Behavior and Design of Prestressed Concrete Beams With ...

Behavior and Design ofPrestressed Concrete BeamsWith Large Web Openings

George B. BarneyStructural EngineerStructural Development SectionPortland Cement AssociationSkokie, Illinois

John M. HansonVice-President of OperationsWiss, Janney, Elstner and AssociatesNorthbrook, Illinois

W. Gene CorleyDirectorEngineering Development DepartmentPortland Cement AssociationSkokie, Illinois

Richard A. ParmeleeProfessor of Civil EngineeringNorthwestern UniversityEvanston, Illinois

George B. Barney W. Gene Corley

John M. Hanson Richard A. Parmelee

32

Reports the results of tests on 18 full-size precastprestressed tee beam (13 long-span beams and 5short-span beams) to 1etermine the effects of largeweb openings on the erformance of prestressedconcrete members.The results showed th t large web openings can beaccommodated in pre tressed concrete memberswhile maintaining their full strength. Also,serviceability requirements can be satisfied.A design procedure b sed on the results of thesetests is presented and a fully worked numericalexample illustrates the applicability of the proposeddesign method.

T he trend in recent years to andthe systems approach to building

has generated a need for web open-ings in structural members. Mec an-ical and electrical services in 4iostbuildings are carried in the s acewithin the floor-ceiling sandwich.

Passing these services thro ghopenings in the webs of the foorbeams eliminates a significant am untof dead space and results in a norecompact and often more econorricaldesign. However, the effect of theopenings on the strength and er-viceability of the floor beams mus beconsidered.

Background

Only limited research has bendone to determine the effects of ebopenings in prestressed concrete teebeams, a member widely used in heprecast concrete industry. Ragan tndWarwaruk' at the University of 1-

berta were the first to conduct a seriesof tests on prestressed concrete teebeams with multiple web openings.They found that sizeable web open-ings could be accommodated withoutsacrificing strength. Deflections forbeams with openings were not signif-icantly greater than those for beamswithout openings.

Tests2 were also conducted to de-termine the effect of both vertical andlongitudinal reinforcement in teebeams with multiple openings. It wasfound that increasing the vertical rein-forcement in the posts between open-ings increased shear capacity of thespecimens. Additional tests of pre-stressed beams with multipleparallelogram-shaped web openings3indicated that these beams werestronger than similar beams with rec-tangular openings.

The investigation reported in thispaper was carried out at the StructuralDevelopment Laboratory of the Port-land Cement Association.

PCI JOURNAL/November-December 1977 33

Design Recommendations

Based on the results of the investi-gation reported here, a design proce-dure is recommended for prestressed,pretensioned concrete beams withlarge rectangular web openings. Theprocedure is applicable for beamshaving straight strands.

To avoid slip of the prestressingstrands, openings must be locatedoutside the required strand embed-ment length. This length may be cal-culated using the provisions of Sec-tion 12.11 of the 1971 ACI BuildingCode.4 The value of the force to betransferred by each strand can be es-timated as its breaking strength.

Vertical stirrups must be providedadjacent to both sides of all web open-ings. These stirrups should be propor-tioned to carry the total shear force atthe section where they are located.

The analytical procedure describedlater in this report may be utilized todetermine axial forces, shear forces,and moments in the struts above andbelow openings. The capacity of thestruts to resist flexure and axial loadsmay conveniently be determined frominteraction curves.

Slenderness effects in the compres-sive strut should he considered in ac-cordance with Section 10.10 of theAC! Code.4 When the section beinganalyzed is a tee beam, the effectivewidth of the flange in determiningproperties and capacities of the com-pressive strut should not exceed thelimits established in Section 8.7 of theACI Code.4

Axial forces should be accounted forin the shear design of the struts. Theshear capacity of concrete in the com-pressive strut can be determined fromthe provisions of Section 11.4.3 of theACI Code' for members subjected toaxial compression.

Design of the tensile strut for sheardepends on whether the net axial

force is tension or compression. Fornet axial compression, the shearcapacity of the concrete section with-out web reinforcement can be deter-mined from the provisions of Section11.5.2 of the ACl Code. 4 Eq. (11-11)will usually govern the design. Interms of notation used in this report,this equation becomes:

v = 0.6 f,` ^ (1)*

wherea = minimum width of tensile

strutd = distance from extreme com-

pressive fiber to centroid ofprestressed reinforcementbut not less than 0.8h

fc = compressive strength of con-crete

h = overall depth of tensile strutM. = maximum moment in tensile

strut at section considereddue to superimposed loads.Note that M. _ V 1/2 where lis the effective strut length

M me. = bending moment causingflexural cracking at sectionconsidered due to superim-posed loads

vcz = shear stress at diagonal crack-ing due to all design loads,when such cracking is resultof combined shear and mo-meat

F k = shear force in tensile strut

and

cr = _. ^]' p `

Yt d.A1,t IIn Eck. (2)A = gross area of tensile strutcis = distance between centroidal

axis at tensile and compres-sive struts

*Note that in SI units, ^}' psi = 008304 y f MPa.

34

fpe = compressive stress In on-crete due to prestress myafter all losses, at extr mefiber of section at which en-sile stresses are cause byapplied loads. Note hatfieP(ds + A d) /(d3 gt)where P is the effective re-stress force and d d is the is-tance of the effective re-stress force resultant be owthe centroidal axis of the en-sile strut

I^ = moment of inertia ofcracked section transfor:to concrete

M = moment at center of openy t = distance from centroidal

of uncracked section totreme fiber in tension

When the tensile strut is in net a ialtension, the shear capacity of concr tomay be determined from the pro i-sions of Section 11.4.4 o€ the A ICode.4

Shear reinforcement in the s ts,when required, should be prop r-tioned using the provisions of Sect on11.6 of the ACI Code.4

Results of the tests indicate t atcracking in the struts occurred in m stspecimens prior to reaching serviceload. However, this did not appear tohave a significant effect on deft c-tions. In view of this finding, it is r c-ommended that the allowable tensilestress of 6 , specified in Secti n18.4.2(b) of the ACI Code 4 be i -creased to 7.5 VT psi (0.62 3 1l a)for concrete in the struts..

Furthermore, it is likely that eprovisions of Section €8.4.2(e) w 1Ioften apply for prestressed bea swith web openings. The a Iowab etensile stress would then be increas dto IZ,1 psi (1.00 V MPa). Allow -ble tensile stresses at sections aw yfrom openings should not be i -creased above those allowed in thACI Code.4

PC[ JOURNAL/November-December

SynopsisResults of tests on 18 full-

size prestressed, pretensionedconcrete tee beams represent-ing one-half of a structuraldouble tee section are re-ported. The variables investi-gated were opening size, loca-tion of opening along the span,type and amount of web shearreinforcement, and amount ofprimary flexural reinforcement.

Behavior of beams withopenings was found to be simi-lar to that of a Vierendeeltruss. For the sizes of open-ings studied, distribution ofshear force above and belowan opening was dependent onthe relative flexural stiffnessesof the struts. Based on thesefindings, a method of analysiswas established. Criteria forstrength design are presentedand a fully worked numericalexample is included to illus-trate the application of the pro-posed design method.

The tests indicate that largeweb openings can be accom-modated in pretensioned dou-ble tees while maintaining re-quired strength and servicea-bii'cte. However, the openingsmust be located outside therequired strand embedmentlength and adequate shearreinforcement must be pro-vided adjacent to the open-ings.

The experimental work wascarried out at the PortlandCement Association.

35

Table 1. Variables and measured material properties for short-span beams.

Opening**Concrete Properties

Equivalent % ofSpecimen Location f, fsp Ec Load Calculated

ft psi psi ksi kips/ft.+

Ultimate Load

Pl_P* 6 6040 570 3570 5.36 61

P2-P 3 5990 570 3530 3.69 42

P3-W 6 5920 550 3290 8.52 97

P4-R 3 6040 560 3350 5.38 61

P5-W 6 6200 590 3750 8.25 94

*Symbols indicate reinforcement as follows:

P - no web reinforcementW - welded wire fabric to provide minimum shear reinforcementR - welded wire fabric and stirrups near openings

**Distance shown as X in Figure la

+Ultimate load for flexure calculated with = 1.0

1 ft = 0.305 m

1 psi = 0.006895 MPa

1 ksi = 6.895 MPa

1 kip/ft = 14.6 kN/m

Experimental Investigation

The experimental investigation wasconducted to determine whether pre-stressed concrete beams could ac-commodate large web openings whilemaintaining adequate strength andserviceability. Details and results ofthe investigation are presented below.

Short-span beam testsA series of short-span beams tested

at the Structural Development Labo-ratory of the Portland Cement Associ-ation indicated that slip of the pre-stressing strands limited strength ofbeams containing openings in the re-gion required for strand embedment.

In this investigation, five 26-in. (660mm) deep tee beams, each containingfive 1 -in. (12.7 mm) diameter, 7-wire,prestressing strands with a breakingstrength of 286 ksi (1972 MPa) weretested. Forces were applied to simu-late conditions in a uniformly loadedbeam. Note that the cross section ofthese beams was similar to that for thelong-span beams, except that theflange was monolithic.

Rectangular openings 10-in. (254mm) deep and 30-in. (762 mm) longwere placed symmetrically aboutmidspan. These openings were cen-tered at either 6 ft (1.8 m) or 3 ft (0.9m) from each end of the 18-ft (5.5 m)span.

Welded wire fabric and U-shapedNo. 3 stirrups with yield strengths of80.6 and 67.3 ksi (556 and 464 MPa),respectively, were used as shear rein-forcement in three specimens. Theother two specimens contained noshear reinforcement. Variables andconcrete material properties of theshort-span beam specimens are shownin Table 1.

Strand slip occurred in all fivespecimens causing a premature loss ofstrength. Vertical stirrup reinforce-ment along each side of the openingsdid not delay the occurrence of strandslip in these beams. However, Speci-men P3-W with welded wire fabric forminimum shear reinforcement andwith openings centered at 6 ft (1.8 m)from each support, carried a load cor-responding to 97 percent of the calcu-lated flexural capacity for a beamwithout openings.

36

Table 2. De ails of test specimens.

Specimen

Beam Concrete To ping Concrete openingSize

Opening**Locationf^ fsp Sc fc fsp Ec

psi psi ksi psi psi ksi in. ft.

B1-W* 6820 590 4150 2990 370 3150 10x45 6

B2-W 7610. 650 4260 2820 370 2680 - -

B3-W 7600 650. 4320 2840 380 3080 10x45 12

B4-W 7820 690 4340 2800 360 3100 10x45 9

B5-W 7410 630 4320 2830 380 3110 10x60 9

B6-W 7500 650 4230 2860 360 3050 10x30 9

B7-R 8020 660 4220 2860 390 3010 10x60 9

B8-P 7660 650 4190 3110 440 3220 "10x60` 9

B9-W 8140 670 4300 3680 420 3330 10x60 15

B10-W 7910 670 4310 3710 380 3440 10x60 12

B11-R 8000 670 4530 3760 500 3740 10x60 9 & 15

B12-R+ 7170 690 3770 3460 400 3580 10x60 9

B13-R+ 7460 680 4130 3110 400 3250 14x60 9

*Symbols indicate reinforcement as follows:P - no web reinforcementW - welded wire fabric to provide minimum shear reinforcementR - welded wire fabric an stirrups near openings

**Distance shown as X in Figure la+Specimens B12 and 513 contained 2 strands, all others contained 3 strands

1 psi 0.006895 MPa

1 ksi = 6.895 MPa

1 in. = 25.4 mm

1 ft = 0.305 m

Long-span beam testsTest beams in the long-span s ries

represented one-half of a stan anddouble tee section. Dimensions anddetails of the test specimen are sh wnin Figs. 1 and 2.

The web and the lower 2 in. ( 0.8mm) of the flange were cast with or-mal weight concrete designed to avea compressive strength of 4000 psi(27.6 MPa) at the time of initial pre-stress. After transfer of prestre s, a2-in. (50.8 mm) topping of no malweight unreinforced concrete wascast. The topping had a design om-pressive strength of 3000 psi (0.7MPa) at 14 days, normally the a e attime of testing.

A span of 36 ft (11.0 m) was sele tedso that openings could be place I atseveral locations outside the reqi firedstrand embedment length. This wasintended to decrease the likeliho d ofbond failures that were observe I in

the short-span test series. The overalllength of each specimen was 37 ft(11.3 m). Specimen details and con-crete material properties are shown inTable 2.

Prestressing was provided with1/2-in. (12.7 mm) diameter, 270KGrade, 7-wire straight strand with abreaking strength of 278 ksi (1917MPa). Most specimens containedthree strands spaced at 4 in. (102 mm),as shown in Fig. 2b. This arrangementpermitted 10-in. (254 mm) deep open-ings to be placed in the web whilestill maintaining adequate concretecover. One specimen was tested withonly the top and bottom strands inplace. A specimen containing 14-in.(356 mm) deep openings was testedwith only the two bottom strands.

Web shear reinforcement, whenprovided, was of two types. Minimumshear reinforcement as specified bythe 1971 ACI Building Code 4 was


a) FIVE POINT LOADS

I I I

Sot 6'-0" 30'-d" 3tOe

I" = 254 mmb) SIX POINT LOADS 0.305m

Fig. 1. Loading for long-span beams.

----

Topping

•

2622

I" = 25.4 mm

3 a) DIMENSIONS

—Welded wire fabric0 (4x4-10/IO)

Welded wire fabric(6x6-8/8) —I/2Dja. 270 K Strand

b) REINFORCEMENT

Fig. 2. Properties of long-span beams.

38

Fig. 3. Test setup for prestressed beam with four openings.

provided by welded wire fabri inmost specimens with three stra ds.This type of reinforcement had ea-sured yield strengths between 8.2and 75.2 ksi (470 and 519 M a).Specimens with two strands, alth ughhaving a lower minimum shear ein-forcement requirement, were ro-vided with the same amount of ebsteel.

One additional U-shaped No. 3 stir-rup was placed at each side and dia-cent to each opening in four s eci-mens. Distance from the centroi ofthe stirrups to the edge of the ope ingwas 1 in. Welded wire fabric was ro-vided in the flange of each speci wnto satisfy temperature reinforc itrequirements.

The tests were limited to be mswith rectangular-shaped openi gs.Opening depths in 12 beams wer 10in. (254 mm). Specimen B13 had14-in. (356 mm) deep openings. heopenings were long enough, 30, 45,and 60 in. (762, 1143, and 1524 m),to cause failures in the struts of s me

test specimens. Consequently, infor-mation was obtained for the develop-ment of strength design criteria.

Openings were placed symmetri-cally about midspan with the distancefrom the supports to the center of theopenings, shown as "X" in Fig. la,being either 6, 9, 12, or 15 ft (1.8, 2.7,3.7, or 4.6 m). Web shear reinforce-ment in most specimens consisted ofat least the minimum amount requiredby the 1971 ACI Building Code.4Some specimens contained additionalstirrup reinforcement adjacent to theopenings. Specimen B8 was testedwith no shear reinforcement.

Specimen B2 was tested with noopenings and with minimum shearreinforcement. This provided a stan-dard with which to compare the re-sults of other specimens. The beamwas designed to be under-reinforced.

Specimen B11 contained four open-ings. This beam was tested to deter-mine the effect of closely spacedopenings on forces in the struts (seeFig. 3).

PCI JOURNAL/November-Decembor 1977 39

Test procedure—The test setup isshown in Fig. 3. Concentrated loadslocated directly over the web wereapplied at 6-ft (1.8 m) intervals alongthe span to simulate conditions in auniformly loaded beam.

To avoid having load points directlyover openings, a five-point loadingscheme was used for specimens hav-ing openings at 9 and 15 ft (2.7 and 4.6m) from supports, and a six-point load-ing scheme was used for specimenshaving openings at 6 and 12 ft (1.8 and2.7 m) from supports. Positions ofloads are shown in Fig. 1.

During each test, the design serviceload was reached with the applicationof seven to twelve equal load incre-ments. The beams were then un-loaded in one increment. Next, theservice load was reapplied in one in-crement. The beams were then un-loaded before being tested to destruc-tion. This sequence provided data inthe service load range both before andafter the specimens cracked.

Instrumentation—To determine thedistribution of forces in the vicinity ofthe openings, parallel lines of electri-cal resistance strain gages were at-tached to the compressive struts overthe openings. Additional strain gageswere placed on the prestressingstrands in the tensile strut.

Gage points for a Whittemore me-chanical strain gage were attached tothe test specimens at the level of theprestressing steel at the center of thespan. These gage points were used todetermine strains due to prestressboth after transfer and at the time oftesting.

In addition to strain readings, de-flections were measured at midspanand at points directly under the edgesof one of the openings. Load cellsmeasured the applied forces and reac-tions. Dial gages placed at the tip ofthe strands extending from the ends ofthe beam were used to detect slip.

Test Results

Observed behaviorObserved behavior of the test

specimens can be assigned to threedifferent categories. Examples of eachare shown in Fig. 4.

Specimens with adequate strengthat the openings, and the standardspecimen with no openings, reachedtheir capacity in flexure. These testsended when the prestressing strandsfractured at midspan. Fig. 4a illus-trates this behavior.

The capacity of several specimenshaving openings in high shear regionswas limited by an unrestrained shearcrack extending from the low momentside of an opening toward the support.These cracks normally propagatedalong the prestressing strands. Insome beams the cracks extended intothe region required for strand em-bedment causing the strand to slip.

The capacity of the tensile strut tocarry shear was reduced as the cracklengthened. As a result, additionalshear was transferred to the compres-sive strut. Capacity was reached whena hinging mechanism formed in thecompressive strut. An example of thisbehavior is shown in Fig. 4b.

Some tests ended with fracture ofthe prestressing strands beneath anopening. For specimens having open-ings in high shear regions, strand frac-ture frequently coincided with theformation of a hinging mechanism inboth struts. This mechanism is illus-trated in Fig. 4c.

FindingsThe tests clearly established that

large web openings can be accommo-dated in prestressed concrete beamswithout decreasing their strength.However, this is possible only whencracking at an opening is not allowedto extend into the required strandembedment length.

40

(a) Specimen B2 just before strand (fracture at midspan.

•ss

(b) Specimen B1 just before strand

(c) Specimen B4 just before strand racture beneath opening.

Fig. 4. Behavior of beams un er overload (Specimens B2, B1, and B4).

To satisfy this requirement,ings must be located outside t]quired embedment length. Ad(ally, vertical stirrup reinforcemust be provided adjacent toings in an amount sufficient tothe full design ultimate shear foi

Principal test results are preyin Table 3. Listed are the ma).load carried by each specime.

en- percentage of design ultimate load at-re- tained, the type and cause of failure,on- and the location of damage.,nt In most specimens with openings,en- cracking occurred in the struts at less.rry than service load. Therefore, design of

these members may be controlled byted serviceability requirements whenurn cracking is not allowed.the In all tests, measured service load


Table 3. Principal test results.

SpecimenEquivalent

Load% of

Design

Failure

TypeInitiated Locationkips/ft Load* by

B1 0.98 77 Shear-secondary Strand OpeningHinging** Slip

B2 1.36 107 Flexure Strand MidspanFracture


B4 1.35 106 Flexure Strand OpeningFracture

B5 1.18 92 Shear-secondary Shear OpeningHinging** Cracking



B8 0.94 73 Shear-secondary Strand OpeningHinging** Slip

B9 1.34 104 Flexure Strand OpeninFracture

310 1.36 105 Flexure Strand OpeningFracture

811 1.33 103 Flexure Strand OpeningFracture

B12 0.91 106 Flexur Strand OpeningFracture


*Ultimate load for flexure based upon = 1.0

**Indicates shear failure of tensile strut followed by hinging incompressive strut

1 kip/ft = 14.6 kN/m

deflections were well within those al-lowed in the 1971 ACI BuildingCode. 4 A comparison of load versusdeflection for Specimen B2 with noopenings and Specimen Bit with fouropenings is shown in Fig. 5. It may beconcluded that the influence of open-ings on deflection is minor in properlydetailed beams.

Behavior of the test specimens wasanalogous to that of a Vierendeeltruss. Analysis of recorded strains in-dicated that points of contraflexureexisted in the compressive struts ofspecimens with openings. For open-ings in high shear regions, the point of

contraflexure was near the midlengthof the strut.

Forces determined from strain read-ings indicated that shear in the com-pressive and tensile struts was carriedin proportion to their flexural stiffnes-ses. It was also found that cracking inthe struts had a significant effect onthe distribution of shear. This is illus-trated in Fig. 6. Axial forces in thestruts were close to those calculatedon the basis of a Vierendeel trussanalogy. Cracking had little effect onaxial forces, as seen in Fig. 7.

Specimen B11 was tested with10 x 60-in. (254 x 1524 mm) openings

MIESPAN DEFLECTION,mm100 200

Coiculaled Flexural opocily1:0

APPLIEDLOAD

kipsl U.

0.5Service Load

30020

APPLIEDLOADKN/m

10

0J0 5 10 15

Ml SPAN DEFLECTION, in.

Fig. 5. Load versus flecti n for beams with and without openings.

—CALCULATED FROM MEASURED 30B6 STRAINSB4 °— CALCULATED ASSUMING Vc V

s , ---- CALCULATED BY VIERENDEEL TRUSS8

SECTIONS ASSUMING UNCRACKED83

25BII (exi)

5 810

8 B5 20

We 4 B12 Vckips

3 15 kN3

!02 BIIOnf)

/JJ/ J J Jj::/

0APPLIED HEAR AT OPENING, V

Fig. 6. Shear f rce in compressive strut.

PCI JOURNAL/November-Decembers 1977 43

100

80

C 60kips_

BI40

20/

0

T I I I I I I I I I- CALCULATED FROM MEASURED STRAINS--- CALCULATED BY VIERENDEEL TRUSS ANALOGY

BIoB3

B9 I

I 4 IIB6 87 I I7ji ii

I IB5 I I I elll(int) I BII(ext)I i I I I

B8 I I I B12

II I I !I f '^ 5/13

MOMENT AT OPENING

Fig. 7. Axial force in compressive strut.

400

300VckN

200

Ite200 k4-m-

k-ft

centered at 9 and 15 ft (2.7 and 4.6 m)from each support. The openingswere separated by 1-ft (0.31 m) thickweb elements, referred to here asposts. Forces in the struts of Speci-mens B7 and B9, having isolatedopenings centered 9 and 15 ft (2.7 and4.6 m) from supports, respectively,were compared to correspondingforces in Specimen B11. It was de-termined that cracking of the postsaffected strut forces. Nominalshear stresses in excess of 7 J f psi(0.58 f MPa) were calculated fromexperimental data. However, thecracks in the posts did not reduce thestrength of the specimen. Further re-search is needed to identify the be-havior of posts between closelyspaced openings.

Effect of variablesOf the variables considered in this

investigation, those having thegreatest effect on specimen strengthand behavior were the location of theweb openings along the span and the

amount of web shear reinforcement.The effect of web reinforcement andopening length on beam strength isshown in Fig. 8a. Only specimenswith openings centered 9 ft (2.7 m)from the supports are compared inthis figure.

Minimum shear reinforcement pro-vided adequate strength for speci-mens having opening lengths of 45 in.(1143 mm) or less. For specimenswith 60-in. (1524 mm) openings, addi-tional stirrup reinforcement at theopenings was required to preventstrength reduction. Stirrup forces forSpecimen B7 are shown in Fig. 9.

The effect on strength of openinglocation and opening size for speci-mens containing minimum shear rein-forcement is shown in Fig. 8b. No losswas found for specimens with open-ings centered 12 ft (3.7 m) or morefrom the supports. However, 10 x 60-in. (254 x 1524 mm) openings located9 ft (2.7 m) from the supports de-creased the strength of Specimen B5.

Specimen B1, with 10 x45-in.

44

82 x86 xB4 0B7

100 x— 8I

FLEXURAL q B8CAPACITY,

50

q NO WEB REINFO CEMENTx WELDED WIRE FA IRICo WELDED WIRE FA bRIC PLUS STIRRUPS NEAR OPENINGS

0

0 15 30 45 60OPENIN LENGTH, in

a) EFFECT OF WEB REINFORCEMENT AND OPENING LENGTH

q B6 x83

100 B8 °BIO xEo B5

BI I

FLEXURAL REQUIRED X I

CAPACITY, EMBEDMENT ZO E I

50 I

q IOX30 in. IX IOX45 in. Io IOX60 in.

00 3 6 9 12 15(support) OPENING LOCATION, ft.

b) EFFECT OF OPENI G LOCATION AND SIZE

x812100

x613

FLEXURALCAPACITY,

sn

I° = 25.4 mm

1^ =0.305m

10 14OPENIN DEPTH, in.

c) EFFECT OF OPENI G DEPTH

Fig. 8. ffect of variables.

(254 x 1143 mm) openings center d 6ft (1.8 m) from the supports, exhil iteda substantial loss of strength. Cr cksextending from the openings into theregions required for strand em ed-ment in this specimen caused thestrands to slip. This led to a prem urefailure.

A comparison of test results for

Specimens B12 and B13 provides anindication of the effect of openingdepth on behavior. Increased openingdepth in Specimen B13 was providedby decreasing the depth of the tensilestrut.

As shown in Fig. 8c, Specimen B12with an opening depth of 10 in. (254mm) carried a load 6 percent greater


APPLIED SHEAR,kN

10 20 30 40

8

30

6

Strain Gage Location

20

4

ST 2STIRRUP

STIRRUP FORCEFORCEkips

Sit

ILJ2

I I

15 3.0 4.5 6.0 7.5 B.0APPLIED SHEAR, kips

Fig. 9. Stirrup forces in Specimen B7.

than that corresponding to its calcu-lated flexural capacity. The capacity ofSpecimen B13 with an opening depthof 14 in. (356 mm) was 9 percentbelow its calculated flexural strength.

Varying the amount of primaryflexural reinforcement did, not signifi-cantly change specimen behavior.Specimens B7 and B12 were similarexcept for the number of prestressingstrands. Since Specimen B7 had threestrands, its calculated flexural capacitywas greater than that of Specimen B12with only two strands.

Therefore, the shear at the openingfor Specimen B7 was much more se-vere. However, the behavior of thetwo specimens was virtually identical.Both carried loads exceeding theircalculated flexural capacities. Frac-ture of the prestressing strands be-neath the openings occurred in bothspecimens at ultimate load.

Analytical Procedure

In this section, an analytical proce-dure for determining forces in thestruts of prestressed beams with largeweb openings is presented.

Analysis of recorded strain data in-dicated that the behavior of the testspecimens was similar to that of a Vie-rendeel truss. Cracking was observedto have a significant effect on theshear distribution in the struts. Beforecracking, shear was distributed to thestruts in proportion to their gross mo-ments of inertia.

Forces acting on a beam with anopening are illustrated in Fig. 10.Loads on the beam produce. shear, V,and moment, M. Moment is resistedby the two struts acting together as in-tegral parts of a beam. This results inthe primary stress condition indicated.

46

Primary SecondaryStresses Stresses

tresses at opening.

CombinedStresses

Fig. 10. 1

Each strut also carries a staticallyindeterminate portion of the t talshear force acting at the section. heshear carried by the compressive s rutis designated V e and that carried bythe tensile strut is designated Vt.These produce secondary flex ralstresses in the struts.

At some section near the left edgeof the opening, the secondary stre sesbefore cracking are similar to ti oseshown. Superposition of these wo

states of stress results in a combinedstress condition as shown.

Distribution of forces

Simplified method An idealizedmodel of a beam with an opening isshown in Fig. Ila. The length of thestruts, shown as 1, is conservativelytaken as the distance between verticalstirrups on each side of the opening.In practice, these stirrups must beprovided to contain cracking.

Ac/Rigid

Abutment

7ds

Ltd'--^

StrrruQ Stirrup';

HingM LM 1MM

VC

P

c) IDEALIZED MODEL

__'HVt b) FREE BODY DIAGRAM

free body diagram at beam opening.

P/2

AM(1vStirrup

Fig. 11. Idealized model

PC[ JOURNAL/November- December+ 1977 47

The compressive and tensile strutsare assumed to frame into rigid abut-ments on each side of the opening. Toreflect the Vierendeel truss action ob-served in the tests, hinges are as-sumed at the midlength of each strut.Moments of inertia for the compres-sive and tensile struts are shown as I,and I, respectively.

Shear, moment, and prestress areintroduced into the system throughthe rigid abutments. For strength de-sign, the shear, V, and moment, M, atthe center of the opening are deter-mined from beam forces at ultimate.

When the opening length,1, is smallcompared to the span length of thebeam, V can be assumed constant overthe length of the opening. Momentthen varies linearly across the open-ing from M – AM to M + OM, where:

AM ° l(3)2 and AM denotes the change in mo-ment over one-half of the strut length.

The prestress force, P, acts at a dis-tance Ad below the centroidal axis ofthe tensile strut. The distance be-tween the centroidal axes of the strutsis shown in Fig. ha as d8.

Forces acting at a section throughthe center of the opening are shownin Fig. Jib. With respect to theapplied loads, the axial forces in thestruts are calculated as:

C= M –P(Ad) (4)d8

T= M– P(d8 +Ad) (5)d8

For design purposes a simplifiedprocedure for estimating shear forcesin the struts has been derived. Whenno cracking has occurred in the struts,shear is carried in proportion to theuntracked moments of inertia. Thiswill often be the case at transfer ofprestress and at service load.

Once cracking has occurred, a redis-

tribution of forces takes place in thestruts. For this case, the design proce-dure is dependent on the extent ofcracking in the tensile strut. WhenEq. (5) results in:

(6)a crack extending the full depth of thetensile strut is likely to have occurred.For this condition, it is recommendedthat the struts be designed for:

V c =V (7)

Vt _ 0 (8)For values of T satisfying the condi-

tion:

T < 6 A g, (9)a full-depth crack has not occurred.For this case the tensile strut must bedesigned to carry some of the shear.The recommended design forces are:

V0 =V I` (10)^ i;+ lt(cr)

Vt =VI I +It 1 (11)It

whereI, = moment of inertia of un-

cracked compressive strutI t = moment of inertia of un-

cracked tensile strutIt(cr) = moment of inertia of fully

cracked tensile strutThe use of this simplified method

for determining strut shear forces atultimate load results in a conservativedesign. This should be satisfactory formost design applications.

Iterative method When an accu-rate determination of strut shearforces is required, analysis using themodified idealized model shown inFig. 12 is recommended. Variable ef-fective moments of inertia I ci, 'c2, I tl,and I t2 are assumed in each strut seg-ment to allow the effects of crackingto be included in the analysis.

Axial forces in the struts are calcu-lated from Eqs. (4) and (5).

48

between the loads in the untoppedsystem and loads in the compositesystem. Dead load and prestressforces are initially resisted by the un-topped system. However, once crack-ing occurs, some of the dead load andprestress forces along with all of theload applied after the topping is castare redistributed to the composite sys-tem.

The two systems must be analyzedseparately to satisfy compatibility.However, the effects of forces in bothsystems must be considered togetherin determining the properties of thestruts after they have cracked. Moredetail on the aspects of this analysisare contained elsewhere.' A computerprogram for this analysis is available.

The analysis applies only when thestruts behave primarily as flexuralmembers. As a guide in this regard,the analysis is not recommendedwhen overall length-to-depth ratios ofthe struts are less than 2.5. Fur-thermore, to ensure that the posts be-have rigidly, it is recommended thatadjacent web openings be separatedby web elements (posts) having over-all width-to-height ratios of at least2.0 where the width of the posts is thedistance between adjacent stirrups. Alimit on nominal total design shearstress, v,,., of 2f J psi (0.17fj MPa) isadvised for the posts.

AcknowledgmentsThe investigation described in this re-

port was carried out in the Structural De-velopment Laboratory of the Portland Ce-ment Association in Skokie, Illinois. B. W.Fullhart, A. G. Aabey, and W. Hummerich,Jr. of the technician staff of the StructuralDevelopment Section fabricated andtested the beams. Photographic serviceswere provided by P. J. Walusek. Astaire M.Parisi provided the secretarial services.Figures in the report were prepared byLouise S. Masten.

Concluding Remarks

Tests were carried out on 18 full-size tee beams containing large rec-tangular web openings. The beamswere loaded to simulate conditions ina uniformly loaded beam.

Principal variables in the test pro-gram were size and location of open-ings, type and amount of web shearreinforcement, and amount of primaryflexural reinforcement.

The behavior of beams with open-ings was similar to that of a Vieren-deel truss. Test results indicate thatlarge web openings can be placed inprestressed concrete beams withoutsacrificing strength or serviceability.However, openings must be locatedoutside the required strand embed-ment length. Adequate shear rein-forcement must be provided adjacentto openings.

An analytical procedure has beenestablished for determining forces andmoments in struts above and belowopenings. Design criteria have beenpresented.

References1. Ragan, H. S. and Warwaruk, J., "Tee

Members With Large Web Openings,"PCI JOURNAL, V. 12, No, 4, August1967, pp. 52-65.

2. Sauve, Jacques Germain, "PrestressedConcrete Tee Beams With Large WebOpenings," MS Thesis, University ofAlberta, Fall 1970.

3. LeBlanc, Eric P., "ParallelogramShaped Openings in Prestressed Con-crete Tee Beams," MS Thesis, Univer-sity of Alberta, Fall 1971.

4. ACI Committee 318, "Building CodeRequirements for Reinforced Concrete(ACI 318-71)," American Concrete In-stitute, Detroit, Michigan, 1971, 78 pp.

5. Barney, G. B., "Design of PrestressedConcrete Beams with Large Web Open-ings," PhD Thesis, Northwestern Uni-versity, June 1975.

50

APPENDI^C A-NOTATION

A 0 = gross area of tensile strutb z = minimum width of tensile

strutC = axial force in compres ive

strutd = distance from extreme c m-

pressive fiber to centroi ofprestressed reinforcementbut not less than 0.8h

d8 = distance between centro dalaxes of tensile and comp es-sive struts

Ad = distance of effective re-stressing force result ntbelow centroidal axis of n-sile strut

E , = modulus of elasticity of c n-crete

= compressive strength of c n-crete

fYe = compressive stress in c n-crete due to prestress o lyafter all losses, at extre efiber of a section at wh chtensile stresses are cau edby applied loads

f8P = splitting tensile strength ofconcrete

h = overall depth of tensile st tI, = moment of inertia of n-

cracked compressive strutIc,. = moment of inertia of crac ed

strut section transformed toconcrete

1 ,i, 1C2 = effective moments of iner isin compressive strut

'elf = effective moment of inertiIg = moment of inertia of n-

cracked section transform dto concrete

= moment of inertia of un-cracked tensile strut

I t(cr) = moment of inertia of fullycracked tensile strut

Ic,, I t2 = effective moments of inertiain tensile strut

= effective strut lengthM = moment at center of openingAM = change in moment over

one-half of strut lengthMa = maximum moment in strut

segmentMC,. = bending moment causing

flexural cracking at sectionconsidered due to superim-posedloads

M„ = maximum moment in tensilestrut due to superimposedloads

P = effective prestress forceT = axial force in tensile strutv ex = shear stress at diagonal

cracking due to all designloads, when such cracking isresult of combined shear andmoment

v u = nominal total design shearstress

V = shear force at center of open-ing

V, = shear force in compressivestrut

Vt = shear force in tensile strutX = distance from support to

center of opening

lit = distance from centroidal axisof uncracked section to ex-treme fiber in tension

(P = capacity reduction factor


APPENDIX B-DESIGN EXAMPLE

In this section a design example ispresented for a prestressed concretedouble tee beam with a 2-in, thickreinforced concrete topping. Part I ofthe example demonstrates strengthdesign using the simplified methodreferred to in the text.

Part II illustrates procedures forchecking stress and•. deflection re-quirements at service load.

Note: For the convenience of read-ers unfamiliar with the American sys-tem of units, a table of metric (SI)equivalents is included below.

Part 1-Strength designThe simply supported prestressed

concrete double tee beam shown inFig. B1 has been designed withoutweb openings to carry a live load of 50psf. Material properties are as follows:

Concrete:Beam f, = 6000 psiTopping., = 3000 psi

Non-prestressed reinforcement:f 60ksi

Prestressed reinforcement:f = 270 ksi

A requirement of two 10-in, deep by36-n, long web openings for passageof mechanical and electrical serviceshas been introduced. The centers ofthe openings are located 8Yz and 15 ftfrom one support as shown in Fig. Bi.

Design the beam to carry the re-quired loads. Assume that the dis-tance from the vertical edges of theopenings to the centroid of the stirrupreinforcement adjacent to the open-ings is 1 in.

1. Calculate the required embed-ment length t d for 1/a-in, diameterstrand from the provisions of Section12.11 of ACI 318-71. Assume a strandstress of 0.7fg.. immediately aftertransfer and 15 percent losses underservice load conditions from the ef-fects of creep, shrinkage, and strandrelaxation:

1d = (fpff- z/afse)dt,_ { 270 – 2/3 (270) (0.7) (1 – ©.15)} (0.5)= 81.5 tn.

Embedment length provided6+ 102- 18– 1= 89in.> 81.5 in.

2. Calculate the effective flangewidth from Section 8.7.2 of ACI 318-71:

1!a of span length =(1/4) (36) (12) = 108 in. > 48 in.

Allowable overhang8 times slab thickness

= 8 (4) = 32 in.or allowable overhang

= 1!a clear distance to nextbeam

= (1/2) (48 – 5.75)= 21.13 in.

Therefore, full flange width is ef-fective.

52

36 .E `^ 2 { 36.. A

10,;;Hole A Hole 8

8.4

15'-0.

A

L 36 - 0 „ 6

(a) ELEVATION

48

Toppin

Symmetrical Welded Wire FabricAbout Center (4x4-10/f0)Line

2 ' Dia.270 K Strands

Welded Wire Fabric(4x4-10/10) N

1 , = 25.4 mm3

I' 0.305m 5

(b) SECTION A-A

Fig. 81. Elevation and cross secti n of prestressed beam used in example.

3. Check the width-to-height ra 'o 111. = 1/i wLX — ½ w,,.X2

of post: = 1/2 (760) (36) (8.5) -(42 — 2) /10 = 4.0 > 2.0 1,4 (760) (8.5)2

= 88,825 ft-lb = 1066 in.-kips4. Calculate the beam design loads

assuming a uniformly distribut d V. = % w.L — wuXdead load. Since the double tee is = ½ (760) (36) — (760) (8.5)symmetric about its centerline, desi n = 7220 lb = 7.2 kipsfor one-half of a standard double t e 6. Calculate the size of stirrup rein-section. forcement adjacent to the opening:

wu = 1.4wa+ 1.7w1= 1.4(299.7)+ 1.7(50)(4) A„= "= 760 lb/ft

5. Calculate the design mome it _ 7200= 0.14 sq in.

and shear at the center of Hole A. (0.85) (60,000)


Use U-shaped No.. 3 stirrup, i.e.,0.22 sq in. of steel each side ofopening.

7. Calculate the axial forces fromEqs. (4) and (5):

C_M-P(Ad)dg

where d s = 17.57 in. based ontransformed strut properties andOd = 0.25 in.

C = { 1066 - (2) (270) (0.153) (0.7) x(1 - 0.15) (0.25) } /17.57

= 60 kips (compression)

T= M-P(d3+Ad)d,

_ { 1066 - (2) (270) (0.153) (0.7) x(1 - 0.15) (17.57 + 0.25) } /17.57

= 10.8 kips (tension)

8. Check the extent of cracking inthe tensile strut from Eqs. (6) and (9).

The value of A,- is calculated as53.2 sq in.

_ 6(53.2)fOO60

6A gt J 1000

= 24.7 kipsT = 10.8 kips < 24.7 kipsTherefore, the tensile strut is notpenetrated by a crack over its fulldepth.

9. Calculate the strut shear forcesfrom Eqs. (10) and (11). The sectionproperties of the struts are:I, = 214 in. 4 (with topping, based on

transformed section)I, = 32 in.4 (without topping)I t = 642 in.4I t(cr) = 72 in ,4

From Eq. (10) the shear force inthe compressive strut is:

I,Ve=V

I c +I t( .)

_ 214V^-7.2 214+72

= 5.4 kips

To obtain a conservative estimate ofshear force in the tensile strut, assumethat cracking causes no redistributionof dead load shear, V 4. to the com-pressive strut.

Vd = 1/s w dL - wdX= ½ (299.7) (36) - (299.7) (8.5)= 2847 lb = 2.9 kips

ItV t =Vd I +It +

It(Vu -V4)

I ` + It

I 642 f=2.9It 32+6421+

(7.2 - 2.9) 642

214 + 642

= 6.0 kips

10. Calculate the moments at endsof the struts using the free body dia-gram of Fig. B2. Assume strut lengthsof 36 + 2 = 38 in. and hinges at mid-lengths of the struts:

In compressive strutM,= ±5.4(19)

= 102.6 in.-kipsIn tensile strutMt= ±6.0(19)

114.0 in.-kips

11. Determine the magnified mo-ments in the compressive strut to ac-count for slenderness effects. UsingEq. (10-5) from Section 10.11.5 of ACI318-71, for a frame not braced againstsidesway:

s = CmP

1 -"

wlatre

C m = 1.0 for unbraced memberP u = C = 60 kips0 =0.7

54

Calculating dead load sh ar incompressive strut:

V , = 2.9 3232 + 642

= 0.1 kips

/34= 1.4 (0.1) (19)

(5.4) (19)= 0.03

Using Eq. (10-8):E,.I,.

El = 2.51+/3a

(57) (214)2.51.03

= 3.67 x 105 lb-in.2

Using Eq. (10-6):

P^ 7r2EI_

(kl)2

__ir2 (3.67 x 105)(1.0 x 38)2

= 2508 kips

S - 1.0_ 60

1 0.7 (2508)

= 1.04BM,= ±1.04(102.6)

= ± 106.7 in.-kips12. Check the positive and nega ive

flexural capacities of struts with re-spect to interaction curves shown inFigs. B3 and B4 (see next page).

For C = 60 kips(Mc)allow = + 114 in.-kips

- 143 in.-kips ( k)(M t)^Zo,,, _ + 120 in.-kips

- 152 in.-kips (ok)

13. The net axial force in thesile strut is 10.8 kips tension

Fig. B2. Free body diagram.

Step 7). Determine the shear capacityfrom Section 11.4.4 of ACI 318-71.

Vu

`Vb wd

= 6.0(0.85) (3.75) (0.8) (12)

= 0.20 ksi = 200 psi

v =2I 1+0.002^u9

= 2 ( 1+ 0.002 ( 10800) I J,MW53.2

= 92 psi < 200 psi

Hence, additional reinforce-ment is required. Proportionthe shear reinforcement in ten-sile strut according to Eq. (11-13) in ACI 318-71. Maximumspacing by Section 11.1.4(b) ofACI 318-71 is 0.75h = 9 in.

A- - yc) bwsv- f.

_ (200 - 92) (3.75) (9)60000

= 0.06 sq in.Use No. 3 bar single leg stirrupsat 9-in, centers in tensile strutto provide A„ = 0.11 sq in. An-chor stirrups around the strandsusing a 180-deg bend at eachend.

14. Determine the shear capacity ofcompressive strut using the provisions

PCI JOURNAL/November-December h977 55

Ctxir+s) CQsN)-25 4 _100

-20 -101 ____

-200 -150 -100 -500

25 100

20050

300

75

ao

50 100

MtxN;m)

20

ISO 200

Min-kips)

1001

Fig. B3. Interaction curve for compressive strut.

C(kN)

M(in-kips)

Fig. B4. Interaction curve for tensile strut.

56

L

of Section 11.4.3 of ACI 318-71.d is different at each end of thecheck capacity at each end. Eo.end having d = 1;in.

Vuvu-

5.4(0.85) (48) (1)

=0.13 ksi = 130 psi

v, =21 1+0.0005 _ 19

where A g = 165 sq in. isgross area of the compresstrut transformed to 600Cconcrete. Therefore:

v, = 2j 1 + 0.0005 6 00 1 ' 60 0

= 183 psi > 130 psi (ok)

For the end having d = 3 in.,assume the capacity is ov-erned by topping concete av-ing f 3000 psi since this s incompression. Thus, in Eq.(11-6) of ACI 318-71, A 9 = 193sq in.

_ 5.4_ v " (0.85) (48) (3)

= 0.04 ksi = 40 psi

v c =2f 1+0.0005 LrJ

127 psi> 40 psi (ok)

15. Check the horizontal sicapacity between flange and top;using Sections 17.5.3 and 17.5. ofACT 318-71.

V.

vah=^dv

5.4_ (0.85) (48) (3)

= 0.04 ksi = 40 psi<80 psi (allowable)

Therefore, the tensile strut ofHole B is penetrated by a full-depth crack. The distribution ofshear to the struts is determinedfrom Eqs. (7) and (8).

V, = V„ = 2.3 kipsVe = 0

Moments adjusted for slender-ness effects become:

SM c = - 4.3 in.-kipsM = 0

From the interaction diagram inFig. B3, for C = 81 kips:

(M) Qilo _ + 140 in.-kips– 172 in.-kips (ok)

Shear design for the end of thestrut with d = 1 in. results in

v = 56 psiv 1. = 193psi>56psi(ok)

For the end of the strut withd = 3 in.

v, = 17 psiv, = 133 psi > 17 psi (ok)

Checking horizontal shear be-tween topping and flange leadsto:

v dn = 17 psi < 80 psi (ok)

17. The shear stress in the post iscalculated as:

flee 16. Following a similar procedure.rut, for Hole B results in the following:the Design moment at center of

opening:

M,, = 1436 in.-kipsDesign shear at center of open-ing:

VV, = 2.3 kipsStirrup reinforcement adjacentto opening: U-shaped No. 3 bar

C = 81 kips [from Eq. (4)11

T = 31.9 kips {from Eq. (5)]thelive 6Ag = 24.7 kips

<T=31.9 kips

PCI JOURNAL/November-flecembe} 1977 57

vu = ( C)Hole B - (C)Hote A

4b,d_ 81-60

(0.85) (4.84) (42 - 1)

= 0.12 ksi = 120 psi

2/T = 155 psi (ok)

Part II-Stress anddeflection requirements

Check deflection and stressesaround the openings at service load.Assume a strand stress of 0.7f.,.. im-mediately following transfer of pre-stress. Also, assume that the entireprestress loss of 15 percent occursafter the topping is cast.

1. Service load stresses will be cal-culated at the extreme fibers of thestruts at the four points shown in Fig.B5. Stresses will be calculated assum-ing that full dead load, including theweight of the topping, and the fullprestress force are resisted by the un-topped beam. After the topping be-comes an integral part of the beam,additional stresses resulting from liveload and loss of prestress will be cal-culated.

2. Referring to Fig. B2, the forcescaused by dead load and prestress act-ing on the untopped beam at Hole Aare calculated as follows:

P = AJ.

= 2 (0.153) (0.7) (270)= 57.8 kips

M = 1/2 wdLX - 1/2 wdX2.

= 1/2 (199.7) (36) (8.5)- 1/2 (199.7) (8.5)2

= 23340 ft-lb = 280 in.-kips

V = 1/2wdL -wdX= 1/2 (199.7) (36)

- (199.7) (8.5)= 1897 lb = 1.9 kips

I ^,V, = VIcu+It

=1.932

32 + 642

= 0.1 kips

V t = V -Ve= 1.9 - 0.1= 1.8 kips

The calculated values of d $ andAd shown in Fig. ha for the un-topped beam are:

ds = 16.75 in.O d- 0.25 in.

From Eqs. (4) and (5):

C 280 - 57.8 (0.25)16.75

= 15.9 kips

T - 280 - 57.8 (16.75 + 0.25)16.75

= - 42.0 kips

3. For the bottom extreme fiber atPoint 1:

f = -C

VJ ')yb

21e

15900 (100)(19)(1)97 32

= - 223 psi

Stresses at other locations arecalculated using a similar proce-dure.

4. Referring to Fig. 11b, the changein forces caused by live load and 15percent loss of prestress at Hole A arecalculated following the procedure inStep 2.

P = - 2 (0.153) (0.7) (0.15) (270)_ - 8.7 kips

M = 1/2 (50) (4) (36) (8.5)- 1/2 (50) (4) (8.5)2

= 23375 ft-lb = 281 in.-kips

58

I" = 25.4 mm . Q=38

Stirrup .' .

Fig. B5. Location of stresses.

V = i/2 (50) (4) (36) From Eqs. (4) and (5):– (50) (4) (8.5 281 – (– 8.7) (0.25)

= 1900 lb = 1.9 kipsC _

17.57

16.1 kips

VI –V^

1

1 Ic +it 281–(-8.7)(17.57+0.25)

T=17.57

214= 1.9 = 24.8 kips

L2 14 + 6425. Stresses for live load and 15 per-

0.5 kips cent loss of prestress acting on thetopped beam are now calculated from

V t = V – V, first principles. A summary of service= 1.9 – 0.5 load stresses for Hole A is presented= 1.4 kips in Table B1.

Table B1. Str sses (psi) at service load.

Hole A Hole B

Location TopIbeI[

ottom Fiber Top Fiber Bottom Fiber

Calculated

1 -48* -455 -71* -519

2 -92* -241 -112* -421

3 +648 -1052 +451 -279

4 -614 +478 -29 +241

Allowable

1 -1350, +657* - 700, +930 -1350, +657* -2700, +930

2 -1350, +657* - 700, +930 -1350, +657* -2700, +930

3 -2700, +930 - 700, +930 -2700, +930 -2700, +930

4 -2700, +930 - 700, +930 -2700, +930 -2700, +930

*Stresses in topping

PCI JOURNAL/November-Decembe 1977 59

r

t:.l

(b) U-SHAPED STIRRUP (c)SINGLE-LEG STIRRUP

Fig. 86. Details of additional reinforcement.

6. Stresses at service load for HoleB are calculated in a similar manner.A summary of service load stresses forHole B is given in Table B1.

7. Allowable stresses in tension andcompression for 3000- and 6000-psiconcrete from Section 18.4.2 of ACI318-71 are as follows:

For 3000-psi concrete—in compression,

O.45 f e = 1350 psiin tension, 12 J , = 657 psi

For 6000-psi concrete—in compression,

0.45f= 2700 psi

in tension, 12 f 1 = 657 psiModulus of rupture is deter-mined from Section 9.5.2 of ACI318-71 are as follows:For 3000-psi concrete-

.fr =7.5 f 411 psiFor 6000-psi concrete—f1.=7.5 f 581 psi

As indicated in Table B1, no al-lowable stresses are exceeded.However, cracking is indicatedin the top extreme fiber of thetensile strut at Location 3.

8. Estimate the midspan deflectioncaused by live load and 15 percentloss of prestress. Assume that thecomponent of deflection caused bylive load shear at each opening is de-termined from the following expres-sion derived from moment-area prin-ciples.

V (12) 3

s" – 23E,(I,+It)

where I is the opening length of38 in. and E 6 is the concretemodulus. For Hole A, conserva-tively estimate I t = 72 in.4 Thisis the moment of inertia of thefully cracked strut.

Calculate the component ofmidspan deflection caused byloss of prestress from the expres-sion:

_ _ PeL28,, 8E1

where e = 13.27 in. is the eccen-tricity of the prestressing steel.

Conservatively estimate I =12939 in.4, the moment of inertia

M

of the beam at a section throughan opening.Total midspan deflection is cal-culated as:

& =

(&v+ )Hote A + (6v)Hole B + 8v

= 5 (6x12412

384 (4415000) (12939)2 (1900) (19) +

3 (4415000) (214 + 72)2 (600) (19) +

3 (4415000) (214 + 642)

(8700) (13.27) (36 x 12)2

8 (4415000) (12939)= 0.132 + 0.007 + 0.001 + 0.04k= 0.187 in. < L1360 = 1.2 in. (ol)

The above deflection comparesto a calculated deflection of0.168 in. for a similar beam withno web openings. Note that theincrease in deflections caused bythe holes is small. This findingwas verified by the experimentalprogram.

Design summaryI. Use U-shaped No. 3 stirrups ad-

jacent to both edges of eachopening to contain crackingwithin the struts. See reinforce-ment details in Fig. B6.

2. Use single-leg No. 3 stirrups at9-in, centers as additional rein-forcement in the tensile strut ofHole A. See reinforcement de-tails in Fig. B6.

Discussion of this paper is invited.Please forward your comments toPCI Headquarters by May 1, 1978.

PCI JOURNALfNovember-Decembei 1977 61

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