Class Note for Structural Analysis 2strana.snu.ac.kr/lecture/struct2_2020/Notes/Note_all_2020.pdf · Class Note for Structural Analysis 2 Spring Semester, 2020 Hae Sung Lee, Professor

Class Note for Structural Analysis 2

Spring Semester, 2020

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………………………… 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 32

2.1 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 Buckling of Structures 48

3.0 Stability of Structures……………………………………………………………... 493.1 Governing Equation for a Beam with Axial Force………………………………... 503.2 Homogeneous Solutions…………………………………………………………... 513.3 Homogeneous and Particular solution…………………………………………….. 57

Chapter 4 Energy Principles 58

4.1 Spring-Force Systems……………………………………………………………... 594.2 Beam Problems……………………………………………………………………. 604.3 Truss problems…………………………………………………………………….. 644.4 Buckling problems………………………………………………………………… 66

Chapter 5 Matrix Structural Analysis 72

5.1 Truss Problems…………………………………………………………………….. 735.2 Beam Problems……………………………………………………………………. 845.3 Frame Problems………………………………………………………………….... 925.4 Buckling of Beams and Frames…..……………………………………………….. 95

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



2

1.0 Comparison of Flexibility Method and Stiffness Method

Flexibility Method

Remove redundancy (Equilibrium)

Compatibility

21 δ=δ

Pkk

kX

k

XP

k

X

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

EI

PLPL

EI

LB 16

14

)21

1(6

2

0 =×+=δ , EI

LBB 3

2=δ

323

0 00

PLMM

BB

BBBBBB −=

δδ−=→=δ+δ

Stiffness Method

Compatibility

BBCBA θ=θ=θ

Equilibrium

0 , 16

3 =−= fBC

fBA M

PLM , B

BBC

BBA L

EIMM θ== 3

EI

PL

L

EIPL

MMMMM

BB

BBC

BBA

fBC

fBAB

320

616

3

02

=θ→=θ+−

→=+++=

1

+ +

EI

L

EI

L

L/2 P

A B

C

16

3PL

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

MEI

LM

EI

L

MEI

LM

EI

L

→ AA L

EIM θ−= 4

, AB L

EIM θ= 2

ii) 0=Aθ

BA L

EIM θ−= 2

, BB L

EIM θ= 4

Sign Convention for M :Counterclockwise “ +”

0≠θA , 0≠θB

BAB

BAA

L

EI

L

EIM

L

EI

L

EIM

θ+θ=

θ+θ=

42

24

1.1.2 Relative motion of joints

∆

MA MB

A B

Aθ



6

Flexibility Method

LM

EI

LM

EI

LL

MEI

LM

EI

L

BA

BA

∆=+

∆−=+

36

63 →LL

EIM A

∆−= 6 ,

LL

EIM B

∆= 6

or in the new sign convention : LL

EIM A

∆= 6 ,

LL

EIM B

∆= 6

Final Slope-Deflection Equation

LL

EI

L

EI

L

EIM

LL

EI

L

EI

L

EIM

BAB

BAA

∆+θ+θ=

∆+θ+θ=

642

624

In Case an One End is Hinged

LL

EI

L

EI

LL

EI

L

EI

L

EIM

LL

EI

L

EI

L

EI

LL

EI

L

EI

L

EIM

BBAB

BABAA

∆+θ=∆+θ+θ=

∆−θ−=θ→=∆+θ+θ=

33642

320

624

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 f

B −=−=−−= , 0=fAM

1.1.4 Joint Equilibrium

=+−− 0jointmemberfixed FFF

or

=+ jointmemberfixed FFF

MA MB

MA MA/2

AML2

3AM

L2

3

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 a

EIM θ= 21 , BBA a

EIM θ= 41 , BBC b

EIM θ= 41 , BCB b

EIM θ= 21

BABBA a

EIVV θ=−=

211 6

, BCBBC b

EIVV θ==−

211 6

iii) 0=θB , 0≠∆B

BAB a

EIM ∆=

22 6

, BBA a

EIM ∆=

22 6

, BBC b

EIM ∆−=

22 6

, BCB b

EIM ∆−=

22 6

BABBA a

EIVV ∆=−=

322 12

, BCBBC b

EIVV Δ

123

22 =−=

Ba

EI θ4Ba

EI θ2Bb

EI θ4Bb

EI θ2

Ba

EI θ2

6 Ba

EI θ2

6 Bb

EI θ2

6 Bb

EI θ2

6

Ba

EI ∆2

6 Ba

EI ∆2

6 Bb

EI ∆2

6 Bb

EI ∆2

6

Ba

EI ∆3

12Ba

EI ∆3

12Bb

EI ∆3

12Bb

EI ∆3

12

b a EI

P

A B

C



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

ab

a

EI

a

EIMMM BBABABAB 2

2

221 62 =∆+θ=+= ,

Pl

ba

b

EI

b

EIMMM BBCBCBCB 2

2

221 62 −=∆−θ=+=

1.2.2 Analysis of a Two-span Continuous Beam (Approach I)

DOF : Bθ , Cθ

Analysis

i) Fix all DOFs and Calculate FEM.

12

2qLM f

AB = , 12

2qLM f

BA −= , 8

2qLM f

BC = , 8

2qLM f

CB −=

ii) 0≠θB , 0=θC

BAB L

EIM θ= 21 , BBA L

EIM θ= 41 , BBC L

EIM θ= 81 , BCB L

EIM θ= 41

iii) 0=θB , 0≠θC

CBC L

EIM θ= 42 , CCB L

EIM θ= 82


→=++++→= 00 211BCBCBA

fBC

fBA

iB MMMMMM 0412

24

2

=θ+θ+ CB L

EI

L

EIqL

→=++→= 00 21CBCB

fCB

iC MMMM 084

8

2

=θ+θ+− CB L

EI

L

EIqL

qL

EI 2EI L L

q

A B

C



10

2

512

17qL

2

16

1qL 2

8

1qL 2

16

3qL

L16

7

EI

qLB 96

3

−=θ , EI

qLC 48

3

=θ

Member End Forces

22

1

16

12

12qL

L

EIqLMMM BAB

fABAB =θ+=+=

22

1

8

14

12qL

L

EIqLMMM BBA

fBABA −=θ+−=+=

22

21

8

148

8qL

L

EI

L

EIqLMMMM CBBCBC

fBCBC =θ+θ+=++=

084

8

221 =θ+θ+−=++= CBCBCB

fCBCB L

EI

L

EIqLMMMM

Various Diagram - Free Body Diagram

- Moment Diagram

2

16

1qL 2

8

1qL

qL16

7

qL16

9qL

8

5qL

8

3

qL16

19



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 f

AB = , 12

2qLM f

BA −= , 16

3

82

1

8

222 qLqLqLM f

BC =+=

ii) 0≠θB

BAB L

EIM θ= 21 , BBA L

EIM θ= 41 , BBC L

EIM θ= 61

Construct Stiffness Equation

00 11 =+++→= BCBAf

BCf

BAB MMMMM

EI

qL

L

EI

L

EIqLqLBBB 96

06416

3

12

322

−=θ→=θ+θ++−

Member End Forces

22

1

48

32

12qL

L

EIqLMMM BAB

fABAB =θ+=+=

22

1

8

14

12qL

L

EIqLMMM BBA

fBABA −=θ+−=+=

22

1

8

16

16

3qL

L

EIqLMMM 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 , L

Cθ , RCθ , ∆C

Analysis

i) All fixed

12

2qlM f

AB = , 12

2qlM f

BA −=

ii) 0≠θB

BAB l

EIM θ= 21 , BBCBA l

EIMM θ== 411 , BCB l

EIM θ= 21 , BCB l

EIV θ=

21 6

iii) 0≠θL

C

LCBC l

EIM θ= 22 , L

CCB l

EIM θ= 42 , L

CCB l

EIV θ=

22 6

iv) 0≠θRC

RCCD l

EIM θ= 43 , R

CDC l

EIM θ= 23 , L

CCD l

EIV θ−=

23 6

v) 0≠∆C

CCBBC l

EIMM ∆==

244 6

, CDCCD l

EIMM ∆−==

244 6

, CCBCD l

EIVV ∆==

244 12

q

EI EI EI

l l l

A B C

D



13

Construct Stiffness Equation

06 2812

02

2

1 =∆+θ+θ+−→= CLCB

i

i

l

EI

l

EI

l

EIqlM

06 42 02

2 =∆+θ+θ→= CLCB

i

i

l

EI

l

EI

l

EIM

06 4 023 =∆−θ→= C

RC

i

i

l

EI

l

EIM

024666 03222

4 =∆+θ−θ+θ→= CRC

LCB

i

i

l

EI

l

EI

l

EI

l

EIV

Elimination of LCθ and R

Cθ

- 2nd and 3rd equation

)3(22 CB

LC l

EI

l

EI

l

EI ∆+θ−=θ , CRC l

EI

l

EI ∆=θ2

3 2

- 1st equation

03 712

06 )3 (812

62812

2

2

22

2

2

2

=∆+θ+−→=∆+∆+θ−θ+−

=∆+θ+θ+−

CBCCBB

CLCB

l

EI

l

EIql

l

EI

l

EI

l

EI

l

EIql

l

EI

l

EI

l

EIql

- 4th equation

063024)3(3)3(36

24666

3233322

3222

=∆+θ→=∆+∆−∆+θ−θ

=∆+θ−θ+θ

CBCCCBB

CRC

LCB

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EIl

EI

l

EI

l

EI

l

EI

1.2.5 Analysis of a Beam with an Internal Hinge (2 DOFs System)

DOF : Bθ , ∆C

Analysis

i) All fixed

12

2qlM f

AB = , 12

2qlM f

BA −=

q

EI EI EI

l l l

A B C

D



14

ii) 0≠θB

BAB l

EIM θ= 21 , BBA l

EIM θ= 41 , BBC l

EIM θ= 31 , BCBBC l

EIVV θ−=−=

211 3

iii) 0≠∆C

CDCBC l

EIMM ∆=−=

222 3

, CCBBC l

EIVV ∆−=−=

322 3

, CDCCD l

EIVV ∆=−=

322 3


03 712

02

2

1 =∆+θ+−→= CBi

i

l

EI

l

EIqlM

063 0322 =∆+θ→= CB

i

i

l

EI

l

EIV

EI

qlB 66

3

=θ , EI

qlC 132

4

−=∆

EI

ql

lll

EI

l

EI CRC

CRC 882

3)(

32 3

−=∆=θ→∆−−=θ

EI

ql

EI

ql

EI

ql

lC

BLC 2641322

3

1322

3

2

1 333

=+−=∆−θ−=θ

1.2.6 Beam with a Spring Support

Analysis

i) All fixed

12

2qlM f

AB = , 12

2qlM f

BA −=

l l l

D

k

q

EI EI EI A

B C



15

ii) 0≠θB

BAB l

EIM θ= 21 , BBA l

EIM θ= 41 , BBC l

EIM θ= 31 , BCBBC l

EIVV θ==−

211 3

iii) 0≠∆C

CDCBC l

EIMM ∆=−=

222 3

CCBBC l

EIVV ∆−=−=

222 3

, CDCCD l

EIVV ∆=−=

222 3

, CS kV ∆=2


03 712

02

2

1 =∆+θ+−→= CBi

i

l

EI

l

EIqlM

0)6(3 0 322 =∆++θ→= CBi

i kl

EI

l

EIV

EI

qlB 6611/141

1 3

α+α+=θ ,

EI

qlC 132)11/141(

1 4

α+−=∆ where 3

6

l

EIk α=

0→α

EI

qlB 66

3

=θ , EI

qlC 132

4

−=∆

∞→α

EI

qlB 84

3

=θ , 0=∆C

k∆C



16

1.2.7 Support Settlement

DOF : Bθ

Analysis

i) All fixed

ll

EIM f

BA

δ= 6 ,

ll

EIM f

BC

δ−= 3

ii) 0≠θB

BBA l

EIM θ= 41 , BBC l

EIM θ= 31

Construct the Equilibrium Equation

0343601 =θ+θ+δ−δ→= BBi

i

l

EI

l

EI

ll

EI

ll

EIM

lB

δ−=θ→7

3

1.2.8 Temperature Change

T1

T2

lh

TTl

h

TTBA 2

)( ,

2

)( 1212 −α−=θ−α=θ

Fixed End Moment

EIh

TT

L

EI

L

EIM

EIh

TT

L

EI

L

EIM

BAB

BAA

)(42

)(24

12

12

−α−=θ+θ=

−α=θ+θ=

EI EI

l l A B C

δ

A B



17

1.3 Analysis of Frames

1.3.1 A Portal Frame without Sidesway

DOF : Bθ , Cθ

Analysis

i) All fixed

80 Pl

M BC = , 8

0 PlMCB −=

ii) 0≠θB

BAB l

EIM θ= 11 2

, BBA l

EIM θ= 11 4

BBC l

EIM θ= 21 4

, BCB l

EIM θ= 21 2

iii) 0≠θC

CBC l

EIM θ= 22 2

, CCB l

EIM θ= 22 4

CCD l

EIM θ= 12 4

, CDC l

EIM θ= 12 2


02

)44

(8

0 221 =θ+θ++→= CBiB l

EI

l

EI

l

EIPlM

0)44

(2

80 212 =θ++θ+−→= CB

iC l

EI

l

EI

l

EIPlM

8241 2

21

Pl

EIEICB +=θ=θ−

l/2 P

A

B C

D

EI1

EI2

EI1



18

Member End Forces

82422

21

11 Pl

EIEI

EI

l

EIM BAB +

−=θ=

82444

21

11 Pl

EIEI

EI

l

EIM BBA +

−=θ=

824424

8 21

122 Pl

EIEI

EI

l

EI

l

EIPlM CBBC +

=θ+θ+=

824442

8 21

122 Pl

EIEI

EI

l

EI

l

EIPlM CBCB +

−=θ+θ+−=

82444

21

11 Pl

EIEI

EI

l

EIM CCD +

=θ=

82422

21

11 Pl

EIEI

EI

l

EIM CDC +

=θ=

In case 21 EIEI =

24

PlM AB −= ,

12

PlM BA −= ,

12

PlM BC = ,

12

PlMCB −= ,

12

PlMCD = ,

24

PlM DC =

1.3.2 A Portal Frame without Sidesway – hinged supports

DOF : Bθ , Cθ

Analysis

i) All fixed

80 Pl

M BC = , 8

0 PlMCB −=

l/2 P

A

B C

D

EI1

EI2

EI1



19

ii) 0≠θB

BBA l

EIM θ= 11 3

BBC l

EIM θ= 21 4

, BCB l

EIM θ= 21 2

iii) 0≠θC

CBC l

EIM θ= 22 2

, CCB l

EIM θ= 22 4

CCD l

EIM θ= 12 3


02

)43

(8

0 221 =θ+θ++→= CBiB l

EI

l

EI

l

EIPlM

0)43

(2

80 212 =θ++θ+−→= CB

iC l

EI

l

EI

l

EIPlM

8231 2

21

Pl

EIEICB +=θ=θ−

Member End Forces

0=ABM

82333

21

11 Pl

EIEI

EI

l

EIM BBA +

−=θ=

823324

8 21

122 Pl

EIEI

EI

l

EI

l

EIPlM CBBC +

=θ+θ+=

823342

8 21

122 Pl

EIEI

EI

l

EI

l

EIPlM CBCB +

−=θ+θ+−=

82333

21

11 Pl

EIEI

EI

l

EIM CCD +

=θ=

0=DCM

In case of 21 EIEI =

0=ABM , PlM BA 40

3−= , PlM BC 40

3= , PlMCB 40

3−= , PlMCD 40

3= ,

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 l

EIM θ= 21 , BBA l

EIM θ= 41

BBC l

EIM θ= 31 , BBA l

EIV θ=

21 6

iii) 0≠∆

∆=2

2 6

l

EIM AB , ∆=

22 6

l

EIM BA

∆=3

2 12

l

EIVBA

Construct the stiffness equation

06

)34

( 02

=∆+θ+→= l

EI

l

EI

l

EIM B

iB

Pl

EI

l

EIPV B

i =∆+θ→= 32

126

EI

Pl

EI

PlB 48

7 ,

8

32

=∆−=θ

Member end forces

Pll

EI

l

EIM BAB 8

5622

=∆+θ= , Pll

EI

l

EIM BBA 8

3642

=∆+θ= , Pll

EIM 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

l

PabM BC = , 2

20

l

bPaMCB −=

ii) 0≠θB

BAB l

EIM θ= 21 , BBA l

EIM θ= 41

BBC l

EIM θ= 41 , BCB l

EIM θ= 21 , BBA l

EIV θ=

21 6

iii) 0≠θC

CBC l

EIM θ= 22 , CCB l

EIM θ= 42

CCD l

EIM θ= 42 , CDC l

EIM θ= 22 , CCD l

EIV θ=

22 6

a P

A

B C

D



22

iv) 0≠∆

∆==2

33 6

l

EIMM BAAB , ∆==

233 6

l

EIMM DCCD ,

∆==3

33 12

l

EIVV CDBA


06

28

022

2

=∆+θ+θ+→= l

EI

l

EI

l

EI

l

PabM cB

iB

06

82

022

2

=∆+θ+θ+−→= l

EI

l

EI

l

EI

l

bPaM cB

iC

02466

0322

=∆+θ+θ→= l

EI

l

EI

l

EIV cB

i

)(4 CB

l θ+θ−=∆

0 22

13

2

2

=θ+θ+ cB l

EI

l

EI

l

bPa

0 2

13

22

2

=θ+θ+− cB l

EI

l

EI

l

Pab

l

ba

EI

PabB

)13(

84

1 +−=θ

l

ba

EI

PabC

)13(

84

1

+=θ

)(28

1 ab

EI

Pab −=∆

1.3.5 A Portal Frame with a Bracing (Vertical Load)


Analysis

i) All fixed

2

20

l

PabM BC = , 2

20

l

bPaMCB −=

4

la =

a P

A

B C

D

0,0 ≠= EAEI



23

ii) 0≠θB

BAB l

EIM θ= 21 , BBA l

EIM θ= 41

BBC l

EIM θ= 41 , BCB l

EIM θ= 21

BBA l

EIV θ=

21 6

iii) 0≠θC

CBC l

EIM θ= 22 , CCB l

EIM θ= 42

CCD l

EIM θ= 42 , CDC l

EIM θ= 22

CCD l

EIV θ=

22 6

iv) 0≠∆

∆==2

33 6

l

EIMM BAAB , ∆==

233 6

l

EIMM DCCD ,

∆==3

33 12

l

EIVV CDBA

22

∆=l

EAABD (Compression)

222

1

22

∆==∆=→l

EAV

l

EAV BDBD


06

28

022

2

=∆+θ+θ+→=l

EI

l

EI

l

EI

l

PabM cB

iB

06

82

022

2

=∆+θ+θ+−→=l

EI

l

EI

l

EI

l

bPaM cB

iC

0)1(2466

0322

=∆α++θ+θ→=l

EI

l

EI

l

EIV cB

i

EI

EAllCB

248 , )(

411 2

=αθ+θα+

−=∆

Solution for ab 3=

EI

PlB

2

)107(2565240

α+α+−=θ ,

EI

PlC

2

)107(2562816

α+α+−=θ ,

EI

Pl3

)107(1283

α+=∆

For a hw× rectangular section and hl 20= , 250=α .

EI

PlB

2

0203.0−=θ , EI

PlC

2

0109.0 =θ EI

Pl34103282.0 −×=∆

∆

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= , 6/6/)2//( 2 hPwhhIM allallallall =σ=σ=

Unbalanced shear force in the columns = Pl

EICB 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

ii)-iv) the same as the previous case

P

A

B C

D

0,0 ≠= EAEI



25


06

28

02

=∆+θ+θ→=l

EI

l

EI

l

EIM cB

iB

06

82

02

=∆+θ+θ→=l

EI

l

EI

l

EIM cB

iC

Pl

EI

l

EI

l

EIV cB

i =∆α++θ+θ→= )1(2466

0322

CB θ=θ , ll CB θ−=θ−=∆3

5

3

5

Solution

EI

Pl

EI

PlCB

32

)4028(35

,)4028(

1θθ

α+=∆

α+−==

For 250=α ,

EI

PlCB

23103501.0 −×−=θ=θ ,

EI

Pl33105835.0 −×=∆

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) 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 without bracing. whP allall σ= , 6/hPM allall =

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



26

1.3.7 A Portal Frame with a Spring

DOF : Bθ , Cθ , ∆ Analysis

iv) 0≠∆

∆==2

33 6

l

EIMM BAAB , ∆==

233 6

l

EIMM DCCD ,

∆==3

33 24

l

EIVV CDBA , ∆= kVS

3


06

28

022

2

=∆+θ+θ+→=l

EI

l

EI

l

EI

l

PabM cB

iB

06

82

022

2

=∆+θ+θ+−→=l

EI

l

EI

l

EI

l

bPaM cB

iC

∆−=∆+θ+θ→= kl

EI

l

EI

l

EIV cB

i322

2466 0

0)24

(66

322

=∆++θ+θ kl

EI

l

EI

l

EIcB

Deformed Shapes ( 41.0=∆∆S )

without a spring with a spring ( )243l

EIk =

k

a P

A

B C

D

k∆



27

1.3.8 A Portal Frame Subject to Support Settlement


Analysis

i) All fixed

δ==2

00 6

l

EIMM CBBC

ii)-iv) the same as the previous problem Construct the Stiffness Equation

06

286

022

=∆+θ+θ+δ→=l

EI

l

EI

l

EI

l

EIM cB

iB

06

826

022

=∆+θ+θ+δ→=l

EI

l

EI

l

EI

l

EIM cB

iC

02466

0322

=∆+θ+θ→=l

EI

l

EI

l

EIV cB

i

A

B C

D

δ



28

1.3.9 A Portal Frame with Unsymmetrical Supports


Analysis

i) All fixed

80 Pl

M BC = , 8

0 PlMCB −=

ii) 0≠θB

BBA l

EIM θ= 31

BBC l

EIM θ= 41 , BCB l

EIM θ= 21

BBA l

EIV θ=

21 3

iii) 0≠θC

CBC l

EIM θ= 22 , CCB l

EIM θ= 42

CCD l

EIM θ= 42 , CDC l

EIM θ= 22

CCD l

EIV θ=

22 6

iv) 0≠∆

∆=2

3 3

l

EIM BA ,

∆==2

33 6

l

EIMM DCCD ,

∆=3

3 3

l

EIVBA , ∆=

33 15

l

EIVCD

l/2 P

A

B C

D



29


03

27

8 0

2=∆+θ+θ+→=

l

EI

l

EI

l

EIPlM cB

iB

06

82

8 0

2 =∆+θ+θ+−→=

l

EI

l

EI

l

EIPlM cB

iC

01563

0322

=∆+θ+θ→=l

EI

l

EI

l

EIV cB

i

)2(5 CB

l θ+θ−=∆

EI

PlB

2

441−=θ ,

EI

PlC

2

89

441

=θ , EI

Pl3

1761−=∆

Load Location that Causes No Sidesway

CBCB

l θ−=θ→=θ+θ−=∆ 20)2(5

- Stiffness equation

0 27

02

2

=θ+θ+→= cBiB l

EI

l

EI

l

PabM , 0

82 0

2

2

=θ+θ+−→= cBiC l

EI

l

EI

l

bPaM

012

2

2

=θ− Cl

EI

l

Pab , 0 4

2

2

=θ+− cl

EI

l

bPa

abl

bPa

l

Pab33

2

2

2

2

=→=

a P

A

B C

D



30

1.3.10 A Frame with a Skewed Member

DOF : Bθ , ∆

Analysis

i) All fixed : PlPlPl

M BC 16

3

1680 =+= , PVBC 16

11220 −=

ii) 0≠θB

BBAB l

EI

l

EIM θ=θ= 2

2

21 , BBA l

EIM θ= 221 , BBC l

EIM θ= 31 , BBA l

EIV θ=

21 3

BBC l

EIV θ−= 2

1

223

iii) 0≠∆

∆=∆==2

22 3

22

6l

EI

ll

EIMM ABBA ∆−=∆−=

2

2

2

23

2

3

l

EI

ll

EIM BC , ∆=

32 23

l

EIVBA ,

∆=3

2

2

3

l

EIVBC

P

A

B C

BBB l

EI

ll

EI

l

EI θ=θ+θ2

32

1)222(

BB l

EI

ll

EI θ=θ 222

3221

3

Bl

EI

ll

EI

l

EI θ=∆+∆322

232

1)33(

∆=∆ 32 23

2

1122

3l

EI

ll

EI



31

Construct the stiffness equation

Pll

EI

l

EIM B

iB 16

3 )

223

3()32(20 2 −=∆−+θ+→=

Pl

EI

l

EIV B

i

1611

22

)23

23()223

(3 0 32 =∆++θ−→=

Pll

EI

l

EIB 1875.0 8787.05.8284

2−=∆+θ

Pl

EI

l

EIB 4861.07426.58787.0

32=∆+θ

EI

PlB

2

0460.0−=θ , EI

Pl3

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

bXaXaXaXa

bXaXaXaXa

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

a

ai 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

bXaXaXa

bXaXaXaXa

=++++

=++++

=++++=+++++

LL

M

LL

M

LL

LL

where 11

11)2(

a

aaaa ji

ijij −= . Again, the second unknown X2 is eliminated from the third equa-

tion by multiplying )2(

22

)2(2

a

ai 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

bXaXaXa

bXaXaXaXa

=

=++

=++++=+++++

MM

L

MMM

LL

LL



34

where 1

1,1

)1(,1

)1(1,)1()(−

−−

−−

−−− −=k

kk

kjk

kkik

ijk

ija

aaaa 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

na

bX =

→=+ −−

−−−

−−−

)1(1

)1(,11

)1(1,1

nnn

nnnn

nnn bXaXa

21,1

)1(,1

)1(1

1 −−−

−−

−−

−

−=

nnn

nn

nnn

nn a

XabX

→=++ )()(,

)( iin

inii

iii bXaXa L

1

1)()(

−

+

=−

=iii

i

nkk

iik

ii

ia

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

a

XaXabX

a

XaXaXaXabX

a

XaXaXabX

a

XaXabX

)(

)(

)(

)(

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.



35

2.1.3 Iterative Method – Gauss-Seidal 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-Seidal Method.

nn

knnnknnkn

ii

kninkiiikiiikiiki

knnkkk

knnkk

a

XaXabX

a

XaXaXaXabX

a

XaXaXabX

a

XaXabX

))()(()(

))()()()(()(

))())(()(

))()(()(

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

2.1.4 Example

Stiffness Equation

0)36

(3

0.4503.133

03

)63

(3.1337.168

=θ++θ++−

=θ+θ+++−

CB

CB

l

EI

l

EI

l

EIl

EI

l

EI

l

EI

For the simplicity of derivation, CCBB l

EI

l

EI 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 4



36

Gauss-Jordan Iteration

))(37.316(9

1)(

))(34.35(9

1)(

1

1

−

−

θ−−=θ

θ−=θ

kBkC

kCkB

Gauss-Seidal Iteration

))(37.316(9

1)(

))(34.35(9

1)( 1

kBkC

kCkB

θ−−=θ

θ−=θ −

Gauss-Seidal Gauss-Jordan GAUSS-SIEDAL ITERATION

======================

***** Iteration 1*****

X(1) = 0.3933334E+01

X(2) = -0.3650000E+02

ERROR = 0.1000000E+01


X(1) = 0.1610000E+02

X(2) = -0.4055556E+02

ERROR = 0.2939146E+00


X(1) = 0.1745185E+02

X(2) = -0.4100617E+02

ERROR = 0.3197497E-01


X(1) = 0.1760206E+02

X(2) = -0.4105624E+02

ERROR = 0.3544420E-02


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

======================


X(1) = 0.3933334E+01

X(2) = -0.3518889E+02

ERROR = 0.1000000E+01


X(1) = 0.1566296E+02

X(2) = -0.3650000E+02

ERROR = 0.2971564E+00


X(1) = 0.1610000E+02

X(2) = -0.4040988E+02

ERROR = 0.9044394E-01


X(1) = 0.1740329E+02

X(2) = -0.4055556E+02

ERROR = 0.2971564E-01


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

L

EI

L

EIL

EI

L

EIM

MDM

L

EI

L

EIL

EI

L

EIM

L

EI

L

EIM

MML

EI

L

EIM

=+

=θ=

=+

=θ=

+=θ→+=θ+=

43

4

4

43

3

3

43 )

43(

1

1

11

- Moment carry over to joint C: BBCBBC

BCCB MD

L

EIM

2

121 =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

L

EI

L

EIL

EI

L

EIM

MDM

L

EI

L

EIL

EI

L

EIM

L

EI

L

EIM

MML

EI

L

EIM

=+

=θ=

=+

=θ=

+=θ→+=θ+=

34

3

3

34

4

4

34 )

34(

2

2

22

- Moment carry over to joint B: 2

122CCBC

BC

BCBC MD

L

EIM =θ=

A B C D

3@30=90

EI 1.5EI EI

30 35.6 4



38

Stiffness Equation in terms of Moment at Joints

→

=++

=++−

02

17.316

02

135.4

CBBC

CCBB

MMD

MDM

BBCC

CCBB

MDM

MDM

2

17.316-

2

14.35

−=

−=

- Gauss-Seidal Approach

kBBCkC

kCCBkB

MDM

MDM

)(2

1316.7 )(

)(2

14.35)( 1

−−=

−= −

- Gauss-Jordan Approach

1

1

)(2

1316.7 )(

)(2

14.35)(

−

−

−−=

−=

kBBCkC

kCCBkB

MDM

MDM

For the given structure

3

2 ,

3

1 ==== 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)(4.353

2

2

17.316)(

2

17.316)(

4.35)(4.35)(2

14.35)(

111

101

−=∆→−−=−−=

=∆→=−=

CBBCC

BcCBB

MMDM

MMDM

- For 1>k

kBBCkCkBBCkC

kBBCkBBCkBBCkC

kcCBkBkcCBkB

kcCBkcCBkcCBkB

MDMMDM

MDMDMDM

MDMMDM

MDMDMDM

)(2

1)()(

2

1)(

)(2

1)(

2

17.316)(

2

17.316)(

)(2

1)()(

2

1)(

)(2

1)(

2

14.35)(

2

14.35)(

1

1

111

121

∆−=∆→∆−=

∆−−−=−−=

∆−=∆→∆−=

∆−−=−=

−

−

−−−

−−−



39

- Iteration 1

0)( 0 =BM , 0)( 0 =CM

4.35)(4.35)(2

14.35)( 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)(5.3284.353

2

2

17.316)(

2

17.316)( 111 −=∆→−=−−=∆−−= cBBCC MMDM

5.109)(5.1094505.32833.0450)()(

0.2198.11)(

0.2198.113.133)()(2

1)(

111

1

111

−=∆→−=×−=∆+=

−=∆

→−+−=∆+∆+=

CDCCDf

CDCD

CB

CCBBBCf

CBCB

MMDMM

M

MDMDMM

- Iteration 2

−=∆=∆−=

∆+∆=∆→−=∆−=∆

+−=

∆+∆=∆

=∆=∆

→=∆−=∆

2.12)()(3.245.36

)()(2

1)(

5.36)(2

1)(

0.735.109

)()(2

1)(

5.36)()(

5.109)(2

1)(

22

222

22

212

22

12

CCDCD

CCBBBCCB

BBCC

BBCcCBBC

BBABA

cCBB

MDM

MDMDMMDM

MDMDM

MDM

MDM

- Iteration 3

−=∆=∆

−=∆+∆=∆→−=∆−=∆

+−=

∆+∆=∆

=∆=∆

→=∆−=∆

4.1)()(

7.21.4)()(2

1)(1.4)(

2

1)(

2.82.12

)()(2

1)(

1.4)()(

2.12)(2

1)(

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)(2

17.316)(

4.35)(4.35)(2

14.35)(

101

101

−=∆→−=−−=

=∆→=−=

CBBCC

BcCBB

MMDM

MMDM

- For 1>k

111

121

111

121

)(2

1)()(

2

1)(

)(2

1)(

2

17.316)(

2

17.316)(

)(2

1)()(

2

1)(

)(2

1)(

2

14.35)(

2

14.35)(

−−−

−−−

−−−

−−−

∆−=∆→∆−=

∆−−−=−−=

∆−=∆→∆−=

∆−−=−=

kBBCkCkBBCkC

kBBCkBBCkBBCkC

kcCBkBkcCBkB

kcCBkcCBkcCBkB

MDMMDM

MDMDMDM

MDMMDM

MDMDMDM

- Iteration 1

0)( 0 =BM , 0)( 0 =CM

4.35)(4.35)(2

14.35)( 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



41

7.316)(7.3167.316)(2

17.316)( 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

M

MDMM

- Iteration 2

−=∆=∆−=

∆+∆=∆→−=∆−=∆

+−=

∆+∆=∆

=∆=∆

→=∆−=∆

9.3)()(9.78.11

)()(2

1)(

8.11)(2

1)(

1.711.106

)()(2

1)(

0.35)()(

1.106)(2

1)(

22

212

12

212

22

12

CCDCD

CCBBBCCB

BBCC

BBCcCBBC

BBABA

cCBB

MDM

MDMDMMDM

MDMDM

MDM

MDM

- Iteration 3

−=∆=∆−=

∆+∆=∆→−=∆−=∆

+−=

∆+∆=∆

=∆=∆

→=∆−=∆

7.11)()(9.236.35

)()(2

1)(

6.35)(2

1)(

7.20.4

)()(2

1)(

3.1)()(

0.4)(2

1)(

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

-116.4

133.3 23.6

-106.1 71.1 -4.0 2. 7

-12.0 8.0

-0.5 0.3

116.4

-133.3 -212.2

11.8 -7. 9 35.6

-23.9 1.4

-0.9 4.0

-2.7 -328.1

450.0 -104.5

-3.9

-11.7

-0.5

-1.3

328.1



42

2.3 Example - MDM for a 4-span Continuous Beam

4.0)5.1

44(4 =+=L

EI

L

EI

L

EIDBA , 6.0)

5.144(

5.14 =+=

L

EI

L

EI

L

EIDBC , 5.0)

5.14

5.14(

5.14 =+=

L

EI

L

EI

L

EIDCB ,

5.0)5.1

45.1

4(5.1

4 =+=L

EI

L

EI

L

EIDCD , 67.0)3

5.14(

5.14 =+=

L

EI

L

EI

L

EIDDC , 33.0)3

5.14(3 =+=

L

EI

L

EI

L

EIDDE

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

0

0.0

4.44

8.88

2466

682

628

C

B

L

EI

L

EI

L

EIL

EI

L

EI

L

EIL

EI

L

EI

L

EI

00

)()(

)(

)(][=

+

∆Θ

∆∆θ∆

∆θθθ P

KK

KK

∆∆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

))(())((

KKK 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 ( L = 30 )

-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 = VAB+VDC = 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 = 8.1, MDC = 10-4.1/2 = 7.9, VAB = -0.47, VDC = -0.46, ∆V = VBA+VCD = -0.93

A

10 20

B C

D

0.5

0.5

0.5

0.5

C

A D

B ∆



46

Shear Equilibrium Condition (the 2nd equation)

No Sidesway (1st MDM) Sidesway Only (2nd MDM)

97.0

93.0

89.0 =−−=−=α

∆V

V P

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

PKLLPLLK =→θ=×θ

Unstable state

PKLLPLLK <→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

qdx

dVqdxVdVV −=→=+−+ 0)(

Equilibrium for moment

Vdx

dwP

dx

dMdx

dx

dwP

dxqdxVdxMdMM =−→=−+−−+ 0

2)(

Elimination of shear force

qdx

wdP

dx

Md −=−2

2

2

2

Strain-displacement relation

dx

duy

dx

wd +−=ε2

2

Stress-strain relation (Hooke law)

dx

duEy

dx

wdEE +−=ε=σ

2

2

Definition of Moment

2

22

2

2

)(dx

wdEIdAy

dx

duEy

dx

wdEydAEydAM

AAA

−=−−=ε=σ=

Beam Equation with Axial Force

qdx

wdP

dx

wdEI =+

2

2

4

4

M M+dM

V V+dV

q

w

dxdx

dw

P

P



51

3.2 Homogeneous Solutions

Characteristic Equation for 0>P

0 ,0)( 224 ie

ewx

x

β±=λ→=λβ+λ=

λ

λ

where EI

P=β2

Homogeneous solution

DCxBeAew ixix +++= β−β

Exponential Function with Complex Variable

xixi

xi

xi

ixee

xxxxee

xi

xi

xi

xi

xi

ixe

xi

xi

xi

xi

xi

ixe

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

DCxxBxA

DCxxBAixBA

DCxxixBxixAw

++β+β=++β−+β+=

++β−β+β+β=

sincos

sin)(cos)(

)sin(cos)sin(cos

Characteristic Equation for 0<Q

0 ,0)( 224 β±=λ→=λβ−λ

=λ

λ

x

x

e

ew where

EI

P=β 2

Homogeneous solution for 0<Q

DCxxBxA

DCxee

BAee

BA

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−=′′=+=

CBLBLwCLLBLw

AAwDAw

− Characteristic Equation

00 =→==== wDCBA (trivial solution) or

L3,2,1 , 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=′=+=

CLBLALw

DCLLBLALw

CBw

DAw

− Characteristic Equation

0)

01cossin

1sincos

010

1001

(

0

0

0

0

01cossin

1sincos

010

1001

=

ββββ−ββ

β→

=

ββββ−ββ

β

LL

LLLDet

D

C

B

A

LL

LLL

1cossin

sincos

10

01cos

1sin

01

01cossin

1sincos

010

1001

LL

LLL

L

LL

LL

LLLββββ−

βββ

−ββ

ββ

=

ββββ−ββ

β

P P

P P


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

2sin

2cos

2(

2sin

cos2

sin22

sin4

sin2cos2

(cos()cos(

2

β−βββ

ββ+β−

=ββ+−ββ−−ββ−−β−

LLLL

LL

L

LLL

L

− Eigenvalues

Symmetric modes

2 0

2sin

β→=β LL

0)0( =+= DAw

0 =→=+ ADA

Anti-symmetric modes

sin2

cos2

LL β−ββ

of Civil & Environmental Eng., SNU

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

2sin

2cos

2or 0

2sin0)

2

2cos

sin)1(cos2

sin2cos2())sin(cos

=β−ββ=β→=

=β=ββ+−β

β+−ββ=β+ββ+β

LLLLL

L

LLL

LLLLL

L3,2,1 ,4

2 2

22

=π=→π= nL

EInPn

L

)( ,0)()0( , 0 +==+β=′=′ CLALwCBLww

)12

(cos −π=→−= xL

nAwD for 0≠A

symmetric modes

2

218.8

2tan

20

2 L

EIP

LLL π=→β=β→=β

Structural Analysis Lab.

53

0

0)sin

=

=βL

0=+ DCL



54

Cantilever Beam


0)(

0)sincos()()(

0)0(

0)0(

3

3

22

=−−=

=ββ−ββ−−=′′−=

=+β=′=+=

dx

dwP

dx

wdEILV

LBLAEILwEILM

CBw

DAw

L3,2,1 ,4

)12(2

22

=π−= nL

EInP

Beam-Columns - Simple Beams with a Uniform Load

− Governing Equation

qdx

wdP

dx

wdEI =+

2

2

4

4

− Particular Solution

2

2)( x

P

qxw p =

− General Solution

P

qxBxAxw

xP

qDCxxBxAxw

+ββ−ββ−=′′

+++β+β=

sincos)(

2sincos)(

22

2

− Boundary Conditions

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

β−=

β=→=+β−−=′′=+=

P

qD

P

qA

P

qAEIwDAw

0)

sincos()()(

02

)1(cos

sin)(

22

2

2

=+ββ−ββ−−=′′−=

=+−ββ

++β=

P

qLBLAEILwEILM

LP

qL

P

qCLLBLw

P P

P P q



55

LP

qC

L

P

qB

L

P

qLLB

LP

qLB

P

qLBL

P

q

2

,

2tan

1

)12

sin21(

2cos

2sin2

)1(cos

sin0

sincos

2

22

22

−=ββ

=

−β−−=βββ

−β−=ββ→=+ββ−β−

− Final Solution

)

sin2

tan

cos()()(

2

2

sin

2tan

)1(cos)( 2

22

P

qx

L

P

qx

P

qEIxwEIxM

xP

qLx

P

qx

L

P

qx

P

qxw

−ββ+β=′′−=

+−βββ

+−ββ

=

− Displacement at the mid-span

)( )8

1

2cos

1(

7884.0

)8

1

/2

cos

1()(7884.0

)8

1

/2

cos

1(

5

384

384

5

)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

2

2

2

22

crst

crcr

crst

P

Ptt

tt

P

P

PPP

P

EI

PL

EIPLLP

EI

EI

qL

LLP

q

LP

qLLL

P

q

LP

qLL

P

qL

P

qLw

=π−−π

δ=

π−−π

δ=

ππ−−

πππ

π=

β−−ββ

=

−−ββ+ββ

=

−βββ

+−ββ

=

− Moment at the Mid-span

)1

2cos

1(

1811.0

)1/

2cos

1(

8

)12

sin2

tan2

(cos8

8

)12

sin2

tan2

(cos)2

(

2

2

2

2

2

−π

=

−ππ

=

−ββ+βππ

=

−ββ+β=

ttM

PPP

PM

LLL

PL

EIqL

LLL

P

qEI

LM

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 s

t

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

dx

wdEI

dx

wdP

dx

wdEI

dx

wdP

dx

wdEI

dx

wwdP

dx

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(

00sincos

1sincos

000

1001

22

2

=+→=

′′

′′+

ββ−ββ−ββ

β−FKX

Lw

Lw

w

w

D

C

B

A

LL

LLL

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

2

1

, 2

1)()(

Equilibrium Equation

k∆e=P

Principle of Minimum Potential Energy for an arbitr ary displacement ∆+∆=∆ e .

etotal

etotal

ee

eee

eee

eetotal

k

kPk

PkkPk

Pkkk

Pk

Π≥∆+Π=

∆+∆−∆=

−∆∆+∆+∆−∆=

∆+∆−∆+∆∆+∆=

∆+∆−∆+∆=Π

2

22

22

22

2

)(2

1

)(2

1)(

2

1

)()(2

1)(

2

1

)()(2

1)(

2

1

)()(2

1

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

wdEIdx

EI

M

00

22

2

int

00

22

2

0

2

int

)(2

1

, )(2

1

2

1


qdx

wdEIq

dx

MdEI

ee

=−=4

4

2

2

or

Principle of Minimum Potential Energy for a virtual displacement www e += .

wdxdx

wdEI

dx

wd

qdxwdxEI

MMdx

dx

wdEI

dx

wd

qdxwdxdx

wdEI

EIdx

wdEIdx

dx

wdEI

dx

wd

qdxwdxdx

wdEI

dx

wddx

dx

wdEI

dx

wd

qdxwdxdx

wdEI

dx

wd

qdxwwdxdx

wwdEI

dx

wwd

el

e

ll ele

ll ele

ll el

le

l ee

le

l eeh

virtualallfor )(2

1

)(2

1

)(1

)()(2

1

)()(2

1

)(2

1

)())()(

(2

1

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

wdEIw

l e

for all virtual displacement w

00

3

3

02

2

002

2

2

2

=−+− lelell e

dx

wdEIw

dx

wdEI

dx

wdqdxwdx

dx

wdEI

dx

wd

00000

2

2

2

2

=−=− qdxwdxEI

MMqdxwdx

dx

wdEI

dx

wd 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

dxEI

MMxwdxxxwdxqwqdxw

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

dx

wdEI

dx

wd

0 10 11002

2

2

2

)()(2

1)(

2

1

- The first-order Necessary Condition

0

)])()([2

1

))()(2

1(

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

2

1

4)2(

2

1)

2()(

2

1 22

2

0

2

00

2 laPlaEI

laPdxaEIwdx

lxPdxwEI

lll

total +=+=−δ−′′=Π

)(1616

04

40 22

xlxEI

Plw

EI

Pla

lPaEIl

atotal −−=→−=→=+→=

∂Π∂

EI

Pllw

64)

2(

3

= , EI

Pl

EI

Pllwe

33

0208.048

)2

( == , Error = 25.0)

2(

)2

()2

(=

−

lw

lw

lw

e

e

ii) with two unknowns

bxawlxbxlxaxw 62)()( 22 +=′′→−+−=

)8

3

4()

336

2244(

2

1

)8

3

4()62(

2

1)

2()(

2

1

3232

22

32

0

2

00

2

lb

laP

lb

lablaEI

lb

laPdxbxaEIwdx

lxPdxwEI

lll

total

++++=

−−−+=−δ−′′=Π

8

3)126(0

4)64(0

332

22

lPblalEI

b

lPbllaEI

a

total

total

−=+→=∂

Π∂

−=+→=∂

Π∂

0 , 16

1 =−=→ bEI

Pla (???)

iii) with three unknowns

23322 1262)()()( cxbxawlxcxlxbxlxaxw ++=′′→−+−+−=

)16

7

8

3

4(

)4

1443

482

245

1443

364(2

1

)16

7

8

3

4()1262(

2

1

)2

()(2

1

432

43252

322

432

0

22

00

2

lc

lb

laP

lbc

lac

lab

lc

lblaEI

lc

lb

laPdxcxbxaEI

wdxl

xPdxwEI

l

ll

total

++

++++++=

−−−−++=

−δ−′′=Π

P



63

16

7)

5

144188(0

8

3)18126(0

4)864(0

4543

3432

232

lPclblalEI

c

lPclblalEI

b

lPclbllaEI

a

total

total

total

−=++→=∂

Π∂

−=++→=∂

Π∂

−=++→=∂

Π∂

=

−=

=

→

EIl

PEI

Pb

EI

Pla

64

5c

32

5

64

1

)(64

5)(

32

5)(

64

1 3322 lxxEIl

Plxx

EI

Plxx

EI

Plw −+−−−=

EI

Pl

EI

Pl

EI

Pllw

333

0205.01024

21)

16

7

64

5

8

3

32

5

4

1

64

1()

2( ==−+−= , Error = 0.0144

iv) with one sin function

xll

awxl

awππ=′′=→π= sin)(sin 2

aPl

lEIaaPxdx

llEIa

aPdxxll

aEIwdxl

xPdxwEI

l

lll

total

+π=+ππ=

−ππ=−δ−′′=Π

2)(

2

1sin)(

2

1

)sin)((2

1)

2()(

2

1

42

0

242

0

22

00

2

xlEI

Plw

EI

PlaP

l

lEIa

atotal π

π=→

π=→=−π→=

∂Π∂

sin22

02

)(03

4

3

44

EI

Pllw

3

7045.48

1)

2( = ,

EI

Pllwe

48)

2(

3

= , Error = 0145.0

v) with two sin function

xll

bxll

awxl

bxl

awππ+ππ=′′=→π+π= 3

sin)3

(sin)(3

sinsin 22

bPaPl

lEIb

l

lEIa

bPaPxdxll

EIb

xdxllll

EIabxdxll

EIa

bPaPdxxll

bxll

aEI

wdxl

xPdxwEI

l

ll

l

ll

total

+−π+π=

+−ππ

+ππππ+ππ=

+−ππ+ππ=

−δ−′′=Π

2)

3(

2

1

2)(

2

1

3sin)

3(

2

1

3sinsin)

3()(sin)(

2

1

)3

sin)3

(sin)((2

1

)2

()(2

1

4242

0

242

0

22

0

242

0

222

00

2



64

)3

sin81

1(sin

2

)3(

20

2)

3(0

20

2)(0

3

4

3

44

3

44

xl

xlEI

Plw

EI

PlbP

l

lEIb

b

EI

PlaP

l

lEIa

a

total

total

π−ππ

=

π−=→=+π→=

∂Π∂

π=→=−π→=

∂Π∂

EI

Pl

EI

Pllw

33

0208.0)0123.01(0205.0)2

( =+= , EI

Pllwe

3

0208.0)2

( = , Error ≅0

4.3 Truss problems

Potential Energy

==

==

+−=Π+Π=Π

+−=Π=Π

njn

i

iiiinmb

k k

kkexttotal

njn

i

iiiiext

nmb

k k

kk

vYuXEA

lF

vYuXEA

lF

11

2

int

11

2

int

)()(

2

1

)( , )(

2

1

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



65

Principle of Minimum Potential Energy

ntdisplaceme virtualallfor )(

2

1

)()(

)(

2

1

)()(

2

1)(

)(

2

1

))()(()(

2

1

1

2

1

2

11

2

111

2

11

2

11

2

enmb

k k

kke

nmb

k k

kkie

injn

i

ie

inmb

k k

ke

k

njn

i

iiiinmb

k k

kke

knmb

k k

kkie

injn

i

ie

inmb

i k

ke

k

iie

iinjn

i

ie

inmb

k k

kke

ktotal

EA

lF

EA

lFvYuX

EA

lF

vYuXEA

lFF

EA

lFvYuX

EA

lF

vvYuuXEA

lFF

Π≥+Π=

++−=

+−+++−=

+++−+=Π

=

===

=====

==

where ekF and kF are the bar force of k-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

kkk vv θ− sin)( 12

kkk uu θ− cos)( 12

kθ kθ

)( 12kk uu −

)( 12kk vv −



66

==

== ==

= ==

+=−θ+−θ

+=θ+θ

=+θ−++θ−

njn

i

iiiinmb

kkkk

ekkkk

ek

njn

i

iiiinjn

i

im

j

ij

ij

iim

j

ij

ij

i

njn

i

iiim

j

ij

ij

iiim

j

ij

ij

vYuXvvFuuF

vYuXFvFu

vYFuXF

11

1212

11

)(

1

)(

1

1

)(

1

)(

1

) ())(sin )(cos(

) ()sin cos(

0))sin( )cos((

==

====

+=

==∆=−θ+−θ

njn

i

iiiinmb

k k

kke

k

nmb

k k

kke

k

k

kknmb

k

ek

nmb

kk

ek

nmb

kkkkkkk

ek

vYuXEA

lFF

EA

lFF

EA

lFFlFvvuuF

11

1111

1212

) (

))(sin )((cos

The principle of virtual work yields the displacement of a joint i in a truss by applying an

unit load at a joint i in an arbitrary direction and by using the reciprocal theorem,

=

=α=α=+nmb

k k

kke

kie

ie

ie

iie

i

EA

lFFvYuX

1

coscos uuX

Since α represnts the angle between the applied unit load and the displacement vector,

αcosieu are the displacement of the joint i in the direction of the applied unit load.

4.4. Buckling Problems

Total Potential Energy

dxwqPdxdx

wdEI

dx

wd ll

−∆−=Π00

2

2

2

2

2

1

dxwdxwdsLLLL

))(2

11()(1

0

2

0

2

0

∆−∆−

′+≈′+==

LdxwdxwdxwLLLLL

<<∆′≈′=∆→′+∆−= ∆−∆−

for )(21

)(21

)(21

0

2

0

2

0

2

dxwqdxdx

dwP

dx

dwdx

dx

wdEI

dx

wd lll

−−=Π000

2

2

2

2

2

1

2

1

∆ Q



67

Principle of the Minimum Potential Energy

−+Π=

−+−++Π=

−+−

−+−−=

+−++−++=Π

le

ll eee

ll

l eele

l eeee

le

l eel eeh

dxdx

wdP

dx

wd

dx

wdEI

dx

wd

dxdx

wdP

dx

wd

dx

wdEI

dx

wddxq

dx

wdP

dx

wdEIw

dxdx

wdP

dx

wd

dx

wdEI

dx

wdqdxw

dxdx

wdP

dx

dw

dx

wdEI

dx

wdqdxwdx

dx

dwP

dx

dw

dx

wdEI

dx

wd

qdxwwdxdx

wwdQ

dx

wwddx

dx

wwdEI

dx

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

)(2

1

)(2

1)(

)(2

1

)()(2

1

)()()(

2

1)()(

2

1

The principle of minimum potential energy holds if and only if

wdxdx

wdP

dx

wd

dx

wdEI

dx

wdl

possible allfor 0)(0

2

2

2

2

>−

The principle of the minimum potential energy is not valid for the following cases.

wdxdx

wdP

dx

wd

dx

wdEI

dx

wdl

somefor 0)(0

2

2

2

2

≤−

The critical status of a structure is defined as

0)(0

2

2

2

2

=−l

dxdx

wdP

dx

wd

dx

wdEI

dx

wd

Approximation

− Approximation of displacement: =

=n

iii gaw

1

− Critical Status

0])[]([0)])([]([)(

)()()()(

00

1 11 1

0

1 1 01 1 0

0 110 11

=−→=−

=−=′′−′′′′

=′′−′′′′

= == == == =

====

GGT

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

Det

aKaaKaadxgQgaadxgEIga

dxgaQgadxgaEIga

KKaKKa



68

Example - Simple Beam

− with a parabola: 2 , 2)( 11 =′′−=′→−= glxglxaxw

%)22error , 86.9:exact( 12

0)3

14(

3

1)12

3

4()44()2(

44

22

2

2

3

33

0

22

0

2

0

00

==π=→=−

=+−=+−=−=′′

==′′′′

l

EI

l

EI

l

EIPlPEIlDet

lldxlxlxdxlxdxgg

EIldxEIdxgEIg

cr

lll

ii

ll

ii

− with one sine curve: sin)( , cossin 211 l

x

lg

l

x

lg

l

xaw

ππ=′′ππ=′→π=

)exact(0)2

)(2

)((

2)(cos)(

2)(sin)(

2

224

2

0

22

0

0

424

0

l

EIP

l

lP

l

lEIDet

l

ldx

l

x

ldxgg

l

lEIdx

l

x

lEIdxgEIg

cr

ll

ii

ll

ii

π=→=π−π

π=ππ=′′

π=ππ=′′′′

Example – Cantilever Beam

− with one unknown:

%)22error , 46.24

:exact( 3

0)3

44(

3

44 , 422

2 , 2

22

2

2

3

3

0

2

000

112

==π=→=−

==′′==′′′′

=′′=′→=

l

EI

l

EI

l

EIPlPEIlDet

ldxxdxggEIldxEIdxgEIg

gxgaxw

cr

ll

ii

ll

ii

P P

P



69

− with two unknowns:

3

0

222

2

0

2112

0

11

5

0

422

4

0

32112

3

0

211

212

2132

1236 , 612 , 44

5

99 ,

4

66 ,

3

44

6 , 2 , 3 , 2

ldxxKlxdxKKldxK

PldxxPKPldxxPKKPldxxPK

xggxgxgbxaxw

lll

lG

lGG

lG

=======

=======

=′′=′′=′=′→+=

32.181or 487.20727.1

or 0829.0

45

27.2226

45

1802626

0455240)456()5412)(404(

30

1 , 0

5412456

4564040)

5445

4540

30

1

126

64(

222228

121226

2864242533

5342

423

54

43

32

2

l

EI

l

EIP

llll

lll

lllllllll

EI

P

llll

llll

ll

llP

ll

llEIDet

cr =→=±=−±=α

=α+α−→=α−−α−α−

=α=α−α−α−α−

→=

−

249.2

l

EIPexact = (error = 1.2%) or

219.22

l

EIPexact = (error = 45%)

Beam-Columns - Simple Beams with a Uniform Load

− with one sine curve:

sin)( , cossin 211 l

x

lg

l

x

lg

l

xaw

ππ=′′ππ=′→π=

π=π=

π=ππ=′′π=ππ=′′′′

lqdx

l

xqqdxg

l

lQdx

l

x

lQdxgQg

l

lEIdx

l

x

lEIdxgEIg

ll

i

ll

ii

ll

ii

2sin

2)(cos)( ,

2)(sin)(

00

2

0

22

00

424

0

tQQEI

qll

EI

Qql

EI

q

lQ

lEI

qa

lqa

l

lQa

l

lEI

a

al

qal

lQa

l

lEI

stcr −

δ=−π

×=

π−

πππ

=π−ππ=

=π

−π−π=∂Π∂

→π

−π−π=Π

1

0038.1

/1

1

5

4384

384

5

)(1

/4)(

14

)()(

/4

022

)( 2

)(

22

)(2

1

2)(

2

1Min

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

ff

L

EAf

L

EAf

Member Stiffness Matrix in Local Coordinate System

e

Ry

Rx

Ly

Lx

e

Ry

Rx

Ly

Lx

L

EA

f

f

f

f

δδδδ

−

−

=

0000

0101

0000

0101

eee )(][)( δkf =

Transformation Matrix

φφφ−φ

=

→

φ+φ=

φ−φ=

y

x

y

x

yxy

yxx

v

v

V

V

vvV

vvV

cossin

sincos

cossin

sincos

)]([)()(][)( VvvV γ=→γ= T

1

2

xv

yv

φ

xV

yV

Rxf

Ryf L

yf

Rxδ

RyδL

yδ

LxδL

xf



74

Member End Force

→

φφφ−φ

φφφ−φ

=

e

Ry

Rx

Ly

Lx

e

y

x

y

x

f

f

f

f

F

F

F

F

cos sin00

sincos00

00cos sin

00sincos

2

2

1

1

eTe )(][)( fF Γ=

Member End Displacement

→

∆∆∆∆

φφ−φφ

φφ−φφ

=

δδδδ

e

y

x

y

x

e

Ry

Rx

Ly

Lx

2

2

1

1

cossin00

sincos 00

00cossin

00sincos

ee )]([)( ∆δ Γ=

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

L

EA

][][

][][

sincossinsincossin

cossincoscossincos

sincossinsincossin

cossincoscossincos

][2221

1211

22

22

22

22

KK

KKK

5.1.2 Global Stiffness Equation Nodal Equilibrium

)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

=

=++++=)(

1

)(2or 1)5(1)4(2)3(1)2(2)1(1 )()()()()()()(mnm

k

kmmmmmmm FFFFFFP

)]([)(][)(

)(

00

0

0

00

0

)(

)(

0

)(

)(

)(

)(

)(

)(

)(11 1

2

1

2

1

)(

1

)(2or 1

)(

1

)(2or 1

)1(

1

)(12or 11

FEFEF

F

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

FEEEE

M

M

LL

m-th row

n-th row

1

2

i)( 2F

i)( 1F

m-th joint

mPi-th member

nP n-th joint



76

Compatibility Condition

=

=→==

n

mi

ii

ni

mi

u

uuu

I0

0I

)(

)()( )( , )( 2

121

∆∆

∆∆∆

)(][000

000

)(

)()(

1

2

1

uC

u

u

u

u

I

I i

q

n

m

i

ii =

=

=

M

M

M

LLL

LLL

∆∆

∆

)]([

][

][

)(

)(

)(1

11

uC

u

u

C

C

=

=

=

qpp

MMM

∆

∆∆

Contragradient

→=⋅→⋅=⋅ )]([)()()()()()()( uCFuPFuP TTTT ∆

][)()()( possible allfor 0)])([)()(( CFPuuCFP TTTT =→=−

][][)]([)(][)( ECFEFCP =→== TT

m-th column

n-th column

i-th member

m

n

um

un

1

2

i)( 1∆

i)( 2∆

Compatibility Matrix



77

Unassembled Member Stiffness Equation

)(

)(

)(

][00

0][0

00][

)(

)(

)( 111

→

=

p

i

p

i

p

i

∆

∆

∆

M

M

LL

MOMNM

LL

MNMOM

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

MOMNM

LL

MNMOM

LL

LL

=

=

=

00000

0][0][0

00000

0][0][0

00000

0][][0

0][][0

00

0

0

00

000

000

][][

][][

00

0

0

00

][][][

2221

1211

2221

1211

2221

1211

ii

ii

ii

ii

ii

iiiiiT

KK

KK

KK

KK

I

I

I

I

KK

KK

I

I

CKC

LLL

LLL

MM

MM

MM

LLL

LLL

MM

MM

MM

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

L

EA

][][

][][

sincossinsincossin

cossincoscossincos

sincossinsincossin

cossincoscossincos

][2221

1211

22

22

22

22

KK

KKK

- Member 1: φ = 45o

−−

−−

−−

−−

=

2

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

1

2][ 1

L

EAK

- Member 2: φ = -45 o

−−

−−

−−

−−

=

2

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

1

2][ 2

L

EAK

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

−

−

=

0000

0101

0000

0101

2][ 3

L

EAK


31226

31225

21124

21123

32112

32111

)()(

)()(

)()(

)()(

)()(

)()(

yy

xx

yy

xx

yy

xx

FFP

FFP

FFP

FFP

FFP

FFP

+=

+=

+=

+=

+=

+=

001010000000

000101000000

000000101000

000000010100

100000000010

010000000001

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

F

F

F

F

F

F

F

F

F

F

F

F

P

P

P

P

P

P

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

000010

000001

100000

010000

100000

010000

001000

000100

001000

000100

000010

000001

u

u

u

u

u

u

u

u

u

u

u

u

y

x

y

x

y

x

y

x

y

x

y

x

C

C

C

)]([)( uC=∆

TTTT][][ ][][ ,][][ ,][][ 332211 CECECECE =→===

Global Stiffness Matrix

333222111 ][][][][][][][][][]][[][][ CKCCKCCKCCKCKTTTT +++== L

u3

u6

u5

u2

u1

u4

[C]1

[C]3

[C]2



81

−−

−−

−−

−−

=

−−

−−

−−

−−

=

−−

−−

−−

−−

=

000000

000000

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

2

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

0000

0000

1000

0100

0010

0001

2

001000

000100

000010

000001

2

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

1

0000

0000

1000

0100

0010

0001

2][][][ 111

L

EA

L

EA

L

EATCKC

−−

−−

−−

−−

=

−−

−−

−−

−−

=

−−

−−

−−

−−

=

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

000000

000000

2

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

0000

0000

1000

0100

0010

0001

2

100000

010000

001000

000100

2

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

12

1

2

1

2

1

2

1

1000

0100

0010

0001

0000

0000

2][][][ 222

L

EA

L

EA

L

EATCKC

−

−

=

000010

000001

100000

010000

0000

0101

0000

0101

0010

0001

0000

0000

1000

0100

2][][][ 333

L

EATCKC



82

−

−

=

−

−

=

000000

010001

000000

000000

000000

010001

2

000000

010001

000000

010001

0010

0001

0000

0000

1000

0100

2 L

EA

L

EA

−−

−+−−

−+−−−

−−+−−

−−

−−−+

=

−

−

+

−−

−−

−−

−−

+

−−

−−

−−

−−

=

22

1

22

1

22

1

22

100

22

1

2

1

22

1

22

1

22

10

2

122

1

22

1

22

1

22

1

22

1

22

1

22

1

22

122

1

22

1

22

1

22

1

22

1

22

1

22

1

22

1

0022

1

22

1

22

1

22

1

02

1

22

1

22

1

22

1

2

1

22

1

000000

010001

000000

000000

000000

010001

2

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

2

1

2

1

2

1

2

100

000000

000000

2

000000

000000

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

002

1

2

1

2

1

2

1

2][

L

EA

L

EA

L

EA

L

EAK



83

Stiffness Equation

−−

−+−−

−−−

−−−

−−

−−−+

=

6

5

4

3

2

1

6

5

4

3

2

1

22

1

22

1

22

1

22

100

22

1

22

21

22

1

22

10

2

122

1

22

1

2

10

22

1

22

122

1

22

10

2

1

22

1

22

1

0022

1

22

1

22

1

22

1

02

1

22

1

22

1

22

1

22

21

u

u

u

u

u

u

L

EA

P

P

P

P

P

P

Application of Support Conditions (Boundary Conditions)

−−

−+−−

−−−

−−−

−−

−−−+

=

6

5

4

3

2

1

6

5

4

3

2

1

22

1

22

1

22

1

22

100

22

1

22

21

22

1

22

10

2

122

1

22

1

2

10

22

1

22

122

1

22

10

2

1

22

1

22

1

0022

1

22

1

22

1

22

1

02

1

22

1

22

1

22

1

22

21

u

u

u

u

u

u

L

EA

P

P

P

P

P

P

Final Stiffness Equation

+−

−

=

→

+−

−

=

−

5

4

3

1

5

4

3

5

4

3

5

4

3

22

21

22

1

22

122

1

2

10

22

10

2

1

22

21

22

1

22

122

1

2

10

22

10

2

1

P

P

P

EA

L

u

u

u

u

u

u

L

EA

P

P

P

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

L

EI

L

EI

L

EI

L

EI

L

MMf

L

EI

L

EI

L

EI

L

EI

L

MMf

L

EI

L

EI

L

EI

L

EIm

L

EI

L

EI

L

EI

L

EIm

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


e

y

y

y

ee

eeee

ee

eeee

e

e

e

y

y

LL

LLLL

LL

LLLL

L

EI

M

F

M

F

Θ

∆

Θ

∆

−

−−−

−

−

=

2

2

1

1

22

22

2

2

1

1

46

26

612612

26

46

612612

or eee )(][)( ∆KF =

LθLm

Lyf

Ryf

Rθ Rm

RyδL

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

FF

FF

FF

FF

F

P

P

P

P

P

P

P

=

=

+

+

+

+

+

=

−

+

−

−−

+

+

−

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

0

0

00

0

)(

)(

0

FEF

F

I

I

F

F=

=

MM

MM

M

M

Compatibility

222

122

21

121

)(

)(

)(

)(

+

+

−

=Θ

=∆

=Θ

=∆

ii

iiy

ii

iiy

u

u

u

u

→

=

=→==

++

1

1

0

0

)(

)()( )( , )(

i

i

Ri

LiiiR

iiL

iu

uI

Iuu

∆∆

∆∆∆

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

)(][000

000

)(

)()(

1

1

1

uC

u

u

u

u

I

Ii

p

i

i

Ri

Li

i =

=

=

+

+

M

M

LL

LL

∆∆

∆

)]([

][

][

)(

)(

)(1

111

uC

u

u

C

C

=

=

=+p

pp

MMM

∆

∆∆

Unassembled Member Stiffness Equation

→

=

p

i

p

i

p

i

)(

)(

)(

][00

0][0

00][

)(

)(

)( 111

∆

∆

∆

M

M

LL

MOMNM

LL

MNMOM

LL

M

M

K

K

K

F

F

F

)]([)( ∆KF =


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

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


=

0000

0][][0

0][][0

0000

][][][2221

1211

LL

MOMMNM

LL

LL

MNMMOM

LL

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

MMVV

MMMVVV

MMMVVV

MMVV

34

34

323

323

212

212

11

11

,

,

,

,

==

+=+=

+=+=

==

)]([

100000000000

010000000000

001010000000

000101000000

000000101000

000000010100

000000000010

000000000001

)(

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

V

M

V

M

V

M

V

M

V

M

V

M

V

M

V

M

V


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



88

)]([

][

][

][

10000000

01000000

00100000

00010000

00100000

00010000

00001000

00000100

00001000

00000100

00000010

00000001

)(

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

uC

C

C

C

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

)]([

)(

)(

)(

][00

0][0

00][

)(

)(

)(

)(

3

2

1

3

2

1

3

2

1

ΔK

Δ

Δ

Δ

K

K

K

F

F

F

F =

=

=


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

[ ]

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

)(

][

][

][

][00

0][0

00][

][ ][ ][)(

333222111

3

2

1

3

2

1

321

uCKCCKCCKCuK

u

C

C

C

K

K

K

CCCP

TTT

TTT

++==

=


++

=

322321

312311222221

212211122121

112111

][][00

][][][][0

0][][][][

00][][

][

KK

KKKK

KKKK

KK

K [ ]

=

00

0

0

00

K

[C]1

[C]3

[C]2



89

- Member 1

−

−−

−

−

=

−

−−−

−

−

00000000

00000000

00000000

00000000

000046

26

0000612612

000026

46

0000612612

00001000

00000100

00000010

00000001

46

26

612612

26

46

612612

0000

0000

0000

0000

1000

0100

0010

0001

11

1211

21

11

1211

21

1

1

11

1211

21

11

1211

21

1

1

LL

LLLL

LL

LLLL

L

EI

LL

LLLL

LL

LLLL

L

EI

- Member 2

−

−−−

−

−

=

−

−−−

−

−

0000000000000000

0046

26

00

00612612

00

0026

46

00

00612612

00

0000000000000000

00100000000100000000100000000100

46

26

612612

26

46

612612

00000000100001000010000100000000

22

2222

22

22

2222

22

2

2

22

2222

22

22

2222

22

2

2

LL

LLLL

LL

LLLL

L

EI

LL

LLLL

LL

LLLL

L

EI



90

- Member 3

−

−−−

−

−

=

−

−−−

−

−

46

26

0000

6126120000

26

46

0000

6126120000

00000000000000000000000000000000

10000000010000000010000000010000

46

26

612612

26

46

612612

10000100001000010000000000000000

33

3233

23

33

3233

23

3

3

33

3233

23

33

3233

23

3

3

LL

LLLL

LL

LLLL

L

EI

LL

LLLL

LL

LLLL

L

EI


−

−−−

−++−

−+−+−−

−++−

−+−+−−

−

−

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

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

E



91

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

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

w

w

w

w

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

L

IL

I

L

I

L

I

L

I

E

M

V

M

V

M

V

M

V

Final Stiffness Equation

θ

θθ

−

−−

−+

+

=

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

1

1

4

4

3

2

4

620

61260

26442

00244

w

L

I

L

I

L

IL

I

L

I

L

IL

I

L

I

L

I

L

I

L

IL

I

L

I

L

I

E

M

V

M

M



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

L

EI

L

EI

L

EI

L

EI

L

MMf

L

EI

L

EI

L

EI

L

EI

L

MMf

L

EI

L

EI

L

EI

L

EIm

L

EI

L

EI

L

EI

L

EIm

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

VV

L

EAf

L

EAf


θ

δ

δ

θ

δ

δ

−

−−−

−

−

−

−

=

Re

Ry

Rx

Le

Ly

Lx

e

e

ee

e

e

e

e

e

e

e

e

e

e

ee

e

e

ee

e

e

e

e

e

e

e

e

e

e

ee

e

R

Ry

Rx

L

Ly

Lx

IL

II

L

IL

I

L

I

L

I

L

IAA

IL

II

L

IL

I

L

I

L

I

L

IAA

L

E

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

φφφ−φ

=

→

=

φ+φ=

φ−φ=

m

vv

M

VV

mM

vvV

vvV

y

x

y

x

yxy

yxx

100

0cossin

0sincos

cossin

sincos

)]([)()(][)( VvvV γ=→γ= T

Member End Force

→

φφφ−φ

φφφ−φ

=

e

R

Ry

Rx

L

Ly

Lx

e

y

x

y

x

m

ffm

ff

M

FFM

FF

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

2

2

2

1

1

1

eTe )(][)( fF Γ=

Member End Displacement

→

θ

∆∆θ

∆∆

φφ−φφ

φφ−φφ

=

θ

δδθ

δδ

e

y

x

y

x

e

Re

Ry

Rx

Le

Ly

Lx

2

2

2

1

1

1

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

ee )]([)( ∆δ Γ=

1

2

xvyv

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

KK

KKK

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 NNNNNNNNw ∆δ NN ==

θ

δ

θ

δ

=θ+δ+θ+δ=

3

3

2

2

1

231

L

x

L

xN +−= ,

2

32

2

2L

x

L

xxN +−= ,

3

3

2

2

3

23

L

x

L

xN −= ,

2

32

4 L

x

L

xN +−=

Total Potential Energy

)()()]([)(2

1

)(][)()](][[][)(2

1

)(][)()](])[[]([][)(2

1

)()()])([]([)(2

1

)()()()()(2

1)()()(

2

1

)()(2

1)(

2

1

2

1

2

1

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

dxqdxdx

dP

dx

ddx

dx

dEI

dx

d

dxqwdxdx

dwP

dx

dwdx

dx

wdEI

dx

wd

dxqwdxdx

dwP

dx

dwdx

dx

wdEI

dx

wd

−=

−=

−−=

−−=

−−=

−−=

−−=Π

==

==

==

===

===

===

∆∆∆

∆∆∆∆∆

−

−−−

−

−

=

46

26

612612

26

46

612612

22

22

0

ii

iiii

ii

iiii

i

ii

LL

LLLL

LL

LLLL

L

EIK ,

−−

−−−

−−

−

=

15

2

10

1

3010

110

1

5

6

10

1

5

63010

1

15

2

10

110

1

5

6

10

1

5

6

ii

ii

ii

ii

Gi

LLLL

LLLL

PK

12 −iu

12 +iu

i-th member

iu2

22 +iu

P



96

Principle of Minimum Potential Energy for crPP <

== =

−=−=Πn

iii

n

ij

n

jiji

TT PuuKu11 12

1)()()]([)(

21

PuuKu

)()]([,,1for 02

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

111

1111

11 11 1

PuK =→==−=−+=

−+=−+=

∂∂

−∂∂

+∂∂

=∂

Π∂

===

====

== == =

nkPuKPuKuK

PuKuKPKuuK

Pu

u

u

uKuuK

u

u

u

kj

n

jkjk

n

jjkjj

n

jkj

k

n

iikij

n

jkjk

n

iikij

n

jkj

n

ii

k

in

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

e

e

e

ee

e

e

e

e

e

e

e

e

e

e

ee

e

e

ee

e

e

e

e

e

e

e

e

e

e

ee

e

R

Ry

Rx

L

Ly

Lx

LLLL

LLLL

P

IL

II

L

IL

I

L

I

L

I

L

IAA

IL

II

L

IL

I

L

I

L

I

L

IAA

L

E

m

f

f

m

f

f

)

15

2

10

10

3010

10

10

1

5

60

10

1

5

60

0000003010

10

15

2

10

10

10

1

5

60

10

1

5

60

000000

46

026

0

6120

6120

0000

26

046

0

6120

6120

0000

(

22

22

e

Rx

Lx

ee LEAP

δ−δ=

Documents

Class Note for Structural Analysis 2strana.snu.ac.kr/lecture/struct2_2020/Notes/Note_all_2020.pdf · Class Note for Structural Analysis 2 Spring Semester, 2020 Hae Sung Lee, Professor