Civil Engineering Reference
In-Depth Information
which becomes, after integration of the first term by parts,
x
2
x
2
x
2
x
1
−
d
y
d
x
x
d
N
i
d
x
d
y
d
x
N
i
x
d
x
−
4
xN
i
d
x
=
0
i
=
1, 2
x
1
x
1
Substituting the element solution form and rearranging, we have
x
2
d
N
1
d
x
y
2
d
x
x
2
x
2
x
1
−
d
N
i
d
x
d
N
2
d
x
d
y
d
x
x
y
1
+
=
N
i
x
4
xN
i
d
x
i
=
1, 2
x
1
x
1
Expanding the two equations represented by the last result after substitution for the inter-
polation functions and first derivatives yields
x
1
−
x
2
x
2
1
(
x
2
−
x
1
)
2
d
y
d
x
x
2
−
x
x
2
−
x
1
x
(
y
1
−
y
2
)d
x
=−
x
1
4
x
d
x
x
1
x
1
x
2
−
x
2
x
2
1
(
x
2
−
x
1
)
2
d
y
d
x
x
−
x
1
x
2
−
x
1
x
(
y
2
−
y
1
)d
x
=
x
2
4
x
d
x
x
1
x
1
Integration of the terms on the left reveals the element stiffness matrix as
k
(
e
)
=
1
x
2
−
x
1
2(
x
2
−
x
1
)
2
−
1
−
11
while the gradient boundary conditions and nodal forces are evident on the right-hand
side of the equations.
To illustrate, a two-element solution is formulated by taking equally spaced nodes at
x
=
1,
1.5, 2 as follows.
Element 1
x
1
=
1
x
2
=
1
.
5
k
=
2
.
5
1
.
5
x
1
.
5
−
x
1
.
5
−
1
F
(1)
1
=−
4
d
x
=−
1
.
166666
...
1
1
.
5
x
−
1
1
.
5
−
1
F
(1)
2
=−
4
x
d
x
=−
1
.
33333
...
1
Element 2
x
1
=
1
.
5
x
2
=
2
k
=
3
.
5
2
x
2
−
1
.
5
2
−
F
(2)
1
=−
4
x
d
x
=−
1
.
66666
...
1
.
5
2
x
x
5
2
−
1
.
5
−
1
.
F
(2)
2
=−
4
d
x
=−
1
.
83333
...
1
.
5
