AASHTO LRFD B id D iAASHTO LRFD Bridge Design ... · PDF file1 AASHTO LRFD B id D iAASHTO LRFD Bridge Design Specifications Prestressed Concrete RICHARD A. MILLER, PhD, PE, FPCI PROFESSOR

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AASHTO LRFD B id D iAASHTO LRFD Bridge Design Specifications

Prestressed Concrete

RICHARD A. MILLER, PhD, PE, FPCIPROFESSORPROFESSOR

UNIVERSITY OF CINCINNATI

AASHTO-LRFD Specification, 4th Edition

This module covers prestressed concrete superstructure elements.

General

Segmental boxes are NOT covered.Topics which are related to reinforced concrete only are covered in another module.Concrete structures are covered in Chapter 5. Chapter 5 uses a unified approach – reinforced concrete and prestressed concrete are covered in the same chapter.

AASHTO-LRFD 2007ODOT Short Course

July 2007

Loads and load combinations related to concrete are covered in Chapter 3.Analysis of concrete structures is covered in Chapter 4.

Do Not Duplicate Prestressed Concrete: Slide #2

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General

LRFD equations are in KSI units!Example Modulus of Rupture:Example Modulus of Rupture:

In most cases, the equations are simply the old Standard Specifications equations converted to ksi units.

..53050005.7

530.0524.0

SPECSTDpsipsif

LRFDksiksif

r

r

==

==


July 2007

ksiksiksiksipsipsi

524.01000

55.71000

510005.7/1000

50005.7===


§ 5.4 – Material Properties

Materials must meet AASHTO LRFD Bridge Construction Specifications.Construction Specifications.Unless specified otherwise, all provisions apply for strengths up to 10 ksi (Art. 5.4.2.1).

Some provisions allow up to 15 ksi.There is an effort to extend all provisions to 18 ksi.If a provision does not allow higher strength, use a


July 2007

maximum of 10 ksi in the calculations.Decks must have a minimum strength of 4 ksi.


3


A current problem with the LRFD Specifications is that some provisions allow strengths up to 18 ksi, but many are limited to 15 ksi or the default of 10 ksi. So what do you do if you are using a high strength concrete and a specific provision does not allow that strength?

Use the highest strength allowed by that provision. For example, assume a 15 ksi strength is specified but a particular provision has not been verified for that strength. For that particular provision, you must use a concrete strength of 10 ksi for your calculations (you may still use 15 ksi concrete in


July 2007

ksi for your calculations (you may still use 15 ksi concrete in the structure, you just cannot take advantage of the additional strength for that particular provision). However, if other provisions allow the use of 15 ksi concrete, you can use 15 ksi for those provisions.



§ 5.4.2.3 – Shrinkage and Creep

For calculation of creep and shrinkage, the engineer may use:

Articles 5.4.2.3.2 and 5.4.2.3.3CEB-FIP Model CodeACI 209

For prestressed concrete – the loss equations include creep and shrinkage.The main use of these provisions for prestressed


July 2007

The main use of these provisions for prestressed concrete is for calculating restraint moments for continuous for live load bridges.These are verified to 15 ksi. The creep equations do not work for strengths over 15 ksi.


4



( ) = −tkkkkttArttCoefficienCreep

itdfhcvsi 9.1,:)2.3.2.4.5.(

118.0ψH = Relative Humidity

t = time from first loading to( )

+=

−=

≥

−=

tk

fk

HkSVk

cif

hc

vs

itdfhcvsi

'15

008.056.1

0.113.045.1

,ψ t = time from first loading totime being considered

ti = time of first loading

V/S = volume to surface

fci = concrete strength at time ofprestress transfer or time offirst load (RC)


July 2007

+−

=tf

kci

td '461


first load (RC). If unknown, assume = 0.8fc’.

Std. Spec did not have a creep coefficient. Previous versions of LRFD use a different equation. It is similar to the ACI equation using ∆t0.6 /(10+ ∆t0.6).



( )−= −xkkkkArtShrinkage

tdfhsvssh 1048.0:)3.3.2.4.5.(

3εH = Relative Humidity

t = time from end of cure to( )

+=

−=

≥

−=

t

fk

HkSVk

cif

hs

vs

tdfhsvssh

'15

014.02

0.113.045.1

t = time from end of cure to time being considered

V/S = volume to surface

fci = concrete strength at time of prestress transfer or time of first load (RC). If unknown, assume = 0.8fc’.


July 2007

+−

=tf

tkci

td '461


Std. Spec. set shrinkage = 0.002. Previous editions of LRFD used an ACI type equation with a term of t/(35+t).

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§ 5.4.2.6 – Modulus of Rupture

There are now 3 defined Moduli of Rupture for normal weight concrete:g

For Arts. 5.7.3.4 (crack control) and 5.7.3.3.2 (Ieff):0.24 √fc’ksi (= 7.5√fc’ in psi units)

For Art. 5.7.3.3.2 (minimum reinforcement):0.37 √fc’ksi (= 11.5√fc’ in psi units)

For Art. 5.8.3.4.3 (shear) (this is new in 2007):√ √


July 2007

0.20 √fc’ksi (= 6 √fc’ in psi units)

Note that the value for Article 5.8.3.4.3 (shear) ONLY applies to the new, “simplified” method.



§ 5.4.2.4 – Modulus of Elasticity & § 5.4.2.5 – Poisson’s Ratio

(5.4.2.4-1)

(5 4 2 5)

1.5c 1 c cE 33,000K w f '

0 2=

µ

Where:K1 = Aggregate factor. Taken as 1.0 unless determined by testing or as approved by a jurisdiction.w = concrete unit weight in kcffc’ = concrete strength ksi

(5.4.2.5)0.2µ =


July 2007

E is basically the old Standard Specifications equation converted to ksi units and with an aggregate correction factor added.

µ is unchanged from Standard Specifications.


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§ 3.4 - Loads and Load Factors

§3.4.1: Load Factors and Load Combinations

For prestressed girders, the following service load combinations are most common:

Service I: Used for compression and transverse tension in prestressed concrete.

Service III: Used for longitudinal tension in prestressed concrete girders.

Service IV: Used for tension in prestressed columns, for crack control.Strength I: Basic load combination. Fatigue : Fatigue of reinforcement does NOT need to be checked for

fully prestressed components designed using Service III (A 3 1)


July 2007

(Art. 5.5.3.1)

Strength II-V and Extreme Event I and II are checked as warranted. Service II is for steel and never applies to prestressed concrete.


DCDD LL Use One of These at a Time



Table 3.4.1-1 Load Combinations and Load Factors

Load Combination

DWEHEVESEL

IMCEBRPLLS WA WS WL FR

TUCRSH TG SE EQ IC CT CV

STRENGTH I(unless noted) γp 1.75 1.00 -- -- 1.00 0.50/1.20 γTG γSE -- -- -- --

STRENGTH II γp 1.35 1.00 -- -- 1.00 0.50/1.20 γTG γSE -- -- -- --

STRENGTH III γp 1.00 1.40 -- 1.00 0.50/1.20 γTG γSE -- -- -- --

STRENGTH IV γp 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --

STRENGTH V γp 1.35 1.00 0.40 1.0 1.00 0.50/1.20 γTG γSE -- -- -- --


July 2007

γp γTG γSE


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Table 3.4.1-1 Load Combinations and Load Factors (cont.)DCDD LL

Use One of These at a Time

Load Combination

DWEHEVESEL


TUCRSH TG SE EQ IC CT CV

EXTREME EVENT I γp γEQ 1.00 -- -- 1.00 -- -- -- 1.00 -- -- --

EXTREME EVENT II γp 0.50 1.00 -- -- 1.00 -- -- -- -- 1.00 1.00 1.00

FATIGUE – LL, IM, & CE ONLY -- 0.75 -- -- -- -- -- -- -- -- -- -- --


July 2007Do Not Duplicate Prestressed Concrete: Slide #13



Table 3.4.1-1 Load Combinations and Load Factors (cont.)DCDD LL Use One of These at a

Ti

Load Combination

DWEHEVESEL


TUCRSH TG SE

Time

EQ IC CT CV

SERVICE I 1.00 1.00 1.00 0.30 1.0 1.00 1.00/1.20 γTG γSE -- -- -- --

SERVICE II 1.00 1.30 1.00 -- -- 1.00 1.00/1.20 -- -- -- -- -- --

SERVICE III 1.00 0.80 1.00 -- -- 1.00 1.00/1.20 γTG γSE -- -- -- --

SERVICE IV 1.00 -- 1.00 0.70 -- 1.00 1.00/1.20 -- 1.0 -- -- -- --



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Service III applies only to LONGITUDINAL TENSION in prestressed girders. The modifier to (LL+IM) is 0.8. Theprestressed girders. The modifier to (LL IM) is 0.8. The modifier is < 1 because it was found that the tensile capacity of prestressed girders is underestimated. This is largely because the loss of prestressing force is usually overestimated and a lower bound is used for the tensile strength (modulus of rupture).



AASHTO LRFDAASHTO-LRFDDistribution Factors for Precast/Prestressed Concrete Elements


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Distribution Factors for Precast/Prestressed Concrete Elements

§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear

The simplified distribution factors may be used if:

Width of the slab is constant Number of beams, Nb > 4Beams are parallel and of similar stiffness Roadway overhang de < 3 ftCentral angle < Article 4.6.1.2

Cross section conforms to AASHTO Table 4.6.2.2.1-1



Note: Multiple presence factors are NOT used with simplified distribution factors.

This is part of Table 4.6.2.2.1-1 showing common precast/prestressed concrete bridgeconcrete bridge types.The letter below the diagram correlates to a set of distribution factors.



factors.

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Beam and Slab Bridges would be a Type “k” bridge.

Moment distribution factors - LRFD Table 4.6.2.2.2b-1:T l l d d


July 2007

Two or more lanes loaded:DFM = 0.075+(S/9.5)0.6 (S/L)0.2 (Kg/12.0Lts

3)0.1

One lane loaded:DFM= 0.06+( S/14 )0.4 ( S/L )0.3 (Kg/12.0Lts

3)0.1




S = girder spacing (ft) 3.5 < S < 16.0L = span length (ft) 20 < L < 240t = slab thickness (in) 4 5 < t < 12 0ts = slab thickness (in) 4.5 < ts < 12.0Nb = Number of Beams Nb > 4Kg = n(Ig + Ageg

2) (in4) 10,000 < Kg < 7,000,000n = Ec,beam/Ec,slabIg = gross moment of inertia, non composite girder (in4)Ag = gross area, non composite girder (in2)eg = distance between centers of gravity of the non composite beam

d l b (i )


July 2007

and slab. (in)

If Nb = 3, use the lesser of the equations above with Nb = 3 and the lever rule.


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Beam and Slab Type “k” bridge

Shear Distribution Factors - LRFD Table 4.6.2.2.3a-1:


July 2007

Two or more lanes loaded:DFV = 0.2 + ( S/12 ) - ( S/35 )2

One lane loaded:DFV = 0.36 + ( S/25 )




3.5 < S < 16.0 ft.20 < L < 240 ft20 < L < 240 ft.4.5 < ts < 12.0 in.Nb > 4

If Nb = 3; use the lever rule.



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Beam and Slab Bridge Type “k” – Exterior – MomentTwo or more lanes loaded:

One lane loaded – use the Lever Rule

LRFD Table 4.6.2.2.2d-1

1.977.0

int

e

ext

de

egg

+=

=


July 2007

g = DFMde = distance from edge of the traffic railing to the exterior web of the exterior beam. The term de is positive when the railing is outboard (shown) and negative when the railing is inboard. -1.0 < de < 5.5 ft.




Beam and Slab Bridge Type “k” – Exterior – ShearTwo or more lanes loaded:Two or more lanes loaded:

One lane loaded – use the Lever Rule

LRFD Table 4.6.2.2.3b-1

106.0

int

e

ext

de

egg

+=

=


July 2007

g = DFV-1.0 < de < 5.5 ft.


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Beam and Slab Bridge – Type “k” Longitudinal Beams on Skewed SupportsAny number of lanes loaded; multiply DFM by: (LRFD Table 4.6.2.2.2c-1)

θ = Angle of skew; 30o < θ < 60o;

( )5.025.0

1

5.11

1225.0

tan1

=

−

LS

LtKc

c

s

θ


July 2007

θ Angle of skew; 30 < θ < 60 ; if θ<30o, c1 = 0; if θ>60o then θ=60o

L = Span, 20 < L < 240 ftS = Beam Spacing, 3.5 < S < 16 ft Nb > 4




Beam and Slab Bridge – Type “k” Longitudinal Beams on Skewed Supportson Skewed Supports

Correlation Factor for Load Distribution Factor for Support Shear atObtuse Corner - (LRFD Table 4.6.2.2.3c-1)

θ = Angle of skew; 0o < θ < 60o;

θtan1220.00.13.0

3

+

g

s

KLt


July 2007

L = Span, 20 < L < 240 ftS = Beam Spacing, 3.5 < S < 16 ft Nb > 4


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Lever Rule: Assume a hinge develops over each interior girder and solve for the reaction in the exterior girder as a fraction of the truck load.



1.2 01.2 1.2

HM Pe RSPe eR DFS S

→ − =

= ∴ =

∑

This is for one lane loaded. Multiple Presence Factors apply 1.2 is the MPF

In the diagram P/2 are the wheel loads; P

1.5’

36k36k


July 2007Do Not Duplicate Loads & Analysis: Slide #27

In the diagram, P/2 are the wheel loads; P is the resultant force. All three loads are NOT applied at the same time.

Note that truck cannot be closer than 2’ from the barrier

8 ft

(3.6.1.3)

Minimum Exterior DFM: (Rigid Body Rotation of Bridge Section)



NL - Number of loaded lanes under considerationNb - Number of beams or girders

E t i it f d i t k l d f CG f tt f

∑

∑+=

b

L

MinExt N

N

Ext

b

L

x

eX

NNDF

2, (C4.6.2.2.2d-1)



e - Eccentricity of design truck or load from CG of pattern of girders (ft.)

x - Distance from CG of pattern of girders to each girder (ft.)XExt - Distance from CG of pattern of girders to exterior girder (ft.)

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Adjacent Box GirdersAdjacent box girders with shear keys and a cast in placeAdjacent box girders with shear keys and a cast-in-place overlay are Type “f” sections.

Adjacent box girders with shear keys, but no cast-in-place deck, are Type “g” sections. Type “g” sections may or may not be laterally post-tensioned.

Lack of lateral post-tensioning causes a reduction of the


July 2007

distribution factor.




Interior Box GirdersThe following distribution factors may be used for a Type g y yp“f” (composite deck) or a Type “g” (non-composite) bridge IF the girders are “sufficiently connected together” – meaning they achieve transverse flexural continuity.

This can be done with lateral post-tensioning of at least 250 psi (Commentary 4.6.2.2.1; paragraph 12).

The Commentary further states that bridges without a t t l l d hi h t i d t


July 2007

structural overlay and which use untensioned transverse rods should NOT be considered as sufficient to achieve transverse flexural continuity, unless demonstrated by testing or experience (Commentary 4.6.2.2.1, paragraph 14).


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Interior Box GirdersType “f” (composite deck) or “g” with lateral PT -Type f (composite deck) or g with lateral PT LRFD Table 4.6.2.2.2b-1Moment:Two lanes loaded

DFM = k ( b/305 )0.6 ( b/12.0L )0.2 ( I/J )0.06

One lane loadedDFM = k(b/33.3L)0.5(I/J)0.25





Interior Box Girdersk = 2 5 ( N ) 0 2 > 1 5k = 2.5 ( Nb )-0.2 > 1.5Nb = number of beams 5 < Nb < 20b = width of beam, in 35< b < 60 in L = span of beam, ft 20< L < 120 ftI = moment of inertia of beam, in4

J = St. Venant torsional constant, in4


July 2007

J St. Venant torsional constant, in

For preliminary design, ( I/J )0.06 = 1.0


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Interior Box GirdersDistribution Factors for Shear - LRFD Table 4.6.2.2..3a-1Two Lanes Loaded:

DFV = (b/156)0.4 (b/12L)0.1 (I/J)0.05(b/48)

One Lane Loaded:DFV = (b/130L)0.15 (I/J)0.05

5 < Nb < 20

These are used for both composite and non-composite; even if the girders are NOT sufficiently connected.


July 2007

5 Nb 2035< b < 60 in 20< L < 120 ft25,000 < J < 610,000 in4

40,000 < I < 610,000 in4




Type “g” box with NO lateral PT

DFV (distribution factor for shear) does not change. It is the same for Type “g” structures with and without lateral PT.DFM is different. For Type “g” structures without lateral PT, the old


July 2007

Standard Specifications equations are used.

NOTE: The Standard Specifications equations were based on wheel loads and the LRFD equations are based on axle loads; so the equations changed by a factor of 2.


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Distribution Factor for Moment - LRFD Table 4.6.2.2.2b-1

DFM = S/DS = width of precast beam (ft)

D = (11.5 - NL)+1.4NL(1-0.2C)2 when C < 5D = (11.5 - NL) when C > 5


July 2007

Where:NL = number of traffic lanes C = K(W/L) < K




For Preliminary DesignBeam Type KN id d t l b 0 7

C = K(W/L) < KWhere:

Nonvoided rectangular beams 0.7Rectangular beams with circular voids: 0.8Box section beams 1.0Channel beams 2.2T-beam 2.0Double T-beam 2.0

JIK )1( µ+

=


July 2007

W = overall width of bridge measured perpendicular to the longitudinal beam (ft)

L = span (ft)µ = Poisson’s ratio = 0.2 for concrete (5.4.2.5)

J


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≈ SA4J2



∑ tSJ

Where:

A = Area enclosed by the centerline of the webs and flanges.l h f b fl li



S = length of a web or flange centerline.t = thickness of the corresponding web or flange.



The bending moment for exterior beams is determined by multiplying the distribution factor for interior beams by amultiplying the distribution factor for interior beams by a factor, e, which accounts for the distribution of load to the exterior girder. Note that this applies to type “g” even if there is no lateral post-tensioning. Lack of lateral post-tensioning is accounted for in the DVM.

Minimum exterior distribution factor based on rigid body



Minimum exterior distribution factor based on rigid body rotation does not apply to adjacent box girders.

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Exterior Box GirdersMultiplier for Moment – Types “f” and “g” - LRFD Table 4 6 2 2 2d-1Multiplier for Moment Types f and g LRFD Table 4.6.2.2.2d 1

Two or more lanes loaded:gext= eginterior

Where:e = 1.04 + ( de / 25 ) > 1 de=distance from edge of the traffic railing to the exterior web of


July 2007

ethe exterior beam. The term de is positive when the railing is outboard (shown) and negative when the railing is inboard.

de < 2.0 UNIT IS FEET!g= DFM




Exterior Box GirderMultiplier for Moment – Types “f” and “g” - LRFD Table 4 6 2 2 2d-1Multiplier for Moment Types f and g LRFD Table 4.6.2.2.2d 1

One lane loaded:gext= eginterior

e = 1.125 + ( de / 30 ) > 1 de < 2.0 ft.

e accounts for the distribution of load to


July 2007

e


the exterior girder

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Exterior Box GirdersMultiplier for Shear – Types “f” and “g” - LRFD Table 4 6 2 2 3b-1Multiplier for Shear Types f and g LRFD Table 4.6.2.2.3b 1Two or more lanes loaded:

148

48

5.0

int

≤

=

b

b

beggext

d < 2 0


July 2007

0.140

0.2121 ≥

−++=

bde

e


de < 2.035 < b < 60 ing = DFV



Multiplier for Shear – Types “f” and “g” - LRFD Table 4.6.2.2.3b-14.6.2.2.3b 1

One lane loaded:gext = eginterior

e = 1.125 + ( de / 20 ) > 1 de < 2.0 ft.



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Skewed Box GirdersMultiplier for Moment LRFD Table 4 6 2 2 2c 1Multiplier for Moment - LRFD Table 4.6.2.2.2c-1

1.05 - 0.25 ( tan θ) < 1.0

θ = skew angleIf θ > 600 use θ = 600


July 2007

This is optional.




When the skew angle of a bridge is small, say, less than 20o, it is often considered safe to ignore the angle of skew and to analyze the bridge as a zero-skew bridge whose span is equal to the skew span. This approach is generally conservative for moments in the beams, and slightly unsafe (<5%) for slab-on-girder decks for longitudinal shears.The LRFD Specifications Table 4.6.2.2.e-1 lists reduction multipliers for moments in longitudinal beams. The previous slide illustrates the multiplier for spread box beams,


July 2007

adjacent box beams with concrete overlays or transverse post-tensioning and double tees in multi-beam decks or Types (b), (c), (f) and (g).


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Correlation Factor for Load Distribution Factor for Support Shear at Obtuse Corner – Types “f” and “g” - (LRFD Table yp g (4.6.2.2.3c-1) – This is mandatory.

0o < θ < 60o

20 < L < 240 ft

θtan90

0.120.1dL

+


July 2007

20 < L < 240 ft.17 < d < 60 in d is depth of the girder35 < b < 60 in b is width of the flange5 < Nb < 20


S TOAASHTO-LRFDFlexure and Axial Loads


24

Flexure and Axial Loads

Definitions of various “d” terms for




AASHTO LRFD now uses the same terminology as ACI 318-05.This is a unified method for prestressed and reinforced concrete members.Article 5.7.2.1 defines 3 states:

Tension ControlledCompression ControlledTransition


July 2007

TransitionIn all cases, extreme fiber compressive strain = 0.003 (Article 5.7.2.1).

Values above 0.003 are allowed for confined cores.


25


§ 5.7.2 Assumptions for Strength and Extreme Event Limit States

Definition of Section Types

Extreme tensile steel strain when the extreme concrete compressive strain = 0.003

Type of section

εt > 0.005 Tension controlled

εt < fy / Es (may use = 0.002) Compression controlled

0.005 > εt > fy / Es Transition



For all prestressing or Grade 60 non-prestressed steel, εt may be assumed = 0.002 in place of fy/Es for compression controlled.

The ACI 318 code, upon which this provision is based, requires flexural members (that is, members with a superimposed axial load of < 0.1fc’Ag) to have εs > 0.004. AASHTO does not impose this requirement.



Definition of strain conditions for determining tension or compression control. Note that tensile strain in the steel closest to the tensile face is used


July 2007

face is used.Balanced condition is when εt = εy. For Grade 60 steel and all prestressing steel, εy may be taken as 0.002. Note that for prestressing steel, εt is the tensile strain which occurs in the steel after the pre-compression in the concrete is lost.


26



For a prestressed beam, it is important to understand theimportant to understand the definition of εt.

Begin by considering the strain condition of the beam at the point where the only loads are the prestressing force and the beam self weight.

dt



In this condition, the top of the beam is usually in tension (due to the prestressing). There is a net tensile strain in the prestressing steel of εp1. This is the initial pull minus any strain lost due to prestress losses. At the level of the steel, there is a compressive strain the concrete, εc.



As load is applied, the strain profile changes, the bottom d d t lldecompresses and eventually reaches a point where the CONCRETE strain at the level of the steel is 0. This is called “decompression”.

If there were no losses (except for elastic shortening), the strain in the steel ε at this point

dt


July 2007

in the steel, εp2 at this point would be the initial pull. The actual strain in the steel, with losses, can be calculated by mechanics.


27



This is the condition at Mn. The compressive strain in the concrete is 0 003 The total strain in theis 0.003. The total strain in the prestressing steel is the sum of the strain in the steel at decompression, εp2, and the strain developed between decompression and the ultimate state, εt.

The specifications only regulate the

dt


July 2007

The specifications only regulate the strain developed between decompression and the ultimate state, εt. The additional strain in the prestressing steel, εp2 is not part of the specification.



§ 5.5.4.2 Resistance Factors

Φ = 0.9 tension controlled reinforced concrete members1 0 tension controlled prestressed concrete members1.0 tension controlled prestressed concrete members0.75 compression controlled members with spirals or

ties (except for members in Seismic Zones 3 & 4)0.90 shear and torsion0.70 shear and torsion lightweight concrete


July 2007

For transition members, use a linear interpolation of the Φfactor based on the extreme tensile steel strain.


28



1

1.05

Prestressed:Strain = 0 004

Prestressed

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

0 0 001 0 002 0 003 0 004 0 005 0 006 0 007

Phi F

acto

r

CompressionControlled Transition

Tension Controlled

Strain = 0.004Phi = 0.92 Reinforced


July 2007

Prestressed Members

Reinforced Members

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007

Extreme Steel Strain


(5.5.4.2.1-1)

(5.5.4.2.1-2)

0.75 0.583 .25 1 1.0

0.75 0.65 .15 1 1.0

≤ = + − ≤ ≤ = + − ≤

t

t

dcdc

φ

φ



Effect of New Resistance FactorsIt is allowable to design flexural members with extremeIt is allowable to design flexural members with extreme fiber steel strains < 0.005. This is done by increasing the area of steel. However, in general, the Φ factor is reduced at a slightly lower rate than moment resistance is gained. There is a slight increase in Mn but it is minimal.Thus there is little effect on the allowable moment by


July 2007

Thus, there is little effect on the allowable moment by increasing the amount of steel above that required to bring the extreme fiber steel strains to 0.005.


29



For tension controlled partially prestressed members:(5 5 4 2 1 3)0 90 0 10PPRφ +

PPR = Partial prestressing ratioAps = Area of prestressing steel

(5.5.4.2.1-3)

(5.5.4.2.1-4)ps py

ps py s y

0.90 0.10PPRA f

PPRA f A f

φ = +

=+


July 2007

Aps Area of prestressing steelfpy = Yield strength of the prestressing steelAs = Area of mild steelfy = Yield strength of the mild steel



The stress block remains the same as Standard Specifications.


July 2007

Analysis of reinforced concrete RECTANGULAR beams is the same as Standard Specifications.HOWEVER, there are some differences with prestressed concrete.


30

AASHTO LRFDAASHTO-LRFDPrestressed Beams with Bonded Tendons


Prestressed Beams with Bonded Tendons

§ 5.7.3 Flexural Members

The value of fps can be found from (if fpe > 0.5fpu):

5 7 3 1 1 1)cf f 1 k pyf

k 2 1 04 (5 7 3 1 1 2)

Then:

(5.7.3.1.1-1)ps pu

p

cf f 1 kd

= −

c ps ps

1

c 1 ps pup

0.85f 'b a A f

a c

c0.85f 'b c A f 1 kd

=

= β

β = −

py

pu

k 2 1.04f

= −

(5.7.3.1.1-2)


July 2007

Stress in the steel, fps, can also be found from strain compatibility analysis.


ps pu

puc 1 ps

p

A fc f

0.85f ' b kAd

=β +

31



pups

ffA

c =

c = depth of neutral axisb = width of compression block Aps= area of TENSILE prestressing steel

p

pupsc df

kAbf +1'85.0 β


July 2007

dp = depth to centroid of tensile prestressing steelfpu = tensile strength of prestressing steelfpy = yield strength of prestressing steelβ1 = stress block factor – same as Std. Spec.




If there is mild (nonprestressed) tensile steel, As and mild compression steel As’ both with a yield stress of fy , the p s y yequation for c becomes:

(5.7.3.1.1-4)

1.85 ' ' ' 1c s y s y ps pup

cf b c A f A f A f kd

β

+ = + −

' 'ps pu s y s yA f A f A fc f

+ −=


July 2007

The engineer must do an analysis to see if the compression steel yields. If the compression steel does not yield, the actual stress is substituted for fy’ into equation 5.7.3.1.1-4.


10.85 ' puc ps

p

ff b kA

dβ +

32



Sometimes, things change for the better!!!!Std. SpecAndLRFD 2005 Interim

Editions 1 th h 3 f

In Editions 1-3 of the LRFD Specifications, the β factor was applied to the flange as well as to the web. This made no sense. It was changed with the 2005 Interim back to the old definition



through 3 of LRFD

back to the old definition. Now it is the same definition as ACI 318 and Std. Spec.



The T beam equation returns to normal:a a

(5.7.3.1.1-1)

( )

2 2

' ' ' 0.85 '2 2 2

n ps ps p s y s

fs y s c w f

a aM A f d A f d

ha aA f d f b b h

= − + − −

− + − −


July 2007

Again the engineer must do an analysis to see if the compression steel yields. If the compression steel does not yield, the actual stress is substituted for fy’ into equation 5.7.3.1.1-1.


33



( ) fn ps ps p s y s s s s c w f

a a a a hM A f d A f d A ' f ' d ' 0.85f ' b b h2 2 2 2 2

= − + − − − + − −

The LRFD Specifications give only this equation:

2 2 2 2 2

a aM A f d A f d +

n ps ps p s y s s s sa a aM A f d A f d A ' f ' d '2 2 2

= − + − − −

If the section is NOT a “T” beam, b = bw and:

If there is no compression steel:


July 2007

n ps ps paM A f d2

= −

n ps ps p s y sM A f d A f d2 2

= − + −


If there is no non-prestressed tensile steel:



For prestressed T- Beams:

bw = web width

(5.7.3.1.1-3)( )

1

' ' 0.85 '

0.85 '

ps pu s y s y c w f

puc w ps

p

A f A f A f f b b hc f

f b kAd

β

+ − − −=

+


July 2007

w

b = flange widthhf = flange thickness


34

AASHTO LRFDAASHTO-LRFDPrestressed Beams with Unbonded Tendons

AASHTO-LRFD Specification, 4th Edition.

Prestressed Beams with Unbonded Tendons


The stress in the prestressing steel can be found from: d

e = effective tendon length

i = length of tendon between anchoragesl

+

=

<

−+=

s

ie

pye

ppeps

N

fcd

ff

22

900

ll

l(5.7.3.1.2-1)

(5.7.3.1.2-2)

l


July 2007

i length of tendon between anchoragesNs = Number of support hinges crossed by the tendon between

anchorages or discretely bonded points.fpe= Effective stress in the steel after losses.


35



For rectangular beams:

For T-beams:

(5.7.3.1.2-4)

1

' '0.85 '

ps ps s y s y

c

A f A f A fc

f bβ+ −

=

( )' ' 0.85 'ps ps s y s y c w fA f A f A f f b b hc

+ − − −=



(5.7.3.1.2-3)10.85 'c wf bβ



For unbonded tendons, the equations for “c” require the value of f , but the equation for f requires the value ofvalue of fps, but the equation for fps requires the value of “c”.The two equations can be solved simultaneously in a closed form, but most people will not do this.Thus, finding fps becomes an iterative procedure.The Commentary (C5.7.3.1.2) gives an equation for a first estimate of f (in ksi):


July 2007

first estimate of fps (in ksi):


(C5.7.3.1.2-1)15ps pef f= +

36

AASHTO LRFDAASHTO-LRFDComponents with Both Bonded and Unbonded Tendons


Components with Both Bonded and Unbonded Tendons


Article 5.7.3.1.3 allows two methods:

Article 5.7.3.1.3a “Detailed Analysis”In this method, a detailed, strain compatibility is used.

Article 5.7.3.1.3b “Simplified Analysis”Shown on the following slide Apsb = area of bonded tendonsA f b d d t d


July 2007

Apsu = area of unbonded tendons


37

Components with Both Bonded and Unbonded Tendons


Simplified Analysis - The stress in the UNBONDED tendons may be conservatively taken as the effective stress after losses: fpe. p

For T-beams:


( )

p

pupswc

fwcysyspepsupupsb

df

kAbf

hbbffAfAfAfAc

+

−−−++=

1'85.0

'85.0''

β


July 2007



10.85 '

psb pu psu pe

puc ps

p

A f A fc f

f b kAd

β

+=

+

S TOAASHTO-LRFDMoment Capacity


38

Moment Capacity

§ 5.7.3.2 Flexural Resistance

For T-beams (where a>hf): aa

For rectangular beams, b=bw; thus equation 5.7.3.2.2-1

( )

−−+

−−

−+

−=

22'85.0

2'''

22

ffwcsys

sysppspsn

hahbbfadfA

adfAadfAM(5.7.3.2.2-1)


July 2007

wbecomes:

−−

−+

−=

2'''

22adfAadfAadfAM syssysppspsn


Moment Capacity

§ 5.7.3.2 Flexural Resistance

In the preceding equations:

dp = distance from the extreme compression fiber to theprestressing steel.

ds = distance from the extreme compression fiber to thenon-prestressed tensile steel.

ds’ = distance from the extreme compression fiber to thenon prestressed compression steel


July 2007

non-prestressed compression steel.fy = yield strength of the non-prestressed tensile steel.fy’ = yield strength of the non-prestressed compression

steel.


39

Moment Capacity

§ 5.7.3.3 Limits for Reinforcement

Minimum reinforcement (Article 5.7.3.3.2):

It is the smaller of:

φMn > 1.2 Mcr – same as in Std. Spec.φMn > 1.33Mu – LRFD added



Moment Capacity


For the minimum reinforcement requirement, the cracking moment Mcris found from:

Sc = composite section modulusfr = modulus of rupture = 0.37√fc’ (ksi units)fcpe = compressive stress in the concrete due to effective

prestressing force, at the extreme tensile fiber for applied

(5.7.3.3.2-1)( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥


July 2007

prestressing force, at the extreme tensile fiber for appliedloads.

Mdnc = Unfactored dead load moment on the non-composite or monolithic section.

Snc = Non-composite section modulus.


40

Moment Capacity


Maximum reinforcement provision was dropped with 2005 InterimInterim

No longer needed with new definitions of tension controlled, compression controlled and transition.LRFD previously used a c/d ratio. This can still be used:

38t

cd≤ Tension Controlled

εt > 0.005

3


July 2007

3 35 8t

cd> >


Compression Controlledεt <0.002

Transition

35t

cd

≥

Moment Capacity


Maximum reinforcement is now controlled by εt.


July 2007

To determine εt , calculate c. Then, using similar triangles:

−

=ccdt

t 003.0ε


41

Moment Capacity


Maximum ReinforcementThis is more restrictive that Std. Specification or previous editions of p pLRFD.

For reinforced sections, 0.75ρbal was used. This was a strain of 0.0037 in the steel. For prestressed, Std. Spec. c/de ratio was limited to 0.42. This corresponded to a strain of 0.0041

0.375t

cd

≤ Tension Controlledεt > 0.005


July 2007

42.0≤edc

0.45e

cd≤


t

Previous Editionsεt >0.0041

Std. Specifications, RC.

Moment Capacity

§ 5.7.3.4 Control of Cracking by Distribution of Reinforcement

css

e df

s −≤ 2700β

γ (5.7.3.4-1)

s = spacing of reinforcement closest to the tension face.γe = exposure factor; 1 for Class 1 and 0.75 for Class 2

ODOT uses 0.75 for decks, 1 for everything elsedc = cover to extreme tension fiberfs = Steel stress @ service limit stateh = overall thickness or depth

( )cc

s dhd−

+=7.0

1β


July 2007

h = overall thickness or depth

Does not apply to slabs designed using the empirical method (ODOT does not allow empirical design).

It applies to all other concrete components where the service tensile stress exceeds 0.8fr = 0.8(0.24)√fc’ = 0.20√fc’


42

Moment Capacity

§ 5.7.3.5 Moment Redistribution

ODOT does not permit moment redistribution



Moment Capacity

§ 5.7.3.6.2 Deflection and Camber

Prestressed members are usually designed as uncracked at service loads. Instantaneous deflections and cambers are then calculated using the gross moment of inertia, Ig.

If the deflection is calculated using Ig, long term deflection can be found by multiplying the instantaneous deflection by 4.


July 2007

For prestressed members, the Commentary (C5.7.3.6.1) allows the multipliers given in the PCI Design Handbookto be used for long term camber/deflection values.


43

Prestressing and Partial Prestressing

§ 5.9.3 Stress Limitations for Prestressing Tendons

Tendon Type

Table 5.9.3-1 Stress Limits for Prestressing Tendonsfpy = yield stress of prestressing steel

Condition

Stress-Relieved Strand and Plain High-

Strength Bars

Low Relaxation

Strand

Deformed High-Strength

Bars

Pretensioning

Immediately prior to transfer (fpbt) 0.70 fpu 0.75 fpu __

At service limit state after all losses (fpe) 0.80 fpy 0.80 fpy 0.80 fpyPost-Tensioning

Prior to seating – short-term f bt may be allowed 0 90 f 0 90 f 0 90 f

fpu = ultimate strength of prestressing steel


July 2007

Prior to seating short term fpbt may be allowed 0.90 fpy 0.90 fpy 0.90 fpy

At anchorages and couplers immediately after anchor set

0.70 fpu 0.70 fpu 0.70 fpu

Elsewhere along length of member away from anchorages and couplers immediately after anchor set

0.70 fpu 0.74 fpu 0.70 fpu

At service limit state after losses (fpe) 0.80 fpy 0.80 fpy 0.80 fpy



§ 5.9.4 Stress Limits for Concrete

Table 5.9.4.2-1 Temporary Tensile Stress Limits in Prestressed Concrete Before Losses, Fully Prestressed Components. (Partial)

Bridge Type Location Stress LimitOther than Segmentally Constructed Bridges

•In precompressed tensile zone without bonded reinforcement•In areas other than the precompressed tensile zone and without bonded reinforcement•In areas with bonded reinforcement (reinforcing bars or prestressing steel) sufficient to resist the tensile force in the

N/A

0.0948√f’ci <0.2(ksi)

0.24√f’ci (ksi)


July 2007

concrete computed assuming an uncracked section, where reinforcement is proportioned using a stress of 0.5 fy, not to exceed 30 ksi.•For handling stresses in prestressed piles 0.158√f’ci (ksi)

Compression Limit at Transfer 0.6 f’ci (ksi)


44

Debonding and Harping

If the tensile stresses at the end of girder are above 0.24√fci’ , then the stress must be reduced either by0.24√fci , then the stress must be reduced either by debonding the strand or harping the strand.If debonding is used, no more than 25% of the total number of strands may be debonded and not more than 40% in any single row may be debonded. (Art. 5.11.4.3)





Location Stress Limit

Table 5.9.4.2.1-1 Compressive Stress Limits in prestressed Concrete at Service Limit State After Losses, Fully Prestressed Components.

Location Stress Limit• In other than segmentally constructed bridges due to the

sum of effective prestress and permanent loads• In segmentally constructed bridges due to the sum of

effective prestress and permanent loads• In other than segmentally constructed bridges due to live

load and one-half the sum of effective prestress and permanent loads

• Due to the sum of effective prestress permanent loads

0.45f’c (ksi)

0.45f’c (ksi)

0.40f’c (ksi)

0 60φ f’ (ksi)


July 2007

• Due to the sum of effective prestress, permanent loads, and transient loads and during shipping and handling

0.60φwf’c (ksi)


45



B id T L ti St Li it

Table 5.9.4.2.2-1 Tensile Stress Limits in Prestressed Concrete at Service Limit State After Losses, Fully Prestressed Components. (Partial)

Bridge Type Location Stress LimitOther than Segmentally Constructed Bridges

Tension in the Precompressed Tensile Zone Bridges, Assuming Uncracked Sections

• For components with bonded prestressing tendons or reinforcement that are subjected to not worse than moderate corrosion conditions

• For components with bonded prestressing tendons or reinforcement that are subjected to severe corrosive conditionsF t ith b d d t i

0.19√f’c (ksi)

0.0948√f’c (ksi)

N T i


July 2007

• For components with unbonded prestressing tendons

No Tension


Again, these are Std. Spec. limits in ksi units.

0.19(1000)0.5 = 6

S TOAASHTO-LRFDLoss of Prestressing Force


46

§ 5.9 – Prestressing and Partial Prestressing

§ 5.9.5 Loss of Prestress

Loss of prestressing force was changed with the 3rd

Edition.Edition.Like creep and shrinkage, the changes are based on the results NCHRP Report 496 “Prestressed Losses in Pretensioned High Strength Concrete Bridge Girders”These provisions are applicable up to 15 ksi concrete





The basic equations:Pretensioned Members:Pretensioned Members:

Post-tensioned Members:

fffff ∆+∆+∆+∆=∆

pLTpESpT fff ∆+∆=∆ (5.9.5.1-1)

(5 9 5 1-2)



pLTpESpApFpT fffff ∆+∆+∆+∆=∆ (5.9.5.1-2)

47



∆fpT = Total loss of prestressing force (ksi).∆f = Loss due to friction (ksi)∆fpF = Loss due to friction (ksi).∆fpA = Loss due to anchorage set (ksi).∆fpES = Loss due to elastic shortening (ksi).∆fpLT = Loss due to long term shrinkage and creep of the

concrete and relaxation of the steel (ksi).


July 2007

∆fpA is usually given by the manufacturer.




Friction losses:

Loss due to friction between an internal tendon and a duct wall:

Loss due to friction between an external tendon and a

( )( )µα+−−=∆ kxpjpF eff 1 (5.9.5.2.2b-1)


July 2007

single deviator pipe:( )( )04.01 +−−=∆ αµeff pjpF


(5.9.5.2.2b-2)

48



fpj= initial jacking stress in the tendon (ksi).x = length of tendon from the jacking point to the pointx = length of tendon from the jacking point to the point

being considered (ft).K = wobble friction coefficient (per ft. of tendon)µ = friction coefficient.α = sum of the absolute value of angular change of

prestressing steel path from jacking end (or nearest


July 2007

prestressing steel path from jacking end (or nearestjacking end if jacked from both ends) to point underconsideration. (radian)




Steel Duct K µTable 5.9.5.2.2b-1 Friction Coefficients for Post-Tensioning Tendons.

Wire or Strand

Rigid or Semi rigid galvanized metal sheathing

0.0002 0.15-0.25

Polyethylene 0.0002 .23

Rigid steel deviator bar for external tendons

0.0002 .25

HS Bar Galvanized metal sheathing 0.0002 .30



Values for K and µ should be found from experimental data. If such data is absent, values from the table above may be used.

49



Elastic Shortening, pretensioned members:

Ect = modulus of elasticity of the concrete at transfer or at time of load

Elastic Shortening, Post-tensioned Members:

=∆ fEE

f cgpct

ppES

E1N

(5.9.5.2.3a-1)



fEE

2N1N∆f cgp

ci

ppES

−= (5.9.5.2.3b-1)



fcgp = concrete stresses at the center of gravity of the prestressing tendons due to prestressing force immediately after transfer (pretensioning) or immediately after jacking (post-tensioning) and the self-weight of the member at the sections of maximum moment (ksi).

In pretensioned members, at transfer, fcgp may be calculated by assuming the stress in the prestressing tendon after release = 0.9fpi; where fpi is the initial prestressing stress (jacking stress) in the tendons.

f ( )


July 2007

Ep = Elastic Modulus of the prestressing strand (ksi).Eci = Elastic Modulus of the concrete at the time of transfer or time of

load application (ksi).N = number of identical strands.


50



Long Term LossesFor standard precast pretensioned members subject toFor standard, precast, pretensioned members subject to normal loading and environmental conditions:

(5.9.5.3-1)

(5.9.5.3-2)

10 12

1 7 0 01

= + +

= −

pi pspLT h st h st pR

g

h

f Af f

A

. . H

∆ γ γ γ γ ∆

γ



( )

(5.9.5.3-3)5

1=

+stcif

γ



fpi = prestressing steel stress immediately PRIOR to transfertransfer.

H = Average annual relative humidity in percent (e.g.70 not 0.7)

∆fpR = 2.5 ksi for LoLax10 ksi for stress relieved

γh = humidity factor


July 2007

h

γst = strength factor


51



To use the ∆fpLT equation, the following criteria must be met:met:

Members are pretensionedNormal weight concrete is usedMembers are moist or steam curedPrestressing is by bar or strand with normal and low relaxation properties


July 2007

Average exposure conditions and temperatures.




This table can be used to estimate time dependent losses in prestressed members which do not have composite slabs and are stressed after attaining a compressive strength of at least 3.5 ksi.

Type of Beam Section

Level For wires and Strands with fpu = 235,250 or 270 ksi

For Bars with fpu = 145 or 160 ksi

Rectangular Upper Bound Average

29.0 + 4.0PPR 19.0 + 6.0 PPR

Box Girder Upper Bound Average

21.0 + 4.0PPR19.9 + 4.0PPR

15.0

Table 5.9.5.3-1 Time-Dependent Losses in ksi.



PPRf

PPRf

c

c

0.60.6

0.6'15.00.10.33

0.60.6

0.6'15.00.10.39

+

−

−

+

−

−PPRf c 0.6

0.60.6'15.00.10.31 +

−

−Single T, Double T, Hollow core and Voided Slab

Upper Bound

Average

PPR is the partial prestressing ratio.

52



Lump Sum Losses:For lightweight concrete the stresses in the table areFor lightweight concrete, the stresses in the table are increased 5 ksi.For low relaxation strand, the values in the table are reduced by:

4 ksi for box girders6 ksi for rectangular beams and solid slabs


July 2007

8 ksi for single T’s, double T’s, hollow core and voided slabs.




For post-tensioned members, the “Refined Method” for estimation of time dependent losses must be used.estimation of time dependent losses must be used. However, this method is based on NCHRP 496, but requires a large amount of calculation.

Since longitudinal post-tensioning is not common in Ohio, the method is not presented here. However, it can be found in Article 5 9 5 4 of the LRFD Specifications


July 2007

be found in Article 5.9.5.4 of the LRFD Specifications.


53

S TOAASHTO-LRFDBond/Development Length


§ 5.11 – Bond and Development Length

§ 5.11.4.1 – Transfer Length

F f ll b d d t d th t f l th fFor fully bonded strands, the transfer length from the end of the girder is assumed to be 60db, where db is the bar or strand diameter.



54


§ 5.11.4.2 – Development Length

Development length for fully bonded strand is given by:

d ps pe b2f f d3

= κ −

l (5.11.4.2-1)





Where:ld = development lengthfps = steel stress at strength limit statefpe = effective prestressing stress after all lossesdb = strand diameterκ =1.0 for pretensioned panels, piles and other

pretensioned members with a depth < 24 inches.1 6 f t i d b ith d th 24 i h



= 1.6 for pretensioned members with a depth > 24 inches = 2.0 for debonded strand

55



In previous editions of the LRFD Specifications, bond stress was assumed linear – e.g, if the bonded length was only ½ the development length, it was assumed that the strand could only develop 0.5fps.

This assumption is still true for TRANSFER LENGTH; e.g at ½ the transfer length it is assumed only 0.5fpe is developed.



However, stress in the steel beyond the transfer length, but less than the development length, can now be calculated by a bilinear formula.



( )60px bpx pe ps pe

df f f f

d−

= + −l ( )60px pe ps ped b

f f f fd−l

Where:fpx = stress at “x” from the end of the girderf = effective stress in the steel after all losses

(5.11.4.2-4)



fpe effective stress in the steel after all lossesfps = stress in the steel at the strength limit statelpx = length were the stress is being calculatedld = development lengthdb = strand diameter

56



Within the transfer length (which is 60db):

60px pe

pxb

ff

d=l

(5.11.4.2-3)







1

S TOAASHTO-LRFDShear


§ 5.8 - Shear and Torsion

§ 5.6 Design Considerations

Important things about the shear sectionThis section has the provisions of the LRFDThis section has the provisions of the LRFD Specifications, through the 2007 changes.This section concentrates the provisions as they apply to prestressed concrete; both pretensioned and post-tensioned.Segmental box girder bridges and spliced girders are NOT covered



NOT covered.Reinforced concrete is covered in another section.

2


§ 5.6.3 Strut-and-Tie Model

Strut and Tie ModelStrut and tie can be used for analysis of anchorageStrut and tie can be used for analysis of anchorage zones and support regions.It is also useful for deep footings, pile caps and sections where the depth is more than ½ the span.This model is covered in Article 5.6.3.Strut and tie will not be discussed as part of this module.



It will be covered in another presentation.


§ 5.8.2 General Requirements

nr VV =φ (5.8.2.1-2)

Vn = nominal shear resistance given in Article 5.8.3.3 (kip)φ = 0.9 normal weight concreteφ = 0 7 lightweight concrete

ru VV ≤



φ = 0.7 lightweight concreteVu = Factored shear at the cross section being

considered. If there is significant torsion present, this term is modified for torsion.

3


§ 5.8.3.3 Nominal Shear Resistance

The nominal shear resistance, Vn, can be assumed to be the sum of three forces, the forces in the stirrups, thethe sum of three forces, the forces in the stirrups, the vertical component of the force in the concrete and the vertical component of any harped or draped prestressing strand. This leads to the basic equation:

Vn = Vc + Vs + Vp (5.8.3.3-1)





Assumptions about Shear Strength:The beam fails when the concrete in the struts reachesThe beam fails when the concrete in the struts reaches its crushing strength.At failure, the beam has shear cracks and the cracks have opened

This would cause the stirrups to yield.The compressive strength of concrete between the

( )



shear cracks (struts) is not fc’. As will be shown, it may be greater than or less than fc’.

4



Assume that the angle of the strut is θ and theof the strut is θ and the distance between the compressive and tensile forces is jd where d is effective depth and j<1. Thus the horizontal distance



is jd/tanθ = jdcotθ.



The stirrup contribution is:Force per stirrup times the number of stirrups.Force per stirrup times the number of stirrups.If the stirrups are spaced at “s”, the number of stirrups in the length jd cotθ is (jd cotθ)/sThe force per stirrup is Avfy so:

v y v y vA f jd cot A f d cotV

θ θ



(Note that if j = 1 and θ = 45o, we get the old, familiar equation: Vs = (Avfy d) / s . Also note that jd = dv)

= =y ysV s s

5



The LRFD Specifications consider the most general case where the stirrups may be inclined at an angle of α from thewhere the stirrups may be inclined at an angle of α from the longitudinal axis. Thus, the equation becomes:

( )+= v y v

s

A f d cot cot sinV

sθ α α

(5.8.3.3-4)



However, in almost all cases, α = 90o ; thus cotα = 0 and sinα = 1. The equation reverts the one shown on the previous slide.



If a line is cut perpendicular toperpendicular to the cracks, it has a length of jdcosθ. It may cross several struts. The total force in the struts will be the



concrete stress times the area.

Fc = fc (jd cosθ) bvwhere fc is the concrete stress and bv web width.

6



The force triangle shows thatthe force along the struts isthe force along the struts is

V / sinθ.

Substituting into the previous equation and assuming Vc is



Substituting into the previous equation and assuming Vc is the shear force carried by the concrete:

Vc = fc (jd cosθ) sinθ bv



( )cos sinc c vV f jd bθ θ=

1, 45

4 '

2 ' ( ) 0.0632 ' ( )

o

c c

c c v c v

Note that if j

and f f

V f b d lbs f b d kips

θ= =

=

= =



This is the ACI 318 equation and the old Standard Specification equation.The Vc equation, in ksi units, may be used for NON-PRESTRESSED concrete members (LRFD 5.8.3.4).

7



The basics of these equations were developed by research done at the University of Illinois in the 1920’s.research done at the University of Illinois in the 1920 s. They found that the actual angle varies along the beam and that the angle can be anywhere from 25 to 65 degrees. While it is possible to calculate the angle, it is difficult. In the days before computers or calculators, it was nearly impossible. Therefore, the value of 45 degrees was



impossible. Therefore, the value of 45 degrees was chosen for simplification. The value of the crushing strength was also chosen as a simplification.



In the 1980’s, Vecchio and Collins (University of Toronto) proposed a method for finding the shear strength of aproposed a method for finding the shear strength of a beam. This method required the calculation of the actual angle, θ, and the crushing strength of the concrete struts. The crushing strength is a function of the strain perpendicular to the strut.The original theory was called “Compression Field Theory”. Later the theory was improved to account for



Theory . Later the theory was improved to account for additional mechanisms, such as aggregate interlock, and was renamed “Modified Compression Field Theory”.

8


Modified Compression Field Theory

The basis of the Modified Compression Field Theory (MCFT) is to determine the point at which the diagonal compressive struts fail and to determine the angle of the struts. From the crushing strength and the angle, the contribution of the concrete, Vc , can be found.




Why isn’t the crushing strength fc’ ? The value of fc’ is for uniaxial load. The concrete fails by cracking parallel to theuniaxial load. The concrete fails by cracking parallel to the load. If a lateral (biaxial) force is applied, it changes the apparent compressive strength. Lateral compression holds cracks together and increases compressive strength. Lateral tension pulls them apart and decreases the compressive strength.



9


Vecchio and Collins proposed that the compressive strength of the strut is a function of both the compressive g pstress along the strut and the tensile stress perpendicular to the strut. They wrote several equations in terms of the applied average shear stress, v = V/bd, the principal tensile strain (perpendicular to strut), ε1, and the angle of the strut, θ.

To use MCFT values of ε and θ are assumed It then



To use MCFT, values of ε1 and θ are assumed. It then takes 17 steps and 15 equations to recalculate ε1 and θ. If these are not close to the assumed values, then iterations are needed.


§ 5.8.3 Sectional Design Model

Sectional Design ModelObviously, no one wanted to use an iterative procedureObviously, no one wanted to use an iterative procedure involving 17 steps and 15 equations. As a result the LRFD Code simplified the method to use a table. This is called the “Sectional Design Model”.Unfortunately, soon after the 1st Edition came out, there was controversy with the Sectional Design Model. The equations provided low values of shear strength. It was found that simplifying the method created inaccuracies



found that simplifying the method created inaccuracies.Editions after the 2nd Ed. still use(d) the Sectional Design Model, but have new equations and tables with more realistic values of shear resistance.

10



The shear strength of the beam is:V = V + V + V (5 8 3 3-1)Vn Vc Vs Vp

Vc = contribution of the concreteVs = contribution of the stirrupsVp = vertical component of the force in harped strands.

Note that there is a limit:V < 0 25f ’ b d + V (5 8 3 3 2)

(5.8.3.3 1)



Vn < 0.25fc’ bv dv + Vpbv = effective web widthdv = effective depth for shear dv = de – a/2 > greater of 0.9de or 0.72 h

(5.8.3.3-2)



According to Articles 5.8.2.5 and 5.8.2.9, the web width, b , must be adjusted for the presence of ducts.bv, must be adjusted for the presence of ducts.bv = Effective web width, defined as the minimum web

width, parallel to the neutral axis, between the compressive and tensile flexure resultants. For circular sections, it is the diameter of the section.

At a particular level, one half the diameter of ungrouted ducts and one quarter the diameter of grouted ducts is



ducts and one quarter the diameter of grouted ducts is subtracted from the web width.

11



The nominal shear resistance is the lesser of:

n c s pV V V V= + + (5.8.3.3-1)

Vc and Vs are defined as:

d is the shear depth = d – a/2

( )sincotcot'0316.0+

=

=

sdfA

V

LdbfV

vyvs

vvcc

ααθβ

p

n c v vV 0.25f 'b d≤

(5.8.3.3-3)

(5.8.3.3-2)

(5.8.3.3-4)



dv is the shear depth = de – a/2the greater of 0.9de or 0.72 hs = stirrup spacingAv = stirrup area.

The 0.0316 converts psi to ksi units.



In all of the preceding equations, the factors β and θ are unknown and must be determined.unknown and must be determined.The LRFD Specifications require a sectional approach. The girder is divided into sections along the length, the factors β and θ are determined at each section.Traditionally, the sections are every 0.1L and importantpoints like harp points, debond points, etc.



The first sections must be the critical section from the face of the support.

12


§ 5.8.3.2 Sections Near Supports

Critical SectionThe critical section is defined in Article 5.8.3.2.

The beam must be checked using Article 5.8.1.2 to determine if it is a deep beam.

The critical section is taken as dv from the face of the support IF the reaction is compressive.

The term dv has limits of the greater of 0.72h and 0.9de .Previous editions defined critical section as the larger of dvand 0 5d cotθ but this made the process iterative



and 0.5dvcotθ, but this made the process iterative. Otherwise it is taken at the face of the support.At interior supports, the critical section on each side of the support must be determined separately based on loading conditions.


§ 5.8.3.2 Sections Near Supports

In 2007, Article 5.8.3.2 introduces a limit on average shear stress, vu, at any design section:

If the value of vu > 0.18fc’, AND the flexural element is NOT integral with the support then strut and tie model (Article 5 6 3) must be used for

−=

dbVV

vvv

puu φ

φ(5.8.2.9-1)



the support, then strut and tie model (Article 5.6.3) must be used for analysis.

13



The first step in the Sectional Design Model is to determine if the section has at least the minimum amount of transverse steel (stirrups).Minimum transverse reinforcing (stirrups) are needed if:

Vu > 0.5φ(Vc + Vp)

However, Vc cannot be determined until β is found from

(5.8.2.4-1)



However, Vc cannot be determined until β is found from tables, but the tables used to find β are different depending on whether there are minimum stirrups or not. It is probably best to put minimum stirrups throughout the entire beam to avoid excessive iterations.



The previous slide shows the first problem with this method. The term θ is found in a table which dependsmethod. The term θ is found in a table which depends on whether or not there are minimum stirrups. However, to find if minimum stirrups are needed, it is necessary to know Vc which depends on θ. Thus, the engineer must make an assumption about whether minimum stirrups are needed to determine which table is needed for θ. The table for θ depends on whether or not minimum



stirrups are PROVIDED, not whether or not they are required. Thus, by always specifying minimum stirrups, iterations between the tables can be avoided.

14


§ 5.8.2.7 Maximum Spacing of Transverse Reinforcement

The maximum spacing of stirrups is, smax is:If vu < 0.125 fc’

−=

VVv pu φ

u csmax = 0.8 dv < 24”

If v > 0.125 fc’smax = 0.4 dv < 12”

The minimum area of stirrups is:

dbv

vvu φ

sb

(5.8.2.9-1)

(5.8.2.7-1)

(5.8.2.7-2)



Note: If torsion must be considered, Vu in the equation for v must be modified for torsion (as given in Eq’ns 5.8.2.1-6 and 7). This will be explained later in the torsion section.

This provision does NOT apply to segmental boxes. Different equations are used.

'0316.0min, =fsbfAy

vcv (5.8.2.5-1)


§ 5.8.2.5 Minimum Transverse Reinforcement

For sections with at least the minimum amount of transverse steel (stirrups):transverse steel (stirrups):

A value of θ is assumed and this is used to find the term εx (the formulae for εx are shown on the following slides).

The LRFD Tables, which are based on vu /fc’ and εx, are used to find values of β and θ. If θ is close to the assumed value, then Vn can be calculated. If it is not close, iteration is



needed.

15



To avoid iteration, it is permissible to assume the term 0.5(V -V )cotθ = (V -V ) in the following equations (i.e.0.5(Vu Vp)cotθ (Vu Vp) in the following equations (i.e. 0.5cotθ = 1). (Commentary – C5.8.3.4.2 paragraph 4).

This means cotθ can be assumed = 2. For cotθ = 2, θ = 26o, the most conservative, reasonable angle.





cot5.05.0 −−++ fAVVNdM

popspuuu θ

(5 8 3 4 2 1)

εx = longitudinal strain at 0.5dv . The initial value should be < 0.001.This equation ASSUMES cracked section and is only forbeams where at least the minimum amount of transversereinforcing (stirrups) is provided.

( )2 +=

AEAEd

pspss

vxε

(5.8.3.4.2-1)



g ( p ) p

Note: If torsion must be considered, Vu in the equation must be modified for torsion (as given in Eq’ns 5.8.2.1-6 and 7). This will be explained later in the torsion section.

Again, this equation is used if at least minimum stirrups are provided, not whether or not they are required.

16



Really Important Definitions:

The flexural tension side of a beam is the (½ h) on the flexural tension side. In all the equations for shear which require a value of the area of the longitudinal tensile steel, As or Aps , ONLY the steel on the flexural tension side counts. Tensile steel on the flexural compression side (the ½ h on the flexural



on the flexural compression side (the ½ h on the flexural compression side) or compression steel is NOT counted for shear strength.





Definition of “flexural tension side”, the term Ac, and the term εx for cases with and without minimum stirrups.

17



popspuuv

u fAVVNdM

−−++ cot5.05.0 θ

The first term in the numerator, Mu / dv , is the tensile force in the flanges due to the moment. The dv is shear depth = de – a/2.

( )pspss

vx AEAE +=

2ε





popspuuv

u fAVVNdM

−−++ cot5.05.0 θ

The second term in the numerator, Nu, is any APPLIED axial force (not prestressing force). It is assumed that ½ of the axial load is taken by each flange. If the load is compressive, Nu is negative.

( )pspss

vx AEAE +=

2ε



18



( )popspuu

v

u

x AEAE

fAVVNdM

+

−−++=

2

cot5.05.0 θε

The third term in the numerator, (Vu – Vp )cotθ, is the axial force component of strut force and the inclined force from any harped tendons, as shown in the force triangle. Half the force is assumed to be taken by the tensile flange and the other half by the compression flange.

( )pspss AEAE +2





popspuuv

u fAVVNdM

−−++ cot5.05.0 θ

The last term in the numerator, Apsfpo corrects for the strain in the prestressing steel due to prestressing. The term fpois the “locked in” stress in the prestressing steel, usually taken as 0.7fpu (LRFD Art. 5.4.8.3.2), unless the section

( )pspss

vx AEAE +=

2ε



pubeing considered is within the transfer length. If the section is within the transfer length, the value of fpo must be reduced to reflect the lack of development (e.g. if the section is at ½ the transfer length, fpo = 0.35fpu).

19



popspuuv

u fAVVNdM

−−++ cot5.05.0 θ

The denominator is the stiffness of the tensile side. Notice that this equation ASSUMES cracking. If the section doesn’t crack (εx is negative), the effect of the uncracked concrete must be considered.

( )pspss

vx AEAE +=

2ε





If Equation 5.8.3.4.2-1 is used and εx < 0; the section has not cracked. The effect of the uncracked concrete must benot cracked. The effect of the uncracked concrete must be considered and the equation becomes:

Ac is the area of concrete on the tension half of the section.

( )2

cot5.05.0

++

−−++=

AEAEAE

fAVVNdM

ccpspss

popspuuv

u

x

θε (5.8.3.4.2-3)



Ac is the area of concrete on the tension half of the section.


20


§ 5.8.3.4 Determination of β and θ

Once the values of vu /fc’ and εx are calculated, use the table in the LRFD Specifications to find θ and β. If the value of θ is close to the original assumption, use the β given. If not, use the table value of θ as the next estimate and repeat the calculations of εx .

22.3 20.4 21.0 21.8 24.3 26.6 30.5 33.7 36.4 40.8 43.96.32 4.75 4.10 3.75 3.24 2.94 2.59 2.38 2.23 1.95 1.6718.1 20.4 21.4 22.5 24.9 27.1 30.8 34.0 36.7 40.8 43.13.79 3.38 3.24 3.14 2.91 2.75 2.50 2.32 2.18 1.93 1.6919.9 21.9 22.8 23.7 25.9 27.9 31.4 34.4 37.0 41.0 43.23.18 2.99 2.94 2.87 2.74 2.62 2.42 2.26 2.13 1.90 1.67

0.1

-0.05 0.00 0.125-0.20 -0.10

0.125

v/f'c

0.075

1.00 1.50 2.00

εx * 1,000

0.25 0.50 0.75≤ ≤ ≤ ≤ ≤ ≤ ≤≤ ≤ ≤ ≤

≤

≤≤

≤

Table 5.8.3.4.2-1 Values of θ and β for Sections with Transverse Reinforcement



21.6 23.3 24.2 25.0 26.9 28.8 32.1 34.9 37.3 40.5 42.82.88 2.79 2.78 2.72 2.60 2.52 2.36 2.21 2.08 1.82 1.6123.2 24.7 25.5 26.2 28.0 29.7 32.7 35.2 36.8 39.7 42.22.73 2.66 2.65 2.60 2.52 2.44 2.28 2.14 1.96 1.71 1.5424.7 26.1 26.7 27.4 29.0 30.6 32.8 34.5 36.1 39.2 41.72.63 2.59 2.52 2.51 2.43 2.37 2.14 1.94 1.79 1.61 1.4726.1 27.3 27.9 28.5 30.0 30.8 32.3 34.0 35.7 38.8 41.42.53 2.45 2.42 2.40 2.34 2.14 1.86 1.73 1.64 1.51 1.3927.5 28.6 29.1 29.7 30.6 31.3 32.8 34.3 35.8 38.6 41.22.39 2.39 2.33 2.33 2.12 1.93 1.70 1.58 1.50 1.38 1.29

0.225

0.25

0.15

0.175

0.2

≤

≤

≤

≤

≤



Notes:It is NOT necessary to interpolate the previous tableIt is NOT necessary to interpolate the previous table. The terms β and θ apply to the range of strains and shear in the table. Taking higher values of εx is acceptable.Example from the Commentary: θ = 34.4o and β=2.26 can be used provided that εx < 0.75x 10-3 and vu/fc’ < 0.125(Commentary – C5 8 3 4 2 paragraph 9)



(Commentary – C5.8.3.4.2 paragraph 9).If 0.5cotθ was assumed = 1, the values of θ and βobtained from the table may be used without further iteration.

21



After finding the value of β and θ :

Vn = Vc + Vs + Vp < 0.25fc’ bv dv + Vp

sdfA

V

dbfV

vyvs

vvcc

θβcot

'0316.0

=

= (5.8.3.3-3)



Vn Vc Vs Vp 0.25fc bv dv Vp

Then Vu < φ Vn



If the section does NOT have at least the minimum required transverse steel (stirrups), two modifications are

d Fi t th t i i th i l it di lmade. First, the strain, εx , is the maximum longitudinal strain in the web. It can be calculated by:

The initial value of εx should < 0.002

( )pspss

popspuuv

u

x AEAE

fAVVNdM

+

−−++=

θε

cot5.05.0



xAs before, the section is assumed to be cracked and0.5cotθ may be taken = 1


22



If the section is not cracked:u fAVVN

M++ θt5050

( )ccpspss

popspuuv

u

x AEAEAE

fAVVNd

++

−−++=

θε

cot5.05.0






The second modification is that a crack spacing parameter, sxe , is used in place of v in the table.parameter, sxe , is used in place of v in the table.

sx = lesser of dv or the spacing of longitudinal steel placed in the web to control cracking. The area of

.8063.0

38.1 ina

ssg

xxe ≤+

=



longitudinal steel in each layer must be at least 0.003 bvsx

ag = maximum aggregate size – inch.

23







Once the values of sxe and εx are calculated, use the table in the LRFD Code for this case to find θ and β. If the valuein the LRFD Code for this case to find θ and β. If the value of θ is close to the original assumption, use the β given. If not, use the table value of θ as the next estimate and repeat the calculations of εx . Iterate (unless 0.5cotθ is assumed = 1). Again, interpolation is not necessary. After finding the value of β and θ :

dbfV vvcc β '0316.0=



Vn = Vc + Vs + Vp < 0.25fc’ bv dv + Vp

Then Vu < φ Vn

sdfA

V vyvs

θcot=

24



Here is the table for beam with less than minimum stirrups:β

* 1 000Table 5.8.3.4.2-2 Values of θ and β for Sections without Transverse Reinforcement

25.4 25.5 25.9 26.4 27.7 28.9 30.9 32.4 33.7 35.6 37.26.36 6.06 5.56 5.15 4.41 3.91 3.26 2.86 2.58 2.21 1.9627.6 27.6 28.3 29.3 31.6 33.5 36.3 38.4 40.1 42.7 44.75.78 5.78 5.38 4.89 4.05 3.52 2.88 2.50 2.23 1.88 1.6529.5 29.5 29.7 31.1 34.1 36.5 39.9 42.4 44.4 47.4 49.75.34 5.34 5.27 4.73 3.82 3.28 2.64 2.26 2.01 1.68 1.4631.2 31.2 31.2 32.3 36.0 38.8 42.7 45.5 47.6 50.9 53.44.99 4.99 4.99 4.61 3.65 3.09 2.46 2.09 1.85 1.52 1.3134 1 34 1 34 1 34 2 38 9 42 3 46 9 50 1 52 6 56 3 59 0

sXE (in)εx * 1,000

-0.20 -0.10 -0.05 0.00 0.125 0.25 0.50 0.75 1.00 1.50 2.00

5

10

15

20

< < < < < < < < < < <

<

<

<

<



34.1 34.1 34.1 34.2 38.9 42.3 46.9 50.1 52.6 56.3 59.04.46 4.46 4.46 4.43 3.39 2.82 2.19 1.84 1.60 1.30 1.1036.6 36.6 36.6 36.6 41.2 45.0 50.2 53.7 56.3 60.2 63.04.06 4.06 4.06 4.06 3.20 2.62 2.00 1.66 1.43 1.14 0.9540.8 40.8 40.8 40.8 44.5 49.2 55.1 58.9 61.8 65.8 68.63.50 3.50 3.50 3.50 2.92 2.32 1.72 1.40 1.18 0.92 0.7544.3 44.3 44.3 44.3 47.1 52.3 58.7 62.8 65.7 69.7 72.43.10 3.10 3.10 3.10 2.71 2.11 1.52 1.21 1.01 0.76 0.62

30

40

60

80

<

<

<

<



Some final notes:The shear must be checked at the critical sections andThe shear must be checked at the critical sections and then at intervals along the beam, usually every 0.1L, and any important points (like harp points) . The values of dv , β and θ must be calculated at each section. As with all concrete members, minimum stirrups are required when Vu > 0.5φ(Vc – Vp)For reinforced concrete members β and θ may be



For reinforced concrete members, β and θ may be taken as 2 and 45o, respectively. Previously, there was a depth limit of 16 inches on this, but this is removed in 2007.

25

S TOAASHTO-LRFDComing in 2007! - Simplified Shear

(or, what goes around, comes around –again, and again and again.)


again, and again and again.)


§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections

Article 5.8.3.4.3 – new in 2007Well, not really new, y

Vci and Vcw return from the Standard Specifications with some modification.

This is the result of a National Co-operative Highway Research Program (NCHRP) study.

Report 549Available on line at www.trb.org – follow the NCHRP links.



Note: Simplified shear has been accepted by the AASHTO Subcommittee on Bridges. However, no change is official until it is actually published. Article and equation numbers are from the proposed article, but these may change for editorial reasons in the final publication.

26



Why the change?According to NCHRP 549:According to NCHRP 549:

Sectional Design Model, as given in the current LRFD Specifications, is still considered too complex.Designers said the process has to be automated.

Automated processes cause the engineers to “lose the feel” of designs.

The old V and V orked ell for man ears



The old Vci and Vcw worked well for many years.Still the ACI 318 method.



Not exactly the old Standard Specifications method.NCHRP 549 suggested 4 changes:NCHRP 549 suggested 4 changes:

Change 1 – The expression for web shear cracking, Vcw, is made more conservative and now also applies to partially prestressed members.



27



Change 2 – The variable angle truss model is used for calculating the contribution of shear reinforcement incalculating the contribution of shear reinforcement in web shear regions. For flexural shear regions where Mu> Mcr, the 45o truss model is used.Change 3 - Maximum shear stress is substantially increased.Change 4 - Minimum shear reinforcement is the same as for the Sectional Design Model



for the Sectional Design Model.



Rules:No significant axial tensionNo significant axial tensionProvide minimum shear reinforcement as given in Art. 5.8.2.5 (same as Sectional Design Model).Take Vp = 0 when finding Vn in Eq’n 5.8.3.3-1.

Then, Vc is the lesser of:Vcw



cw

Vci

As before, the beam is divided into sections and shear is investigated at each section. The critical section is the same as for Sectional Design Model.

28



Vcw

Nominal shear resistance provided by concrete whenNominal shear resistance provided by concrete when inclined cracking results from excessive principal tensions in the web.“Web Shear”

Vci

Nominal shear resistance provided by concrete when



inclined cracking results from combined shear and moment.“Flexural Shear”



A quick reminder.Exactly what are Vci and Vcw?



29



There are two types of shear:Flexural shear where shear cracks grow from flexuralFlexural shear where shear cracks grow from flexural cracks. This is Vci .Web shear where thin webs crack due to high principal tensile stresses. This is Vcw.





Flexural Shear - Vci

A prestressed beam will form a flexural crack when theA prestressed beam will form a flexural crack when the moment at a section reaches Mcre . The shear at the section which exists at the time of cracking is called Vcre . The shear does NOT cause the cracking. The cracking is caused by the moment, Mcre. Vcre is simply the shear which is associated with Mcre.So how is V found?



So how is Vcre found?The simplifying assumption is made that V and Mincrease proportionally.

30



Thus, if V and M increase proportionally, Vcre can be found from this proportionality. Since M is known, it is possiblefrom this proportionality. Since Mu is known, it is possible to find Vu FOR THE LOADING CASE WHICH CAUSES Mu. The equation becomes:

cre

cre

u

u

MV

MV

=



creu

ucre M

MVV =



Experiments have shown that if the shear at the section increases by (0.02√f ’)b d ksi, the flexural crack will growincreases by (0.02√fc )bv dv ksi, the flexural crack will grow into a shear crack.

The flexural shear at the time the crack grows into a shear crack can be written as:

'02.0MdbfVV vvccreci +=



This form of the equation is valid for non-composite members with uniform loads. It is NOT valid for bridges.

)('02.0 ksiVMMdbfV u

u

crevvcci +=

31



Flexural Shear - Vci

It was assumed that the shear and moment increaseIt was assumed that the shear and moment increase proportionally. However, in a composite section or a section with other than uniform loads, the dead load doesn’t increase proportionally, so subtract it out of the proportionality part of the equation.Two new terms are defined:

M = Maximum moment at a section caused by all



Mmax = Maximum moment at a section caused by allFACTORED superimposed loads.

Vi = Shear at the section associated with Mmax.



If the dead load is taken separately, the equation is:MV

Vd = shear due to UNFACTORED dead load – noncomposite section

Mmax = maximum moment at the section due all super-imposed FACTORED loads.

V FACTORED h t th ti di t M

max

'02.0MMVVdbfV crei

dvvcci ++=



Vi = FACTORED shear at the section corresponding to Mmax.bv = minimum web widthdv = effective shear depth

32



There is a lower limit to Vci:

Near simple supports, the Vci equation goes to infinity

'06.0'02.0max

≥++= dbfMMVVdbfV vvc

creidvvcci

(5.8.3.4.3-1)



Near simple supports, the Vci equation goes to infinity because Mmax goes to 0. However, the Vcw equation is finite at supports, so it will control.



How is the cracking moment found?

In a prestressed beam, what will eventually be the tensile fiber will be in compression due to prestressing forces, fcpe. The beam cracks when enough moment is applied to the beam to remove the compressive stress and add enough tension to crack the beam. The usual cracking strength in flexure is the modulus of rupture, fr.



33



Assuming an elastic system:

Where Sc is the section modulus to the tension fiber.

( )rcpeccre

c

crecrercpe

ffSMSM

IcM

ff

+=

==+





Mcre must be adjusted to reflect the fact that the dead load effect has been accounted for. In the LRFD equation, onlyeffect has been accounted for. In the LRFD equation, only the non-composite DL is subtracted:

Mdnc = Moment due to UNFACTORED dead loads applied to the non-composite or monolithic section

12

−+=

SMffSMnc

dnccperccre

(5.8.3.4.3-2)



to the non-composite or monolithic section.Sdnc = Section modulus to the tensile fiber of the non-

composite or monolithic section.In the LRFD equation, Mcre is in inch-k, but Mdnc is in ft-k. The 12 converts feet to inches.

34



Web Shear - VcwIn a beam, there are shear stresses from flexure. The maximum shear ,stress occurs at the neutral axis. For most beams, there is no normal stress at the neutral axis. However, in a prestressed beam there is a normal stress from the P/A term in the stress equation. In a composite beam, the neutral axis of the composite is not the same as in the non-composite. At the neutral axis of the composite section, there will also be normal stresses from bending, caused by the prestressing and the dead load applied to the non-composite section.





Web Shear - VcwThe normal stress, fpc is:

The top sign is used above the non-composite pc

Peff = effective prestressing forceAnc = noncomposite areaInc = noncomposite moment of inertiay = distance between neutral axis of composite and

dl,nc ceff eff cpc

nc nc nc

M yP P eyfA I I

= ±m

neutral axis, the bottom sign is used below the non-composite neutral axis.



yc = distance between neutral axis of composite and noncomposite sections

= 0 for noncomposite beamsMdl,nc= noncomposite dead loadse = eccentricity of prestressing

35



Web Shear - Vcw

V can be calculated using the shearing stress formula:



Vcw can be calculated using the shearing stress formula: v = (Vcw Q)/(It)

where v is the shear stress which causes a maximum principal tensile stress of 4(fc’)1/2 when the normal stress is fpc.



Web Shear - Vcw



An approximate equation is provided to find Vcw:

( )3.0'06.0 ++= VdbffV pvvpcccw (5.8.3.4.3-3)

36



The shear strength of the beam is:V = V + V + V < 0.25f ’ b d + V

(5.8.3.3-1 & 5 8 3 3-2)Vn Vc Vs Vp 0.25fc bv dv Vp

For the simplified method, Vp is taken = 0 in this equation, so:

Vn = Vc + Vs < 0.25fc’ bv dv

V is the lesser of V and V

5.8.3.3-2)



Vc is the lesser of Vci and Vcw .

Vp is taken = 0 only in Equations 5.8.3.3-1 and 5.8.3.3-2 and when the simplified method is used. It is NOT taken = 0 in the equation for Vcw, 5.8.3.4.3-3



'06.0'02.0max

≥++= dbfMMVVdbfV vvc

creidvvcci (5.8.3.4.3-1)

This is the old Vci equation, just adjusted to ksi units, rounded off and

max

12

−+=

SMffSMnc

dnccperccre (5.8.3.4.3-2)



with new notations.

0.02√fc’ ksi = 0.63√fc’; 0.06√fc’ ksi = 1.9√fc’

37



Vd = Shear force at the section from UNFACTOREDdead load (includes DC and DW) (k-in).dead load (includes DC and DW) (k in).

Mcre = Moment causing flexural cracking at a section due to externally applied load (k-in).

Mmax = Maximum factored moment at a section due toexternally applied loads (k-in).

Vi = Shear force at a section due to factored superimposed loads which occurs simultaneously



superimposed loads, which occurs simultaneouslywith Mmax (kip).

Mmax and Vi are found from the load combination causing maximum moment at the section.



fcpe = compressive stress in concrete due to effective prestressing forces only (after loss) at the extremeprestressing forces only (after loss) at the extreme fiber of the section where externally applied loads cause tensile stress.

fr = modulus of rupture. For this provision:

f f

( )ksiff cr '2.0=



Note that this definition of fr is a new bullet in Article 5.4.2.6 (2007).This is the old 6√fc’ just converted to ksi units.

38



Mdnc= total unfactored dead load moment acting on the non composite or monolithic section (k ft)non-composite or monolithic section (k-ft).

Note that this is k-ft. That’s why there’s a 12 in the numerator – converts ft. to in.

Sc = Section modulus to the extreme fiber of the composite section where tensile stress is caused by externally applied loads (in3).S f f



Snc = Section modulus to the extreme fiber of the non-composite or monolithic section where tensilestress is caused by externally applied loads (in3).



For composite members, the commentary allows for a simplification:simplification:

Mmax = Mu – Md Vi = Vu – Vd

(C5.8.3.4.3) 7th Paragraph



Note: The ACI-318 code allows a simplification for non-composite members, however, this simplification was developed for building beams with UNIFORM loads. This simplification should NOT be applied to bridge girders, which are loaded with point (axle) loads.

39



( )3.0'06.0 ++= VdbffV pvvpcccw (5.8.3.4.3-3)

This is an approximate equation for finding the condition where the principal tensile stress is 4√fc’Note that Vp is NOT = 0 in this equation. Vp is only set = 0 when finding Vn in equations 5.8.3.3-1 and 2 and when using the simplified shear method.



using the simplified shear method.Again, this is the old Vcw equation, converted to kip units.



fpc = compressive stress in concrete (after allowance for all prestress loses) at centroid of cross section resistingprestress loses) at centroid of cross section resisting externally applied loads or at the junction of the web and the flange when the centroid lies within the flange (ksi). In a composite section, fpc is the resultant compressive stress at the centroid of the composite section (or at the junction of the web and the flange if the centroid lies in the flange) due to both prestress



and the moments resisted by the precast member acting alone.

40



Stirrups:Recall that if α = 90o (and it almost always is) then:Recall that, if α = 90o (and it almost always is), then:

If Vci < Vcw (in other words, Vci controls), then:

cotθ = 1

sdfA

V vyvs

θcot=



If Vcw < Vci, (Vcw controls) then:

8.1'

30.1cot ≤

+=

ff

c

pcθ (5.8.3.4.3-4)



Simplified Shear - SummaryBasically it is the old V and V method from the StdBasically, it is the old Vci and Vcw method from the Std. Specifications (and ACI 318).The equations are slightly different.

Be sure to use the new version of the equations.The biggest change is needing to find cotθ for finding Vsand longitudinal steel requirements.



If Vci controls, cotθ = 1If Vcw controls, cotθ must be calculated.

41


Lightweight Concrete

This applies to all shear methods, If the splitting strength is known the term √f ’ is replaced by :known the term √fc is replaced by :

If the splitting strength is not known, substitute:

'7.4 cct ff ≤

'75.0 f All lightweight



'

'85.0

75.0

c

c

c

fofplaceIn

f

f g g

Sanded lightweight


Deep Components

Deep Components:Components may be considered as deep components if:Components may be considered as deep components if:

There is a point of zero shear within a distance of 2d from the face of the support.A load causing more than ½ the shear at the support is within 2d of the face of the support (for segmental boxes, the limit is 1/3 the shear).



Design with strut and tie (Article 5.6.3)Detail according to Article 5.13.2.3

42


§ 5.8.3.5 Longitudinal Reinforcement

As shown in the previous slides, the shear forces causeAs shown in the previous slides, the shear forces cause tensile forces in the longitudinal reinforcement. According to the commentary in the LRFD Specifications, this tension becomes larger as θ becomes smaller and Vc gets larger. Therefore, the tensile steel doesn’t only have to resist moment, but it also must resist the tensile component of the shear. It is possible that these tensile forces might be



great enough, when combined with the tensile forces due to moment and axial load, to fail the longitudinal tensile steel. Therefore, a check must be made to assure that there is sufficient tensile steel to resist all the forces.





The longitudinal tensile steel must be able to resist the tension due to bending and axial load, along with the tensile component of shear force in the concrete.

43



≥+

VNM

fAfA yspsps

Note that φ is the appropriate strength reduction factor for that specific load effect (e.g. 1.0 for Mu in prestressed concrete, 0.9 for shear, etc.).

cot5.05.0

−−++ VVVN

dM

spuu

v

u θφφφ (5.8.3.5-1)



There is also a limit of Vs < Vu / φNote: If torsion must be considered, Vu in the equation must be modified for torsion. This will be explained later in the torsion section.



At the inside edge of the bearing area of a simple end support to the section of critical shear:support to the section of critical shear:

cot5.0

−−≥+ VVVfAfA ps

uyspsps θ

φ

(5.8.3.5-2)



44



In Equations 5.8.3.5-1 and 5.8.3.5-2, there is a cotθ term. The value of cotθ depends on the method used. If theThe value of cotθ depends on the method used. If the Sectional Design Model is used, then cotθ is found using the value of θ found from the table.

If the simplified method is used, the value of cotθ depends on which value controls. If Vci controls, then cotθ=1. If Vcwcontrols then cotθ must be calculated:



controls, then cotθ must be calculated: cotθ=1.0+3(fpc/√fc’) < 1.8.



Finally, it is necessary to account for any lack of development of the tensile steel. In the diagram below, thedevelopment of the tensile steel. In the diagram below, the strand/bar may not be fully developed before it reaches the crack. If so, the terms fy and fps must be reduced by the ratio of the actual length/development length.



45



The longitudinal reinforcement does not have to be greater than that required to carry Mu in cases where there is a compressive reaction on the flexural compression face.

In other words – it is not necessary to check this provision at the interior supports of a continuous girder. However, it IS necessary to check this provision for a continuous for live l d i d



load girder.


§ 5.8.4 Interface Shear Transfer – Shear Friction

Interface (horizontal) shear must be considered at:An existing or potential crackAn existing or potential crackAn interface between dissimilar materialsAn interface between two concretes cast at different timesThe interface between different elements of a cross section



This provision appears to be for the vertical interface between flanges and webs of box girders – especially segmental boxes.

46



The factored interface shear resistance. Vri shall be taken as:

ri niV V= φ (5.8.4.1-1)

The design shall satisfy:

ri uiV V≥ (5.8.4.1-2)





Where:

Vni = Nominal Shear Resistance (kip)

Vui = Factored interface shear force due to total load based on the applicable strength and extreme event load combinations in Table 3.4.1-1 (kip)

φ = Resistance factor for shear specified in Article ff f



5.5.4.2.1. In cases where different weights of concrete exist on different sides of the interface, the lower of the two values of φ shall be used.

47



The strength of the interface, Vni, is:

But not greater than the lesser of:

( 8 4 1 )

(5.8.4.1-4)

(5.8.4.1-3)ni cv vf y c

ni 1 c cv

V cA A f P

V K f ' AV K A

= +µ +

≤



(5.8.4.1-5)ni 2 cv

cv vi vi

V K A

A b L

≤

= (5.8.4.1-6)



Vni = Nominal shear resistance (k)A = area of concrete engaged in shear transfer (in2)Acv area of concrete engaged in shear transfer (in )Avf = area of shear reinforcement crossing the shear

plane (in2 )fy= yield strength of reinforcementc = cohesion factorµ = friction factorP = permanent net compressive force normal to the



Pc = permanent net compressive force normal to the shear plane (k). If tensile, Pc = 0.

fc’ = 28 day compressive strength of the WEAKERconcrete

48



bvi = interface width considered to be engaged in shear transfer (inch)transfer (inch)

Lvi = interface length considered to be engaged in shear transfer (inch)

K1 = fraction of the concrete strength available to resist interface shear, as specified in Article 5.8.4.3

K2 = limiting interface shear resistance specified in Article 5 8 4 3 (ksi)



Article 5.8.4.3 (ksi)



Based on consideration of a free body diagram and utilizing the conservative, envelope value of the factored, vertical shear force at

u1ui

vi v

Vvb d

=

the section, Vu1.

(5.8.4.2-1)

Where dv is the previously defined shear depth.

The factored interface shear force in kips/ft for a concrete girder/slab



The factored interface shear force in kips/ft for a concrete girder/slab bridge may be determined as:

ui ui cvV v A= (5.8.4.2-2)

49



If the net (normal) force, Pc , across the interface shear plane is tensile, additional reinforcement shall be provided:additional reinforcement shall be provided:

cvpc

y

PAf

=φ

(5.8.4.2-3)



For beams and girders, the longitudinal spacing of the rows of interface shear transfer reinforcing bars shall not exceed 24 inches.


§ 5.8.4.2 Cohesion and Friction

For concrete placed monolithicallyc = 0.40 ksiµ = 1.4

K1 = 0.25K2 = 1.5 ksi

For normal weight concrete placed against a clean concrete surface, free of laitance and intentionally roughened 0.25 inches

c = 0.24 ksi



µ = 1.0 K1 = 0.25K2 = 1.5 ksi

50


§ 5.8.4.2 Cohesion and FrictionFor concrete anchored to as-rolled structural steel by headed studs or by rebar where all the steel in contact with the concrete is clean and free of paint:

c = 0.025 ksiµ = 0.7K1 = 0.2K2 = 0.8 ksi

For concrete placed against clean, hardened concrete not intentionally roughened but free of laitance and clean

c 0 075 ksi



c = 0.075 ksiµ = 0.6

K1 = 0.2K2 = 0.8 ksi


§ 5.8.4.2 Cohesion and Friction

For lightweight concrete placed against a clean concrete surface, free of laitance and intentionally roughened 0.25 inches

c = 0.24 ksiµ = 1.0 K1 = 0.25K2 = 1.0 ksi

For a cast-in-place concrete slab on clean concrete girder surfaces, free of laitance and intentionally roughened 0.25 inches

c = 0.28 ksi



µ = 1.0 K1 = 0.3K2 = 1.8 ksi – normal weightK2 = 1.3 ksi - lightweight

51



Avf has a minimum:

(5.8.4.4-1)cvvf

y

0.05AAf

≥





For a cast-in-place concrete slab on a clean concrete girder surface, free of laitance:g

The minimum interface shear reinforcement, Avf, need not exceed the lesser of the amount determined from equation 5.8.4.1-1 and the amount needed to resist 1.33Vui /φ as determined using equation 5.8.4.1-3.The minimum reinforcement provisions shall be waived for girder/slab interfaces with surface roughened to an amplitude of 0 25 inches where the



roughened to an amplitude of 0.25 inches, where the factored interface shear stress, vui < 0.210 ksi and all of the vertical shear reinforcement required by Article 5.8.1.1 is extended across the interface and adequately anchored in the slab.

52

S TOAASHTO-LRFDTorsion



§ 5.8.2.1 General

Torsion causes a condition of pure shear, as shown by



element “a”. However, element “a” can be rotated to show principal stresses, as shown in element “c”. For principal stress, two of the normal stresses are tensile and two are compressive.

53


§ 5.8.2.1 General

In a torsion test brittle materials, which are weaker in tension than in shear, will break along surfaces forming a 45 degree angle with the longitudinal axis.



CONCRETE IS A BRITTLE MATERIAL!!!!!!


§ 5.8.2.1 General

Because concrete is brittle and tension weak, torsion forces will crack the member diagonally, perpendicular to the



maximum principal tensile stress. As a result, concrete members under torsional loads tend to ‘unwrap’.

54


§ 5.8.2.1 General

As with shear, compression struts will occur.



Stirrups will arrest the cracks. As with shear, the presence of stirrups (in tension) and compression struts forms a truss, but here the truss is 3-D.


§ 5.8.2.1 General

The important part is this:

Torsion causes shear stresses which are additive to the flexural shear stresses.



55


§ 5.8.2.1 General

General Requirements:

Tu = factored torsional momentTr = factored torsional resistanceTn = nominal torsional resistance given in Article 5.8.3.6

= TT nr φ (5.8.2.1-1)



Tn nominal torsional resistance given in Article 5.8.3.6 (k-in)

φ = 0.9 normal weight concreteφ = 0.7 lightweight concrete


§ 5.8.2.1 General

In many cases, torsional stresses are not significant. Article 5.8.2.1 states that torsional effects may NOT BEArticle 5.8.2.1 states that torsional effects may NOT BE ignored if:

25.0≥ TT cru φ (5.8.2.1-3)

1'12502

+

=fA

fT pccp



(5.8.2.1-4)'125.01125.0 +

=fp

fTcc

ccr

56


§ 5.8.2.1 General

'12501'125.0

2pccp

ccr ffA

fT +

= (5.8.2.1-4)

Tcr = Cracking torsion (k-in)Acp = Total area enclosed by the outside perimeter of

the concrete cross section (in2)pc = length of the outside perimeter of the concrete cross

section (in)f i t i t ( ft ll f

'125.0 ccccr fpf

(5.8.2.1 4)



fpc = compressive stress in concrete (after allowance for all prestress loses) at centroid of cross section resisting externally applied loads or at the junction of the web and the flange when the centroid lies within the flange (ksi). (This is the same as for Vcw).


§ 5.8.2.1 General

2A 789 i



2cp

2 2 2 2c

c

A 789 in

p 26 20 2 8 9 9 23 6 6 8

p 166.4 in

=

= + + + + + + + +

=

57


§ 5.8.2.1 General

For cellular structures:

A 2

A0 = Area enclosed by the shear flow path, including any holes therein.

vc

cp bApA

0

2

2≤ (5.8.2.1-5)




§ 5.8.2.1 General

Torsional Design

For torsion, the area of ADDITIONAL transverse reinforcement is calculated.

The required area of stirrups for shear must be added to the required area of stirrups for the concurrent torsion



(Article 5.8.3.6.1).

58


The Commentary (C5.8.3.6.1) explains the use of the word “concurrent”.word concurrent .It is not appropriate to design for the maximum shear and the maximum torsion (unless they are concurrent).It is appropriate to examine the area of transverse reinforcement required for the maximum shear with the concurrent torsion and the maximum torsion with the concurrent shear Use the largest area required



concurrent shear. Use the largest area required.


When calculating the shear resistance, Vn, several equations require the term Vu. When considering shear q q u gand torsion, the EQUIVALENT factored shear force, Vushall be taken as equal to:

(5.8.2.1-6)2

2

0

:

0.92

h uu

Solid Sections

p TVA

B S i

+



ph = perimeter of the centerline of the closed, transverse torsion reinforcement.

(5.8.2.1-7)

:

2u

uo

Box SectionsT dVA

+

59


§ 5.8.3.6.2 Torsional Resistance

The nominal torsional resistance is:

fAA θ2

At = Area of one leg of closed transverse reinforcement provided for torsion in solid members or the total area of transverse torsion reinforcement in the exterior web of a cellular member.

sfAA

T ytn

θcot2 0= (5.8.3.6.2-1)



CAUTION: The Specifications require that the area of transverse reinforcement for shear be added to that for torsion. However, the transverse reinforcement for shear, Av, includes ALL legs of the stirrups which cross the plane of the shear crack. For torsion, At is the area of ONE leg. Thus, when detailing the reinforcement, it is important to add these areas correctly.


§ 5.8.3.6.3 Longitudinal Reinforcement

The longitudinal steel requirements are modified if torsion must be considered.must be considered.

Solid Sections:

245.05.0cot5.0

22

+

−−++

≥+

ATpVVVN

dM

fAfA

uhsp

uuu

yspsps

φφθ

φφ



2 0

Ad sp

v φφφφ

(5.8.3.6.3-1)

60


§ 5.8.3.6.3 Longitudinal Reinforcement

In box sections, the required amount of ADDITIONAL longitudinal steel is:longitudinal steel is:

2 0

=fApTA

y

hnl (5.8.3.6.3-2)




Design for Shear and Torsion

Step 1Determine if torsion must be consideredDetermine if torsion must be considered.IF Tu < 0.25ΦTcr, torsion may be ignored.

Step 2Determine the maximum factored shear and concurrent factored torsion.Determine the maximum factored torsion and



concurrent factored shear.

61


Design for Shear and Torsion (cont.)

Step 3Modify V to reflect the presence of torsionModify Vu to reflect the presence of torsion.

This is the equivalent factored shear force.

Equations 5.8.2.1-6 or 7For the Sectional Design Model is used for shear, the equivalent factored shear force is used for Vu in the equations for vu and εx.





Step 4Determine the area of transverse shear reinforcementDetermine the area of transverse shear reinforcement needed to resist the maximum value of Vu.Determine the area of transverse shear reinforcement needed to resist the value of Vu concurrent with the maximum torsion.



62



Step 5Determine the area of transverse torsionDetermine the area of transverse torsion reinforcement needed to resist the maximum value of Tu.Determine the area of transverse torsion reinforcement needed to resist the value of Tu concurrent with the maximum shear.





Step 6Add together the areas of transverse reinforcementAdd together the areas of transverse reinforcement required for torsion and shear.

Add the required areas for the cases of maximum shear and concurrent torsion and maximum torsion and concurrent shear. Use the maximum.Remember, the calculated shear area is for ALL the stirrup legs; the calculated torsion area is for ONE leg. Be sure to

dd th tl



add the areas correctly.

63



Check the requirements for longitudinal steel using the equations modified for torsion.equations modified for torsion.

5.8.3.6.3-1 or 2Finally, although the specifications do not say it specifically, it appears that if torsion is present, sectional design model must be used.



1

S TOAASHTO-LRFDContinuous for Live Load


New in 2007

§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous

Article 5.14.1.3 has been extensively revised for 2007.Results of NCHRP Study 12-53.Results of NCHRP Study 12 53.

NCHRP Report 519 (available on the web at TRB.org)This article only applies to bridges intended to be continuous for live load.

This does not apply to bridges designed as simple spans.Some states use “poor boy” continuity. A negative



Some states use poor boy continuity. A negative moment connection is provided in the slab, but no positive moment connection is provided. The bridge is designed as simple spans. 5.14.1.3 does NOT apply to this type of bridge.

2

New in 2007


Construction Sequence



New in 2007


Negative moment reinforcement over a diaphragm



3

New in 2007


A bent strand positive moment connection



New in 2007


Girders carry self weight and slab weight as simple, non-composite spans.composite spans.All superimposed DL and LL carried as continuous, composite spans.Negative moment connection over pier is usually reinforced slab.Creep, shrinkage and temperature may cause girders to camber up causing positive moment



camber up, causing positive moment.Usually in young girdersPositive moment connection required.

4

New in 2007


Over time, creep and shrinkage of the girders may cause additional camber in the girders. This creates a positive moment at the diaphragm which often causes cracking, so positive moment connections are needed. These moments are called “restraint” moments.Experimental evidence shows that this behavior is most prevalent when the girders are very young.



when the girders are very young.When the girders are old, theory says shrinkage of the slab causes the girders to de-camber, resulting in a negative restraint moment at the diaphragm. However, this is not seen in field measurements. Field measurements show the girders camber up until the slab is cast, then every thing “locks up” – no cambering or decambering is seen.

Continuous for Live Load

§ 5.14.1.4.2 Restraint Moments

Methods of analysis are NOT covered in the LRFD Specifications.Specifications.

Many commercial bridge analysis programs will calculate positive moments from creep/shrinkage.PCA EB-14 is a popular hand method.Q-Con Bridge is available for free from WSDOT.

Current analysis methods are questionable.



Creep and shrinkage properties are extremely variable.Analysis results do not match field data for older girders.

5


§ 5.14.1.4.2 Restraint Moments

The MOST important variable is the age of the girders at the time continuity is established (Art 5.14.1.4.4).the time continuity is established (Art 5.14.1.4.4).

If the girders are less than 90 days old when continuity is established:

The engineer must estimate or specify the girder age at continuity.Restraint moments must be calculated.

If the girders are SPECIFIED to be no less than 90



If the girders are SPECIFIED to be no less than 90 days old when continuity is established:

Provide a specified positive moment connectionNo calculations of restraint moments are needed.


§ 5.14.1.4.4 Age of Girder when Continuity is Established

The 90 day specificationAt 90 days approximately 70% of the creep andAt 90 days, approximately 70% of the creep and shrinkage has occurred in the girder. This limits positive moment formation.Experimental evidence shows that girders with a positive moment connection which will resist 1.2Mcrcan still provide continuity even if some cracking is present



present.Using the 90 day rule greatly simplifies design.The 90 day rule is verified by experience in several states.

6


§ 5.14.1.4.4 Age of Girder when Continuity is Established

To use the 90 day rule, the 90 day wait must be in the contract documents.Waiting 90 days may not be practical

Precasters do not want to store for 90 days.Production schedules may be significantly altered if a long lead is needed.In some states, precasters are paid for storage.

The commentary allows the owner to change the 90 day



The commentary allows the owner to change the 90 day wait to the time when ktd = 0.7 (Art. 5.4.2.3.2 and 5.4.2.3.3).


§ 5.14.1.4.5 Degree of Continuity at Various Limit States

When the positive moment connection cracks, some degree of continuity may be lost.

In general, the girders act as simple spans until the cracks close; then act as continuous after the crack closes.The design must consider possible loss of continuity.

If the calculated stress at the bottom of the continuity diaphragm for the combination of superimposed permanent loads, settlement, creep, shrinkage, 50% live load and temperature gradient, if applicable, is compressive, the spans may be considered as fully continuous for all limit states.



If the girders are specified to be at least 90 days old when continuity is established, the spans may be assumed fully continuous for all limit states.Negative moment deck cracking may be neglected.

7


§ 5.14.1.4.6 Service Limit State for Girder Stress Limits

For loads carried as simple spans (including release of prestressing force), the girders must satisfy the tensileprestressing force), the girders must satisfy the tensile stress requirements for prestressed girders (Art. 5.9.4).For the top of the girder at an interior support at service limit state after losses, either:

Treat it as a prestressed girder. Use the prestressed tensile limits and Service III, as applicable.Treat it as a reinforced concrete section



Treat it as a reinforced concrete section.A cast-in-place composite deck slab shall not be subject to the tensile stress limits for the service limit state after losses specified in Table 5.9.4.2.2-1.


§ 5.14.1.4.7 Strength Limit State

The negative moment connection must be able to resist the factored negative moment at the section.the factored negative moment at the section.The positive moment connection must be able to resist the factored restraint moments.



8


§ 5.14.1.4.8 Negative Moment Connections

The most common negative moment connection is a reinforced concrete slab on top the girders.reinforced concrete slab on top the girders.

This is designed as a reinforced concrete section and must meet all applicable provisions. Bars must be properly anchored and splices must be staggered.

Other types of connections are permitted if verified by testing



testing.


§ 5.14.1.4.9 Positive Moment Connections

Positive moment connections resist restraint moments caused by creep and shrinkage of the girders.caused by creep and shrinkage of the girders. Without positive moment connections, the girder/diaphragm interface cracks and continuity is lost. Continuous for Live Load Bridges MUST have positive moment connections.



9



Three types permitted:Leave some of the strand extended from the end ofLeave some of the strand extended from the end of the girder and bend it to a 90o angle.Embed mild steel bars in the end of the girder. These bars have either 90o or 180o hooks into the diaphragm.Any connection verified by analysis/testing to provide adequate resistance Mechanical connections would



adequate resistance. Mechanical connections would be permitted under this section.



The positive moment connection must be designed to resist the factored restraint moments unless the 90 dayresist the factored restraint moments unless the 90 day rule is used.If the connection is designed using restraint moments, the capacity of the connection must be between 0.6 Mcrand 1.2 Mcr.

Mcr is the cracking moment of the gross composite girder cross section at the diaphragm



girder cross section at the diaphragm.Mcr is calculated using the strength of the diaphragm concrete.

10



If the connection is designed using the 90 day rule, the capacity of the connection must be at least 1.2 M .capacity of the connection must be at least 1.2 Mcr.IMPORTANT – The 1.2 Mcr capacity referred to here IS NOT the same 1.2 Mcr referred to in Art. 5.7.3.3.2 (which states that prestressed elements must have a minimum capacity of 1.2 Mcr). Art. 5.7.3.3.2 does NOT apply to positive moment connections.





Bent Bar Type Connection:Connection is made by embedding mild steel in the endConnection is made by embedding mild steel in the end of the girder.Use the provisions for development of straight and bent bar (Art. 5.11) to design the bars. The critical section is the girder/ diaphragm interface.Stagger the ends of the bars in the girder to prevent stress concentrations



stress concentrations.

11



Bent Bar type connection.





Bent Bar Type Connection:Often the bars cannot be installed pre bent (especiallyOften, the bars cannot be installed pre-bent (especially in Bulb-T and I sections). It may be necessary to field bend. Field bend specifications are needed.Embedded bars may increase end zone congestion. To mesh the bars in the diaphragm, the bars must be offset. However, an excessively asymmetrical connection detail will cause uneven bar stress The



connection detail will cause uneven bar stress. The connection should be kept as symmetrical as possible while still allowing meshing.

12



This shows that bent bars extend above the top of the flange. They cannot be installed bent or the forms cannot be closed. They must be installed straight and field bent.





Bent Strand Type Connection:This connection is made by leaving a length of strandThis connection is made by leaving a length of strand extend from the end of the beam.

The strand may be left straight and developed into the diaphragm.The strand may be bent into a 90o hook.

This connection develops the strand for the purposes of



Art. 5.8.3.6.3 (Longitudinal reinforcement).The strands should be symmetrical about the vertical axis of the cross section.

13



Strand stress in bent strand connections is found from:fpsl = (ℓdsh – 8)/0.228 < 150 ksi (5.14.1.4.9-1)psl ( dsh )fpul = (ℓdsh – 8)/0.163

where:ℓdsh = total length of extended strand (IN)fpsl = stress in the strand at the SERVICE limit state.

Cracked section shall be assumed. (KSI)f = stress in the strand at the STRENGTH limit state

(5.14.1.4.9-2)



fpul = stress in the strand at the STRENGTH limit state. (KSI)

Strands shall project at least 8 IN from the face of the girder before they are bent.


§ 5.14.1.4.10 Continuity Diaphragms

The design of continuity diaphragms at interior supports may be based on the strength of the concrete in the precast girders.Precast girders may be embedded into continuity diaphragms. If horizontal diaphragm reinforcement is passed through holes in the precast beam or is attached to the precast element using mechanical connectors, the end precast element shall be designed to resist positive moments caused by superimposed dead loads, live loads, creep and shrinkage of the girders, shrinkage of the deck slab, and temperature effects. Design of the end of the girder shall account for the reduced effect of prestress within the transfer length.



Where ends of girders are not directly opposite each other across a continuity diaphragm, the diaphragm must be designed to transfer forces between girders. Continuity diaphragms shall also be designed for situations where an angle change occurs between opposing girders.

1

AASHTOAASHTO LRFD Bridge Design Specifications –Design Example 1

Simple Span Prestressed Adjacent Box Bridge

RICHARD MILLER


Design Example - Simple-Span Adjacent Box Girder Bridge

Problem Statement and Assumptions

This design example demonstrates the design of a single span, 65 ft. long adjacent box girder bridge with a 30o right forward skew, as shown below. This



References:•“Precast Prestressed Concrete Bridge Design Manual,” Published by Precast/Prestressed concrete Institute

j g g g ,example illustrates the design of typical interior and exterior beams at the critical sections in positive flexure, shear and deflection due to prestressing, dead load, and live load.

2







This problem was chosen to illustrate skew bridge design.

Note: Table 4.6.2.2.2e-1 has an inconsistency. It does not include this type of bridge in the description in the first column, but names it as a cross section type in the second column.

It is assumed the skew factor applies to this structure.



It is assumed the skew factor applies to this structure.

3


1.2.1 Precast Beams

Ohio B33-48 box girder as shownfc’ = 7.0 ksi @ 28 daysf ’ 5 0 k ifci’ = 5.0 ksi

ODOT Bridge Design Manual (BDM) allows a range of strengths.



g gThese are chosen from that range.[BDM 302.5.1.7]


Selecting the Girder Size

The LRFD Specifications were checked against the old Standard Specificationsagainst the old Standard Specifications. LRFD should give a more refined design, but not a radically different design.For prestressed concrete, the difference is usually a few strands one way or the other.



Design tables developed for Standard Specifications can usually be used to approximate the section for initial sizing.

4



When using tables based on the Standard Specifications try to stay in the middle of the designSpecifications, try to stay in the middle of the design range. Sections near either end of the design range may be inadequate.The Precast/Prestressed Concrete Institute (PCI) publishes preliminary design tables in their Bridge Design Manual.



These work when ODOT uses the AASHTO standard section (e.g. Type IV)It will give an approximate section for cases where the ODOT section is not AASHTO standard (boxes).



The B33-48 section was chosen from preliminary design charts in ODOT Design Data Sheets Groupdesign charts in ODOT Design Data Sheets. Group “B” Design (roadway width 36 ft. to 48 ft.). The span of 65 ft is the midrange for this section.The design data sheet suggests using 20 strands, ½” diameter.ODOT requires the use of minimum span to depth



ODOT requires the use of minimum span to depth ratios given in LRFD Article 2.5.2.6.3. For a precast box, the limit is 0.03L = 0.03(65ft)(12in/ft) =23.4 inches < 33 inches OK

5


1.2.3 Prestressing Strand

½ in diameter, low-relaxation ASTM A 415 [ODOT BDM 302.5.1.2a]

ODOT BDM allows either ½ inch or 0.6 inch. Here, ½ inch diameter is chosen.

Area of one strand = 0.153 in2

Ultimate strength, fpu = 270.0 ksi



1.2.4 Reinforcing Bars

GR 60; Yield strength, fy = 60 ksi [BDM 302.5.1.8]Modulus of elasticity, Es = 29,000 ksi


1.2.5 LoadsDiaphragms: 2 - 12” wide at 1/3 points

(ODOT Std. Drawings)Future wearing gsurface: 0.060 ksf (ODOT Std. Drawings)

Barriers: 0.090 k/ft each (ODOT Design Data Sheets)

Truck: HL 93, including dynamic allowance

1 2 6 Bridge Parameters



Single SpanOverall Length: 67 ft.c/c Span: 65 ft.Support: Elastomeric Bearing Pad

1.2.6 Bridge Parameters

6


1.3.1 Non-Composite Section Properties

Area in2 733.5Weight (k/ft) 0.764h (in) 33

yb (in) 16.61yt (in) 16.39I (in4) 108,150Sb (in3) 6,511St (in3) 6,599



1.5133,000 'C C cE K w f= (5.4.2.4-1)

1.533,000 1.0 0.150 5.0 4,300CE ksi= × × =1.533,000 1.0 0.150 7.0 5,072CE ksi= × × =

At Transfer

At Service Loads


Material Properties

It is important to remember that the LRFD Specifications use KSI units The formulaSpecifications use KSI units. The formula given for E is the old E=33w1.5√fc’, just adjusted to ksi units. The K1 factor was added for high strength concrete, but it applies to all concrete. E is heavily influenced by aggregates. At high



y y gg g gstrengths, E is limited by aggregate stiffness. The K1 factor allows the owner or designer to adjust E based on experimental evidence.

7


1.3.2 Assumptions

The current ODOT standard is to tie the girders together with tie rods tightened enough to bring thetogether with tie rods, tightened enough to bring the girders together, but not providing significant lateral post-tensioning. According to the commentary in the LRFD Specifications, for this bridge to be considered to have the girders “sufficiently connected”, a lateral post-tensioning force causing a stress of 0.25 ksi across the keyway is needed Therefore this bridge



across the keyway is needed. Therefore, this bridge will be considered as not being “sufficiently connected”. This changes the distribution factor significantly.


1.4.1 Dead Loads

DC = Dead load of structural components and non structural attachmentsnon-structural attachments

DC Dead Loads carried by the girders:Beam Weight: 0.764 klfDiaphragms: 2 at each 1/3 point

( ) ( ) ( ) ( ) ( )d 2 2

33in 10.5in 48in 11inDC 1 ft 2 diaphragms 0.150kcf 1.75k

− −= =



ODOT specifies a MINIMUM of 3 inches in the Bridge Design Manual, but the Design Data Sheets use a 3.5 inch average to account for camber along the length of beam.

( ) ( ) ( )d 2 2DC 1 ft 2 diaphragms 0.150kcf 1.75k144in / ft

8


1.4.1 Dead Loads

DC Dead Loads carried by the girders (con’t):Asphalt Wearing Surface: at ConstructionAsphalt Wearing Surface: at Construction

DW = future wearing surfaces and future DL

( )( )3.5 4 0.120 0.14012 /ws

inDC ft kcf klfin ft

= =



DW future wearing surfaces and future DLFWS: (0.060 ksf)(4 ft) = 0.240 klf


1.4.1 Dead Loads

An important note on the asphalt wearing surface:

The ODOT standards call for a minimum 3 inch asphalt surface.

However, the ODOT Design Data Sheets call for a 3.5 in surface. Actually, this is the average surface thi k D t b th f b



thickness. Due to camber, the surface may be thicker at the ends of the girder. The surface may be thicker on an individual girder due to differential camber.

9


1.4.1 Dead Loads

fRails: 0.090 klf applied to exterior girders.

In other example problems, barrier/railing loads are distributed equally to all the girders, but Article 4.6.2.2 appears to require a deck to distribute the load equally to all girders. Here, assume the railing load is applied only to the exterior girders



to the exterior girders.


1.4.1.1 DL-Unfactored Shear Forces & Bending Moments

Since this is a simple span beam, the most critical moment is at midspan:s at dspa

( )( )

( )( ) ftk8.1268

ft65klf240.0M

ftk3.515k75.13

ft658

ft65klf140.0klf764.0M

2

DW

2

DC

−==

−=

+

+=



8

10


1.4.2 Live Loads

According to LRFD Article 4.6.1.2.1 vehicular live loading on the roadways of bridges or incidental structures, y gdesignated HL-93, shall consists of a combination of the:

Design truck or design tandem with dynamic allowance. The design truck shall consists of an 8.0 kip front axle and a pair of 32.0 kip back axles. The first and second axle are spaced 14’-0” apart. The space between the rear axles shall be varied between 14.0’ and 30.0’ to produce extreme force effects. The design tandem shall consist of a pair of 25 0 kip axles spaced 4 0’ apart



tandem shall consist of a pair of 25.0 kip axles spaced 4.0 apart. [LRFD Article 3.6.1.2.2 and 3.6.1.2.3]Design lane load shall consist of a load of 0.64 kip/ft uniformly distributed in the longitudinal direction. [LRFD Article 3.6.1.2.4]


1.4.2 Live Loads

Since this is a simple span, the maximum moment from the LANE LOAD occurs when the girder is fully loaded. Thus:LANE LOAD occurs when the girder is fully loaded. Thus:

The HL-93 truck controls for this span length and, since this is a simple span, the maximum moment is:

( )( ) ftk3388

ft65klf640.0M2

Lane,LL −==



896LL,TruckM k ft= −

11


1.4.2 Live Loads

A note on live loads:

The lane load is just a uniform load, so for a simple span the moment is:

M = 0.5wx(L-x)



w = load (klf)L = total spanx = point where moment is calculated.


1.4.2 Live Loads

The HL-93 Truck is treated as a series of axle loads. For a SIMPLE SPAN (only), the maximum moment occurs when th id f th b i ½ b t th lt tthe midspan of the beam is ½ way between the resultant load and the nearest axle load:

The resultant is used only for positioning the loads. It is NOT included in the



analysis.

Don’t you wish you would have paid more attention in Structural Analysis?????

12


1.4.2 Live Loads

The HL-93 has the same axle loads as the old HS-20 truck. The Standard Specifications published moments for simple p p pspans under the old HS-20 loading in Appendix B.

BE CAREFUL – Appendix B gives the moment for the controlling load case which might be either the truck load or the lane load!! Recall that the Standard Specifications use EITHER Lane or Truck; LRFD uses BOTH.



The HS 20 lane load is NOT the same as the HL-93 truck or HL-93 lane!!! (Standard Specification Lane Load has a point load!)


1.4.2.1 Distribution Factors

The live load bending moments and shear forces are determined by using the simplified distribution factor f l [LRFD 4 6 2 2] T th i lifi d li l dformulas [LRFD 4.6.2.2]. To use the simplified live load distribution factor formulas, the following conditions must be met [LRFD 4.6.2.2.1]

Width of deck is constant. OKNumber of beams, Nb > 4. OKOverhang part of the roadway < 3 ft OK

de = 0.23 ftCurvature in plan < Article 4.6.1.2 OK



Curvature in plan Article 4.6.1.2 OKBeam parallel and of same stiffness OKCross Section listed in Table 4.6.2.2.1-1 OK

For a precast concrete box beam with an asphalt surface , the bridge type is (g).

13


1.4.2.1 Distribution Factors

The number of design lanes should be determined by taking the integer part of the ratiodetermined by taking the integer part of the ratio w/12, where w is the clear roadway width in feet between curbs and/or barriers.

w = 48 feetNumber of design lanes = integer part of (48/12) = 4



(3.6.1.1.1)


1.4.2.1.1 Distribution Factors for Bending Moment

DFM = S/DS = width of precast beam (ft)S = width of precast beam (ft)D = (11.5 -NL)+1.4NL(1-0.2C)2 when C < 5D = (11.5 -NL) when C > 5

Range of Applicability:

(Table 4.6.2.2.2b-1)



6LN ≤ 45Skew ≤ °

14



Where:

NL = Number of Lanes = 4C = K(W/L) < KW = Clear width of the bridge = 48 ft.



( )J

I1K µ+=



J is not published for ODOT girders. However, it can be approximated by:can be approximated by:

( )= = =

+ +

∑

2224

4 1180in4AJ 211625inS 27.75in 42.5in 42.5in2t 5.5in 5.5in 5in



A is the area enclosed by the centerline of the box walls.t is the wall thicknessS is the length of the centerline of a box wall.

15



( )+= =

4

4

1 0.2 108150inK 0.783

211625in

( ) ( ) ( )( )

= =

= − + − =

= =

2

211625in48 ftC 0.783 0.57865 ft

D 11.5 4Lanes 1.4 4Lanes 1 0.2 0.578 11.9

S 4 ft 0 336



= = 0.336D 11.9

µ = Poisson’s Ratio = 0.2 [LRFD 5.4.2.5]



Note that for boxes, K can be conservatively taken as 1 The DFM = 0 361 a difference of 8%as 1. The DFM 0.361, a difference of 8%.

Also note that there is only one distribution factor for this case. This is different from other cases where there are factors for one lane loaded and two lanes loaded.



16


1.4.2.1.2 Distribution Factors for Shear Force

Two Lanes Loaded:

DFV = (b/156)0.4 (b/12L)0.1 (I/J)0.05(b/48)

One Lane Loaded:

DFV = (b/130L)0.15 (I/J)0.05



DFV (b/130L) (I/J)

(Table 4.6.2.2.3a-1)



Where DFV = distribution factor for moment for interior beam Provided:interior beam. Provided:

5< Nb < 20 Nb = 12 OK Nb = number of beams

35 < b < 60 b = 48 OK b = beam width, in20 < L < 120 L = 65 OK L = beam span, ft25,000 < J <610,000

J = 211,625 OK



40,000 < I <610,000

I = 108,150 OK

17



For two or more lanes loaded:0 1

For one design lane loaded:

( )

0.10.4 0.0548 48 108150 48 0.456156 12 65 211625 48

= = DFV

0.15 0.0548 108150DFV 0 445



Because I/J is raised to a very small power, assuming I/J = 1 changes the DFV very little. Here, the DFV is about 4% higher if I/J = 1.

( )DFV 0.445

130 65 211625 = =


1.4.2.2 Dynamic Allowance

IM = 33%

Where: IM = dynamic load allowance, applied only to truck load



18


1.4.2.3 Moment Reduction Factor for Skew

For 0 60θ° ≤ ≤ °1.05 0.25 tan 1.0g θ= − ≤

The specifications state that the MOMENT DISTRIBUTION FACTOR in a skewed bridge MAY

( )1.05 0.25 tan 30 0.905= − =og



gbe reduced by this factor. (Table 4.6.2.2.2e-1)

Note: Table 4.6.2.2.2e-1 has an inconsistency. It does not include this type of bridge in the description in the first column, but names it as a cross section type in the second column. It is assumed the skew factor applies to this structure.


1.4.2.4 Unfactored Bending Moments

Unfactored bending moment due to HL-93 truck, per beam:per beam:MLL,Truck= (bending moment per lane)(DFM)(1+IM)(skew factor)= (bending moment per lane)(0.336)(1.33)(0.905)= (bending moment per lane)(0.404) = 896 k-ft (0.404) = 362.3 k-ft



( )

19


1.4.2.4 Unfactored Bending Moments

U f d b di d HL 93 l l dUnfactored bending moment due to HL-93 lane load, per beam:MLL,Lane = (bending moment per lane)(DFM)(skew factor)

= (bending moment per lane)(0.336)(0.905)= 338 k-ft (0.304) = 102.7 k-ft



(Impact is not applied to lane loads.)


1.4.3 Load Combinations

The following limit states are applicable: Service I:

(3.4.1)

Q = 1.00(DC + DW) + 1.00 (LL + IM)Service III:

Q = 1.00(DC + DW) + 0.80(LL + IM)Strength I:

Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)



Fatigue: Does not need to be checked for pretensioned beams designed using the Service III load combination.

20


1.5.1 Service Load Stresses at Midspan

Bottom tensile stress due to applied dead and live loads using load combination Service III:

Where: b

ILLDWDCb S

M8.0MMf +++=

fb = Bottom tensile stresses ksiMDC = Unfactored bending moment due to DC loads kip-ftMDW = Unfactored bending moment due to DW loads kip-ft

MLL+I = Unfactored bending moment due to design vehicular live kip-ft



Box girders are usually controlled by Strength I, but it is difficult to estimate number of strands using Strength I. It is easier to estimate the number of strands using Service III and add a few strands. Final strand patterns can be adjusted, if needed, later.

load including impact,Sb = Section modulus to the bottom fiber in3


1.5.1 Service Load Stresses at Midspan

( ){ }( )515 3 126 8 0 8 362 3 102 7 12k f i / f ( ){ }( )3

515 3 126 8 0 8 362 3 102 7 121 87

6511b

. . . . . k ft in / ftf . ksi

in

+ + + − = =

Remember! For Service III (which applies ONLY to tension in fully prestressed members), the LL factor is 0.8!



21


1.5.2 Tensile Stress Limits for Concrete

'0.19r cf f= (Table 5.9.4.2.2-1)

0.19 7.0 0.503rf ksi= =

1.5.3 Required Number of Strands

The first step is determine the required amount of prestressing stress at the tensile fiber:



( )(1.87 0.503) 1.37

pb b r

pb

f f ff ksi

= −

= − =

prestressing stress at the tensile fiber:



Assume the strands are 2 inches from the bottom of the girder So the strand eccentricity at theof the girder So the strand eccentricity at the midspan is:

If Ppe is the total prestressing force, the stress at the bottom fiber due to prestress is:

( ) (16.61 2.0) 14.61c b bse y y in= − = − =



p

pe pe cpb

b

P P ef

A S= +

22



Now plug in the required recompression stress, fpband solve for P :and solve for Ppe:

kips380

in6511in61.14

in5.7331

ksi37.1P

32

pe =

+

=





Final prestress force per strand ( f t d)(f )(1 l %)= (area of strand)(fpi)(1-losses, %)where fpi = initial prestressing stress before

transfer, ksi = 0.75fpu = 202.5 ksi

Assuming 25% loss of prestress the final



Assuming 25% loss of prestress the final prestressing force per strand after losses is:

(0.153)(202.5)(1 0.25) 23.2kips− =

23



Number of strands required:

This shows a need for at least (18) ½ in

380 16.423.2

=



This shows a need for at least (18) ½ in diameter, 270 ksi, low-lax strands as the strand pattern must be symmetrical.


1.5.4 Strand Pattern

At midspan:



The ODOT design data sheets recommend 20 strands. Use 20 strands.

24



Why 20 strands?

1) Boxes tend to be controlled by strength design, but it is hard to use that for strand estimation. It is easier to use Service III and add a few extra strands.

2) The exterior girders will probably require more strand (maybe starting with the exterior would be a better idea!). It is poor design practice to have the exterior girders have



more strand than the interior. This causes fabrication problems. The interior and exterior girders cannot be made on the same bed at the same time.


2.1 Prestress Losses

Total Prestress Losses:

f f f∆ ∆ ∆

Where:∆fpES = loss due to elastic shortening, ksi∆f = loss due to long-term shrinkage and creep of

(5.9.5.1-1)pT pES pLTf f f∆ = ∆ + ∆



∆fpLT loss due to long term shrinkage and creep of concrete, and relaxation of the steel, ksi

25


2.1.1 Elastic Shortening

Where:

ppES cgp

ct

Ef f

E∆ = (5.9.5.2.3a-1)

Where:

fcgp = The concrete stress at the center of gravity of prestressing tendons due to the prestressing force immediately after the transfer and the self-weight of the member at the section of the maximum moment (ksi).



2g ci i c

cgp

M eP PefA I I

= + −



Ep = Elastic Modulus of the prestressing steel (ksi).

Ect = Elastic Modulus of the concrete at the time of transfer or time of load application (ksi).

Mg = girder self weight at release

( )( )20 764 65 65 1 75 441 4 5300. klf ft ftM k k ft k in = + = =



1 75 441 4 53008 3gM . k . k ft k in= + = − = −

26



In the calculation of Mg c/c bearing is used for length. Some designers use overall length, based on the assumption that th i d ill it it d h l d O ll l ththe girder will sit on its ends when released. Overall length gives “a more accurate Mg”. But consider this:

In this case, the difference in the moment between overall length and c/c bearing is 6%.

M is used for ES losses which includes E E is based on



Mg is used for ES losses, which includes Eci. Eci is based on release strength, which is unknown (what is specified is the MINIMUM; the actual will be above this). The formula for E is accurate to, at best, + 10%.



The ES loss is added to the long term losses and the creep and shrinkage equations used to find the long term lossesand shrinkage equations used to find the long term losses are stated in the commentary to only be accurate + 50%.

The weight of the beam is based on ideal cross section and a UW of 150 pcf. Real concrete has UW varying from 140-160 pcf and there are tolerances in the cross section.



Mg based on c/c bearing is conservative (the Mg term subtracts, so using c/c bearing INCREASES ES) and it will be needed later – so why not just use it here??

27



( )( )( )220 0.9 202.5 0.153 558iP strands ksi in k= =

( ) ( )

( )

2

2 4 4

558 14.61 5300 14.61558 1.15733.5 108150 108150

28500 1 15 7 6

cgp

k in k in inkf ksiin in in

ksif ksi ksi

−= + − =

∆ = =



( )1.15 7.64300pESf ksi ksi

ksi∆ = =



In the calculation of fcgp , the initial stress is assumed to be Thi i i d b A i l 9 2 30.9 fpi . This is permitted by Article 5.9.5.2.3a.

In lieu of this, the commentary permits the calculation of the elastic shortening losses using transformed section. The commentary gives the following equation:

2( )+ −∆ ps pi g m g m g gA f I e A e M A

f



2

( )

( )∆ =

+ +

ps pi g m g m g gpES

g g cips g m g

p

ff A I E

A I e AE

(C5.9.5.2.3a-1)

28


2.1.2 Long-Term Losses

For standard, precast, pretensioned members subject to normal loading and environmental conditions:g

In which:

(5.9.5.3-1)

(5.9.5.3-2)

10 12pi pspLT h st h st pR

g

f Af f

Aγ γ γ γ∆ = + + ∆

1.7 0.01h Hγ = −



(5.9.5.3-3)

h

51 'st

cifγ =

+



H = The average annual ambient relative humidity (%)

γh = Correction factor for relative humidity of the ambient air

γhst = Correction factor for specified concrete strength at time of Prestress transfer to the concrete member

∆fpR = An estimate of relaxation loss taken as 2.5 ksi for



low relaxation strand

29



Assume H = 70%

1 7 0 01(70) 1 00γ

So:

1.7 0.01(70) 1.00hγ = − =

5 0.831 5.0stγ = =+

( )( ) ( ) ( ) ( ) ( ) ( )2202.5 20 0.153

10 1 00 0 83 12 1 00 0 83 2 5ksi in

f∆ = + +



( ) ( ) ( ) ( )210 1.00 0.83 12 1.00 0.83 2.5733.5

7.0 10.0 2.5 19.5

pLT

pLT

fin

f ksi

∆ = + +

∆ = + + =


2.1.3 Total Losses at Service Loads

Total Prestress Losses:f f f∆ ∆ ∆ (5.9.5.1-1)

( )

7 6 19 5 27 1

27 1 100 13 3202 5

202 5 27 1 175 4

pT pES pLT

pT

f f f

f . . . ksi

. ksiLoss % . %. ksi

f ksi ksi ksi

∆ = ∆ + ∆

∆ = + =

= =

= =



Loss is less than the 25% initially assumed, so OK.

202 5 27 1 175 4pef . ksi . ksi . ksi= − =

30


2.2 Compression Stress Limit States

(Table 5.9.4.2.1-1)

Sum of effective prestress + permanent loads

< 0.45fc’

1/2(Sum of effective prestress + permanent loads) + live load

< 0.4 fc’

Sum of effective prestress + permanent < 0 6φ f ’



Sum of effective prestress + permanent loads + transient loads

< 0.6φwfc



So what is this φw term?

It is a modifier for sections with thin webs or flanges. It is actually defined in the section for hollow, rectangular compression members (Art. 5.7.4.7).

It i b d th fl b l th/thi k



It is based on the flange or web length/thickness ratio. Since this is for sections with thin webs/flanges, φw term will usually be = 1 for most beams.

31



uw

Xt

λ = (5.7.4.7.1-1)(5.7.4.7.2c-1)

( )w w

w w w

w w

If 15 1.0If 15 25 1 0.0025 15If 25 35 0.75

λ φλ φ λλ φ

≤ =

< ≤ = − −

< ≤ =

( )

(5.7.4.7.2c-2)

(5.7.4.7.2c-3)



( )uX b lesser of 2z or 2 y= −


2.2.1 Φw

Find the web and flange slenderness ratios:

(5.7.4.7.1-1)X u=λ

Where:

(5.7.4.7.1 1)tw =λ

Xu = the clear length of the constant thickness portion of the wall between other walls or fillets

t = wall thickness

( ) ( )48 2 5 5 2 36 2

in . in in. Bottom Flangeλ

− −= =



The top flange λw < 15 by inspection. If λw < 15, φw = 1.0 (5.7.4.7.2c-1)

( ) ( )

6 25

33 5 5 5 2 32 9

5 5

w

w

. Bottom Flangein

in . in in in. Web

. in

λ

λ− − −

= =

32


2.2.2 Service Load Stresses

Pe =20 strand (0.153in2)(202.5 ksi – 27.1 ksi) = 537 kips

( )[ ]( ) ksi17.1in6599

ft/in12ftk8.1263.515f 3top,cDL =−+

=

( )cp ,top 2 3

537k 14.61in537kf 0.457ksi733.5in 6599in

= − = −



( ){ }( )3

362 3 102 7 120 85

6599cLL,top

. . k ft in / ftf . ksi

in+ −

= =


2.2.3 Service Load Compression Stress Check Service I

( )0 457 1 17 0 713

0 45 0 45 7 3 15cp ,top cDL,top

c

f f . ksi . ksi . ksi

. f ' . ksi . ksi

+ = − + =

< = =( )

0 713 0 85 1 212 2

0 4 7 2 8

c

cp ,top cDL,topcLL,top

f

f f . ksif . ksi . ksi

. ( ksi ) . ksi

++ = + =

< =



Compression stresses OK

( )( )0 713 0 85 1 56

0 6 1 0 7 4cp ,top cDL,top cLL,topf f f . ksi . ksi . ksi

. . ksi

+ + = + =

< = 2. ksi

33


2.3.4 Service Load Tensile Stress Check Service III

The Service III stress at the bottom due to dead and live loads, fb, was calculated previously. The allowable tensileloads, fb, was calculated previously. The allowable tensile stress of 0.530 ksi was also calculated previously.

( )2 3

537 14 61537 1 94733 5 65111 87

1 94 1 87 0 07 0 07

pb

b

k . inkipsf . ksi. in in

f . ksif f k i k i k i k i COMPRESSION

= + =

= −+ +



The section is in COMPRESSION, so the tensile allowable does NOT apply.

1 94 1 87 0 07 0 07pb bf f . ksi . ksi . ksi . ksi COMPRESSION+ = − = + =



Because the bottom of the girder is in compression, check with Service I:

( )2 3

537 14 61537 1 94733 5 65112 04

1 94 2 04 0 1 0 1

= + =

= −

pb

b


f . ksif f k i k i k i k i TENSION

( ){ }( )3

515 3 126 8 362 3 102 7 122 04

6511

+ + + − = =b

. . . . k ft in / ftf . ksi

in



1 94 2 04 0 1 0 1+ = − = − =pb bf f . ksi . ksi . ksi . ksi TENSION

Now it’s in tension, which is Service III ?!?!?!?!

34



So what gives?? Is this a Service III or Service I load gcase??

Actually, it is sort of both. For all intents and purposes, the stress at the bottom of the girder is “0” – and this is a dividing line between Service I and Service III. Because of the 0.8 factor on the LL, there is an inconsistency between



the two load cases. However the stress is so low, that really doesn’t matter – we satisfy all allowables in all cases.


3.1 Factored Moment

Strength I:Q 1 25(DC) 1 50(DW) 1 75(LL IM)Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)Since the truck load and lane load have been distributed and impact is included:

Q = 1.25(DC) + 1.50(DW) + 1.75(Truck + Lane)



( ) ( ) ( )1.25 515.3 1.50 126.8 1.75 362.3 102.71648 19780

u

u

MM k ft k in

= + + +

= − = −

35


3.2 Steel Stress at Strength Limit State

Average stress in prestressing steel when :

Where:

(5.7.3.1.1)1ps pup

cf f kd

= −

fps = Average stress in prestressing steel ksik = 0.28 for low relaxation strands

d Di f i fib



dp = Distance from extreme compression fiber to the centroid of the prestressing tendons = 31 in.

c = Distance between the neutral axis and the compressive face

in.



(5.7.3.1.1-4)' '

'0 85

ps pu s y s y

pu

A f A f A fc f

f b kAβ

+ −=

+

Where:

0.85 pc ps

p

f b kAd

β +

Aps = Area of prestressing steel in2

fpu = Specified tensile strength of prestressing steel = 270 ksi



As = Area of mild steel tension reinforcement = 0.0 in2

fy = Yield strength of tension reinforcement = 60.0 ksi

36



A’s = Area of compression reinforcement = 0.0 in2

f’y = Yield strength of compression reinforcement = 60.0 ksi

f’c = Compressive strength of concrete = 7.0 ksiβ1 = Stress block factor specified in LRFD 5.7.2.2

= 0.70b = Effective width of compression flange = 48 in



b = Effective width of compression flange = 48 in.

To compute c, assume rectangular section behavior, and check if the depth of the equivalent compression stress block, a, is equal to or less than ts: Where a =β1c



( )220 0 153 270 0 03 98 5 5

. in ksic in in

+ −= = <

( )( )( ) ( )( )23 98 5 52700 85 7 0 7 48 0 28 20 0 153

31

3 98270 1 0 28 26031ps

c . in. . in.ksi. ksi . in . . inin

. inf ksi . ksiin

<+

= − =



c is also the neutral axis depth, so the stress block depth, a = β1c = 0.7(3.98) = 2.79 inches. Since c < hf, the stress block is entirely in the flange so the beam may be treated as rectangular.

37


3.3 Flexural Resistance

( ) ff

ha a a aM A f d A f d A' f ' d ' 0.85f ' b b h = − + − − − + − −

The moment equation in the LRFD Specification looks like this:

( )n ps ps p s y s s y s c w fM A f d A f d A f d 0.85f b b h2 2 2 2 2+ +

−−

−+

−=

2'''

22adfAadfAadfAM syssysppspsn

(5.7.3.2.2-1)If the section is rectangular (b=bw), the equation becomes:

If th i i ild t i t l th ti



n ps ps paM A f d2

= −

If there is no compression or mild tension steel, the equation becomes:



Since c < hf, the section may be treated as rectangular.

(5.7.3.2.2-1)

( )

2.79

22 79

n ps ps p

a in

aM A f d

in

=

= −



( )( )2 2.7920 0.153 260 31 235502n

inM in ksi in k in = − = −

38



The nominal flange width of 48 inches was used for “b”. In reality, the flange area is reduced by the shear key cut-out. However, this is often ignored as this would require an iterative procedure. If the area is adjusted for the shear key, the nominal moment, Mn changes by only 0.10%. It may not be appropriate to reduce the area by the shear key cut-out as this will be filled with grout and the grout may act with the base concrete to effectively



the grout may act with the base concrete to effectively provide the complete flange width. All of this is a matter of engineering judgment.


3.4 Determination of Phi

To determine Φ, it is necessary to calculate the steel strain at the level of the extreme tensile steel.at the level of the extreme tensile steel.

c = 3.98 inches (calculated above)dt is the distance to the extreme tensile steel. Since there is only one row of steel, dt = dp.

tt

d c0.003c−

ε =



t

c31in 3.980.003 0.0204

3.98−

ε = =

39


3.4 Determination of Phi

Since εt =0.0204 > 0.005, the section is tension controlled.

Φ = 1.0 (5.5.4.2.1)

(5.7.2.1)

This is a big change from the old ρbalanced method.



However, this now makes the LRFD Specifications consistent with ACI 318. This replaces the maximum reinforcement provisions.


3.5 Determination of Flexural Strength

u nM M≤ Φ

( )( )19,780 1.0 23550k in k in OK− < −



40


3.6 Minimum Reinforcement

For minimum reinforcement, the resistance moment, Mrmust be at least the lesser of 1.2 times the cracking gmoment or 1.33 times the factored applied moment.

1.33Mu = 1.33(19780 k-in) = 26310 k-in

For the cracking moment, find the modulus of rupture:



r cf 0.37 f ' 0.37 7ksi 0.979ksi= = = (5.4.2.6)

Note that this is a new MOR for minimum reinforcement. It is equal to 11.5√fc’ in psi; which is the upper bound for MOR.


3.6 Minimum Reinforcement

Next, determine the stress at the bottom of the box due to effective prestressing force:to effective prestressing force:

( )cpe 2 3

537k 14.61in537kipsf 1.94ksi733.5in 6511in

= + =



41


3.6 Maximum and Minimum Reinforcement

Since this is a non-composite section:

1 2M 1 2(19000k i ) 22800 k i 1 33M

(5.7.3.3.2-1)( )cr b r cpeM S f f= +

( )3crM 6511in 0.979ksi 1.94ksi 19000k in= + = −



1.2Mcr = 1.2(19000k-in) = 22800 k-in < 1.33Mu

Mr = φMn = 1.0(23550) k-in = 23550 k-in > 22800 k-in OK


3.6 Maximum and Minimum Reinforcement

Note: When the number of strands was selected itNote: When the number of strands was selected, it was determined that 18 strands would be needed, but 20 were used. If 18 strands had been used, φMn = 21400 k-in, so 18 strands would NOT meet the minimum requirement.



42


4.1 Steel Stress at Transfer

Assume the stress at transfer is 0.9fpi

Pi = 20 strand(0.153in2)(0.9)(202.5 ksi)=558 kips

Tension:

4.2 Allowable Stress at Transfer (Table 5.9.4.1.2-1)



Tension: 0.0948√fci’ < 0.2 ksi w/o bonded reinforcement 0.24√fci’ w/ bonded reinforcementCompression: 0.6fci’


4.3 End Stress at Transfer

( ) ksi474.0in6599

in61.14k558in5733

kips558f 32pt −=−=

These stresses should be calculated at the end of the

( ) ksi01.2in6511

in61.14k558in5.733

kips558f

in6599in5.733

32pb =+=



transfer length = 60db=30 in. The dead load stresses 30 inches from the support should be added. However, these stresses will not be large so it is conservative to use just the stress due to prestressing. (5.11.4.1)

43


4.3 End Stress at Transfer

fpt = 0.474 ksi tension < 0.24√fci’ = 0.24√5 ksi = 0.537 ksi

OK w/bonded steel

fpb = 2.01 ksi compression < 0.6fci = 0.6(5 ksi) = 3 ksi OK

Because the stress is OK, no debonding is needed. If this calculation had shown debonding was needed, it would have been prudent to recalculate



shown debonding was needed, it would have been prudent to recalculate stresses at the end of the transfer length (include the gravity moment) to see if debonding is still needed. If debonding is needed, no more that 25% of the total number of strands could be debonded and no more than 40% in one row can be debonded.


4.3.1 Bonded Steel

Bonded steel is needed at the top of the girder at the end to take the tensile forces. This steel must resist the total t i i th t fl ith t f th 0 5ftension in the top flange with a stress of no more than 0.5fybut not more than 30 ksi. (Table 5.9.4.1.2-1)

The first step it to find the tension in the flange. This requires the location of the neutral axis to be determined. From the top and bottom stresses at the end, the neutral at the end is:



( ) in30.6ksi01.2474.0in33ksi474.0x =

+=

44


4.3.1 Bonded Steel

The top flange is 5.5 inches, so the stress at the bottom of the top flange is:the top flange is:

( ) ksi0602.0in5.5in3.6in30.6ksi474.0

=−




4.3.1 Bonded Steel

( )( )( )( )0.5 6.30 0.474 5.5 2T in ksi in=

Again, this tension could be reduced by calculating the force at the end of the transfer length (including the gravity

( )( )( )( )

( ) ( )( )0.474 0.060 5.5 48 2 5.52

70.8

ksi ksi in in in

T kips

++ −

=



g ( g g ymoment). Including the gravity moment will reduced the calculated tension, but because bars only come in certain sizes, the reduction may not change the number of bars needed.

45


4.3.1 Bonded Steel

The bonded steel must resist the total tensile force with a stress not exceeding the lesser of 0.5f or 30 ksi.stress not exceeding the lesser of 0.5fy or 30 ksi.

Use 8 #5

(5.9.4.1.2-1)2s in36.2

ksi30kips8.70A ==



The length of the bar is determined by the point where bonded steel is no longer required. Since 0.0948√fci’ = 0.212 ksi > 0.2ksi; find the point where the dead load drops the stress below 0.2 ksi.


4.3.1 Bonded Steel

For simplicity, just consider the beam weight and ignore diaphragms.p g

M = ∆fc St = (0.474 ksi – 0.200 ksi) 6599 in3

= 1808 k-in = 150.7 k-ft

( ) ( )x382.0x83.24ftk7.150

xft65xklf764.05.0ftk7.150M2−=−

−=−=



This is from center of bearing, so extend steel 7.75 ft. from each end and then add development length.

ft25.58;ft75.6xx382.0x83.24ftk7.150

=

46


4.3.1 Bonded Steel

(5.11.2.1.1)yb

c

ybd fd4.0

'ffA25.1≥=l

Where:Ab = Area of the bardb = diameter of bar

f’ = compresive strength of concrete at release

( ) ( )( )2

d

1.25 0.31in 60ksi10.4in 0.4 0.625in 60ksi 15in

5ksi= = < =l



f c = compresive strength of concrete at release

Top bar factor = 1.4 : 1.4(15 inches) = 21 inches

So the minimum bar length = 7’- 9” + 1’ – 9” = 9’ – 6”


4.4 Midspan Stress at Transfer

Mg = 5300 k-in (previously calculated)ink5300

ksi329.0ksi803.0ksi474.0f

ksi814.0in6511

ink5300f

ksi803.0in6599

ink5300f

top

3DL,b

3DL,t

=+−=

−=−−

=

=−

=



By inspection, both are below the compression limit.

ksi20.1ksi814.0ksi01.2f

f

bot

top

=−=

47


5.1 Critical Section - Shear

The critical section is at dv from the face of the support for a section where the reaction force in the direction of thesection where the reaction force in the direction of the applied shear introduces compression into the end region of the member.

For this member with only a single layer of prestressing steel:



(5.8.3.2)v ea 2.79ind d 31in 29.6inches2 2

= − = − =


5.1 Critical Section - Shear

The term dv is not taken less than:

0.9de = 0.9(31 inches) = 27.9 inches < 29.6 inches or 0.72h = 0.72(33 inches) = 23.76 inches < 29.6 inches

Assuming a 1 ft. long bearing pad, the critical section is:



29.6 + 6 = 35.6 inches from center of bearing.

For calculations, use 36 inches = 3 ft. The difference is only a few percent.

48


5.2.1 Basic Shear Forces and Moments at the Critical Section

DC:For beam weight:For beam weight:

For the diaphragm, V = 1.75k (shear is constant),

( ) ( )( )( ) ( )( )( ) ftk0.71ft3ft65ft3klf764.05.0xLwx5.0M

k54.22ft3ft655.0klf764.0xL5.0wV

g

g

−=−=−=

=−=−=



For the diaphragm, V 1.75k (shear is constant), M = 1.75(3) = 5.25k-ft



For the DC wearing surface:

( )( )( )( )( )

0.140 0.5 65 3 4.13

0.5 0.140 3 65 3 13ws

ws

V klf ft ft k

M klf ft ft ft k ft

= − =

= − = −

For the DW wearing surface:( )( )



( )( )( ) ( ) ( )

fws

fws

V 0.240klf 0.5 65 ft 3 ft 7.08k

M 0.5 0.240klf 3 ft 65 ft 3 ft 22.3k ft

= − =

= − = −

49



The shear at x is maximizedLive Load: Consider the influence line for shearThe shear at x is maximized by placing the rear wheel of the truck at x and loading the right part of the beam with the uniform load. (Note that influence lines are NOT used for dead loads.



Obviously, it is not possible to have the DL on only part of the beam!)Now don’t you REALLY wish you wouldn’t have slept in Analysis class?????



Using a standard structural analysis program, at the critical section:section:

VLL,Lane = 18.92 kVLL,Truck = 58.33 kMLL,Lane = 56.76 k-ftMLL,Truck = 175.0 k-ft



LL,Truck

50


5.2.2 Skew Factor

This is a multibeam bridge. The shear at the obtuse corner of each girder MUST be increased by:of each girder MUST be increased by:

Note that this factor applies to the distribution factor.

(Table 4.6.2.2.3c-1)

( )( ) ( ) 20.130tan

in3390ft65121tan

d90L121 =+=+ θ



Since the critical section is only 3 feet from the support, apply the skew factor.


5.2.3 Factored Moments and Shears

As calculated in Section 1.4.2.1.1 of this example:DFV 0 456DFV = 0.456DFM = 0.336

The moment MAY be multiplied by the skew factor for moment, 0.905.



The shear MUST be increased by skew factor, 1.20.

51


5.2.3 Factored Moments and Shears


V 0 456(1 2)[58 33(1 33) 18 92] 52 5 kiVLL+IM = 0.456(1.2)[58.33(1.33)+18.92] = 52.5 kips

Vu = 1.25(22.54k + 1.75k + 4.13 k) + 1.50(7.08 k) + 1.75(52.5 k)= 138.0 kips

MLL+IM = DF(SF)[Truck x IM + Lane]M = 0 336(0 905)[175 k ft(1 33)+56 76] = 88 0 k ft



MLL+IM = 0.336(0.905)[175 k-ft(1.33)+56.76] = 88.0 k-ft

Mu = 1.25(71.0 k-ft + 5.25 k-ft + 13.0 k-ft) +1.5(22.3 k-ft) +1.75(88.0 k-ft) = 299.0 k-ft = 3588 k-in


5.3 Shear Design

For shear design, the shear forces at various points along the girder should be calculated. Normally, this is done atthe girder should be calculated. Normally, this is done at the critical section, at points where strands are debonded or harped and then at every 0.1L.

For this design example, only the shear at the critical section is analyzed. The same procedure for the remaining points would be used



points would be used.

52


5.3 Shear Design

The LRFD Specifications adopted the modified compression field theory for shear design with Version 1. This was called the Sectional Design Model.

In Version 4 (2007), the Simplified Method was added. The Simplified Method restores the old Vci and Vcw from the Standard Specifications.



Both methods will be illustrated in this example.


5.3 Sectional Design Model

The sectional design model requires the calculation of two factors:factors:

Concrete strain at : εx

Average shear stress in the concrete: v

These two values are used to find β and θ; which are then

2vd



These two values are used to find β and θ; which are then used to find the strength of the concrete and the strength of the stirrups.

53


5.3.1.1 Finding εx

Strain at is:2

vd

(5.8.3.4.2-1)

0.5 0.5 ( ) cot0.001

2( )

+ + − −= ≤

+ +

uu u p ps po

vx

s s p ps c c

MN V V A f

dE A E A E A

θε

This equation assumes the section is uncracked. If the



qsection is cracked, Ac in the equation above is =0. This equation also assumes at least minimum stirrups are used.


5.3.1.1 Finding εx

Nu = Applied factored normal force at the specified section = 0

kipssection 0

Vp = Component of the effective prestressing force in the direction of the applied shear = 0

kips

fpo = ksi

Aps = Area of prestressing steel on the flexural t i id f th b 20(0 1 3) 3 06

in2

.70 0.70(270.0) 189puf = =



tension side of the member = 20(0.153) = 3.06

As = Area of nonprestressed steel on the flexural tension side of the member = 0

in2

54


5.3.1.1 Finding εx

Ep = 28,500 ksi

A = Area of concrete on the tension half of the in2Ac Area of concrete on the tension half of the beam2(5.5in)(33in)(0.5) + (48in-11in)(5in) = 366.5

in

dv = 29.6 in

Tension Half of the Box



Tension Half of the Box


5.3.1.1 Finding εx

Note that θ is unknown at this point. However, the commentary allows 0 5cotθ=1 as a simplificationcommentary allows 0.5cotθ=1 as a simplification.

Assuming the section is uncracked, the strain at dv/2 is:

( )

( ) ( )

2

6 32 2

3588 138 3.06 18929.6 82 10 0.08 10

2 28500 3.06 5072 366.5− −

− + −= = − ≈ −

+ x

k in k in ksiin x x

ksi in ksi inε



Negative means “uncracked”, so the assumption of uncracked is correct.

( ) ( )

55


5.3.1.2 Finding vu

u pu

V Vv

b dφ

φ−

= (5.8.2.9)

Where:v vb dφ

vu = Shear stress in concrete Ksibv = Effective web width of the beam = 5.5 inVp = Component of the effective prestressing

force in the direction of the applied shear = 0kips

( )



force in the direction of the applied shear = 0

138 0.469 0.18 ' 1.260.9(2)(5.5)(29.6)

= = < =u cv ksi f ksi


5.3.1.3 β and θ

From LRFD Table 5.8.3.4.2-1:

'

0.469 0.0677.0

u

c

vf

= =

30.08 10x xε −= −



θ = 21.0◦

β = 4.10

56


5.3.1.3 β and θ

v/f'cεx * 1,000

22.3 20.4 21.0 21.8 24.3 26.6 30.5 33.7 36.4 40.8 43.96.32 4.75 4.10 3.75 3.24 2.94 2.59 2.38 2.23 1.95 1.6718.1 20.4 21.4 22.5 24.9 27.1 30.8 34.0 36.7 40.8 43.13.79 3.38 3.24 3.14 2.91 2.75 2.50 2.32 2.18 1.93 1.6919.9 21.9 22.8 23.7 25.9 27.9 31.4 34.4 37.0 41.0 43.23.18 2.99 2.94 2.87 2.74 2.62 2.42 2.26 2.13 1.90 1.6721.6 23.3 24.2 25.0 26.9 28.8 32.1 34.9 37.3 40.5 42.82.88 2.79 2.78 2.72 2.60 2.52 2.36 2.21 2.08 1.82 1.6123.2 24.7 25.5 26.2 28.0 29.7 32.7 35.2 36.8 39.7 42.22.73 2.66 2.65 2.60 2.52 2.44 2.28 2.14 1.96 1.71 1.54

<-0.2 <-0.1

<0.125

<0.15

<0.175

v/f c

<0.075

<0.1

<2<0.25 <0.5 <0.75<-0.05 <0 <0.125 <1 <1.5



24.7 26.1 26.7 27.4 29.0 30.6 32.8 34.5 36.1 39.2 41.72.63 2.59 2.52 2.51 2.43 2.37 2.14 1.94 1.79 1.61 1.4726.1 27.3 27.9 28.5 30.0 30.8 32.3 34.0 35.7 38.8 41.42.53 2.45 2.42 2.40 2.34 2.14 1.86 1.73 1.64 1.51 1.3927.5 28.6 29.1 29.7 30.6 31.3 32.8 34.3 35.8 38.6 41.22.39 2.39 2.33 2.33 2.12 1.93 1.70 1.58 1.50 1.38 1.29<0.25

<0.2

<0.225


5.3.2 Shear Strength of Concrete

The contribution of the concrete to the nominal shear resistance is:resistance is:

(5.8.3.3-3)'0.0316=c c v vV f b dβ

( ) ( )( )cV 0.0316 4.1 7ksi 11in 29.6in 111.6k= =



Since Vu = 138k > φVc = 0.9(111.6k) = 100 k; at least minimum stirrups are needed for strength.The equations for β and θ assumed minimum stirrups.

57


5.3.3 Minimum Stirrups

(5.8.2.7)( )u cv 0.469ksi 0.125f ' 0.125 7ksi 0.875ksi= < = =

smax = 23.7 in.Calculate minimum area of steel using a 12 inch spacing to get area of steel per foot:

( )max vs 0.8d 0.8 29.6in 23.7in 24in= = = < (5.8.2.5)



( )( ) 2vv c

y

11in 12inb sA 0.0316 f ' 0.0316 7ksi 0.184inf 60ksi

≥ = =


5.3.3 Minimum Stirrups

ODOT uses #4 bars with 2 legs as standard (Av = 2(0.2 in2) = 0.4 in2) @ 12 inch o.c.

This is adequate to meet minimum.



58


5.3.4 Shear Strength of the Girder

( )v y vs

A f d cot cot sinV

θ+ α α= (5.8.3.3-4)

The stirrups are perpendicular to the main steel so α = 90o; cotα = 0, sinα=1; θ = 21o

s s

( )

( ) ( ) ( ) ( ) ( )

v y vs

2

A f d cot cot sinV

s0 4in 60ksi 29 6 cot 21 0 1

θ + α α=

+



( ) ( ) ( ) ( ) ( )s

s

0.4in 60ksi 29.6 cot 21 0 1V

12inV 154.2k

+ =

=


5.3.4 Shear Strength of the Girder

n c s pV V V V 111.6k 154.2k 0 265.8k= + + = + + =

( )p

u nV 138k V 0.9 265.8k 239.2k= < φ = =

#4 @ 12 inches is OK. Girder is OK in shear.




59


5.3.5 Maximum Nominal Shear Resistance

The upper limit of Vn, given by following equation, is intended to ensure that the concrete in the webis intended to ensure that the concrete in the web of the beam will not crush prior to yield of the transverse reinforcement.

With Vp=0:

(5.8.3.3-2)'0.25n c v v pV f b d V≤ +



OK

'0.25111.6 154.2 0.25(7)(11)(29.6)265.8 569.8

+ ≤+ ≤≤

c s c v vV V f b d


5.4 Simplified Shear

In the 2007 LRFD Specification, the simplified shear method is introduced.

This method brings back Vci and Vcw from the Standard Specification.

Vcw (web shear) usually controls near the support, so Vcw will be checked at the critical section.



Vci (flexural shear) doesn’t control near the support, so for this example, Vc will be calculated at 0.2L. However, in practice Vc and Vcw must be checked at all appropriate sections.

60


5.4.1 Vcw

( )cw c pc v v pV 0.06 f ' 0.3f b d V= + + (5.8.3.4.3-3)

Where:fpc = compressive stress in concrete (after allowance for

all prestress loses) at centroid of cross section resisting externally applied loads or at the junction of the web and the flange when the centroid lies within the flange (ksi)



within the flange (ksi).

For a composite section, this is the compressive stress in the non-composite section at the composite centroid. For a non-composite section, it is the stress at the centroid.


5.4.1 Vcw

Since this is a non-composite section, the only stress at the centroid is the compressive stress due to the axialthe centroid is the compressive stress due to the axial component of prestressing:

epc 2

P 537kf 0.732ksiA 733.5in

= = =

( )



( )( )( ) ( )cwV 0.06 7ksi 0.3 0.732ksi 11in 29.6in 123.2kips= + =

61


5.4.1 Vcw

The critical section is 29.6 inches from the face of the support. Assuming a 1 ft bearing pad, the critical section issupport. Assuming a 1 ft bearing pad, the critical section is approximately 3.5 feet from the end of the beam. The transfer length is 60 bar diameters = 30 inches. Thus, the critical section is past the transfer length, so fpc does not have to be reduced for lack of bond.

If the critical section is within the transfer length f is



If the critical section is within the transfer length, fpc is reduced linearly.


5.4.1 Vcw

One difference between LRFD and Standard Specifications is that LRFD uses cotθ in the V calculation For V the

(5.8.3.4.3-4)

is that LRFD uses cotθ in the Vs calculation. For Vcw, the term cotθ must be calculated:

pc

c

fcot 1.0 3 1.8

f 'θ = + ≤



o

0.732ksicot 1.0 3 1.83 1.8; so use 1.87ksi

29

θ = + = >

θ =

62


5.4.1 Vcw

The minimum stirrup area and maximum spacing calculated in the Sectional Model still apply here.calculated in the Sectional Model still apply here. Assuming #4 stirrups @ 12 in:

( ) ( ) ( ) ( )2

s

0.4in 60ksi 29.6in 1.8V 106.5k

12in= =



( )uV 138k 0.9 123.2k 106.5k 207k= < + =


5.4.2 Vci

Vci does not control near supports of simply supported beams. It will be calculated at 0.2L = 13 ft from thebeams. It will be calculated at 0.2L 13 ft from the center of the support.

DC:Beam Self-weight:

( ) ( )( )V w 0 5L x 0 764klf 0 5 65ft 13ft 14 9k= = =



( ) ( )( )( ) ( )( )( )

g

g

V w 0.5L x 0.764klf 0.5 65ft 13ft 14.9k

M 0.5wx L x 0.5 0.764klf 13ft 65ft 13ft 258k ft

= − = − =

= − = − = −

63


5.4.2.1 Unfactored Dead Loads

DC:For the diaphragm: V = 1 75 k (shear is constant)For the diaphragm: V = 1.75 k (shear is constant),

M = 1.75(13) =22.8 k-ft

For the wearing surface:

( )( )( ) ( ) ( )

wsV 0.140klf 0.5 65ft 13ft 2.73k= − =



( ) ( ) ( )wsM 0.5 0.140klf 13ft 65ft 13ft 47.3k ft= − = −


5.4.2.1 Unfactored Dead Loads

DW:( )( )V 0 240klf 0 5 65ft 13ft 4 68k

The total UNFACTORED dead load shears and moments are:

( )( )( )( )( )

fws

ws

V 0.240klf 0.5 65ft 13ft 4.68k

M 0.5 0.240klf 13ft 65ft 13ft 81.1k ft

= − =

= − = −



Vd = 14.9k + 1.75k + 2.73k + 4.68k = 24.1kMd = 258.0k-ft + 22.8k-ft +47.3k-ft + 81.1k-ft

= 409.2 k-ft = 4910 k-in

64


5.4.2.1 Factored Dead Loads

The FACTORED shears and moments are:

Vud = 1.25(14.9 k + 1.75 k + 2.73 k) + 1.50(4.68 k) = 31.3 k

Mud = 1.25(258.0k-ft + 22.8k-ft +47.3k-ft) + 1.5(81.1k-ft) 531 8 k ft 6381 k i



= 531.8 k-ft = 6381 k-in


5.4.2.2 Live LoadThis method requires two sets of shears and moments for Live Load. The first is the loading where the shear is maximum and the second is where the moment is maximum.

For the lane load, the shear is maximum when the lane load is on the right 52 ft. of the girder (see the influence line from the sectional model):

VLane1 = 13.3k and MLane1 = 173 k-ft = 2076 k-in



The maximum moment occurs when the lane load is on the entire girder:

VLane2 = 12.5k and MLane2 = 216.3 k-ft = 2596 k-in

65


5.4.2.2 Live Load




5.4.2.2 Live Load



Clearly, the moment is maximum when the lane load is placed along the entire beam. The truck load is less certain. The moment at “X” is the value of the point load times the ordinate of the influence line. Unfortunately, it is not clear where this product will be maximum!

66


5.4.2.2 Live Load

For the truck, it is again necessary to consider two placements:p

Placed for maximum shearPlaced for maximum moment

In this case, it just happens that both are the same – the rear axle placed at 0.2L as shown in the previous slide.



However, this is not always the case. It just happened that way in this example.


5.4.2.2 Live Load

For the truck load, the maximum shear at the section and the maximum moment at the section happen to occurthe maximum moment at the section happen to occur under the same loading – the rear wheel of the truck 13 ft. from the support. In this case, the maximum shear loading and the maximum moment loading are the same, but that is NOT always the case. Be sure to carefully check all reasonable load conditions.



67


5.4.2.2 Live Load

VTruck = 47.2 k and MTruck = 613 k-ft = 7356 k-in

Vu,LL = 1.75[Vtruck(1+IM) + VLane](DFV)Vu,LL = 1.75[47.2k(1.33) + 13.3k]( 0.456) = 60.7k

Note that the skew factor is NOT applied. The skew factor is applied only at the obtuse corner and at 0 2L the section is not at the obtuse corner



and at 0.2L, the section is not at the obtuse corner.


5.4.2.2 Live Load

Mu,LL = 1.75[Mtruck(1+IM) + MLane](DFM)(skew factor)M LL = 1 75[613 k-ft(1 33) + 216 3 k-ft](0 336)(0 905)Mu,LL 1.75[613 k ft(1.33) + 216.3 k ft](0.336)(0.905)

= 549.0 k-ft = MmaxNote that the Skew Factor IS Applied to moment

The shear associated with maximum moment is:Vi = 1.75[47.2k(1.33) + 12.5k]( 0.456) = 60.0 k



Why isn’t Vi = Vu? Vi is the shear associated with maximum moment. For the truck, the same position produced both maximum moment and shear, so Vi for the truck is the same. For the lane, maximum shear occurs with the beam partially loaded, but maximum moment occurs when the beam is fully loaded. Thus, Vi is different for the lane load.

68


5.4.2.3 Determination of Cracking Load for Shear

First, find the modulus of rupture:

(5 4 2 6)f 0 2 f ' 0 2 7k i 0 529k i

Note that LRFD has 3 different MORs – be sure to use the correct one!

Next, determine the stress at the bottom of the box due to

(5.4.2.6)r cf 0.2 f ' 0.2 7ksi 0.529ksi= = =



effective prestressing force:

( )cpe 2 3


= + =



(5.8.3.4.3-2)dnccre c r cpe

12MM S f fS

= + −

Where:

( )ncS

Mdnc=

Unfactored moment due to dead load on the non-composite or monolithic section = 409.2 k-ft (note – in k-ft; 12 in numerator converts to inches)

Snc = non-composite section modulus



Sc = composite section modulus = Snc since this is a non-composite structure

69



( ) ( )3cre 3

cre

12 409.2k ftM 6511in 0.529ksi 1.94ksi

6511inM 11165k in 930.5k ft

− = + −

= − = −




5.4.2.4 Vci

i crei d

V MV 0.02 f 'b d V 0.06 f 'b d= + + ≥ci c v v d c v vmax

V 0.02 f b d V 0.06 f b dM

+ + ≥

(5.8.3.4.3-1)

( )( ) ( )( )ci

60.0k 930.5k ftV 0.02 7ksi 11in 29.6in 24.1k 143.0k

549k ft−

= + + = >−



( )( )549k ft

0.06 7ksi 11in 29.6in 51.7k=

70


5.4.2.5 Check Shear Strength

uV 31.3k 60.7k 92k= + =

Assuming #4 @ 12; It is stated that cotθ=1 for Vci

(5.8.3.4.3)( )( )( )( )2

s

0.4in 60ksi 29.6in 1.0V 59.2k

12in= =



The section is adequate in shear.

( )u nV 92.0k V 0.9 143.0k 59.2k 182.0k= < φ = + =


5.4.2.5 Check Shear Strength

If s=18”

sV 39.5kips=

( )u nV 92.0k V 0.9 143.0k 39.5k 164k= < φ = + =



71


Shear Strength

Why are there different values for Vs ?

Sectional Model: V = 154 2kSectional Model: Vs = 154.2kSimplified Model for Vcw ; Vs = 105kSimplified Model for Vci ; Vs = 59.2k

The answer is the θ angle. For sectional model, θ=21o. For Vcw, θ =29o and for Vci, θ=45o. This affects the number of stirrups which cross the shear crack. The smaller the angle,



p g ,the more stirrups which cross the crack and the higher Vs.


5.6 Minimum Longitudinal Steel

At each section:M 0 5N V

For this example, the minimum longitudinal steel will be checked at the critical section. The critical section 29.6inches from the face of the support. Allowing for a 1 ft. bearing pad and one foot from center of bearing to the end

(5.8.3.5-1)u u ups ps s y p s

v

M 0.5N VA f A f V 0.5V cotd

+ ≥ + + − − θ φ φ φ



of the girder, the critical section is 47.6 inches from the end of the girder. However, it is necessary to see if the strand stress is reduced by lack of development.

72



The development length equation is unchanged for strand from Standard Specifications, except that a factor, κ is dd d Thi f t i th lt f O t b 1988 FHWA

(5.11.4.2)

added. This factor is the result of an October, 1988 FHWA memorandum suggesting the need for this conservative multiplier because of strand/bond problems:

( ) ( )2 21 6 260 175 4 0 5 114 53 3d ps pe bf f d . . . . inκ = − = − =

l



( )The terms fps (steel stress at strength limit) and fpe (effective prestressing stress after losses) were calculated previously.κ = 1.6 for member over 24 inches deep (5.11.4.2).



The critical section occurs at 47.6 inches from the end of the beam, but the development length is 114.5 inches. Thus, the steel stress MUST be reduced to account for lack of development.

( )px bpx pe ps pe

60df f f f

−= + −

l



(5.11.4.2-4)

( )px pe ps ped b60d−l

73



( )px b60df f f f

−+l

( )47 6 30174 5 260 0 174 5 192 0114 5 30

−= + − =

−px. in inf . ksi . ksi . ksi . ksi. in in

( )ppx pe ps pe

d b

f f f f60d

= + −−l





The following values were previously calculated or determined:

A = (0 153in2)(20)(260ksi)= 3 06 in2Aps = (0.153in2)(20)(260ksi)= 3.06 in2

Mu = 3588 k-inVu = 138 kθ = 21o (Sectional Design Model)Vs = 153 k (Sectional Design Model)Nu = Vp = 0



φ = 1 for moment; 0.9 for shearAsfy = assumed 0 (ignore any mild steel)fpe = 175.4 ksifps = 260.0 ksi

74



u u ups ps s y p s


+ ≥ + + − − θ φ φ φ

OK

( )

( ) ( )

23 06 192 0 588

3588 138 0 5 153 21 3211 0 29 6 0 9

=

− > + − =

. in . ksi k

k in k . ( k ) cot k. . in .

vdφ φ φ



OK

Note that before the 2005/06 interim, the steel stress was assumed linear with development length, not bilinear. If the stress were assumed linear here, mild steel would need to be added. Also note that Vs < Vu/φ = 153k



Check the inside face of the bearing pad. Assuming a 12 in pad and one foot from center of bearing to the end, the i id f th d i 12 6 18 i h f th d f th

0 5≥ −up ps s

VA f . V cotθφ

18174 5 104 730

= =

pxinf . ksi . ksiin

inside of the pad is 12+6 = 18 inches from the end of the girder. This is inside the transfer length:



( ) ( )2 1383 06 104 7 320 0 5 153 21 1990 9

= > − =

p p

k. in . ksi k . ( k ) cot k.

φ

OK

75



If the stirrup spacing is increased to 18”, Vs = 103 k

( ) ( )2

0 5

1383 06 104 7 320 0 5 103 21 2650 9

≥ −

= > − =

up ps s

VA f . V cot


θφ



OK


5.7 Anchorage Zone Bursting Stirrups

As in the Standard Specification, LRFD requires bursting stirrups which can resist at least 4% of the

20(0.153)(202.5)(0.04) 24.8rP = =

g pinitial prestressing force, with a stress of no more than 20ksi:

24.8 1.2420

= =sA



This steel must be distributed over h/4 from the end. For this girder, h/4=33/4=8.25 inches. Four #4 double leg stirrups @ 3” provides 1.60 in2 over 8 inches.

76


6.1 Exterior Girder - Moment

The exterior girder takes the railing load (DC):( )20 090 65klf ft( )0 090 65

47 5 5708b

. klf ftM . k ft k in= = − = −

Note: Article 4.6.2.2.1 allows the rail load to be equally distributed to all the girders. However, it does not have to be and, in this case, it is probably more correct to assign the railing to the exterior girder.



g g



The live load moments must be multiplied by the exterior girder factor.

ext int

e

g egde 1.04 125

=

= + >

Two or more lanes loaded:

Since the rail is right at the edge of the box d = half the

(Table 4.6.2.2.2d-1)



Since the rail is right at the edge of the box, de = half the web width = 2.75 inches = 0.23 ft.

0.23e 1.04 1.04925

= + =

77



One lane loaded:ext intg eg= (Table 4 6 2 2 2d-1)

Controls

ede 1.125 130

= + >(Table 4.6.2.2.2d-1)

0.23e 1.125 1.13330

= + =





( ) ( ) ( )( )1 25 515 3 47 5 1 50 126 8 1 75 362 3 102 7 1 1331815 21790

uM . . . . . . . . .M k ft k in

= + + + +

= − = −1815 21790uM k ft k in= =

For the interior box with 20 strands, φMn = 23550 k-in so OK for Mu.

Note that there is only one DFM, so the one lane e is multiplied by the DFM. In the equation above, the truck load (362.3 k-ft) is already



t e equat o abo e, t e t uc oad (36 3 t) s a eadymultiplied by the interior DFM and the impact factor; the lane load (102.7 k-ft) is multiplied by the DFM (no impact on lane load). Thus, it is only necessary to multiply by the increasing factor.

78



Stresses at transfer do not need to be checked as these stress occur during fabrication are independent of the railing load and the live load.

The check performed on the interior girders is sufficient.





Service load stresses should be checked. It is clear by inspection that service load compression stresses are OKinspection that service load compression stresses are OK (see Section 2.3.3). Check Service III:

( ) ( ) ( )( )

3

515 3 47 5 126 8 0 8 362 3 102 7 1 133 1111 1333013330 2 05

6511bottom

M . . . . . . . k ft k ink inf . ksiin

= + + + + = − = −

−= =



fpb = 1.94 ksi compression (previously calculated)

fbottom = 1.94 ksi – 2.05 ksi = -0.110 ksi = 0.110 tension < 0.503 ksi tension OK

79


6.1 Exterior Girder - Shear

This check must be performed at all sections. Only the critical section is shown here. The check is also madecritical section is shown here. The check is also made using Sectional Model.

At the critical section:

( ) ( )( )rV w 0.5L x 0.090klf 0.5 65ft 3ft 2.65k= − = − =



( ) ( )( )( )rM 0.5wx L x 0.5 0.090klf 3ft 65ft 3ft 8.37k ft= − = − = −



Two or more lanes loaded: (Table 4.6.2.2.3b-1)48

ext int

0.5

e

48g egb

bd 212e 1 140

=

+ − = + ≥

0 5



0.5480.23 212e 1 1.234

40

+ − = + =

80



One Lane Loaded: (Table 4.6.2.2.3b-1)ig eg=ext int

e

g egde 1.125 120

= + ≥

0.23e 1.125 1.13720

= + =

Check:Two or more lanes: *DFV 1 234(0 456) 0 562 controls



Two or more lanes: e*DFV = 1.234(0.456) = 0.562 controlsOne Lane: e*DFV = 1.137(0.445) = 0.506

Because there are two DFV, each must be checked!



Vu,LL = 0.562(1.2)[58.33(1.33) + 18.92] = 65.08k

VLL,truck = 58.33kVLL,lane = 18.92kIM = 0.33Skew Factor = 1.2



Vu = 1.25(22.54k + 1.75k + 4.13 k+2.65) + 1.50 (7.08 k) + 1.75(65.08k)= 163.3 k

Using the Sectional Design Model, Mu = 3714k-in, β= 3.24, θ=21.4o, φVn = 215 k, so OK.

81


6.1 Exterior Girder

What about the minimum exterior girder distribution factor? LNfactor?

∑

∑+=

bMinExt N

Ext

b

L

x

eX

NNDF

2,

This DOES NOT apply to adjacent box girder bridges. It only applies to slab/beam



only applies to slab/beam bridges (Types a, e and k) with diaphragms or cross braces.


2.5.2.6.2 Deflection

ODOT invokes Article 2.5.2.6.2,which limits Live Load d fl ti t L/800 f t i l i ddeflection to L/800 for precast, simple span girders.

Camber calculations are not directly addressed in LRFD (They were not directly addressed in the Standard Specifications, either).

The same methods used for finding camber and deflection



The same methods used for finding camber and deflection used for Standard Specifications apply for LRFD Designs.

82



Since this is a limit on FLEXURAL deflection, it is appropriate to use the MDF.

MDF = 0.336(0.905) = 0.304

Lane Load = 0.640(0.304) = 0.194klf

Axle Load (rear) = 32k(1.33)(0.304)=12.9k (includes impact)



Axle Load (front) = 8k(1.33)(0.304) = 3.22k (includes impact)



Here are the live loads positioned for maximum deflection.



Using analysis software: ( )65ft 120.654in 0.975in

800δ = < =

1

AASHTOAASHTO LRFD Bridge Design Specifications –Design Example 2

2 Span Continuous Prestressed I-Girder Bridge

RICHARD MILLER


Design Example - Continuous Two-Span I-Girder Bridge


98’-0”

CL to CL of Bearings CL to CL of Bearings

98’-0”

1’-9”96’-3” 96’-3”

This design example demonstrates the design of a two-span (98 ft. each) AASHTO T pe IV I girder ith no ske as sho n This e ample ill strates the



References:•“Precast Prestressed Concrete Bridge Design Manual,” Published by Precast/Prestressed concrete Institute

AASHTO Type IV – I girder with no skew, as shown. This example illustrates the design of typical interior beam at the critical sections for positive flexure, negative flexure, shear, and the continuity connection.

2



34’-0”

8.5” structural+ 1.0”

4 Spaces @ 8’-0” = 32’-0”2.5’

Type IV

2.5’

wearing

37’-0”



Actual thickness, ts = 9.5 in Structural thickness = 8.5 in.Note that 1.0 in wearing surface is considered to be an integral part of the 8.5 in deck.fc’ = 4.5 ksi @ 28 days Concrete unit weight, wc=0.150 kcf


Precast Beams

AASHTO Type IV girder shownfc’ = 7.0 ksi @ 28 days

8”

6”

1’-8”

fci’ = 4.5 ksiConcrete unit weight, wc=0.150 kcf

4’-6”

1’-11”

9”

8”6”

9”

The ODOT Bridge Design Manual (BDM) gives a range of strengths for the precast. These strengths are chosen from that range The BDM



9

8”

2’-2”

from that range. The BDM also gives the deck strength (302.5.2.8).

3


Prestressing Strand

½ in diameter, low-relaxationArea of one strand = 0.153 in2

Ultimate strength, fpu = 270.0 ksi g , pu

Reinforcing BarsYield strength, fy = 60 ksi Modulus of elasticity, Es = 29,000 ksi (BDM 302.5.2.9)

The ODOT BDM allows ½ inch, ½ inch special or 0.6 inch diameter strand (302.5.2.2a). For this girder, ½ inch diameter is chosen.



LoadsFuture wearing surface: 0.060 ksf (ODOT Std. Drawings)Barriers: 0.640 k/ft eachTruck: HL 93, including dynamic allowance


Non-Composite Section Properties

Area in2 789Weight (lb/ft) 822

LRFD uses ksi units.

h (in) 54yb (in) 24.73yt (in) 29.27I (in4) 260,741Sb (in3) 10,542St (in3) 8,909



1.5133,000 'C C cE K w f= (5.4.2.4-1)

1.533,000 1.0 0.150 4.5 4,067CE ksi= × × =1.533,000 1.0 0.150 7.0 5,072CE ksi= × × =

At Transfer

At Service Loads

4


Effective Flange Width

(1/4) Span = (96.25 ft)(12in/ft)/4 = 289 in 12ts plus the greater of the web thickness or ½ the beam s p gtop flange width:

ts = 8.5 in (slab thickness - use structural thickness only)web thickness = 8 in½ top flange = 0.5(20 in) = 10 in (Greatest)

12(8.5 in) + 10 in = 112 in Average spacing between beams = 8 ft = 96 in



Average spacing between beams = 8 ft = 96 in (CONTROLS)

EFFECTIVE FLANGE WIDTH = 96 in Interior Girder

(4.6.2.6)

96”

76.98”

8.5”


Transformed Section Properties

Transformed flange width = n(effective flange width) =

54”

n(effective flange width) (0.8019)(96) = 76.98 in

Transformed flange area = n(effective flange width)(ts) = (0.8019)(96)(8.5) = 654.35 in2



26”

Note that only the structural thickness of the deck, 8.5 in, is considered. A 2” haunch is assumed for calculating weight but not for finding composite properties (ODOT BDM 302.5.2.3).

5


Properties of Composite Section

Ac = Total area of composite section = 1,443 in2

hc = Overall depth of the composite section = 62.5 inIc = Moment of inertia of the composite section = 666,579 in4

ybc = Distance from the centroid of the composite section to the extreme bottom fiber of the precast beam

= 39.93 in

ytg = Distance from the centroid of the composite section to the extreme top fiber of the precast beam

= 14.07 in

ytc = Distance from the centroid of the composite section to the extreme top fiber of the slab

= 22.57 in.

Sb = Composite section modulus for the extreme bottom fiber of the = 16,694 in3



Sbc Composite section modulus for the extreme bottom fiber of the precast beam

16,694 in

Stg = Composite section modulus for the top fiber of the precast beam = 47,376 in3

Stc = Composite section modulus for extreme top fiber of the deck slab = 29,534 in3


Dead Loads

DC = Dead load of structural components and non-structural attachmentsstructural attachmentsDC Dead Loads carried by the girders:

Beam Weight: 0.822 klfSlab: (96 in)(9.5 in)(0.150 kcf)/(144 in2/ft2) = 0.95 klfHaunch: (2 in)(20 in)(0.150 kcf)/(144 in2/ft2) = 0.042 klf

[ODOT BDM 302.5.2.3]



Note: The actual slab thickness of 9.5” is used in calculating dead loads. The 2” haunch thickness is also used in calculating dead loads.

6


Dead Loads

The intermediate diaphragms are assumed as steel “X” braces These are ignored in these dead loadbraces. These are ignored in these dead load calculations. The weight of each brace is less than 0.3 kips. The moment caused by these braces is << 1% of the total DL moment.




Dead Loads

DC Dead Loads carried by the continuous structure, composite section:According to LRFD Article 4.6.2.2.1 permanent loads may be distributed uniformly amount all beams if the following conditions aredistributed uniformly amount all beams if the following conditions are met:


de = 2.5 ft – 1.5 ft = 1.0 ftCurvature in plan < Specified in Article 4.6.1.2 OKCross Section listed in Table 4 6 2 2 1-1 OK



Cross Section listed in Table 4.6.2.2.1 1 OK

The section meets the criteria, so the loads may be uniformly distributed to the girders.

7


Dead Loads

Future Wearing Surface = 0.060 ksf (0.060 ksf)(34 ft)/5 beams = 0.408 kips/ft/girderODOT Std. Drawings

Partial of Table 4.6.2.2.1-1 - This example is a Type “k”



ODOT Std. Drawings

Barrier = 0.640 klf 2 each (0.640)/5 girders = 0.256 kips/ft/girder


Dead Loads

LRFD Article 4.6.2.2.1 allows the slab weight to be evenly distributed to the girders in the same manner as the wearingdistributed to the girders in the same manner as the wearing surface and the barriers. In this case, the decision has been made to use tributary areas to distribute the slab weight to the girders. Either method is allowable.



8


DL-Unfactored Shear Forces & Bending Moments

LocationBeam Weight [Simple Span]

Deck plus Haunch

[Simple Span]Barrier Weight

[Continuous Span]

Future Wearing Surface

[Continuous Span]Sh M Sh M Sh M Sh M

x ft. x/LShear kips

Mg, kip-ft

Shear kips

Ms, kip-ft

Shear kips

Mb, kip-ft

Shear kips

Mws, kip-ft

0.00 0.00 39.6 0 47.7 0 9.2 7.7 14.7 12.4

9.26 0.10 31.9 331 38.5 399.3 6.8 81.8 10.9 130.5

18.97 0.20 24 602.6 28.9 727 4.3 136 6.9 217

28.69 0.30 16 796.5 19.3 961.1 1.8 166 2.9 264.9

38.41 0.40 8 912.9 9.6 1101.5 -0.6 171.9 -1 274.2

48.13 0.50 0 951.9 0 1148.4 -3.1 153.6 -5 245.1



57.84 0.60 -8 912.9 -9.6 1101.5 -5.6 111.2 -8.9 177.5

67.56 0.70 -16 796.5 -19.3 961.1 -8.1 44.7 -12.9 71.3

77.28 0.80 -24 602.6 -28.9 727 -10.6 -46 -16.9 -73.4

86.99 0.90 -31.9 331 -38.5 399.3 -13.1 -160.8 -20.8 -256.7

96.25 Brg. -39.6 0 -47.7 0 -15.4 -292.7 -24.6 -467.1


Live Loads

According to LRFD Article 4.6.1.2.1 vehicular live loading on the roadways of bridges or incidental structures, designated HL-93, shall consists of a combination of the:,

Design truck or design tandem with dynamic allowance. The design truck shall consists of an 8.0 kip front axle and a pair of 32.0 kip back axles. The first and second axle are spaced 14’-0” apart. The space between the rear axles shall be varied between 14.0’ and 30.0’ to produce extreme force effects. The design tandem shall consist of a pair of



g p25.0 kip axles spaced 4.0’ apart. [LRFD Article 3.6.1.2.2 and 3.6.1.2.3]Design lane load shall consist of a load of 0.64 kip/ft uniformly distributed in the longitudinal direction. [LRFD Article 3.6.1.2.4]

9


Live Loads

For negative moment between inflection points, 90% of the effect of two design trucks (HL-9390% of the effect of two design trucks (HL 93 with 14 ft. axle spacing) spaced at a minimum of 50 ft. combined with 90% of the design lane load.Inflection points are determined by loading all spans with a uniform load.


July 2007

Note: See the Loads Module for a complete explanation of how this is applied.

Do Not Duplicate Loads & Analysis: Slide #17


Distribution Factors

The live load bending moments and shear forces are determined by using the simplified distribution factor formulas [LRFD 4 6 2 2] To use the simplified live loadformulas [LRFD 4.6.2.2]. To use the simplified live load distribution factor formulas, the following conditions must be met [LRFD 4.6.2.2.1]


de = 2.5 ft – 1.5 ft = 1.0 ftCurvature in plan



< Specified in Article 4.6.1.2 OKBeam parallel and of same stiffness OKCross Section listed in Table 4.6.2.2.1-1 OK

For a precast concrete I-girder with cast in place deck, the bridge type is (k).

10



The number of design lanes should be determined by taking the integer part of the ratio w/12, where w is thetaking the integer part of the ratio w/12, where w is the clear roadway width in ft between curbs and/or barriers.

w = 34 feetNumber of design lanes = integer part of (34/12) = 2

N t It ld b d th t thi h ld b d i d th l

(3.6.1.1.1)



Note: It could be argued that this should be designed as a three lane bridge because 3 – 11 ft lanes would fit and the minimum lane width is 10ft. However, the distribution factor is for 2 or more lanes loaded and the number of lanes isn’t in the equation so it doesn’t matter.


Distribution Factors for Bending Moment

For all limit states except for fatigue limit state. For two or more lanes loaded:

Where DFM = distribution factor for moment for interior beam. Provided:

(Table 4.6.2.2.2b-1)0.10.6 0.2

30.0759.5 12

g

s

KS SDFML Lt

= +

3.5 < S < 16.0 S = 8 OK S = Spacing, ft

4.5 < ts < 12.0 ts = 8.5 OK ts = slab thickness, in



20 < L < 240 L = 98 OK L = beam span, ftNb > 4 Nb = 5 OK Nb = number of beams10,000 < Kg <7,000,000

Kg = See next slide

Kg = longitudinal stiffness parameter, in4

11



Where:( )2

g gK n I Ae= + (4.6.2.2.1-1)Where:

n = modular ratio between beam and deck materials

A = cross-section area of the beam (non-composite), in2

= 789

( ) 5,072 1.247( ) 4,067

c

c

E beamE slab

= = =



789I = moment of inertia of the beam (non-composite), in4

= 260,741eg= Distance between the c.g. of beam and slab, in

= (8.5/2+2.0+29.27) = 35.52



( )( )21 247 260 741 789 35 52= +K ( )( )4

1.247 260,741 789 35.52

1,566, 480

= +

=

g

g

K

K in

10,000 < Kg < 7,000,000 OK



g

12



For two or more lanes loaded:

0.10.6 0.2

3

8 8 1,566, 4800.0759.5 98 12*98*8.5

0.665

DFM

DFM

= +

=





For one design lane loaded:0.10.4 0.3 KS S

3

0.10.4 0.3

3

0.0614 12

8 8 1,566, 4800.0614 98 12*98*8.5

0.467

g

s

KS SDFML Lt

DFM

DFM

= +

= +

=



The case of two or more design lanes loaded controls,

DFM = 0.665 lanes/beam

0.467DFM

13


Distribution Factors for Shear Force

For two or more lanes loaded:2S S

Where DFV = distribution factor for moment for interior beam. Provided:

0.212 35S SDFV = + −

(4.6.2.2.1-1)

3.5 < S < 16.0 S = 8 OK S = Spacing, ft



p g,4.5 < ts < 12.0 ts = 8.5 OK ts = slab thickness, in20 < L < 240 L = 98 OK L = beam span, ftNb > 4 Nb = 5 OK Nb = number of beams



For two or more lanes loaded:

28 80.212 35

0.814

DFV

DFV

= + −

=



14




S 0.362580.3625

0.68

SDFV

DFV

DFV

= + = +

=



The case of two or more design lanes loaded controls, DFV = 0.814 lanes/beam


Dynamic Allowance

IM = 33%Where: IM = dynamic load allowance applied only to truckWhere: IM = dynamic load allowance, applied only to truck load



15


Unfactored Shear Force and Bending Moments

Unfactored shear forces and bending moment due to HL-93 truck, per beam:93 truck, per beam:

VLT = (shear force per lane)(DFV)(1+IM)= (shear force per lane)(0.814)(1.33)= (shear force per lane)(1.083) kips

MLT= (bending moment per lane)(DFM)(1+IM)



LT

= (bending moment per lane)(0.665)(1.33) = (bending moment per lane)(0.884) kips-ft



Unfactored shear forces and bending moment due to HL-93 lane load, per beam:93 lane load, per beam:

VLANE = (shear force per lane)(DFV)= (shear force per lane)(0.814) kips

MLANE= (bending moment per lane)(DFM)



LANE

= (bending moment per lane)(0.665) kip-ft

16



Location HL-93 Live Load

Di t S tiMax

ShMax. Positive

M tMax. Negative

M tDistance x ft.

Section x/L

Shear kips

Moment MLL+I, kip-ft

Moment MLL+I, kip-ft

0.00 0.00 89.4 48.5 -5.6

9.26 0.10 76.3 624.6 -83.3

18.97 0.20 62.7 1049.3 -163.4

28.69 0.30 50.1 1300.5 -243.6

38.41 0.40 39.9 1412.4 -323.7

48.13 0.50 -48.3 1386.2 -403.9



57.84 0.60 -60.3 1239.1 -484

67.56 0.70 -72.2 961.1 -564.2

77.28 0.80 -83.8 577.5 -776.2

86.99 0.90 -95 215.9 -877.6

96.25 Brg. -104.6 14.8 -1380.7



Shown in the preceding table are maximum values of shear positive moment and negative moment Theshear, positive moment, and negative moment. The maximum values at a given location are not necessarily from the same load case.



17


Load Combinations


(3.4.1)Service I:



Maximum Q = 1 25(DC) + 1 50(DW) + 1 75(LL + IM)



Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)


Load Combinations

A reminder:

This is a continuous bridge so both maximum and minimum loadThis is a continuous bridge, so both maximum and minimum load combinations must be considered.

Remember, in some cases loads mitigate load effects in other spans, but it is not appropriate to use different load factors for the same analysis. For example, the DC in one span mitigates the positive moment in the other span; but it is not appropriate to use different load factors in this case!



different load factors in this case!

18


Load Combinations

The required number of strands is usually governed by Service III load combination at the section of maximum moment or harp points.

In a continuous for live load structure, the maximum moments do not occur at the same place for each load. The point of maximum moment depends on whether the load was applied to the continuous or simple structure. Thus, each point must be checked for the combinations of loads.

In this structure, the maximum flexural stresses occur at Midspan (48.13)



feet for Service I and Service III (although this is NOT where the continuous load moments are maximum). The Strength I maximum is at 0.4L. It is inappropriate to simply take maximum moments without regard to location along the length of the girder.


Load CombinationsService 1 Service 3 Strength 1 LengthV M V M V Mk k-ft k k-ft k k-ft ft.

200 6 68 6 182 72 58 9 299 125 113 1 Bearing 0200.6 68.6 182.72 58.9 299.125 113.1 Bearing 0192.6 431.7 175.3 393.72 287.45 644.925 Trans. 2.04189.8 549.9 172.7 502.76 283.375 817.925 H/2 2.73164.4 1567.2 149.14 1442.28 246.375 2303.925 0.10L 9.26126.8 2731.9 114.26 2522.04 191.575 3993.775 0.20L 18.9790.1 3489 80.08 3228.9 138.4 5077.725 0.30L 28.6955.9 3872.9 47.92 3590.42 89.575 5615.875 0.40L 38.41

-56.4 3885 -46.74 3607.76 -95.9 5610.625 MidSpan 48.13-92.4 3542.2 -80.34 3294.38 -147.875 5091.675 0.60L 57.84128 5 2834 7 114 06 2642 48 199 95 4041 75 0 70L 67 56



-128.5 2834.7 -114.06 2642.48 -199.95 4041.75 0.70L 67.56-164.2 434 -147.44 589.24 -251.375 -329.31 0.80L 77.28-199.3 -564.8 -180.3 -389.28 -301.825 -1464.58 0.90L 86.99-222.3 -1614.4 -201.94 -1375.8 -334.65 -2795.88 H/2 93.52-224.8 -1742.2 -204.3 -1494.76 -338.2 -2961.82 Trans. 94.21-231.9 -2140.5 -210.98 -1864.36 -348.325 -3482.75 Bearing 96.25

19


Determining Number of Strands from Service Load Stresses at Midspan

At this point it is necessary to determine the neededAt this point, it is necessary to determine the needed number of strands. Box girders tend to be controlled by the Strength Limit State, but “I” girders (this example) tend to be controlled by service load tensions.

The initial estimate of number of strands will be found from the Service III combination. Recall that Service III ONLY



applies to tension in prestressed sections.


Service Load Stresses at Midspan


Where:

(0.8)( )g s b ws LL Ib

b bc

M M M M MfS S

++ + +

= +

fb = Bottom tensile stresses ksiMg = Unfactored bending moment due to beam self-weight, kip-ftMs = Unfactored bending moment due to slab and haunch

i htkip-ft



weights,Mb = Unfactored bending moment due to due to barrier weights, kip-ft

Mws = Unfactored bending moment due to future wearing surface, kip-ftMLL+I = Unfactored bending moment due to design vehicular live

load including impact,kip-ft

20


Service Load Stresses at Midspan

[ ]153.6 245.1 (0.8)(1,386.2) (12)(951.9 1,148.4)(12)10 542 16 694

+ ++= +bf 10,542 16,694

2.39 1.083.47

= +=

b

b

ff ksi

Stress Limits for Concrete

'0.19 cf= (Table 5.9.4.2.2-1)



0.19 7.0 0.503ksi= =Required Compressive Stress From Strands

(3.47 0.503) 2.97pbf ksi= − =


Required Number of Strands

Assume a strand center of gravity at midspan as 8% of the height of the girder.g g

So the strand eccentricity at the midspan is:

0.08(54) 4.32bsy in= =

( ) (24 73 4 32) 20 41b be y y in= − = − =



( ) (24.73 4.32) 20.41c b bse y y in= = =

21



If Ppe is the total prestressing force, the stress at the bottom fiber due to prestress is:p

Now plug in the required recompression stress, fpb and solve form P : ( )

pe pe cpb

b

P P ef

A S= +



solve form Ppe: (20.41)2.97

789 10,542927

pe pe

pe

P P

P kips

= +

=



The required prestressing force after all losses is 927 kips This is after an assumedlosses is 927 kips. This is after an assumed 25% loss. That means the initial prestressing force will be approximately 1240 kips. Check with your local precast producer to ensure the capacity prestressing beds can withstand this force


July 2007

force.


22



Final prestress force per strand = (area of strand)(fpi)(1-losses, %)

h f i iti l t i t b f t f k iwhere fpi = initial prestressing stress before transfer, ksi= 0.75fpu = 202.5 ksi

Assuming 25% loss of prestress the final prestressing force per strand after losses is:

(0.153)(202.5)(1 0.25) 23.2 /F kips strand= − =

927



Number of strands required = strands

Try (40) ½ in diameter, 270 ksi, low-lax strands.

927 39.923.2

=


Strand Pattern

At midspan:

No. Strands

Distance from bottom (in)

7 811 611 411 2



10 Spa.@ 2”

2” 2”

23


Strand Pattern

The distance between the center of gravity of strands and the bottom concrete fiber of the beam is, ybs, is:the bottom concrete fiber of the beam is, ybs, is:

Strand eccentricity at midspan:

[(11)2 (11)4 (11)6 (7)8] 4.7040bsy in+ + +

= =

24 73 4 70 20 0i



24.73 4.70 20.0c b bse y y in= − = − =


Prestress Losses


(5 9 5 1 1)f f f∆ ∆ + ∆

Where:∆fpES = loss due to elastic shortening, ksi∆fpLT = loss due to long-term shrinkage and creep of

concrete, and relaxation of the steel, ksi

(5.9.5.1-1)pT pES pLTf f f∆ = ∆ + ∆



24


Elastic Shortening

∆ = ppES cgp

ct

Ef f

E (5.9.5.2.3a-1)

Where:

2g ci i c M eP Pe

+ −




A I I+ −=

Ep = Elastic Modulus of the prestressing steel (ksi).Ect = Elastic Modulus of the concrete at the time of transfer or

time of load application (ksi).


Elastic Shortening

According to the LRFD Commentary for pretensioned members, the loss due to elastic shortening may bemembers, the loss due to elastic shortening may be determined by the following alternative equation (this is the calculation of elastic shortening loss by transformed section):

2

2

( )

( )

+ −∆ =

+ +


g g ct

A f I e A e M Af A I E

A I A



(C5.9.5.2.3a-1)

2( )+ + g gps g m g

p

A I e AE

25


Elastic Shortening

Aps = Area of prestressing steel, 40(0.153) = 6.12 in2

fpi = Prestressing steel stress immediately prior to transfer, 202.5 ksi

Ag = Gross area of section, 789 in2

Ect = Elastic Modulus of the concrete at transfer, 4,067 ksiEp = Elastic Modulus of the prestressing steel, 28,500 ksiem = Average prestressing steel eccentricity at midspan,

20.0 in



Ig = Moment of inertia of the gross concrete section, 260,741 in4

Mg = Midspan moment due to member self-weight, 951.9(12) = 11,422.8 kip-in


Elastic Shortening

26 12*202 5(260 741 20 0 *789) 20 0*11 422 8*7892

2

6.12*202.5(260,741 20.0 *789) 20.0*11,422.8*789789*260,741*4,0676.12(260,741 20.0 *789)

28,50016.24

+ −∆ =

+ +

∆ =

pES

pES

f

f ksi

Note: If the self weight moment is calculated using total beam length



rather than c/c bearing, the moment becomes 11641 k-in. The elastic shortening loss becomes 16.13 ksi; < 1% different.

26


Long-Term Losses

For standard, precast, pretensioned members subject to normal loading and environmental conditions:g

In which:

(5.9.5.3-1)

(5 9 5 3-2)


g

f Af f

Aγ γ γ γ∆ = + + ∆

1 7 0 01Hγ = −



(5.9.5.3-2)

(5.9.5.3-3)

1.7 0.01h Hγ = −

51 'st

cifγ =

+


Long-Term Losses

Where:

H = The average annual ambient relative humidity (%)γh = Correction factor for relative humidity of the ambient airγhst = Correction factor for specified concrete strength at time

of Prestress transfer to the concrete member∆f = An estimate of relaxation loss taken as 2 5 ksi for low



∆fpR = An estimate of relaxation loss taken as 2.5 ksi for low relaxation strand

27


Long-Term Losses

Assume H = 70%1 7 0 01*70 1 00γ

So:

1.7 0.01*70 1.00= − =hγ

5 0.911 4.5stγ = =+

202.5*6.1210 1.00*0.91 12*1.00*0.91 2.5pLTf∆ = + +



ksi

78914.29 10.92 2.5

27.71

pLT

pLT

pLT

f

f

f

∆ = + +

∆ =


Total Losses at Service Loads


(5.9.5.1-1)

16.24 27.71

43.95

pT pES pLT

pT

pT

f f f

f

f

f

∆ = ∆ + ∆

∆ = +

∆ =



Losses are approximately 22% < 25% OK

202.5 43.95 158.6pef = − =

28


Stresses at Transfer

Force per strand after initial losses:Stress in tendons after transfer:

Force per strand = fpt(strand area) = 186.26(0.153) = 28.50kips

202.5 16.24 186.26pt pi pif f f ksi= − ∆ = − =



Therefore, the total prestressing force after transfer is, Pi = 1,140 kips



In this example, Pi is determined by subtracting the elastic shortening loss from the initial stress.

In the previous example, Pi was found by assuming the stress after transfer was 0.9fpi.

Either method is acceptable. If 0.9fpi is used, Pi = 1115 kips The difference is 2%



kips. The difference is 2%.

29



Compression: 0.60f i’ = 0.60(4.5) = +2.700 ksi (5 9 4 1 1)0.60fci 0.60(4.5) 2.700 ksi

Tension: 1. In areas other than the precompressed tensile zone

and without bonded reinforcement

(5.9.4.1.1)

(5.9.4.1.2)'0.0948 0.2t cif f ksi= ≤



Therefore, 0.200 ksi (CONTROLS)

( )0.0948 4.5 0.2

0.201 0.2tf ksi

ksi ksi= ≤

≤



2. In areas with bonded reinforcement sufficient to resist the tensile force in the concrete computedresist the tensile force in the concrete computed assuming an uncracked section, where reinforcement is proportioned using a stress of 0.5fy, not to exceed 30 ksi.

'0.24 0.24 4.5 0.509= = =t cif f ksi



t cif f

(5.9.4.1.2)

30


Stresses At Transfer Length Section

Stresses at this location need only be checked at release since this stage almost always governs. Also, losses withsince this stage almost always governs. Also, losses with time will reduce the concrete stresses making them less critical.

Transfer length = 60(strand diameter) = 60(0.5) = 30 in = 2.5 ft (5.8.2.3)



The bending moment at a distance 2.5 ft from the end of the beam due to beam self-weight is:

(0.5)(0.822)(2.5)(97.167 2.5) 97.3gM k ft= − = −



Compute top stress at the top fiber of the beam:

1,140 1,140(20.0) 97.3(12)789 8,909 8,909

1 44 2 56 0 13 0 99

= − +

= − +

= − + = −

gi it

t t

t

MP PefA S S

f

f ksi



Tensile stress limit for concrete with bonded reinforcement: 0.509 ksi NG

1.44 2.56 0.13 0.99= + =tf ksi

31



Compute bottom stress at the bottom fiber of the beam:

1,140 1,140(20.0) 97.3(12)789 10,542 10,542

1 44 2 16 0 11 3 49

gi it

b b

t

MP PefA S S

f

f ksi

= + −

= + −

= + − = +



Compressive stress limit for concrete: +2.700 ksi NG

1.44 2.16 0.11 3.49tf ksi= + − = +



Harp 9 strands at the 0.35L points as shown.

At endsAt Midspan At endsNo.

StrandsDistance from

bottom (in)3 523 503 484 88 6

At MidspanNo.

StrandsDistance from

bottom (in)7 8

11 611 411 2



8 411 2

32



4”

9 Strands

2’-6”

48’-7”

50”

14’-7”34’-0”

31 Strands ψ





The distance between the center of gravity of the 9 harped strands at the end of the beam and the top fiber of the precast beam is:

in

The distance between the center of gravity of the 9 harped strands at the harp point and the bottom fiber of the precast beam is:

in

The distance between the center of gravity of the 9 harped strands and

3(2) 3(4) 3(6) 4.009

+ +=

3(4) 3(6) 3(8) 6.009

+ +=



The distance between the center of gravity of the 9 harped strands and the top fiber of the beam at the transfer length section is:

in(54 6 4)4.00 (2.5) 7.2534− −

+ =

33



The distance between the center of gravity of the 31 straight bottom strands and the extreme bottom fiber of the beam is:

in

The distance between the center of gravity of the total number of the strands and the bottom fiber of the precast beam at the transfer length is:

in

11(2) 8(4) 8(6) 4(8) 4.3231

+ + +=

9(54 7.25) 31(4.32) 13.8740

− +=



Eccentricity of the strand group at transfer length is:

in

40

24.73 13.87 10.86− =



The distance between the center of gravity of the total number of the strands and the bottom fiber of the precast beam at the end of the beam is:

in

The eccentricity at the end of the beam is:

in

9(54 4) 31(4.32) 14.6040

− +=

24.73 14.60 10.13− =



34



Recompute top and bottom stresses at the transfer length section using the harped pattern. Concrete stress at the topsection using the harped pattern. Concrete stress at the top fiber of the beam:

1,140 1,140(10.86) 97.3(12)789 8,909 8,909

1.44 1.39 0.13 0.18

t

t

f

f ksi

= − +

= − + = +



Compressive stress limit for concrete: +2.700 ksi OK



At the bottom:

Compressive stress limit for concrete: +2 700 ksi OK

1,140 1,140(10.86) 97.3(12)789 10,542 10,542

1.44 1.17 0.11 2.50

b

b

f

f ksi

= + −

= + − = +




35


Stresses at Harp Points

The strand eccentricity at the harp points is the same as at the midspan,the midspan,

ec = 20.0 in

The bending moment due to beam self-weight at a distance 34.00’ from the end of the beam is:



(0.5)(0.822)(34.00)(97.17 34.00) 882.7gM k ft= − = −



Concrete stress at the top fiber of the beam:

1,140 1,140(20.0) 882.7*12789 8,909 8,909

1.44 2.56 1.19 0.07

= − +

= − +

= − + = +

gi it

t t

t

t

MP PefA S S

f

f



36



Compute bottom stress at the bottom fiber of the beam:

1,140 1,140(20.0) 882.7*12789 10,542 10,542

1 44 2 16 1 00 2 60

gi ib

b b

b

MP PefA S S

f

f

= + −

= + −

= + − = +




1.44 2.16 1.00 2.60bf = + = +


Stresses at Midspan

The bending moment due to beam self-weight at a distance 48’-7” (midspan) from the end of the beam is:


(0.5)(0.822)(48.58)(97.17 48.58) 970.1gM k ft= − = −

= − + gi it

t t

MP PefA S S




1,140 1,140(20.0) 970.1*12 1.44 2.56 1.31 0.19789 8,909 8,909

= − + = − + = +

t t

tf

37


Stresses at Midspan

Compute bottom stress at the bottom fiber of the beam:MP P

Compressive stress limit for concrete: +2 700 ksi OK

1,140 1,140(20.0) 970.1*12 1.44 2.16 1.10 2.50789 10,542 10,542

gi ib

b b

b

MP PefA S S

f

= + −

= + − = + − = +





Hold-Down Forces

Assume that the stress in the strand at the time of prestressing, before any losses, is:prestressing, before any losses, is:

Then, the Prestress force per strand before any losses is:

0.75 0.75(270) 202.5puf ksi= =

' 0.153(202.5) 31.0 /iP k strand= =



Harp angle:1 54 4 6tan 6.2

34(12)ψ − − −= =

o

38


Hold-Down Forces

Therefore, hold-down force per strand = 1 05 (force per strand)(sin ψ)= 1.05 (force per strand)(sin ψ)=1.05(31.0) sin 6.2◦ = 3.5 kips per strand

Note that the factor, 1.05, is applied to account for friction.

Total hold down force = 9 strands(3.5) = 31.6 kips



Total hold down force 9 strands(3.5) 31.6 kips


Hold-Down Forces

ODOT BDM States that the following limits are not to be exceeded:be exceeded:

No. of Draped Strands per Row

PU/Strand(lb)

1 6,0002 4,000



So hold-down force per strand = 3.5 kips per strand OK

3 4,000

39



At transfer, stresses at the end of girder tend to exceed allowables if the strand is straight.

Stresses can be brought within the allowable stress range either by harping or debonding the strand. The question arises as to which is better, harping or debonding?

Boxes tend to use debonding because harping isn’t practical as the strand would go through the void. I and Bulb T girders tend to use harping



harping.

However, not all fabricators have the ability to harp (the bed won’t take the hold down force). Therefore, before deciding to harp, contact probable fabricators or the local PCI section for assistance and advice.


Summary of Stresses at Transfer

Top Stressesft (ksi)

Bottom stressesft (ksi) stressesfb (ksi)

At transfer length section

+0.27 +2.43

At harp points +0.07 +2.60At midspan +0.19 +2.50



Note that the bottom stresses at the harp points are more critical than the ones at midspan.

No Tension! The entire beam is in compression.

40


Summary of Stresses at TransferTop Stress

0.3

0.1

0.15

0.2

0.25

Stre

ss (k

si)

Transfer Length



0

0.05

0 5 10 15 20 25 30 35 40 45 50

Length (ft)

g

HarpPoint

Mid-Span


Summary of Stresses at TransferBottom Stress

3

1

1.5

2

2.5

Stre

ss (k

si)

Transfer Length



0

0.5

0 5 10 15 20 25 30 35 40 45 50

Length (ft)

Length

HarpPoint

Mid-Span

41


Stresses at Service Loads

Total loss of prestress at service loads is

Stress in tendon after all losses

43.95pTf ksi∆ =



202.5 43.95 158.55= −∆ = − =pe pi pTf f f ksi


Stresses at Service Loads

Force per strand = (fpe)(strand area)p

= (158.55)(0.153) = 24.3 kips

The total prestressing force after all losses

Ppe = 24.3(40) = 972.0 kips



42



Compression: (5.9.4.2.1)

Due to permanent loads, for service limit states:

For the precast beam: 0.45fc’ = 0.45(7.0) = +3.150 ksiFor the deck: 0.45fc’ = 0.45(4.5) = +2.025 ksi

Due to one half the permanent loads and live load:



For the precast beam:0.40fc’ = 0.40(7.0) = +2.800 ksiFor the deck: 0.40fc’ = 0.40(4.5) = +1.800 ksi



Compression (con’t): (5.9.4.2.1)

Due to permanent and transient loads for service limit states:

For the precast beam: 0.60Φw fc’ = 0.60(1.0)(7.0) = +4.200 ksi

F th d k 0 60Φ f ’ 0 60(1 0)(4 5) 2 700 k i



For the deck: 0.60Φw fc’ = 0.60(1.0)(4.5) = +2.700 ksi

Note: Φw is a factor for slender webs/flanges. It is not really meant for “I” girders. If the calculations required for Φw are done, Φw=1.

43



Tension:For components with bonded prestressing tendons:p p g

For the precast beam:

'



'0.19 0.19(7.0) 0.503cf ksi= =


Stresses at Midspan

Concrete stress at the top fiber of the beam, three cases:1 Under permanent loads Service I:1. Under permanent loads, Service I:

1

1

( ) ( )

972 972(20.0) (951.9 1,148.4)*12 (153.6 245.1)*12789 8,909 8,909 47,3761 23 2 18 2 83 0 10 1 98

pe pe c g s ws btg

t t tg

tg

P P e M M M MfA S S S

f

f

+ += − + +

+ += − + +

+ + +




1 1.23 2.18 2.83 0.10 1.98tgf = − + + = +

44


Stresses at Midspan

2. One-half permanent loads plus live loads:( )M

2 1

2

2

( )0.5

1,386.2*120.5(1.98)47,376

0.99 0.35 1.34

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +





Stresses at Midspan

3. Under permanent and transient loads:( )M

3

3

3

( )

1,386.2*121.9847,376

1.98 0.35 2.33

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +




45


Stresses at Midspan

Concrete stress at the top fiber of the deck, three cases:1. Under permanent loads:

( )M M( )

(245.1 153.6)*1229,534

0.162

ws btc

tc

tc

tc

M MfS

f

f

+=

+= +

= +




Note that deck stresses under service loads are almost always well below allowable for continuous for LL bridges; but they still must be checked.


Stresses at Midspan

2. One-half permanent loads plus live loads:( )LL IMf f +

2 1

2

2

( )

1,386.2*120.5(0.162)29,534

0.08 0.563 0.64

LL Itc tc

tc

tc

tc

Mf fS

f

f

+= +

= +

= + = +




46


Stresses at Midspan

3. Under permanent and transient loads:

( )

(245.1 153.6 1,386.2)*1229,534

0.73

ws b LL Itc

tc

tc

tc

M M MfS

f

f

++ +=

+ +=

= +




tcf


Stresses at Midspan

Tension stress at the bottom fiber of the beam, Service III:

[ ]

( ) ( ) 0.8

(245.1 153.6) (0.8*1,386.2) *12972 972(20.0) (951.9 1,148.4)*12789 10,542 10,542 16,6941.23 1.84 2.39 1.08 0.40

++ + +

= + − −

+ ++= + − −

= + − − = −

pe pe c g s ws b LL Ib

b b bc

b

b

P P e M M M M MfA S S S

f

f



Tensile stress limit for concrete: -0.503 ksi OK

Service III has the 0.8LL factor!

bf

47


Positive Moment Section

Total Ultimate bending moment for Strength I is:

At point of maximum moment 0.4L:

(Tables 3.4.1-1&2)

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +

1.25( ) 1.5( ) 1.75( )1 2 (912 9 1 101 1 1 9) 1 (2 4 2) 1 (1 412 4)

uM DC DW LL IM= + + +



1.25(912.9 1,101.5 171.9) 1.5(274.2) 1.75(1,412.4)5,615

u

u

MM k ft

= + + + += −



Average stress in prestressing steel:

(5.7.3.1.1)1ps pu

p

cf f kd

= −



48



fps = Average stress in prestressing steel ksi

k =

=

0.28 for low relaxation strands

dp = Distance from extreme compression fiber to th t id f th t i t d

in.

2 1.04 py

pu

ff

−

(Table C5.7.3.1.1-1)



= the centroid of the prestressing tendonsh - ybs = 62.5 – 4.70 = 57.80


in.



' 'A f A f A f+(5.7.3.1.1-4)

'0.85

ps pu s y s y

puc ps

p

A f A f A fc f

f b kAd

β

+ −=

+



49



Aps = Area of prestressing steel = 40 * 0.153 = 6.12 in2

f = Specified tensile strength of prestressing steel = 270 ksifpu = Specified tensile strength of prestressing steel = 270 ksi

As = Area of mild steel tension reinforcement = 0.0 in2

fy = Yield strength of tension reinforcement = 60.0 ksi

As‘ = Area of compression reinforcement = 0.0 in2

fy‘ = Yield strength of compression reinforcement = 60.0 ksi



fc‘ = Compressive strength of deck concrete = 4.5 ksi

β1 = Stress block factor specified in LRFD 5.7.2.2 = 0.83

b = Effective width of compression flange = 96 in.



6.12(270) 0.0 0.0270

+ −=c

a = depth of the equivalent stress block = β1c

2700.85(4.5)(0.83)(96) 0.28(6.12)57.8

5.28

+

=c



0.83(5.28) 4.39 8.5sa in t in= = < =

50



Therefore, the assumption of rectangular section behavior is valid and the average stress in prestressing steel is:valid and the average stress in prestressing steel is:

Nominal flexural resistance:

5.28270 1 0.28 263.357.8psf ksi = − =

4.396 12(263 3) 57 80 −



6.12(263.3) 57.802

2 127,467

n ps ps p

n

aM A f d

M k ft

= − =

= −



M Mφ=Factored flexural resistance:

7 467 5 615M k ft M k ftφ = − > = −

r nM Mφ=

Where Φ= resistance factor = 1.0 for flexure and tension of prestressed concrete



7, 467 5,615n uM k ft M k ftφ = > =

51


Maximum Reinforcement-Positive Moment Section

The old ρmax requirement has been deleted. The LRFD Specifications now require that φ be determined based on whether the section is tension controlled compressionwhether the section is tension controlled, compression controlled or a transition section. In the calculation of Mr, tension control was assumed.

Check the strain in the extreme tensile steel:

td 54.0 8.5 2 60.5= + − =



This is a tension controlled section, so φ = 1.0(5.7.2.1 & 5.5.4.2)

tt

d c 60.5 5.280.003 0.003 0.032 0.005c 5.28− − ε = = = >


Minimum Reinforcement – Positive Moment Section

At any section, the amount of prestressed and nonprestressed tensile reinforcement shall be adequate to p qdevelop a factored flexural resistance, Mr, at least equal to the lesser of:

1.2 times the cracking moment, Mcr, determined on the basis of elastic stress distribution and the modulus of rupture, fr,



1.33 times the factored moment required by the applicable strength load combinations

(5.7.3.3.2)

52


Minimum Reinforcement - Positive Moment Section

(5.7.3.3.2-1)( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥

Where:

'0.37 0.37 7.0 0.979cf = =fr = Modulus of rupture = ksifcpe

=Compressive stress in concrete due to effective prestresss forces only (after allowance for all Prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads

ksi(5.4.2.6)

nc



tensile stress is caused by externally applied loads

972 972(20.0) 1.23 1.84 3.07789 10,542

pe pe c

b

P P eA S

+ = + = + =



Mdnc= Total unfactored dead load moment acting on the non-composite section =

kip-ftthe non composite section Mg+Ms = 951.9+1,148.4 = 2,100.3

Sc= Section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads = 16,694

in3



Snc= Section modulus for the extreme fiber of the noncomposite section where tensile stress is caused by externally applied loads = 10,542

in3

53



ki ft

16,694 16,694 16,694(0.98 3.07) 2,100.3 1 (0.979)12 10,542 12

4 400 1 362

crM

M

= + − − ≥

≥ kip-ft

kip-ft

At midspan, the factored moment required by the Strength I load combination is: Mu = 5,610 kip-ft

Therefore, kip-ft

Since Controls

1.33 7,461uM =

1 2 1 33M M 1 2M

1.2 5,290crM =

4, 400 1,362crM = ≥



Since , Controls

OK

Note: The LRFD Specifications states that this requirement be met at every section.

1.2 1.33cr uM M< 1.2 crM

7, 467 1.2r crM M= >


Design of the Negative Moment Section

Total Ultimate bending moment for Strength I is: (3 4 1-1&2)1 25( ) 1 5( ) 1 75( )M DC DW LL IM= + + +

At the pier section:kip-ft

Notes:1. At the negative moment section, the compression

face is the bottom flange of the beam and is 26 in

(3.4.1-1&2)1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +

1.25( 292.7) 1.5( 467.1) 1.75( 1,380.7) 3, 483uM = − + − + − = −



wide.2. This section is a nonprestressed reinforced concrete

section, thus Φ = 0.9 for flexure.

54



Assume the deck reinforcement is at the mid-height of the deck. A f deck.

'1.7s y

u s yc

A fM A f d

f bφ

= −

fy = Yield strength of compression reinforcement

= 60.0in2

fc‘ = Compressive strength of girder = 7.0 ksi

(5.14.1.2.7j)



54 0.5(8.5) 58.25+ =

c p g gd = Effective depth to negative moment

reinforcement from bottom of girder =in



(60)3 483(12) 0 90 (60) 58 25 sAA

= −

This is the required amount of mild steel reinforcement

2

2

3, 483(12) 0.90 (60) 58.251.7(7.0)(26)

0 10.47 3145 41,796

13.94

s

s s

s

A

A A

A in

=

= − +

=



This is the required amount of mild steel reinforcement required in the slab to resist the negative moment and it is equal to 18 #5 bars and 19 #6 bars.

55


Longitudinal Deck Reinforcement

The longitudinal reinforcement in the deck includes distribution reinforcement and other minimum reinforcementdistribution reinforcement and other minimum reinforcement.



2( ) 5.58s providedA in=


Negative Moment Deck Reinforcement

The additional area of deck reinforcement required: 22

, ' 13.93 5.58 8.35s Add lA in= − =



56



Typical longitudinal deck reinforcement No. 5 @ 12” Top -No. 5 @ 8” Btm.

T l A f l i di l i f 5 58 i 2Total Area of longitudinal reinforcement provided

5.58 in2

Factored negative design moment -3,483 kip-ftTotal area required to resist negative moment

13.93 in2

Additional area of deck reinforcement required

8.35 in2

Addi i l i f id d 19 N 6 B



Additional reinforcement provided 19 No. 6 BarsAdditional area of deck reinforcement provided

8.36 in2

Total As provided 13.94 in2 > 13.93 in2

OK



Location of steel:Top – 8 #5 + 8 #6 with 2” clear

Note: Epoxy coated steel assumed. Min. Top 8 #5 8 #6 with 2 clear

Btm – 10 #5 + 11 #6 with 2 5/8” clear.in218(0.31) 19(0.44) 13.94sA = + =

8(0.31)(2.3125) 8(0.44)(2.375) 10(0.31)(8.5 2.9375) 11(0.44)(8.5 3)13.94

57.96 4 16

x

x

+ + − + −=

= =

cover is 1.5 in.(5.12.4)



We assumed 4.25” from top OKd = 58.34 in

4.1613.94

x

57



Now check Mn:

( )( )( )( )

s y

c

1

A f 13.94 60a 5.41in

0.85f 'b 0.85 7 26a 5.41c 7.72

0.741

= = =

= = =β



( )( )( )r n

r u

5.41M M 0.9 13.94 60 58.342

M 41,880k in 3,490k ft M 3,483k ft

= φ = −

= − = − > = −


Effective Tension Flange Width

The effective tension flange width is the lesser of:

The effective flange width = 96 in CONTROLS

A width equal to 1/10 of the average of adjacent spans between bearings =

(5.7.3.4)

0 10(96 25)(12) 115 5in=



(4.6.2.6)

0.10(96.25)(12) 115.5in

58


Control of Cracking by Distribution Reinforcement

According to LRFD 5.7.3.4 the spacing of the mild steel reinforcement in the layer closest to the tension face shall satisfy equation 5.7.3.4-1.

The tensile stress in mild reinforcement is computed to be:

700 2ec

s s

s dfγ

β≤ −



sls

s

MfA jd

=



fy = Yield strength of reinforcement = 60.0 ksi

Msl = (292.7)+(467.1)+(1,380.7)= 2,140.5 kip-ft

As = Area of negative moment reinforcement = 13.94

in2

d = Effective depth to negative moment reinforcement from bottom of girder =

in



g62.5-4.16 = 58.34

j = 1-(k/3) = 1-(0.275/3) = 0.908

59



Where:22 ( )k n n nρ ρ ρ= + −

Where:

29,000 5 718steelE

ρ =

M d l R i

22(0.00919)(5.718) (0.00919*5.718) 0.00919 *5.7180.275

kk= + −

=

13.94 0.00919(26)(58.34)

sAbd

= =



So:

, 5.7185,072

steel

girderE= =n = Modular Ratio =

2,140.5(12) 34.813.94(0.908)(58.34)sf ksi= =



The previous calculation made the simplifying assumption th t th ti t lthat the section was rectangular.

If this assumption is NOT made, the neutral axis, calculated using working stress concepts, can be calculated as 16.45 inches from the bottom of the beam. The cracked, transformed moment of inertia is 177200 in4. The steel stress is found to be 34 6 ksi which compares to 34 8 ksi using the



is found to be 34.6 ksi which compares to 34.8 ksi using the rectangular assumption.

60


Control of Cracking by Distribution ReinforcementA quick review of working stress:

1) The cracked, transformed section is used.2) Th t l i i t th t i t id

s

c

EnE

=



2) The neutral axis is at the geometric centroid.3) Concrete stress is assumed linear.4) Steel is converted to an equivalent area of concrete by multiplying

by n.5) Tension in concrete is ignored



The maximum concrete stress is:

The steel stress is:

slc

tr

M cfI

=

( )The term M(d-c)/I gives the equivalent concrete



( )sls

tr

M d cf n

I−

=the equivalent concrete stress. It is converted to steel stress by multiplying by n.

61





This is the assumed cracked, transformed section. Note that it is a negative moment section. Based on a previous iteration, the neutral axis, x, is within the tapered section of the Type IV flange.



To determine “x”, the position of the neutral axis, the first moment of inertia of the area about theof inertia of the area about the neutral axis must be = 0. Define the downward direction as positive.

It can be shown that b = 42-2x

( ) 1 x 8−



( )( )( )( ) ( )( )

( ) ( )

1 x 826 42 2x 8 x 4 2 x 8 x 8 x 82 3

xx 42 2x 79.5 58.34 x 02

− − − + − − − −

+ − − − =

62



The equation reduces to:3 20.33x 21x 15.5x 4467.3 0− + + − =

The roots are -13.55, 16.45 and 60.75. The only root which makes any sense is x = 16.45 in. Thus, b = 9.10 inches and x-8 = 8.45 in.

( )( ) ( ) ( )( )3 231 1I 9.10 16.45 2 8.45 8 8.45 8 16.45 43 12

= + + −



( )( ) ( )( )

( )

23

2 4

1 1 8.452 8.45 8.45 8.45 8.45 16.45 836 2 3

79.5 58.34 16.45 177200in

+ + − −

+ − =



( )sls

M d cf n

I−

=

( ) ( )( )

str

s

I2140.5 12 58.34 16.45

f 5.7 34.6ksi177200

−= =

This is lower than the stress found by assuming a rectangular section Since the steel stress in the



rectangular section. Since the steel stress in the denominator of the spacing equation, using the rectangular assumption is conservative (requires a closer spacing) in this case.

63



The spacing of mild steel reinforcement in the layer closest to the tension face shall satisfy the following:to the tension face shall satisfy the following:

Where:

(5.7.3.4-1)700 2e

cs s

s dfγ

β≤ −

γe = Exposure factor = 0.75 for Class 2 exposure condition



fs = Tensile stress in steel reinforcement at the service limit state

ksi

βs = 10.7( )

c

c

dh d

+−



Where:

dc = Thickness of concrete cover measured from extreme tension fiber to center of the flexural reinforcement located closest therto = 2.00+5/8(0.5) = 2.31

in

h = Overall height on the composite section = 62.5

in



64



2.311 1.0550 7(62 5 2 31)sβ = + =

700 0.75 2(2.31) 9.671.055 34.8

s in⋅≤ − =

⋅

0.7(62.5 2.31)−

OK6.0 9.67in in≤



OKFor this example the tensile stress in the mild reinforcement is less than its allowable. Thus, the distribution of reinforcement for control of cracking is adequate.


Maximum Reinforcement – Negative Moment Section

As before, check the strain in the extreme tensile steel:

This is a tension controlled section, so φ = 0.9

tt

d c 59.9 7.720.003 0.003 0.020 0.005c 7.72− − ε = = = >



φ

(5.7.2.1 & 5.5.4.2)

65


Minimum Reinforcement – Negative Moment Section

At any section, the amount of prestressed and nonprestressed tensile reinforcement shall be adequate to p qdevelop a factored flexural resistance, Mr, at least equal to the lesser of:

1.2 times the cracking moment, Mcr, determined on the basis of elastic stress distribution and the modulus of rupture, fr,



1.33 times the factored moment required by the applicable strength load combinations

(5.7.3.3.2)



(5.7.3.3.2-1)( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥

Where:

nc

'0.37 0.37 4.5 0.785cf = =

0g sM M+ =

fr = ksifcpe = 0.0 ksi

Mdnc= kip-ftS = 29 534 in3



Sc= 29,534 in

66



29,534 (0.785)12

=crM

At bearing, the factored moment required by the Strength I load combination is: Mu = -3,483 kip-ft

Therefore, kip-ft1.33 4,631uM =

1,932= −crM k ft

1.2 2,318= −crM k ft



Since , Controls

OKNote: The LRFD Specifications states that this requirement be met at every section.

1.2 1.33cr uM M< 1.2 crM

3,490 1.2 2,318r crM M= > =


Positive Moment Connection

Continuous for live load bridges are covered in Article5.14.1.4.4. Much of this article is new in 2007 (4th Ed.).

One requirement of this article is for a positive moment connection. These positive moments are caused by the upward camber of the prestressed girders due to creep and shrinkage. The positive moment connection is needed to provided continuity at the pier.

Th ti b d ith b t di ild t l t



The connection can be made either by extending mild steel out of the end of the girder into the diaphragm or by leaving strand extend out of the end of the girder into the diaphragm. This example illustrates bent strand connections.

67


Positive Moment Connection

Positive moments develop at the connection between girders at in interior supports due to livebetween girders at in interior supports due to live-load effects (if more than two spans) and restraint caused by temperature, creep, and shrinkage. According to LRFD 5.14.1.4.4, these restraint moments are negligible when continuity is established after 90 days.



y


Development of Extended Strands

The strands are bent up 90° into the diaphragm so that the hook extends 8 inches from the end of the girder. The ends of the girders are placed 10 inches apart With the 8 inch projection this leaves 2are placed 10 inches apart. With the 8 inch projection this leaves 2 inches of clear allowing for construction tolerances. Typically mild steel is placed in the corner of the hooks to enhance the development length of the hooks. These bars should be at least #5.



68


Required Area of Strand

The design moment used for the working stress check is Mcr while the design moment for the strength check is 1 2M According to LRFD 5 14 1 4 9c the stress in the1.2Mcr. According to LRFD 5.14.1.4.9c the stress in the strands used for design as a function of the total length of the strand shall not exceed:

where:

( 8) 1500.288

−= ≤dsh

psllf ksi

( 8)0.163

−= dsh

pullf

(5.14.1.4.9c-1) (5.14.1.4.9c-2)ℓ = total length of extended strand in



ℓdsh = total length of extended strand infpsl = stress in the strand at the service limit state ksi

Cracked section shall be assumedfpul = stress in the strand at the strength limit state ksi



The design moments, parameters, and results for the design of the positive moment connectionthe design of the positive moment connection using bent strand are found in following table. The cracking moment is found using the gross, composite cross section, but assuming that cracking occurs at the diaphragm. Thus the diaphragm concrete strength is used. For these



p g gcalculations the effective width of 96 inches, 0.5 inch strand, and concrete strength of 4.5 ksi were used.

69



When using working stress design the number of strands is assumed to calculate the length of the strand. When using

( )cr c cbM 0.24 f ' S 0.24 4.5 16694 8500k in 708k ft1 2M 850k f

= = = − = −

g gthe strength design method, the length of strand is assumed to calculate the number of strands required. Design iterations are performed to determine the most efficient combination of strand and length.



cr

e dsh

1.2M 850k ftL 8

= −

= −l



Working Stress Design

No of Strand 6 8 10 12 16No. of Strand 6 8 10 12 16

42.29 33.78 29.36 25.83 21.42

As. 0.92 1.22 1.53 1.84 2.45

Moment 708.00 708.00 708.00 708.00 708.00

n 7.00 7.00 7.00 7.00 7.00

d 62.50 62.50 60.50 60.50 60.50

dshl



ρ 45E-6 52E-6 263E-6 317E-6 422E-6

k 0.05 0.05 0.06 0.07 0.08

j 0.98 0.98 0.98 0.98 0.97

fs 150 113.07 93.68 78.22 58.87

70



Strength DesignNo. of Strand 5.18 6.52 8.00 9.27 13.13

42.00 35.00 30.00 27.00 22.00As 0.79 1.00 1.22 1.42 2.01

Moment 849.70 849.70 849.70 849.70 849.70d 62.50 62.50 62.50 60.50 60.50a 0.45 0.45 0.45 0.45 0.47

f 209 166 135 117 86

dshl



* Back calculated based on strand length

fpul 209 166 135 117 86



In this example working stress design governs. Multiple iterations are performed to determine the least length ofiterations are performed to determine the least length of extension of the strand required.

If the results indicate an odd number of strands they are rounded up to an even number to provide symmetry in the connection. It may be more desirable to have a larger number of shorter strands as opposed to fewer longer strands. Girder fabrication may be more difficult with longer strand extensions as this may require excessive space between girders in the bed In addition



require excessive space between girders in the bed. In addition, if a larger number of shorter strands are used the stress can be distributed throughout a larger area.

71



The designer chooses from the previous tables. A reasonable design would be 12 strands extended 26 inches. That would be an 8 inch horizontal extension from the face of the beam and an 18 inch vertical “tail” to the hook. Any 12 strands could be extended, but spacing them out and using different rows makes construction easier and limits stress concentrations.

Also note that, consistent with the design examples in NCHRP Report 519, the haunch has been included.




Shear Design

The area and spacing of shear reinforcement must be determined at regular intervals along the entire length ofdetermined at regular intervals along the entire length of the beam. In this design example, transverse shear design procedures are demonstrated below by determining these values at the critical section near the supports.Transverse reinforcement shall be provided where:

φ



(5.8.2.4-1)0.5 ( )u c pV V Vφ≥ +

72


Shear Design

Vu = Total factored shear force kips

Vc = Shear strength provided by concrete kips

Vp = Component of the effective prestressing force in the direction of the applied shear

kips



Φ = Resistance factor (5.5.4.2.1)


Critical Section

dv = Effective shear depthDistance between resultants of tensile andDistance between resultants of tensile and compressive forces, de – a/2, but not less than 0.9de or 0.72h.

(5.8.2.9)

de = The corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement 58 34

in



reinforcement = 58.34a = Equivalent depth of the compression

block = 5.41in

h = Total height of section = 62.5 in

73


Effective Shear Depth

0 5( ) 58 34 0 5(5 41) 55 63d d a in= = =

Therefore, dv = 55.63 in.

0.5( ) 58.34 0.5(5.41) 55.630.9 0.9(58.34) 52.50.72 0.72(62.5) 45

v e

e

d d a ind inh in

= − = − =

≥ = =

≥ = =



Therefore, dv 55.63 in.


Calculation of Critical Section

The critical section near the support is dv = 55.63 in from the FACE of the supportfrom the FACE of the support.

Note: Assume the length of the bearing pad is 10 inches.



Thus the critical section is:55.63 in + 5 in = 60.63 inches.

74



Using values from previous tables (linearly interpolated), the factored shear force and bending moment at the critical gsection for shear, according to Strength I load combinations.

kips(All shear goes the same way!)

1.25(35.4 42.7 14.1) 1.50(22.6) 1.75(99.4) 323.1uV = + + + + =



0.9(185.2 223.5) 1.25( 219.3) 1.50( 350.0) 1.75( 1,080.9)2,323 27880

u

u

MM k ft k in

= + + − + − + −

= − − = − −



At this point, there are three choices:

1. Ignore the prestressing steelThen, this is a reinforced sectionβ = 2θ = 45◦

(This is VERY conservative)



(This is VERY conservative)

2. Use Sectional Model for RC3. Include PS Steel

75


Contribution of Concrete to Nominal Shear Resistance

1. Ignore prestressing steel:

A #4 h A 0 4 i 2 90 i 1 0

'0.0316 0.0316(2) 7(8)(55.63) 74.4c c v vV f b d kβ= = =

323.1 74.4 284.60.9sV kips= − =



Assume #4 hoops Av = 0.4 in2 α = 90 sin α =1 cot α =0



cot 0 4(60)(55 63)cot 45A f d θin

Use #4@4 in

cot 0.4(60)(55.63)cot 45 4.7284.6

v y v

s

A f ds

Vθ

= = =



Vs = 334 kips

76



2. Use Sectional Model but for RC:M 27 880 kip inMu = 27,880 kip-indv = 55.63 in.

Nu = Applied factored normal force at the specified section = 0 kips

Vu = 323.1kips

As = Area of nonprestressed steel on the flexural tension side of the member = 13.94

in2

A = 0 in2



Ap 0 inEp = 28,500 ksiEs = 29,000 ksi



Vp =

==

Component of the effective prestressing force in the direction of the applied shear(force per strand)(number of draped strands)(sin ψ)0

kips



fpo = 0

77



Assume 0.5 cot θ = 1.

3

27,880 0.5(0) (323.1) 055.63 0.001

2(29,000(13.94))1.0 10 0.001

x

x

ε

−

+ + −= ≤

≤





u pu

V Vv

b dφ

φ−

=v vb dφ

vu = Shear stress in concrete kipsbv =

=Effective web width of the beam8

in


kips

Where:



==

direction of the applied shear(force per strand)(number of draped strands)(sin ψ)0

78



ksi323.1 0.9(0) 0.810 9(8)(55 63)uv −

= =

Use (vu / fc’) < 0.125 and εx < 1 from LRFD Table

0.9(8)(55.63)

'

0.81 0.1157.0

= =

u

c

vf



( u c ) x5.8.3.4.2-1:

θ = 37◦β = 2.13



kips'0.0316 0.0316(2.13) 7(8)(55.63) 79.3c c v vV f b dβ= = =

kips

Use #4 hoops Av = 0.4 in2 α = 90 sin α =1 cot α =0

323.1 0.9(79.3) 280.00.9sV −

= =

cot 0.4(60)(55.63)cot 37.0 6.32280.0

v y v

s

A f ds

Vθ

= = =



So #4 hoops at 6 in Vs = 295.0 kips

OK0.9(79.3 295.0 0) 337r uV k V= + + = >

79



3. Include Prestressing Steel:M 27 880 kip inMu = 27,880 kip-indv = 53.6 in.

Nu = Applied factored normal force at the specified section = 0 kips

Vu = 323.1kips


in2

A = 9(0 153) = 1 38 in2



Ap 9(0.153) 1.38 inEp = 28,500 ksiEs = 29,000 ksi

Note, when the prestressing steel in included, de = 57 inches. The term c = 9.76 in and a = 6.77in. Thus, dv = 53.6 in.



If dv < 60db = 30 in, Vp and fpo must be reduced for lack of bond. d = 53.6 , so the critical section is 70.6 from the endbond. dv 53.6 , so the critical section is 70.6 from the end of the girder > 30 in so:Vp===

Component of the effective prestressing force in the direction of the applied shear(force per strand)(number of draped strands)(sin ψ)24.3(9)(sin 6.2◦) = 23.6

kips

fpo = A parameter taken as modulus of elasticity of ksi



.7 0.7(270.0) 189= =puf

p

=

prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and the surrounding concrete

[LRFD 5.8.3.4.2]

80



Assume 0.5 cot θ = 1.

3

27,880 0.5(0) (323.1 23.6) 1.38(189)53.6 0.001

2(29,000(13.94) 28,500(1.38))0.63 10 0.001

x

x

ε

−

+ + − −= ≤

+

≤





u pu

V Vv

b dφ

φ−

=v vb dφ

vu = Shear stress in concrete kipsbv =

=Effective web width of the beam8

in


kips

Where:



==

direction of the applied shear(force per strand)(number of draped strands)(sin ψ)23.6

81



ksi323.1 0.9(23.6) 0.7820 9(8)(53 6)uv −

= =

Use (vu / fc’) < 0.125 and εx < 0.75 from LRFD

0.9(8)(53.6)

'

0.782 0.1117.0

u

c

vf

= =



( u c ) xTable 5.8.3.4.2-1:

θ = 34.4◦β = 2.26



kips'0.0316 0.0316(2.26) 7(8)(55.63) 84.1c c v vV f b dβ= = =

kips


323.1 0.9(84.1 23.6) 251.30.9sV − +

= =

cot 0.4(60)(53.6)cot 34.4 7.5251.3

v y v

s

A f ds

Vθ

= = =



So #4 hoops at 6 in Vs = 313 kips

OK0.9(84.1 313.0 23.6) 378.6r uV V= + + = >

82


Minimum Reinforcement Requirement

Check which is true:(5.8.2.7)

or

ksiksi

(5.8.2.7-1)

(5.8.2.7-2)

'0.125u cv f<

'0.125u cv f≥

'0.125 0.125(7.0) 0.875cf = =0.81uv =



Since , Then in 24 in CONTROLS

'0.125u cv f< max 0.8 0.8(55.63) 44.5 24.0= = = ≤vs d



Calculate minimum area of steel using a 6 inch spacing to get area of steel:get area of steel:

<0.4 in2 OK( )( ) 2vv c

y


≥ = =

(5.8.2.5)



( )

83


Critical Section – Positive Moment

Critical Section near the supports is at dv. Where:

Where:

dv = Effective shear depthDistance between resultants of tensile and compressive forces, de – a/2, but not less than 0.9de or 0.72h.

(5.8.2.9)

d = The corresponding effective depth from the in



de The corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement = 58.2

in

a = Equivalent depth of the compression block = 3.42 inh = Total height of section = 62.5 in



In this area, the positive moment properties are needed. However, since this section is where the strand is harped, the positive moment properties must be recalculated using 31 strands. Ap = 4.74 in2 and dp = 62.5 - 4.32 = 58.2 inches. The value of 4.32 inches as the centroid of 31 strands was calculated earlier in Section 1.7.2. Refer to Section 1.9.1 for the equations below:

( )( )( )( )( )( ) ( )

4 74 2704 112700 85 4 5 0 83 96 0 28 4 74

58 2

.c . in

. . . . .= =

+



( )( )

58 24 11270 1 0 28 264 858 2

0 83 4 11 3 42

ps

..f . . ksi.

a . . . in

= − =

= =

84



0.5( ) 58.2 0.5(3.42) 56.5d d a in= − = − =


0.5( ) 58.2 0.5(3.42) 56.50.9 0.9(58.2) 52.40.72 0.72(62.5) 45

v e

e

d d a ind inh in

≥ = =

≥ = =





The critical section near the support is dv = 56.5 in from the FACE of the supportfrom the FACE of the support.

Note: Assume the length of the bearing pad is 10 inches.



Thus the critical section is:56.5 in + 5 in ≈ 62 inches.

85



Using values from previous tables, the factored shear force and bending moment at the critical section for shear, gaccording to Strength I load combinations.

It is conservative to take the highest factored moment that will occur at that section, rather than the moment corresponding to maximum V Therefore

1.25(35.4 42.7 7.9) 1.50(12.6) 1.75(82.2) 250.01.25(185.2 223.5 49.6) 1.50(79.1) 1.75(373.9) 1,346

u

u

V kM k in

= + + + + =

= + + + + = −



corresponding to maximum Vu. Therefore,

kipskip-ft

250.0uV =1,346uM =

(5.8.3.4.2)



The contribution of the concrete to the nominal shear resistance is:resistance is:

'0.0316c c v vV f b dβ= (5.8.3.3-3)



86


Strain in Flexural Tension Reinforcement

Strain in the reinforcement is (assuming uncracked):

(5 8 3 4 2 1)M

Where:

(5.8.3.4.2-1)0.5 0.5 cot0.001

2( )

uu u p ps po

vx

s s p ps c c

MN V V A f

dE A E A E A

θε

+ + − −= ≤

+ +

Nu = Applied factored normal force at the specified section = 0 kipsVp =

=

Component of the effective prestressing force in the direction of the applied shear(force per strand)(number of draped strands)(sin ψ)

kips



==

(force per strand)(number of draped strands)(sin ψ)24.3(9)(sin 6.2◦) = 23.6

fpo =

=

A parameter taken as modulus of elasticity of prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and the surrounding concrete

ksi[LRFD 5.8.3.4.2]

.7 0.7(270.0) 189= =puf



f f 2

Where (cont.):Aps

=

=

Area of prestressing steel on the flexural tension side of the member, as shown in LRFD Figure 5.8.3.4.2-1.31(0.153) = 4.74

in2

As = Area of nonprestressed steel on the flexural tension side of the member = 0

in2

Ac= Area of concrete on the flexural tension half. This in2



=

term is calculated as the area on the tension side (bottom in this case) from the tension fiber to h/2.475

87



Note that either θ can be assumed OR 0.5 cot θ can be assumed =1. Assume 0.5 cot θ = 1:assumed 1. Assume 0.5 cot θ 1:

( )( )3

1,346(12) 0.5(0) (250 23.6) 4.74(189)56.5 0.001

2 28,500(4.74) 5072 475

0.07 10 0.001

x

x

ε

−

+ + − −= ≤

+

− ≤



The negative value means the section is uncracked


Shear Stress

u pu

V Vv

b dφ

φ−

=

Where:v vb dφ

vu = Shear stress in concrete kipsbv = Effective web width of the beam = 8 in

Vp =

=

Component of the effective prestressing force in the direction of the applied shear(force per strand)(number of draped strands)(sin ψ)

kips



= 23.6

250 0.9(23.6) 0.5620.9(8)(56.5)uv ksi−

= =

88


Values of β & θ

0.562 0 0803uv = =

Use (vu / fc’) < 0.1 and εx < -0.05 from LRFD Table 5.8.3.4.2-1:θ = 21.4◦

β = 3.24

' 0.08037.0cf

= =




Concrete Contribution

The contribution of the concrete to the nominal shear resistance is: (5 8 3 3 3)resistance is:

kips

(5.8.3.3-3)'0.0316c c v vV f b dβ=

0.0316(3.24) 7.0(8)(56.5) 122.4cV = =



89


Contribution of Reinforcement of Nominal Shear Resistance

Check if: (5.8.2.4-1)

kips

At least minimum stirrups are needed.

( )250 0.5 ( ) 0.5 0.9 (122.4 23.6) 65.7u c pV kips V Vφ= > + = + =





Check which is true: (5.8.2.7)

or

ksi

(5.8.2.7-1)

(5.8.2.7-2)

'0.125u cv f<

'0.125u cv f≥

'0.125 0.125(7.0) 0.875cf = =



ksi

Since , Then in : 24 in CONTROLS

0.562uv =

'0.125u cv f< max 0.8 24.0vs d= ≤

90



Calculate minimum area of steel using a 12 inch spacing to get area of steel per foot:get area of steel per foot:

ODOT uses #4 bars with 2 legs as standard; (Av = 2(0.2 in2) = 0.4 in2)

(5.8.2.5)( )( ) 2vv c

y


≥ = =



#4@ 24 inch o.c.= 0.2 in2/ftThis is adequate to meet minimum.


Maximum Nominal Shear Resistance

The upper limit of Vn, given by following equation, is intended to ensure that the concrete in the web of theintended to ensure that the concrete in the web of the beam will not crush prior to yield of the transverse reinforcement.

Comparing this previous equation with equation LRFD 5 8 3 3 1:

(5.8.3.3-2)'0.25n c v v pV f b d V≤ +



5.8.3.3.-1:'0.25c s c v vV V f b d+ ≤

91



Assume #4 @ 24”:

( )

( )( )( ) ( ) ( )20 4 60 56 5 21 4 0 1

24144 2

v y vs

s

s

A f d cot cot sinV

s. in ksi . cot .

Vin

V . k

θ α α+=

+ =

=



OK( )122.4 144.2 266.6 0.25(7.0)(8)(56.5) 791+ = ≤ =



( )V V V Vφ= + +( )( )0 9 122 4 144 2 23 6 261 2

250

r c s p

r

r u

V V V V

V . . . . . kipsV V kips

φ + +

= + + =

> =



92


Factored Horizontal Shear

It will be assumed that the critical section is the same as for vertical shear Using load combination Strength I:vertical shear. Using load combination Strength I:

kipsin

Both of these values were found in the preceding section.

323.1uV =

55.6vd =



p gThis is shear at the critical section near the pier.


Required Interface Shear Reinforcement

ri niV Vφ= (5.8.4.1-1)The nominal shear resistance of the interface plane is:

Where:(5.8.4.1-3)[ ]ni cv vf y cV cA A f Pµ= + +

c = Cohesion factor ksi [LRFD 5.8.4.3]µ = Friction factor

Acv = Area of concrete engaged in shear transfer = bviLvi in2

A f h i f t i th h l i 2



Avf = Area of shear reinforcement crossing the shear plane in2

Pc = Permanent net compressive force normal to the shear plane kipsfy = Shear reinforcement yield strength ksibvi= Width of area of concrete engaged in shear transfer in

Lvi = Length of area of concrete engaged in shear transfer in

93



For a cast-in-place concrete placed against clean concrete girder surfaces, free of laitance with surface intentionallygirder surfaces, free of laitance with surface intentionally roughened to an amplitude of 0.25 in:

(5.8.4.2)0.28c =1.0µ =





Begin by exploring what happens when the shear reinforcement is the minimum used anywhere in the girder.reinforcement is the minimum used anywhere in the girder. The shear reinforcement was previously calculated to be#4 @ 24 inches minimum. The shear width is bvi = 20inches as this is the width of the top of the girder. If Lvi = 24inches:

( ) 2

[ ]

20 24 480ni cv vf y cV cA A f P

A in

µ= + +

= =



( )( )( ) ( )

( )

20 24 480

0.28 480 1.0 0.4 60 0 158.4

0.9 158.4 142.6

cv

ni

ri ni

A in

V k

V V kφ

= + + = = = =

94



ui ui cvV v A= (5.8.4.2-2)

142 6

(5.8.4.2-1)

142 6 0 297480ui ,max

.v . ksi= =

1uui

vi v

Vvb d

=

( )( )1 0 297 20 55 6 330u ,maxV . . kips= =



Therefore, #4 @ 24 is adequate anywhere that Vu < 330 kips. Note that the critical section, the reinforcement is actually #4 @ 4 inches or #4 @ 6”; depending on the model used. Note that #4 @ 24 would be adequate for horizontal shear, so it is NOT necessary to extend every shear stirrup into the slab.

( )( ),


Minimum Interface Shear Reinforcement

Minimum shear reinforcement, 0 05A

A #4 double leg bar at 24 in spacing is provided from the beam extending into the deck. Therefore, Avf =0.4 in2 every 2 ft.

(5.8.4.1-4)0.05≥ cv

vfy

AAf



OK 0.05(480)0.40 0.4060

≥ =

95


Minimum Interface Shear Reinforcement

Article 5.8.4.4 states that Avf need not exceed that required to resist 1 33V /φ The same article alsorequired to resist 1.33Vui/φ. The same article also states that the minimum reinforcement provisions are waived for girder slab interfaces with surfaces roughened to an amplitude of 0.25 inches where the factored interface shear, vui, found in equation 5.8.4.2-1 is less than 0.210 ksi and all of the vertical (transverse) shear reinforcement required by Article



(transverse) shear reinforcement required by Article 5.8.1.1 is extended and anchored into the slab.



Vni must be less than:' 0 3(4 5)(480) 648K f A k (5 8 4 1 4)

Vni provided = 158.4 k OK

1 0.3(4.5)(480) 648c cvK f A k= =

2 1.8(480) 864cvK A k= =

(5.8.4.1-4)

(5.8.4.1-5)

'1

2

c cv

cv

K f AK A

≤≤



K1 = 0.3 and K2 = 1.8 (for normal weight concrete) are found in Article 5.8.4.3.

2 cv

96


Minimum Longitudinal Reinforcement Requirement

At each section the tensile capacity of the longitudinal reinforcement on the flexural tension side of the memberreinforcement on the flexural tension side of the member shall be proportioned to satisfy:

0.5 0.5 cotu u ups ps s y p s

M N VA f A f V Vd

θφ φ

+ ≥ + + − −



(5.8.3.5-1)

vd φ φ



According to Article 5.8.3.5, it is not necessary to provide any steel beyond that to resistto provide any steel beyond that to resist moment if there is a compressive reaction on the flexural compression face; in other words, in a negative moment zone over a support, the equation in this article does not need to be satisfied. However, it makes an exception for a



continuous for live load bridge; saying that this equation must be checked for a continuous for live load bridge.

97



This provision will be checked at the simply supported end, using positive moment properties. The check at theusing positive moment properties. The check at the continuous end is made in a similar manner.

The development length is:

( ) ( )d ps pe b2 2f f d 1.6 264.8 158.6 0.5 127.3in3 3

= κ − = − =

l



(5.11.4.2)



( )a 3.42d d 62 5 4 32 56 5in= − = − − =( )v pd d 62.5 4.32 56.5in2 2

= − = − − =

So the critical section is 56.5 inches from face of support. Allowing for a 10 inch bearing pad and that the center of bearing is 12 inches from the girder end, the critical section is 56.5+10/2+12=73.5 inches from the end of the girder.



Since this is less than the development length, the stress in the steel must be reduced for lack of development.

98



The stress in the undeveloped steel can be found from:

( )px bpx pe ps pe

d b

60df f f f

60d−

= + −−

l

l(5.11.4.2-4)

( )73.5in 30inf 158 6ksi 264 8ksi 158 6ksi 206ksi−+



( )pxf 158.6ksi 264.8ksi 158.6ksi 206ksi127.3in 30in

= + − =−



0.5 0.5 cot

+ ≥ + + − − u u u

ps ps s y p s

M N VA f A f V V θ

( )( )

( ) ( ) ( )

0.5 0.5 co

4.74 206 977

1346 2500 23.6 0.5 144.2 cot 21.4 4881.0 56.5 0.9

= >

+ + − − =

ps ps s y p sv

f f V Vd

k

k

θφ φ φ

This is OK. Note that Vs may not be taken as greater than



Vu/φ [LRFD 5.8.3.5].

250144 277 80 9

us

V kV k . k.φ

= < = =

99



At the inside edge of the bearing area of a simply supported end:supported end:

The steel is not fully developed. Since the bearing pad is assumed 10 inches and the center of bearing is 12 inches from the end of the girder, this section is 12+10/2 =17 inches from the end of the girder This is within the transfer

(5.8.3.5-2)0.5 cotups ps s y p s

VA f A f V V θφ

+ ≥ − −



inches from the end of the girder. This is within the transfer length, so:

( )158 6 1790

60 30= = =

lpe pxpx

b

f .f ksi

d (5.11.4.2-3)



0.5 cot ≥ − −

ups ps p s

VA f V V θφ

Assume #4 bars will be used.

( )( )

( ) ( )

4.74 90 426

250 23.6 0.5 144.2 cot 21.4 464.60.9

= <

− − =

k

k

φ

( )0 2 60A f

NG



( )

( )( )

0 2 601 25 1 25 5 7

7

0 4 0 4 0 5 60 12

b yd

c

b y

A f .. . . in

f '

. d f . . in

= = =

< = =

l (5.11.2.1)

100



The development length is 12 inches so the bar is fully developed:developed:

Thus:2464 6 426 0 64

60−

= =s.A . in

# # #



4 #4 works. 3 #5 also works as a # 5 needs a 15 inch development length.



Can also add stirrups. Increase to #4 @ 12:( )( )0 4 60 56 5 21 4cot V

Therefore, Vs = 277.8 for this calculation.

( )( )0 4 60 56 5 21 4288 277 8

12u

s

. . cot . VV k . kφ

= = > =

0.5 cot ≥ − −

ups ps p s

VA f V V θφ



( )( )

( ) ( )

4.74 90 426

250 23.6 0.5 277.8 cot 21.4 294.20.9

= >

− − =

k

k

101



In the previous slides, the assumption was made that the center of bearing was 12 inches from the end of the girder.center of bearing was 12 inches from the end of the girder.

What if the bearing pad is placed right at the end of the girder? That is, what if the center of bearing is only 5 inches from the end? What effect does that have on longitudinal steel?





( )a 3.42d d 62 5 4 32 56 5in= − = − − =( )v pd d 62.5 4.32 56.5in2 2

= − = − − =

So the critical section is 56.5 inches from face of support. Allowing for a 10 inch bearing pad, the critical section is 66.5 inches from the end of the girder.



Since this is less than the development length, the stress in the steel must be reduced for lack of development.

102



The stress in the undeveloped steel can be found from:

( )px bpx pe ps pe

d b

60df f f f

60d−

= + −−

l

l(5.11.4.2-4)

( )66.5in 30inf 158 6ksi 264 8ksi 158 6ksi 198 4ksi−+



( )pxf 158.6ksi 264.8ksi 158.6ksi 198.4ksi127.3in 30in

= + − =−




M N VA f A f V V θ

+ ≥ + + − −

( )( )

( ) ( ) ( )

0.5 0.5 co

4.74 198.4 940.4

1346 2500 23.6 0.5 144.2 cot 21.4 4881.0 56.5 0.9

ps ps s y p sv

f f V Vd

k

k

θφ φ φ

= >

+ + − − =

This is OK. Note that Vs may not be taken as greater than



Vu/φ [LRFD 5.8.3.5].

250144 277 80 9

us

V kV k . k.φ

= < = =

103



At the inside edge of the bearing area of a simply supported end:supported end:

The steel is not fully developed. Since the bearing pad is assumed 10 inches, this section is 10 inches from the end of the girder. This is within the transfer length, so:

(5.8.3.5-2)0.5 cotups ps s y p s

VA f A f V V θφ

+ ≥ − −



( )158 6 1052 9

60 30pe px

pxb

f .f . ksi

d= = =

l (5.11.4.2-3)



( )( )

0.5 cotups ps p s

VA f V V θφ

≥ − −


( )( )

( ) ( )

4.74 52.9 250.8

250 23.6 0.5 144.2 cot 21.4 464.60.9

k

k

= <

− − =

( )0 2 601 25 1 25 5 7

7b y

d

A f .. . . in

f '= = =l (5.11.2.1)

NG



( )( )7

0 4 0 4 0 5 60 12c

b y

f '

. d f . . in< = =

104



The development length is 12 inches so:10

The #4 can only develop 50 ksi. Thus:

( )10 60 5012sxf ksi= =

2464 6 250 8 4 350s

. .A . in−= =



This would be 22 #4! Clearly unrealistic!

50



Add stirrups. Increase to #4 @ 12:( )( )0 4 60 56 5 21 4cot V


( )( )0 4 60 56 5 21 4288 277 8

12u

s


= = > =

0.5 cotups ps p s

VA f V V θφ

≥ − −



( )( )

( ) ( )

4.74 52.9 250.8

250 23.6 0.5 277.8 cot 21.4 294.20.9

k

k

= <

− − =

105



This is much more workable:

294 2 250 8

This is 5 #4 bars.

So decrease stirrup spacing from the end of the girder to

2294 2 250 8 0 8750s

. .A . in−= =



So decrease stirrup spacing from the end of the girder to the critical section (this will be 66.5 inches from the end of the girder) to #4 @ 12. Add 5 #4 bars longitudinal in the bottom flange.


Anchorage Zone

The bursting resistance of pretensioned anchorage zones provided by vertical reinforcement in the ends of the pretensioned beams at the service limit state shall be takepretensioned beams at the service limit state shall be take as: (5.10.10.1-1)r s sP f A=

As = Total area of transverse reinforcement located within the distance h/4 from the end of the beam

in2

f = Stress in steel but not taken greater than 20 ksi



40(0.153)(202.5)(0.04) 49.6=

fs = Stress in steel, but not taken greater than 20 ksiPr = Bursting resistance, should not be

less than 4% of fpi

kips

106


Anchorage Zone

Solving for the required area of steel in249.6 2 47A = =Solving for the required area of steel, in2

At least 2.47 in2 of vertical transverse reinforcement should be provided at the end of the beam for a distance equal to one-fourth of the depth of the beam, h/4 = 54/4=13.5 in

2.4720sA = =



Therefore, for a distance of 13.5 in from the end of the member, use 7 #5 bars at 2 inches on center. The reinforcement provided:

OK. 7(2)0.2 2.8 2.47= >


Confinement Reinforcement

For a distance of 1.5d = 1.5(54) = 81 in, from the end of the beam, reinforcement is placed to confine the prestressingbeam, reinforcement is placed to confine the prestressing steel in the bottom flange. The reinforcement should not be less than #3 deformed pars, with spacing not exceeding 6.0 in, and shaped to enclose the strands.



(5.10.10.2)

107

AASHTO LRFD B id D i S ifi iAASHTO LRFD Bridge Design Specifications –Design Example 2

2 Span Continuous Prestressed I-Girder Bridge

EXTERIOR GIRDER


Design Example - Continuous Two-Span Inelastic I-Girder Bridge

Transverse Section

34’-0”

Type IV

8.5” structural+ 1.0” wearing



4 Spaces @ 8’-0” = 32’-0”2.5’ 2.5’

37’-0”

108


Effective Flange Width – Exterior Girder

The effective flange width is taken as one-half the effective width of the adjacent interior girder plus the least of:

One-eighth of the effective span length = 0.125(96.25)(12)= 144 in.

6.0 times the average thickness of the slab, plus the greater of half the web thicknessorone-quarter of the width of the top

= 6.0(8.5) + 0.5(8)=55 in.

= 6.0(8.5) + 0.25(20)

j g p



one quarter of the width of the top flange of the basic girder

6.0(8.5) 0.25(20)= 55 in.

The width of the overhang = 2.5 ft = 30 inches

Therefore, the effective flange width for the exterior girder is:(96/2) + 30 = 78 in.


Exterior Girder Properties

6.5 ft = 78 in

From the previous calculation of beff, the center to center distance controls.

beff Trans = nbeff = (0.8015) 78 in = 62.5 in



2.5 ft 4.0 ft

109


Exterior Girder Properties

yb= 38.22 inI 624512 in4I = 624512 in4

A = 50457 in2

h = 62.5 inyTC = 24.28 inyTG = 15.78 in

S 16340 in3



Sb= 16340 in3

STG = 39576 in3

STC = 25721in3


Dead Loads

Slab Self Weight:78 i (8 5 i )(0 150 k f)/144 0 691 klf78 in (8.5 in)(0.150 kcf)/144 = 0.691 klf

Haunch Weight: (Same as interior girder)0.042 klf

Recall that tributary area was used for the slab weight.



This will DECREASE the dead load moment on the exterior girders.

110


Dead Loads – Deck Plus Haunch – Exterior Girder

Distance x ft.

Shear kips

Moment kip-ft

0 00 35 3 00.00 35.3 0

9.26 28.5 295

18.97 21.4 537

28.69 14.2 710

38.41 7.1 814

48.13 0 849

57.84 -7.1 814

67.56 -14.2 710



77.28 -21.4 537

86.99 -28.5 295

96.25 -35.3 0


Exterior Beams

Exterior Girders:One Lane Loaded:

Lever Rule

Two or More Lanes Loaded:

g= egint

Where:



1.977.0 ede +=

g = DFMext

gint= DFMint

111


Distribution Factor for Moment

Positive Moment Region:Exterior Girder Two or More Lanes Loaded:Exterior Girder – Two or More Lanes Loaded:

DFExt = e DFInt

0.779.1

ede = +



DFExt+ = (0.880) (0.665) = 0.585

1.00.77 0.880

9.1= + =

Lever Rule: Assume a hinge develops over each interior girder and solve for the reaction in the exterior girder as a fraction of the truck


Exterior Beams

load.

1.2 01.2 1.2

HM Pe RSPe eR DF

S S

→ − =

= ∴ =

∑

This is for one lane loaded. Multiple Presence Factors apply 1.2 is the MPF

In the diagram P/2 are the wheel loads; P

1.5’

36k36k



In the diagram, P/2 are the wheel loads; P is the resultant force. All three loads are NOT applied at the same time.

Note that truck cannot be closer than 2’ from the barrier

8 ft

(3.6.1.3)

112



One Lane Loaded: [ ]1.2(36 ) (10.5 3.5) (10.5 9.5)kR

− + −=

Multiple Presence:

MPF = 1.2

N t th t thi l th t k

1.5’

36k36k

72 (8 )0.6 /

k ftR lanes girder=



Note that this only uses the truck.

By dividing by the total truck weight of 72 kips, R is given in lanes/girder

8 ft



Exterior Beams Moment

NL - Number of loaded lanes under considerationN N b f b i d

∑

∑+=

b

L

MinExt N

N

Ext

b

L

x

eX

NNDF

2, (C4.6.2.2.2d-1)



Nb - Number of beams or girderse - Eccentricity of design truck or load from CG of pattern of

girders (ft.)x - Distance from CG of pattern of girders to each girder (ft.)XExt - Distance from CG of pattern of girders to exterior girder (ft.)

113


Distribution Factor - One Lane

2’1.5’ 6’

36k

8’-0” 2.5’

36k 36k e = 12’



16’-0”

Note: Only the truck is used and it cannot be closer than 2’ from the barrier (3.6.1.3)



Minimum Exterior Girder Distribution Factor One Lane:LN

X ∑

( )

,

,

2

2 2

1 16(12)5

Ext Minb

Ext Min

ExtL

Nb

X eN

DFMN x

DFM

= +

= +

∑

∑



( ),

,

2 25 2

0.50

16 8

Ext MinDFM =

+

, ( ) 1.2(0.5) 0.6Ext MinDFM MPF DF= = =

114


Distribution Factor - Two Lanes

2’1.5’ 6’ 6’4’ 2’

12’ Lane 12’ Lane

36k 36k

e1 = 12’

36k 36k

e2 = 18.5’ - 1.5’ - 2’ - 6’ - 4’ - 2’ - 3’ = 0’



Note: Truck cannot be closer than 2’ from the barrier and the truck must be 2 feet from the lane edge.

(3.6.1.3)



Minimum Exterior Girder Distribution Factor Two Lane:N

,

2

2 16(12 0)

L

Ext Minb

N

ExtL

Nb

X eN

DFMN x

DFM

= +

+

∑

∑



,

,

2 2

( )5 2(160.70

8 )Ext Min

Ext Min

DFM

DFM

= +

=

+

, ( ) 1.0(0.7) 0.7Ext MinDFM MPF DF= = = CONTROLS

115



DFMtwo lanes = 0.585 lanes/girderDFMone lane = 0.600 lanes/girder (lever rule)DFMone lane 0.600 lanes/girder (lever rule)DFMminimum = 0.600 lanes/girder (one lanes)DFMminimum = 0.700 lanes/girder (two lanes)

The controlling DFM is the minimum DFM with two lanes loaded DFM = 0.7This is a 5% increase from the interior girder (DFM =


July 2007

This is a 5% increase from the interior girder (DFM = 0.665)


Exterior Girders:One Lane Loaded:


Exterior Beams Shear

One Lane Loaded:

Lever Rule

Two or More Lanes Loaded:

DFM Ext = e DFM Int



DFM,Ext e DFM,Int

1060.0 ede +=

116


Distribution Factor for Shear

Shear:Exterior Girder Two or More Lanes Loaded:Exterior Girder – Two or More Lanes Loaded:

DFExt = e DFInt

0.6101 0

ede = +



DFExt+ = (0.70) (0.814) = 0.570

1.00.6 0.70

10= + =


Distribution Factor for Shear

One Lane Loaded: (Lever Rule)

DFVEXT = 0.6

This is the same as moment calculation.

However, the minimum DF = 0.7 (from possible rigid body



However, the minimum DF 0.7 (from possible rigid body rotation) - THIS CONTROLS.

117


DL-Unfactored Shear Forces & Bending Moments - Exterior Girder

LocationBeam Weight [Simple Span]

Deck plus Haunch

[Simple Span]Barrier Weight

[Continuous Span]


[Continuous Span]Sh M Sh M Sh M Sh M

x ft. x/LShear kips

Mg, kip-ft

Shear kips

Ms, kip-ft

Shear kips

Mb, kip-ft

Shear kips

Mws, kip-ft

0.00 0.00 39.6 0 35.3 0 9.2 7.7 14.7 12.4

9.26 0.10 31.9 331 28.5 295.2 6.8 81.8 10.9 130.5

18.97 0.20 24 602.6 21.4 537.3 4.3 136 6.9 217

28.69 0.30 16 796.5 14.2 710.4 1.8 166 2.9 264.9

38.41 0.40 8 912.9 7.1 814.2 -0.6 171.9 -1 274.2

48.13 0.50 0 951.9 0 848.8 -3.1 153.6 -5 245.1



57.84 0.60 -8 912.9 -7.1 814.2 -5.6 111.2 -8.9 177.5

67.56 0.70 -16 796.5 -14.2 710.4 -8.1 44.7 -12.9 71.3

77.28 0.80 -24 602.6 -21.4 537.3 -10.6 -46 -16.9 -73.4

86.99 0.90 -31.9 331 -28.5 295.2 -13.1 -160.8 -20.8 -256.7

96.25 Brg. -39.6 0 -35.3 0 -15.4 -292.7 -24.6 -467.1



Exterior shear and bending moments

Length LL+IMV M

ft. k k-ft bending moments.

Maximum envelope values shown.

The values shown may not be from the

Bearing 0 76.5 50.9Trans. 2.04 74.0 199.4H/2 2.73 73.2 247.50.10L 9.26 65.3 655.80.20L 18.97 53.7 1101.80.30L 28.69 42.9 1365.50.40L 38.41 34.2 1483.0MidSpan 48.13 -41.3 1455.50.60L 57.84 -51.6 1301.1



may not be from the same load case.

0.70L 67.56 -61.8 1009.20.80L 77.28 -71.7 -815.00.90L 86.99 -81.3 -921.5H/2 93.52 -87.1 -1252.7Trans. 94.21 -87.7 -1299.1Bearing 96.25 -89.5 -1449.7

118


Load Combinations


(3.4.1)Service I:



Maximum Q = 1 25(DC) + 1 50(DW) + 1 75(LL + IM)



Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)


Load Combinations – Exterior Beam

Length Service 1 Service 3 Strength 1V M V M V M

ft. k k-ft k k-ft k k-ftBearing 0 175.3 71.0 160.0 60.8 261.1 117.3Trans. 2.04 168.2 416.2 153.4 376.4 250.8 630.3H/2 2.73 165.8 528.7 151.1 479.2 247.2 797.30.10L 9.26 143.4 1494.4 130.3 1363.2 214.6 2228.50.20L 18.97 110.2 2594.7 99.5 2374.3 166.4 3848.50.30L 28.69 77.8 3303.3 69.3 3030.2 119.5 4878.10.40L 38.41 47.7 3656.2 40.8 3359.6 76.4 5380.4MidSpan 48.13 -49.4 3654.7 -41.2 3363.6 -83.7 5357.40.60L 57.84 -81.2 3316.9 -70.9 3056.7 -129.6 4841.00 70L 67 56 -113 0 2632 0 -100 7 2430 2 -175 4 3812 5



0.70L 67.56 -113.0 2632.0 -100.7 2430.2 -175.4 3812.50.80L 77.28 -144.6 205.5 -130.3 368.5 -220.8 -568.00.90L 86.99 -175.6 -712.8 -159.3 -528.5 -265.4 -1635.0H/2 93.52 -195.9 -1707.1 -178.5 -1456.5 -294.3 -2930.0Trans. 94.21 -198.1 -1829.0 -180.6 -1569.2 -297.5 -3092.5Bearing 96.25 -204.4 -2209.5 -186.5 -1919.6 -306.4 -3603.6

119



Compression: (5.9.4.2.1)

Due to permanent loads, for service limit states:

For the precast girder: 0.45fc’ = 0.45(7.0) = +3.150 ksiFor the deck: 0.45fc’ = 0.45(4.5) = +2.025 ksi




For the precast girder: 0.40fc’ = 0.40(7.0) = +2.800 ksiFor the deck: 0.40fc’ = 0.40(4.5) = +1.800 ksi



Compression (con’t):Due to permanent and transient loads for service limit states:

(5.9.4.2.1)p

For the precast girder:

0.60Φw fc’ = 0.60(1.0)(7.0) = +4.200 ksi

For the deck:

0 60Φ f ’ = 0 60(1 0)(4 5) = +2 700 ksi



0.60Φw fc = 0.60(1.0)(4.5) = +2.700 ksi

Note: Φw is a factor for slender webs/flanges. It is not really meant for “I” girders. If the calculations required for Φw are done, Φw=1.

120



Tension:For components with bonded prestressing tendons:p p g

For the precast girder:

'0.19 0.19(7.0) 0.503cf ksi= =




Stresses at Midspan

Concrete stress at the top fiber of the girder, three cases:1 Under permanent loads Service I:1. Under permanent loads, Service I:

1

1

( ) ( )

972 972(20.0) (951.9 848.8)*12 (153.6 245.1)*12789 8,909 8,909 395761 23 2 18 2 43 0 12 1 60

pe pe c g s ws btg

t t tg

tg


f

f

+ += − + +

+ += − + +

+ + +




1 1.23 2.18 2.43 0.12 1.60tgf = − + + = +

121


Stresses at Midspan

2. One-half permanent loads plus live loads:( )M

2 1

2

2

( )0.5

1, 455*120.5(1.60)39576

0.80 0.44 1.24

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +





Stresses at Midspan


3 1

3

( )

1,455*12(1.60)39576

1 60 0 44 2 04

LL Itg tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +




3 1.60 0.44 2.04tgf = + = +

122


Stresses at Midspan

Concrete stress at the top fiber of the deck, three cases:1. Under permanent loads:

( )( )

(245.1 153.6)*1225271

0.186

ws btc

tc

tc

tc

M MfS

f

f

+=

+= +

= +




Note that deck stresses under service loads are almost always well below allowable for continuous for LL bridges; but they still must be checked.


Stresses at Midspan

2. One-half permanent loads plus live loads:( )LL IM

2 1

2

2

( )0.5

1,455*120.5(0.186)25721

0.09 0.68 0.77

LL Itc tc

tc

tc

tc

Mf fS

f

f

+= +

= +

= + = +




123


Stresses at Midspan

3. Under permanent and transient loads:( )M

C f O

3 1

3

3

( )

1, 455*12(0.186)25721

0.19 0.68 0.87

LL Itc tc

tc

tc

tc

Mf fS

f

f

+= +

= +

= + = +





Stresses at Midspan

Tension stress at the bottom fiber of the girder, Service III:

[ ]

( ) ( ) 0.8

(245.1 153.6) (0.8*1455) *12972 972(20.0) (951.9 848.8)*12789 10,542 10,542 16,3401.23 1.84 2.05 1.15 0.13


b b bc

b

b


f

f

++ + +

= + − −

+ ++= + − −

= + − − = −




124


Stresses at Midspan

GIRDER STRESSES INT EXTGIRDER STRESSES INT EXT

COMP – PERMANENT LOADS 1.98 ksi 1.60 ksi

COMP – ½ PERMANENT LOADS + LL

1.34 ksi 1.24 ksi

COMP – PERMANENT LOADS + LL

2.33 ksi 2.04 ksi


July 2007

LLTENSION 0.40 ksi 0.13 ksi






(Tables 3.4.1-1&2)

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +

,

,

1.25( ) 1.5( ) 1.75( )1.25(912.9 814.2 171.9) 1.5(274.2) 1.75(1,483)

u ext

u ext

M DC DW LL IMM

= + + +

= + + + +



, ,int5380 5,615u ext uM k ft M k ft= − < = −

Since exterior Mu is less than interior Mu, OK

125



The positive moment, under the Strength I limit state, for the exterior girder is less than that for interior girder. Although the LL increases, the DL decreases due to the flange (slab) being narrower.

The interior girder design met all the checks for positive moment design. These were: Nominal Strength, tension controlled, and minimum reinforcement. All of these checks depend on Mu and/or Mn. Since MU,ext<Mu,int, the design for the interior girder for POSITIVE MOMENT is adequate for exterior girder.


July 2007

Stresses at transfer of prestressing force is independent of whether the girder is interior or exterior, so no check is needed.

Loads & Analysis: Slide #249Do Not Duplicate




(3 4 1 1&2)1 25( ) 1 5( ) 1 75( )M DC DW LL IM= + + +

At the pier section:kip-ft

This is 4% greater than the moment for the interior girder. This is because the LL moment increases. At the support, the slab moment is 0 so it has no effect Away from the support the slab moment is

(3.4.1-1&2)1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +

1.25( 292.7) 1.5( 467.1) 1.75( 1, 450) 3604uM = − + − + − = −



0, so it has no effect. Away from the support, the slab moment is positive, so it would mitigate the negative moment. Thus, the smaller slab moment has the effect of INCREASING the negative moment, as compared to the interior girder.

126



(60)3 604(12) 0 90 (60) 58 25 sAA

= −

This is the required amount of mild steel reinforcement required in the slab to resist the negative moment and it is equal to 33 #6 bars.

2

2

3,604(12) 0.90 (60) 58.251.7(7.0)(26)

0 10.47 3145 43248

14.5

s

s s

s

A

A A

A in

=

= − +

=



g qDistributed over a length of 6.5 feet, this would be #6 @ 4 inches top and bottom! Use 16 bars on the bottom and 17 on the top. As = 14.52 in2

Note: Only 13.98 in2 were required for the interior girder.



Location of steel:Top – 17 #6 with 2” clearTop 17 #6 with 2 clearBtm – 16 #6 with 2 5/8” clear.

in233(0.44) 14.52sA = =

17(0.44)(2.375) 16(0.44)(8.5 3)14.52

x + −=


July 2007

We assumed 4.25” from top OKd = 58.6 in

56.48 3.914.52

x = =


127



Now check Mn:

( )( )( )( )

s y

c

1

A f 14.52 60a 5.63in

0.85f 'b 0.85 7 26a 5.63c 8.04

0.763

= = =

= = =β


July 2007

( )( )( )r n

r u

5.63M M 0.9 14.52 60 58.62

M 43740k in 3,645k ft M 3,604k ft

= φ = −

= − = − > = −





Based on the check made for the interior girders (requiring a spacing of 9 inches) #6@ 4 inches will clearly satisfy this requirement Note

700 2ec

s s

s dfγ

β≤ −


July 2007

of 9 inches), #6@ 4 inches will clearly satisfy this requirement. Note that the service level stress will increase, but not enough to bring the requirement below 4 inches.


128


Maximum Reinforcement – Negative Moment Section



tt

d c 59.9 8.040.003 0.003 0.019 0.005c 8.04− − ε = = = >


July 2007

This is a tension controlled section, so φ 0.9

(5.7.2.1 & 5.5.4.2)




(5.7.3.3.2-1)( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥

Where:

nc

'0.37 0.37 4.5 0.785cf = =

0g sM M+ =

fr = ksifcpe = 0.0 ksi

Mdnc= kip-ftS = 16340 in3


July 2007

Sc= 16340 in


129



16340 (0.785)12crM =

At bearing, the factored moment required by the Strength I load combination is: Mu = -3604 kip-ft

Therefore, kip-ft1.33 4793uM =

1069crM k ft= −

1.2 1282crM k ft= −


July 2007

Since , Controls

OKNote: The LRFD Specifications states that this requirement be met at every section.

1.2 1.33cr uM M< 1.2 crM

3,645 1.2 1282r crM M= > =




The design of the exterior section meets all requirements for positive and negative bending under both Service andfor positive and negative bending under both Service and Strength Limit States.


July 2007Loads & Analysis: Slide #258Do Not Duplicate

130


Design of the Section for Shear

This compares Strength I shears and moments for the interior and

Strength ILength Interior Exterior

V M V Mft k k-ft k k-ft for the interior and

exterior girders. Note that the exterior girder shears are LESS than the interior girdershears. Thus, the previous design works for vertical and

ft. k k ft k k ftBearing 0 299.125 113.1 261.0657 117.3438Trans. 2.04 287.45 644.925 250.7524 630.3376H/2 2.73 283.375 817.925 247.1722 797.26250.10L 9.26 246.375 2303.925 214.6325 2228.4850.20L 18.97 191.575 3993.775 166.3629 3848.4510.30L 28.69 138.4 5077.725 119.4571 4878.1260.40L 38.41 89.575 5615.875 76.42157 5380.371MidSpan 48.13 -95.9 5610.625 -83.733 5357.4420.60L 57.84 -147.875 5091.675 -129.581 4841.0080.70L 67.56 -199.95 4041.75 -175.438 3812.453



horizontal shear. The longitudinal steel requirements are also met.

0.80L 77.28 -251.375 -329.31 -220.846 -567.9670.90L 86.99 -301.825 -1464.58 -265.37 -1635.04H/2 93.52 -334.65 -2795.88 -294.34 -2929.99Trans. 94.21 -338.2 -2961.82 -297.47 -3092.54Bearing 96.25 -348.325 -3482.75 -306.435 -3603.56

Box Girder Example To be Used as an Example, Only ODOT Short Course Page 1 of 35

E3GUIDED DESIGN EXAMPLE

Non-composite, Skewed, Adjacent Box Girder Bridge; LRFD Specifications

1.1 INTRODUCTION

This design example demonstrates the design of a single span, 65 ft. long adjacent box girder bridge with a 30o right forward skew, as shown below. This example illustrates the design of typical interior and exterior beams at the critical sections in positive flexure and shear due to prestressing, dead load, and live load.

1.1-1

Longitudinal Section

1.1-2

Transverse Cross Section

1.1-3

Plan View


1.2

MATERIALS

Concrete @ release: fci’ = 5000 psi Concrete @ 28 days: fc’ = 7000 psi ODOT Bridge Design Manual (BDM) allows a range of strengths. These are chosen from that range [ODOT BDM 302.5.1.7]

1.2.1

Precast Beams

Ohio B33-48 box girder. Chosen from preliminary design charts in ODOT Design Data Sheets. Group “B” Design (roadway width 36 ft. to 48 ft.).

ODOT requires the use of minimum span to depth ratios given in LRFD Article 2.5.2.6.3. For a precast box, the limit is 0.03L = 0.03(65ft)(12in/ft) =23.4 inches OK.

1.2.3

Prestressing Strand

AASHTO M203 (ASTM A416) 7 wire, low relaxation, ½ inch dia., Gr. 270. Here, ½ inch strand is chosen, although the BDM allows both ½ inch and 0.6 inch diameter. [ODOT BDM 301.5.1.2a] Area of one strand = 0.153 in2 Ultimate strength, fpu = 270.0 ksi

1.2.4

Reinforcing Bars AASHTO M31 (ASTM A615), Gr. 60 [ODOT BDM 302.5.1.8].

Yield strength, fy = 60 ksi Modulus of elasticity, Es = 29,000 ksi


1.2.5

Loads Diaphragms: Two, 12 inch wide diaphragms at the 1/3 points

[ODOT Std. Drawings] Future wearing surface: 0.060 ksf (ODOT Design Data Sheets) Barriers: 0.090 k/ft each (ODOT Design Data Sheets) Truck: HL 93, including dynamic allowance

1.2.6

Bridge Parameters

Single Span Overall Length: 67 ft. c/c Span: 65 ft. Support: Elastomeric Bearing Pad

1.3

CROSS-SECTION

PROPERTIES FOR A

TYPICAL BEAM

1.3.1 Non-Composite

Section

Area in2 733.5 Weight (k/ft) 0.764 h (in) 33 yb (in) 16.61 yt (in) 16.39 I (in4) 108,150 Sb (in3) 6,511 St (in3) 6,599

1.5c 1 cE 33,000K w f '= [LRFD 5.4.2.4-1]

Units are kips; w is weight in kcf, fc’ is given in ksi. 1.5

cE 33,000(1.0)(0.150 kcf ) 5ksi 4,300 ksi= = - at transfer K1 is an aggregate factor = 1.0 unless specified by the owner.

1.5CE 33,000 1.0 0.150 7.0 5,072 ksi= × × = - service loads

1.3.2

Assumptions The current ODOT standard is to tie the girders together with tie rods, tightened

enough to bring the girders together, but not providing significant lateral post-tensioning. According to the commentary in the LRFD Specifications, for this bridge to be considered to have the girders “sufficiently connected”, a lateral post-tensioning force causing a stress of 0.25 ksi across the keyway is needed. Therefore, this bridge will be considered as not being “sufficiently connected”. In practice, all this does is change the distribution factor.

1.4

SHEAR FORCES & BENDING

MOMENTS 1.4.1

Dead Loads

DC = Dead load of structural components and non-structural attachments

Beam Weight: DCg = 0.764 klf Diaphragms: 2 at each 1/3 point:

( )( ) ( )( )( )d 2 2

33in 10.5in 48in 11inDC 1ft 2 diaphragms 0.150kcf 1.75k

144in / ft− −

= =


Asphalt Wearing Surface – at construction: ODOT specifies a MINIMUM of 3 inches in the Bridge Design Manual, but the Design Data Sheets use a 3.5 inch average to account for camber along the length of beam.

( )( )ws3.5inDC 4ft 0.120pcf 0.140klf

12in / ft= =

Rails – 0.090 klf – applied to exterior girders only (In other examples, barrier/railing loads are distributed equally to all the girders, but Article 4.6.2.2 appears to require a deck to distribute the load equally to all girders).

DW = future wearing surfaces and future DL

( )( )fwsDW 0.060ksf 4ft 0.240klf= =

1.4.1.1 DL-Unfactored

Bending Moments

Since this is a simple span beam, the most critical moment is at midspan.

( )( )

( )( )

2

DC

2

DW

0.764klf 0.140klf 65ft 65ftM 1.75k 515.3k ft8 3

0.240klf 65ftM 126.8k ft

8

+ = + = −

= = −

1.4.2

Live Loads According to LRFD Article 4.6.1.2.1 vehicular live loading on the roadways of bridges

or incidental structures, designated HL-93, shall consists of a combination of the: • Design truck or design tandem with dynamic allowance. The design truck shall

consists of an 8.0 kip front axle and a pair of 32.0 kip back axles. The first and second axle are spaced 14’-0” apart. The space between the rear axles shall be varied between 14.0’ and 30.0’ to porduce extreme force effects. The design tandem shall consist of a pair of 25.0 kip axles spaced 4.0’ apart. [LRFD Article 3.6.1.2.2 and 3.6.1.2.3]

• Design lane load shall consist of a load of 0.64 kip/ft uniformly distributed in the longitudinal direction. [LRFD Article 3.6.1.2.4]

Since this is a simple span, the maximum moment from the LANE LOAD occurs when the girder is fully loaded. Thus:

( )( )2

LL,Lane

0.640klf 65ftM 338k ft

8= = −

The HL-93 truck controls for this span length. Since this is a simple span, there is a simple formula for finding the maximum moment. The position of the resultant load is found and the midspan of the beam is placed halfway between the resultant and the nearest axle load. Note that the resultant is NOT used to find the moment, just the position of the axle loads. Also note that for a simple span, the moment is greatest when the back axles are as close together as possible, thus the minimum spacing of 14 feet is used.


The calculated moment is:

LL,TruckM 896.0 k ft= −

1.4.2.1 Distribution

Factors

The live load bending moments and shear forces are determined by using the simplified distribution factor formulas [LRFD 4.6.2.2]. To use the simplified live load distribution factor formulas, the following conditions must be met [LRFD 4.6.2.2.1]

Width of deck is constant. OK Number of beams, Nb > 4. OK Overhang part of the roadway < 3 ft OK de = 0.23 ft Curvature in plan < specified in Article 4.6.1.2 OK Beam parallel and of same stiffness OK Cross Section listed in Table 4.6.2.2.1-1 OK

For a precast concrete box beam with an asphalt surface, the bridge type is (g). [LRFD 4.6.2.2.1-1]

The number of design lanes should be determined by taking the integer part of the ratio w/12, where w is the clear roadway width in ft between curbs and/or barriers. [LRFD 3.6.1.1.1]

w = 48 ft. Number of design lanes = integer part of (48/12) = 4


1.4.2.1.1 Distribution Factors for

Bending Moment

This bridge is assumed to have no significant lateral post-tensioning. DFM = S/D

Where: S = width of precast beam (ft)

D = (11.5 -NL)+1.4NL(1-0.2C)2 when C < 5 [LRFD Table 4.6.2.2.2b-1] D = (11.5 -NL) when C > 5 Range of Applicability:

6LN ≤ 45Skew ≤ °

Where: NL = number of traffic lanes C = K(W/L) < K

Where: ( )

JI1K µ+

=

J is not published for ODOT girders. However, it can be approximated by:

( )= = =

+ +

∑

2224

4 1180in4AJ 211625inS 27.75in 42.5in 42.5in2t 5.5in 5.5in 5in

Where:

A = the area enclosed by the centerline of the box walls. T = wall thickness S = length of the centerline of a box wall.

µ = Poisson’s Ratio = 0.2 [LRFD 5.4.2.5]

( )

( ) ( ) ( )( )

4

4

2

1 0.2 108150inK 0.783

211625in48 ftC 0.783 0.57865 ft

D 11.5 4Lanes 1.4 4Lanes 1 0.2 0.578 11.9

S 4 ft 0.336D 11.9

+= =

= =

= − + − =

= =

Note that for boxes, K can be conservatively taken as 1. The DFM = 0.361, a difference of 8%. Also note that there is only one distribution factor for this case. This is different from other cases where there are factors for one lane loaded and two lanes loaded.


1.4.2.1.1 Distribution

Factors for Shear Force

Shear forces will be calculated in the section on shear design. The distribution factors will be calculated here. Two Lanes Loaded:

DFV = (b/156)0.4 (b/12L)0.1 (I/J)0.05(b/48) [LRFD Table 4.6.2.2.3a-1] Where:

1.048b≥

One Lane Loaded: DFV = (b/130L)0.15 (I/J)0.05 Range of Applicability:

5 < Nb < 20 Number of beams 35< b < 60 in Beam width 20< L < 120 ft Span 25000 < J < 610000 in4 40000 < I < 610000 in4

Two Lanes Loaded:

( )

0.10.4 0.0548 48 108150 48 0.456156 12 65 211625 48

= = DFV CONTROLS

One Lane Loaded:

( )

0.15 0.0548 108150DFV 0.445130 65 211625 = =

Because I/J is raised to a very small power, assuming I/J = 1 changes the DFV very little. In this example, the DFV is about 4% higher if I/J = 1.

1.4.2.2

Dynamic Allowance

IM = 33%

Where: IM = dynamic load allowance, applied only to truck load

1.4.2.3

Moment Reduction Factor for

Skew

1.05 0.25 tan 1.0g θ= − ≤ For 0 60θ° ≤ ≤ ° [LRFD Table 4.6.2.2.2e-1]

( )1.05 0.25 tan 30 0.905= − =og The specifications state that the MOMENT DISTRIBUTION FACTOR in a skewed bridge MAY be reduced by this factor. Note: Table 4.6.2.2.2e-1 has an inconsistency. It does not include this type of bridge in the description in the first column, but names it as a cross section type in the second column. It is assumed the skew factor applies to this structure.


1.4.2.4 Unfactored

Bending Moments

Unfactored bending moment due to HL-93 truck, per beam: MLL,Truck = (bending moment per lane)(DFM)(1+IM)(skew factor)

=(bending moment per lane)(0.336)(1.33)(0.905) =(bending moment per lane)(0.404)

= 896 k-ft (0.404) = 362.3 k-ft Unfactored bending moment due to HL-93 lane load, per beam: MLL,Lane = (bending moment per lane)(DFM)(skew factor) = (bending moment per lane)(0.336)(0.905) = 338 k-ft (0.304) = 102.7 k-ft (Impact is not applied to lane loads.)

1.4.3 Load

Combinations

The following limit states are applicable: [LRFD 3.4.1] Service I:

Q = 1.00(DC + DW) + 1.00 (LL + IM) Service III:

Q = 1.00(DC + DW) + 0.80(LL + IM) Strength I:


Fatigue: Does not need to be checked for pretensioned beams designed using the Service III load combination.

1.5

ESTIMATE REQUIRED

PRESTRESS


1.5.1

Service Load Stresses at

Midspan


DC DW LL Ib

b

M M 0.8MfS

++ +=

Where: fb = Bottom tensile stresses ksi

MDC = Unfactored bending moment due to DC loads kip-ftMDW = Unfactored bending moment due to DW loads kip-ft

MLL+I = Unfactored bending moment due to design vehicular live load including impact,

kip-ft

Sb = Section modulus to the bottom fiber in3

( ){ }( )b 3

515.3 126.8 0.8 362.3 102.7 k ft 12in / ftf 1.87ksi

6511in

+ + + − = =


Remember! For Service III (which applies ONLY to tension in fully prestressed members), the LL factor is 0.8!


1.5.2

Tensile Stress Limits for

Concrete

According to LRFD Table 5.9.4.2.2-1 the tensile stress limit at service loads is

b cf 0.19 f ' 0.19 7ksi 0.503ksi≤ = =

1.5.3 Required Number

of Strands

The difference between the bottom fiber tensile stress due to applied loads and the tensile stress limit is the required precompression stress.

( )pbf 1.87ksi 0.503ksi 1.37ksi= − = Assume the strands are 2 inches from the bottom of the girder. So the strand eccentricity at the midspan is:

ce 16.61in 2in 14.61in= − =

If Ppe is the total prestressing force, the stress at the bottom fiber due to presstress is:

= +pe pe cpb

b

P P ef

A S

Now plug in the required precompression stress, fpb and solve form Ppe:

pe

2 3

1.37ksiP 380kips1 14.61in

733.5in 6511in

= = +

Final prestress force per strand = (area of strand)(fpi)(1-losses, %) where fpi = initial prestressing stress before transfer, ksi. For Grade 270 strand, fpi = 0.75fpu = 202.5 ksi. Assuming 25% loss of prestress the final prestressing force per strand after losses is:

(0.153)(202.5)(1 0.25) 23.2kips− =

380kips# strands 16.423.2kips

= =

This shows a need for at least (18) ½ in diameter, 270 ksi, low-lax strands as the strand pattern must be symmetrical.



The ODOT Design Data Sheet for Group “B” roadway widths gives 20 strands at 2” from the bottom. Use the strand pattern of 20 strands shown at the midspan:

Using 20 strands allows for the possibility that the Strength Limit State controls. This pattern should work for exterior girders. Recall that the exterior girders will have the guardrail load and increased live load because of the exterior girder factor. It is NOT good design practice to have the exterior girder strand patterns be different than that for the interior girders. By using the same pattern for all girders, the fabricator has the option to fabricate exterior and interior girders in the same bed at the same time.

2.0

SERVICE LOAD LIMIT STATE

2.1 Prestress Losses


∆ = ∆ + ∆pT pES pLTf f f [LRFD 5.9.5.1-1] Where:

∆fpES = loss due to elastic shortening, ksi ∆fpLT = loss due to long-term shrinkage and creep of

concrete, and relaxation of the steel, ksi

2.1.1

Elastic Shortening

ppES cgp

ct

Ef f

E∆ = [LRFD 5.9.5.2.3a-1]


Where: fcgp = The concrete stress at the center of gravity of prestressing tendons due

to the prestressing force immediately after the transfer and the self-weight of the member at the section of the maximum moment (ksi). For the purpose of estimating fcgp, the prestressing force immediately after transfer may be assumed to be equal to 0.9 of the force just before transfer; also, change of concrete stress at the center of gravity of prestressing tendons due to subsequent applied loads, when considered.

2g ci ci

cgp

M ePePfA I I

= + −

Ep = Elastic Modulus of the prestressing steel (ksi). Eci = Elastic Modulus of the concrete at the time of transfer or time of load

application (ksi). Mg =

=

girder self weight at release ( )( )2

g

0.764klf 65ft 65ftM 1.75k 441.4k ft 5300k in8 3

= + = − = −

( )( )( )

( ) ( )

( )

2i

2

cgp 2 4 4

pES

P 20strands 0.9 202.5ksi 0.153in 558k

558k 14.61in 5300k in 14.61in558kf 1.15ksi733.5in 108150in 108150in

28500ksif 1.15ksi 7.6ksi4300ksi

= =

−= + − =

∆ = =

Note: In many example problems, the gravity moment for elastic shortening losses and stresses at release are calculated using the overall length of the girder. The thought here is that the girder will “sit up on its ends” and the span will be the overall length. In this example, the center of bearing to center of bearing span is used rather than overall length. This is done for 3 reasons:

1) This value will be needed later for service load calculations. Using it in this calculation saves a calculation later.

2) It is conservative as it actually results in higher losses and higher stresses in the concrete.

3) It doesn’t make that much of a difference. In this case, using the overall length increases the gravity moment 6% and decreases the loss 4%. The concrete unit weight, modulus of elasticity, modulus of rupture and the strength are not known with an accuracy that justifies being concerned over a few percent differences in the gravity moment.

2.1.2

Long-Term Losses

For standard, precast, pretensioned members subject to normal loading and environmental conditions:


pi pspLT h st h st pR

g

f Af 10 12 f

A∆ = γ γ + γ γ + ∆ [LRFD 5.9.5.3-1]

In which: h 1.7 0.01Hγ = − [LRFD 5.9.5.3-2]

stci

51 f '

γ =+

[LRFD 5.9.5.3-3]

Where: H = The average annual ambient relative humidity (%) γh = Correction factor for relative humidity of the ambient air γhst = Correction factor for specified concrete strength at time of

Prestress transfer to the concrete member ∆fpR = An estimate of relaxation loss taken as 2.5 ksi for low

relaxation strand, 10.0 ksi for stress relieved strand, and in accordance with manufacturers recommendation for other types of strand (ksi)

Assume H = 70%

( )h 1.7 0.01 70 1.00γ = − =

st5 0.83

1 5γ = =

+

So: ( )( )( ) ( ) ( ) ( ) ( )

2

pLT 2

pLT

202.5ksi 20 0.153inf 10 1.00 0.83 12 1.00 0.83 2.5

733.5inf 7.0 10.0 2.5 19.5ksi

∆ = + +

∆ = + + =

2.1.3



( )

pT pES pLT

pe

f f f 7.6 19.5 27.1ksi

27.1ksiLoss 100% 13.3%202.5ksi

f 202.5 27.1 175.4ksi

∆ = ∆ + ∆ = + =

= =

= − =

[LRFD 5.9.5.1-1]

Loss is less than the 25% initially assumed, so OK.

2.2

Compression Stress Limit

Sum of effective prestress + permanent loads < 0.45fc’ 1/2(Sum of effective prestress + permanent loads) + live load < 0.4 fc’ Sum of effective prestress + permanent loads + transient loads < 0.6φwfc’ [LRFD Table 5.9.4.2.1-1]


2.2.1 φw

φw is a modifier for sections with thin webs or flanges. It is actually defined in the section for hollow, rectangular compression members (Art. 5.7.4.7). It is based on the flange or web length/thickness ratio. Since this is for sections with thin webs/flanges, φw term will usually be = 1 for most beams. Find the web and flange slenderness ratios:

tX u

w =λ [LRFD 5.7.4.7.1-1]

Where: Xu = the clear length of the constant thickness

portion of the wall between other walls or fillets t = wall thickness

( ) ( )

( ) ( )w

w

48in 2 5.5in 2 3in6.2 Bottom Flange

5in33in 5.5in 5in 2 3in

2.9 Web5.5in

− −λ = =

− − −λ = =

The top flange λw < 15 by inspection. If λw < 15, φw = 1.0 [LRFD 5.7.4.7.2c-1]

( )2 2uX b lesser of z or y= −

2.2

Service Load

Stresses

Pe =20 strand (0.153in2)(202.5 ksi – 27.1 ksi) = 537 kips

( )cp,top 2 3

537k 14.61in537kf 0.457ksi733.5in 6599in

= − = −

( ) ( )

cDL,top 3

515.3 126.8 k ft 12in / ftf 1.17ksi

6599in + − = =

( ){ }( )

cLL,top 3

362.3 102.7 k ft 12in / ftf 0.85ksi

6599in+ −

= =


2.2.3 Service Load Compression

Stress Check

Service I

( )

( )( )

cp,top cDL,top c

cp,top cDL,topcLL,top

cp,top cDL,top cLL,top

f f 0.457ksi 1.17ksi 0.713ksi 0.45f ' 0.45 7ksi 3.15ksi

f f 0.713ksif 0.85ksi 1.21ksi 0.4(7ksi) 2.8ksi2 2

f f f 0.713ksi 0.85ksi 1.56ksi 0.6 1.0 7ksi 4

+ = − + = < = =

++ = + = < =

+ + = + = < = .2ksi

Compression stresses OK

2.3.4

Service Load Tensile

Stress Check Service III

The Service III stress at the bottom due to dead and live loads, fb, was calculated previously. The allowable tensile stress of 0.530 ksi was also calculated previously

( )pb 2 3

b

pb b


f 1.87ksif f 1.94ksi 1.87ksi 0.07ksi 0.07ksi COMPRESSION

= + =

= −+ = − = + =

No Tensile Stresses!!! Compression obviously OK Because the bottom of the girder is in compression, check with Service I:

Now it’s in tension, which is Service III ? Actually, it is sort of both. For all intents and purposes, the stress at the bottom of the girder is “0” – and this is a dividing line between Service I and Service III. Because of the 0.8 factor on the LL, there is an inconsistency between the two load cases. However the stress is so low, that really doesn’t matter – we satisfy all allowables in all cases.

( )2 3

537 14 61537 1 94733 5 65112 04

1 94 2 04 0 1 0 1

= + =

= −+ = − = − =

pb

b

pb b


f . ksif f . ksi . ksi . ksi . ksi TENSION

( ){ }( )3

515 3 126 8 362 3 102 7 122 04

6511

+ + + − = =b

. . . . k ft in / ftf . ksi

in


3.0 STRENGTH

LIMIT STATE 3.1

Factored Moment

Strength I:

Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)

Q = 1.25(DC) + 1.50(DW) + 1.75(Truck + Lane)

( ) ( ) ( )uM 1.25 515.3 1.50 126.8 1.75 362.3 102.7 1648k ft 19780k in= + + + = − = −

3.2

Steel Stress At Strength Limit State

ps pu

p

cf f 1 kd

= −

[LRFD 5.7.3.1.1-1]

Where: k = 0.28 for low relaxation strands

Assume the section is rectangular:

ps pu s s s s

puc 1 ps

p

A f A f A 'f 'c f

0.85f ' b kAd

+ −=

β + [LRFD 5.7.3.1.1-4]

Where: Aps = Area of prestressing steel in2 fpu =

= Specified tensile strength of prestressing steel 270

ksi

As = =

Area of mild steel tension reinforcement 0.0

in2

fy = =

Yield strength of tension reinforcement 60.0

ksi

As‘ = =

Area of compression reinforcement 0.0

in2

fy‘ = =

Yield strength of compression reinforcement 60.0

ksi

fc‘ = =

Compressive strength of deck concrete 7.0

ksi

β1 = = =

Stress block factor specified in LRFD 5.7.2.2 0.85 – 0.05(f’c – 4.0) > 0.65 for f’c > 4.0 0.70

b = =

Width of compression flange 48

in.

( )

( )( )( ) ( )( )

2

2

ps

20 0.153in 270ksi 0 0c 3.98in. 5.5in.270ksi0.85 7ksi 0.7 48in 0.28 20 0.153in

31in

3.98inf 270ksi 1 0.28 260ksi31in

+ −= = <

+

= − =


c is also the neutral axis depth, so the stress block depth, a = β1c = 0.7(3.98) = 2.79 inches. Since c < hf, the stress block is entirely in the flange so the beam may be treated as rectangular.

3.3

Flexural Resistance

The moment equation in the LRFD Specification looks like this

( ) fn ps ps p s y s s y s c w f

ha a a aM A f d A f d A ' f ' d ' 0.85 f ' b b h2 2 2 2 2

= − + − − − + − −

If the section is rectangular (b=bw), the equation becomes:

' ' '2 2 2n ps ps p s y s s y sa a aM A f d A f d A f d = − + − − −

If there is no compression or mild tension steel, the equation becomes:

n ps ps paM A f d2

= −

Since c < hf, the section may be treated as rectangular.

( )( )

n ps ps p

2n

a 2.79inaM A f d2

2.79inM 20 0.153in 260ksi 31in 23550k in2

=

= −

= − = −

[LRFD 5.7.3.2.2-1]

Note: The nominal flange width of 48 inches was used for “b”. In reality, the flange area is reduced by the shear key cut-out. However, this is often ignored as this would require an iterative procedure. If the area is adjusted for the shear key, the nominal moment, Mn changes by only 0.10%. It may not be appropriate to reduce the area by the shear key cut-out as this will be filled with grout and the grout may act with the base concrete to effectively provide the complete flange width. All of this is a matter of engineering judgment.


3.4 Determination

Of Phi

To determine Φ, it is necessary to calculate the steel strain at the level of the extreme tensile steel. c = 3.98 inches (calculated above) dt is the distance to the extreme tensile steel. Since there is only one row of steel, dt = dp.

tt

d c 31in 3.980.003 0.003 0.0204c 3.98− −

ε = = =

Since εt > 0.005, the section is tension controlled. [LRFD 5.7.2.1] Φ = 1.0 [LRFD 5.5.4.2.1] This is a big change from the old ρbalanced method. However, this now makes the LRFD Specifications consistent with ACI 318. This replaces the maximum reinforcement provisions.

3.5

Determination of Flexural Strength

( )( )u nM M

19,780k in 1.0 23550k in OK≤ Φ

− < −

3.6

Maximum and Minimum

Reinforcement

For minimum reinforcement, the resistance moment, Mr must be at least the lesser of 1.2 times the cracking moment or 1.33 times the factored applied moment.

1.33Mu = 1.33(19780 k-in) = 26310 k-in For the cracking moment, find the modulus of rupture:

r cf 0.37 f ' 0.37 7ksi 0.979ksi= = = [LRFD 5.4.2.6] Note that this is a new MOR for minimum reinforcement. It is equal to 11.5√fc’ in psi; which is the upper bound for MOR. Next, determine the stress at the bottom of the box due to effective prestressing force:

( )cpe 2 3


= + =


( )cr b r cpeM S f f= + [LRFD 5.7.3.3.2-1]

( )3crM 6511in 0.979ksi 1.94ksi 19000k in= + = −


1.2Mcr = 1.2(19000k-in) = 22800k-in < 1.33Mu Mr = φMn = 1.0(23550) k-in = 23550k-in > 22800 k-in OK Note: When the number of strands was selected, it was determined that 18 strands would be needed, but 20 were used. If 18 strands had been used, φMn = 21400 k-in, so 18 strands would NOT meet the minimum requirement.

4.0

STRESSES AT

TRANSFER 4.1

Steel Stress At Transfer

Assume the stress at transfer is 0.9fpi Pi = 20 strand(0.153in2)(0.9)(202.5 ksi)=558 kips

4.2

Allowable Stresses at

Transfer

Tension: 0.0948√fci’ < 0.2ksi w/o bonded reinforcement [LRFD Table 5.9.4.1.2-1] 0.24√fci’ w/bonded reinforcement Compression: 0.6fci’

4.3

End Stress

At Transfer

( )

( )pt 2 3

pb 2 3



= − = −

= + =

These stresses should be calculated at the end of the transfer length = 60db=30 inches. The dead load stresses 30 inches from the support should be added. However, these stresses will not be large and it is conservative to use just the stress due to prestressing.

fpt = 0.474 ksi tension < 0.24√fci’ = 0.24√5ksi = 0.537 ksi OK w/bonded steel fpb = 2.01 ksi compression < 0.6fci = 0.6(5ksi) = 3 ksi OK

Because the stress is OK, no debonding is needed. However, if debonding was needed, no more that 25% of the total number of strands could be debonded and no more than 40% in one row can be debonded. [LRFD 5.11.4.3]

4.3.1

Bonded Steel

Bonded steel is needed at the top of the girder at the end to take the tensile forces. This steel must resist the total tension in the top flange with a stress of no more than 0.5fy but not more than 30 ksi. [LRFD Table 5.9.4.1.2-1] The first step it to find the tension in the flange. This requires the location of the neutral axis to be determined. From the top and bottom stresses at the end, the neutral at the end is:


From the top and bottom stresses at the end, the neutral at the end is:

( )0.474ksi 33in

x 6.30in0.474 2.01ksi

= =+

The top flange is 5.5 inches, so the stress at the bottom of the top flange is:

( )0.474ksi 6.3in 5.5in 0.060ksi6.30in

− =

The total tensile force is:

( )( )( )( ) ( ) ( )( )0.474ksi 0.060ksiT 0.5 6.30in 0.474ksi 5.5in 2 5.5in 48in 2 5.5in2

T 70.8kips

+= + −

=

Again, this tension could be reduced by calculating the force at the end of the transfer length (including the gravity moment). Including the gravity moment will reduced the calculated tension, but because bars only come in certain sizes, the reduction may not change the number of bars needed. The bonded steel must resist the total tensile force with a stress not exceeding the lesser of 0.5fy or 30 ksi. [LRFD Table 5.9.4.1.2-1]

2

s70.8kipsA 2.36in

30ksi= =

Use 8 #5 The length of the bar is determined by the point where bonded steel is no longer required. Since 0.0948√fci’ = 0.212 ksi > 0.2ksi; find the point where the dead load drops the stress below 0.2 ksi. For simplicity, just consider the beam weight and ignore diaphragms. The required moment = ∆fc St = (0.474 ksi – 0.200 ksi) 6599 in3 = 1808 k-in = 150.7k-ft

( ) ( )2

M 150.7k ft 0.5 0.764klf x 65ft x

150.7k ft 24.83x 0.382xx 6.75ft; 58.25ft

= − = −

− = −=


This is from center of bearing, so extend steel 7.75 ft. from each end and then add development length.

b yd b y

c

1.25A f0.4d f

f '= ≥l [LRFD 5.11.2.1.1]

( ) ( )( )2

d

1.25 0.31in 60ksi8.8in 0.4 0.625in 60ksi 15in

7ksi= = < =l

Where: Ab = Area of the bar db = diameter of bar

Top bar factor = 1.4 1.4(15inches) = 21 inches So the minimum bar length = 6’- 9” + 1’ – 9” = 9’ – 6”

4.4

Midspan Stress

At Transfer

Mg = 5300 k-in (calculated in the section on losses - 2.1.1)

t,DL 3

b,DL 3

top

bot

5300k inf 0.803ksi6599in5300k inf 0.814ksi6511in

f 0.474ksi 0.803ksi 0.329ksi

f 2.01ksi 0.814ksi 1.120ksi

−= =

− −= = −

= − + =

= − =

By inspection, both are below the compression limit.

5.0

SHEAR 5.1

Critical Section

The critical section is at dv from the face of the support for a section where the reaction force in the direction of the applied shear introduces compression into the end region of the member. [LRFD 5.8.3.2] For this member with only a single layer of prestressing steel:

v pa 2.79ind d 31in 29.6inches2 2

= − = − =

The term dv is not taken less than: 0.9de = 0.9(31 inches)=27.9 inches < 29.6 inches or 0.72h = 0.72(33 inches) = 23.76 inches < 29.6 inches

Assuming a 1 ft. long bearing pad, the critical section is: 29.6+6 = 35.6 inches from center of bearing. For calculations, use 36 inches = 3 ft. The difference is only a few percent.


5.2 Shear Forces

And Moments At the Critical

Section 5.2.1 Basic Shear

Forces And

Moments At the

Critical Section

DC: For beam weight:

( ) ( )( )( ) ( )( )( )

g

g

V w 0.5L x 0.764klf 0.5 65ft 3ft 22.54k

M 0.5wx L x 0.5 0.764klf 3ft 65ft 3ft 71.0k ft

= − = − =

= − = − = −

For the diaphragm, V = 1.75k (shear is constant), M = 1.75(3) = 5.25k-ft

For the wearing surface:

( )( )( )( )( )

0.140 0.5 65 3 4.13

0.5 0.140 3 65 3 13

= − =

= − = −ws

ws

V klf ft ft k

M klf ft ft ft k ft

DW:

( )( )( )( )( )

fws

ws

V 0.240klf 0.5 65ft 3ft 7.08k

M 0.5 0.240klf 3ft 65ft 3ft 22.3k ft

= − =

= − = −

Live Load: Consider the influence line for shear:

The shear at x is maximized by placing the rear wheel of the truck at x and loading the right part of the beam with the uniform load. (Note that influence lines are NOT used for dead loads. Obviously, it is not possible to have the DL on only part of the beam!) Using a standard structural analysis program, at the critical section:

VLL,Lane = 18.92k VLL,Truck = 58.33k MLL,Lane = 56.76 k-ft MLL,Truck = 175.0 k-ft


5.2.2 Skew

Factor

This is a multibeam bridge. The shear at the obtuse corner of each girder MUST be increased by:

( )( ) ( )12 65ft12L1 tan 1 tan 30 1.20

90d 90 33in+ θ = + = [LRFD Table 4.6.2.2.3c-1]

Note that this factor applies only to the distribution factor. Since the critical section is only 3 feet from the support, apply the skew factor.

5.2.3

Factored Moments

And Shears

As calculated in Section 1.4.2.1.1 of this example: DFV = 0.456 DFM = 0.336

The moment MAY be multiplied by the skew factor for moment, 0.91. The shear MUST be increased by skew factor, 1.20. Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)

VLL+IM = 0.456(1.2)[58.33(1.33)+18.92] = 52.5 k Vu = 1.25(22.54k + 1.75k + 4.13 k) + 1.50 (7.08 k) + 1.75(52.5 k) = 138.0 k

MLL+IM = 0.336(0.905)[175 k-ft(1.33)+56.8] = 88.0 k-ft Mu = 1.25(71.0 k-ft + 5.25 k-ft + 13.0 k-ft) + 1.5(22.3 k-ft) + 1.75(88.0 k-ft) = 299 k-ft = 3588 k-in

5.3

Sectional Design Model

For shear design, the shear forces at various points along the girder should be calculated. Normally, this is done at the critical section, at points where strands are debonded or harped and then at every 0.1L. For this design example, only the shear at the critical section is analyzed. The same procedure for the remaining points would be used. The LRFD Specifications adopted the modified compression field theory for shear design with Version 1. This was called the Sectional Design Model. In Version 4 (2007), the Simplified Method was added. The Simplified Method restores the old Vci and Vcw from the Standard Specifications. Both methods will be illustrated in this example.

5.3.1

Finding β and θ

The sectional design model requires the calculation of two factors:

• Concrete strain at : εx • Average shear stress in the concrete: v

These two values are used to find β and θ; which are then used to find the strength of the concrete and the strength of the stirrups.


5.3.1.1 Finding

εx

The first step is to find the strain at 0.5dv in the cross section. It is assumed the section is uncracked and that at least minimum transverse reinforcement will be used. Note that θ is unknown at this point. However, the commentary allows 0.5cotθ=1 as a simplification. [LRFD C5.8.3.4.2]

[LRFD 5.8.3.4.2-3]

Where:

Nu = Applied factored normal force at the specified section = 0.0

kips

Vp = Strands are not harped = 0.0 kips fpo =

=

A parameter taken as modulus of elasticity of prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and the surrounding concrete .7 0.7(270.0) 189puf = =

ksi [LRFD 5.8.3.4.2]

Aps =

=

Area of prestressing steel on the flexural tension side of the member, as shown in LRFD Figure 5.8.3.4.2-1.20 strands(0.153) = 3.06

in2


in2

Ac =

=

Ac is the area of concrete on the tension half of the beam; it is the area of the bottom half (h/2). 2(5.5in)(33in)(0.5) + (48in-11in)(5in) = 366.5 in2

in2

Ep = 28,500 ksi dv = 29.6 in

( )

( ) ( )

2

6 3x 2 2

3588k in 138k 3.06in 189ksi29.6in 82x10 0.08x10

2 28500ksi 3.06in 5072ksi 366.5in− −

−+ −

ε = = − ≈ − +

Negative means “uncracked”.

5.3.1.2

Finding vu

( )( )( )( )u p

u cv v

V V 138kv 0.469ksi 0.18f ' 1.26ksib d 0.9 2 5.5in 29.6in−φ

= = = < =φ

[LRFD 5.8.2.9] Where:

Vp = 0 φ = 0.9 [LRFD 5.5.4.2.1]

( )ccpspss

popspuuv

u

x AEAEAE2

fAcotVV5.0N5.0dM

++

−−++=

θε


5.3.1.3 β and θ

Using u

c

v 0.469ksi 0.067f ' 7ksi

= = and x 0.08ε = −

From LRFD Table 5.8.3.4.2-1: θ = 21◦ β = 4.1

5.3.2

Shear Strength of Concrete

c c v vV 0.0316 f 'b d= β [LRFD 5.8.3.3-3]

( ) ( )( )cV 0.0316 4.1 7ksi 11in 29.6in 111.6k= = Since Vu = 138k > φVc = 0.9(111.6k) = 100 k; at least minimum stirrups are needed for strength. The equations for β and θ assumed minimum stirrups.

5.3.3

Minimum Stirrups ( )

( )u c

max v

v 0.469ksi 0.125f ' 0.125 7ksi 0.875ksi

s 0.8d 0.8 29.6in 23.7in 24in

= < = =

= = = < [LRFD 5.8.2.7]

smax = 23.75 in.

Calculate minimum area of steel using a 12 inch spacing to get area of steel per foot:

( )( ) 2vv c

y


≥ = = [LRFD 5.8.2.5]

ODOT uses #4 bars with 2 legs as standard (Av = 2(0.2in2) = 0.4in2) @ 12 inch o.c. This is adequate to meet minimum.

5.3.4

Shear Strength of the Girder

( )v y vs

A f d cot cot sinV

sθ+ α α

= [LRFD 5.8.3.3-4]

The stirrups are perpendicular to the main steel so α = 90o; cotα = 0, sinα=1; θ = 21o

( ) ( )( )( ) ( ) ( )2v y v

s

s

0.4in 60ksi 29.6 cot 21 0 1A f d cot cot sinV

s 12inV 154.2k

+θ+ α α = =

=

( )n c s p

u n

V V V V 111.6k 154.2k 0 265.8k

V 138k V 0.9 265.8k 239.2k

= + + = + + =

= < φ = =


5.3.5 Maximum Nominal Shear

Resistance

The upper limit of Vn, given by following equation, is intended to ensure that the concrete in the web of the beam will not crush prior to yield of the transverse reinforcement.

'0.25n c v v pV f b d V≤ + [LRFD 5.8.3.3-2]


With Vp=0:

'0.25111.6 154.2 0.25(7)(11)(29.6)265.8 569.8

c s c v vV V f b d+ ≤+ ≤≤

5.4

Simplified Shear

In the 2007 LRFD Specification, the simplified shear method is introduced. This method brings back Vci and Vcw from the Standard Specification. • Vcw (web shear) usually controls near the support, so Vcw will be checked at the

critical section. • Vci (flexural shear) doesn’t control near the support, so for this example, Vci will be

calculated at 0.2L. However, in practice Vci and Vcw must be checked at all appropriate sections.

5.4.1

Vcw

[LRFD 5.8.3.4.3-3] Where:

fpc = compressive stress in concrete (after allowance for all prestress loses) at centroid of cross section resisting externally applied loads or at the junction of the web and the flange when the centroid lies within the flange (ksi).


epc 2

P 537kf 0.732ksiA 733.5in

= = =

( )( ) ( )( )cwV 0.06 7ksi 0.3 0.732ksi 11in 29.6in 123.2kips= + =

The critical section is 29.6 inches from the face of the support. Assuming a 1 ft bearing pad, the critical section is approximately 3.5 feet from the end of the beam. The transfer length is 60 bar diameters = 30 inches. Thus, the critical section is past the transfer length, so fpc does not have to be reduced for lack of bond. If the critical section is within the transfer length, fpc is reduced linearly. One difference between LRFD and Standard Specifications is that LRFD uses cotθ in the Vs calculation. For Vcw, the term cotθ must be calculated:

pc

c

fcot 1.0 3 1.8

f 'θ = + ≤ [LRFD 5.8.3.4.3-4]

( )cw c pc v v pV 0.06 f ' 0.3f b d V= + +


0.732ksicot 1.0 3 1.83 1.8; so use 1.8

7ksiθ = + = >

θ = 29° The minimum stirrup area and maximum spacing calculated in the Sectional Model still applies here. Assuming #4 stirrups @ 12 in:

( ) ( ) ( ) ( )2

s

0.4in 60ksi 29.6in 1.8V 106.5k

12in= =

( )uV 138k 0.9 123.2k 106.5k 207k= < + =

5.4.2

Vci Vci does not control near supports of simply supported beams. It will be calculated at

0.2L=13 ft from the center of the support.

5.4.2.1 Unfoactored Dead Loads

The equation for Vci requires the calculation of unfactored dead loads. DC:

For beam weight: ( ) ( )( )

( ) ( )( )( )g

g

V w 0.5L x 0.764klf 0.5 65ft 13ft 14.9k

M 0.5wx L x 0.5 0.764klf 13ft 65ft 13ft 258k ft

= − = − =

= − = − = −

For the diaphragm, V = 1.75k (shear is constant), M = 1.75(13) =22.8k-ft For the wearing surface:

( )( )( )( )( )

fws

ws

V 0.140klf 0.5 65ft 13ft 2.73k

M 0.5 0.140klf 13ft 65ft 13ft 47.3k ft

= − =

= − = −

DW:

( )( )( )( )( )

fws

ws

V 0.240klf 0.5 65ft 13ft 4.68k

M 0.5 0.240klf 13ft 65ft 13ft 81.1k ft

= − =

= − = −

The total UNFACTORED shears and moments are:

Vd = 14.9k + 1.75k + 2.73k + 4.68k = 24.1k Md = 258.0k-ft + 22.8k-ft +47.3k-ft + 81.1k-ft = 409.2 k-ft = 4910 k-in

The FACTORED shears and moments are:

Vud = 1.25(14.9k + 1.75k + 2.73k) + 1.50(4.68k) = 31.3 k Mud = 1.25(258.0k-ft + 22.8k-ft +47.3k-ft) + 1.5(81.1k-ft) = 531.8 k-ft = 6381 k-in


5.4.2.2 Live Load

This method requires two sets of shears and moments for Live Load. The first is the loading where the shear is maximum and the second is where the moment is maximum. For the lane load, the shear is maximum when the lane load is on the right 52 ft. of the girder (see the influence line from the sectional model):

VLane1 = 13.3k and MLane1 = 173 k-ft = 2076 k-in The maximum moment occurs when the lane load is on the entire girder:

VLane2 = 12.5k and MLane2 = 216.3 k-ft = 2596 k-in

Clearly, the moment is maximum when the lane load is placed along the entire beam. The truck load is less certain. The moment at “X” is the value of the point load times the ordinate of the influence line. Unfortunately, it is not clear where this product will be maximum!


For the truck, it is again necessary to consider two placements: Placed for maximum shear Placed for maximum moment

In this case, it just happens that both are the same – the rear axle placed at 0.2L as shown in the previously. For the truck load, the maximum shear at the section and the maximum moment at the section happen to occur under the same loading – the rear wheel of the truck 13 ft. from the support. In this case, the maximum shear loading and the maximum moment loading are the same, but that is NOT always the case. Be sure to carefully check all reasonable load conditions. However, this is not always the case. It just happened that way in this example.

VTruck = 47.2 k and MTruck = 613 k-ft = 7356 k-in Vu,LL = 1.75[Vtruck(1+IM) + VLane](DFV) Vu,LL = 1.75[47.2k(1.33) + 13.3k]( 0.456) = 60.7k

Note that the skew factor is NOT applied. The skew factor is applied only at the obtuse corner and at 0.2L, the section is not at the obtuse corner.

Mu,LL = 1.75[Mtruck(1+IM) + MLane](DFM)(skew factor) Mu,LL = 1.75[613 k-ft(1.33) + 216.3 k-ft](0.336)(0.905) = 549.0 k-ft = Mmax

Note that the Skew Factor IS Applied to moment The shear associated with maximum moment is:

Vi = 1.75[47.2k(1.33) + 12.5k]( 0.456) = 60.0 k

5.4.2.3 Determination of

Cracking Load for Shear

First, find the modulus of rupture:

r cf 0.2 f ' 0.2 7ksi 0.529ksi= = = [LRFD 5.4.2.6]

Note that LRFD has 3 different MORs – be sure to use the correct one! Next, determine the stress at the bottom of the box due to effective prestressing force:

( )cpe 2 3


= + =

dnccre c r cpe

nc

12MM S f fS

= + −

[LRFD 5.8.3.4.3-2]


Where: Mdnc = Unfactored moment due to dead load on the non- composite or

monolithic section = 409.2 k-ft (note – in k-ft; 12 in numerator converts to inches)

Snc = non-composite section modulus Sc = composite section modulus = Snc since this is a non-composite

structure

( ) ( )3cre 3

cre

12 409.2k ftM 6511in 0.529ksi 1.94ksi

6511inM 11165k in 930.5k ft

− = + −

= − = −

5.4.2.4

Vci

i creci c v v d c v v

max

V MV 0.02 f 'b d V 0.06 f 'b dM

= + + ≥ [LRFD 5.8.3.4.3-1]

( )( ) ( )( )

( )( )

ci

60.0k 930.5k ftV 0.02 7ksi 11in 29.6in 24.1k 143.0k

549k ft0.06 7ksi 11in 29.6in 51.7

−= + + = >

−=

5.4.2.5

Check Shear Strength

uV 31.3k 60.7k 92.0k= + = Assuming #4@12; cotθ=1 for Vci [LRFD 5.8.3.4.3]

( )( )( )( )2

s

0.4in 60ksi 29.6in 1.0V 59.2k

12in= =

( )u nV 92.0k V 0.9 143.0k 59.2k 182.0k= < φ = + =

The section is adequate in shear. If s=18”

sV 39.5kips=

( )u nV 92.0k V 0.9 143.0k 39.5k 164k= < φ = + =

5.6 Minimum

Longitudinal Steel

At each section: [LRFD 5.8.3.5-1]

u u ups ps s y p s

v


+ ≥ + + − − θ φ φ φ

For this example, the minimum longitudinal steel will be checked at the critical section. The critical section 29.6 inches from the face of the support. Allowing for a 1 ft. bearing pad and one foot from center of bearing to the end of the girder, the critical section is 47.6 inches from the end of the girder. However, it is necessary to see if the strand stress is reduced by lack of development.


The development length equation is unchanged for strand from Standard Specifications, except that a factor, κ is added. This factor is the result of an October, 1988 FHWA memorandum suggesting the need for this conservative multiplier because of strand/bond problems:

( ) ( )d ps pe b2 2f f d 1.6 260 175.4 0.5 114.5in3 3

= κ − = − =

l [LRFD 5.11.4.2]

The terms fps (steel stress at strength limit) and fpe (effective prestressing stress after losses) were calculated previously. κ = 1.6 for member over 24 inches deep The critical section occurs at 47.6 inches from the end of the beam, but the development length is 114.5 inches. Thus, the steel stress MUST be reduced to account for lack of development.

( )6060

px bpx pe ps pe

d b

df f f f

d−

= + −−

l

l [LRFD 5.11.4.2-4]

The following values were previously calculated or determined:

Aps =20(0.153)= 3.06 in2 fps = 260.0 ksi fpe = 174.5 ksi Mu = 3588 k-in Vu = 138 k θ = 21o (Sectional Design Model) Vs = 153 k (Sectional Design Model) Nu = Vp = 0 φ = 1 for moment; 0.9 for shear Asfy = assumed 0 (ignore any mild steel) 60db = 30 inches

( )47 6 30174 5 260 0 174 5 192 0114 5 30px

. in inf . ksi . ksi . ksi . ksi. in in

−= + − =

−

u u u

ps ps s y p sv


+ ≥ + + − − θ φ φ φ

( )

( ) ( )

23 06 192 0 588

3588 138 0 5 153 21 3211 0 29 6 0 9

. in . ksi k

k in k . ( k ) cot k. . in .

=

− > + − =

OK.


Note that before the 2005/06 interim, the steel stress was assumed linear with development length, not bilinear. If the stress were assumed linear here, mild steel would need to be added. Also note that Vs < Vc/φ = 153k Check the inside face of the bearing pad. Assuming a 12 in pad and one foot from center of bearing to the end, the inside of the pad is 12+6 = 18 inches from the end of the girder. This is inside the transfer length:

18174 5 104 730px

inf . ksi . ksiin

= =

( ) ( )2

0 5

1383 06 104 7 320 0 5 153 21 1990 9

up ps s

VA f . V cot


θφ

≥ −

= > − =

OK If the stirrup spacing is increased to 18”, Vs = 103 k

( ) ( )2

0 5

1383 06 104 7 320 0 5 103 21 2650 9

up ps s

VA f . V cot


θφ

≥ −

= > − =

OK

5.7 Anchorage Zone

(Bursting Stirrups)

The bursting resistance of pretensioned anchorage zones provided by vertical reinforcement in the ends of the pretensioned beams at the service limit state shall be take as:

r s sP f A= [LRFD 5.10.10.1-1] Where:


in2

fs = Stress in steel, but not taken greater than 20 ksi Pr = Bursting resistance, should not be less than

4% of Fpi 20(0.153)(202.5)(0.04) 24.8=

kips

Solving for the required area of steel, 24.8 1.2420

= =sA in2

As in the Standard Specification, LRFD requires bursting stirrups which can resist at least 4% of the initial prestressing force, with a stress of no more than 20ksi: This steel must be distributed over h/4 from the end. For this girder, h/4=33/4=8.25 inches. Four #4 double leg stirrups @ 3” provides 1.60 in2 over 8 inches.


6.0 EXTERIOR

GIRDER 6.1

Moment

The exterior girder takes the rail load (DC): ( )2

b

0.090klf 65ftM 47.5k ft 570k in

8= = − = −

Note: Article 4.6.2.2.1 allows the rail load to be equally distributed to all the girders. However, it does not have to be and, in this case, it is probably more correct to assign the railing to the exterior girder. The live load moments must be multiplied by the exterior girder factor. Two or more lanes loaded: [LRFD Table 4.6.2.2.2d-1]

ext int

e

g egde 1.04 125

=

= + >

Since the rail is right at the edge of the box, de = half the web width = 2.75 inches = 0.23 ft. Note that de is in FEET.

0.23e 1.04 1.04925

= + =

One lane loaded:

ext int

e

g egde 1.125 130

=

= + >

0.23e 1.125 1.13330

= + = Controls

Note that there is only one DFM, so the one lane e is multiplied by the DFM. In the equation below, the truck load (362.3 k-ft) is already multiplied by the interior DFM and the impact factor; the lane load (102.7 k-ft) is multiplied by the DFM (no impact on lane load). Thus, it is only necessary to multiply by the increasing factor:

( ) ( ) ( )( )u

u

M 1.25 515.3 47.5 1.50 126.8 1.75 362.3 102.7 1.133M 1815k ft 21790k in

= + + + +

= − = −

For the interior box with 20 strands, φMn = 23550 k-in so OK for Mu Stresses at transfer do not need to be checked as these stress occur during fabrication are independent of the railing load and the live load. The check performed on the interior girders is sufficient.


Service load stresses should be checked. It is clear by inspection that service load compression stresses are OK (see Section 2.3.3). Check Service III:

( ) ( ) ( )( )

bottom 3

M 515.3 47.5 126.8 0.8 362.3 102.7 1.133 1111k ft 13330k13330k inf 2.05ksi

6511in

= + + + + = − = −

−= =

fps = 1.94 ksi compression (previously calculated) fbottom = 1.94 ksi – 2.05 ksi = -0.110 ksi = 0.110 tension < 0.503 ksi tension OK

6.2

Exterior Girder Shear

This check must be performed at all sections. Only the critical section is shown here. The check is also made using Sectional Model. At the critical section:

( ) ( )( )( ) ( )( )( )

g

g

V w 0.5L x 0.090klf 0.5 65ft 3ft 2.65k

M 0.5wx L x 0.5 0.090klf 3ft 65ft 3ft 8.37k ft

= − = − =

= − = − = −

Two or more lanes loaded: [LRFD Table 4.6.2.2.3b-1]

ext int

0.5

e

48g egb

bd 212e 1 140

=

+ − = + ≥

0.5480.23 2

12e 1 1.23440

+ − = + =

One Lane Loaded:

ext int

e

g egde 1.125 120

=

= + ≥

0.23e 1.125 1.13720

= + =


Check: Two or more lanes: e*DFV = 1.234(0.456) = 0.562 Controls One Lane: e*DFV = 1.137(0.445) = 0.506 Because there are two DFV, each must be checked! Vu= 0.562(1.2)[58.33(1.33) + 18.92] = 65.08k VLL,truck= 58.33k VLL,lane = 18.92k IM = 0.33 Skew Factor = 1.2 Vu = 1.25(22.54k + 1.75k + 4.13 k+2.65) + 1.50 (7.08 k) + 1.75(65.08k)= 163.3 k Using the Sectional Design Model, Mu = 3714k-in, β= 3.24, θ=21.4o, φVn = 215 k, so OK.

7.0

CAMBER AND DEFLECTION

Camber calculations are not directly addressed in LRFD (They were not directly addressed in the Standard Specifications, either). The same methods used for finding camber and deflection used for Standard Specifications apply for LRFD Designs. ODOT invokes Article 2.5.2.6.2,which limits Live Load deflection to L/800 for precast, simple span girders. The limit for a Box Girder Bridge is L/800. Since this is a limit on FLEXURAL deflection, it is appropriate to use the MDF. MDF = 0.336(0.905) = 0.304 Lane Load = 0.640(0.304) = 0.194klf Axle Load (rear) = 32k(1.33)(0.304)=12.9k (includes impact) Axle Load (front) = 8k(1.33)(0.304) = 3.22k (includes impact) The live load, positioned for maximum deflection is:


Using a standard analysis software:

( )65ft 120.654in 0.975in

800δ = < = OK

2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 1 of 65

GUIDED DESIGN EXAMPLE

AASHTO Type IV, Two Span, Composite Deck, LRFD Specifications

INTRO

This design example demonstrates the design of a two-span AASHTO Type IV – I girder with no skew, as shown below. This example illustrates the design of a typical interior and exterior beam at the critical sections for positive flexure, negative flexure, shear, and the continuity connection. The superstructure consists of five beams spaced at 8’-0” centers as shown below. Beams are designed to act compositely with the 8.5-in-thick cast-in-place concrete deck slab to resist all superimposed dead loads, live loads, and impact.

Longitudinal

Section

Transverse

Cross Section

MATERIALS

Slab Actual thickness, ts = 9.5 in

Structural thickness = 8.5 in. Note that 1.0 in wearing surface is considered to be an integral part of the 8.5 in deck. fc’ = 4.5 ksi @ 28 days (ODOT Bridge Design Manual (BDM) 302.5.2.8) Concrete unit weight, wc=0.150 kcf

96’-3” 96’-3”

1’-9”

4 Spaces @ 8’-0” = 32’-0” 2.5’

34’-0”

Type IV

2.5’

8.5” structural+ 1.0” wearing

37’-0”


Precast Beams AASHTO Type IV girder shown below fc’ = 7.0 ksi @ 28 days fci’ = 4.5 ksi Concrete unit weight, wc=0.150 kcf The ODOT BDM allows a range of strengths (302.5.2.8). Given strengths are within that range.

Prestressing

Strand ½ in diameter, low-relaxation, ASTM A416

Area of one strand = 0.153 in2 Ultimate strength, fpu = 270.0 ksi The ODOT BDM (302.5.2.2a) allows ½ inch, ½ inch special or 0.6 inch diameter strands. Regular ½ inch diameter is chosen here.

4’-6”

8”

6”

1’-11”

9”

8”

1’-8”

2’-2”

8”6”

9”


Reinforcing Bars

Yield strength, fy = 60 ksi Modulus of elasticity, Es = 29,000 ksi (ODOT BDM 302.5.2.9)

Loads Future wearing surface: 0.060 ksf (ODOT Std. Drawings)

Barriers: 0.640 k/ft each Truck: HL 93, including dynamic allowance

CROSS-

SECTION PROPERTIES

FOR A TYPICAL

INTERIOR BEAM

Non-Composite Section

Area in2 789 Weight (lb/ft) 822 h (in) 54 yb (in) 24.73 yt (in) 29.27 I (in4) 260,741 Sb (in3) 10,542 St (in3) 8,909

1.5

133,000 'C C cE K w f= [LRFD 5.4.2.4-1]

1.533,000 1.0 0.150 4.5 4,067CE ksi= × × = - at transfer

1.533,000 1.0 0.150 7.0 5,072CE ksi= × × = - service loads

Composite

Section Effective flange

Width

(1/4) Span = (96.25 ft)(12in/ft)/4 = 289 in [LRFD 4.6.2.6]

12ts plus the greater of the web thickness or ½ the beam top flange width:

ts = 8.5 in (slab thickness - use structural thickness only)

web thickness = 8 in

½ top flange = 0.5(20 in) = 10 in (Greatest)

12(8.5 in) + 10 in = 112 in

Average spacing between beams = 8 ft = 96 in (CONTROLS)

EFFECTIVE FLANGE WIDTH = 96 in Interior Girder

EFFECTIVE FLANGE WIDTH = 78 in Exterior Girder (overhang is 4 ft).

Modular Ratio ( ) 4,067 0.8019

( ) 5,072c

c

E SlabnE beam

= = =

Transformed Section

Properties

Transformed flange width = n(effective flange width) = (0.8019)(96) = 76.98 in

Transformed flange area = n(effective flange width)(ts) = (0.8019)(96)(8.5) = 654.35in2


Note that only the structural thickness of the deck, 8.5 in, is considered. A 2” haunch is assumed for calculating weight but not for finding composite properties (ODOT BDM 302.5.2.3). Figure below shows the dimensions of the composite section.

Properties of Composite

section

Ac = Total area of composite section = 1,443 in2 hc = Overall depth of the composite section

= 62.5 in

Ic = Moment of inertia of the composite section

= 666,579 in4

ybc = Distance from the centroid of the composite section to the extreme bottom fiber of the precast beam

= 39.93 in

ytg = Distance from the centroid of the composite section to the extreme top fiber of the precast beam

= 14.07 in

ytc = Distance from the centroid of the composite section to the extreme top fiber of the slab

= 22.57 in.

Sbc = Composite section modulus for the extreme bottom fiber of the precast beam

= 16,694 in3

Stg = Composite section modulus for the top fiber of the precast beam

= 47,376 in3

Stc = Composite section modulus for extreme top fiber of the deck slab

= 29,534 in3

96”

76.98”

8.5”

54”

26”


SHEAR FORCES & BENDING

MOMENTS

The self-weight of the beam, haunch, and slab act on the non-composite section as a simple span structure. The weight of the barriers, future wearing surface, and live loads with impact act on the composite section as a continuous structure.

Dead Loads DC = Dead load of structural components and non-structural attachments

DC Dead Loads carried by the girders:

Beam Weight: 0.822 klf

Slab: (96 in)(9.5 in)(0.150 kcf)/(144 in2/ft2) = 0.95 klf

Haunch: (2 in)(20 in)(0.150 kcf)/(144 in2/ft2) = 0.042 klf (ODOT BDM 302.5.2.3)

Note: The actual slab thickness of 9.5” is used in calculating dead loads. The 2” haunch thickness is also used in calculating dead loads. The intermediate diaphragms are assumed as steel “X” braces. These are ignored in these dead load calculations. The weight of each brace is less than 0.3 kips. The moment caused by these braces is << 1% of the total DL moment. DC Dead Loads carried by the continuous structure, composite section: According to LRFD Article 4.6.2.2.1 permanent loads may be distributed uniformly to all beams if the following conditions are met: Width of deck is constant. OK Number of beams, Nb > 4. OK Overhang part of the roadway < 3 ft OK de = 2.5 ft – 1.5 ft = 1.0 ft Curvature in plan < Specified in Article 4.6.1.2 OK Cross Section listed in Table 4.6.2.2.1-1 OK

Partial of Table 4.6.2.2.1-1 - This example is a Type “k”

The section meets the criteria, so the loads may be uniformly distributed to the girders.

Future Wearing Surface = 0.060 ksf = (0.060 ksf)(34 ft)/5 beams = 0.408 kips/ft/girder

Barrier = 0.640 klf = 2 each (0.640)/5 girders = 0.256 kips/ft/girder


LRFD Article 4.6.2.2.1 allows the slab weight to be evenly distributed to the girders in the same manner as the wearing surface and the barriers. In this case, the decision has been made to use tributary areas to distribute the slab weight to the girders. Either method is allowable.

DL-Unfactored Shear Forces &

Bending Moments

For dead loads the length of the span depends on the construction stage:

The shear forces and bending moments are given in the table below:

Location Beam Weight [Simple Span]

Deck plus Haunch [Simple Span]

Barrier Weight [Continuous Span]

Future Wearing Surface [Continuous

Span] Distance

x ft. Section

x/L Shear kips

Moment Mg, kip-ft Shear kips

Moment Ms, kip-ft

Shear kips

Moment Mb, kip-ft

Shear kips

Moment Mws, kip-ft

0.00 0.00 39.6 0 47.7 0 9.2 7.7 14.7 12.4 9.26 0.10 31.9 331 38.5 399.3 6.8 81.8 10.9 130.5

18.97 0.20 24 602.6 28.9 727 4.3 136 6.9 217 28.69 0.30 16 796.5 19.3 961.1 1.8 166 2.9 264.9 38.41 0.40 8 912.9 9.6 1101.5 -0.6 171.9 -1 274.2 48.13 0.50 0 951.9 0 1148.4 -3.1 153.6 -5 245.1 57.84 0.60 -8 912.9 -9.6 1101.5 -5.6 111.2 -8.9 177.5 67.56 0.70 -16 796.5 -19.3 961.1 -8.1 44.7 -12.9 71.3 77.28 0.80 -24 602.6 -28.9 727 -10.6 -46 -16.9 -73.4 86.99 0.90 -31.9 331 -38.5 399.3 -13.1 -160.8 -20.8 -256.7 96.25 Brg. -39.6 0 -47.7 0 -15.4 -292.7 -24.6 -467.1

Live Loads According to LRFD Article 4.6.1.2.1 vehicular live loading on the roadways of bridges or

incidental structures, designated HL-93, shall consists of a combination of the:

• Design truck or design tandem with dynamic allowance. The design truck shall consists of an 8.0 kip front axle and a pair of 32.0 kip back axles. The first and second axle are spaced 14’-0” apart. The space between the rear axles shall be varied between 14.0’ and 30.0’ to porduce extreme force effects. The design tandem shall consist of a pair of 25.0 kip axles spaced 4.0’ apart. [LRFD Article 3.6.1.2.2 and 3.6.1.2.3]

• Design lane load shall consist of a load of 0.64 kip/ft uniformly distributed in the

longitudinal direction. [LRFD Article 3.6.1.2.4]

• For negative moment between inflection points, 90% of the effect of two design trucks (HL-93 with 14 ft. axle spacing) spaced at a minimum of 50 ft. combined with 90% of the design lane load.

• Inflection points are determined by loading all spans with a uniform load.


The live load bending moments and shear forces are determined by using the simplified distribution factor formulas [LRFD 4.6.2.2]. To use the simplified live load distribution factor formulas, the following conditions must be met [LRFD 4.6.2.2.1]


Width of deck is constant. OK Number of beams, Nb > 4. OK Overhang part of the roadway < 3 ft OK de = 2.5 ft – 1.5 ft = 1.0 ft Curvature in plan < Specified in Article 4.6.1.2 OK Beam parallel and of same stiffness OK Cross Section listed in Table 4.6.2.2.1-1 OK

For a precast concrete I-girder with CIP deck, the bridge type is (k) [LRFD 4.6.2.2.1-1] The number of design lanes should be determined by taking the integer part of the ratio w/12, where w is the clear roadway width in ft between curbs and/or barriers. [LRFD 3.6.1.1.1] w = 34 ft. Number of design lanes = integer part of (34/12) = 2

Note: It could be argued that this should be designed as a three lane bridge because 3 – 11 ft lanes would fit and the minimum lane width is 10ft. However, the distribution factor is for 2 or more lanes loaded and the number of lanes isn’t in the equation so it doesn’t matter.

Distribution Factors for

Bending Moment

For all limit states except for fatigue limit state. [LRFD Table 4.6.2.2.2b-1] For two or more lanes loaded:

0.10.6 0.2

30.0759.5 12

g

s

KS SDFML Lt

= +

Where DFM = distribution factor for moment for interior beam. Provided that:

3.5 16.0S≤ ≤ 8.0S OK= S = Spacing, ft 4.5 12.0st≤ ≤ 8.5st OK= ts = slab thickness, in 20 240L≤ ≤ 98L OK= L = beam span, ft

4bN ≥ 5bN OK= Nb = number of beams 10,000 7,000,000gK≤ ≤ gK = See below Kg = longitudinal

stiffness parameter, in4

( )2g gK n I Ae= + [LRFD 4.6.2.2.1-1]

Where: n = modular ratio between beam and deck materials

( ) 5,072 1.247( ) 4,067

c

c

E beamE slab

= = =

A = cross-section area of the beam (non-composite), in2 = 789 I = moment of inertia of the beam (non-composite), in4 = 260,741 eg = Distance be c.g. of beam and slab, in = (8.5/2+2.0+29.27) = 35.52


So: ( )2

4

1.247 260,741 789*35.52

1,566, 480

g

g

K

K in

= +

=

10,000 < Kg < 7,000,000 OK

The haunch is included in this calculation as this results in the most conservative DFM. Using L = 98 ft:

0.10.6 0.2

3

8 8 1,566,4800.0759.5 98 12*98*8.5

0.665

DFM

DFM

= +

=


0.10.4 0.3

3

0.10.4 0.3

3

0.0614 12

8 8 1,566,4800.0614 98 12*98*8.5

0.467

g

s

KS SDFML Lt

DFM

DFM

= +

= +

=

The case of two design lanes loaded controls, DFM = 0.665 lanes/beam

Distribution Factors for

Shear Force

For two or more lanes loaded: [LRFD 4.6.2.2.1-1]

2

0.212 35S SDFV = + −

Where DFV = distribution factor for shear for interior beam. Provided that:

3.5 16.0S≤ ≤ 8.0S OK= S = Spacing, ft 4.5 12.0st≤ ≤ 8.5st OK= ts = slab thickness, in 20 240L≤ ≤ 98L OK= L = beam span, ft

4bN ≥ 5bN OK= Nb = number of beams So:

28 80.212 35

0.814

DFV

DFV

= + −

=

For one design lane loaded


0.362580.3625

0.68

SDFV

DFV

DFV

= + = +

=

The case of two design lanes loaded controls, DFV = 0.814 lanes/beam

Dynamic Allowance

IM = 33% Where: IM = dynamic load allowance, applied only to truck load

Unfactored

Shear Force and Bending

Moments

Unfactored shear forces and bending moment due to HL-93 truck load, per beam:

VLT = (shear force per lane)(DFV)(1+IM)=(shear force per lane)(0.814)(1.33) =(shear force per lane)(1.083) kips MLT = (bending moment per lane)(DFM)(1+IM)=(bending moment per lane)(0.665)(1.33) =( bending moment per lane)(0.884) kips-ft Unfactored shear forces and bending moment due to HL-93 lane load, per beam: VLT = (shear force per lane)(DFV)(1+IM)=(shear force per lane)(0.814) MLT = (bending moment per lane)(DFM)(1+IM)=(bending moment per lane)(0.665) This table, obtained from a structural analysis program, is truck load + lane load, with dynamic effect and distribution factor included.

Location

HL-93 Live Load

Distance x ft.

Section x/L

Max Shear kips

Max. Positive Moment

MLL+I, kip-ft

Max. Negative Moment

MLL+I, kip-ft 0.00 0.00 89.4 48.5 -5.6 9.26 0.10 76.3 624.6 -83.3

18.97 0.20 62.7 1049.3 -163.4 28.69 0.30 50.1 1300.5 -243.6 38.41 0.40 39.9 1412.4 -323.7 48.13 0.50 -48.3 1386.2 -403.9 57.84 0.60 -60.3 1239.1 -484 67.56 0.70 -72.2 961.1 -564.2 77.28 0.80 -83.8 577.5 -776.2 86.99 0.90 -95 215.9 -877.6 96.25 Brg. -104.6 14.8 -1380.7

Shown in this table are maximum values of shear, positive moment, and negative moment. The maximum values at a given location are not necessarily from the same load case.


Load Combinations

The following limit states are applicable: [LRFD 3.4.1] Service I:

Q = 1.00(DC + DW) + 1.00 (LL + IM) Service III:

Q = 1.00(DC + DW) + 0.80(LL + IM) Strength I:

Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM) Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)

ESTIMATE REQUIRED

PRESTRESS

The required number of strands is usually governed by Service III load combination at the section of maximum moment or harp points. In a continuous for live load structure, the maximum moments do not occur at the same place for each load. The point of maximum moment depends on whether the load was applied to the continuous or simple structure. Thus, each point must be checked for the combinations of loads. In this structure, the maximum flexural Service Load stresses occur at 48.13 ft. (although this is NOT where the continuous load moments are maximum). It is inappropriate to simply take maximum moments without regard to location along the length of the girder. At this point, it is necessary to determine the needed number of strands. Box girders tend to be controlled by the Strength Limit State, but “I” girders (this example) tend to be controlled by service load tensions. The initial estimate of number of strands will be found from the Service III combination. Recall that Service III ONLY applies to tension in prestressed sections.

Service 1 Service 3 Strength 1 LengthV M V M V Mk k-ft k k-ft k k-ft ft.

200.6 68.6 182.72 58.9 299.125 113.1 Bearing 0192.6 431.7 175.3 393.72 287.45 644.925 Trans. 2.04189.8 549.9 172.7 502.76 283.375 817.925 H/2 2.73164.4 1567.2 149.14 1442.28 246.375 2303.925 0.10L 9.26126.8 2731.9 114.26 2522.04 191.575 3993.775 0.20L 18.9790.1 3489 80.08 3228.9 138.4 5077.725 0.30L 28.6955.9 3872.9 47.92 3590.42 89.575 5615.875 0.40L 38.41

-56.4 3885 -46.74 3607.76 -95.9 5610.625 MidSpan 48.13-92.4 3542.2 -80.34 3294.38 -147.875 5091.675 0.60L 57.84

-128.5 2834.7 -114.06 2642.48 -199.95 4041.75 0.70L 67.56-164.2 434 -147.44 589.24 -251.375 -329.31 0.80L 77.28-199.3 -564.8 -180.3 -389.28 -301.825 -1464.58 0.90L 86.99-222.3 -1614.4 -201.94 -1375.8 -334.65 -2795.88 H/2 93.52-224.8 -1742.2 -204.3 -1494.76 -338.2 -2961.82 Trans. 94.21-231.9 -2140.5 -210.98 -1864.36 -348.325 -3482.75 Bearing 96.25


Service Load Stresses at

Midspan


(0.8)( )g s b ws LL Ib

b bc

M M M M MfS S

++ + +

= +

Where: fb = Bottom tensile stresses ksi

Mg = Unfactored bending moment due to beam self-weight, kip-ft Ms = Unfactored bending moment due to slab and haunch weights, kip-ft Mb = Unfactored bending moment due to due to barrier weights, kip-ft

Mws = Unfactored bending moment due to future wearing surface, kip-ft MLL+I = Unfactored bending moment due to design vehicular live

load including impact, kip-ft

[ ]153.6 245.1 (0.8)(1,386.2) (12)(951.9 1,148.4)(12)

10,542 16,6942.39 1.083.47

b

b

b

f

ff ksi

+ ++= +

= +=


According to LRFD Table 5.9.4.2.2-1 the tensile stress limit at service loads is '0.19

0.19 7.0 0.503cf

ksi

=

= =

Required

Number of Strands

The difference between the bottom fiber tensile stress due to applied loads and the tensile stress limit is the required precompression stress.

(3.47 0.503) 2.97pbf = − = ksi At this point, the number of rows of strands is unknown. Assume a strand center of gravity at midspan as 8% of the height of the girder.

0.08(54) 4.32bsy = = in So the strand eccentricity at the midspan is:

( ) (24.73 4.32) 20.41c b bse y y= − = − = in If Ppe is the total prestressing force, the stress at the bottom fiber due to prestress is:

pe pe cpb

b

P P ef

A S= +

Now plug in the required recompression stress, fpb and solve form Ppe:

(20.41)2.97

789 10,542927

pe pe

pe

P P

P kips

= +

=


The required prestressing force after all losses is 927 kips. This is after an assumed 25% loss. That means the initial prestressing force will be approximately 1240 kips. Check with your local precast producer to ensure the capacity prestressing beds can withstand this force. Final prestress force per strand = (area of strand)(fpi)(1-losses, %) where fpi = initial prestressing stress before transfer =0.75 fpu = 202.5 ksi Assuming 25% loss of prestress the final prestressing force per strand after losses is:

(0.153)(202.5)(1 0.25) 23.2 /kips strand− ≈

Number of strands required = 927 39.923.2

= strands

Try (40) ½ in diameter, 270 ksi, low-lax strands.

Strand Pattern See figure below for the assumed strand pattern at the midspan:

No.

Strands Distance from

bottom (in) 7 8 11 6 11 4 11 2

10 Spa. @ 2”

2” 2”


The distance between the center of gravity of strands and the bottom concrete fiber of the beam is, ybs, is:

[(11)2 (11)4 (11)6 (7)8] 4.7040bsy + + +

= = in

Strand eccentricity at midspan:

24.73 4.70 20.0c b bse y y= − = − = in PRESTRESS

LOSSES Total Prestress Losses

pT pES pLTf f f∆ = ∆ + ∆ [LRFD 5.9.5.1-1]

Where:

∆fpES = loss due to elastic shortening, ksi ∆fpLT = loss due to long-term shrinkage and creep of concrete, and relaxation of the steel, ksi

Elastic

Shortening p

pES cgpct

Ef f

E∆ = [LRFD 5.9.5.2.3a-1]

Where:


2g ci i c M eP Pe

A I I+ −

Ep = Elastic Modulus of the prestressing steel (ksi).

Eci = Elastic Modulus of the concrete at the time of transfer or time of load application (ksi).

According to the LRFD Commentary for pretensioned member the loss due to elastic shortening may be determined by the following alternative equation (This is the calculation of elastic shortening loss by transformed section):

2

2

( )

( )


g g ctps g m g

p

A f I e A e M AA I E

A I e AE

+ −∆ =

+ + [LRFD C5.9.5.2.3a-1]


Where:

Aps = =

Area of prestressing steel, in2

40(0.153) = 6.12 fpi =

= Prestressing steel stress immediately prior to transfer, ksi 202.5

Ag = =

Gross area of section, in2

789 Ect =

= Elastic Modulus of the concrete at transfer (ksi). 4,067

Ep = =

Elastic Modulus of the prestressing steel (ksi). 28,500

em = =

Average prestressing steel eccentricity at midspan, in 20.0

Ig = =

Moment of inertia of the gross concrete section, in4

260,741 Mg =

= Midspan moment due to member self-weight, kip-in 951.9(12) = 11,422.8

So:

2

2

6.12*202.5(260,741 20.0 *789) 20.0*11,422.8*789789*260,741*4,0676.12(260,741 20.0 *789)

28,50016.24

pES

pES

+ −∆ =

+ +

∆ =

Note: If the self weight moment is calculated using total beam length rather than c/c bearing, the moment becomes 11641 k-in. The elastic shortening loss becomes 16.13 ksi; < 1% different.

Long-Term

Losses For standard, precast, pretensioned members subject to normal loading and environmental

conditions:


g

f Af f

Aγ γ γ γ∆ = + + ∆ [LRFD 5.9.5.3-1]

In which:

1.7 0.01h Hγ = − [LRFD 5.9.5.3-2]

5

1 'stcif

γ =+

[LRFD 5.9.5.3-3]


Where:

H = The average annual ambient relative humidity (%) γh = Correction factor for relative humidity of the ambient air γhst = Correction factor for specified concrete strength at time of

Prestress transfer to the concrete member ∆fpR = An estimate of relaxation loss taken as 2.5 ksi for low

relaxation strand, 10.0 ksi for stress relieved strand, and in accordance with manufacturers recommendation for other types of strand (ksi)

Assume H = 70%

1.7 0.01*70 1.00hγ = − = 5 0.91

1 4.5stγ = =+

So:

202.5*6.1210 1.00*0.91 12*1.00*0.91 2.5789

14.29 10.92 2.5

27.71

pLT

pLT

pLT

f

f

f

∆ = + +

∆ = + +

∆ =



16.24 27.71

43.95

202.5 43.95 158.6

pT pES pLT

pT

pT

pe

f f f

f

f

f

∆ = ∆ + ∆

∆ = +

∆ =

= − =

[LRFD 5.9.5.1-1]

Losses are approximately 22% < 25% OK

STRESSES AT

TRANSFER Force per strand after initial losses:

Stress in tendons after transfer: 202.5 16.24 186.26pt pi pif f f= −∆ = − = ksi

Force per strand = fpt(strand area) = 186.26(0.153) = 28.50 kips

Therefore, the total prestressing force after transfer is, Pi = 1,140 kips

(Note: The LRFD Specifications permit 0.9fpu to be used here; the difference is minimal.)


Compression: 0.60fci’ = 0.60(4.5) = +2.700 ksi [LRFD 5.9.4.1.1]


Tension: [LRFD 5.9.4.1.2] 1. In areas other than the precompressed tensile zone and without bonded

reinforcement '0.0948 0.2

0.0948 4.5 0.20.201 0.2

cif ≤

≤≤

Therefore, -0.200 ksi (Controls)

2. In areas with bonded reinforcement sufficient to resist the tensile force in the concrete computed assuming an uncracked section, where reinforcement is proportioned using a stress of 0.5fy, not to exceed 30 ksi.

'0.24

0.24 4.5cif

-0.509 ksi


Length Section

Stresses at this location need only be checked at release since this stage almost always governs. Also, losses with time will reduce the concrete stresses making them less critical. Transfer length = 60(strand diameter) = 60(0.5) = 30 in = 2.5 ft [LRFD 5.8.2.3] The bending moment at a distance 2.5 ft from the end of the beam due to beam self-weight is:

(0.5)(0.822)(2.5)(97.17 2.5) 97.3gM = − = kip-ft

Compute top stress at the top fiber of the beam:

1,140 1,140(20.0) 97.3(12)789 8,909 8,909

1.44 2.56 0.13 0.99

gi it

b b

t

t

MP PefA S S

f

f

= − +

= − +

= − + = −

Tensile stress limit for concrete with bonded reinforcement: -0.509 ksi NG Compute bottom stress at the bottom fiber of the beam:

1,140 1,140(20.0) 97.3(12)789 10,542 10,542

1.44 2.16 0.11 3.49

gi it

b b

t

t

MP PefA S S

f

f

= + −

= + −

= + − = +

Compressive stress limit for concrete: +2.700 ksi NG


Since the top and bottom concrete stresses exceed the stress limits, harp 9 strands at 0.35L = 34 ft. as shown in the following figures.

At Midspan At ends

No. Strands


No. Strands


7 8 3 52 11 6 3 50 11 4 3 48 11 2 4 8 8 6 8 4 11 2

Compute the center of gravity of the prestressing strands at the transfer length using the harped pattern. The distance between the center of gravity of the 9 harped strands at the end of the beam and the top fiber of the precast beam is:

3(2) 3(4) 3(6) 4.009

+ += in

The distance between the center of gravity of the 9 harped strands at the harp point and the bottom fiber of the precast beam is:

3(4) 3(6) 3(8) 6.009

+ += in


The distance between the center of gravity of the 9 harped strands and the top fiber of the beam at the transfer length section is:

(54 6 4)4.00 (2.5) 7.2534− −

+ = in

The distance between the center of gravity of the 31 straight bottom strands and the extreme bottom fiber of the beam is:

11(2) 8(4) 8(6) 4(8) 4.3231

+ + += in

The distance between the center of gravity of the total number of the strands and the bottom fiber of the precast beam at the transfer length is:

9(54 7.25) 31(4.32) 13.8740

− += in

Eccentricity of the strand group at transfer length is: 24.73 13.87 10.86− = in The distance between the center of gravity of the total number of the strands and the bottom fiber of the precast beam at the end of the beam is:

9(54 4) 31(4.32) 14.6040

− += in

The eccentricity at the end of the beam is: 24.73 14.60 10.13− = in Recompute top and bottom stresses at the transfer length section using the harped pattern: Concrete stress at the top fiber of the beam:

1,140 1,140(10.86) 97.3(12)789 8,909 8,909

1.44 1.39 0.13 0.18

= − +

= − + = +

t

t

f

f

Compressive stress limit for concrete: +2.700 ksi OK At the bottom:

1,140 1,140(10.86) 97.3(12)789 10,542 10,542

1.44 1.17 0.11 2.50

t

t

f

f

= + −

= + − = +



The strand eccentricity at the harp points is the same as at the midspan, ec = 20.0 in The bending moment due to beam self-weight at a distance 34.00 ft. from the end of the beam is:

(0.5)(0.822)(34.00)(97.17 34.00) 882.7gM = − = kip-ft



1,140 1,140(20.0) 882.7*12789 8,909 8,909

1.44 2.56 1.19 0.07

gi it

b b

t

t

MP PefA S S

f

f

= − +

= − +

= − + = +

Compressive stress limit for concrete: +2.700 ksi OK Compute bottom stress at the bottom fiber of the beam:

1,140 1,140(20.0) 882.7*12789 10,542 10,542

1.44 2.16 1.00 2.60

gi it

b b

t

t

MP PefA S S

f

f

= + −

= + −

= + − = +


Stresses at Midspan

The maximum moments due to non-composite loads and composite load do not occur at the same place. In this example, the maximum combined stresses occur at midspan. The bending moment due to beam self-weight at a distance 48’-7” from the end of the beam is:

(0.5)(0.822)(48.58)(97.17 48.58) 970.1gM = − = kip-ft


1,140 1,140(20.0) 970.1*12789 8,909 8,909

1.44 2.56 1.31 0.19

gi it

b b

t

t

MP PefA S S

f

f

= − +

= − +

= − + = +

Compressive stress limit for concrete: +2.700 ksi OK Compute bottom stress at the bottom fiber of the beam:

1,140 1,140(20.0) 970.1*12789 10,542 10,542

1.44 2.16 1.10 2.50

gi it

b b

t

t

MP PefA S S

f

f

= + −

= + −

= + − = +



Hold-Down Forces

Assume that the stress in the strand at the time of prestressing, before any losses, is: 0.75 0.75(270) 202.5= =puf ksi

Then, the Prestress force per strand before any losses is: From previous figure, harp angle:

1 54 4 6tan 6.234(12)

ψ − − −= =

o

Therefore, hold-down force per strand = 1.05(force per strand)(sin ψ) =1.05(31.0) sin 6.2◦ = 3.5 kips per strand

Note that the factor, 1.05, is applied to account for friction. Total hold down force = 9 strands(3.5) = 31.6 kips ODOT BDM States that the following limits are not to be exceeded:

No. of Draped Strands per Row

PU/Strand (lb)

1 6,000

2 4,000

3 4,000 So hold-down force per strand = 3.5 kips per strand OK

Summary of


At transfer, stresses at the end of girder tend to exceed allowables if the strand is straight. Stresses can be brought within the allowable stress range either by harping or debonding the strand. The question arises as to which is better, harping or debonding? Boxes tend to use debonding because harping isn’t practical as the strand would go through the void. I and Bulb T girders tend to use harping. However, not all fabricators have the ability to harp (the bed won’t take the hold down force). Therefore, before deciding to harp, contact probable fabricators or the local PCI section for assistance and advice.

Top Stresses ft, (ksi)

Bottom stresses fb (ksi)

At transfer length section +0.27 +2.43 At harp points +0.07 +2.60 At midspan +0.19 +2.50

Note that the bottom stresses at the harp points are more critical than the ones at midspan.

' 0.153(202.5) 31.0 /iP k strand= =


STRESSES AT SERVICE

LOADS

Total loss of prestress at service loads is 43.95pTf∆ = ksi Stress in tendon after all losses, 202.5 43.95 158.55pe pi pTf f f= −∆ = − = ksi Force per strand = (fpe)(strand area) = (158.6)(0.153) = 24.3 kips

The total prestressing force after all losses, Ppe = 24.3 (40) = 972.0 kips


Compression: [LRFD 5.9.4.2.1]Due to permanent loads, (i.e. beam self-weight, weight of slab and haunch, weight of future wearing surface, and weight of barriers), for service limit states:

For the precast beam:0.45fc’ = 0.45(7.0) = +3.150 ksi For the deck: 0.45fc’ = 0.45(4.5) = +2.025 ksi


For the precast beam:0.40fc’ = 0.40(7.0) = +2.800 ksi For the deck: 0.40fc’ = 0.40(4.5) = +1.800 ksi

Due to permanent and transient loads (i.e. all dead loads and live loads), for service limit states:

For the precast beam:0.60Φwfc’ = 0.60(1.0)(7.0) = +4.200 ksi For the deck: 0.60Φwfc’ = 0.60(1.0)(4.5) = +2.700 ksi

Φw = 1.0 [LRFD 5.7.4.7.2] Note: Φw is a factor for slender webs/flanges. It is not really meant for “I” girders. If the calculations required for Φw are done, Φw=1. Tension:

For components with bonded prestressing tendons: For the precast beam: '0.19 0.19(7.0) 0.503cf = = ksi

Stresses at Midspan-

Compression

Concrete stress at the top fiber of the beam, three cases: 1. Under permanent loads, Service I: Use bending moments given in table in Section 1.4.1.1.

1

1

1

( ) ( )

972 972(20.0) (951.9 1,148.4)*12 (153.6 245.1)*12789 8,909 8,909 47,3761.23 2.18 2.83 0.10 1.98

pe pe c g s ws btg

t t tg

tg

tg


f

f

+ += − + +

+ += − + +

= − + + = +



2. One-half permanent loads plus live loads:

2 1

2

2

( )0.5

1,386.2*120.5(1.98)47,376

0.99 0.35 1.34

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +



3

3

3

( )

1,386.2*121.9847,376

1.98 0.35 2.33

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +

Compressive stress limit for concrete: +4.200 ksi OK Concrete stress at the top fiber of the deck, three cases: 1. Under permanent loads:

1

1

1

( )

(245.1 153.6)*1229,534

0.162

ws btc

tg

tc

tc

M MfS

f

f

+=

+= +

= +

Compressive stress limit for concrete: +2.025 ksi OK Note: Note that deck stresses under service loads are almost always well below allowable for continuous for LL bridges; but they still must be checked. 2. One-half permanent loads plus live loads:

2 1

2

2

( )0.5

1,386.2*120.5(0.162)29,534

0.08 0.563 0.64

LL Itc tc

tg

tc

tc

Mf fS

f

f

+= +

= +

= + = +




( )

(245.1 153.6 1,386.2)*1229,534

0.73

ws b LL Itc

tg

tc

tc

M M MfS

f

f

++ +=

+ +=

= +


Stresses at Midspan-

Tension

Tension stress at the bottom fiber of the beam, Service III:

[ ]

( ) ( ) 0.8

(245.1 153.6) (0.8*1,386.2) *12972 972(20.0) (951.9 1,148.4)*12789 10,542 10,542 16,6941.23 1.84 2.39 1.08 0.40


b b bc

b

b


f

f

++ + +

= + − −

+ ++= + − −

= + − − = −

Tensile stress limit for concrete: -0.503 ksi OK Service III has the 0.8LL factor!

STRENGTH

LIMIT STATE Positive Moment Section

Total Ultimate bending moment for Strength I is: [LRFD Tables 3.4.1-1&2]

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + + The maximum moments for non-composite and composite loads do not occur at the same places. Here, the maximum factored moment occurs at 0.4L (although midspan is only 5k-ft lower). At point of maximum moment, 0.4L:

1.25( ) 1.5( ) 1.75( )1.25(912.9 1,101.5 171.9) 1.5(274.2) 1.75(1,412.4)5,615

u

u

u

M DC DW LL IMMM k ft

= + + += + + + += −

Average stress in prestressing steel when 0.5pe puf f≥ : [LRFD 5.7.3.1.1]

1ps pup

cf f kd

= −


Where:

fps = Average stress in prestressing steel ksi k =

=

2 1.04 py

pu

ff

−

0.28 for low relaxation strands

[LRFD Table C5.7.3.1.1-1]

dp =

=

Distance from extreme compression fiber to the centroid of the prestressing tendons h - ybs = 62.5 – 4.70 = 57.80

in.


in.

To compute c, assume rectangular section behavior, and check if the depth of the equivalent compression stress block, a, is equal to or less than ts: Note: a =β1c

' '

'0.85

ps pu s y s y

puc ps

p

A f A f A fc f

f b kAd

β

+ −=

+ [LRFD 5.7.3.1.1-4]

Where:

Aps = =

Area of prestressing steel 40 * 0.153 = 6.12

in2

fpu = =

Specified tensile strength of prestressing steel 270

ksi

As = =

Area of mild steel tension reinforcement 0.0

in2

fy = =

Yield strength of tension reinforcement 60.0

ksi

As‘ = =

Area of compression reinforcement 0.0

in2

fy‘ = =

Yield strength of compression reinforcement 60.0

ksi

fc‘ = =

Compressive strength of deck concrete 4.5

ksi

β1 = =

Stress block factor specified in LRFD 5.7.2.2 0.83

b = =

Effective width of compression flange 96

in.


6.12(270) 0.0 0.02700.85(4.5)(0.83)(96) 0.28(6.12)

57.805.28

c

c

+ −=

+

=

a = depth of the equivalent stress block = β1c

0.83(5.28) 4.39a = = in. < ts=8.5 in. OK Therefore, the assumption of rectangular section behavior is valid and the average stress in prestressing steel is:

5.28270 1 0.28 263.357.80psf = − =

ksi

Nominal flexural resistance:

24.396.12(263.3) 57.80

212

7,467

n ps ps p

n

n

aM A f d

M

M

= −

− =

=

Factored flexural resistance: r nM Mφ=

Where φ = resistance factor = 1.0 for flexure and tension of prestressed concrete 7, 467rM = kip-ft > 5,615uM = kip-ft OK

NOTE: The equation given above for Mn is not the exact equation 5.7.3.2.2-1. Equation 5.7.3.2.2-1 assumes T-beam behavior, the presence of non-prestressed tensile steel, prestressed tensile steel and non-prestressed compression steel. When the section is rectangular and the non-prestressed reinforcement is ignored, equation 5.7.3.2.2-1 simplifies to the one used above.

Maximum Reinforce-

ment Positive Moment Section

The old ρmax requirement has been deleted. The LRFD Specifications now require that φ be determined based on whether the section is tension controlled, compression controlled or a transition section. In the calculation of Mr, tension control was assumed. Check the strain in the extreme tensile steel:

t

tt

d 54.0 8.5 2.0 60.5d c 60.5 5.280.003 0.003 0.032 0.005

c 5.28

= + − =

− − ε = = = >

This is a tension controlled section, so φ = 1.0 [LRFD 5.7.2.1 and 5.5.4.2]


Minimum

Reinforce- ent Positive

Moment Section

[LRFD 5.7.3.3.2] At any section, the amount of prestressed and nonprestressed tensile reinforcement shall be adequate to develop a factored flexural resistance, Mr, at least equal to the lesser of:

3. 1.2 times the cracking moment, Mcr, determined on the basis of elastic stress distribution and the modulus of rupture, fr,

4. 1.33 times he factored moment required by the applicable strength load combinations

( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥

[LRFD 5.7.3.3.2-1]

Where: fr =

=

Modulus of rupture '0.37 0.37 7.0 0.979cf = =

ksi

[LRFD 5.4.2.6] fcpe =

=

Compressive stress in concrete due to effective prestresss forces only (after allowance for all Prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads

972 972(20.0) 1.23 1.84 3.07789 10,542

pe pe c

b

P P eA S

+ = + = + =

ksi

Mdnc=

=

Total unfactored dead load moment acting on the non composite section

951.9 1,148.4 2,100.3g sM M+ = + =

kip-ft

Sc=

=

Section modulus for the extreme fiber of the composite section where tensile stress is caused by externally applied loads 16,694

in3

Snc=

=

Section modulus for the extreme fiber of the noncomposite section where tensile stress is caused by externally applied loads 10,542

in3

16,694 16,694 16,694(0.98 3.07) 2,100.3 1 (0.979)

12 10,542 124,408 1,362

cr

cr

M

M

= + − − ≥

= ≥

1.2 5,290crM = kip-ft At midspan, the factored moment required by the Strength I load combination is: Mu = 5,610 kip-ft


Therefore, 1.33 7,461uM = kip-ft Since 1.2 1.33cr uM M< 1.2 crM Controls

7, 467 1.2r crM M= > OK

Note: The LRFD Specifications requires that this provision be met at every section.

Design of Negative Moment Section

Longitudinal

Deck Reinforcement

Total Ultimate bending moment for Strength I is: [LRFD 3.4.1-1&2] At the pier section: kip-ft Notes:

1. At the negative moment section, the compression face is the bottom flange of the beam and is 26 in wide.

2. This section is a nonprestressed reinforced concrete section, thus Φ = 0.9 for flexure. Assume the deck reinforcement is at the mid-height of the deck. [LRFD 5.14.1.2.7j]

fy = Yield strength of compression reinforcement = 60.0

in2

fc‘ = Compressive strength of girder = 7.0 ksi

d = Effective depth to negative moment reinforcement from bottom of girder

in

This is the required amount of mild steel reinforcement required in the slab to resist the negative moment and it is equal to 18 #5 bars and 19 #6 bars.

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +

1.25( 292.7) 1.5( 467.1) 1.75( 1,380.7) 3,483uM = − + − + − = −

'1.7s y

u s yc

A fM A f d

f bφ

= −

54 0.5(8.5) 58.25+ =

2

2

(60)3, 483(12) 0.90 (60) 58.251.7(7.0)(26)

0 10.47 3145 41,796

13.94

ss

s s

s

AA

A A

A in

= −

= − +

=


The total area of longitudinal reinforcement provided, ( ) 5.58s providedA = in2.

1.9.2.3


The additional area of deck reinforcement required, ( ) 13.93 5.58 8.35s additionalA = − = in2. The reinforcement layout is shown in the figure below. The additional reinforcement bars are placed between the longitudinal reinforcement.

The table below is a summary of the negative moment continuity calculations.

Typical longitudinal deck reinforcement

No. 5 @ 12” Top = 8 bars No. 5 @ 8” Btm. = 10 bars

Total Area of longitudinal reinforcement provided

5.58 in2

Factored negative design moment

-3,483 kip-ft

Total area required to resist negative moment

13.93 in2

Additional area of deck reinforcement required

8.35 in2

Additional reinforcement provided

19 No. 6 Bars

Additional area of deck reinforcement provided

8.36 in2

Total As provided 13.94 in2 > 13.93 in2 OK


Location of steel: Top – 8 #5 + 8 #6 with 2” clear Bottom – 10 #5 + 11 #6 with 2 5/8” clear Note: Epoxy coated steel assumed. Min. cover is 1.5 in. [LRFD 5.124.]

18(0.31) 19(0.44) 13.94sA = + = in2

8(0.31)(2.3125) 8(0.44)(2.375) 10(0.31)(8.5 2.9375) 11(0.44)(8.5 3)13.94

57.96 4.1613.94

x

x

+ + − + −=

= =

The steel was assumed 4.25” from top OK d = 58.34 in

Now check Mn:

( )( )( )( )

( )( )( )

s y

c

1

r n

r u

A f 13.94 60a 5.41in

0.85f 'b 0.85 7 26a 5.41c 7.72

0.75.41M M 0.9 13.94 60 58.34

2M 41,880k in 3,490k ft M 3,483k ft

= = =

= = =β

= φ = −

= − = − > = −

Effective

Tension Flange Width

The effective tension flange width is the lesser of: [LRFD 5.7.3.4]1. The effective flange width, specified in LRFD Art. 4.6.2.6 = 96 in CONTROLS 2. A width equal to 1/10 of the average of adjacent spans between bearings =

0.10(96.25)(12) 115.5in=


Control of Cracking by Distribution

Reinforcement

According to LRFD 5.7.3.4 the spacing of the mild steel reinforcement in the layer closest to the tension face shall satisfy equation 5.7.3.4-1. The tensile stress in mild reinforcement is computed to be:

sls

s

MfA jd

=

Where: fy =

= Yield strength of compression reinforcement 60.0

ksi

Msl = 292.7 467.1 1,380.7 2,140.5uM = + + = kip-ft As =

= Area of negative moment reinforcement 13.94

in2

d =

=

Effective depth to negative moment reinforcement from bottom of girder 62.5 4.16 58.34− =

in

j =

0.2751 1 0.9083 3k

− = − =

Where:

2

2

2 ( )

2(0.00919)(5.718) (0.00919 *5.718) (0.00919)(5.718)0.275

k n n n

kk

ρ ρ ρ= + −

= + −

=

Where:

ρ =

13.94 0.00919(26)(58.34)

sAbd

= =

n = =

Modular ratio 29,000 5.7185,072

steel

girder

EE

= =

2,140.5(12) 34.8

13.94(0.908)(58.34)sf ksi= =

The previous calculation made the simplifying assumption that the section was rectangular. If this assumption is NOT made, the neutral axis, calculated using working stress concepts, can be calculate as 16.5 inches from the bottom of the beam. The cracked, transformed moment of inertia is 177,600 in4. The steel stress is found to be 34.6ksi which compares to 35.4 ksi using the rectangular assumption.

700 2ec

s s

s dfγ

β≤ −


The spacing of mild steel reinforcement in the layer closest to the tension face shall satisfy the following:

700 2ec

s s

s dfγ

β≤ − [LRFD 5.7.3.4-1]

Where: γe =

= Exposure factor 0.75 for Class 2 exposure condition

fs = Tensile stress in steel reinforcement at the service limit state

ksi

βs = 10.7( )

c

c

dh d

+−

Where:

dc =

=

Thickness of concrete cover measured from extreme tension fiber to center of the flexural reinforcement located closest therto 2.00 5/8 (1/2) 2.31+ =

in

h = =

Overall height on the composite section 62.5

in

2.311 1.0550.7(62.5 2.31)

= + =−sβ

( )( )

700 0.752(2.31) 9.67

1.055 34.8s in≤ − =

6.0 9.67in in≤ OK

For this example the tensile stress in the mild reinforcement is less than its allowable. Thus, the distribution of reinforcement for control of cracking is adequate.

Maximum Reinforce-

ment. Negative Moment Section


tt

d c 59.9 7.720.003 0.003 0.020 0.005c 7.72− − ε = = = >

This is a tension controlled section, so φ = 0.9 [LRFD 5.7.2.1 and 5.5.4.2]

Minimum Reinforcement

Negative Moment Section

[LRFD 5.7.3.3.2] At any section, the amount of prestressed and nonprestressed tensile reinforcement shall be adequate to develop a factored flexural resistance, Mr, at least equal to the lesser of:

2. 1.2 times the cracking moment, Mcr, determined on the basis of elastic stress distribution and the modulus of rupture, fr,

3. 1.33 times he factored moment required by the applicable strength load


combinations

( ) 1ccr c r cpe dnc c r

nc

SM S f f M S fS

= + − − ≥

[LRFD 5.7.3.3.2-1]

Where:

fr = '0.37 0.37 4.5 0.785cf = = ksi [LRFD 5.4.2.6]

fcpe = 0 ksi Mdnc= 0g sM M+ = kip-ft

Sc= 29,534 in3

29,534 (0.785)

121,932

=

= −

cr

cr

M

M k ft

1.2 2,318= −crM k ft At bearing, the factored moment required by the Strength I load combination is: Mu = -3,483 Therefore, 1.33 4,631uM = k-ft. Since 1.2 1.33cr uM M> 1.2 crM Controls

3, 490 1.2 2,318r crM k ft M k ft= − > = − OK Note: The LRFD Specifications states that this requirement be met at every section.

Positive Moment

Connection

Continuous for live load bridges are covered in Article 5.14.1.4.4. Much of this article is new in 2007 (4th Ed.). One requirement of this article is for a positive moment connection. These positive moments are caused by the upward camber of the prestressed girders due to creep and shrinkage. The positive moment connection is needed to provide continuity at the pier. The connection can be made either by extending mild steel out of the end of the girder into the diaphragm or by leaving strand extend out of the end of the girder into the diaphragm. This example illustrates bent strand connections. Positive moments develop at the connection between girders at in interior supports due to live-load effects (if more than two spans) and restraint caused by temperature, creep, and


shrinkage. According to LRFD 5.14.1.4.4, these restraint moments are negligible when continuity is established after 90 days.

Development

Extended Strands

The strands are bent up 90° into the diaphragm so that the hook extends 8 inches from the end of the girder. This distance is required to use the equations in the following section. The ends of the girders are placed 10 inches apart. With the 8 inch projection this leaves 2 inches of clear allowing for construction tolerances. Typically mild steel is placed in the corner of the hooks to enhance the development length of the hooks. These bars should have a minimum area equal to that of the bent strand or bar.

Required Area

of Strand The design moment used for the working stress check is Mcr while the design moment for

the strength check is 1.2Mcr. According to LRFD 5.14.1.4.9c the stress in the strands used for design as a function of the total length of the strand shall not exceed:

( 8) 1500.288dsh

psllf ksi−

= ≤ [LRFD 5.14.1.4.9c-1]

( 8)0.163dsh

pullf −

= [LRFD 5.14.1.4.9c-2]

Where:

ℓdsh = total length of extended strand in fpsl = stress in the strand at the service limit state ksi

Cracked section shall be assumed fpul = stress in the strand at the strength limit state ksi

The design moments, parameters, and results for the design of the positive moment connection using bent strand are found in following table. The cracking moment is found using the gross, composite cross section, but assuming that cracking occurs at the diaphragm. Thus the diaphragm concrete strength is used. For these calculations the effective width of 96 inches, 0.5 inch strand, and concrete strength of 4.5 ksi were used.


When using working stress design the number of strands is assumed to calculate the length of the strand. When using the strength design method, the length of strand is assumed to calculate the number of strands required. Design iterations are performed to determine the most efficient combination of strand and length.

( )cr c cb

cr

e dsh

M 0.24 f ' S 0.24 4.5 16694 8500k in 708k ft1.2M 850k ftL 8

= = = − = −

= −= −l

Thus Le is the length of the extended strand beyond the bend.

Working Stress Design No. of Strand 6 8 10 12 16 ℓdsh 42.29 33.78 29.36 25.83 21.42 As. 0.92 1.22 1.53 1.84 2.45

Moment 708.00 708.00 708.00 708.00 708.00 n 7.00 7.00 7.00 7.00 7.00 d 62.50 62.50 60.50 60.50 60.50

rho 45E-6 52E-6 263E-6 317E-6 422E-6 k 0.05 0.05 0.06 0.07 0.08 j 0.98 0.98 0.98 0.98 0.97 fs 150 113 94 78 59

Strength Design No. of

Strand* 5.18 6.52 8.00 9.27 13.13 ℓdsh 42.00 35.00 30.00 27.00 22.00 As. 0.79 1.00 1.22 1.42 2.01

Moment 849.70 849.70 849.70 849.70 849.70 d 62.50 62.50 62.50 60.50 60.50 a 0.45 0.45 0.45 0.45 0.47

fpul 208.59 165.64 134.97 116.56 85.89 * Back calculated based on strand length

In this example working stress design governs. Multiple iterations are performed to determine the least length of extension of the strand required.

• If the results indicate an odd number of strands they are rounded up to an even number to provide symmetry in the connection.

• It may be more desirable to have a larger number of shorter strands as opposed to fewer longer strands. Girder fabrication may be more difficult with longer strand extensions as this may require excessive space between girders in the bed. In addition, if a larger number of shorter strands are used the stress can be distributed throughout a larger area.


The designer chooses from the tables above. A reasonable design would be 12 strands extended 26 inches. That would be an 8 inch horizontal extension from the face of the beam and an 18 inch vertical “tail” to the hook. Any 12 strands could be extended, but spacing them out and using different rows makes construction easier and limits stress concentrations. Also note that, consistent with the design examples in NCHRP Report 519, the haunch has been included.

SHEAR

DESIGN The area and spacing of shear reinforcement must be determined at regular intervals along

the entire length of the beam. In this design example, transverse shear design procedures are demonstrated below by determining these values at the critical section near the supports. Transverse reinforcement shall be provided where:

0.5 ( )u c pV V Vφ= + [LRFD 5.8.2.4-1]

Where: Vu = Total factored shear force kips Vc = Shear strength provided by concrete kips Vp = Component of the effective prestressing

force in the direction of the applied shear kips

φ = Resistance factor [LRFD 5.5.4.2.1]

Critical Section

Negative Moment

Critical Section near the supports is at dv. [LRFD 5.8.3.2]

Where: dv =

= Effective shear depth Distance between resultants of tensile and compressive forces, ( / 2)ed a− , but not less than 0.9 ed or 0.72h.

[LRFD 5.8.2.9]

Where: de =

=

The corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement 58.34

in

a = =

Equivalent depth of the compression block 5.41

in

h = =

Total height of section 62.5

in



The critical section will occur in the negative moment area, so use the negative bending properties:

0.5( ) 58.34 0.5(5.41) 55.63

0.9 0.9(58.34) 52.50.72 0.72(62.5) 45

v e

e

d d a ind in

h in

= − = − =≥ = =≥ = =

Therefore, dv = 55.63 in

Calculation of

Critical Section The critical section near the support is dv = 55.63 in from the FACE of the support.

Note: Assume the length of the bearing pad is 10 inches. Thus the critical section is 55.63 in + 5 in = 60.63 inches. Using values from previous tables (linearly interpolated), the factored shear force and bending moment at the critical section for shear, according to Strength I load combinations.

1.25(35.4 42.7 14.1) 1.50(22.6) 1.75(99.4) 323.1uV = + + + + = kips (All shear goes the same way!)

0.9(185.2 223.5) 1.25( 219.3) 1.50( 350.0) 1.75( 1,080.9)2323 27880

uMk ft k in

= + + − + − + −= − − = −

At this point, there are three choices:

1. Ignore the prestressing steel Then, this is a reinforced section β = 2 θ = 45◦

(This is VERY conservative) 2. Use Sectional Model for Reinforced Concrete 3. Include prestressing steel

1. Ignore prestressing steel: '0.0316 0.0316(2) 7(8)(55.63) 74.4c c v vV f b d kβ= = =

323.1 74.4 284.60.9sV = − =

Assume #4 hoops Av = 0.4 in2 α = 90 sin α =1 cot α =0 cot 0.4(60)(55.63)cot 45 4.69

284.6v y v

s

A f ds

Vθ

= = = in

Use #4 at 4 in Vs = 334 kips


2. Use Sectional Model for Reinforced Concrete Mu =27,880 kip-in dv =55.63 in. Nu =Applied factored normal force at the specified section = 0 kips

Vu =323.1 kips As =Area of nonprestressed steel on the flexural tension side of

the member = 13.94 in2

Ap =0 in2 Ep =28,500 ksi

If dv < 60db = 30 in, Vp and fpo must be reduced for lack of bond. dv = 55.63 from center bearing, so it is 66.63 from end of girder > 30 in. OK:

Vp =

==

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 0

kips

fpo =0 Assume 0.5cotθ = 1.

3

27,880 0.5(0) (323.1) 055.63 0.001

2(29,000(13.94))1 10 0.001

x

x

ε

−

+ + −= ≤

≤

u p

uv v

V Vv

b dφ

φ−

=

Where: vu = Shear stress in concrete kips bv =

= Effective web width of the beam 8

in

Vp =

= =

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 0

kips

323.1 0.9(0) 0.810.9(8)(55.63)uv −

= = kips


'

0.81 0.1157.0

u

c

vf

= =

Use (vu / fc’) < 0.125 and εx < 1 from LRFD Table 5.8.3.4.2-1:

θ = 37◦ β = 2.13

'0.0316 0.0316(2.13) 7(8)(55.63) 79.3c c v vV f b dβ= = = kips

323.1 0.9(79.3) 280

0.9sV −= = kips


cot 0.4(60)(55.63)cot 37 6.32280

v y v

s

A f ds

Vθ

= = =


( )0.9 79.3 295 337 323.1n uV k k VΦ = + = < =

3. Include Prestressing Steel:

Mu = 27,880 kip-in dv = 53.6 In.

Nu = =

Applied factored normal force at the specified section 0

kips

Vu = 323.1 kips As =

=

Area of nonprestressed steel on the flexural tension side of the member 13.94

in2

Ap =

9(0.153) = 1.38 in2

Ep = 28,500 ksi Es = 29,000 ksi

Note, when the prestressing steel in included, de = 57 inches. The term c = 9.76 in and a = 6.77in. Thus, dv = 53.6 in. If dv < 60db = 30 in, Vp and fpo must be reduced for lack of bond. dv = 55.63 from face of support so this > 30 in from the end of the girder, so:


Vp =

= =

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 24.3(9)(sin 6.2◦) = 23.6

kips

fpo =

=

A parameter taken as modulus of elasticity of prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and the surrounding concrete .7 0.7(270.0) 189= =puf

ksi [LRFD 5.8.3.4.2]

Assume 0.5cotθ = 1.

3

27,880 0.5(0) (323.1 23.6) 1.38(189)53.6 0.001

2(29,000(13.94) 28,500(1.38))0.63 10 0.001

x

x

ε

−

+ + − −= ≤

+

≤

u p

uv v

V Vv

b dφ

φ−

=



in

Vp =

= =

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 23.6

kips

323.1 0.9(23.6) 0.782

0.9(8)(53.6)uv −= = kips

'

0.782 0.117.0

u

c

vf

= =

Use (vu / fc’) < 0.125 and εx < 0.75 from LRFD Table 5.8.3.4.2-1:

θ = 34.4◦ β = 2.26


'0.0316 0.0316(2.26) 7(8)(55.63) 84.1c c v vV f b dβ= = = kips

323.1 0.9(84.1 23.6) 251.30.9sV − +

= = kips


cot 0.4(60)(53.6)cot 34.4 7.5251.3

v y v

s

A f ds

Vθ

= = =


0.9(84.1 313.0 23.6) 378.6r uV V= + + = > OK

Minimum Reinforcement

Requirement

Check which is true: • '0.125u cv f< [LRFD 5.8.2.7-1] Or • '0.125u cv f≥ [LRFD 5.8.2.7-2]

'0.125 0.125(7.0) 0.875cf = = ksi

0.81uv = ksi, max Since '0.125u cv f< , Then max 0.8 0.8(55.63) 44.5 24.0s d= = = ≤ in : 24 in CONTROLS

Calculate minimum area of steel using a 6 inch spacing:

( )( ) 2 28 60 0316 0 0316 7 0 067 0 40

60v

v cy

in inb sA . f ' . ksi . in . inf ksi

≥ = = < OK

[LRFD 5.8.2.5]


Critical Section Positive Moment

Critical Section near the supports is at dv. [LRFD 5.8.3.2]

Where: dv =

= Effective shear depth Distance between resultants of tensile and compressive forces, ( / 2)ed a− , but not less than 0.9 ed or 0.72h.

[LRFD 5.8.2.9]

Where:

de =

=

The corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement 58.2 = dp

in

a = =

Equivalent depth of the compression block 3.42

in

h = =

Total height of section 62.5

in

In this area, the positive moment properties are needed. However, since this section is where the strand is harped, the positive moment properties must be recalculated using 31 strands. Ap = 4.74 in2 and dp = 62.5-4.32 = 58.2 inches. The value of 4.32 inches as the centroid of 31 strands was calculated earlier in Section 1.7.2. Refer to previous section for the equations below:

( )( )( )( )( )( ) ( )

( )( )

4 74 2704 112700 85 4 5 0 83 96 0 28 4 74

58 24 11270 1 0 28 264 858 2

0 83 4 11 3 42

ps

.c . in

. . . . ..

.f . . ksi.

a . . . in

= =+

= − =

= =


0.5( ) 58.2 0.5(3.42) 56.5

0.9 0.9(58.2) 52.40.72 0.72(62.5) 45

v e

e

d d a ind in

h in

= − = − =

≥ = =

≥ = =




The critical section near the support is dv = 56.5 in from the FACE of the support. Note: Assume the length of the bearing pad is 10 inches. Thus the critical section is 56.5in + 5 in ≈ 62 inches. Using values from previous tables, the factored shear force and bending moment at the critical section for shear, according to Strength I load combinations.

1.25(35.4 42.7 7.9) 1.50(12.6) 1.75(82.2) 250.01.25(185.2 223.5 49.6) 1.50(79.1) 1.75(373.9) 1,346

u

u

V kM k in

= + + + + == + + + + = −

It is conservative to take the highest factored moment that will occur at that section, rather than the moment corresponding to maximum Vu, [LRFD 5.8.3.4.2]. Therefore,

250.0uV = kips 1,346uM = kip-ft

The values used to find Vu and Mu are linearly interpolated from the table of shears and moments in previous section.

Contribution of

Concrete to Nominal Shear

Resistance

The contribution of the concrete to the nominal shear resistance is: '0.0316c c v vV f b dβ= [LRFD 5.8.3.3-3]

Strain in Flexural Tension

Reinforcement

Strain in the reinforcement is (assuming uncracked):

0.5 0.5 cot0.001

2( )

uu u p ps po

vx

s s p ps c c

MN V V A f

dE A E A E A

θε

+ + − −= ≤

+ + [LRFD 5.8.3.4.2-1]

Where: Nu =

= Applied factored normal force at the specified section 0

kips

Vp =

= =

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 24.3(9)(sin 6.2◦) = 23.6

kips

fpo =

=

A parameter taken as modulus of elasticity of prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and the surrounding concrete .7 0.7(270.0) 189= =puf

ksi [LRFD 5.8.3.4.2]


Aps =

=

Area of prestressing steel on the flexural tension side of the member, as shown in LRFD Figure 5.8.3.4.2-1. 31(0.153) = 4.74

in2

As =

=

Area of nonprestressed steel on the flexural tension side of the member 0

in2

Ac=

=

Area of concrete on the flexural tension half. This term is calculated as the area on the tension side (bottom in this case) from the tension fiber to h/2. 475

in2

This section is beyond the transfer length, so fpo and Vp do not need to be reduced. Note that either θ can be assumed OR 0.5cotθ can be assumed =1. Assume 0.5cotθ=1:

( )( )3

1,346(12) 0.5(0) (250 23.6) 4.74(189)56.5 0.001

2 28,500(4.74) 5072 475

0.07 10 0.001

x

x

ε

−

+ + − −= ≤

+

− ≤

The negative value means the section is uncracked.

Shear Stress u p

uv v

V Vv

b dφ

φ−

=



in

Vp =

= =

Component of the effective prestressing force in the direction of the applied shear (force per strand)(number of draped strands)(sin ψ) 23.6

kips

250 0.9(23.6) 0.562

0.9(8)(56.5)uv −= = kips

'

0.562 0.08037.0

u

c

vf

= =

Values of β & θ Use (vu / fc’) < 0.1 and εx < -0.05 from LRFD Table 5.8.3.4.2-1:

θ = 21.4◦ β = 3.24


Concrete Contribution

The contribution of the concrete to the nominal shear resistance is: '0.0316c c v vV f b dβ= [LRFD 5.8.3.3-3]

0.0316(3.24) 7.0(8)(56.5) 122.4cV = = kips

Contribution of Reinforcement

of Nominal Shear

Resistance

Check if: [LRFD 5.8.2.4-1]

( )250 0.5 ( ) 0.5 0.9 (122.4 23.6) 65.7u c pV kips V Vφ= > + = + = kips At least minimum stirrups are needed.

Minimum

Reinforcement Requirement

The area of transverse reinforcement should not be less than: '0.0316 v

v cy

b sA ff

≥ [LRFD 5.8.2.5-1]

Check maximum spacing of transverse reinforcement: [LRFD 5.8.2.7] Check which is true:

• '0.125u cv f< [LRFD 5.8.2.7-1] Or • '0.125u cv f≥ [LRFD 5.8.2.7-2]

'0.125 0.125(7.0) 0.875cf = = ksi

0.562uv = ksi Since '0.125u cv f< , Then max 0.8 0.8(56.5) 45.2 24.0s d= = = ≤ in : 24 in CONTROLS

Calculate minimum area of steel using a 12 inch spacing to get area of steel per foot:

( )( ) 2vv c

y


≥ = = [LRFD 5.8.2.5]

ODOT uses #4 bars with 2 legs as standard; (Av = 2(0.2in2) = 0.4in2) #4@ 24 inch o.c. = 0.2 in2 / ft.

This is adequate to meet minimum.

Maximum

Nominal Shear Resistance

The upper limit of Vn, given by following equation, is intended to ensure that the concrete in the web of the beam will not crush prior to yield of the transverse reinforcement.

'0.25n c v v pV f b d V= + [LRFD 5.8.3.3-2] Comparing this previous equation with equation LRFD 5.8.3.3.-1:

'0.25c s c v vV V f b d+ ≤


Assume #4 @ 24”:

( ) ( )( )( ) ( ) ( )20 4 60 56 5 21 4 0 1

24144 2

v y vs

s

. in ksi . cot .A f d cot cot sinV

s inV . k

θ α α ++ = =

=

( )122.4 144.2 266.6 0.25(7.0)(8)(56.5) 791+ = ≤ = kips OK

( )( )0 9 122 4 144 2 23 6 261 2

250

r c s p

r

r u

V V V V

V . . . . . kipsV V kips

φ= + +

= + + =

> =

INTERFACE

SHEAR TRANSFER

Factored Horizontal

Shear

It will be assumed that the critical section is the same as for vertical shear. Using load combination Strength I:

323.1uV = kips 55.6vd = in

Both of these values were found in the preceding section. This is shear at the critical section near the pier.

Required

Interface Shear Reinforcement

ri niV Vφ= [LRFD 5.8.4.1-1] The nominal shear resistance of the interface plane is:

[ ]ni cv vf y cV cA A f Pµ= + + [LRFD 5.8.4.1-3] Where:

c = Cohesion factor ksi [LRFD 5.8.4.3] µ = Friction factor [LRFD 5.8.4.3]

Acv =

=

Area of concrete engaged in shear transfer bviLvi

in2 [LRFD 5.8.4.1-6]

Avf = Area of shear reinforcement crossing the shear plane

in2

Pc = Permanent net compressive force normal to the shear plane

kips

fy = Shear reinforcement yield strength ksi bvi= Width of area of concrete engaged in

shear transfer inch

Lvi = Length of area of concrete engaged in shear transfer

inch


For a cast-in-place concrete placed against clean concrete girder surfaces, free of laitance with surface intentionally roughened to an amplitude of 0.25 in:

0.28c = [LRFD 5.8.4.2] 1.0µ =

Begin by exploring what happens when the shear reinforcement is the minimum used anywhere in the girder. The shear reinforcement was previously calculated to be #4 @ 24 inches minimum. The shear width is bvi = 20 inches as this is the width of the top of the girder. If Lvi = 24 inches:

( )( )( ) ( )

( )

2

[ ]

20 24 480

0.28 480 1.0 0.4 60 0 158.4

0.9 158.4 142.6

ni cv vf y c

cv

ni

ri ni

V cA A f P

A in

V k

V V k

µ

φ

= + +

= =

= + + = = = =

ui ui cvV v A= [LRFD 5.8.4.2-2]

142 6 0 297480ui ,max

.v . ksi= =

1u

uivi v

Vvb d

= [LRFD 5.8.4.2-1]

( )( )1 0 297 20 55 6 330u ,maxV . . kips= =

Therefore, #4 @ 24 is adequate anywhere that Vu < 330 kips. Note that the critical section, the reinforcement is actually #4 @ 4 inches or #4 @ 6”; depending on the model used. Note that #4 @ 24 would be adequate for horizontal shear, so it is NOT necessary to extend every shear stirrup into the slab.

Minimum

Interface Shear Reinforcement

Minimum shear reinforcement, 0.05 cvvf

y

AAf

≥ [LRFD 5.8.4.1-4]

A #4 double leg bar at 24 in spacing is provided from the beam extending into the deck. Therefore, Avf =0.4 in2

0.05(480)0.40 0.4060

≥ = OK


Article 5.8.4.4 states that Avf need not exceed that required to resist 1.33Vui/φ. The same article also states that the minimum reinforcement provisions are waived for girder slab interfaces with surfaces roughened to an amplitude of 0.25 inches where the factored interface shear, vui, found in equation 5.8.4.2-1 is less than 0.210 ksi and all of the vertical (transverse) shear reinforcement required by Article 5.8.1.1 is extended and anchored into the slab.

Maximum

Nominal Shear Resistance

Vni must be less than: '

1 0.3(4.5)(480) 648c cvK f A k= = [LRFD 5.8.4.1-4]

2 1.8(480) 864cvK A k= = [LRFD 5.8.4.1-5]

Vni provided = 158.4k '

1

2

c cv

cv

K f AK A

≤

≤ OK

K1 = 0.3 and K2 = 1.8 (for normal weight concrete) are found in Article 5.8.4.3.

MINIMUM

LONG- ITUDINAL

REIN- FORC-

EMENT REQUIRE-

MENT

At each section the tensile capacity of the longitudinal reinforcement on the flexural tension side of the member shall be proportioned to satisfy: [LRFD 5.8.3.5-1]


v

M N VA f A f V Vd

θφ φ φ

+ ≥ + + − −

According to Article 5.8.3.5, it is not necessary to provide any steel beyond that to resist moment if there is a compressive reaction on the flexural compression face; in other words, in a negative moment zone over a support, the equation in this article does not need to be satisfied. However, it makes an exception for a continuous for live load bridge; saying that this equation must be checked for a continuous for live load bridge. This provision will be checked at the simply supported end, using positive moment properties. The check at the continuous end is made in a similar manner. The development length is:

( ) ( )d ps pe b2 2f f d 1.6 264.8 158.6 0.5 127.3in3 3

= κ − = − =

l [LRFD 5.11.4.2]

( )v pa 3.42d d 62.5 4.32 56.5in2 2

= − = − − =

So the critical section is 56.5 inches from face of support. Allowing for a 10 inch bearing pad and that the center of bearing is 12 inches from the girder end, the critical section is 56.5+10/2+12=73.5 inches from the end of the girder.


Since this is less than the development length, the stress in the steel must be reduced for lack of development. The stress in the undeveloped steel can be found from:

( )px bpx pe ps pe

d b

60df f f f

60d−

= + −−

l

l [LRFD 5.11.4.2-4]

( )( )

( ) ( ) ( )

0.5 0.5 cot

4.74 206 977

1346 2500 23.6 0.5 144.2 cot 21.4 4881.0 56.5 0.9

u u ups ps s y p s

v

M N VA f A f V Vd

k

k

θφ φ φ

+ ≥ + + − −

= >

+ + − − =

This is OK. Note that Vs may not be taken as greater than Vu/φ [LRFD 5.8.3.5].

250144 277 80 9

us

V kV k . k.φ

= < = =

At the inside edge of the bearing area of a simply supported end:

0.5 cotups ps s y p s

VA f A f V V θφ

+ ≥ − −

[LRFD 5.8.3.5-2]

The steel is not fully developed. Since the bearing pad is assumed 10 inches and the center of bearing is 12 inches from the end of the girder, this section is 12+10/2 =17 inches from the end of the girder. This is within the transfer length, so:

( )158 6 1790

60 30pe px

pxb

f .f ksi

d= = =

l [5.11.4.2-3]

( )px73.5in 30inf 158.6ksi 264.8ksi 158.6ksi 206ksi

127.3in 30in−

= + − =−


( )( )

( ) ( )

0.5 cot

4.74 90 426

250 23.6 0.5 144.2 cot 21.4 464.60.9

ups ps p s

VA f V V

k

k

θφ

≥ − − = <

− − =


( )

( )( )

0 2 601 25 1 25 5 7

7

0 4 0 4 0 5 60 12

b yd

c

b y

A f .. . . in

f '

. d f . . in

= = =

< = =

l [LRFD 5.11.2.1]

The development length is 12 inches so the bar is fully developed, thus:

2464 6 426 0 6460s

.A . in−= =

4 #4 works. 3 #5 also works as a # 5 needs a 15 inch development length. Can also add stirrups. Increase to #4 @ 12:

( )( )0 4 60 56 5 21 4288 277 8

12u

s


= = > =


( )( )

( ) ( )

0.5 cot

4.74 90 426

250 23.6 0.5 277.8 cot 21.4 294.20.9

ups ps p s

VA f V V

k

k

θφ

≥ − − = >

− − =

In the previous calculations, the assumption was made that the center of bearing was 12 inches from the end of the girder. What if the bearing pad is placed right at the end of the girder? That is, what if the center of bearing is only 5 inches from the end? What effect does that have on longitudinal steel?


( )v pa 3.42d d 62.5 4.32 56.5in2 2

= − = − − =

So the critical section is 56.5 inches from face of support. Allowing for a 10 inch bearing pad, the critical section is 66.5 inches from the end of the girder. Since this is less than the development length, the stress in the steel must be reduced for lack of development. The stress in the undeveloped steel can be found from:

( )px bpx pe ps pe

d b

60df f f f

60d−

= + −−

l

l [LRFD 5.11.4.2-4]

( )px66.5in 30inf 158.6ksi 264.8ksi 158.6ksi 198.4ksi

127.3in 30in−

= + − =−

( )( )

( ) ( ) ( )

0.5 0.5 cot

4.74 198.4 940.4

1346 2500 23.6 0.5 144.2 cot 21.4 4881.0 56.5 0.9

u u ups ps s y p s

v

M N VA f A f V Vd

k

k

θφ φ φ

+ ≥ + + − −

= >

+ + − − =

This is OK. Note that Vs may not be taken as greater than Vu/φ [LRFD 5.8.3.5].

250144 277 80 9

us

V kV k . k.φ

= < = =

At the inside edge of the bearing area of a simply supported end:

0.5 cotups ps s y p s

VA f A f V V θφ

+ ≥ − −

[LRFD 5.8.3.5.-2]

The steel is not fully developed. Since the bearing pad is assumed 10 inches, this section is 10 inches from the end of the girder. This is within the transfer length, so:

( )158 6 1052 9

60 30pe px

pxb

f .f . ksi

d= = =

l [LRFD 5.11.4.2-3]


( )( )

( ) ( )

0.5 cot

4.74 52.9 250.8

250 23.6 0.5 144.2 cot 21.4 464.60.9

ups ps p s

VA f V V

k

k

θφ

≥ − −

= <

− − =

NG. Assume #4 bars will be used.

( )

( )( )

0 2 601 25 1 25 5 7

7

0 4 0 4 0 5 60 12

b yd

c

b y

A f .. . . in

f '

. d f . . in

= = =

< = =

l

The development length is 12 inches so:

( )10 60 5012sxf ksi= =

The #4 can only develop 50 ksi. Thus:

2464 6 250 8 4 350s

. .A . in−= =

This would be 22 #4! Clearly unrealistic! Add stirrups. Increase to #4 @ 12:

( )( )0 4 60 56 5 21 4288 277 8

12u

s


= = > =


( )( )

( ) ( )

0.5 cot

4.74 52.9 250.8

250 23.6 0.5 277.8 cot 21.4 294.20.9

ups ps p s

VA f V V

k

k

θφ

≥ − −

= <

− − =

This is much more workable:

2294 2 250 8 0 8750s

. .A . in−= =


This is 5 #4 bars. So decrease stirrup spacing from the end of the girder to the critical section (this will be 66.5 inches from the end of the girder) to #4 @ 12. Add 5 #4 bars longitudinal in the bottom flange.

PRE-

TENSIONED ANCHORAG

E ZONE Anchorage

Zone

The bursting resistance of pretensioned anchorage zones provided by vertical reinforcement in the ends of the pretensioned beams at the service limit state shall be take as:

r s sP f A= [LRFD 5.10.10.1-1] Where:


in2

fs = Stress in steel, but not taken greater than 20 ksi Pr = Bursting resistance, should not be less than

4% of Fpi 40(0.153)(202.5)(0.04) 49.6=

kips

Solving for the required area of steel, 49.6 2.4720sA = = in2

At least 2.47 in2 of vertical transverse reinforcement should be provided at the end of the beam for a distance equal to one-fourth of the depth of the beam, h/4 = 54/4=13.5 in Therefore, for a distance of 13.5 in from the end of the member, use 7 #4 bars at 2 inches on center. The reinforcement provided 7(2)0.2 2.8 2.47= > OK. This may be unrealistic, so larger bars may be needed.

Confinement

Reinforcement [LRFD 5.10.10.2]

For a distance of 1.5d = 1.5(54) = 81 in, from the end of the beam, reinforcement is placed to confine the prestressing steel in the bottom flange. The reinforcement should not be less than #3 deformed pars, with spacing not exceeding 6.0 in, and shaped to enclose the strands.

EXTERIOR

GIRDER


Effective Flange Width – Exterior Girder

The effective flange width is taken as one-half the effective width of the adjacent interior girder plus the least of:

One-eighth of the effective span length = 0.125(96.25)(12) = 144 in.

6.0 times the average thickness of the slab, plus the greater of half the web thickness or one-quarter of the width of the top flange of the basic girder

= 6.0(8.5) + 0.5(8) =55 in. = 6.0(8.5) + 0.25(20) = 55 in.

The width of the overhang = 2.5 ft = 30 inches Therefore, the effective flange width for the exterior girder is: (96/2) + 30 = 78 in.

From the previous calculation of beff, the center to center distance controls. beff Trans = nbeff = (0.8015) 78 in = 62.5 in

Exterior Girder

Properties From the previous calculation of beff, the center to center distance controls.

beff Trans = nbeff = (0.8015) 78 in = 62.5 in yb= 38.22 in

I = 624512 in4

A = 50457 in2

h = 62.5 in

yTC = 24.28 in

yTG = 15.78 in

Sb= 16340 in3

STG = 39576 in3

STC = 25721in3


Dead Loads Slab Self Weight: 78 in (8.5 in)(0.150 kcf)/144 = 0.691 klf

Haunch Weight: (Same as interior girder) 0.042 klf

Recall that tributary area was used for the slab weight. This will DECREASE the dead load moment on the exterior girders.

Distance x, ft. Shear, kips Moment, kip-ft

0.00 35.3 0

9.26 28.5 295

18.97 21.4 537

28.69 14.2 710

38.41 7.1 814

48.13 0 849

57.84 -7.1 814

67.56 -14.2 710

77.28 -21.4 537

86.99 -28.5 295

96.25 -35.3 0

Distribution

Factors One Lane Loaded: Lever Rule

Two or More Lanes Loaded: g= egint

Where:

g = DFMext gint= DFMint

0.779.1

ede = +


Distribution Factor for

Moment

Positive Moment Region: Exterior Girder – Two or More Lanes Loaded: DFExt = e DFInt

1.00.77 0.77 0.880

9.1 9.1ed

e = + = + =

DFExt+ = (0.880) (0.665) = 0.585

Lever Rule Assume a hinge develops over each interior girder and solve for the reaction in the exterior

girder as a fraction of the truck load.

1.2 01.2 1.2

HM Pe RSPe eR DF

S S

→ − =

= ∴ =

∑

This is for one lane loaded. Multiple Presence Factors apply 1.2 is the MPF In the diagram, P/2 are the wheel loads; P is the resultant force. All three loads are NOT applied at the same time. Note that truck cannot be closer than 2’ from the barrier

Distribution for

Factor for Moment

One Lane Loaded:

[ ]1.2(36 ) (10.5 3.5) (10.5 9.5)72 (8 )

0.6 /

kR

k ftR lanes girder

− + −=

=

Multiple Presence: MPF = 1.2 Note that this only uses the truck. By dividing by the total truck weight of 72 kips, R is given in lanes/girder



,

2

L

Ext Minb

N

ExtL

Nb

X eNDFN x

= +∑

∑ [LRFD C4.6.2.2.2d-1]

Where:

NL - Number of loaded lanes under consideration Nb - Number of beams or girders e - Eccentricity of design truck or load from CG of pattern of girders (ft.) x - Distance from CG of pattern of girders to each girder (ft.) XExt - Distance from CG of pattern of girders to exterior girder (ft.)

Note: Only the truck is used and it cannot be closer than 2’ from the barrier Minimum Exterior Girder Distribution Factor One Lane:

( )

,

,

,

2

2 2

1 16(12)5 2

0.50

16 8

L

Ext Minb

Ext Min

Ext Min

N

ExtL

Nb

X eN

DFMN x

DFM

DFM

= +

= +

=

+

∑

∑

, ( ) 1.2(0.5) 0.6Ext MinDFM MPF DF= = =


Two Lanes Loaded:

Note: Truck cannot be closer than 2’ from the barrier and the truck must be 2 feet from the lane edge. Minimum Exterior Girder Distribution Factor Two Lane:

,

,

,

2

2 2

2 16(12 0)5 2(160.70

8 )

L

Ext Minb

Ext Min

Ext Min

N

ExtL

Nb

X eN

DFMN x

DFM

DFM

= +

+= +

=

+

∑

∑

, ( ) 1.0(0.7) 0.7Ext MinDFM MPF DF= = = CONTROLS DFMtwo lanes = 0.585 lanes/girder DFMone lane = 0.600 lanes/girder (lever rule) DFMminimum = 0.600 lanes/girder (one lanes) DFMminimum = 0.700 lanes/girder (two lanes) The controlling DFM is the minimum DFM with two lanes loaded DFM = 0.7 This is a 5% increase from the interior girder (DFM = 0.665)

Distribution for

Factor for Shear

One Lane Loaded: Lever Rule

Two or More Lanes Loaded: DFM,Ext = e DFM,Int

0.6610

ede = +


Two or More Lanes Loaded: DFExt = e DFInt

1.00.6 0.6 0.70

10 10ed

e = + = + =

DFExt+ = (0.70) (0.814) = 0.570 One Lane Loaded: (Lever Rule) DFVEXT = 0.6 This is the same as moment calculation. However, the minimum DF = 0.7 (from possible rigid body rotation) - THIS CONTROLS.

Unfactored Shear Forces &

Bending Moments

Dead Loads:

Location Beam Weight [Simple Span]

Deck plus Haunch

[Simple Span]

Barrier Weight [Continuous

Span]


[Continuous Span]

x ft. x/L Shear kips

Mg, kip-ft

Shear kips

Ms, kip-ft

Shear kips

Mb, kip-ft

Shear

kips Mws, kip-ft

0.00 0.00 39.6 0 35.3 0 9.2 7.7 14.7 12.4

9.26 0.10 31.9 331 28.5 295.2 6.8 81.8 10.9 130.5

18.97 0.20 24 602.6 21.4 537.3 4.3 136 6.9 217

28.69 0.30 16 796.5 14.2 710.4 1.8 166 2.9 264.9

38.41 0.40 8 912.9 7.1 814.2 -0.6 171.9 -1 274.2

48.13 0.50 0 951.9 0 848.8 -3.1 153.6 -5 245.1

57.84 0.60 -8 912.9 -7.1 814.2 -5.6 111.2 -8.9 177.5

67.56 0.70 -16 796.5 -14.2 710.4 -8.1 44.7 -12.9 71.3

77.28 0.80 -24 602.6 -21.4 537.3 -10.6 -46 -16.9 -73.4

86.99 0.90 -31.9 331 -28.5 295.2 -13.1 -160.8 -20.8 -256.7

96.25 Brg. -39.6 0 -35.3 0 -15.4 -292.7 -24.6 -467.1


Live Loads: Maximum envelope values shown. The values shown may not be from the same load case.

Load

Combinations The following limit states are applicable: [LRFD 3.4.1]

Service I: Q = 1.00(DC + DW) + 1.00 (LL + IM) Service III: Q = 1.00(DC + DW) + 0.80(LL + IM) Strength I: Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM) Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)

Length LL+IMV M

ft. k k-ftBearing 0 76.5 50.9Trans. 2.04 74.0 199.4H/2 2.73 73.2 247.50.10L 9.26 65.3 655.80.20L 18.97 53.7 1101.80.30L 28.69 42.9 1365.50.40L 38.41 34.2 1483.0MidSpan 48.13 -41.3 1455.50.60L 57.84 -51.6 1301.10.70L 67.56 -61.8 1009.20.80L 77.28 -71.7 -815.00.90L 86.99 -81.3 -921.5H/2 93.52 -87.1 -1252.7Trans. 94.21 -87.7 -1299.1Bearing 96.25 -89.5 -1449.7

Length Service 1 Service 3 Strength 1V M V M V M

ft. k k-ft k k-ft k k-ftBearing 0 175.3 71.0 160.0 60.8 261.1 117.3Trans. 2.04 168.2 416.2 153.4 376.4 250.8 630.3H/2 2.73 165.8 528.7 151.1 479.2 247.2 797.30.10L 9.26 143.4 1494.4 130.3 1363.2 214.6 2228.50.20L 18.97 110.2 2594.7 99.5 2374.3 166.4 3848.50.30L 28.69 77.8 3303.3 69.3 3030.2 119.5 4878.10.40L 38.41 47.7 3656.2 40.8 3359.6 76.4 5380.4MidSpan 48.13 -49.4 3654.7 -41.2 3363.6 -83.7 5357.40.60L 57.84 -81.2 3316.9 -70.9 3056.7 -129.6 4841.00.70L 67.56 -113.0 2632.0 -100.7 2430.2 -175.4 3812.50.80L 77.28 -144.6 205.5 -130.3 368.5 -220.8 -568.00.90L 86.99 -175.6 -712.8 -159.3 -528.5 -265.4 -1635.0H/2 93.52 -195.9 -1707.1 -178.5 -1456.5 -294.3 -2930.0Trans. 94.21 -198.1 -1829.0 -180.6 -1569.2 -297.5 -3092.5Bearing 96.25 -204.4 -2209.5 -186.5 -1919.6 -306.4 -3603.6


Stresses at Midspan

Concrete stress at the top fiber of the girder, three cases: 1. Under permanent loads, Service I:

1

1

1

( ) ( )

972 972(20.0) (951.9 848.8)*12 (153.6 245.1)*12789 8,909 8,909 395761.23 2.18 2.43 0.12 1.60

pe pe c g s ws btg

t t tg

tg

tg


f

f

+ += − + +

+ += − + +

= − + + = +

Compressive stress limit for concrete: +3.150 ksi OK 2. One-half permanent loads plus live loads:

2 1

2

2

( )0.5

1,455*120.5(1.60)39576

0.80 0.44 1.24

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +

Compressive stress limit for concrete: +2.800 ksi OK 3. Under permanent and transient loads:

3 1

3

3

( )

1, 455*12(1.60)39576

1.60 0.44 2.04

LL Itg tg

tg

tg

tg

Mf fS

f

f

+= +

= +

= + = +

Compressive stress limit for concrete: +4.200 ksi OK Concrete stress at the top fiber of the deck, three cases: 1. Under permanent loads:

( )

(245.1 153.6)*1225271

0.186

ws btc

tc

tc

tc

M MfS

f

f

+=

+= +

= +

Compressive stress limit for concrete: +2.025 ksi OK Note that deck stresses under service loads are almost always well below allowable for continuous for LL bridges; but they still must be checked.


2. One-half permanent loads plus live loads:

2 1

2

2

( )0.5

1,455*120.5(0.186)25721

0.09 0.68 0.77

LL Itc tc

tc

tc

tc

Mf fS

f

f

+= +

= +

= + = +

Compressive stress limit for concrete: +1.800 ksi OK 3. Under permanent and transient loads:

3 1

3

3

( )

1, 455*12(0.186)25721

0.19 0.68 0.87

LL Itc tc

tc

tc

tc

Mf fS

f

f

+= +

= +

= + = +

Compressive stress limit for concrete: +2.700 ksi OK Tension stress at the bottom fiber of the girder, Service III:

[ ]

( ) ( ) 0.8

(245.1 153.6) (0.8*1455) *12972 972(20.0) (951.9 848.8)*12789 10,542 10,542 16,3401.23 1.84 2.05 1.15 0.13


b b bc

b

b


f

f

++ + +

= + − −

+ ++= + − −

= + − − = −


GIRDER STRESSES INT EXT

COMP – PERMANENT LOADS 1.98 ksi 1.60 ksi

COMP – ½ PERMANENT LOADS + LL 1.34 ksi 1.24 ksi

COMP – PERMANENT LOADS + LL 2.33 ksi 2.04 ksi

TENSION 0.40 ksi 0.13 ksi


Total Ultimate bending moment for Strength I i: [LRFD Tables 3.4.1&2]

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + +



,

,

, ,int

1.25( ) 1.5( ) 1.75( )1.25(912.9 814.2 171.9) 1.5(274.2) 1.75(1,483)5380 5,615

u ext

u ext

u ext u

M DC DW LL IMMM k ft M k ft

= + + +

= + + + +

= − < = −

Since exterior Mu is less than interior Mu, OK The positive moment, under the Strength I limit state, for the exterior girder is less than that for interior girder. Although the LL increases, the DL decreases due to the flange (slab) being narrower. The interior girder design met all the checks for positive moment design. These were: Nominal Strength, tension controlled, and minimum reinforcement. All of these checks depend on Mu and/or Mn. Since MU,ext<Mu,int, the design for the interior girder for POSITIVE MOMENT is adequate for exterior girder. Stresses at transfer of prestressing force is independent of whether the girder is interior or exterior, so no check is needed.

Negative Moment Section

Total Ultimate bending moment for Strength I is: [LRFD Tables 3.4.1&2]

1.25( ) 1.5( ) 1.75( )uM DC DW LL IM= + + + At the pier section: 1.25( 292.7) 1.5( 467.1) 1.75( 1,450) 3604uM = − + − + − = − kip-ft This is 4% greater than the moment for the interior girder. This is because the LL moment increases. At the support, the slab moment is 0, so it has no effect. Away from the support, the slab moment is positive, so it would mitigate the negative moment. Thus, the smaller slab moment has the effect of INCREASING the negative moment, as compared to the interior girder.

2

2

(60)3,604(12) 0.90 (60) 58.251.7(7.0)(26)

0 10.47 3145 43248

14.5

ss

s s

s

AA

A A

A in

= −

= − +

=

This is the required amount of mild steel reinforcement required in the slab to resist the negative moment and it is equal to 33 #6 bars. Distributed over a length of 6.5 feet, this would be #6 @ 4 inches top and bottom! Use 16 bars on the bottom and 17 on the top. As = 14.52 in2 Note: Only 13.98 in2 were required for the interior girder.


Location of steel: Top – 17 #6 with 2” clear Btm – 16 #6 with 2 5/8” clear. 33(0.44) 14.52sA = = in2

17(0.44)(2.375) 16(0.44)(8.5 3)14.52

56.48 3.914.52

x

x

+ −=

= =

We assumed 4.25” from top OK d = 58.6 in Now check Mn:

( )( )( )( )

( )( )( )

s y

c

1

r n

r u

A f 14.52 60a 5.63in

0.85f 'b 0.85 7 26a 5.63c 8.04

0.75.63M M 0.9 14.52 60 58.6

2M 43740k in 3,645k ft M 3,604k ft

= = =

= = =β

= φ = −

= − = − > = −

Control of

Cracking by Distribution

Reinforcement


700 2ec

s s

s dfγ

β≤ −

Based on the check made for the interior girders (requiring a spacing of 9 inches), #6@ 4 inches will clearly satisfy this requirement. Note that the service level stress will increase, but not enough to bring the requirement below 4 inches.


Maximum Reinforcement

– Negative Moment Section

As before, check the strain in the extreme tensile steel: : [LRFD 5.7.2.1 & 5.5.4.2]

tt

d c 59.9 8.040.003 0.003 0.019 0.005c 8.04− − ε = = = >


Minimum

Reinforcement – Negative

Moment Section

( ) 1c

cr c r cpe dnc c rnc

SM S f f M S fS

= + − − ≥

[LRFD 5.7.3.3.2-1]

Where:

fr = '0.37 0.37 4.5 0.785crf f= = = ksi

fcpe = 0.0 ksi

Mdnc= 0g sM M+ = kip-ft

Sc= 16340 in3

16340 (0.785)12

1069

cr

cr

M

M k ft

=

= −

1.2 1282crM k ft= − At bearing, the factored moment required by the Strength I load combination is: Mu = -3604 kip-ft Therefore, 1.33 4793uM = kip-ft Since 1.2 1.33cr uM M< , 1.2 crM Controls

3,645 1.2 1282r crM M= > = OK Note: The LRFD Specifications states that this requirement be met at every section. The design of the exterior section meets all requirements for positive and negative bending under both Service and Strength Limit States.


Shear This compares Strength I shears and moments for the interior and exterior girders. Note that the exterior girder shears are LESS than the interior girder shears. Thus, the previous design works for vertical and horizontal shear. The longitudinal steel requirements are also met.

Strength ILength Interior Exterior

V M V Mft. k k-ft k k-ft

Bearing 0 299.125 113.1 261.0657 117.3438Trans. 2.04 287.45 644.925 250.7524 630.3376H/2 2.73 283.375 817.925 247.1722 797.26250.10L 9.26 246.375 2303.925 214.6325 2228.4850.20L 18.97 191.575 3993.775 166.3629 3848.4510.30L 28.69 138.4 5077.725 119.4571 4878.1260.40L 38.41 89.575 5615.875 76.42157 5380.371MidSpan 48.13 -95.9 5610.625 -83.733 5357.4420.60L 57.84 -147.875 5091.675 -129.581 4841.0080.70L 67.56 -199.95 4041.75 -175.438 3812.4530.80L 77.28 -251.375 -329.31 -220.846 -567.9670.90L 86.99 -301.825 -1464.58 -265.37 -1635.04H/2 93.52 -334.65 -2795.88 -294.34 -2929.99Trans. 94.21 -338.2 -2961.82 -297.47 -3092.54Bearing 96.25 -348.325 -3482.75 -306.435 -3603.56

AASHTO LRFD B id D iAASHTO LRFD Bridge Design ... · PDF file1 AASHTO LRFD B id D iAASHTO LRFD Bridge Design Specifications Prestressed Concrete RICHARD A. MILLER, PhD, PE, FPCI PROFESSOR

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