Alternative Design Procedure for RC-Braced Long Columns ...

Research ArticleAlternative Design Procedure for RC-Braced Long ColumnsBased on New Moment Magnifiers Matrix

Mohamed Farouk 1 Majed Alzara 2 A Ehab3 and A M Yosri 12

1Civil Engineering Department Faculty of Engineering Delta University for Science and Technology Belkas Egypt2Department of Civil Engineering College of Engineering Jouf University Sakakah Saudi Arabia3Department of Civil Engineering Badr University in Cairo Badr City Egypt

Correspondence should be addressed to Majed Alzara majedzarajuedusa

Received 9 March 2021 Revised 31 July 2021 Accepted 16 August 2021 Published 21 October 2021

Academic Editor Giovanni Minafo

Copyright copy 2021 Mohamed Farouk et al is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

Based on modified methods for the results of first-order analysis of RC columns different codes approximate the second-ordereffects by using equations focusing on the maximum additional moment through the column heightese equations did not referto the additional moments between the column and the connected beam only the effect of the connected beams is taken intoconsideration by dealing with the effective length of the column not the total length Moreover these equations did not take intoaccount the second-order effect which is caused by axial force and the inverse moments due to beam restriction for the columnendsis paper presents a newmoment magnifiers matrix for the additional moments at the connection between braced columnsand the connected beams as a simplified computation that can be used in the design procedureat is through an equation basedon transforming the original long column in second-order analysis to an equivalent isolated column e equivalent column wasrepresented as an element restricted with rotational spring support at its ends and it is subjected to lateral distributed loads thathave the same influence of the second-order effect on the induced additional moments in the long column e suggestedequivalent column can be used to form the additional bending moment diagram also to compute the additional deformations aswell Numerous factors were analyzed linearly by using the presented new moment magnifiers matrix and finite element methodand the results proved the efficiency of the proposed model Although the presented suggested model is based on the isolatedanalysis of the long column the effect of the additional moments in the adjacent long column can be considered by presented twosuggestions to improve the model Also development was proceeded on the model by modifying the flexural rigidity (EI) which isrecommended in ACI to appropriate the time of failure e additional moment values of the developed model were close to thevalues calculated by the ACI equation

1 Introduction

In braced long columns a column is subjected to axial loadand equal or unequal end moments which are caused by theconnected beam loads and deformed laterally due to theexistence of end moments e axial load and occurredlateral deflection cause additional bending moments alongthe column height which is called second order e ad-ditional bending moments cause additional lateral dis-placements and rotations of the column and additionalrotation of the members connecting into the column

is in turn leads to change to the first-order bendingmoments through the column height and at the connectedends of the column with the beams which are computedfrom an elastic frame analysis e equations of equilibriumin a first-order analysis are derived by assuming that thedeflections have a negligible impact on the internal forces inthe members In a second-order analysis the equations ofequilibrium consider the deformed shape of the structureInstability can be investigated only via a second-orderanalysis because it is the loss of equilibrium of the deformedstructure that causes instability[1] e elastic structural

HindawiAdvances in Civil EngineeringVolume 2021 Article ID 9921682 20 pageshttpsdoiorg10115520219921682

analysis of the second-order effect can be performed in thefinite element method by adding the geometric stiffnessmatrix to the elastic linear matrix ldquomechanical stiffnessrdquo forthe beam column element e geometric stiffness as shownin (1) is not a function of the mechanical properties of theelement and is only a function of the elementrsquos length andthe force in the element Hence the term ldquogeometricrdquostiffness matrix is introduced so that the matrix has a dif-ferent name from the ldquomechanicalrdquo stiffness matrix which isbased on the physical properties of the element e geo-metric stiffness exists in all structures however it becomesimportant only if it is large compared to the mechanicalstiffness of the structural system [2] To use this stiffness theelement must be divided into small segments between theload points for more accuracy Several trials must be donewhen using the geometric stiffness till achieving the equi-librium as shown in equation (3)

Fi

Mi

Fj

Mj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

T

30L

36 3L minus 36 3L

3L 4L2

minus 3L minus L2

minus 36 minus 3L 36 minus 3L

3L minus L2

minus 3L 4L2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

vi

θi

vj

θj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

orFG KGv

(1)

Fi

Mi

Fj

Mj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

EI

L3

12 6L minus 12 6L

6L 4L2

minus 6L minus 2L2


minus 6L minus 2L2

minus 6L 4L2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

vi

θi

vj

θj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

orFE KEv

(2)

FT FE + FG

KE + KG1113858 1113859v

KTv

(3)

A nonlinear second-order frame analysis procedure canbe performed to analyze reinforced concrete columns thatare a part of frames In order to account for second-ordereffects due to geometric and material nonlinearities thetheoretical model (computer software) uses classical stiffnessanalysis of linear elastic two-dimensional structural framesthe iterative technique combined with an incrementalmethod for computing load-deflection behavior and failureload of the frame frame discretization to account for columnchord (P-Δ) effects and axial load-bending moment-cur-vature (P-M-ϕ) relationships to account for effects ofnonlinear material behavior [3]

2 Computing of the Additional Moments inDifferent Codes

Design of RC long column must consider the inducedadditional moments in these columns due to the axial loadand occurred lateral deflection Computing the additional

moments for design requires simplified procedure andadequate accuracy ere are many methods that have beenderived from modifying the results of a first-order analysisto approximate the second-order effects as which arerecommended in a lot of codes such as American code ACI[4] and Canadian code CSA [5] ese codes permit the useof a moment magnifier approach to approximate thesecond-order moments due to the axial load acting throughthe lateral deflection caused by the end moments acting ona column In the moment-magnifier analysis unequal endmoments are applied on the column shown in Figure 1(a)e column is replaced with a similar column subjected toequal moments at both ends which is shown in Figure 1(b)e bending moments are chosen where the maximummagnified moment is the same in both columns e ex-pression for the factor Cm was originally derived for use inthe steel beam-columns design and was adopted withoutchange for concrete design

M2 M1 are the larger and smaller end moments of thefirst-order analysis respectively If a single curvaturebending occurred by the moments M1 and M2 M1M2 ispositive However if the moments cause doublecurvatureM1M2 is negative e moment magnifierequation in the cases of no sway according to ACI

Mc δnsM2

Cmδ1M2 geM2(5)

Cm 06 + 04M1

M2 (4)

Chen and Lui [6] explain that Cm and δ1 for pin-endedcolumns subjected to end moments can be derived from thebasic differential equation governing the elastic in-planebehavior of a column ACI Code goes on to define δns asfollows

δns Cm

1 minus 075PPc

(6)

e 075 factor in equation (6) is the stiffness reductionfactor ϕK which is based on the probability of understrength of a single isolated slender column

In Eurocode [7] three methods applied for second-orderimpacts analysis are pointed

Simplified method based on nominal stiffness (MNS)simplified method based on nominal curvature (MNC)and overall method based on nonlinear second-orderanalysis

e method of nominal stiffness is based on the criticalforce due to the buckling computed for the nominal stiffnessof the analyzedmember It is recommended that thematerialnonlinearity creep and cracking which have an effect onthe conduct of the structure members are taken intoconsideration e design moment in the members sub-jected to the bending moment and an axial force whichincludes the impact of the first and second-order effects canbe illustrated as a bending moment boosted by the factordescribed below

2 Advances in Civil Engineering

ME d M0E d + M2

M0E d + M0E dβ

NBNE d( 1113857 minus 1

M0E d 1 +β

NBNE d( 1113857 minus 11113890 1113891

(7)

where M0E d is the 1st order moment including the effectof imperfections M2 is the nominal 2nd order momentNB is the buckling load based on nominal stiffness NE d isthe design value of the axial load β is the factor whichdepends on the distribution of the 1st and 2nd ordermoments

e method of nominal curvature allows for the cal-culation of the second-order moment based on the assumedcurvature distribution (which responds to the first-ordermoment increased by the second-order effects) on the lengthof the membere distribution of the total curvature can beeither parabolic or sinusoidal

e value of the II order moment can be calculated asfollows

M2 NE de2 (8)

where NE d is the design value of the axial load e2 is thedeflection calculated by taking into account such parametersas creep the intensity of reinforced and also distribution ofthe reinforcement over the height of the cross-section

e2 1rl20c

(9)

where c is the factor depending on the curvature distribu-tion l0 is the effective length and 1r is the curvature

According to Egyptian Code [8] (Madd) is induced bythe deflection (δ) given by the following

Madd Pδ (10)

If the column is long in t direction

δt λt

2t

2000

Madd Pδt

(11)

However if the column is long in b direction

δb λb

2b

2000

Madd Pδb

λb He

b

He kH0

(12)

where He is the effective height of the column H0 is clearheight of the column k is length factor which depends on theconditions of the end column and the bracing conditions

e presented equations in the mentioned codes dependon their derivation on the isolated analysis for the longcolumn and computed maximum bending moments in-duced through the height of the column e additionalmoments analysis at the joints between the column andconnected beams did not receive any interest in the differentcodes Only the recommended equations in these codes takeinto account the effect of the connected beams on the ad-ditional moments through the column height by dealingwith the effective length of the column not the total lengthIn this research a new moment magnifiers matrix will bepresented in a derived equation for an equivalent column tocompute the additional moments of the braced long columnincluding the moments at the joints between the column andthe connected beams In this model the additional momentdiagram of a braced long column and its deformations canbe computed taking into consideration the second-ordereffect of the axial load and the inverse moments at theconnection between the columns and the beams Materialnonlinearities will be considered by modifying the elasticflexural rigidity (EI) to effective flexural stiffness computedaccording to ACI (2019)

M2

Max M

M1

(a)

Cm

M2

Cm

M2

Max M

(b)

Figure 1 Equivalent moment factor Cm (a) Actual moments at failure (b) Equivalent moments at failure

Advances in Civil Engineering 3

3 Lateral Displacements in a LongColumn under End Moments

In the first-order analysis the curvature equation for a longcolumn under equal end moments as shown in Figure 2 canbe expressed as follows

d2ydx

2 minusM

EI (13)

where M M0

dy

dx minus

M0

EIx + C1 (14)

δ0 minus1

EI

M0x2

21113890 1113891 + C1x + C2 (15)

By applying the boundary conditions it is found thatC2 0 and C1 M0L2EI

And equation (15) becomes as follows

δ0 1EI

minusM0x

2

2+

M0L

21113874 1113875x1113890 1113891 (16)

Maximum lateral displacement at the mid-span of thecolumn can be expressed as follows

δ0( 1113857max M0L

2

8EI (17)

Due to the second-order effect the lateral displacementof the column increases and it can be expressed as (δo + δa)where δa is the lateral displacement which is caused by theadditional moments To compute the maximum additionallateral displacement at the middle span of the column thevirtual work method can be used As observed that thedeformation shape of the first-order analysis is 2nd curve asshown in equation (16) As a result additional displacementand the additional moment diagram will be 2nd curve whereMadd P(δ0 + δa)

e additional lateral displacement can be found asfollows

δa 1113946L

0MaddM11dx

Madd P δ0 + δa( 1113857

δa 1EI

2lowast23

P δ0 + δa( 1113857 middotL

2middot58

middotL

41113874 1113875

δa δ05PL

248EI1 minus 5PL

248EI

But1Pe

5L

2

48EIwherePe is Euler load

δa δ0PPe

1 minus PPe

(18)

e final maximum displacement will be as follows

δf δ0PPe

1 minus PPe

+ δ0

It can be put η 1

1 minus PPe

So δf δ0η

(19)

emaximum additional moments can be formulated asfollows

Madd Pδoη (20)

When the column is deformed under unequal endmoments the bigger moment can be divided into two partsM M0 + ΔM and the column will be considered underequal end moments (M0) which was illustrated previouslyand one end moment (ΔM)

In the first-order analysis the curvature equation for along column under one end moment can be expressed asfollows

d2ydx

2 minusM

EI (21)

M Rxwhere R is the reaction and equalsR ΔM

L (22)

dy

dx minus

1EI

Rx2

2EI1113888 1113889 + C1 (23)

δΔ0 minus1EI

Rx3

61113890 1113891 + C1x + C2 (24)

where δΔ0 the lateral displacement due to ΔMBy applying the boundary conditions C2 0

C1 RL26EI and put R ΔML

1 kNL4

M01

M02

δa

δ0

P

Figure 2 Deformed shape of a long column under end moments


δΔ0 1EI

minusΔMx

3

6L+ΔMLx

61113890 1113891 (25)

As shown in equation (24) the deflection curve due toΔM is 3rd-degree parabolic curve and the maximum de-flection occurs when dδdx 0

Thusdy

dx

1EI

Mx2

2L+ΔML

61113888 1113889 0 (26)

By solving equation (26) max lateral displacement dueto ΔM will be at x L

3

radicand Max lateral displacement

due to ΔM is given by the following

δΔ0 ΔML

2

93

radicEI

(27)

Also due to the second-order effect the lateral dis-placement of the column increases and it can be expressedas (δΔo + δΔa) where δaΔ is the lateral displacement which iscaused by the additional moments Considering that thedeformation shape of the first-order analysis is 3rd curve asshown in equation (25) also additional displacement and theadditional moment diagram will be 3rd curve Similarly themaximum lateral displacement due to ΔM in second-orderanalysis can be found as the samemanner of the case of equalend moments as in the following equation

δfΔ δΔ0

1 minus PPe

δΔ0η

(28)

e additional moments can be formulated as follows

Madd PδΔoη (29)

4 Equivalent Lateral Load for the Second-OrderEffect in a Long Braced Column

As shown in section (2) when the column is loaded with anequal end moment the deformed shape of the column in thesecond-order analysis was 2nd-degree curve As a result theexpected additional bending moment diagram will be as theinduced bending moment from the regular distributed loadus the long column in the second-order effect can bereplaced in a beam element subjected to an equivalentregular distributed load and the equivalent load can becomputed as follows

weqRL2

8 Pη

M0L2

8EI1113888 1113889 (30)

where weqRL28 max moment due to equivalent regularload Pη(M0L

28EI) max moment due to second-orderanalysis weqR equivalent regular distributed load

weqR PηM0

EI (31)

In similar to the column under equally end moment thedeformed shape in the second-order analysis due to one endmoment as shown in Figure 3 is 3rd-degree curve and theadditional bending moment diagram in the second-orderanalysis will be as the induced bending moment from thetriangular distributed load Also in this case the longcolumn in the second-order effect can be replaced in a beamthat is subjected to equivalent triangular distributed loadand the equivalent load can be computed as follows

weqΔL2

93

radic PηΔML

2

93

radicEI

1113888 1113889 (32)

where weqΔL29

3

radicis the max moment due to equivalent

triangular load Pη(ΔML293

radicEI) is the max moment due

to second-order analysis and weqΔ is the equivalent trian-gular distributed load

weqΔ PηΔMEI

(33)

5 New Moment Magnifiers Matrix of BracedLong Columns

51 Equivalent Column Modeling Based on the equivalentcolumn concept Afefy and El-Tony [9] have shownequivalent pin-ended columns for columns bent in eithersingle or double curvature modes where the impact of endeccentricity ratio was related to the equivalent columnlength ey deduced that the equivalent column conceptcan be generalized to simplify columns bent in single cur-vature modes with different end eccentricities combinationsto pin-ended axially loaded columns Furthermore theequivalent column concept can be carried out for a specificstate of a column bent in double curvature mode

Here in the suggested equivalent column model thecolumn at any structure will be analyzed as an isolatedelement e equivalent column was represented as anelement restricted by a rotational spring support at its endsand it is subjected to lateral distributed loads e lateraldistributed loads have the same influence of the second-order effect on the induced additional moments in the longcolumn Column (1) for example in the shown closedframe in Figure 4 will be analyzed to illustrate the modele column will be modeled as a pin-supported memberrestricted by the connected beams which are as rotationalspring supports Computing the rotational stiffness (Kθ) ofthese beams will be discussed later e second-orderanalysis of the modeled column in Figure 5 can be dividedinto two parts

e first part is concerned with the deformation due tothe end moments of the first-order analysis without theexistence of the reaction moments of the rotational springe induced moments of the second-order effect is equiv-alent to the induced moments of trapezoidal load us thecolumn can be represented as a pin-supported columnsubjected to triangular distributed loads weq1 and weq2


weq1 pηM01

EI (34a)

where weq1 is the triangular equivalent load for the secondeffect due to moment at the column end at the beam ofhigher stiffness and M01 is the the end moment at the beamof higher stiffness due to first order

weq2 pηM02

EI (34b)

weq2 is the triangular equivalent load for the secondeffect due to the moment at the column end at the beamof lower stiffnessM02 is the the end moment at the beam of lowerstiffness due to first order

e second part is concerned with the deformation dueto the reaction moments of the spring rotational supportonly Also the column in the second effect will be repre-sented as a pin-supported column subjected to triangulardistributed loads weqlowast1

and weqlowast2

weqlowast1is the triangular equivalent load for the second

effect due to the reaction moment of spring rotationalsupport at the beam of higher stiffness M1 the ad-ditional moment of spring rotational support at thebeam of higher stiffness

weqlowast1 pη

M1

EI (35)


effect due to the reaction moment of spring rotationalsupport at the beam of lower stiffness M2 the addi-tional moment of spring rotational support at the beamof lower stiffness

weqlowast2 pη

M2

EI (36)

By arranging the linear stiffness matrix of a beam ele-ment for computing the moments of the modeled column inFigure 5 the formula will be as follows

EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)

e terms in equation (37) are as shown below

K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)

e matrix in (37) can be divided into the following

Lb

I (column 2)I (column 1)

L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)

Figure 4 Closed frame as an example

M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI

Figure 3 Equivalent lateral load for the second-order effect of along column


minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)

e final formula to compute the additional moments atthe column ends will be as follows

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)

Equation (40) can be rewritten as follows

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)

where [A] is considered as moment magnifiers matrix forthe end additional moments and it is equal to the following

[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)

Just the additional moments at the column ends werecomputed the final load of the equivalent column will be asshown in Figure 6

RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o

Figure 5 Equivalent column modeling for the restricted column (1) in the closed frame


rough the model in Figure 6 the additional bendingmoment at any section can be computed and an additionalbending moment diagram can be formed Also by using oneof themethods of structural analysis such as the virtual workmethod or area moment method the additional lateraldisplacement and rotations at any point can be calculatede total lateral displacement also can be computed easilyby dividing the additional bendingmoment at any section bythe axial load (δfinal MaddP)

6 Approximate Rotational Stiffnesses for theColumn at Upper and Lower Joints

e connected beams which represent the rotational stiffnessof the upper and lower joints of the column (Kb) can beapproximately computed by applying one unit of the mo-ment toward the end of the connected beams with thestudied column as shown in Figure 7 e opposite end ofthe beam is considered as rotational restricted end by an-other pin column (adjust column to the studied column)Rotation of the beam end (θb) under the unit moment can becalculated en the rotational stiffness will be computed asKb 1θb

e rotation at the loaded end of the adjacent column iscomputed as follows

θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)

where LAcol and IAcol are the length and moment of inertia ofthe adjacent column

e rotational stiffness of the adjacent column to thebeam is as follows

KAcol 3EIcol

Lcol

(44)

Due to the unit moment at the beam end the reactionmomentMlowast at the opposite end of the beam can be found bythe force method as follows

Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)

By using the virtual work the rotation at the loaded endof the beam can be determined as follows

θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)

e rotational rigidity of the connected beam end for thestudied column can be expressed as follows

Kb EIb

Lb (13) minus Lb36( 1113857 middot 1EIbKAcol + Lb3( 1113857 + L2b108 1EIbKAcol + Lb3( 1113857

21113872 1113873

(47)

where Lb and EIb are the length andmoment of inertia of theconnected beam

In fact most of the long columns are connected withbeams that have stiffness bigger than or close to the columnstiffness us the effect of the adjacent column which as

rotational spring for the beams will be slight and (47) can besimplified as follows

Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1

Figure 6 Final load of the equivalent column


7 Computing the Additional Moments by Usingthe Equivalent Column with More Accuracy

As mentioned before the additional moments in a longcolumn can be computed according to equation (41) asisolated column analysis If there are other long columnsadjacent to the studied columns the additional momentsof these columns will affect the additional moments ofthe studied column For more accuracy the effect ofadditional moments of adjacent columns must be con-sidered where a part of these moments will be trans-ferred through the connected beams to the studiedcolumn By one of the following two suggestions theeffect of the adjacent long columns can be taken intoconsideration

71 Suggestion 1 Assume that the studied column is the leftcolumn in the shown closed frame in Figure 4 In thissuggestion the additional moments in each column will becomputed according to equation (41) as a separate analysisof each of them en the transferring ratio of the addi-tional moments between the columns will be found Eachcolumn will be considered as a rotational spring for boththe bottom and top beams e rotational stiffness of thecolumns will be computed in the same manner in section 5equation (44)

By using the force method the transmitting momentfrom the right column to the left studied column at joint 1 asan example can be calculated as follows

Mlowast1 Madd( 11138572 middot

Lb6EIb( 1113857Top

Lcol3EIcol + Lb3EIbTop1113872 1113873

Mlowast1 Madd( 11138572 middot α1

(49)

where Mlowast1 is the transferred moment from the adjacentcolumn (joint (2) to the studied column joint (1) α1 α2factor of transferring ratio by the top beam

(Lb6EIb)Top(Lcol3EIcol + (Lb3EIb)Top) (Ma dd)2 the ad-ditional moment at joint 2 of the adjacent column

Mlowast3 Madd( 11138574

Lb6EIb( 1113857bottomLcol3EIcol + Lb3EIb( 1113857bottom( 1113857


(50)

where Mlowast3 is the the transferred moment from the adjacentcolumn (joint 4) to the studied column joint (3) α3 α4factor of transferring ratio by the bottom beam

(Lb6EIb)bottom(Lcol3EIcol + (Lb3EIb)bottom)

After obtaining the transmitting moment between thetwo columns (41) can be carried out one time for thesecond-order effect of the transmitting moments Also thiscan be considered by modifying equation (41) as follows

M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30

minusM03 + βMadd4( 1113857

30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)

Equation (51) takes into account the transmitting ad-ditional moments between two adjacent columns for onetrial e equation can be carried out for several trials till theratio of transferred additional moment gets close to zero andit can be modified to include the effect of more adjacentcolumns Whereas the deformations in reinforced concretestructures are small thus the additional moments at the endof the long columns will not be large values As a result theexpected transmitting moments will be small and it can beignored or one trial as maximum can be carried out But for

spring roationalsuppport

KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)

Figure 7 Rotational stiffness of the connected beam (a) Studied column (b) Connected beam (c) Adjacent column


more accuracy the effect of the additional moments ofadjacent columns can be considered as in (51) e effect ofadjacent additional moments can be considered schematicmethod as in Figures 8(a) and 9 presented the additionalmomentsrsquo transmission between the columns

If a number of slender columns exist in the structure asshown in Figure 8(b) equations (43) and (51) easily can beformulated as follows

Equation (51) will become as follows

Madd 1113944trialn

trial1Mtrial1 + αMtrial1 + Mtrial2 + αMtrial2 + middot middot middot middot middot middot + Mtrial(n)

(52a)

where n is the trial number which at it the condition of(αMtrial(n) zero) will be achieved

(M1)add Final (M2)add Final

(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m

09 times 03 m09 times 03 m

w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2

w2 = 150kN(mprime)beam 2

Lb

beam 4

Lb

(b)

Figure 8 (a) Schematic method for the transmitting additional moments (b) Multibays frame is an example of a structure that has morethan two adjacent slender columns


Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =

(M3)Final=(M3)trial+(αM4)trial1+(M3)trial2+(M4)Final=(M4)trial1+(αM3)trial1+(M4)trial2+

(M3)trial2 (M3)trial1(M4)trial1

(M4)trial2(M3)trial1

M03 M04

(βM4)trial1 (M4)trial1 (βM3)trial1

(αM4)trial1(αM3)trial1

(αM2)trial1 (αM1)trial1

(αM1)trial1 (M2)trial2

(M1)Final =(M1)trial1+(αM2)trial1+(M1)trial2+ (M2)Final=(M2)trial1+(αM1)trial1+(M2)trial2+

(M1)trial2 (αM2)trial1 (M1)trial1 M01

(M1)trial1 (M2)trial1

(M2)trial1M02

Figure 9 Additional moments transmission between the columns


Mtransfferal (α)lowast [M]ttrail1

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b

1113936(L3EI)elementslowast

(3EIL)coloumn

1113936(3EIL)elaments minus (3EIL)b

(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf

Figure 10 Approximate initial moments at the ends of the beam to calculate the relative rotational stiffness


Lb(6EI)b is for the connected beam between the studiedcolumn and the other column sent the transferred moments(3EIL)coloumn is for the studied column and (3EIL)elements

is for the all connected elements with the studied columnincluding it

[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0

0 0 minus C2 minus4C2L

22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0

0 0 0 0 minus C3 minus4C2L

23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)

If a number of slender columns exist in the structure andthe effect of additional moments of adjacent columns will beconsidered the sequence of solving can be schematized asshown in Figure 8(a) or it can be programmed e shownframe in Figure 8(b) was analyzed by MATLAB program

M-file of programming and the results are shown in Ap-pendix (I)

72 Suggestion 2 In this suggestion the effect of the ad-ditional moments of the adjacent columns on the studied

infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10

Figure 11 Jackson-Moreland Alignment Chart for braced frames [10]


column will be taken into the relative rotational stiffness ofthe connected beams is will be considered by applyingapproximate values of additional moments at the ends of theconnected beams to evaluate the relative rotational stiffnessof these ends for the long columns

As shown in Figure 10 the initial additional momentscan be calculated as follows

Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)

where weq(av) is the average equivalent regular load for thesecond effect due to the end moments of the column Lfeffective length of the long column and it can be calculatedas in Figure 11

e initial moments of the four joints of the shown framein Figure 10 will be calculated as follows

Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)

And by using the virtual workmethod the rotation at theends of the connected beams at joint 1 as an example can becalculated as follows

θ1 1EIB

12Mint1Lb middot

23

+12Mint2 middot Lb middot

13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)

e rotational stiffness of the connected beams can becomputed as follows

Kθ1 Mint1 EIbLb( 1113857Top

Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857

Kθ3 Mint3 EIbLb( 1113857BottomMint33 + Mint46( 1113857


(56)

And by substituting the computed rotational stiffness in(41) additional moments in the long columns in the closedframe will be computed taking into account the approxi-mate effect of the additional moments of each column oneach other

8 Elastic Analysis for Checking the StructuralAnalysis Efficiency of the Equivalent Column

Numerous factors will be studied here through linearanalysis of closed frames by utilizing the new momentmagnifiers matrix and finite element methode reason forthis investigation is to check the structural analysis profi-ciency of the recommended model in a wide range withoutthe restriction of materials failure e factors were a var-iation of the stiffness of connected beams together inducedaxial force to Euler load ratio slenderness ratio and thestiffness of upper beam to lower beame left column is thetarget column in this study e results of the comparisonare shown in Figures 12ndash 16

Solving steps for the equivalent column model

(1) Calculating the terms K0 C Kθ1 and Kθ2 accordingto equations (38-a) (38-b) (48) respectively tosatisfy the moment magnifiers matrix [A]

(2) Applying (41) for each column to compute the ad-ditional moments between the columns and theconnected beams

(3) For more accuracy the effect of the additionalmoments of each column on each other can beconsidered by using a schematic method in Figure 8then find the final additional moments ldquosuggestion1rdquo

(4) By using model of final loads in Figure 6 the ad-ditional moment diagram and additional deforma-tions of the column (lateral displacements androtations) can be computed

From the results it was observed that using (41) in thesuggested equivalent column model gives values of addi-tional moments close to their values calculated by the finite

1400 kN 800 kNW1 = 200 kNmprime

upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m

7m W2 = 150 kNmprime

10m

Figure 12 e analyzed frame


element method e results proved the structural analysisefficiency of the proposed model for analyzing the longcolumn as an isolated element After the satisfaction to theefficiency of the model the model can be developed bymodifying the flexural rigidity (EI) to appropriate the ma-terials case at the moment of computing the additionalmoments as shown in the next section

9 Design Procedures for Computing AdditionalMoments in Long Columns by Using NewMoment Magnifiers Matrix

e calculation of terms (K0 C) and (Kθ1 Kθ2) in themoment magnifiers matrix of (41) involves the use of theflexural rigidity EI of the column and the connected beamsrespectively To use the suggested equation (41) in com-puting the additional moments in a long column the flexuralrigidity for a given column section must be considered at thetime of failure taking into account the effects of crackingand nonlinearity of the stress-strain curves JamesG MacGregor et al [11 12] describe empirical attempts toderive values for EI ACI-318-19 includes two different sets

of stiffness values EI the first set is for the computation of EIin caudation the critical load of an individual column asfollows

EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)

where Ec Es are the modules of elasticity of the concrete andthe steel respectively Ig is the gross moment of inertia of theconcrete section Ise is the moment of inertia of the rein-forcement about the centroidal axis of the concrete sectionand (1 + βdn s) term reflects the effect of creep on the columndeflections

It can be used in equation (57) or equation (58) butequation (57) is more accurate

e second set is for values of the moment of inertia Ifor use in elastic frame analyses or in computing the effectivelength factor k In this set the column and beam stiffnesseswere computed as 07EcIc and 035EcIc respectively

beam section 07times03m

0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu

beam section 06times03 m

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)

Figure 13 (a-d) Using equivalent column with varying of beam stiffnesses


According to the concepts of the previous recommen-dations in ACI-318-19 for using the approximate flexuralrigidity in different cases it is appropriate to use (EI)according to equation (57) or equation (58) when calculatingthe term (C) in equation (41) Either when calculating theterms (K0) and (K1 K2) in (41) the column and beamstiffnesses will be computed as 07EcIc and 035EcIbrespectively

A long column in closed frames as shown in Figure 10will be analyzed by each of the equations of (ACI-318-19)(ECP 203-2018) the suggested equation (41) by usingsuggestions (1) (2) was considered the moment transmittingbetween the long columns and by using (41)) with neglectingthe moment transmitting e frames were analyzedaccording to the design requirements and with changing twoparameters slenderness ratio and PPe e main crosssection of the studied column is 50times 30 cm with a rein-forcement ratio in-between 001 to 0035 e slendernessratio will be varied from (98ndash215) and PPe will be variedfrom (023-095) e relationship between the maximumadditional moment through the height of the column andthe changed parameters are shown in Figures 18 17

e results of the analysis show that there is a largeconvergence of the additional moments calculated by (41) in

the equivalent column between each of suggestions (1) and(2) where the effect of transmitted additional moments wastaken into account and by using equation (41) directlywhere the transmitted additional moments were neglectedis is expected because of that the failure limitations of thematerial which makes the additional moments induced in along column are not the large values that strongly affect orget affected by the adjacent columns us for easiness (41)can only be used where a separate analysis of the longcolumn is included without being affected by the additionalmoments of adjacent long columns

Moreover it is observed that (41) gives values of addi-tional moments close to their values which are given by theequation of ACI is means that using the suggestedequation gives good efficiency Also it means the appro-priate use of flexural rigidity (EI) according to Eqs (57) or(58) for computing the term (C) in (41) while using thecolumn and beam stiffnesses

According to 07EIc and 035EIb respectively forcomputing the terms (K0) and (K1 K2) in (41) the term(C) is the concern of the lateral deformations of the longcolumn and using (EI) according to Eqs (57) or (58) inthis term corresponds to using these equations for com-puting Pcr in the equation of ACI While the terms (K0)

PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)

Figure 14 (a-d) Using equivalent column with varying axial force to Euler load


and (K1 K2) are the concern of the additional moments atthe column ends and the effects of these terms in (41) aresimilar to the effects of the effective length in the ACIequation

e additional deformations of the long column (lateraldisplacement and rotations) at any point can be computedeasily as shown in Figures 19 20 as an example

It is clear that there is a big difference in the additionalmoments between each of ACI the suggested equation in theequivalent column and ECP Prab Bhatt et al [13] illustratedthe basic and the assumptions of computing the additionalmoments in British code equation BS8110 1997 [14] whichis the same as the ECP equation He illustrates that thedeformed column curvature will typically vary along thecolumn as a sinusoidal value of(1π2) Figure 21 shows theinteraction diagram between the bending and the normalforce and strain diagram in the ultimate stage (balancedfailure) us the central lateral deflection au will be as-sumed as follows

au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)

e column curvature (1r) is calculated based on thestrain diagram at the balanced failure as follows

1rb

(0003 + 0002)

d (60)

e maximum deflection for the case set out above isgiven in the code by the following expression

au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)

It was noted that the ECP equation was based oncomputing the additional moments in a specific case whichis at the balanced failure and this restriction is difficult toachieve when designing the columns us if the failuremode of the column section is not compatible with thebalanced failure it is supposed that this equation is not validand it will give far values of the additional acting momentsas shown in previous analyzing cases So it seems that theECP equation cannot be used generally to compute theadditional moments and it is for a specific case Where it can

slenderness ratio 11

0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu

(b)slenderness ratio 16

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)

Figure 15 (andashd) Using equivalent column with varying slenderness ratio


EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00

50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)

Figure 16 (andashd) Using equivalent column with varying upper beam to lower beam stiffness

90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15

Slenderness ratio17 19 21 23

ACIEq 41 using sugg1Eq 41 using sugg2Eq 41 neglecting moment transferECP

Figure 17 Additional moments vs slenderness ratio

140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12


Figure 18 Additional moments Vs PPe


be used to evaluate the maximum allowable lateral dis-placement of the column at the balanced failure then themoment capacity of the column section can be checked forrestrained design (balanced failure) is is maybe one of thereasons for the big difference between the results of the ACIequation the suggested equation in the equivalent columnand the EPC equation

Also the ECP equation does not take into account thesecond-order effect which is caused by the axial force and theinverse moments due to beams restriction to the columnends ECP equation considers only the connected beamseffect by dealing with the effective length of the column notthe total length ACI equation is similar to ECP equation atthis point but returning to the original equation of ACI (Eq62) a term (1 + 023PPe) was found which was omitted fromthe final equation to generalize the use of the equationwhere the factor 023 varies as a function of the momentdiagram shape is leads to a decrease in the additionalmoments among the inflected points and that approachesthe results between the ACI equation and the analysis whichtakes into account the second-order effect which is causedby the inverse moments as in the suggested equivalentcolumn

Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)

On the other hand the computed flexural rigidity (EI)according to ACI in Eq (57) remains constant regardless ofthe magnitude of end moments and therefore Pc also re-mains constant As a result the moment magnifier remainsconstant for a given column However Pc is stronglyinfluenced by the effective flexural stiffness (EI) which variesdue to the nonlinearity of the concrete stress-strain curveand cracking along the height of the column among otherfactors [3]

10 Summary and Conclusions

Based on the equivalent column concept a new momentmagnifiers matrix was presented in this paper for computingthe additional end moments in the braced long column eequivalent columnwas an element restricted at its ends by twospring rotational supports and is subjected to lateral dis-tributed loads which have the same influence of the second-order effect in a long column e additional momentsrsquo di-agram and additional deformations (lateral displacementsand rotations) can be computed by using the suggestedequivalent column taking into consideration the second-or-der effect which is caused by the axial load and the inversemoments due to beams restriction for the column ends thiseffect is important although it is neglected in design codese long column in the suggested model was analyzed as anisolated element but by two presented suggestions the effectof the additional moments of other adjacent long columns ifany can be considered e first suggestion took into accountthe effect of adjacent additional moments by computing thetransmitting additional moments among columns throughtransfer coefficients depended on the rigidity of the connectedbeams then the equation of the moment magnifiers matrixwas applied more than once for the transmitted momentsis suggestion can be carried out by schematic method asshown in the paper content In the second suggestion theadjacent additional momentsrsquo effect was considered in therelative rotational stiffness of the connected beams which areas a rotational spring for the long columns is will be

0025 002 0015Additional lateral displacement (m)

Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00

Figure 19 Computing additional lateral displacement by usingequivalent column model for slenderness ratio 112 as an example

0008 0006 0004 0002 00

2

4

6

8

10

12

-0002Additional rotations (rad)

Hei

ght o

f col

umn

(m)

-0004 -0006 -0008

Figure 20 Computing the additional rotations by using anequivalent column model for slenderness ratio 112 as an example

0002

0003

d

Balanced failure

M

N

Figure 21 e interaction diagram between the bending and thenormal force and strain diagram in the ultimate stage (balancedfailure)


considered by applying approximate values of additionalmoments at the ends of the connected beams to evaluate therelative rotational stiffness of these beams Development wascarried out on the model by modifying the flexural rigidity(EI) in each of the connected beams and the long column as itis recommended in ACI to appropriate the time of failureFrom the results presented in this paper the following isconcluded

(1) e suggested equivalent column proved a goodefficiency for analyzed numerous factors linearly byfinite element method and the equivalent columnwas satisfying as a successful structural model

(2) For analyzing many designed long columns in closedframes the results showed that there are small dif-ferences of computed additional moments by ap-plying the suggested equation of moment magnifiersdirectly and by using the two suggestions of con-sidering the adjacent additional moments so theadditional moments of the adjacent columns can beneglected for simplifying

(3) e developed model gave close values of the ad-ditional moments for many analyzed long columnswith ACI equation and it is appropriate to generalizethis model for second-order analysis of long bracedcolumns as an easy-to-use model that yields goodresults

(4) ere was a gap between the values of the additionalmoments computed by the two methods the sug-gested equivalent column and ECP equation ismay be because ECP equation was based on com-puting the additional moments in a specific casewhich is at the balanced failure Moreover the ECPequation did not take into account the second-ordereffect which is caused by the axial force and theinverse moments due to beamsrsquo restriction for thecolumn ends

Appendix

M-File in MATLAB program to solve the frame inFigure 8(b)

e results in Command window of MATLAB program

Data Availability

e data used to support the findings of this study are in-cluded within the article and are available from the corre-sponding author upon reasonable request

Conflicts of Interest

All authors declare that they have no conflicts of interest

Acknowledgments

e authors extend their appreciation to the Deanship ofScientific Research at Jouf University is work was funded

by the Deanship of Scientific Research at Jouf Universityunder grant No (DSR-2021-02-0353)

References

[1] J G Wight and J K MacGregor Reinforced concrete Me-chanics and Design Prentice-Hall Upper Saddle River NJUSA 6th ed edition 2011

[2] R D Cook D S Malkus and M E Plesha Concepts andApplications of Finite Element Analysis John Wiley amp SonsHoboken NJ USA 3rd edition Article ID 0-471-84788-71989

[3] T K Tikka and S A Mirza ldquoEffective length of reinforcedconcrete columns in braced framesrdquo International Journal ofConcrete Structures and Materials vol 8 no 2 pp 99ndash1162014

[4] Aci Committee 318 Building Code Requirements for Struc-tural concrete (ACI 318-19) and Commentary AmericanConcrete Institute Farmington Hills MI USA 2019

[5] Csa Design of concrete Structures Vol A233-04 CanadianStandards Association Mississauga Canada 2004

[6] W F Chen and E M Lui Structural Stability-eory andImplementation Elsevier Science Publishing Company IncNew York NY USA 1987

[7] EN 1992-1-1 Eurocode 2 (EC2) Design of concrete Structures-Part 1-1 General Rules and Rules for buildings EuropeanCommittee for Standardization Brussels Belgium 2004

[8] Ecp Egyptian Code for Design and Construction of ReinforcedConcrete Structures ECP Tamworth UK 2018

[9] H M Afefy and E-T M El-Tony ldquoSimplified design pro-cedure for reinforced concrete columns based on equivalentcolumn conceptrdquo International Journal of Concrete Structuresand Materials vol 10 no 3 pp 393ndash406 2016

[10] L Duan W S King and W F Chen ldquoK-factor equation toalignment charts for column designrdquo ACI Structural Journalvol 90 no 3 pp 242ndash248 1993

[11] J G MacGregor U H Oelhafen and S E Hage ldquoA Reex-amination of the El Value for Slender Columnsrdquo ReinforcedConcrete Columns pp 1ndash40 American Concrete InstituteFarmington Hills MI USA 1975

[12] J G MacGregor J E Breen and E O Pfrang ldquoDesign ofslender columnsrdquo ACI Journal Proceedings vol 67 no 1pp 6ndash28 1970

[13] P Bhatt J T MacGinly and B S Choo ldquoReinforced ConcreteDesign eory and Examplesrdquo CRC Press Boca Raton FLUSA 3rd edition 2006

[14] Bs8110 Structural Use of Concrete Part 1 Code of Practice forDesign and Construction BSI London UK 1997


analysis of the second-order effect can be performed in thefinite element method by adding the geometric stiffnessmatrix to the elastic linear matrix ldquomechanical stiffnessrdquo forthe beam column element e geometric stiffness as shownin (1) is not a function of the mechanical properties of theelement and is only a function of the elementrsquos length andthe force in the element Hence the term ldquogeometricrdquostiffness matrix is introduced so that the matrix has a dif-ferent name from the ldquomechanicalrdquo stiffness matrix which isbased on the physical properties of the element e geo-metric stiffness exists in all structures however it becomesimportant only if it is large compared to the mechanicalstiffness of the structural system [2] To use this stiffness theelement must be divided into small segments between theload points for more accuracy Several trials must be donewhen using the geometric stiffness till achieving the equi-librium as shown in equation (3)

Fi

Mi

Fj

Mj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

T

30L

36 3L minus 36 3L

3L 4L2

minus 3L minus L2


3L minus L2

minus 3L 4L2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

vi

θi

vj

θj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

orFG KGv

(1)

Fi

Mi

Fj

Mj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

EI

L3

12 6L minus 12 6L

6L 4L2

minus 6L minus 2L2


minus 6L minus 2L2

minus 6L 4L2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

vi

θi

vj

θj

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

orFE KEv

(2)

FT FE + FG

KE + KG1113858 1113859v

KTv

(3)

A nonlinear second-order frame analysis procedure canbe performed to analyze reinforced concrete columns thatare a part of frames In order to account for second-ordereffects due to geometric and material nonlinearities thetheoretical model (computer software) uses classical stiffnessanalysis of linear elastic two-dimensional structural framesthe iterative technique combined with an incrementalmethod for computing load-deflection behavior and failureload of the frame frame discretization to account for columnchord (P-Δ) effects and axial load-bending moment-cur-vature (P-M-ϕ) relationships to account for effects ofnonlinear material behavior [3]

2 Computing of the Additional Moments inDifferent Codes

Design of RC long column must consider the inducedadditional moments in these columns due to the axial loadand occurred lateral deflection Computing the additional

moments for design requires simplified procedure andadequate accuracy ere are many methods that have beenderived from modifying the results of a first-order analysisto approximate the second-order effects as which arerecommended in a lot of codes such as American code ACI[4] and Canadian code CSA [5] ese codes permit the useof a moment magnifier approach to approximate thesecond-order moments due to the axial load acting throughthe lateral deflection caused by the end moments acting ona column In the moment-magnifier analysis unequal endmoments are applied on the column shown in Figure 1(a)e column is replaced with a similar column subjected toequal moments at both ends which is shown in Figure 1(b)e bending moments are chosen where the maximummagnified moment is the same in both columns e ex-pression for the factor Cm was originally derived for use inthe steel beam-columns design and was adopted withoutchange for concrete design

M2 M1 are the larger and smaller end moments of thefirst-order analysis respectively If a single curvaturebending occurred by the moments M1 and M2 M1M2 ispositive However if the moments cause doublecurvatureM1M2 is negative e moment magnifierequation in the cases of no sway according to ACI

Mc δnsM2

Cmδ1M2 geM2(5)

Cm 06 + 04M1

M2 (4)

Chen and Lui [6] explain that Cm and δ1 for pin-endedcolumns subjected to end moments can be derived from thebasic differential equation governing the elastic in-planebehavior of a column ACI Code goes on to define δns asfollows

δns Cm

1 minus 075PPc

(6)

e 075 factor in equation (6) is the stiffness reductionfactor ϕK which is based on the probability of understrength of a single isolated slender column

In Eurocode [7] three methods applied for second-orderimpacts analysis are pointed

Simplified method based on nominal stiffness (MNS)simplified method based on nominal curvature (MNC)and overall method based on nonlinear second-orderanalysis

e method of nominal stiffness is based on the criticalforce due to the buckling computed for the nominal stiffnessof the analyzedmember It is recommended that thematerialnonlinearity creep and cracking which have an effect onthe conduct of the structure members are taken intoconsideration e design moment in the members sub-jected to the bending moment and an axial force whichincludes the impact of the first and second-order effects canbe illustrated as a bending moment boosted by the factordescribed below


ME d M0E d + M2

M0E d + M0E dβ


M0E d 1 +β

NBNE d( 1113857 minus 11113890 1113891

(7)




M2 NE de2 (8)


e2 1rl20c

(9)



Madd Pδ (10)


δt λt

2t

2000

Madd Pδt

(11)


δb λb

2b

2000

Madd Pδb

λb He

b

He kH0

(12)



M2

Max M

M1

(a)

Cm

M2

Cm

M2

Max M

(b)





d2ydx

2 minusM

EI (13)

where M M0

dy

dx minus

M0

EIx + C1 (14)

δ0 minus1

EI

M0x2

21113890 1113891 + C1x + C2 (15)



δ0 1EI

minusM0x

2

2+

M0L

21113874 1113875x1113890 1113891 (16)


δ0( 1113857max M0L

2

8EI (17)



δa 1113946L

0MaddM11dx

Madd P δ0 + δa( 1113857

δa 1EI

2lowast23


2middot58

middotL

41113874 1113875

δa δ05PL

248EI1 minus 5PL

248EI

But1Pe

5L

2


δa δ0PPe

1 minus PPe

(18)


δf δ0PPe

1 minus PPe

+ δ0

It can be put η 1

1 minus PPe

So δf δ0η

(19)


Madd Pδoη (20)



d2ydx

2 minusM

EI (21)


L (22)

dy

dx minus

1EI

Rx2

2EI1113888 1113889 + C1 (23)

δΔ0 minus1EI

Rx3

61113890 1113891 + C1x + C2 (24)



1 kNL4

M01

M02

δa

δ0

P



δΔ0 1EI

minusΔMx

3

6L+ΔMLx

61113890 1113891 (25)


Thusdy

dx

1EI

Mx2

2L+ΔML

61113888 1113889 0 (26)


3



δΔ0 ΔML

2

93

radicEI

(27)


δfΔ δΔ0

1 minus PPe

δΔ0η

(28)


Madd PδΔoη (29)



weqRL2

8 Pη

M0L2

8EI1113888 1113889 (30)



weqR PηM0

EI (31)


weqΔL2

93

radic PηΔML

2

93

radicEI

1113888 1113889 (32)

where weqΔL29

3





weqΔ PηΔMEI

(33)






weq1 pηM01

EI (34a)


weq2 pηM02

EI (34b)



and weqlowast2



weqlowast1 pη

M1

EI (35)



weqlowast2 pη

M2

EI (36)


EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)


K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)


Lb


L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)


M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI



minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















ME d M0E d + M2

M0E d + M0E dβ


M0E d 1 +β

NBNE d( 1113857 minus 11113890 1113891

(7)




M2 NE de2 (8)


e2 1rl20c

(9)



Madd Pδ (10)


δt λt

2t

2000

Madd Pδt

(11)


δb λb

2b

2000

Madd Pδb

λb He

b

He kH0

(12)



M2

Max M

M1

(a)

Cm

M2

Cm

M2

Max M

(b)





d2ydx

2 minusM

EI (13)

where M M0

dy

dx minus

M0

EIx + C1 (14)

δ0 minus1

EI

M0x2

21113890 1113891 + C1x + C2 (15)



δ0 1EI

minusM0x

2

2+

M0L

21113874 1113875x1113890 1113891 (16)


δ0( 1113857max M0L

2

8EI (17)



δa 1113946L

0MaddM11dx

Madd P δ0 + δa( 1113857

δa 1EI

2lowast23


2middot58

middotL

41113874 1113875

δa δ05PL

248EI1 minus 5PL

248EI

But1Pe

5L

2


δa δ0PPe

1 minus PPe

(18)


δf δ0PPe

1 minus PPe

+ δ0

It can be put η 1

1 minus PPe

So δf δ0η

(19)


Madd Pδoη (20)



d2ydx

2 minusM

EI (21)


L (22)

dy

dx minus

1EI

Rx2

2EI1113888 1113889 + C1 (23)

δΔ0 minus1EI

Rx3

61113890 1113891 + C1x + C2 (24)



1 kNL4

M01

M02

δa

δ0

P



δΔ0 1EI

minusΔMx

3

6L+ΔMLx

61113890 1113891 (25)


Thusdy

dx

1EI

Mx2

2L+ΔML

61113888 1113889 0 (26)


3



δΔ0 ΔML

2

93

radicEI

(27)


δfΔ δΔ0

1 minus PPe

δΔ0η

(28)


Madd PδΔoη (29)



weqRL2

8 Pη

M0L2

8EI1113888 1113889 (30)



weqR PηM0

EI (31)


weqΔL2

93

radic PηΔML

2

93

radicEI

1113888 1113889 (32)

where weqΔL29

3





weqΔ PηΔMEI

(33)






weq1 pηM01

EI (34a)


weq2 pηM02

EI (34b)



and weqlowast2



weqlowast1 pη

M1

EI (35)



weqlowast2 pη

M2

EI (36)


EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)


K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)


Lb


L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)


M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI



minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References


















d2ydx

2 minusM

EI (13)

where M M0

dy

dx minus

M0

EIx + C1 (14)

δ0 minus1

EI

M0x2

21113890 1113891 + C1x + C2 (15)



δ0 1EI

minusM0x

2

2+

M0L

21113874 1113875x1113890 1113891 (16)


δ0( 1113857max M0L

2

8EI (17)



δa 1113946L

0MaddM11dx

Madd P δ0 + δa( 1113857

δa 1EI

2lowast23


2middot58

middotL

41113874 1113875

δa δ05PL

248EI1 minus 5PL

248EI

But1Pe

5L

2


δa δ0PPe

1 minus PPe

(18)


δf δ0PPe

1 minus PPe

+ δ0

It can be put η 1

1 minus PPe

So δf δ0η

(19)


Madd Pδoη (20)



d2ydx

2 minusM

EI (21)


L (22)

dy

dx minus

1EI

Rx2

2EI1113888 1113889 + C1 (23)

δΔ0 minus1EI

Rx3

61113890 1113891 + C1x + C2 (24)



1 kNL4

M01

M02

δa

δ0

P



δΔ0 1EI

minusΔMx

3

6L+ΔMLx

61113890 1113891 (25)


Thusdy

dx

1EI

Mx2

2L+ΔML

61113888 1113889 0 (26)


3



δΔ0 ΔML

2

93

radicEI

(27)


δfΔ δΔ0

1 minus PPe

δΔ0η

(28)


Madd PδΔoη (29)



weqRL2

8 Pη

M0L2

8EI1113888 1113889 (30)



weqR PηM0

EI (31)


weqΔL2

93

radic PηΔML

2

93

radicEI

1113888 1113889 (32)

where weqΔL29

3





weqΔ PηΔMEI

(33)






weq1 pηM01

EI (34a)


weq2 pηM02

EI (34b)



and weqlowast2



weqlowast1 pη

M1

EI (35)



weqlowast2 pη

M2

EI (36)


EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)


K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)


Lb


L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)


M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI



minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















δΔ0 1EI

minusΔMx

3

6L+ΔMLx

61113890 1113891 (25)


Thusdy

dx

1EI

Mx2

2L+ΔML

61113888 1113889 0 (26)


3



δΔ0 ΔML

2

93

radicEI

(27)


δfΔ δΔ0

1 minus PPe

δΔ0η

(28)


Madd PδΔoη (29)



weqRL2

8 Pη

M0L2

8EI1113888 1113889 (30)



weqR PηM0

EI (31)


weqΔL2

93

radic PηΔML

2

93

radicEI

1113888 1113889 (32)

where weqΔL29

3





weqΔ PηΔMEI

(33)






weq1 pηM01

EI (34a)


weq2 pηM02

EI (34b)



and weqlowast2



weqlowast1 pη

M1

EI (35)



weqlowast2 pη

M2

EI (36)


EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)


K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)


Lb


L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)


M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI



minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















weq1 pηM01

EI (34a)


weq2 pηM02

EI (34b)



and weqlowast2



weqlowast1 pη

M1

EI (35)



weqlowast2 pη

M2

EI (36)


EIL3

4L2 2L

2

2L2 4L

2

⎡⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎦

θ1

θ2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ +

minus weq1L2

20minus

weq2L2

30+

PηM2L2

30EIminus

PηM1L2

20EI

weq1L2

30+

weq2L2

20minus

PηM2L2

20EI+

PηM1L2

30EI

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ (37)


K0 EIL3 (38a)

C PηL

2

EI (38b)

θ1 minus M1

Kθ1 (38c)

θ2 minus M2

Kθ2 (38d)


Lb


L col

I (beam 1)

I (beam 2)

W1 kN (mprime)

W2 kN (mprime)


M0

+ ∆M

M0

P P

δft=(δ

0+ δ∆

0)η

δ0+ δ∆

0

weqR = Pη

M0

EI

weqR = Pη

∆MEI



minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















minus 1 minus4L

2K0

Kθ1+

C

201113890 1113891M1 + minus

2L2K0

Kθ2+

C

301113890 1113891M2

weq1L2

20+

weq2L2

30

minus2L

2K0

Kθ1+

C

301113890 1113891M1 + minus 1 minus

4L2K0

Kθ2minus

C

201113890 1113891M2

minus weq1L2

30minus

weq2L2

20

(39)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(40)


M1

M2

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01

20+

M02

30

minusM01

30minus

M02

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(41)


[A] C

minus 1 minus4L2K0

Kθ1minus

C

20minus2L2K0

Kθ2+

C

30

minus2L2K0

Kθ1+

C

30minus 1 minus

4L2K0

Kθ2minus

C

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minus 1

(42)


RotationalSpring

weq∆ = Pη ∆MEI

weqa = PηM1EI

weqR = PηEI M02

weq2 = PηM2EI

δ0+ δ∆0

δft+( δa+δΔ

P

P

M01 = M02

Ma

M1

M2΄

δΔ1δΔ2η

δΔ2δΔ1η

o







θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References




















θcol 1

EIcol

1113946L

0MoM1

LAcol

3EIAcol

(43)



KAcol 3EIcol

Lcol

(44)


Mlowast

Lb6

EIbKAcol + Lb31113888 1113889 (45)


θb 1

EIb

12Lb

32 minus

L2b

1213

1EIbKAcol + Lb3

+L2b

3613

1EIbKAcol + Lb3

1113890 1113891

2⎡⎣ ⎤⎦ (46)


Kb EIb


21113872 1113873

(47)




Kb 3EIbLb

(48)

weq 1 ndash weq 1

weq 2 ndash weq 2

M2

M1








Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References





















Lb6EIb( 1113857Top



(49)






(50)




M1

M3

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ [A]

M01 + αMadd2( 1113857

20+

M03 + αMadd4( 1113857

30


30minus

M03 + βMadd4( 1113857

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

minusα1Madd2

α3Madd4

⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦

(51)



KB2

KB1P

δo

(a)

1 kNm Kcol

Lb

M0

M11

Mlowast

1

1

(b)

1 kNm1

M0

L col

M11

(c)






Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References



















Madd 1113944trialn


(52a)



(M1)trial3

(M1)trial2

(M1)trial1

M01 M02

(αM2)trial2

(αM2)trial1

(M2)trial3

(M2)trial2

(M2)trial1

(αM1)trial2

(αM1)trial1

Applying Eq (41)

①

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

Applying Eq (41)

+

+

+

+

②

(a)

06 times 03 m

col 2col 1

P1 = 5100kN

col 3

06

times 0

6 m

Lcol

06

x 0

6 m

06

times 0

6 m

06 times 03 m


w1 = 200kN(mprime)P2 = 5750 P3 = 1500kN

beam 3 5beam 1 3

64

1

2


Lb

beam 4

Lb

(b)



Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















Mtrail1

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail1

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M01

M02

M03

M04

M05

M06

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52b)

43

21

weq2

M0p

Elweq =

(α o

r βM

13)

trial

1p El

w eq

=

(α o

r βM

24)

trial

1p El

w eq

=

M0p

Elweq =




M03 M04








(M2)trial1M02




α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References

















α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52c)

Mtrail2

M1

M2

M3

M4

M5

M6

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

trail2

[A]

120

130

0 0 0 0

minus130

minus120

0 0 0 0

0 0120

130

0 0

0 0 minus130

minus120

0 0

0 0 0 0120

130

0 0 0 0 minus130

minus120

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

α1M3

α2M4

α3LM1 + α3RM5

α4LM1 + α4RM6

α5M3

α6M4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52d)

(α) Lb(6EI)b


(3EIL)coloumn


(52e)

Lf

21 LB

LcolLef

2

2

Lf

Lf

2

2

Mint3

weq1

weq1 weq1

weq1

Mint4

3 4

Mint1

Mint2

weq2

weq2

weq2

weq2

Lf

Lf





[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References


















[A]

minus C1 minus4C1L

21K01

Kθ1minus

C21

20minus2C1L

21K01

Kθ2+

C21

300 0 0 0

minus2C1L

21K01

Kθ1minus

C21

20minus C1 minus

4C1L21K01

Kθ2minus

C1

200 0 0 0


22K02

Kθ3minus

C22

20minus2C2L

22K02

Kθ4+

C22

300 0

0 0 minus2C2L

22K02

Kθ3+

C22

30minus C2 minus

4C2L22K02

Kθ4minus

C2

200 0


23K03

Kθ5minus

C23

20minus2C2L

23K02

Kθ6minus

C23

30

0 0 0 0 minus2C2L

23K03

Kθ5+

C23

30minus C3 minus

4C1L21K03

Kθ6minus

C3

20

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(52f)




infin infin

GA

GB

Pin-Ended ColumnK

500100

50

30

20

10

08

06

05

04

03

02

01

0

500100

50

30

20

10

08

06

05

04

03

02

01

0Fix-Ended Column

Rang

e of K

for

Fix-

Ende

d C

olum

nsRa

nge o

f K fo

rPi

n-En

ded

Col

umns

Rang

e of G

B for

is

Stud

y

05

06

07

08

09

10





Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References


















Mint weq(av)

4L minus Lf1113872 1113873

2+

weq(av)

4L minus Lf1113872 1113873 (53)



Mint1 Mint3

p M01 + M03( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

Mint2 Mint4

p M02 + M04( 1113857Lcol

8EIcol

Lcol minus Lf1113872 1113873

(54)


θ1 1EIB

12Mint1Lb middot

23


13

1113874 1113875

θ1 LB

EIB

Mint1

3+

Mint2

61113874 1113875

(55)



Mint13 + Mint26( 1113857


Mint23 + Mint26( 1113857



(56)











upper beam 0603 m

(col

umn

1) 0

50

3m

(col

umn

2) 0

50

3m

Lower beam 0603m


10m







EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References




















EI 02EcIg + EsIse

1 + βdn s

(57)

EI 04EcIg

1 + βdn s

(58)





0

2

4

6

8

10

12

-60 -40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(b)

FEMSuggEqu


0

2

4

6

8

10

12

-150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(c)

FEMSuggEqu


0

2

4

6

8

10

12

-200 -150 -100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

(d)









PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References






















PPe=02

0

2

4

6

8

10

12

-60 -40 -20 0 20 40Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)

PPe=03

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=04

(c)

0

2

4

6

8

10

12

-150 -50-100 500 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

PPe=05

(d)






au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References



















au 1π2

1113888 1113889l2e

1r

1113874 1113875 (59)


1rb

(0003 + 0002)

d (60)


au 00005l

2e

h

au h

2000middot

le

h1113888 1113889

2

λ2

2000middot h

(61)



0

2

4

6

8

10

12

-60 -40 -20 0 20Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(a)


0

2

4

6

8

10

12

-100 -50 0 50 100

Col

umn

leng

th (m

)

Add Moments (kNm)

FEMSuggEqu


0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)


0

2

4

6

8

10

12

-100 -50 0 50 100 150Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)



EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References
















EI (TB)EI(BB)=03

2

4

6

8

10

12

-150 -100 -50 00


Col

umn

leng

th (m

)

FEMSuggEqu

(a)

EI (TB)EI(BB)=055

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(b)

EI (TB)EI(BB)=1

0

2

4

6

8

10

12

-100 -50 0 50 100Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(c)

EI (TB)EI(BB)=24

0

2

4

6

8

10

12

-40 -20 0 20 40 60Add Moments (kNm)

Col

umn

leng

th (m

)

FEMSuggEqu

(d)


90

80

70

60

50

Addi

tiona

l mom

ent

40

30

10

20

07 119 13 15




140

120

100

80

Addi

tiona

l mom

ent

60

40

20

00 02 04 06

PPe08 1 12






Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References


















Mc M0 1 + 023PPe( 1113857

1 minus PPe

(62)





Hei

ght o

f col

umn

(m)

001 0005

12

10

8

6

4

2

00


0008 0006 0004 0002 00

2

4

6

8

10

12


Hei

ght o

f col

umn

(m)

-0004 -0006 -0008


0002

0003

d

Balanced failure

M

N








Appendix



Data Availability




Acknowledgments



References





















Appendix



Data Availability




Acknowledgments



References
















Alternative Design Procedure for RC-Braced Long Columns ...

Documents

Transcript of Alternative Design Procedure for RC-Braced Long Columns ...