Secondary Moment and Redistribution in Continuos Prstressed Concrete Beam.pdf

SECONDARY MOMENT ANDMOMENT REDISTRIBUTIONIN CONTINUOUS PRESTRESSEDCONCRETE BEAMS

T. Y. Lin

Keith ThorntonUniversity of California T. Y. Lin & Associates

Berkeley, California

New York, New York

The subject of moment redistribu-tion in prestressed continuous beamshas received careful study(') forbeams with concordant cables*, butnot for those with non-concordantcables. In the latter case, what hap-pens to the moment produced byprestressing (the secondary moment)in the plastic or elasto-plastic stagehas become a subject of much dis-cussion(2.3^

The purpose of this paper is tobring up the serious nature of theproblem, particularly in the light ofthe 1971 ACI Building Code. Theexistence of secondary moments inthe post-elastic range will be clari-fied. One simple and conservativesolution is suggested, but no exactmethod is proposed.

1971 ACI Building Code'4 ) Iimitsthe percentage of moment redistri-bution in prestressed concrete con-tinuous beams in Sect. 18.12, some-what similar to that for conventional

' A concordant cable is so located to pro-duce a line of compressive force in theconcrete at each section that coincideswith the center of gravity of the steel(c.g.s.).

reinforced concrete in Sect. 8.6. Inaddition, it states, "The effect of mo-ments due to prestressing shall beneglected when calculating the de-sign moments." In the 1971 CodeCommentary( 5 ) to Sect. 18.12, it isfurther explained, "The secondarybending moments produced by theprestress force in a non-concord-ant tendon disappear at the capac-ity at which, due to plastic hinge for-mation, the structure becomes stati-cally determinate. Therefore, the de-sign load moments at the critical sec-tions of a continuous prestressedbeam are only those due to dead andlive loads."

The above statement on the disap-pearance of secondary moments,while correct by itself if properly in-terpreted, has been quite mislead-ing to many engineers, especiallywhen taken together with the limita-tions on moment redistribution.

It is well known that secondarymoments are produced by prestress-ing a continuous beam with a non-concordant c.g.s. line. It is also wellknown that when plastic hinges formin a continuous beam, converting itinto a statically determinate struc-

PCI Journal

Explains the 1971 ACI Building Code provisions for momentredistribution in continuous prestressed concrete beams. Emphasizes,by means of examples, when the secondary moments due toprestress force are to be included in determining the ultimatecapacity of continuous beams.

tare, the moments in the beam canbe computed, taking into accountonly the external dead and live loadsand the moment capacity at the criti-cal sections. To these moments, thesecondary moments due to prestress-ing need not be added. In fact,whether or not to add the secondarymoments will yield the same loadcarrying capacity for the beam, ifcomplete moment redistribution cantake place.

It has been stated(°), "Lineartransformation of the c.g.s. line doesnot change the ultimate load-carry-

ing capacity of a continuous beam."This means a non-concordant c.g.s.line can be linearly transformed in-to a concordant c.g.s. line withoutchanging the ultimate load capacity.Since this beam will now possess aconcordant c.g.s. line, it will have nosecondary moment. This phenome-non led to the statement in the ACIBuilding Code Commentary that thedesign (ultimate) load moments areonly those due to dead and liveloads.

All of the above, however, is predi-cated on full moment redistributionwith complete development of plas-

C- Line due to prestress28'1

c.g

SYMM.^80 , .4-

I8 14

________•[12

4Parabolic c.g.s.

80

Fig. 1. Beam elevation for Example 1January-February 1972

c.9.c.48°

3l

Fig. 2. Beam section for Example 1

tic hinges. If redistribution is notpermitted at all, i.e., if the beam isconsidered to be totally elastic, thenthe elastically computed secondarymoments due to non-concordant ca-bles are obviously there and shouldnot be neglected. However, theCode and its Commentary seem tosay that in calculating the designmoments, full moment redistribu-tion is not permitted and secondarymoments must be neglected at thesame time. This interpretation willlead to grossly incorrect answers,which could be conservative or non-conservative.

Since the problem cannot beeasily discussed in formulas, whichcould become quite lengthy andcomplicated, the nature and magni-tude of the problem will be illustrat-ed by two examples.

EXAMPLE 1

Consider a two-span continuousbeam, Figs. 1 and 2, prestressedwith a non-concordant parabolicc.g.s. line and bonded tendons; beamproperties are:

1. Section propertiesA, = 1440 in.2I = 253,340 in.4yt = 12 in.yb = 36 in.S t = 21,100 in.3S b = 7,030 in.3

2. Steel propertiesA, of tendons = 4.0 in.2

10

Ultimate strength of steelf 8 = 250 ksi

Effective prestress f$e = 150 ksiTotal effective prestress

Fe = 600 k.

3. Concrete propertiesUltimate compressive strength

f =5 ksiTensile stress at cracking =

6'/f ' = 424 psi

4. Secondary moments and reactions.With the c.g.s. line located as.

shown, it can be determined by in-spection s) that the C-line underprestress alone can be located by lin-early transforming the parabolicc.g.s. line 28 in. upward at centersupport, and 14 in. upward at mid-span. This results in a net eccentri-city of 36 in. up at center supportand 18 in. down at midspan. Thesecondary moment due to the effec-tive prestress is

600 x 28/12 = 1400 k. ft.

over center support, and 700 k. ft. atmidspan. Each end reaction due toprestressing is

1400 - 17.5 k. upward

80

and the corresponding secondary re-action at center support is

2 x 17.5 = 35 k. downward

To simplify our discussionthroughout this example, we willconsider the controlling +M at mid-span simultaneously with the —M

over center support. It is realizedthat, to get an accurate computationof the utlimate load, the exact loca-tion of controlling +M away frommidspan should be considered.

5. Balanced load due to prestressing.Since the parabolic c.g.s. has an

effective sag of

PCI Journal

h=32+8/2=36 in. =3ft.

uniform load balanced by prestressis

8Fh 8x600x 3w E = 80= -=2.25k./ft.

Thus, under an external load of 2.25k. /ft., the beam has a uniform corn-

pressive stress of 600,000 = 417 psi1440

along its entire length. Internal mo-

ment = 600 x 12 = —400 k. ft. over

center support and 600 x 322 = 1600

k. ft. at midspan.6. —M section at center support

To obtain zero tension at top fiber,it is necessary to negate the precom-pression of 417 psi by an additionalmoment,

_ _ 417 x 21, 100 =736k. ft.M — f iSa 12,000

which corresponds to a uniform loadof

=

8M = 8x736 = 0.92 k. /ft.w — L`-' 80^

Adding this to the balanced load of2.25 k. /ft., we have

wT= 0.92+2.25=3.17k./ft.

which is the total uniform load pro-ducing zero tension at top fiber overcenter support. The internal moment

—736 — 400_ —1136 k. ft.

To obtain a top fiber tension of424 psi, additional uniform load canbe computed by direct proportion-ing from the above 0.92 k./ft.

0.92 x 417 = 0.93 k./ft.

Thus, the total uniform load atcracking is

wT= 3.17+0.93=4.10k./ft.

and internal moment

= —1136 — 736 x 417

= —1880 k. ft.

To obtain ultimate moment ca-pacity with A8 = 4.0 in-., b = 12 in.,and d = 44 in., compute

4 = 0.756%p 12 x 44

which indicates an over-reinforcedsection. Using 1963 ACI Code

—M,, = 0.25f', bd2

=0.25x5x12x44

= —2420 k. ft.

Since no moment redistribution ispermitted for this over-reinforcedsection, the elastic moment over thecenter support, —M = wL2/8, mustbe used for calculating the ultimateload capacity. If, as per 1971 ACICode, moment due to prestressing isneglected, we will have an ultimateuniform load

8 M,,wu = Lz

8 x 2420= 802= 3.03 k./ft.

Note that this is less than wT = 4.10k./ft. at start of top fiber cracking,and even less than wT = 3.17 k./ft.under zero tension. This is obviouslyan incorrect answer as a result of theabove interpretation of the 1971 ACICode. The mistake results from theinconsistent assumption that, on theone hand, the beam is elastic underexternal load while, on the otherhand, it is not elastic under the effectof prestress.

If it is agreed that secondary mo-ments do exist, since the beam still

January-February 1972 11

SYMM.3520"

1400" 2820"'

_ ^700k'

1210k^ ^'– -2420"'

80 T 80'

(a ) (b)

(a) With no secondary moments (b) With secondary moments

Fig. 3. Ultimate load capacity assuming full moment redistribution

behaves elastically, then the ultimateload capacity should be computedincluding the effect of the secondarymoments. If secondary moment dueto effective prestress is used, wehave

,wu = 8 (Mu1t. + Msec.)L2

8 (2420 + 1400)802

= 4.78 k./ft.which is correct if no moment redis-tribution takes place.7. +M section at midspan

To obtain zero tension at bottomfiber, additional moment is

417 x 7030M = f`Sb = 12,000

= 244 k. ft.which for +M = wL2/16 at mid-span, means a uniform load of

16M 16 x 244W = Lz = 802

= 0.61 k./ft.

Wr = 0.61 + 2.25 = 2.86 k./ft.Internal moment = 1600 + 244

= 1844 k. ft.To obtain a bottom fiber tension of

424 psi, additional uniform load iscomputed

w = 0.61 x 417 = 0.62 k./ft.

PVT= 2.86+0.62=3.48k./ft.aad internal moment

= 1844 + 244 x 417= 2092 k. ft.

To obtain ultimate moment capac-ity with A8 = 4.0 in2., b = 120 in.,and d = 44 in., we have

4 = 0.076%p 120 x 44Using 1963 ACI Code

ts, = f 3 ( 1— 0.5p f$ /

=250 (1 — 0.5 x 0.00076 x 250 )= 245 ksi

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M = ASfsd (1-0.6p f")=4x245x44

x (1— 0.6 x 0.00076 x 2551

= 3520 k. ft.

As far as the ultimate moment ca-pacity of this section is concerned,neglecting both moment redistribu-tion and the secondary moment, i.e.simply using elastic moment coeffi-cients, we would have a theoreticalultimate load of

16MwT = Lz

_ 16 x 3520802

= 8.80 k./ft.

If the secondary moment of —700k. ft. produced by prestressing stillexists, it must be deducted from the

3250 k. ft. to give a theoretical ulti-mate load of

16(3520 — 700)WT = 802

= 7.05 k./ft.

again assuming elastic moments,with no moment redistribution.

8. Ultimate load capacity with fullmoment redistributionIf full moment redistribution were

possible, whether to include second-ary moments does not make any dif-ference. This is shown by the follow-ing two simple calculations, illustrat-ed by Fig. 3 for this two-span con-tinuous beam. The left half, Fig. 3(a), assumes no secondary moments.Corresponding to —M = -2420 k.ft.and +M = 3520 k. ft., the load ca-pacity is

wu — 8(3520 + 1210)802

= 5.91 k./ft.

9

8

7

wu (full redisitribution)6

wu (no redistri.) ^possible plastic behavior 3

5

A cracking4

w (no redistri.u no Msec)

ED zero tension

Moto 3C balanced S x y

seo4load 2os1oMS^\

/Cracking

Bi

i^wu(redistri.)

possible plasticbehavior

gn

e

-3000 -2000 -1000 0 1000 2000 3000 4000

- M at center support + M at midspan

(k/ft) (k/ft)

Fig. 4. Load and moment relationships for Example 1January-February 1972ҟ 13

Table 1. Loads and moments for critical conditions (Example 1)

Stress and Over center At midspanloading support

—M (k. ft.) wT (k. /ft.) +M (k. ft.) WT (k. /ft.)condition

Prestress only +1400 0 + 700 0Balanced load — 400 2.25 +1600 2.25Zero tension —1136 3.17 +1844 2.86

Cracking at424 psi —1880 4.10 +2092 3.48

Ultimate—noredistribution —2420 4.78 +3520 7.05

Ultimate—fullredistribution —2420 5.91 +3520 5.91

Prestress only 0 0 0 0Ultimate—no

°redistribution —2420 3.03 +3520 8.80

Ultimate—fullZ redistribution —2420 5.91 +3520 5.91

The right half, Fig. 3(b), considerssecondary moments, —M = 1400 k.

ft. and +M = 700 k. ft., and the loadcapacity is

8 [2820 + 2420 + 1400

Wu = 802= 5.91 k./ft.

This simple phenomenon justifiesthe ACI Code Commentary that thesecondary moments disappear, but itshould be emphasized •that this istrue only at full moment redistribu-tion. If the elastic moments are notfully redistributed, then the second-ary moments must also remain.

9. Summary of load and moment re-lationshipsThe above values of critical mo-

ments and corresponding uniformload intensity are now summarizedin Table 1 and plotted in Fig. 4.Along the Y-axis of Fig. 4 is plotted

the uniform load w. Along the X-axisare the —M over center support andthe +M at midspan. The solid linesshow the actual elastic moments pro-duced by external loads includingthe effect of prestressing (M ee). Thedotted lines show the elastic mo-ments, without considering the effectof prestressing (these, of course, donot represent the actual internal mo-ments, but are plotted for the pur-pose of comparison).

Now let us look at Fig. 4 and tryto visualize the actual ultimate loadcapacity. Let us assume that second-ary moments do remain, and let ususe elastic moments with no redistri-bution at all, the ultimate load isw2^ = 4.78 k./ft. Then let us assumefull moment redistribution, the ca-pacity is w„ = 5.91 k./ft. (with orwithout secondary moments). Forthis particular example, because ofthe high value of p = 0.756% for the—M section, plastic hinging action

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24 32°12° 4°

36 ^i^\ ^c.g.s 44" 36j

-r;:- -„ k”C-Line due to prestress

SYMM. C^80 , -I► 80

Fig. 5. Beam elevation for Example 2

cannot easily develop, and w,, isprobably only slightly higher than4.78 k./ft. However, if this section isnot so highly over-reinforced, somehinging action will develop and mo-ments will be redistributed. Thus,wu will fall between 4.78 and5.91 k./ft. It can be shown that theultimate load capacity will not fallbelow 4.78 k./ft., since the +M sec-tion is very much under-reinforcedin this beam.

The cracking of the —M and +Msections starts at points D and D' re-spectively. Since the +M sectionwill crack first (at w = 3.48 k./ft.),its moment will be gradually redis-tributed to the —M section startingat point E. Then when the —M sec-tion also starts to crack, momentswill be redistributed back to the+M section. Hence, the load-mo-ment curve will follow the dottedline EDB, approaching point B asa limit, if full redistribution isachieved. If moment redistribution isnot complete, this dotted line willnot end at point B, but at some pointbetween A and B( 7 ). It is just notpossible for this line to dip down-ward toward point C, as an improperinterpretation of the 1971 ACI Codewould indicate.

EXAMPLE 2In order to illustrate a different sit-

uation, when neglecting secondarymoments can result in a mistake onthe non-conservative side, we willnow discuss another 2-span continu-ous beam, Figs. 5 and 6. This beamis the same as the one in Example 1,except with the section upside down,changing into an inverted T-section.The section properties for the beamin Example 1 can be reversed andused for this section and will not belisted again.

1. Secondary moment and reactionsIt can be shown that the second-

ary moment at center support due toa prestress of 600 k. is

600 x 8 = —400 k. ft.

36 48"

8 ^— 120"

Fig. 6. Beam section for Example 2January -February 1972 15

8M 8 x 3520 4.40 k./ft.i°u = Lz 802Secondary exterior reaction = 480= 5 k. downward.2. Balanced load due to prestressing

Effective parabolic sagh=8+ 32 =24 in.

8Fhw = L^

_ 8x600x(24/12)802

= 1.50 k./ft.Thus under an external load of

1.50 k./ft., the beam has a uniformcompressive stress of 417 psi alongits entire length. Internal moment

= 600 X 12 = —1600 k. ft. over cen-

ter support, and 600 x 12 = 400 k. ft.at midspan.3. —M section at center support

Using calculations similar to Ex-ample 1, the following critical mo-ment and load values are obtained.

For zero tension at top fiber, addi-tional moment is

M _ 7030 x 417 = 244 k. ft.12,0008M=--=_---=0.30k./ft. 8 x 244

w L^ 802

Wr= 0.30+1.50=1.80k./ft.Internal moment = —244 — 1600 =—1844 k. ft.

For tension = 424 psi at top fiber

Wr = 1.80 + 0.31 = 2.11 k./ft.Internal moment = -1844 — 248 =—2092 k. ft.

Ultimate moment capacity = 3520k. ft. (from Example 1). If no mo-ment redistribution is permitted andsecondary moment neglected, theultimate load capacity is

If the secondary moment of —400k. ft. is considered, actual capacityleft for load-carrying is 3520 — 400 =3120 k. ft., which, with no redistribu-tion, yields

8x0120= _3.90k./ft.1Uu 802

This indicates that neglecting sec-ondary moment gives a non-con-servative value of 4.40 k./ft. for thisbeam.4. +M section at midspan

For zero tension at bottom fiber,additional moment isM = ftSt _ 417 x 21,100 = 736 k. ft.12,000TV = 16M = 16 x 736 = 1.84 k./ft.

LL 802wi, = 1.50 + 1.84 = 3.34 k./ft.Internal moment = 400 + 736

= 1136 k. ft.For ultimate moment = 2420 k. ft.,

neglecting secondary moment, (con-sidering only this section)

16 x 2420 = 6.05 k./ft.1Uu = 802

If secondary moment of —200 k. ft. isconsidered

16(2420 + 200) = 6.55 k. /ft.wu = 802

5. Ultimate load capacity with fullmoment redistributionSimilar to Example 1, it can be

shown that, assuming full momentredistribution, the ultimate load ca-pacity is

8(2420 + 3520/2) = 523 k./ft.wu = 802

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Table 2. Loads and moments for critical conditions (Example 2)

Stress and Over center At midspan

loading support

—M (k. ft.) WT (k./ft.) +M (k. ft.) WT (k./ft.)condition

Prestress only — 400 0 — 200 0

Balanced load —1600 1.50 + 400 1.50

Zero tension —1844 1.80 +1136 3.34

Cracking at424 psi —2092 2.11 +1884 5.21

0 Ultimate—noredistribution —3520 3.90 +2420 6.55

Ultimate—fullredistribution —3520 5.23 +2420 5.23

Prestress only 0 0 0 0bn

Ultimate—no

aoredistribution -3520 4.40 +2420 6.05

Ultimate—fullZ redistribution —3520 5.23 +2420 5.23

which value is good whether or notthe secondary moments are consid-ered.

6. Summary of load and momentrelationships

The above values of critical mo-ments and corresponding uniformload intensities are now summarizedin Table 2 and plotted in Fig. 7,similar to Fig. 4. From Fig. 7 it canbe observed that:

a. Point C is higher than Point Aindicating that if secondary mo-ments are neglected, a mistake ismade on the non-conservativeside.b. When cracking starts at PointD, moment redistribution will be-gin. Depending upon the devel-opment of both +M and —Mhinges, the ultimate load capacitymay fall anywhere between Point

A and Point B.c. The + M section may pick upmore moments than indicated bythe elastic values, owing to hing-ing at the —M section. Theamount of redistribution is subjectto further study.

This example thus shows that ne-glecting secondary moments mayyield a non-conservative result.

DESIGN METHOD

An approximate method for de-termining the ultimate load capacityof a continuous beam is proposed,based on a load-balancing approachwhich takes into account the second-ary moments produced by the ulti-mate prestress. For convenience, thismethod will neglect moment redis-tribution caused by plastic hingingaction.

This method assumes that, for un-January-February 1972 17

tension possible plastic behavior3

`' o

ooy /e^oy balanced load//

z ;ro

6 f CB wu (full redistri.)

wu(no redistri.)wa(no C possible plastic behaviorredistri.) q `

D ^e/zero te nsion o^rno41. 2_ \ sebalanced load M%sx4,. ^ \

ee

cracking

-4000 -3000 -2000 -1000 0 1000 2000 3000

– M at center support + M at midspan(k/ft) (k/ft)

Fig. 7. Load and moment relationships for Example 2

bonded tendons, the ultimate stresscapacity of the steel is at 1.20 X ef-fective prestress (the coefficient 1.20can be modified as required). It isfurther assumed that cable eccentri-cities in the elastic stage will be usedfor calculation. Hence, the load bal-anced by ultimate prestress is 1.20X load balanced by the effective pre-stress. This load-balancing approachconveniently takes into account theeffect of secondary moments at ulti-mate load. This method is quite ac-curate for unbonded tendons whosestress is more or less uniform alongtheir length. For bonded tendons,high stresses are concentrated atpoints of hinging, and this methodmay not be accurate enough.

This method is applied to theabove two examples.

Example 1

Under effective prestress, the bal-anced load is

Wbal = 2.25 k./ft.

and the internal moments are

18

—M = —400 k. ft. over centersupport

+M = 1600 k. ft. at midspan

If ultimate prestress is 1.2 x effec-tive prestress, the balanced ultimateload is

Wbal = 2.25 x 1.2 = 2.70 k./ft.

and the corresponding internal mo-ments are

—M = —400 x 1.2 = —480 k. ft.over center support

+M = 1600 x 1.2 = 1920 k. ft.at midspan

The moment capacities left for loadsabove the balanced load are (sinceultimate —M capacity = —2420 k. ft.,and +M capacity = 3520 k. ft.)

—2420 + 480 = —1940 k. ft.over center support

3520 — 1920 = 1600 k. ft.at midspan

Obviously, —M controls, and addi-tional load is

q

cracking 0 / wu (full redistri.)

4

PCI Journal

zU = 8 x81940= =, 2.43 k./ft.

Total load capacity at ultimate is

w=2.43+2.70=5.13k./ft.

As would be expected, this is slightlyhigher than the 4.78 k./ft. value pre-viously obtained when consideringthe secondary moments produced byeffective prestress.

Example 2

Similarly to the above, load bal-anced at ultimate prestress is

Wba1 = 1.50 X 1.2=1.80k./ft.

while the corresponding internal mo-ments are

—M = 1600 x 1.2=-1920k.ft.over center support

+M=400 x 1.2=480k. ft.at midspan

Moment capacities left to carry addi-tional load

—M = —3520 + 1920 = —1600 k. ft.

+M = 2420 — 480 = 1940 k. ft.

Again, —ivI controls, and the addi-tional load above balanced load is

8M _8x1600w = L2 =

802 = 2.00 k./ft.

Total load capacity at ultimate is

w = 2.00 + 1.80 = 3.80 k. /ft.

This is slightly lower than the 3.90k./ft. value previously obtainedwhen considering the secondary mo-ments produced by effective pre-stress. This is correct, because in-creasing the secondary moments by20% decreases the load capacity, inthis case.

This method is a conservative ap-proach, since it neglects moment re-distribution completely. It takes into

account the secondary moments dueto prestress without calculating forthem. It correctly considers the sec-ondary moments to increase insteadof disappear at ultimate capacity. Ifit is desired to consider moment re-distribution in the plastic range, ad-justments can be best made by con-sidering the load added beyond thebalanced load. It should be notedthat, under the balanced load, thebeam has no curvature nor deflec-tion anywhere.

CONCLUSIONS

A review of the above leads to thefollowing conclusions with respectto continuous concrete beams pre-stressed with non-concordant cables.While tee and inverted-tee sectionsare used in the examples, rectangu-lar or other sections will obviouslyfollow similar reasoning. In fact, thefollowing will apply generally tostatically indeterminate structuresprestressed with non-concordant ca-bles.

1. If elastic moments are used tocompute ultimate load capacity of acontinuous prestressed beam, sec-ondary moments produced by pre-stressing shall definitely be includedin the calculations.

2. If plastic moments with fullmoment redistribution are used tocompute ultimate load capacity of acontinuous beam, then secondarymoments may be either neglected orincluded, since the results will be thesame.

3. If plastic hinges do not fullydevelop, then ultimate load capacitywill fie between the two values com-puted in Items 1 and 2 above. Theexact capacity can be determined bytheoretical analysis, taking into ac-count the moment-curvature rela-tionships of the entire beam up to

January-February 1972ҟ 19

the ultimate load, although this isnot presented in this paper.

4. Approximate but conservativemethods can be devised to considerthe effect of secondary moments atultimate load, but to neglect momentredistribution. This can be con-veniently done using a load balanc-ing approach.

5. More analytical and experi-mental research should be carriedout concerning the plastic behaviorof prestressed concrete continuousbeams with non-concordant cables.However, it is abundantly clear thatSect. 18.12 of the 1971 ACI BuildingCode, if interpreted to neglectsecondary moments when full mo-ment redistribution does not occur,could lead to erroneous results.

REFERENCES1. Guyon, Y., "Statically Indeterminate

Structures in the Elastic and PlasticStates," General Report, First CongressFIP, 1953.

2. Mattock, Alan H., discussion of "Pro-posed Revision of ACI 318-63: Build-ing Code Requirements for ReinforcedConcrete," ACI Journal, Sept. 1970,Vol. 67, No. 9, p. 710.

3. Bondy, Kenneth, "New ACI BuildingCode (ACI 318-70)," Atlas EngineeringNewsletter, March 1970, pp. 3-4; June1971, pp. 2-5.

4. "Building Code Requirements for Rein-forced Concrete (ACI 318-71)," Ameri-can Concrete Institute, Detroit, Michi-gan, 78 pp.

5. "Commentary on Building Code Re-quirements for Reinforced Concrete(ACI 318-71)," American Concrete In-stitute, Detroit, Michigan, 96 pp.

6. Lin, T. Y., "Design of Prestressed Con-crete Structures," John Wiley & Sons,Inc., New York, New York, Second Edi-tion, 1963, 614 pp.

7. Lin, T. Y., "Strength of ContinuousPrestressed Concrete Beams Under Stat-ic and Repeated Loads," ACI Journal,June 1955, Vol. 26, No. 10, pp. 1037-1059.

Discussion of this paper is invited. Please forward your comments to PCI Headquartersby May I to permit publication in the May-June 1972 issue of the PCI JOURNAL.

20 PCI Journal