flexibility method – temp. and pre-strain

8/13/2019 Flexibility Method Temp. and Pre-Strain

1/32

Lecture 10: Flexibility Method Temp. and Pre-Strain

Effects Of Temperature, Pre-strain & Support Displacement

In the previous sections we have only considered loads acting on the structure. We

would also like to consider the effects of

Temperature changes {DQT}

Prestrain of members {DQP}

These effects are taken in to account by including them in the calculation ofdisplacements (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 eitheruniform changes in temperature or to differential changes in temperature. A

differential change in temperature assumes that the top and the bottom of the member

changes 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.


2/32


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}.

Consider the possibility of known displacements occurring at the restraints (or

supports) of the structure. There are two possibilities to consider, depending onwhether 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 }.

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 to

temperature displacements and prestrains. When restraint displacements occur in the

released structure a new matrix {DQR } is introduced.

QFDDDD QPQTQLQ


3/32


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

QSQ DDFQ 1

The generalized form of the superposition equation becomes

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


4/32



5/32


Joint Displacements, Member End Actions And Reactions

In previous sections we have focused on finding redundants using flexibility (force) methods.

Typically redundants (Q1

, Q2

, , Qn

) specified by the structural engineer are unknown

reactions. However, we have also stipulated, and found, redundants that are internal actions,

i.e., shear forces, bending moments and/or axial forces. Redundants are determined by

imposing displacement continuity at the point in the structure where redundants are applied.

If the redundants specified are unknown reactions then after these redundants were found

other actions in the released structure could be found using equations of equilibrium.

When all actions in a structure have been determined it is possible to compute displacements

by isolating the individual subcomponents of a structure. Displacements in these

subcomponents can be calculated using concepts learned in Strength of Materials. These

concepts allow us to determine displacements anywhere in the structure but usually the

unknown displacements at the joints are of primary interest if they are non-zero.

Instead of following the procedure just outlined we will now introduce a systematic

procedure for calculating non-zero joint displacements, member end actions and reactions

directly using flexibility methods.


6/32


7/32


The principle of superposition is used to obtain the joint displacement vector {DJ}, which is

a vector of displacements that occur in the actual structure. For the structure depicted on

the previous page the rotations in the actual structure at jointsB ( =DJ1) and C( =DJ2) are

required. Initially, when the redundants Q1 and Q2 are found superposition is imposed on

the released structure requiring the displacement associated with the unknown redundantsto be equal to zero. In finding unknown, non-zero, joint displacements in the actual

structure superposition is used again and displacements in the released structure must be

equated to the displacement in the actual structure. Focusing on jointB, superposition

requires

Here

DJ1 = non-zero displacement (a rotation) at jointB in the actual structure, at

the joint associated with Q1

DJL1 = the displacement (a rotation) at jointB associated withDJ1 caused by

the external loads in the released structure.

DJQ11 = the rotation at jointB associated withDJ1 caused by a unit force at

jointB corresponding to the redundant Q1 in the released structure

DJQ12

= the rotation at jointB associated withDJ1

caused by a unit force at joint

Ccorresponding to the redundant Q2 in the released structure

21211111 QDQDDD JQJQJLJ


8/32


Thus displacements in the released structure must be further evaluated for information

beyond that required to find the redundants Q1 and Q2 . In the released structure the

displacements associated with the applied loads are designated {DJL} and are depicted

below. The displacements associated with the redundants are designated [DJQ ] and aresimilarly depicted.

In the figure to the right unit

loads are shown applied at theredundants. These unit loads

were used earlier to find

flexibility coefficients [Fij ].

These coefficients were then

used to determine Q1 and Q2 .Now the unit loads are used to

find the components of [DJQ ].

released structure


9/32


A similar expression can be derived for the rotation at C( =DJ2), i.e.,

Here

DJ2 = non-zero displacement (a rotation) at joint C in the actual structure, at

the joint associated with Q2DJL2 = the displacement (a rotation) at joint C associated withDJ2 caused by the

external loads in the released structure.

DJQ21 = the rotation at joint Cassociated withDJ2 caused by a unit force at jointB

corresponding to the redundant Q1 in the released structure

DJQ22 = the rotation at joint Cassociated withDJ2 caused by a unit force at joint C

corresponding to the redundant Q2 in the released structure

The expressionsDJ1 andDJ2 can be expressed in a matrix format as follows

where

QDDD JQJLJ

2

1

J

J

J D

D

D

2

1

JL

JL

JL D

D

D

2

1

Q

QQ

22212122 QDQDDD JQJQJLJ

2221

1211

JQJQ

JQJQ

JQ DD

DD

D


10/32


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

For the first expression

AM1 = is the shear force atB on memberAB

AML1 = is the shear force atB on memberAB caused by the external loads on

the released structure

AMQ11 = is the shear force atB on memberAB caused by a unit loadcorresponding to the redundant Q1

AMQ12 = is the shear force atB on memberAB caused by a unit load

corresponding to the redundant Q2

The other expressions follow in a similar manner.

21211111 QAQAAA MQMQMLM





11/32


QAAA MQMLM

2

1

Q

Q

Q

4

3

2

1

M

M

M

M

M

AA

A

A

A

4

3

2

1

ML

ML

ML

ML

ML

AA

A

A

A

4241

3231

2221

1211

MQMQ

MQMQ

MQQM

MQMQ

MQ

AAAA

AA

AA

A

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

where

The sign convention for member end actions is as follows:

+ when up for translations and forces+ when counterclockwise for rotation and couples


12/32


In a similar manner we can find reactions via superposition

For the first expression

AR1 = the reaction in the actual beam at A

AR2 = the reaction in the actual beam at A

ARL1 = the reaction in the released structure due to the external loads

ARL2 = the reaction in the released structure due to the external loads

ARQ11 = the reaction at A in the released structure due to the unit actioncorresponding to the redundant Q1ARQ22 = the reaction at A in the released structure due to the unit action

corresponding to the redundant Q2ARQ12 = the reaction at A in the released structure due to the unit action

corresponding to the redundant Q2

21211111 QAQAAA RQRQRLR

22212122 QAQAAA RQRQRLR


13/32


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

where

QAAA RQRLR

2

1

R

R

RA

AA

2

1

RL

RL

RLA

AA

2

1

Q

QQ

2221

1211

RQRQ

RQRQ

RQAA

AAA


14/32


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

the only change in the superposition equation is in the first term. Take

here

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 the 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 theredundants {Q}, values that account for temperature, prestrain and displacement effects.

JRJPJTJLJS DDDDD

QDDD JQJSJ


15/32


Example

PPPP

PLM

PP

3

2

1 2

Consider the two span beam to the left

where it is assumed that the objective isto calculate the various joint

displacementsDJ, member end actions

AM , and end reactionsAR. The beam has

a constant flexural rigidity EI and is acted

upon by the following loads


16/32


Consider the released

structure and the attending

moment area diagrams.

The (M/EI) diagram was

drawn by parts. Each

action and its attending

diagram is presented one at

a time in the figure starting

with actions on the far

right.


17/32


From first moment area theorem

1

2

1 12 1 1.5 0 .5

2 2

12 2

5

4

JL

P L P LD L L

E I E I

P L P L LLE I E I

P L

E I

2

2

1 2 1 3 32

2 2 2 2

1

2 2

13

8

JL

P L P L LD L

E I E I

P L PL LL

E I E I

P L

E I

13

10

8

2

EI

PL

D JL


18/32


PA

PPPA

F

RL

RL

Y

2

2

0

1

1

2

22

3

22

0

2

2

PLA

LPL

PPLL

PA

M

RL

RL

A

Using the following free body diagram of the released structure

Then from the equations of equilibrium


19/32


0

22

0

1

1

ML

ML

Y

A

PPA

F

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

then once again from the equations of equilibrium

23

222

2

0

2

2

PLA

PLPLL

PA

M

ML

ML

B


20/32


0

0

3

3

ML

ML

Y

A

PPA

F

0

4 2

4 2

MB

P LA P L

M L

P LAM L

Using a free body diagram from segmentBCof the entire beam, i.e.,

then once again from the equations of equilibrium


21/32


2

0

2

3

0

PL

PL

AML

2

2

RL

P

APL

Thus the vectorsAML andARL are as follows:

Member end actions in the released structure.

Reactions in the released structure.


22/32


Consider the released beam with a unit load at pointB

EI

L

LEI

LD JQ

2

2

1

2

11

EI

L

LEI

LD JQ

2

2

1

2

21

L


23/32


Consider the released beam with a unit load at point C

EIL

LEI

LD JQ

23

122

1

2

12

EIL

LEI

LD JQ

2

22

2

22

2

1

2L

L


24/32


2 1 3

1 42

JQ

LD

E I

L

LAMQ

0

10

0

11

1 1

2R QA

L L

Thus

In a similar fashion, applying a unit load associated with Q1 and Q2 in the previous

cantilever beam, we obtain the following matrices


25/32


6 9

6 45 6

PQ

Previously (Lecture #6)

with

2 2

2

1 0 1 3 6 9

1 3 1 4 6 48 2 5 6

1 7

51 1 2

J

P L L PD

E I E I

P L

E I

QDDD JQJLJ

then


26/32


0

1 13

0 6 92

0 0 1 6 45 6

0

2

5

2 0

6 45 6

3 6

M

P L

LPA

P L L

LP

L

Similarly, with

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

QAAA MQMLM


27/32


28/32


29/32


7. Determine the other displacements and actions. The following are the four flexibility

matrix equations for calculating redundants member end actions, reactions and joint

displacements

where for the released structure

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

QDDD JQJSJ

QAAA MQMLM

QAAA

RQRLR

JRJPJTJLJS DDDDD


30/32


31/32


MATRIX ORDER DEFINITION

jx1 Jointdisplacementintheactualstructure(j = numberof

joint

displacement)jx1 Jointdisplacementsinthereleasedstructureduetoloads

jx1 Jointdisplacementsinthereleasedstructureduetounit

valuesof

the

redundants

jx1

Jointdisplacementsinthereleasedstructuredueto

temperature,prestrain,andrestraintdisplacements(other

thanthoseinDQ)

jx1

JLD

QL

D

JRJPJT DDD ,,

JRJPJTJLJS DDDDD

JD

JSD


32/32


MATRIX ORDER DEFINITION

mx1 Memberendactionsintheactualstructure

(m = Numberofendactions)

mx1 Member

end

actions

in

the

released

structure

due

to

loads

mxq Memberendactionsinthereleasedstructureduetounit

valuesoftheredundants

rx1 Reactions

in

the

actual

structure

(r

=

numberof

reactions)

rx1 Reactionsinthereleasedstructureduetoloads

rxqReactions

in

the

released

structure

due

to

unit

values

of

the

redundants

MLA

RA

MA

RLA

MQA

RQA

flexibility method – temp. and pre-strain

Documents