DESION OF A REINFORCED concurs BRIDGE «on 1339 " ’ . ' NEAR GRAND LEDOB The.“ in the Bound I I. WCHIOAN STAT! COLLEGE _ . R. 6‘ My”: _ ’ .1941 " 1m nw4»;- - . -. _ .» " '7 ‘ -—-J «QA — _‘ J:- _ _ - ‘ . - . 1* . . . I ' 'I I 1' ' I} ' u - { I ‘5 -'§ Design of a Reinforced Concrete Bridge on M—SQ Near Grand Ledge A Thesis Submitted to The Faculty of MICHIGAN STATE COLLEGE of AGRICULTURE AND APPLIED SCIENCE by R. G. Myers .3 Alina», Canidate for the Degree of Bachelor of Science June 1941 TH 9:539 ("1' f 3 . I” l’ Introduction This thesis covers the complete drawings and design for a bridge on‘M 39 near Grand Ledge. At the present time there is at the site a 25 foot by 18 foot reinforced concrete bridge. It is at the bottom of two comparatively steep slopes and is too narrow for the present volume of traffic on the highway. it is my object in this thesis to design a bridge with improved approach grades and wide enough to carry the present and future volume of traffic. Keeping in mind future road work the grade ed the bridge was established as shown on drawing number 1. No attempt has been made to show the future road work but when the highway is improved it can be tied in very readibr with the designed bridge approach. Both skab and girder bridges were investigated and it was desided that the proper bridge for the existing conditions is a slab bridge. The span of the bridge was controled by the existing conditions and was established as 25 feet clear. "Specifications for The Benign of Highway Bridges" adopted by the Michigan State Highway Department, November, 1936, were used throughout.atAt times it was deemed advisable to vary a small amount from.these specifications. This variation is noted in the desi n. . . g 1&6095 I wish to take this opportunity to thank Hr. martin of the Portland Cennnt Association and Hr.mShuttleworth of the Michigan State Highway Department for their valuable aid and advise in the preparation of this thesis. II Design of Slab Width of Roadway - For four lane highways, the Michigan State Highway Department specifications suggest 44 ft. curb to curb, measured at right angles to the longitudinal centerline of the bridge. .170" 44.15” 42]?" 2 ft. 5 in. from the curb to the edge of the bridge allows 1 ft. 6 in. clear distance from the curb to the railing , 1 ft. assumed for the railing. .Effective Width for the Distribution of Concentrated Wheel Loads on the Slab - The Specifications, Article 41, states: B 8 0.75 f 2 where B 8 Effective width of slab in feet. 8 ' Span of slab, center to center of supports. assume 12 in. abutment wall. S = 25 f l = 28 ft. B = 0.7 x 25 f 2 = 18.2 f 2 = 20.2 ft. max. B = 7 ft (specifications, Article 41) A rticle 31 of the specifications assumes that the traffic lanes will not occupy a position in which the center of the lane is nearer than 4 ft. 8 in. to the roadway face of the curb. assumed effective width of wheel load B With traffic lanes as shown, two wheels can come within 3 ft. of each other, and their effective widths overlap and a new distribution must be calculated. The specifications, Article 41 also gives : effective width 3 §§_Q where C 3 distance center to center of loads. effective width = 7 i 3 = 5 ft. 2 distance from curb to edge of distributed wheel load B = 4.5 f 3.0 - 3.5 - 4 ft. assume d = 13% in. depth of slab = 13% f 2 f i - 15 3/4 in. Article 59 calls for a protective coverage of 1% diams. (min. of 1% in.) from the surface of the concrete to the center of the nearest bar, but after investigation it was decided that a protective coverage of 2 in. clear would be better for this bridge deck. Dead Load slab - l%§1§ x 1 x 150 =196.75 future wearing surface (Specs. Art. 30) = 30-00 g in. wearing surface (Specs. Art. 26) 0-5 x 1 x 150 z 6.25 T3— ""223'.'o'0" lbs. per sq. in. wl2 dead load moment = 5— = 1/8 x 223 x 2'53 x 12 = 209,000 in. lbs. Live Load Rear wheel in the middle of the deck. Izzkxa“ i 3 /2* ‘ / 1 ' 5 ca“ 4000” J .5 Try both wheels on the deck. .Afirua 4%gqo f 2.’ " :J'_ /2"6~ J [ 1, Maximum moment occurs when the 12,000 lb. wheel is at the center of the span, and is: - 4 _ 6 000 Live load moment = Impact Article 37 of the specifications states coefficient shall be: Impact coefficient - I 8 L L - Span length I a 25 £20 6 x 25 # 20 - Impact moment Total moment Calculate depth of slab Use: fo = intermediate grade steel fS 170 0.265 x 180,000 = 3,000 lbs. per sq. in. and n = -43-—- x 12.5 x 12.0 - 180,000 in. lbs. that the impact 20 0 0.265 47,700 in. lbs. 209,000 / 180,000 { 27,700 436,700 in.1bs. per ft. width of slab. 10 a 18,000 lbs. per sq. in. yg-Izamfl79' t—rl‘ ‘ was! 3 r r C t b l' 7' L——~ é. II a u ‘&%%§L-=/Eum97%; emu-41.. 1200 3,000 kd = 0.4a jd = d — 3% = 0.8670 T . c : $339. x .hd x 12 = 2880d M = T x jd 436.7000 = 2880d x .867d b36,700 _ 175 02 = 2880 x .867 d - 13.22 in. use d n 13% in. as was assumed As : 2’880 X 13'25 = 2.12 sq. in. 18,000 use 1 in. square bars 9 5.5 in. A - 2.18 sq. in. h k [0000’ , . C so Shear L [4.0 {.3000 T‘fi 11 Live load shear a VL = 12,000 {‘25 3,000 = 13,320 lbs. Impact shear a VI . 0.265 X 13,320 = 3,535 lbs. Dead load shear = VD - 223 x 12.5 . 2790 lbs. - V Dead load unit shear = VD = '—E§I_ 2790 12 x 0.867 x 13.25 20.25 lbs. per sq. in. There have been many attempts to establish a formula for the effective width for shear distribution. Kirkham.in "Highway Bridges", page 118, assumes that the end shear on simple slabs due to live load and impgct may be distributed transversely over 2.5' Caughey in his book "Reinforced Concrete", on page 1&9, gives the results of an experiment at the Iowa Engineering Experiment Station in which the formula was found to be as follows: E = .75 % 1.2t / x vhere t . slab thickness in feet x distance of load from.unsupported edge of slab in feet, never to be assigned a greater value than 1 % 2.ht 2.h X 13.25 x = 1 / 2.4t = 13- 12 = 3.68 ft. which is less than the distance to the edge of the slab for the point at which we are figuring the shear. E then becomes E = .75 % 1.2 x lziéé' / 3.68 = 5.76 ft. The specifications do not limit their effective width of slab formula, to figuring moment so it is assumed that it can be used for shear. There are many other formulas advocated for use which compare favorably with the last two shown above. In view of all this it was decided to use the same distribution for shear that was used for moment, namely 5 feet Live load & impact shear = 16,855 lbs. distributed over 5 ft. transversely V V: i -533 16,855 _ 12 x 5 x 867 x 13.25 2h.h lbs. per sq. in. Live Lead Impact unit shear total unit shear VBVL" VD 24.w ; 20.25 = hh.65 lbs per sq. in allowable V 60 lbs per sq. in. Check Bond V 2" ‘2on /v = 16:55 / 2790 . 6161 lbs. 6161 : A X 2 x 867 X 13.25 = 61.4 lbs per sq in. /” is allowable/a = 150 lbs. per sq. in. Special anchorage is not shown to be necessary but good practice will require a standard hook on the main longitudinal Steele Crown of Roadway 2:6" 2:0; 40-0“ _ 210" 2-4 ”w I ‘ ‘5 % I ”\i Q o a Q (Note - g" wearing surface is not included here.) Bottom of slab edge of effecient width for distributed wheel load. See fig. pagez I .00187R2 (for middle 40 ft. - Specs. Art. 2h) = Crown of roadway in inches Width of roadway in feet =‘ .0018? x 1'02 = 3 in. 02300 II The two feet outside of the middle 60 feet is tangent to the curve with a slope equivalent to 3 inch vertical in 10 foot horizontal. E = x '1?’ _ 1 x - .6 use 2 inch Determine crown 4 feet from the face of the curb. 1+0 "' [iv 2" 36 ft. 0 = .00187 x 362 = 2.42 in.say 2.5 in. 3 - 2.5 = 0.5 in. Temperature Reinforcement "Temperature reinforcement shall be placed at exposed surfaces and shall provide not less than one eigth square inch of reinforcement per foot of width of surface." Specs. Art. 66. 1/8 sq. in. reinforcement per foot of width of surface 3 i in. round bars @ 19 in. Use % in. round @ 18 in. logitudinaly in the top of the slab. There is no restraint of the slab transversely in the middle of the span so that except for tie bars, no temperature reinforcement is necessary. There is, however, restraint transversely near and at the abutmnets and transverse temperature reinforcement should be used. in. round @ 18 in. transversely in the top of the slab feet next to the abutments. and Use for \J‘INIH Use % in. round @ 30 in. transversely in the top of the slab for the middle 15 feet. —— _ 10 _ Transverse Steel in Bottom of Slab There should be steel in the bottom of the slab to take care of the moment perpendicular to the span of the slab. It may be argued that there have been no failures in slabs due to this moment, but this is not due to the fact that the m moment does not exist. Putting steel in a slab does not mean that all of the moment is in the direction of this steel. For a balanced design, steel should be added perpendicular to the main reinforcement. For spans of this length the Michigan State Highway Department uses 15% of the main reinforcement in the outer k of the span and 25% of the main reinforcement in the middle % of the span. 1 in. sq. bars @ 5.5 in. A 2.18 sq. in. S 2.18 x .15 = .327 sq. in. 2.18 x .25 = .5h5 sq. in. Use 2 in sq. bars at 9 in. in the outer 2 span. Use % in sq. bars at 5.5 in. in the middle 8 of the span. Check the Stresses at the Centerline Depth of slab = 15 3/2 x 2% = 18% in,d = 15 3/4 in. _ 11 - . [2,. I l =X I Na» l. .Af9%; .fiL .fl 0' o . As I 2.18 in. Transformed area steel = n x AS - 10 x 2.18 = 21.8 sq. in. Take moments about neutral axis. 12 x X x “%;' = 21.8 x (15.75 ~x) 6X2 = 34h - 21.8X 6x2 / 21.8x - BAA = 0 5615 {62 - hac 2 a flfl "' 21.8 f 12 5.98 id = 15.75 — “‘3‘— = 13.76 in. Dead Load slab - l§f§£~ x 1 x 150 228.00 future wearing surface - Specs. Art. 30 20.00 5 in. wearing surface 2 .2 2 . e. per sq. in. - 13 - w 12 Dead load moment "§"" 1/8 x 254.25 x 252 x 12 3 239,000 in. 1b. Live load and Impact moment = 180,000 / 47,700 = 227,700 in. lb. Total moment - 227,700 # 239,000 - 488,700 in. lb. f M ' 2° x k d x b x j d r 488,700 = g x 5.98 x 12 x 13.78 f - 944 lbs. per sq. in. 'allowable £0 a 1,200 lbs. per sq. in. I fsAB-+xkdxb 944 5.98 x 12 f8 = 3 x 2.18 = 15,550 lbs. per sq. in. allowable f8 - 18,000 lbs. per sq. in. V a 1l§§§§_ 254.25 x 12.5 5 7‘ - 3371 / 3180 .. 8551 lbs. - V . 6551 = 'V' - 533 T§—§_T3.76 39.6 lbs. per sq. in. allowable we a 60 lbs. per sq. in. .ddg = __!___, a 6551 - 54.6 lbs. Zfijd 4 x 12 x 13.76 per sq, in. 5.5 allowable/A, : 150 lbs. per sq. in. - 13 - aback_fiizsssss_al_flnzb .A2” 1 T 8 \ \ . Depth of slab = 15 3/4 - n . 1 1/8 in. ' 4.2150" s 14 5/8 in. d - 12 1/8 in. Transformed area steel = nAg ' 10 x 2.18 a 21.8 sq. in. Take mooments about neutral axis. 12 x x 1: -§- = 21.8 (12.125 - x) 6121‘ 21.81: - 284 = o - z - ‘4— X ___ b {{b 48.0 2 a - 21.9 2‘. [478 ,l 8340 12 = 5.05 in. 5.05 jdad--§g": 13.13-? a 10.45111. Dead logg 63 slab - - 1 x 150 ' 182.75 12" " future wearing surface - specs. Art. 30 20. A in. wearing surface 6.35 209.0 lbs. per sq. in. Dead load moment = 8 1/8 x 209 x 253 x 12 - 196,500 in. lbs. Effective width for concentrated load Max. B is 7 ft., but the distance from a wheel at the curb to the edge of the slab is less than -%-so the femmula; eff. width = -%—¥ distance to edge of slab, must be used. Assume the wheel at the curb. . 7.5 distance to edge of slab a 3-5 A '13?"‘ = 3.125 where -%§§- = distance of center of tire from face of curb in feet. effective width '%-'/ 3.125 = 6.625 ft. Live lead moment = —§f%%%- x 12.5 x 12 ' 136,000 in. lbs. Impact moment = .265:x 136,000 8 36,100 in lbs. Total moment -. 198,500 / 138,000 / 38,100 -- 388,800 in lbs. M ' x kd x b x jd 368,600 a x 5.05 x [2 x 10.45 I" I“ MONO H1 c-= 1165 lbs per sq. in. 13‘s 2 f; x kd x b . 1165 f a ’5‘” X 5-05 I 13 - 16,200 lbs. per sq. in. s 2.18 allowable a 18,000 per sq. in. - 15 - 15.15.55 1! 209 x 12.5 - 5,350 lbs. 6.685 I Y a 51150 2.. a 41 lbs. per sq. in. de 13 x 10-45 allowable v a 60 lbs. per sq. in. 56.5 lbs. per sq. in. allowable/7: 150 lbs per sq. in. v __ 5415 _ : ‘ 3 . ‘ liojd 4 x 5 x 10 45 'HO 01 -16- III Design of Railing /.‘a" k A [54 " / 7f! :qi IJUV7QXUF¥or£¢VQI [ 1-r—5Qbkar l’ ‘ l . Jan 717. Assume a 12 in. railing with % in. round bars @ 16 in. vertically. Check Shear If only reasonable care is taken with the joint at the bane of the rail it may be expected to develop at least 50% of the shearing resistance of monolithic concrete. - 17- 150 As the unit shear is only; ‘33:“ = 1.05 lbs. per sq. in., the railing of course id’safe in shear. Check Railing for over Turning tbout Point A Moment about A = 150 x 3 - 100 x .5 - 150 x 3 x .5 = 450 — 50 - 225 = 175 ft. lbs. Area of steel needed to withstand this moment per ft. of 175 x 12 - railing 3 A8 3 W "' 0117 sq. in. § in. round @ 16 in. have an area of .20 x“%§‘ = .15 sq. in. per ft. of rail Use % in round @ 16 in. vertically in each face. Use 4 - i in. round horizontally for temperature steel in each face. The moment and shear are both less on the curb section and % in. round 0 16 in. vertically may also be used. Use 5 ~ % in. round horizontally as shown. Check the Slab For a Wheel on the Curb /o ”‘ Izaoa" : 24 4' ' 1 9" ,, \—+/4/z _ l8 - Dead load railing 3 X 1 X 150 =- #50 curb and slab 9 l a X l X 150 = 29A 12 load 100 = 100 8th lbs. per ft. of curb w 12 Dead load moment 3 —@T-' 1/8 x 84h x 252 x 12 = 791,000 in. lbs. Live moment = -é§%%9 X 12.5 X 12 = 360,000 in. lbs. Impact moment = .265 X 360,000 - 95,h00 in. lbs. Total moment .= 791,000 % 360,000 % 95,h00 = l,2&6,h00 in. u lbs. ~30 1 3 __ __ 33 U Q -r A, ’Z./6'Z.J*OT¥J'" Transformed area steel - n A3 3 5h.5 sq. in. 30 X X x.% = 56.5 (21 -AX) 15x2 % 5a.5x - 11u5 = 0 ;;A§Z£iZEE-tac 2a = - 51+.5 {[2970 f 6900 - 7.1 in. 30 ‘ . V‘Nr" - WW A _ 19 - Jd 3 2]. - T 3 18.6 in. T=c=—g—x30x7.l r c x jd = 1,2b6,100 = _%_ x 30 x 7.1 x 18.6 f0 = 630 lbs. per sq. in. allowable f0 - 1,200 lbs per sq. in. Asfsxjd= M f _ 1,2h6,hOO . 12,300 lbs per sq. in. s 5.45 X 18.6 allowable fs = 18,000 lbs per sq. in. E: II Check Shear — Dead load shear Vd I 84h X 12.5 = 1,0550 lbs. 11 Live load shear Vl = 12,000 / '5? 3,000 = 13,320 lbs - .31. a —324§ZQ—- - 523 lbs er s in v bjd 2.5 x 867 x 21 ” p q The curb is not designed for a wheel load on it. _ - 10:550—fi‘, - 2 1 lbs er s . in. Dead lwad v - 2.5 X 867 X 21 3 P q The allowable unit shear with special anchorage and web reinforcement = 270 lbs per sq. in. v I 231 lbs per sq. in. v0 . 90 lbs per sq. in. vs I v - v0 = lhl lbs per sq. in. 'v aS fs 3 b s sinoL % in. round bars aS = .20 s = ’20 x 18,000 = 10 / inches 2.5 X l X ltl _ 20 _ Use 9 inch spacing. Try shear at 5 point. V = 8AA X 12.5 - 864 X 6.25 a 5,280 leo v - 5’280 __. = 115 lbs per sq. in. 2.5 X .867 X 21 VS = 115 - 90 = 25 lbs per sq. in. S = .20 X 18,000 _ 50 in. 2.5 X l X 25 maX. s - ZZ7TIU5 "Iaa—a Use % in. round stirrups at 9 in. in outside i of span and 1 2 in. round stirrups at 18 in. in middle half of span. _ 21 _ Design 0d Abutment Walls , j I». K /-’o" :0. Q” . - A 9 d: B a L 6"?" f 15414 53.?" , L 40—1" _ 22 _ Design of stem Jbvvlbcpr ‘5 -.________..___1_fl. firlMM/ P h - - e r- 760 I/ WI// 1’__\_ ___ __ °~ \ b ‘ \\ ‘5 1 ". _t P \3 ; ‘1 L ’0‘!” \ \ ‘ .c 7 Max Pr’JJl/f't' 4" (file/’7. "Retaining walls, abutments and structures built to retain fills shall be designed to resist pressures determined in accordance with "Rankine" theory of pressure distribution in non cohesive granular material provided that no structure shall be designed for an equivalent fluid pressure of less than 30 lbs per sq. ft.", Specifications, Article 39. An equivalent earth surcharge of four feet is used as the live load. Values of Ce 8 .33 and W = 100 lbs per cu. ft. are used by the.Michigan State Highway Department for soil conditions similar to those on this job. . X 150 _ .75 ft. of concrete = 7:00 2 - 1.13 ft. Of earth _ 23 _ 2 w h _ 100 X 21.632 a 7 730 lbs. P - Ce 2 - 033 2 , a .. 3.1.5.93— . 7.21 ft. 100 882 p = .33 g 5. n 570 lbs. .88 b. = 2.3...— ’( 15.75 = 17.71 ft. P-p = 7,730 - 570 - 7,160 lbs. Take moments about A 71120 x 7.21 - 570 x 17.71 _ c . 7,160 _ 6.38 ft. Moment at base = M.= 7160 X 6.38 X 12 = 558,000 in lbs. I Id -KB' d = for fc = 1,200 lbs per sq. in. and fs = 18,000 lbs per sq. in. K=ao_8 d = “/ 5u8,000 - 14.82 say 15 in. 208 X 12 Depth of slab = 15 1 3 = 18 in. M = f3 AS jd l l ___—_— B ___—— - 001+ k ,_. 13" ts 18,000 ‘fifg— l % 10 x 1,200 J = 1 -'§ . .867 A 548,000 = 2.3a sq. in. s a 18,000 X .867 X 15 Use 1 in. round @ A in. AS = 2.36 sq. in. V = (r — p) - 7,160 lbs. _ v _ 7,160 maX. V -' W 12 x .867 X 15 ' h5.5 lbs per sq. in. allowable 60 lbs per sq. in j: _L. 7,160 22“ per sq. in. allowablgjla 150 lbs per sq. in Check Stem.5 ft. Above the Footing I 7;} I’M/l . K3 . Q 52:14.»: la \ be clocl , ’0’, off'éz ‘Q 1 .33 " _ [1,560 lbs. _ 2 16163 = al I 3 5051? ft. 5.88 P - 570 lbs. b1 ar-3- / 10.75 = 12.71 ft. - 25 - P - p = 6,560 - 570 - 3,990 lbs. l C a [1&60 X job-LL " 570 X 1.2.71 9 [+050 ft. 1 3.990 M I 3,990 X 6.50 X 12 = 215,500 in lbs. 215 500 .2 ° . d 208 x 12 a 9 5 in Depth of stem needed - 9.25 f 3 - 12.25 in. Depth of stem available 3 l2 # 61§.%2°72 ' 16-1 in. d - 13.1 in. Assume that 5 the steel will be terminated 5 ft. above the footing. X 12 x X x “'2'" = 1.3.8 2: 10(13.l - X) 6x2 2 11.8 X — 154.5 = 0 - 11.8 5 15139.11? 31700 X = 12 kd = X = h.l7 in. _ 26 - id = 13.1 - 3%?“ 11.71 in. 215,500 = :5; x 1.17 x 12 x 11.71 fc = 735 lbs. per sq. in. allowable fc - 1,200 lbs per sq. in. 1‘3 x 1.18 . 122.1: 1.17 x 12 fS = 15,600 lbs per sq. in. allowable fS = 18,000 lbs per sq. in. Check for anchorage 2 TI'a X 18,000 allowable tension in a round rod = Bond developed in a bar = 11 X a X l X 150 rial X18,000= IIaleSO 1: 30a Anchor 5 of the bars 30 diameters or 30 in. beyond 5 ft. ‘5: above the footing. Temperature Steel in the Stem The specifications call for 1/8 sq. in. of temperature steel per foot of surface. Spacing for 8 in. round bars. ~ 1.1.2.5...- -20 s-1 in. 12 s 9 ’Z' Use 5 in. round @ 18 in. horizontal and vertical in the front face and use 5 in. round @ 18 in. horizontal at 18 in. in the back face. _ 27 - Determine Dimention for the Abutment The cross—section will be analysed using three methods of loading. Cast I - No superstructure load No live load Case II - Superstructure dead load All live load surcharges Case III - Superstructure dead load No surcharges Case I for dimension see sketch udiJLjhlLl§Q_Z_l+iZE Wt. of stem.— 15.75 X l X l X 150 % 2 = 2,360 % 591 = 2,951 lbs. Wt. 03 base a 10.5 X l X 2 X 150 a 3,150 " Wt. of earth = 5.25 x 15.75 x 1 x 100 = 8,275 " Total weight 16,376 lbs. Distance of resultant of vertical forces from point B = X 2.360 2:11.751 591 x 1.08} 31150 x 5.25 11 81275 x 7.87L X: 14376 3 6.63 -28- 'f7—752 Earth pressure = P2 = .33 100 3 ° = 5,200 lbs. a2 = 17575 = 5092 in. #374 "‘ & 622k? ._JZ_ _ 200 5.91 ‘ 14,376 Y - 2.16 ft. Toe distance to resultant = 6.63 - 2.11 I 4.49 ft. Eccentricity = e = 5.25 - hoh9 = .76 ft. 10.5 . -——g——- - 1.75 ft. Resultant passes through the middle third of the base. 16,376 X .4 = 5,200 1.105 14.376 x 6.63 - 3.10 5,200 X 5.92 Sliding factor of safety = f = Overturning factor of safety Earth pressures = p :— ES (1 f 6bR .7; £11712; (15.2.2212 . 10.5 _ 29 _ - 1965 lbs per sq. ft. = 775 lbs per sq. ft. Case II Wt. of deck = 1.5 x 1 x 13.5 x 150 - 3.050 lbs. Wt. of surcharge on deck = A X 1 X 13.5 X 100 = 5,600 " Wt. of stem = 2,360 l 591 = 2,951 " Wt. Of base ; 3,150 n Wt. of earth and surcharge = 21.63 X 5.25 X 100 *ll,350 Total weight 25,891 lbs. Distance of resultant of vertical forces from point B = X X (3.060 I 5.400 f 2.360) 4.756% 591 X 4.08 { 3.50 X 11350 X 7.875 25,891 = 6.17 ft. ‘K——— . I\ 3 fife/mfl1l\:fi _ P ‘37 \ \ 5: ‘a A Q —"\d 7’ .flblzunfffiii‘ '1 f5 _ 3o - = 9,210 lbs. a3 t-Elgél— - 7.88 ft. P - 570 lbs. b3 _ _5..'.%§_; 17.75 - 19.71 ft. P3 - p 2 9,210 - 570 : 8610.118. _ 9,210 x 7.88 - 570 x 19.71 03 ‘ 8,610 I 7010 ft. y a 8,660 7.10 25,891 y = 2.37 ft. Toe distance to resultant = 6.17 - 2.37 = 3.80 ft. C = 5.25 - 3.80 1.65 ft- Sliding factor of safety = 25gziox '5 = 1.20 Overturning factor of safety = 25,891 X 6.17 = 2.61 8,611.0 X 7010 6 X 1. Earth pressures = 21 891 (1 é 10.91+5 h,500 lbs. per sq. ft. = 627 lbs per sq. ft. ’ Case III Wt. of deck = 3,090 lbs. Wt. of stem - 2,360 % 591 = 2,951 " Wt. od base a 3,150 " Wt. of earth = 17.63 X 5.25 X l X 100 : 9,250 n Total weight 18,391 lbs. _ 31 - Distance of resultant of vertical forces from point B IIX = (20401} 2260) 4.751% 591 X 4.08%3150 X 5.25/ 9250 X 7.875 18391 = 6.39 ft. 75/, d/Wd//’;‘}\ ___ 3 ; \ \ . \\ :3 \ ‘Q “WI-5 30/11. off/992 \ 4‘ 100 I913. 2 Ph - .33 X n 6,370 lbs. a =,l2;é2 - 6.54 2 h 3 y : 6370 6.55 18391 y ' 2'27 ft' Toe distance to resultant = 6.39 — 2.27 = 5.12 ft. e - 5.25 - 1.12 - 1.13 ft.‘ 8 . Sliding factoreof safety I l ggéox A I 1-155 Overturning factor of safety I 12336 $2352 g 2.82 0 ' 6 1.1 Earth pressures = p i?%‘%— ( l i ‘_I%_5_42 ) = 2,940 lbs. per sq. ft. 566 lbs. per sq. ft. - 32 - Design of Toe Slab maximum.moment occurs under Case II .a | ' ‘ 1 a ' 1“" “it-17:2:fi2m' I ,, ” I ., 51.729077: 4500 - 627 = h073 lbs per sq. ft. X . 6.75 . 2620 lbs. per sq. ft. #073 10.5 A500 - 2620 - #27 = 1A53 lbs. per sq. ft. Wt. of concrete = 300 lbs. per sq. ft. Moment about a - a = M Mi= 3047 x 3.75 x.2312_ ; 1553 g 375 x 2 x33.z5 - 300 x 3.75 x 3&75 a 26,150 ft. lbs. 26 150 X 12 . . - ’ = 11.2 1n. sa ll. 1n. d ‘ V 208 x 12 y 5 11.5 # 3.5 = 15 in. Depth of toe slab Check §1ear - 111.53 )2: 3075 _ 300 X 3.75 : 13015 lbs. v = 3017 x 3.75 % _ 33 _ v =.11__. = 13015 = 109 lbs. per sq. in. de 12 x .867 x 115 O . . d needed = ;g%-' x 11.5 = 20.9 1n. say 21 in. Depth of toe slab = 26.5 in. The 2A in. assumed checks. fl .q___£ZL____.+ .zx" 1200 C = 2 - X 12 x kd 1200 ML: 26,150 x 12 = -——§-ex 12 x kd (21 - 5%—-) 2100 (kd)2 - 151,200 kd ; 313,800 = 0 . ,1 151.2 {[2,286,1L.L. - 301,2118" 1+8 ‘ 1512 - 1409 = $8 = 2.15 in. kd jd = 20.283 T I C 3 AS fs j a .97 1200 12 x 2.15 _ s : '*-§—fx 18,000 - .86 sq. 1n. - 3h - Use 3/h in. round bars @ 6 in. As - .88 sq. in. Check Bond _ 13015 fl- [+07]. X .97 X 21 allowable/41a 150 lbs. per sq. in. = 135 lbs. per sq. in. Design of Heel Slab Determination of maximum.moment. Case I 15.75 X l x 1 = 1575 lbs. per sq. o ___- -= 121. [fi’~]- ‘1 1875 lbs. per .4 | Sqo fie a ' ,i L____- __ | X’/” 77’%' !,/// 120535? 1965 - 775 = 1190 lbs. per sq. ft. _ 1190 X - —10:5' X 5.25 - 595 lbs. per sq. ft. Moment at a - a = M 2 "———2 M = 1875 x'5725 _ 775 x 5.25 _ 595 x 5.25 x 5525 2 2 2 - 12,110 ft. lbs. _ 35 _ Case II 21.63 x 1 x 100 = i:’ 2163 lbs. per sq. 6? f“*’["“ - 300 ' a 2# 3 lbs. per sq. ' ft. #500 - #27 = #073 lbs. per sq. ft. x = %%Z% x 5.25 = 2036 lbs. per sq. ft. Moment at a-a = M M a 2163 $57252 _ 1.27 g 5.252 _ 2036 3:75.2i X 2.3.5.. M = 18,760 ft. 1bs. Case III 17.63 x l x 100 1763 lbs. per sq. H. 300 2063 lbs. per sq. ff. .2982775" - 36 - 29#O - 56# = 2376 lbs. per sq. ft. x 3 '§%%%- X 5.25 = 1188 lbs. per sq. ft. Moment at a-a 8 1M 2063 x 5.252 564 x 5.252 - 5 5 1M - 2 _ 2 11882]: 5022 X 2 = 15,170 ft. lbs. Maximmm.moment = 18,760 ft. lbs. 18760 x 12: d = 208 x 12 = 9'5 in' Depth of heel slab = 9.5 / 3.5 = 13 in. Check Shear 2036 x 5.25 ’2 V = 2#63 X 5.25 - #27 x 5.25 - = 5370 lbs. 2 537° = . 1b . . ° . v 12 x .867 X 9.5 5A A S per sq 1n Use the same effective depth for the fieel slab as was used for the toe slab - 21 in. I. t” 'I 1:00 x 12 x kd _ 37 - 1200 kd M = 18760 x 12 a -g- X 12 X kd (21 - ‘3‘) 2#00(kd)2 - 151200 kd ; 225120 = 0 = 15120 f y228611100 - 21,611,520 #80 kd «3 1053 in. T=C =AS‘fS 1200 12 X 1.53 .. X As “ 2 18000 .613 sq. in. Use 5/8 round @ 5% in. As - .67 Sq. in. jd = 20.69 J = .97 Check Bond _ 5370 //v- 1.96 leg— X .97 X 21 I 61.7 lbs. per sq. in. 5.5 allowable/Al: 150 lbs. per sq. in. Check factor of safety against sliding *AM I :zeuaa' .21__————n522x7” P ___ I 1‘ - -. .mcauzu' ” Factor of safety against sliding is smallest in Case I f = 1.105. . _ l _ ' _ PaSSlre Ce - 733' - 3 to be safe use Ce _ 2 -—-—2 00 . p = 2 X l 2 7 25 . 5260 lbs. Wt. = 8270 ¢ 2951 / 5.75 x 5.25 x 100 / 3075 = 16266 1bs. _ 16266 x .42! 5260 = 2.26 ‘ 5200 -39... II Bibliography Kirkham, John E., "Highway Bridges" H001 and Johnson, "Handbook of Building Construction" Hool,"Reinforced Concrete Construction" Urqnhart and O'Rourke, "Design of Concrete Structures" Caughey, "Reinforced Concrete" \ Lurneaure and Maurer, "Principles of Reinforced Concrete Construction" American Concrete Institute, "Reinforced Concrete Design Handbook" Peabody, RReinforced Concrete Structures" Portland Cement Association, "Continuous Concrete Bridges" a u u .5 , ; . O o. 71!“ l 3.01.0)2 mqbam CZ_