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Chapter 1
Plastic moment redistribution
Abrham E.
&
Sophonyas A.
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The purpose of any analysis is to know how the
structure responds to a given loading and therebyevaluate the stresses and deformations.
Then the sections will be designed to resist the
internal forces induced by external loads so that
the stresses and deformations developed are
within permissible limits.
Most reinforced concrete structures are designed
for internal forces found by elastic theory withmethods such as slope deflection, moment
distribution, and matrix analysis.
INTRODUCTION
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There is an apparent inconsistency in determining thedesign moments based on an elastic analysis, whiledoing the design based on a limit state designprocedure.
The structural analysis is based on linear elastictheory, whereas the structural design is based oninelastic section behavior.
It should be noted, however, that there is no realinconsistency if the moment-curvature (M- )relationship remains linear even under ultimate loads.
According to EBCS-2, 1995 for the analysis in the
Ultimate Limit State; Plastic,
Non-linear and
Linear elastic theory may be applied.
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In materials obeying Hooke’s law, the load
deformation relationship is linear.
This behavior is simple to analyze.
For instance, the use of the principle of
superposition is allowed only by assuming linearbehavior of structures.
In linear elastic analysis, there are two different
ways of analyzing a statically indeterminatestructure:
1. The displacement or Equilibrium methods
2. The force or compatibility methods
Linear Elastic Analysis of Structures
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The non-linear analysis procedures are more
complex and therefore very time consuming. It is beyond the scope of this course.
Nonlinear structural problems usually fall into one
of the following main categories; Large deformations associated with geometric
nonlinearity ,
Nonlinear material behavior associated with material
nonlinearity , and The combination of geometric and material
nonlinearities.
Non-Linear Analysis of Structures
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Most structural members, under gradually increasing
strain, have an elastic stage & a plastic stage.
When a beam yields in bending, an increase in
curvature does not produce an increase in moment
resistance. Analysis of beams and structures made of
such flexural members is called Plastic Analysis. For a proper determination of the distribution of
bending moments for loading beyond the yielding
stage at any section, inelastic analysis is used.
This is generally referred to as limit analysis, whenapplied to reinforced concrete framed structures, and
plastic analysis when applied to steel structures.
Plastic Analysis of Structures
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Curvature is defined as the angle change per unit
length at a given location.
In ordinary design, the relation between moment
applied to a given beam section and the resulting
curvature, is not needed explicitly. It is important:
to study the ductility of members
to understand the development of plastic hinge, and
to account for the redistribution of elastic momentsthat occurs in most reinforced concrete structures
before collapse.
Moment curvature relationship
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Point ‘O' is the center
of curvature of the
deflected curve.
(rho), radius of
curvature, is the
distance from the
curve to the center of
curvature.
If we take a portion of a beam:
Curvature (kappa) is defined as the reciprocal ofthe radius of curvature.
Thus, =
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And from the geometry
of triangle oab,
*d = ds
=1
=
Point O is located much further than it is located in the
figure, because most beams give very small deflections and
have nearly flat deflection curve.
Curvature is the measure of
how sharply a beam is bent
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To evaluate normal strain in beams subjected to
pure bending, consider a line cd located at a
distance ‘y’ from the neutral axis.
Fig. Deformed beam under pure bending
The longitudinal line cd had
the same length with ab on
the neutral axis beforebending, but after bending
the curve cd will shorten.
And this shortened curve is
at a distance ( - y) fromthe center of curvature.
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Before bending of the beam =
= ∗
After bending the size of remains constant but
′ = .
Therefore,
Δ =
′
=
=
=−
=−
Therefore, the strain curvature relation is:
=
=
For linear elastic materials, by substituting hook’s law foruniaxial stress ( = *E) into the above equation, we get:
σx = E ∗ εx = E ∗ y
ρ= E ∗ κ ∗ y
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From the study of mechanics of materials, we have
the flexural formula
σx =M ∗ y
I
The moment curvature relation can be stated, from
the above equations as follows:
σx
=M ∗ y
I=
E ∗ y
ρ= E ∗ κ ∗ y
σx =M ∗ y
I=
E ∗ y
ρ= E ∗ κ ∗ y
Implies that: κ =
=
This is relation is known as
moment - curvature equation
From the relation the curvature is: directly proportional to the bending moment M and
inversely proportional to the quantity EI, which is
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Drawing Moment - Curvature diagram
The Bernoulli hypothesis of plane sections remaining
plane after bending holds also for analysis based onnon-linear - relationship
It is to be noted that for RC cross sections in state II,it holds true only on the average or “smeared” sense
along the axis of the beam.At a crack location, the hypothesis is violated, but is
not considered in the analysis
We need the following two important equations to
draw the Moment(M) - Curvature(κ ) diagram
=
= Strain - curvature equation
κ =
=
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STATE I:
The stiffness used in linear analysis for Reinforced
Concrete structures is generally determined usinguncracked section
STATE II:
The stiffness may be determined on the basis ofcracked section (linear elastic analysis with reduced
stiffness)
The important points along the moment curvature
diagram at which M and κ are to be calculated areas follows:
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Theoretical Moment-Curvature diagram for R.C.C.
Because the tension stiffening effect of the
concrete is ignored
In reality, the uncracked concrete in
the cracked section shares in
resisting flexural tension, resulting
in what is known as tension
stiffening.
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Just before cracking (Pt. 1):
In this case, concrete also takes tension and the moment isthe cracking moment. We find the second moment of area
of the uncracked transformed section
κ =
∗
Mcr = 1.7 ∗ f ct ∗ Z
Mcr - Theoretical moment which causes cracking
Z - Section modulus
f ctk - Characteristic Tensile strength of Concrete
Just after cracking (Pt. 2):
In this case, concrete does not take any tension and themoment is the cracking moment.
We find the second moment of area of the crackedtransformed section
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κ =
∗
At yielding of the steel (Pt. 3): In this case, tension steel reaches yield point but
concrete does not reach maximum strain - under
reinforced beam.
κ = κ =
=
=
+
At Ultimate Limit State of the R.C.C. section (Pt. 4):
In this case either = 3.5
0
/00
or =10.00
/00
, forconcrete and steel respectively.
κ = κ =
=
= +
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By trial and error, we canfind the strain in the steel
or the maximum strain in
concrete from the relation
C = T.
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The figure below shows the calculated and experimentally
registered curves M- κ for two different cross-sections, with
steel f yk = 560 MPa and concrete C60/75.
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The actual moment curvature relationship measured in
beam tests differ some-what from the theoretical curve
shown above, mainly because, the tension stiffening
effect of the concrete is ignored.
The theoretical moment curvature can be modified to
resemble the actual one by considering the concrete share
in resisting flexural tension.
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The above Moment-Curvature diagram can be simplified
as follows , for the purpose of limit analysis.
With the idealized M – relation, the ultimate moment of
resistance (MuR) is assumed to have been reached at a‘critical’ section in a flexural member with the yielding of
the tension steel.
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From the analysis of a section, with the same section
size and that has different tension reinforcement, the
moment capacity at yield and at ULS is shown below:
Size of
reinforcement
Cross
sectionsize (mm.)
Moment Capacity (kNm.)
When the steelyields (My )
When the section is atULS (MuR)
310
W i d t h = 2 0 0
T o t a l d e p t h = 4 0 0 26.568 27.953
312 37.299 39.322
3
14 49.450 51.953
316 62.794 65.343
320 91.364 92.670
324 Compression failure 112.923
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With the idealized M – relation,
On further straining (increase curvature: > y ), the
moment at the section cannot increase.
However, the section ‘yields’ & the curvature continues
to increase under a constant moment (M = Mu).
Size ofreinforcement
M(kNm)
Curvature (1/mm.)When the steelyields (y )
When the section is at
ULS (u)
310 27.953 6.890E-06 3.273E-05
312 39.322 7.487E-06 3.453E-05
314 51.953 7.487E-06 3.689E-05
316 65.343 8.987E-06 3.061E-05
320 92.670 1.133E-05 1.959E-05
324 112.923 1.447E-05 1.447E-05
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Idealized Moment Curvature relation
The assumption generally
made in limit analysis is
that the moment-curvature relation is an
idealized bilinear elasto-
plastic relation.
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If there is no moment redistribution:
Region 3
Has both the ductile and brittle portion and the boundary
line Connect
= 3.50 0/0
0 and
= 4.3125 0/0
0
Region 2
Ductile Failure
Region 4
Brittle Failure
Moment Curvature curve and ductility
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From the figure above, it is easy to understand that:
If the R.C.C. member is under reinforced section,
It has larger change in angle per unit length or
It has enough bent before failure.
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The main advantage underlying under-reinforced
sections is that they exhibit ductile behavior
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0.00
20.00
40.00
60.00
80.00
100.00
120.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Moment Curvature relation 3 d 10
3 d 12
3 d 14
3 d 16
3 d 20
3 d 24
Fig. Moment curvature diagrams for 20cm x 40cm RC section with differentreinforcement
Size of
reinforcement
Cross section
size
Strain at ULSFailure Mode
s1 (0/00) cm (
0/00)
310
W i d t h = 2 0 0
T o
t a l d e p t h = 4 0
010.00 1.85 STEEL
FAILURE312 10.00 2.46
314 10.00 3.28
316 7.49 3.50CONCRETE
FAILURE320 3.49 3.50
324 1.64 3.50
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Size of
reinforcement
Cross
section size
Strain at ULSFailure Mode
s1 (0/00) cm (
0/00)
310
W i d t h = 2 0
0
T o t a l d e p t h =
4 0 0 10.00 1.85
TENSION
FAILURE
312 10.00 2.46314 10.00 3.28
316 7.49 3.50
320 3.49 3.50
3
24 1.64 3.50 COMPRESSION FAILURE
Size of
reinforcement
Cross
section size
Strain at ULSFailure Mode
s1 (0/00) cm (
0/00)
310
W i d t h = 2 0 0
T
o t a l d e p t h =
4 0 0 10.00 1.85
DUCTILEFAILURE
312 10.00 2.46314 10.00 3.28
316 7.49 3.50
320 3.49 3.50 BRITTLE
FAILURE324 1.64 3.50
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• Exercise : Determine the M- relationship forthe singly reinforced beam cross section
shown C-25 concrete and S-460 steel
Moment Curvature Diagram
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• The steps are as follows
I. choose a value for s1II. take trial NA depth x
III. determine c from s1 and xIV. determine stress resultants Cc and Ts. Are
they in equilibrium? If no, adjust x and go tostep (ii)
V. iterate until Cc = Ts. The corresponding straindistribution represents one point on the M- diagram
Moment Curvature Diagram
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εcm ≤ 2‰ and N.A. within the section
εcm ≥ 2‰ and N.A. within the section
Moment Curvature Diagram
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Determine cracking moment Mcr from:Mcr = (f ctdIut)/(h-x); where Iut is the moment
of inertia of uncracked transformed section
Es= 200 GPa, Ec=29 GPa n = 200/29 = 6.9
(n-1)As = (6.9-1)397.11 =
2343 mm2
200
450
Moment Curvature Diagram
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• x=(2005002502343450)/(2005002343) =254.58 mm from top
• Iut = 2174907745 mm6
• Mcr = (1.0)(2174907745)/(500-254.58) =8861982.5 Nmm = 8.862 kNm
• cr = c1/245.42; where c1=f ctd/Ec = 1.0/29000 =0.0344828 (c1=bottom fiber strain)
• cr = 0.0344828 10-4/(245.4210-3) =
1.4050510-4 (1/m)
Moment Curvature Diagram
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• My (at yielding of reinforcement)
d
Cc
Ts=397.11400 = 158844 N
x
yd = 0.002
c = ?
Moment Curvature Diagram
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kx =x/d xc( ) c=cm/12(
6-cm)kx
Cc =(8-
cm)/(4(
6-cm)
0.4 180 1.333333 0.2074 2111493
0.3 135 0.8752 0.1102 112375
0.35 157.5 1.077 0.1546 157689
0.36 162 1.125 0.1645 167773
0.355 159.75 1.101 0.1595 162684
0.352 158.4 1.086 0.1566 159672
0.351 157.95 1.0817 0.1556 158675 0.123433
d=55.545
Moment Curvature Diagram
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• In the table above, c/x = 0.002/(450-x)
• Formula for c is for parabola only. So check
on values of c 0.002
• My =158844(450-55.545) = 62.66 kNm
• And y = c/x=1.081710-3/(157.9510-3)
=0.006848 (1/m) (rotation of section (rad) per
m length of beam axis)
Moment Curvature Diagram
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• Somewhere b/n cracking moment and My, say
at s1 = 0.001 = 1
d
Ccx
s1 = 0.001
c = ?
Ts=397.11 0.001 200000 = 79422
Moment Curvature Diagram
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kx =x/d xc( ) c=cm/1
2(6-
cm)kx
Cc=(8-
cm)/(4(
6-cm)
0.5 225 1.0 0.208 212438
0.2 90 0.25 0.023958 24430
0.28 126 0.389 0.0509 51918.6
0.35 157.5 0.53846 0.08577 87464
0.32 144 0.47058 0.0939 70755.6
0.33 148.5 0.492537 0.0746 76066.9
0.34 153 0.51515 0.080057 81634
0.366 151.2 0.506024
1
0.07784 79376 0.11458
d=51.56
Moment Curvature Diagram
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• M = 79422(450-51.56) Nmm
• M = 31.64 kNm
•
= 0.5060241
10-3/(151.2
10-3) =
0.00334672 (1/m)
Moment Curvature Diagram
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Somewhere between My and Mu, say ats1 = 0.005 = 5
d
Ccx
s1 = 0.005
c = ?
Ts=397.11 400= 158844 N
Moment Curvature Diagram
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M C Di
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kx =x/d xc( ) c=cm/1
2(6-
cm)kx
Cc=(8-
cm)/(4(
6-cm)
0.25 112.5 1.666666
7
0.15046 153427
0.26 117 1.756756 0.16151 164692
0.254 114.3 1.70241 0.15486 1579120.255 114.75 1.7114 0.155965 159038 0.09348
d=48.07
M = 158844(450-42.07) = 65.798 kNm
= cm/x = 1.711410-3/(114.7510-3) = 0.014914 (1/m)
Moment Curvature Diagram
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• At the ULS strain distribution at ULS
• ult
= (10+2.63) /0.45 = 0.0280667(1/m)
• Using design chart No 1
• Sd,s = 0.143
MSd,s = 0.14311.332004502 = 65.62 kNm
x = kxd = 0.208d = 93.6 mm
s1 = 10
c = -2.63
Moment Curvature Diagram
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Moment Curvature Diagram
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Observations:
For low values of M, cross section is in state I. M-
relationship is linear. The slope is the flexural
stiffness EcIi in state I.
After cracking, cross section is in state II. M- relationship b/n Mcr and My is also approximately
linear. The reduced flexural stiffness is depicted by
the lower slope.
After yielding, the bending stiffness is practicallylost with the almost horizontal slope. The RC section
(under reinforced) shows a highly ductile behavior.
Moment Curvature Diagram
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In an indeterminate structure, once a beam sectiondevelops its ultimate moment of resistance Mu, it
behaves as a plastic hinge resisting a constant
moment of that value.
Further loading must be taken by other parts of the
structure, with the changes in moment elsewhere
being just the same as if a real hinge existed.
Thus, after yielding, moments are redistributed toother cross-sections of the member which are still
elastic.
Moment Redistribution for Linear Elastic
Analysis of structures
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As the load increases further, other sections will
yield and shall develop hinges.
When enough hinges have formed in any span of themember to make it unstable (a mechanism rather
than a flexural member ), the member is considered
to have failed .
The load at which a mechanism forms in any span is
called the ‘limit’ load in the span.
Thus redistribution of Moments is the transfer of
loads after the formation of first plastic hinge at thesection, having the highest bending moment till the
collapse of the structure.
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When a beam is loaded beyond working load, plastic
hinges are formed at certain locations.
If a plastic hinge is formed in determinate structures,
uncontrolled deflection takes place, and the structure
will collapse. Thus a statically determinate system requires the
formation of only one plastic hinge in order to
become a mechanism.
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I th f i d t i t t t t bilit
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In the case of indeterminate structures, stability may
be maintained even though hinges have formed at
several cross sections
Plastic hinges modify the behavior of structures much in the
same way as mechanical hinges.
The only difference is that the plastic hinges permit rotation
offering constant resisting moment Mp to the rotation.
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Plastic hinge: is defined as a zone that yields due to
bending in a structural member, at which large
rotations can take place at a section with a
constant moment, Mp.
The plastic hinges are likely to be formed under the
points of
maximum bending moment,
points of supports, or
under concentrated loads.
Formation of plastic mechanism:
is one of the major Ultimate Limit States
is formed when the reinforcement yields to form
plastic hinges at enough sections to make the
structure unstable.
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A stable structure can
resist displacement .There has to be a force corresponding to
any displacement.
The structure is converted into mechanism and collapse
occurs, if there is displacement without resistance.
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Rotation Requirement
Before failure, reinforced concrete sections are usually
capable of considerable inelastic rotation at nearlyconstant moment.
This permits a redistribution of elastic moments andprovides the basis for plastic analysis of beams, frames andslabs.
Therefore, the designer adopting full limit analysis inconcrete must calculate not only the amount of rotationrequired at critical sections to achieve the assumed degreeof moment redistribution but also the rotation capacity ofthe members at those sections to ensure that it is
adequate. AND: It also requires an extensive analysis of all possible
mechanisms.
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Plastic Moment Redistribution According to
EBCS-2, 1995
To obtain most of the advantages of limit analysis, arestricted amount of redistribution of elastic moments cansafely be made without complete analysis.
This is possible, if the section forming the plastic hinge has
the ability to rotate at constant moment, which dependson the ductility of that section.
Therefore, R.C.C. section must be ductile; this implies itshould be under reinforced.
According to EBCS-2, 1995, a limited amount ofredistribution is permitted, depending upon a roughmeasure of available ductility, without explicit calculationof rotation requirement and capacity.
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is a reduction coefficient for redistribution of moments
depending on the ductility of the section.
Moments obtained from a linear analysis may bemultiplied by provided that the moments are increased
in other sections in order to maintain equilibrium.
For continuous beams & beams in rigid jointed frames
with span/effective-depth ratio not greater than 20,
≥ 0.44 + 1.25
≤
0.44
1.25, ≤ 35
≥ 0.56 + 1.25
≤ 0.56
1.25
, > 35
The neutral axis height, x, is calculated at the Ultimate
Limit State and the term x/d refers to the section where
the moment is reduced.
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For other continuous beams and rigid jointed braced
frames ≥ 0.75
For sway frames with slenderness ratio of columns
less than 25, ≥ 0.90
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(reduction
coefficient )
s1
( 0/00)
cm
(0/00) C C sd Kx
0.70 10.0 -2.626 0.155 0.082 0.142 0.208
0.80 8.6528 -3.50 0.233 0.120 0.205 0.288
0.90 6.0109 -3.50 0.298 0.153 0.252 0.368
1.0 4.3125 -3.50 0.363 0.186 0.295 0.448
μ*
0.143
0.205
0.252
0.295
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Advantages of moment redistribution:
It gives a more realistic picture of the actual load carrying
capacity of the indeterminate structure.
Structures designed considering the redistribution of
moment (though limited) would result in economy as the
actual load capacity is higher than that we determine from
any elastic analysis. The designer will have the freedom to modify, with in
limit , the design bending moments to reduce reinforcing
bars, which are crowded especially at location of high
bending moment, such as beam-column joint . This can be achieved by letting the section to be under-
reinforced, if the moment resistance of adjacent critical
section is increased correspondingly.
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The application of the inelastic moment redistribution
is illustrated by means of the continuous beam shown
in figure (next slide).
Given:
– 2 span continuous beam and design load
(g+q)d = 80 kN/m
Required:
– BMD after inelastic moment redistribution using aredistribution factor = 0.8 applied to the support
moment
57
Inelastic Moment Redistribution
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58
80 kN/m
6.0 m8.0 m
-520 kNm
406.4 kNm
146.9 kNm
(b) BMD based on linear analysis
(a) System and Loading
x x’
Inelastic Moment Redistribution
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Solution:
Determine the BMD of the continuous beam using
linear analysis (result is shown in figure. Students will
check the result)
MS,red = 0.8MS,el = -0.8520 = -416 kNm
Now since the support section has turned into a
plastic hinge, the system has changed from
indeterminate to two determinate simple span
beams
59
Inelastic Moment Redistribution
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reactions and internal action effects can be
determined from conditions of static equilibrium.
60
80 kN/m
Ms,red = -416 kNm
6.0 m
80 kN/m
-416 kNm
8.0 m
A
C
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From FBD of left span: A = 170.67 kN
From FBD of right span: C = 268 kN
B = 681.33 kN
Maximum span moments: – Left span: xo = A/(g+q) = 170.67/80 = 2.13m
– Mf,l = 170.67(2.13)-80(2.13)2/2 = 182.05 kNm (max
span moment after redistribution)
– Right span: x’o measured from right x’o = C/(g+q) =268/80 – 3.35m
– Mf,r = 268(3.35)-80(3.35)2/2 = 448.9 kNm
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-416 kNm
448.9 kNm182.5 kNm
Fig. BMD before and after redistribution
-520 kNm
406.4 kNm
146.9 kNm
Inelastic Moment Redistribution
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Points to be observed For any degree of inelastic moment redistribution,
there must take place some amount of rotation at a
plastic hinge which must be endowed with sufficient
rotation capacity.
Thus such redistributions are typically followed by a
check for sufficiency of plastic rotation capacity.
Such a check is rather involved
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Excercise
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Excercise
1. Determine the moment – curvature diagram of the
x-section (b/h/d) = 300/500/450 RC section if it isreinforced with:
2Φ24 bars
4Φ24 bars
6Φ24 bars
2. Design a 6m beam fixed at both ends to support adesign load of 24kN/m
a) Without redistribution
b) With 20% support moment redistribution Section 200mm x 400mm
C-25 concrete & S-400 steel, class I works.