# Eml 4500 Finite Element Analysis and Design

RAYLEIGH-RITZ METHOD 1. Assume a deflection shape v(x) c1f1(x) c 2 f2 (x)..... c n fn (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 (c1,c 2 ,...c n ) U V 3. Apply the principle of minimum potential energy to determine the coefficients 0 c1 c 2 c n 1 Application to beam problems The differential equation for the beam displacement is d4 v EI 4 p(x) dx For a constant p(x)=p0, the general solution of this differential 4 p x equation is v( x) 0 a3 x 3 a2 x 2 a1 x a0 24 EI When p(x)=0, the exact solution is a cubic polynomial. Concentrated forces and couples cause discontinuities, in third and second derivatives, respectively. 2 EXAMPLE SIMPLY SUPPORTED BEAM p0 Assumed deflection curve

v(x) C sin x L E,I,L Strain energy 2 2 d v 1 C2EI4 U EI 2 dx 2 0 dx 4L3 L Potential energy of applied loads (no reaction forces) L L 2p L x dx 0 C L 0 0 Potential energy EI4 2 2p0L U V 3 C C 4L V p(x)v(x)dx p0Csin PMPE: d EI4 2p0L 3 C 0 dC 2L

4p0L4 C EI5 3 EXAMPLE SIMPLY SUPPORTED BEAM cont. Approximate bending moment and shear force 4p0L4 p0L4 x x v(x) sin 0.01307 sin EI5 L EI L 4p0L2 d2 v 2 x x x 2 M(x) EI 2 EIC 2 sin sin 0.129 0 p L sin 0 dx L L 3 L L 4p0L

d3 v 3 x x x Vy (x) EI 3 EIC 3 cos cos 0.4052 p L sin 0 dx L L 2 L L Exact solutions p0L 3 p0 4 1 p0L3 v(x) x x x EI 24 12 24 p0L3 v(L / 2) 0.01302 EI p0L p x 0 x2 M(L / 2) 0.125p0L2 2 2 pL Vy (x) 0 p0 x Vy (0) 0.5p0L

2 M(x) 4 EXAMPLE SIMPLY SUPPORTED BEAM cont. Deflection 0.8 v(x)/v_max 1.0 0.6 0.4 v-exact 0.2 v-approx. 0.0 0 0.2 0.4 Bending Moment M(x) 0.8 1 0 0.2 0.4 0.6 0.8

1 -0.04 -0.06 -0.08 -0.10 M_exact -0.12 M_approx -0.14 0.6 V_exact 0.4 Shear Force V(x) Shear force Error increases -0.02 0.6 x 0.00 Bending moment x V_approx 0.2 0.0 -0.2 -0.4 -0.6 0

0.2 0.4 x 0.6 0.8 1 5 Quiz-like questions Calculate the Rayleigh-Ritz approximation to the maximum displacement, bending moment and shear force for a simply supported beam under uniform load, when the displacement is assumed of the form v=Cx(L-x) Why is the solution with the sine so much better? The coefficient of x4 is shown in Slide 2 to be 1/(24EI). What boundary conditions does one use to calculate the coefficients of the other terms (the cubic polynomial in Slide 2)? Answers in notes page 6 EXAMPLE CANTILEVERED BEAM p0 Assumed deflection C v(x) a bx c 1x 2 c 2 x 3 E,I,L F Need to satisfy BC v(x) c1x 2 c 2 x 3 v(0) 0, dv(0) / dx 0 L Strain energy U EI 2c 6c x 2 dx

1 2 2 0 Potential of loads L V c1,c 2 p0 v(x)dx Fv(L) C 0 dv (L) dx 4 p0L3 p L 2 3 2 0 c1 FL 2CL c 2 FL 3CL 3 4 7 EXAMPLE CANTILEVERED BEAM cont. Derivatives of U: L U 2EI 2c1 6c 2 x dx EI 4Lc1 6L2c 2 c1 0

L U 6EI 2c1 6c 2 x xdx EI 6L2c1 12L3c 2 c 2 0 PMPE: 0 c1 0 c 2 p0L3 EI 4Lc1 6L c 2 FL2 2CL 3 4 p L EI 6L2c1 12L3c 2 0 FL3 3CL2 4 2 Data: E 100GPa, I 10 7 m4 , L 1m, p0 300 N / m, F 500 N , C 100 Nm Solve for c1 and c2: c 23.75 10 3 , c 5.833 10 3 1 2 Deflection curve: v(x) 10 3 23.75x 2 5.833x3 v(L) 17.92 10 3

Exact solution: 1 v(x) 5400x 2 800x 3 300 x 4 24EI Check exact solution v(L) 17.9110 3 8 EXAMPLE CANTILEVERED BEAM cont. Deflection Error increases Bending moment Shear force 9 Quiz like questions Consider the cantilever beam in the figure Will a cubic polynomial Ritz solution be exact? We use the Rayleigh-Ritz method with v=Cx2. Compare the exact and approximate shear force. Compare the exact and approximate tip deflection. Answers in the notes page 10

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