Civil Engineering Reference
In-Depth Information
Now, from Eqs (16.28) and (16.30)
θ
B
−
10
v
1
2
×
2
.
5
EI
10
3
M
AB
=−
−
12
.
5
(i)
2
θ
B
−
10
v
1
2
×
2
.
5
EI
10
3
M
BA
=−
+
12
.
5
(ii)
In Eqs (i) and (ii) we are assuming that the displacement,
v
1
,
is to the right.
Furthermore
2
EI
20
(2
θ
B
+
M
BC
=−
θ
C
)
−
100
(iii)
2
EI
20
(2
θ
C
+
θ
B
)
M
CB
=−
+
100
(iv)
2
θ
C
+
10
v
1
2
×
2
.
5
EI
10
3
M
CD
=−
θ
D
+
(v)
2
θ
D
+
10
v
1
2
×
2
.
5
EI
10
3
M
DC
=−
θ
C
+
(vi)
From the equilibrium of the member end moments at the joints
M
BA
+
M
BC
=
0
M
CB
+
M
CD
−
54
=
0
M
DC
=
0
Substituting in the equilibrium equations for
M
BA
,
M
BC
, etc., from Eqs (i)-(vi) we
obtain
1
.
25
EI
θ
B
+
0
.
1
EI
θ
C
−
0
.
15
EI
v
1
+
87
.
5
=
0
(vii)
1
.
2
EI
θ
C
+
0
.
1
EI
θ
B
+
0
.
5
EI
θ
C
+
0
.
15
EI
v
1
−
46
=
0
(viii)
EI
θ
D
+
0
.
5
EI
θ
C
+
0
.
15
EI
v
1
=
0
(ix)
Since there are four unknown displacements we require a further equation for a solu-
tion. This may be obtained by considering the overall horizontal equilibrium of the
frame. Thus
S
AB
+
S
DC
−
10
=
0
in which, from Eq. (16.29)
6
×
2
.
5
EI
10
2
12
×
2
.
5
EI
10
3
S
AB
=
θ
B
−
v
1
+
5
where the last term on the right-hand side is
S
AB
(
=+
5 kN), the contribution of the
10 kN horizontal load to
S
AB
. Also
6
×
2
.
5
EI
10
2
12
×
2
.
5
EI
10
3
S
DC
=
(
θ
D
+
θ
C
)
−
v
1
Hence, substituting for
S
AB
and
S
DC
in the equilibrium equations, we have
EI
θ
B
+
EI
θ
D
+
EI
θ
C
−
0
.
4
EI
v
1
−
33
.
3
=
0
(x)
