Civil Engineering Reference
In-Depth Information
Analogous to Equation 3.21, Equations 3.55 and 3.56 can be expressed as
U
(
e
)
1
U
(
e
)
2
U
(
e
)
3
U
(
e
)
4
U
(
e
)
5
U
(
e
)
6
u
(
e
1
u
(
e
2
cos
x
cos
y
cos
z
0
0
0
=
0
0
0
cos
x
cos
y
cos
z
[
R
]
U
(
e
)
(3.57)
where [
R
] is the transformation matrix mapping the one-dimensional element
displacements into a three-dimensional global coordinate system. Following the
identical procedure used for the 2-D case in Section 3.3, the element stiffness
matrix in the element coordinate system is transformed into the 3-D global co-
ordinates via
=
[
R
]
T
k
e
[
R
]
K
(
e
)
=
−
k
e
(3.58)
−
k
e
k
e
Substituting for the transformation matrix [
R
] and performing the multiplication
results in
c
x
c
x
c
y
c
x
c
z
−
c
x
−
c
x
c
y
−
c
x
c
z
c
y
c
y
c
x
c
y
c
y
c
z
−
c
x
c
x
−
−
c
y
c
z
K
(
e
)
=
c
z
c
z
c
x
c
z
c
y
c
z
−
c
x
c
z
−
c
y
c
z
−
k
e
(3.59)
c
x
c
x
−
−
c
x
c
x
−
c
x
c
z
c
x
c
y
c
x
c
z
c
y
c
y
−
c
x
c
y
−
−
c
y
c
z
c
x
c
y
c
y
c
z
c
z
c
z
−
c
x
c
z
−
c
y
c
z
−
c
x
c
z
c
y
c
z
as the 3-D global stiffness matrix for the one-dimensional bar element where
c
x
=
cos
x
c
y
=
cos
y
(3.60)
c
z
=
cos
z
Assembly of the global stiffness matrix (hence, the equilibrium equations),
is identical to the procedure discussed for the two-dimensional case with the ob-
vious exception that three displacements are to be accounted for at each node.
EXAMPLE 3.3
The three-member truss shown in Figure 3.8a is connected by ball-and-socket joints and
fixed at nodes 1, 2, and 3. A 5000-lb force is applied at node 4 in the negative
Y
direction,
as shown. Each of the three members is identical and exhibits a characteristic axial stiff-
ness of 3(10
5
) lb/in. Compute the displacement components of node 4 using a finite
element model with bar elements.
