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Example: Simply supported beam with lateral restraint at load application point

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Example: Simply supported supported beam with lateral restraint restraint at load application point

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EN 1993-1-1

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Exampl Example: e: Simp Simply ly su ppo rted beam beam wit h lateral lateral restraint restraint at lo ad application p oint This worked example deals with a simply supported beam with lateral restraints at supports and at load application point.

The following distributed loads are applied to the beam:

•

self-weight of the beam

•

concrete slab

•

imposed load

5,0 m

5,0 m

1

t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

1 1

1 : Lateral restraint

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 bending resistance, including the exact calculation of the elastic critical moment for lateral torsional buckling,

-

the calculation of shear resistance, including shear buckling resistance,

-

the calculation of the deflection at serviceability limit state.

Partial factors

•

γ G = 1,35

(permanent loads)

•

γ Q = 1,50

(variable loads)

•

γ M0 M0 = 1,0

EN 1993-1-1

•

γ M1 M1 = 1,0

§ 6.1 (1)

EN 1990


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Basic data

Design a non composite floor beam of a multi-storey building according to the data given below. Two secondary beams are connected to the calculated one at mid-span. The beam is assumed to be laterally restrained at mid-span and at the ends

7m

Calculated Beam t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

7m

5m

5m Secondary beam Concrete slab

150 mm

597 mm

5m

Restraints to lateral buckling

5m

10


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Span length :

•

Secondary beam:

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10 m

o

Span length:

7m

o

Bay width :

5m

•

Slab depth :

•

Secondary beams

0,10 kN/m2

•

Partitions and false ceiling:

0,50 kN/m2

•

Imposed load :

2,50 kN/m2

•

Concrete density :

24 kN/m3

•

Steel grade :

15 cm

S355

Weight of the slab : 0,15 × 24 kN/m 3 = 3,6 kN/m2

Try

IPEA 600 – Steel grade S355 Depth

h = 597 mm

Width

b = 220 mm

Web thickness

t w = 9,8 mm

Flange thickness

t f = 17,5 mm

Fillet

r = 24 mm

Mass

108 kg/m

Section area

A = 137 cm2

z

t f

Euronorm 19-57 t w

y

y h

Second moment of area /yy I y = 82920 cm 4 Second moment of area /zz I z = 3116 cm4 Torsion constant

I t = 118,8 cm4

Warping constant

I w = 2607000 cm 6

Elastic modulus /yy

W el,y = 2778 cm3

Plastic modulus /yy

W pl,y = 3141 cm3

z b


CALCULATION SHEET

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Self weight of the beam : qG = (108 × 9,81) × 10 -3 =1,06 kN/m Permanent load : F G = (3,6 + 0,10 + 0,50)× 5,0 × 7,0 = 147 kN Variable load (Imposed load) : F Q = 2,50 × 5,0 × 7,0 = 87,5 kN

EN 1990

ULS Combination :

γ G qG = 1,35 × 1,06 = 1,43 kN/m γ G F G + γ Q F Q = 1,35 × 147 + 1,50 × 87,5 = 329,70 kN

§ 6.4.3.2 (6.10)

Moment diagram t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

M

842,13 kNm

Maximal moment at mid span : M y,Ed = 0,125 × 1,43 × 10,00 2 + 0,25 × 329,70 × 10 = 842,13 kNm Shear force diagram 172 kN

164,85 kN V

Maximal shear force at supports : V z,Ed = 0,50 × 1,43 × 10,0 + 0,50 × 329,70 = 172 kN Maximal shear force at mid-span : V z,Ed = 0,50 × 329,70 = 164,85 kN

Yield strength

Steel grade S355 The maximum thickness is 17,5 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 :

EN 1993-1-1

The parameter ε is derived from the yield strength : ε =

235 f y [N/mm 2 ]

= 0,81

Table 5.2 (sheet 2 of 3)

Outstand flange : flange under uniform compression

c = (b – t w – 2 r ) / 2 = (220 – 9,8 – 2 × 24)/2 = 81,10 mm c/tf = 81,1 / 17,5 = 4,63

≤

9 ε = 7,29

Class 1

Internal compression part : web under pure bending

EN 1993-1-1

c = h – 2 t f – 2 r = 597 – 2 × 17,5 – 2 × 24 = 514 mm c / tw = 514 / 9.8 = 52,45 < 72 ε = 58,32 t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

Class 1

Table 5.2 (sheet 1 of 3)

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.

Moment resistance

The design resistance for bending of a cross section is given by : M c,Rd = M pl,Rd = W pl,y f y / γ M0 = (3141 × 355 / 1,0) / 1000 M c.Rd = 1115 kNm M y,Ed / M c,Rd = 842,13 / 1115 = 0,755 < 1 OK

EN 1993-1-1 § 6.2.5


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Reduction factor for lateral torsional buckling

To determine the design buckling resistance moment of a beam, the reduction factor for lateral torsional buckling must be determined. The following calculation determines this factor by calculation of the elastic critical moment for lateral torsional buckling.

Critical moment for lateral torsional buckling

The critical moment may be calculated from the following expression :

⎧

2

⎫

π 2 E I z ⎪ ⎛ k ⎞ I w (k Lc ) G I t ⎪ ⎜⎜ ⎟⎟ M cr = C 1 + + (C 2 z g ) 2 − C 2 z g ⎬ 2 ⎨ 2 π E I z (k Lc ) ⎪ ⎝ k w ⎠ I z ⎪ 2

⎩


⎭

E is the Young modulus :

E = 210000 N/mm 2

G is the shear modulus :

G = 81000 N/mm2

Lc is the distance between lateral restraints : Lc = 5,0 m

In the expression of M cr , the following assumption should be considered : k = 1

since the compression flange is free to rotate about the weak axis of the cross-section,

k w = 1

since there is no device to prevent the warping at the ends of the beam.

The C 1 and C 2 coefficients depend on the moment diagram along the beam segment between lateral restraints. It can be assumed that the diagram is linear, then : C 1 = 1,77 for k = 1 C 2 z g = 0

Therefore : π 2 E I z

(k Lc )

2

=

π 2 × 210000 × 3116 × 10 4 (5000)

2

×10 −3 = 2583 kN

See NCCI SN005


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⎧

2607000

⎩

3116

M cr = 1,77 × 2583 × ⎨

Sheet

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81000 × 1188000 ⎫ −3 ⎬.10 2583000 ⎭

M cr = 1590 kNm Non-dimensional slenderness

The non-dimensional slenderness is obtained from : λ LT =

W pl,y f y M cr

=

3141000 × 355

EN 1993-1-1 § 6.3.2.2 (1)

= 0,837

1590 × 10 6

EN 1993-1-1

For rolled profiles, λ LT,0 = 0,4 t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

Note :

The value of λ LT,0 may be given in the National Annex. The recommended value is 0,4.

So

λ LT = 0,837 > λ LT,0

§ 6.3.2.3(1)

Reduction factor

For rolled section, the reduction factor for lateral torsional buckling is EN 1993-1-1 calculated by : § 6.3.2.3 (1) ⎧ χ LT ≤ 1.0 1 ⎪ 1 χ LT = but ⎨ χ LT ≤ 2 2 2 ⎪⎩ φ LT + φ LT − β λ LT λ LT

[

(

)

2

where : φ LT = 0,5 1 + α LT λ LT − λ LT,0 + β λ LT

]

α LT is the imperfection factor for LTB. When applying the method for rolled profiles, the LTB curve has to be selected from the table 6.5 : For h/b = 597 / 220 = 2,71 > 2 λ LT,0 = 0,4 and β = 0,75



Curve c 

LT =

0,49

EN 1993-1-1 Table 6.5 Table 6.3


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The values of λ LT,0 and β may be given in the National Annex. The recommended values are 0,40 and 0,75 respectively.

[

We obtain :

φ LT = 0,5 1 + 0,49 (0,837 − 0,4) + 0,75 × (0,837) 2

and :

χ LT =

1 0,870 + (0,870) − 0,75 × (0,837) 2

2

]

= 0,870

= 0,740

Then, we check : χ LT = 0,740 < 1,0 2

and :


χ LT = 0,740 < 1 / λ LT = 1,427

The influence of the moment distribution on the design buckling resistance moment of the beam is taken into account through the f-factor :

[

(

f = 1 − 0,5 (1 − k c )1 − 2 λ LT − 0,8

where : k c =

1 1,33 − 0,33 ×ψ

)] 2

and ψ = 0

Lc

Then :

k c =

So :

f = 1 – 0,5 (1 – 0,752) [1 – 2 (0,837 – 0,8) 2] = 0,876

1,33

§ 6.3.2.3 (2)

but ≤ 1

Simplified moment distribution :

1

EN 1993-1-1

= 0,752

We obtain : χ LT,mod = χ LT / f = 0,740 / 0,876 = 0,845

EN 1993-1-1 Table 6.6


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Design buckling resistance moment

M b,Rd = χ LT,mod W pl,y f y / γ M1 M b,Rd = (0,845× 3141000 × 355 / 1,0) × 10 -6 = 942,22 kNm M y,Ed / M b,Rd = 842,13 / 942,22 = 0,894 < 1 OK

EN 1993-1-1 § 6.3.2.1

EN 1993-1-1

Shear Resistance

In the absence of torsion, the shear plastic resistance depends on the shear area, which is given by:

§ 6.2.6 (3)

Av,z = A – 2 b t f + (t w + 2 r ) t f Av,z = 13700 – 2 × 220 × 17,5 + (9,8 + 2 × 24) × 17,5 = 7011,5 mm 2 But not less than η hw t w = 1,2 × 562 ×9,8 = 6609,12 mm 2 OK t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

EN 1993-1-1

Shear plastic resistance

V pl,z,Rd =

Av,z ( f y / 3 ) γ M0

=

7011,5 × (355 / 3 )/1000 1,0

V z,Ed / V pl,z,Rd = 172 / 1437 = 0,12 < 1

= 1437 kN

§ 6.2.6 (2)

OK

Resistance to shear buckling

Unstiffened webs with hw/tw greater than 72 ε / η should be checked for resistance to shear buckling and should be provided with transverse stiffeners at the supports. The value η may be conservatively taken as 1,0.

hw / t w = (597 – 2 × 17,5) / 9,8 = 57,35 < 72 × 0,81 / 1,0 = 58,3

So the shear buckling resistance does not need to be checked.

EN 1993-1-1 § 6.2.6 (6)


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Serviceability Limit State verification

EN 1990

SLS Combination

q g = 1,06 kN/m

§ 6.5.3

F G + F Q = 147 + 87,50 = 234,50 kN

Deflection due to G +Q :

w=

w= t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e 7 v 0 r 0 e 2 s , e r 1 s 0 t r h g e i r b l l m e a v o t h N g , i y r a y d p o s r c u s h i l T a n i r o t e d a e t m a s i e r h C T

( F G + F Q ) L3 48 E I y

+

5q g L4 384 EI y

234500 × (10000)3 48 × 210000 × 82920×10 4

+

5 × 1,06 × (10000) 4 384 × 210000 × 82920×10 4

w = 28,85 mm The deflection under G+Q is L/347 – OK

Deflection due to Q :

w=

F Q L3 48 E I y

=

87500 × (10000)3 48 × 210000 × 82920 ×10 4

w = 10,47 mm The deflection under Q is L/955 – OK

EN 1993-1-1 § 7.2.1(1)B

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 : Concerning vibrations, the National Annex may specify li mits concerning the frequency.

EN 1993-1-1 § 7.2.3(1)B


Example: Simply supported beam with lateral restraint at load application point SX007a-EN-EU

Quality Record RESOURCE TITLE


Reference(s) ORIGINAL DOCUMENT Name

Compan y

Date

Created b y

Valérie LEMAIRE

CTICM

08/04/2005

Technical content checked by

Alain BUREAU

CTICM

11/05/2005

1. UK

G W Owens

SCI

17.08.2005

2. France

Alain BUREAU

CTICM

17.08.2005

3. Germany

A Olsson

SBI

17.08.2005

4. Sweden

C Muller

RWTH

17.08.2005

5. Spain

J Chica

Labein

17.08.2005

G W Owens

SCI

21.05.2005

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