8. industrial halls - fsv Čvut -- peoplepeople.fsv.cvut.cz/~machacek/prednaskyok3e/ok3-8e.pdf ·...

© 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


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 eachloading combination !!

1. First-order analysis structures (αcr > 10):

10≥=Ed

crcr F

Fα

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

∑∑=

EdH,Ed

Edcr δ

αh

VH

Note: For given loading FEd the αcr results from FEM by common software (e.g. SCIA Engineer).

Ed

y

NfA

,30≥λ

The check of all members with buckling length equal to the system length (between joints) is then conservative (acc. to Eurocode if αcr > 25 then χ = 1).

H1 H2

V1 V2

h

δH , Ed


OK3 3

2. Second-order analysis structures (αcr < 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, no stability 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 ≤ αcr< 10 and sway buckling mode (corresponds to αcr 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.


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 φdue 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 coefficients χ):

hcr = β h

δ Lcr determined similarly as for columns or to usesystem length and increase moments fromhorizontal loadings by about ~ 20%.

ensure stabilityof free flange ! !

given in many references

111

1≥

−crα


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 commonsoftware of corresponding αcr (corresponding to buckled member) as follows:

Edcr

2

cr

2

cr NIE

NIEL

αππ

==

Note: 1) Using αcr from approximate formula (i.e. for sway buckling mode), the minimum

buckling length equals the system length.1) Mind the modification of cross sections after check:

results in different αcr and hence also Lcr.

For symmetricalloading

for Irafter = ∞

for Irafter = ∞


OK3 6

77110518496

235170903 ,

,,NAf

=⋅⋅

⋅==

cr,1

yλ

mm13742410518496

1063662100003

62

cr

y2

y2

cr ,,,

,N

IEN

IEh =

⋅⋅⋅⋅⋅

===π

αππ

Edcr,1

Practical example:

10000

24000

IPE 550

HE 340 B

12 kN/m'

40 kN 40 kN

imp 1

(for calculation of αcr see Complementary note)

Instead of determination of buckling length hcr the direct check using relativeslenderness is preferred:

cr

y

NAf

=λ

... and from tables directly χ

For given example:

αcr,1 = 6,9(αcr,2 = 44,3)

< 10 (2nd order)

Mind the change of Ncr bychanging cross sectionsafter checks !!!


OK3 7

Portal knees

Approximate resolution of internal forces into flanges:

Vb Mb

Nb

21bb N

hMF +≅

22bb N

hMF −≅

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-rigidconnection)

welds forM, N, V

mind a lamellar tearingof the end plate

(check for buckling)


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 webfor shear: ⎟⎟

⎠

⎞⎜⎜⎝

⎛≈

w

2

w

1maxEd tb

F;tb

Fτ

M1

ywwRdb,Rdb,Ed 3 γ

χττ

fth/V ==≤Considering buckling:

F2

h

b

F1τweld for F'2 h'

extended end plate: less desirable (shorter arm h):

σ

tw

F'2


OK3 9

Note: Shear capacity of the web surrounded by flanges and stiffeners may be increasedby “frame effect” (contribution from flanges, creating 4 plastic hinges in the frame):

hMMf Rdst,pl,Rdc,pl,

M1

ywwEd

223

++≤

γ

χτ

plastic capacityof flanges and stiffeners

c) Increase of shear capacity of an unstiffened knee

tw

t

increasing of web thicknesscontinuous transition of flanges

radialstiffeners

stiffening of the shear panel

diagonal check forloading minus strengthof web in shear


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

possiblestiffener

pinned connection

thick end plates(or a stiffening)


OK3 11

Space behaviour of frames

Analysis:a) Space analysis of the building as a whole (demanding);b) Approximate analysis using continuous girder on elastic supports:

Substantial for local loading(e.g. cranes):

< H

H

elastic supports:

δ1

=c

δ1

withoutcooperation

withcooperation

• roof bracing distributes the loading to more mainframes


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 αcr ≥ 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 gableshear fields

edge members loadedby axial force


OK3 13

Příčná vazba

Přípojeplech/vaznice

plech/plechPřípoje

(jedna tabule)Trapézový plech (podélný prvek)

Přípojesmykové spojky

Smyková spojka

(příčný prvek)Vaznice

V

bVa

b

vaV

bVa

sheetingsheeting

(one sheet)

mainframe

purlin

shear connectorsheeting

connectionssheeting–purlin

jointshear connector

joint

Example ofshear 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 sheetingconnections 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.


OK3 14

Overhead cranes weight of crane Qc(without crab)

crab

bridge

hoistloadhoist 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 ...)


OK3 15

Dynamic effects:- introduced approximately by dynamic coefficients ϕ1 up to ϕ7:

e.g.: for vertical actions ϕ1 up to ϕ4, depends on hoisting speed, crane type ...for drive horizontal actions ϕ5 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 xthe Winkler criterion is valid:

However, usually Mmax and Vmax within all girder length is required:

∑ <> L

xRFi

e.g. 4 forces

arithmetic mean load: P3

position for Mmax = M3 position for Vmax

1st crane 2nd crane (heavier)


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

Gtw

H

HT

heHH =T

h

for design of bottom flange

H + HT

15 tw

assign to upperflange

y SG

z

truss may be replaced by a platewith thickness teff of the sameshear stiffenes

HQ


OK3 17

3. Usual design (on unsafe side, torsion neglected):

Main girder: Horizontal girder:

• vertical loading (mind interactionof buckling due to M, N, V, F)

• longitudinal horizontal loading(implicates N, M)

15 tw • transverse horizontalloading

Fatigue of crane runway beamsCheck for equivalent characteristic stress range (γFf = 1,00):

Mf

CE,2Ff

∆∆γσ

σγ ≤

equivalent constant amplitude direct stress range (must be < 1,5 fy including dynamic coefficient ϕfat)

“fatigue strength" for 2.106 cyclesaccording to category detailFor σ :

(similarly for τ) 1,15

HQ


OK318

Equivalent constant amplitude stress range:

σλϕσ ∆∆ fatE,2 =stress range caused by the fatigue loads acc. to EN 1991

damage equivalent factor, corresponding to2×106 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 80fillet welds: IIτ

→ KD 36*

plan view:

KD 90KD 40

r ≥ 150

r

⊥⊥ στ and


OK3 19

Note: For other members (which buckle at other - higher buckling shapes) is slenderness determined from the 1st shape conservative.

Critical length of the checked member (embodying the right boundary conditions in the structure) may be determined from:

Critical length, however, is just historical auxiliary value, enabling determination of reduction factor with help of slenderness λ. Determination of critical lengths (except of basic cases) with help of formulas, graphs etc. is out-of-date at present. More suitable is direct determination of relative slenderness according to the formula above.

Complementary notes:Critical loading of given load case (combination) may be received by software.E.g. SCIA Engineer : choose → calculation, type: buckling eigenmode, number of shapes

(suitable at least 4). Further setting as for static global analysis (choose mesh, design load case, solution). In postprocessor chooseload case, the first natural shape of buckling: k1 = αcr,1, the secondk2 = αcr,2 , etc. and shapes of buckling (results, deformation shapes).

For the checked member (deciding of the whole structure instability in relevant, i.e. particularly the 1st buckling shape) the critical force at the given load case(combination) and relative slenderness are:

(or ), and → χEdcr,1cr NN α= Ed1cr NkN =

cr

2

cr NIEL π

=

cr,1

y

NAf

=λ


OK3 20

Approximate determination of αcr corresponding to sway mode

Buckling shape of a member with one-side elastic support:

From moment equilibrium:

hence for follows:

hHV EdEdH,cr =δ

1 Support rigidity 3

2

hIEc π

<

Ed

crcr V

V=α

EdH,Ed

Edcr δ

α hVH

=

2 Support rigidity 3

2

hIEc π

≥

VcrHEd

HEd = δH, Ed c

δH, Ed

Ecr VV <

sway mode buckling

Vcr

2

2

E hEIV π

=

h buckling without sway (Euler)

V

8. industrial halls - fsv Čvut -- peoplepeople.fsv.cvut.cz/~machacek/prednaskyok3e/ok3-8e.pdf ·...

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