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. www.studymaterialz.in
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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
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.