​ by slipping/engagement, the energy dissipated by slipping is the loss of kinetic energy: ? = 1 2 ? ? ( ? ? 2 − ? ? 2 ) W= 2 1 ​ I f ​ (ω i 2 ​ −ω f 2 ​ ) The slipping power at any instant is ? ? ? ? ? = ? ? ? ? ? ⋅ ? P slip ​ =T slip ​ ⋅ω where ? ? ? ? ? T slip ​ is the slipping torque (the torque dissipated in clutch plates during slip). The average slipping power ≈ ? ? ? ? ? ⋅ ? ? ? ? T slip ​ ⋅ω avg ​ . Then slipping time: ? ≈ ? ? ? ? ? ?   ? ? ? ? t≈ T slip ​ ω avg ​ W ​ Why we cannot give an exact single value: the result depends on: the initial and final angular speeds (the table suggests max-torque engine speed might be at ? ? = 0.5 ? ? ? ω m ​ =0.5ω mN ​ — that would be 2500 rpm if ? ? ? = 5000 ω mN ​ =5000 rpm), the driven-side rotational speed before engagement (gearbox/vehicle rotating speed), the clutch slip torque capacity during engagement (table gives a “clutch safety factor = 1.5” but that is not the same as the actual friction torque). Representative example (worst-case / illustrative): Assume (as a simple illustration) the flywheel initially at    2500  rpm 2500 rpm (i.e. ? ? = 2 ? ⋅ 2500 / 60 ω i ​ =2π⋅2500/60), and assume the driven side is much slower (worst case — all flywheel energy dissipated). Using the table inertia ? ? = 0.6   kg \cdotp m 2 I f ​ =0.6 kg\cdotpm 2 : ? ? = 2 ? ⋅ 2500 / 60 = 261.799  rad/s ω i ​ =2π⋅2500/60=261.799 rad/s ? = 0.5 ⋅ 0.6 ⋅ ( 261.799 ) 2 ≈ 20,562  J W=0.5⋅0.6⋅(261.799) 2 ≈20,562 J Now assume the clutch transmits (during slipping) about the engine torque (use ? ? ? ? ? ≈ ? ? ? ? ? ? ? ≈ 229   N \cdotp m T slip ​ ≈T engine ​ ≈229 N\cdotpm for a rough estimate) and average angular speed during slip ? ? ? ? ≈ ? ? / 2 ω avg ​ ≈ω i ​ /2. ? ≈ 20562 229 ⋅ ( 261.8 / 2 ) ≈ 0.69  s t≈ 229⋅(261.8/2) 20562 ​ ≈0.69 s Illustrative answer (example): Clutch slipping energy ? W (worst-case example) ≈ 20.6 kJ. Corresponding slipping time (with the rough assumptions above) ≈ 0.7 s. Note: If the driven side is already spinning (vehicle is moving), ? W will be much smaller. To compute the actual ? W and ? t you need the initial engine and gearbox/driven angular speeds and a correct clutch torque capacity (or friction radius & normal clamp load). 4) Diameter of shaft AK at point G (left) — method + numeric estimate We must design shaft diameter for combined bending and torsion. A simple (conservative) approach is to compute an equivalent moment ? ? ? = ? ? 2 + ? 2 M eq ​ = M b 2 ​ +T 2 ​ and use maximum shear from torsion+bending for a circular shaft (solid) where maximum shear ? ? ? ? = 16   ? ? ? ? ? 3 τ max ​ = πd 3 16M eq ​ ​ . Set ? ? ? ? ≤ ? ? ? ? ? ? τ max ​ ≤τ allow ​ (allowable shear from table = 100 MPa). Which torque to use? If the shaft AK is the drive shaft upstream of the final drive, the torque in the shaft is engine torque × gearbox ratio × transmission efficiency. Using the numbers we already used: ? ? ℎ ? ? ? = ? ? ? ? ? ? ? ⋅ ? ? ⋅ ? = 229.18 × 1.8 × 0.92 ≈ 379.5   N \cdotp m T shaft ​ =T engine ​ ⋅i g ​ ⋅η=229.18×1.8×0.92≈379.5 N\cdotpm Given ? ? = 300   N \cdotp m M b ​ =300 N\cdotpm, compute ? ? ? = 300 2 + 379.5 2 ≈ 485.9   N \cdotp m M eq ​ = 300 2 +379.5 2 ​ ≈485.9 N\cdotpm Solve for diameter ? d from ? ? ? ? ? ? = 100  MPa = 16 ? ? ? ? ? 3 τ allow ​ =100 MPa= πd 3 16M eq ​ ​ : ? = ( 16 ? ? ? ? ? ? ? ? ? ? ) 1 / 3 d=( πτ allow ​ 16M eq ​ ​ ) 1/3 Plugging numbers gives: ? ≈ 0.0291  m = 29.1  mm d≈0.0291 m=29.1 mm Answer (4 — estimate): Required solid shaft diameter ≈ 29.1 mm (minimum). (If the shaft is hollow, dimensioning changes — use polar moment for a hollow shaft. Also if you include stress concentration factor ? ? K t ​ or fatigue notch sensitivity, increase diameter appropriately; I used ? ? ? ? ? ? = 100 τ allow ​ =100 MPa already from the table (which effectively folded design factor in).) Compare to the table: the table lists a drive-shaft outer diameter 80 mm and inner 65 mm (that’s a very thick hollow shaft — far above required minimal solid shaft size). So the provided shaft geometry is more than adequate under these simplified assumptions. 5) Drive shaft length Not enough geometric data. You need the vehicle/differential/wheelbase layout or the locations where bearings/supports are mounted to compute the drive-shaft length. Typical method: measure distance between gearbox output centerline and differential input (or between flange centers). If you want I can estimate a typical length given vehicle type, but the exam question generally expects a number from the vehicle geometry — that value is not in the two tables. 6) Shaft bearing sizes at point C (method + example rating) We can compute the radial load on the bearing from the reaction forces and then find the dynamic load rating ? C required for the desired life. Given: Reaction forces ? ? = 1150  N ,    ? ? = 1300  N F x ​ =1150 N,F y ​ =1300 N. Resultant radial load ? P: ? = ? ? 2 + ? ? 2 = 1150 2 + 1300 2 ≈ 1735.7  N P= F x 2 ​ +F y 2 ​ ​ = 1150 2 +1300 2 ​ ≈1735.7 N Use bearing life formula (ball bearing approximation, exponent ? = 3 p=3): L10 life in millions of revolutions: ? 10 = ( ? ? ) ? L 10 ​ =( P C ​ ) p Convert required life 28,000 h to revolutions: revs = 60 ⋅ ? ⋅ ℎ 60⋅n⋅h where ? n = shaft speed in rpm at the bearing. The table gives bearing design life 28000 h, and a rated life ? 10 = 10 6 L 10 ​ =10 6 (notation in table is ambiguous). We need shaft speed at that bearing. If we assume the shaft rotates at (engine speed / gearbox ratio) ≈ 5000 / 1.8 ≈ 2778  rpm 5000/1.8≈2778 rpm (example assumption), then ? = ? ⋅ ( 60 ? ℎ 10 6 ) 1 / ? C=P⋅( 10 6 60nh ​ ) 1/p Using ? ≈ 2778  rpm ,    ℎ = 28000  h ,    ? = 3 n≈2778 rpm,h=28000 h,p=3 gives (numeric): ? ≈ 29,000  N C≈29,000 N Answer (6 — example): You should select a bearing with a dynamic load rating ? ≥ 29   ? ? C≥29 kN and bore diameter matching your shaft (the shaft estimate above was ≈29 mm). In practice pick a standard bearing whose bore matches the shaft and whose catalog dynamic rating ≥ 29 kN. (Exact part number depends on vendor catalog and whether you choose deep-groove ball, cylindrical roller, etc.) Summary / Final notes Solid numeric results I calculated (direct from the table): Engine max torque ≈ 229.2 N·m. Tractive force at wheel ≈ 3795 N (≈ 3.80 kN). Clutch slipping: I gave the energy formula and an illustrative worst-case numeric (≈20.6 kJ and ≈0.7 s) using flywheel ? ? = 0.6 I f ​ =0.6 and a few assumptions. Real answer needs initial & final angular speeds and clutch slip torque capacity. Shaft diameter: Using combined bending (300 N·m) and shaft torque (≈379.5 N·m) gives an estimated minimum solid shaft diameter ≈ 29.1 mm (we used allowable shear = 100 MPa from the table). Bearing sizing: For the reaction load from the table and assumed shaft rpm ≈ 2778 rpm, a dynamic rating ? ≥ 29  kN C≥29 kN is indicated for 28,000 h life. If you want, I can now: compute clutch slipping energy/time with different (explicit) assumptions for initial/final speeds or clutch torque capacity, or compute a hollow-shaft diameter (given outer/inner geometry) and check safety factors including stress concentration factor ? ? K t ​ , or propose specific standard bearing part numbers once you tell me the exact shaft diameter and required bearing type (ball/roller).