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CHAP 4 FINITE ELEMENT ANALYSIS OFBEAMS AND FRAMES1INTRODUCTION We learned Direct Stiffness Method in Chapter 2– Limited to simple elements such as 1D bars we will learn Energy Method to build beam finite element– Structure is in equilibrium when the potential energy is minimum Potential energy: Sum of strain energy and potential ofapplied loadsPotential of U V Interpolation scheme:applied loadsv( x) N( x) {q}Strain energyBeamInterpolation NodaldeflectionfunctionDOF2

BEAM THEORY Euler-Bernoulli Beam Theory–––––can carry the transverse loadslope can change along the span (x-axis)Cross-section is symmetric w.r.t. xy-planeThe y-axis passes through the centroidLoads are applied in xy-plane (plane of loading)yyNeutral axisPlane of loadingxzALFF3BEAM THEORY cont. Euler-Bernoulli Beam Theory cont.– Plane sections normal to the beam axis remain plane and normal tothe axis after deformation (no shear stress)– Transverse deflection (deflection curve) is function of x only: v(x)– Displacement in x-dir is function of x and y: u(x, y)dvu ( x, y ) u0 ( x) ydxxx u du0 x dxy dvdxy(dv/dx)Neutral axisxLd 2vy 2dxFy dv/dxv(x)4

BEAM THEORY cont. Euler-Bernoulli Beam Theory cont.–––– u du0 x dxd 2vy 2dxStrain along the beam axis: 0 du0 / dxStrain xx varies linearly w.r.t. y; Strain yy 0Curvature: d 2v / dx 2Can assume plane stress in z-dirbasically uniaxial statusxx xx Exx Ed 2vEy 2dx0 Axial force resultant and bending momentd 2vP xx dA E 0 dA E 2 ydAdx AAAM y xx dA E0A ydA EAP EA0d 2vM EI 2dx2d v 2y dAdx 2 AEA: axial rigidityEI: flexural rigidityMoment of inertia I(x)5BEAM THEORY cont. Beam constitutive relation– We assume P 0 (We will consider non-zero P in the frame element)– Moment-curvature relation:d 2vM EI 2dxMoment and curvature is linearly dependent Sign convention M Vyyx P M P Vy– Positive directions for applied loadsyp(x)xC1F1C2F2C3F36

GOVERNING EQUATIONS Beam equilibrium equations 0()f pxdx y Vy M M dM dx dx dx Vy 0dx dVy pdx dVydxdxVy dx 02 p( x)Vy dMdxd 4v– Combining three equations together: EI 4 p ( x)dx– Fourth-order differential equationpVyMdVydxMVydxdMdxdxdx7STRESS AND STRAIN Bending stressd 2v xx Ey 2dx xx ( x, y ) d 2vM EI 2dxM ( x) yIBending stress– This is only non-zero stress component for Euler-Bernoulli beam Transverse shear strain xy u y v v x x v 0 xu ( x, y ) u0 ( x) ydvdx– Euler beam predicts zero shear strain (approximation)VQ– Traditional beam theory says the transverse shear stress is xy Ib– However, this shear stress is in general small compared tothe bending stress8

POTENTIAL ENERGY Potential energy U V Strain energy– Strain energy density1U 0 xx2xx1 E(221 d 2v 1 2 d 2v E y 2 Ey 2 xx ) 2 dx 2 dx 22– Strain energy per unit length21 2 d 2v 1 d 2v U L ( x) U 0 ( x, y, z ) dA Ey 2 dA E 2 22 dx dx AA1 d v U L ( x) EI 2 2 dx 222 y dA2AMoment ofinertia– Strain energy21 L d 2v U U L ( x) dx EI 2 dx02 0 dx L9POTENTIAL ENERGY cont. Potential energy of applied loadsV L0NCNFp( x)v( x) dx Fv( x ) Ci 1iiii 1dv( xi )dx Potential energy21 L d 2v U V EI 2 dx2 0 dx L0NFp( x)v( x) dx Fvi ( xi )i 1NC Cii 1dv( xi )dx– Potential energy is a function of v(x) and slope– The beam is in equilibrium when has its minimum value 0 vv*v10

RAYLEIGH-RITZ METHOD1. Assume a deflection shapev( x) c1 f1 ( x) c2 f 2 ( x). cn f n ( x)––Unknown coefficients ci and known function fi(x)Deflection curve v(x) must satisfy displacement boundary conditions2. Obtain potential energy as function of coefficients (c1 , c2 ,.cn ) U V3. Apply the principle of minimum potential energy to determinethe coefficients11EXAMPLE – SIMPLY SUPPORTED BEAMp0 Assumed deflection curvev( x) C sinxLE,I,L Strain energy21 L d 2v C 2 EI 4U EI 2 dx 2 04 L3 dx Potential energy of applied loads (no reaction forces)LL2 p0 LxV p( x)v( x)dx p0C sin dx CL00EI 4 2 2 p0 L Potential energy U V CC4 L3d EI 4 PMPE: CdC2 L32 p0 L4 p0 L4 0 C EI 512

EXAMPLE – SIMPLY SUPPORTED BEAM cont. Exact vs. approximate solutionsCapproxp0 L4 76.5EICexactp0 L4 76.8EI Approximate bending moment and shear force4 p0 L2d 2v2xxM ( x) EI 2 EIC 2 sinsin dxLLL34 p0 Ld 3v3xxVy ( x) EI 3 EIC 3 coscos 2dxLLL3 Exact solutions v( x) 1 p0 L xEI 24p0 L 3x12p0 4 x24 p0 Lp0 2xx22pLVy ( x) 0p0 x2M ( x) 13EXAMPLE – SIMPLY SUPPORTED BEAM cont.1.0Deflection0.8v(x)/v max 0.60.4v-exact0.2v-approx.0.000.20.4Bending Moment M(x) 0.10M exact-0.12M approxError increases0.00-0.02xx-0.140.6V exact Shear forceShear Force V(x)0.4V approx0.20.0-0.2-0.4-0.600.20.4x0.60.8114

EXAMPLE – CANTILEVERED BEAM–p0 Assumed deflectionCv( x) a bx c1x 2 c2 x3E,I,LF Need to satisfy BCv(0) 0, dv(0) / dx 0v( x) c1 x 2 c2 x3L Strain energy U EI 2c1 6c2 x 2 dx2 0 Potential of loadsV c1 , c2 L p0 v( x)dx Fv( L) C0 p0 L3 c1 32FLdv( L)dx p0 L42CL c2 4 FL3 3CL2 15EXAMPLE – CANTILEVERED BEAM cont.L Derivatives of U: U 2 EI 2c 6c x dx EI 4 Lc 6 L2c 12 1 2 c10 U 6 EI 2c1 6c2 x xdx EI 6 L2c1 12 L3c2 c20L PMPE: 0 c1 0 c2p0 L3EI 4 Lc1 6 L c2 FL2 2CL3p0 L423EI 6 L c1 12 L c2 FL3 3CL242 Solve for c1 and c2: c1 23.75 10 3 , c2 8.417 10 Deflection curve: v( x) 10 Exact solution: v( x) 3 23.75x238.417 x3 15400 x 2 800 x3 300 x 4 24 EI16

EXAMPLE – CANTILEVERED BEAM cont.Deflectionv(x)/v max r increases BendingmomentBending Moment M(x)500.00M exact400.00M approx300.00200.00100.000.00-100.0000.20.4x0.60.81 Shear forceShear Force V(x)600.0500.0400.0300.0V exactV approx200.000.20.4x0.60.8117FINITE ELEMENT INTERPOLATION Rayleigh-Ritz method approximate solution in the entire beam– Difficult to find approx solution that satisfies displacement BC Finite element approximates solution in an element– Make it easy to satisfy displacement BC using interpolation technique Beam element––––––Divide the beam using a set of elementsElements are connected to other elements at nodesConcentrated forces and couples can only be applied at nodesConsider two-node bean elementPositive directions for forces and couplesF2F1Constant or linearlydistributed loadC1xC2p(x)18

FINITE ELEMENT INTERPOLATION cont. Nodal DOF of beam element– Each node has deflection v and slope– Positive directions of DOFsv2– Vector of nodal DOFs {q} {v112}T Scaling parameter s– Length L of the beam is scaled to 1 using scaling parameter sv2v1x x11, ds dx,LLds 1dx Lds, dx L1s 2xLx1s 0x2s 1 Will write deflection curve v(s) in terms of s19FINITE ELEMENT INTERPOLATION cont. Deflection interpolation– Interpolate the deflection v(s) in terms of four nodal DOFs– Use cubic function: v( s) a0 a1s a2 s 2 a3s 3– Relation to the slope:dv dv ds 12 – Apply four conditions:v(0) v1dv(0) dxdx1 ds dxv(1) v2 L(a1 2a2 s 3a3 s )dv(1) dx2– Express four coefficients in terms of nodal DOFsv1 v(0) a01dv (0) a11dxLv2 v(1) a0 a1 a2 a32 1dv(1) (a1 2a2 3a3 )dxLa0 v1a1 L1a2 3v1 2 La3 2v1 L113v22v2LL2220

FINITE ELEMENT INTERPOLATION cont. Deflection interpolation cont.v( s) (1 3s 2 2s3 )v1 L( s 2s 2 s 3 )v( s ) [ N1 ( s )N 2 ( s)N1 ( s ) 1 3s 2L( s 2 s 3 )2v( s) N {q}1.02s3N 2 ( s ) L( s 2 s 2(3s 2 2s 3 )v2v1! !! !N 4 ( s)] " 1 #! v2 !! 2!%N3 ( s) Shape functions1N10.8s3 )N30.6N 3 ( s ) 3s 2 2 s 30.4N 4 ( s ) L( s 20.2s3 )– Hermite polynomials– Interpolation propertyN2/L0.00.00.20.40.6N4/L0.81.0-0.221FINITE ELEMENT INTERPOLATION cont. Properties of interpolation– Deflection is a cubic polynomial (discuss accuracy and limitation)– Interpolation is valid within an element, not outside of the element– Adjacent elements have continuous deflection and slope Approximation of curvature– Curvature is second derivative and related to strain and stress v1! !d 2v 1 d 2v 1! 1![612s,L(46s),612s,L(26s)] " #dx 2 L2 ds 2 L2! v2 !!d 2v 1 2!% Bq{} 22 B:strain-displacementvectordxL 1 4 4 1– B is linear function of s and, thus, the strain and stress2– Alternative expression: d v 1 T Tdx 2 q {B }L2 1 4 4 1– If the given problem is linearly varying curvature, the approximation isaccurate; if higher-order variation of curvature, then it is approximate22

FINITE ELEMENT INTERPOLATION cont. Approximation of bending moment and shear forced 2v EIM ( s) EI 2 2 B {q}dxLLineardMd 3v EIVy EI 3 3 [ 12dxdxL6 L]{q}6 L 12Constant– Stress is proportional to M(s); M(s) is linear; stress is linear, too– Maximum stress always occurs at the node– Bending moment and shear force are not continuous between adjacentelements23EXAMPLE – INTERPOLATIONv1v2 Cantilevered beam Given nodal DOFs{q} {0, 0, 0.1, 0.2}T1L Deflection and slope at x 0.5L Parameter s 0.5 at x 0.5L1L1L Shape functions: N1 ( 12 ) , N 2 ( 12 ) , N3 ( 12 ) , N 4 ( 12 ) 2828 Deflection at s 0.5:v( 12 ) N1 ( 12 )v11 02N 2 ( 12 )1N 3 ( 12 )v2L10v282 Slope at s 0.5:dv 1 dv 1 dN1 v1dx L ds L ds1 v1 ( 6s 6s 2 )L11dN 2ds 1L8v2N 4 ( 12 )2 v22dN3ds4 s 3s 2 v2222L 2 0.0258dN 4 ds 1(6 s 6 s 2 )L2 2 s 3s 2 0.124

EXAMPLE A beam finite element with length LL3v1 0, 1 0, v2 ,3EI2L2 2 EILF Calculate v(s)v( s) N1 ( s)v1 N 2 (s)v( s) (3s 2 2s3 )v2N3 (s)v21L( s 2s3 )N 4 (s)22 Bending momentd 2v EI d 2v EIM ( s ) EI 2 2 2 2 & (6 12s )v2 L( 2 6s )dxL dsLEI (L3L2 ) 2 (6 12s )L( 2 6 s )3EI2 EI L 2' L(1 s ) ( L x) Bending moment cause by unit force at the tip25FINITE ELEMENT EQUATION FOR BEAM Finite element equation using PMPE– A beam is divided by NEL elements with constant sections Strain energy– Sum of each element’s strain energyLTNEL0e 1ex2 NELU U L ( x)dx e U L ( x)dx U e x1e 1– Strain energy of element (e)U e ex2 EI e x121 d 2v EIdx 3 2 L2 dx yC1p(x)x1F1x 1121 d 2v 0 2 ds 2 ds1x 21F2C2C3C4234x12x 22F3x1 3x 23F4x1 4C55F5x 2426

FE EQUATION FOR BEAM cont. Strain energy cont.– Approximate curvature in terms of nodal DOFs2 d 2v d 2v d 2v Te e Tq {} B B {q } ds 2 ds 2 ds 2 1 44 14 11 4 – Approximate element strain energy in terms of nodal DOFsU(e)1 e ( EI {q }T 32 L e 1 e T e e ) e BBqds{} {q } [k ]{q } 0 21T Stiffness matrix of a beam element( 6 12s ) EI 1 L( 4 6 s) e [k ] 3 & 6 12sL 0 6 12 s L( 2 6 s ) L( 4 6 s) 6 12 sL( 2 6 s) ' ds27FE EQUATION FOR BEAM cont. Stiffness matrix of a beam element( 12 EI 6 L e [k ] 3L 12 6L6L4 L26L2 L2126L126L6L )2 L2 6L 4 L2 Symmetric, positive semi-definiteProportional to EIInversely proportional to L Strain energy cont.NELU U(e)e 11 NEL e T e e {q } [k ]{q }2 e 1– Assembly1U {Q s }T [K s ]{Q s }228

EXAMPLE – ASSEMBLYy2EIx3212L{Qs }T {v1Lv1(3 EI 3L 1 [k ] 3L 3 3 L Two elements Global DOFsEIv213L4 L23L2 L2F233L33L(3 3 L EI 3[K s ] 3 L 3 L 0 0F3v22( 12 EI 6 L 2 [k ] 3L 12 6L3L ) v12 L2 1 3L v2 4 L2 23L4 L23L2 L2003L2 L23L8 L26L2 L233L153L126L00126L126Lv212v326L4 L26L2 L2126L126Lv33}36 L ) v22 L2 2 6 L v3 4 L2 30 )0 6L 2 L2 6L 4 L2 29FE EQUATION FOR BEAM cont. Potential energy of applied loadsF1!C !– Concentrated forces and couplesND!! 1 !!V Fi vi Ci i V v1 1 v2 . ND " F2 # {Q s }T {Fs }i 1!!!! CND !%– Distributed load (Work-equivalent nodal forces) !V NEL e 1V(e) L(e)ex2 ex1 NELp( x)v( x) dx V(e)V(e)e 11 p(s) v N111N 2 v2 N 32ex2 e p( x)v( x) dx Lx1(e)1 p(s)v(s) ds0N 4 ds0111 v1 L( e ) p ( s ) N1ds 1 L( e ) p ( s ) N 2 ds v2 L( e ) p (s ) N 3ds 000 (e)(e)(e)(e) v1F1v2 F21C12 C2 (e) 1 2 L p ( s ) N 4 ds 0 30

EXAMPLE – WORK-EQUIVALENT NODAL FORCES Uniformly distributed loadpL00211pL2223C1 pL N 2 ( s ) ds pL ( s 2s s ) ds 001211pLF2 pL N3 ( s) ds pL (3s 2 2s 3 ) ds 00211pL2223C2 pL N 4 ( s ) ds pL ( s s ) ds 001211F1 pL N1 ( s) ds pL (1 3s 2 2s 3 ) ds pL{F} " 2TpL212pL2pL2#12 %ppL/2EquivalentpL/2pL2/12pL2/1231FE EQUATION FOR BEAM cont. Finite element equation for beam( 12 EI 6 LL3 12 6L6L4 L26L2 L2126L126L6 L ) v12 L2 !! 1 !! !! " # "6 L ! v2 ! ! 4 L2 ! 2 !% !pL / 2pL2 /12 !!#pL / 2 !pL2 /12 %!F1! !! C1 !" #! F2 !! C2 !%– One beam element has four variables– When there is no distributed load, p 0– Applying boundary conditions is identical to truss element– At each DOF, either displacement (v or ) or force (F or C) must beknown, not both– Use standard procedure for assembly, BC, and solution32

PRINCIPLE OF MINIMUM POTENTIAL ENERGY Potential energy (quadratic form)1 U V {Q s }T [K s ]{Q s } {Q s }T {Fs }2 PMPE– Potential energy has its minimum when[K s ]{Q s } {Fs }[Ks] is symmetric & PSD Applying BC– The same procedure with truss elements (striking-the-rows andstriking-he-columns)[K ]{Q} {F}[K] is symmetric & PD Solve for unknown nodal DOFs {Q}33BENDING MOMENT & SHEAR FORCE Bending momentd 2v EI d 2v EIM ( s) EI 2 2 2 2 B {q}dxL dsL– Linearly varying along the beam span Shear forcedMd 3v EI 3 Vy ( s ) dxdxEI d 3v EI [ 12L3 ds 3 L36 L 12v1! !! !6 L] " 1 #! v2 !! 2!%– Constant– When true moment is not linear and true shear is not constant, manyelements should be used to approximate itMy Bending stress x I Shear stress for rectangular section1.5Vy 4 y 2 xy ( y ) 1bh h 2 34

yEXAMPLE – CLAMPED-CLAMPEDBEAM Determine deflection &slope at x 0.5, 1.0, 1.5 m Element stiffness matricesv1( 12 6(1)[k ] 1000 12 6( 12 6 121000 6 0 mF2 240 Nv26 ) v12 1 6 v2 4 23( 12 6(2)[k ] 1000 12 62v3646212612636 ) v22 2 6 v3 4 30 ) v1F1! !0 ! 1 ! ! C1 !!! !!!240 !6 v2 " # " 0 #2 ! 2 ! !!!!!F3 !6 v3! ! ! !4 3 % C3 %35EXAMPLE – CLAMPED-CLAMPED BEAM cont. Applying BCv2 0.01( 24 0 ) v 22401000 "#"# 0 8 2 % 0 % At x 0.51dN31v2L(1)dsN 2 ( 12 ) v2 N3 ( 12 )2N 4 ( 12 ) 0.01 N3 ( 12 ) 0.005m 0.015rads 12 At x 1.0either s 1 (element 1) or s 0 (element 2)v(1) v2 N 3 (1) 0.01 N 3 (1) 0.01m(1) 0.0s 0.5 and use element 1v( 12 ) v1 N1 ( 12 )( 12 ) 2dN 31v(1) 2Lds 0.0 rads 1v(0) v2 N1 (0) 0.01 N1 (0) 0.01m(0) 1dN1v(2) 2Lds 0.0 rads 0Will this solution be accurate or approximate?36

EXAMPLE – CANTILEVERED BEAMp0 120 N/m One beam element No assembly required Element stiffness( 12 6[K s ] 1000 12 66462126126EI 1000 N-m2L 1m6 ) v12 1 6 v2 4 2C –50 N-m Work-equivalent nodal forces1 3s 2 2 s 3F1e!!! !!1 !( s2s 2 s3 ) L !! C1e !!" # p0 L 0 "# ds p0 L "23Fss32! 2e !!!!! C2 e !%! ( s 2 s 3 ) L !%! 1/ 260L /12 !! !! 10 !!# "#1/ 2 ! ! 60 !L /12!% ! 10!%37EXAMPLE – CANTILEVERED BEAM cont. FE matrix equation( 12 61000 12 66462 Applying BC(121000 61261266 ) v1F1 602 !! 1 !! !! C1 10 !! " # "#6 ! v2 ! ! 60 ! 4 ! 2 !% ! 10 50 %!6 ) v260 " # "#4 2 % 60 %v2 0.01m2 0.03 rad Deflection curve: v( s) 0.01N3 ( s) 0.03N 4 ( s) 0.01s 3 Exact solution: v( x) 0.005( x 4 4 x3 x 2 )38

EXAMPLE – CANTILEVERED BEAM cont. Support reaction (From assembled matrix equation)1000 12v21000 6v26222 F1 C1F1 120 N60C1 10 N m10 Bending momentEI B {q}L2EI 2 & ( 6 12 s)v1 L( 4 6 s) 1 (6 12 s)v2L 1000[ 0.01(6 12s ) 0.03( 2 6 s)] 60s N mM ( s) L( 2 6 s )2' Shear forceEI&12v1 6 L 1 12v2 6 L 2 'L3 1000[ 12 ( 0.01) 6( 0.03)] 60 NVy 39EXAMPLE – CANTILEVERED BEAM cont. Bending moment-50-60Shear force-12000.20.4x0.60.8100.20.4x0.60.8140

PLANE FRAME ELEMENT Beam– Vertical deflection and slope. No axial deformation Frame structure– Can carry axial force, transverse shear force, and bending moment(Beam Truss) Assumptionv1– Axial and bending effectsare uncoupled– Reasonable when deformationu2is small 3 DOFs per nodev2u1u212pFv2122u2v232{ui , vi , i } Need coordinate transformation like plane truss2u131v1u14v11141PLANE FRAME ELEMENT cont. Element-fixed local coordinates x y Local DOFs {u , v , }Local forces { f x , f y , c } Transformation between local and global coord.f x1( cos *! f ! sin *! y1 ! ! c1 ! 0" # ! fx 2 ! 0! fy2 ! 0! ! c2 % 0sin * 0cos * 001000000000cos *sin *000 ) f x100 ! f y1 ! ! !00 ! c1 ! " #sin * 0 ! f x 2 !cos * 0 ! f y 2 ! ! !01 c2 %Local coordinatesy{f } [T]{f }{q} [T]{q}v1yu11xv2u2x2*21Global coordinates42

PLANE FRAME ELEMENT cont. Axial deformation (in local coord.)1) u1f x1 " # " #1 u2 % f x 2 %EA ( 1L 1 Beam bending( 12 EI 6 LL3 12 6L6L4 L26L2 L2126L126L6 L ) v1f y1! !2 L2 !! 1 !! ! c1 ! " # " #6 L ! v2 ! ! f y 2 ! 4 L2 ! 2 !% ! c2 !% Basically, it is equivalent to overlapping a beam with a bar A frame element has 6 DOFs43PLANE FRAME ELEMENT cont. Element matrix equation (local coord.)( a100a100012a26 La2012a26 La206 La24 L2 a206 La22 L2 a2a100a100012a26 La2012a26 La20 ) u1f x1! ! ! !6 La2 ! v1 ! ! f y1 ! 2 L2 a2 ! 1 ! ! c1 ! " # " #0 ! u2 ! ! f x 2 !6 La2 ! v2 ! ! f y 2 ! ! ! ! !4 L2 a2 2 % c2 %EALEIa2 3La1 [k ]{q} {f } Element matrix equation (global coord.)[k ][T]{q} [T]{f }[T]T [k ][T]{q} {f }[k ]{q} {f }[k ] [T]T [k ][T] Same procedure for assembly and applying BC44

PLANE FRAME ELEMENT cont. Calculation of element forces––––Element forces can only be calculated in the local coordinateExtract element DOFs {q} from the global DOFs {Qs}Transform the element DOFs to the local coordinate {q} [T]{q}Then, use 1D bar and beam formulas for element forcesAE u2 u1 LEI– Bending moment M ( s) 2 B {q}LEI– Shear force Vy ( s ) 3 [ 126 L 12L Other method: V6L( 12y1!!2 ! M 1 ! EI 6 L 4 L"# 3V! y 2 ! L 12 6 L 2! M 2 !% 6L 2L– Axial force P 6 L] q,126L126L6 L ) v12 L2 !! 1 !! " #6 L ! v2 ! 4 L2 ! 2 !%45

xy VQ Ib 9 POTENTIAL ENERGY Potential energy Strain energy – Strain energy density – Strain energy per unit length – Strain energy UV 2222 22 0 22 11 1 1 22 2 2xx xx xx dv dv U