Civil Engineering Reference
In-Depth Information
M
A
B
θ
A
θ
B
δ
1
C
(a)
M
A
B
θ
C
δ
2
C
(b)
W
A
B
C
F
IGURE
15.25
Model
analysis of a fixed beam
M
A
M
B
(c)
Thus
θ
A(b)
=
θ
B
It follows that the rotation at A due to
M
B
atBis
M
B
M
θ
B
θ
A(c),1
=
(i)
where (b) and (c) refer to (b) and (c) in Fig. 15.25.
Also, the rotation at A due to a unit load at C is equal to the deflection at C due to a
unit moment at A. Therefore
θ
A(c),2
W
=
δ
1
M
or
W
M
δ
1
θ
A(c),2
=
(ii)
in which
θ
A(c),2
is the rotation at A due to
W
at C. Finally the rotation at A due to
M
A
at A is, from Fig. 15.25(a) and (c)
M
A
M
θ
A
θ
A(c),3
=
(iii)
The total rotation at A produced by
M
A
at A,
W
at C and
M
B
at B is, from Eqs (i), (ii)
and (iii)
M
B
M
θ
B
+
W
M
δ
1
+
M
A
M
θ
A
=
θ
A(c),1
+
θ
A(c),2
+
θ
A(c),3
=
0
(iv)
