Class Note for Structural Analysis 2
Fall Semester, 2010
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 332.1 Solution Method for Linear Algebraic Equations………………………………… 342.2 Moment Distribution Method……………………………………………………... 382.3 Example - MDM for a 4-span Continuous Beam…………………………………. 432.4 Direct Solution Scheme by Partitioning…………………………………………... 452.5 Moment Distribution Method for Frames………………………………………… 46
Chapter 3 Buckling of Structures 493.0 Stability of Structures… 503.1 Governing Equation for a Beam with Axial Force 513.2 Homogeneous Solutions………… 523.3 Homogeneous and Particular solution… 58
Chapter 4 Energy Principles 594.1 Spring-Force Systems……………………………………………………………... 604.2 Beam Problems……………………………………………………………………. 614.3 Truss problems……………………………………………………………………. 674.4 Buckling problems
Chapter 5 Matrix Structural Analysis 735.1 Truss Problems…………………………………………………………………… 745.2 Beam Problems…………………………………………………………………… 855.3 Frame Problems…………………………………………………………………... 935.4 Buckling of Structures 965.5 Beam Columns 985.6 Nonlinear Truss
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1
Chapter 1
Slope Deflection Method
A B
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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
A B
C
163PL
EI
L
EI
L
L/2 P
A B
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 6 BCB bEIM ∆−= 2
2 6
BABBA aEIVV ∆=−= 3
22 12 , BCBBC bEIVV ∆=−= 2
22 12
baEI
P
A B
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
A B
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
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
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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24
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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Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
25
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.
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
26
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
School of Civil, Urban & Geosystem Eng., SNU
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28
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
School of Civil, Urban & Geosystem Eng., SNU
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29
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
School of Civil, Urban & Geosystem Eng., SNU
Structural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac.kr
30
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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32
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