9- Ctr 11119 - Notes - Structural Steelwork - Frame Stability - EC3 (1)

M.Sc. Advanced Structural Engineering

Lecture 9Frame Stability - 2nd order effects in sway framesFrame Stability 2 order effects in sway frames

Dr. W.M.C. McKenzie 1

Sway frames – 2nd order effects

V

Consider a cantilever with horizontal and vertical loads applied at the free end:

VP

δ1 V

P

δ2V

PP

h

1st order effects: 2nd order effects:Applied load onδ1 due to P aloneM = (P × h)

δ2 due to P and VM = (P × h) + (V × δ2)

Applied load on the cantilever

2Dr. W.M.C. McKenzie

Figure 9.1

Sway frames – 2nd order effects (cont.)Second-order effects apply to both isolated members and to complete

structures.structures.

Local bow effect in isolated member Global effect due to sway of frame

Fi 9 2Figure 9.2Clause 5.2.1(2)EC3-1-1

‘The effects of the deformed geometry (second-order effects) should be consideredThe effects of the deformed geometry (second-order effects) should be considered

if they increase the action effects significantly or modify significantly the structural

b h i ’


behaviour.’

Sway frames: 2nd order effects – Elastic methods of analysis

The in-plane second-order effects can be allowed for by using a first-order analysis

in one of the following three methods:in one of the following three methods:

1. the amplified sway moment method,

2 the iterative method or2. the iterative method or

3. the sway-mode buckling method.

Method 1 is the simplest and most straight for ard method the principles of hichMethod 1 is the simplest and most straight forward method the principles of which

are given in Clause 5.2.2 of EN 1993-1-1.

h i f i i i i f i i b kli i d h ldThe imperfections existing in frames sensitive to buckling in a sway mode should

be allowed for in a frame analysis by means of ‘equivalent imperfections’ as

follows:

1. relative initial local bow imperfections of members for flexural buckling and

2. global initial sway imperfections.


Sway frames: 2nd order effects – local bow imperfections

Clause 5.2.2(3)(b)EC3-1-1

The relative local bow imperfections of members for flexural buckling can be

estimated from:estimated from:

e0/L (EN 1993-1-1: Equation 5.6)

where L is the member length

Notes:

The value of e0/L may be given in The National Annex, recommended values are

given in Table 5.1.given in Table 5.1.

It is unlikely that sway frame columns will be sufficiently slender to require

id ti f l l b i f ticonsideration of local bow imperfections.


Sway frames: 2nd order effects – local bow imperfections (cont.)

Clause 5.3.2 (6)EC3-1-1

The sensitivity of a frame to local bow imperfections is determined in accordance

with Clause 5.3.2(6) of EN 1999-1-1, i.e.

Local bow imperfections should be considered when:

1 at least one moment resisting joint is present at one end and when1. at least one moment resisting joint is present at one end and when

2. (EN 1993-1-1: Equation 5.8)y

Ed

0,5AfN

λ >

where:

NEd is the design value of the compression force,

Ed

is the in-plane non-dimensional slenderness calculated for the member

considered as hinged at its ends.λ g


Sway frames: 2nd order effects- equivalent horizontal forces (cont.)

Initial bow imperfections can be represented by equivalent horizontal forces asindicated in Clause 5.3.2.(7)

NEd NEd

wL/2Clause 5.3.2(7)EC3-1-1

Consider the moment at the mid-height

position:position:

NEde0,d = − (w × L/2 × L/4) + (w/2 × L/2)2/8

e0,d w kN/mL

= + wL2/8

∴w = (8NEde0,d)/L2

and

wL/2 = (4NEde0,d)/LN NEd

WL/2

Figure 9.3NEd

Ed


Sway frames: 2nd order effects - sway imperfections

Clause 5.3.2(4)BEC3-1-1

For building frames sway imperfections may be disregarded where:

Clause 5.3.2(4)B

HEd ≥ 0,15VEd (EN 1993-1-1: Equation 5.7)

where:where:

HEd is the design value of the horizontal reaction at the bottom of the storey to the

horizontal loads and fictitious horizontal loads,

VEd is the total design vertical load on the structure on the bottom of the storey.


Sway frames: 2nd order effects- sensitivity to sway

Clause 5.2.1(3)EC3-1-1

First-order analysis may be used if the following criterion is satisfied:

For elastic analysis - αcr = 10crF≥For elastic analysis αcr

For plastic analysis - αcr =EdF

15cr

Ed

FF

≥(EN 1993-1-1: Equation 5.1)

where:

αcr is the factor by which the design loading would have to be increased to cause

EdF

αcr is the factor by which the design loading would have to be increased to cause

elastic instability in a global mode,

F i th d i l d th t t dFEd is the design load on the structure and

Fcr is the elastic critical buckling load for the global instability mode based on

initial elastic stiffness.


Sway frames: 2nd order effects- sensitivity to sway (cont.)

Portal frames with shallow roof slopes (i.e. < 26°) and beam-and-column type

plane frames in buildings may be checked for sway mode failure with first orderplane frames in buildings may be checked for sway mode failure with first order

analysis if the criteria given in Equation (5.1) are satisfied for each storey.

I th b i t d f hEdH h⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟In these cases αcr may be approximated from: αcr = where:

HEd is the design value of the horizontal reaction at the bottom of the storey to

Ed

Ed H,EdV δ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

the horizontal loads and fictitious horizontal loads,

VEd is the total design vertical load on the structure on the bottom of the storey,

δH,Ed is the horizontal displacement at the top of the storey, relative to the bottom

of the storey, when the frame is loaded with (horizontal loads) and fictitiousof the storey, when the frame is loaded with (horizontal loads) and fictitious

horizontal loads which are applied at each floor level,

h i th t h i ht (Neglect: see NCCI SN004 in (access steel))h is the storey height (Neglect: see NCCI - SN004 in (access steel))



NCCI (SN004a-EN-EU): (EC3: Equation (5.2))Edcr

Ed H,Ed

H hV

αδ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

This NCCI from www.access-steel.com sets out the basis for the calculationof α Three alternatives are given for the direct application of

,⎝ ⎠⎝ ⎠

of αcrit. Three alternatives are given for the direct application ofEquation (5.2) as given in EN 1993-1-1.

i l l d ( ) δ δ δδP4

H = P

Horizontal loads (P1 to P4)(e.g. wind + fictitious loads)

δH,Ed = δ3 − δ2

δH,Ed = δ4 − δ3

δ3

δ4

P3

P4

HEd,3 = P4

HEd,2 = P4 + P3δH,Ed = δ2 − δ1δ2

P2

HEd,1 = P4 + P3 + P2 δH,Ed = δ1δ1P1

h


HEd,0 = P4 + P3 + P2 + P1 Figure 9.4


1. Use Equation (5.2) with HEd determined by the fictitious horizontal

loads from the initial s a imperfections in 5 3 2(7) alone and ithloads from the initial sway imperfections in 5.3.2(7) alone and with

δH,Ed as the displacements arising from these fictitious horizontal loads

(i.e. exclude the effects of any other horizontal loads, such as wind

loads).

2. Calculate αcr by computer by finding the first sway-mode from an

i l l i Wh i thi t f l i it i i t t teigenvalue analysis. When using this type of analysis, it is important to

study the form of each buckling mode to see if it is a frame mode or a

local column mode.

3. Fcr may be found from design charts appropriate to the type of building


3. cr ay be ou d o des g c a ts app op ate to t e type o bu d g

considered.

Sway frames: 2nd order effects- global initial sway imperfections

Clause 5.3.2(3)(a)EC3-1-1

The global initial sway imperfections are estimated by assuming an initial sway of

h f d φ h b lthe frame due to φ as shown below:

h hh

φ φφ

Figure 9 5


Figure 9.5

Sway frames: 2nd order effects- global initial sway imperfections

The global initial sway imperfections due to the sway angle φ are estimated from:

φ φ (EN 1993 1 1 E ti 5 5)φ = φ0αhαm (EN 1993-1-1: Equation 5.5)

where:

φ0 is the basic value: φ0 = 1/200,

αh is the reduction factor for height h applicable to columns:h g pp

where αh = but

h i h l h i h f h i

2h h

2 1,03

α≤ ≤

h is the total height of the structure in metres,

αm is the reduction factor for the number of columns in a row =10,5 1m

⎛ ⎞+⎜ ⎟⎝ ⎠

m is the number of columns in a row including only those columns which carry a

vertical load NEd ≥ 50% of the average of the column in the vertical plane

⎝ ⎠

Ed g p

considered.14Dr. W.M.C. McKenzie

Sway frames: 2nd order effects- equivalent horizontal forces

Initial sway imperfections can be represented by equivalent horizontal forces asindicated in Clause 5.3.2.(7)

NEd NEd

PxClause 5.3.2(7)EC3-1-1

x = φ h

P

Consider rotational force about the base: h

NEd x = NEd φh = Ph

∴ P = φNEdφ

N N

P

Figure 9.6NEd NEd


Methods of Global Structural Analysis: Sway imperfections

1. For unbraced frames with low slenderness, i.e.10 0 for elastic analysisα ≥

1st order analysis only is required.cr 10,0 for elastic analysisα ≥

2. For unbraced frames with moderate slenderness, i.e.cr10,0 3,0 for elastic analysisα> ≥

1st order analysis combined with an amplifying factor for horizontal

loads equal to may be used.11 1⎛ ⎞

−⎜ ⎟loads equal to may be used.

3. For unbraced frames with high slenderness, i.e.

cr1 1

α⎜ ⎟⎝ ⎠

A more accurate 2nd order analysis is required.cr 3,0 for elastic analysisα <


Sway frames: Example 9.1 – Unbraced, two-storey frame

Using the data given, determine the design loads required for a first-order analysis

which allows for second order sway-effects of the frame indicatedwhich allows for second order sway effects of the frame indicated.

3,5

mm

4,0

m

6,0 m 6,0 m


Figure 9.7

Sway frames: Example 9.1 – frame data

Outer columns: 203 x 133 x 30 UB

Second moment of area about major axis I = 2896 cm4Second moment of area about major axis Iyy 2896 cm

Gross Cross-sectional area Ag = 38,2 cm2

Inner columns: 254 x 102 x 28 UBInner columns: 254 x 102 x 28 UB

Second moment of area about major axis Iyy = 4005 cm4

G C ti l A 36 1 2Gross Cross-sectional area Ag = 36,1 cm2

Roof beams 406 x 178 x 67 UB



Floor beams: 457x 191 x 74 UB




Sway frames: Example 9.1 – design data

Frame spacing 8,0 m

Fl l dFloor loads

characteristic permanent load (gk,1) 5,65 kN/m2

characteristic variable load (qk,1) 2,5 kN/m2

Roof loads

characteristic permanent load (gk2) 5,0 kN/m2

characteristic variable load due to snow (qk,2) 1,5 kN/m2

Wind loads

windward side of building (wk,windward) 0,45 kN/m2,

leeward side of building (wk,leeward) 0,20 kN/m2

Assume the altitude of the site is greater than 1000 m above mean sea level.g


Sway frames: Example 9.1 – Solution

Fundamental Combination

(i.e. excluding accidental and seismic effects)

Clause 6.4.3.2(3)NA –EC0 EN 1993-1-1: Equation (6.10)

G,j k,j Q,1 k,1 Q,i 0,i k,i1 1j i

G Q Qγ γ γ ψ≥ >

+ +∑ ∑

Permanent Leading Accompanying

j

PermanentActions

LeadingVariableAction

AccompanyingVariableActions

Assume the wind load to be the leading variable action 20Dr. W.M.C. McKenzie

Sway frames: Example 9.1 – Solution (cont.)

Using Equation (6.10) G j k j Q 1 k 1 Q i 0 i k iG Q Qγ γ γ ψ+ +∑ ∑Combination Factors

G,j k,j Q,1 k,1 Q,i 0,i k,i1 1j i

Q Qγ γ γ ψ≥ >∑ ∑

Table NA.A1.2(b) - γG = 1,35

Table NA.A1.2(b) - γQ,f = 1,5

Table NA.A1.2(b) - γQ,s = 1,5

bl A A1 0

Table NA A1 ψ = 0 7

Table NA.A1. - ψ0,variable floor = 0,7

Table NA.A1. - ψ0,snow = 0,7



d G,j k,j Q,1 k,1 Q,i 0,i k,iF G Q Qγ γ γ ψ= + +∑ ∑1 1j i≥ >

gk 1 = permanent floor load = 5 65 kN/m2 γ 1 = 1 35gk,1 permanent floor load 5,65 kN/m γg,1 1,35

gk,2 = permanent roof load = 5,0 kN/m2 γg,2 = 1,35

qk,2 = variable floor load = 2,5 kN/m2 γq,2 = 1,5 ψq,2 = 0,7

i bl l d 1 5 kN/ 2 1 5 0 7qk,3 = variable snow load = 1,5 kN/m2 γq,3 = 1,5 ψq,3 = 0,7

Qk,1 = wwindward load = 0,45 kN/m2 γQ,1 = 1,5, Q,

Qk,2 = wleeward load = 0,20 kN/m2 γQ,1 = 1,5



Leading variable due to wind load

Design loads for frames at 8,0 m centres

Roof load:

Froof = [(1,35 × 5,0) + (1,5 × 0,7 × 1,5)] × 8,0 = 66,60 kN/m

Floor load:

Ffloor = [(1,35 × 5,65) + (1,5 × 0,7 × 2,5)] × 8,0 = 82,02 kN/m

Wind load:

Fwindward = [(1,5 × 0,45) × 8,0] = 5,40 kN/m

Fleeward = [(1,5 × 0,20) × 8,0] = 2,40 kN/m


Sway frames: Example 9.1 – Solution (cont.)Initial Design loads 66,60 kN/m

m3,

582,02 kN/m5,40 kN/m

wind

m

winddirection 2,40 kN/m

4,0

6,0 m 6,0 m


Figure 9.8

Figure 9 9


Figure 9.9


QSE Analysis -results

66,60 kN/m

3,5

m82,02 kN/m5,4 kN/m

i d

4,0

m

2,40 kN/m

winddirection

6,0 m 6,0 m

Figure 9.10

166 kN 169 kN465 kN

Figure 9.11365 kN 385 kN1034 kN



EC3-1-1

Check if local bow imperfection calculations are required

AfClause 5.3.2(6)/Equation 5.8EC3-1-1

yAf Af Af

y

Ed

0,5AfN

λ >

y

cr

AfN

λ = y y

cr Ed

0,5Af AfN N

∴ > Ed cr 0,25N N∴ >

Note: this is the in-plane non-dimensional slendernessOuter columns:

2 2 4

cr cr2 2 3

210000 2896 10 3751 kN 0,25 937,8 kN4000 10

EIN Nl

π π × × ×= = = =

×Inner columns: NEd = 365 kN < 0,25Ncr

2 2 4

cr cr2 2 3

210000 4005 10 5188 kN 0,25 1297 kNEIN Nl

π π × × ×= = = =cr cr2 2 3 ,

4000 10l ×NEd = 1034 kN < 0,25Ncr

Since Equation (5 8) is not satisfied local bow imperfections do not have to be


Since Equation (5.8) is not satisfied local bow imperfections do not have to be considered.


Check need for sway imperfection calculations

Clause 5.3.2(4)BEC3-1-1

when HEd ≥ 0,15VEd (EN 1993-1-1: Equation 5.7)

sway imperfections may be disregarded.sway imperfections may be disregarded.

where HEd and VEd are defined in Clause 5.2.1(4)B i.e.

HEd is the design value of the horizontal reaction at the bottom of the storey to

the horizontal loads.

VEd is the total design vertical load on the structure on the bottom of the

storey.y



Clause 5.3.2(4)BEC3-1-1

HEd = (Fwindward + Fleeward ) = [(5,4 + 2,4) × 7,5] = 58,50 kN

VEd = (Froof + Ffloor ) = [(66,60 + 82,02) × 12,0] = 1783,4 kNEd ( roof floor ) [( , , ) , ] ,

0,15VEd = (0,15 × 1783,4) = 267,51 kN (> HEd)

Since H < 0 15V sway imperfections must be consideredSince HEd < 0,15VEd sway imperfections must be considered.

Check Sensitivity to Sway: αcr

Clause 5.2.1(3)EC3-1-1

First-order analysis may be used if the following criteria is satisfied:

For elastic analysis - αcr = (EN 1993-1-1: Equation 5.1)cr 10FF

≥


EdF


Portal frames with shallow roof slopes (i.e. < 26°) and beam-and-column type

plane frames in buildings may be checked for sway mode failure with first order

analysis if the criteria given in Equation (5.1) is satisfied for each storey. In theseanalysis if the criteria given in Equation (5.1) is satisfied for each storey. In these

cases αcr may be approximated from:

αcr = (EN 1993-1-1: Equation 5.1)Ed

Ed H,Ed

H hV δ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

where:

δH,Ed is the horizontal displacement at the top of the storey, relative to the bottom

of the storey, when the frame is loaded with horizontal loads and fictitious

horizontal loads which are applied at each floor level,pp


Sway frames: Example 9.1 – Solution (cont.)δH,Ed can be determined from a frame analysis using computer software , e.g. QSE

or equivalent. In this case the analysis is carried out using the equivalent horizontalq y g q

loads and the design vertical loads as shown.

EC3 1 1 φ = φ0 αh αm (EN 1993-1-1: Equation 5.5)Clause 5.3.2(3)EC3-1-1

EC3 1 1 66,6 kN/m

H = φNEd

Figure 5.4EC3-1-1

5 m

,

82 02 kN/m

Hroof

φ Ed

3,582,02 kN/m

Hfloor

4,0

m

Figure 9.126,0 m 6,0 m


Figure 9.12


φ = φ0αhαm (EN 1993-1-1: Equation 5.5)

αm is dependent on the number of columns in a row including only those which

carry a vertical load NEd not less than 50% of the average value of the column in they Ed g

vertical plane considered.

NEd,average = [(66,60 + 82,02) × 12]/3

595 20 kN

166 kN 465 kN 169 kN

= 595,20 kN

50% NEd,average = 297,60 kN 365 kN 1034 kN 385 kN

Consider 3 columns for αm


Figure 9.13


φ = φ0αhαm (EN 1993-1-1: Equation 5.5)

φ0 = 1/200,

α = but2 2 0,730= = h

2 1,0α≤ ≤αh = but,7,5h h ,

3

1 10 5 1 0 5 1 0 816⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟αm =

The vertical load N ≥ 50% of the average of the column in the vertical plane

0,5 1 0,5 1 0,8163m

+ = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The vertical load NEd ≥ 50% of the average of the column in the vertical planeconsidered.

φ = φ α α = (0 005 × 0 730 × 0 816) = 0 003φ = φ0αhαm = (0,005 × 0,730 × 0,816) = 0,003


Sway frames: Example 9.1 – Solution (cont.)Equivalent horizontal forces due to sway imperfections:

H φNFi 5 4EC3-1-1

Hroof = [φ × (qroof × L)] = [0,003 × (66,60 × 12,0)] = 2,40 kN

H = φNEdFigure 5.4EC3-1-1

roof [φ (qroof )] [ , ( , , )] ,

Hfloor = [φ × (qfloor × L)] = [0,003 × (82,02 × 12,0)] = 3,05 kN

Portal frames with shallow roof slopes (i.e. < 26°) and beam-and-column type plane

frames in buildings may be checked for sway mode failure with first order analysisframes in buildings may be checked for sway mode failure with first order analysis

if the criteria given in Equation (5.1) is satisfied for each storey. In these cases αcr

may be approximated from: αcr =

(S Cl 5 2 2(6)B f lti t f )

Ed

Ed H,Ed

H hV δ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(See Clause 5.2.2(6)B for multi-storey frames)



QSE Analysis results for the horizontal displacement δH,Ed at each storey due to

equivalent horizontal loadsequivalent horizontal loads.

2,40 kN

3,5

m

3 05 kN

0 m

3,05 kN

4,0

6,0 m 6,0 m

Figure 9 14


Figure 9.14

Figure 9.15


g


Portal frames with shallow roof slopes (i.e. < 26°) and beam-and-column type plane

frames in buildings may be checked for sway mode failure with first order analysis

if the criteria given in Equation (5.1) is satisfied for each storey. In these cases αcrg q ( ) y cr

may be approximated from: αcr = Ed

Ed H,Ed

H hV δ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Relative horizontal displacements for each storey:

δH Ed f = (2 142 – 1 570) = 0 572 mm;δH,Ed,roof (2,142 1,570) 0,572 mm;

δH,Ed,floor = 1,570 mm

VEd,roof = (66,60 × 12) = 799,20 kN

VEd,floor = (82,02 × 12) = 984,24 kN



Upper storey:

α = = = 18 38 ≥ 10EdH h⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟

2,40 3500⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟αcr = = = 18,38 ≥ 10

Ed H,EdV δ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 799,20 0,572⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Lower storey:

αcr = =Ed,roof Ed,floor

Ed,roof Ed,floor H,Ed,floor

H H hV V δ

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

2,40 3,05 4000799,20 984,24 1,57

+⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

= 7,79 < 10

⎝ ⎠⎝ ⎠

If sway effects can be neglected.Clause 5.2.1(3)EC3-1-1 10crcr

Ed

FF

α = ≥

Since the minimum value of αcr = 7,79 i.e. < 10 sway effects cannot be neglected.



Clauses 5.2.2(5)B and (6)BEC3-1-1

The 2nd Order sway effects may be calculated according to 1st Order theory if the

horizontal loads i.e. wind and equivalent horizontal loads are increased by a factor

equal to provided that αcr ≥ 3,0.cr

11 1 α−

In this case αcr ≈ 7,79 and the method can be used.

Multiplying factor = = = 1,147 cr

11 1 α−

11 1 7,79−



Clauses 5.2.2(5)B and (6)BEC3-1-1

The following loads are considered for a 1st order analysis:

Froof = 66,60 kN/m

F = 82 02 kN/mFfloor = 82,02 kN/m

Fwindward = (1,147 × 5,4) = 6,19 kN/m

Fleeward = (1,147 × 2,4) = 2,75 kN/m

Hroof = (1,147 × 2,4) = 2,75 kN/m

Hfloor = (1,147 × 3,5) = 4,01 kN/m



66 60 kN/m

Design loads using 1st order analysis to allow for 2nd order effects

66,60 kN/m

2,75 kN

3,5

m

82,02 kN/m6,19 kN/m

2,75 kN/m

4,01 kN

4,0

m

6,0 m 6,0 m

41Dr. W.M.C. McKenzieFigure 9.16

Figure 9.17


Figure 9.18


Figure 9.19


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9- Ctr 11119 - Notes - Structural Steelwork - Frame Stability - EC3 (1)

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