Flexibility Method w316q

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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 s calculating deflections using virtual work will be reviewed and several examples are presented.

1

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Consider the following arbitrarily loaded beam

Identify M

 M(x)  M oment at any section in the beam due to external loads

m

 m(x)  M oment at any section in the beam due to a unit action my I  Stress acting on dA due to a unit action

~ 

2

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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 Washkewicz College of Engineering

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  dx    dA  E I  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.,

 dW A

Wdiffernetial segment

cT     M m y2    dA  dx     2    cB  E I   cT     Mm  2     2   y dA  dx     EI c B 

  Mm    Mm     2  I  dx    dx EI EI      

4

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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.,  M x  mx   0  EI  dx L

WInternal 

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

1D

WExternal



WExternal

 WInternal

With

1 D  D 



 M  x  m x   0  EI  dx



 M  x  m x   0  EI  dx

L

L

or the deformation (D) of the a beam at the point of application of a unit action (force or 5 moment) is given by the integral on the right.

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Example 6.1

6

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Example 6.2 Flexibility Coefficients by virtual work

7

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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 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.

RB

5   wL 8 10

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

DB

5 w L4   384 EI

Thus by the compatibility equation

DB

  B RB

 0

RB

 

DB

B

5   w L 8

11

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 Washkewicz College of Engineering

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 Washkewicz College of Engineering

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

DQ 2

 DQL 2

 F21Q1

 F22Q2

 0

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

DQ1

 DQL1

 F11Q1

 F12Q2

DQ 2

 DQL 2

 F21Q1

 F22Q2

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Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

D  Q



D  QL



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

D  Q

 DQ1     D Q 2  

D  QL

F

 DQL1     D QL 2  

F   11  F21

F12  F22 

Q

Q1     Q2 

15

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

F  Q Q





D 

D 



Q

F 1  DQ 

QL



 D  QL

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

DQ

0    0 16

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

DQL1

13PL3  24 EI

DQL 2



97 PL3 48EI

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

D  QL



PL3 48EI

26   97 

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Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

F11 

L3 3EI

F21 

5L3 6 EI

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

F12



5L3 6 EI

F22



8L3 3EI

18

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

The flexibility matrix is

F 



L3  2 5  6 EI  5 16

The inverse of the flexibility matrix is

F 1



6 EI 7 L3

 16  5   5 2   

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

Q

Q    1  Q2 

F 1   DQ 

  16  5   0    2   0   5  P  69       64 56     6 EI   3  7L



D   QL

 PL3    48 EI

 26     97  19

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

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Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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 s 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 ading span.

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Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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



DQL1

wL3 24 EI

PL2  16 EI

Similarly,

DQL 2

PL2  16 EI

PL2  16 EI

PL2  8EI

such that

DQL



L2 48EI

2wL  3P    6P   22

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

The flexibility coefficients are determined next. The flexibility coefficient F11 is the sum of two rotations at t B. One in span AB and the other in span BC (not shown below)

F11 

L 3EI



L 3EI



2L 3EI

Similarly the coefficient F21 is equal to the sum of rotations at t C. However, the rotation in span CD is zero from a unit rotation at t B. Thus

F21 

L 6 EI

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Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Similarly



F22

L 3EI F12

 

L 3EI



2L 3EI

L 6 EI

The flexibility matrix is

F



L 6EI

4 1  1 4  

2 EI 5L

 4  1  1 4   

The inverse of the flexibility matrix is

F

1



24

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

Q

Q    1  Q2 

F 1  DQ 



D   QL

 4  1 0 L2   1 4  0  48EI     L  8wL  6 P       120  2wL  21P  2 EI 5L

2wL  3P    6 P  

and

Q1

wL2   15



PL 20

Q2



wL2 60



7 PL 40

25

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Example 6.6

26

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

t 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

D  Q



D  QL



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 ts are of primary interest if they are non-zero. . 27

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Instead of following the procedure just outlined we will now introduce a systematic procedure for calculating non-zero t 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 t displacements DJ1 and DJ2, both rotations, as well as reactions AR1 and AR2. can be computed. We will focus on the t 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 Washkewicz College of Engineering

The principle of superposition is used to obtain the t 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 ts 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 t 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 t B, superposition requires

DJ 1

 DJL1

 DJQ11Q1

 DJQ12Q2

Here DJ1

=

DJL1

=

DJQ11

=

DJQ12

=

non-zero displacement (a rotation) at t B in the actual structure, at the t associated with Q1 the displacement (a rotation) at t B associated with DJ1 caused by the external loads in the released structure. the rotation at t B associated with DJ1 caused by a unit force at t B corresponding to the redundant Q1 in the released structure the rotation at t B associated with DJ1 caused by a unit force at t 29 C corresponding to the redundant Q2 in the released structure

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

DJ 2

 DJL 2

 DJQ 21 Q1  DJQ 22 Q2

Here DJ2

=

DJL2

=

DJQ21

=

DJQ22

=

non-zero displacement (a rotation) at t C in the actual structure, at the t associated with Q2 the displacement (a rotation) at t C associated with DJ2 caused by the external loads in the released structure. the rotation at t C associated with DJ2 caused by a unit force at t B corresponding to the redundant Q1 in the released structure the rotation at t C associated with DJ2 caused by a unit force at t C corresponding to the redundant Q2 in the released structure

31

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

DJ  where

D  JQ



DJL 

D Q JQ

DJ 

D    J1   DJ 2 

DJL 

D    JL1  DJL 2 

   

and

Q which were determined previously



DJQ11

DJQ12

DJQ 21

DJQ 22

  

Q    1 Q2  32

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

In a similar manner we can find reactions via superposition

AR1



ARL1  ARQ11Q1  ARQ12Q2

AR 2



ARL 2  ARQ 21Q1  ARQ 22Q2

For the first expression AR1 AR2 ARL1 ARL2 ARQ11

= = = = =

ARQ22

=

ARQ12

=

the reaction in the actual beam at A the reaction in the actual beam at A the reaction in the released structure due to the external loads the reaction in the released structure due to the external loads the reaction at A in the released structure due to the unit action corresponding to the redundant Q1 the reaction at A in the released structure due to the unit action corresponding to the redundant Q2 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 Washkewicz College of Engineering

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

AR 



ARL 



A Q RQ

where

AR 

Q

A    R1   AR 2 

ARL 

A    RL1   ARL 2 

A  RQ

   

ARQ11

ARQ12

ARQ 21

ARQ 22

  

Q    1 Q2 

34

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

AM 1



AML1



AMQ11Q1



AMQ12Q2

AM 2



AML2



AMQ21Q1



AMQ22Q2

AM 3



AML3



AMQ31Q1



AMQ32Q2

AM 4



AML4



AMQ41Q1



AMQ42Q2

For the first expression AM1 AML1

= =

AMQ11

=

AMQ12

=

is the shear force at B on member AB is the shear force at B on member AB caused by the external loads on the released structure is the shear force at B on member AB caused by a unit load corresponding to the redundant Q1 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 Washkewicz College of Engineering

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

AM 



AML 



A Q MQ

where

AM 

Q



 AM 1  A   M2   A  M3   AM 4  

AML 



 AML1  A   ML 2    A  ML3    AML 4  

A  MQ



 AMQ11 AMQ12     AM Q 21 AMQ 22     AMQ31 AMQ32  A  A MQ 41 MQ 42  

Q    1 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 t 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 P2

 PL  P

P3

 P

37

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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 DJL1



1 PL 2 EI

 2  1 L

 

DJL 2





PL L EI

 

1 2 1 2

PL 1.5  0.5  L EI PL L EI 2

5 PL2 4 EI

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

DJL 



PL2 8EI

10 13  

39

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Consider the released beam with a unit load at point B

L

DJQ11

 

1 L L 2 EI L2 2 EI

DJQ 21

 

1 L L 2 EI L2 2 EI

40

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Consider the released beam with a unit load at point C

L 2L

DJQ12

 

1 L 2  1 L 2 EI 3L2 2 EI

DJQ 22

 

1 2L 2L 2 EI 2 L2 EI

leading to

  DJQ  



L2 2 EI

1 1 

3 4 

41

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Previously in Example 6.4

Q



 69     64  

P 56

with

DJ 



DJL 



D Q JQ

then the displacements DJ1 and DJ2 are

 DJ 

10  13  



PL2 8 EI



17  PL2   112 EI 5



L2 2 EI

1 1 

3  P  69    4  64 56   

42

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Using the following free body diagram of the released structure

Then from the equations of equilibrium

M



0



ARL 2





PL 2



0



ARL1





2P

A

ARL 2

F

Y

ARL1



2P

2P



L 2



P



PL



P

3L 2



P2L

P 43

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

then once again from the equations of equilibrium

F

Y

AML1

M

B

AML 2



0



AML1



0



0



AML 2



3PL 2





2P

2P



2P

L 2



PL 2



2 PL 44

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

then once again from the equations of equilibrium

F

 0

Y



A ML 4



P



P

 0

AML3 M

AML3

B



0



A ML 4 PL  2





PL 2



PL 45

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Thus the vectors AML and ARL are as follows:

AML



 0      3 PL    2       0       PL    2   

ARL

 2P          PL    2  

Reactions in the released structure.

Member end actions in the released structure.

46

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Finally with

AR 



ARL 



A Q RQ

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

 AR 





 2P     PL      2   P 107    56 31L 

P 56

 1  L 

1   69    64  2 L   

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

A  MQ

  ARQ  





1 0   0  0  1  L 

1  L    1   L  1  2 L  

48

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

Similarly, with

AM 



AML 



A Q MQ

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

 AM 





 0   3PL     2      0   PL     2   5    P  20 L    56  64   36 L  

P 56

1  0  0  0

1   69 L      1   64  L 

49

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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.,

D  Q

D  QS



D 

Q

QL







D  QS

F  Q



D  QT

F 1  DQ 

 

D  QP



D   QS

D  QR

50

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

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

DJ 



DJS 



D Q

AM 



AML 



A Q

AR 



ARL 



A Q

JQ

MQ

RQ

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 Washkewicz College of Engineering

MATRIX

ORDER

Q

qx1

Unknown redundant actions (q = Number of redundant)

qx1

Displacements in the actual structure Corresponding to the redundant

qx1

Displacements in the released structure corresponding to the redundants and due to loads

qxq

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

DQ DQL

DJQ

DEFINITION

DQT , DQP , DQR

qx1

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

DQS

qx1

DQS  DQL  DQT  DQP  DQR 52

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

MATRIX

ORDER

DEFINITION

DJ

jx1

t displacement in the actual structure (j = number of t displacement)

jx1

t displacements in the released structure due to loads

jx1

t displacements in the released structure due to unit values of the redundants

DJL

DQL DJT , DJP , DJR

jx1

DJS

jx1

F

qxq

t displacements in the released structure due to temperature, prestrain, and restraint displacements (other than those in DQ)

DJS



DJL  DJT  DJP  DJR

Matrix of flexibility coefficients 53

Section 6: The Flexibility Method - Beams Washkewicz College of Engineering

MATRIX

ORDER

DEFINITION

mx1

Member end actions in the actual structure (m = Number of end-actions)

mx1

Member end actions in the released structure due to loads

mxq

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

AR

rx1

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

ARL

rx1

Reactions in the released structure due to loads

rxq

Reactions in the released structure due to unit values of the redundants

AM AML AMQ

ARQ

54