Design for a 180' Concrete Arch Bridge at Devil's Gate Citation Harris, Tom Johnson (1916) Design for a 180' Concrete Arch Bridge at Devil's Gate. Bachelor's thesis, California Institute of Technology. doi:10.7907/07f5-1j16. https://resolver.caltech.edu/CaltechTHESIS:07172024-224722154 Abstract No abstract. Item Type: Thesis (Bachelor's thesis) Subject Keywords: (Civil Engineering) Degree Grantor: California Institute of Technology Division: Engineering and Applied Science Major Option: Civil Engineering Thesis Availability: Public (worldwide access) Research Advisor(s): Unknown, Unknown Thesis Committee: Unknown, Unknown Defense Date: 1 January 1916 Record Number: CaltechTHESIS:07172024-224722154 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:07172024-224722154 DOI: 10.7907/07f5-1j16 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 16569 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 18 Jul 2024 18:48 Last Modified: 18 Jul 2024 18:50 Thesis Files PDF (Full thesis) - Final Version See Usage Policy. 8MB PDF (Plate 1 Reinforced Concrete Arch General Drawing) - Supplemental Material See Usage Policy. 2MB PDF (Plate 2 Concrete Arch Bridge Influence Lines) - Supplemental Material See Usage Policy. 1MB PDF (Plate 3 Concrete Arch Bridge Influence Lines of Fibre Stress) - Supplemental Material See Usage Policy. 2MB PDF (Plate 4 Reinforced Concrete Arch Thrust Line) - Supplemental Material See Usage Policy. 1MB PDF (Plate 5 Concrete Arch Bridge Map of Bridge Site at Devil's Gate) - Supplemental Material See Usage Policy. 1MB Repository Staff Only: item control page

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LE L= 3600 lo =4#400 18000 0 16000 m 22500 39000 THESIS KEEINFORCED CONCRETE ARCH THRUST LINE THROOP COLLEGE ¥ TECHNOLOGY PASADENA CALIFORMA Scale I= 8 er 30000" e516 QrawneThaced by Jom Yum ' |CONCRETE ARCH BRIDGE THE S/S MAP ¥ BRIDGE SITE # DEVILS GATE THROOP COLLEGE + TECHNOLOGY FASADENA CALIFORNIA Sca/fe /240° ZB6 Drie Paced Ly Som, pf flawrea THESIS. Design for a 180' Concrete Arch Bridge at Devil's Gate . by '11 0m J. Harris. Class of Nineteen Hundred Sixteen. Department of Civil Engineering. THROOP COLLEGE OF TECHNOLOGY Pasadena, California. 1916. 1 The present structure over the Arroyo Seco at Devil's Gate is a steel deck truss of the Pratt type. The maximum allowa ble load on this bridge is a five ton truck. As this bridge is on the main line drive from Pasadena to La Canada and Glendale it would appear to be inadequate for the service which it might be expected to undergo in the very near future. Furthlrmore, the::ce has been some talk of beautifying the Arroyo and making a public park of it. Any such proceedure would certainly necessitate the removal of the present structure at Devil's Gate. It is through the influence then of these two needs that I was induced to design a concrete arch bridge to take the place of the present one and I hope that it will not only satisfy the need of a stronger ·bridge but will fit into any improvement plan of the future. In order to satisfy the esthetic sense I de- cided to design the bridge with open spandrels and three centered arch ring. This type of structure will be appro- priate for any plan of beautification beacuse of the already existing open spandrel arch on Colorado Street. The method usea in the design of the arch was to assume a given ring and then investigate it V': i th the required loading. To better estimate the like)y dimensions of the ring, it is advisable to look up the records of existing arches which are fulfilling conditions somewhat similar to the ones et hand. By such records I found 2 that for tbe span required, i.e. 180', the following dimensions were so mewhere in the vicinity of the correct ones; these dimensions are, rise 45 feet, ring thickness at crown 4 feet, at springing line 12 feet. Steel reinforcement equals 1¼% of area at crown, . 625o/o is placed. at top and bottom, 2 inches from surface. Having assumed these di~ensions the next p roccedure is to investigate them. This investi gation was carried out in a manner as out lined in "Principles of Reinforced Concrete Construction" by Turneaure and 1v1aurer, and in "Concrete, Plain and R€inforced '' by Taylor and Thomson. The method and results follow. 3 The arch ring is divided into 20 divisions of such a length that ds is a constant, where ds = length of divI ision measured along the arch axis, I = moment of inertia of any section= I concrete~ 15 I steel (see pg.92 Turneaure and Maurer). The method of ~king this division is; let i = !, ia = mean value of i found by taking nn" divI isions of equal length, s = half length of arch ring measured along the axis, and n = desired number of di visions in ½ the arch ring. Then ds • sia. The value of ds I n I being known, the length ds can be determined by the cut and try method, until 2 ds = s. ]'allowing are the re- quired data and computations of ds. I Table I. DIVISION OF ARCH RHT8. Assume it~ steel (1¼% of area at crown) placed as shown 2n from surface. Preliminary Divisions. No. of Division. Depth in Feet le bh 3 /12. 15 Is 1 4.00 5.33 2 4.10 3 I- i- le+- 151 8 1/1. 2.52 7.85 .1275 5.75 2.66 8.41 .1185 4.30 6.63 2.95 9.58 .1045 4 4.60 8.13 3.41 11.54 .0865 5 5.00 10.41 4.08 14.49 .0691 6 5.60 14.62 5.21 19.83 .0504 7 6.50 22.85 7.15 30.00 .0333 8 7.70 38.01 13 .30 51.31 .0195 9 9.10 62.82 14.40 77.22 .0129 10 10. 90 107.80 20.90 128.70 .0077 4 Table I (cont) . z i = .6299. Final Divisions. No. of Division. i ds Ho. of Division . i ds 1 . 1290 5.31 6 . 09 70 7.06 2 .1260 5.44 7 .0825 8.07 3 .1215 5.63 8 .0699 9.80 .4 .1150 5.95 9 .0486 14.10 5 .1070 6.40 10 .1066 41.20 ds I = sia n 108.8 10 = X .6299 10 = .685 s = i length of arch ring measured along :r . .Axis = 108. 8 ia • average i n • number of divisions a 10. Note : I= le+ 15 ls= bh 3/12 + 15 x .m125 b'h' x a 2 • where, bh = area of concrete section at point in question. b'b' = " " II at crown . . 0125 = o/o steel. a= distance from N.Axis to center of steel 15 x .0125 x b'h' = .75 5 After the divisions ds/I are laid off on the arch, the next step is to calculate l:i 0 , V0 , and .M 0 (Thrust, shear and moment at crov~n), for load of unit~ placed successively at the load points H, B, C, D, E, and F, at base of span~rel pier. Table II gives the necessary data for the calculations. In Table II the points 1, 2, 3, etc. are the center points of the divisions ds. "X" is the horizontal distance from the crown to the point i.n question. "Y" is the vertical distance from the neutral axis at the crown to the point in question. ation. 11 X2 " and "Y2 11 n€ed no explan- The arch ring is assumed to be cut by a vertical plane passing through the crown perpendicular to the neutral a.xis. Each bali of the ring is thus acting as a cantilever v:ith V 0 , H0 , and. M0 substituted in place of the half arch removed. Now MR and M1 are the moments at the given points 1, 2, 3, etc. on these cantilevers due to all loads between the point in question and the crown. Thus for a load of unity placed at B, the moment MR at point 2 is derived as follows. Let the distance from the crown to the load = X , then MR at :p oint 2 = (X - X1} 1 = 16.4 - 15 = 1.4. 6 Table II. Ca lc ul a t i on s for H0 , i • Mo for Loa d of Unity plac ed successiv ely at A, B, etc. Lo a d at A. Point. X y 1 5. 3 .15 2 16.4 3 ,y 2 ML 28.16 .225 0 .9 0 268.00 . 81 0 27.0 2. 50 729.00 6.25 0 4 3 7.6 4 . 90 1416.0 24 .00 0 5 48.0 8 . 00 2304.0 64 . 00 0 6 58.00 12.20 33 64 .0 149.0 0 7 67.6 17.60 4575.0 3 10 . 0 0 8 76.4 23 .9 0 5 840.0 572.0 0 9 84.4 3 1.10 7120.0 968.0 0 10 91.8 39 . 20 84 50.0 1535 . 0 0 2£ 14 0.45 34 09 3 .2 3 629 . 3 Q)Q) x2 Load at A. Point. M,~ (ML + MR)Y 1 5. 3 .795 2 8 .15 2 1 6 .4 14 . 76 268.0 OJ '7- . 27. 0 67.50 72 9 .0 4 3 7.6 1 84.1 14 15.0 5 48.0 384. 0 23 04 . 6 58.0 70 7.5 33 64. 7 6 7.6 119 0 .0 4 575. 8 76. 4 1 825 . 0 5840 . 9 84. 4 2625.0 7120. 10 91b.8 3 610 .0 8450 . - 512.5 -10 60 8 .7 - 20493 . 2 z: (MR - M1 )X 7 Table II (cont l . Load at B. Point. ~ilR MRY lv1£ ~ 2 1.4 1.26 22.95 3 12.0 30.0 324.0 4 22.6 111.0 849.00 5 23.0 264.0 1585.0 6 43.0 525.0 2490.0 7 52.6 926.0 3560.0 8 61.4 1465.0 4690. 9 69.4 2160.0 5850. 10 76.8 z,010. o 7060. -372.2 -8492.3 -26431. 1 ~ Load at c. Point MR MRY MRX 4 7.6 37.3 286.1 5 18.0 144.0 865.0 6 28.0 341.0 1625.0 7 37.6 661.0 2542.0 8 46.4 1110.0 3545.0 9 54.4 1695. 4685.0 10 61.8 2425. 5675.0 -253.8 -6413. -19123.0 2£ 8 Table II (cont) . Point. MR ,;.. Load at D MRY MRX 5 L/. 24.0 144.0 6 13. 158.5 754.0 7 22.1 3 89.0 1510.0 8 31.4 751.0 2395.0 9 39.4 1225.0 3322.0 10 46.8 1835.0 4~:S OO. 0 -155.7 -4382.5 -12425. ~ Load at E. Point Iv1R MRY 1v1RX 7 7.6 134 .0 514 .o 8 16.4 392.0 1254.0 9 24 .4 758.0 2060.0 10 31.8 1248.0 2920.0 -80.2 -2532.0 -6748.0 ~ Load at F Point MR MRY .MRX 8 1.4 3 3 .5 107. 9 9.4 292.4 791. 10 16.3 639.9 1500.0 -27.1 -965.8 -2400. b Load at G Point MR MRY rvlRX 10 1.8 70.6 165.4 -1.8 -70.6 -165.4 £ 9 Of course as the load advances from the crown to the springing line the computations become i'ewer and fewer. This is e a sily understood when it is remembered that the half arch is assumed to act as a cantilever and the mement on a cantilever is zero between a given load and the free end of the cantilever. From Table II then, we have the necessary quantities to f'ind. H0 , V0 , and M0 d.ue to a unit lead at A, B, C, D, etc. 'l'bese val ueE are found by subs ti tu ting in the following equations. Ho : n Z: MY - g M~ Y 2 ((.2Y) 2_ n.z,Y2J Z ( MB - Mr.)X 2 .2 x 2 2 M + 2 Ho 2 Y 2 n In these equations, the general equations of the elastic arch, ;s Y2 and ~ x 2 are for ½ of the arch 2 .M is for the entire arch and is equal to only. 2. MR ::;SY, -4' .Z ~v1 1 • The summation z tion of the products ( MR - i"11 }x ( MR - M1 )X is a summa- in which MR and M1 are the bending moments at corresponding points in the right and left halves wbich have eoual abscissae X; and the sumwation 2 ..:2 MY (lVIR • M1 ) Y. is for the enti1·e s,rch anct equals Tbe cornputa t ions e_ re given be low. 10 Calculations of Ho. Vo. Mo. for load of unity placed at A, B, C, D, E, F, and G successively. Load at A. Ho_ 10 (-10609) - (-512.5)(140.5) -33186 Vo e - 24Q93 = - .5 2 X 34093 Mo = e = + 1.025 - -51~? - 2 20 1.025 X • ~ Ho X 141 :: ... 11.2 11.2 - + 10.9 1.025 = Load at B. 10 (-8492) - (- 3 72)(141 ) -Z31E6 Ho = - - - - - - - - -·- --"- = Vo -372 Mo = e -- 26431 = -68186.'f ,.388 .98 2 - 20 X X 141 = t 5.77 5.77 = + 5. 89 = Ho ~ = .98 Load at c. Ho - (-253.8)(141) = 10 (-6413) -33186 Mo = - -253.8 - 2 X .854 X 141 20 Vo= -19123 681864 e = _ . 28 = M.o_ = .65 - + .762 Ho • 854 - 4 .854 = + 0.65 ll Calculations of Ho• Vo• Mo for load of uni t"tr placed " at A, B, C , D, E, and F successively. (cont) Load at D. Ho = 10 (-4382.5) - (-155.7)(141) = + • 688 - 33186 , - 12425 = - il825 68186.4 Mo._ -155.7 - 2 x.688 x 141 = _ 1 _95 20 e = = - 1.95 = . 688 - 2.87 Load at E. Ho - (-80.2)(141) = 10 (-2532) - 33186 Vo = - 6748 68186.4 = - .0988 Mo = - -80.2 - 2 X .421 X 141 20 e = -M :::0 Ho Load at = + .421 1.92 = - .421 = - 1.~2 = - 4.56 F. Ho = 10 (-965.8) - (-27.1)(141) -33186 Vo = -24 00 68186. 4 - = + .1765 - .0352 Mo = - - 27.1 - 2 X 20 .1765 X 141 = - 1.135 12 load at F (cont). e _o = -1.135 = - 6.43 =M Ho .1765 Load at G. - (-1.8)(141) Ho = 10 (-70.6) -33186 Vo = - 165.4 = 68186.4 - + .CD136 - .00243 Mo - - -1.8 - 2 X .0204 X 141 = - .1975 20 = Mo = e Ho -.1975 .0136 = - 14.5 These equations and calculations to follow pg. 9. H0 - .Z M 4Y -- n2 211/IY ~~y)2 - il42Y2] Vo = ) ) _.Z (MB - M1)X 2 ) ) 2..1 x 2 See pg. 338 Turneaure and Maurer. ) } .?i M .... 2 HQ .:SY Mo = 2 n 2 Dz Y)2 - M.2Y 2 ] : 2( 140.45 2 - 10 X 3629.3}: -33186. 2 ~ X2 = 2 X 34093. 2 2 I l : 2 X 10: 20. = 68186.4 13 Influence lines of Fibre Stress at Sections A------ G Inclusive. GRAPHICAL SOLUTION. An influence line of fibre etress shows the character and amount of stress at any section due to a load of unity passing over the bridge. In ord.er to s~~e graphically for the fibre stress at any section, the force polygon of plate II was layed with the load line 11N equal to unity. Then in order to determine the directions of the thrust lines, the shear (V 0 ), caused by the load at any point, is laid off on li'1N from N and the corresponding horizlontal thrust (H 0 ) is layed off as the pole distance. The rays drawn from M and N to the end of H0 give the directions of the thrust lines. The distance below or above the neutral axis, at the crown, of this thrust line is given by M0 /H 0 • If thi~ quantity is positive, the thrust line passe s above the neutral axis, if negative bel ow. With these polygons drawn, the bending moment at any section, due to a unit load at any section is given b y the vertical distance from the neutral axis at that section to the proper thrust line, multiplied by the corresponding pole distance H0 • The quantity thus found is the true bending moment at the given s ection. by the The fibre stress at that section can be computed equation fc = Mu/I= N/A, where fc = fibre stress, M equals the true moment at tbe section, u = the distance 14 from the neutral a:xj s to the firire in quest ion. ( :E'or our purpose u = distanc e to extreme fibre.) I s the moment of inertia of the section, N = th e thrustmormal to the section and A= area of the section. The above method is correct but it is too cumbersome. The f-0llcwing method is much shorter. M = Ne, where M = moment, N = thrust and e = eccentricitJ. If r = radius of gyration, Ar2 = I, then the expression f 0 = Mu/It N/A N r2 can be written, f 0 =Neu/It ux u I .. r2 ) u -N ( e_,rI But ( 1) u = d/2 r2 =I/A• d3/12d = d 2 /12 with unit width. d = a length. 6 + r2 is a moment which may be written u This new moment is equal to the thrtst N mult- :. N( as M'. e - iplied by the eccentricity e t the additional distance 2 -r = -d • Then if we take the center of moments at a u 6 distance et d below or above the gravity axis, or at 7' the lower or up~ edge of the middle third according as upper or lower fibre stress is desired we can obtain the fibre stress direct by multiplying this moment M' by u/I for the section. As will be seen from the above, the fibre stress varies directly with the moment M'. Therefore an influence line drawn for such a moment 15 will serve as an influence line for fibre stress. It should be noted that in the actual calculation of the moment id' it is much simpler to measure the vertical a istances b 1 d 1 and b 2 d 2 , from the ecl p:e of the middle third or kern, to the proper thrust line. These distances multiplied by the proper pole distance gives M' for upper and lower fibre stress respectively. For values of M' with loads on the left vhalf of arch consider the load on the right half and the comresponding section on the left half. Thus values of l'v5: ' Cfl,n be computed I point. for any section for loads at any If the thrust line passes above the upper edge of the kern, the values of M' are plotted as positive ordinates showing compression in the upper iibre and tension in the lower. If the thrust line passes through the kern Ho x (b1d1) is positive showing compression in upper fibre, H0 x (b¢1,l is negative showing compression in the lower fibre. If the thrust line passes below the kern both R0 x (b 1d 1 } and H0 x (b 1 d 1 ) are negative showing tension in upper fibre and compression in lower fibre. Table III gives the values of H0 x (b 1 d 1 ) and H0 x (b 1a 1 ) at sections A1B1 ------G for a unit load 16 moving across the arch. Plate III shows these values of M' plotted to scale. A word of explanation may be needed for Plate III. :For any section the value of the ordinates at any point give values of ::.1 1 due to a load of unity at that point on the a rch. The actual stress= M'u/I. Table III SOLUTION BY USE OF INFLUENCE LIUES. Unit Load at A. M S.ection. A 9.133 B 3.70 5.23 3.79 C .30 1.30 0. 308 1.33 D 2.90 4.7 2.97 4.82 E 4.00 6.10 4.10 6.25 F 2. 00 4.80 2.05 4.92 G 10.00 6.10 10.25 6.25 B' 5.00 3.60 5.13 3.69 C' 0.20 1.40 0.20 5 1.44 D' z.oo 4.60 E' 4.00 6.00 4.10 6.15 F' 1.90 4.60 1.95 4.72 10.0 6.10 10. 25 6.25 Unit load at B. 4.72 H0 = .98 M Section. b2d2 A 5.0 6.47 11.50 12.75 B 13.0 11.25 l7 Table III ( 111ont) Unit load at B (cont) M Section b1d1 b2d2 Hobldl Hob2d2 C 6.5 4.70 6.Z7 4.61 D 1.0 0.80 0.98 0.78 E 2.0 4.20 1.96 4.12 :E' 2.0 4.80 1.96 4 .71 1.7: ,.. 6.8 2.70 6.66 2.65 B' 1.5 o.o 1.47 o.o C' 2.0 3.5 1.96 3.43 D' 3.9 5.4 3.82 5.29 E' 3.4 5.4 3.2,3 5.29 I!' ' o.o 2.5 o.o 2.45 14.0 10.5 13.7 10 .3 "' \J \ Unit load at c. Ho = 0.854 A 1.6 o.o 1.36 o.o B 6.8 5.4 5. 81 4.61 C 13.9 12.6 11.85 1 0 .75 D 6.0 4.1 5 .1~. 3 .bO E o.o 2.4 o.o 2. 0 5 I!, 3.2 6.2 2.73 5.29 1.7: 1.1 3.8 .94 3.25 B' 2.6 4 .0 2.22 3.42 C' 5.0 6.6 4.27 5.64 D' 6.0 '7 • 6 5.13 6.49 E' 4.6 6.6 3.93 5.64 F' 0.2 2.6 0.17 2.22 ,rV:. ' 15.3 12.0 13.05 10.25 ('I 18 Table III (cont) Unit load at D. Ho= 0.688 Section b1d1 b2d2 Hob1d1 H0 b,!dz A 2.2 3.7 1.51 2.55 B ,;, 2 •U 0.9 1.58 0.62 C 8.4 7.0 5.78 4.81 D 16.2 14.8 11.15 10 .18 E 6.6 3.8 4.54 2.61 F 1. 3 5.0 8. 89 5 3.44 G 4.4 10.6 3.03 7.0 B' 5.4 7.0 3.72 4.81 f"I .., I 7.0 8.6 4.81 5.91 D' 7.0 8.6 4.81 5.91 E' 4.8 6.6 3 .30 4.54 F' 0.4 2.0 0 .2 8 1. 38 r, 'J ' 17.2 14.0 11.85 <;:. 64 Unit load at E. Ho = 0.421 A 3 .7 5.3 1.55 2.23 B o.o 1.4 o.o 0.59 D 12.5 11.0 5.26 4. 64 C 5.7 4.2 2.40 1.77 E 21.5 20.0 9.05 8 .43 F 4.4 0.6 1.85 0125 r, 19.6 28 .8 8 .25 12.10 B' 6.3 7.8 2.65 3.28 C' 7.3 8.8 3.07 3.70 J 19 Table III (cont) Unit load at E (cont) Ho = 0.421 M Section b1d1 b2d2 Hob1d1 Hob2d2 D' 6.6 8.2 2.78 3 .45 R' D 3.8 5.5 1.60 2.31 ]' I 2.1 0.1 0.885 0.092 (' ' 19.8 16.8 8.33 7.06 J Unit load at F. Ho = 0.176 A 5.7 7.2 1.001 1.27 B 2.0 3.4 0.353 0.60 C 3.3 1.8 0.582 0.218 D 10.2 8.8 1.86 1.55 E 19.1 17.6 3.37 3.11 F z.o. 5 28.9 5.38 5.1 1..7 r, 51.4 71.6 9.06 1 2 .7 B' 8.0 9.6 1.41 1.69 CI 8.9 10.3 1.57 1.82 D' 8.0 9.6 1.41 1.69 1?' J.:I 5.1 6.8 0.90 1.20 F' 0.9 1.4 0.247 0.15 ' 18 .6 15.8 3 .28 2.79 (' J 20 FLOOR DESIGN. The floor system is designed alent to a 20 ton truck. f§o a loading, equiv- This loading gives two con- centrated loads of 13000# each over the rear wheels of tbe truck as shown in the diagram. Since the maximum floor space is 5 feet the floor is designed to carry a concentrated load of 13000# at the center of the span. lo The moment coefficients used are for continuous beams. ( See "Hool' s Concrete Construction", volume II, pages 374-376.) An allowance for impact is given in the coefficents. The beams are designed as simple T beams, span 18', and loading equivalent of 20 ton truck. The girders are designed as continuous rectangular beams. Intermediate spans are 15', end spans are 25'. The complete calculations are given below. Note: For moment coefficients see Hool, pages 374376; floor is de signed for impact due to live load. Int ermed ~~t e span~. Span= 5'- 0". Live load= 13000#, concentrated at center of span, due to twenty ton truck. Dead load = 150ff ft~ ( :H'loor assumed 12" thick) M = .191 x 13000 x 5 x 12 21 M = 149 000 inch pounds due to live loaa. including impact. M = .085 X 150 X 5 X 5 X 12 ~ 3 830 inch polllldS d.ue to dead. load. Total M = 149 000 +. ..,,r;z, 830 = 153 000 inch p ounds. 2 Mc = fc/2 x bkjd = 325 X 12 X .38 X .875 d2 = 153 000. 153 000 325 X 12 X .38 X .875 d = d • 10.85 inches net depth. (12" gross depth,. Steel Area Ms= Afcjd • 153 000 inch pounds. 153 000 A = = .995 square inches of steel. 16 000 X .875 X 11 Use ½" square twisted rods through out floor spaced 3 11 • End Spans Span = 5'. Loading same as for intermediate spans. Live M = .211 x 13 000 x 5 x 12 s Dead M 164 500 inch pounds (includes impact) = .116 x 150 x 5 x 5 x 12 = 5 220 inch pounds Total M = 164 000 + 5 220 = 170 000 inch p ounds Ms: fc x bkjd2: 170 000 a2. 110 ooo 325 X 12 X .38 X .875 d 11.41 inches net depth of floor - 22 Steel Area 170 000 16 000 X .875 X 11 A= BEAM Span = 1.27 square inches. DESIGN. 18'. D€signed as simple T beams for loading equivalent to a twenty ton truck~ Moment coefficients increased in ratio of 191 , for 130 impact (see Hool, page 376). Dead load of floor= 750 lbs. per ft. of beam II IT " beam • 250 II " " " II Live load Moment = 13 000 X 6 X 12 X 191 130 = 1 370 000 inch pounds. Dead load Moment = .13 X 1 000 X 18 X 18 X 12 = 505 000 inch pounds. Total Moment = 1 875 000 inch pounds. bt(d - !t) Mc = !c 2 d = 1 87b 000 + 6. 325 X 30 X 12 = 22 inches, minimum allowable depth. Use 28 inches for 1'1 6 - 11 d" otherwise% steel runs too high. Asfs {d - ½t) 1 875 000 = 5.32 square inches of steel. 16 000 X 22 % steel= 5.32 = .59% SJOO 23 Shear Considerations Allowable unit shear on concrete = 10 0ft ( vvi th stirrups) Area carrying shear= width of stem x (d - Jt) Effective area for beam in question= 12 x 22 = 264 in~ 100 x 264 = 26 400 lbs. shear carried by concrete. Total shear on beam= 22 000 lbs. Use stirrups of ¾-" steel wire, spaced 15 inches. Intermediate Girder Design Span 15'. Note: Girders designed as continuous rectangular beams. Loading = 28 000 lbs. @ 1/3 points ( Impact factor applied= l91. 130 Moment coefficients from Bool's Concrete Construction Volume II, page 376. Assume 16 x 20 Girder. Load per foot of Girder due to its own weight equals 330 pounds. Moment - .122 x 28 000 x 15 x 12 - 615 000 inch lbs. due to beam loads at 1 pts. 3 Dead Load Moment= .086 x 330 x 15 x 15 x 12 = 76,700'f1 in.lbs. due to wt. of gir. _;J I~ Total Homent = 692 000 inch pounds M0 - !.a. x bkjd2 = 692 000 2 24 6~2 000 - - - - - - - - - - - = 400. 325 X 16 X .38 X .875 = 20 inches, net depth d Use 22 gross II below slab. II Ms= Asf 8 jd 692 000 16 000 X .875 X 20 = 2.47 square inches of steel. % Steel• 162 • 47 = .775~ X 20 Shear. Maximum shear= 30 500 pounds. Allowable unit on concrete if stirrups are used= 100# Allowable Total Shear= 100 x 16 x 20 = 32 000# Use¼" wire stirrups spaeed at 10 inches. Girders (End Spa~) Span 25 feet. Assume 20 x 26 Girder. Load from b€ams at l points - 28 OOOff (Includes impact) 5 Moment over support= 4.47 x 28 000 x 12 = 1 500 000 in. lbs. (From beams) (See Hool's Con. Const. p 359) Deed load Moment - .107 x 540 x 15 x 15 x 12 ( Wt • Gi rd er ) = 156 000 inch pounds. Total Moment Mc =½fc bkjd a2 _ d = 1 656 000 inch pounds. 2 1 656 ooo Z25 X 20 X . 38 X .875 = 765. = 27.7 inches Use 28 inches net depth 20 x 30 beam (Overall) 25 Shear. Maximum shear= 62 000 pounds 100 x 20 x 26 ~ 52 000 pounds carried by concrete Use¼" wire stirrups spaced at 14 inches. Stress in stirrups ~ 10000 x 14: 5 000 lbs. • 28 At 120 pounds per square inch concrete will carry total shear. Bridge Loading From a consideration of the influence lines on plate III we see that any loading which will give a maximum fibre stress at G will probably be a safe loading for which to investigate the arch. Consequently live loads were p1acea. so as to give maximum fibre stress at section G, i.e. at ] 11 , E', D', C', B', A and B. Since spand.rel coulm are used to take tbe floor load to the ring it was thought unnecessary to use a concent r ated floor loading when investigating the arch,hence a uniform loading, which would give a column load equal to or greater than the assumed concentrated load was used. The loading as finally de- cided upon was 150 lbs. p e r square foot extending from th e left end of the arch up to point B. Th e total load per column per foot width of arch rin g is shown on plate lV. Now having th e loads at the various points F ' E ' F given, we can find the fibre stress at any section 26 causeo by any load by simply multiplying the proper ord_inate as scaled from the influence line on Plate III, by the corresponding load. The total stress is f ou.nd by sum- ming algebraically the products of the ordinates and the corresponding loads and multiplying the quantit.;y thus found. by ; for the section in qu.eetion. take section C. For example The algebraic sum of the rpoducts of ord inates times 1oaas equals 111 ~940 for upper fibre and 108 : 050 for lower fibre. The stress equals 111 940 X .19,9 = 165 lbs. per in? compression in upper fibre 144 ----·-- ' or 108 050 X .199 144 149 lbs. per in~ compression in 1 c.wer fibre Table IV gives the complete data and calcul&tions for fibre stress at sections A------G inclusive. From these calculations we sec that the maximum cempression stress is at section A with a value of 422 _pounds per square inch. The maximum tensile stress is at G, with a value of 21 pounds per square inch. Both of these values are entirely within safe limits, especially since we have .62510 of steel in top and bottom of arch. 27 Table IV. Not e: Positive signs show tension. Negative " ,, compression. FIBRE STRESS - Section A. Load at Section Ordin. to Influ. Line. Up . Fib. Low . Fib Total Load i~loment = M1 Cpper f ib re Lower fibre + + F' "-1.20 -1.50 44 000 52800 66000 E' +1.65 - 2 . 40 26 000 42800 62400 D' +1.65 -2.6 5 20 000 23000 53000 CI -l. Z5 o.oo 19 000 257 50 B' -6.40 +4.90 18 000 115000 88300 A -11.75 +9.35 18 000 210000 168500 B -6.75 +5.10 16 000 108000 81600 C -1.35 0.00 16 000 21600 0000 D -t-1.65 -2.65 17 000 28100 45100 l<' .L' +l.65 -2 . 40 23 000 38000 55200 F +-1. 20 -1.50 :39 000 46750 5 8 500 2-i 480?50 0000 241450 340200 3384 00 Upper fibre strESS = (480350 - 241450) .255 = 144 422 pounds per square inch compre ssion. Lower fibre stress = (340200 - 3;jf3400) .255 = 144 3 .l9 pounds pto-r square ir:ch compression. FIBRE STRESS - Section B. L oad at Sect:ton Total Ord in.to Influ. Line Loaa.. Up .Fib. Low.Fib. Moment = M' Upper .Jibre Lower Fibre. + T F' +1.40 -1. 65 44 000 61600 72600 E' +2.60 - z.20 26 000 67600 8Z200 D' +3. 75 -4.75 20 000 75000 95000 28 Table IV (Cont) Load at Section FIBRE STP~SS - Section B. (Cont ) . 0rdin.to Influ. Total Moment= M' Line Load. Upper fibre Lower fibre Up.Fib. Low.Fib. ~ ~ C' 4-2.10 B' -1.50 o.oo 18 000 27000 A -5.30 +3. 75 18 000 95500 67500 .5 -12.80 +11.00 16 000 205000 176000 C -6.00 +4.80 16 000 96000 76750 D -:-1.70 -t-0.70 17 000 28900 11900 E 00.00 -0.59 23 000 0000 F + • 353 - . 60 39 000 19 000 ~ 39900 63750 0000 13600 13800 23400 ----- 452400 2b7900 531550 2·32150 Upper Fibre stress= ( 4 52400 - 257900) .2 415 = 144 2 26 pounds per square inct compression. Lower Fibre strees = ( 351550 - 33215 0) .2415 = 144 32.6 pounds per square inch compression. FIBRE STRESS - Section C. Load at Section Ordin.to Influ. Total Moment= M' Load. Upper fibre Lower fibre. Line + Up.fib. Low.fib . ~ .E'' -t-1.57 -1.82 44 000 69100 80100 E' +Z.07 -3.70 26 000 79~00 962 00 l) I +4.81 -5.91 20 000 96000 116000 (;j I +4.27 -5.64 19 000 81 200 1074-00 1' +1.96 -Z.42 18 000 Z,5300 A' - . 308 -1 .40 18 000 B -6.37 +4.61 16 000 102000 73600 C -ll.t15 -1-10.75 16 000 190000 172000 5540 61700 25200 29 Table IV (Cont) Load at Section :FIBRE S'l RESS - Section C (Cont) . 1 Ordin.to Influ. Line Up.fib. Low.fib Total Moment= ~ ' Load. Upper fibre Lower fibre. ~ D -5.78 +4.81 17 000 98000 81600 E -2.40 +1.77 23 000 55200 40750 F' - .582 + .318 39 000 ~ _________ _____ 22700 ....;.._ 12.;;. oo 4 73440 3 61500 488600 380550 -Upper Fi'bre stress= ( 472440 - 3 61500) .199 = 144 165 pounds per square inch compression. Lower fibr e stress= ( 48 8 6 00 - 380550) .199 14 4 149 pounds pe 1· square inch co rripr e ssion. FIBRE S~l:'RESS - Section D. Load at Se ction Crd in. to Influ. Total Aoment = M' Load . Upper fib r e Lower fibre Line Up.fib. Low.fib. T\ I .J.J +1.41 -1.69 44 000 62000 74400 +2.78 -Z.45 26 0 00 72400 RS750 +4.81 -5. 91 20 0 00 961 0 0 118500 J g 000 97800 12 3500 CI B' +3 .82 -5.29 18 00 0 68800 95100 A ;-2.98 -4.72 18 000 53800 85000 B - • 98 - . 78 16 000 15750 C -5.13 +3.50 16 000 82250 56100 D -11.15 +10.18 17 000 1 8 9500 173000 -5.26 +4.64 23 000 121000 106500 -1.80 +1.55 2 9 000 60500 F Z 704 0 0 12500 478900 450 900 598750 3961 00 30 Table IV (cont) FIBRE STRESS - Section D (Cont). Upper Fibre stress= ( 4789 00 - 4 5 0090) .1724 = 144 34.5 pounds per square inch compressiotj, Lower Fibre stress= ( 598750 - 396 lOOJ .1724 = 144 242 pounds per squarE; lnch compression. FIBRE STRESS - Section E. Load at Section Ordin.to Influ. Line Moment= M' Lower fibre Total Load. Upper fibre Up. fib . Low.fib • F' -t -t + • 90 -1.20 44 000 39700 52800 +l. 60 -2.31 26 000 41600 59900 90000 D' +2. z,o -4.50 20 000 66000 C' +3.93 -5.64 19 000 74800 107500 B' -5.29 18 000 60000 A -6.15 18 000 73900 111000 95100 B +l. 96 -4.12 16 000 31400 66000 C O:iOO -2.05 16 000 00000 32800 D -4.54 +2.61 17 000 77100 44400 E -9.05 +8.43 2Z 000 208000 19~000 39 000 lZ-1500 12lfi00 _z; 416600 387400 615100 359900 ) .1250 = 144 24.5 pounds per sauare inch compression. Lower fibre stress= ( 615100 - 359900 1 .1250 = 144 221 pounds per square inch compression. Upper fibre stress= (416600 - 387400 31 Table IV (Cont) Load at Section FIBRE STRESS - Section F. Ordin.to Influ. Line Up.fib. Low. fib. Total i.\.G.ornent = M' Load. Upper i'ibre Lo wer fibre. F' - .247 -0.]E, 44 000 10900 6600 E' - .885 -0.042 26 000 23000 1100 D' - .28 -1.38 20 000 5600 27700 .17 -2.22 190000 3240 42200 B' o.oo -2.45 18 000 0000 44000 A +2.00 -4.90 18 000 36000 88000 B +1.96 -4. 71 . 16 000 ~')1400 75500 C +2.73 -5.29 16 000 43600 84500 D + .895 73.74 17 000 15300 6 ~700 E -1.85 -0.25 2 ,r;;_, 000 :B' -5.38 +5.1 39 000 210000 ("11 t- I.I $ 42600 5750 199000 292100 129540 439050 199000 Upper fibre stress = (292100 - 129540) .0975 = 144 110 :pounds per square inch compression. Lower fibre stress = (439050 - 199000) .0974 = 144 163 pounds per square inch compression. FIB:.='.E STRJi~SS Load at Section - Section G. Ordin.to In flu. Line Up.fib. Low. fib Total MoY!lent = r,o l Load. Upper fH:re Loer fibre. Li. t- + ]' I -2,. 28 -t2.79 44 000 144000 123000 E' -8.2'3 .J..7 .09 26 000 217000 184000 D' -11.85 +9.64 20 0 ('0 237000 193000 C' -13.05 +10.25 19 000 249000 19 4500 B' -12', .70 +10.30 18 000 247000 1850()0 32 Table IV (Cont) Load at Section FIBRE STRESS - Section G (Cont). Ora_in. to In flu. Line Up.fib. Low.fib. Total Moment = M' Load. Upper fibre Lower fibr€. .:.. A -10.25 +6.25 18000 184000 112800 B -6.66 +2.65 16000 106800 42400 C -0. 94 -3.25 16000 15100 D +3.03 -7.30 17000 51600 124000 E t8.25 -12.10 23000 190000 278000 ]' +9.06 -12.70 39000 354000 495000 ~ 1421900 59b600 949000 1033700 52000 Upper fibre stress = ( 1421900 - 5 ~ 5600 ) .0355 144 - 204 pounds per square inch compression. Lower fibre stress = (1069700 - 8Z9000) .0355 = 144 21 pounds per square incb tension. PROPERTIES of SECTIONS A-----G• I = u/I 2.0 1 7.85 .2550 4.2 2.1 8.70 . 2415 C 4.6 2.3 11.54 .1990 D 5.0 2.5 14.50 .1724 E 6.0 3.0 24.00 .1250 F 7.4 3.7 28.00 .0~74 G 12.0 6.0 169.00 .0355 Section Depth u = d/2 A 4.0' B = le - 151 8 33 FINAL THRUST LINE CALCULATIONS. The final calculations of H0 , V0 , and M0 due to assumed loads are given in Table V. ref er to plate IV. The quantities all .i;JIR and M:1 were obtained graphically as explained on page 348 of Turneaure and Maurer. the values of H0 , After X , and M0 were found., the true equilib- rium polygon or thrust line was drawn in as follows. The loads at ]'', E', -------.li are layed off verti1 cally to sea.le as the load. line MN. ]'rom the middle of load A, v0 is layed off towards M (V 0 is positive). From the upper end of V0 , H0 is layed off as a pole distance. The rays are drawn in as usual. The strings of the equil- ibrium polygon are drawn in ptirallel to the proper rays of the force polygon. The eccentricity e of the e quilib- rium _polygon at the crown is given b:y· Mo • Ho It is seen that the thrust line falls within the kern at all but two points, i.e. D and G. According to the analytic solution "G" is the only section at which the thrust falls outside the kern. This seems to show an error but upon inspection it will be seen that at D the thrust is very close to tte lower edge of the kern. This indicates that at this section the upper fibre is in tension, whereas actually there is 34.5 pounds compression in the upper fibre. This discrepancy simply lies in the slight error in the graphival solution. On the whole, ho Beter, the thrust line checks the analytic 34 solution very well, i.e. it lies close to the upper egge of the kern where the compression is large in the upper fibre, ana close to the lower edge of the kern where the compression is large in the lower fibre. This completes the analysis of the arch stresses. Plate V shows the completed structure in elevation, and section.