'm. '3‘ v ‘ . v . . ‘ \1'.- _ ~ ‘I ., ‘ - ' ,. ., , w ' .. __ H '.'.. —_—',‘ .-_-- : —‘."'l_' H .H H ‘ ‘1 . A A . .ul' _ . ._ a” _ a ‘ ' ' h. ..I " l‘ ‘ ' ' 9,: ." . O I ‘ ' .. 0‘ W . \O’: \‘, I c . ‘ , ' I l' . H. 0 ' . u' .' .w - . (J I ‘ .u :l L‘. r' b 1'... ‘l: I ' m— . . .I:"-1?vfy‘u7‘zl\oE,W-T’1Fritf :{gP-Jrgrygmw‘bl . _ . L ‘ - ' ' my... . . ’ I. ‘ R'GlD-i FRAMET’BRlDGE ° " ~ 4 F,0,R.-FARM5FM¢* '- . W . '1' .Thé'si‘g. fit)?” {he chrgc ‘of B. S. ' ._Vincent Vanderburg ‘ ‘ 4937. - . r . .,... 1//1//11.u/1//. “11.1611/1/1 111/ /.1¢/.11/ . 1111111117 . 1.11./ . .1.111...1...11.1. 1/111 ”171/1111M110fl/ u . mm .m I E - e . w m .. a .m u. d rm m .m mm 0 6 0m I u u .9... a a.» m Em .m Wm fl . y at m 0 0 mm m wmemm v... m. r . fir. F. .u .m fh e a M. e e m dr .1 9.. m c mm J .mrro. ma. m...“ U .m as m m m w. v as A E m w m c m TH 5.515 INTRODUCTION Fara.Lane is a public road bisecting the property of’Eichigan State College at East Lansing. The college campus and farms owned or leased comprise a total of more than nineteen hundred acres. When the first of this property was acquired there were few traveled highways and it was probably not intended that the lane through the center of the farm should ever be used as such. It was maintained as a typical farm lane for the purpose of’noving cattle, farm machinery, etc., from one field to another, and from the farm buildings to and from the various fields and pastures. A simplerflone‘fihss.hridge that had been previously used elsewhere was pur- chased and placed in position across the Red Cedar River. With the expansion of college property south of ht. Hope Road the gates across Farm Lane were eliminated. This Opened it to public use and it became the most direct route between East Lansing and a prOSperoue farming section. It has been used for many years by the general public as a public highway. In 1956 the B. P. A. through a cooperative project placed a hard sur- face on the roadway which has tended to increase the amount of traffic. The present bridge was first condemned as unsafe for highway purposes by the Engineering Department in 1895. Its use becomes increasingly pre- carious Iith the passing of the years not only due to the deterioration of time but also due to the high speed with which heavy trucks traverse it. The State Board of Agriculture has repeatedly requested the State Highway Department to build a substantial bridge at this point. In 1929 the legislature passed an enabling act authorizing the highway Department 10832”? ' \Jlmrnum‘ll ‘I .'1.|': ‘ * ‘ ..M‘ a». an rt“. to take over the construction and maintenance of roads and drives at the various State institutions. The Highuy Department has not as yet agreed to undertake this project. be college has placed earning signs on the bridge warning all tref- fic that it uses the bridge at its on risk. But, in spite of these pro-- cautions there is a moral responsibility on the college and the State to provide a safe bridge. be time is not far off when a new and adequate bridge must replace the present one. the college campus now extends to the river at this point. The area east of tern Lane is used for the hater Carnival and other similar pur— poses. Tho new bridge in order to be in keeping with the fine appearance of the college buildings and the beautiful landscaped campus must be of an attractive design, and should combine the maximum beauty with the necessary utility. iith this in nind I an mggesting :1 Rigid Frame Concrete Bridge for this location resembling in general style the beautiful Rigid Frans Con-— crate Bridge in [rape Park, Frooport, Illinois. This type of bridge is economical to build, easy to maintain, stun-w and rigid, and can be made graceful and artistic in appearance. It repre— sents a newer type of construction that should be of interest and value to the engineering departnent of the college for instructional purposes. flies. specifications describe a bridge eighty feet long clear span, eiehtoon foot roedsey with a single eight foot sidewalk with the clear- ance above nor-a1 water level of fifteen feet. An adequate number of borings have been taken to demonstrate that '10 difficulty will be encountered in the construction of the foundation. With this type of construction it is of the utmost importance that careful laboratory checks be made of the concrete poured in the monolithic structure. The specifications used in working these necessary problems were ob— tained from the Michigan State Highway Department. finch valuable informa- tion see obtained from the Portland Cement Association through Br. J. O. Brenna. Grateful acknowledgment is also made to Professor illen of the R. S. 0. Civil Engineering Department, and to J. G. Hartin of the Portland ' Cement Association for their many helpful and valuable suggestions. Amour-m4.“ ’7 “rem-nu Kin—u: i ,rynh . _ 3 ‘~‘ A! At!“ Jud-wagsjv 5&1 ”magnum-ares..- -“a. e$qufl hmd\w .1 I l 40’ b£fi ”"'-——— -———_-’TT .0 52+- C/em" 604/2 80’ _r4||..\.a.m:l.l flare—~— w a ...... we Problem 1. Frame Dimensions. Axis and Coefficients. 2 Stiffness (c is fixed):- 12 x 25%.. = 1.9 one b “(’,;r”’/ c 5 i \5fifi‘neos (”cf/Ls h/hjedyf-987Xg7g‘g = /7/ °( 90 g 80’ , Cocffidenfa 79F 1970/ [447/15 Coe/y/‘b/énl‘s [fir Dec/i:- M- M 3/14" 1 5.3.3 ‘ P75 0/. - 2.33 '955 64.59 [3:64:43 6=/2:00 81,8 /.3-0 #5 55: 7‘43 5/" 6.0 654/ varéj= 9.87’ 4 La , 1, J Deck coefficients:— dl : 5.58 - 2.55 _-_ ’90,, 2.35 From Chart II S : r3; .1. SxL’ 2.0 8.0 Carry over factor, r, equals 2.. = -33. 3 2 88 z 509 .- 3 or proportional to 12.00 1 big). = 1.9 .666 Sb 2 13.0 ra Se 2 4.5 rbSb = 4.5 3 ‘Airin'._‘/I“W. 11W" 'zv“ "_ ‘ ’a - H» J mm .9. ‘1".‘:'-‘.’ ‘ \. . \ I - - 1232):. I=9-87x-I-=9.87 x—l-rrw -15.:11— :— Sb L( a”) ( 1315.9)1. L x 5.555 = 17.06 21.33 The relative stiffness in per cent at "b" is then 1‘9 x 100 = 10. for the deck 1.9 + 17.06 17.05 _ f. . 1.9 ‘ 17.06 x 100 - 90.-or the wall Problem 2. Distribution of Fixed End Moment 0 0 | ,QQI I /O I c O H 00- 00 find and mom. 0. O. 30 " / O. 0/3 fr/bufeo’ mom. 0. 0. 0- O- ' Cor/y - oycr mom. ~66 0. 0 o. O/Sfiy‘bU/‘ed .. + .66 +5.95z 0. + #356 Cor/y- orcr mom. 0. 0. 13.920 - 04356 O/S/r/bufea’ u o. o. 0' 0. @{Zarrj/ ~0/c'r mom. -—0. 0.38 7485‘ O. 9_ o. /Sfr/'bafco/ .. + .0026 74 + 0256 7¢ ”9059 290.89 75/‘4/ Mommfs 15.965876‘ 261965874 a a/ computations - / lst Cycle - 100 x .10 = - 10 in be - 100 x .90 2 ~90 in be total moments at end of let cycle - 90 + 90 at "b” zero at “c“ zero F ~.'.'!_'.A.."G4'il-'-' .1'4. ' ‘ A (t' 1" - v ‘.’.‘,r"_‘.wkm~.~ero‘rv.' ‘ 1 .. £-' c‘. 437‘ {ti Cy W e11 2nd Cycle - Moment carried from b to c equals - 10 x .66 = — 6.6 Distributed moments at ”c" are + 6.6 x .l : + .86 in cb + 6.6 x .9 : + 5.94 in cd total moment at end 2nd cycle — 90 9 90 at b, - 5.94 9’+ 5.94 at c 5rd Cycle + .66 x .66 = .4356 - .4356 x .l : .04556 in be - .4556 x .9 .59204 in be total moment at end of 5rd cycle 90.59204 9 + 90.39204 at b " 5.94 9' '9’ 5-94 at 3 4th chle .04556 x .86 = -.0287495 + .0287493 I .1 +.00287496 in cb + .0287496 1 .9 Z + .025374 in cd If a Fixed l5nd Moment — 100.00 is applied in “be" at "b" (i.e., in the end wall immediately below the corner Joint) and the joints are then released, the final corner moments become:- - 100.00 + 10.00 + 90. - .4556 f .59204 f .04556 - 9.60796 in 'ba' at 'b' + 9.60796 in ”be“ at "b" + 5.965874 and 5.965874 at "C'' (These relative moment values will be helpful for subsequent analysis by eliminating repetition of’conputationsv) .. ' I""".\' -r1 ———"hm"-M T' .'I~ ' - .,. ‘r-M'fl"\ _ H—“A.__._.-‘-...' . . A: ”'5 J: lu'n. “it A gm; «9 ‘1 ~Vu do I Problem 5. Dead Load The weight at the end walls is carried directly down to the foot~ ings and creates no moments. E chigsn State Hirhwey Department Specifications call for an allowa once of 20i/sq. ft. of roadway for additional separate wearing surface; P133 a 'fi‘ additional thickness to provide monolithic wearing surface. 1/64. '- 2/. /3. 71.5” 4.0 ”0 4004/:- 53/50 4 950 - /, /25 600 A50 J” (Concrete weighs approximately 160i 1 cu. ft.) Fixed End Eoment per foot of width:- Unifbrm load:~ 2.57 (150) + 20 = 856 + 20 = 3765/sq. ft. 576 x 802 x .132 : 245,452 ft. lbs. El“ 3.9 Ste u v. Equivalent concentrated loads 5,150 x 80 x .05 12,600 1,950 x 60 x .125 19,500 1,125 x 80 x .16 - 16,200 600 x 60 x .20 = 9,600 150 x 80 x .167 - 2,240 150 x 80 x .14 1,660 600 x 60 x .1 ‘ = 4,600 1,125 x 80 x .04 5,600 1,950 x 80 x .012 1,670 5,150 x 80 x .002 : 500 516,042 ft. lbs. Using values determined in Problem 2, pages 6 and 7, the numerical values of the corner moments at "b'I are:- 518,042 x .9059 due to Fixed End hon. at "b” 318,042 x .05965 ' ' ' ' ' ” "c" Total moment when deck is straight is:— 518,042 x (.9059 + .05965) = 506,000 ft. lbs. and produces tension in the outside corner. Correcting this moment for curvature of deck:— (Raise of deck axis is 1.621) 505,000 I 21.55 4- .5 x 1.62 = 506,000 x 22.14 = 21.55~+ 1.62 22.95 506,000 x .965 = 295,000 ft. lbs. The crown moment for straight deck centerline can now be found by Statics. The total positive moment assuming a simply supported deck is:- _ _ .' .___.._.._..4 m" . nan-o“ xv “Ann n n -- .A 'r' .nvau‘a ...,_.._,, ., " _ " 02'1".“ ‘f‘m\"_.n v'_.-k1-‘ u.-mla 24-11 It. mate] lathe] C0 ’ a £93533- LC? ’38 of 0014:6; Wife: The difference between this moment and the negative corner moment 5,150 1,950 1,125 600 150 1/8 x 576 x 80 x .05 = 12,600 x 80 x .15 = 25,400 x 80 x .25 : 22,500 x 60 x .55 : 16,600 x 60 x .45 : 5,400 x 602 = 500,600 created by the same loading:— 581,500 ft. lbs. 561,500 - 506,000 : 75,500 ft. lbs. is the moment at the crown with straight deck. 10 Correct for curvature of deck and determine the final crown moment (tension in bottom of deck). 75,500 x.£1~55~+ -5 x 1193 = 75,500 x .965 = 72,800 ft. lbs. 2.1.055 + 1062 Checking on the final corner and crown moments will now be made by use of influence lines (Chart 1)- §:fl&3: S 80 Concentrated Loads Uhifbrn Load 5,150 1,950 1,125 600 150 150 600 1,125 5,150 576 .266 (Interpolating between .54 and .18 on Chart I) x 80 x .048 x 80 x .115 x 80 x .182 x 80 x .198 x 80 x .185 x 80 x .158 x 80 x .089 x 80 x .048 x 80 x .02 x 80 x .004 x 602 x .102 1| I! " 12,100 17,620 16,400 9,500 2,220 1,655 4,270 4,520 5,120 1,010 245,4§2 517,667 ft. lbs. _ 1 -‘A9?;:*“.. «rum-u...” ...‘ . 5! C h“ OVA. not i: The treated 13 +12: 5 {$5012 "3 Total moment when the deck is straight is:— 517,667 x (.9059 — .05965) : 317,687 X .96355 : 505,030 ft. lbs. and produces tension in outside of the corner. Correcting this moment fer curvature of the deck, the final corner moment is:- 21055 + e5 1 1.62 - a en! 00 - 505 000 x .9r5 - 3 000 ft. lbs. 505’ O x 21.55 + 1.62 ’ a “ *’ Total positive moment has been previously computed:- (page) 10 — 581,500 ft. lbs. The difference between this moment and the negative corner moment created by the same loading:- 581,500 - 505,000 = 76,500 ft. lbs. is the moment at the crown of the frame with straight deck. Correct for curvature of deck and determine the final crown moment (tension in bottom of deck):- 76,500 x.§1«85;r -§;£_ls§Z.: 76,500 x .965 : 75,600 21. lbs. 21055 * 1.62 Corner Crown From Chart I - 294,000 + 75,800 iron Moment Distribution - 295,000 + 72,800 - Jen ' l I ld’L‘M W’Y ~ 11' 1"“). I mixtw. 3‘“ 7?. 12 Total deed load of the frame, one foot wide, is:— leerinz eurfaoex- 20* x 80 = 1,600 Deckz- 2.57 x 80 x 150 : 27,640 Deck:- .55 x 5.25 x 80 x 150 = 15,000 Cornerez- 5.25 x 5.58 x 2 x 150 . 5,442 lellex- .5 x (5.56 + 5.55) x 18.54 x 2 x 150 = 24,800 hetingsv— (6.0 - 5.55) x 3.55 x 2 x 150 = 2,670 75,152 The vertical reaction on each footing is:- O.5 x 75,152 = 57,576 lb. any 57,500 lb. The horizontal thmet at the footing, whm the deck is curved, is:- 295,000 21.55 = 15,650 lb. The crown thrust also equals 15,850 1b., since all the loads are gravity loads. *Puture [coring Surface - flichigan State Highway Department Specifications. 15 Problem 4. «3/6Z) géTSC) .4/ c5 c5 C> xcn7 [J IIJH‘IILIIIIHTHIW] .376 mm,- /f 72,600 O l_¢gjfik29 (l 5. &9\\ .‘a29a9” zwfififl' GO (Shana/[Lowm/ ML v ' , l qQQ’ r 62500. -Liye loads taken from:- "Theory of hodern Steel Structures" - By Grinter - Page 107 - Article 115. These loads are considered by the American Association of State Eigheay Officials as the loadings designed for the various types of bridges. le have selected according to the specifications of the A. A. S. H. 0. the alternate loading or equivalent loading to take the place of the truck train for long spans. The loads are fer a lane 9 feet wide. 8 15 Alternate loading or equivalent loading is used. (15,500 lbs. for moment (Concentrated load : ( (20,500 lbs. fer shear ( Without Impact( (UnifOrm loading = 480 lb. per linear foot. I f#,_*_aii In 1501 2/4 16 - =.§9.__.:._5.9.=24.5or25 1‘9"“ _I z. + 125 205 i 6 §\\'\. . \\ \\ VN \\ ‘ \ \‘.. an. \‘ |\ .\\h \ ._\‘\ Ben and used values for: .\ .\ \\‘ ‘ Nhhh (16,875 lbs. for accent (Concentrated load : ( ' iith Impact ( (25,625 lbs. for shear \‘ \. :6 x h (Uniform load : 600 lbs. per linear feet (D ' aV°/‘(m* J lllllllllj 646K25\\\ AWL4U 66)AQWBACOGZ/ Man Mom. 02‘ Crown ZRQCL? J ’ . .¢Cy . , a, 600 laxinun moment is produced at the crown when the concentrated load is placed at the midpoint and the uniform load covers the entire span. The first step is the analysisjto determine the fixed and moment co— efficients by entering Chart 11 with d1 s .966. 15 Fixed End.flomentz- (9 foot lane) Uniform loadx- 600 :802 x .101 = §§Z$§$Q 2 45,000 221.999. .-. 25,200 Concentrated loadz- 16,875 x 80 x .168 9 68,200 ft. lbs. By using the values from Problem 2, the corner moment is found to abes- 68,200 x (.90592 + .059658) : 68,200 3 .965578 : 65,600 ft. lbs. (Straight Deck) The total positive crown moment assuming a simply supported deck 18:-- 600:40x.5x40: 9.0.5132 =55,500 .5 x 16,675 x 40 : 5579500 : 57,500 90,800 ft. lbs. The difference between this moment and the negative corner moment created by the same loading. 90,800 - 65,600 = 25,200 ft. lbs. is the moment at the crown of the frame with straight deck. Correct for curvature of deck and determine the final crown moment (tension in bottom of deck): 25,200 x W = 25,200 x .965 : 24,500 r1. lb. 21.55 + 1.62 A check on the final crown moment can be obtained by use of the in- fluence lines in Chart I. 600 x 602 x .021 .. £955.42 = 8,950 16,875 x 80 x .078 = 105,000 :11,680 9 20,650 ft. lb. ‘ \ ‘f‘filq .-“. '. ”3r". 5 16 The moments, thrusts and shears for the position of the live load that gives the laxinue moment at the croIn are shown. The corner moment when the deck is curved is:- M = , , 65,600 x 22.85 61,600 It lb The corresponding horisontal thrust is:- élaig : 2,902 lb. The vertical reaction on each footing is:- 6002.5180+.5116,875: 15.25.421.15,6001b. 5' ' /5,875 LIIJTIIILTTLILLIILIIIIllllj CWOC)‘;oernéhczv"/fixzt 45.9 60.90% 2764“ ,Akéaf/izwvn av‘c3ovvnn" .32“fl9” (SSAflnsncgy/éhucnvéqfl 26225.4; V (l . ‘4CY 636013 laxilul moment at the corner (point 1.0 in Chart II) is produced with uniform lead over the entire span and when the concentrated load is placed ‘\\ Vl|:‘.“_‘ x“..'.'. ‘.. ‘ \ ' . 0 .‘x H.‘ H. \ M \t‘.“ -\ - ‘\'\ \ \‘I fl ' 1‘ Nikki.“ 32""in V3.\\\\§.\\\¢\\§\;-; \ Q 77 I 1 f. "JG-1 .7 1,, . 5/4 5 ,r I,’ a ’z // we \ \\ § .\\§\\ \ 17 at or near point .625. The following values are obtained with the load of’l6,875# at point 0.625. Fixed End homent: (9 foot lane) dl : .966 At point 1.0 16,675 x 60 x .2 : 6193999 : 50,000 155,000 9 At point 0.0 16,875 x 80 x .1 15,000 Corner moment after distribution:— st 1.0 50,000 x .9059 + 15,000 x .0596 27,100 - 894 = 27,994 at 0.0 50,000 x .0596 + 15,000 x .9059 1,789 - 15,560 : 15,569 laximum corner moment (including uniform load) when the deck is straight. 27,994 + 45,000 x (.9059 + .0596) = 27,994 + 41,400 : 69,594 ft. lbs. Final corner moment allowing for curvature of deck. 2.1.55 4' e5 1 1e62 .- 69 9 -—_—-= 69 594 x .965 - 66 900 ft. lb. ’3 4 x 21.55 + 1.62 ’ -**—- Check by Chart I with 16,875 16., at point .625 600 :602 x .095 = 9933999 . 40,600 16,675 x 60 x .167 .2533999 - 26,000 68,600 ft. lb. The moments, thrusts and shears for the position of the live load that gives the maximum moment at the crown are shown:- The corner moment when the deck is curved is:- 66,900 x gé~5§ = 65,000 ft. lbs. The corresponding horizontal thrust is:- QLQQQ a 2,959 lb. 21.55 The vertical reaction on each footing:- 600 x .5 x 60 + .5 x 16,875 =.§£3391 = 5,605 16. M “a..- - ucxvu— .. -'. 1:. ‘1-‘ “I’- Sui In"!!! ' .'.‘ 18 Problem 5. Change in Length of Deck and Horizontal Diaplacenent 1 relative change in length of deck may be either a shortening (temperature drop, shrinkage, outward diaplecement of footings) or a lengthening temperature rise, inward displacement of footings). Assume that the frame in Problem 1 is subject to a deck shortening due to (a) temperature drop of 45° F., (b) shrinkage corresponding to a shrinkage factor of 0.0002, and (c) outward horizontal displacement of the footings equivalent to a contraction coefficient of .0002. The shortening per unit of length is:- ‘5 3 0.0000065 + .0002 + .0002 : .0006925 @: s. n. 6. Specifications) The total shortening in the span of 80 feet is:- .0006925 1 80 : .0554 ft. This is equivalent to an outward displacement of .02771 at "a" and "d“. When analysing the frame by moment distribution, begin by locking the Joints "b" and 'c". Figure below illustrates the static conditions from which the fixed and moment at “b" is determined. A =3/..3.3 Hymn-um; :; 1': ' .- - [fill-i9 ; ‘4 .80 ”L‘.A..A ‘ - "3 v a -.<.‘“-‘A1 K ‘7‘"..3 "I. 'M'Wu .34.. A. 19 It can be shown that:- F. E. h. at 'b' : Eli—Q x Sb 1 %§ 1 (1 - rarb) : (5 x 106 x 122); 0.0277 15 (1/12) x 5.553 1 _ .52 __. 786 000 21.55 x x 21.55 x ( 15.0 x 5.9 ) ’ F. E. E. at "b" = 786,000 ft. lb. According to Problem 2, the formula for both corner and crown moment when deck is straight may be written as:- fioment = 766,000 x:[f(1.0000 - 0.900 - .592) - .05957 : 15,700 ft.lb. .0174 Final ficments - allowing for curvature of deck — equal Corner: 15,700 x («——-§L&§§-——-—)2 11,820 ft. lb. 21.55 a 1.62 Crownz. 15,700 x ( 21-55 ) 12,760 ft. lb. 21.55 + 1.62’ N The effect of changes in the thickness at the bottom of the wall is relatively insignificant. Inserting numerical values in the empirical formula gives:— Corner moment:- 4.55 x 106 x 21-55 1 -0277fi x 2.552 = 24,604 ft.lb. (21.55 + 1.62)“ .00109 Crown momentz- 12,700 x El:§%_§:11§3.: 15,650 ft. lb. 1.075 ULTWG‘H _~ _ .315. .-e~mm::v'\.~ 'f .‘ .2 (a .1"; '.. '-‘ 2C .11 «ma-ngnxx ee- r .‘1 7* t "I ?“.m 12.11! 7 of». '1" \b + kfllélaof éflkyfiez&37 amm/ ' .ZZLAO" z€50V277'.(2¢y2émtznovzf éV‘4’ .CL5ZF¢‘Z% xi76367fi% (1' #fl . V_ Corner moment when the deck is curved is:- 11,820 x 3.14% a: 11,180 ft. lb. 22 85 Corresponding horizontal thrust is:- Mg I 524 lb. 21 55 Problem 6. Earth Pressure Rigid frames should be designed to withstand two groups of influences, (l) the ferces characteristic of continuous structures; and (?) the dead and live loads, tractive forces and earth pressure. The loads of group (2) are identical with those acting on ordinary simple-Span bridges with the exception of the earth pressure on the end walls. Earth pressure on abutments for simple—Span bridges is usually active pressure, produced by the backfill moving toward the abutment. In rigid frame bridges it is possible - at least theoretically - to de— velop some passive earth pressure by a moveaent of the end wall against the backfill. Tests are recorded which indicate that little passive earth pressure is developed; it may ordinarily be disregarded. Problem 7. Dissymmetry and Sidesway If the frame or loading is unsymmetrical, the moment distribution method as discussed and applied in the foregoing gives horizontal thrusts that apparently do not satisfy the statical requirements for equilibrium. Take the frame on page 5 loaded in 16,875 lb. at point 0.625 11th a lane 9 feet wide. The corner moments, determined in Problem 4 for straight deck are:- at point 1.0: 27,994 at point 0.0: 15,569 The correSponding horizontal thrusts are:- .' «don-.41; .- .. 5" . ‘ m ‘22-:- ‘2 ‘ 30’ > w 50' ‘r / 27,993h , .1 .592 . ,/ 0:369 @V~5%5&51n37 A5./?cvemv%z/ 2746”" 4131.1,“ 4 ’ j a/ . 74 43/3 54% 311.2%: a 1,515 lb. at ”a" and w = 721 1b. at "d" The algebraic sum of the horizontal forces is 1,515 - 721 : 592 1b., but it should be zero to satisfy the static requirement that the sum of the projections of all external forces on any line must be zero. This apparent discrepancy will be clarified by the discussion of sidesway Ihioh follows. It is evident that the deck I'bc" in figure shown above .111 tend to move sideiise relative to ”a” and 'd' whenever the frame or the loading is unsymmetrical, and also that a lateral displacement of “be" will set up moments at the corners. Refer to Problems 2 and 4 and observe that no fixed and moment due to displacement of "bc‘I was included in the analysis. t 55mm.mm.7r 2:. -. '- ';11.—~r.‘.nu.xs 2‘“Y.: < ‘4... .LX‘J-é’. ‘1’ l 5£hYAIRA“"1’)'vF '..~ — A — 25 The significance of this omission is that points 9b“ and "e“ have been kept in their original position vertically above 'a" and "d"; or, as it is called, sidesway of the frame has been prevented. It is obvious that an external force must be added in the line ”bc' then horizontal displacement of "be" is to be prevented. The laws of equilibrium require that the force equal 592 lb. The loads, reactions and deflected axis for the frame in which sidesway is prevented are shoun in the above figure. The force of 592 lb. in "be" increases the vertical reaction at 'a' (and decreases the vertical reaction at 'd"), thereby make it equal to 1.31687 x§9+59233l2§§= 9 80 80 1,875 x .625 + 592 x .268 = 1,551 lb. or the two assway and no sidesway - the latter is obviously closer to the actual condition in the ordinary rigid frame for highway bridges.f The assumption that no sway takes place is therefore preferable, especial- ly since it gives the greater corner moment. ‘It shall be illustrated, honever, how readily results obtained by moment distribution.may be ad- justed to allow for the assumption that sidesnny is permitted. Consider, for example, the frame in the figure below analyzed by moment distribution, in which a force of 592 1b. is required to prevent sidesway. Eliminate this ferce by adding another equal but Opposite force in "be”, simultan- eously displacing the deck horisontally in the direction from “b" to '6". Determine the F. E. H. and then by moment distribution - in a manner sim- ilar to that in Problem 5 - the final corner moments. This general pro- cedure can often be simplified. In the frame in the figure below for wiri— _. . - J z'..'-umuzr': "" A “(4.449. «firm .x. .n >|