structural analysis lecture notes

8/10/2019 structural analysis lecture notes

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Introduction


2/112

Process of St

Stage 1 Developin

Stage 2 Investigat

Stage 3 Prelimina

Stage 4 Selection

Stage 5 Reanalysi

Stage 6 Drawing a


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No Yes

nP

uP

n uP P

Stage 1 Developi

Selecting the best location ohas not been predetermined

Legal, financial, sociologica

Structural types, materials

Usually this is done by clien

order of experience, skill, ge

Preliminar

Stage 2 Investigating the loads

Information is generally

provided in the codes and

specifications

Specify the load conditions

and take care of exceptionalcases

Determine the types and magnitudes of loads to be carried

by the structure

Types o

Weight of structure itself an

attached to it.

It has the fixed location and

approximated at first and re

Dead load


4/112

Types of load (2)

Movable load the load which can be transported from one

location to another in the structure without dynamic impact.

Live load

- People, furniture, goods on a building floor (live load),- Snow or ice load

Types oLive load

Moving load the load that m

Sometimes it may be applied s

- Trains or cars on a bridge, wind, eacceleration of vehicles, hydrostat

- In ordinary structural design it is tImpact load --- percentage of liveEarthquake force --- percentage

Types of load (3)

Thermal load

Blast load

Stage 3 Prelimin

It is performed using the princip

It can give the internal forces in

some controlling point

When live loads are involved, on

internal forces in each member.


5/112

Stage 4 Selection of elementsBased on the results from stage 3 and design provisions of the

specification or codes, the sizes and shapes assumed in the stage 2

are checked.

Economical and adequate proportioning of memberscan be achieved using a trial-and-error approach

Stage 5

This step is required on

of members determined

assumed in the prelimin

accuracy of the prelimi

Engineering judgment i

Stage 6 Drawing and detailing

It involves contract drawings, detailing, job

specifications and calculation of final cost.

This information must be accurate to allow

construction to proceed.

Classification of S

Statics vs. dynamics

Plane vs. space

Linear vs. nonlinear

Statically determina

indeterminate stru

Force vs. displacem


6/112

Statics vs. Dynamics

All types of load are treated as the ones having fixed

magnitude and point of application.

It neglects the effect of inertia force

It is the main subject studied in elementary theory of

structure

Statics

Study the dynamic effects on structures considering the

variation of magnitude and location of load on the structure.

Typically considered cases are earthquake, wind, blast,

impact, accelerated moving load.

Dynamics

Dyn


7/112

Plane vs. Space

Typically all structures are three-dimensional, i.e.,

space structureGenerally beam, trussed bridge and roof and rigid

frame buildings are treated as planar structure.

For structures the stresses between members do not

lie in a plane, they must be treated as space

structures.

Idealization of Structures

Y1 Frame

Y2Frame

X1 Frame X2 Frame X3 Frame

Linear vs. Nonl

The material of the structu

law at all points and throug

considered.

The changes in the geome

neglected when the stresse

Linear structures

The above conditions lead tothe applied load and the resul

Principle of superposition is v

-The total effect at a given poin

causes (displacement or force)

sum of the effects for the cause


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Linear vs. Nonl

The material of the structure

points undergoes the inelasticunder the range of applied lo

(Plastic analysis of structures

The material is within the ela

but the geometry of the struc

changes significantly during

application of loads. (Buckli

stability analysis of structure

Nonlinear structures

Statically Determinate vs.Statically Indeterminate Structures

If the structure can be analyzed considering the conditions

for static equilibrium alone, it is statically determinate.

A statically indeterminate

structure is solved by the

equations of static equilibrium

and the equations for the

deformation of the structure.

Force vs. D

Force method (flexi- The forces are treated

express the displaceme

forces.

Displacement meth

- The displacements areand express the forces i

displacements.


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An engineering structure is erected by connecting a number of

structural components and used to support specified loads.

To analyze an engineering structure factors such as load,geometry, connection, material properties and support

conditions must be considered.

The process of transforming a real structure to an ideal

structure suitable for analysis is called modeling. It usually

involves assumptions which is based on experience and in-

depth understanding of real structural behavior.

The results obtained from the ideal structure must be

reasonably as close to the those of real structure.

Idealization (

Idealization (modeling)

Ide

The idealization or

both skills and expe


10/112


Y1 Frame

Y1Frame

X1 Frame X2 Frame X3 Frame

Rigid Zone

Types of Structures Studied

A straight member subjected

only to transverse loads.

It is completely analyzed

when the values of bendingmoment and shear force are

determined

Beam

Members connected byfrictionless pins or hinges.

The external loads are appliat the joints.

Each member is subjected taxial load only and is a twoforce member.

It is completely analyzed wthe values of axial force isdetermined.

Types of Stru

Truss


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They are built of members

connected by rigid joints

capable of resisting moments.

The members are subjected

shear force, axial force and

bending moments.

Types of Structures Studied

Rigid Frame

Structural E

Axial-Force members,

Bracing Struts

Tie Rod

Structural Elements (2)

A horizontal straight member

subjected to vertical load

Primarily designed to resist

bending moment

For short beam carrying large

loads, shear force may govern

the design.

Beam

Beam E

Steel Beam- I section( wide flange beam

- Plate girder

Fabricated by using a plate for wbolting 2 plates to its ends

Used for long span and carryingBuild-up Sections

Concrete Beam- Usually a rectangular cross- Use steel to resist tension a

- Can be cast-in-place or pre

Timber beam (single p


12/112

Structural Elements (3)

Generally a vertical member and resist axial

compressive load.

If a column is subjected to both axial load and a

bending moment is called a beam-column.

Usually is of rectangular or circular cross sections

for concrete and of tubular and wide-flange cross

section for steel column.

Column

Structural E

Usually used for structure

requiring large span and no

concern over depth.

Consists of slender eleme

usually arranged in triangles

Loads that cause the struc

bend are converted into tens

compressive forces in the m

Trusses

Truss Elements

Plane Trusses- Members lies in the same plane

- Used for bridge or roof support

Space Trusses- Members extending in 3-D

- Suitable for derricks and towers

Structural E

Suspen

Cables


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Cable Elements

It is flexible and carries load in tension.

It will not collapse suddenly.

External load is not acting along the axis.

It is commonly used to support bridges

and roof. For such use it has advantage

over beam and truss especially for long-

span structures.

It is limited only by sag, weight and

method of anchorage.

ArcIt is the reversed shape of ca

It must be rigid in order to m

the internal forces of momen

properly designed shape cancompressive forces.


14/112

2Fundamentals

Scalar

()

Vector()

?


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1P

2P

3P

O

A

P

1P

B

O

A

P B

2PO

A

P1P

?

yP

O

P

xP

moment


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moment

verticalforce

momentverticalforce

Equil

The concept of equilibriumis of foremost importance

A structure is said to be in of external forces, it remai

Each part of the structure,structure, called a free bodto the earth under the actiosection and of the external

, F 0

Equations of Equilibrium

Planar structures

Coplanar force systems

Y1Frame

Y2Frame

X1Frame

X2Frame

X3Frame

1F

2F

3F

1M

2M

x F summation of the x comof each force in the syst

summation of the y comof each force in the syst

summation of moments any point a in the plane each force in the system

y F

a M

The balanced force systthree simultaneous equa

y F,x F 0

Equations o


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,x F 0 ,y F 0 a M 0

,y F 0 ,a M 0 b M 0

The line through points a and b is not perpendicular to

the y axis, a and bbeing arbitrarily chosen points and

the y axis being an arbitrarily chosen axis in the plane.

x

y

a

y

a

b

ya b


The three arbitrarily cho

collinear.

y

a

b

a

Equations o

,y F 0 M

,a M 0 M

Planar structures

Coplanar force systems

Concurrent forces

or

or

1F

2F

3F

Imply o M 0

O


,x F 0 ,y F 0 a M 0

y F 0,x F 0

,y F 0 a M 0

,a M 0 b M 0

Parallel forces

or

Equations o

Planar structures

Coplanar force system

,x F 0 y F

,y F 0 M

a M 0 M


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Special Cases of Equilibrium

Two-force member

Two forces must be equal in

magnitude and opposite

aF

bFa

b aF

bF

ab

Three-force member

Three forces must be

at a common point O.

aF

bF

a

b

cFc

Special Cases

SupportsLinking structures to foundations

SupLinking structu

Roller support

Reaction acts normal to thsupporting surface

Cannot resist moment andforce along the surface

rollermovable


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Hinge support

Can resist a general force in any direction Cannot resist the moment of the force

about the connecting point

hinge rotatable

SupportsLinking structures to foundations

Fixed support

Can resist force in any dirmoment of force about the c

Fixed

cannot movecannot rotate

SupLinking structu

Roller

Hinge

Fix

hinge

roller

hinge

1

2

3

Rigid connection

Not allow any relative mobetween members

Used in rigid frames and shear force and axial force f

ConnLinking memb


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Semi-rigid connection

Used in the steel structure between column and beam,allowing partial rotation.

The amount of rotation depends upon the amount ofmoment that the connection transmits.

ConnectionsLinking members to each other

Hinge (pin) connection

Prevent the relative translatibetween members

Transmit force but not mom

Create one or more conditioadditional equations of statics,construction to supplement the

Conn

Truss Frame

Connections

Pin

Rigid

Conn


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Roller connection

Occasionally used in steel structures to provide thefreedom to expand between members

Prevent translation only in the direction normal to themember axis.

0, 0M H

Connections Membe

Truss structure

Pin-connected and pin-lo

(two-force members) One unknown element of

( the axial forc

Members of beams and rigid frames

Internal forces are composed of three unknown elements

the normal force N

the shearing force V

the resisting moment M

N

V

1M

Member Forces

N

V

2M

Stabil ity and

a conc

remain

Determinacy

Stability

an evaluation allows oneto figure out the strategy

for establishing the

solution equations.


22/112

In general the structural stability and determinacy must

be judged by

the number and arrangement of the supports the number and arrangement of the members

the connection of structures

staticallydeterminate

staticallyindeterminate

staticallydeterminate

Stabil ity and Determinacy

Stabil ity and Determinacy of aStructure with Respect to Supports

In studying the stability and determinacy of a structure with

respect to supports, the structure is treated as rigid body and we

must pay attention to the arrangement and number of supports.

Since three equilibrium eq

for the plane problems, tw

supplied by the supports a

stability of a rigid body.

1 2,R R

Two elements of reacti

x F

Unknowns

Equations


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Collinear arrangement

1R

2R

unstableconditionally

stable

Two elements of reactions

Parallel arrangement

1R 2Runstable

Two elements of reactioTwo elements of react

Concurrent arrangement

1R 2R

Two elements of reactionsAt least three elements orestrain a body in stableThere is no inappropriat

1R

2R 3R

1R

2R

3R

The body will neither move hoThe system is called statically

Unknowns

Equations x F 0

1 2, R R


24/112

If there are more than three elements of reaction and noinappropriate arrangement, the structure is more stable andthe number of unknown reaction is greater than the numberof equations of equilibrium. Such a system is called astatically indeterminate with regard to support reactions.

1R

2R 3R 4R

5R

5 3 2 Degree of statical indeterminacy

1R

1R

2R2R

3R

3R4R 4R

5R 5R

It is important to note that

should be at least three is

a sufficient condition for a

External Geo

The instability of strucof arrangement of supp

For a monolithic rigid body, its internal force can

always be determined using equations of equilibrium

once the reactions are completely determined.

Sum

If the number of unknown ele

three, the equations of equiliband the system is said to be s


and if no external geometric

system is statically stable and


three, then the system is statistable provided that no extern

The excess number n of unkndegree of statical indetermina

Fig. 2-13 there are five unkn3=2, which indicates a statica


25/112

General Stability and Determinacy of Beams

without internal

connections

with internalconnections

Treat the beam as a rigid body and use the conditions

described previously to study the stability and determinacy

Provide at least one

additional reaction

to restore the beam

to a statically stable

state

0M

1aR

0M

1aR

Provide at least two

restore the beam to a

0, 0M H

2aR

1aR

2R

1R

Beams with the number of total unknown forces greater

than the number of static equilibrium equations.

The excess number of unknown forces is called the

degree of indeterminacy

Statically Indeterminate Beam

4R

3R

Unknowns : 4

Equations : 3

Degree : 1

Internal hinge me

by applied load.


26/112

An internal connection is introduced to yield an

originally stable beam.

The instability of the structure is not because ofinadequacy of supports, but because of inadequatearrangement of members.

Internal Geometric Instability

Criterion for investigatin

3r c

r

c

: Number of the reacti

: Number of equation

1c

for a hinge ;2c for a roller ;

0c for a beam with

Statically uns

Statically dete

geometric ins

is involved

Statically ind

3r c

3r c

General Stability and

A truss is composed of connected at ends with

The basic form of a trus


27/112

General Stability and

The number of unknowns

is the number of bars,b,

and reactions,r.

The number of equationsof equilibrium is 2j where

thej is the number of

joints.

2b r j Statical

Statical

that it i

Statical

2b r j

2b r j


The value ofr must be equal tfor statical stability of supports.There must be no inadequacy bars so as to avoid both externa


28/112

b r j ( ) ? (2 )b r j

It is stable and determina

and supports.

0M


29/112

General Stability and Determinacy of

Rigid Frames

A frame is built of beams (girders) and columns. :b:r

Number of the beam e

Number of the reactio

Number of jointsNumber of equations

3 3b r j c Static

Static

that it

Static

3 3b r j c

3 3b r j c

:j:c


pin

14 9 13 4

11 9 10 1


30/112

b r j (3 ) ? (3 )b r j c c

Number of equations of condition ?

General Stability a

Rigid

General Stability and Determinacy of

Rigid Frames Unstable

Stable

External geomreactions (num

Internal geommembers (con

Determinate

Indeterminate


31/112

!

?!


32/112

Method of S

=

+

Method of Superposition Method of S


33/112

Three types of trusses

Simple truss

Compound truss

Complex truss

Truss

Starts with three memin a form of triangle andtwo new bars for each n

Simpl

Simple Truss

Two or more simple trussat certain joints by three linconcurrent or by equivalent

Compou


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Trusses that can not be classified assimple or compound.

Complex Truss

Simple

Simple

Com

Co

Analysis of Statically Determinate Truss

Method of joint

0 12 kipsd aM R

aR

0 12 kipsy dF R

dR

2 2

Method of section

Analysis of Statical


35/112

0iM

150 kipsaR

aR

Find the forces in chord members cdand CD and in thediagonal Cd c

M

cdS

0oM

150 200109 kips

275CdV

7.8109 142 kips

6CdS

A compound truss consisconnected by three bars,

3

First, find the reactions:A H


36/112

0CM

120 330 kN

12AFS

0FM

120 3 30 kN

12CDS

2

2

Method of SubThis is a method for solvsupport reactions can be dthe member forces can not

3

3


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Method of Substitute MemberStep 1

Determine which member is to be substituted. It isimportant that the new arrangement of the members

must make the structure stable.

2

Analyze the truss underthe member forces

Method of SubSte

At two ends of the removedmember, apply two equal butopposite unknown forcesdenoted as, for example, X, andthen analyze the truss withoutconsidering the applied force.This unknown force is actuallythe member force of removedmember under the applied load.

The member force of themember at the new location is thesum of step 2 and step 3 whichmust be zero since the memberdoes not exist originally and thevalue of Xcan be obtained.

Method of Substitute MemberStep 3-1

However, to make theanalysis simple usually theunit member force is applied

Method of SubSte


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= +

Member iS i iX i i iS S X

ABBCCDDEEFFA

ACBECF

0AC ACS X

ACS

ACS ACXAC

1 X

1 X

DBS

=

iS

Notes for Complex Truss

Complex trusses may often be arranged so as to begeometrically unstable. However, it is not always possible tosee a critical form just by inspection. Detection is based onthe principle that, if the analysis for the truss yields a uniquesolution, then the truss is stable and statically determinate; onthe other hand, if the analysis fails to yield a unique solution,the truss has a critical form.

The method just described is practical for complex trusseswith only a few members when it is easy to determine whichmember is to be substituted. A more practical method,applicable to all trusses, is the matrix analysis based on themethod of joints described in the next section.

Matrix AStatically Dete

This method consists o

Write all the equationjoints.

Arrange the unknown

Express the equations

matrix form Solve the problem usi


39/112

( )

( )

If the determinant ofmatrix A is zero, thenthe truss is unstable.

AQ R1

Q A R

Q b R (Force transformation equation)

b : force transfer matrix1

0

0

0

0

0

A

A

B

B

a

b

H

V

H

V

S

S

Q


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1 0 0 0 0.36 0.48 0

0 1 0 0 0.48 0.64 0

0 0 1 0 0.64 0.48 0

0 0 0 1 0.48 0.36 0

0 0 0 0 0.6 0.8 0

0 0 0 0 0.8 0.6 10

A

A

B

B

a

b

H

V

H

V

S

S

1Q A R bR

Description o

Descript ion of Bridge Truss Types o


41/112

Types of Trusses

The Forth Rail Brid

Greater New Orleans Br


42/112

Sydney Harbor Bridge Australia (1932)

New York Bay Bridge USA (1931) (

(Vierendeel)

(Truss)


43/112

Tacoma Bridge Seattle, USA (1940)

Reconstruction


44/112

Analysis of Statically DeterminateRigid Frames

In the analysis, the support reactions have to be

determined first.Using the equation of equilibrium and the free-

body diagram, one can obtain the internal forces:

axial force, shear force and bending moment.

The axial force, shear force and bending moment

diagrams can then be plotted along the centroidal

axis.


45/112

54

3

2 8 30.96

10 5

2 8 41.28

10 5

Load

Shear

Moment

Axial force


46/112


47/112

Approximate Method for StaticallyIndeterminate Rigid Frames

For statically indeterminate rigid

frame, assumptions have to bemade in order to render theproblem equivalent to a staticallydeterminate problem

The assumptions are based onexperience and knowledge ofmore exact analyses.

The approximate analysis isuseful for preliminary design andcost estimation.

ApproximaStatically Indeterm

under Uniform

The axial force in each girder is

small and can be neglect.

A point of inflection (zero moment)

occurs in each girder at a point one-

tenth of the span length from the

left end of the girder.

A point of inflection (zero moment)

occurs in each girder at a point one-

tenth of the span length from the

right end of the girder

Approximate Method forStatically Indeterminate Rigid Frames

under Uniform Vertical Loads

Approximate MeIndeterminate Rigid Concentrated Load


48/112

A point of inflection exists at the

center of each column.

The unit axial stresses in the columns

vary as the horizontal distances of the

columns from the center of gravity of

the bent.

All columns are identical in a story,

so the axial forces of the columns in a

story will vary in proportional to the

distances from the center gravity of

the bent.

A point of inflection exists at the center of each girder.

Approximate Method for StaticallyIndeterminate Rigid Frames under LateralConcentrated Loads---Canti lever Method


49/112

Elastic Deformations

The calculation of elasticis of great importance inconstruction of structure

Cantilever method app

The allowable deflectiin the code.

Analysis of statically i

The deformation involvepoints and the rotationaltheir original positions.

Ge

Methods for Calculating the ElasticDeformation

Conjugate Beam

Unit Load Method

Castiglianos Second Theorem

Curvature of

12

s


50/112

12

2 1 s

tan dy

dx

2

2tan

d d y

dx dx

22

2(1 tan )

d d y

dx dx

2 2

2

/

1 /

d d y dx

dx dy dx

?d

dx

d dx

dx ds

12

s

dx

ds

d dx

dx ds

12

2 1 s

2 2

3 / 22

/

1 /

d y d x

dy dx

1 d d dx

ds dx ds

Neglect 2

/dy dx

2

2

d d y

ds dx

Curvature duSmall deflection of beam

Elastic material

Only bending moment considered

Plane section remaining plane aft

d


51/112

2

2

d y M

EIdx

X

Y

2

2

d y M

EIdx

Y

Curvature due to Bending

d

d

cd

cd

d

d

cd

cd

2

2

d y M

EIdx

Conjuga

tandy

dx

Deflection is small.

Mdx

EI

The beam which has the same length as real beam and is subjected

to elastic load is called conjugate beam.

Md dx

EI

My dx dxdx

EI

dy

dx

dV wdx

V wdx

dMV

dx

M Vdx wdxdx

Actual Beam Conjugate Beam2

2

d y M

EIdx

2

2

d Mw

dx

Mdx

EI

Elastic load w

Mdxdx

EI

The slope of a given section

equals the shear in the corres

beam subjected to the elastic

The deflection of a given sec

beam) equals the bending mof the conjugate beam subje

Conjuga

V

y M


52/112

Actual Beam

Elastic Load

Conjugate Beam

w

V wdx

M Vdx

Mw

EI

Units for Con

Sign Convention for Conjugate Beams

( ) ( )V

( ) ( )M y

X

Y

+

(down ) (clockwise )

Elastic Load

( ) ( )

M

w EI w

In order to make the abov

the support conditions of

Fixed end

Simple end

Interior connection


53/112

Actual Beam

Subjected to Applied Load

Conjugate Beam

Subjected to Elastic Load

0 0y

0

0y

0V

0M

Hinge

0V 0M Hinge

Actual Beam


0 0y

0

0y

Actual Beam


Conjugate Beam

Subjected to Elastic Load

0

0y

0

0y

0

0y

0V

0M

hinge

0V

0M pin connection

0V

0M pin connection

pin connection


54/112

Find, by the conjugate beam method, the vertical deflection at

the free end c of the cantilever beam. Assume constantEI.

cM (down )3

6 4

c

wk k

EI

cM

Find for the

beam method. Assume constan

, , andA C C

cM2 ( )

6

c

Pa b a P

EI

Use the conjugate beam method to determine the deflection

and rotation at point b.220,000 kN/cmE

1I 2I

right

2

270 b

EI

1EI

bM1

180 6 4

2b

EI

leftb

V left

1

180 6

2b

EI

rightb

V


55/112

1 2

900 135 0.006 radEI EI

The relative rotation between the left and right sides of b is the

reaction at support b of the conjugate beam.

( ) ( ) ( ) ( )b left b right b left b right bV V R

Determine the maximum defle6200 GPa, 60 10 mmE I

45

EI

63

EI

81

EI27

EI

Work

LTotal work done by F

Total work performed by the applied load during this period

0 0

1

2

PW Fds s ds P

PP

dW Fds

Conservatio

External Wor

(Potential Energ

External Work


56/112

Internal work for a beam1

2IdW Md

d M

ds EI

Md ds

EI

2

2

Mds

EI

2

2

Mdx

EI( )ds dx

2

0 2I

MW dx

EI

Conservation of Energy

External Work = Internal Work

1

2E bW P

M Px

Conservatio

External Work

Internal work for a truss

1

2IdW S dL

S dLE

A dx

SdxdL

EA

2

2

Sdx

EA

2

0 2

L

I

SW dx

EA

L L

S S

Internal Work (Strain Energy)

2

2

S L

EA

2

0 2

lMdx

EI2

2

SL

EA

Beam Truss

Additio

Further deflection occurs,

W

Additional Work


57/112

Method of Virtual Force(Unit Load Method)

1 1 2 21 1 1

2 2 2P P S dL 1

1 1(1)( )

2 2u dL

E IW W1

(1)( )2

11

2u dL

:EW

:IW

1 11 1

2 2P

Method of Virtual(Unit Load Met

1P 2P

1

2

dLS

Unit Load Method

1

2S dL u dL

1 (1)( )2

:EW

:IW

1 1 2 21 12 2

1P P

11

2u dL

Virtual Actual

1 u dL

Unit Loa

1

1

actual sma

virtual force sys

actual sm

virtual force syst

Displacement

Rotation


58/112

Method of Virtual Force(Complementary vir tual work)

The virtual (or imagined) force system in equilibrium

is acting on the structure when the structure is

subjected to a set of actual (or real) deformation.

The deformation system consists of a set small,

compatible deformation.

The principle of virtual work applies to all structure

irrespective of whether the material behaves linearly

or nonlinearly.

Unit Load Method

1

act

virtu1P 2P

1

M

m

Virtual

Actual

Find the deflection and slope at the free end of a cantilever beam

subjected to a uniform load.

1

12 4

1

0 0

( / 2)( )

8b

Mm wx x wdx dx

EI EI EI

(down)

2 32

0 0

( / 2)( 1)

6b

Mm wx wdx dx

EI EI EI

(clockwise)

b

b

2

2

wxM

1m x

2 1m

Find , , and of the loadA C C

Section Origin Limit M

M

2mC


59/112

0 0 0

l a bMm Mm Mmdx dx dx

EI EI EI

10 0 0

( / ) 1 ( / ) ( / )( / )l a bA

Pbx l x lMm Pax l x ldx dx dx

EI EI EI

2 3 3 2 2

2 2

1 2 2 ( )3

2 6 63 3

Pa b Pa b Pab Pab a b Pab l ba

EI l EIl l l EIll l

Section Origin Limit M 1( )Am 2 ( )Cm 3( )Cm

0 0

( / )( / ) (a bC

Pbx l x l Paxdx

EI

3 3

2 2

1 (

33 3

Pa b Pab Pab b

EI El l

0 0

( / )( / ) (a bC

Pbx l bx l Paxdx

EI

3 2 2 3

2 2

1

33 3

Pa b Pa b Pa

EI El l

Section Origin Limit M

Find the deflection at the center of the beam. UseE=30,000 kips/in2

0

l

CMm

E dxI

2

10 10

0 0

1 1 5(10 )2 (5 ) 9.44

1000 2 1500 2

x xx dx dx

1I 2I

M

1

m

Check the units

01

l

CMm

E dxI

2

kips30,000 (1 kips)( )

in.C

Thus3

2

9.44 ft

30,000 in.C

(9.44)(1,728

30,000


60/112

Determine the horizontal, vertical, and rotational deflection

components at end a of the rigid frame. Assume that all members

have the same value ofEI.

F ab bc cd

Mm Mm Mm Mmdx dx dx

EI EI EI EI

1m 3m2mM

horizontal vertical rotational1mM

horizontal

M

1

1m 2m 3mM 1( ) 2( ) 3( ) = horizontal deflection at a d

= vertical deflection at a due

= rotational displacement at a

11

21

31

1mM

horizontal

11

1M m

21

11 21

( );

F F

m dx

EI

Subjected to a unit horizontal f


61/112

= horizontal deflection at a due to a unit vertical force at a

= vertical deflection at a due to a unit vertical force at a

= rotational displacement at a due to a unit vertical force at a

12

22

32

1m 3m2mM

horizontal vertical rotational

12 22 32

2M m

Subjected to a unit vertical force at a2

2 31 2 212 22 32

( ); ;

F F F

m mm m mdx dx dx

EI EI EI

= horizontal deflection at a d

= vertical deflection at a due

= rotational displacement at a

13

23

33

1mM

horizontal

13

3M m

Subjected to a unit couple at a

1 313 23;F F

m m mdx

EI

symmetry

Unit Load Met

1

actual small c

virtual force sy

SdL L L

E EA


62/112

1

11

22

1 2 2m

mm

L

EAS

LS

u u u EA

SL

EAm

1

1m SuL

EA

Matrix form

Unit Load Method for Trusses Find the vertical deflection of joithatL(ft)/A(in.2)=1 and thatE=30

S

v

Actual

1 202

30,000v

SuL SuL

EA E A

0.00673 ft (down)

Find the absolute deflection of jo

and thatE=30,000 kips/in.2 for a

S

Actual


63/112

Find the rotation of member bc.

11

S u

bc

VirtualActual

r

bc bc bc

SuL S

EA

L L L

rSuL

EA

Working formula for finding deflection

due to temperature change

1 u dL

1 u tL

dL tL

where = coefficient of linear thermal expansion

= temperature rise in degrees

t

Find the vertical deflection at joint

of 50oF in the top chordsBCand C

S

b u t L t L

b

b

Actu


64/112


1 2 3( , , ,..., )I E nW W f P P P P

(internal strain energy)IW

The external work is a function of the external loads.

is increased by a differential amount while the internal

work is increased.iP idP

II I I i

i

WW dW W dP

P

External Work = Internal WorkEW 1PiP

nP

1 i

n

1

2I i iW P

id

idP

i

i iP dP

1

2 i iW dP d

1

2i i ij Ij idP P dP W

j

jPothers

1P

nP

iP

iid

idP

1

2i i j jdP P

Castiglianos S

1 2 3( , ,I E W W f P P P

The external work is a functio

is increased by a differen

work is increased.iP

II I I i

i

WW dW W dP

P

I i iW dP


65/112


The displacement of in the direction of isequal to the first partial derivative of strain energy

with respect to .

iP

Ii

i

W

P

PW

P

iP

i

It is a statement regarding the structures compatibility.

The forces must be conservative forces which do work thatis independent of the path and therefore create no energy loss.

Since the forces causing elastic responses are conservative,this method is restricted to linear elastic behavior of thematerial.

1 2

=

11 21

11 P

1 1 1M m P

+

12 22

21 P

2 2 2M m P

M


P

W

P

Internal work for a beam2

0 2

lMW dx

EI

Total bending moment at any section

1 2 1 1 2 2M M M m P m P

21 1

10 0 01 1

( / )

2

l l lM M P MmW Mdx dx dx

P P EI EI EI

: bending moment at any section due to a unit load in place of1 2,m m 1 2,P P

Castiglianos S

Total internal force in any1 2 1 1 2S S S u P u

2

1 1

(

2

S W S L

P P EA

: internal force in any bar du1 2,u u


66/112

Unit Load Method

1 u dL

actual small compatible deformation system

virtual force system in equilibrium1P 2P

1

M

m

myu dA dA

I

MydL dx dx dx

E EI

1 my My

dA dxI EI

2

20 0

l l

A

Mmdx Mmy dA dx

EIEI

Unit Load Met

1

actual small c

virtual force sy

SdL L L

E EA

CastiglianosSecond Theorem

Unit LoadMethod

( / )M M Pdx

EI

Mmdx

EI

( / )S S P L

EA SuL

EA Truss

Beam orFrame

Find the vertical deflection at th

0

( /l b

W M M P

P EI

,M Px M

P

0

1( )( )

l

b Px x dxEI

M P


67/112

Determine the vertical deflection at the free endA.6 4200 GPa, 60 10 mmE I

3Q

0 0

( / ) 1l lA

W M M Q M dx M dx

Q EI EI Q

3)3 (M Qx x Q M

xQ

3 3

0

1( )( ) 0.0833 m

3

l

AQl l

Qx x dxEI EI EI

(down)

Determine the slope at pointB.

3M x

0M

Q

AB BC

0 0

( / ) 1l lB

M M Q dx M

EI EI

5 5

0 0

1( 3 )(0) 3(5x dx

EI

0Q

3M x 0M

Q

AB

x x

3(5 )

+Q

M x 1M

Q

BC

0

( / )lB

M M Qdx

EI

1

M

m

0

l

BMm

dxEI

3M x 0mAB

BC 3(5 )M x 1m

Mm

Q

Determine the vertical displacem2200 GPa, 400 mmE A

Member L S (S Q

AB

AC

BC

8

5

5

22

3Q

5 5

6 2Q

5 5

6 2Q

2

5

2

5

2


68/112

0Q

2

3

5

6

5

6

1

2

3

5

6

5

6

S

Q

u

( / )S S Q L

EA

SuL

EA

Su

Q

Find the vertical and rotational

2

0 0

( / ) (l la

M M Q wdx

EI

1 2, 0Q Q

Find, by Castiglianos theorem, the horizontal displacement

and the rotational displacement at support c. Consider the

bending effect only.

1

2

1 2, 0Q Q


69/112

Find the horizontal

displacement atD.0Q

Maxwells Law of Reciprocal Deflection

1 221 0

lM mdx

EI

2 112 0

lM mdx

EI

1 2 2 1 2 121 120 0 0

( ) ( )l l lPm m Pm m M mdx dx dx

EI EI EI

1 1M Pm 2 2M Pmand

Maxwells Law of R

1 221 0 0

( ) (l l Pm m Pdx

EI

The special case:

When 1P 21

where

= deflection at point 2 resul

= deflection at point 1 alon

load applied at point 2 alon

21

12


70/112

Maxwells Law of Reciprocal Deflection

21 12

where

= rotation at point 2 resulting from a unit couple applied at point 1

= rotation at point 1 due to a unit couple applied at point 2

21

12

The reciprocity extends also to rotational displacement.

For the case of two unit couples applied separately to

any two points of a structure, the law is the rotational

deflection at point 2 on a structure caused by a unitcouple at point 1 is equal to the rotational deflection at

point 1 due to a unit couple at point 2.

Maxwells Law of R

21 12

Maxwells law is perfectl

any type of structure as lo

structural is elastic and fo

Because of virtual force w

deflection at point 2 due t

in magnitude to the linear

original force due to a uni

Determine the horizontal displacement at jointIof the truss due to

a downward vertical load of 15 kips at (a) jointEand (b) joint F.

15 kips

Hint: compute the deflection at two points of a truss for one position of

loading instead of the deflection at one point of the truss for two different

position of loading.

IE EI

IF FI

330 10 ksiE

0.012ei iev v

S SA


71/112

Betti Law

For a linearly elastic structure,

the virtual work done by a P

system of forces and couplesacting through the deformation

caused by a Q system of forces

and couples is equal to the

virtual work of the Q system

acting through the deformation

due to the P system.

Bett

PM

QM

0

1

nL P Q

i Qi

i

M MP dx

EI

01

mL Q P

j Pj

j

M MQ dx

EI

1 1

n m

i Qi j Pj

i j

P Q

PM

QM

Betti Law

W

ij

ji

Bett


72/112

Method of

Consistent Deformation

5A statically indeterminunknown forces includnumber of equilibrium

Gen

1R

2R 3R

Indeterminate to

A statically indeterminate structure can be madedeterminate and stable, called primary or releasedstructure, by removing the extra forces (or restraints)called redundant forces or statical redundant.

1X 2X

1R

2R 3R

Primary or released structure

1 2,X X : redundant forces or statical redundant

General

The number of redundanredundancy.

The choice of redundantbe chosen such that the p

Gen


73/112

21

1 1or 1X X 1

11 21

1

12 22

1 1 11 12

2 2 21 22

1 2

1 2

12

12 2or X 22 2or X

2 2or 1X X

21 1or X

1X 2X

11 1or X

22

11

= displacement at i due to aunit at j, all other points beingassumed unloaded

ij

Flexibility coefficient

21

1 1or 1X X

1 2

1 2

12

12 2or X 22or

2 or 1X X

21or

1X 2X

11 1or X

22

11

The conditional equations or geometric consistenceof the original structure at redundant points are calledcompatibility equations.

1 0 2 0

The compatibility equations can be obtained from theprimary structure by superposition of the deformationscaused separately by the original loads and theredundants.

1 11 1 12 2

2 21 1 22 2

0

0

X X

X X

The method which releasestructure first and then sothe compatibility equationconsistent deformations.

The number of compatibithat of redundants.

Number


74/112

Establish the primary structure

Procedure of Method ofConsistent Deformation

21

1 1or 1X X

1 2

1 2

12 22

12 2or X 22 2or X

2 2or 1X X

21 1or X

1X 2X

11

11 1or X

Set up compatibility equations

1 0 2 0

Compute the deformation ofprimary structure and that due tounit load at each redundant pointand sum them up using principleof superposition as follows.

1 11 1 12 2

2 21 1 22 2

0

0

X X

X X

For the case of n degre

1 21 11 12

2 21 22

1 2

1 2

1 2n n n

X X

X X

X X

Method of Consi

F : structure flexibil

1 11 12 1

2 21 22 2

1 2

n

n

n n n nn

For the case of n degree of indetermancy

1 11 1 12 2 1

2 21 1 22

1

2

1 1 2 2

22

n n

n n

n n n nn n n

X X X

X X X

X X X

In a general form including prescribed displacements,such as support settlement, elastic supports

1 11 12 1 1 1

2 21 22 2 2 2

1 2

n

n

n n n nn n n

XX

X

1( ) ( ) X F

F X

Analyze the propped beam, which is

degree. Assume constant EI.

bX

1 bX

b

b

bb bX


75/112

3

8b

wlX

bX

Moment diagram

0yF

3 5

8 8aV wl wl wl

aV

0aM

2 2 21 3 1

2 8 8aM wl wl wl

aM

aM

1 aM

a

a

aa aM

0a

0c c cc cM

cM

1 cM

0

l

c

Mmdx

EI

2 3

0

/ 4 ( / 2) (2 / )

12

l wlx wx x l dx wl

EI EI

2 2

0 0

(2 / ) 4

3

l l

ccm dx x l dx l

EI EI EI

c

c

cc cM

Compatibility equation0c

3 40

12 3c

wl lM

EI EI

21

16cM wl

Determine the reaction at b (the sp

spring flexibility is f(displacemen

b

wb

b b b


76/112

Determine the reaction at b (the spring force), denoted by . The

spring flexibility is f(displacement per unit force).bX

4

8b

wl

EI

3

3bb

l

EI

b bb b bX f X b

wb

bb bX

1 bX

3If 0,

8bf X wl

If , 0bf X

4 3

08 3 b b

wl lX f XEI EI

3

3 1

8 1 (3 / )bX wl

f EI l

If a beam is provided with n red

spring flexibilities

compatibility equation is1 2, , , nf f f

1 11 1 12

2 21 22 2

1 2n n n

f

f

1 11 12 1

2 21 22 2

1 2n n n n

Find the reactions for the beam with two sections.

0b b bb bR

b

bb bR

bbR

1 bR

2( )

0b bbMm m

dx R dxEI EI

2( )

b

b

b

Mmdx

EIRm

dxEI

,a dR R

1 bR


77/112

The end moments for a fixed-end beam is calledfixed-end moment.

Fixed-end moments are important in slope-deflectionmethod and moment-distribution method.

Fixed-End Moments

AM BM

AM BM

Fixed-end moments of unifoconcentrated load.

AM

0yF

02 2 2

A BM l M lPab

EI EI EI

(1)A BPab

M Ml

0BM

20

2 3 2 3 2 3

A BM l M lPab l b l l

EI EI EI

2

22 (2)A B

Pab PabM M

l l

(1), (2)2 2

2 2,A B

Pab Pa bM M

l l

AM

BM

/ 2Pab EI

2

AM l

EI 2

BM l

EI

Fixed-end moments of unifuniform load.

M


78/112

0y

F

2 20

8 3

wl l Ml

EI EI

21

12M wl

Ml

EI

2

28 3wl lEI

M M

Fixed-end moments of unifconcentrated moment.

AM

BM

0yF 2 2

02 2 2 2

A BM l M lMb Ma

EI EIl EI EIl

2 2

2

( ) (1)A B

M a bM M

l

0BM 2

2

2 2

2 3 2 3 2 3

02 3

A BM l M ll Mb b l

EI EIl EI

Ma abEIl

2

2

[ 2 ( )]2 (2)A B

M a b a bM M

l

AM

BM

2

2

Mb

EIl

2

2

Ma

EIl

2

AM l

EI

2

BM l

EI

2 2(2 ), (2 )A B

Mb MaM a b M b a

l l

Find the reactions components at th

diagram for the entire frame. Assum

Redundants = 3

Analysis of StaticFra


79/112

11 X

21 X31 X

1 2 3, ,

1X

2X

3X

12 22, 32, 13 23, 33,

11 21, 31,

1 11 12 13

2 21 22 23

3 31 32 33

5,000 1,667

7,500 1,000

800 200

1

2

3

1

6 kips

3.33 ft

X

X

X

Apparently the computaconsistent deformation wdegree of redundancy fo

Thus the method of consused for analyzing the rislope-deflection method

method are frequently us

Analysis of StaticFra


80/112

The indeterminateness of truss can be due to redundantsupports or redundant bars or both.

If the indeterminacy is due to supports only, then theapproach described previously can be applied as well.

If the indeterminateness is due to bars, then

The compatibility condition is the relative axial displacementof the two sides of the cut section caused by the combinedeffect of the original loading and the redundants should be zero.

Each of the bars chosen to be redundant forces is cut andreplaced by two equal and opposite axial redundant forcesrepresenting the internal force of the bar.

Analysis of Statically IndeterminateTrusses

Analyze the continuous truss. Assum

(in.2) =1 for all members.

Redundants = 1

c

c

1 cX

0c c cc cX

'S cu

Using virtual force2

0c ccS u L u L

XAE AE

2

c

c

c

S u L

AEXu L

AE

122

3.25

37.5 kips

Bar force:

c cS S u X

Analyze the truss. Assume that E=30

all member.

Redundants = 2


81/112

1X

2X

11 X

21 X

1 1 11 12 1

2 2 21 22 2

0

0

X

X

21 1 1 2

1

222 1 2 2

'

0

' 0

S u L u L u u L

XAE AE AE

XS u L u u L u L

AE AE AE

or

11 1X

1

21

2

21 1X

12 2X

22 2X

'S

1 1u X 2 2u X

S

21 1 1 2

1

222 1 2 2

S u L u L u u L

XAE AE AE

XS u L u u L u L

AE AE AE

1X

2X

11 X

21 X

'S

1 1u X 2 2u X

S

1 1 2 2S S u X u X

Analyze the truss subject to a rise of

Assume =0.0000065 in./in./ , E

for all members.

Redundants = 1

1 F

By virtual force,


82/112

21

1 1( ) 0u L

u t L X AE

14

0.00468 030,000

X

1 35.1 kipsX (tension)

1 1S u X

Each bar force

11 X

It applies only to structure wunyielding supports, and lin

The expressions that the dispequals zero for a loaded strumay be set up by using Cast

11

22

0

0

0n

n

W

X

W

X

W

X

Castiglianos Com(Method of

1X

1 2

0n

W W W

X X X

The redundants must have such value that the total strain energyof the structure is a minimum consistent with equilibrium.

It is sometimes referred to as the theorem of least work.

It cannot be used to determine stresses caused by temperaturechange, support movements, fabrication errors etc.

Castiglianos Compatibility Equation

Total strain energy

2

2

MW dx

EI

Take first derivative with

(

i

M MW

X

Analysis of Stat icBeams and


83/112

For a statically indeterminate beam or rigid framewith n redundants

1

1

2

2

( / )

0( / )

0

0( / )n

n

W M M X dxX EI

W M M X dx

X EI

M M X dxW

EIX

Analysis of Statically IndeterminateBeams and Rigid Frames

Analysis of Stat icTru

Total strain energy

2

2

S W

E

Take first derivative with

i

SW

X

For a statically indeterminate truss with n redundants

1

1

2

2

( / )

0( / )

0

0( / )n

n

W S S X LX EA

W S S X L

X EA

S S X LW

EAX

Analysis of Statically IndeterminateTrusses

Derive a working formula for solvin

beam under general loading.

AV

AM

Apply the method of least wo


84/112

AV

AM

( ) ( )M x M x

( ) AM x M

( ) AM x V x

A AM M M V x

A M M M

whereM Bending moment

1 and A A

M M

M V

AV

AM

0

0

( / )0

( / )0

l A

A

l A

A

M M M dxW

M EI

M M V dxW

V EI

A AM M V x ( )A AM M V x P x a

0 x a a x l

0 0

0 0

( ) [ ( )] 0

( ) [ ( )] 0

l a l

A A A Aa

l a l

A A A A

a

Mdx M V x dx M V x P x a dx

Mxdx M V x xdx M V x P x a xdx

AV

AM

2 2

2 3 2

02 2

( 2 )0

2 3 6

AA

A A

V l PbM l

M l V l Pb a l

2

2

2

3

( 2 )

A

A

PabM

l

Pb l aV

l

Similarly

2

2

2

3

( 2 )

B

B

Pa bM

l

Pa l bV

l

Analyze the frame by taking the inte

mid-span section of the beam as redu

e

e

W

M

W

H

Apply


85/112

3 10 35 0

3 20 45 0

e e

e e

M H

M H

1.0 kip

8.33 ft-kips

e

e

H

M

1.0 kip

3.33 ft-kips

a

a

H

M

eM

eH

Symmetrical

2

5 100 0

10

0

2 (1.2) (1) 15 (1) 02

215 ( ) 0

e e e

e e

xM dx M H x dxEI

M H x x dxEI

Apply the method of l

Analyze the truss. Assume that E=30,000 kips/in.2 and L (ft)/A (in.2)=1

for all members.

1X 2X Apply least work

1

2

( / )0

( / )0

S S X L

EA

S S X L

EA

11

21

( / )78 4 0.6

( / )27.2 0.64

S S X LX

EA

S S X LX

EA


86/112

Find the force in the rod. E=30,000 kips/in.2

X

Internal work in the rod

21

12

X l

EA

Internal work in the beam

2 22

6 12 2

0 62

0.6 10( 6) ( 0.8 )(0.6 )

2 2 2

Xx x X lXxdx dx

EI EI EA

Apply least work 0W

X

2 261

01

(0.6 )

2 2

X l XxW dx

EA EI

The effect of the axial force

small and can be neglected

610

1

2

2

(0.6 )(0.6 )

( 0.8 )( 0.8)

XlW Xx x dX EA EI

X l

EA

15 207.4 3,024 1,94

1/144 1/144

X X

15 207.4 1,080 0.64X X X


87/112

Slope-Deflection Method

Degrees o

When a structure is loadecalled nodes, will underg

These displacements are rfreedom for the structure.

In the displacement methto specify these degrees othe unknowns when the m

Six degrees of freedom for each

node of 3-D beam members

Three degrees of freedom for

each node of 2-D beam members

Degrees of Freedom

The slope-deflection mmethod for beam and ri

The axial deformation o

The deformed configurrotations (slopes) and jo

The basic unknowns aretranslation subjected to the axial deformation o

Ge


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The number of displacement unknowns is usuallyreferred to as displacement degrees of freedom or

kinematic indeterminancy.

The unknowns are solved from equilibriumequations that are equal in number to the unknowns.

General Basic Slope-Def

The basis of the slope deflection equations that express the end mothe end distortions of that membe

( , , , load on span)

( , , , load on span)

ab a b

ba a b

M f

M g

A free-body diagram of a member

Basic Slope-Deflection Equations

The moment due to end rotatioThe moment due to end rotatioThe moment due to a relative dthe member without altering thends.The moment caused by placingwithout altering the existing en

Basic Slope-Def


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abM

abM

abM

baM

baM

baM

FabM

FbaM

abM

baMb

a

b

a 2

2

d y M

EIdx

The moment due to end rotation

0dV

wdx

dMV

dx

abM

a

xy

4

40

d yEI

dx

(0) 0, ( ) 0, '(0) , '( ) 0ay y l y y l B.C.

abM baM

a

xy

The moment due to end rotation while the other end b is fixed.a

2 30 1 2 3( )y x a a x a x a x

2

1 ax

y xl

2

2

d y M

EIdx

M

ab M

IK

l ab M

0x

(stiffness factor)

abM

a

xy

The moment due to end rotation


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The moment due to end rotation while end a is fixed.b

22bab b

EIM EK

l

44bba b

EIM EK

l

abMb baM

(0) 0, ( ) 0, '(0) 0, '( ) by y l y y l B.C.

xy

The moment due to a relative the member without altering thends.

abM

R

(0) 0, ( )y y l B.C.

l

(rigid-body rotation)

x

y

The moment caused by placing the actual loads on the spanwithout altering the existing end distortions.

FabM

FbaM

ab

ba

M

M

I Kl

ab ab ab ab

ba ba ba ba

M M M M

M M M M

ab

ba

M

M

Rl Let


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2 2 3

2 2 3

Fab a b ab

Fba a b ba

M EK R M

M EK R M

Clockwise is takenas positive

abM

abM

abM

baM

baM

baM

FabM

FbaM

a

b

abM

baMb

a

Use the conjugate beam me

deflection equations.

Procedures of Analysis by theSlope-Deflection Method

Identify the joint rotations and joint translations and drawfree-body diagrams for all members.

Write the slope-deflection equation for each member.Simplify the equations with known joint rotations andtranslations.

For each joint write the equation of equilibrium in terms of

the end forces of members connected at that joint.

Solve the equilibrium equations to obtain the values forunknown joint rotations and translations.

The end forces of each member can then be obtained byusing these displacements.

b c

Ste


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2 3

2 3 2520

Fab ab b ab

b

M EK R M

IE R

2 2 3

2 2 3 2520

Fba ab b ba

b

M EK R M

IE R

2 2

2 210

bc bc b c

b c

M EK

IE

2 2

2 210

cb bc c b

c b

M EK

IE

2 2 3

2 2 320

cd cd c

c

M EK R

IE R

2 3

2 320

dc cd c

c

M EK R

IE R

Step 2To simplify the solution wevalues for 2EI/L

2

2 3 2520

ab bI

M E R

2 2 3 2520

ba bI

M E R

2 210

bc b cI

M E

2 210

cb c bI

M E

2 2 320

cd cI

M E R

2 320

dc cI

M E R

2

Ste

Due to horizontal equilibrium

(R)

joint 00b ba bcMM M

joint 00c cb cd MM M

10 520 20

0

ab ba cd dcM M M M

100 0ab ba cd dcM M M M

Due to joint equilibrium

Step 3 (Continue) 6 22 6

3 3 1

b c

b

b c

1.20, 5b c

3 25 5ab bM R

2 3 25ba bM R

2 2bc b cM

2 2cb c bM

2 3 17.cd cM R

3 22.8dc cM R


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abM

abM

abM

baM

baM

baM

FabM

FbaM

abM

baMb

a

b

a

2

2

4 2 6

2 4 6

Fa bab ab

Fa bba ba

EI EI EIM M

l l l

EI EI EIM Ml l l

Fab ab ab ab ab

Fba ba ba ba ba

M M M M M

M M M M M

22 3

22 3

Fab a b ab

Fba a b ba

EIM M

l l

EIM M

l l

Pin-supported end span

0baM 3

2 2a b

32 2

2 2 4

Fa ba

ab a

MRM EK

EK

1

3 (2

F Fa ab bEK R M M


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Analysis of Rigid Frames WithoutJoint Translation By

The Slope-Deflection Method

A B C DAnalysis of Rig

One Degree Of Transl

The Slope-Def


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1 2

sin sin(90

1 2 tan

31 2

2 1 2 1sin sin( ) sin


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Analysis of Rigid Frames WithTwo Degree Of Freedom Of Joint

Translation ByThe Slope-Deflection Method

31 2

1 2sin( ) sin(90

1 2

1 2sin( ) co

31 2 4

1 2 2 1sin( ) sin(90 ) sin(90 )

31 2 4

1 2 2 1sin( ) cos cos


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10

Influences Lines

Concept of I

In designing a structure with live load we have toknow where to place the live load so that it willcause the maximum live stresses.

The maximum live stresses are support reactions,the shear forces, bending moment and axial forcesdepending upon the structure considered.

The position to cause maximum moment may notcause the maximum reactions and so on.

Concept of Influence Line Concept of I

FM


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(10 )(1)1

10 10A

x xR

(1)

10 10k

x xR

AR KR

52

F Kx

M R 5 (5) 1 510 2

F A

x xM R

2

xy 5

2

xy

FM

0 5x 5 10x

Concept of Influence Line

2

xy

Concept of I

AR

An influence line is a curve whose ordinate (y value)gives the value of the function (shear, bendingmoment, reaction, bar force, etc) in a fixed element(member section, support, bar in truss) when a unitload is at the abscissa.

An influence line is only for a selected location of astructure.

Concept of Influence Line Use of Inf

It serves as a criterion istress; that is a guide foof the structure should bthe maximum effect on

It simplified the compu


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Consider a simple beam 10 ft long subjected to the passageof a moving uniform load of 1 kip/ft without limit in lengthand a movable concentrated load of 10 kips that may be

placed at any point of the span. Determine the maximumbending moment at the midspan section C.

CM

CM

Use of Influence Line

CM

0 0( )

l lwdx y wydx w y

(load intensity)

CM

(2.5)(10)(10)(2.5) (1) 25 12.5 37.5 ft-kips

2CM

210 (1)(10)(5) 25 12.5 37.5 ft-kips

2 8CM

By conventional method of computing Mc

Using influence line

CM

Use of Influence Line Method for CInfluence Line --

Apply the equilibrium btaking the appropriate ftravels along the structu

For beam it is often conreaction influence linesshear and moment influ


100/112

2016

BxR

20 41 1

16 16C B

x xR R

BR CR0CM

0 or 0B yM F

( )B LV

( )B RV

BR R

( )B LV

DV

0 8x

8 20x

DV

CRD CV R

DVBR

D BV R

CR

BR

DM

0 8x

8 20x

DM CR

DMBR

BR

CR


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0 4x

8 20x

BMCR

16B CM R

BMBR0BM

BR

BM

BR

CR

Muller-Bresl

It is based on the prin

The desired quantity is released fras the external load. Then a virtuawill have a unit displacement alonshape is then the influence line for

AR

1

As

1

CV

( )( ) (1)( ) 0C C

V s y

C

C

yV

s

1Cs

CV y

Let

1C

s

ah

bh

1a b

h h

a

b

h a

h b

/a

h a

/b

h b

Muller-Breslaus Principle

1

CM

1

( )( ) (1)( ) 0C CM y

C

C

yM

Cs1

a

bC

a h

Muller-Bresl


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For Statically Determinate BeamTo obtain an influence line for the reaction of any statically determinatebeam, remove the support and make a positive unit displacement of its

point of application. The deflected beam is the influence line for thereaction.

To obtain an influence line for the shear at a section of any staticallydeterminate beam, cut the section and induce a unit relative transversesliding displacement between the portion to the left of the section andthe portion to the right of the section, keeping all other constraints (bothexternal and internal) intact. The deflected beam is the influence line forthe shear at section.

To obtain the influence line for the moment at a section of any staticallydeterminate beam, cut the section and induce a unit rotation between theportion to the left of the section and the portion to the right of the section,keeping all other constraints (both external and internal) intact. Thedeflected beam is the influence line for the moment at the section.

Muller-Breslaus Principle Draw influence lines for RMuller-Breslaus principl

AR

1As

DV

Draw influence lines forRA, VD, MD, VE, and MEby

Muller-Breslaus principle.

1D

s

DM

D

Draw influence lines for R

Muller-Breslaus principl


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EV

1E

s

Draw influence lines forRA, VD, MD, VE, and MEby

Muller-Breslaus principle.

Draw influence lines for R

Muller-Breslaus principl

Descript ion of Bridge Truss Description o


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Influence Lines for StaticallyDeterminate Bridge Trusses

Load transferring mechanism for bridge truss.

In the analysis the stringers areassumed to be simply supported.

The load on the deck is transmitted to the stringersThe stringers then transmit the load to floor beams

which are connected to the joints of bridge trusses

Influence LineDeterminate B

Although it is always possible to obtany element for a unit load at each ptime consuming when dealing with a

aid of a computer. Alternatively, we support reactions since they are relatvariable position. After that we can dvery quickly, as can be seen in the fo

l x

l

x

l

Draw the influence lines for

forces in members aB, Bb,

Bc, and bc.

aBS

BbS

BcV

bcS

1 Draw the influence lines forforces in members aB, Bb,

Bc, and bc.


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aBS

BbS

BcV

bcS



Bc, and bc. 1



Bc, and bc.


bar forces (or components) in

members cd, and Cc.

aR

gR

CcV

cdS

gRcdS

a x c 5 21cd gS R

21

5

gcd

RS

aR cdS

d x g 5 15cd aS R

3cd aS R



members cd, and Cc.


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gR

CcV

a x c Cc gV R



members cd, and Cc.

aR

gR

CcV

cdS

aR CcV

d x g Cc aV R



members cd, and Cc.

Moving Loads Truck

0.4w0.1w

0.1w 0.4w

425cm

0.4w

0.4w

V

Moving Lo


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Moving Loads Train Influence Lines aLoad S

AR

1R

2R

1 22 4 12.4R R R

Figure shows a simple beam

subjected to the passage of

wheel loads. We wish to find

the maximum reaction at the

left end A.

maximum

Maximum Shear Fto a Series of Co

CV


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Maximum Shear Fto a Series of Co

Instead of using trial-and-error comthe critical position of the loads cadetermined by computing the chan

V for each possible case.

If a negative change in is obsefirst time, then the location of the pcase is the critical position.

The increment of from positionalong a line with slope s iswherep is the magnitude of load.

The increment of for a load moacross a jump is is the magnitude of load and y2 andordinates of influence diagram.

V

VV p

V( 2 1V p y y


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1 21( 1) (1 4 4)(0.025)(5)

0.125 k

V

2 3

4( 1) (1 4 4)(0.025)(5)

2.875 k

V

0.025s

Maximum Bendingdue to a Series of C

Instead of using trial-and-error comloads can be determined by compufor each possible case.

If a negative change in is obseof the previous case is the critical p

The increment of from positio

M

( 2 1)M p s x x M

wherep is th

1 2

7.5 7.52 (4) (4 3) (4)

10 40 10

1.0 k ft

M

2 3

7.5 7.5(2 4) (6) 3 (6)

10 40 10

22.5 k ft

M

max

(2) 4.5 4 7.5 3(6.0) 57.0 k ft

CM

7.5

40 10s

7.5

10s

1

BcV


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(1) Joint Equilibrium

Force Method & Displacement Method

(2) Member Flexibility/Stiffness

(3) Joint Displacement (Compatibility)

(Flexibility) (Stiffness)

- Determine the internal member forces

- Express the member elongation in term ofmember force

- Express the joint displacement in term ofmember elongation

Force Method

Force Method & Displacement Method(Flexibility) (Stiffness)

(1) (2) (3)

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