By the end of this lesson, you will be able to:
High-precision CNC machines require extremely rigid bed structures to maintain accuracy during cutting operations. Modern CNC beds often use composite construction - combining steel reinforcement with aluminum casting to achieve optimal stiffness, weight, and thermal stability while managing cost.
CNC Machine Bed Components:
During machining operations, the CNC bed experiences:
Engineering Question: How do we analyze the stress distribution in a composite CNC bed where steel reinforcement beams are encased in aluminum, and how do we ensure both materials work together effectively under bending loads?
Consequences of Poor Composite Design:
Benefits of Proper Composite Analysis:
For composite beams with perfect bonding between materials:
Since different materials have different elastic moduli, we transform the composite section into an equivalent single-material section:
🔄 Transformation Ratio Formula
Where:
Physical Meaning: The transformation ratio allows us to convert a composite beam into an equivalent single-material beam by adjusting the width of one material based on stiffness differences.
Transform one material to the reference material:
🎯 Neutral Axis Location
Where:
Physical Meaning: The neutral axis location is found using the weighted average of transformed areas.
Calculate moment of inertia of transformed section:
Using parallel axis theorem
⚡ Composite Beam Stress Formulas
Reference material stress:
Transformed material stress:
Where:
Physical Meaning: Stresses in composite beams are calculated using the transformed section, with the transformed material stress adjusted by the transformation ratio.
Let’s analyze a realistic CNC bed cross-section step by step.
System Parameters:
w = 50 kN uniform distributed loadSafety factor: 2.0Choose aluminum as reference material:
Transform steel section to equivalent aluminum:
Analysis approach:
Analyze transformed section as all-aluminum, then convert steel stresses using transformation ratio
Aluminum region (original):
Steel region (transformed to aluminum):
Combined centroid location:
Neutral axis at geometric center (125 mm from bottom) due to symmetry
Maximum bending moment:
For simply supported beam with uniform load:
Transformed section moment of inertia:
Maximum distance from neutral axis:
c = 125 mm (to top or bottom fiber)
Stress in aluminum regions:
Stress in steel regions:
Safety factor assessment:
Load distribution analysis:
Steel carries 2.86× more stress per unit strain, efficiently using its higher strength
A composite beam consists of a steel I-beam with a concrete slab cast on top, used in building construction.
Given:
Find: Maximum stresses in both steel and concrete.
✓ Calculate transformation ratio
Using steel as reference material: n = E_concrete/E_steel = 25/200 = 0.125 Transformed concrete width: b_t = n × 1000 = 125 mm
✓ Locate neutral axis of composite section
For steel I-beam: A₁ = 4580 mm², ȳ₁ = 100 mm (from bottom) For concrete slab: A₂ = 125 × 100 = 12,500 mm², ȳ₂ = 250 mm (from bottom)
ȳ = (A₁ȳ₁ + A₂ȳ₂)/(A₁ + A₂) = (4580×100 + 12,500×250)/(4580 + 12,500) = 202 mm
✓ Calculate composite moment of inertia
Using parallel axis theorem for transformed section: I_composite = I_steel + A₁d₁² + A₂d₂² I_composite = 40.5×10⁶ + 4580×(202-100)² + 12,500×(250-202)² I_composite = 40.5×10⁶ + 47.6×10⁶ + 28.8×10⁶ = 116.9×10⁶ mm⁴
✓ Calculate maximum bending moment
For simply supported beam: M_max = wL²/8 = 15×6²/8 = 67.5 kN·m
✓ Calculate stresses
Steel (bottom): σ_s = M×c_s/I = 67.5×10⁶×202/116.9×10⁶ = 116.7 MPa Concrete (top): σ_c = (M×c_c/I) = 67.5×10⁶×48/116.9×10⁶ = 27.7 MPa (before transformation correction)
A lightweight beam combines aluminum flanges with a steel web for optimal strength-to-weight ratio in aerospace applications.
Given:
Find: Maximum bending stresses in steel and aluminum components.
✓ Transform aluminum to equivalent steel
Transformation ratio: n = E_aluminum/E_steel = 70/200 = 0.35 Transformed aluminum flange width: b_t = 150 × 0.35 = 52.5 mm
✓ Calculate transformed section properties
Due to symmetry, neutral axis is at mid-height (120 mm from bottom)
✓ Calculate moment of inertia of transformed section
I_transformed = I_web + I_flanges + Ad² terms I_transformed = (8×200³)/12 + 2×[(52.5×20³)/12 + 52.5×20×100²] I_transformed = 5.33×10⁶ + 2×[35,000 + 10.5×10⁶] = 26.4×10⁶ mm⁴
✓ Calculate maximum bending moment
For cantilever: M_max = P × L = 5000 × 1200 = 6×10⁶ N·mm
✓ Calculate stresses in both materials
Steel web (at neutral axis level): σ_steel = 0 MPa Aluminum flanges: σ_aluminum = M×c/I_transformed × n⁻¹ σ_aluminum = (6×10⁶ × 120)/(26.4×10⁶) × (1/0.35) = 78.0 MPa
A high-performance beam uses carbon fiber reinforcement bonded to aluminum structure for maximum stiffness with minimal weight.
Given:
Find: Stress distribution and maximum stresses in each material.
✓ Calculate transformation ratio
Using aluminum as reference: n = E_CF/E_Al = 150/70 = 2.14 Carbon fiber strips transform to equivalent aluminum: b_t = 60 × 2.14 = 128.6 mm
✓ Locate composite neutral axis
Aluminum beam: A₁ = 60 × 80 = 4800 mm², ȳ₁ = 40 mm CF strips: A₂ = 2 × (128.6 × 3) = 771.6 mm², ȳ₂ = 42.5 mm (average of top/bottom)
Due to near symmetry, neutral axis ≈ 40 mm from bottom
✓ Calculate composite moment of inertia
I_aluminum = 60×80³/12 = 2.56×10⁶ mm⁴ I_CF = 2×[128.6×3³/12 + 128.6×3×38.5²] = 2×[96.4 + 57,200] = 114,600 mm⁴ I_total = 2.56×10⁶ + 0.115×10⁶ = 2.675×10⁶ mm⁴
✓ Calculate maximum bending moment
M_max = wL²/8 = 2000 × 0.8²/8 = 160 N·m
✓ Determine maximum stresses
Aluminum: σ_Al = Mc/I = 160×10³ × 40/(2.675×10⁶) = 2.39 MPa Carbon fiber: σ_CF = n × Mc/I = 2.14 × 160×10³ × 41.5/(2.675×10⁶) = 8.46 MPa
In this lesson, you learned to:
Key Design Insights:
Critical Formula:
Coming Next: In Lesson 2.6, we’ll analyze principal stresses and failure criteria for critical stress evaluation in mechatronic joint design using Mohr’s circle analysis.
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