ar231_chap06_analysisofstructures.ppt

8/9/2019 AR231_Chap06_AnalysisofStructures.ppt

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IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311Fall12/13Fall12/13

04.02.15 Dr. Engin Akta 1

Analysis of the StructuresAnalysis of the Structures


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Introduction For the equilibrium of structures made of several

connected parts, the internal forcesas well the external

forcesare considered.

In the interaction between connected parts,

Newtons 3rdLaw states that theforces of

action and reactionbetween bodies in contact

have the same magnitude, same line of action,and opposite sense.

hree categories of engineering structures are considered!

a) Frames! contain at least one one multi"force member, i.e., member acted upon

b# 3 or more forces.b) Trusses! formed from two-force members, i.e., straight members with end point

connections

c) Machines! structures containing moving parts designed to transmit and modif#

forces.


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Definition of a Truss $ truss consists of straight members connected at

%oints. Nomember is continuous through a %oint.

&olted or welded connections are assumed to be

pinned together. Forces acting at the member ends

reduce to a single force and no couple. 'nl# two-

force membersare considered.

(ost structures are made of several trusses %oined

together to form a space framewor). *ach truss

carries those loads which act in its plane and ma#

be treated as a two"dimensional structure.

+hen forces tend to pull the member apart, it is in

tension. +hen the forces tend to compress the

member, it is in compression.


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Definition of a Truss(embers of a truss are slender and not capable of supporting large lateral loads.

oads !ust "e a##lied at t$e %oints.


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Definition of a Truss


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*i!#le Trusses

$ rigid trusswill not collapse underthe application of a load.

$simple trussis constructed b# successivel# adding two

members such as &- and -/ and one connection to the

basic triangular truss.

In a simple truss, m0 1n" 3 where mis the total number of members and n

is the number of %oints.

$simple trussdoes not alwa#s contain triangular elements, it is possible to have

trusses as


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-ismember the truss and create a freebod# diagram for each

member and pin.

he two forces e2erted on each member are equal in

magnitude, have the same line of action, and

opposite sense. Forces e2erted b# a member on the pins or %oints at

its ends are directed along the member and equal in

magnitudeand opposite.

onditions of equilibrium on the pins provide 1nequations for 1nun)nowns. For a

simple truss, 2n= m+ 3. (a# solve for mmember forces and 3 reaction forces atthe supports.

onditions for equilibrium for the entire truss provide 3 additional equations

which are not independent of the pin equations.

Anal+sis of Trusses "+ t$e ,et$od of -oints

f /


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-oints nder *#ecial oading onditions Forces inopposite members intersecting in two straight lines at a %oint are equal.

he forces in two opposite members are

equal when a load is aligned with a third

member. he third member force is equal

to the load including ero load/.

he forces in two members

connected at a %oint are equal if themembers are aligned and ero

otherwise.

4ecognition of %oints under special

loading conditions simplifies a truss

anal#sis.

ero force

!e!"er

IZMIR INSTITUTE OF TECHNO OGYIZMIR INSTITUTE OF TECHNOLOGY D f A hiD t t f A hit t AR23AR2311 F ll12/13F ll12/13


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*#ace Trusses

$n elementary space trussconsists of 5 members

connected at 6 %oints to form a tetrahedron.

$simple space trussis formed and can be

e2tended when 3 new members and 7 %oint are

added at the same time.

*quilibrium for the entire truss provides 5

additional equations which are not independent of

the %oint equations.

In a simple space truss, m0 3n" 5 where mis thenumber of members and nis the number of

%oints. onditions of equilibrium for the %oints provide

3nequations. For a simple truss, 3n0 m8 5 and

the equations can be solved for mmember forcesand 5 support reactions.

IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY D t t f A hit tD t t f A hit t AR23AR2311 F ll12/13F ll12/13


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04.02.15 Dr. Engin Akta 1&

*a!#le ro"le! &.1

9sing the method of %oints,determine the force in each member

of the truss.

:'L9I'N!

-raw the free"bod# diagram of the entiretruss, and using that solve the 3

equilibrium equations for the reactions at

Eand C.

:tart with ;oint A since it is sub%ected to

onl# two un)nown member forces.-etermine these from the %oint

equilibrium requirements. ontinue with solving un)nown member

forces at %oints D, B, and E from %oint

equilibrium requirements.

$ll member forces and support reactions

are )nown at %oint C.


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04.02.15 Dr. Engin Akta 1'

*a!#le ro"le! &.1

N===,7=+=E

:'L9I'N!

-raw the free"bod# diagram of the entire truss, andusing that solve the 3 equilibrium equations for the

reactions atEand C.

( )( ) ( ) ( ) ( )m3m5N7===m71N1===

=

E

MC

+=

=

= N===,7=E

== xx CF = ==xC

++== yy CF N7=,===N7==="N1====

= N>===yC

1000 2000

& ! & !

3 !

& !

3 !

4 !

E

+

Cy >====

IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY D t t f A hit tDepartment of Architect re AR23AR2311 F ll12/13Fall12/13


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04.02.15 Dr. Engin Akta 1(

1000 2000

& ! & !

3 !

& !3 !

4 !

E

+

*a!#le ro"le! &.1

:tart with ;ointAsince it is sub%ected to onl# two

un)nown member forces. -etermine these fromthe %oint equilibrium requirements.

?36

N1=== ADAB FF ==

ompressionF

ensionF

AD

AB

C

T

N1?==

N7?==

=

=

here are now onl# two un)nown member

forces at %oint -.

( ) DADE

DADB

FF

FF

?

31=

=

CF

!F

DE

DB

N3===

N1?==

=

=

IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13


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04.02.15 Dr. Engin Akta 1)

1000 2000

& ! & !

3 !

& !3 !

4 !

E

+

*a!#le ro"le! &.1

here are now onl# two un)nown member

forces at %oint &. $ssume both are in tension.

( )

N3>?=

1?==7====?6

?6

=

==BE

BEy

F

FF

CFBE N3>?==

( ) ( )N?1?=

3>?=1?==7?=== ?3

?

3

+===

BC

BCx

FFF

!FBC N?1?==

here is one un)nown member force at

%ointE. $ssume the member is in tension.

( )

N@>?=

3>?=3====?3

?3

=

++==EC

ECx

F

FF

CFEC N@>?==



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*a!#le ro"le! &.1 $ll member forces and support reactions are

)nown at %oint C. ?=>===

chec)s=@>?=?1?=

?6

?3

=+=

=+=

y

x

F

F

1000 2000

& ! & !

3 !

& !

3 !

4 !

E

+



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Trusses ,ade of *eeral *i!#le Trusses Compound trussesare staticall#

determinant, rigid, and completel#constrained.

31 = nm

russ contains a redundant member

and isstatically indeterminate"31 > nm

Necessar# but insufficient condition

for a compound truss to be

staticall# determinant, rigid, and

completel# constrained,

nrm 1=+

non-rigid rigid

31 < nm

$dditional reaction forces ma# be

necessar# for a rigid truss.

61 < nm



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Anal+sis of Trusses "+ t$e ,et$od of *ections

+hen the force in onl# one member or the

forces in a ver# few members are desired, themethod of sectionswor)s well.

o determine the force in memberBD# pass a

sectionthrough the truss as shown and create

a free bod# diagram for the left side.

+ith onl# three members cut b# the section,

the equations for static equilibrium ma# be

applied to determine the un)nown member

forces, includingFBD.



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*a!#le ro"le! &.3

-etermine the force in membersF$,

%$, and %&.

:'L9I'N!

a)e the entire truss as a free bod#.

$ppl# the conditions for static equilib"

rium to solve for the reactions atAand'.

Aass a section through membersF$,

%$, and %&and ta)e the right"handsection as a free bod#.

$ppl# the conditions for static

equilibrium to determine the desired

member forces.



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*a!#le ro"le! &.3:'L9I'N!

a)e the entire truss as a free bod#.

$ppl# the conditions for static equilib"

rium to solve for the reactions atAand'.

( ) ( ) ( )( ) ( )( )

( ) ( ) ( )( ) ( )

=

++===

+

==

)N?.71

)N1==

)N?.>

m3=)N7m1?)N7m1=

)N5m7?)N5m7=)N5m?=

A

A'F

'

'

M

y

A



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*a!#le ro"le! &.3 Aass a section through membersF$, %$, and %&

and ta)e the right"hand section as a free bod#.

( ) ( ) ( ) ( ) ( )

)N73.73

=m33.?m?)N7m7=)N>.?=

=

+=

=

=

%&

%&

$

F

F

M

$ppl# the conditions for static equilibrium todetermine the desired member forces.

!F%& )N73.73=



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04.02.15 Dr. Engin Akta 2&

*a!#le ro"le! &.3

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

)[email protected]

=m@cos

m?)N7m7=)N7m7?)N>.?

=

=>.1@?333.=m7?

m@tan

=

=+

=

====

F$

F$

%

F

F

M

%'

F%

CFF$ )[email protected]=

( )

( ) ( ) ( ) ( ) ( ) ( )

)N3>7.7

=m7?cosm?)N7m7=)N7

=

7?.63B3>?.=m@

m?tan

31

=

=++

=

====

%$

%$

'

F

F

M

$&

%&

CF%$ )N3>7.7=



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04.02.15 Dr. Engin Akta 2'

Anal+sis of Fra!es Frames and machinesare structures with at least one

multiforcemember. Frames are designed to support loads

and are usuall# stationar#. (achines contain moving parts

and are designed to transmit and modif# forces.

$ free bod# diagram of the complete frame is used to

determine the e2ternal forces acting on the frame.

Internal forces are determined b# dismembering the frame

and creating free"bod# diagrams for each component.

Forces between connected components are equal, have the

same line of action, and opposite sense.

Forces on two force members have )nown lines of action

but un)nown magnitude and sense.

Forces on multiforce members have un)nown magnitudeand line of action. he# must be represented with two

un)nown components.



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04.02.15 Dr. Engin Akta 2(

Fra!es $ic$ ease To 6e 7igid $en

Detac$ed Fro! T$eir *u##orts

:ome frames ma# collapse if removed fromtheir supports. :uch frames can not be treated

as rigid bodies.

$ free"bod# diagram of the complete frame

indicates four un)nown force components which

can not be determined from the three equilibriumconditions.

he frame must be considered as two distinct, but

related, rigid bodies.

+ith equal and opposite reactions at the contactpoint between members, the two free"bod#

diagrams indicate 5 un)nown force components.

*quilibrium requirements for the two rigid

bodies #ield 5 independent equations.



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IZMIR INSTITUTE OF TECHNOLOGYS U O C O OG Department of Architectureepa t e t o c tectu e AR2331Fall12/13a / 3

04.02.15 Dr. Engin Akta 2)

*a!#le ro"le! &.4

(embersACEandBCDare

connected b# a pin at Cand b# the

lin)DE. For the loading shown,determine the force in lin)DEand the

components of the force e2erted at C

on memberBCD.

:'L9I'N!

reate a free"bod# diagram for thecomplete frame and solve for the

support reactions.

-efine a free"bod# diagram for member

BCD. he force e2erted b# the lin)DE

has a )nown line of action but un)nownmagnitude. It is determined b# summing

moments about C.

+ith the force on the lin)DE)nown,

the sum of forces in thexandy

directions ma# be used to find the force

components at C. +ith memberACEas a free"bod#,

chec) the solution b# summing

moments aboutA"



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pp


*a!#le ro"le! &.4:'L9I'N!

reate a free"bod# diagram for the complete frameand solve for the support reactions.

N6@== == yy AF = N6@=yA

( ) ( ) ( )mm75=mm7==N6@== BMA +=== N3==B

xx ABF +== = = N3==xA

== =>.1@tan7?=@=7

Note!



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pp


*a!#le ro"le! &.4 -efine a free"bod# diagram for member

BCD. he force e2erted b# the lin)DEhas a)nown line of action but un)nown

magnitude. It is determined b# summing

moments about C.

( ) ( ) ( ) ( ) ( ) ( )

N?57

mm7==N6@=mm=5N3==mm1?=sin=

=

++==

DE

DEC

F

FM

CFDE N?57=

:um of forces in thexandydirections ma# be used to find the force

components at C.

( ) N3==cosN?57=N3==cos=

+=

+==

x

DExxC

FCFN>B?=xC

( ) N6@=sinN?57=

N6@=sin=

=

==

y

DEyy

C

FCF

N175=yC

60 mm



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pp

*a!#le ro"le! &.4

+ith memberACEas a free"bod#, chec)

the solution b# summing moments aboutA"

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) =mm11=>B?mm7==sin?57mm3==cos?57

mm11=mm7==sinmm3==cos

=+=

+=

xDEDEA CFFM

chec)s/

ar231_chap06_analysisofstructures.ppt

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