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where the elements are segments along the beam. This difference results in a reduction of
the dimensionality of the problem when using a boundary element approach, in that the
solution of a one dimensional beam problem is transformed to a solution of linear equations
involving boundary points, with no consideration of in-span nodes.
The solution of Eq. (9.11) provides values of the unknown boundary variables. A well-
posed beam problem consists of four known boundary variables and the same number of
unknown boundary variabl es . Equation (9.11) can be prepared for solution by moving the
known quantities in V and P together with the corresponding elements in the H and G
matrices to the right-hand side of the equation and the unknown quantities are moved to
the left. A tractable system of linear equations AX
F in terms of the unknown variables
X is obtained and the unknown variables can be calculated. Once the unknown boundary
variables are determined, they can be substituted into Eqs. (9.9) and (9.10) to find the
responses along the beam.
It should be observed that simple manipulation of Eq. (9.11) leads to stiffness and transfer
matrices. For example, multiplication of Eq. (9.11) by G 1 results in
=
P
G 1 HV
G 1 B
=
+
i.e.,
P
P 0
=
KV
(9.12)
where P contains the shear forces a nd moments on both boundaries, V contains the bound-
ary displacements, K
G 1 H , and P 0
G 1 B
This corresponds to Chapter 5, Eq. (5.100).
Thus, Eq. (9.12) is the stiffness equation of the beam problem. It can be verified that the
stiffness matrix K obtained in this manner is the same as that given in Chapter 4. Since only
one element is involved in this beam problem, the global stiffness matrix K of Eq. (9.12) is
equal to the element stiffness matrix k i
=
=
.
.
EXAMPLE 9.1 Beam with Linearly Varying Load
Consider the familiar beam of Chapter 7, Fig. 7.1 with loading p z (
).
The solution should begin with the determination of the unknown boundary variables.
For this beam the known boundary conditions are
x
) =
p 0
(
1
x
/
L
w
= w
=
0 ,
θ
=
0 ,
M L
=
0
(1)
0
L
0
and the boundary variables to be calculated are
L ,V 0 ,M 0 , and V L . Equation (9.11) is to be
used to find these unknown variables. Substitute (1) and p z =
θ
p 0
(
1
x
/
L
)
into Eq. (9.11),
giving
L 3
6 EI
θ L
L 4
20
L 3
4
L
0
0
0 L 2
2 EI
V 0
10
p 0
6 EI
=−
(2)
L 3
6 EI
L 2
2 EI
L 4
5
3 L 3
4
M 0
0
0
L 2
2 EI
L
EI
1
0
V L
A
X
=
F
Solve this system of equations to find
p 0 L 3
120 EI
2
5 p 0 L,
1
15 p 0 L 2 ,
p 0 L
10
V 0 =
M 0 =−
=−
,
θ L =
(3)
L
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