Shear and Torsion Design of Prestressed and Non ... · PDF filecan be used to design...

•'• Shear and TorsionDesign of Prestressed.i.

• and Non -PrestressedI•'' Concrete Beams

Michael P. CollinsProfessorDepartment of Civil EngineeringUniversity of TorontoToronto, Ontario

Denis MitchellAssociate Professor

Department of Civil Engineeringand Applied Mechanics

McGill UniversityMontreal, Quebec

D esign procedures which are based on rational models

rather than empirical equationsenable the engineer to develop abetter understanding of actualstructural behavior. In this regard,the unsatisfactory nature of cur-rent shear and torsion design pro-cedtires is evident if the ACICode' chapter on shear and tor-sion is compared with the ACIchapter on flexure and axial load.In the flexure and axial loadchapter a rational, simple, generalmethod is explained in a few par-agraphs of text.

On the other hand, the shearand torsion chapter consists of acollection of complex, restrictive,empirical equations which, whileleading to safe designs, lacks anunderstandable central philoso-phy. This lack, in the opinion ofthe authors, is the source of manyof the complaints which arise fromthe engineering profession aboutmodern design codes becomingunworkably complicated.

In this paper an attempt is madeto present procedures based onrational models which enablemembers containing web rein-

32

Shear and torsion design recommendations whichare believed to be more rational and more generalthan current code provisions are presented. Theuse of the design recommendations is illustrated bymeans of several design examples. Comparisonswith the results of other design methods are made.

forcement to be designed to resistshear andlor torsion.

In order to illustrate thecharacteristics of a rational modelof structural behavior, the paperfirst briefly reviews the theory forflexure and axial load. Then theprogress made in developingcomparable rational models fortorsion and shear is summarized.The way in which these modelscan be used to design prestressedand non-prestressed concretebeams for torsion and shear is ex-plained.

In addition, design proceduresfor combinations of flexure andshear and flexure combined withshear and torsion are presented.Minimum reinforcement require-ments, diagonal crack control re-quirements and detailing re-quirements are also discussed. Fi-nally, the recommended proce-dures are summarized in a set ofspecific design recommendations,the use of which are illustrated bymeans of several design examples.Derivations of the major equationspresented are included in three ap-pendices at the end of the paper.

Plane Sections Theoryfor Flexure and

Axial Load

The "plane sections" theory whichis capable of predicting the responseof prestressed and non-prestressedconcrete beams loaded in flexure andaxial load is described in several text-books (e.g., Refs. 2, 3 and 4). Thistheory will he briefly illustrated herein order to review concepts which willhe used in developing the models fortorsion and shear_

Assume that it is desired to find themoment-curvature relationship for therectangular prestressed concrete beamshown in Fig. 1. Since it is assumedthat plane sections remain plane onlytwo variables (say the concrete strainat the top, and the depth to the neutralaxis) are required to define the con-crete longitudinal strain distribution.For a chosen value of top concretestrain, a trial value of the depth ofcompression can be selected and theconcrete strain distribution will thenbe fixed.

The longitudinal concrete stressescan be found from the concrete strainsby using the concrete stress-straincharacteristics. Usually, it is assumed

PCI JOURNAUSeptember-October 1 9e0 33

(C } Equivalent Stresses (d) Stress Resultants

fct I

I I

E st I Eo

e 1 Concrete Stress-Strain

fpT 11111

Ecp+GEp

►y1 4Ultimate

Steel yield / ^

(b.lto,rn

lculated point M // ØL 'SM' M C •id

king on

cing on top '— L

ep 0I A6 p IEcp1(f ) Steel Stress-Strain (g) Moment -Curvature

Fig. 1. Plane sections theory for flexure showing various relationships.

that in compression the stress-straincurve obtained from a test cylindercan be used and that in tension theconcrete is not capable of resistingstress after cracking.

Due to the prestressing operationthe strain in the prestressing steel willbe substantially greater than the strain

in the surrounding concrete. Forexample, for a pretensioned beambefore release the concrete strain iszero while the prestressing steel has ahigh tensile strain. This difference instrain, se,,, which is caused by andcan he calculated from the specifics ofthe prestressing operation, is assumed

34

to remain constant throughout the lifeof the beam. For the concrete straindistribution being investigated, thestrain in the concrete surrounding theprestressing steel is known and henceby adding the strain difference, Acv,the total strain in the prestressingsteel, e p , can be determined. From thestress-strain characteristics of the pre-stressing steel, the stress, f, cone-sponding to the strain, e p, can be de-term med.

Knowing the stresses acting on thecross section, the resulting compres-sive force in the concrete and the ten-sile force in the steel can be com-puted. In the case of zero axial load,equilibrium requires that the com-pressive force in the concrete equalsthe tensile force in the steel. If thiscondition is not satisfied, the trialvalue of the depth of compressionmust be adjusted and the calculationsrepeated.

When the correct value of the depthof compression has been found, themoment corresponding to the chosenvalue of top concrete strain can thenbe calculated. This moment alongwith the curvature calculated from thestrain distribution, will give one pointon the moment-curvature plot. Re-peating the calculations for differentvalues of top concrete strain will pro-duce the complete moment-curvaturerelationship shown in Fig. 1.

The moment-curvature relationshippredicted on the basis that the con-crete cannot resist tensile stresses isshown by the solid line in Fig. 1(g).The dashed line in Fig. 1(g) indicatesthe predicted precracking response iftensile stresses in the concrete are ac-counted for. Also shown are thecracking loads for the beam whichwill of course depend on the tensilestrength of the concrete. Since thismember is eccentrically prestressed,the concrete on the top face will crackif the applied moment is too low.

In determining the magnitude and

position of the resultant compressionin the concrete, it is convenient to re-place the actual stress distributionwith an equivalent uniform stress dis-tribution. Thus, the distributionshown in Fig. 1(b) could be replacedby a uniform stress of a,f acting overa depth p lc, Fig. 1(c), where the stressblock factors a, and 13, have been cho-sen so that the magnitude and positionof the resultant compression do notchange. For a constant width of beam,the value of a, and R r will dependonly on the shape of the concretestress-strain curve, and the value ofthe highest concrete strain. The wayin which these factors may heevaluated for a particular concretestress-strain curve is shown in Appen-dix A.

In the AC! Code' the plane sectionstheory is the basis for determining themoment capacity. For this determina-tion the following additional assump-tions are made:

(a) The maximum moment willoccur when the compressivestrain at the extreme fiber is0.003.

(b) The value of the stress blockFactor a 1 is 0.85-

(c) The value of the stress blockfactor fi, is 0.85 for concretestrengths of 4000 psi or less andis reduced continuously by 0.05for each 1000 psi of strength inexcess of 4000 psi but 8 t shallnot be taken less than 0.65.*

These assumptions, of course, applyto both prestressed and non-pre-stressed members. in addition, theAC! Code' permits the use of an ap-proximate expression for the stress inthe prestressing steel at ultimate inlieu of a more accurate determinationbased on strain compatibility.

'For SI units f, shall he taken as 0.85 forstrengths fi LIP to 30 MPa and shall be reducedcontinuously at a rate of 0.08 for each 10 MPa ofstrength in excess of 30 MPa but ,B, shall not betaken less than 0.65.

PCI JOURNAL/September-October 1980 35

Truss Models forShear and Torsion

Early design procedures for rein-forced concrete members in shearwere based on the truss analogy de-veloped at the turn of the century byRitter' (1899) and by M6rsch" (1902).This theory, which assumes that con-crete is not capable of resisting ten-sion, postulates that a cracked rein-forced concrete beam (see Fig. 2) actsas a truss with parallel longitudinalchords and with a web composed ofdiagonal concrete struts and trans-verse steel ties. When shear is appliedto this truss, the diagonal struts go intocompression while tension is pro-duced in the transverse ties and in thelongitudinal chords.

Examination of the free body dia-gram of Fig. 2(b) reveals that theshear, V. is resisted by the verticalcomponent of the compression force,D, in the diagonal struts. The hori-zontal component of the compressionin the struts must be balanced by ten-sion in the longitudinal steel. Themagnitude of this tension will thus begiven by:

AN = V (1)tang

where B is the angle of inclination ofthe diagonal struts. It can he seenfrom Fig. 2(c) that the diagonal com-pressive stress, fd , is given by:

V.1(1= bd,.s iris cos6 (2)

where b, is the effective web widthand d„ is the effective shear depth.

Examination of the free bod y dia-gram of Fig. 2(d) shows that the ten-sion in a transverse tie is given by:

A u.ff = Vs tan8 (3)d„

In discussing the choice of theangle of inclination of the diagonals, 0,MSrschi in 1922 made the followingstatement:

"We have to comment with regardsto practical application that it is abso-lutely impossible to mathematicallydetermine the slope of the secondaryinclined cracks according to whichone can design the stirrups. For prac-tical purposes one has to make a pos-sibly unfavorable assumption for theslope 0 and therefore, with tan20 = w,we arrive at our usual calculation forstirrups which presumes B = 45 deg.Originally this was derived from theinitial shear cracks which actually ex-hibit this slope."

The equation for the amount oftransverse reinforcement neededwhich resulted from Morsch's as-sumption that 0 equals 45 deg becameidentified as the truss equation forshear.

Experience with the 45-deg trussanalogy revealed that the results ofthis theory were typically quite con-servative, particularly for beams withsmall amounts of web reinforcement.As a consequence, in North America itbecame accepted design practice toadd an empirical correction term tothe 45-deg truss equations. In the ACICode this added shear capacity istaken as equal to the shear at thecommencement of diagonal crackingand is often termed the "concretecontribution." As prestressing in-creases the diagonal cracking load, thebeneficial effects of prestress are ac-counted for in the AC! Code by in-creasing the "concrete contribution."

The truss analogy predicts that inorder to resist shear a beam needsboth stirrups and longitudinal steel.The AC! Code,' rather than specifyingthe amount of additional longitudinalsteel required for shear, gives rules forthe extension of the flexural rein-forcement (e.g., "reinforcement shallextend beyond the point at which it is

36

d bv

(o) Reinforced Concrete Beam in Shear

11 i I i t2

\ 4

2

ib) Longitudinal Equilibrium at Zero Moment Section

AN= Vton0

V = shear at section

^d cos 8

Se1l,fi

] Diagonal Stresses

_ p _ V -ttl b, d^cos6 b,d„sinecose

1^ T T f >< ^ 1 t t

td) Transverse EquilibriumII

' d„/toneV Aviv s

s4s s

d,/tan

Fig. 2. Truss model for shear showing various relationships.

no longer required to resist flexure fara distance equal to the effective depthof the member").

The recent CEB Code,' has recog-nized that the angle of inclination ofthe concrete struts is not in general 45deg. This code permits tan8 to bevaried between 3/5 and 5/3. Theselimits are a modification of the em-pirical limits determined by Lampertand Thtirlimann e for beams in torsion.It can be seen from Eqs. (1) and (3)that if a lower angle of inclination ofthe diagonal struts is chosen, then lesstransverse reinforcement but morelongitudinal reinforcement will he re-quired. Even though the CEB Codeallows a range of values for 0, it stillfinds it necessary to include an em-piricaI correction term (a "concrete

contribution") for lightly reinforcedmembers,

The truss analogy equations for tor-sion were first developed by Rausch1°in 1929. As in shear, it is assumed thatafter cracking the concrete can carryno tension and that the beam acts as atruss with longitudinal chords andwith walls composed of diagonal con-crete struts and transverse steel ties(see Fig. 3).

Fig. 3 illustrates that the torsion isresisted by the tangential componentsof the diagonal compression whichproduce a shear flow, q, around theperimeter. This shear flow is relatedto the applied torque by the equilib-riirm equation:

T = 2r1„q (4)

PCI JOUR NAL/September-October 1980 37

(a I Cracked Beam in TorsionDiagonal compressive stressesacting at angle 6

(b) Longitudinal EquilibriumLongitudinal components of diagonal compression,q/tone per unit length

^ T(c) Shear Flow Path,

f`.

T = 2A aq 1^ ^L '` / 1

Shear flow, q (d) Equilibriumper unit length of Corneraround perimeter p,,

0

Fig. 3. Truss model for torsion showing assumed forces acting on element.

where A,, is the area enclosed by theshear flow path.

The longitudinal component of thediagonal compression must be bal-anced by tension in the longitudinalsteel[see Fig. 3(h)], given by:

AN = q x) - – — Po (5)tang 2A Q tang

To balance out the horizontal com-pression in the concrete, the resultanttension force in the steel must act atthe centroid of the perimeter pa.

An examination of the equilibriumof a corner element, shown in Fig.

3(d), indicates that the force in eachhoop is:

A j, = s q tang = T tang (6)2A,,

Rausch, like Morsch, assumed H tobe 45 deg. In addition, he assumedthat the path of the shear flow coin-cided with the centerline of the closedstirrups. The resulting equations be-came identified as the truss equationsfor torsion.

In the ACI Code' the expressionsfor torsional strength consist of amodified form of the 45-deg truss

38

equations. These modifications,primarily based on the work of Hsu"and Mattock,' 2 consist of adding anempirical "concrete contribution" re-lated to the diagonal cracking load andreplacing the "2" in Eqs. (5) and (6)by an empirical coefficient which is afunction of the shape of the beam.While the ACI Code provisions do nottreat prestressed concrete members intorsion, the recent PCI Design Hand-book' 3 includes a torsion design pro-cedure for prestressed concrete whichis an extension of the ACI provisions.This procedure is based primarily onthe work of Zia and McGee."

The CEB Code" recognizes that fortorsion the angle of inclination of thediagonal struts is not always 45 deg.Again, this code permits tanO to varybetween 315 and 5/3. In addition,rather than using the centerline of theclosed stirrups as the shear flow path,the CEB Code, based on the work ofLampert and Thurlimann, 9 uses a pathdefined by a line connecting the cen-ters of the longitudinal bars in thecomers of the closed stirrups.

Comparisons between the amountsof shear and torsion reinforcement re-quired by the ACI and CEB Codes,and the authors' recommendationswill be given later in this paper.

Compression Field Theoryfor Shear and Torsion

Before the equilibrium equations ofthe truss analogy can be used to de-sign a member for shear and/or tor-sion, the inclination of the diagonalcompression struts must be known. In1929, Wagner'" dealt with an analo-gous problem in studying the post-buckling shear resistance of thin-webbed metal girders. He assumedthat after buckling the thin webswould not resist compression and thatthe shear would be carried by a fieldof diagonal tension.

To determine the angle of inclina-tion of the diagonal tension, Wagnerconsidered the deformations of thesystem. He assumed that the angle ofinclination of the diagonal tensilestress would coincide with the angleof inclination of the principal tensilestrain. This approach became knownas the tension field theory.

Applying Wagner's approach toreinforced concrete where it is as-sumed that after cracking the concretecan carry no tension and that the shearis carried by a field of diagonal com-pression results in the following ex-pression for the angle of inclination ofthe diagonal compression:

tan2 0 = Et + Ed (7)E t + @4

whereei = longitudinal tensile strainE, = transverse tensile strained = diagonal compressive strain

This geometric equation can bethought of as a compatibility relation-ship which links the strains in theconcrete diagonals, the longitudinalsteel and the transverse steel.

Using the compatibility condition ofEq. (7), the equilibrium equations ofthe truss, and the stress-strain re-lationships of the concrete and thesteel, the full behavioral response ofreinforced concrete members in shearor torsion can be predicted. This ap-proach is called the compression fieldtheory."

To demonstrate how the compres-sion field theory can be used to pre-dict response, imagine that we wish todetermine the behavior of a givenbeam subjected to a certain magnitudeof shear. The solution could com-mence by assuming a trial value of 0.Knowing 9, the tensile stresses in thelongitudinal and transverse steel andthe diagonal compressive stresses inthe concrete can he determined from

PCI JOURNALSeptember-Qctober 1980 39

Tension in hoop

tv, Compression Compression

in concrete in ConcreteOutside ofconcrete °o.

ryti

I^^ Tension in hoop _ I

UNSPALLED SPILLED

Fig. 4. Spalling of the concrete cover due to torsion.

the truss equilibrium relationships.Knowing the stress-strain characteris-tics of the reinforcement and thestresses in the reinforcement, thestrains and et can be determined,

Similarly, knowing the stress-straincharacteristics of the concrete and thestress in the concrete, the strain, ed,can be determined. The calculatedvalues of the strains can then be usedto check the initial assumption of the

angle of inclination of the diagonalcompression, 9. If the angle calculatedagrees with the estimated angle, thenthe solution would be correct. If itdoes not agree, then a new estimate of6 could be made and the procedurerepeated.

Thus, it can he seen that the com-pression field theory can predict theangle of inclination of the diagonalcompression.

El

a.<<

Fig. 5. Effective wall thickness of a twisted beam.

Members in TorsionIn applying the compression {field

theory to members in torsion, a fewadditional aspects of the behaviormust be taken into account. In resist-ing the torsion, not all of the concreteis effective in providing diagonalcompressive stresses. If the equilib-rium of a corner element for a beam intorsion (Fig. 4) is examined, it can beseen that the compression in the con-crete tends to push off the cornerwhile the tension in the hoops holds it

on. Since concrete is weak in tensionat higher torsions, the concrete out-side of the hoops spalls off. Because ofthis spalling it is assumed that the ef-fective outer surface of the concretecoincides with the hoop centerline.

If the deformed shape of the twistedbeam in Fig. 5 is examined, it can beobserved that the walls of the beam donot remain plane surfaces. Because ofthe curvature of the walls, the diago-nal compressive strains will be amaximum, e,,,, at the surface and will

PCI JOURNAUSeptember-October 1980 41

a,/2

A,

Hoop centerline

A.

ae/2

o, /2

Ao7

Ho opce^^e.l^ne

Fig. 6. Area enclosed by the shear flow for different member cross sections.

decrease linearly with the distancefrom the surface becoming tensile fordepths below a certain distance, ta.Thus, in torsion as in flexure, we havea depth of compression below whichwe may assume that the concrete,being in tension, is ineffective. Theoutside concrete spalls off and the in-side concrete goes into tension;hence, we are left with a tube of ef-fective concrete td thick which liesjust inside the hoop centerline.

The diagonal concrete stresses willvary in magnitude over the thicknessof the effective concrete tube fromzero at the inside to a value fd, corre-

sponding to the strain -,,, at the effec-tive outside surface. As in flexure wecan replace this actual stress distribu-tion by a uniform stress of c &' = fdacting over a depth of p, t d = ao wherethe stress block factors a l and f3, de-pend on the shape of the concretestress-strain curve and the value ofsurface compressive strain, e dd . Thecenterline dimensions of the resultingtube of uniformly stressed concrete ofthickness, a 0 , will define the path ofthe shear flow, q. This path will liea0/2 inside the centerline of the hoopas shown in Fig. 5. Knowing the pathof the shear flow, the terms A„ (the

42

Boo

Measuredhoop yield

600 '\,^9i Predictedhoop yield

C

a

. 400 Crackedprediction

0a=or BEAM P1 —a

Untracked Cover = 1/2 in.

200 prediction f^ 4680psi*3 hoops at 3,8 in, f = 47.5 ksiAj'=0.88 in? f- 475 ksi

A0.718 in? f,e166 ksiAE 0 -0.0059 In =214 ksi

010 05 1.0 1 5 2.0 R 10-3

TWIST {ad/in.)

Fig. 7. Measured and predicted torque-twist response for a prestressed beam.

{4

17^ ,

area enclosed by the shear flow) andp o (the perimeter of the shear flowpath) can he determined. Examinationof Fig. 5 shows that A. may be takenas:

ao (8)Ao"Aan—

j-Ph

where A ar, is the area enclosed by thecenterline of the hoop and pn is thehoop centerline perimeter. Theperimeter of the shear flow path, p.,can be taken as:

po = ph – 4a o (9)

The area enclosed by the shearflow, A o , for a variety of cross-sec-tional shapes is shown in Fig. 6.

As in flexure, the depth of compres-

sion will be a function of the tensileforces in the reinforcement. It can beshown (see Appendix B) that:

_ ,N + Acf` (10)a0 aJ, Ao alfcs

To illustrate the use of the compres-sion field theory for torsion, the pre-diction of the torque-twist curve forthe prestressed concrete beam shownin Fig. 7 will be described. The cal-culations would commence bychoosing a value for the diagonalcompressive strain at the surface ofthe concrete, eds . Knowing e,, and thestress-strain curve of the concrete, thestress block factors a l and I3 1 could bedetermined (Appendix A).

To determine the longitudinal andtransverse strains in the beam which


correspond to the chosen value of fde,

it is convenient to rearrange the basicequations (see Appendix B) to givethe following expressions:

1–L

f a113dr'Aohs – E482phAft

da

E __ r a lY]J r' 'A ahpa — 1 J

E (12)i IIL 2 p, AN r9

The strain in the prestressing steelis determined by adding the straindifference, to the longitudinaltensile strain, el . When Eqs. (11) and(12) have been evaluated, then thetension in the hoop A J, and the ten-sion in the longitudinal steel, AN, forthe chosen value ofet , will be known.Eq. (10) can then be used to calculatethe depth of compression, a 0 , andhence the terrnsA0 and p0.

Solving Eqs. (5) and (6) for the twounknowns, T and 0, gives:

T = 2A o (13)Po S

and

tang = —'f= ° (14)s ^N

Knowing 0, the twist of the beam forthe chosen value of E d„ caul be deter-mined by the geometrical relationshipgiven in Fig. 5. Repeating the abovecalculations for different values of €enables the complete torque-twist re-sponse to be determined.

The torque-twist curve given by thesolid line in Fig. 7 is based on the as-surnption that the concrete cannot re-sist any tensile stress. The dashed linerepresents the predicted precrackingtorque-twist response if tensile stress-es in the concrete are taken into ac-count. The torsion at which crackingoccurs depends, of course, on the ten-

Bile strength of the concrete and thelevel of prestress. An expression forthis cracking torque will he givenlater in the paper.

If the ohserved" experimental be-havior of the beam shown in Fig. 7 iscompared with the two theoreticalpredictions, it can he seen that prior tocracking the behavior closely followsthe uncracked member predictionwhile after cracking the behaviortends towards the fully crackedmember prediction.

To determine the ultimate torsionalstrength, it is not necessary to predictthe complete torsional response. As inflexure, it can be assumed that theload which corresponds to a concretestrain of 0.003 is the maximum loadthe section can carry. When using thisassumption to determine the torsionalcapacity, it is appropriate to use theACI stress block factors.

Members in ShearIn applying the compression field

theory to members in shear, it is againnecessary to take some additional be-havioral aspects into account. As intorsion, the unrestrained concretecover may spall off at higher loads(see Fig. 8). Once again, it is assumedthat after spalling the effective outersurface of the concrete will coincidewith the centerline of the stirrups.

In calculating the diagonal com-pressive stress, fr, it is assumed thatthe magnitude of the shear flow isconstant over the effective sheardepth d. Hence, the maximum valueof the diagonal stress will occur at thelocation of the minimum effectiveweb width, b,,, within the depth, d,,.In the truss analogy, d u is the distancebetween the top and the bottom lon-gitudinal chords. While dy could betaken as the flexural lever arm jd, wewill assume that d, is the vertical dis-tance between the centers of the lon-gitudinal bars which are anchoring theends of the stirrups,

44

< s,^^i ;,6^ r,q Sin zo 5 cZy t

Outside cconcrete

F^moconcrele

aP

sion X11 11

Tension

IF j_._- in stirrup'

/2

UNSPALLED SPALLED

Fig. 8. Spalling of concrete cover due to shear.

PCI JOURNALJSeptember-October 1980 45

f .> IA d

br

Flanges prevent coverfrom spoiling

d I/2 bY d Yd

I1/2 by

Diameter of duct bra bvz b„'da Grouted duct

jj2ibvt— y2dd)+bY,

Fig. 9. Effective shear area of members with various cross sections.

The assumed effective areas resist-ing shear for a variety of cross-sec-tional shapes are shown in Fig. 9. Theactual shear stress distributions for thecross sections shown will of course benon-uniform, However, the use of theeffective shear areas shown in Fig. 9will lead to conservative results. It hasbeen observed" , "" that the presence of

large post-tensioning ducts in thinwebs reduces the shear capacity. Thesuggested reduction in Fig. 9 comesfrom the CEB Code.e

The previously determined trussequations for shear plus the geometricrelationship for 0 can be used to de-termine the response of prestressedand non-prestressed concrete mem-

46

125

100

Predicted failure load

N 75 Uncracked BEAM CF 1prediction Side cover = 1/2 in.

6" Top and bottom cover = I inw =5600 psivim+ 50 24 * 3 stirrup at 6 in.,

Cracked 16 a . fY = 53.2 ksiprediction

A1= 0.66in` f =53 2ksiA p =1.436 in2, {=150 Ksi

25 ^t2" ^.LEp =a oo54 {pY =2iCi 5 Ksi

o Test values

OL0 5.0 Is.o z to -3

SHEAR STRAIN, y

Fig. t 0. Measured and predicted shear-shear strain response for a prestressedconcrete beam.

bers in shear by the procedure alreadyexplained.

The resulting predicted shearforce-shear strain response for a pre-stressed concrete box girder is shownin Fig. 10. The solid line representsthe predicted response based on theassumption that concrete cannot resistany tensile stress. The predicted pre-cracking response is represented bythe dashed line. Once again, the ob=serveds ' experimental behavior fol-lows the uncracked member predic-tion prior to cracking and tends to-wards the fully cracked member pre-diction after cracking.

In predicting the ultimate shearcapacity of members, it has beenfound necessary to limit the maximumcompressive stress, fd . It must be ap-preciated that fd is unlikely to reach

the cylinder crushing stress f, Apartfrom the problems of the actual dis-tribution of the principal compressivestresses (we have assumed a uniformstress distribution), this stress must betransmitted across cracked and se-verely deformed concrete.

Fig. 11 compares the failure condi-tions for the concrete in a test cylinderwith the failure conditions for diago-nally stressed concrete in a crackedbeam loaded in shear. It has beenproposed" that the size of the stresscircle that causes the concrete to fail isrelated to the size of the co-existingstrain circle. As an indicator of theintensity of strain, the maximum shearstrain, y.„ (i.e., the diameter of thestrain circle) is used. It has beensuggestedY2 that the maximum valueof f,1 be taken as:

PCI JOURNAL15eptember-October 1980 47

Cylinder

Y2

EauEa

Strain CircleStress Circle

T

..e a

^fdDiagonally CrackedConcrete S ress Circle

Strain Circle

Fig. 11. Comparison of stress and strain conditions for a test cylinder and fordiagonally cracked concrete.

0.40

0.30

i-nf^

0.20

0.10

00°

B

Fig. 12. Limits on angle of diagonal compression for torsion.

48

where

f = 5.5 f f (15) wheredu 4 + ym/e,rn = T"rah (19)2A,,

*ym = 2€4 + + €+ E l (16)

and en at failure is assumed to be0.002, that is, the strain correspondingto the peak concrete stress. When thecompressive stress f„ reaches thelimiting value, fdU , failure is predicted.For the beam discussed above, the re-sulting predicted failure load is shownin Fig. 10. It can be seen that the pre-diction is conservative.

Before the above expressions can beused, it is necessary to evaluate thearea enclosed by the shear flow, A0.

This area is a function of the depth ofcompression, a,,. By rearranging Eqs.(10), (5), (6), and (8), the following ex-pression fora,, can be obtained:

ao = Ann 1

ph

Design for l - T f tang + fTorsional Strength 0.85fÀ °n 1 ta"B

In many practical situations the sizeand shape of the beam together withthe amount of prestressed or non-pre-stressed longitudinal reinforcementwill already have been chosen tosatisfy other design considerations.The objectives of the torsion designthen become:

(a) Check if the section size is ade-quate to resist the design torsion;

(b) Determine the area of trans-verse reinforcement required to resistthe torsion; and

(c) Evaluate the additional lon-gitudinal reinforcement required toresist the torsion.

If the section size is inadequate, theconcrete will crush before the rein-forcement yields. The transverse andlongitudinal strains at the nominal tor-sional moment capacity T„ can be de-rived from Eqs. (5), (6), (11), and (12),as:

0.85 R_.f,Ao 1L tone –r 1 10-003âk (17)

f 0.85 13 fJA , lE t = L tang- 1 J

0.003rnA nA

(18)

(20)

It can be seen from the above equa-tions that the strain conditions at ulti-mate depend on the angle B.

The tasks of selecting an appropri-ate value of S, and then calculating Er

and e, are made considerably simplerby plotting Eqs. (17) and (18) in theform of the design chart shown in Fig.12. In plotting the chart a value of 0.80forî l was used.

As an example, let us assume thecharacteristics of the reinforcementare such that when e, = 0.002 thetransverse reinforcement will yieldand when er = 0.002 the longitudinalreinforcement will "yield." For thiscase it can be seen from Fig. 12 that ifr„If f equals about 0.3, then both typesof steel will yield only if B is about 45deg, On the other hand, if r,,/fr equalsabout 0.1 both steels will yield for anyvalue oiO between 16 and 74 deg.

The effect of choosing a lower valueof B is that, for a given torque, lesshoop steel but more longitudinal steelwill be required. Since hoop steel istypically more expensive than lon-gituclinal steel, the design engineermay wish to use the lowest possiblevalue of 0.

PC I JOUR NAL'September-October 1980 49

Having chosen an appropriate valueof 0 from Fig. 12, a o can he deter-mined from Eq. (20), the shear flowpath parameters A,, and pa can then befound from Eqs. (8) and (9), and fi-nally the required areas of reinforce-ment can be calculated from Eqs. (5)and (6).

Designing for CombinedTorsion and Flexure

The compression field theory de-scribed above is only strictly validwhen the longitudinal strain e, is con-stant over the whole section which re-stilts in a constant value of 0. A morecomplex version of the compressionfield theory has been developed2which is capable of predicting the re-sponse of reinforced concrete beamsunder combined torsion and flexure.Fortunately, for under-reinforcedbeams a simple superposition proce-dure produces accurate results.

In this superposition procedure thetransverse reinforcement is designedto resist the torsion by using the pro-cedure already explained. The lon-gitudinal reinforcement is then de-signed by the conventional plane sec-tions theory to resist the applied mo-ment plus the equivalent tension, AN,produced by the torsion.

To investigate the accuracy of thesimple superposition method outlinedabove, it was used to predict the tor-sion-flexure interaction relationshipfor a series of uniformly prestressed,symmetrically reinforced concretebeams which had been tested at theUniversity of Toronto. 24 The concretecylinder strength varied somewhatbetween the five beams of the testseries (individual values are shown inFig. 13) so in the calculations the av-erage value of 5.57 ksi (38 MPa) wasused.

The predicted interaction curve wasdetermined in the following manner:

For a chosen torque T the angle 9,which would result in the givenamount of hoop reinforcement, wasdetermined from Eqs. (6), (8), and(10). The axial tension AN, resultingfrom the torsion T with the calculatedangle 0, was then found from Eq. (5).Finally, the moment M which thegiven section could resist in combina-tion with the axial tension AN wasdetermined from the conventionalplane sections theory,

As can be seen from Fig. 13, thesuggested procedure predicts accu-rately the observed strengths of thetested beams.

The predicted and observed crack-ing loads, which are shown in Fig. 13,will be discussed later in the paper.

The theory predicts that as the ratioof torsion to moment decreases, theangle of inclination of the diagonalcompression 0 will increase. Predictedvalues of 0 range from 24 deg for puretorsion to 90 deg for pure flexure.While these predicted values of 0 willnot necessarily coincide with the av-erage inclination of the cracks (theywill coincide with the inclination ofcracks which form just prior to fail-ure), this inclination will provide anindication of the value of 9. If thbcrack patterns for the five testedbeams, which are shown in Fig. 14,are studied, it can be seen that thecrack inclinations are in reasonableagreement with the predicted valuesof 0.

In predicting the strengths of someof the beams shown in Fig. 13, it wasnecessary to apply the theory in re-gions where the longitudinal steel wasnot yielding. For example, in pure tor-sion B = 24 deg and z„Ife = 0.20which, as can he seen from Fig. 12,will correspond to e j = 0.6 X l0-3.

When the capacity of such membersis investigated by using the concept ofan equivalent axial tension, allowancemust he made for the fact that the lon-gitudinal steel will not actually yield

50

3 5"—2#5

276n wiresB 0 i

Concrete cover-1 in. top # 3 at 6 in.and bottom 17 fr=54.5 ksiV2 in. sides •. 3

2 0 5, f,=52.6 ksi8-0.276 in wires

A =24° fa.=I66 ksi—i2" —600 / 1f,=5.85ksi) I' foy=214ksi

• T64 T91 1R„= 247ksi

)f5.16ksi) \ (f6.53ksi)500 6=28^ TB2

400

{kip in) /`\

300 Observed cracking \ Observed ultimate oQ=50o ^•

TB32°D Predicted cracking (f^ =4.88 ks{ )

f'°5.57 ksiPredicted ultimate_

I00 l(f' 5.4 5 k si )165p =90°

0o 200 400 600 800 I OD 1 200 1400 1600 1800 2000 2200

M (kip in)

Fig. 13. Torsion-flexure interaction for a series of prestressed concrete beams.

under torsion while under direct ten-sion the longitudinal steel wouldyield. To allow for this effect, theequivalent axial tension AN as givenby Eq. (5) is increased by the ratio ofthe yield force of the longitudinalsteel to the force in the steel at a strainofez.

Designing for CombinedShear and Flexure

Procedures analogous to those fortorsion and flexure can he used to de-sign beams subjected to shear and

flexure. As in torsion, it is necessary tocheck that the section is of an ade-quate size to resist the applied loadsand then to determine the requiredamounts of transverse and longitudi-nal reinforcement.

To determine that the section size isadequate to resist the applied shear, itis necessary to check the transverseand longitudinal strains at ultimate.The following equations (see nextpage) which relate the strains at thenominal shear capacity, V,,, the angle0, and the applied shear can be ob-tained by rearranging Eqs. (3), (15),(16), and (7),

PCI JOURNAL15eptembes-October 1980 51

ia^^A*.. t^ 24

TB M413 • !^o -h^ •, ^+ ,

TB2T,/M=0.40 t^ s

34

T tl-0-!8 :::

.0 : ^ ^ O d 4 p ii

T8 5 fl Y f t•

Fig. 14. Crack patterns for five prestressed concrete beams.(Specimens TB4: TB1. TB2, T63, and TB5.)

52

5.5 sin0 cos0 – 4 Tn

f l I 0.002 (21)Tn (1 + tan20)

5.5 sines cosO – 4 T"

1 f` – 1 0.002 (22)

ff \ 1 + tan40l

where Tn = VnI(b rda).

Once again, the tasks of selecting anappropriate value of 0 and then de-termining if the section size is ade-quate are made considerably simplerif the above equations are plotted inthe form of a design chart similar toFig. 12. A modified form of this designchart will be presented in the follow-ing section.

Having chosen an appropriate valueof 0, the required area of transversereinforcement can be found from Eq.(3). The longitudinal reinforcementcan then be designed by the planesections theory to resist the appliedmoment plus the equivalent axial ten-sion, AN, given by Eq. (1).

Designing forCombined TorsionShear and Flexure

Although the compression fieldtheory has not yet been fully extendedto the case of beams loaded in com-bined torsion, shear, and flexure, asomewhat more approximate modelcalled the variable angle space truss isavailable Y5 and has been used as thebasis of a computer-aided design pro-cedure. 26 In this paper, an alternativesimplified, conservative design pro-cedure will he developed on the basisof the design charts for shear and tor-sion.

In Fig, 15 the design chart for tor-sion is compared with the design chartfor shear. The design charts can bethought of as defining the limits of Bfor a given level of stress and for givensteel strains. It can be seen that thetwo design charts have somewhatsimilar shapes and that either chartcould be reasonably represented bythe following equation:

10+ T"1f,' 35 < 0(0.42 – 50 el)

rn/J n

< 80 – (0.42 – 50 et) 3 (23)5

where the angle 0 is in degrees.

In applying the above design equa-tion for the case of combined torsionand shear, we will take:

= 1.s + n (24)n 2

A M bed.

In designing a member for com-bined torsion and shear, the nominalshear stress r„ would first be deter-mined. Based on the yield strength of'the reinforcement, appropriate valuesof e i and e1 would then be calculated.Eq. (23) could then he used to calcu-late the range of possible values of B.If the lower limit on 0 is calculated tohe higher than the upper limit on a it

PCI JOURNALJSeplernber-October 1980 53

means that the section size is inade-quate for the applied load.

After choosing an appropriate valueof 0 from within the allowable range(usually a value close to the lowerlimit of B would be chosen), theamount of transverse reinforcementrequired to resist the shear could hedetermined from Eq. (3), while theamount of transverse steel needed toresist the torsion could be determinedfrom Eqs. (6), (8), and (20). The re-quired amount of transverse rein-forcement is assumed to he the sum ofthe amount required for shear and theamount required for torsion.

If the shear alone were acting, thenthe longitudinal tension force re-quired could be determined from Eq.(1). If on the other hand the torsionwere acting alone, then the longitudi-nal tension force required could bedetermined by Eqs. (5), (8), and (20).If these two forces are simply addedfor the combined loading case, the re-suit would be a conservative design.This is because on one face of the

member the torsion and the shearstresses counteract reducing the totallongitudinal force required forequilibrium.

Comparisons with predictions fromthe variable angle space truss suggestthat a simple conservative procedurefor determining the required equiva-lent tension under combined loadingis to take the square root of the sum ofthe squares of the individually calcu-lated tensions. Thus, for combinedloading the longitudinal reinforce-ment is designed to resist the appliedmoment and axial load plus an addi-tional axial tension AN given by:

sAN A = 1 V,2 + f T -I.o (25)

tang ` 2Ao

For members not subjected to anexternally applied axial load, it maybe more convenient to design for anequivalent additional moment ratherthan an additional axial load. The lon-gitudinal reinforcement at a givensection could thus be designed as

0 50

040

0.30

Tnfl'

o 20

0 10

00 10 20 30 40 50 60 70 80

e DEGREES

Fig. 15. Design limits on angle of diagonal compression for shear and torsion.

54

flexural tension reinforcement to re-sist a positive moment of M.+ 1/z d,AN„ and a negative moment of ½ drAN„ – M. In regions of high positivemoment, M,, would exceed '/a d, AN.,indicating that top longitudinal rein-forcement is not required.

Minimum ReinforcementRequirements

To ensure ductile behavior offlexural members, the ACT Code' re-quires that the amount of longitudinalreinforcement provided in prestressedconcrete beams be large enough toensure that the flexural capacity is atleast 1.2 times the cracking moment.If the reinforcement is not capable oftransmitting the cracking load, thenthe member may fail in a brittle man-ner when the first crack forms. Toprevent such brittle failures for non-prestressed members, the ACT Codespecifies a minimum percentage offlexural reinforcement. This minimumreinforcement is necessary unless thereinforcement provided is one-thirdgreater than that required by analysis.

If the ACT Code philosophy forminimum flexural reinforcement isapplied to members subjected tocombined torsion, shear, and flexure,then either the reinforcement shouldbe designed to transmit at least 1.2times the cracking load or the rein-forcement should be designed totransmit four-thirds of the factored de-sign loads.

A simple, approximate procedurefor calculating the cracking loadsunder combined loading can be de-veloped from the following interactionequation:

1 T wrl E * ^ V.xr) • Mocr

1

where Tc ,., V c r , and M e,. are the crack-ing loads under combined loading

while Tar is the pure torsional crack-ing load, Vu,. is the pure shear crack-ing load, and is the pure flexuralcracking load.

The pure torsional cracking load canbe estimated27 as:

A 2Tr= A°4 f, 1+ fn` (27)PC `1fJ[

where p G is the outside perimeter ofthe concrete section, A, is the area en-closed by pc , and fr,,, is the compres-sive stress due to prestress at the cen-troid of the section. In Eq. (27) f, is inpsi units. If MPa units are used for fthe coefficients 4 should be replacedby 0.33.

The pure shear cracking load can beestimated as;

Var = b,nd4 f^ 1+ f^4^f

(28)

where b,m is the unspalled web widthand d is the effective depth of theflexural steel. Once again, f f is in psiunits. If MPa units are used, replacethe 4's by 0.33.

The pure flexural cracking load canbe estimated' as:

M^^ = I (7.5 [ + f) (29)yt

where Ily t is the section modulus ofthe beam, ff is the compressive stressdue to prestress at the extreme fiber ofthe section where tensile stress iscaused by the applied moment, and f^

is in psi. If MPa units are used, re-place 7.5 by 0.6.

The above procedure was used tocalculate the cracking loads for theprestressed concrete beams shown inFig. 13. It can be seen that the ob-served cracking loads agree rea-sonably well with the predictions.

PCI JOURNALSeptember-October 1980 55

Control ofDiagonal Cracking

Excessively wide cracks at serviceloads are undesirable. Obviously,cracking will not be a problem if theservice load is lower than the crackingload. This will often be the case forconcrete members which are pre-stressed. If the service load is greaterthan the cracking load, then it isnecessary to provide an adequateamount of well detailed reinforcementto restrain the opening of the diagonalcracks.

In 1974 ASCE-ACI Shear Commit-tee suggested 28 that "limiting themaximum stirrup strains at workingloads to 0.001... should prevent un-sightly inclined cracks at workingloads." In checking this suggestedlimitation, the compression fieldtheory can be used to predict thestrain in the transverse reinforcementat the specified service loads.

As can be seen in Fig, 16, the com-pression field theory, because it ne-glects concrete in tension, predictsthat the transverse reinforcement willcommence straining as soon as theload is applied. In reality, the trans-verse reinforcement will not begin tobe strained until cracking occurs.After cracking there will be a transi-tion between the uncracked conditionand the fully cracked condition as theloads go above the cracking Ioad_

Faced with an analogous problem indetermining a stiffness which liesbetween the uncracked value and thefully cracked value, the ACI Code'uses an empirical transition formulabased on the work by Branson. 49 Thisequation can he modified for our pur-poses to give the following expres-sion:

r 1 3— t— I Vrr ) I"

1 (30)

in which et. is the expected transversestrain at the shear V and Er is thetransverse strain predicted by the di-agonal compression field theory.

The effect of this transition curvefor a prestressed and a non-pre-stressed concrete beam is illustratedin Fig. 16. It can be seen that even ifthese two members have the same ul-timate load, the prestressed memberwill have much smaller strains at ser-vice load levels. There are two rea-sons for this more desirable behaviorof the prestressed concrete member;the cracking load is higher and thestiffness after cracking is larger.

In designing the section for ultimateIoads, the choice of the angle 0 deter-mines the relative amounts of trans-verse and longitudinal steel. If a verylow value of 0 is chosen, only a verysmall amount of transverse steel willbe supplied which may result in ex-cessive stirrup strains at service loads.Thus, it is possible to think of thecrack control limit as determining alower limit on 0. Rearranging thebasic equations and introducing somesimplifying assumptions (see Appen-dix C) results in the fallowing limit:

tang _( f VSP ] z ( ... L fxl x29 VRI 1 29ff)

i(L)3 JII

is

(31)

where V,,e is the shear at service loadand f,, is the yield stress of the trans-verse steel in ksi units. If MPa unitsare used, the coefficients 29 should bereplaced by 200.

Thus, in choosing the angle 0 re-quired to design the reinforcement atultimate loads, the following two limitstates must be considered:

1. To ensure that at ultimate thereinforcement yields prior to crushingof the concrete, 6 must lie betweenthe lower and tipper limits given byEq. (23).

56

r

theory

RESTRESSE p MEMBER

o ^ -C U y

•a } v

N N

TRANSVERSE STEEL STRAIN

a

0J

vsVcr

TRANSVERSE STEEL STRAIN

Fig. 16. Strain in transverse reinforcement at service load for a prestressed andnon-prestressed member.

00J

us

Vcr

2. To ensure that at service loadsthe strain in the transverse reinforce-ment does not exceed 0.001, 9 must begreater than the limit specified in Eq.(31).

These two limit states are comparedin Fig. 17. In preparing this figure itwas assumed that f,' = 5000 psi (35MPa), V.,IV, = 0.55, bt,dt,= 0.75 bd,

and that the cracking shears for thenon-prestressed and prestressed

beams were 2 V f b m d and 3.5\fh,,d, respectively (in MPa these val-ues would he 0.17 b,.d and 0.29J f bd). It can be seen that forf„ = 40 ksi (300 MPa) the crackinglimit on 8 is not critical. It can also beseen that prestressing the beam makesthe crack width limit ors 0 less restric-tive.

In order to ensure control of diago-nal cracking, it is necessary not only to

PCI JOURNAL1September-October 1980 57

fY=40ksi

Crushinglimit

0.4

0.2

® 0Tn

fc,

90° 0

Non—Prestressed

T

04

0.2

0

AsibIe

0ksi

4 -0

0.4J fY a 40 ksi Crack I fY 60 ksi

width ^7limit /

20.

0 f0 45° 90° 8 0 45° 90°Prestressed, f = 1000 psi

Fig. 17. Crushing limits and crack width limits for prestressed and non-prestressedconcrete beams.

provide a sufficient amount of trans- amount of reinforcement he providedverse reinforcement but also to limit but also it is essential that this rein-the spacing of reinforcement so that forcement be correctly detailed. A ra-an undesirable widening of the cracks tional model which enables the en-between the reinforcement does not gineer to understand the requiredoccur. It is suggested that provided functions of the reinforcement willboth the longitudinal reinforcement help the designer to avoid detailingand the transverse reinforcement have mistakes, particularly in unusual situ-a spacing equal to or less than 12 in. ations not adequately covered by code(300 nom), this requirement will be rules.satisfied. Since this requirement is A cracked beam in shear isconcerned with controlling cracking at idealized in Fig, 18(a) so that theservice loads, these spacing limits can functions of the reinforcement can bebe waived if the specified service visualized more clearly. The primaryloads are less than the cracking load, function of the stirrups is to hold the

beam together in the lateral direction.

Detailing ofThe distribution plates for the trans-verse steel shown in the idealized

Reinforcement model enable the concentrated stirruptensions to be distributed along the

In order that members subjected to length of the beam balancing the out-shear and torsion perform satisfacto- ward thrusts of the diagonal concreterily, not only must an adequate compressions.

58

0 45°

rn

0.4

0.2

0

`^aeiu i..y i^rW 701" r^nv ii =rv^^ar iiIIIUHIHIEnd

distributionplate I)

Longitudinal steel Stirrup anchorage points Transverseanchorage points / distribution plates

! 1

0

(a) Functions of the Reinforcement in anIdealized Truss Model

Longitudinalsteel anchoredin end region

{bl Properly Detailed Beam

Endon0oracfailure

(c} Poorly Detailed Beam Corner pushoutfailure

Fig. 18. Detailing considerations for a beam subjected to shear and/or torsion.

In a properly detailed beam [Fig.18(b)], longitudinal bars at the anchorpoints of the stirrups perform thefunction of the distribution plates. Inan improperly detailed beam [Fig.18(c)] with excessive stirrup spacing,the longitudinal bars will not be capa-ble of distributing the concentratedstimip tensions and hence a prema-ture failure may occur. For beams intorsion where the thrusts from twoadjacent faces must be resisted (seeFig. 4), it is suggested that the diam-

eter of the longitudinal bars in thecorners of the hoops should be at leasts tang/16.

The consequences of large stirrupspacing are further illustrated in Fig,19. The compression field theory as-sumes a uniform distribution of diag-onal compressive stresses over thebeam as shown in Fig. 19(a)_ Withwidely spaced stirrups these diagonalstresses will not he uniform along thelength of the beam but will becomeconcentrated at the stirrup locations

PCI JOURNALiSeptember-October 1980 59

Uniform field of diagonalcompressive stresses

5 3IonG

(a) Small Stirrup Spacing

High diagonal stressesI s'.f ^

A ee $<` Small diagonal stressese „ due to flexibility of

supports between stirrups

f n5 tan 9

lb) Large Stirrup Spacing

Fig. 19. Consequences of large stirrup spacing.

[see Fig. 19(b)]. These local concen-trations of stress may result in pre-mature diagonal crushing. To ensure areasonably uniform distribution ofstress, it is suggested that the stirrupspacing should not exceed do/(3 tan8).An analogous requirement for torsionwould limit the spacing of the hoopsto p h /(8 tang).

The three beams shown in Fig. 20illustrate what happens when thespacing limits suggested above are ex-ceeded. If Eqs. (3) and (23) areapplied to these beams, which wereamong the many hundreds of beamstested at the University of Toronto byG. N. J. Kani, 30 a shear capacity of 14,9kips (66.2 kN) and an angle 0 of 18.7deg would be predicted. Themaximum spacing would thus be dr/(3tan 18.7) = d. Beam 781, which had amaximum spacing of d, did not fail inshear. At a load corresponding to ashear of 13.4 kips (59.6 kN}, the lon-gitudinal steel in the central region ofthe beam yielded and the beam failedin flexure (see Fig. 20). Beam 782,

which had a maximum spacing of 2dand somewhat stronger flexural steel,failed in shear at 14.0 kips (62.2 kN)while Beam 783, with a maximumspacing of 3d, failed in shear at 10.3kips (45.8 kN).

As shown in the idealized model(Fig. 18), the transverse reinforcementmust be properly anchored. The ACICode' gives detailing requirementsfor the development of the transverseshear reinforcement. In applyingthese requirements, it should be keptin mind that at high shear stressesspalling of the unrestrained concretecover may occur. In torsion, becauseof spalling of the concrete cover, con-siderable care must he taken toachieve proper end anchorage of theclosed stirrups."

In the idealized model the primaryfunction of the longitudinal steel is tohold the beam together along its axis.The end distribution plate shown inthe idealized model [Fig. 18(a)] en-ables the concentrated tensile forcesin the longitudinal steel, which must

60

6". (;". I"FP nr;MM rai

5^sx3^,,^ R 13.4 kips 13,4 kips

LELMLI uSEa.r i

BEAM 783

Fil 2 CROSS-SECTIONAL PROPERTIESStirrups "r2 I T = 50 ksiConcrete f,= 3.88 ksi

1218 095• Side cover = 3/4"Longitudinol steel 23 + 28Yield strengths as shownd= 10.7"

.33 _6"6

Fig. 20. Three beams with large stirrup spacings

of course he anchored to the plate, tobe distributed over the end of thebeam. The tensile force in the steelbalances the outward thrusts of the di-agonal concrete stresses which areattempting to push off the end of thebeam.

If the end of an actual beam framesinto an adjacent member, then thismember can act as the end distribu-tion plate. Alternatively, end dia-phragms or end blocks could act asend distribution plates. If none ofthese conditions are met, then caremust be taken to provide proper endanchorage details. One such solution

involving well anchored and welldistributed end longitudinal steel isshown in Fig. 18(b). Fig. 18(c) illus-trates the consequences of not prop-erly detailing the end anchorage re-gion.

The beams shown in Fig. 19 do nothave end blocks or well distributedlongitudinal steel. For such cases thediagonal compressive stresses "fan-out" from the end hearing plate. Thespread of the fan will be defined bythe lowest angle of inclination, 9, andthe highest angle of inclination, 0,, ofthe diagonal compressive stresses,

If Eq. (23) is examined, it can be


imle III ICCIeiciiI

1

ave

ILb Distributed bottom

Reinforcement

- r4 3 6. ^ .^

It

H ^ -^ LAk' tom#e S} F^

d ye r

^^ t

450

tb

(c) Flanged Section

Fig. 21. Fanning of diagonal compression at ends of beams.

seen that for a given shear stress thesum of the lowest allowable angle andthe highest allowable angle is ap-proximately 90 deg. The two angles, Band 9e , defining the fan will usuallyhe the lowest and the highest allow-able angles, respectively, and hence itcan be assumed that 6 + Be = 90 deg.

The presence of the end tans (seeFig. 21) eliminates the need for ten-

sion in the top longitudinal steel nearthe end of the beam.32

While the fan reduces the total lon-gitudinal tensile force required in thesteel at the end of the beam, it in-creases the compressive stresses inthe concrete. In checking for crushingof the concrete in the end region, itcan be assumed that the diagonalcompressive stresses are distributed

62

ai d Y M1,,;..,,. Equwalent additional2 `' tan gmoment due to niear

Fig. 22. Fanning of diagonal compression under concentrated load.

over the depth of the fan, d,,, at theedge of the bearing plate (see Fig. 21).

The depth, d,, should be taken as totan6P = l/tanO, Fig. 21(a), unless amore detailed analysis indicates thatwell-anchored longitudinal steelspread over the depth of the beam,Fig. 21(b), enablesd,,e to be increased.In checking web crushing for flangedmembers, the effective bearinglength, 1r„ can he increased since thecritical section will no longer occur atthe edge of the bearing plate [Fig.21(c)].

If it is assumed that the tensilestrains are negligible in the confinedregion near the bearing plate, thenfrom Eq. (23) crushing will be avoidedif:

yn , (B – 10) 0.42 f (32)b a d, 35

One further effect caused by thefanning out of diagonal compressivestresses from a concentrated load isillustrated in Fig. 22. The tension inthe longitudinal steel caused by sheardecreases as the angle of inclination of

the diagonal compression increases.Directly under the load, where 0 = 90deg, the shear will not cause any in-crease. As a consequence, the area oflongitudinal steel in this region neednot exceed the area required for themaximum flexure.1,32

Distribution ofTransverse Reinforcement

Design procedures to determine therequired spacing of transverse rein-forcement in regions of constant shearand/or torsion have already been ex-plained. In regions of changing shearssome additional factors need to beconsidered.

Fig. 23 compares three beamswhich have the same magnitudes ofapplied loading per unit length andhence the same shear force diagrams.The first beam has loading applied atthe top face, the second has loadingapplied at the middle of the side face,while the third beam has the loadingapplied near the bottom face. For thefree body diagrams shown, vertical

PCI JOURNAL'September-October 1980 63

SHEAR DIAGRAMw„L

2

i Vu3 Vu2 yui

I d„ /2fan8' d v /2 ton &Shift for Shift for

W bottom loading top loadingV

BEAMTopLoading qbAV fr dV

S i 2ton&-VuI

uv/TOn t1 7

wuBEAM 2 } 1SideLoading rt ^A2{^ dv

SZ 21anB Vu2sZ

BEAM 3Bottom W^

Loading ^Avfy dvS3

53 2tanB Vu3

Fig. 23. Required shear strengths for top-loaded, side-loaded, and bottom-loadedbeams.

equilibrium requires that the totaltension in the stirrups crossed by thecut equals the end reaction minus theloading applied to the left of the cut.For Beam 2 this total force equals theshear at the midpoint of the cut. ForBeam 1 this total force which the stir-rups must resist is the reduced shearwhich occurs d,/(2 tan g ) to the right of

the midpoint of the cut. The increasedshear which occurs, d cl(2 tang ), to theleft of the midpoint of the cut must beresisted by the stirrups in Beam 3.

Fig. 24 illustrates the way in whichthe distribution of transverse rein-forcement is determined for a typicalbeam. From the given loading, thefactored shear force diagram can be

64

d„

Stirrups at Stirrups at Minimumispacing s a Spacing 5 b stirrups

! VuyRequiredua

Capacity- -

2 oAyfy dv 1 ^ v Minimumn

^Avfy: dv5 tpn9 V..a

stirrups

5p taIle_ huh - -, :%:

iaan f nA Shift of requireda shear strength

Fig. 24. Required transverse reinforcement for a uniformly loaded beam.

determined. For this top loaded beam,the shears for which the stirrups are tohe designed are found by "shifting"the shear force diagram a distance ofd v/(2 tan g ) towards the support asshown by the dashed line. Over theIength dr,Itanfl equilibrium will besatisfied if the stirrups are designedfor the average shear force over thislength.

The net effect of the shifting andthe averaging of the shear forces for a

top loaded beam is that the transversesteel within a length, d,ItanO, is de-signed for the lowest factored shear,V w , within this length. For a beamloaded near its bottom face, designingthe transverse shear for this "lowestshear" would lead to insufficientreinforcement. For this case, it wouldhe necessary to add additional trans-verse reinforcement capable of trans-mitting this load to the top face of themember.

PROPOSED DESIGN RECOMMENDATIONS FORBEAMS IN SHEAR AND TORSION

The design procedures which havebeen discussed above have beensummarized in the form of specificdesign recommendations. In for-mulating these recommendationssome of the design equations were

generalized so that the recommen-dations could cover a wider range ofpractical problems. in particular, in-clined stirrups, inclined prestressingtendons, and variable depth membershave all been included.

PCI JOURNAL/Sepiember-October 1980 65

1.0 — Notationa, = equivalent depth of compres-

sion in torsionA, = area enclosed by perimeter of

cross section, p,A,, = gross area of concrete cross

sectionA, = area enclosed by shear flow

pathA, = area enclosed by hoop center-

lineA, = cross-sectional area of one leg

of a closed stirrupA, = cross-sectional area of shear

reinforcement within a dis-tance s

hr = minimum effective web widthwithin depth d D (see Section1.6.1)

b. = minimum unspalled webwidth within depth d

d = distance From extreme com-pression fiber to centroid oflongitudinal tension reinforce-ment

d r, = effective shear depth can betaken as flexural lever arm butneed not he taken less thanthe vertical distance betweencenters of bars or prestress-ing tendons in corner of stir-rups

d, = effective shear depth at end ofbeam

E, = modulus of elasticity of steel= specified compressive strength

of'concrete= compressive stress in concrete

at centroid of cross section dueto prestress (after allowancefor all prestress losses)

= compressive stress in concretedue to effective prestress forc-es only (after allowance for allprestress losses) at extreme 11-her of section where tensilestress is caused by externallyapplied loads

= stress in prestressing steelwhen strain in surroundingconcrete is zero

f, = stress in prestressing steel atnominal strength (see Section18.7 of ACI 318-77)

= specified yield strength ofnon-prestressed reinforce-ment

I = moment of inertia of gross con-crete section resisting exter-nally applied factored loads

1, = effective length of bearingarea taken as actual length ofbearing except that for mem-bers with flanges where thebearing area is wider than theweb 1 b is taken as the actuallength of bearing plus the ver-tical distance from the outerface of the flange to the junc-tion of the web and the flange

Meg = cracking moment under com-bined loading

M„ = nominal moment strengthM,. = pure flexural cracking strengthM, = factored flexural momentN = unfactored axial loadN. = factored axial loadAN. = equivalent factored axial load

caused by shear and torsionp, = outside perimeter of concrete

cross sectionph – perimeter of hoop centerlinepo = perimeter of shear flow paths = spacing of shear or torsion re-

inforcement in direction paral-lel to longitudinal axis

1'c, = torsional cracking moment un-der combined loading

T„ = nominal torsional momentstrength provided by circula-tory shear flow

T,c,. = pure torsional crackingstrength

1'. = factored torsional momentV. = cracking shear under com-

bined loadingV. = nominal shear strengthVim,. = pure shear cracking strengthV, = vertical component of effective

prestressing force or for vari-able depth members the sumof the vertical component of

66

the effective prestressing forceand the vertical components ofthe flexural compression andtension

V„ = service load shearV s = factored shear forceeft = distance from centroidal axis of

section to extreme fiber intension

a = angle between inclined stir-rups and longitudinal axis ofmember

= factor accounting for nun-yielding of the longitudinalsteel under shear and/or tor-sion defined in Section 1.9.1

X3 1 = concrete stress block factor &1e-fined in Section 10.2,7 of ACI318-77

E, = tensile strain of longitudinalreinforcement due to shearand/or torsion

eau = yield strain of transverse re-inforcement

6 = angle of inclination to longi-tudinal axis of member (indegrees) of diagonal compres-sive stresses

X = factor to account for light-weight concrete defined inSection 1.4.1

r„ = nominal shear stress= strength reduction factor de-

fined in Section 9.3 of ACI318-77

1.1 —ScopeThese recommendations are con-

cerned with the design of prestressedand non-prestressed concrete beamssubjected to shear or shear combinedwith torsion.

1.2 — General Principles andRequirements1.2.1 — Beams shall be designed tohave adequate strength, adequateductility, and satisfactory performanceat service load levels.1.2.1.1 —To ensure adequatestrength, the transverse reinforcement

shall be designed in accordance withSection 1.8 to resist the applied shearand torsion, the longitudinal rein-torcement shall be designed in accor-dance with Section 1.9 to resist theapplied flexure, shear, and torsion,while the section size shall be pro-portioned in accordance with Section1.6 to avoid diagonal crushing of theconcrete.1.2.1.2 — To ensure adequate ductil-ity, the members shall be designed tosatisfy the minimum reinforcementrequirements of Section 1.3 and themaximum reinforcement requirementsspecified in Section 1.6.1.2.1.3 — To ensure adequate controlof diagonal cracking at service loadlevels, members shall be designed tosatisfy the requirements of Section1.7.1.2.2 — Reinforcement detailing re-quirements of Section 1.10 shall besatisfied.

1.3 — Minimum ReinforcementRequirements1.3.1 — Amount of reinforcement in amember shall he chosen such that areserve of strength exists after initialcracking.1.3.2 — Requirement of Section 1.3.1may be waived if the reinforcement isdesigned to resist factored loadsone-third greater than those deter-mined by analysis.1.3.3 — The requirements of Section1.3,1 will be satisfied if the amount ofreinforcement at any section is suchthat the nominal sectional strengths,Tx , V, and M„ are at least equal to 1.2times the cracking loads, T^„ VMcr , determined in accordance withSection 1.4.1.3.4 — For members not subjected tomoving loads, requirements of Section1.3.3 need to be investigated only atlocations of maximum moments.

1.4 — Cracking Loads1.4.1 — In lieu of more exact analysis,


the cracking loads under combinedflexure, shear, and torsion can be de-termined from the following interac-tion equation.

^ uo,r ^ ^ + ^ V xr ^ ^ +

1 T nar!

= 1

(1-1)

where

M«. T = ('lye) (7.5A .f' + f,,,) (1-2)

(Stresses are in psi units; for MPaunits replace 7.5 by 0.6.)

Vag. = b w d 4X Vfc 1 + fx114X y .=c )

+ VP(1-3)

(Stresses are in psi units; for MPaunits replace 4 by 0.33.)

Tr = (A C-1p,) 4A .,fJ

1 +f/(4x ^.f,) (1-4)

(Stresses are in psi units; for MPaunits replace 4 by 0.33.)where

A = 1.0 for normal weight concreteA = 0.75 for "aIl-lightweight"

concrete andA = 0.85 for "sand-lightweight"

concrete1.4.2 —Eq. (1-4) can be used for hol-low sections provided the least wallthickness is not less than 0.75A,/p,.1.4.3 — In calculating the crackingloads, the ratio of torsion to shear,

can be assumed equal toTb/V,,.1.4.4 — In calculating the crackingloads, the ratio of moment to shear,M cr fV c„ can be assumed equal toM„/V U . However, this ratio shall riotbe taken as less than d.1.4.5 — Influence of axial loads on themagnitude of the cracking loads canbe accounted for by increasing (for

compression) or decreasing (for ten-sion) both f,K. and ff by N/AQ.

1.4.6 — For members not subjected totorsion or to axial load, cracking shearV, need not be taken as less than2A Y T b,,d (stresses in psi units; forMPa units replace 2 by 0.17).

1.5 — Consideration of Torsion1.5.1 — if the magnitude of the fac-tored torsional moment, T., as deter-mined by an analysis using uncrackedstiffness values exceeds (0.25T.,),then torsional reinforcement designedin accordance with Sections 1.8 and1.9 shall be provided. Otherwise, tor-sional eflects may be neglected.1.5.2 — In a statically indeterminatestructure where reduction of torsionalmoment in a member can occur due toredistribution of internal forces, themaximum factored torsional moment,T, tnay be reduced to O(0.67T...)provided that the corresponding ad-justments to the moments and shearsin adjoining members are made.

1.6 --- Diagonal Crushing of theConcrete1.6.1 Nominal shear stress shall becomputed as:

T rt _ R - V P) + I T I ] h (1 -5)b r. dr AA

1.6.1.1 — In determining b, the tmre-strained concrete cover down to thecenterline of the outer transversereinforcing bar shall be assumed tohave spalled off; however, b = need notbe taken less than one-half of theminimum unspalled web width, b.1.6.1.2—In determining theminimum effective web width, b x , thediameters of ungrouted ducts or one-half the diameters of grouted ductsshall be subtracted from the webwidth at the level of these ducts.1,6.2 —Cross-sectional propertiesshall be chosen such that the trans-

68

verse reinforcement will yield prior todiagonal crushing of the concrete.1.6.3 — Requirements of Section 1.6,2may he considered satisfied if it ispossible to choose an angle 0 withinthe following limits:

10+ 35(;n jf,) <a(0.42– 50e1)

<80– 35(Tn1fc) (1-6)

(0.42 -- 65et„)

where the value of E t can be chosen.However, the selected value of e t mustalso be used in satisfying the re-quirements of Section 1.9.1.

1.7 — Control of DiagonalCracking1.7.1 —Cross-sectional propertiesshall be chosen to ensure adequatecontrol of diagonal cracking at serviceloads.1.7.2 — Requirements of Section 1.7.1may be considered satisfied if thecracking loads as determined by theprocedures of Section 1.4 exceed theservice Ioads.1.7.3 — For uniformly loaded simplysupported beams, crack control re-quirements need only be investigatedfor sections one-quarter of the spanlength from the supports.1.7.4 — Requirements of Section 1.7.1ma y be considered satisfied if thefollowing three conditions are met:

(a) Calculations show that the strainin the transverse reinforcementat service loads does not exceed0.001.

(h) Spacing of transverse reinforce-ment does not exceed 12 in. (300mm).

(c) Spacing between longitudinalreinforcing bars at the crackedfaces of the member does notexceed 12 in. (300 mm).

1.7.4.1 — Requirements of Section1.7.4(a) may be considered satisfied if

in calculating the area of transversereinforcement either the value of fa, istaken as not greater than 40 ksi (300MPa), or the value of 0 is such that;

V l 2 _tangy 29 V p 1 29f[][

xJ 1–( . ) J

E (1-7)L S/

where f„ is in ksi units (for MPa unitsreplace 29 by 200).1.7.4.2 — Bonded prestressing ten-dons may he considered equivalent tolongitudinal reinforcing bars in satis-fying the requirements of Section1.7.4(c).

1.8 — Design of TransverseReinforcement1.8.1 — Transverse reinforcementprovided for shear shall satisfy thedetailing requirements of Section 1.10and may consist of:

(a) Stirrups perpendicular to theaxis of the member;

(Ib) Welded wire fabric with wireslocated perpendicular to the axisof the member;

(c) Stirrups making an angle of 45deg or more with the longitudi-nal axis of the member.

1.8.2 — Transverse reinforcementprovided for torsion shall satisfy thedetailing requirements of Section 1.10and may consist of:

(a) Closed stirrups perpendicular tothe axis of the member;

(b) A closed cage of welded wirefabric with wires located per-pendicular to the axis of themember;

(c) Spirals.1.8.3 — Transverse reinforcementprovided shall be at least equal to thesum of that required for shear and thatrequired for torsion.1.8.4 — In determining the requiredareas of transverse reinforcement, the


values of 8 chosen must satisfy thelimits specified in Eq. (1-6).1.8.5 —Transverse reinforcement re-quired for shear shall be determinedfrom the requirement that V _- 0 V,,.1.8.6 —When shear reinforcementperpendicular to the axis of themember is used;

(1-8)V s tang + V

1.8.7 —When inclined stirrups areused as shear reinforcement:

V _ A rf„ sinsR – d1, + coca 1 + Vp

S tang(1-9)

1.8.8 —Transverse reinforcement re-quired for torsion shall be determinedfrom the requirement that T,, = 0 T,,.1.8.9 — Nom inal torsional momentstrength shall be computed by:

Tn = A efv 2A4 (1-10)s tang

1.8.10 — The area enclosed by thetorsional shear flow, A„ may be com-puted as A,, – a.p5/2.1.8.11 — The equivalent depth ofcompression in torsion, a 0 , may hecomputed as:

ao =A°h 1–Ph

I– T,, ph f tang + __2.__ 110.85f r 20 ` tang

(1-11)

1.8.12 — For hollow sections in tor-sion, the smallest thickness from thecenterline of the stirrup to the insideface of the wall shall not be less than

1.8.13 — For uniformly loaded beams

where V. and T. change linearly, thetransverse reinforcement requiredwithin a length d0Itan8 may be deter-mined by using the lowest values ofV,and T. which occur within this lengthprovided that 9 is chosen to satisfy Eq.(1-6) by using in determining r R thehighest values of V u and T. whichoccur within this length.1.8.14 — For those beams for whichthe provisions of Section 1.8.13 do notapply, the transverse reinforcementrequired within a length d,/tan6 maybe determined by using the averagevalues ofA als andA tis calculated fromEqs. (1-8) or (1-9) and (1-10).1.8.15 — In regions near supportswhich introduce direct compressioninto the member, the transverse rein-forcement may be designed by usingthe values of V. and T which occur ata distance de,/(2 tang ) from the face ofthe support provided that 9 is chosento satisfy Eq. (1-6) by using in deter-mining T. the values of V. and T.which occur at the face of the support.1.8.16 —When downwards load isapplied at the bottom face of a beamadditional transverse reinforcement,capable of transmitting in tension theapplied load to the opposite face ofthe beam, shall be provided.1.8.17 -- When a downwards load isapplied to the side faces of a beam,then additional transverse reinforce-ment shall be provided. The amountof additional transverse reinforcementrequired may be assumed to varylinearly with the position of load onthe side face going from zero whenthe load is at the top to the amountspecified in Section 1.8.16 when theload is at the bottom.

1.9 — Design of LongitudinalReinforcement1.9.1 — The longitudinal reinforce-ment shall be designed by the planesections theory described in Chapter10 of AC! 318-77 to resist the factoredmoment, M y , and the factored axial

70

load, N., together with an additionalfactored axial tension, ON acting atmid-depth of the stirrups and givenby:

(V, — 4 Vn)2+ 1 reT^o

tan8 (2Ao

(1-12)

where S„ is a function of the value of E,

used in satisfying Eq. (1-6) and 0 isthe value used in the design of thetransverse reinforcement.1.9.1.1 — For non-prestressed beams,,8, can be taken as fr,l(E,E,) but not lessthan one.1.9.1.2 — For prestressed beams, /3.can be taken as f„al(f^ + E,e) but notless than one.1.9.1.3 — Perimeter of shear flowpath, p, may be computed as)Pn – 4uQ.1.9.2 —The ratios of longitudinalreinforcement shall be such that therequirements of Sections 10.3.3 and18.8.1 of ACI 318-77 are satisfied.

1.9.3 — For members not subjected toaxial load (N,, = 0), the requirementsof Section 1.9.1 will be satisfied if thesection is capable of resisting a fac-tored moment equal to M,, ± '12 d^,AN„.1.9.4 — When an interior support or aconcentrated load responsible formore than 50 percent of the shear atits location introduces direct compres-sion into the flexural compression faceof a member, then for those sectionscloser than d„ltanO to the support orthe load, the area of longitudinalreinforcement on the tension sideneed not exceed the area required toresist the M , and N,, which exist at thenearest section where maximum mo-ments occur. However, if torsion ispresent, then the sections must be ca-pable of resisting an additional ten-sion force of Typ 0/(2A o tanO) at themid-depth of the stirrups.

1.9.5 — At the loaded free ends ofcantilevers or at the ends of simplysupported beams where the loads orreactions introduce compression intothe end regions, the longitudinalreinforcement in these regions shallbe designed according to the follow-ing provisions.1.9.5.1 - The longitudinal reinforce-ment near the flexural tension faceshall be anchored such that at theinner edge of the bearing area a fac-tored moment of Mu + 1/2 d„4N„ canhe resisted.1.9.5.2 — For those sections closerthan dt.Itane to the inner edge of thebearing area, the longitudinal rein-forcement near the flexural compres-sion face shall be designed such thatat each section a factored moment of:

do ^r. Tu po — M2 tang 2A o

M.

causing tension on the flexural com-pression face can be resisted.

1.9.5.3 — Axial tensile loads, N,applied to ends of cantilevers orbeams shall he resisted by appropri-ately anchored additional longitudinalreinforcement.

1.9.5.4 —The cross-sectional proper-ties and bearing area dimensions shallbe such that at the inner edge of thebearing area:

Vn d Vi + Azih''-- 0.012 (0 - 10)f,r re o

(1-13)

where 0 is the angle used in deter-mining the amount of transversereinforcement. If 0 is greater than 45deg, the term (0 – 10) shall be re-placed by (80– 0). The term d„e shallbe taken as lbltan0 unless a more de-tailed analysis indicates that well an-chored longitudinal reinforcementspread over the depth of the beam en-ables d. to be increased.


1.10 — Reinforcement Details1.10.1 — Transverse shear and torsionreinforcement shall be anchored ac-cording to Section 12.14 of ACI 318-77except that in regions of unconfinedconcrete cover, torsion reinforcementshall be anchored by means of 135-deg standard hooks.1.10.2 — A longitudinal reinforcingbar or prestressing tendon shall beprovided at each interior corner oftransverse reinforcement. The nomi-nal diameter of this bar or tendonshall be at least equal to the diameterof the stirrup but shall not be less thanV2 in. (12 mm).1.10.2.1 — Nominal diameter of thebar or tendon provided in each cornerof closed transverse reinforcement re-

quired for torsion shall not be lessthan s tanG116.1.10.3 — Where shear reinforcementis required, its spacing shall not ex-ceed do/(3 tang).1.10.4 — Where torsion reinforcementis required, its spacing shall not ex-ceed p,,!(8 tan&).1.10.5 — Except at supports of simplespans and at free ends of cantilevers,the longitudinal reinforcement shallextend beyond the point at which it isno longer required to resist any stressfor a distance equal to twelve timesthe bar diameter. The additional re-quirement of Section 12.11.3 of ACI318-77 to extend this reinforcementfor a distance equal to the effectivedepth of the member may be waived.

Comparisons With OtherDesign Procedures

Fig. 25 compares the predictions ofthe compression field theory (CFT),the ACI Code, and the CEB Code forfour different series of beams. Thecompared predictions relate theamount of transverse reinforcement(expressed by the non-dimensionalratio, A rff,I(bsf,) to the resultingshear strength (expressed by thenon-dimensional ratio, r„If, ).

In preparing this figure, it was as-sumed that ff = 5000 psi (35 MPa),V./V n = 0.55, b y = 0.8 b,n , d„ = 0.9 d,

and that the cracking shears for thenon-prestressed and prestressedbeams were 2Jf"b,4 and 3.5VJ'b ,4 ,respectively (0.17 \iffb Wd and 0.29.Jffb,g1 for MPa units)_

To facilitate comparisons betweenthe ACI and the CEB approaches, thematerial strength reduction factors

used in the CEB Code were taken asunity. Further, the crack control re-quirements of the CEB Code whichare expressed in terms of the stirrupspacing were not applied.

It can be seen that for lightly rein-forced beams, the three methods givequite similar results. However, formembers subjected to very high shearstresses there is a considerable di-vergence between the predictions ofthe three methods. It is worth notingthat the ACI upper limit was primarilyintended to control cracking at serviceloads. For the non-prestressed beamswith the higher strength reinforce-ment, where the compression fieldtheory predicts that cracking at serviceloads will control, this ACI upperlimit looks appropriate.

Further comparisons between thereinforcement required by the com-pression field theory and that requiredby other design methods are given inthe three design examples at the end

72

Tn

0,4{-

0,4

0.2

3 bws{s 0

Non— Prestressed

T^

04—

0.4

CE8CFT

0.2ACI i

0 {r=40ks1n o.l 0.2 O.

ty=60ksi A y fYI0.1 0.2 0.3 bsf^

0.2 0.2

flyf,= 4o ks A f or fy=60ksi

w Y

of this paper. From these examples itcan be seen that the amount of rein-forcement required by the compres-sion field theory is in general compa-rable to the amounts required byexisting empirical design procedures.

Concluding Remarks

This paper has presented designrecommendations, based on the com-pression field theory, which enableprestressed and non-prestressed con-crete beams to he designed for shearand for torsion.

The compression field theory,which is a development of the tra-ditional truss model for shear and tor-sion, considers in addition to the trussequilibrium conditions, geometriccompatibility conditions and material

stress-strain relationships. Like theplane sections theory for flexure andaxial load, the compression fieldtheory is capable of predicting notonly the failure load but also the com-plete load-deformation response.

The design recommendations de-veloped from the compression fieldtheory have the following features:

1. The expressions that relate therequired areas of reinforcement to thedesign loads are the classical trussequilibrium equations.

2. Rather than setting the angle ofdiagonal compression, 0, to 45 deg asin traditional North American practiceor choosing it within the empiricalrange of 31 to 59 deg as in recentEuropean practice, the range of allow-able angles is calculated based onconsiderations of concrete crushingand control of diagonal cracking. For


highly stressed members, H will be re-stricted to a narrow range of valuesclose to 45 deg while for lightlyloaded members the range of allow-able angles will be very wide (in thelimit 10 deg _- e -_ 80 deg).

3. In determining shear and torsioncapacities, the tensile strength of theconcrete is ignored. Because 0 canvary within such wide limits, it is notnecessary to introduce an empiricalcorrection term (a "concrete contribu-tion") to account for the strength oflightly reinforced members.

4, Prestressed concrete beams,non-prestressed concrete beams, andpartially prestressed concrete beamshaving a wide variety of cross-sec-tional shapes can all he designedusing the same basic expressions. Thebeneficial effects of prestressing areaccounted for in the design by allow-ing lower values of a to be used whichwill result in less transverse rein-forcement.

The compression field theory as-sumes that diagonal compressivestresses can be transmitted throughcracked concrete. Transmitting diago-nal compressive stresses across a crackwhich is inclined to these stresses willrequire shear stresses to be transmit-ted from one face of the crack to theother. These interface shear stressesare transmitted by aggregate inter-lock4 and by the dowel action of thereinforcement crossing the crack.

If an excessively wide crack forms

across the whole member depth, themechanisms of "interface sheartransfer" will deteriorate and maycause a premature failure_ For thisreason the design recommendationsshould not be applied to sectionssubjected to reversed cyclic loading ofsuch an intensity that excursions intothe yield range of the flexural rein-forcement are expected.

In certain situations the presentrecommendations may lead to overlyconservative designs. Further re-search will probably result in refine-ments of the design recommendationsin the following areas:—Local effects caused by concen-

trated forces;—Minimum reinforcement require-

ments;—Effectiveness of large unrestrained

concrete covers;—Influence of variable web width;—Effects of indirect loading; and—Torsional resistance provided by re-

strained warping stresses.The authors believe that the rec-

ommendations presented in this paperare more rational and more generalthan the shear and torsion provisionsof current North American codes.Hopefully, the design method pre-sented will enable engineers to de-velop a better understanding of thebehavior of beams in shear and tor-sion, an understanding which shouldresult in more efficient and economi-cal designs.

74

DESIGN EXAMPLESTo illustrate the proposed design

method, three numerical examples arepresented: (1) single tee, (2) spandrelbeam, and (3) bridge girder.

DESIGN EXAMPLE 1—PCIStandard Single Tee

The single tee floor beam shown inFig. 26 has been constructed withsand-lightweight concrete weighing120 pcf and topped with 2½ in. ofnormal weight concrete. The beamsupports a live load of 75 psf andspans 70 ft.

1. Determine factored loads

Dead load:

570x 120 + 2=5 x 150 x 8144 12

= 475 + 250 = 0.725 kips/ft

Live load

8 x 75 = 0.600 kips/ft

Therefore, factored uniform load perunit length:

ru„ = 1.4 x 0.725 + 1.7 x 0.600=2.035 kips/ft

2. Check minimum reinforcementrequirements (Section 1.3)

Since this member is not subjectedto moving loads, it is only necessary tocheck Section 1.3.3 at midspan. As-sume the effective prestress equals150 ksi. Therefore, prestressing force:

P = 13 x 0.153 x 150 = 298 kips

Hence, bottom fiber stress due toprestress:

fPe = A + ZPe

b

298 298 x 23.01=-+570 2650

= 3.110 ksi

Note that untopped section proper-ties have been used in determiningjam . The dead load of the topping plusthe dead load of the precast beam willcause a moment at midspan of

O.725< 702 x 12 = 5330 kip-in.S

which will reduce the bottom fiber

2 viz topping 8-0'

BLST 36 I 'I7 3i$ in. cover to stirrups="lsupport)

36" d=3322

" midspond

l TOPPING-normal weight7"long x 8"wide _ f, =4000 psibearing pad 8L ST 36-sand-lightweight (120 pct}

L ______________ 000 psiPRESTRESSING- 13-1/2 in. diameter

'=5

270 K strandsSingle depression

Span = 70' 0=1^- 8 at midspanp40,000 40,000 psiL.L. = 75 psf

Fig. 26. Design Example 1--Design of a lightweight single-tee member.


precompression by 5330/2650 = 2.011 Factored shear force:ksi.

To crack the beam the bottom fiber V u = 2.035 x 35 = 71.2 kipsmust reach a tensile stress of

7.5A \lfc = 7.5 x 0,85 x ^ 5000=0,451ksi

Nominal shear strength:

V,, = 71.2/0.85 = 83.8 kips

Hence, additional moment to crack Minimum effective web width:beam:

3142(3.110-2.011+0.451)=4870 b„=8-2x 8 – 8kip-in,

(unrestrained cover neglected)(Note that 3142 in. 9 isZ 5 of composite Effective shear depth:section.)

Thus, cracking moment under com- dt = 36 – 2 – 1 = 33 in.bined loading:

From Eq. (1-5), nominal shearMe ,. = 5330 + 4870 = 10,200 kip-in. stress:

Factored flexural moment:

M =wLz = 2.035x702x12

"+ 8 8

= 14,960 kip-in.

Hence, to provide the flexuralstrength required, the nominal mo-ment strength:

M„ : M„f4, = 14,96010.9= 16,620 kip-in.

Therefore, the nominal to crackingmoment ratio:

M nl M,r = 16,620110,200= 1.63> 1.20

Hence, minimum reinforcement re-quirements will be satisfied.

3. Check cross-sectional size andchoose angle 0 (Section 1.6)

Calculate the nominal shear stressat face of support. For simplicity,neglect the vertical component of theprestressing force, Vp, at this stage.

,rn = V, = 83.8 – 0.368 ksib u d,. 6.9 x 33

Therefore, r.1f,' = 0.36815 = 0.074.Forf„ = 40 ksi, yield strain oftransverse reinforcement:

Eau = 40/29,000 = 0.00138Use €,, = ct,.

To avoid crushing of the concrete,the angle of inclination, B, must satisfythe limits given by Eq. (1-6). Thus:

10 + 35 (T„Ifs) <

0.42 – 50e,

< 80 – 35 (r,,/ f) f )0.42

35 x 0.07410+

<00.42 – 50 x 0.00138

< 80 – 35 x 0.0740.42 – 65 x 0.00138

17.4 deg < d < 72.2 deg

Therefore, the section size is ade-quate. Choose 9 = 20 deg.

76

Although different values of 8 couldbe used in designing different regionsalong the length of the beam, it willtypically he more convenient to usethe same value of 0 over the length ofthe beam.

While the smallest allowable valueof 0 (in this case 17.4 deg) will resultin the smallest amount of transversesteel, it may result in an excessivelylarge amount of longitudinal steel.Hence, it is usually prudent to choosea value of B somewhat larger than thesmallest allowable angle.

4. Check diagonal crack controlrequirements (Section 1.7)

Compare cracking loads (Section1.4) and service loads at £14 from sup-port (Section 1.7.3).

Calculate the service load shear atL/4 from support:

V. = 0.5 x 35 (0.725 + 0.600)= 23.2 kips

Calculate the shear required tocause diagonal cracking. For simplic-ity use Section 1.4.6:

Vcr = 2x Ifsb d

=2x0.85,i5000<8x

(22+ 332

28.9 kips

Since V cr > V, the crack controlrequirements are satisfied.

5. Design of transverse reinforcement(Section 1.8)

For a given shear the transversereinforcement required can be deter-mined by rearranging Eq. (1-8) togive:

Afr^ n - Vus tan8 – V:.

If Grade 40 #3 U stirrups are used,then:

A,,f„ d _ 0.22 x 40 x 33 _ 798s tang s tan20 s

For this uniformly loaded beam thetransverse reinforcement within eachlength of d„Itano is designed for thelowest value of shear within thislength (Section 1.8.13).

_!L _ 33 = 90.7 in. (or 7.6 ft)tang tan20

This results in the "stepped" re-quirement for transverse steel shownin Fig. 27. The stirrup spacings cho-sen to satisfy the requirements arealso shown in Fig. 27.

Note that maximum spacing of shearreinforcement (Section 1.10.3) is

d„1(3 tan8) = 33/(3>< tan20) = 30.2 in.

6. Design of longitudinalreinforcement (Section 1.9)

To show that the longitudinal rein-forcement is adequate, we must dem-onstrate that at each section the beamcan resist a factored moment equal toM. + ' d, rAN„ where AN. is theequivalent axial tension produced bythe shear (Section 1.9.3). The calcula-tions for these equivalent additionalfactored moments and for the nominalflexural capacities along the length ofthe beam are shown in Table 1.

The resulting required values ofM,,are compared with the values of M,provided by the prestressing strand inFig. 27. It can be seen that the lon-gitudinal reinforcement is adequateexcept near the end support.

To cover the 2000 kip-in. momentdeficiency near the support, addi-tional reinforcing bars welded to ananchor plate will be provided. If3—#6 Grade 40 bars are used the ad-ditional moment capacity providedwill he approximately:

PC€ JOURNALISeptember-October 1980 77

100 -6^

#3I a 3@2D'

—3®3O"SHEAR Vu(kips 1 — Q.85

SHEAR

sa I VA CapacitypIAGRAM

provided

Support L c.face

Required capacitya-- —

tonA 76_-7,6

•=7.6 35-0"

Beam Support Laceend —

U

0.9

Deficiencyat support LN,^face ? 2000 kipin 0.5 d„ 0 a5 0000

Required Mh

MMOMENTDIAGRAM

(kip my

M„ provided20 000

Fig. 27. Design Example 1—Shear and moment diagrams for design of stirrups andlongitudinal reinforcement.

3 x 0.44 X 40 X 37.5 = 1980 kip-in.

V„ _- 0.012 (8 – 10) f,'b d,

If the length of the bearing pad is 70.5 <0.012 (20 – 10) 5too small, then additional well an- (0 85 x 8) dux

shored longitudinal steel spread somedistance up the web will have to be Therefore, d 17.3 in.provided to avoid local crushing of the The 7 in. long bearing pad will pro-concrete (Section 1.9.5.3). Check vide an effective shear depth of:bearing pad length by Eq. (1-13) suchthat at the inner edge of the bearing do/tang = 7ltan20 = 19.2 in,area: which is satisfactory.

78

Table 1. Additional factored moments required to be resisted and nominal momentcapacities available.

Distance from end of beans 7 in. 25 in. 5 ft 3'/z in. IS ft 3% in. 25 ft 3½in. 35 ft 3 1/2 in.

d (in.) 24.6 25.1 26.2 29.3 32.5 35.5

f. [ Fig. 3.9.9, PCI Designhandbook and Eq. (18-3), 42.0 150 227 264 264 265AC1318-77) (ksi)

f va=^n = r

f v+ 1.0 1.0 1.0 1.0 1.15 1.15 1.16 0.7 x 270 + 40

Section 1.9.1.2)

V,, (kips) 70.6 67.6 61.1 40.7 20.4 0

N (degrees) (vatin chosen) 20 211 211 20 20 20

` tanO 194 186 165 129 64 0

[Eq. (1-12)] (kips)

Va d,, AN. (Section 1.9.3) 3200 3070 2770 2130 1060 0(kfp-in.)

%=A,fp,(d-alt)= 13 x 0.153f. 2040 7350 11520 14960 16640 18290

13x0.153xf'p,d —

2x0.85x 4xr96^(kip-in.)

Note: I It = 0.305 in; I i,,. - 25.4 mm; 1 ksi =' 8.895 MPa; I kip-in. = 113 N • m

Table 2. Comparison of web reinforcement requirements by compression fieldtheory and ACI 318-77 methods.

DDesign method

Location

Near About 12 ft About 24 ft Near support from support from support midspan

Compression field theory #3 4 12 in. #3 (&- 20 in. #3 @ 30 in. 0ACI 318-77 #3 @ 20 in. #3 @ 11 in, #3 @ 20 in. 0\nte; I in. — 25.1 mm.

7. Comparison with PCI DesignHandbook results

It is of interest to compare the re-sults given above with the solution tothe same design problem given by the

PCI Design Handbook.' 3 The Hand-book uses the shear provisions of ACI318-77. Table 2 compares the webreinforcement required by the twoapproaches.


DESIGN EXAMPLE 2—Parking Dead load of beam:Garage Spandrel Beam

1.4x 0.725 = 1.02 kips/ftThe typical precast, prestressed

concrete parking garage spandrel Dead load of deck:beam shown in Fig. 28 spans betweencolumn corbels. The double-tee floor 1.4 x 0.0895 x 60/2 x 4 = 15.04 kipsframing system spans 60 ft, carries a per stemlive load of 50 psf, and is supported onthe ledge of the spandrel beam. Live load:

1. Determine factored loads on the 1.7 x 0.050 x 30 x 4 = 10.2 kips perspandrel stem

24"x 24" column A

Angle welded to plate in column

24„

10 x 8y neo p rene pod LjA

24"x 24"columnConnection angle

SECTION A-Ae 3,.

75 e„ fif 5000 psi, normal weight 24,f^, = 40 ksi V80T

6-1/2 in. dia., 270 K strands d =69 in.Cover to stirrups = I %4 in.

Corbel

Fig. 28. Design Example 2—Structural framing of a precast prestressed concreteparking garage spandrel beam.

80

The loads and the resulting values loading M e,. at midspan. Assume ef-of M, V,, and T, are shown in Fig. fective prestress equals 150 ksi.29 . Therefore, prestressing force:

2. Check minimum reinforcementrequirements (Section 1.3)

To check Section 1.3.3, calculatethe cracking moment under combined

P = 6 x 0.153 x 150 = 138 kips

Therefore, bottom stress due to pre-stress:

,^4'-o" —a-o" ^^

2 kips

I4 2' H

u^Ln k+p ft

101 100

75.3 71.2

46.041.9

12.6 v#€s7 u

kips

58.8

Tu&4 kip ft

Fig. 29. Design Example 2-Moment shear, and torsion diagrams forspandrel beam.


e A Zs

138 138 x 27.2696 10990

= 0.540 ksi

Therefore, cracking moment:

Mrr = Z (7.5 J' +.fire)

_ (10,990/12) (0.530 + 0.540)= 980 kip-ft

Since the maximum value of M„(578 kip-ft) is considerably smallerthan 1.2 M r , we will use Section 1.3.2and satisfy the minimum reinforcingrequirements by increasing all designloads by one-third,

3. Check cross-sectional size andchoose angle 0 (Section 1.6)

We wish to check crushing of theconcrete in the web of the beam at theface of' the support. The load which isapplied by the stem which is almostdirectly over the corbel can be trans-ferred through the ]edge of the beamto the support without causing shearstresses in the web of the spandrelbeam. Hence:

V, = 75.3 X 4/3 100.4 kipsV„ = 100.4/0.85 = 118.1 kipsT„ = 42.0 x 12 x 413 = 672

kip-in.T„ = 672/0,85 = 791 kip-in.b„ = 8 - 2 x 1.5 = 5.0 in.(unrestrained cover neglectedand #4 stirrups assumed)d =75-2x2=71 in.(2 in, assumed from outsideconcrete surface to center ofcorner reinforcement)A c,, = 5x 72+ 8x 9=432 in.2ph =5±72+13+9+8±63

= 170 in.

From Eq. (1-5), the nominal shearstress:

=--+in ph

bvdv Aon

118.1 791 x 170_ +

5 x 71 4322

= 1.053 ksir/f =1.053i5=0.211

Forf, = 40 ksi:Et„ = 40/29,000 = 0.00138

Use E =From Eq. (1-6):

10+ 35 (Tn/fc) <00.42 - 50er

< 80 - 35 (r /f) )0.42 - 65E1,

10+ 35x 0.211 <e0.42 - 50 x 0.00138

<80- 35 x 0.2110.42 - 65 x 0.1)0138

31.0 deg< 0< 57.6 deg

Therefore, section is adequate.Choose 0 = 35 deg.

4. Check diagonal crack controlrequirements (Section 1.7)

Since f„ = 40 ksi, the requirementsof Section 1.7 will be satisfied by re-stricting the spacing of both the lon-gitudinal and transverse reinforce-ment to 12 in. This spacing restrictioncould he relaxed for regions of thebeam where calculations indicate thatthe cracking load exceeds the serviceload.

5. Design of transverse reinforcement(Section 1.8)

Design values of V. and T. nearsupports (Section 1.8.14) to be calcu-lated at d „/(2 tanO) from support face.

d„ 71__= 50.7 in. (or 4.2 11)

2 tanO 2 tan 35

82

V„ = 45.9 x 4/3 = 61.2 kips T„ – 25.2 x 12 x 4/3 -- 403 kip-in.V„ = 61.2/0.85 = 72.0 kips .' 7'„ = 403/0.85 = 474 kip-in.

Calculate torsional depth of compression, a., by Eq. (1-11):

^ 0 = nh 1 1 — T, p 5 1h ^tan8 + J

AP S 0.85f-VAoh tan J 9

432 474 x 170

170 1– – 0.85x 5x 4322 (tan35 + tan35) I

= 0.29 iri.

A n =A M – 1/2a o p h 6. Design of longitudinal= 432 – '/2 x 0.29>< 170 reinforcement (Section 1.9)= 407 in.2

Calculate transverse reinforcementfor torsion by Eq. (1-10):

Ar _ T„tan8

s 2Agf„

474 X tan352 407 x 40

0,0102 in.2/in.

Calculate transverse reinforcementfor shear by Eq. (1-8):

A n _ V. tang

5' d,,.fr

72.0) tan35

7140

-- 0.0178 in .2/in.

Thus,

A—° + 2- = 0.0381 in.2/in.$ s

(for two legs)

Therefore, use #4 closed stirrups at10-in. centers.

PCI JOURNAL^September-October 1980

The longitudinal steel requirementswill be satisfied if the member is ca-pable of resisting the positive factoredmoments given by M,^ + '/2 d„AN„and the negative factored momentsgiven by r/a da AN„ – M. (Section1.9.3). However, in the end regions(within a distance d„/tanO = 71/tan35= 8.45 ft), the negative moment that isto be resisted need not exceed:

d r f3 ` T" p, – M, (Section 1-9.5.2)2 tan() 2A,

The resulting required moment ca-pacities are shown in Fig. 30.

Table 3 summarizes the calculationsof required additional factored mo-ment for torsion and shear.

The longitudinal reinforcement thatis continued along the entire length ofthe beam consists of 15—#4 bars andthe 6—lei-in. diameter prestressingstrands (see Fig. 32). This reinforce-ment results in the flexural capacitiesshown in Fig. 30. These capacitiesdecrease near the ends of the beambecause in these regions the strandsare not fully developed.

From Fig. 30 it can be seen thatboth top and bottom additional lon-gitudinal steel is required near the

83

10 in Available negative moment strength

—Required negative moment strength

beam end ,

M„0.9

Mn 10 D000 d ON„ kip in.

0.85Required positivemoment strength

.120000Available positive moment strength kip in.

Fig. 30. Design Example 2—Required moment capacities along length ofspandrel beam.

Table 3. Summary of calculations giving required additional factored moment fortorsion and shear.

Distance from end of heath 10 in. 4 ft 1.1 in. S $ 11 in. 12 tt I I in. Units

las I f' ig . 3.9.9, PCI DesignHandbook and Eq. (18-3), ACI 318-

60 218 258 258 ksi77]

J w pe

---

— .ln = fm + E,E, 0.7 X 2701 40

1.0 1.0 1.13 1.13d; 1.0 (Section 1.9.1.2)

X 4/3 100.4 94.9/61.3" 55.9122.3* 16.8 kips7' x 4/3 672 672/403 403/134 134 kip-in.O 35 35 35 35 deg.

=Y^^rs + uNo x

tang2A0

I Eq. (1-12)) 245 2411148 162/58 52 kipsPo = p h — 4aa = 170— 4 x 0.29= 169 in.

o = 407 in.2

'/s d, AN 0 (Section 1.9.3) 8700 8555!5250 5750/2060 1850 kip-iii

*The two v:Jues refer to the sections either side iif the applied Shin load.Note; 1 ft = 0.305 ft; 1 in. = 25.4 in.; 1 in. 2 = 64.5. hf mms; 1 kip = 4.448 kN; 1 kip-in. = 113 N • m;1 ksi = 6.895 MPa.

84

ends of the member. The additionalbottom steel must provide an addi-tional moment capacity of 2720 kip-in.and the additional top steel must pro-vide 3320 kip-in. If 3-46 Grade 40bars are welded to corner angles at thetop and at the bottom the resultingadditional moment capacity will beapproximately:

3 x 0.44 x 40 x 70 = 3695 kip-in.

Use 4-ft long bars to cover the mo-ment diagram.

Check end anchorage details of lon-gitudinal steel (Section 1.9.5.3), Eq.(1-13):

Un + f- R ` 0.012(8- 10).fe1: c ye doh

100.4672 x 170+ (0.85 x 8)d,e 0.85 x 4322

0.012 (35 - 10) 5

Therefore, d„ P y 18.9 in.

The 10 in. long bearing pad willprovide an effective shear depth ofl bltanO = 10/tan35 = 14.3 in. There-fore, an additional 4.6 in. of effectivedepth is required. Hence, provide a5 x 5-in, corner angle to anchor the3 - #6 welded bars.

Note that in checking web crushingat the face of the bearing, it was as-sumed that the bearing pad and theledge would prevent spalling of thecover, hence b z = 8 in.

7. Check reinforcement details(Section 1.10)

Maximum spacing of shear rein-forcement:

d ti/(3tanO) = 711(3tan35) = 33.8 in.

Maximum spacing of torsion rein-forcement:

p 5/(8tanO) = 170/(8tan35) = 30.4 in.

Therefore, spacing of bars at 10 in.is satisfactory.

8. Transfer of forces from ledge

It will be assumed that the loadfrom the stem of the tee is transferredfrom the ledge to the bottom of theweb of the spandrel by means of thecompression strut shown in Fig. 31.This strut will be inclined at an angle0, where tanO = 9/8 (i.e., 0 = 48.4deg).

Crushing of the concrete can bechecked by Eq. (1-13) with b= 33.'4

in. and d, 1.5 + 6/tan48.4 = 6.8 in.

0.012(80- B)f

33.6 _ 0,012(80 - 48.4)50.85 x 3,75 x 6.8

1.55 ksi < 1.90 ksi (.•_ ok)

Tension tie force required:

33.6 x 8 = 31.6 kips

0.9 0.9 (9 + 1.5)

Area of steel required:

31.6140 = 0.79 in.2

Required diameter of longitudinal Therefore, use 4 - #4 closed stir-bars in corners of stirrups: rugs in ledge near each stem.

s tanOl16 = 10 x tan35/16 = 0.44 in. Calculate additional stirrups in webrequired to "hang-up" the load from

Therefore, #4 comer bars are satis- the ledge (Section 1.8.17).factory. Reinforcement must be capable of


Table 4. Comparison of shear and torsion reinforcement byvarious methods.

Design method Web Longitudinal reinforcement reinforceinent

Compression field theor y #4 at 10 in. 1,5—#4PCI Design Handbook #3 at 12 in. 14—#3Zia and Hsu #4 at 9 in. 12—#4

Note: 1 in. = 25.4 mm.

transmitting in tension a load of: The resulting reinforcement isshown in Fig. 32.

33.675-12 X = 31.4 kips0.9 75 9. Comparisons with results of otherdesign methods

Required area of steel: It is of interest to compare the solu-tion given above with solutions to es-

31.4/40 = 0.78 in.' sentially the same design problemgiven by the PCI Design Handbook

Therefore, at each stem location and by Zia and Hsu, 34 Table 4 corn-provide two double legged #4 closed pares the shear and torsion reinforce-stirrups in addition to the shear and ment calculated by the three differenttorsion reinforcement, approaches.

25.2 x 4/3 = 33.6 kips

4 i Tension Ge

d ye Compressiono o 9n strut

o -fa n ^'

,1.5

CROSS SECTION ELEVATION

Fig. 31. Design Example 2—Transfer of forces from ledge.

86

*4 closed stirrups

0 15 *4 longitudinal

*4 closed stirrups

ff6- $'2 in diameter270 K strand

4spaces @10 ' 4 spaces @ 10" 4 spaces @ 10"-j TYPICAL CROSS SECT ION3spaces=8" 3spaces=8 3spaces=8"

Reinforcing bars - grade 40ELEVATION f5000 psi

Fig. 32. Design Example 2—Spandrel beam reinforcement details.

DESIGN EXAMPLE 3—PrecastBridge Girder

The composite beam and slab twolane highway bridge described in Fig.33 is to be designed to resist HS 20-44loading, 35 The design of an interiorgirder will be described.

1. Determine loads on the girderDead Loads:

The dead loads are determined asshown in the accompanying table.Live Loads:

Fraction of wheel load applied to

Determination of dead loads.

Dead loads supported by naked girder:Girder685 X 150/144 .............................= 0,714 kipslftSlab 12< 8.20X 7.5............ = 738Haunch 22x3 .................... = 66

804X 150/144 = 0.838 kips/tt

Diaphragms at 1/3 points 0.75 3.66 7.5 0.150= 3.09 kips

Dead loads supported by composite section:3 in. asphalt 37.72 x 0.25 x 150 = 1415 lbs/ftBarrier walls 322,5 x 2 x 150/144 = 672 lbs/ftRails (approx.) = 48 lbs/ftEach composite girder takes 2135/5 = 0.427 kips/ft


HS 20 Truck Loading14 - O ' --T-- 14'-O" —^ }

32 kips 32,kips Skips Q Barrier wall

18wide x 10'longg pods Diaphragms at L/3bearing

Span =82= 0"-4 F

ELEVATION

CPCI LY

L39' 2 2.0 mf3,7.5 m lone(6.56230) 1 7i^slab

3'aspholt

20' ----- — 8.20' Variable depth haunchto allow for comber3" at (L of bearings

CROSS SECTION

22"

9'-t"–LY

I MATERIALS.

294 Concrete girders f^'= 6000 psi

54

Strandslab

f^ = 4 000 psi

112"24.6" q5 270 K low rela:otion

Rebors

H 26"H fY= 60 ksi

CPCI ] GIRDER

Fig. 33. Design Example 3—Composite beam slab highway bridge,

88

each girder determined from Table1.3.1(B) of AASHTO-1977.35

Girder spacing in ft _ 8.2= 1.491

5.5 5.5

Therefore, fraction of truck loadapplied to each girder:

t x 1.491 = 0.745

Impact allowance for truck loadmoments from AASHTO Section1.2.12:

I = 50/(L + 125)50/(82 + 125)

= 24.2 percent

B"

1bJ

^I q I ..•.•

/Zip

BetweenAt ends hold —down points

q points

Fig. 34, Design Example 3—Strandpattern in CPCI IV Girders.

Note that for shear, I is a function ofposition along the span.

The maximum moments and shearsin the girder which result from the Based on the allowable stresses at

application of the HS 20 loading are transfer and at service loads designshown in Table 5. the required prestressing. The full

details of these standard calculations2. Design prestressing (Fig. 34) and will not he given. It was found, how-determine cracking loads ever, that if the stress in strands be-

Table 5. Calculated moments and shears for bridge girder.

Distance from support 0 0.1 L 0.2 L 0.3 L 0.1 IL 0.5 L

sen+iceLoads

DL on naked girder 0 495 886 1172 1336 1388

DL on composite 0 129 230 301 345 359b19

kip-ft Total DL 0 624 1116 1.173 1681 1747

LL+ I 0 4311 750 961 1063 1107

Total DL 84.2 68.0 5 [.7 35.5 16.2 0

1, percent 24.2 25.2 26.2 27.4 28.7 30 kips

LL + 1 59.0 52.8 46-:1 40. I 33.6 27.0

FactoredLoads

k^I " 1.3 [ D + 5 (L + 1)] 0 1743 3076 3997 4488 4670p-

V1.31D+ a (L + 1)1 237 203 168 133 94 59

kips 3

Note: I kip = 4.448 kN; 1 kip-it = 1.356 kN • m.


fore transfer (fm ) was 204 ksi, then thestress in the strands after all losses(f) would be 164 ksi. Using hold-down points at the third points of thespan, 28 — t -in, diameter strands inthe pattern described in Fig. 34 werechosen,

The stresses due to the chosen pre-stressing and the resulting calculated

cracking loads are shown in Table 6.To simplify the calculations, the depthof the haunch which actually variesalong the span was assumed constantat 1.5 in.

By comparing the calculated crack-ing shears, V^r, with the service loadshears, Vr 1. (D + L + 1), and recalling

that V r need not be taken less than2 v jr' b,r d (Section 1.4.6) which isabout 64 kips, it can be seen that di-agonal cracks will not occur at serviceloads; hence the requirements of Sec-tion 1.7 will be satisfied.

To satisfy the minimum reinforce-ment requirements of Section 1.3.3, itis necessary that the nominal sectionalstrength V„ is at least equal to 1.2 Vcr.For our girder this requirement willgovern only for the section 0.1 L fromthe support.

3. Design transverse andlongitudinal reinforcement

The design calculations for the

Table 6. Stresses due to prestress and cracking loads for bridge girder.

Distance from support centerline 5 in.'" 0.1 L 0.2 L 0.3 L 0.4 1, 0.3 L.

Strand stress f, (ksi) 85 164 164 164 164 164

Prestress force, P (kips) 364 703 703 703 703 703

e = eccentricity ofP (in.) 10.6 13.6 16.7 19.9 20.9 20.9

Vertical component, Vp (kips) 11.6 22.4 22.4 22.4 0 0

Concrete stress, fD (ksi) 0.531 1.026 1.026 1.026 1.026 1.026

Concrete stress, ff (ksi) 0.923 1.996 2.217 2.445 2.516 2.516

Tensile stress due to DL onnaked girder, fde (ksi)

0 0.602 1.078 1.426 1.626 1.689

Additional moment to crack=Z&t(7..5f; + f —f2)(kip-ft)

1935 2542 2214 2059 1893 1812

Total moment to crack,Mir (kip-0)

1935 30.37 3100 3231 3229 3200

Va t Eq. (1-3) (kips) 187 257 271 285 267 267

1 2 f M„lVVrr = 1 1 J(_-) + L

ter) (kips) 171 208 144 101 66 40

LV. 1.2 V,.r (kips)263 2,51) 187 148 104 66

* Taken at inner edge of hearing, 13 in. from end ofbeam,t Z, for composite section taken as 15440 in.1# d taken as 7% + 1½-f 29.4 + e in.§ ¢ = 0.9 for shear and 1.0 for flexure. AASHTO-77, Clause 1.6.5.Note; 1 in. = 25.4 mm; 1 kip = 4.448 kN; l kip-it = 1.356 kN • m; 1 ksi = 6.l95 MPa.

90

transverse and longitudinal steel aresummarized in Table 7. In determin-ing d, it was assumed that the trans-verse steel would be anchored 2 in.above the bottom face of the girderand 5 in. above the top face of thegirder. The 5-in, extension above thegirder plus a standard hook will ena-ble the #3 transverse bars to he fullydeveloped at the interface. The con-tact surface at the interface is assumedto be clean and intentionallyroughened. The transverse steel cho-sen to satisfy the design requirementsis shown in Fig. 35.

With regard to longitudinal rein-forcement, it can be seen from Table 7that the prestressing strands alone areadequate (1471 - 1475) to resist theapplied moment in addition to theequivalent moment caused by theshear.

Check crushing of web near support(Section 1.9.5.3).

- ln "-_ 0.012 (9 - 10).fc

251 `0.012(22- 10)67 x d,

I Table 7. Design of transverse and longitudinal reinforcement for bridge girder.

Distance from support 0 0.1 L 0.2 L 0.31. 0.4 L 0.5 L

V„ - V, (kips) 251 228 165 126 104 66

T Vn -V Vx -V a- 0.105 0.095 0.069 0.053 0.043 0.028

.ff b ^*dJ 7 x 57 x 6

Range of0 Eq. (1-6) (deg) 21.6 20.5 17.6 15.9 14.8 13.0E, = e, R, = 60/29000 (deg) -67.1 --+68.4 -7L.5 -73.5 -+74.7 -+76,6

Design choice of (deg) 22 22 22 22 22 22

`° (V ,,) (tang) (in. 2 /in.) 0.0297 0.0269 0.0195 0.0149 0.0123 0.007$R d r.l r

d (in.) 49.0 52.0 55.1 58,3 59.3 59.3

fvs [ACI318-7712.10.1+ Eq. (18-3)1 65 262 263 263 263 263(ksi)

1.0 1.0 1.0 1.0 1.0 1.0 1.0 204 + 60

AN. 9^ (V n -V v)_ (kips) 621 564 408 312 257 163(b tan6

1AN ° (kip-ft)d

1475 1341) 969 741 610 3882 0

t A,LfA+ d (kip-ft) 1475 3083 404,5 4738 5098 5058

rh 2 d>

i4x=A,,fp,(d- a/2)28 x 0.153 X fv,. I28 x 0.153 f,,,+ d- 1471 4707 5015 5316 5410 5410

2x 0.85x 4x 8.2x 12)(kip-ft)

*b, = b^, since flanges restrain concrete cover.Note; Lin. = 25.4 mm; I kip = 4.448 kN; 1 kip-lt = 1.36 kN • m.


Table 8. Comparison of transverse reinforcement for bridgegirder by various methods.

Various methods LocationNear support Near 0.25L

Compression field theory #3 at 8 in. #3 at 12 in.

ACI 318-77 #3 at 10 in. #3 at 24 in.

AASHTO-77 #3 at 12 in. #3 at 12 in.

Note: 1 in. = 25.4 mm.

r3 *5 tAASHTO 1.6.151I— '3@8" *3@ 12"— - -'

0.030Av

5iin 2/fn } I "; Provided0.020 k End_of

girde

SupportO.OID ^ face

Hegwred'^dv j

2rangI.

0 OIL 0.2L 0,3L 0.4L 0.'

DISTANCE FROM SUPPORT

Fig. 35. Design Example 3—Design of transverse reinforcement.

from which d„p=54+5– 13.5= 45.5 in. > 41.5 in. (.•. ok)

d,, r 41.5 in. 4. Comparisons with other design

For this flanged section: methods

It is of interest to compare theI b = 10 + 13.5 = 23.5 in. transverse reinforcement obtained

above with the transverse reinforce-dtt = I6/tan9 = 23.5Itan22 = 58 in., meat that would have been obtained

by satisfying the requirements of thewhich exceeds the available depth ACI Code (with AASIITO load factorsabove the junction of the web and and yb factors) or by following theflange. AASHTO recommendations (see

Hence, available: Table 8).

92

ACKNOWLEDGMENTS

The recommendations presented in thispaper are the result of a research programinitiated at the University of Toronto in1969.

This research was made possible bya series of grants from the Natural Sciencesand Engineering Research Council ofCanada. This continuing support is grate-fully acknowledged.

The significant contributions of Dr. PaulLampert during the initial stages of the tor-sion and shear research program deservespecial mention. The individual contribu-tions of the following engineers, each ofwhom tested at least one series of beams intorsion and/or shear, are gratefully ac-knowledged: Petar Krpan, Basile Rabbat,and Winston Onsongo (former doctoralstudents); Ersin Ocerdinc, Selahattin To-prak, Marcus Aregawi, SithambaramChockalingain, Jamil Mardukhi, BenjaminArbesman, Mario Kaiii, Chris Sadler, Frank

Vecchio, Jacek Morawski and MilindJogllekar (former Master's students).

Thanks are also extended to the SteelCompany of Canada who over the yearshave donated the reinforcing steel and pre-stressing wire used in the test beams.

The 1922 statement of Morsch quoted inthe paper was brought to the authors' at-tention by a paper by Kuyt. a' The transla-tion of MOrsch's quote given in the paperwas provided by Mario Kani.

Design Example 3 was based on aworked example by Kris Bassi and HidGrouni of the Ontario Ministry of Trans-portation and Communications which waspresented at an Ontario Bridge CodeSeminar in October, 1979.

Finally, the authors wish to express theirappreciation to the many individuals whoreviewed this paper and for their veryhelpful suggestions in strengthening thevalue of the article.

Metric (SI) UnitEquivalents

1 in.=25.4 mm1 in. 2 =645.16 mm'1 ft=0,3048 m1 ft2 =0.0929 mz1 psf=47.88 NJm21 pst=47.88 Pa1 psi=0.006895 MPa1 pcf=16.02 kg/m31 kip=4.448 RN1 kiplft=14.594 kN,,m1 kip-in. = 113N • m1 kip-ft=1.356 kN • m

NOTE: A References section andfour appendices follow.

Discussion of this paper is invited.Please forward your comments toPCI Headquarters by May 1, 1981.


REFERENCES

1, ACI Committee 318, "Building CodeRequirements for Reinforced Concrete(ACI 318-77)," American Concrete In-stitute, Detroit, 1977.

2. Gerstle, K. H., Basic Structural De-sign, McGraw-Hill, New York, 1967,405 pp.

3. Khachaturian, N., and Gur€`inkel, C.,Prestressed Concrete, McGraw-Hill,New York, 1969, 460 pp.

4. Park, R., and Pau lay, T., ReinforcedConcrete Structures, Wiley-Intersci-ence, New York, 1975, 769 pp.

5. Ritter, W., Die Bauuaeise Hennebique,Schweizerische Bauzeitung, Zurich,February 1899.

6. M6rsch, E., Concrete-Steel Construc-tion, English Translation by E. P.Goodrich, McGraw-Hill Book Com-pany, New York, 1909, 368 pp.(Translation from third edition of DerEi.senbetonbau, first edition, 1902.)

7. Mi)rsch, E., Der Eisenbetonhau, Ver-lag von Konrad Wittwer, Stuttgart,1922, 460 pp. (quote is from page 128).

8. CEB-FIP, Model Code for ConcreteStructures, CEB-FIP InternationalRecommendations, Third Edition,Comit6 Euro-International du Beton,Paris, 1978, 348 pp.

9. Lampert, P., and Thurlimann, B., "Ul-timate Strength and Design of Rein-forced Concrete Beams in Torsion andBending," International Associationfor Bridge and Structural Eengineer-ing, Publication 31-1, 1971, pp. 1.07-131.

10. Rausch, E., Berechnung des Eisenbe-tons gegen Verdrehung (Torsion) andAbscheren, Julius Springer-Verlag,Berlin, 1929, 51 pp.

11. Hsu, T. T. C., "Torsion of StructuralConcrete-Behavior of Reinforced Con-crete Rectangular Members," Torsion

of Structural Concrete, SP-18, Ameri-can Concrete Institute, Detroit, 1968,pp. 261-306.

12. Mattock, A. H., "How to Design forTorsion," Torsion of Structural Con-crete, SP-18, American Concrete In-stitute, Detroit, 1968, pp. 469-495.

13. PCI Design Handbook—Precast Pre-stressed Concrete, Second Edition,

Prestressed Concrete Institute,Chicago, 1978.

14. Zia, P., and McGee, W. D., "TorsionDesign of Prestressed Concrete," PCIJOURNAL, V. 19, No. 2, March-April1974, pp. 46-65.

15. Wagner, H., "Ebene Blechwandtragermit sehr dunnem Stegblech," Zeits-chrift fur Flugtechnik and Motorlufts-chiffahr, V. 20, Nos. 8 to 12, Berlin,1929.

16. Mitchell, D., and Collins, M. P., "Di-agonal Compression Field Theory—ARational Model for Structural Concretein Pure Torsion," ACI Journal, V. 71,August 1974, pp. 396-408.

17. Collins, M. P., "Towards a RationalTheory for RC Members in Shear,"Journal of the Structural Division,American Society of Civil Engineers,V. 104, April 1978, pp. 649-666.

18. Mitchell, D., and Collins, M. P., "In-fluence of Prestressing on TorsionalResponse of Concrete Beams," PCIJOURNAL, V. 23, No. 3, May-June,1978, pp. 54-73.

19. Leonhardt, F., "Shear and Torsion inPrestressed Concrete," Proceedings,the Sixth Congress, Federation Inter-nationale de la Precontrainte, Prague,1970, pp. 137-155.

20. Campbell, T. I., Batchelor, B. D., andChitnuyanondh, L., "Web Crushing inConcrete Girders with PrestressingDucts in the Web," PCI JOURNAL, V.24, No. 5, Sept.-Oct. 1979, pp. 70-88.

24. Mardukhi, J., "The Behaviour of Uni-formly Prestressed Concrete BoxBeams in Combined Torsion and 32Bending," MA Sc Thesis, University ofToronto, Toronto, 1974, 73 pp.

21. Rabhat, B. C., "A Variable AngleSpace Truss Model for Structural Con-crete Beams,' PhD Thesis, Universityof Toronto, Toronto, 1975, 236 pp.

22. Collins, M. P., "Investigating theStress-Strain Characteristics of Diago-nally Cracked Concrete," IABSEColloquium on Plasticity in Rein-forced Concrete, Copenhagen, May1979, V. 29, pp. 27-34.

23. Onsongo, W. M., "The Diagonal Com-pression Field Theory for ReinforcedConcrete Beams Subjected to Com-bined Torsion, Flexure and AxialLoad," PhD Thesis, University of To-ronto, Toronto, 1978, 246 pp.

25. Rabbat, B. C., and Collins, M. P., "AVariable Angle Space Tniss Model forStructural Concrete Members Sub-jected to Complex Loading," DouglasMcHenriy International Symposium onConcrete and Concrete Structures,ACE SP-55, American Concrete Insti-tute, Detroit, 1978, pp. 547-587.

26. Rabbat, B. G., and Collins, M. P., "TheComputer Aided Design of StructuralConcrete Sections Subjected to Com-bined Loading," Computers andStructures, V. 7, No. 2, PermagonPress, April 1977, pp. 229-236.

27. Collins, M. P„ "Design for Shear andTorsion," Chapter 2 of Metric DesignHandbook for Reinforced ConcreteElements, Canadian Portland CementAssociation, Ottawa, 1978, pp. 2-1-2-68.

28. ACI-ASCE Committee 426, "TheShear Strength of Reinforced ConcreteMembers," Journal of the StructuralDivision, American Society of Civil

Engineers, V. 100, August 1974, pp.1543-1591.

29. Branson, D. E., "Instantaneous andTime-Dependent Deflections on Sim-ple and Continuous Reinforced Con-crete Beams," HPR Report No. 7, Part1, Alabama Highway Department,Bureau of Public Roads, August 1965,78 pp.

30. Kani, M. W., Huggins, M. W., andWittkopp, R. R., Kani on Shear inReinforced Concrete, Department ofCivil Engineering, University of To-ronto, 1979, 225 pp.

31. Mitchell, D., and Collins, M. P., "De-tailing for Torsion," ACt Journal, V.73, No. 9, September 1976, pp. 506-511.

Grob, J., and ThUrlimann, B., "Ulti-mate Strength and Design of Rein-forced Concrete Beams Under Bend-ing and Shear," Publications, Interna-tional Association for Bridge andStructural Engineering, V. 36 -I1, 1976,pp. 107.

33. Kuyt, B., "Oeer de Dwarskrachtsterktevan Slanke Balken met VerticaleBeugels. (I) en (11)," Cement, Amster-danm, The Netherlands, V. 24, No. 6,1972, pp. 243-251, and V. 24, No. 9,1972, pp. 346-353 (English translationavailable as Technical Translation1822, Canada Institute for Scientificand Technical Information, NationalResearch Council, Ottawa, Canada.)

.34. Zia, P., and Hsu, T. 1'. C., "Design forTorsion and Shear in Prestressed Con-crete," Preprint 3424, American Soci-ety of Civil Engineers Convention,Chicago, October 1978, 17 pp.

35. AASHTO, Standard Specifications forHighway Bridges, American Associa-tion of State Highway and Transporta-tion Officials, Washington, 1977, 496PP.

APPENDIX A-DERIVATION OF STRESS BLOCK FACTORS

For the beam shown in Fig. 1, the b cY+`^_

resultant compression in the concrete f I f`f`dE^

is given by: _ (A7)bc rct

f dEcC= + fib dy (Al) IE`E °

Jo

For the equivalent uniform stressAs the strain distribution is linear: distribution, the distance from the

neutral axis up to the resultant corn-

= E (A2) pression force will be:C Ec,

=c(I–pi,/2) (A8)Using Eq. (A2) and assuming that

the width of the compression zone is Hence, for the position of the re-constant, Eq. (Al) can be rewritten as: sultant compression force to remain

the same:fct

C = C [fde, (A3) fir r

Ecf10 J ,I€d€c(1–f3/2)= ° (A9)

If the equivalent uniform stress Ec,IF`f„dE„distribution is used, Fig. 1(d), then theresultant compression in the concreteis given by:

If the stress-strain curve of the con-crete is known, then for a given value

C = a,f3 1 b c f^

(A4) of maximum compressive strain, e,,,the stress block factors a, and /3, can

Hence, for the magnitude of the re- be determined from Eqs. (A9) andsultant compression force to remain (A5)•the same: If the following parabolic concrete

stress-strain curve is assumed, seeFig. 1(e):

EC,

a is, — r 1 +fd€, (A5)JcEcrJO

The position of the resultant con-crete compression force can be deter-mined by taking moments about theneutral axis. The distance up to theresultant force will be:

J cfcb ydyy = (A6)

1cf,bdy°

Eliminating y by using Eq. (A2)leads to:

Y

f^ = 2 Eo – 1 Eo(A10)

Then Eqs. (A9) and (A5) give:

(All)6 – 2 E,,/Ea

Yai1Q ^ = E^, – 1 ) (Al2)

E^ 3 Ep )

Note: When using Eqs. (All) and(Al2) for beams in torsion replace e,by €,

96

AN tanO

a 1fe Po Sing GUSH

2 AN tanO(B10)

a 1f^ p, sin20

S

APPENDIX B—DERIVATION OF EXPRESSIONSFOR a,, e t AND E,

For a beam in torsion the equivalent longitudinal and transverse lines inuniform concrete stress distribution the plane of the hoop centerline.

consists of a uniform diagonal corn- From Mohr's circle of strain (seepressive stress of a lf,' acting over a Fig. B1):depth of a ° . These diagonal stresses

acting at an inclination of o will pro- __ 2(€ + Er) (B7)

duce a shear flow [ see Eq. (2)] of: l` tanO

q = a j, a ° sines cos9 (BI) and

From Eq. (5) and Eq. (B1): yr = 2 (ed, + e t ) tarn6 (B8)

LAN =sin( cosH (B2) [Note: eliminating )'u from thesetan@ two equations results in Eq. (7)1.

Substituting from Eqs. (B7) and (B6)

From Eq. (6) and Eq. (B1): into Eq. (B5) gives:

fl f € A°h tanO

p•j= a, sine cosO tang (B3) a, = Q^ (B9)

s p (€ + €j1) sin2Q

Adding Eqs. (B2) and (B3) gives: But from Eq. (B2):

X14' + ArftaLfC Po ajc S

_ a o sing cosO I 1 + tanO I = a.,`tans J

(B4)

which completes the derivation of Eq.(10).

From F'iv S it ran hR gF+ pn tf^at-

p E de = eds yltao — N1 t Yrt — ! 1b /3d! q sin2H 2

(BS) i-

where iJr is the twist of the beam. Thetwist can be determined23 as:

= (B6)hYu ZA,h

where yrr is the shear strain between Fig. B1. Mohr's circle of strain.


Comparing Eqs. (B9) and (B10) wesee that:

l2 EdsAan _ 2 AN (B11)Y1 Ph ( Edit + E!) a1J c pa

Rearranging this equation gives thetensile strain of longitudinal rein-forcement due to shear andlor torsion:

E! _ f ^1f f A oh^}o _ 1 J Ed,]L 2 1 ,,, AN

(B12)

which completes the derivation of Eq.(12). By substituting from Eqs. (B8)and (B6) into Eq. (B5) and comparingthe resulting expression for a o withEq. (B3), Eq. (11) can be derived.

APPENDIX C-DERIVATION OF CRACKCONTROL LIMIT ON 0

We wish to find the lower limit on 0which will ensure that when V = V,,the strain in the transverse steel doesnot exceed 0.001. Thus, from Eq. (30):

s

1_(Vc

r) }Et -_0.001 (Cl)Vae )

For a beam subjected to a shear V,0,the strain e r can be calculated fromEq. (3) as:

E, = V' 5 tan8$ (C2)E a A„dO

where 0, is the inclination of the diag-onal stresses at the service load shearV,,.

The area of shear reinforcement A,,is determined from Eq. (3) by using:

A,, =y - tang (C3)f

Substituting Eq. (C3) into Eq. (C2)gives:

% tang„ (C4)

` E„ V. tang

In design we will choose 0 andhence to find e, at service load weneed a relationship between 0,, theangle of inclination of the diagonals atservice load, and 8, the angle of incli-nation of the diagonals at the nominalcapacity.

In developing an approximate re-lationship between 8r and 0 we shouldrecognize that the higher the estimateof 8 ,, the higher will be the estimate ofe, [ Eq. (C4)] . Hence, for our purposesoverestimating 0. will be conservative.

For a known value of V„ and a givenchoice of 0, the required areas oftransverse and longitudinal rein-forcement could he calculated. Theresponse of the beam at service loadcould then be determined to find thevalue of 8,. It will be found that for agiven value of8, the higher the valueof V, the higher the value of 0,.Hence, to determine a conservativevalue of B, for a given value of 8, thehighest possible value of V„ should beused, that is, the value which justsatisfies the concrete crushing limit ofEq. (23) should be used.

The calculated relationships be-tween 0, and 0 for a non-prestressedand for a prestressed series of beams

98

LO

Non prestressed

taf9S

,PrefY

tones= lane('-30 fPc `

Accurate relationshipbetween Ian 95 and tong

fy = 60 ksi

010 05 .0

tang

Fig. C1. Relation between a, and t for prestressedand non-prestressed concrete beams,

It can be seen that this equationrepresents reasonably well the re-lationship between 0, and 8.

Substituting from Eqs. (C5) and(C2) into Eq. (C1) results in Eq. (31):

ztang = 2

. V__ } 1 - 29 f 1 x

n fc f

tanO — tan0 1 — tU —"`8 30 .fc 1 _ Vcr z (31)

(C5) L V

APPENDIX D—NOTATION

Symbols not defined in the notation a = equivalent depth of compres-section of the Design Recommen- siondations are defined below: b = width of beam

0,5

are shown in Fig, C1. In calculatingthese relationships it was assumedthat VEe/V„ = 0.55. Additionally, forthe prestressed beams it was assumedthat f= = 150 ksi and AE = 6.5 x 10-3.

Also shown in Fig. C1 is the approxi-mate relationship.

Al = area of longitudinal reinforcingbars

Ap = area of longitudinal prestress-ing steel

C = resultant compressive force inconcrete

c = neutral axis depthD = diagonal compressive force in

concrete

PCI JOURNALISeptember- October 1980 99

d – effective depth to flexural rein-forcement

d, = distance from extreme com-pression fiber to neutral axis

dd = duct diametere = eccentricity of prestressing

= compressive stress in concretecorresponding to strain €,

fCt = concrete stress correspondingto the strain Ear

fd = equivalent uniform compres-sive stressfdaz = limit of principal diagonal com-pressive stress in diagonallycracked concrete

f, = stress in longitudinal prestress-ing steel

faQ = tensile stress due to dead loadon naked girder (Example 3)

= "yield" stress of prestressingsteel; taken as 0.2 percent off-set stress

= effective stress in prestressedreinforcement (after allowancefor all prestress losses)

f^ = stress in hoop reinforcement= stress in shear reinforcement

I = impact fraction (Example 3)jd = lever ann for flexural resistanceL = length of'memberM = flexural momentMa = service load flexural momentM. = factored flexural momentAN = tensile force in longitudinal

steel produced by shear andtorsion

P = prestressing Forceq = shear flowT = torsional momenttd = effective wall thickness under

torsional loadingV = sheara s = nominal shear stress at ultimatetv u = factored uniform load per unit

length

y = distance from neutral axisyt = distance from centroidal axis to

extreme fiber in tensiony = distance from neutral axis to re-

sultant compressive forceZ b = cross-sectional section modulus

for bottom fibera l = stress block factor defined in

Eq. (1)y = shear strain

= shear strain between longitudi-nal and transverse lines

ym = maximum shear strainE – normal straine, = compressive strain in concreteE, = strain in concrete at level of

prestressing steelear = maximum concrete compres-

sive strained = diagonal compressive straine& = concrete diagonal compressive

strain at effective surface of'beam in torsion

eo = strain in concrete correspond-ing to maximum compressivestress

= strain in prestressingDe,, = difference in strain between

prestressing steel and sur-rounding concrete

E, = strain in transverse reinforce-ment

efe = expected transverse strain atservice load level

B = angle of diagonal compressivestresses in concrete

8e = value of B at end of beam (sec.Fig. 21)

= value of 6 at service loadd) = curvature(bd = curvature of walls due to tor-

sionifi = twist per unit lengthrr = normal stressT = shear stress

I,

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