example_simply supported beam with lateral restraint at load application point

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Document Ref: SX007a-EN-EU Sheet 1 of 10Title

CALCULATION SHEET

Example: Simply supported beam with lateral restraint at

load application point

Eurocode Ref EN 1993-1-1

Made by Valrie LEMAIRE Date April 2005

Checked by Alain BUREAU Date April 2005

Example: Simply supported beam with lateralrestraint at load application point

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

1 1

5,0 m 5,0 m

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 exactcalculation of the elastic critical moment for lateral torsional

buckling,

- the calculation of shear resistance, including shear bucklingresistance,

- the calculation of the deflection at serviceability limit state.

Partial factors

G = 1,35 (permanent loads)

Q = 1,50 (variable loads)

M0 = 1,0

M1 = 1,0

EN 1990

EN 1993-1-1

6.1 (1)

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

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CALCULATION SHEET






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

7m

5m 5m

Secondary beam

Calculated

Beam

Concrete slab

5m 5m

150 mm

597 mm

Restraints to lateral buckling


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CALCULATION SHEET






Span length : 10 m

Secondary beam:

o Span length: 7 m

o Bay width : 5 m

Slab depth : 15 cm

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 : S355

Weight of the slab : 0,15 24 kN/m3 = 3,6 kN/m2

Try IPEA 600 Steel grade S355

Depth h = 597 mm

Width b = 220 mm

Web thickness tw = 9,8 mm

Flange thickness tf= 17,5 mm

Fillet r= 24 mm

Mass 108 kg/m

z

z

y y

tf

tw

b

h

Euronorm

19-57

Section area A = 137 cm2

Second moment of area /yy Iy = 82920 cm4

Second moment of area /zz Iz = 3116 cm4

Torsion constant It = 118,8 cm4

Warping constant Iw = 2607000 cm6

Elastic modulus /yy Wel,y = 2778 cm3

Plastic modulus /yy Wpl,y

= 3141 cm3


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CALCULATION SHEET






Self weight of the beam : qG = (108 9,81) 10-3 =1,06 kN/m

Permanent load :

FG = (3,6 + 0,10 + 0,50) 5,0 7,0 = 147 kN

Variable load (Imposed load) :

FQ = 2,50 5,0 7,0 = 87,5 kN

ULS Combination :

G qG = 1,35 1,06 = 1,43 kN/m

G FG + Q FQ = 1,35 147 + 1,50 87,5 = 329,70 kN

EN 1990

6.4.3.2(6.10)

Moment diagram

M 842,13 kNm

Maximal moment at mid span :

My,Ed = 0,125 1,43 10,002 + 0,25 329,70 10 = 842,13 kNm

Shear force diagram

V

172 kN 164,85 kN

Maximal shear force at supports :

Vz,Ed = 0,50 1,43 10,0 + 0,50 329,70 = 172 kN

Maximal shear force at mid-span :

Vz,Ed = 0,50 329,70 = 164,85 kN

Yield strength

Steel grade S355

The maximum thickness is 17,5 mm < 40 mm, so :fy = 355 N/mm2

Note : The National Annex may impose either the values offy from the

Table 3.1 or the values from the product standard.

EN 1993-1-1

Table 3.1


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CALCULATION SHEET






Section classification :

The parameteris derived from the yield strength : 0,81][N/mm

235

2

y

==f

Outstand flange : flange under uniform compression

c = (b tw 2 r) / 2 = (220 9,8 2 24)/2 = 81,10 mm

c/tf= 81,1 / 17,5 = 4,63 9 = 7,29 Class 1

EN 1993-1-1

Table 5.2

(sheet 2 of 3)

Internal compression part : web under pure bending

c = h 2 tf 2 r= 597 2 17,5 2 24 = 514 mm

c / tw = 514 / 9.8 = 52,45 < 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.

EN 1993-1-1

Table 5.2

(sheet 1 of 3)

Moment resistance

The design resistance for bending of a cross section is given by :

Mc,Rd =Mpl,Rd = Wpl,yfy / M0 = (3141 355 / 1,0) / 1000

Mc.Rd = 1115 kNm

My,Ed /Mc,Rd = 842,13 / 1115 = 0,755 < 1 OK

EN 1993-1-1

6.2.5


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CALCULATION SHEET






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 :

( )( )

++

= g2

2

g2

z

2

t

2

z

w

2

w

2

z

2

1cr )(

zCzCIE

IGLk

I

I

k

k

Lk

IECM c

c

Eis the Young modulus : E= 210000 N/mm2

G is the shear modulus : G = 81000 N/mm2

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

See NCCI

SN005

In the expression ofMcr, the following assumption should be considered :

k= 1 since the compression flange is free to rotate about the weak

axis of the cross-section,

kw = 1 since there is no device to prevent the warping at the ends of

the beam.

The C1 and C2 coefficients depend on the moment diagram along the beam

segment between lateral restraints. It can be assumed that the diagram is

linear, then :

C1 = 1,77 for k = 1

C2zg = 0

Therefore :

( )kN258310

(5000)

103116210000 32

42

2

z

2

=

=

cLk

IE


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CALCULATION SHEET






= 25831,77crM3.10

2583000

118800081000100

3116

2607000

+

Mcr= 1590 kNm

Non-dimensional slenderness

The non-dimensional slenderness is obtained from :

0,837101590

3553141000

6

cr

yypl,LT =

==

M

fW

EN 1993-1-1

6.3.2.2 (1)

For rolled profiles, 0,4LT,0 =

Note : The value of LT,0 may be given in the National Annex. The

recommended value is 0,4.

So LT,0LT 0,837 >=

EN 1993-1-1

6.3.2.3(1)

Reduction factor

For rolled section, the reduction factor for lateral torsional buckling is

calculated by :

+=

2

LT

LT

LT

2

LT2

LTLT

LT1

1.0

but1

where : ( )[ ] 10,5 LTLT,0LTLTLT 2 ++=

EN 1993-1-1

6.3.2.3 (1)

LTis the imperfection factor for LTB. When applying the method for rolled

profiles, the LTB curve has to be selected from the table 6.5 :

Forh/b = 597 / 220 = 2,71 > 2 Curve c LT = 0,49

0,4LT,0 = and = 0,75

EN 1993-1-1

Table 6.5

Table 6.3


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Note : The values of LT,0 and may be given in the National Annex. The

recommended values are 0,40 and 0,75 respectively.

We obtain : ( )[ ] 0,870(0,837)0,750,40,8370,4910,5 2LT =++=

and : 0,740(0,837)0,75(0,870)0,870

1

22LT =

+=

Then, we check : LT = 0,740 < 1,0

and : LT= 0,740

example_simply supported beam with lateral restraint at load application point

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