Civil Engineering Reference
In-Depth Information
Again, note the use of capital letters for global quantities and the superscript
notation to refer to an individual element. As the nodal displacements must be
the same in both coordinate systems, we can equate vector components of global
displacements to element system displacements to obtain the relations
u
(
e
1
U
(
e
1
cos
U
(
e
2
sin
=
+
(3.4a)
v
(
e
1
U
(
e
1
sin
U
(
e
2
cos
=−
+
u
(
e
2
U
(
e
3
cos
U
(
e
4
sin
=
+
(3.4b)
v
(
e
1
U
(
e
3
sin
U
(
e
4
cos
=−
+
As noted, the
v
displacement components are not associated with element stiff-
ness, hence not associated with element forces, so we can express the axial de-
formation of the element as
=
U
(
e
3
U
(
e
1
cos
+
U
(
e
4
U
(
e
2
sin
u
(
e
2
u
(
e
1
(
e
)
=
−
−
−
(3.5)
The net axial force acting on the element is then
f
(
e
)
(3.6)
Utilizing Equation 3.6 for element 1 (Figure 3.3d) while noting that the dis-
placements of element 1 are related to the specified global displacements as
U
(1
1
k
(
e
)
U
(
e
3
U
(
e
1
cos
+
U
(
e
4
U
(
e
2
sin
k
(
e
)
(
e
)
=
=
−
−
U
1
,
U
(1
2
U
2
,
U
(1
3
U
5
,
U
(1
4
=
=
=
=
U
6
, we have the force in element 1 as
f
(1
3
f
(1
1
k
(1)
[(
U
5
−
=−
=
U
1
)cos
1
+
(
U
6
−
U
2
)sin
1
]
(3.7)
and similarly for element 2 (Figure 3.3e):
f
(2
3
f
(2
2
k
(2)
[(
U
5
−
2
]
(3.8)
Note that, in writing Equations 3.7 and 3.8, we invoke the condition that the dis-
placements of node 3 (
U
5
and
U
6
) are the same for each element. To reiterate, this
assumption is actually a requirement, since on a physical basis, the structure
must remain connected at the joints after deformation. Displacement compatibil-
ity at the nodes is a fundamental requirement of the finite element method.
Substituting Equations 3.7 and 3.8 into the nodal equilibrium conditions
(Equations 3.1-3.3) yields
=−
=
U
3
)cos
2
+
(
U
6
−
U
4
)sin
k
(1)
[(
U
5
−
−
U
1
)cos
1
+
(
U
6
−
U
2
)sin
1
]cos
1
=
F
1
(3.9)
k
(1)
[(
U
5
−
−
U
1
)cos
1
+
(
U
6
−
U
2
)sin
1
]sin
1
=
F
2
(3.10)
k
(2)
[(
U
5
−
−
U
3
)cos
2
+
(
U
6
−
U
4
)sin
2
]cos
2
=
F
3
(3.11)
k
(2)
[(
U
5
−
−
U
3
)cos
2
+
(
U
6
−
U
4
)sin
2
]sin
2
=
F
4
(3.12)
k
(2)
[(
U
5
−
U
3
)cos
2
+
(
U
6
−
U
4
)sin
2
] cos
2
k
(1)
[(
U
5
−
+
U
3
)cos
1
+
(
U
6
−
U
4
)sin
1
]cos
1
=
F
5
(3.13)
k
(2)
[(
U
5
−
U
3
)cos
2
+
(
U
6
−
U
4
)sin
2
]sin
2
k
(1)
[(
U
5
−
+
U
1
)cos
1
+
(
U
6
−
U
2
)sin
1
]sin
1
=
F
6
(3.14)
