structural analysis istrana.snu.ac.kr/lecture/struct1_2019/notes/19_note(ch1... · 2019-05-03 ·...

Dept. of Civil and Environmental Eng., SNU

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

Structural Analysis I

Spring Semester, 2018

Hae Sung Lee

Dept. of Civil and Environmental Engineering Seoul National University

yδ

yf

zδ zf

xδ

xf

yM yθ

zM zθ xM xθ



This page is intentionally left blank.



Contents

1. Introduction

2. Reactions & Internal Forces by Free Body Diagra

ms

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 dia

gram 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 inde-

terminate structures.

EI EI

q

Ra Rb Rc

L L

q



3

Fundamentals of Differential Equations

Governing Equations

The governing equations of a (engineering) system are usually defined by a system of diffe-

rential equations, which governs behaviors of the given system within a domain. Notice that

the domain does not include boundaries.

11 qwEI =′′′′ for lx



4

1.2 Mechanics of Material

Governing Equation

– Left span

112

13

1

4

1''''

1 24dxcxbxa

EI

qxwqEIw ++++=→=

– Right Span

222

23

2

4

2''''

2 24dxcxbxa

EI

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

31

4

11111 2402)0( , 0)0( ++=→==′′==

– Right Support

xcxaEI

qxwbwdw 2

32

4

22222 2402)0( , 0)0( ++=→==′′==

x x

y

z

z

y

q

w1 w2



5

– 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

22

2

3

12

1

3

23

2

4

2

13

1

4

1

)32(48

33421 xLLxxEI

qww +−=≡

83

, 8

32 11

211

qLqxwEIVx

qLx

qwEIM +−=′′′−=+−=′′−=

Moment Diagram

Shear Diagram

Reactions

0.375qL

L83

+

-

+

0.625qL

-

0.375qL 0.375qL 1.25qL

0.125qL2

0.070qL2

L8

3

+

-

+



6

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 (compatibility condition)

δ0+ δR=0

1.3.2 Calculation of δ0

Bending Moment

qLxqx

M +−=2

2

q

q

δ0

δR

Rb

q

qL2/2



7

Governing Equation

baxqLxqx

EIwMwEI ++−=→−=′′624

34

00

Boundary (support) Conditions

– Left Support : 0)0(0 =w

– Right Support : 0)2(0 =Lw

Determination of Integration Constant

– Left Support

0 0)0(0 =→= bEIw

– Right Support

EI

LqaLa

LqLqLEIw

24

)2(0)2(

6

)2(

24

)2()2(

334

0 =→=+−=

Deflection

))2()2(2(24

3340 LxLxxEI

qw +−=

EI

LqLLLLL

EI

qLw

384)2(5

))2()2(2(24

)(4

33400 =+−==δ

1.3.3 Calculation of δR

Bending Moment

22,

22

21

1

LRxRM

xRM bbb −=−=

Governing Equation

+++−=

++=→

−=′′−=′′

222

22

32

2

111

31

1

22

11

412

12

bxaLxRxR

EIw

bxaxR

EIw

MwEI

MwEI

bbR

bR

R

R

1x 2x

RbL/2

Rb



8

Boundary (support and mid-span) Conditions

– Left Support & Right Support

0)( , 0)0( 21 == Lww RR

– Mid-span

)( )0()()( , )0()( 221121 LwLwLwLw RRRRRR θ=′=′=θ=

Determination of Integration Constant

– Left Support

0 0)0( 11 =→= bwR

– Right Support & Mid-span

−=

=

−=

→

′==+=′

==+=

=+++−=

6

04

(0)4

)(

(0)12

)(

0412

)(

3

2

2

2

1

221

2

1

221

3

1

22

33

2

LRb

a

LRa

wEIaaLR

LwEI

EIwbLaLR

LEIw

bLaLRLR

LEIw

b

b

Rb

R

Rb

R

bbR

Deflection

)23(12

)3(12

322

322

123

11

LLxxEI

Rw

xLxEI

Rw

bR

bR

+−−=

−=

EI

RLwLw bRRR 48

)2()0()(

3

21 −===δ

Compatibility Condition

δ0+ δR=0 → 048

)2(

384

)2(5 34 =−EI

RL

EI

Lq b → qLRb 810=



9

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

EI

Lqdx

EI

MML R384

)2(5 42

0

00 −== δ , EI

LRdx

EI

MM bL

RRR 48

)2( 32

0

== δ

Solution

δ0+ δR=0 → 048

)2(384

)2(5 34 =+−EI

RL

EI

Lq b → qLRb 810=

RbL/2

Rb

q

q

qL2/2



10

– Moment

– Shear

+

-

+

-

+

+ -

+

-

=

0.070qL2

5qL2/8

Rb

+

=

0.125qL2 L

8

3

+

-

+

qL2/2



11

1.5 Supports

– Supports are used to maintain the position of the structure by constraining free mo-

tions at some discrete points.

– If you want to constrain a free motion of a structure at any part of the structure, some

kind of forces should be applied.

– Since a support prevent free motion of a structure, a force called as a reaction force is

developed in the direction of a constraints.

– The support conditions provide boundary conditions for the governing equation(s).

– The reaction forces are unknown, and should be calculated through the structural

analysis.

– The most popular support types are a fixed support, a hinge support and a roller sup-

port.

δ0δ



12

Fixed Support(고정단)

– Both the horizontal and vertical motion and rotational motion are prevented.

– Reaction forces are developed in three directions.



13

Hinged Support(회전단)

– The horizontal and vertical motion are prevented, but the rotational motion is allowed.

– The horizontal and vertical reaction forces are developed in all three directions

Roller Support(이동단)

– Only the vertical motion is prevented, and the horizontal and the rotational motion is

allowed.

– A reaction force is developed only in the vertical direction.



14

1.6 Two-Dimensional Idealization of a Structure

Real 3-D Structures

– 3 force components and 3 moment components

– 3 displacement components and 3 rotational components

Main Structures : A part of a structure designed mainly to resist external loads.

– Beams, Truss, Frame, Arch…

Secondary Structure : A part of a structure designed to transfer external loads to

main structures or increase stability and/or stiffness of a structure

– Stringer : Small beam that runs parallel to a main structure to support the floor

system.

– Cross beam : Small beam that runs perpendicular to the main structure to support

stringers.

– Bracing (tie) : member that connects two or more main structures to increase mainly

lateral stability of the main structures.

xδ xf

ym yθ zδ

x

y

z

yδ yf

zδ zf

xδ xf ym yθ

zm zθ xm xθ



15

Load transfer path

External load → Slab → Stringer → Cross Beam → Main Structure → Support →

Foundation

– External loads applied on a slab are supported by stringers, which transfer the carried

loads to cross beams. The cross beams, in turn, transfer the carried-over loads from

the stringers to main structures.

주 구조물 (Main Structure)

가로 보 (Cross Beam)

세로 보 (Stringer )

Cross Bracing (Wind Bracing)

지 점 (Support)



16

Two-dimensional Idealization of a Structures

All members are idealized as one-dimensional members, which are represented by

lines passing the centroids of the cross sections, because dimensions in cross-

sectional directions are very small compared to their lengths (usually less than 1/20).

Suppose we somehow know portions of the external loads transferred to each main

structure.

Suppose all members in each main structure lie in one plane, and external loads car-

ried-over from substructures are applied on the same planes.

Then, all structural responses are developed in the same plane, and main load resist-

ing actions occur in the same plane.

If this the case, each main structure in the same plane can be separated from the

whole structure by neglecting three-dimensional effects, and analyzed in the plane

where each main structure is defined for the simplicity of a structural analysis.

Two-dimensional structures are often referred to as plane structures.

However, fundamentals employed in the 2-D analysis of structures are exactly the

same as those in the 3-D analysis.



17

Plane truss idealization

– Structure composed of a number of bars carrying axial forces that lie in one plane.

– Vertical and horizontal force

– Vertical and horizontal displacement

절점

xf xδ

yδ yf



18

Plane beam idealization

– Structure composed of a number of beams carrying bending moments and shear

forces that lies in one plane.

– Shear force and Moment on z-axis

– Vertical displacement and rotational angle with respect to z-axis

11 , wV 22 , wV

33 , wV

11 , θM 22 , θM 33 , θM



19

Plane Frame idealization

– Structure composed of a number of beams carrying bending moments, shear forces

and axial forces that lie in one plane.

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

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

xδ xf

yf yδ zM zθ



20

1.7 Stability of Structures

Internal Stability (내적 안정)

A structure maintains its shape under the actions of all possible loads within a certain limit.

External Stability (외적 안정)

A structure maintains its position under the actions of all possible loads within a certain limit.



21




22

Chapter 2

Reactions & Internal Forces

by Free Body Diagrams



23

2.1 Free Body Diagram

Any portion of a structure can be isolated by passing any desired section through structure.

A free body sketch is a diagram drawn showing this portion acted upon by the external loads

and reactions, together with any forces that may be act on the faces of the members cut by the

isolating section. Or, a diagram drawn by removing some part(s) of a structure and replacing

the removed part(s) with unknown forces.



24

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 -



25

2.2 Reactions

Beams

→

+=−→=

=−+→=

) (clockwise 00

00

LRPaM

PRRF

BA

BAVP

L

aRB = , PL

bRA =

→

+=−+→=

=+→=

) (clockwise 0)(0

0

LRaLPM

PRRF

BA

BAVP

L

aRB )1( += , PL

aRA −=

Truss

PHPHF AAH −=→=+→= 00 ,

00 =−+→= PRRF BAV

0)3(0 =−+→= aRPaPaM BA (Clockwise +)

PRA 31= , PRB 3

2=

P a b

L

P

RAL

RBL

b a

P a

L

P a

L RAL

RBL

RAL

RB P

P

HAL

P

P

[email protected]

a



26

Frame

+=

−=→

=−−=

=−−=

=+=

=−→=−+=−+=

42

2

0422

02

082

24

2 02

qLHR

HR

LqLLH

LRM

LHL

RM

HHF

qLHH

qLH

qLH

qLRRF

AA

BB

AALh

BBRh

BAH

BABABAV

What if there is no internal hinge ?

042

)(2

)(

0422

02

0

02

=−+−−→

=−−+−=

=−−=

=+=

=−+=

LqLLHH

LRR

LqLLH

LRMM

LHL

RMM

HHF

qLRRF

ABBA

AALh

BBRh

BAH

BAV

HB

RA RB

HA

L

L

h

q

→−=

=

16

16qL

H

qLH

B

A

8

8

3

qLR

qLR

B

A

=

=

HB

RA RB

HA

q

M M



27

2.3 Internal Forces in Framed Structures

Axial Force

Shear Force

Bending Moment

Torsion

+

+

+

+



28

2.4 Internal Forces in Simple Beams

Reactions

q

RA=qL/2 Rb= qL/

q

RA=qL/2 RB= qL/2

RA RB



29

Free Body Diagram for Shear and Moment

qxqL

qxRVVqxRF AxxAV −=−=→=−−= 20

22

02

2qxx

qLMM

xqxxRM xxAx −=→=−−=

Differential Equation

baxxq

Mqdx

Md ++−=→−= 22

2

2

Boundary Conditions: 2222

,00)()0( xq

xqL

MqL

abLMM −=→==→==

Shear Force and Moment Diagrams

Deflected Shape

Remember 2

2

dx

wdEIM −= !!!

RA

x

RB

Mx

Vx

+

qL/2

qL2/8



30

2.5 Gerber Systems

A structural system that an unstable structural component is supported by a stable structure.

Sometimes there may be several unstable structures are supported by a stable structure, and

vice versa. An external load applied to an unstable structure is transferred to a stable struc-

ture, but an external load applied to a stable structure is not transferred to an unstable structure.

In the later case, a unstable structure is just deflected as a rigid body without deformation.



31



32

2.5.1 Internal Forces in a Gerber Beam - I

Reactions

PRL

PL

RM HHC 3

20

24

30 =→=×−×→=

PRPRRF CCHv 3

100 =→=−+→=

PRL

RLRM BHBA 6

50

4

50 =→=×+×−→=

PRPRRF ABAv 6

10

3

20 −=→=−+→=

L/4

P

RH

RA RB

RC

P

P/6 5P/6 P/3

P



33

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= Px6

1−



34

ii) LxL2

3≤≤ , Lxx −=′

iii) LxL 22

3 ≤≤ xLx −=′ 2

Deflected Shape

qdx

wdEI =

4

4

or 2

2

dx

wdEIM −=

P/6 5P/6

2P/3

Mx=

xPPL ′+− 2

P/6 5P/6

P

P/3 P/3

Mx= xP ′3

1

+

-

PL/6

PL/6



35

2.5.2 Internal Forces in a Gerber Beam - II

Free Body diagram

Shear

q

2

ql

2

ql

2

ql

2

ql

2

2ql

2

2ql

2

ql

2

ql

+

L

q

L L

2

ql

2

ql

x

qxql

Vx −= 2 2

ql



36

Moment

Deflected Shape

8

2ql

2

2ql 2

2ql

+

xqlql

Mxqlql

M xx 220

22

22

+−=→=−+

2

ql

2

2ql

2

ql

x

2

ql

220

2

2qxx

qLMM

xqxxR xxA −=→=++−



37

2.6 Truss (트러스)

Structures composed of a number of straight bars carrying axial forces.



38

Assumption

1. All joints are hinges.

2. All members are straight.

3. Small deformatiom

4. The external loads are applied only at joints.

The 1st assumption seems to be unreasonable, but it is very reasonable assumption !!!

Because the length of each member is short compared to the total length of a truss, the mo-

ment induced by rigid joints is negligibly small in an engineering sense.

Characteristics of truss

– By the 2nd , 3rd and 4th assumptions

02

2

2

2

=→−=dx

Mdq

dx

Md→ baxM +=

– By the 1st assumption

0 , 000)()0( ≡≡→==→== VMbaLMM

– 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.

– The axial force is constant along a member by assumption 1.

– There is only one unknown per member.

– As the equilibirium of each member is always satisfied, the equilibrium conditions at

joints are to be considered.

– Determinancy

– The number of unknowns : RM NN +

– The number of Equilibrium Equations : JN2

– 02 =−+ JRM NNN : Determinant, 02 >−+ JRM NNN : Indeterminant



39

2.6.1 Internal Forces in Howe Truss

At U1 and U3

At L1

2 ,

22

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

9+3-2×6=0: determinant truss

L2

7

9 1

4

2

3

U1 U2 U3

6

5

L1

P/2 P/2

L3

8

P

[email protected]

a



40

At U2

=+−

−==→

=−−−

=++−−

02

1

2

12

2

02

2

2

2

02

2

2

237

753

8734

PPP

PFF

FFF

FFFF

At L3

Because we already have 3 equilibrium equations to calculate the reactions

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

Tension Compression



41

Equilvalent Beam Action

Axial forces in members generate bending moments and shear forces just like in a beam !!!

Deflected Shape

2.6.2 Internal Forces in Warren Truss

V=P/2

P/2

1 P

3P

32P

2 3

4

5

6

7

8

9

10

11

L1 L2 L3

U1 U2 U3

L4

x

PL/4 M=Px/2

[email protected]

l



42

At L1

PFPFP

F

FF

32

, 3

22

03

222

022

12

2

12

=−=→

=+

=+

At U1

PFPFF

FF

FFF

34

,232

022

22

02

2

2

2

423

32

432

−==−=→

=+

=++−

At L2

PFPF

FFFF

PFF==→

=+−+

=−+65

5361

35

,32

022

22

-

022

22

F2

F1

2P/3

F4

F2 F3

F3

F1 F6

P

F5



43

At U2

PFPF

FF

FFFF

32

,32

022

22

02

2

2

2

87

75

7584

−=−=→

=−−

=+−+−

At L3

3,

32

022

22

022

22

109

97610

97 PFPF

FFFF

FF==→

=+−−

=+

At U3

PF

FF

FFF

32

022

22

022

22

11

119

8119

−=→

=−−

=−+−

F8

F5 F7

F4

F7

F6 F10

F9

F11

F8

F9



44

At L1

OK

Deflected Shape

Equivalent Beam Action

1 2 3 4 5 6

1. PxM32=

2. PxPlM32

32 +=

3. PxPlxlPPlM31

34

)(31 −=−+=

4. PxPlxlPPlM31

)(31

32 −=−+=

5. PxPlxlPPlM31

32

)(31

31 −=−+=

6. PxPlxlPM31

31

)(31 −=−=

P/3

P/3

P32

P

Pl3

4

PxPlM3

2

3

2 += 2P/3

2P/3

4P/3=2P/3+2P/3

2P/3 x

x

2P/3

PxM3

2=



45

2.6.3 Method of Sections

PFL

FLPM L 34

023

2442 −=→=×+×=

PFL

FL

PLPMU =→=×−×−×= 662 02223

32

PFFPPFV 32

022

32

55 =→=×+−=

P

3P

P32

1

2 3

4

5

6

7

8

9

10

11

L1 L2 L3

U1

U

U2 U3

L4

Cut out

P

P32

F5

F4

F6



46

2.7 프레임 (Frame)



47

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

LqH

2

2

L

qH

2

2

L

qH

2

2



48

Axial, Shear and Moment diagram

Deflected Shape

L

qH

2

2

+ - Axial

L

qH

2

2

qH

-

+ Shear

L

qH

2

2

2

2qH

+

Moment

+



49

2.7.2 Internal Forces in a 3-hinged Frame

Reactions (+:Clockwise for mement)

02

=+

=+

BA

BA

HH

qLRR

= 0RhM : BBBB HRLHLR 202 −=→=−−

= 0LhM : 420422qL

HRLqL

LHL

R AAAA +=→=−−

082

24

2

=+

=−→=−+

BA

BABA

HH

qLHH

qLH

qLH

16

16qL

H

qLH

B

A

−=

= →

8

83

qLR

qLR

B

A

=

=

L

L

HA HB

RA RB

h

q



50

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

16qL

- +

- 8

qL

+

83qL

Shear

83qL

8qL

16qL

- -

-

Axial



51

283

16

028

316

22

2

xqqLx

qLM

xqxqLx

qLM

−+−=

=−+−−

Deflected Shape

- -

-

Moment

- -

2

16qL 2

16qL

83qL

16

2qL

M

V



52

2.8 Arches



53

2.8.1 Three Hinged Arch

Arch Curve : )( 22

2xl

l

hy −=

Reactions

2

2P

R

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

222

xlxl

P

xlP

xll

h

h

Pl

xlP

yh

PlM

+=

++−−=

++−=

8maxPl

M −=

HA

RA

HB

RB

P

h

2l

x

y

HA

RA

V

H

M

y

l+x



54

Axial force and Shear Force

θθθ−θ

=

→

θ+θ=θ−θ=

S

A

V

H

SAV

SAH

cossin

sincos

cossin

sincos

θθ−θθ

=

→

V

H

S

A

cossin

sincos

224224

2

2

4

2sin ,

4cos

2tan

xhl

hx

xhl

l

l

hxy

+

−=θ+

=θ

−=′=θ

)2

(4

cos2

sin2

)2

(4

sin2

cos2

2

224

23

224

llx

xhl

PP

h

PlS

h

xhl

xhl

PP

h

PlA

−−+

=θ−θ=

−

+−=θ−θ−=

Deflected Shape

V

H

A S θ



55

2.8.2 Zero Moment Arch I

Reactions

2

2P

R

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



56

2.8.3 Zero Moment Arch II

Reactions

qlR

qlR

B

A

== ,

h

qlH

h

qlH

B

A

2

22

2

=

=

Freebody Diagram

02

)()()( =++−++−− xlxlqxlRyHM AA

)(

02

)()()(

2

222

2

xll

hy

xlxlqxlqly

h

qlM

−=

=++−++−=

HA

RA

HB

RB

q

h

2l

x

y

HA

RA

V

H

M

y

l+x

q



57

Chapter 3

Principle of Virtual Work

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



58

3.1 Beam Problems

3.1.1 Governing Equations

Equilibrium for vertical force

qdx

dVqdxVdVV −=→=+−+ 0)(

Equilibrium for moment

Vdx

dMdxqdxVdxMdMM =→=+−−+ 0

2)(

Elimination of shear force

qdx

Md −=2

2

Strain-displacement relation

A plane section before deformation remains a plane after deformation.

A section perpendicular to the neutral axis before deformation maintain the right

angle to the neutral axis after deformation

ydx

wd2

2

−=ε

Stress-strain relation (Hooke’s law)

ydx

wdEE

2

2

−=ε=σ

Definition of Moment

2

22

2

2

dx

wdEIdAy

dx

wdEydAEydAM

AAA

−=−=ε=σ=

Beam Equation

qdx

wdEI =

4

4

M M+dM

V V+dV

q



59

3.1.2 Limit Process

Equilibrium for vertical force

qdx

dVq

x

Vq

x

VxqVVV

x−=→=+

∆∆→=+

∆∆→=∆+−∆+

→∆0)(lim00)(

0

Equilibrium for moment

0))()(1

(lim0)()()(0

00

=ξξ−∆ξ+∆

+−∆

∆→=ξξ−∆ξ++∆−−∆+ ∆

→∆

∆

dxxqx

Vx

MdxxqxVMMM

x

x

x

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

lim)(1

lim)(lim

)(),(

00

00

00

=×−=ξ

ξ−=∆

−∆−=ξξξ+∆

−ξξ+

ξξξ+=ξ

=ξ→∆

∆

→∆

∆

→∆

xqd

xdQ

x

xQxxQdxq

xdxq

dxqxQ

x

x

x

x

x

3.1.3 Modelling of Concentrate loads - Dirac delta functions

0lim→ε

= = )( ξ−δ x

122

1lim)0

2

10(lim)(

00

00

=εε

=+ε

+=ξ−δ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε ll

dxdxdxdxx

)()(2

)()(lim)(

2

1lim

)0)(2

1)(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

fFFF

dxxf

dxxfdxxfdxxfdxxxfll

ξ

∞

ξ

ε21

2ε

ξ

M M+∆M

V V+∆V

q

∆x



60

3.1.4 Principle of Virtual Work (Beam)

The real(actual) load case A virtual(imaginary) load case

For the real load case

02

2

=+ qdx

Md,

2

2

dx

wdEIM −= for lx



61

Principle of virtual work

dxEI

MMdxM

dx

wd ll

−=00

2

2

dxqwdxEI

MM ll =00

→ extWW δ=δ int

The internal virtual work is equal to the external virtual work if a beam satisfy the

equilibrium for the real and virtual load.

Equilibrium equation for load case q

0)(0

2

2

=+ dxqdxMd

wl

Virtual work expression

dxqwdxEI

MM ll

=00

Betti-Maxwell’s Reciprocal Theorem

dxEI

MMdx

EI

MM ll

=00

→ dxqwdxqwll

=00

Calculation of displacement for the load case q

dxqwdxEI

MM 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

0

0

xwdxxxwdxEI

MM ll =−δ= → dxEIMM

xwl

=0

0 )(



62

3.1.5 Example

A simple beam subject to an uniform load

– Moment of load case q

– Moment of load case q

– Deflection at the center of the span

EI

ql

EI

qllqlql

EI

dxxq

xql

EIdxx

qx

qlx

EI

dxxq

xqlxl

EIdxx

qx

qlx

EIdx

EI

MMlw

ll

l

l

ll

4443

2/

0

322/

0

2

2/

22/

0

2

0

384

5)

256

1

96

1(

2))

2(

4

1

4)

2(

3

1

4(

2

)44

(2

)22

(2

2

)22

)(22

(1

)22

(2

1)

2(

=−=−=

−=−=

−−+−==

or from the integration table,

EI

qlqll

EI

lMM

l

ab

EI

ldx

EI

MMlw

l

3845

84)

41

1(3

)1(3

)2

(42

3120

=+=+==

l/4

1

q

ql2/8



63

Values of Product Integrals L

LU dxMM0



64

3.1.6 Conservation of Energy

Equilibrium and Conservation of Energy

– Equilibrium Equation

qdx

wdEI =

4

4

– External work

int

ll

lll

lll

ll

ext

WdxEI

Mdx

dx

wdEI

dx

wd

MwVdxdx

wdEI

dx

wd

dx

wdEI

dx

dw

dx

wdwEIdx

dx

wdEI

dx

wd

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

2

1

2

1

][2

1

][2

1

2

1

2

1

Conservation of Energy in each load case

dxwqdxEI

M ll

=00

2

21

21

, dxqwdxEI

M ll

=00

2

21

21

Two load cases are applied simultaneously.

dxqwqwdxEI

MM

dxwqqwqwqwdxEI

Mdx

EI

MMdx

EI

M

dxqqwwdxEI

MM

ll

llll

ll

+=

+++=++

++=+

00

00

2

00

2

00

2

)(2

1

)((2

1

2

1

2

1

))((2

1)(

2

1

w

q

w

q



65

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

dxqwdxqwdxqwqwdxEI

MM llll

==+=0000

)(2

1

3.1.7 General Conservation and Equilibrium

Conservation in General

=+⋅−S V

fdVdS 0nv

v n

dS

w

q

w

q

ww +

w

q

w

q

ww +qq + qq +



66

– By divergence theorem,

⋅∇−=⋅−VS

dVdS vnv where ),,(),,(321 xxxzyx ∂

∂∂∂

∂∂=

∂∂

∂∂

∂∂=∇

=+⋅−∇=+⋅∇−=+⋅−VVS VV

dVffdVdVfdVdS 0)( vvnv

– Since the integral equation should hold for all systems,

0=+⋅∇− fv

Potential Problems

– The vector field of a system is defined by a gradient of a scalar function referred to as

a potential function

Φ∇⋅−= kv , ),,(zyx ∂Φ∂

∂Φ∂

∂Φ∂=Φ∇

– The famous Laplace equation for a conservative system.

0)( =+Φ∇⋅⋅∇=+⋅∇− ff kv

– If the system properties are homogeneous and isotropic, Φ∇−= kv

0)( 2 =+Φ∇=+Φ∇⋅∇=+⋅∇− fkfkfv or 02

2

2

2

2

2

=+∂

Φ∂+∂

Φ∂+∂

Φ∂f

zyx

3.1.8 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 =

⋅=σ=3

1jijiji nT nσ ,

=

σσσσσσσσσ

=

3

2

1

333231

232221

131211

σσσ

σ ,

=

3

2

1

n

n

n

n

TT=

E, ν

uu = Su

St

V

x

y

z

bx

by

bz



67

– Divergence Theorem

0)( =+⋅∇=

+⋅∇=+⋅=+

V

ii

V V

ii

S V

ii

S V

ii

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 bx

bzyx

bσ 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

– In case elasticity problems are potential problems

ii

i uC ∇⋅=σ

– However, to maintain symmetry condition of stress,

uC ∇= :σ or = = ∂

∂=σ3

1

3

1k l l

kijklij u

uC

– Equilibrium equation in terms of the potential functions

0):( =+∇⋅∇ buC

– What is σ, and why is σ related to a potential function in such a way?

out of scope of this class !



68

3.1.9 Displacement on boundaries

Virtual Work Expression

llll

MVwdxqwdxEI

MM00

00

θ−+=

llll

MVwdxqwdxEI

MM00

00

θ−+=

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

0000

MlMlVwlVlwdxqwMVwdxqwdxEI

MM lllll θ+θ−−+=θ−+=

By coinciding the positive direction of forces and displacement

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

MlMlVwlVlwdxqwdxEI

MM ll θ+θ+++=

Deflection of a cantilever beam subject to an end load

0)0( , 0)0( , 0)( , 1)( =θ=== wlMlV

q (real) system q (virtual) system

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

MlMlVwlVlwdxqwdxEI

MM ll θ+θ+++=

EI

PlPll

EI

lMM

EI

ldx

EI

MMlw

l

3))((

33)(

3

31

0

=−−===

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

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

MlMlVwlVlwdxqwdxEI

MM ll θ+θ+++=

EI

PlPl

EI

lMM

EI

ldx

EI

MMl

l

2)(1

22)(

2

31

0

−=−××===θ

P

-Pl

1

-l

P

-Pl

1



69

Rotation in the a body (or a structure)

– Modeling of a unit moment applied at x0

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

MlMlVwlVlwdxqwdxEI

MM ll θ+θ+++=

)()]2

()2

([1

lim

))]2

((1

))2

((1

[lim

0000

0

000

0

0

xdx

dwxwxw

dxxxxxwdxEI

MM

xx

ll

θ−=−=ε+−ε−ε

=

ε+−δε

−ε−−δε

=

=→ε

→ε

– by coinciding the positive direction of the rotational angle with that of the applied

moment.

dxEI

MMx

l

=θ0

0 )(

ε 1/ε 1/ε

x0



70

3.2 Principle of Virtual Work in General

3.2.1 3-Dimensional Elastic Body

Rigid Body

Rigid body: A body that does not deform at all. Displacement is constant over the body.

Deformation of a body is caused by the variation of displacements in a 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 bTwbwTw

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 !!!

q-Force System

q -Force System

w

T

T

dVdSij V

ijij

S

σε=⋅ Tw



71

⋅+⋅=δS V

ext dVdSW bwTw

=δV

dVW σε :int

= == =

= == ==

∂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 V

ijij

i j S

jijii S j

jijii S

ii

S

dVx

wx

wdVw

x

dSnwdSnwdSTwdSTw

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

)(2

1

)(2

1

)(2

1

)(

)(

WdVdVx

w

x

w

dVx

wdV

x

w

dVx

wdV

x

wdV

x

w

dVbx

wdVx

w

dVbwdVx

wx

wW

i j V

ijiji j V

iji

j

j

i

i j V

iji

j

i j V

ijj

i

j i V

jii

j

i j V

ijj

i

i j V

ijj

i

i V ji

j

iji

i j V

ijj

i

i V

iii j V j

ijiij

j

iext

δ=σε=σ∂∂

+∂∂=

σ∂∂

+σ∂∂=

σ∂∂

+σ∂∂=σ

∂∂=

+∂σ∂

+σ∂∂=

+∂σ∂

+σ∂∂=δ

= == =

= == =

= == == =

= == =

== =

=⋅+⋅S VV

dVdVdS σε :bwTw



72

3.2.2 General Framed Structures

)( τγ+σε=τγ+σε=σ⋅εe VVVVij V

ijij dVdVdVdVdVee

Internal virtual work by normal stress – bending moment

=−−=

==

==−−=σε

ee

ee

e

e

eeee

ll

ll

A

l

AVVV

dxEI

MMdx

EI

MEI

EI

M

dxdx

wdEI

dx

wddx

dx

wddAEy

dx

wd

dAdxydx

wd

dx

wdEdVy

dx

wd

dx

wdEdVy

dx

wdEy

dx

wddV

00

02

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

FFdx

A

FdA

EA

FdV

A

F

EA

FdV

00

)(

Internal virtual work by shear stress

QyIb

V

)(=τ and Q

yGIb

V

)(=γ where =

a

y

ydAQ

===τγee

eee

ls

l

AVV

VdxVGA

fdAdx

ybI

Q

G

VVQdV

yIb

VQ

yGIb

VdV

0022

2

)()()(

Total displacement

++=e

l ll

s

e ee

dxEA

FFdx

GA

VVfdx

EI

MMxw

0 00

0 )()(



73

3.2.3 Effect of Shear Deformation

For simple beam with a uniform load case

EI

qllwM 384

5)

2(

4

=

Shear Effect

▬ Shear force of load case q

▬ Shear force of load case q

▬ Deflection by shear force

GA

qlfVV

l

GA

fVdxV

GA

fVdxV

GA

fw ss

ls

ls

S

e

822

122 22/

00

====

GA

EI

l

f

EIql

GAqlf

w

w ss

M

s

40384

384/58/

24

2

==

for a rectangle section of bh× with steel

2

2

2

3

5.21240

6.238456

l

h

bhl

bh

w

w

M

s =×

××=

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

1/2

ql/2



74

3.3 Truss Problems

3.3.1 Principle of Virtual Work

From General principle

=++=

α+=α=⋅

ii

ii

e

l ll

s

iii

S

lEA

FFdx

EA

FFdx

GA

VVfdx

EI

MM

vuwdS

e ee

0 00

22

)(

coscoswq

From equilibrium equation

0 , 0)(

1

)(

1

=+−=+− ==

iim

j

ij

iim

j

ij YVXH for njni ,,1L= (

ij

ij

ij

ij

ij

ij FVFH θ=θ= sin , cos )

where m(i), ijH and ijV 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−

njn

i

iiiinjn

i

im

j

ij

ij

iim

j

ij

ij

i

njn

i

iiim

j

ij

ij

iiim

j

ij

ij

vYuXFvFu

vYFuXF

11

)(

1

)(

1

1

)(

1

)(

1

) ()sin cos(

0))sin( )cos((

Each member has two ends, and each end should be connected to two different joints.

Therefore adding all member end forces in a joint-wise fashion is equivalent to adding all

member end forces in a member-wise fashion because the member end forces of a

member

iY

iX

iF1−

ijF−

i imF )(−

iY

iX

pqF−

ijF ijF−

pqF

i-th joint

p-th joint

k-th member

1

2



75

==

== ==

+=−θ+−θ

+=θ+θ

njn

i

iiiinmb

ikkkkkkkk

njn

i

iiiinjn

i

im

j

ij

ij

iim

j

ij

ij

i

vYuXvvFuuF

vYuXFvFu

11

1212

11

)(

1

)(

1

) ())(sin )(cos(

) ()sin cos(

==

====

=+

==∆=−θ+−θ

nmb

k k

kkknjn

i

iiii

nmb

k k

kkk

k

kknmb

kk

nmb

kkk

nmb

kkkkkkkk

EA

lFFvYuX

EA

lFF

EA

lFFlFvvuuF

11

1111

1212

) (

))(sin )((cos

Since α represnts the angle between the applied unit load and the displacement vector, αcosiu are the displacement of the joint i in the direction of the applied unit load.

For vertical displacement For Horizontal displacement

iu αcosiu

kkk vv θ− sin)(12

kk uu − cos)(12

kθ

kθ

)( 12 kk uu −

)( 12 kk vv −By applying Betti-Maxwell reciprocal

theorem,

==

=+nmb

i k

kkknjn

i

iiii

EA

lFFvYuX

11

) (

The displacement of a joint i in a truss is

obtained by applying a unit load at a joint i

in an arbitrary direction.

=

=α=

α=+nmb

k k

kkki

iiiii

EA

lFF

vYuX

1

cos

cos

u

uX



76

3.3.2 Example

Real System Virtual System

Table for calculation of the deflection of a truss

Member EA

l 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

EA

l

EA

L215)260130( =+=δ

1

2

3

4

5

6

7

8

9

10

11

12

13

20 20 20



77

3.3.3 Conservation of Energy (Truss)

Equilibrium and Conservation of Energy

▬ Equilibrium Equation

0 , 0)(

1

)(

1

=+−=+− ==

iim

j

ij

iim

j

ij YVXH for njni ,,1L=

▬ External work

= ====

+=⋅∆=+=njn

i

iim

j

ij

iim

j

ij

njn

iii

njn

i

iiiiext vVuHvYuXW

1

)(

1

)(

111

) (21

21

)(21

P

int

nmb

k k

kkknmb

kkk

nmb

kkkkkkkkk

njn

i

im

j

ij

ij

iim

j

ij

ij

i

njn

i

iim

j

ij

ij

iim

j

ij

ij

njn

i

iim

j

ij

iim

j

ijext

WEA

lFFlF

vvFuuF

FvFu

vFuFvVuHW

==∆=

−θ+−θ=

θ+θ=

θ+θ=+=

==

=

= ==

= === ==

11

1

1212

1

)(

1

)(

1

1

)(

1

)(

11

)(

1

)(

1

2

1

2

1

))(sin )(cos(2

1

)sin cos(2

1

)sin cos(2

1) (

2

1

intext WW =

Conservation of Energy in each load case

==

⋅∆=njn

iii

nmb

k k

kk

EA

lF

11

2

21

21

P , ==

⋅∆=njn

iii

nmb

k k

kk

EA

lF

11

2

2

1

2

1P

Two load cases are applied simultaneously

====

⋅∆+⋅∆=→+⋅∆+∆=+njn

iiiii

nmb

k k

kkknjn

iiiii

nmb

k k

kkk

EA

lFF

EA

lFF

1111

2

)(2

1)()(

2

1)(

2

1PPPP

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

k k

kkk

EA

lFF

111

PP

===

δ=δ=njn

kkk

njn

kkk

nmb

k k

kkk lFlFEA

lFF

111



78

3.4 Frame Problems

++=∆e

l ll

s

e ee

dxEA

FFdx

GA

VVfdx

EI

MM

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

EI

PllPlllPll

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



79

+=δ

2/

00

2 lls

S dxdxGA

f

GA

Plf)

PlPl(

GA

fdx

GA

VVf ss

V

sS 8

3

2

1

224

1

4

2 =××+××== δ

+=

2/

00

2 lls

A dxdxEAδ

EA

PlPlPl

EAdx

EA

AA

V

A 16

9)

4

1

422

1

2(

2 =××+××==δ

))(75.0)(56.11(16

)961(16

223

22

3

l

h

l

h

EI

Pl

EAl

EI

GAl

EIf

EI

Pl sASM

++=

++=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

dxEI

MM

0

×

P/4 1/4

×

P/2 1/2

×

P/4 1/4

×

P/2 1/2



80




81

Chapter 4

Analysis of Statically Indeterminate Beams

Rxθ L

xθ

1



82

4.1 Propped Cantilever Beam

Equilibrium equation

02

0

=−+

=+−−

lRl

qlM

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

qll

qll

EIdx

EIdx

EI

MM ll

8))(

2(

4

1

1 42

00

0 =−−===δ

EI

lll

l

EIdx

EIdx

EI

MM ll

x 3))((

3

1

1 3

00

=−−===δ

q

RA RB

MA

δ0

R

xBR δ

-ql2/2

1

-l



83

Compatibility condition and the final solution

qlREI

lR

EI

qlBB 8

30

38

34

−=→=+ (up)

281

,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



84

– Rotional Angle at the fixed end

EI

qlqll

EIdx

EIdx

EI

MM ll

241

83

1

1 32

00

0 =×===θ

EI

ll

EIdx

EIdx

EI

MM ll

x 311

3

1

1

00

=××===θ

Compatibility condition and the final solution

23

81

0324

qlMEI

lM

EI

qlAA −=→=+

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



85

4.2 Cantilever Beam with Spring Support

Robin BC (The third type BC)

)()( lkwlwEIV −=′′′−=

Primary structure


wbeam(l)=δspring → spring0 δ−=δ+δ xBR

EIkl

klql

kEI

lEI

ql

Rk

R

EI

lR

EI

qlB

BB 38

31

3

838 3

3

3

4

34

+−=

+−=→−=+

As qlRk B 8

3, →∞→ , and As 0,0 →→ BRk

Deflected Shape for 3

100l

EIk =

δ0

BR

xBR δ

)(lkw



86

4.3 Support Settlement

Primary structure


∆=→∆=δ+δ30

3

l

EIRR BxB

Deflected Shape

∆

RB

xBR δ



87

4.4 Temperature Change

Primary structure

Curvature due to temperature change

dxTTdxTThd )()( 0102 −α−−α=θ

2

212 )(

dx

wd

h

TT

dx

d −=−α

=θ

baxxh

TTw ++

−α−= 212

2

)(

For simple beam, 0)()0( == lww → lh

TTa

2

)( 12 −α=

Comaptibility condition

00 =θ+θ xAM → 032)( 12 =+

−αEI

lMl

h

TTA → EIh

TTM A

)(

2

3 12 −α−=

T1

T2

θ0

MA

xAM θ



88

4.5 Shear Effect

Primary Structures – Shear force diagram


0)()( 000 =δ+δ+δ+δ=δ+δSx

MxB

SMxB RR

GA

fqlql

l

GA

fdx

GA

fdx

GA

VVf

llS

2)1)((

2

2

00

0 ====δ

GA

fll

GA

fdx

GA

fdx

GA

VVf

llSx ====δ )1)(1(

00

2

2

2

2

3

24

7801

0411

8

3

31

41

8

3

3

28

)l/h(.

)l/h(.ql

GAl/fEI

GAl/fEIql

GA/flEI/ql

GA/fqlEI/qlRB +

+−=++−=

++−=

– For retangular section

( ) 222

3

2260

125

30126

12

12

5

6)

l

h(.)

l

h(

.

)v(/Ehbl

/Ebh

GAl

fEI =×

+×=+

=

For 201=

l

h

8

300071

002001

002601

8

3

7801

0411

8

32

2 ql.

.

.ql

)l/h(.

)l/h(.qlRB −=+

+−=++−= (0.07 % error)

1

ql 1

+ +



89

For 101=

l

h

8

300261

007801

010401

8

3

7801

0411

8

32

2 ql.

.

.ql

)l/h(.

)l/h(.qlRB −=+

+−=++−= (0.26 % error)

For 51=

l

h

8

301011

031201

041601

8

3

7801

0411

8

32

2 ql.

.

.ql

)l/h(.

)l/h(.qlRB −=+

+−=++−= (1.0 % 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 RxLxBRL M

EI EI q

ql

Rxθ

Lxθ

1

4

2ql 82ql

q

ql

R0θ L0θ



90

EI

qlqlql

EI

qllqll

EI

dxEI

dxEI

dxEI

MM

ll

lRL

33322

00

2

0

00

48

5)

2416(

1)1

831

4)

2

11(

6(

1

1

1

=+=××+×+=

+=

=θ+θ

EI

ldx

EIdx

EI

dxEI

MM

ll

lRx

Lx

3

2

1

1

00

2

0

=+=

=θ+θ

200

325

qlMRx

Lx

RL

B −=++−=

θθθθ

Deflected shape

1



91

4.7 Fixed-Fixed End Beam

4.7.1. Primary Structure type I

0-th load case (X0)

1st load case (X1)

2nd load case (X2)

ijδ : the deflection in the direction of the unit concentrated load in load case i caused by

load case j. In most of engineering problem, the first subscript denotes results while the

second subscript indicates causes.

EI

qlqll

l

EIdx

EI

MMl

8)

2()(

4

1 42

0

0110 =−×−×==δ

EI

qlqll

EIdx

EI

MMl

6)

2(1

3

1 32

0

0220 −=−××==δ

EI

lll

l

EIdx

EI

MMl

3)()(

3

1 3

0

1111 =−×−×==δ

EI

ll

l

EIdx

EI

MMl

21)(

2

1 2

0

212112 −=×−==δ=δ

EI

ll

EIdx

EI

MMl=××==δ 11

1

0

2222

q

RA RB

MA MB

-ql2/2

M0

1

-l

M1

1 1

M2



92

Compatibility condition (Flexibility equation)

=+−−

=−+→

=δ+δ+δ=δ+δ+δ

026

0238

0

0

21

23

2

2

1

34

22212120

21211110

XEI

lX

EI

l

EI

ql

XEI

lX

EI

l

EI

ql

XX

XX

21ql

X −= , 12

2

2

qlX −=

4.7.2. Primary Structure type II

EI

qlqll

EIdx

EI

MMl

2481

3

1 42

0

0110 =××==δ

EI

qlqll

EIdx

EI

MMl

2481

3

1 32

0

0220 =××==δ

EI

ll

EIdx

EI

MMl

3)1()1(

3

1

0

1111 =−×−×==δ

EI

ll

EIdx

EI

MMl

611

61

δδ

0

212112 =××===

EI

ll

EIdx

EI

MMl

311

3

1

0

2222 =××==δ

M0

M2 1

1

M1

1

1 8

2ql



93


=++

=++→

=δ+δ+δ=δ+δ+δ

03624

06324

0

0

21

3

21

3

22212120

21211110

XEI

lX

EI

l

EI

ql

XEI

lX

EI

l

EI

ql

XX

XX

12

2

21

qlXX −==

Reactions and Moment Diagrams

Deflected Shape

24

2ql 12

2ql−

2

ql 2

ql



94

4.8 3-Span Continuous Beam

4.8.1. Uniform load case

Primary structure

)1(248

13

1

81

3

1

2

1

1

32

2

2

1

2

0

0110 I

I

EI

qlqll

EI

qll

EIdx

EI

MMl+=××+××==δ

)1(248

13

1

81

3

1

2

1

1

32

1

2

2

2

0

0220 I

I

EI

qlqll

EI

qll

EIdx

EI

MMl+=××+××==δ

)1(3

113

111

3

1

2

1

121

2

0

1111 I

I

EI

ll

EI

l

EIdx

EI

MMl+=××+××==δ

220

212112 6

116

1δδ

EI

ll

EIdx

EI

MMl =××===

)1(3

113

111

3

1

2

1

112

2

0

2222 I

I

EI

ll

EI

l

EIdx

EI

MMl+=××+××==δ

EI1 EI2

q

EI1

M0

1

M1

1

M2



95


=++++

=++++→

=δ+δ+δ=δ+δ+δ

0)1(36

)1(24

06

)1(3

)1(24

0

0

22

1

11

22

1

1

3

22

12

1

12

1

1

3

22212120

21211110

XI

I

EI

lX

EI

l

I

I

EI

ql

XEI

lX

I

I

EI

l

I

I

EI

ql

XX

XX

2

1

2

1

221

5.11

1

8

1

I

II

I

qlXX+

+−==

In case 21 II = , 2

21 101

qlXX −==

4.8.2. Complicated Load Case

Primary structure

M0

EI EI

q

EI



96

EI

qlqll

EIdx

EI

MMl

2481

3

1 32

0

0110 =××==δ

EI

ql

EI

ql

EI

qlqll

EI

qll

EIdx

EI

MMl

48

5

162441)

2

11(

6

1

81

3

1 333222

0

0220 =+=××++××==δ


=++

=++→

=δ+δ+δ=δ+δ+δ

03

2

648

5

063

2

240

0

21

3

21

3

22212120

21211110

XEI

lX

EI

l

EI

ql

XEI

lX

EI

l

EI

ql

XX

XX

2

1 401

qlX −= , 22 406

qlX −=

Compatibility Condition (Flexibility Equation) in General

=

∆=δ+δn

jijiji X

10



97

Chapter 5

Analysis of Statically Indeterminate Trusses

1 1

1

21− 2

1−

21−



98

5.1 Various Types of Trusses

Determinate Truss

Externally Indeterminate Truss

Internally Indeterminate Truss

Mixed Indeterminate Truss



99

5.2 A Simple Truss

5.2.1 Method - I

+

P

X

P

P

=



100

Primary structure

At L1

0

75.06.0

25.1

18.0

1

31

3

==+

==

F

FF

F

F

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

L1 L2

U1 U2



101

At L2

75.0 , 25.1

18.0

06.0

45

5

54

−===

=+

FF

F

FF

At U1

1 , 75.0

08.0

06.0

21

52

51

−=−==+=+

FF

FF

FF

1

-1.0

1.25

1.25

-0.75

1

-0.75

F4 F5

1

F1

F2

F5



102

Axial force table for primary structure

Mem 0F xF A L LEA

FF x00 =δ

LEA

Fxx

2

=δ

1 0 -0.75 0.5 0.75L 0 LEA5.0

75.0 3

2 -P -1.0 0.5 L EA

PL

5.0 L

EA5.01

3 1.25 P 1.25 1.0 1.25L EA

PL325.1

EA

L325.1

4 -0.75 P -0.75 0.5 0.75L EA

PL

5.075.0 3

LEA5.0

75.0 3

5 0 1.25 1.0 1.25L 0 EA

L325.1

EA

PL79.4

EA

L59.7


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



103

5.2.2 Method - II

P

+

P

X

X

=



104


EA

XLX 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

EA

XL



105


Mem 0F xF A L LEA

FF x00 =δ LEA

FF xxx =δ

1 0 -0.6 0.5 0.75L 0 EA

L54.0

2 -P -0.8 0.5 L EA

PL6.1 EA

L28.1

3 1.25 P 1.0 1.0 1.25 L EA

PL56.1 EA

PL25.1

`4 -0.75 P -0.6 0.5 0.75 L EA

PL68.0

EA

PL54.0

5 - - 1.0 - - -

EA

PL84.3 EA

L61.3

EA

XLX x −=δ+δ0 → XEA

LX

EA

L

EA

PL 25.161.384.3 −=+

PPX 79.086.4

84.3 −==

PXH 63.08.02 −==



106

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 - 2P - 2P - 22 P

1

1

1

2

1− 2

1−

2

1−

2

1−



107


Mem F 0 xF EAL

LEA

FF x00 =δ LEA

FF xxx =δ

1 P 0 1 0 0

2 P2− 0 2 0 0

3 P 2

1− 1 2P−

2

1

4 P2 2

1− 1 22P−

2

1

5 P2− 1 2 P2− 2

6 - - 2 - -

7 P− 21− 1

2

P

2

1

8 P3 2

1− 1 23P−

2

1

9 P2 0 1 0 0

10 P22− 0 2 0 0

-5.54P 3.41


LEA

XXx 20 −=δ+δ

PXXXP 15.141.141.354.5 =→−=+−

X



108

Temperature change and fabrication error

)22(0 fx LTLEA

XX ∆+∆α+−=δ+δ

In case of no external loads

)2(21.0)2(82.4 ff LTL

EAXLTX

EA

L ∆+∆−=→∆+∆−= αα

5.4 Truss with Two Hinge Supports

Primary structure and compatibility condition

0

2

22212120

121211110

=δ+δ+δ

−=δ+δ+δ

XX

LEA

XXX

3P 2P P

X1

1

2

3

4

5

6

7

8

9

structural analysis istrana.snu.ac.kr/lecture/struct1_2019/notes/19_note(ch1... · 2019-05-03 ·...

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