MN I | 100 343 THS A COMPAMSQN Q? RENRDRSEB CGHCRE‘TE AND PRESTRESSED GMCRETE SLAB DESiGHS Thasis {or 'i‘he Degree cf M. S. MICHEGAN STATE COLLEGE Zaki Alemdargil 1950 ’ THESKS Certification of Acceptance of Master's Thesis Michigan State College This is to certify that thesis A COMPARISON OF REINFORCED CONCFETE AND PRESTRESSED CONCRETE SLAB DESIGNS Presented By ZEKI ALEMDARGIL Candidate for degree of Master of Science in Civil Engineering Has been found worthy of acceptance. 02/4. 2 @1611- U, ajor Adviser Date June 22J 1959 A COMPARISON OF REINFORCED CONCRETE AND PRESTRESSED CONCRETE SLAB DESIGNS By Zeki.élendargil A THESIS submitted to the School of Graduate Studies of Michigan State College of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Department of Civil Engineering I950 I. II. III. V. VI: M Introduction Purpose of Thesis Notations Prestressed Concrete Design Theory Fundamental Conditions Depth Bond Shear and Diagonal Tension Design Procedure Design of a Highway Bridge Slab with Reinforced Concrete Specifications Design Nozl SH-2O Loading? Design Nole “H-lS Loading“ Design No:III“H-1o Loading“ Design of a Highway Bridge Slab withPrestressed Concrete Specifications Design No:I "Hp20 Loading" Design Nole “3-15 Loading“ Design No:III“H-1o Loading“ Conclusion - “ Bibliography 237374 Page ~1de ll 12 l3 14 15 16 17 23 28 33 34 35 42 54 AC KNOWLED GEL‘VEENT The author wishes.to express his indebtedness to many individual and firms for criticism and councel, but he desires particularly to acknowledge his thanks to Prof. C. A. Miller of Michigan State College, who Originated the idea of comparison between Prestressed and Reinforced Concrete Slab designs and read the man- uscript; Introduction Application of a virtually unexplored concept-- prestressing--to concrete construction makes possible remarkable advances in structural engineering. Eugene Freyssinet's five flat arch bridges over the Marne near Meaux, for instance which have a depth .at the crown of only 57' for a span of 245. Roebling's 14,000 square foot Joinuqh:s_glahgin a Chicago warehouse, which is only 3' thick yet does not show even a hair crack, though it has been sub- Jected to heavy loads and large temperature varia- tions for two and one-half years. Gustave Magnel's hangars at Brussels airport with 165-foot beams, and Preload Corporation's dome at Kan- sas City spanning 180 recto-though only 5' thick. The secret lies in applying forces to the concrete to overcome the tensile stresses caused by loads, volume changes and impact. This creates concrete capable of taking tension without cracking, and makes it an effi- cient building material with many applications apart from the spectacular ones cited. The great shortcoming of concrete is that its strength in tension is much smaller than in compression. To overcome this weakness, the usual procedure has been to reinforce the concrete and put in steel rods to take tension. The technique has proved very successful with respect to safety, but it has not prevented concrete from cracking. In recent years, however, great strides have been made in improving the quality and strength of both con- crete and steel. Reliable concrete with a compressive strength of 9000 to 12000 psi and suitable steel with a tensile strength of over 200,000 psi can be attained readily. However, these materials cannot be used to full advantage in reinforced concrete beams or slabs. In 'the first place, the added expense of obtaining high- strength concrete is not warranted; secondly, if high- strength steel is used, the strain will be several times that of mild steel and wide cracks will br formed in the beam under working loads. But full use can be made of the best concrete and highest strength steel obtainable if the concrete is prestreesed. Since the steel is stretched independently to com- press the concrete, steel with high tensile strength is an asset. The reasons for this are that large pre- stress can be applied with a small amount of steel, and the stress losses in the steel, due to plastic flow and shrinkage of the concrete become proportionally smaller as the applied tension is increased. High strength concrete is advantageous because the amount of prestress that can be applied increases with the compressive strength of the concrete. And obviously the greater the prestress the greater the B. M. that a prestreesed con- crete slab can carry. Prestressing converts into an elastic material-- one for which the stress and deformations can be ac- curately computed. Since the entire concrete section carries bending moment, a prestreesed beam can carry much heavier loads without cracking than a reinforced concrete beam of the same section. Or it can span much greater distances than has so far been possible with conventionally-designed beams. In fact, prestressing places concrete in full economic competition with struc- tural steel. One of the most impressive characteristics of pre- stressing is that it enables long-span beams to be con- structed of short, precast units. Much like a group of books can be lifted from a shelf by applying longidu- dinal pressure to the and pair, such a beam is given a monolithe character by the prestress. Another big advantage is that if heavy loads induce large deflections and cause slight cracks, a prestress beam will recover its original shape and the cracks will be effectively sealed when the loads are removed. Tests were made by the John P. Roeblings Sons Co. in which a 10' long, 1-5/8' thick prestreesed concrete slab was deflected as much as 3' or l/40th span. Despite this large deflection, the slab returned to its original shape on removal of the load, no cracks being apparent. The Pugpose o This Thesis Prestressed concrete holds the promise of provid- ing a new structural material by making more efficient use of the physical pronerties of.steel and concrete in combination. The purpose of this thesis is to show the design of prestreesed concrete and reinforced concrete and thus making a comparison of the two. Notations A : cross sectional area of a slab ' AS : cross sectional area of reinforcing steel b : width of the slab b = as a subscript denotes the bottom fiber of a slab c : as a subscript denotes concrete d : depth of slab (effective) d = a subscript denotes "dead load" e = eccentricity of the steel from the neutral axis Ec : Modulus of elasticity for concrete Es = Modulus of elasticity for steel F1 = Initial prestress force supplied by the steel. f : denotes a fiber unit stress fc = Allowable fiber unit stress for concrete in compression rot or ftc : Allowable fiber unit stress for concrete in tension fdu : upper fiber stress due to the dead load upper fiber stress due to the live load H) C. l--’ ll fdl é flb = lower or bottom fiber stress due to the live load I = moment of inertia Moment or total bending moment '3 Cur ll _ Moment due to the live load on beam 3 l-‘ I Moment due to dead load on the beam '3: Q. n q _ prOportion of F1 that remains permanently, usually has a value of .85 n = ratio of ES/EC percentage of steel in the entire cross section Radius of gyration of the concrete section A subscript denoting steel A subscript denoting tOp fiber Total shear Unit shear stress 7 Design Theory Fundamental conditions: The most important difference between a structural unit of reinforced concrete and one of prestressed concrete is that in the latter, all the concrete in any cross section must be kept in compression. In general, a prestressed concrete beam subject to bending only must resist two moments. The first is produced while the prestressed is being established and is the result of the moment due to the dead load plus the moment caused by the prestress force. These two moments are in opposition to each other and result in smaller total stress. The second moment is produced by loading the beam in actual use. This moment is generally called the live load moment and includes impact. Next, consider a beam of cross-section A, a steel force Fl, eccentricity of the steel, e, and the distance from the neutral axis to the upper and lower fibers, yl and y2, reapectively as shown in fig. Assume that the bending moments are such that the t0p fiber is in compression for a positive moment. Then the section of the beam at the point of maximum bending stress must satisfy-the following conditions. In tOp fiber 1) As soon as the prestress has been established, the combination of stress in the concrete due to the beam weight, Wd, plus the stress due to prestress force, F1 must not exceed osmium. 023 $4522. 522 43? CS. 576 manned 55 10 the permissible concrete tensile stress, fof‘ for is equal to zero as minimum compression value. 2) After a period of time, the beam is acting under wd, the dead load and W1, live loads combination. The com- pressive stress in the concrete must not exceed the permissible concrete stress, fc. In the bottom fiber 1) As soon as the pre-stress has been established the stress in the concrete due to combined effects of the pre- stressed moment and the dead load moment must not exceed the allowable compressive stress in the concrete, fc. 2) After a period of time under the combined live and dead loads plus the initial prestress, the tensile stresses in the concrete must not exceed rte- Using the general fiber stress equation f : F/A.$ Mc/I now it is possible to find the fiber stresses due to prestress acting alone. f : F/A I MC/I M : Fie I = Are Then using column analogy method, in the tOp fiber £110 3 (F/A) - (Fyle)/I In the bottom fiber fbc = (F/A) + (Fyge/I) (F/A)(l - eyl/I) Tension (F/A)(l + ey2/I) Compression ll So, four conditions described can be expressed mathematically as below: In tOp fiber: 1) -(F1/a)(ey1/I‘2 - 1) + fat ‘5 rte 2) -r‘(F1/'A)(ey1/r2 - l) + fdt + fat ‘5 f0 Bottom fiber: 1) -(F1/A)(l 4 eye/r2) + fdb é fC 2) -n(F1/A)(l + eye/r2) 4 fdb + fab g rte Damn Subtracting 1 from 2, we get M a I/yl Zrtc + rc - (1 - nHFl/Awayl/n‘ - 17 An evaluation of the terms from various researches are n = .85 F/A = not larger than .5fc eyl/r2 : 2 usually and that gives I/yl Z Ml/(ftC - .9251}? Adding 3 to 4 M I l 2 /y2 ftc + fc — (1 - anFl/A)(l — ey2/r 7 Evaluating the terms I/y2 g Ml/(ftc . .775rc) and this gives larger section modulus. 1/0 = bd2/6 I/y2 = bd2/6 : M1/.775fc d =\/(m16/bx.775fc) = 2.78% In this equation: M1 is in inches; b is in inches; fC is in #/in2; the depth will then be in inches. 12 @239 In prestress concrete, bond is of importance only in cases where the prestress is transferred from the steel to the concrete through the bond. Where the reinforcement is embedded in the concrete under initial tensile stress, a very strong bond is produced as follows: Upon release of the force producing tension in the steel, the tension in the steel is reduced due to elastic shortening of the concrete and subsequent volume changes, such as shrinkage. As this happens, the wire reinforce- ment deve10ps a tendency to enlarge its diameter slightly. As may be readily seen, this action is similar in effect to that of a "forced fit," where considerable resistance to sliding is deve10ped in Spite of smooth contact surfaces. Thus, the bond deve10ped in prestressed concrete is greater than is deve10ped in reinforced concrete due to the "force fit" action. Rosov deve10ps an equation which he suggests may be used to check the bond stresses. It is: u = v x Pn(q-k)/Czod Where: u ; the unit bond stress psi V = Total shear at any section p a percentage of steel in the entire cross section n : ratio of ES/Ec a constant : (1/12) — (% - k)2 - (n - 1)p - (k - Q)2 l3 2% = Total perimeter of the steel d depth of a slab or beam (id/d) - a Shear and diagonal tension k In conventional reinforced concrete design, diagonal tension is of great importance, and to offset it, an intri- cate system of special reinforcement must be designed. Since the longitudinal stress is always compressive for a pre-stressed beam, and since the maximum unit shear stress does not occur at the same place in the beam as the maximum fiber stress, the diagonal compression will never be much in excess of the allowable. However, if shear exists, there will always be a diagonal tensile stress existing. The diagonal tension at any point on a beam, can be checked by using the equations: fd = v2/r1 f1 : MC/I From the equations above it is seen that the value for diagonal tension is much smaller than for a conventional reinforced beam where the diagonal tension is computed from v : 8V/7bd. Note: There has not been any attempt made for the reinforced concrete design theory, as everyone is familiar with the theory. 14 Design Procedures The time required for making a design is of importance Simplicity is also a major consideration.A structural engineer's work encompasses many kinds of design problems and the design of prestressed concrete is only one sub_ Ject among many.It is therefore important that the prin_ ciples of prestressed concrete design be made as easy as possible, to remember and use.Such considiration have influenced the choice and scope of the procedures ill_ ustrated in the design. ' The slab considered in the design examples is simply supported and has a 32ft. span length 46 ft. span width (ctnsisting of four IO ft.lanes plus 2ft 3 in.side walks and rails, to carry highway loadings H—20,H—IS and H-IO in current AASHO specifications. The individual slabs are rectangular shaped which is economical shape for prestressed and reinforced concrete.The most suitable and economical layout can not be selected from books nor computed solely by application of equations but a .nust largly based on experience and judgment. 15 DESIGN OF A HIGHWAY BRIDGE SLAB WITH REINFORCED CONCRETE 16 Specifications: Span Length: 32'00" Span Width: #6100" (consisting of four 10' lanes and 2'3" sidewalks and rails) Span Loading: ASSHO H-2O . HAO . HAS Allow Stresses: fs fc Impact 22000 psi for main reinforcing 16000 psi for stirrup .45 x 2500 e 1125 psi % 2 x 2500 = 50 without web reinforcing or Special anchorage. % 6 x 2500 = 150 with web reinforcing but without special end anchorage. % 5 x 2500 e 125 psi 10 : 50/(L-125) : % 30 but 25% will be sufficient, since concrete stress was somewhat on the safe side. fc 1125 fc + (rs/n) = 1125 + (22000/12) - .38 1 - (k/3) = .873 (fc/2)kJ = 186.5 17 Design No. 1 - For H-2O Loading 32! |¢ ‘3“ 115 f V ‘4’ Moment @ A 8 x 14 = 40X x = (8 x 14)/40 X = 2.8 X : the distance from A to the Resultant Force 44.5’ 3“" 44' t a B1 c r E,‘ m‘ZSK 7'2122‘015K Maximum Moment and shear_at critical sections Maximum Moment: M @ B (1n.6)(18.25) = 266.u50 ft - kip Moment per foot _ 26.645 ft - #/ft of width Reactions: M @ A (14.6)(32) + (28.6)(8) . 32 R2 R2 21.75k #0 + 21.75 = 18.25k R1 18 Maximum Shear 3:“ “4' %§- 183 I f" 2.: 36.5L ‘lute 3.5L Lane Loading for Maximum Shear Reactions: M @ B 32R1 e 32x32 + 8 x 18 = 0 32131 ; 1024 - 141+ 32R1 = 1168 Maximum shear : 36.5 kips Shear Diagram Check Shear V = V/bjd v a (3650)/(l2x.873x13.35) a 26.3 psi v = 26.3 psi (Allow 50 psi) Allow. (50 psi) 0.x. Stirrups are not required. 19 Maximum Live Load Moment: Live Load Moment + % 25 Impact 266u5 + .25(26645) = 33306 ft - #/ft. Required Depth (1 “/8786 = W33306x127/(1865x127 : 13.35" Rf Depth of 13%" Protective cover 2%" h {5 72 Use an overall depth of 16" «V II Reinforcement: , As = M/fsid a (33306x12)/22000x.873x13.35) T27" z (33306x6)/(11000x.873x13.35) (33306x6)/(9603x13.35) : 1.555 d'/ft. As : 1.555 d'/ft. Try %"— ¢ A : .20 d7“ Use 8%"— <(> A Spacing (.20xl2)/(l.60) : 1.5" 1.60 :Ei/ft. Use a spacing of 1%fl Check spacing = (2%) (bar diameter) = 111:" Spacing 0.K. Check bond stress: ‘1 = V/zni u a (3650)/(l2.6x.873x13.35) : 24.8 psi Bond 0.x. (Allow = 125 psi) 20 22 Temperature and shrinkage reinforcing Abs = .002(15.5)(12) = .360 ,g/rt. “Try 3/8" 4: A : .11 o" Use 3/8" 4D ’4 - 3/8" c.c. A .-. .440" s a (12x.ll)/(.360) a 3.66 Use Baa/’4" Spacing 23 Design No. 2 for AASHO - H—l5 Loading 24K *3! bl | k I .‘LS p..— t4' “fij M @ A 6xll+=30x x = (6x14)/30 X = 2.8 x = the distance from A to the resultant force MK x 4445' ' Lg' :11 ’ B k S:[::]t> a! R: ‘—' 43.63k r1cla=469511, Maximum Moment and Shear at Critical Sections Maximum Moment: Reactions (14.6)(24) + (28.6)(6) : 32R2 R2 = l6.3lk R1 : 30 - 16.33 : 13.69k1 Maximum Moment (14.6)(13.69) : 201,000 ft. - # Moment per foot e 201,000/10 = 20,100 ft - #/ft. 24 Maximum Shear: 24K , 6L ' as .l sail A, , E5 12, a 27.31 v. 32’ 12,-. 2.651 Reactions: M @ B (amuse) . (6)08) = 32R, 32R1 = 768 + 108 32R1 : 876 R1 = 27.37k = v Maximum Shear V = 2737 #/ft. Maximum Live Load Moment: Live Load Moment + %25 Impact 20,100 + (.25)(20,100) = 5025 = 25,125 ft. #/ft. MLlL : 25,125 ft.-#/ft. Required Depth. d =\/'('M'7fib = \/(25,125)(12)/(186.57(127 :2'11'82" _ ‘h——4L——___.1 Depth 12" Protective Cover 2%" . 42 Use an overall Depth 14%" 44%" _i 25 26 27 Reinforcement :1:- l - (M)/(fSJd) e (25,125xl2)/(22,000x.873xll.82) (25,l25x6)/(ll,000x.873xll.82) = 1.32 Eg/ft. 1.32 n"/ft. Tryé"-

A = .11 D" Use 3 - 3/8"<(>0.c. A = .3311754 s = (12 x .11)/( 288) = 4.6 ' Use 443/4" Spacing (v)/(de) e. (1825)/(12 x .873 x 9.48) 18.4 psi 33 DESIGN OF A HIGHWAY BRIDGE SLAB WITH PRE-STRESSED CONCRETE 34 Specifications: Live Span Length - 32'00" Span Width - 46'00" (consisting of four 10 ft. Lanes and 2'3" sidewalks and rails) Span Loading ASSHO H-20, H—15, and H-lO Allowable stress in the concrete ' Compression, f 1500 psi c, Compression ultimate 4500 psi Tension fct : 0 psi n = .85 (Loss in prestress with age) Allowable steel tension - 120,000 psi That requires a steel tensile strength of 200,000 psi Use a wire of .2 inch diameter for reinforcing Load ASSHO H-20 Impact = (50)/(L - 125) = 50/167 a 30% But %25 will be sufficient as we are on conservative side. 35 Design No. l - For H—20 Loading Maximum live load: 33,306 ft.-# Maximum v = 3650 #/ft. Computations: d : 2.78 /(M1)/(bfs) : 2.78 ((33306 x 12)/(12 x 1500) d . 2.78\/22.—2 : 2.78 x 4.71 a 13.1" Assume a section to satisfy I/Y2 = (M1)/(.775fc + rte) = (33306 x 12)/(1162.5) I/Y2 : 342 111.3 Check the depth 7. I/C e (bd2)/6 4+" T---—------- : (12 x 14 x 14)/6 = 28 X 14 . = 392 in3 4,, ~ Depth is O.K. The Dead Load Weight Weight of concrete 150#/ft3 wo = 150(14/12) e 175 #/ft2 Md e 1/8w12 = (175 x 12 x 32 x 32)/8 = 1536 x 175 Md = 268500 in #/ft Concrete stresses 3 = (6M)/(bd2) . (268500 x 6)/(12 x 14 x 14) : 686 #/in2 (33306 x 6 x 12)/(12 x 14 x 14) e 1020 fdu = fdb ful 3 f1b 36 Determination e and F1 P/A 3; (Pe/I)y rte z 0 e = r2/y r2 e. I/A I :— (1/12)bd3 = (l/i2)(12)(14)3 I = 196 x 14.. 2744 in4 A=l4xl2 =l68 in2 r2 a 2744/168 {£1.65 y = 6 e ==l6.3/7-=-2.33" For safety use an e = 3.25" Mtotal = ML.L.+ Md.L. rte =_. Fl/A -y(Fie - Mfg/(1).: 0 F1 2:. (Mt)/(I‘2/IY + 9) Thereiis a loss in the concrete due to creep and shrink- age. Therefore a 20"loss is estimated a reasonable reductioniin prestress for this problem. Then F1 becomes F, e. (Mt)/(,80)(r2/y -+e) Mt = 268500 (33306 x 12) = 668172 F, = 668172/( .80) (2.31 + 3.25) .=-. 150000454 a, = 150000 ,4! A " 37 Steel Area’ A ' Fl/Allowable steel stress 150000/120000 l.25"/Ft Try .2" (#5 American wire size) A a .0314 0" A No. of wires l.25/.O§i4 = 39.91“ Use 40 wires Spacing of wires 4 3 rows of 14 wires each /Ft Wire spacing 12/14 .85" Use a spacing bf l" apart Note: There is a reduction in the concrete area because“ of the wires.For practical purposes it has been omitted. 38 Check Concrete Stresses F1 = 150000 #/ft e : 3" 1) -(F1/A)(9Y1/I’2 " 1) 4* fdt 9. ft .0 -(150000/166.75)(3x6.98/16.4 — 1) + 686 = rt ~900(1.273 - 1) + 686 = rt -900(.273) + 686 = +441 : rt ft : 0 O.K. + compression 2) -n(F1/A)(ey1/r2 " 1) * fdt " f‘at 4; f“c -.85(150000/166.75)(.273) 4 686 + 1021 €- 1500 psi -.85(900)(.273) 4 686 4 1021 E‘ 1500 psi -208 . 686 4 1021 E- 1500 psi 1499 psi 4 1500 psi O.K. 3) -(F1/A)(l + ey2/r2) 4 tab s: re 1500 psi -900(1 + 3(7.02)/16.3) + 686 fc —2050 4 686 = 1350 r 1500 9.; O.K. 4) -n(Fl/A)(l + ey2/r2) + rdb + fab :1 rt rt : 0 -.85(900)(2.282) 4 686 4 1030 <1 0 -l742.5 . 1723 < 0 ~19.5 = 0 O.K. Concrete Stresses 0K Check Bond u“ = V x Pn(q-b)/c %d v e 3650 #/ft Aw/Ac : 1.25/168 -_- .00745 (1/3)d/14 : 10/14 = .714 .0 ll 39 N d = 12 ‘% a 3.14 x .2 x 38 c = (1/12) _ (3 - a)? — (n - 1)p - (k - qt)2 0 a (1/12) - 0 - 5(.0081)(.75 - .5) 0 2 .083 - .006090 C = .077 u = V x Pn(q-b)/C 206 u e 2737 x (.0081 x 6 x .75-12)/(.077 x 23.864) u = 71 #/in2 3 Allow. bond 80 71 1500 O.K. : 171.0 compression 1385 psi ‘SHL r —————————————————— 7: 4 ____d ‘2 I 44.6 47 EON. SE 38.: D 59;; A #ww a: 48 Design No. 37- For H-lO Loading Maximum Live Load M = 16645 ft-# Maximum V = 1825 #/ft. a) b) 0) Reg. depth d z 2.78‘/(M7bfc) = 2.78(/(16645 x 12)7(12 x 1500) d = 10.61" Assume a section to satisfy equation I/Y2 = Ml/(.775fc - rte) a (16645 x 12)/1162.5 : 143 in3 Check the depth I/C e bd2/6 : (12 x 112)/6 = 242 in3 ‘1 Section O.K. Dead Load Weight Weight of concrete : 150 #/ft3 “ wd = 150 x (11/12) = 137.5 #/ft. . Md : (1/8)wd12 e (1/8)(137.5)(32)2 : 176500 in-#/ft. Concrete stresses 8 = 6M/bd2 rdu : (176500 x 6)/12 x 112) : 176500/242 : 730 psi (16645 x 6 x 12)/(12 x 112) a 825 psi ful Determination of e and F1 F/A.- (Fe/I)y e 0 e : r2/y r2 : I/A : (1/12 x 12 x 113)/(12 x 11) r2 a 10.08" c = 22/12 : 1.831" Use an e = 2.25" rte : Fl/A -y(Ffi - Mt)/I : 0 F1 : (Mt/(Pe/y & 8) 49 Mtotai = 176.500 t 199742 = 376242 ft-# Fi = (376242)/(-80)(1.831 + 2.25) = 115200 # F1 I 115200 # Compute Wire Area (F1)/(Allow. Steel Stress) : 115200/120000 : .962 "/ft. Try .2" (#/5 American Size) A : .031413" No. of wires 2 .962/.0314 = 30.6" Use 32 wires Spacing of Wires 2 rows of 16 wires each/ft. Wire Spacing : 16/12 : 1.32" Use a Spacing of 1%" apart Stresses due to Loads (176500 x 5 5)/1331 = 728 7‘“ H.) O. Cf ll fdb = fdt 1 : (1/12)bd3 I = (1/12)(12)(11)3 I : 1331 in4 (16645 x 12 x 5.5)/1331 e 825 pa: fat fat = fab Check Concrete Stresses F1 : 115200# e = 2.25" 1) -(F1/'A)(ey1/r2 - 1) 4 rdt 5; ft ft 2 0 —(115200/l32)(2.25x5.5/10.08 - 1) 4 730 5- ft -(875 x .23) r 730 +204 — 730 = -526 compression O.K. 50 2) -q(Fl/A)(ey1/r2 - 1) 4 fat 4 rat 5; re -(.85)(115200/132)(2.25x5.5/10.8 — 1) 4 730 4 825 §~ r0 -(.85)(875)(.23o) 4 730 4 825 = 1383 s43 1383 > 1500 0.x. 3) -(F1/A)(1 - eye/r2) 4 rdb s r, -(115200/132)(1 - 2.25x5.5/10.8) 4 730 f0 875(2.23) + 730 -1950 4 730 z 1220 g; 1500 O.K. 4) —q(F1/A)(l 4 ey2/r2) + fdb 4 fab g. ft .85(115200/132)(1 - 2.25x5.5/10.8) 4 825 4 730 .§2 ft -111 0 0.x. Check Bond u : V x np(q-b)/C od v = 1825 # p = .962/132 z .0072 q = 8/11 : .728 k e 5.5/11 2 .5 1% e 3.14 x .2 x 32 = 20.01 c = .083 - 6(.0072(.728 - .5) c = .083 - .00985 C : .07318 u = 1825 (6 x .0072 x .728-12)/(.0731 x 20.01 x 11) : 52 psi u : 52 psi 52 80 O.K. Allow min. bond = .04f'c e 2000 x .04 : 80 psi Diagonal Tension rd 2 v2/f1 f1 : 1500 psi V/bjd = VQ/Ib Q = 6 x 5.5 x 5.5 a 181.5 in3 (1825 x 181 5)/(1331 x 12) = 20.6 #/in2 V V 51 f1 = Mc/I : Fee/I = (115200 x 2.25 x 5.5)/1331 r1 = 1071 Safety factor 3 f1 = 537 rd : (20.6)3/537 = .795 psi Allow. 24 O.K. Section at the End (2/3)d1 = 7.5 dl : 11.2" min. Use d1 : 12" 'L5” d. . . .. . 12* A z 1.44 1112 P2 : 12 e - 2" r, 2 (Fl/AMeyl/r2 - 1) = (115200/144)(2x6/12 - 1) - (115200/144)(0) = 0 O.K. fb - -(ll5200/144)(2x6/12 4 1) 2 -(115200/144)(2) : ~1600 1600 1500 Not O.K. Try a section 13" A = 169 in2 f.” 0‘ I II II II 52 (115200/169)(2x6.5/13 - 1) 115200/169 x .081 55.2 psi O.K. _ (115200/169)(2x6.5/13 4 1) 1420 psi 1420 ;> 1500 O.K. 53 .300 “dawn: Hui/mg; we. 54 CONCLUSION Comparative Table For The Designed Slabs: Design Max.L.L.Mom. vMax Depth SteeleArea Ft-lb lb in' in Nozl Reinforced 33 306 3550 16 1.555 Prestressed‘ 33 306 3650 14 1.250 No:II Reinforced 25125 2737 14.5 13320 Prestressed 25125 2737 12 1.167 No:III Reinforced 16 645 1825 12 1.095 Prestressed 16 645 1825 11 0.962 As can be seen from the above table the major difference between the two slabs are.in materials. In prestressed concrete economical use of both materials {Concrete and Steel" have been Obtained since2the total crosSectional area of c6ncrete is effective in comression. That is prestressed concrete slab, requires less concrete , less steel atdthe;site. This is an important factor in that reduction of materials is always a desirable point. Purpose of all design: is to construct at the lbwest cost with the use of minimum amount of materials. 55. After careful consideration of the above facts and data presented I came to the. conclusion that the prestressed con- crete slab would be most economical and best to use for this designs . 1. 2. 5. 6. 7. 9. __Bibliographx_. Sutherland and Reese; "Reinforced Concrete" Portland Cement Assoc.;"Modern.Development; in Reinforced Concrete? No:25 “ ' ' Schorer,R., “Prestressed Concrete, Design Principles and Reinforcing Unitsf; American Concrete.InstituteeProciding v.14,p.493-528,June 1943 Abeles,P.W., ”Beams in Prestressed Reinforced Concrete I II and III; CSncrete and Constructional Engineering,v.4l, pp.l47,l9l,225:l946. Anon.,SA Cemparison of Partially and Fully Prestressed Concrete Beams”; Concrete and Constructional Engineering, v.42,p.ll-l5, danuary 1947. Anon.,”Prestressed Reinforcement of Fine‘Wires Gives Springato Precast Beams",Gbncrete, v.47,p.8,November, 1939. Abeles,P.W., SSaving Reinforcement by Prestressing",Con— ‘ cretezand ConStructional Engineering,v.35,p.328-333, Gueritte,T.J., ”Recent Developments of Prestressed Cone crete Construction‘with.Resulting Economy in the Usesof ,Steel”; Structural Engineer,v.18,p.626,July 1940. :Magnei,e., "Prestressed Concrete9,London 1948. “ A 10. ll. 12. 13. 14. 15. 16. 17.. 18. 19. Magnel,G., ’Creep of Stoel'and.Concrete in Relation to PrestressedAConcreteF; American Concrete Institute Procidings, v.44,p.485-5oo, February 1948. Mautner,K.W},§Rebutta1 Saving Reinforcement by Prestress- ing” ; Concrete and Constructional Engineering, v.36,p.73- 95,‘Februery 1941.. Evans,Rsfl.,PRelative Merits owaire and Bar Reinforcement in Prestressed Concrete Beams: Institution of Civil engi- neers of London Journal,v.18,§.315-329,February 1942. Evans,R.H. and Vilson,G.,”Influence of Prestressing Rein- forced Concrete Beams on Iheir Resistance to Shear", Structural Engineer, v.20,p.lO9-122, August 1942. A M9Ilmoyle,R.H.,fPrestressed Concrete Bridge Beams Being Tested in England9,Railway Age,v.123,p.478, Sept.l947. Schorer,H.,¢Analysis and Design of Elementary Prestressed Concrete Members9,American Concrete Institute Procidings v.93,p.49-87, Sentember 1946. Abeles,P.W.,”Formulae for the Design of.Prestressed con- crete Beams§:Concrete and Constructional Engineering,v. 43,p.115-12§ ,April 1948. Abeles,P.W},”Examples of the Design of Prestressed Cone crete Beams'IConcrete and Constructional Engineering, v. 43 ,p.149-1156,May 1948. Coff,L.,’Prestressed Concrete ,’a New Frontier?, Engi- neering Revs Record,v.l43,p.187-190,§eptember i949, Peabody,”Reinforced Cencrete Design? . Q A ROOM use ONLY - _ __ . .1- v" ‘.'_ v". '1"... I" {A l. . , " <1! -. .8484, ; 'x, '1‘... "WT-1:94;} ‘34 A.'+;--'+'b.:y;,’.~;~‘za“-23,”. a. . ""894’94 f ' ' . ‘ V. o ' ‘e We. ". ‘J ' Jr 4‘ Q .. 413.1 2‘. e -. }' Q O i 0‘ MVERSWYIJBRARES III III” I 3056 0829 TATE ""Il’il'filifli'uml 293 In 0