Moment Distribution and Moment Re-distribution

8/3/2019 Moment Distribution and Moment Re-distribution

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Moment distribution

Moment re-distribution

J. P. Forth

(Acknowledgements: Dr A. N. Fried and theConcrete Centre)


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Moment distribution

Simple hand calculation method of analysingframes by distributing the fixed end moments

according to their relative stiffness. Used mainly for analysing sub frames orcontinuous beams where a simplified analysis isrequired.

However, most structural analysis is undertakenusing convenient and efficient softwarepackages.


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The stages of momentdistribution are

Calculate member stiffness

Calculate distribution factors Calculate fixed end moments Distribute moments according to stiffness

by an iterative process until balance isachieved.


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Member stiffness.

Stiffness k of a member, defined as 4 x productof elastic modulus and its second moment ofarea divided by the length ; thus k = 4EI/L

Greater the stiffness of a member, the more

moment it will attract. Stiffness is modified depending on the degree offixity at the ends of a member.

k =k =

k =k =

k =k =

k=


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k = 0.75(4EI/L)F

k = 0.75(4EI/L)E

k = 0G

k = 4EI/LC

k=4EI/LB

k = 0.75(4EI/L)

Dk = 4EI/LA

Effect of end fixity on stiffness

Built in at one end and continuous at other k = 4EI/L {element A} Continuous at both ends k = 4EI/L {elements B and C} Continuous at one end and simply supported at the other k = 0.75(4EI/L) {elements D

and E} Built in at one end and simply supported at the other k = 0.75(4EI/L) {elements F} Simply supported at both ends k = 0 {element G}


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Distribution factors

Use member stiffness to find proportion of totalstiffness a member provides at a connection.

Distribution factors enable the proportioning ofthe stiffness.

Total stiffness at connectionk1 + k2 + k3 + k4 = kt

k3k1

k2

k4

Member stiffnesses

f1 = k1/ktf2 = k2/ktf3 = k3/ktf4 = k4/kt

f3f1

f2

f4

Distribution factors

f1 + f2 + f3 + f4 = 1.0


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Distribution factors and carry over

Out of balance moments are redistributedat the joint using distribution factors.

As part of re-distribution half of theunbalanced moment is carried over toother end of the beam.


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Fixed end moments.

These are the moments at the fixed ends of amember due to a given vertical load.

The values for a uniformly distributed load (UDL)

acting on the whole of a span and a point loadare shown. The formulae for other loadingsituation can be found in many reference books.

Sign convention.

Clockwise moments are considered to bepositive, anticlockwise moments negative.


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Common Fixed end moments

Fixed endmoment

Loadingconfiguration

Fixed endmoment

wL2/12 wL2/12

WL/8 WL/8

(Wab2)/L2 (Wba2)/L2

w/unit length

L

W

L/2

W

a b


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Moment distribution Example.

Three span beam has fixed ends. Outer spans are 5.33m in length, and the inner

span 8.0m long.

Beam is of the same section throughout and soEI is a constant hence we only require relativestiffnesses.

For this purpose assume 4EI = 100, say (4EI

cancels out through all the terms so 100 ischosen simply for convenience).

A udl of 36kN/m exists over the beam.


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Beam stiffnesses

Beam stiffnesses = 4EI/L Outer span (2 No.) = 100/5.33 = 18.76

Inner span = 100/8 = 12.5


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k = 18.76k = 12.5k = 18.76

Setting up the problem

8.0m 5.33m5.33m

1233.536

2

85.23-85.23 85.23-85.23192.0-192.0

12

836 2Fixed end moments

DistributionFactors5.1276.18

76.18

5.1276.18

5.12

00 0.6 0.4 0.60.4

No distributionat fixed end


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Moment distribution Iteration 1

85.23 + 64.06= 149.29

85.23-85.23 85.23-85.23192.0-192.0

00 0.6 0.4 0.60.4

Step 1. Start at left support.

Distributed moment= - 85.23 x 0 = 0

Step 2. Next support alongOut of balance moment = (85.23 192.0) = -106.77Distribution to left of support = 106.77 x 0.6 = 64.06Distribution to right of support = 107.77 x 0.4 = 42.71Carry over to left = 0.5 x 64.06 = 32.03Carry over to right = 0.5 x 42.71 = 21.35

-85.23 + 32.03 = -53.02 192 + 21.35 = 213.35

149.29-53.02

Step 3. Next support alongOut of balance moment = (213.35 85.23) = 128.12Distribution to left of support = -128.12 x 0.4 = -51.25

Distribution to right of support = -128.12 x 0.6 = -76.87Carry over to left = 0.5 x -51.25 = -25.62Carry over to right = 0.5 x -76.87 = -38.43

213.15 51.25 = 162.10

213.35

162.10

-85.23 -76.87 = -162.10

-162.10

85.23 -38.43 = 46.80

-149.29 25.62 = -174.91

-192 + 42.71= -149.29

46.80

Boxed numbers below the beam only indicate the method and are to be omitted.

Step 4.Distributed moment = 46.80 x 0 = 0

-149.29

-174.91


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Iteration 2

85.23-85.23 85.23-85.23192.0-192.0

00 0.6 0.4 0.60.4

162.1-162.1213.35149.29-53.02 -149.29 46.80

-174.91-45.33167.2

164.7

165.2

-165.3

-165.68

-164.745.27


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Moment diagram

Draw support moment diagram Superimpose free moment diagrams on

support moment diagram.


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Fixed end moments

-165.32165.32165.32-45.18 45.18-165.32

Free body moments

Mss = 36 x 82/8

= 288 kNm

Mss = 36 x 5.332/8

= 127.1 kNmMss = 36 x 5.33

2/8= 127.1 kNm


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Superimpose diagrams

165.32165.3245.18

127.8 127.8288.0


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Shear Force Diagrams.

The shear force along a beam is thesummation of the individual effects of endmoments and loads.


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Final shear force diagram

144

118.495.9

-22.5

118.4

144

73.4


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Shear force diagrams and reactions.

Using virtual work or similar Reactions at A, B, C and D can be found. They are the same as the sum of the shear

forces at the ends of the beams spanning ontothe support :

Reaction at A = 73.4kN

Reaction at B = 144 + 118.4 = 262.4kN Reaction at C = 118.4 + 144 = 262.4kN Reaction at D = 73.4Kn


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Moment dist. (Practical applications)

Usually analysis of elements is undertaken using some form ofcomputer package.

Using moment distribution on the sub frames in a 10 storey buildingwould be most unusual these days.

Computers reduce errors

Check moments using wL2

/8, or wL2

/12 depending on the fixity toensure you are not an order of 10 out. Engineers call this having a feel for the structure. In reality it is

unlikely that any project which is big would not be checked at leastfour or five times.

With a number of load cases, number of computations rapidly

expands. Consequently, if at all possible use electronic methods to assist theanalysis.


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Redistribution of moments.

In BS 8110 considerable information onmoment redistribution is included. Verylittle is given in the European Code. Theentire clause on this aspect of concretedesign covers about 1.5 pages.


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Moment re-distribution - background

Moment re-distribution is an optionalprocedure and may be applied to frames,sub-frames and continuous beams.


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Interior spancontinuous rect. beamof length L & subjectto a udl of w/m

Beam but with ends

fully fixedBeam but simply

supported.

The overall amount of moment between span and supports remains the

same, its just the distribution of the moment that changes, the supportand span moments could be equal.

Relationship between a fully fixed, simplysupported and continuous beams

moments.

wL2/8

wL2/12

wL2/8

wL2/16

wL2/8


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Case for moment re-distribution

A beam fixed at both ends supports anever increasing UDL.

Bending strengths are equal at the spanand supports.

If the ultimate bending moment at asupport is reached at a UDL of 30kN/m,what will be the value of the UDL atcollapse.


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Solutionw kN/m

LLoading diagram

u kN/m Additional load

uL2/8

Additional moment

for hinges at each end

At collapse Mu (at centre span) = wL2/12

= wL2/24 + uL2/8

So u = w/3 and collapse load = 30 + 30/3 = 40 kN/m

wL2/12wL2/12

wL2/24

Bending moment diagram


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Benefits of Moment Re-distribution.

A more balanced distribution of momentsresults in terms of the magnitude of spanand support moments.

A reduction in the range of the momentenvelope may occur.

A corresponding reduction in thereinforcement requirements leading toeasier placement and fit.


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Summary of Code Guidance Moments calculated from a linear elastic analysis may be re-distributed in a

continuous beam or slab where the ratio of adjacent spans is between 0.5 and 2.0. Equilibrium between internal and external forces must be maintained under all

appropriate combinations of design ultimate load. When using high ductility steel (Class B and Class C Characteristic yield strength

between 400 600MPa ) sections may be designed for moments down to 70% of themaximum moment obtained from a linear elastic analysis at that section for any loadcase.

For normal ductility steel the figure is reduced to 80%. When the maximum moment at a support or in the span is reduced the sectionshould be designed so that :-

Where, xbal is the depth of the neutral axis. The suffix bal is used

when the beam is balanced.

d is the effective depth is the ratio of the redistributed moment to the momentbefore distribution

According to the UK annex, k1= 0.4 k2= 0.6 + 0.0014/cu3 assuming a rectangular stress

block.

21/)( kdkxbal


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Flexure and Shear design of beams.

zbal = d 0.8xbal/2

Moment of resistance of concrete= Mbal = Fc x zbal = (0.453fckbxbal)zbal And Kbal = Mbal/bd2fck = 0.453 xbal zbal/d2 or Kbal = 0.453(xbal/d)(zbal/d)

The following table summarises this data in adesign friendly way.

2/)( 21 ECfromkdkxbal


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The effect or moment re-distribution on Kbal

Redistribution(%) x

bal/d zbal/d Kbal d/d

0 1.0 0.45 0.82 0.167 0.171

10 0.9 0.45 0.82 0.167 0.17115 0.85 0.45 0.82 0.167 0.171

20 0.8 0.4 0.84 0.152 0.152

25 0.75 0.35 0.86 0.137 0.13330 0.7 0.3 0.88 0.120 0.114

k1 = 0.4, k2 = 1.0. UK National Annex case

Moment re distrib tion


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Moment re-distributionConceptual example.

A three span continuous beam is analysed usingconventional elastic analysis. Three load cases are

assessed and these are given in the Figure below

min max min

max min max

max max max

L1 L2 L3


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240

144

375

349358

Bending moment diagrams

550

452


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Explanation

Load case which produces the maximum supportmoments of 452 and 550kNm does not produce themaximum span moment of 375kNm.

If maximum support moments were reduced to the

maximum support moments produced by the maximumspan moment case, namely 358 and 349kNm, we wouldproduce an identical moment diagram because both loadcases have maximum loading on the span. Identical inthis context means of the same shape and overall value,

but obviously in a different vertical positions. Therefore it is possible to reduce the maximum support

moments without increasing the maximum spanmoment.


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Redistribution in accordance with the code.

For high ductility steel - code permits 30% reduction inmoment :

Our example : support moments may become :- 452 x 0.7 = 316.4kNm

550 x 0.7 = 385kNm Problem Reducing 452 below 358 [value for min max

min case has no advantage. Support moment willincrease.

Therefore, only reduce LH moment by 21% to 358. If we then superimpose the free bending moment fromthis load case on support moments of 358 and 385, thediagram shown below is obtained.


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Final Bending Moment

349240358 = 0.79 x 458

144

375

550

452

FINAL BENDING MOMENT ENVELOPE AFTER RE-

DISTRIBUTION

385 = 0.7 x 550

Free body diagrambetween two redistributedsupport moments


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Solving a continuous beam.

Span above (centre) has been considered inisolation to explain the process of re-distributionand to satisfy the conditions given in the code.

In practice, consider the effects of moment re-distribution on all spans,

For example reducing a support moment for agiven load case, not only increases the spanmoment in this span but also in the adjacentspan.


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Moment re-distribution

Example


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Example Moment re-distribution

Design the main tensile reinforcement in abeam which is to carry a moment of400kNm after 30% re-distribution. Use

25mm diameter bars, a cover of 30mm,and take the beam section as 350 x550mm deep. Characteristic strength of

concrete = 40N/mm2.


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Solution Maximum bending moment = 400kNm

Effective depth d = 550 12.5 30 -10 = 497.5mm

Limiting depth to neutral axis xbal = ( k1)d/k2 From the National Annex, k1 = 0.4 and k2 = 1.0 So limiting xbal = (0.7 0.4)497.5/1.0 = 149.25mm And zbal = d 0.4xbal = 437.6mm

K = M/(bd2fck) = (400 x 106)/(350 x 497.52 x 40) = 0.115 Kbal = 0.453(xbal/d)(zbal/d) = 0.453 x (149.25/507.5)(437.6/497.5) = 0.12 K Kbal so singly reinforced beam

As = M/(0.87 fyk z) = (400 x 106)/(0.87 x 500 x 442.8 = 2076mm2

Use 5 No. 25mm diameter bars, Area = 2454mm2 Shear design as previous

133.1/25.05.0 Kdz

dmmddz 95.08.44289.0133.1/115.025.05.0


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Solution continued

Max. and min. reinforcement areas

EN 1992-1-1:2004 9.2.1.1). Clause 9.2.1.1

As,min 0.26 (fctm/fyk) bd and 0.13%bd

fctm = 0.3 x fck2/3=0.3 x 402/3= 3.55N/mm2 or see Table 3.1 code

As,min 0.26 x (3.55/500) x (350 x 507.5) = 328mm2

As,min 0.13% = 0.13/100(350 x 507.5 = 230.9mm2

As,max 0.04 x bd = 0.04 x (350 x 507.5) = 7105mm2

And A

s,min= 328 2454 7105 = A

s,maxso OK.


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Ultimate capacity of beam

Set K= 0.12

Mu = Kbalbd2fck = 0.12 x 350 x 497.52 x 40 x 10-6= 415.8kNm

dmmddz 95.043888.0133.1/12.025.05.0


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Example Moment re-distribution

Design the main tensile reinforcement in abeam which is to carry a moment of600kNm after 30% re-distribution. Use

25mm diameter bars, a cover of 30mm,and take the beam section as 350 x550mm deep. Characteristic strength of

concrete = 40N/mm2.

Solution


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Solution Maximum bending moment = 600kNm

Effective depth d = 550 12.5 30 -10 = 497.5mm

Limiting depth to neutral axis xbal = ( k1)d/k2 From the National Annex, k1 = 0.4 and k2 = 1.0 So limiting xbal = (0.7 0.4)497.5/1.0 = 149.25mm And zbal = d 0.4xbal = 437.8mm

K = M/(bd2fck) = (600 x 106)/(350 x 497.52 x 40) = 0.17 Kbal = 0.453(xbal/d)(zbal/d) = 0.453 x (149.25/497.5)(437.8/497.5) = 0.12

K Kbal so doubly reinforced beam

Compressive steel = (K Kbal)fckbd2/(fsc(d-d) d = 52.5mm, so steel compressive strain = 0.0035x (149.25 52.5)/149.5 = .0023, Therefore, steel has yielded (>0.00217) so use fyk = 500N/mm2 Area comp steel = (0.17 0.12)40 x 350 x 497.52 / [0.87 x 500 x (497.5 - 52.5)]

= 895mm2 [3 H25 1470mm2]

Area of tens steel = [Kbalfckbd2]/[0.87fykz] + As= [0.12 x 40 x 350 x 497.52]/[0.87 x 500 x 437.8] + [895]

= 3078mm2. 7 H25 bars [3436mm2]

(As,provAs,req) (As,prov- As,req) so (1470 895) = 575mm2 (3436 3078) = 358mm2so OK


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Solution continued

Max. and min. reinforcement areas

EN 1992-1-1:2004 9.2.1.1). Clause 9.2.1.1

As,min 0.26 (fctm/fyk) bd and 0.13%bd

fctm = 0.3 x fck2/3

=0.3 x 402/3

= 3.55N/mm2 or see Table 3.1 code

As,min 0.26 x (3.55/500) x (350 x 507.5) = 328mm2

As,min 0.13% = 0.13/100(350 x 507.5 = 230.9mm2

As,max 0.04 x bd = 0.04 x (350 x 507.5) = 7105mm2

And A = 328 3436 7105 = A so OK.

Moment Distribution and Moment Re-distribution

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