STATE-OF-THE-ART PAPER

The Practice ofWarping Double Tees

Paul Mack, P.E., FPCIRocky Mountain PrestressDenver, Colorado

Gregory Force, P.E., FPCIVice President and General Manager

Tindall CorporationPetersburg, Virginia

This paper presents the results of a study ofwarping (twisting) in precast concrete double teesand draws conclusions on what is beingsuccessfully done within the industry to minimizeor eliminate longitudinal cracking through decksdue to warping. Conclusions are given based onthe results of an industry survey, an assessment oftorsion parameters, theoretical calculation ofcracking point, and empirical data from plantcontrolled testing and field experiences. Data arepresented for field-topped tees and pretopped teesof various lengths and widths. Recommendationsare given on the best practice to follow incontrolling warping in double tees.

Charles MagnesloVice PresidentJVI, Inc.Lincoinwood, Illinois

Kp Bryan, RE.Project Manager

FDG, Inc.Arvada, Colorado

Double tees are perhaps the most widely used member in the precast, prestressed concrete industry (seeFig. 1). In general, double tees have performed ex

tremely well over the past 50 years. However, occasionally, a double tee may experience longitudinal cracks at theinterface of stem and deck elements. The cracks may rangein length from 5 to 30 percent of the tee length.

Under close scrutiny, these cracks appear to develop ontop of the deck on one stem and on the underside of theother stem. The cause and significance related to long-termperformance of double tees experiencing these cracks havenot been universally agreed upon within the industry. Nevertheless, one purpose of this paper is to bring all availableknowledge on this subject together so that a consensus canbe reached.

A PCI committee report in the January-February 1983PCI JOURNAL1states that the cause of longitudinal cracks

32 PCI JOURNAL

is generally warping or twisting thathas been imposed on the double teemember.

The most common occurrence ofwarping is on decks of parking structures. Therefore, many discussions inthis paper are directed toward this application. In general, cracking iscaused by:• Twisting that occurs when the tee is

stripped (removed) from its castingbed (e.g., uneven lifting duringstripping from binding in the form,uneven sling lengths, or use of twocranes).

• Racking from improper storage inthe manufacturing plant storageyard (e.g., dunnage placed on soft oruneven ground).

• Twisting during handling or hauling(e.g., a trailer running over a curb).

• Torsion being a condition of the design (geometry) of the structurecontaining the erected tee (e.g.,sloped/warped for drainage in aparking structure).Typically, this type of crack is non-

structural and has a negligible effecton the load carrying capacity of thetee. If the tee is covered with a composite concrete topping (field-topped),

no repair is necessary unless for cosmetic reasons.

A pretopped tee (one in which theentire deck is cast monolithically withthe tee stems during manufacture) experiencing cracks may need remedialattention after the tee is erected, depending on the crack width and location. If water penetration is a concern,a wide range of weather resistant materials is available to fill and seal thecracks, such as epoxies, caulking materials, and other sealants.

This paper presents the results andconclusions of a study of warping(twisting) in double tees. Conclusionsare based on the results and dissemination of an industry survey, theoretical calculation of cracking point andtorsion parameters of the cross section, and empirical data extractedfrom plant controlled testing and fieldobservations. Data are presented forfield-topped tees and pretopped tees ofvarious lengths.

Note that double tees may crackeven if the guidelines presented hereare followed closely. Recommendations provided in this article, therefore, should be considered with engineering judgment. There is no

guarantee that adhering to these recommendations will entirely eliminatethe condition, but it should be borne inmind that concrete cracking is not unusual and that not all cracks necessitate remedial action.

The first three bulleted items identified as causes of longitudinal crackingare generally unpredicted, but thefourth is an intentional design condition wherein twisting and warping isknown to occur. To achieve durabilityin parking structures, one of the mostimportant recommendations made inthe American Concrete Institute publication ACI 362.1R-972 is to slopedecks in order to drain water from thedeck surface.*

While direct rain or snow may notenter all areas of a parking structure,wind blown rain and/or vehicles carrying ice/slush/water will distribute

* Other components of durability include (1) lowwater-cementitious material ratios (0.40 or less), (2) airentrained concrete (6V2 percent, 11/2 percent tolerance),(3) coffosion inhibiting admixtures, (4) good cornicetions, (5) high quality joint sealants, (6) cover over reinforcement of 1/2 to 2 in. (38 to 51 mm), and (7) awell-planned maintenance program. Other less effective measures like surface scalers and traffic bearingmembranes are sometimes employed.

January-February 2003 33

water throughout the facility. If thewater is contaminated with salts fromdeicing materials or other sources,corrosion of reinforcement may be accelerated due to chloride penetration.As a result, the life of the structuremay be shortened.

Frequent floor wash-downs that arepart of a maintenance program arealso a source of water throughout thefacility. If the floor is not adequatelysloped for proper drainage, pools ofwater or sheets of ice can form, promoting the potential for deteriorationof the concrete slab due to freeze-thawcycles. If a floor is to be sloped, howmuch slope is necessary? Parking industry guidelines are cited in the following paragraphs.

Floor Slope Recommendations —

Standard References

Recommendations on the amount ofslope necessary to drain a floor hasbeen based mainly upon experience.Many references on the subject areavailable, and the following excerptsare taken from the most often citedsources:

1. Parking Structures: Recommended Practice for Design and Construction, PCI publicaton MNL-129-98:

“Parking Decks should be pitchedfor draining. Slopes of 11/2 percentare common with 1 percent beingthe minimum acceptable field limitafter construction tolerances areconsidered.”2. The Dimensions of Parking,

Third Edition, ULI & NPA:4“The recommended slope to drainsis 2 percent minimum in all directions. Normally, in precast parkingstructures, the main slope (typicallyin the span direction) is 2 percentand the cross slope (perpendicular

to the direction of the main slope) islimited to about 1 percent or evenless. If a traffic membrane is used, 1percent slopes are acceptable, but 2percent slopes are still preferable.”3. Guide for the Design of Durable

Parking Structures, ACT 362.1R-97:2“A minimum slope in any directionof 11/2 percent is recommended with2 percent being preferred.”4. Parking, ENO Foundation for

Transportation:5“Storage floors of cast-in-placefloor garages should be slopedabout 2 percent (approximately 1/4

in. per ft). Precision manufactured(and installed) precast floors canprovide positive drainage with 1percent (approximately 1/8 in. per ft)slopes, although some designersprefer a 2 percent slope.”5. Parking Structures, Planning, De

sign, Construction, Maintenance andRepair, Third Edition:6

“For drainage, the absolute minimum slope should be 1/8 in. per ft orabout 1 percent preferred slope is

u16 in. per ft or about 1/2 percent.”6. The Parking Professional, May

2000, “Parking Design Checklist2000”:

“Slopes: 2 percent preferable (1 percent is used typically in precast concrete and is based on the use ofquality sealer on surface and thedenser type of concrete normally associated with precast concrete).”

Warping of Double Tees

The PCI document, Parking Structures: Recommended Practice for Design and Construction, MNL 129-98,presents a cursory discussion of warping of double tees in Chapter 4, Section 4.4. Mention is made of warpingfor both pretopped and field toppedmembers, and a general comment is

made about the flexibility of the members. A sketch indicates how a typicalbay of a parking structure could besloped for drainage, but warping is notshown.

Cross-bay drainage is achieved bylowering one end of the double teemembers. Transverse drainage is typically achieved by one of three meth

1. Pitching the supporting girder orspandrel at the low end only.

2. Sloping the structural topping.3. Pitching both the high and low

ends of the member by using slopedledges, variable dap depths in the double tee stems, corbels at varyingheights on the supporting spandrel,and/or sloping of the entire supportingelement.

The cause of warping (twisting) indouble tee members in situ is due totransverse floor slopes (Method No.1). Many times the other two methodsare neither practical nor cost effective.Sloping the field placed topping addscost for the additional concrete and itsweight can require more prestressingforce (usually, more strands), therebyadding even more cost to the project.

Using sloped ledges, varying heightdaps, or placing corbels at varyingheights on the face of the spandrel allrequire particular attention to engineering detailing and to setup withinthe form so as to ensure proper orientation and dimension. Higher material,labor and forming costs are introduced. It should also be noted that aesthetic considerations often precludeutilization of sloped spandrels on theexterior of a parking deck.

Using varying height daps on theexterior may create a conditionwherein the ends of the stems are visible from the outside, which may alsobe objectionable. Accommodatingvarying height ledges or corbels on thesupport spandrel may require that thedepth of the spandrel be increased byat least the change in elevation fromhighest to lowest stem support withina given span.

This may create a problem in thatthe deck must contain a certainamount of open area (i.e., the spacebetween the top of one spandrel andthe bottom of the spandrel above) onits perimeter in order to be considered

Amount of warping

idisplacement jAJ

Fig. 2. Warping of double tee section. ods:

34 PCI JOURNAL

an open parking structure. Referenceshould be made to the governingmodel building code for specific requirements.

For these reasons, designers oftenmaintain a constant elevation for thedeck surface at the exterior perimeterof the deck and slope beams or wallson the interior gridline to achieve thetransverse slope to the drains. A twistor warp (see Fig. 2) is thereby introduced at the low end of the double tee.This twist propagates along the lengthof the tee.

The purpose of this paper is to provide a rationale to designers on thelimits of warping that can be reasonably expected of typical double teemembers currently used in parkingdeck construction.

As a matter of definition, warpingoccurs in a double tee when the fourstem bearing points are not co-planarand the ends of the tee are rotated withrespect to each other. The differencein the slopes at the ends of the tee defines the amount of warp (twist angle).The warp can be expressed as a percent slope (equal to 100 times twistangle) or as a twist with dimensions ofin. per ft across the cross section.

INDUSTRY SURVEYOver the years, the PCI Parking

Structures Committee has received numerous questions regarding the process of warping double tees, with particular reference to allowable warpand stresses that will initiate cracking.To provide guidance to the engineering community, the committee conducted a survey of PCI producer members to determine the industrycommon practice and collect feedbackregarding testing or other means of analyzing the limits of twisting andcracking. Fifty-eight surveys were returned to the committee from producers in the United States and Canada.

Survey Results

The survey requested producers toprovide section properties, deck reinforcement and the amount of warpingthey recommend, given variouslengths of double tees. Producers werealso asked to provide a rationale fortheir recommendations. All survey re

sponses contained information for teeslonger than 50 ft (15.2 m).

The information received (see Tables 1 and 2) regarding the amount ofwarping (A/Span Length) diminishedas the double tee lengths decreased.Some respondents reported that shorttees are so stiff that they would notsettle into their desired (warped) position without forcing the stem downwith weights. Blank responses werenot considered in the statistical evaluations.

Producer Comments —

Amount of Warping:

• Double tees with widths of 10 ft(3.05 m) or less and with a 2 in. (51mm) thick deck have been warpedas much as 0.6 in. per ft (A 3 in.).The decks cracked but are functioning well.

• Pretopped double tees with widthsof 10 ft (3.05 m) or less with a 4 in.(102 mm) thick deck have beenwarped as much as 0.4 in. per ft (A

2 in.) and had cracks.• Three producers reported testing a

10 ft (3.05 m) wide, 60 ft (18.30 m)long double tee. Two found that thetees began to crack at 0.2 in. per ftwith a 4 in. (102 mm) thick deck.The third producer conducted a finite element analysis on a 10 ft(3.05 m) wide, 60 ft (18.30 m) longdouble tee with a 43/4 in. (121 mm)thick deck. The tee cracked at 0.175in. per ft of warp. Test results confirmed their theoretical analysis.

Producer Comments —

Regarding Cracking:

• Double tees are less susceptible totorsion cracking if the transition

from stem to deck is a radius ratherthan a chamfer; this is becausestress concentration is less.

• Using a ‘8 in. (9.5 mm) diameterlongitudinal prestressing strand(parallel to the stems) at the stem locations in pretopped decks helpscontrol cracks. The strand also provides a surface for supporting thedeck reinforcement, ensuring properplacement. Typically, the strand isstressed to a nominal value of about5 kips (22.3 kN).

• It is possible to initiate warping inthe plant by shimming one leghigher than the other while the tee isin storage and the shear modulus issmaller. By doing this, the warpingis taking place while the concrete iscuring and allows the double tee towarp more without developingcracks. Care must be taken to warpthe tee in the proper direction.

• Allowing a double tee to settle intoits warped position may increase thetotal amount a double tee can bewarped versus forcing the doubletee down to make the deck connections.

• There was insufficient data to determine whether or not the type of aggregate in the concrete mix (normalweight or lightweight) influencedthe amount double tees can bewarped without cracking. However,engineers should keep in mind that,in the end, it is the shear modulusthat is important.

• The depth of the tee or the spacingof tee stems appears to have littleeffect on the amount double teescan be warped without cracking. Noreason for this observation wasgiven. It is shown later in this articlethat the depth of the tee stems has a

Table 1. Field-topped double tees [2 in. (51 mm) average deck thickness].

Span length < 20 ft 20 to 30 ft J 31 to 40 ft

Average warp (twisQ 0.068 in/ft 0.100 in./ft 0.151 in.Ift 0.224 in/ft

Standard deviation 0.026 in/ft 0.035 in./ft 0.043 in/ft 0.128 in/ft

Note: 1 ft = 0.3048 m; 1 in. = 25.4 mm.

Table 2. Pretopped double tees [4 in. (102 mm) average deck thickness].

Note: 1 ft = 0.3048 m; 1 in. = 25.4 mm.

Span length 1 <20 ft 20 to 30ft 31 to 40 ft I > 50ft -

Average warp (twist) 1 0.076 in./ft 0.082 in./ft 0.128 in./ft 0.160 in/ft

Standard deviation J 0.039 in./fi 0.040 in./ft 0.056 in/ft 0.074 inJft

January-February 2003 35

Fig. 3. Trailerrocker beamsupporting adouble tee.

Table 3. Double tee geometry.

Tee size Tee width Stem height Deck thickness

8DT22+2 8 ft 22 in. 2 in. - Field topped

8DT22+4 8 ft 22 in. 4 in. - Pretopped

8DT30+2 8 ft 30 in. 2 in. - Field_topped

8DT30+4 8 ft 30 in. 4 in. - Pretopped

100T22+2 10 ft 22 in. 2 in. - Field topped

10DT22+4 10 ft 22 in. 4 in. - Pretopped

1ODT3O±2 J lOft 30 in. 2iit-Field topped

1ODT3O+4 10 ft 30 in. 4 in. - Pretopped

12DT26+2 12 ft 26 in. 2 in. - Field topped

I 2DT26+4 12 ft 26 in. 4 in. - Pretopped

12DT30+2 12 ft 30 in. 2 in. - Field topped

12DT30+4 12 ft 30 in. 4 in. - Pretopped

cause shear stresses. An analysis ofstandard double tee sections subjectedto pure torsion was carried out to isolate the effects of torsion. The complex interaction of stresses caused bygravity and axial loads are not addressed in this paper.

The purposes of the study were to:(1) Compare rotational stiffness andmaximum torsion stresses for the various tee sections, and (2) Determine thepercentage of total torsion resisted bythe deck and by the stems. It was alsoexpected that computer modelingwould reveal regions on the sectionwhere stress concentrations are likelyto occur.

Tee sections and their propertieswere taken from Chapter 2 of the FifthEdition of the PCI Design Handbook.9The sections are somewhat “historical” in that tee widths have increasedover time. For a brief period, producers manufactured 6 ft (1.83 m) widemembers. The 8 ft (2.44 m) wide teebecame the industry standard for abouttwo decades, followed by 10 ft (3.05m) tees, then 12 ft (3.66 m) wide tees.

Currently, 15 ft (4.57 m) wide teesare fabricated in some locations. TheDesign Handbook presents data for 8,10, and 12 ft (2.44, 3.05, and 3.66 m)sections with varying stem heights anddeck thicknesses. Table 3 summarizesthe sections used in this analysis.

Two methods were used to calculatethe torsion constant K. First, the double tee sections were modeled usinglow-order finite differences to solvethe general partial differential equation for cross sections without openings:

+ (1)8x2 3y2

whereH = -2G0G = shear modulus of elasticity(9 = twist angle per unit lengthEq. (1) is derived by using the

Prandtl stress function, where is afunction of the orthogonal coordinatesX and Y.

The differential equation accountsfor stresses due to pure rotation aswell as the out-of-plane warping thatoccurs for non-circular cross sections.The stress function for torsion has

Note: 1 ft = 0.3048 m; tin. = 25.4 mm.

dramatic effect on the warping behavior of a double tee.

• There was too much variation indeck reinforcement to conclude howor if this affects cracking or theamount double tees can be warpedprior to cracking.

• The data indicate that the thicker thedeck, the less double tees can bewarped. The reason for this is thatthe torsion constant is proportionalto the cube of thickness. There wasinsufficient data to determine successful warping criteria for deckthicknesses other than 2 and 4 in.(51 and 102 mm).

• Using rocker-beam assembliesunder the front end of hauling trailers can mitigate cracks resultingfrom transportation of tees withinplant storage areas or to jobsites.Fig. 3 shows a trailer fitted with arocker beam. The double tee stemssat on rubber softeners in the twobrackets located on top of the steel

tube section. Without the rockerbeams, twisting transmitted fromthe trailer to the tee can be considerably more than the amount designedinto the structure to accommodatefloor drainage. Truck drivers cannotalways avoid curbs and potholes.

RATIONAL METHOD TODETERMINE TORSION

STIFFNESSDescribed below is a simplified ap

proach for analyzing torsion and a discussion of some practical results obtained from using this methodology.

Method of Analysis

Tension and shear stresses in doubletee sections that are warped are influenced by a number of stress combinations. Prestress forces and flexurecause tension and compressionstresses, whereas flexure and torsion

36 PCI JOURNAL

b

Fig. 4. Typicaldouble tee section.

the same form as the differential equation describing the deflection of anelastic membrane subjected to pressure. It is also analogous to the shapeof the deflected membrane.

The tangent to the membrane parallel to the ZX-plane is equal to the puretorsion shear stress acting in the Y-direction. The tangent to the membraneparallel to the ZY-plane is equal to thetorsion shear stress acting in the X-direction. These tangents give the following equations:

T7 =

ay

T =—

8x

The volume under the elastic membrane is exactly equal to one-half ofthe applied torsion moment. Thederivation of Eqs. (1), (2) and (3) isbeyond the scope of this paper; however, it can be found in most advancedmechanics of materials textbooks.

A fine mesh size [0.10 in. (2.54mm)] was utilized so that stress concentrations could be discerned. Thisresulted in solving more than 125,000equations for the larger tee cross sections. Stresses and bending momentswere numerically integrated over thecross section to check equilibrium requirements and the accuracy of eachmodel.

As a verification, twist angles andstresses were also checked and foundto be within 1 percent of known closedform solutions for elliptical, triangularand rectangular cross sections. Oncethe computer program had generatedthe membrane surface, the volume

was calculated. After the volume wascalculated, the torque on the doubletee was easily obtained by multiplyingthe volume by 2.

The relationship between the angleof twist and the torque is taken fromReference 9:

QLKG

(4)

in whicho = twist angle per unit lengthM1 = calculated torsion moment

(2) G = shear modulus of elasticityK = torsion constant for sectionIn Eq. (4), the only unknown quan

tity is the torsion constant K.(3) The second method of calculating

K, using a simplified approach, is described in the following paragraphs.

Table 4. Constant c based on ratio bit.bit

or Constant

(Ii + 2t)It avg c

1.0 0.141

1.5 0.196

2.0 0.229

2.5 0.249

3.0 L 0.263

4.0 0.281

5.0 29i

10.0 0.312

= 0.333

Fig. 4 shows a cross section of a typical double tee member.

Eq. (5) gives the St. Venant’s torsion constant K in terms of the thickness (t) and width (b) for a rectangularsection (see Reference 9):

ts—top

ts_avg —

Jtsbot -

A - bitsavg =

2h

I t

A = cross section area II h

C

Double Tee Size

Fig. 5. Torsion stiffness vs. double tee size with 30 in. (762 mm) stem depth.

January-February 2003 37

KDeck = c1bt3 (6)

Fig. 6. Contribution of stems to torsion stiffness for 2 in. (51 mm) deck.

I 22” Stem 26” Stem 32” Stem I

Fig. 7. Contribution of stems to torsion stiffness for 4 in. (102 mm) deck.

Ktem = c2(h + 2t)(ts3avg) (7)

The torsion constant for the complete double tee section was obtainedby summing the values of K for thedeck and the two stems as shown inEq. (8):

Ksjmpie Kieci +2Ktem (8)

All of the double tee cross sectionslisted in the Fifth Edition of the PCIHandbook that are practical to be usedfor typical parking deck spans wereanalyzed for stiffness and stresses. A15 ft (4.57 m) wide tee was also analyzed to evaluate an upper bound.

A shear modulus G = EI(2)(1 +

in which E is the modulus of elasticityand i is Poisson’s ratio, was calculated to be 2006 ksi (13.8 GPa). Thisvalue was held constant for all tee sections. [E = 4696 ksi (32.4 GPa) forf’= 6000 psi (41.4 MPa) and = 0.17for concrete].

Two sets of analyses were done foreach section, assuming a 60 ft (18.30m) tee length. The first assumed a A =

11/2 in. (38 mm) warp while the second assumed a 2 in. (51 mm) warpmeasured over the full width of the tee[i.e., 10 ft (3.05 m)1.

Unit twist angles of 0 = 0.000018radians per in. and 0 0.000023 radians per in. were used for the 11/2 and 2in. (38 and 51 mm) warp cases, respectively. For reference, the twistangle of 0.000023 radians per in. for a10 ft (3.05 m) wide x 60 ft (18.30 m)long tee corresponds to a warping displacement of:

A = (0.000023)(10)(12)(60)(12)= 1.987 in. (50.5 mm)

K = cbt3

In this equation, c is a constantwhose value depends on the bit ratioas shown in Table 4.

Table 4 gives the values for the constant c for corresponding bit ratios.

Double tee sections are subdividedinto three rectangular sections, twostems and one deck. For double tees, band t are the long and short sides ofthe deck and stem rectangles, respec

(5) tively. The contribution of the stems tothe total torsion constant K was modified by adding two times the deckthickness to the length.

The reason for this modification willbe discussed later. By doing so, thesimplified value of K matched themore accurate value from the finitedifference method. The torsion contribution to constant K for the deck andthe stems are calculated using the following equations:

Tables 5 and 6 show the results ofthe analysis for 8, 10 and 12 ft (2.44,3.05 and 3.66 m) wide double tees.

Column (2) lists the theoreticaltwisting moment M1 needed to causethe assumed angle of twist.

Column (3) lists the torsion stiffnessconstant K calculated by Eq. (4).

Column (4) lists the K calculated bythe proposed simple method.

Column (5) shows the margin oferror of K by using the simple method.

22” Stem 26” Stem 32” Stem

=

‘U

C

8-0” DT 10-0” DT

DoubieTee Size

12-0” DT

C

(U

=

80.0%

70.0%

60.0%

50.0%

40.0%

30.0%8-0” DT 10-0” DT

Double Tee Size

12-0” DT

38 PCI JOURNAL

Table 5. Torsion stiffness of double tee members: 11/2 in. warp. C = 2006 ksi; Maximum recommended principaltensile stress = 4Iöb = 309 psi; 0 = 0.000018 radians per in. (11/2 in. warp over 10 ft width for 60 ft span).

Double tee M actual KA, K Error KDek/KSjm t, KAf relative to

size (in.-kips) (in.4/rad) (in.4/rad) (percent) (percent) (percent) (psi) (psi) 10DT22+4 (percent)

8DT22+2 64.2 1946.9 1937.0 —0.5 12.7 87.3 199 2.57 40.5

8DT22+4 135.3 4104.4 3897.4 —5.0 49.7 50.3 258 3.33 85.3

8DT30+2 173.4 5257.9 5203.5 —1.0 4.7 95.3 238 3.07 109.3

8DT30+4 257.4 7804.8 7506.2 —3.8 25.8 74.2 314 4.05 162.2

10DT22+2 69.1 2095.4 2094.0 —0.1 14.8 85.2 205 2.65 43.6

10DT22+4 - 158.6 4810.7 4499.7 —6.5 54.0 46.0 254 3.28 100.0

10DT30+2 179.0 5428.1 5415.0 —0.2 5.7 94.3 3.27 112.8

10DT30+4 282.4 8564.9 8166.4 —4.7 29.8 70.2 309 3.99 178.0

12DT26+2 158.2 4796.5 4842.5 1.0 7.8 92.2 251 3.24 99.7-. - ---___

12DT26+4 276.1 8371.8 8046.5 —3.9 36.4 63.6 305 3.94 174.0

12DT30+2 181.1 5492.7 5480.4 —0.2 6.9 93.1 253 I 3.27 114.2- if - --- - - - —-- -

12DT30+4 299.5 9081.7 8665.7 —4.6 33.8 66.2 309 3.99 188.8

Note; I ft = 0.3048 m; 1 in. = 25.4 ann; 1 psi =0.006895 MPa; 1 kip=4.448 kN; 1 in.-kip=0.113 kN-m.

Table 6. Torsion stiffness of double tee members: 2 in. warp. C = 2006 ksi; Maximum recommended principaltensile stress = 4Jöö = 309 psi; 0 = 0.000023 radians per in. (2 in. warp over 10 ft width for 60 ft span).

Double tee M, actual KAC.S K Error KDkIKs,,, Ks5,,/Ks1,,, tma:

KAj relative to

size (in..kips) (in.4lrad) (in.4/rad) (percent) (percent) (percent) (psi) (psi) 10DT22+4 (percent)

8DT22±2 - 82.0 1946.9 1937.0 —0.5 12.7 87.3 255 3.29 83.0

8DT22+4 172.9 4104.5 3897.4 —5.0 49.7 50.3 1 330 4.26 175.0. - I - - -____ -

8DT30+2 221.6 5258.0 5203.5 —1.0 I 4.7 95.3 I 304 3.92 224.2. - I - - I - -. -

8DT30+4 328.9 7804.9 7506.2 —3.8 25.8 74.2 401 5.18 332.8

10DT22+2 88.3 2095.4 — 2094.0 0.1 J 14.8 85.2 262 3.38 89.4

10DT22+4 202.7 4810.7 4499.7 —6.5 54.0 46.0 325 4.20 205.1

IODT3O+2 228.7 5428. 5415.0 —0.2 5.7 94.3 324 4.18 231.5

IODT3O+4 360.9 8564.9 8166.4 -4.7 29.8 70.2 395 5.09 250.1

12DT26+2 202.1 4796.6 4842.4 0.9 7.8 92.2 321 4.14 204.5

12DT26+4 352.8 8372.0 8046.4 =3.8 36A 63.6 390 5.03 357.0

12DT30+2 231.4 5492.8 5480.4 —0.2 6.9 93.1 324 4.18 234.2

12DT30+4 382.7 9081.8 8665.7 -4.5 33.8 66.2 395 5.10 387.3

Note; I ft=0.3048 m; I in. = 25.4 mm; I psi = 0.006895 MPa; 1 kip = 4.448 leN; 1 in.-kip=0.113 kN-m.

Column (6) lists the percentage oftorsion moment resisted by the deck.

Column (7) lists the percentage ofthe torsion moment resisted by the twostems.

Column (8) lists the maximum resultant torsion shear stress found anywhere on the cross section.

Column (9) shows how the calculated shear stresses compare to themaximum tensile stress as recommended in ACT 31899,b0 Section11.6, Commentary.

Column (10) compares the torsionstiffness for each section to the10DT22+4.

Discussion of Results

The torsion stiffness constant (K) ofa double tee is dependent upon the two

stems and the deck (flange). For allstem heights and deck thicknessesshown in the preceding tables, thecontribution of stem and deck thickness for a given tee width is shown.

Fig. 5 shows the variation of the torsion stiffness constant (K) for all threedouble tee widths with a constant 30in. (762 mm) stem depth.

The figure shows two interestingitems; first, as the width of the doubletee is increased the stiffness is increased moderately and, second, as thedeck thickness increases the stiffnessincreases significantly.

The contribution of the deck in relation to that of the stems varies as thedimensions of the double tees change.Figs. 6 and 7 show the contribution ofthe stems to the total stiffness of thesection. A comparison of the double

tee sections with 2 in. (51 mm) decksshows that if the stem depth is heldconstant, the contribution of the deckdecreases as the width increases.

The same trend holds true for 4 in.(102 mm) decks. The difference between 2 and 4 in. (51 and 102 mm)decks is that for the 2 in. (51 mm)decks the contribution of the stems tothe total stiffness is between 85 and 95percent while for the 4 in. (102 mm)decks it is between 45 and 75 percent.

Note that all of the sections in Tables 5 and 6 have fillets. The effect of3 in. (76 mm) chamfers was also investigated. Table 7 shows a comparison of 10 ft (3.05 m) wide double teeswith 3 in. (76 mm) fillets and 3 in. (76mm) chamfers.

Table 7 shows that both the torsionconstant and maximum stress for a

January-February 2003 39

Table 7. Comparison of 3 in. fillet to 3 in. chamfer for 10 ft wide double tee sections:11/2 in. warp; G = 2006 ksi; maximum recommended principal tensile stress = 4Jöö = 309 psi;0 = 0.000023 radians per in. (2 in. warp over 10 ft width for 60 ft span).

Double lee M, actual KA, KsI,,.,,k K Error KDk/KSjm K5im/Ksj, r,,,.,,/.j K,15 relative to

size (in.-kips) (in.4 per radian) (in.4 per radian) (percent) (percent) (percent) (psi) (psi) 10DT22+4 (percent)

I 0DT22+2. 69.1 2095.4 2094.0 —0.1 14.8 85.2 205 2.65 43.6

3 in. fillet

10DT22+2 t I. 77.3 2345.0 2373.7 1.2 13.1 86.9 309 3.98 48.7

3 in. chamfer

10DT22+4. 158.6 4810.7 4499.7 —6.5 54.0 46.0 254 3.28 100.0in. fillet

10DT22+4 I ‘—

. . 176.3 5346.5 4823.6 —9.8 50.4 49.6 372 4.81 111.13 in. chamfer

IODT3O+2. 179.0 5428.1 5415.0 —0.2 5.7 94.3 254 3.27 112.8

3 in._fillet

1ODT3O+2 —

. 189.2 5737.4 5861.8 2.2 5.3 94.7 375 4.84 119.3In. chamfer

IODT3O+4. 282.4 8564.9 8166.4 -4.7 29.8 70.2 309 3.99 178.0

3 in._fillet

IODT3O+4. 304.5 9234.4 8667.5 —6.1 28.1 71.9 459 5.92 192.0

3 in. chamfer

I —‘ote: 1 ft = 0.3048 m; 1 in. = 25. mm; 1 psi = 0.006895 MPa; 1 kip=4.448 kN; I in.-kip=0.113 kN-m.

double tee with a chamfer is largerthan that of a double tee with a fillet.However, for 10 ft (3.05 m) wide double tees, the torsion constant increasedby about 10 percent but the maximumstress increased by over 45 percent.

Torsion Stress Behavior

The pure torsion shear stresses wereextremely well behaved and consistent. A single discontinuity always occuffed in the form of a short “crease”when chamfers were provided at thestem-to-deck interface. Fig. 8 shows athree-dimensional plot of, the stressmembrane that represents the solutionto the differential Eq. (1) by the finitedifference method.

Torsion stresses are obtained fromthis membrane. Theory indicates thatthe steeper the curve, the higher thestress. Maximum shear stresses occurin the deck over each stem and alongthe exterior edges of the stems justbelow the deck, i.e., at the fillets orchamfers on the underside of the deck.Fig. 9 shows contour lines of stresseson a double tee section with fillets andchamfers, respectively.

Stresses induced by chamfers weremuch greater than tees that had fillets.Fillets were substituted for chamfersto provide consistency in modelingand stress comparisons. Fillet sizes

Fig. 8. Torsion membrane for double tee section.

Double Tee wI3 in. (76 mm)

Fillet

Double Tee w/3 in. (76 mm)

Chamfer

Fig. 9. Shear stresses caused by pure torsion on double tees.

40 PCI JOURNAL

500

E -*

g:Trendne• A 4Deck

.

4 Deck Trenciflne4*fcA1/2

5*ftAlI2

.

6*fcA1I2

I I I

r_A

.._

• ÷ 2 Deck—2’ Deck Trendline

: A 4”Deck.:: 4” Deck Trendline

• 4*fcA1I2

6*fc1!2

Fig. 10. Relationshipbetween maximumtorsion stress andtorsion stiffness forstandard PCI doubletee sections: warpof approximately11/2 in. (38 mm) fora lOft (3.05 m) widedouble tee over a60 ft (18.30 m) span.

Fig. 11. Relationshipbetween maximumtorsion stress andtorsion stiffness forstandard PCI doubletee sections: warp ofapproximately2 in. (51 mm) for alOft(3.05 m)widedouble tee over a60 ft (18.30 m) span.

from 2 to 10 in. (51 to 254 mm) wereanalyzed for a few of the tees but werefound to have little impact on the magnitude of the maximum torsionstresses.

The portion of deck outside of thestems behaves much like that of longthin rectangles subjected to torsion.Concentric stress contours are parallelto the deck surfaces with the higheststresses occurring on the exterior surfaces. The stress contours diminish tozero at the interior of the deck. Thestems also reveal concentric stresscontours that are parallel to the vertical axis rather than parallel to the tapered sides of the tee stems. The overlapping stem and deck regions show asmooth transition of horizontal deck

contours and vertical stem contours.With the exception of the overlap

ping stem and deck regions, the plotted stress functions show, as expected,that the deck and stems tend to behaveas independent torsion resisting components. However, the classical approach of summing the stiffness ofcomponent rectangles to calculate thetorsion constant of a built-up sectionwas found to consistently underestimate the theoretical stiffness.

This behavior is caused by the increased resistance found in the overlapping stem and deck regions (seeFig. 8). A good correlation with thetheoretical stiffness was found by simply increasing the depth of the stemsby two times the deck thickness.

It was noted that torsion stiffness increases at a much greater rate than thecross-sectional area. For instance, thetorsion stiffness for a 12DT30+4 isnearly twice that of a 1 0DT22+4. It isalso interesting to note that increasingthe deck widths from 8 to 10 ft (2.44to 3.05 m), while holding the stem anddeck thickness constant had absolutelyno effect on the maximum shearstresses.

The relationship between the torsionconstant K and the torsion stress canbe used as a guideline for designers toassess any particular double tee crosssection for its potential for cracking.By using the least squares fit theory,the data from Tables 5 and 6 were reduced to the simple curves shown in

400

• 3000.

x

+ 200

100

00 5000 K - (in4lradian) 10000 15000

500

400

300

j 200

100

00 5000 10000 15000

K - (in.4lradian)

January-February 2003 41

Figs. 10 and 11. The first figure usesdata for a warp of 11/2 in. (38 mm)across the full width of the double teesection, while the second curve is for a2 in. (51 mm) warp.

Note the following assumptions:1. Elastic analysis assumes 3 in. (76

mm) deck/stem fillets — deck stemchamfers tend to increase stress.

2. Plots are for a unit twist angle of0.000018 and 0.000023 radians per in.This is equivalent to about 1’/2 and 2in. (38 and 51 mm) of warp fora lOft(3.05 m) width over a 60 ft (18.30 m)span for Figs. 10 and 11, respectively.

3. Plotted shear stresses are maximum resultant ;. and occur on ornear the deck/stem chamfers.

4. Linear interpolation is used tocalculate Tmax for other twist angles.

The three horizontal lines on Figs.10 and 11 represent stresses equal to 4,5, or These allow the designerto exercise judgment as to how muchstress is acceptable for the particulardesign conditions.

Stresses are sensitive to the degreeof warp. When the warp is 11/2 in. (38mm), all but one of the data points fallunder the 4JI line, but when the teeis warped to 2 in. (51 mm), thestresses shift to the 5J state ofstress.

THEORETICALCALCULATION OFCRACKING POINT

An in-depth research project on thedurability of precast, prestressed parking structures was carried out by PCIin 1995 by the Consulting EngineersGroup of San Antonio, Texas.” Part

of the focus of this research was thedetermination of warping withoutcracking of pretopped double teeswith 4 in. (102 mm) thick decks(flanges).

To determine the theoretical limitsof warping for these members, a finiteelement analysis was performed usinga 1 in. (25.4 mm) grid mesh modelinga system with three supported elements (stems) and the fourth supported on a spring with a constant K.The model assumed 5 ft (1.52 m) between the stems [10 ft (3.05 m) overallwidth] and analyzed tee lengths of 40and 60 ft (12.2 and 18.3 m).

The model varied the degree ofwarp to determine the maximum displacement between the stems at themaximum tension stress as defined bythe modulus of rupture for a specificconcrete. Note that ACT 31810 recommends using a value of fr in the rangeof 0.1f to 0.15f.

The maximum tensile stress occursin the top of the flange near the unwarped (non-spring-supported) stemat the end that is warped. The stressdistribution across the flange is similarto a classical bending stress distribution. However, the stress distributionat the top surface is similar to thatcaused by a concentrated load. Oncethe maximum tensile stress is reached,the crack length is approximately 3 in.(76.2 mm).

For a high quality concrete with service compressive strength of 6500 psi(44.8 MPa) and corresponding modulus of rupture of 975 psi (6.72 MPa)(equal to 0.1 5f’), the permissible warp(measured between stems) would beon the magnitude of 0.45 in. (11.4

mm) for a 40 ft (12.2 m) tee lengthand 0.50 in. (12.7 mm) for a 60 ft(18.3 m) length member. Fig. 12shows the maximum stresses in thedeck with respect to stem-to-stemwarping. The program write-up cautions that the values chosen forfr are afunction of the acceptability of noncritical cracking.

The finite analysis model was alsoused to predict crack propagation. Theanalysis revealed that the crack propagates along the flange rather thanthrough the flange. This is an important point when considering the crackcontainment afforded by welded wirefabric reinforcement typical in thistype of member. For more informationregarding this study, see Reference 11.

EMPIRICAL DATAThe program referred to in the pre

vious section also included full scaletesting of two double tee members, 10ft (3.05 m) wide by 28 in. (711 mm)deep with 4 in. (102 mm) thick flanges(10DT24+4). The last 2 ft (0.61 m) oflength at each end of the members washeld down to a 2 in. (51 mm) thickness as though a pour strip was to beadded in the field.

One double tee was 47 ft (14.3 m)long and the other was 60.5 ft (18.4m) long. Warping was induced in thetee by setting the tee stems on levelshim stacks and then removing shimsincrementally from beneath one stem.

Based on the test results, the following observations can be made:

The 60.5 ft (18.4 m) long memberwithstood a warp of 3/4 in. (19 mm)between the stems before cracking,while the 47 ft (14.3 m) long member had a very small crack start at 5/

in. (16 mm) differential between itsstems. The shorter member was difficult to test because it tended to riseup on opposite corners as shimswere removed.

• Cracking in the top flange progressed diagonally across the flangeand the bottom cracking remained inthe flange to stem intersection. Fig.13 shows the crack pattern developed in the 60.5 (18.4 m) longmember.

• The maximum crack width for 1/8in. (43 mm) differential between the

18001600

1400

1200ioco80O

600

400

2000

o o.i 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2

Degree 01 Warp (Stem to Stem), in

Fig. 12. Maximum elastic tension stresses in warped double tee flanges.

42 PCI JOURNAL

stems was 0.010 in. (0.25 mm) inthe longer member. The cracklength extended approximately 6 ft(1.83 m) from the end of the member.As the warping increased, crackingoccurred at the juncture of the stemand the flange bottom for the legthat was lowered. Cracking occurredin the top of the flange above theadjacent stem on the same end. Nocracking occurred at the end withboth stems at the same elevation.

• Maximum crack widths increased aswarping increased.

• Crack widths were less at the bottom of the flange than at the top.The load tests substantiated the re

sults predicted by the finite elementstudy. Both studies confirmed that observable cracking occurred when theoverall warp exceeded 11/2 in. (38mm).

SUMMARY OF RESULTS

Results are summarized for the industry survey and theoretical analysis.

Industry Survey

Industry practice and associated experience gained therefrom indicatethat warping of twin stemmed teemembers in parking structures is notonly ongoing, but also commonplacein order to create transverse slopes.When double tees are to be warped, allparties should be well informed regarding the potential for cracks in themembers.

The study indicates that a 60 ft (18.3m) long double tee element can betwisted to the amounts shown in Tables1 and 2 without cracking. Warping toproduce floor slopes of about 11/4 percent is routinely done within the industry with no cracking, and as much as21/2 percent with minor cracking.

As previously stated, not all flangecracking is attributable to the in-placecondition of the tee. Stripping, handling, and, probably most significantly,hauling the product to the jobsite induce torsion stresses in the element.Because prestressed concrete producers have been facing the challenge ofmanufacturing plant-cast 4 in. (102mm) thick flange double tees for park-

(Note: 1 in. = 25.4 mm)

ing deck applications for over twentyyears, special procedures for handlingthese members have been developed.

One of the methods successfullyused by several precast concrete manufacturers involves using a specially designed rocker beam, which was discussed earlier in the paper (see Fig.14). Other precautions include providing adequate jobsite access for deliverytrucks because the jobsite typically hasthe worst terrain encountered en route.

Theoretical Analysis

The following is a list and discussion of the conclusions that can bedrawn from the theoretical analysis.

Narrower tees have a lower torsionstiffness constant (K) than widertees.Results from both the finite differ

ence method and simplified methodconfirm that as the width of the deckdecreases, the torsion constant K alsodecreases.

Double tees with 2 in. (51 mm)decks are more flexible than thosewith 4 in. (102 mm) decks.By reviewing Tables 5 and 6, it is

seen that double tees with 2 in. (51mm) decks have a much lower K

value than double tees with 4 in. (102mm) decks. By looking at the stiffnessof a 1ODT3O+2 and 1ODT3O+4 it isseen that the contribution of the stemsKtem remains constant and the contribution of the deck KDeCk changes.• In general, the stems contribute

more to the total stiffness than thedeck.As the ratio of the area of the stems

to the total area of the double tee increases so does the ratio of the stiffness of the stems to the total stiffness.• At the junction of the tee stems and

the deck, chamfers cause higherstress concentrations in the doubletees than do fillets.The relationship between the torsion

stiffness, K, and the maximum stressesis not evident from the analysis. However, it is clear that the use of chamfers rather than fillets affects the maximum torsion stress much more than itaffects the torsion constant (K).• Results confirm that the simplified

method for calculating the torsionconstant K is quite accurate whencompared to the classical solution.The torsion constant calculated by

the simplified method differs from themore accurate method by only 21/2 percent on average for all cross sections.

TOP VIEW

1) /8 in. warp no cracks

2) 3/4 in. warp 0.002 in. width

BOTTOM VIEW

3) lV8 in. warp 0.005 in. @PT1 and 0.002 in. @PT3

4) 1/g in. warp 0.010 in. @PT1 and 0.005 in. @PT3

Fig. 13. Crack pattern for 10 ft (3.05 m) wide 60.5 ft (18.4 m) long tee.

January-February 2003 43

• A relationship exists between thetorsion constant and the torsionstress.As a tee becomes stiffer, the stress

developed in the cross section increases. The degree of warp greatlyaffects the state of torsion stress in thedouble tee. For standard tees (as contained in the PCI Design Handbook8),stresses are less than 4’Z when thewarp is 11/2 in. (38 mm), but stressesrise to about 5/ when the warp isincreased to 2 in. (102 mm).• Curves were developed to aid de

signers in evaluating torsion cracksin typical double tee sections.Using simple equations, a designer

can calculate the torsion stiffness of adouble tee shape and enter a curve todetermine the state of torsion stressthat will be induced in the tee when itis warped.• Lightweight concrete double tees

are less flexible than normal weighttees.The shear modulus (G) for

lightweight concrete is smaller thanthat for nonnal weight concrete. Looking at Eq. (4), if M, and 0 are heldconstant, as G decreases K increases.This combined with the fact thatlightweight concrete cracks at lowerstresses than normal weight concretemeans that double tees constructed oflightweight concrete are more susceptible to warping cracks.

CONCLUSIONSBased on the results of this study,

the following conclusions can bedrawn:

1. Double tees have an inherentability to sustain a certain amount oftwist, thereby providing the designertwo-way slope capability utilizing either field-topped or pretopped elements. While surface cracking can becontrolled, it should be noted thatmost surface cracks require no furtherremedial action than to be coveredwith a sealer to prevent water penetration. The double tee cast in factorycontrolled conditions with high performance concrete is one of the besttools in the designer’s arsenal for highquality, cost effective parking deckconstruction. Key aspects of high performance concrete are low watercementitious material ratio (< 0.40),high strength [>6000 psi (41.4 MPa)],and sufficient air entrainment (6’/2percent ± 1 ‘/2 percent).

2. It should not be assumed that thepresence of a crack in the top surfaceof a double tee flange is cause forstructural or long-term durability concern. Reports issued by ACI, PCI, andothers within the industry indicate thatcracking on the magnitude of 0.0 15 in.(0.38 mm) wide or less is generally insignificant. The depth of the crack,when compared to the depth of reinforcement (as opposed to cracking inline with the reinforcement), is nottypically cause for concern. The nature of the cracking described earlierin this paper falls into that category.The significance of a crack in the surface becomes more pronounced whenthat crack occurs in an area wherethere is ponded water, perhaps contaminated with chloride ions that leadto corrosion of reinforcement. This is

why proper drainage is imperative inthe design of the deck.

3. The information contained hereinis informational and not a directive forwarping double tee elements. Individual owners and producers need to establish what they are willing to acceptregarding the amount double teesshould be warped. Also, cracks indouble tee elements can originate fromother sources such as shipping andhandling.

Needs for Future Study

This article presents a comprehensive analysis of double tee sectionssubjected to torsion only. It excludesother stress combinations. Prestressforces and flexure cause tension andcompression stresses, while shearstresses are caused by flexure and torsion. The complex interaction of gravity and axial loads are not addressedhere. Further analysis taking these factors into account will help verify theresults derived from the torsion onlymodel or may give new informationregarding the initiation of cracking atthe interface of the stem and the flangeas influenced by member geometryand applied loads.

Controlled member testing could bedone to provide actual data that couldbe used to expand the knowledge ofdouble tee warping. A prescribedmethod of testing and results documentation could be prepared to assistthose willing to do the tests and thuscreate a database for empirically derived warping limits. As of this writing, one producer is performing teststo determine the relation betweenmember width and the initiation oftorsion cracking, as well as the implications of a “starter” crack to the development of torsion cracks. Thisstudy will be made available to PCIupon completion.

Closer scrutiny of product throughits various cycles (i.e., production, cutting of strand, stripping, storage, shipping, and erection) will expand the experience database and provide bettercorrelation of theoretical and observedperformance of warped double tees. Anumber of producers are actively collecting this information. A centraldatabase would be helpful in assessing

Fig. 14. Trailer rocker beam supporting a double tee.

44 PCI JOURNAL

where cracking is most likely to occurbased on the highest number of observations available.

ACKNOWLEDGMENTThe authors are grateful to the

many individuals who contributedtechnical comments and guidancethroughout the various stages of this

project. They are Bob Corry, TomD’Arcy, John Garlich, Degan Hambacher, Roger Kaness, Joe Miller, andLeo Whiteley.

Particular thanks go to John Hanlonfor his understanding and applicationof numerical methods and advancedmathematics used to develop a rational method to determine torsion stiff

ness. His computer modeling and

methodology in integrating graphic results into custom AutoCAD drawingsmade concepts and data presentationsmuch easier to read and understand.

The authors are also appreciative ofthe excellent technical contributionsmade by the PCI JOURNAL reviewers. They are Alex Aswad, Ken Baur,

Suresh Gami, Walter Korkosz, GeorgeNasser, and Kim Seeber.

REFERENCES

1. PCI Committee on Quality Control Performance Criteria,“Fabrication and Shipment Cracks in Prestressed Hollow-CoreSlabs and Double Tees,” PCI JOURNAL, V. 28, No. 1, January-February 1983, pp. 18-39.

2. ACI Committee 362, “Guide for the Design of Durable Parking Structures (ACI 362.1R-97),” Section 3.1 Drainage, American Concrete Institute, Farmington Hills, MI, 1997.

3. Parking Structures: Recommended Practice for Design andConstruction, MNL-129-98, Chapter 4, Section 4.4.1 Slope,PrecastlPrestressed Concrete Institute, Chicago, IL, 1997, pp.4-16.

4. The Dimensions of Parking, Third Edition, Chapter 18,Drainage, ULI-the Urban Land Institute & NPA-the NationalParking Association, Washington, DC, 1993.

5. Weant, R. A., and Levinson, H. S. (Editors), Parking, Chapter9, Garage Design, ENO Foundation for Transportation, Inc.,Westport, CT, 1990.

6. Chrest, A. P., Smith, M. S., Bhuyan, S., Monahan, D. R., andIqbal, M., Parking Structures: Planning, Design, Construction,

Maintenance and Repair, Third Edition, Chapter 12, Section12.2.3.2, Drainage, Kiuwer Academic Publishers, Boston/Dordrecht/London.

7. Lew, I. P., Gutman, A., Hoffman, R. G., “Parking DesignChecklist 2000,” The Parking Professional, May 2000, 25 pp.

8. PCi Design Handbook — Precast and Prestressed Concrete,Fifth Edition, Precast/Prestressed Concrete Institute, Chicago,IL, 1999.

9. Ugural, A. C., and Fenster, S. K., Advanced Strength and Applied Elasticity, Chapter 1 and 2, Elsevier North-Holland Publishing Company, Inc., 1977.

10. ACI Committee 318, “Building Code Requirements for Structural Concrete (318-99) and Commentary (31 8R-99),” American Concrete Institute, Farmington Hills, MI, 1999.

11. PCI Research and Development Committee, Durability ofPrecast Prestressed Concrete Structures, Prepared by the Consulting Engineers Group, Inc. for the Precast/Prestressed ConcreteInstitute, Chicago, IL, 1995.

January-February 2003 45

APPENDIX A — NOTATION

b = long side of deck rectanglec = constant whose value depends on bit ratioE = modulus of elasticity

f = specified compressive strength of concrete

fr = modulus of rupture of concreteG = shear modulus of elasticityH =-2G0K = torsion constant for sectionKDeck = torsion contribution to constant K for deckKtem = torsion contribution to constant K for stemKsjmpie torsion constant for complete double tee section

M = calculated torsion momentt = short side of stem rectangleT = tangent to membrane parallel to ZX plane

Tz), = tangent to membrane parallel to ZY plane= vertical warp= function of orthogonal coordinates X and Y= Poisson’s ratio

B = twist angle per unit length= torsional stress in XZ plane= torsional stress in YZ plane

Tmax = maximum torsional stress

APPENDIX B — EXAMPLE OF WARPING FOR DRAINAGE

Fig. B 1 shows a typical framing plan for a two-bay, singlehelix parking structure that has been designed in generalconformance with the floor slope recommendations of Section 1. The west bay includes a 5.83 percent ramp, while theeast bay is flat (not ramped). Typical east bay double teesare longer than typical west bay tees [64 ft 6 in. and 59 ft 6in. (19.66 and 18.14 m), respectively] to accommodate a 4 ft(1.2 m) wide walkway needed to satisfy the client’s requirement for Americans with Disabilities Act (ADA) spaces.The shorter tees are I 0DT22+2 members while the longertees are 10DT26+2 and 10DT26+4.

The tees from Gridlines 3 to 8 are pretopped with a 4 in.(102 mm) thick deck. The short tees by stair cores, by thesoutheast notch, and the tees from Gridlines 2 to 3 and 9 to10 are field-topped with a constant 3 in. (76 mm) thicknessof cast-in-place concrete placed over the precast members.

The rationale for field-topping these tees were to eliminate differential cambers between adjacent short/long teesand the need to slope these tees greater than recommendedvalues for pretopped members.

In the flat (east) bay, drains are located about every 60 ft(18.29 m) along Grid B at the wash line. Tees are set at aconstant elevation around the perimeter of the structure

Table Bi. Double tee warping summary.

(top-of-tee at elevation zero). Typically then stem bearingsat the west end of tees are lowered and tees warped in a 30ft (9.14 m) pattern that allows a ridge/valley arrangement toestablish a suitable drainage scheme. Basically, tees arelowered 16 in. (406 mm) at valley lines and 12 in. (305 mm)at ridge lines so that over each 30 ft (9.14 m) bay, a 4 in.(102 mm) rise or fall along Grid B is developed.

Floor slopes resulting from this geometry are 2.07 percentalong valley lines and 1.55 percent along ridge lines. Theseslopes are adequate to drain the floor properly. Becausesome tees by the elevator/stair core are shorter than typicaltees, slopes along the diagonals vary slightly, but are generally about 1.9 percent. Adjustments were necessary in theend bays (turning radius bays), to accommodate notches andnot-typical drain locations. On the ramp bay, drains are located in only two places as shown because water will rundown the ramp along washes.

To accomplish the drainage pattern shown, tees in the flatbay were warped as shown in Fig. Bi. Table Bi shows theanticipated stress created in each section given the specifiedamount of warp. This example was chosen because it usesdouble tee sizes that were not included in the analysis. Theexample shows how the information presented in this paper

Double tee Location Length Warp Angle of Ksjj r Tensile stress range

section (Grids) (ft) (in.) twist (0) (in.4 per radian) (psi) (psi) 6j7 (psi)

10DT22+2 1-2&B-C 45.25 1.90 0.000029 2094 344 310 465

10DT26+2 2-3 &B-C 64.33 2.90 0.000031 5003 404 310 465

100T26+4 3-9 & B-C 64.33 1.33 0.000014 7837 230 310 465

10DT26+2 9-10&B-C 64.33 3.00 0.000032 5003 421 310 465

Note: 1 jn.= 25.4 mm; 1 psi = 0.006895 MPa.

46 PCI JOURNAL

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can be used to help predict warping behavior of varioussizes of double tees.

The following is an explanation of how each item in TableBi was obtained using the 10DT26+2 at Grids 2-3 and B-Cas an example.

Double tee section, location, length and warp are obtainedfrom Fig. B1.

Angle of Twist (9):

Warp

Width x Length

Torsion Constants (K):

Ksjmpie is calculated using the method described in the“Rational Method to Determine Torsion Stiffness” portionof this article. See Fig. 4, Table 4 and Eqs. (6), (7) and (8).

577 sq in. (645.2 mm2)10 ft (3.05 m)2 in. (51 mm)26 in. (660.4 mm)6.48 in. (164.6 mm)

A =

b =

h =

tsavgbit =60(h + 2t)it avg = 4.63c1 = 0.324 (Table 4)

= 0.287 (Table 4)KDcck = 311 in.4 per radian [Eq. (6)]Kgtem 2346 in.4 per radian [Eq. (7)1

KTotal = 5003 in.4 per radian [Eq. (8)]

Calculated Shear Stress (-r):

From Fig. 11, read off the stress for a 2 in. (51 mm) warpfor the given torsion constant (Kgjmpie = 5003 in.4 per radian)and deck thickness (t = 2 in.):

warp = 300 psi (2.07 MPa)

Multiply the stress obtained from Fig. 11 by the angle oftwist (8 = 0.000031) and divide it by the angle of twist associated with a 2 in. (51 mm) warp (2” warp = 0.000023).

T2var xQ 300 psi x 0.00003 1= T

°2,varp 0.00023

= 404 psi (2.79 MPa)

All the stresses in this example fall between 4L and6,/j. This does not mean that the tees will not crack, ratherit simply gives a range of stresses for comparison. Whenusing the information presented in this article, it is left to thedesigner’s judgment as to whether the calculated stresses areacceptable for each individual project.

This example shows that setting double tee elevations requires a careful study regarding drainage. This is particularly true when the designer-of-record desires to minimizethe number of floor drains by placing them far apart. Particular attention must be given to the camber of the doubletees. Designers should not depend on camber of the prestressed members to provide drainage slopes.

2.90 in.

(l0ftxl2in.ift)(64.33ftx12 in./ft)

= 0.000031 radians per in.

48 PCI JOURNAL