🎯 Engineering Scenario
Application: Industrial robotic gripper handling steel rods in a manufacturing assembly line
Object: Steel cylinder (φ50mm, 2.5kg) - typical hydraulic piston rod
Challenge: Analyze bending stresses in cantilever gripper jaws during maximum grip force
Objective: Ensure gripper jaws can safely handle the required gripping force without failure
Step 1: Define System Parameters
| Parameter | Value | Description |
|---|
| Gripper Length (L) | 80 mm | Cantilever span from fixed support to tip |
| Cross-section | 15 × 8 mm | Rectangular gripper jaw cross-section |
| Material | Aluminum 6061-T6 | σ_yield = 270 MPa, E = 70 GPa |
| Steel Cylinder | φ50mm, 2.5kg | Object being gripped |
| Grip Force | 600 N | Required force to securely hold cylinder |
| Safety Factor | 3.0 | Required safety margin for industrial application |
Step 2: Calculate Maximum Bending Moment
For a cantilever beam with concentrated load at the tip, the maximum moment occurs at the fixed support:
M_max = F × L
M_max = 600 N × 0.08 m = 48 N·m
Maximum Bending Moment: 48 N·m
Step 3: Calculate Section Properties
For rectangular cross-section (15mm × 8mm):
Second Moment of Area: I = (b × h³)/12
I = (15 × 8³)/12 = 640 mm⁴ = 640 × 10⁻¹² m⁴
Distance to extreme fiber: c = h/2 = 4 mm
Section Modulus: S = I/c = 160 × 10⁻⁹ m³
Section Modulus: 160 × 10⁻⁹ m³
Step 4: Calculate Maximum Bending Stress
Using the flexural formula:
σ_max = M_max / S
σ_max = 48 N·m / (160 × 10⁻⁹ m³)
σ_max = 300 × 10⁶ Pa = 300 MPa
Maximum Bending Stress: 300 MPa
Step 5: Safety Assessment
Check safety factor against yield strength:
Safety Factor = σ_yield / σ_max
SF = 270 MPa / 300 MPa = 0.9
⚠️ CRITICAL FAILURE RISK: Safety Factor = 0.9 < 3.0 required
The current design will fail under the specified load!
Step 6: Design Improvement
To achieve SF = 3.0, we need to reduce stress to:
σ_allowable = σ_yield / SF = 270 MPa / 3.0 = 90 MPa
Required Section Modulus: S_req = M_max / σ_allowable
S_req = 48 N·m / (90 × 10⁶ Pa) = 533 × 10⁻⁹ m³
Redesign with 15mm × 12mm cross-section:
I_new = (15 × 12³)/12 = 2160 mm⁴
S_new = 2160 × 10⁻¹² / (6 × 10⁻³) = 360 × 10⁻⁹ m³
σ_new = 48 / (360 × 10⁻⁹) = 133 MPa
SF_new = 270 / 133 = 2.03
Still insufficient! Try 15mm × 15mm cross-section:
I_final = (15 × 15³)/12 = 4219 mm⁴
S_final = 4219 × 10⁻¹² / (7.5 × 10⁻³) = 563 × 10⁻⁹ m³
σ_final = 48 / (563 × 10⁻⁹) = 85 MPa
SF_final = 270 / 85 = 3.18
✅ DESIGN APPROVED: 15mm × 15mm cross-section provides SF = 3.18 > 3.0 required