Learn combined stress analysis through robotic wrist joints, covering principal stress theory, equivalent stress calculations, and failure prediction under simultaneous bending and torsion loading.
By the end of this lesson, you will be able to:
Robotic wrist joints represent one of the most complex loading scenarios in mechatronics. The wrist must simultaneously handle bending moments from tool and payload weights, torsional moments from tool rotation and orientation changes, and dynamic loads from rapid positioning movements.
6-DOF Robotic Wrist Components:
During industrial operations, the wrist joint simultaneously experiences:
Engineering Question: How do we analyze a robotic wrist shaft that experiences both 150 N·m bending moment and 200 N·m torque simultaneously, and determine if this combination will cause failure?
Consequences of Inadequate Analysis:
Benefits of Proper Combined Analysis:
When multiple loads act simultaneously, the stress state becomes more complex:
📊 Combined Normal Stress Formula
Individual components:
Combined normal stress:
Physical Meaning: Normal stresses from different loading types are algebraically combined, with tension positive and compression negative.
🌀 Combined Shear Stress Formula
Individual components:
Combined shear stress:
Physical Meaning: Shear stresses from different sources are vectorially combined at each point in the cross-section.
🔱 Principal Stress Formula
Where:
Physical Meaning: Principal stresses represent the maximum and minimum normal stresses at a point, occurring on planes with zero shear stress.
🌀 Maximum Shear Stress
In-plane maximum shear:
Absolute maximum shear:
Physical Meaning: Maximum shear stress occurs on planes oriented 45° from the principal stress directions. For 3D analysis, consider all three principal stresses.
Maximum Normal Stress Theory: σ₁ ≤ σ_allowable
Maximum Shear Stress Theory (Tresca): τ_max ≤ τ_allowable = σ_yield/2
Distortion Energy Theory (von Mises): σ_eq ≤ σ_yield
⚙️ Von Mises Equivalent Stress
General form:
For combined bending and torsion:
Physical Meaning: Von Mises stress represents the equivalent uniaxial stress that produces the same distortion energy as the actual multi-axis stress state.
⚡ Maximum Shear Stress Theory
Physical Meaning: Tresca theory predicts failure when maximum shear stress reaches the shear strength of the material (typically yield strength/2).
Let’s analyze a realistic wrist joint under combined loading.
System Parameters:
σ_yield = 415 MPa, τ_yield = 240 MPa)Safety factor: 2.5M = 150 N·m (from tool weight and cutting forces)T = 200 N·m (from joint rotation)P = 1000 N (from joint preload)Cross-sectional area:
Second moment of area:
Polar moment of inertia:
Axial stress from preload:
Bending stress at outer fiber:
Combined normal stress:
Torsional shear stress:
Von Mises equivalent stress:
Maximum shear stress theory:
For pure shear and normal stress state:
Von Mises safety factor:
Tresca safety factor:
Design assessment:
Both safety factors exceed the required SF = 2.5 ✅
A motor drive shaft in an industrial conveyor system experiences both torsional loads from power transmission and bending moments from belt tension.
Given:
Find: Von Mises equivalent stress and safety factor verification.
✓ Calculate section properties
For solid circular cross-section (d = 40 mm):
✓ Calculate normal stress from bending
Maximum bending stress at outer fiber: σ = Mc/I = (150 × 10³ × 20)/125,700 = 23.9 MPa
✓ Calculate shear stress from torsion
Maximum torsional shear stress at outer fiber: τ = Tr/J = (300 × 10³ × 20)/251,300 = 23.9 MPa
✓ Apply von Mises failure criterion
Von Mises equivalent stress: σ_eq = √(σ² + 3τ²) = √(23.9² + 3 × 23.9²) = √(571.2 + 1,713.5) = 47.8 MPa
✓ Calculate safety factor
SF = σ_yield/σ_eq = 250/47.8 = 5.23 Required SF = 2.0, Actual SF = 5.23 ✓ (Very safe design)
A robotic wrist joint experiences simultaneous bending and twisting during manipulation tasks.
Given:
Find: Compare von Mises and maximum shear stress theories.
✓ Calculate hollow section properties
For hollow circular cross-section:
✓ Calculate stress components
Normal stress: σ = MR/I = (80 × 10³ × 25)/230,700 = 8.66 MPa Shear stress: τ = TR/J = (120 × 10³ × 25)/461,400 = 6.50 MPa
✓ Apply von Mises theory
σ_eq = √(σ² + 3τ²) = √(8.66² + 3 × 6.50²) = √(75.0 + 126.8) = 14.2 MPa SF_vonMises = 500/14.2 = 35.2
✓ Apply maximum shear stress theory (Tresca)
Principal stresses: σ₁ = σ/2 + √((σ/2)² + τ²) = 4.33 + √(18.7 + 42.3) = 12.1 MPa σ₂ = σ/2 - √((σ/2)² + τ²) = 4.33 - 7.81 = -3.48 MPa σ_eq,Tresca = σ₁ - σ₂ = 12.1 - (-3.48) = 15.6 MPa
✓ Compare safety factors
SF_Tresca = 500/15.6 = 32.1 Both theories show SF >> 3.0 required (very conservative design)
A flexible coupling shaft connects an electric motor to a pump, experiencing both torque transmission and misalignment bending.
Given:
Find: Maximum equivalent stress and failure theory comparison.
✓ Calculate section properties
For d = 30 mm solid circular shaft:
✓ Calculate individual stress components
Bending stress: σ = Mc/I = (75 × 10³ × 15)/39,760 = 28.3 MPa Torsional shear: τ = Tr/J = (200 × 10³ × 15)/79,520 = 37.7 MPa
✓ Von Mises equivalent stress
σ_eq = √(σ² + 3τ²) = √(28.3² + 3 × 37.7²) = √(800.9 + 4,265.5) = 71.2 MPa
✓ Maximum shear stress equivalent
Principal stresses: σ₁ = 14.15 + √(14.15² + 37.7²) = 14.15 + 40.2 = 54.4 MPa σ₂ = 14.15 - 40.2 = -26.0 MPa σ_eq,Tresca = 54.4 - (-26.0) = 80.4 MPa
✓ Safety factor assessment
SF_vonMises = 210/71.2 = 2.95 SF_Tresca = 210/80.4 = 2.61 Both theories show adequate safety (SF > 2.5 for rotating machinery)
In this lesson, you learned to:
Key Design Insights:
Critical Formula:
Coming Next: In Lesson 2.5, we’ll analyze composite and built-up beam systems in CNC machine beds, exploring how different materials work together to resist complex loading conditions.
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