chapter 7 design of prestresed concrete bridges

7/27/2019 CHAPTER 7 Design of Prestresed Concrete Bridges

1/28

CHAPTER 7: DESIGN OF REINFORCED CONCRETE BRIDGES

Bridge deckconsists of reinforced flooring and a system of longitudinal beams:

Concrete Steel timber- 3000 to 4000psi compressive strength- in-thick integral concrete, or 1 to 2in thick asphalt concrete or latex-

modified concrete

- (AASHTO,1992a:CRSI,1993;PCI,1975)- 20 t0 60ft span range for concrete bridges(NBI)- 60 to 120ft span range for steel bridges(NBI)

MATERIALS OF THE CONSTRUCTION

The two materials used for the construction of reinforced concrete bridges are

concrete and reinforcing steel.

Various reinforced concrete bridge deck sections:

a) Slabb) Voided slabc) T-beamd) A three-cell boxe) A four-cell girder

DESIGN METHODS

Structural design of reinforced concrete bridges can be performed by the service

load design load (AASHTO 8.15) or by the strength (or load factor) design method

(AASHTO 8.16). This design approach is stipulated in AASHTO 8.14. AASHTO

8.14.1.3 further stipulates that the strength and serviceability requirements of the strength

design are satisfied by the service load design if the service load stresses are limited to

values prescribed by AASHTO 8.15 (discussed in the following sections).

For analysis of a cracked rectangular reinforced concrete section, the neutral axis

factor, k, is given by:

(6.1)

Where = As/bd (reinforcement ratio)


2/28

n = Es/Ec (modular ratio)

(lever-arm factor)The California DOT specifications (CALTRANS, 1993a) stipulate the following values

of n:

fc = 2000-2400 psi n = 15

= 2500-2900 = 12

= 3000-3900 = 10

= 4000-4900 = 8

= 5000 or more = 8

Allowable concrete stresses:

Mc =1/2fckjbd2

(6.2)

Allowable steel stresses:

Mc=Asfsjd (6.3)

1. The depth of the slab or of a rectangular beam

(6.4)

2. Area of reinforcing steel required: (6.5)

Where b = 12 in. for slab

(6.6) (6.7)

Where fc = allowable stress in concretefs = allowable stress in steel reinforcement

Slab thickness, h , should be varied in increments of in (AASHTO 8.9.2)

Allowable stresses (AASHTO 8.15): Flexure

Allowable stresses in concrete

Extreme fiber stresses in compression 0.40fc

Extreme fiber stress in tension for plain concrete 0.21fr


3/28

The modulus of rupture, fr, should be obtained from the tests. In the absence of test

results, the following values of the modulus of rupture should be used:

Normal-weight concrete 7.5Sand-lightweight concrete 6.3

Bearing stress, fb 0.30 A higher value of the bearing stress is permitted when the supporting surface is wider on

all sides than the loaded area; in such a case the allowable bearing stress on the loaded

area may be multiplied by a factor of but not one greater than 2.Al lowable stress, fs, in steel reinf orcement

Grade 40 reinforcement 20,000psi

Grade 60 reinforcement 24,000psi

Certain states or agencies may specify their own allowable stress values, which may be

lower than those specified previously. For example, the California Department of

Transportation specifies that the following stresses(CALTRANS, 1993a):

Extreme fibre stress in compression transversely reinforced deck slabs:

1,200psi

Grade 60 reinforcement for transversely reinforced deck slabs:

20,000psi

Allowable stresses (AASHTO 8.15.2): Shear

Shear stress in a reinforced concrete beam can be calculated from Eq.6.8:

(6.8)Allowable stresses and design for shear are covered AASHTO 8.15.2. In members

subjected to flexure and shear, the allowable shear is given by Eq.6.9:

(6.9)A more detailed calculation of the allowable shear stress can be made from Eq.6.10:

(6.10)where M = design moment occurring simultaneously with V at the section being

considered, and Vd/M 1.0

AASHTO specifications stipulate two conditions for providing shear reinforcement:


4/28

1. Shear reinforcement is to be provided when the design stress, V, exceeds Vc(AASHTO 8.15.5.3)

2. Minimum shear reinforcement is to be provided when the design shear exceedsone-half the allowable shear stress (AASHTO 8.19).

Design for shear. When the design exceeds the allowable shear stress, the excess shear

must be carried by shear reinforcement. However, this excess shear is limited in such a

way that the excess shear stress, , does not exceed (AASHTO 8.15.5.3.9);if it does, a larger beam cross section is indicated. The required shear reinforcement,

perpendicular to the beams longitudinal axis (vertical stirrups), can be calculated from

Eq. 6.11:

(6.11)where Av = cross-sectional area of all legs of stirrup

s = spacing of stirrups

AASHTO 8.19.3 stipulates that the stirrup spacing should not exceed the smaller of

d/2 or 24in. These spacings are to be reduced by half, to the smaller of d/4 or 12 in.,

if the excess shear stress, , exceeds .The minimum shear reinforcement is calculated by the following expression, with

stirrup spacing limited to d/2 or 24 in.:

(6.12)Strength Design Method (Load Factor Design Method)

The load factor design method calls for designing reinforced concrete members by the

strength design method, for which the relevant design requirements are covered in

AASHTO 8.16, and the design assumptions are stipulated in AASHTO 8.16.2.

In spite of similarities in the application of strength design principles to buildings and

bridges, different load factors are used in two cases because of differences in the

variability of loads involved. For example, for the two cases because of differences in the

variability of loads are 1.4 and 1.7, respectively (ACI,1995), whereas those for a bridge

are to be taken as 1.3 for the dead load and 6.5/3 (1.3 x 5/3) for live load plus

impact(AASHTO, 1992a):


5/28

Factors (AASHTO, 1992a)

a Flexure 0.9

b Shear 0.85

c Axial compression with

spirals 0.75ties 0.7

d Bearing on concrete 0.7

Design Parameter

AASHTO Eq.8.17 (6.15a) AASHTO Eq.8.17 (6.15b)

The moment equilibrium yields

(6.16)Equation 6.16 gives the nominal strength of the section, which, multiplied with , the

strength reduction factor, is equated to the factored design loads, Mu (AASHTO

Eq.8.16):

(6.17)In Eq.6.17, one substitutes the value of a from Eq.6.15a to yield

(6.18)The constant 0.588 in the right-hand-side parenthetical term in Eq.6.18 is rounded off to

0.59 in the ACI building code (ACI,1995) and to 0.6 in AASHTO Eq.8.15. Thus

(6.19)The value of the balanced steel ratio,, obtained from compability of balanced

strains in concrete (c = 0.003, AASHTO 8.16.3.1.2) and reinforcement ( ) at ultimate conditions, is given by AASHTO Eq.8.18: (6.22)

Since is clearly a function of the given design parameters, namely, , , and(which again is a function of ), its value is known in advance. For computationalconvenience and efficiency in the design process, values of are tabulated for variouscombinations of and , as given in Appendix A (Table A.18). This table also givesvalues of 0.5 , a value which may be used for the initial design trial. Note that the depthof Whitneys rectangular stress block, a, is related to the depth of the neutral axis, c, by


6/28

the relationship a = , where is a coefficient whose value, subject to a minimum of0.65, is given by Eq. 6.23 (AASHTO 8.16.2.7):

(6.23)Service load stresses

Stresses under service load conditions should not exceed the allowable stresses,

as presented in the earlier sections.

Fatigue stresses limits

AASHTO 8.16.8.3 stipulates that the range between a maximum tensile stress

and the minimum stress in straight reinforcement should not exceed ff , given by

AASHTO Eq. 8.60:

(6.24)Whereff = stress range, ksi

fmin = algebraic minimum stress level, ksi (tension positive and compression

negative)

r/h = ratio of base radius (r) to height (h) of rolled-on traverse deformations, equal

to 0.3 when the actual value ofr/h is not known.

This fatigue limit is waived when the deck slab, with primary reinforcement

perpendicular to traffic, is designed in accordance with AASHTO 3.24.3, Case A.

Distribution of flexural reinforcement

To control flexural cracking of concrete, tension reinforcement should be well

distributed within the maximum flexural zones. This requirement is critical only when

the yield strength of flexure reinforcement exceeds 40,000 psi. In such case, the bar sizes

and their spacing at the maximum positive and negative sections should be so chosen

that the stress in reinforcement steel under service load conditions does not exceed the

following value (AASHTO 8.16.4, Eq.8.61):

(6.25)where A = effective area, in.2, of concrete surrounding the flexure tension reinforcement

and

having the same centroid as the reinforcement, divided by the number of bars or

wires.


7/28

dc = thickness of concrete cover measured from extreme tension fibre to the

center of

the closest bar or wire, in., limited to 2 in. maximum for calculations

It is instructive to note that this requirement is similar to that required for the reinforced

concrete buildings (ACI, 1995).

The Gerly-Lutz formula for crack width is expressed in a modified form

(Nawy,1972,1985; Lutz, 1974):

(6.26)where number of bars if all are of the same diameter, or the total area ofreinforcing steel divided by the area of the largest bar if more than one size is used.

AASHTO Eq.6.61 is the Gergely-Lutz formula (Gergely and Lutz, 1968) derived

from the general formula based on a statistical study of test data of several investigations

(Nawy,1985):

(6.27)

Two bars Three bars Four bars

where crack width in units of 0.001 in.

depth factor, average value = 1.20

fs = maximum stress (ksi) in steel at service load level, 0.6fy to be used if no

computations are available

Note:fs= 0.6fy

For computational simplicity, a parameter,z, is introduced in Eq. 6.27 (Nilson

and Winter, 1986):

(6.28)Thez-w relationship of Eq.6.28 can be expressed as

(6.29)


8/28

Note that the units ofzare kips per inch. The AASHTO specifications (AASHTO,

1992a) prescribe the following upper and lower limits on the value ofz:

z = 170 kips/in. for members in moderate exposure condition

z = 170 kips/in. for members in severe exposure condition

REINFORCEMENT

Compression reinforcement used to increase the strength of flexural members is

to be enclosed by ties or stirrups of at least #3 bar for longitudinal bars that are #10 or

smaller, and at least #4 bar for #11, #14, #18, and bundled longitudinal bars. The

minimum size of bars should be #5(AASHTO 8.18.1.2). The use of #3 bar is generally

not practical for longitudinal reinforcement in slab construction.

Minimum reinforcement

AASHTO 8.17.1 requires that any section of a flexural member where tension

reinforcement is required by analysis, the actual reinforcement provide should be

sufficient to develop a moment at least 1.2 times the cracking moment, calculated on the

basis of modulus of rupture for normal weight concrete (AASHTO 8.15.2.1):

(6.30) (6.31)where S = section modulus, based on uncracked section

Spacing limits

Spacing limits for reinforcing bars for bridges, different and generally more

conservative than those for reinforced concrete buildings recommended by the

ACI(1995), are covered in AASHTO 8.20:

1. For cast-in-place concrete, the clear distance between parallel bars in a layershould not be less than 1 times the bar diameters, 1 times the coarse

aggregate, or 1 in.

2. Where bars are place into two or more layers, the bars of the upper layers shouldbe placed directly above the ones in the lower layers. The clear distance between

the bars into two layers should not be less than 1in.

3. Bundling of bars in a group is permitted. For bars #10 and smaller, a maximum offour bars can be bundled; for bars #11, #14 and #18, the number of bars in a

bundle is limited into two. When the spacing limitation are based on the diameter,


9/28

a unit of bundle bar should be treated as a single bar of a diameter derived from

the equivalent total area (AASHTO 8.21.5).

Protection against corrosion

AASHTO 8.22 requires the following minimum clear cover (in inches), outside

of the outermost steel, for protection of reinforcing bars from the environment:

Concrete cast against and permanently exposed to earth 3

Concrete exposed to earth or weather

Primary reinforcement 2

Stirrups, ties, and spiral 1

Concrete deck slabs in mild climatesTop reinforcement 2

Bottom reinforcement 1

Concrete deck slab that have no positive corrosion protection

and are frequently exposed to deicing salts

Top reinforcement 2

Bottom reinforcement 1

Concrete not exposed to weather or in contact with ground

Primary reinforcement 1

Stirrups, ties, and spirals 1

Concrete piles cast against or permanently exposed to earth 2


10/28

Transverse Reinforcement

AASHTO 3.24.10 stipulates that all reinforced concrete slabs be provided with

distribution reinforcement, near the bottom fibre, placed perpendicularly to the main

reinforcement, to distribute loads laterally. The amount of such reinforcing steel is

governed by the orientation of the reinforcement with the respect of the direction of the

traffic:

For main reinforcement parallel to traffic (AASTO Eq.3.21):Percentage of reinforcement but 50% (6.32)

where S= effective length.

For main reinforcement perpendicular to traffic (AASTO Eq.3.21):Percentage of reinforcement but 67% (6.33)


11/28

CONSTRUCTION DETAILS

Diaphragms

In T-beam construction, they are provided between the beams at the ends and at

some intermediate points along the span, and they are cast integrally with the slab and

the webs. At the intermediate points, they provide lateral support to beams, and when the

abutment does not extend above the bridge-seat level, the end diaphragms support the

backfill.

Deck Joints

To provide for expansion and contraction of the deck due to temperature and the

other causes, it is common practice to provide expansion joints in a bridge deck at the

expansion ends and at other desirable locations. Details for the expansion andconstruction deck joints are spelled out in AASHTO Div. II, 8.9. In humid climates in

areas subjected to freezing, joints should be sealed to prevent erosion, filling with debris,

and freezing-induced spalling.

Bearings

Bearings are devices provided at the abutments o piers and positioned between

the bottom flanges of the beam and the top of the bridges seats. Basically, they serve the

following important functions:

1. Uniformly distribute the concentrated horizontal and vertical loads due to beamreactions over bearing stress to eliminate highly localized stress and resulting

structural damage

2. Allow movements between the superstructure and the substructure (abutmentsand piers), and minimize the effects of loads due to volume changes resulting

from shrinkage, creep and temperature

3. For longer spans, allow rotations at the support that will be caused by thedeflection of the loaded beam

Sidewalks, Curbs, Parapets and Railings

Sidewalks are usually provided for pedestrian traffic on bridges, generally

carrying urban express ways, and wherever necessary. They should be in such width as

required by the controlling and concerned agencies, but preferably not less than 4ft (ACI,

1969). When sidewalks are provided, they should be separated from the bridge roadway

by the use of a suitable railing (AASHTO 2.7.3).


12/28

Curbs and parapets are safety barriers provided parallel to longitudinal axis of the

bridge on both sides of the roadway, designed to prevent a moving vehicle from leaving

the roadway. But maybe designed to form a combination curb and gutter section. Curbs

are of two general types: parapets (nonmountable) and mountable curbs (also called

brush curbs). Curbs are generally not less than 2ft 6in. high, and may be used

combination with the curbs.

Curbs and sidewalks are poured after the hardening of the deck. Both should have

vertical slits or other provisions of discontinuity to prevent them from resting deck

bending moments; this avoids any structural cracking.

Railings are safety devices provide to contain the vehicles. AASHTO

specifications (1989, 1992b) provide current guide specifications for highway bridge

railings.

Medians

Lanes carrying the opposing traffics should preferably be carried on two separate

structures. However, when the width limitations dictate the use of the same bridge for

lanes of opposing traffic, they should be separated by traffic separators known as

medians, usually of parapets sections 12 to 27 in. high.

Drainage

The drainage should be both traverse and longitudinal, should be provided on a

bridge roadway. Traverse drainage is provided by a suitable crown in a deck (about 1/8

in. ft), and the longitudinal drainage by camber or gradient (AASHTO 1.5). For

longitudinally horizontal bridges, a camber of 0.5 to 1 percent is generally provided to

carry the water in the gutters to the ends of the span.

Roadway Width

The width of the roadway on a bridge is defined as the clear width measure at the

right angles to the longitudinal centreline of the bridge between the bottom of the curbs,

or in their absence, the distance between the nearest faces of the bridge railing

(AASHTO 2.1.2). The roadway widths are at least the distances between the approach

guardrails, if guardrails are provided, or they equal the length of the roadway section,

including shoulders. Where a curbed roadway sections approaches a bridge, the same

section is carried across the bridge (AASHTO, 1992a, Article 2.3.1).


13/28

DESIGN OF SLAB BRIDGES

Slab bridges are characterized by simple or continuous concrete slab spanning in

the direction of traffic. The wearing surface, say 25 to 35 lb/ft2, is added to obtain the

total dead load. The live load moment and the shear is calculated according to the

methods describe by Chapter 4 and are multiplied by the factor (1 +I) to account for the

effect of impact. The differences of reinforcing for continuous slab bridges designed for

HS15 or HS20 loading, having 20-ft, 30-ft, and 40-ft end spans are only 0.3, 0.6 and 1.5

lb/ft2 of the deck area, respectively. It is recommended for short span to design for HS20

loading (PCI, 1975).

Moments and shear at intermediate points. It is customary to determine the maximum

moment and shears at several points along the span, usually at every tenth points (L/10).

Example: in this 40 ft span bridge, the maximum moment occurs at 17.67 ft from one ofthe supports.

16k 16k 16k

14 14

5 35

4.375 2.625 0.875

Influence line for Mmax at 5ft from the left support.

Solution:

MD = 1.052 x x 4.375 x 40 = 92.05 k-ft

ML = 16(4.375 + 2.625) + 4 x 0.875 = 115.5 k-ft

ML+I = 115.5 x 1.3 = 150.15 k-ft

ML = 1.3(92.05 + 1.67 + 150.15) = 445.64 k-ft

16k 16k 16k

14 14

10 30

7.5 4.0 0.5


14/28


Solution:

MD = 1.052 x x 7.5 x 40 = 157.8 k-ft

ML = 16(7.5 + 4.0) + 4 x 0.875 = 186.0 k-ft

ML+I = 186.0 x 1.3 = 241.8 k-ft

ML = 1.3(157.8+ 1.67 + 241.8) = 730.1 k-ft

Flexural reinforcement will be cut off, calculate of the nominal moment capacity of the

T-section with only 3bars in the bottom layer and with only six bars in the in the bottom

two layers. All bars however will be extende to full development.

Mn with 3 # 11 barsd = 29.54 in. As = 0.85ab = Asfya =

= 0.765 inMn = Asfy = = 456.7 k-ft

Mu = Mn = 0.9 x 456.7 = 411.03 k- ft

Mn with 6 # 11 bars

d = 27.96 in. As = 0.85ab = Asfya =

= 1.53 in

Mn = Asfy = = 849.39 k-ftMu = Mn = 0.9 x 849.39 = 764.45 k- ft

The bars can be cut off as follows:

1. At 10 ft from the support Mu = 730.1 k-ft, whereas the Mu of the section with six# bars is 764.45 k-ft . hence, out of the eight total # 11 bars, the top two # 11 bars

can be theoretically stopped at this point.


15/28

2. At 5ft from the support, Mu = 445.64 k-ft, whereas the Mu of the section withthree # 11 bars in the bottom layer is only 411.03 k-ft (smaller); hence,the

additional three #11 bars cannot be cut off at this point. Calculate the maxmum

moment at 4 ft from the support and check if the bars can be cut off at this point.

16k 16k 16k

14 14

4 36

3.6 2.2 0.8


Solution:

MD = 1.052 x x 3.6 x 40 = 75.7 k-ft

ML = 16(3.6 + 2.2) + 4 x 0.8 = 96.0 k-ft

ML+I = 96.0 x 1.3 = 124.8 k-ft

ML = 1.3(75.7 + 1.67 + 124.8) = 369.3 k-ft

This is the smaller than the (411.03 k-ft) of the section with three # 11 bars. Hence, of

the six #11, the three #11 bar in the bottom layer will be extended the full length of the

beam to satisfy the requirements of AASHTO 8.242.2.1 the least one third shall extend

along the same face of the member into the support.

Development lengths. The required development length of flexural reinforcement are

governed by AASHTO 8.24 and 8.25

*the basic development length (AASHTO 8.25.1)Ldb is

Ldb = = = 39.46 in.

0.0004db fy = 0.0004 x 1.14 x 40,000 = 22.56 in. or 12 in (AASHTO 8.25.4) Ldb =

39.46 in say 40 in---governs. The two #11 bars must be extended 40 in. in the directionof decreasing moment, beyond the point of maximum stress.


16/28

*According to AASHTO 8.24.1.2.1

15 bar diameter, 15 db = 15 x 1.14 = 21 in

1/20 of clear span, L/20 = 40 x 12/20 =24 in

Effective depth, d = 26.69, say 27 ingoverns

*AASHTO 8.24.2.2 continuing reinforcement (6#11 bars) must have development length

at least equal to ld (40 in) beyond the point where they no longer required to resist

tension.

* AASHTO 8.24.2.3 requires that the bar size should be such that ld computed for fy per

AASHTO 8.25 satisfies the following equation.

Ldb 1.3 (M/V) + la (AASHTO eq. 8.65, modified)Where:

M= computed moment capacity, assuming all positive tension reinforcement at

the section as fully stressed

V= maximum shear force at the section

La= embedment length beyond the center of support

1.3= coefficient giving 30 percent increase in the value of M/V in the

development length limitation, to account for the confinement of the end reinforcement

by compressive reaction, which tends to prevent splitting and bond failure along the bars

Flexural capacity of three #11 bars (Mu = 411..03 k-ft) and maximum shear force at the

support (Vu = 120.3 kips) had been calculated earlier.

The value of la is kept 6 in (AASHTO 8.24.2.1)

Ldb 1.3 (M/V) + la= 1.3[(411.03 x 12 ) / 120.3] + 6

=59.3 in

The actual of 40 in. meets this restriction.*when flexural reinforcement is cut off in some points in the tension zone of beam

provision of AASHTO must be satisfied to prevent premature flexural cracks in the

vicinity of the cut ends.


17/28

a. extra stirrups will be provided along each terminated bar over a distance from

the termination point equal to of the effective depths of the member, 0.75 x 26.69 = 20

in.

b. the excess stirrup area Av, required is not less than. Av =

c. the spacing s, should not exceed d/8, where is the ratio of the area ofreinforcement cut off to the total area of the reinforcement at the section. In present

section d= 26.69 in, the values of d/8 are as follows:8#11 bars, As= 12.5 6#11 bars, As= 9.37

=(12.59.37)/12.5 = 0.21, d/8

= 15.9 in

3#11 bars, As= 4.68 = 5.68/9.37 = 0.5, d/8 = 7.4 inDistribution of flexural reinforcement. To control the flexural cracking of concrete,

tension reinforcement should be well distributed within the maximum flexural zones.

exceeds 40000 psi provision of AASHTO 8.16.8.4 must be satisfied.Design of exterior girders.

Exterior girder have the same cross section as the interior ones. By AASHTO

3.23.2.3.1.4

= 307.1 k-ft= 255.8 k-ft= 1.3[D + (L+l)]= 1.3(1.0 x 307.1 + 1.67 x 255.8) = 954.6 k-ft

The required nominal capacity of the beam,

=

/

= 954.6/0.9 = 1060. k-ft

Interior beam, = 1069.6 k-ft (calq earlier)= wL = x 1.557 x 40 = 31.14 kipsMaximum liveload shear in the exterior girder due to service load plus impact was

calculated earlier as 34.0 kips. Therefore,

= 1.3[D + (L+l)]= 1.3(31.14 + 1.67 x 34.0) = 114.3 kips

Interior was calculated to be 120.3 kips which is greater than the value of for theexterior girder.


18/28

DESIGN OF BOX GIRDER BRIDGES

1937- Construction of the first reinforced concrete bridge in the United States.

The popularity of box girder bridges( reinforced and prestressed) has increase in western

states particularly in California where 90 percent of the bridges built on the state

highway system are concrete box girder and about 80 percent of them are prestressed.

Although cast-in-place box girder have been built for span of 460 ft, for simple span are

economical 95 to 140 ft range, limited by excessive dead load deflections

[Degenkolb,1977]. This make them ideally suited for highway interchange structures for

which the general span range is 50 to 150 ft. for longer span, prestressed concrete box

are more economical.Lightweight concrete is used to reduce the deadweight of the superstructure in cases

where the normal weight concrete is too heavy from the practical standpoint.

The differences in the physical properties of normal weight ad lightweight aggregates

cause their design to vary (Modulus of rapture , modular ratio n, for deflection and themultiplier for their shear strength).

Advantages

1. The relative shallow depth requirement of a box girder bridge is a definiteadvantage whe headroom is limited, a condition frequently encountered in urban

areas.

2. Monolithic construction of the superstructure and the substructure offers structureadvantage as well as enhanced aesthetics.

3. They provide ideal space for utilities such as gas and water pipelines; powertelephone and cable ducts; storm drains; and sewer which safely placed inside the

large cells and hidden from view.

4. High torsional stiffness.5. Easy aesthetic treatment through smooth finishing of the soffit and the sides.

6.8.2 General Design Consideration

6.8.2.1 Structural behaviour

A reinforced concrete box girder is essentially a T-beam with the transverse bottom

flange similar to the top flange, resulting in a closed, torsionally stiff multicell


19/28

configuration. Figure 6.8 shows configurations of several commonly used multicell box

girder bridges [Libby and Perkins, 1976]

Top deck. Supported on webs (also referred as girders).

interior web. Resist shear and often only a small portion of moment. Vertical moment

while

exterior web. May be vertical, inclined curve or otherwise profiled as shown in fig.

6.6.and if inclined their sloped should preferably be 1:2

6.8.2.2 Proportion

Depth. for constant depth of box girders, AASHTO 8.9.2 minimum depth span ratio of0.060S for simple span and 0.055S for continuous span.

Top slab thickness. No recommendation for minimum thickness of the top slab except

that they satisfy the design requirements. However many designer use greater deck

thickness, varying from 6 to 9 in as required by the standards and manual in use. Cover

in AASHTO 8.22.

Bottom slab thickness. Bottom slab or soffit essentially performs three important

functions:

1.) It contains reinforcement for positive moment.

2.) It function as compression flange for the negative moments for continuous

box girders,

3.) It provides the desired architectural features of the girder.

Web. There is no provision in the AASHTO specification governing the spacing and

thickness of web (or girder). The primary purpose is to resist shear and Resist shear and

often only a small portion of moment.

Fillet. Requirements for fillet are covered by AASHTO 9.8.2.3, which requires that

adequate fillet be provided at the intersection of all surfaces within the cell of a box

girder, except the junction of the web at the bottom flange. The usual practice is to

provide 4 in by 4 in.


20/28

Reinforcement. By AASHTO 8.17.2.3

1. A minimum distributed reinforcement of 0.4 percent of the flange are should beplaced in the bottom slab parallel to the girder span. Spacing should not exceed in

18 in.

2. A minimum distributed reinforcement of 0.5 percent of the cross-sectional area ofthe slab, based on the least thickness, should be placed in the bottom slab

transverse to the girder span, and distributed on both faces of the slab. Spacing

should not exceed in 18 in.

3. Girder reinforcement most commonly used sized. #11 bar when it is design byservice load method. #14 bar and #18 bar may be warranted if the number of

smaller bars causes undesirable congestion. #10 and #11 bar when it is design by

load factor design method

Diaphragm. Diaphragm are not required in box girder unless the box girders are sharply

curved. AASHTO 8.12.3 stipulates that diaphragm are not required for straight box

girder bridges and curve if the inside radius of 800 ft or greater.

Construction. In the case of multispan structures, two or more span is made continuous

over the intermediate supports. Two alternatives are in use: 1.) Pour the soffit and web

together without a construction joint between them, 2.) have the soffit poured first,

followed by the pouring of the webs and the top slab.

6.8.3 Design Procedure and Example

Example: Design a reinforced concrete box girder for a simple s[pan of 100 ft for a two

lane highway bridge for HS20 loading. The typical cross section is shown in fig. E6.3a.


21/28

Use = 43000 psi and grade 60 reinforcement.Calculation:

The minimum depth satisfy AASHTO 8.9.2

= 0.055S = 0.055 x 100 = 5.5 ftThe bottom flange thickness, t, is computed according to AASHTO 8.11.2:

Clear span between girder wens = 9.51.0 = 8.5 ft.

t = = 6.375 in or 6 in.

Reinforcement parallel to girder. (To satisfy the requirement of AASHTO 8.17.2.3.1)

the width between the outside faces of thbe exterior web is 29.5 ft.

= 0.004 x 6.375 ( 29.5 x 12 ) = 9.03 DESIGN OF SLAB BRIDGES

Method of design:

1. The service load method2. The load factor design

EXAMPLE: DESIGN OF A TYPICAL SINGLE-SPAN REINFORCED

CONCRETE SLAB BRIDGE

Design a two-lane reinforced concrete slab bridge for a clear span of 20ft to carry HS20

loading. A provision of 30 lb/ft2 of dead load should be made in the design for the future

wearing surface. The following data are given:

fc= 3000psi Grade 40 reinforcement clear width =

40 ft

Calculations.The design of a slab bridge essentially consists of two parts:

1. Design of the slab2. Design of the longitudinal edge beams.

Design by the service load method: Design of slab. Assume 6 in. to the center of the

bearings from the face of the abutments:

L = S = 20 + 2 x (6/12) = 21 ft

The minimum slab thickness is given by AASHTO 8.9.3 (Table 5.3 in this text) for

deflection control:

tmin = =

= 1.24 ft


22/28

Therefore, try a 15 in. thick slab

Self-weight of slab = 1 x 1.25 x 150 = 187.5 lb/ft2

Wearing surface = 30.0 lb/ft2

Total load = 217.5 lb/ft

2

Dead-load momentMD = wL2/8 = 0.2175 x (21)2/8 = 11,990 lb-ft

To calculate live-load moment, the weight of one rear wheel of the HS20 truck (16 kips)

is distributed over a width, E, per AASHTO 3.24.3.2:

E = 4 + 0.06S = 4 + 0.06 X 21 = 5.26 < 7.0 ft

The weight on a unit width of slab =

= = 3,042 lb

This 3042-lb load is treated as a concentrated load on a span of 21 ft, and the

corresponding live-load moment is

Compare this with moment due to HS20 lane loading. The lane load (both the uniform

and the concentrated) is distributed over a width of 2E; thus,

The live-load moment due to this load is

The concentrated load is

The moment due to this load is given by

The total live-load moment due to lane loading is

ML = 3.352 + 8.98 = 12.332 kip-ft < 15.97 kip-ft (truck loading)

Therefore, the truck loading governs. UseML = 15.97 kip-ft (the larger value):


23/28

Impact: Hence, useI= 0.3 (maximum value). Thus,

MI = 0.3 x 15.97 = 4.79 kip-ft

Mtotal = 11.99 + 15.97 + 4.79 = 32.75 kip-ft

For fc = 3000 psi, n = 9 (Table A.14, Appendix A). For the design of rectangular

sections, k= 0.351 andj = 0.883 (Table A.16, Appendix A). Thus, from Eq. 6.4,

Assuming #8 for main reinforcing bar, and 1 in. as clear cover, the total required slab

thickness is h = 13.2 + 0.5 + 1 = 14.7 in. Use h = 15 in., required for deflection control.

Then, the effective depth is d= 150.51.0 = 13.5 in. From Eq. 6.5,

Therefore, provide #9 @ 7 in. o.c.,As = 1.71 in.

2 (Appendix A, Table A.19).

Distri bution reinf orcement (AASHTO 3.24.10.4)

Percentage of reinforcement = 100/

= 100/

= 21.82% < 50% max.

Therefore,As = 0.2182 x 1.65 = 0.36 in.2 Provide #6 @ 15 in. o.c.,As = 0.36 in.

2, directly

above and perpendicular to the main reinforcement.

Temperature rein for cement (AASHTO 8.20.1)

in each direction. Provide #4 @ 18 in. o.c. in each direction bellow the top face of the

slab. All reinforcement details are shown in Fig. E6.1a.

Shear and bond (AASHTO 3.24.4).Slab designed for bending by the previous method is

considered satisfactory in shear and bond; hence, no further checking is necessary.

Check for min imum rein forcement (AASHTO 8.17.1.1)


24/28

Fatigue stress limits (AASHTO 8.16.8.3).Fatigue stress limits are to be checked for the

service load conditions as provided by AASHTO Eq. 8.60.

The minimum stress level, fmin, is caused by dead loads only. From service load design

calculations,MD = 11.99 kip-ft.

From Table A.15 (Appendix A), forn= 9 and = 0.0106 (by interpolation), k= 0.351

andj = 0.883. Therefore, the minimum stress,fmin, is

The maximum stress is caused by the live load plus impact; the corresponding moment,

including the dead-load moment, isMtotal = 32.75 kip-ft. The corresponding stress in the

main reinforcement is

Thus, the actual stress range is ff = 19.287.05 =12.23 ksi. According to AASHTO Eq.

8.60,

Hence, the fatigue stress limit requirements are satisfied.

Design of longitudinal edge beams. AASHTO 3.24.8 stipulates that edge beams be

provided for all slabs having primary reinforcement parallel to traffic. Three alternatives

are suggested for providing edge beams:

1. A slab section additionally reinforced2. A beam integral with and deeper than slab3. An integrally reinforced section of slab and curb

The first alternativean additionally reinforced section of the slabas the longitudinal

edge beam is 24 in. x 15 in. Provide a curb 10 in. high (standard depth), which gives the

total height of the beam as 25 in. (for computing the self-weight).

Self-weight (including the curb) =

= 625 lb/ft

Weight of railing (assumed) = = 15 lb/ft

Total dead load = 640 lb/ft


25/28

The live-load moment in a longitudinal edge beam is given by AASHTO 3.24.8.2

The required beam depth, with b =24., can now be calculated from Eq. 6.4. With k

=0.351 and j = 0.883 for n = 9 (Table A.16, Appendix A),

With 1 in. cover and #8 as the main reinforcing bar, hrequired = 13.53 + 1 + 0.5 = 15.03 in.,

Say 15 in., as provided. The required reinforcement can now be calculated from Eq. 6.5:

Therefore, provide four #9 bars, As = 4.0 in.

2 > 3.47 in.2 (Table A.20, Appendix A).

Check for min imum rein forcement (AASHTO 8.17.1.1).From Eq. 6.30:

Then, from Eq. 6.31:

Fatigue stress l imit (AASHTO 8.16.8.3). Fatigue stress limits will be checked for theservice load conditions. The permissible stress range is given by AASHTO Eq. 8.60:

The minimum stress level,fmin , is caused by dead loads only. From service load design

calculations,MD = 35.28 kip-ft.


26/28

From Table A.15 (Appendix A), for n = 9 and = 0.0123 (by interpolation), k = 0.373, j

= 0.875. The minimum stress, therefore, is (from Eq. 6.5)

The maximum stress is caused by the live load plus impact; the corresponding moment,including the dead load moment, isMtotal = 68.88 kip-ft. The corresponding stress in the

main reinforcement is

Thus, the actual stress range, According to AASHTO Eq.8.60, the permissible stress range is

. Hence, the fatigue stress limit requirements are satisfied.

FIGURE E6.1b

Reinforcement details (slab and the edge beams).

Reinforcement details are shown in Fig.E6.1b. #5 dowels @ 18 in. o.c., along

with two #5 hanger bars, are provided for the entire lengths of longitudinal edge

beams.

Design by the load factor method: Design of slab. The following service load

moments were calculated earlier:


27/28

MD = 11.99 kip-ft

ML = 15.97 kip-ft

MI= 4.79 kip-ft

These moments will be appropriately factored for a Group I loading combination perAASHTO 3.22.1 (see Table 3.12, Chapter 3).

[ ]Where:

Thus, [ ] = 60.66 kip-ft. for flexure, = 0.9(AASHTO 8.16.1.2). For a slab, b = 12 in., d= 13.5 in. (based on the minimum depth

requirements, determined earlier). Thus,

From Table A.17 (Appendix A), for

= 369.0 lb/in.2 (close enough to 369.82

lb/in.2), = 0.0100, from which As = 0.010 x 12 x 13.5 = 1.62 in.2 Therefore, provide

#9 @ 7 in. o.c., As = 1.71 in.2 (as in the case of the service load design method).

Distri bution rein forcement. This requirement is the same as in the case of service

load design. Provide #6 @ 12 in. o.c. directly above and perpendicular to the main

reinforcement.

Temperature and shr inkage reinforcement.This requirement is the same as in the

case of service load design. Provide #4 @ 18 in. o.c. n each direction under the top

face of the slab.

Check for min imum rein forcement (AASHTO 8.17.1.1). Calculating are the same as

provided earlier for the service load design.

Fatigue stress l imi ts (AASHTO 8.16.8.3). Fatigue stress limits are to be checked for

the service load conditions. Calculations are the same as provided earlier for the

service load design.

Design of longitudinal edge beams. From calculations for the service load design,

one has the following values of design moments:

MD= 35.28 kip-ft

ML= 33.60 kip-ft


28/28

Thus,

[ ] [ ]

From Table A.17 (Appendix A), for

= 362.2 lb/, = 0.0098, from whichAs= bd= 0.0098 x 24 x 13.5 = 3.17 in.

2 Therefore, provide four #9 bars, As = 4.0

in.2 >3.17 in.2 (Table A.15, Appendix A).

Check for min imum rein forcement (AASHTO 8.17.1.1).These calculations are the

same as provided earlier for the service load design.

Fatigue stress limits (AASHTO 8.16.8.3). A fatigue stress limit is to be checked for

the service load conditions. Calculations for this are the same as provided earlier for

the service load design. Since there has been no change in any type of reinforcement

as a result of design by the load factor design method, the reinforcement details

remain the same as those for design by the service load method (Fig.E6.1b).

chapter 7 design of prestresed concrete bridges

Documents