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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
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