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2
3
2
3
2
3
2
3
w(ξ)
=
(
1
−
3
ξ
+
2
ξ
)w
+
(
−
ξ
+
2
ξ
−
ξ
)θ
+
(
3
ξ
−
2
ξ
)w
+
(ξ
−
ξ
)θ
a
a
b
b
=
(ξ )w
+
(ξ ) θ
+
(ξ ) w
+
(ξ ) θ
H
1
H
2
H
3
H
4
(4.47b)
a
a
b
b
The terms in brackets are Hermitian
2
polynomials. Note that due to scaling,
G
of Eq. (4.45)
differs from
G
of Eq. (4.47).
The goal in this subsection is to derive the element stiffness matrix using an assumed
series. Although it will be illustrated below that for beams this can be accomplished directly,
the more useful and more important technique is to employ the principle of virtual work.
Direct Evaluation of the Stiffness Matrix
Thus far, the deflection has been expressed in terms of the end displacements using
w
=
N
u
Gv
.
Since we seek a stiffness relationship of the form
p
=
kv
, we still need to
relate the end forces
p
to the deflection
A force-displacement relationship is provided
by the material law [Chapter 1, Eq. (1.106)]
M
w.
w
,
which relates the
internal moments to the second derivative of the deflection. Use
=
EI
κ
=−
EI
w
=
N
u
Gv
,
so that
M
=−
EI
N
u
Gv
.
With
N
u
=
[1
ξξ
2
ξ
3
] and
N
u
=
(
d
N
u
/
d
ξ)(
d
ξ/
dx
)
,
N
u
becomes
N
u
=
2
[0026
ξ
]
/
(4.48)
Define
N
u
=
B
u
(4.49)
Then
M
=−
EI
B
u
Gv
(4.50)
It is a simple matter to use the conditions of equilibrium to find the shear force
V
in
terms of the displacements and thereby to complete the derivation of the stiffness matrix.
The moment
M
of Eq. (4.50) is an internal moment which adheres to Sign Convention
1 as shown in Fig. 4.9. Let a superscript 1 indicate Sign Convention 1. Then the internal
bending moments at
a
and
b
are
M
x
=
a
=
M
b
,
respectively. It can be
observed in Fig. 4.9 that the moments defined according to the two sign conventions
are related by
M
a
=−
M
a
and
M
x
=
b
=
Furthermore, from the element equilibrium
condition of the summation of moments being zero first about
a
and then about
b,
it is
found that
V
b
=
(
M
a
and
M
b
=
M
b
.
M
b
)/.
Thus, the set of equations relating the end moments and shear forces in Sign Convention
2 to the internal end moments is
M
a
+
M
b
)/
=
(
−
M
a
+
M
b
)/
and
V
a
=
(
−
M
a
−
M
b
)/
=
(
M
a
−
V
a
M
a
V
b
M
b
1
−
1
p
a
M
a
M
b
=
=
1
−
0
(4.51)
−
11
0
p
b
Finally, evaluate
M
a
and
M
b
in Eq. (4.51) by setting
M
a
=
M
ξ
=
0
and
M
b
=
M
ξ
=
1
in Eq.
(4.50). Substitution of these values for
M
a
and
M
b
into the expression of Eq. (4.51) gives
the desired stiffness relation
p
=
kv
.
2
Charles Hermite (1822-1901) was a great French algebraist, probably the leading French mathematician of the
second half of the 19th century. From 1869, he was a professor of mathematics at the Sorbonne University. His
work and that of his students exercised a profound influence on contemporary mathematics. Henri Poincare, his
student, said, “Talk with Hermite: he never evokes a concrete image; yet you soon perceive that the most abstract
entities are for him like living creatures.”
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