aashto lrfd b id d iaashto lrfd bridge design ... · pdf file1 aashto lrfd b id d iaashto lrfd...
TRANSCRIPT
1
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
2
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
==
==
AASHTO-LRFD 2007ODOT Short Course
July 2007
ksiksiksiksipsipsi
524.01000
55.71000
510005.7/1000
50005.7===
Do Not Duplicate Prestressed Concrete: Slide #3
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007
maximum of 10 ksi in the calculations.Decks must have a minimum strength of 4 ksi.
Do Not Duplicate Prestressed Concrete: Slide #4
3
§ 5.4 – Material Properties
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #5
§ 5.4 – Material Properties
§ 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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #6
4
§ 5.4 – Material Properties
§ 5.4.2.3 – Shrinkage and Creep
( ) = −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)
AASHTO-LRFD 2007ODOT Short Course
July 2007
+−
=tf
kci
td '461
Do Not Duplicate Prestressed Concrete: Slide #7
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).
§ 5.4 – Material Properties
§ 5.4.2.3 – Shrinkage and Creep
( )−= −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’.
AASHTO-LRFD 2007ODOT Short Course
July 2007
+−
=tf
tkci
td '461
Do Not Duplicate Prestressed Concrete: Slide #8
Std. Spec. set shrinkage = 0.002. Previous editions of LRFD used an ACI type equation with a term of t/(35+t).
5
§ 5.4 – Material Properties
§ 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):√ √
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #9
§ 5.4 – Material Properties
§ 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 =
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #10
6
§ 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)
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #11
DCDD LL Use One of These at a Time
§ 3.4 - Loads and Load Factors
§3.4.1: Load Factors and Load Combinations
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 -- -- -- --
AASHTO-LRFD 2007ODOT Short Course
July 2007
γp γTG γSE
Do Not Duplicate Prestressed Concrete: Slide #12
7
§ 3.4 - Loads and Load Factors
§3.4.1: Load Factors and Load Combinations
Table 3.4.1-1 Load Combinations and Load Factors (cont.)DCDD LL
Use One of These at a Time
Load Combination
DWEHEVESEL
IMCEBRPLLS WA WS WL FR
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 -- -- -- -- -- -- -- -- -- -- --
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #13
§ 3.4 - Loads and Load Factors
§3.4.1: Load Factors and Load Combinations
Table 3.4.1-1 Load Combinations and Load Factors (cont.)DCDD LL Use One of These at a
Ti
Load Combination
DWEHEVESEL
IMCEBRPLLS WA WS WL FR
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 -- -- -- --
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #14
8
§ 3.4 - Loads and Load Factors
§3.4.1: Load Factors and Load Combinations
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-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #15
AASHTO LRFDAASHTO-LRFDDistribution Factors for Precast/Prestressed Concrete Elements
AASHTO-LRFD Specification, 4th Edition
9
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #17
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #18
factors.
10
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #19
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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 )
AASHTO-LRFD 2007ODOT Short Course
July 2007
and slab. (in)
If Nb = 3, use the lesser of the equations above with Nb = 3 and the lever rule.
Do Not Duplicate Prestressed Concrete: Slide #20
11
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
Beam and Slab Type “k” bridge
Shear Distribution Factors - LRFD Table 4.6.2.2.3a-1:
AASHTO-LRFD 2007ODOT Short Course
July 2007
Two or more lanes loaded:DFV = 0.2 + ( S/12 ) - ( S/35 )2
One lane loaded:DFV = 0.36 + ( S/25 )
Do Not Duplicate Prestressed Concrete: Slide #21
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #22
12
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
+=
=
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #23
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
+=
=
AASHTO-LRFD 2007ODOT Short Course
July 2007
g = DFV-1.0 < de < 5.5 ft.
Do Not Duplicate Prestressed Concrete: Slide #24
13
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #25
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
L = Span, 20 < L < 240 ftS = Beam Spacing, 3.5 < S < 16 ft Nb > 4
Do Not Duplicate Prestressed Concrete: Slide #26
14
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.
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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)
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #28
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.)
15
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
distribution factor.
Do Not Duplicate Prestressed Concrete: Slide #29
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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).
Do Not Duplicate Prestressed Concrete: Slide #30
16
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #31
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
J St. Venant torsional constant, in
For preliminary design, ( I/J )0.06 = 1.0
Do Not Duplicate Prestressed Concrete: Slide #32
17
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007
5 Nb 2035< b < 60 in 20< L < 120 ft25,000 < J < 610,000 in4
40,000 < I < 610,000 in4
Do Not Duplicate Prestressed Concrete: Slide #33
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #34
18
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
Where:NL = number of traffic lanes C = K(W/L) < K
Do Not Duplicate Prestressed Concrete: Slide #35
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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+
=
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #36
19
≈ SA4J2
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
∑ tSJ
Where:
A = Area enclosed by the centerline of the webs and flanges.l h f b fl li
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #37
S = length of a web or flange centerline.t = thickness of the corresponding web or flange.
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #38
Minimum exterior distribution factor based on rigid body rotation does not apply to adjacent box girders.
20
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #39
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
e
Do Not Duplicate Prestressed Concrete: Slide #40
the exterior girder
21
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
0.140
0.2121 ≥
−++=
bde
e
Do Not Duplicate Prestressed Concrete: Slide #41
de < 2.035 < b < 60 ing = DFV
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #42
22
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
This is optional.
Do Not Duplicate Prestressed Concrete: Slide #43
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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,
AASHTO-LRFD 2007ODOT Short Course
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).
Do Not Duplicate Prestressed Concrete: Slide #44
23
Distribution Factors for Precast/Prestressed Concrete Elements
§ 4.6.2.2.2 Distribution Factor Method for Moment and Shear
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
+
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #45
S TOAASHTO-LRFDFlexure and Axial Loads
AASHTO-LRFD Specification, 4th Edition
24
Flexure and Axial Loads
Definitions of various “d” terms for
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #47
Flexure and Axial Loads
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #48
25
Flexure and Axial Loads
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #49
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.
Flexure and Axial Loads
§ 5.7.2 Assumptions for Strength and Extreme Event Limit States
Definition of strain conditions for determining tension or compression control. Note that tensile strain in the steel closest to the tensile face is used
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #50
26
Flexure and Axial Loads
§ 5.7.2 Assumptions for Strength and Extreme Event Limit States
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #51
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.
Flexure and Axial Loads
§ 5.7.2 Assumptions for Strength and Extreme Event Limit States
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #52
27
Flexure and Axial Loads
§ 5.7.2 Assumptions for Strength and Extreme Event Limit States
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #53
Flexure and Axial Loads
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007
For transition members, use a linear interpolation of the Φfactor based on the extreme tensile steel strain.
Do Not Duplicate Prestressed Concrete: Slide #54
28
Flexure and Axial Loads
§ 5.5.4.2 Resistance Factors
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
Prestressed Members
Reinforced Members
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007
Extreme Steel Strain
Do Not Duplicate Prestressed Concrete: Slide #55
(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
Flexure and Axial Loads
§ 5.5.4.2 Resistance Factors
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #56
29
Flexure and Axial Loads
§ 5.5.4.2 Resistance Factors
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 = +
=+
AASHTO-LRFD 2007ODOT Short Course
July 2007
Aps Area of prestressing steelfpy = Yield strength of the prestressing steelAs = Area of mild steelfy = Yield strength of the mild steel
Do Not Duplicate Prestressed Concrete: Slide #57
Flexure and Axial Loads
The stress block remains the same as Standard Specifications.
AASHTO-LRFD 2007ODOT Short Course
July 2007
Analysis of reinforced concrete RECTANGULAR beams is the same as Standard Specifications.HOWEVER, there are some differences with prestressed concrete.
Do Not Duplicate Prestressed Concrete: Slide #58
30
AASHTO LRFDAASHTO-LRFDPrestressed Beams with Bonded Tendons
AASHTO-LRFD Specification, 4th Edition
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007
Stress in the steel, fps, can also be found from strain compatibility analysis.
Do Not Duplicate Prestressed Concrete: Slide #60
ps pu
puc 1 ps
p
A fc f
0.85f ' b kAd
=β +
31
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #61
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
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
+ −=
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #62
10.85 ' puc ps
p
ff b kA
dβ +
32
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #63
through 3 of LRFD
back to the old definition. Now it is the same definition as ACI 318 and Std. Spec.
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
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
= − + − −
− + − −
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #64
33
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
( ) 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:
AASHTO-LRFD 2007ODOT Short Course
July 2007
n ps ps paM A f d2
= −
n ps ps p s y sM A f d A f d2 2
= − + −
Do Not Duplicate Prestressed Concrete: Slide #65
If there is no non-prestressed tensile steel:
Prestressed Beams with Bonded Tendons
§ 5.7.3 Flexural Members
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
+ − − −=
+
AASHTO-LRFD 2007ODOT Short Course
July 2007
w
b = flange widthhf = flange thickness
Do Not Duplicate Prestressed Concrete: Slide #66
34
AASHTO LRFDAASHTO-LRFDPrestressed Beams with Unbonded Tendons
AASHTO-LRFD Specification, 4th Edition.
Prestressed Beams with Unbonded Tendons
§ 5.7.3 Flexural Members
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
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #68
35
Prestressed Beams with Unbonded Tendons
§ 5.7.3 Flexural Members
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
+ − − −=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #69
(5.7.3.1.2-3)10.85 'c wf bβ
Prestressed Beams with Unbonded Tendons
§ 5.7.3 Flexural Members
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):
AASHTO-LRFD 2007ODOT Short Course
July 2007
first estimate of fps (in ksi):
Do Not Duplicate Prestressed Concrete: Slide #70
(C5.7.3.1.2-1)15ps pef f= +
36
AASHTO LRFDAASHTO-LRFDComponents with Both Bonded and Unbonded Tendons
AASHTO-LRFD Specification, 4th Edition
Components with Both Bonded and Unbonded Tendons
§ 5.7.3 Flexural Members
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
Apsu = area of unbonded tendons
Do Not Duplicate Prestressed Concrete: Slide #72
37
Components with Both Bonded and Unbonded Tendons
§ 5.7.3 Flexural Members
Simplified Analysis - The stress in the UNBONDED tendons may be conservatively taken as the effective stress after losses: fpe. p
For T-beams:
For rectangular beams:
( )
p
pupswc
fwcysyspepsupupsb
df
kAbf
hbbffAfAfAfAc
+
−−−++=
1'85.0
'85.0''
β
AASHTO-LRFD 2007ODOT Short Course
July 2007
For rectangular beams:
Do Not Duplicate Prestressed Concrete: Slide #73
10.85 '
psb pu psu pe
puc ps
p
A f A fc f
f b kAd
β
+=
+
S TOAASHTO-LRFDMoment Capacity
AASHTO-LRFD Specification, 4th Edition
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007
wbecomes:
−−
−+
−=
2'''
22adfAadfAadfAM syssysppspsn
Do Not Duplicate Prestressed Concrete: Slide #75
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
non-prestressed compression steel.fy = yield strength of the non-prestressed tensile steel.fy’ = yield strength of the non-prestressed compression
steel.
Do Not Duplicate Prestressed Concrete: Slide #76
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #77
Moment Capacity
§ 5.7.3.3 Limits for Reinforcement
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
= + − − ≥
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #78
40
Moment Capacity
§ 5.7.3.3 Limits for Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
3 35 8t
cd> >
Do Not Duplicate Prestressed Concrete: Slide #79
Compression Controlledεt <0.002
Transition
35t
cd
≥
Moment Capacity
§ 5.7.3.3 Limits for Reinforcement
Maximum reinforcement is now controlled by εt.
AASHTO-LRFD 2007ODOT Short Course
July 2007
To determine εt , calculate c. Then, using similar triangles:
−
=ccdt
t 003.0ε
Do Not Duplicate Prestressed Concrete: Slide #80
41
Moment Capacity
§ 5.7.3.3 Limits for Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
42.0≤edc
0.45e
cd≤
Do Not Duplicate Prestressed Concrete: Slide #81
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
AASHTO-LRFD 2007ODOT Short Course
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’
Do Not Duplicate Prestressed Concrete: Slide #82
42
Moment Capacity
§ 5.7.3.5 Moment Redistribution
ODOT does not permit moment redistribution
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #83
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.
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #84
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
AASHTO-LRFD 2007ODOT Short Course
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
Do Not Duplicate Prestressed Concrete: Slide #85
Prestressing and Partial Prestressing
§ 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)
AASHTO-LRFD 2007ODOT Short Course
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)
Do Not Duplicate Prestressed Concrete: Slide #86
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #87
Prestressing and Partial Prestressing
§ 5.9.4 Stress Limits for Concrete
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007
• Due to the sum of effective prestress, permanent loads, and transient loads and during shipping and handling
0.60φwf’c (ksi)
Do Not Duplicate Prestressed Concrete: Slide #88
45
Prestressing and Partial Prestressing
§ 5.9.4 Stress Limits for Concrete
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
• For components with unbonded prestressing tendons
No Tension
Do Not Duplicate Prestressed Concrete: Slide #89
Again, these are Std. Spec. limits in ksi units.
0.19(1000)0.5 = 6
S TOAASHTO-LRFDLoss of Prestressing Force
AASHTO-LRFD Specification, 4th Edition
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #91
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
The basic equations:Pretensioned Members:Pretensioned Members:
Post-tensioned Members:
fffff ∆+∆+∆+∆=∆
pLTpESpT fff ∆+∆=∆ (5.9.5.1-1)
(5 9 5 1-2)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #92
pLTpESpApFpT fffff ∆+∆+∆+∆=∆ (5.9.5.1-2)
47
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
∆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).
AASHTO-LRFD 2007ODOT Short Course
July 2007
∆fpA is usually given by the manufacturer.
Do Not Duplicate Prestressed Concrete: Slide #93
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007
single deviator pipe:( )( )04.01 +−−=∆ αµeff pjpF
Do Not Duplicate Prestressed Concrete: Slide #94
(5.9.5.2.2b-2)
48
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
prestressing steel path from jacking end (or nearestjacking end if jacked from both ends) to point underconsideration. (radian)
Do Not Duplicate Prestressed Concrete: Slide #95
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #96
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
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #97
fEE
2N1N∆f cgp
ci
ppES
−= (5.9.5.2.3b-1)
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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 ( )
AASHTO-LRFD 2007ODOT Short Course
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.
Do Not Duplicate Prestressed Concrete: Slide #98
50
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #99
( )
(5.9.5.3-3)5
1=
+stcif
γ
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
h
γst = strength factor
Do Not Duplicate Prestressed Concrete: Slide #100
51
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
Average exposure conditions and temperatures.
Do Not Duplicate Prestressed Concrete: Slide #101
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #102
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
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
8 ksi for single T’s, double T’s, hollow core and voided slabs.
Do Not Duplicate Prestressed Concrete: Slide #103
§ 5.9 – Prestressing and Partial Prestressing
§ 5.9.5 Loss of Prestress
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
be found in Article 5.9.5.4 of the LRFD Specifications.
Do Not Duplicate Prestressed Concrete: Slide #104
53
S TOAASHTO-LRFDBond/Development Length
AASHTO-LRFD Specification, 4th Edition
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #106
54
§ 5.11 – Bond and Development Length
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #107
§ 5.11 – Bond and Development Length
§ 5.11.4.2 – Development Length
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #108
= 1.6 for pretensioned members with a depth > 24 inches = 2.0 for debonded strand
55
§ 5.11 – Bond and Development Length
§ 5.11.4.2 – Development Length
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #109
However, stress in the steel beyond the transfer length, but less than the development length, can now be calculated by a bilinear formula.
§ 5.11 – Bond and Development Length
§ 5.11.4.2 – Development Length
( )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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #110
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
§ 5.11 – Bond and Development Length
§ 5.11.4.2 – Development Length
Within the transfer length (which is 60db):
60px pe
pxb
ff
d=l
(5.11.4.2-3)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #111
§ 5.11 – Bond and Development Length
§ 5.11.4.2 – Development Length
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #112
1
S TOAASHTO-LRFDShear
AASHTO-LRFD Specification, 4th Edition.
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #2
NOT covered.Reinforced concrete is covered in another section.
2
§ 5.8 - Shear and Torsion
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #3
It will be covered in another presentation.
§ 5.8 - Shear and Torsion
§ 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 ≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #4
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 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #5
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
( )
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #6
shear cracks (struts) is not fc’. As will be shown, it may be greater than or less than fc’.
4
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #7
is jd/tanθ = jdcotθ.
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #8
(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
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #9
However, in almost all cases, α = 90o ; thus cotα = 0 and sinα = 1. The equation reverts the one shown on the previous slide.
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #10
concrete stress times the area.
Fc = fc (jd cosθ) bvwhere fc is the concrete stress and bv web width.
6
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #11
Substituting into the previous equation and assuming Vc is the shear force carried by the concrete:
Vc = fc (jd cosθ) sinθ bv
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
( )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= =
=
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #12
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
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #13
impossible. Therefore, the value of 45 degrees was chosen for simplification. The value of the crushing strength was also chosen as a simplification.
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #14
Theory . Later the theory was improved to account for additional mechanisms, such as aggregate interlock, and was renamed “Modified Compression Field Theory”.
8
§ 5.8 - Shear and Torsion
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #15
§ 5.8 - Shear and Torsion
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #16
9
§ 5.8 - Shear and Torsion
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #17
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 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #18
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
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #19
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)
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #20
ducts and one quarter the diameter of grouted ducts is subtracted from the web width.
11
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #21
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.
§ 5.8 - Shear and Torsion
§ 5.8.3.3 Nominal Shear Resistance
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #22
The first sections must be the critical section from the face of the support.
12
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #23
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 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #24
the support, then strut and tie model (Article 5.6.3) must be used for analysis.
13
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #25
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.
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #26
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 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #27
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 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #28
needed.
15
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #29
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #30
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
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #31
on the flexural compression side (the ½ h on the flexural compression side) or compression steel is NOT counted for shear strength.
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #32
Definition of “flexural tension side”, the term Ac, and the term εx for cases with and without minimum stirrups.
17
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #33
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #34
18
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
( )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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #35
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #36
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
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #37
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #38
Ac is the area of concrete on the tension half of the section.
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.
20
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #39
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
≤
≤
≤
≤
≤
§ 5.8 - Shear and Torsion
§ 5.8.3.4 Determination of β and θ
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #40
(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
§ 5.8 - Shear and Torsion
§ 5.8.3.4 Determination of β and θ
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #41
Vn Vc Vs Vp 0.25fc bv dv Vp
Then Vu < φ Vn
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #42
xAs before, the section is assumed to be cracked and0.5cotθ may be taken = 1
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.
22
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
If the section is not cracked:u fAVVN
M++ θt5050
( )ccpspss
popspuuv
u
x AEAEAE
fAVVNd
++
−−++=
θε
cot5.05.0
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #43
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.
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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 ≤+
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #44
longitudinal steel in each layer must be at least 0.003 bvsx
ag = maximum aggregate size – inch.
23
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #45
§ 5.8 - Shear and Torsion
§ 5.8.3.4 Determination of β and θ
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=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #46
Vn = Vc + Vs + Vp < 0.25fc’ bv dv + Vp
Then Vu < φ Vn
sdfA
V vyvs
θcot=
24
§ 5.8 - Shear and Torsion
§ 5.8.3.4 Determination of β and θ
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
< < < < < < < < < < <
<
<
<
<
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #47
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
<
<
<
<
§ 5.8 - Shear and Torsion
§ 5.8.3 Sectional Design Model
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #48
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.)
AASHTO-LRFD Specification, 4th Edition.
again, and again and again.)
§ 5.8 - Shear and Torsion
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #50
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #51
The old Vci and Vcw worked well for many years.Still the ACI 318 method.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #52
27
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #53
for the Sectional Design Model.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #54
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #55
inclined cracking results from combined shear and moment.“Flexural Shear”
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
A quick reminder.Exactly what are Vci and Vcw?
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #56
29
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #57
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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?
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #58
So how is Vcre found?The simplifying assumption is made that V and Mincrease proportionally.
30
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #59
creu
ucre M
MVV =
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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 +=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #60
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #61
Mmax = Maximum moment at a section caused by allFACTORED superimposed loads.
Vi = Shear at the section associated with Mmax.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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 ++=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #62
Vi = FACTORED shear at the section corresponding to Mmax.bv = minimum web widthdv = effective shear depth
32
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #63
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.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #64
33
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
Assuming an elastic system:
Where Sc is the section modulus to the tension fiber.
( )rcpeccre
c
crecrercpe
ffSMSM
IcM
ff
+=
==+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #65
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #66
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #67
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #68
yc = distance between neutral axis of composite and noncomposite sections
= 0 for noncomposite beamsMdl,nc= noncomposite dead loadse = eccentricity of prestressing
35
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
Web Shear - Vcw
V can be calculated using the shearing stress formula:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #69
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.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
Web Shear - Vcw
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #70
An approximate equation is provided to find Vcw:
( )3.0'06.0 ++= VdbffV pvvpcccw (5.8.3.4.3-3)
36
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #71
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
'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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #72
with new notations.
0.02√fc’ ksi = 0.63√fc’; 0.06√fc’ ksi = 1.9√fc’
37
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #73
superimposed loads, which occurs simultaneouslywith Mmax (kip).
Mmax and Vi are found from the load combination causing maximum moment at the section.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #74
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #75
Snc = Section modulus to the extreme fiber of the non-composite or monolithic section where tensilestress is caused by externally applied loads (in3).
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
For composite members, the commentary allows for a simplification:simplification:
Mmax = Mu – Md Vi = Vu – Vd
(C5.8.3.4.3) 7th Paragraph
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #76
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
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
( )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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #77
using the simplified shear method.Again, this is the old Vcw equation, converted to kip units.
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #78
and the moments resisted by the precast member acting alone.
40
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #79
If Vcw < Vci, (Vcw controls) then:
8.1'
30.1cot ≤
+=
ff
c
pcθ (5.8.3.4.3-4)
§ 5.8 - Shear and Torsion
§ 5.8.3.4.3 Simplified Procedure for Prestressed & Nonprestressed Sections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #80
If Vci controls, cotθ = 1If Vcw controls, cotθ must be calculated.
41
§ 5.8 - Shear and Torsion
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #81
'
'85.0
75.0
c
c
c
fofplaceIn
f
f g g
Sanded lightweight
§ 5.8 - Shear and Torsion
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).
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #82
Design with strut and tie (Article 5.6.3)Detail according to Article 5.13.2.3
42
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #83
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.
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #84
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
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
≥+
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #85
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.
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #86
44
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #87
controls, then cotθ must be calculated: cotθ=1.0+3(fpc/√fc’) < 1.8.
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #88
45
§ 5.8 - Shear and Torsion
§ 5.8.3.5 Longitudinal Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #89
load girder.
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #90
This provision appears to be for the vertical interface between flanges and webs of box girders – especially segmental boxes.
46
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #91
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #92
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
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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 +
≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #93
(5.8.4.1-5)ni 2 cv
cv vi vi
V K A
A b L
≤
= (5.8.4.1-6)
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #94
Pc = permanent net compressive force normal to the shear plane (k). If tensile, Pc = 0.
fc’ = 28 day compressive strength of the WEAKERconcrete
48
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #95
Article 5.8.4.3 (ksi)
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #96
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
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #97
For beams and girders, the longitudinal spacing of the rows of interface shear transfer reinforcing bars shall not exceed 24 inches.
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #98
µ = 1.0 K1 = 0.25K2 = 1.5 ksi
50
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #99
c = 0.075 ksiµ = 0.6
K1 = 0.2K2 = 0.8 ksi
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #100
µ = 1.0 K1 = 0.3K2 = 1.8 ksi – normal weightK2 = 1.3 ksi - lightweight
51
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
Avf has a minimum:
(5.8.4.4-1)cvvf
y
0.05AAf
≥
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #101
§ 5.8 - Shear and Torsion
§ 5.8.4 Interface Shear Transfer – Shear Friction
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #102
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
AASHTO-LRFD Specification, 4th Edition.
§ 5.8 - Shear and Torsion
§ 5.8.2.1 General
Torsion causes a condition of pure shear, as shown by
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #104
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 - Shear and Torsion
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #105
CONCRETE IS A BRITTLE MATERIAL!!!!!!
§ 5.8 - Shear and Torsion
§ 5.8.2.1 General
Because concrete is brittle and tension weak, torsion forces will crack the member diagonally, perpendicular to the
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #106
maximum principal tensile stress. As a result, concrete members under torsional loads tend to ‘unwrap’.
54
§ 5.8 - Shear and Torsion
§ 5.8.2.1 General
As with shear, compression struts will occur.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #107
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 - Shear and Torsion
§ 5.8.2.1 General
The important part is this:
Torsion causes shear stresses which are additive to the flexural shear stresses.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #108
55
§ 5.8 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #109
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 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #110
(5.8.2.1-4)'125.01125.0 +
=fp
fTcc
ccr
56
§ 5.8 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #111
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 - Shear and Torsion
§ 5.8.2.1 General
2A 789 i
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #112
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 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #113
§ 5.8 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #114
(Article 5.8.3.6.1).
58
§ 5.8 - Shear and Torsion
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #115
concurrent shear. Use the largest area required.
§ 5.8 - Shear and Torsion
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
+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #116
ph = perimeter of the centerline of the closed, transverse torsion reinforcement.
(5.8.2.1-7)
:
2u
uo
Box SectionsT dVA
+
59
§ 5.8 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #117
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 - Shear and Torsion
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #118
2 0
Ad sp
v φφφφ
(5.8.3.6.3-1)
60
§ 5.8 - Shear and Torsion
§ 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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #119
§ 5.8 - Shear and Torsion
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #120
concurrent factored shear.
61
§ 5.8 - Shear and Torsion
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #121
§ 5.8 - Shear and Torsion
Design for Shear and Torsion (cont.)
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #122
62
§ 5.8 - Shear and Torsion
Design for Shear and Torsion (cont.)
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #123
§ 5.8 - Shear and Torsion
Design for Shear and Torsion (cont.)
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #124
add the areas correctly.
63
§ 5.8 - Shear and Torsion
Design for Shear and Torsion (cont.)
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #125
1
S TOAASHTO-LRFDContinuous for Live Load
AASHTO-LRFD Specification, 4th Edition.
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #2
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
§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous
Construction Sequence
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #3
New in 2007
§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous
Negative moment reinforcement over a diaphragm
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #4
3
New in 2007
§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous
A bent strand positive moment connection
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #5
New in 2007
§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #6
camber up, causing positive moment.Usually in young girdersPositive moment connection required.
4
New in 2007
§ 5.14.1.4 Bridges Composed on Simple Span Precast Girders Made Continuous
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #7
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #8
Creep and shrinkage properties are extremely variable.Analysis results do not match field data for older girders.
5
Continuous for Live Load
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #9
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.
Continuous for Live Load
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #10
present.Using the 90 day rule greatly simplifies design.The 90 day rule is verified by experience in several states.
6
Continuous for Live Load
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #11
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).
Continuous for Live Load
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #12
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
Continuous for Live Load
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #13
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.
Continuous for Live Load
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #14
8
Continuous for Live Load
§ 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #15
testing.
Continuous for Live Load
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #16
9
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #17
adequate resistance. Mechanical connections would be permitted under this section.
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #18
girder cross section at the diaphragm.Mcr is calculated using the strength of the diaphragm concrete.
10
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #19
Continuous for Live Load
§ 5.14.1.4.9 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #20
stress concentrations.
11
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
Bent Bar type connection.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #21
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #22
connection detail will cause uneven bar stress. The connection should be kept as symmetrical as possible while still allowing meshing.
12
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #23
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #24
Art. 5.8.3.6.3 (Longitudinal reinforcement).The strands should be symmetrical about the vertical axis of the cross section.
13
Continuous for Live Load
§ 5.14.1.4.9 Positive Moment Connections
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #25
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.
Continuous for Live Load
§ 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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Prestressed Concrete: Slide #26
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
AASHTO-LRFD Specification, 4th Edition.
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #2
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
Design Example - Simple-Span Adjacent Box Girder Bridge
Problem Statement and Assumptions
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #3
Design Example - Simple-Span Adjacent Box Girder Bridge
Problem Statement and Assumptions
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #4
It is assumed the skew factor applies to this structure.
3
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #5
g gThese are chosen from that range.[BDM 302.5.1.7]
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #6
Design tables developed for Standard Specifications can usually be used to approximate the section for initial sizing.
4
Design Example - Simple-Span Adjacent Box Girder Bridge
Selecting the Girder Size
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #7
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).
Design Example - Simple-Span Adjacent Box Girder Bridge
Selecting the Girder Size
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #8
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #9
1.2.4 Reinforcing Bars
GR 60; Yield strength, fy = 60 ksi [BDM 302.5.1.8]Modulus of elasticity, Es = 29,000 ksi
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #10
Single SpanOverall Length: 67 ft.c/c Span: 65 ft.Support: Elastomeric Bearing Pad
1.2.6 Bridge Parameters
6
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #11
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #12
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #13
across the keyway is needed. Therefore, this bridge will be considered as not being “sufficiently connected”. This changes the distribution factor significantly.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
− −= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #14
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #15
DW future wearing surfaces and future DLFWS: (0.060 ksf)(4 ft) = 0.240 klf
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #16
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #17
to the exterior girders.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
−==
−=
+
+=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #18
8
10
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #19
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]
Design Example - Simple-Span Adjacent Box Girder Bridge
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 −==
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #20
896LL,TruckM k ft= −
11
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #21
w = load (klf)L = total spanx = point where moment is calculated.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #22
analysis.
Don’t you wish you would have paid more attention in Structural Analysis?????
12
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #23
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!)
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #24
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #25
(3.6.1.1.1)
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #26
6LN ≤ 45Skew ≤ °
14
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.1 Distribution Factors for Bending Moment
Where:
NL = Number of Lanes = 4C = K(W/L) < KW = Clear width of the bridge = 48 ft.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #27
( )J
I1K µ+=
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.1 Distribution Factors for Bending Moment
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #28
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
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.1 Distribution Factors for Bending Moment
( )+= =
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #29
= = 0.336D 11.9
µ = Poisson’s Ratio = 0.2 [LRFD 5.4.2.5]
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.1 Distribution Factors for Bending Moment
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #30
16
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #31
DFV (b/130L) (I/J)
(Table 4.6.2.2.3a-1)
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.2 Distribution Factors for Shear Force
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #32
40,000 < I <610,000
I = 108,150 OK
17
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.1.2 Distribution Factors for Shear Force
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #33
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 = =
Design Example - Simple-Span Adjacent Box Girder Bridge
1.4.2.2 Dynamic Allowance
IM = 33%
Where: IM = dynamic load allowance, applied only to truck load
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #34
18
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #35
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #36
( )
19
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #37
(Impact is not applied to lane loads.)
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #38
Fatigue: Does not need to be checked for pretensioned beams designed using the Service III load combination.
20
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #39
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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!
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #40
21
Design Example - Simple-Span Adjacent Box Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #41
( )(1.87 0.503) 1.37
pb b r
pb
f f ff ksi
= −
= − =
prestressing stress at the tensile fiber:
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.3 Required Number of Strands
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= − = − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #42
p
pe pe cpb
b
P P ef
A S= +
22
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.3 Required Number of Strands
Now plug in the required recompression stress, fpband solve for P :and solve for Ppe:
kips380
in6511in61.14
in5.7331
ksi37.1P
32
pe =
+
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #43
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.3 Required Number of Strands
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #44
Assuming 25% loss of prestress the final prestressing force per strand after losses is:
(0.153)(202.5)(1 0.25) 23.2kips− =
23
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.3 Required Number of Strands
Number of strands required:
This shows a need for at least (18) ½ in
380 16.423.2
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #45
This shows a need for at least (18) ½ in diameter, 270 ksi, low-lax strands as the strand pattern must be symmetrical.
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.4 Strand Pattern
At midspan:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #46
The ODOT design data sheets recommend 20 strands. Use 20 strands.
24
Design Example - Simple-Span Adjacent Box Girder Bridge
1.5.4 Strand Pattern
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #47
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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∆ = ∆ + ∆
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #48
∆fpLT loss due to long term shrinkage and creep of concrete, and relaxation of the steel, ksi
25
Design Example - Simple-Span Adjacent Box Girder Bridge
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).
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #49
2g ci i c
cgp
M eP PefA I I
= + −
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.1 Elastic Shortening
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 = + = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #50
1 75 441 4 53008 3gM . k . k ft k in= + = − = −
26
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.1 Elastic Shortening
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #51
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%.
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.1 Elastic Shortening
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #52
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
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.1 Elastic Shortening
( )( )( )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
−= + − =
∆ = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #53
( )1.15 7.64300pESf ksi ksi
ksi∆ = =
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.1 Elastic Shortening
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #54
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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 = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #55
(5.9.5.3-3)
h
51 'st
cifγ =
+
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.2 Long-Term Losses
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #56
low relaxation strand
29
Design Example - Simple-Span Adjacent Box Girder Bridge
2.1.2 Long-Term Losses
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∆ = + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #57
( ) ( ) ( ) ( )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
∆ = + +
∆ = + + =
Design Example - Simple-Span Adjacent Box Girder Bridge
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
∆ = ∆ + ∆
∆ = + =
= =
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #58
Loss is less than the 25% initially assumed, so OK.
202 5 27 1 175 4pef . ksi . ksi . ksi= − =
30
Design Example - Simple-Span Adjacent Box Girder Bridge
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 ’
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #59
Sum of effective prestress + permanent loads + transient loads
< 0.6φwfc
Design Example - Simple-Span Adjacent Box Girder Bridge
2.2 Compression Stress Limit States
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #60
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
Design Example - Simple-Span Adjacent Box Girder Bridge
2.2 Compression Stress Limit States
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #61
( )uX b lesser of 2z or 2 y= −
Design Example - Simple-Span Adjacent Box Girder Bridge
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
− −= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #62
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #63
( ){ }( )3
362 3 102 7 120 85
6599cLL,top
. . k ft in / ftf . ksi
in+ −
= =
Design Example - Simple-Span Adjacent Box Girder Bridge
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
++ = + =
< =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #64
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= + =
= −+ +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #65
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+ = − = + =
Design Example - Simple-Span Adjacent Box Girder Bridge
2.3.4 Service Load Tensile Stress Check Service III
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
k . inkipsf . ksi. in in
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #66
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
Design Example - Simple-Span Adjacent Box Girder Bridge
2.3.4 Service Load Tensile Stress Check Service III
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #67
the two load cases. However the stress is so low, that really doesn’t matter – we satisfy all allowables in all cases.
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #68
( ) ( ) ( )1.25 515.3 1.50 126.8 1.75 362.3 102.71648 19780
u
u
MM k ft k in
= + + +
= − = −
35
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #69
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
3.2 Steel Stress at Strength Limit State
(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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #70
As = Area of mild steel tension reinforcement = 0.0 in2
fy = Yield strength of tension reinforcement = 60.0 ksi
36
Design Example - Simple-Span Adjacent Box Girder Bridge
3.2 Steel Stress at Strength Limit State
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #71
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
Design Example - Simple-Span Adjacent Box Girder Bridge
3.2 Steel Stress at Strength Limit State
( )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
<+
= − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #72
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #73
n ps ps paM A f d2
= −
If there is no compression or mild tension steel, the equation becomes:
Design Example - Simple-Span Adjacent Box Girder Bridge
3.3 Flexural Resistance
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
=
= −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #74
( )( )2 2.7920 0.153 260 31 235502n
inM in ksi in k in = − = −
38
Design Example - Simple-Span Adjacent Box Girder Bridge
3.3 Flexural Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #75
the grout may act with the base concrete to effectively provide the complete flange width. All of this is a matter of engineering judgment.
Design Example - Simple-Span Adjacent Box Girder Bridge
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 =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #76
t
c31in 3.980.003 0.0204
3.98−
ε = =
39
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #77
However, this now makes the LRFD Specifications consistent with ACI 318. This replaces the maximum reinforcement provisions.
Design Example - Simple-Span Adjacent Box Girder Bridge
3.5 Determination of Flexural Strength
u nM M≤ Φ
( )( )19,780 1.0 23550k in k in OK− < −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #78
40
Design Example - Simple-Span Adjacent Box Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #79
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= + =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #80
41
Design Example - Simple-Span Adjacent Box Girder Bridge
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= + = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #81
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #82
42
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #83
Tension: 0.0948√fci’ < 0.2 ksi w/o bonded reinforcement 0.24√fci’ w/ bonded reinforcementCompression: 0.6fci’
Design Example - Simple-Span Adjacent Box Girder Bridge
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 =+=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #84
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #85
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #86
( ) in30.6ksi01.2474.0in33ksi474.0x =
+=
44
Design Example - Simple-Span Adjacent Box Girder Bridge
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
=−
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #87
Design Example - Simple-Span Adjacent Box Girder Bridge
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
++ −
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #88
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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 ==
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #89
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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−=−
−=−=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #90
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #91
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”
Design Example - Simple-Span Adjacent Box Girder Bridge
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
=+−=
−=−−
=
=−
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #92
By inspection, both are below the compression limit.
ksi20.1ksi814.0ksi01.2f
f
bot
top
=−=
47
Design Example - Simple-Span Adjacent Box Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #93
(5.8.3.2)v ea 2.79ind d 31in 29.6inches2 2
= − = − =
Design Example - Simple-Span Adjacent Box Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #94
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
−=−=−=
=−=−=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #95
For the diaphragm, V 1.75k (shear is constant), M = 1.75(3) = 5.25k-ft
Design Example - Simple-Span Adjacent Box Girder Bridge
5.2.1 Basic Shear Forces and Moments at the Critical Section
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:( )( )
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #96
( )( )( ) ( ) ( )
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
Design Example - Simple-Span Adjacent Box Girder Bridge
5.2.1 Basic Shear Forces and Moments at the Critical Section
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #97
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?????
Design Example - Simple-Span Adjacent Box Girder Bridge
5.2.1 Basic Shear Forces and Moments at the Critical Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #98
LL,Truck
50
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #99
Since the critical section is only 3 feet from the support, apply the skew factor.
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #100
The shear MUST be increased by skew factor, 1.20.
51
Design Example - Simple-Span Adjacent Box Girder Bridge
5.2.3 Factored Moments and Shears
Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #101
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #102
points would be used.
52
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #103
Both methods will be illustrated in this example.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #104
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #105
qsection is cracked, Ac in the equation above is =0. This equation also assumes at least minimum stirrups are used.
Design Example - Simple-Span Adjacent Box Girder Bridge
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 = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #106
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #107
Tension Half of the Box
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #108
Negative means “uncracked”, so the assumption of uncracked is correct.
( ) ( )
55
Design Example - Simple-Span Adjacent Box Girder Bridge
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
( )
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #109
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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 −= −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #110
θ = 21.0◦
β = 4.10
56
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #111
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #112
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #113
( )( ) 2vv c
y
11in 12inb sA 0.0316 f ' 0.0316 7ksi 0.184inf 60ksi
≥ = =
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #114
58
Design Example - Simple-Span Adjacent Box Girder Bridge
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=
+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #115
( ) ( ) ( ) ( ) ( )s
s
0.4in 60ksi 29.6 cot 21 0 1V
12inV 154.2k
+ =
=
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #116
#4 @ 12 inches is OK. Girder is OK in shear.
59
Design Example - Simple-Span Adjacent Box Girder Bridge
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≤ +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #117
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #118
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #119
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.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= = =
( )
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #120
( )( )( ) ( )cwV 0.06 7ksi 0.3 0.732ksi 11in 29.6in 123.2kips= + =
61
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #121
If the critical section is within the transfer length, fpc is reduced linearly.
Design Example - Simple-Span Adjacent Box Girder Bridge
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 = + ≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #122
o
0.732ksicot 1.0 3 1.83 1.8; so use 1.87ksi
29
θ = + = >
θ =
62
Design Example - Simple-Span Adjacent Box Girder Bridge
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= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #123
( )uV 138k 0.9 123.2k 106.5k 207k= < + =
Design Example - Simple-Span Adjacent Box Girder Bridge
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= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #124
( ) ( )( )( ) ( )( )( )
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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= − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #125
( ) ( ) ( )wsM 0.5 0.140klf 13ft 65ft 13ft 47.3k ft= − = −
Design Example - Simple-Span Adjacent Box Girder Bridge
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
= − =
= − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #126
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #127
= 531.8 k-ft = 6381 k-in
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #128
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
Design Example - Simple-Span Adjacent Box Girder Bridge
5.4.2.2 Live Load
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #129
Design Example - Simple-Span Adjacent Box Girder Bridge
5.4.2.2 Live Load
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #130
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #131
However, this is not always the case. It just happened that way in this example.
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #132
67
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #133
and at 0.2L, the section is not at the obtuse corner.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #134
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #135
effective prestressing force:
( )cpe 2 3
537k 14.61in537kipsf 1.94ksi733.5in 6511in
= + =
Design Example - Simple-Span Adjacent Box Girder Bridge
5.4.2.3 Determination of Cracking Load for Shear
(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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #136
Sc = composite section modulus = Snc since this is a non-composite structure
69
Design Example - Simple-Span Adjacent Box Girder Bridge
5.4.2.3 Determination of Cracking Load for Shear
( ) ( )3cre 3
cre
12 409.2k ftM 6511in 0.529ksi 1.94ksi
6511inM 11165k in 930.5k ft
− = + −
= − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #137
Design Example - Simple-Span Adjacent Box Girder Bridge
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−
= + + = >−
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #138
( )( )549k ft
0.06 7ksi 11in 29.6in 51.7k=
70
Design Example - Simple-Span Adjacent Box Girder Bridge
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= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #139
The section is adequate in shear.
( )u nV 92.0k V 0.9 143.0k 59.2k 182.0k= < φ = + =
Design Example - Simple-Span Adjacent Box Girder Bridge
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 = + =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #140
71
Design Example - Simple-Span Adjacent Box Girder Bridge
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,
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #141
p g ,the more stirrups which cross the crack and the higher Vs.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #142
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
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #143
( )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).
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #144
(5.11.4.2-4)
( )px pe ps ped b60d−l
73
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
( )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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #145
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #146
φ = 1 for moment; 0.9 for shearAsfy = assumed 0 (ignore any mild steel)fpe = 175.4 ksifps = 260.0 ksi
74
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
u u ups ps s y p s
M 0.5N VA f A f V 0.5V cotd
+ ≥ + + − − 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #147
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
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #148
( ) ( )2 1383 06 104 7 320 0 5 153 21 1990 9
= > − =
p p
k. in . ksi k . ( k ) cot k.
φ
OK
75
Design Example - Simple-Span Adjacent Box Girder Bridge
5.6 Minimum Longitudinal Steel
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
k. in . ksi k . ( k ) cot k.
θφ
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #149
OK
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #150
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #151
g g
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Moment
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #152
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
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Moment
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
= + =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #153
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Moment
( ) ( ) ( )( )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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #154
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
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Moment
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #155
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Moment
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
= + + + + = − = −
−= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #156
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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= − = − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #157
( ) ( )( )( )rM 0.5wx L x 0.5 0.090klf 3ft 65ft 3ft 8.37k ft= − = − = −
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #158
0.5480.23 212e 1 1.234
40
+ − = + =
80
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #159
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!
Design Example - Simple-Span Adjacent Box Girder Bridge
6.1 Exterior Girder - Shear
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #160
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
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #161
only applies to slab/beam bridges (Types a, e and k) with diaphragms or cross braces.
Design Example - Simple-Span Adjacent Box Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #162
The same methods used for finding camber and deflection used for Standard Specifications apply for LRFD Designs.
82
Design Example - Simple-Span Adjacent Box Girder Bridge
2.5.2.6.2 Deflection
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #163
Axle Load (front) = 8k(1.33)(0.304) = 3.22k (includes impact)
Design Example - Simple-Span Adjacent Box Girder Bridge
2.5.2.6.2 Deflection
Here are the live loads positioned for maximum deflection.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #164
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
AASHTO-LRFD Specification, 4th Edition.
Design Example - Continuous Two-Span I-Girder Bridge
Problem Statement and Assumptions
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #2
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
Design Example - Continuous Two-Span I-Girder Bridge
Problem Statement and Assumptions
34’-0”
8.5” structural+ 1.0”
4 Spaces @ 8’-0” = 32’-0”2.5’
Type IV
2.5’
wearing
37’-0”
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #3
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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #4
9
8”
2’-2”
from that range. The BDM also gives the deck strength (302.5.2.8).
3
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #5
LoadsFuture wearing surface: 0.060 ksf (ODOT Std. Drawings)Barriers: 0.640 k/ft eachTruck: HL 93, including dynamic allowance
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #6
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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #7
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”
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #8
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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #9
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
Design Example - Continuous Two-Span I-Girder Bridge
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]
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #10
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
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #11
Design Example - Continuous Two-Span I-Girder Bridge
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:
Width of deck is constant. OKNumber of beams, Nb > 4. OKOverhang part of the roadway < 3 ft OK
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #12
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
Design Example - Continuous Two-Span I-Girder Bridge
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”
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #13
ODOT Std. Drawings
Barrier = 0.640 klf 2 each (0.640)/5 girders = 0.256 kips/ft/girder
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #14
8
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #15
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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #16
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
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007
Note: See the Loads Module for a complete explanation of how this is applied.
Do Not Duplicate Loads & Analysis: Slide #17
Design Example - Continuous Two-Span I-Girder Bridge
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]
Width of deck is constant. OKNumber of beams, Nb > 4. OKOverhang part of the roadway < 3 ft OK
de = 2.5 ft – 1.5 ft = 1.0 ftCurvature in plan
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #18
< 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
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #19
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #20
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
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Bending Moment
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
= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #21
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
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Bending Moment
( )( )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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #22
g
12
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Bending Moment
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
= +
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #23
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Bending Moment
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
= +
= +
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #24
The case of two or more design lanes loaded controls,
DFM = 0.665 lanes/beam
0.467DFM
13
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #25
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
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Shear Force
For two or more lanes loaded:
28 80.212 35
0.814
DFV
DFV
= + −
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #26
14
Design Example - Continuous Two-Span I-Girder Bridge
Distribution Factors for Shear Force
For one design lane loaded:
S 0.362580.3625
0.68
SDFV
DFV
DFV
= + = +
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #27
The case of two or more design lanes loaded controls, DFV = 0.814 lanes/beam
Design Example - Continuous Two-Span I-Girder Bridge
Dynamic Allowance
IM = 33%Where: IM = dynamic load allowance applied only to truckWhere: IM = dynamic load allowance, applied only to truck load
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #28
15
Design Example - Continuous Two-Span I-Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #29
LT
= (bending moment per lane)(0.665)(1.33) = (bending moment per lane)(0.884) kips-ft
Design Example - Continuous Two-Span I-Girder Bridge
Unfactored Shear Force and Bending Moments
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #30
LANE
= (bending moment per lane)(0.665) kip-ft
16
Design Example - Continuous Two-Span I-Girder Bridge
Unfactored Shear Force and Bending Moments
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #31
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
Design Example - Continuous Two-Span I-Girder Bridge
Unfactored Shear Force and Bending Moments
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #32
17
Design Example - Continuous Two-Span I-Girder Bridge
Load Combinations
The following limit states are applicable: Service I:
(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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #33
Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)
Design Example - Continuous Two-Span I-Girder Bridge
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!
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #34
different load factors in this case!
18
Design Example - Continuous Two-Span I-Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #35
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #36
-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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #37
applies to tension in prestressed sections.
Design Example - Continuous Two-Span I-Girder Bridge
Service Load Stresses at Midspan
Bottom tensile stress due to applied dead and live loads using load combination Service III:
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #38
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
Design Example - Continuous Two-Span I-Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #39
0.19 7.0 0.503ksi= =Required Compressive Stress From Strands
(3.47 0.503) 2.97pbf ksi= − =
Design Example - Continuous Two-Span I-Girder Bridge
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= − = − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #40
( ) (24.73 4.32) 20.41c b bse y y in= = =
21
Design Example - Continuous Two-Span I-Girder Bridge
Required Number of Strands
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= +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #41
solve form Ppe: (20.41)2.97
789 10,542927
pe pe
pe
P P
P kips
= +
=
Design Example - Continuous Two-Span I-Girder Bridge
Required Number of Strands
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
force.
Do Not Duplicate Loads & Analysis: Slide #42
22
Design Example - Continuous Two-Span I-Girder Bridge
Required Number of Strands
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #43
Number of strands required = strands
Try (40) ½ in diameter, 270 ksi, low-lax strands.
927 39.923.2
=
Design Example - Continuous Two-Span I-Girder Bridge
Strand Pattern
At midspan:
No. Strands
Distance from bottom (in)
7 811 611 411 2
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #44
10 Spa.@ 2”
2” 2”
23
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #45
24.73 4.70 20.0c b bse y y in= − = − =
Design Example - Continuous Two-Span I-Girder Bridge
Prestress Losses
Total 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∆ = ∆ + ∆
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #46
24
Design Example - Continuous Two-Span I-Girder Bridge
Elastic Shortening
∆ = ppES cgp
ct
Ef f
E (5.9.5.2.3a-1)
Where:
2g ci i c M eP Pe
+ −
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).
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #47
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).
Design Example - Continuous Two-Span I-Girder Bridge
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
( )
( )
+ −∆ =
+ +
ps pi g m g m g gpES
g g ct
A f I e A e M Af A I E
A I A
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #48
(C5.9.5.2.3a-1)
2( )+ + g gps g m g
p
A I e AE
25
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #49
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
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #50
rather than c/c bearing, the moment becomes 11641 k-in. The elastic shortening loss becomes 16.13 ksi; < 1% different.
26
Design Example - Continuous Two-Span I-Girder Bridge
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 01Hγ = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #51
(5.9.5.3-2)
(5.9.5.3-3)
1.7 0.01h Hγ = −
51 'st
cifγ =
+
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #52
∆fpR = An estimate of relaxation loss taken as 2.5 ksi for low relaxation strand
27
Design Example - Continuous Two-Span I-Girder Bridge
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∆ = + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #53
ksi
78914.29 10.92 2.5
27.71
pLT
pLT
pLT
f
f
f
∆ = + +
∆ =
Design Example - Continuous Two-Span I-Girder Bridge
Total Losses at Service Loads
Total Prestress Losses:
(5.9.5.1-1)
16.24 27.71
43.95
pT pES pLT
pT
pT
f f f
f
f
f
∆ = ∆ + ∆
∆ = +
∆ =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #54
Losses are approximately 22% < 25% OK
202.5 43.95 158.6pef = − =
28
Design Example - Continuous Two-Span I-Girder Bridge
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= − ∆ = − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #55
Therefore, the total prestressing force after transfer is, Pi = 1,140 kips
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Transfer
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%
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #56
kips. The difference is 2%.
29
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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= ≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #57
Therefore, 0.200 ksi (CONTROLS)
( )0.0948 4.5 0.2
0.201 0.2tf ksi
ksi ksi= ≤
≤
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #58
t cif f
(5.9.4.1.2)
30
Design Example - Continuous Two-Span I-Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #59
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= − = −
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #60
Tensile stress limit for concrete with bonded reinforcement: 0.509 ksi NG
1.44 2.56 0.13 0.99= + =tf ksi
31
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
= + −
= + −
= + − = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #61
Compressive stress limit for concrete: +2.700 ksi NG
1.44 2.16 0.11 3.49tf ksi= + − = +
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #62
8 411 2
32
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
4”
9 Strands
2’-6”
48’-7”
50”
14’-7”34’-0”
31 Strands ψ
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #63
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
+ +=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #64
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
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
− +=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #65
Eccentricity of the strand group at transfer length is:
in
40
24.73 13.87 10.86− =
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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− =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #66
34
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
= − +
= − + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #67
Compressive stress limit for concrete: +2.700 ksi OK
Design Example - Continuous Two-Span I-Girder Bridge
Stresses At Transfer Length Section
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
= + −
= + − = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #68
Compressive stress limit for concrete: +2.700 ksi OK
35
Design Example - Continuous Two-Span I-Girder Bridge
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #69
(0.5)(0.822)(34.00)(97.17 34.00) 882.7gM k ft= − = −
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Harp Points
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #70
36
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Harp Points
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
= + −
= + −
= + − = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #71
Compressive stress limit for concrete: +2.700 ksi OK
1.44 2.16 1.00 2.60bf = + = +
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Midspan
The bending moment due to beam self-weight at a distance 48’-7” (midspan) from the end of the beam is:
Concrete stress at the top fiber of the beam:
(0.5)(0.822)(48.58)(97.17 48.58) 970.1gM k ft= − = −
= − + gi it
t t
MP PefA S S
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #72
Compressive stress limit for concrete: +2.700 ksi OK
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
Design Example - Continuous Two-Span I-Girder Bridge
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
= + −
= + − = + − = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #73
Compressive stress limit for concrete: +2.700 ksi OK
Design Example - Continuous Two-Span I-Girder Bridge
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= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #74
Harp angle:1 54 4 6tan 6.2
34(12)ψ − − −= =
o
38
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #75
Total hold down force 9 strands(3.5) 31.6 kips
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #76
So hold-down force per strand = 3.5 kips per strand OK
3 4,000
39
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Transfer
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #77
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #78
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
Design Example - Continuous Two-Span I-Girder Bridge
Summary of Stresses at TransferTop Stress
0.3
0.1
0.15
0.2
0.25
Stre
ss (k
si)
Transfer Length
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #79
0
0.05
0 5 10 15 20 25 30 35 40 45 50
Length (ft)
g
HarpPoint
Mid-Span
Design Example - Continuous Two-Span I-Girder Bridge
Summary of Stresses at TransferBottom Stress
3
1
1.5
2
2.5
Stre
ss (k
si)
Transfer Length
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #80
0
0.5
0 5 10 15 20 25 30 35 40 45 50
Length (ft)
Length
HarpPoint
Mid-Span
41
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Service Loads
Total loss of prestress at service loads is
Stress in tendon after all losses
43.95pTf ksi∆ =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #81
202.5 43.95 158.55= −∆ = − =pe pi pTf f f ksi
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #82
42
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #83
For the precast beam:0.40fc’ = 0.40(7.0) = +2.800 ksiFor the deck: 0.40fc’ = 0.40(4.5) = +1.800 ksi
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #84
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
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
Tension:For components with bonded prestressing tendons:p p g
For the precast beam:
'
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #85
'0.19 0.19(7.0) 0.503cf ksi= =
Design Example - Continuous Two-Span I-Girder Bridge
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
+ += − + +
+ += − + +
+ + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #86
Compressive stress limit for concrete: +3.150 ksi OK
1 1.23 2.18 2.83 0.10 1.98tgf = − + + = +
44
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #87
Compressive stress limit for concrete: +2.800 ksi OK
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #88
Compressive stress limit for concrete: +4.200 ksi OK
45
Design Example - Continuous Two-Span I-Girder Bridge
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
+=
+= +
= +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #89
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #90
Compressive stress limit for concrete: +1.800 ksi OK
46
Design Example - Continuous Two-Span I-Girder Bridge
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
++ +=
+ +=
= +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #91
Compressive stress limit for concrete: +2.700 ksi OK
tcf
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #92
Tensile stress limit for concrete: -0.503 ksi OK
Service III has the 0.8LL factor!
bf
47
Design Example - Continuous Two-Span I-Girder Bridge
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= + + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #93
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
= + + + += −
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
Average stress in prestressing steel:
(5.7.3.1.1)1ps pu
p
cf f kd
= −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #94
48
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #95
= the centroid of the prestressing tendonsh - ybs = 62.5 – 4.70 = 57.80
c = Distance between the neutral axis and the compressive face
in.
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
' '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
+ −=
+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #96
49
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #97
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.
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #98
0.83(5.28) 4.39 8.5sa in t in= = < =
50
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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 −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #99
6.12(263.3) 57.802
2 127,467
n ps ps p
n
aM A f d
M k ft
= − =
= −
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #100
7, 467 5,615n uM k ft M k ftφ = > =
51
Design Example - Continuous Two-Span I-Girder Bridge
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= + − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #101
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 = = = >
Design Example - Continuous Two-Span I-Girder Bridge
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,
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #102
1.33 times the factored moment required by the applicable strength load combinations
(5.7.3.3.2)
52
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #103
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
+ = + = + =
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement - Positive Moment Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #104
Snc= Section modulus for the extreme fiber of the noncomposite section where tensile stress is caused by externally applied loads = 10,542
in3
53
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement - Positive Moment Section
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 = ≥
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #105
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 Example - Continuous Two-Span I-Girder Bridge
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 = − + − + − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #106
wide.2. This section is a nonprestressed reinforced concrete
section, thus Φ = 0.9 for flexure.
54
Design Example - Continuous Two-Span I-Girder Bridge
Design of the Negative Moment Section
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #107
54 0.5(8.5) 58.25+ =
c p g gd = Effective depth to negative moment
reinforcement from bottom of girder =in
Design Example - Continuous Two-Span I-Girder Bridge
Design of the Negative Moment Section
(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
=
= − +
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #108
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
Design Example - Continuous Two-Span I-Girder Bridge
Longitudinal Deck Reinforcement
The longitudinal reinforcement in the deck includes distribution reinforcement and other minimum reinforcementdistribution reinforcement and other minimum reinforcement.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #109
2( ) 5.58s providedA in=
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
The additional area of deck reinforcement required: 22
, ' 13.93 5.58 8.35s Add lA in= − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #110
56
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #111
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
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #112
We assumed 4.25” from top OKd = 58.34 in
4.1613.94
x
57
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #113
( )( )( )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 = −
= − = − > = −
Design Example - Continuous Two-Span I-Girder Bridge
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=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #114
(4.6.2.6)
0.10(96.25)(12) 115.5in
58
Design Example - Continuous Two-Span I-Girder Bridge
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≤ −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #115
sls
s
MfA jd
=
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #116
g62.5-4.16 = 58.34
j = 1-(k/3) = 1-(0.275/3) = 0.908
59
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #117
So:
, 5.7185,072
steel
girderE= =n = Modular Ratio =
2,140.5(12) 34.813.94(0.908)(58.34)sf ksi= =
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #118
is found to be 34.6 ksi which compares to 34.8 ksi using the rectangular assumption.
60
Design Example - Continuous Two-Span I-Girder Bridge
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
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #119
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
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
The maximum concrete stress is:
The steel stress is:
slc
tr
M cfI
=
( )The term M(d-c)/I gives the equivalent concrete
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #120
( )sls
tr
M d cf n
I−
=the equivalent concrete stress. It is converted to steel stress by multiplying by n.
61
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #121
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.
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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−
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #122
( )( )( )( ) ( )( )
( ) ( )
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
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
= + + −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #123
( )( ) ( )( )
( )
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
+ + − −
+ − =
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
( )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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #124
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
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #125
fs = Tensile stress in steel reinforcement at the service limit state
ksi
βs = 10.7( )
c
c
dh d
+−
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #126
64
Design Example - Continuous Two-Span I-Girder Bridge
Control of Cracking by Distribution Reinforcement
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≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #127
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.
Design Example - Continuous Two-Span I-Girder Bridge
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 = = = >
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #128
φ
(5.7.2.1 & 5.5.4.2)
65
Design Example - Continuous Two-Span I-Girder Bridge
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,
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #129
1.33 times the factored moment required by the applicable strength load combinations
(5.7.3.3.2)
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement – Negative Moment Section
(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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #130
Sc= 29,534 in
66
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement – Negative Moment Section
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #131
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= > =
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #132
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
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #133
y
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #134
68
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #135
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
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #136
p g gcalculations the effective width of 96 inches, 0.5 inch strand, and concrete strength of 4.5 ksi were used.
69
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #137
cr
e dsh
1.2M 850k ftL 8
= −
= −l
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #138
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
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #139
* Back calculated based on strand length
fpul 209 166 135 117 86
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #140
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
Design Example - Continuous Two-Span I-Girder Bridge
Required Area of Strand
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #141
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #142
(5.8.2.4-1)0.5 ( )u c pV V Vφ≥ +
72
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #143
Φ = Resistance factor (5.5.4.2.1)
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #144
reinforcement = 58.34a = Equivalent depth of the compression
block = 5.41in
h = Total height of section = 62.5 in
73
Design Example - Continuous Two-Span I-Girder Bridge
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
= − = − =
≥ = =
≥ = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #145
Therefore, dv 55.63 in.
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #146
Thus the critical section is:55.63 in + 5 in = 60.63 inches.
74
Design Example - Continuous Two-Span I-Girder Bridge
Calculation of Critical Section
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 = + + + + =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #147
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
= + + − + − + −
= − − = − −
Design Example - Continuous Two-Span I-Girder Bridge
Calculation of Critical Section
At this point, there are three choices:
1. Ignore the prestressing steelThen, this is a reinforced sectionβ = 2θ = 45◦
(This is VERY conservative)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #148
(This is VERY conservative)
2. Use Sectional Model for RC3. Include PS Steel
75
Design Example - Continuous Two-Span I-Girder Bridge
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= − =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #149
Assume #4 hoops Av = 0.4 in2 α = 90 sin α =1 cot α =0
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #150
Vs = 334 kips
76
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #151
Ap 0 inEp = 28,500 ksiEs = 29,000 ksi
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #152
fpo = 0
77
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
−
+ + −= ≤
≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #153
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
u pu
V Vv
b dφ
φ−
=v vb dφ
vu = Shear stress in concrete kipsbv =
=Effective web width of the beam8
in
Vp = Component of the effective prestressing force in the direction of the applied shear
kips
Where:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #154
==
direction of the applied shear(force per strand)(number of draped strands)(sin ψ)0
78
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #155
( u c ) x5.8.3.4.2-1:
θ = 37◦β = 2.13
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #156
So #4 hoops at 6 in Vs = 295.0 kips
OK0.9(79.3 295.0 0) 337r uV k V= + + = >
79
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
As = Area of nonprestressed steel on the flexural tension side of the member = 13.94
in2
A = 9(0 153) = 1 38 in2
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #157
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.
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #158
.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
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
−
+ + − −= ≤
+
≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #159
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
u pu
V Vv
b dφ
φ−
=v vb dφ
vu = Shear stress in concrete kipsbv =
=Effective web width of the beam8
in
Vp = Component of the effective prestressing force in the direction of the applied shear
kips
Where:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #160
==
direction of the applied shear(force per strand)(number of draped strands)(sin ψ)23.6
81
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #161
( u c ) xTable 5.8.3.4.2-1:
θ = 34.4◦β = 2.26
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
kips'0.0316 0.0316(2.26) 7(8)(55.63) 84.1c 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(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
= = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #162
So #4 hoops at 6 in Vs = 313 kips
OK0.9(84.1 313.0 23.6) 378.6r uV V= + + = >
82
Design Example - Continuous Two-Span I-Girder Bridge
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 =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #163
Since , Then in 24 in CONTROLS
'0.125u cv f< max 0.8 0.8(55.63) 44.5 24.0= = = ≤vs d
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement Requirement
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
8in 6inb sA 0.0316 f ' 0.0316 7ksi 0.067inf 60ksi
≥ = =
(5.8.2.5)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #164
( )
83
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #165
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
Design Example - Continuous Two-Span I-Girder Bridge
Critical Section – Positive Moment
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
. . . . .= =
+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #166
( )( )
58 24 11270 1 0 28 264 858 2
0 83 4 11 3 42
ps
..f . . ksi.
a . . . in
= − =
= =
84
Design Example - Continuous Two-Span I-Girder Bridge
Critical Section – Positive Moment
0.5( ) 58.2 0.5(3.42) 56.5d d a in= − = − =
Therefore, dv = 56.5 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
≥ = =
≥ = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #167
Design Example - Continuous Two-Span I-Girder Bridge
Calculation of Critical Section
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #168
Thus the critical section is:56.5 in + 5 in ≈ 62 inches.
85
Design Example - Continuous Two-Span I-Girder Bridge
Calculation of Critical Section
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
= + + + + =
= + + + + = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #169
corresponding to maximum Vu. Therefore,
kipskip-ft
250.0uV =1,346uM =
(5.8.3.4.2)
Design Example - Continuous Two-Span I-Girder Bridge
Contribution of Concrete to Nominal Shear Resistance
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #170
86
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #171
==
(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
Design Example - Continuous Two-Span I-Girder Bridge
Strain in Flexural Tension Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #172
=
term is calculated as the area on the tension side (bottom in this case) from the tension fiber to h/2.475
87
Design Example - Continuous Two-Span I-Girder Bridge
Strain in Flexural Tension Reinforcement
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
−
+ + − −= ≤
+
− ≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #173
The negative value means the section is uncracked
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #174
= 23.6
250 0.9(23.6) 0.5620.9(8)(56.5)uv ksi−
= =
88
Design Example - Continuous Two-Span I-Girder Bridge
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
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #175
Design Example - Continuous Two-Span I-Girder Bridge
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 = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #176
89
Design Example - Continuous Two-Span I-Girder Bridge
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= > + = + =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #177
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement Requirement
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 = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #178
ksi
Since , Then in : 24 in CONTROLS
0.562uv =
'0.125u cv f< max 0.8 24.0vs d= ≤
90
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement Requirement
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
8in 12inb sA 0.0316 f ' 0.0316 7ksi 0.134inf 60ksi
≥ = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #179
#4@ 24 inch o.c.= 0.2 in2/ftThis is adequate to meet minimum.
Design Example - Continuous Two-Span I-Girder Bridge
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≤ +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #180
5.8.3.3.-1:'0.25c s c v vV V f b d+ ≤
91
Design Example - Continuous Two-Span I-Girder Bridge
Maximum Nominal Shear Resistance
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+=
+ =
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #181
OK( )122.4 144.2 266.6 0.25(7.0)(8)(56.5) 791+ = ≤ =
Design Example - Continuous Two-Span I-Girder Bridge
Maximum Nominal Shear Resistance
( )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 + +
= + + =
> =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #182
92
Design Example - Continuous Two-Span I-Girder Bridge
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 =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #183
p gThis is shear at the critical section near the pier.
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #184
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
Design Example - Continuous Two-Span I-Girder Bridge
Required Interface Shear Reinforcement
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 =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #185
Design Example - Continuous Two-Span I-Girder Bridge
Required Interface Shear Reinforcement
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= + +
= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #186
( )( )( ) ( )
( )
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
Design Example - Continuous Two-Span I-Girder Bridge
Required Interface Shear Reinforcement
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= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #187
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.
( )( ),
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #188
OK 0.05(480)0.40 0.4060
≥ =
95
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #189
(transverse) shear reinforcement required by Article 5.8.1.1 is extended and anchored into the slab.
Design Example - Continuous Two-Span I-Girder Bridge
Maximum Nominal Shear Resistance
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
≤≤
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #190
K1 = 0.3 and K2 = 1.8 (for normal weight concrete) are found in Article 5.8.4.3.
2 cv
96
Design Example - Continuous Two-Span I-Girder Bridge
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
+ ≥ + + − −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #191
(5.8.3.5-1)
vd φ φ
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #192
continuous for live load bridge; saying that this equation must be checked for a continuous for live load bridge.
97
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #193
(5.11.4.2)
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
( )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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #194
Since this is less than the development length, the stress in the steel must be reduced for lack of development.
98
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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−+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #195
( )pxf 158.6ksi 264.8ksi 158.6ksi 206ksi127.3in 30in
= + − =−
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #196
Vu/φ [LRFD 5.8.3.5].
250144 277 80 9
us
V kV k . k.φ
= < = =
99
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
+ ≥ − −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #197
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)
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #198
( )
( )( )
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
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
The development length is 12 inches so the bar is fully developed:developed:
Thus:2464 6 426 0 64
60−
= =s.A . in
# # #
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #199
4 #4 works. 3 #5 also works as a # 5 needs a 15 inch development length.
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #200
( )( )
( ) ( )
4.74 90 426
250 23.6 0.5 277.8 cot 21.4 294.20.9
= >
− − =
k
k
101
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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?
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #201
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
( )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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #202
Since this is less than the development length, the stress in the steel must be reduced for lack of development.
102
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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−+
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #203
( )pxf 158.6ksi 264.8ksi 158.6ksi 198.4ksi127.3in 30in
= + − =−
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
0.5 0.5 cotu u ups ps s y p s
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #204
Vu/φ [LRFD 5.8.3.5].
250144 277 80 9
us
V kV k . k.φ
= < = =
103
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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
+ ≥ − −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #205
( )158 6 1052 9
60 30pe px
pxb
f .f . ksi
d= = =
l (5.11.4.2-3)
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
( )( )
0.5 cotups ps p s
VA f V V θφ
≥ − −
Assume #4 bars will be used.
( )( )
( ) ( )
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #206
( )( )7
0 4 0 4 0 5 60 12c
b y
f '
. d f . . in< = =
104
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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−= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #207
This would be 22 #4! Clearly unrealistic!
50
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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 cotups ps p s
VA f V V θφ
≥ − −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #208
( )( )
( ) ( )
4.74 52.9 250.8
250 23.6 0.5 277.8 cot 21.4 294.20.9
k
k
= <
− − =
105
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Longitudinal Reinforcement Requirement
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−= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #209
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #210
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
Design Example - Continuous Two-Span I-Girder Bridge
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 = =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #211
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= >
Design Example - Continuous Two-Span I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #212
(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
AASHTO-LRFD Specification, 4th Edition.
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Transverse Section
34’-0”
Type IV
8.5” structural+ 1.0” wearing
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #214
4 Spaces @ 8’-0” = 32’-0”2.5’ 2.5’
37’-0”
108
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #215
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.
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #216
2.5 ft 4.0 ft
109
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #217
Sb= 16340 in3
STG = 39576 in3
STC = 25721in3
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #218
This will DECREASE the dead load moment on the exterior girders.
110
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #219
77.28 -21.4 537
86.99 -28.5 295
96.25 -35.3 0
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Exterior Beams
Exterior Girders:One Lane Loaded:
Lever Rule
Two or More Lanes Loaded:
g= egint
Where:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #220
1.977.0 ede +=
g = DFMext
gint= DFMint
111
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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 = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #221
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
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #222
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
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Distribution Factor for Moment
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=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #223
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
Minimum Exterior DFM: (Rigid Body Rotation of Bridge Section)
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #224
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
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Distribution Factor - One Lane
2’1.5’ 6’
36k
8’-0” 2.5’
36k 36k e = 12’
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #225
16’-0”
Note: Only the truck is used and it cannot be closer than 2’ from the barrier (3.6.1.3)
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Distribution Factor for Moment
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
= +
= +
∑
∑
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #226
( ),
,
2 25 2
0.50
16 8
Ext MinDFM =
+
, ( ) 1.2(0.5) 0.6Ext MinDFM MPF DF= = =
114
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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’
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #227
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)
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Distribution Factor for Moment
Minimum Exterior Girder Distribution Factor Two Lane:N
,
2
2 16(12 0)
L
Ext Minb
N
ExtL
Nb
X eN
DFMN x
DFM
= +
+
∑
∑
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #228
,
,
2 2
( )5 2(160.70
8 )Ext Min
Ext Min
DFM
DFM
= +
=
+
, ( ) 1.0(0.7) 0.7Ext MinDFM MPF DF= = = CONTROLS
115
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Distribution Factor for Moment
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 =
AASHTO-LRFD 2007ODOT Short Course
July 2007
This is a 5% increase from the interior girder (DFM = 0.665)
Do Not Duplicate Loads & Analysis: Slide #229
Exterior Girders:One Lane Loaded:
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
Exterior Beams Shear
One Lane Loaded:
Lever Rule
Two or More Lanes Loaded:
DFM Ext = e DFM Int
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #230
DFM,Ext e DFM,Int
1060.0 ede +=
116
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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 = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #231
DFExt+ = (0.70) (0.814) = 0.570
1.00.6 0.70
10= + =
Design Example - Continuous Two-Span Inelastic I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #232
However, the minimum DF 0.7 (from possible rigid body rotation) - THIS CONTROLS.
117
Design Example - Continuous Two-Span I-Girder Bridge
DL-Unfactored Shear Forces & Bending Moments - Exterior Girder
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 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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #233
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
Design Example - Continuous Two-Span I-Girder Bridge
Unfactored Shear Force and Bending Moments
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #234
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
Design Example - Continuous Two-Span I-Girder Bridge
Load Combinations
The following limit states are applicable: Service I:
(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)
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #235
Maximum Q = 1.25(DC) + 1.50(DW) + 1.75(LL + IM)Minimum Q = 0.90(DC) + 0.65(DW) + 1.75(LL + IM)
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #236
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
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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
Due to one half the permanent loads and live load:
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #237
For the precast girder: 0.40fc’ = 0.40(7.0) = +2.800 ksiFor the deck: 0.40fc’ = 0.40(4.5) = +1.800 ksi
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #238
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
Design Example - Continuous Two-Span I-Girder Bridge
Stress Limits for Concrete
Tension:For components with bonded prestressing tendons:p p g
For the precast girder:
'0.19 0.19(7.0) 0.503cf ksi= =
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #239
Design Example - Continuous Two-Span I-Girder Bridge
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
P P e M M M MfA S S S
f
f
+ += − + +
+ += − + +
+ + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #240
Compressive stress limit for concrete: +3.150 ksi OK
1 1.23 2.18 2.43 0.12 1.60tgf = − + + = +
121
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #241
Compressive stress limit for concrete: +2.800 ksi OK
Design Example - Continuous Two-Span I-Girder Bridge
Stresses at Midspan
3. Under permanent and transient loads:
3 1
3
( )
1,455*12(1.60)39576
1 60 0 44 2 04
LL Itg tg
tg
tg
Mf fS
f
f
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #242
Compressive stress limit for concrete: +4.200 ksi OK
3 1.60 0.44 2.04tgf = + = +
122
Design Example - Continuous Two-Span I-Girder Bridge
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
+=
+= +
= +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #243
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.
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #244
Compressive stress limit for concrete: +1.800 ksi OK
123
Design Example - Continuous Two-Span I-Girder Bridge
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
+= +
= +
= + = +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #245
Compressive stress limit for concrete: +2.700 ksi OK
Design Example - Continuous Two-Span I-Girder Bridge
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
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
++ + +
= + − −
+ ++= + − −
= + − − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #246
Tensile stress limit for concrete: -0.503 ksi OK
124
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007
LLTENSION 0.40 ksi 0.13 ksi
Do Not Duplicate Loads & Analysis: Slide #247
Design Example - Continuous Two-Span I-Girder Bridge
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( )1.25(912.9 814.2 171.9) 1.5(274.2) 1.75(1,483)
u ext
u ext
M DC DW LL IMM
= + + +
= + + + +
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #248
, ,int5380 5,615u ext uM k ft M k ft= − < = −
Since exterior Mu is less than interior Mu, OK
125
Design Example - Continuous Two-Span I-Girder Bridge
Positive Moment Section
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.
AASHTO-LRFD 2007ODOT Short Course
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
Design Example - Continuous Two-Span I-Girder Bridge
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
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 = − + − + − = −
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #250
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
Design Example - Continuous Two-Span I-Girder Bridge
Design of the Negative Moment Section
(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
=
= − +
=
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #251
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.
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
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 + −=
AASHTO-LRFD 2007ODOT Short Course
July 2007
We assumed 4.25” from top OKd = 58.6 in
56.48 3.914.52
x = =
Loads & Analysis: Slide #252Do Not Duplicate
127
Design Example - Continuous Two-Span I-Girder Bridge
Negative Moment Deck Reinforcement
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
AASHTO-LRFD 2007ODOT Short Course
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 = −
= − = − > = −
Loads & Analysis: Slide #253Do Not Duplicate
Design Example - Continuous Two-Span I-Girder Bridge
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.
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≤ −
AASHTO-LRFD 2007ODOT Short Course
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.
Loads & Analysis: Slide #254Do Not Duplicate
128
Design Example - Continuous Two-Span I-Girder Bridge
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 8.040.003 0.003 0.019 0.005c 8.04− − ε = = = >
AASHTO-LRFD 2007ODOT Short Course
July 2007
This is a tension controlled section, so φ 0.9
(5.7.2.1 & 5.5.4.2)
Loads & Analysis: Slide #255Do Not Duplicate
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement – Negative Moment Section
(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
AASHTO-LRFD 2007ODOT Short Course
July 2007
Sc= 16340 in
Loads & Analysis: Slide #256Do Not Duplicate
129
Design Example - Continuous Two-Span I-Girder Bridge
Minimum Reinforcement – Negative Moment Section
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= −
AASHTO-LRFD 2007ODOT Short Course
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= > =
Loads & Analysis: Slide #257Do Not Duplicate
Design Example - Continuous Two-Span I-Girder Bridge
Design of the Negative Moment Section
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.
AASHTO-LRFD 2007ODOT Short Course
July 2007Loads & Analysis: Slide #258Do Not Duplicate
130
Design Example - Continuous Two-Span I-Girder Bridge
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
AASHTO-LRFD 2007ODOT Short Course
July 2007Do Not Duplicate Loads & Analysis: Slide #259
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 2 of 35
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 3 of 35
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− −
= =
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 4 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 5 of 35
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 6 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 7 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 8 of 35
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:
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.
1.5
ESTIMATE REQUIRED
PRESTRESS
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.
1.5.1
Service Load Stresses at
Midspan
Bottom tensile stress due to applied dead and live loads using load combination Service III:
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
+ + + − = =
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 9 of 35
Remember! For Service III (which applies ONLY to tension in fully prestressed members), the LL factor is 0.8!
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.
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 10 of 35
1.5.4 Strand Pattern
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
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
2.1.1
Elastic Shortening
ppES cgp
ct
Ef f
E∆ = [LRFD 5.9.5.2.3a-1]
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 11 of 35
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:
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 12 of 35
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
Total Losses at Service Loads
Total Prestress Losses:
( )
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]
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 13 of 35
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+ −
= =
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 14 of 35
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
537k 14.61in537kipsf 1.94ksi733.5in 6511in
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
k . inkipsf . ksi. in in
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 15 of 35
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
+ −= = <
+
= − =
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 16 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 17 of 35
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
537k 14.61in537kipsf 1.94ksi733.5in 6511in
= + =
Since this is a non-composite section:
( )cr b r cpeM S f f= + [LRFD 5.7.3.3.2-1]
( )3crM 6511in 0.979ksi 1.94ksi 19000k in= + = −
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 18 of 35
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
558k 14.61in558kipsf 0.474ksi733.5in 6599in
558k 14.61in558kipsf 2.01ksi733.5in 6511in
= − = −
= + =
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:
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 19 of 35
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
= − = −
− = −=
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 20 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 21 of 35
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 22 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 23 of 35
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
As = Area of nonprestressed steel on the flexural tension side of the member = 0.0
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 24 of 35
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
11in 12inb sA 0.0316 f ' 0.0316 7ksi 0.184inf 60ksi
≥ = = [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 = =
#4 @ 12 inches is OK. Girder is OK in shear.
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]
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 25 of 35
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).
Since this is a non-composite section:
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= + +
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 26 of 35
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
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 27 of 35
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!
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 28 of 35
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
537k 14.61in537kipsf 1.94ksi733.5in 6511in
= + =
dnccre c r cpe
nc
12MM S f fS
= + −
[LRFD 5.8.3.4.3-2]
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 29 of 35
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
M 0.5N VA f A f V 0.5V cotd
+ ≥ + + − − 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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 30 of 35
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
M 0.5N VA f A f V 0.5V cotd
+ ≥ + + − − 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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 31 of 35
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
k. in . ksi k . ( k ) cot k.
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
k. in . ksi k . ( k ) cot k.
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:
As = Total area of transverse reinforcement located within the distance h/4 from the end of the beam
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 32 of 35
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.
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 33 of 35
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
= + =
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 34 of 35
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:
Box Girder Example To be Used as an Example, Only ODOT Short Course Page 35 of 35
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”
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 2 of 65
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”
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 3 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 4 of 65
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”
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 5 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 6 of 65
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.
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 7 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 8 of 65
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
= +
=
For one design lane loaded:
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 9 of 65
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 10 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 11 of 65
Service Load Stresses at
Midspan
Bottom tensile stress due to applied dead and live loads using load combination Service III:
(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
+ ++= +
= +=
Stress Limits for Concrete
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
= +
=
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 12 of 65
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”
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 13 of 65
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:
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 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
( )
( )
ps pi g m g m g gpES
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 14 of 65
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:
10 12pi pspLT h st h st pR
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 15 of 65
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
∆ = + +
∆ = + +
∆ =
Total Losses at Service Loads
Total Prestress Losses:
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.)
Stress Limits for Concrete
Compression: 0.60fci’ = 0.60(4.5) = +2.700 ksi [LRFD 5.9.4.1.1]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 16 of 65
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
Stresses at Transfer
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 17 of 65
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
Distance from bottom (in)
No. Strands
Distance from bottom (in)
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 18 of 65
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
= + −
= + − = +
Compressive stress limit for concrete: +2.700 ksi OK
Stresses at Harp Points
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 19 of 65
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
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
= + −
= + −
= + − = +
Compressive stress limit for concrete: +2.700 ksi OK
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
Concrete stress at the top fiber of the beam:
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
= + −
= + −
= + − = +
Compressive stress limit for concrete: +2.700 ksi OK
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 20 of 65
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
Stresses at Transfer
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= =
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 21 of 65
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
Stress Limits for Concrete
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
Due to one half the permanent loads and live load:
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
P P e M M M MfA S S S
f
f
+ += − + +
+ += − + +
= − + + = +
Compressive stress limit for concrete: +3.150 ksi OK
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 22 of 65
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
+= +
= +
= + = +
Compressive stress limit for concrete: +2.800 ksi OK
3. Under permanent and transient loads:
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
+= +
= +
= + = +
Compressive stress limit for concrete: +1.800 ksi OK
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 23 of 65
3. Under permanent and transient loads:
( )
(245.1 153.6 1,386.2)*1229,534
0.73
ws b LL Itc
tg
tc
tc
M M MfS
f
f
++ +=
+ +=
= +
Compressive stress limit for concrete: +2.700 ksi OK
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
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!
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
= −
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 24 of 65
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.
c = Distance between the neutral axis and the compressive face
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 25 of 65
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 26 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 27 of 65
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
= −
= − +
=
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 28 of 65
The total area of longitudinal reinforcement provided, ( ) 5.58s providedA = in2.
1.9.2.3
Negative Moment Deck Reinforcement
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 29 of 65
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=
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 30 of 65
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≤ −
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 31 of 65
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
As before, check the strain in the extreme tensile steel:
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 32 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 33 of 65
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 34 of 65
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 35 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 36 of 65
Effective Shear Depth
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- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 37 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 38 of 65
'
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
Use #4 hoops Av = 0.40 in2 α = 90 sin α =1 cot α =0
cot 0.4(60)(55.63)cot 37 6.32280
v y v
s
A f ds
Vθ
= = =
So #4 hoops at 6 in Vs = 295.0 kips
( )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:
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 39 of 65
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−
=
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) 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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 40 of 65
'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
Use #4 hoops Av = 0.40 in2 α = 90 sin α =1 cot α =0
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 = 313.0 kips
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 41 of 65
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
= =+
= − =
= =
Effective Shear Depth
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
= − = − =
≥ = =
≥ = =
Therefore, dv = 56.5 in.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 42 of 65
Calculation of Critical Section
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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 43 of 65
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−
=
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) 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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 44 of 65
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
8in 12inb sA 0.0316 f ' 0.0316 7ksi 0.134inf 60ksi
≥ = = [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+ ≤
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 45 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 46 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 47 of 65
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]
0.5 0.5 cotu u ups ps s y p s
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 48 of 65
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−
= + − =−
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 49 of 65
( )( )
( ) ( )
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
≥ − − = <
− − =
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 [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
. . cot . VV k . kφ
= = > =
Therefore, Vs = 277.8 for this calculation.
( )( )
( ) ( )
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?
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 50 of 65
( )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]
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 51 of 65
( )( )
( ) ( )
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
. . cot . VV k . kφ
= = > =
Therefore, Vs = 277.8 for this calculation.
( )( )
( ) ( )
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−= =
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 52 of 65
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:
As = Total area of transverse reinforcement located within the distance h/4 from the end of the beam
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 53 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 54 of 65
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 = +
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 55 of 65
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- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 56 of 65
Minimum Exterior DFM: (Rigid Body Rotation of Bridge Section)
,
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= = =
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 57 of 65
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 = +
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 58 of 65
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]
Future Wearing Surface
[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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 59 of 65
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
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 60 of 65
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
P P e M M M MfA S S S
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- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 61 of 65
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
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
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
Positive Moment Section
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= + + +
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 62 of 65
At point of maximum moment 0.4L:
,
,
, ,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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 63 of 65
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
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.
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 64 of 65
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 = = = >
This is a tension controlled section, so φ = 0.9
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.
2- Span Continuous Example July 2007 To Be Used as an Example Only ODOT Short Course Page 65 of 65
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