Structural Analysis I - SNUstrana.snu.ac.kr/lecture/struct1_2015/Notes/15_Note_All.pdf · Dept. of Civil and Environmental Eng., SNU . Structural Analysis I . Spring Semester, 2015

Dept. of Civil and Environmental Eng., SNU

Structural Analysis I

Spring Semester, 2015

Hae Sung Lee

Dept. of Civil and Environmental Engineering Seoul National University

yδ yf

zδ zf

xδ xf

yM yθ

zM zθ xM xθ

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


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





Dept. of Civil and Environmental Eng., SNU 1

Chapter 1

Introduction



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 RRLRLRLqLLqLM

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



1.2 Mechanics of Material

Governing Equation

– Left span

112

13

1

4

1''''

1 24dxcxbxa

EIqxwqEIw ++++=→=

– Right Span

222

23

2

4

2''''

2 24dxcxbxa

EIqxwqEIw ++++=→=

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 13

1

4

11111 2402)0( , 0)0( ++=→==′′==

– Right Support

xcxaEI

qxwbwdw 23

2

4

22222 2402)0( , 0)0( ++=→==′′==

x x

q

w1 w2



– Center Support

==

−==→

+=+

−−−=++

=++=

=++=

EIqLcc

EIqLaa

LaEI

qLLaEI

qL

cLaEI

qLcLaEI

qL

LcLaEI

qLLw

LcLaEI

qLLw

48

483

62

62

36

36

024

)(

024

)(

3

21

21

2

2

1

2

22

2

3

12

1

3

23

2

4

2

13

1

4

1

)32(48

33421 xLLxx

EIqww +−=≡

83 ,

83

2 112

11qLqxwEIVxqLxqwEIM +−=′′′−=+−=′′−=

Moment Diagram Shear Diagram Reactions

0.375qL

L83

+ -

+

0.625qL

-

0.375qL 0.375qL 1.25qL

0.125qL2

0.070qL2

L83

+ -

+



1.3 Mechanics of Material + α

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

δ0+ δR=0

(compatibility condition) 1.3.2 Calculation of δ0

Governing Equation

dcxbxaxEI

qxwqEIw ++++=→= 234

0''''

0 24

q

q

δ0

δR

Rb



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

+=

−=→

=+

=++→

=′′=

EILqc

EILqa

LaEILq

LcLaEILq

LwLw

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

3340 LxLxx

EIqw +−=

EILqLLLLL

EIqLw

384)2(5))2()2(2(

24)(

4334

00 =+−==δ

1.3.3 Calculation of δR

Governing Equation

dcxbxaxwEIw RR +++=→= 23'''' 0

Boundary (support) Conditions

– Left Support

0)0()0( , 0)0( =′′−== RRR wEIMw – Mid-span

2

)()( , 0)()( bRRR

RLwEILVLwL −=′′′−==′=θ


– Left Support

00)0( , 0 0)0( =→=′′=→= bwdw RR Structural Analysis Lab.

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


– Mid-span

)3(12

123

12

26

03

2)(

0)( 23

2

2

xLxEI

Rw

EILR

c

EIR

aR

aEI

caLR

LwEI

Lwb

Rb

b

bbR

R

−=→

−=

=→

=

=+→

=′′′

=′

EIRL

EIRL

Lw bbRR 48

)2(122

)(33

−=−==δ

1.3.4 Final Solution

Reaction at Supports

qLRRR cba 2=++

qLRR cb 22 =+

δ0+ δR=0 → 048

)2(384

)2(5 34

=−EI

RLEILq b → qLRb 8

10=

qLRR ca 83

==

Moment

xqLxqxqLxLxqwEIwEIMMM RR 83

285)2(

222

00 +−=−+−=′′−′′−=+=

Shear

8

38

5)(00qLqxqLLxqwEIwEIVVV RR +−=−+−=′′′−′′′−=+=



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

δ0+ δR=0

Principle of Virtual Work

EILqdx

EIMML

R

384)2(5 42

0

00 −== ∫δ ,

EILRdx

EIMM b

LRR

R 48)2( 32

0

== ∫δ

Solution

δ0+ δR=0 → 048

)2(384

)2(5 34

=+−EI

RLEILq b → qLRb 8

10=

RbL/2

Rb

q

q

qL2/2



– Moment

– Shear

+

-

+

-

+

+ -

+ -

=

0.070qL2

5qL2/8

Rb

+

=

0.125qL2 L83

+ -

+

qL2/2



1.5 지 점 (Supports)

고정단 (fixed support)



회전단 (hinge support)

이동단 (roller support)



1.6 구조물의 2차원 이상화

주 구조물 (Main Structure)

가로 보 (Cross Beam)

세로 보 (Stringer )

Cross Bracing (Wind Bracing)

지 점 (Support)



Truss

Beam

절점(Joint)



Frame



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

yδ yf

zδ zf

xδ xf

yM yθ

zM zθ xM xθ

22 , wV

22 , θM

11 , wV

11 , θM

33 , wV

33 , θM



Plane Frame Idealization

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

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

xδ xf

yδ yf

xδ xf

yδ yf zM

zθ



1.8구조물의 안정 (Stability of Structures)

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






Chapter 2

Reactions & Internal Forces

by Free Body Diagrams



2.1 Free Body Diagram



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 -



2.2 Reactions

Beams

0 0 =−+→=∑ PRRF BAV

00 =−→=∑ LRPaM BA (Clockwise +)

PLaRB = , P

LbRA =

PRRF BAV =+→=∑ 0

0)(0 =−+→=∑ LRaLPM BA (Clockwise +)

PLaRB )1( += , P

LaRA −=

L

P

a b

RA RB

L

P

a

RA RBaL



Truss

PHPHF AAH −=→=+→=∑ 00 ,

00 =−+→=∑ PRRF BAV

0)3(0 =−+→=∑ aRPaPaM BA (Clockwise +)

PRA 31

= , PRB 32

=

Frame

0

02

=+=

=−+=

∑∑

BAH

BAV

HHF

qLRRF

∑ =−−= 02

LHLRM BBRh

∑ =−−= 0422LqLLHLRM AA

Lh BB HR 2−= ,

42 qLHR AA +=

822

42 qLHHqLHqLH BABA =−→=−+

16

16qLH

qLH

B

A

−=

= →

8

83

qLR

qLR

B

A

=

=

P RA RB

P

HA

L

L

HA HB

RA RB

q



2.3 Internal Forces in Framed Structures

Axial Force

Shear Force

Bending Moment

Torsion

+

+

+

+



2.4들보 (Beam)

Reactions

q

RA=qL/2 Rb= qL/2 Structural Analysis Lab.



Free Body Diagram for Shear and Moment

qxqLqxRVVqxRF AxxAV −=−=→=−−=∑ 20

22

02

2qxxqLMMxqxxRM xxAx −=→=−−=∑

Shear Force and Moment Diagrams

q

RA=qL/2 RB= qL/2

RA RB

RA

x

RB

Mx

Vx

+

qL/2 qL2/8



Deflected Shape

2.5 Gerber Systems





2.5.1 Internal Forces in a Gerber Beam - I

Free Body Diagram

PRLPLRM HHC 320

2430 =→=×−×→=∑

PRPRRF CCHv 3100 =→=−+→=∑

PRLRLRM BHBA 650

450 =→=×+×−→=∑

PRPRRF ABAv 610

320 −=→=−+→=∑

L/4 P

RH

RA RB

RC

P

P/6 5P/6 P/3

P



Shear Force i) Lx ≤≤0

ii) LxL23

≤≤

iii) LxL 223

≤≤

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



ii) ,

iii)

Deflected Shape

P/6 5P/6

2P/3

Mx=

P/6 5P/6

P

P/3 P/3

Mx=

+ -

PL/6

PL/6



2.5.2 Internal Forces in a Gerber Beam - II Free Body diagram Shear

Moment

q

2ql

2ql

2ql

2ql

2

2ql

2

2ql

2ql

2ql

+

L

q

L L

8

2ql

2

2ql 2

2ql

+



2.6 트러스 (Truss)



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 the 2nd , 3rd and 4th assumptions

– By the 1st assumption

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

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



2.6.1 Internal Forces in Howe Truss

At U1 and U3

At L1

2 ,

22

022

2

022

23

3

23 PFPFPF

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



At U2

=+−

−==→

=−−−

=++−

021

21

22

022

22

022

22

37

753

8734

PPP

PFF

FFF

FFFF

At L3

Axial Force Diagram

2P

2

P

F5

F4 F8

F3 F7

2P

0

P/2

P/2

P/2 P/2

P 0

0 0

0

P P/2 P/2



Equilvalent Beam Action

Deflected Shape

2.6.2 Internal Forces in Warren Truss

V=P/2

P/2

1 P

2 3

4

5

6

7

8

9

10

11

L1 L2 L3

U1 U2 U3

L4

x

PL/4 M=Px/2



At L1

PFPFPF

FF

32 ,

322

03

222

022

12

2

12

=−=→

=+

=+

At U1

PFPFFFF

FFF

34,2

32

022

22

022

22

423

32

432

−==−=→

=+

=++−

At L2

PFPFFFFF

PFF==→

=+−+

=−+65

5361

35

,32

022

22-

022

22

F2

F1

2P/3

F4

F2 F3

F3

F1 F6

P

F5



At U2

PFPFFF

FFFF

32,

32

022

22

022

22

87

75

7584

−=−=→

=−−

=+−+−

At L3

3,

32

022

22

022

22

109

97610

97 PFPFFFFF

FF==→

=+−−

=+

At U3

PFFF

FFF

32

022

22

022

22

11

119

8119

−=→

=−−

=−+−

F8

F5 F7

F4

F7

F6 F10

F9

F11

F8

F9



At L1

OK

Deflected Shape Equivalent Moment

1 2 3 4 5 6

1.

2.

3.

4.

5.

6.

P/3

P/3

P



2.6.3 Method of Sections

PFLFLPM L 340

232

442 −=→=×+×=∑

PFLFLPLPMU =→=×−×−×=∑ 662 0222

332

PFFPPFV 320

22

32

55 =→=×+−=∑

P

3P

P32

1

2 3

4

5

6

7

8

9

10

11

L1

L2 L3

U1

U2 U3

L4

Cut out

P P

32

F5

F4

F6



2.7 프레임 (Frame)



2.7.1 Internal Forces in a Frame

Reactions

Freebody Diagram

H

qH

L

LqH2

2

q

LqH2

2

2

2qH

qH

LqH2

2

LqH2

2

2

2qH

LqH2

2

LqH2

2

LqH2

2

LqH2

2



Axial, Shear and Moment diagram

Deflected Shape

+ - Axial

-

+ Shear

+

Moment

+



2.7.2 Internal Forces in a 3-hinged Frame

Reactions (+:Clockwise for mement)

02

=+

=+

BA

BA

HH

qLRR

∑ = 0RhM : BBBB HRLHLR 20

2−=→=−−

∑ = 0LhM :

420

422qLHRLqLLHLR AAAA +=→=−−

082

24

2

=+

=−→=−+

BA

BABA

HH

qLHHqLHqLH

16

16qLH

qLH

B

A

−=

= →

8

83

qLR

qLR

B

A

=

=

L

L

HA HB

RA RB

h

q



Freebody Diagram

Axial, Shear and Moment diagram

16qL

16qL

8qL

16qL

83qL

16qL

8qL

83qL

8qL

16qL

16qL

16

2qL 16

2qL

16

2qL 16

2qL

16qL 16

qL

- +

- 8

qL

+

83qL

Shear

83qL

8qL

16qL

- -

-

Axial



Deflected Shape

- -

-

Moment

- -

M

V



2.8 Arches



2.8.1 Three Hinged Arch

Arch Curve : )( 22

2 xllhy −=

Reactions

2

2PR

PR

B

A

=

= ,

hPlH

hPlH

B

A

2

2

=

=

Freebody Diagram

hPlH

PV

2

2

=

=

0)( =++−− xLRyHM AA

)(2

)(2

)(2

)(22

2

222

xlxl

P

xlPxllh

hPl

xlPyh

PlM

+=

++−−=

++−=

8maxPlM −=

HA

RA

HB

RB

P

h

2l

x

y

HA

RA

V

H M

y

l+x



Axial force and Shear Force

Deflected Shape

V

H

A S θ



2.8.2 Zero Moment Arch I

Reactions

2

2PR

PR

B

A

=

= ,

hPlH

hPlH

B

A

2

2

=

=

Freebody Diagram

hPlH

PV

2

2

=

=

0)( =++−− xLRyHM AA

)(

0)(22

xllhy

xlPyh

PlM

+=

=++−=

HA

RA

HB

RB

P

h

2l

x

y

H

HA

RA

V M

y

l+x

x y



2.8.3 Zero Moment Arch II

Reactions

2

2qlR

qlR

B

A

=

= ,

hqlH

hqlH

B

A

2

22

2

=

=

Freebody Diagram

02

)()()( =+

+−++−−xlxlqxlRyHM AA

)(

02

)()()(2

222

2

xllhy

xlxlqxlqlyh

qlM

−=

=+

+−++−=

HA

RA

HB

RB

q

h

2l

x

y

HA

RA

V

H

M

y

l+x

q



Chapter 3

Principle of Virtual Work

The principle of virtual work is the most important subject in the area of the structural analysis !!!!


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


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

M M+dM

V V+dV

q




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




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




– Moment of load case

– Deflection at the center of the span

or from the integration table,

Values of Product Integrals

l/4

1




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




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 +

v n

dS




∫∫∫ ∫∫ =+⋅−∇=+⋅∇−=+⋅−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




– 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




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 applied moment.

dxEIMMx

l

∫=θ0

0 )(

ε 1/ε 1/ε

x0




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




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

1fwvw

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 )()(




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




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 ,,1=

where m(i), ijH and i

jV are the number of member connected to joint i, the horizontal

component and the vertical component of the bar force of j-th member connected to joint i, respectively.

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

lFFEA

lFFEA

lFFlFvvuuF

11

1111

1212

) ()(

))(sin )((cos

iY iX

iF1−

ijF−

iimF )(−

iY iX




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 a unit load at a joint k in an 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

iθ iθ

)( 12 uu −

)( 12 vv −

ku αcosku




3.3.2 Example

Real System Virtual System

Table for calculation of the deflection of a truss

Member EAl F F EA

lFF

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




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 ,,1=

▬ 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

iii

EAlFF

EAlFF

1111

2

)ΔΔ(21)()ΔΔ(

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




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 virtual system.

Moment Shear Axial

+=δ ∫∫

2/

00

2 ll

M dxdxEI

EIPllPlllPll

EIM 16

3)

446443(2

=××+××=δ

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 l/4

×

Pl/4 l/4




+=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 negligibly small compared to that caused by the bending moment. If this is the case, the displacement 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








Chapter 4

Analysis of Statically Indeterminate Beams

1




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.

EI

qllqllEI

dxEI

dxEIMM 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

xBR δ

-ql2/2

1

-l




Compatibility condition and the final solution

(up)

Moment Diagram

Deflected shape

4.1.2. The second idea

=

+


-ql2/2

3ql2/8

-ql2/8 -

3l/8

+ =

θ0

MA




– Rotional Angle at the fixed end

EIqlqll

EIdx

EIdx

EIMM 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




4.2 Cantilever Beam with Spring Support

Robin BC (The third type BC)

Primary structure

Compatibity Condition

wbeam(l)=δspring →

As , and As

Deflected Shape for

δ0

Ra




4.3 Support Settlement

Primary structure


Deflected Shape

∆

RB




4.4 Temperture Change

Primary structure

Curvature due to temperature change

For simple beam, →

Comaptibility condition

→ →

T1

T2

θ0

MA




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

+ +




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 EI q

ql

Rxθ L

xθ

1

4

2ql 8

2ql

q

ql

R0θ L

0θ




Deflected shape

1




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

RA RB

MA MB

-ql2/2

M0

1

-l

M1

1 1

M2




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

61δδ

0

212112 =××=== ∫

EIll

EIdx

EIMMl

311

31

0

2222 =××==δ ∫

M0

M2

1

1

M1

1

1 8

2ql





Reactions and Moment Diagrams

Deflected Shape




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

1δδEIll

EIdx

EIMMl

=××=== ∫

)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





In case ,

4.8.2. Complicated Load Case

Primary structure

M0

EI EI

q

EI




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

2

1 401 qlX −= , 2

2 406 qlX −=

Compatibility Condition (Flexibility Equation) in General

∑=

∆=δ+δn

jijiji X

10




Chapter 5

Analysis of Statically Indeterminate Trusses

1 1

1

21

− 21

−

21

−




5.1 Various Types of Trusses

Determinate Truss

Externally Indeterminate Truss

Internally Indeterminate Truss

Mixed Indeterminate Truss

Structural Analysis Lab.




5.2 A Simple Truss

5.2.1 Method - I

+

P

X

P

P

=




Primary structure

At L1

075.06.0

25.118.0

1

31

3

==+

==

FFF

FF

P -1.0

1.25

0

-0.75 0

1 0.75 0.75

F1 F3

0.75

1

①,②,④:0.5A ③,⑤:A

P ②

①

③

④

⑤

L

0.75L

0.75L L1 L2

U1 U2




At L2

75.0 , 25.1

18.006.0

45

5

54

−===

=+

FFF

FF

At U1

1 , 75.0

08.006.0

21

52

51

−=−==+=+

FFFFFF

1

-1.0

1.25

1.25

-0.75

1

-0.75

F4 F5

1

F1

F2

F5




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




5.2.2 Method - II

P

+

P

X

X

=





AEXLX x −=δ+δ0

Primary structure

-1.0P P

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





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 EA

PL6.1 EAL28.1

3 1.25 P 1.0 1.0 1.25 L EA

PL56.1 EAPL25.1

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

EAPL54.0

5 - - 1.0 - - -

∑ EAPL84.3 EA

L61.3

AEXLX x −=δ+δ0 →

AEXLX

AEL

AEPL 25.161.384.3 −

=+

PPX 79.086.484.3

−==

PXH 63.08.02 −==




5.3 A Truss with 1 Roller Support

Primary Structures

1

2 3

4

5

6

7

8

9

10

3P

P 2P 3P

-P

2P P 2P

P 3P - 2 P - 2 P - 22 P

1

1

1

21

− 2

1−

21

−

21

−





Mem F 0 xF

EAL 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


LEAXXx 20 −=δ+δ

PXXXP 15.141.141.354.5 =→−=+−

X




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

3P 2P P

X1

1

2 3

4

5

6

7

8

9

X1

X2

X1

X1




– F0

– F1

– F2

1

1

1 1 1

2P P 3P

-P

2P 2P P

P 3P - 2 P - 2 P - 22 P

21

−

1 1

1

21

− 21

−

21

−




Axial force table for the primary structure

Mem 0F 1F 2F EAL 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 P3 22

− 0 1 P2

23− 0 2

1 0 0

4 P2 22

− 1 1 P2

22− P2 2

1 22

− 1

5 P2− 1 0 2 P2− 0 2 0 0

6 P− 22

− 0 1 P22 0 2

1 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

−==

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


Chapter 6

Analysis of Statically Indeterminate Frames

EI,l

w




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

RC l

0.75L

l





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




Moment Diagram

+

=

Deflection Shape

Pl/2

P

Pl/2

-

-

P

0.45P

+

0.45Pl

0.45P

0.45Pl

+

0.225Pl

+

-

- 0.05Pl






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





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

+




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




Moment Diagram

Deflected Shape


-3wl2/28

11wl2/196

-wl2/28

M0

wl2/8




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

+

+




6.3 Portal Frame subject to Horizontal Load

6.3.1. Primary Structure type I

1

w

M0

-wl2/2

-

EI,l

w

M2

l

+

+

M3

1

1

+

+

+

M1

-l

- -

-

1




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

3

2

1

22

233

233

→




Reactions

Moment Diagram

Deflected Shape

w

5wl/24 19wl/24

wl /7 wl /7

59wl2/252 31wl2/252





EIwldx

EIdx

EIMM ll

6 1 3

00

0110 −===δ ∫∫

EIwldx

EIdx

EIMM ll

8 1 3

00

0220 ===δ ∫∫

EIwldx

EIdx

EIMM ll

8 1 3

00

0330 −===δ ∫∫

EIldx

EIdx

EIMM ll

34) (1

0

22

0

1111 =+==δ ∫∫

1

1 +

+ M1

+

+

-

1

M2

+ +

1

1

M3

M0 w

-wl2/2




2100

2112 3

) (1δ=−=+==δ ∫∫ EI

ldxEI

dxEIMM ll

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




6.4 Portal Frame subject to Vertical Load

1

1 +

+ M1

+

+

-

1

M2

+ +

1

1

M3

EI,l

a P

M0

4Pab




Matrix Form

Sidesway :

Deflected Shape




6.5 Order of Indeterminacy

# of unknowns

# of member × # of internal force per member +

# of reactions - # of known quantities

# of equations

# of member × # of E.E. per member +

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

# of Indeterminacy = # of unknowns - # of equations

Order of Indeterminacy of the frame shown above

6×10 +3×3 – (3×10 +3×9) = 69 – 57 = 12

6.5.1. Order of Indeterminacy – Beam

Number of Internal Forces in a Member : 4

Number of Equilibrium Equations in a Member : 2

Number of Equilibrium Equations at a Joint : 2

Simple Beam

- # of unknowns : 4×1 + 1×2 = 6 or 4×1 + 2×2 – 2= 6

- # of equations : 2×1 + 2×2 = 6




- Order of Indeterminacy : 6 – 6 = 0

Overhanged Beam

- # of unknowns : 4×2 + 1×2 – 2 = 8

- # of equations : 2×2 + 2×2 = 8


Gerber Beam

- # of unknowns : 4×3 + 1×3 – 2 = 13

- # of equations : 2×3 + 2×4 – 1= 13


Continuous Beam

- # of unknowns : 4×3 + 1×4 = 16

- # of equations : 2×3 + 2×4 = 14


6.5.2. Order of Indeterminacy - Truss







Determinate Truss

- # of unknowns : 1×21 + 3 = 24

- # of equations : 2×12 = 24



- # of unknowns : 1×25 + 3 = 28

- # of equations : 2×12 = 24



- # of unknowns : 1×25 + 5 = 30

- # of equations : 2×12 = 24





6.5.3. Order of Indeterminacy - Frame




Internally Indeterminate Frame

- # of unknowns : 6×3 + 3 = 21

- # of equations : 3×3 + 3×4 = 21


Frame with Hinges

- # of unknowns : 6×4 + 4 – 2 = 26

- # of equations : 3×4 + 3×5 – 1 = 26


Portal Frame with Fixed Supports

- # of unknowns : 6×3 + 6 = 24

- # of equations : 3×3 + 3×4 = 21





6.5.4. Selecting a Primary Structure for a Complicated Frame

Order of Indeterminacy of the frame

- # of unknowns : 6×10 + 3×3 = 69

- # of equations : 3×10 + 3×9 = 57


By Replacing a Rigid Joint with a Hinge, we can reduce

- # of unknowns by the number of members at the joint

- # of equations by one

In the Primary Structure

- Reduction in unknowns : 1 + 1 + 3 + 4 + 3 + 2 + 3 = 19

- Reduction in equations : 5

- Reduction in Indeterminacy : 17 – 5 = 12




6.6 General Frame

Primary Structure

M1, M2, M3

EI

L

L/2

2EI

q qL

EI

L L

EI

ql2/8 4

2ql M0

M1 M2

M3




EIqlql

EIldx

EIMMl

2481

3

32

0

0110 =××==δ ∫

EIqlqlldx

EIMMl

164)1()

211(

6

32

0

0220 −=×−×+==δ ∫

00

0330 ==δ ∫

l

dxEIMM

EIldx

EIMMl

30

1111 ==δ ∫ , 01312 =δ=δ

EIldx

EIMMl

30

2222 ==δ ∫ , 02321 =δ=δ

EIldx

EIMMl

620

3333 ==δ ∫ , 03231 =δ=δ

ql2/8 4

2ql

+ 1

- 1

+ 1

M0 M1

M2 M3





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 compatibility equations and one equilibrium equation, which leads to

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




6.7 General Joint Compatibility


nδ==δ=δ 21

j

k

njijj

n

jijii MM ∑∑

+==

δ+δ+δ=δ11

0 ni 1for =

One Additional Equilibrium Equation

021 =+++ nMMM

Joint i

M1

Mn








Chapter 7

Influence Lines for

Determinate Structures




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

1 ξ

I(xp,ξ)

xp

ξ q(x)

R(xp)

q(ξ)dξ




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

P1 ξ1

I(xp,ξ)

ξn

Pn

RA=(l - ξ)/l

ξ

RB=ξ/l

P = 1

RA=(l - ξ)/l

ξ P = 1 Mx




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




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

=

105

104 LL

≤ξ≤

RA=(l - ξ)/l

Vx

+

1/2

+

P/2 ξ

P

L/10

2)2/( ξ

=LIM

22)2/( ξ

−=LLI M




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/2 L

RA=1

ξ P=1

Vx=0

RA=1

Vx=1

+

1




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 −=−=

ξ ξ−=xM

)2( ξ′−−= LM x

ξ′

Lξ′2

Lξ′

−21

-

L )23()2( ξ−−=ξ′−− LL −ξ

- L

L/4




– 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




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

l Ra Rb

x

l0

l/2

P =1

l

Ra Rb

x (l-l0)/2

l/2

P =1




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




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




Influence 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≤≤ )

61(0)

61(

axFFaa

ax

−=→=−×−

+

-

264

2

61

axRa 6

1−=

A

65

+




Chapter 8

Influence Lines for Indeterminate Beams




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

1/2 1/2




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

L

EI

L L

ξ P = 1

EI EI

M = 1 dxb

θbb

Structural Analysis Lab.




– 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 = 1 dxb

θbb

M = 1 dxb

M = 1

x = 1




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−=




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




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




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




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




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

χ

P = 1

Mb

0.203L

-0.048L

0.423L

× =≤− )2

(22 LxLx

L

0.406

-0.096

0.423L

-0.594




8.2.3. Influence line of 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

dxb dbb

L/2

1 1

1

2 1




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

=→+−=

−=→=++−→=++−




v) Deflection of the left span

– Suspended span : )5(6

16

233

xLxEI

axEIxwS −−=+−=

– Overhanged span : )1334128(48

1 3223 LxLLxxEI

wO +−+−=

– Right span : ( )xLLxxEI

wR223 23

61

+−=


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 I - SNUstrana.snu.ac.kr/lecture/struct1_2015/Notes/15_Note_All.pdf · Dept. of Civil and Environmental Eng., SNU . Structural Analysis I . Spring Semester, 2015