Chapter 5 Continuous composite beams.pdf

8/10/2019 Chapter 5 Continuous composite beams.pdf

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Chapter 5 : Continuous Composite BeamsSummary:

Continuous beams are an alternative to simply supported beams and their use is justified byconsiderations of economy. In the hogging bending regions at supports, concrete will be in tension and the steel bottom

flange in compression, leading to the possibility of onset of local buckling. This is taken up bythe classification of cross sections.

Rigid-plastic design may be performed for beams with Class 1 cross-sections. Plastic momentresistance of cross-sections can be used for Class 1 and 2 cross-sections.

For Class 3 sections, elastic analysis and elastic cross-section resistance must be used. The principles of calculation of cross-section resistance, either plastic or elastic, are similar to

the case of sagging bending. The tension resistance of concrete is neglected. Lateral-torsional buckling is a special phenomenon which can be prevented by conforming to

certain detailing rules. The design of the shear connection in the case of continuous beams is more complex than for

simply supported beams. Serviceability checks include deflection and vibration control as well as that of concrete

cracking. This latter is specific to continuous beams because tension in concrete at the hoggingmoment regions may cause unacceptable cracks, while in simply supported beams cracking isonly due to shrinkage of concrete and is therefore lower in magnitude.

Objectives:The student should: Appreciate the advantages of continuous composite beams and be aware of their disadvantages. Understand the methods of plastic and elastic design of continuous beams. Understand the methods of calculation of elastic and plastic cross-section resistance for hogging

bending moment, shear resistance and resistance against lateral-torsional buckling. Understand the way shear connection is designed for class 1 and class 2 cross-sections. Be aware the need for serviceability checks for cracking in the hogging moment region.


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5.2. Effective Width, B e

For simply supported beam Be is taken as the sum of effective width b e ofthe portion of concrete flange on each side of the steel section, where b e =L/8, and L = span length of the simply supported beam.

For sagging moment regions of a continuous beam, the effective width is proportioning to the effective length Lo between the point ofcontraflexture. This length will depend on the type and magnitude of theloading on the various spans of the continuous beam, and will change inaccordance with the different load cases.

An approximation may be made as follows:For end span Lo = 0.8LInternal span Lo = 0.7LL = distance between supports for the span concerned.

Over an internal supportLo = 0.25(L1 +L 2)


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5.3. Moment of Resistance in Hogging Bending

The reinforcement bars within the effective width are assumed to bestressed to their design yield strength, y.

The concrete slab may be assumed to be cracked.

The tensile resistance of the reinforcement, R r , within the effective widthof the slab under negative moment is given by:

R r = yAr

Ar = area of the reinforcement within the effective width.

The axial resistance of the web is

R w = dtpy

d = depth between the steel section flange.t = Web thickness

The negative plastic resistance moment M c can be determined byconsidering moment of each rectangular stress block about the neutralaxis.


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For compact section:

Case 1: R r < R w (PNA lies in web)vr R /R |1

76td

+ (Web compact)

Moment about the center of the beamMc = Ms + R r (0.5D + Dr ) + R r x

X =2d

R R

d /R 2R

t p2R

v

r

v

r

y

r ==

Mc = Ms + R r (0.5D + Dr ) + (R r 2d)/4R w whereMs = the plastic resistance moment of the steel section aloneD = overall depth of the steel sectionDr = the distance from the top of the steel beam to the centroid of the reinforcement.


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Case 2: R s > R r R w (PNA lies in steel flange)

38td (Web compact)

2T

R R R

B p2R R

xf

r s

y

r s ==

Moments about op of the steel flange

Mc = 0.5R sD + R r D - (R s - R r )2T/4R f whereR s = tensile resistance of the steel section, = yA. R f = resistance of the steel flange = BTy.T = thickness of the steel flange

Light mesh reinforcement in the slab is neglected when calculating M c.

If no reinforcement is provided then M c = M p.

Case 3: R r > R s (PNA outside the steel beam)

Mc = R s (0.5D + Dr )


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5.4. Shear Connections in Negative Moment Regions

The required number of shear connectors, N n, is

Nn = R r /Qn

Qn = design capacity of a shear connector in negative moment regionsQn = 0.6Qk compared with 0.8Qk for positive bending, because of the

influence of cracking of concrete.

A suitable spacing can be determined by calculating the total number ofconnectors N p + Nn needed between the point of maximum moment and

each adjacent support.

The total number of shear connectors may be spaced uniformly along the beam between the point of maximum positive moment and the adjacentsupport.


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5.5. Analysis MethodsThree analysis methods are recommended for determining the moments incontinuous beams:Simplified methodElastic global analysis

plastic hinge analysis

5.5.1 Simplified Method The method is based on a Table of coefficients as given below. Certainrestrictions are placed on this method (5.2.2):1. The steel beam should be of uniform section with equal flanges and

without any haunches.

2. The steel beam should be of the same section in each span.3. The loading should be uniformly distributed.4. the unfactored imposed load should not exceed 2.5 times the unfactored

dead load.5. No span should be less than 75% of the longest.6. End span should not exceed 115% of the length of the adjacent span.7. There should not be any cantilevers.

The coefficient in the Table should be multiplied by the free bendingmoment WL/8, where W is the total factored load on the span L.The values in the Table already allow for pattern loads and forredistribution. No further redistribution should be carried out.


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Table: Simplified table of moment coefficients


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5.5.2 Elastic Global Analysis

Two methods of elastic global analysis are available for ultimate limitstate design:

(a) cracked section method(b) uncracked section method.

Cracked Section Method.For a length of 15% of the span on each side of the internal support, thesection properties are those of the cracked section for negative moments.

The second moment of area of the cracked section is calculated using asection comprising the steel section together with the effectively anchoredreinforcement located within the effective width of the concrete flange atthe support.

Outside the 15% length, the section properties are those of theuncracked section, this being calculated using the mid-span effectivewidth for the concrete flange but ignoring any longitudinal reinforcement.

The continuous beam can be analysed using a standard program orformulae. The forces obtained from the analysis can be used to checkagainst the capacity at various critical locations along the beam.

Uncracked Section Method The properties of the uncracked section are used throughout. The analysiscan be carried out without prior calculation of the cross-section.


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Global Analysis based on Cracked Section Method

Uncracked Section (B.3.1)

I IB D D AB D D D D D

B D Dg xe s p

ee s p s p

e e s p= +

+

+ ++

( ) ( )( )[A ( )]

3 2

12 4

Cracked Section, Negative Moment (B.3.2)

)AA(a)D2D(AA

IIr

2r r

xn ++=


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5.5.3 Redistribution of Moments for Elastic Analysis

Design codes permit negative moment (hogging) at the supports to bereduced, except at cantilevers, by redistribution to mid-span. The extentof the redistribution is dependent on the method of analysis and sectionclassification, as shown in the following Table.

Limits to redistribution of hogging moments

Class of section in

hogging region

1 2 3 4

For cracked analysis 30% 20% 10% 0

For un-cracked analysis 40% 30% 20% 10%


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Elsatic Global AnalysisMoments are calculatd for two load cases:


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5.6 Serviceability Limit State

Cracking may not need to be controlled in composite beams. If the slab is constructed ascontinuous, uncontrolled cracking is permitted by design codes provided that this does not impairthe functioning of the structure. Most interiors of buildings for offices have low air humidity andcrack width has no influence on durability. Appearance requires a floor finish with ductile behaviour or provision of a covering. Even so, British Standards and Eurocodes specify minimumareas of reinforcement to prevent fracture of the bars or the formation of very wide cracks underservice loading. To avoid visible cracks where hard finishes are used, crack control joints should beconsidered.

This part of the lecture therefore addresses deflections only. These are influenced by: pattern loading cracking of concrete shakedown effects.

DeflectionFor uniformly distributed or symmetrical loading, the deflection at mid-span for a continuous beamis given approximately as

c = o{1-0.6(M1+M2)/Mo}whereo = deflection of a simply supported composite beam under thensame loading conditions (see

chapter 4).Mo = maximum moment in a simply supported beam subject to the same loading.M1 and M2 = moments at the adjacent supports of the continuous beam (following redistributionetc).

The support moment M1 and M2 may be determined approximately using an elasticanalysis assuming uncracked section.


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Pattern loading

BS5950: Part 3.1 allows the beam to be analysed with the imposed load on each

span. Except adjacent to cantilevers, the moment at each support is reduced to allowfor absence of imposed load on the adjacent spans.

To allow for the effect of pattern loading as shown in the figure, the moments atsupports (not adjacent to cantilevers) are reduced by 30% for beams carrying normalloads or 50% for storage loads, to allow for pattern loading.

For other non-symmetric load cases it is more accurate to calculate the deflectionsfrom the bending moment diagram at serviceability.

Shakedown effectsIf the beam has been designed for ultimate moments determined by plastic analysis, or by elasticanalysis with substantial redistribution, then irreversible deformation may have occurred at a support.To allow for this, BS5950 recommends that the support moment used in deflection calculations isreduced as follows.

The support moment is calculated using the un-cracked section throughout, normally under deadload plus 80% of the imposed load. If this moment exceeds the plastic moment resistance of thesection in negative (hogging) bending, the difference (termed the excess moment) is calculated.


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The support moment used in deflection calculations should be reduced by the excess moment, aswell as by the reduction due to pattern loading.

Stresses

Determine the stresses in steel beam in the positive moment region based on the bending momentdiagram used to calculate the service deflections. For unpropped beams add the steel stresses tothose calculated for the steel section due to self weight of the floor. Check that the total stress doesnot exceed py. No further checks are required in the negative moment regions.

5.7 Summary: - Design Procedure

1) Loading and momentsObtain the factor loads through suitable combination of load factors, and calculate the free bending moment on the beam ignoring continuity .

2) Initial selection of beam size Use the simplified table of moment coefficient, and obtain thedesign moments in the negative and positive moment regions by multiplying the free bendingmoment by the coefficients. Select the steel section so that it can resist the negative momentobtained in the above without the need of reinforcement bars. Further refinement of section sizemay be made by including additional reinforcement in the slab (see Step 5).

3) Perform section classification, and determine the analysis and design method.4) Global Analysis

Select the following methods of analysisa) Simplified table of moment coefficient b) Elastic global analysis -uncracked sectionc) Elastic global analysis -cracked section d) Plastic hinge analysis

If method (a) is used, no further refinement is needed.

5) Check moment capacity at positive moments.

6) Check moment capacity at negative moments.7) The section size may be re-selected depending on the results of this global analysis.

8) Check interaction of moment and shear

9) Check shear connectors10) Check construction stage11) Check stability of the lower flange over the internal supports.12) Provide transverse reinforcement


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13) Check service load deflection and stresses.14) Check natural frequency


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EXAMPLEDesign a two-span composite beam with single span length 12m as shown inFigure. Assuming that there is no reinforcement at the intermediate support.

Design data:Steel: Grade 50Concrete: Grade 30 light-weightSlab thickness Ds = 130mmShear studs: 19mm diameter, 100mm length, use 2 studs per troughMetal decking perpendicular to the steel beam:Profile depth D p = 50mm, thickness t = 1mm , average trough width b r =130mm, trough spacing = 300Unfactored Dead Load = 8.1 kN/mUnfactored Imposed load = 18 kN/m

Design Load = 1.4DL + 1.6ILw = 1.4 x 8.1 + 1.6 x 18 = 40kN/m

Check assumption for simplified table:

(Unfactored I L) / (Unfactored DL) = 18/8.1=2.22 < 2.5 OK!

12m

w = 40 kN/m

12m

w= 40 kN/m


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Continuous Beam ABCL = 12m, w = 40kN/m

For hogging moment

Select 406x178x60 UB Grade 50Section is plastic in bending (NA in the web)

Mcx = 426kNm > 361kNm OK!

Classfication for bending, S355 steel: Plastic

Check shear at the intermediate support

Fv = 6x 40 + 361/12= 271kNPvx = 675kNShear is OK0.6Pvx = 405 > Fv = 271kN i.e., low shear

Note that high Shear does not coincide with the maximum moment.

Check sagging resistance

Bo = 3000mmBe = 0.8L/4 = 0.8 x 12000/4 = 2400mm (Control!)Rc = 0.45f cuBe (Ds-D p)R kNc = =045 30 2400 130 50 10 25903. ( )R s = 2700kNR R s c> PNA is not in the concrte slabR w = R s 2R f = 1084kNR c > R w PNA is in the steel flange

0.5wL2/8 = 361kNm

0.79wL2/8 =5715kNm

FF 12m

M = 361 kNm

W = 40 kN/m


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4T

R )R R (

2DD

R 2DR M

f

2cs ps

csc

++=

Mc = 781kNm>571kNm

Shear ConnectionR c = 2590kNR s = 2700kNSmaller of R c and R s is 2590kN.

Capacity of shear connector (19mm diameter and 95mm long) inlightweight concrete

Q kN kNk = =0 9 100 90. Design capacity Q = 0.8Q kNk =72 Reduction factor for deck profile

k bDh

Dr

p p=

060 1 08. . for two studs per rib br = Average trough width = 150 mmh = overall height of the stud = 95 mm

k =

= >060150

50

95

501 162 08. . .

k = 0.8Resistance of a shear connector = 0.8 x 72 = 57.6kNFor full composite, no. of connectors required= 2590/(57.6) = 45

Evaluate the x distance between points of zero moment

0.5wL2

/8 = 361kNm

0.79wL2/8 =5715kNm

x

F = 271kNF 12m

M = 361 kNmW = 40 kN/m


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Take moment about the zero moment pointM = 0 = Fx wx2/2X = 2F/w = 10.5mSince the trough spacing is 300 mm,no. of connectors that can be accommodated in half span (x/2), assumingtwo connectors per trough= 2 x (10.2/2)/300 = 35.

R q = 35x57.6 = 2016 kN < R c i.e., partial composite

Degree of partial composite = 35/45 = 0.78Calculate reduced moment using simplified formulaeM = Ms + k (Mc- Ms)

= 426 + 0.78 (781-426)= 703kNm > 571 kNm OK

( More exact value for Mc is 746 kNm > 571 kNm )

Check deflection For Unfactored Imposed Load OnlyUnfactored imposed load, w = 18 kN/m e s s= + ( ) s = =10 25 for lightweight concreteLong term loading: Dead load 8.1kN/m

1/3 Imposed Load 6 kN/m

12m 12m

10.5m

35 + 35studs

10.5m

35 + 35studs


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14.1 kN/m Total Loading = DL + IL = 26.1 kN/m

= =141261 054

.

. .

Compute composite section propertiese = 18.1l =0.54I I

B D D AB D D D D DA B D Dg x

e s p

e

e s p s p

e e s p

= +

+ + +

+ ( ) ( )( )

[ ( )]

3 2

12 4 for uncracked section

I g = 60129 cm4

Uncracked analysis

M1 reduces by 30% to allow for patterned lodingM

1 = 0.7 x 324 = 227kNm

c = o{1-0.6(M1+M2)/Mo}

o = 5384 39 44wl

EI mmg= . =deflection of simply supported composite beam

s =5

384 10974wl

EI mms= . =deflection of simply supported steel beam

M wlo = 2 8/

M1 =0.125wL2 = 324kNm

IL w = 18kN/m

M1 = 227

12m

M2 = 0


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To account for partial composite connectiono = o + 0.3(1-Na/N p) (s - o)

= 39.4 + 0.3 (1-0.78) ( 109.7- 39.4)= 44mm

c = o{1-0.6(M1+M2)/Mo} = 44(1-0.6x0.7)= 26mm < L/360 = 33mm OK

Check stresses in concrete and steel!!


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HOMEWORK Composite ConstructionQuestion 3(a) Design the composite beam AB assuming that it is simply supported with a span length 12m asshown in Fig. Qa. Check the moment and shear resistances, and determine the number of shearconnectors required for the full length of the beam.(b) Design the two-span composite beam CDE with a single span length 12m as shown in Fig. Qb.Assuming that there is no reinforcement at the intermediate support, check the shear resistance atthe intermediate support and the moment resistances at the hogging and sagging moment regions.Determine the number of shear connectors required for each span length.The following information should be used for the above design:Design data:Steel: S275Concrete: Grade 35 light-weightSlab thickness Ds = 150mmShear studs: 19mm diameter, 100mm length, use 2 studs per troughMetal decking perpendicular to the steel beam: Profile depth D p = 60mm,thickness t = 1mm , average trough width br = 130mmUnfactored uniformly distributed loads:Dead Load = 11 kN/mImposed load = 10 kN/mDesign constrainst:Height of steel beam must be less than 400mm.Beam spacing = 3000mm

Ds

Dp

300mm

130mm

1mm thicksteel deck

95mm

19mm diameter stud

12m

w kN/m

A B

12m 12m

w kN/mC

D

E

Fi . a

Fi . b

2 studs

Fi . c


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Q4 A two-span composite beam of L1 = 10.5m and L2=12m is shown in Fig. Q4. The hoggingsection at the intermediate support is reinforced by deformed bars of 12mm diameter spaced at150mm. Using the simplified table in BS5950:Part3, check the moment resistances at the hoggingand sagging regions assuming full composite action.The following information should be used for the above design:

Design data:Steel beam: Grade S275, UB 457 x 191 x 74Concrete slab: Grade 30, light-weight, slab thickness Ds = 150mmShear studs: 19mm diameter, adequately spaced for full composite actionRe-bars: 12mm diameter, f y = 460N/mm2 Unfactored uniformly distributed loads:Dead Load = 9.5 kN/mImposed load = 15.0 kN/m

Beam spacing = 3000mm

Fi . 4

10.5m 12m

150m30m Be

150m

UB 457 x 191 x74

12mm re-bars

Section at intermediate su ort


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Q 5 The following figure indicates the proposed composite steel frame structure to be built over anexisting building. The floors to the new building are of reinforced concrete slab.

Design data Roof

Insulated roof decking 0.1 kN/m2

Purlins 0.1 kN/m2 Services 0.15 kN/m2 Self weight of plate girders (estimated) 1.0 kN/m2 Imposed load (Roof) 0.75 kN/m2

Typical FloorFinish 0.1 kN/m2

150 mm concrete slab 2.5 kN/m2 Suspended ceiling 0.15 kN/m2 Services 0.15 kN/m2

Weight of steel beams (Approx.) 0.4 kN/m2

Partition (superimposed dead load) 1.00 kN/m2 Imposed load (Floor) 4.0 kN/m2

External cladding 4.00 kN/m2

S275 steel and Grade 30 normal weight concrete should be used.

(a) Design beam A as a two-span continuous composite beam and check for moment and shear.Determine the number of shear studs for full composite design. Check the total beamdeflection assuming propped construction

(b) Determine suitable section sizes for hanger B (deigned as tension member).(a) Design column C, assuming simple construction.


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8000

3500 (clear height)

3500

5000

Hanger 'B'

Suspended Ceiling

Suspended Ceiling

Suspended Ceiling

8000

150 mm concretefloo

150 mm concretefloo

Plate Girder 'D'

fal fal

Existing Floor

Existing Floor

Existing Building

SECTION A-A

A

A

Denotes span ofreinforcedconcrete slab

Vertical Bracing

Vertical Bracing

Hanger 'B'

Column 'C'

Beam 'A'

6000 6000 6000 6000 Outline of building

8000

8000

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