SA 2 CHAP 11

8/20/2019 SA 2 CHAP 11

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

Chapter 11:Chapter 11:Displacement Method of Analysis: SlopeDisplacement Method of Analysis: Slope--Deflection EquationsDeflection Equations

Displacement Method of Analysis: GeneralDisplacement Method of Analysis: General

ProceduresProcedures

•• Disp method requires satisfying eqm eqn for theDisp method requires satisfying eqm eqn for thestructuresstructures

•• The unknowns disp are written in terms of theThe unknowns disp are written in terms of theloads by using the loadloads by using the load--disp relationsdisp relations

•• These eqn are solved for the dispThese eqn are solved for the disp

•• Once the disp are obtained, the unknown loadsOnce the disp are obtained, the unknown loads

are determined from the compatibility eqn usingare determined from the compatibility eqn usingthe load disp relationsthe load disp relations

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th Edition

Chapter 11: Displacement Method of Analysis: Slope-Deflection Equations

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Displacement Method of Analysis: GeneralDisplacement Method of Analysis: GeneralProceduresProcedures

•• When a structure is loaded, specified points on itWhen a structure is loaded, specified points on itcalled nodes, will undergo unknown dispcalled nodes, will undergo unknown disp

•• These disp are referred to as the degree ofThese disp are referred to as the degree offreedomfreedom

•• The no. of these unknowns is referred to as theThe no. of these unknowns is referred to as thedegree in which the structure is kinematicallydegree in which the structure is kinematicallyindeterminateindeterminate

•• We will consider some e.g.sWe will consider some e.g.s




Displacement Method of Analysis: GeneralDisplacement Method of Analysis: General

ProceduresProcedures

•• Any load applied to the beam will cause node A to Any load applied to the beam will cause node A torotaterotate

•• Node B is completely restricted from movingNode B is completely restricted from moving

•• Hence, the beam has only one unknown degree ofHence, the beam has only one unknown degree offreedomfreedom

•• The beam has nodes at A, B & CThe beam has nodes at A, B & C

•• There are 4 degrees of freedomThere are 4 degrees of freedom θθ A A ,, θθBB,, θθCC andand ∆∆CC




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SlopeSlope--deflection equationdeflection equation

•• Slope deflection method requires less work both toSlope deflection method requires less work both towrite the necessary eqn for the solution of awrite the necessary eqn for the solution of aproblem& to solve these eqn for the unknown dispproblem& to solve these eqn for the unknown disp& associated internal loads& associated internal loads

•• General CaseGeneral Case

•• To develop the general form of the slopeTo develop the general form of the slope--deflectiondeflectioneqn, we will consider theeqn, we will consider thetypical span AB of thetypical span AB of the

continuous beam whencontinuous beam whensubjected to arbitrary loadingsubjected to arbitrary loading





•• Angular Disp Angular Disp

•• Consider node A of the member to rotateConsider node A of the member to rotate θθ A A whilewhileits while end node B is held fixedits while end node B is held fixed

•• To determine the moment MTo determine the moment M AB AB needed to cause thisneeded to cause thisdisp, we will use the conjugate beam methoddisp, we will use the conjugate beam method




8/20/2019 SA 2 CHAP 11







03

2

2

1

32

1

0

03

2

2

1

32

1

0

'

'

=+

−

=∑

=

−

=∑

L

L

L EI

M L

L EI

M

M

L L

EI

M L L

EI

M

M

A

AB BA

B

BA AB

A

θ



•• From which we obtain the following:From which we obtain the following:

•• Similarly, end B of the beam rotates to its finalSimilarly, end B of the beam rotates to its finalposition while end A is held fixedposition while end A is held fixed

•• We can relate the applied moment MWe can relate the applied moment MBA BA to theto theangular dispangular disp θθBB & the reaction moment M& the reaction moment M AB AB at theat thewallwall




2

4

A BA A AB

L

EI M

L

EI M θ θ ==

B AB B BA

L

EI M

L

EI M θ θ

2

4==

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•• Relative linear dispRelative linear disp

•• If the far node B if the member is displaced relativeIf the far node B if the member is displaced relativeto A, so that the cord of the member rotatesto A, so that the cord of the member rotatesclockwise & yet both ends do not rotate then equalclockwise & yet both ends do not rotate then equalbut opposite moment and shear reactions arebut opposite moment and shear reactions aredeveloped in the memberdeveloped in the member

•• Moment M can be related to the disp usingMoment M can be related to the disp usingconjugate beam methodconjugate beam method





•• Relative linear dispRelative linear disp

•• The conjugate beam is free at both ends since theThe conjugate beam is free at both ends since thereal member is fixed supportreal member is fixed support

•• The disp of the real beam at B, the moment at endThe disp of the real beam at B, the moment at endBB’’ of the conjugate beam must have a magnitude ofof the conjugate beam must have a magnitude of∆∆ as indicatedas indicated




( ) ( )

∆−

===

=∆−

−

=∑

2

'

6

032

1

3

2

2

1

0

L

EI M M M

L L

EI

M L L

EI

M

M

BA AB

B

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•• Fixed end momentFixed end moment

•• In general, linear & angular disp of the nodes areIn general, linear & angular disp of the nodes arecaused by loadings acting on the span of thecaused by loadings acting on the span of themembermember

•• To develop the slopeTo develop the slope--deflection eqn, we mustdeflection eqn, we musttransform these span loadings into equivalenttransform these span loadings into equivalentmoment acting at the nodes & then use the loadmoment acting at the nodes & then use the load--disp relationships just deriveddisp relationships just derived





•• SlopeSlope--deflection eqndeflection eqn

•• If the end moments due to each disp & loadings areIf the end moments due to each disp & loadings areadded together, the resultant moments at the endsadded together, the resultant moments at the endscan be written as:can be written as:




0322

0322

=+

∆−+

=

=+

∆−+

=

BA A B BA

AB B A AB

FEM L L

I E M

FEM L L

I E M

θ θ

θ θ

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•• SlopeSlope--deflection eqndeflection eqn

•• The results can be expressed as a single eqnThe results can be expressed as a single eqn




[ ]

supportendnearat themomentendfixed

displinearatoduecorditsof rotationspan

supportsat thespantheof dispangularorslopesendfarandnear,

stiffnessspan&elasticityof modulus,

spantheof endnearat themomentinternal

0322

=

=

=

=

=

=+−+=

N FEM

F N

k E

N M

N F N N FEM Ek M

ψ

θ θ

ψ θ θ


•• Pin supported end spanPin supported end span

•• Sometimes an end span of a beam or frame isSometimes an end span of a beam or frame issupported by a pin or roller at its far endsupported by a pin or roller at its far end

•• The moment at the roller or pin is zero provided theThe moment at the roller or pin is zero provided theangular disp at this support does not have to beangular disp at this support does not have to bedetermineddetermined




[ ][ ] 03220

322

+−+=

+−+=

ψ θ θ

ψ θ θ

F N

N F N N

Ek

FEM Ek M

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•• Pin supported end spanPin supported end span

•• Simplifying, we get:Simplifying, we get:

•• This is only applicable for end span with far endThis is only applicable for end span with far endpinned or roller supportedpinned or roller supported




[ ] N N N

FEM Ek M +−= ψ θ 3

Draw the shear and moment diagrams for the beam where EI isconstant.

Example 11.1Example 11.1




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Slope-deflection equation

2 spans must be considered in this problem

Using the formulas for FEM, we have:

SolutionSolution




mkN wL

FEM

mkN wL

FEM

CB

BC

.8.1020

)6(6

20)(

.2.730

)6(6

30)(

22

22

===

−=−=−=


•Note that (FEM)BC is –ve and

•(FEM) AB = (FEM)BA since there is no load on span AB

•Since A & C are fixed support, θ A = θC =0

•Since the supports do not settle nor are they displaced up ordown, ψ AB = ψBC = 0

SolutionSolution




[ ]

[ ]

(1) 4

0)0(3)0(28

2

322

B

B AB

N F N N

EI

I E M

FEM L

I E M

θ

θ

ψ θ θ

=

+−+

=

+−+

=

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SolutionSolution




(4) 10.83

(3) 7.2-3

2

(2) 2

Similarly,

+=

=

=

BCB

B BC

B BA

EI

M

EI M

EI M

θ

θ

θ

Equilibrium eqn

The necessary fifth eqn comes from the condition of momentequilibrium at support B

Here MBA & MBC are assumed to act in the +ve direction to beconsistent with the slope-deflection eqn

SolutionSolution




(5) 0=+ BC BA

M M

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Equilibrium equation

SolutionSolution




kNm M kNm M kNm M kNm. M

EI

.θ

CB BC

BA AB

B

86.12 ;09.3 ;09.3 ;541

:gives(4)to(1)eqnintovaluethissub-Re

176

:gives(5)eqninto(3)and(2)eqnSub

=−= ==

=

Equilibrium equation

•Using these results, the shears at the end spans aredetermined.

•The free-body diagram of the entire beam & the shear &moment diagrams are shown.

SolutionSolution




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Analysis of Frames: No sidesway Analysis of Frames: No sidesway

•• A frame will not sidesway to the left or right A frame will not sidesway to the left or rightprovided it is properly restrainedprovided it is properly restrained

•• No sidesway will occur in an unrestrained frameNo sidesway will occur in an unrestrained frameprovided it is symmetric wrt both loading andprovided it is symmetric wrt both loading andgeometrygeometry




Determine the moments at each joint of the frame. EI is constant.





8/20/2019 SA 2 CHAP 11


Slope-deflection eqn

3 spans must be considered in this case: AB, BC & CD

SolutionSolution




0and0that Note

.8096

5)(

.8096

5)(

2

2

=====

==

−=−=

CD BC AB D A

CB

BC

mkN wL

FEM

mkN wL

FEM

ψ ψ ψ θ θ


We have

SolutionSolution




[ ]

C DC

C CD

BC CB

C B BC

B BA

B AB

N F N N

EI M

EI M

EI EI M

EI EI M

EI M

EI M

FEM L

I E M

θ θ

θ θ θ θ

θ θ

ψ θ θ

1667.0333.0

8025.05.08025.05.0

333.01667.0

322

==

++=−+=

==

+−+

=

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•The remaining 2 eqn come from moment equlibrium at joints B& C, we have:

•Solving for these 8 eqn, we get:

SolutionSolution




0

0

=+

=+

CDCB

BC BA

M M

M M

8025.0833.0

8025.0833.0

−=+

=+

BC

C B

EI EI

EI EI

θ θ

θ θ


Solving simultaneously yields:

SolutionSolution




kNm M kNm M

kNm M kNm M

kNm M kNm M

EI

DC CD

CB BC

BA AB

C B

9.22 ;7.457.45 ;7.45

7.45 ;9.22

1.137

−=−=

=−=

==

=−= θ θ

8/20/2019 SA 2 CHAP 11


Analysis of Frames: Sidesway Analysis of Frames: Sidesway

•• A frame will sidesway when it or the loading acting on it is A frame will sidesway when it or the loading acting on it isnonsymmetricnonsymmetric

•• The loading P causes an unequal moments at joint B & CThe loading P causes an unequal moments at joint B & C

•• MMBCBC tends to displace joint Btends to displace joint Bto the rightto the right

•• MMCBCB tends to displace joint Ctends to displace joint Cto the leftto the left

•• Since MSince MBCBC > M> MCBCB, the net result, the net resultis a sidesway of both joint B & Cis a sidesway of both joint B & Cto the rightto the right





•• When applying the slopeWhen applying the slope--deflection eqn to eachdeflection eqn to eachcolumn, we must consider the column rotation,column, we must consider the column rotation, ψψas an unknown in the eqnas an unknown in the eqn

•• As a result, an extra eqm eqn must be included in As a result, an extra eqm eqn must be included inthe solutionthe solution

•• The techniques for solving problems for framesThe techniques for solving problems for frames

with sidesway is best illustrated by e.g.with sidesway is best illustrated by e.g.




8/20/2019 SA 2 CHAP 11



•• When applying the slopeWhen applying the slope--deflection eqn to eachdeflection eqn to eachcolumn, we must consider the column rotation,column, we must consider the column rotation, ψψas an unknown in the eqnas an unknown in the eqn

•• As a result, an extra eqm eqn must be included in As a result, an extra eqm eqn must be included inthe solutionthe solution

•• The techniques for solving problems for framesThe techniques for solving problems for frameswith sidesway is best illustrated by e.g.with sidesway is best illustrated by e.g.




Explain how the moments in each joint of the two-story frame. EIis constant.





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We have 12 equations that contain 18 unknowns

SolutionSolution





No FEMs have to be calculated since the applied loading acts atthe joints

Members AB & FE undergo rotations of ψ1 = ∆1 /5

Members AB & FE undergo rotations of ψ2 = ∆2 /5

Moment eqm of joints B, C, D and E, requires

SolutionSolution




(16) 0

(15) 0

(14) 0

(13) 0

=++

=+=+

=++

ED EB EF

DE DC

CDCB

BC BE BA

M M M

M M

M M

M M M

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Similarly, shear at the base of columns must balance the appliedhorizontal loads

SolutionSolution




(18) 055

120

0VV80400

(17) 055

40

0VV400

FEABx

EDBCx

=++++

=−−+⇒=∑

=+

++

+

=−−⇒=∑

FE EF BA AB

DE EDCB BC

M M M M

F

M M M M

F


•Sub eqn (1) to (12) into eqn (13) to (18)

•These eqn can be solved simultaneously

•The results are resub into eqn (1) to (12) to obtain themomentsat the joints

SolutionSolution




SA 2 CHAP 11

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