Section 6: The Flexibility Method - Beams

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Preliminaries: 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 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.

1 Section 6: The Flexibility Method - Beams

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Consider the following arbitrarily loaded beam

Identify M  M(x)  Moment at any section in the beam due to external loads m  m(x)  Moment at any section in the beam due to a unit action m y ~  I  Stress acting on dA due to a unit action 2 Section 6: The Flexibility Method - Beams

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The force acting on the differential area dA due to a unit action is ~ f  ~ dA  m y     dA  I  The stress due to external loads is M y   I The displacement of a differential segment dA by dx along the length of the beam is

   dx       dx  E   M y     dx  E I 

3 Section 6: The Flexibility Method - Beams

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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 ~ dW  f   m y   M y     dA   dx  I   E I   M m y2     dA dx  2   E I  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.,

c  T  M m y2   dW    dA dx   2      E I  A  cB   

 cT    Mm  2  Wdiffernetialsegment    2  y dA  dx   EI     cB    Mm    Mm     2  I  dx    dx   EI    EI  4 Section 6: The Flexibility Method - Beams

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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.,

L  M x mx W  dx Internal   0  EI 

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

WExternal  1D With

WExternal  WInternal L  M x mx 1 D    dx 0  EI  L  M x mx D    dx 0  EI  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. 5 Section 6: The Flexibility Method - Beams

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Example 6.1

6 Section 6: The Flexibility Method - Beams

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Example 6.2

Flexibility Coefficients by virtual work

7 Section 6: The Flexibility Method - Beams

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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 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 Section 6: The Flexibility Method - Beams

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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 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 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering Example 6.3

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 at the

redundant is set equal to DB.  5  RB    w L  8  10 Section 6: The Flexibility Method - Beams

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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 L3   B 48EI

The displacement at B caused by RB is δB RB. The displacement caused by the uniform load w acting on the released structure is 5 w L4 D   B 384 EI Thus by the compatibility equation D  5  D   R  0 R   B  w L B B B B   11  B  8  Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering Example 6.4

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 Section 6: The Flexibility Method - Beams

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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 Section 6: The Flexibility Method - Beams

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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

DQ1  DQL1  F11Q1  F12Q2  0

DQ2  DQL2  F21Q1  F22Q2  0

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

DQ1  DQL1  F11Q1  F12Q2

DQ2  DQL2  F21Q1  F22Q2

14 Section 6: The Flexibility Method - Beams

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

DQ  DQL   FQ 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

DQ1  DQL1  Q1  DQ     DQL     Q    D D  Q2   QL2  Q2 

F11 F12 F    F21 F22 15 Section 6: The Flexibility Method - Beams

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

FQ  DQ   DQL 

1 Q  F DQ    DQL 

To see how this works consider the previous beam with a constant flexural rigidity EI. If we identify actions on the beam as

P1   2P M  PL P2   P P3  P

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

0 DQ    0

16 Section 6: The Flexibility Method - Beams

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The vector [DQL] represents the displacements in the released structure corresponding to the redundant loads. These displacements are

13PL3 97PL3 D  D  QL1 24EI QL2 48EI

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

PL3 26 DQL     48EI 97

17 Section 6: The Flexibility Method - Beams

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The flexibility matrix [F ] is obtained by subjecting the beam to unit load corresponding to Q1 and computing the following displacements

L3 5L3 F  F  11 3EI 21 6EI

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

5L3 8L3 F  F  12 6EI 22 3EI

18 Section 6: The Flexibility Method - Beams

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The flexibility matrix is L3  2 5  F    6EI  5 16

The inverse of the flexibility matrix is

1 6EI  16 5  F  3   7L 5 2 

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

Q1  1 Q     F  DQ   DQL   Q2   6EI  16  5   0  PL3  26      3           7L  5 2   0  48EI  97  P  69       56  64 19 Section 6: The Flexibility Method - Beams

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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 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 Section 6: The Flexibility Method - Beams

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Example 6.5

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 Section 6: The Flexibility Method - Beams

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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

wL3 PL2 D   QL1 24EI 16EI

Similarly, PL2 PL2 PL2 D    QL2 16EI 16EI 8EI such that

L2 2wL  3P DQL    48EI  6P  22 Section 6: The Flexibility Method - Beams

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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)

L L 2L F    11 3EI 3EI 3EI

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

L F  21 6EI 23 Section 6: The Flexibility Method - Beams

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Similarly L L 2L F    22 3EI 3EI 3EI

L F  12 6EI The flexibility matrix is L 4 1 F    6EI 1 4

The inverse of the flexibility matrix is

1 2EI  4 1 F    5L 1 4 

24 Section 6: The Flexibility Method - Beams

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

Q1  1 Q     F  DQ   DQL   Q2  2EI  4 1 0 L2 2wL  3P         5L 1 4  0 48EI  6P  L  8wL  6P      120  2wL  21P and

wL2 PL wL2 7PL Q    Q   1 15 20 2 60 40

25 Section 6: The Flexibility Method - Beams

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Example 6.6

26 Section 6: The Flexibility Method - Beams

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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

DQ  DQL   FQ

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 Section 6: The Flexibility Method - Beams

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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 Section 6: The Flexibility Method - Beams

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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

DJ1  DJL1  DJQ11Q1  DJQ12Q2 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 29 Section 6: The Flexibility Method - Beams

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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 released structure 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 ].

30 Section 6: The Flexibility Method - Beams

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

DJ 2  DJL2  DJQ21 Q1  DJQ22 Q2

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

31 Section 6: The Flexibility Method - Beams

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

DJ   DJL   DJQ Q where DJ1  DJ     DJ 2 

DJL1  DJL     DJL2 

 DJQ11 DJQ12  DJQ      DJQ21 DJQ22  and

Q1  Q    Q2  which were determined previously 32 Section 6: The Flexibility Method - Beams

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In a similar manner we can find reactions via superposition

AR1  ARL1  ARQ11Q1  ARQ12Q2

AR2  ARL 2  ARQ 21Q1  ARQ 22Q2

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 ARQ22 = the reaction at A in the released structure due to the unit action corresponding to the redundant Q2 ARQ12 = the reaction at A in the released structure due to the unit action corresponding to the redundant Q2 33 Section 6: The Flexibility Method - Beams

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

AR   ARL   ARQ Q where

ARL1   ARQ11 ARQ12  ARL     ARQ     AR1  ARL 2  ARQ 21 ARQ 22 AR       AR2 

Q1  Q    Q2 

34 Section 6: The Flexibility Method - Beams

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

AM1  AML1  AMQ11Q1  AMQ12Q2

AM 2  AML2  AMQ21Q1  AMQ22Q2

AM 3  AML3  AMQ31Q1  AMQ32Q2

AM 4  AML4  AMQ41Q1  AMQ42Q2 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 corresponding to the redundant Q1 AMQ12 = is the shear force at B on member AB caused by a unit load corresponding to the redundant Q2

The other expressions follow in a similar manner. 35 Section 6: The Flexibility Method - Beams

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

AM   AML  AMQQ where

 A   A  AMQ11 AMQ12  M1 ML1       A A AM 2  AML2   M Q21 MQ22 AM     AML    AMQ    AM 3  AML3  AMQ31 AMQ32  A  A     M 4   ML4  AMQ41 AMQ42 

Q1  Q    Q2 

The sign convention for member end actions is as follows: + when up for translations and forces

+ when counterclockwise for rotation and couples 36 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering Example 6.7

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

P1  2P M  PL

P2  P

P3  P

37 Section 6: The Flexibility Method - Beams

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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 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering From first moment area theorem 11PL PL DLL2  1  1.5  0.5 JL1 22EI EI PL1 PL L L EI22 EI 5PL2  4EI

1 2PL  1  3 PL  3 L DLJL2  2   2EI  2  2 EI  2 PL1 PL L L EI22 EI 13PL2  8EI

PL2 10 DJL     8EI 13 39 Section 6: The Flexibility Method - Beams

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Consider the released beam with a unit load at point B

L

1 L 1 L D  L DJQ21  L JQ11 2 EI 2 EI 2 L2 L   2EI 2EI

40 Section 6: The Flexibility Method - Beams

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Consider the released beam with a unit load at point C

L 2L

1 L 1 2L D  2 1 L D  2L JQ12 2 EI JQ22 2 EI 3L2 2L2   2EI EI leading to

L2 13  DJQ   2EI 14 41 Section 6: The Flexibility Method - Beams

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Previously in Example 6.4 P 69 Q   56 64 with

DJ   DJL   DJQ Q

then the displacements DJ1 and DJ2 are

PL2210  L  1 3  P  69  DJ       8EI13  2 EI  1 4  56  64  PL2 17   112EI 5

42 Section 6: The Flexibility Method - Beams

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Using the following free body diagram of the released structure

Then from the equations of equilibrium

 M A  0 L 3L  A  2P  PL  P  P2L RL 2 2 2 PL A   RL 2 2

 FY  0

 ARL1  2P  P  P

ARL1  2P 43 Section 6: The Flexibility Method - Beams

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

then once again from the equations of equilibrium

 FY  0

 AML1  2P  2P

AML1  0

 M B  0 L PL  A  2P   2PL ML2 2 2 3PL A  44 ML2 2 Section 6: The Flexibility Method - Beams

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

then once again from the equations of equilibrium

 FY  0

 AML3  P  P

AML3  0

 M  0 B PL A   PL ML4 2 PL A  45 ML4 2 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering

Thus the vectors AML and ARL are as follows:

 0       3PL  2P  2     A  AML    RL  PL  0     2    PL   Reactions in the released structure.  2 

Member end actions in the released structure.

46 Section 6: The Flexibility Method - Beams

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Finally with

AR   ARL   ARQ Q

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

2P  P 1 1   69  AR PL     56 LL 2    64  2 P 107   56 31L

47 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering

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

1 1    0 L  AMQ    0 1 0  L

11  ARQ   LL2

48 Section 6: The Flexibility Method - Beams

Washkewicz College of Engineering

Similarly, with

AM   AML  AMQ Q

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

0 3PL 11   2 P 0L  69 AM       0 56  0 1  64 PL  0  L   2 5  P 20L   56 64 36L 49 Section 6: The Flexibility Method - Beams

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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.,

DQ   DQS   FQ

DQS   DQL   DQT   DQP   DQR 

1 Q  F DQ   DQS  50 Section 6: The Flexibility Method - Beams

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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

DJ   DJS   DJQ Q

AM   AML  AMQ Q

AR  ARL   ARQ Q

where for the released structure

DJS   DJL   DJT   DJP   DJR 

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

51 Section 6: The Flexibility Method - Beams

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MATRIX ORDER DEFINITION

Q q x 1 Unknown redundant actions (q = Number of redundant) Displacements in the actual structure Corresponding to the DQ q x 1 redundant

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

Displacements in the released structure corresponding to the DJQ q x q redundants and due to unit values of the redundants

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

DQS q x 1 DQS  DQL  DQT  DQP  DQR

52 Section 6: The Flexibility Method - Beams

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MATRIX ORDER DEFINITION

D j x 1 Joint displacement in the actual structure (j = number of joint J displacement) j x 1 Joint displacements in the released structure due to loads DJL

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

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

DJS j x 1 DJS  DJL  DJT  DJP  DJR

F q x q Matrix of flexibility coefficients 53 Section 6: The Flexibility Method - Beams

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MATRIX ORDER DEFINITION m x 1 Member end actions in the actual structure AM (m = Number of end-actions) m x 1 Member end actions in the released structure due to loads AML m x q Member end actions in the released structure due to unit AMQ values of the redundants r x 1 Reactions in the actual structure (r = number of reactions) AR

r x 1 Reactions in the released structure due to loads ARL

Reactions in the released structure due to unit values of the

ARQ r x q redundants

54