4 NON-LINEAR ANALYSIS

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1 4 NON-INEAR ANAYSIS arge displacement elasticity theory, principle of virtual work arge displacement FEA with solid, thin slab, and bar models Virtual work density of internal forces revisited 4-1

2 SOURCES OF NON-INEARITY Geometry: Equilibrium equations should be satisfied in deformed geometry depending on displacement. Strain measures of large displacements are always nonlinear. the source to be considered here! Material: Constitutive equation g(,) u 0 may be non-linear. Near reference geometry, truncated Taylor series g ( g /)( /) 0 g u u gives a useful approximation. Contact and follower forces: Constitutive equation of external forces may be nonlinear. Even the simplest contact condition is one sided (inequality) and therefore nonlinear (linear combination of two solutions is not a solution). 4-

EXAMPE. Consider the structure of figure and determine the relationship between the vertical displacement of node (positive upwards) and force F acting on node.

3 EXAMPE. Consider the structure of figure and determine the relationship between the vertical displacement of node (positive upwards) and force F acting on node. Assume that the force-length relationship (small strains and large displacements) is given by N EA( h / h 1) (EA is constant and h is the length when N 0). y x F y h Y h x α X Answer F EA 1 a sin a 1 (sin a) 0, where 1 a sin a u h a Y 4-3

4 Strain definition should not induce stress under rigid body motion of motion of a bar. Strain measure e h / h 1, based on the relative length change, satisfies the criterion. At the deformed geometry, when displacement is u Y, h h cos( I sin) h 1u aj sin h a Y h sin a h u u Y Y uy 1 a sin a h N EA( 1)( 1EA a sin a 1) h, where a u Y h. Virtual work expressions of external and internal forces for one bar element, written at the deformed geometry with length h, are ext W Fu Y and W int N h. As 4-4

5 the structure consists of two bars (internal parts of the bars are same by symmetry), virtual work expression of the structure 1 a sin a 1 W [ F (sin EA a) ] u Y 1 a sin a. Principle of virtual work and the fundamental lemma of variation calculus are valid also in large displacement analysis F 1 a sin a 1 (sin EA a) 0 1 a sin a. The remaining mathematical problem is to find a solution or solutions to the nonlinear algebraic equation. 4-5

6 FORCE-DISPACEMENT REATIONSHIP / 3 non-physical (unstable) F / EA a Numerical method may give mathematically correct but non-physical solutions in nonlinear elasticity. In practice, displacement (solution) may not depend continuously on the force (data). 4-6

m 0 : J p F : f 0 p F : n t r V r V t M : c r V Z A constitutive equation f (,) u 0 Y deformed body u()

7 BAANCE AWS OF MECHANICS; OCA FORMS Given the solution, u, V (usually u 0 ), the goal is to find a new solution, u, V when e.g. external given forces are changed. m 0 : J p F : f 0 p F : n t r V r V t M : c r V Z A constitutive equation f (,) u 0 Y deformed body u() brings the material details into the model. For an unique solution, a displacement boundary condition is needed somewhere on X O z y P V x V fdv tda z P V. O y x 4-7

The principle of virtual work PRINCIPE OF VIRTUA WORK W int ext W 0 u U 4-8 holds at the equilibrium. Therefore, the domain V corresponds to the deformed geometry.

8 The principle of virtual work PRINCIPE OF VIRTUA WORK W int ext W 0 u U 4-8 holds at the equilibrium. Therefore, the domain V corresponds to the deformed geometry. In practice, a mapping (depending on u ) is used so that integrals are over domain V of the reference geometry. A tda W int ( :)( c:) V dv V S E c dv V P W ext V ()() f u dv g u dv u V V gdv P W () t u da (not needed here) ext A A Also (in particularly) strain-displacement and stress-strain relationships of a more precise theory are non-linear! u 0

9 STRAIN MEASURES A rigid body motion should not induce strains! A proper strain measure with this respect is always non-linear in displacement components (small strain h h h) h inear strain 1 h Fc F () I u u c Green-agrange E 1 h [() 1] h ()() E Fc F I u u c u u c Superscript refers to the initial geometry and subscript c denotes conjugate dyad (kind of transpose). Deformation gradient F x I u is one of the key quantities of large c displacement theory. Material coordinate system of initial geometry is usually assumed to be Cartesian so that i / x j / y k / z. 4-9

10 GREEN-AGRANGE STRAIN MEASURE inear strain measure is just a simple example of the numerous useful definitions. The Green-agrange strain has the components (in the basis of the initial geometry) Exx ux, x ux, x u u u u u u E yy uy, y u y, y u u u u u u Ezz 1 uz, z uz, z 1 u u u u u u E xy ux, y u y, x u u u u u u E yz uy, z uz, y u u u u u u Ezx uz, x ux, z u u u u u u x, x x, x y, x y, x z, x z, x x, y x, y y, y y, y z, y z, y x, z x, z y, z y, z z, z z, z x, x x, y y, x y, y z, x z, y x, y x, z y, y y, z z, y z, z x, z x, x y, z y, x z, z z, x. Green-agrange E gives zero strain in a rigid body motion, whereas linear strain does not. inear strain can be taken as an approximation to E valid when strains and rotations of material elements are small! 4-10

11 INEARY EASTIC MATERIA Under the assumption of large displacement and small strains the Green-agrange strain measure does not differ much from the linear setting with small displacements and small strains. Constitutive equations Exx 1 Sxx 1 E yy 1 Syy C E 1 zz Szz and Exy Sxy 1 Eyz S yz, G Ezx Szx with material parameters C (which replaces E),, and G C /( ) are same as those of the linear case, are assumed to simplify the setting. Also, the uni-axial and two-axial (plane) stress and strain relationships follows just by using strains instead of engineering strains and C instead of E. 4-11

12 EXAMPE. Consider a bar whose left end is fixed and right end is free to move. Assuming that the displacement of the right end with respect to the horizontal initial geometry is given by ux (1) cos 1 sin x u y (1)sin cos 1 y y Y α h h x X (rotation with simultaneous length increase h h), determine the linear strain component xx and the Green-agrange strain component E xx. Answer 1 Exx when 1 and xx (1)cos 1 when 1 4-1

13 Partial derivatives of the displacement components are ux, x (1) cos 1 u y, x (1)sin and ux, y sin uy, y cos 1 inear and Green-agrange axial strain components xx ux, x (1)cos 1 and 1 1 Exx ux, x () ux, x uy, x. The former depends strongly on the rotation angle even when is small although pure rotation should not cause any strains. The latter does not depend on the rotation at all. Also, for small length changes, the Green-agrange strain is close to the relative change of length h / h. 4-13

14 KINEMATICS OF ARGE DISPACEMENT (GRADIENTS) Displacement... r u() Deformation gradient... F r u I u c Green-agrange... ()()() E Fc F I u u c u u c Variation... E F F c Domain element... dv JdV Jacobian... J det F 1 Nanson... nda JF nda c or da JF da 1 c 4-14

15 Piola-Kirchhoff 1... J P F Piola-Kirchhoff... J F S F KINETICS OF ARGE DISPACEMENTS c c (F S P ) Force element... df tda n da nda PndA int Virtual work... wv S : Ec : cj Simple material... S tr() E I E c Analysis may use PK stress. Cauchy (true) stress is obtained from the relationship between the quantities. In practice, the simple constitutive equation applies to isotropic material subjected to small deformations (displacements may be large). 4-15

16 4.1 NON-INEAR ANAYSIS BY FEM Model the structure as a collection of beam, plate, etc. elements. Derive the element contributions e W and express the nodal displacement and rotation components of the material coordinate system in terms of those in the structural coordinate system. Sum the element contributions to end up with the virtual work expression of the e T structure W W. Re-arrange to get W a F() a ee Use the principle of virtual work W 0 a, fundamental lemma of variation calculus for n a to deduce the system equations F() a 0. Find a physically meaningful solution by some of the standard numerical methods for non-linear algebraic equation systems. 4-16

17 BAR Virtual work expression can be expressed in a concise form in terms of initial and deformed lengths of a bar element (linear approximations to the nodal displacements) int h 1 h [() W h CA 1], h h W ext T ux1 1 1 hag. ux 1 u x1 u y1 u z1 z g C, A h u y u x u z x ength squared ()()() x x1 y y1 z z1 h h u u u u u u of the deformed element depends also on the nodal displaments in the y and z directions. Transformation into the components of the structural system follows the lines of linear theory. 4-17

Use the principle of virtual work and assume the constitutive relationship S xx CE xx, in which Green-agrange

18 EXAMPE 4.1. Consider the bar structure shown subjected to large displacements. Determine the relationship between vertical displacement of node ( positive upwards) and force F acting on node. Use the principle of virtual work and assume the constitutive relationship S xx CE xx, in which Green-agrange strain constant. Cross-sectional of initial geometry is A. Exx [( h /) h 1] / and C is y x F y Y x X Answer F CA 1 (sin a)(a sin a) 0 where u h a Y 4-18

19 FORCE-DISPACEMENT REATIONSHIP 3 F F / CA / EA a Solutions with strain definitions e (blue) and () h/ h h. If is small, strain definitions do not differ much E [((1) 1]/ (black) in which

20 In (geometrically) non-linear analysis, equilibrium equations are satisfied at the deformed geometry. Virtual work expressions of internal forces of the bar element and the point force are int h 1 h [() W h CA 1] h h and ext W Fu Y. For element 1, the relationship between the displacement components in the material coordinate system are u u sin and u u cos x Y y Y giving ( Y sin)( cos) Y + sin Y Y h u u u u, E xx 1 h 1 [() 1] asin a h (sin a) a where E xx a u Y. 4-0

21 Then, virtual work expression of the structure becomes (the contribution for bar is the same due to the symmetry). 1 W uy (sin a)(a CAsin a) F uy. Principle of virtual work and the fundamental lemma of variation calculus give 1 W [ F (sin a)(a CAsin a)] u0 Y 1 F (sin a)(a CAsin a)

22 EXAMPE 4.. Determine the nodal displacement uz and uz 3 of the bar structure shown. Use non-linear bar elements and linear approximations. Cross-sectional areas and length of the initial geometry are C 100Nm and external force F 0.05N. Answer uz and uz A 0.01m and 1m 1 Z x z 3 X 1. Elasticity parameter z x 4 F x z

23 The physically correct solution is just one of the mathematically correct solutions to the nodal displacements (in this case the number of solution is 6). The solver for nonlinear analysis returns a real solution with the minimal norm. In the example, when 1m, A 0.01m, C E 100Nm, and F 1/ 0 N: 4-3

24 4. EEMENT CONTRIBUTIONS Virtual work expressions for the elements combine virtual work densities of the model and an approximation depending on the element shape and type. To derive the expression for an element: Start with the large displacement versions of the virtual work densities ext w of the formulae collection. int w and Represent the unknown functions by interpolation of the nodal displacement and rotations (see formulae collection). Substitute the approximations into the density expressions. Integrate the virtual work density over the domain occupied by the element at the initial geometry to get W. 4-4

25 EEMENT APPROXIMATION In MEC-E8001 element approximation is a polynomial interpolant of the nodal displacements and rotations in terms of shape functions. In non-linear analysis, shape functions depend on x, y, and z. Approximation u T N a always of the same form! Shape functions N {( 1,,)(,,)(,,)} n N x y z N x y z N x y z T Parameters a {a a a } 1 n T Nodal parameters a { u, u, u,,, } x y z x y z may be just displacement or rotation components or a mixture of them (as with the Bernoulli beam model). 4-5

26 SOID The model does not contain kinetic or kinematic assumptions in addition to those of nonlinear elasticity theory. T T Exx E E xx xy Exy int V yy [ ] yy yz 4 yz w E C E E G E, E E E E zz zz zx zx T ux gx ext V y y w u g. u g z z u 0 V P gdv A u P tda The solution domain can be represented, e.g, by tetrahedron elements with linear interpolation of the displacement components u( x, y,) z, v( x, y,) z, and w( x, y,) z 4-6

Determine the equilibrium equation for the displacement uz 3 of node 3 with one tetrahedron element and linear approximation.

27 EXAMPE 4.3. A tetrahedron of edge length, density, and elastic properties C and is subjected to its own weight on a horizontal floor. Determine the equilibrium equation for the displacement uz 3 of node 3 with one tetrahedron element and linear approximation. Assume that u 3 u 3 0 and that the bottom surface is fixed and that X Y the geometry and density described is concerned with the initial geometry (gravity omitted). 3 Z,z g Answer: 1 (1 a)a(1 a) F 0 where F 1 1 g u and a Z C Y,y X,x 4-7

28 inear shape functions can be deduced directly from the figure N1 x /, N y /, N z, and N4 1 x / y / z /. Only the shape function of node 3 is 3 / actually needed as the other nodes are fixed. Approximations to the displacement components are u x z 1 uy 0 and uz uz 3, giving uz, x uz, y 0 and uz, z uz 3. When the approximation is substituted there, the non-zero Green-agrange strain component takes the form 1 1 E zz u Z 3 u 3 Z 1 1 Ezz uz 3 u Z 3 uz

29 Virtual work densities of the internal and external forces simplify to (we assume that the material is described by the constitutive equation of linear elasticity theory in which the Young s modulus E is replaced by elasticity parameter C) C() w E S u u u u (1)(1 ) int V :()() Z 3 Z 3 Z 3 Z 3, z w u g g u ext V Z 3. Virtual work expressions are obtained as integrals of densities over the volume occupied by the body at the initial geometry. With a uz 3 / 3 int int int 1 1 V V Z 3(1 a)(a a) V W w dv w C u 6 6 (1)(1 ), 4-9

30 ext 3 ext V V Z 3. W w dv g u 4 Finally, principle of virtual work W 0 with equilibrium equation int ext W W W implies the 3 ( 1) 1 C (1 a)(a a) g 0 6 (1)(1 ) 4. In terms of F 1 1 g 4 1 C the physically meaningful solution is given by 1 a 1 where 1/3 / /3 ( 9F 3 1 7) F. 4-30

31 BAR With the assumptions of the bar model u u()()() x i v x j w x k S S i i etc. in, xx the generic expressions for large displacement analysis for the solid model simplify to int xx xx, w E A CE u x1 f x u x ext w u f x, where Exx u u v w. u y1 u z1 z C, A h u y u z x In FEM, the solution domain (a line segment) is represented by line elements and the displacement components u() x, v() x, w() x by their interpolants. 4-31

32 et us start with the kinematical assumption u u()()() x i v x j w x k (in linear displacement analysis only the first term is accounted for). The kinetic assumption is S S i i. Green-agrange strain ( u u,x etc.) xx Exx u u v w E xx u u u v v w w. Assuming the constitutive equation S xx CE xx, virtual work density of internal forces per unit length of the initial domain becomes (expression is integrated over the cross section) int xx xx and int w E A CE w u f x. 4-3

33 VIRTUA WORK EXPRESSION; BAR Virtual work expression can be expressed in a concise form in terms of initial and deformed lengths of a bar element (linear approximations to the nodal displacements) int h 1 h [() W h CA 1], h h W ext T ux1 1 1 hag. ux 1 u x1 u y1 u z1 z g C, A h u y u x u z x ength squared ()()() x x1 y y1 z z1 h h u u u u u u of the deformed element depends also on the nodal displaments in the y and z directions. Transformation into the components of the structural system follows the lines of the linear theory. 4-33

34 inear approximations to the displacement components give constant values to the derivatives u, v, and w and the Green-agrange strain component relative difference in the squares of lengths: Exx is simply the E xx 1() h 1h h [() 1] () h h and h h Exx. h h As virtual work density of internal forces is constant and the approximation linear Virtual work of internal and external forces int int h 1 h [() W w h h CA 1] h h, W ext T ux1 1 1 hag ux

35 It is noteworthy that PK does not represent the true stress in bar. The constitutive equation for the (true) axial force in terms of Green-agrange strain follows from the relationship between the Cauchy stress and PK stress. Here the relationship J F S F c simplifies to J FSF in which J V / V ha / ha and F h / h giving S h h ha h () A h N h h ha ha ha h h N A S A CE h h. Using the axial force N and the variation h (at deformed geometry) int h 1 h [() W N h h CA 1] h h (the same as earlier). 4-35

36 EXAMPE 4.4. Determine the nodal displacement uz and uz 3 of the bar structure shown. Use non-linear bar elements and linear approximations. Cross-sectional areas and material properties are 1m, A 1/100m, C 100 Nm and F 1/ 0 N. Answer uz 0.085m and uz m 1 Z x z 3 X 1 z x 4 F x z

37 For bar 1, the nodal displacement components of material coordinate system are u x1 uz1 0, u u 3 /, and u 1 u 3 /. As approximations are linear, derivatives are x Z z Z u 3 1 (0) Z uz 3 u u, v 0, 3 1 (0) Z u w Z 3. E xx 1 u 1 u 1 u u. Z 3 (1) Z 3 E Z 3 (1) Z xx 3 When the approximations are substituted there, virtual work expression of internal forces simplifies to (density is constant) W 3 (1)() u CA u u. 4 1 u Z 3 Z 3 Z 3 Z 4-37

38 For bar, the nodal displacement components are ux3 uz 3, ux uz and u z1 uz 0. As approximations are linear, derivatives and the Green-agrange strains take the forms ux ux3 uz 3 u u Z, v 0, and w 0. E xx u u 1 u u (1) u u u u. Z 3 Z Z 3 Z E Z 3 Z (1) Z 3 Z xx When the approximations are substituted there, virtual work expression of internal forces simplifies to u ()(1)(1) Z uz uz uz uz 3 u W u Z Z 3 uz CA. 4-38

39 For bar 3, the nodal displacement components are u 4 u 4 0, u u /, and u u /. As approximations are linear, derivatives and the Green- z Z agrange strain take the forms u 1 () Z uz u u, v 0, and 1 () Z u w Z. x z x Z E xx 1 u 1 u 1 u u. Z ( 1) Z E Z ( 1) Z xx When the approximations are substituted there, virtual work expression of internal forces simplifies to (density is constant) W CA u u u. 4 3 u ( 1)( ) Z Z Z Z 4-39

40 Virtual work expression is sum of the element contributions. By taking into account also the point force contribution 4 W u F Z u (1)()()(1) Z CA uz uz uz u W u Z Z uz 3 uz 4 u Z 3 u Z 1 u (1)( 1)( Z 3 u ) Z u Z Z CA u Z u CA u Z uz F. 4 Principle of virtual work and the fundamental lemma of variation calculus give a nonlinear algebraic equation system for the active dofs uz and u Z 3. However, finding an analytical solution in terms of the parameters of the problem is not possible. Mathematica code of the course gives a real valued solution with the minimal norm for selections 1m, A 1/100m, E C 100 Nm and F 1/ 0 N (that is likely to be the physically meaningful solution when the initial displacement is zero) 4-40

41 4-41

42 EXAMPE 4.5. A bar truss is loaded by a point force having magnitude F as shown in the figure. Determine the equilibrium equations according to the large displacement theory. At the initial (non-loaded) geometry, cross-sectional area of bar 1 is A and that for bar A /. Find also the solution for 1m, F 1/ 0 N. A 1/100m, C 100 Nm and 3 X Answer u m and u 0.5m X Z Z x F 1 x

43 For bar 1, the nodal displacement components of material coordinate system are u x1 uz1 0, ux ux, and uz uz. As the approximations are linear u X u Z u, v 0, and w and the virtual work expression (density is constant) of internal forces simplifies to 1 u 1 1 () [()() ] X uz ux ux u W u Z X ux uz CA. For bar, the nodal displacement components of material coordinate system are u, ux () u/ X uz and uz () / ux uz (notice the use of x3 uz3 0 initial geometry). As the approximations are linear u u u u w u X Z, Z X 4-43

44 and the virtual work expression (density is constant) of internal forces simplifies to u [()()()()] X uz uz u W u X X uz ux uz uz ux u 1 1 [()()() X u ] Z ux uz uz u CA X. Virtual work expression of the point follows from definition of work 3 W F u Z. Virtual work expression is sum of the element contributions. After a considerable amount of manipulations, the standard form with notations a 1 ux / and a u / Z 4-44

45 W EA u 8 u 8 [a(1 5a) a(1 5a) a( 3a 5a ] EA T (1 a)(10a 1 1 5a 1 a 5a) X F Z 1 1 ) 0. Principle of virtual work and the fundamental lemma of variation calculus give a nonlinear algebraic equation system ( a 1 uz 3 / and a uz / ) (1 a)(10a 1 1 5a 1 a 5a) F 8 [a(1 5a) a(1 5a) a( 3a 5a)] EA It is obvious that finding an analytical solution in terms of the parameters of the problem become impossible even when the truss is very simple when the number of active dofs exceeds one. Mathematica code of the course gives as the real valued 4-45

46 solution with the minimal norm (that is likely to be the physically meaningful solution when the initial displacement is zero) ( 1m, A 1/100m, E C 100 Nm and F 1/ 0 N. 4-46

MEC-E8001 FINITE ELEMENT ANALYSIS

MEC-E8001 FINITE ELEMENT ANALYSIS MEC-E800 FINIE EEMEN ANAYSIS 07 - WHY FINIE EEMENS AND IS HEORY? Design of machines and structures: Solution to stress or displacement by analytical method is often impossible due to complex geometry,