Civil Engineering Reference
In-Depth Information
the displacements produced by axial force and to assume that they are caused solely
by bending.
The point D in the primary structure will suffer vertical and horizontal displacements,
D,V
and
D,H
. Thus if we designate the redundant reactions as
R
1
, and
R
2
, the
equations of compatibility are
D,V
+
a
11
R
1
+
a
12
R
2
=
0
(i)
D,H
+
a
21
R
1
+
a
22
R
2
=
0
(ii)
in which the flexibility coefficients have their usual meaning. Again, as in the preceding
examples, we employ the unit loadmethod to calculate the displacements andflexibility
coefficients. Thus
M
0
M
1,V
EI
D,V
=
d
x
L
in which
M
1,V
is the bending moment at any point in the frame due to a unit load
applied vertically at D.
Similarly
M
0
M
1,H
EI
D,H
=
d
x
L
and
M
1,V
EI
M
1,
H
EI
M
1,V
M
1,H
EI
a
11
=
d
xa
22
=
d
xa
12
=
a
21
=
d
x
L
L
L
We shall now write down expressions for bending moment in the members of the
frame; we shall designate a bending moment as positive when it causes tension on the
outside of the frame. Thus in DC
M
0
=
0
M
1,V
=
0
M
1,H
=−
1
x
1
In CB
x
2
2
2
x
2
M
0
=
4
x
2
=
M
1,V
=−
1
x
2
M
1,H
=−
3
In BA
M
0
=
4
×
3
.
5
×
1
.
75
+
10
x
3
=
24
.
5
+
10
x
3
M
1,V
=−
3
.
5
M
1,H
=−
1(3
−
x
3
)
Hence
3
.
5
3
0
−
1
EI
489
.
8
EI
2
x
2
)d
x
2
+
D,V
=
(
−
(24
.
5
+
10
x
3
)3
.
5d
x
3
=−
0
3
.
5
3
0
−
1
EI
241
.
0
EI
6
x
2
)d
x
2
+
D,H
=
(
−
(24
.
5
+
10
x
3
)(3
−
x
3
)d
x
3
=−
0
