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© 8 Prof. Ing. Josef Macháček, DrSc. 1
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8. Industrial hallsClassification (first and second order) structures, frame haunches, space
behaviour of halls, design of crane runway beams.
Cross sections of portal frames
At present usually:
• pinned based columns (or ”erection stiff”),
• site connections mostly with end plates and pretensioned bolts (instead of splices),
• haunched rafters and columns.
One-bay (portal) frame: span up to 80 m
Two-bay frame: span up to 2x80 m
Three-bay frame: span up to 3x70 mFour-bay frame: span up to 4x70 m
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For sway mode failure approximately
At the same time the slenderness of all
members must fulfil:
In plane frames this shall be applied at each
floor level, the lowest value decides.
Classification of frames and complex multistorey structuresClassifications depends on both geometry and loadings → different for each
loading combination !!
1. First-order analysis structures (
cr
> 10):
10Ed
cr cr
F
F
⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛
∑
∑
=
EdH,Ed
Ed
cr δ
h
V
H
Note: For given loading FEd the cr results from FEM
by common software (e.g. SCIA Engineer).
Ed
y
N
f A,30≥λ
The check of all members with buckling length equal to the system length
(between joints) is then conservative (acc. to Eurocode if cr > 25 then χ = 1).
H1 H2
V1 V2
h
δH , Ed
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2. Second-order analysis structures ( cr < 10):
In general three methods may be used:
a) Geometrical non-linear analysis with imperfections (GNIA).
Second order effects considering global and member imperfections are then
included in resulting internal forces and moments. Check of individual
members is done for simple compression or bending (without χ , χ LT, nostability check is necessary). The solution is demanding on software,
introduction of imperfections and evaluation of results.
b) Geometrical non-linear analysis (GNIA) with global imperfection only (using
frame sway or equivalent horizontal forces). Members shall be checked on
buckling (i.e. 2nd order effect and influence of imperfections), taking the
system length as buckling length (e.g. h, L /2).
If 3 ≤ cr < 10 and sway buckling mode (corresponds to cr determined from
approximate relation above) the 2nd order effects from sway may be
evaluated approximately in accordance with following method b1):
hcr ≤ h
fictitious support for subsequence check of membersfor buckling
Note: for small slopes (up to 15º or flat rafters)
the Lcr equals distance of columns.
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© 8 Prof. Ing. Josef Macháček, DrSc. 4
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b1) Second order sway effects due to vertical loads may be calculated by
increasing the horizontal loads HEd (e.g. wind) and equivalent loads VEd φ
due to imperfections and other possible sway effects according to first
order theory by second order factor:
c) Frequently (classical method) is used first order theory without any
imperfections and members are checked with equivalent global buckling
lengths (using relevant reduction coefficientsχ
):
hcr = h
δ
Lcr determined similarly as for columns or to use
system length and increase moments from
horizontal loadings by about ~ 20%.
ensure stability
of free flange ! !
given in
many
references
11
1
1 ≥
cr
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Typical global buckling lengths (for sway buckling mode):
Global buckling lengths are given in tables or formulas in literature.
They may be preferably determined from critical loading Ncr by common
software of corresponding cr (corresponding to buckled member) as follows:
Edcr
2
cr
2
cr N
IE
N
IEL
π
=
Note:
1) Using α cr from approximate formula (i.e. for sway buckling mode), the minimumbuckling length equals the system length.
1) Mind the modification of cross sections after check:
results in different cr and hence also Lcr .
For symmetrical
loading
for Irafter =∞
for Irafter =∞
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77110518496
235170903
,,,N
Af =
⋅
⋅
=
cr,1
yλ
mm137424
10518496
1063662100003
62
cr
y2
y2
cr ,
,,
,
N
IE
N
IEh =
⋅
⋅
=
π
π
Edcr,1
Practical example:
10000
24000
IPE 550
HE 340 B
12 kN/m'
40 kN 40 kN
imp 1
(for calculation of α cr see
Complementary note)
Instead of determination of buckling length hcr the direct check using relative
slenderness is preferred:
cr
y
N
Af =
... and from tables directly χ
For given example:
cr,1 = 6,9( cr,2 = 44,3)
< 10 (2nd order)
Mind the change of Ncr by
changing cross sections
after checks !!!
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Portal knees
Approximate resolution of internal forces into flanges:
Vb Mb
Nb
21
bb N
h
MF
22
bb NhMF
1) Unhaunched portal knees
a) Knee stiffened for compression
F1
F2
hD
D
b
compression diagonal welded connection bolted connection
thick end plates
(otherwise semi-rigid
connection)
welds for
M, N, V
mind a lamellar tearing
of the end plate
(check for buckling)
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b) Knee with unstiffened shear panel
cover plate with flush and end plate (more expensive)
available for shear V
loaded in shear F1 (friction-grip bolts to avoid slip)
Check of the knee web
for shear: ⎟
⎠
⎞
⎜
⎝
⎛
≈ w
2
w
1
maxEd tb
F
;tb
F
τ
M1
yww
Rdb,Rdb,Ed3 γ
χ
τ
f th/V =Considering buckling:
F2
h
b
F1
weld for F'2 h'
extended end plate: less desirable (shorter arm h):
σ
tw
F'2
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Note:
Shear capacity of the web surrounded by flanges and stiffeners may be increased
by “frame effect” (contribution from flanges, creating 4 plastic hinges in the frame):
hMMf Rdst,pl,Rdc,pl,
M1
ywwEd 22
3
γ
χ
τ
plastic capacity
of flanges and stiffeners
c) Increase of shear capacity of an unstiffened knee
tw
t
increasing of web thicknesscontinuous transition of
flanges
radial
stiffeners
stiffening of the shear panel
diagonal check for loading minus strength
of web in shear
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Note: Site connection may also be offset from column face (column with a cantilever).
2) Hauched portal knees
Portal apexes - similarly:
F2
weld for force M/h
F1
F'1
F'2
I
cutting of I
stiffener
h
shear
tension
compression
cutting of I profile
possible
stiffener
pinned connection
thick end plates
(or a stiffening)
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2. Stressed skin design
stiff cladding (trapezoidal sheeting, monolithic deck):
- acts as a web of high girder, the flanges of which are purlins
(in side-walls rails);
- unloads mainframes, transfers the transvers horizontal
loading to stiff gables;
- usually changes classification of frames for cr ≥ 10.
2 high web girders:
Requirements:- during assembly the structure is non-stiff, secure by temporary bracings, props ...
- the cladding must be effective all the structure life (mind fire, rebuilding ...)
- suitable for short industrial buildings (L /B < 4), with stiff gables.
transfer to stiff gable
shear fields
edge members loaded
by axial force
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Př íčná vazba
Př ípoje plech/vaznice
plech/plechPř ípoje
(jedna tabule)Trapézový plech
(podélný prvek)
Př ípojesmykové spojky
Smyková spojka
(př íčný prvek)Vaznice
V
b
Va
b
v
aV
b
Va
sheeting
sheeting
(one sheet)
mainframe
purlin
shear connector
sheeting
connectionssheeting–purlin
joint
shear connector
joint
Example of
shear field:
Design progress (demanding, usually for repeated use only):
- design of cladding for common bending loading,
- global analysis of non-sway frame (supported by stiff roof plane),
- subdividing the roof into shear fields (diaphragms),
- determination of shear strength and rigidity of the shear field including sheeting
connections and joints (for design procedure see e.g. guideline ECCS No.88),
- determination of cladding effects (unloading of internal frames and design of the
high web girder),
- design of gables.
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Overhead cranesweight of crane Qc
(without crab)
crab
bridge
hoistload
hoist weight + crane loadActions of overhead cranes (EN 1993-3):
• selfweight of the crane Qc• variable:
- vertical action of cranes QH (hoist load given in crane tables)
- horizontal actions acts at rail vertex:
from crane acceleration
(starting, braking)
from crane skewing from crab acceleration
(starting, braking)
crab
- further loading (buffer loads, tilting loads, test loading ...)
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Dynamic effects:
- introduced approximately by dynamic coefficients ϕ 1 up to ϕ7:
e.g.: for vertical actions ϕ 1 up to ϕ 4, depends on hoisting speed, crane type ...
for drive horizontal actions ϕ 5 according to drive, etc.
SLS:
Generally is checked vibration.Practical calculation consists in determination of deflections (δmax < L/600 ≤ 25 mm).
Global analysis
In case of moving loading the influence lines should be used. E.g. for Mmax in section x
the Winkler criterion is valid:
However, usually Mmax and Vmax within all girder length is required:∑
<
> L
x
RFi
e.g. 4 forces
arithmetic mean load: P3
position for Mmax = M3 position for Vmax
1st crane 2nd crane (heavier)
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Example:
s
V V
Design of a crane runway beam
1. Correct design: - requires space (3D) calculation, incl. torsion
(resulting internal forces N, My, Mz, B, Vy, Vz, Tt, Tw)
(necessary to try numerically)
2. Approximate (conservative) introduction of H:
≈+=e
H
G
tw
H
HT
h
eHH =T
h
for design of bottom flange
H + HT
15 tw
assign to upper
flange
yS
G
z
truss may be replaced by a plate
with thickness teff of the sameshear stiffenes
HQ
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Equivalent constant amplitude stress range:
σ∆∆ fatE,2 =stress range caused by the fatigue loads acc. to EN 1991
damage equivalent factor, corresponding to
2 106
cycles (given by EN 1991-3 acc. to crane category)
Structural details (requirement: prevent notches)
max. 100
(buckling)
acc.need
KD 80
KD 80
KD 80
KD 80
KD 45 up to KD 90
KD 112 (for manual weld KD 100)
KD 112 (for manual weld KD 100)
For web to flange → KD 80
fillet welds: II
→ KD 36*
plan view:
KD 90KD 40
r ≥ 150
r
⊥
and
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Approximate determination of cr corresponding to sway mode
Buckling shape of a member with one-side elastic support:
From moment equilibrium:
hence for follows:
hHV EdEdH,cr =
1 Support rigidity 3
2
h
IEc π
<
Ed
cr cr
VV
=
EdH,Ed
Edcr
δ
hVH
=
2 Support rigidity 3
2
h
IEc π
≥
Vcr HEd
HEd = δH, Ed c
δH, Ed
Ecr VV <
sway mode buckling
Vcr
2
2
Eh
EIV
π
=
h
buckling without
sway (Euler)
V