CHAPTER 4 – BEAM ELEMENT
Introduction
Beam element has six degrees of freedom at each node
y
v
jy
jy
v
jx
i x
jz
j
jx
z v
jz
Beam element is a slender structure
Has uniform cross section.
The element is unsuitable for structures that have complex
geometry, holes, and points of stress concentration.
The stiffness constant of a beam element is derived by
combining the stiffness constants of a beam under pure
bending, a truss element, and a torsion bar.
A beam element can represent a beam in bending, a truss
element, and a torsion bar.
In FEA it’s a common practice to use beam elements to
represent all or any of these three loads.
ME 273 Lecture Notes © by R. B. Agarwal
2
Derivation of Stiffness Equation for a Beam
Element Under Pure Bending in 2D
A beam, under pure bending (without axial loads or torsion
loads), has twodegrees of freedom at any point.
F
v
A beam element in pure bending has a total of four degrees
of freedom, two at each node.
The size of the stiffness matrix of a beam element has the size
4 x 4.
Stiffness matrix equation is derived using the Stiffness
Influence Coefficient Method.
For a twonode beam element, there are two deflections and
two rotations, namely, v
1
,
1
, v
2
, and
2
.
Force and influence coefficient relationship is established by
setting each of the four deflection values to unity, with the
remaining deflection values equal to zero. The procedure
follows.
ME 273 Lecture Notes © by R. B. Agarwal
3
Consider a beam element, loaded in such a way that it has the
deflection values: v
i
= 1,
i
= 0, v
j
= 0,
j
= 0
i j
v
i
,
i
v
j
,
j
The above deflections can be produced by a combination of load
conditions, shown in figure 4.4.
The deflection relationships for loading can be found in any
Machine Design Handbook, and is given as,
v
max
v
max
= (FL
3
)/(3EI)
y
=  (FL
2
)/(2EI)
i L j x
F
(a)
ME 273 Lecture Notes © by R. B. Agarwal
4
y
M
i
x
i L M
j
v
max
=  (ML
2
)/(2EI)
v
max
j
= (ML)/(EI)
(b)
Applying these relationships to the beam, we get,
1 = v
i
= (v
i
)
F
+ (v
i
)
M
1 = v
i
= (F
i
L
3
)/3 EI  (M
i
L
2
)/2EI
(4.1)
and = 0 = ()
F
+ ()
M
0 =  (F
i
L
2
)/2EI + (M
i
L)/EI
(4.2)
Solving Equations (4.1) and (4.2), we get,
F
i
= (12EI)/L
3
(A)
F
j
=  F
i
= (12EI)/L
3
(B)
M
i
= (6EI)/L
2
(C)
ME 273 Lecture Notes © by R. B. Agarwal
5
From the above Figures,
M
j
= F
i
L  M
i
= (12EI)/L
2
 (6EI)/ L
2
= (6EI)/ L
2
(D)
Writing equations (A) through (D) in a matrix form we get,
F
i
(12EI)/L
3
1 (12EI)/ L
3
0 0 0 1
M
i
(6EI)/ L
2
0
(6EI)/ L
2
0 0 0 0
= =
F
j
(12EI)/ L
3
0
(12EI)/ L
3
0 0 0 0
M
j
(6EI)/ L
2
0
(6EI)/ L
2
0 0 0 0
Using a similar procedure and setting the following deflection
values:
v
i
= 0,
i
= 1, v
j
= 0,
j
= 0, we get,
F
i
(6EI)/L
2
0 0 (6EI)/ L
2
0 0 0
M
i
(4EI)/ L 1
0
(4EI)/ L
0 0 1
= =
F
j
(6EI)/ L
2
0
0 (6EI)/ L
2
0 0 0
M
j
(2EI)/ L
0 0 (2EI)/ L
0 0 0
ME 273 Lecture Notes © by R. B. Agarwal
6
Similarly, setting v
j
= 1 and ,
j
= 1, respectively, and keeping all
other deflection values to zero, we get the final matrix as,
F
i
(12EI)/L
3
(6EI)/ L
2
(12EI)/ L
3
(6EI)/ L
2
1
M
i
(6EI)/ L2 (4EI)/ L
(6EI)/ L
2
(2EI)/ L 1
= (4.7)
F
j
(12EI)/ L
3
(6EI)/ L
2
(12EI)/L
3
(6EI)/ L
2
1
M
j
(6EI)/ L
2
(2EI)/ L (6EI)/ L
2
(4EI)/ L
1
Note that, the first term on the RHS of the above equation is the
stiffness matrix and the second term is the deflection. In the case
where deflections are other than unity, the above equation will
provide an element equation for a beam (in bending), which can be
written as,
F
i
(12EI)/L
3
(6EI)/ L
2
(12EI)/ L
3
(6EI)/ L
2
v
i
M
i
(6EI)/ L2 (4EI)/ L
(6EI)/ L
2
(2EI)/ L
i
=
F
j
(12EI)/ L
3
(6EI)/ L
2
(12EI)/L
3
(6EI)/ L
2
v
j
M
j
(6EI)/ L
2
(2EI)/ L (6EI)/ L
2
(4EI)/ L
j
Where F
i
, M
i
, F
j
, M
j
are the loads corresponding to the deflections
v
i
,
i
, v
j
,
j
.
ME 273 Lecture Notes © by R. B. Agarwal
7
The above equation can be written in a more solution friendly form
as,
F
i
12 6L
12 6L v
i
M
i
6L 4L
2
6L 2L
2
i
F
j
= EI/L
3
12 6L 12 6L v
j
M
j
6L 2 L
2
6L 4L
2
j
The above equation is the equation of a beam element, which
is under pure bending load (no axial or torsion loads).
The stiffness matrix is a 4 x 4, symmetric matrix.
Using this equation, we can solve problems in which several
beam elements are connected in an uniaxial direction.
The assembly procedure is identical to the truss elements.
However, if the beam elements are oriented in more than one
direction, we will have to first transform the above equation
into a global stiffness matrix equation (analogues to the
procedure used for truss elements).
For a beam element, transformation of a local stiffness matrix
into a global equation involves very complex trigonometric
relations.
ME 273 Lecture Notes © by R. B. Agarwal
8
Example 1
For the beam shown, determine the displacements and slopes at the
nodes, forces in each element, and reactions at the supports.
3 m 3 m 50 kN
E = 210 GPa, I = 2x10
4
m
4
K = 200 kN/m
Solution
The beam structure is descritized into three elements and 4nodes,
as shown.
[1] [2]
3
1 2
[3]
4
First, we will find the element stiffness matrix for each element,
next we will assemble the stiffness matrices, apply the boundary
conditions, and finally, solve for node deflection. Internal forces
and reactions are calculated by backsubstituting the deflections in
the structural equation.
ME 273 Lecture Notes © by R. B. Agarwal
9
Element 1
1 2
EI/L
3
= (210 x 10
9
) x (2x10
4
)/(3)
3
= 15.55 x 10
5
The general equation of a stiffness matrix is given by equation 4.7.
Writing this equation by placing the common factor EI/L
3
outside
the matrix, we get
Element 1
12 6L 12 6L v
1
6L 4L
2
6L 2 L
2
1
[K
e
]
(1)
= (EI/L
3
)
12 6L 12 6L v
2
6L 2 L
2
6L 4 L
2
2
[2]
Element 2 2 3
12 6L 12 6L v
2
6L 4L
2
6L 2 L
2
2
[K
e
]
(2)
= (EI/L
3
)
12 6L 12 6L v
3
6L 2 L
2
6L 4 L
2
3
ME 273 Lecture Notes © by R. B. Agarwal
10
3
Element 3
[3]
4
[K
e
]
(3)
= K K v
3
K K v
4
To get the global stiffness matrix, we will use the same procedure
used for assembling truss element stiffness equations. In terms of
E, L, and I the assembled global stiffness matrix is,
v
1
1
v
2
2
v
3
3
v
4
v
1
12 6L 12 6L 0 0 0
1
4L
2
6L 2 L
2
0 0 0
v
2
24 0 12 6L 0
x (EI) /(L
3
)
2
8L
2
6L 2L
2
0
v
3
12 +K’
6L  K’
3
4L
2
0
v
4
SYMMETRY
K’
Where K’ = (K) x [L
3
/ (EI)] = 200 x 10
3
/(15.55 x10
5
) = .1286
ME 273 Lecture Notes © by R. B. Agarwal
11
Our next step is to write the structural equation; however, we can
reduce the size of the stiffness matrix by applying the given
boundary conditions:
v
1
=
1
= 0 node 1 is fixed
v
2
= 0
node 2 has no vertical deflection, but it’s free to
rotate.
V
4
= 0 node 4 is fixed.
The reduced stiffness matrix is
8L
2
6L 2L
2
K
G
= EI / (L
3
) 6L 12+K’ 6L
2L
2
6L 4L
2
Substituting the values of E, L, and I the structural equation can be
written as,
0 72 18 18
2
100 = 15.55 x 10
5
18 12.1 18 v
3
0 18 18 36
3
2
=  0.0025 rad
Solving, we get v
3
=  0.0177 m
3
= 0.0076 rad
ME 273 Lecture Notes © by R. B. Agarwal
12
Derivation of a Plane (2D) Beam Element an an
Arbitrary angle
We will derive the 2D beam element equation that has an arbitrary
orientation with axial and bending loads.
F
M
θ
The stiffness equation will be derived in three steps
1. Pure Beam element arbitrarily oriented in space
2. Local Beam stiffness with axial and bending loads
3. Arbitrarily oriented beam with axial and bending loads.
ME 273 Lecture Notes © by R. B. Agarwal
13
Arbitrarily Oriented 2D Beam Element
The stiffness equation for an arbitrarily oriented beam element can
be derived with a procedure similar to the truss element.
d
2y
x
y
2
y
d
1y
d
1y
1
d
1x
x
d
1y
= d
1y
cosθ  d
1x
sinθ = d
1y
C  d
1x
S =  d
1x
S + d
1y
C
d
2y
= d
2y
cosθ – d
2x
sinθ = d
2y
C – d
2x
S = – d
2x
S + d
2y
C
and
1
=
1
,
2
=
2
Note: The underscored terms represent local coordinate values.
Thus, x and y are local coordinates and x and y are global
coordinates.
The above equations can be written in a compatible matrix form,
by introducing 0s where necessary,
d
1x
d
1y
s c 0 0 0 0 d
1y
1
= 0 0 1 0 0 0
1
d
2y
0 0 0 s c 0 d
2x
2
0 0 0 0 0 1 d
2y
2
ME 273 Lecture Notes © by R. B. Agarwal
14
s c 0 0 0 0
Let T
= 0 0 1 0 0 0 (A)
0 0 0 s c 0
0 0 0 0 0 1 , the transformation matrix.
Thus, {d} = [T] {d}
Global
Local
Note that angle
is independent of the coordinate systems, and
1
=
1
,
2
=
2
As derived in the case of the truss element, relationship between
local and global stiffness matrices is given as
[k
g
] = [T]
T
[k] [T]
Where, [k
g
] = Global stiffness matrix of an element
[T] = Transformation matrix
[k] = Local stiffness matrix of the element
Substituting the values of [T] and [k], we get the global equation of
a beam element oriented arbitrarily at an angle
猬
†ㄲ
2
ㄲ千 ⴶ䱓ⴱ㉓
2
ㄲ千 ⴶ䱓
† ‱2C
2
‶䱃 ㄲ千1ⴱ㉃
2
‶䱃
欠㴠䕉⽌
3
‴
2
‶䱓 ⴶ䱃㉌
2
‱㉓
2
ㄲ千 㙌匠
⁓祭 整特e ㄲ1
2
㙌
† † L
2
ME 273 Lecture Notes © by R. B. Agarwal
15
This is the equation of a beam element (without axial or torsional
load, and oriented at an angle
⸠
䅬獯Ⱐ匠㴠獩A
Ⱐ䌠㴠捯,
渠瑨攠慢潶攠敱畡瑩潮⸠
ME 273 Lecture Notes © by R. B. Agarwal
16
Stiffness Equation of a Beam Element with
Combined Bending and Axial loads
First, we will derive the stiffness matrix in local coordinates and
then convert it in to global coordinates.
The stiffness equation for the combined bending and axial load can
be written by superimposing the axial stiffness terms over the
bending stiffness.
For axial loading, the structural equation is,
f
1x
1 1 d
1x
f
1x
f
2x
= AE/L
f
2x
1 1 d
2x
Truss Element
And for bending, the structural equation is,
f
1y
12 6L 12 6L d
1y
m
1
6L 4L
2
6L 2L
2
φ
1
= AE/L
3
f
2y
12 6L 12 6L d
2y
m
2
6L 2L
2
6L 4L
2
φ
2
f
1y
f
2y
m
1
m
2
Beam Element
ME 273 Lecture Notes © by R. B. Agarwal
17
Therefore, the combined loading equation is
d
1x
d
1y
φ
1
d
2x
d
2y
φ
2
f
1x
C
1
0 0  C
1
0 0 d
1x
f
1y
0 12 C
2
6C
2
L 0 12 C
2
6C
2
L d
1y
m
1
0 6 C
2
L 4C
2
L
2
0 6C
2
L 2C
2
L
2
φ
1
=
f
2x
C
1
0 0 C
1
0 0 d
2x
f
2y
0 12 C
2
6C
2
L 0 12 C
2
6C
2
L d
2y
m
2
0 6 C
2
L 2C
2
L
2
0 6C
2
L 4C
2
L
2
φ
2
Where,
C
1
= AE/L
C
2
= EI/L
3
The first matrix on the RHS in the above equation gives the local
Stiffness matrix k as
d
1x
d
1y
φ
1
d
2x
d
2y
φ
2
C
1
0 0  C
1
0 0
0 12 C
2
6C
2
L 0 12 C
2
6C
2
L
0 6 C
2
L 4C
2
L
2
0 6C
2
L 2C
2
L
2
k =
C
1
0 0 C
1
0 0
0 12 C
2
6C
2
L 0 12 C
2
6C
2
L
0 6 C
2
L 2C
2
L
2
0 6C
2
L 4C
2
L
2
ME 273 Lecture Notes © by R. B. Agarwal
18
The local stiffness matrix k has 3DOF at each node, representing
bending and axial loads.
For axial loading:
u
2y
u
2
u
1y
2
u
2x
1
u
1x
u
1
u
1
= u
1x
cos + u
1y
sin = c u
1x
+ s u
1y
u
2
= u
2x
cos + u
2y
sin = c u
2x
+ s u
2y
where, cos = c, and sin = s
Which can be written as,
u
1x
u
1
c s 0 0 u
1y
= u
2x
(3.2)
u
2
0 0 c s u
2y
Where
c s 0 0
T =
(3.2)
0 0 c s
is the transformation Matrix
Since the transformation matrix T is not a square matrix, it can’t be
inverted, which will be incompatible with the matrix operations.
ME 273 Lecture Notes © by R. B. Agarwal
19
Let us convert the matrix T into a compatible form by placing 0s as
necessary for compatibility.
u
1x
u
1y
u
2x
u
2y
u
1
c s 0 0 u
1x
v
1
= s c 0 0 u
1y
u
2
0 0 c s u
2x
v
2
0 0 s c u
2y
Where
,
u
1x
u
1y
u
2x
u
2y
c s 0 0
T
= s c 0 0
0 0 c s
0 0 s c
For bending loading:
The stiffness equation for an arbitrarily oriented beam element can
be derived with a procedure similar to the truss element.
u
2y
y x
y
2
u
1y
u
1y
1
u
1x
x
u
1y
= u
1y
cosθ  u
1x
sinθ = u
1y
C  u
1x
S =  u
1x
S + u
1y
C
u
2y
= u
2y
cosθ – u
2x
sinθ = u
2y
C – u
2x
S = – u
2x
S + u
2y
C
ME 273 Lecture Notes © by R. B. Agarwal
20
and
1
=
1
,
2
=
2
Note: The underscored terms represent local coordinate values.
Thus, x and y are local coordinates and x and y are global
coordinates.
The above equations can be written in a compatible matrix form,
by introducing 0s where necessary,
u
1x
u
1y
s c 0 0 0 0 u
1y
1
= 0 0 1 0 0 0
1
u
2y
0 0 0 s c 0 u
2x
2
0 0 0 0 0 1 u
2y
2
s c 0 0 0 0
Let T
= 0 0 1 0 0 0
0 0 0 s c 0
0 0 0 0 0 1 ,
the transformation matrix.
Combining the axial and bending transformations, we get,
C S 0 0 0 0
S C 0 0 0 0
T = 0 0 1 0 0 0
0 0 0 C S 0
0 0 0 S C 0
0 0 0 0 0 1
ME 273 Lecture Notes © by R. B. Agarwal
21
Using the relation, [k] = [T]
T
[k][T] and simplifying, we get,
I
C
L
I
C
L
I
ASsymmetry
S
L
I
CS
L
I
AS
L
I
AC
IC
L
I
S
L
I
I
C
L
I
C
L
I
ASCS
L
I
AC
L
I
C
L
I
AS
S
L
I
CS
L
I
AS
L
I
ACS
L
I
CS
L
I
AS
L
I
AC
L
E
K
4
612
61212
2
66
4
61212612
6121261212
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
This is the stiffness equation for a 2D beam element at an
arbitrary orientation and under axial and bending loads.
ME 273 Lecture Notes © by R. B. Agarwal
22
2D Beam Element with combined loading
Bending, Axial, and Torsion (θ = 0)
A similar procedure can be used to find the equation of a beam
under general loading of axial, bending and torsion loads.
The torsional loads are m
1x
and m
2x
, and the corresponding
deflections are,
x1
湤
x2
周攠瑯牳楯湡氠獴牵捴畲慬煵慴楯渠楳㨠
ㅸ
†ㄠ ⴱ φ
1x
= JG/L
m
1x
1 1 φ
2x
These terms can be superimposed on the stiffness equation derived
previously for the combined bending and axial loads.
d
y
φ
y
3D Beam Element: d
x
φ
z
φ
x
d
z
A 3D beam element has 6 DOF at each node, and 12 DOF for
each element. The stiffness matrix can be derived by super
imposing the axial, bending, and torsion loadings in the XY, XZ,
and YZ planes. The equation is,
ME 273 Lecture Notes © by R. B. Agarwal
23
The stiffness equation in local coordinate is:
^
2
^
2
^
2
^
2
^
2
^
2
^
1
^
1
^
1
^
1
^
1
^
1 zyxzyxzyxzyx
dddddd
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
GJ
L
GJ
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
GJ
L
GJ
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
K
zzzz
yyyy
yyyy
zzzz
zzzz
yyyy
yyyy
zzzz
4
000
6
0
2
000
6
0
0
4
0
6
000
2
0
6
00
0000000000
0
6
0
12
000
6
0
12
00
6
000
12
0
6
000
12
0
0000000000
2
000
6
0
4
000
6
0
0
2
0
6
000
4
0
6
00
0000000000
0
6
0
12
000
6
0
12
00
6
000
12
0
6
000
12
0
0000000000
22
22
2323
2323
22
22
2323
2323
Similarly, we can find the transformation matrix for Axial,
bending, and torsional loading and then use the equation,
]][[][
^
TKTK
T
The equation gets very complicated, so we will not go in to the
derivation any further.
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