AASHTO Prestressed Beams AASHTO Examples

Author
alioeztuerk 
Category
Documents

view
86 
download
14
Embed Size (px)
Transcript of AASHTO Prestressed Beams AASHTO Examples

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 51
Prestressed IBeam Design Example
This example illustrates the design of a pretensioned IBeam for a two span bridge without skew. The 130'0" spans are supported with Mn/DOT 72" beams. Mn/DOT standard details and drawings for diaphragms (B406, B812), railings (Fig. 5397.117), and beams (Fig. 5397.517) are to be used with this example. This example contains the design of a typical interior beam at the critical sections in positive flexure, shear, and deflection. The superstructure consists of six beams spaced at 9'0" centers. A typical transverse superstructure section is provided in Figure 5.7.2.1. A framing plan is provided in Figure 5.7.2.2. The roadway section is composed of two 12' traffic lanes and two 12' shoulders. A Type F railing is provided on each side of the bridge and a 9" composite concrete deck is used. End diaphragms (B812) are used at each end of the bridge and interior diaphragms (B406) are used at the interior third points and at the pier.
Figure 5.7.2.1

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 52
Figure 5.7.2.2

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 53
A. Materials The modulus of elasticity for high strength concrete suggested by ACI Committee 363 is used for the beam concrete. The composite deck is assumed to have a unit weight of 0.150 kcf for dead load computations and 0.145 kcf for elastic modulus computations. The beam concrete is assumed to have a unit weight of 0.155 kcf for dead load computations. The material and geometric parameters used in the example are shown in Table 5.7.2.1: Table 5.7.2.1 Material Properties
Material Parameter Prestressed Beam Deck
cif at transfer 7 ksi 
cf at 28 days 8.0 ksi 4 ksi
ciE at transfer ( ) 1000cif1265 +
ksi4347= 
Concr
ete
cE at 28 days ( ) 1000cf1265 +
ksi4578= ( ) cf5.1145.0000,33
ksi3644=
yf for rebar 60 ksi 60 ksi
puf for strand 270 ksi 
sE for rebar 29,000 ksi 29,000 ksi
pE for strand 28,500 ksi 
Ste
el
Strand type 0.6 inch diameter
270 ksi, low relaxation 
The beams are designed to act compositely with the deck on simple spans. The deck consists of a 7 inch thick concrete slab with a 2 inch wearing course. For simplicity and in order to be conservative, the beams are designed assuming the full 9 inches of thickness is placed in a single pour. A 1/2 inch of wear is assumed. A thickness of 81/2 inches is used for composite section properties. The haunch or stool is assumed to have an average thickness of 21/2 inches for dead load computations and 1
B. Determine CrossSection Properties for a Typical Interior Beam
1/2 inches for section property computations.
[4.6.2.6.1] The effective flange width, , is the smallest of: eb
1) 1/4 x Effective Span Length in0.3901213041 ==
2) 12 x Slab Thickness + 1/2 Top Flange Width
in0.117230
5.812 =+=

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 54
3) Average Beam Spacing = 108.0 in GOVERNS The modular ratio of the deck concrete to the beam concrete is:
796.045783644
EE
ncbeam
cdeck ===
This results in a transformed effective flange width of:
( ) in0.860.108796.0betrans == Properties for an interior beam are given in Table 5.7.2.2. Table 5.7.2.2 CrossSection Properties
Parameter Noncomposite Section Composite Section
Height of section, h 72 in 82.0 in
Deck thickness  8.5 in
Average stool thickness  1.5 in (section properties)
2.5 in (dead load)
Effective flange width, eb  108.0 in (deck concrete)
86.0 in (beam concrete)
Area, A 786 in 2 1553 in2
Moment of inertia, I 547,920 in 4 1,235,000 in 4
Centroidal axis height, y 35.60 in 56.29 in
Bottom section modulus, bS 15,390 in3 21,940 in3
Top section modulus, tS 15,050 in3
48,040 in3 (beam concrete)
60,350 in3 (deck concrete)
Top of prestressed beam 15,050 in3 78,610 in3
Three load combinations will be considered; Strength I, Service I, and Service III. As a result of the simple span configuration, only maximum
C. Shear Forces and Bending Moments

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 55
p values need to be considered. Load effects related to settlement, thermal effects, water load, or stream pressure will not be considered. It shall be assumed that traffic can be positioned anywhere between the barriers. Number of design lanes = 4
1248
width lane designbarriers between distance ==
[3.6.2] Dynamic load allowance IM %33=
[4.6.2.2] 1. Determine Live Load Distribution Factors
Designers should note that the approximate distribution factor equations include the multiple presence factors.
[4.6.2.2.2] Distribution Factor for Moment Interior Beams LRFD Table 4.6.2.2.11 lists the common deck superstructure types for which approximate live load distribution equations have been assembled. The cross section for this design example is Type (k). To ensure that the approximate distribution equations can be used, several parameters need to be checked.
1) ft0.16ft0.9 spacing beam ft5.3 = OK 2) in0.12in5.8 thickness slab in5.3 = OK 3) ft240ft130 length span ft20 = OK 4) 6 beams ofnumber 4 = OK
The distribution factor equations use a Kg factor that is defined in LRFD Article 4.6.2.2.1.
256.136444578
E
E
deckc
beamc ===
( ) ( ) in15.4260.3575.77 centroid beam centroiddeck eg === ( )[ ] ( )[ ] 442.215.42786920,547256.1eAIK 22gg =+=+= x 106
One design lane loaded:
1.0
3s
3.04.0
tL12
KLS
14S
06.0gM
+=
1.063.04.0 10x442.2900.9 35.81301213014
06.0gM +=

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 56
473.0gM = lanes/beam Two or more design lanes loaded:
1.0
3s
g2.06.0
tL12
K
LS
5.9S
075.0gM
+=
1.062.06.0 10x442.2900.9075.0gM +=
35.8130121305.9
698.0gM = lanes/beam
[4.6.2.2.2d] Distribution Factor for Moment  Exterior Beams LRFD Table 4.6.2.2.2d1 contains the approximate distribution factor equations for exterior beams. Type (k) crosssections have a deck dimension check to ensure that the approximate equations are valid. The distance from the inside face of barrier to the centerline of the fascia beam is defined as . For the example this distance is: ed
( ) ft50.10.95.224de == The check to use the approximate equations is:
ft5.5ft50.1dft0.1 e = OK One design lane loaded:
Figure 5.7.2.3

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 57
The lever rule shall be used to determine the live load distribution factor for one lane. The fascia beam live load distribution factor is found by summing reactions about the first interior beam:
lanes5.0WW 21 ==
== 0.92.1S2.1gM2211 + + 5.25.05.85.0LWLW
733.0gM = lanes/beam Two or more design lanes loaded: The distribution factor is equal to the factor e multiplied by the interior girder distribution factor for two or more lanes
ft5.15.220.24de ==
935.01.9
77.01.9
77.0e e =+=+=
5.1d
653.0698.0935.0gegM int === lanes/beam
[4.6.2.2.2e] Skew Factor No correction is necessary for a skew angle of zero.
[4.6.2.2.3] [4.6.2.2.3a]
Distribution Factor for Shear Interior Beams LRFD Table 4.6.2.2.3a1 can be used. One design lane loaded:
720.00.250.9
36.00.25
S36.0gV =
+=
+= lanes/beam
Two or more design lanes loaded:
884.035
0.912
0.92.0
35S
12S
2.0gV22
=
+=
+= lanes/beam
[4.6.2.2.3b] Distribution Factor for Shear Exterior Beams
One Design Lane Loaded: The lever rule shall be used which results in the same factor that was computed for flexure and is equal to 0.733 lanes/beam Two or more design lanes loaded:

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 58
750.010
5.16.0
10d
6.0e e =
+=
+=
The exterior beam shear distribution factor for two or more design lanes is determined by modifying the interior distribution factor:
663.0884.0750.0gegV int === lanes/beam
[4.6.2.2.3c] Skew Factor No correction is necessary for a skew angle of zero. Distribution Factor for Deflection [2.5.2.6.2]
[Table 3.6.1.1.21] The distribution factor for checking live load deflections assumes that the entire cross section participates in resisting the live load. The minimum Multiple Presence Factor (MPF) used by Mn/DOT when checking live load deflection is 0.85. The deflection distribution factor is:
( ) ( )( ) 567.06 85.04lines beam of # MPFlanes of #gD === lanes/beam Table 5.7.2.3 contains a summary of the live load distribution factors. Table 5.7.2.3 Distribution Factor Summary (lanes per beam)
Loading Flexure Shear
One Design Lane 0.473 0.720
Two or More Design Lanes 0.698 0.884
Inte
rior
Bea
m
Deflection 0.567 
One Design Lane 0.733 0.733
Two or More Design Lanes 0.653 0.663
Ext
erio
r Bea
m
Deflection 0.567 
[1.3.3 1.3.5] 2. Load Modifiers
The following load modifiers will be used for this example:
Strength Service Fatigue
Ductility D 1.0 1.0 1.0 Redundancy R 1.0 1.0 1.0 Importance I 1.0 n/a n/a
= IRD 1.0 1.0 1.0

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 59
3. Dead and Live Load Summary
Beam Selfweight ( ) ( ) kip/ft 846.0kip/ft 155.0144/786 3 == Stool Weight ( ) ( ) ( ) kip/ft 078.0kip/ft 150.0ft208.0ft5.2 3 ==
( ) ( ) ( )Deck Weight kip/ft 013.1kip/ft 150.0ft75.0ft0.9 == 3 ( ) ( ) ( )2 kip/ft 160.06/1ft48kip/ft 020.0 == Future Wearing Surface
Barrier Weight ( ) ( ) kip/ft 146.06/1kip/ft 439.02 == Diaphragm Weight ( ) ( ) ( )[ ]0149.00103.020.9 +
( ) ( ) kip 561.0490.01212
17.42 =+5.017
The bending moments and shears for the dead and live loads were obtained with a line girder model of the bridge. They are summarized in Tables 5.7.2.4 and 5.7.2.5.
Table 5.7.2.4 Shear Force Summary (kips/beam)
Load Type/Combination Brg CL
(0.0')
Brg Face
(0.63')
Trans Point
(2.38')
Critical Shear Point (5.8')
0.1 Span Point
(13.0')
Strand Dev Point
(13.6')
0.2 Span Point
(26.0')
0.3 Span Point
(39.0')
0.4 Span Point
(52.0')
0.5 Span Point
(65.0')
Selfweight 55 54 53 50 44 44 33 22 11 0
Stool 5 5 5 5 4 4 3 2 1 0
Deck 66 65 63 60 53 52 40 26 13 0
FWS 10 10 10 9 8 8 6 4 2 0
Barrier 9 9 9 9 8 8 6 4 2 0
Diaphragms 1 1 1 1 1 1 1 1 0 0
Dea
d L
oad
s
Total 146 144 141 134 118 113 89 59 29 0
Uniform Lane 37 36 35 34 30 30 24 18 13 9
Truck with DLA 79 78 77 75 70 70 62 53 45 36
Live
Load
s
Total 116 114 112 109 100 100 86 71 58 45
Strength I Load Comb
)LL75.1DL25.1( + 386 380 372 358 323 316 262 198 138 79
Service I Load Comb
)LL00.1DL00.1( + 262 258 253 243 218 213 175 130 87 45
Service III Load Comb
)LL80.0DL00.1( + 239 235 231 221 198 193 158 116 75 36

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 510
Table 5.7.2.5 Bending Moment Summary (kipft/beam)
Load Type/Combination Brg CL
(0.0')
Brg Face
(0.63')
Trans Point
(2.38')
Critical Shear Point (5.8')
0.1 Span Point
(13.0')
Strand Dev Point
(13.6')
0.2 Span Point
(26.0')
0.3 Span Point
(39.0')
0.4 Span Point* (52.0')
0.5 Span Point
(65.0')
Selfweight 0 34 128 305 643 670 1144 1501 1716 1787
Stool 0 3 12 28 59 62 105 138 158 165
Deck 0 41 154 365 770 802 1370 1798 2054 2140
Diaphragms 0 0 1 3 7 8 15 22 24 24
DC1
Total DC1 0 78 297 701 1479 1542 2634 3459 3952 4116
Barrier 0 6 22 53 111 116 197 259 296 308
FWS 0 7 24 58 122 127 216 284 324 338
DC2
Total DC2 0 13 46 111 233 243 413 543 620 646
Dea
d L
oad
s
Total (DC1+DC2) 0 91 343 812 1712 1785 3047 4002 4572 4762
Uniform Lane 0 18 68 161 340 354 604 793 906 944
Truck with DLA 0 39 145 343 719 749 1265 1638 1857 1912
Live
Load
s
Total 0 57 213 504 1059 1103 1869 2431 2763 2856
Strength I  Load Comb
)LL75.1DL25.1( + 0 214 802 1897 3993 4162 7080 9257 10,550 10,951
Service I  Load Comb
)LL00.1DL00.1( + 0 148 556 1316 2771 2888 4916 6433 7335 7618
Service III Load Comb
)LL80.0DL00.1( + 0 137 513 1215 2559 2667 4542 5947 6782 7047
* Drape point for strands.
Typically the tension at the bottom of the beam at midspan dictates the required level of prestressing.
D. Design Prestressing
1. Estimate Required Prestress Service III load combination shall be used Bottom of beam stress:
+
+
=
cb
LL
cb
2DC
gb
1DC
S8.0M
SM
SM
ksi81.4940,21940,21390,15
=++=8.012285612646124116

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 511
As a starting point, the total prestress losses will be assumed to be 30%. This results in an effective prestress of
( ) ksi8.14170.027075.030.01f75.0f pupe === Strands are typically placed on a 2" grid. The bottom flange of a 72" beam can hold a maximum of 48 strands. The centroid of a 48 strand pattern would be
( ) ( )( )
=strands of # total
strands of strands of #ystr
( ) ( ) ( )in79.5
151311975328464210 = ++++++++++=48
Using the centroid of this group as an estimate of the strand pattern eccentricity results in
in81.2979.560.3579.5ye g48 === The area of a 0.6" diameter 7wire strand is 2in217.0 The axial compression produced by the prestressing strands is
( ) ( ) ( )8.141217.0strands of #fAP pes == The internal moment produced by the prestressing strands is
( ) 81.298.141217.0strands of #efAM 48pess/p == The allowable tension after losses ksi54.0819.0f19.0 c === This moment and the axial compression from the prestress must reduce the bottom flange tension from 4.81 ksi tension to a tension of 0.54 ksi or
Required ksi27.454.081.4fpe ==
Using the fact that SA
fpe += MP
One can estimate the required number of strands:

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 512
( ) ( ) strands2.438.141217.08.141217.0 =+ =+ 390,1581.29
7861
27.4
S81.29
A1
27.4
gbg
+
+
A strand pattern with 44 strands should be tried. After reviewing Bridge Details Part II Figure 5397.517, a 44 strand draped strand pattern was selected. Also, the drape points were chosen to be at 0.40L = 52.0 ft from the centerline of bearing locations. The trial strand pattern is shown in Figure 5.7.2.4. The properties of this strand pattern at midspan are:
( ) ( )in27.5
44131198753264210
ystrand =
+++++++++=
in33.3027.560.35yye strandbstrand === Section Modulus at the strand pattern centroid is
3
strand
ggps in065,1833.30
920,547e
IS ===
Figure 5.7.2.4

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 513
[5.9.5] 2. Prestress Losses Prestress losses are computed using the refined method.
[5.9.5.4.4b] Initial Relaxation Loss It shall be assumed that the prestress is transferred 18 hours after stressing
days75.02418
t ==
ksi50.20227075.0f75.0f pupj === ( )
pjp
pj1pR f55.0f
f 40
t24logf
=
( )ksi80.150.20255.0
50.20275.024logf = =
2709.0401pR
[5.9.5.2.3] Elastic Shortening Loss
The alternative equation presented in the LRFD C5.9.5.2.3a shall be used. )(
( )p
ciggg
2mgps
ggmgmgpbtpspES
E
EIAAeIA
AMeAeIfAf
+++=
2
( ) ( ) 2ps in55.9217.044area strandstrands of #A ===ksi70.20080.150.202fff ===
1pRpjpbt
in33.30ee strandm ==
( ) ( ) 6p
cigg in764,687,65500,28E
== 4347920,547786EIA
( ) [ ( ) ( )] 62g2mgps in748,137,1278633.30920,54755.9AeIA =+=+ ( ) ( ) ( ) ( )
ksi73.24764,687,65748,137,12
fpES =+=78612178733.30748,137,1270.200
[5.9.5.4.2] Shrinkage Loss
Use an average humidity for North Dakota of 70%.
( ) ( ) ksi05.673150.00.17H150.00.17fpSR ===
[5.9.5.4.3] Creep Loss Noncomposite dead load moment excluding selfweight
( ) 232917874116M SW1DC == kipft

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 514
Composite dead load moment, kipft 646M 2DC =
++
=
cg
strandgcg2DC
g
strandSW1DCcdp I
eyyM
Ie
Mf
000,235,1920,547 ++ = 33.3060.3529.5612.64633.3012.2329
ksi87.1=
gps
sw
gps
strandi
g
icgp SSA
f += MePP
( ) ( ) ( ) kips168155.973.2480.150.202AfffP pspES1pRpji === ( ) ( )
ksi77.318065065,18786
fcgp =+= 12178733.3016811681
( ) ( ) ksi15.3287.1777.312f7f12f cdpcgppCR ===
[5.9.5.4.4c] Relaxation Loss After Transfer
( )[ ]pCRpSRpES2pR ff2.0f4.02030.0f += ( )[ ] ksi74.015.3205.62.073.244.02030.0 =+=
[5.9.5.1] Total Losses
ksi67.6374.015.3205.673.24ffffTL 2pRpCRpSRpES =+++=+++= ksi83.13867.6350.202TLff pjpe ===
prestress loss percentage %4.3110050.202
100fpj
=== 67.63TL
[5.9.4.1] 3. Stresses at Transfer (compression +, tension )
Stress Limits for P/S Concrete at Release Compression in the concrete is limited to:
ksi20.40.760.0f60.0 ci == Tension in the concrete is limited to: The minimum of ksi25.00.70948.0f0948.0 ci ==
or ksi20.0 Tension limit = 0.20 ksi

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 515
Check Release Stresses at Drape Point (0.40 Point of Span)
985,5033.301681eP strandi == kipin
Top stress due to P/S =
=050,15786SA gt
strandi
g
i 985,501681ePP
ksi25.1=
Bottom Stress due to P/S +=
+=390,15786SA gb
strandi
g
i 985,501681ePP
ksi45.5= Selfweight moment at drape point 1716M 40.0sw == kipft
Top stress due to selfweight ksi37.1050,15Sgt
40.0sw ==
= 121716M
Bottom stress due to selfweight ksi34.1121716M 40.0sw = =
=
390,15Sgb ksi20.0ksi12.037.125 kipft OK
[5.7.3.3.1] 6. Limits of Reinforcement Maximum Reinforcement The depth of the flexural compressive block is compared to the depth of the steel centroid to verify adequate ductility.
42.010.073.7602.8
dc
p

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 519
[5.7.3.3.2] Minimum Reinforcement
ksi68.00.824.0f24.0f cr ===
gb
strande
g
epeb SAf += ePP
ksi30.433.3013261326 =+=
( )390,15786
Sgb1DCcgbpebrcr += 1SMSffM cgb
( ) ( )
240,881940,21
124116940,2130.468.0 = += 390,15 7353=
kipin
kipft
294,15882473532.1M2.1 cr

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 520
With this approximation to the strand centroid, can be computed: pd
( ) in57.6343.185.85.172yhd [email protected] =++== From the flexural strength computations, a = 6.82 in
in16.60282.6
57.632a
dd pv === But the effective shear depth need not be less than vd
in04.59h72.0dv = or
pev d9.0d9.0d = ( ) in21.5757.639.0 == Therefore take in16.60dv = Based on the assumed , the critical section location is:
( ) ( ) (26cot16.605.05.7cotd5.05.7d vcritv )+=+= ft8.5in17.69 == GOVERNS
or in63.6716.605.7d5.7d vcritv =+=+=
Determine Shear Stress From Table 5.7.2.4 the Strength I design shear at 5.8 ft is
kips358Vu = The amount of force carried by the draped strands at their effective prestress level is:
kips5.36183.138217.012P d12 == The inclination of the draped strands is:
( )reesdeg887.4
63.5212/862
arctan =
=
The vertical prestress component is:
( ) ( ) kips8.30887.4sin5.361sinPV d12p === The maximum shear capacity of the section
kips7.7528.300.616.600.825.0Vbdf25.0V pvvcn =+=+= The maximum design shear the section can have is:
kips358kips4.6777.75290.0Vnv >>==

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 521
The shear stress on the section is:
ksi017.116.60690.0
8.3090.0358db
VVv
vvv
pvu =
=
=
The ratio of the shear stress to the compressive strength is:
127.00.8
017.1fv
c==
Determine Longitudinal Strain x It shall be assumed that minimum transverse reinforcement will be provided in the cross section.
psA shall first be determined. Note that computed here is different than the Aps computed earlier. This Aps includes only the area of prestressing steel found on the flexural tension side of the member. Near the end of the beam, Aps must also be reduced for development
psA
Development length is: dl
bpepsd df32
fK
=l
( ) ( )2 in8.1626.083.1383
10.2626.1 = =
Transfer length is: trl
( ) in0.366.060d60 btr ===l At the critical section in14.69dcritv = from the beam end, the strand development fraction
+=pu
pe
trd
trcritv
pu
pedev f
f1
df
fF ll
l
641.0270
10.368.162270
= +=83.1380.3614.6983.138

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 522
The flexural tension side of the member is defined as:
in0.412
0.822
hcomp ==
At none of the draped strands fall on the flexural tension side.critvdTherefore, ( ) ( ) ( )devps Farea strandstr. straight #A =
( ) ( ) ( ) 2in451.4641.0217.032 ==
[5.8.3.4.3] Equation 5.8.3.4.21 shall be used to compute the strain:
( ) ( )( )
= psp
popspuv
x AE2d
+u fAcotVV5.0M
( ) ( ) ( )( )
+
=451.4500,282
27070.0451.426cot8.303585.016.6012.1897
000502.0707,253
4.127 =
=
Because the value is negative, equation three should be investigated with the additional concrete term: From Figure 5.4.6.1, A 2c in431=
( ) ( )( )
+= psp
popspuv
x AEAcEc2d
+u fAcotVV5.0M
( ) ( ) ( )( )
++
=451.4500,2843145782
27070.0451.426cot8.303585.016.6012.1897
000030.0943,199,44.127 =
=
With the strain and shear stress to cf ratio determined, interpolate to find and in LRFD Table 5.8.3.4.21. = 23.2 degrees = 2.90

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 523
Since the original assumption for (26 degrees) does not match the computed angle, do an iteration assuming 2.23= degrees. Because new angle is flatter than original assumption, do not revise location of critical section for shear (conservative). For degrees, 2.23= 000019.0x =
Then with 126.0fv
c= and 000019.0x = , interpolate to get:
4.23= degrees (close enough to assumed angle) 89.2=
The concrete contribution:
kips2.9316.6068896.20316.0dbf0316.0V vvcc === The required steel contribution is:
kips8.2738.302.9390.0
358VV
VVVVV pc
v
upcns ====
The required spacing of double leg #13 stirrups:
( ) ( )in2.12
8.2734.23cot16.606020.02
V
cotdfAs
s
vyv ===
Double leg stirrups at a 12 inch spacing at the end of the beam shall be provided.
ft/in040A 2v = kips9.277Vs = Other sections are investigated similarly.
[5.8.4] 2. Interface Shear Transfer Top flange width in30bv = The Strength I vertical shear at the critical shear section due to all superimposed loads is:
( ) ( ) 3.21310975.19925.1Vu =++= Interface shear force is:
55.316.603.213
2a
d
VdV
V
p
u
e
uh ==
== kip/in

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 524
Required nominal interface design shear is:
94.390.055.3V
Vv
hn === kip/in
The interface area per 1 inch length of beam is:
2cv in0.30130A == /in
The upper limits on nominal interface shear are:
kip/in94.3kip/in0.240.3042.0Af2.0 cvc >== OK and
kip/in94.3kip/in0.240.308.0A8.0 cv >== OK The amount of interface shear carried by cohesion shall be determined. A note on the Bridge Details II Fig. 5397.517 requires the top flanges of the beam to be roughened. Consequently use a cohesion (c) of 0.100 ksi and a friction factor ( ) of 1.0. The nominal interface shear resistance is:
( )cvfcvn PfAAcV ++= kip0.0Pc =
The required interface shear steel is:
( ) 2y
cvvf in016.0600.1
0.301.094.3fAcV
A ==
= /in /ft 2in19.0=
[Eng. 5.8.4.14] The minimum shear steel that needs to be provided is:
2
y
vminvf in025.060
3005.0f
b05.0A === /in /ft 2in30.0=
The minimum requirement controls. Vertical shear reinforcement /ft /ft at the critical section for shear. Therefore, no additional reinforcement is required for interface shear.
2v in40.0A = 2in30.0>
Other sections are investigated similarly.

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 525
[5.8.3.5] 3. Minimum Longitudinal Reinforcement Requirement A diagonal section shall be checked with a crack starting at the inside edge of the bearing sole plate. Straight strands cross the crack at:
sole plate ( ) ( ) in82.244.23cot25.415coty32 =+=+ The transfer length for 0.6" strands is:
in366.060d60 s == To find the tensile capacity of the straight strands at the crack interpolate:
kips6.66436
82.24217.03283.138
lengthtransfer crack to length
AfT psper === The force to carry is:
( )
= cotVV5.0V
T psv
u
( )4.23cot8.309.2775.0358 =90.0
kips6.664kips9.526

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 526
This steel shall be located at the end of the beam within a distance of:
in0.18472
4h ==
The number of #16 double legged stirrups necessary to provide this area is:
4.531.02
36.3A2
A
b
s ==
The first set of stirrups is located 2 inches from the end of the beam. Provide six sets of #16 stirrups spaced at 3 inch centers. ( )in18in17352

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 527
Downward deflection due to selfweight
( )in28.2
920,547434738412
IE384Lw5 4
sw ===
12130846.0
5 4
Camber at release in56.328.284.5swpsrel === To estimate camber at the time of erection the deflection components are multiplied by standard PCI handbook multipliers. (See Figure 4.6.3 of the 3rd Edition.) They are:
Release to Erection Multipliers: Prestress = 1.80 Selfweight = 1.85
Camber and selfweight deflection values at erection are:
Prestress: in51.1084.580.1 = Selfweight: ( ) in22.428.285.1 = Diaphragm DL: in03.0 Deck and stool DL: in84.2 Parapet: in17.0
The values to be placed in the camber diagram on the beam plan sheet are arrived at by combining the values above. Initial Total Camber
in26.603.022.451.10 = say 61/4 in Est. Dead Load Deflection
in01.317.084.2 =+ say 3 in Est. Residual Camber
61/4 3 = 31/4 in
Live Load Deflection The deflection of the bridge is checked when subjected to live load and compared against the limiting values of L/800 for vehicle only bridges and L/1000 for bridges with bicycle or pedestrian traffic. Deflection due to lane load is:
( )in73.0
000,235,14578384
1213012
5
IE384Lw5 4
lane =
=
=
64.0 4

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 528
Deflection due to a truck with dynamic load allowance is found using hand computations or computer tools to be:
in46.1truck = Two deflections are computed and compared to the limiting values; that of the truck alone and that of the lane load plus 25% of the truck. Both deflections need to be adjusted with the distribution factor for deflection.
in83.046.1567.0DF truck1 === ( ) ( ) in62.046.125.073.0567.025.0DF trucklane2 =+=+=
There is no bicycle or pedestrian traffic on the bridge, the deflection limit is:
21 orthanin95.180012130
800L >>== OK
H. Detailing Items Approximate weight of each beam is:
tons5.55kips2ton1
ft
kips155.0
/ftin144
in786ft25.131yAL
322
2==
Initial prestress force at jacking is:
kips193327075.0217.044 = Figure 5.7.2.5 shows the detailed beam sheet for the bridge.

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 529
Figure 5.7.2.5