Precast and Prestressed Concrete. PCI Design Handbook. (Third Edition, MNL )

Transcription

1 Precast and Prestressed Concrete PCI Design Handbook (Tird Edition, MNL )

2 Sect. Cange Te Left at ERRATA PCI Design Handbook- Precast and Prestressed Concrete Tird Edition MNL pp troug 2-43 p p p pp and 4-29 p p p pp and 4-74 Cange Fig Upper cart, bottom curve sould be kf/r = 150. Eq Te quantity r/ r sould be multiplied by te lateral load, W. Fig. dimension on eac sketc from Widt varies to 4 O , p. 4-62, is corrected Example is affected as indicated on te corrected p In Example 4.3.9, if non-prestressed reinforcement is present (as in Example 4.2.6), te force in te bars, Af or sould be included in te calculation of T or C. Left column, cange bottom equation to T = / 0.4(53.1) + ko )) 2 = 186 in.-kips calculations in rigt column cange accordingly as indicated on te corrected p Rigt column, line 11, cange to >0.25 Tocr 0.85 (0.25) (81.8) ft-kips 0.8 /ö (3680) coefficients Kes and Kc, are for pretensioned members as are Kcir and KS. coefficient in Eq and in equation for icr in eading for table, Fig from 1.67 to 1.6. Tabular values for C in Fig are based on coefficient of 1.6 and are correct as given. p Tabulated cambers and deflections (top of page) are incorrect. A corrected p is attaced. Cange numbers in Example as follows: Final long-time camber = 1.24 in. = ( ) in < 3.50 in. OK p p p In Te In equations 1 (Design), and 3 (Analysis), cange d to d. tabular values for concrete strengts of 5000 psi and greater are incorrect. Also, in sketc, cange d to d; second note, cange > to > ; in table eading, Cal. 1, cange f to f. A corrected table (p. 4-62) is attaced. equation, cange d in denominator of 1st term to d. p. 5-4 Top of left column, below Fig sould read = f/6.7 p.5-15 In Fig , cange eading, Col. 2, to: Reactions and vertical sear in Col. 3, cange tird line to: Wc B and C p p pp. 6-5 and 6-49 p. 6-7 M = 2 Rigt column Units for w sould be lb/in Left In 2 (or lb/in/in). column, line 6, cange less to more. effective troat dimensions of 0.2 db or 0.3 db are valid only wen te space is filled flus to te solid section of te bar. Fig , Case 1, in te expression for P1, cange e to p. 6-8 column, in Eq and in line immediately following Eq , cange f to f.

3 Rigt Rigt Rigt Fig. sould In p p pp. 6-19, 6-20 and 6-30 p p In Equations In Left column, line 6 cange to ASTM A-36. Fig , Section No. 4, expression for, cange minus sign to plus sign in denominator , and complete equations by adding: (values in Table 6.7.1) Table 6.7.1, rigt column, cange eading to read: Maximum V, (= V/4)), lb column, te expression for X sould read: = fi6.7/vt p pp troug 6-67 p In Table , middle table for Tension on External Ancor Bolts, missing values for b in Col. 1 vary from 12 to 28 in., same as in lower table. Te tables do not contain te limitation of Eq , 800 bd. Tis will frequently control, especially on small corbels. Table Add to criteria: a =0.75 d = 1.25 V =4)V 0.84)bd Cange tabular values accordingly. A corrected set of tables is attaced. Beam No. 24, te expression for M sould read: p column, line 6 read: Use 2 #3 stirrups = 0.44 sq in. max p p p pp troug 6-58 Left column, top: Te value sown for fb is an upper limit. column, line 10 Cange Ref. 7 to Ref. 13. Values For (a), missing dimension is ee. For (c), ee is to end of embedment, not full widt of column; cange strain and stress diagrams accordingly. in te tables do not include 4) as indicated. Cange 4) PC in eadings to P1. Example 6.5.2, p. 6-9, is affected by te above, as indicated on te corrected pp. 6-9 and p In Table , cange values in tables for in, strand: For 270-ksi, 7-wire strand For 250-ksi, 7-wire strand p Cange to 16,387. Cange to 416,231. For explanation of te canges and background on new material in te PCI Design Handbook, see Explanatory Discussion on PCI Design Handbook, Tird Edition by PCI Committee on Industry Handbook, PCI JOURNAL, May June 1988, pp Reprints available from Prestressed Concrete Institute at $3.00 per copy (Code No.: JR 344) member price.

4 /2) Example Use of Fig.4.1O.2 for determination of prestressing steel require ments bonded strand Given: PCI standard rectangular beam 16RB24 Applied factored moment, M = 600 ft-kips f 5 = 6000 psi normal weigt concrete f = 270 ksi, low-relaxation strand Concrete: f = 5000 psi normal weigt concrete Prestressing steel: 10-3/8 diameter 250K stress-relieved strand = 10(0.080) = sq in. Section properties: A Zb Yb =218in = 381 in in. Problem: Find design flexural strengt, 4M Determine Cw,,, for te section: = C + (u ) Problem: Find te required amount of prestressing steel. Solution: Referring to Fig : terefore M = K bd/1 2,000 M5 (12,000) 600(12,000) Req d K 0 bd 2 16(21)2 =1020 for( = 0.23, K = , K = = 0.23 (0.01) = A = bv c PS = 1.73 sq. in (16)(21)(6) 270 Use 12 1/2 diameter strands; = 1.84 sq in. Example Use of Fig values off5 by stress-strain relationsip bonded strand Given: 3-4 x 8 ollow-core slab 40. = 1¼ since w = = (0.8)(250) C0 = = (40)(7)(5) Entering Fig wit tis parameter and an assumed effective stress, f55 = 150 ksi gives a value of: = or f5 = 0.965(250) = 241 ksi Determine te flexural strengt: 4M5 = a/2) a = (A5f. +A5f)/(0.85f 0b) Since A = 0: +A5f5(d a/2)] 0.800(241) a = In. = 0.85(5)(40) 4M = 0.9[0.8 (241) (7 + 0] = 1116 in-kips 93.0 ft-kips Ceck te ductility requirement, 4M5> 1.2M 5r P = se Aps = 150 (0.80) = 120 kips 1.2Mcr = 1.2(P/A + Pe/Zb \/Tc)Zb / (2.98) = V\ ) 381 = 923 in.-kips = 76.9 ft-kips <93.0 ft-kips OK 4 10 PCI Design Handbook

5 bd = T Span of spandrel beam = 30 ft clear 4. Determine torsion reinforcement require = 5000 psi, normal weigt concrete ments: Reinforcement f = 60,000 p1 T= 4T = I(T d=69in. or Loads (kips/ft): T 376 T = -- = T) D.L.: = 256 in.-kips Precast floor 60 psf (20ft) 1.2 (1.4) = 1.68 Topping 25 (20) = 0.5 (1.4) 0.70 Superimposed 10 (20) = 0.2 (1.4) = 0.28 By Sect (ACI ) Window = 0.50 (1.4) = 0.07 T 4T = 4(186) = 744 OK Spandrel = 0.63 (1.4) 0.88 By Eq of ACI L.L.: 50 psf (20) 1.00 (1.7) =.LZQ T =A1ax1y1f w=5.31 s Problem: or Determine torsion reinforcement requirements Ta(s) = cx1yf Solution: 1. Compute torsion moment (Ta) at critical sec tion, assumed to be 5 -O from face of support: Assume x1 4 in.; y1 = 70 in. = V = w(15 5) = 5.31(10) = 53.1 kips (y = w,, for torsion = (70/4) = 6.44, use a = kips/ft 256(12) Eccentricity = 2/3 (8) = 8.62 in. A = 0.12 sq in.ift = 1.5(4)(70)(60) T w, (a) (e/2 5) 4.36 (8.62) (10) =0.01 sq in./in. = 376 in.-kips 2. Determine if torsion effects must be considered. If T 4 (0.5 \/i x2y) must consider torsion. Zx 3 2y = 62(72) + 62(8)(2) + 82(8) = 3680 in. 4(O.5 fx2y) = 376> 110.6, consider torsion 1/x1) s 1.5 Tis is te required area of steel in eac leg of te closed stirrup for torsion only. Te sear steel requirement must be added to A. Te minimum area of closed stirrups is: A + 2A = 50bs/f 0.85(0.5)/ (3680) Placement of closed ties in a 6-in, web is dif 1000 ficult. Consider re-design wit greater web in.-kips tickness, or arrange reinforcement as follows: 3. Determine te torsion moment strengt pro vided by concrete 0.8\/f Zx2y T = I 1 + /0.4V CT )2 were: 6(69) T = 0.8 (3680) / 0.4 (53.1) ( (376) )2 = 186 in.-kips 4 30 PCI Design Handbook

6 1.24) (1) Release Multiplier (2) Erection Multiplier (3) Final Prestress 4.35 t 1.80 x (1) X (1) t Wd x (1) x (1) t t Wd x (2) t 1.12 t w, Example Use of multipliers for determining long-time cambers and deflec tions Given: 8DT24 of Examples 4.2.9, 4.6.1, and Non-structural elements are attaced, but not likely to be damaged by deflections (ligt fixtures, etc.). Problem: Estimate te camber and deflection and deter mine if it meets te requirements of Table 9.5(b) of te Code (see Table 4.6.1). Solution: Calculate te instantaneous deflections caused by te superimposed dead and live loads. 5we 384 E0l 4 = 0.48 in., 5(0.080) (70x 384 (4287)(20,985) = 2.36 in. (see Example 4.6.3) For convenience, a tabular format is used (above). Te estimated critical cambers and deflections would ten be: At erection of te member after WSd is applied Final long-time camber 1.80 in in. Te deflection limitation of Table 9.5(b) for te above condition is e/240. (70 x 12)/240 = 3.50 in. Total deflection occurring after attacment of non-structural elements: ( in. < 3.50 in. OK 4.7 Compression Members Precast and prestressed concrete columns and load bearing wall panels are usually proportioned on te basis of strengt design. Stresses under service conditions, particularly during andling and erection (especially wall panels) must also be con sidered. Te procedures in tis section are based on Capter 10 of te Code and on te recommen dations of te PCI Committee on Prestressed Con crete Columns (referred to in tis section as te Recommended Practice ) Strengt Design of Precast Concrete Compression Members Te capacity of a reinforced concrete compres sion member wit eccentric loads is most easily determined by constructing a capacity interaction curve. Points on tis curve are calculated using te compatibility of strains and solving te equa tions of equilibrium as prescribed in Capter 10 of te Code. Solution of tese equations is illus trated in Fig ACI waives te minimum vertical rein forcement requirements for compression mem bers if te concrete is prestressed to at least an average of 225 psi after all losses. In addition, te Recommended Practice permits te elimination of column ties, if te nominal capacity is multiplied by Interaction curves for typical prestressed square columns and wall panels are provided in Part 2. Construction of an interaction curve usually fol lows tese steps: Step 1: Determine P0 for M0 = 0. (See Fig (c)) Step 2: Determine M0 for P = 0. Tis is normally done by neglecting te reinforcement above te neutral axis and determining te moment capacity by one of te met ods described in Sect Step 3: For non-prestressed columns, Pflb and Mflb at te balance point may be determined (see Fig (d)). For prestressed col umns, te yield point of te prestressed reinforcement is not well defined and te stress-strain relationsip is non-linear over a broad range (see Fig ). PCI Design Handbook 4 47

7 (0.59w)()] FLEXURE Fig Coefficients, K, for determining flexural design strengt bonded prestressing steel Procedure: A5f = 1. Determine _2_ bd f a 2. Find K from table Determine 4) M = K bd (ft kips) 12,000 Basis: K :u C()[1 7 Note: K from tis table is approximately equiv alent to 4) K f from Table Table values are based on a strain compatibility analysis, using a stress-strain curve for prestress ing strand similar to tat sown in Fig As terisk(*) indicates o > 0.363, and 4)M = 44fbd (0.36 l i)] Values of K f psi * 670* 670* 670* 670* psi * 894* 894* 894* 894* psi * 1066* 1066* 1066* 1066* 1066* 1066* psi * 1215* 1215* 1215* 1215* 1215* 1215* 1215* 1215* psi * * 1341* 1341* 1341* 1341* 1341* 1341* 1341* 1341* 1341* psi * 1441* 1441* * 1441* 1441* 1441* 1441* 1441* 1441* 1441* 1441* 1441* 4 62 PCI Design Handbook

8 1.35/2 Problem: Find te maximum sear strengt. Solution: Ab = 0.20 sq in. (a) IV,, = 0.85(800)(0.20)(1)Vô= 9617 lb/stud (b) For de = 2 in. 0.85(2)(3.14)(2)2 (1)\/ = 1510 lb (c) 4V0 For de = 3.5 in. = 0.85(2)(3.14)(3.5)2 (1)\/U 4624 lb Example Capacity of welded eaded studs Given: Bracket on column as sown. f 5000 psi (normal weigt) Factored load on bracket 75 kips (d) For de = 5 in. = 0.85(2)(3.14)(5)2 (1)\/ 9436 lb Maximum capacity of te group: 1. 10(1510) = 15,000 lb 2. 4(1510)(3) = 18,120 lb or 2(4624)(3) = 27,744 lb 3. 4(9436) = 37,744 lb Tus condition 1 controls. Te design sear strengt as governed by steel strengt is: = 0.75 Ab fs = 45,000 Ab (Eq ) were4 1.0 Table tabulates te maximum capacities from te above equations. Combined sear apd tension Te design strengt of studs under combined tension and sear sould satisfy te following in teraction equations: Concrete: [ were 4 = 0.85 Steel: (PS\2 PC) [(P)2 + ()2] 1.0 (Eq ) (Eq ) Problem: Determine if studs are adequate to resist te loads sown. Solution: (a) Ceck concrete strengt: Tension (top group of studs) from Table : de 4P0 = 5 in., e = 6 in., % in. studs = 6(27.4) = kips Tis is te cumulative capacity of six individual cones, reduced for edge distance. It can also be determined from Eqs a and Or P1 from Table (Case 3): y = 3 in., x 16 in. ee 6 in kips 4P 01 = 0.85 (67.5) = 57.4 kips Tis is te capacity of a truncated pyramid ac counting for te stud spacing and controls te de sign. A moment-resisting couple is formed: C = T = 0.85 f ba = 57.4 kips were4 = 1.0 P and V, are te factored tension and sear loads. Plate tickness Tickness of plates to wic studs are attaced sould be at least 2/3 of te diameter of te stud. comp. block, a jd 11 o.81o) = 10.3 in. = T + N = M/jd + N = 75(6)/ =S57kips Ceck sear (all studs): 1.35 in. PCI Design Handbook 6 9

9 2.34/2 From Table : Deformed Bar Ancors f = 5000 psi, d0 > 9 in., 5/8 in. studs Deformed bar ancors are automatically welded = 12(14.7) = kips to steel plates, similar to eaded studs. Tey are ancored to te concrete by bond, and te devel To satisfy Eq : opment lengt can be taken te same as Grade 60 reinforcing bars (see Table ). = 2rd A (not critical) Bolts and Treaded Connectors V,, = 176.4/0.85 = kips In most connections, bolts are sipped loose and treaded into inserts. Occasionally a precast Combined capacity: concrete member will be cast wit a treaded connector projecting from te face. Tis is usually From Eq : undesirable because of possible damage during andling. Wen embedded in suc a manner, de sign for concrete strengt is similar to tat for 2 fv 2] 1 F1p\ j-j + j J 1.0 studs. I P L \ c/ \Vc/ Hig strengt bolts are used infrequently in pre 2 2 cast concrete connections because it is question _L_ F + ( 75 I 1 able as to weter te tension can be eld wen 0.85 L \67.5) \207.5) ] tigtened against concrete. Wen used, AISC rec ommendations sould be followed < 1.0 OK Table gives allowable working and de sign strengts for most commonly used treaded fasteners. (b) Ceck steel strengt: Tension in top group of studs: From Table for % in. studs: P = 6 (16.6) = 99.6 kips (Could also be determined from Eq ) C = T = 0.85 f ba 99.6 kips Hig strengt treaded rods Rods wit treads and specially designed nuts and couplers are available wit properties similar to Grade 60 reinforcing bars and post-tensioning bars. Design information is give in Part 11. comp. block, a Inserts Cast in Concrete = 0 85(5)(10) = 2.34 in. Loop inserts of te type sown in Fig can id 11 = 9.83 in. be investigated in a manner similar to tat for welded studs, using Eqs , and for = M/jd + N = 75(6)/ te concrete tensile and sear strengts. Te 57.8 kips strengt as controlled by steel can be taken from manufacturers catalogs, or calculated based on Sear in studs: wire strengts sown in Table or te strengt of te bolt or treaded rod sown in Ta- From Table for % in. studs: bles and V = 12 (13.8) kips An evaluation of publised test results leads to certain caracteristics common to most of te (Could also be determined from Eq ) available inserts: Combined capacity: 1. Controlling strengt conditions of various types of inserts are similar. From Eq : 2. Pullout strengt decreases wit decreasing unit /57.8\2 \2 weigt of concrete. / ) + 3. For inserts located in zones of potential flexural k165 6) cracking, te pullout strengt sould be re = < 1.0 OK duced by about 10%. 6 i 0 PCI Design Handbook

10 CONNECTIONS Table Design strengt of concrete brackets, corbels, or aunces projection Design strengt by Eqs or for following criteria: N = psi : L V, = O.8bd L_, 4VU D Values of 4, V,,, kips 4 Projection 6 Projection 8 Projection A, # # # # # # # # # # # # b 2 # # # # # # # # # # # II 2 # # # # # # # PCI Design Handbook 663

11 CONNECTIONS Table (continued) Design strengt of concrete brackets, corbels, or aunces 10 Projection 12 Projection 14 Projection A, # # # # # # A, # CO 2 # # # # # A, # a 2 # # # # # A, # # # # # # # # # # # # I I 6 64 PCI Design Handbook

12 CONNECTIONS Table (continued) Design strengt of concrete brackets, corbels, or aunces 4 Projection 6 Projection 8 Projection 2 # # # # # # # # # # # # # # v 3 # # # # # # # # A, 4, # # # Co 2 # # # # # # # # # # # # # # # # # # # # PCI Design Handbook 6 65

13 C- -- CONNECTIONS Table (continued) Design strengt of concrete brackets, corbels, or aunces 10 Projection 12 Projection 14 Projection I.0 2-#4 2-#5 2-#6 2-#7 2-#8 2-#9 3-#4 3-#5 3-#6 3-#7 3-#8 3-# #6 2-#7 2-#8-2-#9 3-#6 3-#7 3-#8 3-#9 4-#6 4-# # I-.0 2-#6 2 #7 2-#8 2-#9 3-#6 3-#7 3-# #6 4 #7 4-# # #8 2-# # # # PCI Design Handbook

14 8 CONNECTIONS Table (continued) Design strengt of concrete brackets, corbels, or aunces 4 Projection 6 Projection - Projection 2 # # # # # # # # # # # # # # # # # # # # # # # Projection 12 Projection 14 Projection A, # # # # J 3 # # # # # # # # A, # # # # # # # # # # # PCI Design Handbook 6 67