Chapter 3: BASIC ELEMENTS. solid mechanics)
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1 Chapter 3: BASIC ELEMENTS Section 3.: Preliminaries (review of solid mechanics) Outline Most structural analsis FE codes are displacement based In this chapter we discuss interpolation methods and elements based on displacement interpolations ti Stiffness matri formulations will be presented Shortcomings and restrictions of the elements due to the interpolations used will be discussed We will review the governing equations (for solids elastic bodies) to help us understand the solution methods and accurac
2 3 Review of Solid Mechanics The analsis of an solid elastic bod has to define and develop the following quantities and/or relations Stress Strain (strain-displacement relations) Constitutive Properties (Stress-Strain Strain relations) Compatibilit Equilibrium Equations Boundar Conditions 4 Stress Stresses are distributed internal forces that result from eternall applied forces B F F F F ΔF F 3 F F 5 5 F 4 Note: There are two tpes of forces: Surface forces that act on an area of eternal surface and bod force that acts on the volume of the bod
3 5 Stress/Force acting on a surface A force acting on a surface can be resolved into two components: One tangential to the surface (shear force) and the other normal to the surface τ S Force is a st order tensor (vector) Stress is a nd order tensor. Wh is it a tensor? tˆ t nˆ σ Needs two vectors to specif it. One is the vector of the internal force and the other is the outward normal of the defining area 6 Representation of stress at a point In 3-D space the stress at a point is denoted b the stresses acting on three mutuall perpendicular planes at that point Often we use a simple infinitesimal rectangular solid and indicate the stresses on the faces of that solid Stress terms have two indices. The first indicating the plane on which it acts The second indicating the direction in which it acts
4 Stress at a point 7 State of stress at a point in 3-D Cartesian Coordinates σ Matri representation of state of stress τ τ τ τ σ τ τ τ τ σ τ τ τ σ τ σ τ σ Sign Conventions: Normal stresses are positive when acting outward from a surface (tension) Shear stresses are positive when the act in the +ve direction on a positive face and ve direction on a ve face Stress on an arbitrar plane (-D) We often need to enforce stress boundar conditions on surfaces that are not alwas rectangular Y Let cos(θ) l and sin(θ) m C N If length of side BC A, then length of sides OCAl and OB τ α Am If we write the force equilibrium in X and Y-directions, we have π/ α X A σ ( Al ) + τ ( Am ) Which simplifies to X σ + σ l τ m Y σ m + τ l O τ σ 8 B X
5 9 Stress on an arbitrar plane (3-D) The forces (per unit area) in X, Y and Z-directions on an arbitrar plane defined d b its normal vector N is given b N B X lσ + mτ + nτ Y lτ + mσ + nτ Z lτ + mτ + nσ C O A Where l, m and n are the direction cosines of the normal vector of the arbitrar plane l cos N, m cos N, n cos N 0 Stress transformations in 3D Stress transformation in 3D space can be defined using the directions cosines as follows. σ l σ + m σ + n σ + mlτ + mnτ + nlτ ' ' l + m + n + m l + m n + n l ' l3 + m3 + n3 + ml m3n3 + nl 3 3 σ σ σ σ τ τ τ σ σ σ σ τ τ τ τ ' ' ll σ + mm σ + nn σ + ( lm ml ) τ + ( mn nm ) τ + ( nl ln ) τ τ ' ' ll 3σ + mm 3σ + nn 3σ + ( lm 3 ml 3) τ + ( mn 3 nm 3) τ + ( nl 3 ln 3) τ τ llσ + mmσ + nnσ + ( lm 3 ml 3 ) τ ' ' ( mn nm) τ + ( nl ln) τ
6 Direction cosines in 3D The direction cosines l.m and n between the new coordinate aes,, and and the original coordinate sstem, and are defined as follows Since the transformation is orthogonal, the direction cosines must satisf the following properties l m n ' l m n l + m + n l 3 m n l + l + l 3 3 Where, l cos, m cos l l + m m l m + l m 3 + nn + l m Equilibrium Equations (-D) The -D stresses are shown on a volume given of length d and d in X- and Y-directions and unit thickness in the Z- direction σ τ d d σ τ + F Y dτ F X σ + dσ Summation of forces in X-direction τ ( σ + dσ ) d σ d + ( τ + dτ ) d τ d + F X 0
7 3 Equilibrium of forces in X-direction ( σ + d σ ) d σ d + ( τ + d τ ) d τ d + F dd 0 ( σ + dσ ) d σ d + ( τ + dτ ) d τ d + F dd dd 0 X X d σ σ dτ τ d d dσ dτ dd + dd + FX dd d d σ τ + + F F X dd D Equilibrium Equations The force equilibrium provide the relations shown below referred to as differential equation of equilibrium σ τ + τ σ + + X 0 + Y 0 Establishing moment equilibrium b the same method will provide the condition for smmetr of the stress tensor τ τ
8 5 Strain Wh do we need the strain measures? Will displacement not suffice? Strain better quantifies the deformation of the bod and eliminates rigid bod motion/ rotation Strain in ver general terms is a measure of relative deformation Relative to what? Undeformed bod : Lagrangian strain Deformed bod: Eulerian strain 6 Strain-Displacement Relations For uniaial (-D) case: l Δl Δl l d P d Q P Q Q P, P Q du Q P u P u ui +vj +wk PP(,,) P( P P (+u,+v,+w) dv Q
9 7 Strain -D : Normal strain d P d Q P Q d d d u v d d + + d d P, P Q du d dv Q u u v du dv u d v d u u + + v + u u v + + O d d Q P STRAIN: Shear Strain u O d d Q B O d θ v v Q d P d Q d 8 γ θ + θ v P d A θ u d P
10 9 Shear strain With a bit of trigonometr (see for eample, Allen and Haisler, Introduction to Aerospace Structural Analsis, p.60) γ v u u v u u For small displacements The normal and shear strains are epressed as u v γ v u + Engineering Strains v u + Tensor strains Shear strain definitions are of two forms. The above form is referred to as engineering strains. The alternate form referred as tensorial strains have a factor of ½ applied to engineering strains. To appl coordinate transformations need the tensor form.
11 Compatibilit Deformation must be such that the pieces fit together without an gaps or overlap. Wh is this an issue? In -D we require onl displacements u, and v to describe deformation, but have three strain quantities,, and γ. This implies onl two of the three strain terms are independent. u v v u γ + u v ( γ ) + + Stress-Strain Strain Relations The stress-strain relations in solid mechanics is often referred to as the Hooke s Law Hooke s law of proportionalit stated as etension is proportional to the force refers to the aial etension of a bar under an aial force This can be etended to 3-D stress/strain state referred to as the Generalied Hooke s Law relates the components of the 3-D stress state to 3-D strains as follows. σ c c c 3 c 4 c 5 c 6 σ c c c3 c4 c5 c6 σ c3 c3 c33 c34 c35 c36 τ c4 c4 c43 c44 c45 c46 γ τ c5 c5 c53 c54 c55 c56 γ τ c6 c6 c63 c64 c65 c66 γ
12 Generalied Hooke s Law In the most general form the generalied Hooke s Law requires 36 constants to relate the terms of a 3-D Stress state to its corresponding 3-D strain state for an elastic material However, from smmetr of the strain energ terms, it can be shown that c ij c ji This reduces the number of unknown constants to σ c c c3 c4 c5 c6 σ c c c3 c4 c5 c6 σ c3 c3 c33 c34 c35 c36 τ c4 c4 c34 c44 c45 c46 γ τ c5 c5 c35 c45 c55 c56 γ τ c6 c6 c36 c46 c56 c66 γ 3 4 Hooke s Law: Orthotrop and Isotrop If we assume the, and coordinates provide the planes of smmetr we can further reduce the number of constants to 9. σ c σ c σ c τ 0 τ c c c τ c c c c c γ 0 γ c γ This corresponds to an full orthotropic material Isotrop assumes that there is no directional variation on propert. Using this argument we can obtain c c c c c c c55 c66 c
13 Hooke s law : Engineering Elastic Constants 5 The two engineering elastic constants used to relate stress to strain for isotropic materials are the Elastic modulus, E and the Poisson s ration ν. For uniaial loading, strain in the loading direction obtained from Hooke s law, states σ E σ Transverse to loading direction ν ν E The relation between the shear stress component and its corresponding shear-strain strain component is called the modulus of rigidit or modulus of elasticit in shear and is denoted b the letter G. τ E G μ G γ +ν ( ) Generalied Hooke s Law from Isotropic Materials 6 The generalied Hooke s law epressed in engineering elastic constants τ ( + ν ) γ [ σ ν ( σ σ )] + τ E G E τ [ ( )] ( + ν ) σ ν σ + σ γ τ E G E τ ( + ν ) [ σ ν ( σ + σ )] γ τ E G E
14 7 Plane state of stress There are a large class of problems for which the stresses normal to the plane of the solid are absent or negligibl ibl small. If we assume that the stresses are restricted to the - plane, then σ τ τ 0 This simplifies the stress strain relationship to the form as shown below. E [ σ νσ ] σ [ + ν ] E ν E σ + ν ν [ σ νσ ] [ ] E ν + E ( σ + σ ) τ γ G τ Gγ 8 Plane strain Strains that deform the bod normal to the reference plane are absent or are negligible γ γ This indicates that the stress normal to the plane of strain is dependent on the stresses in the plane of the strain [ σ ν ( σ + σ )] 0 σ ν ( σ + σ ) E Substituting σ into other strain epressions we obtain [ ( ν ) σ νσ ] + ν E + ν ( ν ) σ νσ E [ ] 0 ( ν ( ) + ν )( ν ) [ ν ] σ E + σ E ( )( ) ( ν ) + ν ν [ + ν ] ν E σ ( )( ) [ ] + + ν ν γ τ G τ Gγ
15 Conversion from plane strain to plane stress and vice-versa 9 The solution obtained for the stress and strains in plane stress and plane strain states are qualitativel similar. To use a plane strain solution for a plane stress or vice versa, we simpl interchange the appropriate constants as shown below For plane stress the epressions in E, ν For plane stress the epressions in E *, ν * * E ν * + ν ν E, ν or E E and ν * * ν ν + ( + ν ) ν 30 Solution of 3D Elasticit Problems 3 Equilibrium equations 3 Equilibrium equations 6 Strain-displacement 3 Compatibilit equations 6 Hooke s law equations 6 Hooke s law equations σ,σ, σ, τ, τ, τ σ,σ, σ, τ, τ, τ,,,,, γ γ γ,,, γ, γ, γ u,v,w 3 Equilibrium equations 6 Strain displacement equations σ, σ, σ, τ, τ, τ u, v, w 3 Equilibrium equations in displacements u,v, w 3 Equilibrium equations 3 Compatibilit equations epressed in stress σ,σ, σ, τ, τ τ 6,
16 3 Principle of superposition Effects of several forces acting together are equal to the combined effect of the forces acting separatel. This is valid onl when The stresses and displacements are directl proportional to the load The geometr and loading of the deformed object does not differ significantl from the undeformed configuration 3 Energ Principles Strain Energ Densit: When an elastic bod is under the action of eternal forces, the bod deforms and work is done b these forces. The work done b the forces is stored internall b the bod and is called the strain energ. Let us consider the unit element of volume ddd with onl the normal stress σ acting on it. The work done, or work stored in the element is C C D D σ σ B B A A σ σ σ 0 σ σ u σ d u + d dd d σ σ σ u σ ddd σ 0 σ 0 ( u)dd
17 33 Strain Energ Using Hooke s law u σ E σ σ Work done σ E σ E dσ ddd σ 0 ddd du For shear stresses, it can be similarl shown that the work done is τ G ddd The strain energ stored in an element ddd under a general three dimensional stress sstem is calculated as ( σ + σ + σ + τ γ + τ γ + τ γ )ddd 34 Strain Energ Densit The strain energ densit refers to strain energ per unit volume du 0 ( σ + σ + σ + τ γ + τ γ + τ γ ) Using principal stresses and strains, this can be epressed as du 0 3 du 0 I ( +ν ) I E I σ + σ + σ 3 I σ σ + σ σ + σ σ ( σ + σ + σ ) ( ) 3 3 3
18 35 3. Interpolation Functions 36 Introduction to Interpolation Functions Interpolation is to devise a continuous function that satisfies prescribed conditions at a finite number of points. In FEM, the points are the nodes of the elements & the prescribed conditions are the nodal values of the field variable Polnomials l are the usual choice for FEM
19 37 Polnomial interpolation The polnomial function φ() is used to interpolate a field variable based on its values at n-points n i φ( ) ai or φ X a i00 {} n X... and {} a a0 a a... an The number of terms in the polnomial is chosen to match the number of given quantities at the nodes. T With one quantit per node, we calculate a i s using the n-equations resulting from the epressions for φ i at each of the n-known points n i φ ( j ) ai j { e } [ A ]{ a } i 0 φ { a } [ A ] { φ } φ e 38 Shape Functions or Basis Functions Traditional interpolation takes the following steps. Choose a interpolation function. Evaluate interpolation function at known points 3. Solve equations to determine unknown constants φ X {} a { φe} [ A]{} a {} a [ A] { } φ e φ X {} a In FEM we are more interested in writing φ in terms of the nodal values φ X { a} { φ e } [ A]{} a { a} [ A] { } X [ A ] { } ϕ N { φ } N X [ A] e ϕ φ e φ e
20 Degree of Continuit 39 In FEM field quantities φ are interpolated in piecewise fashion over each element This implies that φ is continuous and smooth within the element However, φ ma not be smooth between elements An interpolation function with C m continuit provides a continuous variation of the function and up to the m- derivatives at the nodes For eample in a -D interpolation of f() C 0 continuit indicates that f is continuous at the nodes and f, is not continuous. If the displacement u() is C 0 then displacements are continuous between elements, but the strains are not (bar elements) 40 Degree of Continuit Function φ is C 0 continuous while φ is C continuous φ() φ φ C 0 C dφ d dφ d C 0 dφ d C 0 0
21 Eample: Deriving a D linear interpolation shape function { } From φ N φ e each interpolation function is ero at all the dofs ecept one. This can allow us to derive interpolation functions one at a time. For linear interpolation ti between and, N ( ), N ( )0, N a +a. So obviousl, N -(- )/L, L -. 4 C 0 Interpolation -D Element 4 ( )( 3 ) N ( )( ) 3 3 φ φ φ N N N φ φ φ ( )( 3 ) N ( )( ) 3 3 φ φ N 3 φ 3 ( )( ) ( )( ) 3 φ N φ φ 3 3
22 43 Lagrange Interpolation Formula Shape functions shown for the C 0 interpolations are special forms of the Lagrangian interpolation functions f ( ) n k N k f k N k ( )( )... [ k ]... ( n ) ( )( ) [ ] ( ) k k k k n k In above epressions for N k the terms in square brackets are omitted Properties of C 0 Interpolation Shape Functions 44 All shape functions N i, along with function φ are polnomials of the same degree For an shape function N i, N i at node i ( i ) and ero at all other nodes j, ( j i ) C 0 shape functions sum to one n N k k
23 C Interpolation Also called Hermitian interpolation (Hermite polnomials) Use the ordinate and slope information at the nodes to interpolate 45 φ φ φ, C 0 interpolation curve φ, C interpolation curve φ φ φ Hermitian interpolation used for beam elements 46
24 47 -D and 3-D Interpolation The -D and 3-D shape functions follow the same procedure as for -D We now have to start with shape functions that have two or more independent terms. For eample a linear interpolation in -D from 3 nodes will require an interpolation function T 3 f (, ) a a a If there are two or more components (e.g., u, v and w displacements) then the same interpolation function is used for all components 48 Principle of Virtual Work The principle of virtual work states that at equilibrium the strain energ change due to a small virtual displacement is equal to the work done b the forces in moving through the virtual displacement. A virtual displacement is a small imaginar change in configuration that is also a admissible displacement An admissible displacement satisfies kinematic boundar conditions Note: Neither loads nor stresses are altered b the virtual displacement.
25 Principle of Virtual Work 49 The principle of virtual work can be written as follows T T T { δ} { σ} dv { δu} { F} dv + { δu} { Φ} ds The same can be obtained b the Principle of Stationar Potential Energ The total potential energ of a sstem Π is given b δ Π δ U δ W δ U + δ V 0 U is strain energ, W is work done, or V is potential of the forces Π U W 50 Element and load derivation Interpolation { u } [ ]{ d } { u } Strain displacement Virtual Constitutive law N u v w { } [ B]{ d} [ B] [ ][ N] T T T T T T { δu} { δd} [ N] and { δ} { δd} [ B] { σ} [ E]{ } Altogether T T T T { δd } ( [ ] [ ][ ] { d } [ ] [ ]{ 0 } + [ ] { σ 0 } T T [ N] { F} dv [ N] {} φ ds ) 0 B E B dv B E dv B dv
26 5 Stiffness matri and load vector Equations of equilibrium k d r e Element stiffness matri Element load vector [ ]{ } { } T [ ] [ ] [ ][ ] k B E B dv T T T T { er } [ B ] [ E ]{ 0} dv [ B ] { σ0} dv+ [ N ] { F } dv+ [ N ] {} φds ) Loads due to initial strain, initial stress, bod forces and surface tractions Plane Problems: Constitutive Equations Constitutive equations for a linearl l elastic and isotropic material in plane stress (i.e., σ τ τ 0): 5 Initial thermal strains 0 0 αδt, γ 0 0 In matri form, where
27 53. Plane Problems: Strain-Displacement Relations Plane Problems: Displacement Field Interpolated 54 From the previous two equations, where B is the strain-displacement matri.
28 Constant Strain Triangle (CST) 55 The node numbers sequence must go counter clockwise Linear displacement field so strains are constant! CST ELEMENT Constant Strain Triangular Element Decompose two-dimensional domain b a set of triangles. Each triangular element is composed b three corner nodes. Each element shares its edge and two corner nodes with an adjacent element Counter-clockwise or clockwise node numbering Each node has two DOFs: u and v displacements interpolation using the shape functions and nodal displacements. Displacement is linear because three nodal data are available. Stress & strain are constant. v 3 3 u 3 v u v u 56
29 Displacement Interpolation CST ELEMENT cont. Since two-coordinates are perpendicular, u(,) and v(,) are separated. u(,) ( ) needs to be interpolated t in terms of u, u, and u 3, and v(,) in terms of v, v, and v 3. interpolation function must be a three term polnomial in and. Since we must have rigid bod displacements and constant strain terms in the interpolation function, the displacement interpolation must be of the form u (, ) α+ α+ α3 v (, ) β+ β+ β3 The goal is how to calculate unknown coefficients α i and β i, i,, 3, in terms of nodal displacements. u (, ) N ( u, ) + N ( u, ) + N ( u, ) Displacement Interpolation CST ELEMENT cont. -displacement: Evaluate displacement at each node u (, ) u α + α + α 3 u (, ) u α+ α + α3 u ( 3, 3) u3 α+ α3 + α33 In matri notation u α α u α u Is the coefficient matri singular? 58
30 Displacement Interpolation CST ELEMENT cont. α u f f f3 u α u b b b u 3 α A u3 c c c3 u3 where f, b, c f, b, c f3, b3, c3 Area: A det α ( fu + fu + fu 3 3 ) A α ( bu + bu + bu 3 3 ) A α 3 ( cu + cu + cu 3 3 ) A CST ELEMENT cont. Insert to the interpolation equation u (, ) α + α + α 3 [ ( fu + fu + fu 3 3 ) + ( bu + bu + bu 3 3 ) + ( cu + cu + cu 3 3 ) ] A ( f + b + c ) u A + ( f A + b + c ) u + f A + b + c ) u N (,) N (,) N 3 (,) 60
31 Displacement Interpolation CST ELEMENT cont. A similar procedure can be applied for -displacement v(, ). u (, ) [ N N u N 3] u u3 v (, ) [ N N v N 3] v v3 N(, ) ( f+ b + c) A N (, ) ( f + b + c ) A N3(, ) ( f3 + b3 + c3) A Shape Function N, N, and N 3 are linear functions of - and -coordinates. Interpolated displacement changes linearl along the each coordinate direction. 6 Displacement Interpolation CST ELEMENT cont. Matri Notation u v u N 0 N 0 N3 0 u { u} v 0 N 0 N 0 N3 v u 3 v3 { u(, )} [ N(, )]{ q } [N]: 6 matri, {q}: 6 vector. For a given point (,) within element, calculate [N] and multipl it with {q} to evaluate displacement at the point (,). 6
32 Strain Interpolation CST ELEMENT cont. differentiating the displacement in - and -directions. differentiating shape function [N] because {q} is constant u N i bi Ni(, ) ui ui ui i i i A v N i ci Ni (, ) vi vi vi i i i A γ 3 3 u v ci bi + + ui vi i A i A 63 Strain Interpolation CST ELEMENT cont. u u v b 0 b 0 b3 0 v u {} 0 c 0 c 0 c3 [ B]{} q A v c b c b c3 b3 u v u + 3 v3 [B] B matri is a constant matri and depends onl on the coordinates of the three nodes of the triangular element. the strains will be constant over a given element 64
33 Constant Strain Triangle (CST): Stiffness Matri Strain-displacement relation, Bd 65 A is the area of the triangle and ij i - j. (tetbook has results for a coordinate sstem with aligned with side - T B EBtA From the general formula where t: element thickness (constant) k CST ELEMENT cont. Strain Energ: U C h ( e ) T ( e ) A { } [ ]{ } da h { q } [ B] [ C] [ B] { q } ( e) T T dA ( e) A { ( e) } T [ ( e) ] ( ) 6 6 { e q k q } ( e) Element Stiffness Matri: [ k ] ha[ B] [ C][ B] Different from the truss and beam elements, transformation matri [T] is not required in the two-dimensional element because [k] is constructed in the global coordinates. The strain energ of the entire solid is simpl the sum of the element strain energies NE NE assembl ( e) ( e ) T ( e ) ( e ) T U U { q } [ k ]{ q } U { Qs} [ Ks ]{ Qs } e e T 66
34 BEAM BENDING EXAMPLE -F m σ is constant along the -ais and linear along -ais Eact Solution: σ 60 MPa Ma deflection v ma m σ 5 m F Ma v Linear Strain Triangle (LST) 68 The element has si nodes and dof.
35 Linear Strain Triangle (LST) The quadratic displacement field in terms of generalied coordinates: 69 The linear strain field: Bilinear Quadrilateral (Q4): CQUAD4 in NASTRAN 70 Q4 element has four nodes and eight dof. Can be quadrilateral; but for now rectangle. Displacement field: So, u and v are bilinear in and. Because of form, sides are stiffer than diagonals-artificial anisotrop!
36 7 Strain field: Q4: The strain fields Observation : f() Q4 cannot eactl model the beam where Q4: Behavior in Pure Bending of a Beam 7 Observation : When β 4 0, varies linearl in - desirable characteristic for a beam in pure bending because normal strain varies linearl along the depth coordinate. But γ 0 is undesirable because there is no shear strain. Fig. (a) is the correct deformation in pure bending while (b) is the deformation of Q4 (sides remain straight). Phsical interpretation: applied moment is resisted b a spurious shear stress as well as fleural (normal) stresses.
37 Q4: Interpolation functions Eas to obtain interpolation ti functions 73 where matri N is 8 and the shape functions are Q4: The Shape (Interpolation) Functions 74 N, N N 3 N 4 0 at node, -a, -b, so u N u u at that node. Similarl N i while all other Ns are ero at node i. Fig 3.6- See Eqn for strain-displacement matri ( NdBd). All in all, Q4 converges properl with mesh refinement and works better than CST in most problems.
38 Coarse mesh results Q4 element is over-stiff in bending. For the following problem, deflections and fleural stresses are smaller than the eact values and the shear stresses are greatl in error: 75 BEAM BENDING PROBLEM cont. S Plot Ma v Stress is constant along the -ais (pure bending) linear through the height of the beam Deflection is much higher than CST element. In fact, CST element is too stiff. However, stress is inaccurate. 76
39 Caution: BEAM BENDING PROBLEM cont. In numerical integration, we did not calculate stress at node points. Instead, we calculate stress at integration points. Let s calculate l stress at the bottom surface for element in the beam bending problem. 4(0,) 3(,) Nodal Coordinates:(0,0), (,0), 3(,), 4(0,) Nodal Displacements: u [0, , , 0] v [0, , , 0] (0,0) 0) (,0) Shape functions and derivatives N ( )( ) N / ( ) N / ( ) N N 3 ( ) N4 ( ) N / ( ) N / 3 N / 4 N / N / 3 N / ( ) 4 77 BEAM BENDING PROBLEM cont. At bottom surface, N/ N/ N / N / N3/ 0 N3/ N4/ 0 N4/ ( ) Strain 4 N I u I I 4 N I v I γ Stress: I 4 NI NI v + u I I I { σ} [ C]{ } {4.44,.33,.55} 0 u [0, , , 0] v [0, , , 0] 7 78
40 RECTANGULAR ELEMENT -normal stress and shear stress are supposed to be ero. σ Plot is a linear function of alone if a linear function of alone γ is a linear function of and 4 N I u I I 4 N N I v I I N / ( ) N / ( ) N / 3 τ Plot N / ( ) N / N 4 / N4 / ( ) N / Discussions RECTANGULAR ELEMENT Can t represent constant shear force problem because must be a linear function of. Even if can represent linear strain in -direction, the rectangular element can t represent pure bending problem accuratel. Spurious shear strain makes the element too stiff. u α + α + α + α 3 4 v β + β + β + β 3 4 Eact α + α 4 β + β 3 4 γ α β α β ( 3+ ) α4 0 Rectangular element 80
41 Two-Laer Model σ v ma RECTANGULAR ELEMENT 8 BEAM BENDING PROBLEM cont. Distorted Element Ma v As element is distorted, the solution is not accurate an more. 8
42 BEAM BENDING PROBLEM cont. Constant Shear Force Problem Ma v S is supposed to change linearl along -ais. But, the element is unable to represent linear change of stress along -ais. Wh? Eact solution: v m and σ 6e7 Pa. 83 BEAM BENDING PROBLEM cont. Higher-Order Element? 8-Node Rectangular Element u (, ) a + a + a + a + a Strain u (, ) a + a+ a a + a + a + a+ a Can this element accuratel represent pure bending and constant shear force problem? 6 84
43 BEAM BENDING PROBLEM cont. 8-Node Rectangular Elements Tip Displacement m, Eact! 85 BEAM BENDING PROBLEM cont. If the stress at the bottom surface is calculated, it will be the eact stress value. S S 86
44 Q6: Additional Bending Shape Functions Q4: Artificial shear deformation under pure bending Additional shape functions to solve the issue 4 i(,) i ( ) ( ) i u N s t u + s a + t a 4 i(,) i ( ) 3 ( ) 4 i v N s t v + s a + t a Bubble modes Strain can var linear along -dir. Shear strain γ can vanish for pure bending Nodeless DOFs, a, a, a 3, and a 4, are condensed in the element level (total DOFs) 87 Modeling Bending with the Q6 Element 88 Kdd Kda d fd Kad Kaa a fa aa a ad { a} K { f K d} dd da aa ad d da aa a [ K K K K ]{ d} { f K K f } Modeling the previous bending problem with Q6 elements gives the following stresses:
45 Quadratic Quadrilateral (Q8) corner nodes and 4 side nodes and 6 nodal dof. Quadratic Quadrilateral (Q8): Displacement 90 The displacement field, which is quadratic in and : Two tpes of shape functions (ξ/a, η/b) : The edges ±a deform into a parabola (i.e., quadratic displacement in ) (same for ±b)
46 Quadratic Quadrilateral (Q8): Strains 9 The strain field: Strains have linear and quadratic terms. Hence, Q8 can represent man strain states eactl. For eample, states of constant strain, bending strain, etc Nodal loads Consistent (work-equivalent) loads W dr ufdv + uφds T T T e Mechanical loads: concentrated loads, surface traction, bod forces.
47 93 Eample: Beam under uniform loads Normal forces are obvious. For moments M θ L L 3 qn θ d q ( + ) d L L 0 0 ql 3 4 ql ( + ) Work equivalent (consistent) normal loads Normal surface traction on a side of a plane element whose sides remain straight (q is force/length): 94
48 95 Loads on Quadratic sides * Distributed Shear Traction Shear traction on a side of a plane element whose sides remain straight (q is force/length): 96 In (b), a Q4 element and two LSTs share the top midnode so that the nodal loads from Q4 and the right LST are combined.
49 Uniform Bod Force Work-equivalent nodal forces corresponding to weight as a bod force: 97 LST has no verte loads and verte loads of Q8 are upwards! The resultant in all cases is W, the weight of the element. Connecting beam and plane elements Since the previous plane elements have translational dof onl, no moment can be applied to their nodes. The connection (a) of a beam and plane elements cannot transmit a moment and the beam can freel rotate. t (Singular K!) 98 In (b). the beam is etended. Rotational dof at A, B and C are associated with the beam elements onl. A plane element with drilling dof would also work but is not recommended.
50 Elements with Drilling DOF Drilling dof: rotational dof about ais normal to the plane. A CST with these added to each node has 9 dof. This dof allows twisting and bending rotations of shells under some loads to be represented. See Section Stress calculation 0 0 Combining Hooke s law with strain-displacement equation σ ν 0 αt E σ E( Bd 0 ) eg.. σ ν 0 B αt d ν τ 0 0 ( ν)/ 0 Stresses are most accurate inside elements
51 Improving stresses at nodes and boundaries 0 One common technique is averaging, g, but There are interpolation and etrapolation techniques that we will stud later Eamples of poor meshing 0 Do not create unnecessar discontinuities!
52 Eample of discontinuities in un- averaged stresses 0 3
Bilinear Quadrilateral (Q4): CQUAD4 in GENESIS
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