Beam Elements

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 105 m
Cross-sections
6 cm
5 cm

similar documents