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By:school.edhole.com

school.edhole.com

5.1 ANALYSIS OF INDETERMINATE STRUCTURES BY FORCE METHOD - AN OVERVIEW

5.2 INTRODUCTION 5.3 METHOD OF CONSISTENT DEFORMATION 5.4 INDETERMINATE BEAMS 5.5 INDETRMINATE BEAMS WITH MULTIPLE DEGREES OF

INDETERMINACY 5.6 TRUSS STRUCTURES 5.7 TEMPERATURE CHANGES AND FABRICATION ERRORS

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5.2 Introduction While analyzing indeterminate structures, it is necessary to

satisfy (force) equilibrium, (displacement) compatibility and force-displacement relationships

(a) Force equilibrium is satisfied when the reactive forces hold the structure in stable equilibrium, as the structure is subjected to external loads

(b) Displacement compatibility is satisfied when the various segments of the structure fit together without intentional breaks, or overlaps

(c) Force-displacement requirements depend on the manner the material of the structure responds to the applied loads, which can be linear/nonlinear/viscous and elastic/inelastic; for our study the behavior is assumed to be linear and elastic

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Two methods are available to analyze indeterminate structures, depending on whether we satisfy force equilibrium or displacement compatibility conditions - They are: Force method and Displacement Method

Force Method satisfies displacement compatibility and force-displacement relationships; it treats the forces as unknowns - Two methods which we will be studying are Method of Consistent Deformation and (Iterative Method of) Moment Distribution

Displacement Method satisfies force equilibrium and force-displacement relationships; it treats the displacements as unknowns - Two available methods are Slope Deflection Method and Stiffness (Matrix) method

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Solution Procedure: (i) Make the structure determinate, by releasing the extra

forces constraining the structure in space (ii) Determine the displacements (or rotations) at the

locations of released (constraining) forces (iii) Apply the released (constraining) forces back on the

structure (To standardize the procedure, only a unit load of the constraining force is applied in the +ve direction) to produce the same deformation(s) on the structure as in (ii)

(iv) Sum up the deformations and equate them to zero at the position(s) of the released (constraining) forces, and calculate the unknown restraining forces

Types of Problems to be dealt: (a) Indeterminate beams; (b) Indeterminate trusses; and (c) Influence lines for indeterminate structures

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5.4.1 Propped Cantilever - Redundant vertical reaction released

(i) Propped Cantilever: The structure is indeterminate to the first degree; hence has one unknown in the problem.

(ii) In order to solve the problem, release the extra constraint and make the beam a determinate structure. This can be achieved in two different ways, viz., (a) By removing the vertical support at B, and making the beam a cantilever beam (which is a determinate beam); or (b) By releasing the moment constraint at A, and making the structure a simply supported beam (which is once again, a determinate beam).

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(a) Release the vertical support at B:

The governing compatibility equation obtained at B is,

fBB = displacement per unit load (applied in +ve direction)

x

y

PP

BC

L/2L/2L

C

B= +

B B

RB

BB=RB*fBB

B + '

B B = 0

BBBB

BBBB

fR

fR

/

0)()(

F r o m e a r l i e r a n a l y s e s ,

)/)(48/5(

)16/()24/(

)2/()]2/()2/([)3/()2/(

3

33

23

EIPL

EIPLEIPL

LEILPEILPB

)3/(3 EILf BB

PEILEIPLR BB )16/5()]3/(/[)]/)(48/5([ 33

Applied in +ve direction

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G o v e r n i n g c o m p a t i b i l i t y e q u a t i o n o b t a i n e d a t A i s , )()( AAAA M , AA = r o t a t i o n p e r u n i t m o m e n t

AA

AAM

F r o m k n o w n e a r l i e r a n a l y s i s , )16(

2

EI

PLAA [ u n d e r a c e n t r a l c o n c e n t r a t e d

l o a d ])]3/()[1( EILAA

T h i s i s d u e t o t h e f a c t t h a t + v e m o m e n t c a u s e s a – v e r o t a t i o n

PL16)/(3

EI)]L/(3/[EI)]/(16PL[M 2A

5.4 INDETERMINATE BEAM (Cont’d)

5.4.2 Propped cantilever - Redundant support moment released

L

PL/2

(b) Release the moment constraint at a:

A B

A

=

A BP

Primary structure

+ BA

MA A=MAAA

Redundant MA applied

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To recapitulate on what we have done earlier,I. Structure with single degree of indeterminacy:

(a) Remove the redundant to make the structure determinate (primary structure)

(b) Apply unit force on the structure, in the direction of the redundant, and find the displacement

(c) Apply compatibility at the location of the removed redundant

A BRB

A BBo

fBB

5.4.3 OVERVIEW OF METHOD OF CONSISTENT DEFORMATION

B0 + fBBRB = 0

P

P

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(a) Make the structure determinate (by releasing the supports at B, C and D) and determine the deflections at B, C and D in the direction of removed redundants, viz., BO, CO and DO

AB C D E

RB RC RD

B0 C0D0

w/u.l

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(b) Apply unit loads at B, C and D, in a sequential manner and

determine deformations at B, C and D, respectively.

AB C D E

fBBfCB fDB

1

AB C D E

fBCfCC fDC

AB C D E

fBDfCD fDD

1

1

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(c ) Establish compatibility conditions at B, C and D

BO + fBBRB + fBCRC + fBDRD = 0

CO + fCBRB + fCCRC + fCDRD = 0

DO + fDBRB + fDCRC + fDDRD = 0

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5.4.2 When support settlements occur:

Compatibility conditions at B, C and D give the following equations:

BO + fBBRB + fBCRC + fBDRD = B

CO + fCBRB + fCCRC + fCDRD = C

DO + fDBRB + fDCRC + fDDRD = D

AB C D E

B C D Support settlements

w / u. l.

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(a) (a) Remove the redundant member (say AB) and make the structure

a primary determinate structure

The condition for stability and indeterminacy is:

r+m>=<2j,

Since, m = 6, r = 3, j = 4, (r + m =) 3 + 6 > (2j =) 2*4 or 9 > 8 i = 1

C

80 kN

60 kN

A B

D

C

80 kN

60 kN

A B

D

1 2

Primary structure

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5.5 Truss Structures (Cont’d)

(b)Find deformation ABO along AB:

ABO = (F0uABL)/AE

F0 = Force in member of the primary structure due to applied loaduAB= Forces in members due to unit force applied along AB

(c) Determine deformation along AB due to unit load applied along AB:

(d) Apply compatibility condition along AB:

ABO+fAB,ABFAB=0

(d) Hence determine FAB

AE

LABu

ABABf

2

,

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(e) Determine the individual member forces in a particular member CE by

FCE = FCE0 + uCE FAB

where FCE0 = force in CE due to applied loads on primary structure (=F0), and uCE = force in CE due to unit force applied along AB (= uAB)

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Temperature changes affect the internal forces in a structure

Similarly fabrication errors also affect the internal forces in a structure(i) Subject the primary structure to temperature

changes and fabrication errors. - Find the deformations in the redundant direction

(ii) Reintroduce the removed members back and make the deformation compatible

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