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Lecture 6: The Flexibility Method - Beams

Flexibility Method

method for indeterminate structures. His method was based on considering deflections, but

the presentation was rather brief and attraced little attention. Ten years later Otto Mohr

independently extended Maxwells theory to the present day treatment. The flexibility

method will sometimes be referred to in the literature as Maxwell-Mohr method.

With the flexibility method equations of compatibility involving displacements at each of

needed for solution. This method is somewhat useful in analyzing beams, framse and

trusses that are statically indeterminate to the first or second degree. For structures with a

high degree of static indeterminacy such as multi-story buildings and large complex trussesstiffness methods are more appropriate. Nevertheless flexibility methods provide an

understanding of the behavior of statically indeterminate structures.

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The fundamental concepts that underpin the flexibility method will be illustrated by the

study of a two span beam. The procedure is as follows

1. Pick a sufficient number of redundants corresponding to the degree ofindeterminacy

2. Remove the redundants

3. Determine displacements at the redundants on released structure due to external or

mpose act ons

4. Determine displacements due to unit loads at the redundants on the released

structure

5. Employ equation of compatibility, e.g., if a pin reaction is removed as a redundant

the compatibility equation could be the summation of vertical displacements in the

re ease s ruc ure mus a o zero.

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The beam to the left isstatically indeterminate to

the first degree. The

reaction at the middle

supportRB is chosen as theredundant.

The released beam is also

shown. Under the external

loads the released beam.

A second beam is

considered where the

released redundant is treatedas an external load and the

corresponding deflection at

the redundant is set equal to

B.LwRB

=

8

5

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A more general approach consists in finding the displacement atB caused by a unit load in

the direction ofRB. Then this displacement can be multiplied byRB to determine the total

displacement

Also in a more general approach a consistent sign convention for actions and displacementsmust be adopted. The displacements in the released structure atB are positive when they are

, . ., .

The displacement atB caused by the unit action is

L3

The displacement atB caused byRB is BRB. The displacement caused by the uniform load

EIB

48=

EI

LwB

384

5 4=

T us y t e compat ty equat on

LwRRB

BBBBB

=

==+

8

50

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If a structure is statically indeterminate to more

than one degree, the approach used in thepreceeding example must be further organized

.

Consider the beam to the left. The beam is

statically indeterminate to the second degree. A

by releasing two redundant reactions. Four

possible released structures are shown.

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The redundants chosen are atB and C. The

.

The released structure is shown at the leftwith all external and internal redundants

s own.

DQL1 is the displacement corresponding to Q1and caused b onl external actions on the

released structure

DQL2 is the displacement corresponding to Q2cause y on y ex erna ac ons on e

released structure.

Both displacements are shown in their

assumed positive direction.

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We can now write the compatibility equations for this structure. The displacements

corresponding to Q and Q will be zero. These are labeledD andD respectively

021211111 =++= QFQFDD QLQ

In some casesDQ1 andDQ2 would be nonzero then we would write

022212122 =++= QFQFDD QLQ

21211111 QFQFDD QLQ ++=

22212122 QFQFDD QLQ ++=

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The equations from the previous page can be written in matrix format as

where:

QLQ +=

{DQ } - matr x o actua sp acements correspon ng to t e re un ant

{DQL } - matrix of displacements in the released structure corresponding to the

redundant action [Q] and due to the loadsF - flexibilit matrix for the released structure corres ondin to the redundant

actions [Q]

{Q} - matrix of redundant

{ }

=

2

1

Q

Q

Q D

DD { }

=

2

1

QL

QL

QL D

DD { }

=

2

1

Q

QQ

=

2221

1211

FF

FFF

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The vector [Q] of redundants can be found by solving for them from the matrix equation

on the previous overhead.

[ ] [ ] [ ] [ ]QLQ DDQF =

To see how this works consider the previous beam with a constant flexural rigidityEI. If

[ ] [ ] [ ] [ ]( )QLQ DDFQ = 1

we identify actions on the beam as

PPPPPLMPP ==== 321 2

Since there are no displacements corresponding to Q1 and Q2, then

0

=

0Q

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The vector [D ] represents the displacements in the released structure corresponding to

the redundant loads. These displacements are

PLPL 9713 33

EIEIQLQL

482421 ==

The positive signs indicate that both displacements are upward. In a matrix format

263

PL

=9748EI

QL

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The flexibility matrix [F ] is obtained by subjecting the beam to unit load corresponding

1

LL 5 33

EIEI 632111 ==

2

following displacements

33

EIF

EIF

362212 ==

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The flexibility matrix is

=

523L

The inverse of the flexibility matrix is

1656EI

[ ]

=

25

516

7

63

1

L

EIF

As a final step the redundants [Q] can be found as follows

=

=

11

DDF

Q

Q

=

97

26

480

0

25

516

7

6 3

3

2

EI

PL

L

EI

=

64

69

56

P

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The redundants have been obtained. The other unknown reactions can be found from

.

the released structure and imposing the compatibility equations.

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Example

t ree span eam s own at t e e t s

acted upon by a uniform load w and

concentrated loadsPas shown. The

beam has a constant flexural ri idit EI.

Treat the supports atB and Cas

redundants and compute theseredundants.

In this problem the bending moments atB

and Care chosen as redundants to

indicate how unit rotations are applied toreleased structures.

Each redundant consists of two moments,

one acting in each adjoining span.

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The displacements corresponding to the two redundants consist of two rotations one for

each adjoining span. The displacementD L1 andD L2 corresponding to Q1 and Q2.

These displacements will be caused by the loads acting on the released structure.

The displacementDQL1 is composed of two parts, the rotation of endB of memberAB

EI

PL

EI

wL

DQL 1624

23

1 +=

Similarly,

PLPLPL

222

EIEIEIQL

816162

such that

+=

Pw

EILDQL

648

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The flexibility coefficients are determined next. The flexibility coefficientF11 is the sum

of two rotations at joint B. One in spanAB and the other in spanBC(not shown below)

EI

L

EI

L

EI

LF

3

2

33

11 =+=

Similarly the coefficientF21 is equal to the sum of rotations at joint C. However, the

rotation in span CD is zero from a unit rotation at jointB. Thus

EI

LF

621 =

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LLL 2

Similarly

EIEIEI 33322 ==

LF12 =

The flexibility matrix is

14L

=416EI

The inverse of the flexibility matrix is

=

41

14

5

21

L

EIF

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As a final step the redundants [Q] can be found as follows

[ ] [ ] [ ] [ ]( )

( ) +

=

=

PwLLEI

DDF

Q

Q QLQ

320142 2

1

2

1

( ) +

=

=

PwLL

PEIL

68

6480415

and

2015

2

1

PLwLQ =

40

7

60

2

2

PLwLQ =

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