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
2
5
3
2
2
2
2
2
2
10
1
96
2
2
0
2
2
0
2
2
IGL
E
bh
EI
L
P
U
U
U
y
h
I
V
dV
G
U
L
x
Px
M
dx
EI
M
U
s
b
xy
xz
V
xz
xy
s
L
b
2
2
3
2
3
5
6
1
96
12
GL
Eh
EI
L
P
U
U
U
bh
I
s
b
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
M
F
F
M
Determining Equivalent Loads
Goal is to ensure equivalent loads produce same
strain energy
2
4
2
3
1
2
1
1
)
(
)
(
)
(
)
(
)
(
x
N
v
x
N
x
N
v
x
N
x
v
2
3
2
4
2
2
3
3
3
2
3
2
2
2
2
3
3
1
1
1
)
(
3
2
)
(
2
1
)
(
1
3
2
)
(
x
L
x
L
x
N
x
L
x
L
x
N
x
x
L
x
L
x
N
x
L
x
L
x
N
2
2
1
1
0
4
2
0
3
2
0
2
1
0
1
1
0
2
4
0
2
3
0
1
2
0
1
1
0
2
4
2
3
1
2
1
1
0
0
12
2
1
12
2
1
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
L
v
L
v
qL
W
dx
x
N
dx
x
N
v
dx
x
N
dx
x
N
v
q
W
dx
x
N
q
dx
v
x
N
q
dx
x
N
q
dx
v
x
N
q
W
dx
x
N
v
x
N
x
N
v
x
N
q
W
dx
x
v
q
qdx
x
v
W
L
L
L
L
L
L
L
L
L
L
L
Equivalent Loads (continued)
2
1
2
1
M
v
v
F
W
2
2
1
1
2
1
2
1
12
2
1
12
2
1
L
v
L
v
qL
M
v
v
F
W
12
2
2
qL
M
qL
F
M
F
F
M
Putting Two Elements Together
M
F
F
M
M
F
F
M
M
F
F
F
2F
M
An Example
Consider a beam of length D divided into 4
elements
Distributed load is constant
For each element, L=D/4
192
12
8
2
2
2
qD
qL
M
qD
qL
F
qD/8
qD/4
qD/4
qD/8
qD/4
qD
2
/192
qD
2
/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
)
EI
qL
v
8
4
max
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
3 m
2 m
Cross

sections
6 cm
5 cm
m
v
I
J
I
R
R
I
xx
i
o
5
max
4
4
10
59
.
2
2
4
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