secondary composite example

8/13/2019 Secondary Composite Example

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Document Ref: SX014a-EN-UK Sheet 1 of 11Title

CALCULATION SHEET

Example: Simply supported secondary composite beam

Eurocode Ref EN 1994-1-1

Made by Laurent Narboux Date Oct 2006

Checked by Charles King Date Oct 2006

Localized resource for UK

Example: Simply supported secondarycomposite beam

This example deals with a simply supported secondary composite beam

under a uniformly distributed loading.

The following distributed loads are applied to the beam.

self-weight of the beam

concrete slab

imposed load

The beam is a UKB profile in bending about the strong axis. This example

includes :

- the classification of the cross-section,

- the calculation of the effective width of the concrete flange,- the calculation of shear resistance of a headed stud,

- the calculation of the degree of shear connection,

- the calculation of bending resistance,

- the calculation of shear resistance,

- the calculation of longitudinal shear resistance of the slab,

- the calculation of deflection at serviceability limit state.

This example does not include any shear buckling verification of the web.

Partial factors

G= 1.35 (permanent loads)

Q= 1.50 (variable loads)

M0= 1.0

M1= 1.0

V= 1.25

C= 1.5

EN 1990

EN 1993-1-1

6.1 (1)

EN 1994-1-1

6.6.3.1

EN 1992-1-1

Example: Simply supported secondary composite beam (GB)

CreatedonFriday,

January09,

2009

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2/12


CALCULATION SHEET





Basic data

Design a composite floor beam of a multi-storey building according to the

data given below. The beam is assumed to be fully propped during

construction.

The profiled steel sheeting is transverse to the beam.

Span length : 7.50 m

Bay width : 3.00 m

Slab depth : 12 cm

Partitions : 0.75 kN/m2

Imposed load : 2.50 kN/m2

Reinforced Concrete density : 25 kN/m3

Steel grade : S355

Try UKB 254x146x37

Depth ha= 256.0 mm

Width b= 146.4 mm

Web thickness tw= 6.3 mm

Flange thickness tf= 10.9 mm

Fillet r= 7.6 mm

Mass 37.0 kg/m

z

z

y y

tf

tw

b

ha

BS4

Corus

Advance

Section area Aa= 47.2 cm2

Second moment of area /yy Iy= 5537 cm4

Elastic modulus /yy Wel,y= 432.6 cm3

Plastic modulus /yy Wpl.y= 483.2 cm3

Modulus of elasticity of steel Ea= 210000 N/mm2


CreatedonFriday,

January09,

2009




This is a bit of a copout. Most UKconstruction isunpropped and would

require constructioncondition (non-composite) calculationstoo. See primary beamexample.


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CALCULATION SHEET





Profiled steel sheeting

Thickness of sheet t= 0.75 mm

Slab depth h= 120 mm

Overall depth of the profiled steel sheeting hp= 58 mm

b1= 62 mm b2= 101 mm e= 207 mm

Connectors

Diameter d= 19 mm

Overall nominal height hsc= 100 mm

Ultimate tensile strength fu= 450 N/mm2

Number of shear connectors studs n= 7500 / e= 36

Number of studs per rib nr= 1

0,5hp

hp

hsc

h

h0

b1

b2

e

Concrete parameters : C 25/30

Value of the compressive strength at 28 days fck= 25 N/mm2

Secant modulus of elasticity of concrete Ecm= 31 476 N/mm2

EN 1992-1-1

3.1.3

Table 3.1

To take into account the troughs of the profiled steel sheeting, the weight of

the slab is taken as :

25 3.0 (0.12 5 2

062.0101.0 + 0.058) = 7.2 kN/m

Self weight of the beam : (37 9.81) 10-3=0.363 kN/m


CreatedonFriday,

January09,

2009





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CALCULATION SHEET





Permanent load :

G= 0.363 + 7.2 + 0.75 3.0 = 9.81 kN/m

Variable load (Imposed load) :

Q= 2.5 3.0 = 7.50 kN/m

ULS Combination :

GG+ QQ= 1.35 9.81 + 1.50 7.50 = 24.49 kN/m

EN 1990

6.4.3.2

Moment diagram

M

172.19 kNm

Maximum moment at mid span :

My,Ed= 0.125 24.49 7.502= 172.19 kNm

Shear force diagram

V

91.84 kN

Maximum shear force at supports :

Vz,Ed= 0.5 24.49 7.50 = 91.84 kN

Yield strength

Steel grade S355

The maximum thickness is 10.9 mm < 40 mm, so :fy= 355 N/mm2

Note: The National Annex may impose either the values of fy from the

Table 3.1 or the values from the product standard.

EN 1993-1-1

Table 3.1


CreatedonFriday,

January09,

2009





5/12


CALCULATION SHEET





Section classification:

The parameter is derived from the yield strength : 0.81][N/mm

235

2

y

==f

Note: The classification is made for the non composite beam. For the

composite beam the classification is more favourable.

EN 1993-1-1

Table 5.2

(sheet 2 of 3)

Outstand flange: flange under uniform compression

c= (b tw 2 r) / 2 = (146.4 6.3 2 7.6)/2 = 62.45 mm

c/tf= 62.45 / 10.9 = 5.73 9 = 7.29 Class 1

Internal compression part:

c= h 2 tf 2 r= 256 2 10.9 2 7.6 = 219.0 mm

c/ tw= 219.0 / 6.3 = 34.76 < 72 = 58.32 Class 1

The class of the cross-section is the highest class (i.e the least favourable)

between the flange and the web, here : Class 1

So the ULS verifications should be based on the plastic resistance of the

cross-section since the Class is 1.

EN 1993-1-1

Table 5.2

(sheet 1 of 3)

Effective width of concrete flange

At mid-span, the total effective width may be determined by :

+= ei0eff,1 bbb

b0is the distance between the centres of the outstand shear connectors, here

b0= 0 ;

beiis the value of the effective width of the concrete flange on each side of the

web and taken as bei=Le/ 8 but bi= 3.0 mbeff,1= 0 + 7.5 / 8 = 0.9375 m, then beff= 2 0.9375 = 1.875 m < 3.0 m

EN 1994-1-1

5.4.1.2

(figure 5.1)

At the ends, the total effective width is determined by :

+= eii0eff,0 bbb

Withi= (0.55 + 0.025Le/ bei) but 1.0

= (0.55 + 0.025 7.5 / 0.9375) = 0.75

beff,0= 0 + 0.75 7.5 / 8 = 0.703 m, then beff= 2 0.703 = 1.406 m < 3.0 m

EN 1994-1-1

5.4.1.2

(figure 5.1)


CreatedonFriday,

January09,

2009




Only applicable ifthere is more than

one shearconnector for eachrib

Not needed forsimple calcs. Seeclause 6.1.2(2)
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CALCULATION SHEET





Design shear resistance of a headed stud

The shear resistance should be determined by :

=

V

cmck

2

V

2

utRd

29.0;

4/8.0Min

EfddfkP

hsc/ d= 100 / 19 = 5.26 > 4, so = 1

EN 1994-1-1

6.6.3.1

Reduction factor(kt)

For sheeting with ribs transverse to the supporting beam, the reduction factor

for shear resistance is calculated by :

= 1

7.0

p

sc

p

0

r

th

h

h

b

nk but ktmax

EN 1994-1-1

6.6.4.2

Table 6.2

Where : nr= 1

hp= 58 mm

b0= 82 mm

hsc= 100 mm

So, 717.0158

100

58

82

1

7.0t =

=k ktmax= 0.75

for profiled sheeting with holes.

=

25.1

314762519129.0;

25.1

4/194508.0Min717.0

22

Rd

P

310

( )kN29.74;kN66.81Min717.0 =

PRd= 53.27 kN


CreatedonFriday,

January09,

2009




Shear stud resistancecalculation is semi-empirical with lots of'fudge' factors derivedfrom experiments.
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CALCULATION SHEET





Degree of shear connection

The degree of shear connection is defined by :

fc,

c

N

N=

Where : Nc is the design value of the compressive normal force in the

concrete flange

Nc,f is the design value of the compressive normal force in the

concrete flange with full shear connection

EN 1994-1-1

6.2.1.3(3)

At mid-span :

The compressive normal force in the concrete flange represents the total

connection.

Acis the cross-sectional area of concrete, so at mid-spanAc= beffhc

with hc= h- hp= 120 58 = 62 mm,Ac= 1875 62 = 116250 mm2

So, ==== 3

c

ckccdcfc, 10

5.1

2511625085.085.085.0

fAfAN 1647 kN

The resistance of the shear connectors limits the normal force to not more

than :

Nc= 0.5 nPRd= 0.5 36 53.27 = 959 kN

So, 582.01647

959

fc,

c ===N

N

The ratio is less than 1.0 so the connection is partial.

Verification of bending resistance

Minimum degree of shear connection

The minimum degree of shear connection for a steel section with equal

flanges is given by :

( ey

min 03.075.0355

-1 L-f

= )withLe25

EN 1994-1-1

6.6.1.2


CreatedonFriday,

January09,

2009




Effectivewidth ascalculatedabove.

Number of shear connectors.0.5 as peak moment at mid-span. Note even number ofstuds on beam. If an oddnumber, stud at mid-spanwould carry no load andwould be ignored, as inprimary beam example.


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CALCULATION SHEET





Leis the distance in sagging bending between points of zero bending momentin metres, for our example :Le= 7.5 m

So, min= 1 (355 / 355) (0.75 0.03 7.50) = 0.475

Then, min= 0.475 < = 0.582 OK

Plastic Resistance Moment at mid span

The design value of the normal force in the structural steel section is given

by :

Npl,a

=Aaf

y/

M0= 4720 355 10-3/ 1.0 = 1675 kN

So,Npl,a>Nc= Nc,f= 959 kN

EN 1994-1-1

6.2.1.2and

6.2.1.3

With the ductile shear connectors and the cross-section of the steel beam in

Class 1, the resistance moment of the critical cross-section of the beamMRdat

mid span is calculated by means of rigid-plastic theory except that a reduced

value of the compressive force in the concrete flange Ncis used in place of

the forceNcf.

Here, the plastic stress distribution is given below :

MRd

+

-

Nc= Nc,f= 959 kN

Na= 1316 kNhn

hp 364 kN

9

The position of neutral axis is : hn= 249 mm

Then the design resistance for bending of the composite cross section is :

MRd= 298.6 kNm

So, My,Ed/MRd= 172.2 / 298.6= 0.58 < 1 OK


CreatedonFriday,

January09,

2009




I make this 242.2mmand the othernumbers a littledifferent

MG


9/12


CALCULATION SHEET





Shear Resistance

The shear plastic resistance depends on the shear area of the steel beam,

which is given by:

Av,z=A 2 btf+ (tw+ 2 r) tf

Av,z= 4720 2 146.4 10.9 + (6.3 + 2 7.6) 10.9 = 1762 mm2

But not less than hwtw

conservatively taken equal to 1.0

hwtw= 1.0 234.2 6.3 = 1475 mm2 < 1762 mm2 OK

EN 1993-1-1

6.2.6(3)

Shear plastic resistance

613100.1

)3/355(1762

)3/( 3

M0

yzv,

Rdz,pl, =

== -fA

V

kN

Vz,Ed/ Vpl,z,Rd= 91.84 / 361 = 0.254 < 1 OK

EN 1994-1-1

6.2.2.2

Nota that the verification to shear buckling is not required when :

hw/ tw72 /

may be conservatively taken as 1.0

hw/ tw= 234.2 / 6.3 = 31.2 < 72 0.81 / 1.0 = 58.3 OK

EN 1993-1-1

6.2.6(6)

Longitudinal Shear Resistance of the Slab

The plastic longitudinal shear stresses is given by :

xh

Fv

=

f

dEd

Where x = 7.5 / 2 = 3.75 m

EN 1992-1-1

6.2.4

(figure 6.7)

The value for x is half the distance between the section where the moment iszero and the section where the moment is maximum and we have two areas

for the shear resistance.

Fd=Nc/ 2 = 952/ 2 = 476 kN

hf= h- hp= 120 58 = 62 mm

=

=

=

375062

10476 3

f

dEd

xh

Fv 2.05 N/mm2


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January09,

2009



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10/12


CALCULATION SHEET





To prevent crushing of the compression struts in the concrete flange, thefollowing condition should be satisfied :

ffcdEd cossin fv < with [ ]250/16.0 ckf= and f= 45

5.45.05.1

25

250

2516.0Ed =


11/12

Document Ref: SX014a-EN-UK Sheet 11 of 11Title Example: Simply supported secondary composite beam



CALCULATION SHEET


The deflection under (G+Q) isL/444

Note1: The limits of deflection should be specified by the client. The

National Annex may specify some limits. Here the result may be

considered as fully satisfactory.

Note 2: The National Annex may specify limits concerning the frequency. of

vibration. Here the total deflection is low and the mass fairly high

and by experience there is no problem of vibration.

EN 1994-1-1

7.2.1

EN 1993-1-1

7.2.3


CreatedonFriday,

January09,

2009





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SX014a-EN-UK

Quality Record

RESOURCE TITLE Example: Simply supported laterally unrestrained beam

Reference SX014a-EN-GB

LOCALISED RESOURCE DOCUMENT

Name Company Date

Created by Laurent Narboux SCI Oct 2006

Technical content checked by Charles King SCI Oct 2006

Editorial content checked by D C Iles SCI 19/2/07


CreatedonFriday,

January09,

2009




secondary composite example

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