Composite Structures Chap 3

7/25/2019 Composite Structures Chap 3

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3.1 Introduction

Composite slabs are structural plates made ofsteel profiled sheeting and reinforcedconcrete. (The sheeting is very thin foreconomic reasons, usually between 0.8 mmand 1.2 mm).

During construction the permanent steel

formwork serves as a working platform, andafter the concrete has hardened as bottomreinforcement for the slab

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3.2 Design of composite slab

Area of the steel sheeting that is neededfor the construction phase often providesmore than enoughbottom reinforcement

for the composite slab. It is usual to design the slabs as simplysupported

The concrete is continuous over thesupporting beams, so are also thealternate sheets(sheets are typically 1mwide up to 6m long)

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The simply supported "slabs require toplongitudinal reinforcement at their supports, tocontrol crack widths

EC4 specifies 0.2% of the concrete area abovethe steel ribs for unpropped construction and0.4% for propped construction

Long span slabs are sometimes designed as

continuous To avoid local buckling, the breadth/thickness

ratio does not exceed 35.

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3.3 Resistance of composite slabs to positive

bending

The total depth htis required by EC4 nottobe less than 80 mm, and the toppingnot lessthan 40 mm

Normally the topping is 60 mm thick or more,to provide sufficient sound or fire insulation,

and resistance to concentrated loads

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Except where the sheeting is unusually deep,the NA liesin the concretewhere there is fullinteraction

For partial interaction there is always NAwithin the steel section

Local buckling of compressed sheeting hasto be considered

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This is done by using effective widths of flatregions of sheeting which are allowed to beup to twice the limits given for class 1steelweb plates in beams, because the concrete

prevents the sheeting from buckling

The effective area per meter width, Ap, and

the height of the center of area, e, are usuallybased on tests. These usually show that ep,the height of the PNA, is different from e.

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Because local buckling is allowed in this way, thebending resistance can be calculated by simpleplastic theory.

compressive force in the concrete component, Ncf axial load capacity of the steel component, Npa

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There are three casesi) Neutral axis above the sheeting and full

shear connection

The compressive force in the concrete

component, Ncf, is equal to the axial loadcapacity of the steel component, Npa

Horizontal equilibrium for pure bending

Ncf= Npa= Apfyp/px= Ncf/(b(0.85fck/c))

MP,Rd= Ncf(dp0.5x)

Where fyp= the characteristic yield strength of the sheeting

MP,Rd The design resistance to bending, x=NA depth 10


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ii) Neutral axis within the sheeting, and full

shear connection

Now the axial load capacity of the steelcomponent N

pa

is greater than the capacity ofthe concrete, and neglecting the compressionin the concrete within the ribs, Ncfis given by:

Ncf= bhc(0.85fck/c) . . . (f stands for full interaction)

There is no simple method of calculating x,because of the complex properties of profiledsheeting

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the following approximate method is usedin EC4

The tensile force in the sheeting isdecomposedinto a force at the bottom equalto Nac(the compressive force in the sheeting)and a force Na, where the following criteria issatisfied:

Na= Ncf

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The equal and opposite forces Nacprovide aresistance moment Mpr, equal to the plasticmoment resistance for the sheeting Mpa,reducedby the effect of the axial force Na.

the value represented by the symbol NC,fdepends on the ratio x/hc.

The relationship Mpr

/Mpa

and Ncf

/Npa

dependson the profile, but is typically as shown by thedashed line ABC in figure.

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This is approximated in EC4 by the equation:Mpr=1.25Mpa(1-Ncf/Npa)Mpa

which is shown by as ADC

Thus, the resistance moment is then given by:

Mp,Rd= Ncfz + Mpr (see previous figure)

Check the two points E and F for validity of theextreme cases:

The lowest level that the NA can reach is thecentroidal axis, because NA below that requirestension in the concrete for horizontal equilibrium.

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Now investigate the situation as the NA enters theprofile

Point F corresponds to the beginning where theNA is at the interface at the top of the sheeting

here, Ncf= NpaNac= 0 Mpr= 0 The design moment resistance

MpR,d= Ncf(dp0.5x) with x =hc The lever arm z is found by the approximation

shown by line EF.

z = dp0.5hc= ht e 0.5hcas given by point F.

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To check point E, we assume that Ncfis nearlyzero (e.g. if the concrete is too weak) Na0and MprMpa

The NA for Mpaalone is at the height epfromthe bottom of the sheet, and the lever arm is:

z = ht e 0.5hcas given by point F.

This method has been validated by tests

The line EF is given by: z = ht 0.5hcep=(epe)Ncf/Npa

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iii) Partial shear connection

The compression force in the slab, Nc, is nowless than the compressive strength of theslab, as is the tensile force Naless than Npa.

The forces Nc, and Npaare determined by thestrength of the shear connection

The NA depth x in the concrete component is

determined by: x = Nc/(b(0.85fck/c)


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There is a second NA within the steelsheeting, and the stress blocks are as shownin the previous figure.

The calculation of MpR,dis as for method (ii),except that Ncfis replaced by Nc, Npaby Ncf,and hcby x, so that:

z = ht 0.5x ep+(epe)Nc/Npa Mpr=1.25Mpa(1-Nc/Ncf)Mpa

Mp,Rd= Ncz + Mpr

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3.4 Resistance of composite slabs to

longitudinal shear

For profiled sheeting that relies on frictionalinterlock to transmit longitudinal shear, thereis no satisfactory conceptual model.

This led to the development of the shear-bond test, and the empirical m-k method of

design where the shear resistanceis given bythe EC4 equation:

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Vl,Rd= bdp(mAp/bLs+k)/vs (refer chapter 2)

Where m and k are constants


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For uniformly distributed load on a span L, the length Ls, istaken as L/4 (from equivalent area of SFDs for UDL andtwo point loading in the experiment)

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Defects of the m-k method (i)The m-k method is not based on a

mechanical model, so that conservativeassumptions have to be made in design when

the dimensions, materials, or loading differfrom those used in the tests (e.g. calculationof Ls)

(ii) many additional tests are needed before

the range of application can be extended; fore.g., to include end anchorage or the use oflongitudinal reinforcing bars

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(iii) The method of evaluation of test data isthe same, whether the failure is brittle orductile. The use in EC4 of a penalty factor of0.8 for brittle behavior does not adequately

represent the advantage of using sheetingwith good mechanical interlock, because thisincreases with span

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Partial interaction

The method based on shear-bond testsaccording to Bode is described first

The method based on slip-block tests, takesmore specific account of the effects offriction near supports and can be moreeconomical for short spans

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For composite slabs of given x-sectionandmaterials, the result of each shear-bond teston a profile with ductile behavior enables thedegree of partial shear connectionin that test

to be calculated This gives the compressive force Nc

transferred from the sheeting to the slabwithin the shear span of known length L

s

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The ultimate uniformly distributed (assumed)shear stress uover the shear span Lscan then bedetermined from the following equation:

u= (

testN

cf)/(bL

s)

There is usually a mid-span region where fullshear connection is achieved, and Mp,Rdisindependent of x.

For safe design, this curve of Mp,Rdas a functionof x (the resistance diagram) must at all points lieabove the BMD for the applied loading.

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3.5 Resistance of composite slabs to vertical

shear

Tests show that resistance to vertical shear isprovided mainly by the concrete ribs.

For open profiles, their effective width boistaken as the mean width, though the width atthe centroidal axis is accurate enough.

Design methods are based on those for in RCT-beams

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EC4 gives the design shear resistance of acomposite slab, Vv,Rdwith ribs of effectivewidth boat spacing b as:

Vv,Rd=(bo/b)dpRdkv(1.2 + 40) per unit width

Where:

dp = the depth to the centroidal axis

Rd= the basic shear strength of the concrete

Rd = 0.25fctk0.05/c dpin m. allows for highershear strength of shallow

members.

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Kv= (1.6 dp)


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Resistance to vertical shear is most likely tobe critical in design where span/depth ratiosare low, as is the case for beams.

Punching Shear

Where a thin composite slab has to bedesigned to resist point loads, resistance to

punching shear should be checked.

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Critical perimeter

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Serviceability limit states for composite slabs

ASSIGNMENT

Consider different codes and discuss

Deflection Cracking of concrete and

Fire resistance and temperature

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Cracking of concrete

EC4 specifies minimum amounts ofreinforcement above internal supports;

0.2% and 0.4% of the area of concreteabove the sheeting for unpropped andpropped construction respectively.

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EC4 specifies that the deflection of thesheeting due its own weight and wetconcrete slab should not exceed L/180

or 20 mm, where L is the effectivespan. This span can be reduced to anydesired level by using propped

construction-but cost increases For service: the maximum deflectionshould not exceed L/250

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All buildings are vulnerable to damage fromfire, which is usually the first accidentaldesign situation to be considered in design

thermal insulation criterion, denoted I,

concerned with limiting the transmission ofheat by conduction, and

integrity criterion, denoted E, concerned withpreventing the passage of flames and hotgases into an adjacent compartment.

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Partial safety factors for fire =1)

Design action effects for fire

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Design Example: Composite Slab

NEXT CLASS

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The calculation of MpR,dis as for method (ii),except that Ncfis replaced by Nc, Npaby Ncf,and hcby x, so that:

z = ht 0.5x ep+(epe)Nc/NpaMpr=1.25Mpa(1-Nc/Ncf)Mpa

Mp,Rd= Ncz + Mpr

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These enable M

Rd

to be calculated for

any value of between zero and 1.0


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V=0.23*112*1000/3.3=25.7kN/3.3= 7.78kN

Assume there are3.3 studs (19mm)

per meter.

per meter

Composite Structures Chap 3

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