Nice — this is a classic “is the pin & yoke safe?” homework / design check. I don’t have your SolidWorks dimensions, so I’ll make conservative, explicit assumptions and show the full analytical work (equations, numbers, and sources for the stress-concentration and Hertz contact equations). At the end I give FEA setup guidance, compare analytic vs FEA expectations, and concrete design recommendations you can act on (or re-run with your actual dimensions). 0) Assumptions (stated explicitly so you can replace them) Pin (journal) is solid circular, diameter d = 25 mm. Yoke bears the pin on two opposing arms; bearing (contact) length per arm L = 20 mm (I also show results for L = 40 mm for comparison). Fillet radius at the pin shoulder / bore root r = 1.5 mm. Material: Alloy steel, E = 210 GPa, ν = 0.30, Yield σ_y = 550 MPa. (You gave these — used directly.) Loads: steady torque T = 110 N·m, compressive axial load F = 8 kN (service) and F = 16 kN (shock). Load shared by two yoke arms → each arm reaction ≈ F/2. Contacts treated elastic (Hertz) for contact stresses; all contacts otherwise ideal (rigid mating) for strength checks. Conservative stress concentration factors are used from Peterson / Mechanical Design charts — I cite the references and explain chosen values below. Replace d, L, r with your real CAD values to get exact numbers. I used these because they’re typical for small driveshaft universal-joint pins. 1) Simple section-strength checks (pin under combined torsion + axial compression) 1.1 Equations (nominal stresses) Axial (compressive) normal stress: ? ? = ? ? = ? ? ? 2 / 4 σ a = A F = πd 2 /4 F . Torsional shear (solid circular shaft, max shear at surface): ? ? = 16 ? ? ? 3 τ t = πd 3 16T . Combined von Mises (axial + torsion): ? ? ? = ? ? 2 + 3 ? ? 2 σ vm = σ a 2 +3τ t 2 . Apply a geometric stress-concentration factor ? ? K t (fillet) to the local peak stress: ? ? ? , ? ? ? = ? ? ⋅ ? ? ? σ vm,max =K t ⋅σ vm . 1.2 Numbers (with d = 25 mm) (Work shown step-by-step, calculated exactly.) Cross-sectional area ? = ? ? 2 / 4 = ? ( 0.025 2 ) / 4 = 4.91 × 10 − 4 m 2 . A=πd 2 /4=π(0.025 2 )/4=4.91×10 −4 m 2 . Axial stress (service): ? ? ( ? ) = 8000 / ? = 16.30 MPa . σ a (s) =8000/A=16.30 MPa. Axial stress (shock): ? ? ( ? ℎ ) = 16000 / ? = 32.59 MPa . σ a (sh) =16000/A=32.59 MPa. Torsional shear (same for both cases): ? ? = 16 ? / ( ? ? 3 ) = 35.85 MPa . τ t =16T/(πd 3 )=35.85 MPa. Nominal von Mises (service): ? ? ? = 16.30 2 + 3 ( 35.85 ) 2 ≈ 64.2 MPa . σ vm = 16.30 2 +3(35.85) 2 ≈64.2 MPa. Nominal von Mises (shock axial): ≈ 70.1 MPa . ≈70.1 MPa. 1.3 Choose a conservative stress-concentration factor ? ? K t For a shoulder fillet on a round shaft the charts in Peterson / Mechanical Engineering Design give ? ? K t values that depend on ? / ? r/d and step ratio. For small fillet ratios (r/d ≈ 0.06) a conservative ? ? ≈ 2.0 K t ≈2.0 (bending/torsion sensitive regions) is reasonable to use for a worst-case local peak at the fillet root. (Sources: Peterson charts and summary tables). CloudFront IIT Delhi Web Services 1.4 Apply Kt and safety factors Max local von Mises (service): ? ? ? , ? ? ? ≈ 2.0 × 64.2 ≈ 128.4 MPa . σ vm,max ≈2.0×64.2≈128.4 MPa. Max local von Mises (shock): ? ? ? , ? ? ? ≈ 2.0 × 70.1 ≈ 140.3 MPa . σ vm,max ≈2.0×70.1≈140.3 MPa. Safety factors vs yield (static yielding): Service: ? ? = ? ? / ? ? ? , ? ? ? = 550 / 128.4 ≈ 4.3. SF=σ y /σ vm,max =550/128.4≈4.3. Shock: ? ? ≈ 550 / 140.3 ≈ 3.9. SF≈550/140.3≈3.9. Conclusion (section strength): with the chosen geometry (d = 25 mm) and a conservative ? ? = 2 K t =2, the pin cross-section + fillet region has a comfortable static safety factor (~4) both for service and shock. So at the shaft/fillet level the material will not plastically yield under the given loads. (Reference: Peterson for Kt charts). CloudFront 2) Bearing / contact stress (Hertz) at the pin — the critical risk The pin transfers the axial compressive load to the yoke via a small contact area — this is where Hertz contact stresses (very high local pressures) govern. 2.1 Hertz line-contact (parallel cylinders) equations used For two parallel cylinders pressed with force per unit length ? P (N/m): reduced radius: 1 ? = 1 ? 1 + 1 ? 2 R 1 = R 1 1 + R 2 1 (for cylinder–cylinder). reduced modulus: 1 ? ∗ = 1 − ? 1 2 ? 1 + 1 − ? 2 2 ? 2 . E ∗ 1 = E 1 1−ν 1 2 + E 2 1−ν 2 2 . contact half-width: ? = 4 ? ? ? ? ∗ a= πE ∗ 4PR . maximum contact pressure (centerline): ? 0 = 2 ? ? ? . p 0 = πa 2P . for steel-to-steel ( ? = 210 GPa , ? ≈ 0.3 ) (E=210 GPa,ν≈0.3) these formulas are directly applicable. (Sources: standard Hertzian contact references / tutorials). Optics at Arizona e6.ijs.si 2.2 How I applied them (numbers) Each arm load (service): ? ? ? ? = ? / 2 = 4000 N F arm =F/2=4000 N. If the bearing length is ? L (m), the line-load ? = ? ? ? ? / ? P=F arm /L (N/m). I compute two cases: L = 20 mm (compact yoke) and L = 40 mm (longer bearing). Pin radius ? 1 = ? / 2 = 12.5 R 1 =d/2=12.5 mm. For a tight circular bore (yoke contacting pin) use ? 2 = ? 1 R 2 =R 1 (cylinder–cylinder). If the yoke inner surface is effectively flat relative to the pin, use ? 2 = ∞ R 2 =∞ (cylinder-on-flat) — I show the conservative cylinder–cylinder case (smaller effective radius, higher pressure). Case A — L = 20 mm, two identical cylinders (R1=R2=12.5 mm) Computed p₀ ≈ 1.08 × 10⁹ Pa = 1,084 MPa (maximum contact pressure at center). Approximate maximum subsurface shear (commonly used approximation for cylinder line contact): ? ? ? ? ≈ 0.3 ? 0 ≈ 325 MPa . τ max ≈0.3p 0 ≈325 MPa. Case B — L = 40 mm (same geometry otherwise) Computed p₀ ≈ 542 MPa and ? ? ? ? ≈ 163 MPa . τ max ≈163 MPa. (Details: I used the equations above and the standard formulae from Hertzian contact references.) Optics at Arizona andrew.cmu.edu 2.3 Yielding comparison (contact) For yielding under shear the material shear yield ≈ ? ? / 3 ≈ 550 / 3 ≈ 318 MPa . σ y / 3 ≈550/ 3 ≈318 MPa. Case A (L = 20 mm): ? ? ? ? ≈ 325 MPa > 318 MPa τ max ≈325 MPa>318 MPa → contact is at or slightly above shear yield → plasticity at contact / local indentation very likely (bad). Case B (L = 40 mm): ? ? ? ? ≈ 163 MPa < 318 MPa τ max ≈163 MPa
Question:
Nice — this is a classic “is the pin & yoke safe?” homework / design check. I don’t have your SolidWorks dimensions, so I’ll make conservative, explicit assumptions and show the full analytical work (equations, numbers, and sources for the stress-concentration and Hertz contact equations). At the end I give FEA setup guidance, compare analytic vs FEA expectations, and concrete design recommendations you can act on (or re-run with your actual dimensions). 0) Assumptions (stated explicitly so you can replace them) Pin (journal) is solid circular, diameter d = 25 mm. Yoke bears the pin on two opposing arms; bearing (contact) length per arm L = 20 mm (I also show results for L = 40 mm for comparison). Fillet radius at the pin shoulder / bore root r = 1.5 mm. Material: Alloy steel, E = 210 GPa, ν = 0.30, Yield σ_y = 550 MPa. (You gave these — used directly.) Loads: steady torque T = 110 N·m, compressive axial load F = 8 kN (service) and F = 16 kN (shock). Load shared by two yoke arms → each arm reaction ≈ F/2. Contacts treated elastic (Hertz) for contact stresses; all contacts otherwise ideal (rigid mating) for strength checks. Conservative stress concentration factors are used from Peterson / Mechanical Design charts — I cite the references and explain chosen values below. Replace d, L, r with your real CAD values to get exact numbers. I used these because they’re typical for small driveshaft universal-joint pins. 1) Simple section-strength checks (pin under combined torsion + axial compression) 1.1 Equations (nominal stresses) Axial (compressive) normal stress: ? ? = ? ? = ? ? ? 2 / 4 σ a = A F = πd 2 /4 F . Torsional shear (solid circular shaft, max shear at surface): ? ? = 16 ? ? ? 3 τ t = πd 3 16T . Combined von Mises (axial + torsion): ? ? ? = ? ? 2 + 3 ? ? 2 σ vm = σ a 2 +3τ t 2 . Apply a geometric stress-concentration factor ? ? K t (fillet) to the local peak stress: ? ? ? , ? ? ? = ? ? ⋅ ? ? ? σ vm,max =K t ⋅σ vm . 1.2 Numbers (with d = 25 mm) (Work shown step-by-step, calculated exactly.) Cross-sectional area ? = ? ? 2 / 4 = ? ( 0.025 2 ) / 4 = 4.91 × 10 − 4 m 2 . A=πd 2 /4=π(0.025 2 )/4=4.91×10 −4 m 2 . Axial stress (service): ? ? ( ? ) = 8000 / ? = 16.30 MPa . σ a (s) =8000/A=16.30 MPa. Axial stress (shock): ? ? ( ? ℎ ) = 16000 / ? = 32.59 MPa . σ a (sh) =16000/A=32.59 MPa. Torsional shear (same for both cases): ? ? = 16 ? / ( ? ? 3 ) = 35.85 MPa . τ t =16T/(πd 3 )=35.85 MPa. Nominal von Mises (service): ? ? ? = 16.30 2 + 3 ( 35.85 ) 2 ≈ 64.2 MPa . σ vm = 16.30 2 +3(35.85) 2 ≈64.2 MPa. Nominal von Mises (shock axial): ≈ 70.1 MPa . ≈70.1 MPa. 1.3 Choose a conservative stress-concentration factor ? ? K t For a shoulder fillet on a round shaft the charts in Peterson / Mechanical Engineering Design give ? ? K t values that depend on ? / ? r/d and step ratio. For small fillet ratios (r/d ≈ 0.06) a conservative ? ? ≈ 2.0 K t ≈2.0 (bending/torsion sensitive regions) is reasonable to use for a worst-case local peak at the fillet root. (Sources: Peterson charts and summary tables). CloudFront IIT Delhi Web Services 1.4 Apply Kt and safety factors Max local von Mises (service): ? ? ? , ? ? ? ≈ 2.0 × 64.2 ≈ 128.4 MPa . σ vm,max ≈2.0×64.2≈128.4 MPa. Max local von Mises (shock): ? ? ? , ? ? ? ≈ 2.0 × 70.1 ≈ 140.3 MPa . σ vm,max ≈2.0×70.1≈140.3 MPa. Safety factors vs yield (static yielding): Service: ? ? = ? ? / ? ? ? , ? ? ? = 550 / 128.4 ≈ 4.3. SF=σ y /σ vm,max =550/128.4≈4.3. Shock: ? ? ≈ 550 / 140.3 ≈ 3.9. SF≈550/140.3≈3.9. Conclusion (section strength): with the chosen geometry (d = 25 mm) and a conservative ? ? = 2 K t =2, the pin cross-section + fillet region has a comfortable static safety factor (~4) both for service and shock. So at the shaft/fillet level the material will not plastically yield under the given loads. (Reference: Peterson for Kt charts). CloudFront 2) Bearing / contact stress (Hertz) at the pin — the critical risk The pin transfers the axial compressive load to the yoke via a small contact area — this is where Hertz contact stresses (very high local pressures) govern. 2.1 Hertz line-contact (parallel cylinders) equations used For two parallel cylinders pressed with force per unit length ? P (N/m): reduced radius: 1 ? = 1 ? 1 + 1 ? 2 R 1 = R 1 1 + R 2 1 (for cylinder–cylinder). reduced modulus: 1 ? ∗ = 1 − ? 1 2 ? 1 + 1 − ? 2 2 ? 2 . E ∗ 1 = E 1 1−ν 1 2 + E 2 1−ν 2 2 . contact half-width: ? = 4 ? ? ? ? ∗ a= πE ∗ 4PR . maximum contact pressure (centerline): ? 0 = 2 ? ? ? . p 0 = πa 2P . for steel-to-steel ( ? = 210 GPa , ? ≈ 0.3 ) (E=210 GPa,ν≈0.3) these formulas are directly applicable. (Sources: standard Hertzian contact references / tutorials). Optics at Arizona e6.ijs.si 2.2 How I applied them (numbers) Each arm load (service): ? ? ? ? = ? / 2 = 4000 N F arm =F/2=4000 N. If the bearing length is ? L (m), the line-load ? = ? ? ? ? / ? P=F arm /L (N/m). I compute two cases: L = 20 mm (compact yoke) and L = 40 mm (longer bearing). Pin radius ? 1 = ? / 2 = 12.5 R 1 =d/2=12.5 mm. For a tight circular bore (yoke contacting pin) use ? 2 = ? 1 R 2 =R 1 (cylinder–cylinder). If the yoke inner surface is effectively flat relative to the pin, use ? 2 = ∞ R 2 =∞ (cylinder-on-flat) — I show the conservative cylinder–cylinder case (smaller effective radius, higher pressure). Case A — L = 20 mm, two identical cylinders (R1=R2=12.5 mm) Computed p₀ ≈ 1.08 × 10⁹ Pa = 1,084 MPa (maximum contact pressure at center). Approximate maximum subsurface shear (commonly used approximation for cylinder line contact): ? ? ? ? ≈ 0.3 ? 0 ≈ 325 MPa . τ max ≈0.3p 0 ≈325 MPa. Case B — L = 40 mm (same geometry otherwise) Computed p₀ ≈ 542 MPa and ? ? ? ? ≈ 163 MPa . τ max ≈163 MPa. (Details: I used the equations above and the standard formulae from Hertzian contact references.) Optics at Arizona andrew.cmu.edu 2.3 Yielding comparison (contact) For yielding under shear the material shear yield ≈ ? ? / 3 ≈ 550 / 3 ≈ 318 MPa . σ y / 3 ≈550/ 3 ≈318 MPa. Case A (L = 20 mm): ? ? ? ? ≈ 325 MPa > 318 MPa τ max ≈325 MPa>318 MPa → contact is at or slightly above shear yield → plasticity at contact / local indentation very likely (bad). Case B (L = 40 mm): ? ? ? ? ≈ 163 MPa < 318 MPa τ max ≈163 MPa
Asked by: vara prasad
Created at: 2025-08-31 02:05:24
Final Answer:
Question:
Nice — this is a classic “is the pin & yoke safe?” homework / design check. I don’t have your SolidWorks dimensions, so I’ll make conservative, explicit assumptions and show the full analytical work (equations, numbers, and sources for the stress-concentration and Hertz contact equations). At the end I give FEA setup guidance, compare analytic vs FEA expectations, and concrete design recommendations you can act on (or re-run with your actual dimensions). 0) Assumptions (stated explicitly so you can replace them) Pin (journal) is solid circular, diameter d = 25 mm. Yoke bears the pin on two opposing arms; bearing (contact) length per arm L = 20 mm (I also show results for L = 40 mm for comparison). Fillet radius at the pin shoulder / bore root r = 1.5 mm. Material: Alloy steel, E = 210 GPa, ν = 0.30, Yield σ_y = 550 MPa. (You gave these — used directly.) Loads: steady torque T = 110 N·m, compressive axial load F = 8 kN (service) and F = 16 kN (shock). Load shared by two yoke arms → each arm reaction ≈ F/2. Contacts treated elastic (Hertz) for contact stresses; all contacts otherwise ideal (rigid mating) for strength checks. Conservative stress concentration factors are used from Peterson / Mechanical Design charts — I cite the references and explain chosen values below. Replace d, L, r with your real CAD values to get exact numbers. I used these because they’re typical for small driveshaft universal-joint pins. 1) Simple section-strength checks (pin under combined torsion + axial compression) 1.1 Equations (nominal stresses) Axial (compressive) normal stress: ? ? = ? ? = ? ? ? 2 / 4 σ a = A F = πd 2 /4 F . Torsional shear (solid circular shaft, max shear at surface): ? ? = 16 ? ? ? 3 τ t = πd 3 16T . Combined von Mises (axial + torsion): ? ? ? = ? ? 2 + 3 ? ? 2 σ vm = σ a 2 +3τ t 2 . Apply a geometric stress-concentration factor ? ? K t (fillet) to the local peak stress: ? ? ? , ? ? ? = ? ? ⋅ ? ? ? σ vm,max =K t ⋅σ vm . 1.2 Numbers (with d = 25 mm) (Work shown step-by-step, calculated exactly.) Cross-sectional area ? = ? ? 2 / 4 = ? ( 0.025 2 ) / 4 = 4.91 × 10 − 4 m 2 . A=πd 2 /4=π(0.025 2 )/4=4.91×10 −4 m 2 . Axial stress (service): ? ? ( ? ) = 8000 / ? = 16.30 MPa . σ a (s) =8000/A=16.30 MPa. Axial stress (shock): ? ? ( ? ℎ ) = 16000 / ? = 32.59 MPa . σ a (sh) =16000/A=32.59 MPa. Torsional shear (same for both cases): ? ? = 16 ? / ( ? ? 3 ) = 35.85 MPa . τ t =16T/(πd 3 )=35.85 MPa. Nominal von Mises (service): ? ? ? = 16.30 2 + 3 ( 35.85 ) 2 ≈ 64.2 MPa . σ vm = 16.30 2 +3(35.85) 2 ≈64.2 MPa. Nominal von Mises (shock axial): ≈ 70.1 MPa . ≈70.1 MPa. 1.3 Choose a conservative stress-concentration factor ? ? K t For a shoulder fillet on a round shaft the charts in Peterson / Mechanical Engineering Design give ? ? K t values that depend on ? / ? r/d and step ratio. For small fillet ratios (r/d ≈ 0.06) a conservative ? ? ≈ 2.0 K t ≈2.0 (bending/torsion sensitive regions) is reasonable to use for a worst-case local peak at the fillet root. (Sources: Peterson charts and summary tables). CloudFront IIT Delhi Web Services 1.4 Apply Kt and safety factors Max local von Mises (service): ? ? ? , ? ? ? ≈ 2.0 × 64.2 ≈ 128.4 MPa . σ vm,max ≈2.0×64.2≈128.4 MPa. Max local von Mises (shock): ? ? ? , ? ? ? ≈ 2.0 × 70.1 ≈ 140.3 MPa . σ vm,max ≈2.0×70.1≈140.3 MPa. Safety factors vs yield (static yielding): Service: ? ? = ? ? / ? ? ? , ? ? ? = 550 / 128.4 ≈ 4.3. SF=σ y /σ vm,max =550/128.4≈4.3. Shock: ? ? ≈ 550 / 140.3 ≈ 3.9. SF≈550/140.3≈3.9. Conclusion (section strength): with the chosen geometry (d = 25 mm) and a conservative ? ? = 2 K t =2, the pin cross-section + fillet region has a comfortable static safety factor (~4) both for service and shock. So at the shaft/fillet level the material will not plastically yield under the given loads. (Reference: Peterson for Kt charts). CloudFront 2) Bearing / contact stress (Hertz) at the pin — the critical risk The pin transfers the axial compressive load to the yoke via a small contact area — this is where Hertz contact stresses (very high local pressures) govern. 2.1 Hertz line-contact (parallel cylinders) equations used For two parallel cylinders pressed with force per unit length ? P (N/m): reduced radius: 1 ? = 1 ? 1 + 1 ? 2 R 1 = R 1 1 + R 2 1 (for cylinder–cylinder). reduced modulus: 1 ? ∗ = 1 − ? 1 2 ? 1 + 1 − ? 2 2 ? 2 . E ∗ 1 = E 1 1−ν 1 2 + E 2 1−ν 2 2 . contact half-width: ? = 4 ? ? ? ? ∗ a= πE ∗ 4PR . maximum contact pressure (centerline): ? 0 = 2 ? ? ? . p 0 = πa 2P . for steel-to-steel ( ? = 210 GPa , ? ≈ 0.3 ) (E=210 GPa,ν≈0.3) these formulas are directly applicable. (Sources: standard Hertzian contact references / tutorials). Optics at Arizona e6.ijs.si 2.2 How I applied them (numbers) Each arm load (service): ? ? ? ? = ? / 2 = 4000 N F arm =F/2=4000 N. If the bearing length is ? L (m), the line-load ? = ? ? ? ? / ? P=F arm /L (N/m). I compute two cases: L = 20 mm (compact yoke) and L = 40 mm (longer bearing). Pin radius ? 1 = ? / 2 = 12.5 R 1 =d/2=12.5 mm. For a tight circular bore (yoke contacting pin) use ? 2 = ? 1 R 2 =R 1 (cylinder–cylinder). If the yoke inner surface is effectively flat relative to the pin, use ? 2 = ∞ R 2 =∞ (cylinder-on-flat) — I show the conservative cylinder–cylinder case (smaller effective radius, higher pressure). Case A — L = 20 mm, two identical cylinders (R1=R2=12.5 mm) Computed p₀ ≈ 1.08 × 10⁹ Pa = 1,084 MPa (maximum contact pressure at center). Approximate maximum subsurface shear (commonly used approximation for cylinder line contact): ? ? ? ? ≈ 0.3 ? 0 ≈ 325 MPa . τ max ≈0.3p 0 ≈325 MPa. Case B — L = 40 mm (same geometry otherwise) Computed p₀ ≈ 542 MPa and ? ? ? ? ≈ 163 MPa . τ max ≈163 MPa. (Details: I used the equations above and the standard formulae from Hertzian contact references.) Optics at Arizona andrew.cmu.edu 2.3 Yielding comparison (contact) For yielding under shear the material shear yield ≈ ? ? / 3 ≈ 550 / 3 ≈ 318 MPa . σ y / 3 ≈550/ 3 ≈318 MPa. Case A (L = 20 mm): ? ? ? ? ≈ 325 MPa > 318 MPa τ max ≈325 MPa>318 MPa → contact is at or slightly above shear yield → plasticity at contact / local indentation very likely (bad). Case B (L = 40 mm): ? ? ? ? ≈ 163 MPa < 318 MPa τ max ≈163 MPa
Asked by: vara prasad
Created at: 2025-08-31 02:06:53
Final Answer:
Question:
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).
Asked by: vara prasad
Created at: 2025-08-31 02:14:11
Final Answer:
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :
VIPSolutions