Design Of Post-Tension Continuous Slab

STUDENT’S NAME

1. Sarah Mohammed Shamal

2. Israa Ali Hussein

A Project Report Submitted in Partial Fulfilment of the

requirements for the award of the degree of

Bachelor of Civil Engineering

Civil Engineering Department

Engineering College

University of Misan

Iraq

2018-2019
‫سورة يوسف‬

‫‪II‬‬
 DECLARATION

I hereby declare that this project report is based on my original work

except for citations and quotations which have been duly acknowledged.

Signature : ____________________

Name : Sarah Mohammed Shamal

Date :

Signature :_______________________

Name : Israa Ali Hussein

Date :

III
 APPROVAL FOR SUBMISSION

I certify that this project report “Design Of Post-Tension Continuous Slab”

was prepared by Sarah Mohammed Shamal and Israa Ali Hussein has met
the required standard for submission in partial fulfilment of the
requirements for the award of Bachelor of Civil Engineering at University of
Misan.

Approved by,

Signature : _________________________

Supervisor : Dr. Nasir Hakeem Toma

Date : _________________________

IV
 ACKNOWLEDGEMENTS

I would like to thank everyone who had contributed to the successful

completion of this project. I would like to express my gratitude to my research
supervisor, Dr. Nasir Hakeem Toma for his invaluable advice, guidance
and his enormous patience throughout the development of the research.

In addition, I would also like to express my gratitude to my loving parent who

had helped and given me encouragement......

V
 ABSTRACT

The purpose of this project is studying the design of post-tensioned continuous

slab. Post-tensioning method is widely now a days increasing, due to its
application in construction of many projects around the world. By using post-
tensioning method one can design the most economic and the safe design. Post
tension slabs (PT) are one of the most widely used system which are very effective
method take into consideration the ultimate and serviceability limit states. While
using this method more precautions has to be made for shear criteria for the slabs.
This project concentrates on the analysis and design of two-way flat plate slabs.
The analysis of post- tensioned flat slab can be done by using load balancing and
equivalent frame method. In the load balancing method the 60 to 80% of the dead
load is carried by the tendon itself. So that there is an upward deflection due to
tendon profile resulting the reduction in overall deflection. The design procedures
are based upon the ACI-318 of post-tensioned two –way slab. The effects of
different parameters are investigated in this project. The different parameters such
as include concrete thickness of the flat slab, grade of concrete, pre-stress losses,
normal reinforcement, reinforcement for the shear, stressing force per tendon, and
number of tendons. For the application of design procedure one storey building
ignoring column stiffness in equations for simplicity of hand calculations
considered as a case I study and four stories building as a case II study. The design
procedure was done based on the conventional methods of concrete text books.
The computer programs of STAAD Pro. Was used as aid for the analysis works.
The design works were done by hand calculations and provided aid based on the
specification of ACI-318-14.

VI
 LIST OF CONTENTS

Contents Page

DECLARATION ................................................................................................ III

APPROVAL FOR SUBMISSION ..................................................................... IV
ACKNOWLEDGEMENTS .................................................................................. V
ABSTRACT ........................................................................................................ VI
LIST OF CONTENTS....................................................................................... VII
LIST OF FIGURES AND TABLES ................................................................. IX
LIST OF EQUATIONS...................................................................................... XI

Chapter One .......................................................................................................... 1

Introduction
1.1Background ................................................................................................... 1
1.2Principle of Pre-stressing .............................................................................. 3
1.3 Methods of pre-stressing concrete ............................................................... 5
1.4 Benefits of post-tensioning .......................................................................... 8
1.5Types of post-tensioning systems ............................................................... 10
1.6 Strands ........................................................................................................ 12
1.7 Two-Way Slab Systems ............................................................................. 14

Chapter Two ....................................................................................................... 17

Literature review

Chapter Three ..................................................................................................... 23

Methodology
3.1 Introduction ................................................................................................ 23

VII
 3.2 The Load Balancing Method ..................................................................... 23
3.3 Load Balancing in Continuous Structures ................................................. 27
3.4 Introduction to Hyperstatic (Secondary) Forces ........................................ 30
3.5 Pre-stress Losses ........................................................................................ 35
3.6 ACI 318 Requirements: ............................................................................. 40
3.7 Minimum Bounded Reinforcing: ............................................................... 45
3.8 Flexural Strength ........................................................................................ 48
3.9 Punching Shear .......................................................................................... 50
3.10 Transfer of moments at columns.............................................................. 55
3.11 Tendon layout .......................................................................................... 58

CHAPTER FOUR
HAND CALCULATION .................................................................................... 61
4.1 Case I One story building........................................................................... 62
4.2 case II Four Floors Building....................................................................... 80

Chapter Five ..................................................................................................... 125

Conclusions

References ......................................................................................................... 126

VIII
 LIST OF FIGURES AND TABLES

Figure page

1.1 Several examples of structures resisted only compressive forces…………......1

1.2 Comparison of Reinforced and Pre-stressed Concrete Beams……..………..4
1.3 Stages of pre-tensioning……………………………………………………...5
1.4 Stages of post-tensioning……………………………………………………6
1.5 Reinforced concrete floor and post-tensioned slab………………………… 9
1.6 The components of a typical multi strand grouted system………………....10
1.7 The typical components and construction sequence for an

unbonded system…………………………………………………..............11

1.8 Steel tendons used in Post Tensioning of Slab…………………………….13

1.9 Types of slabs……………………………………………………………...15
2.1 Pre-stressing equivalent load……………………………………………...18
2.2 stress–strain distribution at each loading stage……………….………...….20
3.1 simple span beam with a draped tendon with force………………………...24
3.2 Parabolic Tendon Drape……………………………………….…………..25
3.3 Equivalent Tendon Loads Applied to Beam……………….....………..….25
3.4 Harped Tendon Drape..................................................................................26
3.5 Usual Parabolic Tendon Drape ……………………………..…….………27
3.6 Equivalent Loads for a Reverse Parabolic Tendon Drape…………………27
3.7 Idealized Continuous Parabolic Tendon Drape…………...……………….28
3.8 Equivalent Tendon Loads Applied to Continuous Beam………..……….29
3.9 Post-tensioning induced Reaction……………..…………………………30
3.10 Theoretical Deflected Shape due to Post-Tensioning….………...………31
3.11 Hyperstatic Moment Diagram for a 2-Span Beam……………..………..31

IX
3.12 Hyperstatic Moment Diagram for a 3-Span Beam…………………...……31
3.13 Equivalent Loading Due to Post-Tensioning…………………………..…..33
3.14 Moment Diagram due to balance load…………………… …………. ……33
3.15 primary moment diagram………………………………..…………………34
3.16 Hyperstatic Moment Diagram…………………………………………….34
3.17 Act is graphically defined for several typical cross sections………………45
3.18 tension force in the concrete ………………………………………….......46
3.19 Minimum Lengths of Bonded Reinforcing……………….……………....47
3.20 The rectangular compression block………………………….…………….48
3.21 Failure surface defined by punching shear………………………………..51
3.22 The location of the critical section for various conditions………………..51
3.23 Some techniques to provide adequate punching shear strength ……….…54
3.24 Eccentricity of shear…………………………………………….…..........55
3.25 Calculation of Modulus of critical section………………………..............57
3.26 Principle options of tendon layout………………………………………..59
3.27 Section though the distributiondirection at support………………………60

4.1 final design sketch………………………………………………………79

4.2 minimum bonded reinforcement –East-West interior………………….102

4.3 minimum bonded reinforcement –North –south interior………………122

4.4 Final design sketch of an interior bay……………………………………123

Table

1.1 Common specifications of strands according to ASTM A416……………13

3.1 Typical Mechanical Properties of Standard Pre-stressing Steels…………39

3.2 concrete compressive stress limits immediately after transfer prestress….41
3.3 concrete tensile stress limits immediately after transfer of prestress…......41
3.4 concrete stress limits at service loads………………………………..……43
3.5 calculation of 𝑣𝑐 for two –way shear…………………………………..…..53

X
 LIST OF EQUATIONS

Equation page

3.1…………………………………………………………………………..….24
3.2…………………………………………………………………………..….32
3.3……………………………………………………………………………...42
3.4……………………………………………………………………………...42
3.5…………………………………………………………………………..….42
3.6……………………………………………………………………………...42
3.7……………………………………………………………………………...43
3.8…………………………………………………………………………..….43
3.9…………………………………………………………………………..….43
3.10………………………………………………………………………….…43
3.11………………………………………………………………………….…45
3.12………………………………………………………………………….…46
3.13………………………………………………………………………….…46
3.14……………………………………………………………………….……46
3.15……………………………………………………………….……………47
3.16………………………………………………………………………….…49
3.17…………………………………………………………………………….49
3.18…………………………………………………………………………….49
3.19…………………………………………………………………………….49
3.20…………………………………………………………………………….49
3.21…………………………………………………………………………….49
3.22…………………………………………………………………………….52
3.23………………………………………………………………………...…..52
3.24………………………………………………………………………….....52
3.25……………………………………………………………………….........52
3.26…………………………………………………………………………….56
3.27……………………………………………………………...…………......56
3.28…………………………………………………………………..………...56

XI
 LIST OF SYMBOLS / ABBREVIATIONS

a = depth of equivalent rectangular stress block.

a = drape of tendon measured at centre of profile between points of inflection.

Ab = area of an individual bar or wire.

Ac = area of concrete section resisting shear transfer.

Acf = greater gross cross-sectional area of the slab-beam strips of the two orthogonal
equivalent frames intersecting at a column of a two -way slab.

Act = area of that part of cross section between the flexural tension face and
centroid of gross section.

Ag = gross area of concrete section.

Aps = area of prestressed longitudinal tension reinforcement.

ANC= Anchorage slip.

As,min = minimum area of flexural reinforcement.

As = area of nonprestressed longitudinal tension reinforcement.

Ast = total area of nonprestressed longitudinal reinforcement including bars or

steel shapes, and excluding prestressing reinforcement.

b = width of member.

bo = perimeter of critical section for two-way shear in slabs.

b1 = dimension of the critical section b o measured in the direction of the span for
which moments are determined.

XII
b2 = dimension of the critical section b o measured in the direction perpendicular
to b1.

c1 = dimension of rectangular or equivalent rectangular column, measured in the

direction of the span for which moments are being determined.

c2 = dimension of rectangular or equivalent rectangular column, measured in the

direction perpendicular to c 1.

CR=Creep losses.

db = nominal diameter of bar, wire, or prestressing strand.

dp= distance from extreme compression fiber to centroid of prestressing

reinforcement.

e = The eccentricity of the post-tensioning force with respect to the neutral axis of
the member.

Ec= modulus of elasticity of concrete.

Es= modulus of elasticity of reinforcement and structural steel, excluding

prestressing reinforcement.

ES= Elastic shortening losses.

FR= Frictional losses

ƒc′ = specified compressive strength of concrete.

ƒci′ = specifed compressive strength of concrete at time of initial prestress.

𝑓𝑐𝑑𝑠 = stress in concrete at steel level due to superimposed dead loads applied to the
member after transfer of pre-stress.

XIII
ƒcir= concrete stress at level of steel immediately after transfer.

ƒpc = compressive stress in concrete, after allowance for all prestress losses, at
centroid of cross section resisting externally applied loads or at junction of
web and flange where the centroid lies within the flange.

ƒpe = compressive stress in concrete due only to effective prestress forces, after
allowance for all prestress losses, at extreme fiber of section if tensile stress
is caused by externally applied loads.

ƒps = stress in prestressing reinforcement at nominal flexural strength.

ƒpu = specified tensile strength of prestressing reinforcement.

ƒpy = specified yield strength of prestressing reinforcement.

ƒs = tensile stress in reinforcement at service loads,excluding prestressing

reinforcement.

ƒse = effective stress in prestressing reinforcement, after allowance for all

prestress losses.

ƒy = specified yield strength for nonprestressed reinforcement.

h = overall thickness, height, or depth of member.

I = moment of inertia of section about centroid axis.

M1= primary moment due to the eccentricity of the post-tensioning force with
respect to the neutral axis of the member.

MHPY= Hyperstatic Moment in indeterminate.

MBAL= Balanced load moment.

Msc = factored slab moment that is resisted by the column at a joint.

XIV
Mu= factored moment at section.

Nc = resultant tensile force acting on the portion of the concrete cross section that
is subjected to tensile stresses due to the combined effects of service loads
and effective prestress.

Po = nominal axial strength at zero eccentricity.

RE= Relaxation losses

νc = stress corresponding to nominal two-way shear strength provided by

concrete.

νu = maximum factored two-way shear stress calculated around the perimeter of

a given critical section.

νug = factored shear stress on the slab critical section for two-way action due to
gravity loads without moment transfer.

β = ratio of long to short dimensions.

γf = factor used to determine the fraction of M sc transferred by slab flexure at

slab-column connections.

γv = factor used to determine the fraction of M sc transferred by eccentricity of

shear at slab-column connections

αs = constant used to calculate ν c in slabs.

μ = coefficient of friction.

ρp = ratio of Aps to bdp .

ϕ = strength reduction factor.

Ɵ= Creep coefficient.

XV
C HAPTER ONE
INTRODUTION
Chapter one Introduction

Chapter One
Introduction

1.1Background

Concrete as a building material has been around for thousands of years. Unlike
other isotropic building materials such as steel, wood, and aluminum, concrete and
masonry have a high compressive strength as compared to their relatively weak
tensile strength. Therefore, until the advent of reinforced concrete in the 1800s,
concrete and masonry structures mainly resisted only compressive forces. These
structures generally consisted of columns, arches, and domes to take advantage of
their compressive capacity while eliminating any tensile demand. Several
examples include the following:

Roman Aqueduct Brunelleschi's Dome Stari Most Bridge

Segovia, Spain Basilica di Santa Maria del Fiore Mostar, Herzegovina
(1st Century) Florence, Italy (1461 ) (1567)

Figure (1-1) Several examples of structures resisted only compressive forces.

1
Chapter one Introduction

In the middle of the 1800s, the idea of adding iron to concrete to resist tensile
stresses was first developed. Joseph Monier exhibited this invention at the Paris
Exposition in 1867. With the invention of steel in the later part of the 1800s, the
use of steel reinforcing bars to resist tensile forces in concrete structures quickly
became widespread. Thus, "mild" reinforcing steel is strategically placed within,
and continuously bonded to, concrete members to resist tensile forces to which
they may be subjected. Mild steel reinforcing is also commonly used in
combination with concrete to resist compressive and shear forces.
In the early 1900s, the idea of tightening the reinforcing bars to compensate for the
shrinkage of the concrete was first suggested. Embedded high strength steel rods
were coated to prevent bond with the concrete. “Pre-stressed Concrete ”soon
became the single most significant new direction in structural engineering. This
unique concept gave the engineer the ability to control the actual structural
behavior while forcing him or her to dive more deeply into the construction process
of the structural material. It gave architects as well as engineers a new realm of
reinforced concrete design pushing not only the structural but also the architectural
limits of concrete design to a level that neither concrete nor structural steel could
achieve.
Ordinary reinforced concrete could not achieve the same limits because the new
long spans that pre-stressed concrete were able to achieve could not be reached
with reinforced concrete. Those longer spans required much deeper members,
which quickly made reinforced concrete uneconomical. Additionally, steel
structures weren’t able to create the same architectural forms that the new pre-
stressed concrete could.

2
Chapter one Introduction

1.2Principle of Pre-stressing

The function of pre-stressing is to place the concrete structure under compression

in those regions where load causes tensile stress. Tension caused by applied loads
will first have to cancel the compression induced by the pre-stressing before it can
crack the concrete. Figure [1.2(a)] shows a plainly reinforced concrete simple span
beam and fixed cantilever beam cracked under applied load. Figure [1.2(b)] shows
the same unloaded beams with pre-stressing forces applied by stressing post-
tensioning tendons. By placing the pre-stressing low in the simple-span beam and
high in the cantilever beam, compression is induced in the tension zones; creating
upward camber. Figure [1.2(c)] shows the two pre-stressed beams under the action
of post-tensioning and applied loads. The loads cause both the simple span beam
and cantilever beam to deflect down, creating tensile stresses in the bottom of the
simple-span beam and top of the cantilever beam. The designer balances the effects
of load and pre-stressing in such a way that tension from the loading is
compensated by compression induced by the pre-stressing. Tension is eliminated
under the combination of the two and tension cracks are prevented. As a result,
durability is increased and more efficient, cost effective construction is realized.

3
Chapter one Introduction

Figure (1-2) Comparison of Reinforced and Pre-stressed Concrete Beams

4
Chapter one Introduction

1.3 Methods of pre-stressing concrete

Pre-stressed concrete includes both pre-tensioned and post-tensioned concrete.

Pre-tensioning. The prefix "pre" means that the pre-stressing steel is stressed before
the concrete is cast. This method consists of first stressing high-strength steel strands
or wires between buttresses, and then casting the concrete around the steel. Once
the concrete has reached a certain specified strength, the steel is cut between the
ends of the members and the buttresses to transfer the pre-stressing forces to the
concrete This process typically takes place at a precast plant and requires the
completed pre-tensioned concrete member to be trucked out to the job site and then
assembled.

Figure (1-3) Stages of pre-tensioning

5
Chapter one Introduction

The other method of pre-stressing concrete is called post-tensioning. The prefix

"post" means that the pre-stressing ' steel is stressed after the concrete is cast.
Instead of stressing the high-strength steel between buttresses at a precast plant.
The steel is simply installed on the job site after the contractor forms up the
member. The high-strength steel is housed in a sheathing or duct that prevents it
from bonding to the concrete. The steel is attached to the concrete at the ends of
the member by specially designed anchorage devices. Once the concrete has cured
(hardened), the steel is stressed to induce forces in the concrete.

Figure (1-4) Stages of post-tensioning

6
Chapter one Introduction

Post-tensioning is an attractive method of reinforcing concrete slabs and beams

due to its versatility and economy. Post-tensioning is also more versatile than pre-
stressing due the ability to cast members in place, including many situations where
pre-stressing is not feasible.

Post-tensioning is a highly efficient structural system that offers many benefits in

a wide range of construction, repair, and rehabilitation applications.
Post-tensioning has been successfully used for small as well as large projects. The
efficiency stems from being able to use high strength materials, to structurally
utilize the entire cross section. To vary the force and location of the reinforcing to
best resist applied loads, and to control the timing of when the pre-stressing force
is applied to the structure.

Post-tensioning offers a perfect balance of two materials which complement each

other. Concrete is strong in compression and relatively weak in tension. The tensile
strength of concrete is about 10% of its compressive strength. Pre-stressing steel,
on the other hand, has a very high tensile strength (270,000 psi for strand) which
is about four times that of common reinforcing bars. By combining the two, a
structural member can resist both compressive and tensile forces caused by various
loads. This results in greater efficiency in resisting tensile as well as compressive
stresses resulting from the applied loads.
Post-tensioning can be used in all facets of construction from buildings and bridges
to highway pavements, slabs-on-ground and ground anchors. It has also been used
for rehabilitation and retrofit applications.

7
Chapter one Introduction

1.4 Benefits of post-tensioning

Post-tensioned concrete is used in commercial buildings, residential apartments,

high-rise condominiums, office buildings, parking structures, and mixed-use
facilities such as hotels and casinos. Benefits of post-tensioning include:
 A significant reduction in the amount of concrete and reinforcing steel required.
 Thinner structural members as compared to non-pre-stressed concrete, resulting
in lower overall building heights and reduced foundation loads.
 Aesthetically pleasing structures that harness the benefits of cast in place
structures with curved geometries, and longer, slender members with large
spaces between supports.
 Superior structural integrity as compared to precast concrete construction
because of continuous framing and tendon continuity.
 Monolithic connections between slabs, beams, and columns that can eliminate
troublesome joints between elements.
 Profiled tendons that result in balanced gravity loads (typically a portion of dead
load only), significantly reducing total deflection.
 Better crack control, which results from permanent compressive forces applied
to the structure during pre-stressing.
 Post-tensioning reduces overall building mass, which is important in zones of
high seismicity.

Post-tensioning also offers the following construction advantages as compared to

steel, non-pre-stressed concrete and precast construction:
 Faster floor construction cycle
 Lower floor weight

8
Chapter one Introduction

 Lower floor-to-floor height

 Larger spans between columns
 Reduced foundations

High early-strength concrete allows for faster floor construction cycles. The use of
standard design details of the post-tensioned elements, minimum congestion of pre-
stressed and non-pre-stressed reinforcement, and earlier Stripping of formwork
after tendon stressing can also significantly reduce the floor construction cycle.
Greater span-to-depth ratios are allowed for post-tensioned members as compared
to non-pre-stressed members. This results in a lighter structure and a reduction in
floor-to-floor height while maintaining the required headroom.

Figure (1-5) Reinforced concrete floor and post-tensioned slab

9
Chapter one Introduction

1.5Types of post-tensioning systems

In most post-tensioned construction, the pre-stressing tendons are embedded in the

concrete before the concrete is cast. The posttensioned systems can be either
bonded or unbonded.

1.5.1Bonded Post-Tensioning Systems

Bonded post-tensioning systems consist of tendons with multiple strands or bars.

The strands or bars are placed in corrugated galvanized steel, high density
polyethylene (HDPE) or polypropylene (PP) ducts. Depending on the site
conditions and system used the strands may be installed before the concrete is
placed or the ducts may be installed without the strands. The strands are then pulled
or pushed through the ducts. Once the concrete has hardened the tendons are
stressed and the ducts filled with grout. Inlets and outlets are provided at high/low
points to ensure that the grout fills the ducts completely.Figure(1.6) shows the
components of a typical multistrand grouted system.The grout provides an alkaline
environment and protects the prestressing strands from corrosion.It also bonds the
strands to the surrounding concrete.

Figure(1-6) The components of a typical multistrand grouted system.

10
Chapter one Introduction

1.5.2Unbonded Post-Tensioning Systems

The tendons in an un-bonded system typically consist of single-strands that are

coated with a corrosion-inhibiting coating and protected by extruded plastic
sheathing. This allows the strand to move inside the plastic sheathing and prevents
ingress of water. The strands are anchored to the concrete using ductile iron anchors
and hardened steel wedges. The tendon is supported by chairs and bolsters along
its length to maintain the desired profile. Figure (1-7). Depending on the exposure
of the single-strand unbonded system it can be classified as a standard or an
encapsulated system. Encapsulated systems are required for aggressive
environments where there is a possibility of tendon exposure to chlorides or other
deleterious substances. Encapsulated tendons are designed to prevent any ingress
of water during and after construction.

Figure (1-7) the typical components and construction sequence for an unbonded
system.

11
Chapter one Introduction

1.6 Strands

Strand for post-tensioning is made of high tensile strength steel wire. A strand is
comprised of 7 individual wires, wrapping six wires around a central straight
wire.All strand should be Grade 1860 Mpa (270 ksi) low relaxation, seven-wire
strand conforming to the requirements of ASTM A416 “Standard Specification for
Steel Strand, Uncoated Seven Wire Strand for Prestressed Concrete.” ASTM
A416 provides minimum requirements for mechanical properties (yield, breaking
strength, elongation) and maximum allowable dimensional tolerances. Strand is
most commonly availablein two nominal sizes,12.7mm (0.5in) and 15.2mm
(0.6in)diameter, with nominal cross sectional areas of 99mm2and 140mm2 (0.153
and 0.217 square inches), respectively.Though the majority of post-tensioning
hardware and stressing equipment is based on these sizes, the use of 15.7mm
(0.62in) diameter strand has been increasing.Strand size tolerances may result in
strands being manufactured consistently smaller than, or larger than nominal
values. Recognizing this,“Acceptance Standards for Post-Tensioning Systems”
(Post-Tensioning Institute, 1998) refers to the “Minimum Ultimate Tensile
Strength” (MUTS), which is the minimum specified breaking force for a strand.
Strand size tolerance may also affect strand-wedge action leading to possible
wedge slip if the wedges and strands are at opposite ends of the size tolerance
range.Strand conforming to ASTM A416 is relatively resistant to stress corrosion
and hydrogen embrittlement due to the cold drawing process. However, since
susceptibility to corrosion increases with increasing tensile strength, caution is
necessary if strand is exposed to corrosive conditions such as marine environments
and solutions containing chloride or sulfate,phosphate, nitrate ions or similar.

12
Chapter one Introduction

Consequently, ASTM A416 requires proper protection of strand throughout

manufacture, shipping and handling. Protection during the project, before and after
installation, should be specified in project drawings and specifications.

Figure (1-8) Steel tendons used in Post Tensioning of Slab

Table(1-1) Common specifications of strands according to ASTM A416

13
Chapter one Introduction

1.7 Two-Way Slab Systems

Structural two way slabs may be classified as follows:

1.7.1 Flat Plate: Span not exceedi ng 6.0 to 7.5 m and live oad not exceeding
3.5 to 4.5 kN/m2

Advantages

1. Low cost formwork

2. Exposed flat ceilings

3. Flexibility in column arrangement

Disadvantages

1. Low shear capacity

2. Low Stiffness (notable deflection)

3. Need of special formwork for drop panels and capitals

1.7.2 Waffle Slab: Span up to 14 m and live load up to 7.5 kN/m2

Advantages

1. Carries heavy loads

2. Attractive exposed ceilings

Disadvantages

1. Formwork with panels is expensive

14
Chapter one Introduction

1.7.3 Slab with beams: Span up to 10 m

Advantages

1. Versatile

2. Framing of beams with columns

Disadvantages

1. Visibility of drop beams in ceilings

(a) Flat Plate (b) Flat slab withdrop panels

(c) Slab with beams (d) Waffle Slab

Figure (1-9) Types of slab

15
 CHAPTER TWO
LITERATURE REVIE
`
Chapter two literature review

Chapter Two

Literature review

The capitals and/or drop panels are generally applied to strengthen the slab-
column conjunction and improve its punching shear resistance. Introducing these
strengthening detailing in two-way post-tensioned slab structures changes the
stiffness and moment for using the equivalent frame analysis and causes two
additional pre-stressing equivalent loads: upward point load and concentrated
moment at the interfaces between columns and/or capitals and drop panels. Wei
Zhou [3] used a moment-area procedure to analyze a slab-beam with step
haunches, rotational stiffness, carry over and fixed-end moments of a flat slab
with capitals and/or drop panels. The design factors that derived represent in
graphical tables and used Equivalent Frame Method (EFM) for analyzing and
designing two-way post-tensioned slabs. Both of equivalent pre-stressing loads: a
point load and/or a concentrated moment were evaluated for the stiffer column-
slab conjunction region due to tensioning pre-stressing tendons. The expressions
of two equivalent loads due to pre-stress were derived in result of the dimension
of member end and pre-stressing effect. Compared to a prismatic member, a stiffer
column-slab joint region increases the stiffness and carryover factor, and
influences fixed-end moment of slab under gravity and pre-stressing loads. Herein
five kinds of equivalent loads were required to analyze the inner force due to post-
tensioning the tendons, including partially uniform loads, horizontal compression
loads applied at both of ends, additional pre-stressing equivalent load.

17
Chapter two literature review

Figure (2-1) Pre-stressing equivalent load.

Faria , Lúcio and Pinho Ramos[4] they studied a new flat slab strengthening
technique based on post-tensioning with anchorages by bonding using an epoxy
adhesive. The main advantages of this technique over the traditional pre-stress
strengthening systems that use mechanical anchorages are that it did not need
external permanent anchorages, meaning that the forces are introduced into the
concrete gradually instead of being localized, thereby preserving aesthetics and
useable space. They used the seven tested slab models show that this technique met
its objective as it is able to reduce reinforcement strains at service loads by up to
80% if the strengthening technique is applied in two directions and slab
deformations by up to 70%, consequently making crack widths smaller. It can also
increase punching load capacity by as much as 51% when compared to non-
strengthened slabs.

18
Chapter two literature review

Mohamed H.AbuGazia, Mahmoud El-Kateb, Tamer Elafandy and Amr

Abdelrahman [5], they proposed the strengthening continuous slabs using
external pre-stressing .It is one of the widely used methods of enhancing the
flexural strength of the Reinforced Concrete (RC) slabs since it is fast, effective,
and economic compared to other strengthening techniques. An experimental
investigation was conducted on two-span continuous slabs strengthened with
externally pre-stressed steel strands using different profiles. The ultimate load of
the strengthened slabs increased significantly due to using external pre-stressing.
Stitching the external pre-stressing strand in the slab at the zones of high stresses
(at the intermediate support and mid span regions) resulted in a significant increase
of the flexural capacity with less values of deflection and cracks.

The un-bonded post-tension (UPT) method has been often applied to continuous
flexural members (e.g., slabs and beams), where tendons are typically placed
continuously through the supports. Most of the previous studies, however, focused
on simply supported UPT flexural members, whereas little research has been
performed on continuous UPT members. Moreover, as the few existing studies on
continuous beams mainly aimed at strength prediction, the results of these studies
are not applicable to examine the service load behavior of UPT members. They
also did not reflect the moment redistribution phenomenon in continuous members,
which is quite important to accurately predict the flexural strength and behavior of
continuous UPT members. Therefore, a flexural behavior model for continuous
UPT members has been proposed by Kang Su Kim and Deuck Hang Lee [6] ,
which is a nonlinear analysis model that reflects the moment redistribution. The
accuracy of the proposed analysis model for flexural behavior of the continuous
UPT members was consistent, regardless of the

19
 Chapter two literature review

tendon profiles, the types of the applied load, and the shapes of the section. It was
also confirmed that the proposed analysis model could also be applied to the cases
of the internal and external post-tension method and of the various reinforcement
indices of the members. The effect of the loading types on the flexural behavior of
continuous UPT members was reflected by using the curvature distribution in the
maximum moment region and the area of bending moment diagram, which were
considered to be relatively simple and rational, based on the analysis results, also
reflected the moment redistribution phenomenon utilizing the bending moment
distribution along the member and the flexural stiffness ratio.

Figure (2.2) stress–strain distribution at each loading stage.

Dezsõ Hegyi - András Árpád Sipos 7] They designed a post-tensioned concrete

cantilever slab with a 6.50 m free span supported by columns for a villa near Pécs.
The shape and number of the bonded strands were determined to balance the dead
load of the structure by the transversal component of the pre-stressing force. The
deflections measured on finished structure were in good agreement with the
approximated values of the structural calculation. During the design they
investigated some other structural possibilities as well and found that the 350 mm
thick slab is the most efficient solution for the problem. The final solution is

20
Chapter two literature review

definitely the thinnest compared to the other variants namely the system of steel
beams (HEA members, the thickness is 400 mm) or a reinforced concrete slab
strengthened by beams (the thickness is at least 550 mm).

Erez[8] studied the factors affecting the increase in tendon stress at nominal
strength, ∆ƒps, in un-bonded partially pre-stressed continuous concrete members. It
described a nonlinear numerical model that capable of predicting the response up
to failure of un-bonded, partially pre-stressed continuous concrete beams and it
presented a comparison of results from the model against test data. Loading pattern,
type of loading and degree of concrete confinement are shown by means of a
parametric study to have a significant effect on the tendon stress at ultimate.
Finally, modifications were suggested to the current A23.3-94 Canadian Code
equation for predicting the tendon stress at ultimate in concrete members pre-
stressed with un-bonded tendons, in order to consider the contribution from all
plastic hinges likely to develop under a particular pattern of loading. A comparison
between test data and predictions according to provisions in the Canadian and the
American Codes (A 23.3-94and ACI-318) revealed a poor agreement. The pattern
of loading appeared to have a significant influence on the value of, ∆ƒ ps, because
the increase in tendon stress is directly related to the maximum number of plastic
hinges that can be developed under a given pattern of loading. The larger the
number of hinges, the larger is the increase in tendon stress. The change in tendon
stress was influenced by the type of loading. Significantly higher values of, ∆ƒ ps
were predicted when the member was subjected to a two-point loads per span or a
uniformly distributed load, than when a single-point load per span was applied.

21
CHAPTER THREE
METHODOLOGY
Chapter three Methodology

Chapter Three
Methodology

3.1 Introduction
This chapter covers the fundamentals of post-tensioned concrete design for
building structures and learn about the load balancing concept, hyperstatic
moments, pre-stress losses, the basic requirements of the American Concrete
Institute’s Building Code Requirements for Structural Concrete, and nominal
flexure and shear capacities of post-tensioned members.

3.2 The Load Balancing Method

The load balancing method is the most widely used technique to design post-
tensioned concrete beams and slabs. In the load balancing method, a portion of the
design load is selected to be "balanced", or carried, by the action of the tendons.
The balanced load is commonly taken as 60 to 80% of the dead load. The required
force in the tendons to carry the balanced load is easily calculated using statics. The
concrete member is then analyzed using conventional structural analysis techniques
with the equivalent set of tendon loads acting on the member in combination with
other externally applied design loads, such as dead load and live load.

Let's consider the free body diagram at mid-span of the following simple span beam
with a draped tendon with force P . Note that the common simplifying assumption
made in post-tensioned concrete analysis is that the tendon force acts in the
horizontal direction at the ends of the member and the small vertical component, if

23
Chapter three Methodology

any, can be ignored or is transferred directly to the support. Note that the shear force
at the right side of the free body diagram is zero since this occurs at the mid-span
of a simple span beam with a uniformly distributed load.

Figure(3-1) simple span beam with a draped tendon with force.

If we sum the moments about the force P at the left support, we get:

∑𝑀=0
𝑤𝑙 𝑙
× =𝑃 ×𝑎
2 4

𝑤𝑙 2
=𝑃 Eq(3.1)
8𝑎

where

L: distance between points of inflection

a:drape of tendon measured at centre of profile between points of inflection.

P = average prestessing force in tendon

24
Chapter three Methodology

The load balancing concept is further illustrated in the figure below, which shows
a simply supported beam and a tendon with a parabolic profile. The beam shown
in the first figure below may be analyzed with an equivalent set of tendon loads
acting on the member as shown in the second figure. Thus, the equivalent loads
acting on the beam consist of the axial force P, an upward uniform load of w, and
a clockwise moment M at the left end due to the eccentricity of the tendon with
respect to the neutral axis of the beam.

Figure (3-2) Parabolic Tendon Drape

Figure (3-3) Equivalent Tendon Loads Applied to Beam.

25
Chapter three Methodology

Note that reactions are induced at both ends to keep the system in equilibrium. If
we sum the moments about the left support, we get:

∑𝑀= 0

8𝑃𝑎 𝐿
× 𝐿 × − 𝑃𝑒 − 𝑅𝑟 × 𝐿 = 0
𝐿2 2
4Pa M
− = Rr
L L

And if we sum the vertical forces we find the left reaction is:

4𝑃𝑎 𝑀
+ = 𝑅𝐿
𝐿 𝐿

Note that the vertical component of the applied pre-stressing force is neglected.
This is practical since the tendons are customarily horizontal, or very nearly
horizontal, at the end of the members, and the vertical component is usually small.
As we have seen, a draped tendon profile supports, or balances, a uniformly
distributed load. Now let's consider a beam that is required to support a
concentrated load. In the case of a concentrated load on a beam, a concentrated
balancing load would be ideal. This can be achieved by placing the pre-stressing
tendons in a harped profile. This concept using harped tendons is illustrated in the
figure below.

Figure (3-4) Harped Tendon Drape

26
Chapter three Methodology

3.3 Load Balancing in Continuous Structures

Let's now turn our attention to the load balancing concept applied to continuous
structures, post-tensioning in continuous structures induces secondary , or so-called
hyperstatic, forces in the members. Consider the figure below. In real continuous
structures, the tendon profile is usually a downward parabola at the supports and
an upward parabola between supports such that the tendon drape is a smooth curve
from end to end. The tendon curve changes at an inflection point. This tendon
configuration actually places an equivalent downward load on the beam near the
supports between inflection points while an upward equivalent load acts on the
beam elsewhere, as shown in the figure below. This type of reverse loading can be
taken into account in computer analysis, where a more rigorous approach to the
structural analysis can be accommodated. However, for hand and more
approximate calculations, tendon drapes are idealized as a single upward parabolic
drape in each span.

Figure (3-5) Usual Parabolic Tendon Drape

Figure (3-6) Equivalent Loads for a Reverse Parabolic Tendon Drape

27
Chapter three Methodology

Now let's consider the figure below, which shows a two span continuous beam with
a cantilever on the right end. Each span has a different tendon drape as shown.

Figure (3-7) Idealized Continuous Parabolic Tendon Drape

As in the previous examples, the continuous beam shown above may be analyzed
with an equivalent set of tendon loads acting on the member. Thus, the equivalent
loads acting on the each beam span due to the pre-stressing force in the tendons
consist of the axial force P and an upward uniform load. Since the tendon force, P
, acts at the neutral axis at the ends of the beam in this example, there are no end
moments induced due to the eccentricity of the tendon. The diagram on the
following page shows the equivalent set of tendon loads acting on the beam for the
diagram above. Note that the drape "a" in span 1 is not equal to the drape "b" in
span 2. Drape "c" in the right cantilever is also different. If we assume that the
tendons are continuous throughout all spans of the beam, then the post-tensioning
force, P , is also constant throughout all spans. Therefore, for a given post-
tensioning force, P , we may balance a different amount of load in each span,
depending on the drape in each span and the span length, according to the equation
(3.1).

28
Chapter three Methodology

Figure (3-8) Equivalent Tendon Loads Applied to Continuous Beam

However, it is not usually desirable or practical to balance different loads in each

span. Nor is it practical to apply a different post-tensioning force in each span to
balance the same load in each span, except in end spans or in spans adjacent to a
construction joint, where tendons can be added. Therefore, the drape can be
adjusted in each span to balance the same amount of load in each span. Thus, for a
given post-tensioning force and balanced load, we can find the required drape
𝒘𝒍𝟐
𝒂=
𝟖𝒑

29
Chapter three Methodology

3.4 Introduction to Hyperstatic (Secondary) Forces

Hyperstatic, or secondary, forces are the forces generated in a statically

indeterminate structure by the action of the post-tensioning. Generally, hyperstatic
forces are generated due to support restraint. Hyperstatic forces are not generated
in a statically determinate structure. It is important to isolate the hyperstatic forces
as they are treated with a separate load factor when considering ultimate strength
design.

Let's consider the two span post-tensioned beam in the following illustration. We
know from previous examples that the tendon force will create an upward
uniformly distributed load acting on the beam as shown in the figure. If the center
support were not there, the beam would deflect upward due to the post-tensioning
force. Since the center support is there, and it is assumed that it resists the upward
deflection of the beam, a downward reaction is induced at the center support solely
due to the post tensioning force.

Figure (3-9) Post-tensioning induced Reaction

30
Chapter three Methodology

The next figure shows the deflected shape of the two-span beam due to the
posttensioning force (ignoring self-weight) as if the center support were not there.
In order to theoretically bring the beam back down to the center support, a force
equal to the center reaction would have to be applied to the beam. This induces a
hyperstatic moment in the beam. The hyperstatic moment diagram is illustrated
below for the two span beam in this example. Note that the hyperstatic moment
varies linearly from support to support. An example of a three-span hyperstatic
moment diagram is also illustrated below.

Figure (3-10) Theoretical Deflected Shape due to Post-Tensioning

Figure (3-11) Hyperstatic Moment Diagram for a 2-Span Beam

Figure (3-12) Hyperstatic Moment Diagram for a 3-Span Beam

31
Chapter three Methodology

Note that the above example assumes that the beam is supported by frictionless pin
supports and therefore no moments can be transferred into the supports by the
beam. However, in real structures, the post-tensioned beam or slab is normally built
integrally with the supports such that hypersatic moments are also induced in
support columns or walls.

Hyperstatic moments in support members are normally ignored in hand and

approximate calculations, but can and should be accounted for in post-tensioning
computer software. Thus, when designing supporting columns or walls in
posttensioned structures, it is important to take into account hyperstatic moments
induced by post-tensioning forces. The above illustrations serve to introduce the
concept and source of hypersatic forces in continuous post-tensioned structures.
The hyperstatic moment at a particular section of a member is defined as the
difference between the balanced load moment and the primary moment. We refer
to the primary moment as M1 and this is the moment due to the eccentricity of the
post-tensioning force with respect to the neutral axis of the member at any given
section. In equation form,

MHYP=MBAL - M1 Eq. (3.2.a)

MHYP=MBAL - P×e Eq. (3.2.b)

Consider the two-span post-tensioned beam shown below. As we have seen

previously, the beam can be analyzed with equivalent loads due to the tendon force.
The draped tendons with force P may be replaced with an equivalent upward acting
uniform load of 8Pa/L2. When the two-span beam is analyzed

32
Chapter three Methodology

using this load, a moment diagram is developed as illustrated below. This is called
the balanced moment diagram. Recall that the sign convention results in a negative
moment when tension is in the top.

Figure (3-13) Equivalent Loading Due to Post-Tensioning

Figure (3-14) Moment Diagram due to balance load

Using our sign conventions, we can construct a primary moment diagram for
momentsM1. The primary moment is obtained from the product of the post
tensioning force times its eccentricity with respect to the neutral axis of the beam
at any given section. Thus, referring to the diagram above, the primary moment at

33
Chapter three Methodology

the center support is P × e1 and the primary moment at the mid-span of each span
is P × e2 According to our sign conventions, the primary moment at the center
support is positive.

Figure (3-15) primary moment diagram

Now we can determine the hyperstatic moments using the equation (3.3) at the
center support, we obtain

MHYP= Pa-Pe1

Near the mid-span of each span we have

4Pa
MHYP=− + Pe2
7

Figure (3-16) Hyperstatic Moment Diagram

34
Chapter three Methodology

3.5 Pre-stress Losses

When an unbonded tendon is stressed, the final force in the tendon is less than the
initial jacking force due to a number of factors collectively referred to as prestress
losses. When a hydraulic jack stresses an unbonded tendon, it literally grabs the
end of the tendon and stretches the tendon by five or six or more inches. As the
tendon is stretched to its scheduled length, and before the jack releases the tendon,
there is an initial tension in the tendon. However, after the tendon is released from
the jack and the wedges are seated, the tendon loses some of its tension, both
immediately and over time, due to a combination of factors, such as seating loss,
friction, concrete strain, concrete shrinkage, concrete creep, and tendon relaxation.
Prior to the 1983 ACI 318, pre-stress losses were estimated using lump sum values.
For example, the loss due to concrete strain, concrete creep, concrete shrinkage,
and tendon relaxation in normal weight concrete was assumed to be 25,000 psi for
posttensioned concrete. This did not include friction and tendon seating losses. This
was a generalized method and it was subsequently discovered that this method
could not cover all situations adequately. Since the 1983 ACI 318, each type of pre-
stress loss must be calculated separately. The effective pre-stress force, that is the
force in the tendon after all losses, is given on the design documents and it is
customary for the tendon supplier to calculate all prestress losses so that the number
of tendons can be determined to satisfy the given effective prestress force. Even
though most design offices do not normally calculate prestress losses, it is
informative and relevant to understand how losses are calculated. Therefore, each
type of pre-stress loss is discussed in more detail below.

35
Chapter three Methodology

• Seating Loss

When an unbonded tendon is tensioned, or stretched, to its full value, the jack
releases the tendon and its force is then transferred to the anchorage hardware, and
thereby into the concrete member. The anchorage hardware tends to deform
slightly, which allows the tendon to relax slightly. The friction wedges deform
slightly, allowing the tendon to slip slightly before the wires are firmly gripped.
Minimizing the wedge seating loss is a function of the skill of the operator. The
average slippage for wedge type anchors is approximately 0.1 inches. Seating
losses are more significant on shorter tendons. There are various types of anchorage
devices and methods, so the calculation of seating losses is dependent on the
particular system used.

• Friction and Wobble

Loss of pre-stress force occurs in tendons due to friction that is present between the
tendon and its surrounding sheathing material as it is tensioned, or stretched.
Friction also occurs at the anchoring hardware where the tendon passes through.
This is small, however, in comparison to the friction between the tendon and the
sheathing (or duct) throughout its length. This friction can be thought of as two
parts; the length effect and the curvature effect. The length effect is the amount of
friction that would occur in a straight tendon - that is the amount of friction between
the tendon and its surrounding material. In reality, a tendon cannot be perfectly
straight and so there will be slight "wobbles" throughout its length. This so called
wobble effect is rather small

36
Chapter three Methodology

compared to the curvature effect. The amount of loss due to the wobble effect
depends mainly on the coefficient of friction between the contact materials, the
amount of care and accuracy used in physically laying out and securing the tendon
against displacement, and the length of the tendon. The loss in the pre-stressing
tendons due to the curvature effect is a result of the friction between the tendon and
its surrounding material as it passes through an intentional curve, such as drape, or
a change in direction, such as a harped tendon. The amount of loss due to the
curvature effect depends on the coefficient of friction between the contact
materials, the length of the tendon, and the pressure exerted by the tendon on its
surrounding material as it passes through a change in direction.

 Elastic Concrete Strain

When a concrete member is subjected to a compressive force due to pre-stressing

tendons, it will shorten elastically. If this compressive force were removed, the
member would return to its original length. Although ACI 318 does not specifically
give procedures or requirements for calculating losses due to elastic strain, there
are references available that provide some guidance. In general, the elastic strain
shortening is a simplified computation involving the average net compressive stress
in the concrete due to pre-stressing and the modulus of elasticity of the pre-stressing
steel and the concrete at the time of stressing.

 Concrete Creep
A well know phenomenon of concrete in compression is that it creeps, or shortens,
over time. The creep rate diminishes over time. Although ACI 318 does not
specifically give procedures or requirements for calculating losses due to creep,

37
Chapter three Methodology

there are references available that provide some guidance. In general, the creep
shortening is a simplified computation involving the average net compressive stress
in the concrete due to pre-stressing and the moduli of elasticity of the pre-stressing
steel and the 28-day concrete strength. For post-tensioned members with unbonded
tendons, for example, creep strain amounts to approximately 1.6 times the elastic
strain.

 Concrete Shrinkage:
The hardening of concrete involves a chemical reaction called hydration between
water and cement. The amount of water used in a batch of concrete to make it
workable far exceeds the amount of water necessary for the chemical reaction of
hydration. Therefore, only a small portion of the water in a typical concrete mix is
consumed in the chemical reaction and most of the water evaporates from the
hardened concrete. When the excess mix water evaporates from a particular
concrete member, it loses volume and therefore tends to shrink. Reinforcing steel
and surrounding construction can minimize concrete shrinkage to some extent, but
nonetheless shrinkage stresses are developed. Factors that influence concrete
shrinkage include the volume to surface ratio of the member, the timing of the
application of pre-stressing force after concrete curing, and the relative humidity
surrounding the member.

 Tendon Relaxation:
As a pre-stressed concrete member shortens, the tendons shorten by the same
amount, thus relaxing some of their tension. The concrete member shortens due to
the above three sources –elastic strain, creep, and shrinkage. Thus, the total tendon

38
Chapter three Methodology

relaxation is a summation of these three concrete shortening sources. There may

also be some relaxation in the pre-stressing steel over time, similar to concrete
creep. This steel relaxation is a function of the type of steel used and on the ratio
of actual steel stress to the specified steel stress.

Table (3-1) Typical Mechanical Properties of Standard Pre-stressing Steels

As mentioned earlier, most design offices only show the required effective pre-
stress force and the location of the center of gravity of the tendons on the
construction documents. Remember that the effective pre-stress force is the force
in the tendons after all losses have been accounted for. Calculating pre-stress losses
for the design office can be very tedious and may not be exact, and therefore it is
customary for the tendon supplier to calculate all pre-stress losses based on their
experience and the specific tendon layout. Then, the number of tendons are
determined to satisfy the given effective pre-stress force.

39
Chapter three Methodology

3.6 ACI 318 Requirements:

We will now review some of the requirements contained in the 2014 edition of the
Building Code Requirements for Structural Concrete ACI 318.

3.6.1 The allowable extreme fiber tension stress at service loads:

ACI 318 places limits on the allowable extreme fiber tension stress at service loads
according to the classification of a structure. Class U members are assumed to
behave as uncracked sections and therefore gross section properties may be used in
service load analysis and deflection calculations. Class C members are assumed to
be cracked and therefore cracked section properties must be used in service load
analysis, and deflection calculations must be based on an effective moment of
inertia or on a bilinear moment-deflection relationship. Class T members are
assumed to be in a transition state between cracked and uncracked, and the Code
specifies that gross section properties may be used for analysis at service loads, but
deflection calculations must be based on an effective moment of inertia or on a
bilinear moment-deflection relationship.
 The allowable extreme fiber tension stresses in flexural members at service
loads according to ACI (24.5.2.1) are as follows:

(a)Class U: 𝒇𝒕 ≤ 0.62 √𝒇′𝒄

(b) Class T: 0.62√𝒇′𝒄 < 𝒇𝒕 ≤ 1.0 √𝒇′𝒄

(c) Class C: 𝒇𝒕 > 1.0 √𝒇′𝒄

Pre-stressed two-way slabs must be designed as Class U and the extreme fiber
tension stress must not exceed 0.50√𝒇′𝒄 .

40
Chapter three Methodology

3.6.2 Stages of loading:

ACI 318 stipulates two cases of serviceability checks. The first is a check of the
concrete tension and compression stresses immediately after transfer of pre-stress.
The concrete stresses at this stage are caused by the pre-stress force after all short
term losses, not including long term losses such as concrete creep and shrinkage,
and due to the dead load of the member. These limits are placed on the design to
ensure that no significant cracks occur at the very beginning of the life of the
structure. The initial concrete compressive strength, f' ci is used in this case. f'ci is
normally taken as 75% of the specified 28-day concrete compressive strength.

 The maximum permissible concrete stresses at force transfer (before time-

dependent prestress losses) according to ACI (24.5.3.1) and (24.5.3.2) for
compressive and tensile stress respectively shall not exceed the following:

concrete compressive stress limits immediately after transfer of prestress

Location Concrete compressive stress limits

End of simply-supported member 0.7 𝒇′𝒄𝒊
All other location 0.6 𝒇′𝒄𝒊

Table (3-2) concrete compressive stress limits immediately after transfer prestress

concrete tensile stress limits immediately after transfer of prestress, without additional
bonded reinforcement in tension zone

Location Concrete tensile stress limits

End of simply-supported member 0.50 √𝒇′𝒄𝒊

All other location 0.25 √𝒇′𝒄𝒊

Table (3-3) concrete tensile stress limits immediately after transfer of prestress

41
Chapter three Methodology

To check the slab stress in top and bottom fibers with the allowable stress, use the
follow equation

Mid span Stresses

−MDL +Mbal P
𝑓𝑡𝑜𝑝 = − Eq.(3-3)
S A
MDL −Mbal P
𝑓bot = − Eq.(3-4)
S A

Support Stresses
(𝑀𝐷𝐿 −𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑡𝑜𝑝 = − Eq.(3-5)
𝑆 𝐴
(−𝑀𝐷𝐿 +𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑏𝑜𝑡 = − Eq.(3-6)
𝑆 𝐴

If the above stresses are exceeded at force transfer, additional bonded

reinforcement shall be provided in the tensile zone to resist the total tensile force
in concrete computed with the assumption of an uncracked section.

The second serviceability check is a check of the concrete tension and compression
stresses at sustained service loads (sustained live load, dead load, superimposed
dead load and pre-stress) and a check at total service loads (dead load,
superimposed dead load, and pre-stress) these checks are to preclude excessive
creep deflection and to keep stresses low enough to improve long term behavior.
Note that the specified 28-day concrete compressive strength is used for these stress
check.
 The maximum permissible concrete stresses at the service load ((based on
uncracked section properties, and after allowance for all prestress losses) state
according to ACI (24.5.4.1) shall not exceed the following:

42
Chapter three Methodology

concrete stress limits at service loads

Load condition concrete stress limits

compression due to pre-stress plus sustained load 0.45ƒ′𝒄

tension due to pre-stress plus sustained load 0.5√ƒ′𝒄

Table (3-4) concrete stress limits at service loads

To check the slab stress in top and bottom fibers with the allowable stress, use the
follow equation

Mid span Stresses

(−𝑀𝐷𝐿 −𝑀𝑙𝐿 +𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = − Eq.(3-7)
𝑆 𝐴

(𝑀𝐷𝐿 +𝑀𝑙𝑙 −𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = − Eq.(3-8)
𝑆 𝐴

Support Stresses

(+𝑀𝐷𝐿 +𝑀𝑙𝐿 −𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = − Eq.(3-9)
𝑆 𝐴

(−𝑀𝐷𝐿 −𝑀𝑙𝑙 +𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = − Eq.(3-10)
𝑆 𝐴

If the above stresses are exceeded, then additional bounded reinforcement shall
be provided in the tensile zone to resist the total tensile force. Also note that the
above stress checks only address serviceability. Permissible stresses do not ensure
adequate structural strength.

43
Chapter three Methodology

3.6.3 Final effective force

Tensile stress in Post-tensioning tendons, at anchorage devices and couplers,

immediately after force transfer according to ACI (20.3.2.5.1) shall not
exceed…………… 0.70 ƒpu .

In the above, ƒpu is the specified tensile strength of the pre-stressing steel. The most
commonly used pre-stressing steel in the United states is grade 270, low-relaxation,
seven wire strand, Defined by ASTM416. therefore, for this common pre-stressing
steel, ƒpu =270Ksi(1860Mpa).

3.6.4 Average pre-compression

For prestressed slabs, the effective prestress force 𝐴𝑝𝑠 ƒ𝑠𝑒 shall provide a
minimum average compressive stress of 0.9 MPa on the slab section tributary
to the tendon or tendon group. For slabs with varying cross section along the slab
span, either parallel or perpendicular to the tendon or tendon group, the minimum
average effective prestress of 0.9MPa is required at every cross section tributary to
the tendon or tendon group along the span to address punching shear concerns of
lightly reinforced slabs. For this reason, the minimum effective prestress is required
to be provided at every cross section.

44
Chapter three Methodology

3.7 Minimum Bounded Reinforcing:

All flexural members with unbonded tendons require some amount of bounded
reinforcing. For beams and one-way slabs, this bonded reinforcing is required
regardless of the services load stresses. For two-way slabs, the requirement for this
bonded reinforcing depends on the services load stresses. This bonded reinforcing
is intended to limit crack width and spacing in case the concrete tensile stress
exceeds the concrete’s tensile capacity at service loads. ACI requires that this
bonded reinforcement be uniformly distributed and located as close as possible to
the tension face.

 ACI (7.6.2.3) For beams and one-way slabs, the minimum area of bonded
reinforcing As is:

As=0.004Act Eq. (3-11)

Act is defined as the area of the part of the cross section between the flexural tension
face and the center of gravity of the cross-section.

Figure (3-17) Act is graphically defined for several typical cross sections.

45
Chapter three Methodology

 ACI (8.6.2.3) For two-way post-tensioned slabs with undonded tendons, bonded
reinforcing is not required in positive moment areas (In the bottom of the slab)
if the extreme fiber tension at service loads, after all pre-stress losses, does not
exceed the following:

ƒ t ≤ 0.17√ƒ′𝐜 Eq.(3-12)

Recall that the maximum extreme tension fiber stress in a two-way slab is 0.5√ƒ′𝐜 ,
so a minimum amount of bottom steel is required for tensile stresses in the range
of:

0.17 √ƒ′𝐜 < ƒ t≤ 0.5√ƒ′𝐜 Eq.(3-13)

When it is required per the above equation for two-way post-tensioned slabs with
unbonded tendons, the minimum area of bonded reinforcement in positive moment
regions is
𝐍𝐜
𝐀𝐬 = Eq. (3-14)
𝟎.𝟓ƒ𝐲

Where Nc is the tension force in the concrete due to unfactored (service) dead puls
live load and is illustrated below.

Figure (3-18) tension force in the concrete

46
Chapter three Methodology

 For two-way post-tensioned slabs with unbonded tendons, the minimum area of
bonded reinforcement required in negative moment regions at column supports
is

𝐴𝑠 = 0.00075𝐴𝑐f Eq. (3-15)

Where 𝐴𝑐f is defined as the area of the larger gross cross-sectional area of the slab
beam strips in two orthogonal equivalent frames intersecting at the column.

 ACI (8.7.5.5.1) Length of deformed reinforcement required by shall be in

positive moment areas, length of reinforcement shall be at least ℓn/3 and be
centered in those areas and in negative moment areas, reinforcement shall extend
at least ℓn/6 on each side of the face of support.

Figure (3-19) Minimum Lengths of Bonded Reinforcing

ACI (8.7.5.3) The bonded reinforcing required in negative moment regions at

columns shall be distributed between lines that are 1.5h outside both sides of the
face of the column support, shall be a minimum of four bars, and shall be spaced
no more than 300 mm on center.

47
Chapter three Methodology

3.8 Flexural Strength

Let us now begin our investigation into the ultimate flexural strength of a pre-
stressed member. The design moment strength may be computed using the same
methodology used for non-prestressed members. That is, a force couple is
generated between a simplified rectangular compression block and the equal and
opposite tensile force generated in the reinforcing steel, which is in our case pre-
stressing steel.

we will narrow our focus to post-tensioned members with unbonded tendons, and
we will use the approximate values for the nominal stress in the pre-stressing steel,
ƒps, instead of using strain compatibility. For more accurate determinations of the
nominal stress in the pre-stressing steel, and for flexural members with a high
percentage of bonded reinforcement, and for flexural members with prestressing
steel located in the compression zone, strain compatibility should be used.

Figure (3-20) The rectangular compression block

The rectangular compression block has an area equal to a times b. Equating the
compression resultant, C, to the tensile resultant, T, the nominal moment capacity
can be written as
𝐚 𝐚
Mn=T(dp- ) = 𝐂(dp- )
𝟐 𝟐

48
Chapter three Methodology

𝐚
ϕMn= ϕAps ƒps (dp- ) Eq.(3-16)
𝟐

Where Aps is the area of pre-stressed reinforcement, ƒps is the stress in the pre-
stressed reinforcement at nominal moment strength, and ϕ is the strength reduction
factor (0.90 for flexure). ACI 318 (20.3.2.4.1) defines the approximate value for
ƒps the lesser of the following, depending on the span-to-depth ratio
span
For ≤ 35 , the lesser of
depth

ƒ′c
ƒps= ƒse +70+ ≤ ƒpy or ƒse +420 Eq.(3-17)
100𝛒𝐩

span
For > 35 , the lesser of
depth

ƒ′c
ƒps= ƒse +70+ ≤ ƒpy or ƒse +210 Eq.(3-18)
300𝛒𝐩

where ƒse is the effective stress in the pre-stressing steel after all losses. The depth
of the compression block is defined as

𝐴𝑝𝑠 ׃𝑦
a= Eq.(3-19)
0.85 ƒ′c b

Let's now consider including the contribution of the bonded reinforcing steel to the
nominal moment capacity. The tensile component of the moment couple becomes
the sum of both pre-stressing steel and the bonded reinforcement can be written as
follows
𝐚
ϕMn= ϕ(Aps ƒps + As ƒy) (dp- ) Eq.(3-20)
𝟐

The depth of the compression block then becomes:

𝐴𝑝𝑠 ׃𝑝𝑠 +𝐴𝑠 ׃𝑦

a= ′ Eq.(3-21)
0.85 ƒc b

49
Chapter three Methodology

3.9 Punching Shear

Punching shear, or two-way shear, at column supports is often the controlling factor
in establishing a slab thickness. This failure mechanism occurs when the column
below the slab literally “punches through” the slab due to overload, insufficient
shear strength, or construction defects. This type of failure tends to be catastrophic
and therefore the shear strength of the slab at column supports must be examined
very carefully. ACI 318 also requires that one-way shear, or beam shear, be
checked, but this condition does not usually control in two-way slabs and will not
be covered here in this project. ACI 318 also requires, in most situations, two
structural integrity tendons to pass through the column core in each direction in
case of a punching shear failure. These integrity tendons are intended to prevent a
total collapse or a progressive collapse.

The first step in analyzing two-way shear in slabs is to define the critical section.
The critical section is established by an imaginary line around the slab support
along which the support is assumed to punch through the slab. ACI 318 defines the
critical section to be located so that its perimeter, b0, is a minimum but need not be
closer to the face of the support than d/2, where d is the depth to the centroid of the
tension reinforcing. Thus the critical section is b0 times d. Note that the critical
section is an area over which the shear stress is assumed to be distributed. The
dashed line in the following sketches illustrate the location of the critical section
for various conditions see figure (3-22). It should also be mentioned that openings
through the slab in the vicinity of the column tends to reduce the punching shear
strength of the slab. They also can reduce the flexural capacity of the slab in one or
both directions. ACI 318 gives requirements to account for these openings, but they
are beyond the scope of this project.

50
Chapter three Methodology

Figure (3-21)Failure surface defined by punching shear.

(a) Square column (b) column with drop panel

(c) Rectangular column (d) L-shape column

Figure (3-22) The location of the critical section for various conditions.

51
Chapter three Methodology

At columns of two-way pre-stressed slabs, the nominal punching shear strength of

the slab 𝑣c according to ACI(22.6.5.5) shall be permitted to be the lesser of (a)
and (b):
𝑉𝑝
(a) 𝑣c= 0.29 λ√𝑓′𝑐 +0.3ƒpc+ Eq.(3-22 )
𝑏𝑜 𝑑

𝛼𝑠 𝑑 𝑉𝑝
(b) 𝑣c=0.083(1.5+ ) λ√𝑓′𝑐 +0.3ƒpc+ Eq.(3-23)
𝑏𝑜 𝑏𝑜 𝑑

Where:
αs = 40, 30, and 20 for interior, edge, and corner columns, respectively.
λ= Lightweight Concrete Factor, 1.0 for Normal Weight Concrete.
𝑉𝑝 =vertical component of all effective pre-stress forces.
ƒpc = is the average value in the two directions and shall not exceed 3.5 MPa

√𝑓′𝑐 =shall not exceed 5.8 MPa.

In typical buildings with post-tensioned two-way slabs, normal weight concrete is

customarily used and the vertical component of the pre-stress force can be ignored
in preliminary design since it is small due to the relatively flat drapes in two-way
slabs. However, in real structures where reverse parabolic tendon drapes are
normally used, the vertical component of the pre-stress should be included. This is
accounted for is computer software packages. Assuming normal weight concrete
and neglecting the vertical component of the pre-stress, the nominal punching shear
strength of two-way prestressed slabs at columns in typical situations can be
simplified to:

(a) 𝑣c= 0.29 √𝑓′𝑐 +0.3ƒpc Eq(3-24)

𝛼𝑠 𝑑
(b) 𝑣c=0.083(1.5+ ) √𝑓′𝑐 +0.3ƒpc Eq(3-25)
𝑏𝑜

52
Chapter three Methodology

The above equation applies at interior columns of two-way post-tensioned slabs,

and at perimeter columns only if the slab edge extends past the face of the column
by at least 4 times the slab thickness h. If the slab does not extend far enough past
the face of the column, the pre-stress force in the slab ƒpc , is not fully effective, in
which case the nominal shear strength calculation reverts back to the smallest of
the following non-prestressed two-way nominal shear strength according ACI
(22.6.5.2) equations:

𝑣𝑐

0.33 λ√𝑓′𝑐 (a)

Least of (a),(b)and (c) 2 (b)
0.17(1+ ) λ√𝑓′𝑐
𝛽

𝛼𝑠 𝑑 (c)
0.083(2+ ) λ√𝑓′𝑐
𝑏𝑜

Table (3-5) Calculation of 𝑣𝑐 for two –way shear.

where β is the ratio of long side to short side of the column.

In most typical buildings, the slab does not usually extend 4h past the outside face
of the columns and so the above equations for non-prestressed slabs would apply.

There are various techniques to provide adequate punching shear strength in two-
way slabs. These include sufficient flat plate thickness, drop panels, column
capitals, reinforcing bars, shear head reinforcement, and headed stud shear
reinforcement. In this project, we will only be studying flat plate slabs (slabs
without drop panels) without shear reinforcement. For example, the use of a drop
panel merely increases the slab thickness at the column, providing more shear
strength at the column, but also creates a second critical section just outside the

53
Chapter three Methodology

drop panel that must also be checked. (Drop panels also influence the equivalent
frame stiffness and the distribution of moments and shears). Headed stud shear
reinforcement is quite commonly used and is usually included in computer design
software.

a) Shear stud reinforcement for concrete slab

b) shear head reinforcement

Figure (3-23) Some techniques to provide adequate punching shear strength

54
Chapter three Methodology

3.10 Transfer of moments at columns

In two-way continuous slab systems, slab moments must be transferred to the

column support through the slab-column joint. Slab moments to be transferred may
be caused by lateral and/or gravity loads. A portion of the moment is required to be
transferred from the slab to the column through flexure, and the remainder is
required to be transferred by “eccentricity of shear.” This concept is illustrated
below.

Figure (3-24) Eccentricity of shear

From the above figures, by statics, we can write the following equations for the
unit shear stress, vu,AB and vu,CD, at the critical section according to ACI(8.4.4.2.3):

55
Chapter three Methodology

𝛾𝑣 𝑀𝑠𝑒 𝑐AB
𝑣 u,AB =𝑣𝑢𝑔 + Eq(3-26)
𝐽𝑐

𝛾𝑣 𝑀𝑠𝑒 𝑐CD
𝑣𝑢,𝐶𝐷 =𝑣𝑢𝑔 − Eq(3-27)
𝐽𝑐

The above equations are analogous to P/A +/- Mc/I. The total shear demand on the
critical section is thus the sum of the direct shear (VUg) and the eccentricity of shear
(γvMUc/J) due to moment. The terms in the above expressions are defined as:

Ac = The area of concrete resisting shear, or b0 times the effective depth d, or the
critical section

J = A property of the critical section analogous to the polar moment of inertia.

Notes on ACI gives a very convenient tabular summary of the section properties of
criticalmsections for rectangular and circular columns.

γv = fraction of moment to be transferred by eccentricity of shear = 1 – γf

1
γf = fraction of moment to be transferred by flexure γf= 2 𝑏
Eq(3-28)
1+( )√ 1
3 𝑏2

One study found that, for square columns, approximately 60% of the moment is
transferred by flexure about the critical section and approximately 40% of the
moment is transferred by eccentricity of shear about the centroid of the critical
section. ACI provides equation(3-28) to calculate the fraction of moment
transferred by flexure and shear for column shapes other than square . These
equations are conveniently graphed in Notes on ACI and, in general, are not very
sensitive for aspect ratios up to about 2.

From Notes on ACI, we find the following equations for the properties of the
critical section based on the moment acting to the edge and position of column:

56
Chapter three Methodology

Figure (3-25) Calculation of Modulus of critical section

57
Chapter three Methodology

3.11 Tendon layout

When designing two-way, post-tensioned slab systems, in addition to the Code

provisions and various checks, there is an important consideration that simplifies
the construction process. Imagine a multi-span two-way slab system with tendons
draped in two perpendicular directions. Installing tendons in this configuration
would be very difficult, and would be like weaving a basket. Thus, there are several
possible arrangements for the layout of the tendons in each design strip. Figure (3-
26) illustrates the alternatives. Note that the tendons in each direction may be
arranged in banded, distributed, or a mixed layout. In the banded direction all of
the tendons of a design strip are grouped in a number of flat bundles and placed
parallel to one another with a relatively small gap separating the constituent
bundles. The tendons form a narrow band, typically up to or slightly larger than
1.20m in width, following the support line. Tendons in the distributed direction are
placed in bundles of one to 4 strands, spread over the entire width of the design
strip with essentially equal spacing between the bundles.

 Banded tendons in one direction, and distributed in the other direction.

 Banded in both directions.
 Distributed in both directions.
 Mixed banded and distributed in both directions.
All of the stated four options are deemed to provide equal strength capacity. The
choice of layout is generally governed by constructability. The option of banded in
both directions (Figure (3-26(a)) is not permitted by the ACI Code. The
constructability advantage of this scheme is that it does not require interweaving
of tendons in different directions. The distributed tendons directly over the support

58
Chapter three Methodology

(see Figure 3-27) are placed and secured in position first, followed by placement
of all banded tendons. Then the rest of the distributed tendons are placed over the
bands. Most other tendon layout schemes require some interweaving. One other
advantage of the banded distributed option, from a design standpoint, is that both
directions can be designed with the maximum permissible tendon drape.
Banded and distributed tendons generally do not cross at their high or low points,
with the exception of two distributed tendons over the supports (see Fig. (3-27)).
Therefore, the bulk of the strands can be placed with the maximum allowable drape
without interference from tendons in the perpendicular direction.

Figure (3-26) Principle options of tendon layout

59
Chapter three Methodology

Figure (3-27) Section though the distributiondirection at support

60
 CHAPTER FOUR
HAND CALCULATION
Chapter four Hand calculation

4.1 Case I One story building

The following example illustrates the design method presented. The example
presented here is for Two-tensioned design

Loads:

Framing dead load =self weight

Superimposed dead load=1.2 KN/m2

partitions, M/E,misc

Live load =3 KN/m2 Laboratories

Material:

Concrete normal weight =24 KN/m3

𝑓′𝑐 =35 N/mm2

𝑓′𝑐𝑖 =21 N/mm2

Rebar 𝑓𝑦 =414 N/mm2

PT: unbonded tendons

1.27 cm ∅,7-wire

strands,A= 99mm2

ƒpu=1860 N/mm2

Determine preliminary slab Thickness

Start with L/h=45

Longest span =9.14m

62
Chapter four Hand calculation

(9.14m)(100)
h=
45

=20.3cm

Take t=22cm preliminary slab thickness

DESIGN OF EAST-WEST INTERIOR FRAME

Ignore column stiffness in equations for simplicity of hand calculation

Total bay width between centerlines = 7.62m

No pattern loading required, since LL/DL<3/4 ACI (6.4.3.3)

Estimated prestress losses ACI (20.3.2.6)

1) Elastic shorting=0 (simultaneous tensioning and anchorage)

2) Shrinkage:
0.0002
Ɛsh =
log10(t+2)
0.0002
= = 1.35×10-4
log10(28+2)

SH =Ɛsh × Es
=(1.35×10-4)(2×105)=27MPa
27
SH% = × 100 = 1.45%
1860

3)Relaxation=2.5%
4) anchorage slip:
∆ = 0.2 in
∆𝑙
∆𝑓𝑠𝑙𝑖𝑝 = 𝐸𝑠
𝑙
5.08
= × 2 × 105
25.6×1000

=40 MPa

63
Chapter four Hand calculation

40
Anchorage slip%= × 100 = 2.15%
1860

5) creep losses:

𝐶𝑅 = 𝜃𝑛(𝑓𝑐𝑖𝑟 − 𝑓𝑐𝑑𝑠 )

𝑃° 𝑃° × 𝑒 2 𝑀𝐺 × 𝑒
𝑓𝑐𝑖𝑟 =− − +
𝐴𝑔 𝐼 𝐼

𝑏ℎ3 7.62×0.223
𝐼= = = 6.76 × 10−3 𝑚4
12 12

Ag =bh

=7.62×0.22=1.6764 𝑚2

𝑃° = 𝐴 × 𝑓𝑝𝑢

= 99 × 1860 =184140N= 184.14KN

22 1.27
e= − 2.5 − =7.87cm
2 2

Wd=ɤ×t

Wd =24×0.22=5.28KN/m2

Wd=5.28×7.62=40.23KN/m

64
Chapter four Hand calculation

Max. moment =303.09 KN.m

184.14 184.14×0.07872 303.09×0.0787

𝑓𝑐𝑖𝑟 = − − + = 3.25 𝑀𝑝𝑎
1.6764 6.76×10−3 6.76×10−3

Wcds=1.2×7.62=9.14KN/m

max. moment due to superimpose dead load =68.86 KN.m

𝑀×𝑐 68.86 ×0.0787
ƒcds= = =0.802MPa
𝐼 6.76×10−3

𝜃 = 1.6 (28 𝑑𝑎𝑦𝑠 )

𝐸𝑠
n=
𝐸𝑐

2×105
= = 7.2
4700√35

CR=1.6×7.2(3.25 − 0.802)= 28.2MPa

28.2
CR%= × 100 =1.5%
1860

6) friction losses:

65
Chapter four Hand calculation

The following properties of parabolas are used. For segment (1-2), the parabola in
the sketch below is used.

The change in slope from the origin to the end of the parabola is same as the slope
at the end of the tendon which is α= 2e/L,

α1-2= α 6-7 =2e/L

2×0.0787
= = 0.046
3.365

For segments 2-3 and 3-4 and subsequent pairs of segments, the following property
is used.

For the two parabolic segments joined at the inflection point as shown in the sketch
above, the slope at the inflection point

α = 2(e1+ e2)/λL

2(0.0787+0.0787)
α2-3= α5-6= =0.065
4.865

2(0.0787+0.0787)
α3-4= α4-5= = 0.069
4.57

For extruded seven-wire strand:

Curvature coefficient (µ) =0.05

66
Chapter four Hand calculation

Wobble coefficient (K) = 0.00984/m

Segments cumulative Wobble Curvature Length uα+kL Exp(-uα-kL) stress@

angle coefficient coefficient end of segment
α K, per m µ m
1-2 0.046 0.00984 0.05 3.365 0.0354116 0.965208055 0.965208055PO
2-3 0.065 0.00984 0.05 4.865 0.0511216 0.950163124 0.9171051 PO
3-4 0.069 0.00984 0.05 4.57 0.0484188 0.952734698 0.873757851 PO
4-5 0.069 0.00984 0.05 4.57 0.0484188 0.952734698 0.832459423 PO
5-6 0.065 0.00984 0.05 4.865 0.0511216 0.950163124 0.790972245 PO
6-7 0.046 0.00984 0.05 3.365 0.0354116 0.965208055 0.763452782 PO

Stress@ end of segment =0.763452782×184.14=140.58 Mpa

140.58
FR%= × 100 = 7.6%
1860

Total losses=ES+SH+RE+ANC+CR+FR
Total loss = 0+27+46.5+40+28.2+140.58=282.28Mpa
ƒse=0.7 ƒpu - Total losses
ƒse=0.7(1860)-282.28=1019.72Mpa ACI(20.3.2.5.1)
Peff=A׃se= (99)(1019.72)= 100952.28 N= 100.95kN

67
Chapter four Hand calculation

Calculate section properties

Two-way slab must be designed as class U (ACI 24.5.2.1), Gross-Sectional Properties

allowed

S=bh2/6=7.62x 0.222/6=0.0615m3

Set Design parameters

Allowable stresses: class U(ACI 24.5.2.1)

At time of jacking (ACI (24.5.3.1) and (24.5.3.2) for compressive and tensile stress
respectively)

𝑓′𝑐𝑖 =21 N/mm2

compression =0.6 𝑓′𝑐𝑖 =0.6(21)=12.6 MPa

Tension =0.25√𝑓𝑐𝑖′ =0.25×√21 =1.15 MPa

At service loads (ACI 24.5.4.1)

𝑓′𝑐 = 35MPa

Compression = 0.45 ƒ'c = 0.45(35) = 15.75MPa

Tension = 0.5√𝑓′𝑐 = 0.5√35 = 3MPa

Average precompression limits(3.6.4)

𝑃
= 0.9MPa min. (ACI 8.6.2.1)
𝐴

Target load balances:

60%-80% of DL(self weight) for slabs (good approximation for hand calculation)

For this example: 0.75 WDL = 0.75(5.28) = 3.96 KN/m2

68
Chapter four Hand calculation

Tendon profile:

Parabolic shape;

For a layout with spans of similar length, the tendons will be typically be located at
the highest allowable point at the interior columns, the lowest possible point at the
mid spans, and the neutral axis at the anchor locations. This provides the maximum
drape for load-balancing.

Tendon Ordinate Tendon (CG) Location*

Exterior support 22/2=11

Interior support-top 22-2.5-1.27/2=18.865

Interior span – bottom 2.5+1.27/2=3.135

End span – bottom 2.5+1.27/2=3.135

(CG) = center of gravity

*Measure from bottom of slab

aINT = 18.865-3.135=15.73cm

(11+18.865)
aEND = − 3.135=11.798cm
2

eccentricity, e, is the distance from the center to tendon to the neutral axis; varies
along the span

Prestress Force Required to Balance 75% of selfweight DL

Since the spans are of similar length, the end span will typically govern the maximum
required post-tensioning force. This is due to the significantly reduced tendon drape,
aEND.

wb = 0.75 WDL = 0.75(5.28)(7.62)=30.18 KN/m

69
Chapter four Hand calculation

Force needed in tendons to counteract the load in the end bay:

𝑊𝑏
P= 𝐿2
8𝑎𝑒𝑛𝑑

(30.18)(8.23)2
= = 2165.8𝐾𝑁
8×(11.798/100)

Check Interior Span Force

(30.18)(9.14)2
P= = 2003.5 𝐾𝑁<2165.8𝐾𝑁
8×(15.73/100)

Less force is required in the center bay

Check Precompression Allowance

Determine number of tendons to achieve 2165.8𝐾𝑁

2165.8
No. of tendons= =21.45
100.95

Use 21 tendons

Actual force for banded tendons

Pactual = (21 tendons) (100.95) = 2119.95kN

The balanced load for the end span is slightly adjusted

2119.95
wb= × 30.18 = 29.5 𝑘𝑁/𝑚
2165.8

Determine actual Precompression stress

Pactual /A = ( 2119.95) / (1.6764)

=1.26 MPa > 0.9 MPa ………………………………………….. OK

For this example, continue the force required for the end spans into the interior span
and check the amount of load that will be balanced:
2119.95×8×0.1573
Wb= = 31.93𝐾𝑁/𝑚
9.14 2

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Chapter four Hand calculation

𝑤𝑏 31.93
= × 100 = 79.4% This value is less than 100%; acceptable for this
𝑤𝐷𝐿 5.28×7.62

design.

East-West interior frame:

Effective prestress force, Peff = 𝟐𝟏𝟏𝟗. 𝟗𝟓 kN

Check Slab Stresses

Separately calculate the maximum positive and negative moments in the frame for the
dead, live, and balancing loads. A combination of these values will determine the slab
stresses at the time of stressing and at service loads.

Dead Load Moments

WDL=9.14+40.23=49.4KN/m

Live load moment

WLL=(3)(7.62)=22.86KN/m

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Chapter four Hand calculation

Total Balancing moment

Wb= -30.31kN/m (average of 3 bays)

Stage 1: Stresses immediately after jacking (DL + PT) (3.6.2)

Mid span Stresses

−MDL +Mbal P
𝑓𝑡𝑜𝑝 = −
S A

MDL − Mbal P
𝑓bot = −
S A

Interior Span

(−144.62 +88.43)×106
𝑓𝑡𝑜𝑝 = − 1.26 = −2.17𝑀𝑃𝑎 compression <0.6 𝑓𝑐𝑖′
0.0615×109

=12.6Mpa ..……………………………………………………..Ok

(144.62 −88.43)×106
𝑓𝑏𝑜𝑡 = − 1.26 = −0.35𝑀𝑃𝑎 compression <0.6 𝑓𝑐𝑖′
0.0615×109

=12.6Mpa………………………………………………………Ok

End span

(−232.75 +142.59)×106
𝑓𝑡𝑜𝑝 = − 1.26 = −2.73𝑀𝑃𝑎 compression <0.60𝑓𝑐𝑖′
0.0615×109

=12.6Mpa……………………………………………………...Ok

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Chapter four Hand calculation

(232.75 −142.59)×106
𝑓𝑏𝑜𝑡 = − 1.26 = 0.21𝑀𝑃𝑎 tension <0.25√𝑓𝑐𝑖′
0.0615×109

=1.15 MPa……………………………………………….…Ok

Support Stresses

(𝑀𝐷𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(−𝑀𝐷𝐿 + 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑏𝑜𝑡 = −
𝑆 𝐴
(371.72−228.08)×106
𝑓𝑏𝑜𝑡 = − 1.26 = 1.08 𝑀𝑃𝑎 Tension <0.25 √𝑓𝑐𝑖′
0.0615×109

=1.15MPa…………………………………………………......................Ok

(− 371.72+228.08)×106
𝑓𝑏𝑜𝑡 = − 1.26= -3.6MPa compression < 0.60𝑓𝑐𝑖′
0.0615×109

=12.6Mpa……………………………………………………………….Ok

Stage 2: Stresses at service load (DL + LL + PT)

Mid span Stresses

(−𝑀𝐷𝐿 − 𝑀𝑙𝐿 + 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(𝑀𝐷𝐿 + 𝑀𝑙𝑙 − 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

Interior Span

(−144.62 −66.698+88.43)×106
𝑓𝑡𝑜𝑝 = − 1.26
0.0615×109

= -1.38 MPa compression < 0.45 𝑓𝑐′ = 15.75𝑀𝑝𝑎 … … … … … … . . … . . 𝑂𝑘

(144.62 + 66.698 − 88.43) × 106

𝑓𝑏𝑜𝑡 = − 1.26
0.0615 × 109

73
Chapter four Hand calculation

= 0.74 MPa tension < 0.5√𝑓′𝑐 = 3𝑀𝑝𝑎 … … … … … … … … … … . 𝑂𝑘

End Span
(−232.75 − 107.54 + 142.59) × 106
𝑓𝑡𝑜𝑝 = − 1.26
0.0615 × 109

= -4.47 MPa compression <0.45 𝑓𝑐′ =15.75MPa…………………Ok

(232.75 + 107.54 − 142.59) × 106

𝑓𝑏𝑜𝑡 = − 1.26
0.0615 × 109

= 1.95 MPa tension < 0.5√𝑓′𝑐 = 3𝑀𝑝𝑎 … … … … … … … … … … … 𝑂𝑘

Support Stresses

(+𝑀𝐷𝐿 + 𝑀𝑙𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(−𝑀𝐷𝐿 − 𝑀𝑙𝑙 + 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

(371.72 + 172.02 − 228.08) × 106

𝑓𝑡𝑜𝑝 = − 1.26
0.0615 × 109

= 2.87𝑀𝑃𝑎 Tension < 0.5√𝑓′𝑐 MPa……………………………Ok

(− 371.72 − 172.02 + 228.08) × 106

𝑓𝑏𝑜𝑡 = − 1.26
0.0615 × 109

= -6.39MPa compression < 0.45 𝑓𝑐′ =15.75 MPa……………..Ok

All stresses are within the permissible code limits

Ultimate Strength

Determine factored moments

The primary post-tensioning moments, M1, vary along the length of the span.

M1 = P × e

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Chapter four Hand calculation

e = 0 m at the exterior support

e = 0.0787𝑚 at the interior support (neutral axis to the center of tendon)

M1 = (2119.95)(0.0787) = 166.84KN.m

The secondary post-tensioning moments, Msec, vary linearly between supports(3.4).

Msec = Mbal - M1

= 228.08-166.84

= 61.24kN.m at the interior supports

The typical load combination for ultimate strength design is

Mu = 1.2 MDL + 1.6 MLL + 1.0 Msec

At mid span: Mu = 1.2 (232.75) + 1.6 (107.54) + 1.0 (30.62) = 481.98 KN.m

At support: Mu =1.2 (-371.72) + 1.6 (-172.02) + 1.0 (61.24) =-660.06 KN.m

Determine minimum bonded reinforcement: to see if acceptable for ultimate

strength design.(3.7)

Positive moment region:

Interior span: 𝑓𝑡 = 0.74 MPa < 0.17√𝑓′𝑐 = 0.17√35= 1 MPa

No positive reinforcement required (ACI 8.6.2.3a)

Exterior span: 𝑓𝑡 = 1.95 MPa > 0.17 √𝑓′𝑐 =1 MPa

Minimum positive moment reinforcement required (ACI 8.6.2.3)

75
Chapter four Hand calculation

𝑓𝑡
y= ×ℎ
(𝑓𝑡 +𝑓𝑐 )

1.95
= × 220
(1.95+4.47)

= 66.82mm

𝑀𝐷𝐿+𝐿𝐿
Nc = × 0.5 × 𝑦 × 𝑙2
𝑆

(232.75+107.54)×1000
= × 0.5 × 66.82 × 7620
0.0615×109

= 1408.66KN
𝑁𝐶 1408.66×1000
As = = = 6805.12mm2
0.5𝑓𝑦 0.5×414

Distribute the positive moment reinforcement uniformly across the slab-beam width
and as close as practicable to the extreme tension fiber.
6805.12
As, min =
7.62

= 893.06 mm2/m

Use Ø 25 (Ab=510mm2) Bottom

Minimum length shall be 1/3 clear span and centered in positive moment region (ACI
8.7.5.5.1)

Negative moment region:

As, min = 0.00075Acf (ACI 8.6.2.3)

Interior supports:
9.14+8.23
Acf = max. (0.22)[ ,7.62]
2

=max.(0.22)[8.685,7.62]

= 1.91 m2

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Chapter four Hand calculation

As, min = 0.00075(1.91)

= 0.00143 m2

= 3Ø25 Top (1530mm2)

Exterior supports:
8.23
Acf = max. (0.22)[ ,7.62]
2

=max.(0.22)[4.115,7.62]

= 1.676m2

As, min = 0.00075(1.676)

= 0.001257m2

=3Ø25 Top (1530mm2)

Must span a minimum of 1/6 the clear span on each side of support (ACI 8.7.5.5.1)

At least 4 bars required in each direction (ACI 8.7.5.3)

Place top bars within 1.5h away from the face of the support on each side (ACI
8.7.5.3)

=1.5h

=1.5(220)

=330mm

Maximum bar spacing is 300mm (ACI 8.7.5.3)

Check minimum reinforcement if it is sufficient for ultimate strength (3.8)

𝑎
Mn = (Asfy + Apsfps) (d - )
2

Aps = 99×(number of tendons)

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Chapter four Hand calculation

= 99×(21 tendons)

= 2079 mm2

𝑓′ 𝑐
𝑓𝑝𝑠 = 𝑓𝑠𝑒 + 70+ ≤ 𝑓𝑝𝑦 𝑜𝑟 𝑓𝑠𝑒 + 210 for slabs with L/h > 35 (ACI 20.3.2.4.1 )
300𝜌𝑝

(𝐴𝑠 𝑓𝑦 + 𝐴𝑝𝑠 𝑓𝑝𝑠 )

a=
(0.85𝑓′𝑐 𝑏)

At supports
1.27
d= 22 -2.5-( )=18.865 cm
2

𝐴𝑝𝑠 2079
ρ = = = 1.45 × 10-3
𝑏𝑑 7620×188.65

35
𝑓𝑝𝑠 = 1019.72 + 70 + ≤ 1860 𝑜𝑟 1019.72 + 210
300×1.45× 10−3

𝑓𝑝𝑠 = 1170.18MPa ≤ 1860 𝑜𝑟 1229.72 MPa

𝑓𝑝𝑠 =1170.18MPa

1530×414+2079×1170.18
a= = 13.5𝑚𝑚
0.85×35×7.62×103

13.5
ϕMn=0.9(1530 × 414 + 2079 × 1170.18)(188.65- )
2

=502KN.m<660.06KN.m Reinforcement for ultimate strength requirements governs

As,reqd=3612mm2

8-Ø25Top at interior supports

3-Ø25Top at exterior supports

When reinforcement is provided to meet ultimate strength requirements, the minimum

lengths must also conform to the provision of ACI 318

78
Chapter four Hand calculation

At mid span (end span)

6805.12×414+2079×1170.18
a= = 23.16mm
0.85×35×7.62×103

23.16
ϕMn=0.9(6805.12 × 414 + 2079 × 1170.18)(188.65- )
2

=836.68 KN.m> 481.98 KN.m Minimum reinforcement ……………..Ok

Ø25 oc. Bottom at end span.

Figure (4-1) final design sketch

79
Chapter four Hand calculation

4.2 case II Four Floors Building

Loads:

Framing dead load =self weight

Superimposed dead load=1.2 KN/m2

partitions, M/E,misc

Live load =3KN/m2

column size 500mm×500mm
story height 3m
Material:

Concrete normal weight =24 KN/m3

𝑓′𝑐 =35 N/mm2

𝑓′𝑐𝑖 =21 N/mm2

Rebar 𝑓𝑦 =414 N/mm2

PT: unbonded tendons

1.28 cm ∅,7-wire strands,

A= 99mm2,ƒpu=1860 N/mm2

80
Chapter four Hand calculation

Determine preliminary slab Thickness

Start with L/h=45

Longest span =9.14m

h=(9.14m) (100)/45

=20.3cm

Take t=22cm preliminary slab thickness

Estimated prestress losses ACI(20.3.2.6)

1) Elastic shorting=0 (simultaneous tensioning and anchorage)

2) Shrinkage:

0.0002
Ɛsh =
𝑙𝑜𝑔10 (t+2)
0.0002
= = 1.35×10-4
𝑙𝑜𝑔10 (28+2)

SH = Ɛsh × Es
= (1.35×10-4)(2×105)=27MPa
27
SH% = ×100=1.45%
1860

3)Relaxation=2.5%

4) anchorage slip:
∆ = 0.2 in
∆𝑙
∆𝑓𝑠𝑙𝑖𝑝 = 𝐸𝑠
𝑙
5.08
= × 2 × 105
25.6×1000

=40 MPa
40
Anchorage slip%= × 100 = 2.15%
1860

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Chapter four Hand calculation

5) creep losses:

𝐶𝑅 = 𝜃𝑛(𝑓𝑐𝑖𝑟 − 𝑓𝑐𝑑𝑠 )

F° F° ×e2 MG ×e
𝑓𝑐𝑖𝑟 =- - +
AG I I

bh3 7.62×0.223
𝐼= = =6.76×10-3 m4
12 12

AG =bh =7.62×0.22=1.6764 𝑚2

𝑃° = 𝐴 × 𝑓𝑝𝑢

= 99 × 1860 =184140N= 184.14KN

22 1.27
e= − 2.5 − =7.87cm
2 2

Wd=ɤ×t

Wd =24×0.22=5.28KN/m2

Wd=5.28×7.62=40.23KN/m

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Chapter four Hand calculation

Max. moment =277.967 KN.m

184.14 184.14×0.07872 277.967×0.0787

𝑓𝑐𝑖𝑟 =- - + = 2.96𝑀𝑃𝑎
1.6764 6.76×10-3 6.76×10-3

Wcds=1.2×7.62 =9.14KN/m

max. moment due to superimpose dead load =63.152 KN.m

𝑀×𝑐 63.152 ×0.0787
ƒcds= = =0.74MPa
𝐼 6.76×10−3

𝜃 = 1.6 (28 𝑑𝑎𝑦𝑠 )

𝐸𝑠 2×105
n= = = 7.2
𝐸𝑐 4700√35

CR=1.6×7.2(2.96 − 0.74)= 25.57MPa

25.57
CR%= × 100 =1.37%
1860

6) friction losses:

P=PO ×e-(µα+KL)

83
Chapter four Hand calculation

Segments cumulative Wobble Curvature Length uα+kL Exp(-uα-kL) stress@

angle coefficient coefficient end of segment
α K, per m µ m
1-2 0.046 0.00984 0.05 3.365 0.0354116 0.965208055 0.965208055PO
2-3 0.065 0.00984 0.05 4.865 0.0511216 0.950163124 0.9171051 PO
3-4 0.069 0.00984 0.05 4.57 0.0484188 0.952734698 0.873757851 PO
4-5 0.069 0.00984 0.05 4.57 0.0484188 0.952734698 0.832459423 PO
5-6 0.065 0.00984 0.05 4.865 0.0511216 0.950163124 0.790972245 PO
6-7 0.046 0.00984 0.05 3.365 0.0354116 0.965208055 0.763452782 PO

84
Chapter four Hand calculation

Stress @ end of segment =0.763452782×184.14=140.58Mpa

FR% = (140.58)/1860×100=7.6%

Total losses=ES+SH+RE+ANC+CR+FR

Total losses =0+27+46.5+40+25.57+140.58=279.65Mpa

ƒse=0.7(1860)-279.65=1022.35Mpa (ACI 20.3.2.5.1)

Peff=A׃se = (99) ( 1022.35)= 101.21 kN

Calculate section properties

Two-way slab must be designed as class U (ACI 24.5.2.1), Gross-Sectional

Properties allowed

S=bh2/6=7.62x 0.222/6=0.0615m3

Set Design parameters

Allowable stresses: class U (ACI 24.5.2.1)

At time of jacking (ACI 24.5.3.1) and (ACI 24.5.3.2) for compressive and
tensile stress respectively)

𝑓′𝑐𝑖 =21 N/mm2

compression =0.6 𝑓′𝑐𝑖 =0.6(21)=12.6 MPa

Tension =0.25√𝑓𝑐𝑖′ =0.25×√21 =1.15 MPa

At service loads (ACI 24.5.4.1)

𝑓′𝑐 = 35MPa

Compression = 0.45 ƒ'c = 0.45(35) = 15.75MPa

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Chapter four Hand calculation

Tension = 0.5√𝑓′𝑐 = 0.5√35 = 3MPa

Average pre-compression limits(3.6.4)

𝑃
= 0.9MPa min. (ACI 8.6.2.1)
𝐴

Target load balances:

60%-80% of DL (self weight) for slabs (good approximation for hand

calculation)

For this example: 0.75 WDL = 0.75(5.28) = 3.96 KN/m2

Tendon profile:

Parabolic shape;

For a layout with spans of similar length, the tendons will be typically be
located at the highest allowable point at the interior columns, the lowest
possible point at the mid spans, and the neutral axis at the anchor locations.
This provides the maximum drape for load-balancing.

Tendon Ordinate Tendon (CG) Location*

Exterior support 22/2=11
Interior support-top 22-2.5-1.27/2=18.865
Interior span – bottom 2.5+1.27/2=3.135
End span – bottom 2.5+1.27/2=3.135

(CG) = center of gravity

*Measure from bottom of slab

aINT = 18.865-3.135=15.73cm

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Chapter four Hand calculation

(11+18.865)
aEND = − 3.135=11.798cm
2

eccentricity, e, is the distance from the center to tendon to the neutral axis;
varies along the span

Pre-stress Force Required to Balance 75% of self weight DL

Since the spans are of similar length, the end span will typically govern the
maximum required post-tensioning force. This is due to the significantly
reduced tendon drape, aEND.

wb = 0.75 WDL = 0.75(5.28)(7.62)=30.18 KN/m

Force needed in tendons to counteract the load in the end bay:

𝑊𝑏
P= 𝐿2
8𝑎𝑒𝑛𝑑

(30.18)(8.23)2
= = 2165.8𝐾𝑁
8×(11.798/100)

Check Interior Span Force

(30.18)(9.14)2
P= = 2003.5 𝐾𝑁<2165.8𝐾𝑁
8×(15.73/100)

Less force is required in the center bay

Check Precompression Allowance

Determine number of tendons to achieve 2165.8𝐾𝑁

2165.8
No. of tendons= =21.399
101.21

Use 21 tendons

Actual force for banded tendons

Pactual = (21 tendons) (101.21) = 2125.41kN

The balanced load for the end span is slightly adjusted

87
 Chapter four Hand calculation

2125.41
wb= × 30.18 = 29.6 𝑘𝑁/𝑚
2165.8

Determine actual Precompression stress

Pactual /A = (2125.41) / (1.6764)

=1.27MPa > 0.9 MPa…..………………………………………… OK

For this example, continue the force required for the end spans into the interior
span and check the amount of load that will be balanced:
2125.41×8×0.1573
Wb= = 32𝐾𝑁/𝑚
9.14 2

𝑤𝑏 32
= × 100 = 79.5% This value is less than 100%; acceptable for this
𝑤𝐷𝐿 5.28×7.62

design.

East-West interior frame:

Effective prestress force, Peff = 2125.41kN

Check Slab Stresses

Separately calculate the maximum positive and negative moments in the frame
for the dead, live, and balancing loads. A combination of these values will
determine the slab stresses at the time of stressing and at service loads.

Dead Load Moments

WDL=9.14+40.23=49.4KN/m

88
Chapter four Hand calculation

Live load moment

WLL=(3)(7.62)=22.86KN/m

Total Balancing moment Wb= -30.4kN/m (average of 3 bays)

89
 Chapter four Hand calculation

The final slab service moment, at the column centerline are as follows

Service load Exterior Exterior First Interior Center

Moment column Mid span column Midspan
MDL (KN.m) -241.763 +146.659 -341.326 +174.531
MLL(KN.m) -111.877 +67.867 -157.950 +80.765
MBAL(KN.m) +148.777 -90.252 +210.04 -107.404

Stage 1: Stresses immediately after jacking (DL + PT) (3.6.2)

Mid span Stresses

−MDL +Mbal P
𝑓𝑡𝑜𝑝 = −
S A

MDL − Mbal P
𝑓bot = −
S A

Support Stresses

(𝑀𝐷𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

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 Chapter four Hand calculation

(−𝑀𝐷𝐿 + 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

Transfer stresses Exterior Exterior First Interior Center

column Mid span column Mid span
MDL+MBAL (KN.m) -92.986 +56.407 -131.1286 67.127
M/S (MPa) +/- 1.5 +/-0.92 +/-2.13 +/-1.09
P/A (MPa) -1.27 -1.27 -1.27 -1.27
𝑓𝑡𝑜𝑝 (MPa) +0.23 -2.19 +0.86 -2.360
𝑓bot (MPa) -2.77 -0.35 -3.4 -0.18

The allowable concrete stresses at transfer are:

𝑓𝑐 <0.6 𝑓𝑐𝑖′ =12.6MPa compression

𝑓𝑡 < 0.25√𝑓𝑐𝑖′ =1.15 MPa tension

From the above table, the maximum tension and compression values are in the
shaded boxes (+0.86 MPa and -3.4MPa, respectively), and we can see these
stresses are well below the allowable values and therefore the stresses at
transfer are acceptable.

Stage 2: Stresses at service load (DL + LL + PT) (3.6.2)

Mid span Stresses

(−𝑀𝐷𝐿 − 𝑀𝑙𝐿 + 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(𝑀𝐷𝐿 + 𝑀𝑙𝑙 − 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

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 Chapter four Hand calculation

Support Stresses
(+𝑀𝐷𝐿 + 𝑀𝑙𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑡𝑜𝑝 = −
𝑆 𝐴
(−𝑀𝐷𝐿 − 𝑀𝑙𝑙 + 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

Service load stresses Exterior Exterior First Interior Center

column Mid span column Mid span
𝑀𝐷𝐿 + 𝑀𝑙𝐿 + 𝑀𝑏𝑎𝑙 (KN. m) -204.763 +124.274 -289.236 +147.892
M/S (MPa) +/- 3.33 +/-2.02 +/-4.70 +/- 2.40
P/A (MPa) -1.27 -1.27 -1.27 -1.27

𝑓𝑡𝑜𝑝 (MPa) +2.06 -3.29 +3.43 -3.67

𝑓bot (MPa) -4.6 +0.75 -5.97 +1.13

The maximum permissible service loads are:

Tension ˂ 0.5√35= 3 MPa

compression < 0.45 𝑓𝑐′ =15.75 MPa

From the above table, the maximum tension and compression values are in the
shaded boxes (+3.43 MPa and -5.97 MPa, respectively), and we can see the
maximum compressive stress is acceptable. but the tensile stress exceeds the
allowable value at the column. The tensile stress at the bottom of the slab at
mid span (+1.13Mpa) also exceeds the allowable value of 0.17√𝑓′𝑐 .=1Mpa
and therefore requires bonded reinforcing. When the tensile stresses exceed the
limit, then the minimum area of bonded reinforcing is:

𝑓𝑡
y= ×ℎ
(𝑓𝑡 +𝑓𝑐 )

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 Chapter four Hand calculation

3.43
= × 220= 80.28mm
(3.43+5.97)

𝑀𝐷𝐿+𝐿𝐿
Nc = × 0.5 × 𝑦 × 𝑙2
𝑆

(341.326+157.950)×1000
= × 0.5 × 80.28 × 7620
0.0615×109

= 2483KN
𝑁𝐶 2483×1000
As, min = = = 11995.2 mm2
0.5𝑓𝑦 0.5×414

Distribute the positive moment reinforcement uniformly across the slab-beam

width and as close as practicable to the extreme tension fiber.
11995.2
As, min = 1574 mm2 /m
7.62

use Ø25 @ 3080mm oc. Bottom

Determine the hyperstatic moments (secondary moment)(3.4):

The primary post-tensioning moments, M1, vary along the length of the span.

M1 = P × e

The secondary post-tensioning moments, Msec, vary linearly between supports.

Msec = Mbal - M1

We can tabulate the hyperstatic moments as follows:

93
Chapter four Hand calculation

Exterior Exterior First Interior Center

column Mid span column Mid span
MBAL (KN.m) +148.777 -90.252 +210.04 -107.404
e (cm) 0 -7.87 +7.87 -7.87
M1 (KN.m) 0 -167.27 +167.27 -167.27
Msec (KN.m) +148.777 +77.018 +42.77 +59.87

The typical load combination for ultimate strength design is

Mu = 1.2 MDL + 1.6 MLL + 1.0 Msec

Exterior Exterior First Interior Center
column Mid span column Mid span
MUDL (KN.m) -290.12 +175.99 -409.59 +209.44
MULL(KN.m) -179 +108.587 -252.720 +129.224
Msec(KN.m) +148.777 +77.018 +42.77 +59.87
MU(KN.m) -320.343 +361.595 -619.54 +398.534
Determine minimum bonded reinforcement: to see if acceptable for ultimate
strength design. (3.7)

positive moment areas:

Exterior Mid span ƒt=0.75MPa˂ 0.17√𝑓′𝑐 .= 0.17√35=1MPa

Thus, minimum bonded steel is not required

Center Mid span: ƒt=1.13MPa˃0.17√𝑓′𝑐 .= 0.17√35=1MPa

Thus, minimum bonded steel is required

𝑓𝑡
y= ×ℎ
(𝑓𝑡 +𝑓𝑐 )

1.13
= × 220
(1.13+3.67)

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Chapter four Hand calculation

=51.79mm
𝑀𝐷𝐿+𝐿𝐿
Nc = × 0.5 × 𝑦 × 𝑙2
𝑆

(174.531+80.765)×1000
= × 0.5 × 51.79 × 7620
0.0615×109

=819.12KN
𝑁𝐶 819.12×1000
As = = = 3957mm2
0.5𝑓𝑦 0.5×414

Distribute the positive moment reinforcement uniformly across the slab-beam

width and as close as practicable to the extreme tension fiber.
3957
As, min = =519.29 mm2/m
7.62

Use Ø25 Bottom (Ab=510mm2)

Minimum length shall be 1/3 clear span and centered in positive moment region
(ACI 8.7.5.5.1)

In negative moment areas (tension in the top of the slab), the minimum area of
bonded reinforcing in each direction is:

First Interior column

As, min = 0.00075Acf

9.14+8.23
Acf = max. (0.22)[ ,7.62]
2

=max.(0.22)[8.685,7.62]

= 1.91 m2

As, min = 0.00075(1.91)

= 0.00143 m2

=3Ø25 Top (1530mm2)

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 Chapter four Hand calculation

Exterior column:
8.23
Acf = max. (0.22)[ ,7.62]
2

=max.(0.22)[4.115,7.62]

= 1.676m2

As, min = 0.00075(1.676)

= 0.001257m2

=3Ø25 Top (1530mm2)

Must span a minimum of 1/6 the clear span on each side of support (ACI
8.7.5.5.1)

At least 4 bars required in each direction (ACI 8.7.5.3)

Place top bars within 1.5h away from the face of the column on each side (ACI
8.7.5.3))

=1.5h

=1.5(220)

=330mm

Maximum bar spacing is 300mm (ACI 8.7.5.3))

Check minimum reinforcement if it is sufficient for ultimate strength

𝑎
Mn = (Asfy + Apsfps) (d - )
2

Aps = 99×(number of tendons)

= 99×(21 tendons)

= 2079 mm2

96
Chapter four Hand calculation

𝑓′ 𝑐
𝑓𝑝𝑠 = 𝑓𝑠𝑒 + 70+ ≤ 𝑓𝑝𝑦 𝑜𝑟 𝑓𝑠𝑒 + 210 for slabs with L/h > 35 (ACI
300𝜌𝑝

20.3.2.4.1)

(𝐴𝑠 𝑓𝑦 + 𝐴𝑝𝑠 𝑓𝑝𝑠 )

a=
(0.85𝑓′𝑐 𝑏)

At First Interior column

1.27
d= 22 -2.5-( )=18.865 cm
2

𝐴𝑝𝑠 2079
ρ = = = 1.45 × 10-3
𝑏𝑑 7620×188.65

35
𝑓𝑝𝑠 = 1022.35 + 70 + ≤ 1860 𝑜𝑟 1172.8 + 210
300×1.45× 10−3

𝑓𝑝𝑠 = 1172.8MPa ≤ 1860 𝑜𝑟 1382.8 MPa

𝑓𝑝𝑠 =1172.8MPa

1530×414+2079×1172.8
a= =13.5mm
0.85×35×7.62×103

13.5
ϕMn=0.9(1530 × 414 + 2079 × 1172.8)(188.65- )
2

=502.86KN.m<619.54 KN.m Reinforcement for ultimate strength

requirements governs

As,req’d=3252mm2

7 Ø25Top at interior column

3 Ø25Top at exterior supports

When reinforcement is provided to meet ultimate strength requirements, the

minimum lengths must also conform to the provision of ACI 318.

97
 Chapter four Hand calculation

At Center Mid span

3957×414+2079×1172.8
a= = 18mm
0.85×35×7.62×103

18
ϕMn=0.9(3957 × 414 + 2079 × 1172.8)(188.65- )
2

=635KN.m> 398.534 KN.m Minimum reinforcement …………..OK

Ø25 oc. Bottom at end span

We must now investigate the shear stress in the slab at the columns in the
direction of the analysis. Let's begin by determining the critical section, and
then calculate the nominal shear capacity using the appropriate equation. After
that, we will calculate the shear stress demand.

For the exterior column:

we can see that the slab does not extend at least 4h past the face of the columns,
and therefore we cannot take advantage of the post-tensioning to calculate the
shear capacity and must revert to the non pre-stressed equation:

ϕ 𝑣 c = ϕ0.33λ√𝑓′𝑐 =0.75×0.33×1×√35 =1.46Mpa

To calculate the punching shear demand, we need to determine the factored

shear forces and the transfer moments at each column. The factored moments
are the algebraic sum of the moments due to external loading (sustained dead
loads, live loads, etc.) plus the hyperstatic moments.

The load factor combination for moment

is 1.2MD +1.6ML + 1.0MHYP.

𝑣 u= [1.2(1.2+5.28) +1.6(3)](4.115×7.62)=394.34KN

And from the table above MU= -320.343 KN.m

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 Chapter four Hand calculation

Since we have a square column, we will transfer 40% of the unbalanced

moment by eccentricity of shear. The exterior
188.65
b1=(500+ )=594.33mm
2

b2 =(500+188.65)=688.65mm

b0=2b1+ b2 =1877.31mm

Thus, the area of the critical section is:

Ac= b0 ×d=1877.31×188.65=0.354m2

For a rectangular edge column with the moment acting perpendicular to the
edge, we find the following equations for the properties of the critical section:

𝐽 2𝑏12 𝑑(𝑏1 + 2𝑏2 ) + 𝑑 3 (2𝑏1 + 𝑏2 )

=
𝑐𝐴𝐵 6𝑏1

2×594.332 ×188.65(594.33+2×688.65)+188.653 (2×594.33+688.65)

= =7.72×107 mm3
6×594.33

J 2b12 d(b1 + 2b2 ) + d3 (2b1 + b2 )

=
cCD 6(b1 + b2 )

2 3
2×594.33 ×188.65(594.33+2×688.65)+188.65 (2×594.33+688.65)
= 6(594.33+688.65)
= 3.58×107mm3

Since we have a square column, we will transfer 40% of the unbalanced

moment by eccentricity of shear

γv Mse cAB
𝑣 u,AB =vug +
Jc

394.34×103 0.4×320.343 ×106

= + =4.69MPa
0.354×106 7.72×107

𝛾𝑣 𝑀𝑠𝑒 𝑐CD
𝑣𝑢,𝐶𝐷 =𝑣𝑢𝑔 −
𝐽𝑐

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 Chapter four Hand calculation

394.34×103 0.4×320.343×106
= − = −2.47MPa
0.354×106 3.58×107

Thus, the shear demand at the exterior column is greater than the allowable
shear stress of 1.46 MPa. Therefore, we need to increase the shear capacity of
the slab at the exterior column. it is probably most practical and economical to
add headed stud reinforcing.

For the interior column:

(a) ϕ 𝑣c= ϕ(0.29 λ√𝑓′𝑐 +0.3ƒpc)

= 0.75(0.29√35 +0.3×1.27)=1.57MPa
b1=b2= 500+188.65=688.65mm

b0= 2 b1+2 b2=2754.6mm

Ac= b0 ×d=2754.6×188.65=0.52m2

𝛼𝑠 𝑑
(b) ϕ 𝑣c= ϕ [0.083(1.5+ ) √𝑓′𝑐 +0.3ƒpc]
𝑏𝑜

40×188.65
=0.75 [0.083(1.5+ ) √35 +0.3×1.27]= 1.85MPa
2754.6

use the less value of ϕ 𝑣c =1.57 MPa

And from the table above MU=54.087KN.m

𝐽 J 𝑏1 𝑑(𝑏1 + 3𝑏2 ) + 𝑑3
= =
𝑐𝐴𝐵 cCD 3

688.65×188.65×(688.5+3×688.5)+188.653
= =12.15×107mm3
3

𝑣 u= [1.2(1.2+5.28)+1.6(3)](8.685×7.62)=832.28KN

𝛾𝑣 𝑀𝑠𝑒 𝑐AB 832.28×103 0.4×54.087×106

𝑣 u,AB =𝑣𝑢𝑔 + = + =1.78MPa
𝐽𝑐 0.52×106 12.15×107

100
Chapter four Hand calculation

𝛾𝑣 𝑀𝑠𝑒 𝑐CD 832.28×103 0.4×54.087×106

𝑣𝑢,𝐶𝐷 =𝑣𝑢𝑔 − = − =1.42MPa.
𝐽𝑐 0.52×106 12.15×107

Thus, the shear demand at the exterior column is greater than the allowable
shear stress of 1.57 MPa. Therefore, we need to increase the shear capacity of
the slab at the interior column. it is probably most practical and economical to
add headed stud reinforcing.

The last thing we need to check in this direction of analysis is the unbalanced
moment transfer by flexure at the columns. We need to transfer 60% of the
unbalanced moment by flexure. Thus, we need to transfer (0.6)(320.343)=-
192.2KN.m and(0.6)(54.08 )=32.448KN.mat the exterior and interior columns,
respectively. This moment transfer must occur in a slab width equal to the
column face dimension plus 1.5h on both sides of the column,500+2(1.5×220)
=1160mm.

0.85𝑓′𝑐 2.353𝑀𝑢
𝜌= [1 − √1 − ]
𝑓𝑦 ∅ 𝑓 ′ 𝑐 𝑏𝑑 2

0.85×35 2.353×192.2×106
=
414
[1 − √1 − 0.9×35×1160×188.652 ] =0.0138

As= 𝜌 𝑏𝑑

=0.0138×1160×188.65 = 3019.9mm2

Based on this demand, we can see that the minimum area of bonded mild

reinforcement of 6Ø25.

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 Chapter four Hand calculation

Figure(4-2)minimum bonded reinforcement –East-West interior

To summarize the analysis in this direction, we have used a slab thickness of

0.22mm with an effective pre-stress force of 279KN per meter. So, in the 7.62
wide equivalent frame, we need a total effective pre-stress of 2125.41KN.
Using 101.21kN per 1.27cm tendon, we need 21tendons (each will be stressed
slightly less than 101.21kN) placed in a narrow band on the column line in the
east west direction, with at least two of them passing through the column core
to act as integrity tendons.

Now, let's turn our attention briefly to the analysis in the perpendicular,
or north-south direction.
Estimated prestress losses

2) Elastic shorting=0 (simultaneous tensioning and anchorage)

2) Shrinkage:
0.0002
Ɛsh =
log10(t+2)
0.0002
= = 1.35×10-4
log10(28+2)

SH =Ɛsh × Es

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 Chapter four Hand calculation

=(1.35×10-4)(2×105)=27MPa
27
SH% = × 100 = 1.45%
1860

3)Relaxation=2.5%
4) anchorage slip:
∆ = 0.2 in
∆𝑙
∆𝑓𝑠𝑙𝑖𝑝 = 𝐸𝑠
𝑙
5.08
= × 2 × 105
30.48×1000

=33.3 MPa
33.3
Anchorage slip%= × 100 = 1.79%
1860

5) creep losses:

𝐶𝑅 = 𝜃𝑛(𝑓𝑐𝑖𝑟 − 𝑓𝑐𝑑𝑠 )

𝐹° 𝐹° × 𝑒 2 𝑀𝐺 × 𝑒
𝑓𝑐𝑖𝑟 =− − +
𝐴𝐺 𝐼 𝐼

𝑏ℎ3 8.685×0.223
𝐼= = = 7.71 × 10−3 𝑚4
12 12

AG =bh =8.685×0.22=1.91 𝑚2

𝑃° = 𝐴 × 𝑓𝑝𝑢

= 99 × 1860 =184140N= 184.14KN

22 1.27
e= − 2.5 − =7.87cm
2 2

Wd=ɤ×t

Wd =24×0.22=5.28KN/m2

Wd=5.28×8.685=45.86KN/m

103
 Chapter four Hand calculation

Max. moment =236.05 KN.m

184.14 184.14 × 0.07872 236.05 × 0.0787

𝑓𝑐𝑖𝑟 =− − + = 2.17𝑀𝑃𝑎
1.91 7.71 × 10−3 7.71 × 10−3

Wcds=1.2×8.685=10.422 KN/m

max. moment due to superimpose dead load =53.644 KN.m

104
 Chapter four Hand calculation

𝑀×𝑐 53.644 ×0.0787

ƒcds= = =0.55MPa
𝐼 7.71×10−3

𝜃 = 1.6 (28 𝑑𝑎𝑦𝑠 )

𝐸 2×105
n= 𝑠 = = 7.2
𝐸𝑐 4700√35

CR=1.6×7.2(2.17 − 0.55)= 18.66 MPa

18.66
CR%= × 100 =1 %
1860

6) friction losses:

P=PO×e-(µα+KL)

105
Chapter four Hand calculation

Segments cumulative Wobble Curvature Length uα+kL Exp(-uα-kL) stress@

angle coefficient coefficient end of segment
α K, per m µ m
1-2 0.00984 0.05 0.028178 0.972215297 0.972215297PO
0.05 3.175
2-3 0.00984 0.05 0.042592 0.958302298 0.931676153 PO
0.066 5.075
3-4 0.00984 0.05 0.0372936 0.963393242 0.897570509 PO
0.088 3.81
4-5 0.00984 0.05 0.0372936 0.963393242 0.864713362 PO
0.088 3.81
5-6 0.00984 0.05 0.0372936 0.963393242 0.8330559009PO
0.088 3.81
6-7 0.00984 0.05 0.0372936 0.963393242 0.802563419 PO
0.088 3.81
7-8 0.00984 0.05 0.042592 0.958302298 0.716550609 PO
0.066 5.075
8-9 0.00984 0.05 0.028178 0.972215297 0.696641464 PO
0.05 3.175

Stress @ end of segment =0.696641464×184.14=128.3Mpa

FR%=(128.3 )/1860×100=6.897%

Total losses=ES+SH+RE+ANC+CR+FR

106
Chapter four Hand calculation

Total losses =0+27+46.5+33.3+18.66+128.3=253.76 Mpa

ƒse=0.7(1860)-253.76=1048.24 Mpa (ACI 20.3.2.5.1)

Peff=A׃se = (99)(1048.24)= 103.78kN

Calculate section properties

Two-way slab must be designed as class U (ACI 24.5.2.1), Gross-Sectional

Properties allowed

S=bh2/6=8.685 x 0.222/6=0.07m3

Target load balances:

60%-80% of DL (self weight) for slabs (good approximation for hand

calculation)

For this example: 0.75 WDL = 0.75(5.28) = 3.96 KN/m2

Tendon profile:

Parabolic shape;

For a layout with spans of similar length, the tendons will be typically be
located at the highest allowable point at the interior columns, the lowest
possible point at the mid spans, and the neutral axis at the anchor locations.
This provides the maximum drape for load-balancing.

Tendon Ordinate Tendon (CG) Location*

Exterior support 22/2=11
Interior support-top 22-2.5-1.27/2=18.865
Interior span – bottom 2.5+1.27/2=3.135
End span – bottom 2.5+1.27/2=3.135

107
Chapter four Hand calculation

(CG) = center of gravity

*Measure from bottom of slab

aINT = 18.865-3.135=15.73cm

(11+18.865)
aEND = − 3.135=11.798cm
2

eccentricity, e, is the distance from the center to tendon to the neutral axis;
varies along the span

Pre-stress Force Required to Balance 75% of self weight DL

Since the spans are of similar length, the end span will typically govern the
maximum required post-tensioning force. This is due to the significantly
reduced tendon drape, aEND.

wb = 0.75 WDL = 0.75(5.28)(8.685)=34.39 KN/m

Force needed in tendons to counteract the load in the end bay:

𝑊𝑏
P= 𝐿2
8𝑎𝑒𝑛𝑑

(34.39)(7.62)2
= = 2115.6𝐾𝑁
8×(11.798/100)

Check Interior Span Force

(34.39)(7.62)2
P= = 1586.8 𝐾𝑁<2115.6𝐾𝑁
8×(15.73/100)

Less force is required in the center bay

Check Precompression Allowance

Determine number of tendons to achieve 2115.6𝐾𝑁

2115.6
No. of tendons= =20.4
103.78

108
 Chapter four Hand calculation

Use 20 tendons

Actual force for banded tendons

Pactual = (20 tendons) (103.78) = 2075.6 kN

The balanced load for the end span is slightly adjusted

2115.7
wb= × 34.39 = 35.1 𝑘𝑁/𝑚
2075.6

Determine actual Precompression stress

Pactual /A = (2075.6) / (1.91)

=1.1 MPa > 0.9 MPa ……………… OK

For this example, continue the force required for the end spans into the interior
span and check the amount of load that will be balanced:

2075.6 ×8×0.1573
Wb= = 44.98𝐾𝑁/𝑚
7.622

𝑤𝑏 44.98
= × 100 = 98.1% This value is less than 100%; acceptable
𝑤𝐷𝐿 5.28×8.685

for this design.

north-south interior frame

Effective prestress force, Peff = 2075.6 kN

Check Slab Stresses

Separately calculate the maximum positive and negative moments in the frame
for the dead, live, and balancing loads. A combination of these values will
determine the slab stresses at the time of stressing and at service loads.

109
Chapter four Hand calculation

Dead Load Moments

WDL=10.422+45.86=56.3KN/m

Live load moment

WLL=(3)(8.685)=26.055 kN/m

110
Chapter four Hand calculation

Total Balancing moment

Wb= -40 .04 kN/m (average of 4 bays)

111
 Chapter four Hand calculation

The final slab service moment, at the column centerline are as follows
Service load Exterior Exterior First Center Interior
moment column Mid span interior column span
column
MDL (kN.m) -233.148 +147.160 -289.788 -271.139 +135.556
MLL (kN.m) -107.898 +68.104 -134.111 -125.480 +62.734
M bal (kN.m) +165.813 -104.659 +206.094 +195.581 -96.406

Stage 1: Stresses immediately after jacking (DL + PT) (3.6.2)

Mid span Stresses

−MDL +Mbal P
𝑓𝑡𝑜𝑝 = −
S A

MDL − Mbal P
𝑓bot = −
S A

Support Stresses

(𝑀𝐷𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(−𝑀𝐷𝐿 + 𝑀𝑏𝑎𝑙 ) 𝑃
𝑓𝑏𝑜𝑡 = −
𝑆 𝐴
Transfer stresses Exterior Exterior First Interior interior Center
column Mid span column Mid span column
MDL+MBAL (KN.m) -67.34 42.501 -83.694 39.15 -75.558
M/S (MPa) +/- 0.962 +/-0.607 +/-1.196 +/-0.559 +/-1.08
P/A (MPa) -1.1 -1.1 -1.1 -1.1 -1.1
𝑓𝑡𝑜𝑝 (MPa) -0.138 -0.493 +0.096 -1.659 -0.02
𝑓bot (MPa) -2.062 -1.71 -2.296 -0.541 -2.18

112
 Chapter four Hand calculation

The allowable concrete stresses at transfer are:

𝑓𝑐 <0.6 𝑓𝑐𝑖′ =12.6MPa compression

𝑓𝑡 < 0.25√𝑓𝑐𝑖′ =1.15 MPa tension

From the above table, the maximum tension and compression values are in the
shaded boxes (+0. 0.096 MPa and - 2.296MPa, respectively), and we can see
these stresses are well below the allowable values and therefore the stresses at
transfer are acceptable

Stage 2: Stresses at service load (DL + LL + PT)

Mid span Stresses

(-MDL -MlL +Mbal ) P

𝑓𝑡𝑜𝑝 = -
S A

(𝑀𝐷𝐿 + 𝑀𝑙𝑙 − 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

Support Stresses

(+𝑀𝐷𝐿 + 𝑀𝑙𝐿 − 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑡𝑜𝑝 = −
𝑆 𝐴

(−𝑀𝐷𝐿 − 𝑀𝑙𝑙 + 𝑀𝑏𝑎𝑙 ) 𝑃

𝑓𝑏𝑜𝑡 = −
𝑆 𝐴

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 Chapter four Hand calculation

Service load stresses Exterior Exterior First interior Center

column Mid span Interior Mid span column
column
𝑀𝐷𝐿 + 𝑀𝑙𝐿 -175.233 +110.605 -217.805 +101.884 -201.038
+ 𝑀𝑏𝑎𝑙 (KN. m)
M/S (MPa) +/- 2.5 +/-2.02 +/-3.11 +/- 1.46 +/- 2.87
P/A (MPa) -1.1 -1.1 -1.1 -1.1 -1.1

𝑓𝑡𝑜𝑝 (MPa) +1.4 -3.29 +2.01 -2.56 +1.77

𝑓bot (MPa) -3.6 +0.92 -4.21 +0.36 -3.97

The maximum permissible service loads are:

Tension < 0.5√35= 3 MPa

compression < 0.45 𝑓𝑐′ =15.75 MPa

From the above table, the maximum tension and compression values are in the
shaded boxes (+2.01MPa and -4.21MPa, respectively), and we can see these
stresses are well below the allowable values and therefore the stresses at service
loads are acceptable.

Determine the hyper-static moments (secondary moment):

The primary post-tensioning moments, M1, vary along the length of the span.

M1 = P × e

The secondary post-tensioning moments, Msec, vary linearly between supports.

Msec = Mbal - M1

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Chapter four Hand calculation

We can tabulate the hyperstatic moments as follows:

Moment Exterior Exterior First Interior Center interior

column Mid span column column Mid span
MBAL (KN.m) +165.813 -104.659 +206.094 +195.581 -96.406
e (cm) 0 +7.87 -7.87 -7.87 +7.87
M1 (KN.m) 0 +163.35 -163.35 -163.35 +163.35
Msec (KN.m) +165.813 +58.691 +42.744 +32.231 +66.944

The typical load combination for ultimate strength design is

Mu = 1.2 MDL + 1.6 MLL + 1.0 Msec

Moment Exterior Exterior First Interior Center interior

column Mid span column column Mid span
MUDL (KN.m) -279.78 +176.59 -347.74 -325.37 +162.67
MULL(KN.m) -172.64 +108.97 -214.58 -200.77 +100.37
Msec(KN.m) +165.813 +58.691 +42.744 +32.231 +66.944
MU(KN.m) -286.61 +344.25 -519.58 -493.91 +329.984

Determine minimum bonded reinforcement: to see if acceptable for ultimate

strength design.

positive moment areas:

Exterior Mid span ƒt=0.92 MPa˂ 0.17√𝑓′𝑐 .= 0.17√35=1MPa

Thus, minimum bonded steel is not required

interior Mid span: ƒt=0.36 MPa˃0.17√𝑓′𝑐 .= 0.17√35=1MPa

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Chapter four Hand calculation

Thus, minimum bonded steel is not required

In negative moment areas (tension in the top of the slab), the minimum area of
bonded reinforcing in each direction is:

First Interior column

As, min = 0.00075Acf

7.62+7.62
Acf = max. (0.22)[ ,8.685]
2

=max.(0.22)[7.62,8.685]

= 1.91 m2

As, min = 0.00075(1.91)

= 0.00143 m2

=3Ø25 Top (1530mm2)

Exterior column :
7.62
Acf = max. (0.22)[ , 8.685]
2

=max.(0.22)[3.81 , 8.685] = 1.91m2

As, min = 0.00075(1.91)

=0.00143 m2

= 3Ø25 Top (1530mm2)

Must span a minimum of 1/6 the clear span on each side of support. At least
4 bars required in each direction, Place top bars within 1.5h away from the face
of the column on each side.

=1.5h

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 Chapter four Hand calculation

=1.5(220)

=330mm

Maximum bar spacing is 300mm

Check minimum reinforcement if it is sufficient for ultimate strength

𝑎
Mn = (Asfy + Apsfps) (d - )
2

Aps = 99×(number of tendons)

= 99×(20 tendons)

= 1980 mm2

𝑓′ 𝑐
𝑓𝑝𝑠 = 𝑓𝑠𝑒 + 70+ ≤ 𝑓𝑝𝑦 𝑜𝑟 𝑓𝑠𝑒 + 210 for slabs with L/h > 35 (ACI 18.7.2)
300𝜌𝑝

(𝐴𝑠 𝑓𝑦 + 𝐴𝑝𝑠 𝑓𝑝𝑠 )

a=
(0.85𝑓′𝑐 𝑏)

At First Interior column

1.27
d= 22 -2.5- ( )=18.865 cm
2

𝐴𝑝𝑠 1980
ρ = = = 1.21 × 10-3
𝑏𝑑 8685×188.65

35
𝑓𝑝𝑠 =1048.24 + 70 + ≤ 1860 𝑜𝑟 1172.8 + 210
300×1.21× 10−3

𝑓𝑝𝑠 = 1214.8 MPa ≤ 1860 𝑜𝑟 1382.8 MPa

𝑓𝑝𝑠 =1214.8MPa

1530×414+1980×1214.8
a= = 11.76mm
0.85×35×8.685×103

11.76
ϕMn=0.9(1530 × 414 + 1980 × 1214.8)(188.65- )
2

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 Chapter four Hand calculation

=500KN.m<519.58 KN.m Reinforcement for ultimate strength requirements

governs

As,req’d= 1817.5 mm2

4Ø25Top at interior column

3Ø25Top at exterior supports

When reinforcement is provided to meet ultimate strength requirements, the

minimum lengths must also conform to the provision of ACI 318.

We must now investigate the shear stress in the slab at the columns in the
direction of the analysis. Let's begin by determining the critical section, and
then calculate the nominal shear capacity using the appropriate equation. After
that, we will calculate the shear stress demand.

For the exterior column:

we can see that the slab does not extend at least 4h past the face of the columns,
and therefore we cannot take advantage of the post-tensioning to calculate the
shear capacity and must revert to the non pre-stressed equation:

ϕ 𝑣 c = ϕ0.33λ√𝑓′𝑐 =0.75×0.33×1×√35 =1.46Mpa

To calculate the punching shear demand, we need to determine the factored

shear forces and the transfer moments at each column. The factored moments
are the algebraic sum of the moments due to external loading (sustained dead
loads, live loads, etc.) plus the hyperstatic moments. The load factor
combination for moment is 1.2MD +1.6ML + 1.0MHYP.

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 Chapter four Hand calculation

𝑣 u=[1.2(1.2+5.28)+1.6(3)](3.81×8.685)=416.14 KN

And from the table above MU=-286.61 KN.m

Since we have a square column, we will transfer 40% of the unbalanced

moment by eccentricity of shear. The exterior
188.65
b1=(500+ )=594.33mm
2

b2 =(500+188.65)=688.65mm

b0=2b1+ b2 =1877.31mm

Thus, the area of the critical section is:

Ac= b0 ×d=1877.31×188.65=0.354m2

For a rectangular edge column with the moment acting perpendicular to the
edge, we find the following equations for the properties of the critical section:

𝐽 2𝑏12 𝑑(𝑏1 + 2𝑏2 ) + 𝑑 3 (2𝑏1 + 𝑏2 )

=
𝑐𝐴𝐵 6𝑏1

2×594.332 ×188.65(594.33+2×688.65)+188.653 (2×594.33+688.65)

=
6×594.33

=7.72×107 mm3

𝐽 2𝑏12 𝑑(𝑏1 + 2𝑏2 ) + 𝑑 3 (2𝑏1 + 𝑏2 )

=
𝑐𝐴𝐵 6(𝑏1 + 𝑏2 )

2×594.332 ×188.65(594.33+2×688.65)+188.653 (2×594.33+688.65)

=
6(594.33+688.65)

=3.58×107 mm3

Since we have a square column, we will transfer 40% of the unbalanced

moment by eccentricity of shear

119
Chapter four Hand calculation

𝛾𝑣 𝑀𝑠𝑒 𝑐AB
𝑣 u,AB =𝑣𝑢𝑔 +
𝐽𝑐

416.14 0.4×286.61
= + = 1.18𝑀𝑃𝑎
0.354 7.72×107

𝛾𝑣 𝑀𝑠𝑒 𝑐CD
𝑣𝑢,𝐶𝐷 =𝑣𝑢𝑔 −
𝐽𝑐

416.14 0.4×286.61
= − = 1.18MPa
0.354 3.58×107

Thus, the shear demand at the interior column is less than the allowable shear
stress of 1.46Mpa and therefore is acceptable.

For the interior column:

use the minimum

(a) ϕ 𝑣c= ϕ(0.29 λ√𝑓′𝑐 +0.3ƒpc) = 0.75(0.29√35 +0.3×1.1)=1.53MPa

b1=b2= 500+188.65=688.65mm
b0= 2 b1+2 b2=2754.6mm
Ac= b0 ×d=2754.6×188.65=0.52m2
𝛼𝑠 𝑑
(b) ϕ 𝑣c= ϕ [0.083(1.5+ ) √𝑓′𝑐 +0.3ƒpc]
𝑏𝑜
40×188.65
= 0.75 [0.083(1.5+ ) √35 +0.3×1.1]= 1.81MPa
2754.6

use the less value of ϕ 𝑣c =1.53Mpa

And from the table above MU=133.16 KN.m

b0= 2 b1+2 b2=2754.6mm

Ac= b0 ×d=2754.6×188.65=0.52m2

𝐽 𝐽 𝑏1 𝑑(𝑏1 + 3𝑏2 ) + 𝑑 3
= =
𝑐𝐴𝐵 𝑐𝐶𝐷 3

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Chapter four Hand calculation

688.65×188.65×(688.5+3×688.5)+188.653
=
3

=12.15×107 mm3

v u=[1.2(1.2+5.28)+1.6(3)](7.62×8.685)=832.3 KN
𝛾𝑣 𝑀𝑠𝑒 𝑐AB 832.3 0.4× 133.16
𝑣 u,AB =𝑣𝑢𝑔 + = + = 1.6 𝑀𝑃𝑎
𝐽𝑐 0.52 12.15×107

𝛾𝑣 𝑀𝑠𝑒 𝑐CD 832.3 0.4× 133.16

𝑣𝑢,𝐶𝐷 =𝑣𝑢𝑔 − = − = 1.6 𝑀𝑃𝑎
𝐽𝑐 0.52 12.15×107

For the center column:

ϕ 𝑣c= ϕ(0.29 λ√𝑓′𝑐 +0.3ƒpc) = 0.75(0.29√35 +0.3×1.1)=1.53MPa

And from the table above MU=0 KN.m

b0= 2 b1+2 b2=2754.6mm

Ac= b0 ×d=2754.6×188.65=0.52m2

v u=[1.2(1.2+5.28)+1.6(3)](7.62×8.685)=832.3 KN
𝛾𝑣 𝑀𝑠𝑒 𝑐AB 832.3
𝑣 u,AB =𝑣𝑢𝑔 + = = 1600.6𝑘𝑝𝑎 = 1.6 𝑀𝑃𝑎
𝐽𝑐 0.52

𝛾𝑣 𝑀𝑠𝑒 𝑐CD 832.3

𝑣 u,cd =𝑣𝑢𝑔 − = = 1600.6𝑘𝑝𝑎 = 1.6 𝑀𝑃𝑎
𝐽𝑐 0.52

we need to increase the shear capacity of the slab at the interior and center
column. it is probably most practical and economical to add headed stud
reinforcing.

The last thing we need to check in this direction of analysis is the unbalanced
moment transfer by flexure at the columns. We need to transfer 60% of the
unbalanced moment by flexure. Thus, we need to transfer (0.6)( 286.61)=-172

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Chapter four Hand calculation

KN.m and (0.6)( 133.16 )=80KN.m at the exterior and interior columns,
respectively. This moment transfer must occur in a slab width equal to the
column face dimension plus 1.5h on both sides of the column,500+2(1.5×220)
=1160mm.

0.85𝑓′𝑐 2.353𝑀𝑢
𝜌= [1 − √1 − ]
𝑓𝑦 ∅ 𝑓 ′ 𝑐 𝑏𝑑 2

0.85×35 2.353×80×106
=
414
[1 − √1 − 0.9×35×1160×188.652 ] =0.0054

As= 𝜌 𝑏𝑑
=0.0054×1160×188.65 = 1181.7mm2
Based on this demand, we can see that the minimum area of bonded mild
reinforcement of 3Ø25.

Figure (4-3)minimum bonded reinforcement –north –South interior

To summarize the analysis in this direction, we have used a slab thickness of

0.22mm with an effective pre-stress force of 239KN per meter. So, in the 8.685
wide equivalent frame, we need a total effective pre-stress of 2075.6KN. Using

122
Chapter four Hand calculation

103.78kN per 1.27cm tendon, we need 20tendons (each will be stressed slightly
less than 103.78kN) placed in a narrow band on the column line in the east west
direction, with at least two of them passing through the column core to act as
integrity tendons.

Figure (4-4) Final design sketch of an interior bay.

123
CHAPTER FIVE
CONCLUSIONS
Chapter five Conclusions

Chapter Five
Conclusions
5.1 Conclusion

The main conclusion can be drawn from this project:

1.know something about the historical background of post-tensioned concrete

and the difference between post-tensioned members and pre-tensioned
members.

2. Compute effective pre-stress force Peff for a given drape and balanced load.

3.Understand allowable stresses at transfer and service transfer according to

ACI 318-14.

4. Understand pre-stress losses.

5. Compute the balanced and hyperstatic moments for a continuous structure.

6. Determine the minimum amount of flexural and shear reinforcing required.

7. Calculate the nominal moment capacity φMn and punching shear capacity
of a two-way slab system.

8.Check the moment transfer capacity of the slab at the support.

125
References

[1] Building Code Requirements for Structural Concrete, ACI 318-14,

American Concrete Institute, 2014.

[2] John P.Miller,2018, "Fundamentals of Post-Tensioned Concrete Design

for Buildings" Part One and Part three.

[3] Wei Zhou ,2017, "Analysis of Non-Prismatic Two-Way Post-Tensioned

Slabs ".

[4] Duarte M.V. Faria, Válter J.G. Lúcio, A. Pinho Ramos,2011,”

Strengthening of flat slabs with post-tensioning using anchorages by bonding”.

[5] Mohamed H. AbuGazia, Mahmoud El-Kateb , Tamer Elafandy

and Amr A.Abdelrahman, “BEHAVIOR OF RC CONTINUOUS SLABS

STRENGTHENED BY EXTERNAL PRESTRESSING STEEL STRANDS”.

[6] Kang Su Kim, Deuck Hang Lee,2012,” Nonlinear analysis method for
continuous post-tensioned concrete members with unbonded tendons”.

[7] Dezsõ Hegyi - András Árpád Sipos “A post-tensioned concrete slAb

cAntilevering 6.50 m”.

[8] Erez N. Allouche, Ell, T. Ivan Campbell, Mark F. Green and Khaled
A. Soudki” Tendon Stress in Continuous Unbonded Prestressed Concrete
Members Part 2: Parametric Study”.

126