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Vw fir... . .1 .. .1 —.~... . «A! . ..\.v1o..: . .; . u M. .. .l...) f . « . . . «m «’5’.» : . . : . .. . . 1 ...?w h. .. .. . . ....3 . . .....Bze ..«Q.34. . .J; l I. « 4Ler 7 1. .3: I « l .. . . .. . .. . . .x. ...m? .v . 1 . 3 .. L. - I.u«/ |.-, «A .r . :« .ar. . . «III \ I v SUDN r:I ”VARY Nit-n LHIHL IN BACK OF BOOK Design of 3 Rigid Frame Concrete Highway Bridge A Thesis Submitted to The Faculty of MICHIGAN STATE COLLEGE or AGRICULTURE AND APPLIED SCIENCE '0! Wayne A. Groesbeek N Candidate for the Degree of Bachelor of Science June 1941 'l'l"|Ei1'3"a (ff/J. / INTRODUCTION The site chosen for this bridge is on U.S.~12 three- fourths of a mile east of Marshall, where the highway crosses Rice Creek. The highway is one of the oldest in theState and as in many cases, the roadway and its appur- tenances have lagged behind the ever increasing demands of modern traffic. The present bridge is of concrete slab-girder construc- tion, has a twenty-four foot roadway, and was built nearly twenty years ago. In the last few years the accidents and fatalities occurring at this bridge have mounted steadily. National defense, too, will make its demands, since Michigan's Fort Custer is located on U.S.-12. The heavy military traffic which now uses this road requires wider and heavier roads and bridges. Rider roads and bridges mean the discarding of the present bridge and it is for this particular site that the following design is preposed. The rigid frame concrete bridge was chosen here for two reasons, (1) because of its economy, (2) because it offers design principles different from those covered in any of the courses in concrete design. The economy afforded by the rigid frame cannot be ig- nored. It requires only about 60 per cent of the material required of a bridge having members of uniform cross—sec- tion. The abutments required by fixed arches of the same span are massive and require, in addition to more concrete, 1313.1“?0 greater excaVation. The shallow bridge floor lowers the overpass roadway and thus lowers the grade line. This reduces the amount of approach fill. Not only are rigid frames low in first cost, but also maintenance and replacement charges are considerably less. Widening of rigid frame bridges can be done easily, with slight alteration of the existing structure. AC KNO ‘éz-LEDG as. EN T The author wishes to take this Opportunity to thank Professor 0. A. miller of the Civil Engineering Depart- ment for the valuable assistance and advice he has given.” And also, to thank those in the Bridge Design Division of the michigan state Highway Department, and those in the local office of the Portland Cement Association who have extended their very courteous assistance. “I QQWhansU «QWQ w 881K . as Sow, sub ..‘N 5 1. DESIGN OF THE FRAME A. Frame Dimensions, Axes, and Coefficients (ZJvV3 ~5flzyhear(fIsfahqyx/5229snryo ‘=flCZ7001%i£%4 : CW— 2. .S/é§29&s(3'Aslééya(Zfikzea4 gaéEZ'flfvsioc¢95125 (:ofi£?0£mbvf§ fGr.£ku/IVEA/ Iéb~ [Decf' I’I' I £50- 2.00 , 350—550 C/:' 2500 -';K’ 4/- Ihfb 6.20 4.52:4.60 J‘= A100 /4.00 7.; =4.60 r5= moo (/-I;.I;)= la64 5/sfl9 autos. Lc/JL- JL 42,15" T Deck Coefficients: - d' : ggso - 1.50 a 1.53 1.50 From Chart II 3 8 15000 r8 8 11000 S x l, or proportional to 15.00 x $1.502:5 = 1.00 L 50.50 Carry over factor, r, equals: £§ = 11.00 8 15.00 5"." 0067 End hall Coefficients: - d. = 5.50 " 2000 = 0075 2.00 Sb x (1 - the 10.64 10.64 x The relative 1.00 x l.OO+5.16 Parb) : 14.00 x (1 -_L4.60 12 ) x I 14 e OOX6 e 20 T, X tflH or preportional to (2.00)3 g 5.15 16.625 stiffness in per cent at "b" is then 100 a 16.24 for the deck 5016 X 100 = 83076 for the wall 1.00+5.16 The charts used in selecting the coefficients are found in "Analysis of Rigid Frame Concrete Bridges", a publication of the Portland Cement Association. 8. Distribution of Fixed End Moment fl 41.67', éia4 C [{Fh'reo’ Em/ Moaeo/s o 3 o [JahvaaAk/ " 0 GD .0 O 0 (any 0Ver " —/o. 90 o o 0 Z Deathly/ed " '. /. 77 *9/3 0 «(/9 Carry aver " o 0 —/.oo - ./9 3(0/lr/ria'u/ed " o o C) 0 Cbhcy'Chmv' " ~.A3 ‘O 0 0 4(0/3/riéu/d " + .02 4 .l/ “6’4. 76 + 64.76 ' 1 Term. MOMENTJ - 9.24 + 3.24- a.. ..a’ Computations lst Cycle - 100 x .1624 a - 16.24 in be - 100 x .8576 ; - 83.76 in be Total moments at end of let cycle - 85.76, +‘83.76 at "b"; at “c“ zero 2nd Cycle Moment carried from b to c equals - 16.24 x .67 a - 10.90 Distributed Moments at "c" are -r10.90x.1624 s -rl.77 in be +10.90x.8376 : -+9.15 in cd Total moment at end of 2nd cycle - 83.76,+ 85.76 at b -+ 9.13, - 9.13 at 0 5rd Cycle +'1.77 x .67 a +'1.19 - 1.19 x .1624 a -0.19 in be - 1.19 x .8576 : - 1.00 in be Total moment at end of 3rd cycle - 84.76, +~84.76 at b - 9.13, +-9.15 at 0 4th cycle " 019 x 067 :3 0013 +-013 X 01624 I +’0002 in be +-.1s x .8376 - #‘O.ll in ad If a fixed end moment of - 100.00 is applied in "ab" at "b” (1.6., in the end wall immediately below the corner Joint) and the joints are then released, the final corner moments become: - lO0.00-+-83.76-+ 1.00 : - 15.24 in "ab" at “b"; + 16.24 - 1.19+.19 = 4—15.24 in ”be” at "b"; + 9.24 and - 9.24 at ”c" (These relative moment values will be useful in the subsequent analyses by eliminating repetition of computa- t10n8)e C. Dead Load Michigan State Highway Department Specifications (1936) call for the following in regard to dead loads: ”For structures with concrete slab floors without sap rate wearing surface, a minimum allowance of 20 #/sq.ft. of roadway shall be made, in addition to the weight of any monolithically placed concrete wearing surface, to provide for future wearing surface." ‘ (The weight of concrete, plain or reinforced = 160#/c.f. in addition to the dead load of the frame an additional con« crete thickness of %” was allowed for an integral roadway wearing surface. No moment is created by the weight of the end walls since it is carried directly down to the footings. The longitudinal section through the deck is divided into an area 50 ft. long and depth of 1 ft. 6% in., weighing 230 p.s.f. The remaining area is subdivided as follows: 10 315/ .95 .65 .75 .6 5 . 55 area c226 5.05 2. 63 /. 0/ .2 2 Load IZJ’O 760 400 /.5’0 33 _ l i t l 1? .2 AL P: .af.3 _ - r Fixed end moment per foot of width Uniform lead: 250 + 20 = 250 p.8.f. 250 x 50.52 x .105 2 66900 Equivalent concentrated loads 1250 x 50.5 x .060 n 3165 760 x 50.5 x .130 = 4980 400 x 50.5 x .185 a 5740 160 x 50.5 x .210 1590 11 55 x 50.5 x .195 a 525 55 x 50.5 x .145 s 242 150 x 50.5 x .087 : 658 400 x 50.5 x .040 2 808 760 x 50.5 x .012 : 460 : 126 1250 x 50.5 x .002 - 82984 ft.1b. Using the values of 84.76 and 9.24 per cent determined in Part B the numerical values of the corner moments at “b" are 82984 x .8476 due to F.E.M. at "b" 82984 x .0924 due to F.E.M. at "c" The total moment when the deck is straight is: 82984 x (.8476 + .0924) = 78000 ft. lb. and produces tension in the outside of the corner Correcting for curvature of the deck (rise of deck centerline = l ft.l% in.) the final corner moment is £50625 + 1.1253.5 : 78000 x 16.625 + 1.125 The crown moment for straight deck centerline can be 75500 ft.1bo found by statics. The total positive moment assuming a simply supported deck is: 1250 x .05 x 50.5 = 5155 760 x .15 x 50.5 = 5750 400 x .25 x 50.5 = 5050 150 x .55 x 50.5 = 2650 35 x .45 x 50.5 3 750 250 x 25.25 x .25 x 50.5 = 79600 96955 ft. lb. 12 The difference between this moment and the negative corner moment created by the same loading is the moment at the crown of the frame with straight deck and is: 96955 - 78000 = 18955 ft. lb. Correcting for curvature of deck the final crown moment (tension in bottom of deck) is: 15.625 + 1.125x.5 - 18955x 16.625 + 1.125 ‘ 18550 ft. lb. Check the final crown and corner moments Corner Crown From Chart 1 ~75448 7'14140 From Moment Distribution ~75500 -t18550 The total dead load of the frame (1 ft. wide) is: bearing surface: (20+—6.25) 54.5 = 1450 Back: 1.50 x 50.5 x 150 3 11560 Deck: .555 x 2 x 50.5 x 150 = 5040 Corners: 2 x 5.5 x 2 x 150 = 2100 halls: .5 x (5.25-k2) x 14.75 x 2 x 150 = 11610 Footings: (5.00 - 2.00) x 5 x 2 x 150 =__gZQQ_ 54240 1b. say, 55000 1b. The vertical reaction on each footing is .5 x 55000 = 17500 1b. The horizontal thrust at the footing, when the deck is curved, is m : 4540 lbs. 16.625 The crown thrust also equals 4540 lbs. since all loads are ymw'fy leads. 15 D. Live Load Live load shall be the standard H-RO truck train: (michigan State Highway Dept. Spec.) Impact: I : L-+ 20 : 70.5 = .218 6L-+ 20 525 Load per foot of width due to front axle: 8000 x 1.218 a 1100 lbs. 9 Load per foot of width due to rear axle: 9 Maximum moment is produced at the crown with the con- centrated loads at the center of the span. [loo “"4!” I pita 1 ‘!J piao zoar ' 7 liflr L+——~zza: zdukr———a— p.32..— . Live Laaa’ /or Max. Ckvwwt ’WhM»on¥ Fixed and moment per foot of width: I! at pt. 1.0: 4400 x 50.5 x .160 55600 46000 ft. 1b. 0". O O on ‘ x .04 O O '1' 41750 43970 ft. lb. 14 Corner moment after distribution (straight deck): at 1.0; 46000 1.8476't43970 x.0924 - 590004‘4060 a 45060 at 0.0; 46000 1.0924-t43970 x.8476 = 4260-r57250 = 41500 The corresponding horizontal thrusts (straight deck) are: at 1.0; 46060 . 2690# 16.625 at 0.0- 4 6 . 2500# ’ 161625 ”an {km 4U«¢\ l I 4awo k. ‘ ‘ 1 l .Lfic'load'lév‘ I1M;C;umn fl¢bw. zfib *fl” 'M A 32 ca 2300 The laws of equilibrium require a force of 2590 - 2500 n 90 lb. The vertical reaction at "a" then becomes: 4400 x 25.66.11100 x 57.85.+90 x 16.625 3 3200 lb. 50.50 50.50 50.50 Vertical reaction at "d": 50.50 50.50 50.50 4400 x 26.65_+1100 x 12.65 - 90 x 16.625 - 2500 lb. [too 4400 5” 4a”;\ 4nmo L. J ‘f 5' 45 4Lém'loau’/Q» Max. Crown AM. 2 500 25.90 L H 2 J 0 0 The total positive crown moment assuming a simply sup- ported deck: 2905 x 25.25 - 1100 x 12.60 = 75500 - 15860 = 59440 ft. lb. The crown moment (straight deck) is: +-5944O - 45060 T 45.0 x 16.625 g-fl7128 Correcting for curVature the crown moment becomes: 17128 x 16.6251'1.125 x .5 a 17128 x .968 _ 16600 ft.1b. 16.625i'1.125 The final corner moment, after curvature correction, is: The horizontal thrust at "a" is: 16.625 or; 2590 x .968 a 2510 lb. Check the final crown and corner moments: Corner Crown From Chart I 40680 15042 From Moment Distribution 41700 16600 Maximum moment at the corner is produced when the concen- trated loads are at point .625 (pt. 0.0 being at the right hand corner). #00 4100 825‘ L'F—3" sax Fl 302: 32:9 Jfiéi—————a4 -4(2 ° 4(h4!.(oav’ fQ-fi4hc C2vwar I460» 16 Fixed and moment per foot of width: at pt. 1.0; 1100 x 50.5 x .110 6100 at pt. 1.0; 4400 x 50.5 x .208 46150 52250 ft. lb. at pt. 0.0; 1100 x 50.5 x .011 = 610 27010 ft. lb. The values of "k” for the 825 1b. load aporoach zero so closely that the load was not considered when computing the fixed end momenta. Corner moment after distribution (straight deck): at 1.0; 52250 x .8476+27010 x .0924 1' 44300+2500 a 46800 at 0.0 ; 52250 x .09247‘27010 x .8476 3 4830*?22900 = 27730 The corresponding horizontal thrusts (straight deck) are: at 1.0; 46800 2 2820 lb. 16.585 at 0.0; 27730 ; 1670 lb. 16.625 [/00 «00 { 468°0\ ; £ ‘27730 K 1’ 4Lfimvlaailflw- lime Gmmmw_f¢hn £680 . zero H ) 17 The laws of equilibrium require a force of 2820 - 1670 3 1150 lb. The vertical reaction at "a” then becomes: 4400 x 50.2 +1100 1 _4_4.‘2+825 .2 +1150 x 16.625 50.5 50.5 50.5 50.5 8 3975 lb. Vertical reaction at "d": 4400 x 20.5-+llOO 5.5-f825 50.3 - 1150 16.625 50.5 {1005 50.5 5005 \ [/00 M00 (- sz‘bo ‘ ‘ ffnmp, 377" 7 :7! Lu}: (and /~—- ”OX. Carnal“ Mom. 20“ ‘ I670 392: 2"” The total positive crown moment, assuming a simply sup- ported deck: 5596 x 25.25 — 1100 x 18.95 - 4400 x 4.95 = 90800 - 42650 a 48150 ft.1b. The crown moment, (straight deck) is: -+48150 - 46800i-575 x 16.625 x 10900 ft. lb. Correcting for curvature, the crown moment becomes: 10900 x .968 - 10560 ft. lb. 18 The final corner moment, after curvature corrections, is: 46800 x .968 a 45400 ft. lb. The horizontal thrust at “a" is: 16.625 or: 2820 x .968 = 2725 it. Check the final crown and corner moments: Corner Crown From Chart I 46560 10120 From Moment Distribution 45400 10580 19 E. Change in Length of Deck and Horizontal Displacement A relative change in length of deck may be either a shortening, due to temperature drop, shrinkage, and outward displacement of footings, or a lengthening due to tempera- ture rise and an inward displacement of the footings. Assume the frame subject to a deck shortening due to (a) temperature drop of 60° F, (b) shrinkage corresponding to a shrinkage factor of .0002, and (c) outward displace- ment of the footings equivalent to a contraction coeffi- cient of .0002. The coefficient of thermal expansion is .0000065. The shortening per unit of length is: 60 x .0000065-+.0002-+.0002 = .00079 The total shortening in the span of 50.5 feet is: .00079 x 50.5 3 .0399 ft. This is equivalent to an outward displacement of .0200 ft. at the footings. when analyzing the frame by moment distribution begin by locking the Joints "b” and "c”. The static conditions from which the fixed and moment at "b" is determined are illustrated by the following sketch. //‘//// L A sleazy ' 1' ~21? .JL it can be shown that: Fixed end moment at “b“ = E % D x Sb x %; x (l-rarb) £§x106x122)x.02 x 14.0 x (1[12)x25 x (1 - 4.62} ) : = 264500 According to Part "B” the formula for both corner and crown moments, when the deck is straight, may be written as: Moment 3 264500 1 I(1.0000 - .8576 - .0100) - .0924.1 8 264500 x .0600 = 15880 ft. lb. Final moments, correcting for curvature: Corner: 15880 x 16.625 32 = 15920 ft. lb. {16.625't1.125 Crown: 15880 x“ <16.625 : 14880 ft. lb. 16.625-rl.125 Or the moments can be obtained by substitution in the empirical formula: Corner moment I 4.35 x 106 x HXDfi 1 (Crown thicknese)2 (H 1154 4.35 x 106 x 16.625 x .02 x 1.5 (16.625+-1.125)2 10520 ft. lb. 2 Crown moment 3 10520 x 16.625jjl.125 = 11020 ft. lb. 16.625 The values obtained by moment distribution will be used. The horizontal thrust is: 16.655 82" [”0 7‘0 400 4,1 4 21 [6‘0 3 3 l i 7 I fi 2“ ‘7' mm 4540 5 73.91")" K‘ A ' ’ x 17’9" I6 72' 4:40 aJL ' H 26".? .7 17.5w Dead Load 25/0 3209 (I've l 000’ /or Max. Crown Mamm/ (510/6qu Pram M] [/00 4400 16600 ) , 5466 r 22 [/00 . k75fifl/ "—_‘ 6 are 2 7‘! 3376’ (lbs [000’ /or- Max. (om er I490: 00/— 6538er Pmen/edj [3920 24 14660 ) ea 8 038 Deck 55 ar/en I?) and Roz/0'3; DIE/o/acemeawl F. Earth fressure Moment distribution gives satisfactory results, for a frame subject to earth-pressure, only when the walls are relatively high and slender compared with the deck, (when the wall height is greater than .4 of the span length). Passive earth pressure can be developed, in addition to active pressure, by flexure under live load. Tests have been run on actual bridges and the results showed that very little passive pressure was developed. The earth pressure moments are usually a very small percentage of the moments caused by the dead plus the live load and can be disregarded. The dead and live loads cre- ate tension in the intrados, while tension is created in the extrados by the earth pressure. It is, therefore, more conservative to disregard the earth pressure moment at the crown. G. Dissymmetry and Sidesway Under unsymmetrical loading the moment distribution method gives horizontal thrusts which apparently do not satisfy the statical requirements for equilibrium. For instance, in Part D, when the live loads are at the center of the span, the corner moments for straight deck are: at point 1.0: 45060 ft. lb. at point 0.0: 41500 ft. lb. The corresponding horizontal thrusts are: at 1.0: 2590 lb. at 0.0: 2500 lb. The algebraic sum of the horizontal forces is not zero, which means a tendency for deck "be" to move side- wise. If the frame is to remain with the points "b" and 0 I!" vertically above a and "0“, respectively, sidesway must be prevented. This can be done by the application of a horizontal force of 90 lb. in the line "be". The force of 90 lb. will increase the vertical reaction at "a" and decrease that at "d" by an amount equal to: 90 x 16.625 : 295 lb. 50.50 The condition approached most closely by the ordinary rigid frame highway bridge is that in which there is no sidesway. Greater corner moment is given by the assumption of no sidesway. 37 The following computations illustrate how the results obtained by moment distribution can be adjusted for the assumption that sidesway is permitted. The added force of 90 lb. creates the same horizontal thrust at both footings, it being 45 lb. The total horizontal thrust at both footings is then: 2590 - 45 : 2500-?45 = 2545 lb. The corner moments at Joints ”a" and "b" are: 2545 x 16.625 3 42400 ft. lb. The maximum negative corner moment when sidesway is prevented is: 45060 ft. lb. 88 B. Shears The maximum shear and unit stresses at (l) the crown, (2) the corner, and (5) the top of the footing were com- puted in the following section. (1) Crown. The shear due to dead load, deck shorten- ing, and symmetrical earth pressure is zero. Sidesway was assumed to be permitted, making the shears in the deck equal to the shears in a simply supported beam of span equal to 50.5 ft. Shear due to live load was found as follows: — 50.5 e \\\\\\\\\\\\\“r\\l J 2 ‘\\\\\\\\\\\\\“\\1 fiyaérnre [has {Qr.52awecul (bnlhr N\\ It can be seen by reference to the influence line, that the maximum negative shear will occur with the 4400 lb. load at a “d1" distance to the left of the midpoint. The maximum positive shear will occur with the 1100 lb. load a distance of "dx" to the right of the midpoint. 29 Maximum positive shear: +—1100 x %-+4400 x 11.25 x 6 . see-+980 = 1530 lb. 25.25 Maximum negative shear: 1100 x 11.25 x 6 - 4400 x a g ~246 - 2200 = 25.25 - 2446 lb. The corresponding unit stress is: V a .1. s 2446 = 14.6 p.s.i. id 12x2x 16 8 Investigation of the shear based on the assumption of side- sway prevented is unwarranted. (2) Corner. The total dead load from face to face of end walls is: Wearing surface: (20+‘6.25)48 - 1260 Deck: 1.50 x 150 x 48 = 10800 Deck: .635 x 2 x 48 x 150 a uéfiQQ 16860 lb. The maximum shear due to dead load is: 16860 3 8430 lb. 2 Deck shortening and symmetrical earth pressure produce no vertical shears. 50 Shear due to live load was found as follows: ‘1 5.003 " hill/end? [II/)6 [or 549m" (7/ (8// [00/ It can be seen from the influence line that the maxi- mum shear will occur with the 4400 lb. load a distance "dx" to the right of the left end. The shear will be: +4400 x l+825 1; 20.51.5300 1: 6.5 a 5160 lb. 50.5 50.5 The total dead and live load shear is: 84301-5160 3 13590 1b. The corresponding unit shearing stress is: V a V a 15590 - 35 p.8.1. :- 12 x 1.x 5 8 (3) Tap of footing. The maximum shear equals the horizontal thrust at the support. The dead load thrust is: '75500 : 4540 lb. 16.625 Maximum live load thrust is produced by the same load arrangement that produces maximum corner moment, and is: 16.625 51 Since the horizontal thrusts produced by earth pres- sure and deck shortening counteract the thrusts due to dead and live load, it is more conservative to disregard the earth pressure and deck shortening and take the maximum shear as: 4540+~2725 s 7265 lb. The corresponding unit stress is: V 3 V 8 7265 = 2808 p.801. jd I§xfix23 8 Allowable I 60 13.3.1. 52 I. Stresses at Crown and Corner The tensile and compreSSive reinforcement was computed using a 5000 lb. concrete, and the corresponding unit stresses were checked to see whether the allowable working stresses, (fc z 1200 p.s.i., fs = 20000) were exceeded. Moments and thrusts at the midpoint of the deck are: Axial homent Thrust Dead Load +'18650 4‘4540 Live Load 4-16600 -+2466 Change in Deck Length +-;4889 :_§§§ Total +-49830 +6168 Eccentricity with respect to centerline: 49860 g 8.08 ft., say, 97 inches 6168 The tensile steel area for this moment and thrust must be somewhat less than that required when the axial thrust is disregarded; namely: A3 - Ms 3 49850 x:1g_ : 2.14 sq. in. fsjd 20000x2x16 8 A tensile reinforcement of l in. square bars spaced at 6 in. will be chosen. sith this reinforcement, axial thrust still disregarded, the extreme fiber stress in concrete is less than 900 p.s.i. This stress will be raised by the addition of axial thrust, and compressive steel of l in. square bars at 12 in. will be chosen. The depth to the neutral axis in the concrete section equals d (r2pn-I-Gm)? - pn) -.-.-. 5.82 when d a 16 in., A’ . 2.00 squ. in., n a 10, and s is infinity. Addition of the axial thrust will tend to increase the effective depth to, say, 6.20 in.. The sec- tion coefficients with the estimated value of Z a 6.20 will be: Estimated Section Coefficients Cor- Corrected rection i Value A = 12x6.2+(10-l)x 1.004'10x2.00 : 103.40 -4.67 § 98.75 Q . %x1216.22+9.00 x 2-r20.0316 : 569.00 r4.67x6.01 £540.90 1 . izlsxe.25+9.00x22+20.0x162 = 6109.00 -4.6716.012 3 5940.20 i [33 I 9700 " 05X18 I 8800 If Z were correctly chosen, it should satisfy the equation: Z : I+Ex§ g 6109.0+88x569.0 : 56209 : 5.81 Q+£XA 569.0+881103.4 9669 Using the second value of Z a 5.81, correct A, Q, and I used above for the discrepancy in effective concrete area, which equals 12 x (6.20 - 5.81) g 4.57 sq. in. the center of which is at a distance of 5.81-ké-(6.2O - 5.81) a 6.01 in. below the extreme concrete fiber. The final value of Z is: Z 3 $94002+ 88 I. 54009 8 5081 in. 54 s~2 9 2 540.9 - 5.48 A 98.7 e : 3+8 2'.’ 88000 ~fl5o48 = 95048 Ig g I -132 = 5940.2 - 5g0.92 = 3015.2 A . 1'8 0 7 then; fc : P x e x Z 3 6168 x 95.48 x 5.81 g 1110 lg 3015 fs" = P x e x (z-c n g 6168 x 93.48 x 3.81 x 10 g 726 Is 3015 fs' 3 P x e x (d-Z)n s 6168 L 2;.é8 a 19.12“; 19 lg 5015 : 19500 Checking against numerical errors: kd : ed = 1110 x 16 g 5.81 3 Z fc+fs n 11104-1950 Moments and thrusts at the corner are Axial Moment Thrust Dead Load -78500 -+l7500 Live Load ~45400 +‘3975 Total «120900 -t21475 Eccentricity with respect to center line: 120900 3 5063 ft. or 6705 in. 21475 If the axial thrust is disregarded, the steel area required in tension is: 20000 :91 x 37 8 55 Carry the 1 in square bars at 12 in. from the top of the deck around the corner and add in each interval between these bars one 1-1/8 in.square bar, making a total tensile steel area of 2.2780. in. per lin. ft. at the corner. No compression reinforcement will be included in the following stress determination. The depth to the neutral axis in the concrete section equals d (rfigpnrt(pn)2' - pn ) g 10.1 in. when d a 37 in., A g 2.27 sq. in., n z 10, and E is infinity. Adding the axial thrust will tend to increase the effective depth to, say, 13.5. The section coefficients with the estimated value of Z 3 15.5 will be: Estimated Section Coefficients Correction Corrected Value A .... 1211505+1012027 = 184.7 "' 1308 170.9 2 Q - %x12x15.5 +22.7137 3 1935.0 - 15.8x12.95 1756.5 1x12x15.53+22.7x372 : 40940.0 - 13.8x12.932 58630.0 3 3 6705 " .5 X 37 2 4900 E3 Second value of Z is: z = 309401-49.0 x 1955 : 155640 = 12.85 1935-r49.0 x 185 10995 Correct A, Q, and 1: 12 x (13.5 - 12.35) : 13.8 sq. in. 12.35+—% (13.5 - 12.55) = 12.93 in. The final value of Z is: Z : 38860—f49.0 x 1751.: 12.5 l757+—49.0 x 171 56‘ 171 e : 49.0't10.27 - 59.27 lg a 58650 - 11512 = 20580 171 fc a 21475 x 59.27 x 12.3 = 761 20580 fs' : 21475 x 59.27 x 24.7 x 10 : 15500 20580 Checking the value of Z kd : 761 x 37 : 12.5 : k 761+'1550 The unit stresses of £0 = 1110, fc 3 761, fs = 19500 and fa a 15500 are conservative, since they include the combined effects of dead load, live load, temperature drop, shrinkage and spreading of footings. They will be reduced by the effect of symmetrical earth pressure, which has been disregarded. II DESIGN OF THE FOOTINGS A 5000 lb. concrete was used in the design of the footings. The bearing power of the soil was taken as 2 T Vper sq. ft. and the net pressure was assumed to be uniform. Load: Dead load reaction (Part C) per foot of width - 17500 I Maximum live load reaction " " u " z 4585 Estimated footing weight 3 1900 Total 25785 Area footing : 25785 ; 5.95 sq. ft. 4000 Bending homent: The footing is 6 ft. long and projects 24.0 in. on each side of the frame as a cantilever beam. Net pressure p = 21885 : 5650 lb./ sq. ft. The section at the edgg.gg the frame is a rectangular beam 1 ft. Wide. Maximum.moment = 312 g 3650 x124)2 : 87600 in.1b. 2 12 x 2 __g 87600 d 2 Kb = I 197 x 12 2 6009 inc, Say, 601 in. The steel is at the bottom of the footing. Assuming that projecting stones, clay, muck puddles, etc., may spoil ‘2 in. of concrete, and adding 3 in. for damp proofing, and 12 in. for the key on the bottom of the frame, the minimum depth becomes: h 3 d+'17 : 25 in. 38 Shear: At the face of the frame: Maximum shear V - wl : 3550 x 2.00 - 7300 lb. Maximum shear stress v = _1_ = I 7500 : 114 . bjd 12x1;6.1 8 Allowable shear v z .03 fc' ; 90 p.s.i. (special anchorage of long. bars) Diagonal Tension: The first failure crack to extend com- pletely through the footing will start at the face of the frame and extend outward at an angle of 45°. The diagonal tension will be computed at a distance of d : 6.1 in. from the face of the frame. The shear is due to the net soil pressure on the extension of the footing beyond this section. , V = V 3 §§5Q_L_llia ’ 85 3d 1211;6.1 ’ s Allowable v a .05 fc' : 90 lb./ sq. in. (special anchorage) height: The footing weighs 23/12 x 6.00 x 150 z 1725 lb. Assumed wt. 3 1900 lb. Steel: As : H : 87600 : .82 sq. in. fsjd 20000 1‘1 x 6.1 8 39 Use 3/4 in. round bars at. 6 in. A .88 Bond at the face of the frame: u g V 3 7300 : 290 253d 4.711236.1 8 Allowable u : .080 fc' z 240 lb./sq. in. (special anchorage) (Change d to 8.0 in. making h = 25 in. 'The new values, using d = 8 in., are: Shear : v : 87.0 Allowable g 90 (Special anchorage) Diagonal Tension: v = 57.8 Allowable : 60 (ordinary anchorage) height: 25/12 x 6.00 x 150 g 1875 lb. Steel: As : .627 sq. in. Use 5/8 in. round bars at 5 in. A : .745 u a 222 Allowable g 240 (special anchorage) Bars are to be 6 ft. plus 12.5 in. or 6 ft. 14 in. long. 40 E 095(9): 0/ I59 Mbjwa/A’ T I , =3000 A :Zaooo paw/3 1?: A97 J"?! ”“ ’0 [6:3u Fir—(“z [load t-IJ 1 E?” Earl-A Pru: are .' l—-———8 :1~ // L 4L5: ‘Lmd ‘ ‘ 30‘ 41 ab 3 8 :2 15085 ft., say 14 ft. ten 500 52000 x 1.215 g 5540 1b./ft. of width (live load) 11 Combined dead and live load pressure on 11' slab : 55404-125 (10 in. slab) = 5655 lb./ft. 5665 x 11 z 2060 1b./eq. ft. at 14 ft. down 19 0 3 0°, ¢ . 50°, w z 120 lb./c.f. Ce : Cos 6 Cos 9 okCosZQ- 0083 B Cos 9+V00529 - 0032 d 3 1. llzskflEZEEZ. : l -.5 : .553 . .1+yT?7fiT— 1+ue Side P. . 2060 x .555 a 682 lb./sq. ft. Earth Pressure : ce whz : .555 120x(18.25)2 : 2 2 =6660 1b. Pressure: on PWm” _._T§§§2_ 2900 4.87 ‘31 1 42 9560 'i' : 6660 x 18.25+ 2900 x 4.25 T “2"— 3': 4.67 ft. Vertical Forces: Soil W‘e of earth = 11067 X 17017 I 120 = 24050 It. stem 3 1.55 x 17.17 x 150 2. 5420 wt. base slab : 14.x 1.08 x 150 : 2270 29740 1b80 Pt. of application or vert. forces: y .-.- 2270 x 7+5420 x 1.67+24050 x a. 7 3 210100 “W 2 40 29745 : 7.34 ft. X = 4. 87 x g 10 56 85-60 W43" Toe diet. 8 7.54 - 1556 a 5.78 ft. e 3 7.00 --6.78 = 1.22 ft. Pressure: ‘=§(1:§—-)~ : 29740 2125 (1.523) a 5255 lbs. 8 2125 (.477) a 1015 lbs. Sliding: f 8 29740 x .33 = 1.03 43 Overturning: f = 29740 x 7.33 a 4.7 9560 x 4.87 To overcome the small factor of safety against sliding a friction key was put on the bottom of the base slab. Friction Key: 7* 6 A08 4.5 Ce :: 006 0 006 6 +}’0osz 6 -1 cosgé - 1+.5 : 3.00 003 9 -7“;;;2€ - 006 fi -. Total Passive Earth Presure: p 3 Ce “32 = 5.0 120 x (5.623 I 5650 ' ' 2 Pressure at 1.08 ft.: 2 P 3 3.0 120 X (1.08) 8 ’2'? $11.38. Total Active Earth.Pressure: P ' .533 120 X (22.7542 = 10550 lbs. 2 44 Sliding: f 8 (29740+—1 x 4.5 x 150}.35‘t5440 . 1.50 0350 Resultant of passive earth pressures: 5650 x 5.6 . 210 x 4.91—5440'56 x = 10560 - 1060 : 1.75 n. W4 Bending Moment: BM - 5440 x 2.75 - 15000 ft. lb. Effective Depth: ' 7500, dz}??? I I 157 II 8.73, sayQin. D: 9¥3¢ 12 in. Shear: V's V a 5440 g 67.7 533 nggxg Allowable 8 60 (ordinary anchorage) Area Steel: As 3 pbd 3 .0115 x 12 x 9 2 1.22 sq. in. Use‘g ” round at 4“ A.- 1.80 This particular size and spacing was used to per- mit the use of the stem steel as the steel for the key. Bond: u . V g 5440 g 85.7 0 0253? X ‘3 Allowable I 120 (plain bars, ordinary anchorage) Stem: 45 Earth Pressure: = 0. she a .55)! 120 x (1mm)2 : 5640 1b. ”'2' “—2—” Bending Moment: - 5840 x 17.17’: 63400 ft. lb. ._1§_. Effective Depth: 3% : 13.0 D a 151'5 a 16 in. Area Steel: I pbd 8 .0115 x 12 x 15 a 1.76 sq. in. Use ‘3' in. round at 4 in. A I 1.80 Bond: a V g 5840 : 62.2 gold .. x x Allowable : 120 Shear: : v - 5840 g 42.3 STE 1W x '6’ Allowable = 60 Economy can be effected by using bars of several dif- ferent lengths. Cut off 2 bars out of 5. The area remaining is 1.80 8 .60 sq. in. This occurs at a distance of 10 ft. from 46 the top. The bars will end at 10 ~ 40 x 7/8 = 7 ft. 1 in., say, 7 ft. down from the ton. Cut off half of the remaining bars. As a 1469 = .5 sq. in. This occurs at a distance of 7 ft.62 in. from the top. The bars will end at 4 ft. 5 in. from the top. The remaining bars run to the top and are spaced at 2 ft. apart. Temp. Steel: In front of wall As : pbd : .0025 x 12 x 15 : .59 Use % in. round at 6 in. A - .40 In back Use % in. round at 12 in. A — .20 t - Toe Slab: Ffi—421;~{ 11’53 l g Fawuwuny'on 739'Jlm6 T ‘0 l3 1 3255 l Maximum Bending Moment: BMab:507611x%+1591%—x22/591651 1x 3.; = 1509 47 Effective Depth: 1509 d a 197 a 2.76 use 5 in. D:3+3+2:81n. Area Steel: As : pbd : .0115 x 12 x 5 : .41 because the toe slab is so small, the heel slab will be computed and the same d used for the toe slab. Heel Slab: plenum on #93/ J/ql ,L; M‘ 7 :4- C b 1 [87.5“ i l T 13" a! J. 1 [d5 2 .942: Maximum Bending Moment: BMcd : 1875 x 11.67 x 11.67 - 1015 x 11.67 x 11.67 2 2 - 1910 x 11.67 x 11.67 a 15500 2 5 Effective Depth: 15500 d 3 197 g 8.8, say 9 in. 9+5+2 : 14 in. 48 Area Steel: As a pbd : .0115 x 12 x 9 a 1.22 sq. in. Use 7/8 in. round at 5-5/4 in. A a 1.25 Shear: Shear due to 1875 lb. load : 1875x " “ ” 1015 " " = 1015x " " " 0 to 1910 15. load . 82x2 vX-x a 18751 - 1015x - 8212 01.: 860 - 164x d1 X ‘ 5025 ft. v 860 x 5.25 - 62(5.25)2 = 4520 - 2260 2260 v 3 V = 2260 a 23.9 bjd 12:: Z x 9 8 Allowable s 50 Bond: 2260 - 50.5 u ___l__~ - ' 25%67 ‘ 5.711x9 8 Allowable a 120 Too Slab: Area Steel: As . pbd . .0115 x 12 1 9 . 1.22 sq. in. Use 7/8 in. round at 5-5/4 in.‘ A = 1.25 Shear: v z 165 x l-5076x1-159 x 6- = 2995 v -._E_ : f 2993 a 31.7 bjd 12x1z9 8 Allowable - 60 Bond: u = V : 2993 3 66.7 iojd 5.7xZ;9 8 Allowable a 120 49 1. 2. 3. 4. 5. 6. 7. 8. G! 0 Bibliography "Specifications for the Design of Highway Bridges,” Michigan State Highway Department, Jan. 1936 “Analysis of Rigid Frame Concrete Bridges," Port- land Cement Aesociation, 1936 “The Rigid Frame Bridge,I Arthur 0. Hayden “Reinforced Concrete Design Handbook," American Concrete Institute “Reinforced Concrete Structures.“ Dean Peabody, Jr. “Retaining Walls," George Paaswell “Specifications for Design of Highway Structures,“ Ohio Dept. of Highways. 1935 "Theory and Practice of Reinforced Concrete,” Clarence w. Dunham "V. I Qfl'. Lao-IO” .- llllllln lull‘itlbl .wlr. ‘ . W “In Vllhw hrlllh‘lu ‘ AFI..I.. .'.| ‘I m. m We .2. .1 c... n M mm a .a/ _ ma 1 m ., Jr 5 a _ Mo 5 _ m” “m ..6 e .. _ mo u a 5. _ _ _;,- TJ !ll|”l._ _ ._ lllllll . _. _..l_ - E L 6. m E D K .... E R C E a .m B _IJ lllllll _ _ _ ... _ _|L _ _ , _ . _ W _ q _ _ _ W ._ W _ w _ O ||||||||||| I— W - fi‘ I; - f4 . " . quthL _ \g «— z ’—a-1 . ' 2 ’60:; COFFEE 0 . Varsesrop “\2 \ _ . 1 \ '? ( JOINT Darn/am ’6‘” X I f Ia ’ ‘ MNGh/ALl 4' FRAMe' 2-” ["27 1| 1| 41—! if‘r _F H ! 3"" _—1~| I 3 at 54:: re ANCHOR -fi===d= T[ l ' / 4M4“ re FRAME 'I c" . C“ ’-_—-————-.-O . O O O _ - . \‘4 ___3. . . . /é‘¢@/e' l602. .CoPPe‘I? ,5" Wartrsrop .121 5% 6) 6 231' u OI ; 23’9" 2—— 16'6 ’1 Li’s"— o ...]... _. —-—...—- —M-_.._.—.- _«O— ‘ ...... -h J‘s" 1L 1k ’4,’ . . ./ MCH/GAIV .5 ran: COLLEGIE’ DEPT. OF CIVIL ENG. WINGk/ALL 6:05:43 sex, HA 3cm: / "3' Dar: $7£q/¥/ —'"‘-~~<1'*-m~——-—-—— —. .1... _ 1. - ‘ .... _ v-9“-..r.“ “M’_ ‘_' _-_‘ , ~ _. , ‘ , . 7 .1 _ _ __ ...,» __ _ dfi -... 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I 1 aefif’ = ‘ 56672 515 3-} I} 542: G) 3 ” I Memo.“ 5 r47: (04 1. £6: Depr. or Cum. Elva. BRIDGE Lemur 4‘ Bot/um; DE 774/1 3 564“ A: SHOW Dar: Ski/4V 64201-305“ )h/A. ““W’m‘m'” .--- ,4- __._._..._.1. -_ —_ 'v—‘—1”WO—Ow~‘ Wfifl——«A-H-_ _. __ 1. -_ -fi' ,- __..__..—1 - ~ - - ...... -o-v- ..- - — 1. __, ----..“r- VI MICHIGAN . . . i DVI . . n'uu.a.t .... , .47.. v \»\r ..4 . . .M§ . “cut". ‘. ..aalu avg-uh . ... . h « 5.x. .. gawk. I». 13--‘.-Iy_| I-v": q)” w'v-I' . ."' “y. '1'! VI“ ..‘I'.‘ LI' . 1%., , ..' I, ' '3' . _ . . ' W V." . ‘ r‘ ' --v- I w , , , . I.“ .I .I I . ; _..;17..,.Il "III :1 I6-"- II .l__.I 7.4‘7‘I II. "Z'I {‘J'JP'II 74“" .... MICHIGAN STATE UNIVERS SITY LIBRARIES II III I II III III I .I I I; II' II I I“ I 312930360 8795 ‘JA_O‘ 5