Analysis of Cracked Prestressed Concrete Sections: a Practical Approach
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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.