Scholars' Mine
Masters Theses Student Theses and Dissertations
1939
Design study of a three span continuous tied-arch bridge
George Perry Steen
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Recommended Citation Steen, George Perry, "Design study of a three span continuous tied-arch bridge" (1939). Masters Theses. 4754. https://scholarsmine.mst.edu/masters_theses/4754
This thesis is brought to you by Scholars' Mine, a service of the Missouri S&T Library and Learning Resources. This work is protected by U. S. Copyright Law. Unauthorized use including reproduction for redistribution requires the permission of the copyright holder. For more information, please contact [email protected]. Design Study of a Three Span Continuous Tied-Arch Bridge
By George Perry Steen
A Thesis
Submitted to the Faculty of the School of Mines and Metallurgy of The University of Missouri In partial fulfillment of the work required for the Degree Of Master of Science in Civil Engineering
Rolla, Missouri 1939
Approved by: A~~ Professor of Structrual Engineering.
- 1 - ACKNOWLEDGEMENT
To Professor E. W. Carlton, for his valuable criticism, and to Mr. Howard Mullins, for his help and suggestions, the writer owes an expression of appreciation.
- 2 - TABLE OF CONTENTS
Page Acknowledgment ...... 2 Synopsis ••••••••••.•••.••••••••••••••••••• 5 List of Illustrations •••••••••••••••••••••• 6 A General Discussion of Statically Indeterminate Structures ...... 7 A Brief History of Arch History...... 11 Description of Type ••••••••••••••••••••••• 14 The Principle of Least Work ••••••••••••••• 16 Direct Stresses ...... 18 Notation ...... 19 Influence Lines ...•...... • 21 Stress Analysis ...... 22 Diagram For Main Span Equations ...... 25
Equations For Design - Main Span • • • ••••• • • .26
Diagram For Side Span Equations • • • • •• • • • • • 35
Equations For Design - Side Span •••••••••• 36
.s~c~ edx Evaluation of ± •••••••••••• II; A sec.~¢dx Evaluation of f • • • ••• • • • • • • o At TABLE OF CONTENTS
Page
Computations - Design No.1. 46 Conclusions ••••.•••••••••••••.••••.•..••• 54 Bibliography...... 57 Tables and Illustrations ••••••••••••..••• 59
4 Synopsis
In this study of the continuous tied-arch bridge~ having one main span and two equal side-spans, the follow ing items are endeavored: 1. To set up the total internal work equations and to find the values of the unknown forces or redundants. 2. To compute to maximum moments for design and to complete the design of the super-structure for one bridge. Details, such as the gusset plates, rivets, connection angles, etc., are not design ed; the effect of such being allowed in the design for dead load as a certain per cent for details. 3. To set up hypothetical conditions for other bridges of different span lengths and to con struct the influence lines of the unknown forces of each. 4. To compare the different bridges upon the basis of their influence diagrams.
- 5 - List of Illustrations
Diagram for Main Span Equations Plate A Diagram For Side Span Equation Plate B
Evaluation of Equations Design No.1. Plate 1 Evaluation of Equations Design No.2. Plate 2 Evaluation of Equations Design No.3. Plate 3 Evaluation of Equations Design No.4. Plate 4 Influence Diagrams Design No.1. Plate 5 Influence Diagrams Design No.2 Plate 6 Influence Diagrams Des,ign No. 3 Plate 7 Influence Diagrams Design No. 4 Plate 8 Influence Ordinates For Live - Load Moment Design No. 1 Plate 9 Maximum Positive and Negative Live-Load Moment Design No. 1 Plate 10 Influence Ordinates for Dead - Load Moment and Maximum Moment for Design No. 1 Plate 11 Section Summary Design No. 1 Plate 12 Section and Weight Summary Design No. 1 Plate 12A
_. 6 - A General Discussion of Statically Indeterminate Structures
The statically indeterminate type of struc- ture is not generally favored by the American Engineer. Until recently this type has been actually shunned be cause of the lack of information and the lack of train- ing in the design of such. The majority of structural engineers prefer to use types with which they are famil iar, making many estimates based on judgment and experi ence. They contend, too, that for every indeterminate structure that it is possible to design a determinate type that will carry the same load with as little or less material. A few citations follow, giving the com parative amount of material needed for a type of bridge, stepping from a statically determinate one to one of in determinateness in the first degree or from one toanoth er type that is statically indeterminate in a higher degree. trFor the Sciotoville Bridge over the Ohio river at Sciotoville, Ohio, consisting of two spans at 775 feet, the saving in favor of continuous spans over two simple spans was found to be about 15 per cent." ftAccording to comparative studies reported by Winkler, the economy ef continuous spans of two, three,
- 7 - four spans is 16, 19, and 21 per cent, respectively, when the span length is about 325 feet; and 20, 24, and 28 per cent, respectively, when the length is about 500 feet." 1 "Ierriman and Jacoby investigated the result of placing a hinge at the crown of a 550-foot two-hinged spandrel-braced arch bridge over the Niagara River and found that if just one hinge were placed in the upper chord the weight of the arches alone would be decreased 11.8 per cent." 2 "Professor W. Dietz of Munich states that a very careful comparison of material required for a two hinged and three-hinged arch for the 110-foot span of the Hacker Bridge in Munich, indicated that 11.3 per cent excess was required for the latter type." 2
A chief objection to the use of a continuous bridge or of a hingeless arch has been that it is sensi tive to settlements of foundations. Needless to say,
1. Hool and Kline -- Movable and Long-Span Steel Bridges, p. 200.
2. Parcel and Maney Statically Indeterminate Con- struction, p. 405.
- 8 - this type of bridge would not be considered unless rock foundation was available, and any slight settle ment of the foundation would cause very little change in the stress. Another objection has been that the continu ous bridge would be seriously affected by temperature changes. The effect of the expansion and contraction of intermQdiate piers will cause little changes in stresses according to several studies reported. 3 It has also been determined that the effect of unequal temperature changes in different parts of the span can safely be neglected providing the members of the span have the same coefficient of expansion. A continuous bridge of the indeterminate type has advantages in rigidity and economy and in the fewer number of expansion joints. Its use permits a decrease in the width of intermediate piers. In addition, a continuous bridge and especially a continuous-arch span has all the advantages of a cantilever bridge for loca- tions over navigable waters where clearance must be maintained during the construction period. The contin- uous structure eliminates the detailing and expense of hinges. There is no question but that the design of an
Hool and Kline -- Movable and Long-Span Steel Bridges, p. 201.
- 9 - indeterminate structure involves a great deal more work and calculations. It is usually necessary to make more than one trial design. However, if a small per cent of material and weight is saved, the extra time spent in design is warranted. In the final analysis, the results given by engineers, both pro and con, on the statically indeter minate structure are largely theoretical and are not based on extensive experiments. There are not enough actual monuments of indeterm~ate structures that have approximately the same site conditions of the many sim ple structures. Until more such bridges are built affording us more reliable comparisons, the advantages of one or of the other will be debated largely on the basis of theory. It is probable that the statically determinate structure will be favored for relatively short spans and the indeterminate bridge will replace them for longer spans•.
- 10 - A Brief History of Arch Bridges
The modern period in arch-bridge building be gan in the year 1716 when the French ~Departement des Ponts et Chaussies" was formed although the first real emploYment of the true-arch type was as early as 600 B.C. The first arch span of the monolithic type was a wrought iron bridge built in 1808 over the River Crow at St. Den nis. However, it was not until several years later, in 1879, that Weyrauch developed the fundamental equations upon which the elastic-arch theory is based. Arches of iron have given way to structural steel spans, which type in the form of ribs (hinged and fixed), truss design, or braced spandrel construction is being utilized to a very large extent in the field today. The concrete arch con stitutes the latest development in the arch and is a very important type, especially for short-span structures, in which it has nearly an exclusive field. Weare more particularly concerned with the long-span structure, in which the steel arch supersedes the use of concrete for material. It is pertinent and interesting to follow the several examples of arch spans that have been erected. One of the first in which structural steel was used ex-
- 11 - tensively is the Eads Bridge erected in 1874 across the Mississippi River at st. Louis. It consists of two side spans of 502 feet and a center span of 520 feet with four lines of fixed circular-arch trusses with parallel chords. The structure carries an upper deck of a 52-foot highway and a lower deck 32 feet wide for railway traffic. In 1889, the Washington Arch Bridge was built over the Harlem River at New York. Two. spans were erect ed at 509 feet and had 6 lines of circular plate-girder arches 13 feet deep and a 90 foot rise. Each girder has two hinges. The bridge carries a highway 80 feet wide. The bridge Alexander III was constructed in 1899 over the River Seine at Paris, France. Spans of 352 feet with 15 lines of circular plate-girder arches 3 feet deep and a 21 foot rise were used. A highway deck was supported on top with a width of 131 feet. During 1897-1898, the Rhine Bridge at Bonn, Germany was erected. It was a through arch having a highway floor 46 feet wide. It contained two lines of arch trusses 614 feet center to center of hinges with a rise of 97 feet to the bottom chord.
The Wupper Railway Viaduct was completed in 1897 at Mungsten, Germany. The spans were hingeless arch trusses 588 fe·et long with a rise of 219 feet and
~ 12 - carried a deck of 28 feet in width. The trusses were
16 feet apart at the c~ovvn and 84 feet apart at the springing line. One of the first bridges of the type which is under study was a three-span continuous truss with a tied arch in the middle span built over the Angora River at Irkutsk, Russia. The structure consisted of two side spans 278 feet and a center span of 500 feet with the arch rising 78 feet. In 1933, the Lachine Bridge was designed and later built across the St. Lawrence River near Montreal. It consisted of a continuous structure of several spans, but the main span was a continuous tied-arch truss 400 feet long. This structure was designed by the Lake St. Louis Bridge Corporation and constructed by the Dominion Bridge Company, Ltd. There is no bridge known that consists ofa three-span continuous plate girder with a tied-arch rib in the middle span. This scheme was suggested by the last competition at Mannheim, Germany, but it is not known whether any has been designed and erected. The tied arch itself has been used frequently in Europe but no examples can be cited. Description of Type
The type under study is a three-span continu ous plate-girder bridge with a tied-arch rib in the mid dle span. The structure is statically indeterminate in the third degree. In an ordinary arch inclined reactions are exerted on the bearings, but in a tied arch the hori zontal component of the reaction is resisted by the tie connecting the two bearings, the tie acting as a beam as far as the reactions are concerned. The stresses, how ever, are determined as in an arch. The bridge is a through structure, the roadway being suspended by hangers from the arch rib. One end of the arch will be fixed with the remaining bearings consisting of rollers.
The arch itself will be parabolic in shape, the rise of the arch being about 1/6 of the span length. Ample rigidity must be secured by making the width be tween the arch spans at least 1/20 of the span length. The side spans are plate girders with the roadway fasten ed to the bottom of the girders. The one exception is in Design No.4. The length of the side spans do not allow the use of a plate girder, the weight and section requir ed being excessive for economy. The bridge structure must be braced against
- 14 - side-sway and windloads in the lateral direction by ample bracing both in the upper chord and in the sec tion of the tie. It is generally agreed that the Warren truss system is the most economical for such lateral bracing. Types similar to this are generally designed on the basis of elastic centers or on the basis of least work. The "method of least work" will be follow ed in this study. This method, commonly known as Cas tigliano's second theorem, was discovered independently by A Castigliano and was developed in his treatise on the t1Theorie de l'equilibredes systemes elastiquestl • A detailed explanation of the method will follow.
- 15 - The Principle of Least Work
"The fundamental conception upon which the method of least work is based may be presented in an ideally simple statement, namely, that a structure will deform in a manner consistent with physical limitations, and so that the internal work of deformation will be a minimum. nl In order to apply the above me.thod, the structure must be severed so as to obtain a statically determinate condition, the unknown forces being called redundants. Then, an expression for the total internal work is set up in terms of these redundants. The total internal work is equal to one half the product of the square of the moment and the sub-division dx divided by the product of the moment of inertia and the modulus of elasticity.2 Partial derivatives of the internal work are taken in respect to each redundant and the de- rivatives are set equal to zero. The unknowns may be determined by solving the resulting simultaneous equa tions. A mimimum condition is obtained in setting the
1. Grunter -- Theory of Modern Steel Structures, p. 64.
2. See Plate A for the mathematical development of the total internal work.
- 16 - derivative to zero. "In every case of statical inde termination where an indefinite nrunber of different values of the redundant forces X will satisfy all stat ical requirements, the true values are those which rend er the total internal work of deformation a minimum. D) The indeterminate forces or redundants are chosen at points where .any displacements are prevented, such as supports. Hence, the partial derivative is equal to zero.
Parcel and Maney - An Elementary Treatise of Stati cally Indeterminate Stresses, p. 123.
-n- Direct Stresses
The tension stress in the. horizontal tie produces a direct stress in the arch ring equal to H Sec e. There is also a direct stress produced in the tie itself. It is affected by the camber in the tie, but the angle of inclination is so small that the secant of the angle is nearly unity. There fore, the direct stress is equal to H. The camber of the tie also causes a direct stress in the hangers, but the value of the stress is negligible.
- 18 - Notation
H = horizontal thrust in tie Ml= moment at left end of arch span over pier M2= moment at right end of arch span over pier I = moment of inertia at any cross-section of arch 1,= moment of inertia of side span cross-section A = arch section, in square feet Ah= hanger section, in square feet At= horizontal tie section, in square feet X = distance from end of arch span to any section Xl= distance from end of side span to any section
Y =: vertical distance between arch and tie axes at any point ~ = inclination of arch rib ¢ = inclination of horizontal tie section AS = Linear increment in arch span, in feet
..£1 Sl = linear increment in side span, in feet L = length of arch span, in feet Ll = length of side span, in feet
E =: modull1s of elasticity of structural steel Eh= modulus of elasticity of hanger if other than steel. f = camber or horizontal tie, in feet
- 19 - A h = length of any ha.nger, in. feet A = spacing of hangers~ or panel length, in feet do = depth of section over pier, in feet d = depth of arch, in feet
dl = depth of side span section, in feet k = position ratio of any load (x...!. 1) •
- 20 - Influence Lines
An influence line may be described as a curve any ordinate of which gives the value of the func tion (shear, moment, horizontal thrust, stress, etc.) for which the curve is drawn when a load of unity is at the ordinate. An influence line for shear or moment records graphically the value of the function at a sin gle section for loads at all sections, which contrasts greatly with a simple shear or moment curve that records graphically the value of the function at all sections of a structure under a fixed loading. From its manner of construction it rol1ows that any influence line gives the effect ofa concentrated. load, or of a series of such loads, by the multiplication of each load intensity by the value of the influence ordinate at the load.
- 21 - Stress Analysis
Ordinarily, the first step in the analysis of a continuous structure is the construction of an influence diagram for one of the unknown reactions known as the elastic curve. The influence lines for all members of the structure are constructed by draw ing straight lines across the curve. In the case of the structure under study, not only are the vertical reactions unknown but the moments at the piers and the horizontal thrust in the tie as well. In such a case, influence lines must be obtained for the three redun- After these values are found, the stresses may be determined by statics. The sections are proportioned for a uniform live load in combination with one concentrated load placed where its effect isa maximum. This is in ac cordance with the bridge specifications of the American Association of State Highway Officials.l This load is chosen of such magnitude as to give an effect equal to that produced by the actual vehicle on the structure.
1. The American Association of State Highway Officials Standard Specifications for Highway Bridges, p. 174.
- 22 - Live loads can be used as a uniform load across the span, and the value of the function can be obtained by multiplying the area of the influence diagram by the uniform load per foot. This has been varied slightly by breaking up the load into a series of con centrated loads assumed to be acting at the panel points. The live-load moment is then computed by mul tiplying the panel load by the ordinate at that point. After the simultaneous equations are solved by determinants, and the values of H, Ml, and M2, are found, we tabulate the values for each at the various load points.2 By taking these values and substitut ing in our initial moment equations, we calculate the influence ordinates for the moment at each load point.3 The dead load moments are computed directly from our original influence ordinates for H, Ml, and M2, as the dead loads are stationary and will always be acting.4 For our maximum negative and positive live load moments, we must load the influence diagrams whose ordinates are given on Plate 10.5 For the maximum positive moment,
2·See Plate 5, p. 63 3·See Plate 9, p. 67 4·See Plate 11, p. 69 5·See Plate 10, p.68
- 23 - the length of the structure showing positive moment ordinates is loaded with our live load. The maximum negative moment is obtained in like manner by loading the length of the structure showing negative moment. After these are obtained the maximum dead and live load moments are combined and the maximum for design is found. The moments are secured by multiplying the load at the point in question by the ordinate at the same point and the results added algebraically.
- 24 - DIAGRAM FOR MAIN 5 PAN EQUATIONS PLATE A.
.U
z A (r"C.~IN~'f'O"'-0 .. TI~) o , ~ ~ Q, l'J C1l I fi 5r~tiS/£ DEVELOPMENT OF INTERNAL WORK EQUATION 1M ...,.--,n d\:£ TF as 15 SUFF/C,ENTl-Y SMALL, THE AVERAGE NOTATION r C VAlUE OF' M'" (f'1I1+r'1L)+Z. THt:Ie£FORE, W ~ TOTAL. INTERNAL woRK M:: BENDING MOMEN'"f MI. U \ THf; woRK OF THE: SENDING MOMe:rlT IS \1 Mit 0( = ANGULAR. DISTORTION 'I~ (. I I , .j W= Me( S":: TOTAL. LINEAR O/:5TORTIO"l C:: DISTANCE TO r::-XTR£ME FIBER \ FROM MeCHAN IC.s: 0( = 57 C S: lds-;- E -!" UNIT STR5i:SS IN EXTREMe: FIBER \ i"'Mc+I .'. 0(= Mds~IE E -:: MOOUl-US OF" ELASTIC ITY , \ L.AMIr4~ o ,~ 50 W:. ±N;P FO/l! THE AND Setting up the values for moment, we have the fol lowing for loads in the main span only: For a unit load at a distance kl from the end of the main span, we have, When x< kl) M x = (\- k)x - Hy + M, (~) + M2. t ----- CD When X>kl J Mx=kCl-x)-Hy +M,(?)+M,f -----0 Side Span: For the left side span Mx -= M b ------t l, ® For the right side span Mx = Ml.1.'. ------@ - 26 - The total work will be, since W=.1..JM2.dx , 2. EI Where the last three terms represent the work done by the arch, tie, and hangers in resisting the di- rect stresses. Differentiating equation 5 with respect to H, Ml, and ~2 which are the unknown forces: .. 27 - kl fk.l.~· kl J-. li-k){t-J<.),.d", -.tl (l-"'hldx t M.!. S \.-x'1 dx. + Mi- rn-x.)xdx. E\.~ E1 f0 I E.'l 0 I 0 Ell. 0 1. t. 1d + 11!. )(., x E.l1. I Io I aW. 8 at{ Setting the above first derivatives equal to zero, multiplying by E, an9. rearranging, we have - 28 - ® 1 kt l. M2. Jr..'1 d x. - [U-k)f Y-'fdx + k r (l-1.)'ldx. ] =0 t 0 I 0 I J~ I H f1X'Idx. - M. Jl(\-xhdx - M2.( ,\2. r1~ldx -+ -\Sl'xfdx. 1 I lz. I J I ' 1, 0 I, 00 0 These three equations may be expressed in the fol lowing form by sUbstituting singular coefficients for one or more terms. - 29 - Instead of actually integrating in order to ob- tain the various values for a structure, we may divide the structure into increments or 65 distances, obtain the values for the various subdivisions, and make a summation of all individual values in order to obtain an evaluation for the entire structure. Theoretically, these subdivisions should be infinitesimally small, but for practical purposes we make each division equal to a panel length, or, in case of the arch ring, the amount subtended by ,a panel length. The main reason for so doing is to allow computations to be made in tabular form. Therefore, we substitute the subdivision ~s for the term dx and the sign of summation, ~ , for the integral sign, f. . For a symmetric.al structure we write the following. In computing these values we have made the substitution that the moment of inertia varies d o )3 as the cube of the depth. For I we substitute ( ~ where do is the depth over the pier and d is at any sec tion. - ~O - (\-~,,65 02.= b,= t t ------@ I 1 1 ) ~ 1. 2- b :: c ~ -1. [L " l1s +(-) ~ X'~51] ------@ Z. 3 1 0 I 1. 0 I I " k't \ A = (\-k)'E J<.'ltls +k~ n-XbJ6S ------@ , 0 I u I I k't A \ z. A ~ - [(l-k)~ (\-)(.)~us + k Z (\-x) AS] ------® 2. 1 0 I k\ I kl \ ~ )(~ ~ -x~ A 3 : t[(,-k) s k t n As] ------@ - 31 - It should be noted that the second term of is antisymmetrical to the first term, is antisymmetrical to the and A) is antisymmetrical to A2. The word antisymmetrical is applied to the condi tion in which the total sum of the values for a certain term is equal to that of another term, but as the values for one are increasing or decreasing, the values for the other are increasing or decreasing but in the reverse or- der. For example, the values of x are increasing from a minimum at the left side of the main span to a maximum at the right pier. At the same time, the values for (l-x) are increasing from a minimum at the right end to a maximum at the left pier where x was a minimum. The summation for both terms is equal. Inter-changing some of the coefficients for their equal in order to have as few different coefficients as possible, we have: Ha.-M,Qz. - Mzaz. =At ------@ ... 32 - Solving for the unknown values by determinants, we have: A,- Q2. - Qz. A2.- b 2. - 03 A3-b~-b2.. A,( b~- b:)+A2-(a1.b,-Chbl.) + A3(Q1.b~- a b;!.) H = = 2 ~I - Q2. - al. Cl\(b~- b~) + a~{2b3-2b:z.) . ~ Q1.. PI. - b 3 a 2. - b~ - b" Simplifying and cancelling the term we obtain: ------@ Q,+A,-a2. Qz.+A2..- b 3 Q31-A 3 -bl. M ,:=- 1----- ® ell - Qz. - ell. D Q1.- b z.- b 3 Ok- b.3 - b 2 .~ 33 - Where D is the same as for H. The influence ordinates for M2 are antisymmetrical to those of MI. That is the ordinates de~ease or in crease from the left side to the right in the same man ner and value as the ordinates for Ml decrease or in crease from the right to the left and the summation of the ordinates is the same. DIAGRAM FOR 51DE -'sPAN EQUATIONS PLATE B. o ~ I. .kt ~ ,.3I- Z -;) z ~I -kt, (f) 1;;}J • A D tv CJl I ..Q., t Q. DEVELOPMENT OF REACTION:) & MOMENTS ."'-"- I1* I. I -.II:-}J(. "1 USING UNITY LOAD AND TAKING MOMENTS fiT R~~ ,m n Z.MRa. : R,)(~ - I (l-k).Q." 0 R. x I R2, :. R,: (1-1<.) R~ ~ .Jl>- x WHEN )<.<~.tl Mrn=(I-Je.)X J- WHf:N)( >~~, Mn~(I-Je)x-l(x-.k.t)",.k.(.Q-)() Equations for Design Side Span: Developing the parallel theory for the side spans we have the following: For a unit load at a distance kl, from the end of the side span, we have: Loaded Side Span when ~< kl'l Mil. == (l-k)x, + M, ~' ------@ I Unloaded Side Span Mx= M~~'------@ \. MJI. ~ M, (1- t) + M2. l - Hy ------® The total work will be: - 36 - W=- ® Where the last three terms, as before, represent the work done by the arch, tie, and hangers in resist ing the direct stresses. Differentiating the equation for total work with respect to the unknown values H, Ml' and MZ respective ly, we have for partial derivatives: -.MJl\)('l-)(" ydx -&it~'Jd)(. +.Iifl.:l~ + ll.st sec:l. e dx El 0 I El. 0 1: E. 0 1:. E G A .... 37 - aw ={ M2. r\~dx +~ (n-x.h.dx + M2.S\':Ld X -11 ItX'IdX } ® El~ II E1 ~ I E\ I dM1. Jo J0 I 0 0 Setting the above ~irst derivatives equal to zero, multiplying through by E, and rearranging, we have: ----@ -M !. r\(t-l<.)yd~ - M xydx "= 0 \\J. I 2.1.1.1\. I o 0 - 3.8 - These three equations may be expressed in the fol- lowing manner: Ha,-M,b,-Mz.c,=O ------® Hal-M,b3-M;tC3=O ------® In the same manner as before, we substitute ~s for dx and the E sign for the sign of integration. For a symmetrical structure we have: \ "\. '\. l..E \. :L Q ""h~ + L: sec"l.fA5 + L sec-:l.0A5 +frl£._ 1: A f).s - --- @ \ 0 I 0 0 A..., \.4 Eh 0 A;; \. = a =l ~ l \-x.)v~s ------@ b , 1 \.L.r I o l c -: a - 1- ~ "''I6os , ~- l.. I ------® 't "l..£i 'l "J. t. 2 ] b =C :=.l.1.[L Cl- x ) "5 +/_) L. ~,~s, ------@ 2. 3 l 0 I\ t, 0 II .,.. 39 - Again we have the term for b, and aZ antisymmetri cal to that for c and a3. Changing some of the coefficients for their equals, we have: Hal. - M,b 3 - M2.bl. : 0------® Solving for the unknown value by the method of determinan~s) we have: 0- Q2. -Q;t A~-b1.-b3 o - b 3 - b1. H = ~------+ = a,-G\1.-Q~ Ct1. - b 1 - b 3 01 - b a - bz. Multiplying through by -1 and cancelling terms of (~Z - 0-3), we have: A4-t------® - 40 - a,+O -02. C12. + A4 -b 3 Q2.+0-b:z. M= Q I - Q2. - 02 0 1 - b2. - b 3 Q2.-b~ -b:z. Where D has the same value as for H. Q.- C1 2 - Q:2,. Q1- b2. - b 3 Q:2,.-b3 -b2. For the values of the various unknowns we will use a tabular form. - 41 - t Evaluation of '"~ sec}-edx. o A The arch axis is assumed to be in the form of a parabola. If so, the vertical ordinates, or y distances, varies as the square of the horizontal measurement, or x distances, from the vertex of the curve, or at the crown of the arch. In accordance, we assume that the equation of the parabola is x"1.= lay • At the springing line, when x = 110 and y = 36.67, which is the total use of the arch, we have by substitution in the equation ~~~lay l\OZ. :: 2.Q )(. 36.67 a -.:: \2.\00 = \65 78."31 Then, the equation becomes The slope of the curve at any point is the tangent or ~ value of the curve equation From the trigOm.Qlb.et:eie relation$lli~s,we 1lavethe - 42 - which becomes When the value of x at any point is substituted in this equation, the value of sec.2.g is obtained. For example, at the mid-point of the fifth panel, x =20. sec:J.e-= \+{~oo5Y= \+(O.\2.12.Y· = \.0\47 . In tabulating the valies, we substitute ISs . the length arch axis subtended by the panel. feet and A = 108 sq. in. or 0.75 sq. ft., we have sec1 8As:= 1.0141,. 20.5 :. 2.7.75 A 0.75 Securing the values for e~ch panel, the summation across the entire span gives us the value of ~ sec.~eAs o A Since we multiplied all our terms by d: we must \ multiply our summation of ~ sec.:J.e /15 by to ab- o d: tain the final result. - 43 - Evaluation of J\ sec 0 dx. o AT The horizontal tie between the bearings of the arch is cambered, but the angle of inclination is so small that the secant of the angle approaches unity. For all practical purposes, it is safe to call the value of sec"¢ as one. The cross-section of the tie is constant, so the term may be integrated direct- ly and evaluated. If At = 40 sq. in. or 0.278 ft., and the limits are from 0 to 220 ft., we have This value must also be multiplied by d: to obtain the final correct result. 6~f2. E "}?~h Evaluation of ~ - E ~ Ah h The ratioE/Eh is used so that the value will apply to all conditions. If the hangers were compos ed of a different material, the moduli of elasticity would not be the same, and the ratio would be used. -44- In this particular design, we have assumed that the hanger and arch section are of identical material, structural steel, so the ratio is equal to unity. If we evaluate the term at panel point L,o where the tie is cambered 0.25 feet, A = 0.21 sq. ft., the panel spacing is .20 feet, and the length of the hang er is 36.14 feet, we have the following: 0.0000\\5 The term is negligible as far as affecting the results of the design. Even after multiplying by the term d; , we obtain (7.5)3 " 0.0000115= 0.00485 the value is still very small and can safely be neg- lected. - 45 - Design No.1. Data: Main span = 220' -0" center to center of bearings (11 panels at :20' -on) Two side spans = 100' -0" (5 panels at 20 1 -on) Rise ratio o~ parabolic arch = 1 . 6 Theoretical height o~ arch = 36' -8" Thiclmess of arch at piers = 7' -6" Thiclmess of arch at crown = 3 1 -6" Clear width of roadway = 24' -0" Impact = 50 1+125 , Floor slab = 8" slab crowned to IOU at center With 25 lb. per sq. ft. wearing surface totaling 150 Ib.p~r sq.ft. Stringers assumed to weigh 40# per ft. and spaced 4' -0" c. to c. 40,000 = bending moment due to concentrated load 0.345 - 50 = impact ratio 20+125 13,800 = 40,000 x 0.345 = bending moment due to impact _.2 32,000 = 150x.4+40 x 10 = bending moment due to dead 2 load - 46 - 85,800 = total bending moment 12 x 85,SOO = 18,000 S S = 57.2 (a 16' - 40# wf beam is ample) Use all stringers as of this section. Design of Floor Beams: (Spaced 28' -0" c. to c.) Maximum concentration on floor beam at stringer point nearest the center comes from 15-ton truck when one 12,000# wheel load is over the floor beam and a 3000# load 14 feet away. This makes a concentration of 12,000 + 3000 x ~ = 12,300# 20 I = 50 = 0.327 ,28+125 RL = 12,300 (5.75 + 11.75 + 14.75 + ,20.75) + 28 =.23,300# 234,900 = (23,300 x 13 •.25 - 12,300 x 6) = bending moment due to live load. 71,800 = (234,900 x 0.327) = bending moment due to impact. 3352 = (150 + 1&) 20 + ].5.2 = equivalent dead load per 4 foot. (152# section assumed) 3.27,000 = (3352 ,£. 2) (13 •.25 x 14.75) bending moment due • to dead load. - 47 - 633,700 = total bending moment 12 x 633,700 = 18000 S S = 425 (a 33 tr -152# wf beam is sufficient) Equivalent live load = (480 x 12 x 0.94) = 600 lb. 9 per foot. Total panel load = 600 x 20 = l2,000 lb. Impact = 0.30 x 600 = 180 lb. per foot. Total panel load due to impact = 3,600 lb. Total 15,600 lb. Concentrated load for moment = 13,500 x 12 x 0.94 = 16,900 lb. 9 Concentrated load for shear = 19,500 x 16 x 0.94 = 24,400 lb. 9 Dead load from each floor beam = 3352 x .28 = 46,900 lb. "2 Lateral Forces: Assume 150# per linear foot on unloaded chord Assume 300# per linear foot on loaded chord Assume a lateral force of 200 lb. per linear foot acting 6 ft. above the floor due to moving live load and wind pressure upon the load. Load per panel = 500 x .20 = 10,000 lb. on loaded chord Load per panel = 150 x 20 = 3000 lb. on unloaded chord - 48 - 1st coefficient = 1+2+3+4+5+6+7+8+9+10 = 2i 11 11 Bottom Chord: Wind Load Stresses p= /0, ooo#. tan ()= ~o :::- 0.834-; sec f)-= I. 305; ,: Psec e.:= 1/80:f! / A~ / Pian e =: 8340# Stress in chords = coefficient x P tane Stress in webs = coefficient x P sec e A double system of bracing will be used in each panel, but each diagonal will be assumed capable of resisting tension only. The maximum shear in any panel occurs when every panel point to the right of the section is loaded. The total wind load = 110 kips. Design: Diagonals Maximum unsupported length may be taken as Ii feet less than the diagonal distance c. to c. of panel points be cause of the support which is provided by flanges of main girders and gusset plates at connection. 1 = (1.305 x ,24 x 12) - 18 = 357 in. - 49 - This unsupported length is fairly large so it will be considered as each member developing tension only. The net area required is 65,.200/ 16000 = 4.07 Q n using 7/8" rivets at 3" spacing. Panel ab Net area furnished by two 5" x 3t" X 3/8" angles, with 5fl legs back to back. Net area furnished is 80% x 6.10 = 4.88 lJ n Number of rivets = 65,200 = 11 rivets (Field driven bOlO rivet value in single shear) 12 will be used. 6 in each angle. Net area for panel bc 53,400/16000 = 3.34 D " Area furnished by two 4 f1 X 3" X 3/8" angles is (.2 x 2.48). .80 = 3.97 D n 53400/6010 = 9 rivets (10 will be used - 5 in each angle). Panel cd 42,700/16000 = .2.67 0 n Area furnished by two 3!" x ,2!" x 3/8" angles is (2 x 2.11). .80 = 3.38 0 n No. of rivets = 42,700/6010 = 7 (8 will be used - 4 in each angle). - 50 - Panel de 33,200 ;: 16000 =: .2.08 a" Area furnished by two 3" x :2-~" X 5/l6n angles is (.2 x 1.6.2). .80 =: 2.59 c" Number of rivets = 33,.200} 6010 =: 6 rivets (8 will be used - 4 in each angle). This same size angle will be used in panels remaining as it is the minimum angle allowed for bracing. Chords: Panel AB 41,700/ 16000 =: .2.60 an required A 10" [15.3# will be used although furnishing more area than necessary. (4.47 a nG .) (4.47-2xO.24) = 3.99 °nN. Panel Be 75,100/ 16000 = 4.70 0" required A 1011 [.20# will be used (5.86 °nG.)(5.86-2xO.38) = 5.08 On N. fane1 CD 100,100/16000 = 6.25 Dn required A 10"[ .25# will be used (7.33 - .2 x 0.53) = 6.27 sq.in. N. Panel DE 125.1/16000 =: 7.82 !J 0 n required ... 51 - A 10" [30# will be used (8.80 0" G.) (8.80 - .2 x 0.67) = 7.46 sq. in. N. Panel EF 141.8/16000 = 8.85 LJ " A 10" [ 35# will be used (10.27 [J" G.) (10.27-2 x 0.82 = 8.63 Q" N• Panel FE' Pn 150.1/16000 = 938 • A 10" [35# will be used (10.27 On G.) (10.27 - .2 x 0.82) = 8.63 sq. in. N. See diagram No. 12 for sections. Design of Side Span: Stringers and floor beams of the same section as for the main span will be used. Lateral Bracing The lateral bracing in the plane of the lower flanges must provide for all the lateral forces due to the wind and to the effect of sway, including the wind force which is normally assigned to the upper flanges. Wind force = moving load equal to 30# per a, on one and one-half times area in elevation. It x 30 x 5 x I = 225# per linear foot. - 52 - Also a 200# load asa lateral force is acting against the live load. Total horizontal force = 425# per linear foot. The lateral system moving panel load is 425 x20 = 8500 lb. A double system of bracing will be used, but each diagonal will be assumed capable of resisting tension only. First coefficient = 1 + 2 + 3 + 4 = 10 = 2 5 5 P = 8500#; tan-&= 20 = 0.834; sec-tt = 1.305; 1/5 P sec 24 ~ = 2220#; P tanf!r-;707o# • Stresses in chords = coefficient x P sec~ The total wind load is 42.5 kips. A minimum size angle of 3" x .2t" x 5/16n was used for all diagonals which furnished a net area of 2.59 sq. in. - 53 - Conclusions In combining our influence diagrams for dead load moment, our initial equations are used. The first term (l-k) x or k(l-x) in the equations for the main span is equivalent to the moment at any section, treating the arch as though it were a simple beam. The sum of the values for the last two terms Ml(l-x) and 1 M2 ~ produces a straight line diagram for a symmetri- 1 cal design since Ml and M2 are antisymmetrical and the totals are identical. In superimposing the diagram for the simple beam moment upon that for the end moment, we obtain points of zero stress at the intersections of the two diagrams. It is evident that there can be a considera- ble error in the values for direct stress and the de- sign will not be materially affected. The tension in the tie causes direct stress in the hangers, but the values are negligible. We can have considerable error in assuming the areas of the arch ring and horizontal tie and our design will not be affected a great deal. The arch ring area may be in error as much as 50% and yet the values of the redundants be affected but by a small percent. - 54 - The method of evaluating the equations for the redundant forces may be simplified a great deal by assuming panel lengths in the side span equal to those in the main span. Then, the panel length as a whole may be used as unity. The values of moment ordinates must then be multiplied by the length of the panel in feet. The values for the horizontal thrust in the tie is correct in either case. The values of thrust do not depend upon the span of the structure, but the thrust is analagous to the shear in a beam. This is evident upon inspecting the H-in fluence lines for the different designs. The values vary only slightly and are entirely independent of the span lengths. The change in span lengths has a decided effect upon the value of the redundants Ml and M2. This is in accordance with the fact that moment is dependent upon the length of the structure. As the length of the side span is made to approach the main span length, the end moment at the intermediate pier produced by a load in the side span approaches the end moment at the same point produced by a load in the main span. This increase in length of the side span tends to increase the negative end moment and so de crease the positive moment ~t the arch crown. - ~5 - Increasing the side span past a certain point affects the type of section used. A plate gird er section may be used economically up to lengths of 125 feet. Above that box girders must be used. Above a certain point box girders would prove uneconomical, and it would be feasible to use a truss for the side span. In Design No.4·, the main span is 325 feet and the side spans are 200 feet in length. The side span section would of necessity be composed of a box gird er or of a truss. Above the point where a box girder section proves economical and a truss section must be used, it would prove more satisfactory to use a truss for the main span also. The design as followed would be equally applicable. - 56 - BIBLIOGRAPHY McCullough, C. B. and Thayer, E. S. Elastic Arch Bridges. New York City. John Wiley & Sons, Inc., 1931. pp. 8-14; 33-42; 125-141. Spofford, Chas. M. The Theory of Continuous Struc tures and Arches. New York and London. Me Graw Hill Book Company, Inc., 1937. pp. 1, 57, 58; 184-.203. Krivoshein, G. G. Simplified Calculation of Static ally Indeterminate Bridges. Prague, Czechoslov akia. Published by author, 1930. pp. 77-80; 86-95; 110-113. Parcel, J. I. and Maney, G. A. An Elementary Treatise on Statically Indeterminate Stresses. New York and London. John Wiley & Sons, Inc.; Chapman & Hall, Ltd., 1936. pp. 121-122; 320-338; 403-420. Cross, Hardy and Morgan, N. D. Continuous Frames Of Reinforced Concrete. New York and London. John Wiley & Sons, Inc.; Chapman & Hall, Ltd., 193.2. pp. 247-270. Sutherland, Hale and Bowman, H. L. An Introduction to Structural Theory and Design. New York and London. John Wiley and Sons, Inc.; Chapman and Hall, Ltd., 1930. pp. 184;·217-18; 231-37. Seely, Fred B. Advanced Mechanics of Materials. New York City. John Wiley and Sons, Ind., 1932. pp• .283-.284. The American Association of State Highway Officials. Standard Specifications for Highway Bridges. Wash ington, D.C., pUblished by the Association, 1935. pp. 165...,205. Kunz, F. C. Design of Steel Bridges. New York and London. McGraw Hill Book Company, Inc., 1915. pp. 312-366. - 57 - BIBLIOGRAPHY Hool, G. A. and Kinne, W. S. Movable and Long-Span Steel Bridges. New York and London. McGraw-Hill Book Company, Inc., 1923. pp • .218-258; 393-482. Hool, G. A. and Kinne, W. S. Structural Members and Connections. New York and London. McGraw-Hill Book Company, Inc., 1923. pp. 199~203. - 58 - £.VALUATION or EQUATIONS DES (GN NO. 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SI o. .t8 /.1.0 3. ..,. 005 /1.0 0.6 o.oq /1:0 1.1 0.111 12.0 J.'l 0./8 11..0 .l.t. 0./6 12.0 / q I 5~ ~ =¥~6~,i'-IJ~I====1===5""'iloeo8·.;a;=t.=;5:=:So"':7~. ='==f.=2:;='1:;=3""i.Cf~=7..,.",'7i'tJir =i,c1==;8~.="~9 -~ 10~. I~W9-1 1871*='1='1'+.t.o=oF,.,=/=.=.2,=7=*=====- hl -'1-5 .7 ¥- = 3"'*=2=0=.'=2.=1=.:;=2=.=1==6+= .. 61.'l 6 " i 3 I 83/. 0-' I Pa..,el ~,,. t£., Panel PDUlt LB' Pa/Je/ /?/",t L9 ~he/ Po, n t L,.. t M.s.I'7 Spa/? ~~~:_, Or cI'l"af~j~:: IkP #8.7 0,85 12.0 10.)., 1'1 .,.H no 5'7 I ~ 77\ I /20 51..t.1 1'1.31 "-.0 11;'./ 6.~7 IJ..O 'IS.1. 3.0~ J1..o 39.' /~.O i lo 8.80 _ J1.o 105-" 1 oJ.'!7 I 12.0, 1f7." 3.06 I /:2..0 36.7 /.3.:l9 1.z.O 15'I.S 'I. 75 93.0 L'IO 8.8" 120 106.~ 16.7' 13." I 80.. :l·bo 1:J.0 31·~ .1.71' /2,0 "I'.' 7. '15 12.0 '3.0 L'9 ;,bO I~·O .,.,.2. !"./O ,ul 73.l ,3.6:3 Un 93.h 0.;z8 1;1.(1 ~.-f- 3.03 12.0 3'. If L'I! ... ~·A 1.2.0 51.t; ~.~-fI}.1.D 51/.5 3,.5~1 l:l.o -'1;,.5 0,1/:0 JU) //.1) 0·8$ 12.0 lo.:t I',: -t 7'1 1).0 -'1.' I 3./8 1.t.0 ~8..2. l:t. 79 I~.I) 3~.~ /.-98 /:<.0 11.8 0.5.3 n.o I, o.·/,l. 1:(0 1/.0 ' I 1.17 IUJ J~.'f I· DZ '''.0 I"-.Z 0.50 1:1.0 h.D 0·10 11..0 .L~ o.~'1 /10 '1.1 0.13 I~.o 8.8 0./13 11.0 7.b ~.~" /:l.O 3., 0.05 I~·O /.'1 " ,:".' 1:1,1 9.'1 0.9h 1:Z.O 11.1 0.83 /~.o /0..0 0.'1'0 IA.D· -9.8 o.o~ 1.z·0 o.f "i- .'.i,Q I~.O 8.3 lo.83 , 1/2.0 10.0 O·7~ 1:l.D 8.8 o.J5 11.0 '!.:It 0.06 12.0 0·7 " I" 39 I I1.C '1.7 i o· 'llB j i lUI 5.8 0.1I~ /2.0 6.L 0.2./ 17..0 t.5 0.06 1:1. 0 o.b _, I" .,,9:l.b '~"I2.1; ~"J~/-2.1.61r !I.n.7 o¥08.-1f 3~.f6 18."1'1' -'8lf'''' .U8./~~.lJ'f 3.'10 JIJ"I.7 iii;!! H.Bo 1.32 - _. ,2eZ,.ooi==rr=/h.=O=l INFLUENCE ORO\NATES FOR DEAD LOAD MOME.NT AND MAXIMUM MOMENT FOR D'E.s \GN DE5\GN No. \. PLATE I' P"rH~L H ORDINATES I DE AD ST~E'sSlKIf') LI'IE" 5T~E'S (KIPS)I M, 0 RDIN"Tl'~ Mol''IliHT M, LOI'ID \. 0 AD pO'NT -,-- I == + - 'N l<'P. + - ." WI.' + - --+ - + - :1-- AR.C,H A ..... G SEC.TION L, 0.0,"2. 55.0 3.11 12.." 0·7/f JI.l~ t:JZ. , , L~ O.IO~ ~b.' 6.1'" 11.0 1.30 '!.'IO '111. 1 I L3 o.I%.If 5b.1 1.tJ5 /2.0 ,.49 8. 'So ~3.'J , I L.., 0.0'1" 55.0 5.'" /2.0 ,.13 h.4f 3~.%.1 O.I6'!i I 'Jt.3 ~.lf" I 11.0 /.86 I Q.5.t. "11./ L" I I L'1 ~ I 63.1- 2~.oa /2.0 JB7 I 15./13 %/.3 o. '" ! 0 I Lg 0.')70 I b:l.5 35.bO 11- bBIt , 11,.92- o57.S ~ 6/.3 17.1.0 11.0 ~r1..f I 'tb •.., "'l' L9 0.110 /5.1" I\' /o,Cf3 L,o o. Cf / I I 5'1.<) 5'/.15 /2,0 11.80 7/2.' I L',o o.q" I 5'5 5f./5 /1.0 ,0.'IJ 5.'0 35/·' I 0.710 bL3 "I1.~o 11. 0 '1.1'/ I.b6 103.0 L9 1- .. -r1 L' e o. ''''0 .2.') J5.00 12.0 ".8 II D.H ;Ud 1--18~ --l L' /:,3.'1 23.08 /2.0 f.37 0.8h 5lfS , 1 o 3"4 t..'~ 0/55 j, 'f. 3 q.% 11.0 1.86 o.bB ,/3:1 , MIN. S EC"-'ON = 85 DIOG. O.oq-t 55.0 5,1"- 1~.0 I. I .3 0.35 leu L'I MA~.5Ec..TION =: ll%"G. L'3 O./:l.4! 5/:d '7.05 /2.0 l.¥t 0.'110 :lo.2. L''). 0.108 5b// 1>.14- Ilo 1.30 0.,/0 ~2.8 0.01.1., 55.0 3.'11 /'LO 0.14 01' lA.7 '" L', ~- T 2'1fo.5 5 'I. ,(" 11lf.5 "'33.3 (7'£ IVS10N) (re illS/ON.) DIAec.r 1"1 ••. L.l.I"'}o",.~r 1'1A-. """.~. 0" ro ;Ji'II"[L , 5'MPL[ . ~ (, ~. Mil;'. D. L.I"I."f"T co...e ••· L040 1I"'lP"ClT M,u. l"Io",.,.r s-.-,",~. 8[ A 1'1 M, -- 11 • 7,- M, M, f-I pO,Nr f"lOMc,.,r .I. r .,. ~SEte ~'TIVa: PO~lrIV. - 1 i Po ""''"CoA rIvE ,Jlil ",r, 'I' PoS,T,,,. + - - + - + - I - --- L, + :2.l38 1.4'1 /121 101/ ~:?O.9 , 8'7.8 25h.0 5'1. , I %"0 85.' 1955.Q I 3Ylb ~5.l 37"1.8 1/'1.5 I 220/." Ll. Y'91:l 1150 qn 355.5 3%8.0 140'.0 T I 151.~ L~ • ~ 37" 1'1-1 3bB5 30'/ 5'18.1 55'7.3 JI'l.S 11/.5 ~"U· l:tall.' I . L't • tH8 991. I -Mol> .He8 :tIfJ.'1 7'i8.3 ISH .t1'l0 I I 'f7.lt> 337.6 158'.., I ~ Lb t to.lto I 145 l.IH mil }507 10'4.3 813.0 I :Jo.iJ 171.7 MS 5:l.S '1/18/i ... 34'\·0 ,,~., ~ I.t. blH 19'1-1 H.l./ S'fl./ .:u'l.l. /f89 I'#~. 3 25~0 I 3a.Q.O L7 ./1/51' , S 6ii10 hi 3/H1 '#08., 2.50.8 /1'1.0 218.7 180.• q,~a. +31~O Le f 1"'810 I 115 "'3' 8&2" Lt5 b13~ /ooe3 10'10 ~89." :z.'1S.-l .HQ.7 b/.'f %1'.5 II°.! S."i "3050 Lq • /11..31, I /:1.5 bIB 1078/f ISo'l ~7 Alhfl 21.«S .t5.0 ~/O.o 1-11-8 81(t.2.. t ;l4l8.0 =_J..'o 1t 16'1Zb 5ECTlON SUMMARY DESIGN NO.1 PLATE 12 .sPEt'I~'CArIONS: A;"ERICAIII HS'O'JI. Sr4rl H'~"~IlY OF,-.e·,-.LS \: lr 1ft Q UN,T STRESSlS·. TCNS,ON_ IB,Ooo-/""; COMP.-.I5.000-~(+-)%-tI/D" " :1 "i ' t 300 --1Ft. LOAD£" C I ~ .",. "'Clr" DutaL., l..O"I~~"'~·"'AL lATI!:~IIIL' LOAD 0 .. ~OO·lrTAcrINC • H; Ae."" ROADWAY LONI;"';.:':!NIll- C r"",r I l''l~AGTr 50/.L ... /:l5(R"y.o) .J ~ • , 0 ""r ~ ) ~ l. ~o en qo of. I ~O'-Oll ~ALE I" : ~ I S ... ~/. ",' ">'" :t/.o' :lo.5 ' oI0.a. ' ft. 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