CONSTRUCTION OF DESIGN AIDS FORBIAXIAL BENDING OF
LONGRECTANGULAR REINFORCEDCONCRETE COLUMNSByMARC LeROY
GULLISONBachelor of Architectural. EngineeringOklahoma State
UniversityStillwater, Oklahoma1969Submitted to the Faculty of
theGraduate College of theOklahoma State Universityin partial
fulfillment ofthe requirements forthe Degree ofMASTER OF
ARCHITECTURAL ENGINEERINGJuly, 1975Ij'lS-CC{f:,7c.CONSTRUCTION OF
DESIGN AIDS FORBIAXIAL BENDING OF LONGRECTANGULAR
REINFORCEDCONCRETE COLUMNSThesis Approved:nn --Dean of the Graduate
College923486ii UNIVERSITYt/.fjRARYPREFACEThis study presents a
refined approach to the analysisand design of rectangular tied
reinforced concrete columnssubjected to axial thrust and biaxial
bending. One of theseveral techniques for the design of'concrete
columns cur-rently in use is thoroughly examined, organized into a
log-ical procedure and converted into graphical form to be usedas
design aids. Due to the character of the resulting chartsthe scope
of this study is limited to the construction andillustration of
design charts only so far as to convey theprocess by which they
were formulated. It is intended forthe future that a complete set
of design charts be construct-ed for use over a large range of
design parameters to serveas functional design aids for the
structural engineer.I wish to express my appreciation to my
principal ad-viser, Professor Louis o. Bass, for his guidance,
advice,and assistance during this study.And in special recognition,
sincere gratitude is ex-tended to my wife, Janet, for her respect,
andmany sacrifices.iii'TABLE OF CONTENTSChapter PageI .
INTRODUCTION. . . . . . . . . . . . . . . .. 1II. STATEMENT AND
PURPOSE OF STUDY. . 5III. ASSUMPTIONS AND CODE PROVISIONS. . . . .
. 10IV. SLENDERNESS EFFECTS ON COLUMNS. . . . . .. 14V UNIAXIAL
BENDING OF COLUMNS. . . .. 21VI. BIAXIAL BENDING OF COLUMNS .
28VII. TRANSFORMATION INTO GRAPHICAL FORM 44VIII. APPLICATION AND
USE OF DESIGN GRAPHS . 62BIBLIOGRAPHY. . . . . . . . 81APPENDIX.
83ivLIST OF FIGURESFigure Page1.2 3.4.Resisting Forces of Column
Section UnderPure Thrust. . . . . . . .Resisting Forces of Column
Section UnderEccentric Load . . . . . . .Resisting Forces of Column
Section UncerEccentric Load Load-Moment Interaction
Diagram.222325275. Compression Area of Rectangular Secti.onUnder
Biaxial Bending . 286. Compression Area ,of Circular SectionUnder
Biaxial Bending . 287. Biaxial Load-Moment Interaction Diagram..8.
Section Through Biaxial Load-MomentInteraction Diagram at Constant
Load9. Relation of Uniaxial Capacities to BiaxialLoad-Moment
Interaction Diagram. . . . . . .10. Equivalent Eccentricities of
Axial Load ..11. Failure Surface for Load vs Eccentricity12.
Failure Surface for Reciprocal of Load vsEccentricity . . . . ..
....30313133343513. Bresler's Approximation of the FailurePlane lip
vs e . . . .. .... 3614. Load Contour of Biaxial Interaction
Surface .. 3915. Approximation of Load Contour From
BiaxialInteraction Surface ..........v39 .16.17.18.Load Contour for
Square Section with EqualReinforcement in all Faces . . . . . .
.Load Contour for Rectangular Sectionwith Symmetrical Reinforcement
..Flow Chart for Conventional Design ofReinforced Concrete Columns
....4142455519. Graphical Representation of P/(1/5EcIc + EsIs )20.
Graphical Representation of(1 + 6 d) x p/ (1/5EcIc + EsIs). . . . .
. . . .. 5721. Graphical2Representation of(klu/x) x P(l +
Sd)/(1/5Eclc + EsIs'>. . . . .. 5822. Graphical Representation
of8. = Cm/ (1 - p / 2 - 2[ 1 (0 0 3 ) (cb- d I )+ - d' cb- Asfy [d
- .003 dcb = .003 + .0020726(5.12b)For all values of c smaller than
cb the stress in the tensionsteel is fy and Fs2 = Asfy With this
value for Fs2 substi-tuted into equations (5.10) and (5.11), they
will yield thefailure loads and for the range C < cb .Over the
first range from the minimum eccentricity tothe point where
balanced loading exists, failure in the sec-tion is controlled by
compressive stress. As the eccentric-ity increases beyond balanced
loading, failure is controlledby tension. Also, as e increases,
decreases and in-creases up to some maximum value and then
decreases. This isbest illustrated in a load moment-interaction
diagram as inFigure 4.If is plotted versus (or the curve
willrepresent the locus of the maximum theoretical allowableaxial
loads and moments and for any eccentricity e asshown in Figure 4.
This curve is unique for some percentageand configuration of
reinforcement. Normally the diagram isa family of curves for
various percentages of steel. Thistype of diagram is to be used as
the basis for the designaids in the appendix. Its application will
be discussed ingreater detail in Chapters VII and VIII and can be
found in27most texts dealing with the design of reinforced
concretecolumns.,~e/ M1~ 'CA, U.Figure 4. Load-Moment
InteractionDiagram.Another method of design better suited to
longhand cal-culations is to construct a portion of an ipteraction
dia-gram with several values of c giving allowable loads nearthe
design loads. 3The capacity of the column can then betaken directly
from the curve.3Riceand Hoffman, pp. 265-274.CHAPTER VIBIAXIAL
BENDINGIn comparison to bending about one axis of a
reinforcedconcrete column, biaxial bending presents an entirely
differ-ent and more complex situation. As a second bending momentis
introduced, the neutral axes are no longer parallel to
thecentroidal axes of the section, but lie at some angle e
fromthem. Due to the rectangular shape of the cross section, ase
increases, the area of the cross section under compressiony
OF'4ru=1V'rrr:Figure 5. Compression Areaof RectangularSection
UnderBiaxial Bending.28Figure 6. Compression Areaof CircularSection
UnderBiaxial Bending.29becomes triangular as shown in Figure 5. If
this area main-tained the same shape as a changed, as in the case
of a cir-cular cross section, it would be a simple matter to find
therelationship of one moment to the other, as illustrated inFigure
6. However, the shape of the cross section does varywith a, but no
simple and exact relationship is to be foundbetween a and the load
capacity of the column. It is there-fore necessary to rely on
empirical relationships developedfrom biaxial bending tests on
rectangular concrete columns.In recent years several methods for
designing biaxiallyloaded columns have been published. Most methods
have incommon some form of an interaction surface as shown inFigure
7. The curve ABeD represents the load-moment inter-action diagram
for the X axis of the column and the curveDEFG represents the
load-moment interaction diagram for theY axis. For any given value
of axial load p ~ a horizontalplane may be passed through the
interaction surface definingthe maximum allowable moment the column
will withs.tand whenapplied simultaneously with P ~ . For example
we shall assumethat point H in Figure 7 represents a value of axial
load tobe applied to a column. A horizontal plane passed
throughthis point is represented by plane BHF. If this plane wereto
be removed from the diagram and viewed in plan it wouldappear as in
Figure 8. For this value of p ~ , if a momentoccurred about only
the X axis, the maximum allowable momentthe column could withstand
would be represented by point M ~ x .Likewise, if a moment occurred
only about the Y axis, the30IPu.cFigure 7. Biaxial Load-Moment
InteractionDiagram.maximum moment capacity would be given by point
M ~ y . How-ever, if the resultant moment applied to the column is
aboutan axis at some angle a from the Y axis, its maximum
allow-able value is given by M ~ . Ideally, point M ~ would lie
onthe dashed elliptical curve, but in the case of
rectangularcolumns, the fact that the compression area becomes
triangu-lar, as in Figure 5, alters the boundary similar to the
solid31Figure 8. Section Through BiaxialLoad-Moment Interaction
Diagramat Constant Load P.~ lFigure 9. Relation of Uniaxial
Capacitiesto Biaxial Load-MomentInteraction Diagram.32curve in
Figure 8. If the maximum allowable moment isdivided into X and Y
components and it becomes asimpler matter to formulate a
relationship between the axialload and each of the moment
components.This problem is expanded in Figure 9 where is
theultimate concentric load with no eccentricity. At points and an
equivalent moment may be produced by theaxial load acting at
eccentricities ex and ey from theX and Y axes, respectively. If the
eccentricity ex werefixed and the load allowed to vary, the maximum
allowableaxial load would be If no bending occurred about the
Yaxis, the maximum axial load would be If the analysisof Figure 9
is approached from the opposite direction, thatis if piex and ey
are known, a relationship can beoy'found between and One of the
simplest relation-ships was by Bresler.1Tests and investigations
ofbiaxial bending have shown his equation to be
satisfactorilyaccurate under most of the range of axial load and
bendingmoments. If a given cross section as in Figure 10 is
sub-jected to bending moments and and an axial load as shown, the
system of forces can be reduced to a singleload acting at
equivalent eccentricities obtained from =PulBoris Bresler, "Design
Criteria For Reinforced ColumnsUnder Axial Load and Biaxial
Bending, II ACI Journal (1970),pp. 481-490.33yxe'j '1'
t)(
- --yFigure 10. Equivalent Eccentricities ofAxial Load.Bresler's
equation is_1_P'u= 1 + 1 1
o(6.1)where is the ultimate axial capacity of the cross
sectionand is ultimate axial capacity under a concentricload, is
the ultimate axial capacity if moment occursonly about the X axis
and is the ultimate axial capacityif moment occurs only about the Y
axis. can of course befound by simple statics once a column size
and reinforcement34is assumed. The capacity is determined by
consideringthe column subjected only to and and "accounting
forslenderness effects if applicable. Then can be
similarlydetermined. Finally, a value is obtained for and
comparedwith the actual load applied to the column.The origin of
Bresler's equation stems from a failuresurface obtained by plotting
the failure load as a func-tion of eccentricities x and y as shown
in Figure 11. Thevalues of x and y also serve to illustrate the
relationshipof and the bending moment components and As canbe seen
in the diagram as the eccentricities increase,p-- FAILUR,E,
:xFigure 11. Failure Surface for Load vsEccentricity.35bending
moments increase and the failure load decreases tosome limit at the
bottom of the curve where axial load be-comes negligible and the
section is considered to be in purebending. If the reciprocal of
the failure load is plottedas a function of the eccentricities, a
surface such as thatin Figure 12 will be obtained. It is this
surface from whichBresler's equation is actually derived. The
surface beingsomewhat flat resembles a slightly warped plane. For a
given,p/////IIIIIII y'------ F Figure 12. Failure Surface for
Reciprocalof Load vs Eccentricity.36column, at least three points
on the surface are known forsome particular value of and are
coordinates for the fail-ure load for pure axial load the ex
corresponding to thefailure load were moment to occur only about
the Y axis,and the ey similarly corresponding to the failure load
P' .oyThese points can be plotted as (l/Po', 0,0), (lip' , e ,D)ox
xand ey ' 0) as shown in Figure 13. If a plane werepassed through
the three points, any point on the failureI /////////_--.//" /\" I
//_____\v////////Figure 13. Bresler's Approximation of the
FailurePlane lip vs e.37surface ex' ey ) can be approximated by a
point ex' ey) on the vertical projection to the plane. Ifsome point
were defined by the three coordinates ex' ey)' see Figure 13, the
location of fallsvery near the intersection of the failure surface
and theplane where the error in approximation is 2ero.Since the
plane is unique in that each value of and will yield unique values
for ex and ey , the errorin the approximations will very nearly be
the same for allpositions of the plane. The error will increase
slightly,however, for very large values of and Resultsof the
approximation were compared with theoretical resultsin Bresler's
paper and found to be in excellent agreement,the average error
being 3.3 percent.In order to apply the approximation the plane
must bedefined by the three known points. The equation of the
planefor some eccentricities and0 lip' 1 lip' e 1 e 0 1ox ox x xe
lip' 1 e' + lip' 0 1 e' + 0 ey 1 lip' =y oy x oy y u0 lip' 1 lip' 0
1 a 0 10 0ex 0 o ey o 0 where ey) are the coordinates of the
failure loadfor biaxial bending. Simplifying the equation we
obtain:e'(_l __1)+x p' p'o oxe' ey xex ( 1 1 ) ( 1 1 ) _pr- - pi +
ex pi - pi - 0o oy u 038For biaxial bending the eccentricities will
be the same asthose for uniaxial bending, see Figure 10. Therefore,
ife ~ = ex and ey= ey , then....LP'u= 1 + 1 1p ~ x P ~ y - p ~which
is Bresler's formula. It is perhaps the simplest andmost widely
applicable relationship that has been developed.Another method
introduced by Bresler is of the formwhere Mx = Puy, Mox = PuYo when
x = My = 0: and My = Pux,May = Puxo when y = Mx = O. Looking at the
failure surfaceformed by the load-moment interaction diagram"
Figure 14, thesurface is formed by a family of curves at constant
values ofPu . Bresler refers to these curves as load contours. If
aplan bounded by a load contour at some Pu is examined as inFigure
15, and if the load contour is assumed to be astraight line, the
equation of the load contour is given byThe equation can be written
as.+ 1p39Figure 14. Load Contour of Biaxial
InteractionSurface.(a)Figure 15. Approximation of Load Contour
FromBiaxial Interaction Surface.40If the load contour is curved
instead of straight its equa-tion is approximated byin which a and
S are dependent on the dimension of the col-umn, steel reinforcing,
stress-strain behavior of the mate-rials, the concrete cover and
lateral ties. Tests showedthat this equation provided good
approximations of analyticalresults but no one value of a or 8 can
be assigned to accur-ately represent the load contour for all
cases.2Therefore,the determination of a and S would add undesired
complexityto the design procedure.Other approximations have been
derived, among them amethod by Pannell based on a failure surface
as in Figure14. 3These methods are either less accurate or require
addi-tional functions making the design procedure more complex.For
this reason the first method by Bresler will be used inthe design
charts to be constructed here.The one limitation of Bresler's
equation is its appli-cable range. For small axial loads tension
has more influ-ence on a column's capacity. The relationships on
whichBresler's equation is based are no longer valid due to the3F
N. Pannell, "Biaxially Loaded Reinforced ConcreteColumns,"
Proceedings, ASCE, Vol. 85, ST6 (June, 1959),pp. 47-54.41absence of
the failure surface, Figure 13, in this range.Therefore, the
equation will not be used for of Puless than Po/10.4Another
equation must be used for thislower range of axial loads.If a load
contour of Figure 14 is examined for a squarecolumn with equal
steel in all faces at some Pu ' the biaxialload-moment capacity
will be represented by a near circularcurve, Figure l6a, where Mox
and Moy are the uniaxial moment(b) (a.)M",'j 1---------.-... Figure
16. Load Contour for Square Section WithEqual Reinforcement in All
Faces.capacities for the X and Y axes, respectively, and are
equal.The design moments Mux and Muy are bounded by their
inter-section with the load contour Mu . An approximation of
thislimit can be made by assuming a straight line between Mox4 Rice
and Hoffman, p. 290.42and Moy as in Figure l6b. Then Mu will lie on
the line andinside the curve for any combination of Mux and Muy
giving amore conservative solution. The design moment Mu is
simplythe vector sum of Mux and Muy but should not be larger
thanMox5This equation is suitable(6.2)for a square column with
equal reinforcement in all faces.However, if the reinforcement is
not symmetrical or the col-umn is rectangular, Mox and Moy are not
equal and the loadcontour will resemble Figure l7a. A similar
approximationcan be made in this case as shown in Figure l7b. If
equation(a) (b)Figure 17. Load Contour for Rectangular SectionWith
Symmetrical Reinforcement.5Ibid ., pp. 286-289.43(6.2) is written
asMux + Muy ~ 1Max May(6.3)where Mox and M are assumed equal to M,
the relationshipoy ualso represents the approximation in Figure
l7b. Since Maxwas the upper limit of equation (6.2), when the
equation isdivided by Max' one is the upper bound of equation
(6.3).This equation was given by the previous code (ACI
318-63,Section 1407c, Eqn. 14-14) and limited to situations
wheretension controls the design.Because of its simplicity,
equation (6.3) will be usedto determine biaxial capacity for the
design charts to bepresented here. Although the conservative error
is signifi-cant, simplicity is considered to be important and since
mostcolumns encountered are not controlled by tension, the use
ofequation ,(6.3) Should not provide unreasonable design as faras
economy is concerned. Further discussion of the errorsinvolved may
be found .in Paul F. Rice and Edward S. Hoffman,Structural Design
Guide to the ACI Building Code, New York,1972, Chapter "lO.CHAPTER
VIITRANSFORMATION INTO GRAPHICAL FORMThe preceding chapters have
presented all the equationsand proce"dures necessary for the
complete design of a cornmanreinforced concrete column. In those
areas having severalaccepted techniques and theories, one method
was selected foruse in constructing the design aids. The equations
used inthis chapter will be repeated and referenced to their
intro-duction in the text.An outline of the series of operations
required to de-sign a column is given in Figure 18. The flow chart
followsthe column design procedure from the initial assumptions
tofinal sizing. This procedure represents the fundamental ap-proach
to column design. Numerous short cuts and approxima-tions have been
introduced in many texts for preliminarychecks to make the trial
and error process less cumbersome.The reader is referred to Phil M.
Ferguson, ReinforcedConcrete Fundamentals, New York, 1973, for some
of the morecommon methods of approximating column design.Any method
of column design requires some initial as-sumptions as to column
size, reinforcement, or loading condi-tions. The procedure used
here requires the selection of atrial size and percentage of
reinforcement for a column. The44 eJ( 45 l F' OR.. COt\lT[
DSTE.R.M'NE- IF Or ;S\sLE..c.:T NSW ,0 1,D'+'ox 15ADt;QUATE. __
COMPu,S k.lu. 40 TO ZFigure L8 Continued.47capacity of the column
is then determined and compared to thegiven load. Once the
dimensions have been selected, the re-maining unknown parameters
lend themselves very well tographical presentation. As can be seen
from Figure 18 thesection is first checked for uniaxial load-moment
capacity.Then slenderness effects are checked for each axis and
final-ly the biaxial capacity is determined. The general form ofthe
design charts in the appendix follows these three steps.Design
charts for column slenderness have been presentedbased on the
previous code (ACI 318-63) and are assembled interms of
dimensionless parameters.lThis requires that sev-eral preliminary
calculations be made prior to using thecharts. The sections of the
charts dealing with slendernesseffects also require more than
simple calculations. Thecharts to be constructed here will follow a
similar patternexcept that relationships will conform to the
current codeand the parameters w'ill be separated so that only
simple cal-culations will be necessary. Dimensionless parameters
willnot be used, but rather a chart for each given column size.Also
the relationships within the charts will be extended toinclude
biaxial relationships. It is possible that since theexactness of
the equations is to be sacrificed for the con-venience of graphical
solutions, some accuracy will be lost.However, the equations
themselves are but empirical approxi-mations and any errors in the
solution will depend primarilylRichard w. Furlong, IIColumn
Slenderness and Charts forDesign, II ACI Journal (1971), pp.
9-17.48on judgement and accuracy of plotting. A simple check
bystatics at the end of the design process should be suffi-cient to
identify any significant errors made during the de-sign
procedure.The basis of the design aids is the load-moment
inter-action diagram as shown in Figure 4. With the equationsgiven
in Chapter V, a diagram may be constructed for eachaxis of a given
column size and percentage of steel. If thedimensions, reinforcing,
load and properties are known and ifan interaction diagram is
available, it will be obviouswhether failure is controlled by
tension or compression andthe columnls capacity for any combination
of load and momentmay be found instantly. The upper boundary of the
diagram,point P ~ ' is the capacity of the section under pure a x i
~ lload and from statics is given by equation (5.2).As moment is
introduced the position of the neutral axis,distance c from the
compression edge of the column, shiftstoward the compression edge.
As the eccentricity increases,c decreases. The capacity for axial
load also decreases, butthe moment resisting capacity increases.
This curve betweenp ~ and Ph is defined bypiU ( 5 10)M'.u49=
O.36125fcAnc(h- .S5c) + -d') .003(c-d')c- A E (d- h/2) .003(d-l)sSe
(5.11)When the neutral axis falls outside the section on the
ten-sion side (.85c >h), equations (5.5) and (5;6) must be
used.As the balanced loading condition Pb is passed, the mo-ment
capacity begins to decrease. This portion of the curveis given
byM'u(7.1)( 7 .2)The code provides that when the load capacity is
less than the safety factor may be increased from 0.7 to 0.9as
decreases to zero.2This puts an outward bend in thebottom of the
interaction curve.Since any column may contain several
configurations andpercentages of steel, a family of interaction
curves may bedrawn, one for each size bar group possible within the
samesize column. This will eliminate interpolation between
per-centages found on other diagrams. Lines of constant
eccen-tricity are also plotted on the charts to aid in theselection
of reinforcement. The uppermost line on the2Section 9.2.1.2, p.
26.50interaction diagram represents the minimum eccentricity
al-lowed by the code, h/10.The family of interaction curves is
located in the upperright quadrant on each design chart. Another
curve is super-imposed on the graph and will be explained later in
the chap-ter. The upper left quadrant consists of three families
ofcurves. These three groups determine the ratio of appliedload to
critical load P / ~ P c r . In order to determine thisratio, Pcr
must be found from equation (4.1). This equationcan be separated
into a product of two terms in which the= (4.1)term EI is given by
equation (4.2). If the equation for EIis separated into two terms,
then the term (Eclc/5 + EsIs )EI = 1/5 EcI c + EsI s1 + Sd (4.2)has
a unique value for each bar size group in a given sizecolumn. This
term becomes one of the three slenderness par-ameters for equation
(4.1). The second term 1/(1 + Sd) isthe second parameter since its
value is independent of thecolumn properties and is a function of
loading. The thirdslenderness parameter is x2j(k1u) 2from equation
(4.1). Theproduct of these three parameters will yield Per' but P /
~ P c ris required to find the moment magnifier 8 given by
equation(4.5). In order to obtain the ratio of P to Pcr' the
param-eters given above may be combined as shown in equation
(7.3).o =1 - pepPcr51(4.5)= p(7 .3)Each of the three families of
curves in the upper left quad-rant of the design charts is the
graphical representation ofone of the three terms in equation
(7.3). The lower leftquadrant is a plot of P/epPcr vs 0 for several
values of emwhich is obtained from equation (4.6). In order to
eliminatethe calculation of Cmthe actual parameter used will
beMl/M2 and the family of curves is then defined by equation(7.4) .
Of course the code requires that C must not be lessmthan 0.4 and
that o. must not be less than 1.0.em = 0.6 + 0.4( : ~ ) ~ 0.40.6 +
0.4 (:1)a = 21.0P ~1 ---epPcr(4. 6)(7.4)The lower right quadrant
corrects the initial designmoment Mx by the moment magnifier 2fy
..use Fsi = 120 k( 003) (cb - d l- s) AIFs 2 = 2- Escb 2= (. 003)
(9. 2. 2. -- 2. 44 - 4. 3 7) (41_ ( 2 9x1 03 ) =(9.21) (2)= (.003)
(d - S - Cb) As2 Es= (.003) {IS.S8 - 4.37 - 9.22) (4)
(29Xl03)(9.21) (2)Fs 3 = -38 kFs4 = -120 k6845 kPh = 301 + 120 + 45
- 120 313 kM'bM'b= Fe - + Fs1 - d ' ] + - - Fs4 - d ' ]= (306) [128
- (.8S)J9.22)] + (120) [\8 - 2.44]+ (45) [4.237] - (-38) [4'237] -
(-120) [128 - 2.44]= 3311 "k= 3311313 = 10.57"Since the balanced
eccentricity is greater than the actualeccentricity, compression
controls the design. Now the ca-pacity of the column in compression
must be found. The dis-tance to the neutral axis must be found by
trial and error.As shown in Figure 26 the reactive force of the
concrete isassumed to act between P and the center line of the
column.The neutral axis is assumed to lie as shown. Since
thestrains in the steel caused by Fsi and Fs4 may exceed themaximum
steel strain of .00207 in/in, a check may be made to69e pFigure 26.
Resisting Forcesof Section,Example 1.determine whether the forces
given below are valid. SinceFc = 85 f ~ { .85cb - 3Ar )Fe = ( .85)
(4) l(85) (12) c - 6] = 34.68c - 20.4Fsl = (.OO3)(c - d')A Ee r
sFsl = (.OO3)(c - 2. .44) (2) (2.9xlO3)cFsl = 174 - 424.56cFs 2. =
(.OO3)(c - s - d ')ArEscFs 2 = (.OO3)(c - 4.37 - 2.44)(2)
(2.9xl03)cFs 2 = 174 - 1184.94cc70F = (.003)(c - d + s)A E83 c r
sFs3 = ( 003) (c - 15.58 + 4. 37) (2) (2 9x1 03)cFs3 = 174 -
1950.54c( 003) (d - c) A Ec r s{ 003} U5. 58 - c} ( 2) (2 9x103)cF
1 74 - 2710.92s4 =Fs4 is close to the neutral axis, it will not be
critical.The c required for Fs1 to produce a strain of 0.00207
isgiven by'c = d l1 - 0.002070.003= 2.441 - .00207.003= 7.87"Since
c is assumed to be greater tha.n 7.87 ", Fsl will prod\.lcea strain
greater than .00207 and the maximum stress in thesteel is 60 ksi.
Therefore,Fsl =fyAr = (60) (12) = 120 kAssuming a trial value for c
of 15 i n c h e s ~Fc = 500 kFs 1 = 120 kFs 2 = 95 kFs 3 = 44 kFs4
= -7 k71and = 762. k = Fc - 81C] + -d I} + Fs2 - - Fs4 - d l] =
(500) (9 - 6.38) + (120) (9 - 2.44) + (95) (2.19)(44) (2..19) -
(-7) (9 - 2.44) = 2278 "ke = = 2.96"The e obtained from c = 15" is
larger than e so a larger cxwill be tried. With c = 15.49", e =
2..75' which is correct. = 517 + 120 + 98 + 48 - 1 = 782 kPox = 0.7
= 547 kMax = 517(9 - 6.49) + 12.0(9 - 2.44) + 98(2.19)- 48(2.19) +
(9 - 2.44) = 2152 "kMox = 0.7 Max = 1506 "kPox and Mox represent
the maximum design loads that may beapplied to the column with an
eccentricity of 2.75".The next step in the problem is to determine
the momentof inertia of the section with respect to the major
axis.From equation (4.2.)1EI = SEcIc + EsI s1 + Sd72Ie = bh3= (12)
(18)3 = 5832 in412 12Is = 4 {'7t(04,564)4 + (1) (9 - 2.44)2]+ 4 [3t
(04,5 64) 4 + (1) (4 "/ 7 ) 2] = 1 91 86 in4t (3 64 x 106) (5832) +
( 2 9 :x 106) (191. 87)EI =1 + 0.6EI = 6.131 x 109#-in2The critical
buckling load is then=(1000) [(.75) (14) (12)] 2= 3811 kThe moment
magnifier is computed from equations (4.5) and(4.6) .Cm = 0.6 +
0.4(780 ) = 0.8811000 = Cm= 0.88= 1.041 _P_-1 400- -
2.1311239Compression will control the design. Figure 27 shows
theassumed locations and directions of the resisting forcesGcepI I
I ,,,It.-_~ . 5 b "-- .-.- 12 " ....- ..Figure 27. Resisting
Forces, MinorAxis, Example 1.Another trial and error procedure is
required to determinethe distance to the neutral axis. Having been
illustratedonce, the iteration is not given here. A value for c
of10.52" was found to be correct.P ~ y = (.85) (4) [(.85) (18)
(10.52) - 8] + 240+ (.003) (10.52 - 9.56) (4) (29 x 103)10.5275P ~
y = 520 + 240 + 32 = 792 kPay = 0.7 P ~ y = 554 kM ~ y = (52.0) (6
- 4.48) + (240) (6 - 2.44)+ (-32) (6 - 2.44) = 1535 "kMoy = 1075
Ilk(18) (l2)3 = 2592 in412.e = 1535792= bh3=12= 1.94"EI4= 8 lJt( .
~ 6 4 ) + (1) (6 2.44) 2] = 102.2 in4= t (3 64 x 106) (25 92) +
(102 02) (29 x 106)1 + 0.6EI = 3.028 x 109#-in2The critical
buckling load is then= Jt2(3.029 x 109) =(1000) l(. 75) (14) (12)]
21883 k= 0.6 + 0.4(310) = 0.76775= 1.114000.771 - (.7) (1883)== 8M
= (775) (1.11) = 860 "kYAgain a new eccentricity is calculated ase
= 860 = 2 15"400 ..76It was found that a c of 10.13" satisfied the
eccentricityof 2.15 11andP ~ y = 760 kMay = 1631 "kPoy = 532 kMoy =
1142 "kBoth values are greater than the design loa:3s.Biaxial
bending will now be considered. The valuesrequired are summarized
below.P = 400 k, P ~ x = 766 k, P ~ y = 760 kThe third term in
Bresler's biaxial equation is the recipro-cal of the axial capacity
of the column in the absence ofbending. From equation (5.2)P' =
(.85) (4) (12) (18) - 8 + (60) (8) = 1187 k01 1 1 1pi +P ~ y ' pi =
piox 0 u1 _1__ 1 1766 + =562 760 1187Pu = 0.7 (562) = 393 k 400
kThe Pu obtained from Bresler's equation is just less than
thedesign load, but the error is less than two percent:400 - 393393
= 0.018 or 1.8%This is close enough to be satisfactory and verifies
theresults of the design conducted with the design charts.77As a
second example, assume the loads are a 300 k axialload, moments
about one axis at either end are E40 "k and430 Ilk, and moments
about the other axis are 500 "k and-125 "k. The eccentricity for
the larger moment will be2.13". The eccentricity for the other axis
is 1.67". Forthe first case S = 0.9 and S = 0.4 for the second.A
square section 14 inches deep will be tried, withfour number eight
bars in the corners. Assume kill is 11 feetfor both axes. The
ratiQs of end moments are 0.67 for thelarger moments and -0.25 for
the smaller moments. UsingFigure 30 connect 300 on the load scale
to the "bar size 8 11line, then up to a point midway between the S
= 0.8 and theB = 1.0 lines, then right to the "klu = 11" line and
finallydown to the IIP/4>Pcr" scale reading 0.305.
Interpolatingbetween Ml/M2 = 0.6 and 0.8, connect the previous
point onthe P/JstT L\NE..0 CUR..VE.. t::lLDN,4 e.,SE..rLE.CT Fbfl-
-51ZS \
