Civil Engineering Reference
In-Depth Information
k
00000
k
0
0 000
∆
∆
∆
P
P
P
M
M
M
P
P
P
M
M
M
11
,
17
,
ix
ix
0
k
0 00 0
k
k
0 00
k
22
,
26
,
28
,
212
,
iy
iy
00 0
k
k
0 00
k
0
k
0
33
,
35
,
339
,
311
,
iz
iz
000
k
0 0000 00
k
q
q
q
44
,
410
,
ix
ix
00 0
k
k
0 00 0
k
k
0
53
,
55
,
59
,
511
,
iy
iy
0
k
,
000
k
0 000
k
k
iz
6
2
6 6
,
6 8
,
6 12
,
iz
=
k
00000
k
0
0 000
∆
71
,
77
,
jx
jx
0
k
0 00 0
k
k
0000
k
∆
∆
jy
82
,
86
,
88
,
812
,
jy
00 0
k
k
0 00 0
k
k
0
93
,
95
,
99
,
911
,
jz
jz
000
k
0 0000 00
k
q
q
q
10 4
,
1010
,
jx
jx
00
k
0
k
0 00 0
k
k
0
11 3
,
115
,
11 9
,
1111
,
jy
jy
0
k
0 00 0
k
k
0 00
k
12 2
,
126
,
12 8
,
1212
,
jz
jz
(5.1)
Example 5.1 ∆
iz
end release
Derive the local member stiffness for a ∆
iz
member end release using the
conjugate beam method.
A free-body diagram of the released beam is shown in Figure 5.1.
Since the beam is allowed to move at the
i
i-end in the
z
direction, the
reaction
P
iz
is equal to zero. The loaded conjugate beam is also shown in
Figure 5.1. Note that the shear in the conjugate beam is equal to the rota-
tion in the real beam and the moment in the conjugate beam is equal to the
deflection in the real beam.
θ
iy
EI
y
EI
y
∆
iz
=0
∆
iz
=0
θ
iy
θ
iy
=0
Figure 5.1.
Example 5.1 ∆
iz
end release.
If a motion ∆
iz
is imposed, there is no resistance and therefore no
forces. The resulting forces are the stiffness values due to the motion. The
following are forces and stiffness due to ∆
iz
:
