Structural Analysis Chapter 10.ppt

Structural Analysis 7Structural Analysis 7thth Edition in SI Units Edition in SI UnitsRussell C. HibbelerRussell C. Hibbeler

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

MethodMethod

Statically Indeterminate StructuresStatically Indeterminate Structures

• Advantages & DisadvantagesAdvantages & Disadvantages• For a given loading, the max stress and For a given loading, the max stress and

deflection of an indeterminate structure are deflection of an indeterminate structure are generally smaller than those of its statically generally smaller than those of its statically determinate counterpartdeterminate counterpart

• Statically indeterminate structure has a Statically indeterminate structure has a tendency to redistribute its load to its tendency to redistribute its load to its redundant supports in cases of faulty designs redundant supports in cases of faulty designs or overloadingor overloading

© 2009 Pearson Education South Asia Pte LtdStructural Analysis 7th EditionChapter 10: Analysis of Statically Indeterminate Structures by the Force Method


• Advantages & DisadvantagesAdvantages & Disadvantages• Although statically indeterminate structure Although statically indeterminate structure

can support loading with thinner members & can support loading with thinner members & with increased stability compared to their with increased stability compared to their statically determinate counterpart, the cost statically determinate counterpart, the cost savings in material must be compared with savings in material must be compared with the added cost to fabricate the structure the added cost to fabricate the structure since often it becomes more costly to since often it becomes more costly to construct the supports & joints of an construct the supports & joints of an indeterminate structureindeterminate structure

• Careful of differential disp of the supports as Careful of differential disp of the supports as wellwell



• Method of AnalysisMethod of Analysis• To satisfy equilibrium, compatibility & force-To satisfy equilibrium, compatibility & force-

disp requirements for the structuredisp requirements for the structure• Force MethodForce Method• Displacement MethodDisplacement Method


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

• From free-body diagram, there would be 4 From free-body diagram, there would be 4 unknown support reactionsunknown 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 supportscompatibility of disp at one of the supports• Choose one of the support reactions as Choose one of the support reactions as

redundant & temporarily removing its effect redundant & temporarily removing its effect on the beamon the beam



• This will allow the beam to be statically This will allow the beam to be statically determinate & stabledeterminate & 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 downwardB 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 upwardat B to be displaced upward



• 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:eqn at the rocker as:


BByB

BByBBBBB

fB

fB

BBfyBBB

0

''0

:get we,eqnfirst intoeqn second sub

tcoefficieny flexibilitlinear Bat reaction unknown

Bat disp upward '


• Using methods in Chapter 8 or 9 to solve for Using methods in Chapter 8 or 9 to solve for BB and f and fBBBB, B, Byy can be found can be found

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

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



• The moment at A can be determined directly The moment at A can be determined directly by removing the capacity of the beam to by removing the capacity of the beam to support moment at A, replacing fixed support moment at A, replacing fixed support by pin supportsupport by pin support

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

• The rotation at A The rotation at A caused by the caused by the redundant Mredundant MAA at at A is A is ’’AAAA




AAAA

AAAAA

M

M

0:requiresity Compatibil

' Similarly,

Maxwell’s Theorem of Reciprocal Maxwell’s Theorem of Reciprocal Disp: Betti’s LawDisp: Betti’s Law

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

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


ABBA ff

Maxwell’s Theorem of Reciprocal Maxwell’s Theorem of Reciprocal Disp: Betti’s LawDisp: Betti’s Law

• The theorem also applies for reciprocal The theorem also applies for reciprocal rotationsrotations

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


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

Example 10.1Example 10.1


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

SolutionSolution


Compatibility equation

SolutionSolution


kNBEI

BEI

EImf

EIkNm

f

fB

yy

BBB

BBB

BByB

6.1557690000

:(1)eqn into Sub

576 ;9000

table.standard using obtainedeasily are and

eqn(1) 0 )(

33

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



Principle of SuperpositionSince 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.

SolutionSolution


Compatibility eqnReference to points A & B requires

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

SolutionSolution


(2)eqn 0

(1)eqn 0

BBBBAAB

ABBAAAA

MM

MM

EIEIEI

EIEI

BAABBBAA

AA

1 ;2 ;2

1.118 ;9.151

Compatibility eqn

SolutionSolution


,1.28 ;9.61

211.1180

129.1510

:gives(2) and (1)eqn into Sub

kNmMkNmM

EIM

EIM

EI

EIM

EIM

EI

BA

BA

BA

Composite Structures

• Composite structures are composed of some Composite structures are composed of some members subjected only to axial force while members subjected only to axial force while other members are subjected to bendingother members are subjected to bending

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


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.



Principle of superpositionThe 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.

SolutionSolution


Compatibility eqnWith 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 fBDBD

SolutionSolution


(1) 0 BDBDBDBD fF

Compatibility eqnFor 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.

SolutionSolution


mm

AE

AEEIdxxx

AEnNLdx

EIMm

o

o

L

BD

326.0

)45cos/8.1)(1)(0(

)30cos/8.1)(816.0)(3.69()0)(333.330(

3

0

3

0

Compatibility eqnFor fBDBD we require application of a real unit load & a virtual unit load at the cut ends of member BD.

SolutionSolution


kNm

AEAEEIdx

AELndx

EImf

oo

L

BD

/)10(092.1

)45cos/8.1()1()30cos/8.1()816.0()0(

5

223

0

2

0

22

Compatibility eqnSub into eqn (1) yields

SolutionSolution


)(9.29

)10)(092.1(0003264.00

0

5

CkNF

F

fF

BD

BD

BDBDBDBD

Additional remarks on the force Additional remarks on the force method of analysismethod of analysis

• Flexibility coefficients depend on the Flexibility coefficients depend on the material and geometrical properties of the material and geometrical properties of the members and members and notnot on the loading of the on the loading of the primary structureprimary structure

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


0......

0......

0......

2211

22221212

12121111

nnnnnn

nn

nn

RfRfRf

RfRfRf

RfRfRf

Additional remarks on the force Additional remarks on the force method of analysismethod of analysis

BDBD 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 differences or fabrication errors in the membersmembers

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

• Note that Note that ffijij=f=fjiji

• Hence, the flexibility matrix will be symmetricHence, the flexibility matrix will be symmetric© 2009 Pearson Education South Asia Pte LtdStructural Analysis 7th EditionChapter 10: Analysis of Statically Indeterminate Structures by the Force Method

fR

Symmetric StructuresSymmetric Structures

• A structural analysis of any highly A structural analysis of any highly indeterminate structure or statically indeterminate structure or statically determinate structure can be simplified determinate structure can be simplified provided the designer can recognise those provided the designer can recognise those structures that are symmetric & support structures that are symmetric & support either symmetric or antisymmetric loadingseither symmetric or antisymmetric loadings

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



• Here the horizontal reaction at the pin is zero, Here the horizontal reaction at the pin is zero, so both these structures will deflect & produce so both these structures will deflect & produce the same internal loading as their reflected the same internal loading as their reflected counterpartcounterpart

• As a result, they can As a result, they can be classified as being symmetricbe 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 sideon its left & right side



• A symmetric structure supports an A symmetric structure supports an antisymmetric loading as shownantisymmetric 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 have to be performed on half the members of the structure since the same or opposite of the structure since the same or opposite results will be produced on the other halfresults will be produced on the other half



• 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 structurethe structure


Influence lines for Statically Influence lines for Statically Indeterminate BeamsIndeterminate Beams

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

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

• Reaction at AReaction at A• To determine the influence line for the To determine the influence line for the

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

• At each point, the reaction at A must be At each point, the reaction at A must be computedcomputed



• Reaction at AReaction at A• A plot of these results yields the influence A plot of these results yields the influence

lineline• The reaction at A can be determined by the The reaction at A can be determined by the

force methodforce 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:



• Reaction at AReaction at A• A plot of these results yields the influence A plot of these results yields the influence

lineline• The reaction at A can be determined by the The reaction at A can be determined by the

force methodforce 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


AAADyAAyAD ffAfAf /0

DAAA

yDAAD ff

Aff

1


• Shear at EShear at E• Using the force method & Maxwell’s theorem Using the force method & Maxwell’s theorem

of reciprocal disp, it can be shown that of reciprocal disp, it can be shown that


DEEE

E ff

V

1


• 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 Edetermined by placing a pin or hinge at E• Applying a +ve unit couple moment, the Applying a +ve unit couple moment, the

beam then deflects to the dashed positionbeam then deflects to the dashed position• Using the force method & Maxwell’s theorem Using the force method & Maxwell’s theorem

of reciprocal disp, it can be shown thatof reciprocal disp, it can be shown that


DEEE

E fM

1


• 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 Edetermined by placing a pin or hinge at E• Applying a +ve unit couple moment, the Applying a +ve unit couple moment, the

beam then deflects to the dashed positionbeam then deflects to the dashed position• Using the force method & Maxwell’s theorem Using the force method & Maxwell’s theorem

of reciprocal disp, it can be shown thatof reciprocal disp, it can be shown that


Qualitative Influence lines for FramesQualitative 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 moment at the center I of girder FG of the frame is shown by the dashed linesframe is shown by the dashed lines

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


Draw the influence line for the vertical reaction at A for the beam. EI is constant. Plot numerical values every 2m



The capacity of the beam to resist reaction Ay 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.

SolutionSolution


SolutionSolution


EIEIEIM

M

M

DD

D

BB

67.3432)2(2

21)2(18

0

0

'

'

'

,D'For

B'at beam conjugate on the existsmoment no since ,B'For

SolutionSolution


EIM

EIEIEIM

M

AA

CC

C

72

33.6134)4(4

21)4(18

0

'

'

'

,A'For

,C'For

•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

SolutionSolution


x Ay

A 1C 0.852D 0.481B 0

Structural Analysis Chapter 10.ppt

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