Capstone project

Norwich University
Direct Connector between I-10 and Highway-99
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0
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Prepared by: Dawit Bogale Submitted to: Dave Mukerman
1919 lenora ct. katy Tx. 77493
8/12/2012

Capstone project 2012
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Direct Connector between I-10 and Highway 99
Table of Contents
Acknowledgement ….......................................................................................................I
Nomenclature.....………………………………………………………………….........II
List of Tables………………………………………………………………………......III
List of Figures……………………………………………………………………........ IV
Abstract…...................................................................................................................... VI
1. Introduction ........................................................................................................10
1.1 Background…………………………………………………………………...10
1.2 Objective ……………………………………………………………………...11
1.3 Thesis Content ………………………………………………………………..11
1.4 Application of the study and limitations………………………………….......11
2. Bridge Structure .................................................................................................13
2.1. General……………………………………………………………………......13
2.2. Precast pre-stressed concrete Bridge……………………………………........13
2.2.1. Description…………………………………………………………….13
2.3. Importance of Bridges in Architecture…………………………………..….14
2.4 Design task ....................…………………….. …………….………..………15
3. Structural Analysis and Design of Pre-stressed Concrete Bridges........................18
3.1. Bridge loading……………………………………………………………..…18
3.1.1. Dead Load.....……….…………... …………………….………………19
3.1.2. Live Load...................……………....…….…………………………….19

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3.1.3. Dynamic load allowance..............................……..........……………….20
3.1.4. Wind load..………................................………….…………………….20
3.1.5. Earth quake Load ………………………………………………………21
3.1.6 Earth pressure…........................................................................................21
3.2. Bridge design and analysis…………………………………………………...22
3.2.1Super-structure Analysis and Design……………………………….……22
3.2.1.1 Slab Analysis and Design………………………………………....24
3.2.1.2 Girder Analysis and Design……………………………………....33
3.2.2 Sub-structure Analysis and Design……………………………………...64
3.2.2.1 Abutment Analysis and Design.............. …………………………64
3.2.2.1.1 Design Abutment Heel…………….………………..…….82
3.2.2.1.2 Design Abutment Toe…………………………………….84
3.2.2.1.3 Design Abutment Stem ………………..…………………87
4. Project Management……………………….................................................................97
4.1. Project Scope Management…………………………………………………...97
4.1.1 Define Scope……………………………………………………………99
4.1.2 Create Work Break Down Structure……………………………………100
4.2. Project Time Management. ………..................................................................102
4.2.1 Define Activities, Sequence Activities, and Develop Schedule………..102
4.2.2 Schedule Changes and Thresholds….………………………………….104
4.2.3 Schedule Control…………….…………………………………………104
4.3. Project Cost Management................................................................................107
4.3.1 Estimate Cost ………………………………………………………….107

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4.3.2 Determine Budget………………………………………………………107
4.3.3 Project Cost and Schedule Performance………………………………..111
4.4. Project Quality Management …………………………………………………115
4.4.1Project quality measurement……………………………………………..115
4.5. Project Human Resource Management……………………………………….115
4.6. Project Communication Management………………………………………..116
4.7. Risk Management.............................................................................................116
4.8. Project Procurement Management……………………………………………117
5 Conclusions and Summary................................................................................................118
6. References…....................................................................................................................119
7. Appendix A......................................................................................................................120
8. Appendix B.......................................................................................................................129

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I. Acknowledgements
First of all, I would like to thank God for each and every success in my life. Then I would like to
extend my love and heart felt appreciation to Meskerem Eshete, Elias Manyazewal, and my
families not only for their encouragement but also for their being with me in all ups and downs.
My deepest gratitude goes to my thesis advisor Mr. Dave Mukerman, Mr. Nick Marianos, and all
the staff of Norwich University for their professional, genuine guidance and valuable advice to
accomplish the thesis.
At last, but not least, I would like to express my profound and special thanks to those engineers
who have collaborated during site visit.

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II. Nomenclature
Symbol Definition
⍺ Depth of equivalent rectangular stress block in concrete
As* Area of pre-stressing steel (in2
)
Ast Total area of longitudinal reinforcement (in2
)
A’s Area of compression reinforcement (in2
)
Av Area of shear reinforcement (in2
)
E Earth pressure (lbf/ft2
)
D Dead load (lbf)
B Width of footing (ft.)
ρ Ratio of steel reinforcement
ρ’ Ratio of compression reinforcement
ρmax Maximum permitted reinforcement ratio
β Coefficients applied to actual loads for service load and load factor designs
β1 Factor for concrete strength
I Moment of inertia (in4
)
I Live load impact (lbf)
Tf Temperature force due to friction at bearing (lbf).
FSo Factor of safety against overturning
FSs Factor of safety against sliding
g Centroid of pre-stressing strand pattern (in)
R Reaction (lbf)

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III. List of Tables
TABLES Page
Table 1 Section Properties of AASHTO Type IV Girder.........................................................36
Table 2 Properties of Composite Section..................................................................................38
Table 3 Summary of service loads and moments…………………..........................................77
Table 4 Stability and Bearing Pressures…………………………………................................77
Table 5 Factored Vertical Loads………………………………...............................................79
Table 6 Moment resulted from Factored Vertical Loads.........................................................79
Table 7 Factored Horizontal Loads .........................................................................................79
Table 8 Moment resulted from Factored Horizontal Loads………........................................80
Table 9 Factored Bearing Pressures…….................................................................................80
Table 10 Factored Vertical Loads……...................................................................................89
Table 11 Factored horizontal loads………………………………………………………….89
Table 12 Factored Moment at stem base resulting from vertical and horizontal Loads…….89
Table 13 Factored Vertical Loads……................................................................................. 90
Table 14 Factored Horizontal Loads………………………………………………………..90
Table 15 Factored Moment at stem base resulting from vertical and horizontal Loads…….90
Table 16 Project Budget…………………………………………………………………….108
Table 17 Project Cost Performance………………………………………………………….113
Table 18 Cash Flow………………………………………………………………………….129

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IV. List of Figure
Figure Page
Fig. 1 Project Location……………....................................................................................... 10
Fig. 2 Plan view of Bridge…….............................................................................................. 16
Fig. 3 Sectional view of Bridge……………………………………………………………...23
Fig. 4.Sectional view of slab, curb and parapet, and slab reinforcement detail….…………..32
Fig. 5 Section Geometry AASHTO Type IV Girder………………………………………...35
Fig. 6 Sectional view of Composite Section and properties…………………....................... 40
Fig. 7 Maximum live load shear……………………………………………………………..47
Fig. 8 Sectional view and strand arrangement of Girder…………………………………….52
Fig. 9 Sectional view of Abutment…………………………………………………………..65
Fig. 10 Sectional view of Live load and Earth pressure on Abutment………………………69
Fig. 11 Sectional view of pressure on Footing………………………………………………81
Fig. 12 Reinforcement detail of Abutment.............................................................................95
Fig. 13 Summary of Project Schedule…………….................................................................105
Fig. 14 Project Schedule…………..........................................................................................121

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V. Abstract
The existing signalized intersection and travel corridor between IH 10 and SH 99 is insufficient
to cycle the anticipated volume of traffic. Hence, the purpose of this project is to improve traffic
flow by providing a Direct Connector between IH 10 and SH 99. It connects northbound and
southbound movements along SH 99 from IH 10 and westbound and eastbound movements
along IH 10 from SH 99.
The scope of work for this project will focus both on structural engineering and Project
Management from initiation to execution of the proposed direct connectors. This project will
perform the design of Pre stressed Concrete IV Girder, Abutment, and Concrete deck slabs. The
design is guided by the American Association of State Highway and Transportation Officials
(AASHTO) Standard Bridge Design Specifications and Texas department of Transportation’s
manuals. The analysis will be performed using traditional method. In addition scheduling, Gantt
bar chart, cost estimation is performed using Microsoft project.
In conclusion, in addition to design and analysis of concrete slab, concrete girders, and abutment,
this project clearly demonstrates the advantage of using precast concrete products to construct
cost effective, complex long span structure where aesthetic and urban geometric are significant
design consideration.

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CHAPTER ONE:
INTRODUCTION

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1. Introduction
1.1. Background
The project is located at the intersection of State Highway (SH) 99 and Interstate Highway (IH)
10 in Harris County, Texas. These two high-speed highways intersect at a grade separated
interchange. The existing frontage roads provide northbound and southbound movements along
SH 99 from IH 10 and westbound and eastbound movements along IH 10 from SH 99. This
project would provide fully directional direct connectors (DCs) between these two facilities in
addition to the existing frontage road system. In order to connect between IH 10 and SH 99,
motorists are required to travel through at grade signalized intersection. The existing signalized
intersection and travel corridor is insufficient to cycle the anticipated volume of traffic. Hence,
there is a delay and interrupted connection that reduces local and regional transportation system
mobility [8]. Highway 99 N
I-10 W I-10 E
Highway 99 S
Fig. 1 Project Location

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1.2. Objective
The importance of this project is to improve traffic flow by providing a Direct Connector that
enhances transportation continuity regionally and locally. It helps to connect northbound and
southbound movements along SH 99 from IH 10 and westbound and eastbound movements
along IH 10 from SH 99. The main objective of the thesis is to develop analysis and design of
slab, girder, and abutment of concrete bridge.
1.3. Thesis Content
The study of the thesis mainly focuses on application of the knowledge gained through different
courses on design and analysis of bridges. Based on the main objective of the thesis, the study
has focused on developing analysis and design of pre-stressed concrete bridge between I-10 and
Highway 99 using AASHTO Standard Bridge Specification. In addition to this, the project
management concept like project time, cost, quality, risk, human resource, and procurement
management will be included. Beside the above objective, the thesis helps to promote the
importance of precast pre-stressed concrete bridge for beauty of the city.
1.4. Applications and Limitations
The study shall benefit the client, public, and constructors as these bridges would only take a
short time to produce and assemble. This enables the quick restoration of traffic. The scope of
the study has been limited to the preparation of analysis and design for slab, girder, and abutment
only. This research may be used as a basis for future study to include the design of other parts of
the bridge so that the whole bridge could be designed and analyzed.

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CHAPTER TWO:
BRIDGE STRUCTURE

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2. Bridge Structure
2.1. General
A bridge structure is divided into super structure and substructure. The superstructure (upper
part), which consists of the slab, the floor system, the girders, and the substructure (lower part),
which are piers, footings, piles and abutments. The super structure provides horizontal spans
such as deck and girders and carries the traffic loads directly where as the function of sub
structure is to support the superstructure of the bridge. The following factors are taken into
account and the type that is most economical and can give maximum service is designed. Some
of the factors considered to be the main criteria for the selection of type of bridge studied in the
thesis are:
 Existing structures.
 Availability of fund.
 Time available for construction of the bridge.
 Appearance of bridge from aesthetic point of view.
2.2. Pre-stressed Concrete Bridge
2.2.1. Description
Pre-stressed concrete structures are shallower in depth than concrete reinforced structure for the
same span and loading condition. In conventional reinforced concrete, the high tensile strength
of steel is combined with concrete's great compressive strength to form a structural material that
is strong in both compression and tension. The principle behind pre-stressed concrete is that
compressive stresses induced by high-strength steel tendons in a concrete member before loads
are applied will balance the tensile stresses imposed in the member during service. Pre-stressing

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removes a number of design limitations conventional concrete places on span and load and
permits the building of bridges with longer unsupported spans. This allows designing and
building lighter and shallower concrete structures without sacrificing strength. [2]
Pre-stressed concrete structure can be produced through pre-tensioning or post-tensioning. In
pre-tensioning, the steel is stretched before the concrete is placed. High-strength steel tendons
are placed between two abutments and stretched to percentage required of their ultimate strength.
Concrete is poured into molds around the tendons and allowed to cure. Once the concrete reaches
the required strength, the stretching forces are released. As the steel reacts to regain its original
length, the tensile stresses are translated into a compressive stress in the concrete. In post-
tensioning, the steel is stretched after the concrete hardens. Concrete is cast around, but not in
contact with steel. In many cases, ducts are formed in the concrete unit using thin walled steel
forms. Once the concrete has hardened to the required strength, the steel tendons are inserted and
stretched against the ends of the unit and anchored off externally, placing the concrete into
compression. [2]
2.3 Importance of Bridges in Archtecture
The fact that bridges last many generations and become symbols of a particular city makes them
special for their designers and users. During the design, sizes and aesthetic appearance are
chosen carefully because it shows the culture of the time and the ambition of the civilizations
which built them. Bridges have a significant place in human civilization. Hence, the design of
bridge is not as easy as the design of an ordinary structure. The success of the bridge comes not
only with its function but its visual beauty. Bridges are highly visible elements in city life. The
function, stability, stiffness and strength were the primary concerns of the designer. However,
with the advancement of civilization, aesthetic parameters started to be considered. When an

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engineer builds a bridge, he/she creates a visible object in the environment. The location and size
of the major structural features of the bridge, its piers, girders, and abutments, are its major
aesthetic features; because of their size, they have a major role in the aesthetic impact of the
bridge.
2.4 Design Task
In this capstone project, the task focuses both on structural engineering and Project Management
from initiation to execution of the proposed direct connectors.
Structural design and analysis: This project has precast pre-stressed Concrete girder, Concrete
deck slabs, and Abutment. It is designed using the American Association of State Highway and
Transportation Officials (AASHTO) Standard Bridge Design Specifications. Most of the data are
assumed by visiting the site.
Project Management: The project management concept like project time, cost, quality, risk,
human resource, and procurement management will be included.
The project Management section includes all connectors: Bridge on highway 99 bypass over I-10
and Direct Connector A, B, C, D, E, F, G, and H. However, the structural design and analysis
section includes only bridge on highway 99 bypass over I-10. Look Fig. 2 in the next page.

PRODUCED BY AN AUTODESK EDUCATIONAL PRODUCT
PRODUCEDBYANAUTODESKEDUCATIONALPRODUCT

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CHAPTER THREE:
STRUCTURAL ANALYSIS AND DESIGN
OF PRESTRESSED CONCRETE BRIDGE

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3. Structural Analysis and Design of pre-stressed concrete Bridges
3.1. Bridge Loading
The design of any component of bridge is based on a set of loading conditions which the
component must withstand. The various types of loading which need to be considered can
broadly be classified as permanent, or temporary.[1] Permanent loads are those due to the weight
of the structure itself and of any other immovable loads that are constant in magnitude and
permanently attached to the structure. They act on the bridge throughout its life. Temporary
loads are those loads that vary in position and magnitude and act on the bridge for short period of
time such as live loads, wind loads and water loads etc. [4]. Some of these are:
1. Permanent loads
 dead load of structure
 superimposed dead loads
2. Temporary loads
 vehicular live loads
 pedestrian live loads
 impact loads
 wind loads
 earth quake loads
In order to form a consistent basis for design, AASHTO has developed a set of standard loading
condition, which is taken in to account while designing a bridge. These loads are factored and
combined to produce extreme adverse effect on the member being designed.

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3.1.1 Dead Load
The dead load from superstructure is the aggregate weight of all structural elements and
nonstructural parts of the bridge above the bearing. This would include the main supporting
girders, the deck, parapets and road surfacing. The dead load from substructure is the weight of
all structural elements of the bridge between the lowest levels of the structure till the bearing.
This would include the abutment, foundation, and pier.
3.1.2 Live Loads
The live load for bridges means a load that moves along the length of the span that consists of
the weight of the applied moving load of vehicles and pedestrians. The traffic over a highway
bridge consists of a multitude of different types of vehicles. To form a consistent basis for
design, standard loading conditions are applied to the design model of structure. These standard
loadings are specified in Standard specification of American Association of State Highway and
transportation Officials (AASHTO).
The highway live loadings on bridge consist of standard trucks or lane loads that are equivalent
to truck trains. Two systems of loading are provided: H-loading and HS loading. The number
after H or HS indicates the gross weight in tons of the truck or tractor. The design truck is
designated as HS 25 consisting of 10 kip front axle and two 40 kip rear axles. The design
tandem consists of a pair of 32-kip axles spaced 4 ft. apart. However, for spans longer than 40 ft.
the tandem loading do not govern, thus only the truck load is investigated [1].
The lane load consists of a load of 0.8klf uniformly distributed in the longitudinal direction. The
design of the deck slab is based on HS 25 loading; hence, 20 kip wheel load will govern the
design.

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3.1.3 Dynamic Load Allowance (Impact)
The truckloads on bridges are applied not gently and gradually but rather violently, causing
stress increase. In order to account the dynamic effect of sudden loading of vehicle on to a bridge
structure, additional loads called impact loads must be considered. These are taken into account
by increasing the static effects of design truck or tandem, with the exceptions of centrifugal and
breaking forces, by the Dynamic Load Allowance. The factor to be applied to the static load shall
be taken as: (1 + IM/100). The dynamic load allowance shall not be applied to pedestrian loads
or to the design lane load. [3]
3.1.4 Wind Load
Wind forces are extremely complicated, but through a series of simplifications are reduced to an
equivalent static force applied uniformly over the exposed faces of the bridge (both super and
sub-structures) that are perpendicular to the longitudinal axis. AASHTO specifies that the
assumed wind velocity should be 100 mph. For a common slab-on-stringer bridge this is usually
a pressure of 50psf, and a minimum of 300p/lf. These forces are applied at the center of gravity
of the exposed regions of the structure.
AASHTO recommends the following for common slab-on-stringer bridges:
1) Wind force on structures (W): a) transverse loading = 50psf b) longitudinal loading = 12psf
2) Wind force on live load (WL): a) transverse loading = 100psf b) longitudinal loading = 40psf
The transverse and longitudinal loads are placed simultaneously for both the structure and the
live load (AASHTO 3.15.2.1.3).
For the usual girder having span lengths less than 125ft, the transverse wind loading on the
superstructure can be taken as 50lbf/ft2, and the longitudinal wind loading can be taken as
12lbf/ft2. [AASHTO 3.15.2.1.3]

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Wind forces are resisted by the bracing systems for a through bridge. The bracing systems are
neither analyzed nor designed in this thesis since the load is considered insignificant.
3.1.5 Earthquake Loading
When earthquakes occur, bridges can be subject to large lateral displacements from the ground
movement at the base of the structure. In many areas of the United States, the risk of earthquakes
is low. Since city of Houston lies not within the seismic zone, the risk of earthquakes is less.
Bridge earthquake loads depend on a number of factors, including the earthquake magnitude, the
seismic response of soil at the site, and the dynamic response characteristics (stiffness and
weight distribution) of the structure. Hence, it can be ignored for this capstone project.
3.1.6 Earth Pressure
In this capstone project, earth pressure is the lateral pressure generated by fill material acting on
abutments. The magnitude of earth pressure depends on the physical properties of the soil, the
interaction at the soil-structure interface, and the deformations in the soil-structure system. From
AASHTO 5.5.2, an equivalent fluid weight of 35lb/ft3 is more commonly used (sandy backfill
with a unit weight of approximately 120lb/ft3). The earth pressure acting on abutment is
increased when vehicle live loads occur in the vicinity of the structure. When vehicle traffic can
come within a horizontal distance from the top of a retaining structure equal to one-half its
height, a live load surcharge of 2 feet of fill is added to compensate for vehicle loads (AASHTO
3.20.3). The resulting load distribution on the structure is trapezoidal is as shown in drawing no.
10. [1]

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3.2. Bridge design and analysis
The Direct Connector between I-10 and highway 99 projects has eight connectors with different
span length. In this project, structural design of slab, girder, and abutment that connects north
and south of I-10 on highway 99 is considered. The calculation used in design of girder in this
project connects north and south of I-10 on Highway 99. In this project, to obtain optimum
design longest span of 80 ft. single span bridge is considered. The design is based on the
AASHTO Standard Bridge Design Specifications. The sectional view of the bridge is shown in
the next page, Fig. 3.

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3.2.1. Super-structure Analysis and Design
DESIGN PARAMETERS
The bridge considered for this design has a span length of 80 ft. (center-to-center (c/c) pier
distance), a total width of 42 ft. and total roadway width of 39 ft. The bridge superstructure
consists of slab design, five AASHTO IV Girders spaced 9 ft. center-to-center, designed to act
compositely with 9 in. thick cast-in-place (CIP) concrete deck. The wearing surface thickness is
0.5 in., which includes the thickness of any future wearing surface.
3.2.1.1 Slab Analysis and Design
The overall width of the bridge is 42ft. The clear road way width is 39ft. The road way is a
concrete slab 9 in thick, with a concrete strength of fc’=4 kips/in2 and steel reinforcement equal
to Fy= 60 kips/in2. The top width of girder spaced 9 ft. apart is 20 in. The future wearing surface
is 0.03 kips/ft3.
Determine the effective slab span length:
The effective slab span length S is the clear span plus one half the stringer top widths
[AASHTO 3.24.1.2(b)] S = 9 ft. + 10/12 ft. =9.83 ft.
Determine minimum thickness of the slab:
[AASHTO Table 8.9.2] tmin = (S+10)/30 = (9.83+10)/30 *12 in/ft. = 7.9 in
Assumed slab thickness is t= 7.9 +.5=8.4
Use 9 in

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Determine factored load
Group loading combinations for load factor design are
[AASHTO Table 3.22.1A] Group I =ɤ*(βD*D + βL(L+I))
= 1.3(1*D+1.67(L+I))
= 1.3D + 2.17(L+I)
Determine the factored dead loads
WD= 1.3(deck slab + FWS)
= 1.3(9in*(1/12)*.15kip/ft3
+ 0.03kip/ft2
)
= 0.185kip/ft2
WD = 0.185kip/ft per foot of width of slab
WC+P = 1.3(curb and parapet)
= 1.3*3.37ft2
*(0.15kip/ft3
)
= 0.657 kip/ft.
Determine the factored live plus impact loads.
The design truck is designated as HS 25 consisting of 10 kip front axle and two 40 kip rear axles.
The design tandem consists of a pair of 32-kip axles spaced 4 ft. apart. However, for spans
longer than 40 ft. the tandem loading does not govern, thus only the truck load is investigated.

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The lane load consists of a load of 0.8klf uniformly distributed in the longitudinal direction. The
design of the deck slab is based on HS 25 loading; hence, 20 kip wheel load will govern the
design.[1]
[AASHTO 3.8.2] The live load impact is:
I= 50/ (L+125) = 50/ (9ft + 125) = 0.37
The maximum impact load allowed is 0.3.
The factored wheel plus impact load is
PL+I = 2.17(L+I) = 2.17*(20kips + 0.3*20kips) = 56.4 kips
Analyze for factored moment.
For continuous spans, the factored positive and negative dead load moments are assumed to be
MD= (WD S2
)/10 = ((0.185 Kip/ft2
)*(9.83ft)2
)/10
MD =1.79ft-kips/ft. width of slab
Based on AASHTO 3.24.3.1 in slabs continuous over three or more supports, a continuity factor
of 0.8 is applicable. The factored positive and negative live load plus impact moments are:
[AASHTO 3.24.3.1 & Equation 3-15 ] ML+I= O.8*((S+2)/32)*PL+I
= 0.8((9.83+2)/32)*56.4
= 16.68 ft.-kips/ft. width of slab

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The total factored positive and negative moments are:
Mu = MD+ML+I
= 1.79ft-kips + 16. 68 ft.-kips
[AASHTO3.24.5.] For cantilever spans, the factored negative dead load moment is;
MD= ((WDS2
)/2) + MC+P*L
AASHTO 3.24.1.2 S= 3ft – (10/12) ft. = 2.17ft
L= 2.17ft-0.66ft = 1.51ft
MD= (((0.185 kips/ft2
)*(2.17ft)2
)/2) +(0.657 kip/ft2
)*1.51ft
[AASHTO 3.24.2.1] The center line of the wheel will be placed 1 ft. from the face of the curb.
[AASHTO 3.24.5] Each wheel on the slab perpendicular to traffic is distributed over a width of
[AASHTO Equation 3.17] E= 0.8X + 2.17 = (0.8*(0.58ft.)) + 2.17 ft. = 2.63 ft.
Where X= distance in feet from wheel load to point of support
X= I [2.17ft. - ((12+2+7+12)/12)] I = 0.58 ft.
[AASHTO 3.24.5.1] The factored negative wheel load plus impact moment is
ML+I = PL+ I(X/E)

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= (56.4 kip)*(0.58ft/2.63)
= 12.44ft.-kips/ft. width of slab
The total factored negative moment is
Mu= MD + ML+ I
= 1.43 ft.-kips + 12.44 ft.-kips
= 13.87ft.-kips/ft. width of slab
Due to the short overhang, the live load acting at a distance on one foot from the barrier face
does not act on the overhang. Therefore, this case need not be investigated.
Design for moment
The compressive strength of the concrete at 28 days is f’c = 4000lbf/in2
. The specified minimum
yield point of the steel is Fy = 60000lbf/in2
. Determine the maximum and minimum steel
reinforcement needed.
[AASHTO 8.16.3.1 & 8.16.3.2] The maximum ratio of tension reinforcement is
[AASHTO 8.16.3.1.1] ρ max = 0.75ρb = 0.75* ρb
Where, β =0.85-0.025= 0.825; fc’=4000lbf/in2
[AASHTO 8.16.2.7]
ρb =( (0.85β1fc’)/Fy)*(87000/(87000+Fy) [AASHTO 8.16.3.2.2]
ρb =( (0.85*0.825*4000)/60000)*(87000/(87000+60000)

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ρb =0.028
ρ max = 0.75ρb = 0.75*0.028 = 0.021
Determine the areas of positive and negative steel. [AASHTO 8.16.3.2 & Equation 8-16]
ΦMn = ΦAs Fy(d- (⍺/2))
⍺= (As*Fy)/ (0.85*fc’*b)
⍺= (As*60kips/in2
)/ (0.85*4kips/in 2
*12in)
⍺= 1.47*As
Assume No. 7 steel rebar,
d= 9in – 0.5in for integral wearing surface – 2.0 in for cover – 0.44in
d= 6.06in
[AASHTO 8.16.1.2] Φ=0.9 for flexure
Mu= Φ*Mn for continuous spans
Mu= 18.47*12(in/ft.)= Φ*Mn = 0.9*As*(60kips/in2
)*(6.06in-(1.47/2)*As)
221.64= 327.24As - 39.69As2
As2
-8.245*As + 5.584 = 0
As=0.745 in2
/ft. of slab width

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Use #7@9in (As= 0.8in2
/ft.)
ρ= As/bd = 0.8in2
/(12in*6.06in)= 0.011< ρmax=0.021 ok
Check moment capacity
⍺=(As*Fy)/(0.85*fc’*b)= (0.8in2
*60kips/in2
)/(0.85*4 kips/in2
*12in)
= 1.18in
⍺/2 = 0.588in
ΦMn= 0.9*0.8*60*(6.06in-0.588in)(1ft/12in)
= 19.7 ft-kips/ft of slab width
Mu= 18.47 ft-kips/ft of slab width [ < ΦMn, so ok]
Check minimum steelIn a flexure member where tension reinforcement is required by analysis,
the minimum reinforcement provided shall be adequate to develop a moment capacity at least
1.2times the cracking moment. [AASHTO 8.17.1]
[AASHTO Equation 8-62] ΦMn≥1.2 Mcr
[AASHTO 8.13.3 & Equation 8-2] Mcr= fr*Ig/yt
fr=modulus of rupture= 7.5(√fc') for normal weight concrete
fc= 4000lbf/in2
Ig= moment of inertia

31
Yt= distance from centroidal axis to extreme fiber in tension
ΦMn≥ 1.2Mcr
Mcr= fr*Ig/yt= (7.5 ) *((12in)(9in)3
)/4.5in)*(1ft/12in)(1kip/ft)
Mcr= 6.404 ft.-kips/ft.
1.2 Mcr=1.2*6.404ft-kips = 7.68ft-kips/ft. [≤ΦMn=19.47 ft.-kips/ft. ok
Distribution Reinforcement
Reinforcement traverse to main steel reinforcement is placed in the bottom of all slabs. The
amount shall be a percentage of the main reinforcement required as determined in the following
formula. [AASHTO 3.24.10]
[AASHTO 3.24.10.2 & Equation 3-22] The percentage is 220/ (√S), with a maximum of 67%
= 220/ (√9.83ft) =70.17% [67%maximum allowed]
As= 0.67*0.7in2
= 0.47 in2
/ ft.
Use #6@9in (As= 0.59in2
/ft.) in the bottom and perpendicular to the main reinforcement in the
middle half of the slab span. 50% of the specified distribution reinforcement is used in the outer
quarters of the slab span. [AASHTO 3.24.10.3]
5. Design for shear and bond
Slabs designed for bending moment in accordance with AASHTO sec.3.24.3 (wheel loads) are
considered satisfactory in bond and shear. Fig. 4 shows slab and curb detail in the next page.

33
3.2.1.2 Girder Analysis and Design
Considering the bridge connects north and south of highway 99 that pass over I-10 as simply
supported span 80 ft. There are five pre-stressed AASHTO standard IV beams with compressive
strength f’c= 5500 psi at initial pre-stress and f’cg = 6500 psi at 28 days. The roadway is 9 in
slab three lanes wide. The dead load of curb and parapet is 0.506 kips/ft. Pre-stress is to be
provided by ½ in strand(seven-wire) steel with an area of 0.153in2 and ultimate stress of pre-
stressing steel of f’s= 270 kips/in2. For HS 25 loading, use load factor design method to design
AASHTO standard IV beams.
Span Length (c/c piers) = 80 ft.-0 in.
Overall girder length = 80'-0" – 2(2") = 79'-8" = 79.67 ft.
Design Span = 80'-0" – 2(9") = 78'-6" = 78.5 ft. (c/c of bearing)
Cast-in-place slab:
Thickness, ts = 9.0 in.
Concrete strength at 28 days, fc′ = 4000 psi
Thickness of asphalt wearing surface (including any future wearing surface), tw = 0.5 in.
Unit weight of concrete, wc = 150 pcf
Precast girders:
Concrete strength at release, fc′i = 5500 psi

34
Concrete strength at 28 days, fc′g = 6500 psi
Concrete unit weight, wc = 150 pcf
Pre-tensioning strands: ½ in. diameter, seven wire low relaxation
Area of one strand = 0.153 in.2
Ultimate stress, fpu = 270,000 psi
Yield strength, fpy = 0.9fpu = 243,000 psi
Stress limits for prestressing strands:
Before transfer, fpi ≤ 0.75 fpu = 202,500 psi
At service limit state (after all losses) fpe ≤ 0.80 fpy = 194,400 psi
Modulus of Elasticity, Ep = 28,500 ksi
Non pre-stressed reinforcement:
Yield strength, fy = 60,000 psi
Modulus of Elasticity, Es = 29,000 ksi
Look the cross sectional drawing of Type IV Girder Fig. 5 in the next page.

36
A.2.4 CROSS SECTION PROPERTIES FOR A TYPICAL GIRDER
A.2.4.1 NON-COMPOSITE SECTION
The section properties of an AASHTO Type IV girder are provided in the following table.
Table 1 Section Properties of AASHTO Type IV Girder
yt (in) yb (in) Area(in2) I (in4) Wt/lf (lbs)
29.25 24.75 788.4 260403 821
Where:
I = Moment of inertia about the centroid of the non-composite precast girder = 260,403 in.4
yb = Distance from centroid to the extreme bottom fiber of the non-composite precast girder =
24.75 in.
yt = Distance from centroid to the extreme top fiber of the non-composite precast girder = 29.25
in.
Sb = Section modulus referenced to the extreme bottom fiber of the non-composite precast
girder, in.3 = I/yb = 260,403/24.75 = 10,521.33 in.3
St = Section modulus referenced to the extreme top fiber of the non-composite precast girder,
in.3 = I/ yt = 260,403/29.25 = 8902.67 in.3
COMPOSITE SECTION
Effective Flange Width
[AASHTO 9.8.3.2] The effective flange width is lesser of:

37
Case 1: 0.25 span length of girder:
= 80(12 in. /ft.)/ 4 = 240 in.
Case 2: 12 × (effective slab thickness) + (greater of web thickness or one half top flange width):
= 12(9) + 0.5(20) = 118 in.
[0.5 × (girder top flange width) = 10 in. > web thickness = 8 in.]
Case 3: Average spacing of adjacent girders:
= (9 ft.)(12 in. /ft.) = 108 in. (controls)
Effective flange width = 108 in.
A.2.4.2.2Modular Ratio between Slab and Girder Concrete
The modular ratio between the slab and girder concrete is used for service load design
calculations. For the flexural strength limit design, shear design, and deflection calculations, the
actual modular ratio based on optimized concrete strengths is used.
n = Ecs for slab/ Ecg for girder
where n is the modular ratio between slab and girder concrete, and
Ec is the elastic modulus of concrete.
Slab concrete fcs’ = 4000lbf/ft2
,
[AASHTO Equation 10-68] Ecs = Wc1.5
*33*√fcs'

38
Ecs = (150lbf/ft3)1.5
*33*√ (4000lbf/in2
) = 3.83 x 106
lbf/in2
Girder fcg’ = 6500lbf/ft2, fci’ = 5500lbf/ft2
,
Ecg = Wc1.5
*33*√fcg' = (150lbf/ft3)1.5*
33*√ (6500lbf/in2) = 4.89 x 106
lbf/in2
n= Ecg/Ecs = 4.89/3.83 = 1.28
A.2.4.2.3 Transformed Section Properties
Transformed flange width = n × (effective flange width) = (1.28)(108) = 138.24 in.
Transformed Flange Area = n × (effective flange width)(ts) = (1.28)(108)(9) = 1244.16 in.2
Table 2 Properties of Composite Section.
Transformed
Area( in2)
yb
( in)
A*yb A(ybc – yb)2
I
(in4
)
I + A(ybc - yb)2
(in4
)
Girder 788.4 24.75 19512.9 336517.19 260403 596920.19
Slab 1244.16 58.5 72783.36 213184.45 6561 219745.45
Total 2032.56 92296.26 816665.64
Note
 Ac = Total area of composite section
Ac= Ab + flange area = 788.4 + 1244.16 = 2032.56 in.2
 hc = Total height of composite section
hc = 54 + 9 = 63 in.
 Ic = Moment of inertia about the centroid of the composite section

39
= 260403+788.4*(45.41-24.75)2
+6561+ 1244.16*(8.59+4.5)2
= 260403+336517.19+6561+213184.45 in.4
= 816665.64 in4
 ybc = Distance from the centroid of the composite section to extreme bottom fiber of the
precast girder, in. = 92296.26/2032.56 = 45.41 in.
 ytg = Distance from the centroid of the composite section to extreme top fiber of the
precast girder, in. = 54 – 45.41 = 8.59 in.
 ytc = Distance from the centroid of the composite section to extreme top fiber of the slab
= 63 – 45.41 = 17.59 in.
 Sbc = Section modulus of composite section referenced to the extreme bottom fiber of the
precast girder, in.3 = Ic/ybc = 816665.64 /45.41 = 17984.27 in.3
 Stg = Section modulus of composite section referenced to the top fiber of the precast
girder, in.3 = Ic/ytg = 816665.64 /8.59 = 95071.67in.3
 Stc = Section modulus of composite section referenced to the top fiber of the slab, in.3
= Ic/ytc = 816665.64 /17.59 = 46427.84 in.3
Look the composite sectional drawing shown in the next page Fig. 6.

41
SHEAR FORCES AND BENDING MOMENTS
The self-weight of the girder and the weight of the slab act on the non-composite simple span
structure, while the weight of the barriers, future wearing surface, live load, and dynamic load
act on the composite simple span structure. [2]
Shear Forces and Bending Moments due to Dead Load
Dead Loads
Dead loads acting on the non-composite structure:
Self-weight of the girder = 0.821 kip/ft.
Weight of cast-in-place deck on each girder = (0.150kcf) (9 in.) (9 ft.)/ (12 in./ft.)
= 1.012 kips/ft.
Total dead load on non-composite section = 0.821 + 1.012 = 1.833 kips/ft.
Superimposed Dead Loads
The superimposed dead loads placed on the bridge, including loads from railing and wearing
surface can be distributed uniformly among all girders.
Weight of 0.5 in. wearing surface = 0.03 kips/ft2
. This load is applied over the entire clear
roadway width of 39 ft.-0 in.
DFWS= (FWS)*9ft
= (0.03kip/ft2
)*9ft

42
= 0.27 kip/ft. /girder
WC+P = (curb and parapet) for one girder
= 3.37ft2*(0.15kip/ft3)*2/5
= 0.202 kip/ft. / girder
Total superimposed dead load = 0.202(Wc+p) + 0.27(WD) = 0.472 kips/ft.
Wind load
Wind Forces (W and WL).
sub-structures) that are perpendicular to the longitudinal axis.
AASHTO specifies that the assumed wind velocity should be 100 mph. For a common slab-on-
stringer bridge this is usually a pressure of 50psf, and a minimum of 300p/lf. These forces are
applied at the center of gravity of the exposed regions of the structure.

43
Determine the wind load, Wsuper, on the superstructure transmitted to the substructure. For the
usual girder having span lengths less than 125ft, the transverse wind loading on the
superstructure can be taken as 50lbf/ft2
, and the longitudinal wind loading can be taken as
12lbf/ft2
. [AASHTO 3.15.2.1.3]
The height exposed to wind is
= 54 in (girder) + 9in (slab) + 32in (curb and parapet)
=95 in or 7.92ft
The longitudinal wind loading is
= 80 ft. * 7.92 ft. * 0.012 kip/ft2
= 7.6kips (at 7.92/2= 3.96 ft.)
The longitudinal (horizontal) wind loading of the superstructure is
WH= 7.6 kips/80ft
= 0.095 kips/ft. (including wind load on the girder)
The vertical wind loading of the superstructure is
= 7.6 kips*3.96ft/80ft
= 0.376 kips (including wind load on girder)
WV = 0.376 kips/ 80 ft. = 0.0047 kip/ft. [negligible]

44
= 54 in (girder) + 9in (slab) + 6ft (wind load on vehicle acts)
=11.25 ft.
= 80 ft. * 0.04 kip/ft.
= 3.2kips (at 11.25/2= 5.625 ft.)
The longitudinal (horizontal) wind loading of the superstructure is
WLH= 3.2 kips/80ft
= 0.04 kips/ ft. (including wind load on the girder)
WLV = 3.2 kips*3.96ft/80ft
the maximum lateral moment due to the factored wind loading is computed as follows:
M = WL2
/10 =
A longitudinal force of 5% of the live load in all lanes is located 6ft above floor slab. For HS 25
loading, the longitudinal force is
= 3 lanes*[80ft*(0.64kip/ft.*1.25) +(26kips*1.25)]*0.05

45
= 14.475 kips
The horizontal force is
LFH = 14.475 kip/80 ft. =0.18 kip/ft.
Determine the live load distribution
[AASHTO 3.12.1]
The reduction in load intensity for three traffic lanes loaded is 90%. The transverse distribution
of wheel loads for beam design for an interior beam is
= S/5.5 [AASHTO Table 3.23.1]
= 9ft. /5.5 = 1.45
This distribution factor will be applied for HS loading.
Determine impact load [AASHTO 3.8.2]
I= 50/ (L+125) = 50/(80+125) = 0.244 [3]
Determine service loads moment and shear
The moment at mid span due to the weight of the beam is
M0 = WL2
/8 = (0.821 * 802
)/8 = 656.8 ft.-kips
The moment at mid span due to the slab dead load per beam is
MD = (WD*L2
)/8 = (1.012*802
)/8 = 809.6 ft.-kips

46
The moment mid span due to the superimposed dead load per beam is
Ms = WSL2
/8 = (0.472*802
)/8 = 377.6 ft.-kips
The maximum live load moment due to the HS 25 truck can be determined through the influence
line method.
The live load reaction at A is
= 40kips*(28.33ft. +42.33ft.) / 80ft. + (10kips*56.33 ft.) /80 ft.
= 42.37 kips
The live load moment, ML, is
= 42.37 kips*(14ft. + 23.67 ft.) - (10kips*14 ft.)
= 1456.07 ft.-kips for all wheels
The maximum live load moment with impact for each beam is
ML+ I = (1456.07 ft.-kips/2 wheel lines)*1.45*1.244
= 1313.23 ft.-kips
The maximum dead load shear is
VD= (0.821kip/ft. + 1.012kip/ft. + 0.472kip/ft.)*40ft.
= 92.12 kips at each support
The maximum live load shear due to the HS 25 truck is

47
40kips 40kips 10kips
14ft. 14 ft. 52ft.
Fig. 7 Maximum live load shear
The live load reaction at A is
= 40 kips + (40kips*(52 ft. + 14ft.) + 10kips*(52ft.))/80ft.
= 79.5 kips
For an HS lane loading
W=0.64 kips/ft *1.25 = 0.8 kips/ft
Axel load= 26 kips*1.25 = 32.5 kips
Max. R= 32.5 kips + 0.8*80/2 = 64.5 kips
Therefore, HS 25 truck loading controls the maximum reaction, which is 79.5kips/lane.
The maximum live load shear with impact for each beam is
VL+I = (79.5 kips/2 wheel lines)*1.45*1.244
= 71.7 kips
Factored loads are used for designing structural members using the load factor concept. Group
loading combinations for load factor design are given by [AASHTO 3.22.1A; and foot notes]

48
Group I =1.3(1D + (1.67) (L+I))
Group II =1.3(1D + 1W)
Group III =1.3(1D + 1L+0.3W+1WL)
Group IV =1.3(1D + 1L)
Group V =1.25(1D + 1W)
Group VI =1.25(1D + 1L+0.3W+1WL)
For the strength limit state, wind on the structure is considered for the Strength III and Strength
V Limit States. Due to the magnitude of the live load stresses, Strength III will clearly not
control for this design. Therefore, for this design, group I control the design factor load.
Calculate the factored moment and shear
Mu = 1.3(1D + (1.67) (L+I))
= 1.3*[656.8+809.6+377.6+ (1.67)*(1313.23)]
=5248.22 ft.-kips
Vu = 1.3(1D +1.67*(L+I))
= 1.3(92.12 kips + 1.67*(71.7kips))
= 275.42 kips
Allowable concrete stresses for I-beam girders [AASHTO 9.15.2]

49
[AASHTO 9.15.2.1 & 9.15.2.2]
Compression (pretension members) before losses due to creep and shrinkage,
fci=0.6 f’ci= 0.6*5500lbf/in2
= 3300psi
Tension (with no bonded reinforcement) before losses due to creep and shrinkage,
fti =200lbf/in2
or )
[AASHTO 9.15.2.2]
Compression stresses after losses, fcs= 0.4f’cg= 0.4*6500lbf/in2
= 2600lbf/in2
Tension after losses, fts = ) = √ ) = 483.7 lbf/in2
For severe corrosive exposure conditions, fts= ) = 241.87 lbf/in2
Calculate pre-stress force and eccentricity
The temporary stress limits before losses due to creep and shrinkage are as follows.
For compression,
fcs= 0.4f’cg= 0.4*6500lbf/in2
= 2600lbf/in2
For tension,
fts= ) = √ )= 556 lbf/in2
Allowable stresses after losses are as follows.

50
For compression,
fcs= 0.4f’cg= 0.4*6500lbf/in2
= 2600lbf/in2
For tension,
fts= )= √ )= 242 lbf/in2
N.B: A 25% loss of pre-stress force is assumed in the straight tendons. It will be assumed that the
critical stresses are the initial tensile stress at the top of the beam at the bearing and the final
tensile stress at the bottom of the beam at mid-span.
For initial tensile stress at the top of beam at bearing,
fti= Pi/Ab + Pie/St
Substituting values for the stresses and section properties gives
300 lbf/in2
= -Pi/788.4 + Pi *e/8902.67 -------------------------------------------------------Equation 1.
For final tensile stress at the bottom of the beam at midspan,
fts = (-Pe/Ab) – (Pe* e/Sb) +( MO+D/Sb) +(( Ms+ML+I)/Sbc)
242= (-0.75Pi/ 788.4) – (0.75Pi*e/10,521.33) + [((656.8 ft.-kips+809.6 ft.-kips)/ 10,521.33) +
((377.6 kip- ft. +1313.23 kip- ft.) / 17984.27 in.3)]*12*1000 -----------------------------Equation 2.
Solving Equation 1 and 2 simultaneously gives
Pi=1,348,630 lbf; e= 13.27 in
[AASHTO 9.15.1] The amount of force taken by one ½ in, seven-wire strand at 70% of ultimate

51
stress f’s is
F=As*fsi
Where, fsi = 0.7*f’s
F= O.153 in2
*0.7*270000 = 28920lbf
The number of strands required is
N = 1348630/28920 = 46.63 ≅ [USE 47 STRANDS]
For 47 strands, the initial pre-stressing force is
Pi = 28920lbf*47 = 1359240lbf
Try the following pattern
Number of strands 5 5 5 5 3 3 3 2 2 2 2 2 2 2 2 2
Distance from
bottom fiber
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32
Strand eccentricity at mid-span after strand arrangement
ec = 24.75 –[(5(2 + 4 + 6+8) +3(10+12+14)+ 2(16+18+20+22+24+26+28+30+32))/47]
= 24.75- 13.62 in. = 11.11 in
Look Fig. 8 cross sectional view drawing no.8 strand arrangement in the next page.

53
Check the assumed pre-stress loss
The loss of pre-stress will be determined using the modified Bureau of Public Roads formula.[1]
Total loss of pre-stress is
=6000lbf/in2 + 16fcs + 0.04fsi
fsi = 0.7*f’s = 0.7*270000 = 189,000 lbf/in2
fcs= Pi/Ab + (Pi*e2 – Mo*e)/IB
fcs=[1359240lbf / 788.4in2] +[ [(1359240kips*11.112
) – (656.8 ft.-kips*12000*11.11in)]/
260403in4
]
fcs=1724.05 + 308.02 = 2032.07lbf/in2
Total loss of pre-stress is
=6000lbf/in2 +16*2032.07lbf/in2 +0.04*189000lbf/in2 =46073.1lbf/in2
The percentage loss is =(46073.1/189000)*100% = 24.4% < 25% assumed; therefore minimum
loss of 25% assumed is conservative.
The effective pre-stress loss force after losses is
Pe = (1-0.25)*1359240lbf = 1019430lbf
Determine the critical stresses at support and mid-span
Initial stresses before losses at support for the basic beam section are

54
Pi/Ab = 1359240lbf/788.4 in2
= -1724.47lbf/in2 [compression]
Pi*e/St = 1359240lbf * 11.11in/8902.67 in3
m= 1696.25lbf/in2 [tension]
Pi*e/Sb = 1359240lbf*11.11in / 10521.33 in3
Check the allowable stresses
[AASHTO 9.15.2.1 &9.15.2.2]
For compression at the beam base, allowable concrete stress = fci
Fci= 0.6*f’ci = 0.6*5500lbf/ in2 = 3300lbf/in2 [> -3159.76 lbf/in2 = (-1724.47-1435.29) lbf/in2,
ok]
For tension, to be conservative, allowable concrete stress fti lesser of 200lbf/in2 or
√ = 222.5 lbf/in2
fti=200lbf/in2 [>-28.22 lbf/in2 =( -1724.47+1696.25) lbf/in2]
Final stresses after losses at mid span are as follows.
The stress in the basic beam section at girder base is
= -Pe/Ab – Pe*e/Sb
= -1019430/788.4 – 1019430*11.11/10521.33
= -1293.04 – 1076.47
= -2369.51 lbf/in2 [compression]

55
The stress at the top of the girder is
f= -Pe/Ab + Pe*e/St
= -1293.04 + 1019430*11.11/8902.67
Due to the beam and slab weight, the stress at the top of the girder is
=-MO+D/St = -((656.8 +809.6)/8902.67)*12*1000 = -1976.58lbf/in2 [compression]
The stress at the girder base is
= MO+D/Sb = -((656.8 +809.6)/10521.33)*12*1000 = 1672.49lbf/in2 [tension]
Due to the superimposed dead load (parapet and curb) plus live load, the stress in the composite
section at slab top is
f= -Ms+ML+I/Stc = - ((377.6 +1313.23)/ 46427.84)*12*1000 = -437.02lbf/in2 [compression]
The stress in the composite section at the girder base is
f=Ms+ML+I/Sbc = ((377.6 +1313.23)/ 17984.27)*12*1000 = 1128.21lbf/in2 [compression]
Check the allowable stresses in the girder after losses.
[AASHTO 9.15.2.2] For compression at the beam base, allowable concrete stress =
fcs= 0.4*f’cg = 0.4*5500lbf/ in2 = 2600lbf/in2 [> -2210.85lbf/in2 , ok]
[AASHTO 9.15.2.2] For tension,

56
fts= 6√(f’cg = 6√(6500 lbf)/in2 = 483.7lbf/in2(>431.19lbf/in2 ok)
Check the stress in the slab
[AASHTO 8.15.2.1.1] fcs allowable compressive stress in concrete (slab)
fcs= 0.4*f’cs= 0.4*4000 = 1600 lbf/in2
fslab = (Ms +ML+I)/Stc = -((377.6 +1313.23)/ 46427.84)*12*1000 = -437.02 lbf/in2
[compression]
[<1600lbf/in2, so ok]
Check moment capacity by Load factor design. First, determine whether the beam section is
flanged or rectangular.
[AASHTO 9.17.2] ⍺= As*f*su/0.85fc,slab b
f’c,slab= f’cs=4000lbf/in2
[AASHTO 9.17.4.1] f*su=average stress in pre-stressing steel at ultimate load
f*su=f’s(1-(ɤ/β1)(ρ*f’s/f’c))
[AASHTO 9.1.2] ɤ*= 0.4 for steel relieved steel [AASHTO 9.1.2]
[AASHTO 8.16.2.7] β1=0.85-0.05*(500/1000)=0.825 [AASHTO 8.16.2.7]
f’s=270000lbf/in2
ρ*=As/bd = 47*0.153/(108*(63-13.62)) = 0.00135

57
f*su=(270000lbf/in2)*(1-[(0.4/0.825)*(0.00135*270000/4000)])
= 258070.91lbf/in2
⍺= (47*0.153*258070.91)/(0.85*4000*108)
= 5.05 in
Since ⍺<9 in, the effective thickness of the slab, the beam is rectangular.
The moment capacity by load factor design is
ΦMn= Φ(A*sf*su d(1-0.6(ρ*fsu*/f’c))) [AASHTO 9.1.7.2]
[AASHTO 9.14] Φ=1 for flexure in pre-stressed concrete members for load factor design
ΦMn= 1(47*0.153*258070.91*(63-13.63)(1-0.6(0.00135*258070.91*/4000)))
= 104420290.04in-lbf
= 8701.69ft-kips
Determine the design moment by the load factor method (Group I loading)
Mu=1.3D + 2.17(L+I)
= 1.3(Mo+MD+Ms) + 2.17(ML+I)
= 1.3(656.8+809.6+377.6) + 2.17(1313.23)
= 5246.91 ft-kips [<ΦMn =8701.69ft-kips ok]
[AASHTO 9.18.1] The pre-stress steel’s reinforcement index cannot exceed 0.36β1.

58
Ρ*f*su/f’c = 0.00135*258070.91/4000
= 0.087 [<0.36β1=0.36*0.825=0.297, ok]
Determine the minimum amount of pre-stressing steel that will be necessary.
[AASHTO 9.18.2] The total amount of pre-stressing reinforcement shall be adequate to develop
an ultimate moment at the critical section of at least 1.2 times the cracking moment M*cr.
ΦMn≥ 1.2 M*cr
M*cr= (fr+fpe)Sbc – Md/nc((Sc/Sb) -1)
Md/nc = non-composite dead load moment
= Mo + MD + Ms
= 656.8+809.6+377.6
= 1844 ft-kips
[AASHTO 9.15.2.3] fr= √ = = 604.67lbf/in2
fpe= Peffect/Ab + Peffect*e/Sb – Mo/Sb
Peffect = Pi*0.75 = (0.7f’sA*s)*0.75
= 0.7*270*0.153*42*0.75
= 910.89kips
fpe = 910.89/788.4 + 910.89*11.11/10521.33 – 656.8*12/10521.33

59
= 1.155 + 0.962 -0.749
= 1.37kips/in2
M*cr. = (0.6047+1.37)* 17984.27 – 1844*((17984.27 /10521.33) -1)
= 2959.46 – 1307.98
= 1651.48 ft-kips
1.2M*cr. = 1.2*1651.48 = 1981.776 ft.-kips [<ΦMn = 8701.69ft-kips, ok]
Design for shear
Shear design for straight fully bonded strands pre-stressed beams will be in accordance with
Article 9.20 of the AASHTO Specifications.
Vu ≤ Φ*(Vc + Vs) [AASHTO Equation 9-26]
The shear stress at the support is
The maximum live load and impact load shear is
VL+I= (79.5 kips/ 2 wheel lines)*1.45*1.244= 71.7kips
The maximum dead load shear is
VO+D+S = (0.821kip/ft. + 1.012kip/ft. + 0.472kip/ft.)*40ft.
= 92.12 kips at each support
Calculate shear stress at the quarter point

60
[AASHTO 9.20.2.1 & 9.20.2.2] The shear strength of concrete, Vc, shall be the lesser of Vci or
Vcw.
Vci= 0.6(√’fc) b’ d + Vd + ViMcr/Mmax
b’= 8in
d= yt+e= 29.25 in + 11.11 in = 40.39 in
From shear force diagram, since dead load is uniform, the shear force at quarter point equals half
of the support shear force.
Vd= ½(92.12) = 46.06 kips
The factored shear force due to externally applied loads occurring simultaneously with Mmax is
Vi= 1.3*1.67(L+I)
= 1.3*1.67*53.69
= 116.56 kips
Mcr = (Ic/yb)( 6√f’cg + fpe – fd)
Moments at quarter point are (Wo+WD = WL). For beam weight (W) and slab weight (WD),
MO+D = (WL/2)(L/4) – (WL/4)(L/8)
= WL2
/8 – WL2
/32
= 3(WL) 2
/32

61
= (3/32)*(0.821 + 1.012)*802
= 1099.8 ft.-kips
For the superimposed dead load, Ws(parapet/curb),
Ms= (3/32)*(Ws) 2
= (3/32)*(0.472)* 802
= 283.2 ft-kips
fpe= 1.37 kips.in2
The stress due to the un-factor dead load at the quarter point in the beam span is
fd= (MO+D/Sb) + (Ms/Sbc)
= (1099.8/10521.33 + 283.2/17984.27)*12in/ft
= 1.44 kips/in2
Mcr = (816665.64 /45.41)*[( 6√6500 *1/1000) +1.37 – 1.44]*1/12
= 620.06 ft- kips
The maximum factored moment due to externally applied load at quarter point is
Mmax = ML+I
= 1.3*1.67(L+I)
= 2.17*(1012.73)

62
= 2197.62 ft.-kips
Vci= (0.6*(√6500)* 8*40.39)*1/1000 + 46.06 + 116.56*620.06/2197.62
= 15.63 + 46.06 + 32.89
= 94.58 kips
Vc = Vci = 94.58 kips
Vc, min = 1.7*√f’c *b’ *d
= 1.7*((√6500)*1/1000)*8*40.39
= 44.27 kips
d≥ 0.8h = 0.8*54 in = 43.2 in [ d=40.39 ≅ 0.8h, hence so close ok]
Vu ≤ Φ(Vc +Vs)
Vu= 1.3(1D +1.67(L+I))
= 1.3(46.06 + 1.67(53.69))
= 176.44 kips
For load factor design, Φ=0.9 for shear[AASHTO 9.14]
Vu/Φ = 176.44/0.9 = 196.04 kips
Vs= Vu/Φ – Vc = 196.04 – 94.58 = 101.46 kips
[AASHTO 9. 20.3.1] Check Vs ≤ 8*√f’cg b’ *d = 8*(√6500* 1/1000)*8*40.39 = 208.41 kips

63
Vs= 101.46 kips ≤ 208.41kips ok
For grade 60 No. 4 vertical stirrups, the spacing required
[AASHTO Equation 9-30] S = (Av * fsy * d) / Vs
= 0.4*60*40.39/ 101.46
= 9.55 in [use 10 in spacing]

64
3.2.2. Sub-structure Analysis and Design
DESIGN PARAMETERS
The abutments are a part of the substructure or foundation of the bridge. They act at end
supports. Abutments provide vertical support to the bridge and lateral support to the soil at the
ends of the roadway. The abutment has 16 ft. height and 42 ft. length, and it has spread footing
and no back wall. Assuming it has placed on gravel and sandy soil with a safe bearing capacity
of 8.5 kips/ft2
. The density of the compacted earth fill is 120lbf/ft3
, and the lateral earth pressure
35lbf/ft3
. The load that must be considered in this analysis are dead load, live load, earth
pressure, wind load on structure, wind load on live load, longitudinal force from live load, and
longitudinal friction load due to temperature. Centrifugal force does not exist for straight
structure. Whereas impact load and earthquake load are not considered. A live surcharge of 2ft.
of soil is placed on the bridge approach (AASHTO 3.20.3 and 5.5.2). Service loads are used in
determining if the abutment is safe against overturning about the toe of the footing, against
sliding on the footing base, and against crushing of the foundation material at the point of
maximum pressure.
3.2.2.1 Reinforced concrete Abutment and Footing
From AASHTO load combination for service load design for abutment and footing, group I, II,
III, IV, V, and VI will be the only ones considered in checking for stability and bearing pressure
for the abutment. Group I, II, III, IV, V, and VI factored load group are considered in designing
structural members using the load factor concept. In the next page Fig. 9 shows sectional view of
Abutment.

66
Dead loads of super structure on the abutment:
Self-weight of the girder = 0.821 kip/ft.
Weight of cast-in-place deck on each girder = (0.150kcf) (9 in.) (9 ft.) / (12 in./ft. )
= 1.012 kips/ft.
For the end block and diaphragm,
= 42ft*2.75ft*1.5ft*0.15kip/ft3
= 25.99 kip
Future wearing surface
DFWS= (FWS)*9ft
= (0.03kip/ft2)*9ft
= 0.27 kip/ft./girder
WC+P = (curb and parapet) for one girder
= 3.37ft2*(0.15kip/ft3)*2/5
= 0.202 kip/ft. / girder
Total dead load of super structure
=([0.821(beam) + 1.012(slab) + 0.27(wearing surface) + 0.202(curb and parapet)]*5*80/2) +
25.99(end block and diaphragm)

67
=486.99 kips
The dead load weight per foot of the abutment is
D= 486.99/42 =11.6 kips/ft.
Determine the live load weight on the superstructure. Use HS 25 loading.
The maximum live load shear due to the HS 25 truck (HS 20 loading x 1.25) is
40kips 40kips 10kips
14ft. 14 ft. 52ft.
The maximum live load reaction occurs due to truck loading
= 40 kips + (40kips*(52 ft. + 14ft.) + 10kips*(52ft.))/80ft.
= 79.5 kips/lane
For an HS lane loading
W=0.64 kips/ft. *1.25 = 0.8 kips/ft.
Axel load= 26 kips*1.25 = 32.5 kips
Max. R= 32.5 kips + 0.8*80/2 = 64.5 kips
Therefore, HS 25 truck loading controls the maximum reaction, which is 79.5kips/lane.
The maximum live load per foot of abutment

68
= (3lanes * 79.5 kips/lane)/42ft.
=5.68 kips/ft.
Determine the lateral earth pressure
An equivalent fluid weight of 35lbf/ft3
is assumed for determining the lateral earth pressure. The
effect of passive pressure due to soil in front of the abutment is neglected.
[AASHTO 3.20.3 &5.5.2] Alive load surcharge pressure equal to 2 ft. of earth will be added to
the approach.[1]
= 2ft*35lbf/ft3
= 70lbf/ft2
A lateral pressure due to the earth back fill is
= (5.37ft + 16ft)*35lbf/ft3
= 747.95lbf/ft2
The load due to earth and live load surcharge is as follows
Ls= 0.07*21.37 =1.5 kips/ft.
E = 0.5*0.748*21.23 = 7.98 kips/ft.
In the next page, Fig. 10 shows Sectional view of Live load and Earth pressure on Abutment.

70
Wind load (W and WL).
sub-structures) that are perpendicular to the longitudinal axis.
AASHTO specifies that the assumed wind velocity should be 100 mph. For a common slab-on-
stringer bridge this is usually a pressure of 50psf, and a minimum of 300 plf. These forces are
applied at the center of gravity of the exposed regions of the structure.
Determine the wind load, Wsuper, on the superstructure transmitted to the substructure. For the
usual girder having span lengths less than 125ft, the transverse wind loading on the
superstructure can be taken as 50lbf/ft2, and the longitudinal wind loading can be taken as
12lbf/ft2. [AASHTO 3.15.2.1.3]
= 54 in (girder) + 9in (slab) + 32in (curb and parapet)
=95 in or 7.92ft

71
The distance of the center of gravity of the superstructure area above the top of the abutment is
= (7.92ft/2) +0.12(bearing) = 4.08 ft
= 80 ft. * 7.92 ft. * 0.012 kip/ft2
= 7.6kips (at 7.92/2= 3.96 ft.)
This force transmitted to the substructure through the bearings.
[AASHTO 3.15.2.1.3] The longitudinal (horizontal) wind loading of the superstructure is
W super (H) = 7.6 kips/42ft abutment width
= 0.18 kips/ft
= 7.6 kips*4.08ft/80ft
Wsuper(V) = 0.388 kips/ 42 ft. = 0.009 kip/ft. [negligible]
[AASHTO 3.15.2.2] Determine the wind load on sub-structure, Wsub.
A wind loading of 40ft/ft2 will act perpendicular to the exposed stem of the abutment. The
horizontal wind loading at 10ft(= (12ft/2)+2ft+2ft) above the base of the abutment footing is
Wsub(H)= (0.04kip/ft2)*12ft = 0.48kip/ft. [exposed stem=12ft]

72
[AASHTO 3.15.3] Determine the upward wind load, Wup.
The upward force will be 20 lbf/ft2
of horizontal area for group II and V combinations and 6
lbf/ft2 for group III and VI combinations.
For group II and V, the uplift is
Wup= (80ft*39ft*0.02kip/ft2)/(2*42) = 0.743kip/ft
Whereas, roadway width =39ft
Slab width= 42ft = abutment width
Overall beam length = 80ft
For groups III and VI, the uplift is
Wup= (80ft*39ft*0.006kip/ft2)/(2*42) =0.222 kip/ft
[AASHTO 3.15.2.1.2 & 8.3.15.2.1.3] Determine the wind load, WL, transmitted to the
substructure by the wind load on the moving live load.
The longitudinal wind loading on the live load is taken as 0.04 kip/ft and acts at a point 6ft.
above the deck.
= 54 in (girder) + 9in (slab) + 6ft (wind load on vehicle acts)+ 0.12(bearing)
=11.37 ft

73
= 80 ft. * 0.04 kip/ft
= 3.2kips (at 11.37/2= 5.685 ft.)
The longitudinal (horizontal) wind loading of the superstructure transmitted through bearing.
The reaction at support
= 3.2kips*11.37/80ft = 0.455kip
The horizontal wind loading at the top of the abutment,
WLH= 3.2 kips/42ft
= 0.076 kips/ ft. (including wind load on the girder)
The vertical wind loading at the top of the abutment is
WLV = 0.455 kips/42ft
= 0.011 kips/ft width
[AASHTO Fig. 3.7.6B & 8.3.9] A longitudinal force of 5% of the live load in all lanes is located
6ft above floor slab. For HS 25 loading, the longitudinal force is
= 3 lanes*[80ft*(0.64kip/ft*1.25)+(26kips*1.25)]*0.05
= 14.475 kips
The horizontal force is

74
LFH = 14.475 kip/42 ft. =0.345 kip/ft.
The reaction at support is
= 14.475*11.37/80 = 2.057kips
The vertical force at the top of the abutment is
LFV= 2.057/42 = 0.049 kip/ft width [negligible]
[AASHTO 3.16] Determine the temperature force, Tf, due to friction.
The longitudinal force due to friction at expansion bearings, which is transmitted to both
abutments through the superstructure, is assumed to be 10% (coefficient of friction) of the dead
load reaction.
The friction force is
Tf =( 0.1)*(6.55 kips/ft)= 0.655kip/ft.
The service loads and moments due to the loads are summarized in Table 1.
Perform a stability analysis and bearing pressure check (service load design).
[AASHTO 3.3.6] The density of the normal weight concrete is 150lbf/ft3. The density of the
compacted earth fill is 120lbf/ft3. To compensate for incidental field adjustments in the location
of bearings, a 2 in longitudinal eccentricity from the theoretical centerline of bearing will be
used.
Ls= lateral pressure from 2 ft of soil for live load surcharge

75
= 2 ft.*0.035kip/ft3 * 21.37ft
=1.5kips/ft.
E= lateral earth pressure
= 21.37ft*0.035kip/ft3*21.37ft*0.5
= 7.98kips/ft.
To create the maximum pressure under the toe and the minimum pressure under the heel of the
footing, an eccentricity of 2in to the left for vertical forces will be used.
Check against sliding and overturning.
The factor of safety against sliding is
[AASHTO 5.5.5 & Table 5.5.2B]
FSs= f*∑V/∑H= 0.6*∑V/∑H≥1.5 [Assuming friction factor of 0.6]
The minimum factor of safety against sliding is 1.5.
The factor of safety against overturning is 2 for footings set on soil.
The location of the resultant soil pressure from the toe is given by
Ẋ= ((∑Mv,toe - ∑MH)/∑V)
The eccentricity of the resultant soil pressure is given by
eB= (B/2)-Ẋ

76
For eB < B/6, the pressure under the footing is given by
qt=(∑V/B)(1+(6*eB/B)) at toe
qt=(∑V/B)(1-(6*eB/B)) at heel
[AASHTO 5.5.5 & 4.4.7] For footing set on soil, the location of the resultant soil pressure should
be with in the middle one-third of the base.
The safe bearing capacity of the gravel sand foundation material is assumed to be 8.5 kips/ft2.
Table 3 shows service loads and service load moments at the toe of the footing in the next page.
Table 4 shows stability and bearing pressure in the next pressure.
The minimum values of the factor of safety against sliding and overturning from Table 4 are:
FSs, min= 2.11 from group V loading [>1.5, ok]
FSo, min= 2.43 from group VI loading [>2, ok]
The eccentricity of the resultant soil pressure, ∑V, is
eB, max = 2.24 ft from group VI loading [> B/6 = 11/6=1.83FT]
Check soil pressure qt, max = 8.16 kips/ft2 from group VI loading
[<safe bearing capacity = 8.5 kips/ft, ok]

V Lever arm from toe Mv, toe
calculation kips/ft ft ft-kips/ft
1.00 concrete D 11*2*0.15 3.30 5.50 18.15
2.00 concrete D 14*1.5*0.15 3.15 3.75 11.81
3.00 concrete D 0.5*14*1.17*0.15 1.23 4.89 6.01
4.00 soil E 0.5*14*1.17*0.12 0.98 5.28 5.19
5.00 soil E 14*5.33*0.12 8.95 8.33 74.59
6.00 soil E 6.5*5.37*0.12 4.19 7.75 32.46
7.00 live load surcharge Ls 6.5*2*0.12 1.56 7.75 12.09
8.00 superstructure D 11.60 3.75 43.50
9.00 live load L 5.68 3.75 21.30
10.00 wind upward Wup group II &V -0.74 3.75 -2.79
0.3Wup group III &VI -0.22 3.75 -0.84
load type H
Lever arm from
footing base Mv, toe
kips/ft ft ft-kips/ft
11.00 LS Live load surcharge 1.50 10.69 16.03
12.00 E 7.98 7.12 56.82
13.00 LF 0.35 16.00 5.52
14.00 Tf 0.66 16.00 10.48
15.00 WLH 0.08 16.00 1.22
16.00 Wsuper(H) group II &V 0.18 16.00 2.88
17.00 0.3Wsuper(H) group III &VI 0.05 16.00 0.86
18.00 Wsub group II &V 0.48 10.00 4.80
19.00 0.3Wsub group III &VI 0.14 10.00 1.44
Group
loading ∑V ∑H ∑Mv, toe ∑M H FSs Fso Ẋ eb qt qh
kips kips ft-kips/ft ft-kips/ft
I 40.64 9.48 225.10 72.85 2.57 3.09 3.75 1.75 7.23 0.16
II 32.66 8.64 188.92 64.50 2.27 2.93 3.81 1.69 5.71 0.23
III 40.42 10.10 224.26 81.89 2.40 2.74 3.52 1.98 7.64 -0.29
IV 40.64 10.14 225.10 83.33 2.41 2.70 3.49 2.01 7.75 -0.36
V 32.66 9.30 188.92 74.98 2.11 2.52 3.49 2.01 6.23 -0.29
VI 40.42 10.75 224.26 92.37 2.26 2.43 3.26 2.24 8.16 -0.81
Item
Number
Item
Number
Table 1. Summary of sercice loads and moments
(at the toe of footing, Mv and MH)
Table 2 Stability and Bearing Pressures
Vertical Loads
Description
load type
Horizontal Loads

78
3.2.2.1 Analyze and design for footing (Load factor design)
Factored loads and moments resulting from the factored loads are given in Tables 5-8 .The
factored bearing pressure under the footing at toe qt and heel qh for various group loading
combinations are summarized in Table 9.
Group I loading is critical for the toe of the footing. The factored shear and moment at the front
of the stem for this loading are assuming d=21.37in or 1.63 ft
Vu= (((10.94+9.55)/2)-0.39)*(3-1.63)
= 13.5kips/ft [at a distance d from the face of the stem]
Mu= (((7.9-0.39)*3*1.5 + 0.5*(10.94-7.9)*3*2
= 52.035 ft-kips/ft
Design the footing using strength criteria for concrete and steel of f’c=3000lbf/in2 and Fy =
60000lbf/in2.
[AASHTO 8.16.2.7 & 8.16.3] Calculate the maximum and minimum steel reinforcement
ρb= ((0.85β1f’c/Fy)*(87000/(87000+Fy))
= ((0.85*0.85*3000/60000)*(87000/ (87000+60000))
= 0.02138
ρmax= 0.75*ρb = 0.75*0.02138 = 0.016

V
Lever arm
from toe
calculation kips/ft ft I II III IV V VI
1.00 concrete D 11*2*0.15 3.30 5.50 4.29 4.29 4.29 4.29 4.13 4.13
2.00 concrete D 14*1.5*0.15 3.15 3.75 4.10 4.10 4.10 4.10 3.94 3.94
3.00 concrete D 0.5*14*1.17*0.15 1.23 4.89 1.60 1.60 1.60 1.60 1.54 1.54
4.00 soil E 0.5*14*1.17*0.12 0.98 5.28 1.28 1.28 1.28 1.28 1.23 1.23
5.00 soil E 14*5.33*0.12 8.95 8.33 11.64 11.64 11.64 11.64 11.19 11.19
6.00 soil E 6.5*5.37*0.12 4.19 7.75 5.45 5.45 5.45 5.45 5.24 5.24
7.00
live load
surcharge Ls 6.5*2*0.12 1.56 7.75 3.39 2.03 2.03 1.95
8.00 superstructure D 11.60 3.75 15.08 15.08 15.08 15.08 14.50 14.50
9.00 live load L 5.68 3.75 12.33 7.38 7.38 7.10
10.00 wind upward Wup group II &V -0.74 3.75 -0.97 -0.29 -0.93 -0.28
∑V 59.14 42.46 52.55 52.84 40.83 50.53
V
Lever arm
from toe
calculation kips/ft ft I II III IV V VI
1.00 concrete D 11*2*0.15 3.30 5.50 23.60 23.60 23.60 23.60 22.69 22.69
2.00 concrete D 14*1.5*0.15 3.15 3.75 15.36 15.36 15.36 15.36 14.77 14.77
3.00 concrete D 0.5*14*1.17*0.15 1.23 4.89 7.81 7.81 7.81 7.81 7.51 7.51
4.00 soil E 0.5*14*1.17*0.12 0.98 5.28 6.75 6.75 6.75 6.75 6.49 6.49
5.00 soil E 14*5.33*0.12 8.95 8.33 96.97 96.97 96.97 96.97 93.24 93.24
6.00 soil E 6.5*5.37*0.12 4.19 7.75 42.20 42.20 42.20 42.20 40.58 40.58
7.00
live load
surcharge Ls 6.5*2*0.12 1.56 7.75 26.25 0.00 15.72 15.72 0.00 15.11
9.00 live load L 5.68 3.75 46.24 0.00 27.69 27.69 0.00 26.63
10.00 wind upward Wup group II &V -0.74 3.75 0.00 -3.62 -1.09 0.00 -3.48 -1.04
∑V 321.71 245.60 291.54 292.63 236.16 280.33
Table 3. Factored Vertical Loads
Description
load type
Table 4. Moment resulted from Factored Vertical Loads
Description
Item
Number
Factored V(kips/ft)
Item
Number
Factored Mv, toe(ft-kips/ft)
load type

load type service lever arm I II III IV V VI
11.00 LS Live load surcharge 1.50 10.69 3.26 1.95 1.95 1.88
12.00 E 7.99 7.12 10.39 10.39 10.39 10.39 9.99 9.99
13.00 LF 0.35 16.00 0.45 0.43
14.00 Tf 0.66 16.00 0.85 0.82 0.82
15.00 WLH 0.08 16.00 0.10 0.10
16.00 Wsuper(H) group II &V 0.18 16.00 0.23 0.07 0.23 0.07
18.00 Wsub group II &V 0.48 10.00 0.62 0.19 0.60 0.18
∑H 13.64 11.25 13.14 13.19 11.63 13.46
load type service lever arm I II III IV V VI
12.00 E 7.98 7.12 73.96 73.96 73.96 73.96 71.11 71.11
13.00 LF 0.35 16.00 7.18 6.90
14.00 Tf 0.66 16.00 13.62 13.10 13.10
15.00 WLH 0.08 16.00 1.58 1.52
16.00 Wsuper(H) group II &V 0.18 16.00 3.74 1.12 3.60 1.08
18.00 Wsub group II &V 0.48 10.00 6.24 1.87 6.00 1.80
∑MH 108.75 83.94 106.54 108.42 93.81 115.55
Group
loading ∑V ∑H ∑Mv, toe ∑M H FSs Fso Ẋ eb qt qh
kips kips ft-kips/ft ft-kips/ft
I 59.14 13.64 321.71 108.75 2.60 2.96 3.60 1.90 10.95 -0.19
II 42.46 11.25 245.60 83.94 2.27 2.93 3.81 1.69 7.42 0.30
III 52.55 13.14 291.54 106.54 2.40 2.74 3.52 1.98 9.93 -0.38
IV 52.84 13.19 292.63 108.42 2.40 2.70 3.49 2.01 10.08 -0.47
V 40.83 11.63 236.16 93.81 2.11 2.52 3.49 2.01 7.79 -0.36
VI 50.53 13.46 280.33 115.55 2.25 2.43 3.26 2.24 10.20 -1.02
Table 7 Factored Bearing Pressures
Item
Number
Factored Mv, toe(ft-kips/ft)
Item
Number
Factored V(kips/ft)
5. Factored Horizontal Loads
6. Moment resulted from Factored Horizontal Loads
description
description

82
3.2.2.1.1 Design for the toe.
It will be safe and conservative to neglect compression reinforcement in the design calculation.
For a singly reinforced section, the ultimate moment is
ΦMn = ΦAsFy(d-(⍺/2))
[AASHTO 8.16.1.2.2] Φ=0.9 for flexure
⍺= ((As*Fy)/ (0.85*f’c*b))
[AASHTO 8.22.1] Use d= 24in- 4 in cover – 0.5 in for 0.5 diameter of steel assumed =
19.5 in.
⍺= ((As*60)/(0.85*3*12))= 1.96 As
⍺/2 = 0.98As
Mu = ΦMn
(52.035 ft-kip/ft)*12 in/ft = 0.9 As*60(19.5in- 0.98As)
As2
– 19.9As + 11.8 = 0
As=19.29 in2 or 0.61 in2 [Since As should be less than 1, As=0.61 in2 is correct answer]
ρ= As/bd = 0.61/12*19.5 = 0.0026 [<ρmax = 0.016 ok]
Use #8@15 in( As= 0.62in2)
Check minimum steel
[AASHTO 8.17.1] The minimum reinforcement provided shall be adequate to develop a factored
moment capacity at least 1.2 times the cracking moment, unless the area of reinforcement
provided is at least one third greater than that required by analysis.

83
ΦMn ≥ 1.2Mcr
⍺= ((0.62*60)/(0.85*3*12)) = 1.216 in
⍺/2 = 0.608 in
ΦMn = 0.9*0.62*60*(19.5-0.608)/12 =52.71ft-kips/ft
The cracking moment, Mcr, is determined by the transformed area method.
Mcr = fr*I/ c
fr=modulus of rupture= 7.5(√fc') for normal weight concrete wc=150 lbf/ft3
fc= 4000lbf/in2
I= Icg= moment of inertia
c= distance from centroid axis to extreme fiber in tension
[AASHTO 8.7.1 & 8.7.2] Ec = 33*wc 1.5
(√fc')
= 33 1501.5
(√3000)
= 3,320,560.95lbf/in2
Es = 29,000,000 lbf/in2
n= 29,000,000 / 3,320,560.95 = 8.7 ≅ 9
The transformed area is
(n-1)*As= (9-1)*0.62 = 4.96 in2
Ac +As = 12*24 + 4.96 = 292.96 in2
Taking moment about the bottom and solving the equation,
292.96*yb = 288*12 +4.96*4.5
yb = 11.87in from bottom
The moment of inertia about the neutral axis is

84
Icg= (1/12)(12*243
) + 288*(12-11.87)2
+ 4.96(11.87 – 4.5)2
= 14100 in4
[AASHTO 8.15.2.1.1] fr=modulus of rupture= 7.5(√fc') = 7.5(√3000) =410.79lbf/in2
Mcr = fr*I/ c = (0.411*14100/11.87) /(12) = 40.68 ft.-kips/ft.
1.2Mcr = 1.2*40.68 = 48.82 ft.-kips/ft. [< ΦMn = 52.71 ft- kips/ft., ok]
Use #8@ 15 in.
[AASHTO 8.16.1.2.2 & Equation 8-49] Check shear at a distance d from the face of the stem.
The ultimate shear capacity without shear reinforcement is
ΦVc=Φ*2(√fc') *bw*d
= 0.85*2(√3000) *12*19.5 in
= 21788.4 lbf [> Vu from group I loading (=13.5kips/ft.) at a distance d from face of stem,
ok]
3.2.2.1.2 Design for the heel
Group VI loading is critical for the heel of the footing. The factored shear and moment at the
back of the stem for this loading are as follows.
w=1.25(21.35ft)(0.12) + 2ft* 0.15= 3.506
Vu = (3.506kips/ft2)*5.33ft – 0.5*4.424*4.341
= 9.082 kips/ft
Mu =3.506*4.341*2.67 – 0.5*4.424*4.341*(4.341/3)
= 26.74 ft-kips/ft
[AASHTO 8.22.1] For the heel steel design, use d= 24-3-0.5in for ½ diameter of steel
assumed=20.5in
⍺= ((As*60)/(0.85*3*12)) = 1.96As
⍺/2 = 0.98As

85
The moment in the heel at the back of the stem from group VI loading was 26.74ft-kips/ft
Mu = ΦMn = ΦAsFy(d-(⍺/2))
26.74ft-kips*12in/ft = 0.9*As*60*(20.5-0.98As)
As2
– 20.918As + 6.063 = 0
As= 20.624 in2 or 0.294 in2
As= 0.294 in2/ft
ρ= 0.294/12*20.5 = 0.0012[<ρmax = 0.016 ok]
Try 7@16in, As= 0.45 in2
Check minimum steel
⍺= ((0.45*60)/(0.85*3*12)) = 0.882 in
⍺/2 = 0.441 in
ΦMn = 0.9*0.45*60*(20.5-0.441)/12 =40.62ft-kips/ft
(n-1)*As= (9-1)*0.45 = 3.6 in2
Ac +As = 12*24 + 3.6 = 291.6 in2
Taking moment about the top and solving the equation,
291.6*yt = 288*12 +3.6*3.5
yb = 11.895 in from top
Icg= (1/12)(12*243
) + 288*(12-11.895)2
+ 3.6(11.895 – 3.5)2
= 14080.89 in4
fr=modulus of rupture= 7.5(√fc') = 7.5(√3000) =410.79 lbf/in2
Mcr = fr*I/ c = (0.411*14080.89/11.895) / (12) = 40.54 ft-kips/ft
1.2Mcr = 1.2*40.54 = 48.65 ft-kips/ft [> ΦMn = 40.62 ft- kips/ft, not ok, increase the amount of

86
stee]
Try #8@ 15 in., As =0.62 in2
⍺= ((0.62*60)/(0.85*3*12)) = 1.216 in
⍺/2 = 0.608 in
= 0.9*0.62*60*(20.5-0.608)
= 55.5 ft-kips/ft
(n-1)*As= (9-1)*0.62 = 4.96 in2
Ac +As = 12*24 + 4.96 = 292.96 in2
Taking moment about the top and solving the equation,
292.96*yt = 288*12 +4.96*3.5
yb = 11.86 in from top
Icg= (1/12)(12*243
) + 288*(12-11.86)2 + 4.96*(11.86 – 3.5)2 = 14176.3 in4
Mcr = fr*I/ c = (0.411*14176.3/11.86) /(12) = 40.94 ft-kips/ft
1.2Mcr = 1.2*40.94 = 49.13 ft-kips/ft [< ΦMn = 55.5 ft- kips/ft, ok]
Use 8@15 in, As= 0.62 in2
Check shear in the heel at the back of the stem. The ultimate shear capacity without shear
reinforcement is [AASHTO 8.16.1.2.2]
= 0.85*2(√3000) *12*20.5 = 22900lbf/ft = 22.9 kips/ft [> Vu from group VI loading

87
(=9.082kips/ft) at a distance d from face of stem, ok]
3.2.2.1.3 Design for stem
Analyze for the stem
The stem will be designed for combined axial load and bending. To compensate for incidental
field adjustments in the location of bearings for for vertical loads, a 2in longitudinal eccentricity
from the theoretical centerline of bearing will be used. This 2 in eccentricity will produce the
maximum or minimum moment at the stem base.[1]
Determine the minimum axial load and maximum moment with a 2 in eccentricity toward the
stem front face from the bearing point as follows.
Factored loads are used for designing structural members using the load factor concept. Group
loading combinations for load factor design are given by AASHTO 3.22.1A; and foot notes
βD = 0.75 because of designing members for minimum axial load and maximum moment, and
βE = 0.75 for lateral earth pressure.
Group I =1.3(0.75*D +0.75EVERTICAL+ 1.67(L+LS) +1.3ELATERAL)
Group II =1.3(0.75*D +0.75EVERTICAL+ 1W +1.3ELATERAL)
Group III =1.3(0.75*D +0.75EVERTICAL+ 1L+LS+0.3W+1WL+1.3ELATERAL +1LF)
Group IV =1.3(0.75*D +0.75EVERTICAL +1L+LS +1T+1.3ELATERAL)
Group V =1.25(0.75*D +0.75EVERTICAL+ 1W+1W +1T+0.75ELATERAL)
Group VI =1.25(0.75*D +0.75EVERTICAL+ 1L+LS +0.3W+1WL+1T+0.75ELATERAL +1LF)
The minimum axial load and corresponding moment are
∑V= Pu= 15.68 kips/ft see Table
∑M = Mu= 95.51 ft- kips/ft see Table
The maximum moment and corresponding axial load are

88
The maximum shear is
∑H= Vu= 14.06 kips/ft see Table
Determine the maximum axial load and minimum moment with a 2 in eccentricity toward the
stem rear face from the bearing point. [AASHTO Table 3.22.1A]
βD = 1 because of designing members for maximum axial load and minimum moment, and
βE = 1 for vertical earth pressure
βE = 0.75 for lateral earth pressure.
Group I =1.3(1D +1EVERTICAL+ 1.67(L+LS) +1.3ELATERAL)
Group II =1.3(1D +1EVERTICAL+ 1W +1.3ELATERAL)
Group III =1.3(1D +1EVERTICAL+ 1L+LS+0.3W+1WL+1.3ELATERAL +1LF)
Group IV =1.3(1D +1EVERTICAL +1L+LS +1T+1.3ELATERAL)
Group V =1.25(1D +1EVERTICAL+ 1W +1T+1.3ELATERAL)
Group VI =1.25(1D +1EVERTICAL+ 1L+LS +0.3W+1WL+1T+1.3ELATERAL +1LF)
The maximum axial load and corresponding moment are
The minimum moment and corresponding axial load are
The maximum shear is
∑H= Vu= 14.06 kips/ft see Table

V
Lever arm from center
of stem base
Description kips/ft ft I II III IV V VI
2 concrete D 14*1.5*0.15 3.15 0.58 3.07 3.07 3.07 3.07 2.95 2.95
3 concrete D 0.5*14*1.17*0.15 1.23 -0.56 1.20 1.20 1.20 1.20 1.15 1.15
4 soil E 0.5*14*1.17*0.12 0.98 -0.94 0.96 0.96 0.96 0.96 0.92 0.92
5 soil E 5.37*1.17*0.12 0.75 -0.75 0.74 0.74 0.74 0.74 0.71 0.71
7
live load
surcharge Ls 1.7*2*0.12 0.28 -0.75 0.61 0.37 0.37 0.35
8 superstructure D 11.60 0.75 11.31 11.31 11.31 11.31 10.88 10.88
9 live load L 5.68 0.75 12.33 7.38 7.38 7.10
10 wind upward Wup group II &V -0.74 0.75 -0.97 -0.29 -0.93 -0.28
∑V 30.21 16.31 24.73 25.02 15.68 23.78
service lever arm I II III IV V VI
11 LS Live load surcharge 1.36 9.69 2.95 1.77 1.77 1.70
12 E Lateral 6.57 6.46 11.10 11.10 11.10 11.10 10.68 10.68
13 LF 0.35 14.00 0.45 0.43
14 Tf 0.66 14.00 0.85 0.82 0.82
15 WLH 0.08 14.00 0.10 0.10
16 Wsuper(H)group II &V 0.18 14.00 0.23 0.07 0.23 0.07
18 Wsub group II &V 0.48 8.00 0.62 0.19 0.60 0.18
∑H 14.06 11.96 13.68 13.72 12.32 13.97
V
of stem base
kips/ft ft I II III IV V VI
2 concrete D 14*1.5*0.15 3.15 0.58 1.78 1.78 1.78 1.78 1.71 1.71
3 concrete D 0.5*14*1.17*0.15 1.23 -0.56 -0.67 -0.67 -0.67 -0.67 -0.64 -0.64
4 soil E 0.5*14*1.17*0.12 0.98 -0.94 -0.90 -0.90 -0.90 -0.90 -0.87 -0.87
5 soil E 5.37*1.17*0.12 0.75 -0.75 -0.55 -0.55 -0.55 -0.55 -0.53 -0.53
7
live load
surcharge Ls 1.7*2*0.12 0.28 -0.75 -0.46 -0.27 -0.27 -0.26
8 superstructure D 11.60 0.75 8.48 8.48 8.48 8.48 8.16 8.16
9 live load L 5.68 0.75 9.25 5.54 5.54 5.33
10 wind upward Wup group II &V -0.74 0.75 -0.72 -0.22 0.00 -0.70 -0.21
∑V 16.93 7.42 13.19 13.41 7.13 12.68
11 LS Live load surcharge 1.36 9.69 28.61 17.13 17.13 16.47
12 E Lateral 6.57 6.46 71.73 71.73 71.73 71.73 68.97 68.97
13 LF 0.35 14.00 6.28 6.04
14 Tf 0.66 14.00 11.92 11.46 11.46
15 WLH 0.08 14.00 1.38 1.33
16 Wsuper(H)group II &V 0.18 14.00 3.28 0.98 3.15 0.95
18 Wsub group II &V 0.48 8.00 4.99 1.50 4.80 1.44
∑H 100.34 80.00 99.00 100.78 88.38 106.66
∑Total Moment stem 117.27 87.41 112.19 114.19 95.51 119.34
Table 8. Factored vertical loads
Table 9. Factored Horizontal Loads
Table 10. Factored Moment at stem base resulting from Vertical and Horizontal Loads
Description
Description
Item
Number
Factored Moment V(kips/ft)
Item
Number
Factored V(kips/ft)
Item
Number
Factored H(kips/ft)
Item
Number

V
of stem base
2.00 concrete D 14*1.5*0.15 3.15 0.58 4.10 4.10 4.10 4.10 3.94 3.94
3.00 concrete D 0.5*14*1.17*0.15 1.23 -0.56 1.60 1.60 1.60 1.60 1.54 1.54
4.00 soil E 0.5*14*1.17*0.12 0.98 -0.94 1.28 1.28 1.28 1.28 1.23 1.23
5.00 soil E 5.37*1.17*0.12 0.75 -0.75 0.98 0.98 0.98 0.98 0.94 0.94
7.00
live load
surcharge Ls 1.7*2*0.12 0.28 -0.75 0.61 0.37 0.37 0.35
9.00 live load L 5.68 0.42 12.33 7.38 7.38 7.10
∑V 35.97 22.06 30.49 30.78 21.22 29.32
12.00 E 6.57 6.46 11.10 11.10 11.10 11.10 10.68 10.68
13.00 LF -0.35 14.00 -0.45 -0.43
14.00 Tf -0.66 14.00 -0.85 -0.82 -0.82
15.00 WLH -0.08 14.00 -0.10 -0.10
16.00 Wsuper(H) group II &V -0.18 14.00 -0.23 -0.07 -0.23 -0.07
18.00 Wsub group II &V -0.48 8.00 -0.62 -0.19 -0.60 -0.18
∑H 14.06 10.25 12.07 12.02 9.03 10.78
V
of stem base
2.00 concrete D 14*1.5*0.15 3.15 0.58 2.38 2.38 2.38 2.38 2.28 2.28
3.00 concrete D 0.5*14*1.17*0.15 1.23 -0.56 -0.89 -0.89 -0.89 -0.89 -0.86 -0.86
4.00 soil E 0.5*14*1.17*0.12 0.98 -0.94 -1.20 -1.20 -1.20 -1.20 -1.15 -1.15
5.00 soil E 5.37*1.17*0.12 0.75 -0.75 -0.74 -0.74 -0.74 -0.74 -0.71 -0.71
7.00
live load
surcharge Ls 1.7*2*0.12 0.28 -0.75 -0.46 -0.27 -0.27 -0.26
9.00 live load L 5.68 0.42 5.18 3.10 3.10 2.98
10.60 5.47 8.58 8.71 5.26 8.25
H
11.00 1.36 9.69 28.61 0.00 17.13 17.13 0.00 16.47
12.00 E 6.57 6.46 71.73 71.73 71.73 71.73 68.97 68.97
13.00 LF 0.35 14.00 -6.28 -6.04
14.00 Tf 0.66 14.00 -11.92 -11.46 -11.46
15.00 WLH 0.08 14.00 -1.38 -1.33
16.00 Wsuper(H) group II &V 0.18 14.00 -3.28 -0.98 -3.15 -0.95
18.00 Wsub group II &V 0.48 8.00 -4.99 -1.50 -4.80 -1.44
100.34 63.46 78.72 76.94 49.56 64.23
110.94 68.93 87.30 85.64 54.82 72.48
∑H moment
∑Total Moment stem
Table 11. Factored Vertical Loads
Description
Description
Description
Description
Table 12. Factored Horizontal Loads
Table 13. Factored Moment at stem base resulting from Vertical and Horizontal Loads
∑V moment
Item
Number
Factored Moment H(kips/ft)
Item
Number
Factored V(kips/ft)
Item
Number
Factored H(kips/ft)
Item
Number

91
Design for stem f’c= 3000 lbf/in2
and Fy= 60000lbf/in2
Reinforcement for stem will be designed for bending only as a singly reinforced beam. The
section will then be checked for combined axial force and bending, neglecting the front faces
reinforcing steel.
Use d= 32in – 3 in cover – 0.5 in for ½ steel assumed= 28.5 in
d’’ =(32/2)-3.5 = 12.5 in
⍺= ((As*60)/(0.85*3*12)) = 1.96As
⍺/2 = 0.98As
119.34ft-kips*12in/ft = 0.9*As*60*(28.5-0.98As)
As2
– 29.08As + 27.06 = 0
As= 28.12 in2
or 0.96 in2
As= 0.96 in2/ft
Try #8@9 in, As= 1.05 in2
ρ= 0.96/12*28.5 = 0.0028[<ρmax = 0.016 ok]
Use, #8@ 9 in, As= 1.05 in
Check for the combined axial force and bending.
The AASHTO equations for pure compression and balanced conditions neglect the slight change
of location for plastic centroid when compression steel is used. The footing and wing walls brace
the abutment stem, so slenderness effects will not have to be considered.
[AASHTO 8.16.4.2.1, Equation 8-31 & 8-30]
For pure compression (As=Ast),

92
ΦPo= Φ(0.85*f’c*(Ag-Ast)+Ast*Fy)
= Φ (0.85*f’c*(Ag-Ast)+Ast*Fy)
= 0.7[0.85*3[12*32-1.05] + 1.05*60]
= 715.49 kips/ft
ΦPn(max)= 0.8*715.49 = 572.39 kips/ft [0.8 is the minimum eccentricity for a tied column]
[AASHTO 8.16.4.2.3, Equation 8-32 & 8-33]
For balanced conditions,
ΦPb= Φ(0.85*f’c*b*ab +Asf’s - Ast*Fy)
ΦMb= (0.85*f’c*b*ab[d - d’’- (ab/2)] +A’s*f’s[d- d’ – d’’] + As*Fy*d’’)
ab= (87000/(87000+Fy))β1d [AASHTO Equation 8-34]
fs= 87000(1-(d’/d)[(8700+Fy)/87000]) ≤ Fy
As= 1.05 in2
A’s= f’s=0 [assumed a singly reinforced stem section]
ab = (87000/(87000+60000)*0.85*28.5
= 14.34 in
ΦPb= 0.7*(0.85*3*12*14.34 +0 – 1.05*60)
= 263.06 kips/ft.
ΦMb= (0.85*f’c*b*ab[d - d’’- (ab/2)] +A’s*f’s[d- d’ – d’’] + As*Fy*d’’)
= 0.7(0.85*3*12*14.34*(28.5-12.5- (14.34/2)) + 1.05*60*12.5)
= 3263.5in- kips/ft
[AASHTO 8.16.4.2.2, Equation 8-16 &17]
For pure bending,

93
⍺= AsFy/0.85*f’c*b = 1.05*60/0.85*3*12 = 2.06 in
⍺/2 = 1.03 in
ΦMn = 0.9*1.05*60*(28.5-0.75)/12
= 131.12 ft-kips/ft
Check minimum steel for bending
(n-1)*As= (9-1)*1.05 = 8.4 in2
Ac +As = 12*32 + 8.4 = 392.4 in2
Taking moment about the bottom and solving the equation,
392.4*yb = 384*16 + 8.4*3.5
yb = 15.73 in from bottom
Icg= (1/12)(12*323
) + 384*(16-15.73)2
+ 4.96*(15.73 – 3.5)2
= 34052.41 in4
Mcr = fr*I/ c = (0.411*34052.41/15.73) /(12) = 74.14 ft-kips/ft
1.2Mcr = 1.2*74.14 = 88.97 ft-kips/ft [< ΦMn = 131.12 ft- kips/ft, ok]
Use #8@9 in, As= 1.05 in2
Check for shear in the stem. The ultimate shear capacity without shear reinforcement is
= 0.85*2(√3000) *12*28.5 in
= 31844.59 lbf/ft
= 31.84 kips/ft

94
[> ∑H for group I loading (=14.06 kips/ft), ok]
Determine temperature and shrinkage reinforcement
[AASHTO 8.20.1] Reinforcement should be provided near exposed surfaces not otherwise
reinforced. The area provided shall be As (temperature and shrinkage)= 0.125 in2/ft. in each
direction.
Use # 4@ 15 in (As= 0.16 in2
).
Fig. 12 shows the abutment reinforcement detail.

96
CHAPTER FOUR:
PROJECT MANAGEMENT

97
Project Management
“Project management has five process groups as mentioned in PMBOK Gide, namely: project
initiation, project planning, project execution, project monitoring and control, and project
closure.” [7] Success of any project depends upon how best the activities are managed from
conception till completion. To be successful in a project management, a project manager must
achieve the project objectives by managing the four basic elements of a project: resource, time,
money, and scope. Hence, to manage this direct connector project, the assigned project manager
has to manage by combing effectively the following key elements:
 Resources: people, equipment, material
 Time: task durations, dependencies, critical path
 Money: costs, contingencies, profit
 Scope: project size, goals, requirements
4.1. Project Scope Management
Project scope management clearly defines all roles and responsibilities. For this direct connector
project the role and responsibilities of the Executives Sponsors, Resource Manager, Program
Manager, Project Manager, Customer Service, Team Leader, and Project Team is defined as
shown below which plays key roles in managing the scope of this project. All the teams and
responsible personnel must be aware of their responsibilities in order to ensure that work
performed on the project is within the established scope throughout the entire duration of the
project.
The following table summarizes the roles and responsibilities of the scope management of this
project.

98
Role Role and Responsibilities
Executive sponsor  Oversees project delivery from a business perspective.
 Signs off on results during project delivery, project planning, and
quality assurance reporting.
 Approve or deny scope change requests as appropriate
 Evaluate need for scope change requests
 Accept project deliverables
Resource manager  Assign resources to projects.
Program manager  Works to ensure project requirements are being met.
 Responsible for business planning and administration of the
overall project.
 Responsible for coordinating, monitoring and reporting status of
the project and deliverables to appropriate parties.
 Oversees project delivery from a customer business process
service perspective.
Project Manager  Responsible for coordinating and reporting project status
 Measure and verify project scope
 Facilitate scope change requests
 Facilitate impact assessments of scope change requests
 Organize and facilitate scheduled change control meetings
 Communicate outcomes of scope change requests
 Update project documents upon approval of all scope changes

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