# Lec.18.pptx - 18. Tensor Transforma2on of Stresses I...

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Unformatted text preview: 10/23/11 18. Tensor Transforma2on of Stresses I Main Topics A Objec2ve B Approach C Deriva2on D Example 10/23/11 GG303 1 17. Mohr Circle for Trac2ons •  From King et al., 1994 (Fig. 11) •  Coulomb stress change caused by the Landers rupture. The leT ­lateral ML=6.5 Big Bear rupture occurred along doYed line 3 hr 26 min aTer the Landers main shock. The Coulomb stress increase at the future Big Bear epicenter is 2.2 ­2.9 bars. 10/23/11 hYp://earthquake.usgs.gov/research/modeling/papers/landers.php GG303 2 1 10/23/11 18. Tensor Transforma2on of Stresses II Objec2ve Lecture 16 Transforma2on Lecture 18 Stresses to trac2ons Stresses to stresses Number of arbitrary planes 1 plane 2 perpendicular planes Stresses accounted for Normal and shear stresses 10/23/11 Normal stresses only GG303 3 18. Tensor Transforma2on of Stresses III Approach Vectors Tensors vi ′ = ai ′j v j σ i ′j ′ = ai ′k a j ′lσ kl Number of subscripts in quan2ty being converted 1 2 Number of direc2on cosines in equa2on 1 2 Equa2on Key concept Total value of each stress component in one reference frame is the sum of the weighted contribu2ons from all the components in another frame 10/23/11 GG303 4 2 10/23/11 18. Tensor Transforma2on of Stresses V Deriva2on A Descrip2on of terms Term Meaning Ax’, Ax, Ay Sides of prism σxx/σxy Normal/shear stress on Ax σyy/σyx Normal/shear stress on Ay σx’x’ /σx’y’ Normal/shear stress on Ax’ θx’x Angle from x’ to x axis θx’y Angle from x’ to y axis θy’x Angle from y’ to x axis θy’y Angle from y’ to y axis 10/23/11 GG303 5 18. Tensor Transforma2on of Stresses IV Deriva2on B Contribu2on of σxx to σx’x’ w = dimensionless weigh2ng factor 1 σ x ′x ′ = w(1)σ xx + w( 2 )σ xy + w( 3)σ yx + w( 4 )σ yy ⎛ A F (1) ⎞ F (1) ⎛ A F ( 2 ) ⎞ Fy( 2 ) ⎛ Ay Fx(′3) ⎞ Fx( 3) ⎛ Ay Fx(′4 ) ⎞ Fy( 4 ) = ⎜ x x(′1) ⎟ x + ⎜ x x(′2 ) ⎟ + +⎜ ⎟ Ax ′ ⎝ Ax ′ Fx ⎠ Ax ⎝ Ax ′ Fy ⎠ Ax ⎜ Ax ′ Fx( 3) ⎟ Ay ⎝ Ax ′ Fy( 4 ) ⎠ Ay ⎝ ⎠ 2 F x′ 3 10/23/11 σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy GG303 6 3 10/23/11 18. Tensor Transforma2on of Stresses B Contribu2on of σxx to σx’x’ Start with the deﬁni2on of a stress vector: 1  σx’x’(1) = Fx’(1))/ Ax’ First ﬁnd Fx(1) associated with σxx 2 Fx(1) = σxx Axx Find Fx’ (1), the component of Fx(1) in the x’ ­direc2on 3  Fx’ (1) = Fx(1) cos θx’x Now ﬁnd Ax in terms of Ax’ Ax = Ax’ cos θx’x 4 Ax’ = Ax/cos θx’x 10/23/11 GG303 7 18. Tensor Transforma2on of Stresses B Contribu2on of σxx to σx’x’ (cont.) 3  Fx’ (1) = Fx(1) cos θx’x 4  4 Ax’ = Ax/cos θx’x Now subs2tute: 5a σx’x’ (1) = Fx’ (1) / Ax’ 5b σx’x’ (1) = Fx(1) cos θx’x/ (Ax/cos θx’x) 5c σx’x’ (1) = cos θx’x cos θx’x (Fx(1)/ Ax) 5d σx’x’ (1) = ax’x ax’x σxx Weigh2ng factor w(1) 6 w(1) = ax’x ax’x 10/23/11 GG303 8 4 10/23/11 18. Tensor Transforma2on of Stresses IV Deriva2on C Contribu2on of σxy to σx’x’ w = dimensionless weigh2ng factor 1 σ x ′x ′ = w(1)σ xx + w( 2 )σ xy + w( 3)σ yx + w( 4 )σ yy ⎛ A F (1) ⎞ F (1) ⎛ A F ( 2 ) ⎞ Fy( 2 ) ⎛ Ay Fx(′3) ⎞ Fx( 3) ⎛ Ay Fx(′4 ) ⎞ Fy( 4 ) = ⎜ x x(′1) ⎟ x + ⎜ x x(′2 ) ⎟ + +⎜ ⎟ Ax ′ ⎝ Ax ′ Fx ⎠ Ax ⎝ Ax ′ Fy ⎠ Ax ⎜ Ax ′ Fx( 3) ⎟ Ay ⎝ Ax ′ Fy( 4 ) ⎠ Ay ⎝ ⎠ 2 F x′ 3 10/23/11 σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy GG303 9 18. Tensor Transforma2on of Stresses C Contribu2on of σxy to σx’x’ Start with the deﬁni2on of a stress vector: 1  σx’x’(1) = Fx’(2))/ Ax’ First ﬁnd Fy(2) associated with σxy 2 Fy(2) = σxy Ax Find Fx’ (2), the component of Fy(2) in the x’ ­direc2on 3  Fx’ (2) = Fy(2) cos θx’y Now ﬁnd Ax in terms of Ax’ Ax = Ax’ cos θx’x 4 Ax’ = Ax/cos θx’x 10/23/11 GG303 * * 10 5 10/23/11 18. Tensor Transforma2on of Stresses C Contribu2on of σxy to σx’x’ (cont.) 3  Fx’ (2) = Fy(2) cos θx’y 4 Ax’ = Ax/cos θx’x Now subs2tute: 5a σx’x’ (2) = Fx’ (2) / Ax’ 5b σx’x’ (2) = Fy(2) cos θx’y/ (Ax/cos θx’x) 5c σx’x’ (2) = cos θx’x cos θx’y (Fy(2)/Ax) 5d σx’x’ (2) = ax’x ax’y σxy Weigh2ng factor w(2) 6 w(2) = ax’x ax’y 10/23/11 GG303 11 18. Tensor Transforma2on of Stresses IV Deriva2on D Contribu2on of σyx to σx’x’ w = dimensionless weigh2ng factor 1 σ x ′x ′ = w(1)σ xx + w( 2 )σ xy + w( 3)σ yx + w( 4 )σ yy ⎛ A F (1) ⎞ F (1) ⎛ A F ( 2 ) ⎞ Fy( 2 ) ⎛ Ay Fx(′3) ⎞ Fx( 3) ⎛ Ay Fx(′4 ) ⎞ Fy( 4 ) = ⎜ x x(′1) ⎟ x + ⎜ x x(′2 ) ⎟ + +⎜ ⎟ Ax ′ ⎝ Ax ′ Fx ⎠ Ax ⎝ Ax ′ Fy ⎠ Ax ⎜ Ax ′ Fx( 3) ⎟ Ay ⎝ Ax ′ Fy( 4 ) ⎠ Ay ⎝ ⎠ 2 F x′ 3 10/23/11 σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy GG303 12 6 10/23/11 18. Tensor Transforma2on of Stresses D Contribu2on of σyx to σx’x’ Start with the deﬁni2on of a stress vector: 1  σx’x’(3) = Fx’(3))/ Ax’ First ﬁnd Fx(3) associated with σyx 2 Fx(3) = σyx Ay Find Fx’(3), the component of Fx(3) in the x’ ­direc2on 3  Fx’(3) = Fx(3) cos θx’x Find Ay in terms of Ax’ Ay = Ax’ cos θx’y 4 Ax’ = Ay/cos θx’y 10/23/11 * * GG303 13 18. Tensor Transforma2on of Stresses D Contribu2on of σyx to σx’x’ (cont.) 3  Fx’ (3) = Fx(3) cos θx’x 4 Ax’ = Ay/cos θx’y Now subs2tute: 5a σx’x’(3 ) = Fx’ (3) / Ax’ 5b σx’x’(3 ) = Fx(3) cos θx’x/ (Ay/cos θx’y) 5c σx’x’(3) = cos θx’y cos θx’x (Fx(3)/Ay) 5d σx’x’(3) = ax’y ax’x σyx Weigh2ng factor w(3) 6 w(3) = ax’y ax’x 10/23/11 GG303 14 7 10/23/11 18. Tensor Transforma2on of Stresses IV Deriva2on E Contribu2on of σyy to σx’x’ w = dimensionless weigh2ng factor 1 σ x ′x ′ = w(1)σ xx + w( 2 )σ xy + w( 3)σ yx + w( 4 )σ yy ⎛ A F (1) ⎞ F (1) ⎛ A F ( 2 ) ⎞ Fy( 2 ) ⎛ Ay Fx(′3) ⎞ Fx( 3) ⎛ Ay Fx(′4 ) ⎞ Fy( 4 ) = ⎜ x x(′1) ⎟ x + ⎜ x x(′2 ) ⎟ + +⎜ ⎟ Ax ′ ⎝ Ax ′ Fx ⎠ Ax ⎝ Ax ′ Fy ⎠ Ax ⎜ Ax ′ Fx( 3) ⎟ Ay ⎝ Ax ′ Fy( 4 ) ⎠ Ay ⎝ ⎠ 2 F x′ 3 10/23/11 σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy GG303 15 18. Tensor Transforma2on of Stresses E Contribu2on of σyy to σx’x’ Start with the deﬁni2on of a stress vector: 1  σx’x’(4) = Fx’(4) /Ax’ First ﬁnd Fy(4) associated with σyy 2 Fy(4) = σyy Ay Find Fx’(4), the component of Fx(4) in the x’ ­direc2on 3  Fx’(4) = Fy(4) cos θx’y Find Ay in terms of Ax’ Ay = Ax’ cos θx’y 4 Ax’ = Ay/cos θx’y 10/23/11 GG303 * * 16 8 10/23/11 18. Tensor Transforma2on of Stresses E Contribu2on of σyy to σx’x’ 3  Fx’ (4) = Fy(4) cos θx’y 4 Ax’ = Ay /cos θx’y Now subs2tute: 5a σx’x’(4) = Fx’(4)/ Ax’ 5b σx’x’(4) = Fy(4) cos θx’y/ (Ay/cos θx’y) 5c σx’x’(4) = cos θx’y cos θx’y (Fy(4)/Ay) 5d σx’x’(4) = ax’y ax’y σyy Weigh2ng factor w(4) 6 w(4) = ax’y ax’y 10/23/11 * * GG303 17 18. Tensor Transforma2on of Stresses IV Deriva2on F Formulas for σx’x’, σx’y’, σy’x’, and σy’y’ 1 σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy 2 σ x ′y ′ = ax ′x ay ′xσ xx + ax ′x ay ′yσ xy + ax ′y ay ′xσ yx + ax ′y ay ′yσ yy 3 σ y ′x ′ = ay ′x ax ′xσ xx + ay ′x ax ′yσ xy + ay ′y ax ′xσ yx + ay ′y ax ′yσ yy 4 σ y ′y ′ = ay ′x ay ′xσ xx + ay ′x ay ′yσ xy + ay ′y ay ′xσ yx + ay ′y ay ′yσ yy 10/23/11 GG303 18 9 10/23/11 18. Tensor Transforma2on of Stresses V Example ⎡ Find σ = ⎢ ⎢ i ′j ′ given σ ij σ x ′x ′ σ x ′y ′ ⎤ ⎥ σ y ′x ′ σ y ′y ′ ⎥ ⎣ ⎦ ⎡ σ xx = −4 MPa σ xy = −4 MPa =⎢ ⎢ σ yx = −4 MPa σ yy = −4 MPa ⎣ ⎤ ⎥ ⎥ ⎦ θ x ′x = −45 ,θ x ′y = 45 ,θ y ′x = −135 ,θ y ′y = −45 10/23/11 GG303 19 18. Tensor Transforma2on of Stresses V Example (cont.) σ x ′x ′ = ax ′x ax ′xσ xx + ax ′x ax ′yσ xy + ax ′y ax ′xσ yx + ax ′y ax ′yσ yy σ x ′x ′ = ( −2 MPa ) + ( −2 MPa ) + ( −2 MPa ) + ( −2 MPa ) = −8 MPa σ x ′y ′ = ax ′x ay ′xσ xx + ax ′x ay ′yσ xy + ax ′y ay ′xσ yx + ax ′y ay ′yσ yy σ x ′y ′ = ( 2 MPa ) + ( −2 MPa ) + ( 2 MPa ) + ( −2 MPa ) = 0 MPa σ y ′x ′ = ay ′x ax ′xσ xx + ay ′x ax ′yσ xy + ay ′y ax ′xσ yx + ay ′y ax ′yσ yy σ y ′x ′ = ( 2 MPa ) + ( 2 MPa ) + ( −2 MPa ) + ( −2 MPa ) = 0 MPa σ y ′y ′ = ay ′x ay ′xσ xx + ay ′x ay ′yσ xy + ay ′y ay ′xσ yx + ay ′y ay ′yσ yy σ y ′x ′ = ( −2 MPa ) + ( 2 MPa ) + ( 2 MPa ) + ( −2 MPa ) = 0 MPa 10/23/11 GG303 20 10 10/23/11 18. Tensor Transforma2on of Stresses V Example (values in MPa) σxx =  ­ 4 τxn =  ­ 4 σx’x’ =  ­ 8 τx’n =  ­ 8 σxy =  ­ 4 τxs =  ­ 4 σx’y’ = 0 τx’s = 0 σyx =  ­ 4 τyn = + 4 σy’x’ =  ­0 σyy =  ­ 4 τys =  ­4 σy’y’ = 0 n τy’s = +0 τy’n = 0 s n s 10/23/11 GG303 21 18. Tensor Transforma2on of Stresses V Example Matrix form ⎡ σ x ′x ′ ⎢ ⎢ σ y ′x ′ ⎣ σ x ′y ′ ⎤ ⎡ ax ′x ′ ⎥=⎢ σ y ′y ′ ⎥ ⎢ a y ′x ′ ⎦⎣ T ⎡σ i ′j ′ ⎤ = [ a ] ⎡σ i ′j ′ ⎤ [ a ] ⎣ ⎦ ⎣ ⎦ ax ′y ′ ⎤ ⎡ σ xx ⎥⎢ ay ′y ′ ⎥ ⎢ σ yx ⎦⎣ σ xy ⎤ ⎡ ax ′x ′ ⎥⎢ σ yy ⎥ ⎢ ay ′x ′ ⎦⎣ ax ′y ′ ⎤ ⎥ a y ′y ′ ⎥ ⎦ T This expression is valid in 2D and 3D! 10/23/11 GG303 22 11 10/23/11 18. Tensor Transforma2on of Stresses V Example Matrix form/Matlab ⎡σ i ′j ′ ⎤ = [ a ] ⎡σ i ′j ′ ⎤ [ a ] ⎣ ⎦ ⎣ ⎦ T >> sij = [ ­4  ­4; ­4  ­4] sij =  ­4  ­4  ­4  ­4 >> a=[sqrt(2)/2 sqrt(2)/2;  ­sqrt(2)/2 sqrt(2)/2] a = 0.7071 0.7071  ­0.7071 0.7071 >> sipjp = a*sij*a’ sipjp =  ­8.0000 0 0 0 10/23/11 GG303 23 12 ...
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## This note was uploaded on 12/05/2011 for the course GEOLOGY 300 taught by Professor Stephenmartel during the Fall '11 term at University of Hawaii, Manoa.

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