Structural Analysis Istrana.snu.ac.kr/lecture/struct1/Notes/12_Note.pdf · 2012. 3. 7. · Dept. of Civil and Environmental Eng., SNU Structural Analysis Lab. Prof. Hae Sung Lee,

Dept. of Civil and Environmental Eng., SNU

Structural Analysis Lab.

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

Structural Analysis I

Spring Semester, 2012

Hae Sung Lee

Dept. of Civil and Environmental Engineering

Seoul National University

y

yf

z zf

x

xf

yM y

zM

z

xM

x




This page is intentionally left blank.




Contents

1. Introduction

2. Reactions & Internal Forces by Free Body Diagrams

3. Principle of Virtual Work

4. Analysis of Statically Indeterminate Beams

5. Analysis of Statically Indeterminate Trusses

6. Analysis of Statically Indeterminate Frames

7. Influence Lines for Determinate Structures

8. Influence Lines for Indeterminate Structures








1

Chapter 1

Introduction




2

1.1 Mechanics of Material - Structural Mechanics

Problem

Calculate the reaction force at each support and draw the moment and shear force diagram for

the two-span beam shown in the figure.

Solution

Equilibrium Equation

qLRRRF cbay 20

qLRRLRLRLqLM cbcba 220220

0022

0 cacab RRLRLRL

qLL

qLM

qLRRLRLRLqLM babac 220220

Since there are three unknowns in two independent equations, we cannot determine a unique

solution for the given structure, and thus we need one more equation to solve this problem.

The main issue of this class is how to build additional equations to analyze statically

indeterminate structures.

EI EI

q

Ra Rb Rc

L L

q

q




3

1.2 Mechanics of Material

Governing Equation

– Left span

11

2

1

3

1

4

1

''''

124

dxcxbxaEI

qxwqEIw

– Right Span

22

2

2

3

2

4

2

''''

224

dxcxbxaEI

qxwqEIw

Boundary Conditions

– Left support

0)0()0( , 0)0( 111 wEIMw

– Center support

)()( , )()( , 0)()( 212121 LwLwLwLwLwLw

– Right support

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

Since there are eight unknowns with eight conditions, we can solve this problem.

Determination of Integration Constant

– Left Support

xcxaEI

qxwbwdw 1

3

1

4

1111124

02)0( , 0)0(

– Right Support

xcxaEI

qxwbwdw 2

3

2

4

2222224

02)0( , 0)0(

x x

q

w1 w2




4

– Center Support

EI

qLcc

EI

qLaa

LaEI

qLLa

EI

qL

cLaEI

qLcLa

EI

qL

LcLaEI

qLLw

LcLaEI

qLLw

48

48

3

62

62

36

36

024

)(

024

)(

3

21

21

2

2

1

2

2

2

2

3

1

2

1

3

2

3

2

4

2

1

3

1

4

1

)32(48

334

21 xLLxxEI

qww

8

3 ,

8

3

211

2

11

qLqxwEIVx

qLx

qwEIM

Moment Diagram

Shear Diagram

Reactions

0.375qL

L8

3

+

-

+

0.625qL

-

0.375qL 0.375qL 1.25qL

0.125qL2

0.070qL2

L8

3

+

-

+




5

1.3 Mechanics of Material +

1.3.1 Main idea

Original Problem

Case I (Removal of the center support)

Case II (Application of the reaction force)

Original Problem = Case I + Case II

0+R=0

(compatibility condition)

1.3.2 Calculation of 0

Governing Equation

dcxbxaxEI

qxwqEIw 23

4

0

''''

024

q

q

0

R

Rb




6

Boundary (support) Conditions

– Left Support

0)0()0( , 0)0( 000 wEIMw

– Right Support

0)2()2( , 0)2( 000 LwEILMLw


– Left Support

00)0( , 0 0)0( 00 bwdw

– Right Support

EI

Lqc

EI

Lqa

LaEI

Lq

LcLaEI

Lq

Lw

Lw

24

)2(

12

)2(

0)2(62

)2(

0)2()2(24

)2(

0)2(

0)2(

32

34

0

0

))2()2(2(24

334

0 LxLxxEI

qw

EI

LqLLLLL

EI

qLw

384

)2(5))2()2(2(

24)(

4334

00

1.3.3 Calculation of R

Governing Equation

dcxbxaxwEIw RR 23'''' 0

Boundary (support) Conditions

– Left Support

0)0()0( , 0)0( RRR wEIMw

– Mid-span

2

)()( , 0)()( b

RRR

RLwEILVLwL


– Left Support

00)0( , 0 0)0( bwdw RR




7

– Mid-span

)3(12

12

3

12

26

03

2)(

0)( 23

2

2

xLxEI

Rw

EI

LRc

EI

Ra

RaEI

caL

RLwEI

Lwb

R

b

b

bb

R

R

EI

RL

EI

RLLw bb

RR48

)2(

12

2)(

33

1.3.4 Final Solution

Reaction at Supports

qLRRR cba 2

qLRR cb 22

0+R=0 048

)2(

384

)2(534

EI

RL

EI

Lq b qLRb8

10

qLRR ca8

3

Moment

xqL

xq

xqL

xLxq

wEIwEIMMM RR8

3

28

5)2(

2

22

00

Shear

8

3

8

5)(00

qLqx

qLLxqwEIwEIVVV RR




8

1.4 Structural Mechanics

Original Problem

Case I (Removal of the center support)

Case II (Application of the reaction force)

Original Problem = Case I + Case II

0+R=0

Principle of Virtual Work

EI

Lqdx

EI

MML

R

384

)2(5 42

0

00 ,

EI

LRdx

EI

MM b

L

RRR

48

)2( 32

0

Solution

0+R=0 048

)2(

384

)2(5 34

EI

RL

EI

Lq b qLRb8

10

RbL/2

Rb

q

q

qL2/2




9

– Moment

– Shear

+

-

+

-

+

+

-

+

-

=

0.070qL2

5qL2/8

Rb

+

=

0.125qL2

L8

3

+

-

+

qL2/2




10

1.5 지 점 (Supports)

고정단 (fixed support)




11

회전단 (hinge support)

이동단 (roller support)




12

1.6 구조물의 2차원 이상화

주 구조물

(Main Structure)

가로 보

(Cross Beam)

세로 보

(Stringer )

Cross Bracing

(Wind Bracing)

지 점

(Support)




13

Truss

Beam

절점(Joint)




14

Frame




15

1.7 Force and Displacement

Real 3-D Structures

– 3 force components and 3 moment components

– 3 displacement components and 3 rotational components

Beam Idealization

– Vertical force and Moment on z-axis

– Vertical displacement and rotational angle w.r.t. z-axis

Plane Truss Idealization

– Vertical and horizontal force

– Vertical and horizontal displacement

x

y

z

y

yf

z zf

x

xf

yM y

zM

z

xM

x

22 , wV

22 , M

11 , wV

11 , M

33 , wV

33 , M




16

Plane Frame Idealization

– Vertical, horizontal force and moment w.r.t. z-axis

– Vertical, horizontal displacement rotational angle w.r.t. z-axis

x xf

y yf

x

xf

y

yf zM

z




17

1.8구조물의 안정 (Stability of Stuctures)

내적 안정 (Internal Stability)

어느 한계 내의 크기의 어떠한 하중의 작용을 받더라도 형상이 허물어 지지

않는 구조물의 상태

외적 안정 (External Stability)

어느 한계 내의 크기의 어떠한 하중의 작용을 받더라도 구조물이 움직이지 않

는 상태




18





19

Chapter 2

Reactions & Internal Forces

by Free Body Diagrams




20

2.1 Free Body Diagram




21

It is impossible to draw too many free-body diagrams.

Time spent in doing so is never wasted

- C. H. Norris & J. B. Wilbur & S. Utku -




22

2.2 Reactions

Beams

0 0 PRRF BAV

00 LRPaM BA (Clockwise +)

PL

aRB , P

L

bRA

PRRF BAV 0

0)(0 LRaLPM BA (Clockwise +)

PL

aRB )1( , P

L

aRA

L

P

a b

RA RB

L

P

a

RA RBaL




23

Truss

PHPHF AAH 00 ,

00 PRRF BAV

0)3(0 aRPaPaM BA (Clockwise +)

PRA3

1 , PRB

3

2

Frame

0

02

BAH

BAV

HHF

qLRRF

02

LHL

RM BB

R

h

0422

LqLLH

LRM AA

L

h BB HR 2 ,

42

qLHR AA

822

42

qLHH

qLH

qLH BABA

16

16

qLH

qLH

B

A

8

8

3

qLR

qLR

B

A

P

RA RB

P

HA

L

L

HA HB

RA RB

q




24

2.3 Internal Forces in Framed Structures

Axial Force

Shear Force

Bending Moment

Torsion




25

2.4들보 (Beam)

Reactions

q

RA=qL/2 Rb= qL/2




26

Free Body Diagram for Shear and Moment

qxqL

qxRVVqxRF AxxAV 2

0

22

02

2qxx

qLMM

xqxxRM xxAx

Shear Force and Moment Diagrams

q

RA=qL/2 RB= qL/2

RA RB

RA

x

RB

Mx

Vx

qL/2

qL2/8




27

Deflected Shape

2.5 Gerber Systems




28




29

2.5.1 Internal Forces in a Gerber Beam - I

Free Body Diagram

PRL

PL

RM HHC3

20

24

30

PRPRRF CCHv3

100

PRL

RLRM BHBA6

50

4

50

PRPRRF ABAv6

10

3

20

L/4

P

RH

RA RB

RC

P

P/6 5P/6 P/3

P




30

Shear Force

i) Lx 0

ii) LxL2

3

iii) LxL 22

3

Bending Moment

i) Lx 0

P/6

V= -P/6

P/6 5P/6

V= 2P/3

P/6 5P/6

P

V= -P/3

+

- -

2P/3

P/3 P/6

P/6

P/6

Mx= Px/6




31

ii) LxL2

3 , Lxx

iii) LxL 22

3 xLx 2

Deflected Shape

qdx

wdEI

4

4

P/6 5P/6

2P/3

Mx= xPPL

3

2

6

-Px/6

P/6 5P/6

P

P/3 P/3

Mx= xP 3

1

+

-

PL/6

PL/6




32

2.5.2 Internal Forces in a Gerber Beam - II

Free Body diagram

Shear

Moment

q

2

ql

2

ql

2

ql

2

ql

2

2ql

2

2ql

2

ql

2

ql

+

L

q

L L

8

2ql

2

2ql 2

2ql

+




33

2.6 트러스 (Truss)




34

Assumption

1. All joints are hinges.

2. All members are straight.

3. Small deformatiom

4. The external loads are applied only at joints.

Characteristics of truss

– By by the 2nd

, 3rd

and 4th

assumptions

02

2

2

2

dx

Mdq

dx

Md

baxM

– By by the 1st assumption

0 , 000)()0( VMbaLMM

– No bending moment and shear force are induced in all members in a truss strcture.

– Only axial forces are the internal forces in a truss.




35

2.6.1 Internal Forces in Howe Truss

At U1 and U3

At L1

2 ,

2

2

022

2

02

2

23

3

23 PFPF

PF

FF

At L2

2 , 0- , 0 265625

PFFPFFFPF

F1 F3

F2

P/2

F5

F2 F6

P

F4=0

F1=0

F8 =0

F9=0

U2

H 1

2

3

4

5

6

7

9

L1 L2

L3

U1 U3

P P/2 P/2

8




36

At U2

02

1

2

1

2

2

02

2

2

2

02

2

2

237

753

8734

PPP

PFF

FFF

FFFF

At L3

Axial Force Diagram

2

P

2

P

F5

F4 F8

F3 F7

2

P 0

P/2

P/2

P/2 P/2

P 0

0 0

0

P P/2 P/2




37

Equilvalent Beam Action

Deflected Shape

2.6.2 Internal Forces in Warren Truss

V=P/2

P/2

1 P

3

P

3

2P

2 3

4

5

6

7

8

9

10

11

L1 L2 L3

U1 U2 U3

L4

x

PL/4 M=Px/2




38

At L1

PFPFP

F

FF

3

2 ,

3

22

03

2

2

2

02

2

12

2

12

At U1

PFPFF

FF

FFF

3

4,2

3

2

02

2

2

2

02

2

2

2

423

32

432

At L2

PFPF

FFFF

PFF

65

5361

35

,3

2

02

2

2

2-

02

2

2

2

F2

F1

2P/3

F4

F2 F3

F3

F1 F6

P

F5




39

At U2

PFPF

FF

FFFF

3

2,

3

2

02

2

2

2

02

2

2

2

87

75

7584

At L3

3,

3

2

02

2

2

2

02

2

2

2

109

97610

97 PFPF

FFFF

FF

At U3

PF

FF

FFF

3

2

02

2

2

2

02

2

2

2

11

119

8119

F8

F5 F7

F4

F7

F6 F10

F9

F11

F8

F9




40

At L1

OK

Deflected Shape

Equivalent Moment

1 2 3 4 5 6

1. PxM3

2

2. PxPlM3

2

3

2

3. PxPlxlPPlM3

1

3

4)(

3

1

4. PxPlxlPPlM3

1)(

3

1

3

2

5. PxPlxlPPlM3

1

3

2)(

3

1

3

1

6. PxPlxlPM3

1

3

1)(

3

1

P/3

P/3

P3

2

P




41

2.6.3 Method of Sections

PFL

FLPM L3

40

23

2442

PFL

FL

PLPMU 662 0222

3

3

2

PFFPPFV3

20

2

2

3

255

P

3

P P

3

2

1

2 3

4

5

6

7

8

9

10

11

L1

L2 L3

U1

U2

U3

L4

L3

L2

L1

U1

U2 U3

L4

Cut out

P

P3

2

F5

F4

F6




42

2.7 프레임 (Frame)




43

2.7.1 Internal Forces in a Frame

Reactions

Freebody Diagram

H

qH

L

L

qH

2

2

q

L

qH

2

2

2

2qH

qH

L

qH

2

2

L

qH

2

2

2

2qH

L

qH

2

2

L

qH

2

2

L

qH

2

2

L

qH

2

2




44

Axial, Shear and Moment diagram

Deflected Shape

L

qH

2

2

+ - Axial

L

qH

2

2

qH

-

+

Shear

L

qH

2

2

2

2

qH

+

Moment

+




45

2.7.2 Internal Forces in a 3-hinged Frame

Reactions (+:Clockwise for mement)

0

2

BA

BA

HH

qLRR

0R

hM : BBBB HRLHL

R 202

0L

hM : 4

20422

qLHR

LqLLH

LR AAAA

0

822

42

BA

BABA

HH

qLHH

qLH

qLH

16

16

qLH

qLH

B

A

8

8

3

qLR

qLR

B

A

L

L

HA HB

RA RB

h

q




46

Freebody Diagram

Axial, Shear and Moment diagram

16

qL 16

qL

8

qL

16

qL 8

3qL

16

qL

8

qL 8

3qL 8

qL

16

qL 16

qL

16

2qL 16

2qL

16

2qL 16

2qL

16

qL 16

qL

- +

- 8

qL

+

8

3qL

Shear

8

3qL 8

qL

16

qL

- -

-

Axial




47

28

3

16

028

3

1622

2

xqqLx

qLM

xqxqLx

qLM

Deflected Shape

- -

-

Moment

- -

2

16

qL 2

16

qL

8

3qL

16

2qL

M

V




48

2.8 Arches




49

2.8.1 Three Hinged Arch

Arch Curve : )( 22

2xl

l

hy

Reactions

2

2

PR

PR

B

A

,

h

PlH

h

PlH

B

A

2

2

Freebody Diagram

h

PlH

PV

2

2

0)( xLRyHM AA

)(2

)(2

)(2

)(22

2

22

2

xlxl

P

xlP

xll

h

h

Pl

xlP

yh

PlM

8max

PlM

HA

RA

HB

RB

P

h

2l

x

y

HA

RA

V

H

M

y

l+x




50

Axial force and Shear Force

S

A

V

H

SAV

SAH

cossin

sincos

cossin

sincos

V

H

S

A

cossin

sincos

224224

2

2

4

2sin ,

4cos

2tan

xhl

hx

xhl

l

l

hxy

)2

(4

cos2

sin2

)2

(4

sin2

cos2

2

224

23

224

llx

xhl

PP

h

PlS

h

xhl

xhl

PP

h

PlA

Deflected Shape

V

H

A S




51

2.8.2 Zero Moment Arch I

Reactions

2

2

PR

PR

B

A

,

h

PlH

h

PlH

B

A

2

2

Freebody Diagram

h

PlH

PV

2

2

0)( xLRyHM AA

)(

0)(22

xll

hy

xlP

yh

PlM

HA

RA

HB

RB

P

h

2l

x

y

H

HA

RA

V

M

y

l+x

x

y




52

2.8.3 Zero Moment Arch II

Reactions

2

2

qlR

qlR

B

A

,

h

qlH

h

qlH

B

A

2

22

2

Freebody Diagram

02

)()()(

xlxlqxlRyHM AA

)(

02

)()()(

2

22

2

2

xll

hy

xlxlqxlqly

h

qlM

HA

RA

HB

RB

q

h

2l

x

y

HA

RA

V

H

M

y

l+x

q


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

53

Chapter 3

Principle of Virtual Work

The principle of virtualwork is the most importantsubject in the area of thestructural analysis !!!!



54

3.1 Beam Problems

3.1.1 Governing Equations

Equilibrium for vertical force

qdxdVqdxVdVV −=→=+−+ 0)(

Equilibrium for moment

Vdx

dMdxqdxVdxMdMM =→=+−−+ 02

)(

Elimination of shear force

qdx

Md−=2

2

Strain-displacement relation

ydx

wd2

2

−=ε

Stress-strain relation (Hooke law)

ydx

wdEE 2

2

−=ε=σ

Definition of Moment

2

22

2

2

dxwdEIdAy

dxwdEydAEydAM

AAA

−=−=ε=σ= ∫∫∫

Beam Equation

qdx

wdEI =4

4

MM+dM

V V+dV

q



55

Modelling of Concentrate loads - Dirac delta functions

0lim→ε

= = )( ξ−δ x

1221lim)0

210(lim)(

00

00

=εε

=+ε

+=ξ−δ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε ∫∫∫∫ll

dxdxdxdxx

)()(2

)()(lim)(21lim

)0)(21)(0)((lim)()(

00

00

0

ξ=ξ′=ε

ε−ξ−ε+ξ=

ε=

+ε

+=ξ−δ

→ε

ε+ξ

ε−ξ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε

∫

∫∫∫∫

fFFFdxxf

dxxfdxxfdxxfdxxxfll

3.1.2 Principle of Virtual Work (Beam)

Equilibrium equation in an integral form

0)(0

2

2

=+∫ dxqdx

Mdwl

Integration by part twice

dxqwdxMdx

wdMdxwd

dxdMw

dxqwdxdx

dMdxwd

dxdMw

llll

lll

∫∫

∫∫

−=+−

−=−

002

2

00

000

llll

MVwdxqwdxMdx

wd00

002

2

θ+−−= ∫∫

In case w is a displacement field of the same structure caused by another load case q ,

then the boundary terms vanish since either displacement or reaction should be zero at a

boundary (support).

ξ

∞

ξ

ε21

2ε



56

Principle of virtual work

dxEIMMdxM

dxwd ll

∫∫ −=00

2

2

dxqwdxEIMM ll

∫∫ =00

→ extWW δ=δ int

Equilibrium equation for load case q

0)(0

2

2

=+∫ dxqdx

Mdwl

Virtual work expression

dxqwdxEIMM ll

∫∫ =00

Betti-Maxwell’s Reciprocal Theorem

dxEIMMdx

EIMM ll

∫∫ =00

→ dxqwdxqwll

∫∫ =00

Calculation of displacement for the load case q

dxqwdxEIMM ll

∫∫ =00

In case q system represents a single unit concentrated load applied at the position where

you want to calculate the displacement for q system.

)()( 00

00

xwdxxxwdxEIMM ll

=−δ= ∫∫ → dxEIMMxw

l

∫=0

0 )(

3.1.3 Example

A simple beam subject to an uniform load

– Moment of load case q

q

ql2/8



57

– Moment of load case q

– Deflection at the center of the span

EIql

EIqllqlql

EI

dxxqxqlEI

dxxqxqlxEI

dxxqxqlxlEI

dxxqxqlxEI

dxEIMMlw

ll

l

l

ll

4443

2/

0

322/

0

2

2/

22/

0

2

0

3845)

2561

961(2))

2(

41

4)

2(

31

4(2

)44

(2)22

(2

2

)22

)(22

(1)22

(2

1)2

(

=−=−=

−=−=

−−+−==

∫∫

∫∫∫

or from the integration table,

EIqlqll

EIlMM

lab

EIldx

EIMMlw

l

3845

84)

411(

3)1(

3)

2(

42

3120

=+=+== ∫

Values of Product Integrals ∫L

LU dxMM0

l/4

1



58

3.1.4 Conservation of Energy

Equilibrium and Conservation of Energy

– Equilibrium Equation

qdx

wdEI =4

4

– External work

int

ll

lll

lll

ll

ext

WdxEIMdx

dxwdEI

dxwd

MwVdxdx

wdEIdx

wd

dxwdEI

dxdw

dxwdwEIdx

dxwdEI

dxwd

dxdx

wdwEIwqdxW

===

θ+−=

−+=

==

∫∫

∫

∫

∫∫

0

2

02

2

2

2

000

2

2

2

2

02

2

03

3

02

2

2

2

04

4

0

21

21

][21

][21

21

21

Conservation of Energy in each load case

dxwqdxEIM ll

∫∫ =00

2

21

21 , dxqwdx

EIM ll

∫∫ =00

2

21

21

Two load cases are applied simultaneously.

dxqwqwdxEIMM

dxwqqwqwqwdxEIMdx

EIMMdx

EIM

dxqqwwdxEI

MM

ll

llll

ll

∫∫

∫∫∫∫

∫∫

+=

+++=++

++=+

00

00

2

00

2

00

2

)(21

)((21

21

21

))((21)(

21

w

q

w

q



59

External work for sequential loading (q first)

dxwqdxwqdxqwqwdxwq

dxwqqwqwqwdxqwqdxwwqdx

dxqqwwdxqwdxwqwqdx

llll

lll l

lll l

∫∫∫∫

∫∫∫ ∫

∫∫∫ ∫

=→+=

+++=++

++=++

0000

000 0

000 0

)(21

)((21

21

21

))((21

21

21

Principle of Virtual work

dxqwdxqwdxqwqwdxEIMM llll

∫∫∫∫ ==+=0000

)(21

3.1.5 General Conservation and Equilibrium

Conservation in General

∫ ∫ =+⋅−S V

fdVdS 0nv

– By divergence theorem,

∫∫ ⋅∇−=⋅−VS

dVdS vnv where ),,(),,(321 xxxzyx ∂

∂∂∂

∂∂

=∂∂

∂∂

∂∂

=∇ .

w

q

w

q

ww +

vn

dS



60

∫∫∫ ∫∫ =+⋅−∇=+⋅∇−=+⋅−VVS VV

dVffdVdVfdVdS 0)( vvnv

– Since the integral equation should hold for all systems,

0=+⋅∇− fv

– In a potential problem, the vector field of a system is defined by a gradient of a scalar

function referred to as a potential function

Φ∇−= kv

– The famous Laplace equation for a conservative system.

02 =+Φ∇=+Φ∇⋅∇=+⋅∇− fffv or

02

2

2

2

2

2

=+∂

Φ∂+

∂Φ∂

+∂

Φ∂ fzyx

Equilibrium in General

– Force Equilibrium: ∑∑∑ === 0zyx FFF or 0=∑F

∫ ∫ =+S V

dVdS 0bT or ∫ ∫ =+S V

ii dVbdST 0 for i = 1,2,3

Suppose nT ⋅= σ or ∑=

⋅==3

1jijiji nT nσσ

=

σσσσσσσσσ

=

3

2

1

333231

232221

131211

σσσ

σ ,

=

3

2

1

nnn

n

Divergence Theorem

0)( =+⋅∇=

+⋅∇=+⋅=+

∫

∫ ∫∫ ∫∫ ∫

Vii

V Vii

S Vii

S Vii

dVb

dVbdVdVbdSdVbdST

σ

σσ n for i = 1,2,3

Since the integral equation should hold for all systems in equilibrium,

03

1

321 =+∂

σ∂=+

∂σ∂

+∂σ∂

+∂σ∂

=+⋅∇ ∑=

ij j

iji

iiiii b

xb

zyxbσ for i = 1,2,3 or

0

0

0

3333231

2232221

1131211

=+∂σ∂

+∂σ∂

+∂σ∂

=+∂σ∂

+∂σ∂

+∂σ∂

=+∂σ∂

+∂σ∂

+∂σ∂

bzyx

bzyx

bzyx



61

– Moment Equilibrium: 0=∑ iM for i=1, 2, 3 or 0=∑M

0=+×+× ∫∫∫VVS

dVdVdS mfxvx

211231133223 , , σ=σσ=σσ=σ

– What is σ, and how is σ related to a potential function? : out of scope of this class !

3.1.6 Displacement on boundaries

lll

llll

MVwdxqwMVwdxqwdxEIMM

000

0000

θ−+=θ−+= ∫∫∫

)0()0()()()0()0()()(00

0

MlMlVwlVlwMVwdxEIMM ll

l

θ+θ−−=θ−=∫

By coinciding the positive direction of forces and displacement

dxEIMMMlMlVwlVlw

l

∫=θ+θ++0

)0()0()()()0()0()()(

Deflection of a cantilever beam subject to an end load

0)0( , 0)0( , 0)( , 1)( =θ=== wlMlV

q (real) system q (virtual) system

EIPlPll

EIlMM

EIldx

EIMMlw

l

3))((

33)(

3

310

=−−=== ∫

Or, you can obtain the same answer by assuming the unit concentrate load is applied at

just left side of the boundary.

Rotation of a cantilever beam subject to an end load

0)0( , 0)0( , 1)( , 0)( =θ=== wlMlV

q (real) system q (virtual) system

P

-Pl

1

-l

P

-Pl

1



62

EIPlPl

EIlMM

EIldx

EIMMl

l

2)(1

22)(

2

310

−=−××===θ ∫

Rotation in the a body (or a structure)

– Modeling of a unit moment applied at x0

)()]2

()2

([1lim

))]2

((1))2

((1[lim

0000

0000

0

0

xdxdwxwxw

dxxxxxwdxEIMM

xx

ll

θ−=−=ε

+−ε

−ε

=

ε+−δ

ε−

ε−−δ

ε=

=→ε

→ε ∫∫

– by coinciding the positive direction of the rotational angle with that of the appliedmoment.

dxEIMMx

l

∫=θ0

0 )(

ε1/ε1/ε

x0



63

3.2 Principle of Virtual Work in General

3.2.1 3-Dimensional Elastic Body

Rigid Body

If a real q-force system is acting on a rigid body is in equilibrium and remains in

equilibrium as the body is given any small displacement , the virtual work done by

the q-force system is equal to zero.

0)(∫ ∫∫ ∫ =+⋅=⋅+⋅=δS VS V

ext dVdSdVdSW fvwfwvw

Deformable Body

If a deformable body is in equilibrium under a real q force system while it is subjected to

small and compatible displacement caused by a virtual q force system, the external

virtual work done by the real q force system is equal to internal virtual work done by

the internal q stress !!!

∫ ∫ ⋅+⋅=δS V

ext dVdSW fwvw

∑∑∫∑∑∫

∑∑∫∑∫ ∑∑∫∫

= == =

= == ==

∂

σ∂+σ

∂∂

=σ∂∂

=

σ=σ==⋅

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

)()(i j V j

ijiij

j

i

i j Viji

j

i j Sjiji

i S jjiji

i Sii

S

dVx

wxw

dVwx

dSnwdSnwdSvwdSvw

q-Force System q -Force System



64

int

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

3

1

)(21

)(

)(21

)(

)(

WdVdVxw

xw

dVxw

dVxw

dVxw

dVxw

dVxw

dVfx

wdVxw

dVfwdVx

wxw

W

i j Vijij

i j Vij

i

j

j

i

i j Vij

i

j

i j Vij

j

i

j i Vji

i

j

i j Vij

j

i

i j Vij

j

i

i V ji

j

iji

i j Vij

j

i

i Vii

i j V j

ijiij

j

iext

δ=σε=σ∂

∂+

∂∂

=

σ∂

∂+σ

∂∂

=

σ∂

∂+σ

∂∂

=σ∂∂

=

+∂

σ∂+σ

∂∂

=

+∂

σ∂+σ

∂∂

=δ

∑∑∫∑∑∫

∑∑∫∑∑∫

∑∑∫∑∑∫∑∑∫

∑∫ ∑∑∑∫

∑∫∑∑∫

= == =

= == =

= == == =

= == =

== =

∫ ∑∑∫∫= =

σε=⋅+⋅S i j V

ijijV

dVdVdS3

1

3

1

fwvw

3.2.2 Framed Structures

)(∑ ∫∫∫∫∑∫ τγ+σε=τγ+σε=σ⋅εe VVVVij V

ijij dVdVdVdVdVee

Internal virtual work by normal stress – bending moment

∫∫∫∫ ∫

∫ ∫∫∫∫

=−−===

==−−=σε

eeee

e

e

eeee

llll

A

l

AVVV

dxEIMMdx

EIMEI

EIMdx

dxwdEI

dxwddx

dxwddAEy

dxwd

dAdxydx

wddx

wdEdVydx

wddx

wdEdVydx

wdEydx

wddV

0002

2

2

2

02

22

2

2

0

22

2

2

22

2

2

2

2

2

2

2

2

)()()(

)()(

Internal virtual work by normal stress – Axial Force

∫∫ ∫∫∫ ===σεee

eee

ll

AVV

dxEA

FFdxAFdA

EAFdV

AF

EAFdV

00

)(

Internal virtual work by shear stress

QyIb

V)(

=τ and QyGIb

V)(

=γ where ∫=a

y

ydAQ

∫∫ ∫∫∫ ===τγee

eee

ls

l

AVV

VdxVGAf

dAdxybI

QGVVQdV

yIbVQ

yGIbVdV

0022

2

)()()(

Total displacement

∑ ∫ ∫∫ ++=e

l ll

s

e ee

dxEA

FFdxGA

VVfdxEIMMxw

0 000 )()(



65

3.2.3 Effect of Shear Deformation

For simple beam with a uniform load case

EIqllwM 384

5)2

(4

=

Shear Effect

▬ Shear force of load case q

▬ Shear force of load case q

▬ Deflection by shear force

GAqlf

VVlGA

fVdxV

GAf

VdxVGAf

w ssl

sl

sS

e

822122 22/

00

==== ∫∫

GAEI

lf

EIqlGAqlf

ww ss

M

s

40384

384/58/

24

2

==

for a rectangle section of bh× with steel

2

2

2

3

5.21240

6.238456

lh

bhlbh

ww

M

s =×

××=

For small h/l, the effect of shear deformation can be neglected.

1/2

ql/2



66

3.3 Truss Problems

3.3.1 Principle of Virtual Work

From General principle

∑∑ ∫ ∫∫

∫

=++=

α+=α=⋅

ii

ii

e

l ll

s

iiiS

lEA

FFdx

EAFFdx

GAVVfdx

EIMM

vuwdS

e ee

0 00

22

)(

coscoswq

From equilibrium equation

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 jointi, respectively.

0])( )[(1

)(

1

)(

1=+−++−∑ ∑∑

= ==

njn

i

iiim

j

ij

iiim

j

ij vYVuXH

∑∑

∑∑ ∑∑

∑ ∑∑

==

== ==

= ==

+=−θ+−θ

+=θ+θ

=+θ−++θ−

njn

i

ii

ii

nmb

iiii

eiiii

ei

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

n

i

iiiinmb

i i

iie

i

nmb

i i

iie

i

i

iinmb

i

ei

nmb

ii

ei

nmb

iiiiiii

ei

vYuXEA

lFF

EAlFF

EAlF

FlFvvuuF

11

1111

1212

) ()(

))(sin )((cos

iY iX

iF1−

ijF−

iimF )(−

iY iX



67

or, by applying Betti-Maxwell reciprocal theorem

∑∑==

=+nmb

i i

iie

injn

i

iiii

EAlFFvYuX

11 )() (

The displacement of a joint k in a truss is obtained by applying an unit load at a joint k inan arbitrary direction.

∑=

===+nmb

i i

iie

ikkkkkk

EAlFFvYuX

1 )(coscos αα 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.

For vertical displacement For Horizontal displacement

ivv θ− sin)( 12

iuu θ− cos)( 12

iθiθ

)( 12 uu −

)( 12 vv −

kuαcosku



68

3.3.2 Example

Real System Virtual System

Table for calculation of the deflection of a truss

Member EAl

F F EAlFF

1 1 30 0.75 22.5

2 2 -30 2 -0.75 2 45 2

3 1 30 0.75 22.5

4 1 40 0.50 20

5 2 -10 2 0.25 2 -5 2

6 1 -30 -0.75 22.5

7 1 20 0 0

8 1 40 0.5 20

9 2 -10 2 -0.25 2 5 2

10 1 -30 -0.25 7.5

11 1 30 0.25 7.5

12 1 30 0.25 7.5

13 2 -30 2 -0.25 2 15 2

∑ 130+60 2

EAl

EAL 215)260130( =+=δ

1

2

3

4

5

6

7

8

9

10

11

12

13

20 20 20



69

3.3.3 Conservation of Energy

Equilibrium and Conservation of Energy

▬ Equilibrium Equation

0 , 0)(

1

)(

1=+−=+− ∑∑

==

iim

j

ij

iim

j

ij YVXH for njni ,,1L=

▬ External work

∑ ∑∑∑∑= ====

+=⋅∆=+=njn

ii

im

j

iji

im

j

ij

njn

iii

njn

iiiiiext vVuHvYuXW

1

)(

1

)(

111) (

21

21)(

21 P

int

nmb

i i

ie

ie

inmb

ii

ei

nmb

iiii

eiiii

ei

njn

i

im

j

ij

iji

im

j

ij

iji

njn

ii

im

j

ij

iji

im

j

ij

ij

njn

ii

im

j

iji

im

j

ijext

WEA

lFFlF

vvFuuF

FvFu

vFuFvVuHW

==∆=

−θ+−θ=

θ+θ=

θ+θ=+=

∑∑

∑

∑ ∑∑

∑ ∑∑∑ ∑∑

==

=

= ==

= === ==

11

1

1212

1

)(

1

)(

1

1

)(

1

)(

11

)(

1

)(

1

21

21

))(sin )(cos(21

)sin cos(21

)sin cos(21) (

21

intext WW =

Conservation of Energy in each load case

∑∑==

⋅∆=njn

iii

nmb

i i

ii

EAlF

11

2

21

21 P , ∑∑

==

⋅∆=njn

iii

nmb

i i

ii

EAlF

11

2

21

21 P

Two load cases are applied simultaneously

∑∑∑∑====

⋅∆+⋅∆=→+⋅∆+∆=+ njn

iiiii

nmb

i i

iinjn

iiiii

nmb

i i

ii

EAlFF

EAlFF

1111

2

)(21)()(

21)(

21 PPPP

External work for sequential loading (P first)

∑∑∑∑

∑∑∑∑

====

====

⋅∆=⋅∆→⋅∆=⋅∆+⋅∆

∆+⋅∆+⋅∆=+⋅∆+∆

njn

iii

njn

iii

njn

iii

njn

iiiii

njn

iii

njn

iii

njn

iii

njn

iiiii

1111

1111

)(21

21

21)()(

21

PPPPP

PPPPP

∑∑∑===

⋅∆=⋅∆=njn

iii

njn

iii

nmb

i i

ii

EAlFF

111PP

∑∑∑===

δ=δ=njn

iii

njn

iii

nmb

i i

ii lFlFEA

lFF111


Pro

70

3.4 Frame Problems

∑ ∫ ∫∫ ++=∆e

l ll

s

e ee

dxEA

FFdxGA

VVfdxEIMM

0 00

)(

where ∆ is the displacement in the direction of applied unit concentrate load in the virtualsystem.

Moment Shear Axial

+=δ ∫∫

2/

0

2 ll

M dxdxEI

l

l

HA HB=P/4

RA=P/2 RB=P/2

P

-

Pl/4

+

P/2

-

-+

P/4

- -

-P/2

P/4

×

Pl/4

×

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

0

EIPllPlllPll

EIM 16

3)

446443(2

=××+××=δ

l/4 Pl/4 l/4



71

+=δ ∫∫

2/

00

2 ll

sS dxdx

GAf

EAPlfPlPl

GAf

dxGA

VVf ss

V

sS 8

3)21

2241

4(

2=××+××==δ ∫

+= ∫∫

2/

00

2 ll

sA dxdx

EAδ

EAPlPlPl

EAdx

EAAA

VA 16

9)41

4221

2(2

=××+××==δ ∫

))(75.0)(56.11(16

)961(16

223

22

3

lh

lh

EIPl

EAlEI

GAlEIf

EIPl s

ASM

++=

++=δ+δ+δ=δ

In most cases, the deformation caused by the shear force and the axial force negligiblysmall compared to that caused by the bending moment. If this is the case, thedisplacement of a frame can be approximated by considering only the bending moment.

∑∫=∆e

le

dxEIMM

0

×

P/4 1/4

×

P/2 1/2

×

P/4 1/4

×

P/2 1/2



72




73

Chapter 4

Analysis of StaticallyIndeterminate Beams

RxθL

xθ

1



74

4.1 Propped Cantilever Beam

Equilibrium equation

02

0

=−+

=+−−

lRlqlM

qlRR

BA

BA

4.1.1. The first idea

=

+

Compatibility condition

00 =δ+δ xBR

– The end displacements of the cantilever beam for two loads cases are calculated by

the principle of virtual work.

EIqllqll

EIdx

EIdx

EIMM ll

8))(

2(

41 1 42

000 =−−===δ ∫∫

EIllll

EIdx

EIdx

EIMM ll

x 3))((

31 1 3

00

=−−===δ ∫∫

q

RA RB

MA

δ0

RBa

xBR δ

-ql2/2

1

-l



75

Compatibility condition and the final solution

qlREIlR

EIql

BB 830

38

34

−=→=+ (up)

2

81 ,

85 qlMqlR AA −==

Moment Diagram

Deflected shape

4.1.2. The second idea

=

+


00 =θ+θ xAM

-ql2/2

3ql2/8

-ql2/8 -

2

1289 ql

3l/8

+=

θ0

MA

xAM θ



76

– Rotional Angle at the fixed end

EI

qlqllEI

dxEI

dxEIMM ll

241

831 1 32

000 =×===θ ∫∫

EIll

EIdx

EIdx

EIMM ll

x 311

31 1

00

=××===θ ∫∫


23

810

324qlM

EIlM

EIql

AA −=→=+

Other reactions by a free body diagram

=

ql2/8 1

1

ql/2 ql/2 ql/8

ql2/8

ql/8

+

3ql/8 5ql/8

ql2/8



77

4.2 Cantilever Beam with Spring Support

Robin BC (The third type BC)

)()( lkwlwEIV −=′′′−=

Primary structure

Compatibity Condition

wbeam(l)=δspring → spring0 δ−=δ+δ xBR

EIklklql

kEIl

EIql

Rk

REIlR

EIql

BB

B 383

13

838 3

3

3

4

34

+−=

+−=→−=+

As qlRk B 83, →∞→ , and As 0,0 →→ BRk

Deflected Shape for 3100lEIk =

δ0

Ra

xBR δ



78

4.3 Support Settlement

Primary structure


∆=→∆=δ+δ 303lEIRR BxB

Deflected Shape

∆

RB

xBR δ



79

4.4 Temperture Change

Primary structure

Curvature due to temperature change

dxTTdxTThd )()( 0102 −α−−α=θ

2

212 )(

dxwd

hTT

dxd

−=−α

=θ

baxxh

TTw ++

−α−= 212

2)(

For simple beam, 0)()0( == lww → lh

TTa2

)( 12 −α=

Comaptibility condition

00 =θ+θ xAM → 032

)( 12 =+−α

EIlMl

hTT

A → EIh

TTM A)(

23 12 −α

−=

T1

T2

θ0

MA

xAM θ



80

4.5 Shear Effect

Primary Structures – Shear force diagram


0)()( 000 =δ+δ+δ+δ=δ+δ Sx

MxB

SMxB RR

GAfqlqll

GAfdx

GAfdx

GAVVf

llS

2)1)((

2

2

000 ====δ ∫∫

GAfll

GAfdx

GAfdx

GAVVf

llSx ====δ ∫∫ )1)(1(

00

2

2

2

2

3

24

)/(56.11)/(1.21

83

31

41

83

3

28lhlhql

GAlfEI

GAlfEI

ql

GAfl

EIql

GAfql

EIql

RB ++

−=+

+−=

+

+−=

– For retangular section

22

2

3

2 )(52.0)(125

6.212

)1(2

1256

lh

lh

hblv

E

bhE

GAlfEI

=××

=

+

=

For 201

=lh

830014.1

039.010053.01

83

)/(56.11)/(1.21

83

2

2 qlqllhlhqlRB −=

++

−=++

−= (0.14 % error)

1

ql 1

+ +



81

For 101

=lh

830053.1

0156.01021.01

83

)/(56.11)/(1.21

83

2

2 qlqllhlhqlRB −=

++

−=++

−= (0.53 % error)

For 51

=lh

8302.1

0624.01084.01

83

)/(56.11)/(1.21

83

2

2 qlqllhlhqlRB −=

++

−=++

−= (2.0 % error)

You may neglect the effect of the shear deformation in most cases !!

4.6 2-Span Continuous Beam

Primary structure

Compatibility

)( 00RxB

RLxB

L MM θ+θ−=θ+θ → 0)(00 =θ+θ+θ+θ Rx

LxB

RL M

EI EIq

ql

RxθL

xθ

1

4

2ql8

2ql

q

ql

R0θL

0θ



82

EIqlqlql

EIqllqll

EI

dxEI

dxEI

dxEIMM

ll

lRL

3332200

2

000

485)

2416(1)1

831

4)

211(

6(1

1 1

=+=××+×+=

+=

=θ+θ

∫∫

∫

EIldx

EIdx

EI

dxEIMM

ll

lRx

Lx

32 1 1

00

2

0

=+=

=θ+θ

∫∫

∫

200

325 qlM R

xLx

RL

B −=++

−=θθθθ

Deflected shape

1



83

4.7 Fixed-Fixed End Beam

4.7.1. Primary Structure type I

EIqlqlll

EIdx

EIMMl

8)

2()(

41 42

0

0110 =−×−×==δ ∫

EIqlqll

EIdx

EIMMl

6)

2(1

31 32

0

0220 −=−××==δ ∫

EIllll

EIdx

EIMMl

3)()(

31 3

0

1111 =−×−×==δ ∫

EIlll

EIdx

EIMMl

21)(

21 2

0

212112 −=×−==δ=δ ∫

EIll

EIdx

EIMMl

=××==δ ∫ 111

0

2222

q

RARB

MA MB

-ql2/2

M0

1

-l

M1

11

M2


StrProf. Hae Sung Lee, http://strana.snu.ac.k

84

Compatibility condition (Flexibility equation)

=+−−

=−+→

=δ+δ+δ=δ+δ+δ

026

0238

00

21

23

2

2

1

34

22212120

21211110

XEIlX

EIl

EIql

XEIlX

EIl

EIql

XXXX

21qlX −= ,

12

2

2qlX −=

4.7.2. Primary Structure type II

EIqlqll

EIdx

EIMMl

2481

31 42

0

0110 =××==δ ∫

EIqlqll

EIdx

EIMMl

2481

31 32

0

0220 =××==δ ∫

EIll

EIdx

EIMMl

3)1()1(

31

0

1111 =−×−×==δ ∫

EIll

EIdx

EIMMl

611

61

0

212112 −=××==δ=δ ∫

EIll

EIdx

EIMMl

311

31

0

2222 =××==δ ∫

M0

M2

1

M1

1

18

2ql

1

uctural Analysis Lab.r



85


=++

=++→

=δ+δ+δ=δ+δ+δ

03624

06324

00

21

3

21

3

22212120

21211110

XEIlX

EIl

EIql

XEIlX

EIl

EIql

XXXX

1221qlXX −==

Reactions and Momenr Diagrams

Deflected Shape

24

2ql12

2ql−

2ql

2ql



86

4.8 3-Span Continuous Beam

4.8.1. Uniform load case

Primary structure

)1(248

13

18

13

1

2

1

1

32

2

2

1

2

0

0110 I

IEI

qlqllEI

qllEI

dxEIMMl

+=××+××==δ ∫

)1(248

13

18

13

1

2

1

1

32

1

2

2

2

0

0220 I

IEI

qlqllEI

qllEI

dxEI

MMl

+=××+××==δ ∫

)1(3

113

1113

1

2

1

121

2

0

1111 I

IEIll

EIl

EIdx

EIMMl

+=××+××==δ ∫

220

212112 6

116

1EIll

EIdx

EIMMl

−=××==δ=δ ∫

)1(3

113

1113

1

2

1

112

2

0

2222 I

IEIll

EIl

EIdx

EIMMl

+=××+××==δ ∫

EI1 EI2

q

EI1

M0

1

M1

1

M2



87


=++++

=++++

→

=δ+δ+δ=δ+δ+δ

0)1(36

)1(24

06

)1(3

)1(24

00

22

1

11

22

1

1

3

22

12

1

12

1

1

3

22212120

21211110

XII

EIlX

EIl

II

EIql

XEIlX

II

EIl

II

EIql

XXXX

2

1

2

1

221

5.11

1

81

II

II

qlXX+

+−==

In case 21 II = , 221 10

1 qlXX −==

4.8.2. Complicated Load Case

Primary structure

M0

EI EI

q

EI



88

EIqlqll

EIdx

EIMMl

2481

31 32

0

0110 =××==δ ∫

EIql

EIql

EIqlqll

EIqll

EIdx

EIMMl

485

162441)

211(

61

81

31 333222

0

0220 =+=××++××==δ ∫


=++

=++→

=δ+δ+δ=δ+δ+δ

032

6485

063

224

00

21

3

21

3

22212120

21211110

XEIlX

EIl

EIql

XEIlX

EIl

EIql

XXXX

21 40

1 qlX −= , 22 40

6 qlX −=

Compatibility Condition (Flexibility Equation) in General

∑=

∆=δ+δn

jijiji X

10



89

Chapter 5

Analysis of StaticallyIndeterminate Trusses

1 1

1

21−2

1−

21−



90

5.1 Various Types of Trusses

Determinate Truss

Externally Indeterminate Truss

Internally Indeterminate Truss

Mixed Indeterminate Truss



91

5.2 A Simple Truss

5.2.1 Method - I

+

P

X

P

P

=

Dept. of Civil and Environmental Eng., SNU 92

Primary structure

At L1

075.06.0

25.118.0

1

31

3

==+

==

FFF

FF

P

0

15

F1 F3

①,②,④:0.5A③,⑤:A

P②

①

③

④

⑤

L

0.75L

L1 L2

U1 U2

1

Prof. Hae Sung Lee

0.75

-1.0

, http://str

1.25

an

0

a.snu.a

-0.75

0.75
0.7
Structural Analysis Lab.c.kr


Prof. Hae Sung Lee

93

At L2

At U1

1 , 75.0

08.006.0

21

52

51

−=−==+=+

FFFFFF

1

F4F5

-1.0

, http://st

1

F1

1.25

1.25

rana.snu.a

8.05

4

F

F

F2

F5

-0.75
-0.75
Structural Analysis Lab.c.kr

75.0 , 25.11

06.0

4

5

5

−===

=+

FF

F

1



94

Axial force table for primary structure

Mem 0F xF A L LEA

FF x00 =δ L

EAFx

x

2

=δ

1 0 -0.75 0.5 0.75L 0 LEA5.0

75.0 3

2 -P -1.0 0.5 L EAPL5.0

LEA5.0

1

3 1.25 P 1.25 1.0 1.25L EAPL325.1

EAL325.1

4 -0.75 P -0.75 0.5 0.75L EAPL

5.075.0 3

LEA5.0

75.0 3

5 0 1.25 1.0 1.25L 0 EAL325.1

∑ EAPL79.4

EAL59.7

Compatibility Condition

00 =δ+δ xX PX 63.0−=→

Final Solution

Mem 0F xF xXF xXFF +0

1 0 -0.75 0.47P 0.47P

2 -P -1.0 0.63P -0.37P

3 1.25 P 1.25 -0.79P 0.46P

4 -0.75 P -0.75 0.47P -0.28P

5 0 1.25 -0.79P -0.79P



95

5.2.2 Method - II

P

+

P

X

X

=



96


AEXLX x −=δ+δ0

Primary structure

-1.0PP

1.25P-0.75P

P

0

0.75P 0.75P

-0.8

1.0-0.6

0.8

-0.6

0.8

1

1

X

= +

F0 Fx

AEXL



97


Mem. 0F xF A L LEA

FF x00 =δ L

EAFF xx

x =δ

1 0 -0.6 0.5 0.75L 0 EAL54.0

2 -P -0.8 0.5 L EAPL6.1

EAL28.1

3 1.25 P 1.0 1.0 1.25 L EAPL56.1

EAPL25.1

4 -0.75 P -0.6 0.5 0.75 L EAPL68.0

EAPL54.0

5 - - 1.0 - - -

∑ EAPL84.3

EAL61.3

AEXLX x −=δ+δ0 →

AEXLX

AEL

AEPL 25.161.384.3 −

=+

PPX 79.086.484.3

−==

PXH 63.08.02 −==



98

5.3 A Truss with 1 Roller Support

Primary Structures

1

23

4

5

6

7

8

9

10

3P

P 2P3P

-P

2PP 2P

P3P - 2P- 2P - 22 P

1

1

1

21

−2

1−

21

−

21

−



99


Mem F 0 xFEAL L

EAFF x0

0 =δ LEA

FF xxx =δ

1 P 0 1 0 0

2 P2− 0 2 0 0

3 P 21

− 1 2P

−21

4 P2 21

− 1 22P

−21

5 P2− 1 2 P2− 2

6 - - 2 - -

7 P− 21

− 1 2P

21

8 P3 21

− 1 23P

−21

9 P2 0 1 0 0

10 P22− 0 2 0 0

-5.54P 3.41

Comaptibility condition

LEAXXx 20 −=δ+δ

PXXXP 15.141.141.354.5 =→−=+−

X



100

Temperature change and fabrication error

)22(0 fx LTLEAXX ∆+∆α+−=δ+δ

In case of no external loads

)2(21.0)2(82.4 ff LTL

EAXLTXEAL

∆+∆−=→∆+∆−= αα

5.4 Truss with Two Hinge Supports

Primary structure and compatibility condition

0

2

22212120

121211110

=δ+δ+δ

−=δ+δ+δ

XX

LEAX

XX

3P2P P

X1

1

23

4

5

6

7

8

9

X1

X2

X1

X1



101

– F0

– F1

– F2

1

1

1 11

2P P3P

-P

2P2P P

P3P- 2P- 2P- 22 P

21

−

1 1

1

21

−21

−

21

−


Structural Analysis Lab.http://strana.snu.ac.kr

102

Axial force table for the primary structure

Mem 0F 1F 2FEAL L

EAFF 10 L

EAFF 20 L

EAFF 11 L

EAFF 12 L

EAFF 22

1 P2 0 1 1 0 P2 0 0 1

2 P22− 0 0 2 0 0 0 0 0

3 P322

− 0 1 P2

23− 0 2

10 0

4 P2 22

− 1 1 P2

22− P2 2

122

− 1

5 P2− 1 0 2 P2− 0 2 0 0

6 P− 22

− 0 1 P22

0 21

0 0

7 P 22

− 0 1 P22

− 0 21

0 0

8 P 0 1 1 0 P 0 0 1

9 P2− 0 0 2 0 0 0 0 0

∑ -5.54P 5P 3.41 -0.71 3

0371.0541.171.041.354.5

21

121

=+−−=−+−

XXPXXXP

→ PXX

PXX5371.0

54.571.082.4

21

21

−=+−=−

→ PX

PX44.1

94.0

2

1

−==



103

Chapter 6

Analysis of StaticallyIndeterminate Frames

EI,l

w



104

6.1 Γ-shaped Frame-I

Equilibrium equation

02

0

=−−

=++

lRPlM

PRR

CA

CA

6.1.1 Primary Structure type I

+

δ0

P 1

Rcδx

P

RA

MA

RCl

l



105


0δ δ0 =+ xCR

End Displacements

333

2

0000

4829)

485

2(1)}

2)(2

2(

61

2))(

2({1

1 1

PlEI

PlPlEI

PlllllPllEI

dxEI

dxEI

dxEIMM

lll

=+=−−−×+−−=

+==δ ∫∫∫

333

000

34)

31(1)})((

3))(({1

1 1

lEI

llEI

llllllEI

dxEI

dxEI

dxEIMM lll

x

=+=−−+−−=

+==δ ∫∫∫


03

448

29 33 =+ lREI

PlEI C

PPRC 45.06429

−=−= , PPRA 55.06435

−=−= , PlM A 643

−=

P

-

-

l

1

l

-

Pl/2

P

Pl/2

- 1



106

Moment Diagram

+

=

Deflection Shape

Pl/2

P

Pl/2

-

-

P

0.45P

+

0.45Pl

0.45P

0.45Pl

+

0.225Pl

+

-

-0.05Pl



107



00 =θ+θ xBM

Rotation Angle

EIPlPll

EIdx

EIdx

EIMM ll

161

423

61 1 2

000 =××××===δ ∫∫

EIlll

EIdx

EIdx

EIMM ll

x 34)

3(1) (1

0

22

0

=+=+==δ ∫∫

P

0θ

P

P/2

P/2

Pl/4

+

1

+

+ 1/l

1/l

MB

xBM θ



108


034

16

2

=+ BMEIl

EIPl →

643PlM B −=

6.2 Γ-shaped Frame-II

6.2.1 Primary Structure type I

w

RA

MA

RB

EI,l

HB

EI,l

-wl2/2

M0 -

l

M1

1

+

+

M2

1

l

+



109

EIwllwll

EIdx

EIdx

EIMM ll

6))(

2(

31 1 42

00

0110 −=−===δ ∫∫

EIwllwll

EIdx

EIdx

EIMM ll

8))(

2(

41 1 42

00

0220 −=−===δ ∫∫

EIlllllll

EIdx

EIdx

EIMM ll

34)})((

3))(({1) (1 3

0

22

0

1111 =+=+==δ ∫∫

EIllll

EIdx

EIdx

EIMM ll

2))((

21 1 3

00

212112 ====δ=δ ∫∫

EIllll

EIdx

EIdx

EIMM ll

3))((

31 1 3

0

2

0

2222 ====δ ∫∫

Compatibility condition (Flexibility Equation)

=++−

=++−→

=δ+δ+δ=δ+δ+δ

0328

023

46

00

2

3

1

34

2

3

1

34

22212120

21211110

XEIlX

EIl

EIwl

XEIlX

EIl

EIwl

XXXX

281wlX −= ,

73

2wlX =

Reactions

w

wl/28

4wl/7

3wl2/28

wl/28

3wl/7



110

Moment Diagram

Deflected Shape


-3wl2/28

11wl2/196

-wl2/28

M0

wl2/8



111

EIwldx

EIdx

EIMM ll

24 1 3

00

012010 ===δ=δ ∫∫

EIll

EIdx

EIdx

EIMM ll

3)1)(1(

31 1

0

2

0

1111 ====δ ∫∫

EIldx

EIdx

EIMM ll

6 1

00

212112 ===δ=δ ∫∫

EIldx

EIdx

EIMM ll

32) (1

0

22

0

2222 =+==δ ∫∫


=++

=++→

=δ+δ+δ=δ+δ+δ

032

624

06324

00

21

3

21

3

22212120

21211110

XEIlX

EIl

EIwl

XEIlX

EIl

EIwl

XXXX

28

2

1wlX −= ,

283 2

2wlX −=

M1

1

+

M2

1

+

+


Prof. Hae Sung

112

6.3 Portal Frame subject to Horizontal Load

6.3.1. Primary Structure type I

w

M0

-wl2/2

-

EI,l

w

M2

l

+

+

M3

1

1

+

+

+

M1

-l

--

-

1

1

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



113

EIwldx

EIdx

EIMM ll

24 1 4

00

0110 ===δ ∫∫

EIwldx

EIdx

EIMM ll

6 1 4

00

0220 −===δ ∫∫

EIwldx

EIdx

EIMM ll

6 1 3

00

0330 −===δ ∫∫

EIldx

EIdx

EIMM ll

35) (21 3

0

22

0

1111 =+×==δ ∫∫

EIldx

EIdx

EIMM ll 3

00

212112 21

−=×==δ=δ ∫∫

EIldx

EIdx

EIMM ll 2

0

2

0

313113

2) (21−=+×==δ=δ ∫∫

EIldx

EIdx

EIMM ll

34) (1 3

0

22

0

2222 =+==δ ∫∫

EIldx

EIdx

EIMM ll

23 ) (1 2

0

2

0

323223 =+==δ=δ ∫∫

EIldx

EIdx

EIMM ll 3 3 1

0

2

0

3333 =×==δ ∫∫


→

=δ+δ+δ+δ=δ+δ+δ+δ

=δ+δ+δ+δ

00

0

33323213130

32322212120

31321211110

XXXXXX

XXX

=++−−

=++−−

=−−+

03232

6

023

34

6

0235

24

32

2

1

23

3

2

2

3

1

34

3

2

2

3

1

34

XEI

lXEIlX

EIl

EIwl

XEIlX

EIlX

EIl

EIwl

XEIlX

EIlX

EIl

EIwl

Matrix form

−

−−=

−

−

−−

EIwlEI

wlEI

wl

XXX

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

6

6

24

3232

23

34

235

3

4

4

2

2

1

22

233

233

→



114

=

−

−

−

−

−−

−=

−

25231

7

24

6

6

24

3232

23

34

235

23

4

41

22

233

233

2

2

1

wl

wl

wl

EIwlEI

wlEI

wl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

XXX

Reactions

Moment Diagram

Deflected Shape


w

5wl/2419wl/24

wl /7 wl /7

59wl2/252 31wl2/252


Prof. Hae Sung

115

EIdx

EIMM ll

1

00

0110 ==δ ∫∫

EIdx

EIMM ll

1

00

0220 ==δ ∫∫

EIdx

EIMM ll

1

00

0330 ==δ ∫∫

EIdx

EIMM ll

(1

0

2

0

1111 ==δ ∫∫

00

2112 (1

==δ ∫∫ EIdx

EIMM ll

1+

+ M1

+

+

-

1

M2

M0w

-wl2/2

1

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

EIwldx6

3

−=

EIwldx8

3

=

EIwldx8

3

−=

EIldx

34) 2 =+

213) δ=−=+

EIldx

++

1

1

M3



116

3100

3113 2

1δ====δ ∫∫ EI

ldxEI

dxEIMM ll

EIldx

EIdx

EIMM ll

=×==δ ∫∫0

2

0

2222 31

3200

3223 6

) (1δ=−=+==δ ∫∫ EI

ldxEI

dxEI

MM ll

EIldx

EIdx

EIMM ll

32 2 1

0

2

0

3333 =×==δ ∫∫


00

0

33323213130

32322212120

31321211110

=δ+δ+δ+δ=δ+δ+δ+δ

=δ+δ+δ+δ

XXXXXX

XXX

Matrix Form

−

−

−=

−

−−

−

EIwlEI

wlEI

wl

XXX

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

8

8

6

32

62

63

2334

3

3

3

3

2

1

−=

−

−

−

−

−=

−

−

−

−−

−

−=

−

25231

50443

50429

8

8

6

2144

212

2116

212

2123

215

2116

215

2123

8

8

6

32

62

63

2334

2

3

3

3

3

3

31

3

2

1

wl

EIwlEI

wlEI

wl

lEI

EIwlEI

wlEI

wl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

EIl

XXX


Prof. Hae Sung L

117

6.4 Portal Frame subject to Vertical Load

1+

+ M1

+

+

-

1

M2

EI,l

a P

M0

4Pab

1

Structural Analysis Lab.ee, http://strana.snu.ac.kr

++

1

1

M3


Prof. Hae Sung Lee

118

lba

EIPab

lPab

lbldx

EIdx

EIMM ll 2

61)1(

6 1

00

0110

+=××+===δ ∫∫

lba

EIPab

lPab

laldx

EIdx

EIMM ll +

=××+===δ ∫∫2

61)1(

6 1

00

0220

00

0330 ==δ ∫

l

dxEIMM

Matrix Form

−−

+

+

−=

+

+

−

−

−=

ba

ba

ba

lPab

lba

lba

lPab

XXX

104

1117

1711

42

0

2

2

2144

212

2116

212

2123

215

2116

215

2123

6 2

3

2

1

Sidesway : )(281 ab

EIPab

−=∆

Deflected Shape

l

Structural Analysis Lab., http://strana.snu.ac.kr

4a =



119

6.5 Order of Indeteminancy

# of unknowns

# of memebr × # of internal force per member +

# of reactions - # of known quantities

# of equations

# of memebr × # of E.E. per member +

# of joints × # of E.E. per joint - # of used equations

# of Indeterminancy = # of unknowns - # of equations


Str ctural Analysis Lab.Prof. Hae Sung Lee, http://strana.snu.ac

120

6.6 General Frame

Primary Structure

M1, M2, M3

EI

L

L/2

2EI

q qL

EI

L L

EI

ql2/84

2ql

M0

M1 M2

M3

u
.kr


Pr

121

dxEIMMl

30

0110 ==δ ∫

ldxEI

MMl

60

0220 ==δ ∫

00

0330 ==δ ∫

l

dxEIMM

Edx

EIMMl

30

1111 ==δ ∫

dxEI

MMl

30

2222 ==δ ∫

dxEIMMl

620

3333 ==δ ∫

ql2/8 4

2ql

+1

M0 M1

M2

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

EIqlql

EIl

2481

32

=××

EIqlql

164)1()

211(

32

−=×−×+

Il , 01312 =δ=δ

EIl , 02321 =δ=δ

EIl , 03231 =δ=δ

-1

+1

M3



122


321 δ=δ=δ

1

3

313212111101 324M

EIl

EIqlMMM +=δ+δ+δ+δ=δ

2

3

323222121202 316M

EIl

EIqlMMM +−=δ+δ+δ+δ=δ

3333232131303 6M

EIlMMM =δ+δ+δ+δ=δ

One Additional Equilibrium Equation

0321 =++ MMM → 213 MMM −−=

Final Compatibility Condition

062246324 21

3

31

3

=++=−+ MEIlM

EIl

EIqlM

EIlM

EIl

EIql

026166316 21

3

32

3

=++−=−+− MEIlM

EIl

EIqlM

EIlM

EIl

EIql

21 64

9 qlEI

M −= , 22 64

11 qlEI

M = , 23 64

2 qlEI

M −=

In case n members are connected to a joint, and a hinge is used to release moment at the

joint you, have n-1 compatiblity equations and one equilibrium equation, which leads to

total of n-1 compatibitlity equations with n-1 unknowns.



123

6.7 General Joint Compatibility


nδ==δ=δ L21

j

k

njijj

n

jijii MM ∑∑

+==

δ+δ+δ=δ11

0 ni L1for =

One Additional Equilibrium Equation

021 =+++ nMMM L

Joint i

M1

Mn



124




125

Chapter 7

Influence Lines for

Determinate Structures



126

7.1 Influence Function

Influence function

Convolution integral – Superposition

ξξξ dqxIxdR pp )(),()( =

∫=l

pp dqxIxR0

)(),()( ξξξ

Dirac delta functions

0lim→ε

= = )( ξ−δ x

1221lim)0

210(lim)(

00

00

=εε

=+ε

+=ξ−δ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε ∫∫∫∫ll

dxdxdxdxx

)()(2

)()(lim)(21lim

)0)(21)(0)((lim)()(

00

00

0

ξ=ξ′=ε

ε−ξ−ε+ξ=

ε=

+ε

+=ξ−δ

→ε

ε+ξ

ε−ξ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε

∫

∫∫∫∫

fFFFdxxf

dxxfdxxfdxxfdxxxfll

ξ

∞

ξ

ε21

2ε

1ξ

I(xp,ξ)

xp

ξ q(x)

R(xp)

q(ξ)dξ



127

Concentrated loads of intensity P at )( ξ−δ=ξ xP

Responses by several concentrated loads

∑

∑∫∫ ∑∫

=

==

ξ=

ξξ−ξδξ=ξξ−ξδξ=ξξξ=

n

iipi

n

i

l

iip

l n

iiip

l

pp

xIP

dPxIdPxIdqxIxR

1

1 00 10

),(

)(),()(),()(),()(

7.2 Influence Line for Simple Beams

7.2.1 Moment

2

0 l≤ξ≤

2

0)2

(12

ξ=→=ξ−×−−

ξ−xx MlMl

ll

P1ξ1

I(xp,ξ)

ξn

Pn

RA=(l - ξ)/l

ξ

RB=ξ/l

P = 1　

RA=(l - ξ)/l

ξ P = 1

Mx



128

ll≤ξ≤

2

2

02

ξξ −=→=−

− lMMll

lxx

Influence line

7.2.2. Shear Force

2

0 l≤ξ≤

lVV

ll

xxξ

−=→=++ξ−

− 01

RA=(l - ξ)/l

ξ P = 1　

Vx

RA=(l - ξ)/l

ξ

RB=ξ/l

P = 1　

+L/4

RA=(l - ξ)/l

Mx



129

ll≤ξ≤

2

llVV

ll

xxξξ −

=→=+−

− 0

Influence line

7.2.3 Maximum Moment in a Simple Beam

010

≤ξ≤−L

20)()

10(

21)( max

PLxMLPxM pp =→+= ξ

104

1020 LLL

=−≤ξ≤

104at

207)(

2043)

10(

21

22)( max

LPLxMPLPLPPxM pp =ξ=→+ξ

=+ξ+ξ

=

RA=(l - ξ)/l

Vx

+

1/2

+

P/2ξ

P

L/10

2)2/( ξ

=LIM22

)2/( ξ−=

LLI M



130

105

104 LL

≤ξ≤

104at

207)(

209

4))

10(

21

2(

22)( max

LPLxMPLPLLPPxM pp =ξ=→+ξ

−=+ξ−+ξ

=

109

105 LL

≤ξ≤

105at

4013)(

2014

43))

10(

21

2()

22(

2)( max

LPLxMPLPLLPLPxM pp =ξ=→+ξ

−=+ξ−+ξ

−=

7.3 Influence Line of a Gerber Beam

7.3.1. Shear Force at x= L/2

2

0 L≤ξ≤

LL≤ξ≤

2

LL 5.1≤ξ≤

Influence line

ξ′

Lξ ′2

Lξ′

−21

Lξ′

−21

L/2L

RA=1

ξP=1

Vx=0

RA=1

Vx=1

+

1



131

7.3.2. Moment at the fixed end

L≤ξ≤0

LL 5.1≤ξ≤

Influence line

7.3.1 Maximum Moment in the Gerber Beam

∫∫++

==4/4/

)()(L

M

L

M dxxIqdxxqIMξ

ξ

ξ

ξ

– L430 ≤ξ≤

)4

2(84

)4

(21 LqLLLqM +ξ−=+ξ+ξ×−= → 22

max 2188.0327 qLqLM −=−=

ξξ−=xM

)2( ξ′−−= LM x

ξ′

Lξ′2

Lξ′

−21

-

L)23()2( ξ−−=ξ′−− LL−ξ

-L

L/4



132

– LL ≤ξ≤43

)358

13(2

)27

232

821(

2

)43)(2

25(

2))((

2222222 ξ−ξ+−−=ξ+ξ+ξ−−ξ−−=

−ξξ−+−ξ−ξ+−=

LLqLLLLq

LLLqLLqM

LLqM650)65(

2=ξ→=ξ−−=′

2

2222max

2292.014433

72150300117

2))

65(3

655

813(

2qL

qLqLqLLLLqM

−=

−=−+−

−=−+−−=

– LL 25.1≤ξ≤

22max 1875.0

163)4

23(

84)2

22(

2qLqLMLqLLLLqM −=−=→ξ′−−=ξ′−+ξ′−−=

– For 025.0 ≤ξ≤− L or LL 5.125.1 ≤ξ≤

The maximum moment should be smaller than any of the above cases. Therefore,the maximum moment is

2max 2292.0 qLM −= at L

65

=ξ

)2( ξ′−− L )22

( ξ′−−L

)225( ξ−− L−ξ

ξ−L LLL43)

4( −ξ=−+ξ



133

7.4 Indirect Load

Equivalent to

10 =+→=∑ bav RRF

0)22

()22

)(1(0 0

0

0

0

=−++−−→=∑ lRlllxll

lxM ba

lx

lllRb +

−=

20 ,

lx

lll

RR ba −+

=−=2

1 0

In case the unit load is applied directly on the simple beam,

10 =+→=∑ bav RRF

0)22

(0 0 =−+−→=∑ lRxllM ba

Ra Rb

l0

1-x/l0 x/l0

lRa Rb

x

l0

l/2

P =1

l

RaRb

x(l-l0)/2

l/2

P =1



134

lx

lllRb +

−=

20

,

lx

lllRR ba −

+=−=

21 0 (Statically equivalent to the indirect load)

0)2

)(2

()1(0 00

0

=ξ+−

−+

+ξ−−−→=∑ ξ

lllx

lll

lxMM

ξ−−

+−

−+

=ξ+−

−+

+ξ−−= )12(22

)2

()2

)(2

()1(0

00000

0 lx

lllll

lx

lllll

lx

lll

lxM

7.4.1 Influence line at the mid-span

422222)12(

22)

2( 000000

0

000 llllllll

llll

lx

lllll

lx

lll

M−

=−

−−+

=−−

+−

−+

=

7.4.2 Truss Case

Ra Rb

1-x/l0 x/l0

ξ

Ra Rb

1-x/l0 x/l0



135

Raax

61−=

Rbax

6=

x

7.5 Influence Line of Truss

7.5.1 Diagonal Member

ax ≤≤0

axF

axF

6201

61

22

−=→=−−+−

axa 2≤≤

)651(202

61

22

axF

axa

axF −−=→=

−−−+−

axa 62 ≤≤

)6

1(206

122

axF

axF −=→=−+−

x a

xaa

ax −=

−−

21



136

Influnece line for the diagonal member

axFax

62 0 −=≤≤

)651(2 2

axFaxa −−=≤≤

)6

1(2 62axFaxa −=≤≤

7.5.2. Bottom Member

Take moment about point A (clockewise +).

ax ≤≤0

axFFaxaa

ax

650)(1)

61( =→=−−×−×−

axa 6≤≤

)6

1(0)6

1(axFFaa

ax

−=→=−×−

+

-

264

261

axRa 6

1−=

A

65

+



141

Chapter 8

Influence Lines forIndeterminate Beams



142

8.1 Influence Lines at Supports

8.1.1. Reaction Force

By the Flexibility Method

– Compatibility Condition

bb

bbbbbb d

dRddR ξ

ξ −=→=+× 0

– Betti-Maxwell’s Reciprocal Theorem

ξξξξ ==→−δ=ξ−δ ∫∫ bLb

L

x

L

xb ddddxdLxdxdx2

0

2

0

)()(

– Influence Line : bb

b

bb

bb d

ddd

R ξξ −=−=

Moment Diagram

EIL

EILLL

EILMM

EILdbb 48

)2(6223

2322

33

==×=×=

L L

ξ

Rb = ??

P = 1

ξ P = 1

dxξ

dbξ

P = 1

dxb

dbb

P = 1P(2L)/4= L /2

1/21/2



143

Calculation of Deflection

baxxEI

wxMwEI ++−=→−=−=′′12

121 3

– Boundary conditions

EILaaL

EILw

bw

40

410)(

00)0(22

=→=+−→=′

=→=

– Deflection of the Beam

)3(12

1 23 xLxEI

wdxb +−==

Influence Line

)](3)[(21

6/)3(

121 3

323

Lx

Lx

EILxLx

EIdd

Rbb

bb −=+−−=−= ξ

8.1.2. Moment

By the force method

1

ξ P = 1

θbξ

dxξ , θxξ

M = 1dxb

θb

L

EI

L L

ξ P = 1

EI EI



144

– Compatibility Condition

bb

bbbbbb MM

θθ

−=→=θ+θ× ξξ 0


ξξξξ θ=θ=→θ−δ=ξ−δ ∫∫ bLb

L

x

L

xb ddxLxdxdx3

0

3

0

)()(


b

bb

bb

dM

θ−=

θθ

−= ξξ


i) Left span

baxLEIxw

LxMwEI LL ++−=→−=−=′′

6

3


EILaaL

EILLw

bw

L

L

60

60)(

00)0(2

=→=+−→=

=→=

– Deflection of the left span

)(6

1 23 xLxLEI

wd Lxb +−==

EILL

bb 3−=θ (counterclockwise)

ii) Analysis of Center and Right Span

M = 1dxb

θbb

M = 1dxb

M = 1

x = 1



145

EIL

cc 32

=θ , EIL

cb 6=θ

– Compatibility condition: 410 −=

θθ

−=→=θ+θcc

cbcccccb MM

– Moment Diagram

iii) Deflection of Center span

baxxxLEI

wxL

MwEI cc ++−=→+−−=−=′′ )224

5(1)145(

23

EILaaLLL

EILw

bw

c

c

2470)

2245(10)(

00)0(22

=→=+−→=

=→=

)7125(24

1 223 xLLxxLEI

wcxb +−==δ

EILwc

Rbb 24

7)0( =′=θ (Clockwise)

EIL

EIL

EILR

bbLbbbb 8

5247

3=+=θ+θ=θ

iv) Deflection of Right Span

baxxL

xEI

wLxMwEI RR +++−=→−−=−=′′ )

824(1)

41

4(

23

EILaaLLL

EILw

bw

R

R

120)

824(10)(

00)0(22

−=→=++−→=

=→=

)23(24

1 223 xLLxxLEI

wRxb +−−==δ

Final Influence Line

i) left span :

)(15

485/)(

61 23

223 xLx

LEILxLx

LEIwM

bb

Lb −=+−−=

θ−=



146

LLxLxL

Mb 577.0310)3(

154 22

2 ==→=−=′

LLLMb 103.0)1577.0)(1577.0(577.0154)577.0( −=−−=

ii) Center Span :

))(75(15

)7125(15

185/)7125(

241

2223

2

223

LxLxLxxLLxx

L

EILxLLxx

LEIwM

bb

Cb

−−−=+−−=

+−−=θ

−=

LLxLxLxL

Mb 384.015

39120)72415(15

1 222 =

−=→=+−−=′

LLLMb 080.0)1384.0)(7384.05(15384.0)384.0( −=−−×−=

iii) Right Span:

))(2(15

1)23(15

185/)23(

241

2223

2

223

LxLxxL

xLLxxL

EILxLLxx

LEIwM

bb

Rb

−−=+−=

+−=θ

−=

LxLxLxL

Mb 423.03

330)263(15

1 222 =

−=→=+−=′

LLMb 026.0)1423.0)(2423.0(15423.0

=−−=

0.577L 0.384L 0.423L

-0.103L-0.080L

-0.026L



147

8.2. Inflence Lines in Members

8.2.1. Moment

By the Flexibility Method

– Compatibility Condition:bb

bbbbbb MM

θθ

−=→=θ+θ× ξξ 0


ξξ

ξξ θ=θ=→θ−δ=ξ−δ ∫∫ bLb

L

x

L

xb ddxLxdxdx2

2

0

2

0

)2/()(


b

bb

bb

dM

θ−=

θθ

−= ξξ


i) Moment Diagram

EIL

bb 38

=θ

L L

ξ Mb = ?? P = 1

ξ P = 1

dxξ , θxξ θbξ

dxb

θbb

M = 1

1

2



148

ii) Suspended span

baxLEIxw

LxMwEI SS ++−=→−=−=′′

32 3


??224

)0()2

(

00)0(2

=+−→=

=→=

LaEI

LwLw

bw

OS

S

– Deflection of the suspended span

axLEIxwS +−=

3

3

iii) Overhanged span

ecxxL

xEI

wLxMwEI OO +++−=→+−=−=′′ )

23(1)21(

23


026

0)2

(

224)

2()0(

2

2

=++−→=

+−=→=

eLcEILLw

LaEI

LeLww

O

SO

iv) Right span

gfxxL

xEI

wLxMwEI RR +++−−=→−−=−=′′ )

3(1)22( 2

3


EILffLLL

EILw

gw

R

R

320)

3(10)(

00)0(

22

=→=++−→=

=→=

– Deflection

)23(3

1 223 xLLxxLEI

wR +−=

v) Determination of a, c, e

EILc

EILcLL

EIEILL

RO 1217

32)

24(1

32)0()

2( =→=++−→=θ=θ

EILaLa

EIL

EIL

EILee

EIL

EILeLc

EIL

−=→+−=−

−=→=++−→=++−

2242413

24130

2417

60

2622

2222



149

vi) Deflection of the left span

– Suspended span: )3(3

13

233

xLxLEI

axLEIxwS +−=+−=

– Overhanged span: )1334128(24

1 3223 LxLLxxLEI

wO +−+−=


i) Suspended span

)3(81

38/)3(

31 23

223 xLx

LEILxLx

LEIwM

bb

Sb +=+=

θ−=

LLLLL

LMb 203.06413)

23)

2((

81)

2(

33

2 ==+=

ii) Overhanged span

)1334128(64

1 32232 LxLLxx

LwM

bb

Ob +−+=

θ−=

iii) Right span

)23(81 223

2 xLLxxL

wMbb

Rb +−−=

θ−=

LxLLxxL

Mb 423.00)263(81 22

2 =→=+−−=′ , LLMb 048.0)423.0( −=

0.203L

-0.048L

0.423L



150

8.2.2. Influence Line of Shear Force using the Influence Line of Moment

i) 2L

≤ξ

)5(412)3(

4122

012

233

233 xLx

LLxxLx

LLx

LM

VMxLV bbbb −=−+=−=→=−×+×

ii) LL≤ξ≤

2 (Overhanged span)

)1334128(32

120

23223

3 LxLLxxLL

MVMLV b

bbb +−+==→=−×

iii) ξ≤L (Right span)

)23(412

02

2233 xLLxx

LLM

VMLV bbbb +−−==→=−×

EI

L

EI

L

ξ Vb = ?? P = 1

Mb

Vb

χP = 1

Mb

0.203L

-0.048L

0.423L

× =≤− )2

(22 LxLx

L

0.406

-0.096

0.423L

-0.594



151

8.2.3. Influence lineof Shear Force by Müller –Breslau’s Principle

Remove Redunduncy and Apply an Unit Load

bb

xbb d

dV −=

Free Body Digram and Moment Diagram

L

EILdbb 6

4 3

=

L

EI

L

ξ Vb = ?? P = 1

1

1

dxbdbb

L/2

11

1

2 1



152

Deflection of the Beam

i) Suspended span

baxEIxwxMwEI SS ++−=→−=−=′′

6

3


??8

)0()2

(

00)0(2

=+−→θ=θ

=→=

aEILL

bw

OS

S

– Deflection of the suspended span

axEIxwS +−=

6

3

ii) Overhanged span

ecxxLxEI

wxLMwEI OO +++−=→+−=−=′′ )46

(1)2

( 23


0212

0)2

(

8)

2()0(

2

2

=++−→=

+−=→θ=θ

eLcEI

LLw

aEILcL

O

SO

iii) Right span

gfxxLxEI

wxLMwEI RR +++−−=→−−=−=′′ )26

(1)( 23


EILffLLL

EILw

gw

R

R

30)

26(10)(

00)0(233

=→=++−−→=

=→=

– Deflection

)23(6

1 223 xLLxxEI

wR +−=

iv) Determination of a, c, e

EILc

EILcLL

EIEILL

RO 2417

3)

48(1

3)0()

2(

22222

=→=++−→=θ=θ

EILaa

EIL

EIL

EILee

EIL

EILeLc

EIL

65

82417

48130

4817

120

212332

2333

=→+−=

−=→=++−→=++−



153

v) Deflection of the left span

– Suspended span : )5(6

16

233

xLxEI

axEIxwS −−=+−=

– Overhanged span : )1334128(48

1 3223 LxLLxxEI

wO +−+−=


i) Suspended span : )5(41

64/)5(

61 23

3

323 xLx

LEILxLx

EIdwV

bb

Sb −=−=−=

ii) Overhanged span : )1334128(32

1 32233 LxLLxx

LdwV

bb

Ob +−+=−=

iii) Right span : )23(41 223

3 xLLxxLd

wVbb

Rb +−−=−=

0.406

-0.096

0.423L

-0.594

Documents

Structural Analysis Istrana.snu.ac.kr/lecture/struct1/Notes/12_Note.pdf · 2012. 3. 7. · Dept. of Civil and Environmental Eng., SNU Structural Analysis Lab. Prof. Hae Sung Lee,