Spring 2008 Beam Elements These are Line Elements with 2 nodes 6 DOF per node 3 translations and 3 rotations Bending modes are included along with torsion tension and compression there also are 2D beam elements with 3 DOFnode 2 translations and 1 rotation ID: 1019025
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1. Beam ElementsJake BlanchardSpring 2008
2. Beam ElementsThese are “Line Elements,” with2 nodes6 DOF per node (3 translations and 3 rotations)Bending modes are included (along with torsion, tension, and compression)(there also are 2-D beam elements with 3 DOF/node – 2 translations and 1 rotation)More than 1 stress at each point on the element
3. Shape functionsAxial displacement is linear in xTransverse displacement is cubic in xCoarse mesh is often OKFor example, transverse displacement in problem pictured below is a cubic function of x, so 1 element can give exact solutionF
4. Beam Elements in ANSYSBEAM 3 = 2-D elastic beamBEAM 4 = 3-D elastic beamBEAM 23 = 2-D plastic beamBEAM 24 = 3-D thin-walled beamBEAM 44 = 3-D elastic, tapered, unsymmetric beamBEAM 54 = 2-D elastic, tapered, unsymmetric beamBEAM 161 = Explicit 3-D beamBEAM 188 = Linear finite strain beamBEAM 189 = 3-D Quadratic finite strain beam
5. Real ConstantsAreaIZZ, IYY, IXXTKZ, TKY (thickness)Theta (orientation about X)ShearZ, ShearY (accounts for shear deflection – important for “stubby” beams)
6. Shear Deflection ConstantsshearZ=actual area/effective area resisting shearGeometryShearZ6/510/9212/5
7. Shear Stresses in BeamsFor long, thin beams, we can generally ignore shear effects.To see this for a particular beam, consider a beam of length L which is pinned at both ends and loaded by a force P at the center.PL/2L/2
8. Accounting for Shear EffectsKey parameter is height to length ratio
9. Distributed LoadsWe can only apply loads to nodes in FE analysesHence, distributed loads must be converted to equivalent nodal loadsWith beams, this can be either force or moment loadsq=force/unit lengthMFFM
10. Determining Equivalent LoadsGoal is to ensure equivalent loads produce same strain energy
11. Equivalent Loads (continued)MFFM
12. Putting Two Elements TogetherMFFMMFFMMFFF2FM
13. An ExampleConsider a beam of length D divided into 4 elementsDistributed load is constantFor each element, L=D/4qD/8qD/4qD/4qD/8qD/4qD2/192qD2/192
14. In-Class ProblemsConsider a cantilever beamCross-Section is 1 cm wide and 10 cm tallE=100 GPaQ=1000 N/mD=3 m, model using surface load and 4 elementsD=3 m, directly apply nodal forces evenly distributed – use 4 elementsD=3 m, directly apply equivalent forces (loads and moments) – use 4 elementsD=20 cm (with and without ShearZ)
15. NotesFor adding distributed load, use “Pressure/On Beams”To view stresses, go to “List Results/Element Results/Line elements”ShearZ for rectangle is still 6/5Be sure to fix all DOF at fixed end
16. Now Try a FrameF (out of plane)=1 N3 m2 mCross-sections6 cm5 cm