Metal Structures

Lecture X

Bracing systems

Contents

Introduction → #t / 3

Types of bracings → #t / 12

Calculations → #t / 31

Example 1 → #t / 60

Example 2 → #t / 75

Example 3 → #t / 86

Example 4 → #t / 90

Conclusions → #t / 93

Examination issues → #t / 94

Plate is stiff member of big

resistance (in plane); it can

bears even big loads. But

without additional support

in perpendicular direction is

completely unstable (out of

plane).

Introduction

Photo: Author

UnstableStable

Generally steel structures consist on flat frame (of big stiffness and big resistance in

plane), but unstable in perpendicular direction. Additional sub-structures is needed: various

types of bracings.

Photo: setrometalgroup.com

Photo: traskostal.pl.

Very important is specific shape of system bracings-frames. Rectangular shape is not

geometrically unchanged and not prevent instability. Only triangles are geometrically

unchanged.

Photo: Author

Wall girts, purlins, roof bracings and side wall

bracings make specific system for wind action.

Front wall housing columns must be connected

with purlins and roof bracings at one point. The

same, girts on front and side walls.

Photo: steelconstruction.info

Photo: greenterrahomes.com

→ #8 / 43

Wind acts on housing (p, [kN / m2]). Housing is support on

wall girts; loads from housing act on girts as continous

loads (q, kN / m). Girts are under bending (mono- or bi-

axial). Loads from girts act on main frames as forces,

applied in points of connection girts - main columns.

Photo: Author

p

q = p a

al

F = q l

Wind acts on front wall: the same way of recalculation p → q → F. Forces are applied to

main frames (perpendiculary to theirs plane) and to housing columns. In case of doors (in

front / side wall), wind action from door is applied to girts and housing columns around

doors.

Photo: Author

→ #8 / 44

Loads from housing columns finally act on bases of housing

column and main frames (main columns, roof girders),

perpendiculary to theirs planes. It potentially makes bi-axial

bending in main frames.

Photo: Author

Main frames are supported in

perpendicular direction by bracings,

purlins and side wall girts. It prevent

from bi-axial bending.

Photo: Author

Roof bracings and purlins make horizontal truss. Roof girders are

chords of truss. The effect is, that loads perpendicular to main

frames make additional axial forces in roof girders. Additionally,

there are axial forces in purlins.

→ #8 / 45

Roof: loads are transported through longitudinal broof bracings.

Wall: loads are transported by side wall girts.

Photo: Author

Photo: Author

Finally, loads act on vertical bracings on side walls, vertical trusses. Main columns are

chords of truss. Depending on location of girts on side walls, there is possible bi-axial

bending in these four of columns (loads out of nodes of truss).

→ #8 / 46

We should avoid too many vertical bracings to

ensure open spaces inside buildings.

Photo: muratorplus.pl

Photo: vmc21.com

Photo: dreamstime.com

Photo: lekkaobudowa.pl

Bracings in plane and

out of plane can be

applied only in outer

walls of structure.

In situation when bracing inside building are needed, should be applied rather portal

bracing than X bracing. Portal bracing didn't block internal communication.

Photo: i.wnp.pl

Photo: dreamstime.com

Types of bracings

Photo: Author

Roof bracings Floor diaphragms

Crane bracing Wall in-plane bracing

Wall out-plane bracing

Reduction buckling

length

(#t / 14 – 19, 54)

Taking forces

perpendicular to main

plane of structure

(#t / 7 - 9)

Increase sway stiffness

(second order analysis)

(#t / 20 - 21)

Roof bracing C C

Wall in-plane bracing C C

Wall out-plane bracing C C

Floor diaphragms C C

Crane bracing C

Types of bracings and reasons of their mountable

Sometimes massive wall girts are named „wind bracings”. But there is

only common name, these members are not bracings. Their rule is only

support housing under wind action.

Photo: Author

Reduction buckling length – one of the most important rule of bracings is change

critical length of members during various type of global buckling (flexural, torsional,

flexural-torsional and lateral). Examples of this rule were presented on Lec #5, Lec #10

and Des #1.

Photo: enterfea.com

Flexural buckling of chords:

Top chords in compression; buckling out of plane: critical length = distance between

horizontal bracings

Photo: Author

→ Des #1 / 47

→ #5 / 47

Photo: Author

Example 1

C 300pS235 → fy = 235 MPaL = 3,00 mE = 210 GPaG = 81 GPaA = 52,5 cm2

Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 33,9 cm4

a = 3,12 cme = 2,89 cmiy = 12,1 cmiz = 3,01 cmys = a + e = 6,01 cmIn this case: zs = ys = 6,01 cm

NEd = 650 kN

I class of cross-section

A fy = 1 233,750 kN

χ A fy = 548,241 kN

NEd = 650 kN

NEd / A fy = 0,527

OK.

NEd / χ A fy = 1,186

Wrong, buckling, destruction!

→ #5 / 52

Proposition: other distance between

supports on y-direction → change of

critical length for z-buckling

Photo: Author

→ #5 / 53L0z = 2,00 m

Ncr, y = 4 398,554 kN

Ncr, z = 2 450,725 kN

Ncr, T = 1 633,427 kN

Ncr, zT = 1 333,190 kN

λy = √(A fy / Ncr, y) = 0,530

λz = √(A fy / Ncr, z) = 0,710

λT = √(A fy / Ncr, T) = 0,869

λzT = √(A fy / Ncr, zT) = 0,962

χ = min(χy ; χz ; χT ; χzT) = 0,562

A fy = 1 233,750 kN

χ A fy = 693,930 kN

NEd = 650 kN

NEd / A fy = 0,567

OK.

NEd / χ A fy = 0,944

OK.

→ #5 / 54

For calculations, new value of horizontal force is applied: VEd* = VEd α*

First- and second-order analysis

There is additional bending moment from axial force for very flexible structures

→ #3 / 86

Photo: Author

Wall in-plane bracing

↓ ↓

df / db-f ≤ 5

→ Non-braced frame

δf / δb-f > 5

↓

↓

Braced frame - no need second-order analysis Second-order analysis

→ Lecture #13

When we must make second-order analysis (PN B 03200)Photo: Author

Type of bracing systems:

Bars

Plates:corrugated sheetsconcrete plates

Photo: nexus.globalquakemodel.orgPhoto: tatasteelconstruction.com

Photo: nexus.globalquakemodel.org

Photo: lekkaobudowa.pl

Requirements for bar bracings:

Additional forces, indicated by bracing, in chords / girders and purlins must be takeninto account;

Horizontal distance between both ends of bracing ≤ 6,00 m → dead weight of bracingcan be neglected (otherwise – bending moments from dead weight);

Additionally, for non-rigid bar bracings:

Rigging screws should be mounted;

Only tensed part of bracing could be analised;

Bracings - recommended cross-sections

(→ Lec # 15)

Photo: stalhart.pl

Photo: calgor.com.pl

Photo: rafstal-inox.pl

Photo: rafstal-inox.pl

Photo: EN 1993-1-1 fig. 6.13

→ #8 / 47

Rigid bracings

Recommended cross-section: RHS, CHS.

There is taken into account entire structure, i.e. tensed and compressed bracing

members. There is long critical length for compression, especially in direction

perpendicular to bracings plane (vertical direction). There are massive cross-sections of

bracings, because of their flexural buckling.

Photo: Author

Non-rigid bracings

Recommended cross-section: C-sections, L-section, round bar.

Bar bracings under compression.

We accepted flexural buckling for part of bracings. Bracings will be mounted in both

directions (X-shape), but for calculations there are taken into account only tensed

bracing members. We have different static scheme than for rigid bracings. Compressed

bracings are under flexural buckling and not cooperate with rest structure.

Photo: Author

Photo: Author

Thermal isolation Factory-made

connecting latch

Anti-buckling

protection for purlins

according to EN

J J L

L J L

L L J

(per 5 - 10 years from

erection)

Photo: steelprofil.pl

Photo: amarodachy.pl

Photo: pruszynski.com.pl

→ #8 / 21

During its life, structure works under various loads and actions. Non-rigid bracings are

subject to alternating buckling. The result is increasing deformations of bracings.

For corrugated sheets important is bearing of bolts and corrosion around holes of bolts.

Photo: Author

Because of buckling, we should mount

rigging screws (or other type of system to

regulate) for reduction of deformations.

Photo: Author

Photo: encrypted-tbn3.gstatic.com Photo: encrypted-tbn3.gstatic.com

Photo: previews.123rf.com

Photo: homeguides.sfgate.com

The result of bearing and corrosion is,

that after few years diameter of hols for

bolts is bigger than diameter of cap of

bolts. Because of this there is no

cooperation between sheet and rest part

of structure. Sheet can't prevent structure

from buckling and should be replaced by

new.

Calculations

There are many various methods of calculations for braing systems. It depends on:

• type of protection (concrete plate, corrugated sheet, bar system);

• type of instability (flexural buckling, lateral buckling);

• position in structure (various types of roof bracings, wall bracings, crane bracings, floor

diaphragm).

Generally, three elements must be calculated:

• equivalent forces (which are result of anti-buckling protection) act on bracings;

• resistance of bracing under these forces;

• behaviour of protected member: no instability (sufficient efficiency of bracings), few types

of instability only (partial efficiency of bracings), each types of instability (insufficient

efficiency of bracings or no bracings).

Sometimes there is no need to calculate equivalent forces; resistance or efficiency of bracings

is calculated based on gemometry of structure only.

Concrete plate:

• there is assumption of total efficiency of protection against each form of instability for

beams;

• no calculations are needed on this situation;

• there is calculation of equivalent forces for concrete plates as floor diaphragms;

• calculation of resistance concrete plate under these forces are need (→ Concrete

Structures).

Floor diaphragms: forces from sway

imperfection of columns act on plate.

Photo: EN 1993-1-1 fig 5.7

Photo: nexus.globalquakemodel.org

Corrugated sheet:

• two separated ways of calculation for protection of beam (purlin, girt) for flexural

buckling and for lateral buckling;

• no calculation of equivalent forces;

• calculation efficency of corrugated sheet as anti-buckling protection (enough or not

enough) based on geometry of beam and corrugated sheet;

• buckling of beam is totally omitted (enough efficency), must be taken into account

partially (partial efficiency) or must be taken into account each type of instability (not

enough efficency);

Photo: i.warosu.org

Corrugated sheets - prevetnion flexural buckling of purlins

Scs ≥ 70 ( E Jw π2 / l2 + G Jt + 0,25 E Jz h π2 / l2) / h2

[N] Scs = 1000 √(t3) [50 + 10 3√(broot)] s / hw [mm]

EN 1993-1-3 (10-1a, 10.1b)

Photo: Author

Corrugated sheets - prevetnion lateral buckling of purlins

(this algorithm is rather dedicated to cold-formed purlins)

Ccs ≥ Mpl2 KD KU / E Jz

KU = 0,35 (elastic analysis)

KU = 1,00 (plastic analysis)

KD → #t / 36

Ccs ≈ k E Jeff / s

Jeff = Jx, roofing / 1 [m]

EN 1993-1-1 BB.2.2;

EN 1993-1-3 (10.16)

Photo: EN 1993-1-3 fig. 10.7

EN 1991-1-1 tab BB.1

Case Moment distribution KD

Without

translational

resistant

With

translational

resistant

1 4,0 0,0

2a

3,5

0,12

2b 0,23

3 2,8 0,0

4 1,6 1,0

5 1,0 0,7

Bar system:

• various technical solutions for protection of beam for

flexural buckling and for lateral buckling;

• various ways of calculation for various position in

structure (various types of roof bracings, wall bracings).

• generally there is need to calculate of equivalent

forces;

• in specific situation equivalent forces can be not

calculated;

• there are additional loads act on structure (not only

equivalent forces), as a effect of cooperation of bar

bracing systems and rest part of structure;Photo: steelconstruction.info

Technical solutions

Photo: Author

Photo: Author

Bracings against flexural buckling

should be applied to centre of gravity of

cross-section, perpendicullary to weak

axis of inertia.

Presentation #t / 39 - 52

Bracings against torsional, flexural-

torsional and lateral buckling should

prevent for torsional deformation of

cross-section

Presentation #t / 53 - 56

L

L

Roof horizontal upper transversal

bracings;

For truss and I-beam girders;

Every eighth band or in the distance of

80,0 m each other;

At the ends of structure;

Next to dilatations;

Loads perpendicular to plane of

structure

Photo: Author

Position in structure

Roof horizontal upper longitudinal

bracings;

For truss and I-beam girders;

Next to eaves and valley

Loads perpendicular to plane of structure

Can be omitted in halls without gantries

Photo: Author

Roof vertical longitudinal bracings;

For truss only;

Next to eaves, ridge and valley, under

skylights or in the distance on 15,0 m

each other;

Loads perpendicular to plane

of structure in montage stage, protection

of bottom chord from buckling

Photo: Author

Roof horizontal lower transversal bracings;

For truss;

Every eighth band or in the distance of 80,0

m each other;

At the ends of structure;

Next to dilatations;

For structure with gantries or for big value

of wind suction only;

Photo: Author

Roof horizontal upper longitudinal

bracings;

For truss;

Next to eaves and valley;

For structure with gantries or for big

value of wind suction only;

Photo: Author

VIIIth example of calculations – roof bracing

Way of calculation - 2D vs. 3D - is very important for algorithm.

Photo: mesilo.pl

3D model in FEM calculations: we have full information about cooperation between trusses,

purlins and bracing bars right away. Calculation is made in two steps:

• Initial calculation: dead weight, climatic actions, live loads etc.

• Equivalent forces: for bracing bars additionally important are equivalent forces from

imperfections of trusses and from prevention of instability of trusses. Values of these

forces are calculated based on forces in trusses after initial calculations. Cumulative

effect: loads from initial calculations and equivalent forces is final result of static

calculations.

→ Des 1 examp / 51

Rys: Autor

Purlin: bi-axial bending

Flat frame

Roof bracings

Loads as in initial calculations

Equivalent forces

Cooperation frame-bracings:

additional forces in purlins,

additional forces in frame.

Purlin: bi-axial bending +

compressive axial force

(recalculation)

Frame: additional axial forces

from cooperation with

bracings (recalculation)

2D model (method of equivalent flat frames) is much more complicated:

→ Des 1 examp / 52

Roof horizontal transversal bracings

View from the top:

Fi - forces perpendicular to truss plane (wind, etc.)

Bracings are calculated as additional horizontal truss

Photo: Author

There is important, how many bands of bracing exist in structure and how many griders are on one band.

g - total number of girders;

tb - total number of bands;

m = g / tb

αm = √[ 0,5 (1 + 1 / m)]

EN 1993-1-1 5.3.2

Photo: Author

Equivalent compressive force in roof girder:

NEd* = max (NEd, comp ; MEd / h ; NEd, comp / 2 + MEd / h)

NEd, comp - important for truss roof girder;

MEd / h or NEd, comp / 2 + MEd / h - important for I-beam roof girder;

Calculation of equivalet loads from imperfection qimperf for roof bracings is complicated.

According to EN 1993-1-1 (5.13), total eqiuvalent load (from wind and imperfection) is

calculated as:

qimperf = qd = Σ [8 NEd, c (e0 + δq) / L2]

δq – deformation comes from load qd

Total load qd depends on δq ,but δq is the effect of qd . Such problem can be solved by iteration

procedure:

δq0 = δ(qw-t) = δw-t

qd(0) = qd

(0)(e0 + δw-t)

δq(1) = δq

(1)(qd(0))

qd(1) = qd

(1)(e0 + δq(1) + δw-t)

δq(2) = δq

(2)(qd(1))

qd(2) = qd

(2)(e0 + δq(2) + δw-t)

…

→ Des 1 examp / 61

Photo: Author

As a result of iterations and calculations, we have Fi = max (Fimperf-wind ; Fbuck-wind) → axial

forces in brace bars → calculations of cross-sections

Additionally, we have axial forces in purlins and additional axial forces in chords. We should

analyze influences of these forces on purlins (bi-axial bending → bi axial bending + axial

force) and truss (change of axial force).

Fi = max (Fimperf-wind ; Fbuck-wind)

Fimperf-wind = a qd

qd = Σ [8 NEd* (e0 + δq) / L

2]

e0 = αm L / 500

Fbuck-wind = Fbuck + Fwind

Fbuck = αm NEd* / 100

Horizontal longitudinal bracings

Generally, the same cross-section as for horizontal transversal bracings.

Photo: Author

Vertical longitudinal bracings

Wall bracings

Under transverse roof bracings in central part of hall between expansion joints;

Transmission of loads from wall (wind on front wall, prevention of girder buckling, imperfections of columns) to bases;

Photo: Author

Loads: perpendicular to

plane of structure and

sway imperfection of

columns.

Horizontal crane bracings (surge girder)

Photo: konar.eu

Photo: Author

→ Metal Structures II, IInd step of study;

Bar bracing against lateral buckling of roof girders and of primary beams

Purlin

Roof girderRoof girder

Photo: EN 1993-1-1 fig 6.5

Photo: builderbill-diy-help.com

In case, when top flange of roof girder / primary

beam is compressed, the protection against lateral

buckling is a purlin-bracing system (system

connected to upper flange).

In case, when bottom flange is compressed, it is

necessary to provide additional anti-buckling

supports connecting bottom flange to purlin-bracing

system.

Photo: steadmans.co.uk

L1

Top part compressed

Bottom part compressed

Photo: Author

L2

L3

L4

L5

L6

L1 length of girder, for flexural buckling in plane;

L2 distance between bracings, for flexural buckling out

of plane, lateral buckling for top part compressed;

L3 length of bottom part compressed;

L4 distance from eaves joint

to first purlin in top part

compressed with additional

anti-buckling support;

L5 distance from eaves joint to first

purlin with additional anti-buckling

support;

L6 distance between the

closest purlins with

additional anti-buckling

supports;

max (L5 ; L6) for lateral buckling for bottom part

compressed; L4 if additional anti-buckling supports

are not applied;

L1 – for flexural buckling in-plane

of frame;

L2 – for flexural buckling out-of-

plane of frame;

L2 – for flateral buckling for top part

compressed;

For calculation value of compressive axial force in this type of bracings can be used

accuracy method or simplified method.

As accuracy method can be adopted method, presented in EN 1993-1-1 6.3.5.2

Force in bracing = additional force acts on purlins and girder:

FEd, bracing = max ( 1,5 αm NEd* / 100 ; Fpurlin)

NEd* = max (NEd ; MEd. / h ; NEd / 2 + MEd. / h)

Fpurlin - force, which acts from purlin to bracing because of

change of static scheme for purlin;

NEd, MEd - axial force and bending moment for girder;

FEd, bracing is inclined to axis of purlins, because of this it produces additional axial force

in purlin (bi-axial bending and axial force in purlin).

Photo: Author

Simplified method, elastic analysis

EN 1993-1-1 6.3.2.4

Members with discrete lateral bracings to the compression flange are not susceptible to latreal

torsional buckling, if distance between bracing LC satisfies formula as follow:

cw

cw / 3

LC kc / ( if, z λ1) ≤ λc0 Mc, Rd / My, Ed

My, Ed - the maximum value of bending moment

within the restraint spacing

Mc, Rd = Wy, c, f fy / γΜ1

kc according to #5 / 72

λ1 = 93,9 ε

λc0 = 0,5

if, z = √ [ Jeff, f, z / (Aeff, f + Aeff, w) ]

Photo: Author

There are two possibilities of modelling of bracing. Bacing members can be connected

each other in half of span, or can pass off in two various planes.

Photo: Author

Photo: spantec.com.au

Photo: halfen.com

Photo: steelconstruction.info

Sometimes bracings are applied per every second field between purlins. In calculation

model only purlins cooperated with bracings are taken into consideration - rest of purlins

are in no contact with bracings. Purlins and bracings are in two various planes.

Purlin - over chords or flanges of roof girders.

Bracings - in axis of chords or flanges of roof girders.

Photo: Author

Photo: Author

Not recommended type of longitudinal bracings. Bracing members are connected to purlin.

It completely change static scheme of purlin (two-span beam, not one-span). Additionally,

on bracing members act loads from purlins.

Example 1

Corrugated sheet, prevention for buckling of purlin.

This example is analised based on Lec #5 example #2

Photo: Author

IPE 300

S235 → fy = 235 MPa

L = 6,00 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 603,8 cm4

Wy = 557,1 cm3

Wpl, y = 628,4 cm3

Jw = 125 900 cm6

JT = 20,12 cm4

iy = 12,46 cm

iz = 3,35 cm

ys = 0,0 cm

MEd = 80 kNm-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L

L

Example 1a

Corrugated sheet, prevention of flexural buckling of purlin

Purlins: IPE 300

h = 300 mm

b = 150 mm

tf = 10,7 mm

tw = 7,1 mm

Jz, el = 604 cm4

Jw = 125 900 cm6

Jt = 20,7 cm4

S 235

One span, l = 6,0 m

Distance between purlins s = 2,0 m = 2 000 mm

Width of roof broof = 14,0 m = 14 000 mm

Corrugated sheet T 18

t = 0,88 mm

h = 10 mm

Photo: W. Bogucki, M. Żyburtowicz, „Tablice do projektowania

konstrukcji metalowych”, Arkady 1996

Girders (truss or I-beam)

PurlinsCorrugated sheet

Photo: Author

Requirement:

Scs ≥ 70 ( E Jw π2 / l2 + G Jt + 0,25 E Jz h π2 / l2) / h2

[N] Scs = 1000 √(t3) [50 + 10 3√(broot)] s / hw [mm]

70 ( E Jw π2 / l2 + G Jt + 0,25 E Jz hI π2 / l2) / hI2 = 3 451 kN

[N] Scs = 1000 √(t3) [50 + 10 3√(broot)] s / hw [mm] =

= 1000 √(0,883) [50 + 10 3√(14 000)] 2 000 / 10 =

= 1000 ∙ 0,826 (50 + 10 ∙ 24,101) 200 =

= 48 074 852 [N] = 48 075 kN

48 075 kN > 3 451 kN

OK., purlin are protected from flexural buckling

Previus result is true only if the sheeting is connected to beam at every rib.

If the sheeting is connected to beam at every second rib only, we must take into

consideration 0,20 Scs Photo: Author

Example 1b

Corrugated sheet, prevention of flexural buckling of purlin

70 ( E Jw π2 / l2 + G Jt + 0,25 E Jz hI π2 / l2) / hI2 = 3 451 kN

[N] 0,20 Scs = 0,20 ∙ 1000 √(t3) [50 + 10 3√(broot)] s / hw [mm] =

= 0,20 ∙ 1000 √(0,883) [50 + 10 3√(14 000)] 2 000 / 10 =

= 0,20 ∙ 1000 ∙ 0,826 (50 + 10 ∙ 24,101) 200 =

= 9 614 970 [N] = 9 614,970 kN

9 614,970 kN > 3 451 kN

OK., purlin are still protected from flexural buckling even if sheeting is connected

to beam at every second rib only.

Example 1c

Corrugated sheet, prevention of lateral buckling of purlin

Purlins: IPE 300

Wy, pl = 628,4 cm3

Jz, el = 604 cm4

S 235

One span, l = 6,0 m

Distance between purlins s = 2,0 m = 2 000 mm

Corrugated sheet T 18

t = 0,88 mm

h = 100 mm

Photo: W. Bogucki, M. Żyburtowicz, „Tablice do projektowania

konstrukcji metalowych”, Arkady 1996

Jx,roofing = 3,7 cm4

Jeff = Jx,roofing / 1 m = 0,037 cm3

Roofing:

Ccs ≈ k E Jeff / s

k = 2 (minumum value)

Ccs ≈ 0,078 kN

Purlin:

Mpl = fy Wy, pl = 147,674 kNm

KU = 0,35 (elastic analysis)

KD = 3,5 (as for single-span beam)

Mpl2 KD KU / E Jz = 21 575 kN

Ccs < Mpl2 KD KU / E Jz

Wrong, purlins are not protected.

Photo: Author

Photo: Author

Photo: Author

„Not protected” means buckling factor χ < 1,0. Is possible, than despite χ < 1,0 corrugated

sheet increase stiffness sufficency to prevent instbility.

Of course, according to results from Lec #5 example #2, beam is prevented from lateral

buckling by bracing bars (3x4,0m) and its own stiffness. This beam no need additional

protection by corrugated sheet:

Wpl, y fy = 147,674 kNm

χLT, mod = 0,712

χLT, mod Wpl, y fy = 105,155 kNm

MEd = 80 kN

MEd / χLT, mod Wpl, y fy = 0,761 OK.

But in initial situation (braing bars 2x6,0m), I-beam is not protcected for lateral buckling and

corrugated sheet prevent only for flexural buckling and not for lateral buckling.

Corrugated sheet is protection for: Conclusion for protected beam

flexural buckling lateral buckling

Yes

(χy = 1,0)

Yes

(χLT = 1,0)

Beam is totally protected.

No

(χy < 1,0)

Yes

(χLT = 1,0)

Such result is very very unlikely, there is a

high probability of mistake in calculations.

Yes

(χy = 1,0)

No

(χLT < 1,0)

Beam is partially protected; there is need to

calculate lateral buckling under conditions

of forced rotation axis

No

(χy < 1,0)

No

(χLT < 1,0)

Beam is not protected, interaction between

flexural buckling and lateral buckling must

be calculated (→ #13)

h/ 2

Photo: Author

Existence of corrugated sheet changes behavior of beam. Lateral buckling

should be calculated as for forced rotation axis, formulas d, #5 / 76:

Mcr = (is2 Ncr, T + cy

2 Ncr, z) / [C1 (cy - by) + C2 (cy - as)]

Ncr, z = 675,654 kN

is = 12,90 cm

Ncr, T = 1 813,849 kN

Geometry (#5 / 74):

ys - position of shear centre, for I-beam = 0

a0 - distance between shear centre and point of applying load,

for this situation = h/ 2 = 150 mm

rx (#5 / 74, I-beam, #5 / 41) = 0

Bending parameter: by = ys - rx / 2 = 0 – 0 / 2 = 0

cy - distance between centre of gravity and position of bracings,

for this situation = h/ 2 = 150 mm

C1 C2 (#5 / 77)

Information about C1 and C2 in Old Polish Standard is not completely clear. According to

literature, table could be presented in dependence of supports:

Photo: Author

Photo: Konstrukcje stalowe, K. Rykaluk,

Dolnośląskie Wydawnictwo Edukacyjne

Wrocław 2001

Photo: Konstrukcje stalowe, K. Rykaluk,

Dolnośląskie Wydawnictwo Edukacyjne

Wrocław 2001

Supports A

Supports BSupports C

Second important parameter is shape of bending moment:

M Supports Old Polish Standard Literature

C1 C2 C1 C2

A 2,00 0,00 2,00 0,00

B 2,00 0,00

A 2,00 0,00 1,13 0,00

B 2,00 0,00

A 0,93 0,81

1,74 0,81B 1,43 0,61

C 0,15 0,91

A 0,60 0,81

1,41 0,81B 1,00 0,81

C 0,00 1,62

Information is presented for initial supports of beam:

C1 = 1,43

C2 = 0,61

(approximation: case B for continous load)

Results for such situation are as follows:

Mcr = 146,230 kNm

Wy, pl fy = 147,674 kNm

λLT = 1,005

ΦLT = 1,429 (general formula, #5 / 84)

χLT = 0,603

χLT, mod = 0,613 (#5 / 85)

χLT, mod Wpl, y fy = 90,503 kNm

MEd / χLT, mod Wpl, y fy = 0,884 OK

Photo: Author

L

L

1. Full protection, χLT, mod =1,0

(concrete plate, small distance

between bracing bars or massive

corrugated sheet); 147,674 kNm

0,0 kNm 50,0 kNm 100,0 kNm150,0 kNm

MEd = 80,0 kNm

2. No protection

(no bracing bars, no

corrugated sheet, no

concrete plate);

33,582 kNm

3 4

5

6

Conclucions: influence of various types of bracings

on loadbearing. Even partial protection from

corrugated sheet could be enough in case of lateral

buckling.

3. Partial protection, #5

example 2a, bracing bars

2x6,00 m; 72,473 kNm

4. Partial protection, #5

example 2b, bracing bars

3x4,00 m; 105,155 kNm

5. Partial protection,

Corrugated sheet only,

partial protection, forced

rotation axis; 73,117 kNm

6. Partial protection, #t example 1c,

corrugated sheet, partial protection,

forced rotation axis, cooperation with

bracing bars 2x6,00m; 90,503 kNm

2

1

Example 2

Horizontal upper transversal bracings for truss roof

Photo: Author

Purlin: one-span simple-supported

IPE 210

MEd, y = 26,865 kNm

MEd, z = 2,687 kNm

Truss: top / bottom chord: O 159 / 8,8

Web members: O 88,9 / 11

J (based on approximated formula →

#9 / 11) = 0,0129 m4

NEd, max, top chord = 603,000 kN

Wind pressure qw = 0,3 kPa

Wind action on front wall (wind

pressure at first front wall + wind

suction at the opposite front wall):

qw-fw = 0,8 kPa

Photo: Author

Total length of hall: 60,00 m

g - total number of girders = 10

tb - total number of bands = 2

m = 10 / 2 = 5

αm = √[ 0,5 (1 + 1 / m)] = 0,775

e0 = αm d / 500 = 31 mm

NEd* = 603,0 kN

Two cases will be analysed: rigid bracings and non-rigid

bracings:

• there are X-shape bracings for both situation;

• for rigid bracings, compressed and tensed part are taken into

calculations;

• for non-rigid bracings, only tensed part are taken into

calculations;

There must be applied front wall structure: wall girts and front wall columns; there must be

place for gates in this structure.

Girts are one-span simple-supported beams, supported by front wall columns.

Photo: Author

Front wall columns are one-span

simple-supported beams, supported

by every other purlins (for example)

and their own foundations.

Area of front wall, attributable to one

column, is equal 2 times distance

between purlin, 2 · 2,5 = 5,0 m.

Roof bracings and front wall

column at the same points.

Wind action from bottom half of wall acts on foundations of front wall columns, from top half acts on purlins and bracing system.

Photo: Author

Area of front wall, attributable to one

column (and purlin) is not completely

the same, but little differences from

inclination of roof can be neglected.

A = a · b = (2 · 2,5) · (4,5 + 0,5) = 25 m2 F = A qw-fw = 20 kN a = 5,0 m b = 5,0 m

A A AA / 2 A / 2

Photo: Author

A (for roof bracings) = 835 m2

A + A (for wall bracings) = 1027 m2

h

d

h / 2

h / 2

L1

L1 = L – min (2d ; 4h)

EN 1991-1-4 tab 7.10

F = A qw cfr = 10,020 kN

F + F= (A + A) qw cfr = 12,368 kN

Surface cfr

steel, smooth concrete 0,01

rough concrete, tar-boards 0,02

ripples, ribs, folds 0,04

Friction of wind (EN 1991-1-4 7.5)

Wind total Fi = F + F = 30,020 kN

qi = Fi / d = 1,510 kN / m

Steps of iteration:

dq0 = d(qw-t) = dw-t = 8,31 mm

qd(0) = S [8 NEd, c (e0 + dw-t) / L

2] =

= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm]) / (22,2 [m])2 = 1,332 kN / m

Static callculation of deformation

dq(1) = dq

(1)(qd(0)) = 2,78 mm

qd(1) = S [8 NEd, c (e0 + dq

(1) + dw-t) / L2] =

= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm] + 2,78 [mm]) / (22,2 [m])2 = 1,429 kN / m

Static callculation of deformation

dq(2) = dq

(2)(qd(1)) = 2,99 mm

qd(2) = S [8 NEd, c (e0 + dq

(2) + dw-t) / L2] = 1,435 kN / m

Static callculation of deformation

dq(3) = dq

(3)(qd(2)) = 3,00 mm

qd(3) = S [8 NEd, c (e0 + dq

(3) + dw-t) / L2] = 1,436 kN / m

Very small difference, end of iteration.

Photo: Author

Photo: Author

Photo: Author

→ Des 1 examp / 63

The same way of iteration

procedure, but, of course, other

values of loads

Photo: Author

Additional compressive axial force in purlin NEd, purlin = 66,0 kN (bi axial bending → bi-

axial bending and compression → Lec #13);

Max axial force in chord of truss up from 603 to 603 + 116 = 719 kN → recalculation of

roof girder;

Compressive axial force in bracing NEd = 93,0 kN.

Purlins must be recalculated because of new values of cross-sectional forces.

Top chord of roof truss girder must be recalculated because of new values of cross-

sectional forces.

Horizontal distance between ends of bar bracing is equal 6,5 m > 6,0 m. Because of this

bar bracing must be calculated for compression and bending from deadweight. Interaction

between bending moment and axial force will be in detail presented on lecture #13.

First assumption about cross-section of bracing: O 38 / 4.

SLS of bracings are not presented in Eurocodes. It can be assumed, that horizontal

deformation of bracing truss can’t be bigger than:

min (limit for horizontal deformation of columns ; limit for vertical deformation for

roof girders)

Calculations for non-rigid bracings

Bracings are applyied in both direction (X bracings), but only tensed part are taken into

consideration → static scheme is completely different, than for rigid bracings.

Photo: Author

Photo: Author

Additional compressive axial force in purlin NEd, purlin = 124,0 kN (nearly 2 times greater

than for rigid bracings);

Axial force in chord of truss up from 603 to 603 + 106 = 709 kN (similar to rigid

bracings) → recalculation of roof girder;

Tensile axial force in bracing NEd = 114,0 kN (about 125 % value for rigid bracings).

Purlins must be recalculated because of new values of cross-sectional forces.

Top chord of truss must be recalculated because of new values of cross-sectional forces.

Horizontal distance between ends of bar bracing is equal 6,5 m > 6,0 m. Because of this

bar bracing must be calculated for compression and bending from deadweight. Interaction

between bending moment and axial force will be in detail presented on lecture #13.

First assumption about cross-section of bracing: round bar φ 26.

SLS of bracings are not presented in Eurocodes. It can be assumed, that horizontal

deformation of bracing truss can’t be bigger than:

min (limit for horizontal deformation of columns ; limit for vertical deformation for

roof girders)

Example 3

Horizontal upper transversal bracings I-beam girders

Photo: Author

Purlin: one-span simple-supported

IPE 210

MEd, y = 26,865 kNm

MEd, z = 2,687 kNm

Roof girder HEA 550,

hHEA 500 = 0,54 m

Wind action on front wall (wind

pressure at first front wall + wind

suction at the opposite front wall):

qw = 0,8 kPa

MEd, max = 932,2 kNm

NEd, comp, max = 140,0 kN

Equivalent compressive force in roof girder:

NEd* = max (NEd, comp ; MEd / h ; NEd, comp / 2 + MEd / h) =

= max (140,0 ; 932,2 / 0,54 ; 140,0 / 2 + 932,2 / 0,54) = 140,0 / 2 + 932,2 / 0,54 =

= 1796, 3 kN

Photo: Author

Calculations of roof bracings in central part of roof is completely the same as in Example 2.

There is only two differences:

• new value of NEd* = 1796, 3 kN;

• chords of truss-bracings are flanges of roof girders;

Photo: Author

Other situation is for part next to eaves. There are

need additional bracings between bottom flange of

roof girder and purlins.

Photo: builderbill-diy-help.com

Photo: EN 1993-1-1 fig 6.5

Photo: Author

Photo: Author

These bracings change static scheme of purlins and produce additional axial force in

purlins. Total value of loads, which act on horizontal truss, is sum of wind-buckling or

wind-imperfection (→ #t / 48 - 49) and loads from inclined bracings (→ #t / 53 - 55).

Photo: Author

Both of these forces are applied to

purlins and horizontal bracings

which cooperate with purlins.

Example 4

Vertical wall bracings

Wind from this

part acts on right

wall roof bracings

Wind from this

part acts on left

wall roof bracings

F = Fwind + Fcolumn-imperf

Photo: Author

Generally, wind action from bottom half of wall acts on

foundations, from top half acts on purlins and vertical bracing

system (→ #t / 78). But there is assumption, that during

calculation of side wall bracings, total wind action acts on

bracing system.

Loads:

Imperfections:

Axial force in column NEd = 160 kN

Number of columns in one wall m = 11

High of column h = 6,0 m

Fcolumn-imperf = NEd Φ0 αh αm

Φ0 = 1 / 200

αh = max{ 2 / 3 ; min[ (2 / √h) ; 1,0]} = 0,814

h – heigh of column [m]

αm = √[ 0,5 (1 + 1 / m)] = 0,739

Fcolumn-imperf = 0,481 kN

Wind action (2 wall bracings, on left and

right side walls):

Fwind = (pressure + suction of both front walls

+ total wind friction) / 2 = 82,184 kN

F = Fwind + Fcolumn-imperf = 82,665 kN

Horizontal force in transported to bases by bracings only - we

analyse only one field between columns; columns-bracing system.

When we assume rigid bracings (tensile and compressive forces

allow), we must analyse this type of truss.

Compressive force in bracing is equal 65,5 kN. Distance between

ends of bracing bar is not greater than 6,0 m; we no need analyse

bending moment from dead-weight of bracings.

Additional compressive force in column 42,1 kN

When we assume non-rigid bracings (tensile force allows only), we

must analyse this type of truss.

Tensile force in bracing is equal 116,91 kN. Distance between ends

of bracing bar is not greater than 6,0 m; we no need analyse

bending moment from dead-weight of bracings.

Additional compressive force in column 82,67 kNPhoto: Author

Conclusions

Equivalent forces (for example → #t / 32, 45, 47, 48, 49, 55, 76, 80, 87, 89, 91), important

for calculation of bracings, are taken based on initial static analysis of structure.

Additional axial forces (for example → #t / 45, 44, 81, 84, 88, 89, 92) which are effects of

co-operation between structure and bracings and change effects of initial static analysis

must be taken into consideration.

Effect of additional forces is analised in various ways for 2D and 3D model (→ #t / 44, 45).

Types of bracings

Role and position varoius roof bracings

Bracings anti lateral buckling and anti flexural buckling - similiarities and differences

Way of calculations for corrugated sheet

Way of calculation for bar bracings

Examination issues

Bracing system - stężenia

Tract - połaćRidge - kalenica

Hood - okap

Rigging screw - śruba rzymska

Valley - kosz

Skylight - świetlik