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DRAFT - 1
CHAPTER 4
Beam Element
4.1 Introduction
Beam element is a very versatile line-element, it has six degrees of freedom at each node,
which include, translations and rotations along the x, y, and z directions, respectively.
Figure 4.1 shows the positive directions of these displacements.
y
v
jy
θ
jy
v
jx
i
x
θ
jz
j
θ
jx
z
v
jz
Figure 4.1 Beam Element with six degrees of freedom at each node
Beam element is employed to simulate a slender structure that has an 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. Thus, 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.
We will derive the element stiffness equation for a beam element by first deriving the
stiffness equation of a beam in bending, and then superimposing the stiffness of a truss
and a torsion bar element.
DRAFT –1
Chapter 4 Beam Element
4.1 Derivation of a Stiffness Equation for a Beam Element
Under Pure Bending
A beam, such as, a cantilever beam, under pure bending (without axial loads or torsional
loads), has two-degrees of freedom at any point, transverse deflection v and rotation
θ
, as
shown in Figure 4.2.
F
θ
v
Figure 4.2 Cantilever Beam with it’s DOF, v and
θ
A beam element has a total of four degrees of freedom, two at each node. Since there are
four degrees of freedom, the size of the stiffness matrix of a beam element has the size 4
x 4.
We will derive the stiffness matrix equation using a simple method, known as Stiffness
Influence Coefficient Method. In this procedure, a relationship between force and the
coefficients that influence stiffness is established. For a beam element, these coefficient
consist of: the modulus of elasticity, moment of inertia, and length of the element. For a
two-node beam element, there are two deflections and two rotations, namely, v
1
,
θ
1
, v
2
,
and
θ
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.
Consider a beam element, loaded in such a way that it has the deflection values: v
i
= 1,
θ
i
= 0, v
j
= 0,
θ
j
= 0
i
j
v
i
,
θ
i
v
j
,
θ
j
Figure 4.3 Beam Element
The above deflections can be produced by a combination of load conditions, shown in
figure 4.4.
ME 273 Lecture Notes © by R. B. Agarwal
4-2
DRAFT –1
Chapter 4 Beam Element
The following deflection relationships for loading of Figures 4.4 (a) and (b) can be found
in any Machine Design Handbook, and is given as,
v
max
v
max
= (FL
3
)/(3EI)
y
θ
θ
= - (FL
2
)/(2EI)
i
L
j
x
F
(a)
y
M
i
L M
j
v
max
= - (ML
2
)/(2EI)
i
j
v
max
x
θ
= (ML)/(EI)
(b)
Figure 4.4
Applying these relationships to the beam of Figure 4.3, 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
θ
= 0 = (
θ
)
F
+ (
θ
)
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)
From Figure 4. 4 (a) and (b),
ME 273 Lecture Notes © by R. B. Agarwal
4-3
DRAFT –1
Chapter 4 Beam Element
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
1
(6EI)/ L
2
0
0
0
0
= =
F
j
-(12EI)/ L
3
1
-(12EI)/ L
3
0 0 0
0
M
j
(6EI)/ L
2
1
(6EI)/ L
2
0 0 0
0
Using a similar procedure and setting the following deflection values:
v
i
= 0,
θ
i
= 1, v
j
= 0,
θ
j
= 0, we get,
F
i
(6EI)/L
2
1
0
(6EI)/
L
2
0 0 0
M
i
(4EI)/ L
1
0
(4EI)/ L
0 0 1
=
=
(4.6)
F
j
-(6EI)/ L
2
1
0 -(6EI)/
L
2
0 0 0
M
j
(2EI)/ L
1 0 (2EI)/ L
0 0 0
Similarly, setting v
j
= 1 and ,
θ
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
ME 273 Lecture Notes © by R. B. Agarwal
4-4
DRAFT –1
Chapter 4 Beam Element
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
θ
i
=
(4.7)
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
θ
j
Where F
i
, M
i
, F
j
, M
j
are the loads corresponding to the deflections v
i
,
θ
i
, v
j
,
θ
j
.
Equation (4.7) 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 uni-
axial 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 (4.7) in to 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, and therefore, we will defer the
derivations at this time. However, Equation (4.7) can be used for solving a beam
problem, loaded under bending loads. In order to understand the application of this
equation, we will apply it to solve some statically indeterminate problems.
ME 273 Lecture Notes © by R. B. Agarwal
4-5