That is, we start with a general matrix: And end with a simpler matrix:

Transcription

1 DIAGON ALIZATION OF THE STR ESS TEN SOR INTRO DUCTIO N By th us of Cauchy s thorm w ar abl to rduc th numbr of strss componnts in th strss tnsor to only nin valus. An additional simplification of th strss can b achivd through diagonalization of th strss tnsor. Diagonalization of th strss tnsor rducs th numbr of componnts to only thr. Many squar matrics can b diagonalizd. In this simplifid (diagonalizd) vrsion of th strss tnsor, th principal plans hav no strss along thm and th principal axs ar th only dirctions along which w hav any strss. Th principal axs dirctions ar orthogonal to th principal plans. That is, w start with a gnral matrix: ij And nd with a simplr matrix: σ ij σ σ σ33 Th last matrix has bn diagonalizd. Th rsultant matrix is asir to handl. For xampl, th squar of th diagonalizd matrix is simply th squar of ach of th componnts. Rcall Cauchy s Thorm that stats: T n i ji j Whr T is th traction/strss vctor at a point on a plan with normal vctor n T th strss tnsor is symmtric. Dfinition A squar matrix, M, can bcom diagonalizd into anothr matrix, D by drivingc, so that

2 -1 D = C MC In th cas that M is th strss tnsor, D bcoms a dscription of th sam strss fild from th prspctiv of a nw, rotatd co-ordinat systm. From th point of viw of this nw strss matrix M is th strss dscribd in th old co-ordinat systm. So, in diffrnt words diagonalization givs th componnts of strss in a nw, rotatd coordinat systm. That is, w start with a gnral matrix: σ σ σ σij σ σ σ σ σ σ and nd with a simplr matrix: σ ij σ σ σ33 Componnts ar not th sam bfor and aftr th diagonalization. Th variabl nam is th sam but th valus ar diffrnt. Whn a matrix is diagonalizabl, it mans that th thr basis vctors in th nw Cartsian coordinat systm ar paralll (in th gnral cas) to thr non-basis traction vctors in th old coordinat systm. Ths spcial thr vctors formrly in th old coordinat systm hav nw componnts in th nw coordinat systm. Ths nw componnts ar th rows of th nw (diagonalizd) strss tnsor. Diagonalization rquirs us to find ths vctors. Eignvalus and Eignvctors In othr words, this mans that: ˆx 1 U and ˆx V and ˆx 3 whr th basis vctors ar in th nw coordinat systm (prims) and ar paralll to vctors that wr formrly in th old coordinat systm. Th nw basis vctors and thir corrsponding formr slvs ar calld ignvctors. W can xprss this anothr way: U 1 U, W

3 V V, W 3 W whr and ar constants, known as ignvalus. Lt s tak a simpl xampl and considr only vctors in a -coordinat systm that ar actd upon by som matrix that w wish to diagonaliz (our targt). U 5 U 1 1 U U U 1 U1 1 U U Also, Th simultanous solution to this problm is givn by Cramr s Rul: Now, So, 5U U U U U U U U U U to obtain solutions to U 1 This is also known as th charactristic quation of matrix M You should work out that can hav two valus: 6 or 1 Whn =1 thn w hav Ths ar

4 DIAGO NALIZATION O F THE STRAIN TENSO R, AN EXAMPLE W will look at th diagonalization of strain instad of th cas of strss as this will lad us in to th nxt sction. Vrsion of 5 Jun 01 Corrctions and Suggstions to Vinc_Cronin@baylor.du How to find th ignvalus and ignvctors of a symmtric x matrix Introduction W will lav th thortical dvlopmnt of ignsystms for you to rad in txtbooks on linar algbra or tnsor mathmatics, or from rliabl sourcs on th wb such as thos listd in th rfrncs sction at th nd of this documnt. Hr, w accpt that for any givn strss or strain tnsor, a coordinat systm can b idntifid in which all of th shar strsss or shar strains hav zro valu, and th only non-zro valus ar along th diagonal of th tnsor matrix. Th ignvctors ar paralll to thos spcial coordinat axs, and th ignvalus ar th valus along th diagonal. Anothr way of charactrizing thm is that th ignvctors ar along th principal dirctions of th strss or strain llipsoids, and th ignvalus ar th magnituds of th principal strsss or strains. Lt s call th squar matrix w ar analyzing matrix M. M = é ê ëê W want to find all possibl valus for a variabl w will call that satisfy th following charactristic quation: d f g û dt(m I) = 0, or using anothr common symbol for dtrminant, M - li = 0, Matrix I in th prcding quations is th idntity matrix

5 é (I = ê ë and variabl is an ignvalu. Thn û if M is a x matrix) é d (M I) = ê f ëê Th charactristic quation is g û - é l 0 é ê = ê ë 0 l û ëê (d - l) f (g - l) û (d - l) f (g - l) = 0 This dtrminant can b unpackd as th quation or, simplifying, ((d )(g )) (f ) = 0 d +gd gf Th rsult is a quadratic quation. You might rmmbr how to solv a quadratic quation from dp in your childhood. In cas you don t, th gnral solution for th quadratic quation is (a x ) + (b x) + c = 0 x = -b ± b - 4ac This xprssion, which is known as th quadratic function, yilds two answrs for x that satisfy th quadratic quation. In our cas, th quivalnt trms in th quadratic function ar noting that [(d + g)] = (d + g), and a a = 1 b = (d + g), c = ((d g) f )) Consquntly, th ignvalus ar givn by th quadratic function

6 (d + g) ± l = (d + g) - 4(dg - f) or, somwhat simplifid, (d + g) ± l = 4 ( ** f ) + d - g ( ) W now hav two valus of that satisfy our quadratic quation, and ths ar th two ignvalus of our x matrix. W will rfr to th largr ignvalu as 1, and th smallr ignvalu is. Now w nd to find th ignvctors that corrspond to 1 and, rspctivly. Rturning to our xampl using matrix M, w hav th following quation to solv to find th ignvctor associatd with 1. é ê ë 0 0 é = ê û ëê d - l 1 f g - l 1 é ê û ëê x 1 y 1 û, which can b rstatd as ((d 1 ) x 1 ) + ( y 1 ) = 0 and (f x 1 ) + ((g 1 ) y 1 ) = 0. Th bst w can do with ths two quations and thir two unknown valus (x 1 and y 1 ) is to dtrmin how on of th unknowns rlats to th othr. In othr words, w can dtrmin th slop of th vctor. If w arbitrarily choos x 1 = 1, w can rarrang ithr or both of th prvious quations to dtrmin th valu of y 1 whn x 1 = 1: or y 1 = - ( d - l 1) y 1 = - f ( ) g - l 1 ( = l - d ) 1 An ignvctor that corrsponds with ignvalu 1 has th following coordinats: Th lngth or norm of that vctor is ( ) l 1, 1 - d î ý

7 ( ) æ l d Rcalling that a unit vctor is a vctor whos lngth is 1, w can find th unit vctor associatd with any vctor of arbitrary lngth by dividing ach componnt of th initial vctor by that vctor s lngth or norm. Th unit ignvctor that corrsponds to ignvalu 1 is 1 æ ( l d) î, ( l 1 - d) æ ( l 1 - d) 1 + W rpat th procss to find th coordinats of th unit ignvctor that corrsponds to ignvalu 1. ý Workd xampl. Givn th matrix M blow, calculat th ignvalus and th corrsponding unit ignvctors. é M = ê ë û Solution. Th ignvalus ar æ (6 + 4) + ( 4 *3* 3) + ( 6-4) = and æ (6 + 4) - ( 4 *3* 3) + ( 6-4) = W rfr to th largr ignvalu (8.168) as 1, and th smallr (1.8377) is. An ignvctor associatd with 1 is

8 ( ) , ý. î 3 Th unit ignvctor associatd with 1 is found by dividing th individual componnts of th prvious vctor by th vctor s lngth î æ ( ) 3, ( ) 3 æ ( ) 1 + An ignvctor associatd with is 3 ( ) 1, ý î 3 and th corrsponding unit ignvctor associatd with is ý = {0.8114, } î 1 ( ) æ , ( ) 3 æ ( ) ý = { , } In th spcific cas in which w hav a -D Lagrangian strain tnsor é ij = ê ëê û th ignvalus ar givn by l = ( 11 + ) ± ( ) + ( 11 - ) Th largr ignvalu is 1, and th smallr ignvalu is. corrsponding to 1 is. An ignvctor

9 and an ignvctor associatd with is 1, l ý, î 1 1, l - 11 ý. î 1 Th unit ignvctors can thn b dtrmind by dividing ach of th componnts of ths vctors by thir lngth or norm. Th unit ignvctor associatd with ignvalu 1 is 1 æ ( l ) 1 î, ( l 1-11 ) 1 æ ( ) l Th unit ignvctor associatd with ignvalu is 1 ý î 1 ( ) æ l , ( l - 11 ) 1 æ ( ) l ý. Rfrncs Frguson, J., 1994, Introduction to linar algbra in gology: London, Chapman & Hall, 03 p., ISBN Onlin rsourcs Eignvalu calculator, accssd 5 Jun 01 via Wisstin, E.W., Eignvalu: MathWorld A Wolfram wb rsourc, accssd 5 Jun 01 via Wisstin, E.W., Eignvctor: MathWorld A Wolfram wb rsourc, accssd 5 Jun 01 via Wisstin, E.W., Quadratic quation: MathWorld A Wolfram wb rsourc, accssd 5 Jun 01 via

10 Eignvalu calculator, accssd 5 Jun 01 via