Carte Mixte Draft 2.0

8/10/2019 Carte Mixte Draft 2.0

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COMPOSITE STRUCTURES

COURSE NOTES AND WORKED EXAMPLES

Authors:

Gelu Danku, PhD, Eng.

Adrian Ciutina, PhD, Senior Lecturer, Eng.


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Chapter 1 - INTRODUCTION

1.1. Composite beams and slabs

The design of structures for buildings and bridges is mainly concerned with the

provision and support of load bearing horizontal surfaces. Despite the great advantages of

reinforced concrete ( a good combination of low cost and high strength, etc.) at spans ofmore than about 10 m steel beams become cheaper than concrete beams, to support the

concrete slab.

Fig.1. Composite beams layout

By about 1950 the development of shear connectors (conectori de forfecare) had

made practicable to connect the concrete slab to the steel beam. The term "compositebeams" used in this course refers to this type of structure. To illustrate the concept of

composite beam we consider a beam consisting of two parts acting first separately (a) or

compositely (b).

(a) (b)Fig.2. Non-composite and composite beam

For the non-composite beam (a) the load will be shared between the two parts

with each deforming in bending, generating separately the typical linear variation of

elastic strain over its own depth (height). Now consider the same beam but with

continuity preserved along the longitudinal interface (b), both parts respond now as a

unit. Bending strains will vary also linearly, but over the entire depth of the beam, with

the neutral axis to the combined section (steel - concrete) corresponding to the position of

0 strains. Moreover since no horizontal slip will occur at the interface vertical lines drawn


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on the depth of the section before loading will remain as single an continuous lines after

deformation.The composite beams are more efficient structurally because they develop smaller

deflections and strains than the non-composite beams.

Fig3. Horizontal slip prevented by the connectors, in composite beam

The deflection in the composite beam c is only 25 % from the non-composite

beam:

c=0.25sand in terms of strain:

c 0.5 s:= In composite beams the steel beam is designed to act with a part of the slab,

preventing the slip at the interface, using the shear connectors.

The main typical cross-sections of composite beams are:

a) composite beam with full slab

b) composite beam with ribbed slab having the ribs perpendicular to the steelbeam

c) composite beam with ribbed slab, having the ribs parallel to the steel beam

The composite floor slabs are constructed from ribbed profiles and concrete slab.The slabs are cast on permanent steel formwork which acts first as a working platform

and than after hardening of concrete it works as a bottom reinforcement for slab. This

formwork is called profiled steel sheeting and these floors are called composite slabs.

1.2. Composite columns

First the columns from steel frames were encased in concrete only to protect them

from fire, without considering the influence of this concrete at the column strength. Then

it was realized that the encasement reduces the effective slenderness of the column and

increases the buckling load of the column.

Eurocode 4 deals with 3 main typical cases of composite columns:a) partially encased in concrete


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- SLS - serviceability limit state, associated to excessive deformations, vibrations,

cracking of concrete, etc.

Practically in ULS is made the strength calculation, while in SLS is made therigidity (deformations) calculation.

In ULS : G= 1.35

Q= 1.5

In SLS : G= Q= 1.0

2.2. Checks performed for an ultimate limit state

The general verification relation is:

where Sdis the internal force or moment (N); in other words it is the internal stress fromthe static calculus. Rd is the corresponding design resistance; the stress from strength

calculus.

with

The resistance Rd is calculated using the design values of properties of the material:

where Xk is the characteristic value of the property and Ma partial safety factor of thematerial.

MaterialStructural

steel

Reinforcement

steel

Profiled steel

sheetingConcrete

Shear

connectors

Property fy fsk fyp fck fsk

Symbol for

Ma s ap c v

ULS 1.10 1.15 1.10 1.5 1.25

SLS 1.00 1.00 1.00 1.00 1.00

Where fy, fyp- yield strengthsfsk- characteristic yield strengthfck- characteristic compressive strength measure on cylinder

Example: - for mild steel

Sd Rd

N

A:=

NRd A R:=

Xd

Xk

M:=

XX

y 235:= N

mm2

a 1.1:=

d

f

ya

:= d 2351.1

:= d 213.636= Nmm

2

Rfy

a

:=fyfy


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2.3. Combinations of actions

Principles - permanent actions are present in all combinations;

- each variable action is chosen in term to be the "leading action" and it is

combined with lower combination values of other relevant variable action;

- the design action in fact is the most unfavorable of those calculated by

this process.

Example: It is assumed that a member is subjected to bending, the design bending

moment Mdis given by: its own weight G, an imposed vertical load Q1and the wind load

Q2.The fundamental combinations of actions are:

a) G x Gk+ Q,1 x Qk,1 + Q,2 x 0,2 x Qk,2

b) Gx Gk+ Q,1 x 0,1x Qk,1 + Q,2 x Qk,2

Where G, Q,1 are partial safety factors for actions;

0,i is a combination factor.

In practice it is obvious which combination will govern:- for low-rise buildings wind is rarely critical so combination (a) will be used;

- for long span lightweight roof wind is important, so combination (b) will be

considered, taking both positive and negative wind pressure.

Eurocode 4 permits the use of simplified combinations. For the example above, assuming

Q1is more adverse that Q2these combinations are:Gx Gk+ Q,1 x Qk,1

and

Gx Gk+ 0,9 ( Q,1 x Qk,1+ Q,2 x Qk,2)

CHAPTER III - MATERIALS USED IN COMPOSITE

CONSTRUCTIONS

Any member must be calculated from strength and rigidity point of view. The

strength calculation is made with the so-called global analysis, which means the

determination of internal stresses (bending moments, shear forces, axial forces etc.) to use

this analysis all three main materials: concrete, structural steel and reinforcing steel, are

assumed to behave in a linear-elastic manner.

3.1. Concrete

Strength classes Eurocode 4 works with the strength classes given in Eurocode2. (e.g. a typical strength class for concrete used in composite structures is

C25/30, where the first number is cylinder strength of concrete at compression


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3.2. Reinforcing Steel

In all calculations we use the characteristic yield strength for reinforcing fskgivenseparately for ribbed bars and for welded mesh. The modulus of elasticity Es=Ea=210

kN/mm2.

3.3. Structural steel

The yield strength fy and the ultimate tensile strength fu are given function the

steel grades. For our examples we shall work with the mild steel having fy=235 N/mm2

and fu=360 N/mm2. The modulus of elasticity Ea=210 kN/mm

2.

3.4. Profiled steel sheeting

EUROCODE 4 recommends a minimum thickness for sheeting of 0.75 mm

(usually approx. 1mm). The yield strength for sheeting is fyp=235...460 N/mm2. The

sheets are protected from corrosion by a Zn coating on each face. All the elastic

properties are identical as for structural steel.

CHAPTER IV COMPOSITE BEAMS

The concept of composite beams refers to the members composed from a concreteslab (full or ribbed) connected with shear connectors to the steel beam.

4.1. Effective width of concrete flange

We assume we have a composite floor composed of steel beams of multiple spans

and situated at distance Bi, and a concrete slab.

Fig.3. Distribution of stresses in continuous composite beams


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As we have seen at theory of elasticity the compressive stresses in the concrete

slab aren't uniform in the longitudinal section A-A. The stresses are maximum at thesupports and decrease towards the middle bay. To study the floor as an ensemble of

independent T beams it must be considered an effective width of slab where an uniformlydistribution of compressive stresses can be accepted.

For the simply supported beam L0=L.For the continuous beam - both positive and negative effective widths should be

calculated function the distances L0 given in the following figure:

Fig.4. Distribution of bending moments assumed in the calculation of the effective width of the composite

cross-section

Examples:

b1 b2+ beff:=b1 b2+

The effective width is beff b i b1 b2+( )=:= b ib i where biis the effective width ofthe concrete slab on each side from

the center line of the steel web.bi minBi

2

L0

8,

:=

Bi

L is the distance between 2 consecutive points of 0 bending moment.

Bi is the bay.


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2

Plastic, Mpl Rd

Limited

3

Elastic, Mel Rd

-

4

Elastic, Mel Rd

-

The class of a composite CS is determined function the least favorable class of itssteel elements in compression. This must be made for each section of positive

respectively negative bending. The steel flanges in compression are classified in

EUROCODE 4 according to the maximum width-to-thickness ratio for steel outstand in

compression. The steel web is classified function the width-to-thickness ratio d/d w. If the

steel compressed stand is effectively attached to the concrete slab by shear connectors,for positive bending moment it may be assumed to be in class 1. If the slab is compressedunder positive moment and the steel beam is in class 1 the neutral axis position imposes

the class of the composite beam.

- if NA is in the slab or in the compressed flange the entire section is in class 1

- if NA passes through the steel web the entire CS is in class 2

The plastic (non-linear) analysis is permitted only for CS having class 1 or 2.The elastic (linear) analysis may be applied to CS of any class.

Generally the plastic analysis is permitted for composite beams while a total

connection is considered between the concrete slab and the steel profile. Anyway anelastic calculus is necessary in SLS state.

Studies on composite beams with critical sections in Class 3 or Class 4 have

shown that provided at least 10% of the span is cracked, as is likely in practice, the

reduction in support moment due to cracking will exceed 8%. It is reasonable to assume

therefore that in round terms the difference between an 'uncracked' and a 'cracked'

analysis with such beams is equivalent to 10% redistribution of the 'uncracked' support

moments.


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In a Class 2 section the full plastic moment resistance can be developed. It has

been proposed that a redistribution of 30% be permitted from an "uncracked" analysis toallow for local yielding at the supports and cracking of concrete. Comparisons with test

results made during the assessment of Eurocode4 confirm the latter figure as appropriatefor sections which can attain the plastic resistance moment at the supports.

A beam with Class 2 (or Class 1) sections at supports will typically have a relatively low

neutral axis, in order to meet the restrictions on the depth of the web in compression

required in such sections. Hence only light tensile reinforcement can be provided and the

ratio of "uncracked" to "cracked" flexural stiffness (I1/I2) can exceed 3.0. For such beams,

the bending moment at the internal support from "cracked" analysis may then be less than

70% of the value from "uncracked" analysis and is almost always less than 85% of the

"uncracked" value. It follows that for Class 2 and Class 1 sections a 15% differencebetween "uncracked" and "cracked" analysis is more appropriate than the 10% difference

adopted for beams with sections in Class 3 or Class 4.

Finally, a Class 1 section is one which can, not only attain the plastic resistancemoment, but also sustain this level of moment whilst rotation occurs. In steel structures,

the limits on flange and web slenderness which define a 'plastic' section are sufficientlyrestrictive to permit plastic global analysis without further checks on rotation capacity.

This is not true for composite beams, partly because the degree of redistribution needed

to attain a plastic hinge mechanism will be higher due to the greater relative moment

resistance at mid-span.

!!! The class of a composite section is determined function the least favorable

class of its steel elements in compression. This must be done for each section of positive

respectively negative bending.

The steel flanges in compression are classified in EUROCODE 4 according to themaximum width-to-thickness ratio for steel outstands in compression.

The steel web is classified function of the width-thickness ratio. If the compressed

steel flange is effectively attached to the concrete slab by shear connectors (for positivebending) it may be assumed to be in class 1.

If the slab is compressed (under positive moment) and the steel beam is inclass 1, the NA's position imposes the class of the entire composite beam:

- if NA is in the slab or in the compressed flange, the entire section is class 1;- if NA passes through the steel web, the entire CS is class 2;

!!! The plastic (nonlinear) analysis is permitted only for CS having class 1 or 2. The

elastic (linear) analysis may be applied to CS of any class.

4.3. Resistance of CS of composite beams

Generally the plastic analysis is permitted for composite beams, while a total

connection is considered between the concrete slab and the steel profile. Anyway an

elastic analysis will be presented, necessary in SLS calculus.


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4.3.1. Elastic Resistance Moment Mel.Rd

4.3.1.1. Basis of design assumptions:

- the connection between slab and beam is considered to be a full shear (total)connection, which prevents the longitudinal slip between these 2 main

elements.- the linear diagram of the normal stress can be considered- plane CS remains plane after deformation- both steel and concrete are considered to have an elastic behavior- the tensile strength of concrete is neglectedBased on these assumptions, a composite CS is calculated considering an

equivalent CS, all in steel, replacing the area of concrete Acby an equivalent area of steel

Ac/n using the modular ratio n. So the equivalent area in steel for a CS will be:

A1=Aa+As+Ac/n

(Aa=steel section area, As=reinforcement area, Ac=concrete area).

4.3.1.2. Composite section under positive elastic moment (sagging bending)The geometrical characteristics and the strength verification depends on the

neutral axis (NA) position.

a) NA is in the steel beam

Ga, Gc, Gm are the centroids of the steel profile, the concrete area and the

equivalent section from steel.

!!! When the concrete slab is compressed the reinforcement area Asis neglected.

To determine the position of ENA (elastic neutral axis) we write the equality ofeach static moment, taking into account that the equivalent area is:


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1

1

2

2

2

22

A

hzA

dh

zAAn

Ad

dh

zAdn

A

dh

zdh

zdd

dAdn

A

n

hbA

n

AAA

c

aa

c

c

aaa

c

c

c

c

aac

c

c

c

aa

c

aca

aac

c

ceff

ac

a

+

=

+=

+

+=

+=+=+

=

+=+=

The distances from ENA to the extreme fibers (where the strength verification iscompulsory) are:

2

cc c

a c

hv d

v h v

= +

=

To make the strength verification we write also the equivalent moment of inertiaconsidering the entire area made of steel which is:

2 2

1c c

a a a c

I AI I A d d

n n= + + +

The strength verification in the extreme fibers will be:

- in the compressed top fiber, from concrete:

1

1

0.85ck

c c

c

c

fM Mz

I W

n IW

v

= =

=

Wc=strength modulus of concrete

fck=characteristic cylinder strength of concrete

c= 1.5 partial safety factor

Only 85% from fckis considered in calculus because there must be considered the

differences between the standard cylinder test and the real behavior of concrete instructural members in time.

- in the inferior fiber from the steel profile:

1

1

y

a a

a

a

fM Mz

I W

n IW

v

= =

=

fy= yield limit for structural steel

a= partial safety factor for structural steel


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b) Neutral axis is in the concrete slab

!!! The tension in concrete is neglected.

1

2

2

( )2

( ) 02

eff ccc

a a

ccc a a

cc

a a c c

c eff ca a c c

eff

c a c a a c

b vA

A A An n

Ad A d

n

vd

d z h v

v b vA z h v

n

bv A v A z h

n

= + = +

=

=

= +

= +

+ + =

2 4 ( )2

2

2

21 ( ) 1

eff

a a a a c

ceff

effac a c c

eff a

bA A A z h

nv

b

n

bA nv z h h

b n A

+ +

=

= + +


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The equivalent area:

2

2

2

( )2

( ) ( )2

( )2

a s

a a c s

cs s

cs s a a c s

cs a s s c a c

c

s c a c

s

a s

A A A

d z h v

hd v

hA v A z h v

hA A v A A z h

hA A z h

vA

v h v

= += +

=

= +

+ = + +

+ +=

=

The equivalent moment of inertia:2 2

2 a a a s sI I A d A d= + +

Where Is0.The strength verification is:

- in steel:

2

y

a a

a

a

fM

W

IW

v

=

=, where

- in reinforcement:

2

sk

s s

s

s

fM

W

IW

v

=

=

, where


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And fsk and s are the characteristic strength and the partial safety factor for

reinforcement, where s=1.15.

Elastic resistance in bending(Bearing capacity)a)section under positive moment

( )(1) (2), , ,

(1) 1,

2 2

1

(2 ) 1,

,

min ;

0.85

sd

el Rd el Rd el Rd

y

el Rd

a a

c c

a a a c

ckel Rd

c c

el Rd

M M M

fIM

v

I AI I A d d

n n

fnIM

v

M M

=

=

= + + +

=

b)section under negative moment

( )(1) (3), , ,

(1) 2,

2 2

2

(3) 2,

,

min ;

sd

el Rd el Rd el Rd

y

el Rd

a a

a a a s s

sel Rd

s s

el Rd

M M M

fIM

v

I I A d A d

fIM

v

M M

=

=

= + +

=

4.3.2. Plastic Resistance Moment

4.3.2.1. Basis of design

To evaluate the plastic moment Mpl,Rd the following assumptions should be

considered:

- there is full interaction between structural steel, reinforcement and concrete,so that in each element the maximum strength should be reached;

- the entire section of the steel profile is plasticized (both tension andcompression zones) the stresses having a rectangular block distribution equal

to fy/a; a=1.1;- the tensile strength of concrete is neglected;- in the compressed concrete the stresses have also a uniform distribution equal

to: 0.85fck/c; c=1.5;- the reinforcements have the design yield stress fsk/s; s=1.15 being neglected

in sections under positive moments (when concrete is compressed).


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4.3.2.2. Composite section under positive plastic moment

a) Plastic neutral axis (PNA) is in the concrete slab

The tensile concrete is neglected.

Fcis the resultant of the compression stresses from concrete:0.85 0.85ck ck c cc c eff

c c

f fF A e b

= =

Fa is the resultant of the tension stresses from steel profile:

aa a

a

fF A

=

The PNA (plastic neutral axis) position is determined from the condition that the

tension resultant must be equal to the compression resultant.

1

0.850.85

a c

y ya c

c acka eff ck a

eff

c

a c

F F

f fA

e A f b fb

e h e

=

= =

=

The plastic moment:

,

,

2

2

pl Rd a

ca c

y cpl Rd a a c

a

M F d

ed z h

f eM A z h

=

= +

= +

b) PNA is in the steel profile


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2

(1)2 2

(2)

(1) & (2)

sk

s s

s

y

at at

a

y

a a

a

a s sk a

s at a at

s y

at s s w w

w

fF A

fF A

fF A

A A fF F F A

f

A b t z t

z

=

=

=

+ = =

= +

The plastic negative moment:

,"

2 2

c cpl Rd a a at

h hM F z F h

= + +

For any of the cases presented above, the general verification is MsdMpl, Rd.

4.3.3. Vertical shear resistance

All relations presented by Eurocode 4 for shear checking are applied to composite

beams having a hot-rolled or welded steel profile, having a solid web withoutlongitudinal stiffeners.

The shear checking is made in accordance to EUROCODE 3, considering that the

entire shear force is taken by the steel profile (thus the concrete slab is neglected in

shear). Similarly to the relation used in bending checks, the relation for checking shear

will be:

Vsd Vpl.Rd

where VSd

- shear force from static calculus

Vpl.Rd

- plastic shear resistance, given in Eurocode 3

Vpl.Rd Av

fy

3

1

a

:=Av

fy

3 design strength at shear of structural steel

For common mild steel: fy 23:= N

mm2


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If this relation isn't checked we have to consider the influence of the shear force

on the plastic moment Mpl.Rd. The influence is considered using the interaction betweenthe shear force and the plastic moment, in evaluating Mpl.Rd. The following interactioncurve is considered:

Fig.5. Interaction curve V-M

The influence of the shear force V on the plastic moment Mpl.Rd is considered

calculating a reduced plastic moment Mpl.Rd interpolating between the values of Mpl.Rd and

Mf.Rd. This reduced plastic moment will be calculated with the formula:

fy3 a

123.343= Rf( )N

mm2

a 1.1:=

fy

a

213.636= N

mm2

R( )

Avis the shear area for the steel member, which is:

- for welded profiles Av d tw:=d

- for hot-rolled I, H profiles, Avis the web area plus a part of the flanges

Av Aa 2b tf tw 2 r+( )tf+:=Aa In practice the inequality written above must take a more severe form:

Vsd 0.5 Vpl.Rd

Mpl.Rdred

Mf.Rd Mpl.Rd Mf.Rd( ) 1 2Vsd

Vpl.Rd

1

2

+ Ms:=Mpl.Rd

redMpl.Rd

red


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In the above relation Mf.Rd is the plastic moment of the entire composite section,

calculated considering that the steel profile is made only from flanges (the steel web isneglected).

Besides the shear verification, the shear buckling resistance of the steel web mustbe checked if:

- for unstiffened and unencased web

If these 3 inequalities are valid we check the shear buckling resistance, with theformula:

d

tw

69> 235

fy

:=

- for unstiffened and encased webd

tw

124>

- for a stiffened web (with transversal stifners)

where k is a buckling factor given in EUROCODE 3:

d

tw

30 k>

a

d1 k 4

5.34

a

d

2+:=

a

if

k 5.344

a

d

2+:=

aa

a

d1> if

a = longitudinal distance between stiffners

Vb.Rd d tw ba

M1

:=d

Vb.Rd VS

where M1 a:= ba

simple post-critical shear strength, given in EUROCODE 3 function the web

slenderness

w

d

tw

37.4 k:=

k

w 4

- for w 1. ba

fyw

3:=

fyw

- for 1.5 w< 3< ba

fyw

3

3

w

0.2 w+ 1.3

:=

ww


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4.4. Design Of Shear Connection in Composite Beams

4.4.1. Basis of design

Shear connectors and transverse reinforcement shall be provided throughout thelength of the beam, to transmit the longitudinal shear force between the concrete slab and

the steel beam, ignoring the effect of bond between these 2 elements.

Shear connectors shall be capable of providing resistance to uplift of the concrete

slab thats why the connectors are designed for a tensile force perpendicular to the plane

of the steel flanges having a value of at least 0.1 from the design shear resistance of theconnector Prd.

Eurocode 4 works with 2 concepts regarding - full connection;- partial connection, which is applied

only in case of composite beams where a plastic analysis is permitted. (section of class1

& 2).

A span of a beam has full shear connection when increase in the number of shear

connectors would not increase the design bending resistance of the member. Otherwise

the shear connection is partial.

4.4.2. Classification of shear connectors

Function the deformation capacity of the connectors Eurocode 4 classifies them

into:

- ductile (elastic) connectors;- non-ductile (rigid) connectors.Ductile connectors are those which have sufficient deformation capacity to justify

the assumption of ideal plastic behavior of the connection. The push-out tests show that

the headed stud connectors are ductile connectors if:

h4d16d22

4.4.3. The longitudinal shear force Vl

As the shear connectors have the role of transmitting this force from the concrete

slab to the steel beam, Vlis calculated using the bearing capacity of each of these 2 mainelements.

For full shear connection: Vlmust be calculated separately on each critical length.The longitudinal force Vlto be resisted by shear connectors between a simple end

support and a point of positive maximum moment (lcr1):

- for 3 w 4 ba

fyw

3

0.9

w

:=

ww

yw yield limit of the steel web


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25/60

if

3 4 0.2 1

4 1

h h

d d

hd

= +

> =

Block connectors

The connectors are drawn in the recommended position with respect to thedirection of thrust.

The design resistance of a block connector isck

rd fi

c

fP A

=

Where Af1 frontal area of connection

ck

c

f

- design compressive strength

2

1

f

f

A

A=

2.5 for normal weight concrete

If

2 for lightweight concrete

Af2 area of the surface of a connector, enlarged at a slope of 15 to the rear surface of

the adjacent connector.

Angle connectors

In solid slabs the angle connectors are positioned with respect to the thrust

directions.

The design resistance of an angle connector is:23

341

10rd ck

v

P bh f

=

Where b, h length and the width of the angle

v partial safety factor for connector

v =1.25


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4.4.5. Dimensioning of shear connectors with ductile connectors

Knowing the longitudinal shear force Vlwhich must be transmitted by connectorsand the design resistance of one connector Prd, the number of connectors will be:

l

rd

VN

P=

The number of connectors must be calculated on each critical length.

A critical length is the length between 2 critical sections, which can be:

- section of max. bending moment

- sections of support- sections where important concentrated forces act (forces which increase M with

more than 20 %; sections where diminishing of the CS exist.

4.4.6. Some design recommendations for the stud connectors

See Eurocode 4 for connector spacing in solid slabs and in ribbed slabs.

4.4.7. Transverse reinforcement

If the longitudinal reinforcement is taken from constructive point of view, the

transverse reinforcement must be designed so that premature longitudinal shear failure or

longitudinal splitting is prevented.The design longitudinal shear/unit length VSd for any surface of shear failure

should not exceed the design resistance to longitudinal shear VRd of the consideredsection.VSdVRd

The potential surfaces of shear failure are:

The design longitudinal shear/unit length is:

rdSd

PV

s=

N

mm

Where Prd design shear resistance of 1 connector,

S longitudinal distance between connectorsThe design resistance to longitudinal shear is:


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( )(1) (2)

(1)

(2 )

min ;

2.5

0.2

rd rd rd

sk

rd cv rd c pd s

ckrd cv

c

V V V

f

V A A V

fV A

=

= + +

=

Where Acv area of the concrete shear surface/unit length (section a-a, b-b, c-c)

factor depending on the concrete self-weight:

- for normal weight concrete (=2400 kg/m3) => = 1,

- for light weight concrete (=2400 kg/m3) => 0.3 0.7

2400

= +

rd concrete shear resistance

0.050.25 ctkrd

c

f

=

- fctk0.05 characteristic tensile strength 5 % fractile;- ex. for C25/30 fctk0.05=1,8 N/mm

2

Ac sum. Of the CS areas of transverse reinforcement/unit length

b tc

A AA

s

+=

2mm

mm

fsk char. strength for reinforcement

s safety factor for reinforcement

s = 1.15

Vpd contribution of eny steel sheeting (for ribbed slabs) (Par 6.6.3 Eurocode 4)

4.5. Deflection control in serviceability limit states

The composite beams should be designed so that their deformations should not

affect the use, efficiency or appearance of the composite structure.

For floor and roofs in buildings, the deflection limits are taken from Eurocode 3.

Deflections due to loading applied to the composite members are calculated using an

elastic analysis also in accordance to Eurocode 3. The difference is that in the rigidity of

the composite members (E x I) we consider the equivalent moment of inertia I1,

respectively I2.The total deflection of a composite beam will be:

max 1 2 0 = +

Where 1 deflection due to char. permanent actions in state (1)

2 deflection due to char. variable actions in state (2)

0 eventual precamber of the composite beam in the unloaded state (0)

Eurocode 4 imposes the limit deflections, for floors in general:


28/60

max

2

250

300

L

L

4.6. Design Example of a composite beam

1. The geometry

Check the continuous beam of an industrial platform, knowing that the beams aredisposed to a distance B=3 m, and having 2 spans, 10 m each. The concrete slab is a full

slab with hc=150 mm, being connected to the structural steel by welded shear stud

connectors, realizing a full interaction.

2. Material properties

a) concrete C25/30 ck 25:= N

mm2

Ecm 30.:= kN

mm2 Ec'

Ecm

2:=

c 2400:= kg

m3

c 1.:=

b) reinforcing steel

sk 400:= N

mm2

Es 210:= kN

mm2

s 7850:= kg

m3

s 1.1:=

c) Structural steel

y 23:= N

mm2

u 360:= N

mm2

Ea 210:= kN

mm2

a 7850:= kg

m3

a 1.1:=


29/60

3. Actions

The permanent actions from the steel beam, concrete slab and a floor should be

considered. In what concerns the variable loads, an imposed load of 1000 kg/mp is

considered.

a) Characteristic actions- permanent actions:

- steel beams weight

- concrete slab

- floor of 5 cm thickness, made of concrete mortar:

- variable actions:

- imposed load:

For the steel beam, the geometrical characteristics are:

Aa 2067:= mm2

Ia Iy:= Iy

Ia 11.0923 108

:= mm4

d) Ductile shear connectors (stud-connectors)

y 380:= N

mm2

u 450:= N

mm2

v 1.25:=

7850 20675 10 6 162.299= kgm

3 1.623 kN

m

2400 3 0.15 1.08 103

= kg

m 10.8

kN

m

2100 0.05 3 315= kg

m 3.15

kN

m

gk 1.623 10.8+ 3.15+ 15.573=:= kN

m

1000 3 3 103

= kg

m 30

kN

m

qk 30:= kN

m


30/60

b) Design actions

4. Internal forces and moments

5. Effective width of concrete flange (beff)

- beffshould be calculated for section under positive moment, respectively undernegative moment.

a) Section under positive moment M.p(sagging bending)

b) Section under negative moment M.n(hogging bending)

G 1.35:= q 1.:=

- permanent: gd gkG 21.024=:= kN

m

- variable: qd qkq 45=:= kN

m

Tmax Vmax:=Vmax

Vmax1 0.375 gd 0.4375 qd+( )L:=

Vmax1 275.713= kN

Vmax2 0.625 qd gd+( ) L:=

Vmax2 412.647= kN

Mmax.p 0.07 gd 0.095qd+( )L2

:=

Mmax.p 574.665= kNm

Mmax.n 0.125 gd qd+( ) L2

:=

Mmax.n 825.294= kNm

L0 0.8 L:= B 3:= m

beff.p min 2L0

8 B,

:=

L0 8= m

beff.p min 28

8 3,

2=:= m

L1 10:= m L2 10:= m

L0 0.25 L1 L2+( ) 5=:= m

beff.n min 2L0

8 B,

1.25=:= m


31/60

6. Resistance of CS in sagging bending Mp

6.1. Plastic resistance Mpl.Rd.p

hc 150:= mm beff.p 200:= mm

Ac beff.phc 3 105

=:= mm2

Aa 2.067 104

=

Fc Ac

0.85 fck

c

10

3:=

Fc 4.25 103

= kN

Fa Aa

fy

a

10

3:=

Fa 4.417 103

= kN

Fa F> => plastic neutral axis is in the steel profile

Fac Aac

2fy

a

:=Aac

Fac Aac427.3:=Aac

Fa 4.417 103

= kN

Fc Fac+ Fa:=Fc Fac+

4250 103

427.3 Aac+ 4417 103

:=4250 103

427.3 Aac+


32/60

As PNA is in the superior compressed flange, the CS class is 1.

6.2. Elastic resistance Mel.Rd.p

First we compute the modular ratio n, which transforms the concrete area into an

equivalent steel area.

Aac4417 10

3 4250 10

3

427.3:=

Afs 24 250 6 103

=:= Fac 167:= kN

Aac zf 250:=zf

Aac 390.826=

mm2

mm2

Aas Afs<

za 305.5:= mm => PNA is in the superior flange

zf 1.56:= mm

The plastic resistance moment is written with respect to Gc:

kNmMpl.Rd.p Fa za

hc

2+

Fac

hc

2

zf

2+

10

3 1.668 10

3=:=

Ec'30.5

215.25=:=

nEa

Ec'

13.77=:=


33/60

The equivalent moment of inertia:

Aa 2.067 104

:= mm2

AadaA

cn

d:=Aada

da dc+ za

hc

2+:=da dc+ da dc+ 380.5:=da dc+ mm

dc 380.5 da:= da

AadaAc

n

380.5 da( ):=Aada

2.067 104

da

2000 150

13.77

380.5 da( ):=2.067 104

da

da 195.2:= mm

dc 185.3:= mm

Aa 2.067 104

:= hc 150:= mm beff.p 2000:= mm n 13.77 :=

Ic

beff.phc3

125.625 10

8=:=

mm

4

I1 Ia Aada2

+Ic

n

+

Ac

n

dc

2+:=

I1 11.0923 108

20675 195.22

+ 2000 150

3

12 13.77

+

2000 150

13.77

185.3

2+ 2.686 10

9=:=

mm4

The elastic moment:

- in the inferior tension fiber:

va 439.7:= mm

Mel.Rd1

fy

a

I1

va

10

6 1.305 10

3=:= kNm

- in the superior compressed fiber:

Mel.Rd2

n I1

vc 0.85

fck

c

:=Mel.Rd2

vc 260.3:= mm

Mel.Rd2 0.85

f

ckc

n I

1

vc

10 6 2.013 103=:= kNm


34/60

The CS class:Compressed flange: as between the concrete slab and the steel profile a full

interaction is considered, the flange is in class 1.

Compressed web:

7. Resistance of CS in sagging bending Mn

7.1. Plastic resistance Mpl.Rd.n

The tensile concrete from slab is neglected, instead we take into account the

longitudinal reinforcement 12/150.

Under negative moment :

On this width we consider 9 bars on each row, so totally we have 18 bars 12with the area:

Mel.Rd min Mel.Rd1Mel.Rd2,( ) 1.305 103

=:= kNm

hat

hac

:=hat

439.7

110.33.986=:=

For the compressed web is in class III => the entire cross-section is in class III.

beff.n 1250:= mm

As 18 12

2( )4

2.036 10

3=:= mm

2


35/60

In this case PNA is always situated in the steel profile.

The plastic moment:

The CS class:

Fa Aa

fy

a

10

34.416 10

3=:= kN

Fat Aat2fy

a

:=Aat Fat 427.3 Aat:= Aat

Fs As

fsk

s

10

3 708.088=:= kN

Fs Fat+ Fa:=Fs Fat+Fs Fat+

Aat4417 10

3 708.088 10

3( )

427.38.68 10

3=:= mm

2

Aat 24 250 12.5 zwt+ 8680=:= zwt mm2

zwt 214.4:= mm

Fat 3709:= kN

kNmMpl.Rd.n Fa103

za

hc

2+

Fat10

3 48.8

hc

2+

10

6 1.221 10

3=:=

MSd.n Mpl.Rd.n< checked !

The compressed flange:

c 350 12.5( )

2

168.75=:=

tf 24:= mm


36/60

7.2. Elastic resistance Mel.Rd.n

The equivalent area:

c

tf

7.031= 7.031 9< => the compressed flange is in class 1.(a)

The compressed web:

d 301.6:= mm (length of compressed part of the web)

d

d0.626=:= 0.626 0.>

If >0.5 => we have to check ifd

tw

396( )

13 1

d 482:= mm

tw 12.:= mm

d

tw

38.56=

396( )

13 155.505= 38.56 55.505< => the compressed web is also in class 1.(b)

From (a), (b) => the entire cross-section is in class 1.

mm2A2 Aa As+ 2.271 10

4=:=

Aada Asds:=Aada


37/60

8. Check to vertical shear

20675 da 2036 380.5 da( ):= da 41.5:= mm=>ds 33:= mm

da ds+ za

hc

2+ 380.5=:=da ds+

ds 380.5 da:=

I2 Ia Aada2

+ Asds2

+:=

I2 13.78 10:= mm4

The elastic moment:

- in the inferior fiber:

M

el.Rd1I2

va

f

ya

:=

vs 414:= mm

va 286:= mm

Mel.Rd1

fy I2( )va a

10

6 1.029 10

3=:= kNm

- in the superior fiber:

Mel.Rd2

I2

vsfsk

s

:=

vs 414:= mm

va 286:= mm

Mel.Rd2

fsk I2( )vs s

10

6 1.158 10

3=:= kNm

kNmMel.Rd min Mel.Rd1Mel.Rd2,( ) 1.029 103

=:= Mel.Rd MSdneg

>

Vpl.Rd Av

fy

3

1

a

:=Av

Av d tw 6.025 103

=:= mm2

Vpl.Rd Av

fy

3

1

a

10

3 743.142=:= kN

VSd.max 412.5:= kN


38/60

Vpl.Rd VSd.max>

but VSd.max

=412.5 kN > 0.5*Vpl.Rd

=371.5 kN => we have to take into account the influence of

shear on the plastic moment Mpl.Rd.n.

Mpl.Rd.red Mf.Rd Mpl.Rd.n Mf.Rd( ) 1VSd

0.5Vpl.Rd

1

2

+:=Mf.RdMf.Rd

Fs As

fsk 10 3

s 708.088=:= kN

Fat Aat2fy

a

427.3Aat=:=Aat and Aat 250 zt:= zt

Fac 24 250 24 350+( )235 10

3

1.1 3.076 10

3=:= kN

Fs Fat+ Fac:=Fs Fat+ => zt 22.1:= mm

zt tf<

Aat 250 zt 5.525 103=:=

Fat 427.3 Aat 10 3

2.361 103

=:=

Mf.Rd Fac za

hc

2+

10

3 Fat

zt

2

hc

2

+

10

3 967.407=:= kNm

Mpl.Rd.n 1221.4:= kNm

Mpl.Rd.red Mf.Rd Mpl.Rd.n Mf.Rd( ) 1VSd

0.5Vpl.Rd

1

2

+ 1218=:=VSd

kNm

Mpl.Rd.red Msdneg

>


39/60

9. The design of connection with headed stud connectors

The dimensioning is made for each critical length L1=4m, L2=6m.The longitudinal shear force that must be transmitted by connectors is Vl.

For L1=4m:

Fcf min Aa

fy

a

Ac0.85fck

c

,

1 10

3 4250=:= kN

Vl.1 Fcf 1 103

425=:= kN

For L2=6m

Vl.2 Fcf Asfsk

s

+:=

Vl.2 4958.2:= kN

The design resistance of 1 connector:

PRd min PRd.1PRd.2,( ):= PRd.1

PRd min 0.8 fu d

2

4

1

v

0.29 d2

fck Ecm 1

v

,

:=

PRd.1 0.8 450 22

2

( )

4

1

v

10 3 109.478=:= kN

PRd.2 0.29 1 222

25 30.5 103

1

1.25

10 3

98.051=:= kN

h

d5.68 4>:=

h

d => 1:=

PRd min PRd.1PRd.2,( ) 98.051=:= kN

The number of connectors for each critical length:

-for L1: N4250

PRd

43.345=:= => 44 connectors.

-for L2: N4958.2

PRd

50.567=:= => 52 connectors.


40/60

10. Design of transverse reinforcement

For L1:

For the potential section of shear failure a-a:

It is calculated from the checking condition VSd


41/60

mmNV

mmNV

mm

mm

s

AAA

mmNfCclassFor

mmNf

concreteweightnormalfor

mm

mmlmmA

Rd

Rd

bt

e

ctk

c

ctk

Rd

aacv

5005.1

2511502.0

8.53615.1

40022.13.011505.2

22.1185

4

122

8.130/25

3.05.1

8.125.025.0

1

1501

)2(

)1(

2

2

2

2

2

==

=+=

=

=+

=

=

===

=

==

VRd=500 N/mm < VSd=> we have to increase the concrete slabs depth to hc=160mm.

!5335.1

2511602.0)2( okVmmNV SdRd >==

For section b-b:

!530713

1283

1.71315.1

40022.13.013855.2

385

3851002

3521252

)2(

)1(

2

okVV

mmNV

mmNV

mm

mmA

l

SdRd

Rd

Rd

cv

bb

=>=

=

=+=

=

=++=

For L2:

For section a-a:

!2.417533

5335.1

2511602.0

544

)2(

)1(

okVV

mmNV

mmNV

SdRd

Rd

Rd

=>=

==

=

For section b-b:

!2.417713 okVV SdRd =>=


42/60

11. Deflections checking in SLS

!!! We must work with the characteristic actions.

A. Characteristic permanent actions:- weight of the steel beam: gk1=1.623 kN/m- weight of the concrete slab: gk2=10.8 kN/m- weigth of the floor finishing: gk3=3.15 kN/m

B. Characteristic variable actions: qk=30 kN/m

The maximum deflection is computed as follows:

max=1+2-0

0 = 0. (precamber=contrasageata);

1= defelction from permanent actions;

2= deflection from variable actions.

1=1+1

where 1

is the deflection from gk1, gk2(for the unhardened wetconcrete) calculated with the moment of inertia of the steel

beam Ia;

- 1 is the deflection from gk3(after the concrete is hardened),

considering the equivalent moment of inertia I1.

2= deflection from qk, with the equivalent moment of inertia I1.

For this type of structure (continuous beam), we have:

!40250

84.5

!3.33300

77.2

84.5

77.210686.2210000

1000030

384

2

07.3

29.010686.2210000

1000015.3

384

2

78.2100923.11210000

10000)8.1062.1(

384

2

384

2

max

2

max

9

4

2

1

9

4"

1

8

4'

1

4

okL

okL

mm

mm

mm

mm

mm

EE

IE

Lq

a

=


43/60

CHAPTER V COMPOSITE SLABS WITH PROFILED STEEL

SHEETING

5.1. Generals

Definition: a composite slab is one in which profiled steel sheets are used:

- As permanent shuttering capable of supporting the wet concrete,reinforcement and construction loads;

- After the concrete is hardened, as tensile reinforcement in the finished floor.Having these two roles, a composite slab must be calculated in two different

phases:

- First when the steel sheet is calculated, considering that it acts as a formwork;- Than, the composite slab is calculated, considering the composite action of the

concrete and the steel sheet.C11 lipsa 1 ora

M+pl.Rd

hp60 mmhc50 mm

The resultant of stresses from the compressed slab:

c

ck

c

fxbN

=

85.0

The resultant of stresses from the steel sheeting:

ap

yp

pp

fAN

=

where fyp yield strength for steel sheeting;

ap partial safety factor for sheeting ap=1.1.

ck

c

ap

ypp

pc

f

f

b

Ax

PNApositionaxisneutralNN

=

=>=

85.0

)('

The positive plastic moment:

RdplSd

pppl

MM

xdNM

.

2

=


44/60

B) PNA in the steel sheeting only when hp>60 mm

!!! The compressed concrete from ribs is neglected.

The real distribution of stresses with rectangular blocks is replaced by a simplified

distribution.

Na is the resultant of the tensile stress from sheeting which equalizes the

compressive stresses from slab.

Nc=Na

The plastic resistance moment:

( )

a

yp

p

c

pp

c

c

ck

cc

prcRdpl

fA

Neee

hhz

fhbN

MzNM

+=

=

+=

2

85.0

.

Mpr is the reduced plastic moment from sheet:

pa

ap

yp

p

c

papr MfA

NMM

=

125.1

Where Mpais the plastic moment capacity of the steel sheeting (from Eurocode 3).

5.3.2. Plastic resistance of section under negative bending

A) PNA in the concrete slab


45/60

( ) ccpcc

ck

c

ap

yp

p

s

sk

scpcs

ap

yp

pp

c

ck

cc

s

sk

ss

AAhxbAA

f

fA

fAANNN

fAN

fAN

fAN

=+=+

==>+=

=

=

=

221

85.0

85.0

x.

Din x => zc, zp.The plastic moment:

ccppRdpl zNzNM += .

B) PNA in the steel sheeting this case appears only for small loads.

!!! The steel sheeting is neglected; the computing is made as for a reinforced

concrete element, considering that the compressive stresses are overtaken only by

concrete.

zNM

zxNN

ribsfromconcretetheofwidthmediumb

fxbN

fAN

sRdpl

cs

c

ck

c

s

skss

==>=>=

=

=

.

0

0

.

85.0

5.3.3. Resistance to longitudinal shear

Eurocode 4 presents two analysis ways:

Steel sheeting without anchorage Steel sheeting with end anchorage

Longitudinal shear for slabs without end anchorage


46/60

This calculation is applied to composite slabs with mechanical and frictional

interlock. The maximum design vertical shear V for a width of slab b, should notexceed the design shear resistance Vl.Rd.

VVl.Rd

Vs

s

p

pRdl

kLb

Am

dbV

+

=.

b, dpin mm; Apin mm2.

m, k: empirical factors obtained from tests (Eurocode 4: par. 10.3.1.)

5.3.4. Resistance to vertical shear

The maximum vertical shear V, should not exceed the vertical shear resistance

Vv.Rd, over a width equal to the distance between centers of ribs b.

( ) 402.10.

.

+=

VRdpRdV

RdV

kdbV

VV

Where: b0 the average width of the concrete from the ribs;

Rd the basic shear strength;


47/60

][,1)6.1(

02.0

5.1;25.0

0

05.0

minddk

db

A

f

ppV

p

p

c

c

ctk

Rd

=

cr=0.2(z-r),

where

)(

.

signtheirwithtakenmomentsendbetweenratiotheisr

N

N

cr

Rpl

=

Also, the second order effects must be taken into account if: 1.0cr

Sd

N

N

2

2 )(

l

EIN ecr

=

For simplification the second order effects in a column may be considered

increasing the greatest 1storder moment MSdIby a correction factor k:


53/60

44.044.066.0

1

1

+=

=

rwhereN

Nk

cr

Sd

The second order moment will be thenMSdII=kMSdI

6.2.2.3. Resistance of members in combined compression and unit axial bending

To check this resistance we start from the interaction curve (M-N) presented

previously, modified like in the figure below. The design procedure is:

Rdpl

Rdpn

pn

Rpl

Sd

d

dn

Rdpl

Sd

N

N

N

N

r

N

N

.

.

.

.

4

)1(

1

=

=

=

=

is the buckling coefficient for centric compression

We take a horizontal line passing through until this line intersects the interaction

curve. The corresponding value for bending R is determined from the buckling

coefficient , which takes into account that the buckling from bending is due to the

geometrical imperfections and residual stresses.

These imperfections decrease linearly from point P to point n.Under the influence of the axial stress NSdthe member has sufficient resistance in

bending ifMSd0.9Mpl.Rd

Where: MSd the maximum bending moment within the column length taking into

account the second order effect (MSd

=MSdII

)

Mpl.Rd is the plastic moment resistance calculated using the stress distribution

corresponding to point B in the interaction curve.

6.2.2.4. Resistance to combined compression and bi-axial bending

The imperfections should be considered only in the plan in which failure isexpected to occur. (e.g. Z-axis). For the other plan of bending we shouldnt take into

account the imperfections. (e.g. Y-axis).

The verification for biaxial bending is made:

Separately for each bending moment:My.Sd0.9yMy.pl.Rd


54/60

Mz.Sd0.9zMz.pl.Rd

Taking the global effect of both moments:

1..

.

..

.

+ RdplzzSdz

Rdplyy

Sdy

M

M

M

M

6.2.2.5. The limits of application. The simplified method of design

This method may be applied only to composite columns having a double

symmetrical cross-section and a constant cross-section along the columns length.

Design prescriptions

The steel contribution ratio to the cross-section resistance:

Rdpl

a

y

a

N

f

A

.

= must be 0.2 0.9.

The non-dimensional slenderness:

2.

=cr

Rpl

N

N

The area of longitudinal reinforcement Asshould be:

0.3%AcAs4%Ac

With Ac the concrete area.

For totally encased sections the thickness of concrete cover must be:40mm cz0.3h

czb/6.

40mm cy0.4b

6.3. Design example of a composite column with partial concrete encasement

Objective - check the composite column having a buckling length of 6m and the

following cross-section:


55/60

The column is laterally supported on the other direction. The axial force has the

value of: 3000 kN.The bending moment diagram on the column height is:

1. Physical and geometrical characteristics:

a. Structural steel: HE360A profileAa 14300:= mm

2

Iya 3.309 108

:= mm4

fy 235:= N

mm2

Iza 0.789 108

:= mm4

fu 360:= Ea 210:=

kN

mm2

a 1.1:=

b. Concrete: C25/30 (normal weight concrete)

Ac 90700:= mm2

Iyc 7.40975 108:= mm4

Izc 2145

3315( )

277.5

245350( )+

1.505 10

9=:=

fck 25:= N

mm2

Ecm 30.5:= kN

mm2 c 1.5:=

c. Reinforcement: 420

As 1256:= mm2

Iys 2 628 1252

1.963 107

=:= mm4

Es 210:= kN

mm2

fsk 400:= N

mm2

s 1.15:=

2. Conditions for the application of the simplified method

The steel ratio:

Aa

fy

a

Npl.Rd

:=Npl.Rd


56/60

The plastic resistance to axial compression:

Npl.Rd Aafy

a

Ac 0.85

fck

c

+ As

fsk

s

+

10

3 4.777 10

3=:= kN

Aa

fy

a

10

3

Npl.Rd

0.64=:=

0.2 < 0.9< ok!

3. The elastic rigidity; non-dimensional slenderness

For this calculus only: c 1.35:=

EIey Ea Iya( ) 0.8Ecm

c

Iyc+ Es Iys( )+ 8.7 10

10=:= kNmm

2

The non-dimensional slenderness :

Npl.R

Ncr.y

:=Npl.R

l 6000:= m

a 1:= c 1:= s 1:=

Npl.R Aa

fy

a

Ac 0.85

fck

c

+ As

fsk

s

+

10

3 5.79 10

3=:= kN

Ncr.y

2

EIey

l2

2.385 104

=:= kN

Npl.R

Ncr.y0.493=:= 0.493 2< (for the application of the simplified method)

as 0.8< the column rigidity shouldn't be recalculated (from Eurocode 3).

From EC3, buckling curve "b" the buckling factor: 0.88:=

4. Checking to centric compressionNSd 3000:= kN

NSd Npl. this is true


57/60


58/60


59/60

The interaction curve has the points coordinates:

A: NA Npl.Rd 4.777 103

=:= kN

MA 0:=

B: NB 0:=

MB Mpl.Rd 437.025=:= kNm

C: NC Npm.Rd 1.285 103

=:= kN

MC Mpl.Rd 437.025=:= kNm

D: ND1

2

Npm.Rd 642.458=:= kN

MD Mmax.Rd 459.318=:= kN

With these points, the non-dimesnional interaction curve is built, with the following coordinates:

A: x 0:= C: x 1:=

y 1:= y p:= p pm

Npm.Rd

Npl.Rd

0.269=:=

B: x 1:= D: x

Mmax.Rd

Mpl.Rd

1.051=:= y 0:=

y pm

20.134=:=

The interaction curve is drawn, as follows:


60/60

We still calculate:

d

NSd

Npl.Rd0.628=:=

n 1 r( )

40.308=:=

As .d>.pm, the factor of reducing the moment , will be:

d( ) 1 pm( )

0.392=:=

The checking in bending:

MSd.II 0.9 Mpl.R true => check ok!

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