Buckling Strength of Tie-Struts and Application to Frame Girders

1. INTRODUCTIONThe so-called tie-struts are those members subjectpartly to axial compression and partly to axial tensionwithin the two end supports. There are three typicalmembers with different patterns of force distribution asshown in Figure 1. In the first type, the member isdivided into two portions: one tensile and onecompressive. In the other two types, the member isdivided symmetrically into three portions, and theforces in the two end portions are of the same sign. InFigure 1, Nt denotes tension and Nc compression. So,the middle portion is subject to tension for type 2, andto compression for type 3.

Figure 2 depicts a lattice girder of rigid frames. Forout-of-plane buckling calculation, the bottom chordbelongs to type 1 when braced at mid point B [Figure 2(b)], and to type 2 when unbraced [Figure 2(c)].In case the lattice girder is a spatial one, which does not

Buckling Strength of Tie-Struts and Application to

Frame Girders

Shaofan Chen*

Xi’an University of Architecture & Technology, Xi’an 710055, China

(Received: 7 January 2009; Received revised form: 12 October 2009; Accepted: 16 October 2009)

Abstract: Tie-struts are members subject partly to axial compression and partly toaxial tension within the two end supports and appear in the chords of lattice girders ofrigid frames. There are three types of tie-strut with different patterns of forcedistribution. The bearing capacity of these members has to be determined by stabilitycalculation. Although the current Chinese code of cold-formed thin-wall steelstructures GB50018 provides a formula for calculating tie-strut chords, it can only beused for the stability check of one of the three types. In this study, analytical equationsof elastic critical load are firstly derived, based on those of struts subject tocompression varying by step and then parametric analyses were carried out andpractical formulas for member effective length are obtained through curve fitting foreach of the three types. The recommended formulas have good accuracy and can beused in design work. The second part of this study covers the application of tie-strutbuckling calculation to solid-web frame girders. Similar to strut analogy in beambuckling, a tie-strut analogy is developed for frame plate girders and this approach isespecially useful for examining buckling induced by lower flange compression whenthe upper flange is restrained against lateral translation.

Key words: tie-strut, buckling, effective length, frame lattice girder, frame plate girder.

*Corresponding author. E-mail: [email protected]; Tel: +86-29-82202199.Editor. K.F. Chung.

Advances in Structural Engineering Vol. 13 No. 4 2010 525

require any bracing for lateral restraint, the upper andlower chords are latticed tie-struts of type 3 and 2respectively.

The buckling strength of tie-struts has scarcely beendiscussed in the engineering literature. To the writer’sknowledge, the Chinese code for cold-formed thin-wallsteel structures GB50018(2002) is the sole documentgiving instructions upon this subject where thefollowing formula for effective length is provided forchecking the out-of-plane stability of the chordmembers of a lattice girder of rigid frames

(1)

where l is the distance between adjacent laterally bracedpoints, within which some panels are subject tocompression and some others to tension; and are average tensile and compressive forces respectively,

NcN t

l N Nn

nl0 1 5 0 5= +( ). . /t c

c

Buckling Strength of Tie-Struts and Application to Frame Girders

526 Advances in Structural Engineering Vol. 13 No. 4 2010

with compression as positive; n and nc are total numberof panels and number of compressive panelsrespectively. The force distribution pattern relevant toEqn 1 is not mentioned in the code provisions. However,analysis shows that this equation is not a universal oneand is applicable solely to tie-struts of type 1. Theobjective of this study is firstly to present theoretical aswell as practical formulas for all three types, and,secondly, to apply these formulas to frame plate girdersin addition to latticed ones.

Considering the similarity between plate and latticedgirders, a tie-strut analogy approach is proposed forinvestigating buckling problems induced by top flangecompression, as well as by compression in the endportions of the lower flange of frame girders.

2. THEORETICAL BACKGROUND OFSTABILITY CAPACITY

In some monographs on structural stability, formulas ofcritical load are given for struts subject to step variationsin compression, such as those shown in Figure 3. Formembers comprising two portions [Figure 3(a)], thecritical condition given by Timoshenko and Gere (1961)and Pflüger (1964) is

(2)

where v1, v2, v3 are parameters:

l is the member

length; l1 and l2 are length of portions subject to force P1

and P1 + P2 respectively.

ν ν ν11

22

31 2, , ;= = = +P

EI

P

EI

P P

EI

νν

νν

ν νν ν

ν22

12

22

32

12

22

1

1 1 1

32

tan− − + −l l

l

ll l

l

− =νν ν

22

2

3 3 2

0tan

Eqn 2 is also valid for the case when one of the forcesis tensile or of negative sign. When P2 < 0 and P1 < |P2|,the right hand portion of strut will be in tension, and thecompressive and tensile forces are Nc = P1 and

respectively. Substituting ,

and into

Eqn 2, it becomes

(3)

By solving this equation for v1, the critical value of Nc

can be obtained:

This critical load applies to fully elastic membersdevoid of any imperfection, and has to be transformedinto an expression for an effective length factor for usein routine design. From the equation

the effective length factor is given by

µ = π /(v1l). (4)

πµ

ν2

2 12EI

lEI

( )=

N P EIc = =1 12ν

νν

νν

ν νν ν

ν22

12

22

32

12

22

1

1 1 1tan+ +

−+

l l

l332

22

2

3 3 2

0l l

ltanh

−=

νν ν

tan tanhν3 22 1

2l iP P

EIl=

−ν3

2 1=−

iP P

EI

ν22= i

P

EIN P Pt = −2 1

Nc NcNc

Nc

lc lc/2 lt/2 l t/2lc/2lt lt lcl

(a) Type 1 (b) Type 2 (c) Type 3

l l

Nt Nt Nt Nt

P1 P1 P1 P1 P1P2 P2 P2 P2P2P2 − P1

Figure 1. Force distribution in typical tie-struts

A B

1

1C

M

B− −

+(a)

(b) (c) (d)

Section 1-1

Figure 2. Tie-strut in lattice girder of rigid frames

l1 l2

l

(a)

(b)

P1P2 P1 + P2

l1 l12l2

l

P1P2 P2 P1

Figure 3. Struts under varying compression

Shaofan Chen


The critical condition of three-portion membersshown in Figure 3(b) can be found in Pflüger (1964) asfollows

(5)

where v1 , v2 , v3 are the same parameters as in Eqn 2; l1and l2 are the lengths of the end portion and the half-length of the middle portion respectively.

When P1 > 0, P2 < 0 and P1 < |P2|, the middle portionis subject to tension, as for type 2 in Figure 1. Thecompressive and tensile forces are Nc = P1 and

respectively. Whereas when P1 < 0, P2 > 0 and P1 < |P2 |, the two end portions will be under tension,as for type 3 in Figure 1. The compressive and tensileforces are respectively and Nt = P1. The critical conditions become respectively

For type 2 (6)

For type 3 (7)− ⋅ − =νν

ν ν1

31 1 3 2 1 0tanh tanl l

νν

ν ν1

31 1 3 2 1 0tan tanhl l⋅ − =

N P Pc = −2 1

N P Pt = −2 1

νν

ν ν1

31 1 3 2 1 0tan tanl l⋅ − =

The corresponding effective length factors arerespectively

µ = π /(v1l ) and µ = π /(v3l) (8, 9)

3. COMPARISON BETWEEN THE THREETYPICAL CASES

The effective length factor calculated by Eqns 4, 8 and9 for various λc = lc /l and various η = Nt /Nc arecompared in Table 1, where also given are resultscalculated from Eqn 1. It is quite obvious that for thesame η and the same λc, the effective length factors µare not identical for the three types, but differ betweenthemselves by a large amount. The µ–factor is largestfor type 2, showing the lowest bearing capacity, andsmallest for type 3 with highest bearing capacity. Thissituation indicates that tension at member ends is veryeffective in restraining the buckling tendency, whereascompression at the ends is more critical than when it isacting on the middle portion. Eqn 1 of GB50018matches type 1 fairly well only in some ranges andoverestimates the buckling strength for small values ofλc. Figure 4 gives a clear picture of the situation for λc = 0.5.

Table 1. Comparison of -factor

c = lc/l = Nt/Nc Type 1 Type 2 Type 3 Eqn 1

0.5 0.575 0.665 0.214 0.3750.3 1.0 0.466 0.583 0.200 0.300

1.5 0.384 0.528 0.192 0.300

0.5 0.591 0.873 0.364 0.6250.5 1.0 0.499 0.730 0.335 0.500

1.5 0.457 0.684 0.320 0.500

0.5 0.690 0.970 0.566 0.8750.7 1.0 0.637 0.961 0.498 0.700

1.5 0.608 0.952 0.463 0.700

Note: lc is the length of the portion under compression`

ηηλλ

µµ

0.8

0.6

0.4

0.20.5 1.0 1.5

= Nt / Ncη

µ

c = 0.5λ

Type 2

Type 1

Type 3

Eqn. 1

Figure 4. Comparison of µ–factor for λc = 0.5



4. SIMPLIFIED PRACTICAL FORMULASEqns 3, 6 and 7 are transcendental ones, complicatedto solve, and only apply to perfect members in theelastic range. So, simplified formulas for calculatingµ–factor are deemed necessary for design use. Toachieve this aim, performed first is a parametriccalculation of µ–factors for various η and λc, thenplotted are the family of curves of µ–versus η and λc,and finally the simplified expressions are formulatedthrough curve fitting.

For type 1 tie-strut, µ–factor is plotted against η forvarious values of λc as shown in Figure 5. These curveshave a similar trajectory and can be approximated by theone expression

where ξ and γ are factors depending on λc. ξ can berecognized as the µ-factor of a particular tie-strut withη = 0, which is actually a partially loaded strut.Referring to Figure 3(a) and letting P1 = 0, this µ or ξ

µ ξ γη= −( )1 0 5.

factor may be solved using the following equation(Pflüger 1964)

(10)

where the notations are identical to those of Eqn 2.Table 2 presents the values of ξ thus obtained, comparedwith the approximate expression

(11)

It can be seen that Eqn 11 errs somewhat on the safeside, but the discrepancy is slight except for λc ≤ 0.2.The factor γ , as determined for each curve in Figure 5,varies approximately as a linear function of λc

γ = 0.4375 − 0.375λc

Finally, the simplified formula for a type 1 tie-struttakes the form

(12)

For a type 2 tie-strut, the µ-factor is plotted againstλc for various values of η [Figure 6(a)]. Examinationof this family of curves leads to the followingexpression

(13)

where α is a function of η as follows

when (13a)

when (13b)0 1 1 2 7 3 1. .� η α η< = + −( )3

1 2 2 2 0 5 22< = + −( )η α η� . .

µ λ α= − −( )1 1 c

µ λ λ γη= + −( ) + −( )

−0 75 1 0 3 0 5 0 7 0 5 1

2 0 5. . . . . .c c (( ).

ξ λ λ. . . . .= + −( ) + −( )0 75 1 0 3 0 5 0 7 0 52

c c

lv

l

l

l v

v l12

22

1

1 2

2 231 0

tan.− − − =

Table 2. Comparison of -factor

c 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

ξ, Eqn 10 0.58 0.65 0.70 0.71 0.73 0.74 0.78 0.83 0.91ξ, Eqn 11 0.73 0.73 0.73 0.73 0.75 0.78 0.82 0.87 0.92

λλ

ξξ

Figure 5. Curves of µ–factor for type 1 tie-strut

1.0

0.9

0.7

0.8

0.6

0.5

0.4

0.3

0.20 0.5 1.0 1.5

η

µ

2.0

0.8

0.7

0.6

0.50.40.30.20.1

c = 0.9λ

Shaofan Chen


versus λc curves, which are fitted by the followingexpressions

when λc � 0.2 µ = 0.85λc (14a)

when λc > 0.2 (14b)

where β = 1 + 0.8η when η � 0.5 (15a)

β = 1.2 + 0.4η � 1.8 when η > 0.5 (15b)

Tables 3, 4 and 5 give the theoretical µ–factor valuesobtained from Eqns 3, 6 and 7 respectively, in parallelwith approximate ones calculated by the simplifiedformulas (Eqns 12, 13 and 14). Comparison reveals thatthe suggested formulas, as a whole, have fairly goodaccuracy. Only a few entries err on the unsafe side, i.e.µ is less than the theoretical value, and within a smallpercentage in any case. Those with noticeablediscrepancy are all on the safe side mostly for smallvalues of λc. Hence, the preferable application range ofthese simplified formulas lies within λc ≥ 0.3 and η ≥1.5. But they may also be safely used for 0.1 ≤ λc < 0.3and 1.5 < η ≤ 2.0 at the expense of accuracy.

The simplified formulas recommended above areaimed at tie-struts with axial forces constant for eachportion as shown in Figure 1. In actual lattice girders,however, each portion may comprise several panelswith different forces and it is conceivable to use theaverage force in the portion as Nt or Nc.

µ λ= +

−

0 17 0 830 2

0 8. .

.

.c

β

Figure 6. Curves of µ–factor for type 2 tie-strut and curve of

exponent α

1.0

0.8

(a)

(b)

0.6

0.4

0.2

0 0.1 0.2

= 0.1

0.3

0.3

0.4 0.5

0.51.0

1.5 2.0

cλ

η

α

µ η

0.6 0.7 0.8

0.5 1.0 1.5 2.0

0.9

10

2

3

4

5

Table 3. Effective length factor of type 1 tie-strut

c

η 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.25 0.415 0.577 0.639 0.654 0.656 0.678 0.729 0.799 0.8890.595 0.598 0.608 0.628 0.656 0.695 0.745 0.806 0.878

0.50 0.355 0.514 0.575 0.585 0.591 0.627 0.690 0.770 0.8700.524 0.543 0.559 0.584 0.617 0.661 0.715 0.781 0.859

0.75 0.317 0.465 0.517 0.522 0.538 0.589 0.660 0.746 0.8530.486 0.501 0.522 0.550 0.588 0.635 0.692 0.762 0.844

1.00 0.291 0.426 0.466 0.470 0.499 0.562 0.637 0.725 0.8370.446 0.465 0.490 0.522 0.563 0.613 0.673 0.746 0.832

1.50 0.257 0.365 0.384 0.398 0.457 0.529 0.608 0.695 0.8090.380 0.406 0.437 0.475 0.520 0.576 0.641 0.715 0.811

2.00 0.234 0.318 0.326 0.366 0.435 0.510 0.589 0.674 0.7860.323 0.356 0.392 0.435 0.485 0.544 0.614 0.697 0.793

Note: For each η value, the upper line gives theoretical µ and the lower line calculated by Eqn 12

λλ

These last two expressions are depicted in Figure 6(b),where α values determined by curves in Figure 6(a) aremarked by sign ×.

As for type 2, simplified formulas for the µ–factor oftype 3 tie-struts are obtained through a family of µ




c

η 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.1 0.414 0.618 0.748 0.840 0.905 0.949 0.978 0.993 0.9990.403 0.664 0.825 0.918 0.966 0.989 0.999 1.000 1.000

0.3 0.310 0.549 0.705 0.813 0.889 0.941 0.974 0.992 0.9990.325 0.565 0.736 0.851 0.925 0.967 0.989 0.998 1.000

0.5 0.255 0.490 0.665 0.787 0.873 0.933 0.970 0.991 0.9990.277 0.496 0.666 0.792 0.881 0.940 0.975 0.993 0.999

0.7 0.226 0.445 0.629 0.763 0.859 0.924 0.966 0.990 0.9990.254 0.462 0.629 0.758 0.854 0.922 0.965 0.989 0.998

1.0 0.200 0.399 0.583 0.730 0.838 0.913 0.961 0.988 0.9980.248 0.453 0.618 0.748 0.846 0.916 0.961 0.987 0.998

1.5 0.177 0.355 0.528 0.684 0.806 0.894 0.952 0.985 0.9980.217 0.405 0.564 0.695 0800 0.881 0.939 0.976 0.995

2.00.164 0.329 0.492 0.647 0.779 0.878 0.944 0.982 0.9990.207 0.388 0.544 0.675 0.782 0.867 0.929 0.971 0.994

Note: For each η value, the upper line gives theoretical µ and the lower line calculated by Eqn 13

λλ


c

η 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.1 0.084 0.168 0.256 0.352 0.455 0.562 0.671 0.780 0.8900.085 0.170 0.258 0.356 0.458 0.563 0.670 0.778 0.888

0.3 0.076 0.152 0.228 0.308 0.396 0.500 0.616 0.741 0.8700.085 0.170 0.233 0.319 0.416 0.521 0.633 0.751 0.875

0.5 0.071 0.143 0.214 0.286 0.364 0.455 0.566 0.699 0.8440.085 0.170 0.215 0.289 0.380 0.485 0.600 0.725 0.858

0.7 0.069 0.139 0.208 0.278 0.350 0.432 0.535 0.671 0.8310.085 0.170 0.208 0.277 0.364 0.468 0.584 0.712 0.851

1.0 0.067 0.133 0.200 0.267 0.335 0.408 0.498 0.627 0.8030.085 0.170 0.200 0.260 0.343 0.444 0.561 0.694 0.840

1.5 0.064 0.128 0.192 0.256 0.320 0.387 0.463 0.573 0.7580.085 0.170 0.190 0.238 0.312 0.408 0.526 0.665 0.823

2.0 0.062 0.124 0.187 0.249 0.311 0.374 0.444 0.538 0.7170.085 0.170 0.190 0.238 0.312 0.408 0.526 0.665 0.823

Note: For each η value, the upper line gives theoretical µ and the lower line calculated by Eqn 14.

λλ

5. TIE-STRUT ANALOGY OF FRAMEGIRDERS

5.1. Strut Analogy of Simple Beams

It is well known that the elastic buckling of beams canbe treated as that of a T-section strut in somecircumstances. This T-section is composed of thecompression flange and one sixth of the web plate ofthe beam. The critical moment of a beam subject touniform bending is given by the expression

(16)MEI

l

h l GI

EIy

crt= ⋅ +

ππ

2

2

2

221

ω

where EIy, EIω and GIt are the flexural rigidity about theweak axis, the warping rigidity and the St. Venanttorsional rigidity respectively; l and h are the spanlength and section depth of the beam respectively. Whenthe St. Venant rigidity term is neglected for simplicity,the corresponding critical stress is given by (Gaylord et al. 1992)

(17)

where iT is the radius of gyration of the above-mentioned T-section about the y-axis, and σcr in Eqn 17

σ π πcr

f w

T= ⋅+

=2

2

2 2

2

2

6

E

l

I

A A

Ei

ly /

/

Shaofan Chen


is just identical to the critical stress of a strut having thisT-section and unsupported length l. The omission of St.Venant rigidity renders the critical stress somewhatconservative, more so for girders of compact section.But, this simplifying measure has the prospect of usagein continuous beams and frame girders whose lowerflange is partly under compression and less effective inproviding restraint to lateral deflection of the upperflange and vice versa.

5.2. Analogy Between Trame Girder

and Tie-Strut

The strut analogy for beams can be extended to therelationship between unbraced plate girder of framesand tie-struts with T-section just as above. It is alsovalid for frame girders braced at the top flange.

Designers often worry about buckling of a framegirder induced by lower flange compression at the girderends, and so far there has been no in-depth investigationon this subject. When the upper flange of the framegirder is not braced against lateral displacement,buckling of the member is governed by upper flangecompression. The normal routine stability verificationprevents buckling of the girder as a whole and there isno need to worry about the lower flange, except itsresistance against local buckling.

On the other hand, when adequate bracing isprovided for the upper flange and no routine stabilityverification is to be performed for the girder, the role oflower flange compression in inducing buckling has to beof concern. One way of treating this problem is to takethe distance between the end support and the inflexionpoint as the effective length for stability calculation.This is certainly not correct because the compressiveportion is a constituent part of the whole lower flangeand of the whole girder as well. Tie-strut analogy maybe useful in this situation.

Figure 7 depicts a simply supported beam subject touniform negative bending moment, where bracingsrestraining the lateral displacement of the upper flangeare spaced closely. When this beam buckles, it rotatesabout the longitudinal axis through the mid point A of theupper flange without flexure, and as a result, this problemcan be solved by a single differential equation of torsionalequilibrium as follows (Timoshenko and Gere 1961):

(18)

where θ is the rotation angle of the cross section and uis the displacement of the shear center along the x-axis.Replacing u by hθ/2, Eqn 18 becomes

(18 a)EI GI Mhω ′′′ − ′ − ′ =θ θ θt

1

20

EI GI Muω ′′′ − ′ − ′ =θ θt 0

For a simply supported beam, the solution of Eqn 18(a)is θ = Asin(nπz/l ). Substituting the derivatives of θ intoEqn 18(a) and knowing that M has a negative sign, oneobtains

which leads to the minimum value of the criticalmoment for n = 1, as follows

(19)

Comparing with Eqn 16, it can be seen that thecritical moment is amplified by the factor

(20)

owing to translational restraint at the upper flange.Again, the term with GIt can be neglected, the factor ρreduces to unity and the critical stress is the same as thatgiven by Eqn 17. Thus the strut analogy can be extendedto a girder braced against lateral translation at the top,and a tie-strut analogy can be developed for solvingconservatively the buckling problem induced bycompression in the end portions of the lower flange.

The feasibility of the tie-strut approach for framegirders is first demonstrated through bucklingcalculation of unbraced girders subject to uniformloading on the upper flange. Figure 8(a) shows thebending moment diagram of such a girder. The lateralbuckling of this girder without any bracing is examinedin parallel by two approaches, namely the design codeprovision and the tie-strut analogy.

ρ = +12

2

l GI

EIt

π ω

MEI

l

h l GI

EIcrt= ⋅ +

ππ ω

2

2

2

221y

n EI

lGI

Mh2 2

2 20

π ω + − =t

M M h

C ux

A

θ

y

y

l

zO

Figure 7. Simple beam subject to uniform negative

bending moment



The overall stability factor, ϕb = σcr / fy, of a simplysupported I-beam is given by the following formula inthe Chinese design code GB50017 (2003)

(21)

where ly is the slenderness ratio about the minor axis y; A and Wx are sectional area and section modulus

about the major axis x respectively;h and t1 are girder depth and thickness of the

compression flange respectively; fy is the yield strength of the material in MPa;βb is a factor of equivalent critical moment for non-

uniform moment distribution, equal to 1.13 forparabolic diagram in a simply supported beam, and isdenoted as Cb in the American design code.

ϕb values larger than 0.6 shall be replaced by

(21a)ϕϕb

b

′ = −1 070 282

..

ϕ βλ

λb b

y x

y

y

= ⋅ +

⋅4320

14 4

2352

1

2Ah

W

t

h f.

As there is no instruction regarding βb of framegirder in the GB50017 code, the following formula ofCb provided in the American design code ANSI/AISC360-05 (2005) is used

(22)

where Mmax is absolute value of maximum moment inthe unbraced segment;

M1/4 and M3/4 are absolute value of moment at the leftand right quarter point of the unbraced segmentrespectively;

M1/2 is absolute value of moment at the centre line ofthe unbraced segment.

This formula does not take account of the effect ofload position, and, for loading above the shear centre,additional factor less than unity should be adopted.

The first step in applying the tie-strut analogyapproach is to transform the varying stress in the threeportions of the girder into constant stresses. In the twoend portions, the upper flange is tensile, with a nearlylinear variation, and 0.5σt.max is taken as the averagevalue so that Nt ∝ 0.5 × 0.6 = 0.3. The middle portionis compressive with stress varying parabolically, andthe effective mean stress is taken as σc.max /1.13, where1.13 is the βb factor mentioned earlier, and Nc ∝ 0.4 ×1.13 = 0.36.

The second step consists of the application of Eqn 14for the type 3 tie-strut to find the effective length factorµ, which leads to the slenderness ratio

(23)

For non-uniform moment distribution, the reducedslenderness ratio

(23a)

should be used.Finally, based on , the stability factor

ϕ = σcr /fy can be obtained from the relevant table in thedesign code. In the GB50017 code, three ϕ–curves a, b,c are specified relating to the intensity of compressiveresidual stress.

The results calculated using the two approaches aresummarized in Table 6. Girders of five different crosssections have been investigated. Two of them are rolledand three others welded. They are all of steel gradeQ235, with yield strength equal to 235MPa. The

λ Tr yf /235

λ λTr T b= / β

λ TT

=µl

i

βb =+ + +

12 5

2 5 3 4 31 4 1 2 3 4

.

.max

max / / /

M

M M M M

0.6

M0

0.184l0

M0

0.632l0 0.184l0l0=l

0.4

M0

(a)

(b)

(c)

Nt¡�0.3

Nt¡�0.267

Nc¡�0.3

Nc¡�0.343

For upper flange

For lower flange

A BC

Figure 8. Girder moment and axial force of the analogous tie-strut

Shaofan Chen


designations and dimensions (in mm) of these sectionsare as follows

R1 588 × 300 × 12 × 20

R2 600 × 200 × 11 × 17

W1 800 × 300 × 12 × 20

W2 1200 × 600 × 16 × 36

W3 1200 × 400 × 18 × 25

All the specimens are subject to uniformly distributedloading acting at the top of the upper flange, and the βb

factor given by Eqn 22 is reduced by multiplying by0.8/1.13 = 0.71 for rolled sections and by 0.73/1.13 =0.65 for welded sections according to Chen (2008). 0.8and 0.73 being the counterpart of 1.13 for loading on theupper flange.

It can be seen from Table 6 that factor ϕb or ϕb′ forthe girders is in good agreement with ϕT,a for sectionsR1, R2 and W2, with ϕT,b for W3, and in between ϕT,a

and ϕT,b for section W1. So, the tie-strut analogyapproach is not conservative for unbraced framegirders, and, to guarantee the required reliability, it isdesirable to use stability factor curve “a” for rolledgirders and curve “b” for welded ones.

If design codes other than the GB50017 are used, acalibration should be carried out to ascertain whetherany adjustment is required. For example, the girderswith section R2 and W2 are calculated according to theAmerican code ANSI/AISC 360-05. As there is no Cb

factor provided in the latter to take account of thedestabilizing effect of the loading on the upper flange,

the reduction factors 0.71 and 0.65 used above areadopted. At the same time, and

are adjustment factors to be applied to the slenderness ratios of the T-section tie-strut. Theresults obtained are summarized in Table 7. The σcr / fyratios from tie-strut approach are in general less than thosefrom girder approach, except for welded section W2with span length l = 18h. A tentative suggestiontherefore is to reduce the calculated strength of the tie-strut by 10% to 15% for girders with l/b ≥ 36, where bis the flange width.

As for frame girders with a closely braced upperflange, owing to the inherent safety margin arising fromthe simplification ρ = 1, there is no need for specificmeasures to reduce the buckling strength obtained fromthe tie-strut analogy.

To apply the tie-strut analogy to buckling induced bylower flange compression (BILFC in brief), thetransformation of the varying stresses into constant stressesis different to the above. For the moment diagram shownin Figure 8(a), the effective mean compressive stress in theend portions is taken as σc.max/1.75 so that Nc ∝ 0.6/1.75 = 0.343 (1.75 being the βb factor for atriangular moment diagram). The average tensile stressequals (2/3)σ t.max for the middle portion and Nt ∝ 0.4 × 2/3= 0.267. At the same time, Eqn 13 for the type 2 tie-strutis used to calculate the effective length factor µ.

The calculation is based on the moment distributionshown in Figure 8(a) and the axial forces in the threeportions shown in Figure 8(c). Results of specimens R2and W2 are given in Table 8.

1 0 65 1 24/ . .=

1 0 71 1 19/ . .=

Table 7. Stability factors according to ANSI/AISC 360-05

Section R2 W2

l/h 12 15 18 12 15 18

σcr /fy, girder 0.82 0.60 0.47 0.95 0.83 0.59σcr /fy, tie-strut 0.70 0.58 0.45 0.86 0.79 0.72

Table 6. Calculation results of stability factors

Section

l/h R1 R2 W1 W2 W3

12 0.93 0.75 0.77 0.91 0.66ϕb or ϕb′ 15 0.87 0.63 0.65 0.84 0.48

18 0.83 0.51 0.51 0.76 0.38

12 0.91, 0.84 0.75, 0.65 0.81, 0.72 0.90,.0.84 0.73, 0.63ϕ T,a, ϕ T,b 15 0.86, 0.77 0.58, 0.51 0.68, 0.59 0.85, 0.76 0.56, 0.49

18 0.79, 0.69 0.44, 0.39 0.54, 0.47 0.77, 0.68 0.42, 0.38

Note: ϕ T,a and ϕ T,b are stability factors of T-strut relevant to ϕ–curve a and b respectively.



If 1.75 times ϕ T is larger than unity, the design willbe governed by cross section strength rather thanbuckling, because the end section will start to yield.Conversely, when 1.75 ϕ T < 1.0 or ϕ T < 0.57,measures have to be taken to prevent buckling. FromTable 8, it can be seen that girders of section R2 withdepth-width ratio of 3, are weaker than those ofsection W2 with depth-width ratio of 2. The latter issafe against buckling for l/h equal to 15 or less.

In order to prevent BILFC, the common measure is toprovide lower flange bracing at the two inflexion points.But, a simpler way is to install bracing at mid-span onlywhich can reduce the effective length to less than onehalf because the type 1 tie-strut has smaller µ–factor(see Figure 4). Figure 9 gives the scheme of bendingmoment within the unsupported length of l/2 and theaxial force of the analogous tie-strut. The latter belongsto type 1 and Eqn 12 is used to obtain the effectivelength factor µ. The ϕT factor calculated for girders ofsection R2 with span length 12h, 15h and 18h are 0.92,

0.89 and 0.84 respectively, all of them much larger than0.57. Thus, mid-span bracing is quite effective forenhancing the resistance to BILFC.

6. CONCLUDING REMARKSThe chords of lattice girder of rigid frames act as tie-struts when buckling out of the plane of the frame whenlongitudinal braces are scarce. There are threepossible types of tie-strut with different patterns offorce distribution. Whereas the formula provided in theChinese code GB50018 (2002) is only applicable totwo-portion tie-struts, the simplified formulas suggestedin this article are applicable to all the three respectivetypes and have fairly good accuracy.

As for the simple beam strut analogy, bucklingproblems of frame plate girders can be solved by a tie-strut analogy approach which is especially useful forinvestigating BILFC. Calculation reveals that bracing ofthe lower flange at mid-span is effective in guaranteeingstability in place of bracings at the inflexion points.

ACKNOWLEDGEMENTThe author thanks Professor Mingzhou Su for reviewingthe manuscript and former graduate student Bo Wu forcalculation of the theoretical data in Tables 3, 4 and 5.

REFERENCESANSI/AISC360-05 (2005). Specification for Structural Steel

Buildings, American Institute of Steel Construction.

Chen, S.F. (2008). “Overall stability of unbraced simply supported

beams with doubly symmetric I-section”, Steel Construction,

Vol. 23, No. 110, pp. 6–13. (in Chinese)

Gaylord, E.H. Jr., Gaylord, C.N. and Stallmeyer, J.E. (1992). Design

of Steel Structures, 3rd Edition, MacGraw-Hill, New York.

GB50017 (2003). Code for Design of Steel Structures, Ministry of

Construction PRC. (in Chinese)

GB50018 (2002). Technical Code of Cold-Formed Thin-Wall Steel

Structures, Ministry of Construction PRC. (in Chinese)

Pflüger, A. (1964). Stabilitätsprobleme der Elastostatik, zweite

Auflage, Springer-Verlag, Berlin.

Timoshenko, S.P. and Gere, J.M. (1961). Theory of Elastic Stability,

2nd Edition. McGraw-Hill, New York.

NOTATIONAf flange areaAw web areab flange widthCb factor of equivalent critical momentE modulus of elasticityfy yield strengthG shear modulus of elasticityIy moment of inertia about minor axis

0.6

M0

0.4

M0

l 0 = l /2

0.632l00.368l0

AC

Nt¡�0.267

Nc¡�0.343

Figure 9. Axial forces in analogous tie-strut of girders braced at

mid-span

Table 8. T values for BILFC

l/h

Section 12 15 18

R2 0.47 0.32 0.23W2 0.79 0.65 0.50

ϕϕ

Shaofan Chen


It St. Venant torsion constantIω warping constantiT radius of gyration of T-section about the

symmetric axish section depthl span length, distance between braced pointsl1, l2 portion lengthlc length of compressive portionlt length of tensile portionM bending momentNc compressionNt tensionP1, P2 axial loadWx elastic section modulus about the major axisα calculation factor for type 2 tie-strutβ calculation factor for type 3 tie-strutβb factor of equivalent critical moment, same as

Cb

γ calculation factor for type 1 tie-strutη ratio of Nt/Nc

θ rotation angleλc ratio lc/lλT slenderness ratio of T-sectionλTr reduced slenderness ratio of T-sectionλ y slenderness of I-section about the minor

axisµ effective length factorv1, v2, v3 parameter relevant to load P1, P2 and

(P1+P2), respectivelyξ calculation factor for type 1 tie-strutρ factor relevant to St. Venant torsional rigidityσcr critical stressϕ stability factor of axially compressed

memberϕb, ϕb′ lateral-torsional stability factor of beamϕT stability factor of T-section strut

Documents

Buckling Strength of Tie-Struts and Application to Frame Girders