'l '1. MI I l“ MIMI". *l’fl! .5 W“: II 00—: com ll) #4 ) THS ANALYSIS OF BOX CG?.‘~:’ER?S Thesis far the Degree of M. S, MICHIGAN STATE COLLEGE Raman D. Pam? 1953 \Y._‘.~ lf: 0', -:.. “1.". a; ,41’, - «a? , 1 riot; 'u e K. V, I .\ «if , '. 3“; ‘ .:’ , ' J‘ .-.~ . l -' A ' ark“ v _"V J ,V" |<'r- _ r v 'c-o This it to certify that the L thesis entitled ANALYSIS OF BOX CULVERTS presented by Ramanbhai Dahyabhai Pate]. has been accepted towards fulfillment ‘ of the requirements for Mes. degree in Civil Engineering Major prdfxuor r!‘ a Datej’l7‘ ’95.} -__—;--o ANALYSIS OF BOX CULVERTS 35’ Raman D. Patel A THESIS Submitted to the School of Graduate Studies of Michigan State College of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Department of Civil Engineering 1953 THESIS \33 U‘ (I u I. INTRODUCTION . A. Introductory Statement CONTENTS B. Purpose and Scope . . C. Acknowledgment . . . II. GENERAL CONSIDERATIONS . III. “Illxitjflwfi> IV. DESIGN TABLES FOR BOX CULVERTS ANALYSIS OF BOX CULVERTS . Virtual Work Method . . Moment Area Method . v. NUMERICAL EXAMPLES .‘ . . VI. CONCLUSION . . BIBLIOGRAPHY 303859 . Column Anology Method . Mbment Distribution Method Principle of Least Work Method SlOpe-deflection method . . . to m mm-PWUNI'OO? (D I. INTRODUCTION A. Introductory Statement. In the beginning of the twentieth century a new material of construction, reinforced concrete, came into existence. This material, being naturally continuous, demanded the study of Statically Indeterminate Structures. Hence, engineers were forced to learn more about the analysis and design of statically indeterminate structures. Accordingly various methods for the analysis of such structures have been developed. It is quite worthwhile to get well acquainted with these various methods. To achieve this purpose, box culvert which is one of the numerous statically indeterminate structures is taken as an example and various methods used to analyse it. Out of numerous methods only six commonly used methods have been used to analyse this problem. These six methods are: l. Moment Area Method. 2. Principle of Least Work Method (Castiglino's Theorem). 3. Virtual Work Method. 4. Slope-Deflection Method. 5. Moment Distribution Method. 6. Column Anology Method. 3/ B. Purpose and Scope. The purpose of this study is two-fold. First is to analyse the box culverts by various methods used in the analysis of the Statically Indeterminate Structures thereby getting well acquainted with methods used in this problem. End moments of the box culverts have been determined by six widely used methods. Second is to introduce design tables showing moments, shear and normal thrust at critical sections of the box culverts. These tables are prepared for designer or checker and the use of the tables herein will eliminate calculations, thereby saving time. Study has been restricted up to two cells a-symmetrical rectangular box culverts. But by going through this it can be seen that the box culverts having three or more cells can be analysed without much trouble. C. Acknowledgement. The author wishes to express his sincere thanks to Dr. R. H. J. Pian for his invaluable advise and suggestions without which this work would have been quite impossible. 4/ II. GENERAL CONSIDERATIONS A. Use of Culverts. Culvert is a traverse drain or waterway under a road, railroad, canal, or channel. In ordinary engineering usage, however, culvert refers to only short structures through roadway or railroad embankments serving as passageways for water and normally not acting under hydrostatic pressure. In this discussion culverts shall be refered as defined in the latter manner. Bridges were used to span Openings of considerable size and importance while culverts were relegated to minor openings. This distinction no longer applies arbitrarily as it is recognized that the efficiency and low cost of culverts make them desirable for a wide range of conditions. From a structural standpoint, culverts are advantageous because of their continuity. Un- expected loads or other unusual conditions are better resisted by culverts as all component parts contribute helpful restrain. B. Culvert Loads. Live Loads. These include moving concentrated super loads as truck wheel loads, with or without impact. For design purposes, the maximum. live load on culvert is taken as that produced by r heavy truck. In accordance with the common practice, 5/ the standard truck train loading of the American Association of State Highway Officials can be adopted. The H-lO loading is used for heavily traveled secondary highways, and even for primary highways in most localities. Pressures from wheel loads are assumed to be uniformly distributed on the culvert slabs gs there is, in general, an intervening earth fill. Impact effect may be included in this uniformly distributed load. Dead Loads. These include weight of embankment material on culvert, weight of the culvert and of contained water, lateral pressure on sides of the culvert. The total vertical load per lineal foot on a projecting culvert due to an embankment may be determined by the formula, P = Cc.w(B‘c)2 In which, w Weight of fill material per cubic foot. Bc - Outside width of the culvert in feet. Coefficient depending on the Cc depth of fill above culvert; width Bc; projection of the culvert above natural surface, and the character of the fill material. Weight of the empty culvert can be 6/ calculated very easily by assuming section. Lateral pressure. Lateral pressure in the embankments are of two distinct types: active pressure and passive pressure. Active pressure on culverts is caused by the action of embankment in attempting to assume its natural condition of repose-- that is its natural s10pe or angle of repose. Passive pressure is induced by the movement of structure against the supporting material. Its magnitude is a function of the amount of movement and also of the soil characteristics. Because of its indeterminate nature and changeable, uncertain magnitude, passive pressure is rarely depended upon to relieve stresses from more positive causes. Active pressure is direct function of vertical pressure, however, and may be estimated by fair accuracy (Rankine's Theory can be used). Active lateral pressures are assumed to be symmetrical on both sides of a culvert as care in back filling will permit large unbalanced pressures. Lateral pressure is computed in terms of equivalent fluid pressure. Its intensity depends on several factors, but is usually taken as one-third (angle of repose of the material 30 degrees) the vertical pressure at the point because it is more precise and safer usually. The load diagram is trapazoidal in shape and can be separated into uniform and triangular loading diagrams. 7/ Above loading conditions can be classified as follows: Uniform vertical load-total embankment and truck load in pounds per foot length of culvert. Uniform latertal load (symmetrical on both sides) equal to lateral pressure at top of the culvert in pounds per square foot. Triangular lateral load (symmetrical on both sides) equal to equivalent lateral unit pressure in pounds per square foot. A combination of II and III will give any desired trapazoidal type of lateral loading. 8/ III. ANALYSIS OF BOX CULVERT BY DIFFERENT HETHODS A. Virtual Work Nethod (a) Rectangular Box Culvert with single cell: Dimensions, moment of inertias of the component parts and loading are as shown in Fig. la. 4““ ”“2 prAx I RB’LIT RAY 987' 207 £~v + R L; at RC x C-H’ go :4: fW9.,/ap Due to symmetrical loading MA : MB and MC : Mb also from Fig. 1b. _ ‘ 2 _ - - RA xh _ Wh/3 + wh /2 4 Mb it and sAy _ RBy _ wL/2 Now relative change in lepe at each corner will be zero, therefore, CA = M.m.ds/EI = O which when simplified turns out to be as, Because horizontal deflection at any corner is zero, Ah = MIm.ds/EI - 0 which can be easily simplified as, - - 2 3 3 Mb(2hIl - 3LI2) + MAhIl - 4Wh 11/15 4 wL 12/4 # wh 11/4 . . . . (2) Solving equations (1) and (2) simultaneously MA and M6 can be determined. If instead of rectangular box culvert there be a square one, then by putting L = h and I1 - 12 two modified equations can be obtained which when i solved will give the corner moments. (b) Rectangular Box Culvert with two cells: Dimensions, moment of inertias of the component parts of the culvert and loading are as shown in the Fig. is. t'.’ lbs“. phi“ font ' 3.35... ”with" "s" + ” Fig. 3b It can be seen from the Fig. 3b = V4 } V7 Now, 2 v = i e " 111 MA I w I"1/2 Ms. . 2 v = .. 3L1 MD + w.L1 /2 MC Therefore, M-Ms‘M-M=o..........(1) A D c s 11/ Also, “ 2 VL =M 4w.L 2- 52 E (2/ NF 2 7L2 H+ Le/ G And, 2 v7 : , - 2Fl M%,4 w L1/2 MA - 2 Therefore, L(M in -M -M =L M in -M -M 2 B C A D ) l( E H F G) oeooeoooee(2) Also, v91: ; MA 4 w.h2/2 J. I'm/3 - MD vth = MF 4 W.h2/2 J, Wh/j - Me Now there are eight unknowns and the six additional equations may be expressed in terms of elastic energy. Since the relative horizontal displace- ment of the point A, at the left of the beam.AB, and the point A, at the top of the member AD. is zero, Therefore, (see Fig. 4a & 4b) A AA = fM.m.ds/EI = O X 44 Therefore, h L1 h oM.y.dy/EI2 4 £h.M.dy/EII t {away/axe = c 12/ Therefore, f-MB y. dy/E12 - (1/EI1 )f(MC h - wx 2h/s 4 v 4mm): - (l/EI2)Jr(MAy - w. h 2y/2 - W.y 4/3h 4 v9.y2 )dy = 0 Which when simplified turns out to be, 2 2 I 3 1121) M4 MhLI v! 1121 vn (‘12(A %)+Cl/l+4L1/1+9/312 3 - w.L1h/EI1 - w.h4/412 -‘Wh:/15I2 = e o o o o o o o o c c e o (3) Now the relative vertical displace- ment of the point A, will be zero, Therefore from Fig. 5a and 5b. A y A Apply a unit vertical load at Point A, Therefore,8 fluids/E11 i fM..L1.2ds/EI 4 fM. 2:. ds/EIl Therefore, 2 MALI/211 - w .L1/811 4v 1.1.1/311 4 ME L 1211/: .. w.L1/811 4 McLl/2I1 4 V4L1/311 Therefore, M 1.2/21: 4 M L2 4 L h/I 4 v 3/3I A1 1 01/le M21 2 411 1 3 = 4 t VIE/311 w.Ll/4Il ...........(4) +1“). ii). (164*: lb. A B E A 5 E W B E D C H D C H D C H Fig. 4 a. Fig. 5a. Fig. 6a. x «I LL Y -2 K Fig. 4b F79- 5’2 Frg. 6b. A mb. 1"?“5 E F174 5 E F e: F H G c H H 0 Fig. 7a. Frg 8a FI'Q- 96L x! *1 L2 c—q 7 Fl'q 76. 1579 8!). Fig. 9b. 13/ Now the relative change in the slope at point A, is zero. Therefore, A 0 = fM.m.ds/EI = 0‘ A A Apply a unit moment at point A, from Fig. 6a and 6b Therefore, M.ds/EIl 4 M.ds/EI2 4 M.ds/EIl 4 M.ds/E12 = 0- Therefore, JM L3/I 4 w.L2/6I 4 V‘L2/2I 4 h/I 4 L /1 All 11 111M32—.M011 3 - 2 1 4 w.L1/611 - V4L1/212 0 Therefore, (Ll/Il)(MA 4 MC) 4 MBh/Ig 4 (Li/2I1HV1 4 V4) 3 a w. 1/3I1 . . . . . . . . (5) Since the relative horizontal and vertical displacement and relative change in slope of the point F, at the right of the beam EF, and the point F, at the top of the member GF, are each separately equal to zero, therefore, following three equations can be obtained. F AF . fM.m.ds/EI - o F x <3 H E fM.y.ds/EI2 4 h Imds/EI} 4 fM.y.cIs/EI2 .-.- o F G ,H 14/ Therefore, from Fig. 7a and 7b. _ 2 4 3 3 MFh /2I2 4 w.h /812 4 W.h /1512 - 10h /3I2 - M L h/I 4 L3 /6 VLah/2I - h2/2I a 0‘ ‘e2 2 w2h I3"82 3 ME 2 ‘ Therefore, 2 I 3 2 (h /212)(MF 4 Ma) 4 MGLzh/13 4 Vloh /312 4 v8 L2h/2I3 4 3 3 . w.h /812 4 w.L2n/6I3 4 Th /1512 ............... (6) Now, F AF ., JM.m.ds/EI .-.- 0 Y F Therefore, from Fig. 8a and 8b. 4/ E F fM.x.dx/EII 4 L2 jM.ds/EIZ 4 M.x.dx/EIl = o' G- E H Therefore, 2 4 3 - ,GL2/213 4 w.L2/813 - Vale/313 - IOIEl-th/I2 4 ,2 .2 - - ,L -'V7L I : 0 Therefore, (L:_/2213)(MGr 4 g) 4 MELzh/Ie 4 (Lg/31¢ (W5 - vs) 4 w.L2/413 .......... . . (7) New, 0F F fM.m.ds/EI - 0‘- F 15/ Therefore, from Fig. 9a and 9b. F G- H E [MAS/Ell 4 fM.ds/E12 4 fM.ds/EIl 4 fM.ds/E12 = o E F. G H Therefore, 3 2 3 - + - — "' o MELz/IB w.L2/6I3 V‘5L2/213 MFh/I2 4 w h /612 Wh2/12I v 112 L / 4 3 1.2/21 4 2 - 10 /2I2 - Mala I3 w.L2/'613 - vb 2 3 - h I = O LIE/2 Therefore, 2 (1.2/13me 4 ME) 4 (11/12) (MF 4 ME) 4 (L2 /2I3)(v5 4 v8) 2 . 3 3 _ 2 4-va0h /’212 = w.L2/'313 4 w.h /’512 4‘wh /1212 O O O O O O O O O O O O O O (8) By solving the above eight equations simultaneously, bending moment at each corner can be determined. In this problem loading is general one and if any of these loading is missing then putting for that particular loading equal to zero new modified eight equations can be obtained from the above equations, and when these new equations will be solved simultaneously they will give moment at each corner of the culvert for that particular type of loading. 16/ B. Mement Area Method (a) Rectangular Box Culvert with single cell: Dimensions, moment of inertias of component parts and loading are as shown in Fig. la. 7--.]... .- MA “a L J I JEN: 93"}. RAV 987 4? 7 ?CV Li—DR L; x RCthJ’ NW) ”C i .F/y. la. Due to symmetrical loading, MA = M3 and MC = Mb. Now relative change in slope at any corner will be zero, therefore, the sum of the area under M/EI curve will be zero, which gives - — 2 3 3 MA 4 Mc (WM 11 4 wL 12 4 wh Il)/6(L12 4 1111) . . . (1) Also relative deflection at any corner is zero, therefore, statical moment of the area under M/EI curve about the corner will be zero. This will give _ 2 3 3 MC(3L12 4 21111) 4 MAhll - Wu 11/15 4 wL 12/4 4 wh 11/4 0 O O (2) Solving equations (1) and (2) simultaneously MI and M A C can be determined. U/ (b) Rectangular Box Culvert with Two Cells: Dimensions, moment of inertias of the component: parts of the culvert and loading are as shown in the Fig. 3a -< f. V -. mm @‘hw‘fifihs fi:g;vhn§ g4g=ytg. New, mh-54mémdg BH'%‘“@@4% Therefore, -%4%-%.o..n...Jp 18/ 2 v2L1-MB4wml/2-MA V4nlamciw.Li/2-MD Therefore, LAMB”. -M ->--L. -L1 .. (Le/513M145 ; 2MP MG 2MB 0. . . . . . (7 From the Figures 9a and 9b. p 6) 3 Im6.M.ds/M.EI F F That is area of moment diagram divided by I is ll 0 equal to zero. 21/ Which will turn out as follows, (h/2I2BHM #MC -ME-MF-M- MH)4Wh223/121 -(L2121N)(ME M? {. Mfi' Mk) 4 w .h:/12I2 # w.L :'=/6I3 .........(8) By solving the above eight equations simultaneously bending moments at each corner can be determined. In this problem general loading is taken but if any of these loadings is missing then putting for that particular loading equal to zero new modified eight equations can be obtained from these above eight equations, which when solved simultaneously will give bending moments at each corner of the culvert for that particular type of loading. 22/ C. Principle of Least Work. (a) Rectangular Box Culvert with Single Cell: Dimensions, moment of inertias and load are as shown in the Fig. l. 1W3. JL Total internal work in a structure are given by U7: (As. is 4 A1. Y1 4 As.2)/II The constant term As. Ye./EI will drop out at the time of differentiation and, hence, can be dropped from the very beginning due to symmetrical loading MA 3 MB and Mb = MD. Therefore, U = A1 Yi/EI 4 AsE/EI 23/ Now, AiYi : um?L 4 MbZL 4 Mhzh 4 (MA MC - 2MB 4 Mb2)h I 211 212 12 312 As? 3 WL3 MA 4 Wh3 (la 4 Mb) 4 Wh?(7MA } sue) 31 121 901 1 2 2 9Q ; O‘ which when solved will be, OMA 3 MA(3L12 4 2hIl) 4 Mth - wh312 - wh 11 1 - 7 Wh212 : o . . . . . . . . . . (1) ‘36— Also, g2 = O, which when solved will be, nd Mb(3L12 4 2h11) + MA hIl - w£2§2 - 1%231 - 8 ngg; ; o . . . . . . . . . . (2) 30 By solving these two equations Rm.and No can be determined. If the culvert is square instead of rectangular, then replacing h by L and 12 by 11 , moment at each corner can be very easily determined. .‘Ub. Fin 75 lb A 16n.lb. 1-'A B E A B E A B E D C Hg D C H D c H Fig 4 a. Fig. 5a. Fig. 6a. M :1— L1. 1 7 Y .2 x fig 4b Flab'b. Fig. 6b. 4Mb 1mm _ F B E rm 5 E F B E C H G C H (J i H a. Fig. 7a. Fig 8a. Flg- 94- ~ :L 1 L2 - H 7 7 r ’ 3L _‘ ‘ # F/g. 8b. Fig. 9b._ 24/ (b) Rectangular Box Culverts with Two Cells: Case 1: Dimensions and sections and loading are as shown in the Fig. 2. I‘x nqnWMHL _ .. "4: C v-vm L,g1:’::l::':‘"'1‘ Al'r‘l .‘ n" Lil‘gtg l.‘| "7 'l | l ." "-r'fi “. . ' l ",7- 1.14““ v ’ ' 'u' ' ;.;a fart“ Kai ' ‘ II _ ‘ C??? “E 11 l .. ‘3 . 3' . 'tl -, l vw ,L‘ Total internal work in structure is given by U = (As{§s 4 Ai{§1 4 As{Z)EI (See Appendix) Dropping the constant term.As{§1/EI and E, the total internal work then will be, U = (mil/I - ALE/I) Now, A1.3r'1/I = file/212 .1 (L2/612HMAMB 4 Mi - 21425) + film/211 4 (Ll/fiwmpmn + M? - 2%) * fill/211 * c 4GB) 4 74.1150. MCD = - 2.E.K1(2&c 4, an) 4 w.L2/12. "no = - 2.E.ml(2<9D 406) '- w.L /12. EDA = - 2.E.K2(2&D 4 67A) 4111/10. MAD = - 2.8'.K2(2 0A ”’13) — 47.11/15. Now, MAB4MAD=0 .........(1) Mm4MBC=o .........(2) MCB4MCD=0 .........(43) MDA4MDC=0 .........(4) When the values from the s10pe- deflections equations will be substituted in the above four equations and when simplified, they will be as follows: 29/ w.L.2/12 - 47.11/15. .........(1) fifth/15 - w.L2/12. .........(2) 2E [2.ec(xl 4 K2) 4 Klan 4. K263 2 1112/12 - 17.11/10. .........(3) 23 [zonal 4 K2) 4 K190 4 K291] = 17.11/10 - w.L2/l2 ' .........(4) From these above four equations QE[2-91(K1 4. K2) 4 K193 4 3291;] 23 [2.eB(K1 4 K2) 4 1:16A 41260] values of 9A, SB, 90, 9D, can be determined and from the values of the above quantities, moments at the corner of culvert can be calculated. If there is uniform lateral load w pounds per foot height of the culvert instead of the above discussed loading, then replacing w.h/15 and 44.11/10 each by w.h2/12 with proper signs, and putting w, the uniform vertical load equal to zero, four revised equations can be obtained and from which end moments can be determined the same way in the above case. If instead of a rectangular box L and culvert it is a square one, then by making h I1 = 12 in the above four equations, unknown "Thetas" can be obtained and,hence, corner moments of the culvert. 19/ Rectangular Box Culvert with two cells: Dimensions, moment of inertias of the component parts of culvert and loading are as shown in the Fig. 2. ‘- | I. < ‘1 ‘ o ' .4 l: 4; - - \ . 1 ‘4 a I ' _ '» ' '1 ' -9-2L‘ Fixed end moments can be easily determined and they are as follows with proper signs: n13 = w.L§/12 aha = w.L§/12 mm. = 11.1.2/12 Mm = 1113/12 Mfg =‘w.h/15 Md? = ‘w.h/1o MGH = w.L§/12 311 = w.L§/12 an - w.L§/12 “be w.L§/12 MhA =‘w.h/1o Fin 2 - w.h/15 K - values will be: xi = I1/11 ’ ‘1 = Iz/h ; K3 = IB/Lz 2&1 Therefore, the slope-deflection equations will be, n13 s-2EK1(2.9A 4 eé) 4 w.L:/lz MEA =-2EKi(2.GB 4 sA) — w.L1/12 NMNNNNNNMNNN .E.K3(2.9E 4 GE) 4 w.L§/12 .E.K3(2.9F 4 oi) - w.L:/12 .E.K2(2.GF 4 6G) -‘W.h/15 .E.K2(2.eG 4 SF) 4 W.héio .E.K3(2.6G 4 9H) - w.L§/12 .E.I:’3(2.eH 4 so) 4 w.L2/12 .E.K1(2.GC 4 on) 4 w.L§/12 .E.Kl(2.9D 4 so) - w.L§/12 .E.K2(2.8D 4 GA) 4 f.h/10 .E.K2(2.9h 4 9D) - w.h/15 .E.K2(2.SB 4 90) .E.K2(2.GC 4 SB) 4 : MAB ' 34D “b1 * ”so “41 * “11 = NEE * MFG = 411 * Men = 0 I. O O O O A [0 31/’ 32/ When the values from the 810pe- deflection equations will be substituted in these above eight equations, they will be: .. 2 — 2.3.[2.9A(K1 4 K2) 4 K163 4 K2913] - w.L1/12 - 14.11/15 0 o o o c o o o c (l) I 1 , = __ 2 2.1:. [2.sD(K1 4 K2) 4 K100 4 Kesfl 44.11/10 w.L1/12 O C O O O O O O O (2) — 2 20B. [200 = w. - 0 F012 4 m5) 4 K29]; 4 K3SG] 11/15 w L2/12 O O O O O O O O O (3) - 2 — 2.E.[2.9G(K2 4 K3) 4 1:st 4 KBGH] - 1112/12 - W.h/lO (4) 143(223 4 9R] 1' w.L2/12 - mafia? 2.E.[2.SC(K1 4 K2) 4 Klan 4 12915931213 éGSJ' '; w.L§/12 - w.L:/12 .........(6) 2.E.[2.QB(K14K2) 4 K1911 4 K290 4 And, -sE .........(7) —-% o o o o c o c o o (8) From the above eight equations when a? Q | solved simultaneously, eight unknown Theta-values can be obtained. The values of lepe thus obtained can be used in finding the moments at each corner of the culvert. 33/ If the load is uniform lateral load w pounds per foot height of the culvert, the moment at each corner can be obtained by replacing 'W.h/lO and W.h/15 by w.h?/l2 with proper signs and putting w-the uniform vertical load equal to zero, and then solving new equations thus obtained, to obtain slopes which can be used to find corner moments. If instead of two cell rectangular box culvert, there be two cell square one, then putting L1, L2 and h each equal to L and also 11' 12, and I3 each equal to I, above eight equations can be modified and these new modified equations can be used in finding moments at the corners of this type of culvert as usual. 34/ E. Mbment Distribution Method. Sign Obventions Used in Solving This Problem: Mbments which indicate tension on the inside face are taken as positive or negative otherwise. Shears that indicate summation of the forces at the left of the section acting outward when viewed from within the culvert are considered as positive. Square Box Culvert having symmetrical uniform vertical load w per foot length of the culvert and the same moment of inertia for all component parts. Distributing factors for this particular case shall be 1/2. Assuming 100 units of fixed end moments and distributing these moments according to above distributing factors, which is not very difficult, it was found that bending moment at each corner comes to 50% of the assumed fixed and moment. Similarly the bending moments at each corner due to uniform lateral load shall be 50% of the fixed end moment. J HUI iflhfllmj F/g.1. 35/ Same culvert as in the previous case with triangular load varying from zero at one end to w pounds per foot of culvert length at the other end as shown in the following Fig. Fixed end moment due to this type of loading will be at A and B equal to WQL/lS and at C and D, to WL/io, where W is total load on one side of the culvert due to above type of loading and it comes to w.L/2. Assuming fixed end moment at A equal to 100 units, consequently fixed end moment at C will be equal to 150, and distributing these moments according to the distributing factors equal to 1/2, the final bending moments at the corners A and B were found as 56% and the corner 0 and D, were 43% of the assumed fixed end moment at reSpective corner. 26/ Rectangular Cell Box Culvert: Dimensions and sections of the component parts are as shown in the Fig. 3. Load is uniform vertical load w pounds per foot length of the culvert. Distributing factors: Taking one foot width of culvert, 11 = 1(1.5t)3/12 and, * 1M11n1w :1- ‘“I‘1 1 12 = 1(1.t)3/12 LAMMMIHUWM .tst = I /L(1 5) r ‘ I“ + j Ki 1 ' -utflc»rw ~IAIe-tn-+4£~e = 2.25t3/12L and, T i K2 - t3/12L “)If‘J = 2025K v—n+ -1 .5 ‘ KL 2 MI . ,TIWTWT WII 41$an UH M! Therefore, . F79 5- Distributing factors will be, 1/1 4 2.25 and 2.25/1 4 2.25 v = 0.31 and 0.59. Due to symmetry the final bending moments at all the four corners shall be the same. Fixed end moment of 100 units was assumed and when distributed by the above distributing factors final moment at each corner came equal to 31% of the assumed fixed end moment. 38/ Box Culverts with Two Square Cells: Case 2. Dimensions and sections of the component parts are the same as in the previous case. Loading is uniform lateral load w pounds per foot length of the culvert height. Distribution factor will be the same as in the previous case. Fixed end moments are assumed as 100 units and these moments then were distributed according to prOper distributing factors. Moments at the four end corners turned out to be 66% and at central diaphram to be 33% of the assumed fixed end moments. 39/ Box Culverts with Two Square Cells: Case 2. Dimensions, sections and distributing factors are the same as in the previous cases. Loading is triangular lateral load. Fixed end moment at A and B were assumed to be 100 units and consequently fixed end moments D and G comes equal to 150 units, because fixed end moments at A and F due to triangular lateral load varying from zero at one end to w per foot height of the culvert at the other end, 11111 be w.h/15, and D and G will be W.h/lO, where W is the total lateral load. When these fixed end moments were distributed according to proper distributing factors, the final moments turned out to be: At A and F 71%: at D and G 62%; at B-E 36% and at 0-H 31% of the assumed fixed end moments. ['1 H «4 504‘ 1 +q.‘—- I I 4 l Ffigynfi 40/ Rectangular Box Culverts with Two Cells: Case 1. Dimensions, moment of inertias of component parts of the culvert and loading are as shown in the Fig.6. (1.5t)3/12 and K1 = 2.25t3/12.L 113/12 and K2 - t3/12.L I 1 I2 Therefore, distribution factors at all the outer corners will be .31 and .69 as shown in the sketch, and the distribution factors at the diaphram will be .41, .41, and .18. Assumed fixed end moments were 100 units and when these moments were distributed according to above factors, the final moments at corners A, F, G and D turned out to be 20% of the assumed fixed end moments. The moment at BrE and 0-H comes to 140%. Procedure is shown on the next page. H I; 1 W H NW 4 952‘ 1m _ | ~c-f m— 4— l 5‘“ ~o—wrf4- 1.1.... -— l——p4f-v 1 lot :4 W 4 HI I I W ¢¢I 0&7. MN: \MIW .M.]. m + .mwr m\.. b.+ Q l W I mT+ mk+ 0%.. b + co\‘ 90‘: s: mfl - . o + o L o__:_,__ 1 m s. £236.? h. cc: : L92 cc. 3:. 0k c5. $5 I me i m\‘. air .mmi. tax] 3 .. 0.: M+. .m+ N.. &N+. QN$ ov\+ 065$ H+ nNI! I N.. nw+ mu+ D I ~w‘ mi- wm+ h + l v; + aw I .N/M 93+. +“+_~__: 00:“ . / . ”.6? 3:39er Gr. 0 mw.T. ,3. we. cowl oo\t b i. em I *0 + .V\ $ Mn n ‘ mu\+ : l 4A. s + .m‘w “Will. I. o$<.. 42/ Case 2. Everything the same as in the previous case but loading is uniform lateral load. Assumed fixed end moments were 100 units, when distributed according to prOper distributing factors, the final moment at A, F, G and D turned out to be 81% and at B-E and 0-H 41%. Procedure is shown on the next page. Case 2. Everything same as in the first case except loading is triangular lateral loading. Assumed fixed and moments at.A and F were 100 units and conseq- uently at Gland D their value will be 150 units. When these moments were distributed according to proper distributing factors, the final moment at A.and F turned out to be 87% and at G and D 78%; at.B-E 43% and at 0-H 39% of the fixed end assumed moments. Procedure is shown on the next page. l5; 7+ 3.: 59+ tu— 00/5 TE- “ ‘1\“1‘\‘°'° “3“» 7“" (sue H «a a H' 000/"‘ I + + + \o ' ‘83 -— 4 24+ .33; I; 79" .9. :57 O .21 3 o 5N” “ H N N‘ do F NV:- H . :5]!leth 7 l9"?_+l +.,_ ; wfiflz‘QEflJ glwslng * *‘ ++ ‘4. A” I +‘ (' ' §+ 31 0 {9+ 3.. .211: G— 9/+ lg;- MQQ‘O‘O 0 % mg n N :‘H'lwui; rs? ?l?l‘l“°9 ,ré-| 00/— s/ - 99+ 9 _ 2+ Ht! ‘4 Q) I ,_45_ +45 Wm M \\ VG“ +l ON \\ ++ .55.: 27.. 9 + g _ I§+ bl“ 09/- (:3 ’9‘ 130””) E33; ' 00/+ \+ +\' 92" '2 52+ I: .61.: z+ T- L: 98+ 3.. :5: £1: 6+ 25.: 0;!— 3"- Wm“? - _ EFLLHHL ' 9/-53+ nh ¥ 9/- o\‘q\vn\ $3313: 12L 9" m * + ' L -—-- i 9'— 9+ {7+ Tf' 7+ 0 9//+ I: r.— b— _€_‘; 99- 51+ §5NWNH§ ongm 00/“ IF 4 #2“- b + z.— T- 1+ ‘9 T ‘0“ mt“ “N “1 ‘7 ‘0 Q n I than?» ++ 1 m I 4. 1) 45/ Box Culverts with Two Rectangular Cells: Case 1. Dimensions and sections of the component parts are as shown in the Fig. EB. Distributing Factors: 11 = t3/12. 12 = (1.25t)3/12 and I3 = (1.25t)3/12 xi = 2.25.t3/12L Ké = t3/12L K3 = 1.252t3/12L. Distributing factor for member BA = KEA/ K = 1.56/1-1.56-2.25 = 0.32. Similarly other distributing factors can be easily determined and their values are as shown in the Fig. 13. For uniform vertical load w per foot length of the culvert length, final moments after distributing assumed fixed end moment equal to 100 units for the span EF turned out to be as follows: At.A and D equal to 14% 'At F and G equal to 21% At B-E and C-H equal to 120%. so: /b.s///- i I 1 (at ~ = 13 L4 E ' 411‘; ‘5; {LE—”J’s:- '1'!»— :—.1 '4 C r G ?1. E L, l HIIIIIIHIIH IIIHHII __;E/g.g3 ‘4 + +l +4 +— 70 aloo «59 qn‘ IO 1. NM ‘1'! 1 N 4. '3 323‘? - +| 4| 2‘4. WM? H W _ ' * fl 4 fl ‘4 4. l [‘21: 3‘3 t+ 47/ Box Culverts with Two Rectangular Cells: Case 2. Dimensions, sections and distributing factors are same as in the previous case. Loading is uniform lateral load w per foot height of the culvert. Assumed fixed end moments at A and F were 100 units and at D and C were 100 units. These moments were distributed according to the prOper distributing factors and the final moments turned out to be as follows: At A and D equal to 75% At F and G equal to 81% At B-E and C-H.equal to 39% of the assumed fixed end moments. ((1 IIHIlIl/IA’LI/‘ICIEIHIIHII N l 'v~4 I + [\W I~+ '3. W5 H' 9' w €3P'fl‘ [72"? +1 ++ 29.: .1; f'+ Q .— gm 6/— 00/- . ITE- mfl mkiwhmst 00/+ $Hl**'+' 6? —— Q/+ z, —— ..— 7+ .12; 28+ 01 I.-. L: 9+ 9+ N ‘ 9—- ? SlmfanHWi . l‘ I 4 |+ . +1 7Z-I 3% amps; :[E t+ 9 + 8+ T” O .99! b,- g -r gt— oZ-r H 001% ~4 kg m o '1 ' G ‘ L 6? . 62- 00/- 68 f— 22'— Tl'* t .— T _. T— 7Ei: r-i 48/ Box Culverts with Two Rectangular Cells: Case 5. Dimensions, sections and distributing factors are same as previous cases. Loading is triangular lateral load varying from zero at one end to w per foot height of culvert at the other end. Fixed and moments at A and F were assumed to be 100 units and at D and G to be 150. These moments were distributed according to proper distributing factors and final moments turned out to be as follows: At.A equal to 79% At B-E equal to 37% At F equal to 55% .At G equal to 85% At C-H equal to 38% .At D equal to 74% of the assumed fixed end moments. i 8.3!; ”I; 21:. 9 + *b .— 951‘ 29- 09/- _ 7.9 '- mwummo WI :9 “‘2 99.31% |*|+H-00/+ ;+I++‘:‘ b9 " 9‘ 92+ _§__: Q + L: L”. ___.._— L 99" z_ ;:: -::= ‘4. 1.1. (.1 OI+ ”- mo ‘0 IV; .. o~ " 7Nq1\7+\\1*\‘“l IZZ‘YV-p1 4‘+ |++ *+++ l4. (“é—é ‘ +‘ +\ \ t+ ‘ Cf- r+ ‘f r+ c— ’1' f): 017+ ” —-—-——1‘ L: ', Li + S.— 5-4 d +‘ 29- 02+ fi 0\ .. .. «m‘ mqe? w :7“: ‘9 +7! 4- 1+ + I69“> \ Ox t\ l 49/ F. Column Anology. Sign Conventions Used in Solving This Problem: Bending moments in the beams will be considered positive if they produce tension on the inside of the elastic ring. Shears are positive if they accompany positive rate of change of bending moment, increase up or to the right. Square box culvert symmetrical loading and same moment of inertia for all component parts. Loading is uniform vertical load w per foot length of the culvert. ‘th'Sai/fto C? At~ L J; f7§i4L E1 is same for all members and is taken as unity. Area of Anologus Column, A = 4L. L.L 4 2L.ng = L/2. L 2L.(L/2)2 4 2.L?/12. 5.9/12. 5? Ix: = Iyy 50/ Load on Anologus Column: w = 2.w.L?L/3.8. = w.L3/l2, Total load = 2.w.L3/12. = w.L3/6. Therefore, fA = MA = w.L3/6.4.L. = w.L2/24. (Mix = Hyy = 0; because loading on the anologus column is symmetrical) Due to symmetry the moments at all corners will be the 88.1116 . Rectangular Cell Box Culvert: Case 1. Dimensions and sections of the component parts are as shown in the fig. (a) below. Load is uniform vertical load w per foot length of the culvert. 51/ Rectangular Cell Box Culvert: Case 2. Dimensions and sections of the component parts are the same as in the previous case. Load is triangular loading, uniformly increasing from one end to the other end according to the law w.y/h. Therefore, Total load 7 =- (w.h/h)(h/2) '= w.h/2. .~—*..q <~ ' V 3 1/‘12 w /é5//‘{- FWgaj' Area of the section of the anologus column, A = 2(1..I2 .1 h.Il)/Il.12 Load on the column will be area of moment diagram due to load alone, 11 elf/12.12 1' Vang/612 52/ E‘ (WIS/612H8/15 - l/2)h = W.h3./18012 Mary = 0 Ixx = 2.(L/’Il) > 5” I :4, . . I" . 1 _ ..T‘ g f , .. , i.._\ .\ t' .‘K . ._... .~ 9 , 3-: . , H... . . ‘ ,N . ' h‘.’.- l t Il§r".". _' {‘0‘ .‘ 5: ‘. - "' . I .. 'l .r‘ “ . i ' ' ." ‘n . , - ' fi‘ . p. '.;'| ‘og "v .. \ ' _ a ..s ‘ ' ' , 1 , 'g ' H .. ', ' ‘. ' " _ ' ." ‘u .I ' 7 . Q . I . ' r . . ' .‘ I '9 H V ‘ ' . h' I. ' t- ' " i ' ‘ ' ‘ ‘ o l I ' ~ I ' - T . a 1 . ' ' .1: - ' M ”L: \I .1 \fi'v.' I ..l L . ‘ 9 ,_ ' ' '. O " p -. . ~ ' ‘. ' * V . . v . $1 . .t"..f‘.‘, ' .’ . ,“ (\ I, . '0, ‘ Y . s ""\' “ ' Av‘ _. ‘ . ’ , 0 -_ ‘ . . .. 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