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FIGURE 4.5
Element for an alternative (reduced) form of the stiffness matrix.
To find
g
, apply the conditions of equilibrium to the beam element of Fig. 4.1b, giving
M
x
=
b
=
0
→
V
a
=−
(
M
a
+
M
b
)/
M
x
=
a
=
(4.30)
0
→
V
b
=
(
M
a
+
M
b
)/
Thus,
V
a
V
b
M
a
M
b
−
1
/
−
1
/
M
a
M
b
=
1
/
1
/
(4.31)
1
0
0
1
g
T
p
=
p
R
where the definition of
p
has been altered from that used previously. The corresponding
independent displacements
v
R
are (from
v
R
=
gv
)
w
a
w
b
θ
a
θ
b
−
1
/
1
/
10
v
R
=
−
1
/
1
/
01
(w
b
−
w
a
)/
+
θ
a
(w
b
−
w
a
)/
+
θ
b
θ
ab
θ
ba
=
=
(4.32)
where it was necessary to rearrange the elements of
v
to correspond to those of
p
.Itis
observed that the newly defined displacement variables are tangent angles of the element.
Flexibility and Stiffness Matrices for a Cantilevered Beam Element
As mentioned, the first case of the section, in which the rigid body displacements of the
left end were removed, corresponds to a beam element fixed on the left end. Since
M
b
and
V
b
are independent non-zero variables, the right end can be treated as being free.
The displacements
θ
R
) are to be related to the forces
M
b
and
V
b
. The
stiffness coefficients are thus relating the variables at end
b
.
w
θ
w
R
and
and
(
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