Machine Design

Machine Design involves selecting and sizing mechanical components to transmit power, handle loads, and function reliably. This brings together mechanics, materials, and manufacturing knowledge.

Design Process

1. Define Requirements (loads, speeds, life, environment)
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2. Conceptual Design (sketches, layout)
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3. Preliminary Design (calculations, selection)
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4. Detailed Design (CAD, specifications)
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5. Analysis (FEA, testing)
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6. Manufacturing Documents (drawings, tolerances)

Power Transmission

Power, Torque, and Speed

Fundamental relationship:

P = Tω = T × (2πN/60)

Where:
- P = power (W)
- T = torque (N⋅m)
- ω = angular velocity (rad/s)
- N = rotational speed (rpm)

Convert between units:

P [W] = T [N⋅m] × N [rpm] × 2π/60
P [kW] = T [N⋅m] × N [rpm] / 9549

Or:
P [hp] = T [lb⋅ft] × N [rpm] / 5252

Example: Motor delivers 50 N⋅m at 1500 rpm. Find power.

P = (50 × 1500) / 9549 = 7.85 kW

Shafts

Shafts transmit rotational power and support rotating elements.

Shaft Design Criteria

1. Torsional stress (from torque)
2. Bending stress (from radial loads)
3. Deflection (avoid excessive bending)
4. Critical speed (avoid resonance)

Torsional Stress in Shafts

For solid circular shaft:

τ = 16T / (πd³)

Where:
- τ = shear stress (Pa)
- T = torque (N⋅m)
- d = shaft diameter (m)

Design equation (with factor of safety):

d = ∛(16T × FOS / (π × τ_allow))

Combined Loading

Shafts often experience both bending and torsion:

Equivalent torque (von Mises):

T_eq = √(M² + T²)

Where:
- M = bending moment (N⋅m)
- T = torque (N⋅m)

Then design for T_eq using torsion formula.

Example: Shaft Sizing

A shaft transmits 10 kW at 1000 rpm, allowable shear stress 50 MPa, FOS = 2.

T = P × 9549 / N = 10 × 9549 / 1000 = 95.49 N⋅m

τ_design = τ_allow / FOS = 50 / 2 = 25 MPa = 25 × 10⁶ Pa

d = ∛(16 × 95.49 / (π × 25 × 10⁶))
d = ∛(0.0000195) = 0.027 m = 27 mm

Use standard size: 30 mm diameter shaft

Keys and Keyways

Keys prevent relative rotation between shaft and hub (gears, pulleys).

Types of Keys

Key Shear Stress

τ = 2T / (d × w × L)

Where:
- T = torque (N⋅m)
- d = shaft diameter (m)
- w = key width (m)
- L = key length (m)

Design: ensure τ < τ_allow with appropriate FOS.

Gears

Gears transmit power between rotating shafts with specific speed ratios.

Gear Ratio

Gear Ratio (GR) = N_driver / N_driven = D_driven / D_driver = T_driven / T_driver

Where:
- N = speed (rpm)
- D = pitch diameter
- T = number of teeth

Speed reduction: GR > 1 (driven slower than driver)
Speed increase: GR < 1 (driven faster than driver)

Example: Gear Train

Input: 1500 rpm, 20 teeth
Output gear: 60 teeth

GR = 60/20 = 3:1
N_output = N_input / GR = 1500 / 3 = 500 rpm

If input torque = 30 N⋅m:
T_output = T_input × GR = 30 × 3 = 90 N⋅m (ideally, neglecting losses)

Types of Gears

Gear Tooth Stress

Bending stress (Lewis equation):

σ = W_t / (F × m × Y)

Where:
- W_t = tangential force on tooth (N)
- F = face width (mm)
- m = module (mm)
- Y = Lewis form factor (from tables)

Contact stress (Hertz): Complex formula, use gear design software or standards (AGMA).

Gear Materials

• Steel: Most common (high strength)
• Cast iron: Low cost, good for low speed
• Bronze: For worm wheels (low friction)
• Plastics: Quiet, low load

Bearings

Bearings support rotating shafts and reduce friction.

Types of Bearings

Rolling Element Bearings:

Plain Bearings (Sleeve/Journal):

• Simpler, lower cost
• Requires lubrication
• Good for high loads, low speed

Bearing Life

Rolling bearings have finite fatigue life:

L₁₀ = (C/P)ᵖ

Where:
- L₁₀ = rating life (millions of revolutions)
- C = dynamic load rating (N, from catalog)
- P = equivalent dynamic load (N)
- p = 3 for ball bearings, 10/3 for roller bearings

Life in hours:

L_h = (10⁶ × L₁₀) / (60 × N)

Where N = speed (rpm)

Example: Bearing Selection

Bearing load = 5000 N, speed = 1000 rpm, desired life = 10,000 hours.

L₁₀ = (L_h × 60 × N) / 10⁶ = (10,000 × 60 × 1000) / 10⁶ = 600 million rev

C = P × ∛(L₁₀) = 5000 × ∛600 = 5000 × 8.43 = 42,150 N

Select bearing from catalog with C > 42,150 N

Lubrication

Essential for bearing life:

• Grease: Common, simple, retains in place
• Oil: Better cooling, continuous supply needed
• Solid lubricants: High temp, space applications

Fasteners

Bolts and Screws

Tensile stress in bolt:

σ = F / A_tensile

Where A_tensile = π d_minor²/4 (based on minor diameter of threads)

Torque-tension relationship:

T = K × F × d

Where:
- T = applied torque (N⋅m)
- K = nut factor (≈ 0.2 for lubricated steel)
- F = clamping force (N)
- d = nominal diameter (m)

Bolt Preload

Tightening bolt creates clamping force. Properly preloaded joint prevents:

• Fatigue failures
• Loosening under vibration
• Gapping under external load

Typical preload: 75-90% of bolt proof strength for static loads.

Thread Standards

Springs

Store and release mechanical energy.

Compression Spring

Spring rate (stiffness):

k = Gd⁴ / (8D³N_a)

Where:
- k = spring rate (N/m)
- G = shear modulus (Pa)
- d = wire diameter (m)
- D = mean coil diameter (m)
- N_a = number of active coils

Force-deflection:

F = k × δ

Where δ = deflection from free length (m)

Shear Stress in Spring

τ = 8FD / (πd³) × K_w

Where K_w = Wahl correction factor (accounts for curvature)

Spring Design

Select:

1. Wire diameter (d)
2. Coil diameter (D) or spring index (C = D/d, typically 4-12)
3. Number of coils (N)
4. Free length

Check: stress, solid length, buckling

Couplings and Clutches

Rigid Couplings

Connect aligned shafts permanently:

• Sleeve coupling
• Flange coupling

Flexible Couplings

Accommodate misalignment:

• Universal joint (Hooke's joint)
• Oldham coupling
• Rubber/elastomer coupling

Clutches

Engage/disengage power transmission:

• Friction clutch (automotive)
• Dog clutch (positive engagement)
• Electromagnetic clutch

Practice Problems

Problem 1: Power and Torque

A motor outputs 15 kW at 3000 rpm. Find torque.

<details> <summary>Solution</summary>

T = P × 9549 / N = 15 × 9549 / 3000 = 47.7 N⋅m

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Problem 2: Gear Ratio

Driver gear (30 teeth) at 1200 rpm drives gear with 90 teeth. Find output speed and torque if input torque is 50 N⋅m.

<details> <summary>Solution</summary>

GR = 90/30 = 3
N_output = 1200/3 = 400 rpm
T_output = 50 × 3 = 150 N⋅m

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Problem 3: Shaft Diameter

Solid shaft transmits 20 kW at 500 rpm, τ_allow = 40 MPa. Find minimum diameter.

<details> <summary>Solution</summary>

T = 20 × 9549 / 500 = 382 N⋅m
d = ∛(16T / (π × τ)) = ∛(16 × 382 / (π × 40 × 10⁶))
d = ∛(0.0000486) = 0.0365 m = 36.5 mm

Use 40 mm standard shaft

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Key Takeaways

✓ Power transmission: P = Tω, fundamental to all rotating machinery
✓ Shafts: Design for torsion, bending, and combined loads
✓ Gears: Change speed and torque, maintain power (minus losses)
✓ Bearings: Support loads, have finite fatigue life
✓ Fasteners: Proper preload critical for joint integrity
✓ Springs: k = F/δ, design for stress and deflection

Next Steps

Design requires understanding manufacturing:

• Chapter 08: Manufacturing Processes: how parts are made
• Learn CAD (SolidWorks, Fusion 360)
• Study machine design standards (AGMA, ISO, DIN)

Pro Tip: Always use standard sizes for shafts, bearings, fasteners. This simplifies procurement and reduces cost.