CIE 536
Design
of
University Ave. Parking Garage
By
Kalpesh Parikh
Acknowledgment
Our deepest g
ratitude goes to Prof Riyad S.
Aboutah for his c
ontinuous and constructive
advic
e and
follow up. His successive advisories and comments were the pillars in our every step during the
Project
.
We are thankful to him for the fact that he has inspired and helped us know about the
Prestressed
Concrete Structural System.
Introduction
Location
The University Ave. Parking Garage building is located in Syracuse University Campus and it is surrounded by
Harrison St., University ave , Walnut Place and East Adams St. It is a Ideal place because in its vicinity University
offices and College is located. From planning point of view it is a best place for a Parking.
Advantages
In recent decades, precast prestressed concrete (PC) structures have become more and
more prevalent in the
construction industries. A precast
concrete structure is composed of
Individual prefabricated members with different
types of connections.
Prestressed concrete structures have better corrosion resistance than reinforced concrete
structures.
Structural Aspects
Generally for the parking gar
agesection which are ideally used are
Double Tee Section
and
Tee Girder
below is
the Snap shot for how it looks.
Figure Showing 3

D View of Double tee and Inverted Tee Section
Structural System
and Materials
Double Tee Section
Figure Showing
15LDT26
The double tee girder flange is designed to carry the load of the cast

in

place slab, the flange weight and a 50 lb./sq.
ft. construction dead load. (or a 300
0
lb. concentr
ated construction load per IBC)
.
When determining the section properties of
the double tee girder the center of gravity is based upon the total area
(excluding corner fillets) of the entire exterior or interior girder and not on the individual stems. Moment of inertia
and dead loads to each stem are figured to the center of the f
lange between the stems. The non

composite dead load
of the external double tee girder is divided according to the following sketch.
Figure showing different member stem and flange
Inverted Tee Girder
Figure Showing 28IT44
The Section was decided
on the basis of the total depth
of the 26” (double tee ) and bearing pad which is should be
l/180 or 3 inch whichever is maximum so in our case it is coming out to be 3 inch so on that basis section was
examine for various check.
The members are like flang
e and web. The main purpose of providing the inverted tee
that Double tee section easily sits on to it. Loads coming into
the
tee girder as shown below.
Materials
Light Weight Concrete
Light weight concrete is preferred as a concrete
materials for our design purpose.
The majority of regular concrete produced is in the density range of 150 pounds per cubic foot (pcf). The last decade
has seen great strides in the realm of dense concrete and fantastic compressive strengths (up to 20,000
psi) which
mix designers have achieved. Yet regular concrete has some drawbacks. It is heavy, hard to work with, and after it
sets, one cannot cut or nail into it without some difficulty or use of special tools. Some complaints about it include
the percep
tion that it is cold and damp. Still, it is a remarkable building material

fluid, strong, relatively cheap, and
environmentally innocuous. And, it is available in almost every part of the world.
Regular concrete with microscopic air bubbles added up to 7
% is called air entrained concrete. It is generally used
for increasing the workability of wet concrete and reducing the freeze

thaw damage by making it less permeable to
water absorption. Conventional air entrainment admixtures, while providing relatively
stable air in small quantities,
have a limited range of application and aren't well suited for specialty lightweight mix designs.
Lightweight concrete begins in the density range of less than 120 pcf. It has traditionally been made using such
aggregates a
s expanded shale, clay, vermiculite, pumice, and scoria among others. Each have their peculiarities in
handling, especially the volcanic aggregates which need careful moisture monitoring and are difficult to pump.
Decreasing the weight and density produces
significant changes which improves many properties of concrete, both
in placement and application. Although this has been accomplished primarily through the use of lightweight
aggregates, since 1928 various preformed foams have been added to mixes, furthe
r reducing weight. The very
lightest mixes (from 20 to 60 pcf) are often made using only foam as the aggregate, and are referred to as cellular
concrete. The entrapped air takes the form of small, macroscopic, spherically shaped bubbles uniformly dispersed
in
the concrete mix. Today foams are available which have a high degree of compatibility with many of the admixtures
currently used in modern concrete mix designs. Gecko Stone of Hawaii is currently experimenting with one such
foam.
Foam used with either
lightweight aggregates and/or admixtures such as fly ash, silica fume, synthetic fiber
reinforcement, and high range water reducers (aka superplasticizers), has produced a new hybrid of concrete called
lightweight composite concrete, or LWC.
.
http://www.geckostone.com/lwc.html
Steel for Prestressing
Here we have chosen high strength
steel that is s
Strands (group of twisted wires)
.
Low relaxation Strands are
chosen because it has good stress

strain res
ponse of seven wire strands
.
Structural welded wire reinforcement
(WWR)
is selected for the top flange of the double tee .It has good ultimate strength (
250 ksi.
) Wires are used at
the top of the flange for double tee section.
Bearing Pads
Bearing pad selected chloroprene (Neoprene) purpose is to absorb the energy produced by moving vehicle .
Neoprene allows beam rotation at the bearing point because of deflection.
Neoprene Bearing Pads, It is manufacture
them in our purpose built plant usin
g high

grade virgin Neoprene Polychloroprene Grade

2 elastomer

mixed with
carbon black and rubber chemicals to form into a rubber compound to match AASHTO M 251 Specification under
controlled heat and pressure.
Properties
1
Having a minimum durometer
hardness of 50 and utilizing 100% chloroprene as the elastomer.
2 Short and Long

Duration Compression Test up to 4200 kN with horizontal shear up to 400kN can be carried
out.
Bearing size is decided on the basis of the minimum in place distance from th
e face of the support to the end of the
double tee in the direction of the span should be (length of Girder)/180 where l is the clear span , butit should be
greater than 3”
http://pretread.com/bridgebearing.html
Summary of the Structural System
1 Girder (Length = 30 ft) selected Inverted tee section
28IT44
2
Double Tee Section (Length =60 ft) selected Section
15LDT26
3
Bearing Pad
(Neoprene)
Selected =
3 inch
Durability and Green engineering aspect in the design
In the Design we can con
sidered green engineering from the point of view of Materials used for designing that is
Steel and Concrete. So main focus on Concrete In today world sustainability is in talk and
lot of structures is
now
being used as green engineering .From the Point of
view Green Engineering in the materials now Prestressed
Structure are concentrated on the Concrete Mixed which are used.
As we know, c
oncrete which is used in order to achieve green engineering the cementitious materials such as use of
Fly ash or Natural P
ozzalano use of Ground Granulated Blast
furnace
and use of Silica Fume. Blast
Furnace
and Fly
ash which are waste can be used as
cementitious materials
achieve the strength required.
The mixing of cementious
materials depends on the concrete strength requi
red. Selection of use of cementitious
materials in the concrete mix,
use of fly ash or silica fumes would may affect the color of the finished concrete but from the stand point of the
strength and durability and optimized use of waste can be made and some
saving in the cost can be added. Apart
from that Carbon dioxide can be reduce which is helpful from the enviormental point of view.
Apart from that we are using Light weight concrete which gives more prestress losses but are beneficial as
a
concrete
materials.
The Precast also contains rebars , the rebars can be made from recycle steel which we can use
as a Green materials.
Design a Double

Tee Precast
–
Pretensioned
Analysis and Design of Double Tee Section
A 60 feet Span Simply Supported Light Weight concrete double tee beam is subjected to un factored live load of 50
psf as it is pre

topped there is 2 inch topping is already considered into the flange. In order to design the double tee
we have selected the
section and procedure are carried and check are made for that it satisfies the condition.
Double Tee Section
Load Analysis
:
Live Load = 50 psf
Span Length = 60 ft
Load Combination
Total Load = 1.2 Dead Load + 1.6 Live Load
Assuming a trial Section
15 LDT
26
Material Properties
Concrete: Light weight Concrete
f’
ci
= Compressive Strength of the concrete at the initial prestress
f’
ci
= 4500 psi
f’
c
= Specified Strength of the concrete.
f’
c
= 8000 psi
w= Density of the concrete =115 lb/ft
3
Prestressing Steel
½” φ seven wire strands
Low Relaxation Grade 270
f
pu
= Specified Tensile strength of prestressing tendons = 270 ksi
Section Properties
Section
15 LDT26
A
c
= 7.486 ft
2
I = 53280 in
4
Y
b
= 19.82 in
Y
t
= 6.18 in
S
b
= 2688 in
3
S
t
= 8618 in
3
Self Wt = 861 plf
V/
S =2.38 in
Number of Strands Provided = 20
All strands are straight
Figure Showing Pretopped Double Tee Cross Section
Dead Load = 0.861 k/ft
Live Load = 50x 15/1000 =0.75 kip/ft
Total Load = 1.2 Dead Load + 1.6 Live Load
Total Load = 2.2332 kip/ft
Calculating Stresses and Losses as Stage Progress
Stages
1. At Transfer
2. After Transfer
Type of Prestressing
1.
Bonded Member Pretensioned.
Losses Stages
Type of Prestressing
Before Transfer
At Transfer
After Transfer
Pre tensioned
Relaxtion losses
Elastic
Shortening
Relaxation Losses
Creep Losses
Shrinkage Losses
Permissible Stresses for the Prestressed concrete flextural member
Description
Stresses At Transfer
Stresses Under Service Load
Extreme Fibre in Compression

0.7 f’
ci

0.45f’
c
Extreme Fibre in Tension
6√f’
c
12√f
c
Calculation of Stresses At Transfer
Figure Showing Beam with its own self weight
M@B =(25.83 x 2.5)
–
(0.861x 0.5 x 2.5
2
) =61.88 kip/ft
M@centre = wl
2
/8 = 387.45 kip/ft
At Transfer we know tension is at the top and
compression is at the bottom. Since we do not know the eccentricity at
this time but we know the Pre

stressed force and allowable stress at extreme fibers, we can easily calculate that.
f
top
= 6√f’
c
= 402.5 psi
f
bottom
=

0.7 f’
ci
=

3150 psi
Fi = Aps*fps
= 619.65k
f
top
=
f
top
=
e = 14.75”
f
bottom
=
e=12.4”
Minimum is taken for calculation ahead for losses.
Calculate Losses:
Elastic
Shortening:
E.S = Kes*Es
Where: Kes=1 (Pre tensioned)
Calculation of young modulus E
c
and E
ci
E
c
= (40000√f’
c
+10
6
) x
)
1.5
=3.23 x 10
6
psi
E
ci
= (40000√f’
ci
+10
6
) x
)
1.5
=2.601 x 10
6
psi
f
cir
= K
cir
(
f
cir
= 0.9(
= 1.96 ksi
The elastic Shortening losses will be: E.S =
= 21.6 ksi
Stress Level at After transfer
Steel initial Stresses = 202.5 ksi
Elastic Shortening and Relaxation = 21.6 ksi
Net Stress after transfer = 202.5
–
21.6 = 180.9 ksi
f
ps
= 180.9 ksi
F
i
= f
ps
* A
ps
= 555.19 kips
Since we revised the Fi value by accounted stress losses so we need to calculate new “e” value with this force b
y
keeping the same max allowable stress at extreme ends.So the new “e” is calculated below:
Now calculating Stresses at top and bottom
f
top
=
0.4025=
e = 15.75”
f
bottom
=
e=14.13”
Minimum is taken for calculation ahead for losses.
I have assumed “e= 14.13” but the value of Fi reduced to Fi=555.19 k so, the max stresses at extreme fiber ends are
under allowable limit state as per PCI code
so STRESSES are………O.K
Stage 2 : Under Service Load (Due to Double Tee beam)
Fi =555.19 kips
e = 14.13”
Calculate losses:
1. Creep loss
C.R=Kcr
(f
cir
–
f
cds
)
Where: f
cir
= Kcir(
f
cir
=(
= 2.398 ksi
f
cds
= 0
The creep loss will be C.R=
1.6
(2.398) = 33.26 ksi
2. Shrinkage loss
S.H = 8.2x10
6
Ksh*Es (1

0.06
)(100

R.H)
Where: Ksh = 1
V/S=2.38
R.H = 75
The shrinkage loss will be =
5.9 Ksi
3. Steel relaxation
R.E = [Kre

J(S.H+C.R+E.S)]c
Where: Kre=5000psi
J= 0.04………….Table4.5.1 (PCI)
C= 1…………….Table4.5.2 (PCI)
The relaxation loss will be = [5

0.04(5.9+33.26+21.6)]*1 = 4.756 ksi
f
T
= Total loss = C.R+R.E+S.H = 33.26+4.756+5.9= 43.916 ksi
Stress left in strands after losses = f
pi
–
f
T
=180.9

43.916= 137.47 ksi
Revise Fi with new f = 3.06*137.47 = 419.17 ksi
Total Loss at all Stage= 65.516 ksi
% of Total Loss = 65.516/202.5 = 32.306 %
Check stresses under Service Load
Allowable stress at super imposed loads is
Compression side =

0.45fc’ =

0.45*8000 =

3.6 ksi
Tension side = 12
√
= 12
√
= 1.073 ksi
f
top
=
f
top
=
=

0.241 ksi <

3.6 ksi
f
bot
=
f
bottom
=
=

2.502 ksi< 1.073 ksi
Stresses at the top and bottom are coming safe when check with
the allowable stresses.
Summary of Stresses
Stresses Level at Various Stages
Steel Stress, ksi
Percentage
After Tensioning (0.75 f
pu
)
202.5
100
Elastic Shortening

21.6

10.67
Creep Loss

33.26

16.42
Shrinkage Loss

5.9

2.91
Relaxation Losses

4.756

2.34
Final net Stress
137.47
67.66
Checking for ultimate Flexural capacity
Mn
1.2
Check stress at Mu.
Ultimate moment calculation
–
Bonded tendons considering no reinforcement
Assume a< h
f
Check fse >0.5f
pu
satisfies the check
we can proceed by these below equation
f
ps
= f
pu
(
)
For Low Relaxation we have
Ѵ
p
= 0.28
f’
c
> 4000 psi where f’
c
= 8000 psi
β
1
= 0.85

0.05
(
)
= 0.65 which is ok
ρ
p
=
= (3.06)/(15*12*20.31) = 8.37 x 10

4
f
ps
= 269.99 ksi
T
ps
= A
ps
F
ps
= 3.06*269.99 =826.16 Kips
T
ps
= 826.16 Kips
C= T
ps
a=
=0.67
a< h
f
hence assumption is correct
M
n
= T
ps
=
16502.54
M
n
= 1375.21 Kips

ft
Figure Showing Total factored load on the beam
Mmax=
wl
2
/8 = 1004.4 kips

ft
>M
u……………………..
Hence ok
Cracking Moment
F
r
= 7.5√f’c = 670.82 psi
M
cr
= f
(
)
F = 419.17 Kips
e=14.13 inch
Mcr =730.96 kips

ft
=
=
1.69>1.2 ………………………………………………….ok
Now Checking Flexural Strength for Flange part
Assume Ast WWR in Precast = W6 @ 6 inch = 0.12 in
2
/ft…………..From Design Aid 11.21
The cantilever flange controls the design since the negative moment over the stem reduces the
moment between the
stems. Construct the strain diagram.
Try C= 0.25 inch
=
=
=2.14 *10^

3
=0.033
Check,
<
........... Hence ok
a=
= 0.1625
C= 0.85 f’
c
b a=13.26 kips/ft
T=A
st
F
y
=0.12 *60 =7.2 kips/ft <13.26 kip/ft
Therefore try,
a=
=0.088”
C=
=0.135
Check,
= 0.064 >0.00214…… Hence ok
Therefore reinforcement yield and analysis is valid
= 0.135 < 0.375 from figure 4.2.1.3 PCI handbook
Therefore tension controls, and
Check minimum and maximum reinforcement
d=3 inch
b=12 inch
A
smin
=
√
=0.16 inch
2
Provide minimum steel reinforcement
As provided
W8
0.319 in
2
@6 inch Spacing
Calculation of development length
L
d
=(f
se
/3)d
b
+ (f
ps

f
se
)d
b
=17.5”
Calculate allowable load,
flange self weight = 4*115/12 =38.33
w
d
(flange section) = 38.33 *12*3.375
2
*0.5
=261.96 lb

ft/ft
M
l
= 785.83

261.96 =523.873 lb

ft/ft.
Calculation of Live load and checking with the actual live load
W
l
=
=344.93 psf < 50 psf ………… Hence safe
Shear Design
The shear design of precast concrete members is designed according to new provision of ACI 318

08.The shear
resistance of precast concrete members must meet the requirement:
Where: Vn = Vc+Vs
Vc = nominal shear strength of concrete
Vs = nomina
l shear strength of shear reinforcement
= 0.75
Since f
se
(137.47ksi)
40% f
pu
(108ksi),used below formula to calculate shear at each point along longitudinal length.
Vc = [0.6
√
+700(
)]b
w
dp…………11.9 ACI Code
Where:
1
; 2
√
b
w
dp
5
√
b
w
dp
b
w
= 16.25 inch
dp >=0.8h so use dp=0.8*h=20.3 inch
Figure Showing Total factored load
80
60
40
20
0
20
40
60
80
0
3
6
9
12
15
18
21
24
27
30
33
36
39
42
45
48
51
54
57
Shear Force in kip
Span Length in ft
Vu
Vu
0
5000
10000
15000
0
3
6
9
12
15
18
21
24
27
30
33
36
39
42
45
48
51
54
57
60
BMD in Kip

inch
Span Length in ft
Mu
Mu
Length
Vu
Mu
Vc
Check For
(Vu*dp/Mu)<1
Vc min
Vc Max
Vu/φ
0
66.96
0
0
0
45.34746
113.3686
89.28
3
60.264
2290.032
2.290032
0.547368421
45.34746
113.3686
80.352
6
53.568
4339.008
4.339008
0.256790123
45.34746
113.3686
71.424
9
46.872
6146.928
6.146928
0.158605664
45.34746
113.3686
62.496
12
40.176
7713.792
7.713792
0.108333333
45.34746
113.3686
53.568
15
33.48
9039.6
9.0396
0.077037037
45.34746
113.3686
44.64
18
26.784
10124.35
10.124352
0.055026455
45.34746
113.3686
35.712
21
20.088
10968.05
10.968048
0.038095238
45.34746
113.3686
26.784
24
13.392
11570.69
11.570688
0.024074074
45.34746
113.3686
17.856
27
6.696
11932.27
11.932272
0.011672278
45.34746
113.3686
8.928
30
0
12052.8
12.0528
0
45.34746
113.3686
0
0
50
100
150
200
250
300
0
5
10
15
20
25
30
35
V Shear Force
Length
Web Steel Envelope for Uniformly Loaded
Prestressed Girder
Vc
Vcmin
Vcmax
Vu/
φ
Check for Shear Reinforcement
Length
Vu
φVc/2
Check
0
66.96
93.75
Shear
Reinforcement
is required as
Vu>φVc/2 as Per
mCf=Cod攠
=
P
=
SMKOS4
=
OOKRVV
=
S
=
RPKRSU
=
OMKMUU
=
V
=
4SKUTO
=
1TKRTT
=
ㄲ
=
4MK1TS
=
1RKMSS
=
ㄵ
=
PPK4U
=
1OKRRR
=
ㄸ
=
OSKTU4
=
1MKM44
=
㈱
=
OMKMUU
=
TKRPP
=
㈴
=
1PKPVO
=
RKMOO
=
㈷
=
SKSVS
=
OKR11
=
㌰
=
M
=
M
=
Table Showing Area Check which is
required to decide on the stirrup
Length
Vs
Smax
Av
Av min=
50bw S/Fy
A 1min=
A2 min
Comment
0

160.72
19.5

1.0864672
0.2640625
0.000177504
0.35427702
Av min=
50bw S/Fy for
reinforced
3

62.759606
19.5

0.424254937
0.2640625
0.000177504
0.35427702
6

2.936780785
19.5

0.019852638
0.2640625
0.000177504
0.35427702
9
11.36566221
19.5
0.076831877
0.2640625
0.000177504
0.35427702
12
14.33209576
19.5
0.096884967
0.2640625
0.000177504
0.35427702
15
12.80879946
19.5
0.086587484
0.2640625
0.000177504
0.35427702
18
9.088503166
19.5
0.061438281
0.2640625
0.000177504
0.35427702
21
4.166429092
19.5
0.028165061
0.2640625
0.000177504
0.35427702
24

1.444163501
19.5

0.009762545
0.2640625
0.000177504
0.35427702
27

7.43789863
19.5

0.050280195
0.2640625
0.000177504
0.35427702
30

13.60423758
19.5

0.091964646
0.2640625
0.000177504
0.35427702
* A2 min= (0.75√fc'bw.S/Fy)
* A1 min= (Aps fpu S/80 Fy d)*√(d/bw)
Table showing at each Station Stirrups Provided and actual area of the stirrups
Length
Av Provided
Stirrups Designation
Spacing in
inch
0
0.395061728
#4

U 2 legged
19.5
3
0.395061728
#4

U 2 legged
19.5
6
0.395061728
#4

U 2 legged
19.5
9
0.395061728
#4

U 2 legged
19.5
12
0.395061728
#4

U 2 legged
19.5
15
0.395061728
#4

U 2
legged
19.5
18
0.395061728
#4

U 2 legged
19.5
21
0.395061728
#4

U 2 legged
19.5
24
0.395061728
#4

U 2 legged
19.5
27
0.395061728
#4

U 2 legged
19.5
30
0.395061728
#4

U 2 legged
19.5
Deflection Check
Most precast, pre stressed concrete flexural members will have a upward camber at the of pre

stress, caused by the
eccentricity of the pre

stressing force. This camber may increase of decrease with time, depending upon the stress
distribution across the me
mber under sustained load, superimposed dead load even during the time of erection. So
must need to calculate the possible deflection would at these stages and all should be under the maximum
permissible deflection according to the PCI handbook.
At Transf
er
Initial , E
ci
= 2.601 x 10
6
psi
Before 28 days E
c
= 3.23 x 10
6
psi
Due to initial Prestress only,
∂=
) + (
)
e
e
=14.75” , I=53280 in
4
P
i
= 619.65 Kips
∂
I
=
=

4.2 “
Self weight intensity = w = 0.861 k/ft = 861/12 lb/in
Self weight ∂
d
=
…………..For Un cracked Section.
∂
d
=
=1.81”
Thus the net camber at transfer =

4.201+ 1.81 =

2.39 inch (↑)
Immediate Se
rvice Load deflection
f
r
= 7.5*
√
= 670.82 psi
f
bottom
under service load =

2.052 ksi
d
p
= 20.31”
ρ
p
=
= (3.06)/(15*12*20.31) = 8.37 x 10

4
n=
= 8.67
Icr = n Aps d
2
p(1

1.6
√
ρ
= 9452 in
4
Moment, Mcr
due to that portion of live load that causes cracking are
Mcr = S
b
(7.5
√
=2842.82 kips

ft
Ma= Unfactored maximum live load moment
Mmax = 337.5 k

ft
=
=8.42
hence we can say use Ig
for check of the deflection
∂
l
=
Wu
l
= 50*15 =0.75 k/ft =750/12 =6.25
∂
l
=
= 1.27” ↓
Long Term Deflection (camber) by PCI multipliers
Load
Transfer
∂
p
inch
PCI Multiplier
∂ (inch) final
Prestress
Camber at
Transfer

4.2
1.8

7.56
Deflection due to own
weight
1.81
1.85
3.3485
Net

2.39↑
=
=
J
4.2115↑
=
䑥a汥捴楯n=du攠瑯=汩l攠汯慤
=
=
=
1.27↓
=
Final
∂

2.39↑

2.941↑
The Result are good because less loading
condition there is not much
deflection due to live load hence check for
allowable deflection.
Maximum permissible computed deflection PCI handbook pg 4

88 Table 4.8.1
l/480 = (60*12/480) = 1.5”
.
Design of Inverted Tee Girder
Loads Analysis:
Loads acting
over inverted tee beams are:

Concentrated loads acting due to the cumulative effect of Self weight of Double tee beam and uniformly
concentrated live load are:
Uniformly distributed live load = 50psf = 0.05k/ft
Width of slab = 15ft
Total live load = 15*
0.05 = 0.75k/ft
Factored Live load = 0.05*15*1.6 =
1.2k/ft
Density of TT = 115 lb/ft
3
Area of TT = 7.86ft
2
Self weight = 0.115*7.86 = 0.86k/ft
Factored self weight = 1.2*0.86 =
1.032k/ft
Therefore, the total load coming from Double tee = 1.2 + 1.032 =
2.232k/ft
Reaction =
= 33.48 k
Figure Showing Load coming from Double tee beam
Self weight of Inverted Tee Beam
Area = 5.44ft
2
Density = 115 lb/ft
3
Self weight = 0.115*5.44 = 0.626k/ft
Factored self weight =
0.75k/ft
Figure Showing Load on
Inverted
tee beam
Section Selected
:

28IT44 (PCI handbook Pg 2

45)
Section Properties:
Designation
h (in)
h1/h2
A(in)
I(in)
Y
b
S
b
S
t
No.Strands
28IT44
44
28/16
784
124437
17.43
7139
4683
20

Straight
Materials Properties:
Pre

stressing steel
Concrete (Light weight concrete)
f
pu
=270ksi
f
c
’= 8000psi
½” diameter
f’
ci
= 4500psi
Low

relaxation strands
w = 115lb/ft
3
f
pi
= 0.75f
pu
= 202.5ksi
Eci’= 377280ksf
Es = 28000ksi
Ec’ = 465120ksf
Aps= 20*0.153=3.06 in
2
Check Stresses At transfer:
Since we do not know the eccent
ricity at this time but we know the Pre

stressed force and allowable stress at
extreme fibers, we can easily calculate that.
Allowable stress at super imposed loads
are
Compression side = 0.7fci’ = 0.7*4500 =

453.6ksf
Tension side = 6
√
= 6
√
=
+57.95ksf
Fi = Aps*fps = 619.65k
f
top
=
(Under Tension)
f
top
=
e = 9.5”
With f
bot
(Under compression)I was getting e = 27.27” but we supposed to considered the small value.
Calculate Losses
Elastic Shortening
E.S = Kes*Es
Where: Kes=1 (Pre tensioned)
f
cir
= Kcir(
f
cir
= 0.9(
= 165.15 ksf
The elastic Shortening losses will be: E.S =
= 1764.96 ksf
=
12.25
ksi
Steel relaxation
R.E = [Kre

J(S.H+C.R+E.S)]c
Where: Kre=5000psi
J= 0.04………….Table4.5.1 (PCI)
C= 1…………….Table4.5.2 (PCI)
The relaxation loss will be = [5000

0.04(0+0+12256.6)]*1 = 4509.73psi =
4.51
ksi
f
T
= Total loss = E.S+R.E = 12.25+4.51=
16.76ksi
Stress left in strands after losses = f
pi
–
f
T
=202.5

16.76
= 185.74ksi
Revise Fi with new f = 3.06*185.74=
568.36k
Since we revised the Fi value by accounted stress losses so we need to calculate new “e” v
alue with this force by
keeping the same max allowable stress at extreme ends.So the new “e” is calculated below:
Fi = Aps*fps = 568.36k
f
top
=
f
top
=
e = 9.85” > 9.5 (old value)
But I in further calculation I have assumed “e= 9.5” but keep Fi=568.36k
Since I have assumed “e= 9.5” but the value of Fi reduced to Fi=568.36k so, the max stresses at extreme fiber ends
are under allowable limit state as per PCI code so STRESSES are………O
.K
At Superimposed dead load
(Due to Double Tee beam)
Fi =568.36k
e = 9.5in
Calculate losses:
Creep loss
C.R=Kcr
(f
cir
–
f
cds
)
Where: f
cir
= Kcir(
f
cir
=0.9(
= 152.76 ksf
f
cds
=
=
0.792 = 52.85ksf = 0.367ksi
Msd is only due to the self wt of Double tee beam, there will be no live load considered
The creep loss will be C.R=
1.6
(1.107
–
0.367) = 9.02 ksi
Shrinkage loss
S.H = 8.2x10
6
Ksh*Es (1

0.06
)(100

R.H)
Where: Ksh = 1
V/S=5.44
R.H = 75
The shrinkage loss will be =
8.2x10
6
*1*28000 (1

0.06*5.44) (100

75) = 3.866 ksi
S
teel relaxation
R.E = [Kre

J(S.H+C.R+E.S)]c
Where: Kre=5000psi
J= 0.04………….Table4.5.1 (PCI)
C= 1…………….Table4.5.2 (PCI)
The relaxation loss will be = [5000

0.04(9.01+3.866+0)*1000]*1 = 4484.96psi =
4.5
ksi
f
T
= Total loss = C.R+R.E+S.H = 9.02+3.866+4.5=
17.386ksi
Stress left in strands after losses = f
pi
–
f
T
=185.74

17.386
= 168.35ksi
Revise Fi
with new f = 3.06*168.35.74=
515.16k
Check stresses
Allowable stress at super imposed loads is
Compression side = 0.45fc’ = 0.45*8000 =

518.4ksf
Tension side = 12
√
= 12
√
= +154.56ksf
f
top
=
(Under compression)
f
top
=
=

94.7ksf <

518.4ksf
f
bot
=
(Under compression)
f
top
=
=

76.21ksf <

518.4ksf
Stresses at the top and bottom are
coming safe……………O.K
Check ultimate flexural strength
In this section our two main objectives are:
Mn
1.2
Ultimate flexural strength
Estimate the steel stress at ultimate by f
ps
equation which is valid to use since fse=168.35ksi >
0.5fpu=135 ksi.
p
=
=
= 0.00707
f
ps
= 270000 (1

0.5
p
=
270000 (1

0.5*0.00707*
= 266.778ksi
T’=A
ps
f
ps
= 3.06*266.778= 816.34
k
C=0.85*f
c
’b*a
But T’=C therefore
a=
= 10in
M
n
= T’(d

a/2) =
816.4(36.07

10/2) = 25363.684 k

in = 2113.64k

ft
Mn = 0.9x2113.64=
1902.27 k

ft
Mu= 147.41x15’

66.96x11’

66.96x3.5’

0.9x15’x7.5’=
1139.13k

ft
Hence
Mn
(1902.27 k

ft)
> Mu
(1139.13k

ft) ………..condition satisfied.
Cracking Moment
Cracking has to be
check at the tension side which is basically at the bottom side of the girder and it is calculated by
following equation:
Mcr = Fi*e +
=
515.16*0.792 +
=
1174.3k

ft
Therefore;
= 1.62 >
1.2……………….condition satisfied.
Check stress at Mu:
Allowable stresses at Mu are:
Compression side = 0.45fc’ = 0.45*8000 =

518.4ksf
Tension side = 12
√
= 12
√
= +154.56ksf
f
top
=
(Under compression)
f
top
=
=

364 ksf <

518.4ksf
f
bot
=
(Under Tension)
f
bot
=
= 82 ksf < 154.56 ksf
Stresses at the top and bottom are coming safe……………O.K
Stresses at each stage
At Transfer(ksf)
Stress at service load(ksf)
At Compression
0.7fci’ =
4RPKS
0.45fc’ =
R1UKPS
At Tension
6
√
=
57.95
12
√
=
145.26
Shear Design
The shear design of precast concrete members is designed according to new
provision of ACI 318

08.The shear
resistance of precast concrete members must meet the requirement:
Where: Vn = Vc+Vs
Vc = nominal shear strength of concrete
Vs = nominal shear strength of shear reinforcement
= 0.75
Since f
se
(168.35ksi)
40% f
pu
(108ksi),used below formula to calculate shear at each point along longitudinal length.
Vc = [0.6
√
+700(
)]b
w
d…………11.9 ACI Code
Where:
1
; 2
√
b
w
d
5
√
b
w
d
Figure Showing Shear and bending Moment calculated From Staad Pro
Table Showing Calculation of Shear Strength
Distance(ft)
Vu(k)
Vu/Ф
Vc(k)
Vc

min
Vc

max
Vs=Vc

Vu/Ф (k)
0
146.187
194.916
320.41
58.07
145.17

49.77
2.5
144.142
192.19
320.41
58.07
145.17

47.02
5
75.138
100.184
124.84
58.07
145.17
24.656
7.5
73.093
97.457
98.37
58.07
145.17
0.913
10
71.049
94.732
81.97
58.07
145.17

12.76
12.5
2.044
2.725
19.968
58.07
145.17
55.345
15
0
0
17.421
58.07
145.17
58.07
The above table shows the comparison between Vc, Vu, Vs. The
negative
values in the last column
showing that we need shear reinforcement Vs at respective stations along the length of girder.
50
0
50
100
150
200
250
300
350
400
0
2
4
6
8
10
12
14
16
Shear force(kips)
Distance along span (ft)
Distribution of shear along span
Vu/
Ф
Vc
Vcmin
Vcmax
Calculate Av at 0 and 2.5ft
Av =
Where: S
max
=
¾*h= ¾*44=33”<24”
Spacing
= 24”
Av =
= 0.55in
2
Provide #5@24” (Av=0.6in
2
)
Av =
Av =
= 0.55in
2
Provide #5@24” (Av=0.6in
2
)
The next shear reinforcement needed at 10’ from support so it’s better to provide the same shear reinforcing till
12’from support and since we do not need any reinforcing at mid span
(
as you can see in below diagram of SFD and
BMD
)
but I have provided mini
mum reinforcing (#3@24”) at the portion 3’ from centre of beam on each side.
Shear strength of ledge portion: The design shear strength of beam ledge ,supporting concentrated loads from stem
of double tee beam can be determined by the lesser of two equati
on below:
The above figure shows detailing of ledge part of inverted beam and it also shows the all parameters which has
been used in equations used below.
For s (24”)
>
b
t
+h
l
(16+7.25=23.25”)
=
√
h
l
[2(b
l

b)+ b
t+
h
l
]……………..4.5.1.1
PCI Hand
book
= 3x0.75x0.75
√
x16[2(20

12)+7.25+16]
= 94.78 Kips ……………gover
n
s
=
√
h
l
[2(b
l

b)+ b
t+
h
l
+2de]………….4.5.1.2
1
PCI Hand book
= 0.75x0.75
√
x16[2(20

12)+7.25+16+2x45] = 101.04 kips
So the the
= 94.78k
and Vu(
reaction from each stem) =
66.96k
Therefore,
is satisfied.
Design of Transverse (Cantilever) Bending of ledge
: Transverse bending of the ledge requires flexural
reinforcement ,As, which is computed by following equation .
As =
(
)
……………..4.5.2.1
=
(
)
= 0.564 in
2
Provide #5@16” (Av=0.6in
2
)
Note
: This reinforcement may be uniformly spaced over the width of 96in on either side of the bearing. Spacing
should not exceed h
l
(16”) or 18”.
Design of Longitudina
l Bending of Ledge:
Longitudinal reinforcing should be placed in both top and bottom of the
ledge portion of beam.
A
l
= 200(b
l

b)d
l
/f
y
= 200(20

12)14.19/60000 = 0.38
in
2
Provide 1

#4 top and 1

#4 bottom” (Av=0.4in
2
)
NOTE: I have considered the one side as ledge for analysis and design of shear reinforcing. Since we have ledge at
both sides because it is inverted tee beam, i have provided the same reinforcing for that.
Deflection and Camber Check
Most precast, pre s
tressed concrete flexural members will have a upward camber at the of pre

stress, caused by the
eccentricity of the pre

stressing force. This camber may increase of decrease with time, depending upon the stress
distribution across the member under sustaine
d load, superimposed dead load even during the time of erection. So
must need to calculate the possible deflection would at these stages and all should be under the maximum
permissible deflection according to the PCI handbook.
Deflection at Transfer of P
re

stressing force:
=
=
= 0.2683in (
Where Fi = Pre stressing at transfer
) =
=
= 0.042in (
Estimate Deflection at Transfer =

)
= 0.2683

0.042 =
0.2263in
At Erection: Using multipliers at erection from table……………
= 0.2683x1.85 = 0.496in
) = 0.042x1.8= 0.0756in (
Estimate Deflection at Erection= 0.496

0.0756=
0.4204in
At Final stage:
= 0.2683x2.45 = 0.657in (
=
0.425
(
) = 0.042x2.7 = 0.1134 (
Estimate Deflection at Erection= 0.657

0.425

0.1134=
0.1186in
Figure showing D
eflection due to only superimposed load is calculated in STAAD.Pro.
Stages
Deflection
Permissible
Condition
At Transfer
0.2263
L/480= 0.75in
Good
At Erection
0.4204
Good
At Final
0.1186
Good
Drawings
References
Books
ACI Code
International building Code
–
IBC Load calculation & Specification
James G Macgregor ,“Reinforced Concrete Mechanics and Design”, (2
nd
edition) by
Prentice hall, A Simon and Schuster company, Upper saddle river ,NJ
–
07458
PCI design handbook

6
th
Edition
Dr. Edward G Nawy, Prestressed Concrete a Fundamental approach

2
nd
Edition
T.Y.Lin, Design of Prestressed Concrete Structures.
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