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Class Note for Structural Analysis 2
Fall Semester, 2013
Hae Sung Lee, Professor
Dept. of Civil and Environmental Engineering
Seoul National University
Seoul, Korea
Contents
Chapter 1 Slope Deflection Method 11.0 Comparison of Flexibility Method and Stiffness Method………………………… 21.1 Analysis of Fundamental System………………………………………………..... 51.2 Analysis of Beams………………………………………………………………… 81.3 Analysis of Frames………………………………………………………………... 17
Chapter 2 Iterative Solution Method & Moment Distribution Method 322.1 Solution Method for Linear Algebraic Equations………………………………… 332.2 Moment Distribution Method……………………………………………………... 372.3 Example - MDM for a 4-span Continuous Beam…………………………………. 422.4 Direct Solution Scheme by Partitioning…………………………………………... 442.5 Moment Distribution Method for Frames………………………………………… 45
Chapter 3 Energy Principles 473.1 Spring-Force Systems……………………………………………………………... 483.2 Beam Problems……………………………………………………………………. 493.3 Truss problems……………………………………………………………………. 53
Chapter 4 Matrix Structural Analysis 564.1 Truss Problems…………………………………………………………………… 574.2 Beam Problems…………………………………………………………………… 684.3 Frame Problems…………………………………………………………………... 76
Chapter 5 Buckling of Structures 795.0 Stability of Structures……………………………………………………………... 805.1 Governing Equation for a Beam with Axial Force………………………………... 815.2 Homogeneous Solutions…………………………………………………………... 825.3 Homogeneous and Particular solution…………………………………………….. 855.4 Energy Method……………………………………………………………………. 865.5 Approximation with the Homogeneous Beam Solutions…………………………. 895.6 Nonlinear Analysis of Truss………………………………………………………. 91
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
1
Chapter 1
Slope Deflection Method
A B
School of Civil, Urban & Geosystem Eng., SNU
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2.0 Comparison of Flexibility Method and Stiffness Method
Flexibility Method
Remove redundancy (Equilibrium)
Compatibility21 δ=δ
Pkk
kXk
XPkX
21
1
21 +=→
−=
Stiffness Method
Compatibility
δ=δ=δ 21
Equilibrium
→=δ+δ Pkk 2121 kk
P+
=δ
P
k1 k2
P
X
P
k1 k2
P
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Flexibility Method
Remove redundancy (Equilibrium)
Compatibility
EIPLPL
EIL
B 161
4)
211(
6
2
0 =×+=δ , EIL
BB 32
=δ
3230 0
0PLMM
BB
BBBBBB −=
δδ
−=→=δ+δ
Stiffness Method
Compatibility
BBCBA θ=θ=θ Equilibrium
0 , 16
3=−= f
BCf
BA MPLM , BBBC
BBA L
EIMM θ==3
EIPL
LEIPL
MMMMM
BB
BBC
BBA
fBC
fBAB
3206
163
02
=θ→=θ+−
→=+++=∑
1
++
EI
L
EI
L
L/2 P
AB
C
163PL
EI
L
EI
L
L/2 P
AB
C
θB
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Flexibility Method
1. Release all redundancies.
2. Calculate displacements induced by external loads at the releasedredundancies.
3. Apply unit loads and calculate displacements at the releasedredundancies.
4. Construct the flexibility equation by superposing the displacementbased on the compatibility conditions.
5. Solve the flexibility equation.
6. Calculate reactions and other quantities as needed.
Stiffness Method
1. Fix all Degrees of Freedom.
2. Calculate fixed end forces induced by external loads at the fixedDOF.
3. Apply unit displacements and calculate member end forces at theDOFs.
4. Construct the stiffness equation by superposing the member endforces based on the equilibrium equations.
5. Solve the stiffness equation.
6. Calculate reactions and other quantities as needed.
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2.1 Analysis of Fundamental System
2.1.1 End Rotation
Flexibility Methodi) 0=θB
036
63
=+
θ−=+
BA
ABA
MEILM
EIL
MEILM
EIL
→ AA LEIM θ−=
4 , AB LEIM θ=
2
ii) 0=Aθ
BA LEIM θ−=
2 , BB LEIM θ=
4
Sign Convention for M :Counterclockwise “+”
0≠θA , 0≠θB
BAB
BAA
LEI
LEIM
LEI
LEIM
θ+θ=
θ+θ=
42
24
2.1.2 Relative motion of joints
MA MB
∆
A B
Aθ
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Flexibility Method
LM
EILM
EIL
LM
EILM
EIL
BA
BA
∆=+
∆−=+
36
63 →LL
EIM A∆
−=6 ,
LLEIM B
∆=
6
or in the new sign convention : LL
EIM A∆
=6 ,
LLEIM B
∆=
6
Final Slope-Deflection Equation
LLEI
LEI
LEIM
LLEI
LEI
LEIM
BAB
BAA
∆+θ+θ=
∆+θ+θ=
642
624
In Case an One End is Hinged
LLEI
LEI
LLEI
LEI
LEIM
LLEI
LEI
LEI
LLEI
LEI
LEIM
BBAB
BABAA
∆+θ=
∆+θ+θ=
∆−θ−=θ→=
∆+θ+θ=
33642
320624
2.1.3 Fixed End Force
Both Ends Fixed
∆
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One End Hinged
+
Ex.: Uniform load case with a hinged left end
8243
2412
2222 qLqLqLqLM fB −=−=−−= , 0=f
AM
2.1.4 Joint Equilibrium
∑∑∑ =+−− 0jointmemberfixed FFF
or
∑∑∑ =+ jointmemberfixed FFF
MA MB
MA MA/2
AM23
AM23
Joint i
Fjoint
Fmember
Ffixed
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2.2 Analysis of Beams2.2.1 A Fixed-fixed End Beam
DOF : Bθ , ∆B
Analysis
i) All fixed : No fixed end forces
ii) 0≠θB , 0=∆B
BAB aEIM θ=
21 , BBA aEIM θ=
41 , BBC bEIM θ=
41 , BCB bEIM θ=
21
BABBA aEIVV θ=−= 2
11 6 , BCBBC bEIVV θ==− 2
11 6
iii) 0=θB , 0≠∆B
BAB aEIM ∆= 2
2 6 , BBA aEIM ∆= 2
2 6 , BBC bEIM ∆−= 2
2 6BCB b
EIM ∆−= 22 6
BABBA aEIVV ∆=−= 3
22 12 , BCBBC bEIVV ∆=−= 2
22 12
baEI
P
AB
C
BaEI
θ4
BaEI
θ2
BbEI
θ4
BbEI
θ2
BaEI
θ26
BaEI
θ26
BbEI
θ26
BbEI
θ26
BaEI
∆26
BaEI
∆26
BbEI
∆26
BbEI
∆26
BaEI
∆312
BaEI
∆312
BbEI
∆312
BbEI
∆312
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Construct the Stiffness Equation
→=+++→=∑ 00 2211BCBABCBA
iB MMMMM 0)11(6)11(4 22 =∆−+θ+ BB ba
EIba
EI
→=+++→=∑ PVVVVPV BCBABCBAi
B2211 P
baEI
baEI BB =∆++θ− )11(12)11(6 3322
PEIl
baabB 3
22
2)( −
−=θ , PEIl
baB 3
33
3=∆
Pl
abaEI
aEIMMM BBABABAB 2
2
221 62
=∆+θ=+= ,
Pl
babEI
bEIMMM BBCBCBCB 2
2
221 62
−=∆−θ=+=
2.2.2 Analysis of a Two-span Continuous Beam (Approach I)
DOF : Bθ , θC
Analysis
i) Fix all DOFs and Calculate FEM.
12
2qLM fAB = ,
12
2qLM fBA −= ,
8
2qLM fBC = ,
8
2qLM fCB −=
ii) 0≠θB , 0=θC
BAB LEIM θ=
21 , BBA LEIM θ=
41 , BBC LEIM θ=
81 , BCB LEIM θ=
41
iii) 0=θB , 0≠θC
CBC LEIM θ=
42 , CCB LEIM θ=
82
Construct the Stiffness Equation
→=++++→=∑ 00 211BCBCBA
fBC
fBA
iB MMMMMM 0412
24
2
=θ+θ+ CB LEI
LEIqL
→=++→=∑ 00 21CBCB
fCB
iC MMMM 084
8
2
=θ+θ+− CB LEI
LEI
LqL
qL
EI 2EI
LL
q
AB
C
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2
51217 qL
2
161 qL 2
81 qL 2
163 qL
L167
EIqL
B 96
3
−=θ , EI
qLC 48
3
=θ
Member End Forces
22
1
4832
12qL
LEIqLMMM BAB
fABAB =θ+=+=
22
1
814
12qL
LEIqLMMM BBA
fBABA −=θ+−=+=
22
21
8148
8qL
LEI
LEIqLMMMM CBBCBC
fBCBC =θ+θ+=++=
0848
221 =θ+θ+−=++= CBCBCB
fCBCB L
EILEIqLMMMM
Various Diagram
- Freebody Diagram
- Moment Diagram
2
161 qL 2
81 qL
qL167 qL
169 qL
85
qL83
qL1619
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2.2.3 Analysis of a Two-span Continuous Beam (Approach II)
DOF : Bθ
Analysis
i) Fix all DOFs and Calculate FEM.
12
2qLM fAB = ,
12
2qLM fBA −= ,
163
821
8
222 qLqLqLM fBC =+=
ii) 0≠θB
BAB LEIM θ=
21 , BBA LEIM θ=
41 , BBC LEIM θ=
61
Construct Stiffness Equation
00 11 =+++→=∑ BCBAf
BCf
BAB MMMMM
EIqL
LEI
LEIqLqL
BBB 96064
163
12-
222
−=θ→=θ+θ++
Member End Forces
22
1
4832
12qL
LEIqLMMM BAB
fABAB =θ+=+=
22
1
814
12qL
LEIqLMMM BBA
fBABA −=θ+−=+=
22
1
816
163 qL
LEIqLMMM BBC
fBCBC =θ+=+=
qL
EI 2EI
LL
q
A B
C
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2.2.4 Analysis of a Beam with an Internal Hinge (4 DOFs System)
DOF : Bθ , LCθ , R
Cθ , ∆C
Analysis
i) All fixed
12
2qlM fAB = ,
12
2qlM fBA −=
ii) 0≠θB
BAB lEIM θ=
21 , BBCBA lEIMM θ==
411 , BCB lEIM θ=
21 , BCB lEIV θ= 2
1 6
iii) 0≠θLC
LCBC l
EIM θ=22 , L
CCB lEIM θ=
42 , LCCB l
EIV θ= 22 6
iv) 0≠θRC
RCCD l
EIM θ=43 , R
CDC lEIM θ=
23 , LCCD l
EIV θ−= 23 6
v) 0≠∆C
CCBBC lEIMM ∆== 2
44 6 , CDCCD lEIMM ∆−== 2
44 6 , CCBCD lEIVV ∆== 2
44 12
q
EI EI EI
l l l
A B C
D
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Construct Stiffness Equation
06 0 2812
0 2
2
1 =∆++θ+θ+−→=∑ CLCB
i
i
lEI
lEI
lEIqlM
06 0 42 0 2 2 =∆++θ+θ→=∑ CLCB
i
i
lEI
lEI
lEIM
06 4 0 2 3 =∆−θ→=∑ CRC
i
i
lEI
lEIM
024666 0 3222 4 =∆+θ−θ+θ→=∑ CRC
LCB
i
i
lEI
lEI
lEI
lEIV
Elimination of LCθ and R
Cθ
- 2nd and 3rd equation
)3(2 2 CBLC l
EIl
EIl
EI∆+θ−=θ , C
RC l
EIl
EI∆=θ 23 2
- 1st equation
03 712
06 )3 (812
62812
2
2
22
2
2
2
=∆+θ+−→=∆+∆+θ−θ+−
=∆+θ+θ+−
CBCCBB
CLCB
lEI
lEIql
lEI
lEI
lEI
lEIql
lEI
lEI
lEIql
- 4th equation
063024)3(3)3(36
24666
3233322
3222
=∆+θ→=∆+∆−∆+θ−θ
=∆+θ−θ+θ
CBCCCBB
CRC
LCB
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
2.2.5 Analysis of a Beam with an Internal Hinge (2 DOFs System)
DOF : Bθ , ∆C
Analysis
i) All fixed
12
2qlM fAB = ,
12
2qlM fBA −=
q
EI EI EI
l l l
A B C
D
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ii) 0≠θB
BAB lEIM θ=
21 , BBA lEIM θ=
41 , BBC lEIM θ=
31 , BCBBC lEIVV θ−=−= 2
11 3
iii) 0≠∆C
CDCBC lEIMM ∆=−= 2
22 3 , CCBBC lEIVV ∆−=−= 2
22 3 , CDCCD lEIVV ∆=−= 2
22 3
Construct the Stiffness Equation
03 712
0 2
2
1 =∆+θ+−→=∑ CBi
i
lEI
lEIqlM
063 0 322 =∆+θ→=∑ CBi
i
lEI
lEIV
EIql
B 66
3
=θ , EI
qlC 132
4
−=∆
EIql
lllEI
lEI CR
CCR
C
3
2643
23)(32
−=∆
=θ→∆
−−=θ
EIql
EIql
EIql
LBLC 2641322
31322
321 333
=+−=∆
−θ−=θ
2.2.6 Beam with a Spring Support
Analysis
i) All fixed
12
2qlM fAB = ,
12
2qlM fBA −=
k
q
EI EI EI
l l l
A B C
D
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ii) 0≠θB
BAB lEIM θ=
21 , BBA lEIM θ=
41 , BBC lEIM θ=
31 , BCBBC lEIVV θ==− 2
11 3
iii) 0≠∆C
CDCBC lEIMM ∆=−= 2
22 3
CCBBC lEIVV ∆−=−= 2
22 3 , CDCCD lEIVV ∆=−= 2
22 3 , CS kV ∆=2
Construct the Stiffness Equation
03 712
0 2
2
1 =∆+θ+−→=∑ CBi
i
lEI
lEIqlM
0)6(3 0 322 =∆++θ→=∑ CBi
i klEI
lEIV
EIql
B 6611/1411 3
α+α+
=θ , EI
qlC 132)11/141(
1 4
α+−=∆ where 3
6lEIk α=
0→α
EIql
B 66
3
=θ , EI
qlC 132
4
−=∆
∞→α
EIql
B 84
3
=θ , 0=∆C
k∆C
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2.2.7 Support Settlement
DOF : Bθ
Analysis
i) All fixed
llEIM f
BAδ
=6 ,
llEIM f
BCδ
−=3
ii) 0≠θB
BBA lEIM θ=
41 , BBC lEIM θ=
31
Construct the Equilibrium Equation
0343601 =θ+θ+δ
−δ
→=∑ BBi
i
lEI
lEI
llEI
llEIM
lBδ
−=θ→73
2.2.8 Temperature Change
T1
T2
lh
TTlh
TTBA 2
)( , 2
)( 1212 −α−=θ
−α=θ
Fixed End Moment
EIh
TTLEI
LEIM
EIh
TTLEI
LEIM
BAB
BAA
)(42
)(24
12
12
−α−=θ+θ=
−α=θ+θ=
A B
EI EI
l l A B C
δ
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2.3 Analysis of Frames
2.3.1 A Portal Frame without Sidesway
DOF : Bθ , Cθ
Analysis
i) All fixed
80 PlM BC = ,
80 PlMCB −=
ii) 0≠θB
BAB lEIM θ= 11 2 , BBA l
EIM θ= 11 4
BBC lEIM θ= 21 4 , BCB l
EIM θ= 21 2
iii) 0≠θC
CBC lEIM θ= 22 2 , CCB l
EIM θ= 22 4
CCD lEIM θ= 12 4 , CDC l
EIM θ= 12 2
Construct the Stiffness Equation
02)44(8
0 221 =θ+θ++→=∑ CBiB l
EIlEI
lEIPlM
0)44(28
0 212 =θ++θ+−→=∑ CBiC l
EIlEI
lEIPlM
8241 2
21
PlEIEICB +
=θ=θ−
l/2 P
A
B C
D
EI1
EI2
EI1
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Member End Forces
82422
21
11 PlEIEI
EIlEIM BAB +
−=θ=
82444
21
11 PlEIEI
EIlEIM BBA +
−=θ=
824424
8 21
122 PlEIEI
EIlEI
lEIPlM CBBC +
=θ+θ+=
824442
8 21
122 PlEIEI
EIlEI
lEIPlM CBCB +
−=θ+θ+−=
82444
21
11 PlEIEI
EIlEIM CCD +
=θ=
82422
21
11 PlEIEI
EIlEIM CDC +
=θ=
In case 21 EIEI =
24PlM AB −= ,
12PlM BA −= ,
12PlM BC = ,
12PlMCB −= ,
12PlMCD = ,
24PlM DC =
2.3.2 A Portal Frame without Sidesway – hinged suppoorts
DOF : Bθ , Cθ
Analysis
i) All fixed
80 PlM BC = ,
80 PlMCB −=
l/2 P
A
B C
D
EI1
EI2
EI1
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ii) 0≠θB
BBA lEIM θ= 11 3
BBC lEIM θ= 21 4 , BCB l
EIM θ= 21 2
iii) 0≠θC
CBC lEIM θ= 22 2 , CCB l
EIM θ= 22 4
CCD lEIM θ= 12 3
Construct the Stiffness Equation
02)43(8
0 221 =θ+θ++→=∑ CBiB l
EIlEI
lEIPlM
0)43(28
0 212 =θ++θ+−→=∑ CBiC l
EIlEI
lEIPlM
8231 2
21
PlEIEICB +
=θ=θ−
Member End Forces
0=ABM
82333
21
11 PlEIEI
EIlEIM BBA +
−=θ=
823324
8 21
122 PlEIEI
EIlEI
lEIPlM CBBC +
=θ+θ+=
823342
8 21
122 PlEIEI
EIlEI
lEIPlM CBCB +
−=θ+θ+−=
82333
21
11 PlEIEI
EIlEIM CCD +
=θ=
0=DCM
In case of 21 EIEI =
0=ABM , PlM BA 403
−= , PlM BC 403
= , PlMCB 403
−= , PlMCD 403
= , 0=DCM
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2.3.3 A Frame with an horizontal force
DOF : Bθ , ∆
Analysis
i) All fixed : None fixed end momentii) 0≠θB
BAB lEIM θ=
21 , BBA lEIM θ=
41
BBC lEIM θ=
31 , BBA lEIV θ= 2
1 6
iii) 0≠∆
∆= 22 6
lEIM AB , ∆= 2
2 6lEIM BA
∆= 32 12
lEIVBA
Construct the stiffness equation
06)34( 0 2 =∆+θ+→=∑ lEI
lEI
lEIM B
iB
PlEI
lEIPV B
i =∆+θ→=∑ 32126
EIPl
EIPl
B 487 ,
8
32
=∆−=θ
Member end forces
PllEI
lEIM BAB 8
5622 =∆+θ= , Pl
lEI
lEIM BBA 8
3642 =∆+θ= , Pl
lEIM BBC 8
33−=θ=
P
A
B C
EI
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2.3.4 A Portal Frame with an Unsymmetrical Load
DOF : Bθ , Cθ , ∆
Analysis
i) All fixed
2
20
lPabM BC = , 2
20
lbPaMCB −=
ii) 0≠θB
BAB lEIM θ=
21 , BBA lEIM θ=
41
BBC lEIM θ=
41 , BCB lEIM θ=
21 , BBA lEIV θ= 2
1 6
iii) 0≠θC
CBC lEIM θ=
22 , CCB lEIM θ=
42
CCD lEIM θ=
42 , CDC lEIM θ=
22 , CCD lEIV θ= 2
2 6
a P
A
B C
D
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iv) 0≠∆
∆== 233 6
lEIMM BAAB , ∆== 2
33 6lEIMM DCCD ,
∆== 333 12
lEIVV CDBA
Construct the Stiffness Equation
06 28 0 22
2
=∆+θ+θ+→=∑ lEI
lEI
lEI
lPabM cB
iB
06 82 0 22
2
=∆+θ+θ+−→=∑ lEI
lEI
lEI
lbPaM cB
iC
02466 0 322 =∆+θ+θ→=∑ lEI
lEI
lEIV cB
i
)(4 CBl
θ+θ−=∆
0 22
13 2
2
=θ+θ+ cB lEI
lEI
lbPa
0 2
1322
2
=θ+θ+− cB lEI
lEI
lPab
lba
EIPab
B)13(
841 +
−=θ
lba
EIPab
C)13(
841 +
=θ
)(281 ab
EIPab
−=∆
2.3.5 A Portal Frame with a Bracing (Vertical Load)
DOF : Bθ , Cθ , ∆
Analysis
i) All fixed
2
20
lPabM BC = , 2
20
lbPaMCB −=
4la =
a P
A
B C
D
0,0 ≠= EAEI
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23
ii) 0≠θB
BAB lEIM θ=
21 , BBA lEIM θ=
41
BBC lEIM θ=
41 , BCB lEIM θ=
21
BBA lEIV θ= 2
1 6
iii) 0≠θC
CBC lEIM θ=
22 , CCB lEIM θ=
42
CCD lEIM θ=
42 , CDC lEIM θ=
22
CCD lEIV θ= 2
2 6
iv) 0≠∆
∆== 233 6
lEIMM BAAB ,
∆== 233 6
lEIMM DCCD ,
∆== 333 12
lEIVV CDBA
22∆
=l
EAABD (C) 222
122
∆==
∆=→
lEAV
lEAV BDBD
Construct the Stiffness Equation
06 28 0 22
2
=∆+θ+θ+→=∑ lEI
lEI
lEI
lPabM cB
iB
06 82 0 22
2
=∆+θ+θ+−→=∑ lEI
lEI
lEI
lbPaM cB
iC
0)1(2466 0 322 =∆α++θ+θ→=∑ lEI
lEI
lEIV cB
i
EIEAll
CB 248 , )(
411 2
=αθ+θα+
−=∆
Solution for ab 3=
EIPl
B
2
)107(2565240
α+α+
−=θ , EIPl
C
2
)107(2562816
α+α+
−=θ , EIPl3
)107(1283
α+=∆
For a hw× rectangular section and hl 20= , 250=α .
EIPl
B
2
0203.0−=θ , EIPl
C
2
0109.0 =θ EIPl3
4103282.0 −×=∆
∆
2∆
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Performance
Responsewith Bracing( 250=α )
w/o bracing( 0=α ) Ratio(%)
θB )/( 2 EIPl× -0.0203 -0.0223 91.03
θC )/( 2 EIPl× 0.0109 0.0089 122.47
∆ )/( 3 EIPl× 0.3282×10-4 0.0033 0.99
ΜΑΒ (Pl) -0.0404 -0.0248 162.90
ΜΒΑ (Pl) -0.0810 -0.0694 116.71
ΜCD (Pl) 0.0438 0.0554 79.06
ΜDC (Pl) 0.0220 0.0376 58.51
ΜP (Pl) 0.1158 0.1216 95.23
ABD (P) 0.0788 - -
Pmax (Pall)* 0.0720 0.0685 105.1
Pmax/vol. 0.0163 0.0228 71.5
*) whP allall σ= , 6/hPM allall =
Unbalanced shear force in the columns = PlEI
CB 0564.0)(62 =θ+θ
The bracing carries 99 % of the unbalanced shear force between the two columns.
2.3.6 A Portal Frame with a Bracing (Horizontal Load)
DOF : Bθ , Cθ , ∆
Analysis
i) All fixed: No fixed end forces
ii)-iv) the same as the previous case
P
A
B C
D
0,0 ≠= EAEI
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Construct the Stiffness Equation
06 28 0 2 =∆+θ+θ→=∑ lEI
lEI
lEIM cB
iB
06 82 0 2 =∆+θ+θ→=∑ lEI
lEI
lEIM cB
iC
PlEI
lEI
lEIV cB
i =∆α++θ+θ→=∑ )1(2466 0 322
CB θ=θ , ll CB θ−=θ−=∆35
35
Solution
EIPl
EIPl
CB
32
)4028(35,
)4028(1θθ
α+=∆
α+−==
For 250=α ,
EIPl
CB
23103501.0 −×−=θ=θ ,
EIPl3
3105835.0 −×=∆
Performance
Responsewith Bracing( 250=α )
w/o bracing( 0=α ) Ratio(%)
θB )/( 2 EIPl× 3103501.0 −×− 1103571.0 −×− 0.98
θC )/( 2 EIPl× 3103501.0 −×− 1103571.0 −×− 0.98
∆ )/( 3 EIPl× 3105835.0 −× 1105952.0 −× 0.98
ΜΑΒ (Pl)2102801.0 −× 0.2857 0.98
ΜΒΑ (Pl)2102101.0 −× 0.2143 0.98
ΜCD (Pl)2102101.0 −× 0.2143 0.98
ΜDC (Pl)2102801.0 −× 0.2857 0.98
ABD (P) 1.4004 - -
Pmax(Pall)* 0.7141 0.0292 2448
Pmax/vol. 0.1617 0.0097 1670
*) Governed by ABD for the structure with bracing, and by MDC for the structure withoutbracing. whP allall σ= , 6/hPM allall =
The bracing carries about 99% of the external horizontal load.
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2.3.7 A Portal Frame with a Spring
DOF : Bθ , Cθ , ∆ Analysis
iv) 0≠∆
∆== 233 6
lEIMM BAAB , ∆== 2
33 6lEIMM DCCD ,
∆== 333 24
lEIVV CDBA , ∆= kVS
3
Construct the Stiffness Equation
06 28 0 22
2
=∆+θ+θ+→=∑ lEI
lEI
lEI
lPabM cB
iB
06 82 0 22
2
=∆+θ+θ+−→=∑ lEI
lEI
lEI
lbPaM cB
iC
∆−=∆+θ+θ→=∑ klEI
lEI
lEIV cB
i322
2466 0
0)24(66 322 =∆++θ+θ klEI
lEI
lEI
cB
Deformed Shapes ( 41.0=∆∆S )
without a spring with a spring ( )24 3lEIk =
k∆
k
a P
A
BC
D
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2.3.8 A Portal Frame Subject to Support Settlement
DOF : Bθ , Cθ , ∆
Analysis
i) All fixed
δ== 200 6
lEIMM CBBC
ii)-iv) the same as the previous problem
Construct the Stiffness Equation
06 286 0 22 =∆+θ+θ+δ→=∑ lEI
lEI
lEI
lEIM cB
iB
06 826 0 22 =∆+θ+θ+δ→=∑ lEI
lEI
lEI
lEIM cB
iC
02466 0 322 =∆+θ+θ→=∑ lEI
lEI
lEIV cB
i
A
B C
D
δ
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2.3.9 A Portal Frame with Unsymmetrical Supports
DOF : Bθ , Cθ , ∆
Analysis
i) All fixed
80 PlM BC = ,
80 PlMCB −=
ii) 0≠θB
BBA lEIM θ=
31
BBC lEIM θ=
41 , BCB lEIM θ=
21
BBA lEIV θ= 2
1 3
iii) 0≠θC
CBC lEIM θ=
22 , CCB lEIM θ=
42
CCD lEIM θ=
42 , CDC lEIM θ=
22
CCD lEIV θ= 2
2 6
iv) 0≠∆
∆= 23 3
lEIM BA ,
∆== 233 6
lEIMM DCCD ,
∆= 33 3
lEIVBA , ∆= 3
3 12lEIVCD
l/2 P
A
B C
D
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Construct the Stiffness Equation
03 278
0 2 =∆+θ+θ+→=∑ lEI
lEI
lEIPlM cB
iB
06 828
0 2 =∆+θ+θ+−→=∑ lEI
lEI
lEIPlM cB
iC
01563 0 322 =∆+θ+θ→=∑ lEI
lEI
lEIV cB
i
)2(5 CBl
θ+θ−=∆
EIPl
B
2
441
−=θ , EIPl
C
2
89
441 =θ ,
EIPl3
1761
−=∆
Load Location that Causes No Sidesway
CBCBl
θ−=θ→=θ+θ−=∆ 20)2(5
- Stiffness equation
0 27 0 2
2
=θ+θ+→=∑ cBiB l
EIlEI
lPabM , 0 82 0 2
2
=θ+θ+−→=∑ cBiC l
EIlEI
lbPaM
0122
2
=θ− ClEI
lPab , 0 4
2
2
=θ+− clEI
lbPa
abl
bPal
Pab 33 2
2
2
2
=→=
a P
A
B C
D
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2.3.10 A Frame with a Skewed Member
DOF : Bθ , ∆
Analysis
i) All fixed : PlPlPlM BC 163
1680 =+= , PVBC 16
11220 −=
ii) 0≠θB
BBAB lEI
lEIM θ=θ= 22
21 , BBA lEIM θ= 221 , BBC l
EIM θ=31 , BBA l
EIV θ= 21 3
BBC lEIV θ−= 2
1
223
iii) 0≠∆
∆=∆
== 222 3
226
lEI
llEIMM ABBA ∆−=
∆−= 2
2
223
23
lEI
llEIM BC , ∆= 3
2 23lEIVBA ,
∆= 32
23
lEIVBC
P
A
B C
BBB lEI
llEI
lEI
θ=θ+θ 2321)222(
BB lEI
llEI
θ=θ 2223
2213
BlEI
llEI
lEI
θ=∆+∆ 322 2321)33(
∆=∆ 32 23
211
223
lEI
llEI
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31
Construct the stiffness equation
PllEI
lEIM B
iB 16
3 )2
233()32(20 2 −=∆−+θ+→=∑
PlEI
lEIV B
i
1611
22)
2323()2
23(3 0 32 =∆++θ−→=∑
PllEI
lEI
B 1875.0 8787.05.8284 2 −=∆+θ
PlEI
lEI
B 4861.07426.58787.0 32 =∆+θ
EIPl
B
2
0460.0−=θ , EIPl3
0917.0=∆
Results
- Deformed shape
- Moment diagram
- Shear force diagram
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Chapter 3
Iterative Solution Method&
Moment Distribution Method
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3.1 Solution Method for Linear Algebraic Equations
3.1.1 Direct Method – Gauss Elimination
nibXa
bXaXaXaXa
bXaXaXaXa
bXaXaXaXabXaXaXaXa
i
n
jjij
nnnnininn
ininiiiii
nni
nnii
L
LL
M
LL
M
LL
LL
1for 1
2211
2211
222222121
111212111
==→
=+++++
=+++++
=+++++=+++++
∑=
or in a matrix form
)()]([ bXA =
By multiplying 11
1
aai to the first equation and subtracting the resulting equation from the i-th
equation for ni ≤≤2 , the first unknown X1 is eliminated from the second equation as fol-
lows.
)2()2()2(22
)2(2
)2()2()2(2
)2(2
)2(2
)2(2
)2(22
)2(22
111212111
nnnninn
ininiiii
nni
nnii
bXaXaXa
bXaXaXa
bXaXaXabXaXaXaXa
=++++
=++++
=++++=+++++
LL
M
LL
M
LL
LL
where 11
11)2(
aaa
aa jiijij −= . Again, the second unknown X2 is eliminated from the third equa-
tion by multiplying )2(22
)2(2
aai to the second equation and subtracting the resulting equation from
the i-th equation for ni ≤≤3 . The aforementioned procedures are repeated until the last
unknown remains in the last equation.
)()(
)()()(
)2(2
)2(2
)2(22
)2(22
111212111
nnn
nnn
iin
iini
iii
nni
nnii
bXa
bXaXa
bXaXaXabXaXaXaXa
=
=++
=++++=+++++
MM
L
MMM
LL
LL
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where 11,1
)1(,1
)1(1,)1()(−
−−
−−
−−− −= k
kk
kjk
kkik
ijk
ij aaa
aa njik ≤≤ , , and ijij aa =1 . Once the system matrix is tri-
angularized, the solution of the given system is easily obtained by the back-substitution.
→= )()( nnn
nnn bXa )(
)(
nnn
nn
n ab
X =
→=+ −−
−−−
−−−
)1(1
)1(,11
)1(1,1
nnn
nnnn
nnn bXaXa 2
1,1
)1(,1
)1(1
1 −−−
−−
−−
−
−= n
nn
nn
nnn
nn a
XabX
→=++ )()(,
)( iin
inii
iii bXaXa L 1
1)()(
−
+
=∑−
= iii
i
nkk
iik
ii
i a
XabX for 11 −≤≤ ni
3.1.2 Iterative Method – Gauss-Jordan Method
A system of linear algebraic equations may be solved by iterative method. For this purpose,
the given system is rearranged as follows.
nn
nnnnnn
ii
niniiiiiiiii
nn
nn
aXaXab
X
aXaXaXaXab
X
aXaXaXab
X
aXaXab
X
)(
)(
)(
)(
11,11
11,11,11
22
232312122
11
121211
−−
−+−−
++−=
+++++−=
+++−=
++−=
L
LL
L
L
Suppose we substitute an approximate solution 1)( −kX into the right-hand side of the above
equation, a new approximate solution k)(X , which is not the same as 1)( −kX , is obtained.
This procedure is repeated until the solution converges.
))((1)( 11
−
≠=
∑−= k
n
jij
jijiii
ki Xaba
X
where the subscript k denotes the iterational count.
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3.1.3 Iterative Method – Gauss-Siedal Method
When we calculate a new Xi value in the k-th iteration of Gauss-Jordan iteration, the values of
11 ,, −iXX L are already updated, and we can utilize the updated values to accelerate conver-
gence rate, which leads to the Gauss-Siedal Method.
nn
knnnknnkn
ii
kninkiiikiiikiiki
knnkkk
knnkk
aXaXab
X
aXaXaXaXab
X
aXaXaXab
X
aXaXab
X
))()(()(
))()()()(()(
))())(()(
))()(()(
11,11
1111,11,11
22
12132312122
11
11121211
−−
−−++−−
−−
−−
++−=
+++++−=
+++−=
++−=
L
LL
L
L
))()((1)( 11
1
11
−
<+=
−
>=
∑∑ −−= k
n
niij
jijk
i
ij
jijiii
ki XaXaba
X
3.1.4 Example
Stiffness Equation
0)36(30.4503.133
03)63(3.1337.168
=θ++θ++−
=θ+θ+++−
CB
CB
lEI
lEI
lEI
lEI
lEI
lEI
For the simplicity of derivation, CCBB lEI
lEI
θ→θθ→θ , . The stiffness equation becomes
0937.316 0394.35
=θ+θ+=θ+θ+−
CB
CB
A B C D
3@30=90EI 1.5EI EI
30 35.6 4klf
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Gauss-Jordan Iteration
))(37.316(91)(
))(34.35(91)(
1
1
−
−
θ−=θ
θ−=θ
kBkC
kCkB
Gauss-Siedal Iteration
))(37.316(91)(
))(34.35(91)( 1
kBkC
kCkB
θ−=θ
θ−=θ −
Gauss-Seidal Gauss-JordanGAUSS-SIEDAL ITERATION======================***** Iteration 1***** X(1) = 0.3933334E+01 X(2) = -0.3650000E+02 ERROR = 0.1000000E+01***** Iteration 2***** X(1) = 0.1610000E+02 X(2) = -0.4055556E+02 ERROR = 0.2939146E+00***** Iteration 3***** X(1) = 0.1745185E+02 X(2) = -0.4100617E+02 ERROR = 0.3197497E-01***** Iteration 4***** X(1) = 0.1760206E+02 X(2) = -0.4105624E+02 ERROR = 0.3544420E-02***** Iteration 5***** X(1) = 0.1761875E+02 X(2) = -0.4106181E+02 ERROR = 0.3937214E-03
****** MOMENT ******
MBA =-115.84375 MBC = 115.82707 MCB =-326.81460 MCD = 326.81458
GAUSS-Jordan ITERATION======================***** Iteration 1***** X(1) = 0.3933334E+01 X(2) = -0.3518889E+02 ERROR = 0.1000000E+01***** Iteration 2***** X(1) = 0.1566296E+02 X(2) = -0.3650000E+02 ERROR = 0.2971564E+00***** Iteration 3***** X(1) = 0.1610000E+02 X(2) = -0.4040988E+02 ERROR = 0.9044394E-01***** Iteration 4***** X(1) = 0.1740329E+02 X(2) = -0.4055556E+02 ERROR = 0.2971564E-01***** Iteration 5***** X(1) = 0.1745185E+02 X(2) = -0.4098999E+02 ERROR = 0.9812154E-02
****** MOMENT ******
MBA =-116.34444 MBC = 115.04115 MCB =-326.88437 MCD = 327.03004
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3.2 Moment Distribution Method
At the Joint B
- Moment distribution
BBCB
BC
BC
AB
AB
BC
BC
BBC
BCBC
BBAB
BC
BC
AB
AB
AB
AB
BAB
ABBA
BC
BC
AB
AB
BBBCBAB
BC
BC
AB
ABB
MDM
LEI
LEI
LEI
LEI
M
MDM
LEI
LEI
LEI
LEI
M
LEI
LEI
MMMLEI
LEIM
=+
=θ=
=+
=θ=
+=θ→+=θ+=
43
44
43
33
43 )
43(
1
1
11
- Moment carry over to joint C: BBCBBC
BCCB MD
LEI
M2121 =θ=
At the Joint C
- Moment distribution
CCDC
CD
CD
BC
BC
CD
CD
CCD
CDCD
CCBC
CD
CD
BC
BC
BC
BC
CBC
BCCB
CD
CD
BC
BC
CCCDCBC
CD
CD
BC
BCC
MDM
LEI
LEI
LEI
LEI
M
MDM
LEI
LEI
LEI
LEI
M
LEI
LEI
MMM
LEI
LEI
M
=+
=θ=
=+
=θ=
+=θ→+=θ+=
34
33
34
44
34 )
34(
2
2
22
- Moment carry over to joint B: 2122
CCBCBC
BCBC MD
LEI
M =θ=
A B C D
3@30=90EI 1.5EI EI
30 35.6 4klf
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Stiffness Equation in terms of Moment at Joints
→
=++
=++−
0217.316
02135.4
CBBC
CCBB
MMD
MDM
BBCC
CCBB
MDM
MDM
217.316-
214.35
−=
−=
- Gauss-Siedal Approach
kBBCkC
kCCBkB
MDM
MDM
)(21316.7 )(
)(214.35)( 1
−−=
−= −
- Gauss-Jordan Approach
1
1
)(21316.7 )(
)(214.35)(
−
−
−−=
−=
kBBCkC
kCCBkB
MDM
MDM
For the given structure
32 ,
31 ==== CBBCCDBA DDDD
Incremental form for the Gauss-Siedal Method
- For 1=k
0)()( 00 == CB MM because we assume all degrees of freedom are fixed for step 0.
5.328)(4.3532
217.316)(
217.316)(
4.35)(4.35)(214.35)(
111
101
−=∆→−−=−−=
=∆→=−=
CBBCC
BcCBB
MMDM
MMDM
- For 1>k
kBBCkCkBBCkC
kBBCkBBCkBBCkC
kcCBkBkcCBkB
kcCBkcCBkcCBkB
MDMMDM
MDMDMDM
MDMMDM
MDMDMDM
)(21)()(
21)(
)(21)(
217.316)(
217.316)(
)(21)()(
21)(
)(21)(
214.35)(
214.35)(
1
1
111
121
∆−=∆→∆−=
∆−−−=−−=
∆−=∆→∆−=
∆−−=−=
−
−
−−−
−−−
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- Iteration 1
0)( 0 =BM , 0)( 0 =CM
4.35)(4.35)(214.35)( 101 =∆→=−= BCCBB MMDM
6.23)(6.233.1334.3567.03.133)()(
8.11)(8.117.1684.3533.07.168)()(
111
111
=∆→+=×+=∆+=
=∆→+−=×+−=∆+=
BCBBCf
BCBC
BABBAf
BABA
MMDMM
MMDMM
5.328)(5.3284.3532
217.316)(
217.316)( 111 −=∆→−=−−=∆−−= cBBCC MMDM
5.109)(5.1094505.32833.0450)()(
0.2198.11)(
0.2198.113.133)()(21)(
111
1
111
−=∆→−=×−=∆+=
−=∆
→−+−=∆+∆+=
CDCCDf
CDCD
CB
CCBBBCf
CBCB
MMDMM
M
MDMDMM
- Iteration 2
−=∆=∆−=
∆+∆=∆→−=∆−=∆
+−=
∆+∆=∆
=∆=∆
→=∆−=∆
2.12)()(3.245.36
)()(21)(
5.36)(21)(
0.735.109
)()(21)(
5.36)()(
5.109)(21)(
22
222
22
212
22
12
CCDCD
CCBBBCCB
BBCC
BBCcCBBC
BBABA
cCBB
MDM
MDMDMMDM
MDMDM
MDM
MDM
- Iteration 3
−=∆=∆
−=∆+∆=∆→−=∆−=∆
+−=
∆+∆=∆
=∆=∆
→=∆−=∆
4.1)()(
7.21.4)()(21)(1.4)(
21)(
2.82.12
)()(21)(
1.4)()(
2.12)(21)(
33
33333
323
33
23
CCDCD
CCBBBCCBBBCC
BBCcCBBC
BBABA
cCBB
MDM
MDMDMMDM
MDMDM
MDM
MDM
- Final Moments
∑
∑
∑
∑
=−−−=∆+=
−=−+−+−+−=∆+=
=+−++−++=∆+=
−=+++−=∆+=
kkCD
fCDCD
kkCB
fCBCB
kkBC
fBCBC
kkBA
fBABA
MMM
MMM
MMM
MMM
9.3264.12.125.1090.450)(
9.326)7.21.4()3.245.36()0.2198.11(3.133)(
4.116)2.82.12()0.735.109(6.233.133)(
3.1161.45.368.117.168)(
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40
0.33 0.66 0.67 0.33
-168.7
11.8
36.5
4.1
0.5
-115.8
133.3
23.6
-109.5
73.0
-12.2
8.2
-1.4
0.9
115.9
-133.3
11.8
-219.0
36.5
-24.3
4.1
-2.7
-326.9
450.0
-109.5
-12.2
-1.4
326.9
Incremental form for the Gauss-Jordan Method
- For 1=k
0)()( 00 == CB MM because we assume all degrees of freedom are fixed for step 0.
7.316)(7.316)(217.316)(
4.35)(4.35)(214.35)(
101
101
−=∆→−=−−=
=∆→=−=
CBBCC
BcCBB
MMDM
MMDM
- For 1>k
111
121
111
121
)(21)()(
21)(
)(21)(
217.316)(
217.316)(
)(21)()(
21)(
)(21)(
214.35)(
214.35)(
−−−
−−−
−−−
−−−
∆−=∆→∆−=
∆−−−=−−=
∆−=∆→∆−=
∆−−=−=
kBBCkCkBBCkC
kBBCkBBCkBBCkC
kcCBkBkcCBkB
kcCBkcCBkcCBkB
MDMMDM
MDMDMDM
MDMMDM
MDMDMDM
- Iteration 1
0)( 0 =BM , 0)( 0 =CM
4.35)(4.35)(214.35)( 101 =∆→=−= BCCBB MMDM
6.23)(6.233.1334.3567.03.133)()(
8.11)(8.117.1684.3533.07.168)()(
111
111
=∆→+=×+=+=
=∆→+−=×+−=+=
BCBBCf
BCBC
BABBAf
BABA
MMDMM
MMDMM
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7.316)(7.3167.316)(217.316)( 101 −=∆→−=−=−−= cBBCC MMDM
5.104)(5.1044507.31633.0450)()(
2.212)( 2.2123.1337.31667.03.133)()(
111
1
11
−=∆→−=×−=+=
−=∆→−−=×−−=+=
CDCCDf
CDCD
CB
CCBf
CBCB
MMDMM
MMDMM
- Iteration 2
−=∆=∆−=
∆+∆=∆→−=∆−=∆
+−=
∆+∆=∆
=∆=∆
→=∆−=∆
9.3)()(9.78.11
)()(21)(
8.11)(21)(
1.711.106
)()(21)(
0.35)()(
1.106)(21)(
22
212
12
212
22
12
CCDCD
CCBBBCCB
BBCC
BBCcCBBC
BBABA
cCBB
MDM
MDMDMMDM
MDMDM
MDM
MDM
- Iteration 3
−=∆=∆−=
∆+∆=∆→−=∆−=∆
+−=
∆+∆=∆
=∆=∆
→=∆−=∆
7.11)()(9.236.35
)()(21)(
6.35)(21)(
7.20.4
)()(21)(
3.1)()(
0.4)(21)(
33
333
23
323
33
23
CCDCD
CCBBBCCB
BBCC
BBCCCBBC
BBABA
CCBB
MDM
MDMDMMDM
MDMDM
MDM
MDM
0.33 0.66 0.67 0.33
-168.711.8
35.0
1.3
4.0
0.2
0.5-115.9
133.323.6
-106.171.1-4.02. 7
-12.08.0
-0.50.3
-1.40.9
115.9
-133.3-212.2
11.8-7. 935.6
-23.91.4
-0.94.0
-2.70.20.1
-328.0
450.0-104.5
-3.9
-11.7
-0.5
-1.3
0.1328.2
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3.3 Example - MDM for a 4-span Continuous Beam
4.0)5.144(4 =+=LEI
LEI
LEIDBA , 6.0)5.144(5.14 =+=
LEI
LEI
LEIDBC , 5.0)5.145.14(5.14 =+=
LEI
LEI
LEIDCB ,
5.0)5.145.14(5.14 =+=LEI
LEI
LEIDCD , 67.0)35.14(5.14 =+=
LEI
LEI
LEIDDC , 33.0)35.14(3 =+=
LEI
LEI
LEIDDE
Gauss-Siedal Approach
0.0 0.0 125 -125 0.0 0.0 93.8-25.0 -50.0 -75 -37.5
40.6 81.3 81.3 40.6-45.0 -90.0 -44.3
11.3 22.5 22.5 11.3-10.4 -20.8 -31.1 -15.6
3.9 7.8 7.8 3.9-5.1 -10.2 -5.0
1.3 2.6 2.6 1.3-1.1 -2.1 -3.1 -0.9 -0.436.5 -72.9 72.9 -63.9 64.1 -44.0 44.1
0.4 0.6 0.5 0.5 0.67 0.33
EI, L
10050
A CB D EEI, L1.5EI, L 1.5EI, L
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Gauss-Jordan Approach
0.0 0.0 125 -125 0.0 0.0 93.8-50 -75 62.5 62.5 -62.8 -31.0
-25.0 31.3 -37.5 -31.4 31.3-12.5 -18.8 34.5 34.5 -21.0 -10.3
-6.3 17.3 -9.4 -10.5 17.3-6.9 -10.4 10.0 10.0 -11.6 -5.7
-3.5 5.0 -5.2 -5.8 5.0-2.0 -3.0 5.5 5.5 -3.4 -1.6
-1.0 2.8 -1.5 -1.7 2.8-1.1 -1.7 1.6 1.6 -1.9 -0.9
-0.6 0.8 -0.9 -1.0 0.8-0.3 -0.5 1.0 1.0 -0.5 -0.3
-36.4 -72.8 72.8 -64.4 64.7 -44.0 44.0
0.4 0.6 0.5 0.5 0.67 0.33
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3.4 Direct Solution Scheme by Partitioning
Slope deflection (Stiffness) Equation
→=
−+
∆θθ
00.04.448.88
2466
682
628
C
B
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
00
)()()(
)(][=
+
∆Θ
∆∆θ∆
∆θθθ PKKKK
∆∆θ
−θθ
−θθ∆θθθ Θ+Θ=∆−−=Θ→∆−−=Θ )()()(][)(][)()()()]([ 11 PKKPKKPK
0))(())(( =∆+Θ+Θ ∆∆∆
θ∆θ∆ KKK P
Direct Solution Procedure by Partitioning
- Assume ∆ = 0 and calculate (Θ)P.
- Assume an arbitrary α∆
=∆ and calculate ∆Θ)( .
- By linearity, α
Θ=Θ
∆∆ )()(
- Calculate α by the second equation by ∆ and ∆Θ)( .
∆θ∆∆∆
θ∆
Θ+∆Θ
−=α))((
))((KK
K P
- Obtain (Θ) by ∆Θα+Θ=Θ )()()( P .
L/320
A
B C
D
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3.5 Moment Distribution Method for Frames
Solution Procedure
- Assume there is no sideway and do the MDM.
- Perform the MDM again for an assumed sidesway.
- Adjust the Moment obtained by the second MDM to satisfy the second equation.
- Add the adjusted moment to the moment by the first MDM.
First MDM – with no Sidesway
-44.4
-8.3
-0.6
-53.3
88.8
-44.4
16.6
8.3
1.1
-0.6
53.3
-44.4
-22.2
33.3
-4.2
2.1
-0.3
0.2
-35.5
33.3
2.1
0.2
35.5
MAB = -53.3/2 = -26.7, MDC = -35.5/2 = 17.8, VAB = 2.67, VDC = -1.78, VP = 0.89
Second MDM – with an Arbitrary Sidesway
10.0
-5.0
1.0
0.1
6.1
-5.0
-1.9
1.0
-0.2
0.1
-6.0
-2.5
-3.8
0.5
-0.3
-5.9
10.0
-3.8
-0.3
5.9
MAB = 10 – 3.9/2.0 = 8.1, MDC = 10 – 4.1/2.0 = 7.9, VAB =-0.47, VDC = -0.46, ∆V = -0.93
C
A D
B
A
L/3 20B C
D
0.5
0.5
0.5
0.5
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Shear Equilibrium Condition (the 2nd equation)
No Sidesway (1st MDM) Sidesway Only (2nd MDM)
α = -0.89/(-0.93) = 0.97
Total Moment = 1st Moment + α 2nd Moment
26.7
53.32.67
17.8
35.51.78
0.89
8.1
6.1
0.47
7.9
5.9
0.46
0.93
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47
Chapter 4
Energy Principles
Principle of Minimum Potential Energy and
Principle of Virtual Work
hMgP
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48
Read Chapter 11 (pp.420~ 428) of Elementary Structural Analysis 4th Edition by C .H. Norris
et al very carefully. In this note an overbarred variable denotes a virtual quantity. The vir-
tual displacement field should satisfy the displacement boundary conditions of supports if
specified. For beam problems, displacement boundary conditions include boundary condi-
tions for rotational angle. Variables with superscript e denote the exact solution that satisfies
the equilibrium equation(s).
4.1 Spring-Force Systems
Total Potential energy
The energy required to return a mechanical system to a reference status
∆−∆=Π+Π=Π
∆−=Π∆=−∆=−∆−=Π ∫∫∆
∆
Pk
Pkduukduuk
exttotal
ext
2int
2
0
0
int
21
, 21)()(
Equilibrium Equation
k∆e=P
Principle of Minimum Potential Energy for an arbitrary displacement ∆+∆=∆ e .
etotal
etotal
ee
eee
eee
eetotal
k
kPk
PkkPk
Pkkk
Pk
Π≥∆+Π=
∆+∆−∆=
−∆∆+∆+∆−∆=
∆+∆−∆+∆∆+∆=
∆+∆−∆+∆=Π
2
22
22
22
2
)(21
)(21)(
21
)()(21)(
21
)()(21)(
21
)()(21
In the above equation, the equality sign holds if and only if 0=∆ . Therefore the total
potential energy of the spring-force system becomes minimum when displacement of
spring satisfies the equilibrium equation.
∆
P
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4.2 Beam Problems
Potential Energy of a Beam
∫∫
∫∫∫
−=Π+Π=Π
−=Π==Π
ll
exttotal
l
ext
ll
qwdxdxdx
wdEI
qwdxdxdx
wdEIdxEIM
00
22
2
int
00
22
2
0
2
int
)(21
, )(21
21
Equilibrium Equation
qdx
wdEIqdx
MdEIee
=−= 4
4
2
2
or
Principle of Minimum Potential Energy for a virtual displacement www e += .
wdxdx
wdEIdx
wd
qdxwdxEIMMdx
dxwdEI
dxwd
qdxwdxdx
wdEIEIdx
wdEIdxdx
wdEIdx
wd
qdxwdxdx
wdEIdx
wddxdx
wdEIdx
wd
qdxwdxdx
wdEIdx
wd
qdxwwdxdx
wwdEIdx
wwd
el
e
ll ele
ll ele
ll el
le
l ee
le
l eeh
virtualallfor )(21
)(21
)(1)()(21
)()(21
)(21
)())()((21
02
2
2
2
0002
2
2
2
002
2
2
2
02
2
2
2
002
2
2
2
02
2
2
2
002
2
2
2
002
2
2
2
Π≥+Π=
−++Π=
−−−++Π=
−+
+−=
+−++
=Π
∫
∫∫∫
∫∫∫
∫∫∫
∫∫
∫∫
Since the equation in the box represents the total virtual work in a beam, the total potential energy
of a beam becomes minimum for all virtual displacement fields when the principle of virtu-
al work holds. In the above equation, the equality sign holds if and only if 0=w .
w we
wany type of support any type of support
q
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50
Principle of Virtual Work
If a beam is in equilibrium, the principle of the virtual work holds for the beam,.
0)(0
4
4
=−∫ dxqdx
wdEIwl e
for all virtual displacement w
00
3
3
02
2
002
2
2
2
=−+− ∫∫lelell e
dxwdEIw
dxwdEI
dxwdqdxwdx
dxwdEI
dxwd
00000
2
2
2
2
=−=− ∫∫∫∫ qdxwdxEIMMqdxwdx
dxwdEI
dxwd ll ell e
In case that there is no support settlement, the boundary terms in above equation vanishes
identically since either virtual displacement including virtual rotational angle or corre-
sponding forces (moment and shear) vanish at supports. The principle of virtual work
yields the displacement of an arbitrary point x~ in a beam by applying an unit load at x~
and by using the reciprocal theorem.
∫∫∫∫ ==−δ==l elll
dxEIMMxwdxxxwdxqwqdxw
0000
)~()~(
Approximation using the principle of minimum potential energy
- Approximation of displacement field
∑=
=n
iii gaw
1
- Total potential energy by the assumed displacement field
∫∑∫ ∑∑∫∫===
−′′′′=−=Πl n
iii
l n
jjj
n
iii
llh qdxgadxgaEIgawqdxdx
dxwdEI
dxwd
0 10 11002
2
2
2
)()(21)(
21
- The first-order Necessary Condition
0
)])()([21
))()(21(
101 0
00 10 1
0 10 11
=−=−′′′′=
−′′′′+′′′′=
−′′′′∂∂
=∂Π∂
∑∫∑∫
∫∫ ∑∫ ∑
∫∑∫ ∑∑
==
==
===
n
ikiki
l
k
n
ii
l
ik
l
k
l
k
n
iii
l n
ijjk
l n
iii
l n
iii
n
iii
kk
h
faKqdxgadxgEIg
qdxgdxgEIgadxgaEIg
qdxgadxgaEIgaaa
or fKa =
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51
Example
i) with one unknown
awlxxalxaxw 2)()( 2 =′′→−=−=
44
21
4)2(
21)
2()(
21 2
22
0
2
00
2 laPlaEIlaPdxaEIwdxlxPdxwEIlll
total +=+=−δ−′′=Π ∫∫∫
)(1616
04
40 22
xlxEI
PlwEI
PlalPaEIlatotal −−=→−=→=+→=
∂Π∂
EIPllw
64)
2(
3
= , EIPl
EIPllwe
33
0208.048
)2
( == , Error = 25.0)
2(
)2
()2
(=
−
lw
lwlw
e
e
ii) with two unknowns
bxawlxbxlxaxw 62)()( 22 +=′′→−+−=
)8
34
()3
362
244(21
)8
34
()62(21)
2()(
21
3232
22
32
0
2
00
2
lblaPlblablaEI
lblaPdxbxaEIwdxlxPdxwEIlll
total
++++=
−−−+=−δ−′′=Π ∫∫∫
83)126(0
4)64(0
332
22
lPblalEIb
lPbllaEIatotal
total
−=+→=∂
Π∂
−=+→=∂
Π∂
0 , 161
=−=→ bEIPla ???
iii) with three unknowns23322 1262)()()( cxbxawlxcxlxbxlxaxw ++=′′→−+−+−=
)167
83
4(
)4
1443
482
245
1443
364(21
)167
83
4()1262(
21
)2
()(21
432
43252
322
432
0
22
00
2
lclblaP
lbclaclablclblaEI
lclblaPdxcxbxaEI
wdxlxPdxwEI
l
ll
total
++
++++++=
−−−−++=
−δ−′′=Π
∫
∫∫
P
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52
167)
5144188(0
83)18126(0
4)864(0
4543
3432
232
lPclblalEIc
lPclblalEIb
lPclbllaEIa
total
total
total
−=++→=∂
Π∂
−=++→=∂
Π∂
−=++→=∂
Π∂
=
−=
=
→
EIlPEIPb
EIPla
645c
325
641
)(645)(
325)(
641 3322 lxx
EIlPlxx
EIPlxx
EIPlw −+−−−=
EIPl
EIPl
EIPllw
333
0205.01024
21)167
645
83
325
41
641()
2( ==−+−= , Error = 0.0144
iv) with one sin function
xll
awxl
aw ππ=′′=→
π= sin)(sin 2
aPll
EIaaPxdxll
EIa
aPdxxll
aEIwdxlxPdxwEI
l
lll
total
+π
=+ππ
=
−ππ
=−δ−′′=Π
∫
∫∫∫
2)(
21sin)(
21
)sin)((21)
2()(
21
42
0
242
0
22
00
2
xlEI
PlwEIPlaPl
lEIa
atotal π
π=→
π=→=−
π→=
∂Π∂
sin2202
)(03
4
3
44
EIPllw
3
7045.481)
2( = ,
EIPllwe
48)
2(
3
= , Error = 0145.0
v) with two sin function
xll
bxll
awxl
bxl
aw ππ+
ππ=′′=→
π+
π=
3sin)3(sin)(3sinsin 22
bPaPll
EIbll
EIa
bPaPxdxll
EIb
xdxllll
EIabxdxll
EIa
bPaPdxxll
bxll
aEI
wdxlxPdxwEI
l
ll
l
ll
total
+−π
+π
=
+−ππ
+ππππ
+ππ
=
+−ππ
+ππ
=
−δ−′′=Π
∫
∫∫
∫
∫∫
2)3(
21
2)(
21
3sin)3(21
3sinsin)3()(sin)(21
)3sin)3(sin)((21
)2
()(21
4242
0
242
0
22
0
242
0
222
00
2
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53
)3sin811(sin2
)3(20
2)3(0
202
)(0
3
4
3
44
3
44
xl
xlEI
Plw
EIPlbPl
lEIb
b
EIPlaPl
lEIa
a
total
total
π−
ππ
=
π−=→=+
π→=
∂Π∂
π=→=−
π→=
∂Π∂
EIPl
EIPllw
33
0208.0)0123.01(0205.0)2
( =+= , EIPllwe
3
0208.0)2
( = , Error ≅0
4.3 Truss problems
Potential Energy
)()(
21
)( , )(
21
11i
2
int
11i
2
int
ii
njn
iii
nmb
iii
exttotal
ii
njn
iiiext
nmb
iii
vYuXEA
lF
vYuXEA
lF
+−=Π+Π=Π
+−=Π=Π
∑∑
∑∑
==
==
where nmb and njn denotes the total number of members and the total numbers of joints
in a truss.
Equilibrium Equations
0 , 0)(
1
)(
1=+−=+− ∑∑
==
iim
j
ij
iim
j
ij YVXH for njni ,,1L=
where m(i), ijH and i
jV are the number of member connected to joint i, the horizontal
component and the vertical component of the bar force of j-th member connected to joint
i, respectively.
iY
iX iF1−
ijF−
iimF )(−
iY iX
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54
Principle of Minimum Potential Energy
fieldsnt displaceme virtualallfor )(
21
)()(
)(21
)()(
21)(
)(21
))()(()(
21
1i
2
1i
2
11i
2
11i1i
2
11i
2
11i
2
enmb
i
iie
nmb
i
iiii
njn
iii
nmb
i
ie
i
iii
njn
ii
nmb
i
iie
inmb
i
iiii
njn
iii
nmb
i
ie
i
iiii
njn
iii
nmb
i
iie
itotal
EAlF
EAlF
vYuXEA
lF
vYuXEA
lFFEA
lFvYuX
EAlF
vvYuuXEA
lFF
Π≥+Π=
++−=
+−+++−=
+++−+
=Π
∑
∑∑∑
∑∑∑∑∑
∑∑
=
===
=====
==
where eiF and iF are the bar force of i-th member induced by the real displacement of joints
and virtual displacement induced by the virtual displacement of joints. Since the equation in the
box represents the total virtual work in a truss, the total potential energy of a truss becomes
minimum for all virtual displacement fields when the principle of virtual work holds. In
the above equation, the equality sign holds if and only if the virtual displacements at all
joints are zero.
Virtual Work Expression
If a truss is in equilibrium, the principle of the virtual work holds for the truss,.
0))(- )((1
)(
1
)(
1=+++−∑ ∑∑
= ==
njn
i
iiim
j
ij
iiim
j
ij vYVuXH
ivv θ− sin)( 12
iuu θ− cos)( 12
iθiθ
)( 12 uu −
)( 12 vv −
School of Civil, Urban & Geosystem Eng., SNU
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55
∑∑
∑∑ ∑∑
∑ ∑∑
==
== ==
= ==
+=−θ+−θ
+=θδ+θ
=+θ++θ−
njn
iiiii
nmb
iiii
eiiii
ei
njn
iiiii
njn
i
im
jj
ij
iim
jj
ij
i
njn
iii
im
jj
ijii
im
jj
ij
vYuXvvFuuF
vYuXFvFu
vYFuXF
11
1212
11
)(
1
)(
1
1
)(
1
)(
1
) ())(sin )(cos(
) ()sin cos(
0))sin(- )cos((
∑∑
∑∑∑∑
==
====
+=
==∆=−θ+−θ
n
iiiii
nmb
i i
iie
i
nmb
i i
iie
i
i
iinmb
i
ei
nmb
ii
ei
nmb
iiiiiii
ei
vYuXEA
lFF
EAlFF
EAlF
FlFvvuuF
11
1111
1212
) ()(
)()())(sin )((cos
The principle of virtual work yields the displacement of a joint k in a truss by applying anunit load at a joint k in an arbitrary direction and by using the reciprocal theorem.
∑=
=α=α=+nmb
i i
iie
ikkkkkk EA
lFFvYuX
1 )(coscos uuX
Since α represnts the angle between the applied unit load and the displacement vector,
αcosku are the displacement of the joint k in the direction of the applied unit load.
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
56
Chapter 5
Matrix Structural Analysis
Mr. Force & Ms. Displacement
Matchmaker: Stiffness Matrix
School of Civil, Urban & Geosystem Eng., SNU
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57
5.1 Truss Problems
5.1.1 Member Stiffness Matrix
Force – Displacement relation at Member ends
0
)(
)(
==
δ−δ=
δ−δ−=
Ry
Ly
Lx
Rx
Rx
Lx
Rx
Lx
ffL
EAf
LEAf
Member Stiffness Matrix in Local Coordinate Systeme
Ry
Rx
Ly
Lx
e
Ry
Rx
Ly
Lx
LEA
ffff
δδδδ
−
−
=
0000010100000101
eee )(][)( δkf =
Transformation Matrix
φφφ−φ
=
→
φ+φ=
φ−φ=
y
x
y
x
yxy
yxx
vv
VV
vvV
vvVcossinsincos
cossin
sincos
)]([)()(][)( VvvV γ=→γ= T
①
xvyv
φ
②xV
yV
Rxf
RyfL
yf
Rxδ
RyδL
yδ
LxδL
xf
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
58
Member End Force
→
φφφ−φ
φφφ−φ
=
e
Ry
Rx
Ly
Lx
e
y
x
y
x
ffff
FFFF
cos sin00sincos0000cos sin00sincos
2
2
1
1
eTe )(][)( fF Γ=
Member End Displacement
→
∆∆∆∆
φφ−φφ
φφ−φφ
=
δδδδ
e
y
x
y
xe
Ry
Rx
Ly
Lx
2
2
1
1
cossin00sincos 00
00cossin00sincos
ee )]([)( ∆Γ=δ
Member Stiffness Matrix in Global CoordinateeeTeeTeTe )]([][][)(][][)(][)( ∆δ ΓΓ=Γ=Γ= kkfF
eee )(][)( ∆KF =
=
φφφφ−φφ−φφφφθ−φ−φ−φφ−φφθφφ−φ−φθφ
= ee
eee
LEA
][][][][
sincossinsincossincossincoscossincos
sincossinsincossincossincoscossincos
][2221
1211
22
22
22
22
KKKK
K
5.1.2 Global Stiffness Equation
Nodal Equilibrium
∑=
=++++=)(
1
)(2or 1)5(1)4(2)3(1)2(2)1(1 )()()()()()()(mnm
k
kmmmmmmm FFFFFFP
)1(1)( mF−
)2(2 )( mF−
)3(1)( mF−)4(2 )( mF−
)5(1)( mF−
mP
mPm-th jointi-th member
n-th joint
School of Civil, Urban & Geosystem Eng., SNU
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59
)]([)(][)()(
00
0
0
00
0
)(
)(
0
)(
)(
)(
)(
)(
)(
)(11 1
2
1
2
1
)(
1
)(2or 1
)(
1
)(2or 1
)1(
1
)(12or 11
FEFEFF
I
I
F
F
F
F
F
P
P
P
P ==
=
=
=
= ∑∑ ∑
∑
∑
∑
== =
=
=
=
p
i
iip
i
p
ii
i
i
i
qnm
k
kq
mnm
k
km
nm
k
k
q
m
MM
MM
MM
M
M
M
M
M
M
M
[ ]
==
p
ipi
)(
)(
)(
)( , ][][][][
1
1
F
F
F
FEEEEM
M
LL
m-th row
n-th row
1
2
i)( 2F
i)( 1F
m-th jointmP
i-th member
nP n-th joint
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
60
Compatability Condition
=
=→==
n
mi
ii
ni
mi
uu
uuI00I
)()(
)( )( , )( 2
121
∆∆
∆∆∆
)(][000000
)()(
)(
1
2
1
uC
u
u
u
u
II i
q
n
m
i
ii =
=
=
M
M
M
LLL
LLL
∆∆
∆
)]([][
][
)(
)()(
111
uCu
u
C
C=
=
=
qpp
MMM
∆
∆∆
Contragradient
→=⋅→∆⋅=⋅ )]([)()()()()()()( uCFuPFuP TTTT
][)()()( possible allfor 0)])([)()(( CFPuuCFP TTTT =→=−
][][)]([)(][)( ECFEFCP =→== TT
i-th member
m
n
um
un
1
2
i)( 1∆
i)( 2∆
m-th column
n-th column
Compatibility Matrix
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
61
Unassembled Member Stiffness Equation
)(
)(
)(
][00
0][0
00][
)(
)(
)( 111
→
∆
∆
∆
=
p
i
p
i
p
i
M
M
LL
MLMLM
LL
LMLM
LL
M
M
K
K
K
F
F
F
)]([)( ∆= KF
Global Stiffness Equation
)](][[][)]([][)(][)( uCKCKCFCP TTT =∆==
)]([)( uKP = where ]][[][][ CKCK T=
Direct Stiffness Method
[ ]
pppiii
p
i
p
ipiT
TTT
TTT
][][][][][][][][][
][
][
][
][00
0][0
00][
][][][]][[][][
111
11
1
CKCCKCCKC
C
C
C
K
K
K
CCCCKCK
++++=
==
LL
M
M
LL
MLMLM
LL
LMLM
LL
LL
=
=
=
000000][0][0000000][0][000000
0][][00][][0
00
0
00
000000
][][][][
00
0
00
][][][
2221
1211
2221
1211
2221
1211
ii
ii
ii
ii
ii
iiiiiT
KK
KK
KKKK
I
I
II
KKKK
I
ICKC
LLL
LLL
MM
M
MM
MM
LLL
LLL
MM
M
MM
MM
m-th row
n-th row
m-th column n-th column
School of Civil, Urban & Geosystem Eng., SNU
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62
5.1.3 Example
Member Stiffness Matrix
=
φφφφ−φφ−φφφφφ−φ−φ−φφ−φφφφφ−φ−φφφ
= ee
eee
LEA
][][][][
sincossinsincossincossincoscossincos
sincossinsincossincossincoscossincos
][2221
1211
22
22
22
22
KKKK
K
- Member 1: φ = 45o
−−
−−
−−
−−
=
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
2][ 1
LEAK
- Member 2: φ =-45 o
−−
−−
−−
−−
=
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
2][ 2
LEAK
1
1
2
3
2L
L
P3, u3
P4, u4
P6, u6
P5, u5
P1, u1
P2, u2
2
3
School of Civil, Urban & Geosystem Eng., SNU
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63
- Member 3: φ = 0o
−
−
=
0000010100000101
2][ 3
LEAK
Equilibrium Equation
31226
31225
21124
21123
32112
32111
)()(
)()(
)()(
)()(
)()(
)()(
yy
xx
yy
xx
yy
xx
FFP
FFP
FFP
FFP
FFP
FFP
+=
+=
+=
+=
+=
+=
001010000000000101000000000000101000000000010100100000000010010000000001
3
2
2
1
1
2
2
2
1
1
1
2
2
1
1
6
5
4
3
2
1
=
y
x
y
x
y
x
y
x
y
x
y
x
FFFF
FFFF
FFFF
PPPPPP
)]([=)]]([ ][ ][[=)( 221 FEFEEEP
P3
P4
P6
P5P1
P2
[E]1 [E]2 [E]3
School of Civil, Urban & Geosystem Eng., SNU
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64
Compatibility Condition
[ ][ ][ ]
=
=
∆∆∆∆
∆∆∆∆
∆∆∆∆
6
5
4
3
2
1
3
2
1
6
5
4
3
2
1
3
2
2
1
1
2
2
2
1
1
1
2
2
1
1
000010000001100000010000100000010000001000000100001000000100000010000001
uuuuuu
uuuuuu
y
x
y
x
y
x
y
x
y
x
y
x
CCC
)]([)( uC=∆
TTTT ][][ ][][ ,][][ ,][][ 332211 CECECECE =→===
Global Stiffness Matrix333222111 ][][][][][][][][][]][[][][ CKCCKCCKCCKCK
TTTT +++== L
u3
u6
u5
u2
u1
u4
[C]1
[C]3
[C]2
School of Civil, Urban & Geosystem Eng., SNU
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65
−−
−−
−−
−−
=
−−
−−
−−
−−
=
−−
−−
−−
−−
=
000000000000
0021
21
21
21
0021
21
21
21
0021
21
21
21
0021
21
21
21
2
0021
21
21
21
0021
21
21
21
0021
21
21
21
0021
21
21
21
000000001000010000100001
2
001000000100000010000001
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
000000001000010000100001
2]][[][ 111
LEA
LEA
LEAT CKC
−−
−−
−−
−−
=
−−
−−
−−
−−
=
−−
−−
−−
−−
=
21
21
21
2100
21
21
21
2100
21
21
21
2100
21
21
21
2100
000000000000
2
21
21
21
2100
21
21
21
2100
21
21
21
2100
21
21
21
2100
000000001000010000100001
2
100000010000001000000100
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
21
100001000010000100000000
2]][[][ 222
LEA
LEA
LEAT CKC
−
−
=
000010000001100000010000
0000010100000101
001000010000000010000100
2]][[][ 333
LEAT CKC
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
66
−
−
=
−
−
=
000000010001000000000000000000010001
2000000010001000000010001
001000010000000010000100
2 LEA
LEA
−−
−+−−
−+−−−
−−+−−
−−
−−−+
=
−
−
+
−−
−−
−−
−−
+
−−
−−
−−
−−
=
221
221
221
22100
221
21
221
221
2210
21
221
221
221
221
221
221
221
221
221
221
221
221
221
221
221
221
0022
122
122
122
1
021
221
221
221
21
221
000000010001000000000000000000010001
2
21
21
21
2100
21
21
21
2100
21
21
21
2100
21
21
21
2100
000000000000
2
000000000000
0021
21
21
21
0021
21
21
21
0021
21
21
21
0021
21
21
21
2][
LEA
LEA
LEA
LEAK
School of Civil, Urban & Geosystem Eng., SNU
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67
Stiffness Equation
−−
−+
−−
−−−
−−−
−−
−−−+
=
6
5
4
3
2
1
6
5
4
3
2
1
221
221
221
22100
221
2221
221
2210
21
221
221
210
221
221
221
2210
21
221
221
0022
122
122
122
1
021
221
221
221
2221
uuuuuu
LEA
PPPPPP
Application of Support Conditions (Boundary Conditions)
−−
−+
−−
−−−
−−−
−−
−−−+
=
6
5
4
3
2
1
6
5
4
3
2
1
221
221
221
22100
221
2221
221
2210
21
221
221
210
221
221
221
2210
21
221
221
0022
122
122
122
1
021
221
221
221
2221
uuuuuu
LEA
PPPPPP
Final Stiffness Equation
+−
−
=
→
+−
−
=
−
5
4
3
1
5
4
3
5
4
3
5
4
3
2221
221
221
221
210
2210
21
2221
221
221
221
210
2210
21
PPP
EAL
uuu
uuu
LEA
PPP
KnownKnown
Known
UnknownUnknownUnknown
UnknownUnknown
Unknown
KnownKnownKnown
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
68
5.2 Beam Problems5.2.1 Member Stiffness Matrix
Force-Displacement Relation at Member Ends
Ry
e
eLy
e
eR
e
eL
e
e
e
Re
LeR
y
Ry
e
eLy
e
eR
e
eL
e
e
e
Re
LeL
y
Ry
e
eLy
e
eR
e
eL
e
eR
Ry
e
eLy
e
eR
e
eL
e
eL
LEI
LEI
LEI
LEI
LMMf
LEI
LEI
LEI
LEI
LMMf
LEI
LEI
LEI
LEIm
LEI
LEI
LEI
LEIm
δ+δ−θ−θ−=+
−=
δ−δ+θ+θ=+
=
δ−δ+θ+θ=
δ−δ+θ+θ=
3322
3322
22
22
121266
121266
6642
6624
Transformation Matrix is not required
Ff → , Mm → , ∆→δ , Θ→θ
Member Stiffness Matrixe
y
y
y
ee
eeee
ee
eeee
e
e
e
y
y
LL
LLLL
LL
LLLL
LEI
M
F
M
F
Θ
∆
Θ
∆
−
−−−
−
−
=
2
2
1
1
22
22
2
2
1
1
4626
612612
2646
612612
or eee )(][)( ∆KF =
LθLm
Lyf
Ryf
RθRm
RyδL
yδ
School of Civil, Urban & Geosystem Eng., SNU
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69
5.1.2 Global Stiffness Matrix
Nodal Equilibrium
Li
Ri
i
iii
iyiyi
MMM
FFV)()(
)()(
)()(11
12
11
2
FFP +=→
+=
+=−
−
−
[ ] )]([
)(
)(
)(
][][][
)(
)()(
)()(
)()(
)()(
)()(
)(
1
1
1
1
1
12
21
1
1
1
1
2
1
FE
F
F
F
EEE
F
FF
FF
FF
FF
FF
F
PP
PPP
PP
=
=
+
+
+
+
+
=
−
+
−
−−
+
+
−
p
ipi
Rp
Lp
Rp
Li
Ri
Li
Ri
Li
Ri
LR
L
p
p
i
i
i
M
M
LL
M
M
M
M
iiRi
Li
Ri
Li )(][
)()(
00
00
00
0
)()(
0
FEFF
II
FF
=
=
MM
MM
M
M
Compatibility
222
122
21
121
)(
)(
)(
)(
+
+
−
=Θ
=∆
=Θ
=∆
ii
iiy
ii
iiy
u
u
u
u
→
=
=→==
++
11
00
)()(
)( )( , )( i
i
Ri
LiiiR
iiL
i uu
II
uu∆∆
∆∆∆
V i
M i
i-th Member(i-1)-th Member
(i+1)-th rowi-th row
i-th Member(i-1)-th Member
12 −iuiu2
12 +iu22 +iu
School of Civil, Urban & Geosystem Eng., SNU
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70
)(][
000000
)()(
)(
1
1
1
uC
u
uu
u
II
i
p
i
i
Ri
Li
i =
=
=
+
+
M
M
LL
LL
∆∆
∆
)]([][
][
)(
)()(
1
111
uCu
u
C
C=
=
=+p
pp
MMM
∆
∆∆
Unassembled Member Stiffness Equation
→
∆
∆
∆
=
p
i
p
i
p
i
)(
)(
)(
][00
0][0
00][
)(
)(
)( 111
M
M
LL
MLMLM
LL
LMLM
LL
M
M
K
K
K
F
F
F
)]([)( ∆= KF
Global Stiffness Equation
)](][[][)]([][)(][)( uCKCKCFCP TTT =∆==
)]([)( uKP = where ]][[][][ CKCK T=
Direct Stiffness Method
=
0
00
0
00000
][][0][][0
00000
][][][2221
1211
M
M
MMMMM
LL
LLMMMMM
ii
iiii
Ti KK
KKCKC
i-th column
(i+1)-th column
i-th rowi+1-th row
i+1-th columni-th column
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71
5.2.3 Example
Equilibrium Equation
RR
LRLR
LRLR
LL
MMVV
MMMVVVMMMVVV
MMVV
44
44
323
323
212
212
11
11
,
, ,
,
==
+=+=
+=+=
==
)]([
100000000000010000000000001010000000000101000000000000101000000000010100000000000010000000000001
)(
3
3
3
3
2
2
2
2
1
1
1
1
4
4
3
3
2
2
1
1
FEP =
=
=
R
R
L
L
R
R
L
L
R
R
L
L
M
V
M
VM
V
MV
M
VM
V
MVMVMVMV
Compatability Condition
43
43
33
33
32
32
22
22
21
21
11
11
, , ,
, , ,
, , ,
θ=θ=θ=θ=
θ=θ=θ=θ=
θ=θ=θ=θ=
RRLL
RRLL
RRLL
wwww
wwww
wwww
[E]1 [E]2 [E]3
11 ,θM
11 , wV
22 ,θM
22 , wV
33 ,θM
33 , wV
44 ,θM
44 , wV
School of Civil, Urban & Geosystem Eng., SNU
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72
)]([
][][][
100000000100000000100000000100000010000000010000000010000000010000001000000001000000001000000001
)(
4
4
3
3
2
2
1
1
3
2
1
4
4
3
3
2
2
1
1
3
3
3
3
2
2
2
2
1
1
1
1
uCCCC
∆ =
θ
θ
θ
θ
=
θ
θ
θ
θ
=
θ
θ
θ
θ
θ
θ
=
w
w
w
w
w
w
w
w
w
w
w
w
w
w
R
R
L
L
R
R
L
L
R
R
L
L
Unassembled Member Stiffness Matrix
)]([)()()(
][000][000][
)()()(
)(
3
2
1
3
2
1
3
2
1
∆K∆∆∆
KK
K
FFF
F =
=
=
Global Stiffness Equation
)]([)](][[][)](][[)]([)( uKuCKCuKEFEP ==== T
[ ]
))( ][][][][][][][][][ ()]([
)(][
][][
][000][000][
][ , ][ , ][)(
333222111
3
2
1
3
2
1
321
uCKCCKCCKCuK
uCCC
KK
KCCCP
TTT
TTT
++==
=
[C]1
[C]3 [C]2
School of Civil, Urban & Geosystem Eng., SNU
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73
- Member 1
−
−−
−
−
=
−
−−−
−
−
00000000000000000000000000000000
00004626
0000612612
00002646
0000612612
00001000000001000000001000000001
4626
612612
2646
612612
00000000000000001000010000100001
11
1211
21
11
1211
21
1
1
11
1211
21
11
1211
2
1
1
LL
LLLL
LL
LLLL
LEI
LL
LLLL
LL
LLLL
LEI
- Member 2
−
−−−
−
−
=
−
−−−
−
−
0000000000000000
00462600
0061261200
00264600
00612612000000000000000000
00100000000100000000100000000100
4626
612612
2646
612612
00000000100001000010000100000000
22
2222
22
22
2222
22
2
2
22
2222
22
22
2222
22
2
2
LL
LLLL
LL
LLLL
LEI
LL
LLLL
LL
LLLL
LEI
School of Civil, Urban & Geosystem Eng., SNU
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74
- Member 3
−
−−−
−
−
=
−
−−−
−
−
46260000
6126120000
26460000
612612000000000000000000000000000000000000
10000000010000000010000000010000
4626
612612
2646
612612
10000100001000010000000000000000
33
3233
23
33
3233
23
3
3
33
3233
23
33
3233
23
3
3
LL
LLLL
LL
LLLL
LEI
LL
LLLL
LL
LLLL
LEI
Global Stiffness Matrix
−
−−−
−++−
−+−+−−
−++−
−+−+−−
−
−
3
323
3
3
323
3
23
333
323
333
3
3
323
3
3
3
2
223
322
2
2
222
2
23
333
323
322
233
332
222
232
2
2
221
2
2
2
2
122
221
1
1
121
1
22
232
222
221
132
231
121
131
1
1
121
1
1
121
1
21
131
121
131
1
4
62
6 0 0 0 0
612
6120 0 0 0
2
644 662 6 0 0
6
12661212 612 0 0
0 0 2 644 662 6
0 0 6
12661212
612
0 0 0 0 264 6
0 0 0 0 6126
12
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
E
School of Civil, Urban & Geosystem Eng., SNU
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75
Application of support Conditions
θ
θ
θ
θ
−
−−−
−++−
−+−+−−
−++−
−+−+−−
−
−
=
4
4
3
3
2
2
1
1
3
323
3
3
323
3
23
333
323
333
3
3
323
3
3
3
2
223
322
2
2
222
2
23
333
323
322
233
332
222
232
2
2
221
2
2
2
2
122
221
1
1
121
1
22
232
222
221
132
231
121
131
1
1
121
1
1
121
1
21
131
121
131
1
4
4
3
3
2
2
1
1
4
62
6 0 0 0 0
612
6120 0 0 0
2
644
662
6 0 0
6
12661212
612 0 0
0 0 2
644
662
6
0 0 6
12661212
612
0 0 0 0 264
6
0 0 0 0 6126
12
w
w
w
w
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
E
MVMVMVMV
Final Stiffness Equation
θ
θ
θ
−
−−
−+
+
=
4
4
3
2
3
323
3
3
3
23
333
323
3
3
323
3
3
3
2
2
2
2
2
2
2
2
2
1
4
4
3
2
4
620
61260
26442
00244
w
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
E
MVMM
School of Civil, Urban & Geosystem Eng., SNU
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76
5.3 Frame Problems5.3.1 Member Stiffness Matrix
Force-Displacement Relation at Member Ends
- Beam action
Ry
e
eLy
e
eRe
e
eLe
e
e
e
Re
LeR
y
Ry
e
eLy
e
eRe
e
eLe
e
e
e
Re
LeL
y
Ry
e
eLy
e
eRe
e
eLe
e
eR
Ry
e
eLy
e
eRe
e
eLe
e
eL
LEI
LEI
LEI
LEI
LMMf
LEI
LEI
LEI
LEI
LMMf
LEI
LEI
LEI
LEIm
LEI
LEI
LEI
LEIm
δ+δ−θ−θ−=+
−=
δ−δ+θ+θ=+
=
δ−δ+θ+θ=
δ−δ+θ+θ=
3322
3322
22
22
121266
121266
6642
6624
- Truss action
0
)(
)(
==
δ−δ=
δ−δ−=
Re
Le
Lx
Rx
Rx
Lx
Rx
Lx
VVL
EAf
LEAf
Member Stiffness Matrix
θ
δ
δ
θ
δ
δ
−
−−−
−
−
−
−
=
Re
Ry
Rx
Le
Ly
Lx
ee
ee
e
e
e
e
e
e
e
e
e
e
ee
ee
ee
e
e
e
e
e
e
e
e
e
e
ee
e
R
Ry
Rx
L
Ly
Lx
ILI
ILI
LI
LI
LI
LI
AA
ILI
ILI
LI
LI
LI
LI
AA
LE
m
f
f
m
f
f
46
026
0
6120
6120
0000
26
046
0
6120
6120
0000
22
22
or eee )(][)( δkf =
LLm θ,
Ly
Lyf δ,
RRm θ,
Ry
Ryf δ,
Lx
Lxf δ, R
xR
xf δ,
School of Civil, Urban & Geosystem Eng., SNU
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77
Transformation Matrix
①②
θθθ−θ
=
→
=
θ+θ=
θ−θ=
m
vv
M
VV
mMvvV
vvV
y
x
y
x
yxy
yxx
1000cossin0sincos
cossin
sincos
)]([)()(][)( VvvV γ=→γ= T
Member End Force
→
θθθ−θ
θθθ−θ
=
e
R
Ry
Rx
L
Ly
Lx
e
y
x
y
x
m
ffm
ff
M
FFM
FF
1000000cossin0000sincos0000001000000cossin0000sincos
2
2
2
1
1
1
eTe )(][)( fF Γ=
Member End Displacement
→
θ
∆∆θ
∆∆
θθ−θθ
θθ−θθ
=
θ
δδθ
δδ
e
y
x
y
x
e
Re
Ry
Rx
Le
Ly
Lx
1
2
2
1
1
1
1000000cossin0000sincos0000001000000cossin0000sincos
ee )]([)( ∆Γ=δ
θ
①
xV
yV
②
xv
yv
School of Civil, Urban & Geosystem Eng., SNU
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78
Member Stiffness Matrix in Global CoordinateeeTeeTeTe )]([][][)(][][)(][)( ∆δ ΓΓ=Γ=Γ= kkfF
eee )(][)( ∆KF = where
= ee
eee
][][][][
][2221
1211
KKKK
K
Nodal Equilibrium & Compatibility
The same as the truss problems.
Global Stiffness Matrix
)](][[][)]([][)(][)( uCKCKCFCP TTT =∆==
)]([)( uKP = where ]][[][][ CKCK T=
Direct Stiffness Method
The same as the truss problems.
School of Civil, Urban & Geosystem Eng., SNU
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79
Chapter 6
Buckling of Structures
P
Lθ
P
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80
6.0 Stability of Structures
Stable state
QKLLQLLK >→θ>×θ
The structure would return its original equilibrium position for a small perturbation in θ.
Critical state
QKLLQLLK =→θ=×θ
Unstable state
QKLLQLLK <→θ<×θ
The structure would not return its original equilibrium position for a small perturbation in
θ.
Q
L
θ
Q
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81
6.1 Governing Equation for a Beam with Axial Force
Equilibrium for vertical force
qdxdVqdxVdVV −=→=+−+ 0)(
Equilibrium for moment
VdxdwQ
dxdM
dxdwPdxqdxVdxMdMM =−→=−+−−+ 0
2)(
Elimination of shear force
qdx
wdQdx
Md−=− 2
2
2
2
Strain-displacement relation
dxduy
dxwd
+−=ε 2
2
Stress-strain relation (Hooke law)
dxduEy
dxwdEE +−=ε=σ 2
2
Definition of Moment
2
22
2
2
)(dx
wdEIdAydxduEy
dxwdEydAEydAM
AAA
−=−−=ε=σ= ∫∫∫
Beam Equation with Axial Force
qdx
wdQdx
wdEI =+ 2
2
4
4
M M+dM
V V+dV
q
w
dxdxdw
Q
Q
School of Civil, Urban & Geosystem Eng., SNU
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82
6.2 Homogeneous Solutions
Characteristic Equation for 0>P
0 ,0)( 224 ieew
x
x
β±=λ→=λβ+λ
=λ
λ
where EIQ
=β2
Homogeneous solution
DCxBeAew ixix +++= β−β
Exponential Function with Complex Variable
xixixixiixee
xxxxee
xixixixixiixe
xixixixixiixe
ixix
ixix
ix
ix
sin2)!7!5!3
(2
cos2)!6
1!4
1!2
11(2
!6)(
!5)(
!4)(
!3)(
!2)(1
!6!5!4!3!21
753
642
66
55
44
33
22
66
55
44
33
22
=+−+−=−
=+−+−=+
+−
+−
+−
+−
+−
+−=
+++++++=
−
−
−
L
L
L
L
xixexixe ixix sincos , sincos −=+= −
Homogeneous solution
DCxxBxADCxxBAixBA
DCxxixBxixAw
++β+β=++β−+β+=
++β−β+β+β=
sincossin)(cos)(
)sin(cos)sin(cos
Characteristic Equation for 0<P
0 ,0)( 224 β±=λ→=λβ−λ
=λ
λ
x
x
eew
where EIQ
=β2
Homogeneous solution for 0<P
DCxxBxA
DCxeeBAeeBA
DCxBeAewxxxx
xx
++β+β=
++−
−++
+=
+++=β−ββ−β
β−β
sinhcosh2
)(2
)(
School of Civil, Urban & Geosystem Eng., SNU
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83
Simple Beam
− Boundary Condition
00sin)( , 0sin)(00)0( , 0)0(
2
2
==→=ββ−=′′=+β=
=→=β−=′′=+=
CBLBLwCLLBLwAAwDAw
− Characteristic Equation
00 =→==== wDCBA (trivial solution) or
L3,2,1 , 2
22
=π
=→π=β nL
EInQnL
xLnBxBw π
=β= sinsin
Fixed-Fixed Beam
− Boundary Condition
0cossin)( 0sincos)(
0)0( 0)0(
=+ββ+ββ−=′=++β+β=
=+β=′=+=
CLBLALwDCLLBLALw
CBwDAw
− Characteristic Equation
0)
01cossin1sincos0101001
(
0000
01cossin1sincos0101001
=
ββββ−ββ
β→
=
ββββ−ββ
β
LLLLL
Det
DCBA
LLLLL
1cossinsincos
10
01cos1sin01
01cossin1sincos0101001
LLLLL
LLL
LLLLL
ββββ−ββ
β−
βββ
β=
ββββ−ββ
β
School of Civil, Urban & Geosystem Eng., SNU
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84
02
sin2
cos2
or 02
sin0)2
sin2
cos2
(2
sin
2cos
2sin2
2sin4
sin)1(cos2sin2cos20)sin2cos2())sin(cos()cos(
2
=β
−ββ
=β
→=β
−βββ
=ββ
β+β
−
=ββ+−β=ββ+−β=ββ+−ββ=β+ββ+ββ−−ββ−−β−
LLLLLLLL
LLLLLLLLLL
LLLLLLL
− Eigenvalues
Symmetric modes
L3,2,1 ,42
02
sin 2
22
=π
=→π=β
→=β n
LEInQnLL
0)( ,0)()0( , 0)0( =++==+β=′=′=+= DCLALwCBLwwDAw
)12(cos0 −π
=→−=→=+ xLnAwDADA for 0≠A
Anti-symmetric modes
2
218.82
tan2
02
sin2
cos2 L
EIQLLLLL π=→
β=
β→=
β−
ββ
School of Civil, Urban & Geosystem Eng., SNU
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85
Cantilever Beam
− Boundary Condition
0)(
0)sincos()()(0)0(
0)0(
3
3
22
=−−=
=ββ−ββ−−=′′−=
=+β=′=+=
dxdwP
dxwdEILV
LBLAEILwEILMCBw
DAw
L3,2,1 ,4 2
22
=π
= nL
EInQ
6.3.Homogeneous and Particular solution
pph wDCxxBxAwww +++β+β=+= sincos
qdx
wdQ
dxwd
EI
dxwd
Qdx
wdEI
dxwd
Qdx
wdEI
dxwwd
Qdx
wwdEI
pp
pphhphph
=+=
+++=+
++
2
2
4
4
2
2
4
4
2
2
4
4
2
2
4
4 )()(
Four Boundary Conditions for Simple Beams
0))(sincos()()(
0 )(sincos)(
0(0))()0((0) , 0(0))0(
22
2
=′′+ββ−ββ−−=′′−=
=+++β+β=
=′′+β−−=′′−==++=
LwLBLAEILwEILM
LwDCLLBLALw
wAEIwEIMwDAw
p
p
pp
00
)()()0()0(
00sincos1sincos0001001
22
2
=+→=
′′
′′+
ββ−ββ−ββ
β−FKX
LwLw
ww
DCBA
LLLLL
p
p
p
p
The homogenous solution is for the boundary conditions, while the particular solution is
for the equilibrium.
School of Civil, Urban & Geosystem Eng., SNU
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86
6.4.Energy Method
Total Potential Energy
dxwqdxdxdw
dxdwQdx
dxwdEI
dxwd lll
∫∫∫ −−=Π000
2
2
2
2
21
21
dxwdxwdsLLLL
))(211()(1
0
2
0
2
0∫∫∫
∆−∆−∆−
′+≈′+==
LdxwdxwdxwLLLLL
<<∆′≈′=∆→′+∆−= ∫∫∫∆−∆−
for )(21)(
21)(
21
0
2
0
2
0
2
Principle of the Minimum Potential Energy
∫
∫∫
∫∫
∫∫∫
∫∫∫
−+Π=
−+−++Π=
−+−
−+−−=
+−++
−++
=Π
le
ll eee
ll
l eele
l eeee
le
l eel eeh
dxdxwdQ
dxwd
dxwdEI
dxwd
dxdxwdQ
dxwd
dxwdEI
dxwddxq
dxwdQ
dxwdEIw
dxdxwdQ
dxwd
dxwdEI
dxwdqdxw
dxdxwdQ
dxdw
dxwdEI
dxwdqdxwdx
dxdwQ
dxdw
dxwdEI
dxwd
qdxwwdxdx
wwdQdx
wwddxdx
wwdEIdx
wwd
02
2
2
2
02
2
2
2
02
2
4
4
02
2
2
2
0
02
2
2
2
002
2
2
2
0002
2
2
2
)(21
)(21)(
)(21
)()(21
)()()(21)()(
21
The principle of minimum potential energy holds if and only if
wdxdxwdQ
dxwd
dxwdEI
dxwdl
possible allfor 0)(0
2
2
2
2
>−∫
The principle of the minimum potential energy is not valid for the following cases.
wdxdxwdQ
dxwd
dxwdEI
dxwdl
somefor 0)(0
2
2
2
2
≤−∫
∆Q
School of Civil, Urban & Geosystem Eng., SNU
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87
The critical status of a structure is defined as
0)(0
2
2
2
2
=−∫l
dxdxwdQ
dxwd
dxwdEI
dxwd
Approximation
− Approximation of displacement: ∑=
=n
iii gaw
1
− Critical Status
0)(0))(()(
)()()()(
1 11 11 1 01 1 0
0 110 11
=−→=−
=−=′′−′′′′
=′′−′′′′
∑∑∑∑∑∑ ∫∑∑ ∫
∫ ∑∑∫ ∑∑
= == == == =
====
GTGT
j
n
i
n
j
Giji
n
ij
n
jijij
n
i
n
j
l
jii
n
ij
n
j
l
iii
l n
jjj
n
iii
l n
jjj
n
iii
Detaa
aKaQaKaadxggaQadxgEIga
dxgagaPdxgaEIga
KKKK
Example - Simple Beam
− with a parabola: 2 , 2)( 11 =′′−=′→−= glxglxaxw
%)22error , 86.9:exact( 120)314(
31)12
34()44()2(
422
22
2
23
33
0
22
0
2
0
00
==π
=→=−
=+−=+−=−=′′
==′′′′
∫∫∫
∫∫
lEI
lEI
lEIQlQEIlDet
lldxlxlxdxlxdxgEIg
EIldxEIdxgEIg
cr
lll
ii
ll
ii
− with one sine curve: sin)( , cossin 211 l
xl
glx
lg
lxaw ππ
=′′ππ=′→
π=
)exact(0)2
)(2
)((
2)(cos)(
2)(sin)(
2
224
2
0
22
0
0
424
0
lEIQl
lQl
lEIDet
ll
dxlx
ldxgEIg
ll
EIdxlx
lEIdxgEIg
ll
ii
ll
ii
π=→=
π−
π
π=
ππ=′′
π=
ππ=′′′′
∫∫
∫∫
Q
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
88
Example – Cantilever Beam
− with one unknown:
%)22error , 46.24
:exact( 30)344(
344 , 422
2 , 2
22
2
23
3
0
2
000
112
==π
=→=−
==′′==′′′′
=′′=′→=
∫∫∫∫
lEI
lEI
lEIQlQEIlDet
ldxxdxgEIgEIldxEIdxgEIg
gxgaxw
cr
ll
ii
ll
ii
− with two unknowns:
3
0
222
2
02112
011
5
0
422
4
0
32112
3
0
211
212
2132
1236 , 612 , 44
599 ,
466 ,
344
6 , 2 , 3 , 2
ldxxKlxdxKKldxK
ldxxKldxxKKldxxK
xggxgxgbxaxw
lll
lG
lGG
lG
=======
=======
=′′=′′=′=′→+=
∫∫∫
∫∫∫
32.181or 487.20727.1or 0829.045
27.222645
1802626
0455240)456()5412)(404(
301 , 0
5412456456404
0)54454540
301
12664
(
222228
121226
2864242533
5342
423
54
43
32
2
lEI
lEIQ
lllllll
lllllllll
EIQ
llllllll
llll
Qllll
EIDet
cr =→=±
=−±
=α
=α+α−→=α−−α−α−
=α=α−α−α−α−
→=
−
249.2lEIQexact = (error = 1.2%) or 219.22
lEIQexact = (error = 45%)
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
89
6.5 Approximation with the Homogeneous Beam Solutions
Homogeneous Solution of Beam
))(())((),,,( 43214321 ii
Rz
Ry
Lz
Ly
Rz
Ry
Lz
Ly
i NNNNNNNNNw ∆=∆=
θ
δ
θ
δ
=θ+δ+θ+δ= N
3
3
2
2
1231Lx
LxN +−= , 2
32
22
Lx
LxxN +−= , 2
32
323Lx
LxN −= , 2
32
4 Lx
LxN +−=
6.5.1 Beam Analysis
Total Potential Energy
)()()]([)(21
)(][)()](][[][)(21
)(][)()](])[[]([][)(21
)()())([]([)(21
)()()()()(21)()()(
21
)()(21)(
21
21
21
11
11
0
11
0
1 01 01 02
2
2
2
1 01 01 02
2
2
2
1 01 01 02
2
2
2
PuuKu
fCuuCKCu
fCuuCKKCu
fKK
NNNNN
TT
p
ii
Ti
Tp
iii
Ti
T
p
ii
Ti
Tp
ii
Gii
Ti
T
p
ii
Ti
p
ii
Gii
Ti
p
i
l
iTT
i
p
ii
lTT
i
p
ii
lTT
i
p
i
l
iT
i
p
i
l iTi
p
i
liTi
p
i
l
ii
p
i
lii
p
i
lii
dxqdxdxd
dxdQdx
dxdEI
dxd
dxqwdxdx
dwdx
dwQdx
dxwd
EIdx
wd
dxqwdxdxdw
dxdw
Qdxdx
wdEI
dxwd
−=
−=
−−=
∆−∆−∆=
∆−∆∆−∆∆=
−−=
−−=Π
∑∑
∑∑
∑∑
∑ ∫∑ ∫∑ ∫
∑∫∑ ∫∑ ∫
∑∫∑ ∫∑ ∫
==
==
==
===
===
===
−
−−−
−
−
=
4626
612612
2646
612612
22
22
0
ii
iiii
ii
iiii
i
ii
LL
LLLL
LL
LLLL
LEI
K ,
−−
−−
−
−
=
152
101
30101
101
56
101
56
30101
152
101
101
56
101
56
ii
ii
ii
ii
Gi
LLLL
LLLL
QK
12 −iu 12 +iu
i-th member
iu2 22 +iu P
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
90
Principle of Minimum Potential Energy for crQQ <
∑∑ ∑== =
−=−=Πn
iii
n
ij
n
jiji
TT PuuKu11 12
1)()()]([)(21 PuuKu
)()]([,,1for 021
21
21
21
21
21
21
21
111
1111
11 11 1
PuK =→==−=−+=
−+=−+=
−∂
∂+
∂∂
=∂
Π∂
∑∑∑
∑∑∑∑
∑∑ ∑∑ ∑
===
====
== == =
nkPuKPuKuK
PuKuKPKuuK
Puuu
KuuKuu
u
kj
n
jkjk
n
jjkjj
n
jkj
k
n
iikij
n
jkjk
n
iikij
n
jkj
n
iii
n
i k
jn
jiji
n
ij
n
jij
k
i
k
L
Calculation of the Critical Load
0])[]([])([ 0 =−= GDetDet KKK
Frame Members
θ
δ
δ
θ
δ
δ
−−
−−
−
−
−
−
−−−
−
−
=
Re
Ry
Rx
Le
Ly
Lx
ee
ee
ee
ee
ee
ee
e
e
e
e
e
e
e
e
e
e
ee
ee
ee
e
e
e
e
e
e
e
e
e
e
ee
e
R
Ry
Rx
L
Ly
Lx
LLLL
LLLL
P
ILII
LI
LI
LI
LI
LI
AA
ILII
LI
LI
LI
LI
LI
AA
LE
m
f
f
m
f
f
)
152
1010
301010
101
560
101
560
0000003010
10152
1010
101
560
101
560
000000
460260
61206120
0000
260460
61206120
0000
(
22
22
e
Rx
Lx
LEAP δ−δ
=
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
91
6.6 Nonlinear Analysis of Truss
Force – Displacement relation at Member ends
Rx
Ly
RyL
yR
y
Lx
Rx
Rx
Lx
Rx
Lx
fl
ff
LEAf
LEAf
δ−δ=−=
δ−δ=
δ−δ−=
)(
)(
Member Stiffness Matrixe
Ry
Rx
Ly
Lx
Lx
Rx
e
Ry
Rx
Ly
Lx
llEA
ffff
δδδδ
−
−δ−δ+
−
−
=
1010000010100000
0000010100000101
eeg
eeeeg
Rx
ee pf ))(][]([))(][]([)( 00 δδ kkkkf +=+=
Equilibrium Analysisei
eg
ei
eei p ))(][]([)( 0 δkkf +=
eeeeg
ei
eTeeTeTe p )(][)]()[][]([][)(][][)(][)( 0 ∆∆δ KkkkfF =Γ+Γ=Γ=Γ=
Successive substitutionee
iee
gei
eTei p )(][)]()[][]([][)( 1101 ∆∆ −−− =Γ+Γ≈ KkkF
iGeip ))()]([]([)( 10 uKKP −+=
Rxf
Ryf
Lyf
Rxδ
Ryδ
Lyδ
LxδL
xf
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
92
Newton-Raphson Method
ei
eeg
ei
eei
ei
eg
ei
ei
eg
ei
eei
eg
ei
e
ei
ei
eg
ei
ei
eg
ei
eei
eg
ei
e
ei
ei
eg
ei
ei
e
ei
eg
ei
eei
p
ppp
ppp
pp
p
δ
δδδ
δδδδ
δδ
δ
∆+++=
∆+∆+++≈
∆+∆+∆+++=
∆+∆++=
+=
σ−−
−−−−
−−−−
−−
)][][]([)(
][)][]([)][]([
)(][)][]([)][]([
))(])[(]([
))(][]([)(
101
110110
110110
110
0
kkkf
kkkkk
kkkkk
kk
kkf
ei
ei
ei
ei
eeg
ei
e p )()()()][][]([ 110 fffkkk ∆=−=∆++ −σ− δ
iGeip )()][)]([]([ 10 PuKKK ∆=∆++ σ−
( )
ei
e
e
i
Ry
Rx
Ly
Lx
ei
Ly
Ry
e
iRy
Rx
Ly
Lx
e
i
Ly
Ry
Ry
Ly
Lx
Rx
e
iRy
Rx
Ly
Lx
ei
eg
ei
lEA
lEA
llEAp
δ
δ
∆=
δ∆δ∆δ∆δ∆
−
−δ−δ=
δ∆δ∆δ∆δ∆
−
δ−δ
δ−δ=
δ∆−δ∆
δδδδ
−
−=∆
σ
−
−
−
−
][
0101000001010000
)(
01010
0
1010000010100000
][
12
1
2
1
1
k
k