GG303 Lecture 19 11/5/0 1 CAUCHY'S FRMULA AN EIGENVAULES (PRINCIPAL STRESSES) IN 3- I II Mai Topics A Cauchy s formula Pricipal stresses (eigevectors ad eigevalues) Cauchy's formula A Relates tractio vector compoets to stress tesor compoets (see Figures 19.1, 19., 19.3 for derivatio) τi = σ ji j 1 Meaig of terms a τ i =tractio vector compoet: r r r r τ = τ1i + τj + τ3k b σij = stress compoet c =uit ormal. The compoets j of the uit ormal are the directio cosies betwee ad the coordiate aes. Fi Fi A1 Fi A Fi A3 d = + + A A A A A A A 1 3 This represets the physics directly 3 The tractio compoet that acts i the i-directio reflects the cotributio of the stresses that act i that directio. 4 Note that the j's "cacel out" i equatio II. 5 Note that the subscripts o the τ ad the differ 6 σ is symmetric (σij =σ ji ), so C τi = σij j Stadard form of Cauchy s formula 1 The subscript j's still "cacel out" The subscripts o the τ ad the still differ 3 Easier(?) to remember tha Full epasio τi = σji j τi = σ ij j τ1 = σ11 1 + σ1 + σ31 3 = σ11 1 + σ1 + σ13 3 τ = σ1 1 + σ + σ3 3 = σ1 1 + σ + σ3 3 τ3 = σ13 1 + σ3 + σ33 3 = σ31 1 + σ3 + σ33 3 E Matri form τ1 σ11 σ1 σ3 1 τ1 σ11 σ1 σ13 1 τ = σ 1 σ σ3 τ = σ1 σ σ3 τ 3 σ13 σ3 σ 33 3 τ 3 σ31 σ3 σ 33 3 Stephe Martel 19-1 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 erivatio of Cauchy's Equatio (-) 19.0 σ 11 σ 1 φ 1 θ τ τ A A 1 θ 1 φ τ1 A σ 1 σ ΣF 1 = 0, so (ΣF 1 )/A = 0. τ 1 (A/A) = (σ 11 )(A 1 /A) + (σ 1 )(A /A). Similarly, (ΣF )/A = 0, so τ (A/A) = (σ 1 )(A 1 /A) + (σ )(A /A). A 1 φ 1 θ θ 1 A A φ A1/A = cosφ1 = cosθ1 = 1 A/A = cosφ = cosθ = The ratio of the areas is give by the directio cosies of the ormal to plae A. This holds i - ad i 3- also. Substitutig for the area ratios: τ 1 = (σ 11 )( 1 ) + (σ 1 )( ). τ = (σ 1 )( 1 ) + (σ )( ). τ i = σ ji j Stephe Martel 19- Uiversity of Hawaii
GG303 Lecture 19 11/5/0 3 erivatio of Cauchy's Equatio C 19.1 C Area A 3 Area A 1 C σ 3 σ 1 σ 11 τ τ σ 33 σ 31 σ 13 τ 3 τ 1 σ Area A σ 3 σ 1 Area A 3 ΣF 1 = 0, so (ΣF 1 )/A = 0. τ 1 (A/A) = (σ 11 )(A 1 /A) + (σ 1 )(A /A) + (σ 31 )(A 3 /A). Similarly, ΣF = 0 ad ΣF 3 = 0, so τ (A/A) = (σ 1 )(A 1 /A) + (σ )(A /A) + (σ 3 )(A 3 /A). τ 3 (A/A) = (σ 13 )(A 1 /A) + (σ 3 )(A /A) + (σ 33 )(A 3 /A). Stephe Martel 19-3 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 4 erivatio of Cauchy's Equatio C 19. C Area A 3 Area A 1 C P''' P' P''' ω P' P ω 1 ω 3 P'' P is ormal to C Area A P'' Area A 3 Note that C of area A projects oto the - plae as C, oto the - 3 plae as C, ad oto the 3 - plae as. P' is perpedicular to C, ad because C is a lie i C, P' is perpedicular to C. Similarly, CP'' is perpedicular to, so CP'' is perpedicular to C. The itersectio of P' ad CP' is perpedicular to C, ad that itersectio is P. ω 1, ω, ω 3, are agles betwee P ad,, ad 3, respectively. A 1 = 1/ (base C)(height C) = (C)(P') = P A 1/ (base C)(height C) (C)(P') P Stephe Martel 19-4 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 5 erivatio of Cauchy's Equatio P' P' 19.3 P ω 1 P ω 1 1 θ ω 1 1 θ Triagles P ad P' are similar right triagles; they both have agle P (i.e., θ) i commo. Therefore, agle P'o = ω 1. A 1 = P' = cos ω 1 = 1 A P' Similarly, A = P'' = cos ω = A P'' ad A 3 = P''' = cos ω 3 = 3 A P''' so τ 1 (A/A) = (σ 11 )(A 1 /A) + (σ 1 )(A /A) + (σ 31 )(A 3 /A) τ 1 = (σ 11 )( 1 ) + (σ 1 )( ) + (σ 31 )( 3 ). τ (A/A) = (σ 1 )(A 1 /A) + (σ )(A /A) + (σ 3 )(A 3 /A) τ = (σ 1 )( 1 ) + (σ )( ) + (σ 3 )( 3 ), τ 3 (A/A) = (σ 13 )(A 1 /A) + (σ 3 )(A /A) + (σ 33 )(A 3 /A) becomes Similarly, becomes ad becomes τ 3 = (σ 13 )( 1 ) + (σ 3 )( ) + (σ 33 )( 3 ). So τ i = σ ji j, but σ ij = σ ji, so τ i = σ ij j.. Stephe Martel 19-5 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 6 III Pricipal stresses from tesor ad matri perspectives Cosider a plae with a ormal vector defied by directio cosies 1,, ad 3. The compoets of tractio τ o the plae, by Cauchy s formula, are τi = σij j. They also are simply the compoets of τ: τ1=τ1, τ=τ, ad τ3=τ3. The compoets ca be equated: σ11 σ1 σ13 1 1 (1) σ1 σ σ 3 = T. σ31 σ3 σ33 3 3 The right side of (1) ca be subtracted from the left side to yield: σ11 T σ1 σ13 1 () σ1 σ T σ 3 = 0. σ31 σ3 σ33 T 3 Equatio () ca be rewritte (3) [σ -I T] [] =0, where I is the idetity matri 1 0 0 (4) I = 0 1 0. For ay square matri [A], [A][I] = [A]. 0 0 1 Accordig to theorems of liear algebra, equatio () ca be solved oly if the determiat σ -IT equals zero: σ11 T σ1 σ13 (5) σ1 σ T σ3 = 0 σ σ σ T 31 3 33 I may cases the compoets of σ are kow but T is must be solved for. Problems of the form of equatio (3) are commo i may braches of mathematics, egieerig, ad physics, ad they have a special ame: eigevalue problems. The values of T (i.e., T, the pricipal values) that solve the equatio are called eigevalues, ad the vectors (the pricipal directios) that give the directios of T are called eigevectors. ecause these problems are so commo, may mathematics packages, icludig Matlab, have special routies to solve for eigevalues ad eigevectors. Stephe Martel 19-6 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 7 Solvig (5) by had requires fidig the roots of a cubic equatio (ot easy), so we cosider the easier - case, which yields a quadratic equatio. (6) σ σ 11 T 1 σ1 σ T = 0 Note: a c b = ad bc d (7) ( σ11 T)( σ T) ( σ1)( σ1) = 0 (8) T T( σ + σ ) + ( σ )( σ ) ( σ )( σ ) = 11 11 1 1 0 (9a) T T( σ + σ ) + [( σ )( σ ) ( σ ) ] = or (9b) T T( I ) + [ T ] = 11 11 1 0 1 0 The term T i equatio (9) is solved usig the quadratic formula: ( σ + σ ) ± ( σ + σ ) 4[( σ )( σ ) ( σ )] I ± I 4I (10) T = = 11 11 11 1 1 1 ( σ + σ ) ± ( σ + σ σ + σ ) 4[( σ )( σ ) ( σ )] (11) T = 11 11 11 11 1 ( σ + σ ) ± ( σ σ σ + σ ) + 4[ σ ] (1) T = 11 11 11 1 ( σ + σ ) ± ( σ σ ) + 4[ σ ] I ± I 4I (13) T = = 11 11 1 1 1 Stephe Martel 19-7 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 8 (14) T + c r I I = ± + I = []±[]= σ σ σ σ 11 11 ± σ = σ σ 1 1 1, 1 A ispectio of the diagram below shows that the first term i brackets i equatio (14) is the mea ormal stress (i.e., the ceter of the Mohr circle) ad the secod term i brackets is the maimum possible shear stress (i.e., the radius of the Mohr circle). So the pricipal stresses lie at the ed of a horizotal diameter through the Mohr circle. The terms c, r, I, ad I are called ivariats ad are idepedet of the frame of i referece. τ s (σ 11, σ 1 ) σ σ 1 + σ σ 1 τ (σ 11, -σ 1 ) Stephe Martel 19-8 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 9 Eample Suppose the stress state at a poit is give by (15) σ ij = 10 3, where dimesios are i MPa. 3 Solvig for the pricipal values usig eq. (14) yields (16) ( 10 + ) λ = ± 10 + 3 = 6± 5 = 11 ad 1 Now we substitute these back ito (3) 10 11 3 (17a) 1 0 3 11 0 1 3 1 0 3 9 0 for T = σ1 = 11MPa. 10 1 3 (17b) 1 0 3 1 9 3 0 1 0 3 1 0 for T = σ = 1MPa. These relatios yield (18a) -1 + 3 = 0 (σ1 = 11MPa) (18b) 31 + = 0 (σ = 1MPa). From (18a), for a eigevalue (pricipal value) of 11 MPa, 1 = 3. From (18b), for a eigevalue (pricipal value) of 1 MPa, = -3 1. MPa ' 1 MPa 3 MPa 11 MPa 3 MPa ' 10 MPa 1 1 3 θ θ 1 1 = cos θ 1 = cos θ = si θ 1 So / 1 = ta θ 1 θ 1 = ta -1 ( / 1 ) = ata(, 1 ) -3 For σ1 = 11MPa* θ ta 1 ta 1 ta 1 ta 1 1 = 18. 5, ormal = 3 3 1 = 1 = = o For σ = 1Mpa* θ = ta 1 ta 1 ta 1 3 1 ta 1 3 71. 5, ormal = 1 = = ( o )= 1 1 The two eigevectors are perpedicular, as they are supposed to be. * I the first epressio for θ, the ormal directio is the 1 directio, ad 1 ad are the directio cosies for a uit vector alog 1. I the first epressio for θ, the ormal directio is the directio, ad 1 ad are the directio cosies for a uit vector alog. Stephe Martel 19-9 Uiversity of Hawaii
GG303 Lecture 19 11/5/0 10 V Matri treatmets of stress trasformatio I matri form, σ ij = a ik a jl σ kl becomes (Mal & Sigh, 1991, p. 37) (19) σ' = [a] [σ] [at], where a a a 11 1 13 a (0) a = a a a ad (1) 1 3 a T a a 11 1 31 = a a a 1 3 a 31 a 3 a 33 a 13 a 3 a 33 The proper order of matri multiplicatio is essetial i order to reproduce the epasios of lecture 17: [a] [σ] [at] [at] [σ][a]! I MATLA, equatio (19) would be writte: sigmaprime = a * sigma * a' The term a' sigifies [at]. Matlab also has a fuctio eig to fid eigevectors (give i terms of the directio cosies) ad eigevalues. [V,] = eig (sigma) Eample»sigmay = [10 3;3 ] sigmay = 10 3 3»a = [3/sqrt(10) 1/sqrt(10);-1/sqrt(10) 3/sqrt(10)] a = 0.9487 0.316 The first row of matri a is the egative -0.316 0.9487 of the first colum of matri V below. The secod row of matri a is the egative»sigmaprime = a*sigmay*a' of the secod colum of matri V below. sigmaprime = 11.0000-0.0000-0.0000 1.0000»[V,] = eig(sigmay) V = -0.9487 0.316 Colum 1 i V relates to colum 1 i -0.316-0.9487 Colum i V relates to colum i = 11 0 0 1 Stephe Martel 19-10 Uiversity of Hawaii