lecture 5: preliminary concepts of structural analysis

Lecture 5: PRELIMINARY CONCEPTS OF STRUCTURAL ANALYSIS

Principle Of Superposition

Mathematically the principle of superposition is stated asMathematically, the principle of superposition is stated as

( ) ( ) ( ) LL ++=++ 2121 aGaGaaG

or for a linear structural system, the response 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.additive map.

Consider a linear spring where

DKA =and K is the linear spring constant.

DKA


For the initial load on the spring

11 DKA =

Now increase the deflection on the system by an amount ∆D. The second additional load on the spring is

DKA ∆=∆

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

( )DKDKAA ∆+=∆+ 11

with( )DDK ∆+= 1

AAA ∆+= 12

and

AAA ∆+12

DDD ∆+= 12 12


then

22 DKA =

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 th f t bt i th d fl ti lti f b th f b i

22

another force to obtain the deflection resulting from both forces being applied;

• the order of loading is not important (∆D 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 to 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.


Consider the beam in figure (a) subject to external actions A1 and A2. These actions

d i i dproduce various reactions and displacements throughout the structure. Reactions are developed at the supports. A displacement is produced at the mid span.p p p

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 1 pA2.

From the figure it becomes obvious that the following equations can be developed g q pthrough the use of superposition:

MMMRRR AAAAAA ′′+′=′′+′=DDDRRR BBB ′′+′=′′+′=


Next consider the same beam subjected to displacements i e the support at B isto displacements, i.e., the support at B is translated down an amount ∆ and rotated counterclockwise an amount θ. Again various reactions and displacements are i d d i hinduced in the structure.

Reactions and displacements with single primes are associated with ∆. Those

ith d bl i i t d ithwith double primes are associated with θ.

Superposition can be invoked for a linear t if th i t i bl i tisystem if the input variable is an action,

or if the input variables are displacements.


Action And Displacement Equations

The relationship between actions and displacements play an important role in structuralThe relationship between actions and displacements play an important role in structural analysis. A convenient way to see this relationship is through a linear, elastic spring

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

FAD =

In this equation F is the flexibility of the spring, and this quantity is defined as the displacement produced by a unit value of the action A.


This relationship can also be expressed as

Here S (earlier it was K) is the stiffness of the spring and is defined as the action required to

SDA =

( ) p g qproduce a unit displacement in the spring. The flexibility and stiffness of the spring are inverse to one another.

1 1

11

1 1

−==

==−

FF

S

SS

F

F


The relationship that holds for a spring holds for any structural component. Consider the

AL3

component. Consider the simple beam subjected to an action A that produces a translation D.

EIALD

48

3

=The action and displacement equation holds if the flexibility F and stiffness S are determined as shown

( ) FEI

LAD ==

=48

1 3

as shown.

The action and displacement equation given on the previous slide is valid only when one EI48slide is valid only when one action is present and we are looking for one displacement within the structure. More

( ) ( ) SL

EIDA ==

= 3

481than one action and one displacement requires a matrix format.


Let’s consider a general example where a beam is subjected to three actions, i.e., two forces (A1 and A ) d (A ) h di i f hA2) and a moment (A3). The directions for the actions are assumed positive.

The deflected shape is given in figure (b) and di l t D D d D d t A Adisplacements D1, D2 and D3 correspond to A1, A2, and A3.

By using superposition each displacement can be d th f di l t d texpressed as the sum of displacements due to

actions A1 through A3

DDDD ++=

In a similar manner expressions for D2 and D3 are

1312111 DDDD ++=

DDDD

3332313

2322212

DDDDDDDD

++=++=


With

1111 AbycausedAatdeflectionD −

3113

2112

AbycausedAatdeflectionDAbycausedAatdeflectionD

−−

and the fact that

onlyAtoalproportiondirectlyisDonlyAtoalproportiondirectlyisD 111

th

onlyAtoalproportiondirectlyisDonlyAtoalproportiondirectlyisD

313

212

then

21212

11111

AFDAFDAFD

==

31313 AFD =


We can express the equations for the deformations D1, D2 and D3 as

AFAFAFD ++

3332321313

3232221212

3132121111

AFAFAFDAFAFAFD

AFAFAFD

++=++=

++=

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)

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 A3figure (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.


Instead of expressing displacements in terms of actions, it is possible to express actions in terms of di l idisplacements, i.e.,

3232221212

3132121111

DSDSDSADSDSDSA

++=++=

Thi t f ti b bt i d f th

3332321313

3232221212

DSDSDSADSDSDSA

++=++

This system of equations can be obtained from the displacement system of equations under suitable conditions.

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 themust be provided. These restraints are shown in the figure by simple supports corresponding to actions A1, A2 and A3.


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 thethe 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 quiteThe 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

tnotes.

The primary purpose of this discussion is for the student to visualize what flexibility and stiffness coefficients represent physically.


Flexibility and Stiffness MatricesWe 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:

nn

AFAFAFDAFAFAFD +++= L 12121111

nnnnnn

nn

AFAFAFD

AFAFAFD

+++=

+++=

L

MLLLM

L

2211

22221212

In matrix format these equations become

n AFFFD L 1112111

=

nnnnn

n

n A

A

FFF

FFF

D

DM

L

MLMM

L

M2

21

222212

or

{ } [ ]{ }AFD =


The action equations with n actions applied to the structure are

nn DSDSDSA +++= L 12121111

nnnnnn

nn

DSDSDSA

DSDSDSA

+++=

+++=

L

MLLLM

L

2211

22221212

In matrix format these equations become

nnnnnn 2211

n DSSSA L 1112111

=

nnnn

n

D

D

SSS

SSS

A

AM

L

MLMM

L

M

2

21

222212

or nnnnnn DSSSA 21

{ } [ ]{ }DSA{ } [ ]{ }DSA =


Since the actions Ai and displacements Di correspond to one another in both formats, it follows the flexibility matrix F and the stiffness matrix S are related to each other Takingfollows the flexibility matrix Fij and the stiffness matrix Sij are related to each other. Taking the matrix inverse of

{ } [ ]{ }AFD =

yields

{ } [ ] { }DFA 1−=

With

{ } [ ] { }

{ } [ ]{ }DSA =then

[ ] [ ] 1−= FS


In a similar fashion one can show that

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

[ ] [ ] 1−= SF

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 setgeometry of the structure only. The matrices are directly related to the geometry and the set of actions and displacements under consideration.


Example

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.


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


(continued)


Then the flexibility coefficients are as follows:

LLL 23

Th di l t

EILF

EILFF

EILF ==== 22211211 23

The displacements are

21

2

22

2

1

3

1 223A

EILA

EILDA

EILA

EILD +=+=

The flexibility matrix becomes

212211 223 EIEIEIEI

[ ]

=LLEIL

EIL

F 232

23

EI

LEIL

2


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


Then the stiffness coefficients are as follows:

Th ti

LEIS

LEISS

LEIS 4612

2222112311 =−===

The actions are

21222213143612 D

LEID

LEIAD

LEID

LEIA +−=−=

The stiffness matrix becomes

212222131 LLLL

[ ]

−

= EIEILEI

LEI

S 46

61223

−

LL2


When the flexibility matrix and the stiffness matrix are multiplied together, theresult is the identity matrix:

[ ][ ]

−

= 2346

612

2

23

23

LLEIL

EIL

EIEILEI

LEI

FS

=

−

1001

22 EIEILL

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

0


Equivalent Joint Loads

The calculations of displacements in larger more extensive structures by the means of the matrix methods derived later requires that the structure be subject to loads applied only at the joints. Thus in general, loads are categorized into those applied at joints, and those that are

t L d th t t li d t j i t t b l d ith t ti ll i l t l dnot. Loads that are not applied to joints must be replaced with statically equivalent loads. Consider the statically indeterminate beam with a distributed load between joints A and B, and a point load between joints B and C:


First one must identify joints, and here we select points A, B, and C for convenience. By superposition the beam can be separated into two beams, one with loads located at the joints, and a second with loads between the joints:and a second with loads between the joints:


To transfer the loads that act on the members to the joints, the joints of the structure are restrained against all displacements. This produces two fixed end beams:

When these fixed end beams are subjected to the member loads, a set of fixed end actions is d d Th fi d d ti h i th f ll i fi h thproduced. The same fixed-end actions are shown in the following figure where they are

depicted as restraint actions in the restrained structure.


If the restraint actions are reversed in direction, they constitute a set of forces and couples that are statically equivalent to the member loads. These equivalent joint loads, when added to the actual joint loads produce the combined joint loads shown in the next figureactual joint loads produce the combined joint loads shown in the next figure.

The combined loads are then used in carrying out the structural analysis. Will the unknown displacements at nodes be correct using this statically equivalent system? Will the displacements computed along segment AB in the figure above be equivalent to thedisplacements computed along segment AB in the figure above be equivalent to the displacements along segment AB given the original load configuration?


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

( )DADADADAW ++++= L21

g y g pvalues at the same time, the work done during this period of time will be the average, hence

( )nnDADADADAW ++++= 33221121

In a matrix format but both A and D are column vectors by definition so to perform thismatrix multiplication we must use the transpose of one or the other column vectors. Thus

( ){ } { } ( ){ }{ }TT DADAW 2121 ==

Recall that

{ } [ ]{ }AFD =

Now, substitute this in the above equation.


This substitution leads to

11

In addition the following relationship holds from matrix algebra

{ } { } { } [ ]{ }AFADAW TT

=

=

21

21

In addition, the following relationship holds from matrix algebra

{ } [ ]{ }( )TT AFD =

Substituting this relationship in the equation from the previous slide yields

[ ] { }TT AF=

Substituting this relationship in the equation from the previous slide yields

{ }{ } { }[ ] { }TTT AFADAW

=

=

11 { }{ } { }[ ] { }AFADAW

22


Equating the two equations for work we obtain

{ } [ ]{ } { }[ ] { }TTT AFAAFA 2121 ={ } [ ]{ } { }[ ] { }AFAAFA 2121 =

{ } [ ]{ } { }[ ] { }TTT AFAAFA =

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

{ }( ) { } { } [ ]{ } { }( ) { } { }[ ] { }TTTTT AFAAAAFAAA 1111 −−−−{ }( ) { } { } [ ]{ } { }( ) { } { }[ ] { }AFAAAAFAAA =

{ }( ) { } { } { }( )[ ] { }( ) { } { } { }( )[ ]TTTTT FAAAAFAAAA 1111 −−−−=

[ ][ ][ ] [ ][ ][ ]TFIIFII =

[ ] [ ]TFF =


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

Substituting this in the equation for work

{ } [ ]{ }DSA =

Substituting this in the equation for work

{ }{ } [ ]{ }{ }TT DDSDAW

=

=

21

21

In addition, the following relationship holds from matrix algebra

22

{ } [ ]{ }( )TT DSA =

{ } [ ] { }TTT DSA ={ } [ ] { }DSA =


Substituting this in the above equation for work

{ } { } [ ] { } { }TTT

11

Equating these two relationships for work

{ } { } [ ] { } { }DDSDAW TTT

=

=

21

21

[ ]{ }{ } [ ] { } { }DDSDDS TTT

=

21

21

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

[ ]{ }{ } [ ] { } { }DDSDDS TTT =

{ }( ) { } { } [ ]{ } { }( ) { } { }[ ] { }TTTTT DSDDDDSDDD 1111 −−−−=


Further manipulation yields

{ }( ) { } { } { }( )[ ] { }( ) { } { } { }( )[ ]TTTTT SDDDDSDDDD 1111 −−−−={ }( ) { } { } { }( )[ ] { }( ) { } { } { }( )[ ]SDDDDSDDDD

[ ][ ][ ] [ ][ ][ ]TSIISII =

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

[ ] [ ]TSS =

ysymmetric 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.

lecture 5: preliminary concepts of structural analysis

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al proportion directly

fixed end beams

directly related

structural analysisthen

structural analysis

matrix format

preliminary concepts

segment ab