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CE8502 STRUCTURAL ANALYSIS I
UNITI STRAIN ENERGY METHOD
Determination of Static and Kinematic Indeterminacies – Analysis of continuous
beams, plane frames and indeterminate plane trusses by strain energy method (up to
two degree of redundancy).
UNITII SLOPE DEFLECTION METHOD
Slope deflection equations – Equilibrium conditions - Analysis of continuous beams
and rigid frames – Rigid frames with inclined members - Support settlements-
symmetric frames with symmetric and skew-symmetric loadings.
UNITIII MOMENT DISTRIBUTION METHOD
Stiffness and carry over factors – Distribution and carryover of moments - Analysis of
continuous Beams- Plane rigid frames with and without sway – Support settlement -
symmetric frames with symmetric and skew-symmetric loadings.
UNITIV FLEXIBLITY METHOD
Primary structures - Compatibility conditions – Formation flexibility matrices -
Analysis of indeterminate pin- jointed plane frames, continuous beams and rigid
jointed plane frames by direct flexibility approach.
UNITV STIFFNESS METHOD
Restrained structure –Formation of stiffness matrices - equilibrium condition -
Analysis of Continuous Beams, Pin-jointed plane frames and rigid frames by direct
stiffness method.
TOTAL: 45 PERIODS
TEXTBOOKS:
1. Bhavikatti, S.S,Structural Analysis,Vol.1,& 2, Vikas Publishing House
Pvt.Ltd.,NewDelhi-4, 2014.
2. Bhavikatti, S.S, Matrix Method of Structural Analysis, I. K. International
Publishing House Pvt.Ltd.,New Delhi-4, 2014.
3. Vazrani.V.N And Ratwani, M.M, Analysis of Structures, Vol.II, Khanna
Publishers, 2015.
4. Pandit G.S.andGupta S.P.,Structural Analysis–AMatrix Approach, Tata McGraw
Hill Publishing Company Ltd.,2006
REFERENCES:
1. Punmia. B.C, Ashok Kumar Jain & Arun Kumar Jain, Theory of structures, Laxmi
Publications, New Delhi, 2004.
2. William Weaver, Jrand James M.Gere, Matrix analysis of framed structures, CBS
Publishers & Distributors, Delhi,1995
3. Hibbeler, R.C.,Structural Analysis, VII Edition, Prentice Hall, 2012.
4. Reddy.C.S, “Basic Structural Analysis”,Tata McGraw Hill Publishing
Company,2005.
5. Rajasekaran. S, & G. Sankarasubramanian., “Computational Structural Mechanics”,
PHI Learning Pvt. Ltd, 2015
6. Negi L.S.and Jangid R.S.,Structural Analysis, Tata McGraw Hill Publishing
Co.Ltd.2004.
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UNIT 2 SLOPE DEFLECTION METHOD
(1). A beam ABC, 10m long, fixed at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.
Solution.
(a) Fixed end moments
Treating each span as a fixed beam, the fixed end moments are as follows:
(b) Slope deflection equations
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The end rotations A and C are zero since the beam is fixed at A and C. hence there
is only o ne unknown, B. the ends do not settle and
hence for each span is zero. Let us assume B to be positive. The result will indicate
the correct sign. The slope deflection equations are as follows:
For span AB,
For span BC,
(c) Equilibrium equation
Since there is only one un known, i.e. B, one equilibrium equation is sufficient. For
the joint B, we have
MBA + MBC = 0
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(0.8 EI B + 3.6) + (0.8 EI B + 5.0) = 0 1.6 EI B = 1.4
The plus sign indicates that B is positive (i.e. rotation of tangent at B is clockwise).
(d) Final moments
Substituting the values of EI B in Eqa. (1) to (4), we get
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(2) A beam ABC, 10m long, hinged at ends A and B is continuous over joint B and is loaded as shown in Fig. Using the slope deflection method, compute the end moments and plot the bending moment diagram. Also, sketch the deflected shape of the beam. The beam has constant EI for both the spans.
SOLUTIONS
(a) Fixed end moments
These are the same as calculated in the previous problem:
MFAB = -2.4 KN-m ; MFBA = +3.6 KN-m
MFBC = -5.0 KN-m ; MFCB = +5.0 KN-m
(b) Slope deflection equations.
(c) Equilibrium equations
Since end A is freely supported,
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(d) Final moments : Substituting the values of EI A and EI B inEq. (2), we get
The bending moment diagram and the deflected shape of the beam are shown in the
Fig. Note. The beam is statically indeterminate to single degree only. This
problem has also been solved by the moment distribution method (example
10.2) treating the moment at B as unknown. However, in the4 slope- deflection
method, the slope or rotations are taken as unknowns, and due to this the
problem involves three unknown rotations A, B and C. hence the method of
slope deflection is not recommended for such a problem.
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(3) A continuous beam ABCD consists of three spans and is loaded as shown in
fig. ends A and D are fixed. Determine the bending moments at the supports and
plot the bending moment diagram.
a)Fixed end moments
(b) Slope deflection equation
A and D are zero since ends A and D are fixed.
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(c) Equilibrium equations
At join B, MBA + MBC = 0
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At join C, MCB + MCD = 0
From (I) and (II), we get EI B = -2.03 kN-m and EI C = + 1.26kN-m
(d) Final moments
Substituting this values in Eqs. (1) to (6), we get
The bending moment diagram and the deflected shape are shown in Figure.
4) A continuous beam ABC is supported on an elastic column BD and is loaded
as shown in figure . Treating joint B as rigid, analyze the frame and plot the
bending moment diagram and the deflected shape of the structure.
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(a)Fixed end moments
MFBD = MFDB = 0
(b)Slope deflection equations.
The slopes A and D are zero since ends A and D are fixed.
For span AB
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(c) Equilibrium equations
At join B, MBA + MBC + MBD = 0
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(d) Final moments
Substituting this values in Eqs. (1) to (6), we get
The bending moment diagram and the deflected shape are shown in Figure.
(5) Analyze the rigid frame shown in figure
(a) Fixed end moments
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(b) Slope deflection equations.
(c) Equilibrium equations
For the equilibrium joint B, MBA + MBD + MBC = 0
(2EIB + 2.67) + (EIB -2) + (-4) = 0
3EIB = 3.33
EIB = 1.11
(d) Final moments
Substituting this value of EIB in Eqs. (1) to (4), we get
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The bending moment diagram and
The deflected shapes are shown in Figure.
(6) A portal frame ABCD is fixed at A and D, and has rigid joints at B and C.
The column AB is 3m long. The beam BC is 2m long, and is loaded with uniformly
distributed load of intensity 6 kN/m. The moment of inertia is 2.1 and that of BC
and CD is I (Fig). Plot B.M. diagram and sketch the deflected shape of the frame.
(a) Fixed end moments
Let the joints B and C move horizontally by
(b)Slope deflection equations.
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(c) Equilibrium equations. At joint B,
d )Shear equation
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e) Final moments
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(7) A portal frame ABCD is hinged at A and fixed at D and has stiff joints at B
and C. the loading is as shown in figure. Draw the bending moment diagram and
deflected shape of the frame.
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Unit 3 - MOMENT DISTRIBUTION METHOD
INTRODUCTION AND BASIC PRINCIPLES
Introduction
(Method developed by Prof. Hardy Cross in 1932)
The method solves for the joint mo ments in continuous beams and rigid frames
by successive approxi mation
Statement of Basic Principles
Consider the continuous beam ABC D, subjected to the given loads,
as shown in Figure below. Assume that only rotation of joints occur
at B, C and D, and that no support d isplacements occur at B, C and
D. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D
In order to solve the problem in a su ccessively approximating manner,
it can be visualized to be made up of a continued two-stage problems
viz., that of locking and releasing the joints in a continuous sequence.
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The joints B, C and D are locked in position before any load is applied on the b
eam ABCD; then given loads are applied on the bea m. Since the joints of beam
ABCD are locke d in position, beams AB, BC and CD acts as ind ividual and
separate fixed beams, subjected to the applied loads; these loads develop fixed
ennd moments.
In beam AB
Fixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.m
Fixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m
In beam BC
Fixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62
= -112.5 kN.m
Fixed end moment at C = + (Pab2)/l2 = + (150)(3)(3)2/62
= + 112.5
In beam AB
Fixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.m
Fixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m
Since the joints B, C and D were fix ed artificially (to compute the the fixed-end
moments), now the joints B, C and D are released and allowed to rotate. Due to
the joint release, the joints rotate maintaining the continuous nature of the beam.
Due to the joint release, the fixed en d moments on either side of joints B, C and
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D act in the opposite direction now, and cause a net un balanced moment to occur
at the joint.
These unbalanced moments act at th e joints and modify the joint moments at B,
C a nd D, according to their relative stiffnesses at the respective joints. The joint
moments are distributed t o either side of the joint B, C or D, according to their
reelative stiffnesses. These distributed moments al so modify the moments at the
opposite side of the beam span, viz., at joint A in span AB, at joints B and C in
span BC and at joints C and D in span CD. This modification is dependent on
the carry-over factor (which is equal to 0.5 in this case);
The carry-over moment becomes the unbalanced moment at the joints to
whic h they are carried over. Steps 3 and 4 are repeated t ill the carry-over
or distributed moment beco mes small.
Sum up all the moments at each of the joint to obtain the joint moments.
SOME BASIC DEFINITIONS
In order to understand the five steps mentioned in section 7.3, some words need
to be defined and relevant derivations made.
1Stiffness and Carry-over Factors
Stiffness = Resistance offered by m ember to a unit displacement or rotation at a
point, for given support constraint conditions
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A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Fin
d A and MB.
Using method of consistent defor mations
Considering moment MB,
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MB + MA + RAL = 0
MB = MA/2= (1/2)MA
Carry - over Factor = 1/2
2 Distribution Factor
Distribution factor is the ratio according to which an externally applied
unbalanced moment M at a joint is apportioned to the various m embers mating
at the joint
M = MBA + MBC + MBD
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Modified Stiffness Factor
The stiffness factor changes when t he far end of the beam is simply-supported.
As per earlier equations for deforma tion, given in Mechanics of Solids text-
books.
Solve the previously given proble m by the moment distribution method
Fixed end moments
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Stiffness Factors (Unmodified Stifffness
Distribution Factors
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Computation of Shear Forces
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UNIT IV FLEXIBLITY METHOD
INTRODUCTION
These are the two basic methods by which an indeterminate skeletal structure
is analyzed. In these methods flexibility and stiffness properties of members are
employed. These methods have been developed in conventional and matrix forms.
Here conventional methods are discussed.
suitable number of releases. The number of releases required is equal to
staticalindeterminacy s. Introduction of releases results in displacement
discontinuities at these releases under the externally applied loads. Pairs of unknown
biactions (forces and moments) are applied at these releases in order to restore the
continuity or compatibility of structure.
The computation of these unknown biactions involves solution of? linear
simultaneous equations. The number of these equations is equal to
staticalindeterminacy s. After the unknown biactions are computed all the internal
forces can be computed in the entires tructure using equations of equilibrium and
free bodies of members. The required displacements can also be computed using
methods of displacement computation.
Inflexibility methods inceunknowns are forces at the releases the method is
also called force method. Since computation of displacement is also required at
releases for imposing conditions of compatibility the method is also called
compatibility method. In computation of displacements use is made of flexibility
properties, hence, the method is also called flexibility method.
EQUILIBRIUM and COMPATABILITY CONDITIONS
Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesa
ndmom ents at anyjoint should beequal to zero.
i.e.?H=0;?V=0;?M=0
Forces should be in equilibrium
i.e.?FX=0;?FY=0;?FZ=0 i.e.?MX=0;?MY=0;?MZ=0
Displacement of a structure should be compatable
The compatibility conditions for the supports can be given as
1.Roller Support ?V=0 2.Hinged Support ?V=0, ?H=0
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3.Fixed Support ?V=0, ?H=0, ??=0
DETERMINATE AND INDETERMINATE STRUCTURAL SYSTEMS
If skeletal structure is subjected to gradually increasing loads, without
distorting the initial geometry of structure, that is, causing small displacements, the
structure is said to be stable. Dynamic loads and buckling or instability of structural
system are not considered here. Ifforthestable structure it is possible to find the
internal forces in all the members constituting the structure and supporting reactions
at all the supports provided from statically equations of equilibrium only, the
structure is said to be determinate.
If it is possible to determine all the support reactions from equations of
equilibrium alone the structure is said to be externally determinateelse externally
indeterminate. If structure is externally determinatebutitis not possible to determine
all internal forces then structure is said to be internally indeterminate. There
foreastructural system may be:
(1)Externally indeterminate but internally determinate
(2)Externally determinate but internally indeterminate
(3)Externally and internally indeterminate
(4)Externally and internally determinate
DETERMINATEVs INDETERMINATESTRUCTURES.
Determinate structures can be solving using conditions of equilibrium alone
(?H=0;?V=0 ;?M=0). No other conditions are required.
Indeterminate structures cannot be solved using conditions of equilibrium
because (?H?0; ?V?0;?M?0). Additional conditions are required for solving such
structures. Usually matrix methods are adopted.
INDETERMINACYOF STRUCTURAL SYSTEM
The indeterminacy of a structure is measured as statically (?s) or kinematical
(?k) Indeterminacy.
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?s= P (M - N + 1) - r = PR- r ?k= P (N - 1) + r - s+?k= PM -c P = 6 for space frames
subjected to general loading
P = 3 for plane frames subjected to inplane or normal to plane loading. N = Numberof
nodes in structural system.
M=Numberofmembersofcompletelystiffstructurewhichincludesfoundationas singly
connected system of members.
Incompletely stiff structure thereisnorelease present. In singly connected
system of rigid foundation members there is only one route between any two
points in which tracks are not retraced. The system is considered comprising of
closed rings or loops.
R = Number of loops or rings in completely stiff structure. r = Number of releases
in the system.
c = Number of constraints in the system. R = (M - N + 1)
For plane and space trusses ?sr educes to:?s=M- (NDOF)N+ P
M= Number of members in completely stifftruss.
P = 6 and 3 for space and plane truss respectively
N= Number of nodes in truss.
NDOF = Degrees of freedomat node which is 2 for plane truss and 3 for space truss.
For space truss?s=M- 3N+ 6
For plane truss?s= M- 2 N+ 3
Test for static indeterminacy of structural system
If ?s> 0 Structure is statically indeterminate
If ?s= 0 Structure is statically determinate
and if?s<0 Structure is a mechanism.
It may be noted that structure may be mechanism even if ?s >0 if thereleases
are present in such away so as to cause collapse as mechanism. The situation of
mechanism is unacceptable.
Statically Indeterminacy
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It is difference of the unknown forces (internal forces plusexternal reactions)
and the equations of equilibrium.
Kinematic Indeterminacy
It is the number of possible relative displacement softhenodes in the directions
of stress resultants.
PRIMARY STRUCTURE
A structure formed by the removing the excess or redundant restraints from an
indeterminate structure making it statically determinate is called primary structure.
This is required for solving indeterminate structures by flexibility matrix method.
Indeterminate structure Primary Structure
ANALYSIS OF INDETERMINATE STRUCTURES :BEAMS
1Introduction
Solve statically indeterminate beams of degree more than one.
To solve the problem in matrix notation.
To compute reactions at all the supports.
To compute internal resisting bending moment at any section of the
continuous beam.
Beams which are statically indeterminate to first degree, were considered. If the
structure is statically indeterminate to a degree more than one, then the approach
presented in the force method is suitable.
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Problem 1.1
Calculate the support reactions in the continuous beam ABC due to loading as shown
in Fig.1.1 Assume EI to be constant throughout.
Select two reactions vise, at B(R1 ) and C(R2 ) as redundant, since the given beam
is statically indeterminate to second degree. In this case the primary structure is a
cantilever beam AC. The primary structure with a given loading is shown in Fig. 1.2
In the present case, the deflections (? L)1 and (? L) 2 of the released structure at B
and C can be readily calculated by moment-area method. Thus
(? L) 1 = ? 819.16 / EI
(? L) 2 = ? 2311.875/ EI (1)
For the present problem the flexibility matrix is,
a11= 125/3EI ,a21= 625/6EI
a12= 625/6EI , a22 = 1000/3EI (2)
In the actual problem the displacements at B and Care zero. Thus the
compatibility conditions for the problem may be written as, a11 R1+ a12 R2 + (? L)
1 = 0
a21 R1+ a22 R2+ (? L) 2 = 0(3)
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Substituting the value of E and I in the above equation,
R1 = 10.609 KN and R2 = 3.620 KN
Using equations of static equilibrium, R3 = 0.771 KN m and R4 = ?0.755KN m
Problem 1.2
A Fixed beam AB of constant flexural rigidity is shown in Fig.1.3 The beam is
subjected to auniform distributed load of w moment M=wL2 kN.m. Draw Shear force
and bending moment diagrams by force method.
Fig 1.3 Fixed Beam with R1 and R2 as Redundant
Select vertical reaction (R1)and the support moment(R2) at B as the redundant.
The primary structure in this case is acantilever beam which could be obtained by
releasing the redundant R1 andR2.
The R1 is assumed to positive in the upward direction and R2 is assumed to be
positive in the counterclockwise direction. Now, calculate deflection at B duetoonly
applied loading. Let ( L ) be the transverse deflection at1 B and( L 2 bethe slope
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at B due to external loading. The positive directions of the selected redundant are
shown in Fig.8.3b.
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The deflection(? L1)and(? L2)of the released structure can be evaluated from unit
load method. Thus,
(? L1) =wL4/8EI - 3wL4/8EI = ?wL4/2EI - (1)
(? L2) = wL3/6EI - wL3 /2EI = ? 2wL3/3EI --- -(2)
The negative sign indicates that ( L )is downwards and rotation( is 1 L2) clockwise.
Problem 1.3.
A continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in
addition to a concentrated load of 10kN as shown in Fig.7.5a, Draw bending moment
and shear force diagram. Assume EI to be constant for all members.
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It is observed that the continuous beam is statically indeterminate to first degree.
Choose the reaction at B,RBy as the redundant. The primary structure is a simply
supported beam as shown in Fig.1.11. Now, compute the deflection at B, in the
released structure due to uniformly distributed load and concentrated load. This is
accomplished by unit load method. Thus,
In thenextstep, apply a unit load at B in the direction of
RBy(upwards)and
Calculate the deflection at B of the following structure. Thus(seeFig.7.5c),
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Now, deflection at B in the primary structure due to redundant RB is,
In the actual structure, the deflection at B is zero. Hence, the compatibility equation
may be written as
L+ B=0(4)
The other two reactions are calculated by static equilibrium equations (videFig.
1.13)
RA =7.8125kN
RB =2.8125kN
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UNIT V STIFFNESS METHOD
INTRODUCTION
The given indeterminate structure is first made kinematic ally determinate by
introducing constraints atthenodes. The required number of constraints is equal to
degrees of freedom at the nodes that is kinematic indeterminacy ?k. The kinematic
ally determinate structure comprises of fixed ended members, hence, all nodal
displacements are zero. These results in stress resultant discontinuities at these
nodes under the action of applied loads or in other words the clamped joints are not
in equilibrium.
Inorder to restore the equilibrium of stress resultants at the nodes the nodes
are imparted suitable unknown displacements. The number of simultaneous
equations represen ting joint equilibrium of forces is equal to kinematic
indeterminacy ?k. Solution of these equations gives unknown nodal displacements.
Using stiffness properties of members the memberend forces are computed and
hence the internal forces through out the structure.
Since nodal displacements are unknowns, the method is also called
displacement method. Since equilibrium conditions are applied at the joints the
method is also called equilibrium method. Since stiffness properties of members
are used the method is also called stiffness method.
In the displacement method of analysis the equilibrium equations are written
by expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of freedom
of the structure) are calculated by solving equilibrium equations. The slope -
deflection and moment - distribution methods were extensively used before the high
speed computing era. After the revolution in computer industry, only direct stiffness
method is used.
PROPERTIES OFTHESTIFFNESS MATRIX
The properties of the stiffness matrix are:
It is asymmetric matrix
The sum of elements in any column must be equal to zero.
It is an unstable element therefore the determinant is equal to zero.
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ELEMENT AND GLOBAL STIFFNESS MATRICES
Local co ordinates
In the analysis for convenience we fix the element coordinates coincident with
the member axis called element (or) local coordinates (coordinates defined along the
individual member axis )
Global co ordinates
It is normally necessary to define a coordinate system dealing with the entire
structure is called system on global coordinates (Common coordinate system dealing
with the entire structure)
Transformation matrix
The connectivity matrix which relates the internal forces Q and the external
forces R is known as the force transformation matrix. Writing it in a matrix form,
{Q} =[b]{R}
Where Q=member force matrix/vector, b=force transformation matrix R =
external force/load matrix/ vector
ANALYSIS OF CONTINUOUS BEAMS
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ANALYSIS OF PIN JOINTED PLANE FRAMES
An introduction to the stiffness method was given in the previous Page. The
basic principles involved in the analysis of beams, trusses were discussed. The
problems were solved with hand computation by the direct application of the basic
principles. The procedure discussed in the session (page)
though enlighteningarenot suitable for computer programming. It is
necessary to keesphand computation to a minimum while implementing this
procedure on the computer.
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In this session a formal approach has been discussed which may be
readily programmed on a computer. In this less on the direct stiffness method as
applied to planar truss structure is discussed.
Planetrusses are made up of short thin members inter connected a thin gesto form
triangulated patterns. Ahinge connection can only transmit forces from one member
to another member but not the moment. For analysis purpose, the truss is loaded at
the joints. Hence, a truss member is subjected to only axial forces and the forces
remain constant along the length of the member. The forces in the member at its two
ends must be of the same magnitude but actin the opposite directions for equilibrium
as shown in Fig.2.8
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STIFFNESS MATRIX METHOD
1. What are the basic unknowns in stiffness matrix method?
In the stiffness matrix method nodal displacements are treated as the basic
unknowns for the solution of indeterminate structures.
2. Define stiffness coefficient kij.
Stiffness coefficient 'kij' is defined as the force developed at joint 'i' due to
unit displacement at joint 'j'while all other joints are fixed.
3. What is the basic aim of the stiffness method?
The aim of the stiffness method is to evaluate the values of generalized coor
dinates 'r' knowing the structure stiffness matrix 'k' and nodal loads 'R' through
the structure equilibrium equation.
{R} =[K]{r}
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4. What is the displacement transformation matrix?
The connectivity matrix which relates the internal displacement 'q' and the
external displacement 'r' is known as the displacement transformation matrix 'a'.
{q} =[a]{r}
5. How are the basic equations of stiffness matrix obtained?
The basic equations of stiffness matrix are obtained as:
Equilibrium forces
Compatibility of displacements
Force displacement relationships
6. What is the equilibrium condition used in the stiffness method?
The external loads and the internal member forces must be in equilibrium at the
nodal points.
7. What is meant by generalized coordinates?
For specifying a configuration of a system, a certain minimum no of indepen dent
coordinates are necessary. The least no of independent coordinates that are
needed to specify the configuration is known as generalized coordinates.
8. What is the compatibility condition used in the flexibility method?
The deformed elements fit together at nodal points.
9. Write about the force displacement relationship.
The relationship of each element must satisfy the stress-strain relationship of the
element material.
10. Write the element stiffness for a truss element.
The element stiffness matrix for a truss element is given by
11. Write the element stiffness matrix for a beam element.
The element stiffness matrix for a beam element is given by
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12. Compare flexibility method and stiffness method.
Flexibility matrix method
The redundant forces are treated as basic unknowns.
The number of equations involved is equal to the degree of static indete
rminacy of the structure.
The method is the generalization of consistent deformation
method. Different procedures are used for determinate and indeterminate
structures
Stiffness matrix method
The joint displacements are treated as basic unknowns
The number of displacements involved is equal to the no of degrees of
freedom of the structure
The method is the generalization of the slope deflection method.
The same procedure is used for both determinate and indeterminate
structures.
13. Is it possible to develop the flexibility matrix for an unstable structure?
In order to develop the flexibility matrix for a structure, it has to be stable and
determinate.
14. What is the relation between flexibility and stiffness matrix?
The element stiffness matrix 'k' is the inverse of the element flexibility
matrix 'f' and is given by f=1/k or k =1/f.
15. What are the type of structtures that can be solved using stiffness matrix
method?
Structures such as simply supported, fixed beams and portal frames can be solved
using stiffness matrix method.
16. Give the formula for the size of the Global stiffness matrix.
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The size of the global stiffness matrix (GSM) =No: of nodes x Degrees of free
dom per node.
17. List the properties of the stiffness matrix
The properties of the stiffness matrix are:
It is a symmetric matrix
The sum of elements in any column must be equal to zero.
It is an unstable element there fore the determinant is equal to zero.
18Why is the stiffness matrix method also called equilibrium method or
displacement method?
Stiffness method is based on the superposition of displacements and hence is also
known as the dispalcement method. And since it leads to the equilibrium
equations the method is also known as equilibrium method.
19 If the flexibility matrix is given as
20 Write the n stiffness matrix for a 2D beam element. The stiffness matrix for a 2
D beam element is given by
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