160

Reinforced concrete Bridges

Bridges shall be designed for specified limit states to achieve the objectives of constructability

safety and serviceability and also taking into account the economy and aesthetics

The design of reinforced concrete bridges is based on the AASHTO specifications (American

Association of State Highway and Transportation Officials) The bridges in general can be

classified as follows

1 According to the materials of the construction of the super structure reinforced concrete

steel pre-stressed concrete composite etc

2 According to the functions road over a river Road or railway over a valley highway and

pipe line bridge

3 According to the form of super structure

Permanent Loads

1048708DC ndash Structural Components and Attachments

1048708DW - Wearing Surfaces and Utilities

1048708EH - Horizontal Earth Pressure

1048708EL - Locked-In Force Effects Including Pretension

1048708ES - Earth Surcharge Load

1048708EV - Vertical Pressure of Earth Fill

Transient Loads

1048708 BR ndash Veh Braking Force

1048708CE ndash Veh Centrifugal Force

1048708CR - Creep

1048708CT - Veh Collision Force

1048708CV - Vessel Collision Force

1048708EQ - Earthquake

1048708FR - Friction

1048708IC - Ice Load

1048708LL - Veh Live Load

1048708IM - Dynamic Load Allowance

1048708LS - Live Load Surcharge

1048708PL - Pedestrian Live Load

1048708SE - Settlement

1048708SH - Shrinkage

1048708TG - Temperature Gradient

1048708TU - Uniform Temperature

1048708WA - Water Load

1048708WL - Wind on Live Load

1048708WS - Wind Load on Structure

161

Load Factors and Load Combinations (AASHTO 2012 Section 34)

341 Load Factors and Load Combinations

1048708Strength I Basic load combination relating to the normal vehicular use of the bridge without

wind

1048708Strength II Load combination relating to the use of the bridge by Owner-specified special

design vehicles evaluation permit vehicles or both without wind

1048708Strength III Load combination relating to the bridge exposed to wind in excess of 55 mph

1048708Strength IV Load combination relating to very high dead load to live load force effect ratios

(Note In commentary it indicates that this will govern where the DLLL gt7 spans over 600rsquo

and during construction checks)

1048708Strength V Load combination relating to normal vehicular use with a wind of 55 mph

1048708Extreme Event I Load combination including earthquakes

1048708Extreme Event II Load combination relating to ice load collision by vessels and vehicles

and certain hydraulic events with a reduced live load

1048708Fatigue Fatigue and fracture load combination relating to repetitive gravitational vehicular

live load and dynamic responses under a single design truck

1048708Service I Load combination relating to normal operational use of the bridge with a 55 mph

wind and all loads at nominal values Compression in precast concrete components

1048708Service II Load combination intended to control yielding of steel structures and slip of slip-

critical connections due to vehicular load

1048708Service III Load combination relating only to tension in prestressed concrete superstructures

with the objective of crack control

1048708Service IV Load combination relating only to tension in prestressed concrete columns with

the objective of crack control

Common Load Combinations for Prestressed Concrete

1048708Strength I 125DC+ 150DW + 175(LL+IM)

1048708Strength IV 150DC+ 150DW

1048708Fatigue 075(LL+IM)

Common Load Combinations for Reinforced Concrete

1048708 Strength I 125DC+ 150DW + 175(LL+IM)

1048708Strength IV 150DC+ 150DW

1048708Fatigue 075(LL+IM)

3612mdashDesign Vehicular Live Load

36121mdashGeneral

Vehicular live loading on the roadways of bridges or incidental structures designated HL-93

shall consist of a combination of the

bull Design truck or design tandem and

bull Design lane load

162

36122 Design Truck

The weights and spacingrsquos of axles and wheels for the design truck shall be as specified in Figure

36122-1 A dynamic load allowance shall be considered as specified in Article 362 Except

as specified in Articles 36131 and 36141 the spacing between the two 145kN axles shall be

varied between 430m and 90m to produce extreme force effects HL-93 or IL-120 Design

Truck (Structure Design Manual ILLINOIS STATE TOLL HIGHWAY March 2013)

Figures attached

36123 Design Tandem

The design tandem shall consist of a pair of 110kN axles spaced 120m apart The transverse

spacing of wheels shall be taken as 180m A dynamic load allowance shall be considered as

specified in Article 362

36124 Design Lane Load

104870893kNm is applied SIMULTANEOUSLY with the design truck or design tandem over a

width of 30m within the design lane

36133 Design Loads for Decks Deck Systems and the Top Slabs of Box Culverts

When the Approximate Strip Method is Used

1048708Where the slab spans primarily in the transverse direction

1048708only the axles of the design truck or design tandem of shall be applied to the deck slab or the

top slab of box culverts

1048708Where the slab spans primarily in the longitudinal direction

1048708For top slabs of box culverts of all spans and for all other cases (including slab-type bridges

where the span does not exceed 460m only the axle loads of the design truck or design tandem

shall be applied

1048708For all other cases (including slab-type bridges where the span exceeds 460m) the entire HL-

93 loading shall be applied

3616 Pedestrian Loads A pedestrian load of 360kNm2 shall be applied to all sidewalks wider than 600mm and considered

simultaneously with the vehicular design live load in the vehicle lane Where Vehicles can mount the

sidewalk sidewalk pedestrian load shall not be considered concurrently If a sidewalk may be removed in

the future the vehicular live loads shall be applied at 300mm from edge-of-deck for design of the

overhang and 600mm from edge-of-deck for design of all other components The pedestrian load shall

not be considered to act concurrently with vehicles The dynamic load allowance need not be considered

for vehicles Bridges intended for only pedestrian equestrian light maintenance vehicle andor bicycle

traffic should be designed in accordance with AASHTOrsquos LRFD Guide Specifications for the Design of

Pedestrian Bridges

362 Dynamic Load Allowance

3621 General

Unless otherwise permitted in Articles 3622 and 3623 the static effects of the design truck or

tandem other than centrifugal and braking forces shall be increased by the percentage specified

in Table 3621-1 for dynamic load allowance The factor to be applied to the static load shall be

taken as (1 + IM100) The dynamic load allowance shall not be applied to pedestrian loads or to

the design lane load

163

Table 3621-1mdashDynamic Load Allowance IM

Component IM

Deck JointsmdashAll Limit States 75

All Other Components

bull Fatigue and Fracture Limit State

bull All Other Limit States

15

30 1048708The Dynamic Load Allowance is applied only to the truck load (including fatigue trucks) not to

lane loads or pedestrian loads

Section 4 Structural Analysis and Evaluation

462mdashApproximate Methods of Analysis

4621mdashDecks

46211mdashGeneral

An approximate method of analysis in which the deck is subdivided into strips perpendicular to

the supporting components shall be considered acceptable for decks other than

fully filled and partially filled grids for which the provisions of Article 46218 shall apply

and

top slabs of segmental concrete box girders for which the provisions of 46294 shall apply

Where the strip method is used the extreme positive moment in any deck panel between girders

shall be taken to apply to all positive moment regions Similarly the extreme negative moment

over any beam or girder shall be taken to apply to all negative moment regions

46212mdashApplicability

The use of design aids for decks containing prefabricated elements may be permitted in lieu of

analysis if the performance of the deck is documented and supported by sufficient technical

evidence The Engineer shall be responsible for the accuracy and implementation of any design

aids used For slab bridges and concrete slabs spanning more than 4600mm and which span

primarily in the direction parallel to traffic the provisions of Article 4623 shall apply

46213mdashWidth of Equivalent Interior Strips

The width of the equivalent strip of a deck may be taken as specified in Table 46213-1 Where

decks span primarily in the direction parallel to traffic strips supporting an axle load shall not be

taken to be greater than 1000mm for open grids and not greater than 3600mm for all other decks

where multilane loading is being investigated For deck overhangs where applicable the

provisions of Article 36134 may be used in lieu of the strip width specified in Table 46213-

1 for deck overhangs The equivalent strips for decks that span primarily in the transverse

direction shall not be subject to width limits The following notation shall apply to Table

46213-1

S = spacing of supporting components (mm)

h = depth of deck (mm)

L = span length of deck (mm)

P = axle load (N)

Sb = spacing of grid bars (mm)

+M = positive moment

minusM = negative moment

X = distance from load to point of support (mm)

164

Table 46213-1mdashEquivalent Strips

Type of Deck

Direction of Primary Strip

Relative to Traffic

Width of Primary Strip (mm)

Concrete

bull Cast-in-place

bull Cast-in-place with stay-in-

placeconcrete formwork

bull Precast post-tensioned

Overhang

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

1140 + 0833X

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

46214 Width of Equivalent Strips at Edges of Slabs

46214amdashGeneral

For the purpose of design the notional edge beam shall be taken as a reduced deck strip width

specified herein Any additional integral local thickening or similar protrusion acting as a

stiffener to the deck that is located within the reduced deck strip width can be assumed to act

with the reduced deck strip width as the notional edge beam

46214bmdashLongitudinal Edges

Edge beams shall be assumed to support one line of wheels and where appropriate a tributary

portion of the design lane load Where decks span primarily in the direction of traffic the

effective width of a strip with or without an edge beam may be taken as the sum of the distance

between the edge of the deck and the inside face of the barrier plus 300mm plus one-quarter of

the strip width specified in either Article 46213 Article 4623 or Article 46210 as

appropriate but not exceeding either one-half the full strip width or 1800mm

46214cmdashTransverse Edges

Transverse edge beams shall be assumed to support one axle of the design truck in one or more

design lanes positioned to produce maximum load effects Multiple presence factors and the

dynamic load allowance shall apply The effective width of a strip with or without an edge

beam may be taken as the sum of the distance between the transverse edge of the deck and the

centerline of the first line of support for the deck usually taken as a girder web plus one-half of

the width of strip as specified in Article 46213 The effective width shall not exceed the full

strip width specified in Article 46213

46215mdashDistribution of Wheel Loads

If the spacing of supporting components in the secondary direction exceeds 15 times the spacing

in the primary direction all of the wheel loads shall be considered to be applied to the primary

strip and the provisions of Article 9732 may be applied to the secondary direction This Article

attempts to clarify the application of the traditional AASHTO approach with respect to

continuous decks If the spacing of supporting components in the secondary direction is less than

15 times the spacing in the primary direction the deck shall be modeled as a system of

intersecting strips The width of the equivalent strips in both directions may be taken as specified

in Table 46213-1 Each wheel load shall be distributed between two intersecting strips The

distribution shall be determined as the ratio between the stiffness of the strip and the sum of

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

161

Load Factors and Load Combinations (AASHTO 2012 Section 34)

341 Load Factors and Load Combinations

1048708Strength I Basic load combination relating to the normal vehicular use of the bridge without

wind

1048708Strength II Load combination relating to the use of the bridge by Owner-specified special

design vehicles evaluation permit vehicles or both without wind

1048708Strength III Load combination relating to the bridge exposed to wind in excess of 55 mph

1048708Strength IV Load combination relating to very high dead load to live load force effect ratios

(Note In commentary it indicates that this will govern where the DLLL gt7 spans over 600rsquo

and during construction checks)

1048708Strength V Load combination relating to normal vehicular use with a wind of 55 mph

1048708Extreme Event I Load combination including earthquakes

1048708Extreme Event II Load combination relating to ice load collision by vessels and vehicles

and certain hydraulic events with a reduced live load

1048708Fatigue Fatigue and fracture load combination relating to repetitive gravitational vehicular

live load and dynamic responses under a single design truck

1048708Service I Load combination relating to normal operational use of the bridge with a 55 mph

wind and all loads at nominal values Compression in precast concrete components

1048708Service II Load combination intended to control yielding of steel structures and slip of slip-

critical connections due to vehicular load

1048708Service III Load combination relating only to tension in prestressed concrete superstructures

with the objective of crack control

1048708Service IV Load combination relating only to tension in prestressed concrete columns with

the objective of crack control

Common Load Combinations for Prestressed Concrete

1048708Strength I 125DC+ 150DW + 175(LL+IM)

1048708Strength IV 150DC+ 150DW

1048708Fatigue 075(LL+IM)

Common Load Combinations for Reinforced Concrete

1048708 Strength I 125DC+ 150DW + 175(LL+IM)

1048708Strength IV 150DC+ 150DW

1048708Fatigue 075(LL+IM)

3612mdashDesign Vehicular Live Load

36121mdashGeneral

Vehicular live loading on the roadways of bridges or incidental structures designated HL-93

shall consist of a combination of the

bull Design truck or design tandem and

bull Design lane load

162

36122 Design Truck

The weights and spacingrsquos of axles and wheels for the design truck shall be as specified in Figure

36122-1 A dynamic load allowance shall be considered as specified in Article 362 Except

as specified in Articles 36131 and 36141 the spacing between the two 145kN axles shall be

varied between 430m and 90m to produce extreme force effects HL-93 or IL-120 Design

Truck (Structure Design Manual ILLINOIS STATE TOLL HIGHWAY March 2013)

Figures attached

36123 Design Tandem

The design tandem shall consist of a pair of 110kN axles spaced 120m apart The transverse

spacing of wheels shall be taken as 180m A dynamic load allowance shall be considered as

specified in Article 362

36124 Design Lane Load

104870893kNm is applied SIMULTANEOUSLY with the design truck or design tandem over a

width of 30m within the design lane

36133 Design Loads for Decks Deck Systems and the Top Slabs of Box Culverts

When the Approximate Strip Method is Used

1048708Where the slab spans primarily in the transverse direction

1048708only the axles of the design truck or design tandem of shall be applied to the deck slab or the

top slab of box culverts

1048708Where the slab spans primarily in the longitudinal direction

1048708For top slabs of box culverts of all spans and for all other cases (including slab-type bridges

where the span does not exceed 460m only the axle loads of the design truck or design tandem

shall be applied

1048708For all other cases (including slab-type bridges where the span exceeds 460m) the entire HL-

93 loading shall be applied

3616 Pedestrian Loads A pedestrian load of 360kNm2 shall be applied to all sidewalks wider than 600mm and considered

simultaneously with the vehicular design live load in the vehicle lane Where Vehicles can mount the

sidewalk sidewalk pedestrian load shall not be considered concurrently If a sidewalk may be removed in

the future the vehicular live loads shall be applied at 300mm from edge-of-deck for design of the

overhang and 600mm from edge-of-deck for design of all other components The pedestrian load shall

not be considered to act concurrently with vehicles The dynamic load allowance need not be considered

for vehicles Bridges intended for only pedestrian equestrian light maintenance vehicle andor bicycle

traffic should be designed in accordance with AASHTOrsquos LRFD Guide Specifications for the Design of

Pedestrian Bridges

362 Dynamic Load Allowance

3621 General

Unless otherwise permitted in Articles 3622 and 3623 the static effects of the design truck or

tandem other than centrifugal and braking forces shall be increased by the percentage specified

in Table 3621-1 for dynamic load allowance The factor to be applied to the static load shall be

taken as (1 + IM100) The dynamic load allowance shall not be applied to pedestrian loads or to

the design lane load

163

Table 3621-1mdashDynamic Load Allowance IM

Component IM

Deck JointsmdashAll Limit States 75

All Other Components

bull Fatigue and Fracture Limit State

bull All Other Limit States

15

30 1048708The Dynamic Load Allowance is applied only to the truck load (including fatigue trucks) not to

lane loads or pedestrian loads

Section 4 Structural Analysis and Evaluation

462mdashApproximate Methods of Analysis

4621mdashDecks

46211mdashGeneral

An approximate method of analysis in which the deck is subdivided into strips perpendicular to

the supporting components shall be considered acceptable for decks other than

fully filled and partially filled grids for which the provisions of Article 46218 shall apply

and

top slabs of segmental concrete box girders for which the provisions of 46294 shall apply

Where the strip method is used the extreme positive moment in any deck panel between girders

shall be taken to apply to all positive moment regions Similarly the extreme negative moment

over any beam or girder shall be taken to apply to all negative moment regions

46212mdashApplicability

The use of design aids for decks containing prefabricated elements may be permitted in lieu of

analysis if the performance of the deck is documented and supported by sufficient technical

evidence The Engineer shall be responsible for the accuracy and implementation of any design

aids used For slab bridges and concrete slabs spanning more than 4600mm and which span

primarily in the direction parallel to traffic the provisions of Article 4623 shall apply

46213mdashWidth of Equivalent Interior Strips

The width of the equivalent strip of a deck may be taken as specified in Table 46213-1 Where

decks span primarily in the direction parallel to traffic strips supporting an axle load shall not be

taken to be greater than 1000mm for open grids and not greater than 3600mm for all other decks

where multilane loading is being investigated For deck overhangs where applicable the

provisions of Article 36134 may be used in lieu of the strip width specified in Table 46213-

1 for deck overhangs The equivalent strips for decks that span primarily in the transverse

direction shall not be subject to width limits The following notation shall apply to Table

46213-1

S = spacing of supporting components (mm)

h = depth of deck (mm)

L = span length of deck (mm)

P = axle load (N)

Sb = spacing of grid bars (mm)

+M = positive moment

minusM = negative moment

X = distance from load to point of support (mm)

164

Table 46213-1mdashEquivalent Strips

Type of Deck

Direction of Primary Strip

Relative to Traffic

Width of Primary Strip (mm)

Concrete

bull Cast-in-place

bull Cast-in-place with stay-in-

placeconcrete formwork

bull Precast post-tensioned

Overhang

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

1140 + 0833X

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

46214 Width of Equivalent Strips at Edges of Slabs

46214amdashGeneral

For the purpose of design the notional edge beam shall be taken as a reduced deck strip width

specified herein Any additional integral local thickening or similar protrusion acting as a

stiffener to the deck that is located within the reduced deck strip width can be assumed to act

with the reduced deck strip width as the notional edge beam

46214bmdashLongitudinal Edges

Edge beams shall be assumed to support one line of wheels and where appropriate a tributary

portion of the design lane load Where decks span primarily in the direction of traffic the

effective width of a strip with or without an edge beam may be taken as the sum of the distance

between the edge of the deck and the inside face of the barrier plus 300mm plus one-quarter of

the strip width specified in either Article 46213 Article 4623 or Article 46210 as

appropriate but not exceeding either one-half the full strip width or 1800mm

46214cmdashTransverse Edges

Transverse edge beams shall be assumed to support one axle of the design truck in one or more

design lanes positioned to produce maximum load effects Multiple presence factors and the

dynamic load allowance shall apply The effective width of a strip with or without an edge

beam may be taken as the sum of the distance between the transverse edge of the deck and the

centerline of the first line of support for the deck usually taken as a girder web plus one-half of

the width of strip as specified in Article 46213 The effective width shall not exceed the full

strip width specified in Article 46213

46215mdashDistribution of Wheel Loads

If the spacing of supporting components in the secondary direction exceeds 15 times the spacing

in the primary direction all of the wheel loads shall be considered to be applied to the primary

strip and the provisions of Article 9732 may be applied to the secondary direction This Article

attempts to clarify the application of the traditional AASHTO approach with respect to

continuous decks If the spacing of supporting components in the secondary direction is less than

15 times the spacing in the primary direction the deck shall be modeled as a system of

intersecting strips The width of the equivalent strips in both directions may be taken as specified

in Table 46213-1 Each wheel load shall be distributed between two intersecting strips The

distribution shall be determined as the ratio between the stiffness of the strip and the sum of

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

162

36122 Design Truck

The weights and spacingrsquos of axles and wheels for the design truck shall be as specified in Figure

36122-1 A dynamic load allowance shall be considered as specified in Article 362 Except

as specified in Articles 36131 and 36141 the spacing between the two 145kN axles shall be

varied between 430m and 90m to produce extreme force effects HL-93 or IL-120 Design

Truck (Structure Design Manual ILLINOIS STATE TOLL HIGHWAY March 2013)

Figures attached

36123 Design Tandem

The design tandem shall consist of a pair of 110kN axles spaced 120m apart The transverse

spacing of wheels shall be taken as 180m A dynamic load allowance shall be considered as

specified in Article 362

36124 Design Lane Load

104870893kNm is applied SIMULTANEOUSLY with the design truck or design tandem over a

width of 30m within the design lane

36133 Design Loads for Decks Deck Systems and the Top Slabs of Box Culverts

When the Approximate Strip Method is Used

1048708Where the slab spans primarily in the transverse direction

1048708only the axles of the design truck or design tandem of shall be applied to the deck slab or the

top slab of box culverts

1048708Where the slab spans primarily in the longitudinal direction

1048708For top slabs of box culverts of all spans and for all other cases (including slab-type bridges

where the span does not exceed 460m only the axle loads of the design truck or design tandem

shall be applied

1048708For all other cases (including slab-type bridges where the span exceeds 460m) the entire HL-

93 loading shall be applied

3616 Pedestrian Loads A pedestrian load of 360kNm2 shall be applied to all sidewalks wider than 600mm and considered

simultaneously with the vehicular design live load in the vehicle lane Where Vehicles can mount the

sidewalk sidewalk pedestrian load shall not be considered concurrently If a sidewalk may be removed in

the future the vehicular live loads shall be applied at 300mm from edge-of-deck for design of the

overhang and 600mm from edge-of-deck for design of all other components The pedestrian load shall

not be considered to act concurrently with vehicles The dynamic load allowance need not be considered

for vehicles Bridges intended for only pedestrian equestrian light maintenance vehicle andor bicycle

traffic should be designed in accordance with AASHTOrsquos LRFD Guide Specifications for the Design of

Pedestrian Bridges

362 Dynamic Load Allowance

3621 General

Unless otherwise permitted in Articles 3622 and 3623 the static effects of the design truck or

tandem other than centrifugal and braking forces shall be increased by the percentage specified

in Table 3621-1 for dynamic load allowance The factor to be applied to the static load shall be

taken as (1 + IM100) The dynamic load allowance shall not be applied to pedestrian loads or to

the design lane load

163

Table 3621-1mdashDynamic Load Allowance IM

Component IM

Deck JointsmdashAll Limit States 75

All Other Components

bull Fatigue and Fracture Limit State

bull All Other Limit States

15

30 1048708The Dynamic Load Allowance is applied only to the truck load (including fatigue trucks) not to

lane loads or pedestrian loads

Section 4 Structural Analysis and Evaluation

462mdashApproximate Methods of Analysis

4621mdashDecks

46211mdashGeneral

An approximate method of analysis in which the deck is subdivided into strips perpendicular to

the supporting components shall be considered acceptable for decks other than

fully filled and partially filled grids for which the provisions of Article 46218 shall apply

and

top slabs of segmental concrete box girders for which the provisions of 46294 shall apply

Where the strip method is used the extreme positive moment in any deck panel between girders

shall be taken to apply to all positive moment regions Similarly the extreme negative moment

over any beam or girder shall be taken to apply to all negative moment regions

46212mdashApplicability

The use of design aids for decks containing prefabricated elements may be permitted in lieu of

analysis if the performance of the deck is documented and supported by sufficient technical

evidence The Engineer shall be responsible for the accuracy and implementation of any design

aids used For slab bridges and concrete slabs spanning more than 4600mm and which span

primarily in the direction parallel to traffic the provisions of Article 4623 shall apply

46213mdashWidth of Equivalent Interior Strips

The width of the equivalent strip of a deck may be taken as specified in Table 46213-1 Where

decks span primarily in the direction parallel to traffic strips supporting an axle load shall not be

taken to be greater than 1000mm for open grids and not greater than 3600mm for all other decks

where multilane loading is being investigated For deck overhangs where applicable the

provisions of Article 36134 may be used in lieu of the strip width specified in Table 46213-

1 for deck overhangs The equivalent strips for decks that span primarily in the transverse

direction shall not be subject to width limits The following notation shall apply to Table

46213-1

S = spacing of supporting components (mm)

h = depth of deck (mm)

L = span length of deck (mm)

P = axle load (N)

Sb = spacing of grid bars (mm)

+M = positive moment

minusM = negative moment

X = distance from load to point of support (mm)

164

Table 46213-1mdashEquivalent Strips

Type of Deck

Direction of Primary Strip

Relative to Traffic

Width of Primary Strip (mm)

Concrete

bull Cast-in-place

bull Cast-in-place with stay-in-

placeconcrete formwork

bull Precast post-tensioned

Overhang

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

1140 + 0833X

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

46214 Width of Equivalent Strips at Edges of Slabs

46214amdashGeneral

For the purpose of design the notional edge beam shall be taken as a reduced deck strip width

specified herein Any additional integral local thickening or similar protrusion acting as a

stiffener to the deck that is located within the reduced deck strip width can be assumed to act

with the reduced deck strip width as the notional edge beam

46214bmdashLongitudinal Edges

Edge beams shall be assumed to support one line of wheels and where appropriate a tributary

portion of the design lane load Where decks span primarily in the direction of traffic the

effective width of a strip with or without an edge beam may be taken as the sum of the distance

between the edge of the deck and the inside face of the barrier plus 300mm plus one-quarter of

the strip width specified in either Article 46213 Article 4623 or Article 46210 as

appropriate but not exceeding either one-half the full strip width or 1800mm

46214cmdashTransverse Edges

Transverse edge beams shall be assumed to support one axle of the design truck in one or more

design lanes positioned to produce maximum load effects Multiple presence factors and the

dynamic load allowance shall apply The effective width of a strip with or without an edge

beam may be taken as the sum of the distance between the transverse edge of the deck and the

centerline of the first line of support for the deck usually taken as a girder web plus one-half of

the width of strip as specified in Article 46213 The effective width shall not exceed the full

strip width specified in Article 46213

46215mdashDistribution of Wheel Loads

If the spacing of supporting components in the secondary direction exceeds 15 times the spacing

in the primary direction all of the wheel loads shall be considered to be applied to the primary

strip and the provisions of Article 9732 may be applied to the secondary direction This Article

attempts to clarify the application of the traditional AASHTO approach with respect to

continuous decks If the spacing of supporting components in the secondary direction is less than

15 times the spacing in the primary direction the deck shall be modeled as a system of

intersecting strips The width of the equivalent strips in both directions may be taken as specified

in Table 46213-1 Each wheel load shall be distributed between two intersecting strips The

distribution shall be determined as the ratio between the stiffness of the strip and the sum of

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

163

Table 3621-1mdashDynamic Load Allowance IM

Component IM

Deck JointsmdashAll Limit States 75

All Other Components

bull Fatigue and Fracture Limit State

bull All Other Limit States

15

30 1048708The Dynamic Load Allowance is applied only to the truck load (including fatigue trucks) not to

lane loads or pedestrian loads

Section 4 Structural Analysis and Evaluation

462mdashApproximate Methods of Analysis

4621mdashDecks

46211mdashGeneral

An approximate method of analysis in which the deck is subdivided into strips perpendicular to

the supporting components shall be considered acceptable for decks other than

fully filled and partially filled grids for which the provisions of Article 46218 shall apply

and

top slabs of segmental concrete box girders for which the provisions of 46294 shall apply

Where the strip method is used the extreme positive moment in any deck panel between girders

shall be taken to apply to all positive moment regions Similarly the extreme negative moment

over any beam or girder shall be taken to apply to all negative moment regions

46212mdashApplicability

The use of design aids for decks containing prefabricated elements may be permitted in lieu of

analysis if the performance of the deck is documented and supported by sufficient technical

evidence The Engineer shall be responsible for the accuracy and implementation of any design

aids used For slab bridges and concrete slabs spanning more than 4600mm and which span

primarily in the direction parallel to traffic the provisions of Article 4623 shall apply

46213mdashWidth of Equivalent Interior Strips

The width of the equivalent strip of a deck may be taken as specified in Table 46213-1 Where

decks span primarily in the direction parallel to traffic strips supporting an axle load shall not be

taken to be greater than 1000mm for open grids and not greater than 3600mm for all other decks

where multilane loading is being investigated For deck overhangs where applicable the

provisions of Article 36134 may be used in lieu of the strip width specified in Table 46213-

1 for deck overhangs The equivalent strips for decks that span primarily in the transverse

direction shall not be subject to width limits The following notation shall apply to Table

46213-1

S = spacing of supporting components (mm)

h = depth of deck (mm)

L = span length of deck (mm)

P = axle load (N)

Sb = spacing of grid bars (mm)

+M = positive moment

minusM = negative moment

X = distance from load to point of support (mm)

164

Table 46213-1mdashEquivalent Strips

Type of Deck

Direction of Primary Strip

Relative to Traffic

Width of Primary Strip (mm)

Concrete

bull Cast-in-place

bull Cast-in-place with stay-in-

placeconcrete formwork

bull Precast post-tensioned

Overhang

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

1140 + 0833X

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

46214 Width of Equivalent Strips at Edges of Slabs

46214amdashGeneral

For the purpose of design the notional edge beam shall be taken as a reduced deck strip width

specified herein Any additional integral local thickening or similar protrusion acting as a

stiffener to the deck that is located within the reduced deck strip width can be assumed to act

with the reduced deck strip width as the notional edge beam

46214bmdashLongitudinal Edges

Edge beams shall be assumed to support one line of wheels and where appropriate a tributary

portion of the design lane load Where decks span primarily in the direction of traffic the

effective width of a strip with or without an edge beam may be taken as the sum of the distance

between the edge of the deck and the inside face of the barrier plus 300mm plus one-quarter of

the strip width specified in either Article 46213 Article 4623 or Article 46210 as

appropriate but not exceeding either one-half the full strip width or 1800mm

46214cmdashTransverse Edges

Transverse edge beams shall be assumed to support one axle of the design truck in one or more

design lanes positioned to produce maximum load effects Multiple presence factors and the

dynamic load allowance shall apply The effective width of a strip with or without an edge

beam may be taken as the sum of the distance between the transverse edge of the deck and the

centerline of the first line of support for the deck usually taken as a girder web plus one-half of

the width of strip as specified in Article 46213 The effective width shall not exceed the full

strip width specified in Article 46213

46215mdashDistribution of Wheel Loads

If the spacing of supporting components in the secondary direction exceeds 15 times the spacing

in the primary direction all of the wheel loads shall be considered to be applied to the primary

strip and the provisions of Article 9732 may be applied to the secondary direction This Article

attempts to clarify the application of the traditional AASHTO approach with respect to

continuous decks If the spacing of supporting components in the secondary direction is less than

15 times the spacing in the primary direction the deck shall be modeled as a system of

intersecting strips The width of the equivalent strips in both directions may be taken as specified

in Table 46213-1 Each wheel load shall be distributed between two intersecting strips The

distribution shall be determined as the ratio between the stiffness of the strip and the sum of

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

164

Table 46213-1mdashEquivalent Strips

Type of Deck

Direction of Primary Strip

Relative to Traffic

Width of Primary Strip (mm)

Concrete

bull Cast-in-place

bull Cast-in-place with stay-in-

placeconcrete formwork

bull Precast post-tensioned

Overhang

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

Either Parallel or

Perpendicular

1140 + 0833X

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

+M 660 + 055S

minusM 1220 + 025S

46214 Width of Equivalent Strips at Edges of Slabs

46214amdashGeneral

For the purpose of design the notional edge beam shall be taken as a reduced deck strip width

specified herein Any additional integral local thickening or similar protrusion acting as a

stiffener to the deck that is located within the reduced deck strip width can be assumed to act

with the reduced deck strip width as the notional edge beam

46214bmdashLongitudinal Edges

Edge beams shall be assumed to support one line of wheels and where appropriate a tributary

portion of the design lane load Where decks span primarily in the direction of traffic the

effective width of a strip with or without an edge beam may be taken as the sum of the distance

between the edge of the deck and the inside face of the barrier plus 300mm plus one-quarter of

the strip width specified in either Article 46213 Article 4623 or Article 46210 as

appropriate but not exceeding either one-half the full strip width or 1800mm

46214cmdashTransverse Edges

Transverse edge beams shall be assumed to support one axle of the design truck in one or more

design lanes positioned to produce maximum load effects Multiple presence factors and the

dynamic load allowance shall apply The effective width of a strip with or without an edge

beam may be taken as the sum of the distance between the transverse edge of the deck and the

centerline of the first line of support for the deck usually taken as a girder web plus one-half of

the width of strip as specified in Article 46213 The effective width shall not exceed the full

strip width specified in Article 46213

46215mdashDistribution of Wheel Loads

If the spacing of supporting components in the secondary direction exceeds 15 times the spacing

in the primary direction all of the wheel loads shall be considered to be applied to the primary

strip and the provisions of Article 9732 may be applied to the secondary direction This Article

attempts to clarify the application of the traditional AASHTO approach with respect to

continuous decks If the spacing of supporting components in the secondary direction is less than

15 times the spacing in the primary direction the deck shall be modeled as a system of

intersecting strips The width of the equivalent strips in both directions may be taken as specified

in Table 46213-1 Each wheel load shall be distributed between two intersecting strips The

distribution shall be determined as the ratio between the stiffness of the strip and the sum of

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

165

stiffnessrsquos of the intersecting strips In the absence of more precise calculations the strip

stiffness ks may be estimated as

ks =EIs S3

where

Is = moment of inertia of the equivalent strip (mm4)

46216mdashCalculation of Force Effects

The strips shall be treated as continuous beams or simply supported beams as appropriate Span

length shall be taken as the center-to-center distance between the supporting components For the

purpose of determining force effects in the strip the supporting components shall be assumed to

be infinitely rigid The wheel loads may be modeled as concentrated loads or as patch loads

whose length along the span shall be the length of the tire contact area as specified in Article

36125 plus the depth of the deck The strips should be analyzed by classical beam theory The

design section for negative moments and shear forces where investigated may be taken as

follows

bull For monolithic construction closed steel boxes closed concrete boxes open concrete boxes

without top flanges and stemmed precast beams ie Cross sections(b) (c) (d) (e) (f) (g) (h)

(i) and (j)from Table 46221-1 at the face of the supporting component

bull For steel I-beams and steel tub girders ie Cross-sections (a) and (c) from Table 46221-1

one-quarter the flange width from the centerline of support

bull For precast I-shaped concrete beams and open concrete boxes with top flanges ie Cross-

sections(c) and (k) from Table 46221-1 one-third the flange width but not exceeding 380mm

from the centerline of support

4622mdashBeam-Slab Bridges

46221mdashApplication

The provisions of this Article may be applied to straight girder bridges and horizontally curved

concrete bridges as well as horizontally curved steel girder bridges complying with the

provisions of Article 46124 The provisions of this Article may also be used to determine a

starting point for some methods of analysis to determine force effects in curved girders of any

degree of curvature in plan Except as specified in Article 46225 the provisions of this Article

shall be taken to apply to bridges being analyzed for

bull A single lane of loading or

bull Multiple lanes of live load yielding approximately the same force effect per lane

If one lane is loaded with a special vehicle or evaluation permit vehicle the design force effect

per girder resulting from the mixed traffic may be determined as specified in Article 46225

For beam spacing exceeding the range of applicability as specified in tables in Articles

46222and 46223 the live load on each beam shall be the reaction of the loaded lanes based

on the lever rule unless specified otherwise herein

The provisions of 36112 specify that multiple presence factors shall not be used with the

approximate load assignment methods other than statically moment or lever arm methods

because these factors are already incorporated in the distribution factors

Bridges not meeting the requirements of this Article shall be analyzed as specified in Article

463The distribution of live load specified in Articles 46222 and 46223 may be used for

girders beams and stringers other than multiple steel box beams with concrete decks that meet

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

166

the following conditions and any other conditions identified in tables of distribution factors as

specified herein

bull Width of deck is constant

bull Unless otherwise specified the number of beams is not less than four

bull Beams are parallel and have approximately the same stiffness

bull Unless otherwise specified the roadway part of the overhang de does not exceed

910mm

bull Curvature in plan is less than the limit specified in Article 46124 or where

distribution factors are required in order to implement an acceptable approximate or

refined analysis method satisfying the requirements of Article 44 for bridges of any

degree of curvature in plan and

bull Cross-section is consistent with one of the cross sections shown in Table 46221-1

Live load distribution factors specified herein maybe used for permit and rating vehicles whose

overall width is comparable to the width of the design truck

AASHTO 2017 Section 5 concrete structures

- Concrete compressive strength (119891prime119888 ) 5421

16 MPalt119891 prime119888lt 70 MPa

119891prime119888ge 28 MPa for prestressed concrete and decks

- 119864119888 = 4800 radic119891prime119888 for normal density c5424

- Ѵ =02 poissonrsquos Ratio

- Modulus of rupture ( fr ) 5426

fr = 063 radic119891prime119888 for confers of cracking by dist Reinf of defl

fr = 097 radic119891prime119888 for min reinf

- Steel yield stress ( fy ) 543

fyle 520 MPa

for fylt 420 MPa shall be used with the approval of the owner

- Es = 200 000 MPa

- Prestressed steel

For strand

1 Grade 250 ( 1725MPa ) fpu = 1725 MPa fpy asymp ( 085 ndash 09 ) fpu

2 Grade 270 ( 1860MPa ) fpu = 1860 MPa fpy = ( 085 ndash 09 ) fpu

Eps = 197 000 MPa

Limit states AASHTO 2012 Section 55

Structural components shall be proportioned to satisfy the requirements at all appropriate

service fatigue strength and extreme event limit states

Service limit state actions to be considered at the service limit state shall be cracking

deformations and concrete Stresses

Strength limit state the strength limit state issues to be considered shall be those of

strength and stability

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

167

Resistance factor ᵩshall be taken as

- Flexure and tension of RC helliphelliphelliphelliphelliphelliphelliphelliphellip 09

- Flexure and tension of prestressed con helliphellip 10

- Shear and torsion (normal structural concrete ) helliphelliphellip 09

- Axial comp helliphelliphelliphelliphelliphelliphelliphelliphelliphellip 075

- Bearing on concrete helliphelliphelliphelliphelliphellip 07

- Comp in Anchorage zone ( NSt Con ) hellip 08

- Tension in steel in Anchorage helliphelliphelliphelliphellip10

- Resistance during pile driving helliphelliphelliphelliphellip 10

For comp members with flexure ᵩ increased from 075 rarr 09

For partially prestressed concrete ᵩ = 09 + 01 PPR

PPR = 119860119901119904119891119901119910

119860119901119904119891119901119910+119860119904119891119910

PPR partial prestress ratio if PRP lt 50 consider RC

Design for flexure and axial force effect 570

Rectangular stress dist β1 = 065 ndash 085 for normal con

β1 avg = sum 119891prime119888119860119888119888β1

sum 119891prime119888119860119888119888 for composite

Acc area of concrete in comp

Flexural members

Mr = φ Mu

1 Bonded tendons of prestress of fpe ge 05 fpu

fps = fpu( 1 ndash k 119888

119889119901 )

k = 2 ( 104 - 119891119901119910

119891119901119906 )

- T section

c =

119860119901119904119891119901119906+119860119904prime119891119910primeminus 085 β1 119891prime119888(119887minus119887119908)ℎ119891

085 119891prime119888β1 b + k Aps fpud

2 Unbounded tendons

fps = fpe + 6300 ( 119889119901minus119888

ℓ119890 ) le fpy

ℓ119890 = ( 2 ℓ119894

2+119873119904 )

ℓ119890 effective tend Length mm

ℓ119894 length of tendons between anchorages

119873119904 No of supp hinges or bonded points

119891119901119890 effective stress

119891 prime119910 when gtfy or 119891 prime119910 = fs

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

168

- Flanged section (tendons bonded amp comp flange depth ltc

Mn = Aps fps( dp ndash a2)+As fy (ds ndash a2) ndash As fy (d ndash a2)+085119891 prime119888 (b-bw)β1 hf(a2-hf2)

- Rectangular section of flanged comp flange depth ge C b = bw

Mn = Aps fps( dp ndash a2) + As fy (ds ndash a2) ndash As fy (d ndash a2)

Limits of reinforcement 57-33

1 Maximum reinforcement

cde le 042 under reinforced otherwise compression reinforcement required (over

reinforced )

119889119890 =119860119901119904119891119901119904119889119901 + 119860119904119891119910119889119904

119860119901119904119891119901119904 + 119860119904119891119910

S max le 15 t amp 450 mm ( 51032 )

2 Minimum reinforcement

Amount of prestress amp non prestress reinforcement shall be adequate to develop a

factored flexural resistance Mr at least equal to the lesser of

a 12 Mcr

Mcr = Sc ( fr + fcpe ) ndash Mdnc ( 119878119888

119878119899119888minus 1 ) ge Sc fr fr = 097 radic119891prime119888

Sc section modulus ( mm3 ) of composite sec due to DL

Snc Section modulus for monolithic or non-comp

fcpe comp stress in conc Due to effective prestress force

Mdnc total unfactored DL for non-composite section

b 133 M factored reqd strength load combinations

According to (51031)

3 Control of cracking by dist Of reinforcement

Spacing (S) of tens Reinforcement shall satisfy the followings

S le123000 Ɣ119890

120573119904119891119904ndash 2 dc

Ɣ119890 = 10 for class 1 exposure

= 075 for class 2 exposure

dc cover

fs asymp 06 fy or fs = 119872

119860119904119895119889119904

βs = 1 +119889119888

07 ( ℎ minus 119889119888 )

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

169

Shrinkage and temperature reinforcement

Ash ge 011 Ag fy Ag = 1000 t

The steel shall be equally distributionOn both faces with spacing le 3t or 450 mm

If de gt 900 mm long skin reinf shall be uniformly distb Along both sides faces of the

component for a distance d2 nearest the flexural reinf

Ask ge 0001 ( de ndash 760 ) le 119860119904+119860119901119904

1200 mm2m

- Amount for each face le 1

4( As+ Aps )

- S le de6

le 300 mm

Shear and Torsion AASHTO 2012 sec 580

Vr = φ Vn Tr = φ Tn

Torsion design reqd when Tu ge 025 φ Tcr

Tcr = 0328 radic119891prime119888119860119888119901^2

119875119888 lowast radic1 +

119891119901119888

0328 radic119891prime119888

Transverse reinforcement shall be provided where

Vu gt 05 φ ( Vc + Vp )Or Where consideration of torsion is required

Vp component of prestressing force in the direction of shear force

Minimum Transverse reinforcement(Ar)min ge 0083 radic119891prime119888119887119907119904

119891119910

Maximum spacing of transverse reinforcement

If Vu lt 0125 119891prime119888rarr Smax = 08 dv le 600 mm

If Vu ge 0125 119891prime119888rarr Smax = 04 dv le 300 mm

dv =119872119906

119860119904119891119910 + 119860119901119904119891119901119904ge 09 de

ge 072 h

The nominal shear resistance Vn determined as lesser of

Vn = Vc + Vs + VpOrVn = 025 119891prime119888 bv dv + Vp

Where

Vc = 0083 β radic119891prime119888 bv dv

Vs =119860119907119891119910119889119907(119888119900119905120563+119888119900119905120563)+119904119894119899120572

Ѕ for 120572=90o

rarr sin120572 = 1

Vs =119860119907119891119910119889119907119888119900119905120563

Ѕ 120563 = 45o β asymp 20

Provisions for structural types AASHTO 2012 sec514

Precast beam dimensions Thickness of any part of precast concrete beams shall not be less than

- Top flange 50mm

- Web non post tension 125 mm Web post tensioned 165mm

- Bott Flange 125mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

170

Example 1

Design the reinforced concrete bridge shown for the following given data and according to

AASHTO 2012 Wearing surface = 10 KNm2 119891prime119888 = 30 MPa fy = 420 MPa

HL-93 standard truck or IL-120 design truck to be consider in addition to tandem amp lane loading

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

171

Solution

1 Design of reinforced concrete slab

The slab is continuous over five supports main reinforcement Traffic

Assume thickness of slab = 200 mm

Self-weight of the deck = 02 times 24 = 48 KNm2

Unfactored self-weight positive or negative moment = 1

10 DL ℓ2

ℓ girder spacing = 22 m

Unfactored Mdl = 1

10times 48times222 = 232 kNmm

Unfactored future wearing surface (FWS) = 1

10times10 times 222 = 0484 kNmm

Live load moments are taken from AASHTO LRFD table A4-1 These values already

corrected for multiple presence factors and impact loading

M pos = 2404 kNmm for S= 22 m cc of girders

M strength I = 125 MDC + 15 MDW + 175 MLL+IM

= 125 times 232 + 15 times 0484 + 175 times 2404

= 4570 kNmm

M service I = 10 MDC + 10 MDW + 10 MLL+IM

= 232 + 0484 + 2404= 2684 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ (As fy (d - 119886

2 )) ge M strength I

Assume φ = 090 then check assumption limits of reinforcement Assume fs = fy then

assumption valid and assume using φ12 mm As1 = 113 mm2

d = 200 ndash 25 ndash 122 =169 mm b = 10 m

R = 119872119904119905119903119890119899119892119905ℎ119868

120593 1198871198892 = 457times106

09times1000times169sup2 = 1778 MPa

m = 119891119910

085 119891prime119888 =

420

085times30 = 1647

ρ = 1

119898 (1 - radic1 minus

2 119898119877

119891119910 ) = 0004392

500mm

22 22 22 22 12 12

135

02

06 06 10m

Section 2-2

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

172

As = ρ b d = 742 mm2m φ12150mm

(As) provide = 113 times 1000150 = 753 mm2m

Check

c =119860119904119891119910

085119891prime119888119887β1rarr =

753lowast420

085times085times1000times30 = 1459 mm

cd = 1459169 rarr = 0086 lt 042 ok tension failure φ = 090

Check for crack control

S le 123000timesƔ119890

120573119904times119891119904 ndash 2dc Ɣe = 075 for class 2 exposure dc = 25 + 122 = 31 mm h=200 mm

βs = 1 + 119889119888

07 ( ℎminus119889119888 )rarr = 1+

31

07times(200minus31) = 126

fs = 119872119904119890119903119907119894119888119890119868

119860119904119895119889 M service = 2684 kNmm As = 753 mm2m d = 169 mm

j = 1- k3

k = 119891119888

119891119888+119891119904119899 or k = radic(120588119899)2 + 2 120588119899 - ρn

ρ = Asbd rarr = 7531000times169 = 000446 n = EsEcrarr = 200 000

4800radic119891prime119888 = 76

k= radic(000446 times 76)2 + 2 times 000446 times 76 - 000446times76= 0248

j = 1 ndash 02483 rarr = 0917

fs = 230 MPa or use fs = 06 fy asymp 252 MPa

S = 123 000times075

126times230 ndash 2times31 = 256 mm

S= 150 mm lt 256 mm ok cc spacing are adequate for crack control

Check limits of reinforcement

Check maximum reinforcement

ℇt = 0003times( 119889minus119888 )

119888 c = 1459 mm d = 169 mm

ℇt = 00317 gt 0005 rarr φ = 09 ok

For 0002 ltℇtlt 0005 rarr φ = 065 + 015 ( 119889

119888minus 1)

For ℇtlt 0002 rarr φ = 075

Check Minimum reinforcement

1 Mr ge 12 Mcr

Mcr = S timesfr

S = 1

6times b h2

rarr = 1

6times 1000 times 200 2 = 6667 times 10 6 mm3

fr = 097 timesradic119891prime119888 = 531 Nmm2

Mcr = 6667times103times 531 times 10 -3 =354 kNm

12timesMcr = 4248 kNm

Mr = φMu = 09 (As fy (d-a2))

= 09 times (753times420 times (169 - 085times1459

2 )) rarr = 4634 kNmm

Mr = 4634 kNmgt 12 Mcr = 4248 kNm ok

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

173

2 133 times M factored = 133 times 457 = 6078 kNmm

The lesser one is 12 Mcr

Check max Spacing S = 150 mm lt 15 times 200 = 300 hellip ok

lt 450

Check min spacing S = 150 gt 15 times12= 18 mm 15 times 20 = 30 mm 38 mm Ok

Negative moment reinforcement

The design section for negative moments and shear forces may be taken as follows

(AASHTO 46216)

Monolithic construction cross ndash section bcdefghij from table 4622 1-1 at the

face of the supporting component

For precast I ndash shape concrete beams sections c amp k 1

3times flange width le 380 mm from centerline of supp

So unfactored dead loads MDC = 48 times (22 ndash 05)210 = 139 kNmm

MDW= 1times (22 ndash 05)210 = 029 kNmm

Distance from L of girder to design section for negative moment for case e is 250 mm

For Mll in table A4-1 for S=2200 mm either consider 225mm from support

(Mll=1674) or by interpolation find for 250 mm or use 120053 = S = 22 m

face M strength I=157kNmm

Center Mll = 2767 kNmm

Face M strength I = 10 (125 MDC + 15 MDW + 175 MLL+IM)= 2965 kNmm

Center M strength I = 125 times232 + 15 times 0484 + 175 times 2767 = 521 kNmm

Note for more safety consider 521 kNmm in design for this example

Face M service = 139 + 029 + 157 = 174 kNmm

Center M service = 232 + 0484 + 2767 = 305 kNmm

Design for ultimate moment capacity

Mr = φ Mn = φ Asfy (d- a2) geM strength I

For M strength I = 521 kNmm

d= 169 mm b = 10 m R = 2027 MPa m = 1647 ρ = 000504

As = ρ bd rarr = 000504 times1000 times 169 = 851 mm2m φ12125 mm

(As) provide = 113 times 1000125 = 904 mm2m

S = 125 mm le 15 t = 300 S gtSmin Ok

Check

C = As fy 085119891prime119888 b = 904 times 420 085 times 28times1000 = 1595 mm

cd = 0094 lt 04 hellip ok fs = fy

Check for crack control

s le 123 000 Ɣ119890

120573119904119891119904- 2 dc

Ɣe = 075

dc = 31 mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

174

120573119904 = 126

k = 02474 ρ = 000535 n = 76 j = 0918

fs = 305times106

904times0918times169 = 2175 MPa

s = 125 mm lt123 000times075

126times2175 ndash 2times31 = 2746 mm ok

Check limits of reinforcement

Maximum reinforcement

ℇt = 0003 119889minus119888

119888 = 00288 gt 0005 rarr φ = 09 Ok

Check minimum reinforcement

1 12 timesMcr = 4248 kNmm

2 133 times M factores = 133 times 521 = 6929 kNm m

Mr = φ Asfy (d-a2)

= 09 times 904times420 (169 - 085times1595

2 ) 10 -6

Mr = 5543 kNmm gt 12 Mcr = 4248 hellip Ok

Shrinkage and Temperature Reinforcement

Ash = 011 Agfy

= 011 times (1000 times200)420= 524 mm2m

Ashface = 262 mm2m use φ10200 mm (Ash) pro = 400 mm2m gt (Ash) req OK

Design of beams (Girders)

- Live load distribution Factors (4622)

From table 46221-1 bridges with concrete decks supported on cast in place concrete

designated as cross ndash section (e) Other tables in 46222 list the distributionfor interior amp

exterior girders including cross ndash section (e)

Required informations

- bf = 2200 mm Girder spacing S = 22 m Span length L = 175+06 = 181 m

Non-composite beam area Ag = 1015times106 mm2

- Non-composite beam moment of inertia Ig = 23909times1010 mm4

- Deck slab thickness ts = 200 mm

- Modulus of elasticity of slab ED = 2629times103 Nmm2

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

175

- Modulus of elasticity of beam EB = 2629times103 Nmm2

- Modulus ratio between slab amp beam n = 119864119861

119864119863= 10

- CG to top of the basic beam = 4824 mm CG to bottom of the basic beam = 8676 mm

- eg the distance between the CG of non-composite beam and the deck eg = yt ndash ts2

For non-composite eg = yt ndashts2 = 4824 ndash 100 = 3824 mm

Kg the longitudinal stiffness parameter

Kg = n (Ig + Agtimeseg2) = 10 (23909 times 1010 = 1015times106times38242) = 3875 times 109 mm4

Interior Girder

Calculate moment distribution factor for two or more design lanes loaded (table 46222b-1)

DM = 0075 + ( 119878

2900 ) 06 (

119878

119871) 02 (

119870119892

119871times1199051199043)01

1100 lt S = 2200 lt 4900 mm 110 ltts = 200 lt 300 mm

6 m lt L = 181 m lt 73 m 4times109lt Kg= 3875times109lt 3000 times 109

Two or more lanes

DM = 0075 + ( 2200

2900 ) 06 (

2200

18100) 02 (

3875times109

18100times2003) 01rarr = 0688 lanesgirder

One design lane load

DM = 006 + ( 119878

4300 ) 04 (

119878

119871) 03 (

119870119892

119871times1199051199043) 01rarr = 0507 lanegirder

Shear distribution factor (table 46223a-1 )

One design lane

Dv = 036 + S7600 = 065

Two or more design lanes

Dv = 02 + 119878

3600 ndash (

119878

10700 )2rarr = 0769

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

176

02

12 11

06 17

23

115m

P P

04 18 06 06

22 12

Exterior Beam

Moment distribution factor

Single lane loaded (table 4622d-1) using the lever rule

DM = (12 + 22)22

= 1545 wheels2 = 077 lane

Two or more design lanes

DM = e timesDM int e = 077 + de2800

de distance from centerline of exterior girder to the inside face of the curb de = 06 m lt 17 m

Ok

So e = 077+6002800 = 0984

DM = 0984 times 0688 = 0677 lane

Shear distribution Factor (Table 46223b-1)

One ndash lane load Dv = 077

Two or more lanes

e = 06 + de3000 rarr = 06 + 6003000 = 080

Dv = 08 times 0769 = 0615 lane

Summary

Load case DM Int DM Ext DV Int DV Ext

Dist Fact Multiple 0688 0677 0769 0615

Single 0507 077 065 077

Design value 0688 077 0769 077

Dead load calculations

Interior Girder

DC = Ag timesɤc= 1015 times 10 -6times 24 = 2436 kNm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

177

Exterior Girder

DC = Ag timesƔc

Ag = (02times23 ) + ( 02 times 06 ) + ( 05 + 115 ) =1155 m2

DC = 1155 times 24 = 2772 kNm

Future wearing surface

Dw = 10 times 22 = 22 kNm int girder

Dw = 10 times (23 ndash 06) = 17 kNm ext girder

Dead load moments

i Interior girders

MDC = 1

8timesDc times1200012

rarr= 1

8times 2436times1812 = 9976 kNm

MDw= 1

8times 22 times 1812

rarr = 901 kNm

ii Exterior girder

MDC = 1

8times 2772 times 1812

rarr = 11351 kNm

MDw = 1

8times 17 times 1812

rarr = 696 kNm

Mx = (DL (L ndash X) 2) 2

Dead load shear force

The critical section for shear from L of bearing is=hc2 + 05(bearing width)=135

2 +

04

2 =0875 m

V = W (05 L ndash x)

x = 0875

Interior Girder

V Dc = 2436 (05times181 ndash 0875) = 1991 kN

V Dw = 22 (05 times 181 ndash 0875) = 18 kN

Exterior girder

V Dc = 2772 (05times181 ndash 0875) = 2266 kN

V Dw = 17 (05 times 181 ndash 0875) = 139 kN

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

178

Live load moments

Calculate the maximum live load moment due to IL-120 standard truck The maximum live load

in a simple span is calculated by positioning the axle loads in the following locations

P1 = 30 kN

P2 = 60 kN

P3 = 50 kN

R = P1 + 2P2 + 3P3 = 300 kN

sumMo = 0

R X = -P2times 12 ndash P1times 42 + P3times 67 + P3times 79 + P3times 91

X = 329 m

sum M about R2= 0 rarr R1 = 12273 kN

uarrsumfy = 0 rarr R2 = 17727 kN

2757

521

6715

3933

1645

O

X

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3205 1595 1

R1 R

2

181m

R

1645

7108

4283

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

179

( Mll) O = R1times (3205 + 3 + 12) ndash P2times 12 ndash P1 times 42 = 7108 kNm

(Mll) x = R1X X le 3205

= R1 X ndash P1 (X ndash 3205) 3205 lt X le 6205

= R1 X ndash P1 (X ndash 3205) P2 (X ndash 6205) 6205 lt X le 7405

Moment induced by lane load

M lane (X) = 119882119897119886119899119890119871times119883

2 -

119882119871times119883^2

2

M max X= L2 rarr M lane = 119882119897119886119899119890119871^2

4minus

119882119871times119871^2

8rarr =

1

8timesW lane L2

So Wlane = 93 kNmSection36124

L = 181 m

M lane = 1

8times 93 times 1812 = 3809 kNm

Live load moment for truck (including impact) and lane load

Mll(x) = M truck(x) times IM + M lane(x)

IM = 30 table 36211

(Mll) max = 7108times13 + 3809 = 1305 kNm

Distribution factor for int girder = 0688

SoMll = 0688 times 1305 = 8978 kNm

Dist factor for ext girder = 077

Mll = 077 times 1305 = 1005 kNm

P2 P

1

R1

O

Mo

3205 3 12

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

180

Shear due to truck and lane loading

1- Shear due to truck

Clockwise sum M2 = 0 rarr R1 = 1108 kN

uarrsumfy = 0 rarr R2 = 1892 kN

Vll for truck = 1892 kN

2-Shear induced by lane load

V lane (X) = 119882119897119886119899119890119871

2 ndash W lane times X rarr X = 0875

V lane = 1

2times 93 times 181 ndash 93 times 0875 = 76 kN

Total V truck+lane = V trucktimes IM + V lane = 1892 times13 + 76 = 3217 kN

DV dist fact for int girder = 0769

DV dist fact ext girder = 077

(Vll) max for int = 0769 times 3217 = 2474 kN

(Vll) max for ext = 077 times3217 = 2477 kN

Load combinations

Strength I 125 DC + 15 DW + 175 (LL+IM )

1 Design moments

i Interior girder Mu = 125 times 9976 + 15times901 + 175 times8978 rarr = 29533 kNm

ii Ext girder Mu = 125 times 11351 + 15 times696 + 175 times 1005 rarr = 3282 kNm

2 Design shear force

i Int girder Vu = 125 times 1991 + 15 times 18 +175 times 2474 rarr = 7088 kN

ii Ext girder Vu = 125 times 2266 + 15 times139 + 2477times175 rarr = 7376 kN

P1 P

2 P

2 P

3 P

3 P

3

67 3 12 12 12 3925 0875 1

R1 R

2

181m

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm

181

Design Interior beam

i Design for bending

Mu = 2953 kNm

b = 500 mm h = 1350 mm rarr d = 1350 ndash 40 - 12 ndash 252 ndash 875 = 1198 mm

m = fy085 119891prime119888 = 1647

R = 119872119906

120593119887119889^2 = 4573 MPa

ρ = 1

119898( 1 - radic1 minus

2 119898119877

119891119910 ) = 00121 gt ρ min and lt ρ max

As = ρ b d = 7247 mm2 (15 φ 25 mm)

Detail

5 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 0083 β radic119891prime119888 b d

= 0083times2 timesradic30 times 500 times1198 times 10 -3 = 5446 kN

Vu = 7088 kNgt 05 φ Vc = 05 times 09 times 5446 =245 kN shear reinforcement Required

Vs = Vn ndash Vc

= 7088

09 - 5446 = 243 kN

S = 119860119907119891119910119889

119881119904rarr =

226times420times1198

243times10^3 = 468 mm gt 300

lt 04 dv = 479

Use φ 12300 mm gt (Av) min

Note de gt 900 Ask = 0001 (1198 ndash 760) = 0438 mm2m lt 119860119904

1200 rarr φ10 300 mm

Design Exterior Beam

i Design for bending

Mu = 3282 kNm b = 500 mm d = 1198 mm m = 1647 R = 508 MPa

ρ = 0013624

As = ρ b d = 8161 mm2 (16 φ 25 mm)

6 φ 25 continuous with L =185m 5 φ 25 cut with L = 12 m amp 5 φ 25 cut with L = 6m

ii Design for shear

Vc = 5446 kN

Vu = 7376 gt 05 φ Vc = 245 kN shear reinforcement required

Vs = Vn ndash Vc

= 119881119906

09 ndash Vc = 275 kN

S = 119860119907119891119910119889

119881119904 = 414 kN Φ12 300

For Ask Φ10 300mm