MEC-E8001 FINITE ELEMENT ANALYSIS

Transcription

1 MEC-E800 FINIE EEMEN ANAYSIS 07 -

2 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, heterogeneous material etc. ack of the exact solution to an approximate problem is not an issue in engineering work. Finite element method is the standard of solid mechanics: Commercial codes in common use are based on the finite element method. A graphical user interface may make living easier, but a user should always understand what the problem is and in what sense it is solved! Finite element method has a strong theory: Although approximate solution is acceptable, knowing nothing about the error is not acceptable. -

3 SRUCURE ANAYSIS Stress analysis according to the linear elasticity theory may not entirely explain behavior displacement? of a structure! h h F vibration? h h w d 0 X 6 8 buckling? 4 t strength? 9 Z l Y

4 BAANCE AWS IN MECHANICS Balance of mass (def. of a body or a material volume) Mass of a body is constant Balance of linear momentum (Newton ) he rate of change of linear momentum within a material volume equals the external force resultant acting on the material volume. Balance of angular momentum (Cor. of Newton ) he rate of change of angular momentum within a material volume equals the external moment resultant acting on the material volume. Balance of energy (hermodynamics ) Entropy growth (hermodynamics ) -4

5 PROGRAMMER S VIEWPOIN IERAE INEAR IERAE INEAR EEMEN update Δs a = +Δa F = +Fe SEP n n s s s S extend a g g p P END n p p s = +Δs END n u u K = +Ke e E Δa = K-F +Fe END -5

6 SMA DISPACEMEN ANAYSIS Balance laws in mechanics, engineering paradigm, structural parts (elements), nodal displacements and rotations, principle of virtual work, Displacement analysis by FEM, element contributions, fundamental lemma of variation calculus, equilibrium equations Solid-, beam-, plate-, and rigid body elements. -6

7 BAANCE AWS IN MECHANICS 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 : u p F : f 0 r V p F : n t r Vt M : tda c r V y V x O Y z X Z V P fdv z P O y x u ( ) A constitutive equation f (, u ) 0 brings the material details into the model. For an unique solution, a displacement boundary condition is needed somewhere on V. -7

8 INEARY EASIC MAERIA he generalized Hooke s law of an isotropic homogeneous material can be expressed as: xx xx Strain-stress: yy yy E zz zz xy xy xy xx yy and yz yz yz G zz zx zx zx xy xy u x, y u y, x xx xx u x, x Strain-displacement: yy yy u y, y and yz yz u y, z u z, y u u u x, z zz zz z, z zx zx z, x in which E is the Young s modulus, the Poisson s ratio, and G E / ( ) the shear modulus. Using these, one may deduce the uni-axial, and two-axial (plane) stress and strain relationships. -8

9 MAERIA PARAMEERS ρ [ kg / m3 ] E [ GN / m ] ν [] Steel Aluminum Copper Glass Granite Birch Rubber Concrete Material -9

10 PRINCIPE OF VIRUA WORK Principle of virtual work W W int W ext 0 u U is just one form of the balance laws of continuum mechanics. It is important due to its wide applicability and physical meanings of the terms. W int wvint dv V V ( : c )dv ( f u )dv WAext wext A da ( t u )da WVext wvext dv V A V A A V P fdv u u 0 he details of the expressions vary case by case, but the principle itself does not! -0 P tda

11 VIRUA WORK DENSIIES Virtual work densities of the internal forces, external volume forces, and external surface forces are (subscripts V and A denote virtual work per unit volume and area, respectively) xx Internal forces: wvint yy zz fx External forces: wvext f y fz xx xy yy yz zz zx xy yz zx u x t x u x ext u w t u and y y y. A u t u z z z Virtual work densities consist of terms containing kinematic quantities and their work conjugates! -

12 . DISPACEMEN ANAYSIS Model the structure as a collection of beam, plate, etc. elements. Derive the element contributions W e 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 structure W e E W e. Re-arrange to get the standard form W a (Ka F). Use the principle of virtual work W 0 a, fundamental lemma of variation calculus for a n, and solve for the unknown nodal displacement and rotation components from the system equations Ka F 0. -

13 MINIAURE MODE OF CRANE mv ml A 6 03 A 80 mv 4.50 kg ml.04 kg E 06 GN/m B 7850 kg/m B 80 3 M 0.3 M.46 kg A-A : B-B : 40 0

14 ENGINEERING PARADIGM ml g 4 ( A)g ( A)g 3 5 X mv g Z 3 SRUCURE,,3 Beam 4,5,6 Rigid body 7 String 7 4 ( A)g 6-4 Mg

15 SRUCURA AND MAERIA SYSEMS In a particle model, the particles are identified by natural numbers (or labels). In a continuum model, the particles are identified by coordinates ( x, y, z ) of the material coordinate system which moves and deforms with the body. y structural 4 ri O X rp i Y ra Z material x A z AP P body Structural ( X, Y, Z ) coordinate system is needed in the description of geometry. -5

16 SIGN CONVENIONS AND NOAIONS In MEC-E800 displacements, rotations, forces and moments are treated as vectors and, therefore, components are positive in the directions of the chosen coordinate axes. he convention may differ from that used in mechanics of materials courses (be careful). Displacement Force Rotation Moment Material ux, u y, uz Fx, Fy, Fz x, y, z Mx,M y,mz Structural u X, uy, uz FX, FY, FZ X, Y, Z M X, MY, M Z he basis vectors of the material and structural systems are i, j, k and I, J, K, respectively! -6

17 EXAMPE.. A bar truss is loaded by a point force having magnitude F as shown in the figure. Determine the nodal displacements. Cross-sectional area of bar - is A and that for bar 3-8A. Young s modulus is E and weight is omitted. 3 X Z x F x Answer u X F u EA Z -7 3

18 he relationships between the nodal displacement components in the material and structural systems are u x 0 and u x u X. Element contribution W to the virtual work expression of the structure is (see formulae collection) 0 EA 0 0 EA W ( ) u X u X. u X 0 u X For element, u x 3 0 and u x (u X uz ) /. Element contribution takes the form 0 0 E 8 A 0 W ( ) u X uz u X u Z 0 W EA ( u X uz )(u X uz ). -8

19 Virtual work expression of the point force follows from the definition of work W 3 uz F. Virtual work expression of a structure is obtained as the sum of the element contributions W EA EA u X u X ( u X uz )(u X uz ) uz F u X EA u X 0 W ( ). u u Z F Z standard form Using the principle of virtual work W 0 a and the fundamental lemma of variation calculus -9

20 EA u X 0 0 uz F u X F. uz EA In Mathematica code of the course, the problem description is given by -0

21 EXAMPE.. Consider the beam truss of the figure. Determine the displacements and rotations of nodes and 4. Assume that the beams are rigid in the axial directions so that the axial strain vanishes. Cross-sections and lengths are the same and Young s modulus E is constant. x x z X t 4 z z x f 3 3 y t z Z Answer u X u X 4 3 f4 9 f3 5 f3, Y, and Y 4 EI 008 EI 008 EI -

22 Only the bending in XZ-plane needs to be accounted for. he dofs of the structure are u X, Y, and Y 4. As the axial strain of beam vanishes, axial displacements satisfy ux 4 ux EI W ( ) 3 6 u X u X 6 Y Y (u z u X, y Y ) Y Y EI 6 4 W ( 3 ) Y Y 4 ( y Y, y 4 Y 4 ) -

23 6 6 u X u X EI f Y4 W 3 Y 4 ( 3 ) ( u z 4 u X ) Virtual work expression of the structure is 4 6 u X EI 3 W W W W Y ( Y4 6 6 u 6 X f Y 0 ) Y 4 8 Principle of virtual work W 0 a and the fundamental lemma of variation calculus give -3

24 4 6 EI u 6 u X X f3 f Y 0 0 Y 008 EI Y4 8 Y In the Mathematica code calculation, horizontal displacements of nodes and 4 are forced to be same ( u X 4 u X ) -4

25 . EEMEN CONRIBUIONS Virtual work expressions for solid, beam, plate elements combine virtual work densities representing the model and a case dependent approximation. o derive the expression for an element: Start with the virtual work densities w and w of the formulae collection (if not int ext available there derive the expression in the manner discussed in MEC-E050). 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 to get W. -5

26 EEMEN APPROXIMAION In MEC-E800 element approximation is a polynomial interpolant of the nodal displacement and rotations in terms of shape functions. In displacement analysis, shape functions depend on x, y, z and the nodal values are parameters to be evaluated by FEM. always of the same form! Approximation u N a Shape functions N {N ( x, y, z ) N ( x, y, z ) N n ( x, y, z )} Parameters a {a a a n } Nodal parameters a {u x, u y, u z, x, y, z } may be just displacement or rotation components or a mixture of them (as with the Bernoulli beam model). -6

27 EEMEN GEOMERY η η η Ω ξ Ω Ω ξ η η Ω Ω ξ ζ ξ Ω ξ Ω ξ ξ ζ -7

28 SOID MODE he model does not contain assumptions in addition to those of linear elasticity theory. u, x u, x u, y v, x u, y v, x int wv v, y [ E ] v, y v, z w, y G v, z w, y, w w w u w u,z,z,z,x,x,z u wvext v w fx u tx ext f w v and y t y. A w f tz z V fdv u u 0 tda A u he 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 ) -8

29 EXAMPE.3. A tetrahedron of edge length, density, and elastic properties E and is subjected to its own weight on a horizontal floor. Calculate the displacement uz 3 of node 3 with one tetrahedron element and linear approximation. Assume that u X 3 uy 3 0, and that the bottom surface is fixed. Z,z 3 g g Answer: uz 3 4 E 4 X,x -9 Y,y

30 inear shape functions can be deduced directly from the figure N x /, N y /, N 3 z /, and N 4 x / y / z /. However, only the shape function of node 3 is needed as the other nodes are fixed. Approximations to the displacement components are u 0, v 0, and w z uz 3, giving w, x w, y 0 and w, z uz 3. When the approximation is substituted there, the virtual work densities of the internal and external forces simplify to 0 E wvint 0 w ( )( ),z E ( ) uz 3 uz 3 0 ( )( ) w, z

31 wvext u v w fx 0 0 z 0 fy 0 g uz 3. z / u g Z3 fz Virtual work expressions are obtained as integrals of densities over the volume: W int W ext wvint dv V wvext dv V wvint 3 EuZ 3 uz 3, 6 6 ( )( ) 3 g u Z 3. 4 Finally, principle of virtual work W 0 with W W int W ext implies that g uz 3. 4 E -3

32 In Mathematica code of the course, the problem description is given by -3

33 BEAM MODE he bar, torsion bar, and bending modes of a beam are connected unless the first and cross moments (off-diagonal terms of the matrix) of the cross-section vanish: A u, x int w v, xx E S z w S, xx y ext w u v w Sz I zz I zy f x f w y,x v f z,x S y u, x I yz v, xx, x GI rr, x I yy w, xx traction on the end surfaces mx u X FX X M X ext m W u F M ( y Y Y Y Y ) u F M m Z Z Z Z z In FEM the solution domain (a line segment) is represented by line elements and the displacement and rotation components u( x ), v ( x ), w( x ), and ( x ) by their interpolants. -33

34 BAR MODE Assuming that v ( x ) w( x ) ( x ) 0 and a linear interpolation of the axial displacement at the endpoints for u( x ), virtual work expressions of the internal and external forces take the forms fx W int u x EA u x, u x h u x u W ext x u x ux u x EA f x h. z x h Above, f x, E, and A are assumed to be constants. he relationship between the axial displacement component and the displacement components in the structural coordinate system is u x i u i X u X iy uy iz uz. -34

35 ORSION MODE Assuming that u( x ) v ( x ) w( x ) 0 and a linear interpolation of the axial end-point rotations for ( x ), virtual work expressions of the internal and external forces take the forms mx x GI rr x int W, x h x x GIrr m x h W ext x. x z x x h Above, mx, E, and I rr are assumed to be constants. he relationship between the axial rotation component and the rotation components in the structural coordinate system is x i u i X X iy Y iz Z. -35

36 BENDING MODE (xz-plane) Assuming that u( x ) v ( x ) ( x ) 0 and a cubic interpolation for w( x ) in terms of point displacements u z, u z and rotations y, y. W int uz y EI yy 3 u z h y 6h 6h u z 6h h y 6h 4h 6h 6h u z 6h h 6h 4h y fz W ext uz 6 y f z h h u z 6 y h EIzz y Above, f z, I yy and E are assumed to be constants. -36 y u z z x h uz

37 BENDING MODE (xy-plane) Assuming that u( x ) w( x ) ( x ) 0 and a cubic interpolation for v ( x ) in terms of point displacements u y, u y and rotations z, z W int W ext u y EI zz z 3 u y h z 6h 6h u y 6 h 4 h 6 h h z 6h 6h u y 6h h 6h 4h z u y 6 z f y h h u y 6 h z fy EIyy z u y h y Above, f y, I zz and E are assumed to be constants. -37 x z u y

38 EXAMPE.4. he Bernoulli beam of the figure is loaded by its own weight and a point force acting on the right end. Determine the displacement and rotation of the right end starting with the virtual density of the Bernoulli beam model. he x-axis of the material coordinate system is placed at the geometric centroid of the rectangle cross-section. Beam properties A, I zz I, and E are constants. ga F Z,y ga3 F Answer: u X and Y 48 EI EA -38 X,x

39 Bernoulli beam element of the Mathematica code requires the orientation of the y axis unless y and Y axes are aligned. Orientation is given by additional parameter defining the components of j in the structural coordinate system: -39

40 PAE MODE Virtual work densities combine the plane-stress and plate bending modes. Assuming that the material coordinate system is placed so that S y S z I yz 0 int w u,x v, y t[ E ] u v,x,y u w,x, xx t3 [ E ] v, y w, yy u, y v, x w, xy w, xx w, yy, w, xy ext ext w uf x vf y wf z and w ut x vt y wt z. he planar solution domain (reference-plane) can be represented by triangular or rectangular elements. Interpolation of displacement components should be continuous and w( x, y ) should have also continuous derivatives at the element interfaces. -40

41 EXAMPE.5. Consider the thin triangular structure shown. Young s modulus E, Poisson s ratio, and thickness t are constants. Distributed external force vanishes. Assume plane-stress conditions, XY plane deformation and determine the displacement of node when the force components acting on the node are as shown in the figure. y,y 3 u X F ( )( ) Answer: u Et Y F -4 F x,x

42 Nodes are 3 are fixed and the active dofs are u X and uy. inear shape functions N ( x y ) /, N x / and N3 y / are easy to deduce from the figure. herefore u x y u X u v Y u, y u, x u X u X and. uy uy v, y v, x Virtual work density of internal forces is given by ux te int w uy u u X Y 0 ux u 0. Y 0 0 ( ) / u X uy Integration over the triangular domain gives (integrand is constant) -4

43 ux te W uy u u X Y u X te W u Y ux u 0 Y 0 0 ( ) / u X uy u X 3 uy Virtual work expression for the point forces follows from the definition of work u X F W u Y F Principle of virtual work in the form W W W 0 a and the fundamental lemma of variation calculus give -43

44 u X te W ( u Y 4 te u X u X F ) 0 u 3 uy Y u X F 0 3 uy u X F ( ). te uy he point forces acting on a thin slab should be considered as equivalent nodal forces i.e. just representations of tractions acting on some part of the boundary. Under the action of an actual point force, displacement becomes non-bounded. In practice, numerical solution to the displacement at the point of action increases when the mesh is refined. -44

45 In Mathematica code of the course, the problem description is given by -45

46 .5 POIN FORCE/MOMEN External point forces and moments are assumed to act on the joints. hey are treated as elements associated with one node only. Virtual work expression is usually simplest in the structural coordinate system: u X W uy u Z FX X FY Y F Z Z M X MY M Z X Y Z u X, uy, uz X, Y, Z Above, FX, FY, FZ and M X, M Y, M Z are the given components. A rigid body can be modeled as a particle at the center of mass connected to the other joints of the body by rigid links! -46