displacement method of analysis
TRANSCRIPT
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Slope Deflection Method for the
Analysis of Indeterminate Structures
ByProf. Dr. Wail Nourildean Al-Rifaie
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All structures must satisfy:
oad-displacement relationship
!"uili#rium of forces
$ompati#ility of displacements
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%sin& the principle of superposition #yconsiderin& separately the moments
de'eloped at each support of a typical
prismatic #eam (AB) sho*n in +i&. ,(a) of acontinuous #eamdue to each of the
displacements and the
applied loads. Assume cloc*ise moments
are /i'e.
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,. Assume ends A and B are fi0ed i. e. the
rotations . 1his means that *e
ha'e to apply countercloc*ise moment at
end A and cloc*ise moment at end B
due to the applied loads to cause 2ero
rotation at each of ends A and B. 1a#le (,)
&i'es for different loadin& conditions.
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1a#le (,)
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3. Release end A a&ainst rotation at end A (rotates to
its final position ) #y applyin& cloc*ise
moment *hile far end node B is held fi0ed as
sho*n in +i&. ,.
4. No* the cloc*ise moment - rotation
relationship is:
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5. 1he carry o'er moment at end B is:
6. In a similar manner if end B of the #eam rotates to
its final position *hile end A is held fi0ed. 1he
cloc*ise moment 7 rotation relationship
is:
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8. 1he carry o'er moment at end A is:
9. If node B is displaced relati'e to as sho*n in +i&.
(,) so that the cord of the mem#er rotates
cloc*ise i. e. positi'e displacement and yet #oth
ends do not rotate then e"ual #ut anticloc*ise
moments are de'eloped in the mem#er as sho*n
in the fi&ure.
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Slope-Deflection !"uation
oad-displacement relationship
If the end moments due to each displacement and
the loadin& are added toðer the resultant
moments at the ends may then #e *ritten as:
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+or prismatic #eam element e"uation (,) may #e
*ritten as:
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1he slope deflection e"uations (, or 3) relatethe unno*n moments applied to the nodes
to the displacements of the nodes for
any span of the structure. 1o summari2e application of the slope-
deflection e"uations consider thecontinuous #eam sho*n in +i&. (3) *hich has
four de&rees of freedom. No* e"uation (3) can #e applied to each of
the three spans.
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+i&. (3)
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+rom +i&.(3):
!"uili#rium conditions $ompati#ility conditions
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1hese e"uations *ould in'ol'e the four unno*n
rotations . Sol'in& for these four unno*n rotations. It may #e
noted that there is no relati'e deflection #et*een the
supports so that 1he 'alues of the o#tainedrotations may then #e su#stituted in to the slopedeflection e"uations to determine the internal
moments at the ends of each mem#er. If any of the results are ne&ati'e they indicate
countercloc*ise rotation.
DCBA ,,,
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!0ample (,)
Solution
Dra* the shear and moment dia&rams for the #eam
sho*n in +i&.(4). !I is constant.,. %sin& the formulas for the ta#ulated in 1a#le
(,) for the &i'en loadin&s:
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+i&. (4)
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3. 1here are t*o slopes at B and $ i. e.
are unno*ns. Since end A is fi0ed Also
since the supports do not settle nor are they
displaced up or do*n
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No* #y applyin& the e"uili#rium conditions:
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Su#stitutin& the computed 'alues in to momente"uations (a) (#) (c) and (d):
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+i&. (5) Shearin& +orce = Bendin& Moment Di&rams
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!0ample (3)
Determine the internal moments at the supports of thebeam shown in Fig. (5). The support at B is displaced
(settles) 12 mm.
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Solution
1. Two spans must be considered. FEs are determined
using Table (1).
4
4
101.0
12
012.00
10667.618
0012.0
=
=
=
=
x
x
BC
AB
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3. %sin& e"uation 3:
)(0)101.032()12
2(
)()101.032()12
2(
)()10667.632()18
2(
)()10667.63()18
2(
4
4
4
4
lxxEI
M
kxxEI
M
jxxEIM
ixxEI
M
BCBCCB
CBBCBC
BABBA
BABAB
=++=
++=
=
=
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!. E"uilibrium condition#
00 == CB MandM
)(0)101.032()122(
)(0)101.032()12
2()10667.632()
18
2(
)()10667.63()18
2(
4
44
4
pxxEI
nxxEI
xxEI
mxxEI
M
BCBC
CBBCBAB
BABAB
=++
=+++
=
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$n order to obtained the rotations e"uations
(n) % (p) ma& then be sol'ed simultaneousl& it ma& be
noted that since is fi*ed support. Thus
CB and
0=A
.1047647.2.1065294.4 44
radxandradxCB
=+=
+ubstituting these 'alues into e"uations (i to l) &ields
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!0ample (4)
$f end in e*ample (1) is simpl& supported and b&appl&ing the compatibilit& condition their will be three
un,nown rotations
-ow
),,(CBA
)(62.1)2()6.3
2(
)(62.1)2()6.3
2(
)(5.4)2()4.2
2(
)(5.4)2()4.2
2(
dEI
M
cEI
M
bEI
M
aEIM
BCBCCB
CBBCBC
BABBA
BAABAB
++=
+=
+=
+=
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Applyin& the e"uili#rium conditions:
)3(062.1)2()6.3
2(
)2(062.1)2()6.3
2(5.4)2()
4.2
2(
)1(05.4)2()4.2
2(
=++=
=+++
=+=
BCBCCB
CBBCBAB
BAABAB
EIM
EIEI
EIM
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By sol'in& e"uations (, 3 = 4) for and
su#stitute the 'alues into e"uations (a # c d):
CBA ,,
0
.158.4
.158.40
=
=
+=
=
CB
BC
BA
AB
M
iseanticlockwmkNM
clockwisemkNMM
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+hearing force and bending moment diagrams are shown
in the following figure.
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E*ample ()
Determine the moments at each /oint of the frame shown in
Fig.(0). E$ is constant.
Fig. (0)
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MkNFEM
MkNFEM
CB
BC
.8096
)8)(24(5)(
.8096
)8)(24(5)(
2
2
+=+=
==
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Because ends and D are fi*ed supports.
and since no sideswa& will occur.
0=== CDBCAB
CCCDDC
CCCDCD
BCBCBCCB
CBCBBCBC
BBABBA
BBABAB
EIl
EIM
EIEI
M
EIEI
EI
M
EIEIEI
M
EIEI
M
EI
EI
M
1667.0)()2
(
3334.0)2()12
2(
8025.05.080)2()8
2
(
8025.05.080)2()8
2(
3334.0)2()12
2(
1667.0)()12
2
(
==
==
++=++=
+=+=
=+=
=+=
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E"uilibrium 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
EI
OrEIEIEI
MM
EIEI
Or
EIEIEI
MM
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+ol'ing simultaneousl& &ields
EIand
EI CB
1.1371.137==
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Therefore
seanticlokwimkNM
iseanticlockwmkNM
clockwisemkNM
iseanticlockwmkNM
clockwisemkNM
clockwisemkNM
DC
CD
CB
BC
BA
AB
.9.22
.7.45
.7.45
.7.45
.7.45
.9.22
=
=
=
=
=
=
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Fig.()
The bending moment diagram is shown in Fig.().
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E*ample (5)
Determine the internal moments at each of the frame
shown in Fig.().
+olution
Fig.()
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1. Fi*ed end moments#
12)(,
12)(:
12
)5.1()(,
12
)5.1()(:
22
22
wLFEM
wLFEMBCSpan
LwFEM
LwFEMABSpan
CBBC
BAAB
+==
+==
8)(,8)(:
Pl
FEM
Pl
FEMBDSpan BDDB +==
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2. 3oint 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
nPL
L
EIM
mPL
L
EIM
lwL
L
EIM
kwL
L
EIM
jLw
L
EIM
iLwL
EIM
BDBDDB
DBBDBD
BCBCCB
CBBCBC
ABABBA
BAABAB
+=
++=
++=
+=
++=
+=
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!. E"uilibrium conditions#
+ol'ing e"uations (12!) simultaneousl& &ields
)4(8
)()2
(
)3(012
)2()2
(
)2(08
)2()2
(
12)2()
2(
12
)5.1()2()
5.1
2(
0
)1(012
)5.1(
)2()5.1
2
(
2
22
2
PL
L
EIM
wL
L
EIM
PL
L
EI
wL
L
EILw
L
EI
M
Lw
L
EI
M
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB
=
=++=
=++
+++++
=
=+=
)4(8
)()2
(
)3(012
)2()2
(
)2(08
)2()2
(
12)2()
2(
12
)5.1()2()
5.1
2(
0
)1(012
)5.1(
)2()5.1
2
(
2
22
2
PL
L
EIM
wL
L
EIM
PL
L
EI
wL
L
EILw
L
EI
M
Lw
L
EI
M
BBDDC
BCBCCB
DBBD
CBBCABAB
B
BAABAB
=
=++=
=++
+++++
=
=+=
CBA ,,+ol'ing e"uations (12!) simultaneousl& &ields
+ubstituting the rotation 'alues into e"uations (i to n) to
determine the /oint moments.
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E*ample (4)
Determine the /oint internal moments of the frame shown in
Fig.(1) both ends and D are fi*ed.ssume 5.1)(1)()( === CDBCAB
L
EIand
L
EI
L
EI
Fig.(1)
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+olution
1. Fi*ed 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
+=+=
==
=+=
==
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$t is assumed that the a*ial deformation is neglected so that== OO CCBB as shown in the following figure.
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$t ma& be noted that
2. 3oint 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
lMkM
jL
EIM
iM
ABC
ABCDC
ABC
ABCCD
BCCB
CBBC
ABBABBA
ABBAB
=
=
=
=
++=
+=
+=
=
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!. E"uilibrium conditions#
)2(96.1275.65
0:int
)1(96.434
0:int
=+
=+
=+
=+
ABCB
CDCB
ABCB
BCBA
Or
MMCo
Or
MMBo
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+ince a hori6ontal displacemententire frame in the *7direction. This &ields
occurs the summing forces on the
)3(667.1075.425.2
0
6.34.53
1010
6.3
4.53
10
:
010:0
=++
=+
++
++
+=
++=
=+=+
ABCB
DCCDBAAB
DCCDD
BAABA
DA!
Or
MMMM
MM"
and
MM"
w#ic#In
""F
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+ol'ing e"uations (1 2 !) &ields9194.1,565.0,8208.2 +==+= ABCB
B& substituting these 'alues into moment e"uations (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
=
=
+=
=
+=
=
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Bending moment diagram is plotted in the following figure.