-
DESIGN EXAMPLE 1 This example illustrates the design of an
interior and exterior beam of a precast prestressed concrete beam
bridge using fully prestressed beams with harped bonded strands in
accordance with the AASHTO LRFD Bridge Design Specifications, Third
Edition, Customary US Units and through the 2005 Interims. The
bridge consists of a 120-foot simple span. The bridge profile is
shown in Figure 1 and the typical section is shown in Figure 2. The
concrete deck is 9 inches thick and the abutments are not
skewed.
120'-0"
Figure 1 Profile
Centerline bridge1'-0" 18'-0" 18'-0" 1'-0"
AASHTO Type VI Beam
4'-9" 9'-6" 4'-9" 4'-9" 9'-6" 4'-9"
Near AbutmentNear Midspan
10" thick concrete diaphragm
Figure 2 Typical Section
DESIGN PARAMERS Corrosion exposure condition for the stress
limit for tension in the beam concrete: severe Provisions for a
future wearing surface: 0.025 k/ft2Rail dead load per each rail:
0.278 k/foot Diaphragm dead load per each 10-inch thick diaphragm:
5.590 k Deck concrete 28-day strength: 5 ksi fc
' = 5 ksi ( )( )E w fc c= = =33000 33000 0145 5 40741 5 1 5. '
.. ksi (5.4.2.4-1) Beam concrete 28-day strength: 8 ksi fc
' = 8 ksi ( )( )Ec = =33000 0145 8 51541 5. . ksi
Example 1, July 2005 DRAFT 1
-
Beam concrete strength at release: 7 ksi fci
' = 7 ksi ( )( )Eci = =33000 0145 7 48211 5. . ksi
Non-prestressed reinforcement: Grade 60 fy = 60 ksi (5.4.3.2) E s =
29 000, ksi
sii
Prestressing steel: 0.6-inch diameter, 270-ksi low relaxation
strand fpu = 270 ksi (Table 5.4.4.1-1) ( )( )f kpy = = =0 90 0 90
270 243. .fpu (5.4.4.2) Eps = 28 500, ks SECTION PROPERTIES The
non-composite and composite section properties are summarized in
Table 1. Although the haunch between the top of the girder and the
bottom of the deck slab is not included in the composite section
properties, it is included in the dead loads. In order to calculate
the composite section properties, first calculate the effective
flange width. Effective Flange Width (4.6.2.6.1) Calculate the
effective flange width for the interior beam first. For the
interior beam, the effective flange width may be taken as the least
of: a) One-quarter of the effective span length: 120 feet for
simple spans (0.25)(120)(12) = 360 in b) Twelve time the average
thickness of the slab, 9 inches, plus the greater of: The web
thickness: 8 inches One-half of the top flange of the girder: 42
inches
(0.5)(42) = 21 in The greater of these two values is 21 inches
and: (12)(9) + 21 = 129 in c) The average spacing of adjacent
beams: 9.5 feet (9.5)(12) = 114 in The least of these is 114 inches
and therefore, the effective flange width is 114 inches. For the
exterior beam, the effective flange width may be taken as one-half
the effective flange width of the adjacent interior beam, 114
inches, plus the least of: a) One-eight of the effective span
length: 120 feet (0.125)(120)(12) = 180 in b) Six times the average
thickness of the slab, 9 inches, plus the greater of: One-half of
the web thickness: 8 inches
(0.5)(8) = 4 in One-quarter of the top flange of the girder: 42
inches
(0.25)(42) = 10.5 in The greater of these two values is 10.5
inches and: (6)(9) + 10.5 = 64.5 in
Example 1, July 2005 DRAFT 2
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c) The width of the overhang: 4.75 feet (4.75)(12) = 57 in The
least of these is 57 inches and the effective flange width is:
(0.5)(114) + 57 = 114 in Composite Section Properties A = area of
non-composite beam or deck (in2) d = distance between the centers
of gravity of the beam or deck and the composite section
(in) Io = moment of inertia of non-composite beam or deck (in4)
Icomp = moment of inertia of composite section (in4) Sb = section
modulus of non-composite section, extreme bottom beam fiber (in3)
Sbc = section modulus of composite section, extreme bottom beam
fiber (in3) Sslab top = section modulus of composite section,
extreme top deck slab fiber (in3) St = section modulus of
non-composite section, extreme top beam fiber (in3) Stc = section
modulus of composite section, extreme top beam fiber (in3) yb =
distance from the center of gravity of the non-composite section to
the bottom of the
beam (in) ybc = distance from the center of gravity of the
composite section to the bottom of the beam
(in) yslab top = distance from the center of gravity of the
composite section to the top of the deck slab
(in) yt = distance from the center of gravity of the
non-composite section to the top of the beam
(in) ytc = distance from the center of gravity of the composite
section to the top of the beam (in) w = weight of the non-composite
beam (k/ft) For the composite section, the modular ratio, n, to
account for different concrete strengths in the beam and deck slab
is:
nEE
deck
beam= = =4074
51540 7906.
The transformed deck area is: ( )( )( )A = =114 0 7906 9 81112.
. 2 in
The moment of inertia of the transformed deck area is: ( )( )(
)
Io = =114 0 7906 912 54753.
in3
A yb Ayb d Ad2 Io Io + Ad2
Deck 811.12 76.50 62051.00 22.96 427499 5475 432974Beam 1085.00
36.38 39472.30 17.16 319590 733320 1052910Total 1896.12 101523.30
1485884
yAyAbc
b= = =10152330189612
5354.
.. in S
Iybc
c
bc= = =1485884
535427751
. in3
Example 1, July 2005 DRAFT 3
-
y h ytc beam bc= = =72 5354 18 46. . in S Iytcc
tc= = =1485884
18 4680503
. in3
y y tslab top tc slab= + = + =18 46 9 27 46. . in ( )( )SI
y nslab topc
slab top= = =1485884
27 46 0 790668443
. . in3
(114)(0.7906) = 90.12"
C. G. Composite Section
53.5
4"
C. G. Slab
72"
9 "
76.5
0"
22.9
6"
C. G. Beam
36.3
8"17
.16"
18.4
6"
Figure 3 Composite Section, Interior and Exterior Beams
Table 1 Section Properties Non-composite Section Composite
Section
Property Type VI Beams Property Interior Beam Exterior Beam A
(in2) 1085 Icomp (in4) 1485884 1485884 I (in4) 733320 ybc (in)
53.54 53.54 yb (in) 36.38 ytc (in) 18.46 18.46 yt (in) 35.62 yslab
top (in) 27.46 27.46
Sb (in3) 20157 Sbc (in3) 27751 27751 St (in3) 20587 Stc (in3)
80503 80503 w (k/ft) 1.130 Sslab top (in3) 68443 68443
DEAD LOADS The rail and future wearing surface allowance loads
are distributed equally to all beams. Interior Beam The dead loads,
DC, acting on the non-composite section are:
Beam: 1.130 k/foot Slab: (9.5)(0.75)(0.150) = 1.069 k/ft Haunch:
0.025 k/foot Diaphragms: 5.590 k at midspan
Example 1, July 2005 DRAFT 4
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The dead load, DC, acting on the composite section is: Rail: (
)( )0 278 2
40139
..= k / ft
The dead load, DW, acting on the composite section is: Future
wearing surface allowance (FWS):
( )( )0 025 36
40 225
..= k / ft
Exterior Beam The dead loads, DC, acting on the non-composite
section are:
Beam: 1.130 k/foot Slab: (9.5)(0.75)(0.150) = 1.069 k/ft Haunch:
0.025 k/foot Diaphragms: 2.795 k at midspan
The dead load, DC, acting on the composite section is: Rail:
0.139 k/foot
The dead load, DW, acting on the composite section is: FWS:
0.225 k/foot
DISTRIBUTION OF LIVE LOAD Use the approximate formulas found in
Article 4.6.2.2 for cross section k, a concrete slab on concrete
beams. Kg = longitudinal stiffness parameter (in4) L = span length
(ft) Nb = number of girders S = girder spacing (ft) ts = deck slab
thickness (in) Interior Beam LONGITUDINAL STIFFNESS PARAMETER
(4.6.2.2.1) eg = the distance between the centers of gravity of the
basic beam and the deck (in)
e yt
g ts= + = + =
23562
92
4012. . in
The modular ratio for calculating the longitudinal stiffness
parameter is:
nEE
B
D= = =5154
407412651.
The longitudinal stiffness parameter is: ( ) ( ) ( )( )[ ]K n I
Aeg g= + = + =2 212651 733320 1085 4012 3137123. . 4 in
Example 1, July 2005 DRAFT 5
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DISTRIBUTION OF LIVE LOAD FOR MOMENT (4.6.2.2.2b) Check the
range of applicability. 35 16 0. . S S = 9.5 feet O.K. 4 5 12 0 9.
. =ts t inches O.K.s 20 240 L L = 120 feet O.K. 10 000 7 000 000
3137 123, , , , , =Kg K in O.K.g 4 Nb =4 4 N O.K.b For one design
lane loaded:
( )( )( )gS S
LK
Ltg
s= +
= +
=0 06 14 12 0 0 06
9 514
9 5120
313712312 120 9
05060 4 0 3
3
0 1 0 4 0 3
3
0 1
..
.. .
.. . . . . .
For the fatigue limit state, remove the multiple presence
factor.
g = =050612
0 422..
.
For two or more design lanes loaded:
( )( )( )gS S
LK
Ltg
s= +
= +
=0 075 9 5 12 0 0 075
9 59 5
9 5120
313712312 120 9
0 7410 6 0 2
3
0 1 0 6 0 2
3
0 1
.. .
...
..
. . . . . .
DISTRIBUTION OF LIVE LOAD FOR SHEAR (4.6.2.2.3a) Check the range
of applicability. 35 16 0. . S S = 9.5 feet O.K. 4 5 12 0 9. . =ts
t inches O.K.s 20 240 L L = 120 feet O.K. 10 000 7 000 000 3137
123, , , , , =Kg K in O.K.g 4 Nb =4 4 N O.K.b For one design lane
loaded:
gS= + = + =0 36
2500 36
9 5250
0 740..
...
.
For the fatigue limit state, remove the multiple presence
factor:
g = =0 74012
0 617..
.
For two or more design lanes loaded:
gS S= +
= +
=0 2 12 35 0 2
9 512
9 535
0 9182 0 2 0
. .. .
.. .
Exterior Beam DISTRIBUTION OF LIVE LOAD FOR MOMENT (4.6.2.2.2d)
For one design lane loaded, use the lever rule and apply the
multiple presence factor, m, because the truck is manually
positioned on the bridge cross section. When manually positioning
the trucks, the first design lane is placed immediately adjacent to
the face of the traffic rail and subsequent lanes are placed
immediately adjacent to the previous. The closest truck wheel is
placed no closer than two feet from the edge of its design lane.
When using the lever rule,
Example 1, July 2005 DRAFT 6
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assume hinges at all interior beams and solve for the reaction
at the exterior beams. This reaction, when expressed in terms of
lanes, is a lane fraction for the exterior beams.
2'6' 1'
9.5' 4.75'
5.25'assumed hinge
12' design lanecenterline of bridge
P1 P1
Figure 4 Lever Rule
Solving for the reaction, R, at the exterior beam:
R = + =525 11259 5
1737. .
.. wheels = 0.868 lanes
g = mR = (1.2)(0.868) = 1.042 For the fatigue limit state, g =
0.868 For two or more design lanes loaded, the lane fraction is
found by applying a multiplier, e, to the interior beam lane
fraction for two or more design lanes loaded. This multiplier is a
function of the distance from the exterior web of the exterior beam
to the interior edge of the curb or traffic barrier, de. de = 4.75
1.00 0.3333 = 3.4167 ft Check the range of applicability. =10 55
34167. . .de d feet O.K.e e
de= + = + =0 7791
0 7734167
9111455.
..
..
.
g = e ginterior = (1.1455)(0.741) = 0.848 DISTRIBUTION OF LIVE
LOAD FOR SHEAR (4.6.2.2.3b) For one design lane loaded, use the
lever rule. However, this is the same lane fraction for moment,
1.042, and for the fatigue limit state the lane fraction is 0.868.
For two or more design lanes loaded, apply a multiplier, e, to the
lane fraction for two or more design lanes loaded. Check the range
of applicability. =10 55 34167. . .de d feet O.K.e e
de= + = + =0 610
0 634167
100 9417. .
..
g = e ginterior = (0.9417)(0.918) = 0.864
Example 1, July 2005 DRAFT 7
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DISTRIBUTION OF LIVE LOAD FOR MOMENT AND SHEAR Apply the
additional investigation provision, which is for moment
(4.6.2.2.2d) and shear (4.6.2.2.3b). Once again, when manually
positioning the trucks, place the design lanes in the same manner
as for the lever rule and apply the multiple presence factor, m.
Use the suggested commentary equation, C4.6.2.2.2d-1.
2' 6' 1'
center of gravity design truck 1
4.75'4.75'
14.25' 4.75'
center of gravity girder pattern
12' design lane 1
P1 P1
6'13'
Figure 5 Rigid Body Rule, One Design Lane Loaded
2' 6' 6' 1'
4.75'4.75'
14.25' 4.75'
center of gravity design truck 2
12' design lane 112' design lane 2
P1P2 P1P2 2'
6'13'
5'1'
center of gravity girder pattern center of gravity design truck
1
Figure 6 Rigid Body Rule, Two Design Lanes Loaded
Example 1, July 2005 DRAFT 8
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For one design lane loaded: ( )( )
( ) ( ) ( )[ ]R NN x exLB ext= + = + + = + = 2 2 214 14 25 132 4
75 14 25 14 1852545125 0 661.. . .. . g = mR = (1.2)(0.661) = 0.793
For the fatigue limit state, g = 0.661 For two design lanes
loaded:
( )( )R = + + =2
414 25 13 1
451250 942
..
.
g = mR = (1.0)(0.942) = 0.942 ADJUSTMENT IN LANE FRACTIONS FOR
SKEWED BRIDGES Since the skew angle is zero degrees, there is no
adjustment for skew. SUMMARY OF LANE FRACTIONS The calculated live
load lane fractions are summarized in Table 2. The controlling lane
fractions are the largest value for moment and shear for the
interior and the exterior beams. The controlling lane fractions are
then applied to the appropriate live load envelope for one lane of
live load to calculate the design live load moments and shears.
Table 2 Lane Fractions Moment Shear Interior Beam
One lane loaded 0.506 0.740 Two lanes loaded 0.741 0.918 One
lane loaded - fatigue 0.422 0.617
Exterior Beam One lane loaded 1.042 1.042 Two lanes loaded 0.848
0.864 One lane loaded - fatigue 0.868 0.868 Additional
investigation One lane loaded 0.793 0.793 Two lanes loaded 0.942
0.942 One lane loaded - fatigue 0.661 0.661
Table 3 Controlling Lane Fractions
Moment Shear Interior Beam
Service, Strength Limit States 0.741 0.918 Fatigue Limit State
0.422 0.617
Exterior Beam Service, Strength Limit States 1.042 1.042 Fatigue
Limit State 0.868 0.868
Example 1, July 2005 DRAFT 9
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LIMIT STATES Limit states are groups of events or circumstances
that cause structures to be unserviceable. The four limit states
are Service, Fatigue, Strength, and Extreme Event. The service
limit state (5.5.2) places restrictions on stresses, deformations,
and crack width under service conditions. The fatigue and fracture
limit state (5.5.3) places restrictions on stress range in
reinforcement and prestressing tendons under service conditions.
The intent is to limit crack growth under repetitive loads and
prevent fracture. The strength limit state (5.5.4) is intended to
ensure strength and stability and to resist statistically
significant load combinations. The extreme event limit state
(5.5.5) is intended to ensure structural survival during extreme
events as appropriate to the site and use. Includes earthquake, ice
load, collision by vessels and vehicles, and certain hydraulic
events (major floods). The most common limit states for prestressed
concrete beam design are: Strength I - Basic load combination
relating to the normal vehicular use of the bridge without
wind.
1.25 DC + 1.5 DW + 1.75 (LL + IM) Strength II - Load combination
relating to the use of the bridge by Owner-specified special design
vehicles, evaluation permit vehicles, or both without wind.
1.25 DC + 1.5 DW + 1.35 (LL + IM) Service I - Load combination
relating to the normal operational use of the bridge with a 55 MPH
wind and all loads taken at their nominal values. Also to control
crack width in reinforced concrete structures.
1.0 DC + 1.00 DW + 1.00 (LL + IM) Service III - Load combination
relating only to tension in prestressed concrete superstructures
with the object of crack control.
1.0 DC + 1.00 DW + 0.80 (LL + IM) Fatigue - Fatigue and fracture
load combination relating to repetitive gravitational vehicular
live load and dynamic response under a single design truck. 0.75
(LL + IM) BEAM STRESSES In order to determine the number of
required strands, first calculate the maximum tensile stress in the
beam for Service III Limit State. The number of required strands is
usually controlled by the maximum tensile stress. Provide enough
effective prestress force so that the tensile stresses in the beam
meet the tensile stress limit. For simple spans beams, the maximum
tensile force is at midspan at the extreme bottom beam fibers. The
tension service stress at the bottom beam fibers, can be calculated
using:
fM M
SM M M
Sbottombeam slab
b
rail FWS LL
bc= + + + 08. (Service III Limit State)
Example 1, July 2005 DRAFT 10
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Table 4 Beam Moments at Midspan (k-ft)
Dead Load Live Load plus Dynamic Load allowance Non-composite
Composite Composite Beam
Beam Slab/Diaph Rail FWS HL-93 Interior 2034 2137 250 405 2728
Exterior 2034 2053 250 405 3837
Interior Beam Stresses due to dead load and live load.
( )( ) ( )( )[ ]( )fbottom = + + + = 2034 2137 1220157
250 405 08 2728 1227751
3710.
. ksi (t)
Exterior Beam Stresses due to dead load and live load.
( )( ) ( )( )[ ]( )fbottom = + + + = 2034 2053 1220157
250 405 08 3837 1227751
4 044.
. ksi (t)
PRELIMINARY STRAND ARRANGEMENT The development of a strand
pattern is a cyclic process. Two design parameters need to be
initially estimated: the total prestress losses and the
eccentricity of the strand pattern at midspan. The total required
prestress force can be calculated using:
Pf f
Ae
S
eten bottom
bnc
=
1
ften = the tensile stress limit (ksi) fpT = the estimated total
loss in the prestressing steel stress (ksi) fpj = the stress in the
prestressing steel at jacking (ksi) fpe = the effective stress in
the prestressing steel, after all losses (ksi) Pe = the effective
prestress force, after all losses (k) e = the eccentricity of the
strand pattern (in) The concrete stress limit for tension, all
loads applied, after all losses, and subjected to severe corrosion
conditions, Table 5.9.4.2.2-1, is: f ften c= = =0 0948 0 0948 8 0
268. . .' ksi fpT = 40 ksi (estimated) fpj = (0.75)(270) = 202.50
ksi fpe = 202.50 40 = 162.50 ksi Pe = (fpe)(Aps) = (162.50)(0.217)
= 35.3 k (for one strand) e = -32 inches (estimated)
Example 1, July 2005 DRAFT 11
-
Interior Beam ( ) ( )
( )Pe =
=0 268 37101
108532
20157
13718. .
. k
The number of strands required is: 13718
35338 9
..
.= Exterior Beam
( ) ( )( )Pe =
=0 268 4 0441
108532
20157
1504 9. .
. k
The number of strands required is: 1504 9
35342 6
..
.= Try 42 strands. The preliminary strand pattern is shown in
Figure 8 and the preliminary strand profile including the
eccentricities at the span tenth points is shown in Figure 9. At
the midspan of the beam the distance from the bottom of the beam to
the center of gravity of the prestressing strands is:
( )( ) ( )( ) ( )( ) ( )( )y = + + + =13 2 13 4 13 6 3 8
424 29. in
The eccentricity of the prestressing strands at the midspan is:
e y ybnc= = = 4 29 36 38 32 09. . . in At the ends of the beam the
distance from the bottom of the beam to the center of gravity of
the prestressing strands, with 12 strands harped at the 0.4 span
point, is:
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )y = + + + + + +
=10 2 10 4 10 6 3 64 3 66 3 68 3 70
4222 00. in
The eccentricity of the prestressing strands at the ends of the
beam is: e = = 22 00 36 38 14 38. . . in PRESTRESS LOSS LOW
RELAXATION STRAND In order to check the final beam stresses,
calculate the effective prestress force in the prestressing
strands. The effective prestress force is the force in the
prestressing strands after all prestress losses have occurred. This
example assumes that the prestressing strands are jacked to an
initial stress of 0.75fpu or 202.50 ksi.
Example 1, July 2005 DRAFT 12
-
At beam ends At beam centerline
4 sp
a @
2"
4 sp
a @
2"
3 sp
a @
2"
Figure 7 Strand Pattern
C. G. beam
12'-0"
10 strands 3 strands
3 strands3 strands10 strands
10 strands
3 strands
48'-0"6"
e =
-14.
38 i n
e =
- 14 .
56 i n
e =
- 18.
95 in
e =
- 23 .
33 in
e =
- 27 .
71 i n
e =
-32 .
09 i n
e =
- 32 .
09 in
centerline bearing Figure 8 Strand Profile
GIRDER CREEP COEFFICIENTS b(tf,ti) = girder creep coefficient at
final time due to loading introduced at transfer per
Equation 5.4.2.3.2-1 b(td,ti) = girder creep coefficient at time
of deck placement due to loading introduced at
transfer per Equation 5.4.2.3.2-1 b(tf,td) = girder creep
coefficient at final time due to loading at deck placement per
Equation
5.4.2.3.2-1 H = relative humidity (%) kvs = factor for the
effect of the volume-to-surface ratio of the component kf = factor
for the effect of concrete strength khc = humidity factor for creep
ktd = time development factor
Example 1, July 2005 DRAFT 13
-
t = maturity of concrete (Day), defined as age of concrete
between time of loading for creep calculations, or end of curing
for shrinkage calculations, and time being considered for analysis
of creep or shrinkage effects
ti = age of concrete when load is initially applied, age at
transfer (Days) td = age at deck placement (Days) tf = final age
(Days) V/S = volume-to-surface ratio f'ci = specified compressive
strength of concrete at time of prestressing for pretensioned
members and at time of initial loading for nonprestressed
members. If concrete age at time of initial loading is unknown at
design time, f'ci may be taken as 0.80 f'c (KSI)
Girder creep coefficient for final time due to loading at
transfer H = 70% V/S = 5.7 in ti = 2 Days tf = 20000 days t = 20000
- 2 = 19998 days
( )( )k V Svs = = = 145 013 145 013 57 0 7090 0 0. . ( / ) . . .
. . use 0.7090 ( )( )k Hhc = = =156 0 008 156 0 008 70 10000. . . .
. k
ff ci= + = + =
51
51 7
0 6250' .
( )( )ktf ttd ci
= +
= + =61 4
1999861 4 7 19998
0 9984' .
( )( )( )( )( )( ) ( , ) . . . . . . .. .t t k k k k tf i vs hc
f td i= = 19 19 0 7090 10000 0 6250 0 9984 2 0 77460 118 0 118 =
Girder creep coefficient at time of deck placement due to loading
introduced at transfer td = 180 days t = 180 - 2 = 178 days
( )( )ktf ttd ci
= +
= + =61 4
17861 4 7 178
08436' .
( )( )( )( )( )( ) ( , ) . . . . . . .. .t t k k k k td i vs hc
f td i= = 19 19 0 7090 10000 0 6250 08436 2 0 65450 118 0 118 =
Girder creep coefficient at final time due to loading at deck
placement t = 20000 - 180 = 19820 days
( )( )ktf ttd ci
= +
= + =61 4
1982061 4 7 19820
0 9983' .
( )( )( )( )( )( ) ( , ) . . . . . . .. .t t k k k k tf d vs hc
f td i= = 19 19 0 7090 10000 0 6250 0 9983 180 0 45540 118 0 118 =
DECK CREEP COEFFICIENTS Deck creep coefficient at final time due to
loading at deck placement
Example 1, July 2005 DRAFT 14
-
V/S = 5.5 in t = 20000 - 180 = 19820 days
( )( )k V Svs = = = 145 013 145 013 55 0 7350 0 0. . ( / ) . . .
. . use 0.7350 k
ff ci= + = + =
51
51 5
08333' .
( )( )ktf ttd ci
= +
= + =61 4
1982061 4 5 19820
0 9979' .
( )( )( )( )( )( ) ( , ) . . . . . . .. .t t k k k k tf d vs hc
f td i= = 19 19 0 7350 10000 08333 0 9979 180 0 62920 118 0 118 =
TRANSFORMED SECTION COEFFICIENTS Kid = transformed section
coefficient that accounts for time-dependent interaction
between
concrete and bonded steel in the section being considered for
time period between transfer and deck placement
Kdf = transformed section coefficient that accounts for
time-dependent interaction between concrete and bonded steel in the
section being considered for time period between deck placement and
final time
Transformed section coefficient for time period between transfer
and deck placement epg = eccentricity of strands with respect to
centroid of girder e y ypg b= = =36 38 4 29 32 09. . . in
( )K
EE
AA
A eI
t tid
p
ci
ps
g
g pg
gb f i
=+ +
+
1
1 1 1 0 72
. ( , )
( )( ) ( )( )( )Kid =
+
+
+
=11
285004821
91141085
11085 32 09
7333201 0 7 0 7746
083802. .. .
.
Transformed section coefficient for time period between deck
placement and final time epc = eccentricity of strands with respect
to centroid of composite section Ac = area of section calculated
using the net composite concrete section properties of the
girder and the deck and the deck-to-girder modular ratio Ic =
moment of inertia of section calculated using the net composite
concrete section
properties of the girder and the deck and the deck-to-girder
modular ratio at service Ac = 1896.12 in2Ic = 1485884 in4
e y ypc bc= = =5354 4 29 49 25. . . in
Example 1, July 2005 DRAFT 15
-
( )K
EE
AA
A eI
t tdf
p
ci
ps
c
c pc
cb f i
=+ +
+
1
1 1 1 0 72
. ( , )
( )( ) ( )( )( )Kdf =
+
+
+
=11
285004821
91141896
11896 49 25
14858841 0 7 0 7746
084782. .. .
.
Exterior Beam ELASTIC SHORTENING, fpES (5.9.5.2.3a) In order to
calculate fcgp, the prestress force at transfer needs to be known.
However, the prestress force at transfer depends on the loss that
occurs at transfer: elastic shortening. The value of fcgp can be
either taken as 0.90fpi or calculated using an iterative process.
fcgp = sum of concrete stresses at the center of gravity of the
prestressing tendons due to the
prestressing force at transfer and the self weight of the member
at the sections of maximum moment (ksi)
fpt = the stress in the prestressing tendons at transfer, equal
to fpj minus the loss at transfer, elastic shortening (ksi)
Pt = the prestressing force at transfer (k) For the first
iteration, calculate fcgp using a stress in the prestressing steel
equal to 0.90 of the stress just before transfer. Pt =
(0.90)(202.50)(9.114) = 1661.0 k Mbeam = (2034)(12) = 24408 k-in (
) ( )( )[ ]( ) ( )( )f
PA
P e yI
M yIcgp
t t beam= + + = + + =166101085
16610 32 09 32 09733320
24408 32 09733320
2 5026. . . . .
. ksi
( ) f EE
fpESp
cicgp= =
=
285004821
2 5026 14 79. . ksi
fpt = 202.50 - 14.79 = 187.71 ksi For the second iteration,
calculate fcgp using a stress in the prestressing steel of 187.71
ksi. Pt = (187.71)(9.114) = 1710.8 k
( )( )[ ]( ) ( )( )fcgp = + + =171081085
17108 32 09 32 09733320
24408 32 09733320
2 9111. . . . .
. ksi
( ) fpES = =
285004821
2 9111 17 21. . ksi
fpt = 202.50 - 17.21 = 185.21 ksi For the third iteration,
calculate fcgp using a stress in the prestressing steel of 185.21
ksi. Pt = (185.21)(9.114) = 1688.7 k
( )( )[ ]( ) ( )( )fcgp = + + =1688 71085
1688 7 32 09 32 09733320
24408 32 09733320
2 8597. . . . .
. ksi
Example 1, July 2005 DRAFT 16
-
( ) fpES = =
285004821
2 8597 16 91. . ksi
fpt = 202.50 - 16.91 = 185.59 ksi For the fourth iteration,
calculate fcgp using a stress in the prestressing steel of 185.59
ksi. Pt = (185.59)(9.114) = 1691.5 k
( )( )[ ]( ) ( )( )fcgp = + + =169151085
16915 32 09 32 09733320
24408 32 09733320
2 8662. . . . .
. ksi
( ) fpES = =
285004821
2 8662 16 94. . ksi
fpt = 202.50 - 16.94 = 185.56 ksi For the fifth iteration,
calculate fcgp using a stress in the prestressing steel of 185.56
ksi. Pt = (185.56)(9.114) = 1691.2 k
( )( )[ ]( ) ( )( )fcgp = + + =169121085
16912 32 09 32 09733320
24408 32 09733320
2 8655. . . . .
. ksi
( ) fpES = =
285004821
2 8655 16 94. . ksi
LOSSES: TIME OF TRANSFER TO TIME OF DECK PLACEMENT SHRINKAGE OF
GIRDER CONCRETE, fpSR (5.9.5.4.2a) bid = concrete shrinkage strain
of girder between the time of transfer and deck placement
per Equation 5.4.2.3.3-1 Strain due to shrinkage khs = humidity
factor for shrinkage (%) ( ) ( )( )k Hhs = = =2 00 0 014 2 00 0 014
70 10200. . . . .
( )( )( )( )( ) bid vs hs f tdk k k k x= = =0 48 10 0 7090 10200
0 6250 08436 0 00048 0 000183. . . . . . . in / in ( )( )( ) f E
KpSR bid p id= = = 0 00018 28500 08380 4 30. . . ksi CREEP OF
GIRDER CONCRETE, fpCR (5.9.5.4.2b)
( ) ( )( )( ) f EE f t t KpCR pci cgp b d i id= = = , . .
.285004821
2 8655 0 6545 08380 9 29 ksi.
RELAXATION OF PRESTRESSING STRANDS, fpR1 (5.9.5.4.2c) fpt =
stress in prestressing strands immediately after transfer, taken
not less than 0.55fpy in
Equation 4 KL = 30 for low relaxation strands and 7 for other
prestressing steel, unless more accurate
manufacturer's data are available The relaxation loss, fpR1, may
be assumed equal to 1.2 KSI for low-relaxation strands.
f fK
ffpR
pt
L
pt
py1 055
1855630
18556243
055 132=
=
=.
. .. . ksi
Example 1, July 2005 DRAFT 17
-
LOSSES: TIME OF DECK PLACEMENT TO FINAL TIME SHRINKAGE OF GIRDER
CONCRETE, fpSD (5.9.5.4.3a) bdf = shrinkage strain of girder
between time of deck placement and final time, calculated
by Equation 5.4.2.3.3-1, as bdf = bif - bidStrain due to
shrinkage at final time
( )( )( )( )( ) bif vs hs f tdk k k k x= = =0 48 10 0 7090 10200
0 6250 0 9984 0 00048 0 000223. . . . . . . in / in bdf bif bid= =
=0 00022 0 00018 0 00004. . . in / in ( )( )( ) f E KpSD bdf p df=
= = 0 00004 28500 08478 0 97. . . ksi CREEP OF GIRDER CONCRETE,
fpCD (5.9.5.4.3b) fcd = change in concrete stress at centroid of
prestressing strands due to long-term losses
between transfer and deck placement, combined with deck weight
and superimposed loads
fcd can be calculated using: ( ) f M y
IM M y
Icdslab rail WS comp
comp= +
Mslab = (2053)(12) = 24636 k-in Mrail = (250)(12) = 3000 k-in
MWS = (405)(12) = 4860 k-in y y yb= = = 4 29 36 38 32 09. . . in y
y ycomp bc= = = 4 29 5354 49 25. . . in
( )( ) ( )( ) fcd = + =24636 32 097333203000 4860 49 25
148588413386
. .. ksi
( ) f EE f t t t t K EE f t t KpCD pci cgp b f i b d i df pc cd
b f d df= + ( , ) ( , ) ( , ) .0 0 ( )( )( ) ( )( )( ) fpCD =
+
=
285004821
2 8655 0 7746 0 6545 08478285005154
13386 0 4554 08478 4 58. . . . . . . . ksi
RELAXATION OF PRESTRESSING STRANDS, fpR2 (5.9.5.4.3c) f fpR pR2
1 132= = . ksi SHRINKAGE OF DECK CONCRETE, fpSS (5.9.5.4.3d) fpSS =
prestress loss due to shrinkage of deck composite section fcdf =
change in concrete stress at centroid of prestressing strands due
to shrinkage of deck
concrete ddf = shrinkage strain of deck concrete between
placement and final time per Equation
5.4.2.3.3-1 Ad = area of deck concrete Ecd = modulus of
elasticity of deck concrete
Example 1, July 2005 DRAFT 18
-
ed = eccentricity of deck with respect to the transformed gross
composite section, taken negative in common construction
Ad = (114)(9) = 1026 in2ed = ybc ybdeck = 53.54 (72 + 4.50) =
-22.96 in Strain due to shrinkage
( )( )( )( )( ) ddf vs hs f tdk k k k x= = =0 48 10 0 7350 10200
08333 0 9979 0 00048 0 000303. . . . . . . in / in
( ) f A Et t Ae e
Icdfddf d cd
d f d c
pc d
c= + +
1 0 7
1. ( , )
( )( )( )( )( )
( )( ) fcdf = +
+
=
0 00030 1026 40741 0 7 0 4554
11896
49 25 22 961485884
0 2221.
. .. .
. ksi
( ) ( )( ) ( )( )( ) f EE f K t tpSS pc cdf df b f d= + = + = 1
0 7285005154
0 2221 08478 1 0 7 0 4554 137. ( , ) . . . . . ksi TOTAL LOSSES
fpLT = (fpSR + fpCR + fpR1)id + (fpSD + fpCD + fpR2 + fpSS)dffpLT =
(4.30 + 9.29 + 1.32) + (0.97 + 4.58 + 1.32 1.37) = 20.41 ksi fpT =
fpLT + fpES = 20.41 + 16.94 = 37.35 ksi Interior Beam ELASTIC
SHORTENING, fpES (5.9.5.2.3a) The loss due to elastic shortening
only depends on the moment due to the member weight. Since this is
the same for the interior and exterior beams, the loss due to
elastic shortening is the same for both beams. fpES = 16.94 ksi
LOSSES: TIME OF TRANSFER TO TIME OF DECK PLACEMENT SHRINKAGE OF
GIRDER CONCRETE, fpSR (5.9.5.4.2a) The loss due to shrinkage
depends on the relative humidity, the amount of time between the
time of transfer and the time of deck placement, and the
transformed section coefficient for time period between transfer
and deck placement. Since these are the same for the interior and
exterior beams, the loss due to shrinkage of girder concrete is the
same for both beams. fpSR = 4.30 ksi CREEP OF GIRDER CONCRETE, fpCR
(5.9.5.4.2b) The loss due to creep depends on the sum of concrete
stresses at the center of gravity of the prestressing tendons due
to the prestressing force at transfer and the self weight of the
member at the sections of maximum moment, the amount of time
between the time of transfer and the time of deck placement, and
the transformed section coefficient for time period between
transfer and deck placement. Since these are the same for the
interior and exterior beams, the loss due to creep of girder
concrete is the same for both beams. fpCR = 9.29 ksi
Example 1, July 2005 DRAFT 19
-
RELAXATION OF PRESTRESSING STRANDS, fpR1 (5.9.5.4.2c) The loss
due to relaxation of prestressing strands depends on the stress in
prestressing strands immediately after transfer and the type of
prestressing strands used. Since these are the same for the
interior and exterior beams, the loss due to relaxation of
prestressing strands is the same for both beams. fpR1 = 1.32 ksi
LOSSES: TIME OF DECK PLACEMENT TO FINAL TIME SHRINKAGE OF GIRDER
CONCRETE, fpSD (5.9.5.4.3a) The loss due to shrinkage depends on
the relative humidity, the amount of time between the time of deck
placement and the final time, and the transformed section
coefficient for time period between deck placement and final. Since
these are the same for the interior and exterior beams, the loss
due to shrinkage of girder concrete is the same for both beams.
fpSD = 0.97 ksi CREEP OF GIRDER CONCRETE, fpCD (5.9.5.4.3b) Mslab =
(2137)(12) = 25644 k-in ( )( ) ( )( ) fcd = + =25644 32
09733320
3000 4860 49 251485884
13827. .
. ksi
( )( )( ) ( )( )( ) fpCD = +
=
285004821
2 8655 0 7746 0 6545 08478285005154
13827 0 4554 08478 4 68. . . . . . . . ksi
RELAXATION OF PRESTRESSING STRANDS, fpR2 (5.9.5.4.3c) The loss
due to relaxation of prestressing strands between the time of deck
placement and the final time is equal to the loss between the
transfer time and the time of deck placement. Since these are the
same for the interior and exterior beams, the loss due to
relaxation of prestressing strands is the same for both beams. fpR2
= 1.32 ksi SHRINKAGE OF DECK CONCRETE, fpSS (5.9.5.4.3d) The loss
due to shrinkage of the deck concrete depends on the change in
concrete stress at centroid of prestressing strands due to
shrinkage of deck concrete, section properties, the amount of time
between the time of deck placement and the final time, and the
transformed section coefficient for time period between deck
placement and final. Since these are the same for the interior and
exterior beams, the loss due to shrinkage of deck concrete is the
same for both beams. fpSS = 1.37 ksi TOTAL LOSSES fpLT = (fpSR +
fpCR + fpR1)id + (fpSD + fpCD + fpR2 + fpSS)dffpLT = (4.30 + 9.29 +
1.32) + (0.97 + 4.68 + 1.32 1.37) = 20.51 ksi fpT = fpLT + fpES =
20.51 + 16.94 = 37.45 ksi
Example 1, July 2005 DRAFT 20
-
STRAND ARRANGEMENT Now that the total loss in the prestressing
steel stress has been calculated, check if the preliminary strand
pattern and eccentricity are still valid. The calculated effective
stress in the prestressing steel is: fpe = 202.50 37.45 = 165.05
ksi for the interior beam fpe = 202.50 37.35 = 165.15 ksi for the
exterior beam Interior Beam
( ) ( )( )Pe =
=0 268 37101
108532 09
20157
1369 3. .
.. k
The effective prestress force in one strand is: (165.05)(0.217)
= 35.8 k The number of strands required is: 1369 3
35838 2
..
.= Exterior Beam
( ) ( )( )Pe =
=0 268 4 0441
108532 09
20157
1502 2. .
.. k
The effective prestress force in one strand is: (165.15)(0.217)
= 35.8 k The number of strands required is: 1502 2
35842 0
..
.= The preliminary strand arrangement with 42 strands is still
good. SERVICE LIMIT STATE Now that the release and effective
prestress forces are known, the critical beam stresses can be
checked for Service I and III Limit States. Check the stresses in
the interior and exterior beams for the release and service
conditions. The release condition includes loads due the release or
transfer of the prestress force and the self-weight of the beam.
The service condition includes loads due to the effective prestress
force, dead load (beam, concrete deck slab, rails, diaphragms, and
future wearing surface allowance, and live load. Calculate the beam
stress limits for two conditions: temporary before losses for the
release condition and after all losses for the service condition.
For the temporary before losses condition (5.9.4.1): Tension: 0
0948 0 0948 7 0 251. . .'fci = = ksi > 0.2 ksi, use 0.2 ksi
Compression: ( )( )0 60 0 60 7 4 2. . .'fci = = ksiAt the service
limit state after all losses (5.9.4.2): Tension, severe corrosion
conditions:
Example 1, July 2005 DRAFT 21
-
0 0948 0 0948 8 0 268. . .'fc = = ksi Compression: Due to
effective prestress and permanent loads: ( )( )0 45 0 45 8 36. .
.'fc = = ksi Due to live load and one-half the sum of effective
prestress and permanent loads: ( )( )0 40 0 40 8 32. . .'fc = = ksi
Due to effective prestress, permanent loads, and transient loads: 0
6. ' w cf The reduction factor, w, is equal to 1.0 if the flange
slenderness ratio is not greater than
15. The slenderness ratio is the ratio of the flange width to
depth:
flange widthflange depth
= =1149
12 7.
Since this value is not greater than 15, w is equal to 1.0 and
the stress limit is: ( )( )( )0 6 0 6 10 8 4 8. . . .' w cf = =
ksiThe stress limits for the concrete deck slab at the service
limit state: Compression: Due to effective prestress and permanent
loads: ( )( )0 45 0 45 5 2 25. . .'fc = = ksi Due to effective
prestress, permanent loads, and transient loads: ( )( )( )0 6 0 6
10 5 30. . . .' w cf = = ksiThe total area of prestressing steel,
the initial or transfer prestressing force, and the effective
prestressing forces are: Aps = (0.217)(42) = 9.114 in2Pt =
(fpt)(Aps) = (185.56)(9.114) = 1691.2 k Pe = (fpe)(Aps) =
(165.05)(9.114) = 1504.3 k for the interior beam Pe = (fpe)(Aps) =
(165.15)(9.114) = 1505.2 k for the exterior beam STRESSES AT
TRANSFER The release stresses are the same for the interior and
exterior beams and can be calculated using:
fPA
P eS
MSbottom
t t
bnc
beam
bnc= (Service I Limit State)
fPA
P eS
MStop
t t
tnc
beam
tnc= + + (Service I Limit State)
At the harp point due to the initial prestress force and the
beam dead load Mbeam = (1986.7)(12) = 23840 k-in (for 121 foot
casting length) ( )( )fbottom = =169121085
16912 32 0920157
2384020157
3068. . .
. ksi (c)
( )( )ftop = + + =169121085
16912 32 0920587
2384020587
0 081. . .
. ksi (c)
Example 1, July 2005 DRAFT 22
-
At end of the transfer length due to the initial prestress force
and the beam dead load The transfer length is 60 strand diameters
or 36 inches. Therefore, the end of the transfer length is 36
inches from the end of the beam or 30 inches from the centerline of
bearing, the 0.0208 point. The eccentricity of the prestressing
force is: e = 14.56 + (18.95 14.56)(0.208) = 15.47 in Mbeam =
(200.0)(12) = 2400 k-in (for 121 foot casting length) ( )( )fbottom
= =169121085
16912 154720157
240020157
2 738. . .
. ksi (c)
( )( )ftop = + + =169121085
16912 154720587
240020587
0 404. . .
. ksi (c)
STRESSES AT SERVICE CONDITION The service stresses, assuming
tension at the bottom beam fibers, can be calculated using:
fPA
P eS
M MS
M M MSbottom
e e
bnc
beam slab
bnc
rail WS LL
bc= + + + 08. (Service III Limit State)
fPA
P eS
M MS
M M MStop
e e
tnc
beam slab
tnc
rail WS LL
tc= + + + + + + (Service I Limit State)
Exterior Beam At end of the transfer length due to the effective
prestress force and dead load Mbeam = (166.0)(12) = 1992 k-in Mslab
= (164.2)(12) = 1970 k-in Mrail = (20.4)(12) = 245 k-in MWS =
(33.0)(12) = 396 k-in ( )( )fbottom = + + =150521085
15052 154720157
1992 197020157
245 39627751
2 323. . .
. ksi (c)
( )( )ftop = + + + + + =150521085
15052 154720587
1992 197020587
245 39680503
0 457. . .
. ksi (c)
At midspan due to the effective prestress force and dead
load
( )( )fbottom = + + =150521085
15052 32 0920157
24408 2463620157
3000 486027751
1067. . .
. ksi (c)
( )( )ftop = + + + + + =150521085
15052 32 0920587
24408 2463620587
3000 486080503
1521. . .
. ksi (c)
At midspan due to the effective prestress force, dead load, and
live load MLL = (3837)(12) = 46044 k-in ( )( ) ( )( )fbottom = + +
+ = 150521085
15052 32 0920157
24408 2463620157
3000 4860 08 4604427751
0 260. . . .
. ksi (t)
( )( )ftop = + + + + + + =150521085
15052 32 0920587
24408 2463620587
3000 4860 4604480503
2 093. . .
. ksi (c)
Example 1, July 2005 DRAFT 23
-
At midspan due to the live load and one-half the effective
prestress force and dead load ( )( )
ftop = + + + + + + =
12
150521085
15052 32 0920587
24408 2463620587
3000 486080503
4604480503
1332. . .
. ksi (c)
Interior Beam At end of the transfer length due to the effective
prestress force and dead load Mslab = (167.7)(12) = 2012 k-in ( )(
)fbottom = + + =1504 31085
1504 3 154720157
1992 201220157
245 39627751
2 319. . .
. ksi (c)
( )( )ftop = + + + + + =1504 31085
1504 3 154720587
1992 201220587
245 39680503
0 459. . .
. ksi (c)
At midspan due to the effective prestress force and dead
load
( )( )fbottom = + + =1504 31085
1504 3 32 0920157
24408 2564420157
3000 486027751
1015. . .
. ksi (c)
( )( )ftop = + + + + + =1504 31085
1504 3 32 0920587
24408 2564420587
3000 486080503
1571. . .
. ksi (c)
At midspan due to the effective prestress force, dead load, and
live load MLL = (2728)(12) = 32736 k-in ( )( ) ( )( )fbottom = + +
+ =1504 31085
1504 3 32 0920157
24408 2564420157
3000 4860 08 3273627751
0 071. . . .
. ksi (c)
( )( )ftop = + + + + + + =1504 31085
1504 3 32 0920587
24408 2564420587
3000 4860 3273680503
1977. . .
. ksi (c)
At the midspan due to the live load and one-half the effective
prestress force and dead load
( )( )ftop = + + + + +
+ =
12
1504 31085
1504 3 32 0920587
24408 2564420587
3000 486080503
3273680503
1192. . .
. ksi (c)
Check the stress at the top fibers of the deck slab, interior
and exterior beams, at midspan for the concrete compression limit.
The composite section properties and loads are used. The service
stresses, assuming compression in the extreme top slab fibers, can
be calculated using:
fM M M
Sslab toprail WS LL
slab top= + + (Service I Limit State)
Interior Beam At midspan due to the composite dead load
fslab top = + =3000 486068443 0115. ksi (c)
Example 1, July 2005 DRAFT 24
-
At midspan due to the composite dead load and live load
fslab top = + + =3000 4860 3273668443 0593. ksi (c) Exterior
Beam At midspan due to the composite dead load
fslab top = + =3000 486068443 0115. ksi (c) At midspan due to
the composite dead load and live load
fslab top = + + =3000 4860 4604468443 0 788. ksi (c)
Table 5 Stress Summary, Interior and Exterior Beams Maximum
Limit BEAM STRESSES
Release Tension none 0.200 ksi Compression 3.068 ksi 4.2 ksi
Effective prestress, dead loads Tension none 0.268 ksi
Compression 2.323 ksi 3.6 ksi
Effective prestress, dead loads, live load Tension 0.260 ksi
0.268 ksi Compression 2.093 ksi 4.8 ksi
Live load and half the sum of effective prestress and dead loads
Compression 1.332 ksi 3.2 ksi
DECK SLAB STRESS Composite dead loads
Compression 0.115 ksi 2.25 ksi Composite dead loads, live
load
Compression 0.788 ksi 3.0 ksi FATIGUE LIMIT STATE (5.5.3.1) For
fully prestressed concrete components there is no need to check
fatigue when the tensile stress in the extreme fiber at the Service
III Limit State after all losses meets the tensile stress limits.
Since this is the case, fatigue does not need to be checked.
STRENGTH LIMIT STATE The Strength Limit State includes checks on
the nominal flexural resistance and the amount of prestressed and
non-prestressed reinforcement. For practical design the rectangular
stress distribution can be used.
Example 1, July 2005 DRAFT 25
-
NOMINAL FLEXURAL RESISTANCE MIDSPAN The beams do not have any
non-prestressed tension reinforcement or compression reinforcement.
The stress block factor is based on the compressive strength of the
deck concrete, assumes that the neutral axis remains within the
deck slab. Since this is a strength limit state check, the width of
the compression flange, b, is not reduced. dp = the distance from
the extreme compression fiber to the centroid of the
prestressing
tendons (in) 1 = the stress block factor 1. Calculate the
factored moments for the Strength I Limit State, Mu Mu =
(1.25)(2034 + 2053 + 250) + (1.5)(405) + (1.75)(3837) = 12744 k-ft
(exterior beam) Mu = (1.25)(2034 + 2137 + 250) + (1.5)(405) +
(1.75)(2728) = 10908 k-ft (interior beam) 2. Calculate the depth of
the compression block (5.7.3.1.1) b = 114 inches dp = e + yt +
tslab = 32.09 + 35.62 + 9 = 76.71 in
k = 0.28 for low relaxation strand 1 = 0.80 for the 5 ksi deck
concrete In order to calculate the depth of the compression block,
a, first calculate the depth to the
neutral axis, c, assuming rectangular section behavior.
( )( )
( )( )( )( ) ( )( )c
A f A f A f
f b kAfd
ps pu s y s y
c pspu
p
= + +
= + +
=
' '
'.
.
. . . ..
.085
9114 270 0 0
085 5 080 114 0 28 9114270
76 71
6 201
in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = 1c = (0.80)(6.20) = 4.96 in 3. Calculate the stress in the
prestressing steel at the nominal flexural resistance (5.7.3.1.1)
For components with bonded tendons:
( ) ( )f f k cdps pu p
=
=
=1 270 1 0 28
6 2076 71
26389...
. ksi
4. Calculate the factored flexural resistance (5.7.3.2.1)
( )( )M A f d an ps ps p= =
=2 9114 26389 76 71
4 962
112
14878. . ..
k - ft
Mr = Mn = (1.0)(14878) = 14878 k-ft > 12744 k-ft for the
exterior beam O.K. > 10908 k-ft for the interior beam O.K.
REINFORCEMENT LIMITS MIDSPAN Check the reinforcement limits,
maximum and minimum. The maximum amount of prestressed and
non-prestressed reinforcement (5.7.3.3.1) should satisfy:
c
de 0 42.
Example 1, July 2005 DRAFT 26
-
c = the distance from the extreme compression fiber to the
neutral axis (in) de = the corresponding effective depth from the
extreme compression fiber to the centroid of
the tensile force in the tensile reinforcement (in) ds = the
distance from the extreme compression fiber to the centroid of the
non-prestressed
tensile reinforcement (in)
( )( )( )( )( )d
A f d A f dA f A feps ps p s y s
ps ps s y= ++ =
++ =
9114 26389 76 71 09114 26389 0
76 71. . .
. .. in
cde
= = 1884 k-ft O.K.
Example 1, July 2005 DRAFT 27
-
Interior Beam Mdnc = 2034 + 2137 = 4171 k-ft = 50052 k-in f fr
c= = =0 24 0 24 8 0 6788. . .' ksi (5.4.2.6)
( )( )f
PA
P eScpe
e e
nc= = =1504 3
10851504 3 3309
2015737813
. . .. ksi
( )( ) ( )Mcr = +
=27751 0 6788 37813 50052
2775120157
11
128743. . k - ft
Scfr = (27751)(0.6788) = 18837 k-in = 1570 k-ft Since Mcr is to
be less than or equal to Scfr, 1.2 Mcr = (1.2)(1570) = 1884 k-ft
1.33 Mu = (1.33)(10908) = 14508 k-ft The lesser of these two values
is 1884 k-ft and the factored flexural resistance is 17382 k-ft.
17382 k-ft > 1884 k-ft O.K. PRETENSIONED ANCHORAGE ZONE
(5.10.10.1) The bursting resistance provided by vertical
reinforcement in the ends of pretensioned beams at the service
limit state should resist a force not less than 4% of the prestress
force at transfer. The total reinforcement is located within a
distance h/4 from the end of the beam. The stress in the steel, fs,
is not to exceed 20 ksi. Bursting resistance provided by the
vertical reinforcement: Pr = fsAs As = the total area of vertical
reinforcement located within the distance h/4 from the end of
the
beam (in2) h = the overall depth of the precast member Pt = the
prestressing force at transfer (k) Pr = (Pt)(0.04) = (1691.2)(0.04)
= 67.65 k h = 72 inches h4
724
18 0= = . in Solve for the area of vertical reinforcement
required:
APfs
r
s= = =67 45
20338
.. in2
Using pairs of No. 4 bars, As = 0.40 in2 and the number of pairs
of bars required is: 3380 40
85..
.= With 9 pairs of No. 4 bars: As = (9)(0.40) = 3.60 in2
DEVELOPMENT AND TRANSFER LENGTH The development length, ld, in
prestressing strand is the length of strand over which there is a
gradual buildup of strand force, Figure 9. It consists of two
components, the transfer and the
Example 1, July 2005 DRAFT 28
-
flexural bond lengths. The prestress force varies linearly over
the transfer length. At the end of the transfer length the stress
in the strand is the effective prestress stress. The prestress
force then varies in a parabolic manner and reaches the tensile
strength of the strand at the end of the development length. The
transfer and development lengths are calculated. The flexural bond
length can be calculated by taking the difference between the
development and transfer lengths. This example calculates the
development lengths using the stress in the prestressing steel
previously calculated at midspan. The transfer length only depends
on the strand diameter and therefore is the same for the interior
and exterior beams, sixty strand diameters. The development length
for bonded strands:
l d ps pek f f d
23 b
db = the nominal strand diameter (in) fps = the average stress
in the prestressing steel at the time for which the nominal
resistance of
the member is required (ksi) k = 1.6 for precast prestressed
beams
f
f
ld
ps
pe
Distance from free end of strand
Stee
l stre
ss transferlength
Figure 9 Development Length
TRANSFER LENGTH (5.11.4.1) 60 db = (60)(0.6) = 36 in DEVELOPMENT
LENGTH (5.11.4.2) Exterior Beam
( ) ( ) ( )k f f dps pe b =
=
23
16 2638923
16015 0 6 1508. . . . . in
Interior Beam
( ) ( ) ( )k f f dps pe b =
=
23
16 2638923
160 05 0 6 150 9. . . . . in
Example 1, July 2005 DRAFT 29
-
WEB SHEAR DESIGN (5.8) Determine the required transverse
reinforcement spacing at the critical section and at the inside
edge of the elastomeric bearing. Verify that the longitudinal
reinforcement on the flexural tension side of the member is
properly proportioned at these locations. The calculation of the
critical section is an iterative process because the criteria used
to calculate the location depends on the value of the effective
shear depth at the location of the critical section. Vc = the
nominal shear resistance of the concrete (k) Vp = the component in
the direction of the applied shear of the effective prestressing
force (k) Vn = the nominal shear resistance of the section (k) Vs =
the nominal shear resistance of the shear reinforcing steel (k) dv
= the effective shear depth taken as the distance, measured
perpendicular to the neutral
axis, between the resultants of the tensile and compressive
forces due to flexure (in) s = the spacing of the shear
reinforcement (in) = the factor indicating the ability of
diagonally cracked concrete to transmit tension = the angle of
inclination of the diagonal compressive stresses (deg) bv = the
effective web width, taken as the minimum web width within the
depth dv (in) h = the overall depth of the member Av = the area of
shear reinforcement within a distance s (in2) Ac = the area of
concrete on the flexural tension side of the member (in2) Exterior
Beam 1. Calculate the effective shear depth, dv (5.8.2.9) One way
to calculate the effective shear depth is to calculate the depth of
the
compression block and then the distance between the middle of
the compression block and the center of gravity of all the tensile
reinforcement. Since non-prestressed reinforcement is not used, de
is equal to dp. Starting with an estimated location of the critical
section at the 0.05 point:
e = 14.56 + (18.95 14.56)(0.5) = 16.76 in de = dp = e + yt +
tslab = 16.76 + 35.62 + 9 = 61.38 in Assuming rectangular section
behavior:
( )( )
( )( )( )( ) ( )( )c
A f A f A f
f b kAfd
ps pu s y s y
c pspu
p
= + +
= + +
=' '
'.
.
. . . ..
.085
9114 270 0 0
085 5 080 114 0 28 9114270
6138
6171
in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = 1c =(0.80)(6.17) = 4.94 in The calculated effective shear
depth is:
d da
v e= = =2 61384 94
258 91.
.. in
But dv does not need to be less than the greater of: 0.9de =
(0.9)(61.38) = 55.24 in 0.72h = (0.72)(81) = 58.32 in Since the
calculated value is greater than both of these values, dv = 58.91
in
Example 1, July 2005 DRAFT 30
-
2. Calculate the location of the critical section (5.8.3.2) The
critical section is located a distance dv from the face of the
support. Since the
location of the critical section is measured from the face of
the support, add a distance equal to one-half the width of the
bearing pad or 4 inches. Therefore, the distance from the center of
bearing to the critical section is:
58.91 + 4.00 = 62.91 in (~ 0.044 point) Since the revised
location of the critical section is not close enough to the
previous value,
repeat. 3. Calculate the effective shear depth, dv (5.8.2.9) At
the revised estimated location of the critical section, the 0.044
point: e = 14.56 + (18.95 14.56)(0.44) = 16.49 in de = dp = e + yt
+ tslab = 16.49 + 35.62 + 9 = 61.11 in Assuming rectangular section
behavior:
( )( )
( )( )( )( ) ( )( )c
A f A f A f
f b kAfd
ps pu s y s y
c pspu
p
= + +
= + +
=
' '
'.
.
. . . ..
.085
9114 270 0 0
085 5 080 114 0 28 91142706111
6171
in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = 1c =(0.80)(6.17) = 4.94 in The calculated effective shear
depth is:
d da
v e= = =2 61114 94
258 64.
.. in
But dv does not need to be less than the greater of: 0.9de =
(0.9)(61.11) = 55.00 in 0.72h = 58.32 in Since the calculated value
is greater than both of these values, dv = 58.64 in 4. Calculate
the location of the critical section (5.8.3.2) The critical section
is located a distance dv from the face of the support. Therefore,
the
distance from the center of bearing to the critical section is:
58.64 + 4.00 = 62.64 in (~ 0.044 point) Since the revised location
of the critical section is close enough to the previous value,
continue with the next step. 5. Calculate the factored loads for
the Strength I Limit State
Table 6 Beam Moments and Shears, 0.044 Point Dead Load
DC DW Load Beam Slab/Diaph Rail FWS
Live Load plus Dynamic Load
Allowance Moment, k-ft 342 339 42 68 663
Shear, k 61.8 61.3 7.6 12.3 125.4 Mu = (1.25)(342 + 339 + 42) +
(1.5)(68) + (1.75)(663) = 2166 k-ft Vu = (1.25)(61.8 + 61.3 + 7.6)
+ (1.5)(12.3) + (1.75)(125.4) = 401.3 k
Example 1, July 2005 DRAFT 31
-
Nu = 0 k 6. Calculate the prestress tendon shear component = the
inclination of the harped strands (deg)
= =
tan .162
5826 08o
Vp = (Aps)(fpe)(sin) = (12)(0.217)(160.15)(0.1059) = 44.16 k
Since the critical section is located past the transfer length, use
the full effective stress. 7. Calculate the shear stress in the
concrete and the shear stress ratio (5.8.2.9) bv = 8 inches The
shear stress in the concrete is:
( )( )
( )( )( )vV V
b duu p
v v= = =
4013 0 9 44160 9 8 58 64
0856. . .. .
. ksi
The shear stress ratio is:
vf
u
c'
..= =0856
80107
8. Calculate the strain in reinforcement on the flexural tension
side of the member (5.8.3.4.2)
Aps = the area of prestressing steel on the flexural tension
side of the member (in2) As = the area of non-prestressed steel on
the flexural tension side of the member at the
section under investigation (in2) fpo = a parameter taken as the
modulus of elasticity of prestressing tendons multiplied
by the locked in difference in strain between the prestressing
tendons and the surrounding concrete, for usual levels of
prestressing, 0.70 fpu will be appropriate
Mu = the factored moment not to be taken as less than Vudv Aps =
(0.217)(30) = 6.510 in2 (30 strands) As = 0 in2 Ac = 584 in2 fpo =
(0.7)(270) = 189 ksi
Vudv = (401.3)(58.64) = 23532 k-in = 1961 k-ft Since this value
is less than the actual applied factored moment, use the actual
applied
factored moment. Assume an initial value for of 23.7, which is
based on the shear stress ratio of 0.107 and an assumed value for x
of zero. The initial value of x should not be taken greater than
0.001.
( )
( )
x
u
vu u p ps
s s ps ps
Md
N V V A f
E A E A=
+ + +
05 05
2
. . cot po
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
2166 1258 64
0 05 4013 4416 237 6510 189
2 0 28500 651010250.
. . . cot . .
.. x 10 in / in-3
Example 1, July 2005 DRAFT 32
-
Since x is negative:
( )
( )
x
u
vu u p ps
c c s s ps ps
Md
N V V A f
E A E A E A=
+ + + +
05 05
2
. . cot po
( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )( )[ ] x =+ +
+ + = 2166 12
58 640 05 4013 4416 237 6510 189
2 5154 584 0 28500 65100 0595.
. . . cot . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain from Table
5.8.3.4.2-1, 22.8. Since this does not agree with the initial
value, repeat.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
2166 1258 64
0 05 4013 4416 22 8 6510 189
2 0 28500 65100 9765.
. . . cot . .
.. x 10 in / in-3
Since the revised x is negative:
( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )( )[ ] x =+ +
+ + = 2166 12
58 640 05 4013 4416 22 8 6510 189
2 5154 584 0 28500 65100 0567.
. . . cot . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain , 22.8. Since this
agrees with the previous value, continue.
9. Obtain the final theta and beta factors from Table
5.8.3.4.2-1 = 22.8
= 2.94 10. Calculate the shear resistance provided by the
concrete, Vc (5.8.3.3) ( )( )( )( )( )V f b dc c v v= = =0 0316 0
0316 2 94 8 8 58 64 12327. . . .' k. 11. Check if the region
requires transverse reinforcement (5.8.2.4)
Transverse reinforcement is required if: Vu > 0.5(Vc + Vp)
0.5(Vc + Vp) = (0.5)(0.9)(123.27 + 44.16) = 75.34 k Since Vu, 401.3
k, is greater than this value, transverse reinforcement is required
12. Calculate the shear resistance that the transverse
reinforcement needs to provide, Vs
VV
V Vsu
c p= = =40130 9
12327 4416 278 46.
.. . . k
13. Calculate the required transverse reinforcement spacing Av =
0.40 in2 (2 No. 4 bars)
( )( )( )( )
sA f d
Vv y v
s= = =cot . . cot .
..
0 40 60 58 64 22 8278 46
12 0 in
Based on reinforcement placed at 90 to the longitudinal axis of
the beam
Example 1, July 2005 DRAFT 33
-
14. Calculate the maximum allowable spacing of the transverse
reinforcement (5.8.2.7) ( )( )0125 0125 8 1000. . .'fc = = ksi
Since v fcu 0.856 ksi= < 0125. ' s dvmax . . 08 24 0 in 0.8dv =
(0.8)(58.64) = 46.9 in > 24.0 in Therefore, the maximum spacing
is 24 inches. Use 2 No. 4 bars at a 12-inch spacing. 15. Check if
the minimum amount of transverse reinforcement is provided
(5.8.2.5)
( )( ) ( )( )A f b sfv c vy= =
=0 0316 0 0316 8
8 1260
014. .' in2.
Vp
Since the transverse reinforcement provided, 0.40 in2, is
greater than this, the minimum transverse reinforcement provision
is satisfied.
16. Check the nominal shear resistance (5.8.3.3) The nominal
shear resistance is the lesser of: V f b dn c v v +0 25. ' Vc + Vs
+ Vp
( )( )( )( )V
A f dss
v y v= = =cot . . cot . . 0 40 60 58 64 22 812
279 00 k
( )( )( )( )0 25 0 25 8 8 58 64 4416 982 40. . . .'f b d Vc v v
p+ = + = k. Vc + Vs + Vp = 123.27 + 279.00 + 44.16 = 446.43 k The
lesser of these two values is 446.43 k and the factored shear
resistance is: Vr = Vn = (0.9)(446.43) = 401.79 k > 401.3 k O.K.
17. Check the longitudinal reinforcement requirement (5.8.3.5) At
each section the tensile capacity of the longitudinal reinforcement
on the flexural
tension side of the member shall be proportioned to satisfy:
A f A fMd
N VV Vs y ps ps
u
v
u us p+ + +
05 05. . cot
Any lack of full development shall be accounted for. Aps = the
area of prestressing steel on the flexural tension side of the
member (in2) As = the area on non-prestressed steel on the flexural
tension side of the member (in2)
Vs = the shear resistance provided by the transverse
reinforcement at the section under investigation, given by Equation
5.8.3.3-4, except Vs shall not be taken as greater
than Vu
Consider the lack of full development of the prestressing steel
using the plot of the stress in the prestressing steel versus the
distance from the end of the strand. At the end of the transfer
length, 36 inches, the stress in the prestressing strand is fpe,
160.2 ksi. At the end
Example 1, July 2005 DRAFT 34
-
of the development length, 151 inches, the stress in the
prestressing strand, fps, is 263.9 ksi. Using a straight-line
interpolation between the end of the transfer length and the
development length the stress in the prestressing strand at 68
inches from the end of the beam is approximately:
( )fps +
=160 2
2639 160 2151 36
68 36 189.. .
ksi
Calculate the tensile capacity provided by the longitudinal
reinforcement: Asfy + Apsfps = 0 + (6.510)(189) = 1230 k Calculate
the required tensile force where Vs is not to be greater than:
Vu = =
40130 9
44589.
.. k
Since the nominal shear resistance provide, 279.00 k, is less
than this value, use the nominal shear resistance provided.
( )( )( )( ) ( )( ) ( )A f A fs y ps ps+ + +
=
2166 1258 64 10
040130 9
05 279 00 4416 22 8 1067. .
..
. . . cot . k
Since 1230 k > 1067 k O.K. Interior Beam 1. Calculate the
effective shear depth, dv (5.8.2.9) Starting with an estimated
location of the critical section at the 0.05 point: e = 16.76 in de
= dp = 61.38 in Assuming rectangular section behavior:
( )( )
( )( )( )( ) ( )( )c = +
+
=9114 270 0 0085 5 080 114 0 28 9114
2706138
617.
. . . ..
. in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = 1c = (0.80)(6.17) = 4.94 in The calculated effective depth
is:
dv = =6138 4 942 58 91..
. in
But dv does not need to be less than the greater of: 0.9de =
(0.9)(61.38) = 55.24 in 0.72h = 58.32 in Since the calculated value
is greater than both of these values, dv = 58.91 in 2. Calculate
the location of the critical section (5.8.3.2) The critical section
is located a distance dv from the face of the support. Therefore,
the
distance from the center of bearing to the critical section is:
58.91 + 4.00 = 62.91 in (~ 0.044 point) Since the revised location
of the critical section is not close enough to the previous
value,
repeat.
Example 1, July 2005 DRAFT 35
-
3. Calculate the effective shear depth, dv (5.8.2.9) At the
revised estimated location of the critical section, the 0.044
point: e = 14.56 + (18.95 14.56)(0.44) = 16.49 in de = dp = e + yt
+ tslab = 16.49 + 35.62 + 9 = 61.11 in Assuming rectangular section
behavior:
( )( )
( )( )( )( ) ( )( )c
A f A f A f
f b kAfd
ps pu s y s y
c pspu
p
= + +
= + +
=
' '
'.
.
. . . ..
.085
9114 270 0 0
085 5 080 114 0 28 91142706111
6171
in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = 1c =(0.80)(6.17) = 4.94 in The calculated effective shear
depth is:
d da
v e= = =2 61114 94
258 64.
.. in
But dv does not need to be less than the greater of: 0.9de =
(0.9)(61.11) = 55.00 in
0.72h = 58.32 in Since the calculated value is greater than both
of these values, dv = 58.64 in 4. Calculate the location of the
critical section (5.8.3.2) The critical section is located a
distance dv from the face of the support. Therefore, the
distance from the center of bearing to the critical section is:
58.64 + 4.00 = 62.64 in (~ 0.044 point) Since the revised location
of the critical section is close enough to the previous value,
continue with the next step. 5. Calculate the factored loads for
the Strength I Limit State
Table 7 Beam Moments and Shears, 0.044 Point Dead Load
DC DW Load Beam Slab/Diaph Rail FWS
Live Load plus Dynamic Load
Allowance Moment, k-ft 342 346 42 68 472
Shear, k 61.8 62.7 7.6 12.3 110.5 Mu = (1.25)(342 + 346 + 42) +
(1.5)(68) + (1.75)(472) = 1841 k-ft Vu = (1.25)(61.8 + 62.7 + 7.6)
+ (1.5)(12.3) + (1.75)(110.5) = 377.0 k Nu = 0 k 6. Calculate the
prestress tendon shear component Vp = (Aps)(fpe)(sin) =
(12)(0.217)(160.05)(0.1059) = 44.14 k Since the critical section is
located past the transfer length, use the full effective stress. 7.
Calculate the shear stress in the concrete and the shear stress
ratio (5.8.2.9) The shear stress in the concrete is:
Example 1, July 2005 DRAFT 36
-
( )( )
( )( )( )vu = =377 0 0 9 4414
0 9 8 58 640 799
. . .. .
. ksi
The shear stress ratio is:
vf
u
c'
..= =0 799
80100
8. Calculate the strain reinforcement on the flexural tension
side of the member (5.8.3.4.2) Vudv = (377.0)(58.64) = 22107 k-in =
1842 k-ft
Since this value is greater than the actual applied factored
moment, use this value. Assume an initial value for of 22.5, which
is based on the shear stress ratio of 0.100 and an assumed value
for x of zero. ( )( ) ( )( )( ) ( )( )
( ) ( )( )[ ] x =+
+ = 1842 12
58 6405 377 0 4414 22 5 6510 189
2 0 28500 651012172.
. . . cot . .
.. x 10 in / in-3
Since x is negative:
( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )( )[ ] x =+ +
+ + = 1842 12
58 640 05 377 0 4414 22 5 6510 189
2 5154 584 0 28500 65100 0707.
. . . cot . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain from Table
5.8.3.4.2-1, 210.4. Since this does not agree with the initial
value, repeat.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + =
1842 1258 64
05 377 0 4414 214 6510 189
2 0 28500 651011555.
. . . cot . .
.. x 10 in / in-3
Since the revised x is negative:
( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )( )[ ] x =+ +
+ + = 1842 12
58 640 05 377 0 4414 214 6510 189
2 5154 584 0 28500 65100 0671.
. . . cot . .
.. x 10 in / in-3
Use the shear stress ratio and the x to obtain , 21.4. Since
this agrees with the previous value, continue.
9. Obtain the final theta and beta factors from Table
5.8.3.4.2-1
= 21.4 = 3.24
10. Calculate the shear resistance provided by the concrete, Vc
(5.8.3.3) ( )( )( )( )( )Vc = =0 0316 324 8 8 58 64 13585. . . . k
11. Check if the region requires transverse reinforcement (5.8.2.4)
0.5(Vc + Vp) = (0.5)(0.9)(135.85 + 44.14) = 81.00 k Since Vu, 377.0
k, is greater than this value, transverse reinforcement is
required.
Example 1, July 2005 DRAFT 37
-
12. Calculate the shear resistance that the transverse
reinforcement needs to provide, Vs
Vs = =377 00 9 13585 4414 238 90.
.. . . k
13. Calculate the required transverse reinforcement spacing
( )( )( )( )
s = =0 40 60 58 64 214238 90
150. . cot .
.. in
14. Calculate the maximum allowable spacing of the transverse
reinforcement (5.8.2.7) ( )( )0125 0125 8 1000. . .'fc = = ksi
Since v fcu 0.799 ksi= < 0125. ' s dvmax . . 08 24 0 in 0.8dv
= (0.8)(58.64) = 46.9 in > 24.0 in Therefore, the maximum
spacing is 24 inches. Use 2 No. 4 bars at the 12-inch spacing used
in the exterior beam.
13. Check if the minimum amount of transverse reinforcement is
provided (5.8.2.5) Since using the same transverse reinforcement
spacing as the exterior beam, the
minimum transverse reinforcement provision is satisfied. 14.
Check the nominal shear resistance (5.8.3.3)
( )( )( )( )
Vs = =0 40 60 58 64 21412 299 26. . cot .
. k
( )( )( )( )0 25 0 25 8 8 58 64 4414 982 38. . . .'f b d Vc v v
p+ = + = k. Vc + Vs + Vp = 135.85 + 299.26 + 44.14 = 479.25 k The
lesser of these two values is 479.25 k and the factored shear
resistance is: Vr = Vn = (0.9)(479.25) = 431.33 k > 377.0 k O.K.
15. Check the longitudinal reinforcement requirement (5.8.3.5)
Consider the lack of full development of the prestressing steel
using the plot of the stress in the prestressing steel versus the
distance from the end of the strand. At the end of the transfer
length, 36 inches, the stress in the prestressing strand is fpe,
160.1 ksi. At the end of the development length, 151 inches, the
stress in the prestressing strand, fps, is 263.9 ksi. Using a
straight-line interpolation between the end of the transfer length
and the development length the stress in the prestressing strand at
68 inches from the end of the beam is approximately:
( )fps +
=1601
2639 1601151 36
68 36 189.. .
ksi
Calculate the tensile capacity provided by the longitudinal
reinforcement: Asfy + Apsfps = 0 + (6.510)(189) = 1230 k Calculate
the required tensile force where Vs is not to be greater than:
Vu = =
377 00 9
41889.
.. k
Example 1, July 2005 DRAFT 38
-
Since the nominal shear resistance provided, 299.26 k, is less
than this value, use the nominal shear resistance provided.
( )( )( )( ) ( )( ) ( )A f A fs y ps ps+ + +
=
1841 1258 64 10
0377 00 9
05 299 26 4414 214 951. .
..
. . . cot . k
Since 1230 k > 951 k O.K. At the inside edge of the bearing
area of simple end supports to the section of critical shear, the
longitudinal reinforcement on the flexural tension side of the
member shall satisfy:
A f A fV
V Vs y ps psu
s p+
05. cot
The inside edge of the elastomeric bearing is the sum of the
distance from the end of the beam to the centerline of bearing and
one-half of the bearing pad.
Table 8 Beam Shears, Centerline of Bearing (k) Dead Load
DC DW Beam Beam Slab/Diaph Rail FWS
Live Load plus Dynamic Load
Allowance Interior 67.8 68.4 8.3 13.5 117.4 Exterior 67.8 67.0
8.3 13.5 133.2
Exterior Beam inside edge of bearing area 1. Calculate the
effective shear depth, dv (5.8.2.9) Using the values from at the
centerline of bearing: de = dp = 14.56 + 35.62 + 9 = 59.18 in
Assuming rectangular behavior:
( )( )
( )( )( )( ) ( )( )c = +
+
=9114 270 0 0085 5 080 114 0 28 9114
2705918
616.
. . . ..
. in
Since the depth to the neutral axis is less than the slab
thickness, the assumed rectangular section behavior is correct. The
depth of the compression block is:
a = (0.80)(6.16) = 4.93 in The calculated shear depth is:
dv = =5918 4 932 56 72..
. in
But dv does not need to be less than the greater of: 0.9de =
(0.9)(59.18) = 53.26 in 0.72h = 58.32 in Since the calculated value
is less than one of these values, dv = 58.32 in 2. Calculate the
factored shear for the Strength I Limit State, Vu Using the shears
from at the centerline of bearing: Vu = (1.25)(67.8 + 67.0 + 8.3) +
(1.5)(13.5) + (1.75)(133.2) = 432.2 k
Example 1, July 2005 DRAFT 39
-
3. Calculate the prestress tendon shear component
( )fps = =
1036
16015 44 49. . ksi
Vp = (Aps)(fps)(sin) = (12)(0.217)(44.49)(0.1059) = 12.27 k
Since the inside edge of the bearing area is located within the
transfer length, the stress in
the prestressing strands is a proportion of the effective
prestressing stress. 4. Calculate the shear stress in the concrete
and the shear stress ratio (5.8.2.9) The shear stress in the
concrete is:
( )( )
( )( )( )vu = =432 2 0 9 12 27
0 9 8 58 321003
. . .. .
. ksi
The shear stress ratio is:
vf
u
c'
..= =1003
80125
5. Calculate the strain in reinforcement on the flexural tension
side of the member (5.8.3.4.2)
( )( )fpo = =0 7 270
1036
52 50. . ksi
Vudv = (432.2)(58.32) = 25206 k-in = 2100 k-ft Since this value
is greater than the actual applied factored moment, use this value.
Also,
since the inside edge of the bearing area is located within the
transfer length, the value of fpo is a proportion of the full value
of 189 ksi. Assume an initial value for of 23.7, which is based on
the shear stress ratio of 0.125 and an assumed value for x of
zero.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
2100 1258 32
0 05 432 2 12 27 237 6510 52 50
2 0 28500 651015325.
. . . cot . . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain from Table
5.8.3.4.2-1, 37.0. Since this does not agree with the initial
value, repeat.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
2100 1258 32
0 05 432 2 12 27 37 0 6510 52 50
2 0 28500 65100 9943.
. . . cot . . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain , 37.0. Since this
agrees with the previous value, continue.
6. Calculate the shear resistance provided by the transverse
reinforcement, Vs (5.8.3.3) Using the transverse reinforcement
spacing previously calculated, 12 inches:
( )( )( )( )
Vs = =0 40 60 58 32 37 012 154 79. . cot .
. ksi
7. Check the longitudinal reinforcement requirement (5.8.3.5)
Calculate the tensile capacity of the longitudinal reinforcement:
Asfy + Apsfps = 0 + (6.510)(44.49) = 290 k Calculate the required
tensile force where Vs is not to be greater than:
Example 1, July 2005 DRAFT 40
-
Vu = =
432 20 9
480 22.
.. k
Since the nominal shear resistance provided, 154.79 k, is less
than this value, use the nominal shear resistance provided.
( )( ) ( )A f A f V V Vs y ps ps u s p+ =
= 05
432 20 9
05 154 79 12 27 37 0 518. cot.
.. . . cot . k
However, 290 k < 518 k N.G. Try decreasing the transverse
reinforcement spacing so that the maximum value allowed
for Vs, 480.22 k, controls. The required tensile force is
now:
( )( ) ( )A f A f V V Vs y ps ps u s p+ = =
= 05
432 20 9
05 480 22 12 27 37 0 302. cot.
.. . . cot . k
However, 290 k < 302 k N.G. Add non-prestressed
reinforcement, No. 5 bars The basic development length for No. 5
bars is:
125.
'
A f
fb y
c
But is not less than: 0.4dbfy
( )( )( )125 125 0 31 60
88 22
. . ..
'
A f
fb y
c
= = in 0.4dbfy = (0.4)(0.625)(60) = 15 in The larger of these
two values is 15 inches and therefore, the development length is
15
inches. Since the inside edge of the bearing area is located
within the development length, 10 inches from the end of the beam,
consider the lack of full development. The tensile capacity
provided by the longitudinal reinforcement is now:
( )( ) ( )( )A f A f As y ps ps s+ = + =1015 60 6510 44 49 302.
. k Solving for the area of reinforcement required: As = 0.31 in2
Add 2 No. 5 bars, As = 0.62 in2
( )( ) ( )( )A f A fs y ps ps+ = + =
1015
0 62 60 6510 44 49 314. . . k
Since 314 k > 302 k O.K. Calculate the revised transverse
reinforcement spacing.
( )( )( )( )
s = =0 40 60 58 32 37 0480 22
387. . cot .
.. in
Use 3.75-inch transverse reinforcement spacing.
Example 1, July 2005 DRAFT 41
-
Interior Beam inside edge of bearing area 1. Calculate the
effective shear depth, dv (5.8.2.9) The effective shear depth for
the interior beam is the same as the exterior beam, same
strand pattern and section properties, 58.32 inches. 2.
Calculate the factored shear for the Strength I Limit State, Vu
Using the shears from at the centerline of bearing: Vu =
(1.25)(67.8 + 68.4 + 8.3) + (1.5)(13.5) + (1.75)(117.4) = 406.3 k
3. Calculate the prestress tendon shear component
( )fps = =
1036
160 05 44 46. . ksi
Vp = (Aps)(fps)(sin) = (12)(0.217)(44.46)(0.1059) = 12.26 k Once
again the stress in the prestressing strands is reduced because the
inside edge of the
bearing area is located within the transfer length. 4. Calculate
the shear stress in the concrete and the shear stress ratio
(5.8.2.9) The shear stress in the concrete is:
( )( )
( )( )( )vu = =406 3 0 9 12 26
0 9 8 58 320 941
. . .. .
. ksi
The shear stress ratio is:
vf
u
c'
..= =0 941
80118
5. Calculate the strain in reinforcement on the flexural tension
side of the member (5.8.3.4.2) Vudv = (406.3)(58.32) = 23695 k-in =
1975 k-ft Since this value is greater than the actual applied
factored moment, use this value. Once
again the value of fpo is reduced because the inside edge of the
bearing area is located within the transfer length. Assume an
initial value for of 23.7, which is based on the shear stress ratio
of 0.118 and an assumed value for x of zero.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
1975 1258 32
0 05 406 3 12 26 237 6510 52 50
2 0 28500 651013836.
. . . cot . . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain , 37.0. Since this
does not agree with the initial value, repeat.
( )( ) ( )( )( ) ( )( )( ) ( )( )[ ] x =
+ + + =
1975 1258 32
0 05 406 3 12 26 37 0 6510 52 50
2 0 28500 651008787.
. . . cot . . .
.. x 10 in / in-3
Use the shear stress ratio and x to obtain , 37.0. Since this
agrees with the previous value, continue.
6. Calculate the shear resistance provided by the transverse
reinforcement, Vs (5.8.3.3) Using the transverse reinforcement
spacing calculated for the interior beam, 3.75 inches:
Example 1, July 2005 DRAFT 42
-
( )( )( )( )
Vs = =0 40 60 58 32 37 0375 49532. . cot .
.. k
7. Check the longitudinal reinforcement requirement (5.8.3.5)
Calculate the tensile capacity provided by the longitudinal
reinforcement:
( )( ) ( )( )A f A fs y ps ps+ = + =
1015
0 62 60 6510 44 46 314. . . k
Calculate the required tensile force where Vs is not to be
greater than:
Vu = =
406 30 9
45144.
.. k
Since the nominal shear resistance provided, 495.32 k, is
greater than this value, use the limit on nominal shear
resistance.
( )( ) ( )A f A f V V Vs y ps ps u s p+ =
= 05
406 30 9
05 45144 12 26 37 0 283. cot.
.. . . cot . k
Since 314 k > 283 k O.K. INTERFACE SHEAR TRANSFER (5.8.4) For
composite prestressed concrete beams, interface shear transfer is
considered across the plane at the interface between two concrete
cast at different times. The cross-section area, Avf, of the
reinforcement crossing the shear plane is calculated per unit
length along the beam or girder. The maximum longitudinal spacing
of the rows of reinforcing bars is 24 inches. If the width of the
contact surface exceeds 48.0 inches, a minimum of four bars per
rows should be used (commentary provision). Evaluate interface
shear transfer at the critical section. Acv = the area of concrete
engaged in shear transfer (in2) Avf = the area of shear
reinforcement crossing the shear plane (in2) bv = the width of the
interface (in) c = the cohesion factor (ksi) = the friction factor
Pc = the permanent net compressive force normal to the shear plane
(k) de = the distance between the centroid of the steel on the
tension side of the beam to the center
of the compression block in the deck (in) = the specified 28-day
compressive strength of the weaker concrete (ksi) fc
'
Exterior Beam 0.044 Point 1. Calculate the factored vertical
shear force for the Strength I Limit State, composite loads, Vu Vu
= (1.25)(7.6) + (1.5)(12.3) + (1.75)(125.4) = 247.4 k 2. Calculate
the distance, de The distance from the bottom of the beam to the
centroid of the steel on the tension side
of the beam is:
( )( ) ( )( ) ( )( )
y = + + =10 in2 10 4 10 630
4 00.
Example 1, July 2005 DRAFT 43
-
The distance from the top of the deck slab to the center of the
compression block is:
a2
4 942
2 47= =. . in
d h t ya
e beam slab= + = + =2 72 9 4 00 2 47 74 53. . . in 3. Calculate
the horizontal shear per unit length (5.8.4.1)
The horizontal shear per unit length can be calculated
using:
VVdh
u
e=
Vh = =247 474 53 332.
.. k / in
VV
nh= = =
3320 9
369..
. k / in
4. Calculate the required transverse reinforcement spacing The
required transverse reinforcement spacing can be calculated
using:
( )
sA f PV cb
vf y c
n v= +
Avf = 0.40 in2 (2 No. 4 bars) For normal weight concrete placed
against clean, hardened concrete with surface
intentionally roughened to an amplitude of 0.25 inch: c = 0.100
ksi
= 1.0 = (1.0)(1.0) = 1.0 Pc = 0 k
( ) ( )( )[ ]
( )( )s =+
= 10 0 40 60 0369 0100 42
471. .. .
. in
Use 2 No. 4 bars at the 12-inch spacing used for web shear. The
area of transverse reinforcement provided is:
Avf = =0 4012 0 033.
. in / i2 n
cv
v
5. Check the upper limit on the amount of nominal shear
resistance used in design (5.8.4.1) The nominal shear resistance
used in design shall be the lesser of: V f An c 0 2. ' V A n c 08.
Acv = 42 in2
( )( )( )0 2 0 2 5 42 42 0. . .'f Ac cv = = k / in ( )( )08 08
42 336. . .Acv = = k / in The least of these two values is 33.6 k /
in and the nominal resistance of the interface
plane is:
Example 1, July 2005 DRAFT 44
-
( ) ( )( ) ( ) ( )( )[ ]V cA A f Pn cv vf y c= + + = + + = 0100
42 10 0 033 60 0 618. . . . k / in
c
6.18 k / in < 33.6 k / in
The factored shear resistance is: Vr = Vn = (0.9)(6.18) = 5.56 k
/ in > 3.32 k / in O.K. 6. Check the minimum reinforcement
requirements (5.8.4.1) The cross-sectional area, Avf, of the
reinforcement per unit length of the beam or girder
should satisfy that required by the following: ( )V cA A f Pn cv
vf y= + + A
bfvf
v
y 0 05.
The minimum reinforcement requirements may be waived if:
VA
n
cv< 0100. ksi
+n
For the first requirement, set Vn equal to the applied nominal
horizontal shear and solve for Avf.
( )( ) ( ) ( )( )[ ]369 0100 42 10 60 0. . .= + Avf Avf = 0 009.
in / i2 For the second requirement:
( )( )0 05 0 05 42
600 035
. ..
bf
v
y= = in / in2
However:
VA
n
cv= =369
420 088
.. ksi
Since this is less than 0.100 ksi, the minimum reinforcement
requirements m