Slope Deflection Method for Analysis of Indeterminate Beams

1. Subject Name: Theory of Structures
Topic Name: Example onAnalysis of
Indeterminate Beams by Slope
Deflection Method
Lecture No: 16
Dr.Omprakash Netula
Professor & HOD
Department of Civil Engineering
9/28/2017 Lecture Number, Unit Number 1

2. Example
Determine the internal moments at each of the frame shown in Fig.(9).
Solution
Fig.(9)

3. 1. Fixed end moments:
12
)(,
12
)(:
12
)5.1(
)(,
12
)5.1(
)(:
22
22
wL
FEM
wL
FEMBCSpan
Lw
FEM
Lw
FEMABSpan
CBBC
BAAB


8
)(,
8
)(:
Pl
FEM
Pl
FEMBDSpan BDDB 

4. 2. Joint moments:
)(
8
)2()
2
(
)(
8
)2()
2
(
)(
12
)2()
2
(
)(
12
)2()
2
(
)(
12
)5.1(
)2()
5.1
2
(
)(
12
)5.1(
)2()
5.1
2
(
2
2
2
2
n
PL
L
EI
M
m
PL
L
EI
M
l
wL
L
EI
M
k
wL
L
EI
M
j
Lw
L
EI
M
i
Lw
L
EI
M
BDBDDB
DBBDBD
BCBCCB
CBBCBC
ABABBA
BAABAB













5. 3. Equilibrium conditions:
Solving equations (1,2,3) simultaneously yields
)4(
8
)()
2
(
)3(0
12
)2()
2
(
)2(0
8
)2()
2
(
12
)2()
2
(
12
)5.1(
)2()
5.1
2
(
0
)1(0
12
)5.1(
)2()
5.1
2
(
2
22
2
PL
L
EI
M
wL
L
EI
M
PL
L
EI
wL
L
EILw
L
EI
M
Lw
L
EI
M
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB













)4(
8
)()
2
(
)3(0
12
)2()
2
(
)2(0
8
)2()
2
(
12
)2()
2
(
12
)5.1(
)2()
5.1
2
(
0
)1(0
12
)5.1(
)2()
5.1
2
(
2
22
2
PL
L
EI
M
wL
L
EI
M
PL
L
EI
wL
L
EILw
L
EI
M
Lw
L
EI
M
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB













CBA  ,,Solving equations (1,2,3) simultaneously yields
Substituting the rotation values into equations (i to n) to
determine the joint moments.

6. Example (6)
Determine the joint internal moments of the frame shown in Fig.(10),
both ends A and D are fixed.
Assume 5.1)(1)()(  CDBCAB
L
EI
and
L
EI
L
EI
Fig.(10)

7. Solution
1. Fixed end moments:
mkNFEM
mkNFEMBCSpan
mkNFEM
mkNFEMABSpan
CB
BC
BA
AB
.96.12
12
)2.7(3
)(
.96.12
12
)2.7(3
)(:
.0.8
)4.5(
)6.3)(8.1(10
)(
.0.4
)4.5(
)8.1)(6.3(10
)(:
2
2
2
2
2
2





8. It is assumed that the axial deformation is neglected so that
 OO CCBB as shown in the following figure.

9. It may be noted that
2. Joint moments:
05.1  DABCABCD and 
)(75.65.1
)5.4(5.1
)(75.63
)5.42(
)(96.12)2(5.1
)(96.12)2(
)(8)2()
5.1
2
(
)(4)3(
n
M
m
M
lM
kM
j
L
EI
M
iM
ABC
ABCDC
ABC
ABCCD
BCCB
CBBC
ABBABBA
ABBAB

















10. 3. Equilibrium conditions:
)2(96.1275.65
0:int
)1(96.434
0:int






ABCB
CDCB
ABCB
BCBA
Or
MMCJo
Or
MMBJo



11. Since a horizontal displacement
entire frame in the x-direction. This yields
occurs, the summing forces on the
)3(667.1075.425.2
0
6.34.53
10
10
6.3
4.53
10
:
010:0










 
ABCB
DCCDBAAB
DCCD
D
BAAB
A
DAX
Or
MMMM
MM
H
and
MM
H
whichIn
HHF


12. Solving equations (1, 2, 3) yields
9194.1,565.0,8208.2  ABCB 
By substituting these values into moment equations (i to n):
mkNM
mkNM
mNkM
mkNM
mkNM
mkNM
DC
CD
CB
BC
BA
AB
.8035.13
.6509.14
..6509.14
.8833.7
.8833.7
.9374.6







13. Bending moment diagram is plotted in the following figure.