Design Report of University Ave Garage - WordPress.com

determinedenchiladaUrban and Civil

Nov 25, 2013 (3 years and 11 months ago)

64 views



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
=

=
4MK1TS
=
1RKMSS
=

=
PPK4U
=
1OKRRR
=

=
OSKTU4
=
1MKM44
=

=
OMKMUU
=
TKRPP
=

=
1PKPVO
=
RKMOO
=

=
SKSVS
=
OKR11
=

=
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
Vc-min
Vc-max
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.