Post on 30-Apr-2018
Slope Deflection Method for the Analysis of Indeterminate Structures
ByProf. Dr. Wail Nourildean Al-Rifaie
All structures must satisfy: Load-displacement relationship Equilibrium of forces Compatibility of displacements
Using the principle of superposition by considering separately the moments developed at each support of a typical prismatic beam (AB) shown in Fig. 1(a) of a continuous beam, due to each of the displacements , , and the applied loads. Assume clockwise moments are +ive.
1. Assume ends A and B are fixed, i., e., the rotations . This means that we have to apply counterclockwise moment at end A and clockwise moment at end B
due to the applied loads to cause zero rotation at each of ends A and B. Table (1) gives for different loading conditions.
Table (1)
2. Release end A against rotation at end A (rotates to its final position ) by applying clockwise moment while far end node B is held fixed as shown in Fig. 1.
3. Now, the clockwise moment - rotation relationship is:
4. The carry over moment at end B is:
5. In a similar manner, if end B of the beam rotates to its final position , while end A is held fixed. The clockwise moment – rotation relationship is:
6. The carry over moment at end A is:
7. If node B is displaced relative to as shown in Fig. (1), so that the cord of the member rotates clockwise i., e., positive displacement and yet both ends do not rotate, then equal but anticlockwise moments are developed in the member as shown in the figure.
Slope-Deflection Equation
Load-displacement relationship If the end moments due to each displacement and
the loading are added together, the resultant moments at the ends may then be written as:
For prismatic beam element, equation (1) may be written as:
The slope deflection equations (1 or 2) relate the unknown moments applied to the nodes
to the displacements of the nodes for any span of the structure.
To summarize application of the slope-deflection equations, consider the
continuous beam shown in Fig. (2) which has four degrees of freedom.
Now equation (2) can be applied to each of the three spans.
Fig. (2)
From Fig.(2):Equilibrium conditions Compatibility conditions
These equations would involve the four unknown rotations .
Solving for these four unknown rotations. It may be noted that there is no relative deflection between the supports, so that The values of the obtained rotations may then be substituted in to the slope deflection equations to determine the internal moments at the ends of each member.
If any of the results are negative, they indicate counterclockwise rotation.
DCBA ,,,
Example (1)Solution
Draw the shear and moment diagrams for the beam shown in Fig.(3). EI is constant.
1. Using the formulas for the tabulated in Table (1) for the given loadings:
Fig. (3)
2. There are two slopes at B and C, i., e., are unknowns. Since end A is fixed, Also,
since the supports do not settle, nor are they displaced up or down
Now, by applying the equilibrium conditions:
Substituting the computed values in to moment equations (a), (b), (c), and (d):
By considering the values of support moments and the applied loads, the support reactions may then be determined:
RA = 8.3625 kN RB = 10.2042 kN
RC = 1.8333 kN
Shearing force and bending moment digrams are shown in Fig. (4).
Fig. (4) Shearing Force & Bending Moment Digrams
Example (2)
Determine the internal moments at the supports of the beam shown in Fig. (5). The support at B is displaced (settles) 12 mm.
Solution1. Two spans must be considered. FEMs are determined
using Table (1).
4
4
101.012012.00
10667.618
0012.0
x
x
BC
AB
2. Using equation 2:
)(0)101.032()122(
)()101.032()122(
)()10667.632()182(
)()10667.63()182(
4
4
4
4
lxxEIM
kxxEIM
jxxEIM
ixxEIM
BCBCCB
CBBCBC
BABBA
BABAB
3. Equilibrium condition:
00 CB MandM
)(0)101.032()122(
)(0)101.032()122()10667.632()
182(
)()10667.63()182(
4
44
4
pxxEI
nxxEIxxEI
mxxEIM
BCBC
CBBCBAB
BABAB
In order to obtained the rotations equations (n) & (p) may then be solved simultaneously, it may be noted that since A is fixed support. Thus,
CB and
0A
.1047647.2.1065294.4 44 radxandradx CB
Substituting these values into equations (i to l) yields
Example (3)
If end A in example (1) is simply supported, and by applying the compatibility condition, their will be three unknown rotations,
Now,),,( CBA
)(62.1)2()6.3
2(
)(62.1)2()6.3
2(
)(5.4)2()4.2
2(
)(5.4)2()4.2
2(
dEIM
cEIM
bEIM
aEIM
BCBCCB
CBBCBC
BABBA
BAABAB
Applying the equilibrium conditions:
)3(062.1)2()6.3
2(
)2(062.1)2()6.3
2(5.4)2()4.2
2(
)1(05.4)2()4.2
2(
BCBCCB
CBBCBAB
BAABAB
EIM
EIEI
EIM
By solving equations (1, 2 & 3) for and substitute the values into equations (a, b, c, d):
CBA ,,
0.158.4.158.4
0
CB
BC
BA
AB
MiseanticlockwmkNM
clockwisemkNMM
Shearing force and bending moment diagrams are shown in the following figure.
Example (4)Determine the moments at each joint of the frame shown in Fig.(7). EI is constant.
Fig. (7)
MkNFEM
MkNFEM
CB
BC
.8096)8)(24(5
)(
.8096)8)(24(5
)(
2
2
Because ends A and D are fixed supports.
and since no sidesway will occur.
0 CDBCAB
CCCDDC
CCCDCD
BCBCBCCB
CBCBBCBC
BBABBA
BBABAB
EIlEIM
EIEIM
EIEIEIM
EIEIEIM
EIEIM
EIEIM
1667.0)()2(
3334.0)2()122(
8025.05.080)2()82(
8025.05.080)2()82(
3334.0)2()122(
1667.0)()122(
Equilibrium conditions:
)2(08025.08334.0
0803334.025.05.0
0
)1(08025.08334.0
08025.05.03334.0
0
BC
CBC
CDCB
CB
CBB
BCBA
EIOr
EIEIEI
MM
EIEIOr
EIEIEI
MM
Solving simultaneously yields
EIand
EI CB1.1371.137
Therefore,
seanticlokwimkNMiseanticlockwmkNM
clockwisemkNMiseanticlockwmkNM
clockwisemkNMclockwisemkNM
DC
CD
CB
BC
BA
AB
.9.22
.7.45.7.45.7.45.7.45.9.22
Fig.(8)
The bending moment diagram is shown in Fig.(8).
Example (5)Determine the internal moments at each of the frame shown in Fig.(9). Solution
Fig.(9)
1. Fixed end moments:
12)(,
12)(:
12)5.1()(,
12)5.1()(:
22
22
wLFEMwLFEMBCSpan
LwFEMLwFEMABSpan
CBBC
BAAB
8)(,
8)(:
PlFEM
PlFEMBDSpan BDDB
2. Joint moments:
)(8
)2()2(
)(8
)2()2(
)(12
)2()2(
)(12
)2()2(
)(12)5.1()2()
5.12(
)(12)5.1()2()
5.12(
2
2
2
2
nPLLEIM
mPLLEIM
lwLLEIM
kwLLEIM
jLwL
EIM
iLwL
EIM
BDBDDB
DBBDBD
BCBCCB
CBBCBC
ABABBA
BAABAB
3. Equilibrium conditions:
Solving equations (1,2,3) simultaneously yields
)4(8
)()2(
)3(012
)2()2(
)2(08
)2()2(
12)2()2(
12)5.1()2()
5.12(
0
)1(012)5.1(
)2()5.12(
2
22
2
PLLEIM
wLLEIM
PLLEI
wLLEILw
LEI
M
LwL
EIM
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB
)4(8
)()2(
)3(012
)2()2(
)2(08
)2()2(
12)2()2(
12)5.1()2()
5.12(
0
)1(012)5.1(
)2()5.12(
2
22
2
PLLEIM
wLLEIM
PLLEI
wLLEILw
LEI
M
LwL
EIM
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB
CBA ,,Solving equations (1,2,3) simultaneously yieldsSubstituting the rotation values into equations (i to n) to
determine the joint moments.
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 LEIand
LEI
LEI
Fig.(10)
Solution1. Fixed end moments:
mkNFEM
mkNFEMBCSpan
mkNFEM
mkNFEMABSpan
CB
BC
BA
AB
.96.1212)2.7(3)(
.96.1212)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
It is assumed that the axial deformation is neglected so that OO CCBB as shown in the following figure.
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()5.12(
)(4)3(
nM
mM
lMkM
jL
EIM
iM
ABC
ABCDC
ABC
ABCCD
BCCB
CBBC
ABBABBA
ABBAB
3. Equilibrium conditions:
)2(96.1275.65
0:int
)1(96.434
0:int
ABCB
CDCB
ABCB
BCBA
Or
MMCJo
Or
MMBJo
Since a horizontal displacement entire frame in the x-direction. This yields
occurs, the summing forces on the
)3(667.1075.425.2
06.34.53
1010
6.3
4.5310:
010:0
ABCB
DCCDBAAB
DCCDD
BAABA
DAX
Or
MMMM
MMH
and
MMH
whichIn
HHF
Solving equations (1, 2, 3) yields
9194.1,565.0,8208.2 ABCB
By substituting these values into moment equations (i to n):
mkNMmkNMmNkM
mkNMmkNMmkNM
DC
CD
CB
BC
BA
AB
.8035.13.6509.14..6509.14
.8833.7.8833.7.9374.6
Bending moment diagram is plotted in the following figure.