Rational Analysis and Design of Prestressed Concrete Beam … Journal... · 2018-11-01 · Rational Analysis and Design of Prestressed Concrete Beam Columns and Wall Panels Noel D

Rational Analysis andDesign of PrestressedConcrete Beam Columnsand Wall Panels

Noel D. NathanProfessor of Civil EngineeringUniversity of British ColumbiaVancouver, British ColumbiaCanada

Precast concrete columns are ofteneast and shipped in lengths extend-

ing over two, three, or more stories. Inconsequence, they are frequently longand slender, at least until jointing iscomplete, so that handling stresses maygive rise to problems.

Similarly, precast loadbearing wallpanels for light industrial or warehousestructures are generally required to bevery tall, again leading to high slender-ness ratios. The roof loads, distributedalong the wall, are usually quite light, sothat stress conditions are often domi-nated by the bending moments arisingfrom wind loads and load eccentricity.

For the above reasons, precast col-umns and loadbearing walls are fre-quently prestressed.

Codes of practice usually provide em-

pirical formulas for the effects of slen-derness, but limit the range of slender-ness for which they may be used. Forexample, ACI 318-83' requires that ifthe slenderness ratio L/r exceeds 100,design shall be based on analysis which"shall take into account influence ofaxial loads and variable moment of in-ertia on member stiffness and fixed-endmoments, effect of deflections on mo-ments and forces, and the effects of du-ration of loads." No direct guidance,however, is given on how to make suchan analysis.

Precast prestressed concrete columnsand wall panels frequently fall withinthe range of slenderness that excludesuse of the empirical formulas. Further-more, it has been found that the empiri-cal formulas are not as precise for the

82

very low percentages of steel that aregenerally present in precast members,even when the slenderness ratios aresmall.

This paper will address these prob-lems by reviewing the analysis for sec-ondary effects in columns and walls.Methods involving varying degrees ofapproximation will be briefly discussed.The effect of prestressing will be re-viewed, and it will be suggested that ra-tional analysis of components shouldgenerally be the preferred approach forprecast prestressed elements.

The theory of such a rational analysiswill be developed. The construction of acomputer program typical of those usedby researchers for this purpose will bebriefly described. A listing and docu-mentation of the program can be ob-tained at cost of reproduction from thePrestressed Concrete Institute.

The computer program may be usedto determine the design moments for agiven set of loads and end conditions, orto develop design curves of acceptableloading for given member cross sectionsand will predict material or instabilityfailure. The cross section may have anypolygonal shape and may have anynumber of mild steel bars and/or pre-stressing tendons. The concrete andsteel components may have any stress-strain laws, specified in functional formor in the form of points on an experi-mentally determined curve.

Any form of lateral loading can behandled, specified by the primarybending moment arising therefrom. Theaxial load may he applied with differenteccentricities at the two ends. Initialcurvature may be included.

Note that the boundary conditionsmust be known or an effective lengthmust be estimated. A load-deflectionanalysis is used to account for lateraldisplacement of the joints if such dis-placement is permitted.

It is recommended that precast con-crete manufacturers use computer pro-grams such as this to develop design

SynposisSecondary effects of axial loads on

frame members are discussed. Meth-ods of analysis for these forces arereviewed. Characteristics of pre-stressed concrete members whichaffect the method of analysis are de-scribed, and it is noted that "rationalanalysis" is often required. Construc-tion of a computer program to performsuch an analysis of component mem-bers is discussed, and a program isdescribed in detail in an appendix.Several examples are included.

curves for standardized products such ascolumns, double tees, piles, and othercustomized units. The author's programhas been written in very elementaryFORTRAN and therefore should beeasily adaptable to specific needs.

SECONDARY EFFECTSIN FRAMES

This section contains a brief discus-sion of the problems associated withaxial loads in frame elements and of themethods of dealing with them.

The presence of axial compression ina frame member has two effects:

1. The stiffness of the member is re-duced.

2. Secondary bending moments, notaccounted for in the primary analy-sis, are generated when the line ofaction of the axial force is no longercoincident with the centerline ofthe member.

It is possible to make analyses whichtake all these factors into account, butdifficulties are involved,2 3 particularlyin the design stage of reinforced or pre-stressed concrete frames. However, thefirst effect, namely, the reduction in

PCI JOURNALJMay-June 1985 83

stiffness, is negligible if the axial load isa reasonably srnall proportion of theEuler load for the member.'•°

This condition is generally satisfied inpractical building frames, so that theeffect of stiffness Ioss on the distributionof internal forces may effectively he ig-nored. A method of accounting for lossof stiffness, given in Refs. 5 and 6, isreferred to below.

The secondary bending momentsarise from changes in the geometry ofthe structure which are not taken intoaccount in the primary analysis. Theseeffects may again be separated into twoparts:

1. Relative motion of the joints withrespect to one another, in thetransverse direction to the axialloads. Erection tolerances maycontribute to this effect.

2. Departure of the member center-line from the straight line betweenthe joints, due to bending of themember along its length. Thermalbowing, manufacturing tolerances,and camber may be significant inthis respect.

The response of a structure to loads,accounting for these effects, may beevaluated by a variety of procedures:

Method 1An equivalent pin-ended member is

assumed for each column. The "effec-tive length" of this equivalent memberdepends upon the end conditions.?When sidesway is not prevented, thefirst of the effects noted in the previousparagraph is accounted for if the deflec-tions are measured from the thrust line,and the column is given an imaginaryextension, until it recrosses the thrustline at an imaginary point of inflection.

The response of this equivalent pin-ended member is then obtained by mod-ifying the primary bending momentcausing sway by a magnification factor.The moments not associated with swayare also magnified to account for the

second effect, using the braced effectivelength.

Method 2An approximate analysis is made to

determine the horizontal displacementsof the joints, and the so-called P-( load-deflection) moments arisingtherefrom are directly calculated. Themagnification factor is then used to ac-count for the additional secondary mo-ments due to deflection of the columncenterline, using the effective length fora braced column.

Method 3The procedures of Methods 1 or 2 are

used, but the response of the equivalentcolumn is determined by some rationalprocedure instead of by means of thesemi-empirical magnification factor.

Method 4The entire analysis accounting for all

the nonlinear secondary effects is carriedout by a rational procedure, as discussedin Refs. 2 and 8. As mentioned above,this process is far too complex for practi-cal use in the design stage of concretebuildings.

COMMENTS ONANALYTICAL PROCEDURES

In this section, further comments aremade on the individual steps involvedin the above procedures, namely, thedetermination of effective Iength, thecalculation of P-A moments, and theuse of magnification factors.

Effective LengthThe bifurcation buckling load for an

individual column with pure axial loadis determined by the Euler load for ahinged column of effective length kL,

84

^ fPoints of inflection

J J

BRACED CASEw

Points of Inflection

J

~ 1 t 61

r _^

-J SWAY CASE \//

Fig. 1. Effective lengths of columns in a frame (after Ref. 9).

the distance between the points of in-flection at the instant of buckling. Inframed members (see Fig. 1) the effec-tive length depends upon the relativestiffnesses of the columns and the re-straining beams.

When moments or lateral loads arepresent, as is usually the case, instead ofbifurcation buckling, magnified mo-ments (i.e., added secondary moments)

occur, leading eventually to materialfailure or to instability. The magnifica-tion, however, also depends upon theeffective length and therefore on the re-straint conditions at the ends of the col-umn.

The determination of effective lengthis discussed in Refs. 3, 7, and 9. In prac-tice, the Jackson-Moreland alignmentcharts9 • 10 are generally used for this pur-

PCI JOURNALIMay-June 1985 85

400

20 k5 ^, !0

o IEI Columns=8000EI Beom=30000

(0)

+19.7 +19.7 8.42Ir-M1

A = 0,0521°_'Holding Force

IT 400x0,0521cv = =2.08

4.21 10 4.21

(e)

400

-343.1 +249.5 14.03 +4.88 +4.88 2.08

r) "' m in v v A= 0.0429+

r

Holding Force,r, .. 400 x0.0129=0.52

I 0+ 51.46 37.43 1.04 ; ___

(b) (f)

+798 +79.8 34,03 +1.21 +1.21 0.52m _

CO(4I NA= .5 yr - 4=0.0032

Holding Forceo

0210

M1 =400x0.0032 _0.13 N-

o "' 1017.02 17.02 a' 026,--Neglect- 0.26 1

(c) (g)

4

20 -263,3 +329.3 to

-KM of

Lo o =0.2105 N)Holding Force

400 X 0.2105 N10 =8.42 u)

m 34.44 54.44 N

(d)

4Q0

20 -237,5 +355.1-In

IL) m

6=0.279 r?

+ Final Solution mm d

+ 36.14 56.06

LO (h)

Fig. 2, Iterative load-deflection (P- A) analysis.

86

pose; k is often taken as unity whensway is not permitted.

The derivation of these charts is dis-cussed by Galambos, Kavanagh, 7 andChu and Chow." MacGregor andHage,'u however, assert that the chartsare "based on highly idealized and quiteimpractical cases," and they show thatthey lead to unrealistic results — oftenhighly conservative but occasionallyunconservative.

The difficulties in using the effectivelength approach are associated with theevaluation of the restraining effects ofthe beams on the column, particularly inunsymmetrical structures, when side-sway takes place. It is for this reason thatthe P-A method4"2 is advocated for theseparate evaluation of the momentsarising from side sway.

Evaluation of P- MomentsThe application of various procedures

for computing the effects of sideswayare excellently set forth in Refs. 5 and12, and will be briefly discussed here.

An iterative procedure 4 '5• '2 will be il-lustrated by means of the simple portalframe shown in Fig. 2a. In Fig. 2b a firstorder analysis is made of the gravityloads. Sidesway is prevented by meansof a holding force found to be 14.03. Fig.2c shows the removal of that holdingforce together with the application ofthe lateral load of Fig. 2a.

The sum of Solutions 2b and 2c,shown in Fig. 2d, would complete theusual first order analysis. But now, thedeflection (A = 0.2105) is computed andit is deduced that the columns are actedupon by a moment !PA (i.e., 400 x0.2105) which is not, as yet, balanced.Therefore, in order to maintain equilib-rium, there must be a remaining holdingforce of:

V A/h = 400 x 0.2105110 = 8.42

The effect of removing this force isshown in Fig. 2e; however, there is thena further deflection of 0.0521 and

therefore, by the same reasoning, aholding force of 400 x 0.0521110 = 2.08.The effect of removing this force isshown in Fig. 2f, and it is seen that thereis a further deflection of 0.0129 and aholding force of 0.52.

This leads to Fig. 2g, with deflection0.0032 and holding force 0.13, This isconsidered negligible, and the solutionis assumed to be the sum of Figs, 2d, e, f,and g. If the process does not rapidlyconverge, it indicates that the structureis probably too flexible with respect tosway.

Note that the solutions for Figs. 2e, f,and g are simply prorated from that ofFig. 2c. The whole procedure is usuallyan elastic analysis performed with re-spect to the factored loads; however,since it is an elastic analysis, it can oftenhe obtained by factoring up the compo-nents of a service load analysis (similarto Figs. 2b and 2c) performed to checkservice load drifts. Note that if there arechanges in the stiffnesses due to crack-ing as the factored loads are approached,this should be accounted for.'E

The process illustrated in Fig. 2 iseasily extended to the general case ofmultiple stories and multiple bays. Ithas accounted for the secondary mo-ments arising from horizontal motion ofjoints, but moments arising from dis-placement of the member centerlinebetween joints have yet to be accountedfor by means of the magnification factor(based on the braced effective length),or by further rational analysis. Note that,even when side sway is restrained, thisprocess leads to the values of the forcesin the wall or bracing providing the re-straint.

If the increments of deflection in thepreceding method are written out insymbolic form, it will be found that theyform a geometric series whose sum isgiven by:'R

A,= HIK = ^ 1 (1^1— 1— ^'

Kh Hh

PC! JOURNAL May-June 1985 87

400

20 -263.3 +329.3K) M

PRIMARY ANALYSIS FnN

+ Î = 0.2105

O Z 0.2797 ieq.1)

Holding Force N

400 x 0.2797 'r)m 0 F1.19

10N

(a}

+26.2 +26.2 11.19N NtD ^DN N

SECONDARY MOMEN T S

N

N1

N

{b)

-237.1 +355.5NN)

II)LnN N)

+

FINAL RESULT

mv v

Itu^ C%1+

Cc)

Fig. 3. Secondary analysis using series summation for deflection A.

or whereAl = primary deflection (Fig, 2c)

Q s _ 0.2105= 0.2797 A2 = final total deflection

1 _ (400) (0.2105) H = sway force

(34.03) (10) K — lateral stiffness of frame

88

Fig. 4. Imaginary bracing providing secondary forces.13

Note that if the numerator is equal toor less than zero, the series does notconverge and the structure is unstable.In this way, the total deflection is ob-tained without iteration, and the totalholding force from Figs. 2e, f, g followsimmediately. One more elastic analysisunder this force gives the final result(see Fig. 3).

This procedure is strictly valid onlyfor single story frames, but it can beused for multistory structures in whichthere are points of inflection in the col-umns (relatively stiff beams) and inwhich the magnification is of the orderof 1.5 or less .5

Finally, an ingenious method suitablefor computer analysis of large frames,presented by Nixon, Beaulieu, andAdams, 13 will be described.

Suppose, as before, that the final lat-eral deflection of the structure in a givenstory is A., giving rise to total columnmoments IPA. As shown above, an ad-ditional horizontal force ZPAslh in thedirection of A Q would cause these mo-ments.

Now suppose that an imaginary brac-ing member is added as shown in Fig. 4with stiffness such as to cause the force- 'PA2 1h when there is relative lateral

displacement a$ . (Negative because themember must exert a force in the samedirection as its deformation.) The stiff-ness of the bracing member with respectto extension along its axis is AE/L, sothat the stiffness with respect to hori-zontal displacement is (AE/L) costa.Thus:

AEL cosaa D2

= - li 2(2)

or

APL 1A = - Eh costa

where L is the length of the bracingmember.

That is, if imaginary bracing membersare inserted in each story, with negativeareas as given by this equation, they willlead to correct moments in the columnsand beams. The shears and axial forceswill be somewhat in error due to theforces in the bracing, but these effectswill be quite small3 ° (the negative areasrequired are, in fact, very small), andthey can be minimized by making thebracing angle a as flat as possible. Onebracing member per story can stretchacross all the bays of the frame. This

PCI JOURNAL May-June 1985 89

3

2

a

EA (Bracing) _- PL • } 2 [eq.(2)]h cos a

400(18.028)310 . 12

'-1041.65

k = E cos t a =-40

11 200 4000 -480K (Structure) 4000 11200 -480

-480 -480 (192-40)

888.9Generalized Forces R = - 444.4

20.0

116Displacements = K - ^ = R = -.069

.2797

Knee Moments = 237.1 , 355.5

Fig. 5. Second order analysis using bracing with negative area.

procedure, shown in Fig. 5, can he usedin a standard frame analysis programwithout modification.

The loss of stiffness in the column

members can be accounted for in each ofthe foregoing P- analyses by includinga "flexibility factor" in the term P. Thisfactor, developed in Ref. 5, is given by:

y = 1 + 0.22 4('PA - qI ) + ( ALA+3) (q,u+3) {3)[( +2) (tPB +2) - 112

90

10

7654

2

1.5

i4iz

.9B.7.6.5.4

Iyov

.3

2 ! â

1.5

1.0B

0.1 .15 .2 .3 .4 .56.7.8.91 1.5 2 3 4 5 678910

*1

(I E11L)columns at End i of column.Fig, 6. Flexibility factor y of Eq. (3). r = (Y E1!L)beams

where

= 1^ L '/ \`°` at End A or B of column

1 L^ !beam

The t/j factor varies between 1 andabout 1.2, and values are given in Fig. 6.Inclusion of this factor gives a good es-timate of the lateral deflections, al-though the redistribution of the mo-ments is not properly accounted for,

Magnification FactorIt is shown by Galambos (pp. 246-7 of

Ref. 3) that the moments in a columnbent in single curvature are magnified,due to displacements with respect to thethrust line, by the factor:

1S = 1 – PIPS

(4)

Eq. (4) is essentially the same formulaused in the ACI Building Code.' Whenthe effective length for the sway case asillustrated in Fig. 1 is used, this magni-fication factor includes the P -A effectsdiscussed above, since the displacementfrom the thrust line includes the A de-flection. However, it will be observedthat the additional moments of Figs. 2d,e, f, and g all arise from the conditions ofFig. 2c, and are quite independent ofFig. 2b (except insofar as the holdingforce of 14.03 is concerned).

Thus, the magnification factor associ-ated with the sway-permitted effectivelength applies only to the momentsarising from sway forces (Fig. 2c). Thisis made clear in Ref. 1, where the mag-

PCI JOURNAL/May-June 1985 91

nified factor moment in a column is:

Mc = Sn Man + S. t. (5)

in whichMZD = value of larger factored end mo-

ment on compression mem-bers due to loads which resultin no appreciable sidesway,calculated by conventionalelastic analysis

Meg = value of larger factored end mo-ment on compression memberdue to loads which result in ap-preciable sidesway calculatedby conventional elastic frameanalysis

Note that 8h and 8,, discussed below,are based on the effective lengthfor the braced and unbraced cases,respectively.

It is recognized that the two compo-nents of M, in Eq. (5) do not necessarilyoccur at the same point in the column,and are therefore not directly additive.Nevertheless, the true maximum cannotbe greater than their sum, so, for thesake of simplicity, this conservativeform is accepted.

The same problem arises moreacutely when braced columns are sub-ject to double curvature. In that case, themaximum secondary moment certainlyoccurs at a point remote from themaximum primary moment at the end ofthe column. The magnification factorshould then include a factor C,,, _- 1. Thetheoretical value of C. for elastic col-umns is given in Ref. 3, p. 246, togetherwith various approximations, includingthat used in Ref. 1:

Cm = 0.6+0.4M2u0.4 (6)

Finally, noting that, due to diaphragmaction of the floors, the displacements ofall columns are essentially equal at floorlevels, so that the sway magnificationfactors are the same for all columns ofone story:

Sb = Cm . 1 (7a)1 – PIbP^

(7b)1- Y.P110

Y-P.

where the summations are over all thecolumns of the story.

AEIP^ _ (kLy ($)

In calculating P, , the value of k is ob-tained from the Jackson and Morelandcharts for the braced and unbracedcases. Approximate formulas are given'for the rigidity El which represent be-havior as the ultimate load is ap-proached.

An interesting result from Ref. 14shows that the critical load for any floorof a frame is:

Hh (9)

Y

where0 is the deflection from a first-order

analysis.y depends upon the deflected shape

of the columns, Its numerical valuelies between 1 and 1.22. [See Eq.(3) and Fig. 6.1

If one takes y = 1, then:

S $ = 1 (10)_

1 Hi

The deflection A should be obtainedfrom a primary analysis of the sway ef-fects, presumably using the rigidity Elspecified for use in P.

MacGregor and Hage' 2 present step-by-step procedures for the use of thesemethods in the design process. For ex-ample, one may begin with S, from Eq.(10) based on the permissible drift indexMIh and end with column size selectionto ensure that the permissible index isnot exceeded.

92

00J

MOMENT

Fig. 7. Instability failure and material failure.

APPLICATION TOPRESTRESSED CONCRETE

It has been shown" that, since theygenerally have a very low steel ratio,prestressed concrete members havedifferent moment-curvature relation-ships from normal reinforced concretemembers, As a result of this, they aregoverned by instability rather than bymaterial failure for almost the entirerange of slenderness values. Further-more, they are often used at higherslenderness ratios, which compoundsthe problem.

The P-A analyses discussed previ-ously are not directly applicable wheninstability governs. It is assumed inthose analyses, in removing the holdingforces (Figs. 2d, e, f, g), that a linear

analysis at the tangent stiffness of thestructure is possible. The total moment,including the P -A contribution and thebraced column magnification factor, iscompared with short column capacity.

If the columns become unstable be-fore reaching material failure the analy-sis becomes more complicated (see Fig.7). MacGregor and Hage 12 cautionagainst this possibility, giving criteria bywhich its likelihood may he assessed;but they note that it is seldom encoun-tered in reinforced concrete buildingframes. When it is, the analysis of Fig. 2must be modified to account for the lossof stiffness along the lines indicated inRefs. 16 and 17, or the y factor of Eq.(3)can be included in the magnificationfactors to correct the deflections.

The susceptibility to unstable behav-


+00

&o

60

Li±40

20

{,1) io

6

4

2

IPNo

'o

pro

0.02 .04 06 .08 .10 .f 2 .14 .18 30 .40 .50.16 20

(P'0)

Fig. 8. Design aid for X adjusted for use with ACI Code strength reduction factor and loadduration factor 0 -_ ,3u ^ 0.5. Sections with no compression flange.

for in members with very low steel ra-tios also leads to problems in evaluatingthe moment magnification due to de-flection of the member centerline be-tween joints.

Attempts have been made to extendthe range of applicability of the ACImagnification procedure' to memberswith low steel ratios and high slen-derness ratios. Using the theoreticallyaccurate loads generated by the com-puter program detailed below, an effec-tive rigidity El was back-calculated foruse in the computation of P:

El = E,I,1X(1 + 63,) (11)

withX =i63.213d = ratio of maximum factored dead

load moment to maximum fac-tored total load moment (alwayspositive)

= 2.5 + Pl where 6 r^ 70

35 for sections with

Llr – 0.09 compression flange

B=27 for sections with no

– 0.05 compression flange

Design charts charts for ,, are shown in Figs.S and 9.

94

100

60

60

40

2S

20

r0O

(A) r0

8

6

4

2

0.02 04 .06 .0 8 .10 .12 .14 J9 .30 .40 .5016 .20

(P,/PO)

Fig. 9. Design aid for x adjusted for use with ACI Code strength reduction factor and loadduration factor 0 _ {3 d a 0.5. Sections with compression flange.

The artificial nature of these factorsmust be emphasized. In the first place,sections with highly nonlinear mo-ment-curvature relationships are beingrepresented by a formula derived fromlinear material behavior (see Ref. 15,Fig. 14). Further difficulties arise fromthe application of the strength reductionfactor 0, and with the long-term loadfactor f3,.

Together, these two factors accountfor variations in Young's modulus (dueto creep and accidental variations), andfor accidental variations in the momentof inertia and strength of the cross sec-tion. With regard to slenderness effects,

all the section properties are reflected inthe moment-curvature relationship.

It is presumed that the effects of 4 and/3d should be to modify the moment/cur-vature relationship as shown in Fig. 10.For reinforced concrete columns withsteel ratios of at least 1 percent and axialloads not too far below the "balanced"value, the moment-curvature relation-ship tends to be of Type A in Fig. 10.The influence of the ¢ and [ factors isthen accurately represented by applyingthem as in Eqs. (7) and (11). For themembers with low steel ratios and axialloads presently under discussion, themoment-curvature relationships are of


Type B in Fig. 10, and the influence ofthe 0 and fad factors is not properly ac-counted for when they are inserted asshown in Eqs. (7) and (11).

In order to retain the traditional formof these equations, therefore, it is neces-sary to modify the quantities 71 and B stillfurther to the forms given in Eq. (11)above. These are to be applied with theACI Code value of the ¢ factor and withi3 from 0 to 0.5. To predict the actualcapacity of a cross section with 0 = 1and /3d = 0, the expression given in Ref.15, p. 68, should be used.

CURVm (1 +dal

Since it is intended to cover a widerange of cross sections and design pa-rameters, Eq. (11) is often very conser-vative. However, prestressed elements,particularly when precast, usually in-volve a good deal of repetition, sincethat is often the economic justificationfor their use.

This fact, coupled with the growingavailability and declining cost of com-puters, suggests that a rational analysiswould be appropriate for the develop-ment of the magnified moments. Man-ufacturers of standardized items couldeasily supply load capacity charts thatinclude slenderness effects, and spe-cialized items can be quite econom-ically analyzed if reliable programs are

Fig. 10. Application of strength reduction factor and load duration factor to moment-curvature relationships. Heavily reinforced sections (A) and lightly reinforced sections (B).

96

piles and building members is that theyare often statically determinate, or havereasonably well-defined boundary con-ditions, at least during the application ofthe dead loads (when they are oftenmost vulnerable). Thus, one can fre-quently apply the effective lengthmethod, whether or not sidesway is pre-vented.

In summary, the application of the P-Amethods to prestressed concrete shouldhe made with caution, in that stiffnessreduction may have to be considered. Inapplying magnification factors, whetherto account for both P-A moments anddeformation of the member betweenjoints, or only for the latter, the usualprocedures have been found to he lessreliable than they are for normal rein-forced concrete members. However, therepetition associated with precast mem-bers and their simpler boundary condi-tions favor the application of effectivelength methods with computer analysisof magnification effects.

The remainder of this paper will be

concerned with the rational analysis ofsecondary moments in prestressedmembers with known effective lengthsor boundary conditions.

OUTLINE OF RATIONALANALYSIS PROCEDURE

In this section, the essential steps inthe rational analysis of a beam columnwith known boundary conditions (Ref.3, p. 254) are defined.

The external moment at any point in abeam column (Fig. 11) is given by theprimary moment (.M,) arising from theend moments and lateral loads, plus thesecondary moment given by the axialload times the centerline displacement:

M(r) = M 0(x) + Pu

The internal moment of resistancedepends upon the axial load and thecurvature, namely, the P-M-0 relation-ship. This is a complex function of thematerial properties, including the in-

Q I Q2

P M^ M2 P

Pv Secondary Moment

Primary Moment M.due to M and Q

Fig. 11. Primary and secondary moments in a loaded beam column.

PCI JOURNAVMay-June 1985 97

Column

ABMax. End EccentricityC Column L 1 MaterialFailure)

Max End EccentricityColumn L 2 (lnsiobility Failure)

C

r8

Thrust Line .A

L,2

Lz

Fig. 12. Column deflection curves — Equal end eccentricities, single curvature.

elastic range up to failure, and themember cross section. The curvature, inturn, is a function of the centerline dis-placement and its derivatives, and, forsmall displacements, may be approxi-mated by u". Thus:

MW(x) =f= f(P, v", material, section) for

small displacements v

For equilibrium, therefore:

Mea" (x) — Mw (x)

or

Mo (x) + Pv = f (P, v", material,section) (12)

The solution of this differential equa-tion is usually required in one of thefollowing forms:

• The maximum moment in themember, given the applied lateraland axial loads.

• The maximum axial load the mem-ber can sustain, given the appliedlateral loads.

• The maximum eccentrically ap-

plied axial load the member cansustain.

In any case, the first step is the evalu-ation of the PM-4 relationship for thegiven cross section and the appropriatematerial properties (having regard forduration of loading, for example), givingthe function f (P, v", material, section).

It is recommended here that the solu-tion of the differential equation thenproceed by means of the numerical pro-cedure set forth in Ref. 3, p. 279, or Ref.18, p. 171. Starting with the prescribedboundary conditions at one end (usuallydisplacement and moment), the re-maining condition (usually slope) is as-sumed, and the solution curve is evalu-ated at successive nodes along themember by assuming a circular curva-ture within each short segment. Thestarting slope is adjusted and the proce-dure repeated until the prescribedboundary conditions at the far end aresatisfied.

When the maximum sustainable axialload is the sought-for quantity, it may bereached when material failure occurs inthe extreme fibers of the cross section;

98

0C-0cMz

C-C

CD

CDCD

Q E D C 8 A0 O E D, _-C^^B AJ J

COLUMN OF LENGTH L2

COLUMN OF LENGTH L + 'A^ A

B C- A C 0 E

MOMENT MOMENT

Fig. 13. Possible load paths in load-moment space leading to column deflection curves of Fig. 12.COCD

or it may he reached because at somevalue of the load the bending moment ata point within the length of the memberbegins to increase without bound. Inpractice, this means that:

• If there is no possibility of loadshedding, the member would sud-denly how and material failurewould occur, although the last cal-culable value of the moment couldhe well below the failure value.

• If there is a possibility of loadshedding, the axial load on the af.fected member would begin to de-crease, although deflections andinterior moments would continueto increase.

For this reason, the search for themaximum sustainable eccentrically ap-plied load, in the basic case of equal endeccentricities and single curvature, isbest conducted by the generation of setsof column deflection curves (Ref. 3, p.273) as shown in Fig. 12. Starting at themidheight of the column, with slopeequal to zero and moment equal to thatcausing material failure (or with theequivalent eccentricity from the thrustline), the column deflection curve isgenerated by integration of the differ-ential equation as indicated above (LineA, Fig. 12). The starting moment (or ec-centricity) at the midheight of the col-umn is then reduced in small steps, andnew column deflection curves (Lines B,C, D, E, Fig. 12) are generated.

These curves serve for columns of anylength, as shown in the figure; the curvegiving the greatest eccentricity at theend of the column is the governing one:either it corresponds to material failureat midheight (Column L,, Fig. 12) or itcorresponds to instability (Column L2,Fig. 12).

If the load were to be increased atconstant end eccentricity correspondingto A, B, C, D, or E in Fig. 12, the mid-height cross section would follow loadpaths such as those shown in Fig. 13.For the column of length L 2 , there issimply no equilibrium configuration

with loadP and eccentricity greater thanC. At eccentricity C, the practical col-umn bows sharply to failure at load P(unless the load is reduced as the col-umn deflects under it, to give the de-scending branch of the load curve).

At higher eccentricity, this unstablebehavior is exhibited before the load Pcan be reached. For the column oflength L I , on the other hand, the eccen-tricity can be increased to A, whereuponthe column cross section fails at Ioad P.

All the column deflection curves ofFig. 12 are at one value of the axial load,and they generate one point an a load-moment interaction curve for eachlength of column. Repeating the processfor different levels of axial load com-pletes the toad-moment curves, and al-lows the identification of maximum endeccentricities (or moments) for anygiven axial load, for the chosen columnlengths.

ESTABLISHING THEP-M-4 RELATIONSHIP

Turn now to a more detailed exam-ination of these steps, beginning withthe calculation of the moment-curvaturerelationship. Strictly speaking, this de-pends upon the exact loading history; ifthis were known, it would be possible toalter the computational procedure tosuit it. However, it is simplest to makethe computations as though the axialloads were applied and held constantwhile the moments or lateral loads areincreased to their total or failure values.

This is believed to cause no seriouserror. For example, as shown in Fig. 14,the points on a beam column loadedmonotonically at constant eccentricitywould follow Load Paths OA, OB, OC,OD; in the calculations, they are as-sumed to follow Paths PA, PB, PC, PD,but the final results should be unaf-fected by this. Had the actual load pathnot been monotonic (either of thedashed paths), the outcome would have

0a POJ

0 MOMENT

Fig. 14. Load paths in load-moment space.

been influenced, it is believed, in aminor way.

The following additional assumptionsare made:

1. Sections originally plane and nor-mal to the neutral surface remainso. There is no twisting or bendingout of the plane of loading.

2. The material laws are known andare not path dependent. The fibersin the beam column will obey thestress-strain laws indicated by thechosen uniaxial relationship.

Given particular values of the curva-ture 0 and the ordinate of the unstrainedfiber yo , the strain at any point is thendefined'" by:

E = 4(w — i0 (13)

Or, alternatively, if the curvature and

top fiber strain are given:

f=E1 – (Jl 00 (14)

The stress distribution follows fromthe material law, and the load and mo-ment are given by:"

P = A -dA = fYb &t body (15)J

M fA o-

ydA = J b hY rdy (16)

Since the stress-strain law is generallynot known in functional form, the inte-grations must be carried out numer-ically. This is easily done by dividingthe cross section into narrow horizontalstrips, determining the strain and hencethe stress at the centroid of each, andevaluating and summing the contribu-


1200

1000

800

Y

0 600Q0J

400

200

0-300 0 300 600 900 1200 1500 1800

MOMENT ( kips ins)

Fig. 15. Curvature contours in load- moment space.

tions of each strip to the load and mo-ment,

Thus, for selected values of the cur-vature and the ordinates of the un-strained fiber related sets of P-M-4 val-ues can be generated. In this way, con-tours of equal curvature can he devel-oped to cover the entire load-momentspace on an interaction diagram for thecross section, as shown in Fig. 15.

Finally, for a selected load path suchas PA on Fig. 15 (which corresponds tothe moment increasing monotonicallyfrom zero at constant load P) the mo-ment-curvature relationship can be ob-tained by the intersections of the loadpath with the curvature contours (Fig.16).

The next step is to solve the differ-ential equation of column equilibrium.

SOLUTION OF THEDIFFERENTIAL EQUATION

Once the moment-curvature relation-ship for the given load has been estab-lished, the differential equation ofequilibrium [Eq. (12)] is defined, albeitnumerically.

If the displacement v(x) and the slopev (x) are expanded in Taylor series, oneobtains,' 9 after truncation:

v(xa + AX) = t3rxo) + a(x0)Ax+ 1/24(x„)Ax 2(17)

a^(x0 + 1]x) = r^(xo) + k(x0)Ox (18)

wherea = v', the slopedA = v", assumed to be the curvature

102

•^r

a.

5002w02

0 0 0.0005 0.0010 00015 0.0020 0.0025CURVATURE (ins')

Fig. 16. Moment-curvature relationship for monotonic moment increase at constant loadobtained from intersection of load path with curvature contours on Fig. 15.

Note that the truncation of Eqs. (17)and (18) is equivalent to assuming con-stant or circular curvature between xaand xo + Ax0 ; therefore, it is actuallymore accurate to use 0 at xo + Ax/2 in-stead of at x, on the right hand sides ofEqs. (17) and (18).

Now if all the quantities of interest areknown or assumed at x0 , Eq. (12) can beused to determine 0 at the center of thenext segment: first evaluate the momentat that point:

M= Ma +PIv + aAX (19)

whereMo is at x0 + Ax/2

v and a are at xo

The curvature 0 is then obtained fromthe P-M- relationship for the givenload, and Eqs. (17) and (18) are used toevaluate all the necessary quantities at

the next station x„ + Ax.Consider the case of a beam column

with a given lateral load. Suppose thatthe maximum possible eccentricity at agiven load is to he determined in order tofind a point on the load/end momentinteraction curve in the presence of thestated lateral load. The eccentricity at x= L is K times that at x = 0. (Other prob-lems, such as the magnified moment fora given lateral load, axial load, and ec-centricity are simpler specializations ofthis case.)

The object is to find the starting ec-centricity (at x = 0) such that the mem-her is just on the point of failure, eitherby instability or by material failure. Thisis done by first seeking an eccentricityso large that the member does fail; it isthen reduced in steps of 10 percent untilthe member does not fail. This estab-lishes the range of values within whichthe answer lies. This range is then ex-

FOCI JOURNAL7May-June 1985 103

Set startingaccent.. $

IC0=0

425

Set target endaccent, = K 8

;onstruCt columndeflection curve

(call TEN)

Double start.accent. andrestart count

♦70

ie7ma9.fact. Yesrequired for

\

van a 50

7

No 160

Doescolumn

fail

No150

Yes

K'ss the Nost trial' 10

woes Yescolumn

fail

Print CDCar Max_Mom.as required.

Return.

Increaseotart.ecc. I No

by 0.9Reduceslopeby 20%

Set ICO=1

No20

IsIC0=1

(Startingeccent-beingreduced by

\? i sIc0=1\

Reduce starting No (Starting

accent_ byS/10 eccent.being80 educed by

8/100)

Yes

35

leit art inp Ye9ecc.=0

Yea PrintCDC

150 Return. No 85

250

Print 'Column cannot carrythis load'Return.

f 30

Reduce startingIccent. by 8/100

Fig. 17. Flow chart for control of boundary conditions in construction of column deflectioncurves (Subroutine MSIX).

104

C

Fig. 1a. Flow chart for construction of a column deflection curve (Subroutine TEN).See also Figs. 17 and 19.

M

C

0


rz

Mate*+al Failure TARGETw

IfAeCO)u TRIAL I Z(ABCE HI(L)J TRIAL 2va_i MATERIAL FAILURE^ 4}° THRUST LINE

Material Failurezww TRIAL TARGET

TRIALTRIAL 2TRIAL 3 2(ABCEHIMO)

a. J (ABCEHKP )0 (b} INSTABILITY FAILURE {ABCO)

THRUST LINE

Material Failurezwz 3(AB000IMO)

I TARG}

ud

TRIALTRIAL2

GET(ABCF)

TRIAL 3 TRIAL 5 ACCEPTED (ABCEHIMO)v^ TRIAL 4 I (ABCO)q (C) TRIAL 5 THRUST LINE

zw TRIAL 3{ASUD)w ^p

L 4SfASTYX)

4 TRIAL I TRIAL 3 ^tARGETJ TRIAL 2 4(ASTO)au_ TRIAL 5 ACCEPTED (ASVO)

° )IfASaO)

THRUST LINE

zw TARGET

TRIAL 3TRIAL 2 2(ASVO}

a TRIAL 3(ASwZIu_+ INSTABILITY FAILURE t (ASg4)° (e) THRUST LINE

Fig. 19. Examples of construction of column deflection curves,keyed to the flow chart of Fig. 18.

106

plored in steps of! percent of the initialeccentricity until, again, the first valuewhich does not cause failure is found.This is accepted as the answer.

The process is illustrated by the flowchart of Fig. 17, where the node num-bers correspond to statement numbersin the program. In each step of this pro-cedure, it was necessary to construct thecolumn deflection curve for the giveneccentricities, or to determine that themember failed. This will he discussedin the next paragraph. In the programreferred to, it is performed in a subrou-tine labelled TEN.

Now turn to the problem of con-structing the deflection curve for a col-umn with given lateral and axial loads,where the starting and ending eccentric-ities are fixed. The aim is to vary thestarting slope until the correct condi-tions are reached at the far end. Thismay not be possible because the bend-ing moment corresponding to failure isreached at some intermediate point, orbecause, as the starting slope is in-creased, the eccentricity (or deflectionfrom the thrust line) at the far end ap-proaches the target value and then re-cedes without reaching it.

This indicates instability failure: noequilibrium position exists having therequired eccentricities. The situation issimilar to that for the column of lengthL2 illustrated in Figs. 12 and 13.

This problem can he handled by theprocedure outlined in the flow chart ofFig. 18, which is seen to be fairly com-plicated. The easiest way to explain itmight be to follow through some spe-cific cases, which will be explained indetail below, in conjunction with theflow chart. Before entering into the de-tails, a very brief explanation of eachcase will be given.

The examples are illustrated in Fig.19, which shows the deflected shapes ofthe member. For each case, the axialload, lateral load, and the eccentricity atthe left hand end are held constant at thegiven values, and the slope at the left

hand end is varied until the target ec-centricity is reached (if possible) at theright hand end.

In Case I (Fig. 19a), the target eccen-tricity cannot he reached because mate-rial failure occurs within the span.

In Case II (Fig. 19b), the eccentricityat the right hand end reaches a maxi-mum short of the target, and then beginsto reduce with increasing starting slope(although interior deflections continueto increase). This indicates instabilityfailure.

In Case III (Fig. 19c), two configura-tions are found giving the correct targeteccentricity. The one with the smallerstarting slope (and smaller midspan de-flection) is the stable one. The other onerepresents the case where the memberhas snapped through and is on the de-scending load branch. Fig. 20 illustratesthis case.

In Case IV (Fig. 19d) the startingslope is increased in large steps forTrials 1, 2, and 3. Since Trial 3 over-shoots the target, the program returns toTrial 2 and increases the slope in smallsteps until the target is again reachedwith Trial 5, which is accepted. Case Vis similar to Case II, but the solution isreached by a different path through theflow chart.

These cases will now be described indetail, with an indication of the pathfollowed by the program logic. Thereader not interested in the programlogic may move to the end of this sec-tion.

Case I (see Fig. 19a)On the first attempt, with a trial start-

ing slope, material failure is reached atsome point in the column. The attemptis immediately abandoned, and thestarting slope is reduced in small stepsuntil the far end is reached without ma-terial failure. The end eccentricity isthen below the target value. This mayindicate that the target value cannot bereached because of material failure, or

PCI JOURNAIJMay -June 1985 107

the design may be on a descendingbranch of the load path; therefore, stillsmaller values of the starting slope mustbe tried.

The program is directed along PathABCO on Fig. 18, to make a second trialwith reduced starting slope. On the sec-ond trial, the end eccentricity decreases,Path ABCEHKL is followed, and the re-sult is recorded as material failure; thedesign is on a stable branch of the loadpath, since increased starting slopeleads to increased end eccentricity, butmaterial failure intervenes before thetarget end eccentricity can he reached.

Case II (see Fig. 19b)Again, after reducing the slope to

avoid material failure, the end eccen-tricity is below the target value. Theprogram is directed along ABCO as inCase 1, but on the second trial with re-duced starting slope, the end eccentric-ity increases (but does not reach the tar-get), indicating that the design is on anunstable load path; the program is di-rected along ABCEHIMO for a thirdtrial at a further reduced starting slope.This time the end eccentricity also re-duces, indicating that the design hasmoved back along the stable path,reaching the peak before the target, andinstability failure (Path ABCEHKP) oc-curs.

Case Ill (see Fig. 19c)The first two trials are exactly as in

Case II (Paths ABCO and ABCEHIMO).On the third trial, the end eccentricityagain increases and exceeds the targetvalue. The unstable path has to be fol-lowed up to the peak, and then down thestable branch until the target is reached.The program is, therefore, directedalong Path ABCDGIMO for another trialat a still further reduced starting slope.When the end eccentricity begins to de-crease, the process is redirected alongABCDCJNO, until it falls below the

target value again, when ABCF is foI-lowed, and the result is accepted.

Case IV (see Fig. 19d)On the first trial, material failure is not

encountered, but the far end eccentric-ity is below the target value (PathASQO). The starting slope is increasedby a large amount and a second trial ismade; the far end eccentricity increases,but is still below the target (ASVO). Thestarting slope is gradually increaseduntil the target is exceeded (ASUO),when one large decrease in the startingslope is made, whereafter it is increasedagain in small steps. Subsequent trialsfollow Path ASTO until the target isagain exceeded, when Path ASTYX isfollowed and the result is accepted ascorrect.

Case V (see Fig. 19e)The first two trials are as for Case IV

(Paths ASQO and ASVO) but eventually,before the target is reached the end ec-centricity begins to decrease. PathASWZ is followed to indicate instabilityfailure.

Case VIThe first trial is as for Case IV (Path

ASQO), but on the second trial with alarger starting slope, the end eccentric-ity decreases. This indicates that the de-sign is on an unstable load path, andPath ASBCO is followed and the processof Case II or Case III is repeated to workback up this unstable load path to thepeak and beyond, if necessary.

Other cases can be traced through Fig.18 in a similar way.

In the special but most common caseof columns loaded with equal end ec-centricities, causing single curvaturebut no lateral load, an entirely differentprocedure is carried out as described inthe previous section on the solution of

108

UNSTABLEEQUILIBRIUM

STABLE CONFIGURATIONEQUILIBRIUMCONFIGURATION

PP —►

ISTABLE EQUILIBRIUMPOSITION ON ASCENDPATH

UNSTABLE EQUILIBRIUMPOSITION ON OESCENCINGPATH

LOAD PATH AT END X = L

LOAD PATH AT END X = D

LOAD PATH AT MIDSPAN POINT

MOMENT

Fig. 20. Load paths at ends and center of eccentrically loaded column,leading to stable and unstable equilibrium configurations.

Qd POJ

the differential equations, as shown inFig. 12. This set of curves allows one todetermine the magnification factor for agiven eccentricity, or the maximum pos-sible eccentricity (whether governed byinstability or material failure) for alllengths ofthe column for the given load.

APPLICATION OFSTRENGTH REDUCTION ANDLOAD DURATION FACTORS

In the ACI Code column design, thestrength reduction factor is applied attwo points in the calculation. It appearsfirst in the moment magnification pro-

cedure, where it accounts for the influ-ence of analytical and construction inac-curacies on the Euler load and hence onthe moment magnification itself. This isaccomplished in the ACI Code methodby, in effect, reducing the rigidity EI ofthe cross section, which is the slope ofthe moment-curvature relationship.

When the moment-curvature relation-ship has a "yield plateau," the momentcapacity should presumably also be re-duced as shown in Fig. 10. It would besubject to variations from the samecauses as the rigidity, and its value isnow relevant to the moment magnifica-tion. The program is written to performthe modifications to the moment-cur-


vature relationship indicated in Fig. 10,The value of may be entered, or theACI Code values (with the transitionbelowP = 0.1 99 may be requested.

The strength reduction factor appearsa second time when the magnified mo-ment is compared with the nominal ca-pacity of the member without magnifi-cation; the latter is reduced by the ca-pacity reduction factor. This operation isperformed by the program; the value of0 is entered independently of that usedin the magnification procedure. A dif-ferent value may he used if desired, oragain, the ACI Code values may be se-Iected.

The load duration factor fad is also ap-plied in the ACI Code magnificationprocedure in such a way as to reduce therigidity of the cross section by the in-verse of (1 + /3d ). This is again achievedin the program by stretching the curva-ture axis of the moment-curvature re-lationship as shown in Fig. 10.

REVIEW OF ASSUMPTIONSThe assumptions made in the forego-

ing analysis have been stated as theywere made, but they will be collectedtogether here for reference:

1. If sway is permitted, the P-A mo-ments due to sway have been estimatedand the effective length factor is beingset at 1 for the effects of column center-line deflection; or the sway-permittedeffective length factor, greater than 1,has been estimated to account for bothP- moment and centerline deflection.

2. If sway is not permitted, thesway-prevented effective length factorhas been estimated or conservatively setat unity.

3. There is no torsional buckling ofthe section.

4. There is no local buckling of partsof the section, such as the flanges.

5. Deflections are small enough forv"to represent curvature, so that Ely " = M.

6. Sections originally plane and nor-

mal to the neutral surface remain so.7. The stress-strain laws of the mate-

rials are known and are the same inbending as they are in uniaxial stress.

8. The stress-strain laws reflect theduration of loading or other time effects;or the load duration factor /3, correctlymodels the creep.

9. There is perfect bond betweensteel and concrete.

10. Material failure of the cross sec-tion is due to the concrete reachingsome limiting strain.

11. The result is not significantly de-pendent upon the precise load path inthe load-moment space.

Note that Assumption 9 implies thatthe prestressing tendons and other re-inforcement are fully developed at thepoints of high moment. The reduced ca-pacity of the section in the developmentregion of pretensioned members shouldhe considered separately.

CONCLUSIONThe stability analysis of frame struc-

tures has been discussed. Generally, therecommended procedure involves asecond order elastic analysis (with stiff-ness modified to account for cracking) todetermine the P-A effects due to jointtranslations, followed by magnificationof moments based on braced effectivelengths to account for centerline deflec-tions.

Prestressed concrete members aremore subject to instability rather than tomaterial failure than are conventionallyreinforced concrete members; loss ofstiffness in column members maytherefore be of greater significance inthe secondary analysis for P-A effects;and the semi-empirical moment magni-fication procedures need modification,and are in any case less reliable.

However, the large scale productionand standardization associated with pre-cast prestressed members suggests theuse of computer programs for the me-

merit magnification solution; in fact, ra-tional analysis is required by codes ofpractice for the high slenderness ratiosoften encountered in these members.The preparation of a typical program toperform this task has been discussed,and is described in detail in AppendixB.

The program may be used for the ra-tional analysis of slenderness effects inprestressed and/or reinforced concretecolumns, wall panels, or piles. It is use-ful for developing the load-moment re-lationships necessary in the design pro-cess, with or without slenderness; thesolution of particular cases or the prepa-ration of design charts for standardizeditems may be quite easily and inexpen-sively carried out.

The difficulty of establishing the ef-fective length for the sway case or, al-ternatively, of carrying out the second-

order analysis for P-ti effects, remains.Further work is necessary to study theloss of stiffness in prestressed concretemembers in the presence of axial load inthe inelastic range, and to develop reli-able P-A analyses when individualmembers are subject to stability ratherthan to material failures.

ACKNOWLEDGMENTSThis work was funded by the Natural

Sciences and Engineering ResearchCouncil of Canada. Computing facilitieswere made available during the author'ssabbatical by the University of theWitwatersrand, Johannesburg, SouthAfrica. Encouragement and help havebeen received from the PCI PrestressedConcrete Columns Committee, andfrom other members of the PCI.

NOTE: Discussion of this paper is invited. Please submityour comments to PCI Headquarters by January 1, 1986.


REFERENCES1. ACI Committee 318, "Building Code

Requirements for Reinforced Concrete(ACI 318-83)," American Concrete In-stitute, Detroit, Michigan, 1983.

2. MacGregor, J. G., "Stability of Multi-Story Concrete Buildings," ASCE-IABSE Conference on Tall Buildings,Bethlehem, Pennsylvania, SoA-Rep 23-3Planning and Design of Tall Buildings,V. III, 1972, pp. 517-536.

3. Galambos, T. V., Structural Membersand Frames, Prentice-Hall, Inc., Engle-wood Cliffs, N.J., 1968.

4. Wood, B. R., Beaulieu, D., and Adams,P. F., "Column Design by P DeltaMethod,"Journal of the Structural Divi-sion, ASCE, V. 102, No. ST2, Proceed-ings Paper 11936, February 1976, pp.411-427.

5. Lai, S.-M. A., and MacGregor, J. G.,"Geometric Non-Linearities in On-braced Multi-Story Frames," journal ofthe Structural Division, ASCE, V. 109,No. ST11, November 1983, pp. 2528-2545,

6. Lai, S.-M. A., MacGregor, J. C., andHellesland, J., "Geometric Non-Lin-earities in Nonsway Frames,"Jou rnal ofthe Structural Division, ASCE, V. 109,No. ST12, December 1983, pp. 2770-85.

7. Kavanagh, T. C., "Effective Length ofFramed Columns," Transactions, ASCE,V. 127 (1962), Part I1. pp. 81-101.

8. Bleich, F., Buckling Strength of MetalStructures, Engineering Society Mono-graph, McGraw-Hill, New York, N.Y.,1952.

9. Johnston, B. G. (Editor), Guide to Stabil-ity Design Criteria for Metal Structures,3rd Edition, Column Research Council,John Wiley and Sons, Inc., New York,N.Y., 1976.

10. ACI Committee 318, "Commentary onBuilding Code Requirements for Rein-forced Concrete (ACI 318-83)," Ameri-can Concrete Institute, Detroit, Michi-gan, 1983.

11. Chu, K. H., and Chow, H. L., "EffectiveColumn Length in UnsymmetricalFrames," Publication, International As-sociation of Bridge and Structural Engi-neering (IABSE), V. 29-1, 1969.

12. MacGregor, J. G., and Hage, S. E., "Sta-bility Analysis and Design of ConcreteFrames," Journal of the Structural Divi-sion, ASCE, V. 103, No, ST10, Proceed-ings Paper 13280, October 1977, pp.1953-1970.

13. Nixon, D., Beaulieu, D., and Adams,P. F., "Simplified Second-Order FrameAnalysis," Canadian Journal of CivilEngineering, V. 2, No, 4, December1975, pp. 602-605.

14. Rosenhlueth, E., "Slenderness Effects inBuildings,"Journal of the Structural Di-vision, ASCE, V. 91, No. STI, February1965, pp. 229-252.

15. Nathan, N. D., "Slenderness of Pre-stressed Concrete Columns," PCIJOURNAL, V. 28, No. 2, March-April1983, pp. 50-77.

16. Gere, J. M., and Weaver, W., Analysis ofFramed Structures, Van Nostrand, Co.Inc., Princeton, New Jersey, 1965, pp.428-431.

17. Ghali, A., and Neville, A. M., StructuralAnalysis, Chapman and Hall, London,1977, Chapter 15.

18. Timoshenko, S. P., and Gere, J. M.,Theory of Elastic Stability, Second Edi-tion, McGraw-Hill, New York, N.Y.,1961.

19. Nathan, N. D., "Slenderness of Pre-stressed Concrete Beam-Columns," PCIJOURNAL, V. 17, No. 6, November-De-cember 1972, pp. 45-57.

20. Salmons, J. R., and McLaughlin, D. G.,"Design Charts for Proportioning Rec-tangular Prestressed Concrete Col-umns," PCI JOURNAL, V. 27, No. 1,January-February 1982, pp. 120-143.

21. Wang, P. T., Shah, S. P., and Naaman,A. E., "Stress-Strain Curves of Normaland Lightweight Concrete in Compres-sion," PCI JOURNAL, V. 75, No. 11,November 1978, pp. 603-611.

22. Busch, H., "Researches Towards a Gen-eral Flexural Theory for Structural Con-crete," ACI Journal, V. 32, No. 1, July1960, pp. 1-28.

23. Sheppard, David A,, "Seismic Design ofPrestressed Concrete Pilings," PCIJOURNAL, V.28, No. 2, March-April1983, pp. 29-49.

112

APPENDIX A - NOTATIONA = cross section area yb = ordinate of bottom fiberh = width of cross section at ordinate yi = ordinate of top fiber

Y a = slope of column deflection curveC. = modifier to magnification factor a = angle between imaginary brac-

for unequal end eccentricities ing and horizontalE = Young's modulus /3 – ratio of maximum factored deadE, = E of concrete load moment to maximum Eac-h = story height tored total load moment (alwaysH = lateral force positive); the load duration fac-1, = moment of inertia of concrete tor

cross section y = flexibility factor[see Eq. (3)]k = effective length factor S = magnification factorK = lateral stiffness of frame 8y = magnification factor (bracedL = unbraced length case)M = moment 8, = magnification factor (sway case)M. = primary moment 8 = first trial value of starting eccen-

= larger factored end moment tricity(braced case) A = deflection of joint normal to axis

b12a = larger factored end moment of beam column(sway case) Dl = A calculated by first order analy-

fi, = magnified column moment sisMI. = external moment ^ = A including secondary effectsM,,,, = internal moment e = strainP = axial load E, = strain in top fibersPa = maximum pure axial load 7) = factor used in calculation of P,P, = critical load [Eq. (11)]P5 = Euler load B = factor used in calculation of P,P. = ultimate load [Eq. (11)]r = radius of gyration K = ratio of eccentricity at bottom toV = transverse deflection of member eccentricity at top

centerline A = factor used in calculation of P,x = distance along member center- [Eq. (11)1

line o• = stressy = ordinate of point in cross section th = curvatureyo = ordinate of unstrained fiber (h = strength reduction factor

PCI JOURNAUMay-June 1985 113

APPENDIX B - A COMPUTER PROGRAM FORRATIONAL ANALYSIS OF BEAM COLUMNS

What follows is a detailed descriptionof a computer program to perform theanalysis described in the paper. It waswritten by an unsophisticated pro-grammer with the intention of making iteasy to understand and to modify ratherthan to make it efficient from a compu-tational viewpoint; running costs aregenerally not a large factor. Because itwas intended to be used for researchpurposes, the program as presented hereis capable of handling a variety of casesand all the computed curves are gener-ated from an extremely dense array ofpoints.

For practical purposes, the scopecould be more limited and the arrayscould be reduced in size, It is hoped thatthe processes of familiarization, adapta-tion, and modification have been madeas easy as possible. The assumptionsand theories used are as set forth in thepaper.

Equations referred to by number inthis Appendix are those in the paper. Alisting of the FORTRAN source programis available at cost of reproduction fromPCI Headquarters.

CapabilitiesThe problem to be solved is illus-

trated in Fig. BI. The cross section mayhave any polygonal shape (or any curvedshape which can be adequately approx-imated by a polygon) with any arrange-ment of prestressing tendons and/ornonprestressed reinforcing bars.

The axial load P can be applied at anyeccentricity e at one end, and at any ec-centricity Ke at the other. Note that K

may he positive, negative, or zero. Ini-tial crookedness in the shape of a sinecurve is provided for: the amplitudemust be supplied by the user. Camberdue to prestress is automatically ac-counted for. The primary bending mo-

ment due to lateral loads may have anyshape; it is entered by means of its tenthpoint values.

For given values of F, e, K, and pri-mary moment, the maximum value ofthe magnified moment within the spancan be computed; or, while the other pa-rameters are held constant, the eccen-tricity e can be increased until failure.

Failure may be due to material failureat some interior section where the mag-nified moment reaches the short columncapacity, as for the load P, applied at ec-centricity e,, or due to instability, as forthe load Pz applied at eccentricity e2(Fig. BI). The output will indicatewhich has occurred.

By entering several loads and gener-ating several points such as A,, A E , onecan develop the line through them,which is an interaction curve of load Pversus end moment Pe for a given slen-derness ratio.

The program can also be used merelyto compute the short column interactioncurve for complex cross sections, or themoment-curvature relationships atgiven loads. In this mode it generatesdesign information such as that given bySalmons and McLaughlin, 20 for anyshape of member with any arrangementof reinforcement and prestressing ten-dons. Length effects can be included byrunning the whole program.

General ProcedureThe behavior of the section as the

strain is uniformly increased to failure isfirst established. This is the load-mo-ment relationship at zero curvature; itmay lie along the zero moment load axis,but only for symmetrically reinforcedsections (Fig. B2). In fact, it is the mo-ment necessary to remove the camberdue to prestress in the presence of thegiven axial load.

114

.-e

L//)o

a0J

Pz

P,--f - e

PRIMARY BENDING MOMENTMOMENT

P

Fig. B1. Definition of the problem to be solved.

Then the load-moment curve for theshort column corresponding to the max-imum concrete strain is calculated, to-gether with the associated curvatures(Fig. B2). These values are used to de-termine the range of curvatures andmoments which will be encounteredwith the particular section under inves-tigation.

A set of intermediate curvatures isthen selected and contours of equal cur-vature are calculated, covering the en-tire feasible area of the load-momentdiagram. For each of the desired loads,values of moment and curvature are ob-tained from the intersection of the loadcoordinate with these contours, estab-lishing the moment-curvature relation-ship for that load (Figs. 15, 16 of themain part of the paper).

This moment-curvature relationshipis then used to obtain the column de-flection curve for the given load and

boundary conditions by a numericalprocedure.

Each of these steps is performed in aseparate subroutine named MONE,MTWO, MTHREE, etc., which arecalled by a short control program. Vari-ous arithmetic procedures which are re-peatedly used by these main subrou-tines are executed in another set of sub-routines named ONE, TWO, etc. Theresults are printed out by a set of rou-tines named WRITE1, WRITE2, etc.,any of which can be suppressed if onlyspecified data are required. The inputdata, from Subroutine INPUT, areshared by the main subroutines in com-mon statements, but the computed vaI-ues are generally transferred in thecalling statements for the subroutines.All input data referring to the materiallaws are grouped in a separate commonstatement.

Detailed descriptions will now be


10001

Wfn_ ÎY 7..-. III^ }I

o UIJof

N

0.00In

$TRAIN STRESS

may,LJ

W^â

ao

^¢m = 0.007to

000047-00008 = 0.0000

/ \0= o.00ie8(up to such 30 points)

0

4600

MOMENT C hips ins)

Fig. B2. Short column load-moment relationship with curvatures on boundaryof load-moment space, for the section of Fig. B3.

given of each of these routines, illus-trated by means of the actual outputfrom the section of Fig, B3.

MAIN ProgramThe principal variable names are

listed in comment statements. The sub-routines to carry out the steps describedabove are called in turn, each followedby the appropriate output routine. Pro-vision is made for stopping after gener-ation of the short column data, or afterthe moment-curvature relationships.Any of the output segments may also hesuppressed.

INPUT

Details may be found in the program

documentation; this section will providea general description of the optionsopen to the user.

The coordinates of the concrete crosssection are entered with respect to a setof axes set up so that the neutral axis willhe parallel to the x-axis. The originshould be set outside the section, so thatthe entire section lies in the first (posi-tive) quadrant. Any polygonal shapemay he entered, with up to 20 sides.Thus a circular section can be approxi-mated by a twenty-sided polygon, if re-quired.

Data for material laws will be dis-cussed below. Arrays are dimensionedfor up to 20 nonprestressed bars whichmay have different areas and coordi-nates. (Since plane sections are assumedto remain plane and the neutral axis is to

116

60

rn

O oI-

-6000 0001 0.002 0.003 -0.004

STRAIN(N4 TENSION)

270

N

REINF. inSTEEL

o0 0.004 0

STRAIN

e . - 0"

PRESTRESSINGSTRAND

0.02

0.04STRAIN

k N

5

4

3

v2

5.75°

3-#t3

4-112°0 strands

3.75.,

Fig. B3. Illustrative example: double tee used as loadbearing wall unit.

be parallel to the x-axis, the x-coordinateof bars is actually immaterial, and barswith a common y-coordinate might wellbe lumped together.) At present, thesebars must all have the same materiallaw.

Up to 20 prestressing tendons can beincluded, with different areas, cen-troids, and prestress forces. The pre-stress can be entered in the form of astress, a force, or a strain, correspondingto zero strain in the surrounding con-crete, i.e., the value of stress, strain, orforce must be that existing after alllosses except elastic shortening. Again,x-coordinates are immaterial and ten-dons with equal q-coordinates can belumped; and, again, all tendons musthave the same material law.

Up to 20 axial loads may be investi-gated; they may be entered as absolutevalues or as ratios of the maximum axialload. Up to 20 column lengths can bestudied. If there is lateral loading, the

primary bending moment must be en-tered for each length by giving the val-ues at the tenth points and the two ends.The amplitude of the initial crookedness(assumed to be a sine curve) must beentered for each length, if desired.

MONEThis is a subroutine to determine the

area and centroid of the concrete sec-tion, and the maximum load. The load-moment relationship for zero curvatureis also computed; this will be a straightline coinciding with the load axis on theIoad-moment diagram for symmetricsections. But, if the reinforcement is notsymmetrically placed with respect to theplastic centroid,* it may be a curvingline. The data generated by this subrou-

"The centmid of forces when a uniform strain corre-sponding to the peak concrete stress is applied tothe entire cross section.


tine are not necessary for the final result,but they are used in the program for se-lecting other parameters.

The area and centroid of the concreteare calculated in Subroutine FIVE, andthe prestressing is converted to strain ifit was given in some other form. Thestrain in the concrete is then uniformlyincreased in 30 steps from zero to fail-ure, while compatibility is maintainedwith the steel elements. The load andmoment corresponding to each step arecalculated, defining 30 points on theload-moment-zero curvature relation-ship (see Fig. B2).

MTWO

This is a subroutine to determine theload-moment-curvature interactioncurve for the nominal load capacity ofthe short column.

The depth of the section is calculatedand divided into 30 parts. The neutralaxis is then placed at each of these posi-tions in turn, with the extreme fiberstrain of the concrete set to equal thevalue defined as failure. The total forceand moment resulting from stresses onthe concrete section are calculated inSubroutine THREE, using Eqs. (15) and(16).

The compatible strains in the steelelements are computed, and the corre-sponding stresses, from Subroutine SIX,used to obtain the forces which are in-cluded in the total force and moment.The curvature is calculated by dividingthe failure strain in the concrete by theneutral axis depth [Eq. (13)]. See Fig.B2.

MTHREEThis is a subroutine to calculate con-

tours of equal curvature covering theload-moment diagram.

The curvature cam at zero load is ex-tracted from the data generated byMTWO; this is the maximum curvature.The value of curvature 0' at the bal-

anced load and moment is also deter-mined. If 0„, > 2 0', the interval be-tween 0' and 4 m is divided into 10 partsand the interval between 0 and 0' is di-vided into 20 parts. If ¢„, -- 2 ', theinterval from 0 to 0,,, is divided into 30parts. These divisions define the cur-vature values for which contours will bedeveloped; the procedure is found togive a reasonable coverage of the load-moment diagram (see Fig. B2). Forpractical purposes, the number of con-tours could be greatly reduced.

A contour of equal curvature is thengenerated for each of these curvatures.The extreme fiber strain is set equal tothe failure value for the concrete, andthe neutral axis depth is set to give thedesired curvature (curvature = extremefiber strain/neutral axis depth). Subse-quently, the neutral axis depth is de-creased in steps of 'loo of the memberdepth, while the curvature is kept con-stant. For each neutral axis depth theforce and moment arising from the con-crete stresses are calculated by Subrou-tine THREE.

The compatible strains and hence thestresses in the steel elements are thencomputed, and their contributions areadded to the load and moment values.Thus, up to 30 pairs of values of load andmoment are computed to define each of30 contours of equal curvature. See Fig.15 of the paper. The number of pairs ofvalues could also he reduced for practi-cal purposes.

MFOURThis is a subroutine to develop mo-

ment-curvature relationships for thegiven loads. For each of the curvaturecontours obtained in MTHREE, a valueof moment is interpolated (by means ofSubroutine SEVEN) at each of the givenloads. On the zero curvature relation-ship, the value of moment at zero cur-vature for the given load is interpolated,and the maximum moment and the cor-responding curvature are obtained by

118

Fig. B4. Interpolation for curvature as a function of moment.

interpolation on the short column re-lationships. Thus up to 32 pairs of valuesof moment and curvature are obtainedfor each load, defining the moment-cur-vature relationship for that load.

Generally, the moment increasesmonotonically with curvature up to fail-ure, or up to some maximum, whereafterthere is a steadily descending branch tomaterial failure (see Fig. B4). When in-terpolation is made to find the curvaturefor a given moment, the interpolationroutines automatically reject the de-scending branch of the curve; interpo-

lation is made only on the ascendingbranch and the maximum moment rep-resents failure.

Occasionally, for example with I-shaped sections where tension is al-lowed in the concrete and there is verylittle reinforcement, there may be a localmaximum moment before the overallmaximum is reached. This also is shownon Fig. B4. In that case, to avoid prob-lems with the interpolation routine, thecurve is smoothed by removal of thefirst, local maximum, as shown in Fig.B4. This is slightly conservative, but is


believed to deviate very Iittle from thetrue final result.

If there is to be reverse curvature (i.e.,if the end eccentricities are such as toproduce double curvature) the mo-ment-curvature relationship must beextended into the negative range. If thecross section is symmetric about an x-axis, the extended portion of the curve ismerely the mirror image of the portionalready calculated. Subroutine NINE isused to carry out this operation.

If the cross section is not symmetricabout an x-axis, Subroutine EIGHT isused to reverse all the coordinates (i.e.,to turn the section upside down, as itwere) and the complete process, in-volving Routines MTWO, MTHREE,and MFOUR, is repeated. SubroutineEIGHT is again used to merge the posi-tive and negative moment-curvaturerelationships.

MFIVE

This routine deals with the specialcase where there is no lateral load butequal end eccentricities, and the maxi-mum eccentricity is required at givenload levels.

The procedure consists of starting atmidheight of the column, with the max-imum possible moment at that point.This maximum moment is obtained fromthe high point on the moment-curvaturerelationship for the given load. At themidheight point, the starting slope iszero, and the deflection (measured fromthe thrust line) is given by the startingmoment divided by the load.

The column deflection curve is nowconstructed using Eqs. (17), (18), and(19). This process is continued, as de-scribed in the paper, until a completequarter wave of the column deflectioncurve has been generated, or until thehalf lengths of all the columns to be in-vestigated have been exceeded. Theend deflections of all the columns arenoted. (See Fig. 12 in the paper.)

Returning to midheight, the starting

moment and therefore deflection are re-duced by a small amount chosen by theuser (the default value is 0.05 times themaximum moment) and another columndeflection curve is generated. The pro-cess is repeated until the end deflectionis decreasing for all lengths; the maxi-mum end deflection is the requiredvalue; the end eccentricity to cause fail-ure. If it was obtained from the first col-umn deflection curve, which startedwith the maximum moment at mid-height, it represents material failure; if itwas obtained from a subsequent curvewith lower midheight moment, it repre-sents instability failure.

MSIX

This routine develops the column de-flection curves for the more general casein which the end eccentricities are notequal and there is applied lateral load.In contrast to MFIVE, each length casemust he handled separately, since thecolumn deflection curves are con-structed from the end rather than fromthe midheight, as described in the paperunder "solution of the differentialequation."

The segment length is first adjusted ifnecessary so that there will he an exactnumber of segments. Then the startingslope and deflection (or eccentricity) areselected. The problem is to determinethe order of magnitude of slope and de-flection appropriate to the column understudy, since these may vary widely fromcase to case.

The approximate initial slope of themoment-curvature relationship is de-termined by dividing the fifth positivevalue of the moment by the corre-sponding curvature. This gives the ini-tial EI value. Now, the slope at the endof a simply supported beam carrying auniform load is ML2I(3EI), where M isthe maximum moment.

Therefore, the order of magnitude ofthe slope in a given column is expectedto be:

120

aMCL2

5EI

whereM, = maximum moment capacity of

cross sectionEI = initial value of El as determined

aboveThe first starting value of the slope is

set equal to a.The initial eccentricity is set equal to:

& = M,–M"P

where Mp is the maximum value of theprimary moment.

Increments of the starting eccentricityare first set to 8/10, and then to 8/100 forfine adjustments. If the program is beingused to find the magnification factor fora given eccentricity, of course, the initialvalue is set to that quantity and no in-crementation is required.

The procedure to be followed fromthis point requires the establishment ofa starting eccentricity which will lead tofailure. Thus, with the initial value set toS as defined above, a column deflectioncurve is calculated by the SubroutineTEN.

If the column does not fail, subse-quent trials are made with 26, 46, etc.,until failure is achieved. The startingvalue is then reduced in steps of 8110until failure does not occur. At that stagethe last step is retraced, with the initialeccentricity now being reduced in stepsof E/100, until failure again does notoccur; this last trial is taken to give theconfiguration of the column which is"just safe."

The starting slope is set to a, as de-fined above, when TEN is called for thefirst time. After that, it is left equal to thevalue returned from TEN, except whenthe last step in eccentricity is about to beretraced, when it is reduced by 20 per-cent.

Fig. 17 of the paper shows the partialflow chart for MSIX.

ONEIn the numerical integration of

stresses over the concrete area, the latteris divided into narrow strips parallel tothe neutral axis, and this subroutine isused to define the extent of such a strip.(A constant stress will then be assumedover each strip.)

Given the coordinates of the apices ofa polygon, and the y-coordinates defin-ing the upper and lower boundaries ofthe intersecting strip, the subroutinewill determine the coordinates of thesub-polygon which lies within the strip.(If the whole of the original area lieswithin the strip, the original corner co-ordinates will he returned.)

The end points of each side of thegiven polygon are examined in turn; it isestablished whether the side being ex-amined lies wholly outside or wholly in-side the intersecting strip, or whether itis entering or leaving that space. Sub-routine TWO is then used to evaluatethe coordinates of the point where a lineactually crosses one of the strip bound-aries. The comers of the new sub-poly-gon consist of these points together withthe original ones which lay within thestrip.

TWOThis short subroutine determines the

coordinates of the point at which a givenline crosses a line of constant y.

THREEThis subroutine is used to integrate

the concrete stresses to give the totalforce and moment. The coordinates ofthe corners of the concrete area and they-coordinate of the neutral axis aregiven. The strain distribution is implic-itly given, either in the form of the ex-treme fiber strain or as the curvature(equal to the extreme fiber strain di-vided by the neutral axis depth). Theareas above and below the neutral axisare each divided into 20 strips of equal

PCI JOURNAL1May-June 1985 121

depth, parallel to the neutral axis. (If thematerial law for the concrete is from 1 to4, it is assumed that tension is not al-lowed and the area below the neutralaxis is not considered.)

Each of the 40 strips is considered inturn; the coordinates of the comers arefound using ONE, and the area andcentroid of each strip are found usingFIVE, The material law is then used inSIX to determine the stress at the cen-troid and the force in the strip (assumingit is uniformly stressed) and its momentabout Y = 0 are calculated.

FOUR

This is a short subroutine to deter-mine the biggest and smallest of a list ofnumbers. It is used, for example, infinding the overall depth of a cross sec-tion.

FIVE

This is a subroutine to find the areaand centroid of a polygon with n apicesdefined by coordinates x and y. They aregiven by:

Area = 1 1 y j (xc+^ – x_1)1=1

MomentMoment of area (about y = 0)

lk fY3 (x1 +, – Z) (J 1 + yt+t Ya + y )

The apices are taken in cyclic order ineither direction.

SIX

This redirects the program to the ap-propriate subroutine for the assumedmaterial law, which is defined by theparameter MATLAW. The routinesMATT, MAT2, etc., then determine thestress corresponding to the given strainor vice versa. (The sought-for quantity isset to zero in the calling program.)

MAT! interpolates in a curve defined

by pairs of points entered with theinput. It is assumed that there is no ten-sile stress. The routine is intended torepresent concrete.

MAT2 represents the stress-strain lawof concrete in the functional form:

_ 2 1 E/Ea j fC I + (/)2

wherewheref,' = cylinder strengthE = strainEo = strain at peak stressAgain, it is assumed that there is no

tensile stress.MATS contains stored pairs of stress-

strain values representing points on anexperimental curve taken from the liter-ature. El It is entered with the peak stressand the corresponding strain for the casebeing considered, and the stored valuesof stress and strain are scaled up ordown to pass through this point. Thecurrent stress or strain is found by inter-polation. This curve seems more appro-priate for very high concrete strengths(about 7 ksi).

MAT4 gives the stress-strain law inthe functional form proposed in Ref. 21,

MATS is similar to MATT. It is in-tended for steel, and tension may hepermitted.

MATE is a bilinear stress-strain curvefor the elastic/perfectly plastic case. It isintended for mild steel, and the Young'smodulus and yield stress must be en-tered with the data.

MATT is similar to MAT5. It is in-tended for prestressing steel.

MATS contains stored pairs of stress-strain values for a typical seven-wirestrand with f,. = 270 ksi. If it is enteredwith a value off,,, other than 270, all thestress values are scaled up or down ap-propriately.

MAT9 uses the bilinear stress-strainlaw defined in the British Code of Prac-tice CP110 for steel reinforcement. It

122

must be entered with the yield stress inM Pa.

MATI.O is similar to MATT, with ten-sion allowed. It is intended for use withconcrete where MAT5 and MATT areused for mild and prestressed steel.

MATH is similar to MATS, using acurve taken from Ref. 22. It appears tobe more appropriate for the usual con-crete strengths, up to 5 or 6 ksi.

Other stress-strain laws can easily beentered as MATH, MAT12, etc.

SEVENThis subroutine contains a linear in-

terpolation procedure which is used re-peatedly throughout the computationsfor finding values of stress in terms ofstrain, curvature in terms of moment,etc. Cubic interpolation has been usedfor research purposes, but the differencein the end result appears to be neglible.

Several points defining the relation-ship between x and y are given. Theroutine first investigates the relation-ship to determine the range over whichit is single valued in y, i.e., the numberof points for which x is increasing; theremaining points are discarded (see Fig.B4).

The position of the point i for which yis desired is determined. y is calculatedby linear interpolation. If z lies outsidethe range of x, the value of is linearlyextrapolated from the last two points anda message is printed to that effect. If $has a value that cannot be reached by x,such as C on Fig. B4, the value of y cor-responding to the nearest valid x value,such as that at B, is given, and an appro-priate message is printed. This may in-dicate a serious error, or it may only bethat the round off error has led to thevalue of i (NN) being slightly overes-timated.

EIGHTThis subroutine is used when double

curvature is present in unsymmetrical

sections. The previously generated pos-itive branch of the moment-curvaturerelationship is stored in a temporary lo-cation and the section is then inverted.All the relevant coordinates are trans-formed to read from a new origin on theopposite side of the neutral axis. Theprogram then returns to calculate a newmoment-curvature relation, which willbe the negative branch.

NINEThis subroutine merges the positive

and negative branches of the moment-curvature relationships when doublecurvature is present. In the unsymmet-ric case, both branches have been cal-culated, and it is a matter of retrievingthe positive branch from temporary stor-age and of merging the two. In the sym-metric case, the negative branch is con-structed as the mirror image of the posi-tive branch.

TEN

The column deflection curve for thegeneral case is computed here, whenthis subroutine is called by MSIX. Thestarting eccentricity (DeflectionDEFL(I) measured from the thrust line)and the target eccentricity at the otherend, as well as the starting slope for thefirst trial, are given by MSIX. The objectis to vary the starting slope until thetarget eccentricity at the other end isreached.

This may not be possible because thebending moment corresponding to faiI-ure is reached at some intermediatepoint, or because, as the starting slope isvaried, the eccentricity at the far endapproaches and then recedes from thetarget value without reaching it. Thissignifies instability failure.

Starting with the values obtained fromMSIX, the column deflection curve iscomputed node by node as describedfor MFIVE, except that the bendingmoment now contains a contribution


from the primary moment, obtained byinterpolation.

The starting slope is increased insteps of a/10 (after an initial reduction, ifrequired, because of material failurewithin the span) until the end deflectionexceeds the target value. The sub-routine then retreats to the previousstarting slope and increases it in steps ofa/100 until the end deflection again ex-ceeds the starting value. This is taken tohe the correct value.

If the end deflection approaches thetarget and then recedes without reach-ing it, while the starting slope is con-tinuously increased, this signifies insta-bility failure, If the end deflection im-mediately begins to decrease as thestarting slope is increased, this meansthe member is in an unstable configura-tion, representing post-buckling be-havior; the starting slope is then con-tinuously reduced by x1100 until theend deflection begins to reduce also. If,at this stage, the end deflection is lessthan the target, instability failure is

present. If the end deflection is greaterthan the target, the starting slope isgradually reduced until the target isreached from above, indicating a stableequilibrium configuration.

If, during the construction of a columndeflection curve, the moment at aninterior node reaches the maximumvalue, corresponding to material failure,there is an immediate return to the start,with the starting slope reduced byx1100. This is repeated until the end isreached without attaining the failuremoment. If the end deflection is thenbelow the target value, it means thetarget cannot be reached, and materialfailure is present unless this was the firsttrial. In that case, the member may bein a post-buckling configuration, andthis must be tested by reducing thestarting slope still further by a/100, toinvestigate the trend in the end deflec-tion as described above.

The flow chart of Fig. 18 and Fig. 19of the paper, and the accompanying dis-cussion, illustrate these points.

NOTE: A detailed listing and documentation of thecomputer program are available at cost ofreproduction from PCI Headquarters.

124

APPENDIX C - NUMERICAL EXAMPLES

In this section five numerical exam-ples are given to show the application ofthe computer program.

EXAMPLE 1

Ref. 23 gives the results of severaltests in which prestressed concrete pileswere put under axial load and thenloaded laterally to failure. The testshown in Fig. C1 was used as an exam-ple. The program is used to analyticallycheck the stability of one of the pilestested and compare the results with theexperimental values.

InputAll units are in kips and inches.1972 Santa Fe Pomeroy 16 in, Pile Test (Title)

131111010 Program control0.0 Controls on length intervals

to be used in calculation, etc.— default values called for

4 No. of comers to concrete section0.016.0 x-coordinates of corners16.00.00.00.0 y-coordinates of corners16.0

16

Experimental result'15 Failing lateral load = 15K

at deflection = 7.6'

0 14 Comp uted result:

_j Failing lateral load =I3.95Kof deflection = 2.24'

aCrW 13I-4

z

N 2 Z 16'50. Pi I ^' ll- 7/16 ,A 270 K strandsW3.5 Spiral (f, =6200psiat time o1 test)ps =0.002 P=15Kat failure

II 600K K43,-2.,

_I

IO0 I 2 3

MIDSPAN DEFLECTION (ins)

Fig. C1. Example 1. 1972 Santa Fe Pomeroy 16 in. pile test from Ref. 23.


16.0 Output2 Concrete stress-strain law8.2 Cylinder strength of concrete The deflected shape of the member0.002 Strain at peak stress plus the bending moments are printed0.003 Strain at failure out.0 No. of reinforcing bars6 No. of prestressing tendons0.115 Results0.230 Areas of strands - One The results are shown on Fig. Cl.0.230 strand plus five pairs Failure was reached at a lateral load of 0.230 with equal y-coordinates

13.95 kips due to instability. The de-0.230o ,3o flection in the preceding step (13.902.5 kips) was 2.24 in. The experimental re-3.373 sults gave failure at 15 kips and 7.8 in.5.715 y-coordinates of strands However, this would indicate a midspan8,783 moment of:11.60213.277 M - (600)(7.8)/12 + (15)(43.17)142 Control indicating prestress is = 552 kip-ft

to be given as stress at zero This moment seems well beyond theconcrete strain, i.e., after all capacity of the cross section, so it islosses except elastic shortening,

probable that failure occurred at a lower146.6146.6 deflection, the actual value of which 146.6 would have been difficult to measure as146.6 Prestress the member became unstable.146.6146.6 Use of Formulas8 Stress-strain law for prestressing270.0 Ultimate stress of prestressing To compute the actual strength of this

strand member, without 0 or f3,, factors, the ex-1 No. of axial loads pressions ofRef, 15, p. 68 would have to600.0 Axial load be used, They are found to give X = 1.5,5 Nn. of

LengthLength8 = 7.59, and maximum lateral load 40 .0

0.0 kips. The unmodified ACI Code formula259.0 would give P,, = 415 kips < 600 kips.518.0 Primary bending moment at tenth777.0 points for lateral load of 10 kips.1036.0 These will be increased in steps EXAMPLE 21295.0 to failure.1036.0 The load-moment relationships for six777.0 lengths and ten load values were corn-518.0 puted for the double tee section of Fig.259.0 B3, with equal end eccentricities and no0.0 lateral load. The input was similar to0.0 Amplitude of initial crookedness that of Example 1, with different num-0.0 Eccentricity at starting end hers in the program control card for the0.0 Ratio of eccentricity at far end to equal end eccentricity, failure loadthat at starting end1.0 Strength reduction factor in case.

slenderness effects The ACI Code strength reduction1.0 Strength reduction factor on final factor was used both in calculating slen-

results derness effects and in reducing the short0.0 Loading duration factor,8, column capacity. The ratio )9d was set at

126

400

350

300e

250

PDouble Tee of Fiq,B3Load Duration factor Rid-0-2

- Enter with factored load andtp IN cV factored primary moment = Pe

200 Na ^D fiN^ n Conservative for wind loadmomentCoservati m

tod

150

O

50

0

0 5 10 15 20 25 30 35

(DM n (kipsft)

Fig. C2. Example 2, Design curve for double tee of Fig. B3. ACI strength reductionfactors and load duration factor (3d = 0.2. Strands are assumed fully developedat points of high moment.

0.2. Fig, C2 shows the results in a formthat might be provided by a precastmanufacturer. The figure is a designchart for this member, similar to thosegiven in Ref. 20, but including lengtheffects.

The chart is entered with the factoredloads and the factored primary moment:the interaction lines for the statedlengths define the safe values. If theprimary moment arose from a parabolicwind load moment diagram instead of aconstant Fe moment diagram, the resultswould be somewhat conservative.

EXAMPLE 3

Make the stability calculations for thecolumn of Fig. C3; try the section shownin that figure.

Total factored load = 100 kipsEccentricity at starting end (bottom) =20 kip ft x 12 = 2.4 in.

100 kipsEccentricity at far end (top) = 0Lateral load = 0Strength reduction factor: ACI Codevalue


D.L. = 65 kips (Factored)LL. = 35 kips (Factored

ii

L.L.M0M =13 kips ft (Factored)D.L.MOM.= 7kipsft (Factored?

4 j:• —_ -^n270 ksi ^,

strands

13"

Column braced against sway 0.5 0 30 20 10 0

fc = 6000 psi DEFLECTION BENDING MOMENT(ins) (kips ft )

Fig. C3. Example 3. Braced prestressed concrete column.

Sustained load factor 13 d = 7/20 = 0.35Effective length factor = 1

The data are entered as for Example 1,and the column is found able to sustainthe load with a maximum moment of21.4 kip-ft.

The deflected shape and moment dia-gram are shown in Fig. C3. The columnis apparently able to carry the factoredloads with a substantial margin, but re-peating the calculations with increasingload indicates that instability occurs asshown in Fig. C4,

Use of Formulas

A formula that covers the full range ofsection properties, slenderness values,and load durations must necessarily beconservative in most cases, if it is to he

simple yet safe. A penalty must be paidfor the use of simplified calculations.(Hence, the need for more accurate ra-tional analyses as advocated herein.)

In this case, Eq. (11) indicates alarger, 15 in. square section. (Withhigher load duration factor Fa d the ap-proximate procedure becomes still moreconservative.) Then:

P,, = 1087 kipsP = 100 kipsPIP,= 0.092L/r=96r^=2.5+ 1.6/(P/P.) = 19.9B = 271(Llr) – 0.05 = 0.231A = 4.60El = E, I, A(I + $d ) = 3000000 kip-in!P, = 159 kips0 = 0.75C,,, = 0.6

128

500

400

300

n

0--8-

200

• •

00 20 40 60 80 100

0 M n (kips ft }

Flg. C4. Column of Example 3: proximity of instability failure belies appearance ofexcessive safety at magnified factored moment.

8 = C m /1 – PLOP, = 3.72M = 20(3.72) = 74.4 kip-in.This moment is satisfactory for the 15

in. square section, but it is clear that thiscalculation greatly exaggerates the mag-nification of the moment. However, thisdiscrepancy is intentional, reflecting the

impending failure in instability asshown on Fig. C4.

The problem arises from the fact that alinear model is used to describe nonlin-ear behavior as discussed below Eq.(11), and as considered in detail in Ref.15, in reference to Fig. 14 of that paper.

PCI JOUANAUMay-June 1985 129

EXAMPLE 4

Make the stability calculations for theportal frame of Fig. C5, using the sectionof Fig. C3, with modified strand layoutas indicated in Fig. C5.

Dead load on roof = 539 lbsllin. ft

Live load on roof = 320 lbs/lin. ft

Loads on walls:

P1) = 25(0.539) = 13.48 kips/panelPL = 25(0.320) = 8 kips/panel

The two walls cantilever from thebase, with the roof forming a pinnedstrut between them, carrying the com-pression necessary to equalize the deflections. Length changes in the roofunit could be included but will be as-sumed to be zero here, for clarity.

• Primary Analysis at Service Loads:

Let compression in roof member =F lbs

Deflection at roof level = -L4 ± FL

8EI 3EI(Use E = 4000 ksi, I = 2872 in.')Equate deflections at tops of walls:

160L' FL 3 128L 4 FL38EI 3EI 8EI 3E1

F = 1441bs

_ 160L 4 _ 144L3

8EI 3EI

= 0.90 in.

• Secondary Analysis at ServiceLoads:

HoldingforceH = IPA`L

HL3

_ EP_,L 2

6EI

= 0.05 in.

No further iterations are necessary:3 = 0.95 in.

• Design Loads:

Case 1. U = 1.4D + 1.7LP = 32.5 kipsNo wind loads

Case 2, U = 0.75 (1.4D + 1.7L +1,7W)

P = 24.4 kipsW = 204 plf (pressure)

163.2 plf (suction)Case 3. U = 0.9D + 1.3W

P = 12.13 kipsW = 208 plf (pressure)

166.4 plf (suction)Clearly, Case 2 governs.Suppose that, under Case 2 loads, the

moment of inertia of the section will bereduced by cracking to one-half of theinitial value: say 1440 in .4 (This is thedifficult part of this analysis; it shouldbe done iteratively, with the assumedvalue being confirmed by calculationafter the secondary moments are found.)

• Primary Analysis at Design Loads:Equate deflections at tops of walls(Use E = 4000 ksi, I = 1440

204L4 FL 3 = 163.2L 4 + FL38EI 3EI 8EI 3EI

F = 183.6 lbs

_ 204L4 _ 183.6L-'' 8EI 3EI

= 2.28 in.

• Secondary Analysis at Design Loads:

Holding force H = IPA,L

HL3

31EI

_ PAIL26EI

= 0.27 in.

A = Y.PA :L26E1

= 0.03 in.

Accept: A = 2.58 in.

130

50,

8 DT 16

N

2" Normal WT.

ToppingBearing platescentered on centroidof section

UNBRACED CANTILEVER WALLS

Wind load 7 20 psf ( pressure )16 psi ( suction)

Live load on roof = 40 psfBDT12 wolls with 4-l/2 '4 strands

d - 2 = 10'Otherwise as FIG. B2 . I = 2872

Fig. C5. Example 4. Portal frame with unbraced double tee cantilever wall units.

Moment at base of wall =(0.204)(24)212 – (0.184)(24) +(24 .4)(2.58)/12= 59.6 kip-ft

Moment capacity of section 4M atd?P„ = 24.4 kips, assuming strands arefully developed at the point of maximummoment, is calculated to be 63.3 kip-ft.

• Moment Magnification

It would ordinarily be necessary toaccount for additional moments due todeparture of the column from the linejoining the ends. This would he done byapplying the moment magnificationfactor to the braced effective length, orby rational analysis, with the primaryplus P- moments. However, in the caseof a cantilever column, the maximummoment is obviously at the base and thisis not necessary.

Alternative Solution to Example 4Using Program

The problem can be solved entirelyby means of the computer program.

The load P is assumed to act verticallythrough the centroid of the concretesection at the top. At the bottom, it is setto some trial eccentricity. The data arethen entered as for Example 1, with theprimary moment due to the wind loadand the calculated compression in theroof member.

The bottom is used as the starting end,and the slope there is examined. Exam-ination of the curvature contours fromthe first run of the program shows thatthe member is cambered, with curvatureat zero moment equal to 0.000032. Itfollows easily that the camber is 0.33 in.and the slope at the base of the unitwould depart by 0.0046 from the verti-


300

200

I00

0

1000

900

800

700677 —

60001a

s

0< 5000JJ

400

0 500 1000 f500(083

BENOING MOMENT (kips—ins)

Fig. C6. Design data generated for Example 5. 1fi in. square column withno length effects.

cal, if the unit were erected with the topof the section vertically above the bot-tom. The base eccentricity is varied insubsequent runs until this slope isachieved.

When the process is repeated for theleeward side, it is found that the tip de-flections are not equal. This necessitatesa change in the compression force as-

sumed to act in the roof unit, to 50 lbstension. The required balance is thenobtained, with tip deflections equal to2.20 in., and the maximum base mo-ments equal to 64.4 kip-ft (windward)and 50.3 kip-ft (leeward).

It should he recalled that the manualP-A calculation above depended uponthe value adopted for the moment of in-

132

ertia of the cracked section, whichshould have been calculated and cor-rected after the first trial_ Thus, the ap-parent agreement is in part fortuitous,and in part due to the fact that the axialload is relatively small.

Note that this problem could usuallybe solved quite simply by using the ef-fective length concept: by doubling thecolumn length to include the mirrorimage below the foundation, with theaxial load concentrically applied. TheP-A effects would emerge naturally fromsuch a calculation.

The difficulty in this case arose fromthe unsymmetric section with the con-sequent camber in the unloaded state.This means the mirror image column ofeffective length 2L, would have had aslope discontinuity at midheight. Al-though manual calculations can ignorethe camber, there does not appear to beany way of avoiding it in the more exactcalculations.

Use of Formulas

Again, with use of the unbraced ef-fective length of 2L, moment magnifi-cation formulas should lead to the P-Amoment. However, the L/r ratio for thismember is 576/3.16 = 182, which is out-side the range of applicability of anysuch formula. Thus, in this case, rationalanalysis of some form is mandatory.

EXAMPLE 5

A square prestressed concrete columnis to he designed using the followingdata:

P,, = 550 kips f, = 270 ksiM,, = 875 kip-in. J, = 154.9 ksif, = 6 ksi =0.7The minimum cover to the center of

prestressing steel is 2 in.This is Example 1 from Ref. 20, p. 138.

As in that reference, a 16 in. square trialsection was selected. The solution wasobtained by running the program only asfar as the ultimate strength interactioncurve, since no length effects are in-volved.

Fig. C6 was obtained after a few min-utes spent coding and running the pro-gram. As will be seen, it was concludedthat the column was safe with four 3-in.diameter strands, giving a steel area of0.34 in. ! At the given eccentricity, a loadof 677 kips could be safely carried, witha moment of 1083 kip-in. These are ex-actly the results obtained in Ref. 20.

Note that the functional form used forthe concrete stress-strain curve in Ref.20 was not included in the program;nine points on the curve were calculatedand entered, giving this result.

Slenderness effects could have beenincluded by running the whole program.The design curves of Fig. C6 wouldthen include length effects, as do thoseof Example 2 (Fig. C2).

Metric (Si) Conversion Factors

i It = 0.305 rn1in.=25.4mm1 in. 2 = 645.2 mm21 in .4 = 416231 mm"1 lb = 4.448 N

1 kip = 4.446 kN1 ksi = 6.895 MPa1 kip-in. = 113 N-m1 kip-ft = 1356 N-m1 psf = 0.0479 kPa


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