Transcript

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

yf

zδ zf

xf

yM y

θ

zM z

θ x

M xθ

Dept. of Civil and Environmental Eng., SNU

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

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Dept. of Civil and Environmental Eng., SNU

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

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

Dept. of Civil and Environmental Eng., SNU

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

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

1

Chapter 1

Introduction

Dept. of Civil and Environmental Eng., SNU

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

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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

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

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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 <<0

Boundary Condition

– Suppose you have an n-th order differential equation, then n boundary conditions

should be specified to determine a unique solution because there exist n integration

constants in the general solution of your differential equation(s).

– The boundary conditions are specified with up to (n-1)-th derivatives of the unknown

functions.

w , w′ , w ′′ (Moment), w ′′′ (shear Force)

– The same number of the boundary conditions should be specified at each boundary for

structural problems, which means that the governing equation of a structural problem

should an even order differential equation.

– Only half of n-possible boundary conditions is specified at each boundary, and the

other half is unknown, and should be calculated through the structural analysis.

– The derivatives of the unknown function lower than n/2 order is called “displace-

ment” and the derivatives of the unknown function equal to or higher than n/2 order

represent force terms in your structural system

– How to Determine Integration Constants in the General Solution of the given D.E.???

EI

l

q

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

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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

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– 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 xLLxx

EI

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

+

-

+

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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

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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 LxLxx

EI

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

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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 b

RRR 48

)2()0()(

3

21 −===δ

Compatibility Condition

δ0+ δR=0 → 048

)2(

384

)2(5 34

=−EI

RL

EI

Lq b → qLRb 8

10=

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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

MMLR

384)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 8

10=

RbL/2

Rb

q

q

qL2/2

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

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– Moment

– Shear

+

-

+

-

+

+ -

+

-

=

0.070qL2

5qL2/8

Rb

+

=

0.125qL2 L

8

3

+

-

+

qL2/2

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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

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

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Fixed Support(고정단)

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

– Reaction forces are developed in three directions.

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

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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.

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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

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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)

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

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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.

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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

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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

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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

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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.

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

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Chapter 2

Reactions & Internal Forces

by Free Body Diagrams

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

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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.

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

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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 -

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

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2.2 Reactions

Beams

+=−→=

=−+→=

) (clockwise 00

00

LRPaM

PRRF

BA

BAVP

L

aRB = , P

L

bRA =

+=−+→=

=+→=

) (clockwise 0)(0

0

LRaLPM

PRRF

BA

BAVP

L

aRB )1( += , P

L

aRA −=

Truss

PHPHF AAH −=→=+→= 00 ,

00 =−+→= PRRF BAV

0)3(0 =−+→= aRPaPaM BA (Clockwise +)

PRA 31= , PRB 3

2=

P

a b

L

P

RA

LRB

L

b a

P a

L

P a

L

RA

LRB

L

RA

LRB

P

P

HA

L

P

P

3@a

a

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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

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

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2.3 Internal Forces in Framed Structures

Axial Force

Shear Force

Bending Moment

Torsion

+

+

+

+

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

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2.4 Internal Forces in Simple Beams

Reactions

q

RA=qL/2 Rb= qL/

q

RA=qL/2 RB= qL/2

RA RB

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

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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: 2

222,00)()0( x

qx

qLM

qLabLMM −=→==→==

Shear Force and Moment Diagrams

Deflected Shape

Remember 2

2

dx

wdEIM −= !!!

RA

x

RB

Mx

Vx

+

qL/2

qL2/8

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

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.

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

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Dept. of Civil and Environmental Eng., SNU

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

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

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

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= Px

6

1−

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

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

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

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

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

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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 −=→=++−

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

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2.6 Truss (트러스)

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

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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

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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

2@a

a

Dept. of Civil and Environmental Eng., SNU

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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

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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

3@2l

l

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

42

At L1

PFPFP

F

FF

32

, 322

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

Dept. of Civil and Environmental Eng., SNU

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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

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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=

Dept. of Civil and Environmental Eng., SNU

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45

2.6.3 Method of Sections

PFL

FLPM L 34

023

2442 −=→=×+×=

PFL

FL

PLPMU =→=×−×−×= 662 0222

332

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

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46

2.7 프레임 (Frame)

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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

L

qH

2

2

L

qH

2

2

L

qH

2

2

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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

+

Dept. of Civil and Environmental Eng., SNU

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49

2.7.2 Internal Forces in a 3-hinged Frame

Reactions (+:Clockwise for mement)

02

=+

=+

BA

BA

HH

qLRR

= 0RhM : BBBB HRLH

LR 20

2−=→=−−

= 0LhM :

420

422qL

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

Dept. of Civil and Environmental Eng., SNU

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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

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51

283

16

028

316

22

2

xqqLx

qLM

xqxqLx

qLM

−+−=

=−+−−

Deflected Shape

- -

-

Moment

- -

2

16qL 2

16qL

83qL

16

2qL

M

V

Dept. of Civil and Environmental Eng., SNU

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52

2.8 Arches

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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

+=

++−−=

++−=

8max

PlM −=

HA

RA

HB

RB

P

h

2l

x

y

HA

RA

V

H

M

y

l+x

Dept. of Civil and Environmental Eng., SNU

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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

)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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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

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

)()()( =++−++−− xl

xlqxlRyHM 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

Dept. of Civil and Environmental Eng., SNU

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

57

Chapter 3

Principle of Virtual Work

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

Dept. of Civil and Environmental Eng., SNU

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

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

Dept. of Civil and Environmental Eng., SNU

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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

=

+=ξ−δ

→ε

ε+ξ

ε−ξ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε

fFFF

dxxf

dxxfdxxfdxxfdxxxfll

ξ

ξ

ε21

ξ

M M+∆M

V V+∆V

q

∆x

Dept. of Civil and Environmental Eng., SNU

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

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 <<0

For a virtual load case

02

2

=+ qdx

Md,

2

2

dx

wdEIM −= for lx <<0

Equilibrium equation in an integral form

0)(0

2

2

=+ dxqdx

Mdw

l

for all possible w

To multiply some numbers to zero and to add the results always become zero.

Integration by part twice

dxqwdxMdx

wdM

dx

wd

dx

dMw

dxqwdxdx

dM

dx

wd

dx

dMw

llll

lll

−=+−

−=−

002

2

00

000

llll

MVwdxqwdxMdx

wd00

002

2

θ+−−=

As w is the displacement of the same structure caused by another load case q , the

boundary terms vanish since either displacements or reactions should be zero at a

boundary (support).

q w w

q

Dept. of Civil and Environmental Eng., SNU

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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

=+ dxqdx

Mdw

l

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.

)()( 0

0

0

0

xwdxxxwdxEI

MM ll

=−δ= → dxEI

MMxw

l

=0

0)(

Dept. of Civil and Environmental Eng., SNU

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

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

Dept. of Civil and Environmental Eng., SNU

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

63

Values of Product Integrals L

LU dxMM0

Dept. of Civil and Environmental Eng., SNU

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

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

Dept. of Civil and Environmental Eng., SNU

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

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 +

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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

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 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

– 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 !

Dept. of Civil and Environmental Eng., SNU

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

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 lll

ll

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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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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

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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

Dept. of Civil and Environmental Eng., SNU

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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 )()(

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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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= ( i

jij

ij

ij

ij

ij FVFH θ=θ= sin , cos )

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−

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−

iimF )(−

iY

iX

pqF−

ijF

ijF−

pqF

i-th joint

p-th joint

k-th member

1

2

Dept. of Civil and Environmental Eng., SNU

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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

)( 12kk uu −

)( 12kk 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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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79

+=δ

2/

00

2 ll

sS dxdx

GA

f

GA

Plf)

PlPl(

GA

fdx

GA

VVf ss

V

sS 8

3

2

1

224

1

4

2 =××+××== δ

+=

2/

00

2 ll

sA dxdx

EAδ

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

Dept. of Civil and Environmental Eng., SNU

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80

This page is intentionally left blank.

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81

Chapter 4

Analysis of Statically Indeterminate Beams

Rxθ L

1

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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83

Compatibility condition and the final solution

qlREI

lR

EI

qlBB 8

30

38

34

−=→=+ (up)

2

81

,85

qlMqlR AA −==

Moment Diagram

Deflected shape

4.1.2. The second idea

=

+

Compatibility condition

00 =θ+θ xAM

-ql2/2

3ql2/8

-ql2/8 -

2

1289

ql

3l/8

+ =

θ0

MA

xAM θ

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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85

4.2 Cantilever Beam with Spring Support

Robin BC (The third type BC)

)()( lkwlwEIV −=′′′−=

Primary structure

Compatibility Condition

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

Dept. of Civil and Environmental Eng., SNU

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86

4.3 Support Settlement

Primary structure

Compatibility condition

∆=→∆=δ+δ30

3

l

EIRR BxB

Deflected Shape

RB

xBR δ

Dept. of Civil and Environmental Eng., SNU

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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

baxxh

TTw ++

−α−= 212

2

)(

For simple beam, 0)()0( == lww → lh

TTa

2

)( 12 −α=

Comaptibility condition

00 =θ+θ xAM → 032

)( 12 =+−α

EI

lMl

h

TTA → EI

h

TTM A

)(

2

3 12 −α−=

T1

T2

θ0

MA

xAM θ

Dept. of Civil and Environmental Eng., SNU

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88

4.5 Shear Effect

Primary Structures – Shear force diagram

Compatibility condition

0)()( 000 =δ+δ+δ+δ=δ+δ Sx

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

+ +

Dept. of Civil and Environmental Eng., SNU

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

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 Rx

LxB

RL M

EI EI q

ql

Rxθ L

1

4

2ql 8

2ql

q

ql

R0θ L

Dept. of Civil and Environmental Eng., SNU

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

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

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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 111

0

2222

q

RA RB

MA MB

-ql2/2

M0

1

-l

M1

1 1

M2

Dept. of Civil and Environmental Eng., SNU

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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

21

qlX −= ,

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

Dept. of Civil and Environmental Eng., SNU

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93

Compatibility condition (Flexibility equation)

=++

=++→

=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

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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95

Compatibility condition (Flexibility equation)

=++++

=++++→

=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 = , 221 10

1qlXX −==

4.8.2. Complicated Load Case

Primary structure

M0

EI EI

q

EI

Dept. of Civil and Environmental Eng., SNU

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

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

Compatibility condition (Flexibility equation)

=++

=++→

=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 40

6qlX −=

Compatibility Condition (Flexibility Equation) in G eneral

=

∆=δ+δn

jijiji X

10

Dept. of Civil and Environmental Eng., SNU

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

97

Chapter 5

Analysis of Statically Indeterminate Trusses

1 1

1

21− 2

1−

21−

Dept. of Civil and Environmental Eng., SNU

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

98

5.1 Various Types of Trusses

Determinate Truss

Externally Indeterminate Truss

Internally Indeterminate Truss

Mixed Indeterminate Truss

Dept. of Civil and Environmental Eng., SNU

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

99

5.2 A Simple Truss

5.2.1 Method - I

+

P

X

P

P

=

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

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

L

0.75L

L1 L2

U1 U2

Dept. of Civil and Environmental Eng., SNU

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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

Dept. of Civil and Environmental Eng., SNU

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102

Axial force table for primary structure

Mem 0F xF A L LEA

FF x00 =δ L

EA

Fxx

2

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

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

Dept. of Civil and Environmental Eng., SNU

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103

5.2.2 Method - II

P

+

P

X

X

=

Dept. of Civil and Environmental Eng., SNU

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104

Compatibility condition

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

Dept. of Civil and Environmental Eng., SNU

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105

Axial force table for primary structure

Mem 0F xF A L LEA

FF x00 =δ L

EA

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 → X

EA

LX

EA

L

EA

PL 25.161.384.3 −=+

PPX 79.086.4

84.3 −==

PXH 63.08.02 −==

Dept. of Civil and Environmental Eng., SNU

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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−

Dept. of Civil and Environmental Eng., SNU

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107

Axial force table for primary structure

Mem F 0

xF EA

L L

EA

FF x00 =δ L

EA

FF xxx =δ

1 P 0 1 0 0

2 P2− 0 2 0 0

3 P 2

1− 1 2

P− 2

1

4 P2 2

1− 1 2

2P− 2

1

5 P2− 1 2 P2− 2

6 - - 2 - -

7 P− 2

1− 1 2

P

2

1

8 P3 2

1− 1 2

3P− 2

1

9 P2 0 1 0 0

10 P22− 0 2 0 0

-5.54P 3.41

Compatibility condition

LEA

XXx 20 −=δ+δ

PXXXP 15.141.141.354.5 =→−=+−

X

Dept. of Civil and Environmental Eng., SNU

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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

X1

X2

X1

X1

Dept. of Civil and Environmental Eng., SNU

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109

– F0

– F1

– F2

1

1

1 1 1

2P P 3P

-P

2P 2P P

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

2

1−

1 1

1

2

1− 2

1−

2

1−

Dept. of Civil and Environmental Eng., SNU

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110

Axial force table for the primary structure

Mem 0F 1F 2F

EA

L L

EA

FF 10 LEA

FF 20 LEA

FF 11 LEA

FF 21 LEA

FF 22

1 P2 0 1 1 0 P2 0 0 1

2 P22− 0 0 2 0 0 0 0 0

3 P3 2

2− 0 1 P223− 0

2

1 0 0

4 P2 2

2− 1 1 P222− P2 2

1

2

2− 1

5 P2− 1 0 2 P2− 0 2 0 0

6 P− 2

2− 0 1 P2

2 0

2

1 0 0

7 P 2

2− 0 1 P2

2− 0 2

1 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

0371.05

41.171.041.354.5

21

121

=+−−=−+−

XXP

XXXP →

PXX

PXX

5371.0

54.571.082.4

21

21

−=+−=−

→ PX

PX

44.1

94.0

2

1

−==

Dept. of Civil and Environmental Eng., SNU

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

111

Chapter 6

Analysis of Statically Indeterminate Frames

EI

w

Dept. of Civil and Environmental Eng., SNU

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

112

6.1 Γ-shaped Frame-I

Equilibrium equation

02

0

=−−

=++

lRPl

M

PRR

CA

CA

6.1.1 Primary Structure type I

+

δ0

P 1

Rcδx

P

RA

MA

RC l

l

Dept. of Civil and Environmental Eng., SNU

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113

Compatibility condition

0δ δ0 =+ xCR

End Displacements

333

2

000

0

48

29)

48

5

2(

1)}

2)(2

2(

6

1

2))(

2({

1

1

1

PlEI

PlPl

EI

Pll

lll

Pll

EI

dxEI

dxEI

dxEI

MMl

ll

=+=−−−×+−−=

+==δ

333

000

3

4)

3

1(

1)})((

3))(({

1

1

1

lEI

llEI

lll

lllEI

dxEI

dxEI

dxEI

MM lll

x

=+=−−+−−=

+==δ

Compatibility condition and the final solution

03

44829 33 =+ lR

EIPl

EI C

PPRC 45.06429 −=−= , PPRA 55.0

6435 −=−= , PlM A 64

3−=

P

-

-

l

1

l

-

Pl/2

P

Pl/2

- 1

Dept. of Civil and Environmental Eng., SNU

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114

Moment Diagram

+

=

Deflection Shape

Pl/2

P

Pl/2

-

-

P

0.45P

+

0.45Pl

0.45P

0.45Pl

+

0.225Pl

+

-

- 0.05Pl

Dept. of Civil and Environmental Eng., SNU

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115

6.1.2. Primary Structure type II

Compatibility Condition

00 =θ+θ xBM

Rotation Angle

EI

PlPll

EIdx

EIdx

EI

MM ll

161

42

3

6

1

1 2

00

0 =××××===δ

EI

lll

EIdx

EIdx

EI

MM ll

x 3

4)

3(

1) (

1

0

22

0

=+=+==δ

P

P

P/2

P/2

Pl/4

+

1

+

+ 1/l

1/l

MB

xBM θ

Dept. of Civil and Environmental Eng., SNU

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

116

Compatibility condition and the final solution

034

16

2

=+ BMEI

l

EI

Pl → 643Pl

M 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

-

M2

1

l

+

l

M1

1

+

+

Dept. of Civil and Environmental Eng., SNU

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

117

EI

wll

wll

EIdx

EIdx

EI

MM ll

6))(

2(

3

1

1 42

00

0110 −=−===δ

EI

wll

wll

EIdx

EIdx

EI

MM ll

8))(

2(

4

1

1 42

00

0220 −=−===δ

EI

lll

llll

EIdx

EIdx

EI

MM ll

3

4)})((

3))(({

1) (

1 3

0

22

0

1111 =+=+==δ

EI

lll

l

EIdx

EIdx

EI

MM ll

2))((

2

1

1 3

00

212112 ====δ=δ

EI

lll

l

EIdx

EIdx

EI

MM ll

3))((

3

1

1 3

0

2

0

2222 ====δ

Compatibility condition (Flexibility Equation)

=++−

=++−→

=δ+δ+δ=δ+δ+δ

0328

023

4

60

0

2

3

1

34

2

3

1

34

22212120

21211110

XEI

lX

EI

l

EI

wl

XEI

lX

EI

l

EI

wl

XX

XX

281

wlX −= ,

7

32

wlX =

Reactions

w

wl/28

4wl/7

3wl2/28

wl/28

3wl/7

Dept. of Civil and Environmental Eng., SNU

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

118

Moment Diagram

Deflected Shape

6.2.2. Primary Structure type II

-3wl2/28

11wl2/196

-wl2/28

M0

wl2/8

Dept. of Civil and Environmental Eng., SNU

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

119

EI

wldx

EIdx

EI

MM ll

24

1 3

00

012010 ===δ=δ

EI

ll

EIdx

EIdx

EI

MM ll

3)1)(1(

3

1

1

0

2

0

1111 ====δ

EI

ldx

EIdx

EI

MM ll

6

1

00

212112 ===δ=δ

EI

ldx

EIdx

EI

MM ll

3

2) (

1

0

22

0

2222 =+==δ

Compatibility condition (Flexibility Equation)

=++

=++→

=δ+δ+δ=δ+δ+δ

03

2

624

06324

0

0

21

3

21

3

22212120

21211110

XEI

lX

EI

l

EI

wl

XEI

lX

EI

l

EI

wl

XX

XX

28

3 2

1

wlX −= ,

28

2

2

wlX −=

M1

1

+

M2

1

+

+

Dept. of Civil and Environmental Eng., SNU

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120

6.3 Portal Frame subject to Horizontal Load

6.3.1. Primary Structure type I

1

w

M0

-wl2/2

-

EI

w

M2

l

+

+

M3

1

1

+

+

+

M1

-l

-

- -

1

Dept. of Civil and Environmental Eng., SNU

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

121

EI

wldx

EIdx

EI

MM ll

24

1 4

00

0110 ===δ

EI

wldx

EIdx

EI

MM ll

6

1 4

00

0220 −===δ

EI

wldx

EIdx

EI

MM ll

6

1 3

00

0330 −===δ

EI

ldx

EIdx

EI

MM ll

3

5) (2

1 3

0

22

0

1111 =+×==δ

EI

ldx

EIdx

EI

MM ll 3

00

212112 2

1 −=×==δ=δ

EI

ldx

EIdx

EI

MM ll 2

0

2

0

313113

2) (2

1 −=+×==δ=δ

EI

ldx

EIdx

EI

MM ll

3

4) (

1 3

0

22

0

2222 =+==δ

EI

ldx

EIdx

EI

MM ll

2

3 ) (

1 2

0

2

0

323223 =+==δ=δ

EI

ldx

EIdx

EI

MM ll 3 3

1

0

2

0

3333 =×==δ

Compatibility condition (Flexibility Equation)

=δ+δ+δ+δ=δ+δ+δ+δ

=δ+δ+δ+δ

0

0

0

33323213130

32322212120

31321211110

XXX

XXX

XXX

=++−−

=++−−

=−−+

03

232

6

02

3

3

4

6

02

3

5

24

32

2

1

23

3

2

2

3

1

34

3

2

2

3

1

34

XEI

lX

EI

lX

EI

l

EI

wl

XEI

lX

EI

lX

EI

l

EI

wl

XEI

lX

EI

lX

EI

l

EI

wl

Matrix form

−−=

−−

EI

wlEI

wlEI

wl

X

X

X

EI

l

EI

l

EI

lEI

l

EI

l

EI

lEI

l

EI

l

EI

l

6

6

24

3

2

322

3

3

4

2

3

5

3

4

4

3

2

1

22

233

233

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122

=

−−

−=

252

317

24

5

6

6

24

3

2

322

3

3

4

2

3

5

23

4

41

22

233

233

3

2

1

wl

wl

wl

EI

wlEI

wlEI

wl

EI

l

EI

l

EI

lEI

l

EI

l

EI

lEI

l

EI

l

EI

l

X

X

X

Reactions

Moment Diagram

Deflected Shape

w

5wl/24 19wl/24

wl /7 wl /7

59wl2/252

31wl2/252

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123

6.3.2. Primary Structure type II

EI

wldx

EIdx

EI

MM ll

6

1 3

00

0110 −===δ

EI

wldx

EIdx

EI

MM ll

8

1 3

00

0220 ===δ

EI

wldx

EIdx

EI

MM ll

8

1 3

00

0330 −===δ

EI

ldx

EIdx

EI

MM ll

3

4) (

1

0

22

0

1111 =+==δ

1

1 +

+ M1

+

+

-

1

M2

+ +

1

1

M3

M0 w

-wl2/2

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124

21

00

2112 3

) (1 δ=−=+==δ EI

ldx

EIdx

EI

MM ll

31

00

3113 2

1 δ====δ EI

ldx

EIdx

EI

MM ll

EI

ldx

EIdx

EI

MM ll

=×==δ 0

2

0

2222 3

1

32

00

3223 6

) (1 δ=−=+==δ EI

ldx

EIdx

EI

MM ll

EI

ldx

EIdx

EI

MM ll

3

2 2

1

0

2

0

3333 =×==δ

Compatibility condition (Flexibility Equation)

0

0

0

33323213130

32322212120

31321211110

=δ+δ+δ+δ=δ+δ+δ+δ

=δ+δ+δ+δ

XXX

XXX

XXX

Matrix Form

−=

−−

EI

wlEI

wlEI

wl

X

X

X

EI

l

EI

l

EI

lEI

l

EI

l

EI

lEI

l

EI

l

EI

l

8

8

6

3

2

62

63

233

4

3

3

3

3

2

1

−=

−=

−−

−=

252

31504

43504

29

8

8

6

21

44

21

2

21

1621

2

21

23

21

521

16

21

5

21

23

8

8

6

3

2

62

63

233

4

2

3

3

3

3

3

31

3

2

1

wl

EI

wlEI

wlEI

wl

l

EI

EI

wlEI

wlEI

wl

EI

l

EI

l

EI

lEI

l

EI

l

EI

lEI

l

EI

l

EI

l

X

X

X

Dept. of Civil and Environmental Eng., SNU

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125

6.4 Portal Frame subject to Vertical Load

1

1 +

+ M1

+

+

-

1

M2

+ +

1

1

M3

EI,l

a P

M0

l

Pab

Dept. of Civil and Environmental Eng., SNU

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126

l

ba

EI

Pab

l

Pab

l

bldx

EIdx

EI

MM ll 2

61)1(

6

1

00

0110

+=××+===δ

l

ba

EI

Pab

l

Pab

l

aldx

EIdx

EI

MM ll +=××+===δ 2

61)1(

6

1

00

0220

00

0330 ==δ

l

dxEI

MM

Matrix Form

−−

+

+

−=

+

+

−=

ba

ba

ba

l

Pab

l

bal

ba

l

Pab

X

X

X

104

1117

1711

42

0

2

2

21

44

21

2

21

16212

2123

215

2116

215

2123

6 2

3

2

1

Sidesway : )(28

1ab

EI

Pab −=∆

Deflected Shape

4

la =

Dept. of Civil and Environmental Eng., SNU

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127

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

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128

- Order of Indeterminacy : 6 – 6 = 0

Overhanged Beam

- # of unknowns : 4×2 + 1×2 – 2 = 8

- # of equations : 2×2 + 2×2 = 8

- Order of Indeterminacy : 8 – 8 = 0

Gerber Beam

- # of unknowns : 4×3 + 1×3 – 2 = 13

- # of equations : 2×3 + 2×4 – 1= 13

- Order of Indeterminacy : 13 – 13 = 0

Continuous Beam

- # of unknowns : 4×3 + 1×4 = 16

- # of equations : 2×3 + 2×4 = 14

- Order of Indeterminacy : 16 – 14 = 2

6.5.2. Order of Indeterminacy - Truss

Number of Internal Forces in a Member : 1

Number of Equilibrium Equations in a Member : 0

Number of Equilibrium Equations at a Joint : 2

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129

Determinate Truss

- # of unknowns : 1×21 + 3 = 24

- # of equations : 2×12 = 24

- Order of Indeterminacy : 24 – 24 = 0

Internally Indeterminate Truss

- # of unknowns : 1×25 + 3 = 28

- # of equations : 2×12 = 24

- Order of Indeterminacy : 28 – 24 = 4

Internally Indeterminate Truss

- # of unknowns : 1×25 + 5 = 30

- # of equations : 2×12 = 24

- Order of Indeterminacy : 30 – 24 = 6

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130

6.5.3. Order of Indeterminacy - Frame

Number of Internal Forces in a Member : 6

Number of Equilibrium Equations in a Member : 3

Number of Equilibrium Equations at a Joint : 3

Internally Indeterminate Frame

- # of unknowns : 6×3 + 3 = 21

- # of equations : 3×3 + 3×4 = 21

- Order of Indeterminacy : 21 – 21 = 0

Frame with Hinges

- # of unknowns : 6×4 + 4 – 2 = 26

- # of equations : 3×4 + 3×5 – 1 = 26

- Order of Indeterminacy : 26 – 26 = 0

Portal Frame with Fixed Supports

- # of unknowns : 6×3 + 6 = 24

- # of equations : 3×3 + 3×4 = 21

- Order of Indeterminacy : 24 – 21 = 3

Dept. of Civil and Environmental Eng., SNU

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131

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

- Order of Indeterminacy : 69 – 57 = 12

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

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132

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

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133

EI

qlql

EI

ldx

EI

MMl

2481

3

32

0

0110 =××==δ

EI

qlqlldx

EI

MMl

164)1()

2

11(

6

32

0

0220 −=×−×+==δ

00

0330 ==δ

l

dxEI

MM

EI

ldx

EI

MMl

30

1111 ==δ , 01312 =δ=δ

EI

ldx

EI

MMl

30

2222 ==δ , 02321 =δ=δ

EI

ldx

EI

MMl

620

3333 ==δ , 03231 =δ=δ

ql2/8 4

2ql

+ 1

- 1

+ 1

M0 M1

M2 M3

Dept. of Civil and Environmental Eng., SNU

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134

Compatibility Condition

321 δ=δ=δ

1

3

313212111101 324M

EI

l

EI

qlMMM +=δ+δ+δ+δ=δ

2

3

323222121202 316M

EI

l

EI

qlMMM +−=δ+δ+δ+δ=δ

3333232131303 6M

EI

lMMM =δ+δ+δ+δ=δ

One Additional Equilibrium Equation

0321 =++ MMM → 213 MMM −−=

Final Compatibility Condition

062246324 21

3

31

3

=++=−+ MEI

lM

EI

l

EI

qlM

EI

lM

EI

l

EI

ql

026166316 21

3

32

3

=++−=−+− MEI

lM

EI

l

EI

qlM

EI

lM

EI

l

EI

ql

21 64

9qlM −= , 2

2 64

11qlM = , 2

3 64

2qlM −=

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.

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135

6.7 General Joint Compatibility

Compatibility Condition

nδ==δ=δ L21

j

k

njijj

n

jijii MM

+==

δ+δ+δ=δ11

0 ni L1for =

One Additional Equilibrium Equation

021 =+++ nMMM L

Joint i

M1

Mn

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136

This page is intentionally left blank.

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

137

Chapter 7

Influence Lines for

Determinate Structures

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138

7.1 Influence Function

Influence function

Convolution integral – Superposition

ξξξ= dqxIxdR pp )(),()(

ξξξ=l

pp dqxIxR0

)(),()(

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

=

+=ξ−δ

→ε

ε+ξ

ε−ξ→ε

ε+ξ

ε+ξ

ε−ξ

ε−ξ

→ε

fFFF

dxxf

dxxfdxxfdxxfdxxxfll

ξ

ξ

ε2

1

1 ξ

I(xp,ξ)

xp

ξ q(x)

R(xp)

q(ξ)dξ

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139

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

20

l≤ξ≤

2

0)2

(12

ξ=→=ξ−×−−ξ−xx M

lM

l

l

l

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

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140

ll ≤ξ≤2

2

02

ξ−=→=−ξ− lMM

l

l

lxx

Influence line

7.2.2. Shear Force

20

l≤ξ≤

lVV

l

lxx

ξ−=→=++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

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141

ll ≤ξ≤2

l

lVV

l

lxx

ξ−=→=+ξ−− 0

Influence line

7.2.3 Maximum Moment in a Simple Beam

010

≤ξ≤− L

0at 20

)()10

(2

1)( max =ξ=→+ξ= PL

xML

PxM pp

10

4

1020

LLL =−≤ξ≤

10

4at

20

7)(

204

3)

10(

2

1

22)( max

LPLxM

PLPLP

PxM pp =ξ=→+ξ=+ξ+ξ=

10

5

10

4 LL ≤ξ≤

10

4at

20

7)(

20

9

4))

10(

2

1

2(

22)( max

LPLxM

PLPLLP

PxM pp =ξ=→+ξ−=+ξ−+ξ=

RA=(l - ξ)/l

Vx

+

1/2

+

P/2 ξ

P

L/10

2)2/(

ξ=LIM22

)2/(ξ−= L

LI M

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142

10

9

10

5 LL ≤ξ≤

10

5at

40

13)(

20

14

4

3))

10(

2

1

2()

22(

2)( max

LPLxM

PLPLLP

LPxM pp =ξ=→+ξ−=+ξ−+ξ−=

LL ≤ξ≤

10

9

10

9at

40)(

44)

22(

2)( max

LPLxM

PLPLPxM pp =ξ=→+ξ−=ξ−=

7.3 Influence Line of a Gerber Beam

7.3.1. Shear Force at x= L/2

20

L≤ξ≤

LL ≤ξ≤2

LL 5.1≤ξ≤

Influence line

ξ′

L

ξ′2

L

ξ′− 2

1

L

ξ′− 2

1

L/2 L

RA=1

ξ P=1

Vx=0

RA=1

Vx=1

+

1

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143

7.3.2. Moment at the fixed end

L≤ξ≤0

LL 5.1≤ξ≤

Influence line

7.3.1 Maximum Moment at the fixed end

ξ

ξ

==ξ4/4/

)()()(L

M

L

M dxxIqdxxqIM

– L4

30 ≤ξ≤

)4

2(84

)4

(2

1)(

LqLLLqM +ξ−=+ξ+ξ×−=ξ → 22

max 2188.032

7qLqLM −=−=

-

L

)23()2( ξ−−=ξ′−− LL −ξ

-

L

L/4

ξ

ξ ξ−=xM

)2( ξ′−−= LM x

ξ′

L

ξ′2

L

ξ′− 2

1

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144

– LL ≤ξ≤4

3

)358

13(

2)

2

7

2

32

8

21(

2

)4

3)(2

2

5(

2))((

2)(

222222 ξ−ξ+−−=ξ+ξ+ξ−−ξ−−=

−ξξ−+−ξ−ξ+−=ξ

LLq

LLLLq

LLLq

LLq

M

LLq

M6

50)65(

2)( =ξ→=ξ−−=ξ′

2

2222max

2292.0

144

33

72

150300117

2))

6

5(3

6

55

8

13(

2

qL

qLqLqL

LLLq

M

−=

−=−+−−=−+−−=

– LL 25.1≤ξ≤

22max 1875.0

16

3)4

2

3(

84)2

22(

2)( qLqLML

qLLLL

qM −=−=→ξ′−−=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

6

5=ξ

)2( ξ′−− L )22

( ξ′−− L

)22

5( ξ−− L −ξ

ξ−L LLL

4

3)

4( −ξ=−+ξ

Dept. of Civil and Environmental Eng., SNU

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145

7.4 Indirect Load

Equivalent to

10 =+→= bav RRF

0)22

()22

)(1(0 0

0

0

0

=−++−−→= lRll

l

xll

l

xM ba

l

x

l

llRb +−=

20 ,

l

x

l

llRR ba −

+=−=

21 0

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

10 =+→= bav RRF

0)22

(0 0 =−+−→= lRxll

M 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

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146

l

x

l

llRb +−=

20

,

l

x

l

llRR ba −+=−=

21 0 (Statically equivalent to the indirect load)

0)2

)(2

()1(0 00

0

=ξ+−

−+

+ξ−−−→= ξll

l

x

l

ll

l

xMM

ξ−−

+−

−+

=ξ+−

−+

+ξ−−= )12

(22

)2

()2

)(2

()1(0

00000

0 l

x

l

llll

l

x

l

llll

l

x

l

ll

l

xM

7.4.1 Influence line at the mid-span

422222)1

2(

22)

2( 000000

0

000 lll

l

llll

l

lll

l

x

l

llll

l

x

l

llM

−=

−−

−+=−

−+

−−

+=

7.5 Influence Line of Truss

7.5.1 Truss Case

Ra Rb

1-x/l0 x/l0

ξ

Ra Rb

1-x/l0 x/l0

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147

Raa

x

61−=

Rba

x

6=

x

7.5.2 Diagonal Member

ax ≤≤0

a

xF

a

xF

6201

61

22 −=→=−−+−

axa 2≤≤

)65

1(202

61

22

a

xF

a

xa

a

xF −−=→=−−−+−

axa 62 ≤≤

)6

1(206

122

a

xF

a

xF −=→=−+−

x

a

xa

a

ax −=−− 21

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148

Influence line for the diagonal member

a

xFax

62 0 −=≤≤

)65

1(2 2a

xFaxa −−=≤≤

)6

1(2 62a

xFaxa −=≤≤

7.5.3 Bottom Member

Take moment about point A (clockewise +).

ax ≤≤0

a

xFFaxaa

a

x

65

0)(1)6

1( =→=−−×−×−

axa 6≤≤ )

61(0)

61(

a

xFFaa

a

x −=→=−×−

+

-

264

2

61

a

xRa 6

1−=

A

65

+


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