indeterminate analysis force method1 -...

1

Indeterminate AnalysisForce Method1

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

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

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

1Also see pages 10 – 43 in your class notes.

2

Description of the Force Method Procedure

1. Determine the degree of static indeterminacy.

Number of releases* equal to the degree of static indeterminacy are applied to the structure.

Released structure is referred to as the primary structure.

Primary structure must be chosen such that it is geometrically stable and statically determinate.

Redundant forces should be carefully chosen so that the primary structure is easy to analyze

* Details on releases are given later in these notes.

3

Force Method – con’t

2. Calculate “errors” (displacements) at the primary structure redundants. These displacements are calculated using the method of virtual forces.

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

4

Force Method – con’t

4. Calculate redundant forces to eliminate displacement errors.

Use 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 the fact that there is zero relative displacement at each release.

5

Force Method – con’t

These compatibility equations guarantee a final displaced shape consistent with known support conditions, i.e., the structure fits together 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 the released structure.

6

Flexibility Analysis

R1R2

(1)

=

1 (R1)

1 (R2)

+

+

f21 (x R1)f11 (x R1)

f12 (x R2)

f22 (x R2)

D1D2

(2)

(3)

(3)

7

11 1 12 2 1

21 1 22 2 2

(4

f R f R D 0

f R f 0

)

R D

Solve for R1 and R2.

Using matrix methods:

[F] {R} = -{D}

{R} = -[F]-1 {D}

8

[F] = 11 12

21 22

f f

f f

flexibilitymatrix

[F]-1 ( inverse flexibility matrix)

22 12

21 1111 22 12 21

f f1f ff f f f

1

2

D{D}

D

primarystructuredisplacementvector

9

det [F] =

(5)

With R1 and R2 known, remaining structure is statically determinate.

1

2

R{R}

R

redundantforcevector

1 22 1 12 2

2 21 1 11 2

R f D f D1R f D f Ddet[F]

11 22 12 21f f f f

10

Releases

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

One release only breaks a single type of continuity.

Figure 1 shows several types of releases.

Common release is the support reaction, particularly for continuous beams.

11

12

Flexibility Equations

Primary structure displacements at the releases are related to the unknown redundant forces via

i ij jD f R (1)

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

i jf

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

13

1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

D f R f R f R

D f R f R f R

D f R f R f R

(2a)

Matrix form of (2a)

-{D} = [F] {R}

{D} = = <D1 D2 D3>T

= displacement vector at theredundant degrees of freedom

(2b)

1

2

3

D

D

D

14

{R} = = <R1 R2 R3>T

= redundant force vector

11 12 13

21 22 23

31 32 33

f f f

f f f

f f f

[F] =

= flexibility matrix

1

2

3

R

R

R

Displacement Calculations – Method of Virtual Forces

i i i iD p d m d v= + φ +∫ ∫ ∫ dy (3)

subscript i ⇒ direction of at release i iR

d = differential axial displacement

dφ = differential rotation displacement

dy = differential shear displacement

Flexibility Coefficients – Method of Virtual Forces

a b sij ij ij ijf f f f (4)

ja

ij ip

f p dxEA(x)

axial influence coefficient

jb

ij im

f m dxEI(x)

bending influence coefficient

js

ij is

vf v dx

GA (x)

shear flexibility influence coefficient

Nonmechanical Loading [F]{R} ({D} {D })∆= − + (5)

T1 2 n{D } D D D∆ ∆ ∆ ∆= < >… = relative dimen-

sional change displacements, calculated us-

ing principle of virtual forces

Displacements due to dimension changes are

all relative displacements, as are all displace-

ments corresponding to releases. They are

positive when they are in the same vector direc-

tion as the corresponding release.

Structure Forces

Once the redundant forces are calculated from

Eq. (5), all other support reactions and internal

member forces can be calculated using static

equilibrium along with the appropriate free

body diagrams. This is possible since the force

method of analysis has been used to determine

the redundant forces or the forces in excess of

those required for static determinacy.

28

Mathematical Expressions

piA

Calculation of the non-redundant forces Ai (support reactions, internal shears and moments, truss member forces) can be expressed using superposition as

RNp

i ui j jij 1

A A (A ) R

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

Ai on the primary structure due to a unit virtual force at redundant

Rj and NR = number of redundants.

ui j(A )

Displacement Calculations

Displacements for the statically indeterminate

structure can be calculated using the exact

member deformations for a truss or exact shear

and moment expressions along with the virtual

force expressions on the primary structure.

For a truss structure, calculation of a joint dis-

placement ∆ using the principle of virtual

forces results in

1 (∆) =

m

i i

i 1

p ∆

=

δ + δ∑

=

mi i

ii

i 1

F LpEA

∆

=

+ δ∑ (6)

ip = primary structure member forces due to

the application of a unit virtual force at the

joint for which the displacement ∆ is desired

and in the direction of ∆

∆δ = primary structure displacement at de-

sired displacement ∆

iδ = exact member displacements that are ob-

tained for the statically indeterminate struc-

ture using the calculated redundant forces to

determine all the member forces within the

truss structure

For a frame structure, in which shear and axial

deformations are ignored, the displacements

are calculated as

1( )∆ =

Lmi

ii

i 1 0

Mm dxEI

∆ ∆

=

+ ∆∑ ∫ (7a)

1( )θ =

Lmi

ii

i 1 0

Mm dxEI

θ ∆

=

+ θ∑ ∫ (7b)

im , m∆iθ

∆

= primary structure virtual moments

based on the desired displacement ∆ or rotation

θ

,∆∆ θ = primary structure displacements at ∆

or rotation θ

In Eqs. (7a) and (7b) the moment expressions

are exact based on the statically indetermi-

nate structure subjected to the external loads

with the redundant forces known from the

flexibility analysis.

iM

Equations (6), (7a), and (7b) are correct only

because exact real member forces are used in

the calculation of the desired displacements.

17

Force Method Examples

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

L L

w

1 (x R1)

Primary Structure w/ Load

=

+

Primary Structure w/ Redundant

18

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

w

L L

w

R2

Primary Structure w/ Load

=

+

R1

Primary Structure w/ Redundant Forces

19

Prismatic Member Displacements

20

21

22

3. Calculate the support reactions for the two-span continuous beam using the internal moment at B as the redundant force, IAB = 2I and IBC = I; E = constant.

Primary Structure w/ Loading

PL2

23

Primary Structure w/ Redundant

MB

DB = __________________

fBB = _________________

MB = _________________

24

4. Calculate the bar forces for the statically indeterminate truss.

Statically Indeterminate

Truss

Statically Determinate

Released Truss

(Redundant X)

25

VAC AC,AC

F FLF /f 20736/829.44

E A

= 25 kips

Mem L (in) F FV FVFL

AB 192"40 -4/5 -6144

BC 144" 0 -3/5 0

CD192" 0 -4/5 0

DA 144" 30 -4/5 -2592

AC 240" 0 1 0

BD 240" -50 1 -12000

Truss Calculations

29

Example Beam Problem –Nonmechanical Loading

E = 30,000 ksi

I = 288 in4

(a) Given structure

(b) Primary structure

30

The interesting point of this example is that the flexibility equation will have a nonzero right hand side since the redundant displacement is prescribed to equal 0.72” downward. Thus the flexibility equation is

fBB RB = dB - (7)

where

dB = prescribed displacementat redundant B

= -0.72" since RB is positive upward

= -0.24"relative displacementat redundant B

BD

BD

B Bd D

31

Truss Example –Nonmechanical Loading

For the truss structure on the next page, compute the redundant bar EC member force if 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.

32

Truss Example

C

D

E F

A

B

15’

3 @ 20’ = 60’

A 1

B C

D

E F

Primary Structure Subjectedto FCE = 1

33

Mem L FV FV FVL

AB 240" 0 0

AE 300" 0 0

BC 240" -4/5 153.6

BE 180" -3/5 64.8

BF 300" 1 300

CD 240" 0 0

CE 300" 1 300

CF 180" -3/5 64.8

DF 300" 0 0

EF 240" -4/5 153.6

Truss Example Calculations

34

m

CE,CE Vi Vi ii 1

1f F F L

EA

m

CE Vi ii 1

D F

CE,CE CE CEf F D 0

EF EF EF

BF BF

T L 0.072"

0.3"

41

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

40 k

A B

CD

R1

CD

BA 1

Primary Structure Subjected to Virtual Loading

12’

16’

42

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

w

L L

Primary Structure w/ Virtual Load at Desired Displacement

Location

L L

1

R1 R2

43

Influence Lines for Statically Indeterminate

Structures

We will utilize the force

method principles developed in the

last chapter to calculate and

construct influence lines for

statically indeterminate structures.

Recall that an influence line is a

graph of a response function of a

structure as a function of position

of a downward unit load moving

across the structure.

44

Calculating the response function

values simply involves computing

the values of the desired response

function(s) for various positions

of a unit load on the structure.

Constructing the influence line(s)

simply involves plotting the

calculated values.

Earlier in this course, we produced

qualitative influence lines for

statically indeterminate structures

using the Muller-Breslau principle.

45

Influence functions for

statically indeterminate structures

are piecewise cubic for piecewise

prismatic beam members. This

is based on the solution of the

differential equation

4y

4

d uEI 0

dx (b)

where uy = transverse beam

displacement.

46

Truss structure members remain

piecewise linear since the

governing differential equation is

2x

2d u

EA 0dx

(t)

where ux = axial truss member

displacement.

47

Beam and Truss Structure Influence Lines

Consider the continuous beam in

Fig. 1(a) – calculate and draw

the influence line for the

redundant vertical reaction at B.

NOTE: Once the redundant

force(s) is (are) known, the

remaining forces can be

obtained from static equilibrium

or from superposition (Eq. 6).

Figure 1. Influence Line Construction for Single Redundant DOF Beam

The influence line for the reaction By re-

quires that we calculate By as a function of the

unit load position x. Using the force method of

analysis:

BX BB y

BXy

BB

f f B

fB

f

+ =

⇒ = −

0

(1)

where flexibility coefficient fBX denotes the de-

flection of the primary structure (beam) at B

due to a unit load at X (Fig. 1(b); and flexibility

coefficient fBB denotes the deflection at B due

to a unit value of the redundant By.

An efficient procedure for the solution of

(1) involves using Maxwell’s Law of Recipro-

cal Deflections, i.e. the deflection at B due to a

unit load at X must equal the deflection at X

due to a unit load at B. For our problem, this is

stated mathematically as fBX = fXB. Thus, (1)

can be rewritten as

XBy

BB

fB

f= − (2)

which represents the equation for the influence

line for B. We only need to determine fXB and

fBB.

Equation (2) is more convenient that (1)

since (2) shows that the unit load only needs to

be applied at B on the primary (statically de-

terminate) structure and the corresponding dis-

placement is calculated at X. Closed-formed

solutions of (b) for a number of statically de-

terminate structures subjected to point loads are

included in your notes. Also, refer to the table

solutions included in your force method of

analysis notes. The influence line of Fig. 1(d)

is obtained by plotting the solution of (2) for

various values of X; ordinates = -fXB/fBB.

The influence line equation as represented

in (2) shows the validity of Muller-Breslau’s

principle:

The influence line for a response function is given by the deflected shape of the released structure due to a unit displacement (or rota-tion) at the location and in the direction of the response function.

The ordinate of the influence line at any point

X is proportional to the deflection fXB of the

primary structure at that point due to the unit

load at B. Furthermore, this equation indicates

that the influence line for By can be obtained by

multiplying the deflected shape of the primary

structure due to the unit load at B by the scaling

factor -1/fBB.

Consider next a beam structure with multiple

degrees of freedom as shown in Fig. 2. The

procedure is the same you simply have more

redundant degrees of freedom. In Fig. 2, the

vertical reactions at B and C are taken to be re-

dundant forces leading to the equations

(3) BX BB y BC y

CX CB y CC y

f f B f C 0

f f B f C 0

+ +

+ +

=

=

Figure 2. Influence Line Construction for

Multi-Redundant DOF Beam

Using Maxwell’s reciprocal principle, (3) is

B f C 0

f f B f C 0

+ =

+ + = (4)

which facilitates the solution of the problem.

ng

rewritten as

XB Bf f+ B y BC y

XC CB y CC y

Thus, the unit load needs to be placed succes-

sively only at points B and C and the deflec-

tions fXB and fXC at a number of points X alo

the beam are computed.

Procedure for Analysis

tic indetermi-

2. sing Max-

3. ength

4. for the redun-

ce

1. Determine the degree of sta

nacy and select the redundants.

Solve the redundant equations u

well’s law of reciprocal deflections.

Select a number of points along the l

of the structure at which the numerical

values of the ordinates of the influence

lines will be evaluated.

Once the influence lines

dants have been determined, the influen

lines for other response functions can be

generated using static equilibrium.

61

Beam Example

Using the force method of analysis, calculate the ordinates for the positive internal moment at C.

(a) Beam Structure

(b)Qualitative Influence Line Diagram

EI = constant

62

This beam structure is the same as the two redundant dof qualitative example. Thus, taking By and Dyas the redundant forces (see Ghali and Neville, 2003 – “Prismatic Member Displacements”):

BB 1

2 2

f f ( 80 ', x 24 ', b 24 ', P 1)

(80 24) 242(80)(24) (24) (24)

6(80) EI

7526.4

EI

BD DB 1

2 2

f f f ( 80 ', x 56 ', b 24 ', P 1)

24(80 56)2(80) (56) (56) (24)

6(80) EI

6297.6

EI

63

DD 1

2 2

BB

f f ( 80 ', x 56 ', b 56 ', P 1)

(80 56)562(80)(56) (56) (56)

6(80)EI

7526.4f (due to symmetry)

EI

XB 1

2 2

2

f f ( 80 ', b 24 ', P 1)

(80 24) x2(80)(24) (24) x

6(80)EI

7x3264 x for x b

60EI

XB 1

2 2

2

f f ( 80 ', b 24 ', P 1)

24(80 x)2(80)x (24) x

6(80)EI

(80 x)576 160x x for x b

20EI

64

XD 1

2 2

2

f f ( 80 ', b 56 ', P 1)

(80 56) x2(80)(56) (56) x

6(80) EI

x5824 x for x b

20EI

XD 1

2 2

2

f f ( 80 ', b 56 ', P 1)

56(80 x)2(80)x (56) x

6(80) EI

7(80 x)3136 160x x

60EIfor x b

65

1

y

y

XB

XD

R [F] {D(x)}

B EI

D 16,986,931.2

f7526.4 6297.6

f6297.6 7526.4

y XB XD

XB XDy

B 4.4307 f 3.70732fEI3.70732f 4.4307 fD 10,000

y y yA

y y y y y

1M 0 E x 24 B 56 D

80

F 0 A 1 B D E

From equilibrium:

66

Influence Lines forSupport Reactions

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80

ByDyAyEy

67

Influence Line for Mc

-2

-1

0

1

2

3

4

5

6

0 8 16 24 32 40 48 56 64 72 80

Mc

(ft-

kip

s)

C y y

C y y

M 40A 16B (40 x); 0 x 40

M 40A 16B ; 40 x 80

From equilibrium:

68

Truss Example

See example 15.4 in your textbook.

1x

C D

E F

A B

15’

3 @ 20’ = 60’

69

Primary Structurew/ Loading

1x

C D

E F

A B

15’

3 @ 20’ = 60’

70

Primary Structurew/ Unit Redundant

Load

1

C D

E F

A B

15’

3 @ 20’ = 60’

1

71

Mem L A FV FVL/A

AB 240" 6 0 0

AE 300" 6 0 0

BC 240" 6 -4/5 -32

BE 180" 4 -3/5 -27

BF 300" 4 1 75

CD 240" 6 0 0

CE 300" 4 1 75

CF 180" 4 -3/5 -27

DF 300" 6 0 0

EF 240" 6 -4/5 -32

Truss Example Calculations

72

Mem FB FC

AB -8/9 -4/9

AE 10/9 5/9

BC -4/9 -8/9

BE -2/3 -1/3

BF -5/9 10/9

CD -4/9 -8/9

CE 0 0

CF 0 -1

DF 5/9 10/9

EF 8/9 4/9

73

mVi i

CE,CE Viii 1

F L1f F

E A

mB BVi iCE i

ii 1

F L1D F

E A

mC CVi iCE i

ii 1

F L1D F

E A

i iCE,CE CE CEf F D 0;

i B, C

74

Mem

AB 0 0 0

AE 0 0 0

BC 25.6 128/9 256/9

BE 16.2 162/9 81/9

BF 75 -375/9 750/9

CD 0 0 0

CE 75 0 0

CF 16.2 0 243/9

DF 0 0 0

EF 25.6 -256/9 -128/9

CCEE DB

CEE DCE,CEE f

75

CE,CE1

f (25.6 16.2 75E

16.2 25.7 6)

233.

E

5

6

BCE

1 128 162 375 256D

E 9 9 9 9

341

E

CCE

1 256 81 750 243 256D

E 9 9 9 9 9

1202

E

76

ii CECE

CE,CE

BCE

CCE

DF ; i B, C

f

F 1.46

F 5.15

77

Calculate forces in members BE, BC and EF. Note: Member forcesAB, AE, DC, and DF are statically determinate for this truss structure as are the support reactions.

1x

C D

E F

A B

15’

3 @ 20’ = 60’