Civil Engineering Reference
In-Depth Information
Simultaneous solution gives the displacement values as
U
4
=
47974 (10
−
5
)
in.
U
5
=−
2
.
74704 (10
−
4
)
in.
1
.
94058 (10
−
4
)
rad
As usual, the reaction components can be obtained by substituting the computed dis-
placements into the six constraint equations.
For the beam element with axial capability, the stress computation must take into
account the superposition of bending stress and direct axial stress. For element 1, for
example, we use Equation 4.63 with
=
/
2
to compute the element displacement as
U
6
=−
9
.
u
1
v
1
1
u
2
v
2
2
010000
−
U
1
U
2
U
3
U
4
U
5
U
6
0
0
0
100000
001000
000010
000
=
=
−
1
.
74704(10
−
4
)
100
000001
−
−
2
.
47974(10
−
5
)
−
9
.
94058(10
−
4
)
The bending stress is computed at nodes 1 and 2 via Equations 4.33 and 4.34 as
5(10)(10
6
)
6
94058 (10
−
4
)
2
20
(
47974 )(10
−
5
)
x
(
x
=
0)
=±
0
.
20
2
(
−
2
.
−
−
9
.
=±
495
.
2 psi
5(10)(10
6
)
6
94058 )(10
−
4
)
2
20
(2)(
47974 )(10
−
5
)
x
(
x
=
L
)
=±
0
.
20
2
(2
.
+
−
9
.
=±
992
.
2 psi
and the axial stress is
10(10
6
)
−
1
.
74704 (10
−
4
)
20
axial
=
=−
87
.
35 psi
Therefore, the largest stress magnitude occurs at node 2, at which the compressive axial
stress adds to the compressive portion of the bending stress distribution to give
=
1079
.
6 psi
(compressive)
4.8 A GENERAL THREE-DIMENSIONAL
BEAM ELEMENT
A general three-dimensional beam element is capable of both axial and torsional
deflections as well as two-plane bending. To examine the stiffness characteristics
of such an element and obtain the element stiffness matrix, we first extend the
beam-axial element of the previous section to include two-plane bending, then
add torsional capability.
Figure 4.15a shows a beam element with an attached three-dimensional ele-
ment coordinate system in which the
x
axis corresponds to the longitudinal axis
