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Slope-Deflection
Method: Frames 1) Without Side-sway
2) With Side-sway
Slope-Deflection Method:
Frames Without Frames Without Frames Without Frames Without SideSideSideSide----sssswaywaywayway� In frames axial deformations are much smaller than the
bending deformations and are neglected in the analysis.
� With this assumption some frames will not side-sway
(the frames will not be displaced to the right or left).
� Frames do not side-sway if:
� They are restrained against side-sway.
� The frame geometry and loading is symmetrical.
� The analysis of such rigid frames by slope deflection
equation essentially follows the same steps as that of
continuous beams.
1. Degrees of freedom
� The frame is kinematically indeterminate to first
degree. Only one joint rotation ϕϕϕϕb is unknown.
Slope-Deflection Method: Frames Without Side-sway,
Example 1
a
2
EI = const.
2 2
b
c
F1 = 5 kN
d
F2 = 10 kN
4
Slope-Deflection Method: Frames Without Side-sway,
Example 1
2. Fixed end moments are calculated referring to the table.
2
2
1
8
1
8
0
bd bd
db bd
cb bc
M F l
M F l
M M
= − ⋅ ⋅
= + ⋅ ⋅
= =
a
2
EI = const.
2 2
b
c
F1 = 5 kN
d
F2 = 10 kN
4
Slope-Deflection Method: Frames Without Side-sway,
Example 1
a
2
EI = const.
2 2
b
c
F1 = 5 kN
d
F2 = 10 kN
4
3. Express internal end moments
by slope-deflection equations.
( )
( )
( )
( )
22
22
22
22
cbcb c bcb
bcbc b ccb
bdbd b dbd
dbdb d bbd
EIM M
l
EIM M
l
EIM M
l
EIM M
l
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
= + ⋅ +
= + ⋅ +
= + ⋅ +
= + ⋅ +
Slope-Deflection Method: Frames Without Side-sway,
Example 1
a
2
EI = const.
2 2
b
c
F1 = 5 kN
d
F2 = 10 kN
4
4. Equilibrium equations
� End moments are expressed in
terms of unknown rotation ϕb.
Now, the required equation to
solve for the rotation ϕb is the
moment equilibrium equation
at rigid joint b.
b
211 ⋅= FM F
b bdM
bcM
10 : 0Fb bc bd bM M M M= + + =∑
5. End moments
� After evaluating ϕb, substitute it
to evaluate beam end moments.
Slope-Deflection Method: Frames Without Side-sway,
Example 16. Shear forces
� Now, reactions at supports are evaluated using equilibrium
equations. Shear forces are equal to plus/minus this reactions.
( )
( )
1
1
cb cb cb bcbc
bc bc cb bcbc
V R M Ml
V R M Ml
= = − −
= − = − + +
2
2
1
2
1
2
bdbd bd bd db
bd
bddb db bd db
bd
lV R F M M
l
lV R F M M
l
= = ⋅ − −
= − = − ⋅ + +
4c b
Mcb Mbc
Rcb=Vcb Rbc=-Vbc
2b d
Mbd Mbd
Rbd=Vbd Rdb=-Vdb
F2 = 10 kN
2
Slope-Deflection Method: Frames Without Side-sway,
Example 1
7. Normal forces
� The required equations to
solve for normal forces are
the force equilibrium
equations at rigid joint b.
0 : 0xb bd bcF N V= − =∑
10 : 0zb bc bdF N V F= + + =∑b
1F Vbd
Nbd
Vbc
Nbc
a
2
EI = const.
2 2
b
c
F1 = 5 kN
d
F2 = 10 kN
4
Slope-Deflection Method: Frames Without Side-sway,
Example 1
8. Draw normal and shear force diagrams.
Ncb
= N
bc
Nbd = Ndb0
-F1
Vcb
= V
bc
Vbd
Vdb
Slope-Deflection Method: Frames Without Side-sway,
Example 1
8. Draw bending moment diagram
Mcb
-Mbc
-F1·2 Mbd -Mbd
MF
Slope-Deflection Method:
Frames With Frames With Frames With Frames With SideSideSideSide----sssswaywaywayway� In frames axial deformations are much smaller than the
bending deformations and are neglected in the analysis.
� The frames will side-sway (the frames will be displaced
to the right or left or will be displaced up or down).
� Frames side-sway if:
� They are not restrained against side-sway.
� The frame geometry and loading is not symmetrical.
1. Degrees of freedom
� Frame is kinematically
indeterminate to third
degrees (three unknown
displacements - two
rotations ϕϕϕϕb, ϕϕϕϕc and
one translation ub = uc).
Slope-Deflection Method: Frames With Side-sway,
Example 2
a
2I
4
b c
F=
12
kN
d
3
I
I
3
Slope-Deflection Method: Frames With Side-sway,
Example 2
2. Fixed end moments are calculated referring to the table.
1
8
1
8
0
0
ab ab
ba ab
bc cb
cd dc
M F l
M F l
M M
M M
= − ⋅ ⋅
= + ⋅ ⋅
= =
= =
a
2I
4
b c
F=
12
kN
d
3
I
I
3
Slope-Deflection Method: Frames With Side-sway,
Example 2
3. Express internal end moments
by slope-deflection equations.
( )abbaab
ababab l
EIMM ψϕϕ ⋅−+⋅+= 32
2
bar rotation
tan
tan
b c
ab abab
cd cdcd
u u
l
l
ψ ψ
ψ ψ
= = ∆
∆≈ =
∆≈ = a
b c
d
∆∆∆∆ ∆∆∆∆
ψψψψab
ψψψψcd
chords of
the elastic curve
Slope-Deflection Method:
Frames With Side-sway,
Example 23. Express internal end moments
by slope-deflection equations.
( )
( )
( )
( )
( )
( )
22 3
22 3
22 3
22 3
22 3
22 3
ababab a b ab
ab
abbaba b a ab
ab
bcbcbc b c bc
bc
bccbcb c b bc
bc
cdcdcd c d cd
cd
cddcdc d c cd
cd
EIM M
l
EIM M
l
EIM M
l
EIM M
l
EIM M
l
EIM M
l
ϕ ϕ ψ
ϕ ϕ ψ
ϕ ϕ ψ
ϕ ϕ ψ
ϕ ϕ ψ
φ φ ψ
= + ⋅ + − ⋅
= + ⋅ + − ⋅
= + ⋅ + − ⋅
= + ⋅ + − ⋅
= + ⋅ + − ⋅
= + ⋅ + − ⋅
a
2I
4
b c
F=
12
kN
d
3
I
I
3
Slope-Deflection Method: Frames With Side-sway,
Example 2
4. Equilibrium equations
� End moments are expressed in terms of unknown rotations ϕb, ϕc
and translation ∆. Now, the required equation to solve are the moment equilibrium equations at rigid joints b and c and then so-called sideway equation.
0 : 0b ba bcM M M= + =∑
a
2I
4
b c
F=
12
kN
d
3
I
I
3
0 : 0c cb cdM M M= + =∑
Slope-Deflection Method: Frames With Side-sway,
Example 2
4. Equilibrium equations
� Sideway equation Sideway equation Sideway equation Sideway equation (the horizontal equilibrium)
0 : 0x ba cdF V V= + =∑
b c
baV cdV
( )
1
2
1
abba ab ba
ab
cd cd dccd
lV F M M
l
V M Ml
= − ⋅ + +
= − −
a
2I
4
b c
F=
12
kN
d
3
I
I
3
Part of the frame containing
the joints with the same
translation ∆ “cut off”.
Slope-Deflection Method: Frames With Side-sway,
Example 2
4. Solving equilibrium equations
EI
EI
EI
EIEIEI
EIEIEI
EIEIEI
d
c
cb
cb
cb
99,14
88,4
76,2
6556,067,033,0
067,033,25,0
933,05,033,2
=∆
=
−=
=∆⋅⋅+⋅⋅−⋅⋅−=∆⋅⋅−⋅⋅+⋅⋅
−=∆⋅⋅−⋅⋅+⋅⋅
ϕ
ϕ
ϕϕϕϕϕϕ
Slope-Deflection Method: Frames With Side-sway,
Example 2
a
2I
4
b c
F=
12
kN
d
3
I
I
3
kNm75,63
2
3
2
kNm50,33
2
3
4
kNm50,32
kNm32,02
kNm32,033
49
kNm84,1533
29
−=∆⋅⋅−⋅⋅=
−=∆⋅⋅−⋅⋅=
=⋅+⋅=
−=⋅+⋅=
=∆⋅−⋅+=
−=∆⋅−⋅+−=
EIEIM
EIEIM
EIEIM
EIEIM
EIEIM
EIEIM
cdc
ccd
bccb
cbbc
bba
bab
ϕ
ϕ
ϕϕ
ϕϕ
ϕ
ϕ
5. End moments
� After evaluating ϕb, ϕc, ∆, substitute them to evaluate end moments.
Slope-Deflection Method: Frames With Side-sway,
Example 2
6. Shear forces
( )
( )
( )
( ) kN79,050,332,04
1
1
kN79,050,332,04
1
1
−=+−−=
++−=−=
−=−+=
−−==
cb
cbbcbc
cbcb
bc
cbbcbc
bcbc
V
MMl
RV
V
MMl
RV
( )
( ) kN41,332,084,153126
1
2
1
kN59,832,084,153126
1
2
1
−=+−⋅−=
++⋅−=−=
=−+⋅=
−−⋅==
ba
baabab
abbaba
ab
baabab
ababab
V
MMl
Fl
RV
V
MMl
Fl
RV
4b c
Mbc Mcb
Rbc=Vbc Rcb=-Vcb
3a b
MabMba
Rab=Vab Rba=-Vba
F = 12 kN
3 3c d
Mcd Mdc
Rcd=Vcd Rdc=-Vdc
( )
( )
( )
( ) kN41,375,650,33
1
1
kN41,375,650,33
1
1
=−−−=
++−=−=
=++=
−−==
dc
dccdcd
dcdc
cd
dccdcd
cdcd
V
MMl
RV
V
MMl
RV
Slope-Deflection Method: Frames With Side-sway,
Example 2
7. Normal forces
� The required equations to solve for normal forces are the
force equilibrium equations at rigid joint b and c.
kN41,3
0
0
−==
=−=∑
bc
babc
babc
xb
N
VN
VN
F
kN79,0
0
0
=−=
=+=∑
ba
bcba
bcba
zb
N
VN
VN
F
kN41,3
0
0
−=−=
=−−=∑
cb
cdcb
cdcb
xc
N
VN
VN
F
kN79,0
0
0
−==
=−=∑
cd
cbcd
cbcd
zc
N
VN
VN
F
b cVbc
Vba
Vcb
Vcd
Nbc
Nba
Ncb
Ncd
Slope-Deflection Method: Frames With Side-sway,
Example 2
8. Draw normal and shear force diagrams
Nab
= N
ba=
0,7
9
Nbc= Ncb = -3,41 Ncd =
Ndc =
-0,79
Vcd =
Vdc =
3,41
V ab=
8,5
9
Vba
= -
3,41
Vbc= Vcb = -0,79
Slope-Deflection Method: Frames With Side-sway,
Example 2
8. Draw bending moment diagram
Mcd =
-3,50
Mab
= -
15,8
4Mbc= -0,32
kNm92,9359,884,15
3
=⋅+−=⋅+=
F
ababF
M
VMM
-Mcb = -3,50
-Mba
= -
0,32
-Mdc =
6,75MF
= 9
,92
Slope-Deflection Method: Frames With Side-sway,
Example 3
1. Degrees of freedom
� Two unknown displacements
- one rotation ϕϕϕϕb- one translation ua = ub = ud.
� Support rotation ϕϕϕϕa is not necessary to solution because
the moment Ma is known (Ma = 0) ⇒ beam portion ab is taken as hinged-fixed.
a
EI = const.
24
b
c
d
F2 = 5 kN
4
q = 6 kN/m
F1 = 10 kN
Slope-Deflection Method: Frames With Side-sway,
Example 3
2. Fixed end moments are calculated referring to the table.
27
120
0
0
ba ab
bc
cb
M q l
M
M
= + ⋅ ⋅
=
=
a
EI = const.
24
b
c
d
F2 = 5 kN
4
q = 6 kN/m
F1 = 10 kN
Slope-Deflection Method: Frames With Side-sway,
Example 3
3. Express internal end moments by slope-deflection equations.
tan
a b d
bc bcbc
u u u
lψ ψ
= = = ∆
∆≈ =
∆∆∆∆∆∆∆∆
ψψψψbc
chords of
the elastic curve
ab
c
d
∆∆∆∆
( )
( )abaab
ababab
abbaab
ababab
l
EIMM
l
EIMM
ψϕ
ψϕϕ
−+=
⋅−+⋅+=
3
322
fixed end rotation
Slope-Deflection Method: Frames With Side-sway,
Example 3
3. Express internal end
moments by slope-
deflection equations.
( )
( )
( )
3
22 3
22 3
abbaba b ab
ab
bcbcbc b c bc
bc
bccbcb c b bc
bc
EIM M
l
EIM M
l
EIM M
l
ϕ ψ
ϕ ϕ ψ
ϕ ϕ ψ
= + −
= + ⋅ + − ⋅
= + ⋅ + − ⋅
a
EI = const.
24
b
c
d
F2 = 5 kN
4
q= 6 kN/m
F1 = 10 kN
Slope-Deflection Method: Frames With Side-sway,
Example 3
a
EI = const.
24
b
c
d
F2 = 5 kN
4
q = 6 kN/m
F1 = 10 kN
4. Equilibrium equations
� End moments are expressed in terms of unknown rotations ϕband translation ∆. Required equation to solve are the moment
equilibrium equations at rigid joints b and then sideway equation.
20 : 0Fb ba bc bM M M M= + − =∑
b
222 ⋅= FM F
b
baMbcM
Slope-Deflection Method: Frames With Side-sway,
Example 34. Equilibrium equations
� sideway equation
(the horizontal equilibrium)
10 : 0x bcF F V= − =∑
( )1bc bc cb
bc
V M Ml
= − −
a b dF1 = 10 kN
bcV
a
EI = const.
24
b
c
d
F2 = 5 kN
4
q= 6 kN/m
F1 = 10 kN
Part of the frame containing the joints
with the same translation ∆ “cut off”.
Slope-Deflection Method: Frames With Side-sway,
Example 3
4. Solving equilibrium equations?
1,75 0,375 4,4
?0,375 0,1875 10
bb
b
EI EI EI
EI EIEI
ϕϕ
ϕ
=⋅ ⋅ − ⋅ ⋅∆ =
⋅ ⋅ − ⋅ ⋅∆ = − ∆ =
5. End moments
� After evaluating ϕb, ∆, substitute them to evaluate end moments.
35,6
4
3
8
3
2 8
ba b
bc b
cb b
EIM
EIM EI
EI EIM
ϕ
ϕ
ϕ
= + ⋅
= ⋅ − ⋅∆
= ⋅ − ⋅ ∆
Slope-Deflection Method: Frames With Side-sway,
Example 3
6. Shear forces
( )
( )
1
1
bc bc bc cbbc
cb cb bc cbbc
V R M Ml
V R M Ml
= = − −
= − = − + +
21
2 3
1
2 3
ab abab ab ab ba
ab
ab abba ba ab ba
ab
q l lV R M M
l
q l lV R M M
l
⋅ ⋅ = = ⋅ − −
⋅ = − = − ⋅ + +
4b c
Mbc Mcb
Rbc=Vbc Rcb=-Vcb
a b
Mab Mba
Rab=Vab Rba=-Vba
q = 6 kN/m
4
Slope-Deflection Method: Frames With Side-sway,
Example 3
7. Normal forces
� The required equations to solve for normal forces are the
force equilibrium equations at rigid joint b.
0 : 0xb ba bcF N V= − − =∑
20 : 0zb bc baF N V F= − + =∑b
Vba
Vbc
Nba
Nbc
F2
Slope-Deflection Method: Frames With Side-sway,
Example 3
8. Draw normal and shear force diagrams.
Nbc =
Ncb =
-14,975
Nab= Nba = -10
0
Vbc =
Vcb =
10
F2= 5Vab= 2,025
Vba= -9,975
2°
xn= 3,647
m647,36
42975,92=⋅⋅=
⋅⋅=
q
lVx abba
n
Slope-Deflection Method: Frames With Side-sway,
Example 3
kNm35,046
647,36647,3975,99,23
63
3
=⋅
⋅−⋅+−=
⋅⋅−⋅−−=
max
max
ab
nnbabamax
M
M
l
xqxVMM
8. Draw bending moment diagram.M
bc =-13,9
-F2·2 = -10
-Mba= -23,9
3°
-Mcb =
26,1
Mmax= 0,35
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