Civil Engineering Reference
In-Depth Information
the FEMs
M
DE
(
20 kNm) and
M
DC
, balance the beam at D, carry over to C and
then leave the beam at D balanced and pinned; again the stiffness coefficient,
K
DC
,is
modified to allow for this (Case 2).
=−
The FEMs are again calculated using the appropriate results from Table 16.6. Thus
12
×
14
M
AB
=−
M
BA
=−
=−
21 kNm
8
8
2
4
2
7
×
4
×
7
×
8
×
M
BC
=−
M
CB
=−
−
=−
18
.
67 kNm
12
2
12
2
×
22
12
M
CD
=−
M
DC
=−
=−
22 kNm
12
M
DE
=−
5
×
4
=−
20 kNm
The DFs are calculated as follows
K
BA
K
BA
+
3
EI
/
14
3
EI
/
14
DF
BA
=
K
BC
=
4
EI
/
12
=
0
.
39
+
Hence
DF
BC
=
1
−
0
.
39
=
0
.
61
K
CB
K
CB
+
4
EI
/
12
4
EI
/
12
DF
CB
=
K
CD
=
3
EI
/
12
=
0
.
57
+
Hence
DF
CD
=
1
−
0
.
57
=
0
.
43
The solution is completed as follows:
A
C
D
E
B
DFs
1
0.39
0.61
0.57
0.43
1.0
0
-
FEMs
21.0
21.0
18.67
18.67
22.0
22.0
20.0
0
Balance A and D
21.0
2.0
Carry over
10.5
1.0
Balance
5.0
7.83
2.47
1.86
Carry over
1.24
3.92
Balance
0.48
0.76
2.23
1.69
Carry over
1.12
0.38
Balance
0.44
0.68
0.22
0.16
Carry over
0.11
0.34
Balance
0.04
0.07
0.19
0.15
Final moments
0
25.54
25.54
19.14
19.14
20.0
20.0
0
The support reactions are now calculated in an identical manner to that in Ex. 16.17
and are
R
A
=
4
.
18 kN
R
B
=
15
.
35 kN
R
C
=
17
.
4kN
R
D
=
16
.
07 kN
