Example 4 - Cleveland State University · Section 4: Matrix Structural Analysis (MSA) -...

Section 4: Matrix Structural Analysis (MSA) -

FundamentalsWashkewicz College of Engineering

1

Example 4.1



Principle Of Superposition

Mathematically, the principle of superposition is stated in the following fashion

Relative to a linear structural system, the mathematical expression above infers the

displacement at a given point in the system caused by two or more loads is the sum of the

responses which would have been caused by each load individually. Since the addition

function is preserved this is sometimes referred to as an additive map.

Consider a linear spring where

and K is the linear spring constant.

21

2121 ,,

ADAD

AADAAD

DKA

2



For the initial load on the spring

Now increase the deflection on the system by an amount DD. The additional load

on the spring is

The final force on the spring is from the additional deflection is

with

and

11 DKA

DKA DD

DDK

DKDKAA

D

DD

1

11

AAA D 12

DDD D 12

3



then

This result, although deceptively obvious, indicates that for a linear spring system

• the deflection caused by a force can be added to the deflection caused by

another force to obtain the deflection resulting from both forces being

applied;

• the order of loading is not important (DD or D1 could be applied first);

The is the Principle of Superposition – For a structure with a linear response, the

load effects caused by two or more loads are the sum of the effects caused by each

load applied separately.

For the principle to be applicable in a structural analysis the material the structure

is fabricated from must be linear elastic. To guarantee this we typically require the

structure to undergo small deformations.

22 DKA

4



Consider the beam in figure (a) subject to

external actions A1 and A2. These actions

produce various reactions and

displacements throughout the structure.

Reactions are developed at the supports. A

displacement is produced at the mid span.

The effects of A1 and A2 are shown

separately in (b) and (c). A single prime is

associated with A1 and a double prime with

A2.

From the figure it becomes obvious that the

following equations can be developed

through the use of superposition:

DDDRRR

MMMRRR

BBB

AAAAAA

5



Next consider the same beam subjected

to displacements, i.e., the support at B is

translated down an amount D and rotated

counterclockwise an amount q. Again

various reactions and displacements are

induced in the structure.

Reactions and displacements with single

primes are associated with D. Those

with double primes are associated with

q.

Focusing on the reactions, superposition

can also be invoked for a linear system if

the input variables are displacements,

i.e.,

6 2121 DADADDA



Action And Displacement Equations

The relationship between actions and displacements play an important role in structural

analysis. A convenient and simplistic way to see this relationship is through the use of a

linear, elastic spring

The action A will compress (translate) the spring an amount D. This can be expressed

through the simple expression:

In this equation F (in Example 4.1 we used d) is the flexibility of the spring. The spring

flexibility is defined as the displacement produced by a unit value of the (force) action A.

FAD

7



This relationship can also be expressed as

Here S is the stiffness of the spring and is defined as the action (force) required to produce a

unit displacement in the spring. The flexibility and stiffness of the spring are inverse to one

another. To see this consider

Thus

or

This requires a formal discussion of matrix division which occurs in Section 5 of the class

notes.

11

SS

F

SDA

8

FS

AFSA

DSA

1

11 FF

S



3

3 48

48 L

EIDA

EI

ALD

EI

LAF

48

1 3

3

481

L

EIDS

The relationship that holds for a

spring holds for any structural

component. Consider the

simple beam subjected to an

action A that produces a

translation D.

The action and displacement

equation holds if the flexibility

F and stiffness S are determined

as shown.

The action and displacement

equation given on the previous

slide is valid only when one

action is present and we are

looking for one displacement

within the structure. More

than one action and/or more

than one displacement

requires a matrix format.9



Consider a general example where a beam is subjected to

three actions, i.e., two forces (A1 and A2) and a moment

(A3). The directions for the actions are assumed positive.

The deflected shape is given in figure (b) and displacements

D1, D2 and D3 correspond to A1, A2, and A3.

By using superposition each displacement can be expressed

as the sum of displacements due to actions A1 through A3

In a similar manner expressions for D2 and D3 are

1312111

312111

32113211 ,,

DDDD

ADADAD

AAADAAAD

333231

3323133

232221

3222122

DDD

ADADADD

DDD

ADADADD

10



Now it is quite obvious that

and if we note the fact that

We can extend the notation and introduce the concept of a matrix of flexibility coefficients

via the following relationships

3113

2112

1111

AbycausedAatdeflectionD



onlyAtoalproportiondirectlyisD



313

212

111

31313

21212

11111

AFD

AFD

AFD

1132323

22222

12121

AFD

AFD

AFD

33333

23232

13131

AFD

AFD

AFD



Example 4.2

12



Example 4.3

13



We can express the equations for the deformations D1,

D2 and D3 as

Each term on the right-hand side of the equations is a

displacement written in the form of a coefficient times

the action that produces a deformation represented by

the coefficient. The coefficients are called flexibility

coefficients. The physical significance of the flexibility

coefficients are depicted in figures (c), (d) and (e)

3332321313

3232221212

3132121111

AFAFAFD

AFAFAFD

AFAFAFD

All the flexibility coefficients in the figures have two subscripts (Fij). The first subscript

identifies the displacement (Di) associated with an action (Aj). The second subscript

denotes where the unit action is being applied. Figure (c) is associated with action A1,

figure (d) is associated with action A2, and figure (e) is associated with action A3.

Flexibility coefficients are taken as positive when the deformation represented by the

coefficient is in the same direction as the ith action.14



Instead of expressing displacements in terms of

actions, it is possible to express actions in terms of

displacements, i.e.,

Here S is a stiffness coefficient and represents an

action due to a unit displacement. To impose these

unit displacements requires that artificial restraints

must be provided. These restraints are shown in the

figure by simple supports corresponding to actions

A1, A2 and A3.

3332321313

3232221212

3132121111

DSDSDSA

DSDSDSA

DSDSDSA

15



Each stiffness coefficient is shown acting in its assumed positive direction, which is

the same direction as the corresponding action. If the actual direction of one of the

stiffness coefficients is opposite to that assumption, then the stiffness coefficient will

have a negative value.

The calculations of the stiffness coefficients for the beam shown can be quite

lengthy. However, analyzing a beam like the one shown previously by the stiffness

method can be expedited by utilizing a special structure where all the joints of the

structure are restrained. We will get into the details of this in the next section of

notes.

The primary purpose of this discussion is for the student to visualize what flexibility

and stiffness coefficients represent physically.

16



Example 4.4

17



Example 4.5

18



Example 4.6

19



Example 4.7

20



Flexibility and Stiffness Matrices

We can now generalize the concepts introduced in the preceding section. If the number of

actions applied to a structure is n, the corresponding equations for displacements are:

In matrix format these equations become

or

nnnnnn

nn

nn

AFAFAFD

AFAFAFD

AFAFAFD

2211

22221212

12121111

nnnnn

n

n

n A

A

A

FFF

FFF

FFF

D

D

D

2

1

21

22221

11211

2

1

AFD 21



The action equations with n actions applied to the structure with a corresponding n

displacements are

In matrix format these equations become

or

nnnnnn

nn

nn

DSDSDSA

DSDSDSA

DSDSDSA

2211

22221212

12121111

nnnnn

n

n

n D

D

D

SSS

SSS

SSS

A

A

A

2

1

21

22221

11211

2

1

DSA 22



Since the actions Ai and displacements Di correspond to one another in both formats, it

follows the flexibility matrix Fij and the stiffness matrix Sij are related to each other. Taking

the matrix inverse of

yields

With

then

How to formulate the inverses appearing above is discussed in the next section of the notes.

AFD

DFA 1

DSA

1 FS

23



In a similar fashion one can show that

Thus the stiffness matrix is the inverse of the flexibility matrix and vice versa provided that

the same set of actions and displacements are being considered in both equations

Note that a flexibility matrix or stiffness matrix is not an array that is determined by the

geometry of the structure only. The matrices are directly related to the geometry and the set

of actions as well as displacements under consideration.

1 SF

24



Example 4.8

25



Example 4.9

The cantilever beam shown in the figure below is subjected to a force (A1) and moment (A2)

at the free end. Develop the flexibility matrix and the stiffness matrix for assuming

displacements D1 and D2 are of interest.

26



Making use of Case #7 and Case #8 from the following table

27



(continued)

28



Then the flexibility coefficients are as follows:

The displacements are

The flexibility matrix becomes

EI

LF

EI

LFF

EI

LF 22

2

2112

3

1123

21

2

22

2

1

3

1223

AEI

LA

EI

LDA

EI

LA

EI

LD

EI

L

EI

LEI

L

EI

L

F

2

232

23

29



In order to develop the stiffness matrix consider the following beam reactions due to

applied displacements:

30



Then the stiffness coefficients are as follows:

The actions are

The stiffness matrix becomes

L

EIS

L

EISS

L

EIS

46122222112311

1 1 2 2 1 23 2 2

12 6 6 4EI EI EI EIA D D A D D

L L L L

L

EI

L

EIL

EI

L

EI

S46

612

2

23

31



When the flexibility matrix and the stiffness matrix are multiplied together, the

result is the identity matrix:

This infers but does not prove that the two matrices are inverses of one another.

10

01

2

2346

612

2

23

2

23

EI

L

EI

LEI

L

EI

L

L

EI

L

EIL

EI

L

EI

FS

32



nnDADADADAW 33221121

Symmetry – Flexibility & Stiffness Matrices

If the loads on a structure are zero and gradually increase such that all loadings hit peak

values at the same time, the work done during this period of time will be summation of

the area under each individual load deflection curve, i.e.,

In a matrix format but both {A} and {D} are vectors by definition so to perform this

matrix multiplication we must use the transpose of one or the other column vectors. Thus

Recall that

Now, substitute this in the first equation for W immediately above.

T

T

DA

DAW

21

21

AFD

33



This substitution leads to

In addition, the following relationship holds from matrix algebra

Substituting this relationship in the equation from the previous slide yields

AFA

DAW

T

T

2

1

2

1

TT

TT

AF

AFD

TT

T

AFA

DAW

2

1

2

1

34



Now

Multiplying both sides by ({A}T)-1 and {A}-1 we obtain

TTTAFAAFA 2121

TTTAFAAFA

TTTTTAFAAAAFAAA

1111

TTTTTFAAAAFAAAA

1111

TFIIFII

TFF

35



Thus the flexibility matrix must be symmetric. To prove the stiffness matrix is symmetric

recall that

Substituting this in the equation for work

In addition, the following relationship holds from matrix algebra

DSA

T

T

DDS

DAW

2

1

2

1

TT

TT

DS

DSA

36



Substituting this in the above equation for work

Equating these two relationships for work

Multiplying both sides by [D]-1 and [DT]-1 we obtain

DDS

DAW

TT

T

2

1

2

1

DDSDDS

DDSDDS

TTT

TTT

2

1

2

1

TTTTTDSDDDDSDDD

1111

37



Further manipulation yields

Hence the stiffness matrix is symmetric. Of course the fact that the stiffness matrix is

symmetric could have been concluded from the fact that the flexibility matrix is symmetric

and the stiffness matrix is the inverse of the flexibility matrix. But this has not been formally

proven.

TTTTTSDDDDSDDDD

1111

TSIISII

TSS

38

Example 4 - Cleveland State University · Section 4: Matrix Structural Analysis (MSA) -...

Documents

Transcript of Example 4 - Cleveland State University · Section 4: Matrix Structural Analysis (MSA) -...