MEEN 618: ENERGY AND VARIATIONAL METHODS

Size: px

Start display at page:

Download "MEEN 618: ENERGY AND VARIATIONAL METHODS"

Caroline Shepherd
6 years ago
Views:

JN Reddy - 1 MEEN 618: ENERGY AND VARIATIONAL METHODS WORK, ENERGY, AND VARIATIONAL CALCULUS Read: Chapter 4 CONTENTS Work done External and internal work done Strain energy and strain energy density

1 JN Reddy - 1 MEEN 618: ENERGY AND VARIATIONAL METHODS WORK, ENERGY, AND VARIATIONAL CALCULUS Read: Chapter 4 CONTENTS Work done External and internal work done Strain energy and strain energy density Complementary strain energy Strain energy and complementary strain energy of Trusses, Torsional members, and beams The principle of minimum total potential energy Virtual work done Elements of variational calculus

2 JN Reddy - Work done Magnitude of the force multiplied by the magnitude of the displacement in the direction of the force: W Fu F u A du F B dw Fdu W B A Fdu

3 JN Reddy - 3 Energy is the capacity to do work Fs = ke k k m e e m k F F Fs = ke Work done F F W Fe F s f F s de F de, F ke e s s e 1 1 E Fs de ke ke (strain energy) e e F s e de e Note that de W de e e

4 JN Reddy - 4 EXTERNAL AND INTERNAL WORK IN A DEFORMABLE BODY Work done by external forces f d d t d d WE ( ) ( ) d ( s) ( sd ) fx ux t u In calculating the external work done, the applied (external) forces (or moments) are assumed to be independent of the displacements (or rotations) they cause in a body. JN Reddy Work and Energy 4

5 JN Reddy - 5 Work done by internal forces (1D) A STRAIN ENERGY DENSITY AND STRAIN ENERGY u1 u1 u du = dx = dx x1 σ11 11 σ x 1 dx 1 A d dx d ( Adx ) U * U du ( Adx ) U ( ) du, U U d JN Reddy d Work and Energy 5

6 JN Reddy - 6 JN Reddy Strain energy of a 3D solid STRAIN ENERGY DENSITY AND STRAIN ENERGY U U ( ) d, U U d ij ij ij 1 ( x): ( x)d ij ij U ij d ij d For a linear elastic body, we have If the only energy stored in the body is the strain energy, we write U W I Work and Energy 6

7 COMPLEMENTARY STRAIN ENERGY DENSITY AND COMPLEMENTRY STRAIN ENERGY Complementary strain energy for 1D JN Reddy - 7 U * d U A d dx d ( Adx ) * 1 U ( ) du, U U d * * * * 11 du ( Adx ) Complementary strain energy for 3D JN Reddy ij U ( ) d, U U d * * * ij ij ij Work and Energy 7

8 JN Reddy - 8 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES A truss is a collection of uniaxial members, each of which can only carry axial force (compressive or tensile). The members are connected through pins that allow relative rotation. The strain energy and complementary strain energy of a truss with N members each having ist own length, area of cross section, and modulus are N N ( i) ( i) ( i) i i i1 i i1 U U d A LU, U ( ) i i i d i N N i * *( i) *( i) *( i) i i U i1 i i1 U U d A LU, ( ) d i i i

9 AN EXAMPLE K,, K, (a) Pin connections b C B 1 a = 3 K = material constant A i = cross-sectional area of the i th member C Find the (1) strain energy and () complementary strain energy of the truss JN Reddy ( 1) BO BO ( ) a u v a u v au u u BO a a a a 1 1 ( ) CO CO ( a u) ( b v) a b u v ( bv au) 1 1 CO a b a b 1 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES - AN EXAMPLE (1) From Fig. (b), we have the following strains: bv au bv au 3v 3u 1 1 a b a b 4a O P F F 1 P O (b) b+ v B 1 a u O u O v

10 JN Reddy - 1 (1) continued [note that the stress in member 1 is compressive and it is tensile in member ; see part (b) to confirm this] The strain energy densities of each member is 1 ( 1) ( 1) 3 3 ( 1) ( 1) ( 1) ( 1) ( 1) K ( 1) K u 3 U d K d ( ) ( ) 3 ( ) ( ) ( ) ( ) ( ) K ( ) K 3v 3u U d K d The total strain energy of the truss is STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES - AN EXAMPLE a 3 3 4a, U U d U d A LU A LU V ( 1) ( ) ( 1) ( ) V KA a u 4KA a 3v 3u 3 a 3 3 4a

11 JN Reddy - 11 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES - AN EXAMPLE () From Fig. (a), we have Fsin P, F1F cos, 3, F1 3P, F P The complementary strain energy densities of each member are ( 1) ( 1) ( 1) *( ) ( ) ( ) ( ) 1 ( ) 3P U d 3 d, K 3K AK 1 ( ) ( ) ( ) *( ) ( ) ( ) ( ) ( ) 1 8P U d d 3 K 3K 3 AK The total complementary strain energy of the truss is 3 3 * *( 1) *( ) 1 3 3Pa 16Pa AK 1 3AK U U AL U AL

12 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TORSION OF CIRCULAR SHAFTS The shear stress in a shaft subjected to torque Tr x( r) J The strain energy density and strain energy are r TL x x, L GJ JN Reddy - 1 d = diameter of the shaft J = polar moment of inertia G U d 4 G d G L A R x x x x x x x x G 1 L d/ r 1 L x U d G dr rd dx GJ dx L L The complementary strain energy density is x * 1 U xdx x G 1 1 L d/ * 1 Tr 1 L T U x d dr rddx dx v G G J GJ A r T x + r R = 5. d x dr ( r)

13 JN Reddy - 13 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF STRAIGHT E-B BEAMS The strain energy density and strain energy of the Euler-Bernoulli (E-B) beam are (linear elastic material) xx xx E E du d w U xx dxx Ex ( ) xx dxx xx z dx dx L E du d w 1 L du d w U Ud z dadx A xx D xx dx v A dx dx dx dx A E( x) da EA, D z E( x) da EI xx A The complementary strain energy density and complementary strain energy of the Euler-Bernoulli beam are xx A xx xz * xx 1 xz N 1 VQ U xx dxx xz d xz E G E A G Ib L * * 1 LN M fv s A U UdAdx dx, fs Q ( z) da A EA EI GA I b A

14 STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF E-B BEAMS - AN EXAMPLE AN EXAMPLE: Determine the complementary strain energy of the (determinate) frame structure. a B JN Reddy - 14 A b D P v P h N P, V P, BA BA v v BA M P bp x All members have the same E, A, and I N P, V P, M P h DB h DB v DB v h P h x P v P h P h Pb v P v Pb+ P a v P v Pb v x h P v P h V = P h b 3 * Ph 1 fp s v Pb h Pb v fpb s v UDB Pv x dx, EA EI GA EA 6EI GA a * Pv 1 fp s h UBA Pvb Phx dx EA EI GA Pa v fpa s h Pab v PPab v h Pa h EA EI 3 GA N P h V = P v = h P v x v x M= Px Pb v P h N = P v v P v M= Pb+ P x h P

15 JN Reddy - 15 TOTAL POTENTIAL ENERGY AND COMPLEMENTRY ENERGY The potential energy of a 3D solid ( W U, W V ) I E E 1 ( u) U VE ( ): d d d σε ε fu tu The complementary energy of a 3D solid 1 * ( σ) : ( )d d d σ εσ fu tu For an E-B beam, they are L L 1 du d w ( u, w) EA EI dx fu qwdx... dx dx L L 1 N M fv * s ( xx, xz ) dx fu qwdx... EA EI GA

16 JN Reddy - 16 THE PRINCIPLE OF MINIMUM TOTAL POTENTIAL ENERGY Minimum nature of PE: L L 1 du d w ( u, w) EA EI dx fu qwdx dx dx L L 1 du d w EA EI dx fu qwdx dx dx L L 1 du d w EA EI d dx dx x fuqwdx L du du d w d w EA EI dx dx dx dx dx L 1 du d w ( u, w) EA EI dx ( ) ( ) dx dx ( uw, ) ( uw, ) because u and w satisfy the equilibrium equations

17 JN Reddy - 17 VIRTUAL WORK Virtual displacements are those which satisfy the homogeneous form of the specified kinematic boundary conditions, but otherwise arbitrary. dw u( ) = uˆ =, w( ) = wˆ, ˆ dx = zw, Set of admissible displacements ŵ u( x) uˆ ax, w( x) wˆ ˆ xbx Set of admissible virtual displacements u ax, w bx ; u axa x, w bx bx Virtual work done by actual forces respective displacements is L ( qf, ) ˆ q x= F xu, in moving through their W qs ( ) ws ( ) dsfcos ul ( ) Fsin wl ( )

18 JN Reddy - 18 COMPLEMENTARY VIRTUAL WORK Virtual forces are those which satisfy the self-equilibrium conditions, but otherwise arbitrary. u( ) = w( ) = dw dx x = = z, w x, u qx ( ) L f( x) F P The set ( P, F) is clearly is in self-equilibrium. P z, w x, u L F P The virtual work done M = F L F by virtual forces in moving through actual Displacements is * W Pu( L) F w( L) Pu( ) F w( ) M ( ) Pu( L) F w( L)

19 JN Reddy - 19 VIRTUAL WORK DONE Internal virtual work done in a 3D body W Ud ( ): δd σ ( ε ) δε d I ij kl ij Internal complementary virtual work done W U d ( ): δd ε ( σ ) δσ d * * I ij kl ij External virtual work done in a 3D body W f ud t ud E External complementary virtual work done * W f ud t ud E Total virtual virtual work done W W W ; W W W * * * I E I E u

20 JN Reddy - VIRTUAL WORK DONE FOR E-B BEAMS Internal virtual work done L du dw WI N M dx dx dx Internal complementary virtual work done L * ( ) ( 1) ( 1) I xx xx xz W N M V dx External virtual work done in a 3D body b d WE f ( x) u( x) dx q( x) w( x) dx VWpointforces a c External complementary virtual work done b d * WE f ( x) u( x) dx q( x) w( x) dx VW a c

21 JN Reddy - 1 COMPLEMENTARY VIRTUAL WORK DONE: AN EXAMPLE L * * N M V WI U N M V dx, EA EI K sga W V vp up * * E E v h N P, N P, V P, V P DB h BA v DB v BA h M P x, M P bp x DB v BA v h b * NDB MDB V DB U DB NDB MDB V DB dx EA EI K sga Pb Pb Pb Pv, EA EI K GA 3 h v v P h 3 s b * NBA MBA V BA U BA NBA M BA V BA dx EA EI K sga Pa v 1 a Pb v a Pb h Pv EA EI 1 3 a a Pa h Pb v P h h EI 3 K sga P δ P h δn = δp h δv δp h δ P v x B b a A δ P v δv = δp v δm v D δ P v h δ P h δpb+ δp a x = δpx v δ P h δ Pb v δn = δp v δ P h δ P v = δ = δ h + δ v M P x Pb

22 JN Reddy - FIRST VARIATION and VARIATIONAL SYMBOL The delta operator used in conjunction with virtual quantities has special importance in variational calculus. The operator is called the variational operator because it is used to denote a variation (or change) in a given quantity. u uv, v u; ( u) ( v) ( v) ( u) FFxu (, vu, v) Fxuu (,, ) F F F( xuu,, ) v v u u ( v) F ( v)( v) F F( xuu,, )! u! uu F F v v O( ) u u

23 JN Reddy - 3 FIRST VARIATION OF A FUNCTION OF A DEPENDENT VARIABLE Define lim F F F F v v u u Alternatively, F F F F v v u u u u u u df( u v, uv) F d F ( uv) F ( uv) ( uv) ( uv) F F F F v v u u u u u u

24 JN Reddy - 4 ANALOGY BETWEEN TOTAL DIFFERENTIAL AND VARIATIONAL OPERTOR Analogy Properties F F F df dx du du x u u F F F F x u u x u u () 1 ( FF) FF, ( ) ( FF) FFFF F FFF F ( 3),( 4) ( F) nf ( ) F F F n n (1) () du dv d d ( ) ( v) ( u) dx dx dx dx ( ) a a a a udx vdx vdx udx

25 JN Reddy - 5 FIRST VARIATION OF A FUNCTIONAL A functional A functional F is a mapping (or operator) from a vector space U into the real number field R. Thus, if uu (i.e. u is an element of U), then Fu ( ) is a real number. The First Variation of a Functional b I( u) F( x, u, u) dx, u a du dx b b b F F I( u; u) F( x, u, u ) dx F dx u u dx a a a u u b F d F F udx u a u dx u u xb xa

26 JN Reddy - 6 FUNDAMENTAL LEMMA OF VARIATIONAL CALCULUS AND EULER EQUATIONS Lemma: If G is an integrable function and is arbitrary in a< x <b and is arbitrary, then the statement implies that ( a) a b G( x) ( x) dx B( a) ( a) ( x) Gx ( ) axband Ba ( ) which are called the Euler equations. If I and u is arbitrary in( ab, ) and at xa and xb, then xb b F d F F I( u; u) udx u a u dx u ux a xb F d F F axb; u dx u u x a

27 JN Reddy - 7 EULER EQUATIONS OF FUNCTIONAL IN D, INVOLVING TWO DEPENDENT VARIABLES Given the functional u I( u, v) F( x, y, u, v, ux, vx, uy, vy) dxdy, ux, etc. x I find the Euler equations if. We have F F F F F F u u u v v v dxdy x y x y u ux uy v vx vy F F F F F F u v u x u x y u y v x v x y v dxdy y F F F F nx n y u nx n y v dxdy ux u y vx v y

28 JN Reddy - 8 EULER EQUATIONS OF FUNCTIONAL IN D, INVOLVING TWO DEPENDENT VARIABLES u If is arbitrary in and on, then the Euler Equations are as follows: F F F u x u x y u y in F F F v x v x y v y F F nx ny ux u y on F F nx ny vx vy See the textbook for examples (some will be discussed in the class)

29 Problems with Constraints-1 Problem 1: Find the minimum of the function Fxy (, ) with no constraints. Necessary condition F F df = dx + dy = x y Since dx and dy are arbitrary and independent, we have F F = and = x y

30 Problems with Constraints- Problem : Find the minimum of the function Fxy (, ) subjected to the constraint Gxy (, ) = Necessary condition F F df = dx + dy = x y But dx and dy are not independent of each other because of the constraint, we cannot set F F = and = x y

31 Problems with Constraints-3 Lagrange Multiplier Method: Introduce new function F ( x, y, l) = F( x, y) + lg( x, y) L where l is the Lagrange multiplier. Set F F F = + + l = x y l L L L dfl dx dy d Now dx, dy, and dl are independent of each other; so, we can set FL F G FL F G = + l =, = + l =, x x x y y y F l L = Gxy (, ) =.

32 Problems with Constraints-4 Penalty Function Method: Introduce new function g FP ( x, y, l) = F( x, y) + G( x, y) where g is the penalty parameter. Set F F = + = x y P P dfp dx dy [ ] Sincedx and dyare independent of each other, we can set FP F G FP F G = + gg =, = + gg =. x x x y y y

33 Comparison of the two methods Lagrange Multiplier Met hod: FL F G FL F G = + l =, = + l = x x x y y y. Penalty Function Method: FP F G FP F G = + gg =, = + gg =. x x x y y y We find that in the penalty function method we can compute l= ggx (, y) g g

34 AN EXAMPLE: ALGEBRAIC PROBLEM F (x, y) =x + y 8x + y +1, G(x, y) x y = Lagrange Multiplier Method 4x 8+λ =, y +1 λ =, x y = x =.5, y =1., λ =3. Penalty Function Method 4x 8+γ(x y) =, y +1 γ(x y) = x γ = 8+3γ 4+6γ, y γ = 3γ 1 +3γ Clearly, as γ,wehave lim x γ =.5 =x, γ lim y γ =1. =y γ

35 Penalty Function Method (Example - continued) Table: A comparison of the penalty solution with the exact for various values of the penalty parameter γ. γ x γ y γ G(x γ,y γ ) λ γ λ γ = γ G ( x γ, yγ )

36 AN EXAMPLE: CONTINUUM PROBLEM-1 Problem 3: Find the minimum of the functional Iuv (, ) subjected to the constraint Guv (, ) =. Minimize b Iuv (, ) = ò Fxuu (,,, vv, ) dx a subjected to Gu (, u, v, v ) = Lagrange Multiplier Method æ b F F F F ö = di = ò a du du dv dv dx ç è u u v v ø G G G G = dg= du+ du + dv+ dv u u v v

37 AN EXAMPLE: CONTINUUM PROBLEM- b é F F F F = òa du+ du + dv+ dv ê ë u u v v æ G G G G öù + l du+ du + dv+ dv dx è ç u u v v ø û ú ìé b F G d æ F Göù = òa í ï + l - l d u u u dx + ç u u ïîë ïê è øû ú é F G d æ F Göù ü + + l - l d v ï v v dx + ýdx ê ç v v ë è øú û ï ïþ Select l such that the coefficient of dv is zero. Then the coefficient of du is zero because du is independent.

38 AN EXAMPLE: CONTINUUM PROBLEM-3 b I ( u, v, l) º I( u, v) + ò lg( u, u, v, v ) dx L a b = òa ( F + lg) dx éæ b F Gö æ F Gö di = L ò êç a + l du+ + l du ê ç u u ç u u ëè ø è ø æ F Gö æ F Gö ù + + l dv+ + l dv + dlg dx è ç v vø è ç v v ø ú û b ìé ï F G d æ F G öù = òa í ê + l - + l u ïê ïîë u u dx d ç è u u øú û é F F d æ F Göù ü + + l - + l d v G dlï + ý dx ê v v dx ë èç v v ø û ú ï ïþ

39 AN EXAMPLE: CONTINUUM PROBLEM-3 Penalty Function Method 1 b I ( u, v) º I( u, v) + a (,,, ) P ò g ég u u v v ù ë û dx éæ b F Gö æ F Gö di = P ò êç a gg du gg du ç u u ç u u ë ê è ø è ø æ F Gö æ F Gö ù + gg dv gg dv ú dx ç è v v ø çè v v ø ú û b ï ìé F G d æ F Göù = òa í ê + gg - + gg du ïê ïîë u u dxè ç u u ø ú û é F G d æ F Göù ü + + gg - gg dvïdx v v dx + ý ëê çè v v øû ú ïþ ï l = ggu (, u, v, v ) g g g g g

Chapter 5 Structural Elements: The truss & beam elements

Institute of Structural Engineering Page 1 Chapter 5 Structural Elements: The truss & beam elements Institute of Structural Engineering Page 2 Chapter Goals Learn how to formulate the Finite Element Equations