calcul de alpha critique

8/20/2019 Calcul de Alpha Critique

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© 8 Prof. Ing. Josef Macháček, DrSc. 1

OK3 1

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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OK3 2

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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OK3 3

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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OK3 4

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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OK3 5

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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OK3 6

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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OK3 7

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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OK3 8

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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OK3 9

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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OK3 10

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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OK3 12

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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OK3 13

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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OK3 14

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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OK3 15

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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OK3 16

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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OK318

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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OK3 20

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

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