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Mechanical Design 2

School of Aerospace, Mechanical and Manufacturing Engineering 1

MIET2072 2013/Ver. 1

MIET2072

Mechanical Design 2

Topic 8

College of Science, Engineering and Health Learning Package

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Topic 8: INTRODUCTION TO THE DESIGN OF

STEEL COLUMNS OR COMPRESSION

MEMBERS

Learning Outcomes

Upon successful completion of this topic, you will be able to:

•

determine the proportions of supporting columns including consideration ofbuckling, yield, eccentric load and initial crookedness.

Introduction to the Topic

Pressure vessels are supported in a number of ways, one of which is with lugsresting on columns.

Columns are a common part of engineering structures, the basic theory of whichyou have studied in the prior course “Solid Mechanics 3”.

Columns or struts i.e. structural members loaded compressively, differ fromtension loaded members in that they have the additional mode of failure ofbuckling. You have studied the basics of this in the prior course.

Real column design builds on this basic theory but has to add to it consideration ofimperfections such as the load being applied off centre and the column havingsome initial crookedness. Both of these increase the tendency of the column tobend.

Background Skills and Knowledge

Students will require the following:

•

Familiarity with basic column theory.

•

Familiarity with normal stress and bending stress.

• Familiarity with the solution of quadratic equations.

•

Familiarity with the solution of 2nd order differential equations.

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Acti vi ty 8A - Video of buckl ing

Please view the short video of buckling to reacquaint yourself with this extramode of failure that engineers must be conscious of when designing componentssubject to compressive loads.

Topic 8 Buckling video 1

Note: Right click (Google Chrome browser) and choose Save Video As… to download video

Acti vi ty 8B - AS3990 Mechanical Equipment- Steelwork

In your studies on this topic you will be making frequent reference to somepages of AS3990 Mechanical Equipment- Steelwork (previously known asAS1250 “SAA Steel Structures Code”, published by Standards Australia). Printout pages 20, 21, 22, 23, 40, 42, 54 and 55 and have them to hand while you arereading these notes. You will also need to refer briefly to AS1210 Pressure

Vessels.

This Australian Standard can be accessed via the RMIT Library - SAI Global linkto the Australian Standards.2

NOTE: You will need to login with your student login and password.

Acti vi ty 8C - Reading

Read all of Chapter 8 below.

1 https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic_8_Buckling.m4v2 https://login.ezproxy.lib.rmit.edu.au/login?url=http://www.saiglobal.com/online/autologin.asp

https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic_8_Buckling.m4v


https://login.ezproxy.lib.rmit.edu.au/login?url=http://www.saiglobal.com/online/autologin.asp







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Chapter 8: INTRODUCTION TO THE DESIGN OF

STEEL COLUMNS OR COMPRESSION

MEMBERS

Section 8.1 Concentrically Loaded Columns or CompressionMembers

The fundamental theory of columns as developed by Euler has been studied inSolid Mechanics 3, and can be found in most texts on the Mechanics of Solids. Somekey points of this theory will be revised below then consideration will be given tothe behaviour of real columns whose behaviour deviates from the ideal due toeffects such as unintended inaccuracy in the concentricity of the load, initialcrookedness of the column, inhomogeneity and residual stresses. An equation willbe developed that reasonably predicts the failure load for real columns, which, withthe application of a factor of safety, may be used for the design of “concentrically”loaded columns.

Consider now the two columns shown in Figures 8.1a & b, one squat and the otherslender, supported on pinned joints, both subjected to an axial compressive load ofP Newtons.

Fig. 8.1a and b, Axial compressive load on Columns © RMIT University, 2013 (Dixon C., Marchiori G.)

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If the slender column is given a temporary lateral load that deflects it sideways,creating a deflection y at height x as shown in Figure 8.1c(ii), then the columnexperiences a bending moment P x y at this point and generates within itself, due to

its elasticity, a resisting or restoring moment2

2

dx

yd EI − .

Fig. 8.1c Pin jointed column given temporary sideways deflection

© RMIT University, 2013 (Dixon C., Marchiori G.)

If the restoring moment is greater than the bending moment then the column willreturn to its original straightness, and it was therefore in stable equilibrium beforethe temporary disturbance. If the load P is such that the bending moment is equalto the restoring moment then the column will remain in the deflected position afterremoval of the temporary lateral load. This load is referred to as the critical loadPcrit. If the load P is > Pcrit then if the column is deflected sideways the bendingmoment P x y will be greater than the restoring moment and the column willcollapse by bending. Therefore the situation that existed before the temporary

sideways load was applied was one of unstable equilibrium.

The critical situation is therefore when:

Bending Moment Pcrit y =2

2

dx

yd EI − the restoring moment

∴ 2

2

dx

yd + 0= y

EI

Pcrit .

The solution to this differential equation is:

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y = x EI

PCosC x

EI

PSinC crit crit

21 +

The boundary conditions are y = 0 at x = 0 and at x = L, the first of which yieldsC 2=0 and the second gives:

01 = L EI

PSinC crit

This means that either (i) C 1=0, which is a trivial solution corresponding to therebeing no lateral displacement imposed (i.e. no temporary lateral load applied to teststability), or (ii) that:

0= L EI

PSin crit

∴

Π= n L EI Pcrit where n = 1,2,3 etc.

∴ Pcrit =2

22

L

EI n Π n = 1,2,3 etc.

The buckled shapes at two of the critical loads are shown below in Figure 8.1d.

Fig. 8.1d Buckled shape at two of the critical loads © RMIT University, 2013 (Dixon C, Marchiori G.)

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Therefore the lowest value of the critical load Pc is for n = 1

Pc =2

2

L

EI Π eqn. C1

So far it has been assumed that buckling took place in the plane of the page, andthis assumption would be met if there were lateral supports for the column normalto the page. However if there are no such lateral supports and it is only supportedat its ends, it is free to buckle in any direction, and will in fact bend about theprincipal centroidal axis having the lowest value of the second moment of area I .

The analysis above was for a column with pinned ends. Other end supports alsooccur such as built-in at one or both ends, one end free, etc. It is found for thesethat:

Pc =

( )

2

2

2

2

l

EI

Lk

EI

e

Π=

Π eqn. C2

where: l = the effective length = keL

and ke, the effective length factor, depends on the method of end support.

Activ ity 8D

Effective length factor determination

Step 1: View effective length factor for several basic column configurations.

Table 6.2 in the extract at the following e-reserve link shows the effective lengthfactor for several basic column configurations. The effective length of the columnis equal to the distance between points of inflexion of its buckled shape. Pleaserefer to Table 6.2 on page 120 of the Steel designers handbook (PDF 88KB). 3

Step 2: Consider pressure vessel support columns

Figures 3.24(a) and (b) in AS1210 show column supports on pressure vessels.Please go to AS1210 and look at those diagrams. Further column supportarrangements may also be seen in Figs 5.1, 2 and 3 on pages 144 of the Pressure

Vessel Design handbook by Bednar (PDF 39KB). 4

3 Gorenc, B. E. & Tinyou, R., 1981, Compression members, Steel designers handbook, 5th ed., New South WalesUniversity Press, Kensington, N.S.W., pp. 120. viewed on 28th August 2013 <https://equella.rmit.edu.au/rmit/file/817df72d-47bc-43c5-9cf2-5fe6c4db74c3/1/130823_3_063.pdf>

4 Bednar https://equella.rmit.edu.au/rmit/file/11437b2d-663d-4213-a869-

5ffd10b146b8/1/130821_3_036.pdf >

https://equella.rmit.edu.au/rmit/file/817df72d-47bc-43c5-9cf2-5fe6c4db74c3/1/130823_3_063.pdf



https://equella.rmit.edu.au/rmit/file/75996096-53fa-418e-84d7-a093735499b3/1/130821_3_035.pdf






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Step 3: Consider the effective length factor for pressure vessel supportcolumns.

With regard to the effective length factor of column supports on pressure

vessels, Bednar makes useful comments in the extract on pages 148-149 of thePressure vessel design handbook (PDF 122KB) 5 and concludes with a suggestedvalue of 1.5.

Step 4: Optional further reading

Detailed consideration of the effect on effective length of adjoining frameworkand connection type thereto is beyond the scope of this course. The interestedreader is referred to Appendix E of Australian Standard AS3990 “MechanicalEquipment – Steelwork”. This appendix applies in particular to columns withelastically restrained ends. The ratio of the stiffness of columns to the stiffness of

beams is computed at both ends of the column being considered. Furtherreading on the effect on effective length of adjoining framework and connectioncan be found in the following references.

• Trahair, N. S., 1977, The behaviour and design of steel structures, 1st ed.,Chapman and Hall, London

• Gorenc, B. E. & Tinyou, R., 1981, Compression members, Steel designershandbook, 5th ed., New South Wales University Press, Kensington,N.S.W., ISBN 0 86840 248 6. For example the extract 6.3.1 Interactionbetween the various component members on page 121 of the Steel

designers handbook (PDF 88KB) 6 gives an indication of some of theissues involved.

5 Bednar H 1986, Pressure vessel design handbook 2nd ed. Van Nostrand Reinhold, New York. p.148- 149,viewed on 28th August 2013 < https://equella.rmit.edu.au/rmit/file/11437b2d-663d-4213-a869-5ffd10b146b8/1/130821_3_036.pdf>

6 Gorenc, B. E. & Tinyou, R., 1981, Compression members, Steel designers handbook, 5th ed., New South WalesUniversity Press, Kensington, N.S.W., pp. 121. viewed on 28th August 2013 <

https://equella.rmit.edu.au/rmit/file/817df72d-47bc-43c5-9cf2-5fe6c4db74c3/1/130823_3_063.pdf>

https://equella.rmit.edu.au/rmit/file/11437b2d-663d-4213-a869-5ffd10b146b8/1/130821_3_036.pdf














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Returning to equation C2:2

2

l

EI Pc

Π=

The second moment of area I may be expressed as:

I=Ar 2 where A is the cross sectional area, and r is the radius of gyration. Substituting thisinto equation C2 gives:

Pc =( )2

2

r l

EAΠ

The elastic buckling stress, or Euler critical stress, A

PF c

oc = corresponding to this

elastic critical load is therefore:

F oc =( )2

2

r l

E Π eqn. C3

The termr

l is called the slenderness ratio.

For slender columns, i.e.r

l is large, the Euler critical stress is less than the yield

stress and lateral buckling is therefore of more concern than gross yielding.However for squat columns, such as is shown in Fig. 8.1a the greater concern isgross yielding that will occur when f ac, the average axial compressive stress, reaches

a value equal to the yield stress F Y . These two limits are shown by the Euler curveand squash limit in Fig. 8.1e.

Fig. 8.1e Column Buckling and Gross yield limits © RMIT University, 2013 (Dixon C., Marchiori G.)

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Another possible mode of failure in columns, or beams on their compressive side, isthat of local buckling in a thin wall or flange such as illustrated in Figure 8.1f.

Fig. 8.1f Local Plate Buckling in a rectangular hollow section having excessive breadth to thickness ratio © RMIT University, 2013 (Dixon C., Marchiori G.)

Detailed treatment of local plate buckling is beyond the scope of this course. Theinterested reader is referred to chapter 4 “Local Buckling of Thin Plate Elements” in‘The Behaviour and Design of Steel Structures’ by Trahair, N. S., 1977, 1st ed.,Chapman and Hall, London . There it is shown that for a plate of length L , breadth b and thickness t, simply supported at all four edges and subjected to a uniformlydistributed load P at each end, the critical load for plate buckling Pcp is

b

Pcp = k ( )2

2

32

112

b

Et

−Π

ν eqn. C4

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where the buckling co-efficient k depends on the ratiob

L and the number of wave

nodes or points of inflexion. The lowest value of the buckling co-efficient k for thesimply supported plate is 4, which is appropriate for the high values of moststructural steel members.

If a different method of edge support applies then values for k will be different.

The quantity( )2

3

112 ν −

Et is called the flexural rigidity of the plate often given the

symbol D , and is analogous to the quantity EI used in beams and columns. Thisquantity D has been met in the previous topic of the bending of circular plates.

It can be seen that equation C4 for the buckling of plates is similar to that for thebuckling of columns:

Pc =( )2

2

Lk

EI

e

Π eqn. C2

Returning to equation C4, the critical value of stress to cause plate buckling (a caseof which is local buckling in the wall of a hollow column) is:

olF =bt

Pcp = k ( )

( )22

2

112

t b

E

ν −Π eqn. C5

which is similar to the Euler critical stress equation for columns.

ocF = A

Pc =( )2

2

r l

E Π eqn. C3

Looking at equation C5 it can be seen that ift

b is large then the stressolF to cause

local buckling will be low, and local buckling could be of more concern than yield.

Recall also that F ol is affected, via k, by the method of edge support (as is F oc incolumns affected by the method of end support).

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At some limiting value of width to thickness ratiot

b the plate buckling stress

olF is

equal to the yield stress F Y . For example:

a) For flanges supported along both edges as in box section beams or columns,k in equation C5 is 4 and if E is 200,000 MPa, ν is 0.3 and F Y is expressed inMPa then F ol < F Y if:

t

b >

( ) Y Y F F

851

0.3-112

000,200x42

2

=

Π

(i) This may be simplified to the advice to keep

Y F t

b 800< if local buckling is to be

avoided in as-rolled and stress-relieved flanges which is implicit in part of Rule4.3.2 in “Australian Standard AS3990 Mechanical Equipment – Steelwork”. For a

steel whose yield strength is 250 MPa this limit corresponds to a width-thickness

ratiot

b of approximately 50.

(ii) For welded plates which are not stress relieved.

t

b should be <

Y F

560

(iii) And for cold form hollow section members:

t b should be <

Y F 635

(b) For flanges supported along one edge and free along the other, as in an I section,

t

b should be <

Y F

256 which is implicit in rule 4.3.1 of AS3990. Excess widths

beyond the limits of b from the above four equations shall be neglected whencalculating the effective geometrical properties of the section. [See Trahair, N.S.,“The Behaviour and Design of Steel Structures” and “Australian Standard

AS3990 –Mechanical Equipment – Steelwork” for more detail.]

Many commercial structural sections do keept

b less than the limits shown above

as advised and tabulate their values oft

b in catalogues along with other sectional

properties. This being the case yield of the flange rather than local buckling is theconcern. So for a concentrically loaded straight column consisting in essence of

plates joined at various edges whoset

b values are not too large, the maximum load

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should be that which causes yield for squat columns (r

l small) and gross lateral

buckling for slender columns (r

l large).

Acti vi ty 8E - Torsional buckl ing

Buckling by twisting about a longitudinal axis

Another type of buckling however is also possible for open section compressionmembers, which may buckle by twisting about a longitudinal axis as shown inFig.3.27 on page 79 of ‘The behaviour and design of steel structures‘ (PDF104KB)7 or by combined bending and twisting. For most rolled steel sections the

critical load for this mode of failure is greater than the critical load Pc for lateralbuckling about the weaker axis so the possibility of torsional buckling can beignored.

Refer back to Fig. 8.1e where the squash limit (gross yield) and Euler curve (grossbuckling) are shown. Early workers found that for real columns, supposedlyinitially straight and concentrically loaded, that were of slender proportions, thecritical buckling stress was less than the Euler equation predicts for elastic

buckling. This was found to be due to initial crookedness, non-concentricity of theload (i.e. an unintended eccentricity) lack of homogeneity, residual stresses andinelastic buckling (at intermediate slenderness ratios). Perry, [Ayrton, W.E., andPerry, J., “On Struts” in “The Engineer”, 10th December and 24th December 1886.]who was one of the significant early workers in this field, analysed theoretically thecombined effects of initial crookedness and unintended initial eccentricity, andsome of their results will be considered later in this chapter.

You have previously studied the effect of eccentricity on column behaviour in Solid

Mechanics 3. This will be revised now and compared with the effect of initialcrookedness. A column pin jointed at each end, with a length L will be considered,for which the effective length factor ke is 1. For other end support types the effectivelength l=keL would take the place of L.

7 Trahair, N. S., 1977, The behaviour and design of steel structures, 1st ed., Chapman and Hall, London, p. 79,viewed on 28th August 2013 <https://equella.rmit.edu.au/rmit/file/3995dd92-9269-490d-811b-

15f2f5860b02/1/130823_3_057.pdf>

https://equella.rmit.edu.au/rmit/file/3995dd92-9269-490d-811b-15f2f5860b02/1/130823_3_057.pdf



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Initial crookedness Eccentric load

Fig. 8.1g Column with initial crookedness


Initial shape as shown In Fig.8.1g(i)

could be described by L

xaSin y

o

Π=

Final ordinates of the deflected shape

are given by y, deflection from the initialshape is y1 as shown in Figure 8.1g(ii)

∴ the bending moment due to P is

( )1 y yP M o +=

The elastic restoring moment is

2

1

2

dx

yd EI −

For moment equilibrium

( )12

1

2

y yPdx

yd EI o +=−

Fig. 8.1h Column with eccentric load


Deflection from the initial shape is y asshown in Fig.8.1h(ii):

∴ the bending moment due to P is

( ) yeP M +=

The elastic restoring moment is

2

2

dx

yd EI −

For moment equilibrium

( ) yePdx

yd EI +=−

2

2

L

xaSin

EI

P y

EI

P

dx

yd Π−=+∴ 12

1

2

Finding the general solution for y1 andthen considering the boundaryconditions yields

L

xaSin y

Π

−

=α

α

11

e EI

P y

EI

P

dx

yd −=+∴

2

2

Finding the general solution for y and thenconsidering the boundary conditions yields

−+= 1

2

Ltan

1

x EI

PCos x

EI

PSin

EI

Pe y

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where22 / L EI

P

Π=α which is

cP

P

Dividing top and bottom of the latter

equation by cross sectional area A.

The maximum deflection occurs at the

middle of the column where2

L x = .

Substituting this into the above equation

yields

oc

ac

F

f ==⇒

stresscriticalEuler

stressecompressivaverageα

The final ordinates of the deflectioncurve

L

xaSin

L

xaSin y y y

o

Π

−+

Π=+=

α

α

11

L xaSin y Π

−=

α 11

Therefore the maximum deviation fromthe vertical at the middle of the columnwhich is

also the maximum moment arm is

a yα −

=1

1max

−= 1

2max

L

EI

PSece y

Bringing L under the and

multiplying and dividing by Π

yields

−

Π

Π= 1

/222max

L EI

PSece y

−

Π= 1

2max

cP

PSece y

PuttingcP

P=α gives

−

Π= 1

2max α Sece y

The maximum moment arm, which is at

the centre is e y +max

∴ Moment armmax = aα −1

1

Thus the initial deviation a at the middleof the

column is amplified by the factorα −1

1

when the axial compressive load P isapplied. Note that when P approaches Pc, and hence α approaches one, thisamplification factor approaches infinity.

∴ Moment armmax α 2

Π= eSec

Noting thatθ

θ Cos

Sec1

= and the

magnitude of θ Cos is from 0 to 1 it canbe seen that the moment arm e at the topof this column

has been amplified at the middle of the

column by the factor α 2

ΠSec when

the axial compressive load P is applied.Note that when P approaches Pc andhence α approaches 1, this

amplification factor approaches infinity.

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The bending moment at the middle of

the column maxmax Py M =

α −=

1

1max Pa M eqn. C6

The maximum compressive stress is thesum of the average axial compressivestress and the bending compressivestress

i.e. I

c M

A

P maxmax +=σ

where c is the distance from the neutralaxis to the “outer fibre”. Recalling

2 Ar I =

2max1

1

Ar

cPa

A

P α σ −+=⇒

α σ

−+=∴

1

12max

r

ac

A

P

A

P eqn. C8

α σ

−+=∴

1

12max

r

ac f f acac

The term2

r

ac is sometimes called the

imperfection ratio.

Recalling thatoc

ac

F

f =α and if it is

decided that the yield stress F Y is the

limiting value of maxσ and giving the

symbol F L to the consequent limit valueof the average compressive stress f ac gives

The bending moment at the middle of

the column maxmax armmomentxP M =

α 2

max

Π= PeSec M eqn. C7

The maximum compressive stress is thesum of the average axial compressivestress and the bending compressivestress

i.e. I

c M

A

P maxmax +=σ

where c is the distance from the neutralaxis to the “outer fibre”. Recalling

2 Ar I =

2max

2

Ar

cPeSec

A

P

Π

+=⇒

α

σ

α σ 22max

Π+=∴ Sec

r

ec

A

P

A

P eqn. C9

α σ 22max

Π+= Sec

r

ec f f acac

The term2

r

ec is sometimes called the

eccentricity ratio.

Noting thatoc

ac

cc F

f

AP

AP

P

P===

/

/α , and if

it is decided that the yield stress F Y is

the limiting value of maxσ , and giving

the symbol F L to the consequent limitvalue of the average compressive stress

f ac gives

oc

L

L LY

F

F r acF F F

−+=

112 eqn. C10

Figure 8.1i shows plots of F L for

different values of2

r

ac

oc

L L LY

F F Sec

r ecF F F

22 Π+= eqn. C11

Figure 8.1j shows plots of F L for

different values of2r

ec

Refer to Fig. 8.1i Refer to Fig. 8.1j

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Fig. 8.1i Initial crookedness © RMIT University, 2013 (Dixon C., Marchiori G.)

Fig. 8.1j Eccentric load © RMIT University, 2013 (Dixon C., Marchiori G.)

Comparing the two graphs (Fig.8.1i and 8.1j) and using the effective length l, it can

be seen that for a given value of slenderness ratio r

l

, if 22 r

ec

r

ac=

, the values of F L for

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each are very similar. When 0=r

l both formulas C10 and C11 give the same value

of F L (the value of A

P that will cause yield to start at the outer fibre on the concave

side). When 0>r

l, the curves of F L for the non-concentric case are slightly lower

than those for the initially crooked case. It is not surprising that these two situationsshould produce similar results since the eccentrically loaded column might bethought of as a column which first has an end moment Pe applied causing an initialcurvature like the initial crookedness, and then the axial load P is applied whichamplifies the deviation from vertical. Indeed comparing equations C8 and C9 it canbe seen that the two differences are that equation C8 has initial crookedness “a”

instead of eccentricity “e” and an amplification factorα −1

1 instead of α

2

ΠSec .

Perry and Ayrton [Ayrton, W.E., and Perry, J., “On Struts” , pgs 464 to 465 of “TheEngineer”, 10th December and 24th December 1886] compared these twoamplification expressions which is done in the table following.

=

cP

Pα

α −1

1 α

2

ΠSec

0 1.00 1.00

0.1 1.11 1.14

0.2 1.25 1.31

0.3 1.43 1.53

0.4 1.67 1.83

0.5 2.00 2.25

0.6 2.50 2.88

0.7 3.33 3.94

0.8 5.00 6.06

0.9 10.00 12.42

1 ∞ ∞

They observed that the amplification factor for the eccentric case

Πα

2Sec is

slightly larger by a factor of approximately5

6 than that for the initially crooked

case. This greater (i.e. more detrimental) amplification factor for the eccentric caseis the reason that the limiting F L curves for that case are lower than those for theinitially crooked case.

Based on their observations from the table, Ayrton and Perry suggested that thecombined effects of the unintended eccentricity called here “eo” and initialcrookedness “a” could be expressed as a modified initial crookedness which is

given the symbol here of “ a′ ”, such that:

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a′ =o

ea6

5+ eqn. C12

They further suggested that inhomogeneity may be accounted for by a term of the

same kind. Putting a′ instead of a in equation C10 gives:

F Y =

α F

F r

caF F

L

L L

−

′+

1

12

This is in fact a quadratic in F L which can be re-arranged in the more conventionalform:

012

2=+

′++− ocY LocY L F F F F

r caF F

This will be recognised as the quadratic form:

02 =++ c b a L L F F for which the roots are:

( )a

c a b b

2

42−±

−= LF

The smaller root of this quadratic provides the critical value of F L at which yieldingwill occur at the outer fibre on the concave side of the column. The smaller root is:

F L =

−

′++

−

′++

ocY

ocY ocY

F F

F r

caF F

r

caF

2

22

2

1

2

1

eqn. C12

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Important unknown parameters at this stage are the initial crookedness a andunintended eccentricity eo, which are combined in the term a′ as in equation C12.Another problem is that c and r 2 are terms referring to a specific cross-sectional

shape. The symbol η is given to2

r

ca′ and called the imperfection factor. Robertson

(Robertson 1925, p.55)8 suggested that it be assumed that η varied linearly with the

slenderness ratior

l which bypassed the problems just mentioned. He further

suggested the value ofr

l003.0=η would produce a graph for F L versus slenderness

ratio that would pass beneath the great majority of column test results thenavailable. The equation for F L has therefore become essentially an empiricalequation to fit the data although its origin was in the theoretical analysis just given.Godfrey (Godfrey 1962, p.97-112)9 later suggested the imperfection factor bechanged to:

η =2

00003.0

r

l eqn. C13

and this is the value used in AS3990.

Equation C12 thus becomes:

F L =( ) ( )

−

++−

++ocY

ocY ocY F F F F F F

2

2

1

2

1 η η eqn. C14

The graph of this equation is shown in Figure 8.1k, as the PERRY-ROBERTSONCURVE .

8 Robertson, A., 1925, The Strength of Struts, I.C.E. Selected Engineering Papers, No. 28, p.55, The Institution ofCivil Engineers, London.,

9 Godfrey, G.B., 1962, The Allowable Stresses in Axially-Loaded Steel Struts, Structural Engineer Vol. 40., No. 3,p.97-112,.The Institution of Structural Engineers, United Kingdom.,

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Fig. 8.1k Column Perry-Robertson limit © RMIT University, 2013 (Dixon C., Marchiori G.)

This F L then is the value of average axial compressive stress

=

A

P f ac

at which

yielding is predicted to occur on the concave side of the column. For a safe design

the actual value of this average stress f ac should beΩ

≤ LF

where Ω is a factor of

safety. In AS3990

==Ω

6.0

167.1 . The maximum permissible average compressive

stressΩ

LF is given the symbol F ac. Thus f ac is to be acF ≤ where:

F ac =

( ) ( )

−

++

−

++

ΩocY

ocY ocY

F F

F F F F 2

2

1

2

11 η η

eqn. C15

which is equation 6.1.1 in AS3990.

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Table 6.1.1 on page 42 of AS3990 conveniently gives values of F ac [MPa] for a rangeof yield strengths FY [MPa] and slenderness ratios l/r . A sample of this data is givenbelow.

FY 230 240 480 Foc

l/r 0 5 0 5 0 5 0 5

50 125 122 130 127 249 235 790 653

60 118 114 123 118 218 200 548 467

70109

F ac l/r 70 104

F ac l/r 75 113 107 181 164 403 351

80 98 92 101 95 148 133 308 273

For example if a concentrically loaded column is made from steel with a yieldstrength of 230MPa and is of proportions such that it has a slenderness ratio l/r of70, then its permitted average axial compressive stress is 109MPa. If it has aslenderness ratio l/r of 75, then its permitted average axial compressive stress is104MPa.

The two right hand columns of the table give values of Euler critical stress F oc forvarious slenderness ratios. Recall that:

F oc =( )2

2

r l

E Π

Acti vi ty 8F - Graphical representation of F ac

Graphs of the maximum permissible average axial compressive stress F ac for a

range of yield strengths and slenderness ratios may be viewed at Permissiblestresses in compression members (PDF 143KB) 10.

Note that the vertical axis has a change of scale when 100=r

l.

10 Trahair, N. S., 1977, Permissible stresses in compression members, Compression members, The behaviourand design of steel structures, Chapman and Hall, London, pp. 65, viewed on 28th August 2013<https://equella.rmit.edu.au/rmit/file/e3770396-30b1-4301-8623-b978a9de13d1/1/130823_3_055.pdf>

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F ac is the usual maximum allowed value of the average axial compressive stress incolumns or struts, however extra restrictions are placed in Clauses 6.1.2 & 3 ofAS3990 on this stress for built up columns or struts, or on ones made from angle (toavoid torsional buckling in the case of the angle).

Returning now to the equation for F ac in AS3990 (equation C15 in these notes). Thedesigner of a concentrically loaded column will typically know the load P that hasto be carried, the effective length l, for the set up, the yield strength (FY) andYoung’s modulus (E) of the steel proposed for use, and will probably have someidea of a suitable cross sectional shape (e.g. square hollow section, I section, circularhollow section, etc.)

Acti vi ty 8G– Sections used for Columns

Step 1: Typical sections used for columns and struts

Refer to figure 6.3 on page 118 of The Steel Designers Handbook (PDF 80KB)11

for an overview of typical sections used for columns and struts.

Step 2: Safe axial load for various sections

Refer to table 6.1 on page 119 of The Steel Designers Handbook (PDF 80KB) 12

The table compares the safe axial load for various sections, each having the samecross sectional area, and operating with the same effective length.

11 Gorenc, B. E. & Tinyou, R., 1981, figure 6.3, Steel designers handbook, 5th ed., New South Wales UniversityPress, Kensington, N.S.W., p.118, viewed on 28th August 2013 <https://equella.rmit.edu.au/rmit/file/d817860a-baa5-40b2-9722-ed3e38630d2e/1/130823_3_064.pdf>

12 Gorenc, B. E. & Tinyou, R., 1981, Table 6.1, Steel designers handbook, 5th ed., New South Wales UniversityPress, Kensington, N.S.W., p.119, viewed on 28th August 2013 <

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Having made some trial decision on shape, what the designer has to find next is theactual section size that can carry the load P without generating an average axialcompressive stress in excess of F ac.

i.e.

= A

P

f ac must be acF ≤

i.e.( ) ( )

−

++−

++

Ω≤ ocY

ocY ocY

s

F F F F F F

A

P2

2

1

2

11 η η

where η = 0.00003 ( )2r

l

and F oc =

( )2

2

r l

E Π

Looking at these three equations, it can be seen that the unknown quantities are theeffective cross sectional area As and the appropriate radius of gyration . [If l isdifferent in one plane of buckling to another, the plane that has the larger value of

slenderness ratior

l is the one in which buckling is likely to take place so the r

value is the one appropriate to that direction of buckling. Clearly if l is the same forboth planes of buckling the appropriate radius of gyration is the minimum one as itwill produce the larger value of slenderness ratio.]

One possible approach is to guess a size perhaps based on experience, and check

that f ac is acF ≤ . If for the size guessed f ac is << F ac then the size chosen is wasteful

and a smaller section could be checked for suitability. In view of the fact thatcolumn bases are often in moist environments and therefore prone to corrosion itcould be argued that having extra metal thickness to compensate for such corrosionover the life of the structure would be prudent. A sound maintenance regime ofcourse aims to prevent such corrosion.

A second approach, which tries to reduce the number of trial selections, makes useof charts such as those shown in Activity 8H which follows.

Safe load tables for concentrically loaded columns as are provided by manymanufacturers; please see Appendix 8 for some examples.

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Acti vi ty 8H –Design charts to aid column s ize selection

Step 1: Design chart for selecting size of Australian universal sectioncompression members

Refer to Fig. 3.15 on page 66 of ‘The behaviour and design of steel structures’ (PDF 132KB) 13 to view an example of a design chart for selecting size ofuniversal section compression members.

Using this chart one could enter the vertical axis with the desired axial workingload P, and the horizontal axis with the effective length and hence find thesmallest universal section size that could carry the load.

Step 2: Determination of approx. values for F ac for columns of various sections

Refer to Figure 6.10 on page 136 of ‘Steel designers handbook’ (PDF 84KB) 14

In Figure 6.10 one could enter the horizontal axis with the parameter combining

effective length and desired working loadP

l2

, go to the sloping line

representing the section type being considered then move across to the verticalaxis to find the approximate maximum permitted average axial compressive

stress *

acF . The approximate required effective cross sectional area can then be

found from:

f ac = ac

s

F A

P ≤

∴ As ≈ *

acF

P

13 Trahair, N. S., 1977, The behaviour and design of steel structures, 1st ed., Chapman and Hall, London, p.66,viewed on 28th August 2013 < https://equella.rmit.edu.au/rmit/file/929639e4-619e-4279-ac44-9f927f920a5c/1/130823_3_056.pdf>

14 Gorenc, B. E. & Tinyou, R., 1981, Figure 6.10, Steel designers handbook, 5th ed., New South Wales UniversityPress, Kensington, N.S.W., p.136, viewed on 28th August 2013 <

https://equella.rmit.edu.au/rmit/file/2277a2a2-1f72-47c8-ba66-2dcf359af72f/1/130823_3_065.pdf>

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One would then find a section of the desired shape that had an effective area ≥ thatpredicted as being required. Having found this size one would then determine f ac using its effective area As, and F ac using its effective radius of gyration r , and check

that acac F f ≤ . Recall also that Clauses 6.1.2 and 6.1.3 list special limiting provisions on

the maximum allowed average compressive stress for built-up columns or struts, andcolumns or struts made of angle.

Note that the effective cross sectional area As is calculated from the specified sizesof the member cross section deducting:

a) All holes other than bolt holes (which are not deducted for members incompression).

b)

Excess width as given in Clause 4.3 of AS3990. This deduction is related to

plate with excessive width to thickness ratiost

b being prone to local

buckling before it reaches the yield stress and therefore the excess width is

not regarded as useful. Cross sectional areas are often manufactured witht

b

proportions such that this is not a concern. Manufacturers’ tables usuallyindicate the value of the effective and gross cross sectional areas, and mayalso indicate where they differ (see Appendix 8).

Acti vi ty 8I – Excess width determinat ion

To aid in the determination of excess widths Gorenc and Tinyou have preparedsome helpful diagrams in Fig. 6.11 of their ‘Steel designers handbook’(PDF 81KB)15 The sketches there show excessive section areas for compression members withlarge width to thickness ratios. The small calculations shown are for steel with F Y of250 MPa (e.g. for the flange on the I beam in the left most sketch. Clause 4.3.1(a)

says in essence projection b1 greater thanY F T 1256 shall be neglected when

calculating the effective properties of the section; 11 16250256 T T ≈ ).


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There are two noteworthy effects of the effective cross sectional area being less thanthe actual cross sectional area:

• firstly the calculated average axial compressive stress

=

s

ac A

P f increases;

•

secondly, the effective radius of gyration r changes and may decrease,particularly when the neglected area is far from the centroid, and if r is less

then the effective slenderness ratior

l increases and therefore the allowed

maximum value of average compressive stress F ac is reduced.

All things considered it would seem better, as regards use of metal, to avoid use ofsections with excessive plate width to thickness ratios.

With regard to the appropriate effective radius of gyration of the chosen size of

section, it must not be so small that the resulting slenderness ratio exceeds thelimits allowed by Clause 4.6 of AS3990.

Section 8.2 Eccentrically Loaded Columns or CompressionMembers

It is often not possible for a vertical load to be transmitted to a column such that itacts axially through the centroid. The column is therefore subjected to an eccentricload which generates a bending moment. This situation is shown schematically inFigure 8.2a.

Fig. 8.2a: Column with eccentric load © RMIT University, 2013 (Dixon C., Marchiori G.)

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Refer to Clause 6.4 of AS3990 for comments on appropriate choices for the amountof eccentricity to be included in calculations in certain structural situations.

A column may also be subjected to end couples form some other source than aneccentric load, and be simultaneously subjected to an axial load. When studying“concentrically” loaded columns consideration was given to the effect of smallunintended eccentricity and initial crookedness both of which finally ended up as

part of the imperfection parameter η in the expression for the allowed average axialcompressive stress F ac. One of the important observations in that topic was theamplifying effect the axial load had on the bending moment. It is helpful now torevisit some of the main results of that study and to look at the amplifying effectthe axial load can produce in some other situations. This is done in the table on thenext page, which summarises results from analyses done by Timoshenko and Gere.

(Timoshenko & Gere 1963) 16

Regarding the amplification, caused by the axial load, of the central deflection,Timoshenko & Gere also observed (Timoshenko & Gere 1963) that the three

amplification factors λ (u), χ (u) and η (u) could be approximated with goodaccuracy by an amplification factor

cP

P−1

1 eqn. C16

if the ratiocP

P

is not large.

16 Timoshenko, S., Gere, J M., 1963, Beam-columns, Theory of elastic stability, 2nd ed., International student ed.,

McGraw-Hill, Auckland.

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Initially crooked plus axialcompressive load

Equal and opposite couples at eachend plus axial compressive load

Central concentrated lateral loadplus axial compressive load

Uniformly distributed lateral loadplus axial compressive load

Central deflection beforeaxial load applied

a 2

8

Ml

EI

3

48

Ql

EI

45

384

ql

EI

Amplification factor forcentral deflection caused

by axial load

1

1c

P

P−

( )

2

2

1

2 c

Sec uu

Pwhere u

P

called uλ

− ∏

=

( )

33 Tan u uu

called u χ

−

( )

2

412 2 25

Sec u uu

called uη

− −

Angular deflection at endsbefore axial load applied

a

l∏

2

Ml

EI

2

16

Ql

EI

3

24

ql

EI

Amplification factor forangular deflection at ends

caused by axial load

1

1c

P

P−

Tan u

u ( )uλ ( )u χ

Central bending momentbefore axial load applied

O M Ql

u

2

8

ql

Central bending momentafter axial load applied

1

1c

PaP

P

−

M Sec u Ql Tan u

u u

2

2

21

8

qlSec u

u −

Amplification factor forcentral bending moment

caused by axial load not applicable Sec u

Tan u

u ( )uλ

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They note that for values ofcP

P less than 0.6, the error in the approximate expression for the

three amplification factors λ (u), χ (u) and η (u) is less than 2%.

Regarding the amplification, caused by the axial load, of the central bending moment,the three amplification factors [ ]

−= 1

2 )(and

2 Sec u

uu

u

uTanu,Sec λ may also be

approximately represented by

cP

P−1

1, if the ratio

cP

P is not large as may be seen in the

table below which shows various amplification factors as a function of 2u, where

cP

Pu

2

Π= . Data in the last three columns is drawn from Timoshenko and Gere (1963)

It is noteworthy that as P approaches Pcrit the amplification factors approach infinity.

2u cP

P

22

Π=

u c

P

P−1

1

22

1

1

Π−

=u

Sec u

u

uTan ( )uλ

( )122

−= uSecu

( )u χ

[ ]uuTanu

−33

( )uη

[ ]2

422

5

12uuSec

u−−

0 0 1.000 1.000 1.000 1.000 1.000 1.000

0.2 0.004 1.004 1.005 1.003 1.004 1.004 1.0040.4 0.016 1.016 1.020 1.014 1.016 1.016 1.016

0.6 0.036 1.038 1.047 1.031 1.038 1.037 1.037

0.8 0.065 1.069 1.086 1.057 1.073 1.068 1.070

1.0 0.101 1.113 1.139 1.093 1.117 1.111 1.114

1.20 0.146 1.171 1.212 1.140 1.176 1.169 1.173

1.40 0.199 1.248 1.307 1.203 1.255 1.245 1.250

1.60 0.259 1.350 1.435 1.287 1.361 1.346 1.354

1.80 0.328 1.489 1.609 1.400 1.504 1.482 1.494

2.00 0.405 1.681 1.851 1.557 1.704 1.672 1.690

2.20 0.490 1.962 2.205 1.786 1.989 1.949 1.962

2.40 0.584 2.402 2.760 2.143 2.441 2.382 2.400

2.60 0.685 3.174 3.738 2.771 3.240 3.144 3.181

2.80 0.794 4.863 5.883 4.141 4.938 4.808 4.822

2.90 0.852 6.762 8.299 5.681 6.940 6.680 6.790

3.00 0.912 11.350 14.137 9.401 11.670 11.201 11.490

∏ 1 ∞ ∞ ∞± ∞ ∞± ∞

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The approximate substitution of

cP

P−1

1 for Sec u and

u

uTan is not as accurate as it is

for [ ])(and )(and )( uuu η χ λ . For example whencP

P is 0.6 the expression

cPP−1

1

underestimates Sec u by 13%, overestimatesu

uTan by 12%, and underestimates λ

(u) by 1.8%.

It is worth recalling at this point the observation made by Ayrton and Perry(Ayrton, W.E., and Perry, J., “On Struts” in “The Engineer”, 10th December and 24th December 1886.), that was discussed earlier when considering the effects of initial

crookedness and unintended eccentricity, that:

c

c

P

PSec

P

P 26

5

1

1 Π≈

−

(i.e. uSec 6

5)

The reader is referred to Chapter 1 of ‘Theory of Elastic Stability’ (Timoshenko &Gere 1963) for discussion and data on the amplifications that occur when acompressive axial load is applied in beam-column situations with: couples at eachend that differ in size and sense; several concentrated lateral loads; and variabledistributed load.

Brief consideration will be given here to a column with moments of equalmagnitude at each end and an axial compressive load.

For a column such as is shown again in Figure 8.2b(i), loaded with equaleccentricities at each end, the magnitude of the moment at each end is Pe and theirsense is opposite (i.e. one clockwise and one anti-clockwise). They produce singlecurvature bending.

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Fig. 8.2b: Effect of equal and opposite sense end moments © RMIT University, 2013 (Dixon C., Marchiori G.)

This eccentric loading could also be represented by a couple M of magnitude Pe ateach end, and an axial compressive load as shown in Figure 8.2b(ii).

As has been previously shown, the maximum deflection is at the centre and is

[ ]1−uSece which could be written [ ]1−uSecP

M . The maximum bending moment is at

the centre and is formed from:

M max = deflectionmaximumxP M +

= [ ]1x −+ uSecP

M P M

∴ M max = uSec M

& ∴ M max ≈

cP

P M

−1

1

i.e. the axial compressive load has caused amplification of the bending moment by a factor of

approximately

cP

P

−1

1

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If the moments at each end are of equal magnitude and both of the same sense (e.gboth clockwise) then the column has a double curvature shape as is shown in Fig 8.2c.

Fig. 8.2c: Effect of equal and same sense end moments © RMIT University, 2013 (Dixon C., Marchiori G.)

Instead of a single maximum deflection at mid position there are now two; one at ¼and one at ¾ position. Furthermore it will be appreciated that if the moments are ofthe same magnitude as was the case in Fig 8.2b then the maximum deflections in Fig8.2c, will be less, and consequently the moment amplification factor will have a value

less than

cP

P−1

1 ..

The moment amplification factor for beam columns such as shown in Figure 8.2b and

8.2c can thus be represented as

c

m

P

PC

−1

1 where C m is a number ≤ 1.

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If the moments at each end are different in magnitude and/or sense then the shapetaken by the beam column is more complicated, and the maximum deflection from thecentre line may not be at mid height, as shown in Activity 8J .

Optional Activity 8J– Further reading on moment amplification in Beam-columns

In-plane bending moment distribution in a beam-column where moments ateach end are different in magnitude and/or sense

Refer to Fig 7.3 on page 239 of the ‘The behaviour and design of steel structures’ (PDF 68KB) 17

The extra symbol β used here can have any value between –1 (single curvaturebending) and +1 (double curvature bending). So in Figure 8.2b β = –1. The endmoments shown in the Figure 7.3 in this reference, where the column has pin

jointed ends, are applied end moments not reactive end moments from built-inends and the like (which affect effective length and hence critical load).

For situations where the end moments differ it can be shown (Trahair 1977) that:

(i) M max = ( )[ ]2 221 uCosuCosec M ++ β

when uCos 2−< β ,and the point of maximum moment lies in the span

(ii) M max = M

when uCos 2−> β ,

The expression in (i) simplifies to 1when

1

1 max −=

−

≈= β

cP

P M Secu M M ,

which is the maximum value M max can have.

17 Trahair, N. S., 1977, Beam-columns and frames, The behaviour and design of steel structures, Chapman andHall, London, p. 239, viewed on 28th August 2013 < https://equella.rmit.edu.au/rmit/file/4dee5467-6448-

4047-87f6-cfa947033e08/1/130823_3_059.pdf>

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Obviously since the bending moment has been amplified by the presence of the axialcompressive load so too will the resulting bending stress be amplified compared towhat it would be were the axial load not present. Thus:

bc f M

M

= max

max bendingσ

where f bc is the bending stress that would exist if the axial compressive loadwas not present

∴ f bc Z

M = where Z is the appropriate section modulus

Therefore when the end moments are equal and opposite:

max bendingσ ≈ bc

c

f

P

P

−1

1

and when the end moments are other than equal and opposite σ bending max will not beamplified as much, hence:

max bendingσ =bc

c

f

P

P

−

≤

1

1x

momentsend

theof senseand

magnituderelative

on thedepending

1fractionA

In the case of a beam without an axial compressive load the bending compressivestress f bc would ordinarily be compared to the maximum permissible stress due tobending (in a member not subjected to axial force) F b, and the maximumpermissible compressive stress due to bending (in a member not subjected to axialforce) F bc.

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However in a beam-column there is an axial force, the major effects of which are:

(i) to cause the bending moment and consequently the bending stress to beamplified

(ii) to cause an average axial compressive stress

=

AreaEffective

P f

ac

which is

superimposed on the bending compressive stress. [Recall that in a concentricallyloaded column this stress would be compared to F ac, the maximum average axialpermissible compressive stress in an axially loaded strut not subjected to bending.]

AS3990 provides equations in Clause 8.3.1 “Axial compression and bending” whichenable stresses to be checked in the situation where there are the combinedinteracting effects of bending about two axes, and an axial load imposed on thebeam-column.

One such equation is this interaction or “unity” equation:

1 6.0

1

1

6.0

1

1

≤

−

+

−

+

bcy

bcy

ocx

ac

my

bcx

bcx

ocx

ac

mx

ac

ac

F

f

F

f C

F

f

F

f C

F

f

eqn. 8.3.1(a)1 AS3990

The first group of termsac

ac

F

f expresses as a fraction the comparison of the average

axial compressive stress f ac to the lesser maximum permissible average axialcompressive stress F ac about either principal axis as found in Section 6 of AS3990(and discussed in these notes when considering concentrically loaded columns).

The second and third group of terms deal with bending stress about the x and y axis respectively.

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Considering the second group of terms: f bcx is the “maximum” bending stress aboutthe x axis, occurring either at or between braced points along the beam-column,calculated without inclusion as yet of the amplifying effect of the axial compressive

force. The bending stress f bc is calculated using the properties of the section basedon the effective cross sectional area.

The factor

−ocx

ac

F

f

6.01

1 could be written as

−cxP

P

6.01

1 which makes it

recognisable as the approximate amplification factor hitherto discussed with theinclusion of a load factor of 0.6 which tends to increase the amplification. [The

author’s interpretation of this 0.6 is that when f ac = 0.6 F ocx, i.e. P = 0.6 Pcx, then P isat the maximum value allowed for a theoretically perfect column (imperfection

parameter η = 0) if the factor of safety is6.0

1, and there is hence no spare capacity

in the column to deal with any bending stress. The amplification term as shown

then becomes ∞ so that the inclusion of any bending stress f bc would make itimpossible to satisfy the “unity” equation.]

As will be recalled application of the full amplification factor was appropriate

when there are equal and opposite end moments (and hence single curvature), butfor other situations with end moments this can be moderated by a fraction 1≤ depending on the relative magnitude and sense of the end moments. C mx is such amoderating fraction. The standard gives a variety of values for differentcircumstances, and the reader is referred thereto; the range of C mx and C my variesfrom 0.4 to 1.0. For the circumstance where the member is in a frame wheresideways is prevented, and there is no lateral (transverse) loading between themembers’ supports in the plane of bending, then linearised expressions for C m maybe used. These equation are shown in the following optional activity. They givevalues of C m between 0.4 and 1 . Note that if the column is not braced against

sidesway then C m should be taken as 0.85 for all the ratios of bending moments.

However for simplicity the conservative approach of putting Cm=1 may be taken;and this is the approach recommended by Bednar (1986)18 for pressure vesselsupport columns.

18 Bednar H 1986, Pressure vessel design handbook 2nd ed. Van Nostrand Reinhold, New York, p.148.

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Optional Activity 8K–Linearised expression for C m

Step 1: Variation of Cm with End Moment Condition

Refer to Fig 131.1, Variation of Cm with moment on page 46 of ‘Source book forthe Australian Steel Structures Code’ (PDF 73KB) 19

Step 2: Values of the co efficient Cm for columns subject to bending with axialcompressive force

Refer to Fig 6.14 on page 141 of the ‘Steel designers handbook’ (PDF 44KB) 20

Figure 6.14 provides information for obtaining C m for a variety of loading

configurations.

The terms F bcx and F bcy are the maximum permissible compressive bending stressesfor bending about each axis. These are obtained from Section 5 of AS3990. Inparticular for use in equation 8.3.1(a)(1) F bcx and F bcy are derived using Clauses 5.2,5.3 and 5.4 (if applicable).

Values for Fbc for some commercial square hollow sections may be found on the lastpage of the extract in Appendix 8.

Clause 5.2 is based on considerations of yield, Clause 5.3 on consideration of localbuckling, and Clause 5.4 on considerations of flexural-torsional buckling that mayoccur when a beam is bent about its axis of maximum strength.

19 Lay, M. G., 1982, Fig 131.1 variation of Cm with movement condition, Source book for the Australian SteelStructures Code, AS 1250, 3rd ed., Australian Institute of Steel Construction, Milsons Point, N.S.W, p. 46.,viewed on 28th August 2013 < https://equella.rmit.edu.au/rmit/file/b0a21c25-b0b2-4522-9b88-569ae6750d09/1/130823_3_053.pdf>


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Clause 5.2 limits the tensile or compressive bending stresses F b to 0.66 F Y for allbeams other than solid round and square bars and solid rectangular bars bentabout the axis of minimum strength for which the limit is 0.75 F Y .

With respect to the limit of 0.66 F Y , it is noteworthy that this is higher than the limitset as the average axial compressive stress in stocky columns (small

r l ) of 0.60 F Y

which was putting a factor of safety of6.0

1 against yielding. In the case of beams

the limit 0.66 F Y set on the outer fibre stress corresponds approximately to a factor

of safety of6.0

1 against yielding right across the section, i.e. the beam being fully

plastic (i.e. a “plastic hinge”). For most rolled sections the plastic section modulus S is 1.1 to 1.2 times the elastic section modulus Z so taking conservatively 1.1 the

fraction 0.66 = 1.1 x 0.6 is arrived at. Similarly the limiting value of 0.75 F Y for solidbars conservatively reflects the ratio of plastic section modulus to elastic sectionmodulus for these.

The values of F b may not however be the upper limit permissible, the limit havingto be lowered if local buckling or flexural-torsion buckling are issues, as dealt withby Clauses 5.3 and 5.4 respectively.

Earlier in this chapter local buckling was discussed and it was shown that plate

elements loaded in compression were liable to buckle locally before yielding if their

width to thickness

t

b ratios were larger than certain limits. Limits were therefore

imposed on this ratio and widths that caused these limits to be exceeded wereregarded as excessive and the excess portion of width was not to be included in thecalculation of effective cross sectional area.

If a flange has a width that is right on the limit then the compressive stress if largeenough would cause yield and exposure to the likelihood of local buckling to occur

simultaneously. Such flanges therefore have the permitted compressive stresslimited to 0.60 F Y . If the flange

t

b ratio is less than the upper limit then the

compressive stress may be more than 0.60 F Y , the allowed value gradually

increasing to 0.66 F Y ast

b gets less. This 0.66 F Y limit is allowed once the

t

b ratio

has fallen to the point where were the factor of “safety” 1.0 the compressive stresscould theoretically get high enough without local buckling happening to allow thebeam to yield right through [i.e. a “plastic hinge”]. These limits on compressive

stress as affected byt

b ratio, are dealt with in Clause 5.3 in AS3990.

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Optional Activi ty 8L - Extra reading on limits to permitted bendingcompressive stress as affected by b/t ratio.

These limits on permitted compressive bending stress F bc as affected byt

b ratio, are

illustrated graphically for the case of the flange on an I section (i.e. plate elementsupported on one side only) in Fig 45 of the following extract:Reduction of Permitted Compressive Bending Stress due to local buckling for anI section (PDF 102KB) 21

For flanges in box sections, i.e. in which the plate is supported on both sides, thescale on the horizontal axis of Fig 45 is altered by a factor of:

256560 for welded flanges or plates that have not been stress relieved

256

800 for those that have been stress relieved

256

635 for cold formed rectangular hollow sections

F bc would have to fall below 0.60 F Y if the bending stress f bc were calculated based

on the full cross sectional area rather than the “effective” cross sectional area, once

thet

b limit associated with local elastic buckling is exceeded. This falling away is

described by( )

( )t b

t bF F Y bc

limit60.0= and is illustrated in Figure 4 in the following

extract:

Reduction in Maximum Permissible Compressive Stress in Bending F bc (PDF120KB) 22

21 Lay, M. G.,1982, Comments 41-88, Source book for the Australian Steel Structures Code, AS 1250, 3rd ed.,Australian Institute of Steel Construction, Milsons Point, N.S.W., p.18, viewed on 28th August 2013 <https://equella.rmit.edu.au/rmit/file/9d342f52-3ab6-4eca-b7ea-ba3d314853ff/1/130823_3_052.pdf>

22 Australian Institute of Steel Construction 1974, Flange local buckling, Part 1 - Design if beams, design ofcolumns (metric units) : Steel design course : a short course on design of steel structures with special referenceto AS 1250 SAA steel structures code 2nd ed., Australian Institute of Steel Construction, Sydney, p. 4-5, viewedon 28th August 2013 < https://equella.rmit.edu.au/rmit/file/55ba3166-d3f5-449d-b376-9a6373d60125/1/130823_3_054.pdf>

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The limit on compressive stresses for rolled sections, as calculated on effectivesection properties that may have had to be lowered from 0.66 FY to 0.60 FY toguard against local buckling, may have an even lower limit imposed on it ifflexural-torsional buckling is a potential mode of failure. Flexural-torsionalbuckling of beams (sometimes referred to as lateral buckling of beams) is thetwisting sideways that can occur when a beam is bent about the axis of majorstrength.

Acti vi ty 8M - Flexural -tors ional buckl ing of beams

View an example of flexural-torsional buckling of beams which is illustrated in

Figure 8 in the following extract:Torsional flexural buckling of a simply supported beam without immediatelateral restraints (PDF 53KB) 23

Clause 5.4 of AS3990 deals with this issue. A maximum allowed bending stress isdetermined that will, with an appropriate factor of safety, guard against this modeof failure. The calculation of this permitted stress takes into consideration the ratios:

web) to (parallel

thicknessflange

sectionof depth andstrengthminimumof axisabout thegyrationof radius

lengtheffective .

Obviously if a section having equal strength about each axis, such as a squarehollow section, is chosen for use as a beam column then the issue of this mode offailure would not arise.

23 Hosking A K & Harris M, 1981, Structural design to AS 1250-1981, Applied mechanical design 2nd ed., H andH Publishing, Forest Hill, Vic. p.12.7, viewed on 28th August 2013 <

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Returning now to the “unity” of equation 8.3.1(a)(1), we see that the second and thirdgroup of terms are essentially comparing for each axis of bending, the appropriatelyamplified compressive bending stress to the allowed compressive bending stress, saidcomparison being expressed as a fraction. Thus for the whole equation it can be seenthat:

stressecompressivaxial

average permitted maximum

stressecompressiv

axialaverage

+

axis

about bendingforstress bending

ecompressiv permitted maximum

axisabout bendingforstress bending

ecompressivamplified elyappropriat

x

x +

axis

about bendingforstress bending

ecompressiv permitted maximum

axisabout bendingforstress bending

ecompressivamplified elyappropriat

y

y

1≤

If the average axial compressive stress is comparatively low

< 15.0

ac

ac

F

f then the

amplification of bending stress that the axial load produces would be small and isdisregarded, giving:

1≤++bcy

bcy

bcx

bcx

ac

ac

F

f

F

f

F

f eqn. 8.3.1(a)(2)

Equations 8.3.1(a)(1) and (2), as appropriate, are used for checking the member as awhole, using the maximum values of f bcx and f bcy that occur along it. A further check isalso done at the support, using the values of f bcx and f bcy at the support:

160.0

≤++bcy

bcy

bcx

bcx

Y

ac

F

f

F

f

F

f eqn. 8.3.1(b)

For this equation F bcx and F bcy are derived using Clauses 5.2 and 5.3 only (i.e. flexural-

torsional buckling is not limiting the allowed compressive stress at the support endsof the beam-column).

Unity equations such as the three above are sometimes thought of as:

1axisabout bendingtodevoted

strengthscolumn'of fraction

axisabout bendingtodevoted

strengthscolumn'of fraction

load ecompressivaxialtodevoted

strengthscolumn'of fraction≤++

y x

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The designer must therefore choose a size and type of steel section whoseproperties i.e. yield strength, effective cross sectional area, effective sectionmodulus, and radii of gyration are such that in combination with the loads andeffective length of the set-up, the stresses generated within the section are such asto satisfy the interaction “unity” equations.

As earlier shown in Activity 8H, charts are available to help make trial selections ofcolumn section area to safely carry axial loads, however where the column issubject to end moments as well that section area would need to be increased. Theprocess typically involves trial selection aided by past experience and subsequentchecking as described above to make sure the stresses generated within the trialsection are such as to satisfy the interaction “unity” equations

For the case of the columns to be used to support the lugs on the pressure vessel inQuestion 6(b) of Project Part B it is suggested you look at the relative proportions ofthe column and lug widths in Figure 3.24(a) in AS1210 and similarly in Fig 8.2dbelow. In your response to Question 6(a) the lugs have already been selected to beof adequate strength, and their associated gusset spacing, shown as “g”in fig 8.2d,is hence known. Selecting a trial column breadth “B” to match “g”will providecolumn walls aligned for good weight transfer from the lug.

Fig. 8.2d: Column to lug trial proportionality© RMIT University, 2013 (Dixon C., Marchiori G.)

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Appendix 8 gives sample commercial catalogue data of square hollow sections thatcan be used to select such a trial column.

Safe load tables for axial loads are also given in that sample data, however it must

be remembered that the loading situation for this capped column is not simpleaxial, rather it is eccentric loading and must be analysed using the relevantinteraction unity equation. The worst possible position for the vertical load on thecolumn cap is that shown in Fig 8.2e producing biaxial bending, hence all threeadditive terms need to be included in the relevant interaction unity equation.

Fig. 8.2e: Possible eccentricity on column cap under lug © RMIT University, 2013 (Dixon C., Marchiori G.)

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Appendix 8A - Sample Commercial data of square hollow sections

Refer to the sample commercial data of square hollow sections on pages 2, 9, 16and 24 of the ‘Tubeline: safe load tables’ from Tubemakers of Australia Limited.

(PDF 404KB).24 Use it as appropriate in Question 6b of Project Part B.

24 Tubemakers of Australia Limited, Steel Pipe Division, 1984, Tubeline : safe load tables 2nd ed., Milsons Point,N.S.W. : Steel Pipe Division Tubemakers of Australia, p. 2, 9, 16, 24, viewed on 28th August 2013 <

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Activi ty 8N - Reading, calcu lat ions and drawing

Step 1: Refer to Project Part B in AssessmentRefer to Project Part B in the Assessment Section. Read Question 6 (b) and (c).

Step 2: Have to hand relevant parts of AS3990

If you have not already done so, print out pages 20, 21, 22, 23, 40, 42, 54 and 55 ofAS3990 “Mechanical Equipment – Steelwork This Australian Standard can beaccessed via the RMIT Library - SAI Global link to the Australian Standards. 25

NOTE: You will need to login with your student login and password.

Step 3: Plan.

Discuss with your partner the strategy for tackling Question 6 (b) and (c).

The essence of Question 6 (b) is checking that your proposed column satisfies theappropriate “unity” inter-active equation. One such equation is 8.3.1(a)( 1) inAS3990. There are many terms in this equation, which have been discussed inchapter 8. Some tips are: Values for Fbc for some commercial square hollowsections may be found on the last page of the extract accessible in Appendix 8.Fac and Foc may be found in Table 6.1.1. in AS3990. Cmx and Cmy may be taken as1.0; this is a conservative approach. fac is the axial compressive stress in your

column which you must calculate. fbcx and fbcy are the unamplified bendingcompressive stresses induced in your column by the eccentricity of the load,which you must calculate.

Step 4: If you are the partner responsible for completing Question 6 (b), do so,consulting with your partner where appropriate.

Step 5: If you are the partner responsible for completing Question 6 (c), do so,consulting with your partner where appropriate.

Feedback:

Feedback will be provided on your submitted project documentation by theengineering lecturer/tutor responsible for marking it.

25 https://login.ezproxy.lib.rmit.edu.au/login?url=http://www.saiglobal.com/online/autologin.asp





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Summary and Outcome Checklis t

Tick the box for this statement if you agree with it:

I can determine the proportions of supporting columns includingconsideration of buckling, yield, eccentric load and initial crookedness.

Assessment

This topic will be assessed as part of the Project Part B and the end of semesterexamination (see: Assessment section of the Course Introduction for more detail).

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