Chapter 5 Continuous composite beams.pdf

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Summary: Summary:

••

Continuous beams are an alternative to simply supported beams and their use is justified byContinuous beams are an alternative to simply supported beams and their use is justified by considerations of economy.

considerations of economy.

••

In the hogging bending regions at supports, concrete will be in tension and the steel bottomIn 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 flange in compression, leading to the possibility of onset of local buckling. This is taken up by the classification of cross sections.

the classification of cross sections.

••

Rigid-plastic design may be performed for beams with Class 1 cross-sections. Plastic momentRigid-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

resistance of cross-sections can be used for Class 1 and 2 cross-sections.2 cross-sections.

••

For Class 3 sections, elastic analysis and For Class 3 sections, elastic analysis and elastic cross-section resistance must be used.elastic cross-section resistance must be used.

••

The principles of calculation of cross-section resistance, either plastic or elastic, are similar toThe 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

the case of sagging bending. The tension resistance of concrete is neglected.neglected.

••

Lateral-torsional buckling is a special phenomenon which can be prevented by conforming toLateral-torsional buckling is a special phenomenon which can be prevented by conforming to certain detailing rules.

certain detailing rules.

••

The design of the shear connection in the case of continuous beams is more complex than forThe design of the shear connection in the case of continuous beams is more complex than for simply supported beams.

simply supported beams.

••

Serviceability checks include deflection and vibration control as well as that of concreteServiceability 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 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 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.

only due to shrinkage of concrete and is therefore lower in magnitude.

Objectives: Objectives: The student should: The student should:

••

Appreciate the advantages of continuous composite beams and be aware of their disadvantages.Appreciate the advantages of continuous composite beams and be aware of their disadvantages.

••

Understand the methods of plastic and Understand the methods of plastic and elastic design of continuous beams.elastic design of continuous beams.

••

Understand the methods of calculation of Understand the methods of calculation of elastic and plastic elastic and plastic cross-section resistance for hoggingcross-section resistance for hogging bending mom

bending moment, shear resistance and resistanent, shear resistance and resistance against lateral-torsional bce against lateral-torsional buckling.uckling.

••

Understand the way shear connection is designed for class 1 and cUnderstand the way shear connection is designed for class 1 and class 2 cross-sections.lass 2 cross-sections.

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CHAPTER 5 - CONTINUOUS COMPOSITE BEAMS

5.1. INTRODUCTION

Continuous beams offer the following advantages over simply supported

beam constr

beam construction:

uction:

1. 1. greater load capacity due to redistribution of moments, and

greater load capacity due to redistribution of moments, and

2. 2. greater stiffness and

greater stiffness and therefore reduce deflection and vibration.

therefore reduce deflection and vibration.

The disadvantages associated with continuous beams

The disadvantages associated with continuous beams are:

are:

1. 1. increase in complexity in design, and

increase in complexity in design, and

2. 2. susceptible to buckling in the negative moment region over internal

susceptible to buckling in the negative moment region over internal

supports.

supports. Two forms

Two forms of buckling

of buckling may occur:

may occur: (i) local

(i) local buckling of

buckling of the

the

web and/or bottom flange (ii) lateral torsional bukling.

At the supports, th

At the supports, the sections ar subject to ho

e sections ar subject to hogging mom

gging moments.

ents. The concrete

The concrete

slab will

slab will tend to

tend to crack in

crack in tension.

tension. Reinforcement bars may

Reinforcement bars may be placed

be placed in

in

the slab above the steel beam section to

the slab above the steel beam section to increase the moment resistance.

increase the moment resistance.

Continuity of beam can be achieved by spanning secondary beams on top

of the

of the perpendicula

perpendicular to

r to the primary beams.

the primary beams. This concept is

This concept is adopted in the

adopted in the

parallel

parallel beam

beam construction

construction (see

(see figure)

figure) which

which allows

allows continuity

continuity in

in all

all

beams and r

(3)

2

5.2. Effective Width, B

_e

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

L

_o

= 0.8L

Internal span

L

o

= 0.7L

L = distance between supports for the span concerned.

Over an internal support

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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,

ρ

y

.

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

=

ρ

y

A

r

A

r

= area of the reinforcement within the effective width.

The axial resistance of the web is

R

_w

= dtp

_y

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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4

For compact section:

Case 1: R

_r

< R

_w

(PNA lies in web)

v r

/

R

|

1

76 t

d

+

ε

≤

(Web compact)

Moment about the center of the beam

M

_c

’ = M

_s

+ R

_r

(0.5D + D

_r

) + R

_r

x

X =

2 d R R d / R 2 R t p 2 R v r v r y r

₌

M

c

’ = M

s

+ R

r

(0.5D + D

r

) + (R

r 2

d)/4R

w

where

M

s

= the plastic resistance moment of the steel section alone

D = overall depth of the steel section

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Case 2: R

_s

> R

_r

R

_w

(PNA lies in steel flange)

ε

≤

38 t

d

(Web compact)

2 T

R

B

p

2 R

R

x

f r s y r s

−

₌

−

=

Moments about op of the steel flange

M

_c

’ = 0.5R

_s

D + R

_r

D - (R

_s

- R

_r

)

2

T/4R

_f

where

R

_s

= tensile resistance of the steel section, =

ρ

_y

A

_.

R

f

= resistance of the steel flange = BT

ρ

y

.

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)

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6

5.4. Shear Connections in Negative Moment Regions

The required number of shear connectors, N

n

, is

N

n

= R

r

/Q

n

Q

_n

= design capacity of a shear connector in negative moment regions

Q

n

= 0.6Q

k

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

+ N

_n

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

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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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8

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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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10

Global Analysis based on Cracked Section Method

Uncracked Section (B.3.1)

I

_g

I

_x

B D

e

s

D

p

AB D

D

_{B D}

D D

_D

D

e

s

p

s

p

e

s

p

= +

(

−

)

+

(

−

₊

)(

+

₋

+

)

[A

(

)]

3

2

12

α

4

α

Cracked Section, Negative Moment (B.3.2)

)

A

(

a

)

D

2 D

(

AA

I

r 2 r r x n

+

−

+

=

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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

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12

Elsatic Global Analysis

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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

δ

c =

δ

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

where

δ

o = 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 M

1

and M

2

may be determined approximately using an elastic

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14

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.

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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

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16 13) Check service load deflection and stresses.

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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 D

_s

= 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!

12m

w = 40 kN/m

12m

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18

Continuous Beam ABC

L = 12m, w = 40kN/m

For hogging moment

Select 406x178x60 UB Grade 50

Section is plastic in bending (NA in the web)

M

cx

= 426kNm > 361kNm OK!

Classfication for bending, S355 steel: Plastic

Check shear at the intermediate suppo rt

F

_v

= 6x 40 + 361/12= 271kN

P

_vx

= 675kN

Shear is OK

0.6P

vx

= 405 > F

v

= 271kN i.e., low shear

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

Check sagging resistance

B

o

= 3000mm

B

e

= 0.8L/4 = 0.8 x 12000/4 = 2400mm (Control!)

Rc = 0.45f

cu

B

e

(D

s

-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

0.5wL

2

/8 = 361kNm

0.79wL

2

/8 =5715kNm

F

12m

M = 361 kNm W = 40 kN/m

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19 4 T R ) R R ( 2 D D R 2 D R M f 2 c s p s c s c

−

+

=

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 = 0.8Q

_k

=

72 kN

Reduction factor for deck profile

k

b

_D

r

_D

h

p p

=

060 .

_⎝

⎜

−

1 _⎠

⎟ ≤

08 .

for two studs per rib

b

_r

= Average trough width = 150 mm

h = overall height of the stud = 95 mm

k

=

060 .

150 ₅₀

⎛

_⎝

_{50 1}

95 −

_{⎠ =}

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

0.5wL

2

/8 = 361kNm

0.79wL

2

/8 =5715kNm

x

M = 361 kNm W = 40 kN/m

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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 = M

_s

+ k (M

_c

- M

_s

)

= 426 + 0.78 (781-426)

= 703kNm > 571 kNm OK

( More exact value for M

c

is 746 kNm > 571 kNm )

Check deflection For Unfactored Imposed Load Only

Unfactored imposed load, w = 18 kN/m

α α ρ α α

_e

=

_s

+

_ι

(

_ι

−

_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

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

ρ

_ι

=

141 =

261

054 .

.

Compute composite section properties

α

_e

= 18.1

ρ

_l

=0.54

I I B D D AB D D D D D A B D D g x e s p e e s p s p e e s p

= +

−

+

−

+

−

( ) ( )( ) [ ( )] 3 2

12

α

4

α

for uncracked section

I

_g

= 60129 cm

4

Uncracked analysis

M

₁

reduces by 30% to allow for patterned loding

M

1

= 0.7 x 324 = 227kNm

δ

_c

=

δ

’o

{1-0.6(M

1

+M

2

)/M

o

}

δ

_o

=

5

384 39 4

4

wl

EI

_g

=

.

mm

=

deflection of simply supported composite beam

δ

_s

=

₃₈₄

5 wl

_EI

4

1097

mm

s

=

.

=

deflection of simply supported steel beam

M

_o

=

wl

2

_/

₈

M

1

=0.125wL

2

= 324kNm

IL w = 18kN/m

M1 = 227

12m

M2 = 0

(23)

22

To account for partial composite connection

δ

’ o

=

δ

o

+ 0.3(1-N

a

/N

p

) (

δ

s

-

δ

o

)

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

= 44mm

δ

_c

=

δ

’

_o

{1-0.6(M

₁

+M

₂

)/M

_o

} = 44(1-0.6x0.7)

= 26mm < L/360 = 33mm OK

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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 D_s D_p 300mm 130mm 1mm thick steel deck 95mm 19mm diameter stud 12m w kN/m A B 12m 12m w kN/m C D E Fi . a Fi . b 2 studs Fi . c

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24 Q4 A two-span composite beam of L1 = 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 Ds = 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 Fi . 4 10.5m 12m 150m 30m Be 150m UB 457 x 191 x 74 12mm re-bars

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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 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 Floor

Finish 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 beam deflection 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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26 8000 3500 (clear height) 3500 5000 Hanger 'B' Suspended Ceiling Suspended Ceiling Suspended Ceiling 8000 150 mm concrete floo 150 mm concrete floo Plate Girder 'D' fal fal Existing Floor Existing Floor Existing Building SECTION A-A A A Denotes span of reinforced concrete slab Vertical Bracing Vertical Bracing Hanger 'B' Column 'C' Beam 'A' 6000 6000 6000 6000 Outline of building 8000 8000