Tools › Structures › Damage Tolerance › Center Crack › Example
Inclined centre crack, β = 30°, plane strain: K_I = σcos²β√(πa), K_II = σ sinβ cosβ√(πa), G and Irwin's plastic zone
| Tool | Center Crack |
| Source | Anderson, T. L. (2017). Fracture Mechanics: Fundamentals and Applications, 4th ed. CRC Press, Boca Raton, FL. Chapter 2, mixed-mode loading of an inclined through crack and the Irwin plastic-zone estimate; Irwin, G. R. (1957). "Analysis of Stresses and Strains Near the End of a Crack Traversing a Plate." Journal of Applied Mechanics, Vol. 24, pp. 361–364. |
| Level | Graduate |
Problem
The same 0.5 in through crack is inclined 30° to the x axis in a 0.1 in plate carrying 10,000 lb/in in y (E = 10 Msi, ν = 0.33, yield strength 60 ksi) under plane strain. Find K_I, K_II, the effective K, the T-stress, the energy release rate and Irwin's plastic-zone radius.
Inputs to Enter
- N_xx = 0, N_yy = 10000 lb/in, N_xy = 0, t = 0.1 in
- a = 0.25 in, β = 30°; plane strain, E = 10 Msi, ν = 0.33, F_ty = 60 ksi
- r = 0.3 in, θ = 90°
Answers
- Crack-frame far field: σ_y′y′ = 75 ksi, τ_x′y′ = 43.30 ksi, σ_x′x′ = 25 ksi
- K_I = 66.47 ksi√in, K_II = 38.37 ksi√in, K_eff = √(K_I² + K_II²) = 76.75 ksi√in
- T-stress = σ_x′x′ − σ_y′y′ = −50 ksi
- G = (K_I² + K_II²)(1 − ν²)/E = 524.9 lb/in; r_p = (1/6π)(K_I/F_ty)² = 0.0651 in
Notes
Enter N_yy alone, β = 30°, plane strain and F_ty = 60 ksi (the yield strength the changelog's hand values imply; confirmed by a harness run on 2026-11-07).