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to a unit load at the right support is
δ
M
=
L
−
x
.
The unit load method formula of Eq. (3.17)
appears as
R
L
dx
0
0
δ
MM
EI
p
0
(
L
−
x
)
4
2
w
=
0
=
d
(
L
−
x
)
=−
(
L
−
x
)
−
(1)
R
6
L
L
L
This gives
R
L
=
The left-hand reactions are found from the conditions of equilib-
rium for the beam configuration of Fig. 3.11.
p
0
L
/
10
.
2
5
p
0
L,
1
15
p
0
L
2
R
0
=
M
0
=−
EXAMPLE 3.14 Stiffness Matrix for a Beam
The unit load method provides a straightforward technique for establishing a stiffness
matrix. As with the other complementary virtual work theorems, flexibility relationships
are obtained initially. These relationships are reorganized in stiffness matrix form. Return
to the beam element of Fig. 3.9. To find the deflection at
a
in terms of
M
a
and
V
a
,
the unit
load method uses the moment at
x
due to a unit (downwards) force at
a
.
This moment is
of magnitude
−
x
.
The moment at
x
due to
M
a
and
V
a
is
M
=−
M
a
−
V
a
x
.
Thus,
0
−
2
3
1
EI
M
a
V
a
w
=
x
(
−
M
a
−
V
a
x
)
dx
=
+
(1)
a
2
EI
3
EI
The calculation of the slope at
a
involves the moment at
x
due to a unit counterclockwise
moment at
a
.
This moment is equal to
−
1
.
Then
0
(
−
2
1
EI
M
a
V
a
θ
=
1
)(
−
M
a
−
V
a
x
)
dx
=
+
(2)
a
EI
2
EI
From (1) and (2),
V
a
M
a
3
2
/
3
EI
/
2
EI
w
a
=
(3)
θ
2
/
2
EI
/
EI
a
Relationships such as this which relate forces to displacements are the
flexibility
equations.
If (3) is solved for
V
a
and
M
a
we obtain the stiffnesses
k
aa
as
V
a
M
a
12
EI
w
k
aa
w
3
2
/
−
6
EI
/
a
a
=
=
(4)
2
−
6
EI
/
4
EI
/
θ
θ
a
a
From the conditions of equilibrium and (4),
12
EI
6
EI
V
b
=−
V
a
=−
w
+
θ
a
a
3
2
6
EI
2
EI
M
b
=−
V
a
−
M
a
=−
2
w
a
+
θ
a
or
V
b
M
b
−
w
w
3
2
12
EI
/
6
EI
/
a
a
=
=
k
ba
(5)
2
−
6
EI
/
2
EI
/
θ
θ
a
a
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