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Chapter 5 : Continuous Composite Beams Summary:

• Continuous beams are an alternative to simply supported beams and their use is justified by • • • • • • •

considerations 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 by the classification of cross sections. Rigid-plastic design may be performed for beams with Class 1 cross-sections. Plastic moment resistance 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 hogging moment regions may cause unacceptable cracks, while in simply supported beams cracking is only 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, moment, shear resistance and resistance resistance against lateral-torsional buckling. buckling. • Understand the way shear connection is designed for class 1 and class c lass 2 cross-sections. • Be aware the need for serviceability checks for cracking in the hogging moment region.

5-1

CHAPTER 5 - CONTINUOUS COMPOSITE BEAMS 5.1. INTRODUCTION

Continuous beams offer the following advantages over simply supported beam construction: construction: 1. greater load capacity due to redistribution of moments, and 2. greater stiffness and therefore reduce deflection and vibration. The disadvantages associated with continuous beams are: 1. increase in complexity in design, and 2. susceptible to buckling in the negative moment region over internal supports. Two forms of buckling may occur: (i) local buckling of the web and/or bottom flange (ii) lateral torsional bukling. At the supports, the the sections ar subject to hogging hogging moments. moments. The concrete slab will tend to crack in tension. Reinforcement bars may be placed in the slab above the steel beam section to increase the moment resistance. Continuity of beam can be achieved by spanning secondary beams on top of the perpendicular perpendicula r to the primary beams. This concept is adopted in the parallel beam construction (see figure) which allows continuity in all beams and reducing reducing the need of expensive expensive beam-to-beam beam-to-beam conections. conections.

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5.2. Effective Width, Be

For simply supported beam B e is taken as the sum of effective width b e of the 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 L o between the point of contraflexture. This length will depend on the type and magnitude of the loading on the various spans of the continuous beam, and will change in accordance with the different load cases. An approximation may be made as follows: For end span Lo = 0.8L Internal span Lo = 0.7L L = distance between supports for the span concerned. Over an internal support Lo = 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 be stressed to their design yield strength, ρy. The concrete slab may be assumed to be cracked. The tensile resistance of the reinforcement, R r , within the effective width of the slab under negative moment is given by: R r = ρyAr 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 by considering moment of each rectangular stress block about the neutral axis.

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For compact section: Case 1: Rr < Rw (PNA lies in web)

d 76ε ≤ (Web compact) t 1 | + R r / R v

Moment about the center of the beam Mc’ = Ms + R r (0.5D + Dr ) + R r x X = R r = R r = R r d 2 p y t

2R v / d R v 2

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

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Case 2: Rs > Rr Rw (PNA lies in steel flange) d ≤ 38ε (Web compact) t

x=

R s − R r R s − R r T = 2 p y B R f 2

Moments about op of the steel flange Mc’ = 0.5R sD + R r D - (R s - R r )2T/4R f where R s = tensile resistance of the steel section, = ρyA. R f = resistance of the steel flange = BT ρy. 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 ) 5

5-6

5.4. Shear Connections in Negative Moment Regions The required number of shear connectors, N n, is Nn = R r/Q n Qn = design capacity of a shear connector in negative moment regions Qn = 0.6Qk compared with 0.8Q k for positive bending, because of the influence of cracking of concrete. A suitable spacing can be determined by calculating the total number of connectors 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 adjacent support.

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5.5. Analysis Methods Three analysis methods are recommended for determining the moments in continuous beams: Simplified method Elastic global analysis plastic hinge analysis 5.5.1 Simplified Method

The method is based on a “Table of coefficients” as given below. Certain restrictions 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 bending moment WL/8, where W is the total factored load on the span L. The values in the Table already allow for pattern loads and for redistribution. 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 limit state 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, the section properties are those of the cracked section for negative moments. The second moment of area of the cracked section is calculated using a section comprising the steel section together with the effectively anchored reinforcement located within the effective width of the concrete flange at the support. Outside the “15% length”, the section properties are those of the uncracked section, this being calculated using the mid-span effective width for the concrete flange but ignoring any longitudinal reinforcement. The continuous beam can be analysed using a standard program or formulae. The forces obtained from the analysis can be used to check against the capacity at various critical locations along the beam. Uncracked Section Method

The properties of the uncracked section are used throughout. The analysis can 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)

Be (Ds − D p )3 ABe (Ds − D p )(D + Ds + D p )2 + Ig = I x + 12αe 4[Aαe + Be (Ds − D p )] Cracked Section, Negative Moment (B.3.2) AA r (D − 2D r ) 2 In = Ix + a (A + A r )

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5.5.3 Redistribution of Moments for Elastic Analysis Design codes permit negative moment (hogging) at the supports to be reduced, except at cantilevers, by redistribution to mid-span. The extent of the redistribution is dependent on the method of analysis and section classification, 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%

11

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Elsatic Global Analysis Moments 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 as continuous, uncontrolled cracking is permitted by design codes provided that this does not impair the functioning of the structure. Most interiors of buildings for offices have low air humidity and crack 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 minimum areas of reinforcement to prevent fracture of the bars or the formation of very wide cracks under service loading. To avoid visible cracks where hard finishes are used, crack control joints should be considered. This part of the lecture therefore addresses deflections only. These are influenced by: • pattern loading • cracking of concrete • shakedown effects. Deflection

For uniformly distributed or symmetrical loading, the deflection at mid-span for a continuous beam is given approximately as

δc = δo{1-0.6(M 1+M2)/Mo} where δo = 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 redistribution etc).

The support moment M1 and M2 may be determined approximately using an elastic analysis 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 allow for absence of imposed load on the adjacent spans. To allow for the effect of pattern loading as shown in the figure, the moments at supports (not adjacent to cantilevers) are reduced by 30% for beams carrying normal loads or 50% for storage loads, to allow for pattern loading. For other non-symmetric load cases it is more accurate to calculate the deflections from the bending moment diagram at serviceability.

Shakedown effects

If the beam has been designed for ultimate moments determined by plastic analysis, or by elastic analysis 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 is reduced as follows. The support moment is calculated using the un-cracked section throughout, normally under dead load plus 80% of the imposed load. If this moment exceeds the plastic moment resistance of the section in negative (hogging) bending, the difference (termed the “excess moment”) is calculated. 14

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

Stresses

Determine the stresses in steel beam in the positive moment region based on the bending moment diagram used to calculate the service deflections. For unpropped beams add the steel stresses to those calculated for the steel section due to self weight of the floor. Check that the total stress does not exceed py. No further checks are required in the negative moment regions. 5.7 Summary: - Design Procedure

1) Loading and moments Obtain 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 the design moments in the negative and positive moment regions by multiplying the free bending moment by the coefficients. Select the steel section so that it can resist the negative moment obtained in the above without the need of reinforcement bars. Further refinement of section size may 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 analysis a) Simplified table of moment coefficient b) Elastic global analysis -uncracked section c) 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 connectors 10) Check construction stage 11) Check stability of the lower flange over the internal supports. 12) Provide transverse reinforcement 15

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

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

Design data: Steel: Grade 50 Concrete: Grade 30 light-weight Slab thickness Ds = 130mm Shear studs: 19mm diameter, 100mm length, use 2 studs per trough Metal decking perpendicular to the steel beam: Profile depth D p = 50mm, thickness t = 1mm , average trough width b r = 130mm, trough spacing = 300 Unfactored Dead Load = 8.1 kN/m Unfactored Imposed load = 18 kN/m Design Load = 1.4DL + 1.6IL w = 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!

w = 40 kN/m

12m

w= 40 kN/m

12m

17

Continuous Beam ABC L = 12m, w = 40kN/m

5-18

0.5wL2/8 = 361kNm

0.79wL2/8 =5715kNm For hogging moment

Select 406x178x60 UB Grade 50 Section is plastic in bending (NA in the web) Mcx = 426kNm > 361kNm OK! M = 361 kNm Classfication for bending, S355 steel: Plastic

W = 40 kN/m

Check shear at the intermediate suppo rt

F

12m

F

Fv = 6x 40 + 361/12= 271kN Pvx = 675kN Shear is OK 0.6Pvx = 405 > Fv = 271kN i.e., low shear Note that high Shear does not coincide with the maximum moment.

Check sagging resistance

Bo = 3000mm Be = 0.8L/4 = 0.8 x 12000/4 = 2400mm (Control!) Rc = 0.45f cuBe (Ds-D p) R c = 0.45 × 30 × 2400 × (130 − 50) × 10 −3 = 2590 kN R s = 2700kN R s > R c PNA is not in the concrte slab R w = R s – 2R f = 1084kN R c > R w PNA is in the steel flange 18

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D s + D p (R s − R c ) 2 T D − M c = R s + R c 2 2 R f 4

Mc = 781kNm>571kNm Shear Connection

R c = 2590kN R s = 2700kN Smaller of R c and R s is 2590kN. Capacity of shear connector (19mm diameter and 95mm long) in lightweight concrete Q k = 0. 9 × 100 kN = 90 kN Design capacity Q = 0.8 Q k = 72 kN Reduction factor for deck profile

b h k = 060 . Dr ⎜ D − 1⎟ ≤ 08 . for two studs per rib p

⎝

p

⎠

b r = Average trough width = 150 mm h = overall height of the stud = 95 mm

150 ⎛ 95 k = 060 . 50 ⎝ 50 − 1⎠ = 162 . > 08 . ∴k = 0.8

Resistance of a shear connector = 0.8 x 72 = 57.6kN For full composite, no. of connectors required = 2590/(57.6) = 45 Evaluate the x distance between points of zero moment

0.5wL2/8 = 361kNm

W = 40 kN/m

M = 361 kNm

0.79wL2/8 =5715kNm

x 19

F

12m

F = 271kN

5-20

Take moment about the zero moment point M = 0 = Fx – wx 2/2 X = 2F/w = 10.5m Since the trough spacing is 300 mm, no. of connectors that can be accommodated in half span (x/2), assuming two 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 = 0.78 Calculate reduced moment using simplified formulae M = Ms + k (Mc- Ms) = 426 + 0.78 (781-426) = 703kNm > 571 kNm OK ( More exact value for M c is 746 kNm > 571 kNm ) 35 + 35studs

35 + 35studs

10.5m

12m

10.5m

12m

Check deflection For Unfactored Imposed Load Only Unfactored imposed load, w = 18 kN/m α e = α s + ρι (α ι − α s ) α s = 10 α ι = 25 for lightweight concrete Long term loading: Dead load 8.1kN/m 1/3 Imposed Load 6 kN/m 20

5-21

Total Loading

14.1 kN/m = 26.1 kN/m

= DL + IL 141 .

. ρι = 261 = 054 .

Compute composite section properties αe = 18.1 ρl =0.54 Ig = I x +

Be ( Ds − D p ) 3 12αe

+

ABe ( Ds − D p )( D + Ds + Dp ) 2 4[Aαe + Be ( Ds − Dp )]

for uncracked section

I g = 60129 cm 4 Uncracked analysis

M1 =0.125wL2 = 324kNm IL w = 18kN/m

M1 reduces by 30% to allow for patterned loding M1 = 0.7 x 324 = 227kNm M = 0 2

M1 = 227

12m

δc = δ’o{1-0.6(M1+M2)/Mo} 5wl 4 δo = 384EI = 39.4mm =deflection of simply supported composite beam g 5wl 4 . mm =deflection of simply supported steel beam δs = 384EI = 1097 s M o = wl 2 / 8 21

5-22

To account for partial composite connection δ’o = δo + 0.3(1-Na/N p) (δs - δo) = 39.4 + 0.3 (1-0.78) ( 109.7- 39.4) = 44mm δc = δ’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 Construction Question 3 (a) Design the composite beam AB assuming that it is simply supported with a span length 12m as shown in Fig. Qa. Check the moment and shear resistances, and determine the number of shear connectors 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 at the 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: S275 Concrete: Grade 35 light-weight Slab thickness Ds = 150mm Shear studs: 19mm diameter, 100mm length, use 2 studs per trough Metal decking perpendicular to the steel beam: Profile depth D p = 60mm, thickness t = 1mm , average trough width br = 130mm Unfactored uniformly distributed loads: Dead Load = 11 kN/m Imposed load = 10 kN/m Design constrainst: Height of steel beam must be less than 400mm. Beam spacing = 3000mm

w kN/m A

Fi . a

B 12m D

w kN/m

C

E 12m

19mm diameter stud

Fi . b

12m

2 studs

300mm

Ds Dp

95mm 1mm thick steel deck 130mm

Fi . c 23

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Q4 A two-span composite beam of L 1 = 10.5m and L2=12m is shown in Fig. Q4. The hogging section at the intermediate support is reinforced by deformed bars of 12mm diameter spaced at 150mm. Using the simplified table in BS5950:Part3, check the moment resistances at the hogging and 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 74 Concrete slab: Grade 30, light-weight, slab thickness D s = 150mm Shear studs: 19mm diameter, adequately spaced for full composite action Re-bars: 12mm diameter, f y = 460N/mm2 Unfactored uniformly distributed loads: Dead Load = 9.5 kN/m Imposed load = 15.0 kN/m Beam spacing = 3000mm 12mm re-bars

150m 30m

Be

150m 12m

10.5m

Fi . 4

UB 457 x 191 x 74

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 an existing building. The floors to the new building are of reinforced concrete slab. Design data Roof

Insulated roof decking Purlins Services Self –weight of plate girders (estimated) Imposed load (Roof) Typical Floor

0.1 kN/m 2 0.1 kN/m 2 0.15 kN/m2 1.0 kN/m2 0.75 kN/m2

Finish 150 mm concrete slab Suspended ceiling Services Weight of steel beams (Approx.) Partition (superimposed dead load) Imposed load (Floor)

0.1 kN/m2 2.5 kN/m2 0.15 kN/m2 0.15 kN/m2 0.4 kN/m 2 1.00 kN/m 2 4.0 kN/m2

External cladding

4.00 kN/m2

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

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

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6000

6000

6000

6000

A

Outline of building Vertical Bracing

Column 'C'

Beam 'A'

8000 Hanger 'B'

8000

Vertical Bracing

A

8000

Denotes span of reinforced concrete slab

8000 fal

fal Plate Girder 'D'

Existing Building

Suspended Ceiling

Existing Floor

Hanger 'B'

3500 (clear height)

150 mm concrete floo

Suspended Ceiling

Existing Floor

150 mm concrete floo

3500

Suspended Ceiling

5000

SECTION A-A

26

View more...
• Continuous beams are an alternative to simply supported beams and their use is justified by • • • • • • •

considerations 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 by the classification of cross sections. Rigid-plastic design may be performed for beams with Class 1 cross-sections. Plastic moment resistance 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 hogging moment regions may cause unacceptable cracks, while in simply supported beams cracking is only 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, moment, shear resistance and resistance resistance against lateral-torsional buckling. buckling. • Understand the way shear connection is designed for class 1 and class c lass 2 cross-sections. • Be aware the need for serviceability checks for cracking in the hogging moment region.

5-1

CHAPTER 5 - CONTINUOUS COMPOSITE BEAMS 5.1. INTRODUCTION

Continuous beams offer the following advantages over simply supported beam construction: construction: 1. greater load capacity due to redistribution of moments, and 2. greater stiffness and therefore reduce deflection and vibration. The disadvantages associated with continuous beams are: 1. increase in complexity in design, and 2. susceptible to buckling in the negative moment region over internal supports. Two forms of buckling may occur: (i) local buckling of the web and/or bottom flange (ii) lateral torsional bukling. At the supports, the the sections ar subject to hogging hogging moments. moments. The concrete slab will tend to crack in tension. Reinforcement bars may be placed in the slab above the steel beam section to increase the moment resistance. Continuity of beam can be achieved by spanning secondary beams on top of the perpendicular perpendicula r to the primary beams. This concept is adopted in the parallel beam construction (see figure) which allows continuity in all beams and reducing reducing the need of expensive expensive beam-to-beam beam-to-beam conections. conections.

1

5-2

5.2. Effective Width, Be

For simply supported beam B e is taken as the sum of effective width b e of the 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 L o between the point of contraflexture. This length will depend on the type and magnitude of the loading on the various spans of the continuous beam, and will change in accordance with the different load cases. An approximation may be made as follows: For end span Lo = 0.8L Internal span Lo = 0.7L L = distance between supports for the span concerned. Over an internal support Lo = 0.25(L1 +L 2)

2

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5.3. Moment of Resistance in Hogging Bending The reinforcement bars within the effective width are assumed to be stressed to their design yield strength, ρy. The concrete slab may be assumed to be cracked. The tensile resistance of the reinforcement, R r , within the effective width of the slab under negative moment is given by: R r = ρyAr 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 by considering moment of each rectangular stress block about the neutral axis.

3

5-4

For compact section: Case 1: Rr < Rw (PNA lies in web)

d 76ε ≤ (Web compact) t 1 | + R r / R v

Moment about the center of the beam Mc’ = Ms + R r (0.5D + Dr ) + R r x X = R r = R r = R r d 2 p y t

2R v / d R v 2

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

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Case 2: Rs > Rr Rw (PNA lies in steel flange) d ≤ 38ε (Web compact) t

x=

R s − R r R s − R r T = 2 p y B R f 2

Moments about op of the steel flange Mc’ = 0.5R sD + R r D - (R s - R r )2T/4R f where R s = tensile resistance of the steel section, = ρyA. R f = resistance of the steel flange = BT ρy. 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 ) 5

5-6

5.4. Shear Connections in Negative Moment Regions The required number of shear connectors, N n, is Nn = R r/Q n Qn = design capacity of a shear connector in negative moment regions Qn = 0.6Qk compared with 0.8Q k for positive bending, because of the influence of cracking of concrete. A suitable spacing can be determined by calculating the total number of connectors 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 adjacent support.

6

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5.5. Analysis Methods Three analysis methods are recommended for determining the moments in continuous beams: Simplified method Elastic global analysis plastic hinge analysis 5.5.1 Simplified Method

The method is based on a “Table of coefficients” as given below. Certain restrictions 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 bending moment WL/8, where W is the total factored load on the span L. The values in the Table already allow for pattern loads and for redistribution. No further redistribution should be carried out.

7

5-8

Table: Simplified table of moment coefficients

8

5-9

5.5.2 Elastic Global Analysis Two methods of elastic global analysis are available for ultimate limit state 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, the section properties are those of the cracked section for negative moments. The second moment of area of the cracked section is calculated using a section comprising the steel section together with the effectively anchored reinforcement located within the effective width of the concrete flange at the support. Outside the “15% length”, the section properties are those of the uncracked section, this being calculated using the mid-span effective width for the concrete flange but ignoring any longitudinal reinforcement. The continuous beam can be analysed using a standard program or formulae. The forces obtained from the analysis can be used to check against the capacity at various critical locations along the beam. Uncracked Section Method

The properties of the uncracked section are used throughout. The analysis can be carried out without prior calculation of the cross-section.

9

5-10

Global Analysis based on Cracked Section Method

Uncracked Section (B.3.1)

Be (Ds − D p )3 ABe (Ds − D p )(D + Ds + D p )2 + Ig = I x + 12αe 4[Aαe + Be (Ds − D p )] Cracked Section, Negative Moment (B.3.2) AA r (D − 2D r ) 2 In = Ix + a (A + A r )

10

5-11

5.5.3 Redistribution of Moments for Elastic Analysis Design codes permit negative moment (hogging) at the supports to be reduced, except at cantilevers, by redistribution to mid-span. The extent of the redistribution is dependent on the method of analysis and section classification, 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%

11

5-12

Elsatic Global Analysis Moments are calculatd for two load cases:

12

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5.6 Serviceability Limit State Cracking may not need to be controlled in composite beams. If the slab is constructed as continuous, uncontrolled cracking is permitted by design codes provided that this does not impair the functioning of the structure. Most interiors of buildings for offices have low air humidity and crack 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 minimum areas of reinforcement to prevent fracture of the bars or the formation of very wide cracks under service loading. To avoid visible cracks where hard finishes are used, crack control joints should be considered. This part of the lecture therefore addresses deflections only. These are influenced by: • pattern loading • cracking of concrete • shakedown effects. Deflection

For uniformly distributed or symmetrical loading, the deflection at mid-span for a continuous beam is given approximately as

δc = δo{1-0.6(M 1+M2)/Mo} where δo = 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 redistribution etc).

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

13

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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 allow for absence of imposed load on the adjacent spans. To allow for the effect of pattern loading as shown in the figure, the moments at supports (not adjacent to cantilevers) are reduced by 30% for beams carrying normal loads or 50% for storage loads, to allow for pattern loading. For other non-symmetric load cases it is more accurate to calculate the deflections from the bending moment diagram at serviceability.

Shakedown effects

If the beam has been designed for ultimate moments determined by plastic analysis, or by elastic analysis 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 is reduced as follows. The support moment is calculated using the un-cracked section throughout, normally under dead load plus 80% of the imposed load. If this moment exceeds the plastic moment resistance of the section in negative (hogging) bending, the difference (termed the “excess moment”) is calculated. 14

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

Stresses

Determine the stresses in steel beam in the positive moment region based on the bending moment diagram used to calculate the service deflections. For unpropped beams add the steel stresses to those calculated for the steel section due to self weight of the floor. Check that the total stress does not exceed py. No further checks are required in the negative moment regions. 5.7 Summary: - Design Procedure

1) Loading and moments Obtain 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 the design moments in the negative and positive moment regions by multiplying the free bending moment by the coefficients. Select the steel section so that it can resist the negative moment obtained in the above without the need of reinforcement bars. Further refinement of section size may 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 analysis a) Simplified table of moment coefficient b) Elastic global analysis -uncracked section c) 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 connectors 10) Check construction stage 11) Check stability of the lower flange over the internal supports. 12) Provide transverse reinforcement 15

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

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

Design data: Steel: Grade 50 Concrete: Grade 30 light-weight Slab thickness Ds = 130mm Shear studs: 19mm diameter, 100mm length, use 2 studs per trough Metal decking perpendicular to the steel beam: Profile depth D p = 50mm, thickness t = 1mm , average trough width b r = 130mm, trough spacing = 300 Unfactored Dead Load = 8.1 kN/m Unfactored Imposed load = 18 kN/m Design Load = 1.4DL + 1.6IL w = 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!

w = 40 kN/m

12m

w= 40 kN/m

12m

17

Continuous Beam ABC L = 12m, w = 40kN/m

5-18

0.5wL2/8 = 361kNm

0.79wL2/8 =5715kNm For hogging moment

Select 406x178x60 UB Grade 50 Section is plastic in bending (NA in the web) Mcx = 426kNm > 361kNm OK! M = 361 kNm Classfication for bending, S355 steel: Plastic

W = 40 kN/m

Check shear at the intermediate suppo rt

F

12m

F

Fv = 6x 40 + 361/12= 271kN Pvx = 675kN Shear is OK 0.6Pvx = 405 > Fv = 271kN i.e., low shear Note that high Shear does not coincide with the maximum moment.

Check sagging resistance

Bo = 3000mm Be = 0.8L/4 = 0.8 x 12000/4 = 2400mm (Control!) Rc = 0.45f cuBe (Ds-D p) R c = 0.45 × 30 × 2400 × (130 − 50) × 10 −3 = 2590 kN R s = 2700kN R s > R c PNA is not in the concrte slab R w = R s – 2R f = 1084kN R c > R w PNA is in the steel flange 18

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D s + D p (R s − R c ) 2 T D − M c = R s + R c 2 2 R f 4

Mc = 781kNm>571kNm Shear Connection

R c = 2590kN R s = 2700kN Smaller of R c and R s is 2590kN. Capacity of shear connector (19mm diameter and 95mm long) in lightweight concrete Q k = 0. 9 × 100 kN = 90 kN Design capacity Q = 0.8 Q k = 72 kN Reduction factor for deck profile

b h k = 060 . Dr ⎜ D − 1⎟ ≤ 08 . for two studs per rib p

⎝

p

⎠

b r = Average trough width = 150 mm h = overall height of the stud = 95 mm

150 ⎛ 95 k = 060 . 50 ⎝ 50 − 1⎠ = 162 . > 08 . ∴k = 0.8

Resistance of a shear connector = 0.8 x 72 = 57.6kN For full composite, no. of connectors required = 2590/(57.6) = 45 Evaluate the x distance between points of zero moment

0.5wL2/8 = 361kNm

W = 40 kN/m

M = 361 kNm

0.79wL2/8 =5715kNm

x 19

F

12m

F = 271kN

5-20

Take moment about the zero moment point M = 0 = Fx – wx 2/2 X = 2F/w = 10.5m Since the trough spacing is 300 mm, no. of connectors that can be accommodated in half span (x/2), assuming two 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 = 0.78 Calculate reduced moment using simplified formulae M = Ms + k (Mc- Ms) = 426 + 0.78 (781-426) = 703kNm > 571 kNm OK ( More exact value for M c is 746 kNm > 571 kNm ) 35 + 35studs

35 + 35studs

10.5m

12m

10.5m

12m

Check deflection For Unfactored Imposed Load Only Unfactored imposed load, w = 18 kN/m α e = α s + ρι (α ι − α s ) α s = 10 α ι = 25 for lightweight concrete Long term loading: Dead load 8.1kN/m 1/3 Imposed Load 6 kN/m 20

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Total Loading

14.1 kN/m = 26.1 kN/m

= DL + IL 141 .

. ρι = 261 = 054 .

Compute composite section properties αe = 18.1 ρl =0.54 Ig = I x +

Be ( Ds − D p ) 3 12αe

+

ABe ( Ds − D p )( D + Ds + Dp ) 2 4[Aαe + Be ( Ds − Dp )]

for uncracked section

I g = 60129 cm 4 Uncracked analysis

M1 =0.125wL2 = 324kNm IL w = 18kN/m

M1 reduces by 30% to allow for patterned loding M1 = 0.7 x 324 = 227kNm M = 0 2

M1 = 227

12m

δc = δ’o{1-0.6(M1+M2)/Mo} 5wl 4 δo = 384EI = 39.4mm =deflection of simply supported composite beam g 5wl 4 . mm =deflection of simply supported steel beam δs = 384EI = 1097 s M o = wl 2 / 8 21

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To account for partial composite connection δ’o = δo + 0.3(1-Na/N p) (δs - δo) = 39.4 + 0.3 (1-0.78) ( 109.7- 39.4) = 44mm δc = δ’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 Construction Question 3 (a) Design the composite beam AB assuming that it is simply supported with a span length 12m as shown in Fig. Qa. Check the moment and shear resistances, and determine the number of shear connectors 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 at the 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: S275 Concrete: Grade 35 light-weight Slab thickness Ds = 150mm Shear studs: 19mm diameter, 100mm length, use 2 studs per trough Metal decking perpendicular to the steel beam: Profile depth D p = 60mm, thickness t = 1mm , average trough width br = 130mm Unfactored uniformly distributed loads: Dead Load = 11 kN/m Imposed load = 10 kN/m Design constrainst: Height of steel beam must be less than 400mm. Beam spacing = 3000mm

w kN/m A

Fi . a

B 12m D

w kN/m

C

E 12m

19mm diameter stud

Fi . b

12m

2 studs

300mm

Ds Dp

95mm 1mm thick steel deck 130mm

Fi . c 23

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Q4 A two-span composite beam of L 1 = 10.5m and L2=12m is shown in Fig. Q4. The hogging section at the intermediate support is reinforced by deformed bars of 12mm diameter spaced at 150mm. Using the simplified table in BS5950:Part3, check the moment resistances at the hogging and 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 74 Concrete slab: Grade 30, light-weight, slab thickness D s = 150mm Shear studs: 19mm diameter, adequately spaced for full composite action Re-bars: 12mm diameter, f y = 460N/mm2 Unfactored uniformly distributed loads: Dead Load = 9.5 kN/m Imposed load = 15.0 kN/m Beam spacing = 3000mm 12mm re-bars

150m 30m

Be

150m 12m

10.5m

Fi . 4

UB 457 x 191 x 74

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 an existing building. The floors to the new building are of reinforced concrete slab. Design data Roof

Insulated roof decking Purlins Services Self –weight of plate girders (estimated) Imposed load (Roof) Typical Floor

0.1 kN/m 2 0.1 kN/m 2 0.15 kN/m2 1.0 kN/m2 0.75 kN/m2

Finish 150 mm concrete slab Suspended ceiling Services Weight of steel beams (Approx.) Partition (superimposed dead load) Imposed load (Floor)

0.1 kN/m2 2.5 kN/m2 0.15 kN/m2 0.15 kN/m2 0.4 kN/m 2 1.00 kN/m 2 4.0 kN/m2

External cladding

4.00 kN/m2

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

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

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6000

6000

6000

6000

A

Outline of building Vertical Bracing

Column 'C'

Beam 'A'

8000 Hanger 'B'

8000

Vertical Bracing

A

8000

Denotes span of reinforced concrete slab

8000 fal

fal Plate Girder 'D'

Existing Building

Suspended Ceiling

Existing Floor

Hanger 'B'

3500 (clear height)

150 mm concrete floo

Suspended Ceiling

Existing Floor

150 mm concrete floo

3500

Suspended Ceiling

5000

SECTION A-A

26

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