ch 1 inelastic moment redistribution

8/19/2019 Ch 1 Inelastic Moment Redistribution

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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

called flexural rigidity.15-Dec-13


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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


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


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


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εcm ≤ 2‰ and N.A. within the section

εcm ≥ 2‰ and N.A. within the section


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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


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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)


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• My (at yielding of reinforcement)

d

Cc

Ts=397.11400 = 158844 N

x

yd = 0.002

c = ?


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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


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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)


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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


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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


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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)


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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


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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)


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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


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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.


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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’


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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

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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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62

-416 kNm

448.9 kNm182.5 kNm

Fig. BMD before and after redistribution

-520 kNm

406.4 kNm

146.9 kNm


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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.

ch 1 inelastic moment redistribution

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