[Eng]Advanced Training Scaffolding 2009.0.1

7/30/2019 [Eng]Advanced Training Scaffolding 2009.0.1

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All information in this document is subject to modification without prior notice. No part or this manualmay be reproduced, stored in a database or retrieval system or published, in any form or in any way,electronically, mechanically, by print, photo print, microfilm or any other means without prior writtenpermission from the publisher. Scia is not responsible for any direct or indirect damage because ofimperfections in the documentation and/or the software.

Copyright 2008 Scia. All rights reserved.


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Contents

1. General ................................................................................................................................ 72. Materials .............................................................................................................................. 93. Cross sections.................................................................................................................. 114. Construction ..................................................................................................................... 135. Linear Stiffness of connections and supports .............................................................. 156. Load cases ........................................................................................................................ 177. Linear Load Combinations .............................................................................................. 258. Results .............................................................................................................................. 279. Steel Code SLS Check EN 1993-1-1 ............................................................................... 2910. Scaffolding SLS Check EN 12811-1-1 ......................................................................... 3111. Steel Code Check ULS Check EN 1993-1-1 ................................................................ 33

Section Check .................................................................................................................................. 33Stability Check ................................................................................................................................. 37

12. Scaffolding ULS Check EN 12811-1-1 ......................................................................... 4313. Non-Linear Combinations ............................................................................................... 4514. The second order calculation ......................................................................................... 51

Timoshenko ..................................................................................................................................... 51Newton Raphson ............................................................................................................................. 53

15.

Stability ............................................................................................................................. 57

Linear Stability ................................................................................................................................. 57Buckling Shape ............................................................................................................................... 59

16. Input Hinge Types and non-linear stiffness .................................................................. 6717. Scaffolding Coupler Check .......................................................................................... 7318. Scaffolding Non linear Check ......................................................................................... 7719. Aluminium ......................................................................................................................... 79

Aluminium grades ........................................................................................................................... 79Initial bow imperfection e0 .............................................................................................................. 81Initial shape ...................................................................................................................................... 82Section check .................................................................................................................................. 85

20. Document .......................................................................................................................... 8921. Template ............................................................................................................................ 9122. User Blocks....................................................................................................................... 9223. References ........................................................................................................................ 95Annex A: Wind pressure versus Wind force ........................................................................... 97


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1. General

The objective of this manual is to show a way of modelling a scaffolding (class 3) using Scia Engineer. Inthis manual a simple example is elaborated.

The following scaffolding will be treated in this course:

When entering the project, the following project data are chosen:

Frame XYZProject Level: AdvancedMaterial: S235National Code: EC-EN

and the following functionalities:

Non-linearity (+ all options for non linearity on the right hand side)Scaffolding


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2. Materials

For the materials, S235 is generally used. However, floor boards and toeboards will be inserted asmembers. The average weight of these toeboards and floor boards differs from the weight of S235.That is why is chosen to insert an extra material in Scia Engineer, at which the weight will be adapted.

This weight can be determined as shown below:

FLOOR BOARDS

A distinction is made between floor boards of 19cm and 32cm. For each of them an average weightis calculated:

Name Weight

[kg]

Length

[m]

[kg/m] Average

Steel board32/307

23.2 3.07 7.56

Steel board32/257

19 2.57 7.39

Steel board32/207

15.7 2.07 7.58

Steel board32/157

12.2 1.57 7.77 8.20kg/m

Steel board32/140

10.8 1.4 7.71

Steel board32/109

10.4 1.09 9.54

Steel board32/73

7.2 0.73 9.86

Steel board19/307

18.2 3.07 5.93

Steel board19/257

15.5 2.57 6.03

Steel board19/207

12.7 2.07 6.14 6.18kg/m

Steel board19/157

10 1.57 6.37

Steel board19/109

7 1.09 6.42

It is assumed that the floor boards have a thickness of 4 cm. The weights of the floor boards are:

Floor boards 0.32m: 363,64004,032,0

2,8

mkg

mm

mkg

=

=

Floor boards 0.19m: 316,81304,019,0

18,6

mkg

mm

mkg

=

=


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TOEBOARDS

A similar calculation is applied for the toeboards:

Name Weight

[kg]

Length

[m]

[kg/m] Average

Toeboard 3.07 6.3 3.07 2.05

Toeboard 2.57 5.7 2.57 2.22

Toeboard 2.07 4.3 2.07 2.08

Toeboard 1.57 3.5 1.57 2.23 2.32 kg/m

Toeboard 1.09 2.5 1.09 2.29

Toeboard 0.73 1.5 0.73 2.05

Toeboard 0.30 1.0 0.30 3.33

The toeboards are assumed to have a width of 2 cm and a height of 15 cm. The weight of thesetoeboards is 773.33 kg/m.

Toeboards: 333,77315,002,0

32,2

mkg

mm

mkg

=

=

When entering the geometry, the toeboard profiles are entered on a level 7.5 cm above the ledgers ofthe floors.

Three new materials are entered in Scia Engineer (Floor board19, Floor board32 and Toeboards).These materials are entered as an existing steel quality, at which only the weight is adapted:


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3. Cross sections

Below all cross sections used in the construction are displayed:

Cross section Material

Column RO48.3x3.2 S235Ledger RO48.3x3.2 S235

Guardrail RO48.3x3.2 S235Bracing RO48.3x2.3 S235

Floor board 0.32m RECT (40; 320) VL 0.32Floor board 0.19m RECT (40; 190) VL 0.19

Toeboard RECT (150; 20) ToeboardTube RO48.3x3.2 S235

Rule Truss RO48.3x3.2 S235Bracing Truss SHSCF25/25/2.5 S235


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4. Construction

The construction will be entered as a 3D Frame for the first level. Afterwards this level is copiedupward. The type of all members in this frame is entered correctly:

- Columns: column- Ledgers: beam- Bracings: vertical wind bracing-

INPUT OF FLOOR BOARDS

The input of floor board elements depends on the length of the ledgers. In the picture below all thepossibilities are displayed. The total length of the ledger is shown at the bottom. The distance of thepoints, on which the floor board has to be entered, is displayed at the top.

INPUT OF TOEBOARDS

The toeboards are entered on a level 7.5 cm above the ledgers of the floors. The total height of thetoeboard is 15 cm.

INPUT OF BRACINGS

The wind bracings are entered with an eccentricity in the y direction of 48.3 mm (width of the profile): ey=+/- 48.3mm. This way the wind bracings are truly on top of the other members and not in between:


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Without eccentricity With eccentricity


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5. Linear Stiffness of connections and supports

The rigidities are entered through hinge on member.

CONNECTIONS

Standard Standard connections

The overlap length between two columns is 200 mm > 150 mm, so the columns are rigid in the x -direction.

Moreover, the margin between two columns is 3.9 mm (=48.3 mm 2 x 3.2 mm 38 mm), this isless than 4 mm. Because of this, also the connection in the y and z direction are rigid.

According to the code, also the following applies: According to code EN12811-1, the connection column - column can be considered rigid

in the modelling

A hinge will not be entered at the extremities on the standard

Standard Ledger and Standard Guardrail connections

- y direction: 0.15 MNm/rad- z direction: 0.004 MNm/rad

These rigidities are obtained from data from the supplier. In this case, the rigidity for the y direction istaken from the supplier and a smaller value is taken for the z direction. Of course, this is not acorrect method. Both should be obtained from data from the supplier.

Standard Toeboard and Ledger Floor board connections

- y direction: Hinged- z direction: Rigid

Standard Bracing connections

- y direction: Hinged- z direction: Rigid

Summary:As a summary, the following rigidities can be entered in Scia Engineer:

Standard-ledger: y direction: 0.15 MNm/radz direction: 0.004 MNm/rad

Standard-guardrail: y direction: 0.15 MNm/radz direction: 0.004 MNm/rad

Standard-bracing: y direction: Freez direction: Rigid

Standard-toeboard: y direction: Freez direction: Rigid

Ledger-floor board: y direction: Freez direction: Rigid


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SUPPORTS

There are two types of supports: the base jacks at the bottom and the anchorages in the wall.

For the base jacks, translations in all directions are fixed (Translations X, Y and Z fixed). For theanchorages, only the translations according to the x and y direction are fixed (Translations X and Yfixed).

The anchorages are introduced according to the anchorage scheme: from 4m high, every standardhas to be anchored every 4 m upward. Over 20 m high, this needs to be performed every 2 m upward.

Note

Please note that the input described above is only a simplified input method of rigidities. If these rigidities areentered correctly, they have to be entered and calculated with non-linear functionalities.

END INPUT OF CONSTRUCTION

After entering the construction, it is recommended to check the input by using the command CheckStructure Data. Through this function the geometry is checked on errors.

After the check, the option Connect nodes/edges to members is applied to the entire construction. With thisfunction the different parts are connected to each other.


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6. Load cases

There are three main types of loads which need to be considered [EN12811-1, 6.2.1.]:

a) Permanent loads: these shall include the self weight of the scaffold structure, including allcomponents, such as platforms, fences, fans and other protective structures and any ancillarystructures such as hoist towers.

b) Variable loads: these shall include service loads (loading on the working area, loads on the sideprotection) and wind loads and, of appropriate, snow and ice loads.

c) Accidental loads

The permanent loads are inputted as the self weight in Scia Engineer.

The only accidental load specified in the European Standard is the downward loading on the sideprotection [more info in EN12811-1, 6.2.5.1].

The variable loads can be considered as the service loads and the wind load.

The service loadsare considered in Table 3 of EN 12811-1:

Each working area is capable of supporting the various loadings q1, F1and F2, separately but notcumulatively.

q1: Uniformly distributed service load (EN12811-1, 6.2.2.2)

Each working area is capable of supporting the uniformly distributed loads, q1 as specified in thetable above.

F1 and F2: Concentrated load (EN12811-1, 6.2.2.3)

Each platform unit is capable of supporting the load F1 uniformly distributed over an area of 500mmx 500mm and, but not simultaneously, F2uniformly distributed over an area of 200mm x 200mm.

The position of each load is chosen to give the most unfavourable effect.


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When a platform unit is less than 500mm wide, the load may be reduced for this unit in proportion toits width, except that in no case shall the loading be reduced to less than 1,5kN.

q2: Partial area load (EN12811-1, 6.2.2.4)

This load has to be applied only for classes 4, 5 and 6. In those cases each platform is capable of

supporting a partial area loading q2on an area Aq2:

With:

l The length

w The width

ap Coefficient of Table 3

The dimensions and position of the partial area are chosen to give the most unfavourable effect. Oneexample is shown below:

Wind load

Following 6.2.7.4 of EN12811-1, two wind loads have to be calculated: the maximum wind load and theworking wind load.

Maximal wind load

When the European Standard for wind loads is available it is used for the calculation of the maximalwind load.To make allowance for equipment or materials which are on the working area, a nominal reference

area is assumed at its level over its full length. This area is 200mm and includes the height of thetoeboard. (En 12811-1, 6.2.7.4.1)


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Note: For the purposes of structural design of faade scaffolds made of prefabricated components,design velocity pressures are given in EN 12810-1, 8.3.

Working wind load

A uniformly distributed velocity pressure of 0,2 kN/m is taken into account.To make allowance for equipment or materials which are on the working area, a nominal referencearea is assumed. This area is 400mm and includes the height of the toeboard. (En 12811-1, 6.2.7.4.2)

Area of wind loading

The wind load is inputted as an area load. The amount of this load, depends on the amount of theloaded area itself, so without cladding a smaller wind load will be inserted than with cladding or netting.

LOAD CASES IN SCIA ENGINEER

The following load cases will be inputted in Scia Engineer:

Load Case 1: Self Weight

In this load case the complete self weight of the structure is included, including the toeboards, floorboard elements, ...

This load case is automatically calculated by Scia Engineer.

Load Case 2: Service load Main floor Full

This load case represents the service load that operates over the entire main floor. Main floormeans the most important/crucial floor of the scaffolding. If the load is put on this floor, it leads to themost critical values.

In this example, a class 3 scaffolding has been inserted, so a divided load of 200 kg/m, or 2 kN/m(EN12811-1 Table 3). This is transferred to floor boards.

Load Case 3: Service load Secondary floor Full

Analogous to load case 2, the service load in this load case is entered on the complete secondaryfloor. Secondary floor refers to the working area at the first level above or below the main floor.

According to the code EN12811-1, 50% of the service load has to be put on the secondary floor.

Load Case 4: Service load Main floor Partial

This load case is completely analogous to load case 2, but here the service load is only put on half ofthe main floor. By executing this load case, the structure is eccentrically loaded, so effects thatcounterbalance each other in a symmetrical load, are revealed here.

Load Case 5: Service load Main floor Partial

This load case is completely analogous to load case 3, but partially (see also Load case 4).


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Load Case 6: Service load 25% Full

This load case represents an accumulation of materials and equipments for the complete mainworking area, when the scaffolding is subject to the maximal wind load.

Load Case 7: Service load 25% Partial

This load case represents an accumulation of materials and equipments for only half of the mainworking area, when the scaffolding is subject to the maximal wind load.

Load Case 8: Maximal Wind load Perpendicular to Facade

Since the arithmetic method for the calculation of a wind load from code EN12811-1 is not valid forall scaffoldings, provided by nets that completely surround the construction, the code EN 1991-1-4 is

adopted.

There are three cases for calculating a reference height (EN1991-1-4, Figure 7.4)


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In the example discussed in this course, the height is 11m and the building face is 12,35m. So in thiscase clearly 11m < 12,35m and thus h < b. So the wind only has to be calculated for ze =11m.

In this example, the wind is calculated for a construction without nets, situated in Belgium with aterrain category of III.

The terrain category is determined as follows (EN 1991-1-4, Table 4.1):

Terrain category z0 [m] zmin [m]

0 Sea or coastal area exposed to the open sea 0,003 1

I Lakes or flat and horizontal area with negligible vegetation andwithout obstacles

0,01 1

II Area with low vegetation such as grass and isolated obstacles (trees,buildings) with separations of at least 20 obstacle heights

0,05 2

III Area with regular cover of vegetation or buildings or with isolatedobstacles with separations of maximum 20 obstacle heights (such as

villages, suburban terrain, permanent forest)

0,3 5

IV Area in which at least 15% of the surface is covered with buildingsand their average height exceeds 15m

1,0 10

For Belgium, vb,0 equals 26.2m/s following the EC-EN. In addition Cdir and Cseason are all equal to 1.From this vb can be calculated (EN 1991-1-4 (4.1)):

vb = 0,bseasondir vCC

= 26,2m/s

In terrain class III the mean velocity vm are calculated using the following formula (EN 1991-1-4(4.3)):

vm = bor vzczc )()(

where

co(z) is the orography factor taken as 1,0 (unless specified otherwise in EN 1991-1-4 4.3.3.)

and

cr(z) is the roughness factor given by formula (EN 1991-1-4 (4.4)):

=

0

ln)(z

zkzc rr

Formaxmin zzz

( )min)( zczc rr = for

minzz


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And

07,0

,0

019,0

=

II

rz

zk

(EN 1991-1-4 (4.5))

The evaluations of these formulas for this example and category III gives:

215,005,0

3,019,019,0

07,007,0

,0

0 =

=

=II

rz

zk

776,03,0

11ln215,0ln)(

0

=

=

=

z

zkzc rr

And the mean velocity:

3,202,261776,0)()( === borm vzczcv m/s

Out of these values the peak velocity pressure is calculated by:

2

2

1)()()(

bebep vzcqzczq ==

where

= 1,25 kg/m

[ ] ( ) ( )202

)()()(71)( zczczIzc rve +=

=

0

0 ln)(

)(

z

zzc

kzI lv

formaxmin zzz

( )min)( zIzI vv =

forminzz

kl is the turbulence factor. The recommended value for k l is 1,0.

These formulas evaluated for this example and category III gives:

278,0

3,0

11ln0,1

0,1

ln)(

)(

0

0

=

==

=

z

zzc

kzI lv

[ ] ( ) ( ) [ ] ( ) ( ) 77,10,1776,0278,071)()()(71)( 22202

=+=+= zczczIzc rve


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That way, the total external pressure is obtained by:

( ) /76,0/760/2,2625,12

177,1

2

1)()(

22mkNmNmNvzczq bep ====

Furthermore, we can assume that the members in the plane perpendicular on the wind load take upabout 1/5 (=20%) of the total surface of the construction, as shown on the drawing below:

20% of the total pressure is entered on the walls. This equals a total maximal wind pressure of 0.76kN/m x 0.2 = 0.152 kN/m.

Code EN 12811-1 6.2.7.4.1.:

To make allowance for equipment or materials which are on the working area, a nominalreference area shall be assumed at its level over its full length. This area shall be 200 mm highmeasured from the level of the working area and includes the height of the toeboard. The loadsresulting from the wind pressure on this area shall beassumed to act at the level of the workingarea.

In this example, the toeboards are already calculated in the 20% of the total surface:

In this example the toeboards are 150mm high. The area of EN12811-1; 6.2.7.4.1 is 200mm so it isnecessary to calculate with an extra height of 50mm (200mm-50mm) due to equipment of theworking area. 50mm high on a total length of 2m (= 2000mm) corresponds to 2.5% (=50/2000).

So the the wind pressure is: 0.152 kN/m + (0.025 x 0,76kN/m) = 0.17 kN/m.

Note 1

When inputting netting on the structure the following procedure can be taken into account to calculate the

wind pressure on the structure (Netting = 50%):

- Calculate the wind force

- 50% of the wind will be stopped by the netting and carried by the first columns

- 20% of the other 50% of this wind will be inserted on the first plane

- 20% of the other 50% of this wind will be inserted on the second plane

- ...


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Note 2

In this paragraph the wind has been calculated as a wind pressure. In EN 12810-1a calculation with wind

forces instead of wind pressures. The difference of this method with the EN 1991-1-4 will be discussed in

Annex A.

Load Case 9: Maximal Wind load Parallel to Facade

In an analogous way, the Maximal Wind load is entered parallel to the facade on the structure.

Also here maximal wind pressures of 0.17kN/m are entered.

Load Case 10: Working Wind load Perpendicular to Facade

The code EN12811-1 prescribes that if the scaffolding is into service, it only needs to be loaded withthe so-called working wind load.

This working wind load is calculated analogously to the Maximal Wind load on the scaffolding, but areference wind pressure of 0.2 kN/m is assumed.

The calculation is done analogous to the maximal wind pressure, only the reference wind pressure of0.429 kN/m (=26,2m/s) from item Load Case 8 is now replaced by 0.2 kN/m:

EN 12811-1 6.2.7.4.2:A uniformly distributed velocity pressure of 0,2 kN/m2 shall be taken into account. To makeallowance for equipment or materials being on the working area, a nominal reference areaas defined in 6.2.7.4.1, but 400 mm high, shall be used in calculating working wind loads.

/354,0/2,077,1/2,0)()( mkNmkNmkNzczq ep ===

The toeboards in this example are 150 mm high. In total an extra height of 250 mm (= 400mm 150mm) is calculated with. This corresponds to 12.5% of the total construction (= 250/2000).

Subsequently, the wind pressure is multiplied by 20% (total surface of the members) and 12.5% (dueto accumulation of material).

So the wind pressure is: 0,354 kN/m x (0.2+0,125) = 0,115 kN/m

Load Case 11: Working Wind load Parallel to Facade

In an analogous way, the working wind load is entered parallel to the facade on the structure. In this

case, the wind pressures are also 0.115 kN/m.


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7. Linear Load Combinations

Following EN 12811-1, 6.2.9.2 the load cases have to be combined in two different ways: the servicecondition and the out of service condition

In each individual case the service condition and the out of service condition shall be considered.

a) The service condition consists of:

1) The self weight of the scaffold

2) Uniformly distributed service load (EN 12811-1, Table 3, q1) acting on the working area ofthe most unfavourable decked level

3) 50 % of the load specified in a)2) shall be taken to act on the working area at the next levelabove or below if a working scaffold has more than one decked level.

4) Working wind load

b) The out of service condition consists of:

1) The self weight of the scaffold

2) A percentage of the uniformly distributed service load (EN 12811-1, Table 3, q1) acting onthe working area of the most unfavourable decked level. The value depends on the class:

Class 1: 0% (no service load on the working area)Classes 2 and 3: 25% (representing some stored materials on the working area)Classes 4, 5 and 6: 50% (representing some stored materials on the working area)

3) The maximum wind load

In cases a)2) and b)2), the load shall be taken as zero, if its consideration leads to more favourableresults.

This way the following load case combinations are obtained:

Combination 1: Out of service, Floor fully loaded, Wind perpendicular

BG1 Self weightBG6 Service load 25% fullyBG8 Maximal wind load perpendicular with facade

Combination 2: Out of service, Floor partially loaded, Wind perpendicular

BG1 Self weightBG7 Service load 25% partiallyBG8 Maximal wind load perpendicular with facade


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Combination 3: Out of service, Floor fully loaded, Wind parallel

BG1 Self weightBG6 Service load 25% fullyBG9 Maximal wind load Parallel with Facade

Combination 4: Out of service, Floor partially loaded, Wind parallel

BG1 Self weightBG7 Service load 25% partiallyBG9 Maximal wind load Parallel with Facade

Combination 5: Into service, Floor fully loaded, Wind perpendicular

BG1 Self weightBG2 Service load Main floor fullyBG3 Service load Secondary floor fullyBG10 Working wind load perpendicular with Facade

Combination 6: Into service, Floor partially loaded, Wind perpendicular

BG1 Self weightBG4 Service load Main floor partiallyBG5 Service load Secondary floor partiallyBG10 Working wind load perpendicular with Facade

Combination 7: Into service, Floor fully loaded, Wind parallel

BG1 Self weightBG2 Service load Main floor fullyBG3 Service load Secondary floor completelyBG11 Working wind load parallel with Facade

Combination 8: Into service, Floor fully loaded, Wind Parallel

BG1 Self weightBG4 Service load Main floor partiallyBG5 Service load Secondary floor partiallyBG11 Working wind load parallel with Facade

With the Combinations in Ultimate Limit State, a safety factor F of 1,50 on the load cases is taken,

conformable to the code EN 12811 10.3.2. The safety factor M on the material is 1,10.

For the serviceability F and M shall be taken as 1,00.


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8. Results

It is recommended to first eimported loads and load coreviewed as well.

The bill of material can alsosection is shown.

With these results, each usone member, in order to calconstruction.

Example

Scaffolding.esa

The results (such as normalledgers). That is why it is rebe easier to review the resu

Apart from reviewing the resulfound in Reactions.

27

amine the reactions. This way you can check thbinations. The calculation protocol (in the result

be found in the results menu. In this table the tot

r can divide the total length of each type of crosulate the total number of ledgers, columns, ... n

force, moment, ) can be viewed per profile typommended to put every cross section in differen

lts per cross section type.

s, the anchorage forces have to be checked as

Scaffolding

accuracy of thedialog) should be

al length for each cross

section by the length ofcessary for the

e (columns, bracings,t layers. This way it will

ell. These can be


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9. Steel Code SLS C

Relative deformation

For each beam type, limiting v

> Setup > Relative deformatioWith the option 'Steel > Beamrelative deformations are givecheck related to the limit for th

Example

Relative Deformation.esa

- Set beam type for

- Set system length f

- Set limits for relativ

- Relative deformatio

B20: L = 2,9m limit: 2900/100

Uz = -1,400mm 1,400/29Check: (1/2072) / (1/1000) =

B23: L = 5,0m limit: 5000/500

Uz = -2,200mm 2,200/50Check: (1/2273) / (1/500) =

29

eck EN 1993-1-1

alues for the relative deflections are set, using th

ns'.s > Relative deformation', the relative deformatioas absolute values, relative values related to th

e relative value to the span.

ember B20 & B23: Beam and Rafter

r relative deformation

deformations: Beam 1/1000 and Rafter 1/500

n check on member B20 & B23

= 2,9mm

0 = 1/20720,48

= 10,0 mm

0 = 1/22730,22

Scaffolding

e menu 'Steel > Beams

ns can be checked. Thee span, or as unity


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10. Scaffolding

The deformation check in SLS

is L/100. This limit value is se

- Steel:- Setting => Settin

Example SLS Check

Scaffolding.esa- Choose for Relative def

- The maximum check will

The values of this check

31

LS Check EN 12811-1-1

is a part of EC3. According to the code EN1281-11 t

as follows:

of the Relative Deformation => Bring all values to 1

rmation in the steel menu.

be found for beam B352: a unity check of 0,25.

are displayed below.

Scaffolding

e allowable deformation

00


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11. Steel Code Check ULS Check EN 1993-1-1

Section Check

Section Classification

Four classes of cross sections are defined :

Class 1 (EC3, NEN) or PL-PL (DIN) section

Cross sections which can form a plastic hinge with the rotation capacity required for plastic analysis

Class 2 (EC3, NEN) or EL-PL (DIN) section

Cross sections which can develop their plastic moment resistance, but have limited rotation capacity.

Class 3 (EC3, NEN) or EL-EL (DIN) section

Cross sections in which the calculated stress in the extreme compression fibre of the steel member canreach its yield strength, but local buckling is liable to prevent development of the plastic moment resistance.

Class 4 (EC3, NEN) or Slender section

Cross sections in which it is necessary to make explicit allowance for the effects of local buckling whendetermining their moment resistance or compression resistance.


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This classification depends onFor each intermediary section,performed. The classificationFor each load case/combinatithe stability check. So, the staHowever, for non-prismatic seintermediary section.

Section Check

Example

Scaffolding.esa

- Go to the menu Steel -

- Use the Section check.

- The maximum unity che

- The results are shown b

34

the proportions of each of its compression elemthe classification is determined and the relatedan change for each intermediary point.n, the critical section classification over the me

bility section classification can change for each ltions, the stability section classification is deter

Check.

k will be obtained in beam B352: UC = 0,60

low

Advanced Training

nts.section check is

ber is used to performad case/combination.ined for each


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Stability Check

Buckling data

The different system lengths and sway type have to be introduced. The defaults can be overruled bythe user.

During the non-linear analysis, the sway type can be set by user input, or by non-sway.See Steel > Beam > Setup:


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Buckling Ratio

For the calculation of the buckling ratios, some approximate formulas are used. These formulas aretreated in the Theoretical Background (Ref.[32]).

The following formulas are used for the buckling ratios :

for a non sway structure :

24)+11+5+24)(2+5+11+(2

12)2+4+4+24)(+5+5+(=l/L

21212121

21212121

for a sway structure :

4+x

x=l/L1

2

with L the system length

E the modulus of Young

I the moment of inertia

Ci the stiffness in node IMi the moment in node I

Fi the rotation in node I


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1212

12

21

8+)+(

+4=x

EI

LC= ii

i

ii

M=C

The values for Mi and i are acalculated by load cases whic

The following load cases are c

- load case 1 : on the bon the columns the gl

- load case 2 : on the bon the columns the gl

The used approach gives gooconnections. For other cases,

Stability Check

Example

Scaffolding.esa

- Go to the menu Steel -

- Use for the Stability che

- The maximum unity che

- The results are shown b

39

2

proximately determined by the internal forces angenerate deformation forms, having an affinity

onsidered :

ams, the local distributed loads qy=1 N/m and qbal distributed loads Qx = 10000 N/m and Qy =1

ams, the local distributed loads qy=-1 N/m andbal distributed loads Qx = -10000 N/m and Qy=

results for frame structures with perpendicular rthe user has to evaluate the presented bucking r

Check.

k for Class All UGT.

k will be obtained in beam B24: UC = 0,74

low

Scaffolding

d the deformations,ith the buckling form.

z=-100 N/m are used,0000 N/m are used.

z=-100 N/m are used,-10000 N/m are used.

igid or semi-rigid beamatios.


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12. Scaffolding ULS Check EN 12811-1-1

For a steel check according to EC3, the ULS combinations are used.

If necessary, the buckling settings of the members have to be checked.

It is recommended to deselect the option "Standard unbraced in the menu Steel -> Setting ->Setting check steel members for both directions.

Further settings about the buckling data can be adapted in the property window of every member, byclicking on the three dots behind Buckling and relative lengths".

After all the settings have been set right, the check can be performed through Steel -> Check. Alsohere it is recommended to review the check per profile type.

This is the regular steel check that is displayed, but for scaffolding a supplementary check will beadded. This check is performed with the interaction formulae, shown below.

3

1

,

dplV

V 9,0

3

1

,

Library -> Structure,Analysis -> Initial deformations:


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The Type is chosen as According to code, with a standard imperfection of 1/200. The height of theconstruction is 11m for both inclination functions. There are 2 columns in the x-direction and 6columns in the y-direction. The inclination function for the x-direction (Def_X) is displayed below:


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All combinations are entered twice, once with the inclination according to x and once according to y:

Initial bow imperfection e0The initial bow imperfection is given by:

Where L is the member length.The bow imperfection has to be applied when the normal force NEd in a member is higher than 25% ofthe members critical buckling load Ncr.

Scia Engineer can calculate the bow imperfection according to the code automatically for all neededmembers. But in a scaffolding structure all profiles have the same buckling curve and thus the same


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bow imperfection. This bow imperfection is inputted as simple curvature: the same curvature for allmembers.


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14. The second order calculation

Timoshenko

The first method is the so called Timoshenko method (Th.II.O) which is based on the exactTimoshenko solution for members with known normal force. It is a 2

ndorder theory with equilibrium on

the deformed structure which assumes small displacements, small rotations and small strains.

When the normal force in a member is smaller then the critical buckling load, this method is very solid.The axial force is assumed constant during the deformation. Therefore, the method is applicable whenthe normal forces (or membrane forces) do not alter substantially after the first iteration. This is truemainly for frames, buildings, etc. for which the method is the most effective option.

The influence of the normal force on the bending stiffness and the additional moments caused by thelateral displacements of the structure (the P- effect) are taken into account in this method.

This principle is illustrated in the following figure.

M(x) = HxM(L) = HL

First Order Theory

M(x) = Hx + P + P x /L

M(L)= HL + P Second Order Theory

The local P- effect will be regarded further in this course.

If the members of the structure are not in contact with subsoil and do not form ribs of shells, the finite

element mesh of the members must not be refined.

The method needs only two steps, which leads to a great efficiency. In the first step, the axial forcesare solved. In the second step, the determined axial forces are used for Timoshenkos exact solution.The original solution was generalised in SCIA ENGINEER to allow taking into account sheardeformations.

The applied technique is the so called total force method or substitution method. In each iterationstep, the total stiffness of the structure is adapted and the structure is re-calculated until convergence.This technique is illustrated in the following diagram.


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In this figure, the stiffness K is divided in the elastic stiffness KE and the geometrical stiffness KG. Thegeometrical stiffness reflects the effect of axial forces in beams and slabs. The symbol u depicts thedisplacements and F is the force matrix.

The criterion for convergence is defined as follows:

( ) ( )( )

005.02

,

2

,

2

,

2

1,

2

1,

2

1,

2

,

2

,

2

,

++

++++

iziyix

iziyixiziyix

uuu

uuuuuu

With: ux,i The displacement in direction x for iteration i.uy,i The displacement in direction y for iteration i.uz,i The displacement in direction z for iteration i.

Start

Stop

Calculate KE

KG = 0

K = KE - KG

Solve K.u = F

Convergence

in u?

Calculate KG

No

Yes


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The choice of the Timoshenko Method and the maximal amount of iterations can be specified throughCalculation, Mesh > Solver Setup.

Newton Raphson

The second method is the so called Newton-Raphson method (Th.III.O) which is based on theNewton-Raphson method for the solution of non-linear equations.

This method is a more general applicable method which is very solid for most types of problems. It canbe used for very large deformations and rotations; however, as specified the limitation of small strainsis still applicable.

Mathematically, the method is based on a step-by-step augmentation of the load. This incrementalmethod is illustrated on the following diagram:


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In this figure, the tangential stiffness KT is used. The symbol u depicts the displacements and F is theforce matrix.

The original Newton-Raphson method changes the tangential stiffness in each iteration. There are alsoadapted procedures which keep the stiffness constant in certain zones during for example oneincrement. SCIA ENGINEER uses the original method.

As a limitation, the rotation achieved in one increment should not exceed 5.

The accuracy of the method can be increased through refinement of the finite element mesh and byincreasing the number of increments. By default, when the Newton-Raphson method is used, thenumber of 1D elements is set to 4 and the Number of increments is set to 5.

In some cases, a high number of increments may even solve problems that tend to a singular solutionwhich is typical for the analysis of post-critical states. However, in most cases, such a state ischaracterized by extreme deformations, which is not interesting for design purposes.

The choice of the Newton-Raphson Method, the amount of increments and the maximal amount ofiterations can be specified through Calculation, Mesh > Solver Setup.

Start

Stop

Choose Fu0 = 0

F0 = 0

F = F0 + F

Determine KT at F0

Solve KT. u = F

Convergence

in u?

u0 = u

Determine F0

No

Yes

Any F left?

u = u0 + u

No


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As specified, the Newton-Raphson method can be applied in nearly all cases. It may, however fail inthe vicinity of inflexion points of the loading diagram. To avoid this, a specific method has beenimplemented in SCIA ENGINEER: the Modified Newton-Raphson method.

This method follows the same principles as the default method but will automatically refine the numberof increments when a critical point is reached. This method is used for the Non-Linear Stabilitycalculation and will be looked upon in Chapter 7.

In general, for a primary calculation the Timoshenko method is used since it provides a quicker solutionthan Newton-Raphson due to the fact Timoshenko does not use increments. When Timoshenko doesnot provide a solution, Newton-Raphson can be applied.


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15. Stability

Linear Stability

During a linear stability calculation, the following assumptions are used:

- Physical Linearity.

- The elements are taken as ideally straight and have no imperfections.

- The loads are guided to the mesh nodes, it is thus mandatory to refine the finite element mesh inorder to obtain precise results.

- The loading is static.

- The critical load coefficient is, per mode, the same for the entire structure.

- Between the mesh nodes, the axial forces and moments are taken as constant.

The equilibrium equation can be written as follows:

[ ] FuKK GE =

The symbol u depicts the displacements and F is the force matrix.

As specified in the theory of the Timoshenko method, the stiffness K is divided in the elastic stiffnessKE and the geometrical stiffness KG. The geometrical stiffness reflects the effect of axial forces inbeams and slabs.

The basic assumption is that the elements of the matrix KG are linear functions of the axial forces in the

members. This means that the matrix KG corresponding to a th

multiple of axial forces in the structure

is the th

multiple of the original matrix KG.

The aim of the buckling calculation is to find such a multiple for which the structure loses stability.Such a state happens when the following equation has a non-zero solution:

[ ] 0= uKK GE

In other words, such a value for should be found for which the determinant of the total stiffness matrixis equal to zero:

0= GE KK

Similar to the natural vibration analysis, the subspace iteration method is used to solve this eigenmode

problem. As for a dynamic analysis, the result is a series of critical load coefficients withcorresponding eigenmodes.

To perform a Stability calculation, the functionality Stability must be activated.


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In the results menu, the values can be found under the captionThe number of critical coefficients to be calculated per stability combination can be specified underSetup > Solver.

Note:

- The first eigenmode is usually the most important and corresponds to the lowest critical load coefficient. A

possible collapse of the structure usually happens for this first mode.

- The structure becomes unstable for the chosen combination when the loading reaches a value equal to the

current loading multiplied with the critical load factor.

- A critical load factor smaller than 1 signifies that the structure is unstable for the given loading.

- Since the calculation searches for eigen values which are close to zero, the calculatedvalues can be bothpositive or negative.

A negative critical load factor signifies a tensile load. The loading must thus be inversed for buckling to occur

(which can for example be the case with wind loads).

- The eigenmodes (buckling shapes) are dimensionless. Only the relative values of the deformations are of

importance, the absolute values have no meaning.

- For shell elements the axial force is not considered in one direction only. The shell element can be in

compression in one direction and simultaneously in tension in the perpendicular direction. Consequently, the

element tends to buckle in one direction but is being stiffened in the other direction. This is the reason for

significant post-critical bearing capacity of such structures.

- Initial Stress is the only local non-linearity taken into account in a Linear Stability Calculation.

- It is important to keep in mind that a Stability Calculation only examines the theoretical buckling behaviour

of the structure. It is thus still required to perform a Steel Code Check to take into account Lateral Torsional

Buckling, Section Checks, Combined Axial Force and Bending,


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Buckling Shape

As an alternative to Global and Local imperfections, paragraph 13. Non linear combinations allowsthe use of a buckling shape as a unique imperfection.

To input geometrical imperfections, the functionality Nonlinearity > Initial deformations andcurvature and Stability must be activated.

The calculation of the buckling shape through a stability calculation will be looked upon in Chapter 7.

Since the buckling shape is dimensionless, Eurocode gives the formula to calculate the amplitude initof the imperfection. In Ref.[29] examples are given to illustrate this method. In this reference, theamplitude is given as follows:

cr

cry

crinit

IE

Ne

=

''

max,

0

( )

( )

( )21

2

0

1

1

2,0

= M

Rk

Rk

N

Me for 2,0>

With:cr

Rk

NN

=

= The imperfection factor for the relevant buckling curve.

= The reduction factor for the relevant buckling curve, depending on the

relevant cross-section.NRk = The characteristic resistance to normal force of the critical cross-section, i.e.

Npl,Rk.Ncr = Elastic critical buckling load.


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MRk = Thor

cr = Sh

='' max,cr M

The value of init can then be

Example Stability Imper

In Chapter 6, the use of the buexample, the procedure is illu

The column has a cross-sectiproperties:

E = 210.000 N/mmL = 5000 mmIy = 83560000mm

4

60

characteristic moment resistance of the critical

el,Rk as relevant.

pe of the elastic critical buckling mode.

aximal second derivative of the elastic critical bu

ntered in the field Max deformation.

ection

ckling shape as imperfection according to EC3trated for a column.

n of type IPE 300, is fabricated from S235 and h

fy = 235 N/mm M1 = 1.00A = 5380 mmWpl,y = 628400 mm

3

Advanced Training

ross-section, i.e. Mel,Rk

kling mode.

as discussed. In this

as the following relevant


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Calculation of the buckling shape

First a Stability calculation is done using a load of 1kN. This way, the elastic critical buckling load Ncris obtained. In order to obtain precise results, the Number of 1D elements is set to 10. In addition, theShear Force Deformation is neglected so the result can be checked by a manual calculation.The stability calculation gives the following result:

This can be verified with Eulers formula using the member length as the buckling length:

2

42

2

2

)5000(

83560000

000.210

mm

mmmm

N

l

EINcr

==

=6927,51 kN

The following picture shows the mesh nodes of the column and the corresponding buckling shape:


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Using for example an Excel worksheet, the buckling shape can be approximated by a 4th

gradepolynomial.

A polynomial has the advantage that the second derivative can easily be calculated.

212638412622,7284,2240,7117,2117,2

+= ExExExExEcr 57211''

max, 1,4481,2702,540 += ExExEcr

Calculation of e0

=== 5380

235 mm

mm

NAfN yRk 1264300 N

=== 3628400

235 mmmm

NWfM plyRk 147674000 Nmm (class 2)

Buckling Shape

y = -2,117E-12x4 + 2,117E-08x3 - 7,240E-06x2 - 2,284E-01x - 7,622E-02

-400

-350

-300

-250

-200

-150

-100

-50

0

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Length [mm]

BucklingShape


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===N

NN

N

cr

Rk

68852801264300 0,43

= 0,21 for buckling curve a

( ) ( )[ ] ( ) ( )[ ]( ) ( )2222 2.015,02.015,01

+++++

= =0,945

These intermediate results can be verified through SCIA ENGINEER when performing a Steel CodeCheck on the column:

( ) ( )( )

( )N

Nmm

N

Me M

Rk

Rk

1264300

1476740002.043,021,0

1

12,0

2

1

2

0 =

=

= 5,605 mm

The required parameters have now been calculated so in the final step the amplitude of theimperfection can be determined.

Calculation ofinit

The mid section of the column is decisive x = 2500

cr at mid section = -368,24

''

max,cr at mid section = 1,443E-04

1mm

368,24

11,443E83560000210000

6885280605,5

4-4

''

max,

0

=

=

mmmmmmN

Nmm

IE

Ne cr

cry

crinit

= 5,615 mm


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This value can now be inputted as amplitude of the buckling shape for imperfection.

To illustrate this, the column is loaded by a compression load equal to its buckling resistance.

However, due to the imperfection, an additional moment will occur which will influence the sectioncheck. The buckling resistance can be calculated as follows:

==

==

2355380945,01

, mmNmm

fANN

M

y

RdbEd

1194,76 kN

A non-linear combination is created in which the buckling shape as imperfection is specified.

Using this combination, a non-linear 2nd

Order calculation is executed using Timoshenkos method.

The additional moment can be easily calculated as follows:

kN

kNmmkN

N

NNM

cr

EdinitEdinit

28,6885

76,11941

1615,576,1194

1

1,

=

=

= 8,12 kNm

When performing a Steel Code Check on the column for the non-linear combination, this can be

verified. The critical check is performed at 2,5m and has the following effects:

The additional moment thus corresponds to the moment calculated by Scia Engineer.


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As seen in the diagram, Path 3 is followed: the buckling shape serves as a unique global and localimperfection. This implies that only a section check and Lateral Torsional Buckling need to be checked.Since LTB is negligible with this small bending moment, only a section check is required:

This example has illustrated the use of a buckling shape as imperfection. Depending on the geometryof the structure, this imperfection can have a large influence on the results due to the additionalmoments which occur.

When using this method, it is very important to double check all applied steps: small changes to theloading or geometry require a re-calculation of the buckling shape and amplitude before a non-linearanalysis may be carried out.

As a final note: the buckling shape only gives information about a specific zone of the structure. Theimperfection is applied at that zone and results/checks are only significant for that zone. Othercombinations of loads will lead to another buckling shape thus to each load combination a specificbuckling shape must be assigned and a steel code check should only be used on those members onwhich the imperfection applies. Since the applied buckling shape corresponds to a global mode, failureof these members will lead to a collapse of the structure.


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16. Input Hinge Types and non-linear stiffness

For connections in a scaffolding structure, a check has to be performed on the normal forces, theshear forces and the moments. Various connection types are inserted in Scia Engineer. For thedifferent connections, go to Libraries -> Structure, Analysis -> Hinge type.

In this Hinge type library you can choose following types at the option Hinge type:

Right angle coupler Friction sleeve Swivel coupler


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Base jack Pa

In this connection, not only the rigidiconnections, the maximal allowablecouplers:

The rigidities are filled out and theprovided by the supplier.

68

rallel General Righ

ties are entered in a flexible or non-linear way, bforces can also be inserted, as displayed below

aximal forces have to be obtained from the info

Advanced Training

t angle continuouscoupler

t specifically for theseor the Right angled

f the connections,


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Connection between Guardrail/Ledger and Standard

In this example, a connection of Layher is being adopted (K2000):

For this connection, it is advisable to enter a general connection containing the properties of Zulassung Z-8.22-64. For this purpose, the following maximal forces are displayed:

My,R,d = 101 kNcm (=1,01 kNm)

Mz,R,d = 37 kNcm (=0,37 kNm)

Vy,R,d = 10 kN

Vz,R,d = 26,4 kN

ND,R,d = 31 kN

d[rad] =M/(9140 73,6M)

The equation above for the rotation gives the following rotation curve:


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To enter this correctly, it should be performed using a non-linear spring or a non-linear function in SciaEngineer:

If everything is being filled out in the values for this hinge, the following dialog appears:

It is assumed that the translations are rigid in all directions. Also the rotation around the x-axis is

considered as fixed. This is only a rough estimate. For the rotation around the y-axis, the non-linearfunction (determined earlier) is entered with a linear rigidity of 0,051 MNm/rad. The rotation around the localz-axis will be much smaller than around the y-axis namely 0,0051 MNm/rad.


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For the values of Vyk, the same value is taken as for Vyz. This value should be determined by the supplier aswell.

For the maximal moments the same conclusion can be drawn.

Subsequently these hinges are attributed to both extremities of the ledgers and guardrails.

Connection between wind bracings and Standard

For the connection between the wind bracings and the columns, the rotation around the local y-axis is free.There can be a discussion about the rigidity around the z-axis. In this example, it is set on rigid, although itcannot be completely considered as rigid. If you want to enter the exact value, you have to ask the supplier.The rotation around the x-axis and the translations in the x, y and z direction are rigid.

In the direction of the member (local x direction) there is a certain margin of these wind bracings. This can be

entered by adopting Spacings. For this purpose, all bracings are entered as a non-linearity on themembers, containing the type Spacing.

Subsequently a margin of 1 mm per member is entered:


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Supports

It is also better to enter the supports as non-linear supports. In the vertical direction these supports can onlytake pressure and no tension. This is entered in the z direction with rigid press only. The rotations are takenfree in all directions.

For the degrees of freedom according to the x and y displacements, the code prEN12812:2003, Annex B isapplied. In this, the friction coefficients between various materials are entered. If we suppose that thescaffolding is placed on wood, we can see in this code that the maximal and minimal frictions coefficientbetween wood and steel is 1.2 and 0.5. On average a value of 0.85 is adopted. This value is also entered inScia Engineer.

For C flex a large value is taken. This corresponds to a large rigidity of the support in the x- and the y-direction before the friction is exceeded.


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17. Scaffolding

When checking the allowable stressbracings, ). The maximal stresses

base jacks are checked with this val

For the anchorage forces, the reactiOn the other hand, the anchorage cshear.

For connections this check can be pCoupler check.

This check performs a unity

Example

Scaffolding.esa

- Choose for Scaffolding

- Take a look at the check


73

Coupler Check

es, it is recommended to view the results per procan now be compared to the allowable values o

ue.

on force can be tested to the allowable force of aan also be checked manually on the combined ef

erformed by Scia Engineer itself, with the option

check for the couplers for which a rigidity is give

Coupler Check in the steel menu

for beam B165: a unity check of 0,36.


Scaffolding

file type (standards,the supplier. Also the

perpendicular coupler.fect of tension and

Steel -> Scaffolding -

.


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The coupler reaches a value of 0.36 for the unity check for Fx. This maximal force Fx for thisconnection is 31 kN.


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75

The internal normal force that occurs in this connection is 11,24 kN:

If they are divided by each other, we get a unity check of 0,36 (= 11,24/31,00).


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18. Scaffolding

The results of the ULS and SLcombinations and with the cor

Example SLS Check

Scaffolding.esa

- Choose for Relative def

- The maximum check is f


77

on linear Check

S check are repeated in this chapter, but now forrect stiffness for the connections.

rmation in the steel menu.

und for beam B352: a unity check of 0,25.


Scaffolding

the non-linear


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Example ULS Check

Scaffolding.esa

- Choose for Scaffolding

- The maximum check is f


78

Check in the steel code check.

und for beam B350: a unity check of 0,66.


Advanced Training


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19. Aluminium

Aluminium grades

The characteristic values of the material properties are based on Table 3.2a for wrought aluminium

alloys of type sheet, strip and plate and on Table 3.2b for wrought aluminium alloys of type extrudedprofile, extruded tube, extruded rod/bar and drawn tube.


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In Scia Engi!gneer, the following materials are provided by default:

Initial bow imperfection e0

The values of e0/L may be chosen in the National Annex. Recommend values are given in the

following Table 5.1 Ref.[1]. The bow imperfection has to be applied when the normal force NEd in amember is higher than 25% of the members critical buckling load Ncr.


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Where L is the member length.Scia Engineer can calculate the bow imperfection according to the code automatically for all neededmembers or the user can input values for e0. This is done via Aluminium > Setup > National Annex.

Initial shape

For a cross-section with material Aluminium, the Initial Shape can be defined. For a General Cross-section, the Thinwalled representation has to be used to be able to define the Initial Shape.

The inputted types of parts are used further used for determining the classification and reductionfactors.The thin-walled cross-section parts can have for the following types:

F Fixed Part No reduction is needed

I Internal cross-section part

SO Symmetrical Outstand

UO Unsymmetrical Outstand

A part of the cross-section can also be considered as reinforcement:

none Not considered as reinforcement

RI Reinforced Internal (intermediate stiffener

RUO Reinforced Unsymmetrical Outstand (edge stiffener

In case a part is specified as reinforcement, a reinforcement ID can be inputted. Parts having the samereinforcement ID are considered as one reinforcement.

The following conditions apply for the use of reinforcement:

- RI: There must be a plate type I on both sides of the RI reinforcement.


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- RUO: The reinforcement is connected to only one plate with type I.

For standard cross-sections, the default type and reinforcement can be found in (Ref.[1]). For nonstandard section, the user has to evaluate the different parts in the cross-section.

The Initial Shape can be inputted using Cross-sections > Edit > Reduced Section. Afterwards, theuser can select Edit Reduced Section. In this box also welds (HAZ Heath Affected Zone) can beinputted.

The parameters of the welds (HAZ) are:

- Plate ID- Position- Weld Method: MIG or TIG- Weld Material: 5xxx and 6xxx or 7xxx- Weld Temperature- Number of heath paths

These parameters will be discussed further.


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Section check

Partial safety factors

The partial safety factors may be chosen in the National Annex. Recommend values are given in Table6.1 (Ref.[1]).

Resistance of cross-sections whatever the class is M1 = 1,10

Resistance of member to instability assessed by member checks M1 = 1,10

Resistance of cross-sections in tension to fracture M2 = 1,25

Using the menu 'Aluminium > Setup > Member check, the user can input values for M1 and M2.

Bending moments

According to section 6.2.5.1 Ref.[1], alternative values for 3,u and 3,w can be chosen. In SciaEngineer, the user can input these alternative values using Aluminium > Setup > Member check >

Alternative values.


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Shear

The design value of the shear force VEd at each cross-section shall satisfy (Ref.[1]):

1Rd

Ed

V

V

Where VRd is the design shear resistance of the cross-section.

Slender and non-slender sections

The formulas to be used in the shear check are dependent on the slenderness of the cross-sectionparts.

For each part i the slenderness is calculated as follows:

i

begend

iw

wi

t

xx

t

h

=

=

With: xend End position of plate i .xbeg Begin position of plate i.t Thickness of plate i.

For each part i the slenderness is then compared to the limit 39

With0

250f=

and f0 in N/mm

39i => Non-slender plate

39>i => Slender plate

I) All parts are classified as non-slender

39i

The Shear check shall be verified using art. 6.2.6. Ref.[1]

II) One or more parts are classified as slender

39>i

The Shear check shall be verified using art. 6.5.5. Ref.[1].

For each part i the shear resistance VRd,i is calculated.


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Non slender pa

Formula (6.88

N+(u,HAZ)

For Vy: Anet i

For Vz: Anet i

With: i

xendxbegt

u,HAZ

Slender part

Formula (6.

N+(u,HAZ) i

Formula (6.(for I or U-s=> VRd,i,buckl

=> For thisVRd,i,buckling

For each part VRd,i is then det

=> The VRd of the cross-sectio

=i

RdRd iVV

Note:For a solid bar, round tube an(6.31) is applied.

Example

Aluminium.esa- calculate project- aluminium check, detailed

Part Type

1 RUO 10

2 I 29

3 I 100

87

t

) Ref.[1] is used with properties calculated from t

,y,i = iiHAZuibegend txx 2

, cos)(

,z,i = iiHAZuibegend txx 2

, sin)(

The number (ID) of the plateEnd position of plate iBegin position of plate iThickness of plate i

Haz reduction factor of plate i

Angle of plate i to the Principal y-y axis

88) Ref.[1] is used with properties calculated fro

the same way as for a non-slender part. => VRd

89) is used with a the member length or the distections)

ling

slender part, the resulting VRd,i is taken as the mi

rmined.

n is then taken as the sum of the resistances VRd

hollow tube, all parts are taken as non-slender

output

39 Slender? Avy,i Avz,i VRD,y,yiel

3,07 no 2,9 37,1 0,31

3,07 no 53,9 4,1 5,8

53,9 4,1 5,8

3,07 yes 10,5 139,5 1,134,6 61,5 0,5

Scaffolding

he reduced shape for

the reduced shape for

,i,yield

nce between stiffeners

nimum of VRd,i,yieldand

,i of all parts.

y default and formula

ld,i VRD,z,yield,i

4

0,45

0,45

156,61


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88

10,5 139,5 1,13 15

4 I 31,5 43,07 no 58,5 4,5 6,3 0,48

58,5 4,5 6,3 0,48

5 RUO 9 43,07 no 2,6 33,4 0,28 3,6

- In addition: for the slender part 3- a/b = 6000/200 = 30 with a = 6m and b = 200mm and 1 = 0,280


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20. Document

Example

Scaffolding.esa

Look at the document in att

It is possible to save the docu

With this option it is possible t

89

chment as example

ent as a template and use this document for ot

create a document once and use it in all other

Scaffolding

er projects too.

rojects.


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21. Template

To create a template, the following procedure has to be followed:

1. Create a new project or open an existing one.

2. Define all the properties and even parts of a structure that should be included in the template. Ifan already existing project is used, make any changes that are necessary.

3. When you are satisfied with the result and you think that the current state of the project is whatshould become the template, save it As Template using menu function File > Save As.

4. Browse for the folder where the user templates are stored - this folder is specified in the Setup >Options dialogue.

5. Type the name of the template file.6. Complete the action.

With this possibility, it is possible to predefine the cross sections, the (non-linear) combinations, the nonlinear hinges, the document,

After saving a project as a template, this template can be opened just by making a New Project:


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22. User Blocks

The application of user blocks can be divided into three independent steps. The steps must be carried out inthe given order and all of them must be made.

1. Creating the user block

A user block can be created as a standard project. There are no explicit restrictions to it. Usually,the user will be working on his/her project and either at the end or some time during the designphase s/he decides to make a user block of the current state of the project.

Then the only thing that must be done is save the project to the disk. It may be useful, however notcompulsory, to use function Save As and give the project such a name that gives a hint about thestructure in the project.

2. Storing the user block to the library

In order to be usable as a user block, the project must be stored in the User block library folder(see Program settings > Directory settings). This may be achieved in two ways.

The user specifies the proper path in the Save As dialogue (see paragraph above) andsaves the project directly to the User block library folder.

The user saves the project to his/her common project folder and then copies the file to theUser block library folder. The file may be copied in any file-management tool (e.g.Windows Explorer, Total Commander, My Computer dialogue, etc.)

Tip: The user blocks may be stored not only in the given User block library folder, but they may bearranged in a tree of subfolders. The subfolders may then group user blocks that have something incommon. This arrangement may lead to easier and clearer application of user blocks, especially if along time passes from the time they were created and stored.

When using this tip, all the subfolders can be chosen:


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3. Inserting the user block into another project using the following procedure

1. Open service Structure:a. either by means of tree menu function Structure,b. or by means of menu function Tree > Structure,c. or by means of icon Structure on toolbar Project.

2. Select and activate function User blocks.

3. A User block wizard opens on the screen. Its left hand side window shows the organisation ofthe User block library folder, i.e. it shows any possible subfolders. The right hand side windowthen displays all available user blocks saved in the appropriate folder or subfolder.

4. Select the required folder.5. Select the required User block.6. Click [OK] in order to insert the block to the current project.7. Select the required options for the import (see below).8. Position the user block to the desired place and click the left mouse button to put the block

there.9. If required, repeat the previous step as many times as required or necessary.

Note:If the User block is a parameterised project, the program asks the user to provide all necessaryparameters in order to complete the definition of the user block.


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23. References

[1] DIN 4420

Arbeits- und Schutzgerste

1990

[2] HD 1000

Gevelsteigers bestaande uit prefab onderdelen

1992

[3] EN 12811-1

Scaffolds: Performance requirements and general design

2003

[4] EN 12811-3: Scaffolds: Load Testing, 2003

[5] EN 12810-1

Faade Scaffolds with Prefabricated Components:

Products specifications

2003

[6] EN 12810-2

Faade Scaffolds with Prefabricated Components:

Particular methods of structural design

2003

[7] NBN ENV 1991-2-4

Belasting op draagsystemen: Windbelasting

1995

[8] Eurocode 3:

Design of Steel Structures

2002

EN 1993-1-1

[9] Handbuch des Gerstbaus

Friedrich Nather, Hoachim Lindner, Robert Hertle

2005

[10] Eurocode 9

Design of aluminium structures

Part 1 - 1 : General structural rules

EN 1999-1-1:2007

[11] Zulassung Z-8.22-64


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Annex A: Wind pressure versus Wind force

In paragraph 6. Load cases the wind has been calculated with EN1991-1-4 and calculated as a windpressure.

Following EN 12810-1, it is also possible to calculate the wind as a wind force using the followingformula (EN 12810-1, (1)):

With

Ai is the reference area specified in EN 12810-1, Table 4 (see also below)cf is the aerodynamic force coefficient taken from EN 12810-1, Table 5 (see also below)cs is the site coefficient taken from EN 12810-1, Table 6 (see also below)

qi is the design velocity pressure in accordance with EN 12810-1, Figure 3:

In this example, the height of the structure is 11m, so qion 11m is + 940 N/m and on 0m + 800 N/m.When inserting one value a average of those two can be taken: 870 N/m.

The reference Area Ai:

Cladding condition of thesystem configuration

Reference area Ai

Unclad Area of each component projected in the wind direction

Clad Surface area of the cladding (see A.3 of En 12811-1:2003)

This example was an unclad example. For a component with a diameter of 48,3mm, the area is0,0483m x Lcomponent.

The aerodynamic force coefficient cf:

Cladding condition of thesystem configuration

Force coefficient

Normal to the faade Parallel to the faade

Unclad 1,3 1,3Clad 1,3 0,1


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In our example the structure was unclad. We will take a look at the force parallel to the faade, so cf =1,3.

The site coefficient cs:

Cladding condition of the

system configuration

Site coefficient

Normal to the faade Parallel to the faade

Unclad 0,75 1,0

Clad 1,0 1,0

NOTE: The values for the site coefficients correspond to a faade with a solidity ratio B= 0,4; see also EN12811-1.

For an unclad structure, the wind parallel to the faade, results in cs = 1,0.

For each beam the force can be calculated as a line force as shown below:

1,0 0,0483 1,3 870

54,6 0,055

This load is applied on the beams shown below (the toeboards are not taken into account, because theforce coefficient is only available for tube profiles):

So the total force inserted is:

0,055 kN/m x

[(2+2x0,5)x 2,57m

+ (2x0,5)x 2m +3,257]

= 0,055 kN/m (12,97m)

= 0,71 kN

Documents

[Eng]Advanced Training Scaffolding 2009.0.1