Structure Analysis II
STRUCTURAL ANALYSIS II
CE 1352
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
By
R.REVATHI, M. Tech., STRUCTURAL
 Lecturer
 Department of Civil Engineering

PITS
VI Semester
2011-2012
OBJECTIVES
• This Course aims at teaching the students the concept
of analyzing indeterminate structure using classical
and up to date methods.
• It provides students with an understanding of the
methods of analyzing indeterminate structure:
– The force method of analysis
– The Displacement method
• Slope deflection
• Moment distribution
• Stiffness Method (An Introduction to The Finite Element
Method)
UNIT I FLEXIBILITY METHOD FOR INDETERMINATE FRAMES
Equilibrium and compatibility – Determinate Vs Indeterminate structures –
Indeterminacy – Primary structure – Compatibility conditions – Analysis of
indeterminate pin-jointed plane frames, continuous beams, rigid jointed plane frames
(with redundancy restricted to two).
UNIT II MATRIX STIFFNESS METHOD
Element and global stiffness matrices – Analysis of continuous beams – Co-ordinate
transformations – Rotation matrix – Transformations of stiffness matrices, load vectors
and displacements vectors – Analysis of pin-jointed plane frames and rigid frames.
UNIT III FINITE ELEMENT METHOD
Introduction – Discretisation of a structure – Displacement functions – Truss element –
Beam element – Plane stress and plane strain Triangular elements.
UNIT IV PLASTIC ANALYSIS OF STRUCTURES 9
Statically indeterminate axial problems – Beams in pure bending – Plastic moment of
resistance – Plastic modulus – Shape factor – Load factor – Plastic hinge and
mechanism – Plastic analysis of indeterminate beams and frames – Upper and lower
bound theorems
UNIT V SPACE AND CABLE STRUCTURES 9
Analysis of Space trusses using method of tension coefficients – Beams curved in
plan Suspension cables – Cables with two and three hinged stiffening girders.
L: 45 T: 15 Total: 60
TEXT BOOKS
1. Coates R.C., Coutie M.G. and Kong F.K., “Structural Analysis”, ELBS and
Nelson, 1990.
2. Negi, L.S. and Jangid, R.S., “Structural Analysis”, Tata McGraw-Hill
Publications, 2003.
REFERENCES
1. Ghali, A., Nebille, A.M. and Brown, T.G., “Structural Analysis” A Unified
Classical and Matrix approach”, 5th Edition, Spon Press, 2003.
2. Vazirani Vaidyanathan, R. and Perumail, P., “Comprehensive Structural
Analysis
– Vol. I and II”, Laxmi Publications, 2003
INTRODUCTION
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What is statically DETERMINATE structure?
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When all the forces (reactions) in a structure can be
determined from the equilibrium equations its called statically
determinate structure
Structure having unknown forces equal to the available
equilibrium equations
No. of unknown = 3
No. of unknown = 6
No. of equilibrium equations = 3
No. of equilibrium equations = 6
3 = 3 thus statically determinate
6 = 6 thus statically determinate
INTRODUCTION
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What is statically INETERMINATED structure
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Structure having more unknown forces than available
equilibrium equations
Additional equations needed to solve the unknown reactions
No. of unknown = 4
No. of unknown = 10
No. of equilibrium equations = 3
No. of equilibrium equations = 9
4  3 thus statically Indeterminate
10  9 thus statically Indeterminate
INDETERMINATE STRUCTURE
Why we study indeterminate structure
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Most of the structures designed today are statically
indeterminate
Reinforced concrete buildings are considered in
most cases as a statically indeterminate structures
since the columns & beams are poured as
continuous member through the joints & over the
supports
More stable compare to determinate structure or in
another word safer.
In many cases more economical than determinate.
The comparison in the next page will enlighten
more
CONTRAST
Stress
Deflection
Determinate Structure
Indeterminate Structure
Considerable compared to
indeterminate structure
Generally smaller than determinate
structure
P
P
4
PL3
48EI
1
High moment caused thicker
member & more material needed
PL3
192EI
Less moment, smaller cross section
& less material needed
P
P
2
1
PL
4
PL
8
10
Stability in case of over load
CONTRAST
Determinate Structure
Indeterminate Structure
Support will not develop the
horizontal force & moments that
necessary to prevent total collapse
Will develop horizontal force &
moment reactions that will hold
the beam
No load redistribution
Has the tendency to redistribute
its load to its redundant supports
When the plastic hinge formed
certain collapse for the system
When the plastic hinge formed
the system would be a
determinate structure
P
P
Plastic Hinge
Plastic Hinge
CONTRAST
Differential
Displacement
Temperature
Determinate Structure
No effect & no stress would be
developed in the beam
P
No effect & no stress would be
developed
P
Indeterminate Structure
Serious effect and stress would be
developed in the beam
P
Serious effect and stress would be
developed
P
Distinctive Features of Suspension Bridge
Major element is a flexible cable, shaped and supported in such a way that it •
transfers the loads to the towers and anchorage
This cable is commonly constructed from High Strength wires, either spun in situ •
or formed from component, spirally formed wire ropes. In either case allowable
stresses are high of the order of 600 MPA
The deck is hung from the cable by Hangers constructed of high strength ropes •
in tension
As in the long spans the Self-weight of the structures becomes significant, so •
the use of high strength steel in tension, primarily in cables and secondarily in
hangers leads to an economical structure.
The economy of the cable must be balanced against the cost of the associated •
anchorage and towers. The anchorage cost may be high where foundation
material is poor
Distinctive Features of Suspension Bridge
The main cable is stiffened either by a pair of stiffening trusses or by a system of •
girders at deck level.
This stiffening system serves to (a) control aerodynamic movements and (b) limit •
local angle changes in the deck. It may be unnecessary in cases where the dead
load is great.
The complete structure can be erected without intermediate staging from the •
ground
The main structure is elegant and neatly expresses its function.•
It is the only alternative for spans over 600m, and it is generally regarded as •
competitive for spans down to 300m. However, shorter spans have also been
built, including some very attractive pedestrian bridges
The height of the main towers can be a disadvantage in some areas; for •
example, within the approach road for an AIRPORT
Components of a Suspension Bridge
Anchor Block: Just looking at the figure we can compare it as a dead
man having no function of its own other than its weight.
Suspension girder: It is a girder built into a suspension bridge to
distribute the loads uniformly among the suspenders and thus to reduce
the local deflections under concentrated loads.
Suspenders: a vertical hanger in a suspension bridge by which the road
is carried on the cables
Tower: Towers transfers compression forces to the foundation through
piers.
Saddles: A steel block over the towers of a suspension bridge which
acts as a bearing surface for the cable passing over it.
Cables: Members that take tensile forces and transmit it through
saddles to towers and rest of the forces to anchorage block.
•
•
•
•
•
•
Suspension Bridge
Suspension Bridge
Suspension Bridge
The deck of a suspension bridge is usually suspended by vertical hangers,
though, some bridges, following the example of the Severn bridge, use inclined
ones to increase stability. But the structure is essentially flexible, and great effort
must be made to withstand the effects of traffic and wind. If, for example, there
is a daily flow of traffic across a bridge to a large city on one side, the live load
can be asymmetrical, with more traffic on one side in the morning, and more
traffic on the other side in the evening. This produces a periodic torsion, and the
bridge needs to be strong enough to resist the possible effects of fatigue.
We can list the main parts of Suspension bridge
Two towers
Suspended structure
Two main cables
Many hanger cables
Two terminal piers
Four anchorages
5. ANALYSIS OF INDETERMINATE
STRUCTURES BY FORCE METHOD
5.1 ANALYSIS OF INDETERMINATE
STRUCTURES BY FORCE METHOD - AN
OVERVIEW
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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
5.2 INTRODUCTION
5.2 Introduction
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


While analyzing indeterminate structures, it is necessary to satisfy
(force) equilibrium, (displacement) compatibility and forcedisplacement 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
5.2 INTRODUCTION

(CONT’D)
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 forcedisplacement 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 forcedisplacement relationships; it treats the displacements as unknowns Two available methods are Slope Deflection Method and
Stiffness (Matrix) method
5.3 METHOD OF CONSISTENT DEFORMATION
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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
5.4 INDETERMINATE BEAMS
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).
5.4 INDETERMINATE BEAMS
(CONT’D)
(a) Release the vertical support at B:
y
x
L/2
P
C
P
B
B
B
=
B
+
C
L/2
L
RB
BB=RB*fBB
Applied in +ve
direction
The governing compatibility equation obtained at B is,
B + 'BB = 0
 B  ( RB )  ( f BB )  0
 RB    B / f BB
From earlier analyses,
 B   P ( L / 2) 3 /(3EI )  [ P ( L / 2) 2 /( 2 EI )]  ( L / 2)
  PL3 /( 24 EI )  PL3 /(16 EI )
 (5 / 48)( PL3 / EI )
f BB  L3 /(3EI )
RBB  [(5 / 48)( PL3 / EI )] /[ L3 /(3EI )]  (5 / 16) P
fBB = displacement per unit load (applied in +ve direction)
5.4 INDETERMINATE BEAM (Cont’d)
5.4.2 Propped cantilever - Redundant support moment released
L/2
A
P
L
=
B
(b) Release the moment constraint at a:
P
A
A
B
A
+
MA
Primary structure
B
A=MAAA
Redundant MA applied
Governing compatibility equation obtained at A is,
  ( M )  ( )  ,  = rotation per unit moment
A
A
M A  
AA
AA
A
 AA
From known earlier analysis,
 AA  
PL2
(16 EI )
[under a central concentrated
load]
 AA  (1)[ L /( 3EI )]
This is due to the fact that +ve moment causes a –ve rotation
M A  [ PL2 /(16 EI)] /[  L/(3 EI)]
 (3 / 16) PL
5.4.3 OVERVIEW OF METHOD OF CONSISTENT
DEFORMATION
To recapitulate on what we have done earlier,
I. Structure with single degree of indeterminacy:
P
A
B
(a) Remove the redundantRBto make the structure determinate
(primary structure)
P
A
B
Bo
(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
B0 + fBBRB = 0
5.5 INDETERMINATE BEAM WITH MULTIPLE
DEGREES OF INDETERMINACY
w/u.l
A
B
RB
C
RC
D
E
RD
(a) Make the structure
determinate (by
releasing the supports
B0

D0
C0
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
(b) Apply unit loads at B, C and D, in a sequential manner and
determine deformations at B, C and D, respectively.
A
B
fBB
fCB
C
D
E
D
E
fDB
1
A
B
fBC
fCC
C
1 fDC
A
B
fBD
fCD
C
D
fDD
1
E
(c ) Establish compatibility conditions at B, C and D
BO + fBBRB + fBCRC + fBDRD = 0
CO + fCBRB + fCCRC + fCDRD = 0
DO + fDBRB + fDCRC + fDDRD = 0
5.4.2 When support settlements occur:
w / u. l.
A
B
C
B
D
C
E
Support
D
settlements
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
5.5 TRUSS STRUCTURES
80 kN
60 kN
80 kN
C
60 kN
D
A
C
D
B
A
1
2
B
Primary structure
Remove the redundant member (say AB) and make the structure (a)
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
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
load
uAB= Forces in members due to unit force applied along AB
(c) Determine deformation along AB due to unit load applied along
AB:
u2 L
f
  AB
AB, AB
AE
(d) Apply compatibility condition along AB:
ABO+fAB,ABFAB=0
(d) Hence determine FAB
(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)
5.6 TEMPERATURE CHANGES AND
FABRICATION ERROR
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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