1. Introduction

The use of a constant initial tangent stiffness in the analysis of buckling of bars and rigid frames is most desirable but is ,as yet, very difficult to find in the literature A constant stiffness method, as it is the case of non-linear finite-element analysis demonstrated by Zienkiewich[1] and further studied by Duncan and Johnarry[2] always promises to be easier than those that rely on tangent-stiffness-changing iterations with load stages. Iterations based on stability functions alteration of the stiffness matrix according to the extent of the axial buckling forces is amply demonstrated by Ping-Chun Wang(1965),[3], Wood(1974),[4] relied on stiffness alteration in his stiffness distribution buckling studies. Georgios Mageirou and Charles Gantes(2006),[5] and Messaoud Bourezane(2012),[6] ,among others, relied on Stability functions-based matrices in their buckling research on beams and frames . For example, in the slope-deflection equation, the familiar rotational beam stiffness, 4EI/L=(k₁₁) will change to (k₁₁) + function(c,E,I,L) ;c=compression. There are several other approximate methods for critical loads which by-pass the use of stability functions; the treatment by Horne,1975,[7] ,which relied on elastic story drifts ,stands out for its simplicity as it requires the elastic solution and the employment of a formula on the solution displacements/rotations ;the formula was not completely proven. In recent studies Johnarry[8,9] has been examining the employment of acceleration-deflection resonance in new alternative solutions for plates and re-definition of buckling criterion .( curvature ≡ acceleration)

2. Analysis

2.1. Resonance as a Buckling Criterion

In the presence of fiction ,the Euler equation will be,

(1)

For a pin ended bar , try

Y = Σ A_m Sin m πx/L

Continuing ,

(2)

A_m →∞ as P → EI(m²) π² / L² ;(m=1 for mode-1)

Buckling is sudden appearance of uncontrollably large displacement , by this view. Now, in this case

(Y,_xx)_shape = Y_shape = Sin m πx/L

Can it also be generalized that buckling is a condition of resonance between curvature and displacement ?

Can relative-acceleration be substituted for acceleration ? Clearly in this pin-ended case there is no standing support fixing moments and the relative acceleration is everywhere the same as the acceleration. A relative curvature envelop has an immediate common property with its relevant displacement envelop; both have zero end-values if the end values of displacements are prescribed zero. In the event the shapes of the two envelops are similar there is then resonance ,buckling, vibration.

2.2. Relative Acceleration – Displacement Analysis

Figure 1. Pinned – Clamped Column

As an example, in Fig.1 ,consider the pinned-fixed column. Chose the Y-function to be able to satisfy all geometric boundary conditions; let,

Y = Sin K X +Cos KX + A X + B;( -K²(Cos KX) =0;so,

(3)

(4)

Let the relative curvature be ,

(5)

S=0 from starting conditions

Also, in relative curvature analysis, terminal ends of a line

have zero relative curvatures; so,

At x=L , (Y_,xx)_rel .=0 ;so, R=(K² Sin KL) /L

Compare shapes of ‘Y’ and ‘Y_{,xx-rel-zeroends.’}

(6)

At X=L, Y=0; so A= - (SinKL)/L ;from Eq.3 (BCA)

Y_,xx-rel / Y = K² (-Sin KX + (X/L)Sin KL) / [ N^*(Sin KX –

(7a)

(7b)

AT X=L; Y_,x = 0; so, K CosKL + A =0, and with Eq-7,

(8a)

and,

(8b)

So far in this section, no buckling force-P, has been

mentioned , but from Euler’s buckling relation for a

pin-ended bar,

(9a)

(9b)

(9c)

(Y_,xx-rel . / Y)→ resonant , Y_,xx-rel /Y >> Y_,xx-0/Y ; Y→∞ ,

at buckling

So,

E I Y_,xx-rel = -P Y

(9d)

So, buckling(sudden large displacement) is indicated if relative acceleration and displacement become resonant.

Consistent displacements can be introduced and adjusted until the relative curvature - displacement in Eq.9d is constant at each station and when that happens the force ,P, assumes a critical value- P_cr .

This is the exact solution for the problem. In a computer stiffness matrix solution no displacement function is called for ; the slope deflection equation matrix formulation provides initial consistent nodal displacements.

The point-wise constant relativeacceleration-displacement ratio can also be presented to vibration relation, clearly.

2.2.1. Energy Method Versus New

Acceleration-displacement ratio method

In the energy method the buckling load is found from

(10a)

In the present method it is found from,

(10b)

If (M_rel)_{curve shape} ≠ Y_{curve shape} , then rationalize,

(M_rel) = H Y ; find scalar- H ,so that,

∫ (M_rel) Y / ∫ Y.Y = H

Examine a pin-ended column by the restricted function , Y= X(1-X)

M_fixity =0 ; physical support condition ,so, M_relative = M

M_rel = EI (-2) = H . Y

∫ EI(-2) Y dx = ∫ H Y. Y dx ; H=10 EI

Pcr = 10 EI ; exact = (π² )EI

The energy integral Eq.-10a will give a less approximate

result of P_cr=12EI.Eq.-10b is easier than Eq.-10a in many instances.

Fig.2 gives an indication of how the fixed-fixed column is to be handled(suitable equation Y=Cos Kx +C) .The pin-pin

Column is illustrated in Fig.3 divided into 38-elements and later solved.

Figure 2. fixed – fixed column

Figure 3. Pin-Pin Column, 38-elements, 39-nodes

2.3. Computer Stiffness Solution

The equation Y = J ( Y″ + RX + S ) is solved by successive approximation.

(11)

The right hand side(rhs) of eq-11 is available from stiffness analysis , point-wise.

Start by applying a transverse load (say uniform load of any magnitude)

(12)

(13)

Interpret the bending moments so found as the next set of applied loads and re-solve.

(other acceleration expressions can be tried)

A solution is found when ,point-wise,

(14)

is obtained; the displacement and the relative acceleration curves are then similar or of constant ratio

point-wise.

The basic beam stiffness can be stated as:

(15a)

(15b)

So, find for (N=axial

force,M=moment,V=shear-force;member from point-1 to2); a*=EA/L

(15c)

and conduct routine stiffness analysis . The stiffness matrix once found will remain unaltered in the iterations and produces very fast analysis. The results for the pinned-fixed column, the fixed-fixed column and a portal frame in axial compression against its mirror are presented

2.3.1. Pinned-Fixed Column and Others in Computer Solutions

The fixed-pinned column is ,often, a severe test in buckling studies and so the member is divided into 38 equal elements and 39- nodes. The record “moyope” means (m_rel)/(y)/P_E .;(P_E .= reference Euler load)

The quotient is deficient in the vicinities of zero deflections and rapidly changing bending moments. Table 1 shows that the predominant ratio of 2.08 at many nodes is virtually exact. Where the ratio is much different from this the station weight measured by (y m/mmax) will be very small. Tables-1,2,3 show results for iterations =1,3,20. Table 4 compares iterations and the quality of results leading to the conclusion that three iterations are sufficient in a given situation.

As shown in Table-5 for the pin-pin column(20-elements) the result of 1.002 is ,practically, exact and it must be recalled that the stiffness solution did not assume a sinusoidal Y-variation that, in a manual solution, would have led to the same correct result. The result confirms that a column buckles into a sine-curve .

The result of the fixed-fixed column (38-elements) is given in Table-6 and the result P/Pe=4.006 compares with the exact value of 4.0

Table 1. fix-pin-col; iteration= 1; bars = 38, nodes=39

Table 2. fix-pin –col; iteration= 3, bars = 38

Table 3. fix-pin –col; iteration= 20, bars = 38

Table 4. fixed-pined column iterations and P/PE .comparison

From Table-4 and from experience three iterations are often sufficient.

Table 5. pin-pin-col;itertn =3, bars = 20

Table 6. fixd-fix- col; itertn 3; bars = 38

Table 7. fixed-base portal + mirror, itertn =3; bars= 34;(Ibeam)/(Icol)=0.1

Table 8. fixed-base portal + mirror; itertn =3, bars = 34; Ibeam /(Icol) =1

Table 9. fixed-base portal + mirror; itertn =3, bars = 34;(Ibeam)/(Icol)=100

Table 10. pin-base, 1-cell-portl-dirct itrtn= 5; bars = 34; Ib/Ic=0.01

Table 11. pin-base, 1-cell-portl-dirct itrtn= 5; bars = 34 ; Ib/Ic = 1

Table 12. pin-base, 1-cell-portal-direct itrtn= 20; bars = 34 ; Ib/Ic =100

A fixed-base lone portal frame,Fig.4, is analyzed with its mirror so as to find conservative energy situation, as is the case of the cantilever column. The results are as exact as the isolated columns already examined for various beam-to-column stiffness ratios in Tables 7,8,9 and summarized in Table-13.

In the case of the pin-pin portal frame,Fig.5 there is no need for the mirror part as the displacement curve is conservative directly. Results are given in Tables,10,11,12 for different I-beam /I-col ratios. The exactness of the results is maintained.

Many results of these frames for various I-b/I-c ratios can be easily found as demonstrated . Column strengths and design graphs as found in Wood(1974),[4] and in design Codes,BS-8110,BS-5950[10,11] can be replicated by the present method.; the objective here is to show a very fast buckling solution method without going outside the conventional elastic stiffness analysis that forms the back-bone of bridge and building frames design analyses and studies.

Table 13. P./PE. for portal frame for various Ibeam/Icol

Figure 4. Fixed-base sway frame; (a) alone (b) plus mirror for analysis; θ9= θ26 = O; nodes 1-9,18-26 = mirror

Figure 5. Pin-base single portal-analysis

2.4. Extension to Substitute Frames

The substitute frame method allows a large frame to be reduced to easier- to -handle limited frames in the design process.

In applying the present method to any story the main issue to contend with is the value of the fixity at the base of the story. It is somewhere between fixed and hinge support. The present method is based on elastic analysis and so super-position of two solutions(partial fixity and partial hinge) is studied.

2.4.1. Column-Base Fixity Ratio ,F_R

Beams at joints provide fixity to columns and a fixity ratio is defined as

F_R = (I/L²)_beam / [(I/L²)_beam + (I/L² )_col];

maximum=1.0 whenbeam is infinitely rigid (16)

So, hinge-support ratio = 1 -F_R

The equation is rational and highly effective. It is known that in the buckling environment, moments are proportional to (P.y) or (I/L²) and moments must be balanced at joints of beams and columns.

2.4.2. Examples of Story Analysis in Multi-story Frames

These frames ,Figures 6,7,8,9 were taken out of the paper by Horne[7]; results summarized in Table 14

..(Example-Figure.6.)..3-storys frame, Figure. 6; Pinned-bases

; try bottom story ;fixity=0; so direct analysis with pinned bases .

Enter program with Icol=76470000 , E=201,000, L-col=4000, L-beam=8000 ;Ibeam=350,830,000

find P_cr/P_E =0.2177; P_applied =1088Kn.; P_cr /P_applied . = 3.79, ((alternate result quoted by Horne[7]=3.78));

;;try another story if necessary.

..(Example-Figure.7(same as Fig.6 but fixed feet),)) ; ; try story-2

stp1.. F_R = (350,830,000/8000²) / (350,830,000/8000² +76,470,000/4000²) =0.534

stp2.. Hinge-factor =1- 0.534 = 0.464

stp3.. P_cr-hinge/P_E= 0.2177 –already found above .

st4..With new fixed-fixed boundary conditions ,find ( P_cr /P_E )=0.881

st5..So, (P_cr/P_E)_story = 0.881(0.534)+0.2177(0.466) = 0.572 ; P_cr = 10,828 Kn.

st6..(P_cr/P_applied) = 10847/628 =17.24

(Example-Fig.7,story-1)..Try bottom story to check if more critical; (F_r = 1; P_cr =16,680; P_applied = 1088.

P_cr /p_applied = 15.34; compare exact 14.72 - - Table 14;)

(Example-Figure.8,botm story.).. 8-story frame shown in Figure.8 ;try bottom story

stp1.. F_R = (236000000/3050²) / (2360000000/3050² + 479160000/5080²) = 0.577

stp2.. solve with base fixed-fixed ; Pcr/Pe = 0.573

stp3.. solve with pinned bases ; Pcr/Pe = 0.136

stp4.. ( Pcr /Pe)_combined = (0.577(0.573) + 0.423(0.136)) = 0.3875

stp5.. Pcr = 0.3875(π² )E (479160000 (2legs)) /(5080²) = 28,547 Kn

stp6.. P_applied = 7,120 Kn ; Load-factor Pcr/P_applied = 4.0 ((alternate result quoted by Horne[7]=3.95 ))

(Example-Figure.9,story-2).. 5-story – 2bay frame ;Figure.9 ; try story-two

stp1..first reduce to one bay 5-story frame(i.- add up I-beams ;ii.- add up I-cols and divide by two)

stp2.. find F_R = 0.6591

stp3 solve for fixed bases ; Pcr/Pe = 0.9035

stp4.. solve with pinned bases ;Pcr/Pe = 0.223

stp5.. (Pcr/Pe)_combined = 0.671

stp6.. Pcr/P_applied = 7.51 for E=201,000N/sq-mm ((alternate result quoted by Horne[7] =7.70))

(Example-Figure.9,story-1)..try story-one ; F_R =1 ; bases actually fixed-fixed ,so pin-pin contribution is none.

..stp1 ; Pcr/Pe = 0.893

..stp2 ; Pcr / P_applied = 8.52 > 7.38 for story-2 found above.

Figure 6. pin-pin frame

Figure 7. fixed-fixed frame ,same as Fig.6 but fixed

Figure 8. Frame

Figure 9. Frame

2.4.3. Quality of Results in Frames

The fixity ratio method of mixed-boundary-conditions studied here led to results within 5-percent of the best of other quoted results in the limited experience of the method for multi-bay multi-story frames.

3. Conclusions

Buckling instability has been successfully defined as the invariant point-wise quotient of the relative-moment and the consistent displacement, easily obtained from an iterative load-displacement analysis using the constant elastic tangent stiffness. Examples in beams and frames were tested in this preliminary study. About three iterative solutions are needed for an analysis. Above the ground story , the upper storys operate in partial fixity(between fully fixed and pinned);partial fixity and partial pinned factors were rationally found to produce a super-posed solution; each solution is purely elastic. The critical load of each story was easily found directly.

ACKNOWLEDGEMENTS

This study has been financed by Rootway Academy Services .