The basic function of any structure is to withstand or resist loads with a small and definite deformation. Numerous classical and matrix methods are available for analysis of a indeterminate structure/continuous beam. Methods like Moment distribution and Kani’s method are iterative type and accuracy of results will be obtained through more iterations. Three moment equation ,Slope deflection method, energy method, etc. are based on solution of linear simultaneous equations. Dr .Harshvadan. S. Patel and Dr. Patil introduce Relative Deformation Coefficient method for analysis of continuous beam. Present study intends to compare this Relative Deformation Coefficient method to the results obtained from widely accepted classical methods like Slope Deflection and Moment Distribution method. The results obtained by the Relative Deformation Coefficient Approach are found to be act in accordance with those found out by implementing the classical methods.
2. Comparative Study of Classical Method and Relative Deformation Coefficient Approach For
Continuous Beam Analysis
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1. INTRODUCTION
Behr R.A., Goodspeed C.H.[1] have reviewed the existing methods of approximate
structural analysis described in various literatures and compared with slope deflection
method, Conventional approximate method and revised approximate method. Behr
R.A., Grotton E.J. and Dwinal C.A.[2] have suggested some assumption in selection
of point of inflection in beam and column so as to obtain values closer to the exact
method. Dr. Terje Haukaas [3] guess the location of inflection points in beams and
column. Khuda S.N. and Anwar A.M.[4] have developed design tables for selection
of concrete beam section and reinforcement when design bending moment and shear
are available and perform parametric study for assumed variation of parameters for
analysis of continuous beams for moment coefficients. Okonkwo, aginam and
Chidolue[5] have suggested that it is possible to express the values of the internal
support moments with a mathematical model. Ibrahim Amer M.[6] presents the
results of a parametric study of the flexural behavior of continuous beam prestressed
with external tendon. An overview of various approximate method was briefly done
by Life John and Dr. M.G. Rajendran[7].This paper also intends to compare revised
method of structural analysis to the values obtained from STAAD. pro. P.R. Patil,
M.D.Pidurkar and R.H.Mohankar [8] presented comparative study of end moments
regarding application of rotation contribution method and moment distribution
method for the analysis of portal frame. Dipak J Varia and DR. Harshvadan S
Patel[9]presented parametric study of an Innovative approximate method of
continuous beam.
Structurally a building may consist of load bearing walls and floors. The floor
slabs may be supported on beams which in turn may be supported on wall or columns.
But, for a multistoried structure a building frame either of steel or of reinforced
concrete is made[10]. This frame is designed for all the vertical and horizontal loads
transmitted to it. The openings between the columns, where necessary will be filled
with brick walls. A frame of this type will consist of columns and beams built
monolithically forming a network. This provides rigidity to the connections of
members.
A structure is generally defined as a physical object comprising elements which
are invariably so positioned and interconnected in space as to enable the overall
structure to function as an integral unit in properly transferring anticipated loads
resulting from its use under service conditions to the ground through foundations.
Thus the basic function of any structure is to withstand or resist loads with a small and
definite deformation[11,12].
2. INDETERMINATE STRUCTURE
In structural design problems , the aim is to determine a configuration of a loaded
system and the formulating interconnections, which impart the structure certain load-
carrying attributes. The desired attributes are the conditions of equilibrium,
compatibility and force- displacement relations of the materials. Structures mainly
categorized in two forms Determinate and Indeterminate. Statically indeterminate
structures are economical as compared to statically determinate structures since the
former are subjected to generally lower maximum stresses. Indeterminate structures
generally have higher stiffness, that means smaller deformations than the comparable
determinate structures. The choice between statically determinate and statically
indeterminate structures depends to a large extent upon the purpose for which a
3. Dipak J. Varia and Dr. Harshvadan S. Patel
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particular structure is required[13,14]. Continuous beam considered as indeterminate
structure.
3. METHODS OF ANALYSIS
Numerous classical and matrix methods are available for analysis of a indeterminate
structure/continuous beam. Methods like Moment distribution and Kani’s method are
iterative type and accuracy of results will be obtained through more iteration. Three
moment equation, Slope deflection method, energy method, etc. are based on solution
of linear simultaneous equations. Dr .Harshvadan. S. Patel and Dr. Patil introduce
Relative Deformation Coefficient method for analysis of continuous beam. This
paper intends to compare this Relative Deformation Coefficient method to the results
obtained from classical methods like Slope Deflection and Moment Distribution
method.
4. NEW TERMS OF RELATIVE DEFORMATION COEFFICIENT
APPROACH
The method is dependent on four new terms.
4.1 Corrected Member Stiffness (K)
Corrected member stiffness of a frame member is multiplication of fixity
coefficient(Cf) with relative flexural stiffness(EI/L) of frame member.
K = Cf X EI/L (1)
Where, Cf= Fixity coefficient.
4.2 Relative deformation co-efficient(Cr)
Relative deformation coefficient is defined as the deformation at far end of a frame
member due to unit deformation applied at near end.
C r= K1/2∑(K1+K2) (2)
Where, K1 and K2 are corrected member stiffness of members meeting at joint.
4.3 Fixity Co-efficient(Cf)
Fixity coefficient gives the fixity provided against rotation by far end. The value of Cf
at near end is always taken as unity while the same at far end is dependent on relative
deformation coefficient Cr at far end. This is computed using following relation.
Cf = 1– Cr/2 (3)
4.4 Actual Deformation(Ad)
Actual deformation of joints is deformation of that joint due to some deformation
applied at any joint. Actual deformation of a joint is computed by multiplying actual
deformation of preceding joint with relative deformation coefficient of the joint and it
is expressed in equation form as under.
Adi = - Ad(i-1) X Cri (4)
where i is = joint index.
4. Comparative Study of Classical Method and Relative Deformation Coefficient Approach For
Continuous Beam Analysis
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5. EXAMPLE FOR ILLUSTRATION
The application depicted here, demonstrates each step of computation in detail by
Relative Deformation Coefficient approach(RDCA).Example of Three span
continuous beam as per Fig.1 is selected and results are obtain by RDCA and
compared with existing classical and widely accepted Slope deflection and Moment
Distribution methods.(Table1,2,3,4,5.)
Figure. 1 Example of Three span continuous beam
Computation for support moment at A
Here aim is to compute moment at support A. Therefore, Joint D is the extreme far
fixed end. Hence 0r DC − = & 1f DC − = .
Numerical value of r CC − represent relative rotation at C, if unit rotation applied at B.
The fixity coefficient at joint where unit rotation applied is always considered to be
unity. Using equation (2)
( )2
CB
r C
CB CD
K
C
K K
− =
+
( / )
2 ( / ) ( / )
f B CB
f B CB f D CD
C I L
C I L C I L
−
− −
=
+
( )
1 2 / 5
1/ 3
2 1 2 / 5 1 / 5
r C
x I
C
x I xI
− = =
+
Fixity coefficient is calculated using equation (3).
1 / 2 1 (1/ 3) / 2 5 / 6f C r CC C− −= − = − =
Numerical value of r BC − represent relative rotation at B, if unit rotation applied at A.
The fixity coefficient at joint where unit rotation applied is always considered to be
unity. Using equation (2)
( )2
BA
r B
BA CD
K
C
K K
− =
+
( / )
2 ( / ) ( / )
f A BA
f A BA f C BC
C I L
C I L C I L
−
− −
=
+
( )
1 / 6
1/ 6
2 1 / 6 5 / 6 2 / 5
r B
xI
C
xI x I
− = =
+
5. Dipak J. Varia and Dr. Harshvadan S. Patel
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Fixity coefficient is calculated using equation (3).
1 / 2 1 (1/ 6) / 2 11/12f B r BC C− −= − = − =
Actual deformation (AD), here rotation, of each joint is calculated using the equation
(4)
1D AA − =
1 1/ 6 1/ 6D B D A r BA A xC x− − −= − = − = −
( 1/ 6) 1/ 3 1/18D C D B r CA A xC x− − −= − = − − =
1/18 0 0D D D C r DA A xC x− − −= − = − =
Calculation of Fixed End Moments(FEM)
2 2
/12 2 6 /12 6.0 .AFEM wL x kN m= = =
2 2 2
/12 /BFEM wL Pab L= − +
2 2 2
2 6 /12 5 3 2 / 5 3.6 .x x x kN m= − + = −
2 2
/ / 8CFEM Pba L PL= − +
2 2
5 2 3 / 5 8 5 / 8 1.4 .x x x kN m= − + =
( )A DM FEMxA= ∑
[ ](6 1) ( 3.6 1/ 6) (1.4 1/18) (0 5)x x x x= + − − + + − 6.6778 .AM kN m=
Above calculation are depicted in tabular form as under:
Table 1 Support moment at A
Joint A B C D
rC --- 1/6 1/3 0
fC --- 11/12 5/6 1
DA 1 -1/6 1/18 0
FEM 6.0 -3.60 1.40 -5.0
DFEMxA 6.0 0.60 0.0778 0.0
MA=6.6778kN.m (Ref. Ans: MA=6.7444kN.m)
Computation for support moment at B
To compute MB, a unit rotation at joint B should be computed in two parts. Hence
D BAA − and D BCA − are rotations applied at B. both are in positive direction and sum of
D BAA − and D BCA − should be unity. Here support at A and D are extreme for fixed
supports. Therefore, 0r AC − = , 1f DC − = and 0r DC − = & 1f DC − = .
Numerical value of r CC − represent relative rotation at C, if unit rotation applied at B.
The fixity coefficient at joint where unit rotation applied is always considered to be
unity. Using equation (2)
6. Comparative Study of Classical Method and Relative Deformation Coefficient Approach For
Continuous Beam Analysis
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( )2
CB
r C
CB CD
K
C
K K
− =
+
( / )
2 ( / ) ( / )
f B CB
f B CB f D CD
C I L
C I L C I L
−
− −
=
+
( )
1 2 / 5
1/ 3
2 1 2 / 5 1 / 5
r C
x I
C
x I xI
− = =
+
Fixity coefficient is calculated using equation (3).
1 / 2 1 (1/ 3) / 2 5 / 6f C r CC C− −= − = − =
Rotation at joint B are dependent on corrected stiffness of member and it is calculated
as per following:
1 BA
D BA
BA BC
K
A
K K
− = −
+
( / )
1
( / ) ( / )
f A BA
f A BA f C BC
C I L
C I L C I L
−
− −
= −
+
1 / 6
1 2 / 3
1 / 6 5 / 6 2 / 5
D BA
xI
A
xI x I
− = − =
+
Similarly,
1 BC
D BC
BC BA
K
A
K K
− = −
+
( / )
1
( / ) ( / )
f C BC
f C BC f A BA
C I L
C I L C I L
−
− −
= −
+
5 / 6 2 / 5
1 1/ 3
5 / 6 2 / 5 1 / 6
D BC
x I
A
x I xI
− = − =
+
Actual deformation at other joints:
2 / 3 0 0D A D BA r AA A xC x− − −= − = − =
1/ 3 1/ 3 1/ 9D C D BC r CA A xC x− − −= − = − = −
( 1/ 9) 0 0D D D C r DA A xC x− − −= − = − − =
Calculation of Fixed End Moments(FEM)
Sign convention is reversed if FEMs are left of selected joint.
2 2
/12 2 6 /12 6.0 .AFEM wL x kN m= − = − = −
2 2
/12 2 6 /12 6.0 .BAFEM wL x kN m= = − =
2 2 2 2
/ 5 3 2 / 5 2.4 .BCFEM Pab L x x kN m= = =
2 2
/ / 8CFEM Pba L PL= − +
2 2
5 2 3 / 5 8 5 / 8 1.4 .x x x kN m= − + =
7. Dipak J. Varia and Dr. Harshvadan S. Patel
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/ 8 8 5 / 8 5.0 .DFEM PL x kN m= − = − = −
( )B DM FEMxA= ∑
[( 6 0) (6 2 / 3) (2.4 1/ 3)x x x= − + + (1.4 1/ 9) ( 5 0)]x x+ − + −
4.6444 .AM kN m=
Above calculation are depicted in tabular form as under:
Table 2 Support moment at B
Joint A B C D
rC 0 --- --- 1/3 0
fC 1 --- --- 5/6 1
DA 0 2/3 1/3 -1/9 0
FEM -6.0 6.0 2.4 1.40 5.0
DFEMxA 0.0 4.0 0.8 -0.1556 0.0
MB=4.6444kN.m (Ref. Ans:MB=4.5112kN.m)
Similarly, using present approach, MC & MD are computed and presented in tabular
form as under.
Computation for support moment at C
Table 3 Support moment at C
Joint A B C D
rC 0 6/17 --- --- 0
fC 1 14/17 --- --- 1
DA 0 -2/15 17/45 28/45 0
FEM -6.0 3.6 3.6 5.0 -5.0
DFEMxA 0.0 -0.48 1.36 3.1111 0.0
MC=3.9911kN.m (Ref. Ans:MC=3.5378kN.m)
8. Comparative Study of Classical Method and Relative Deformation Coefficient Approach For
Continuous Beam Analysis
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Computation for support moment at D
Table 4 Support moment at D
Joint A B C D
rC 0 6/17 17/90 ---
fC 1 14/17 163/180 ---
DA 0 1/15 -17/90 1
FEM -6.0 3.60 -1.40 5.0
DFEMxA 0.0 0.24 0.2644 5.0
MD=-5.5044kN.m (Ref. Ans:MD=-5.7311kN.m)
6. SUMMARY OF RESULTS
The same three span continuous beam as per Fig.1 is selected and results are obtain by
existing classical Slope deflection method and compared with RDC method and
results are depicted in Table-5.
Table 5 Comparision of support moments with RDCA
Support moment
Slope deflection
method
Moment
Distribution method
RDC Approach
MA 6.7444 6.7444 6.6778
MB 4.5112 4.5112 4.6444
MC 3.5378 3.5378 3.9911
MD 5.7311 5.7311 5.5044
7. CONCLUSIONS
This study gives an introduction to the newly developed Relative Deformation
Coefficient Approach. The calculation steps for solving three span continuous beam
has been demonstrated by suitable example. RDC approach can be use to solve
continuous beam with more than three spans. The results obtained by RDCA are
compared with more popular and classical Slope deflection and Moment Distribution
methods. The results obtained are act in accordance with the standard solutions. In
Moment Distribution and Slope Deflection methods one has to carry out iterations or
to solve equations which is tedious and most time consuming..Design engineers can
use this method effectively for calculating moment without burdensome procedure of
performing /solving simultaneous equations and iterations. One has to accept that
Relative Deformation Coefficient method is userfriendly, simple ,fast and accurate.
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