Indeterminate AnalysisForce Method1

• The force (flexibility) method• The force (flexibility) method expresses the relationships between displacements and pforces that exist in a structure.

• Primary objective of the force method is to determine the chosen set of excess unknown f d/ lforces and/or couples –redundants.

• The number of redundants is• The number of redundants is equal to the degree of static indeterminacy of the structure.

1

y1Also see pages 115 – 141 in your class notes.

Description of the Force Method ProcedureMethod Procedure

1. Determine the degree of static gindeterminacy. Number of releases* equal to the degree of static indeterminacy are applied to the structure.Released structure is referred to as the primary structure. P i b hPrimary structure must be chosen such that it is geometrically stable and statically determinateand statically determinate.Redundant forces should be carefully chosen so that the primary

2

carefully chosen so that the primary structure is easy to analyze

* Details on releases are given later in these notes.

Force Method – con’t

2. Calculate “errors” (displacements) at th i t t d d tthe primary structure redundants. These displacements are calculated using the method of virtual forces.g

3. Determine displacements in the primary structure due to unit values of redundants (method of virtual forces) These displacements areforces). These displacements are required at the same location and in the same direction as the di l t d t i d idisplacement errors determined in step 2.

3

Force Method – con’tForce Method con t

4 Calculate redundant forces to4. Calculate redundant forces to eliminate displacement errors. Use superposition equations inUse superposition equations in which the effects of the separate redundants are added to the displacements of the released structure. Displacement superposition results in a set of n linear equations (n = number of releases) that express thenumber of releases) that express the fact that there is zero relative displacement at each release.

4

p

Force Method – con’tThese compatibility equations guarantee a final displaced shapeguarantee a final displaced shape consistent with known support conditions, i.e., the structure fits co d o s, .e., e s uc u e stogether at the n releases with no relative displacements.

5. Hence, we find the forces on the original indeterminate structure. They are the sum of the correction forces (redundants) and forces on theforces (redundants) and forces on the released structure.

5

Flexibility Analysis

(1)

R1R2

=

(2)

+

D1D2

(2)

f21 (x R1)f11 (x R1)

(3)

1 (R1)

+f (x R )

6

1 (R2)f12 (x R2)

f22 (x R2)(3)

f R f R D 0+ + =11 1 12 2 1(4

f R f R D 0)

+ + =

21 1 22 2 2

(4f R f 0)

R D+ + =

Solve for R and RSolve for R1 and R2.

Using matrix methods:g

[F] {R} = -{D}

⇒ {R} = -[F]-1 {D}

7

f f⎡ ⎤[F] = 11 12

21 22

f ff f

⎡ ⎤⎢ ⎥⎣ ⎦

≡ flexibilitymatrix

[F]-1 (≡ inverse flexibility matrix)[F] (≡ inverse flexibility matrix)

22 12f f1 −⎡ ⎤22 12

21 1111 22 12 21

f f1f ff f f f

⎡ ⎤= ⎢ ⎥−− ⎣ ⎦

1D{D}

⎧ ⎫= ⎨ ⎬

≡ primary2

{D}D⎨ ⎬

⎩ ⎭ structuredisplacement

t8

vector

1R⎧ ⎫≡ redundant

f1

2

R{R}

R⎧ ⎫

= ⎨ ⎬⎩ ⎭

forcevector

1 22 1 12 2R f D f D1R f D f Ddet[F]

−⎧ ⎫ ⎧ ⎫−=⎨ ⎬ ⎨ ⎬+⎩ ⎭ ⎩ ⎭

det [F] =

2 21 1 11 2R f D f Ddet[F] − +⎩ ⎭ ⎩ ⎭

11 22 12 21f f f f−det [F]

(5)

11 22 12 21f f f f

(5)

With R1 and R2 known, remaining structure is

9

remaining structure is statically determinate.

ReleasesReleases

Release is a break in the continuity ofRelease is a break in the continuity of the elastic (displacement) curve.

One release only breaks a single type of continuity.y

Figure 1 shows several types ofFigure 1 shows several types of releases.

Common release is the support reaction, particularly for continuous

10beams.

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Flexibility Equations y qPrimary structure displacements at th l l t d t ththe releases are related to the unknown redundant forces via

i ij jD f R− = (1)

displacement at release i due i jf ≡to a unit force in the direction of and at release j; flexibility coefficients

j

coefficients.

Equation 1 for the case of three redundant forces is expressed as

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redundant forces is expressed as

1 11 1 12 2 13 3D f R f R f R− = + +1 11 1 12 2 13 3

2 21 1 22 2 23 3

D f R f R f RD f R f R f R

= + +

− = + + (2a)

3 31 1 32 2 33 3D f R f R f R− = + +

Matrix form of (2a)

-{D} = [F] {R} (2b)

1D⎧ ⎫{D} = = <D1 D2 D3>T

1

2

DD

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪

= displacement vector at the3D⎪ ⎪

⎩ ⎭

13

predundant degrees of freedom

1R⎧ ⎫⎪ ⎪

{R} = = <R1 R2 R3>T1

2RR

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

= redundant force 3R⎪ ⎪

⎩ ⎭

vector

11 12 13f f f⎡ ⎤⎢ ⎥

21 22 23f f ff f f

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

[F] =

31 32 33f f f⎢ ⎥⎣ ⎦= flexibility matrix

14

y

Displacement Calculations –Displacement Calculations Method of Virtual Forces

D F d M dφli Vi ViD F d M d= + φl

subscript i direction of Ri at⇒(3)

release i

dl = differential axial displacementp

dφ = differential rotational displ

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Flexibility Coefficients –Method of Virtual ForcesMethod of Virtual Forces

a bij ij ijf f f= + (4)ij ij ijf f f+ ( )

VjFVjaij Vi

Ff F dx

EA(x)=

≡ axial deformation influence coefficient

Vjbij Vi

Mf M dx=ij Vif M dx

EI(x)≡ bending deformation

16

≡ bending deformationinfluence coefficient

Force Method Examples

1 Calculate the support reactions1. Calculate the support reactions for the two-span continuous beam, EI = constant. w

L L

w

=

Primary Structure w/ Load

+

171 (x R1)Primary Structure w/ Redundant

2. Calculate the support reactions ppfor the two-span continuous beam, EI = constant.

w

L L

w

=

Primary Structure w/ Load

+

R2

+

R1

18

R2R1

Primary Structure w/ Redundant Forces

Prismatic Member Displacements

19

20

21

3. Calculate the support reactions 3 Ca cu ate t e suppo t eact o sfor the two-span continuous beam using the internal moment at B as the redundant force, IAB = 2I and IBC = I; E = constant.

PL2

Primary Structure w/ Loading22

Primary Structure w/ Loading

MB

i S / d dPrimary Structure w/ Redundant

DB = __________________

fBB = BB _________________

MB = _________________

23

4. Calculate the bar forces for th t ti ll i d t i tthe statically indeterminate truss.

St ti ll StaticallyStatically Indeterminate

Truss

Statically Determinate

Released Truss24

Truss(Redundant X)

Truss CalculationsMem L (in) F FV FVFL

192" 40 4/5 6144AB 192 40 -4/5 -6144

BC 144" 0 -3/5 0

CD 192" 0 -4/5 0

DA 144" 30 3/5 2592DA 144" 30 -3/5 -2592

AC 240" 0 1 0

BD 240" -50 1 -12000

VF FLF /f 20736/82944⎡ ⎤⎢ ⎥∑

25

VAC AC,ACF /f 20736/829.44

E A= − =⎢ ⎥

⎣ ⎦∑= 25 kips

Nonmechanical LoadingLoading

[F]{R} ({D} {D })Δ=− + (5)[F]{R} ({D} {D })= + ( )

T2 n1{D } D D DΔΔ Δ Δ= < >K2 n1{ } K

= relative dimensional change displacements calculated usingdisplacements calculated using principle of virtual forces

Displacements due to dimension changes are all relative displace-

t ll di l tments, as are all displacements corresponding to releases. They are positive when they are in the same

26

positive when they are in the same vector direction as the corres-ponding release (redundant).

Structure ForcesStructure Forces

Once the redundant forces arel l d f ( ) ll hcalculated from Eq. (5), all other

support reactions and internalmember forces can be calculatedusing static equilibrium along withthe appropriate free body diagrams.

Thi i ibl i h fThis is possible since the forcemethod of analysis has been used todetermine the redundant forces orthe forces in excess of those

27required for static determinacy.

Mathematical E iExpressions

Calculation of the non-redundant f A ( t ti i t lforces Ai (support reactions, internal shears and moments, truss member forces) can be expressed usingforces) can be expressed using superposition as

RN

p

Rpi ui j ji

j 1A A (A ) R

== + ∑ (6)

piA

j 1where = desired action Ai on the primary structure due to the applied loading; = action Ai on the primary structure due to

ui j(A )

28a unit virtual force at redundant Rj and NR = number of redundants.

Example Beam Problem –Nonmechanical Loading

E 30 000 k iE = 30,000 ksiI = 288 in4

(a) Given structure

29(b) Primary structure

The interesting point of this example is that the flexibility equation will have a nonzero right hand side since the

d d t di l t i ib dredundant displacement is prescribed to equal 0.72” downward. Thus the flexibility equation isflexibility equation is

fBB RB = dB - (7)BDΔ

wheredB = prescribed displacementB p p

at redundant B= -0.72" since RB is

positive upward= -0.24"BDΔ

30relative displacementat redundant BB Bd DΔ− =

Truss Example –pNonmechanical Loading

For the truss structure on the next page compute thenext page, compute the redundant bar EC member force if the temperature in bar EF isif the temperature in bar EF is increased 50 oF and member BF is fabricated 0.3 in. too short. EA = constant = 60,000 kips and

= 6x10-6 /oF.α

31

Truss Example3 @ 20’ = 60’

CB

3 @ 20 60

DA 15’

E F

B C

A 1 D

E F

Primary Structure SubjectedPrimary Structure Subjectedto FCE = 1

32

Truss Example CalculationsMem L FV FVL

AB 240" 0 0AB 240" 0 0

AE 300" 0 0

BC 240" -4/5 -192

BE 180" -3/5 -108

BF 300" 1 300

CD 240" 0 0

CE 300" 1 300

CF 180" -3/5 -108CF 180 -3/5 -108

DF 300" 0 0

EF 240" 4/5 19233

EF 240" -4/5 -192

mCE CE Vi Vi i

1f F F L= ∑CE,CE Vi Vi ii 1

f F F LEA =

= ∑

mCE Vi iD FΔ Δ= δ∑

i 1=

T L 0 072"Δδ αΔEF EF EF

BF BF

T L 0.072"

0 3"Δ

δ = αΔ =

δ = Δ = −

Δ

BF BF 0.3δ = Δ = −

CE,CE CE CEf F D 0Δ+ =

34

Displacement C l l ti

Displacements for the statically

Calculations

Displacements for the staticallyindeterminate structure can be

l l t d i th t bcalculated using the exact memberdeformations for a truss or exactshear and moment expressionsalong with the virtual force expres-sions on the primary structure.

F t t t l l tiFor a truss structure, calculation of a joint displacement using theprinciple of virtual forces results in

Δ

35

principle of virtual forces results in

mm

Vi iF Δδ + Δ∑1 (Δ) =

i 1=∑

minti i

Vi iF LFEA

Δ⎛ ⎞+ δ + Δ⎜ ⎟

⎝ ⎠∑= (8)Vi ii

i 1EA

=

⎜ ⎟⎝ ⎠∑

ViF = primary structure member forces due to the application of ppa unit virtual force at the joint for which the displacement i d i d d i h di i

Δ

36

is desired and in the direction of Δ

ΔΔ = primary structure displaceΔ = primary structure displace-ment at desired displacement due to nonmechanical effects

iδ = exact member displacements i that are obtained for the sta-tically indeterminate structure

i th l l t d d d tusing the calculated redundant forces to determine all the member forces within the trussmember forces within the truss structure

intδintiδ = member displacements due

to nonmechanical loading on

37the member

For a frame structure, in which shear and axial deformations are ignored, the displacements are calculated asp

1( )ΔLm

iiMΔ ⎛ ⎞∑ ∫1( )Δ = intiVi i

ii 1 0

MM dxEI

Δ

=

⎛ ⎞+ κ⎜ ⎟

⎝ ⎠∑ ∫i 1 0=

(9a)Δ+ Δ

1( )θ =

Lminti

Vi iMM dxθ ⎛ ⎞

+ κ⎜ ⎟∑ ∫1( )θ = Vi ii

i 1 0

M dxEI

=

+ κ⎜ ⎟⎝ ⎠∑ ∫

Δ38(9b)Δ+ θ

M MΔ θVi ViM , MΔ θ = primary structure virtual

moments based on the desired displacement Δ ordesired displacement Δ or rotation θ

,Δ ΔΔ θ = primary structure displace-ments at Δ or rotation θ duements at Δ or rotation θ due to environmental loads or causes

intiκ = primary structure initialiκ primary structure initial

curvature strain caused by nonmechanical loading

39

In Eqs. (9a) and (9b) the moment expressions are exact based on the statically indeterminate structure ysubjected to the external loads with the redundant forces known fromthe redundant forces known from the flexibility analysis.

Equations (8), (9a), and (9b) are cor-rect only because exact real memberrect only because exact real member forces are used in the calculation of the desired displacementsthe desired displacements.

40

Calculate the horizontal displace-ment at joint B for the staticallyment at joint B for the statically indeterminate truss.

16’40 k C

D 16

A BR1

12’

A

CD

BA 1

i S S bj d41

Primary Structure Subjected to Virtual Loading

Calculate the rotation at the center support for the two-span continuous beam, EI = constant.

w

L LR

1

R1 R2

P i St t / Vi t l

L L

Primary Structure w/ Virtual Load at Desired Displacement

Location

42

Alternatively, you can express the desired displacementthe desired displacement calculations also in matrix form following the usual superpositionfollowing the usual superposition process of the force method of analysis:analysis:

(10){ } [F ]{R} {D } {D }Δδ δ δδ = + +

where {δ} = vector of desired displacements; ={D }, {D }Δ

δ δdisplacements; vector of desired displacements for the primary structure for both

{D }, {D }δ δ

p ymechanical and non-mechanical loadings, respectively; [Fδ] =

43

g p y [ δ]

matrix of displacement influence coefficients at the desiredcoefficients at the desired displacement locations due to unit values of the redundant forcesvalues of the redundant forces {R}. Stated mathematically, the coefficients of [Fδ] arecoefficients of [Fδ] are

(11)jij i R 1Fδ == δ (11)

which simply states that the dis-

jj R 1

p yplacement influence coefficients equal the displacement at desired q pdisplacement i on the primary structure due to a unit force at

44redundant j on the primary structure.