# BandF - FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES FINITE...

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1 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES FINITE ELEMENT METHODS IN DESIGN Palaniappan Ramu 1 RAYLEIGH-RITZ METHOD 1. Assume a deflection shape Unknown coefficients c i and known function f i ( x ) 11 2 2 ( ) ( ) ( ). .... ( ) nn vx cf x cf x cf x  Deflection curve must satisfy displacement boundary conditions 2. Obtain potential energy as function of coefficients 3 Apply the principle of minimum potential energy to determine 12 ( , ,... ) n cc c U V  () 2 3. Apply the principle of minimum potential energy to determine the coefficients

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2 EXAMPLE – CANTILEVERED BEAM F C p 0 E,I,L 23 12 () vx a bx cx cx   (0) 0, (0) / 0 vd v d x  vx cx cx  2 0 26 2 L EI Uc c x d x  Assumed deflection: Satisfy BC: Strain Energy : 3  0 0 34 2 00 ,( ) ( ) ( ) L dv Wcc p vxd x F vL C L dx pL cF L C L c F L C L     Potential of loads: EXAMPLE – CANTILEVERED BEAM contd. . UW   0 c    3 22 0 46 2 EI Lc L c FL CL Derivative of potential: 1 2 0 c  4 3 2 0 3 61 2 3 4 EI L c L c FL CL  • Solving the above equations will yield C1 and C2 values • C1 and C2 can be substituted in the approximate expression for v 4 (transverse deflection) assumed in the beginning Consider numerical values: E=100 Gpa, I = 10 -7 m 4, p o =300N/m, F=500N, C=100N.m
3 EXAMPLE – CANTILEVERED BEAM cont . Deflection: Reason: Error increases -100.00 0.00 100.00 200.00 300.00 400.00 500.00 0 0.2 0.4 0.6 0.8 1 x Bending Moment M(x) M_exact M_approx BM: Actual – Quartic poly Assumed – Cubic poly 5 200.0 300.0 400.0 500.0 600.0 0 0.2 0.4 0.6 0.8 1 x Shear Force V(x) V_exact V_approx SF: FINITE ELEMENT INTERPOLATION • Rayleigh-Ritz method approximate solution in the entire beam • Finite element approximates solution in an element • Beam element Divide the beam using a set of elements F 1 F 2 -Divide the beam using a set of elements -Concentrated forces and couples can only be applied at nodes -Consider two-node beam element -Positive directions for forces and couples -Constant or linearly distributed load 6 C 2 C 1 p ( x ) x

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4 • Nodal DOF of beam element -Each node has deflection v and slope -Positive directions of DOFs -Vector of nodal DOFs: FINITE ELEMENT INTERPOLATION cont . 112 2 {} { } T vv q v 1 v 2 2 1 L x 1 s = 0 x 2 s = 1 x 7 1 1 ,, 1 , x x s ds dx LL ds dx Lds dx L  • Scaling parameter s Length L of the beam is scaled to 1 using scaling parameter s • Deflection interpolation - Interpolate the deflection v ( s ) in terms of four nodal DOFs - Use cubic function: - Relation to the slope: FINITE ELEMENT INTERPOLATION cont . 23 01 2 3 () vs a as as as  2 12 3 1 (2 3) dv dv ds aa s a s dx ds dx L 11 22 (0) (1) (0) dv dv dx dx 10 (0) a • Apply four conditions: • Express coefficients in terms of nodal DOF 8 20 1 2 3 21 2 3 1 (0) 1 ( 2 3 ) dv a dx L vv aaaa dv aaa dx L 211 2 2 3112 2 32 3 av aL L v L L v L   
5 FINITE ELEMENT INTERPOLATION cont .

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BandF - FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES FINITE...

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