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DRAFT –1
Chapter 4 Beam Element
Example 1
For the beam shown, determine the displacements and slopes at the nodes, forces in each
element, and reactions at the supports.
5 ft
5 ft
100 lb
E = 1.4 x 10
6
psi,
I = 2.4 in
4
K = 200 lb/in
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 boundary conditions, and finally, solve for node
deflection. Internal forces and reactions are calculated by back-substituting the
deflections in the structural equation.
[1]
Element 1
1 2
EI/L
3
= (1.4 x 10
6
) x (2.4)/(5x12)
3
= 15.55
The general equation of a stiffness matrix is given as,
ME 273 Lecture Notes © by R. B. Agarwal
4-6
DRAFT –1
Chapter 4 Beam Element
12
6L
-12
6L
v
1
6L
4L
2
-6L 2
L
2
θ
1
[K
e
]
(1)
= (EI/L
3
)
-12
-6L
12
-6L
v
2
6L
2
L
2
-6L 4
L
2
θ
2
[2]
Element 2
2
3
12
6L
-12
6L
v
2
6L
4L
2
-6L 2
L
2
θ
2
[K
e
]
(1)
= (EI/L
3
)
-12
-6L
12
-6L
v
3
6L
2
L
2
-6L 4
L
2
θ
3
3
Element 3
[3]
4
[K
e
]
(3)
=
K
-K
v
3
-K
K
v
4
ME 273 Lecture Notes © by R. B. Agarwal
4-7
DRAFT –1
Chapter 4 Beam Element
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
θ
1
v
2
θ
2
v
3
θ
3
v
4
v
1
12
6L
-12
6L 0
0
0
θ
1
4L
2
-6L 2
L
2
0 0 0
v
2
24
0
-12
6L 0
x (EI) /(L
3
)
θ
2
8L
2
-6L 2L
2
0
v
3
12 +K’
-6L - K’
θ
3
4L
2
0
v
4
SYMMETRY
K’
Where K’ = (K) x [L
3
/ (EI)]
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
=
θ
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,
ME 273 Lecture Notes © by R. B. Agarwal
4-8
DRAFT –1
Chapter 4 Beam Element
0
1152 -72
288
θ
2
-100 = (15.55) -72 16.11 -72
v
3
0
288 -72
576
θ
3
θ
2
= - 0.0032 rad
Solving, we get
v
3
= - 0.4412 in
θ
3
= -0.0095 rad
ME 273 Lecture Notes © by R. B. Agarwal
4-9
DRAFT –1
Chapter 4 Beam Element
4.2 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.
y d
2y
x
y
ϕ
2
d
1y
d
1y
ϕ
1
d
1x
x
d
1y
= d
1y
cos
θ
- d
1x
sin
θ
= d
1y
c - d
1x
s
d
2y
= d
1y
cos
θ
– d
2x
sin
θ
= d
2y
c – d
2x
s
and
ϕ
1
=
ϕ
1
,
ϕ
2
=
ϕ
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 matrix form,
d
1x
d
1y
-s c 0 0 0 0 d
1y
ϕ
1
=
0
0 1 0 0 0
ϕ
1
d
2y
0 0
0 -s c 0 d
2x
ϕ
2
0
0 0 0 0 1 d
2y
ϕ
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.
Thus, {d} = [T] {d}
Global
Local
Note that angle
ϕ
is independent of the coordinate systems, and
ϕ
1 =
ϕ
1
,
ϕ
2
=
ϕ
2
ME 273 Lecture Notes © by R. B. Agarwal
4-10