:_:_E_:__:__.:_:2:23 \ul v" ' ,t. 1- a :- ov/mz‘fi" ,-,;;. t O - n . . ‘ - Y" .U ’4 r s u, ate -i fit 1 certify that the ' .nesis entitled . g , . , f I ( 2 h :ison of Highway Bridge _ 4y to One-Hundred Foot Spans. . “ presented by ntilal Ambalal Patel . has been accepted towards fulfillment . . ‘ of the requirements for Civil Engine ring 1.1. S ‘ degree in ______ . . , . .Il - Major professor ‘ _ ‘. _' a Carl L. dhermer , ~‘. 3; Hi? ‘ y I "O ‘- .‘Cn‘ -** *..:'tn ~A b «Er—u- ECONOMIC COMPARISON OF HIGHWAY BRIDGE TYPES FOR FIFTY TO ONE-HUNDRED FOOT SPANS By Shantilal Ambalal_Patel W 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 SC IENCE Department of Civil Engineering 1954 THESRS Acknowledgments I. II. III. IV. V} Anpendix A. Comnosite Construction for 50' ~ 70' - 80' - 90' - lOO' Snans ........ Introduction ............ TEBLE OF‘CONTENTS FactorSDictatingTyne Selection O......OOOOOOOOOOOOOOOCO0...... A. Stream.Behavior ... B. Requirements Of IJaVigatiOn ...... ..... ......OOOOOOOOOOO C. Traffic Considerations .......... D. ArChitectllral Features ......OOOOIOOOOOOOOOOO0.0.0.... TyT)e selection for 50' to 100' spans 00.00....0.00.00.00.0000... A. comnOSite Construction ......OOOOOOOOOOOOOOOC......OOOOOOOOO 7 1. General ...0.00.00.00.00.00..0.0......OOOOOOOOOOOOOOOOOO 7 2. Floor Slab 5.00mp081teBeam.....OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO B. Prestressed Constmction ......OOOOOOOOOOOOOOOOO00.0.0.0... 1. General ... 2. Floor Slab 3. Prestressed Girder ....O.....OOOOOOOOOO......OOOOOOO... Results 00.0.0.0...O....O........OOOOOOOOOOOOOOOOOO0.00........ Conclusion ....... Annendix B. Prestressed Concrete Construction for 50' - 70' - 80' -90' -100, spans oooooo ooooooo o ccccc 000000000000 Bibliograohy ............. 3312 7 .0...0......0.0.............OOOOOOOOOOOOOOCO 12 15 28 28 51 33 44 47 49 59 69 ACIQ‘IOWLEDGLENI‘S The author wishes to express his sincere thanks to Dr; Carl L. Shermer for his kind guidance and valuable help in preparation of this thesis. He is also greatly indebted to Dr. Richard H. J. Pian for his kind suggestions and assistance. Grateful acknowledgment is also due to.Mr. Rydland of Michigan State Highway Department for his practical advice regarding bridge design. The writer deeply appreciates the unit costs for composite construction supplied by thelflichigan State Department, Lansing. He extends his sincere thanks to L. Goff; Consulting Engineer, specializing in prestressed concrete, New York, for supplying the unit costs for prestressed construction. He also appreciates very much the unit costs for prestressed construction furnished by F. Hurlbut Company, Green Bay. Last,but not the least, the author is highly obliged to Dr. Edward A. Brand, Associate Professor, General Business, in putting this thesis in the specific form. I. INTRODUCTION Highway Bridge Engineering can be divided as follows: A. Bridge location B. Preliminary investigational work and stream study C. Economic type selection. D. Detailed design E. Construction 32 Maintenance and operation In general it may be said that many of the highway bridges span from 50' to 100'. Therefore, itis the purpose of this thesis to make an economic study of highway bridge types for these spans including actual cost comparison. It is needless to say that type selection is the most difficult but the most important feature of Bridge Engineering. Yet, the engi- neers in general have not quite appreciated the importance of correct type selection. As a result, millions of dollars have been wasted through improper type adaption due to unwarranted first costs or unneces- sary maintenance expenses. Correct type selection is, therefore, the very corner stone of economy. .A failure to evaluate correctly the factors governing this problem may frequently result in waste many times greater than any saving anticipated from refinements in stress analysis and design. ... I J.“ F I It”?! (RI 9. II. FACTORS DICTATING TYPE SELECTION It is necessary to discuss the general factors that govern the type selection for any span before discussing fifty to one-hundred—foot spans. The Question of economy in the first cost, maintenance and renewal is naturally a major controlling consideration. It should be kept in mind that there are some factors other than economic considera- tions which Operate to dictate type selection. These may be grouped as follows: A. Stream behavior B. Reouirements of navigation C. Traffic considerations. D. Architectural features This is not the place for an extended discussion of these, however, these will be introduced briefly. A. Stream Behavior Stream behavior in this thesis means the peculiar character- istics of the waterway during periods of high waters as regards erosion of bed and banks, lateral shifting of channels, carriage of drift, ice and debris, etc. These characteristics have direct influence on (1) main channel structure, (2) structural approaches, (3) approach embankment. We know that Q =‘VA where Q = discharge of stream in ft. sec. through the main channel. A - Waterway area of main channel ftg. v = Velocity of flow in ft./sec. Construction of any structure across the strewn tends to obstruct the flow of water. This, naturally,affects the velocity of flow and causes erosion or deposition of bed. As a result this characteristic puts limit on minimum spacing of piers. Meandering of stream in flood plains many times make it neces- sary to carry main channel construction clear across the flood plain, where otherwise, short span approach construction would be entirely adequate and much more economical. In such cases, stabi- lization of channel by means of river training works may sometimes be economical against long span construction over the entire flood plain. This automatically dictates the structural approaches as well as the approach embankment. The necessity to guard against the drift or ice dictates the vertical clearance, although extreme flood elevation alone may at times be the limiting factor. These factors often force the elhni- nation of deck truss construction even though it is the most desirable. Shallow truss construction, enough to provide required clear- ance, may sometimes be more economical than deck truss of ordinary depth which requires raising the general grade of the approaches. Requirements of Navigation The horizontal and vertical clearances required for the main channel span to fulfill navigation needs certainly limit the type selection. Generally moveable spans are used for this purpose, of course, the type of design for the moving leaves is dictated by water traffic. In many of the streams in this country even the flanking spans are held to a certain established minimum as regards these clearances by the United States War Departnent. Thus, they control the spacing of piers regardless of other con- siderations. Traffic Considerations In order to allow adequate sight distance at sharp curved approaches structure such as through steel truss construction should be avoided. The ultimate purpose of bridge structure is to facilitate traffic movement. Hence thorough consideration should be given to (l) the direction of traffic movements over the span, (2) provision for the separation of slow and fast moving traffic. Deck construction is much superior to any other type from a traffic standpoint as it satisfies all the above requirements. If the clearance requirements eliminate this type from consideration on preposed grade line, then, it becomes necessary to investigate the feasibility and cost of a new grade line modified so as to per- mit deck construction and also the feasibility of special shallow truss design. Architectural Features In certain cases architectural requirements play an important role in the selection of Bridge Structures. The following can be the order from the architectural point of view. 1» Masonry arch construction 2. Reinforced concrete deck construction 3. Deck truss or plate girder construction with a concrete deck. 4. Through truss or girder construction. If the align- ment is such that the side elevation of structure is plainly visible from the approaching highway, more attention should be given to a type which gives a pleasing side elevation than if only the roadway is ordinarily visible. Natural scenic setting will also influence the type selection. Furthermore, the loca- tion of the bridge in reference to parks, pleasure resorts, etc. will influence the type selection. In general, all such factors operate to place certain limits on type selection. These limits are generally in regard to the spacing of piers, type of approach construction, choice between deck and through construction, grades, clearances, roadway width, etc. In final analysis when the relative economy of the different types is determined, the unobstructed roadway should be given a great deal of consideration and should be chosen unless cost con- siderations are prohibitive. III. TYPE SELESTION FOR 50' to 100' SPANS Assuming all the factors that dictate the type selection are favor- able to the use of a deck structure, then the present trend is to the type of construction using steel beams with concrete floors for these span lengths. Recent developments to provide mechanical shear developers have caused the discarding of the conventional steel construction. As a result state highway departments are designing composite steel bridges. On the other hand, for 50' to 100' spans the conventional reinforced concrete construction cannot be used owing to its limitations in regard to span length. But recent development has brought about the use of pre- stressed concrete construction. This has increased the previous span limit. Hence, the prestressed concrete bridges can be built for these spans. How far this type of construction can compete with the present composite construction can now be found out by actual cost comparison. However, it is assumed that the subéstructure construction is the same in both the cases. Hence, it is now possible to determine cost dif- ferences for these two types of deck constructions for the 50' to 100' spans. Unit costs for the composite construction were supplied to the author by the Michigan State Highway Department, Lansing,ldichigan. Unit costs of the prestressed construction were supplied by the special- ized firms. The order of the following discussion is: A. Composite construction B. Prestressed concrete construction A. COMPOSITE CONSTRUCTION 1. GENERAL a. Introduction During the last twenty years composite construction has replaced the conventional steel construction in the field of highway bridges. This is because, in the conventional type of steel beam bridge with a reinforced concrete deck, each beam carries the entire load transmitted to it by the concrete slab. While in composite construction, adequate provision of shear con- nection between the beam and the slab has made the section compo- site and hence acts similar to the reinforced concrete T-beams. The main advantages of composite construction over the con- ventional type of steel beam and slab construction are: (1). Greater economy (ii). Shallower construction (iii). Greater stiffness (iv). Greater live load factor of safety b. Composite Beam The composite beam consists of three essential parts: (1). Steel section: this may be a rolled beam or a riveted or welded girder. Composite construction is more economical when the bottom flange of the steel section is larger than the top flange. This (ii). (iii). can be accomplished in the case of a standard rolled beam by welding a cover plate at the bottom. Concrete slab: reinforced concrete slab is designed as in conventional construction. The slab may rest on the steel beam or on concrete haunch increasing the carrying capacity of the composite section due to the increased depth. Shear connectors: to provide adequate bond between the concrete slab and the steel beam, mechanical shear connectors are provided. Among the many, the spirals are the most economical and adaptable. These are normally welded to the top flange of the steel beam and embedded in the concrete to transmit the longi- tudinal shear and to prevent the movement between the slab and the beam. c. methods of Erection Two different methods of erection are in practice: (1). (11). No temporary supports are used under the steel beams during the pouring and setting of concrete slab. So steel section carries the dead load while the composite section carries the live load plus any dead load applied after the slab has set e.g. wearing surface. Effective temporary supports are used under the steel beams during the pouring and setting of the concrete slab, hence the composite section carries both the dead and live loads. However, practice has proved the first method as more economi- cal than the second, though the steel section required for the second method is smaller than that for the first method. This is because the cost of the intermediate supports frequently exceeds the saving in the steel. In the following designs, all of the above main factors that bring about economy are taken into consideration. The method of design of composite beams is based on the trans- formed section method used in the design of reinforced concrete beams. For the trial sections, the properties are used from the tables given in Reference No. 2. Data The general layout of the cross-section is as shown in the following Figure l. The roadway is 28' wide and is flanked by two 2-ft. wide curb walks. The stringer spacing is kept 6 ft. c. to c. for all the spans. The slab is kept the same for all the spans. J_,:gfj _ -rll- ____ ~,.‘ 1.-..“ -:. .‘\ . ¢ -’)-(_’J . I"_ ,.- t . ’ [Cf-‘3'.” ! -- T .-A" (J _ .‘ ...: _ ., —, ‘.L______ o ' r p - 2‘-."' ' .- 3 _ T- _._ l 4— - - +- ‘ l --l” in a 9' l o , w—v lV—‘WV CROSS-SECTION OF DECK Fig-l 10 Live load is HBO-44 Future wearing surface is assumed to be 20 lbs. per square foot. Load and design requirements adhere in general to Standard Specifications for Highway Bridges, sixth edition, 1953 adopted by AASHO. e. Notations t" II I' = S's = 38: MDL MLL Mws Fsdl F811: sts= VLL sz Distance from compression surface of slab to neutral axis in inches. Span length in feet. Cross-sectional area of steel section. Moment of inertia of steel section about its NA. Moment of inertia of composite section about its NA. Tensile section modulus of the steel section. Tensile section of modulus of the composite section. Ratio of modulus of elasticity of steel to that of concrete (Es/EC). Bending moment due to dead load. Bending moment due to live load. Bending moment due to wearing surface. Steel stress due to MDL. (Kai). Steel stress due to MIL (Ksi). Steel stress due to Mws (Ksi). Vertical shear due to dead load. Vertical shear due to live load. Vertical shear due to wearing surface. Horizontal shear per unit length. Pitch of spirals. ll Apl - Cross-sectional area of plate. Qpl : Static moment of cover plate about NA of composite section with n - lO. Q'pl: Static moment of cover plate about NA of steel section. Q”pl- Static moment of cover plate about NA of composite section with n = 30. Q = Static moment of the effective concrete area about NA of composite section. Fw = Shearing force transmitted per pitch.(Kips). Fsa Allowable steel stress a 18 (Kai). 12 2. FLOOR SLAB a. Design Distance between stringers c. to c. = 6 ft. Bending Moment shall be calculated by the methods based on the Westergaard's theory of the distribution of loads and design of concrete section. According to AASHO specifications, Art. 3.3.2. (0) case A, where main reinforcement is perpendicular to traffic and spans are 2-7 ft. distribution of wheel loads: E .- O.ds+ 2.5 where s = effective span length which is the distance be- tween edges of flanges plus 1/2 of the stringer flange width for slabs supported on stringers. .AASHO specifications. Art. 3.3.2. (a) Assuming 9' wide flange, 3 : 6' - O” - 9” + 4.5" = 5.62' E = 0.68 + 2.5 = 0.6 x 5.62 + 2.5 = 3.57 + 2.5 : 5.87' Bending moment for continuous snans: M = t 0.2 x‘Pl x s where P1 = Load on one wheel of 3 single axle. 1‘1 12,000 lbs. for H20 - 44 loading. 3: u t (0.2 X 12,000 x 5.62) 5.87 I 1000 i 2034 k. ft. 15 Impact factor 50 L +‘i55 (L = s a 5.52') 0‘ . 5O - 12 5.62 38.4» But max. allowed is 30/o Design.M = t 5.04 k. ft. for 20,000/10/1200, and M = 3.04 k. ft. d = 4'. Minimum slab thickness : 4 + 1.5 = 5.5” Use 7" thick slab. Hence, d = 7 - 1.5 - 0.57 z 5.2". Reinforcement: = As . M 3.04 x12 fsjd 20 x .867 x 5.2 0.405 sq. in./ft. Provide 5/8" ¢ Bars 0 7” c. to c. at bottom as well as at the top. Distribution Reinforcement: reinforcement shall be placed in the bottom of all slabs normal to main steel to provide for lateral dis- tribution of all loads. The amount shall be the percentage of the main reinforcing steel required for positive moment as given by the following formula: % = 100st = loo/5.625 = 42.2% Area of steel regd. = .422 x .402 = 0.17 Provide 1/2" ¢ Bars<0 10" centers. b. Quantity: concrete/sq. ft. =._Z x l = 0.583 cu. ft. 12 c. Cost: 14 Positive steel/sq. ft. 1.04:3 I 12 = 1.79 lbs. 7 Negative steel/sq. ft. 1.79 lbs. Distribution steel/sq. ft. = .668 x_l§ : 0.8 lbs. 10 Per sq. ft. of slab: concrete 0 $48/cu. yd. : 0.583 x 48 = $1.04 27 steel reinf. e $0.1o/1b. = (1.79 - 1.79 - 0.8) x .10 3.38/.10 = 33.8 cents Cost of slab per sq. ft. 3 $1.378 15 3. COMPOSITS BEAM a. Design The following example illustrates the detailed design of a typical composite beam. The snan is 60' as simply supported. Loading H20 - 44 Stringer spacing 6' Slab thickness 7" Wearing surface 20 lbs./sq. ft. Live Load: (Lane load governs) Distributed load/ft. of lane - 640 lbs. Impact factor = 50_“_ 27% 60 + 125 Impact load/ft. of lane - 174 lbs. Total load/ft. of lane 814 lbs. Per stringer, load/ft. =__§ x 814 lbs. : 488 lbs. 10 i.e. WLL : .488 k Concentrated load for moment/lane : 18 k Impact load : 4.86 k Effective concentrated load for moment/lane : 22.86 k Per stringer concentrated load P : 0.6/22.86 Concentrated load for shear/lane : 26 k Impact load : 7 k Effective concentrated load for shear/lane : 33 k 0.6 x 33 Per stringer concentrated load 19.8 k 16 .488 x egg + 13.7 x 60 8 4 Live Load Moment MLL 423 k ft. Dead load carried by steel section alone: Wt. of slab =._Z x 150 x 6 2 525 lbs. 12 Steel section : 120 lbs. Diaphrams, spirals, welds, = 20 lbs. 1.3. pr 665 lbs./ft. .665 x 603 = 299 k. ft. 8 Dead load Moment MDL Dead load carried by Composite Section (n = 30): for the loads which are superimposed after the concrete has set, stresses shall be computed with n a 30 for superimposed deal load and with n I 10 for live load. This higher value of n is supposed to take care of the com- bined effect of dead load and plastic flow stresses in the concrete and steel. Wt. wearing surface = '20 I 6 120 lbs./ft. 4120 x 602 8 Wearing Surfaceluoment .MWS 54 k. ft. Trial Design: Wherever plastic flow is considered in the design of composite beams, its effect may be neglected in the trial design because of its small influence on the required section. Therefore, loads producing plastic flow are grouped with the live load in the trial design procedure. 17 Steel section modulus required for dead load 3 S'sdl a MDL x 12 s 299 x 12 . 200 in3 fs 18 Steel section modulus required for WLL + Wss = 8811 - (423 + 54)_ x 12 e 477 x 12 18 318 in3 Composite Design with.Cover Plate: To select the required composite section, try values from Trial Design Table V of Reference No. 2 for the section 27wf. @ 94. Total Steel area reqd. : As = S'sdl + 8811 S's7as Ss/As = 200 + 318 8.78 13 x 0.978 22.8 + 25.0 = 47.8 sq. in. Now 27 wf. O 94 give steel area : 26.65 sq. in. Therefore, AS - 27.65 2 Area of Plate = Apl 47.8 - 27.65 = 10.75 SQ. in. 2 Use a slightly larger cover plate than determined as above, since it was assumed in the trial design that the load carried by the composite section with n = 30 was carried by the section with n = 10. Therefore use a 27 win 0.94 with a 12" x 15/15" (Apl = 11.25 sq. in.) cover plate at bottom. 18 Properties of section without cover plate (at supports). *‘ "'1 I A ‘ l \i .1 1 L 4- _ __ CL _ «i! _ “3, l a: N ~_—‘Y - “in ,; («0) : n. " (p, a A/) ”"‘* 1 A 3‘ 2% M , , . h e ”r g ml ‘9 : rs . \l -1.L__ .LiLll_ .lL F/y a Steel Section: Area _ 27.65 sq. in. I'o 3264 in4. ll 8'80 243 1n3. Composite Section (n = 10) Area: A1 = 7.2I7 a 50.4: A2 = steel N M Q 0 CD 0‘ Total.area 78.05 sq. in. Statical Moment: A2 x 20.45 = 565 Total = 742 in3. kd : 742/78.05 9.52” from top of slab. Moment of Inertia: __l x 7.2 x 75 12 50.4 x (6.02)2 WF. 27 27.65 x (10.93)? Total = Io : Modulus of Section = 205 1,870 3,267 3,220 8,662 in4 Sso : 8,662/24.38 a 355 in3 Composite Section (n = 30) Area: 2.4X7 = 16.8 WF 27 : 27.65 Total area . StaticallMoment: 44.45 SQ. in. 16.8 x 3.5 = 59 27.65 x 20.45 = 565 Total : 624 in3 kd : 624 x 44.45 : 14" from top of slab. Moment of Inertia: ._; x 2.4 x 73 : 68.4 12 16.8 x (11.5)2 : 2,220.0 WP 27 : 3,267.0 27.65 x (6.45)2 = 1,150.0 Total = I'o : 5,705.4 in4 Modulus of Section = 8'0 = 6705.4/19.9 = 336 in3 19 Properties of Section with cover plate (at mid-span) __.___ __ ———_— ._____—_—__ ———._- ..5 ..-fi '-—+ C TM ._7/ .11____ . b V {3‘ N _ _flf i A J . A 5255/ T is) F79.3 Change in NA = plate to NA of section without plate) / (Equivalent steel area of entire section). Steel Section: 20 Change in Neutral Axis due to adding a cover plate can be determined by the follow- ing formula: (Area of plate) x (Distance from c.g. of NA = 11.25 x 13.92 : 4.0” 38.9 Moment of inertia: 27 WE section = 3,267 2765 x 43 = 432 11.25 x 9.922 . _;Li§g Total 3 1' = 4,819 in4 Modulus of Section = S's = 4819/10.39 = 472 in:5 Composite Section (n = 10) 78" _gfi NA (By adding slab to L IE steel section ) = —*‘—_~ N /v . “12 50.4 x 20.96 . 11.8- a n "’0 1. ,¢ ,( 893 6 $1 \ :1 1 ‘i A/ ,f,f—7L"” ” g 21 Moment of Inertia __1 x 7.2 x 73 = 205 12 (50.4) x (9.16)2 = 4,250 Steel 1' = 4,863 58 9 x (ll 8)2 - 5,420 O O .- . ' Total - I 2 14,718 in4 Modulus of Section 33 = 14718 = 663 in3 22.19 Composite Section (n = 30) NA (By adding slab to steel section) 16.8 x 20.96 = 6.3” 55.7 781/ 7" -___. “i7 Q V s ; h: a V ‘1 1A /7 0 ‘ 4 3 STfl ~m ‘ ”V 57.9/ ‘ 5 \J‘ « 1 __s___ 1 [391.5 Motion of Inertia: 3 .41 12 16.8 x (14.58)3 = 5,480.0 Steel I = 4,863.0 58.9 x (6.5)2 = 1,540.0 Total = I" = 9,951.4 in4 Modulus of Section : S"s 9951.4 = 585 in:3 16.99 22 Grouping together: Section. Moment of Inertia in4 fig_ Sectioanodulii Steel 1' a 4,819 24.46” 3'8 3 472 in3 10.59" Composite I - 14,718 17.86" Se 2 665 153 (I1 = 10) 22.19" Comoosite I" : 9,951.4 17.86" S"s = 585 in3 16.99" Unit stresses at center: Bottom of steel Ton of steel FSDL = 299 x 10.59 x 12 = 7.85 k/in2 15.9 4819 FSLL = 425 x 12 x 22.19 = 7.75 k/ing 2.0 14718 sts : 54 x 12 x 16.99 a 1.11 kginz 0.77 9951.4 2 Total = 16.71 k/in 16.67 k/in2 Top of Concrete: Due to L.L. 452 x 12 x 12.66 a 0.456 k/in2 10 x 14718 Due to W.s. 54 x 12 x 17.86 = .059 k/in2 50 x 9951.4 Total _ 0.575 k/inz All stresses are within the safe allowable stresses. Hence, the section is safe. Length of Cover Plate: 2 L' Lvrit§99 88 60v 1 — 555 = 60\/ (l - .555) 665 40.9' 23 Use 42' cover plate. End weld on cover plate: The and force in cover plate - Apl x Fsa x 380 Ss 11.25 x 18 x 555/665 113.5 kips. Required length of 5/16" weld to transmit this end force only : 115:5, a 50 in. 2.21 Welds along sides of cover plate: To find horizontal shear per inch at the and of plate: At 9' from left support 0.665 x 21 - 14 k. VDL 0.665 (30 - 9) 'VLL = 0.488 x.§l'x.§l + 19.8 x 51 = 10.6 + 16.8 = 27.4 k. 60 2 60 0.120 (50 - 9) - 0.120 x 21 : 2.52 k. 5, Then horizontal shear per inch can be given as: :11 ll 1' I 11" vpL 9'51 . 'VLL Qpl + 'vws 9'51 14L11.25 x 9.92)+ 27.4(11.25 x 21.7?) + 2.52(11.25 x 16.52) 4819 14718 9951.4 0.324 + 0.456 + 0.047 g 0.827 k/in or 0.415 k/in per side. Try 1/4" intenmittent welds, length of each being a 1 1/2' Then strength = 1.77 x 1.5 = 2.65 k. Therefore: spacing s 2.65 = 6.5” 0.413 24 By AWS specifications the minimum weld length : 1-1/2" and the maximum.clear spacing : Joined = 14 x 11/16 = Maximum center to center spacing could be increased to Spiral Shear Connectors: 965" 14 x thickness of thinner part Spacing = 11.15". Therefore, the 11', near the center of the span. Use 5/8" 0 bars with 4-1/2” mean coil diameter. This gives 7” - (4-1/2 + At end supports: 'V(DL h WS) - VLL Spiral pitch, S Fw I = M VQ (0.665 + 0.12) 0.488 I 30 + 19.8 5/8) = 1-7/8” cover over spirals. x 30 = 23.5 k. = 34.4 k. Total shear 57.9 k. 11.04 x 8662 57.9(50.4)(6.02) At 10 ft. from the support: V(DL + WS) ... 'VLL 0.785 (30 "' 10) = 15.7 k. 0.488 x,§g,x.§g + 19.8 X'ég : 26.7 k. 60 2 60 Total shear : 42.4 k. As the cover plate exists in this zone, the properties of the section with cover plate will be used. 'Spiral pitch, 3 = 11.04 x 14718 a 8.52”, say 8" 42.4 x 50.5 X 9.16 25 At 20 ft. from support: V(DL + WS) = 0.785 x (30 - 20) = 7.85 k. VLL 0.488 x_4g,x.49 + 19.8 x.4g = 19.7 k. 60 2 60 Total shear = 27.55 k. Spiral pitch, 8 - 11.04 x 14718 27.55 (50.4)(9.16) = 12:08" say 12-1/2" Spiral £3323. Spiral length Weight so 5-1/2" 10' - 9” 29.8 lbs. 810 8” 10' - 9" 21.9 lbs. 320 12-1/2" 10' - 9” giggglbs. Total = 68.9 lbs. Average weight of spiral per foot of beam = 68.9 = 23 lbs. 30 Live load deflection: The theoretical live load deflection is computed by using n : 8 and by considering the change in section due to adding a cover plate. However, sufficiently close results are obtained by using n = 10 and the properties of the section at mid-span only. A ;_- 22.5 51.4 + 56 P13 EI El : 22.5 x .488 x 604 + 56 x 15.7 x 603 29,000 x 14,718 29,000 x 14,718 = .332 + .25 = 0.582111. 9 = 0.582 = 1 L 60 x12 1,250 26 This is safe because.maximum.deflection is allowed up to l 800 of span. Factor of Safety: Factor of safety is given for canposite construction based on the bottom flange stresses, which governs the design. Minimum yield point stress 33 ksi Dead load stress 8.96 ksi Therefore, stress available for LL _ 24.04 ksi Factor of Safety LL stress I ’01 d ,H (D 1m 5m 1‘9 15 :1:- 1m ‘6 +_, (D fi 0 '1 is C“ 775 Diaphramn: The diaphragms are generally tentatively provided for the better distribution of the live load to the various girders. They also make the structure more rigid to withstand unknown forces such as traction or sudden braking. Hence the diaphragm shall consist of two Ls 3"x3"x 3/8" with a 3/8" plate. The depth of the diaphragm shall be such as to provide 4" clearance from the top flange of the section and 3" to 4" from the bottom flange. The maximum spacing shall not exceed 20 ft. Hence provide two diaphragms, each having two Ls 3“x3"x 3/8" with l9"x 3/8" plate. b. Quantity: (1) Structural steel: 6 Nos. 27” WEE @ 94 for 60 ft. = 94x6x60 = 33,840.0 lbs. 6 Nos. l2"x 15/16" plate 42' long a 38.3x6x42 s 9,651.6 lbs. 4L3 3"x3"x 3/8” - 30' long : 7.2X4I30 864.0 lbs. 2 Nos. l9”x 3/8" plate - 30' long = 24.2x2x30 l,452.0 lbs. Total 45,807.6 lbs. 27 (ii) Spiral shear developers for 6 beams 60' long 2.3 x 6 x 60 _ 828 lbs. 0. Cost: (1) Structural steel (fabrication, erection, painting) 0 $0.155 per 1b. = 45,807.6 x .155 . 56,184.05 (ii) Spiral shear developers 2 $0.50 per lb. _ 828 x 0.50 a $414.00 (iii) 7" slab 28" x 60' a $1.578 per sq. ft. _ $2,520.00 Therefore, total cost of deck 3 $8,918.03 B. PRESTRESSED CUMRETE CONSTRUCTION 1. GENERAL 8. Introduction Conventional reinforced concrete enjoys an enviable position and can be used successfully under most circumstances. But it has also its limitations in regard to span, load or height. But recent development has brought the prestressed concrete construction into practice. This breaks down previous limitations on spans and loads. Moreover prestressed designs have resulted in great savings of material. The prestressing tends to cancel the tensile stresses in the concrete and so the compressive stresses that occur over the entire cross-section tend to prevent the occurrence of cracks. The arrangement of reinforcement is also simplified considerable, because the diagonal tension is considerably less than in the con- ventional design. All of these advantages over the conventional concrete con- struction have made the prestressed concrete construction economical. With this brief introduction the prestressed decks for the same spans and road width as before will be designed. Data: The general layout of the cross-section of the deck is as shown in the following Figure No. 6. The roadway is 28' wide and flanked by two 2' wide curbs. 29 I 1" (f: —.+ ‘1‘ V K - ————¢1-* - , 2 l . q 1- , _ “2...“... -..--- 1 Tc. ,, ' s , “—1 -+ ’ 1 ,1 ~ 1_\~;.J—~'— 1;, f ”T. - 1” ‘f ’ p 1 .1 . 1 1 l ‘ ’ ~1 ‘ 1 J « 1 1 11m; (faoso-SLCruoN or DECK F196 The deck consists of prestressed girders spaced 3.5 ft. on centers. The cross-section of the girder is of T-shape. The con- ventional T-shape appears to be particularly advantageous from the viewnoint of simple form construction. The cast in place concrete slab is 4 in. thick above the girder flanges and 7 in. thick between them. Stirrups are ex- tended above the top surface of the flanges and are spliced in the slab concrete. This has enabled it to assume composite action between slab and girder. The girders are post-tensioned by cables comprising a group of nontwisted wires with ends anchored mechanically. The cables are placed in metal tubing with 1-1/4” outside diameter. The tubes are finally filled with grout so as to provide sufficient bond. 0. Specifications: Load and design requirements shall adhere in general to 30 Standard Specifications for Highway Bridges, sixth edition, 1953, adonted by.AASHO. But as AASHO specifications have not considered prestressed concrete, these will be supplemented by the "Design Criteria for Prestressed Concrete", published by the Bureau of Public Roads, on March 10, 1952. Live load is H-20-44. Concrete: Compressive strength f'c = 4,000 psi Initial allowable compression in extreme fiber 2 2,000 psi II |~' G 0 Initial allowable tension in extrane fiber psi Final allowable compression in extreme fiber _ 1,600 psi Final allowable tension in extreme fiber = 0 psi Allowable tension in bottom fiber at cracking load a 600 psi cracking load shall be = 1.0 DL + 2.0 LL. Ultimate load shall not be less than 2.25 (DL + LL) or (1.0 DL + 3.5 LL) whichever is greater. Allowable principle tensile stress at design load - 160 psi Allowable principle tensile stress at ultimate load = 300 psi Minimum web reinforcement of 3/8” diameter stirrups spaced at a distance not more than half the depth of girder shall be provided. Steel: Tensile strength of the steel wire 3 250,000 psi Allowable initial prestress 150,000 bsi Allowable final prestress 0.85 (150,000) = 127,500 psi The size of wire shall be between and including 0.192 and 0.276 in diameter. 31 2. FLOOR SLAB a. Design As mentioned before,the slab thickness is tentatively pro- vided 7 in. between the girders. It will be noticed that clear distance between flanges is l' 6" for all spans and the width of the flange is 2 ft. for all girders. 1.5 + %(2) Therefore, s 2.5' = 1.5 + 2.5 = 4 ft. Moment = 2 0.2 P1 8 AASHO 5.5.2. (0) Case A E = : ,Q;§H§"12L000 x 2.5 (Wheel load P1 = 12,000 lb.) 4 . = z 1500 ft. lbs. = 18,000 in lbs. Main Reinforcement - g 18,000 .866 15.2 x 20,000 : 3 0.20 sq. in. This is believed to be too little reinforcenent, hence pro- vide 5/8" diameter bars a 12" c. to c. at top as well as bottom arbitrarily. Distribution reinforcement: 50 percent of bottom steel =-% x 0.54 = 0.17 sq. in. Hence, provide-%" diameter bars 0 c. to c. 12". b. Quantity: Per foot of 28 ft. wide roadway slab: (i) Concrete: 28 x _4_+ 8 x 1.5 x_i§ 12 12 12.33 cu. ft. (ii) Steel: 32 Positive steel : 28 x 1.043 : 28.1 lbs. Negative steel = 28 x 1.043 = 28.1 lbs. Distribution steel : 29 x 0.668 = 19;4_1bs. Total steel : 75.6 lbs. 0. Cost: Concrete 0 $46.00 per cu. yd. = 12.33 x 46 = $21.00 Steel reinf. 0 10¢ per lb. : 75.6 x 0.10 : 7.56 Therefore, cost per ft. of 28 ft. wide slab : $28.56 DO 3. PRESTRESSED GIRDER a. Design The girder is simply supported with a span of 60 feet. The section is selected with dimensions shown in the follow- ing Figure No. 7: 1 Symbols: 1 : ' 1 ‘3 l l 0.0.0. : C.G. of concrete ‘ I’— I / x 1/ ( C.G.S. = C.G. Of steel ! cec i A. a Area .1. E ; Yt . Distance from top to C.G.C. (+) 19‘ ‘54 1 Yb = Distance from bottom to S j CoGoco (-) 1 ’ e a Distance from C.G.S. to I coGoco (-) 1 C65 + ————+—% 294‘ : . I a. Moment of Inertia ( I V1 __[ r2. I/A F117]. 7 Properties of Gross Concrete Section: Area Statical Moment A1 a 16 x 4.5 a 72 Al x 2.25 :- 162 Ag = 8 x 56 = 288 A2 I 18 = 5,200 A3 3 3 x 3 a ___9 A5 x 5.5 = 49.5 A : 369 sq. in. Total = 5,411.5 .L Yt : 5411.51369 3 14.6 in. Yb I Yt - 36 a -2l.4 in. e .1: Yb - 405 = -1609 in. Moment of Inertia: I: __; x 16 x (4.5)3 = 122 12 ,1 x 8 x (14.6)5 = 8,260 5 .1 x 8 x (21.4)3 = 26,200 5 72 x (12.55)2 = 10,800 9 x (9.1)2 = 746 I = 46,128 in4. r2 : I/A = 46,128/569 = 125 ing. Girder Stresses: Girder weight 144 Moment = l§_x 23,000 x 60 - 2,080,000 8 Stress = '2,080,QQQ(+14.6) - +660 psi. 46,128 and : = -920 psi. 2 808 000( 46,128 ) Use 62 wires, 0.196 in diameter. Steel area = 62 x 0.05 = 1.86 ing. Therefore, initial prestress, P = 1.86 x 150,000 = This will produce stresses in the girder. f = I e ...... l + _ x 22§$§QQ(1 + -16.9 x +14.6) 125 .. 9 - ~7§é:00[’0097) a -732 13510 and at bottom f .§Z§i§QQ.(1 + -l6.9 x -21-4) 569 "—125— 278 __§é§QQ (3.88) - +2,920 psi. 569 x 150 x 60 = 25,000 lbs. 34 in lbs. 278,500 lbs. 35 Allowing for creep and shrinkage of concrete, final stresses due to prestressing will be, at top 0.85(-732) = -625 psi. at bottom 0.85(2,920) . +2,480 psi. When prestress comes in effect, the girder deflects upward, the girder tries to deflect downward due to its own weight. Therefore, resultant stresses will be as follows: iPrestress Girder Combined Initial -732 +660 -72 psi +2,920 -920 -20,000 psi Final -625 +660 +35 psi +2,480 -920 +l,560 psi Stresses due to slab: After the girders are erected they will first carry the slab. Slab moment, then, will reduce the compressive stress in the bottom fiber. Slab weight - (I x 3.5 +‘1 x 1.5) 150 n 231 1bs./ft. 5 4 Moment = 12.x 251 x 602 = 1,250,000 in lbs. 8 Stress 2 1,250,000 1 (14.6) . +401 psi at top. 46,128 and - 1,250,000 I (-21.4) = -590 psi at bottom 46,218 Stresses in girder, now, will be as follows: Before Slab Due to Slab Combined + 35 + 401 + 436 psi. +l,560 - 590 + 970 psi. Bending stresses in.Composite Section: 5" 3 3? a?" :4 ~-—~— ~ ewe—J: _,.I. ,_.-___ we ~ 07- -~ —— 4 -+1 , r “"T‘*“'“““_“‘—”“”“““‘”1‘"—“‘7' 1 1 , \RI . i After the slab sets, lt ' _f) ____[~~———a—J acts integrally with the /._9‘"' \ // 0. C66 girder. Hence any subse- -+ An-M__—1 quent load stresses will § be calculated with the ‘T assumed composite section ‘1. rL: as shown in Figure 8. CC) F/y. 8 , ( ‘1} r Properties of gross section: Area Statical Moment A1 : 18 x 7 = 126 A1 x 3.5 = 440.0 Ag : 16 I 8.5 : 136 A2 at 4.25 : 577.5 A5 = 8 x 40 : 580 A3 x 20 : 6,400.0 A4 = 3 x 3 = 9 A4 I 9.5 = 85.5 A5 u 1.86 x 7 : 13.0 A5 x 35.5 :__;g§agl A = 604.0 Total = 7,965.0 ing. sq. in. Therefore, Yt : 7965/604 . + 13.2 in. Yb 3 Yt - 40 : - 26.8 in. 3 : Yt - 4.5 = - 22.3 in. Moment of Inertia: I: .1 x 18 x 73 12 .1 x 16 x (8.5)3 2 H x 8 x (15.2)3 x 8 x (26.8)3 Cfihd oaha 126 x (9.7)2 156 x (8.95)2 9 x (3.7)2 15 x (22.5)2 I Stresses due to live load: From Appendix 'A', AASHO for 60' span. Live Load Moment : 555 k. ft. 27 percent impact Therefore, total Live Load 37 I 515 = 820 6,150 51,200 . 11,850 10,900 123 .5449 87,998 in4. specifications, 1953, page 285, 6,650,000 in lbs. 1,800,000 in lbs. 8,450,000 in lbs. Moment per lane Fraction of moment to each girder '3 5.5 = 0.55 2x5.0 10 Design L.L. moment Stress at top of slab At top of girder At bottom of girder 0.55 x 8,450,000 2,960,000 in lbs. 2 960 000 ' ._L__~L._. = + 445 p81. 87,998 (13’2) 2,960,000(9.2) = + 310 psi. 2,960,000(-26.8) : - 910 psi. 38 Final stress analysis: Before ‘ L.L. Applied L.L. Stress Combined At top of slab 0 + 445 + 445 psi. At top of girder + 436 + 310 + 746 psi. At bottom of girder + 970 - 910 + 60 psi. The girder is quite safe as no tensile stress occurs in the final analysis. Checking of stresses at Cracking Load: Cracking load = 1.0 DL + 2.0 LL This means 100 percent over load. Live Load Moment 2x2x8,450,000 = 3,360,000 in lbs. Curb and railing 2x%§x400x602 _ 45,520,000 in lbs. Surfacing : .12x20128x602 = 5,020,000 in lbs. 8 Total moment on 9 girders = 40,950,000 in lbs. Therefore, moment per girder = 4,560,000 in lbs. Stress at top of slab = 4,560,000 x 13.2 a + 680 psi. 87,998 Stress at top of girder = 4,560,000 x 9.2 = + 475 psi. 87,998 Stress at bottom of girder = 4 560 000 x ~26.8 = -l,375 psi. W Therefore, stresses due to 1.0 DL + 2.0 LL will be as follows: Due to D.L. Due to 2.0 LL Combined At top of slab 0' + 680 + 680 psi. At top of girder + 436 + 475 + 911 psi. At bottom of’girder + 970 -1,375 - 405 psi. 39 The tensile stress at the bottmm fiber is less than permis- sible, hence design at cracking load is also safe. Stress diagrams for different stages are shown in the follow- ing Figure No. 9: +445 . +680 -73 +35 #746 9'9” X HE E: l ' [fl g ---i »--———-~1 _ 'r—s-g j— LA :3 E -——x _—-—__, ; _ Donal use 1 -405 Inn‘ia/ Presfrcss ano/ Presfrws proofross pras’r'ess Pres/fess + D L , + Girdzr' + Gim’zr +G/rdzm Slab +D.L.+L.L. + 2L.L. F 1'9- 9 Ultimate Load Ultimate resisting moment, Mu a As fu jd where As .. steel area fu Ultimate Stress -.-. 250,000 psi. jd is moment - arm. Assume j a 0.9 Theanu = (9xl.86)250,000 x 0.9 x 35.5 : 155,000,000 in lbs. Total girder and slab moment - 9 1 3,330,000 3 29,970,000 in lbs. Moment due to curb, railing surfacing, 7,;40,900 in lbs. Total D.L. Moment $7,310,000 in lbs. Total L .L. Moment 16,900,000 in lbs. D.L. . L.L. 54,210,000 in lbs. 40 Ultimate Resisting'Moment, mu 133,000,000 in lbs. Subtracting D.L. Moment - 37,310,000 in lbs. Balance left for L.L. 95,690,000 in lbs. Ultimate load factors: = 133,000,000 (D.L. + L.L.) 2.28 (D.L. + L.LJ 54,210,000 also 95 690 09-9 (L.L.) : 5065 (L.L.) 16,900,000 These are within allowable limits. Hence, the design is also safe at Ultimate Loading. Eccentricity at Subport: As moment at the support is zero, to make top fiber stress equal to zero, 9 : *TZ/yt -125/l4.6 = -8.55” The profile of the c.g. of the steel is assumed to be para- bola for the beam. Principle stresses: The shearing stress v in an uncracked concrete section is maximum at the centroid. This can be given by, v = VQ/bI. Referring to Figure No. 7, Q, the statical moment of sec- tion on either side of centroid taken about that point = 21.4 x s x 10.7 = 1,830 ins. Now shear at the support %(wt. of girder + wt. of slab) : %(23,000 + 231 x 60) = 18,430 lbs. Vertical component of prestress in wires, = -237,000 x 2 x 8.55/12 x 30 = -11,250 lbs. 41 Therefore, V': 18,430 - 11,250 = 7,180 lbs. Hence,'v : VQ/bI = 7,180 x 1,850/8 x 46,128 : 55.5 psi. From AASHO Specifications, Appendix 'A' for 60 ft. span, maximum shear at the support = 45.2 k. Therefore, shear with impact = 1.27 x 45.2 a 57.5 k. Hence, Design shear = .35 x 57.5 = 20.1 k. Referring back to Figure No. 8, Q will be, 26.8 x 8 x 15.5 + 15 x 22 = 2,860 . 290 = 5,150 1n3. Therefore, V : 20,100 x 3,150/8 x 87,998 = 90 psi. Hence, total V = 55.5 + 90 = 125.5 psi. The prestress force creates a horizontal compressive stress at the centroid of concrete. This has intensity, 3X = P/A = 0.85 x 278,500/369 = 642 psi. Now, the stresses V and 3X produce a principal tensile stress which at the support at the centroid can be calculated by the well known formula, St : fiix/ 4v; + sxf- 5x ) = 5(1/4 x 125.52 . 5422 - e42 ) = %{688 - 642) = 25 psi . Shearing stress at ultimate load will be, (1) 2.25 (DL + LL) - 2.25 x 125.5 _ 287 psi. (2) DL + 3.5 LL : 125.5 + 3.5 x 90 _ 350.5 psi. Hence, principle tensile stress, when v : 350.5 will be, ‘%(1/ 4‘E'550.5Z”+ 642*";‘842' ) = 44950 - 542) - 154 psi. These are within limits, hence design is safe. 42 Deflection: The deflection of prestressed girders can be computed by the formula, D: ”5 Mfimax L, when load is uniformly distributed 48 E1 on the span length, L, and the moment of inertia, I, is con- stant throughout the entire snan, D 5 x 16,90 ,_00~§ L60 x l?) ——.—.—. - --— -..—.—— 48 x 5.5 x 10 x 87,9887??? : 0.33 in. : 1/2,180 span This is much less than allowable live load deflection, hence safe. Diaphragms: These are provided tentatively as follows: Provide 2 Nos. 24" x 9" section with two cables each having an area of 0.40 so. in. Quantity: Concrete 2 (2x.75x24) _ 72 cu. ft. Cables 4 (30 x 1.41) _ 169 lbs. Cost: Concrete 0 $46.00 per cu. yd. : z§_§”g§ = $123 27 Cables @1605 per lb. = 169.2 x .6 = 102 Total = $225 b. Quantity for 9 prestressed beams: (1) Concrete: 9 x 369 x 60 144 1,380 0. ft. (ii) Cable: 9 x 60 x 6.44 : 3,480 lbs. c. Cost of deck: (1) Concrete 0 53.00 per cu. ft. 1,580 x 5 s 54,140 43 (ii) Cables (includes fitting, orestressing, anchoring) 0 60 cents per 1b. = 3480 x 0.6 a 32,088 (iii) Transportation up to 100 miles N ,p H v?- 0 $4 per ton : 103.5 x 4 (iv) Erection 0 $10 per ton = 103.5 x 10 3 414 (v) Slab 28' wide 0 $28.56 per foot : 60 x 28.56 _ 1,713 (vi) Diaphragms 2.25 Total _ $9,615 44 IV. RESULTS The costs of the deck for the two different types of construction are tabulated in the following tables with necessary details. Also, the total costs are plotted on the following graph paper for handy comparison. The cost analysis for 50', 70', 80', 90' and 100' spans are given in Appendix. TABLE I COMPOSITE CONSTRUCTION - - ' -9“ --- ”w'-“"'";.*" 9 ' Structural Steel ' Spiral Steel ' 7” Thick Slab ' Total Span ' 0 50.155 ' c 50.50 ' 0 51.578 ' Cost Length ' per pound ' per pound ' per sq. foot ' in ' ' ' ' Dollars ' Quantity ' Cost 'Quantity' Cost'Qnantity' Cost ' ft. ' lbs. ' 5 ' lbs. 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