CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES INTRODUCTION

Transcription

1 CHAP FINITE EEMENT ANAYSIS OF BEAMS AND FRAMES INTRODUCTION We learned Direct Stiffness Method in Chapter imited to simple elements such as D bars we will learn Energ Method to build beam finite element Structure is in equilibrium when the potential energ is minimum Potential energ: Sum of strain energ and potential of applied loads Interpolation scheme: U V v ( ) N( ) { q} Strain energ Potential of applied loads Beam deflection Interpolation function Nodal DOF

2 Euler-Bernoulli Beam Theor BEAM THEORY can carr the transverse load slope can change along the span (-ais) Cross-section is smmetric w.r.t. -plane The -ais passes through the centroid oads are applied in -plane (plane of loading) Neutral ais Plane of loading z F F A BEAM THEORY cont. Euler-Bernoulli Beam Theor cont. Plane sections normal to the beam ais remain plane and normal to the ais after deformation (no shear stress) Transverse deflection (deflection curve) is function of onl: v() Displacement in -dir is function of and : u(, ) dv u (, ) u( ) Neutral ais u du d v (dv/) dv = dv/ F v()

3 BEAM THEORY cont. Euler-Bernoulli Beam Theor cont. du Strain along the beam ais: / u du Strain varies linearl w.r.t. ; Strain = Curvature: d v/ Can assume plane stress in z-dir basicall uniaial status dv E E E d v Aial force resultant and bending moment dv A A A P da E da E da dv M da EdA E da A A A Moment of inertia I() P EA dv M EA: aial rigidit : fleural rigidit 5 BEAM THEORY cont. Beam constitutive relation We assume P = (We will consider non-zero P in the frame element) Moment-curvature relation: dv M Moment and curvature is linearl dependent Sign convention +V +M +M +P +P +V Positive directions for applied loads p() C C C F F F 6

4 GOVERNING EQUATIONS Beam equilibrium equations dv f p( ) V V dv ( ) p dm M M p V Combining three equations together: Fourth-order differential equation p V M V dv p( ) dv M V dm dm 7 STRESS AND STRAIN Bending stress dv E M dv (, ) M( ) I Bending stress This is onl non-zero stress component for Euler-Bernoulli beam Transverse shear strain u v v v dv u (, ) u( ) Euler beam predicts zero shear strain (approimation) Traditional beam theor sas the transverse shear stress is However, this shear stress is in general small compared to the bending stress VQ Ib 8

5 Potential energ Strain energ POTENTIA ENERGY Strain energ densit dv dv U E( ) E E Strain energ per unit length dv dv U ( ) U(,, z) da E da E da A A A dv Moment of U( ) inertia Strain energ U V dv U U ( ) 9 POTENTIA ENERGY cont. Potential energ of applied loads i V p v Fv i i Ci i i Potential energ N F N C ( ) ( ) ( ) dv( ) N F N dv C U V p( ) v( ) Fv( ) C i i i i i dv( i ) Potential energ is a function of v() and slope The beam is in equilibrium when has its minimum value v v * v

6 . Assume a deflection shape RAYGH-RITZ METHOD v ( ) cf( ) cf( )... cf( ) Unknown coefficients c i and known function f i () Deflection curve v() must satisf displacement boundar conditions. Obtain potential energ as function of coefficients ( c, c,... c ) U V n. Appl the principle of minimum potential energ to determine the coefficients n n EXAMPE SIMPY SUPPORTED BEAM Assumed deflection curve v ( ) Csin Strain energ dv C U Potential energ of applied loads (no reaction forces) Potential energ PMPE: p V p( ) v( ) pcsin C p U V C C d p p dc C C 5 p E,I,

7 EXAMPE SIMPY SUPPORTED BEAM cont. Eact vs. approimate solutions C appro p p Ceact Approimate bending moment and shear force dv p M ( ) C sin sin p V ( ) C cos cos Eact solutions dv p p p v ( ) p p M( ) p V ( ) p EXAMPE SIMPY SUPPORTED BEAM cont. Deflection Bending moment Shear force v()/v_ma Bending Moment M() Shear Force V() v-eact v-appro M_eact M_appro V_eact V_appro Error increases

8 EXAMPE CANTIEVERED BEAM Assumed deflection v ( ) abc c Need to satisf BC Strain energ v(), dv() / v ( ) c c Potential of loads U c 6c dv V c, c p v( ) Fv( ) C ( ) p E,I, p p c F C c F C F C 5 EXAMPE CANTIEVERED BEAM cont. Derivatives of U: PMPE: c c U c 6c c 6 c c U 6 c 6c 6 c c c p c 6 c F C p 6 c c F C Solve for c and c : c.75, c 8.7 Deflection curve: v ( ) Eact solution: v ( ) 5 8 6

9 EXAMPE CANTIEVERED BEAM cont. Deflection v()/v_ma..... v-eact v-appro Bending moment Bending Moment M() M_eact M_appro Error increases Shear force Shear Force V() V_eact V_appro FINITE EEMENT INTERPOATION Raleigh-Ritz method approimate solution in the entire beam Difficult to find appro solution that satisfies displacement BC Finite element approimates solution in an element Make it eas to satisf displacement BC using interpolation technique Beam element Divide the beam using a set of elements Elements are connected to other elements at nodes Concentrated forces and couples can onl be applied at nodes Consider two-node bean element Positive directions for forces and couples Constant or linearl distributed load C F F C p() 8

10 FINITE EEMENT INTERPOATION cont. Nodal DOF of beam element Each node has deflection v and slope Positive directions of DOFs Vector of nodal DOFs Scaling parameter s {} q { v v } T ength of the beam is scaled to using scaling parameter s s, ds, ds ds, s = v v s = Will write deflection curve v(s) in terms of s 9 FINITE EEMENT INTERPOATION cont. Deflection interpolation Interpolate the deflection v(s) in terms of four nodal DOFs Use cubic function: vs () aas as as Relation to the slope: dv dv ds ( a as a ) s ds Appl four conditions: dv() dv() v() v v() v Epress four coefficients in terms of nodal DOFs v v() a dv () a v v() a a a a dv () ( aa a) a a v a v v a v v

11 FINITE EEMENT INTERPOATION cont. Deflection interpolation cont. vs () (s s) vs ( s s) (s s) v ( s s) N () s s s v v( s) [ N( s) N( s) N( s) N( s)] vs () v N{} q Shape functions. N s s s s () ( ) N () s s s N s s s () ( ) N N N / Hermite polnomials Interpolation propert. -. N / FINITE EEMENT INTERPOATION cont. Properties of interpolation Deflection is a cubic polnomial (discuss accurac and limitation) Interpolation is valid within an element, not outside of the element Adjacent elements have continuous deflection and slope Approimation of curvature Curvature is second derivative and related to strain and stress v dv dv [ 6 s, ( 6 s ),6 s, ( 6 s )] ds v dv {} B q B: strain-displacement vector B is linear function of s and, thus, the strain and stress Alternative epression: dv T T { } q B If the given problem is linearl varing curvature, the approimation is accurate; if higher-order variation of curvature, then it is approimate

12 FINITE EEMENT INTERPOATION cont. Approimation of bending moment and shear force dv M() s {} B q inear dm d v V [ 6 6 ]{ q} Constant Stress is proportional to M(s); M(s) is linear; stress is linear, too Maimum stress alwas occurs at the node Bending moment and shear force are not continuous between adjacent elements EXAMPE INTERPOATION Cantilevered beam Given nodal DOFs { q} {,,.,.} T Deflection and slope at =.5 v v Parameter s =.5 at =.5 Shape functions: N(), N(), N(), N() 8 8 Deflection at s =.5: v() N() vn() N() v N() v v Slope at s =.5: dv dv dn dn dn dn v v ds ds ds ds ds v ( 6s6 s ) s s v (6s6 s ) ss.

13 EXAMPE A beam finite element with length v,, v, Calculate v(s) vs () N() svn() s N() sv N() s vs () (s s) v ( s s) F Bending moment dv dv M( s) (6 s) v ( 6 s) ds (6 s) ( 6 s) ( s) ( ) Bending moment cause b unit force at the tip 5 FINITE EEMENT EQUATION FOR BEAM Finite element equation using PMPE A beam is divided b NE elements with constant sections Strain energ Sum of each element s strain energ e T U U ( ) U ( ) U NE Strain energ of element (e) NE e e e e e e dv dv U e ds ds p() C C C C C 5 5 F F F F F 5 6

14 Strain energ cont. FE EQUATION FOR BEAM cont. Approimate curvature in terms of nodal DOFs dv dvdv e T T e { } { } ds ds ds q B B q Approimate element strain energ in terms of nodal DOFs ( e) e T T e e T e e U { q } { } { } [ ]{ } ds B B q q k q Stiffness matri of a beam element 6s e ( 6 s) [ k ] 6 s ( 6 s) 6 s ( 6 s) ds 6s ( 6 s) e 7 FE EQUATION FOR BEAM cont. Stiffness matri of a beam element e [ k ] Smmetric, positive semi-definite Proportional to Inversel proportional to Strain energ cont. U NE NE ( e) U q k q e e Assembl { } T U Qs [ Ks ]{ Qs } e T e e { } [ ]{ } 8

15 EXAMPE ASSEMBY F F Two elements Global DOFs T Q v [ k ] v v { s} { v v v } v 6 6 v 6 6 [ k ] 6 6v [ K s] v v 9 FE EQUATION FOR BEAM cont. Potential energ of applied loads F Concentrated forces and couples C ND T V Fv i ic i i V vv... ND F { Qs} { Fs } i Distributed load (Work-equivalent nodal forces) C ND NE e NE V p( ) v( ) V e e e ( e) ( e) ( e) V p() s v N N v N N ds ( e) ( e) ( e) ( e) e ( e) ( e) e V p( ) v( ) p( s) v( s) ds ( e) ( e) ( e) ( e) v psnds () psnds () v psnds () psnds () vf C vf C

16 EXAMPE WORK-EQUIVAENT NODA FORCES Uniforml distributed load () ( ) F p N s ds p s s ds C p N () s ds p ( s s s ) ds () ( ) F p N s ds p s s ds C p N () s ds p ( s s ) ds { } p p p p F T p/ p p p p p Equivalent p/ p / p / FE EQUATION FOR BEAM cont. Finite element equation for beam 6 6 v p/ F 6 6 p / C 6 6 v p/ F 6 6 p / C One beam element has four variables When there is no distributed load, p = Appling boundar conditions is identical to truss element At each DOF, either displacement (v or ) or force (F or C) must be known, not both Use standard procedure for assembl, BC, and solution

17 PRINCIPE OF MINIMUM POTENTIA ENERGY Potential energ (quadratic form) { } T [ ]{ } { } T U V Qs Ks Qs Qs { Fs } PMPE Potential energ has its minimum when [ K ]{ Q } { F} s s s [K s ] is smmetric & PSD Appling BC The same procedure with truss elements (striking-the-rows and striking-he-columns) [ K]{ Q} { F} [K] is smmetric & PD Solve for unknown nodal DOFs {Q} BENDING MOMENT & SHEAR FORCE Bending moment dv dv M() s {} B q ds inearl varing along the beam span Shear force v dm dv dv V () s [ 6 6 ] ds v Constant When true moment is not linear and true shear is not constant, man elements should be used to approimate it M I Bending stress Shear stress for rectangular section.5v ( ) bh h

18 EXAMPE CAMPED-CAMPED BEAM Determine deflection & slope at =.5,.,.5 m Element stiffness matrices () [ ] k v v 6 6 v v 6 6 m m F = N v v k () [ ] 6 6 v v v F 6 6 C 6 6 v v F 6 6 C 5 EXAMPE CAMPED-CAMPED BEAM cont. Appling BC v 8 v.. At =.5 s =.5 and use element v() v N () N () v N () N (). N ().5m dn ( ) v ().5rad ds s At =. either s = (element ) or s = (element ) v() v N (). N ().m dn () v (). rad ds s v() v N (). N ().m dn () v ().rad ds s Will this solution be accurate or approimate? 6

19 EXAMPE CANTIEVERED BEAM One beam element No assembl required Element stiffness 6 6 v 6 6 [ K s] 6 6v 6 6 Work-equivalent nodal forces p = N/m e C e ( s s s ) p ds p F e s s Ce ( s s ) = N-m C = 5 N-m = m F s s / 6 / / 6 / 7 EXAMPE CANTIEVERED BEAM cont. FE matri equation 6 6 v F C 6 6 v Appling BC 6v v.m.rad Deflection curve: Eact solution: vs ( ). N( s). N( s).s v ( ).5( ) 8

20 EXAMPE CANTIEVERED BEAM cont. Support reaction (From assembled matri equation) v 6 F 6 F N 6v C Bending moment M() s B{} q ( 6 sv ) ( 6 s) (6 sv ) ( 6 s) [.(6 s).( 6 s)] 6s Nm Shear force V v 6v 6 [ (.) 6(.)] 6 N C Nm 9 EXAMPE CANTIEVERED BEAM cont. Comparisons v M Deflection FEM Eact Bending moment FEM Eact V Slope FEM Eact FEM Eact Shear force...6.8

21 Beam PANE FRAME EEMENT Vertical deflection and slope. No aial deformation Frame structure Can carr aial force, transverse shear force, and bending moment (Beam + Truss) Assumption Aial and bending effects are uncoupled Reasonable when deformation is small DOFs per node { ui, vi, i} Need coordinate transformation like plane truss v v u u F v v u p u v v u u PANE FRAME EEMENT cont. Element-fied local coordinates ocal DOFs { uv,, } ocal forces { f, f, c} Transformation between local and global coord. f cos sin f f sin cos f c c f cos sin f f ocal coordinates sin cos f v c c u {} f [ T]{} f {} q [ T]{} q v u Global coordinates

22 PANE FRAME EEMENT cont. Aial deformation (in local coord.) EA u f u f Beam bending 6 6 v f 6 6 c 6 6 v f 6 6 c Basicall, it is equivalent to overlapping a beam with a bar A frame element has 6 DOFs PANE FRAME EEMENT cont. Element matri equation (local coord.) a a u f a 6a a 6a v f 6a a 6a a c a a u f a 6a a 6a v f 6a a 6a a c [ k]{} q {} f Element matri equation (global coord.) a a EA T [ k][ T]{ q} [ T]{ f } [ T] [ k][ T]{ q} { f } [ k]{ q} { f} [ k] [ T] T [ k][ T] Same procedure for assembl and appling BC

23 PANE FRAME EEMENT cont. Calculation of element forces Element forces can onl be calculated in the local coordinate Etract element DOFs {q} from the global DOFs {Q s } Transform the element DOFs to the local coordinate {} q [ T]{} q Then, use D bar and beam formulas for element forces AE Aial force P u u Bending moment M() s B{} q Shear force V () s [ 6 6 ] q Other method: V 6 6 v M 6 6 V 6 6v M 6 6 5