7/29/2019 BEAM ELEMENT USING FEM
1/23
CHAPTER 4 BEAM ELEMENT
Introduction Beam element has six degrees of freedom at each node
y
vjy
jy vjxi x
jz j jx
z vjz
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 bycombining 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 trusselement, and a torsion bar.
In FEA its a common practice to use beam elements torepresent all or any of these three loads.
7/29/2019 BEAM ELEMENT USING FEM
2/23
ME 273 Lecture Notes by R. B. Agarwal 2
Derivation of Stiffness Equation for a Beam
Element Under Pure Bending in 2-D
A beam, under pure bending (without axial loads or torsionloads), has two-degrees of freedom at any point.
F
v
A beam element in pure bending has a total of four degreesof freedom, two at each node.
The size of the stiffness matrix of a beam element has the size4 x 4.
Stiffness matrix equation is derived using the StiffnessInfluence Coefficient Method.
For a two-node beam element, there are two deflections andtwo rotations, namely, v1, 1, v2, and 2.
Force and influence coefficient relationship is established bysetting each of the four deflection values to unity, with the
remaining deflection values equal to zero. The procedure
follows.
7/29/2019 BEAM ELEMENT USING FEM
3/23
ME 273 Lecture Notes by R. B. Agarwal 3
Consider a beam element, loaded in such a way that it has the
deflection values: vi = 1, i = 0, vj = 0, j = 0
i j
vi, i vj, 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,
vmax
vmax = (FL3)/(3EI)
y
= - (FL2)/(2EI)i L j x
F
(a)
7/29/2019 BEAM ELEMENT USING FEM
4/23
ME 273 Lecture Notes by R. B. Agarwal 4
y
Mi x
i L Mj vmax = - (ML2)/(2EI)
vmax j = (ML)/(EI)
(b)
Applying these relationships to the beam, we get,
1 = vi = (vi)F + (vi)M
1 = vi = (Fi L3)/3 EI - (Mi L
2)/2EI
(4.1)
and = 0 = ()F + ()M
0 = - (Fi L
2
)/2EI + (Mi L)/EI(4.2)
Solving Equations (4.1) and (4.2), we get,
Fi = (12EI)/L3
(A)
Fj = - Fi = -(12EI)/L3
(B)
Mi = (6EI)/L2 (C)
7/29/2019 BEAM ELEMENT USING FEM
5/23
ME 273 Lecture Notes by R. B. Agarwal 5
From the above Figures,
Mj = Fi L - Mi= (12EI)/L
2- (6EI)/ L
2
= (6EI)/ L2 (D)
Writing equations (A) through (D) in a matrix form we get,
Fi (12EI)/L3
1 (12EI)/ L3
0 0 0 1
Mi (6EI)/ L
2
0
(6EI)/ L
2
0 0 0 0= =
Fj -(12EI)/ L3
0
-(12EI)/ L3
0 0 0 0
Mj (6EI)/ L2
0
(6EI)/ L2
0 0 0 0
Using a similar procedure and setting the following deflection
values:
vi = 0, i = 1, vj = 0, j = 0, we get,
Fi (6EI)/L2
0 0 (6EI)/ L2
0 0 0
Mi (4EI)/ L 1 0
(4EI)/ L
0 0 1
= =Fj -(6EI)/ L
20 0 -(6EI)/ L
20 0 0
Mj (2EI)/ L
0 0 (2EI)/ L
0 0 0
7/29/2019 BEAM ELEMENT USING FEM
6/23
ME 273 Lecture Notes by R. B. Agarwal 6
Similarly, setting vj = 1 and , j = 1, respectively, and keeping allother deflection values to zero, we get the final matrix as,
Fi (12EI)/L3
(6EI)/ L2
-(12EI)/ L3
(6EI)/ L2
1
Mi (6EI)/ L2 (4EI)/ L
-(6EI)/ L2
(2EI)/ L 1
= (4.7)
Fj -(12EI)/ L3
-(6EI)/ L2
(12EI)/L3
-(6EI)/ L2
1
Mj (6EI)/ L2
(2EI)/ L -(6EI)/ L2
(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,
Fi (12EI)/L3
(6EI)/ L2
-(12EI)/ L3
(6EI)/ L2
vi
Mi (6EI)/ L2 (4EI)/ L
-(6EI)/ L2
(2EI)/ L i=
Fj -(12EI)/ L3
-(6EI)/ L2
(12EI)/L3
-(6EI)/ L2
vj
Mj (6EI)/ L2
(2EI)/ L -(6EI)/ L2
(4EI)/ L j
Where Fi, Mi, Fj, Mj are the loads corresponding to the deflections
vi, i, vj, j.
7/29/2019 BEAM ELEMENT USING FEM
7/23
ME 273 Lecture Notes by R. B. Agarwal 7
The above equation can be written in a more solution friendly form
as,
Fi 12 6L
-12 6L vi
Mi 6L 4L2
-6L 2L2 i
Fj = EI/L3
-12 -6L 12 -6L vj
Mj 6L 2 L2
-6L 4L2 j
The above equation is the equation of a beam element, whichis 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 onedirection, 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 matrixinto a global equation involves very complex trigonometric
relations.
7/29/2019 BEAM ELEMENT USING FEM
8/23
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 kNE = 210 GPa, I = 2x10
-4m
4
K = 200 kN/m
Solution
The beam structure is descritized into three elements and 4-nodes,
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 boundaryconditions, and finally, solve for node deflection. Internal forces
and reactions are calculated by back-substituting the deflections in
the structural equation.
7/29/2019 BEAM ELEMENT USING FEM
9/23
ME 273 Lecture Notes by R. B. Agarwal 9
Element 1
1 2EI/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/L3
outside
the matrix, we get
Element 1
12 6L -12 6L v1
6L 4L2
-6L 2 L2
1
[Ke](1)
= (EI/L3)
-12 -6L 12 -6L v2
6L 2 L2
-6L 4 L2
2
[2]
Element 2 2 3
12 6L -12 6L v2
6L 4L2
-6L 2 L2
2
[Ke](2) = (EI/L3)-12 -6L 12 -6L v3
6L 2 L2
-6L 4 L2
3
7/29/2019 BEAM ELEMENT USING FEM
10/23
ME 273 Lecture Notes by R. B. Agarwal 10
3
Element 3
[3]
4
[Ke](3)
= K -K v3
-K K v4
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,
v1 1 v2 2 v3 3 v4
v1 12 6L -12 6L 0 0 0
1 4L2
-6L 2 L2
0 0 0
v2 24 0 -12 6L 0x (EI) /(L
3)
2 8L2
-6L 2L2
0
v3 12 +K -6L - K
3 4L2
0
v4 SYMMETRY K
Where K = (K) x [L3
/ (EI)] = 200 x 103
/(15.55 x105) = .1286
7/29/2019 BEAM ELEMENT USING FEM
11/23
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:
v1= 1 = 0 node 1 is fixed
v2= 0 node 2 has no vertical deflection, but its free to
rotate.
V4 = 0 node 4 is fixed.
The reduced stiffness matrix is
8L2
-6L 2L2
KG = EI / (L3) -6L 12+K -6L
2L2
-6L 4L2
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 v3
0 18 -18 36 3
2 = - 0.0025 radSolving, we get v3 = - 0.0177 m
3 = -0.0076 rad
7/29/2019 BEAM ELEMENT USING FEM
12/23
ME 273 Lecture Notes by R. B. Agarwal 12
Derivation of a Plane (2-D) Beam Element an an
Arbitrary angle
We will derive the 2-D beam element equation that has an arbitraryorientation with axial and bending loads.
F
M
The stiffness equation will be derived in three steps
1.Pure Beam element arbitrarily oriented in space2.Local Beam stiffness with axial and bending loads3.Arbitrarily oriented beam with axial and bending loads.
7/29/2019 BEAM ELEMENT USING FEM
13/23
ME 273 Lecture Notes by R. B. Agarwal 13
Arbitrarily Oriented 2-D Beam Element
The stiffness equation for an arbitrarily oriented beam element can
be derived with a procedure similar to the truss element.
d2yx
y
2y
d1y d1y
1 d1x x
d1y = d1y cos - d1x sin = d1y C - d1x S = - d1x S + d1y C
d2y = d2y cos d2x sin = d2y C d2x S = d2x S + d2y 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,
d1xd1y -s c 0 0 0 0 d1y
1 = 0 0 1 0 0 0 1d2y 0 0 0 -s c 0 d2x
2 0 0 0 0 0 1 d2y2
7/29/2019 BEAM ELEMENT USING FEM
14/23
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 00 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
[kg] = [T]T
[k] [T]
Where, [kg] = 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 as,
12S2
-12SC -6LS -12S2
-12SC -6LS
12C2
6LC 12SC -12C2
6LC
k = EI/L3
4L2
6LS -6LC 2L2
12S2
-12SC 6LSSymmetry 12C
2-6L
4L2
7/29/2019 BEAM ELEMENT USING FEM
15/23
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 .
Also, S = sin , C = cos in the above equation.
7/29/2019 BEAM ELEMENT USING FEM
16/23
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,
f1x 1 -1 d1x f1x f2x= AE/L
f2x -1 1 d2x Truss Element
And for bending, the structural equation is,
f1y 12 6L -12 6L d1y
m1 6L 4L2
-6L 2L2
1= AE/L
3
f2y -12 -6L 12 -6L d2y
m2 6L 2L2
-6L 4L2
2
f1y f2y
m1 m2Beam Element
7/29/2019 BEAM ELEMENT USING FEM
17/23
ME 273 Lecture Notes by R. B. Agarwal 17
Therefore, the combined loading equation is
d1x d1y 1 d2x d2y 2
f1x C1 0 0 - C1 0 0 d1x
f1y 0 12 C2 6C2L 0 -12 C2 6C2L d1y
m1 0 6 C2L 4C2L2
0 -6C2L 2C2L2
1
=f2x -C1 0 0 C1 0 0 d2x
f2y 0 -12 C2 -6C2L 0 12 C2 -6C2L d2y
m2 0 6 C2L 2C2L
2
0 -6C2L 4C2L
2
2
Where,
C1 = AE/L
C2 = EI/L3
The first matrix on the RHS in the above equation gives the local
Stiffness matrix k as
d1x d1y 1 d2x d2y 2
C1 0 0 - C1 0 0
0 12 C2 6C2L 0 -12 C2 6C2L
0 6 C2L 4C2L2
0 -6C2L 2C2L2
k =
-C1 0 0 C1 0 0
0 -12 C2 -6C2L 0 12 C2 -6C2L
0 6 C2L 2C2L2
0 -6C2L 4C2L2
7/29/2019 BEAM ELEMENT USING FEM
18/23
ME 273 Lecture Notes by R. B. Agarwal 18
The local stiffness matrix k has 3-DOF at each node, representing
bending and axial loads.
For axial loading:
u2y
u2
u1y 2u2x
1 u1x
u1
u1 = u1x cos + u1y sin = c u1x + s u1y
u2 = u2x cos + u2y sin = c u2x + s u2y
where, cos = c, and sin = s
Which can be written as,u1x
u1 c s 0 0 u1y
= u2x (3.2)u2 0 0 c s u2y
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 cant be
inverted, which will be incompatible with the matrix operations.
7/29/2019 BEAM ELEMENT USING FEM
19/23
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.
u1x u1y u2x u2y
u1 c s 0 0 u1x
v1 = -s c 0 0 u1y
u2 0 0 c s u2xv2 0 0 -s c u2y
Where,u1x u1y u2x u2yc s 0 0
T = -s c 0 00 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.
u2y
y xy
2u1y
u1y 1 u1x x
u1y = u1y cos - u1x sin = u1y C - u1x S = - u1x S + u1y C
u2y = u2y cos u2x sin = u2y C u2x S = u2x S + u2y C
7/29/2019 BEAM ELEMENT USING FEM
20/23
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 globalcoordinates.
The above equations can be written in a compatible matrix form,
by introducing 0s where necessary,
u1x
u1y -s c 0 0 0 0 u1y
1 = 0 0 1 0 0 0 1u2y 0 0 0 -s c 0 u2x
2 0 0 0 0 0 1 u2y2
-s c 0 0 0 0
Let T = 0 0 1 0 0 0
0 0 0 -s c 00 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 0T = 0 0 1 0 0 0
0 0 0 C S 0
0 0 0 -S C 00 0 0 0 0 1
7/29/2019 BEAM ELEMENT USING FEM
21/23
ME 273 Lecture Notes by R. B. Agarwal 21
Using the relation, [k] = [T]T[k][T] and simplifying, we get,
I
CL
IC
L
IASsymmetry
S
L
ICS
L
IAS
L
IAC
ICL
IS
L
II
CL
IC
L
IASCS
L
IAC
L
IC
L
IAS
SL
ICS
L
IAS
L
IACS
L
ICS
L
IAS
L
IAC
L
EK
4
612
61212
266
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 2-D beam element at an
arbitrary orientation and under axial and bending loads.
7/29/2019 BEAM ELEMENT USING FEM
22/23
ME 273 Lecture Notes by R. B. Agarwal 22
2-D Beam Element with combined loading
Bending, Axial, and Torsion ( = 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 m1x and m2x, and the corresponding
deflections are,
x1 and x2
The torsional structural equation is:
m1x 1 -1 1x
= JG/Lm1x -1 1 2x
These terms can be superimposed on the stiffness equation derived
previously for the combined bending and axial loads.
dy
y
3-D Beam Element: dx
z x
dz
A 3-D 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,
7/29/2019 BEAM ELEMENT USING FEM
23/23
The stiffness equation in local coordinate is:
^
2
^
2
^
2
^
2
^
2
^
2
^
1
^
1
^
1
^
1
^
1
^
1 yxzyxzyxzyx dddddd
L
EI
L
EI
L
EI
L
EI L
EI
L
EI
L
EI
L
EIL
GJ
L
GJL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
GJ
L
GJL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AEL
AE
K
zzzz
yyyy
yyyy
zzzz
zzzz
yyyy
yyyy
zzzz
4000
60
2000
60
04
06
0002
06
00
0000000000
06
012
0006
012
00
6000
120
6000
120
0000000000
2
000
6
0
4
000
6
0
02
06
0004
06
00
0000000000
06
012
0006
012
00
6000
120
6000
120
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,
]][[][^
TKTKT
The equation gets very complicated, so we will not go in to the
derivation any further.