2CO8 Advanced design of steel and composite structures

© Prof. Ing. Josef Macháček, DrSc.

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2CO8 Advanced design of steel and composite structuresLectures: Machacek (M), Netusil (N), Wald (W), Ungureanu (U)_____________________________________________________________________

• 1 (M) Global analysis. Torsion of steel members• 2 (M) Built-up steel members• 3 (M) Fatigue• 4 (M) Lateral-torsional buckling• 5 (M) Buckling of plates• 6 (M) Composite steel and concrete members• 7 (M) Large-span structures 1• 8 (M) Large-span structures 2• 9 (M) Stainless steel structures 1• 10 (M) Stainless steel structures 2• 11 (N) Membrane structures 1• 12 (N) Membrane structures 2• 13 (W) Aluminium structures 1• 14 (W) Aluminium structures 2• 15 (W) Aluminium structures 3• 16 (U) Shell structures 1• 17 (U) Shell structures 2• 18 (U) Lightweight steel structures 1• 19 (U) Lightweight steel structures 2• 20 (U) Lightweight steel structures 3


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1. Global analysis. Torsion.• Global analysis, imperfections, classification of structures according to stability and

methods of analysis. Torsion of open and closed sections.

φ

e0imperfection e: 1. member imperfection (expressed by initial deflection e0),2. frame imperfection (expressed by initial sway of the frame φ).

geometrically nonlinear analysis GNIA(or simplified 2nd order analysis)

linear analysis(1st order, elastic) strength (resistance)

fibre plasticity

plastic analysis GMNIA

δ

F

collapse

e

Methodsof analysis: elastic


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Imperfections:

3

1. Member imperfections: covered by equivalent geometrical imperfection e0

2. Frame imperfection (initial frame sway): covered by angle φ(or equivalent horizontal forces)

V1

V2

V1

V2φ

φ V1

φ V2≈

Effect of all these imperfections, i.e.• initial deflections e0g, • residual stresses in cross section σr, • structural imperfections e1,

are covered in the analysis by a singleequivalent geometrical imperfectionwith amplitude e0.geometrical material structural

(shape imperf.) (residual stresses) (boundary conditions)

e0g

e1

≈ fy

σ r

e.g. for Icross section

e.g. imperfecthinge

≈ e0++

mh0 ααφφ =20010 /=φ

h2

h =α 0132

h .≤≤ α

⎟⎠⎞

⎜⎝⎛ +=

m. 1150mα

in Eurocode:

where

but

h .. total height [m] m .. number of loaded

columns per plane


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Models according to effect of deflection:1. Geometrically nonlinear (in steel structures the 2nd order analysis is sufficient).

GNA is required always for stability analysis and non-rigid elements (cables). Analysis must cover influence of imperfections (GNIA):- in compression members (using equivalent geometrical imperfection e0),- in storey frames (using frame sway φ, or equivalent horizontal forces),- plates in buckling (imperfections according to Eurocode 1993-1-5).

Drawbacks: superposition is not valid, software necessary (otherwise difficult).

2. Linear (1st order theory) – analysis is approximate only.In structures with rigid members the effect of GNIA need to be covered by reduction factors for compression (χ), bending (χLT), buckling (ρ), derived from 2nd order theory.

Models according to use of material:1. Elastic – always possible, but only up to achieving steel yield point fy.

Not economical, if plastic analysis is possible.

2. Plastic – may be used provided the following is true:a) material: the steel is ductile (min. 15 % → meet by all common steels),b) cross section: cross sections in plastic hinges are plastic (Class 1 cross sections),c) member: lateral torsional buckling of the members in plastic hinges is prevented).


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Design for buckling in compression [bachelor study revision]N N N

L

Lcr= L Lcr= 2L Lcr= 0,5L

member i

FEd

5

Basic cases (for 1st buckling mode):

commonly

χ

cr

Ecr N

NLL == β

cr

y

cr

y

1

cr

1 NAf

NNi/L

====λλ

λλ

In general:

for nth critical mode of the frame (use software):

i.e. (n = 1 up to ∞ )

(1st mode is conservative)

For member i the corresponding mode n isdecisive (but using mode n = 1 is conservative).

Ed

ncr,ncr, F

F=α Edncr,ncr, FF α=

Critical member forces (to determine slenderness ):λ

Edcr,1Edncr,ncr, NNN αα ≥=


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Approximately for sway mode according to picture:

For regular building frames this formula may be used for each storey; decisive is the lowest value.

Note: This approximate value of αcr is valid provided:

(otherwise individual members maybuckle in their system lengths)

Frame classificationThe classification depends of both geometry and loading → i.e. it need to be determine for each loading combination !!

1. Frames which may be solved by 1st order theory (αcr ≥ 10): Note: for arbitrary FEd the αcr results from stability analysis by FEM (various software).

Design of all frame members for this loading may be performed for their system buckling lengths (between the frame nodes) and is very conservative (according to Eurocode for αcr > 25 even the value χ = 1 may be used).

for plasticanalysis ≥ 15

H1 H2

V1 V2

h

δH , Ed

design loading FEdfor given loading+ frameimperfection

φ ΣV

V

10≥=Ed

crcr F

Fα

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛∑∑

=EdH,Ed

Edcr δ

α hVH

Ed

y30NAf

.≥λ


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2. Frames to be solved by 2nd theory (αcr < 10): In general 3 methods are used:

a) Geometrically nonlinear analysis of completely imperfect structure.(GNIA = geometrically nonlinear analysis with imperfections).

2nd order effects and both member and frame imperfections are coveredin resulting internal forces and verification of individual compressionmembers is then for simple compression only, with reduction factors χ = 1.

This analysis is demanding for software, introduction of imperfections and evaluation.

Explanation:for

the effect ofnonlinearity isnegligible:

design FEdfor givenloading

φe0(i)

e0(i)e0(i)

Fcr

δFEd

somewhere here is a resistance

10Ed

crcr ≥=

FFα


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Instead of such simplified GNIA a simplified analysis is possible for 3 ≤ αcr< 10: To solve a frame with sway mode of buckling the 1st order theory may be used, but with all horizontal forces HEd (including imperfections VEd φ) increased by 2nd

order factor:

hcr ≤ h

~ Lcr In this case the stability check suppose fictive sway support, as global sway buckling is covered by the 2nd

order analysis with sway imperfection.

Note: for negligible rafter slope (or horizontal rafter, butin theory for 15º) is Lcr equal to spacing of columns. <≈

b) Geometrically nonlinear analysis with frame imperfection only(i.e. by introduction of initial sway – but commonly with equivalent horizontal forces).

Individual members are then verified for buckling length equal to system length (i.e. between the frame nodes). Thus the 2nd order of frame effectsare covered by the initial sway and of member effects by the relevant member reduction factors.E.g. for the frame column the buckling length h (possibly with reduction for built-in end), forframe rafter L/2.

111

1

cr

≥−α


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c) Frequently (classical method) is used 1st order theory without any imperfections and members are checked with equivalent global buckling lengths (using relevant reduction coefficients χ):

Global buckling length may be determined from critical loading corresponding to buckling of the analysed member and depends on rigidity of the frame members and loading (see next page) !!Nevertheless, global buckling length are given in formulas and tables in many references.

E.g. for frame columns the global buckling length may be of many-multiple system length. For rafters the same procedure may be used, or to take system length and increase moments from horizontal loading for about 20%:

hcr = β h

Lcr≥ L/2

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

ensure stabilityof free flange ! !

given in many references


Typical global buckling lengths (for sway buckling mode):

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Preferably the global buckling length may be determined from critical loading Ncr by common software from corresponding αcr,i (corresponding to buckled member i ) as follows:

F FFF

rafter

rafter

(NEd,i is compression of the relevant member iunder given loading FEd and E Ii its rigidity).

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

equals the system length.2) When using αcr,min (valid in general for buckling of another member) the result will be safe, but

uneconomical. 3) Mind the modification of cross sections after check: results in different αcr and hence also Lcr.

Valid for symmetricalloading

2h ÷ ∞h ÷ 2h

if Iraft = ∞

if Iraft = ∞if Iraft = 0

if Iraft = 0

iEd,icr,

i2

icr,

i2

icr, NE

NEL

αππ II

==


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TorsionCommon is elastic solution. Plastic solution (with rectangular stress distribution - see Strelbickaja, Trahair).

Eurocode combine elastic torsion with plastic bending.Distinguish:

- simple torsion: arise shear stresses only,- warping torsion: arise both shear and direct (normal) stresses.

1. Open cross sections (e.g. I, U, L)a) Simple (Saint Venant’s) torsion (occurs only exceptionally, see later)

τt

τiTt

t

bi = 1

i = 2

i = 3

Only shear arises (max. in tmax):

(influence of rounding of rolled sections, otherwise = 1)

Torsion constant:3

/ 0My(i)

t

tt(i)

γτ

ft

IT

≤=

∑=i

3iit 3

1 tbI αbt

T2t

max 3≈τ


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b) Warping torsion (acc. to Vlasov’s theory)Assumptions: 1. Rigid cross section (right angles remain right).2. Null shear deformation (i.e. shear lag is ignored).

- One part of a torsion moment T is transmitted by simple torsion: Tt- Other part by bending torsion: Tw- In total: T = Tt + Tw

T

S shear centre(bending centre)

σwτt τw

= +++

+

-

-

in warping torsion everythingis related to central line

Internal forces: Tt Tw Bmoment of simple

torsionmoment of bending

torsionbimoment

bending torsion

cross-sectionwarping


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Shear stresses:• simple torsion

• bending torsion

e.g. shear stressesin upper flange of I :

Direct stresses:

Stresses in torsion:

sectorial section modulus

first sectorial moment

warping constant

t

τw,max τt

t

stress through thickness t

+ =

total

T

Note: Neglecting rigidity in simple torsion ( ), the bending analogy for bending part of torsion may be used (differential equation of bending and bending torsion is the same):Bending: vertical uniform loading q Bending torsion: uniform torsion loading m = q eBending moment M (např. 1/8 qL2) Bimoment B (analog. 1/8 mL2)Direct stress σ = M/W Direct stress σw = B/WwShear force V → τ =VS/(tI ) Moment of bending torsion Tw → τw=TwSw/(tIw )Deflection δ (např. 5/384·qL4/(EI) ) Torsion angle θ (např. 5/384·mL4/(EIw) Rotation δ’ Warping θ’

w

www It

ST=τ

www W

BwB==

Iσ

0t ≈τ


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Warping characteristics

• For rolled sections in tables.

• In general from sectorial coordinate:

e.g. cross-section I:

U cross section:

a = a

w SwIn this positionno torsion !!

GS

t

h

b

wSw

rG ≡ S

z

The first sectorial moment for I :

z

2

A

2w 4

dA II hw == ∫

The second sectorial moment for I :(warping constant)

w, Sw, Iw ... see tables

The main sectorial coordinate for I :

Position of shear centre S:

(sectorial product of inertia)

Position of main centre w: 0dA1A

=∫ w

4ds

s

hbrw =∫=

16dA

2

Aw

tbhwS =∫=

0dAdAAA

=∫=∫ wzwy


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Determination of internal forces due to torsion:• Solution of Vlasov’s differential equations, or directly from relevant formulas.• Based on „bending analogy“ (see page 13).• Acc. to Eurocode (EN1993-1-1 Czech NA: see Complementary notes) as follows:

Distribution of torsion internal forces:

Superposition for more complex loading is necessary: - e3e2

e1

Simple support in torsion (= couple of forces).If warping is zero, bimoment is not zero (requirese.g. closed stiffeners at girder web).

F

Me

e

V

M

bending

torsion

T = Ve

B

A

Distribution of torsional moment due to eccentrical force corresponds to distribution of transverse force ateccentricity.

Part transmitted by simple torsion is set aside:

κ … see table of Eurocode Czech NA

F

κeVT ≅t

( )κ−≅ 1w eVT

( )κ−≅ 1eMB


moreover, torsion is usually restricted by cladding rigidity ⇒ torsion may often be ignored.

analysis about original shear centre Sis uneconomical !!! S

S ≡ V e

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Important notes:1. Direct stresses are big, they can not be ignored!!2. Direct stresses (from bending torsion) do not arise:

a) For loading by shear stresses τt only (roughly also due to end T torsion (simple torsion arises only):

b) In sections composed of radial outstands(because w = Iw = 0):

3. In practice usually occurs „torsion about enforced axis“ (V):

Simplified (conservative) solution – neglects simple torsion:

bending of flanges only = bimoment(often adequate: conservative for σw)

T0

T T/h

T/h

h

T T

(shear centre S is in cross point)


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a) Simple torsion (shear stresses only, usual for design)

τt tidi

TtBredt’s shear flow (τt t ) = const.

= 2 As

As(area enclosedby centre line)

Torsion constant:

17

2. Closed sections

Contrary to open cross section the maximal shear τtis in the thinnest plate and along thickness constant !!

b) Warping torsion: - Umanskij’s theory (uses rigid cross section),- Vlasov’s theory (uses non-rigid cross section),- FEM (including influence of cross section distortion, gives

also transversal bending moments in all plates).

The stresses are the same as in open cross sections: τt, τw, σw. However, τw, σw are very small, commonly ignored even for bridges.

(i)t(i) t

Tt

Ωτ =

∫=

tds

2

tΩI


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Interaction of bending and torsion ( My + T )e

Direct stresses (open cross sections only):

M1ywMy γσσσ /f≤+=

M1yw

Ed

yLT

Edy, γχ

/fWB

WM

≤+

Shear stresses:

i.e.

( ) ( ) Rdpl,My

Edw,

M0y

Edt,RdT,pl, V

//f//f,V

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

0332511

γτ

γτ

1≤Rdpl,T,

EdV

VVpl,T,Rd ...design plastic shear resistance of the cross section.

For open sections I and U

For closed sections ( ) Rdpl,My

Edt,RdT,pl, V

//fV

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

031

γτ in U sections only

In general, bending and torsion stresses may be summed and von Mises criterion applied:

132

000

22

0≤⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛

My

Ed

My

Edz,

My

Edx,

M0y

Edz,

My

Edx,

γτ

γσ

γσ

γσ

γσ

/f/f/f/f/f


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Complementary notes


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Internal forces due to torsion according to the EN1993-1-1 (Czech NA)Bimoment, moment of simple torsion and moment of bending torsion:

BEd = MEd e (1 – κ)Tt,Ed = VEd e κTw,Ed = VEd e (1 – κ)

where correcting factor: κ = 1/[ β + ( α / K t )2 ]and dimensionless factor: Kt = L (GIT / EI w)0,5

The coefficients α, β according to type of loading and boundary conditions:

Boundary torsion conditions Torsion loading α β

Girder (both sides)

simple supports(free warping)

Full uniform 3.1 1.00

General 3.7 1.08

built-in support(zero warping)

Full uniform For internal forces in support 8.0 1.25

For maximum within span 5.6 1.00

General 6.9 1.14

Cantilever built-in support General (for internal forces in support) 2.7 1.11

2CO8 Advanced design of steel and composite structures

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