Stress Intensity Factor and Critical Crack Size — Worked Example

1. Problem

A steel plate (width 200 mm, thickness 10 mm) contains a central through-thickness crack of length 2a = 20 mm. The plate is subjected to a remote tensile stress of 250 MPa. Determine the stress intensity factor K_I, compare with the material fracture toughness KIC, and calculate the critical crack size.

2. Given

3. Required

• Stress intensity factor K_I
• Check K_I vs KIC
• Critical crack size a_cr

4. Assumptions

• Linear elastic fracture mechanics (LEFM) applies — plastic zone is small relative to crack and ligament
• Plane strain conditions at the crack tip (valid for thick sections)
• The geometry factor Y = 1.0 is approximate for a/W = 0.05 (exact value from Tada/Murakami handbook is ~1.003)
• The crack is through-thickness and central (Mode I loading)
• No crack closure or residual stress effects

5. Governing Equations

Stress intensity factor (Mode I):
  K_I  =  Y × σ × √(π × a)

Fracture criterion:
  K_I  <  KIC   (safe)
  K_I  =  KIC   (critical — crack propagates)

Critical crack size:
  a_cr  =  (1/π) × (KIC / (Y × σ))²

Plastic zone size (plane strain):
  r_p  =  (1/6π) × (K_I / Fty)²

6. Calculation

K_I  =  1.0 × 250 × √(π × 0.010)
     =  250 × √(0.03142)
     =  250 × 0.1772
     =  44.3 MPa·√m

Check:  K_I = 44.3  <  KIC = 140   →  SAFE

Plastic zone size:
  r_p  =  (1/6π) × (44.3 / 1200)²
       =  (1/18.85) × (0.0369)²
       =  (1/18.85) × 0.00136
       =  0.0000722 m  =  0.072 mm

r_p = 0.072 mm << a = 10 mm   →  LEFM is valid

6. Calculation (continued)

Step 2: Critical crack size.

a_cr  =  (1/π) × (KIC / (Y × σ))²
      =  (1/π) × (140 / (1.0 × 250))²
      =  (1/π) × (0.56)²
      =  (1/π) × 0.3136
      =  0.0999 m  =  99.9 mm

2a_cr  =  200 mm  =  W  (full width — physically the plate would fail by net-section yield before reaching this)

7. Result

K_I = 44.3 MPa·√m, well below KIC = 140 MPa·√m. Critical crack size a_cr = 100 mm (2a_cr = 200 mm = full plate width). The current crack (2a = 20 mm) is far from critical. However, net-section yield would occur before the crack reaches a_cr.

8. Check

• Plastic zone (0.072 mm) << crack length (10 mm) and << ligament (90 mm) — LEFM is valid
• K_I/KIC = 0.32 — the crack is well below critical, with a reserve factor of 3.2 on stress intensity
• Net-section stress: σ_net = σ × W/(W − 2a) = 250 × 200/180 = 278 MPa < Fty = 1200 MPa — net-section is safe
• The critical crack size (100 mm) exceeds half the plate width (100 mm) — net-section yield would occur first
• Dimensional check: MPa × √m = MPa·√m — consistent

9. Interpretation

The current crack (10 mm half-length) is well below the critical size (100 mm). The stress intensity K_I = 44 MPa·√m is only 32% of KIC, giving a reserve factor of 3.2. However, the critical crack size is so large that net-section yielding would occur before the crack reaches a_cr — in practice, the failure mode would be plastic collapse, not brittle fracture.

For damage tolerance assessment, the relevant question is how fast the crack grows from its current size to the critical size under service loading. This requires the Paris law (da/dN = C × ΔK^m) and the stress range Δσ from the service spectrum.

10. Limitations

• LEFM only — for ductile materials or large plastic zones, use elastic-plastic fracture mechanics (J-integral, CTOD)
• Y = 1.0 is approximate — use handbook values for the specific geometry
• No crack growth calculation — Paris law integration is needed for remaining life
• No residual stress or crack closure effects
• KIC is material, temperature and orientation dependent — the value must be measured for the specific condition
• No inspection interval determination — requires combining crack growth with detection capability

11. Related Resources

• Fracture Mechanics Quick Reference (Reference Library)
• How to Interpret Fatigue Life from FEA