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where
N
u
and
G
are taken from Eqs. (4.41) and (4.45), respectively. Assume
k
ω
, the coefficient
for the elastic foundation, is constant. Then
1
x
2
x
3
2
/
2
3
/
3
4
/
4
x
2
x
3
x
4
2
/
2
3
/
3
4
/
4
5
/
5
N
u
k
N
u
dx
=
k
dx
=
w
w
x
2
x
3
x
4
x
5
3
/
3
4
/
4
5
/
5
6
/
6
o
o
x
3
x
4
x
5
x
6
4
/
5
/
6
/
7
/
4
5
6
7
and finally,
G
T
o
N
u
N
u
dx
G
k
w
=
k
w
156
−
22
54
13
2
2
−
22
4
−
13
−
3
k
w
420
=
(15)
54
−
13
156
22
2
2
13
−
3
22
4
4.4.3
Properties of Stiffness Matrices
Some properties of stiffness matrices were derived in Chapter 3. It was shown that the
stiffness matrix should be symmetric, i.e.,
k
ba
k
ij
=
k
ji
and
k
ab
=
(4.75)
Note that
k
i
of Eq. (4.42) is indeed symmetric.
It was also shown that the diagonal elements of a stiffness matrix are positive. Further-
more, it was illustrated in Section 4.3 that a stiffness matrix is singular and, hence, cannot be
inverted. Also, after elimination of rigid body motion, a stiffness matrix is positive definite.
4.4.4
Response of Simple Beams
The stiffness matrix as developed in this chapter can be used to solve problems concerning
uniform beams. In the next chapter, structural systems formed of beam elements will be
treated.
It is shown in Section 4.3 that, in general
k
i
is a singular matrix. However, the application
of the boundary conditions has the effect of rendering
k
i
nonsingular and providing the
desired solution for the response of the beam.
EXAMPLE 4.5 Beam with Linearly Varying Loading
Find the response of the beam of Fig. 4.14. This is the same beam treated in Chapter 3,
Example 3.13 and Fig. 3.11.
The
st
iffness matrix for an Euler-Bernoulli beam is provided by Eq. (4.12), and the loading
vector
p
i
0
can be taken from Table 4.2 with
p
a
=
p
0
and
p
b
=
0, giving
7
p
a
/
20
2
p
a
/
−
20
p
i
0
=
(1)
3
p
a
/
20
2
p
a
/
30
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