Metal Structures Lecture V Stability

Metal Structures

Lecture V

Stability

Contents

Introduction rarr t 3

Flexural buckling rarr t 18

Example 1 rarr t 46

Lateral buckling rarr t 57

Example 2 rarr t 83

Example 3 rarr t 89

Local instability rarr t 94

Prevention rarr t 96

Examination issues rarr t 97

Popular experiment with a compressed ruler

instability = buckling

Introduction

Photo Author

Ultimate Limit States ULS (EN 1990 64)

EQU (equilibrium) - loss of static equilibrium of the structure or any part of it

considered as a rigid body

STR (strength) - internal failure or excessive deformaton of the structure or structural

member

GEO (geotechnics) - failure or excessive deformaton of the ground

FAT (fatigue) - fatigue failure

Serviceability Limit States SLS

rarr 3 12

What is the meaning of various types of Limit States

Photo Author

Counterweight

Most important part

Compression only support

reaction

rarr 3 13

Displacement rotation lifting by wind

Stability of retaining wall

Stability of crane

Photo craneaccidentscom

Photo malaysiaconstructionservicescom

No destruction and no deformation of

structure but dangerous situation for

people or structure EQU LS

Photo Author

rarr 3 14

A dangerous situation for people or

structure STR LS

bull exceeding strength (s gt fy )

bull excessive deformations (s lt fy )

bull buckling (instability s lt fy )

Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point

σHMH = radic[σ112 + σ22

2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10

σHMH = radic[σ2 + 3(τ12 + τ2

2)]

Welds (Ist step of study) Shells fatigue calculations crane supporting structures (IInd step of study)

Types of formulas - different for different level of structure

rarr 3 76

(~ 10 of calculations conditions)

F - geometry of cross-section

R = F fy

E R le 10

Elements nodes - when instability is not important bolts rivets pins

Level of cross-sections

Photo Author

rarr 3 77


Level of elements


χ - instability coefficient (depends on element geometry)

R = χ F fy

E R le 10

Elements nodes - when instability is important

Photo Author

rarr 3 78


LOAD Ist class IInd class IIIrd class IVth class

NEd NcRd (1-3) le 10 NEd NcRd (4) le 10

MEd (1) MRd (1-2) le 10 MEd MRd (1-2) le 10 MEd MRd (3) le

10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10

(or another for IV if interaction between MEd and VEd exists)

Steel - different formulas for different class of cross-section

Formulas of resistance

Photo Author

rarr 4 76

NcRd (1-3) = A fy gM0 NcRd (4) = Aeff fy gM0 MRd (1-2) = Wpl fy gM0 MRd (3) = Wel fy gM0 MRd (4) = Weff fy gM0 NtRd = A fy gM0 VRd = Av fy (gM0 radic3)

rarr 4 80

Buckling Lateral buckling

(other name lateral-

torsional buckling)

Local buckling

(IVth class of

cross-section) Flexural Flexural-torsional Torsional

NEd c MEd sc

Level of element Level of point

Various types of buckling (STR LS)

EQU LS ndash empty tank rotation arount poin 0 translation

or lifting (deadweight is smaller than wind action) 0

Photo Author

Instability in real world

Flexural buckling of rails because of thermal

expansion (STR LS)

Photo ttitamuedu

Flexural buckling of steel trusses (STR LS)

Photo ascelibraryorg

Lateral buckling of beam (STR LS)

Photo civildigitalcom

Photo failuremechanismswordpresscom

Flexural-torsional buckling of

bracings (STR LS)

Instability of shell structure (STR LS silo IInd step of

study specific name for shell structures LS3)

Photo publishuccie

Local buckling of flanges of steel beam

(STR LS)

Photo tatasteelconstructioncom

Global instability of rigid body (EQU LS)


Flexural buckling of reinforced concrete columns

(STR LS)

Photo scedccaltechedu

M(x) = N w(x)

d[w(x)]2 dx2 = -M(x) EJ rarr M(x) = -wrdquo(x) E J

-wrdquo(x) E J = N w(x)

wrdquo(x) = - k2 w(x)

k = radic (N EJ)

Flexural buckling

According to Mechanics of Materials

Photo Author


w(x) = W1 sin (k x) + W2 cos (k x)

w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0

sin (k l) = 0 rarr k l = n p

k = radic (N EJ)

l radic (N EJ) = n p

N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)

W1 = (not important ndash buckling is important not its amplitude)

Photo Author

Experiment the same geometrical characteristics of cross-

section the same values of forces but different length of bars

P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion

Long bars Nmax = Ncr = q l2

Short bars Nmax = A fy

Nmax = min (Ncr A fy)

χ = min (10 Ncr A fy)

Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author

STR LS exceeding the strength

STR LS buckling

Resistance of element the same as resistance of cross-section

Critical resistance of element smaller than resistance of cross-section

Nmax

l

Ncr

A fy

Theoretically we have only two conditions for crush and for buckling For real structures because of imperfections ndash described by buckling curves - we have additional conditions Influence of imperfections will be detaily presented on Lec 6

Photo Author

Nmax

l

Five different curves in Eurocode for flexural buckling

Photo EN 1993-1-1 fig 64

Generalization

Formula was elaborated for five assumptions as follow

1 EJ = const (but what if not)

2 N = const (but what if not)

3 Two hinges (but what if not)

4 Force applied in gravity center (but what if not)

5 Bar has straight axis (but what if not)

Photo Author

1 Generally EJ = const is the most often case Differences are negligible if a le 10o for calculations EJ = min (EJ1 EJ2 ) There are different rules for a gt 10o

these rules will be presented on lecture 12

a

Photo Author

2 Generally changes of the axial force NEd along bars are very small We should adopted to calculations NEd = max (NEd1 NEd2)

Photo Author

3 When we have other supports (not two hinges) we have different modes of instability

Photo wikipedia

Critical length is a way to compare different modes of instability

Critical length lcr ndash theoretical length of one sinusoidal wave which occurs in shape of bar after buckling

Buckling length factor m = lcr l0

Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0

There are different factors for different types of support Photo wikipedia

Conclusions different types of support is important factor affecting the calculation

Ncr = p2 EJ (m l0)2

4 Force applied out of gravity center rarr eccentricities between gravity center and point of

force rarr axial force and bending moment simulateneously rarr such type of interaction during

calculation of resistance and stability will be presented on Lec 13

5a Curve axis of bar rarr influence of imperfections rarr various buckling curves rarr t 28

5b Curve axis of bar rarr arch rarr stability of arches rarr Mechanics of Materials

Photo Author

Modes of bucklings under compressive force

Initial configuration

Torsional buckling Flexural-torsional buckling

Flexural buckling (about y about z)

x

y

z

Buckling initial and after-buckling position of cross-section A-A

Flexural Torsional Flexural-torsional

Jy Jz Jw Jt Jz Jw Jt

Modes of cross-sectionrsquos deformation for various types of buckling

Buckling about y axis rarr translation parallel to z axis

Buckling about z axis rarr translation parallel to y axis

Photo Author

Flexural buckling axis y Ncr y = p2 EJy (my l0y)2

Flexural buckling axis z Ncr z = p2 EJz (mz l0z)2

Torsional buckling Ncr T = [p2 EJw (mT l0T)2 + GJt] is2

Flexural-torsional buckling Ncr z-T = Ncr z + Ncr T - radic [(Ncr z + Ncr T)2 - 4 Ncr z Ncr T x] (2 x)

x = 1 - (m zs2 is

2)

m = min[radic (mz mT) radic (mT mz)]

i0 = radic (iy2 + iz

2)

is = radic (i02 + zs

2)

zs - distance between centre of gravity and shear centre (zs ge 0)

Jy Jz ndash moments of inertia

iy iz ndash radius of inertia

E G ndash Young and Kirchhoff modulus

Jt - torsion constant

Jw - warping constant

Formulas (according to Mechanics of Materials)

Here are values of these geometrical characteristics

Photo europrofillu

If we have cross-section which not exists in tables

rarr Lab 1 61

rarr Lab 1 62

J Żmuda bdquoPodstawy projektowania konstrukcji metalowychrdquo TiT Opole 1992

rarr Lab 1 63

Buckling

Flexural Flexural torsional

flexural-torsional

c = cy = cz

(only if lcr y = lcr z )

c = min( cy cz)

c = min( cy cz cT cz T)

(hot rolled I) (welded I)

The result of calculations is buckling factor c

It is calculated in different way for different cross-sections

Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =

min1[F + radic (F2 - l2)]

10

Flexural

torsional

flexural-

torsional

buckling

l = radic (A(eff) fy Ncr)

l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631

Photo EN 1993-1-1 tab 62


Example 1 C 300p S235 rarr fy = 235 MPa L = 300 m E = 210 GPa G = 81 GPa A = 525 cm2

Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4

a = 312 cm e = 289 cm iy = 121 cm iz = 301 cm ys = a + e = 601 cm In this case zs = ys = 601 cm NEd = 650 kN I class of cross-section

Photo Author

l0 y = 600 m l0 z = 300 m l0 T = 600 m my = 100 mz = 090 ndash 100 (in calculation 100) mT = 100

Supports and modes of instability

l0 y l0 T

2 l0z

mz l0z

Photo Author

Ncr y = p2 EJy (my l0y)2 = p2 210 GPa ∙ 7640 cm4 (10 ∙ 600 m)2 = 4 398554 kN

Ncr z = p2 EJz (mz l0z)2 = p2 210 GPa ∙ 473 cm4 (100 ∙ 300 m)2 = 1 089211 kN


2) = 1247 cm


2) = 1384 cm

Ncr T = [p2 EJw (mT l0T)2 + GJt] is2 =

= [p2 210 GPa ∙ 66 500 cm6 (10 ∙ 600 m)2 + 81 GPa ∙ 339 cm4] (1384 cm)2 = 1 633427 kN

m = min[radic (mz mT) radic (mT mz)] = 0975

x = 1 - (m zs2 is

2) = 0816

Ncr zT = Ncr z + Ncr T - radic [(Ncr z + Ncr T)2 - 4 Ncr z Ncr T x] (2 x) =

= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN

ly = radic(A fy Ncr y) = 0530

lz = radic(A fy Ncr z) = 1064

lT = radic(A fy Ncr T) = 0869

lzT = radic(A fy Ncr zT) = 1178

C 300p rarr tab 61 62 EN 1993-1-1 rarr ay = az = aT = azT = 049

Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435

cy = min1[Fy + radic (Fy2 - ly

2)] 10 = 0827

cz = min1[Fz + radic (Fz2 - lz

2)] 10 = 0504

cT = min1[FT + radic (FT2 - lT

2)] 10 = 0619

czT = min1[FzT + radic (FzT2 - lzT

2)] 10 = 0444

c = min(cy cz cT czT) = 0444

A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186

Wrong buckling destruction

L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN






Proposition other distance between

supports on y-direction rarr change of

critical length for z-buckling

Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK

laced compression members

battened compression members

closely spaced build-up compression members

There are special rules for complex cross-sections (two- or more chords) presented on

EN 1993-1-1 p64



Modes of buckling are very important for these types of structures

Photo Author

m le 10 m ge10

There are two possibilities for steel frames not susceptible or susceptible to

second order effects

Calculations of buckling length factor for them will be presented on lecture 13

Photo Author

Photo myews13com

The most often

15 ge m ge 35

Flexural-torsional buckling

Lateral buckling

Flexural-torsional buckling versus lateral buckling

The same shape of deformations but other reasons

Lateral buckling

Photo Author

Experiment whats happend during bending cantilever in two situations

Photo Author

There are two possibilities of destruction for axial force Situation is more complicated for bending moment

Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq

Two possibilities of destruction

1 stresses on support from bending gt resistance rarr destruction of support rarr destruction of

structure

2 lateral buckling

Photo Author

q0 = 0

The same experiment for bending about weak axis

Photo Author

qi ne 0

But for this situation only one way of destruction

is possible ndash by destruction of support

Why there are one possibility for weak

axis and two for strong axis

Photo Author

Effect of load depends on amount of internal energy

For each point inside bar we can calculate matrix of stresses and matrix of strains Product

of these two tensors is value of internal energy in any point Total internal energy for every

points in bar is equal sum after every points

Ei = int int int [s] [e] dx dy dz

Theoretically there are infinitive number of deformations for each type of loads But in

real structures for one type of load is only one type of deformation this type for which

Ei = min

(Classical Mechanics Mechanics of Materials Structural Mechanics Theory of Elasticity)

Increasing of displacements makes increasing of internal energy Values of energy

depends on direction of load and direction of displacements

Increasing of inner energy

Direction of displacement D

| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis

(bending about weak axis)

D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis

(bending about strong

axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author

During analysis of load parallel to weak axis for big value of loads both deformations

parallel to strong axis (lateral buckling)

parallel to weak axis (deflection)

have similar values of internal energy

Final mode of destruction - buckling or destruction of support - depends on geometrical

characteristics of beam length of beam and type of supports

Photo Author

Buckling

Deflection

Internal energy for load parallel to strong axis is always smaller for displacement paralel to

load than for perpendicular to load There is no displacement perpendicual to load

Photo Author

Deflection

No buckling

Conclusions

bull lateral buckling must be analysed only for bending about strong axis (rarr potential instability about weak axis)

bull there is no need to analysis lateral buckling for bending about weak axis

bull for the same reason flexural-torsional buckling is always interaction between torsional buckling and flexural buckling about weak axis

bull flexural-torsional buckling as interaction between torsional buckling and flexural buckling about strong axis is impossible

bull there is no lateral buckling and no torsional-flexural buckling for cross-section with Jy = Jz rarr no lateral buckling for square and round pipes

Ncr z = p2 EJz (mz l0z)2

Ncr T = [p2 EJw (mT l0T)2 + GJt] is2

Mcr = is radic (Ncr z Ncr T)

1 M = const 2 bi-symmetrical cross-section

3 EJ = const 4 two hinges 5 load applied to shear center 6 straight axis of bar

Formulas are elaborated for six assumptions

(similarly to formula for flexural buckling)

Photo Author

M = const rarr C1 = 10

Formula presented by Acces Steel (based on

pre-Eurocode ENV 1993-1-1)

Photo eurocodesjrceceuropaeu

Mcr = p2 E Jz (l)2 radic [(Jw Jz) + (l2 G JT) (p2 E Jz)] = Ncr z radic [(Jw Jz) + (G JT) (Ncr z)] =

= Ncr z radic [(Ncr z Jw) (Ncr z Jz) + (G JT) (Ncr z)] =

= Ncr z radic [(Ncr z Jw) (Jz) + (G JT)] (Ncr z) =

= Ncr z radic [(p2 E Jz Jw) (l2 Jz) + (G JT)] (Ncr z) =

= Ncr z radic [(p2 E Jw) (l2) + (G JT)] (Ncr z) =

= Ncr z radic [(Ncr T is2 ) Ncr z] =

= radic [(Ncr T is2 Ncr z

2 ) Ncr z] =

= is radic (Ncr z Ncr T)

cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

1 M ne const rarr other shape of bending moment


kc

This question is not completely clear in Eurocode

According to EN 1993-1-1 6322 (2) critical

moment Mcr depends on amount others shape of

bending moment But there is no other information

about this influence except recalculation as above

(recalculation for cLT no Mcr) Maybe this means

that we should use other formula for Mcr than

formula on t 69

1 M ne const (alternative way)

or

2 Not bi-symmetrical cross-section

or

5 Load applied out of shear center

Generally bi-symmetrical cross-section is the most often case for steel

structures

If not we use other formula for Mcr

For example old Polish Standard PN B 03200 appendix 3

Table Z1-1 rarr extended version of this table is presented t 40 - 42

Values of rx and ys are presented in this table

Jz JT JW

Shear center (t 40 ndash 42) Point of load application

Distance shear center ndash

load point

Arm of cross-sectionrsquos assymetry

Bending parameter

Symbols Ny rarr Ncr z Nz rarr Ncr T

One-span beam M = const

hinge on both ends

As previous and additionally

bi-symmetrical cross-section

General formula for one-span

beam

One-span I-beam with lateral plate-type bracings which can change axis of rotation

Cantilever

Distance shear center ndash bracing point

M = const

q = const

Force in point at the end of cantilever

M linear or

constant

q = const

Force applied on the half of span

P - hinge

U - rigid support

Point of application of load

Value of Mcr depends on point of application of load In formulas for general solution of

lateral buckling (rarr t 75 point c or general solution on Access Steel) important is distance

between shear centerand point of application zg


Value of Mcr is smaller for load on top flange bigger fot load on bottom flange (bigger

probability of instability for load on top flange)

Load

Photo Author

MT = Load e

Load

e

e

MT = Load e

Deformation from initial imperfection

Deformation as the result of torsional moment



Various points of loadrsquos application makes various

effects of deformations These deformations from

torsional moment can intensify or weaken the

impact of initial imperfections As a result the

cross-section may lose stability more easily (smaller

Mcr) or more difficult (larger Mcr)

Initial imperfections make eccentricity e and

torsional moment MT as the secondary effect of

load

3 The same situation as for previous type of instability (flexuram buckling)

EJ = const is the most often case Differences are negligible if a le 10o for calculations

EJ = min (EJ1 EJ2 ) There are different rules for a gt 10o these rules will be presented on

lecture 12


Curve axis of bar rarr influence of imperfections rarr various buckling curves rarr t 82

4 Other types of support rarr mT

Generally rules are the same as for flexural buckling At now we must

analysed not only translation but additionally change of rotation of

cross-section and types of support for rotation

m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia

lLT = radic (Wy fy Mcr)

EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2

aLT rarr tab 63 64 EN 1993-1-1

cLT = min

1[FLT + radic (FLT2 - lLT

2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10

lLT le 04 rarr cLT = 10

Algorithm

EN 1993-1-1 632

For lateral buckling the same various buckling curves are taken into consideration



General case


Hot-roled and welded I-sections

Distinction not cleary explained

Example 2

IPE 300

hIPE300 = 300 mm

S235 rarr fy = 235 MPa

L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm

I class of cross-section

Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100

mz = 090 ndash 100 (in calculation 100)

mT = 090 ndash 100 (in calculation 100)

L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm

Wy pl fy = 147674 kNm

lLT = 1278

According to formula dedicated for I-beams

aLT rarr tab 63 65 EN 1993-1-1rarr 034

FLT = 1261

but FLT lt lLT we canrsquot calculate radic (FLT2 - lLT

2)]

According to general formula dedicated for bi-symmetrical

cross-sections

aLT rarr tab 63 64 EN 1993-1-1 rarr 021

FLT = 1429

cLT = 0483

Shape of bending moment rarr recalculation for cLT mod rarr t 71

cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491

Wpl y fy = 147674 kNm

cLT mod Wpl y fy = 72473 kNm

MEd = 80000 kN

MEd Wpl y fy = 0542

OK

MEd cLT mod Wpl y fy = 1104


L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712


MEd cLT mod Wpl y fy = 0761 OK

Proposition other distance between supports rarr

change of critical length for z-buckling and

torisional buckling

Photo Author

Example 3


Influence of point of application

bull to top flange

bull to shear center (bi-symmetrical I-beam = center of gravity)

bull to bottom flange


is analysed for two methods Old Polish Standard (rarr t 75

formula c) and pre-Eurocode ENV 1993-1-1 (rarr t 77)

Distance between shear = gravity center for analysed I-beam is

respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard

Mcr = A0 Ncr z + radic[(A0 Ncr z)2 + B2 is

2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2

ry ndash arm of cross-ectionrsquos assymetry rarr t 40-42 (symbol rx) = 0 for bi symmetrical I-beam

ys ndash difference betwen centre of shear and centre of gravity for bi symmetrical cross-

section = 0

as = zg = according to symblos in ENV

A1 A2 B rarr t 76

Statis scheme presented on t 83-84 constant continious load hinge supports proportion

between ctiritical length L0 y = 2 L0 z = 2 L0 T

Parameters for position

Continiuos load P-P 10 ndash 05

A1 = 123 A2 = 012 B = 131

Mcr was calculated for parameters A1 A2 B dedicated to bar under constant continous load

There is no need to recalculate from cLT to cLT mod cLT in in final value Formulas based on

more general assumptions than formuna on Example 2 results for load applied to shear

center are various

Application of load Mcr [kNm] lLT cLT cLTWpl y fy

[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113

to centre of shear (= gravity for

bi-symmetrical I-beam) 118515 1116 0585 86323 0927

to bottom flange 145635 1007 0661 97509 0820

Access Steel SN003a-EN-EU

Mcr = C1 Ncr z [radic(k kw)2 (Jw Jz) + [(k L0 z)2 G Jt ] (p

2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT


Mcr was calculated for parameters C1 C2 dedicated to bar under constant continous load There

is no need to recalculate from cLT to cLT mod cLT in in final value Formulas based on a little

differnt assumptions than formulas in Polish Standard


[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880

General complex of derivative formulas for lateral buckling is

complicated and in the general case it is solved in an approximate

way after many simplifications various for PN and AS

Photo K Rykaluk Zagadnienia stateczności

konstrukcji metalowych DWE 2012

E Jz hrdquordquo + (My j)rdquo = 0

E Jw jrdquordquo ndash [(2 bz My + G JT) jrsquo]rsquo + qz (ez ndash zs) j + My hrdquo = 0

bz = intA[ z (y2 + z2) dA] ndash zs (2 Jy)

Local instability is analysed in different way for different situations

Welded I-beam

Effective cross-section

Shear buckling of web

Buckling of web under transverse force

Flange buckling

Structural joints

Stiffeners

Cold formed cross-sections (distortion)

Local instability

Local instabilities web buckling flange buckling (welded hot-rolled) distortional

buckling (cold-formed EN 1993-1-3)


Photo helpstud2norodru

Photo fgguni-ljsi

Prevention

We can limit or eliminate flexural torsional lateral and local buckling by using

bracings Different type of bracings can eliminate different types of buckling

More information about eliminate flexural torsional and lateral buckling will be

presented on lecture 10

Local instabilities are limited or eliminated by additional stiffeners More

information will be presented on lecture 21

Photo steelconstructioninfo Photo bgstructuralengineeringcom

Difference between resistance of cross-section and stability of element

The cause of instability

Types of instability

Similiarities and differences for lateral buckling and flexural-torsional buckling

Examination issues

Stability - stateczność

Buckling - wyboczenie utrata stateczności

Flexural buckling - wyboczenie giętne

Flexural-torsional buckling - wyboczenie gietno-skrętne

Torsional buckling - wyboczenie skrętne

Lateral buckling - zwichrzenie

Local buckling - niestateczność lokalna

Shear centre - środek ścinania

Torsion constant - moment bezwładności przy skręcaniu

Warping constant - wycinkowy moment bezwładności

Stiffener - żebro

Laced members - słup wielogałęziowy skratowany

Battened members - słup wielogałęziowy z przewiązkami

Closely spaced build-up members - pręt wielogałęziowy

Bracing - stężenie

Thank you for attention

copy 2021 Tomasz Michałowski PhD

tmichalpkedupl

Contents

Introduction rarr t 3

Flexural buckling rarr t 18

Example 1 rarr t 46

Lateral buckling rarr t 57

Example 2 rarr t 83

Example 3 rarr t 89

Local instability rarr t 94

Prevention rarr t 96

Examination issues rarr t 97



Introduction

Photo Author





member




rarr 3 12


Photo Author

Counterweight

Most important part


reaction

rarr 3 13



Stability of crane






Photo Author

rarr 3 14


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Introduction

Photo Author





member




rarr 3 12


Photo Author

Counterweight

Most important part


reaction

rarr 3 13



Stability of crane






Photo Author

rarr 3 14


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





member




rarr 3 12


Photo Author

Counterweight

Most important part


reaction

rarr 3 13



Stability of crane






Photo Author

rarr 3 14


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author

Counterweight

Most important part


reaction

rarr 3 13



Stability of crane






Photo Author

rarr 3 14


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Stability of crane






Photo Author

rarr 3 14


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


structure STR LS




Photo Author

s

s s

rarr 3 15

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Tσ =

s11 τ12 τ13

τ21 s22 τ23

τ31 τ32 s33

Level of point


2 + σ332 - σ11 σ22 - σ11 σ33 - σ22 σ33 + 3(τ12

2 + τ232 + τ13

2 )]

σHMH fy le 10


2)]



rarr 3 76



R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


R = F fy

E R le 10



Photo Author

rarr 3 77


Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Level of elements



R = χ F fy

E R le 10


Photo Author

rarr 3 78





10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




10

MEd MRd (4) le 10

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

Interaction

MEd harr NEd

NEd NtRd le 10

VEd VRd le 10




Photo Author

rarr 4 76


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


rarr 4 80



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



torsional buckling)

Local buckling

(IVth class of


NEd c MEd sc





Photo Author



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



expansion (STR LS)

Photo ttitamuedu







bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





bracings (STR LS)



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Photo publishuccie


(STR LS)





(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




(STR LS)


M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

M(x) = N w(x)




k = radic (N EJ)

Flexural buckling


Photo Author



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



w(0) = 0 rarr W2 = 0

w(l) = 0 rarr W1 = 0 or sin (k l) = 0


k = radic (N EJ)


N EJ = (n p l)2

Ncr = (n p l)2 EJ

w(x) = W1 sin (k x)


Photo Author



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



P0 = 0

Photo Author

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P1 ne 0

Photo Author

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P2 = P1 + DP

Photo Author

Buckling

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P3 = P2 + DP

Photo Author

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P4 = P3 + DP

Photo Author

Buckling

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P5 = P4 + DP

Photo Author

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

P6 = P5 + DP

Photo Author

Crush

Crush

Crush

Conclusion





Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Nmax = χ A fy

χ = χ (l)

χ le 10

Photo Author


STR LS buckling



Nmax

l

Ncr

A fy


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author

Nmax

l



Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Generalization







Photo Author



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



a

Photo Author


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo wikipedia




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




Photo Author

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

m 10 20 07 05 10 20

lcr 10 l0 20 l0 07 l0 05 l0 10 l0 20 l0



Ncr = p2 EJ (m l0)2






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl






Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Photo Author





x

y

z







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl







Photo Author





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





x = 1 - (m zs2 is

2)



2)


2)









Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo europrofillu


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


rarr Lab 1 61

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

rarr Lab 1 62


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


rarr Lab 1 63

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Buckling


flexural-torsional

c = cy = cz


c = min( cy cz)





Photo Author

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Flexural

buckling

(I II III class of

cross-section)

l = (lcr i) l1

l1 = 939 e

F = [1 + a (l -02) + l2] 2

a rarr EN 1993-1-1

tab 61 62

c =


10

Flexural

torsional

flexural-

torsional

buckling


l le 02 rarr c = 10

Algorithm

EN 1993-1-1 631




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Jy = 7640 cm4

Jz = 473 cm4

Jw = 66 500 cm6

JT = 339 cm4


Photo Author



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



l0 y l0 T

2 l0z

mz l0z

Photo Author




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




2) = 1247 cm


2) = 1384 cm




x = 1 - (m zs2 is

2) = 0816


= 1 206953 kN + 1 633427 kN +

- radic [(1 087211 kN + 1 633427 kN )2 - 4 ∙ 1 087211kN ∙ 1 633427 kN ∙ 0816] (2 ∙ 0816) =

= 888974 kN

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

A fy = 1 233750 kN






Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Fy = [1 + ay (ly - 02) + ly2] 2 = 0721

Fz = [1 + az (lz - 02) + lz2] 2 = 1278

FT = [1 + aT (lT - 02) + lT2] 2 = 1041

FzT = [1 + azT (lzT - 02) + lzT2] 2 = 1435


2)] 10 = 0827


2)] 10 = 0504


2)] 10 = 0619


2)] 10 = 0444


A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

A fy = 1 233750 kN

c A fy = 548241 kN

NEd = 650 kN

NEd A fy = 0527

OK

NEd c A fy = 1186


L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

L0z = 200 m

Ncr y = 4 398554 kN

Ncr z = 2 450725 kN

Ncr T = 1 633427 kN

Ncr zT = 1 333190 kN









Photo Author

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

A fy = 1 233750 kN

c A fy = 693930 kN

NEd = 650 kN

NEd A fy = 0567

OK

NEd c A fy = 0944

OK





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





EN 1993-1-1 p64




Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

m le 10 m ge10




Photo Author

Photo myews13com

The most often

15 ge m ge 35


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Lateral buckling



Lateral buckling

Photo Author


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Photo Author

Crush

Crush

Crush

Buckling

Buckling

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

q0 = 0

Photo Author

q1 ne 0

Deflection

q2 = q1 + Dq

Deflection

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

q3 = q2 + Dq



structure

2 lateral buckling

Photo Author

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

q0 = 0


Photo Author

qi ne 0





Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Photo Author








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl








Ei = min






| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





| | to strong axis

(deformation about

weak axis)

| | to weak axis

(deformation about

strong axis)

Direction of

load q

| | to strong axis


D || q || strong

small amount of

energy

D q || weak

very big amount

of energy

| | to weak axis


axis)

D q || strong

medium amount

of energy

D || q || weak

medium amount

of energy

Photo Author







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl







Photo Author

Buckling

Deflection



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Photo Author

Deflection

No buckling

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Conclusions













Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl








Photo Author












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl












2 ) Ncr z] =


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10



kc








formula on t 69


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


or


or



structures





Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



Jz JT JW



load point


Bending parameter



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



hinge on both ends




beam


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Cantilever


M = const

q = const


M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

M linear or

constant

q = const


P - hinge

U - rigid support








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl








Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Load

Photo Author

MT = Load e

Load

e

e

MT = Load e













load




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




lecture 12







m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





m 10 20 07 05

lcr 10 l0 20 l0 07 l0 05 l0

Photo wikipedia


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


EJ =

const

FLT = [1 + aLT (lLT -02) + lLT2] 2


cLT = min


2)]

10

Hot-

rolled

and

welded

I-beam

FLT =

= [1 + aLT (lLT -04) + 075 lLT2] 2


cLT = min


2)]

1 lLT2

10


Algorithm

EN 1993-1-1 632




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




General case




Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Example 2

IPE 300

hIPE300 = 300 mm


L = 600 m

E = 210 GPa

G = 81 GPa

Jy = 8 356 cm4

Jz = 6038 cm4

Wy = 5571 cm3

Wpl y = 6284 cm3

Jw = 125 900 cm6

JT = 2012 cm4

iy = 1246 cm

iz = 335 cm

ys = 00 cm

MEd y = 800 kNm


Photo Author

L

L

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0 2 4 6 8 10 12

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

L0 y = 1200 m

L0 z = 600 m

L0 T = 600 m

my = 100



L0 y = 2 L0 z = 2 L0 T

my L0 y

mz L0 z = mT L0 T

Photo Author

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Ncr z = 347603 kN

i0 = is = 1290 cm

Ncr T = 1 414345 kN

Mcr = 90467 kNm


lLT = 1278



FLT = 1261


2)]


cross-sections


FLT = 1429

cLT = 0483


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


cLT mod = cLT f

f = min 1 - 05(1-kc)[1 - 2(lLT - 08)2] 10

kc = 094

f = 0938

cLT mod = 0491



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



MEd = 80000 kN

MEd Wpl y fy = 0542

OK



L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

L0 y = 1200 m

L0 z = 400 m

L0 T = 400 m

Ncr z = 782108 kN

Ncr zT = 1 958571 kN

Mcr = 159689 kNm

lLT = 0962

FLT = 1042

cLT = 0692

cLT mod = 0712





torisional buckling

Photo Author

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Example 3



bull to top flange







respectively

bull 0150 m

bull 0000 m

bull -0150 m

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Old Polish Standard


2 Ncr z Ncr T]

A0 = A1 by + A2 as

by = ys ndash ry 2



section = 0


A1 A2 B rarr t 76





A1 = 123 A2 = 012 B = 131




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl




center are various


[kNm]

MEd MRd

to top flange 91408 1271 0487 71854 1113



to bottom flange 145635 1007 0661 97509 0820



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl



2 E Jz) + (C2 zg)2 - C2 zg]

k = mz

kw = mT






[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





[kNm]

MEd MRd

to top flange 75279 1401 0418 61630 1298



to bottom flange 128635 1071 0616 90863 0880










Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl


Welded I-beam




Flange buckling

Structural joints

Stiffeners


Local instability





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





Photo fgguni-ljsi

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl

Prevention












Examination issues











Stiffener - żebro







tmichalpkedupl





Examination issues











Stiffener - żebro







tmichalpkedupl











Stiffener - żebro







tmichalpkedupl



tmichalpkedupl

Date post:	18-Dec-2021
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Metal Structures Lecture V Stability

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