Reference
Formula Sheet
All equations from this lesson on one page. Print this and keep it at your workstation.
Stress, Strain, and Material Properties
Stress
σ = F / A
σ = stress (MPa) | F = force (N) | A = cross-sectional area (mm²)
Strain
ε = ΔL / L₀
ε = strain (unitless) | ΔL = change in length (mm) | L₀ = original length (mm)
Hooke's Law (Elastic Region Only)
σ = E × ε
E = Young's Modulus (MPa) | Slope of stress-strain curve in elastic region
Factor of Safety
FoS = Yield Strength / Applied Stress
FoS < 1.0 = failure predicted | FoS > 2.0 = comfortable margin
Three-Point Bend Test (Section 6)
Flexural Stress at Failure
σ_f = (3 × F × L) / (2 × b × d²)
F = failure force (N) | L = support span (mm) | b = width (mm) | d = depth/thickness (mm) | Result in MPa
Flexural Modulus (from Elastic Region)
E_f = (L³ × m) / (4 × b × d³)
m = slope of force-deflection curve in linear region (N/mm) | Result in MPa
Comparison (Section 7)
Percent Error
% Error = |(Predicted − Actual) / Actual| × 100%
Use absolute value — error is always positive
Unit Conversions
| Quantity | Unit | Equivalents |
|---|---|---|
| Stress / Modulus | MPa | 1 MPa = 1 N/mm² = 1,000,000 Pa |
| Stress / Modulus | GPa | 1 GPa = 1,000 MPa |
| Force | N (Newton) | 1 kg mass × 9.81 m/s² = 9.81 N |
| Force (from grams) | N | mass (g) × 0.00981 = force (N) |
| Length | mm | 1 mm = 0.001 m = 0.03937 in |
| Area | mm² | For rectangular cross-section: A = width × thickness |
Specimen Geometry (ASTM D790 Proportions)
| Dimension | Value |
|---|---|
| Length | 80 mm |
| Width (b) | 10 mm |
| Thickness (d) | 4 mm |
| Support span (L) | 64 mm |
| Span-to-depth ratio | 16:1 |
| Cross-sectional area (A) | 40 mm² |
FDM PLA Material Properties (Custom Profile)
| Property | Value |
|---|---|
| Young's Modulus (E) | 3,300 MPa |
| Tensile Strength (UTS) | 50 MPa |
| Yield Strength | 40 MPa |
| Poisson's Ratio | 0.35 |
| Density | 1,240 kg/m³ |
Worked Example: Flexural Stress
Example Calculation
Given: A PLA test bar (b = 10 mm, d = 4 mm) on supports with L = 64 mm span. It failed at a load of F = 85 N.
Find: Flexural stress at failure.
Solution:
σ_f = (3 × F × L) / (2 × b × d²)
σ_f = (3 × 85 × 64) / (2 × 10 × 4²)
σ_f = 16,320 / 320
σ_f = 51.0 MPa
This is close to the published UTS of 50 MPa for FDM PLA — the result makes sense.