Class Note for Structural Analysis 2strana.snu.ac.kr/lecture/struct2_2017/Notes/Lecture_Note_2017.pdf · Class Note for Structural Analysis 2 . Fall Semester, 2017 . Hae Sung Lee,

Class Note for Structural Analysis 2

Fall Semester, 2017

Hae Sung Lee, Professor

Dept. of Civil and Environmental Engineering

Seoul National University

Seoul, Korea

Contents

Chapter 1 Slope Deflection Method 1 1.0 Comparison of Flexibility Method and Stiffness Method………………………… 2 1.1 Analysis of Fundamental System………………………………………………...... 5 1.2 Analysis of Beams………………………………………………………………… 8 1.3 Analysis of Frames………………………………………………………………... 17

Chapter 2 Iterative Solution Method & Moment Distribution Method 32 2.1 Solution Method for Linear Algebraic Equations…………………………………. 33 2.2 Moment Distribution Method……………………………………………………... 37 2.3 Example - MDM for a 4-span Continuous Beam…………………………………. 42 2.4 Direct Solution Scheme by Partitioning…………………………………………... 44 2.5 Moment Distribution Method for Frames…………………………………………. 45

Chapter 3 Buckling of Structures 48 3.0 Stability of Structures……………………………………………………………... 49 3.1 Governing Equation for a Beam with Axial Force………………………………... 50 3.2 Homogeneous Solutions…………………………………………………………... 51 3.3 Homogeneous and Particular solution…………………………………………….. 57

Chapter 4 Energy Principles 58 4.1 Spring-Force Systems……………………………………………………………... 59 4.2 Beam Problems……………………………………………………………………. 60 4.3 Truss problems…………………………………………………………………….. 64 4.4 Buckling problems………………………………………………………………… 66

Chapter 5 Matrix Structural Analysis 72 5.1 Truss Problems…………………………………………………………………….. 73 5.2 Beam Problems……………………………………………………………………. 84 5.3 Frame Problems………………………………………………………………….... 92 5.4 Buckling of Beams and Frames…..……………………………………………….. 95 5.5 Nonlinear Analysis of Truss………………………………………………………. 97

Department of Civil & Environmental Eng., SNU

Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr

1

Chapter 1

Slope Deflection Method

A B

http://strana.snu.ac.kr/



2

1.0 Comparison of Flexibility Method and Stiffness Method

Flexibility Method

Remove redundancy (Equilibrium) Compatibility

21 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




3

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




4

Flexibility Method

1. Release all redundancies. 2. Calculate displacements induced by external loads at the released

redundancies. 3. Apply unit loads and calculate displacements at the released

redundancies. 4. Construct the flexibility equation by superposing the displacement

based 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 fixed

DOF. 3. Apply unit displacements and calculate member end forces at the

DOFs. 4. Construct the stiffness equation by superposing the member end

forces based on the equilibrium equations. 5. Solve the stiffness equation. 6. Calculate reactions and other quantities as needed.




5

1.1 Analysis of Fundamental System

1.1.1 End Rotation

Flexibility Method i) 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

1.1.2 Relative motion of joints

MA MB

∆

A B

Aθ




6

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

1.1.3 Fixed End Force

Both Ends Fixed

∆




7

One End Hinged

+ Ex.: Uniform load case with a hinged left end

8243

2412

2222 qLqLqLqLM fB −=−=−−= , 0=f

AM

1.1.4 Joint Equilibrium

∑∑∑ =+−− 0jointmemberfixed FFF

or

∑∑∑ =+ jointmemberfixed FFF

MA MB

MA MA/2

AML23 AM

L23

Joint i

Fjoint

Fmember

Ffixed




8

1.2 Analysis of Beams 1.2.1 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 Δ123

22 =−=

b a EI

P

A B

C

BaEI

θ4

BaEI

θ2

BbEI

θ4

BbEI

θ2

BaEI

θ26 Ba

EIθ2

6 BbEI

θ26 Bb

EIθ2

6

BaEI

∆26 Ba

EI∆2

6 BbEI

∆26 Bb

EI∆2

6

BaEI

∆312

BaEI

∆312

BbEI

∆312

BbEI

∆312




9

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=+=

1.2.2 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


→=++++→=∑ 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

LEIqL

qL

EI 2EI L L

q

A B

C




10

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

1612

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 - Free Body Diagram - Moment Diagram

2

161 qL

2

81 qL

qL167 qL

169

qL85

qL83

qL1619




11

1.2.3 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

322

−=→=+++− θθθ

Member End Forces

22

1

1612

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 L L

q

A B

C




12

1.2.4 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 θ21 = , BCB l

EIV θ= 21 6

iii) 0≠θL

C

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




13

Construct Stiffness Equation

06 2812

0 2

2

1 =∆+++−→=∑ CL

CBi

i

lEI

lEI

lEIqlM θθ

06 42 0 2 2 =∆++→=∑ CL

CBi

i

lEI

lEI

lEIM θθ

06 4 0 23 =∆−→=∑ CR

Ci

i

lEI

lEIM θ

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

1.2.5 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




14

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 ∆−=−= 3

22 3 , CDCCD lEIVV ∆=−= 3

22 3


03 712

0 2

2

1 =∆+θ+−→=∑ CBi

i

lEI

lEIqlM

063 0 322 =∆+→=∑ CBi

i

lEI

lEIV θ

EIql

B 66

3

=θ , EI

qlC 132

4

−=∆

EIql

lllEI

lEI CR

CCR

C 8823)(32 3

−=∆

=→∆

−−= θθ

EIql

EIql

EIql

lC

BL

C 26413223

13223

21 333

=+−=∆

−−= θθ

1.2.6 Beam with a Spring Support

Analysis

i) All fixed

12

2qlM fAB = ,

12

2qlM fBA −=

l l l

D k

q

EI EI EI A

B C




15

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


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




16

1.2.7 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

1.2.8 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




17

1.3 Analysis of Frames

1.3.1 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


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




18

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 =

1.3.2 A Portal Frame without Sidesway – hinged supports DOF : Bθ , Cθ

Analysis

i) All fixed

80 PlM BC = ,

80 PlMCB −=

l/2 P

A

B C

D

EI1

EI2

EI1




19

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


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




20

1.3.3 A Frame with a horizontal force DOF : Bθ , ∆

Analysis i) All fixed : None fixed end moment ii) 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




21

1.3.4 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




22

iv) 0≠∆

∆== 233 6

lEIMM BAAB , ∆== 2

33 6lEIMM DCCD ,

∆== 333 12

lEIVV CDBA


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

−=∆

1.3.5 A Portal Frame with a Bracing (Vertical Load)


Analysis

i) All fixed

2

20

lPabM BC = , 2

20

lbPaMCB −=

4la =

a P

A

B C

D

0,0 ≠= EAEI




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


06 28 0 22

2

=∆+++→=∑ lEI

lEI

lEI

lPabM cB

iB θθ

06 82 0 22

2

=∆+++−→=∑ lEI

lEI

lEI

lbPaM cB

iC θθ

0)1(2466 0 322 =∆+++→=∑ αθθlEI

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∆




24

Performance

Response with 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.

1.3.6 A Portal Frame with a Bracing (Horizontal Load)


Analysis

i) All fixed: No fixed end forces

P

A

B C

D

0,0 ≠= EAEI




25

ii)-iv) the same as the previous case




26


06 28 0 2 =∆++→=∑ lEI

lEI

lEIM cB

iB θθ

06 82 0 2 =∆++→=∑ lEI

lEI

lEIM cB

iC θθ

PlEI

lEI

lEIV cB

i =∆+++→=∑ )1(2466 0 322 αθθ

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

Response with 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) 0. 2102801.0 −× 0.2857 0.98

ΜΒΑ (Pl) 1. 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 without bracing. whP allall σ= , 6/hPM allall =

The bracing carries about 99% of the external horizontal load.




27

1.3.7 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


06 28 0 22

2

=∆+++→=∑ lEI

lEI

lEI

lPabM cB

iB θθ

06 82 0 22

2

=∆+++−→=∑ lEI

lEI

lEI

lbPaM cB

iC θθ

∆−=∆+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

B C

D




28

1.3.8 A Portal Frame Subject to Support Settlement


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 =∆+++→=∑ lEI

lEI

lEI

lEIM cB

iB θθδ

06 826 0 22 =∆+++→=∑ lEI

lEI

lEI

lEIM cB

iC θθδ

02466 0 322 =∆++→=∑ lEI

lEI

lEIV cB

i θθ

A

B C

D

δ




29

1.3.9 A Portal Frame with Unsymmetrical Supports


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




30


03 278

0 2 =∆+++→=∑ lEI

lEI

lEIPlM cB

iB θθ

06 828

0 2 =∆+++−→=∑ lEI

lEI

lEIPlM cB

iC θθ

01563 0 322 =∆++→=∑ lEI

lEI

lEIV cB

i θθ

)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




31

1.3.10 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




32

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




32

Chapter 2

Iterative Solution Method &

Moment Distribution Method



33

2.1 Solution Method for Linear Algebraic Equations

2.1.1 Direct Method – Gauss Elimination

nibXa

bXaXaXaXa

bXaXaXaXa

bXaXaXaXabXaXaXaXa

i

n

jjij

nnnnininn

ininiiiii

nni

nnii

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

=++++

=++++

=++++=+++++

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

=

=++

=++++=+++++



34

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 1

1)()(

−

+

=∑−

= iii

i

nkk

iik

ii

i a

XabX for 11 −≤≤ ni

2.1.2 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

−−

−+−−

++−=

+++++−=

+++−=

++−=

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.



35

2.1.3 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 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

−−

−−++−−

−−

−−

++−=

+++++−=

+++−=

++−=

))()((1)( 11

1

11

−

<+=

−

>=

∑∑ −−= k

n

niij

jijk

i

ij

jijiii

ki XaXaba

X

2.1.4 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=90 EI 1.5EI EI

30 35.6 4klf



36

Gauss-Jordan Iteration

))(37.316(91)(

))(34.35(91)(

1

1

−

−

−−=

−=

kBkC

kCkB

θθ

θθ

Gauss-Siedal Iteration

))(37.316(91)(

))(34.35(91)( 1

kBkC

kCkB

θθ

θθ

−−=

−= −

Gauss-Seidal Gauss-Jordan

GAUSS-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



37

2.2 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

LEIM

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=90 EI 1.5EI EI

30 35.6 4klf



38

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-Seidal 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-Seidal 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

∆−=∆→∆−=

∆−−−=−−=

∆−=∆→∆−=

∆−−=−=

−

−

−−−

−−−



39

- 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

MMDMMMMDMM

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

MMDMMM

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)(



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

MMDMMMMDMM



41

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

MMDMMM

MDMM

- 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

-168.7 11.8

35.0

1.3

4.0

0.2

0.5

-115.9

133.3 23.6

-106.1 71.1 -4.0 2. 7

-12.0 8.0

-0.5 0.3

-1.4 0.9

115.9

-133.3 -212.2

11.8 -7. 9 35.6

-23.9 1.4

-0.9 4.0

-2.7 0.2 0.1

-328.0

450.0 -104.5

-3.9

-11.7

-0.5

-1.3

0.1

328.2



42

2.3 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 -0.4 36.5 -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

100 50

A C B D E EI, L 1.5EI, L 1.5EI, L



43

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



44

2.4 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/3 20

A

B C

D



45

2.5 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 20 B C

D

0.5

0.5

0.5

0.5



46

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

53.3 2.67

17.8

35.5 1.78

0.89

8.1

6.1

0.47

7.9

5.9

0.46

0.93



47

This page is left bank intentionally.



48

Chapter 3

Buckling of Structures

P

L θ

P



49

3.0 Stability of Structures Stable state

PKLLPLLK >→θ>×θ

The structure would return its original equilibrium position for a small perturbation in θ.

Critical state

PKLLQLLK =→θ=×θ

Unstable state

PKLLQLLK <→θ<×θ

The structure would not return its original equilibrium position for a small perturbation in

θ.

P

L

θ

P



50

3.1 Governing Equation for a Beam with Axial Force Equilibrium for vertical force

qdxdVqdxVdVV −=→=+−+ 0)(

Equilibrium for moment

VdxdwP

dxdMdx

dxdwPdxqdxVdxMdMM =−→=−+−−+ 0

2)(

Elimination of shear force

qdx

wdPdx

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

wdPdx

wdEI =+ 2

2

4

4

M M+dM

V V+dV

q

w

dxdxdw

P

P



51

3.2 Homogeneous Solutions

Characteristic Equation for 0>P

0 ,0)( 224 ieew

x

x

β±=λ→=λβ+λ

=λ

λ

where EIP

=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

=+−+−=−

=+−+−=+

+−

+−

+−

+−

+−

+−=

+++++++=

−

−

−

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 EIP

=β2

Homogeneous solution for 0<P

DCxxBxA

DCxeeBAeeBA

DCxBeAewxxxx

xx

++β+β=

++−

−++

+=

+++=β−ββ−β

β−β

sinhcosh2

)(2

)(



52

Simple Beam

− Boundary Condition

00sin)( , 0sin)(00)0( , 0)0(

2

2

==→=ββ−=′′=+β=

=→=β−=′′=+=

CBLBLwCLLBLwAAwDAw

− Characteristic Equation

00 =→==== wDCBA (trivial solution) or

3,2,1 , 2

22

=π

=→π=β nL

EInPnL

xL

nBxBw π=β= sinsin

Fixed-Fixed Beam


0cossin)( 0sincos)(

0)0( 0)0(

=+ββ+ββ−=′=++β+β=

=+β=′=+=

CLBLALwDCLLBLALw

CBwDAw

− Characteristic Equation

0)

01cossin1sincos0101001

(

0000


=

ββββ−ββ

β→

=

ββββ−ββ

β

LLLLL

Det

DCBA

LLLLL

1cossinsincos

10

01cos1sin01


LLLLL

LLL

LLLLL

ββββ−ββ

β−

βββ

β=

ββββ−ββ

β

P P

P P



53

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

3,2,1 ,42

02

sin 2

22

=π

=→π=β

→=β n

LEInPnLL

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

EIPLLLLL π=→

β=

β→=

β−

ββ



54

Cantilever Beam


0)(

0)sincos()()(0)0(

0)0(

3

3

22

=−−=

=ββ−ββ−−=′′−=

=+β=′=+=

dxdwP

dxwdEILV

LBLAEILwEILMCBw

DAw

3,2,1 ,4

)12(2

22

=−

= nL

EInP π

Beam-Columns - Simple Beams with a Uniform Load

− Governing Equation

qdx

wdPdx

wdEI =+ 2

2

4

4

− Particular Solution

2

2)( x

Pqxwp =

− General Solution

PqxBxAxw

xPqDCxxBxAxw

+ββ−ββ−=′′

+++β+β=

sincos)(

2sincos)(

22

2

− Boundary Conditions

222 ,0)()0( , 0)0(

β−=

β=→=+β−−=′′=+=

PqD

PqA

PqAEIwDAw

0) sincos()()(

02

)1(cos sin)(

22

22

=+ββ−ββ−−=′′−=

=+−ββ

++β=

PqLBLAEILwEILM

LP

qLPqCLLBLw

P P

P P q



55

LP

qCLPqB

LPqLLB

LPqLB

PqLBL

Pq

2 ,

2tan 1

)12

sin21( 2

cos2

sin2

)1(cos sin0 sincos

2

22

22

−=β

β=

−β

−−=ββ

β

−β−=ββ→=+ββ−β−

− Final Solution

) sin2

tan cos()()(

2

2 sin

2tan )1(cos)( 2

22

PqxL

Pqx

PqEIxwEIxM

xP

qLxP

qxLP

qxP

qxw

−ββ

+β=′′−=

+−ββ

β+−β

β=

− Displacement at the mid-span

)( )8

1

2cos

1(7884.0

)8

1/

2cos

1()(7884.0

)8

1/

2cos

1(5384

3845

)8

1

2cos

1(

8 )1

2sin

2tan

2(cos

8

2sin

2tan 1)1

2(cos)

2(

2

2

22

2

22

2242

24

4

4

22

2

22

222

crst

crcr

crst

PPtt

tt

PP

PPPP

EIPL

EIPLLPEI

EIqL

LLP

q

LP

qLLLP

q

LP

qLLPqL

PqLw

=π

−−π

δ=

π−−

πδ=

ππ

−−π

ππ

π=

β−−

ββ=

−−ββ

+β

β=

−ββ

β+−

ββ

=

− Moment at the Mid-span

)1

2cos

1(1811.0

)1/

2cos

1(8

)12

sin2

tan2

(cos88

)12

sin2

tan2

(cos)2

(

2

2

2

2

2

−π

=

−ππ

=

−ββ

+βπ

π=

−ββ

+β=

ttM

PPPPM

LLLPL

EIqL

LLLPqEILM

st

cr

crst



56

Figure 1. Displacement at the mid-span

Figure 2. Moment at the mid-span

0.0

2.0

4.0

6.0

8.0

10.0

0 0.2 0.4 0.6 0.8 1

δ/δ st

P/Pcr

0.0

2.0

4.0

6.0

8.0

10.0

0 0.2 0.4 0.6 0.8 1P/Pcr

M/M

st



57

3.3 Homogeneous and Particular solution

pph wDCxxBxAwww +++β+β=+= sincos

qdx

wdP

dxwd

EI

dxwd

Pdx

wdEI

dxwd

Pdx

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

LwDCLLBLALwwAEIwEIMwDAw

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.



58

Chapter 4

Energy Principles

Principle of Minimum Potential Energy and

Principle of Virtual Work

h Mg P



59

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 virtual

displacement field should satisfy the displacement boundary conditions of supports if speci-

fied. For beam problems, displacement boundary conditions include boundary conditions for

rotational angle. Variables with superscript e denote the exact solution that satisfies the equili-

brium 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



60

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


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 vir-

tual work holds. In the above equation, the equality sign holds if and only if 0=w .

w we w

any type of support any type of support

q



61

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 corres-

ponding 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 =



62

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 unknowns 23322 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



63

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



64

)3sin811(sin2

)3(20

2)3(0

202

)(0

3

4

3

44

3

44

xl

xlEI

Plw

EIPlbPl

lEIb

b

EIPlaPl

lEIa

atotal

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 ,,1=

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



65

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

iθ

iθ

)( 12 uu −

)( 12 vv −



66

∑∑

∑∑ ∑∑

∑ ∑∑

==

== ==

= ==

+=−+−

+=+

=+++−

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((

θθ

θθ

θθ

∑∑

∑∑∑∑

==

====

+=

==∆=−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

lFFEA

lFFEA

lFFlFvvuuF

11

1111

1212

) ()(

)()())(sin )((cos

The principle of virtual work yields the displacement of a joint k in a truss by applying an unit 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.

4.4. Buckling Problems

Total Potential Energy

dxwqPdxdx

wdEIdx

wd ll

∫∫ −∆−=Π00

2

2

2

2

21

dxwdxwdsLLLL

))(211()(1

0

2

0

2

0∫∫∫

∆−∆−

′+≈′+==

LdxwdxwdxwLLLLL

<<∆′≈′=∆→′+∆−= ∫∫∫∆−∆−

for )(21)(

21)(

21

0

2

0

2

0

2

dxwqdxdxdw

dxdwPdx

dxwdEI

dxwd lll

∫∫∫ −−=Π000

2

2

2

2

21

21

∆ P



67

Principle of the Minimum Potential Energy

∫

∫∫

∫∫

∫∫∫

∫∫∫

−+Π=

−+−++Π=

−+−

−+−−=

+−++

−++

=Π

le

ll eee

ll

l eele

l eeee

le

l eel eeh

dxdxwdP

dxwd

dxwdEI

dxwd

dxdxwdP

dxwd

dxwdEI

dxwddxq

dxwdP

dxwdEIw

dxdxwdP

dxwd

dxwdEI

dxwdqdxw

dxdxwdP

dxdw

dxwdEI

dxwdqdxwdx

dxdwP

dxdw

dxwdEI

dxwd

qdxwwdxdx

wwdPdx

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

wdxdxwdP

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.

wdxdxwdP

dxwd

dxwdEI

dxwdl

somefor 0)(0

2

2

2

2

≤−∫

The critical status of a structure is defined as

0)(0

2

2

2

2

=−∫l

dxdxwdP

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

aKaaKaadxggPaadxgEIga

dxgagaPdxgaEIga

KKKK



68

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

lEIPlPEIlDet

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

lEIPl

lPl

lEIDet

ll

dxlx

ldxgEIg

ll

EIdxlx

lEIdxgEIg

cr

ll

ii

ll

ii

π=→=

π−

π

π=

ππ=′′

π=

ππ=′′′′

∫∫

∫∫

Example – Cantilever Beam

− with one unknown:

%)22error , 46.24

:exact( 30)344(

344 , 422

2 , 2

22

2

23

3

0

2

000

112

===→=−

==′′==′′′′

=′′=′→=

∫∫∫∫

lEI

lEI

lEIPlPEIlDet

ldxxdxgEIgEIldxEIdxgEIg

gxgaxw

cr

ll

ii

ll

ii

π

P

P P



69

− 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

PldxxPKPldxxPKKPldxxPK

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

lEIP

lllllll

lllllllll

EIP

llllllll

llll

Pllll

EIDet

cr =→=±

=−±

=

=+−→=−−−−

==−−−−

→=

−

α

ααααα

ααααα

249.2lEIPexact = (error = 1.2%) or 219.22

lEIPexact = (error = 45%)

Beam-Columns - Simple Beams with a Uniform Load

− with one sine curve:

sin)( , cossin 211 l

xl

glx

lg

lxaw ππ

=′′ππ=′→

π=

ππ

ππππππ

lqdxlxqqdxg

ll

dxlx

ldxggl

lEIdx

lx

lEIdxgEIg

ll

i

ll

ii

ll

ii

2sin

2)(cos)( ,

2)(sin)(

00

2

0

22

00

424

0

==

==′′==′′′′

∫∫

∫∫∫∫

tPPEIql

lEIPql

EIq

lP

lEI

qa

lqall

Pall

EIa

alqall

Pall

EI

stcr −

δ=−π

×=

π−

πππ

=π

−π

π=

=π

−π

−π

=∂Π∂

→π

−π

−π

=Π

10038.1

/11

54384

3845

)(1

/4)(14

)()(

/4

022

)( 2

)(

22

)(21

2)(

21Min

5

4

2

4

24

24

2224

P P q



70

0.0

2.0

4.0

6.0

8.0

10.0

0 0.2 0.4 0.6 0.8 1

Exact SolutionEnergy Method with one sine function

δ/δ st

P/Pcr



71

This page is left blank intentionally.



72

Chapter 5

Matrix Structural Analysis

Mr. Force & Ms. Displacement

Matchmaker: Stiffness Matrix



73

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 System e

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

Rxf

Ryf L

yf

Rxδ

Ryδ L

yδ

Lxδ L

xf

1

2

xv

yv

φ

xV

yV



74

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 Coordinate eeTeeTeTe )]([][][)(][][)(][)( ∆δ ΓΓ=Γ=Γ= kkfF

eee )(][)( ∆KF =

=

φφφφ−φφ−φφφφθ−φ−φ−φφ−φφθφφ−φ−φθφ

= ee

eee

LEA

][][][][

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

mP m-th joint i-th member

n-th joint



75

)]([)(][)()(

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

[ ]

==

p

ipi

)(

)(

)(

)( , ][][][][

1

1

F

F

F

FEEEE

m-th row

n-th row

1

2

i)( 2F

i)( 1F

m-th joint mP

i-th member

nP n-th joint



76

Compatibility 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 =

=

=

∆∆

∆

)]([][

][

)(

)()(

111

uCu

u

C

C=

=

=

qpp

∆

∆∆

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



77

Unassembled Member Stiffness Equation

)(

)(

)(

][00

0][0

00][

)(

)(

)( 111

→

∆

∆

∆

=

p

i

p

i

p

i

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

++++=

==

=

=

=

000000][0][0000000][0][000000

0][][00][][0

00

0

0

00

000000

][][][][

00

0

0

00

][][][

2221

1211

2221

1211

2221

1211

ii

ii

ii

ii

ii

iiiiiT

KK

KK

KKKK

I

I

II

KKKK

I

ICKC

m-th row

n-th row

m-th column n-th column



78

5.1.3 Example Member Stiffness Matrix

=

φφφφ−φφ−φφφφφ−φ−φ−φφ−φφφφφ−φ−φφφ

= ee

eee

LEA

][][][][



][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

2

3

2L

L

P3, u3

P4, u4

P6, u6

P5, u5

P1, u1

P2, u2 1 2

3



79

- Member 3: φ = 180o

−

−

=

0000010100000101

2][ 3

LEAK


31226

31225

21124

21123

32112

32111

)()()()(

)()()()(

)()()()(

yy

xx

yy

xx

yy

xx

FFPFFP

FFPFFP

FFPFFP

+=

+=

+=

+=

+=

+=

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

)]([=)](][ ][ ][[=)( 321 FEFEEEP

P3

P4

P6

P5 P1

P2

[E]1 [E]2 [E]3



80


[ ][ ][ ]

=

=

∆∆∆∆

∆∆∆∆

∆∆∆∆

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 Matrix

333222111 ][][][][][][][][][]][[][][ CKCCKCCKCCKCK TTTT +++==

u3

u6

u5

u2

u1

u4

[C]1

[C]3 [C]2



81

−−

−−

−−

−−

=

−−

−−

−−

−−

=

−−

−−

−−

−−

=

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



82

−

−

=

−

−

=

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



83

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

Known Known

Known

Unknown Unknown Unknown

Unknown Unknown

Unknown

Known Known Known



84

5.2 Beam Problems 5.2.1 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 Matrix e

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 =

Lθ Lm

Lyf

Ryf

Rθ Rm

Ryδ L

yδ



85

5.2.2 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

FFFFFF

FFF

PP

PPP

PP

=

=

+

+

+

+

+

=

−

+

−

−−

+

+

−

p

ipi

Rp

Lp

Rp

Li

Ri

Li

Ri

Li

Ri

LR

L

p

p

i

i

i

iiRi

Li

Ri

Li )(][

)()(

00

00

00

0

)()(

0

FEFF

II

FF

=

=

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 row i-th row

i-th Member (i-1)-th Member

12 −iu

iu2

12 +iu

22 +iu



86

)(][

000000

)()(

)(

1

1

1

uC

u

uu

u

II

i

p

i

i

Ri

Li

i =

=

=

+

+

∆∆

∆

)]([][

][

)(

)()(

1

111

uCu

u

C

C=

=

=+p

pp

∆

∆∆

Unassembled Member Stiffness Equation

→

∆

∆

∆

=

p

i

p

i

p

i

)(

)(

)(

][00

0][0

00][

)(

)(

)( 111

K

K

K

F

F

F

)]([)( ∆= KF


)](][[][)]([][)(][)( uCKCKCFCP TTT =∆==

)]([)( uKP = where ]][[][][ CKCK T=


=

0

00

0

00000

][][0][][0

00000

][][][2221

1211

ii

iiii

Ti KK

KKCKC

i-th column

(i+1)-th column

i-th row i+1-th row

i+1-th column i-th column



87

5.2.3 Example


RR

LRLR

LRLR

LL

MMVVMMMVVVMMMVVV

MMVV

34

34

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

MVMVMVMVMVMV

MVMVMVMV


43

43

33

33

32

32

22

22

21

21

11

11

, , , , , , , , ,

θθθθ

θθθθ

θθθθ

====

====

====

RRLL

RRLL

RRLL

wwwwwwwwwwww

[E]1 [E]2

[E]3

11 ,θM

11 , wV

22 ,θM

22 , wV

33 ,θM

33 , wV

44 ,θM

44 , wV



88

)]([

][][][

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

Δ =

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

ΔKΔΔΔ

KK

K

FFF

F =

=

=


)]([)](][[][)](][[)]([)( uKuCKCuKEFEP ==== T

[ ]

))( ][][][][][][][][][ ()]([

)(][

][][

][000][000][

][ ][ ][)(

333222111

3

2

1

3

2

1

321

uCKCCKCCKCuK

uCCC

KK

KCCCP

TTT

TTT

++==

=

[C]1

[C]3

[C]2



89

- 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



90

- 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


−

−−−

−++−

−+−+−−

−++−

−+−+−−

−

−

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

222

2

2

2

1

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



91

Application of support Conditions

−

−−−

−++−

−+−+−−

−++−

−+−+−−

−

−

=

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

222

2

2

2

1

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

θ

θ

θ

θ

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



92

5.3 Frame Problems 5.3.1 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,



93

Transformation Matrix

θθθ−θ

=

→

=

θ+θ=

θ−θ=

mvv

MVV

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

→

∆∆

∆∆

−

−

=

e

y

x

y

x

e

Re

Ry

Rx

Le

Ly

Lx

2

2

2

1

1

1

1000000cossin0000sincos0000001000000cossin0000sincos

θ

θ

θθθθ

θθθθ

θ

δδθ

δδ

ee )]([)( ∆Γ=δ

1

2 xv

yv

φ

xV

yV

M

m



94

Member Stiffness Matrix in Global Coordinate eeTeeTeTe )]([][][)(][][)(][)( ∆δ ΓΓ=Γ=Γ= kkfF

eee )(][)( ∆KF = where

= ee

eee

][][][][

][2221

1211

KKKK

K

Nodal Equilibrium & Compatibility

The same as the truss problems.


)](][[][)]([][)(][)( uCKCKCFCP TTT =∆==

)]([)( uKP = where ]][[][][ CKCK T=


The same as the truss problems.



95

5.4 Buckling of Beams and Frames

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 +−=

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

liTi

p

i

liTi

p

i

l

ii

p

i

lii

p

i

lii

dxqdxdxd

dxdQdx

dxdEI

dxd

dxqwdxdxdw

dxdwQdx

dxwdEI

dxwd

dxqwdxdxdw

dxdwQdx

dxwdEI

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

Q



96

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

Calculation of the Critical Load

0])[]([])([ 0 =−= GDetDet KKK

Frame Members

−−

−−

−

−

−

−

−−−

−

−

=

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

Q

ILII

LI

LI

LI

LI

LI

AA

ILII

LI

LI

LI

LI

LI

AA

LE

m

f

f

m

f

f

θ

δ

δ

θ

δ

δ

)

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

=



97

5.5 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 Matrix e

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 Analysis ei

eg

ei

eei p ))(][]([)( 0 δkkf +=

eeeeg

ei

eTeeTeTe p )(][)]()[][]([][)(][][)(][)( 0 ∆∆δ KkkkfF =Γ+Γ=Γ=Γ=

Successive substitution ee

iee

gei

eTei p )(][)]()[][]([][)( 1101 ∆∆ −−− =Γ+Γ≈ KkkF

iG

eip ))()]([]([)( 10 uKKP −+=

Rxf

Ryf

Lyf

Rxδ

Ryδ

Lyδ

Lxδ L

xf



98

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

iRy

Rx

Ly

Lx

ei

Ly

Ry

e

iRy

Rx

Ly

Lx

e

iLy

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

Documents

Class Note for Structural Analysis 2strana.snu.ac.kr/lecture/struct2_2017/Notes/Lecture_Note_2017.pdf · Class Note for Structural Analysis 2 . Fall Semester, 2017 . Hae Sung Lee,