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TEMTIS Seminar in Horsens September 11, 2008. The effect of Eurocode 5 on timber structure design in Norway. Kolbein Bell and Kjell Arne Malo NTNU, Norway. Objective. To compare some important passages in the current Norwegian timber code NS 3470-1 (5th ed. July 1999) with

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the effect of eurocode 5 on timber structure design in norway

TEMTIS Seminar in Horsens

September 11, 2008

The effect of Eurocode 5on timber structure design in Norway

Kolbein Bell and Kjell Arne MaloNTNU, Norway

Department of structural engineering

objective
Objective

To compare some important passages in

the current Norwegian timber code

NS 3470-1 (5th ed. July 1999)

with

the current version of Eurocode 5 (EC5)

EN 1995-1-1

including EN 1995-1-1:2004/A1

and

to point out some problem areas

Department of structural engineering

important notes
Important NOTES
  • NS 3470-1 was drafted on the basis of the first draft of EC5 – the ”philosophy” is therefore much the same
  • A corrigendum to NS 3470-1 which will bring theNorwegian code closer to EC5 is about to be made an official part of NS 3470-1 – this in order to ”soften” the effects of the transition to EC5.This corrigendum is disregarded in this presentation

Department of structural engineering

slide5

Load duration

EC5

NS 3470-1

Service class

Department of structural engineering

slide6

NS 3470-1

(from 1,1 to 1,32)

1,0 or 1,1

1,1 or 1,2

for solid timber

EC5

for glulam

Department of structural engineering

slide7

Stability - NS 3470-1

Stability - NS 3470-1

Combined axial compression and bending:

Bending:

Department of structural engineering

slide8

Stability - EC5

Combined axial compression and bending:

Bending:

Department of structural engineering

slide9

The factor km

Rectangle:

(bending about two axes)

this factor is not present in NS 3470-1

Department of structural engineering

slide10

Comparison – simple column – P only

p = 0

NS 3470-1

GL36c

C/2

Pu = 188,2 kN

A/3

Pu = 125,5 kN

EC5

Pu = 178,3 kN

C/2

A/3

Pu = 111,5 kN

Department of structural engineering

slide11

Comparison – simple column – P & p

p = const. = 3,0 kN/m

NS 3470-1

GL36c

C/2

Pu = 120 kN

A/3

Pu = 60,6 kN

EC5

Pu = 139,3 kN

C/2

A/3

Pu = 72,5 kN

Department of structural engineering

slide12

Compression perpendicular to grain

The characteristic strength in NS 3470-1is more than twice that of EC5 (for all strength classes), but NS3470-1 uses the actual contact area in the calculation of The formula and the values of factor are not all that different (providing the requirements of “supplement” A1 are used).More about this later.

Department of structural engineering

shear
Shear

The introduction of an effective width

by EC5 (”supplement” A1), may, dependig on the national choice for the value of , have a significant influence on shear design.

Department of structural engineering

slide14

Special glulam components

For curved and pitched cambered beams EC5 has the following formula for combined shear and tension perpendicular to grain:

(6.53)

where

For glulam:

Department of structural engineering

slide15

NS 3470-1 has no such formula, nor does it have the two factors

and

in particular is troublesome; it is both difficult to determine and it seems to have a very significant (adverse) effect for large components).

Formula (6.53) will have a detrimental effect on typical Norwegian arch bridge designs.

Department of structural engineering

slide16

Example: Glulam arch bridge - loading

Department of structural engineering

slide17

Bending moment (M) and shear force (V)

M

problem

V

V

V

Department of structural engineering

slide18

Prior to formula (6.53) EC5 had the formulation: …. shall be satisfied.

In the recently approved ”supplement” A1 this has been changed to: …. should be satisfied.

Exactly how should the designer interpret this?

Department of structural engineering

slide19

Connections

  • EC5 has similar, but more complex formulas thanNS 3470-1.
  • The most noticeable differences are:
  • in NS 3470-1 the first 6 fasteners are effective
  • NS 3470-1 does not take account of the rope effect
  • NS 3470-1 does not recognize block or plug shear
  • Our experiences so far suggest minor differences, but no systematic bias either way.

Department of structural engineering

slide20

More aboutcompression perpendicular to grain

Comparison EC5(A1) vs NS 3470-1

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slide21

geometric parameters:

design parameters (ULS):

Department of structural engineering

slide22

Strength ultimate limit state (ULS)

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slide23

Design: NS 3470-1 & EC5/A1(2004)

Department of structural engineering

slide24

Solid wood, C24:

Department of structural engineering

slide25

EC 5/A1(2004)

Department of structural engineering

slide26

NS 3470-1

  • For
  • is
  • where:
  • while for
  • is

Department of structural engineering

slide27

Solid wood, Case 1:

  • C24
  • Continous support
  • Vertical column on end of beam

Department of structural engineering

slide28

EC5 / A1(2004)

Capacity pr unit width [ N/mm ]

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slide29

NS 3470-1

Capacity pr unit width [ N/mm ]

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slide30

Strength ratio: EC5(A1) / NS 3470-1

Department of structural engineering

slide31

Solid wood, Case 2:

  • C24
  • Continous support
  • Vertical column on continuous beam

Department of structural engineering

slide32

EC5 (A1)

Capacity pr unit width [ N/mm ]

Department of structural engineering

slide33

NS 3470-1

Capacity pr unit width [ N/mm ]

Department of structural engineering

slide35

Solid wood (C24), Case 3:

  • Vertical load transfer through beam section (h > a)
  • Beam continuous
  • EC5: (?)

Department of structural engineering

slide36

EC5 (A1) NS 3470-1

[ N/mm ]

[ N/mm ]

Department of structural engineering

slide38

Solid wood (C24), Case 4:

  • Vertical load transfer through beam section at the beam end
  • EC5?(cover this case?)

Department of structural engineering

slide39

Ratio EC5 (A1) / NS 3470-1Columns at beam end

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slide41

GLULAM

Examples:

GL32c

Department of structural engineering

slide42

Ratio EC5 / NS 3470-1 GL32c”Column at beam end, cont. sup.”

Department of structural engineering

slide43

Ratio EC5 / NS 3470-1 GL32c”Column internal on beam, cont. sup.”

Department of structural engineering

slide44

Ratio EC5 / NS 3470-1 GL32c”Column connections at beam end”

Department of structural engineering

slide45

Ratio EC5 / NS 3470-1 GL32c”Column connections internal on beam”

Department of structural engineering

slide46

Ratio EC5 / NS3 470-1 GL32c”Columns eccentric internal on beam, discrete. sup.”

Department of structural engineering

slide47

Compression perpendicular to grainConcluding remarks

  • EC5 compared to NS 3470-1 gives for:
    • Solid wood:
      • roughly only 2/3 of the capacities
      • small capacities for vertical load transfer through horisontal beams
      • enhanced capacities for colums at beam ends
      • highest capacities for small contact length due to the effektiv length concept
    • GLULAM:
      • Overall similiar to solid wood, but the difference is smaller
      • Higher capacities for small contact length (< 30 mm)
      • Smaller capacity for vertical load transfer through continuous beams

Department of structural engineering

slide48

Summary – ultimate limit state

Generally speaking, ULS-design of timber structures by EC5 will result in somewhat more conservative designs than NS 3470-1.

We are talking about 5 to 25 %, most of which is caused by and .

In some special cases the effect can be much higher.

Our experiences over the past couple of decades do not seem to warrant this ”extra safety”.

Department of structural engineering

slide49

Some additional problems

Serviceability limit state design, as specified by EC5, is rather complex and error prone.

EC5 is not particularly well suited for more accurate, nonlinear static analyses as basis for design (nor is NS 3470-1). Major issues are:- stiffness parameters (E and G)- shape and size of geometric imperfections- modelling of joints- failure criteria

Department of structural engineering

slide50

Consider

  • E (G) as “computational” parameter(s), accounting for all factors influencing the stiffness of the structural members
  • ultimate load design
  • solid timber and glulam of softwood

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slide51

EC5 - some definitions

Mean value:

Characteristic (fifth percentile) value:

Final mean value:

Design value:

- partial factor for a material property

- factor for quasi-permanent value of an action

- factor for the evaluation of creep deformation

Department of structural engineering

slide52

NOTE:

EC5 specifies:

for solid timber

for glulam

for solid timber

for glulam

for solid timber

for glulam

Department of structural engineering

slide53

EC5 states that the analysis of a structure shall be carried out using the following values for stiffness properties:

1st order linear elastic analysis of structure whose distribution of internal forces is insensitive to stiffness distribution

1st order linear elastic analysis of structure whose distribution of internal forces is affected by the stiffness distribution

2nd order linear elastic analysis of structure

Department of structural engineering

slide54

Case - Simply supported column

Glulam GL36c

strong axis

Department of structural engineering

slide55

EC5 - nonlinear approach

Geometric imperfection in the shape of the 1st buckling mode

e = L / 400

e

L

(EC5 suggests L / 500)

For E = 12000 Mpa

PE = 304,7 kN for L = 4 m

PE = 135,4 kN for L = 6 m

Department of structural engineering

slide56

Capacity (Pult) for long-term load in service class 3

q = 0 L = 4 m e = 10 mm

medium slenderness:

Linear analysis: Pult = 122,7 kN

Nonlinear analysis ( kc,y = kc,z = 1,0 ):

Department of structural engineering

slide57

Capacity (Pult) for long-term load in service class 3

q = 0 L = 6 m e = 15 mm

large slenderness:

Linear analysis: Pult = 56,5 kN

Nonlinear analysis ( kc,y = kc,z = 1,0 ):

Department of structural engineering

slide58

Capacity (Pult) for long-term load in service class 3

q = 1 kN/m L = 6 m e = 15 mm

large slenderness:

Linear analysis: Pult = 43,0 kN

Nonlinear analysis ( kc,y = kc,z = 1,0 ):

Department of structural engineering

slide59

Long-term load in service class 3

q = 0 L = 6 m e = 15 mm

q = 0 L = 6 m e = 15 mm

Department of structural engineering

slide60

q = 0 L = 6 m e = 15 mm

S / 1

L / 3

L / 3

S / 1

Department of structural engineering

slide61

This simple example seems to leave a few questions:

  • size of imperfection?
  • what is the appropriate stiffness?
  • and

are factors , and

and parameter really independent of load duration and service class, or should they be functions of in some ways?

Department of structural engineering

slide62

Thank you

Department of structural engineering