Preliminaries: Beam Deflections Virtual Workacademic.csuohio.edu/duffy_s/511_06.pdf · Section 6: The Flexibility Method - Beams Washkewicz College of Engineering 1 Preliminaries:

Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering

1

Preliminaries: Beam Deflections – Virtual Work

There are several methods available to calculate deformations (displacements and rotations)

in beams. They include:

• Formulating moment equations and then integrating to find rotations and

displacements

• Moment area theorems for either rotations and/or displacements

• Virtual work methods

Since structural analysis based on finite element methods is usually based on a potential

energy method, we will tend to use virtual work methods to compute beam deflections.

The theory that supports calculating deflections using virtual work will be reviewed and

several examples are presented.



2

Consider the following arbitrarily loaded beam

Identify

actionunit a todue on acting Stress

~

actionunit a todue beam in thesection any at Moment

loads external todue beam in thesection any at Moment

dA

I

ym

m(x)m

M(x)M



3

The force acting on the differential area dA due to a unit action is

The stress due to external loads is

The displacement of a differential segment dA by dx along the length of the beam is

dAI

ym

dAf

~~

I

yM

dxIE

yM

dxE

dx



4

The work done by the force acting on the differential area dA due to a unit action as the

differential segment of the beam (dA by dx) displaces along the length of the beam by an

amount is

The work done within a differential segment (now A by dx) due to a unit action applied to

the beam is the integration of the expression above with respect to dA, i.e.,

dxdAIE

ymM

dxIE

yMdA

I

ym

fdW

2

2

~

dxEI

MmdxI

EI

Mm

dxdAyEI

MmW

dxdAIE

ymMdW

T

B

T

B

c

c

segmentaldifferneti

c

cA

2

2

2

2

2



5

The internal work done along the entire length of the beam due to a unit action applied to

the beam is the integration of the last expression with respect to x, i.e.,

The external work done along the entire length of the beam due to a unit action applied to

the beam is

With

or the deformation (D) of the a beam at the point of application of a unit action (force or

moment) is given by the integral on the right.

dx

EI

xmxMW

L

Internal

0

D 1ExternalW

dxEI

xmxM

dxEI

xmxM

WW

L

L

InternalExternal

D

D

0

0

1



Example 6.1

6



Example 6.2

7

Flexibility Coefficients by virtual work



Perspectives on the Flexibility Method

In 1864 James Clerk Maxwell published the first consistent treatment of the flexibility

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

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

independently extended Maxwell’s 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

the redundant forces in the structure are introduced to provide the additional equations

needed for solution. This method is somewhat useful in analyzing beams, frames 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 trusses

stiffness methods are more appropriate. Nevertheless flexibility methods provide an

understanding of the behavior of statically indeterminate structures.

8



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 of

indeterminacy

2. Remove the redundants

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

imposed actions

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

released structure must add to zero.

9



The beam to the left is statically

indeterminate to the first degree.

The reaction at the middle support

RB is chosen as the redundant.

The released beam is also shown.

Under the external loads the

released beam deflects an amount

DB.

A second beam is considered

where the released redundant is

treated as an external load and the

corresponding deflection at the

redundant is set equal to DB.

LwRB

8

5

10

Example 6.3



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

the direction of RB. Then this displacement can be multiplied by RB to determine the total

displacement

Also in a more general approach a consistent sign convention for actions and displacements

must be adopted. The displacements in the released structure at B are positive when they are

in the direction of the action released, i.e., upwards is positive here.

The displacement at B caused by the unit action is

The displacement at B caused by RB is δB RB. The displacement caused by the uniform load

w acting on the released structure is

Thus by the compatibility equation

EI

LB

48

3

EI

LwB

384

5 4

D

LwRRB

BBBBB

DD

8

50

11



If a structure is statically indeterminate to more

than one degree, the approach used in the

preceeding example must be further organized

and more generalized notation is introduced.

Consider the beam to the left. The beam is

statically indeterminate to the second degree. A

statically determinate structure can be obtained

by releasing two redundant reactions. Four

possible released structures are shown.

12

Example 6.4



The redundants chosen are at B and C. The

redundant reactions are designated Q1 and Q2.

The released structure is shown at the left

with all external and internal redundants

shown.

DQL1 is the displacement corresponding to Q1

and caused by only external actions on the

released structure

DQL2 is the displacement corresponding to Q2

caused by only external actions on the

released structure.

Both displacements are shown in their

assumed positive direction.

13



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

corresponding to Q1 and Q2 will be zero. These are labeled DQ1 and DQ2 respectively

In some cases DQ1 and DQ2 would be nonzero then we would write

021211111 QFQFDD QLQ

022212122 QFQFDD QLQ

21211111 QFQFDD QLQ

22212122 QFQFDD QLQ

14



where:

{DQ } - vector of actual displacements corresponding to the redundant

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

redundant action [Q] and due to the loads

[F] - flexibility matrix for the released structure corresponding to the redundant

actions [Q]

{Q} - vector of redundants

2

1

Q

Q

Q D

DD

QFDD QLQ

2

1

QL

QL

QL D

DD

2221

1211

FF

FFF

2

1

Q

QQ

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

15



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

on the previous overhead.

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

we identify actions on the beam as

Since there are no displacements imposed on the structure corresponding to Q1 and Q2,

then

QLQ DDQF

QLQ DDFQ 1

PPPPPLMPP 321 2

0

0QD

16



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

the redundant loads. These displacements are

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

EI

PLD

EI

PLD QLQL

48

97

24

13 3

2

3

1

97

26

48

3

EI

PLDQL

17



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

to Q1 and computing the following displacements

Similarly subjecting the beam to unit load corresponding to Q2 and computing the

following displacements

EI

LF

EI

LF

6

5

3

3

21

3

11

EI

LF

EI

LF

3

8

6

5 3

22

3

12

18



The flexibility matrix is

The inverse of the flexibility matrix is

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

165

52

6

3

EI

LF

25

516

7

63

1

L

EIF

64

69

56

97

26

480

0

25

516

7

6 3

3

1

2

1

P

EI

PL

L

EI

DDFQ

QQ QLQ

19



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

the released structure. Displacements can be computed from the known reactions on

the released structure and imposing the compatibility equations.

Discuss the following sign conventions and how they relate to one another:

1. Shear and bending moment diagrams

2. Global coordinate axes

3. Sign conventions for actions

- Translations are positive if the follow the direction of the applied force

- Rotations are positive if they follow the direction of the applied moment

20



A three span beam shown at the left is

acted upon by a uniform load w and

concentrated loads P as shown. The

beam has a constant flexural rigidity EI.

Treat the supports at B and C as

redundants and compute these

redundants.

In this problem the bending moments at B

and C are chosen as redundants to

indicate how unit rotations are applied to

released structures.

Each redundant consists of two moments,

one acting in each adjoining span.

21

Example 6.5



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

each adjoining span. The displacement DQL1 and DQL2 corresponding to Q1 and Q2.

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

The displacement DQL1 is composed of two parts, the rotation of end B of member AB

and the rotation of end B of member BC

Similarly,

EI

PL

EI

wLDQL

1624

23

1

EI

PL

EI

PL

EI

PLDQL

81616

222

2

P

PwL

EI

LDQL

6

32

48

2

such that

22



EI

L

EI

L

EI

LF

3

2

3311

EI

LF

621

The flexibility coefficients are determined next. The flexibility coefficient F11 is the sum

of two rotations at joint B. One in span AB and the other in span BC (not shown below)

Similarly the coefficient F21 is equal to the sum of rotations at joint C. However, the

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

23



EI

L

EI

L

EI

LF

3

2

3322

EI

LF

612

Similarly

The flexibility matrix is

41

14

6EI

LF

The inverse of the flexibility matrix is

41

14

5

21

L

EIF

24



2015

2

1

PLwLQ

PwL

PwLL

P

PwL

EI

L

L

EI

DDFQ

QQ QLQ

212

68

120

6

32

480

0

41

14

5

2 2

1

2

1

40

7

60

2

2

PLwLQ

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

and

25



26

Example 6.6



Joint Displacements, Member End Actions And Reactions

Previously we focused on finding redundants using flexibility (force) methods. Typically

redundants (Q1, Q2, … , Qn) specified by the structural engineer are unknown reactions.

Redundants are determined by imposing displacement continuity at the point in the structure

where redundants are applied, i.e., we imposed

If the redundants specified are unknown reactions then after these redundants are 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.

.27

QFDD QLQ



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

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

directly using flexibility methods.

Consider the two span beam below where the redundants Q1 and Q2 have been computed

previously in Example 6.4. The non-zero joint displacements DJ1 and DJ2, both rotations, as

well as reactions AR1 and AR2. can be computed. We will focus on the joint displacements

DJ1 and DJ2 first. Keep in mind that when using flexibility methods translations are

associated with forces, and rotations are associated with moments.

Reactions other than redundants will be denoted {AR} and these quantities can be

determined as well. The objective here is the extension of the flexibility (force) method so

that it is more generally applied. 28



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 joints B ( = DJ1) and C ( = DJ2) are

required. When the redundants Q1 and Q2 were found superposition was imposed on the

released structure requiring the displacement associated with the unknown redundants to be

equal to zero. In finding joint displacements in the actual structure superposition is used

again and displacements in the released structure are equated to the displacement in the

actual structure. Focusing on joint B, superposition requires

Here

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

the joint associated with Q1

DJL1 = the displacement (a rotation) at joint B associated with DJ1 caused by

the external loads in the released structure.

DJQ11 = the rotation at joint B associated with DJ1 caused by a unit force at

joint B corresponding to the redundant Q1 in the released structure

DJQ12 = the rotation at joint B associated with DJ1 caused by a unit force at joint

C corresponding to the redundant Q2 in the released structure

21211111 QDQDDD JQJQJLJ

29



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 are

similarly depicted.

In the figure to the right unit

loads are shown applied at the

redundants. 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

30



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 Q2

DJL2 = the displacement (a rotation) at joint C associated with DJ2 caused by the

external loads in the released structure.

DJQ21 = the rotation at joint C associated with DJ2 caused by a unit force at joint B

corresponding to the redundant Q1 in the released structure

DJQ22 = the rotation at joint C associated with DJ2 caused by a unit force at joint C

corresponding to the redundant Q2 in the released structure

22212122 QDQDDD JQJQJLJ

31



The expressions DJ1 and DJ2 can be expressed in a matrix format as follows

where

and

which were determined previously

QDDD JQJLJ

2

1

J

J

JD

DD

2

1

JL

JL

JLD

DD

2

1

Q

QQ

2221

1211

JQJQ

JQJQ

JQ DD

DDD

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 action

corresponding to the redundant Q1





21211111 QAQAAA RQRQRLR

22212122 QAQAAA RQRQRLR

33



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

RQ AA

AAA

34



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

For the first expression

AM1 = is the shear force at B on member AB

AML1 = is the shear force at B on member AB caused by the external loads on

the released structure

AMQ11 = is the shear force at B on member AB caused by a unit load


AMQ12 = is the shear force at B on member AB caused by a unit load


The other expressions follow in a similar manner.

21211111 QAQAAA MQMQMLM




35



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

where

QAAA MQMLM

2

1

Q

QQ

4

3

2

1

M

M

M

M

M

A

A

A

A

A

4

3

2

1

ML

ML

ML

ML

ML

A

A

A

A

A

4241

3231

2221

1211

MQMQ

MQMQ

MQQM

MQMQ

MQ

AA

AA

AA

AA

A

The sign convention for member end actions is as follows:

+ when up for translations and forces

+ when counterclockwise for rotation and couples 36



Example 6.7

PP

PP

PLM

PP

3

2

1 2

Consider 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

AM , and end reactions AR. The beam has

a constant flexural rigidity EI and is acted

upon by the following loads

37



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.

38



From first moment area theorem

1

2

1 12 1 1.5 0.5

2 2

1

2 2

5

4

JL

PL PLD L L

EI EI

PL PL LL

EI EI

PL

EI

2

2

1 2 1 3 32

2 2 2 2

1

2 2

13

8

JL

PL PL LD L

EI EI

PL PL LL

EI EI

PL

EI

13

10

8

2

EI

PLDJL

39



Consider the released beam with a unit load at point B

EI

L

LEI

LDJQ

2

2

1

2

11

EI

L

LEI

LDJQ

2

2

1

2

21

L

40



Consider the released beam with a unit load at point C

EI

L

LEI

LDJQ

2

3

122

1

2

12

EI

L

LEI

LDJQ

2

22

2

22

2

1

2L

L

41

2 1 3

1 42JQ

LD

EI

leading to



69

6456

PQ

Previously in Example 6.4

with

2 2

2

10 1 3 69

13 1 4 648 2 56

17

5112

J

PL L PD

EI EI

PL

EI

QDDD JQJLJ

then the displacements DJ1 and DJ2 are

42



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

43



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

2

3

222

2

0

2

2

PLA

PLPLL

PA

M

ML

ML

B

44



0

0

3

3

ML

ML

Y

A

PPA

F

0

4 2

4 2

MB

PLA PL

ML

PLA

ML

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

then once again from the equations of equilibrium

45



2

0

2

3

0

PL

PL

AML

2

2

RL

P

APL

Thus the vectors AML and ARL are as follows:

Member end actions in the released structure.

Reactions in the released structure.

46



Finally with

2

1 1 69

2 64562

107

3156

R

PP

A PLL L

P

L

QAAA RQRLR

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

47



L

LAMQ

0

10

0

11

1 1

2RQA

L L

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

cantilever beam, we obtain the following matrices

48



0

1 13

0 692

0 0 1 6456

0

2

5

20

6456

36

M

PL

LPA

PL L

LP

L

Similarly, with

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

QAAA MQMLM

49



Summary Of Flexibility Method

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

1. Problem statement

2. Selection of released structure

3. Analysis of released structure under loads

4. Analysis of released structure for other causes

5. Analysis of released structure for unit values of redundant

6. Determination of redundants through the superposition equations, i.e.,

QFDD QSQ

QRQPQTQLQS DDDDD

QSQ DDFQ

1

50



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

51



MATRIX ORDER DEFINITION

q x 1 Unknown redundant actions (q = Number of redundant)

q x 1

Displacements in the actual structure Corresponding to the

redundant

q x 1

Displacements in the released structure corresponding to the

redundants and due to loads

q x q


redundants and due to unit values of the redundants

q x 1


redundants and due to temperature, prestrain, and restraint

displacements (other than those in DQ)

q x 1

QD

QLD

JQD

QRQPQT DDD ,,

QSD QRQPQTQLQS DDDDD

Q

52




j x 1 Joint displacement in the actual structure (j = number of joint

displacement)

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

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

of the redundants

j x 1

Joint displacements in the released structure due to

temperature, prestrain, and restraint displacements (other than

those in DQ)

j x 1

q x q Matrix of flexibility coefficients

JLD

QLD

JRJPJT DDD ,,

JRJPJTJLJS DDDDD

JD

JSD

53

F




m x 1 Member end actions in the actual structure

(m = Number of end-actions)

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

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)

r x 1 Reactions in the released structure due to loads

r x q

Reactions in the released structure due to unit values of the

redundants

MLA

RA

MA

RLA

MQA

RQA

54

Preliminaries: Beam Deflections Virtual Workacademic.csuohio.edu/duffy_s/511_06.pdf · Section 6: The Flexibility Method - Beams Washkewicz College of Engineering 1 Preliminaries:

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