18. Tensor Transformation of Stresses

Transcription

1 I Main Topics A Objective B Approach C Derivation D Eample 10/15/18 GG303 1

2 18. Mohr Circle for Tractions rom King et al., 1994 (ig. 11) Coulomb stress change caused b the Landers rupture. The left-lateral ML=6.5 Big Bear rupture occurred along dotted line 3 hr 26 min after the Landers main shock. The Coulomb stress increase at the future Big Bear epicenter is bars. 10/15/18 GG303 2

3 II 18. Tensor Transformation of Stresses Objective Lecture 16 Lecture 18 Transformation Stresses to tractions Stresses to stresses Number of arbitrar planes 1 plane 2 perpendicular planes Stresses accounted for Normal stresses onl Normal and shear stresses 10/15/18 GG303 3

4 III Approach Equation Number of subscripts in quantit being converted Number of direction cosines in equation Vectors Tensors v i = a i j v j σ i j = a i k a j l σ kl Ke concept Total value of each stress component in one reference frame is the sum of the weighted contributions from all the components in another frame 10/15/18 GG303 4

5 V Derivation A Description of terms Term A, A, A Meaning Sides of prism σ /σ σ /σ Normal/shear stress on A Normal/shear stress on A σ /σ Normal/shear stress on A θ θ θ θ Angle from to ais Angle from to ais Angle from to ais Angle from to ais 10/15/18 GG303 5

6 IV Derivation B Contribution of σ to σ w = dimensionless weighting factor 1 σ = w ( 1) σ + w ( 2) σ + w ( 3) σ + w ( 4) σ 2 = A A A ( 1) 1 + A A A ( 1) ( 2) 2 ( 2) + A A ( 3) 3 A + A A A ( 3) ( 4) 4 ( 4) A 3 σ = a a σ + a a σ + a a σ + a a σ 10/15/18 GG303 6

7 B Contribution of σ to σ Start with the definition of a stress vector: 1 σ = )/ A irst find associated with σ 2 = σ A ind, the component of in the -direction 3 = cos θ Now find A in terms of A A = A cos θ 4 A = A /cos θ 10/15/18 GG303 7

8 B Contribution of σ to σ (cont.) 3 = cos θ 4 A = A /cos θ Now substitute: 5a σ = / A 5bσ = cos θ / (A /cos θ ) 5c σ = cos θ cos θ ( / A ) 5dσ = a a σ Weighting factor w 6 w = a a 10/15/18 GG303 8

9 IV Derivation C Contribution of σ to σ w = dimensionless weighting factor 1 σ = w ( 1) σ + w ( 2) σ + w ( 3) σ + w ( 4) σ 2 = A A A ( 1) 1 + A A A ( 1) ( 2) 2 ( 2) + A A ( 3) 3 A + A A A ( 3) ( 4) 4 ( 4) A 3 σ = a a σ + a a σ + a a σ + a a σ 10/15/18 GG303 9

10 C Contribution of σ to σ Start with the definition of a stress vector: 1 σ = )/ A irst find associated with σ 2 = σ A ind, the component of in the -direction 3 = cos θ Now find A in terms of A A = A cos θ 4 A = A /cos θ * * 10/15/18 GG303 10

11 C Contribution of σ to σ (cont.) 3 = cos θ 4 A = A /cos θ Now substitute: 5a σ = / A 5bσ = cos θ / (A /cos θ ) 5c σ = cos θ cos θ ( /A ) 5dσ = a a σ Weighting factor w 6 w = a a 10/15/18 GG303 11

12 IV Derivation D Contribution of σ to σ w = dimensionless weighting factor 1 σ = w ( 1) σ + w ( 2) σ + w ( 3) σ + w ( 4) σ 2 = A A A ( 1) 1 + A A A ( 1) ( 2) 2 ( 2) + A A ( 3) 3 A + A A A ( 3) ( 4) 4 ( 4) A 3 σ = a a σ + a a σ + a a σ + a a σ 10/15/18 GG303 12

13 D Contribution of σ to σ Start with the definition of a stress vector: 1 σ = )/ A * irst find associated with σ * 2 = σ A ind, the component of in the -direction 3 = cos θ ind A in terms of A A = A cos θ 4 A = A /cos θ 10/15/18 GG303 13

14 D Contribution of σ to σ (cont.) 3 = cos θ 4 A = A /cos θ Now substitute: 5a σ (3 ) = / A 5bσ (3 ) = cos θ / (A /cos θ ) 5c σ = cos θ cos θ ( /A ) 5dσ = a a σ Weighting factor w 6 w = a a 10/15/18 GG303 14

15 IV Derivation E Contribution of σ to σ w = dimensionless weighting factor 1 σ = w ( 1) σ + w ( 2) σ + w ( 3) σ + w ( 4) σ 2 = A A A ( 1) 1 + A A A ( 1) ( 2) 2 ( 2) + A A ( 3) 3 A + A A A ( 3) ( 4) 4 ( 4) A 3 σ = a a σ + a a σ + a a σ + a a σ 10/15/18 GG303 15

16 E Contribution of σ to σ Start with the definition of a stress vector: 1 σ = /A irst find associated with σ 2 = σ A ind, the component of in the -direction 3 = cos θ ind A in terms of A A = A cos θ 4 A = A /cos θ * * 10/15/18 GG303 16

17 E Contribution of σ to σ 3 = cos θ 4 A = A /cos θ Now substitute: 5a σ = / A 5bσ = cos θ / (A /cos θ ) 5c σ = cos θ cos θ ( /A ) 5dσ = a a σ Weighting factor w 6 w = a a * * 10/15/18 GG303 17

18 IV Derivation ormulas for σ, σ, σ, and σ 1 σ = a a σ + a a σ + a a σ + a a σ σ = a a σ + a a σ + a a σ + a a σ σ = a a σ + a a σ + a a σ + a a σ σ = a a σ + a a σ + a a σ + a a σ 10/15/18 GG303 18

19 V Eample ind given σ i j = σ ij = σ σ σ σ σ = 4MPa σ = 4MPa σ = 4MPa σ = 4MPa θ = 45!,θ = 45!,θ = 135!,θ = 45! 10/15/18 GG303 19

20 V Eample (cont.) σ = a a σ + a a σ + a a σ + a a σ σ + ( 2MPa) + ( 2MPa) + ( 2MPa) = 8MPa = 2MPa σ = a a σ + a a σ + a a σ + a a σ σ + ( 2MPa) + ( 2MPa) + ( 2MPa) = 0MPa = 2MPa σ = a a σ + a a σ + a a σ + a a σ σ + ( 2MPa) + ( 2MPa) + ( 2MPa) = 0MPa = 2MPa σ = a a σ + a a σ + a a σ + a a σ σ + ( 2MPa) + ( 2MPa) + ( 2MPa) = 0MPa = 2MPa 10/15/18 GG303 20

21 V Eample (values in MPa) σ = - 4 τ n = - 4 σ = - 8 τ n = - 8 σ = - 4 τ s = - 4 σ = 0 τ s = 0 σ = - 4 τ s = + 4 σ = -0 τ s = +0 σ = - 4 τ s = -4 σ = 0 τ n = 0 s n See slide 19 for identification of the angles n s 10/15/18 GG303 21

22 V Eample Matri form σ σ σ σ = a a a a σ σ σ σ a a a a T σ i j = a [ ] σ i j [ ] T a This epression is valid in 2D and 3D! 10/15/18 GG303 22

23 V Eample Matri form/matlab >> sij = [-4-4;-4-4] sij = >> a=[sqrt/2 sqrt/2; -sqrt/2 sqrt/2] a = >> sipjp = a*sij*a sipjp = /15/18 GG303 23