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Eml 4500 Finite Element Analysis and Design

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Eml 4500 Finite Element Analysis and Design CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES 1 INTRODUCTION • 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 of applied loads UV Potential of • Interpolation scheme: applied loads Strain energy vx()4HXGWNq () x {} Beam Interpolation Nodal deflection function DOF 2 BEAM THEORY • Euler-Bernoulli Beam Theory – can carry the transverse load – slope can change along the span (x-axis) – Cross-section is symmetric w.r.t. xy-plane – The y-axis passes through the centroid – Loads are applied in xy-plane (plane of loading) y y Neutral axis Plane of loading x z A L F F 3 BEAM THEORY cont. • Euler-Bernoulli Beam Theory cont. – Plane sections normal to the beam axis remain plane and normal to the 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) dv udvdu 2 dv uxy(, ) u () x y 0 y 0 dx xx xdxdx2 dx y y(dv/dx) Neutral axis x y = dv/dx L F v(x) 4 BEAM THEORY cont. udvdu 2 • Euler-Bernoulli Beam Theory cont. 0 y xx 2 – Strain along the beam axis: 00 du/ dx xdxdx – Strain xx varies linearly w.r.t. y; Strain yy = 0 – Curvature: dv22/ dx – Can assume plane stress in z-dir basically uniaxial status dv2 EEEy xx xx 0 dx2 • Axial force resultant and bending moment dv2 P dA E dA E ydA PEA OOOxx 0 dx2 0 AAA dv2 dv2 MEI M y dA E ydA E y2 dA dx2 OOOxx 0 dx2 AAA EA: axial rigidity Moment of inertia I(x) EI: flexural rigidity 5 BEAM THEORY cont. • Beam constitutive relation – We assume P = 0 (We will consider non-zero P in the frame element) – Moment-curvature relation: dv2 MEI Moment and curvature is linearly dependent dx2 • Sign convention +Vy +M y +M +P x +P +Vy – Positive directions for applied loads y p(x) x C C C 1 2 3 F F F 1 2 3 6 GOVERNING EQUATIONS • Beam equilibrium equations CSdV y dVy B fpxdxVdxVyyy;0() DT 0 px() EUdx dx CSdM dx dM MDT M dx pdx Vy dx 0 V EUdx 2 y dx dv4 – Combining three equations together: EI p() x dx4 – Fourth-order differential equation p dV Vdxy y dx dM V Mdx M y dx dx 7 STRESS AND STRAIN • Bending stress dv2 dv2 Ey MEI xx dx2 dx2 Mxy() (,xy ) xx I Bending stress – This is only non-zero stress component for Euler-Bernoulli beam • Transverse shear strain uv vv dv 0 uxy(, ) u () x y xy yx xx 0 dx – 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 to the bending stress 8 POTENTIAL ENERGY • Potential energy UV • Strain energy – Strain energy density 2222 1122 1CSdv 1 CS dv UEEyEy0 xx xx() xx DT DT 22 2EUdx22 2 EU dx – Strain energy per unit length 22 11CSdv22 CS dv U() x U (,,) x y z dA Ey22 dA E y dA L OO0 22DTdx22 DT dx O AAEU EU A 2 1 CSdv2 Moment of Ux() EI L DT2 inertia 2 EUdx – Strain energy 2 2 LL1 CSdv UUxdxEIdx() OOL DT2 002 EUdx 9 POTENTIAL ENERGY cont. • Potential energy of applied loads N N L F C dv() x VpxvxdxFvxC ()() BB ( ) i O0 ii i ii11dx • Potential energy 2 2 N N 1 LLCSdv F C dv() x U V EI dx p()() x v x dx BB Fv ( x ) C i OO00DT2 ii i 2 EUdxii11 dx – Potential energy is a function of v(x) and slope – The beam is in equilibrium when has its minimum value 0 v * v v 10 RAYLEIGH-RITZ METHOD 1. Assume a deflection shape vx( ) cf11 ( x ) cf 2 2 ( x )..... cfnn ( x ) – Unknown coefficients ci and known function fi(x) – Deflection curve v(x) must satisfy displacement boundary conditions 2. Obtain potential energy as function of coefficients (cc12 , ,... cn ) U V 3. Apply the principle of minimum potential energy to determine the coefficients 11 EXAMPLE – SIMPLY SUPPORTED BEAM p • Assumed deflection curve 0 x vx( ) C sin L E,I,L • Strain energy 2242 1 L CSdv CEI UEIdx O DT23 240 EUdx L • Potential energy of applied loads (no reaction forces) LL x 2 pL VpxvxdxpCdxC ()() sin 0 OO0 L 00 EI 4 2 pL • Potential energy UV C2 0 C 4L3 dEI 4 24pL pL4 • PMPE: ;CC000 dC2 L35 EI 12 EXAMPLE – SIMPLY SUPPORTED BEAM cont. • Exact vs. approximate solutions pL44 pL CC00 approx76.5EI exact 76.8 EI • Approximate bending moment and shear force dv22 x4 pL2 x M( x ) EI EIC sin0 sin dx22 L L 3 L dv33 x4 pL x V() x EI EIC cos 0 cos y dx33 L L 2 L 3 1 CSpL pL34 p • Exact solutions vx() DT00 x x 0 x EI EU24 12 24 pL p Mx() 00 x x2 22 pL Vx() 0 px y 2 0 13 EXAMPLE – SIMPLY SUPPORTED BEAM cont. 1.0 • Deflection 0.8 0.6 0.4 v(x)/v_max v-exact 0.2 v-approx. 0.0 0 0.2 0.4x 0.6 0.8 1 x Error increases 0.00 0 0.2 0.4 0.6 0.8 1 -0.02 • Bending -0.04 moment -0.06 -0.08 -0.10 M_exact Bending Moment M(x)Bending -0.12 M_approx -0.14 0.6 V_exact 0.4 V_approx • Shear force 0.2 0.0 -0.2 Shear Force V(x) -0.4 -0.6 0 0.2 0.4x 0.6 0.8 1 14 EXAMPLE – CANTILEVERED BEAM –p0 • Assumed deflection C 23 vx() a bxcx 12 cx E,I,L F • Need to satisfy BC vdvdx(0) 0, (0) / 0 23 vx() cx12 cx EI L • Strain energy Uccxdx 26 2 2 O 12 0 • Potential of loads L dv Vcc,()()() p vxdxFvL C L 12O 0 dx 0 CSCS34 pL00232 pL cFLCLcFLCL12DTDT23 EUEU34 15 EXAMPLE – CANTILEVERED BEAM cont. U U L • Derivatives of : 226EI c c x dx EI 4 Lc 6 L2 c c O 12 1 2 1 0 U L 626EI c c x xdx EI 6 L23 c 12 L c c O 12 1 2 2 0 pL3 • PMPE: 0 220 c EI46 Lc12 L c FL 2 CL 1 3 pL4 0 EI612 L23 c L c0 FL 32 3 CL c 12 2 4 cc23.75 10 33 , 8.417 10 • Solve for c1 and c2: 12 • Deflection curve: vx( ) 10 32 23.75 x 8.417 x 3 1 vx( ) 5400 x234 800 x 300 x • Exact solution: 24EI 16 EXAMPLE – CANTILEVERED BEAM cont. 0.0 • Deflection 0.0 0.0 v(x)/v_max 0.0 v-exact v-approx. 0.0 0 0.2 0.4x 0.6 0.8 1 500.00 Error increases 400.00 M_exact M_approx • Bending 300.00 moment 200.00 100.00 Bending Moment M(x)Bending 0.00 -100.00 0 0.2 0.4x 0.6 0.8 1 600.0 • Shear force 500.0 400.0 Shear Force V(x) 300.0 V_exact V_approx 200.0 0 0.2 0.4x 0.6 0.8 1 17 FINITE 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 elements – Elements are connected to other elements at nodes – Concentrated forces and couples can only be applied at nodes – Consider two-node bean element – Positive directions for forces and couples F F – Constant or linearly 1 2 C distributed load 1 C x 2 p(x) 18 FINITE ELEMENT INTERPOLATION cont. • Nodal DOF of beam element – Each node has deflection v and slope – Positive directions of DOFs T – Vector of nodal DOFs {}q {vv112 2 } • Scaling parameter s – Length L of the beam is scaled to 1 using scaling parameter s v1 v2 xx 1 1 1 sdsdx,, x 2 LL ds 1 dx Lds, L dx L x1 x2 s = 0 s = 1 • Will write deflection curve v(s) in terms of s 19 FINITE ELEMENT INTERPOLATION cont. • Deflection interpolation – Interpolate the deflection v(s) in terms of four nodal DOFs 23 – Use cubic function: vs() a01 as as 2 as 3 – Relation to the slope: dv dv ds 1 (2aasas 3)2 dx ds dx L 12 3 – Apply four conditions: dv(0) dv (1) vv(0) vv (1) 1122dx dx – Express four coefficients in terms of nodal DOFs vv10(0) a av dv 1 01 (0) a 11 aL dx L 11 avLvL 32 3 vv20123 (1) aaaa 21122 dv 1 avLvL 22 (1) (aaa 2 3 ) 3112 2 2123dx L 20 FINITE ELEMENT INTERPOLATION cont. • Deflection interpolation cont. 2 3 23 2 3 23 vs() (13 s 2 s ) v1122 Ls ( 2 s s ) (3 s 2 s ) v L ( s s ) IYv1 LL LL vs()[()()()()] Ns Ns Ns NsJZ1 vs() GWN {}q 1234v HX LL2 LL K[ 2 • Shape functions 1.0 23 N 0.8 1 N3 Ns1() 1 3 s 2 s 23 Ns2 () Ls ( 2 s s ) 0.6 23 Ns3 () 3 s 2 s 0.4 23 Ns() L ( s s ) 0.2 4 N2/L 0.0 – Hermite polynomials 0.0 0.2 0.4 0.6 0.8 1.0 N4/L – Interpolation property -0.2 21 FINITE ELEMENT INTERPOLATION cont.
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