International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 1259

ISSN 2229-5518

Effective Length Factor for Column in Frame with

Girders on Elastic Foundation

F.Y. Al-Ghalibi

Abstract— This paper considered the effects of elastic foundation on effective length factor calculations using subassembly model for braced and unbraced frames, the girders far ends condition are modeled as rigid, fixed, or hinged. The derivation is under same assumptions of conven- tional effective length factor (K-factor) using two approaches. The first approach considered the effects of elastic foundation and girders far ends conditions by depending on β1 and β2 parameters, the general modified K-factor equations have been derived for braced and unbraced frame us- ing slop-deflection method. In the second approach, formulae of stiffness modification parameter ‘γ’, which modify the relative columns to beams stiffness factor ‘G’ are calculated with direct use of G in the US National codes alignment chart. Approximate modified K-factor formulae are pro- posed for braced and unbraced frames. The Approximate proposed modified K-factor equations are suitable for practical use. Numerical exam- ples are presented to illustrate the effects of elastic foundation taken into consideration girders far ends conditions. The Results showed that the amount of elastic foundation modulus influenced the modified K-factor values and the stiffness of elastic foundation must be taken into account in

K-factor calculations. Generally, modified K-factor values decrease with increasing of stiffness of the elastic foundation.

Index Terms— Stability, stiffness, elastic foundation, modified K-factor, stability functions, braced-frame, unbraced frame.

1 INTRODUCTION

——————————  ——————————
HE importance of K-factor calculation has increased since the middle of the last century. K-factor formulations and applications are widely used in U.S. specifications such as AISC-LRFD, AISC-ASD, ACI (318), AASHTO and many text- books. In the structural engineering, the calculation process of effective length factors are one of the most important applica- tions spatially in the field of second order analysis and mem- bers slenderness. K-factor is widely used in the field of ad- vanced structural analysis, design, buckling issues and struc- tural stability. The studies, research and many references showed K-factor importance and effects on the behavior of structural analysis and design. The value of K-factor is varia- ble and it’s depending upon many factors, those factors are related to structure’s dimensions, types of supports, frame’s geometry, shape of members, type of member’s material, and loading’s case. The derivation of K-factor equation is based on calculation assumptions, these assumptions are simplified the modeling and calculation of K-factor. For example, the as- sumption of simultaneously buckling of all columns in one story with an idealized subassembly model used to prepare the alignment charts, this assumption was proposed by Julian and Lawrence(1959). LeMessurier(1977)modified Julian and Lawrence’s assumption based on considering the columns in a one story buckle simultaneously and the strong column braced the weak column or that carry high axial load. Depend- ing on this assumption, LeMessurier applied some types of
correction factor to the alignment charts.

————————————————

• University Assistance Lecturer, Structures and Water Resources Department, College of Engineering, Kufa University, AN-Najaf city, AN- Najaf province, Iraq. Tel: +964 781 9877878, E-mail: furatyh@yahoo.com .

Lui(1992)proposed simple and more effective method in eval-
uating the K-factor in sway frame, the method considered both instability of member and frame. Hu and Lai (1986) pro- posed a computer programming method to calculate the K- factors. In that computer program, the modeling of element was considered the effects of axial force for typical offshore structures with ends rotational spring and translational elastic springs at a distance from element end. Yura(1971)studied the K-factor calculation for unbraced frame, his strategy involved reducing the calculation process by modeling the problem as an equivalent pinned-ends braced columns. Duan and Chen(1988)proposed a modification to K-factor’s calculation for braced framed column to increase the accuracy of K- factor’s calculation by presenting a modification to G factors which are used in US National codes alignment charts by con- sidering the effect of columns far-ends condition in the above and below the column under consideration. Also, Duan and Chen (1988) modified K-factor calculation for unbraced framed column by taking into account the effect of columns far ends condition in the above and below the considered col- umn. Their work involved modifying G factors used in US National codes alignment charts. Chen et al(1993a) suggested a new method to calculate K-factor for braced and unbraced column in frame restrained by tapered girder with various girders far end conditions. The model improved the conven- tional G factor by girder stiffness modification parameter α. Chen et al (1993a) discussed the ACI (318-89) simplified equa- tions and indicated some comments and limitations used for simplified ACI-code K-factor equations. Dumonteil (1992)discussed the exact conventional K-factor formulae for braced and unbraced columns and check the accuracy of K- factor approximate equations. Dumonteil (1999)discussed some historical K-factor equations for braced and unbraced columns. He checked the accuracy with the exact form for the French “CM 66” approximate formulae for braced and un- braced columns, Donnell’s approximate formulae and New- mark’s approximate formulae for braced column.

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The analysis of beams on elastic foundation was well estab- lished in the literature. The differential equation approach used by Hetenyi (1946) is practical for engineering purposes. In structural buildings, some elements may be interacting with elastic foundation; such interaction affected the accuracy of K- factor calculation. This paper presents the exact and approxi- mate modified K-factor equations for braced and unbraced column in frame with girders on elastic foundation. The gird- ers far ends are modeled as rigid, fixed, and hinged. The mod- ified K-factor exact formulae derived by using two approach- es, the first approach considered the parameter βn which is derived using subassembly model (Fig.1). The value of pa- rameter βn is varying with frame’s case (braced or unbraced), elastic foundation stiffness parameter λ, and girders far ends condition. The second approach considered the conventional
alignment charts for prismatic girders, so that the original G factor can’t use directly. The present study developed G factor used in US codes alignment charts by depending on stiffness parameter γn which is derived for girders on elastic founda- tion with various far end conditions by dividing the bending stiffness of girder on the elastic foundation on the bending stiffness of ordinary member. The proposed modification to G factor allow to use the US cods alignment charts for column in

Fig. 2. Alignment charts of the US National cods, (A) braced frame, (B)

unbraced frame

2 EXACT MODIFIED K-FACTOR EQUATIONS (EXACT FIRST APPROACH)

The differential equation of girder on elastic foundation can be given as:

d 4 y

frame with girders on elastic foundation and various far end
conditions. The exact modified K-factor formulae are derived
according to the following assumptions:

EI + k y = 0 dx 4

(1)
1. Columns buckle simultaneously.
2. All members are elastic and prismatic cross section.
3. All girders have negligible axial force.

From the solution of Eq.1, the stiffness coefficients of girder on
elastic foundation ( S & Sc ) can be obtained as follow:

4. All column ends are rigid, while the girders far end are modeled as rigid, fixed or hinged.

Sinh( λ ) Cosh( λ ) − Sin( λ ) Cos( λ ) S = 2λ

Sinh 2 ( λ ) − Sin 2 ( λ )

(2)
5. All columns have equal stiffness parameter L

P / EI .

6. For braced frame, angles of rotation at opposite girder

Sin( λ ) Cosh( λ ) − Sinh( λ ) Cos( λ )

ends have an equal value and produce single bending
curvature as shown in Fig.1(B); whereas for unbraced
frame, angle of rotation at opposite girder ends also have

Sc = 2λ

Sinh 2

( λ ) − Sin 2

( λ )

(3)
equal value and but produce reverse bending curvature as shown in Fig.1(C).
Where parameter λ is given by:
7. Distribution of resistance of joint is proportion to two columns in above and below the joint.

I / L of

λ = 4 k s

4 EI

(4)

Where, ks is an elastic foundation parameter

2.1 Braced frame

At the buckling load, the columns have an axial force parame- ter

  2

ρ=  = 

 K 

and the stability functions
SK and SC K
stated else-

Fig.1.Subassembly model for a frame with girders on elastic foundation,

where(Duan and Chen, 1988, Duan and Chen, 1989) as shown
in Eq.5 and Eq.6:

(A) modified model, (B) modified braced frame subassembly model, (C)

 π 

 π  2

 π 

 Sin  −  

Cos 

modified unbraced frame subassembly model

S K =  K   K   K   K 

(5)

 π 

2 − 2Cos  −  Sin 

 K 

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 π  2

 π 

 

−  Sin 

β 2 + S

− (S

+ SC )

SC K =  K   K   K 

(6) K

GB K K K

 π 

 π  β

2 − 2Cos  −  Sin 

1 + S K

SC K

− (S K + SC K ) = 0

(14)

 K 

 K  GA

  π  2 

By using the slop-deflection equations for the subassembly

− ( S K

+ SC K )

− ( S K

+ SC K )

2S K



+ SC K

− 1   

2 K 

model for column in braced frame on elastic foundation with
variable girders far end conditions, the following relation can be achieved:

   

Eq.14 can be written in the following form:

GBSC K

β 2 + GBS K

3(β

+ β )   π 2

  β

β 

 π 2 

= 0 (7)

1 2   

− 2SK − 2SCK  +  1 + 2  (SCK )2 − (SK )2 + SK    +

β1 + GAS K

GASC K

GA GB

 π 2

  K 

 GA



GB 



 K  

Eq. 7 can be written as:

  ((SK )2 − (SCK )2 ) = 0

( 15 )

β β  β β 

 K 

1 2 + S

 1 + 2  + S 2 − SC 2 = 0

(8)

GA GB

 GA

GB  K K

The basic equation for stability analysis of unbraced frame on elastic foundation with variable girder far ends conditions
The basic equation for the stability analysis of sway-prevented
resulted from Eq.15 can be given by:
frame on the elastic foundation with variable girder far ends
conditions resulted from Eq.8 can be given as:

GAGB(π / K )2 − β β

2 − π / K = 0

(16)

(π )

β β



+  β



+ β 



tan(π / K )

 β1





+ β 2 



tan( π

/ K )

GAGB

/ K 2  GA

GB  1

π / K 

 GAGB

1 2  1

2  

(9)

2 tan(π =/ 2K )

π / K

− 1 = 0

The parameter βn for unbraced frame can be given as: For rigid girder far end

∑ EI 

β = S

+ SC

= 2λ

Cosh( λn ) − Cos( λn )

(17)

 

n n n

n Sinh( λ

) − Sin( λ )

Where G =

 L  C

∑ EI 

(10) n n

 

For fixed girder far end

 L  g

The parameter βn for braced frame can be given as: For rigid girder far end:

Cosh( λn ) + Cos( λn )

β n = S n = 2λn

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn ) Sinh 2 ( λn ) − Sin 2 ( λn )

(18)

β n = S n − SCn = 2λn

(11)

For hinged girder far end

Sinh( λn ) + Sin( λn )

2 2 2 2

For fixed girder far end:

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

β n = S n = 2λn

(12)

βn n

Cosh ( λn ) − Cos ( λn )

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

(19)

Sinh 2 ( λn ) − Sin 2 ( λn )

For hinged girder far end:

2.3 Simplified form of parameter βn

For the practical and design purposes, simplified formulae of parameter βn were developed by curve-fitting as following:

S 2 − SC 2

Cosh 2 ( λn ) − Cos 2 ( λn )

β n = n n

S n

= 2λn

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

(13)

2.3.1 When λn< 4

2.3.1.1 The case of column in braced frame

Rigid far ends

2 3 4

2.2 Unbraced frame

By using the slop-deflection equations for the subassembly model for column in unbraced frame with girders on elastic

βn = 2 + 0.32λn − 0.714λn + 0.55λn − 0.074λn

Fixed far ends
(20)

2 3 4

foundation have variable girders far ends condition, the fol-
lowing relation can be obtained:

βn = 4 + 0.2λn − 0.42λn + 0.31λn − 0.039λn

Hinged far ends
(21)

2 3 4

βn = 3 − 0.104λn − 0.112λn + 0.295λn − 0.046 λn

(22)

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2.3.1.2 The case of column in unbraced frame

Rigid far ends
as US National codes equation (AISC, 2012, AISC, 1989, ACI,
2014, AASHTO, 1989) for sway-allowed column.

2 3

β n = 6 + 0.0258λn − 0.066 λn + 0.04584λn

Fixed far ends
(23)

3 MODIFIED ALIGNMENT CHART (EXACT SECOND APPROACH)

The modified G -factor equation is proposed by Chen at el

2 3 4

β n = 4 + 0.2λn − 0.42λn + 0.31λn − 0.039λn

Hinged far ends
(24)
(1993a, 1993b) for the case of column in frame restrained by
tapered girders using stiffness modification parameter. This

paragraph introduces a modified G -factor used in the current

2 3 4

alignment charts of US National codes which enable the de-

βn = 3 − 0.104λn − 0.112λn + 0.3λn − 0.0474λn

2.3.2 When λn ≥ 4

(25)

signer to calculate the modified K-factor by direct use of the alignment charts (Fig. 2).The G factor equation which consid- ered the effect of elastic foundation and girders far end condi- tion is given as:
In this case, the close form solution (Eq.s 11, 12, 13, 17, 18 and
19) gives the following result:

Ec I c / Lc

G =

(30)

Cosh( λn ) + Cos( λn ) ≈ Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn ) ≈

∑g E g I g / L g

Sinh( λn ) + Sin( λn )

Cosh( λn ) − Cos( λn ) Sinh( λn ) + Sin( λn )

Sinh 2 ( λn ) − Sin 2 ( λn )

Cosh 2 ( λn ) − Cos 2 ( λn )

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

≈ 1

(26)
Where g : is girder on elastic foundation stiffness modification parameter.

3.1 Exact form of stiffness modification parameter

The girder stiffness modification parameter g is calculated for

2.3.2.1 Case of column in braced or unbraced frame

The parameter βn can be obtained by according to Eq. 26. Then, Eq.s 11, 12, 13, 17, 18 and 19 can be written as shown in Eq.27:
braced and unbraced frames by dividing the bending stiffness of girder on elastic foundation on the bending stiffness of or- dinary girder not interacted with elastic foundation. The g pa- rameter is considered the effects of girder far end condition and elastic foundation stiffness.

 rigid



β n  fixed



far ends 



far ends  = 2λn



(27)
For braced columns

When girder far end is rigid

hinged

far ends

g = S − SC = λ

Cosh( λn

) + Cos( λn )

(31)
From Eq.s 20 to 25, it can be concluded that, when the parame-
ter λn approach to zero, Eq.s 20 to 25are reducing to the fol-

2 n Sinh( λn ) + Sin( λn )

When girder far end is fixed
lowing form:
For braced column

g = S

= λn

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

(32)

2 for



rigid

far end 



2 Sinh 2

( λn ) − Sin 2

( λn )

β n = 4

3

for

fixed

far end 



(28)
When girder far end is hinged

 for

hinged

far end 

2 2

S − SC

Cosh2 ( λ

) − Cos 2 ( λ )

For unbraced column

g =

= λn

n n

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

(33)

6



β n = 4

3

for for

rigid fixed

far end 



far end 



(29)
For unbraced columns

When girder far end is rigid

 for

hinged

far end 

g = S + SC = λn Cosh( λn ) + Cos( λn )

(34)
Eq.s 28 and 29 considered the effects of far end conditions of

6 3 Sinh( λn ) + Sin( λn )

prismatic girders without interaction between frame and elas- tic foundation. In the case of sway-prevented column with
When girder far end is fixed

g = S = λn Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

(35)
rigid girders far ends, Eq. 28 given that β1 = β2 =2, when substi- 6 3
tute these beta values in Eq. 9 the resulted equation is as same

Sinh 2 ( λn ) − Sin 2 ( λn )

as the US National codes equation (AISC, 2012, AISC, 1989,
When girder far end is hinged
ACI, 2014, AASHTO, 1989).Also,Eq.29 given that β1 = β2 =6 for
sway-allowed frame with rigid girders far ends and when ap-

g = S

−=SC = λ

Cosh 2

( λn ) − Cos 2

( λn )

(36)
ply these beta values inEq.16, the resulted equation is as same

6 S 3

Sinh( λn ) Cosh( λn ) − Sin( λn ) Cos( λn )

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3.2 Simplified form of stiffness modification parameter

For unbraced column
For the practical and design purposes, a simplified formulae of stiffness modification parameter γ can be obtained by compare Eq.s 11,12,13,17,18 and 19 with Eq.s 31 to 36.The following

1



g n = 2 / 3

0

for for

rigid fixed

far end 



far end 



(48)
relation between parameter β and parameter γ can be con- cluded as follow

 .5

for

hinged

far end 

For braced frame g = β

(37)
Eq.s 47 and 48aresame as the stiffness modification parameter
proposed by Chen et al (1993b) for prismatic girders and
SSRC-Guide(Johnston, 1979).Fig.3 shows a graphing of modi-

For unbraced frame g = β

(38)
fication stiffness parameter g n
for exact closed form solution

By using Eq.s 37and 38, the simplified form of stiffness modi- fication parameter can be obtained as:

3.2.1 When λn< 4

3.2.1.1 The case of column in braced frame

(Eq.s 31 to 36) and simplified form solution (Eq.s 39 to 46).
Rigid far end g

= 1 + 0.16 λ

− 0.357λ2 + 0.275λ3 − 0.037λ4

(39)

n n n n n

Fixed far end g

= 2 + 0.1λ

− 0.21λ2 + 0.155λ3 − 0.0194λ4

(40)

n n n n n

Hinged far end g

= 1.5 − 0.052λ

− 0.056 λ2 + 0.1475λ3 − 0.023λ4 (41)

n n n n n

3.2.1.2 The case of column in unbraced frame

Rigid far end g

= 1 + 0.0043λ

− 0.011λ2 + 0.00764λ3

(42)

n n n n

Fixed far end g

= 0.66 + 0.034λ

− 0.071λ2 + 0.052λ3 − 0.0065λ4

(43)

n n n n n

Hinged far end g

= 0.5 − 0.0173λ

− 0.0187 λ2 + 0.05λ3 − 0.0079λ4

n n n n

(44)

3.2.2 When λn ≥ 4

3.2.2.1 The case of column in braced frame

The parameter γ n can be obtained by substitute Eq. 37 in Eq.27

 rigid



g n  fixed

hinged

far end 



far end  = λn

far end 

(45)

3.2.2.2 The case of column in unbraced frame

The parameter γ n can be obtained by substitute Eq. 38 in Eq.27

 rigid



g n  fixed

hinged

far end 



far end  = λn / 3

far end 

(46)

Fig.3. The relation between λ and γ

When parameter λn approach to zero (the case of no interac- tion between girder and elastic foundation) Eq.s 39 to 44is re-
ducing to the following form: For braced column

4 APPROXIMATE MODIFIED K-FACTOR PROPOSED FORMULAE

1



g n = 2

1

for for

rigid fixed

far end 



far end 



(47)
The following formulae involved modifying Newmark formu- la for braced frame and French “CM 66” formula for unbraced frame(Dumonteil, 1992),the modified approximate K-factor

 .5

for

hinged

far end 

proposed equations in term of βn
For braced column

can be given as:

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K = (GA + 0.205β1 ) (GB + 0.205β2 ) (GA + 0.41β1 ) (GB + 0.41β2 )

(49)

1.00

0.95 a

For unbraced column

57.6 GAGB + 24( GA + GB ) + 7.5

0.90

0.85

K = β1β2

β1 β2

(50)

0.80

(50)

6( GA + GB ) + 7.5

0.75

β1 β2

0.70

The exact Eq.s 9, 16 and approximate Eq.s 49 and 50 used when the two adjacent girders in one joint have same value of elastic foundation parameter λ. For the general case when the two adjacent girders in one joint have different value of elastic foundation parameter λ, the following formulae can be used: For braced column

(GA + 0.41) (GB + 0.41)

(GA + 0.82) (GB + 0.82)

0.65

0.60

0.55

0.50

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq.9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

0.95

For unbraced column

0.90 b

0.85

K = 1.6 GA GB + 4( GA + GB ) + 7.5

GA + GB + 7.5

(52)

0.80

K 0.75
Eq.s 51, and 52 can be considered as approximate solution of the following general case equations (Eq.s 53 and 54).
For braced column

0.70

0.65

0.60

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

(π / K )

+   1 −

 + (

( )

− 1 = 0

0.55

Eq. 9, lamda=6

GA GB 2

 GA + GB 

π / K 

2 tan π / 2K

Eq. 9, lamda=4

 

 



tan π / K 

π / K

(53)

0.50

Eq. 9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

GA
For unbraced column

GA GB(π / K )2 − 36

π / K

0.95

6 GA GB

− = 0 tan( π / K )

(54)

0.90 c

0.85

When the two adjacent girders in one joint have same value of elastic foundation parameter λ, Eq.s 49 and 50 can be written in term of γ n as:
For braced column

0.80

K 0.75

0.70

K = (GA + 0.41g 1 ) (GB + 0.41g 2 ) (GA + 0.82g 1 ) (GB + 0.82g 2 )

(55)

0.65

0.60

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

For unbraced column

1.6 GAGB + 4( GA + GB ) + 7.5

0.55

0.50

Eq. 9, lamda=8

K = g 1g 2

g 1 g 2

(56)

0 1 2 3 4 5 6 7 8 9 10 11

GA + GB + 7.5 GA

g 1 g 2

Graphing of Eq.s 9 and 49 for the case of braced frame and Eq.s 16 and 50 for the case of unbraced frame with various girder far end condition are shown bellow.

Fig. 4. The relation between modified effective length factor and GA for braced column, (a) for rigid girders far end, (b) for fixed girders far end, (c) for hinged girders far end, GA=GB, λ1=λ2=λ3=λ4=λ

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1.00

0.95 a

1.00

0.95 a

0.90

0.85

0.80

K 0.75

0.80

K 0.75

0.70

0.65

0.60

0.55

0.50

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq. 9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

0.70

0.65

0.60

0.55

0.50

Proposed, Eq. 49

Eq.9, lamda=0

Eq. 9, lamda=2

Eq.9, lamda=4

Eq. 9, lamda=6

Eq. 9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

0.95

0.90 b

0.95

0.90 b

0.85

0.80

0.75

0.70

0.65

0.60

0.55

0.50

proposed, Eq. 49

Eq. 9, lamda=0

Eq.9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq. 9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

0.75

0.70

0.65

0.60

0.55

0.50

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq. 9, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

1.00

0.95 c

0.90

1.00

0.95 c

0.90

0.85

0.80

K 0.75

0.80

K 0.75

0.70

0.65

0.60

0.55

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq. 9, lamda=8

0.70

0.65

0.60

0.55

Proposed, Eq. 49

Eq. 9, lamda=0

Eq. 9, lamda=2

Eq. 9, lamda=4

Eq. 9, lamda=6

Eq.9, lamda=8

0.50

0 1 2 3 4 5 6 7 8 9 10 11

0.50

0 1 2 3 4 5 6 7 8 9 10 11

Fig.5. The relation between modified effective length factor and GA for braced column, (a) rigid girders far end, (b) fixed girders far end, (c)

hinged girders far end, GA=GB, λ1=λ2=λ/2,λ3 = λ4= λ

Fig. 6. The relation between modified effective length factor and GA for

braced column, (a) for rigid girders far end, (b) for fixed girders far end, (c)

for hinged girders far end, GA=GB, λ1=λ2=λ/4,λ3 = λ4= λ

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2.6

2.5

2.4

2.3

2.2

2.1

2.0

1.9

2.7

2.6 a

2.5

2.4

2.3

2.2

2.1

2.0

K 1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

1.4

1.3

1.2

1.1

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

GA
b

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

2.7

2.6

2.5 b

2.4

2.3

2.2

2.1

2.0

K 1.9

1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

2.6

2.5 c

2.4

2.3

2.2

2.1

2.0

1.9

2.8

2.7

2.6 c

2.5

2.4

2.3

2.2

2.1

K 2.0
K 1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq.16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

1.9

1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

Fig. 7. The relation between modified effective length factor and GA for unbraced column, (a) for rigid girders far end, (b) for fixed girders far end, (c) for hinged girders far end, GA=GB, λ1=λ2=λ3 = λ4= λ

Fig. 8. The relation between modifiedGeAffective length factor and GA for unbraced column, (a) for rigid girders far end, (b) for fixed girders far end, (c) for hinged girders far end, GA=GB, λ1=λ2=λ/2,λ3 = λ4= λ

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2.7

2.6 a

1.3

1.2

1.1

Eq. 16, lamda=6

Eq. 16, lamda=8

This section presents two examples, the first example is used to illustrate the calculation of modified K-factor for braced frame using exact equations and US codes alignment chart, and then check the accuracy and the validity of the modified K-factor proposed equations. The solution is obtained using the exact and approximate β, γ, and G factors. Some girders far end are modeled as fixed and hinged to prevent the side- sway. The second example is provided to illustrate the calcula- tion of modified K-factor for unbraced frame. Exact equations and US codes alignment chart are adopted to check the accu- racy of the approximate proposed equations, also the solution was carried out using β, γ, and G factors for the case of exact and approximate solution.
Example 1

0 1 2 3 4 5 6 7 8 9 10 11

2.8

2.7

2.6

2.5

2.4

2.3

2.2

A braced frame supported by girders on elastic foundation as
shown in Fig. 10. The moment of inertia and element length
are given in Table.1 below. The modulus of elasticity (E) was
taken as a constant value for each element. The elastic founda- tion stiffness parameters are λAB =7, λBC = 7, λHJ =3 and λQS =2.5. Determine the effective length factor for columns DA, FB, HC and QH by using two approaches for exact and ap-
proximate solution.

TABLE 1

LENGTH AND MOMENT OF INERTIA FOR ELEMENTS OF EXAMPLE 1

1.4

1.3

1.2

1.1

Eq. 16, lamda=6

Eq. 16, lamda=8

0 1 2 3 4 5 6 7 8 9 10 11

2.8

2.7

2.6

2.5

2.4

2.3

2.2

2.1

K 2.0

1.9

1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

Proposed, Eq. 50

Eq. 16, lamda=0

Eq. 16, lamda=2

Eq. 16, lamda=4

Eq. 16, lamda=6

Eq. 16, lamda=8

Fig.10. Frame of example 1

Solution : The calculation of exact and approximate values of

0 1 2 3 4 5 6 7 8 9 10 11

Fig. 9. The relation between modified effective length factor and GA for unbraced column, (a) for rigid girders far end, (b) for fixed girders far end, (c) for hinged girders far end, GA=GB, λ1=λ2=λ/4, λ3 = λ4= λ

5 ILLUSTRATIVE EXAMPLES

parameter β using Eq.s 11 to 13 and Eq.s 20, 21, 22, and 27 re-
spectively are illustrated in Table 2, Table 2also shows the ex-
act values of parameter γ using Eq.s 31 to 33 and approximate
γ values by depending on Eq.s39, 40, 41, and 45.

TABLE 2

EXACT AND APPROXIMATE VALUES OF β AND γ OF EXAMPLE 1

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λ Girder far end β Exact β App. γ Exact γ App.
0 Rigid 2 2 1 1
7 Rigid 14 14 7 7
3 Hinged - - 2.982 2.96
2.5 Fixed - - 2.592 2.597

The values of G and G factors for each frame joint are tabulat- ed in Table 3 using Eq. 10 and Eq. 30 respectively. Parameter γ illustrated in Table 2 used in calculations of factors G exact and G approximate.

TABLE 3

G AND G OF EXAMPLE 1

K exact equation 0.689 0.637 0.56 0.65
K alignment chart 0.689 0.636 0.56 0.65
K approximate 0.692 0.637 0.567 0.654
Example 2
For the unbraced frame shown in Fig.11, determine the effec-
tive length factor for columns HA, JB, MC, ND, QF, RH, VN and WP by using two approaches for exact and approximate solution. The elastic foundation stiffness parameters are λAB =6, λBC = 6, λCD =6, λFH = 2 and λNP =3. The values of moment
of inertia and elements length are shown in Table 5 below and the modulus of elasticity (E) of each element is constant.

TABLE 5

LENGTH AND MOMENT OF INERTIA FOR ELEMENTS OF EXAMPLE 2

Joint	G	G Exact	G Approximate
A	0.8	0.1143	0.1143
D	2	2	2
B	0.356	0.0508	0.0508
F	0.945	0.945	0.945
C	0.34	0.0476	0.0476
H	0.458	0.23	0.231
Q	1	0.5568	0.556

For column DA, exact solution using Eq. 9

⇒ 1.6 (

/ K )2 +  2 + 0.8  1 −

/ K 2 tan(π =/ 2K )

( ) + − 1 = 0

π   π 



28  2

14 

tan π / K 

π / K

Using iteration exact K =0.689 or by using Eq. 53 exact K =0.689. By using alignment chart with direct use of G , K = 0.689 .

Approximate solution using Eq. 49

(2 + 0.205 * 2) (0.8 + 0.205 * 14)

K =

(2 + 0.41* 2) (0.8 + 0.41* 14)

= 0.692

Approximate solution using Eq. 51

(2 + 0.41) (0.1143 + 0.41)

K =

(2 + 0.82) (0.1143 + 0.82)

= 0.692

Approximate solution by using Eq.55

(2 + 0.41* 1) (0.8 + 0.41* 7 )

K =

(2 + 0.82* 1) (0.8 + 0.82* 7 )

= 0.692

Fig.11. Frame of example 2

For column FB, exact solution using Eq.9 or Eq.53 K =0.637. By using alignment chart with direct use of G , K = 0.636 . Approximate solution by using Eq. 49 or 51 or 55 K =0.637.

For column HC, exact solution using Eq.53 K =0.56. By using
Solution: The calculation of exact and approximate values of

alignment chart with direct use of G ,

K = 0.56 . Approximate

parameter β according to Eq.s 17 to 19 and Eq.s 23, 24, 25, and
solution using Eq. 51 K = 0.567 .

For column QH, exact solution using Eq.53 K =0.65. By using
27 respectively are illustrated in Table 6, also the table shows
the exact values of parameter γ using Eq.s 34 to 36and approx-

alignment chart with direct use of G , solution using Eq. 51 K = 0.654 .

TABLE 4

K = 0.65 . Approximate

imate γ values according toEq.s42, 43, 44, and 46.

TABLE 6

THE SUMMERY OF THE RESULTS OF EXAMPLE 1

Column DA FB HC QH

EXACT AND APPROXIMATE VALUES OF β AND γ OF EXAMPLE 2

λ Girder far end β Exact β App. γ Exact γ App.

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0 Rigid

0.334 * 0.334(π / K )2 − 6 * 6.15

π / K

= 0

6 Rigid

 6 +



6.15 

( 0.334 + 0.334 )

tan( π / K )

3 Rigid 6.717 6.72 1.12 1.12
2 Rigid 6.15 6.154 1.025 1.025

The values of G and G factors for each frame joint are tabulat- ed in Table 7 using Eq. 10 and Eq. 30 respectively. Parameter γ

 2 

Using iteration exact K = 1.1091 or by using Eq. 54 exact K =1.1091. By using alignment chart with direct use of G , K = 1.1 ,approximate solution using Eq. 50 or 52 or 56 K =1.124
For column RH, exact solution using Eq.54 K =1.215. By using
illustrated in Table 6 used in calculations of factors G exact and approximate.
alignment chart with direct use of G , solution using Eq. 52 K =1.242

K = 1.21 ,approximate

For column VN, exact solution usingEq.54 K =1.272. By using

TABLE 7

G AND G OF EXAMPLE 2

alignment chart with direct use of G , mate solution using Eq. 52 K =1.292

K = 1.27 and approxi-

For column WP, exact solution using Eq. 16

0.334 * 0.334(π / K )2 − 6 * 6.717

π / K

 6 +



6.717 

( 0.334 + 0.334 )

− = 0 tan( π / K )

 2 

Using iteration exact K = 1.1046 or by using Eq. 54, exact K =1.1046. By using alignment chart with direct use of G , K = 1.105

Approximate solution using Eq. 50

57.6 * 0.334 * 0.334 + 24( 0.334 + 0.334 ) + 7.5

6 * 6.717 6 6.717

K =

For column HA, exact solution using Eq.54 K =1.185. By using

6( 0.334

+ 0.334

) + 7.5

= 1.119

alignment chart with direct use of G ,

K = 1.185

6 6.717

approximate solution using Eq. 52 given as:
Approximate solution using Eq. 50 or 52 or 56 K =1.119

K = 1.6 * 0.763* 0.4 + 4( 0.763 + 0.4 ) + 7.5 =

0.763 + 0.4 + 7.5

1.207

TABLE 8

For columns JB and MC, Exact solution using Eq. 16

THE SUMMERY OF THE RESULTS OF EXAMPLE 2

1.086 * 0.686 (π / K )2 − 6 * 11.92

 6 + 11.92 ( 1.086 + 0.686 )

π / K

= 0

tan( π / K )

 

 2 

Using iteration exact K = 1.2137 or by using Eq. 54 exact K =1.2137. By using alignment chart with direct use of G , K = 1.21

Approximate solution using Eq. 50 given as:

57.6 * 1.086 * 0.686 + 24( 1.086 + 0.686 ) + 7.5

K = 6 * 11.92 6 11.92 = 1.244

6( 1.086 + 0.686 ) + 7.5

6 THE SUMMARY AND CONCLUSIONS

6 11.92

Approximate solution using Eq.52 given as:

K = 1.6 * 1.068 * 0.343 + 4( 1.068 + 0.343 ) + 7.5 = 1.244

1.068 + 0.343 + 7.5

Approximate solution using Eq. 56 given as:

1.6 * 1.086 * 0.686 + 4( 1.086 + 0.686 ) + 7.5

K = 1* 1.99 1 1.99 = 1.244

1.086 + 0.686 + 7.5

1 1.99

For column ND, exact solution using Eq.54 K =1.2. By using
This paper considered the determination of the effective length factor for column in braced and unbraced frames with girders on elastic foundation. The girders far ends were mod- eled as rigid, fixed or hinged. The exact formulae of the modi- fied K-factor have been derived using two approaches; in the first solution technique of the modified effective length factor calculations were depended on parameter β. The Exact closed form and the simplified approximate form of parameter β have been derived. In the second solution approach, the calcu- lation described using girder stiffness modification parameter γ which provided the ability of direct use of U.S. National

alignment chart with direct use of G , solution using Eq. 52 K = 1.224

K = 1.2

approximate
codes alignment charts. Exact form and simplified form of
parameter γ have been investigated. For practical use, some
For column QF, exact solution using Eq. 16 given as:
approximate formulae of modified K-factor with high accura-
cy are proposed. Two examples were solved to illustrate the

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ISSN 2229-5518

calculation method and solution accuracy.
Based on the results that obtained in the current study, several
conclusions can be drawn. These conclusions are summarized
as follows:
1- The two solution approaches of calculating the modi-
fied K-factor using parameter β or γ gives same modi-
fied K-factor results.
2- Fig. 3 shows an excellent agreement between the
closed form of parameter γ (Eq.s 31, 32, and 33) and
simplified form (Eq.s 39, 40, 41, and 45) for braced
frame, also, for unbraced frame the figure indicated
an excellent agreement between closed form (Eq.s
34,35, and 36) and simplified form (Eq.s 42, 44, 43 and
46). Thus, simplified form of parameter γ is practical
for design purposes.
3- In the second solution approach, the US National
codes alignment charts gives simple exact solution of
Eq. 53 and Eq. 54 without need iteration.
4- Fig.s 4, 5, and 6 shows excellent agreement between the modified K-factor proposed equation (Eq. 49) and exact solution using Eq. 9 for the case of braced frame. Also, for the case of unbraced frame, Fig.s 7, 8, and 9
show excellent agreement between the exact solution using Eq. 16 and approximate proposed solution us- ing Eq. 50. The percentage decrease in the modified K-factor for both braced and unbraced frame become more significant as the elastic foundation stiffness pa- rameter (λ) increases.
5- Results summary of example 1 and example 2 showed that the percent of error in modified K-factor values between exact and approximate solution for braced frame is less than 1.25 %, while for unbraced frame is less than 2.23 %.

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