Civil Engineering Reference
In-Depth Information
and the element axial strain (utilizing Equation 2.29 and the discretization and
interpolation functions of Equation 2.25) is then
N
2
(
x
)]
u
(
e
1
u
(
e
2
d
u
(
e
)
(
x
)
d
x
d
(
e
)
d
x
(
e
)
ε
=
=
[
N
1
(
x
)
u
(
e
1
u
(
e
2
−
u
(
e
2
u
(
e
1
−
1
L
(
e
)
1
L
(
e
)
=
=
(3.49)
L
(
e
)
where
L
(
e
)
is element length. The element axial stress is then obtained via appli-
cation of Hooke's law as
(
e
)
(
e
)
(3.50)
Note, however, that the global solution does not give the element axial displace-
ment directly. Rather, the element displacements are obtained from the global
displacements via Equations 3.48. Recalling Equations 3.21 and 3.22, the ele-
ment strain in terms of global system displacements is
=
E
ε
U
(
e
1
U
(
e
2
U
(
e
3
U
(
e
4
d
u
(
e
)
(
x
)
d
x
d
d
x
[
N
1
(
x
)
(
e
)
ε
=
=
N
2
(
x
)][
R
]
(3.51)
where [
R
] is the element transformation matrix defined by Equation 3.22. The
element stresses for the bar element in terms of global displacements are those
given by
U
(
e
1
U
(
e
2
U
(
e
3
U
(
e
4
E
d
u
(
e
)
(
x
)
d
x
E
d
(
e
)
d
x
(
e
)
(
e
)
(3.52)
=
E
ε
=
=
[
N
1
(
x
)
N
2
(
x
)][
R
]
As the bar element is formulated here, a positive axial stress value indicates that
the element is in tension and a negative value indicates compression per the usual
convention. Note that the stress calculation indicated in Equation 3.52 must be
performed on an element-by-element basis. If desired, the element forces can be
obtained via Equation 3.23.
EXAMPLE 3.2
The two-element truss in Figure 3.5 is subjected to external loading as shown. Using the
same node and element numbering as in Figure 3.2, determine the displacement com-
ponents of node 3, the reaction force components at nodes 1 and 2, and the element
displacements, stresses, and forces. The elements have modulus of elasticity
E
1
=
E
2
=
10
×
10
6
lb/in.
2
and cross-sectional areas
A
1
=
A
2
=
1.5 in.
2
.
