Structural Design for Reinforced Concrete Culverts
Prepared by: - Consultant Engineer Raad Mohammad Dhyiab M.Sc., Structural Engineering 1
CONTENTS No. 1234567-
Details INTRODUCTION Types of culverts. Box Culverts. Design Reinforcing of Culvert Methods for Structural Analysis Examples Design of Culverts. References
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page 3 3 5 5 10 33 48
1-INTRODUCTION A culvert is a hydraulically short conduit, which conveys stream flow through a roadway embankment or past some other type of flow obstruction. Culverts are constructed from a variety of materials and are available in many different shapes and configurations. Culvert selection factors include roadway profiles, channel characteristics, 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 water course like streams, Nallas across the embankment as road embankment cannot be allowed to obstruct the natural water way. The culverts are also required to balance the 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 reinforced cement 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 of the characteristics of these pipes are reviewed below. It is well known that roads are generally constructed in embankment, which come in the way of natural flow of storm water (from existing drainage channels). As such flow cannot be obstructed and some kind of cross, drainage works are required to be provided to allow water to pass across the embankment. The structures to accomplish such flow across the road are called culverts. It is well known that roads are generally constructed in embankment, which come in the way of natural flow of storm water (from existing drainage channels). As, such flow cannot be obstructed and some kind of cross drainage works are required to be provided to allow water to pass across the embankment. The structures to accomplish such flow across the road are called culverts, small and major bridges depending 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; as box culverts and special shapes; and as field-constructed pipe. Shapes, as shown in Fig. (1) Include round, horizontal and vertical ellipse, and arch configuration. No reinforced concrete pipe is available in diameters from 10 to 90 cm and three strength 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 of reinforced 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 the greatest structural strength; according to required D-load strength (ASTM C 655); or according to a direct wall design. (ASTM C 1417). Wall thickness of reinforced concrete pipe can be varied to meet in field conditions. The standard “class” specifications for pipe give wall thickness according to three distinct types, which vary from Wall A, being the thinnest, to Wall C, being the thickest Steel reinforcing for reinforced concrete pipe can be arranged in many combinations to meet the given structural requirements. Figure (2) Shows some of the steel reinforcement layouts used in manufacturing reinforced concrete pipe (3).
Fig. (1) Concrete pipe is manufactured in five common shapes; regional custom and demand usually determine availability.
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Fig. (2) Concrete pipe culvert reinforcement notation. (From PIPECAR: User and Programmer 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 vertical sidewalls, which serve as the outer form for the bottom and sides. The upper portion of the pipe is cast against an inner arch form as illustrated in Fig. (1) the form is pulled along the trench while concrete is poured into a hopper located above Powered spading 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 are constructed with steel reinforcement. Factory-made boxes are shipped in sections 1 to 3 m. long and joined in the field to make a structure of the required length. As shown in Fig. (1).
4-Design Reinforcing of Culvert (4) Structural design of reinforced concrete culvert and inlet structures is quite different from design for corrugated metal structures. For reinforced concrete inlets, the
Designer typically selects a trial wall thickness and then sizes the reinforcing to meet the design requirements. 5
The method for the design of reinforced concrete pipe and box sections presented below was recently adopted by the American Concrete Pipe Association and has been recommended by the AASHTO Rigid Culvert Liaison Committee for adoption by the AASHTO Bridge Committee. This design method provides a set of equations for sizing the main circumferential reinforcing in a buried reinforced concrete culvert. For additional criteria, such as temperature reinforcing in monolithic structures, the designer should refer to the appropriate sections of AASHTO. Typically, the design process involves a determination of reinforcement area for strength and crack control at various governing locations in a slice and checks for shear strength and certain reinforcement limits. The number and location of sections at which designers must size, reinforce, and check shear strength will vary with the shape of the cross section and the reinforcing scheme used. Figure (3). Shows typical reinforcing schemes for precast and cast-in-place one-cell box sections. The design sections for these schemes are shown in Figure (4). For flexural design of box sections with typical geometry and load conditions, Locations 1, 8, and 15 will be positive moment design locations(tension on inside) and locations 4, 5, 11, and 12 will be negative moment design locations. Shear design is by two methods one is relatively simple, and requires checking locations 3, 6, 10 and 13 which are located at a distance dvd from the tip of haunches. The second method is slightly more complex and requires checking 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. The design 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 Box Sections. 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 ( ɸv 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 Locations Precast 4 , 5 , 11 , 12 1 15 8 -
Steel Area AS1 AS2 AS3 AS4 AS8
Cast – In - Place 5, 11 , 12 1 15 8 4
Typical reinforcing schemes and design locations for two cell box sections are shown in 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 and Flexure 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) is also 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/ Vɸvd) = 3.0 Notes:1-Reinforcing in crown (Asc) will be the same as that use at the invert unless 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 and Flexure 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 complete the structural analysis and design of culverts very efficiently. The method discussed 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 may be used to determine moments, thrusts and shears in the pipe due to earth, pipe and 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 +) We = total weight of earth on unite length of buried structure, Ibs/ft Wp = weight of unit length of structure, Ibs/ft. Wf = total weight of fluid inside unit length of buried structure, Ibs/ft. B' = width, ft. Cm, Cn & Cv = coefficients. Figure (1-7): provides coefficients for earth load analysis of circular pipe with 3 loading conditions β1= 90°, 120° and 180°. In all cases, β2= 360° - β1. These load conditions are normally referenced by the bedding. Angle, β2. The 120° and 90°bedding cases correspond approximately with the traditional Class B and Class C bedding conditions (2, 3). These coefficients should only be used when the side fill is compacted during installation. 11
Compacting the side fill allows the development of the beneficial lateral pressures assumed in the analysis. If the side fills are not compacted (this is not recommended), then a new analysis should be completed using the computer program.
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 top slab thickness. After these dimensions are estimated, the section can then be analyzed as a rigid frame, and moment distribution is often used for this purpose. A simplified moment distribution was developed by AREA (8) for box culverts under railroads. Modifications of these equations are reproduced in Table (1) and Table (2) for one and two cell box culverts respectively. This analysis is based on the 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 the sidewalls. 3-Only boxes with "Standard" haunches or without haunches can be considered. Standard haunches have horizontal and vertical dimensions equal to the top slab thickness. 4-The section is assumed doubly symmetrical, thus separate moments and shears are 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. Any one of these cases can be dropped by setting the appropriate unit weight (soil, concrete or fluid) to zero when computing the design pressures pv and ps. The equations 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 and bottom 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 Thrusts in bottom slab:
Eq.(16)
Thrust in sidewall:
Eq.(17)
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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= 0.0 1-
See Equation (18)for determination of
……………………………………..Eq.(18)
2-
If M8 is negative use AS min. for side wall inside reinforcing, and do not check shear 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 standard haunches
Eq.(24)
Eq.(25)
Eq.(26)
For boxes without haunches
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Eq.(27)
Design Moments Moments at Origin: Eq.(28)
Boxes with standard haunches and uniform wall thickness (HH=HV=TT=TS=TB): Eq.(29)
Eq.(30)
Boxes without haunches (HH=HV=0, TT=TB TS): Eq.(31)
Eq.(32) Moment on bottom slab: Eq.(33)
Moment in sidewall: Eq.(34)
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Design Shears Shear on bottom slab: Eq.(35) Shear in sidewall: Eq.(36)
Design Thrusts Thrust in bottom slab: Eq.(37)
Thrust in side slab; boxes with haunches: Eq.(38)
Thrust in side slab, boxes without haunches: Eq.(39)
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 different thickness than the top and bottom slabs (HH=HV=0, and TT=TB TS) use Equation (31), Equation (32), and Equation (39). 3- See Equation (18) for determination of Xdc. 4- If M8 is negative, use ASmin for sidewall inside reinforcing, and do not check shear at Section 9. 5- Geometry constants F1 through F5 are not required for boxes without haunches.
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, the structure is proportioned to satisfy the following limits of structural behavior. 1- Minimum ultimate strength equal to strength required for expected service loading times a load factor. 2- Control of crack width at expected service load to maintain suitable protection of reinforcement from Corrosion, and in some cases, to limit infiltration or exfiltration 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 this analysis, the section stiffness is usually assumed constant, but it may be varied with stress level, loosed on experimentally determined stiffness of crocked sections at the crown, invert and spring lines in computer analysis methods. Multiplying calculated moments, thrusts, and shears (service conditions) determine ultimate moments, thrusts and shears required for design by a load factor (Lf) as follows:
Mu= LfM ………………….……. Eq. (40) Nu= LfN …………………………. Eq. (41) Vu = Lf V ………………………… Eq. (42) Where: - Lf = Load Factor.
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Load Factors for Ultimate Strength: The minimum load factors given below are appropriate when the design bedding is selected near the poorest extreme of the expected installation, and when the design earth load is conservatively estimated. For culvert or trench installations alternatively, these minimum load factors may be earth 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 with the strength reduction factors given below. The 1981 AASHTO Bridge Specifications (4) specify use of a minimum load factor of 1.3 for all loads, multiplied by βcoefficients of 1.0 for dead and earth load and 1.67 for live load plus impact. Thus, the effective load factors are 1.3 for earth and dead load 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,φ, provide "for the possibility that small adverse variations in material strengths, workmanship, and dimensions, while individually within acceptable tolerances and limits of good practice, may combine to result in understrength" . Table (3) presents the maximum φ factors given in the 1981 AASHTO Bridge Specification. Table (3) Strength Reduction Factors in Current AASHTO Standard Specifications for Highway Bridges.
Flexure Shear
Box Culverts Precast (a) Cast-in-Place (b) 1.0 (d) 0.9 0.9 0.85
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Pipe Culverts Precast (c) 1.0 (d) 0.9
a.Section 1.15.7 b. Section 1.5.30 c. Currently recommended by AASHTO Rigid Culvert Liaison Committee for adoption by 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 Culvert Committee on the basis that precast sections are a manufactured product, and are subject to better quality control than are cast-in-place structures. Because welded wire fabric, the reinforcing normally used in precast box and pipe sections, can develop its ultimate strength before failing in flexure, the use of φ= 1.0 with the yield strength still provides a margin for variations equal to the ratio of the yield strength to the ultimate strength. If hot rolled reinforcing is used in a precast structure, 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 shown in Figure (4), Figure (5) and Figure (6). (a) Reinforcement for Flexural Strength, As. ……..…… 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: - For inside face of pipe:
min. As= (Bi+ h)2/65,000 …………….…. Eq. (47)
- For outside face of pipe:
min. As= 0 75 (Bi+ h)2/65,000 ……..….. Eq. (48)
- For elliptical reinforcement in min. As= 2.0 (Bi+ h)2/65,000 ……….….. Eq. (49) Circular pipe: 23
-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 0.07 square 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 = (Di + 2tb)/2 for circular pipe. The term Frp, is a factor used to reflect the variations that local materials and manufacturing processes can have on the tensile strength (and therefore the radial tension strength) of concrete in precast concrete pipe. Experience within the precast concrete pipe industry has shown that such variations are significant. Frp, may be determined with Equation (52) below when a manufacturer has a sufficient amount of test data on pipe with large amounts of reinforcing (greater than ( As ) by Equation (51) to determine a statistically valid test strength, DLut , using the criteria in ASTM C655 (AASHTO M242), "Standard Specification for Reinforced Concrete D-Load Culvert, Storm Drain and Sewer Pipe.."
………………………….. Eq. (52) Once determined, Frp may be applied to other pipe built by the same process and with the same materials. If Equation (52) yields values of Frp less than 1.0, a value of 1.0 may still be used if a review of test results shows that the failure mode was diagonal tension, and not radial tension. If max. Inside (As) is less than (As) required for flexure, use a greater d to reduce the required As, or use radial stirrups, as specified later. (2) Limited by concrete compression:
………………….. Eq.(53) 24
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 required As, or the member must be designed as a compression member subjected to combined axial load and bending. This design should be by conventional ultimate strength methods, meeting the requirements of the AASHTO Bridge Specification; Stirrups provided for diagonal or radial tension may be used to meet the lateral tie requirements 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. Crack Width 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) should not 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 different reinforcements in0.01 in. crack strength tests of reinforced concrete sections. Crack control coefficients Bl and Cl for the type reinforcements noted below are: 25
Type Reinforcement (RTYPE)
B1
C1
1. Smooth wire or plain bars
1.0
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 any reinforcement with stirrups anchored thereto
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 Fcr using coefficients 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 maximum crack 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 a reinforcement area As. If a limiting value of less than 1.0 is specified for Fcr, the probability of a 0.01-inch crack is reduced. No data is available to correlate values of Fcr with specific crack widths other than 0.01 inches at Fcr=1.0 If the calculated Fcr is greater than the limiting Fcr, increase As by the ratio: calculated Fcr/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 Bridge Specification for shear strength of box sections .Under uniform load, the ultimate concrete strength, φv Vc must be greater than the ultimate shear must be greater than the ultimate shear (Vu) computed at a distance φvd from the face of a support, or from the tip of a haunch with inclination of 45 degrees or greater with horizontal:
………………. Eq. (59) ………………. Eq. (60) 26
Current research indicates that this method may be unconservative in some conditions, most importantly, in the top and bottom slab, near the center wall of two cell box culverts. Thus, Method 2 should also be checked. Method 2: Method 2 is based on research sponsored by the American Concrete Pipe Association and is more complex than Method 1, but it reflects the behavior of reinforced concrete sections under combined shear, thrust and moment with greater accuracy than Method 1, or the current provisions in the reinforced concrete design section of the AASHTO Bridge Specification. Determine Vu at the critical shear strength location in the pipe or box. For buried pipe, this occurs where the ratio M/Vφvd = 3.0, and for boxes, it occurs either where M/Vφv d = 3.0 or at the face of supports (or tip of haunch). Distributed load within a distance φvd from the face of a support may be neglected in calculating Vu, but should be included in calculating the ratio M/Vφvd. (a) For pipe, the location where M/Vφv d = 3.0 varies with bedding and load pressure distributions. For the distributions shown in Figure (12), it varies between about 10 degrees and 30 degrees from the invert. For the Olander bedding conditions (Figure 12), the location where M/Vφv d = 3.0 in a circular pipe can be determined from Figure (13), based on the parameter rm/¢vd. For noncircular pipe or other loading conditions, the critical location must be determined 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 MU /VU φvd = 3.0 is at Xdc from the point of maximum positive moment, determined as follows:
………………………… Eq.(18)
This equation can be nondimensionalized by dividing all terms by the mean span of the section being considered. Figure (14) is a plot of the variation of Xdc /l with l/φvd for several typical values of cm, where,
……………………………. Eq.(61) At sections where M/Vφv 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 Uniformly Distributed 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 Fvp is a factor used to reflect the variations that local materials and manufacturing processes can have on the tensile strength (and therefore diagonal tension strength) of concrete in precast concrete pipe. Experience within the precast 31
concrete pipe industry has shown that such variations are significant. Fvp may be determined with Equation (70) below when a manufacturer has a sufficient amount of test data on pipe that fail in diagonal tension to determine a statistically valid test strength DLut , using the criteria in ASTM C655 *AASHTO M242), "Specifications for Reinforced Concrete D-Load Culvert, Storm Drain and Sewer Pipe."
……………………………………Eq. (70) Once determined, Fvp 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 pipe designs for Class 4 strengths given in ASTM C76, "Standard Specification for Reinforced Concrete Culvert, Storm Drain and Sewer Pipe." Thus, it is appropriate to use Fvp=1.0 for pipe manufactured by most combinations of process and local materials. Available 3-edge bearing test data show minimum values of Fvp of about 0.9 for poor quality materials and/or processes, as well as possible increases up to about(1.1) or more, with some combinations of high quality materials and manufacturing process. For tapered inlet structures, Fvp=0.9 is recommended in the absence of test data. If (φvVb< Vu) either use stirrups, as specified as below, or if (M/Vφvd <3.0), calculate the general shear strength, as given below. Shear strength will be greater than Vb when (M/Vφv d < 3.0) at critical sections at the face of supports or, for members under concentrated load, at the edge of the load application point. The increased shear strength when (M/Vφvd < 3.0), termed the general shear strength, VC, is:
…………………………… Eq. (71) If M/Vφv d ≥3.0, use M/Vφv d = 3.0 in Equation (71). Vc shall be determined based on M/Vφv d at the face of supports in restrained end flexural members and at the edges of concentrated loads. Distributed load within a distance φvd from the face of a support may be neglected in calculating Vu, but should be included for determining M/Vφvd.
5-5 Stirrups Stirrups are used for increased radial tension and/or shear strength. 31
(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 around each inner reinforcement wire or bar, and the anchorage at each end shall develop the ultimate strength, f v, used for design of the stirrups. In addition, fv shall not be greater than fy 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/or wherever shear strength limits are exceeded. (f)
Computer Design of Stirrups: The computer program to design reinforced concrete pipe that as described includes design of stirrups. The output gives a stirrup design factor (Sdf) which may be used to size stirrups as follows:
……………………………… Eq. (77) 32
This format allows the designer to select the most suitable stirrup effective ultimate strength and spacing.
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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 (126.5 N/mm2), for mild steel and (140 N/mm2), in the case of cold drawn steel wires. The thickness of the concrete pipe is designed in such a way that under specified test pressure, the maximum tensile stress in concrete when considered as effective to take stress along with the tensile reinforcement, should not exceed (2N/mm2). The minimum 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 are designated as shown in table (4). The spigot dimensions of (NP-1) are shown in Fig.(15) .The longitudinal reinforcement is designed to support the reinforced concrete culvert pipe as a circular beams loaded with twice the self-weight of the pipe and twice the weight of water to fill the pipe across a span equal to the length of the pipe. Under these loading conditions, the stresses in the reinforcement should not exceed the permissible stresses. Table (4) classification of pipes.
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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 loads but minimum quantity of steel reinforcement are specified or different classes of pipes in IS 458-1971. The typical reinforcement requirements for pipes of class (P-1) as shown in Table (5).
Table (5) Reinforcements Requirement in pipes of class (P-1). Reinforcements Internal Longitudinal mild steel at Spiral hard drawn steel Diameter (mm) permissible stress of (126.5 wire at permissible stress of N/mm2) (kg/m) (140 N/mm2) (kg/m) 100 0.863 0.327 200 0.863 0.575 400 1.00 3.800 600 1.25 8.150 800 1.78 14.50 1000 2.50 22.50 1200 3.36 32.50 The pitch of spiral should neither to more than (100 mm) or four times the thickness of the barrel, whichever is less, not less than the maximum size of aggregate plus the diameter of the bar used. The minimum clear cover for concrete pipes specified in the (IS) Code for different types of pipes are as shown in Table (6).
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Table (6) Cover Requirements. Barrel Thickness (mm)
For Spun Pipe (mm)
25 mm & below Over 25 & including (30) 0ver (30) & below (75) 75 mm & above
8.5 9.0 12.0 18.0
6-3 Design Examples of pipe culvert. Example (1):-
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For Pipes other than Spun Pipes (mm) 12.0 12.0 16.0 18.0
Solving t=36.53 mm, but minimum thickness is not less than (55) mm.Adopted t=60 mm. 5. Longitudinal Reinforcements. Assume the pipe to span over a length of 3m Self-weight of pipe = (π x 1.06 x 0.06 x 24) =4.79 kN/m Weight of water= ((π x L2/4) x10) = 7.85 kN/m Total design load =2(4.79+7.85) =25.28 kN/m I= π/64(D4-d4) = π/64(1.124- 14) =0.028 m4. Mmax. = (25.28 x 32/8) =28.44 kN.m Stress σ = (28.44 x 106 x 560/ 0.028 x 1012) = 0.568 N/ mm2. Stresses are negligibly small. Provide minimum longitudinal reinforcement of 2.5 kg/m, use ɸ6 mm bars. Weight of each bar = {(π x 0.0062/4) x 7800} = 0.22 kg/m. Number of bars required = (2.5/0.22) = 11.363. Spacing = (π x 1000/no. of bars) = (π x 1000/ 14) = 224.285 mm Use 14 bars of ɸ6 mm @ spaced 200 mm along the circumference as longitudinal 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 slope of the embankment is (1.5:1). The Maximum discharge is (5 m3/s). The save velocity is (3 m/s). Class (AA) tracked vehicle is to be considered as live load. Assume bell-mouthed entry .Given Ce=1.5, Cs=0.010 and the unit weight of the soil=20 kN/m3. Solution. 1- Hydraulic design. Discharge through the pipe, Q=KAV Where, K = {1/ (1+Ke+Kf) 0.5} Now, Kf = 0.0033(L/R1.3) Where L is the length of the pipe, which is equal to the base width of the embankment. Therefore, L= 10+ (2 x 1.5 x 6) =28 m. Assume (1) m diameter pipe, we have R=A/P = (π D2/4 ÷ π 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 m2
Area provided by each pipe, = π D2/4 = π x 12/4 = 0.785 m2 Therefore, the no. of pipes required = 2.13/0.785 = 2.71 say 3 pipes. 2- Bedding for the pipes. 37
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 Cs (for NP3 pipes) Internal External Height of Embankment above the pipe (m) diameter diameter 0.1 0.2 0.3 0.4 0.6 0.8 1.0 2.0 (mm) (mm) 500 560 0.246 0.288 0.198 0.169 0.117 0.083 0.060 0.017 600 770 0.247 0.234 0.210 0.182 0.131 0.094 0.068 0.022 700 870 0.247 0.236 0.215 0.186 0.140 0.102 0.075 0.024 800 990 0.249 0.240 0.220 0.196 0.149 0.110 0.083 0.027 900 1100 0.249 0.241 0.225 0.202 0.156 0.117 0.089 0.029 1000 1230 0.249 0.242 0.228 0.205 0.162 0.123 0.095 0.032 1200 1440 0.249 0.242 0.230 0.209 0.171 0.131 0.104 0.036
3.0
4.0
0.008 0.010 0.010 0.013 0.014 0.015 0.020
0.005 0.006 0.006 0.007 0.008 0.010 0.011
As Ce = 1.5, therefore, the load on the pipe owing to each fill, Cewd2= 1.5 x 20 x 1.232= 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), three
edge-bearing strength for a (NP3) pipe of 1000 mm internal diameter is (72 kN/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 earth fill (kN/m) ÷ strength factor (SF)} + {Load owing to wheel load/ factor of safety} Or, (72÷1.5) = (45.4÷SF) + (42÷1.5) ………………… Therefore, SF = 2.30 Hence, concrete cradle bedding may be provided, see Table (8), Table (8) The types of Beddings and their Strength Factors Type of Bedding Strength Factor Earth Bedding 2.0 First Class Bedding 2.3 Concrete Cradle Bedding 3.7
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Table (9) DESIGN AND STRENGTH TEST REQUIREMENTS OF CONCRETE PIPES OF CLASS NP3 REINFORCED CONCRETE, MEDIUM-DUTY, NON-PRESSURE PIPES(6) Reinforcement
N0minal internal diameter of pipe
Barrel Wall Thickness
(1)
(2)
mm
mm
80 100 150 200 225 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1400 1600 1800 2000 2200 2400 2600
25 25 25 30 30 30 40 75 75 75 75 85 85 95 100 115 115 120 135 140 150 170 185 200 215
Longitudinal, Mild Steel of Hard-Drawn Steel (3) Min. number 6 6 6 6 6 6 8 8 8 8 8 6+6 6+6 6+6 6+6 6+6 6+6 8+8 8+8 8+8 12+12 12+12 12+12 12+12 12+12
(4) Kg/ linear meter 0.33 0.33 0.33 0.33 0.33 0.33 0.78 0.78 0.78 0.78 0.78 1.18 1.18 2.66 2.66 2.66 2.66 3.55 3.55 3.55 9.36 9.36 9.36 14.88 14.88
Spiral, HardDrawn Steel (5) Kg/ linear meter 0.15 0.22 0.46 0.81 1.03 1.24 1.80 2.95 3.30 3.79 4.82 7.01 10.27 13.04 18.30 21.52 27.99 33.57 46.21 65.40 87.10 97.90 113.30 146.61 175.76
Strength Test Requirement for Three Edge Bearing Test Load to produce Ultimate (0.25) mm Load Crack (6) (7) kN/ linear kN/ linear meter meter 13.00 19.50 13.00 19.50 13.70 20.55 14.50 21.75 14.80 22.20 15.00 22.50 15.50 23.25 16.77 25.16 19.16 28.14 21.56 32.34 23.95 35.93 28.74 43.11 33.53 50.30 38.32 57.48 43.11 64.67 47.90 71.85 52.69 79.00 57.48 86.22 67.06 100.60 76.64 114.96 86.22 129.33 95.80 143.70 105.38 158.07 114.96 172.44 124.54 186.81
Note 1: The actual internal dia. Is to be declared by the manufacturer and the tolerance is to be applied on the declared 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 size available 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/mm2) at 28 days and a minimum cement content of (400 kg/m3). 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) = ((π x 0.0122 x 7850)/4) (π x 1.1) = 3.068 kg/m Table (10) Reinforcement Requirements for (NP3 Pipes According to (IS 458 – 1988) Spiral Longitudinal steel reinforcement Ultimate three Internal diameter with permissible with permissible edge bearing (mm) stress of 125 MPa stress of 140 MPa strength (kg/m) (kg/m) (kg/m) 350 0.78 2.95 25.16 400 0.78 3.30 28.74 450 0.78 3.79 32.34 500 0.78 4.82 35.93 600 1.18 7.01 43.11 700 1.18 10.27 50.30 800 2.66 13.04 57.48 900 2.66 18.30 64.67 1000 2.66 21.52 71.85 1100 2.66 27.99 79.00 1200 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 = (π x 0. 0062 x 1 x 7850)/4 =0.22 kg/m Providing at 4 kg/m, no, of bars = 4/0.22 = 18.18 Spacing = (π 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/m2). The live load on the culvert is (50 kN/m2). Density of soil at site is (18 kN/m3). Angle of repose (ɸ=300). Adopt M-20 or fc=2o N/mm2 grade concrete mix and Fe = 415 or FY=415 MPa 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/m2. - Live load = 50 kN/m2. - Density of soil = 18 KN/ m3. - Angle of repose = ɸ = 300. - Class of concrete = M-20 or fc=2o N/mm2. 2- Permissible stresses. - σcc = 5 N/ mm2 - σcb = 7 N/ mm2 - σst = 150 N/ mm2 ( water face) - σst = 190 N/ mm2 ( Away from water face) 41
- m = 13 = modular ratio (Es/Ec). - 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 = ts = tw = 300 mm. Effective span = 3300 mm
4-Loads. -
Self-weight of the top = 0.30 x 24 = 7.2 kN/m2 Super imposed dead load = 12.8 kN/m2 Live load = 50 kN/m2 Total load = 70.0 kN/ m2 Weight of vertical sidewalls = 0.30 x 3.3 x 24 = w = 24 kN. Soil pressure = p = wh (1-sin ɸ)/ (1+ sin ɸ) At h = 3.3 m, ɸ = 300, w = 18 kN/m3. The soil pressure =p = 18 x 3.3 x 1/3 = 20 kN/m2. Uniform lateral pressure due to the effect of super imposed dead load and live load surcharge is calculated as, P = (50 + 12.8) {(1-sin ɸ)/ (1+ sin ɸ)} = (62.8 x 1/3 = 21 kN/m2 Uniform lateral pressure due to the effect of super imposed dead load surcharge only is, P = 12.8 {(1-sin ɸ)/ (1+ sin ɸ)} = (12.8 x 1/3) = 4.26 kN/m2 Intensity of water pressure is obtained as, P = wh = (10 x 3.3) = 33 kN/m2
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 in Table (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) are compiled in Table (13). The design forces resulting from the combination of the various 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 case
Fixed End Moments MD = MD'
MA = MA'
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
Section
1:1
B-1 A-2
A-3
E-4 D-5
D-6
C-7 1.5:1
B-1 A-2
A-3
E-4 D-5
D-6
C-7
Factor For M N V M N M N V M N V M N M N V M N V M N M N M N V M N V M N M N V M N V M N
1 WL w w + 0.182 0 - 0.068 0 + 0.500 - 0.068 + 0.500 0 - 0.052 + 0.500 - 0.036 + 0.500 0 - 0.036 0 - 0.500 + 0.088 0 + 0.170 0 - 0.079 0 + 0.500 - 0.079 + 0.500 0 - 0.062 + 0.500 - 0.045 + 0.500 0 - 0.045 0 - 0.500 + 0.079 0
2 wL2 wL wL + 0.083 0 - 0.042 0 + 0.500 - 0.042 + 0.500 0 - 0.042 + 0.500 - 0.042 + 0.500 0 - 0.042 0 - 0.500 + 0.083 0 + 0.075 0 - 0.050 0 + 0.500 - 0.050 + 0.500 0 - 0.050 + 0.500 - 0.050 + 0.500 0 - 0.050 0 - 0.500 + 0.075 0
Loading Case 3 4 WL pL2 w pL w pL + 0.021 + 0.019 0 - 0.167 + 0.021 + 0.019 0 - 0.167 0 0 + 0.021 + 0.019 0 0 0 + 0.167 - 0.042 - 0.043 + 0.500 0 - 0.004 + 0.023 + 1.000 - 0.333 0 0 - 0.104 + 0.023 0 0 - 1.020 - 0.333 + 0.146 + 0.023 0 - 0.333 + 0.018 + 0.015 0 - 0.167 + 0.018 + 0.015 0 - 0.167 0 0 + 0.018 + 0.015 0 0 0 + 0.167 - 0.050 - 0.047 + 0.500 0 - 0.118 + 0.018 + 1.000 0 0 - 0.333 - 0.118 + 0.018 0 - 0.333 - 1.000 0 + 0.132 + 0.018 0 - 0.333
5 pL2 pL pL - 0.019 + 0.167 - 0.019 + 0.167 0 - 0.019 0 - 0.167 + 0.043 0 - 0.023 + 0.33 0 - 0.023 0 + 0.333 - 0.023 + 0.333 - 0.015 + 0.167 - 0.015 + 0.167 0 - 0.015 0 - 0.167 + 0.047 0 - 0.018 0 + 0.333 - 0.018 + 0.333 0 - 0.018 + 0.333
6 pL2 pL pL - 0.042 + 0.500 - 0.042 - 0.500 0 - 0.042 0 - 0.500 + 0.083 0 - 0.042 0 + 0.500 - 0.042 + 0.500 0 - 0.042 + 0.500 - 0.033 + 0.500 - 0.033 + 0.500 0 - 0.033 0 - 0.500 + 0.092 0 - 0.033 0 + 0.500 - 0.033 + 0.500 0 - 0.033 + 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 section acts outward when viewed from within. 3- Positive thrust indicates compression on the section. 45
The maximum positive moments develop at the center of bottom & top slab for the condition that the sides of the culvert not carrying the live load and the culverts running full with water. The maximum negative moments develop at the support sections of the bottom slab for the condition, culvert is empty and the top slab carries the dead & live load. Table (13) Forces Components for Different Cases of Loading Section
Forces
B-1
M N M N V M N V M N M N V M N V M N
A-2
A-3
E-4 D-5
D-6
C-7
Case-2 63.20 0 - 31.60 0 115.50 - 31.60 115.50 0 - 31.6 115.50 - 31.60 115.50 0 -31.60 0 -115.50 63.20 0
Loading Case Case-4 Case-5 6.82 - 4.13 - 18.18 + 11.0 6.82 - 4.13 - 18.18 + 11.0 0 0 6.82 - 4.13 0 0 18.18 - 11.00 - 15.45 - 9.36 0 0 8.26 - 5.00 - 36.26 + 21.90 0 0 8.26 - 5.00 0 0 -36.26 +21.90 8.26 -5.00 - 36.26 +21.90
Case-3 1.66 0 1.66 0 0 1.66 0 0 - 3.32 + 39.60 - 0.317 + 79.20 0 -8.23 0 -79.20 11.56 0
Case-6(a) - 9.60 + 34.65 - 9.60 + 34.65 0 - 9.60 0 - 34.65 +19.20 0 - 9.60 0 +34.65 - 9.60 +34.65 0 -9.60 +34.65
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 / (σst * J *d) = (76.10 x 106 / 150 x 0.86 x 270) = 2234 mm2/m Provide 20Ø @ 140 mm c/c Distribution steel = (0.30 x 300 x 100 / 100) = 900 mm2 Provide 10 Ø @ 150 mm(c/c) on both faces. -Section: D-6, (support section) as shown in Table (14) M = -54.43 KN.m N = 34.65 KN 46
Case-6(b) - 1.92 + 6.93 - 1.92 - 6.93 0 -1,92 0 -6.93 +3.84 0 - 1.92 0 +6.93 - 1.92 + 6.93 0 - 1.92 + 6.93
Ast = (M / (σst * J *d) = (54.43 x 106 / 190 x 0.86 x 270) = 1233mm2/m Provide 16 Ø @ 150 mm (c/c) and distribution bars of 10 Ø @ 150 mm (c/c) Table (14) Design Moments and Forces in Box Culvert Section D-6 A-2 B-1 C-7 E-4
Loading Combination Cases 2+3+5+6(a) 2+3+5+6(a) 2+3+4+5+6(b) 2+3+4+5+6(b) 2+3+4+5+6(b)
Moment M KN.m -54.43 -43.67 65.63 76.10 -55.89
Thrust N (kN) +34.65 -23.65 -1.25 -7.43 +155.10
Shear Force V (kN) -172.80 +115.50 0 0 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 /fck *bD2)= (83.83 x 106/ 20 x 1000 x 3002) = 0.046 with fy= 415 N/mm2 (NU / fck *bD) = 232.5 x 103/ 20 x 1000 x 300) = 0.0387 with (d'/D) = (30/300) =0.10 By referring to interaction curve of SP-16, (p / fck) (7) = 0.02 Where, Asc =Ast = 0.5 (phD/100) = 0.5(0.02 x 20 x 1000 x 300) / 100 = 600 mm2 Hence AS = 1200 mm2. But minimum reinforcement of (0.8%) of cross section has to be provide, AS = (0.8 x 300 x 1000 / 100) = 2400 mm2 Hence, provide 16 Ø @150 mm (c/c) on both faces in the vertical sidewall. Distribution steel of 10 Ø @ 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. FHWANHI-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 the Indian Roads Congress, October-December 2009, pp. (190-812). 3- Darryl Shoemaker, Ph.D., Jack Allen, Margaret Ballard, Stephen David, and George 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,1st edition, New Delhi,1988,pp.370. 6- IS 458-1988: "SPECIFICATION FOR PRECAST CONCRETE PIPES ((WITH AND WITHOUT REINFORCEMENT)", BIS, New Delhi, pp.46. 7- (SP-16): IS- 456-l978:"DESIGN AIDS FOR REINFORCED CONCRETE", BIS, New Delhi, And PP.255. 48
