Page 64 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C02.fm Page 44 Thursday, April 14, 2005 6:28 PM

44 Fracture Mechanics: Fundamentals and Applications

TABLE 2.1
Stress Fields Ahead of a Crack Tip for Mode I and Mode II
in a Linear Elastic, Isotropic Material

Mode I Mode II
θ
θ
θ 
σ K I cos  θ 1 sin− θ   sin  3   − K II sin    2 + cos   cos  3θ  
xx  2    2   2         
2π r   2π r 2  2  2 
θ 
θ 
σ yy K I cos  θ 1 sin+ θ   sin  3   K II sin  θ cos  θ  cos  3 
2π r  2     2   2    2π r  2   2   2 
θ 
θ 
τ xy K I cos  θ sin  θ  cos  3  K II cos  θ 1 sin− θ   sin  3  
2π r  2   2   2  2π r  2     2   2   
0 (Plane stress) 0 (Plane stress)
σ zz νσ xx + σ ) (Plane strain) νσ xx + σ ) (Plane strain)
(
(
yy
yy
τ , τ 0 0
xz yz
Note: υ is Poisson’s ratio.
material can be written as
K
θ
limσ () I = I f () I ( ) (2.37a)
r→0 ij 2π r ij
K
limσ () = II f () ( ) (2.37b)
II
θ
II
r→0 ij 2π r ij
K
θ
limσ (III ) = III f (III ) ( ) (2.37c)
r→0 ij 2π r ij
for Modes I, II, and III, respectively. In a mixed-mode problem (i.e., when more than one loading
mode is present), the individual contributions to a given stress component are additive:
σ ij (total ) σ ij I ( ) σ = ij ( II) + σ + ( ij III) (2.38)
Equation (2.38) stems from the principle of linear superposition.
Detailed expressions for the singular stress fields for Mode I and Mode II are given in Table 2.1.
Displacement relationships for Mode I and Mode II are listed in Table 2.2. Table 2.3 lists the
nonzero stress and displacement components for Mode III.

TABLE 2.2
Crack-Tip Displacement Fields for Mode I and Mode II
(Linear Elastic, Isotropic Material)

Mode I Mode II


u K I r cos  θ   κ −+ 2 θ   K II r sin  θ   κ ++ 2 θ  
12sin
12cos
x  2    2    2    2  
2 µ 2 π   2 µ 2 π  
θ
  

u K I r sin  θ   κ +− 2 θ   − K II r cos    κ −− 2 θ
12cos
12sin
y          
2 µ 2 π 2   2   2 µ 2 π 2  2 
Note: µ is the shear modulus. κ = 3 − 4ν (plane strain) and κ = (3 − ν)/(1 + ν) (plane stress).