MAE4700/5700 Finite Element Analysis for Mechanical and Aerospace Design

Transcription

1 MAE4700/5700 Finite Element Analysis for Mechanical and Aerospace Design Cornell University, Fall 2009 Nicholas Zabaras Materials Process Design and Control Laboratory Sibley School of Mechanical and Aerospace Engineering 0 Rhodes Hall Cornell University Ithaca, NY

2 Mathematical preliminaries Let us consider a domain with interior and boundary. Usually the boundary is defined parametrically as follows: x(), s y() s, where s is the arc length along. For a function f(x,y) on the boundary, we can write: f ( s) = f( x( s), y( s)), s. We denote the primary (scalar) variable in 2D as u(x,y). Until further notice, we assume that all functions used here are smooth enough for the operations shown to be valid. 2

3 Mathematical preliminaries Recall that the gradient uxy (, ) is the rate of change of u as we move from (x,y) to nearby locations: uxy (, ) uxy (, ) uxy (, ) = i+ j x y Many times we will interpret this as a mathematical operation of the differential operator on uxy (, ). = i+ x y j The gradient uxy (, ) determines the rate of change of u at (x,y) in any direction The rate of change of u in a given direction t = cosθ i+ sinθ j is given as: duxy (, ) uxy (, ) uxy (, ) uxy (, ) t = i = cos θ + sin θ dt x y 3

4 Flux σ ( xy, ) σ ( xy, ) Similarly to the gradient uxy (, ) the flux is a vector-valued function or vector field. The flux () s at the boundary can be decomposed in normal and tangential components: σ () s σ n () s n = i () s σ () s = σ() s τ() s τ σ 4

5 Flux Consider an arbitrary part containing point ( x0, y0) of the domain. Consider the normal flux σ () s σ () s n n = i () s across the boundary The total flux across this boundary is ω. Σ = ω ω ω σ () sds n σ ( xy, ) Σ A ω If you divide ω by the area of the subregion, we can view the result as the amount of flowing into per unit area. σ ω 5

6 Divergence of the flux The limit of Σ ω / Aω σ n ( sds ) / Aω as decreases in size always ω containing point ( x0, y0) is called the divergence of the flux at : div P 0 σ ( x, y ) ω = 0 0 Taking as the square region shown, then: Σ ω =Δσ xδ y +Δσ yδx σ ( xy, ) Dividing by the area ΔxΔy and taking the limit Δx, Δy 0 : σ (, ) y ( xy, ) x xy σ divσ ( x, y) = + = i + j ( σ ) xi + σ y j σ x y x y So σ is the density of the net flux per unit area at a point ω 6

7 Gauss divergence theorem The total flux out of the region is then: Σ= σdxdy We can now write using an earlier expression: σdxdy σ nds This is the Gauss divergence theorem. Σ= = 7

8 Constitutive equation and balance law Constitutive equation: We consider problems with the following constitutive equation between the flux σ and the state variable u(x,y). σ ( x, y) = k( x, y) u( x, y) material modulus k( x, y) > 0in Conservation principle: Within any portion of the domain, the net flux across the boundary of that part must be equal to the total quantity produced by internal sources. ω σ nds = ω fdxdy f is source per unit area 8

9 Balance law Using the divergence theorem: ω σ nds = fdxdy ω ( σ f ) dxdy = 0 for all ω. ω Since ω is arbitrary, we conclude that the local form of the balance equation is: σ ( x, y) = f( x, y) To make things more interesting, we assume an additional source term proportional to u(x,y): σ ( xy, ) + bxyuxy (, ) (, ) = f( xy, ) 9

10 Conservation principle at interfaces Consider an interface Γ separating two materials with different modulus k and k 2. Take a thin strip of the body around a point on this interface. The balance law takes the form: s ( ) ( + ) ( ) 0 2 Σ= σ n+ σ n ds= s The local balance at the interface reduces to: σ ( ) ( ) () s σ + n () s n σ = () s n = 0, s Γ 0

11 Boundary conditions Let us assume that in the boundary Γ we apply natural boundary q 2 conditions: () s () s n () n s = () s = = ps ()(() us us ()), σ σ σ s 2 On the boundary Γ, we assume essential u boundary conditions: us () = us (), s

12 Summary of the problem of interest What we are given:, 2 and the interface Γ The source f = f( x, y) in i, i=,2 ki = ki( xy, ),( xy, ) i, i=,2 The material moduli bi = bi( xy, ),( xy, ) i, i=,2 The boundary coefficients ps (),() us in 2 or σ () s in 2 With this data we want to compute u(x,y) in : ( kxy (, ) uxy (, )) + bxyuxy (, ) (, ) = f( xy, ),( xy, ), i=,2 us () ks () = ps ()[() us us ()], s 2 or n us () ks () = σ (), s s n us () = us (), s 0, k u n = s Γ 2 i 2

13 A note regarding the boundary conditions The special case in which b=0 and when only natural us () BC of the form ks () = σ (), s s 2 is applied and when n 2, needs special attention: The solution u(x,y) can only be determined up to a constant For u to exist, the following compatibility condition needs to be satisfied: fdxdy = σds 3

14 Variational problem: Weak form We start by multiplying the residual r(x,y) by a sufficient smooth function w(x,y) and integrate over each domain in which rw is smooth and set the resulting weighted average equal to zero: rxy (, ) = ( kxy (, ) uxy (, )) + bxyuxy (, ) (, ) f( xy, ) ( kxy (, ) uxy (, )) bxyuxy (, ) (, ) f( xy, ) + wxydxdy (, ) + ( kxy (, ) uxy (, )) bxyuxy (, ) (, ) f( xy, ) + wxydxdy (, ) = 0 2 We need to integrate by parts the first terms in each of these integrals 4

15 2 Integration by parts gives: Variational problem ( kxy (, ) uxy (, )) bxyuxy (, ) (, ) f( xy, ) + wxydxdy (, ) + ( kxy (, ) uxy (, )) bxyuxy (, ) (, ) f( xy, ) + wxydxdy (, ) = 0 kxy (, ) uxy (, ) wxy (, ) bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) + dxdy+ kxy (, ) uxy (, ) wxy (, ) bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) + dxdy 2 ( wk u) dxdy ( wk u) dxdy = 0 2 The last 2 integrals can be simplified with use of the divergence theorem. 5

16 2 ( ) ( 2) Boundary of the domain Variational problem kxy (, ) uxy (, ) wxy (, ) bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) + dxdy+ kxy (, ) uxy (, ) wxy (, ) + bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) dxdy u k wds k n Boundary of the domain 2 u wds = 0 n We can separate integration over Γ in the last 2 terms: u u u u k wds k wds= k wds k wds n n n n ( ) ( ) ( ) Γ ( ) Γ 2 2 u u + ( k ) wds + ( k ) 2wds n n Γ Γ evaluated in region evaluated in region 2 6

17 Variational problem u u u u u u k wds k wds= k wds k wds+ ( k ) wds+ ( k ) 2wds n n n n n n ( ) ( 2) ( ) Γ ( 2) Γ Γ Γ evaluated in region evaluated in region 2 Note that the outward normal n to region is the negative of n 2 at each point on Γ. ( + ) ( ) u u ( + ) u ( ) u ( k ) wds+ ( k ) 2wds= ( k + k ) wds n n n n Γ Γ Γ evaluated in region evaluated in region 2 7

18 Variational problem The last integral is nothing else but: w σ () n s, which is 0! Finally, we can summarize our weak form as: 2 ( + ) ( ) u u ( + ) u ( ) u ( k ) wds+ ( k ) 2wds= ( k + k ) wds n n n n Γ Γ Γ evaluated in region evaluated in region 2 kxy (, ) uxy (, ) wxy (, ) bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) + dxdy+ kxy (, ) uxy (, ) wxy (, ) + bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) dxdy u u k wds k wds= 0 n n ( ) Γ ( 2) Γ u kxy (, ) uxy (, ) wxy (, ) bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) + dxdy k wds= 0 n 8

19 Variational problem u k( x, y) u( x, y) w( x, y) b( x, y) u( x, y) w( x, y) f ( x, y) w( x, y) + dxdy k wds = 0 n Substitution of the natural boundary condition us () ks () = ps ()[() us us ()], s Γq 2 gives: n kxy (, ) uxy (, ) wxy (, ) + bxyuxywxy (, ) (, ) (, ) f( xywxy, ) (, ) dxdy+ pu ( uwds ) = 0 for all admissible functions w( x, y) ( w = 0 on,sin ce we require u( s) = u ( s), s ) We write: Boundary of the whole domain 2 p( u u ) wds = puwds puwds = puwds γ wds γ

20 Space of admissible functions: H () Finally the weak form looks like: k( x, y) u( x, y) w( x, y) b( x, y) u( x, y) w( x, y) + dxdy + puwds = f wdxdy + γ wds for all admissible functions w( x, y), w = 0on 2 2 What is the class of admissible functions? x y 2 2 w w w dxdy < We indicate this space of functions as reflecting that first derivatives are square integrable indicates the domain over which these functions are defined. H ( ), 20

21 Summary of weak problem Find a function uxy (, ) H( ), such that us () = us (), s and the following holds: k( x, y) u( x, y) w( x, y) b( x, y) u( x, y) w( x, y) + dxdy + puwds = f wdxdy + γ wds for all functions w x y H with w = on (, ) ( ) We can repeat the above weak statement using matrix notation as follows (more appropriate for FEM implementation) Find uxy (, ) H( ), us () = us (), s, such that wxy (, ) H( ) withw= 0on the following holds: T ( w) k( x, y) u + b( x, y) u( x, y) w( x, y) dxdy + puwds = f wdxdy + γ wds 2 2 2x x2 x 2

22 Matrix form of the weak statement Find uxy (, ) H( ), us () = us (), s, such that wxy (, ) H( ) withw= 0on the following holds: T ( w) k( x, y) u + b( x, y) u( x, y) w( x, y) dxdy + puwds = f wdxdy + γ wds 2 2 2x x2 x The gradient operator here is defined as a column vector: u x x T w w =, u =, ( w) = u x y y y We will be interchanging the 2 notations here hoping that from the context you know which notation is implied. 22

23 Appendices Some extra slides are included here with: mathematics background, proof of equivalence of weak and strong formulations, Examples of deriving weak forms for 2D BVP, Including anisotropy (e.g. generalized Fourier s law), etc. 23

24 Equivalence between the weak and strong problems Consider the following simplified BVP and the corresponding weak form: ( qxy (, )) = f( xy, ),( xy, ) where : q = k us u, () () k s q n = q on Γ q, and u = u on Γu, with Γq Γu Γ( ) n The weak form is: Find uxy H w H withw on (, ) ( ): ( ) = 0 Γu : q wdxdy + fwdxdy = qwdγ Γ q We want to show that this weak form is equilavent to the strong problem (recover the PDE + natural BCs) 24

25 Equivalence between the weak and strong problems q wdxdy + fwdxdy = qwdγ Γ Integration by parts of the first term gives the following: q wq nd Γ w qd+ fwdxdy = qwdγ Γ Γ wq ndγ+ wq ndγ w q f d qwdγ= q ( ) 0 Γ Γ Γ u q q w q n q dγ+ wq ndγ w q f d= 0 w H ( ) with w = 0on Γ ( ) ( ) Γ Γ u q u 25

26 Equivalence between the weak and strong problems q wq n qdγ+ wq ndγ w q f d= 0 w H( ) withw= 0onΓ ( ) ( ) Γ Γ u Deriving the PDE first: Select w as follows: w = ψ( x) q f, withψ = 0on Γ and ψ > 0on ( ) The first equation above then results in: ψ q f d= q f = in ( ) u 26

27 Equivalence between the weak and strong problems Γ q wq n qdγ+ wq ndγ= 0 w H( ) withw= 0onΓ ( ) Γ u We next derive the natural BC: Select w as follows: w = ζ( x) q n q, with ζ = 0on Γ and ζ > 0on Γ ( ) The first equation above then results in: Γ q ζ q n q dγ= q n q= onγ ( ) u q u q 27

28 Generalized Fourier s law In general, the flux and temperature gradient are related with the anisotropic (generalized) Fourier s law: u q qx kxx kxy x [ D] u q = y kyx k = yy u 2x2conductivity matrix 2x y For isotropic materials k 0 [ D] = = k [ I] k 0 2 x 2 identity matrix For a 2D BVP with prescribed flux on Γ q and temperature on Γ u and a source term f, can you show that the weak form in matrix notation is: Find uxy H w H withw on (, ) ( ): ( ) 0 u : = Γ ( ) T w [ D] udxdy = qwdγ+ fwdxdy 2x2 2x Γq x2 28

29 Appendix: Green s theorem 0 If uxy (, ) C int egrable, then: ud= und Γ Γ Recall the one-dimensional version of this (where the boundary is only 2 points): du dx = un Γ = u ( x = L ) u ( x = 0) dx 29

30 Appendix: Divergence theorem 0 If qxy (, ) C int egrable, then: qd= q nd Γ Γ Note that: q x q y q = + x y Explicilty, we can write the divergence theorem as: q x x q d ( qxnx qyny) d Γ y + = + Γ y 30

31 Appendix: Green s theorem This is the analogue of integration by parts in D: w qd= wq ndγ w qd Γ The proof of this is simple if you notice that: w qx w qy ( wq) = ( wqx) + ( wqy) = qx + w + qy + w x y x x y y ( wq) = w q + w q This together with the divergence theorem give: w qd= ( wq) d w qd= wq ndγ w qd Γ 3

32 Appendix: Green s theorem in D w qd= wq ndγ w qd Γ The D analogue of Green s theorem is as follows: q w = Γ x w d wqxnd qxd x x Γ or more explicitly as: q w w w dx= wqn qdx= wq wq qdx ( ) x x Γ x x x= L x x= 0 x x x x Recall that we had used this compact form in D to treat boundary conditions on the left (x=0) and the right (x=l) end points in a unified way 32

33 Example problem on divergence theorem Consider the vector function q= ( qx. qy) = (3 x y+ y,3 x + y ), defined in the domain shown in the figure below. Demonstrate the validity of the divergence theorem, i.e. qd= q nd Γ Γ 33

34 Example problem on divergence theorem From q= ( qx. qy) = (3 x y+ y,3 x + y ), we compute the following: q x q y Using this, we compute qd as follows: 2 2 ( 6 ) (3 ) 6 3 q= + = xy + y = xy+ y x y 2 0.5x qd= (6xy + 3 y ) dydx = [3 x( 0.5 x) + ( 0.5 x) ] dx = y = 0.5x dx 34

35 Example problem on divergence theorem Also: Γ q ndγ= q n d Γ+ q n d Γ + q n d Γ AB (0, ) dx BC 5 5 CA (,0) dy (,2) dx (2) 5 Note that on segment BC: ds = dx and n = (, 2). Thus Γ q ndγ= (3 x+ y ) dx+ (3 x y + y ) + 2(3 x+ y ) ( dx) + ( 3 x y y )( dy) x dx q ndγ= 6 + (3 ( 0.5 ) ( 0.5 ) ) 2(3 ( 0.5 ) ) x x + x + x+ x dx Γ 0 q nd Γ= Γ y = 0.5x 2 2 q ndγ= =.5 ds = dx + dy = dx 2 2 dx 5 + = dx 2 2 Γ qd= q nd Γ Γ 35

36 Example: Deriving the weak form Consider isotropic heat conduction in the domain shown with the boundary conditions indicated. Provide a complete weak statement of the problem. ( kxy (, ) T( x, y)) = f( x, y) We start by multiplying with the weight function w and integrating over the domain: ( kxy (, ) T( x, y)) f( x, y) wxy (, ) dxdy = 0, sufficiently smooth w 36

37 Example: Deriving the weak form ( kxy (, ) T( x, y)) f( x, y) wxy (, ) dxdy = 0, sufficiently smooth w Integration by parts gives: T k T w fw dxdy k wds = 0, w H ( ), with w = 0 on Γ n Γ +Γ q h T The boundary term can be further simplified as: T k wds = ( q) wds + ( h( T T )) wds = qwds htwds + ht wds n Γ +Γ Γ Γ Γ Γ Γ q h q h q h h 37

38 Example: Deriving the weak form u k T w fw dxdy k wds = 0, w H ( ), withw= 0 onγ n Γ +Γ The final weak form is then: q h Find T ( x, y) H ( ), with T = T on ΓT such that + = + = Γ k T w fw dxdy htwds qwds ht wds, w H ( ), with w 0on Γ Γ Γ h q h In matrix form, we can also write the weak form as: T ( w) k T fw dxdy 2 x x x2 + htwds = qwds + ht wds Γ Γ Γ h q h w H ( ), with w = 0on Γ T T T 38