Simply Supported Secondary Composite Beam

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SX014a-EN-EU Sheet 1

of 10Document Ref:

Title

CALCULATION SHEET

Example: Simply supported secondary composite beam

Eurocode Ref EN 1994-1-1

Made by Arnaud LEMAIRE Date Au ust 2005

Checked by Alain Bureau Date August 2005

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 I-rolled 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

• γ G = 1,35 (permanent loads)

• γ Q = 1,50 (variable loads)

• γ M0 = 1,0

• γ M1 = 1,0

• γ V = 1,25

• γ C = 1,5

EN 1990

EN 1993-1-1

§ 6.1 (1)

EN 1994-1-1

§ 6.6.3.1

EN 1992-1-1


C r e a t e d o n W e d n e s d a y ,

M a y 1 5 ,

2 0 1 3

T h i s m a t e r i a l i s c o p y r i g h t - a l l r i g h t s r e s e r v e d .

U s e o f t h i s d o c u m e n t i s s u

b j e c t t o t h e t e r m s a n d c o n d i t i o n s o f t h e A c c e s s S t e e l L i c e n c e A g r e e m e n t

http://www.access-steel.com/discovery/linklookup.aspx?id=EC005







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Title

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 IPE 270

Depth ha = 270 mm

Width b = 135 mm

Web thickness t w = 6,6 mm

Flange thickness t f = 10,2 mm

Fillet r = 15 mm

Mass 36,1 kg/m

z

z

y y

t f

t w

b

ha

Euronorm

19-57

Section area Aa = 45,95 cm2

Second moment of area /yy I y = 5790 cm4

Elastic modulus /yy W el,y = 428,9 cm3

Plastic modulus /yy W pl.y = 484,0 cm3

Modulus of elasticity of steel E a= 210000 N/mm2



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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 h p = 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 f u = 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 f ck = 25 N/mm2

Secant modulus of elasticity of concrete E cm = 33 000 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

06201010 , , +× 0,058) = 7,2 kN/m

Self weight of the beam : (36,1 × 9,81) × 10-3 =0,354 kN/m



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CALCULATION SHEET





Permanent load :

G = 0,354 + 7,2 + 0,75 × 3,0 = 9,80 kN/m

Variable load (Imposed load) :

Q = 2,5 × 3,0 = 7,50 kN/m

ULS Combination :

γ G G + γ Q Q = 1,35 × 9,80 + 1,50 ×7,50 = 24,48 kN/m

EN 1990

§ 6.4.3.2

Moment diagram

M

172,13 kNm

Maximum moment at mid span :

M y,Ed = 0,125 × 24,48 × 7,502 = 172,13 kNm

Shear force diagram

V

91,80 kN

Maximum shear force at supports :

V z,Ed = 0,5 × 24,48 × 7,50 = 91,80 kN

Yield strength

Steel grade S355

The maximum thickness is 10,2 mm < 40 mm, so : f y = 355 N/mm2

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

Table 3.1 or the values from the product standard.

EN 1993-1-1

Table 3.1



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Section classification :

The parameter ε is derived from the yield strength : 0,81][N/mm

235

2

y

== f

ε

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 – t w – 2 r ) / 2 = (135 – 6,6 – 2 × 15)/2 = 49,2 mm

c/tf = 49,2 / 10,2 = 4,82 ≤ 9 ε = 7,29 Class 1

Internal compression part :

c = h – 2 t f – 2 r = 270 – 2 × 10,2 – 2 × 15 = 219,6 mm

c / tw = 219,6 / 6,6 = 33,3 < 72 ε = 58,3 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

b0 is the distance between the centres of the outstand shear connectors, here

b0 = 0 ;

bei is 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

Figure 5.1

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

∑+= eii0eff,0 bbb β

With β i = (0,55 + 0,025 Le / 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

Figure 5.1



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SX014a-EN-EU Sheet 6 of 10Document Ref:

Title

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

γ

α

γ

π E f d d f k P

hsc / d = 100 / 19 = 5,26 > 4, so α = 1

EN 1994-1-1

§ 6.6.3.1

Reduction factor (k t)

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 ≤ k tmax

EN 1994-1-1

§ 6.6.4.2

Table 6.2

Where : nr = 1

h p = 58 mm

b0 = 82 mm

hsc = 100 mm

So, 717,0158

100

58

82

1

7,0t =⎟

⎠

⎞⎜⎝

⎛ −=k ≤ k tmax = 0,75

for profiled sheeting with holes.

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ×××××××=

25,1

310002519129,0;

25,1

4/194508,0Min717,0

22

Rd

π P 3

.10−

kN73,73;kN66,81Min717,0 ×=

P Rd = 52,86 kN



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SX014a-EN-EU Sheet 7 of 10Document Ref:

Title

CALCULATION SHEET





Degree of shear connection

The degree of shear connection is defined by :

f c,

c

N

N =η

Where : N c is the design value of the compressive normal force in the

concrete flange

N c,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.

Ac is the cross-sectional area of concrete, so at mid-span Ac = beff hc

with hc = h - h p = 120 – 58 = 72 mm, Ac = 1875 × 62 = 116300 mm2

So, =××=== −3

c

ck ccdcf c, 10

5,1

2511630085,085,085,0

γ

f A f A N 1647 kN

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

than :

N c = 0,5 n P Rd = 0,5 × 36 × 52,86 = 952 kN

So, 578,01647

952

f c,

c === N

N η

The ratio η 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 :

( e

y

min 030750355

-1 L ,- , f ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ =η ) with Le ≤ 25

EN 1994-1-1

§ 6.6.1.2



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Le is the distance in sagging bending between points of zero bending momentin metres, for our example : Le = 7,5 m

So, η min = 1 – (355 / 355) (0,75 – 0,03 × 7,50) = 0,475

Then, η min = 0,475 < η = 0,578 OK

Plastic Resistance Moment at mid span

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

by :

N pl,a

= Aa f

y / γ

M0 = 4595 × 355 × 10-3 / 1,0 = 1631 kN

So, N pl,a > N c = η N c,f = 952 kN

EN 1994-1-1

§ 6.6.1.2 and

§ 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 beam M Rd at

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

value of the compressive force in the concrete flange N c is used in place of

the force N cf .

Here, the plastic stress distribution is given below :

MRd

+

-

Nc=η Nc,f = 952 kN

Na = 1291 kNhn

hp 339 kN

The position of neutral axis is : hn = 263 mm

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

M Rd = 301,7 kNm

So, M y,Ed / M Rd = 172,2 / 301,7 = 0,57 < 1 OK



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Shear Resistance

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

which is given by:

Av,z = A – 2 b t f + (t w + 2 r ) t f

Av,z = 4595 – 2 × 135 × 10,2 + (6,6 + 2 × 15) × 10,2 = 2214 mm2

EN 1993-1-1

§ 6.2.6 (3)

Shear plastic resistance

845310

01

)3/355(2214

)3/( 3

M0

yzv,Rdz, pl, ,

,

f AV - =

×==

γ

kN

V z,Ed / V pl,z,Rd = 91,80 / 453,8 = 0,202 < 1 OK

EN 1994-1-1

§ 6.2.2.2

Verification to shear buckling is not required when :

hw / t w ≤ 72 ε / η

η may be conservatively taken as 1,0

hw / t w = (270 – 2 × 10,2) / 6,6 = 37,8 < 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

F v

Δ

Δ=

f

dEd

Where Δx = 7,5 / 2 = 3,75 m

EN 1992-1-1

§ 6.2.4

Figure 6.7

The value for Δ x is half the distance between the section where the moment is

zero and the section where the moment is maximum and we have two areas

for the shear resistance.

Δ F d = N c / 2 = 951,56 / 2 = 475,8 kN

hf = h - h p = 120 – 58 = 62 mm

=×

×==

375062

108,475 3

f

dEd

xh

F v

Δ

Δ2,05 N/mm2

To prevent crushing of the compression struts in the concrete flange, the

following condition should be satisfied :

f f cdEd cossin θ θ ν f v < with [ ]250/16,0 ck f −=ν and θ f = 45°



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CALCULATION SHEET





5,45,05,1

25

250

2516,0Ed =××⎥⎦

⎤⎢⎣⎡ −×<v N/mm2 OK

The following inequality should be satisfied for the transverse reinforcement :

Asf f yd / sf ≥ vEd hf / cot θ f where f yd = 500 / 1,15 = 435 N/mm2

Assume the spacing of the bars sf = 250 mm and there is no contribution from

the profiled steel sheeting

Asf ≥ =×

××

0,1435

2506205,2

73,05 mm2

We can take 10 mm diameter bars (78,5 mm2) at 250 mm cross-centres

extending over the effective concrete breadth.

Serviceability Limit State verification

SLS Combination

G + Q = 9,80 + 7,50 = 17,30 kN/m

Deflection due to G+Q :384

)(5

y

4

I E

LQG

w

+=

EN 1990

§ 6.5.3

Where I y depends on the modular ratio (n) depending on the type of loading.

By simplification, we can take :

n0 = E a / E cm = 210 000 / 33 000 = 6,36 for primary effects (Q)

So I y = 24 540 cm4 at mid span

And n = 3 E a / E cm = 19,08 for permanent loads (G)

So I y = 18 900 cm4

1024540

507

1018900

809

210000384

575

88

4

=⎟ ⎠

⎞⎜⎝

⎛

×+

××

×=

−−

, ,

,w 16 mm

The deflection under (G+Q) is L/469

Note 1: 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

§ 7.3.1

EN 1994-1-1

§ 7.3.2



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SX014a-EN-EU

Quality Record

RESOURCE TITLE Example: Simply supported secondary composite beam

Reference(s)

ORIGINAL DOCUMENT

Name Company Date

Created by Arnaud LEMAIRE CTICM 14/06/05

Technical content checked by Alain BUREAU CTICM 14/06/05

Editorial content checked by D C Iles SCI 16/9/05

Technical content endorsed by the

following STEEL Partners:

1. UK G W Owens SCI 16/9/05

2. France A Bureau CTICM 16/9/05

3. Germany A Olsson SBI 15/9/05

4. Sweden C Müller RWTH 14/9/05

5. Spain J Chica Labein 16/9/05

Resource approved by TechnicalCoordinator

G W Owens SCI 10/7/06

TRANSLATED DOCUMENT

This Translation made and checked by:

Translated resource approved by:



M a y 1 5 ,

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