class note for structural analysis 2strana.snu.ac.kr/lecture/struct2_2010/notes/note_part_1.pdf ·...

Class Note for Structural Analysis 2

Fall Semester, 2010

Hae Sung Lee, Professor

Dept. of Civil and Environmental Engineering

Seoul National University

Seoul, Korea

Contents

Chapter 1 Slope Deflection Method 11.0 Comparison of Flexibility Method and Stiffness Method………………………… 21.1 Analysis of Fundamental System………………………………………………..... 51.2 Analysis of Beams………………………………………………………………… 81.3 Analysis of Frames………………………………………………………………... 17

Chapter 2 Iterative Solution Method & Moment Distribution Method 332.1 Solution Method for Linear Algebraic Equations………………………………… 342.2 Moment Distribution Method……………………………………………………... 382.3 Example - MDM for a 4-span Continuous Beam…………………………………. 432.4 Direct Solution Scheme by Partitioning…………………………………………... 452.5 Moment Distribution Method for Frames………………………………………… 46

Chapter 3 Buckling of Structures 493.0 Stability of Structures… 503.1 Governing Equation for a Beam with Axial Force 513.2 Homogeneous Solutions………… 523.3 Homogeneous and Particular solution… 58

Chapter 4 Energy Principles 594.1 Spring-Force Systems……………………………………………………………... 604.2 Beam Problems……………………………………………………………………. 614.3 Truss problems……………………………………………………………………. 674.4 Buckling problems

Chapter 5 Matrix Structural Analysis 735.1 Truss Problems…………………………………………………………………… 745.2 Beam Problems…………………………………………………………………… 855.3 Frame Problems…………………………………………………………………... 935.4 Buckling of Structures 965.5 Beam Columns 985.6 Nonlinear Truss

School of Civil, Urban & Geosystem Eng., SNU

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

1

Chapter 1

Slope Deflection Method

A B



2

2.0 Comparison of Flexibility Method and Stiffness Method

Flexibility Method

Remove redundancy (Equilibrium)

Compatibility21 δ=δ

Pkk

kXk

XPkX

21

1

21 +=→

−=

Stiffness Method

Compatibility

δ=δ=δ 21

Equilibrium

→=δ+δ Pkk 2121 kk

P+

=δ

P

k1 k2

P

X

P

k1 k2

P



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 releasedredundancies.

3. Apply unit loads and calculate displacements at the releasedredundancies.

4. Construct the flexibility equation by superposing the displacementbased on the compatibility conditions.

5. Solve the flexibility equation.

6. Calculate reactions and other quantities as needed.

Stiffness Method

1. Fix all Degrees of Freedom.

2. Calculate fixed end forces induced by external loads at the fixedDOF.

3. Apply unit displacements and calculate member end forces at theDOFs.

4. Construct the stiffness equation by superposing the member endforces based on the equilibrium equations.

5. Solve the stiffness equation.

6. Calculate reactions and other quantities as needed.



5

2.1 Analysis of Fundamental System

2.1.1 End Rotation

Flexibility Methodi) 0=θB

036

63

=+

θ−=+

BA

ABA

MEILM

EIL

MEILM

EIL

→ AA LEIM θ−=

4 , AB LEIM θ=

2

ii) 0=Aθ

BA LEIM θ−=

2 , BB LEIM θ=

4

Sign Convention for M :Counterclockwise “+”

0≠θA , 0≠θB

BAB

BAA

LEI

LEIM

LEI

LEIM

θ+θ=

θ+θ=

42

24

2.1.2 Relative motion of joints

MA MB

∆

A B

Aθ



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

2.1.3 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

2.1.4 Joint Equilibrium

∑∑∑ =+−− 0jointmemberfixed FFF

or

∑∑∑ =+ jointmemberfixed FFF

MA MB

MA MA/2

AM23

AM23

Joint i

Fjoint

Fmember

Ffixed



8

2.2 Analysis of Beams2.2.1 A Fixed-fixed End Beam

DOF : Bθ , ∆B

Analysis

i) All fixed : No fixed end forces

ii) 0≠θB , 0=∆B

BAB aEIM θ=

21 , BBA aEIM θ=

41 , BBC bEIM θ=

41 , BCB bEIM θ=

21

BABBA aEIVV θ=−= 2

11 6 , BCBBC bEIVV θ==− 2

11 6

iii) 0=θB , 0≠∆B

BAB aEIM ∆= 2

2 6 , BBA aEIM ∆= 2

2 6 , BBC bEIM ∆−= 2

2 6 BCB bEIM ∆−= 2

2 6

BABBA aEIVV ∆=−= 3

22 12 , BCBBC bEIVV ∆=−= 2

22 12

baEI

P

A B

C

BaEI

θ4

BaEI

θ2

BbEI

θ4

BbEI

θ2

BaEI

θ26

BaEI

θ26

BbEI

θ26

BbEI

θ26

BaEI

∆26

BaEI

∆26

BbEI

∆26

BbEI

∆26

BaEI

∆312

BaEI

∆312

BbEI

∆312

BbEI

∆312



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=+=

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

DOF : Bθ , θC

Analysis

i) Fix all DOFs and Calculate FEM.

12

2qLM fAB = ,

12

2qLM fBA −= ,

8

2qLM fBC = ,

8

2qLM fCB −=

ii) 0≠θB , 0=θC

BAB LEIM θ=

21 , BBA LEIM θ=

41 , BBC LEIM θ=

81 , BCB LEIM θ=

41

iii) 0=θB , 0≠θC

CBC LEIM θ=

42 , CCB LEIM θ=

82


→=++++→=∑ 00 211BCBCBA

fBC

fBA

iB MMMMMM 0412

24

2

=θ+θ+ CB LEI

LEIqL

→=++→=∑ 00 21CBCB

fCB

iC MMMM 084

8

2

=θ+θ+− CB LEI

LEI

LqL

qL

EI 2EI

LL

q

A B

C



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

4832

12qL

LEIqLMMM BAB

fABAB =θ+=+=

22

1

814

12qL

LEIqLMMM BBA

fBABA −=θ+−=+=

22

21

8148

8qL

LEI

LEIqLMMMM CBBCBC

fBCBC =θ+θ+=++=

0848

221 =θ+θ+−=++= CBCBCB

fCBCB L

EILEIqLMMMM

Various Diagram

- Freebody Diagram

- Moment Diagram

2

161 qL 2

81 qL

qL167 qL

169 qL

85

qL83

qL1619



11

2.2.3 Analysis of a Two-span Continuous Beam (Approach II)

DOF : Bθ

Analysis

i) Fix all DOFs and Calculate FEM.

12

2qLM fAB = ,

12

2qLM fBA −= ,

163

821

8

222 qLqLqLM fBC =+=

ii) 0≠θB

BAB LEIM θ=

21 , BBA LEIM θ=

41 , BBC LEIM θ=

61

Construct Stiffness Equation

00 11 =+++→=∑ BCBAf

BCf

BAB MMMMM

EIqL

LEI

LEIqLqL

BBB 96064

163

12-

222

−=θ→=θ+θ++

Member End Forces

22

1

4832

12qL

LEIqLMMM BAB

fABAB =θ+=+=

22

1

814

12qL

LEIqLMMM BBA

fBABA −=θ+−=+=

22

1

816

163 qL

LEIqLMMM BBC

fBCBC =θ+=+=

qL

EI 2EI

LL

q

A B

C



12

2.2.4 Analysis of a Beam with an Internal Hinge (4 DOFs System)

DOF : Bθ , LCθ , R

Cθ , ∆C

Analysis

i) All fixed

12

2qlM fAB = ,

12

2qlM fBA −=

ii) 0≠θB

BAB lEIM θ=

21 , BBCBA lEIMM θ==

411 , BCB lEIM θ=

21 , BCB lEIV θ= 2

1 6

iii) 0≠θLC

LCBC l

EIM θ=22 , L

CCB lEIM θ=

42 , LCCB l

EIV θ= 22 6

iv) 0≠θRC

RCCD l

EIM θ=43 , R

CDC lEIM θ=

23 , LCCD l

EIV θ−= 23 6

v) 0≠∆C

CCBBC lEIMM ∆== 2

44 6 , CDCCD lEIMM ∆−== 2

44 6 , CCBCD lEIVV ∆== 2

44 12

q

EI EI EI

l l l

A B C

D



13

Construct Stiffness Equation

06 0 2812

0 2

2

1 =∆++θ+θ+−→=∑ CLCB

i

i

lEI

lEI

lEIqlM

06 0 42 0 2 2 =∆++θ+θ→=∑ CLCB

i

i

lEI

lEI

lEIM

06 4 0 2 3 =∆−θ→=∑ CRC

i

i

lEI

lEIM

024666 0 3222 4 =∆+θ−θ+θ→=∑ CRC

LCB

i

i

lEI

lEI

lEI

lEIV

Elimination of LCθ and R

Cθ

- 2nd and 3rd equation

)3(2 2 CBLC l

EIl

EIl

EI∆+θ−=θ , C

RC l

EIl

EI∆=θ 23 2

- 1st equation

03 712

06 )3 (812

62812

2

2

22

2

2

2

=∆+θ+−→=∆+∆+θ−θ+−

=∆+θ+θ+−

CBCCBB

CLCB

lEI

lEIql

lEI

lEI

lEI

lEIql

lEI

lEI

lEIql

- 4th equation

063024)3(3)3(36

24666

3233322

3222

=∆+θ→=∆+∆−∆+θ−θ

=∆+θ−θ+θ

CBCCCBB

CRC

LCB

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

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

DOF : Bθ , ∆C

Analysis

i) All fixed

12

2qlM fAB = ,

12

2qlM fBA −=

q

EI EI EI

l l l

A B C

D



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 ∆−=−= 2

22 3 , CDCCD lEIVV ∆=−= 2

22 3


03 712

0 2

2

1 =∆+θ+−→=∑ CBi

i

lEI

lEIqlM

063 0 322 =∆+θ→=∑ CBi

i

lEI

lEIV

EIql

B 66

3

=θ , EI

qlC 132

4

−=∆

EIql

lllEI

lEI CR

CCR

C

3

2643

23)(32

−=∆

=θ→∆

−−=θ

EIql

EIql

EIql

LBLC 2641322

31322

321 333

=+−=∆

−θ−=θ

2.2.6 Beam with a Spring Support

Analysis

i) All fixed

12

2qlM fAB = ,

12

2qlM fBA −=

k

q

EI EI EI

l l l

A B C

D



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

2.2.7 Support Settlement

DOF : Bθ

Analysis

i) All fixed

llEIM f

BAδ

=6 ,

llEIM f

BCδ

−=3

ii) 0≠θB

BBA lEIM θ=

41 , BBC lEIM θ=

31

Construct the Equilibrium Equation

0343601 =θ+θ+δ

−δ

→=∑ BBi

i

lEI

lEI

llEI

llEIM

lBδ

−=θ→73

2.2.8 Temperature Change

T1

T2

lh

TTlh

TTBA 2

)( , 2

)( 1212 −α−=θ

−α=θ

Fixed End Moment

EIh

TTLEI

LEIM

EIh

TTLEI

LEIM

BAB

BAA

)(42

)(24

12

12

−α−=θ+θ=

−α=θ+θ=

A B

EI EI

l l A B C

δ



17

2.3 Analysis of Frames

2.3.1 A Portal Frame without Sidesway

DOF : Bθ , Cθ

Analysis

i) All fixed

80 PlM BC = ,

80 PlMCB −=

ii) 0≠θB

BAB lEIM θ= 11 2 , BBA l

EIM θ= 11 4

BBC lEIM θ= 21 4 , BCB l

EIM θ= 21 2

iii) 0≠θC

CBC lEIM θ= 22 2 , CCB l

EIM θ= 22 4

CCD lEIM θ= 12 4 , CDC l

EIM θ= 12 2


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 =

2.3.2 A Portal Frame without Sidesway – hinged suppoorts

DOF : Bθ , Cθ

Analysis

i) All fixed

80 PlM BC = ,

80 PlMCB −=

l/2 P

A

B C

D

EI1

EI2

EI1



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

2.3.3 A Frame with an horizontal force

DOF : Bθ , ∆

Analysis

i) All fixed : None fixed end momentii) 0≠θB

BAB lEIM θ=

21 , BBA lEIM θ=

41

BBC lEIM θ=

31 , BBA lEIV θ= 2

1 6

iii) 0≠∆

∆= 22 6

lEIM AB , ∆= 2

2 6lEIM BA

∆= 32 12

lEIVBA

Construct the stiffness equation

06)34( 0 2 =∆+θ+→=∑ lEI

lEI

lEIM B

iB

PlEI

lEIPV B

i =∆+θ→=∑ 32126

EIPl

EIPl

B 487 ,

8

32

=∆−=θ

Member end forces

PllEI

lEIM BAB 8

5622 =∆+θ= , Pl

lEI

lEIM BBA 8

3642 =∆+θ= , Pl

lEIM BBC 8

33−=θ=

P

A

B C

EI



21

2.3.4 A Portal Frame with an Unsymmetrical Load

DOF : Bθ , Cθ , ∆

Analysis

i) All fixed

2

20

lPabM BC = , 2

20

lbPaMCB −=

ii) 0≠θB

BAB lEIM θ=

21 , BBA lEIM θ=

41

BBC lEIM θ=

41 , BCB lEIM θ=

21 , BBA lEIV θ= 2

1 6

iii) 0≠θC

CBC lEIM θ=

22 , CCB lEIM θ=

42

CCD lEIM θ=

42 , CDC lEIM θ=

22 , CCD lEIV θ= 2

2 6

a P

A

B C

D



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

−=∆

2.3.5 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

=∆+θ+θ+→=∑ lEI

lEI

lEI

lPabM cB

iB

06 82 0 22

2

=∆+θ+θ+−→=∑ lEI

lEI

lEI

lbPaM cB

iC

0)1(2466 0 322 =∆α++θ+θ→=∑ lEI

lEI

lEIV cB

i

EIEAll

CB 248 , )(

411 2

=αθ+θα+

−=∆

Solution for ab 3=

EIPl

B

2

)107(2565240

α+α+

−=θ , EIPl

C

2

)107(2562816

α+α+

−=θ , EIPl3

)107(1283

α+=∆

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

EIPl

B

2

0203.0−=θ , EIPl

C

2

0109.0 =θ EIPl3

4103282.0 −×=∆

∆

2∆



24

Performance

Responsewith Bracing( 250=α )

w/o bracing( 0=α ) Ratio(%)

θB )/( 2 EIPl× -0.0203 -0.0223 91.03

θC )/( 2 EIPl× 0.0109 0.0089 122.47

∆ )/( 3 EIPl× 0.3282×10-4 0.0033 0.99

ΜΑΒ (Pl) -0.0404 -0.0248 162.90

ΜΒΑ (Pl) -0.0810 -0.0694 116.71

ΜCD (Pl) 0.0438 0.0554 79.06

ΜDC (Pl) 0.0220 0.0376 58.51

ΜP (Pl) 0.1158 0.1216 95.23

ABD (P) 0.0788 - -

Pmax (Pall)* 0.0720 0.0685 105.1

Pmax/vol. 0.0163 0.0228 71.5

*) whP allall σ= , 6/hPM allall =

Unbalanced shear force in the columns = PlEI

CB 0564.0)(62 =θ+θ

The bracing carries 99 % of the unbalanced shear force between the two columns.

2.3.6 A Portal Frame with a Bracing (Horizontal Load)


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 0 2 =∆+θ+θ→=∑ lEI

lEI

lEIM cB

iB

06 82 0 2 =∆+θ+θ→=∑ lEI

lEI

lEIM cB

iC

PlEI

lEI

lEIV cB

i =∆α++θ+θ→=∑ )1(2466 0 322

CB θ=θ , ll CB θ−=θ−=∆35

35

Solution

EIPl

EIPl

CB

32

)4028(35,

)4028(1θθ

α+=∆

α+−==

For 250=α ,

EIPl

CB

23103501.0 −×−=θ=θ ,

EIPl3

3105835.0 −×=∆

Performance

Responsewith Bracing( 250=α )

w/o bracing( 0=α ) Ratio(%)

θB )/( 2 EIPl× 3103501.0 −×− 1103571.0 −×− 0.98

θC )/( 2 EIPl× 3103501.0 −×− 1103571.0 −×− 0.98

∆ )/( 3 EIPl× 3105835.0 −× 1105952.0 −× 0.98

ΜΑΒ (Pl)2102801.0 −× 0.2857 0.98

ΜΒΑ (Pl)2102101.0 −× 0.2143 0.98

ΜCD (Pl)2102101.0 −× 0.2143 0.98

ΜDC (Pl)2102801.0 −× 0.2857 0.98

ABD (P) 1.4004 - -

Pmax(Pall)* 0.7141 0.0292 2448

Pmax/vol. 0.1617 0.0097 1670

*) Governed by ABD for the structure with bracing, and by MDC for the structure withoutbracing. whP allall σ= , 6/hPM allall =

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



26

2.3.7 A Portal Frame with a Spring

DOF : Bθ , Cθ , ∆ Analysis

iv) 0≠∆

∆== 233 6

lEIMM BAAB , ∆== 2

33 6lEIMM DCCD ,

∆== 333 24

lEIVV CDBA , ∆= kVS

3


06 28 0 22

2

=∆+θ+θ+→=∑ lEI

lEI

lEI

lPabM cB

iB

06 82 0 22

2

=∆+θ+θ+−→=∑ lEI

lEI

lEI

lbPaM cB

iC

∆−=∆+θ+θ→=∑ klEI

lEI

lEIV cB

i322

2466 0

0)24(66 322 =∆++θ+θ klEI

lEI

lEI

cB

Deformed Shapes ( 41.0=∆∆S )

without a spring with a spring ( )24 3lEIk =

k∆

k

a P

A

BC

D



27

2.3.8 A Portal Frame Subject to Support Settlement


Analysis

i) All fixed

δ== 200 6

lEIMM CBBC

ii)-iv) the same as the previous problem


06 286 0 22 =∆+θ+θ+δ→=∑ lEI

lEI

lEI

lEIM cB

iB

06 826 0 22 =∆+θ+θ+δ→=∑ lEI

lEI

lEI

lEIM cB

iC

02466 0 322 =∆+θ+θ→=∑ lEI

lEI

lEIV cB

i

A

B C

D

δ



28

2.3.9 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



29


03 278

0 2 =∆+θ+θ+→=∑ lEI

lEI

lEIPlM cB

iB

06 828

0 2 =∆+θ+θ+−→=∑ lEI

lEI

lEIPlM cB

iC

01563 0 322 =∆+θ+θ→=∑ lEI

lEI

lEIV cB

i

)2(5 CBl

θ+θ−=∆

EIPl

B

2

441

−=θ , EIPl

C

2

89

441 =θ ,

EIPl3

1761

−=∆

Load Location that Causes No Sidesway

CBCBl

θ−=θ→=θ+θ−=∆ 20)2(5

- Stiffness equation

0 27 0 2

2

=θ+θ+→=∑ cBiB l

EIlEI

lPabM , 0 82 0 2

2

=θ+θ+−→=∑ cBiC l

EIlEI

lbPaM

0122

2

=θ− ClEI

lPab , 0 4

2

2

=θ+− clEI

lbPa

abl

bPal

Pab 33 2

2

2

2

=→=

a P

A

B C

D



30

2.3.10 A Frame with a Skewed Member

DOF : Bθ , ∆

Analysis

i) All fixed : PlPlPlM BC 163

1680 =+= , PVBC 16

11220 −=

ii) 0≠θB

BBAB lEI

lEIM θ=θ= 22

21 , BBA lEIM θ= 221 , BBC l

EIM θ=31 , BBA l

EIV θ= 21 3

BBC lEIV θ−= 2

1

223

iii) 0≠∆

∆=∆

== 222 3

226

lEI

llEIMM ABBA ∆−=

∆−= 2

2

223

23

lEI

llEIM BC , ∆= 3

2 23lEIVBA ,

∆= 32

23

lEIVBC

P

A

B C

BBB lEI

llEI

lEI

θ=θ+θ 2321)222(

BB lEI

llEI

θ=θ 2223

2213

BlEI

llEI

lEI

θ=∆+∆ 322 2321)33(

∆=∆ 32 23

211

223

lEI

llEI



31

Construct the stiffness equation

PllEI

lEIM B

iB 16

3 )2

233()32(20 2 −=∆−+θ+→=∑

PlEI

lEIV B

i

1611

22)

2323()2

23(3 0 32 =∆++θ−→=∑

PllEI

lEI

B 1875.0 8787.05.8284 2 −=∆+θ

PlEI

lEI

B 4861.07426.58787.0 32 =∆+θ

EIPl

B

2

0460.0−=θ , EIPl3

0917.0=∆

Results

- Deformed shape

- Moment diagram

- Shear force diagram



32

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