3. 5.1 ANALYSIS OOFF IINNDDEETTEERRMMIINNAATTEE SSTTRRUUCCTTUURREESS
BBYY FFOORRCCEE MMEETTHHOODD -- AANN OOVVEERRVVIIEEWW
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
4. 55..22 IINNTTRROODDUUCCTTIIOONN
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
5. 55..22 IINNTTRROODDUUCCTTIIOONN ((CCoonntt’’dd))
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
6. 5.3 METHOD OOFF CCOONNSSIISSTTEENNTT DDEEFFOORRMMAATTIIOONN
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
7. 5.4 IINNDDEETTEERRMMIINNAATTEE BBEEAAMMSS
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).
8. 5.4 IINNDDEETTEERRMMIINNAATTEE BBEEAAMMSS ((CCoonntt’’dd))
(a) Release the vertical support at B:
x
y
P
P
C B
B B
DB = +
DC
L/2 L/2 L
RB
D¢BB=RB*fBB
Applied in +ve
direction
The governing compatibility equation obtained at B is,
DB
+ D'
BB = 0
R f
D+ ( ) ´ ( ) =
0
B B BB
R /
f
From earlier analyses,
=-D
B B BB
3 2
P L EI P L EI L B
D=- +- ´
( / 2) /(3 ) [ ( / 2) /(2 )] ( / 2)
3 3
PL EI PL EI
=- -
/(24 ) /(16 )
3
PL EI
=-
(5/ 48)( / )
f L3 /(3EI ) BB =
R PL EI L EI P BB =-[-(5/ 48)( 3 / )]/[ 3 /(3 )]=(5/16)
fBB = displacement per unit load (applied in +ve direction)
9. 5.4 INDETERMINATE BEAM (Cont’d)
5.4.2 Propped cantilever - Redundant support moment released
L/2 P
A B
P
Governing compatibility equation obtained at A is,
+( )´( )= A A AA q M a , AA a= rotation per unit moment
q
A
AA
A M
=-
a
From known earlier analysis, (16 )
AA q =- [under a central concentrated
2
EI
PL
load]
(1)[L /(3EI )] AA a =-
This is due to the fact that +ve moment causes a –ve rotation
M =-- [ PL2/(16EI)]/[ -
L/(3EI)]
A
=-
(3/16)PL
L
(b) Release the moment constraint at a:
qA
=
A B
Primary structure
+ A B
MA q¢A=MAaAA
Redundant MA applied
10. 5.4.3 OVERVIEW OF METHOD OF
CONSISTENT DEFORMATION
To recapitulate on what we have done earlier,
I. Structure with single degree of indeterminacy:
A B
RB
(a) Remove the redundant to make the structure determinate
(primary structure)
A B
DBo
(b) Apply unit force on the structure, in the direction of the
redundant, and find the displacement
fBB
(c) Apply compatibility at the location of the removed redundant
DB0 + fBB´RB = 0
P
P
11. 5.5 INDETERMINATE BEAM WWIITTHH MMUULLTTIIPPLLEE DDEEGGRREEEESS
OOFF IINNDDEETTEERRMMIINNAACCYY
A
w/u.l
B C D E
RB RC RD
DB0 DC0
DD0
(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., DBO, DCO and DDO
12. (b) Apply unit loads at B, C and D, in a sequential manner and
determine deformations at B, C and D, respectively.
A
B C D E
fBB
fCB fDB 1
A
B C D E
fBC
fCC fDC
A
1
B C D E
fBD
fCD fDD
1
13. (c ) Establish compatibility conditions at B, C and D
DBO + fBBRB + fBCRC + fBDRD = 0
DCO + fCBRB + fCCRC + fCDRD = 0
DDO + fDBRB + fDCRC + fDDRD = 0
14. 5.4.2 When support settlements occur:
A
B C D E
DB DC DD Support settlements
Compatibility conditions at B, C and D give the following equations:
DBO + fBBRB + fBCRC + fBDRD = DB
DCO + fCBRB + fCCRC + fCDRD = DC
DDO + fDBRB + fDCRC + fDDRD = DD
w / u. l.
15. 55..55 TTRRUUSSSS SSTTRRUUCCTTUURREESS
80 kN
C
60 kN
D
A B
80 kN
C
60 kN
D
A 1 2
B
Primary structure
(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
16. 5.5 Truss Structures (Cont’d)
(b)Find deformation Dalong AB:
ABO D=S (FuL)/AE
ABO 0ABF= Force in member of the primary structure due to applied
0 load
u= Forces in members due to unit force applied along AB
AB(c) Determine fAB uABL
=å
deformation 2
along AB due to unit load applied along
AB:
,
AB
AE
(d) Apply compatibility condition along AB:
DABO+fAB,ABFAB=0
17. (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)
18. 5.6 TEMPERATURE CCHHAANNGGEESS AANNDD FFAABBRRIICCAATTIIOONN
EERRRROORR
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