professor s.liu的经典有限元实用教程fem for beams

1、1The Finite Element MethodFEM FOR BEAMS A Practical Course G. R. Liu and S. S. QuekCHAPTER 5: 2CONTENTSINTRODUCTIONFEM EQUATIONSShape functions constructionStrain matrixElement matrices RemarksEXAMPLE AND CASE STUDYRemarks3INTRODUCTIONThe element developed is often known as a beam element. A beam el

2、ement is a straight bar of an arbitrary cross-section.Beams are subjected to transverse forces and moments.Deform only in the directions perpendicular to its axis of the beam.4INTRODUCTIONIn beam structures, the beams are joined together by welding (not by pins or hinges).Uniform cross-section is as

3、sumed.FE matrices for beams with varying cross-sectional area can also be developed without difficulty.5FEM EQUATIONSShape functions constructionStrain matrixElement matrices6Shape functions constructionConsider a beam elementNatural coordinate system: 7Shape functions constructionAssume thatIn matr

4、ix form:or8Shape functions constructionTo obtain constant coefficients four conditionsAt x= -a or x = -1At x= a or x = 19Shape functions constructionoror10Shape functions constructionTherefore,wherein which11Strain matrixTherefore,where(Second derivative of shape functions)Eq. (2-47)12Element matric

5、esEvaluate integrals13Element matricesEvaluate integrals14Element matrices15RemarksTheoretically, coordinate transformation can also be used to transform the beam element matrices from the local coordinate system to the global coordinate system. The transformation is necessary only if there is more

6、than one beam element in the beam structure, and of which there are at least two beam elements of different orientations. A beam structure with at least two beam elements of different orientations is termed a frame or framework.16EXAMPLEConsider the cantilever beam as shown in the figure. The beam i

7、s fixed at one end and it has a uniform cross-sectional area as shown. The beam undergoes static deflection by a downward load of P=1000N applied at the free end. The dimensions and properties of the beam are shown in the figure. P=1000 N 0.5 m 0.06 m 0.1 m E=69 GPa=0.3317EXAMPLEStep 1: Element matr

8、icesExact solution: P=1000 N 0.5 m E=69 GPa=0.33= -3.355E10-4 mEq. (2.59)18EXAMPLEStep 1 (Contd):Step 2: Boundary conditions P=1000 N 0.5 m E=69 GPa=0.3319EXAMPLEStep 2 (Contd):Therefore, K d = F, wheredT = v2 2 , Step 3: Solving FE equation(Two simultaneous equations)v2 = -3.355 x 10-4 m2 = -1.007

9、x 10-3 rad 20EXAMPLEStep 4: Stress recoveringv2 = -3.355 x 10-4 m2 = -1.007 x 10-3 rad Substitute back into first two equations21RemarksFE solution is the same as analytical solutionAnalytical solution to beam is third order polynomial (same as shape functions used)Reproduction property22CASE STUDYResonant frequencies of micro resonant transducer 23CASE STUDYNumber of 2-node beam elementsNatural Frequency (Hz)Mode 1Mode 2Mode 3104.4058 x 1051.2148 x 1062.3832 x 106204.4057 x 1051.2145 x 1062.3809 x 106404.4056 x 1051.2144 x 1062.3808 x 106604.4056 x