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to obtain
(w
Z
2
)
(θ
Z
2
)(w
Z
3
)
V
2
F
Z
2
M
2
M
Y
2
V
3
F
Z
3
3
EI
3
EI
3
EI
−
−
522
−
5220
−
520
3
2
3
k
2
3
EI
3
EI
3
EI
=
−
=
52 200
5220
(5)
2
2
Symmetric
522
3
EI
3
EI
3
EI
−
3
2
3
W
2
=
0
W
3
=
0
2
For this element, the global and local coordinate system coincide. The loading vector is
given by
V
2
2
F
2
Z
2
M
2
2
M
2
Y
2
V
2
3
F
2
Z
3
=
−
5
/
8
−
37
.
5
p
20
2
−
=
q
/
8
75
.
0
(6)
−
3
/
8
−
22
.
5
Stiffness Matrix of Element 3
Element 3 has only hinges on both ends and no transverse loads and, hence, no bending. This
bar acts like a truss member, influencing the bending of the system by second order effects
only (Fig. 11.24). However, these second order effects contained in the geometric matrix
contribute to the change in the moments in other members. The horizontal inclination of
the top of the truss in the direction of the unknown displacement
U
2
(Fig. 11.21) corresponds
to a horizontal reaction force for element 3. With the value
N
0
it contributes through the
geometric stiffness to the equilibrium of the horizontal forces for the determination of the
unknown
U
2
/
.
(
u
X
3
)
(
u
X
4
)
V
3
F
X
3
V
4
F
X
4
63
1
−
523
.
406
523
.
406
/
4
−
1
/
4
(7)
k
geo
=−
2093
.
=
−
1
/
41
/
4
523
.
406
−
523
.
406
U
3
=
U
2
U
4
=
0
The transformation of the stiffness matrix of element 3 to global coordinates is accomplished
by setting
X
=−
z
.
FIGURE 11.24
Element 3 with local and global coordinates.
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