Post on 22-Nov-2014
description
1
! General Case! Stiffness Coefficients! Stiffness Coefficients Derivation! Fixed-End Moments! Pin-Supported End Span! Typical Problems! Analysis of Beams! Analysis of Frames: No Sidesway! Analysis of Frames: Sidesway
DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS
2
Slope � Deflection Equations
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
3
Degrees of Freedom
L
θΑ
A B
M
1 DOF: θΑ
PθΑ
θΒΒΒΒ
A BC 2 DOF: θΑ , θΒΒΒΒ
4
L
A B
1
Stiffness
kBAkAA
LEIkAA
4=
LEIkBA
2=
5
L
A B1
kBBkAB
LEIkBB
4=
LEIkAB
2=
6
Fixed-End ForcesFixed-End Moments: Loads
P
L/2 L/2
L
w
L
8PL
8PL
2P
2P
12
2wL12
2wL
2wL
2wL
7
General Case
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
8
wP
settlement = ∆j
(MFij)∆ (MF
ji)∆
(MFij)Load (MF
ji)Load
+
Mij Mji
θi
θj
+
i jwPMij
Mji
settlement = ∆jθj
θi
ψ
=+ ji LEI
LEI θθ 24
ji LEI
LEI θθ 42
+=
,)()()2()4( LoadijF
ijF
jiij MMLEI
LEIM +++= ∆θθ Loadji
Fji
Fjiji MM
LEI
LEIM )()()4()2( +++= ∆θθ
L
9
Mji Mjk
Pi j kw Cj
Mji Mjk
Cj
j
Equilibrium Equations
0:0 =+−−=Σ+ jjkjij CMMM
10
+
1
1
i jMij Mji
θi
θj
LEIkii
4=
LEIk ji
2=
LEIkij
2=
LEIk jj
4=
iθ×
jθ×
Stiffness Coefficients
L
11
[ ]
=
jjji
ijii
kkkk
k
Stiffness Matrix
)()2()4( ijF
jiij MLEI
LEIM ++= θθ
)()4()2( jiF
jiji MLEI
LEIM ++= θθ
+
=
F
ji
Fij
j
iI
ji
ij
MM
LEILEILEILEI
MM
θθ
)/4()/2()/2()/4(
Matrix Formulation
12
[D] = [K]-1([Q] - [FEM])
Displacementmatrix
Stiffness matrix
Force matrixwP(MF
ij)Load (MFji)Load
+
+
i jwPMij
Mji
θj
θi
ψ ∆j
Mij Mji
θi
θj
(MFij)∆ (MF
ji)∆
Fixed-end momentmatrix
][]][[][ FEMKM += θ
]][[])[]([ θKFEMM =−
][][][][ 1 FEMMK −= −θ
L
13
L
Real beam
Conjugate beam
Stiffness Coefficients Derivation: Fixed-End Support
MjMi
L/3
θi
LMM ji +
EIM j
EIMi
θι
EILM j
2
EILMi
2
)1(2
0)3
2)(2
()3
)(2
(:0'
−−−=
=+−=Σ+
ji
jii
MM
LEI
LMLEI
LMM
)2(0)2
()2
(:0 −−−=+−=Σ↑+EI
LMEI
LMF jiiy θ ij
ii
LEIM
LEIM
andFrom
θ
θ
)2(
)4(
);2()1(
=
=
LMM ji +
i j
14
Conjugate beam
L
Real beam
Stiffness Coefficients Derivation: Pinned-End Support
Mi θi
θj
LM i
LMi
EIM i
EILMi
2
32L
θi θj
0)3
2)(2
(:0' =−=Σ+ LLEI
LMM ii
j θ
i j
0)2
()3
(:0 =+−=Σ↑+ jii
y EILM
EILMF θ
LEIM
EILM
ii
i3)
3(1 =→==θ
)6
(EI
LM ij
−=θ)
3(
EILM i
i =θ
15
Fixed end moment : Point Load
Real beam
8,0
162
22:0
2 PLMEI
PLEI
MLEI
MLFy ==+−−=Σ↑+
P
M
M EIM
Conjugate beamA
EIM
B
L
P
A B
EIM
EIML2
EIM
EIML2
EIPL
16
2
EIPL4 EI
PL16
2
16
L
P
841616PLPLPLPL
=+−
+−
8PL
8PL
2P
2PP/2
P/2
-PL/8 -PL/8
PL/8
-PL/16-PL/8
-
PL/4+
-PL/16-PL/8-
17
Uniform load
L
w
A B
w
M
M
Real beam Conjugate beam
A
EIM
EIM
B
12,0
242
22:0
23 wLMEI
wLEI
MLEI
MLFy ==+−−=Σ↑+
EIwL8
2
EIwL
24
3
EIwL
24
3
EIM
EIML2
EIM
EIML2
18
Settlements
M
M
L
∆
MjMi = Mj
LMM ji +L
MM ji +
Real beam
2
6LEIM ∆
=
Conjugate beam
EIM
A B
EIM
∆
,0)3
2)(2
()3
)(2
(:0 =+−∆−=Σ+L
EIMLL
EIMLM B
EIM
EIML2
EIMEI
ML2
∆
19
wP
A B
A B+θA θB
(FEM)AB(FEM)BA
Pin-Supported End Span: Simple Case
BA LEI
LEI θθ 24
+ BA LEI
LEI θθ 42
+
)1()()/2()/4(0 −−−++== ABBAAB FEMLEILEIM θθ
)2()()/4()/2(0 −−−++== BABABA FEMLEILEIM θθ
BABABBA FEMFEMLEIM )()(2)/6(2:)1()2(2 −+=− θ
2)()()/3( BA
BABBAFEMFEMLEIM −+= θ
wP
AB
L
20
AB
wP
A B
A BθA θB
(MF AB)load
(MF BA)load
Pin-Supported End Span: With End Couple and Settlement
BA LEI
LEI θθ 24
+ BA LEI
LEI θθ 42
+
L
(MF AB)∆
(MF BA) ∆
)1()()(24−−−+++== ∆
FABload
FABBAAAB MM
LEI
LEIMM θθ
)2()()(42−−−+++= ∆
FBAload
FBABABA MM
LEI
LEIM θθ
2)(
21])(
21)[(3:
2)1()2(2lim AF
BAloadFABload
FBABBAA
MMMMLEIMbyinateE ++−+=
−∆θθ
MA
wP
AB
∆
21
Fixed-End MomentsFixed-End Moments: Loads
P
L/2 L/2
P
L/2 L/2
8PL
8PL
163)]
8()[
21(
8PLPLPL
=−−+
12
2wL12
2wL
8)]
12()[
21(
12
222 wLwLwL=−−+
22
Typical Problem
0
0
0
0
A C
B
P1P2
L1 L2
wCB
P
8PL
8PL w12
2wL
12
2wL
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
L L
23
MBA MBC
A C
B
P1P2
L1 L2
wCB
B
CB
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0
24
A C
B
P1P2
L1 L2
wCB
Substitute θB in MAB, MBA, MBC, MCB
MABMBA
MBC
MCB
0
0
0
0
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
25
A BP1
MAB
MBA
L1
A CB
P1P2
L1 L2
wCB
ByR CyAy ByL
Ay Cy
MABMBA
MBC
MCB
By = ByL + ByR
B CP2
MBCMCB
L2
26
Example of Beams
27
10 kN 6 kN/m
A C
B 6 m4 m4 m
Example 1
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
28
PPL8 wwL2
30FEM
PL8
wL2
20
MBA MBC
B
Substitute θB in the moment equations:
MBC = 8.8 kN�m
MCB = -10 kN�m
MAB = 10.6 kN�m,
MBA = - 8.8 kN�m,
[M] = [K][Q] + [FEM]
10 kN 6 kN/m
A C
B 6 m4 m4 m
0
0
0
0
8)8)(10(
82
84
++= BAABEIEIM θθ
8)8)(10(
84
82
−+= BABAEIEIM θθ
30)6)(6(
62
64 2
++= CBBCEIEIM θθ
20)6)(6(
64
62 2
−+= CBCBEIEIM θθ
0:0 =−−=Σ+ BCBAB MMM
EI
EIEI
B
B
4.2
030
)6)(6(10)6
48
4(2
=
=+−+
θ
θ
29
10 kN 6 kN/m
A C
B 6 m4 m4 m
MBC = 8.8 kN�m
MCB= -10 kN�m
MAB = 10.6 kN�m,
MBA = - 8.8 kN�m,
= 5.23 kN = 4.78 kN = 5.8 kN = 12.2 kN
10 kN�m8.8 kN�m
10.6 kN�m8.8 kN�m
10 kN
10.6 kN�m
8.8 kN�m
A B
ByLAy
2 m
6 kN/m8.8 kN�m
10 kN�mB
CyByR
18 kN
30
10 kN 6 kN/m
A C
B 6 m4 m4 m
10 kN�m10.6 kN�m
5.23 kN 12.2 kN
4.78 + 5.8 = 10.58 kN
V (kN)x (m)
5.23
- 4.78
5.8
-12.2
+-
+
-
M (kN�m) x (m)
-10.6
10.3
-8.8 -10- -
+-
EIB4.2
=θ
Deflected shape x (m)
31
10 kN 6 kN/m
A C
B 6 m4 m4 m
Example 2
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
32
PPL8 wwL2
30FEM
PL8
wL2
20
[M] = [K][Q] + [FEM]
10 kN 6 kN/m
A C
B 6 m4 m4 m
)1(8
)8)(10(8
28
4−−−++= BAAB
EIEIM θθ
)2(8
)8)(10(8
48
2−−−−+= BABA
EIEIM θθ
)3(30
)6)(6(6
26
4 2
−−−++= CBBCEIEIM θθ
)4(20
)6)(6(6
46
2 2
−−−−+= CBCBEIEIM θθ
10
10
0
0
0
308
62:)1()2(2 −=− BBAEIM θ
)5(158
3−−−−= BBA
EIM θ
33
MBA MBC
B
)5(158
3−−−−= BBA
EIM θ
)3(30
)6)(6(6
4 2
−−−+= BBCEIM θ
0:0 =−−=Σ+ BCBAB MMM
EI
EIEI
B
B
488.7
)6(030
)6)(6(15)6
48
3(2
=
−−−=+−+
θ
θ
EI
EIEIinSubstitute
A
BAB
74.23
108
28
40:)1(
−=
−+=
θ
θθθ
Substitute θA and θB in (5), (3) and (4):
MBC = 12.19 kN�m
MCB = - 8.30 kN�m
MBA = - 12.19 kN�m)4(
20)6)(6(
62 2
−−−−= BCBEIM θ
34
10 kN 6 kN/m
A C
B 6 m4 m4 m
MBA = - 12.19 kN�m, MBC = 12.19 kN�m, MCB = - 8.30 kN�m
= 3.48 kN = 6.52 kN = 6.65 kN = 11.35 kN
12.19 kN�m
12.19 kN�m8.30 kN�m
10 kN
12.19 kN�m
A B
ByLAy
2 m
6 kN/m12.19 kN�m
8.30 kN�mC
CyByR
18 kN
B
35
Deflected shape x (m)EIB49.7
=θ
10 kN 6 kN/m
A C
B 6 m4 m4 m
V (kN)x (m)
3.48
- 6.52
6.65
-11.35
M (kN�m) x (m)
14
-12.2-8.3
11.35 kN3.48 kN
6.52 + 6.65 = 13.17 kN
EIA74.23−
=θ
36
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
Example 3
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
37
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI(4)(62)/12 (4)(62)/12 (10)(8)/8(10)(8)/8
15
12
)1(8
)8)(10(8
)2(28
)2(4−−−++= BAAB
EIEIM θθ
)2(8
)8)(10(8
)2(48
)2(2−−−−+= BABA
EIEIM θθ
0 10
10
)2(8
)8)(10)(2/3(8
)2(3:2
)1()2(2 aEIM BBA −−−−=− θ
)3(12
)6)(4(6
)3(4 2
−−−+= BBCEIM θ
38
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI(4)(62)/12 (4)(62)/12(3/2)(10)(8)/8
EIEIMM
B
BBCBA
/091.13151275.2:0
==+−==−−
θθ
15
12
)2(8
)8)(10)(2/3(8
)2(3 aEIM BBA −−−−= θ
)3(12
)6)(4(6
)3(4 2
−−−+= BBCEIM θ
mkNEIM
mkNEI
EIM
mkNEI
EIM
BCB
BC
BA
•−=−=
•=−=
•−=−=
91.10126
)3(2
18.1412)091.1(6
)3(4
18.1415)091.1(8
)2(3
θ
39
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
MBA = - 14.18 kN�m, MBC = 14.18 kN�m, MCB = -10.91 kN�m
14.18
140.18 kN�m
10 kN
A B
ByLAy = 6.73 kN= 3.23 kN
14.18 kN�m
10.91 kN�m
4 kN/m
C
CyByR
24 kN
14.18 10.91
= 11.46 kN= 12.55 kN
40
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
11.46 kN3.23 kN
10.91 kN�m
V (kN)x (m)
3.23
-6.73
12.55
-11.46
+-
+-
2.86
M (kN�m) x (m)
12.91
-14.18
5.53
-10.91
+
-+
-
6.77 + 12.55 = 19.32 kN
θB = 1.091/EIDeflected shape x (m)
41
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
Example 4
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
12 kN�m
42
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
12 kN�m
wL2/12 = 12 wL2/12 = 121.5PL/8 = 15
MBA
MBCB
12 kN�mMBA
MBC
EIB273.3
−=θ
EIA21.7
−=θ
)1(158
)2(3−−−−= BBA
EIM θ
)2(126
)3(4−−−+= BBC
EIM θ
)3(126
)3(2−−−−= BCB
EIM θ
8)8)(10(
8)3(2
8)2(4
++= BAABEIEIM θθ
0 -3.273/EI
012:int =−−− BCBA MMBJo
012)122()1575.0( =−+−−− BEIEI θ
mkNEI
EIM BA •−=−−= 45.1715)273.3(75.0
mkNEI
EIM BC •=+−= 45.512)273.3(2
mkNEI
EIM CB •−=−−= 27.1512)273.3(
43
10 kN
A B
4 kN/m
C
24 kN
17.45 kN�m5.45 kN�m
15.27 kN�m
2.82 kN 13.64 kN10.36 kN7.18 kN
12 kN�m10 kN 4 kN/m
A C
B13.64 kN2.82 kN
15.27 kN�m
17.54 kN
mkNEI
EIM BA •−=−−= 45.1715)273.3(75.0
mkNEI
EIM BC •=+−= 45.512)273.3(2
mkNEI
EIM CB •−=−−= 27.1512)273.3(
44
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
12 kN�m
13.64 kN2.82 kN
15.27 kN�m
17.54 kN
V (kN)x (m)3.41 m+
-+
-
2.82
-7.18
10.36
-13.64
M (kN�m) x (m)+
- -+
11.28
-17.45
-5.45-15.27
7.98
Deflected shape x (m)
EIB273.3
=θEIA
21.7−=θ
EIB273.3
−=θ
EIA21.7
−=θ
45
Example 5
Draw the quantitative shear, bending moment diagrams, and qualitativedeflected curve for the beam shown. Support B settles 10 mm, and EI isconstant. Take E = 200 GPa, I = 200x106 mm4.
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
46
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
[FEM]∆
∆
A
B
2
6LEI∆
2
6LEI∆
∆
BC
2
6LEI∆
2
6LEI∆
P w
[FEM]load
8PL
8PL
30
2wL30
2wL
)1(8
)8)(10(8
)01.0)(2(68
)2(28
)2(42 −−−+++=
EIEIEIM BAAB θθ
)2(8
)8)(10(8
)01.0)(2(68
)2(48
)2(22 −−−−++=
EIEIEIM BABA θθ
)3(30
)6)(6(6
)01.0)(3(66
)3(26
)3(4 2
2 −−−+−+=EIEIEIM CBBC θθ
)4(30
)6)(6(6
)01.0)(3(66
)3(46
)3(2 2
2 −−−−−+=EIEIEIM CBCB θθ
-12
0
0
47
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN� m2 :
)1(8
)8)(10(8
)01.0)(2(68
)2(28
)2(42 −−−+++=
EIEIEIM BAAB θθ
)2(8
)8)(10(8
)01.0)(2(68
)2(48
)2(22 −−−−++=
EIEIEIM BABA θθ
)1(10758
)2(28
)2(4−−−+++= BAAB
EIEIM θθ
)2(10758
)2(48
)2(2−−−−++= BABA
EIEIM θθ
)2(2/12)2/10(10)2/75(758
)2(3:2
)1()2(2 aEIM BBA −−−−−−−+=− θ
16.5
48
+ ΣMB = 0: - MBA - MBC = 0 (3/4 + 2)EIθB + 16.5 - 192.8 = 0
θB = 64.109/ EI
Substitute θB in (1): θA = -129.06/EI
Substitute θA and θB in (5), (3), (4):
MBC = -64.58 kN�m
MCB = -146.69 kN�m
MBA = 64.58 kN�m,
MBA MBC
B
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
MBC = (4/6)(3EI)θB - 192.8
MBA = (3/4)(2EI)θB + 16.5
49
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
64.58 kN�m 64.58 kN�m
= -1.57 kN= 11.57 kN
146.69 kN�m
= 47.21 kN= -29.21 kN
10 kN
A B
ByLAy
12 kN�m64.58 kN�m
2 m
6 kN/m
C
CyByR
18 kN
B
64.58 kN�m
146.69 kN�m
MBC = -64.58 kN�m
MCB = -146.69 kN�m
MBA = 64.58 kN�m,
50
12 kN�m 10 kN 6 kN/m
A C
B6 m4 m4 m
2EI 3EI
V (kN)x (m)
11.57 1.57
-29.21-47.21
-+
M (kN�m) x (m)
1258.29 64.58
-146.69
+
-
47.21 kN
146.69 kN�m
11.57 kN
1.57 + 29.21 = 30.78 kN
10 mmθA = -129.06/EI
θB = 64.109/ EIθA = -129.06/EI
Deflected shape x (m)
θB = 64.109/ EI
51
Example 6
For the beam shown, support A settles 10 mm downward, use the slope-deflectionmethod to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape. (3 points)Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
52
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
)1(3
)2(4−−−= CCB
EIM θ
)2(1005.43
)5.1(23
)5.1(4−−−+++= ACCA
EIEIM θθ
)2(2
122
1002
)5.4(33
)5.1(3:2
)2()2(2 aEIM CCA −−−+++=− θ
)3(1005.43
)5.1(43
)5.1(2−−−+−+= ACAC
EIEIM θθ12
0.01 m
C
AmkN •=
××
1003
)01.0)(502005.1(62
mkN •100MF
∆
10 mm
6 kN/m
A C
5.412
)3(6 2
= 4.5MF
w
53
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
)1(3
)2(4−−−= CCB
EIM θ
)2(2
122
1002
)5.4(33
)5.1(3 aEIM CCA −−−+++= θ
10 mm
MCBMCA
C
0=+ CACB MM
02
122
1002
)5.4(33
)5.48(=+++
+C
EI θ
radEIC 0015.006.15
−=−
=θ
)3(1005.43
)5.1(4)06.15(3
)5.1(212 −−−+−+−
= AEI
EIEI θSubstitute θC in eq.(3)
radEIA 0034.022.34
−=−
=θ
� Equilibrium equation:
54
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
radEIC 0015.006.15
−=−
=θ radEIA 0034.022.34
−=−
=θ
mkNEI
EIEIM CBC •−=−
== 08.20)06.15(3
)2(23
)2(2 θ
mkNEI
EIEIM CCB •−=−
== 16.40)06.15(3
)2(43
)2(4 θ
kN08.203
08.2016.40=
+kN08.20
B C40.16 kN�m20.08 kN�m
6 kN/m
AC
12 kN�m
40.16 kN�m
18 kN
8.39 kN26.39 kN
55
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
6 kN/m
AC
12 kN�m
40.16 kN�m8.39 kN26.39 kN
B C40.16 kN�m20.08 kN�m
20.08 kN20.08 kN
V (kN)
x (m)
26.398.39
-20.08
+-
M (kN�m)
x (m)20.08
-40.16
12
radC 0015.0−=θ
radA 0034.0−=θ
Deflected shapex (m)
radC 0015.0=θ
radA 0034.0=θ
56
Example 7
For the beam shown, support A settles 10 mm downward, use theslope-deflection method to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
57
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
5.412
)3(6 2
=
6 kN/m
AC4.5
mkN •=
××
1003
)01.0)(502005.1(620.01 m
C
A
100
34 CEI∆
∆C
CB
34
3)2(6
2CC EIEI ∆
=∆
)2(3
43
)2(4−−−∆−= CCCB
EIEIM θ
)3(1005.43
)5.1(23
)5.1(4−−++∆++= CACCA EIEIEIM θθ
)4(1005.43
)5.1(43
)5.1(2−−+−∆++= CACAC EIEIEIM θθ
12
)1(3
43
)2(2−−−∆−= CCBC
EIEIM θ
)3(2
122
1002
)5.4(323
)5.1(3:2
)4()3(2 aEIEIM CCCA −−−+++∆+=− θ
CC EIEI
∆=∆
23)5.1(6∆C
C
A
CEI∆
58
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
� Equilibrium equation:
)3
()( CBBCCBy
MMC +−=
6 kN/m
AC
12 kN�m
MCA
18 kN
AyB C
MCBMBC
By3
393
)5.1(1812)( +=
++= CACA
CAyMMC
C
MCB MCA
(Cy)CA(Cy)CB
*)1(0:0 −−−=+=Σ CACBC MMM
*)2(0)()(:0 −−−=+=Σ CAyCByy CCC
)5(15.628333.0167.4*)1( −−−−=∆− CC EIEIinSubstitute θ
)6(75.101167.35.2*)2( −−−−=∆+− CC EIEIinSubstitute θ
mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( −=−=∆−=−=θ
59
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
� Solve equation
radEIC 00255.051.25
−=−
=θ
mmEIC 227.527.52
−=−
=∆
Substitute θC and ∆C in (4)
radEIA 000286.086.2
−=−
=θ
Substitute θC and ∆C in (1), (2) and (3a)
mkNM BC •= 68.35
mkNMCB •= 67.1
mkNMCA •−= 67.1
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
35.68 kN�m
kN
Ay
55.56
68.3512)5.4(18
=
−−=
kNBy
45.12
55.518
=
−=
60
10 mm
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
35.68 kN�m
12.45 kN 5.55 kN
Deflected shape
x (m)
radC 00255.0−=θ
mmC 227.5−=∆
radA 000286.0−=θ
radC 00255.0=θradA 000286.0=θ
mmC 227.5=∆
M (kN�m)
x (m)1214.57
1.67
-35.68
-+
V (kN)
x (m)0.925 m12.45
-5.55
+
61
Example of Frame: No Sidesway
62
Example 6
For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EI
63
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EIPL/8 = 30
PL/8 = 30
� Equilibrium equations
10
MBC
MBA
)3(18366
)2(3−−−++= BBC
EIM θ
*)1(010 −−−=−− BCBA MM
05430310 =−+− BEIθ
EIEIB667.4
)3(14 −
=−
=θ)1(30
6)3(2
−−−+= BABEIM θ
mkNMmkNM
mkNM
BC
BA
AB
•=•−=
•=
33.4933.39
33.25
36/2 = 18
(wL2/12 ) =3636
� Slope-Deflection Equations
)2(306
)3(4−−−−= BBA
EIM θ
Substitute (2) and (3) in (1*)
)3()1(667.4 toinEI
Substitute B−
=θ
64
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EI39.33
25.33
49.33
A
B
40 kN
39.33
25.33
C
12 kN/m
B49.33
17.67 kN
27.78 kN
Bending moment diagram
-25.33
27.7
-39.3
Deflected curve
-49.33
20.58
10
θB = -4.667/EIθB
θB
MAB = 25.33 kN�m
MBA = -39.33 kN�m
MBC = 49.33 kN�m
65
A
B
C
D
E
25 kN
5 m
5 kN/m
60(106) mm4
240(106) mm4 180(106)
120(106) mm4
3 m 4 m3 m
Example 7
Draw the quantitative shear, bending moment diagrams and qualitativedeflected curve for the frame shown. E = 200 GPa.
66
A
B
C
D
E
25 kN
5 m
5 kN/m
60(106) mm4
240(106) mm4 180(106)
120(106) mm4
3 m 4 m3 m0
BAABEIEIM θθ
5)2(2
5)2(4
+=0
BABAEIEIM θθ
5)2(4
5)2(2
+=
75.186
)4(26
)4(4++= CBBC
EIEIM θθ
75.186
)4(46
)4(2−+= CBCB
EIEIM θθ
CCDEIM θ5
)(3=
104
)3(3+= CCE
EIM θ
0=+ BCBA MM
)1(75.18)68()
616
58( −−−−=++ CB EIEI θθ
0=++ CECDCB MMM
)2(75.8)49
53
616()
68( −−−=+++ CB EIEI θθ
EIEIandFrom CB
86.229.5:)2()1( =−
= θθ
PL/8 = 18.75 18.75
6.667 (wL2/12 ) = 6.667+ 3.333
67
MAB = −4.23 kN�m
MBA = −8.46 kN�m
MBC = 8.46 kN�m
MCB = −18.18 kN�m
MCD = 1.72 kN�m
MCE = 16.44 kN�m
Substitute θB = -1.11/EI, θc = -20.59/EI below
0
0BAAB
EIEIM θθ5
)2(25
)2(4+=
BABAEIEIM θθ
5)2(4
5)2(2
+=
75.186
)4(26
)4(4++= CBBC
EIEIM θθ
75.186
)4(46
)4(2−+= CBCB
EIEIM θθ
CCDEIM θ5
)(3=
104
)3(3+= CCE
EIM θ
68
MAB = -4.23 kN�m, MBA = -8.46 kN�m, MBC = 8.46 kN�m, MCB = -18.18 kN�m,MCD = 1.72 kN�m, MCE = 16.44 kN�m
A
B
5 m
C E
20 kN
A
B
5 m
4.23 kN�m
2.54 kN
(8.46 + 4.23)/5 = 2.54 kN
8.46 kN�m
C
25 kN
B 3 m 3 m2.54 kN 2.54 kN
8.46 kN�m(25(3)+8.46-18.18)/6 = 10.88 kN
14.12 kN18.18 kN�m
16.44 kN�m
(20(2)+16.44)/4= 14.11 kN
5.89 kN
0.34 kN
28.23 kN
14.12+14.11=28.23 kN1.72 kN�m
(1.72)/5 = 0.34 kN
10.88 kN
10.88 kN
2.54-0.34=2.2 kN
2.2 kN
69
Shear diagram
-2.54
-2.54
10.88
-14.12
14.11
-5.89
0.34
+
-
-
+-
+
2.82 m
1.18 m
Deflected curve
1.18 m
0.78 m2.33 m1.29 m
1.67m
1.29 m
0.78 m
Moment diagram
1.18 m
3.46
24.18
1.72
-18.18 -16.44
4.23
-8.46-8.46
2.33 m
+
-
+
- -
+
θB = −5.29/EI
θC = 2.86/EI1.67m
70
Example of Frames: Sidesway
71
A
B C
D
3m
10 kN
3 m
1 m
Example 8
Determine the moments at each joint of the frame and draw the quantitativebending moment diagrams and qualitative deflected curve . The joints at A andD are fixed and joint C is assumed pin-connected. EI is constant for each member
72
A
B C
D
3m
10 kN
3 m
1 m
MABMDC
Ax Dx
Ay Dy
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(010:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
� Overview
B- Joint B
MBCMBA
73
)3(3
)(3−−−= BBC
EIM θ
)4(1875.0375.021375.0 −−−∆=∆−∆= EIEIEIM DC
)1(4
64
)1)(3(104
)(222
2
−−−∆
++=EIEIM BAB θ
5.625 0.375EI∆
A
B C
D3m
10 kN
3 m
1 m
(5.625)load
(1.875)load
(0.375EI∆)∆
(0.375EI∆)∆
(0.375EI∆)∆
(0.375EI∆)∆
:0=Σ+ CM
MDC
D
C
4 m
Dx
MAB
MBA
A
B
4 m
Ax
10 kN
∆ ∆
(1/2)(0.375EI∆)∆
:0=Σ+ BM
)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ4
)( BAABx
MMA +=
)6(0468.04
−−−∆== EIMD DCx
)2(4
64
)1)(3(104
)(422
2
−−−∆
+−=EIEIM BBA θ
5.625 0.375EI∆
� Slope-Deflection Equations
74
� Solve equation
*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM
)3(3
)(3−−−= BBC
EIM θ
)4(1875.0 −−−∆= EIM DC
)1(375.0625.54
)(2−−−∆++= EIEIM BAB θ
)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ
)6(0468.0 −−−∆= EIDx
)2(375.0625.54
)(4−−−∆+−= EIEIM BBA θ
Equilibrium Conditions:
Slope-Deflection Equations:
Horizontal reaction at supports:
Substitute (2) and (3) in (1*)
2EI θB + 0.375EI ∆ = 5.625 ----(7)
Substitute (5) and (6) in (2*)
)8(437.8235.0375.0 −−−−=∆−− EIEI Bθ
From (7) and (8) can solve;
EIEIB8.446.5
=∆−
=θ
)6()1(8.446.5 toinEI
andEI
Substitute B =∆−
=θ
MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN
75
A
BC
D
Deflected curve
A
BC
D
Bending moment diagram
MAB = 15.88 kN�m, MBA = 5.6 kN�m, MBC = -5.6 kN�m, MDC = 8.42 kN�m, Ax = 7.9 kN, Dx = 2.1 kN,
A
B C
D
3m
10 kN
3 m
1 m
15.88
5.6
8.42
7.9 kN 2.1 kN
15.88
5.6
8.42
∆ = 44.8/EI ∆ = 44.8/EI
θB = -5.6/EI5.6
7.8
76
B C
DA
3m4 m
10 kN4 m
2 m
pin
2 EI
2.5 EI EI
Example 9
From the frame shown use the slope-deflection method to:(a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw thequalitative deflected shape of the entire frame.
77
B
C
DA
3m4 m
10 kN4 m
2 m
2EI
2.5EI EI
Ax Dx
Ay Dy
MDCMAB
B´
∆
∆
∆ CD C´∆BC
� Overview
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(010:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
B- Joint B
MBCMBA
78
B
C
DA 3m4 m
10 kN4 m
2 m
2EI
2.5EI EI36.87° = ∆ tan 36.87° = 0.75 ∆
= ∆ / cos 36.87° = 1.25 ∆
B´
∆
∆∆BC
∆ CD C´
∆
∆BC
∆CD
C
C´
36.87°
)1(5375.04
)(2−−−+∆+= EIEIM BAB θ
)2(5375.04
)(4−−−−∆+= EIEIM BBA θ
)3(2813.04
)2(3−−−∆−= EIEIM BBC θ
)4(375.0 −−−∆= EIM DC
C
C´
BB´
∆BC= 0.75 ∆
0.375EI∆
6EI∆/(4) 2 = 0.375EI∆
B
A
∆´ B´
(6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆0.5625EI∆
D
C
C´∆CD= 1.25 ∆
(1/2) 0.75EI∆
(1/2) 0.5625EI∆
PL/8 = 5
5
(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆
0.75EI∆
� Slope-Deflection Equation
79
B
C
DA
3m4 m
10 kN4 m
2 m
pin2 EI
2.5 EI EI
Dx= (MDC-(3/4)MBC)/4 ---(6)
MBC
4
Ax = (MBA+ MAB-20)/4 -----(5)
B CMBC
C
D
MBC/4MDC
MBC
4
A
B
MBA
10 kN
MAB
� Horizontal reactions
80
� Solve equations
*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM
Equilibrium Conditions:
Slope-Deflection Equation:
Horizontal reactions at supports:
Substitute (2) and (3) in (1*)
Substitute (5) and (6) in (2*)
From (7) and (8) can solve;
EIEIB56.1445.1 −
=∆=θ
)6()1(56.1445.1 toinEI
andEI
Substitute B−
=∆=θ
MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN
)1(4
654
)(22 −−−∆
++=EIEIM BAB θ
)2(4
654
)(42 −−−∆
+−=EIEIM BBA θ
)3(4
)75.0)(2(34
)2(32 −−−
∆−=
EIEIM BBC θ
)4(5
)25.1)(5.2(32 −−−
∆=
EIM DC
)5(4
)20(−−−
−+= ABBA
xMMA
)6(443
−−−−
=BCDC
x
MMD
)7(050938.05.2 −−−=−∆+ EIEI Bθ
)8(05334.00938.0 −−−=−∆+ EIEI Bθ
81
BC
DA
Deflected shape
θB=1.45/EI
θB=1.45/EI
Bending-moment diagram
BC
DA 11.19
1.91
5.35
5.46
1.91
∆ ∆
MAB = 11.19 kN�m MBA = 1.91 kN�m MBC = -1.91 kN�m MDC = 5.46 kN�m
Ax = 8.28 kN�m Dx = 1.72 kN�m
B
C
DA
3m4 m
10 kN4 m
2 m
pin2 EI
2.5 EI EI11.19 kN�m
8.27 kN
5.46
1.91
1.91
0.478 kN1.73
0.478 kN
82
Example 10
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.
A
BC
D
3 m
3m
4m
20 kN/mpin
2EI
3EI
4EI
83
A
BC
D
3 m
3m
4m
20 k
N/m
2EI
3EI
4EI
[FEM]load
� Overview
∆ ∆
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(060:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
B- Joint B
MBCMBA
84
A
BC
D
3 m
3m
4m
20 k
N/m
2EI
3EI
4EI
[FEM]load
wL2/12 = 15
wL2/12 = 15
A
B C
D
∆ ∆
[FEM]∆∆∆∆
6(2EI∆)/(3) 2= 1.333EI∆
6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2
= 1.5EI∆
1.5EI∆
BBCEIM θ3
)3(3=
∆++= EIEI B 333.115333.1 θ ----------(1)
∆+−= EIEI B 333.115667.2 θ ----------(2)
BEIθ3= ----------(3)
∆= EI75.0 ----------(4)
∆+++= EIEIEIM BAAB 333.1153
)2(23
)2(4 θθ0
∆+−+= EIEIEIM BABA 333.1153
)2(43
)2(2 θθ0
∆+= EIEIM DDC 75.04
)4(3 θ0
(1/2)(1.5EI∆)
� Slope-Deflection Equation
85
C
Dx
MDC
4 m
D
MAB
A
B
60 kN
MBA
1.5 m
1.5 m
Ax + ΣMC = 0:
:0=Σ+ BM
)6(188.04
−−−∆== EIMD DCx
)5(30889.0333.13
)5.1(60
−−−+∆+=
++=
EIEIA
MMA
Bx
ABBAx
θ
� Horizontal reactions
86
� Solve equation
*)1(0 −−−=+ BCBA MM
Equilibrium Conditions
Equation of moment
Horizontal reaction at support
Substitute (2) and (3) in (1*)
Substitute (5) and (6) in (2*)
From (7) and (8), solve equations;
EIEIB67.3451.5
=∆−
=θ
)6()1(67.3451.5 toinEI
andEI
Substitute B =∆−
=θ
)7(15333.1667.5 −−−=∆+ EIEI Bθ
)8(30077.1333.1 −−−−=∆−− EIEI Bθ
*)2(060 −−−=−− xx DA
)1(333.115333.1 −−−∆++= EIEIM BAB θ
)2(333.115667.2 −−−∆+−= EIEIM BBA θ
)3(3 −−−= BBC EIM θ
)4(75.0 −−−∆= EIM DC
)6(188.0 −−−∆= EIDx
)5(30889.0333.1 −−−+∆+= EIEIA Bx θ
mkNM AB •= 87.53mkNM BA •= 52.16
mkNM BC •−= 52.16mkNM DC •= 0.26
Ax = 53.48 kNDx = 6.52 kN
87
A
C
DMoment diagram
D
A
B C
Deflected shape
∆ ∆
A
BC
D
3 m
3m
4m20 k
N/m 16.52 kN�m
53.87 kN�m
53.48 kN
6.52 kN26 kN�m
5.55 kN
5.55 kN
26.
16.52
mkNM AB •= 87.53mkNM BA •= 52.16
mkNM BC •−= 52.16mkNM DC •= 0.26
Ax = 53.48 kNDx = 6.52 kN
53.87
B16.52