Effects Of Temperature, Pre-strain & Support Displacementacademic.csuohio.edu/duffy_s/Class Info on Web/511_09.pdf · ... Flexibility Method ... As before the superposition equation

Lecture 9: Flexibility Method

Effects Of Temperature, Pre-strain & Support Displacement

I th i ti h l id d l d ti th t t WIn the previous sections we have only considered loads acting on the structure. We would also like to consider the effects of

• Temperature changes {DQT}p g { QT}

• Prestrain of members {DQP}

These effects are taken in to account by including them in the calculation of displacements (next page) in the released structure in a manner similar to {DQL} The effects will produce displacements in the released structure, and the displacements are associated with the redundant actions {Q}in the released structure.

The temperature displacements {DQT } in the released structure may be due to either uniform changes in temperature or to differential changes in temperature. A differential change in temperature assumes that the top and the bottom of the member h d h ill d l h i f h lchanges temperature and thus will undergo a curvature along the axis of the structural

component. A uniform change in temperature will increase or decrease the length of the structural component.


When the matrices {DQT } and {DQP}are found they can be added to the matrix {DQL } of displacements due to loads in order to obtain the sum of all displacements in the released structure. By superposition

As before the superposition equation is solved for the matrix of redundants {Q}.

{ } { } { } { } [ ]{ }QFDDDD QPQTQLQ +++=

As before the superposition equation is solved for the matrix of redundants {Q}.

Consider the possibility of known displacements occurring at the restraints (or supports) of the structure. There are two possibilities to consider, depending on

h h h i di l d f h d d i {Q}whether the restraint displacements corresponds to one of the redundant actions {Q}.

If the displacement does correspond to a redundant, its effect can be taken into account by including the displacement in the vector {DQ }.y g p { Q }

In a more general situation there will be restraint displacements that do not correspond to any of the selected redundants. In that event, the effects of restraint displacements must be incorporated in the analysis of the released structure in a manner similar tomust be incorporated in the analysis of the released structure in a manner similar to temperature displacements and prestrains. When restraint displacements occur in the released structure a new matrix {DQR } is introduced.


Thus the sum of all matrices representing displacements in the released structure will be denoted by {DQS} and is expressed as follows

{ } { } { } { } { }QRQPQTQLQS DDDDD +++=

{ } { } [ ]{ }QFDD QSQ +=

The generalized form of the superposition equation becomes

{ } { } [ ]{ }QQSQ

When this expression is inverted to obtain the redundants we find that

{ } [ ] { } { }{ }QSQ DDFQ −= −1


Joint Displacements, Member End Actions And Reactionsp

In the previous sections we have focused on finding redundants using the flexibility method. After redundants were found other actions in the released structure could be found using equations of equilibrium When all actions in a structure have beenfound using equations of equilibrium. When all actions in a structure have been determined it is possible to compute displacements by isolating the individual components of a structure and computing displacements from strength of materials expression. Usually in a structural analysis the displacements of the joints are of primary interest.

Instead of following the procedure just outlined we will now introduce a systematic procedure for calculating joint displacements, member end actions and reactions directly into the flexibility method computationsinto the flexibility method computations.


Consider the two span beam to the left where we will compute the redundants Q1 and Q2 as well as the joint 1 2displacements DJ1 and DJ2, as well as reactions AR1 and AR2.

Joint displacement denoted by {D }Joint displacement denoted by {DJ } can be either a translation of rotation

Member end-actions {AM} are the couples and forces that act at the ends of a member when that component is isolated from the remainder if the structure The sign convention forstructure. The sign convention for member end actions will be:

+ when up for translations and forcesh l k i f i+ when counterclockwise for rotation

and couplesReactions other than redundants will be denoted {AR}.


The principle of superposition will be used to obtain the joint displacements [DJ] in the actual structure. In order to do this we need to evaluate the displacements in the released structure.

QDQDDD ++=

In the released structure the displacements associated with the actual joint displacements are designated {DJL}. The rotations at joints B ( = DJ1) and C ( = DJ2) are required. Consider the expressions

Here

DJ1 = is the displacement desired, in this case the rotation at joint B

21211111 QDQDDD JQJQJLJ ++=

J1 p jDJL1 = is the displacement at joint B caused by the external loads in the released

structure.DJQ11 = is the displacement at joint B caused by a unit load at joint B

corresponding to the redundant Qcorresponding to the redundant Q1DJQ12 = is the displacement at joint B caused by a unit action at joint C

corresponding to the redundant Q2

A i il i b d i d f th t ti t C ( D ) iA similar expression can be derived for the rotation at C ( = DJ2), i.e.,

22212122 QDQDDD JQJQJLJ ++=


The expressions on the previous slide can be expressed in a matrix format as follows

{ } { } [ ]{ }QDDD JQJLJ +=

where

{ }

1JD

D { }

1JLD

D { }

1Q

Q

where

{ }

=

2JJ D

D { }

=

2JLJL D

D { }

=

2QQ

[ ]

=

2221

1211

JQJQ

JQJQJQ DD

DDD


In a similar manner we can find member end actions via superposition

21211111 QAQAAA MQMQMLM ++=



For the first expression

23213133 MQMQMLM


AM1 = is the shear force at B on member ABAML1 = is the shear force at B on member AB caused by the external loads on

the released structurethe released structureAMQ11 = is the shear force at B on member AB caused by a unit load

corresponding to the redundant Q1AMQ12 = is the shear force at B on member AB caused by a unit load

di t th d d t Qcorresponding to the redundant Q2

The other expressions follow in a similar manner.


{ } { } { }{ }QAAA

The expressions on the previous slide can be expressed in a matrix format as follows

A A

{ } { } { }{ }QAAA MQMLM +=

where

{ }

=2

1

QQ

Q{ }

=3

2

1

M

M

M

M AAA

A { }

=3

2

1

ML

ML

ML

ML AAA

A

4

3

M

M

A

4

3

ML

ML

A

AA

{ }

=2221

1211

MQQM

MQMQ

MQ AA

AA

AA

A

4241

3231

MQMQ

MQMQ

AA

AA


In a similar manner we can find reactions via superposition

21211111 QAQAAA RQRQRLR ++=

22212122 QAQAAA RQRQRLR ++=

For the first expression

A i th ti i th t l b t AAR1 = is the reaction in the actual beam at AARL1 = is the reaction in the released structure due to the external loadsARQ11 = is the reaction at A in the released structure due to the unit action

corresponding to the redundant Q1p g Q1ARQ12 = is the reaction at A in the released structure due to the unit action

corresponding to the redundant Q2

The other expression follows in a similar manner.p


The expressions on the previous slide can be expressed in a matrix format as

{ } { } [ ]{ }QAAA{ } { } [ ]{ }QAAA RQRLR +=

where

{ }

=2

1

R

RR A

AA { }

=2

1

RL

RLRL A

AA { }

=2

1

QQ

Q

[ ] 1211 RQRQ AA[ ]

=

2221

1211

RQRQ

RQRQRQ AA

A


When the effects of joint displacements, member end actions and reactions are accounted for the equation of superposition becomes

{ } { } [ ]{ }QDDD JQJSJ +=

here

{ } { } { } { } { }JRJPJTJLJS DDDDD +++=

{ } { } [ ]{ }QDDD JQJSJ +

where

{DJT } = joint displacement due to temperature

{DJP } = joint displacement due to prestrain

{DJR } = joint displacement corresponding to redundants

Hence there is no need to generalize the expression for {AM } and {AR } to account for temperature effects, prestrain and displacement effects. None of these effects will produce any actions or reactions in a statically determinate released structure Instead the releasedany actions or reactions in a statically determinate released structure. Instead the released structure will merely change its configurations to accommodate these effects. The effects of these influences are merely propagated into matrices {AM } and {AR } through the value of the redundants {Q}.


Example

C id th t b t th l ftConsider the two span beam to the left where it is assumed that the objective is to calculate the various joint displacements DJ , member end actions p J AM , and end reactions AR. The beam has a constant flexural rigidity EI and is acted upon by the following loads

PPPLMPP

==1 2

PPPP

==

3

2


Consider the released t t d th tt distructure and the attending

moment area diagrams.

The (M/EI) diagram was drawn by parts Eachdrawn by parts. Each action and its attending diagram is presented one at a time in the figure starting with actions on the far right.


From first moment area theorem

( ) ( ) LPLLPLLPLLPLD 150511121+++= ( ) ( )

EIPL

EIL

EIL

EIL

EID JL

45

225.05.1

212

22

1

=

−++−+=

LEIPLL

EIPLL

EIPLL

EIPLD JL 22

12

323

2122

21

2

2 −

+

−

=

EIPL

813 2

=

102PL[ ]

=

1310

8 EIPLD JL


Using the following free body diagram of the released structure

Th f h i f ilib i

30

LLM A =∑

Then from the equations of equilibrium

2

22

32

2

2

2

PLA

LPLPPLLPA

RL

RL

−=

+−+−=

PPPAFY

20=∑

2

PAPPPA

RL

RL

22

1

1

=+−−=


Using a free body diagram from segment AB of the entire beam, i.e.,

then once again from the equations of equilibrium

022

0

1

1

=+−=

=∑

ML

ML

Y

APPA

F

1ML

22

0

PLPLLPA

M B

+

=∑

23

222

2

2

2

PLA

PLPA

ML

ML

=

−−+=


Using a free body diagram from segment BC of the entire beam, i.e.,

then once again from the equations of equilibrium

0

0

3

=+−=

=∑ML

Y

APPA

F

03 =MLA

0PL

M B =∑

2

2

4

4

PLA

PLPLA

ML

ML

−=

+−=


Thus the vectors AML and ARL are as follows:

3

0

PL

2P

=0

2A ML

−

=

2PL

ARL

−

2PL


Consider the released beam with a unit load at point B

L

LEILD JQ 2

1

2

11 =

L

LEILD JQ 2

1

2

21 =

EIL

2= EI2

=


Consider the released beam with a unit load at point C

( ) LEILD JQ 12

21

12 += LEI

LD JQ 22 2221

=

EIL

23 2

=EIL22

=


{ } 312L

Thus

{ }

=

4131

2 EILD JQ

I i il f hi l i it l d i t d ith Q d Q i th i

11

In a similar fashion, applying a unit load associated with Q1 and Q2 in the previous cantilever beam, we obtain the following matrices

{ }

−−

=

L

LA MQ

010

0

[ ]

−−

A11[ ]

−−

=LL

A RQ 2


{ }

60PQ

Previously (Lecture 5)

{ }

−

=6456

Q

with

{ } { } [ ]{ }QDDD JQJLJ +=

{ }

=172PLD J

then

{ }

− 5112 EIJ


Similarly, with

{ } { } [ ]{ }QAAA MQMLM +=

5

and knowing [AML], [AMQ] and [Q] leads to

{ }

=

L

LPA M

3664

205

56 L36


Finally with

{ } { } [ ]{ }QAAA RQRLR +=

then knowing [ARL], [ARQ] and [Q] leads to

{ }

=L

PA R 31107

56


Summary Of Flexibility Method

The analysis of a structure by the flexibility method may be described by the following steps:

1. Problem statement2. Selection of released structure3 Analysis of released structure under loads3. Analysis of released structure under loads4. Analysis of released structure for other causes5. Analysis of released structure for unit values of redundant6. Determination of redundants through the superposition equations, i.e.,

{ } { } [ ]{ }QFDD QSQ +=

{ } { } { } { } { }QRQPQTQLQS DDDDD +++=

{ } { }{ }[ ] [ ] { } { }{ }QSQ DDFQ −= −1


7. Determine the other displacements and actions. The following are the four flexibility matrix equations for calculating redundants member end actions, reactions and joint displacementsdisplacements

{ } { } [ ]{ }QDDD JQJSJ +=

{ } { } [ ]{ }QAAA +{ } { } [ ]{ }QAAA MQMLM +=

{ } { } [ ]{ }QAAA RQRLR +=

where for the released structure

{ } { } { } { } { }JRJPJTJLJS DDDDD +++=

All matrices used in the flexibility method are summarized in the following tables


MATRIX ORDER DEFINITION

Q 1 U k d d t ti ( N b f d d t)Q q x 1 Unknown redundant actions (q = Number of redundant)

q x 1Displacements in the actual structure Corresponding to the redundantQD

q x 1Displacements in the released structure corresponding to the redundants and due to loadsQLD

q x qDisplacements in the released structure corresponding to the redundants and due unit values of the redundants (Flexibility coefficients)

i l i h l d di h

QQDorF

q x 1Displacements in the released structure corresponding to the redundants and due to temperature, prestrain, and restraint displacements (other than those in DQ)

QRQPQT DDD ,,

q x 1QSDQRQPQTQLQS DDDDD +++=



j 1 J i t di l t i th t l t t (j b fj x 1 Joint displacement in the actual structure (j = number of joint displacement)

j x 1 Joint displacements in the released structure due to loadsJLD

JD

j x 1 Joint displacements in the released structure due to unit values of the redundants

QLD

j x 1Joint displacements in the released structure due to temperature, prestrain, and restraint displacements (other than those in DQ)

JRJPJT DDD ,,

j x 1 JRJPJTJLJS DDDDD +++=JSD



m x 1 Member end actions in the actual structure A (m = Number of end‐actions)

m x 1 Member end actions in the released structure due to loadsMLA

MA

m x q Member end actions in the released structure due to unit values of the redundants

r x 1 Reactions in the actual structure (r = number of reactions)RA

MQA

r x 1 Reactions in the released structure due to loads

RA

RLA

r x qReactions in the released structure due to unit values of the

redundantsRQA

Effects Of Temperature, Pre-strain & Support Displacementacademic.csuohio.edu/duffy_s/Class Info on Web/511_09.pdf · ... Flexibility Method ... As before the superposition equation

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