Analysis of Cracked Prestressed Concrete Sections: A Practical Approach

This paper presents a practical approach for analyzing the elastic behavior of cracked prestressed concrete sections of any shape, using existing section property software. The use of the results for estimating deflection and crack control is presented. The method is applicable to .81!.... ' sections with any degree of prestress, from no prestress to full prestress. Examples are given, including the analysis of cracked composite ,f('' sections. The procedural steps for analyzing cracked prestressed Robert F. Mast, P.E. concrete sections are summarized. Senior Principal BERGER/ABAM Engineers Inc. Fed eral Way, Washington o fully understand the behavior The purpose of this paper is to pre­ of a prestressed concrete mem­ sent an analysis method using conven­ T ber cracked at service load, an tional section property software. The analysis of the cracked prestressed solution requires iteration, but the section should be made. This analysis bulk of the work is done by an exist­ is needed in order to find the change ing section property program. The it­ in steel stress after cracking (for use in eration may be done manually or a evaluating crack control at service small additional program may be writ­ load), and for finding the appropriate ten that will do the iteration, using an flexural stiffness for use in deflection existing section property program to calculations. do the computation inside an iteration The analysis of cracked prestressed loop. sections requires, at best, the solution The iterative procedure consists of 2 3 of a cubic equation.'· • .. The complex­ assuming a depth c of the neutral axis, ity of this solution, requiring the use computing section properties of the of charts, tables, or special software, net cracked section, checking stresses has impeded the use of prestressed at the assumed neutral axis location, concrete members with ten sile stresses and revising c as necessary to make beyond the code limits for nominal the concrete stress equal to zero at the tensile stress. assumed neutral axis location.

80 PCI JOURNAL -~ - Mint Mext of transformed section p () ~ . g. + .. ) a. >- ) Mint = Mexr P(yp) nAps .. p 1.: - a. Forces acting on b. Cracked, transformed c. Forces acting at cross section cross section e.g. of cracked transformed section

Fig. 1. Forces acting on cracked transformed section.

DESIGN IMPLICATIONS fc Today, there is a trend to unify the design of nonprestressed and pre­ stressed concrete members, and to per­ mit designs with any combination of nonprestressed and prestressed rein­ forcement. In order to accompJjsh thls , goal, it will be necessary to replace the ,, nominal tensile stress limits in the cur­ ,, rent ACI Code with requirements lim­ ,t===:::t iting cracking and deflection at service I , I , , load. Thls paper describes a practical __I ,______, __ ,____, _ method of performing the needed cracked section analysis. PIA + MintYII Stresses

SOLUTION STRATEGY Fi g. 2. Stresses in cracked transformed section. The analysis will make use of cracked transformed section proper­ Mint acting about the center of gravity EXAMPLE 1 of the cracked transformed section. ties, like those used in the past days of How the solution strategy works can working stress analysis of ordinary The internal bending moment Mint is best be illustrated by a simple exam­ (nonprestressed) reinforced concrete. the external bending moment Mext re­ ple. Fig. 3 shows the cross section of a The area of steel elements is replaced duced by the amount P times yp, where beam, with design parameters given by a "transformed" area of concrete yp is the distance P moved upward below: equal to n times the actual steel area, from the location of the tendons to the 12 x 32 in . (305 x 813 mm) beam where n is the ratio of the modulus of center of gravity of the cracked trans­ elasticity of steel to that of concrete. formed section, as shown in Fig. 1c. J:= 6000 psi (41.4 MPa) To begin the analysis, assume a trial The forces shown in Fig. lc may be Twelve 1h in. (12.7 mrn) 270K strands depth c of the neutral axis of the applied to the cracked transformed Depth dP = 26 in. (660 mm) cracked transformed section. The section, producing the stresses shown Prestress levelfdc = 162 ksi (1117 MPa) forces acting will be the prestress in Fig. 2. The stress at the neutral axis Span= 40ft (12.2 m) force P acting at the level of the ten­ must be zero. That is: Dead and live loads, and midspan dons and the bending moment Mext fna =PIA- Mint Yna /1 0 moments are given in Table 1. caused by external loads (all loads ex­ = cept prestress) (see Fig. I a). On the first try for the neutral axis P =Ap.,(Jdc) = 1.836 (162) The forces may be resolved into an depth c, fna will doubtless not be zero. = 297.4 kips (1508 kN) axial force P acting at the center of If it is positive (compressive), c must n = Ep,l Ec = 28,500/4415.2 = 6.455 gravity of the cracked transformed sec­ be increased; c must be decreased if At= Aps(n) = 1.836(6.455) 2 tion and an internal bending moment fn a is negative. = 11.85 sq in. (7645 mrn )

July-August 1998 81 Table 1. Dead and live loads, and midspan moments (Example 1 ). 12' w Midspan moments Loadings kips per ft in.·kips kN-m

Self weight 0.413 992 112 Additional dead load 1.000 2400 271 Live load 1.250 3000 339 N (o C') (\J Sum 2.663 6392 722

= 12- 1/2 inch PIA= 297.41218.97 1.358 ksi (c) The stresses are shown graphically in Fig. 6. + Strands Mint= Mext - P(yp) = 6392- 297.4(16.43) Note that as the depth to the neutral = 1506 axis decreased, the PIA stress in­ M;nri(IIyna) = 150611108 = -1.359 ksi (t) creased because the area A decreased. Fig. 3. Example 1 beam. Stress at neutral axis :::::: 0 (ok) Also, the bending stress at the neutral Complete the analysis. axis location decreased because the in­

For a first try, assume the neutral fc = PIA + M; 11 /(IIyt) ternal moment Mint decreased as the axis depth c is 18 in. (457 mm). The = 1.358 + 15061(852419.57) shift in the location of the center of cracked transformed section is illus­ = 3.048 ksi (21.0 MPa) gravity of the composite section trated in Fig. 4. The cracked trans­ Nps =[-PIA+ M;nri(IIYps) ]n (tension +) increased. formed section properties are deter­ = [-1.358 + 15061(85241 Equilibrium may be checked manu­ mined by a section property program. 16.43)] (6.455) ally, without the use of computers.

A = 12 X 18 + 11.85 = 9.97 ksi (68.8 MPa) Refer to Fig. 7. 2 = 227.85 sq in. (147,000 mm ) 4 6 4 I= 9079 in. (3779 x 10 mm ) Yt 9.88 in. (251 mm) = fc IIYna = 907918.12 6 3 r-1 = 1119 cu in. (18.33 x 10 mm ) -.---- r-----,-· The stress at the assumed neutral axis location is checked. VI e.g, Co ~l-+--.--t-- + li PIA= 297.41227.85 = 1.305 ksi (c) M;nr = Mexr - P(yp) =6392- 297.4(16.12) ~I ~I I-:___ _.: " ."" = 1598

M; 11 /(IIy11a) = 159811119 = -1.428 ksi (t) At = 11.85 in2 I Stress at neutral axis = -0.123 ksi (t) I ,___I _ I.______. (-0.848 MPa) The stress at the neutral axis must be zero. Reduce c, to reduce tension at the Fig . 4 . Trial value of c. Fi g. 6. Stresses in cracked section. assumed neutral axis location. After a few more trials (not shown), the solu­ tion at c = 17.26 in. (438 mm) is found. Fig. 5 illustrates the cracked trans­ fc formed section for the correct solution. The calculation for determining the stress at the assumed neutral axis loca­ ~!-+---.-----1-- + tion follows. The cracked transformed section c., ·- ~ ~t ~------'1- properties from the section property 1 program are: I I 2 A= 218.97 sq in. (141,270 mm ) I ------+ T 4 6 4 ... I= 8524 in. (3548 x 10 mm ) I _____ . Yt = 9.57 in. (243 mm) IIYna = 8524/7.69 Fig. 5. Solution at c = 1 7.26 in. 3 = 1108 cu in. (18.16 x 106 mm ) (438.4 mm). Fig. 7. Equilibrium check.

82 PCI JOURNAL c =frf;c/2 = 3.048(12)(17.26)/2 is equal to the effective prestress fs e· stresses can be built up in the nonpre­ = 315.7 kips (1404 kN) The error will almost always be on stressed reinforcement, and that these C acts at the top kern of compression the conservative side when comput­ stresses can have a significant effect zone equals dcf3 for a rectangular area ing crack control and deflections at on the cracking moment and the post­ service loads. cracking behavior. 17.26/3 = 5.75 in. (146 mrn) More precise methods of computing Once the decompression stress (usu­ T = P + Np,CAps) = 297.4 + 9.97 (I .836) prestress losses and the decompression ally compressive) in the nonpre­ 6 7 = 315.7 kips (1404 kN) = C Check stress are available.'· • stressed reinforcement is estimated, it M = C or T x lever arm should be combined with the decom­ = 315.7 X 20.25 NON PRESTRESSED pression tensile force in the tendons to produce a resultant force P and a re­ = 6392 in.-kips (722 kN-m) Check REINFORCEMENT sultant location yp of this force, for use The beam of the example was cho- In the design of cracked nonpre­ in the cracked section calculations. sen for simplicity. It is a rectangular stressed sections, it is the usual prac­ beam with only one level of tendons tice to neglect shrinkage and assume and no unstressed reinforcement. Nev­ DEFLECTION that the decompression stress is zero. ertheless, the method is general and When designing a prestressed mem­ In beams without prestress, the error will work for any section, no matter ber intended to be cracked at service introduced is probably small. But, in how complicated, for which the load, it is necessary to check deflec­ beams with a large amount of pre­ cracked transformed properties may tions. Where the prestress and dead stress, creep due to compressive be calculated. load produce stresses below the crack­ stresses at the level of the nonpre­ ing strength, deflection may be calcu­ stressed reinforcement can magnify lated in the usual manner. The incre­ the error in neglecting the decompres­ THE PRESTRESS FORCE P mental deflection due to live load may sion stress in the nonprestressed rein­ For use in a transformed analysis, be found using the bilinear behavior forcement. the prestress level in pretensioned ten­ method or the effective moment of in­ If the prestress losses are calculated dons should be taken as the stress that ertia method described in Sections by the PCI Design Handbook5 method, would exist in the tendons when the 4.8.3 and 4.8 .4 of the PCI Design the same calculations may be used to stress is zero in the adjacent concrete Handbook.5 estimate the decompression stress in at the same level. This is called the de­ The results of the cracked section the nonprestressed reinforcement. compression stress. The use of the analysis provide another way of com­ Using the notation of the PCI Design transformed area of prestressing steel puting deflection. One normally thinks Handbook, the decompression stress in the section properties will automati­ of deflection as a function of Ml El. in the nonprestressed reinforcement cally account for the fluctuation of But MIEI is equal to the curvature K. could be estimated as a compressive stress in the tendons when the stress in And the curvature K is simply equal to stress equal to the creep losses CR the adjacent concrete is not zero. The the maximum concrete strain fe di­ plus the shrinkage losses SH com­ consideration of decompression stress vided by the depth c to the neutral puted by the method given in the PCI was not apparent in the working stress axis. The computation of curvature K Design Handbook. analysis of nonprestressed concrete, for Example I is shown below: More comprehensive methods of es­ because it was assumed (neglecting timating the decompression stress !de K= scfc shrinkage) that it would be zero. in nonprestressed reinforcement are f c =fc/Ec = 3.048/4415 = 0.000690 How should the decompression given in Refs . 6, 7, and 8. These meth­ K = 0 .00069011 7 .26 stress !de be calculated? It does not ods show that substantial compressive = 0.400 X J0-4/in. (0.157 X J0·5fmm) make sense to calculate it more accu­ rately than the loss calculations used to estimate the effective prestress fse· If the estimate of prestress loss is -----K= 36.7 x 1Q·6fin done using the method described in Section 4.7 of the PCI Design Hand­ Additional Curvature Due to Cracking book,' the decompression stress may Neglecting Tension Stiffening be estimated as the effective prestress Curvature Based on Gross Section f se plus an amount equal to (feir - ---7--...... f eds )EPJ Ee. The quantities f cir and f cds - Lo/ -..; are defined in the PCI Design Hand­ K= 20.1 x 1Q-6/in book, and their difference represents an estimate of the stress in the con­ crete adjacent to the tendons under sustained dead loads. The most simple estimate of the de­ compression stress !de is to assume it Fig. 8. Live load curvature diagram.

July-August 1998 83 In order to obtain the incremental Fi g. 9. curvature due to live load, the curva­ Example 2 beam. ture due to prestress and dead load 12" must be subtracted. This curvature ~I (calculated using gross transformed -.-r------, section, and the same Ec as in the (o : :-- Composite Slab cracked section analysis) was equal to I I -r-- 0.034 x 104 /in. Thus, the net curvature KL due to live load is 0.366 x 10·4/in. (0.144 x 10·5/mm). For an approximate value of - BareBeam midspan deflection, a parabolic curva­ ture diagram may be assumed. The live load deflection is: L1L = 5/48KLU 4 2 = 5/48(0.366 X 10 )(480) -~ + ---t- P=288 k =0.88 in. (22 mm) _.....__L...-+-----' This result is conservative because 4"- the beam is not cracked throughout its length. Fig. 8 shows the curvature dia­ gram for this beam, obtained by calcu­ lating the curvature at 1120 points. In­ tegration of this curvature diagram s ::; 540/fs- 2.5cc ::; 432/fs compared to the fs of 36 ksi (248 produces a live load deflection of 0.69 Eq. (10-5) MPa) used in nonprestressed beams. in. (17.5 mm). Even this number is where Locating one strand in each corner conservative because tension stiffen­ s = center-to-center spacing of flex­ of the tension side easily satisfies ing is neglected. ure tension reinforcement, in. crack control requirements. For comparison, the deflection was f s = calculated stress in reinforce­ calculated using the methods given in ment at service loads, ksi COMPOSITE SECTIONS the PCI Design Handbook. Cracking c = clear cover from nearest surface was found to occur at 79 percent of c For composite sections, the situa­ in tension to surface of flexural full live load and the cracked moment tion is more complicated. The pre­ tension reinforcement, in. of inertia Icr was found to be 5513 in. 4 stress and some dead load bending A similar approach to crack control 4 (2.295 x 109 mm ). The results are are usually applied to the bare non­ for cracked prestressed concrete mem­ given below. composite beam. This creates stress bers is under consideration. For pre­ in the bare beam, but not in the com­ Bilinear behavior: tensioned strands, the above equation posite slab. This causes a discontinu­ L1L = in. (25 mm) 1.00 needs to be modified in two ways: ity in stress and strain at the interface Effective moment of inertia: 1. The change in stress L1fps needs to and this discontinuity remains while L1L = 0.86 in. (22 mm) be substituted for fs· additional loads are applied to the 2. For seven-wire strand, the spacing It may be seen that all of the above composite beam. s needs to be reduced by a factor of methods give a conservative estimate How does one find a section prop­ two-thirds to account for the bond prop­ of the instantaneous live load deflec­ erty of a cracked composite section erties of strand being different from tion. Additional information on deflec­ when some of the forces and moments those of deformed reinforcement. This tion of cracked prestressed beams is were applied to a different bare beam is based on a recommendation of ACI section? The solution is to work with given in Ref. 9. 0 Committee 224.' [If supplementary section properties of the composite mild reinforcing steel is used, the maxi­ beam and apply all forces and mo­ CRACK CONTROL mum spacing for deformed bars may be ments to the composite section. This At present, crack control require­ determined directly from Eq. (10-5).] requires modifying the forces and mo­ ments for cracked prestressed con­ The results of Example 1 give L1fps = ments applied to the bare beam to an crete members are not codified in ACI 9.97 ksi (68.7 MPa). This is substi­ equivalent force and moment applied 318-95. ACI Committee 318 is con­ tuted into a modified Eq. (10-5) as fol­ to the composite beam. This equiva­ sidering modifying the crack control lows, using a clear cover of 1.75 in. lent force and moment must produce requirements of Section 10 .6.4 for (44.5 mm): stresses in the bare beam portion of nonprestressed concrete. A bar spac­ s::; 213(540/.1/ps- 2.5cc)::; 2h(432)/L1fps the composite beam that are equal to ing requirement is proposed to replace s ::; 33 in. (840 mm) the actual stresses in the bare beam. the current z factor requirement. The The maximum spacing turns out to The process can best be illustrated proposed spacing requirement is: be very large because L1fps is so low by Example 2, which is purposely

84 PCI JOURNAL 1.75 C') .25 1.75 I- 1- .25 .------, ~t I \ I\ F- I \ :~ik I \ .50 Mbb = 1728 in.- k Mcb= 3150i~ k

)

1.50 -1 .75 -0.25

a. P and Mbb Applied b. Ficticious Force c. Equivalent Force d. Stresses Due to e. Final Stresses to Bare Beam in Slab on Composite Moment Mcb Applied Stresses in ksi Section to Composite Section

Fig. 10. Composite section analysis.

made simple in order to illustrate the in Fig. lOd due to moments M cb ap­ slab due to bare beam stresses. Ref. 11 process. plied to the composite beam. The ficti­ (pp. 201-204) and Ref. 12 give a more tious stresses in the composite slab are general method of analysis for cracked EXAMPLE 2 subtracted in order to obtain the true composite sections. Consider a 12 x 24 in. (305 x 610 stress in the composite slab. The final mm) beam shown in Fig. 9, subjected stresses are shown in Fig. lOe, and EXAMPLE 3 they are identical to stresses calculated to a prestress force P of 288 kips Example 3 illustrates the analysis of in the usual manner. ( 1281 kN) at an eccentricity of 8 in. a cracked composite beam. The given Of course, when the composite beam (203 mm), and a bare beam bending parameters for Example 3 are identical is uncracked, this procedure is unnec­ moment Mbb of 1728 kip-in. (195.3 to those for the example given by Al­ essary. But, this procedure also works kN-m). This produces stresses in the Zaid and N aaman. '3 The beam is for cracked sections. The analysis of a bare beam as shown in Fig. lOa. shown in Fig. 11 , and the given pa­ cracked composite beam is similar to The beam is then made composite rameters are summarized below. with a 6 in. (152 mm) slab of 12 in . that of a cracked noncomposite beam, Estab 3850 ksi (26,550 MPa) (305 mm) width of the same concrete with the additional step of including = strength. It is now necessary to fi nd the fictitious force in the composite n slab = 0.89535 the equivalent forces and moments ap­ plied to the composite section that will produce the same stresses in the bare beam portion of the composite beam. 81" This operation may be accom­ plished by extending the stress di a­ gram for the bare beam up through the composite slab, as shown in Fig. lOa. This produces a fictitious force F in the slab, as shown in Fig. lOb. This fictitious force is combined with the 8" prestress force P to produce the equiv­ alent force Pe at a resultant location to be applied to the composite section. Aps = 2.47 in2 The magnitude and location of P e • combined with M bb produce the de­ As= 2.73 in2 sired stress in the composite section, 26" as shown in Fig. lOc. The stresses shown in Fig. lOc are then combined with the stresses shown Fi g. 11. Example 3 composite beam.

July-August 1998 85 Table 2. Dead and live loads, and midspan moments (Example 3). of the neutral axis is selected and a w Midspan moments section property program is used to Loadings kips per ft in.-kips kN·m compute the section properties of the cracked transformed section. The Self weight 0.822 7890 892 - equivalent force P e is applied at the Added dead load on bare beam 0.675 6480 732 centroid of the cracked transformed . Added dead load on composite 0.250 2396 271 section, and the internal moment Mint Live load 1.111 10,670 1206 is found by subtracting (Pe multiplied Sum 2.858 27,436 3 100 by the distance it is moved from its lo­ cation to the centroid) from the exter­ nal moment Mext. The stress at the assumed neutral btransf = 72 . ~23 in. (1842 mm) formed section properties for the bare axis is checked and the neutral axis Ebeam 4300 ksi (29,655 MPa) beam are given below: = depth c is adjusted until a solution 2 Eps = 27,000 ksi (186,200 MPa) A = 817.7 sq in. (527,560 mm ) for c is found that produces zero 4 6 4 E5 = 29,000 ksi (200,000 MPa) I= 275,755 in . (114,800 x 10 mm ) stress at the assumed neutral axis lo­ cation. Fig. 13 shows the final itera­ Transformed Areas Y1 = 22.08 in. (561 mrn) tion for a neutral axis depth c of n ps = 6.279 Apst = 15.51 sq in . The stresses in the bare beam are 19.67 in . (500 mm). n5 = 6.744 A = 18.4 1 sq in. shown in Fig. 12. The stress gradient 51 The cracked transformed composite in the bare beam is extended upward Decompression stress in tendons section properties are as follows: to the top of the future composite slab. = 146.4 ksi (1010 MPa) 2 The magnitude and location of the fic­ A= 839.5 sq in . (541 ,600 mm ) Decompression stress in reinforcement 4 6 4 titious force F is computed. This is the I= 109,079 in. (45,401 x 10 mm ) =0 force that would be in the composite c = 19.67 in. (500 mm) P = Apslfdc) = 2.47 (146.4) = 361.6 kips slab, if it existed, without affecting the Y1 = 8.80 in. (224 mm) (1608 kN) stresses in the bare beam. The ficti­ Yna = 10.87 in . (276 mm) Dead and li ve loads, and midspan tious force F is combined with the de­ 6 3 1/y,a = 10,035 cu in. (164.4 x 10 mm ) moments are given in Table 2. compression force P and the magni­ The first step is to do an analysis of tude and location of the equivalent Pe!A = 1146.7/839.5 = 1.365 (c) the bare beam for the moments ap­ force Pe are determined. Mint= Mext- Pe (20.79- 8.80) = 27,436- 1146.7(11.99) plied to that beam. For consistency P: 361.6 at 57.43 20,777 = = 13,690 with the later cracked section analysis, F: 785.1 at 3.89 3.057 = M; /(/fycna) = 13 ,690/10,035 transformed section properties are 11 Pe: 1146.7 23,834 - -1.365 (t) used with the areas of composite slab, tendons, and reinforcement trans­ Yp = 23,834/1146.7 Stress at neutral axis = 0 (ok) 20.79 in. (528 mm) formed to an equivalent area of beam = The results are identical to those (measured from top of composite) concrete. The analysis using gross given by Al-Zaid and Naaman. 13 The transformed section properties shows The next step is the analysis of the equilibrium checks are somewhat that the beam is uncracked at the time cracked transformed composite sec­ more complicated than for Example 1, it becomes composite. Gross trans- tion at full service load. A trial depth but they can still be accompli shed without the use of a computer, as shown in Fig. 14. Note that using the method described in this paper, it is 0> not necessary to idealize the !-beam as three rectangles. The actual shape of the fl anges should pose no problem ____ j 1 ~§ ··l-----...---F = 785.1 k 0>,... for a section property program. 1.246 0 C\l DEFLECTION AND Pe= '\ Mbb= )1146.7 k CRACK CONTROL } 14370 in.- k The curvature K is fo und as c.jc, where C.c at the top of the precast beam Mbb= is equal tofJ Ec = 1.467/4300 = 3.41 X 14370 in. -k l 0-4. This distance from the top of the -0.197 precast beam to the neutral axis is 11 .67 in. (296 mm). Dividing by this, K = 2.92 x 10·5/in. Subtracting the Fi g. 12. Stresses in bare beam, and fictitious fo rce F. dead load curvature of 0. 70 X 10·5,

86 PCl JOURNAL btransf = 72.523"

.-----,..- ---7 / ...... ,.' 1.467 ksi ;" u-,------L'li_c_.g.- ;" -----.... ;" ;" ;" ;" ;" ;" ;" ;" ;" ;" , ;" ;" ;" ;" ;" ;" .... ~'~"'------~ 4.744 x nps = 29.79 ksi 5.068 x ns = 34.18 ksi

a. Cracked Transformed Section b. Stresses

Fig. 13. So lution of cracked transformed composite section.

computed from a gross section analy­ mm). For a building, the allowable The steel stress fs is 34.18 ksi (235.7 sis, the live load ~urvature KL = 2.22 x live load deflection would be L/360 = MPa). Assuming a clear cover to the I0-5/in. (0.874 x I0-6/mm). Substitut­ 2.67 in. (68 mm). Thus, the deflection longitudinal steel of 1.5 in. (38 mrn), ing in the conservative and approxi­ is satisfactory. the maximum spacing of reinforce­ mate formula L\L = (5148)KLU, with L The reinforcement nearest the ten­ ment is given directly by Eq. (10-5) as = 960 in. (24.4 m), L\L = 2. 13 in . (54 sion face consists of deformed bars. 12 in. (306 mrn). A minimum of three

1.012 Actual Stress= 1.012(n) = 1.012 x .895 = 0.906

528.4 k

= 0 ,_ N X N ~ Mext = 528.4(51 .92) ;-;; = 27436 in.-k t-. check (0 Co 0 X

435.1 -.....:..--t·~ 528.4 k 93.3 :I:F = 0 check

528.4 k

a. Stresses and Forces b. Resultants

Fi g. 14. Equil ibrium check.

July-August 1998 87 One such technique is called binary existing section property program, Compression Face search. The author first learned of this which computes secti on properties 14 ..c: j ~ technique in an astronomy magazine. using the transformed area of the steel. Try No. 3 CX) y_ The procedure is best illustrated by an The quantity YP is the previously de­ ::c Try No. 4 example. Having checked to fi nd that termined di stance from the compres­ ~ Try No. 2 ..c: the section is cracked, it is known that sion face to the resultant prestressing ~ ~, ..c: the value of the neutral axis depth c fo rce. The quantity Y is the distance ..c: Try No . mu st lie in a range between zero and from the compression face to the cen­ the overall depth h of the section. ter of gravity of the cracked trans­ The key concept is to di vide that formed section, determined by the sec­ range in half and decide in which half tion property subrouti ne. the solution exists. That half range is The same technique is useful for divided in half, and the process re­ solving other problems requiring itera­ peated. Fig. 15 shows pictori ally the tion, such as strain compatibility anal­ Fi g. 15. Binary search for neutral first few steps of the process. After re­ ysis. The technique works even if the ax is depth. peating the process 20 times, the cor­ possible range of the desired variables rect neutral axis location is known to is minus infinity to plus infinity (-= to 20 bars equally spaced across the tension an accuracy of h/(2 ), or about one +=). In this case, a dummy variable face are required. millionth of the overall depth. Of ANGL is incremented, and the desired course, this degree of accuracy is not variable is equated to the tangent of needed, but a personal computer does ANGL. Then as ANGL is incremented ITERATION PROCESS the calculation in an instant. in the range of (almost) ±90 degrees, In order to fi nd the correct location The decision of whether the neutral the desired variable takes on values of of the neutral axis of a cracked section, axis depth should be increased or de­ (almost) ±=. But, the process works it is necessary to assume a neutral axis creased after each try is done in the only if the desired variable is continu­ location and then check to see if the as­ same manner as was illustrated in Ex­ ally increasing or continuall y decreas­ sumed location is correct. The first try ample 1. If the stress at the assumed ing throughout the range investigated. will usually not be correct. It is then neutral axis location is tensile, c must necessary to assume a new neutral axis be decreased, and conversely if the location and repeat the process. When stress is compressive. MANUAL ITERATION doing this manually, one will naturally The BASIC programming steps are If the iteration is done manually, it is try to make a good guess on the correct shown in Fig. 16, with remarks fol­ desirable to make each guess for the neutral axis location on each try, in lowing the exclamation points. neutral axis depth c as accurate as pos­ order to minimize the number of tries Subroutine 900 modifies the input to sible, in order to minimize the number needed to find the correct location. the section property program so that it of iterations needed. Dr. Alan Mattock Iteration is best done by computer, onl y considers concrete areas above has suggested to the author an efficient usi ng a more mechanical approach. the neutral axis. Subroutine l 000 is an process for adjusting the assumed value

Subroutine for finding neutral axis location

C=H/2 First try for C, depth of na DELTC=H/4 Increment of neutral axis depth FOR J=l TO 20 Binary search for correct C GOSUB 900 Modify input to section property program to include only concrete area abov e assumed neutral axis location GOSUB 1000 compute cracked transformed section properties using section property progr am MINT=MEXT - P*(YP-Y) Internal moment with respect to center of gravity of cracked transformed section YNA=C-Y Distance of assumed na below center of grav ity of cracked section FATNA=P/ A-MINT*YNA/I Stress at assumed na location I F FATNA>O THEN C=C+DELTC Increase c if stress at na is comp ELSE C=C-DELTC Decrease c if stress at na is tens END IF DELTC=DELTC/2 Reduce increment by half NEXT J

Fig. 16. BAS IC programming steps for bina ry search techn ique.

88 PCI JOURNAL of c after the ftrst try. Find the depth of the zero stress ftber, and use this for the trial value of c for the next try. For example, consider the first try Co in Example 1, with c assumed to be Co IX! 0, en 18 in. (457 mm). Fig. 17 shows the ,_.:C\J 0 stress diagram for this condition. The "0 (]) __e.g . of Cracked, Transformed location of the zero stress fiber may E 0 ::l 3: . Cross Section /

July-August 1998 89 REFERENCES

I. Nilson, A. H., "Flexural Stresses After Cracking in Partially 8. Tadros, M. K., Ghali, A., and Dilger, W. H., "Effect of Non­ Prestressed Beams," PCI JOURNAL, V. 21, No.4, July­ Prestressed Steel on Loss and Deflection," PCI JOURNAL, August 1976, pp. 72-81. V. 22, No.2, March-April1977, pp. 50-63. 2. Moustafa, S. E., "Design of Partially Prestressed Concrete 9. Tadros, M. K., Ghali, A., and Meyer, A. W., "Prestress Loss Flexural Members," PCI JOURNAL, V. 22, No.3, May-June and Deflection of Precast Concrete Members," PCI JOURNAL, 1977, pp. 13-29; see also discussion, V. 23, No. 3, May-June V. 30, No. I, January-February 1985, pp. 114-141. 1978, pp. 92-105. 10. ACI Committee 224, "Control of Cracking (ACI 224R-90)," 3. Tadros, M. K., "Expedient Service Load Analysis of Cracked American Concrete Institute, Farmington Hills, MI, 1990, p. Prestressed Concrete Sections," PCI JOURNAL, V. 27, No. 6, 43. November-December 1982, pp. 86-111; see also discussion, 11. Ghali, A., and Favre, R., Concrete Structures: Stresses and V. 28, No.6, November-December 1983, pp. 137-158. Deformations, E & FN Spon (imprint of Chapman and Hall), 4. Boczkaj, B. K., "Section Partially Prestressed - An Exact London and New York, Second Edition, 1994, 464 pp. Solution," PCI JOURNAL, V. 39, No.6, November-December 12. Ghali, A., and Elbadry, M., "Cracking of Composite Pre­ 1994, pp. 99-106. stressed Concrete Sections," Canadian Journal of Civil Engi­ 5. PCI Design Handbook, Fifth Edition, Precast/Prestressed neering, V. 14, No.3, June 1987, pp. 314-319 Concrete Institute, Chicago, IL, 1998. 13. AI-Ziad, R. Z., and Naaman, A. E., "Analysis of Partially 6. Suri, K. M., and Dilger, W. H., "Steel Stresses in Partially Pre­ Prestressed Composite Beams," Journal of Structural stressed Members," PCI JOURNAL, V. 31, No. 3, May-June Engineering, American Society of Civil Engineers, V. 112, 1986, pp. 88-112. No.4, Aprill986, pp. 709-725. 7. Moustafa, S. E., "Nonlinear Analysis of Reinforced and Pre­ 14. Sinnot, R. W., "A Computer Assault on Kepler's Equation," stressed Concrete Members," PCI JOURNAL, V. 31, No. 5, Sky and Telescope, V. 70, No. 2, Sky Publishing Co., September-October 1986, pp. 126-147. Cambridge, MA, August 1985, pp. 158-159.

90 PCI JOURNAL APPENDIX- NOTATION

A = cross-sectional area of transformed section J = a counter used to count repetitions in binary Aps = area of prestressed reinforcement search routine Apst = transformed area of prestressed reinforcement K = curvature

A5 = area of nonprestressed reinforcement KL = curvature due to live load

A,1 = transformed area of nonprestressed reinforcement L = span length

A 1 = transformed area of steel M = service load moment b = width of compression zone Mbb = service load moment applied to bare beam

b1ransf = transformed width of composite slab M eb = service load moment applied to composite beam C = compressive force in concrete Mext = service load moment due to external loads (all CR = loss of prestress due to creep of concrete loads except prestress) c = distance from extreme compression fiber to (as­ Mint = internal bending moment acting about center of sumed) neutral axis gravity of cracked transformed section de = depth of compression zone Mmsp = service load moment at midspan dP = distance from extreme compression fiber to cen­ n = ratio of moduli of elasticity

troid of prestressing steel in tension nps = ratio of Ep5 1Ec

E = modulus of elasticity n5 = ratio of EJEc Ee = modulus of elasticity of concrete nslab = ratio of EslabiEc Eps = modulus of elasticity of prestressed reinforcement P = prestress force at decompression

£5 = modulus of elasticity of nonprestressed reinforce­ Pe = effective axial load, including prestress force at ment decompression and fictitious force in composite E stab = modulus of elasticity of composite slab slab e = eccentricity of prestress force SH = loss of prestress due to shrinkage of concrete F = fictitious force in composite slab, needed in T = tension force in steel cracked section analysis of composite beams w = unfactored load per unit length fe = concrete stress Yo = distance between center of gravity of cracked J; = specified compressive strength of concrete at 28 transformed section and zero stress fiber days Yna = distance between center of gravity of cracked f cds = concrete stress at center of gravity of prestress transformed section and (assumed) location of force due to all permanent (dead) loads not used in neutral axis computingfcir Yr = distance between center of gravity of cracked fcir = concrete stress at center of gravity of prestressing transformed section and location of axial force force immediately after transfer acting on section

!de = decompression stress in prestressed tendons that y1 = distance from top fiber to center of gravity of exists when stress in adjacent concrete at same transformed section level is zero ,1frs = change in stress in prestressed reinforcement be­ Ina = concrete stress at (assumed) neutral axis location tween decompression stress and stress at full ser­ fse = effective stress in prestressing steel after losses vice load h = overall height of member ,1L = deflection due to live load I = moment of inertia of transformed section f c = maximum concrete strain at service load

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