Es95d Flexural Buckling II - Session 7

7/29/2019 Es95d Flexural Buckling II - Session 7

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ES95D STRUCTURAL ENGINEERINGES95D STRUCTURAL ENGINEERING

A conversion moduleA conversion module for nonfor non--civilcivil

een ineerinn ineerin raduatesraduates

Session 2-7: Compression members II

MSc Tunnelling and Underground Space

Option (15 credits)

Slide 2-7-2

Steel StructuresSteel Structures

COMPRESSION MEMBERS

Session 2-7

Illustrative practical examples

Stocky columns fail because of the cross-section resistance (Slides 2-6-3 & 2-

6-5). For slender members the compression resistance will be governed by

the global buckling resistance (Slides 2-7-4 to 2-7-8)

where is the buckling reduction factor and is 1. (see Figures 2-6-5 and 2-7-3)

Design buckling resistance of compression member = x Cross-section resistance


2/15

Slide 2-7-3

Curve fitting to Experimental EvidenceCurve fitting to Experimental Evidence

EHS minor axis buckling

0.8

1.0

1.2

ctor

Material yielding

CHS

EHS major axis buckling

FE

0.2

0.4

0.6

Reductionfa

EC 3

(Curve a) Elastic buckling

AISC 360

AS 4100

.

0.0 0.4 0.8 1.2 1.6 2.0 2.4

Non-dimensional slenderness

cr

y

cr

y

N

Af

N

N Figure 2-7.3

Slide 2-7-4

Design Process to BS EN1993Design Process to BS EN1993--11--1:20051:2005

6.3.1 Uniform members in compression*

6.3.1.1 Buckling resistance

compress on mem er s ou e ver e aga ns uc ng as o ows:

(6.46)

where NEd is the design value of the compression force and Nb,Rd is the design

buckling resistance of the compression member.

01Rdb,

Ed .N

N

(2) We shall not consider non-symmetric Class 4 sections but you need to

recognise thatthey require a specific design process.

* From BS EN1993-1-1:2005


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Slide 2-7-5


(3) The design buckling resistance of a compression member should be taken

as:

for Class 1, 2 and 3 cross-sections * 6.47y

Rdb, fA

N

for Class 4 cross-sections (6.48)

Where is the reduction factor for the relevant buckling mode.4 In determinin A and A holes for fasteners at the column ends need not

M1

M1

yeffRdb,

fAN

be taken in account.

* For the member check, with M1 = 1.0.

Slide 2-7-6


6.3.1.3 Buckling curves

(1) For axial compression in members the values of for the appropriatenon-dimensional slenderness may be determined from

but 1.0 (6.49)

where ,

for Class1, 2 and 3 cross-section

22

220150 ..y

N

Af is an imperfection factorNcr is the elastic critical force for the relevant buckling mode

based on the gross sectional properties.


4/15

Slide 2-7-7


(2) The imperfection factor corresponding to the appropriate buckling

curve should be obtained from Table 6.1 and Table 6.2 (slide 2-7-8).

Table 6.1: Imperfection factors for buckling curves*

(3) Values of the reduction factor for the appropriate non-dimensionalslenderness may be obtained from Figure 6.4. (see Slide 2-7-9 and Page 12 of

Buckling curve ao a b c d

Imperfection factor 0.13 0.21 0.34 0.49 0.76

* Takes account of geometric imperfections, residual stresses,

thickness of plate, yield strength refer to Session 2-6

x rac s rom or es gn o ee ruc ures ava a e w exam na on paper

(4) For slenderness 0.2 or forNEd/Ncr 0.04 the buckling effects may beignored and cross-sectional checks apply. (see EC3 Figure 2-6.4 in Slide 2-7-9)

Slide 2-7-8


Table 6.2: Selection of buckling curve for a cross-section

* Extract from BS EN1993-1-1:2005


5/15

Slide 2-7-9


1.0

1.2

0.2

040Ed .N

0.2

0.4

0.6

.

Reductionfactor ao

a

b

c

d

cr

(6.3.1.2(4))

* As given on Page 59 of BS EN1993-1-1:2005

Figure 6.4: Buckling curves *

0.0

0 0.5 1 1.5 2 2.5 3

non-dimensional slenderness 0.2

Slide 2-7-10

Design Process to BS EN1993Design Process to BS EN1993--11--1:20051:20056.3.1.3 Slenderness for flexural buckling *

(1) The non-dimensional slenderness is given by:

for Class 1, 2 and 3 cross-sections (6.50)

yfA cr 1L

where Lcr is the buckling length in the buckling plane considered (see

Slides 2-7-11 to 2-7-13)

i is the radius of gyration about the relevant axis, determined

using the properties of the gross cross-section

crN 1

E 235

* From BS EN1993-1-1:2005 - This approach corresponds to

current UK practice.

and (see Slide 2-5-10 and Table 5.2 *)

(2) For flexural buckling the appropriate buckling curve should be

determined from Table 6.2. (see Slide 2-7-8)

y

1.

f yf


6/15

Slide 2-7-11

Buckling lengthBuckling length LLcrcrEnd restraint (in the plane under consideration) Buckling

length, Lcr

Effectively held in positionat both ends

Effectively restrained in direction at bothends

0.7L

Partially restrained in direction at both ends 0.85L

Restrained in directions at one end 0.85L

Not restrained in direction at either end 1.0L

One end Other end Bucklinglength, Lcr

Effectively held in positionand restrained in direction

Not held inposition

Effectively restrained indirection

1.2L

Partially restrained indirection

1.5L

.

NominalbucklinglengthsL

cr in the bucklingplaneconsidered for compression members.Note: L is the system length and should be taken as the distance between the points ofeffectiverestraint on eachaxis.Lcr(or LE) is also referred to as theeffective length.

* Table 22 from BS 5950-1:2000 is NOT in BS EN1993-1-1:2005

It is on Page 6 of the Extracts from EC3 for Design of Steel

Structures .

Slide 2-7-12

Lcras reported in Table 22 of BS 5950-1:2000 *

Buckling lengthBuckling length LLcrcr

Commonestcase

* This approach corresponds to current UK practice.


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Slide 2-7-13

Buckling lengthBuckling length LLcrcrLcras reported in Table 22 of BS 5950-1:2000 *

* This approach corresponds to current UK practice.

Slide 2-7-14

A compression member AB of grade S235 is

simply supported about the major and minor

principal axes at each end (i.e. Lcr,y = Lcr,z = 6.0

- -

y z

z

NEd

Illustrative Practical ExampleIllustrative Practical Example -- II

A, .

adequacy of the member if the steel section is

457 191 89 UKB for a factored design axialcompression load NEd of 550 kN.

6 m

y

y z

z

N

B

tf

b

r

z

Compression member ABh

tw

c=cf

c=cw

d

tf

z

y y

Figure 2-7-14


8/15

Slide 2-7-15

Dimensions and properties:

Depth of cross-section h = 463.4 mm

Width of cross-section b = 191.9 mm

Depth of straight portion of web d= 407.6 mm

Web thickness tw = 10.5 mm

Member 457 x 191 x 89 UKBMember 457 x 191 x 89 UKB

ange c ness f= . mm

Radius of root fillet r= 10.2 mm

Second moment of area/zz Iz = 2089 cm4

Minor-axis (z-z) radius of gyration iz= 4.29 cm

Cross-sectional area A = 114 cm2

Modulus of elasticity E= 210000 N/mm2 3.2.6(1)

Assessment: Same displacement boundary conditions for flexure about y-y

(major) and z-z (minor) principal axes. Because Lcr = Lcr,y = Lcr,z bucklingfailure for flexure about the weaker minor axis will govern design.

* See Sections Tables. Dimensions are to BS4 specification.

For section classification the coefficient depending on fy is

Outstand flanges: flange under uniform compression

cf= = = 80.5 mm

Slide 2-7-16

cf

Classification of CrossClassification of Cross--sectionsection

01235

235235

y

.f

2

2-- w rtb 2

21025109191 ...

cw

Width-to-thickness ratio = 4.55.

With this ratio < 9 [= 9] (Slide 2-5-11), the flange outstand in compression is Class 1.Internal compression part: web under pure compression. Width cw is section dimension

d= 407.6 mm


717

580

f

f

.

.

t

c

6407w .c t

wWith this ratio < 42 [= 42] the web part (Slide 2-5-10), subjectto compression, is Class 3. Class of the section is the highest class (i.e. the least

favourable between the flange and web). Section 459 x 191 x 89 UKB, S235 is Class 3. *

w.

* under uniform compression, the cross-section is fully

effective


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6.2.4 Compression

(1) The design value of the compression force NEd at each cross-section

should satisfy: (6.9)

Slide 2-7-17

CrossCross--section Resistance,section Resistance, NNc,Rdc,Rd

01Ed .N

N

Given that the section is Class 3 (Slide 2-7-16) we determine Nc,Rd using

Equ. (6.10) (Slide 2-6-3)

= 2679 kNM0

yRdc,

AfN

,

1.0

23510114 2

Cross-section resistance is okay.20502679Rdc,

Ed .N

Slide 2-7-18

Design Buckling Resistance,Design Buckling Resistance, NNb,Rdb,RdStep 1: Determine (for buckling about minor (z-z) axis): (Slide 2-7-10)

= = 1.49 > 0.2 (member check is required)

z

9931

z

zcr,z

.i

L1993

1

42.9

6000

.Step 2: Determine z: (refer to process in Slide 2-7-6 and 2-7-7 for 6.3.1.2)For rolled sections we use

Where the imperfection factor depends on the (geometric) ratio h/b.

For 457 x 191 x 89 UKB we have = 2.07.

*

220150 ..9191

4463

.

.

b

h . - - .

f

buckling curve is b, and so = 0.34 (Table 6.1** (Slide 2-7-7)). Therefore

z = = 1.828

* Page 6 and **page 5 in Extracts from EC3 for Design of Steel

Structures.

24891204891340150 .....


10/15


11/15


12/15

Slide 2-7-23

CrossCross--section Resistance,section Resistance, NNc,Rdc,RdStep 2 contd.

Reduction factor for flexural buckling is given by

= = 0.188 < 1.022

1

22z1

Calculate value to z is much lower than initial guess of 0.5, must not be agood choice!

Second section choose: Let z = (0.5 + 0.188)/2 = 0.344A (550 103) / (0.344 355) = 4500 mm2 (45 cm2).

... zzz

From section table, try a 203 203 x 46 UKC with geometric propertiesh = 203.2 mm, b = 203.6 mm, A = 58.7 cm2, iz = 5.13 cm, tf= 11.0 mm,

tw = 7.2 mm, r= 10.2 mm.

Slide 2-7-24

Classification of CrossClassification of Cross--sectionsection

For section classification the coefficient depending on fy is

Outstand flanges: flange under uniform compression

cf= = = 88 mm

8140335

235235

y

.f

2-- w rtb 2102276203 ...

Width-to-thickness ratio = 8.

With this ratio < 10 [= 8.14] (Slide 2-5-11), the flange outstand in compression is Class 2.Internal compression part: web under pure compression. Width cw is section dimension

d= (h 2tf 2r) = 203.2 2x11.0 2x10.2) = 160.8 mm


011

88

f

f

.t

c

8160w .c

With this ratio < 33 [= 26.9] the web part (Slide 2-5-10), subject to compression, is Class1. Class of the section is the highest class (i.e. the least favourable between the flange

and web). Section 203 x 203 x 46 UKC , S355 is Class 2. (Fully effective)

27w .t


13/15


14/15

Slide 2-7-27

Example QuestionsExample Questions

1(a). For a pin-ended axially loaded strut, derive from firstprinciples the expression for Euler buckling load, Ncr

= 2EI/Lcr2 where E is the modulus of elasticity, I is

z

buckling axis and Lcr is the buckling (effective) length

in the buckling plane considered. (9 marks)

1(b). The column shown in Figure 2-7-27 is a simply

supported pin-ended column of 4 m height. It is

subjected to a factored design axial compression

load N of 200 kN. The column is he 152 152 30

Column

y

4 m

z

y z

UKC of grade S235. Check the adequacy of the

column in terms of section resistance and member

resistance. The section is Class 1. (8 marks)

yz

Figure 2-7-27

Slide 2-7-28


2. A 203 203 46 UKC in S275 steel is used as a column of length 5 m in a frame ofsimple construction. Determine using Clause 6.3.1 of EC3 (Extracts from EC3) the

design buckling resistance Nb,Rd given that:

there are no end moments.

the end restraints (in the plane under consideration) are effectively held in

position at both ends and partially restrained in direction at both ends.

(Ans: Nb,Rd,z = 916 kN)

3. A circular hollow section (CHS) member is to be used as an internal column in a

multi-storey building. The column has pinned boundary conditions at each end and

the inter-storey height is 4 m. The axial force due to the effect of actions is NEd =

1630 kN. Assess the suitability of a hot-rolled 244.5 19 CHS in grade S275 steelfor this column application. Section properties for this hot-rolled Class 1 sectionare: diameterd = 244.5 mm, wall thickness t = 10.0 mm, Wel,y = 425 cm

3 and Wpl,y =

550 cm3. (Hints: Engineering Databook for Second Moment of Area for CHS, Ans:

Buckling resistance Nb,Rd is acceptable.)


15/15

Slide 2-7-29


4. Assess the design buckling resistance of the compression member 203 203 x 52UKC section in grade S275 steel with the following system lengths. For buckling

about the major (y-y) axis, the system length, L is 6 m while a buckling length,

L due o end restraints has been assessed b he desi ner as 0.85L. For minor,

(z-z) axis buckling, the member is restrained by secondary beams at mid height

and the half length assessed by the designer to be pin ended, i.e. Lcr,z = 3 m.

(Ans: Nb,Rd,y = 1469 kN, Nb,Rd,z = 1356 kN; thereby Nb,Rd = Nb,Rd,z =1356 kN.)

Slide 2-7-30

End of Session 2End of Session 2--77

J. T. Mottram 2012

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Es95d Flexural Buckling II - Session 7

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