Report

Beam Elements Jake Blanchard 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 2-D beam elements with 3 DOF/node – 2 translations and 1 rotation) ◦ More than 1 stress at each point on the element Shape functions Axial displacement is linear in x Transverse displacement is cubic in x Coarse mesh is often OK For example, transverse displacement in problem pictured below is a cubic function of x, so 1 element can give exact solution F Beam Elements in ANSYS BEAM 3 = 2-D elastic beam BEAM 4 = 3-D elastic beam BEAM 23 = 2-D plastic beam BEAM 24 = 3-D thin-walled beam BEAM 44 = 3-D elastic, tapered, unsymmetric beam BEAM 54 = 2-D elastic, tapered, unsymmetric beam BEAM 161 = Explicit 3-D beam BEAM 188 = Linear finite strain beam BEAM 189 = 3-D Quadratic finite strain beam Real Constants Area IZZ, IYY, IXX TKZ, TKY (thickness) Theta (orientation about X) ShearZ, ShearY (accounts for shear deflection – important for “stubby” beams) Shear Deflection Constants shearZ=actual area/effective area resisting shear Geometry ShearZ 6/5 10/9 2 12/5 Shear Stresses in Beams For 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. P L/2 L/2 Accounting for Shear Effects M2 Ub dx 2 EI L Px L 0 x 2 2 xy2 xz2 U s dV 2G V M xz 0 xy V 2I h 2 2 y 2 P 2 L3 bh5 E 1 U Ub U s 2 96EI 10IGL bh3 I 12 P 2 L3 6 Eh2 1 U Ub U s 2 96EI 5GL Key parameter is height to length ratio Distributed Loads We can only apply loads to nodes in FE analyses Hence, distributed loads must be converted to equivalent nodal loads With beams, this can be either force or moment loads q=force/unit length F M F M Determining Equivalent Loads Goal is to ensure equivalent loads produce same strain energy v( x) N1 ( x)v1 N 2 ( x)1 N 3 ( x)v2 N 4 ( x) 2 2 3 N1 ( x) 3 x 3 2 x 2 1 L L 1 2 N 2 ( x) 2 x 3 x 2 x L L 2 3 N 3 ( x) 3 x 3 2 x 2 L L 1 1 N 4 ( x) 2 x 3 x 2 L L L L W v ( x )qdx q v ( x ) dx 0 0 L W q N 1 ( x )v1 N 2 ( x )1 N 3 ( x )v2 N 4 ( x ) 2 dx 0 L L L L 0 0 0 0 W q N 1 ( x ) v1dx q N 2 ( x )1 dx q N 3 ( x )v2 dx q N 4 ( x ) 2 dx L L L L W q v1 N 1 ( x ) dx 1 N 2 ( x ) dx v2 N 3 ( x ) dx 2 N 4 ( x ) dx 0 0 0 0 L 1 L 1 W qL v1 1 v2 2 12 2 12 2 Equivalent Loads (continued) W F v1 v2 M 1 2 L 1 L 1 W F v1 v2 M 1 2 qL v1 1 v2 2 12 2 12 2 qL 2 qL2 M 12 F F M F M Putting Two Elements Together F F M F M F M F M F 2F M F M An Example Consider a beam of length D divided into 4 elements Distributed load is constant For each element, L=D/4 qL qD F 2 8 qL2 qD2 M 12 192 qD/4 qD/8 qD2/192 qD/4 qD/4 qD/8 qD2/192 In-Class Problems Consider a cantilever beam Cross-Section is 1 cm wide and 10 cm tall E=100 GPa Q=1000 N/m 1. D=3 m, model using surface load and 4 elements 2. D=3 m, directly apply nodal forces evenly distributed – use 4 elements 3. D=3 m, directly apply equivalent forces (loads and moments) – use 4 elements 4. D=20 cm (with and without ShearZ) 4 vmax qL 8EI Notes For adding distributed load, use “Pressure/On Beams” To view stresses, go to “List Results/Element Results/Line elements” ShearZ for rectangle is still 6/5 Be sure to fix all DOF at fixed end Now Try a Frame F (out of plane)=1 N 3m I R 4 4 o 2m R 4 i I xx J 2 I vmax 2.59 105 m Cross-sections 6 cm 5 cm