## Slide #1.

Beam Elements Jake Blanchard Spring 2008
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## Slide #2.

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
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## Slide #3.

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
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## Slide #4.

Beam Elements in ANSYS BEAM BEAM BEAM BEAM BEAM beam BEAM beam BEAM BEAM BEAM 3 = 2-D elastic beam 4 = 3-D elastic beam 23 = 2-D plastic beam 24 = 3-D thin-walled beam 44 = 3-D elastic, tapered, unsymmetric 54 = 2-D elastic, tapered, unsymmetric 161 = Explicit 3-D beam 188 = Linear finite strain beam 189 = 3-D Quadratic finite strain beam
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## Slide #5.

Real Constants Area IZZ, IYY, IXX TKZ, TKY (thickness) Theta (orientation about X) ShearZ, ShearY (accounts for shear deflection – important for “stubby” beams)
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## Slide #6.

Shear Deflection Constants shearZ=actual area/effective area resisting shear Geometr y ShearZ 6/5 10/9 2 12/5
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## Slide #7.

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
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## Slide #8.

Accounting for Shear Effects M2 U b  dx 2 EI L M Px 2 0x  2 xy   xz2 U s  2G V L 2 dV  xz 0  xy V  2I P 2 L3  6 Eh 2   1   U U b  U s  2  96 EI  5GL   h 2  2    y    2   P 2 L3 U U b  U s  96 EI bh3 I 12  bh 5 E   1   2   10 IGL  Key parameter is height to length ratio
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## Slide #9.

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
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## Slide #10.

Determining Equivalent Loads Goal is to ensure equivalent loads produce same strain energy v ( x )  N 1 ( x )v1  N 2 ( x )1  N 3 ( x )v 2  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 W q N 1 ( x ) v1dx  q N 2 ( x )1 dx  q N 3 ( x )v2 dx  q N 4 ( x ) 2 dx 0 0 0 0 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
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## Slide #11.

Equivalent Loads (continued) W F  v1  v2   M 1   2  L 1 L  1 W  F  v1  v 2   M  1   2  qL v1  1  v 2  2  12 2 12  2 qL 2 qL2 M  12 F F M F M
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## Slide #12.

Putting Two Elements Together F F M F M F M F M F 2F M F M
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## Slide #13.

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 qD/4 qD/4 qL2 qD 2 qD/8 M   12 192 qD2/192 qD/4 qD/8 qD2/192
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## Slide #14.

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
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## Slide #15.

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
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## Slide #16.

Now Try a Frame F (out of plane)=1 N 3m  4 I   Ro  Ri4  4 I xx J 2 I vmax 2.59 10  5 m 2m Crosssections 6 cm 5 cm
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