04 flexibility - juinio

Chapter 4.0 MEMBER AND STRUCTURE FLEXIBILITY MATRIX 4.1 GENERAL

For statically indeterminate structures, the equations of equilibrium are insufficient to determine the response of the structures. Additional equations must be written based on the compatibility conditions of the structure and the stress-strain relationship of the materials used. For structures that could be modeled as an assemblage of line elements we are concerned primarily with determining the stress-resultants, displacements, and deflections at each point rather than the stresses and strains. Therefore, instead of directly using the stress-strain relations, force-displacement relationships shall be developed. This relation is usually expressed in matrix form as either:

[U] = [F] [P]

or [P] = [K] [U]

Where [P] and [U] are the nodal force and the nodal displacement vectors; and [F] and [K] are referred to as the flexibility and stiffness matrix respectively. In this chapter the flexibility matrix for a line element and for an assemblage of elements shall be developed. The flexibility matrices will be developed considering that the force-displacement relationship of a system depends on the force-displacement relationship of its components. Thus the structure [F] will depend on the flexibility matrix [Fm] of its component members. The members may further be viewed as being made up of individual slices. The slices in turn may be viewed as being made up of differential elements. The flexibility matrix for the differential element will depend on the stress-strain relationship of the material. The development of the force-displacement relationships shall therefore begin with the stress-strain relationship and work up to the structure level.

Institute of Civil Engineering, UP Diliman

4-2 Notes on Matrix Structural Analysis

dx

Figure 4-1 Breakdown of Components of a Structure

4.2 DEGREES of FREEDOM and NODAL POINTS The first thing that must be done before developing the force-displacement relationship is to define the elements of the force and displacement vectors [P] and [U]. These are normally defined as the forces and displacements along the degrees of freedom of the structure or member. The degrees of freedom of a system are defined as the displacement components necessary to define its position in space at any time and under any loading. Based on this definition all structures/member have an infinite number of degrees of freedom since each structure/member may theoretically deform in an infinite number of modes. In dynamic analysis, this complexity is usually avoided by lumping masses at selected points and considering only the finite number of degrees of freedom at these points. For static analysis, the resulting displacements under a given set of loads will be a single forced mode. This enables considerable simplification in that the response may be reduced to a function of a finite number of degrees of freedom. The points whose force/displacement components are used to describe the response of the member or structure are referred to as nodal points. To minimize the size of the problems the number of nodal points, and therefore number of degrees of freedom, is the minimum required. For structures the nodal points normally correspond to the points where one or more members meet. For line elements the end points are normally used as the nodal points as these are where it will connect to other members or supports. Recall that the responses of interest are the resulting displacements and internal stress resultants due to actions. For members, if we know the end forces at either end due to a given set of actions, then the internal forces at any point between the ends and the other end can be determined from equilibrium requirements. Further, knowing the displacements of the ends and the internal stress resultant functions, the displacement at any point between the ends can also be determined. For line elements, the number of degrees of freedom at the nodes depends on the type of element. For an axial force member only two degrees of freedom are required based on the member or local coordinates, see Figure 4-2a. This is because the response quantities of interest are only the axial force and axial

dx dz

dy

dx

Structure

Member

Slice Differential Element

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Member Flexibility Matrix 4 - 3

deformations. The local coordinates are used almost exclusively when developing the member force-displacement relationships.

3 2

4

1

6

5 2

1

4 3

2 1

a) Local Coordinates b) Planar Global Coordinates c) Space Global Coordinates

Figure 4-2 Degrees of Freedom for an Axial Force Member In determining the structure force-displacement relationships, the member relationships may have to be expressed in terms of the structure or global coordinates. The axial force and axial displacement quantities must, in this case, be expressed in terms of its components along the global coordinates. Thus, four degrees of freedom are required in the planar case, and six degrees of freedom in the three dimensions. For planar frame members, there are six (6) degrees of freedom consisting of the three displacement components, two translational and one rotational displacement components, at each end of the member. For a planar grid member, there are one translational and two rotational degrees of freedom at each end. For a space frame member every point has six degrees of freedom (three translational and three rotational) resulting in 12 nodal degrees of free-dom.

3 2

4 5 6

1

6 5

4 3

2 1

a) Plane Frame b) Plane Grid Figure 4-3 Degrees of Freedom for an Planar Member

The degrees of freedom, arrows in the figures, represent both forces and displacement. They are numbered to identify its position in the corresponding vectors. This numbering, by convention, is by node starting with direct forces/translations then moments/rotations. Note that the member degrees of freedom above were defined without consideration of any support conditions. These are referred to as Absolute Degrees of Freedom. Degrees of freedoms corresponding to known displacement components, i.e at supports, are referred to as Support Degrees of Freedom. The known displacement is usually zero, but the definition allows consideration of known non-zero support movements. The degrees of freedom of a member or structure other than the support degrees of freedom are referred to as the Relative Degrees of Freedom. Thus for each of the planar frame members with the different support conditions as shown in Figure 4-4 there are three support (indicated by a slash across the arrows) and three relative degrees of freedom.



a) Fixed Free Supported b) Simply Supported Figure 4-4 Relative Degrees of Freedom for an Planar Member

4.3 STRESS-STRAIN RELATIONSHIP AND DIFFERENTIAL

ELEMENT FLEXIBILITY The deformation laws for structural members are based on the Laws of Elasticity, relating stresses and their corresponding strains. In the following we shall consider only materials that are linearly elastic, homogeneous and isotropic. For this material, two elastic constants, the modulus of elasticity and the Poisson's ratio, are sufficient to relate stresses and strains. In matrix form, the stress strain relationship is given by:

++

+

=

zx

yz

xy

x

x

x

zx

yz

xy

z

y

x

E

)1(2000000)1(2000000)1(2000000100010001

1

or more compactly: ]][[ = ][ C (4-1)

where [] and [] are the strain and stress vectors, and [C] is the material compliance matrix. Note that the above recognizes that the shears and shear deformations occur in conjugate pairs, i.e. xy = yx, xy = yx, etc. Alternatively, the material compliance matrix may be written in terms of the elastic modulus, E, and the shear modulus, G, noting that

E =G ) + 2(1

For structural steel, Es = 200,000 MPa (29,000 ksi), = 0.3 in the elastic range, and G = 77,000 MPa (11,000 ksi). Flexibility Matrix for a Differential Element Considering now an differential element dx,dy,dz subject to a general state of stress in Figure 4-5.

1 2

3 2 3

zx

yz xy

z

y

x dy

dz dx

1

Figure 4-5 Differential Element Stresses (conjugate shears not shown) The resultants of the corresponding stresses are equal to the stress multiplied by the area on which it acts. For example the resultant force along the x-axis due

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to x is equal to x (dydz). All the stress resultants may be combined in a single force vector defined as [Pe]T = [ Pex Pey Pez Pexy Peyz Pezx ] and determined as:

=

zx

yz

xy

z

y

x

ezx

eyz

exy

ez

ey

ex

dxdydxdz

dydzdxdy

dxdzdydz

PPPPPP

or [Pe] = [dA] [] (4-2) from which [ ] = [dA]-1 [Pe] (4-3) The corresponding deformations of the stresses acting on the differential element can also be defined by a vector [Ue]T = [ Uex Uey Uez Uexy Ueyz Uezx ], and determined from the strains as:

=

zx

yz

xy

z

y

x

ezx

eyz

exy

ez

ey

ex

dzdy

dxdz

dydx

UUUUUU

or [Ue] = [dL] [] (4-4) Note that the elements of [Ue] are the displacements along the corresponding force element in [Pe]. For example, as illustrated in Figure 4-6, Uex is the deformation in the direction of Pex, and Uexy is the deformation in the direction of Pexy. x

a) Due to x b) Due to xy. Figure 4-6 Stress Resultants and Displacements

Combining equations (4-4) with (4-1) and (4-3),

[Ue] = [dL] [C] []

[Ue] = [dL] [C] [dA]-1

[Pe] or [Ue] = [Fe] [Pe] (4-5)

with [Fe] = [dL][C] [dA]-1 (4-6)

Equation (4-6) should be recognized as a force-displacement relationship for a differential element derived form the stress-strain relationship of the material. The matrix [Fe] is thus a flexibility matrix.

dx Uex = xdx

Pex = x dydz Pexy = x dydz xy xy Uexy = xy dx dx



4.4 SLICE FLEXIBILITY Considering now a differential slice, dx, of a line element which contains our differential element and on which the stress resultants are given by [Sc]T = [S1 S2 S3 ...]. Note that the number of elements of [Sc] depends on the type of member under consideration, and the element Si may be an axial force, shear, moment or torsion stress resultant. At this point we shall assume that the stress distribution throughout the member section which gives rise to the stress resultants is known and can be expressed as:

[] = [Byz] [Sc] (4-7) recalling equation (4-3) and combining,

[Pe] = [dA] [] = [dA] [Byz] [Sc] or [Pe] = [Hec] [Sc] (4-8) where [Hec] = [dA] [Byz] (4-9) Matrix [Hec] should be recognized as a force transformation matrix as this relates the two force systems [Pe] and [Sc]. Now letting [Uc] be the vector of displacements corresponding to the stress resultants [Sc], we have from the principle of contragradience:

d[Uc] = [Hec]T [Ue] (4-10) The differential operator is applied for [Uc] as equation (4-9) represents the deformation of the slice due to the deformation of a single differential element. Combining equations (4-10), (4-5), (4-8) and (4-9)

d[Uc] = [Hec]T [Ue] = [Hec]T [Fe] [Pe]

= [Hec]T [Fe] [Hec] [Sc]

= [Byz]T [dA]T [dL] [C] [dA]-1 [dA] [Byz] [Sc] (4-11)

Recalling that [dA] is a diagonal matrix such that [dA]T = [dA], and with noting that the matrix product [dA] [dL] = [dV], where [dV] = [I6x6] dxdydz, equation (4-11) reduces to

d[Uc] = {[Byz]T [C] [Byz]} dxdydz [Sc] (4-12)

To get the total deformation of the slice due to the deformability of all differential elements making up the slice, we need to integrate over the whole cross-section. Rearranging and defining dydz as dA, the [U] is given by:

[ ] ][][][][][ cA

yz

T

yzA

cc SdxdABCBUdU

==

or [Uc] = [ fs ] dx [Sc] (4-13)

[Uc] = [Fs] [Sc] (4-14)

where [Fs] = [ fs ] dx (4-15)

and (4-16) [ ] =A

yz

T

yzs dABCBf ][][][

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The matrix [Fs] is denoted as the slice flexibility matrix as this relates the forces and corresponding deformations of a slice. Note that equation (4-13) can be rewritten as

[Uc] = [ fs ] [Sc] dx (4-17)

or [Uc] = [ s ] dx (4-18)

That is, the product [fs][Sc] can be viewed as the "strain" of the slice element. These slice strains and deformations will be further discussed following the following sections.

Slice flexibility - Axial Force Member Consider a slice of a prismatic axial force member with a cross-sectional area, A, made up of a homogeneous, linearly elastic material. The axial force is assumed to act along the local x-axis which is centroidal. The axial stress, x, must be uniformly distributed over the cross-section with the other stresses being zero.

S1

x

z

y

dx

x

With x being the only non-zero stress component, the only strain component of interest is x, and the stress-strain relation reduces to [ x ] = (1/E) x. Thus

[C] = (1/E)

For the relation between the stress and stress resultant we have S1 = x A, or

x = S1 / A

thus [Bxy] = (1/A)

Therefore, the slice flexibility is

dx AE1 = dx dA

EA1 = dx dA

EA1 = F

A22

As

][

or [ fs ] = [1/AE]

Slice flexibility - Plane Frame Member. Consider now a slice of a plane frame member of material which follows Hooke's Law with the local axis defined such that the x-axis is centroidal and the y- and z-axes are principal axes.



z

x

y

S1

S2 S3

x dx xy

y

x

dx

Stress Resultants Stress Distribution If we assume that a state of planar stress exist such that y = z = yz = zx = 0, the non-zero stress components are given as

I yS +

AS =

z

31x

xy2

z = S Q

I b

where: A = cross-sectional Area Iz = moment of inertia about the z-axis Q = first moment of area of portion of cross-section above level

where shear stress is being evaluated b = width at the level the shear is being evaluated

In matrix form the above may be written as

=

3

2

1

00

01

SSS

bIQ

Iy

A

z

zxy

x

therefore

=00

01

bIQ

Iy

AB

z

z][

For the stress-strain relationships, considering only the non-zero stresses and their corresponding strains, this reduces to

=

xy

x

xy

xG

E

//

1001

or

= G

EC //][ 10

01

Substitution into our equation for [ fs ] gives

[ ] =A

yz

T

yzs dABCBf ][][][

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dA

Iy

EAEIy

bIQ

G

AEIy

EA

f

zz

z

z

s

=

210

010

01

2

2

][

A Before performing the integration term by term, recall that by convention the x-axis is the centroidal axis and the y and z-axes are principal axes.

centroidal is x if 0 = dAy since0 = dAy AE1 = f = f

AE1 = dA

EA1 = f

AA1331

A211

dA bQ

IG1 = f

I = dA y nce siIE

1 = dA y IE1 = f

2

2

Az222

2

Az

2

Az233

For f22, simplification if possible only if the particular cross-section is known as Q and b are functions of the cross-section. Consider, for example a rectangular cross-section.

( )bdy = dA constant = b

y16 + bh8 - h 64b = Q

y+ 2h y-

2h

2b = Q

42242

2

,

h/2

h/2

y

y x

GA

1= AG

65

1 = AG1

56 =

IGhb

1201 =

dy b b

)y16 + yh8 - h( 64b

IG1 = f

v2z

5

2

422422h

2h-

2z

22 b

Where Av is referred to as the shear area. For rectangular cross sections this is equal to (5/6) times the cross sectional area, for other cross-sections the shear area may be calculated as

dA bQ

I = A

A

2

2z

v

(4-19)

The use of the shear area is preferred as this result in a slice flexibility whose elements are of similar form as shown below.



=

z

vs

EI

GA

AEf

100

010

001

][ (4-20)

It may also be instructive to calculate the "slice-strain" given as [s] = [ fs ][S]

and setting S1 = P, S2 = V, and S3 = M (to express it in more familiar form)

=

=

z

v

z

vs

EIM

GAV

AEP

EIS

GAS

AES

3

2

1

][

The first term can be recognized as the axial strain of the slice due to the axial force while the third term is the curvature of the slice due to the bending moment. The second term on the other hand can be viewed as the change in vertical displacement per unit length of the slice edges due to the shear force.

a) Axial b) Shear c) Moment

Figure 4-7 Slice Stress Resultant and Displacement

Slice Flexibility - Space Frame Member For a slice in three dimensional cases, there are 6 stress resultant components. Following the same procedure and using the same assumptions as in the planar case, the slice flexibility for a space frame member is can be expanded to that given in Equation (4-21). The only item which needs additional discussion pertains to the torsional slice flexibility coefficient (1/GJ). For circular sections, J is the polar moment of inertia of the cross-section. For non-circular sections, J is the torsion constant.

dx Us1 = s1dx

S1

dx

S2

Us2 = s2 dx s2

dx

S3

Us3 = s3 dx

dx

6

5

4

3

2

1 z

y

x

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=

z

y

vz

vy

s

EI

EI

GJ

GA

GA

AE

f

1

1

1

1

1

1

][ (4-21)

The determination for J may become involved due to the warping of the cross-section during twisting. Theory of elasticity concepts must normally be used for non-circular sections. For example, for a rectangular cross-sections, J depends primarily on the ratio of the side dimensions, (b/t) with b being the longer side. Where

J = k1 b t3 (4-22)

with

=

= K,,tanh

31551 2

1192131

n tbn

nbt

k (4-23)

Values of k1 for representative values of (b/t) are shown below.

Table 4-1 Values of k1 b/t 1 2 3 5 10 k1 0.1406 0.229 0.263 0.291 0.312 1/3

4.5 MEMBER FLEXIBILITY MATRIX FROM SLICE FLEXIBILITY

We are now prepared to find the relationship between the nodal forces and nodal deformations in the form.

[Um] = [Fm] [Pm] (4-24)

Where [Um] and [Pm] are the nodal displacement and force vectors respectively, and Fm is the member flexibility matrix. As will be explained below, the flexibility matrix of a member, or structure, is defined only with respect to relative D.O.F's, i.e. enough supports must have been provided to prevent any rigid body motion. For example for a planar frame member, or structure, at least three support reactions must be provided to prevent the independent rigid body motions, (two translational and one rotational motion). For a space frame member, six possible rigid body motion components must be prevented. Note that these numbers of supports are minimum requirements and flexibility matrices may be determined for a statically indeterminate member or structure. The following derivation for [Fm] is applicable for both determinate and indeterminate systems. Example problems, however, will be limited to determinate cases.



When loads along [Pm] are applied, stress resultants will be generated at the differential slices of the member. From the discussion of equilibrium condition, the stress resultant function [S] due to [Pm] may be expressed as a linear force transformation of the form

[S] = [Hs] [Pm] (4-25)

The slice force [S] will result in slice deformations. This slice deformation may be determined from the force-displacement relationship for the slice. Thus

[Us] = [Fs] [S]

or [Us] = [ fs ] [S] dx

As the slice deforms, nodal displacements are produced from compatibility requirements. We can determine the nodal displacements from compatibility, but with contragradient law we know that the relationship between the slice and nodal forces is given by

d[Um] = [Hs]T [Us] (4-26)

Where d[Um] is the deformation at the nodal point due to the deformation of a particular differential slice. Combining

d[Um] = [Hs]T [ fs ] [S] dx

= [Hs]T [ fs ] [Hs] [Pm] dx (4-27)

To get the total deformation at the node we have to sum the contribution of all the slices. Thus;

(4-28) ][)(][][ P dx Hf H = Ud = U mssTsL

mL

m

and thus [ ] [ ] [ ][ ]dx H f H = F ssTsL

m (4-29)

Example 4-1 Determine [Fm] for a Fixed - Free Prismatic Plane Frame Member Solution: [ ] [ ] [ ][ ]dx H f H = F ssTs

Lm

From equilibrium, the slice force at x is given by [S] = [Hs] [Pm], where

=

10

010

001

)(

][

xL

Hs and the slice flexibility [

=

z

v

EIGA

AEfs

//

/]

100010001

L x

1 2

3

dx

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Substituting

dx

EIEIxL

EIxL

EIxL

GA

AE

F

zz

zzvm

+

10

10

001

2

)(

)()(][

=

L

0

+=

zz

zzvm

EIL

EIL

EIL

EIL

GAL

AEL

F

20

230

00

2

23

][

or alternatively

+=

zz

zzm

EIL

EIL

EIL

EIL

AEL

F

20

2310

00

2

23

)(][ (4-30)

where

+== 22 16

3LA

IGLAEI

v

z

v

z )( = shear deformation constant

Example 4-2 Determine [Fm] for a Simply Supported Prismatic Plane Frame

Member Solution: [ ] [ ] [ ][ ]dx H f H = F ssTs

Lm

The slice flexibility for a plane frame member is still

=

z

v

EIGA

AEfs

//

/][

100010001

Unlike that for the fixed-free case, the equilibrium relationship between slice and nodal forces for the simple support case is not as simply derived. This is because the support reactions due to [Pm] must first be determined. It can be verified that the stress resultant functions is given by [S] = [Hs] [Pm], where

L x

1 2 3

dx



LxLxLLL

//)(0/1/10

001 =Hs ][

Substituting

dx

LEIx

GLALEIxLx

GLA

LEIxLx

GLALEIxL

GLA

AE

F

zvzv

zvzvm

+

+=

2

2

222

222

2

2

110

110

001

)(

)()(][

0

L

+

+=

zvzv

zvzvm

EIL

GLAEIL

GLA

EIL

GLAEIL

GLA

AEL

F

31

610

61

310

00

][

or alternatively

+

+=

zz

zzm

EIL

EIL

EIL

EIL

AEL

F

31

6210

621

310

00

)()(

)()(][

(4-31)

where

+==

3A

22 16 LAI

GLEI

v

z

v

z )(

Notes: Shear Deformation Constant For line elements the shear deformation constant is normally much less than one since by definition the length is much larger than the other dimensions of the member. In other words, the displacements caused by shear deformations are much smaller than that due to flexural deformations. As such, it is usually neglected i.e. is assumed to be zero or equivalent to assuming AvG = . Most of the examples in these notes will use this assumption. Flexibility Matrix for a Prismatic Truss Element For a prismatic axial force member, the flexibility matrix is [Fm] = [L/AE], obtained by deleting the rows and columns for the other DOFs in the two examples.

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4.6 PROPERTIES OF THE FLEXIBILITY MATRIX We start with the definition of the flexibility coefficient, fij.

fij is defined as the displacement along degree of freedom 'i' resulting from the application of a unit force along degree of freedom 'j ' with all the other nodal forces being equal to zero.

Figures 4-8 and 4-9 provides an illustration of this definition for the fixed-free and simply supported cases (with =0). a) P1 = 1, P2 = P3 =0 b) P2 = 1, P1 = P3 =0 c) P3 = 1, P1 = P2 =0

Figure 4-8 Flexibility Coefficients for a Fixed-Free Plane Frame Member a) P1 = 1, P2 = P3 =0 b) P2 = 1, P1 = P3 =0 c) P3 = 1, P1 = P2 =0

Figure 4-9 Flexibility Coefficients for a Simply Supported Plane Frame Member

As shown in the Figures, it could also be said that jth column of [Fm] represents a set of compatible nodal displacements resulting from a unit load along DOF 'j'. From the definition of fij and considering that we are using the same coordinate system for forces and displacements, the diagonal elements, fii, have a positive value. It is a physical impossibility that the resulting deformation along the applied force is in the opposite direction of the force. The reason that the flexibility matrix is only defined for relative degrees of freedom can also be explained from the definition and the fact that loads can only be applied at the nodal degrees of freedom. From the definition, a unit load is applied only along the jth DOF. With no support restraints defined the system is unstable since no forces (reactions) can be developed to keep the system in equilibrium. To be stable, sufficient support reactions (restraints) must be present to prevent any rigid body motion. Note, that this does not preclude the determination of member flexibility matrices for indeterminate systems. For instance we may determine [Fm] for a propped cantilever which is indeterminate to the first degree. However, while the number of relative degrees of freedom is reduced, determination of the slice force due to the nodal

Um1 = f11 = L/AE

P1 = 1

Um2 = f21 = 0 Um3 = f31 = 0

P2 = 1

Um1 = f12 = 0 Um2 = f22 = L3/3EI Um3 = f32 = L2/2EI

Um2

Um3

Um2 Um3

P3 = 1

Um3 = f33 = L/EI

Um1 = f13 = 0 Um2 = f23 = L2/EI

P1 = 1

Um2 = f21 = 0 Um3 = f31 = 0

Um1 = f11 = L/AE

P2 = 1

Um1 = f12 = 0 Um2 = f22 = L/3EI Um3 = f32 = -L/6EI

Um2 Um3 P3 = 1

Um3 = f33 = L/3EI

Um1 = f13 = 0 Um2 = f23 = -L/6EI

Um2

Um3



forces will require the solution of an indeterminate problem. In the following, flexibility matrices for determinate systems are the only cases considered.

L

1 2

Figure 4-10 Propped Cantilever with Corresponding Degrees of Freedom Another observation that may be made is the "uncoupling" of the axial with the shear and bending terms for straight line elements. Uncoupling here refers to the dependence of the axial deformation only to axial forces; and the flexural and/or shear deformations being dependent only on the flexural and/or shear forces. This results from the selection of the local axes and the use of the assumption of small displacements. Note that this is true only for straight line elements where the nodal axial force produces only axial slice forces. The final observation we make, and not the least important, is that the flexibility matrix is symmetric, i.e. fij = fji. The proof of this comes directly from Maxwell's Law of Reciprocal Deformations. Maxwells Law is a special case of Betti's Law which is based on the conservation of energy principle. Betti's law may stated as follows: In any structure whose material follows Hooke's Law and in which the supports are unyielding and the temperature is constant, the external virtual work done by a system of forces, [Pm], during the deformation, [Umn] caused by a system of forces, [Pn], is equal to the external virtual work done by the [Pn], system of forces during the deformation, [Unm] caused by the [Pm] force system. Or in matrix form

[ ] [ ] [ ] [ ]U P = U P nmn Tmnm T (4-32)

The proof of Betti's Law proceeds by considering that for a conservative system, the total work done by applying both the [Pm] and [Pn] set of forces to a structure is unique and is independent of the order these two system were applied.

Figure 4-11 Pm and Pn Forces on a System If [Pm] is applied first then [Pn], them the total work done is

U P + U P21 + U P 2

1 = W mnmT

nnnT

mmmT

1

If [Pn] is applied first then [Pm], we have the corresponding work

Pn3 Pn3

Pm1

Pm2 Pmj

Pn1 Pn2

Pnk

Pm1

Pm2 Pmj

Pn1 Pn2

Pnk

= +

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U P + U P21 + U P 2

1 = W nmnTmmmTnnnT2

where: Umm - deformation along forces Pm due to force system Pm Unn - deformation along forces Pn due to force system Pn Unm - deformation along forces Pn due to force system Pm Umn - deformation along forces Pm due to force system Pn

since the total work is independent of the order of load application

W1 = W2

which implies that [ ] [ ] [ ] [ ]U P = U P nmn Tmnm T

Maxwell's Law of reciprocal deflection is a special case where both Pn and Pm set of forces consist of a single unit force. Thus the above relation may be written as;

P1 12 = P2 21

with P1 =1 and P2 =1, then 112 = 1*21 (4-33)

where 12 is the deformation along force '1' due to a unit load along force '2', and 21 is the deformation along force '2' due to a unit load in along force '1'. This definitions of ij is analogous to that for the flexibility coefficient, fij, then it follows directly that

fij = fji These laws are general, within the conditions set, in that these may be applied to any structure and any pair of load or system of loads. Thus, equation (4-33) applies when both forces are direct forces, when both are couples, and when one is a couple and the other is a direct force.

Note that the equality is on the virtual work which is a scalar quantity. The deflections or flexibility coefficients are therefore equal only in magnitude. For example if P1 is a direct force and P2 is a couple then the deflection 12 is a translational displacement and 21 is a rotational displacement (which cannot be equal as vectors).

4.7 FLEXIBILITY TRANSFORMATIONS BETWEEN SETS OF D.O.F.'S

As a member may be provided with different sets of supports to produce a determinate and stable system, different corresponding flexibility matrices are generated depending on the resulting relative degrees of freedom. Despite the apparent difference of the flexibility matrices of a member with different support conditions, these matrices are related to one another. This relationship results from the requirement that the nodal forces are related based on equilibrium and that there is a unique relationship between the loads acting on the member and the resulting deformations. To illustrate the relationships, first consider the simply supported beam with equal and opposite end moments shown in Figure 4-12a. It should be obvious that the response in terms of internal stresses and deformations are identical to that of the equivalent system shown in Figure 4-12b where we have replaced the supports with the support reactions (zeroes for this case). Consider now Figure



4-12c, which has the same internal stress function as the other two but provided with simple supports. It should be recognized that elastic curve (deflection function) with respect to the x-axis (along the chord connecting the member ends) in Figure 4-12c is identical to the elastic curve with respect to the original undeformed members in Figure 4-12a and 4-12b. It follows that the nodal (end) displacements of c) are related to the nodal displacements of the other two cases (for this case by a rotational coordinate transformation about the A-end).

a) Simply Supported b) FBD w/o Supports c) Fixed-Free

Figure 4-12 Relation of Displacements for Different Supports Based on the discussion above, knowing the flexibility matrix for a particular set of relative degrees of freedom it should be possible to determine the flexibility matrix for the same member but for a different set of relative degrees of freedom by a set of transformations. These transformations shall be based on the equilibrium and compatibility requirements. Let the P-system of forces correspond the RDOF set whose flexibility matrix (force-displacement relationship) has been determined, i.e [FPP] in the equation [UP] = [FPP] [P] is known; and let the Q-system of forces correspond to the RDOF set whose flexibility matrix is required, i.e [FQQ] in the equation [UQ] = [FQQ] [Q] is required The equilibrium requirement may be written as a force transformation relationship, using the Q-force support system, in the form

[P] = [HPQ] [Q] (4-36) To understand the meaning of this transformation between forces referring to two different support conditions we should recognize that the [P] and [Q]-force vectors are simply subsets of the same set of absolute degrees of freedom for the member. For example, for the plane frame member with the ADOF shown in Figure 4-13 define the force vector as [A]T = [A1 A2 A3 A4 A5 A6], the force vectors [P] and [Q] are defined as [P]T = [A4 A5 A6] and [Q]T = [A4 A3 A6].

Figure 4-13 Absolute DOFs for Plane Frame Member For the compatibility relationship, the transformation may be determined by considering the geometry of the member when nodal displacements are imposed. However, given the force transformation relationship given in equation (4-36) we have from the contragradient law

[UQ] = [HPQ]T [UP] (4-37)

Combining equations (4-37) with (4-34) and (4-36)

[UQ] = [HPQ]T [UP] = [HPQ]T [FPP] [P]

M M M M M M

x

1 2

3 4

5 6

ABJ Jr Draft 2011


[UQ] = [HPQ]T [FPP] [HPQ] [Q] (4-38)

or [FQQ] = [HPQ]T [FPP] [HPQ] (4-39)

Example 4-3 Determine Simply Supported [Fm] Given the Fixed-Free Flexibility

Matrix for a Plane Frame Member

P-System, Fixed-Free Q- System Simply Supported

Solution: [FQQ] = [HPQ]T [FPP] [HPQ]

where

+=

zz

zzPP

EIL

EIL

EIL

EIL

AEL

F

20

2310

00

2

23

)(][

For equilibrium, [P] = [HPQ] [Q], the transformation matrix may be generated by column using the unit load method noting that the P-forces are the forces at the B-end of the member. Thus

=

100110001

LLHPQ //][

Substitution and performing the triple product gives

+

+=

zz

zzQQ

EIL

EIL

EIL

EIL

AEL

F

31

6210

621

310

00

)()(

)()(][

Notes on compatibility matrix: We have used contragradience to write the compatibility relationship.

[UQ] = [HQP] [UP] = [HPQ]T [UP]

Alternatively, we can generate the [HQP] using compatibility requirements by applying unit displacements along the P-degrees of freedom and determining the displacements corresponding to the Q-degrees of freedom (in this example the horizontal

1 2

3 1 2 3

Q2 = 1

1/L P2 = -1/L P1 = P3 = 0

Q1 = 1 Q3 = 1

P1 = Q1 = 1 P2 = P3 = 0 P3 = Q3 = 1

P1 = 0 P2 = -1/L



displacement at the B-end and the rotations at the ends with respect to the chord joining the ends of the member).

a) 1st Column, UP1 = 1 b) 2nd Column, UP2 = 1 c) 3rd Column, UP3 = 1

Compatibility Relation between [UQ] and [UP] From the figures

=

110010001

LLH PQ

//][

Which is the transpose of [HQP]. 4.8 FLEXIBILITY MATRIX OF ASSEMBLAGES

Having developed the flexibility matrix for a single member, we are now in the position of generating the flexibility matrix of a system composed of an assemblage of members. The simplest application of this is the determination of the flexibility matrix for a nonprismatic member composed of elements whose flexibility matrices are known particularly prismatic elements for which closed-form solutions have already been developed. The procedure that is to be developed in more general and discussions in its application for plane frame and plane truss members are given below. General Formulation Consider an system consisting of assemblage of m number of elements. It is required to determine the system force-displacement relationship in the form [U] = [F] [P], where [U] and [P] are referred to the system nodal degrees of freedom. It is assumed that the flexibility matrices of these are given or known. The force-displacement relationships for the problem are therefore the set of flexibility equations for the component elements.

[Ui] = [Fi] [Pi] with i = 1, m (4-40)

Where [Fi] is the flexibility matrix for the ith element, and (4-40) indicates that there are m such equations. Necessarily, the members must be numbered for identification. The form of the matrix to be used must also be selected, since it depends on the choice of the set of relative degrees of freedom for the element. For the plane frame member, for example, the fixed-free or simply supported flexibility may be used. Selection of the flexibility matrix also defines the element displacement, [Ui], and force vectors. [Pi]. When forces are applied along the system degrees of freedom, element forces are generally induced at the elements. The relation between the applied system forces and resulting element forces can be

U = 1 P1 = UQ1

UQ2 = 0 U 3 = 0

UP3 = 1 UQ3 = 1

QUQ1 = 0

UQ2 = 0

UP2 = 1 UQ1 = 0

UQ2 = -1/L

UQ3 = -1/L

ABJ Jr Draft 2011


determined from equilibrium requirements; and can be expressed as a force transformation in the form

[Pi] = [Ei] [ P ] with i = 1, m (4-41) The compatibility relationships can again be determined using contragradience given equation (4-41) as

[U]i = [Ei]T [Ui] with i = 1, m (4-42)

The delta operator is applied to recognize that this is the contribution to the system nodal displacement of element i. Combining equations (4-42), (4-41) and (4-40) gives

[U]i = [Ei]T [Fi] [Ei] [ P ] with i = 1, m (4-43) To get the total system nodal displacement, we have to sum the contributions for all the elements, such that

[ ] [ ] [ ] [ ]PEFEUUm

iii

Ti

m

ii

==

== 11

][][ (4-44)

Therefore (4-45) [ ] [ ]=

=m

iii

Ti EFEF

1

][][

An alternative formulation involving combining the m-equations each for the force-displacement, equilibrium and compatibility relations into single matrix operations is possible. The force-displacement relationship can be written, by combing the element forces into one vector, and doing the same for the displacement and flexibility matrices, results in

[Ue] = [Fe] [Pe] (4-46) where

=

m

i

m

i

m

i

P

P

PP

F

F

FF

U

U

UU

M

M

O

O

M

M2

1

2

1

2

1

(4-47)

Similarly, the equilibrium relationships may be combined and expressed as [Pe] = [Ee] [ P ] (4-48)

where [Ee]T = [E1T E2 T Ei T Em T] (4-49)

The compatibility relationship can be expressed, using contragradience as

[U] = [Ee]T [Ue] (4-50)

combining [U] = ([Ee]T [Fe] [Ee]) [P] (4-51)

therefore [F] = [Ee]T [Fe] [Ee] (4-52) Performing the triple product of equation (4-52) will verify that this is identical to equation (4-45). The following examples illustrate the application of these equations to different types of structures.



Straight Chain Members Chain members are those composed of elements connected end on end. The simplest example is a nonprismatic member made up of prismatic elements as shown in Figure 4-14. The member flexibility matrix maybe calculated as in equation 4-45 and referred to as chain flexibility formulation.

Figure 4-14 Examples of Straight Chain Member Another application of chain flexibility is in the determination of the member flexibility of a member with continuously varying section properties such as tapered member in Figure 4-15a. A closed form solution using slice flexibilities may be involved due to the increased number of variables. Numerical integration may be resorted to for a specific sets of dimensions. Alternatively, the [Fm] can be approximated by the flexibility of a member consisting of prismatic elements such as in Figure 4-15b.

a) Tapered Member b) Member for Approximation

Figure 4-15 Approximation for Tapered Member The approximation will be closer to the actual value as the number of prismatic elements is increased but the amount of computations also increases. Example 4-4 Determine the member flexibility matrix for the non-prismatic member

using chain flexibility and fixed-free element flexibilities. Neglect shear deformations.

Solution: [ ] [ ]=

=3

1iii

Tim EFEF ][][

Using Fixed-Free element flexibility matrix

iEIL

EIL

EIL

EIL

AEL

F

zz

zzi

=

20

230

00

2

23

][

=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

322560

25615360

008

2

23

1][

1 2

3 2 1 3

1 2

3

L/2L/4 L/42AE, 8EI

1.5AE, 4EI AE, EI A B C D

1 2

3 1 2

3

ABJ Jr Draft 2011


=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

161280

1287680

006

2

23

2 ][

=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

280

8240

002

2

23

3 ][

The [Ei] matrices are determined from equilibrium internal stress resultants at point C for element 1, at point D for element 2 and at B for element 3.

==

1430010001

1

/][][

LHE CB

==

120010001

2

/][][

LHE DB

[E3] = [HBB] = [ I ] Substituting and performing the indicated matrix operations

=

zz

zzm

EIL

EIL

EIL

EIL

AEL

F

3219

25649

0

25649

1536139

0

002419

2

23

][

Example 4-5 Same problem as Example 4-4 except using simply supported element

flexibilities.

Solution: [ ] [ ]=

=3

1iii

Tim EFEF ][][

Using simply supported element flexibility matrix

iEIL

EIL

EIL

EIL

AEL

F

zz

zzi

=

360

630

00

][

=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

961920

192960

008

1][

=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

481920

192480

006

2 ][

=

zz

zz

EIL

EIL

EIL

EIL

AEL

F

6120

1260

002

3 ][

The [Ei] matrices are determined from equilibrium axial and end moments for each element due to [Pm].



=1430110

001

1

/][E

=12101430

001

2

//][E

=1001210

001

3 /][E

It can be verified that the same results are obtained as in the previous problem. Plane Frame Structures In the flexibility or force method of structural analysis of indeterminate structures, redundant forces and/or reactions are released to produce stable and determinate primary or cut-back structures. Force-displacement relations, flexibility matrices, are determined for the primary structures. Consider the plane frame in Figure 4-16a which is indeterminate to the 18th degree, one possible set of primary structures are shown in Figures 4-16b. Primary structures PS1 and PS3 are chain structures. Consider the primary structure PS2. This is referred to as a tree-structure if you imagine the beams branching out of the columns. The structure flexibility matrix for this structure can be determined as in equation 4-45 or 4-52 (the summation formulation shall be used below as this is more instructive). Note that the relative degrees of freedom for the structure are defined as those at the nodes of the structure shown defined in global coordinates.

a) Indeterminate Plane Frame

A B C

D E F

G H I 6 93 85 2 74 1 PS1 PS2 PS3

b) Primary Structures Figure 4-16 Primary Structures of Indeterminate Plane Frame

Instead of numbering the individual degrees of freedom, the joints have been labeled (in actual application it will usually be numbered rather than be assigned letters). The force and displacement vectors of the structure can

ABJ Jr Draft 2011


therefore be partitioned by joint and [F] can also be correspondingly partitioned as below.

[U] = [ F ] [ P ] (4-53)

where

=

N

J

B

A

NNNJNA

INIJIA

BBBA

ANAJABAA

N

I

B

A

P

P

PP

FFF

FFF

FFFFFF

U

U

UU

M

M

LL

MOMM

LLL

MMOM

MM

LL

M

M

The typical submatrix element [FIJ] is defined similarly as the coefficient fij, in that it represents the displacements along the degrees of freedom along joint Ì due to unit loads applied along the degrees of freedom in joint `J with all other joint loads equal to zero. Note that the units loads are still separately applied resulting in a submatrix rather than a single coefficient. Rather than determine the flexibility matrix for the whole structure, this notes will concentrate in the determination of portions of [F] such as a typical submatrix [FIJ] such that [UI] = [FIJ] [PJ]. It can be verified that the this can be determined using

[ ] [ ]=

=m

iiJi

TiIIJ EFEF

1

][][ (4-54)

where [Pi] = [EiJ] [ PJ ] (4-55)

[UI]i = [EiI]T [Ui] (4-56) such that [Pi] = [EiI] [ PI ] (4-55) One major difference between tree-structures and chain-structures is that when loads are applied at joint `J of a tree-structure internal member forces are not necessarily generated in all members of the structure i.e. their corresponding [EiJ]s = []. Only members in the load-path from point of application of the load to the supports generate internal forces. Only the loaded members deform, which may be determined from the member flexibilities. Note that not all members that deform will contribute to the displacement of joint Ì i.e. their corresponding [EiJ]s = []. For example, using the structure of Figure 4-17 and let joint Ì = `F and `J= `G such that we require the relationship [UF] = [FFG] [PG]

A B C

D E F

G H I 6 3 9

P

U

85 2

74 1

Figure 4-17 Example with [PG] and [UF]



When [PG] is applied, internal forces will be generated only on members 1, 2, 3 and 6 (load path is GHEB). Of these loaded members, only members 1 and 2 will contribute to the displacement of joint `F. To understand effect of member deformation to joint displacement, note that joint `F will only displace due to the deformation of member 8 and/or displacement of joint È. For this case, no member forces exist on member 8, and joint È will displace only due to the deformation of members 1 and 2. Figure 4-18 illustrates the deformation of the structure due to the deformation (due to flexure) of members 2 and 3 (deformations are highly exaggerated actual displacements are still assumed to be small). Note, when considering the effect of the deformation a particular member, all other members are assumed to be rigid. Thus for when considering member 3, joint È will be restrained from displacing due to the rigidity of members 1 and 2 which are connected to a fixed support. Joint `H is free to displace due to the deformation of member 3 because there are no support restraint above the member.

A B C

D E F

G H I

A B C

D E F

G H I

a) Due to Member 2 b) Due to Member 3

Figure 4-18 Displacement of Structure Due to Deformation of a Member Note, that [ F ] is symmetric such that [FIJ] = [FJI]T and the diagonal submatrices [FII] is symmetric. Generally, however, the typical submatrix [FIJ] (with I J) is not symmetric. Example 4-6 For the plane frame with prismatic members shown below has columns

with AE = 2,500 kN and EI = 25,000 kN-m2; and beams with AE = 3,000 kN and EI = 65,000 kN-m2, determine the matrix a) [FFG] relating the displacement at `F due to the forces `G. b) [FGG] relating the displacement at `G due to the forces at`G.

A B C

D E F

G H I 6 3 9

P

U

8

7

5

4

2

1

3 m

3 m

3 m

7 m 7 m

ABJ Jr Draft 2011


Solution: a) From discussion in notes, the displacement of joint `F is due to the deformation of

members 1 and 2 only.

[UF] = [UF1] + [UF2]

= [E1F]T[U1] + [E2F]T[U2]

= [E1F]T[F1][P1] + [E2F]T[F2] [P2]

= [E1F]T[F1][E1G][PG] + [E2F]T[F2][E2G][PG]

thus [FFG] = [E1F]T[F1][E1G] + [E2F]T[F2][E2G] Using the fixed-free member flexibilities:

iEIL

EIL

EIL

EIL

AEL

F

zz

zzi

=

20

230

00

2

23

][

A

B

This is with respect to the local axis of the members which must be defined for the problem. It will be assumed that A-ends of the members will be the end closer to the fixed support. [F1] = [F2] = [Fcol] below. [Fbm] is required in problem b).

3101201800180360000201

= xFcol

....

.][ 310

108037700377075910003332

= xFbm

....

.][

The equilibrium matrices can be generated by columns using the unit load method to get the B-end forces for members 1 and 2 due to forces at `F and `G as

=

173001010

1 ][ FE

=

176001010

1 ][ GE

=

170001010

2 ][ FE

=

173001010

2 ][ GE

thus [FFG] = [E1F]T[U1][E1G] + [E2F]T[U2][E2G]

3102400680144016801360908010720004050405

= xFFG

...

...

...][

Note, it can be verified that the same result will be obtained if we have used the simply supported member flexibilities and the corresponding equilibrium matrices. For the simply supported flexibilities, the equilibrium matrices relates the axial load and end



moments of the member to the structure nodal forces (rather than the end forces at the B-end for the fixed-free case). b) When loads are applied at `G members 1, 2, 3, and 6 will develop internal forces

and when they deform will contribute to the displacement of `G. Note that this degenerates into a chain-flexibility problem. Thus, following the same reasoning as in a)

[UG] = [UG1] + [UG2] + [UG3] + [UG6] and [FGG] = [E1G]T[F1][E1G] + [E2G]T[F2][E2G] + [E3G]T[F3][E3G] + [E6G]T[F6][E6G]

Note that for member 6, using the orientation of the members defined in a), will have its a-end at joint `H and the local axis will be oriented as shown below (selecting `G as the A-end will give the same result provided the proper corresponding equilibrium matrix is used). The equilibrium matrices are generated as

AB

=

176001010

1 ][ GE

=

173001010

2 ][ GE

=

170001010

3 ][ GE

=

100010001

4 ][ GE

and [FGG] = [E1G]T[F1][E1G] + [E2G]T[F2][E2G] + [E3G]T[F3][E3G] + [E6G]T[F6][E6G]

3104680897262018972999223401162013401105312

= xFGG.........

][

The following example illustrates the determination of the force-displacement relationship for frames other than a tree-structure; where the nodal forces and displacements are of different number and do not include all the components at the joint; and the use of the simply supported member flexibilities. Example 4-7 For the portal frame shown, determine the force-displacement

relationship for the nodal forces, P, and displacements, U, shown. Section properties are the same as that for problem 4-6.

7 m

U3 U2 U1 P1 P2

1 2 3

7 m

3 m 3

1 2

ABJ Jr Draft 2011


Solution: The solution may be expressed in a similar form as before as

[ ] [ ]=

=3

1iPii

TiU EFEF ][][

where [EiP] relates the member forces to the forces along P, and [EiU] relates the member forces to the forces along the displacement U.

Using simply supported member flexibilities with member incidences indicated in figures:

iEIL

EIL

EIL

EIL

AEL

F

zz

zzi

=

360

630

00

][

31004002000200400

00201

= xFcol

..

...

][ 310036001800018003600

003332

= xFbm

..

...

][

[EiP] are determined from equilibrium requirements such that [Pi] = [EiP] [P]

=

0300

7173

1

//][ PE

=

00007173

2

//][ PE

=

100300

3 ][ PE

[EiU] are determined from equilibrium requirements such that [Pi] = [EiU] [PU], where [PU] are the forces along the [U] displacements.

=

003000

717173

1

///][ UE

=

000000

717173

2

///][ UE

=

100

013

001

3 ][ UE

Thus, [ ] [ ]=

=3

1iPii

TiU EFEF ][][

610

8778409393

03131631254

093938931231

= xF

..

..

..,

][

Plane Truss The final example is for a planar truss. This shall be solved using the matrix formulation in equation 4-52 rather than the summation formulation of equation 4-45.



Example 4-8 Determine the vertical displacements at the bottom chord due to forces

applied at the top chord of the planar truss as shown on the figure. AE = 300,000 kN for the top and bottom chord members, and AE = 200,000 kN for all other members..

Solution:

[F] = [EeU]T [Fe] [EeP]

Where [Fi] = [L/AE]i, with L = 3.0 m for members 1 to 4, L = 3.1623 m for members 5 to 10, L = 1.0 m for members 11 and 12, and L = 2.0 m for member 13.

610

205

58115

81155410

54105410

541010

1010

10

= xeF

..

..

..

][

=

50010100000058110000003722581179105811791058117910581125251252517505175051

..

..

..

..

..

..

..

..

..

..

..

..

..

][ eUE

=

00500000000000000058115811791058117910581179105811372251750517505125251252

..

..

..

..

..

..

..

..

..

..

..

..

..

][ ePE

The first column of [EeU] represents the member forces due to a positive unit load along U1 and the second column represents those due to a unit load along U2 (positive in the direction shown in the figure). The columns of [EeP] are the member forces due to unit

P1 P2

U1 U2 4 @ 3 m = 12 m

2 m 5 6 7

8 10139 121121 43

ABJ Jr Draft 2011



loads along the P-forces (positive in the direction shown in the figure). The elements of the equilibrium matrices are positive for a tensile member axial force. Thus [F] = [EeU]T [Fe] [EeP]

610057174205125409195057179

= xF ..

..][ m/kN

Notes. A closer examination of the calculation show that the flexibility coefficient, fij, can be calculated as (FPFQL/AE)k where FP is the axial load on member k due to a unit load along the nodal P-force corresponding to j, and FQ is the axial load on member k due to a unit load along the displacement Ui. This form is identical to that used in the method of virtual work for calculating displacements. The matrix formulation allows the determination of several displacement components in one matrix equation.

04 flexibility - juinio

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