Chapter 10: Chapter 10: Analysis of Statically Indeterminate Structures by the Force Method Analysis of Statically Indeterminate Structures by the Force Method

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Structural Analysis 7

Structural Analysis 7^th^th Edition in SI Units Edition in SI Units

Chapter 10:

Analysis of Statically Indeterminate Structures by the Force Method Analysis of Statically Indeterminate Structures by the Force Method

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Statically Indeterminate Structures Statically Indeterminate Structures

• Advantages & DisadvantagesAdvantages & Disadvantages

• For a given loading, the max stress and deflection For a given loading, the max stress and deflection of an indeterminate structure are generally smaller of an indeterminate structure are generally smaller

than those of its statically determinate counterpart than those of its statically determinate counterpart

• Statically indeterminate structure has a tendency to Statically indeterminate structure has a tendency to redistribute its load to its redundant supports in

redistribute its load to its redundant supports in cases of faulty designs or overloading

cases of faulty designs or overloading

Chapter 10: Analysis of Statically Indeterminate Structures by the Force Method

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• Advantages & DisadvantagesAdvantages & Disadvantages

• Although statically indeterminate structure can Although statically indeterminate structure can support loading with thinner members & with support loading with thinner members & with

increased stability compared to their statically increased stability compared to their statically

determinate counterpart, the cost savings in determinate counterpart, the cost savings in

material must be compared with the added cost to material must be compared with the added cost to

fabricate the structure since often it becomes more fabricate the structure since often it becomes more

costly to construct the supports & joints of an costly to construct the supports & joints of an

indeterminate structure indeterminate structure

• Careful of differential disp of the supports as wellCareful of differential disp of the supports as well

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• Method of AnalysisMethod of Analysis

• To satisfy equilibrium, compatibility & force-disp To satisfy equilibrium, compatibility & force-disp requirements for the structure

requirements for the structure

• Force MethodForce Method

• Displacement MethodDisplacement Method

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Force Method of Analysis: General Force Method of Analysis: General

Procedure Procedure

• From free-body diagram, there would be 4 From free-body diagram, there would be 4 unknown support reactions

unknown support reactions

• 3 equilibrium eqn3 equilibrium eqn

• Beam is indeterminate to first degreeBeam is indeterminate to first degree

• Use principle of superposition & consider the Use principle of superposition & consider the compatibility of disp at one of the supports compatibility of disp at one of the supports

• Choose one of the support reactions as redundant Choose one of the support reactions as redundant

& temporarily removing its effect on the beam

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• This will allow the beam to be statically This will allow the beam to be statically determinate & stable

determinate & stable

• Here, we will remove the rocker at BHere, we will remove the rocker at B

• As a result, the load P will cause As a result, the load P will cause B to be displaced downward

B to be displaced downward

• By superposition, the unknown By superposition, the unknown reaction at B causes the beam reaction at B causes the beam

at B to be displaced upward at B to be displaced upward

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• Assuming +ve disp act upward, we Assuming +ve disp act upward, we write the necessary compatibility write the necessary compatibility

eqn at the rocker as:

BB y

B

BB y

BB

BB B

f B f B

fBB ByBB









   





0

' '

0

: get we , eqn first into

eqn second

sub

t coefficien y

flexibilit linear

B at reaction unknown

B at disp upward

'

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• Using methods in Chapter 8 or 9 to solve for Using methods in Chapter 8 or 9 to solve for B_B

and f

and f_BB_BB, B, B_y_y can be found can be found

• Reactions at wall A can then be determined from Reactions at wall A can then be determined from eqn of equilibrium

eqn of equilibrium

• The choice of redundant is arbitraryThe choice of redundant is arbitrary

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• The moment at A can be determined directly by The moment at A can be determined directly by removing the capacity of the beam to support removing the capacity of the beam to support

moment at A, replacing fixed support by pin moment at A, replacing fixed support by pin

support support

• The rotation at A The rotation at A caused by P is

caused by P is A_A

• The rotation at A The rotation at A caused by the

caused by the redundant M

redundant M_A_A at at A is

A is ’’

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AA A

A

AA A

AA

M













0 ity requires : Compatibil

' Similarly,

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Maxwell’s Theorem of Reciprocal Disp:

Betti’s Law Betti’s Law

• The disp of a point B on a structure due to a unit The disp of a point B on a structure due to a unit load acting at point A is equal to the disp of point load acting at point A is equal to the disp of point

A when the load is acting at point B A when the load is acting at point B

• Proof of this theorem is easily demonstrated using Proof of this theorem is easily demonstrated using the principle of virtual work

the principle of virtual work

AB BA f

f 

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Maxwell’s Theorem of Reciprocal Disp:

Betti’s Law Betti’s Law

• The theorem also applies for reciprocal rotationsThe theorem also applies for reciprocal rotations

• The rotation at point B on a structure due to a unit The rotation at point B on a structure due to a unit couple moment acting at point A is equal to the

couple moment acting at point A is equal to the rotation at point A when the unit couple is acting rotation at point A when the unit couple is acting

at point B at point B

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Determine the reaction at the roller support B of the beam. EI is constant.

Example 10.1 Example 10.1

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Principle of superposition

By inspection, the beam is statically indeterminate to the first degree. The redundant will be taken as B_y. We assume B_y acts upward on the beam.

Solution Solution

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Compatibility equation

kN EI B

EI B

EI f m

EI kNm f

f B

y y

BB B

BB y

B

6 . 576 15

0 9000

: (1) eqn into

Sub

576 9000 ;

table.

standard using

obtained easily

are and

eqn(1)

0 ) (

3 3

























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Draw the shear and moment diagrams for the beam. EI is constant. Neglect the effects of axial load.

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Principle of Superposition Since axial load is neglected,

the beam is indeterminate to the second degree. The 2 end

moments at A & B will be

considered as the redundant.

The beam’s capacity to resist these moments is removed by

placing a pin at A and a rocker at B.

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Compatibility eqn

Reference to points A & B requires

The required slopes and angular flexibility coefficients can be determined using standard tables.

(2) eqn

0

(1) eqn

0

BB B

BA A

B

AB B

AA A

A

M M

















EI EI

EI

EI EI

BA AB

BB AA

A A

1 2 ;

2 ;

1 . 118

9 ; . 151







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, 1

. 28

; 9

. 61

2 1

1 . 0 118

1 2

9 . 0 151

: gives (2)

and (1)

eqn into

Sub

kNm M

M EI M EI

EI

M EI M EI

EI

B A











 



 



 



 





 



 



 



 



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Composite Structures

• Composite structures are composed of some Composite structures are composed of some

members subjected only to axial force while other members subjected only to axial force while other

members are subjected to bending members are subjected to bending

• If the structure is statically indeterminate, the If the structure is statically indeterminate, the force method can conveniently be used for its force method can conveniently be used for its

analysis analysis

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The beam is supported by a pin at A & two pin-connected bars at B. Determine the force in member BD. Take E = 200GPa & I = 300(10⁶)mm⁴ for the beam and A = 1800mm² for each bar.

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Principle of superposition

The beam is indeterminate to the first degree. Force in member BD is chosen as the redundant. This member is therefore

sectioned to eliminate its capacity to sustain a force.

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With reference to the relative disp of the cut ends of member BD, we require

The method of virtual work will be used to compute BD and f_BDBD

(1) 0  _BD  F_BD f_BDBD

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For _BD we require application of the real loads and a virtual unit load applied to the cut ends of the member BD. We will consider only bending strain energy in the beam & axial strain energy in the bar.

mm AE

AE EI

dx x

x

AE dx nNL

EI Mm

o

o L

BD

326 .

0

) 45 cos /

8 . 1 )(

1 )(

0 (

) 30 cos /

8 . 1 )(

816 .

0 )(

3 . 69 (

) 0 )(

333 .

3 30

³(

0

3 0







 

 











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For f_BDBD we require application of a real unit load & a virtual unit load at the cut ends of member BD.

kN m

AE AE

EI dx

AE L dx n

EI f m

o o

L BD

/ ) 10 ( 092 .

1

) 45 cos /

8 . 1 ( ) 1 ( )

30 cos /

8 . 1 ( ) 816 .

0 ( )

0 (

5

2 3 2

0

2 0

2 2

 

 













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Sub into eqn (1) yields

) ( 9

. 29

) 10 )(

092 .

1 ( 0003264

. 0 0

0

5

C kN F

F f F

BD

BD BDBD BD

BD















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Additional remarks on the force method of Additional remarks on the force method of

analysis analysis

• Flexibility coefficients depend on the material and Flexibility coefficients depend on the material and geometrical properties of the members and

geometrical properties of the members and notnot on on the loading of the primary structure

the loading of the primary structure

• For a structure having n redundant reactions, we For a structure having n redundant reactions, we can write n compatibility eqn

can write n compatibility eqn

0 ...

2 2 1

1

2 2

22 1

21 2

1 2

12 1

11 1





      









n nn n

n n

R f

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Additional remarks on the force method of Additional remarks on the force method of

analysis analysis

 BD_BD are caused by both the real loads on the are caused by both the real loads on the primary structure and by support settlement &

primary structure and by support settlement &

dimensional changes due to temperature dimensional changes due to temperature

differences or fabrication errors in the members differences or fabrication errors in the members

• The above eqn can be re-cast into a matrix form The above eqn can be re-cast into a matrix form or simply:

or simply:

• Note that Note that ff_ij_ij=f=f_ji_ji

• Hence, the flexibility matrix will be symmetricHence, the flexibility matrix will be symmetric



 fR 

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Symmetric Structures Symmetric Structures

• A structural analysis of any highly indeterminate A structural analysis of any highly indeterminate

structure or statically determinate structure can be structure or statically determinate structure can be

simplified provided the designer can recognise simplified provided the designer can recognise those structures that are symmetric & support those structures that are symmetric & support

either symmetric or antisymmetric loadings either symmetric or antisymmetric loadings

• For horizontal stability, a pin is required to support For horizontal stability, a pin is required to support the beam & truss.

the beam & truss.

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• Here the horizontal reaction at the pin is zero, so both Here the horizontal reaction at the pin is zero, so both these structures will deflect & produce the same

these structures will deflect & produce the same internal loading as their reflected counterpart

internal loading as their reflected counterpart

• As a result, they can As a result, they can

be classified as being symmetric be classified as being symmetric

• Not the case if the fixed support Not the case if the fixed support at A was replaced by a pin since at A was replaced by a pin since

the deflected shape & internal the deflected shape & internal

loadings would not be the same loadings would not be the same

on its left & right side on its left & right side

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• A symmetric structure supports an antisymmetric A symmetric structure supports an antisymmetric loading as shown

loading as shown

• Provided the structure is Provided the structure is symmetric & its loading is symmetric & its loading is

either symmetric or either symmetric or

antisymmetric then a antisymmetric then a

structural analysis will only structural analysis will only

have to be performed on half the members of the have to be performed on half the members of the

structure since the same or opposite results will be structure since the same or opposite results will be

produced on the other half produced on the other half

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• A separate structural A separate structural

analysis can be performed analysis can be performed

using the symmetrical using the symmetrical

& antisymmetrical loading

& antisymmetrical loading components & the results components & the results

superimposed to obtain superimposed to obtain the actual behaviour of the actual behaviour of

the structure the structure

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Influence lines for Statically Influence lines for Statically

Indeterminate Beams Indeterminate Beams

• For statically determinate beams, the deflected For statically determinate beams, the deflected shapes will be a series of straight line segments shapes will be a series of straight line segments

• For statically indeterminate beams, curve will For statically indeterminate beams, curve will result

result

• Reaction at AReaction at A

• To determine the influence line for the reaction at A To determine the influence line for the reaction at A , a unit load is placed on the beam at successive

, a unit load is placed on the beam at successive points

points

• At each point, the reaction at A must be computedAt each point, the reaction at A must be computed

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• A plot of these results yields the influence lineA plot of these results yields the influence line

• The reaction at A can be determined by the force The reaction at A can be determined by the force method

method

• The principle of superposition is appliedThe principle of superposition is applied

• The compatibility eqn for point A is:The compatibility eqn for point A is:

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• A plot of these results yields the influence lineA plot of these results yields the influence line

• The reaction at A can be determined by the force The reaction at A can be determined by the force method

method

• The compatibility eqn for point A is:The compatibility eqn for point A is:

• By Maxwell’s theorem of reciprocal dispBy Maxwell’s theorem of reciprocal disp

AA AD

y AA

y

AD A f A f f

f /

0     

DA y

DA

AD f

A f f

f 







 1

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• Shear at EShear at E

• Using the force method & Maxwell’s theorem of Using the force method & Maxwell’s theorem of reciprocal disp, it can be shown that

reciprocal disp, it can be shown that

DE EE

E f

V f 



 



  1

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• Moment at EMoment at E

• The influence line for the moment at E can be The influence line for the moment at E can be determined by placing a pin or hinge at E

determined by placing a pin or hinge at E

• Applying a +ve unit couple moment, the beam then Applying a +ve unit couple moment, the beam then deflects to the dashed position

deflects to the dashed position

DE EE

E f

M 

 



 

 1

(38)

• Moment at EMoment at E

• The influence line for the moment at E can be The influence line for the moment at E can be determined by placing a pin or hinge at E

determined by placing a pin or hinge at E

• Applying a +ve unit couple moment, the beam then Applying a +ve unit couple moment, the beam then deflects to the dashed position

deflects to the dashed position

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Qualitative Influence lines for Frames Qualitative Influence lines for Frames

• The shape of the influence line for the +ve The shape of the influence line for the +ve

moment at the center I of girder FG of the frame moment at the center I of girder FG of the frame

is shown by the dashed lines is shown by the dashed lines

• Uniform loads would be placed only on girders AB, Uniform loads would be placed only on girders AB, CD & FG in order to create the largest +ve

CD & FG in order to create the largest +ve moment at I

moment at I

(40)

Draw the influence line for the vertical reaction at A for the beam.

EI is constant. Plot numerical values every 2m

(41)

The capacity of the beam to resist reaction A_y is removed. This is done using a vertical roller device. Applying a vertical unit load at A yields the shape of the influence line. Using the conjugate beam method to determine ordinates of the influence line.

(42)

EI EI

M EI M

M

D D

D

B B

67 . 34 3

) 2 2 2 (

2 ) 1

2 18 ( 0

0

' '

'

, D' For

B' at beam conjugate

on the exists

moment no

since ,

B' For

 



 



 



 



 













(43)

M EI

EI EI

M EI M

A A

C C

C

72

33 . 61 3

) 4 4 4 (

2 ) 1

4 18 ( 0

' ' '

, A' For

, C' For





 



 



 



 



 









(44)

•Since a vertical 1kN load acting at A on the beam will cause a vertical reaction at A of 1kN, the disp at A, _A should correspond to a numerical value of 1 for the influence line ordinate at A.

•Thus dividing the other computed disp by this factor, we obtain

x A_y

A 1

C 0.852

D 0.481

B 0