Structuran Design of Reinforced Concrete Culverts.pdf

8/10/2019 Structuran Design of Reinforced Concrete Culverts.pdf

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Structural Design for Reinforced Concrete Culverts

Prepared by: - Consultant Engineer

Raad Mohammad Dhyiab

M.Sc., Structural Engineering


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CONTENTS

No. Details page1- INTRODUCTION 32- Types of culverts. 33- Box Culverts. 54- Design Reinforcing of Culvert 55- Methods for Structural Analysis 106- Examples Design of Culverts. 337- References 48


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1-INTRODUCTION

A culvert is a hydraulically short conduit, which conveys stream flow through aroadway embankment or past some other type of flow obstruction. Culverts areconstructed from a variety of materials and are available in many different shapesand configurations. Culvert selection factors include roadway profiles, channelcharacteristics, and flood damage evaluations, construction and maintenance costs,and estimates of service life (1).

Culverts are required to be provided under earth embankment for crossing of watercourse like streams, Nallas across the embankment as road embankment cannot beallowed to obstruct the natural water way. The culverts are also required to balancethe floodwater on both sides of earth embankment to reduce flood level on one side

of road thereby decreasing the water head consequently reducing the flood menace. Culverts can be of different shapes such as circular, slab and box. These can be constructed with different material such as masonry (brick, stone etc.) or reinforcedcement concrete (2).

2-Types of culverts.

The main types of pipe used in highway construction are concrete pipe, metal pipe(steel or aluminum), and plastic pipe (high-density polyethylene and polyvinyl

chloride). They are available in a wide array of sizes, shapes, and properties some ofthe characteristics of these pipes are reviewed below. It is well known that roads aregenerally constructed in embankment, which come in the way of natural flow ofstorm water (from existing drainage channels). As such flow cannot be obstructedand some kind of cross, drainage works are required to be provided to allow water topass across the embankment. The structures to accomplish such flow across the roadare called culverts.

It is well known that roads are generally constructed in embankment, which come inthe way of natural flow of storm water (from existing drainage channels). As, suchflow cannot be obstructed and some kind of cross drainage works are required to beprovided to allow water to pass across the embankment. The structures toaccomplish such flow across the road are called culverts, small and major bridgesdepending on their span, which in turn depends on the discharge.


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2-1:- pipe culverts:-

Concrete pipe is manufactured as no reinforced, reinforced, or cast-in-place pipe; asbox culverts and special shapes; and as field-constructed pipe. Shapes, as shown inFig. (1) Include round, horizontal and vertical ellipse, and arch configuration.

No reinforced concrete pipe is available in diameters from 10 to 90 cm and threestrength classes. No reinforced concrete pipe is available as round pipe only.Reinforced concrete pipe is available in diameters from 30 to 365 cm. The strength ofreinforced concrete pipe can be specified according to five standard pipe classes(ASTM C 76), with Class I pipe being the most economical and Class V offering thegreatest structural strength; according to required D-load strength (ASTM C 655); oraccording to a direct wall design. (ASTM C 1417). Wall thickness of reinforced

concrete pipe can be varied to meet in field conditions. The standard classspecifications for pipe give wall thickness according to three distinct types, which varyfrom Wall A, being the thinnest, to Wall C, being the thickest Steel reinforcing forreinforced concrete pipe can be arranged in many combinations to meet the givenstructural requirements. Figure (2) Shows some of the steel reinforcement layoutsused in manufacturing reinforced concrete pipe (3).

Fig. (1) Concrete pipe is manufactured in five common shapes; regional custom anddemand usually determine availability.


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Fig. (2) Concrete pipe culvert reinforcement notation. (From PIPECAR: User andProgrammer Reference Manual, FHWA, 1989, with permission).

2-1-1:- Cast-in-Place Pipe.

This type of no reinforced pipe is formed in a trench using a continuous process. First,a trench is excavated so that it has a semicircular bottom and vertical or near verticalsidewalls, which serve as the outer form for the bottom and sides. The upper portionof the pipe is cast against an inner arch form as illustrated in Fig. (1) the form is pulledalong the trench while concrete is poured into a hopper located above Poweredspading mechanisms and variable-speed vibrators aid the flow of the concrete.

3-Box Culverts.

Box culverts are rectangular shapes with flat sides, top, and bottom . These shapes areconstructed with steel reinforcement. Factory-made boxes are shipped in sections 1to 3 m. long and joined in the field to make a structure of the required length. Asshown in Fig. (1).

4-Design Reinforcing of Culvert (4)

Structural design of reinforced concrete culvert and inlet structures is quite differentfrom design for corrugated metal structures. For reinforced concrete inlets, the

Designer typically selects a trial wall thickness and then sizes the reinforcing to meetthe design requirements.


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The method for the design of reinforced concrete pipe and box sections presentedbelow was recently adopted by the American Concrete Pipe Association and has beenrecommended by the AASHTO Rigid Culvert Liaison Committee for adoption by theAASHTO Bridge Committee. This design method provides a set of equations for sizingthe main circumferential reinforcing in a buried reinforced concrete culvert. Foradditional criteria, such as temperature reinforcing in monolithic structures, thedesigner should refer to the appropriate sections of AASHTO. Typically, the designprocess involves a determination of reinforcement area for strength and crack controlat various governing locations in a slice and checks for shear strength and certainreinforcement limits. The number and location of sections at which designers mustsize, reinforce, and check shear strength will vary with the shape of the cross sectionand the reinforcing scheme used. Figure (3). Shows typical reinforcing schemes forprecast and cast-in-place one-cell box sections. The design sections for these schemesare shown in Figure (4). For flexural design of box sections with typical geometry andload conditions, Locations 1, 8, and 15 will be positive moment designlocations(tension on inside) and locations 4, 5, 11, and 12 will be negative momentdesign locations. Shear design is by two methods one is relatively simple, andrequires checking locations 3, 6, 10 and 13 which are located at a distance dvd fromthe tip of haunches. The second method is slightly more complex and requireschecking locations (2, 7, 9, and 14) which are where the M/Vd ratio 3.0 and locations(3, 6, 10 and 13) which are located at a distance vd from the tip of haunches. Thedesign methods will be discussed in subsequent sections.

A-precast box sections


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B- Cast in Place Box Section

Fig. (3) Typical Reinforcing Layout for Single Cell Box Culverts

Fig. (4) Locations of Critical Sections for Shear and Flexure Design in Single Cell BoxSections.

Shear Design Locations :-

Method 1 3, 6, 10, 13

Method 2 2, 3,6,7,9,10,13,14


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Note: - For Method 2 shear design, any distributed load within a distance ( v d )from the tip of the haunch is neglected . Thus the shear strength at a locations (4, 5,11 & 12) are compared to the shear forces at locations (3, 6, 10 & 13) respectively.

Flexural Design LocationsSteel Area Precast Cast In - PlaceAS1

4 , 5 , 11 , 12 5, 11 , 12AS2 1 1AS3 15 15AS4 8 8AS8 - 4

Typical reinforcing schemes and design locations for two cell box sections are shownin Figure (5).

a- Typical Reinforcing Layout.

* See note, Fig. (4)

B Design Locations: two cell box culverts.

Fig. (5) Typical Reinforcing Layout and Location of Design Sections for Shear andFlexure Design of Two Cell Box Culverts.


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A typical reinforcing layout and typical design sections for pipe are shown in Fig. (6)Pipes have three flexure design locations and two shear design locations. Fig. (6) isalso applicable to elliptical sections. The details of flexural and shear for the Fig. (6)are.

A-Flexural Design Locations:-

1, 5: maximum positive moment locations at invert & crown.

3 : maximum negative moment location near spring line.

B Shear Design Locations:-

2,4 : locations near invert & crown where ( M/ V vd) = 3.0

Notes:-

1-Reinforcing in crown (Asc) will be the same as that use at the invertunless mat, quadrant or other special reinforcing arrangements are used.2-Design locations are the same for elliptical section.

Fig. (6) Typical Reinforcing Layout and Locations of Critical Sections for Shear andFlexure Design in Pipe Sections.


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5 - Methods for Structural Analysis

Any method of elastic structural analysis may be sued to determine the moments,thrusts, and shears at critical locations in the structure. Computer can completethe structural analysis and design of culverts very efficiently. The methoddiscussed below are appropriate for hand analysis, or are readily programmable

for a hand-held calculator.

5-1 Concrete Pipe Sections.

Using the coefficients presented in Figures (7 to 9), the following equations maybe used to determine moments, thrusts and shears in the pipe due to earth, pipeand internal fluid loads.

M = (cm1We+ cm2Wp+ Cm3Wf) B'/2 Eq. (1)

N = cn1We+ cn2Wp+ cn3Wf .Eq. (2)V = cv1We+ cv2Wp+ cv3Wf ... Eq. (-3)

Where:-

M =moment acting on cross section of width b, service load conditions,

Inch Ib, (taken as absolute value in design equations, always+)

N = axil thrust acting on cross section of width b, service load condition

(+ when compressive, - when tensile), Ibs.

V = shear force acting on cross section of width b, service load condition,

Ibs (taken as absolute value in design equations, always +)

= total weight of earth on unite length of buried structure, Ibs/fteW

nit length of structure, Ibs/ft.= weight of upW

ft.= total weight of fluid inside unit length of buried structure, Ibs/f W

B' = width, ft.

Cm, Cn & Cv = coefficients.

Figure (1-7) : provides coefficients for earth load analysis of circular pipe with 3loading conditions 1 = 90, 120 and 180. In all cases, 2= 360 - 1. These loadconditions are normally referenced by the bedding. Angle, 2. The 120 and90bedding cases correspond approximately with the traditional Class B and Class

C bedding conditions (2, 3). These coefficients should only be used when the sidefill is compacted during installation.


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Compacting the side fill allows the development of the beneficial lateral pressuresassumed in the analysis. If the side fills are not compacted (this is notrecommended), then a new analysis should be completed using the computerprogram.

Fig. (7) Coefficients for M, N, and V due to Earth Load on Circular Pipe.


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Fig. (8) Coefficients or M, N and V due to Pipe Weight on Narrow Support.


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Fig. (9) Coefficients or M, N and V due to Water Load on Circular Pipe.


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5-2 Concrete Box Sections.

The first step in box section design is to select trail wall haunch dimensions. Typically,haunches are at an angle of 45, and the dimensions are taken equal to the topslab thickness. After these dimensions are estimated, the section can then beanalyzed as a rigid frame, and moment distribution is often used for this purpose.

A simplified moment distribution was developed by AREA (8) for box culvertsunder railroads. Modifications of these equations are reproduced in Table (1) andTable (2) for one and two cell box culverts respectively. This analysis is based onthe following assumptions.

1-The lateral pressure is assumed to be uniform, rather than to vary with depth.2-The top and bottom slabs are assumed to be of equal thickness, as are thesidewalls.3-Only boxes with "Standard" haunches or without haunches can be considered.Standard haunches have horizontal and vertical dimensions equal to the top slabthickness.4-The section is assumed doubly symmetrical, thus separate moments and shearsare not calculated for the top and bottom slabs, since these are nearly identical.

4-1: Design Forces in Single Cell Box Culverts.

Fig. (10) Forces in Single Cell Box Culverts


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The equations cover the load cases of earth, dead and internal fluid loads. Anyone of these cases can be dropped by setting the appropriate unit weight (soil,concrete or fluid) to zero when computing the design pressures p v and p s. Theequations provide moments, shears and thrusts at design sections. These designs forces can then be used in the design equations to size the reinforcing based on the assumed geometry.

Table (1). Design Forces in Single Cell Box Culverts.

Design Pressures.

Eq. (4)

Eq.(5)

Eq.(6)

Design Constants.

Eq.(7)

Eq.(8)

For boxes with no haunches(HH= HV= 0)& G2= G3= G4= 0. Eq.(9)

Eq.(10)


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Design Moments.Moment @ origin:

Eq.(11)

Moment in top andbottom slab: Eq.(12)

Moment in sidewall:Eq.(13)

Design Shear.Shear in top and bottom

slab: Eq.(14)

Shear in sidewall:Eq.(15)

Design Thrusts

Eq.(16)

Thrusts in bottom slab:

Eq.(17)

Thrust in sidewall:


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Notes:-

1-Analysis is for boxes with standard haunches (HH= HV= TT).2-Equations may be used to analyze box sections with no haunches by setting

G2= G3= G4= 0.01- See Equation (18)for determination of

..Eq.(18)

2- If M8 is negative use A S min. for side wall inside reinforcing, and do not checkshear at section (9).

4-2 Design Force in Two Cell Box Culverts.

Fig. (11) Force in Two Cell Box Culverts.


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Table (2) Design Force in Two Cell Box Culverts.

Design Pressures.

Eq.(19)

Eq.(20)

Eq.(21)

Geometry Constants

Eq.(22)

Eq.(23)

For boxes with standardhaunches

Eq.(24)

Eq.(25)

Eq.(26)

For boxes withouthaunches

Eq.(27)


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Design Moments

Eq.(28)

Moments at Origin :

Eq.(29)

Boxes with standard haunches and uniform wall thickness(HH=HV=TT=TS=TB):

Eq.(30)

Eq.(31)

Boxes without haunches (HH=HV=0, TT=TB TS):

Eq.(32)

Eq.(33)

Moment on bottom slab:

Eq.(34)

Moment in sidewall:


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Design Shears

Eq.(35)Shear on bottom slab:

Eq.(36)

Shear in sidewall:

Design Thrusts

Eq.(37)

Thrust in bottom slab:

Eq.(38)

Thrust in side slab; boxes with haunches:

Eq.(39)

Thrust in side slab, boxes without haunches:

Notes:-

1- For boxes with standard haunches and all walls of the same thickness(HH=HV=TT=TS=TB) use Equation (31), Equation (33) and Equation (38).


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2- For boxes with no haunches and sidewalls with the amw or differentthickness than the top and bottom slabs (HH=HV=0, and TT=TB TS) useEquation (31), Equation (32), and Equation (39).

3- See Equation (18) for determination of X dc. 4- If M8 is negative, use A Smin for sidewall inside reinforcing, and do not check

shear at Section 9.5- Geometry constants F1 through F5 are not required for boxes withouthaunches.

5 - Reinforced Concrete Design.

5-1: Limit States Design Criteria .

The concept of limit states design has been used in buried pipe engineering practice,although it generally is not formally defined as such. In this design approach, thestructure is proportioned to satisfy the following limits of structural behavior.

1- Minimum ultimate strength equal to strength required for expected serviceloading times a load factor.

2- Control of crack width at expected service load to maintain suitable protectionof reinforcement from Corrosion, and in some cases, to limit infiltration orexfiltration of fluids.

Moments, thrusts and shears at critical points in the pipe or box section, caused by

the design loads and pressure distribution, are determined by elastic analysis. In thisanalysis, the section stiffness is usually assumed constant, but it may be varied withstress level, loosed on experimentally determined stiffness of crocked sections at thecrown, invert and spring lines in computer analysis methods. Multiplying calculatedmoments, thrusts, and shears (service conditions) determine ultimate moments,thrusts and shears required for design by a load factor (Lf) as follows:

Mu= Lf M .. Eq. (40)

Nu= Lf N . Eq. (41)

Vu = Lf V Eq. (42)

Where: - L f = Load Factor.


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Load Factors for Ultimate Strength: The minimum load factors given below areappropriate when the design bedding is selected near the poorest extreme of theexpected installation, and when the design earth load is conservatively estimated.

For culvert or trench installations alternatively, these minimum load factors may beearth pressure distribution are determined by a soil-structure interaction analysis in

which soil properties are selected at the lower end of their expected practical range.In addition, the suggested load factors are intended to be used in conjunction withthe strength reduction factors given below.

The 1981 AASHTO Bridge Specifications (4) specify use of a minimum load factor of1.3 for all loads, multiplied by coefficients of 1.0 for dead and earth load and 1.67for live load plus impact. Thus, the effective load factors are 1.3 for earth and deadload and 1.3 x 1.67 = 2.2 for live loads. These load factors are applied to the moments,thrusts and shears resulting from the loads determined .

Strength Reduction Factors: Strength reduction factors,, provide "for the possibilitythat small adverse variations in material strengths, workmanship, and dimensions,while individually within acceptable tolerances and limits of good practice, maycombine to result in understrength" . Table (3) presents the maximum factors givenin the 1981 AASHTO Bridge Specification.

Table (3) Strength Reduction Factors in Current AASHTO Standard Specifications forHighway Bridges.

Box Culverts Pipe Culverts

Precast (a) Cast-in-Place (b) Precast (c)Flexure 1.0 (d) 0.9 1.0 (d)Shear 0.9 0.85 0.9


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a.Section 1.15.7b. Section 1.5.30c. Currently recommended by AASHTO Rigid Culvert Liaison Committee for adoptionby AASHTO Bridge Committee.d. The use of a strength reduction factor equal to 1.0 is contrary to the philosophy

of ultimate strength design; however, it has been justified by the Rigid CulvertCommittee on the basis that precast sections are a manufactured product, and aresubject to better quality control than are cast-in-place structures. Because weldedwire fabric, the reinforcing normally used in precast box and pipe sections, candevelo p its ultimate strength before failing in flexure, the use of = 1.0 with theyield strength still provides a margin for variations equal to the ratio of the yieldstrength to the ultimate strength. If hot rolled reinforcing is used in a precaststructure, or if any unusual conditions exist, a strength reduction factor of 0.9,instead of 1.0, should be used in flexural calculations.

5-2 Design of Reinforcement for Flexural Strength.

Design for flexural strength is required at sections of maximum moment, as shownin Figure (4), Figure (5) and Figure (6).

(a) Reinforcement for Flexural Strength, A s.

.. Eq.( 43)

... Eq.(44)

d may be approximated as,

.... Eq.( 45) (b) Minimum Reinforcement.

For precast or cast-in-place box sections: min. As= 0.002 bh . Eq. (46)

For precast pipe sections:

47). (.. Eq /65,0002min. As= (Bi+ h) For inside face of pipe:-

48). (.. Eq../65,0002min. As= 0 75 (Bi+ h) For outside face of pipe:-

49). (.. Eq./65,0002min. As= 2.0 (Bi+ h) For elliptical reinforcement in -

Circular pipe:


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-For pipe 84 cm diameter and min. As= 2.0 (Bi+ h)2/65,000 ... Eq. (50)

Smaller with a single cage of

Reinforcement in the middle

third of the pipe wall:

In no case shall the minimum reinforcement in precast pipe be less than 0.07square inches per linear foot.

(c) Maximum Flexural Reinforcement without Stirrups.

(1) Limited by radial tension (inside reinforcing of curved members only):

Eq. (51)

Where rs is the radius of the inside reinforcement = (D i + 2t b)/2 for circular pipe.The term F rp , is a factor used to reflect the variations that local materials andmanufacturing processes can have on the tensile strength (and therefore the radialtension strength) of concrete in precast concrete pipe. Experience within the precastconcrete pipe industry has shown that such variations are significant. Frp, may bedetermined with Equation (52) below when a manufacturer has a sufficient amountof test data on pipe with large amounts of reinforcing (greater than ( As ) by Equation(51) to determine a statistically valid test strength, DL ut , using the criteria in ASTMC655 (AASHTO M242), "Standard Specification for Reinforced Concrete D-LoadCulvert, Storm Drain and Sewer Pipe.."

.. Eq. (52)

Once determined, F rp may be applied to other pipe built by the same process and

with the same materials. If Equation (52) yields values of F rp less than 1.0, a value of1.0 may still be used if a review of test results shows that the failure mode wasdiagonal tension, and not radial tension. If max. Inside (A s) is less than (As) requiredfor flexure, use a greater d to reduce the required As, or use radial stirrups, asspecified later.

(2) Limited by concrete compression:

.. Eq.(53)


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Where:

.Eq.(54)

0.65 b fc' < g' < 0.85 b fc'

If max As is less than As required for flexure, use a greater d to reduce the requiredAs, or the member must be designed as a compression member subjected tocombined axial load and bending. This design should be by conventional ultimatestrength methods, meeting the requirements of the AASHTO Bridge Specification;

Stirrups provided for diagonal or radial tension may be used to meet the lateral tierequirements of this section if they are anchored to the compression reinforcement,as well as to the tension reinforcement.

5-3 Crack Control Check.

Check flexural reinforcement for adequate crack width control at service loads . CrackWidth Control Factor:

. Eq. (55)

Where: Fcr = crack control factor, see note c.

.. Eq. (56)

Note: If e/d is less than 1.15, crack control will not govern and Equation (55) shouldnot be used.

j = 0.74 + 0.1 e/d .. Eq. (57)

Note: If e/d > 1.6, use j = 0.90.

.. Eq. (58)

B1and C1 are crack control coefficients that define performance of differentreinforcements in0.01 in. crack strength tests of reinforced concrete sections. Crackcontrol coefficients B l and Cl for the type reinforcements noted below are:


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Type Reinforcement (RTYPE) B1 C1

1. Smooth wire or plain bars1.0

2. Welded smooth wire fabric, 8 in.max.

Spacing of longitudinal. 1.0 1.5

3. Welded deformed wire fabric,deformed wire, deformed bars, or anyreinforcement with stirrups anchoredthereto

1.9

Notes:

Use n =1 when the inner and the outer cages are each a single layer.Use n = 2 when the inner and the outer cages are each made up from multiple layers.

a- For type 2 reinforcement having, also check F cr usingcoefficients B1 and C1for type 3 reinforcement, and use the larger value for Fcr.

b- Fcr is a crack control factor related to the limit for the average maximumcrack width that is needed to satisfy performance requirements at service load.When Fcr= 1.0, the average maximum crack width is 0.01 inch for areinforcement area As. If a limiting value of less than 1.0 is specified for Fcr, theprobability of a 0.01-inch crack is reduced. No data is available to correlatevalues of Fcr with specific crack widths other than 0.01 inches at Fcr=1.0 If thecalculated Fcr is greater than the limiting Fcr, increase As by the ratio:calculated F cr/limiting Fcr or decrease the reinforcing spacing.

5-4 Shear Strength Check.

Method 1: This method is given in Section 1.5.35 G of the AASHTO BridgeSpecification for shear strength of box sections .Under uniform load, the ultimateconcrete strength, v Vc must be greater than the ultimate shear must be greater thanthe ultimate shear (Vu) computed at a distance vd from the face of a support, orfrom the tip of a haunch with inclination of 45 degrees or greater with horizontal:

. Eq. (59)

. Eq. (60)


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Current research indicates that this method may be unconservative in someconditions, most importantly, in the top and bottom slab, near the center wall of twocell box culverts. Thus, Method 2 should also be checked.

Method 2: Method 2 is based on research sponsored by the American ConcretePipe Association and is more complex than Method 1 , but it reflects thebehavior of reinforced concrete sections under combined shear, thrust andmoment with greater accuracy than Method 1, or the current provisions in thereinforced concrete design section of the AASHTO Bridge Specification.

Determine Vu at the critical shear strength location in the pipe or box. Forburied pipe, this occurs where the ratio M/Vvd = 3.0, and for boxes, it occurseither where M/Vv d = 3.0 or at the face of supports (or tip of haunch). Distributed load within a distance vd from the face of a support may beneglected in calculating Vu, but should be included in calculating the ratioM/Vvd.

(a) For pipe, the location where M/Vv d = 3.0 varies with bedding and loadpressure distributions. For the distributions shown in Figure (12), it variesbetween about 10 degrees and 30 degrees from the invert. For the Olanderbedding conditions (Figure 12), the location where M/V v d = 3.0 in a circularpipe can be determined from Figure (13), based on the parameter rm/vd. For

noncircular pipe or other loading conditions, the critical location must bedetermined by inspection of the moment and shear diagrams.


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Fig.(12) Distribution of Earth Pressure on Culverts.

Angle from invert, degree

Fig.(13) Critical Shear Location in Circular Pipe for Olander , Earth Pressure

Distribution


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(b) For box sections, the location where M U /VU vd = 3.0 is at X dc from the pointof maximum positive moment, determined as follows:

Eq.(18)

This equation can be nondimensionalized by dividing all terms by the mean span ofthe section being considered. Figure (14) is a plot of the variation of Xdc /l with l/vdfor several typical values of cm, where,

. Eq.(61)

At sections where M/Vv d 3.0, shear is governed by the basic shear strength, Vb,calculated as

Eq.(62)

Where:

. Eq.(63)

. Eq.(64)

. Eq.(65)

. Eq.(66)


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Fig.(14) Location of Critical Shear Section for Straight Members with UniformlyDistributed Load.

- when moment produces tension on the inside of a pipe

Eq. (67)- when moment produces tension on the outside of a pipe

.. Eq. (68)

- .. Eq.(69)

The term F vp is a factor used to reflect the variations that local materials andmanufacturing processes can have on the tensile strength (and therefore diagonaltension strength) of concrete in precast concrete pipe. Experience within the precast


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concrete pipe industry has shown that such variations are significant. Fvp may bedetermined with Equation (70) below when a manufacturer has a sufficient amountof test data on pipe that fail in diagonal tension to determine a statistically valid teststrength DLut , using the criteria in ASTM C655 *AASHTO M242), "Specifications forReinforced Concrete D-Load Culvert, Storm Drain and Sewer Pipe."

Eq. (70)

Once determined, F vp may be applied to other pipe built by the same process and

with the same materials. (Fvp= 1.0) gives predicted 3-edge bearing test strengths in

reasonably good agreement with pipe industry experience, as reflected in the pipedesigns for Class 4 strengths given in ASTM C76, "Standard Specification forReinforced Concrete Culvert, Storm Drain and Sewer Pipe." Thus, it is appropriate touse F vp=1.0 for pipe manufactured by most combinations of process and local

materials. Available 3-edge bearing test data show minimum values of F vp of about0.9 for poor quality materials and/or processes, as well as possible increases up toabout (1.1) or more, with some combinations of high quality materials andmanufacturing process. For tapered inlet structures, Fvp=0.9 is recommended in theabsence of test data. If (vVb< Vu) either use stirrups, as specified as below, or if

(M/V vd


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(a) Maximum Circumferential Spacing of Stirrups:

... Eq. (72)

. Eq.(73)

(b) Maximum Longitudinal Spacing and Anchorage Requirements for Stirrups.Longitudinal spacing of stirrups shall equal SL Stirrups shall be anchored aroundeach inner reinforcement wire or bar, and the anchorage at each end shalldevelop the ultimate strength, f v, used for design of the stirrups. In addition, f v

shall not be greater than f y for the stirrup material.

(c) Radial Tension Stirrups (curved members only):

.. Eq. (74)

(d) Shear Stirrups (also resist radial tension):

Eq. (75) Vc is determined in Equation (71) except use,

Eq. (76)

Avr= 0 for straight members.

(e) Extent of Stirrups:

Stirrups should be used wherever the radial tension strength limits and/orwherever shear strength limits are exceeded.

(f) Computer Design of Stirrups:

The computer program to design reinforced concrete pipe that as describedincludes design of stirrups. The output gives a stirrup design factor (Sdf) whichmay be used to size stirrups as follows:

Eq. (77)


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This format allows the designer to select the most suitable stirrup effective ultimatestrength and spacing.

6- Examples Design of Culverts.

6-1 Design Principales(5)

.Reinforced concrete pipes either spun or cast are designed to with stand the

internal hydrostatic pressure without exceeding the permissible stresses of (126.5N/mm 2), for mild steel and (140 N/mm 2), in the case of cold drawn steel wires. Thethickness of the concrete pipe is designed in such a way that under specified testpressure, the maximum tensile stress in concrete when considered as effective totake stress along with the tensile reinforcement, should not exceed (2N/mm 2). Theminimum thickness of pipe varies with internal diameter and classification of pipes.For pressure pipes, the thickness varies from (25) mm for diameter of (80) mm to(65) mm for a diameter of (1200) mm. The type of (NP-1) class pipes aredesignated as shown in table (4). The spigot dimensions of (NP-1) are shown inFig.(15) .The longitudinal reinforcement is designed to support the reinforcedconcrete culvert pipe as a circular beams loaded with twice the self-weight of thepipe and twice the weight of water to fill the pipe across a span equal to the lengthof the pipe. Under these loading conditions, the stresses in the reinforcementshould not exceed the permissible stresses.

Table (4) classification ofpipes.


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Fig. (15) Spigot dimension of NP-1class R.C.C Pipes.

6-2 Reinforcement in pipes.

The circumferential and longitudinal reinforcement are designed for the loadsbut minimum quantity of steel reinforcement are specified or different classesof pipes in IS 458-1971. The typical reinforcement requirements for pipes ofclass (P-1) as shown in Table (5).

Table (5) Reinforcements Requirement in pipes of class (P-1).

InternalDiameter (mm)

ReinforcementsLongitudinal mild steel at

permissible stress of (126.5N/mm 2) (kg/m)

Spiral hard drawn steelwire at permissible stress of

(140 N/mm 2) (kg/m)100 0.863 0.327200 0.863 0.575400 1.00 3.800600 1.25 8.150800 1.78 14.50

1000 2.50 22.501200 3.36 32.50

The pitch of spiral should neither to more than (100 mm) or four times thethickness of the barrel, whichever is less, not less than the maximum size ofaggregate plus the diameter of the bar used. The minimum clear cover forconcrete pipes specified in the (IS) Code for different types of pipes are asshown in Table (6).


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Table (6) Cover Requirements.

Barrel Thickness (mm) For Spun Pipe (mm) For Pipes other thanSpun Pipes (mm)

25 mm & below 8.5 12.0Over 25 & including (30) 9.0 12.0

0ver (30) & below (75) 12.0 16.075 mm & above 18.0 18.0

6-3 Design Examples of pipe culvert.

Example (1):-


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Solving t=36.53 mm, but minimum thickness is not less than (55) mm.Adoptedt=60 mm.

5. Longitudinal Reinforcements.

Assume the pipe to span over a length of 3m

Self-weight of pipe = ( x 1.06 x 0.06 x 24) =4.79 kN/m

Weight of water= (( x L2/4) x10) = 7.85 kN/mTotal design load =2(4.79+7.85) =25.28 kN/m

I= /64(D 4-d4) = /64 (1.12 4- 14) =0.028 m 4.

Mmax. = (25.28 x 3 2/8) =28.44 kN.m

Stress = (28.44 x 10 6 x 560/ 0.028 x 10 12) = 0.568 N/ mm 2.

Stresses are negligibly small. Provide minimum longitudinal reinforcement of

2.5 kg/m, use 6 mm bars.

Weight of each bar = {( x 0.006 2/4) x 7800} = 0.22 kg/m.

Number of bars required = (2.5/0.22) = 11.363.

Spacing = ( x 1000/no. of bars) = ( x 1 000/ 14) = 224.285 mm

Use 14 bars of 6 mm @ spaced 200 mm along the circumference aslongitudinal reinforcement as shown in Fig. (16)


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Example (2):- Design a pipe culvert through a road embankment of height (6) m.The width of the road is (7.5) m and the formation width is (10) m. The side slopeof the embankment is (1.5:1). The Maximum discharge is (5 m 3/s). The savevelocity is (3 m/s). Class (AA) tracked vehicle is to be considered as live load.

Assume bell-mouthed entry .Given C e=1.5, Cs=0.010 and the unit weight of thesoil=20 kN/m 3.

Solution.

1- Hydraulic design.

Discharge through the pipe, Q=KAV

Where,

K = {1/ (1+Ke+Kf ) 0.5}

Now,

Kf = 0.0033(L/R 1.3)

Where L is the length of the pipe, which is equal to the base width of theembankment. Therefore,

L= 10+ (2 x 1.5 x 6) =28 m.

Assume (1) m diameter pipe, we have

R=A/P = ( D2/4 d) = D/4=1/4=0.25

Therefore

Kf = (0.0033 x 28) / (0.25) 1.3= 0.56

And Ke = 0.08 for bell-mouthed entry.

Therefore, we have

Conveyance factor = {1/ (1+0.08+0.56) 0.5} =0.78

Hence, Q=KAV

5 = A x 0.78 x 3 or A = 2.13 m 2

Area provided by each pipe,

= D2/4 = x 12/4 = 0.785 m 2

Therefore, the no. of pipes required = 2.13/0.785 = 2.71 say 3 pipes.

2- Bedding for the pipes .


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From table (7), for a pipe of internal diameter (1) m, the external diameter is(1.23) therefore, height of the embankment over the pipe = (6 1.23) =4.77 m

Table (7) Influence coefficient C s (for NP3 pipes)Internal

diameter(mm)

External

diameter(mm)

Height of Embankment above the pipe (m)

0.1 0.2 0.3 0.4 0.6 0.8 1.0 2.0 3.0 4.0

500 560 0.246 0.288 0.198 0.169 0.117 0.083 0.060 0.017 0.008 0.005600 770 0.247 0.234 0.210 0.182 0.131 0.094 0.068 0.022 0.010 0.006700 870 0.247 0.236 0.215 0.186 0.140 0.102 0.075 0.024 0.010 0.006800 990 0.249 0.240 0.220 0.196 0.149 0.110 0.083 0.027 0.013 0.007900 1100 0.249 0.241 0.225 0.202 0.156 0.117 0.089 0.029 0.014 0.008

1000 1230 0.249 0.242 0.228 0.205 0.162 0.123 0.095 0.032 0.015 0.0101200 1440 0.249 0.242 0.230 0.209 0.171 0.131 0.104 0.036 0.020 0.011

As Ce = 1.5, therefore, the load on the pipe owing to each fill,

Cewd 2= 1.5 x 20 x 1.23 2= 45.4 kN/m

And load on the pipe owing to wheel load,

4CsIP = 4 x 0.010 x 1.5 x 700 = 42 kN/m

Bedding is chosen based on the strength factor. Referring to (IS 458-1988) (6)

, threeedge-bearing strength for a (NP3) pipe of 1000 mm internal diameter is (72kN/m) as shown in Table (9).Hence the equation to be satisfied is:

{Three edge bearing strength (kN/m) factor of safety} = {Load owing to earthfill (kN/m) strength factor (SF) } + {Load owing to wheel load/ factor of safety}

Or, (72 1.5) = (45.4 SF) + (421.5) Therefore, SF = 2.30

Hence, concrete cradle bedding may be provided, see Table (8),

Table (8) The types of Beddings and their Strength FactorsType of Bedding Strength FactorEarth Bedding 2.0

First Class Bedding 2.3Concrete Cradle Bedding 3.7


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Table (9) DESIGN AND STRENGTH TEST REQUIREMENTS OF CONCRETE PIPES OF(6)PRESSURE PIPES-DUTY, NON-REINFORCED CONCRETE, MEDIUMCLASS NP3

N0minalinternal

diameter ofpipe

Barrel WallThickness

ReinforcementStrength Test

Requirement for ThreeEdge Bearing Test

Longitudinal, Mild Steelof Hard-Drawn Steel

Spiral, Hard-Drawn Steel

Load to

produce(0.25) mm

Crack

UltimateLoad

(1) (2) (3) (4) (5) (6) (7)

mm mmMin.

numberKg/ linear

meterKg/ linear

meterkN/ linear

meterkN/ linear

meter80 25 6 0.33 0.15 13.00 19.50

100 25 6 0.33 0.22 13.00 19.50150 25 6 0.33 0.46 13.70 20.55200 30 6 0.33 0.81 14.50 21.75

225 30 6 0.33 1.03 14.80 22.20250 30 6 0.33 1.24 15.00 22.50300 40 8 0.78 1.80 15.50 23.25350 75 8 0.78 2.95 16.77 25.16400 75 8 0.78 3.30 19.16 28.14450 75 8 0.78 3.79 21.56 32.34500 75 8 0.78 4.82 23.95 35.93600 85 6+6 1.18 7.01 28.74 43.11700 85 6+6 1.18 10.27 33.53 50.30800 95 6+6 2.66 13.04 38.32 57.48

900 100 6+6 2.66 18.30 43.11 64.671000 115 6+6 2.66 21.52 47.90 71.851100 115 6+6 2.66 27.99 52.69 79.001200 120 8+8 3.55 33.57 57.48 86.221400 135 8+8 3.55 46.21 67.06 100.601600 140 8+8 3.55 65.40 76.64 114.961800 150 12+12 9.36 87.10 86.22 129.332000 170 12+12 9.36 97.90 95.80 143.702200 185 12+12 9.36 113.30 105.38 158.072400 200 12+12 14.88 146.61 114.96 172.442600 215 12+12 14.88 175.76 124.54 186.81

Note 1: The actual internal dia. Is to be declared by the manufacturer and the tolerance is to be applied on thedeclared dia. (see also 0.3.2)

Note 2: Minimum thickness and minimum length of collars shall be the same as that for the next higher sizeavailable in (NP2) class pipes corresponding to the calculated inner dia. Of collars.

Note 3: The longitudinal reinforcement given in this table is valid for pipes up to 2 m effective length for internal dia.0f pipe up to 250 mm and up to 3m effective length for higher dia. Pipe.

Note 4: Concrete for pipe above 1800 mm nominal dia. shall have a minimum compressive strength of (35 N/mm 2)at 28 days and a minimum cement content of (400 kg/m 3).

Note 5: If mild steel is used for spiral reinforcement, the weight specified in col. 5 shall be increased to 140/125


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3-Reinforcements.

The minimum reinforcements to provided according to (IS 458-1988) (6) Table(10) are:

Spiral reinforcement = 21.52 kg/m

Longitudinal reinforcement = 2.66 kg/m

Weight of the (12) mm spiral (diameter = 1.1 m)

= (( x 0.012 2 x 7850)/4) ( x 1.1) = 3.068 kg/m

Table (10) Reinforcement Requirements for (NP3 PipesAccording to (IS 458 1988)

Internal diameter(mm)

Longitudinal steel

with permissiblestress of 125 MPa(kg/m)

Spiralreinforcement

with permissiblestress of 140 MPa

(kg/m)

Ultimate three

edge bearingstrength (kg/m)

350 0.78 2.95 25.16400 0.78 3.30 28.74450 0.78 3.79 32.34500 0.78 4.82 35.93600 1.18 7.01 43.11700 1.18 10.27 50.30

800 2.66 13.04 57.48900 2.66 18.30 64.67

1000 2.66 21.52 71.851100 2.66 27.99 79.001200 3.55 33.57 86.22

Providing 30 kg/m 0f spiral, no. of spiral = 30/ 3.068 = 9.77 say 10

Spacing c/c distance = 1000/10 = 100 mm.

Providing (6 mm dia.) Mild steel bars as longitudinal steel and providing(4kg/m.)

Weight of a single bar = ( x 0. 006 2 x 1 x 7850)/4 =0.22 kg/m

Providing at 4 kg/m, no, of bars = 4/0.22 = 18.18

Spacing = ( x 1100) / 18.18 =190.08 mm say 150 mm c/c.

The details of reinforcement as shown in Fig. (17).


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6-4 Design Examples of box culvert.

Example (1): Design a reinforced concrete box culvert having a clear vent way

of 3m by 3m. The super imposed dead on the culvert is (12.8 Kn/m 2). The live loadon the culvert is (50 kN/m 2). Density of soil at site is (18 kN/m 3). Angle of repose(=300). Adopt M-20 or f c=2o N/mm 2 grade concrete mix and F e = 415 or FY=415MPa grade for steel. Sketch the details of reinforcement in the box culvert.

Solution.

1- Data .- Clear span = L = 3m.

- Height of event =h = 3m.- Dead load = 12.8 kN/m 2.- Live load = 50 kN/m 2.- Density of soil = 18 KN/ m 3.- Angle of repose = = 300.- Class of concrete = M-20 or fc=2o N/mm 2.

2- Permissible stresses.- cc = 5 N/ mm 2

- cb = 7 N/ mm2

- st = 150 N/ mm 2 ( water face)- st = 190 N/ mm 2 ( Away from water face)


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- m = 13 = modular ratio (Es/E c).- J = 0.86 (lever arm = J= (1-k/3).- Q = 1.198

3- Dimensions of box culvert.

- Adopting thickness of slab as 100 mm/ m span.- Thickness = t s = tw = 300 mm.- Effective span = 3300 mm

4-Loads.

- Self-weight of the top = 0.30 x 24 = 7.2 kN/m 2 - Super imposed dead load = 12.8 kN/m 2 - Live load = 50 kN/m 2 - Total load = 70.0 kN/ m 2

- Weight of vertical sidewalls = 0.30 x 3.3 x 24 = w = 24 kN.Soil pressure = p = wh (1-sin )/ (1+ sin )

At h = 3.3 m, = 300, w = 18 kN/m 3.

The soil pressure =p = 18 x 3.3 x 1/3 = 20 kN/m 2.

Uniform lateral pressure due to the effect of super imposed dead load and liveload surcharge is calculated as,

P = (50 + 12.8) {(1-sin )/ (1+ sin )} = (62.8 x 1/3 = 21 kN/m 2

Uniform lateral pressure due to the effect of super imposed dead load surcharge onlyis,

P = 12.8 {(1-sin )/ (1+ sin )} = (12.8 x 1/3) = 4.26 kN/m 2

Intensity of water pressure is obtained as,

P = wh = (10 x 3.3) = 33 kN/m 2

3- Analysis of Moments, Shear, and Thrusts.The various loading patterns considered are shown in Fig. (18 & 19) The moments,shears and thrust corresponding to the different cases of loading (case 1 to case 6),the fixed end moments developed for the six different loading cases are compiled inTable (11). For two different ratios of (L/H = 1 and 1.5 Where, L = span of the culvert,H = height of the culvert), evaluated using the coefficients given in Table (12) arecompiled in Table (13). The design forces resulting from the combination of thevarious cases yielding maximum moment and forces at the support and mid span

sections are shown in Table (14).


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Fig. (18) Types of Loading for box Culverts.


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Table (11) Fixed End Moments in Box Culvert.

Loading caseFixed End Moments

MA = MA' MD = MD'1

2

3

4

5

6

Note: positive moment indicates tension on inside face.

Fig. (19) Loading Cases Considered for Box Culvert for Example (1)


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Table (12) Coefficients for Moment, Shear and Thrust.

L:H SectionFactor Loading Case

For 1 2 3 4 5 6M WL wL2 WL pL2 pL2 pL2 N w wL w pL pL pLV w wL w pL pL pL

1:1 B-1 M + 0.182 + 0.083 + 0.021 + 0.019 - 0.019 - 0.042N 0 0 0 - 0.167 + 0.167 + 0.500

A-2 M - 0.068 - 0.042 + 0.021 + 0.019 - 0.019 - 0.042N 0 0 0 - 0.167 + 0.167 - 0.500V + 0.500 + 0.500 0 0 0 0

A-3 M - 0.068 - 0.042 + 0.021 + 0.019 - 0.019 - 0.042N + 0.500 + 0.500 0 0 0 0V 0 0 0 + 0.167 - 0.167 - 0.500

E-4 M - 0.052 - 0.042 - 0.042 - 0.043 + 0.043 + 0.083N + 0.500 + 0.500 + 0.500 0 0 0

D-5 M - 0.036 - 0.042 - 0.004 + 0.023 - 0.023 - 0.042N + 0.500 + 0.500 + 1.000 - 0.333 + 0.33 0V 0 0 0 0 0 + 0.500

D-6 M - 0.036 - 0.042 - 0.104 + 0.023 - 0.023 - 0.042N 0 0 0 0 0 + 0.500V - 0.500 - 0.500 - 1.020 - 0.333 + 0.333 0

C-7 M + 0.088 + 0.083 + 0.146 + 0.023 - 0.023 - 0.042N 0 0 0 - 0.333 + 0.333 + 0.500

1.5:1 B-1 M + 0.170 + 0.075 + 0.018 + 0.015 - 0.015 - 0.033N 0 0 0 - 0.167 + 0.167 + 0.500

A-2 M - 0.079 - 0.050 + 0.018 + 0.015 - 0.015 - 0.033N 0 0 0 - 0.167 + 0.167 + 0.500V + 0.500 + 0.500 0 0 0 0

A-3 M - 0.079 - 0.050 + 0.018 + 0.015 - 0.015 - 0.033N + 0.500 + 0.500 0 0 0 0V 0 0 0 + 0.167 - 0.167 - 0.500

E-4 M - 0.062 - 0.050 - 0.050 - 0.047 + 0.047 + 0.092N + 0.500 + 0.500 + 0.500 0 0 0

D-5 M - 0.045 - 0.050 - 0.118 + 0.018 - 0.018 - 0.033N + 0.500 + 0.500 + 1.000 0 0 0V 0 0 0 - 0.333 + 0.333 + 0.500

D-6 M - 0.045 - 0.050 - 0.118 + 0.018 - 0.018 - 0.033N 0 0 0 - 0.333 + 0.333 + 0.500V - 0.500 - 0.500 - 1.000 0 0 0

C-7 M + 0.079 + 0.075 + 0.132 + 0.018 - 0.018 - 0.033N 0 0 0 - 0.333 + 0.333 + 0.500

Refer to Fig. (18) For details & notations:-Note: 1- positive moment indicates tension on side face.

2- Positive shear indicates that the summation of force at the left of the sectionacts outward when viewed from within.

3- Positive thrust indicates compression on the section.


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The maximum positive moments develop at the center of bottom & top slab for thecondition that the sides of the culvert not carrying the live load and the culvertsrunning full with water. The maximum negative moments develop at the supportsections of the bottom slab for the condition, culvert is empty and the top slab carriesthe dead & live load.

Table (13) Forces Components for Different Cases of Loading

Section ForcesLoading Case

Case-2 Case-3 Case-4 Case-5 Case-6(a) Case-6(b)B-1 M 63.20 1.66 6.82 - 4.13 - 9.60 - 1.92

N 0 0 - 18.18 + 11.0 + 34.65 + 6.93A-2 M - 31.60 1.66 6.82 - 4.13 - 9.60 - 1.92

N 0 0 - 18.18 + 11.0 + 34.65 - 6.93V 115.50 0 0 0 0 0

A-3 M - 31.60 1.66 6.82 - 4.13 - 9.60 -1,92

N 115.50 0 0 0 0 0V 0 0 18.18 - 11.00 - 34.65 -6.93E-4 M - 31.6 - 3.32 - 15.45 - 9.36 +19.20 +3.84

N 115.50 + 39.60 0 0 0 0D-5 M - 31.60 - 0.317 8.26 - 5.00 - 9.60 - 1.92

N 115.50 + 79.20 - 36.26 + 21.90 0 0V 0 0 0 0 +34.65 +6.93

D-6 M -31.60 -8.23 8.26 - 5.00 - 9.60 - 1.92N 0 0 0 0 +34.65 + 6.93V -115.50 -79.20 -36.26 +21.90 0 0

C-7 M 63.20 11.56 8.26 -5.00 -9.60 - 1.92N 0 0 - 36.26 +21.90 +34.65 + 6.93

Note: Moments are in (kN.m) & shear force and thrusts are in (kN).

4- Design of Reinforcements

-Section: C-7 , (mid span of bottom slab) as shown in Table (14)

M = 76.10 KN.m

N = - 7.43 KN (tension)

Ast = (M / (st * J *d) = (76.10 x 10 6 / 150 x 0.86 x 270) = 2234 mm 2/m

Provide 20 @ 140 mm c/c

Distribution steel = (0.30 x 300 x 100 / 100) = 900 mm 2

Provide 10 @ 150 mm(c/c) on both faces.

-Section: D-6, (support section) as shown in Table (14)

M = -54.43 KN.mN = 34.65 KN


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Ast = (M / ( st * J *d) = (54.43 x 10 6 / 190 x 0.86 x 270) = 1233mm 2/m

Provide 16 @ 150 mm (c/c) and distribution bars of 10 @ 150 mm (c/c)

Table (14) Design Moments and Forces in Box Culvert

SectionLoading

CombinationCases

MomentM

KN.m

ThrustN

(kN)

Shear ForceV

(kN)D-6 2+3+5+6(a) -54.43 +34.65 -172.80A-2 2+3+5+6(a) -43.67 -23.65 +115.50B-1 2+3+4+5+6(b) 65.63 -1.25 0C-7 2+3+4+5+6(b) 76.10 -7.43 0E-4 2+3+4+5+6(b) -55.89 +155.10 0

-Section: E-4, (vertical sidewall) as shown in Table (14)

M = -55.89 kN.m MU=1.5 x 55.89 = 83.83 kN.m

N= 155.10 KN NU = 1.5 x 155.10 = 232.5 kN.

(MU /f ck *bD2)= (83.83 x 10 6/ 20 x 1000 x 300 2) = 0.046 with f y= 415 N/mm 2

(NU / f ck *bD) = 232.5 x 10 3/ 20 x 1000 x 300) = 0.0387with (d'/D) = (30/300) =0.10

By referring to interaction curve of SP-16 , (p / f ck) (7) = 0.02

Where, A sc =Ast = 0.5 (phD/100) = 0.5(0.02 x 20 x 1000 x 300) / 100 = 600 mm 2

Hence A S = 1200 mm 2.

But minimum reinforcement of (0.8%) of cross section has to be provide,

AS = (0.8 x 300 x 1000 / 100) = 2400 mm 2

Hence, provide 16 @150 mm (c/c) on both faces in the vertical sidewall.

Distribution steel of 10 @ 150 mm (c/c) is provide on both faces.The details of reinforcement in the box culvert is shown in Fig. (20)


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References:1- Federal Highway Administration," HYDRAULIC DESIGN OF HIGHWAY

CULVERTS", U.S. Department of Transportation, Publication No. FHWA-NHI-01-020, September 2001, pp.367.

2- B.N. Sinha, & R.P. Sharma," RCC BOX CULVERT - METHODOLOGY AND DESIGNS INCLUDING COMPUTER METHOD", Paper No. 555, Journal of theIndian Roads Congress, October-December 2009, pp. (190-812).

3- Darryl Shoemaker, Ph.D., Jack Allen, Margaret Ballard, Stephen David, andGeorge Eliason," HIGHWAY ENGINEERING HANDBOOK", McGraw-Hill Companies Copyright , First Edition. 2004, pp.933.

4- FHWA-IP-83-6," Structural Design Manual for Improved Inlets & Culverts", June 1983, pp.314.

5- Krishna, Raju N.,"Advanced Reinforced Concrete Design", satish kumar,1 st edition, New Delhi,1988,pp.370.

6- IS 458-1988: "SPECIFICATION FOR PRECAST CONCRETE PIPES ((WITH AND

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