This document provides an overview of pre-stressed and precast concrete. It discusses basic concepts like pre-stressing, uses of pre-stressed concrete, materials used including high-strength concrete and steel, and methods of prestressing like pre-tensioning and post-tensioning. It also covers topics like tendon profiles, advantages and disadvantages of pre-stressed concrete, losses in prestressing, types of prestressing steel, properties of prestressing steel, and use of non-prestressed reinforcement. The document is submitted by 5 students and contains 15 chapters with information on concepts, introduction, early introduction, uses, the basic idea, methods, profiles, advantages, disadvantages, losses, materials, types of

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REGISTER
Pre-stress & Precast Concrete Register
SUBMITTED BY:
HAMZA HUSSAIN BSCET01193059
ALI RAZA BSCET01193020
SAQIB RAZA BSCET01193019
SHAHYAR KHAN BSCET01193049
M. ARSLAN BSCET01193068
SUBMITTED TO:
HAFIZ M. SHAHZAD ASLAM
DEPARTMENT OF CIVIL TECHNOLOGY
THE UNIVERSITY OF LAHORE

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CHAPTER NO 1
BASIC CONCEPTS
1. Pre-Stressing:
Prestressing is the process by which a concrete element is compressed, generally by steel wires
or strands. Precast elements may be prestressed during the construction process (pre-tensioning)
or structures may be stressed once completed (post-tensioning). Prestressing compensates for the
tensile stresses introduced when the element is loaded. Hence the concrete generally remains in
compression.
2. Introduction:
 Prestressed concrete is a structural material that allows for
predetermined, engineering stresses to be placed in members to counteract the stresses that
occur when they are subject to loading. It combines the
high strength compressive properties of concrete with the high tensile strength of steel.
 In ordinary reinforced concrete, stresses are carried by the steel reinforcement,
whereas prestressed concrete supports the load by induced stresses throughout the
entire structural element. This makes it more resistant to shock and vibration than
ordinary concrete, and able to form long, thin structures with much smaller sectional areas to
support equivalent loads.
 Prestressed concrete was patented by San Franciscan engineer P.H Jackson in 1886, although
it did not emerge as an accepted building material until 50 years later when a shortage
of steel, coupled with technological advancements, made prestressed concrete the building
material of choice during European post-war reconstruction.
 It is now commonly used for floor beams, piles and railways sleepers, as well
as structures such as bridges, water tanks, roofs and runways. Generally, prestressed

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concrete is not necessary for columns and walls, however, it can be used economically for
tall columns and high retaining walls with high bending stresses.
 As a general rule, traditional reinforced concrete is the most economic method for a span of
up to 6 m. Prestressed concrete is more economical when spans are over 9 m. Between 6 and
9 m, the two options must be considered according to the particular requirements as to which
is the most suitable option.
Highway crossing in Switzerland, continuous over two spans
Segmentally precast post-tensioned rigid frames for the Olympic stadium in Montreal

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3. Early introduction:
The prestressed concrete was patented byP.H. Jackson a San Francisco engineer in 1886, it did
not emerge as an accepted building material until a half-century later. The shortage of steel in
Europe after World War II coupled with technological advancements in high-strength concrete
and steel made prestressed concrete the building material of choice during European post-war
reconstruction. North America's first prestressed concrete structure, the Walnut Lane Memorial
Bridge in Philadelphia, Pennsylvania, however, was not completed until 1951.
4. Uses of prestressed concrete:
It is now commonly used for floor beams, piles and railways sleepers, as well as structures such
as bridges, water tanks, roofs and runways. Generally, prestressed concrete is not necessary for
columns and walls, however, it can be used economically for tall columns and high retaining
walls with high bending stresses.itis also use in school auditoriums, gymnasiums, and cafeterias
because of its properties and its ability to provide long, open spaces.
School Gymnasium
5. The Basic Idea Of Pre-Stressed Concrete (Example On A Box):
The principle behind pre stressing is applied when on a box is moved from place to place.
Instead of stacking the boxes vertically and carrying them, the box may be moved in a horizontal
position by applying pressure to the box at the end of the row. When sufficient pressure is
applied, compressive stresses are induced throughout the entire row, and the whole row can be
lifted and carried horizontally at once.
6. METHODS OF PRESTRESSING:
There are two type of method for pre-stressing:
 Pre-tensioning
 Post-tensioning

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7. Pre-Tensioning:
This process involves the stressing of wires or cables by anchoring them at the end of
a metal form, which may be up to 120 m in length. Hydraulic jacks stress the wire as required,
often adding 10% to accommodate creep and other pre-stress losses that may be incurred.
Side moulds are then fixed and the concrete placed around the tensioned wires.
The concrete hardens and shrinks, gripping the steel along its length, transferring
the tension from the jacks to exert a compressive force in the concrete.
Once the concrete has reached the desired strength, the tensioned wires are released from the
jacks. A typical concrete strength of 28 N/mm2 can be achieved by 24-hour steam curing, as well
as using additives.To create shorter members, dividing plates can be placed at any point along
the member which, when removed, permit the cutting of the wires.
8. Post-Tensioning
This follows the reverse method to pre-tensioning, whereby the concrete member is cast and the
prestressing occurs after the concrete is hardened. This method is often used where stressing is to
be carried out on site after casting an insitu component or where a series of precast
concrete units are to be joined together to form the required member.

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9. Tendon Profiles
The linear transformation of a tendon profile can be defined as the tendon placed on the statically
indeterminate structure that rotates around one end without changing its shape within the span;
the total bending moment (summation of sectional forces) is the same, but the reaction forces
change..
10. Tendon Profile Simply Supported Beams With Concentrated Load:
11. Tendon Profile Simply Supported Beams With UDL:
12. Simply Supported Cantilever Beams:

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13. Tendon Profile Simply Supported Beam With Curved Centroid:
14. Tendon Profile Continuous Beams:
15. Advantages of Prestressed Concrete:
Followings are the advantages of prestressed concrete:
 Longer span length increases untroubled floor space and parking facilities.
 Thinner slabs, that are important for high rise building as with the same amount of cost, it
can construct more slabs than traditional thicker slabs.
 As the span length is larger, fewer joints are needed than traditional RC structures.
 Because of fewer joints, maintenance cost also becomes reduced during the design life as
joints are the major locus of weakness in a concrete building.
 Long-term Durability.
 Better finishing of placed concrete.
 It requires a smaller amount of construction materials.
 It resists stresses are higher than normal RCC structures and is free from cracks.
16. Disadvantages of Prestressed Concrete:
Followings are the disadvantages of prestressed concrete:
 It requires high strength concrete and high tensile strength steel wires.

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 The main disadvantage is construction requires additional special equipment like jacks,
anchorage, etc.
 It requires highly skilled workers under skilled supervision.
 Construction cost is little higher than RCC structures.
INSTANTANEOUS LOSSES IN PRESTRESSING
1. Friction:
In the case of pre-tensioning, there is no loss due to friction as the concrete is not being hardened
at the time of tensioning the tendons. In the case of post-tensioning, the tendons are provided
inside the duct of a precast concrete member. So the loss in prestress occurs due to friction
between the concrete surface and the tendon in the process of tensioning. The loss of friction is
also accompanied by the wobble effect.
2. Elastic shortening of concrete:
In the case of pre-tensioning as we cut the tendons the concrete shortens within a short interval
of time by a certain amount which causes loosening of steel tendons. Hence causing a loss in
prestress. In the case of post-tensioning as the concrete is already hardened, after cutting the
tendons the concrete doesn't shorten further.
3. Anchorage slip:
Anchorage, as the name signifies, is a component that is used to anchor the tendons into the
concrete while terminating the tendons. The main function of anchorage is to transfer
the stressing force to the concrete once the stressing process is completed. In the case of pre-
tensioning, the tendons are monolithically embedded into the concrete.
4. Time-dependent losses in prestressing:
 Creep of concrete:
Deformation due to rearrangement of molecules over a period of time under the application of a
constant load is called creep. The deformation of concrete along the direction of tendons, makes
them loosened thereby causing loss in prestress.
 Shrinkage of concrete:
The volumetric changes of concrete structures due to the loss of moisture by evaporation are
known as concrete shrinkage or shrinkage of concrete. Due to shrinkage of concrete, the tendons
are loosened and loss in prestress occurs.
 Relaxation of steel:
Relaxation of steel is defined as the decrease in stress with time under constant strain. Due to the
relaxation of steel, the prestress in the tendon is reduced with time. Just like creep, relaxation
also depends on time. It depends on the type of steel, initial amount of prestress, and the
temperature.
 Total losses in prestressing:
The sum of all type of losses i.e., instantaneous and Time dependent, may be of the order of 20
to 35 percent of the original jacking force.

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CHAPTER 2
MATERIALS
1. Introduction:
 High strength concrete is necessary for prestress concrete as the material offers highly
resistance in tension, shear bond and bearing. In the zone of anchorage the bearing stresses
being hired, high strength concrete is invariably preferred to minimizing the cost. High
strength concrete is less liable to shrinkage cracks and has lighter modulus of elasticity and
smaller ultimate creep strain resulting in a smaller loss of prestress in steel. The use of high
strength concrete results in a reduction in a cross sectional dimensions of prestress concrete
structural element with a reduced dead weight of the material longer span become
technically and economically practicable.
 Tensile strength of high tensile steel is in the range of 1400 to 2000 N/mm2
and if initially
stress upto 1400 N/mm2
their will be still large stress in the high. tensile reinforcement
after making deduction for loss of prestress. Therefore high tensile steel is made for
prestress concrete.
 The total length changes in the memberΔ/=(Esh+Ecu) /c may be such that it exceed the
stretch in the steel that produced the initial stress, and complete loss of prestress force may
result.
 The reduction in steel stress from these causes depends only on the unit strains in the
concrete associated with shrinkage and creep, and the elastic modulus Es of the steel.
Δfs = (Esh+Ecu) x Es
 First the member is prestessed using ordinary reinforcing steel at an initial stress fsi of 30 ksi
 The modulus of elasticity Esfor all steels is about the same and will be as 29000 ksi
 Initial strain of steel Esi = fsi/Es = 30/29000 = 1.03 x 10-3
 Total Steel elongation Esi /s = 1.0 x 1o-3 /s

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 Sum of shrinkage and creep strain in concrete is 0.90x10-3 and the corresponding length
change is (Esh+Ecu)x /c = 0.90 x 10-3 /c
 fsc = (1.03-0.90)x10-3 X 29 X 103 = 4 ksi
 Esi = 150 / 29000 = 5.17 x 10-3
 Es /s = 5.17 x 10-3 /s
 The length change resulting from the shrinkage and creep effects would be the
same as :
 (Esh+Ecu)x /c = 0.90x 10-3 /c
 Effective steel stress fse = (5.17- 0.90)10-3 x 29x103 = 124 ksi
2. Types of Pre stressing steel:
The development of pre-stressed concrete was influenced by the invention of high strength steel.
It is an alloy of iron, carbon, manganese and optional materials. In addition to pre-stressing steel,
conventional non-pre-stressed reinforcement is used for flexural capacity (optional), shear
capacity, temperature and shrinkage requirements.
3. Wires:
A pre-stressing wire is a single unit made of steel. The nominal diameters of the wires are 2.5,
3.0, 4.0, 5.0, 7.0 and 8.0 mm. The different types of wires are as follows:
 Plain wire: No indentations on the surface.
 Indented wire: There are circular or elliptical indentations on the surface.
4. Strands:
A few wires are spun together in a helical form to form a pre-stressing strand. The different
types of strands are as follows:
 Two-wire strand: Two wires are spun together to form the strand.
 Three-wire strand: Three wires are spun together to form the strand.
 Seven-wire strand: In this type of strand, six wires are spun around a central wire. The
central wire is larger than the other wires.
5. Tendons:
A group of strands or wires are placed together to form a pre-stressing tendon. The tendons are
used in post-tensioned members. The following figure shows the cross section of a typical
tendon. The strands are placed in a duct which may be filled with grout after the post-tensioning
operation is completed
Figure 1: Cross-Section of a typical tendon

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6. Cables:
A group of tendons form a pre-stressing cable. The cables are used in bridges.
7. Bar:
A tendon can be made up of a single steel bar. The diameter of a bar is much larger than that of a
wire. Bars are available in the following sizes: 10, 12, 16, 20, 22, 25, 28 and 32 mm.
Figure 2: Forms of reinforcing and pre-stressing steel
8. Types of Pre-stressing Steel
 The steel is treated to achieve the desired properties. The following are the treatment
processes:
 Cold working (cold drawing) is being done by rolling the bars through a series of dyes. It
re-aligns the crystals and increases the strength.
 Stress relieving is being done by heating the strand to about 350°C and cooling slowly.
This reduces the plastic deformation of the steel after the onset of yielding.
 Strain tempering for low relaxation is being done by heating the strand to about 350°C
while it is under tension. This also improves the stress-strain behavior of the steel by
reducing the plastic deformation after the onset of yielding. In addition, the relaxation is
reduced.
9. Properties of Pre-stressing Steel:
The steel in pre-stressed applications has to be of good quality. It requires the following
attributes:
 High strength
 Adequate ductility
 Bend ability, which is required at the harping points and near the anchorage
 High bond, required for pre-tensioned members
 Low relaxation to reduce loses.
 Minimum corrosion.

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Figure 3(a): Test set-up
Figure 3(b): Testing of tensile strength of pre-stressing strand
Table 1: Cold Drawn Stress-Relieved Wires (IS: 1785 Part 1). The proof stress should not be less than
85% of the specified tensile strength.
Table2: As-Drawn wire (IS: 1785 Part 2). The proof stress should not be less than 75% of the specified
tensile strength.
Table 3: Indented wire (IS: 6003). The proof stress should not be less than 85% of the specified tensile
strength.
The minimum tensile strength of high tensile steel bars according to IS:2090 is 980 N/mm2
. The
proof stress should not be less than 80% of the specified tensile strength.The stiffness of pre-
stressing steel is given by the initial modulus of elasticity. The modulus of elasticity depends on
the form of pre-stressing steel (wires or strands or bars). IS:1343 - 1980 provides the following
guidelines which can be used in absence of test data.

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10. Non - Pre-stressed Reinforcement:
 For both pretensioned and post tensioned members, it is common to provide longitudinal bar
steel to control shrinkage and temperature cracking. Overhanging flanges of Tand I shaped
cross sections are normally reinforced in both the transverse and longitudinal directions
with no tensioned bars. It is often convenient to increase the flexural strength of prestressed
beams using supplementary longitudinal bar reinforcement.
 This bond is provided by the relatively large chemical adhesion that develops at the steel
concrete interface, by the natural roughness of the mill scale on hot rolled Continued
 Large diameter bars with 75,000 and 90,000 psi (517 and 621 MPa) yield are available on
special order, although they find little application in prestressed concrete members.
11. Stress - Strain prosperities of steel:
 The stress-strain curve describes the behavior of steel bars under loads. It is created by
testing steel specimens. A steel specimen is gradually pulled through a testing machine until
it breaks, and stress and corresponding strains are recorded.
 There are different mark points on the stress-strain curve that represent various stages that
steel specimen passes through prior to fracture. It is very crucial to understand the stress-
strain curve in order to be able to understand the response of steel bars when subjected to
loads.

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12. Limit of Proportionality:
This stage is represented on the stress-strain curve from the initial point until point "A". In this
region, the stress is low and does not generate permanent strain. The stress and strain are
proportional to each other, so as the stress is removed the steel bar would regain its original
shape.
 Elastic Limit:
It is located between point "A" and "B" on the curve. When the stress on steel specimen is
further increased, it would create elastic strain. The stress and strain are not proportional to each
other
 Yield Point:
It is the most important point on the stress-strain curve from the design point of view. This point,
denoted by letter B on the curve, is considered as the failure point in the design of reinforced
concrete structure.
 Ultimate Strength:
As the stress is further increased beyond yield point, strain hardening takes place that is
represented from point C to D, beyond which necking starts. During strain hardening, the
material undergoes changes in its atomic and crystalline structure, resulting in increased
resistance of the material to further deformation..
 Rupture Strength:
Rupture strength is the strength of the material at rupture. This is also known as the
breaking strength. It is the point "E" on the stress-strain diagram.
 Steel Relaxation:
A steel tendon stretched between two fixed points gradually loses a part of its tension due to
creep. The loss in tension under constant strain, as in a test in which the length of the tendon is
maintained constant after stretching, will be referred to as the intrinsic relaxation.

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Most codes recognize the fact that the magnitude of relaxation of a tendon in a prestressed
concrete member increases with the increase in initial steel stress and decreases with the increase
of loss due to creep and shrinkage.
fp / fpi = 1 – log t (fpi / fpy - 0.55) / 10
fp / fpi = 1 – (log tn –log tr ) / 10 x (fpi / fpy 0.55)
13. Time-Dependent Deformation of Concrete:
Time-dependent deformation s i n prestressed concrete are attribute d to creep and shrinkage of
the concrete and relaxation of the steel. Each of these is a function of numerous other effects.
Creep and shrinkage are affected by almost every variable evolved in the fabrication and loading
of a member. In addition, exposure to different atmospheric conditions results in different rate s
and magnitudes of time-dependent deformations in the concrete
 Creep:
The rate-of-creep method takes into account the fact that concrete stresses i n a prestressed
concrete member change with time . The relationship between creep strain and time is expressed
as follows:
dC/dT = fc dC/dT
C f ta / t l f dC/dT dt
 Shrinkage:
ε sh (t) = t − t ( c ) 35 + t − t ( c ) γ shε shu
ε sh (t) = shrinkage strain at time, t;
ε shu = ultimate shrinkage strain, 780 x 106 (in/in) (standard conditions);
t = age of concrete in days; and t
tc = age of concrete when drying starts at end of moist curing in days.

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CHAPTER NO 3
FLEXURAL ANALYSIS
1. Introduction:
 Beam can requires analysis or design.
 The flexural analysis of concrete and steel dimension because the magnitude and line of
action of the effective prestress force.
 The loads are stated as the one to find the resulting stresses and compare these against a set
of permissible values.
 Initial prestress when Pi alone may act on the concrete.
 Initial pretress plus self-weight of the member.
 Initial prestress plus full dead load.
 Effective prestress Peafter losses plus service loads consisting of full dead and expected live
loads.
 Ultimate load when the expected service loads are increased by load factor and the member
is at incipient failure.
 At low the service load stage both concrete and steel stresses are usually within the elastic
range.
 The member be overloaded or one or both materials are likely to be stressed into the inelastic
range in this case prediction of ultimate strength must be based on actual nonlinear stress
strain relations.
2. Notation:
 Tensile strain and stresses are taken to be positive (they are linked with the length increase)
and compressive strains and stresses can be negative.
 Strains refer to the top surface of a flexural member are given the subscript 1 and refer to the
bottom the subscript 2.
3. Partial loss of prestress force:
 Jacking tension Pj initial applied to the tendon is reduced at once to what is termed the
initial prestress force Pi
 The loss in jacking tension which is due to friction between a post tensioned tendon and
its encasing conduit occurs before the transfer of the prestress force to the concrete.
 The elastic shortening of concrete is due to slip of post tensioning anchorages and the
wedges the take hold its happen immediately after transfer.
 Losses can happen over extended period and the concrete shrinkage and creep because of
relaxation of the stress in the steel tendon. The prestress force is reduced from Pi to its
final or effective value Pe after all the time dependent losses have taken place.
 The great value of interest to the designer is the initial prestress Pi and the effective
prestress Pe. It is convenient to show the relation between the values is noted to
effectiveness ratio R
 .

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4. Behavior of prestress beams in the elastic range:
 The concrete centroid is the entire uncracked cross section and the steel will be expressed
by it centroidal axis whether it have one tendon or more.
 The eccentricity of a steel centroid is positive if measured downward from the concrete
centrois is e.
 The distance from the concrete centroid to the top and bottom of the member are C1 and
C2 respectively.
5. Elastic stresses:
 The member is subjected only to the initial prestressing force Pi its have expressed that the
compressive resultant acts at the steel centroid.
 The concrete stress f1 at the top face of the member and f2 at the bottom face can be found
by superimposing axial and bend effects
 f1 = - P/ Ac + Pi e c1 / lc
 f2 = Pi / AC – Pi e c2 / lc
 The tendon eccentricity measured downward from the concrete centroid Ac is the area of the
concrete cross section andlc is the moment of inertia of the concrete cross section.
 Substituting the radius of gyration r2
= lc / Ac these equation can be written:
 f1 = - Pi / Ac ( 1 – e c1 / r2
)
 f2 = - Pi / Ac ( 1 + e c2 / r2
)

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 f 1 = - Pi / AC ( 1 – e c1 / r2
) – Mo / S1
 f 2 = - Pi / AC ( 1 – e c2 / r2
) + Mo / S1
 The MO is the bending moment resulting from the self-weight of the member and S1
= lc/c1and S2 = lc / c2are the section moduli with respect to the top and bottom surfaces of
the beam.
 Superimposed dead loads may be placed when the prestress force is still close to its initial
value that is before time dependent losses have occurred.
 Superimposed live load are applied sufficiently late for the greatest part of the loss of
prestress to have occurred.
 The next load stage of interest is the full service load stage when the effective prestress acts
with the moments resulting from self-weight (Mo) superimposed dead load (Md) and
superimposed lived load (Ml).
 f 1 = - Pe / AC ( 1 – e c1 / r2
) – Mt / S1
 f 2 = - Pe / AC ( 1 – e c2 / r2
) + Mt / S1
 The total moment M is:
 Mt = Mo + Md + Ml

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6. Calculation of Section properties:
 The properties of the concrete of the concrete cross section to be used in the previous
equations it should be noted that in post tensioned construction tendons may pass through
conduits of considerable size.
 After grouting the transformed section should be used.
 The holes can be filled with concrete and the steel replaced with its transformed area of
equivalent concrete equal to (np – 1)Apwhere np is the modular ratio Ep / Ecand Apis the
area of the prestressing steel.
 The holes deduction is small and the gross concrete section can provide the basis for all
calculations.
7. Cross section kern or core:
 The prestressing force acts alone cause no tension in the cross section to be acting within
the kern or the core of the cross section.
 To calculate the lower kern dimendion the concrete stress at the top surface is set equal to
zero:
 f 1 = - (Pi / Ac) (1 – e c1 / r2
) = 0
 That tells us that the quantity in parentheses must equal zero
 Solving for that particular eccentricity is expressed as e = k2the lower kern limit is:
 1 – k2c1 / r2
= 0
 K2 = r2
/ c1
 The upper kern limit is calculated by setting the expression for the concrete stress at the
bottom surface equal to zero from which k1 = - r2
/ c2 the minus sign confirming that the
limit dimension is measured upward from the concrete centroid.
8. EXAMPLE:
The simply supported I-beam shown in cross section and elevation is to carry a uniformly
distributed service dead and live load totaling 0.55 kips/ft over the 40 ft span in the addition to its
own weight. Normal concrete having density of 150 lb/ft3
will be used. The beam will be
pretensioned using multiple seven wire strands eccentricity is constant and equal to 5.19 in. the
prestress force Pi immediately after transfer ( after elastic shortening loss) is 169 kips. Time
dependent losses due to shrinkage creep and relaxation total 15 percent of the initial prestress
force.
Find the concrete flexural stresses at mid-span and support sections under initial and final
conditions.

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9. Given data:
 Beam section = I-section
 Span = 40 ft
 Service dead and live load = 0.55 kips / ft
 Eccentricity = 5.19 in.
 Initial prestress force = 169 kips
 Time dependent losses = 15 percent of the initial prestress force.
 Concrete flexural stresses at mid-span and support section under initial and final
condition =?
 For that Section:
 Moment of inertia Ic = 12000 in4
 Concrete area Ac = 176 in2
 Section modulus S1 =S2= 1000 in3
10. Radius of gyration r2
= lc / Ac = 68.2 in2
 f1 = -P1/ Ac (1- ec1 / r2
) = - 169000/ 176 (1 – 5.19 x12/ 68.2) = -83 psi
 f 2 = -P1/ Ac (1- ec1 / r2
) = - 169000/ 176 (1 + 5.19 x12/ 68.2) = - 1837 psi
11. The member dead load is
 Wo = 176 / 144 x 0.150 = 0.183 kips/ft
12. Dead load moment
 Mo = 1/8 x 0.183 x 402
= 36.6 ft-kips
13. Self-weight moment
 f 1 = - Mo / S1 = -36.6 x 12000 / 1000 = - 439 psi
 f 2 = + Mo / S2 = +36.6 x 12000 / 1000 = + 439 psi

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 f 1= -83 -439 = -522 psi
 f 1=-1837 +439 = -1398 psi
14. Time dependent losses are 15 percent of Pi :
 R= Po / Pi = 0.85
 Po =0.85 x169 = 144 kips
15. Top and bottom concrete stresses due to Pe :
 f 1 =0.85 x -83 = -71 psi
 f 2 = 0.85 x -1837 = -1561 psi
16. Flexural Stresses due to self-weight:
 f 1 = -71 -439 = 510 psi ( -3.5 MPa)
 f 1 = -1561 + 439 = 1122 psi ( -7.7 MPa)
17. Mid-span moment due to superimposed dead and live load :
 Md + Ml = 1/8 x 0.55 x 402
= 110 ft-kips
18. Corresponding concrete stresses:
 f 1 = -110x12000/1000 = -1320
 f 2 = +110x12000/1000 = +1320
19. Effective prestress force with moments due to self-weight and superimposed:
 f 1 = -510 -1320= -1830 psi
 f 2= -1122 +1320= -198 psi
20. Stresses immediately after prestress transfer (before time- dependent prestress losses)
shall not exceed the following:
Extreme fiber stress in compression 0.60fci . Extreme fiber stress in tension except as permitted
in c (next) 3√fci. Extreme fiber stress in tension at ends of simply supported members 6√fci
Where computed tensile stresses exceed these values, bonded auxiliary reinforcement (non-
prestressed or prestressed) shall be provided in the tensile zone to resist the total tensile force in
the concrete computed with the assumption of an uncracked section.
21. Stresses at service loads (after allowance for all prestress losses) shall not exceed the
following:
Extreme fiber stress in compression 0.45fc Extreme fiber stress in tension in recompressed
tensile zone 6√fci Extreme fiber stress in tension in recompressed tensile zone of members
(except two-way slab systems) where analysis based on transformed cracked sections and on
bilinear moment deflection relationships shows that immediate and long-time deflections comply
with requirements stated elsewhere in the Code, and where special cover requirements are met.
12√fci The permissible stresses of Sections 1 and 2 may be exceeded if shown by test or analysis
that performance will not be impaired.
21. Tensile stress in prestressing tendons shall not exceed the following:
Due to tendon jacking force but not greater than 0.85fpu or maximum value recommended by
manufacturer of prestressing tendons or anchorage 0.94fpy Immediately after prestress transfer
but not greater than 0.74fpu. 0.82fpy Post-tensioning tendons, at anchorages and couplers,
immediately after tendon anchorage 0.70fpu

22. REGISTER
22
22. Although the importance of cracking has been
Overemphasized in the past, it may be necessary to predict the cracking load for any of the
following reasons:
 Deflection is influenced by the reduction in flexural rigidity that accompanies cracking
 After the beam cracks, cracks, the prestressing prestressing steel is more vulnerable to
corrosion.
 The fatigue resistance of beams is reduced by cracking because of the greater stress range
experienced by the prestressing steel near the cracks.
 Cracks may be visually objectionable in some cases.
 In the case of liquid containment vessels, leaks are more likely after cracking
23. Allowable Stress Design Method.
 According to current practice in the United States, prestressed concrete members are
proportioned using the Allowable Stress Design Method.
 Cross-section dimensions, prestress force, and prestress eccentricity are selected to keep
concrete stresses within specified limits as the member ranges from the unloaded stage to
the full service load stage.
 When the member is unloaded, with initial prestress force Piand self-weight acting,
concrete stress limits are imposed that relate to the Concrete strength f’ciat the time the
prestress force is transferred to the concrete.
 At full service load, with effective prestress force Peacting, plus the actual dead loads and
specified service live loads, other concrete stress limits are imposed that relate to the full
specified concrete strength f’c.

23. REGISTER
23
 The prestress tendon area is chosen, usually based on the required initial prestress force
Pi, and certain allowable stresses for the steel, related to the yield strength and ultimate
strength of the steel.
 Concrete stress limits imposed by the provisions of the ACI Code are summarized in
Table 3.1, and allowable steel stresses are shown in Table 3.2.
 Beams proportioned based on stress limits as just described must also satisfy other
requirements.
 Deflections at full service load, under sustained load, and possibly other load
combinations must be calculated, and the results compared against limit values.
24. Strength Design Method
 By this method, the concrete section dimensions, steel area, and steel centroid location
are selected to provide the required strength at factored loads.
 This approach is similar to that generally used for reinforced concrete.
 It is more difficult to employ for prestressed beams, mainly because the stress in the
tendon at flexural failure fpsis unknown at the start of the design procedure.
 For typically under-reinforced concrete beams, the steel stress is equal to the yield stress
fy.
25. Load Balancing Design Method
 Trial dimensions are selected for the concrete section, and prestress force and eccentricity
are chosen to provide an upward equivalent load that is equal and opposite to a certain
downward load (often the full dead load).
 The factored load stage is then investigated, and if the flexural strength is less than that
required, the strength is increased, usually by adding non-pre stressed bar reinforcement
to supplement the tensile force in the prestressing tendons.
 The resulting design is often a combination of reinforced concrete and prestressed
concrete.
 Flexural tensile cracks are generally present at normal service load, and a check of crack
widths is important.
 While designing beams with constant eccentricity, the requirements on the section
moduli are that:

24. REGISTER
24
• The concrete centroidal stress may be found by Eq. (4.4) and the initial prestress force by
as before.
• The expression for required eccentricity differs, as:
e = (fti – fcci) S1 / Pi
Certain alternative means are available for coping with the problem of excessive concrete
stresses resulting from prestress at the ends of members with constant eccentricity.
The prestress force may be reduced near the ends of the span by encasing some of the tendons in
plastic sheathing, effectively moving the point of application of prestress force inward toward
midspan for a part of the strands. Or
Supplementary non-prestressed bar reinforcement may be used in the end regions to
accommodate the local high stresses.

25. REGISTER
25
ASSIGNMENT NO 1
QNO1
GIVEN DATE:
Fsi = R + 30 ksi
R = 5+9 = 14
Fsi = 14 + 30 = 44 Ksi
SOLUTION
Converting Ksi into MPa:
1 Ksi is equal 6.895 MPa
44 Ksi = 303.29 MPa
Es = 29000 Ksi = 199948 .01 MPa
E = Fsi / Esi
Esi = Fsi / E
Esi = 303.29 / 199948.01
Esi = 1.51 x 10-3
Total steel Elongation = Esi x lc
= 1.51x10-3
lc
Total Concrete Elongation = ( Esh + Ecu ) x lc
= 0.90 x 10-3
lc
Effective Strain = (Steel Elongation – Concrete Elongation) x Es
= (1.51x10-3
–0.90x10-3
)× 199948.01 = 121.96 MPa ANSWER

26. REGISTER
26
QNO 2
GIVEN DATA
Fsi = R + 150 ksi
R = 5+9 =14
Fsi = 14+ 150 = 164ksi
SOLUTION.
Converting Ksi into MPa:
1 Ksi is equal 6.895 MPa
164 Ksi = 1130.78 MPa
Es = 29000 Ksi = 199948 .01 MPa
E = Fsi / Esi
Esi = Fsi / E
Esi = 1130.78 / 199948.01
Esi = 5. 65 x 10-3
Total steel Elongation = Esi x lc
= 5.65x10-3
lc
Total Concrete Elongation = (Esh + Ecu) x lc
= 0.90 x 10-3
lc
Effective Strain = (Steel Elongation – Concrete Elongation) x Es
= (5.65x10-3
–0.90x10-3)
x 199948.01 = 949. 75 MPa

27. REGISTER
27
ASSIGNMENT NO 3
QNO 1.
Given data:
Beam section = I-section (Symmetrical)
Span = 12192 mm
Concrete Density = 2.35 x 10−5
N/mm3
Service dead and live load = 86.99 N/mm
Eccentricity = 258.82 +0.59mm=259.14
Initial prestress force =Pi = 773990.28 N
Time dependent losses = 15.59 percent of the initial prestress force.
Final Pre stress force =Pe = 657504.74 N
Mo = 49424417.28 N / mm
MD + ML = 1504843442 N / mm
Concrete flexural stresses at mid-span and support section under initial and
final condition =?
Section Properties
 Moment of inertia Ic = 4994777107 mm4
 Concrete area Ac = 113548.16 mm2
 Section modulus S1 =S2= `16387064 mm3
 Centroid = C1 = C2 = 3657.6 mm
Initial Loading: ( F1 )
F 1 = - Pi / Ac + Pi e C1 / I
F 1 = - 773990.28 / 113548.16 + 773990.28 X 258.82 X 3657.6 / 4994777107
F 1 = - 6.81 + 146 .69
F 1 = + 139.88 N / mm

28. REGISTER
28
Initial Loading:
F 2 = - Pi / Ac - Pi e C1 / I
F 2 = - 773990.28 / 113548.16 - 773990.28 X 258.82 X 3657.6 / 4994777107
F 2 = - 6.81 – 146 .69
F 2 = - 153.5 N / mm
Final loading:
F 1 = - Pe / Ac + Pe e C1 / I
F 1 = - 657504.74 / 113548.16 + 657504.74 X 258.82 X 3657.6 / 4994777107
F 1 = - 5.79 + 124.61
F 1 = 118.82 N / mm
Final loading:
F 1 = - Pe / Ac + Pe e C1 / I - Mo / S1
F 1 = + 118.82 - 49424417.28 / 16387064
F 1 = + 118 .82 - 3.01
F 1 = + 115.81 N / mm
Final loading:
F 1 = - Pe / Ac + Pe e C1 / I - Mo / S1 - MD + ML / S1
F 1 = + 115.81 - 1504843442 / 16387064
F 1 = + 115.81 - 91.83
F 1 = + 23.98 N / mm
Final loading:
F 1 = - Pe / Ac + Pe e C1 / I
F 2 = - 657504.74 / 113548.16 - 657504.74 X 258.82 X 3657.6 / 4994777107
F 2 = - 5.79 - 124.61

29. REGISTER
29
F 2 = - 130.4 N / mm
Final loading:
F 2 = - Pe / Ac - Pe e C1 / I + Mo / S2
F 2 = - 130.4 + 49424417.28 / 16387064
F 2 = - 130.4 + 3.01
F 2 = - 127.39 N / mm
Final loading:
F 2 = - Pe / Ac - Pe e C1 / I + Mo / S2 + MD + ML / S2
F 2 = -127.39 + 1504843442 / 16387064
F 2 = - 128.68 + 91.83
F2 = - 35.56 N / mm

30. REGISTER
30
ASSIGNMENT NO 4
QNO1
GIVEN DATA
WL = 1000+14lb/ft =1014 lb/ft
WL =
𝟏𝟎𝟏𝟒×𝟒.𝟒𝟒𝟖
𝟏𝟐×𝟐𝟓.𝟒
= 14.797 N/mm
WD = 500+14lb/ft = 514lb/ft
WD =
𝟓𝟏𝟒×𝟒.𝟒𝟒𝟖
𝟏𝟐×𝟐𝟓.𝟒
= 7.500 N/mm
Span = 40'+(14)'= 54'
fc’ = 6000/145.08 = 41.36 Mpa
fci’ = 4200/145.08 = 28.94 Mpa
R = 0.85+0.14 = 0.99
SOLUTION:
STEP 1.
Initialloadingcondition
fci = -0.6(fci)
fci = -0.6(28.94) = - 17.36
fti = 3 √𝑓𝑐𝑖
fti = 3√28.94 = 16.13
Service loading condition
fcs= 0.45fc’
fcs= -0.45(41.36) = -18.612
fts =6 √41.36
fts = 6 √41.36= 38.58
STEP 2.
D.L, L.L, Self weight
MD =
𝑊𝐿²
8
=
514×54²
8
= 187353 lb-ft

31. REGISTER
31
ML =
𝑊𝐿²
8
=
1014×54²
8
= 369603 lb-ft
MD+ML = 187353+369603 = 556956
MO =
𝑊𝑂𝐿²
8
=
3.85×54²
8
= 1403.32lb
STEP 3
S1 ≥
(1−R)M0+MD+ML
Rfti−fcs
S1 ≥
(1−0.99)1403.32+(556956)
0.99(16.13)+17.36
= 16711.39 in3
S2 ≥
(1−R)M0+MD+ML
fts−R(fci)
S2 ≥
(1−0.99)1403.32+(556956)
38.58−0.99(−18.612)
= 9771.06 in3
TRIAL
I1 =
𝑏𝑑3
12
+Ay2
I1 =
12×63
12
+(12×6) (28-3)2
= 45216 in4
I2 =
𝑏𝑑3
12
+Ay2
I2 =
4×443
12
+(4×44) (0)2
= 28394.66 in4
I3 =
𝑏𝑑3
12
+Ay2
I3 =
12×63
12
+(12×6) (28-3)2
= 45216 in4
I = 118826.66 in4
STEP 4
S1 =
𝐈
𝐂𝟏
=
𝟏𝟏𝟖𝟖𝟐𝟔.𝟔𝟔
𝟐𝟖
=4243.80 in3
S2 =
𝐈
𝐂𝟐
=
𝟏𝟏𝟖𝟖𝟐𝟔.𝟔𝟔
𝟐𝟖
=4243.80 in3
Ac = (12×6)+(6×44)+(12×6) = 408 in2

32. REGISTER
32
STEP 5
fcci = fti-c1(fti-fci)
fcci = 16.13-28(16.13+17.36) = -921.59 Mpa
STEP 6
Pi = fcci×Ac = -921.59×408 = - 376008.72 Mpa=376008.72 Mpa
Pe = 376008.72 ×0.99 = -372248.6328 Mpa = 372248.6328 Mpa
STEP 7
e = (fti- fcci) ×
𝑆1
𝑃𝐼
+
𝑀𝑜
𝑃𝐼
e = (16.13-+921.59) ×
4243.80
376008.72
+
1403.32
376008.72
= 10.58 Mpa
Designofbeamwithvariableeccentricity
Pi = - 376008.72 Mpa
STEP 8
Selection of Pre-stress steel will be provided by using tendon
A=πd2
/4 = 0.0491in2
dia = ¼’’
Minimum Tensile = fpu = 240 Ksi
fpy = 0.99 fpu = 0.99×240 = 237.6 Ksi Prestress steel stress
1) 0.82fpy = 0.82(237.6) = 194.40 Ksi
2) 0.74fpu = 0.74(240) = 177.6 Ksi
fp = 19.68 Ksi
STEP 9
G =
𝐹
𝐴
Ap =
Pi
fp
=
376008.72
19.68
= 19106.13 in2
STEP 10
Number of Pre-stress steel wire equal to =
Ap
A
= 19106.13/0.0491
= 389.39=389.39 / 2194.695

33. REGISTER
33
OBE ASSIGNMENT
Q NO 1: Explain the ACI stress limitation for different classes of beam?
ANSWER:
1. fci ’ is the compressive strength of concrete at the time of initial prestress, and fc’ is the
specified i time of initial prestress, and fc’ is the specified compressive strength of the
concrete. Both are expressed in psi units, as are the resulting stresses. stresses immediately
after prestress transfer (before time- dependent prestress losses) shall not exceed the
following:
 Extreme fiber stress in compression 0.60fci
 Extreme fiber stress in tension except as permitted in c (next) 3√fci
 Extreme fiber stress in tension at ends of simply supportedmembers 6√fci
2. Where computed tensile stresses exceed these values, bonded auxiliary reinforcement (non-
prestressed or prestressed) shall be provided in the tensile zone to resist the total tensile force
in the concrete computed with the assumption of an uncracked section. Stresses at service
loads (after allowance for all prestress losses)shall not exceed the following:
 Extreme fiber stress in compression 0.45fc
 Extreme fiber stress in tension in recompressed tensile zone 6√fci
 Extreme fiber stress in tension in recompressed tensile zone of members (except
two-way slab systems) where analysis based on transformed cracked sections and
on bilinear moment deflection relationships shows that immediate and long-time
deflections comply with requirements stated elsewhere in theCode, and where special
cover requirements are met. 12√fci
3. The permissible stresses of Sections 1 and 2 may be exceeded ifshown by test or analysis that
performance will not be impaired. The permissible tensile stresses in prestressing steel
 These are expressed in terms of fpu, the ultimate strength of the steel, and fpy the
specified yieldstrength.
 It is seen that the stress permitted by the Code depends on the stage of loading.
 When the jacking force is first applied, a stress of 0.85fpu or 0.94fpy is allowed,
whichever isLower
4. No limit need be placed on final steel stress after all losses, because that stress will always be
lessthan the steel stress under initial conditions, when an adequate factor of safety must be

34. REGISTER
34
obtained tensile stress in pre stressing tendons shall not exceed the following:
 Due to tendon jacking force but not greater than 0.85fpu or maximum value
recommended by manufacturer of pre stressing tendons or anchor 0.94fpy
 Immediately after prestress transfer but not greater than 0.74fpu 0.82fpy.
 Post-tensioning tendons, at anchorages and couplers, immediately after tendon
anchorage0.70fpu
QNO 2: Evaluate a post-tensioned prestressed concrete beam is to carry an intermittent
liveload of 1500lb/ft and super imposed dead load of 1000lb/ft, in addition to its own weight,
on a 40ft simple span. Normal density concrete will be used with design strength
fc’=6000psi.it is estimatedthat, at the time of transfer, the concrete will have attained 70%
of fc’ or 4200psi.Time dependent losses may be assumed to be 15 percent of the initial
prestress, giving an effectiveness ratio of 0.85.Determine the required concrete dimensions,
magnitude of prestress force and eccentricity of the steel centroid based on ACI stress
limitation for a class i beam?
Given data:
Beam Section = I-Section
Span = 40ft
Dead Load = Wd = 1000lb/ft
Live Load = WL = 1500 lb/ft
FC,
= 6000psi
Fci = 4200psi
R = .85
SOLUTION
STRESS LIMITATIONS
STEP 1:-
 INITIAL LOADING CONDITION
fci = 0.6 × 𝑓𝑐𝑖 ; -0.6 × 4200 = -2520 psi
fti = ; 3√4200 = 194.42 psi
 SERVICE LOADING CONDITION
fcs= 0.45 × ƒc′ ; - 0.45 × 6000 = -2700 psi
fts =6√c′ ; 6√6000 = 464.75 psi

35. REGISTER
35
STEP 2:-
Calculation of Moment with respect to Deal load, Live load &Service load
 Dead load
WD=
𝑊𝐷×𝐿2
8
=
1000×402
8
= 200000 lb-ft
 Live load
WL=
𝑊𝐿×𝐿2
8
=
1500×402
8
= 300000 lb-ft
 Service load
MO=
𝑀𝑂×𝐿2
8
=
250×402
8
= 50000 lb-ft
STEP 3:-
Calculation of Required section Modulus for section
S1 ≥
(𝟏−𝑹)𝑴𝑶+𝑾𝑫+𝑾𝑳
𝑹𝒇𝒕𝒊−𝒇𝒄𝒔
=
(1−0.85)50000+200000+300000
0.85×192.42−(−2700)
=345.51 in3
S2 ≥
(𝟏−𝑹)𝑴𝑶+𝑾𝑫+𝑾𝑳
𝒇𝒕𝒊−𝑹𝒇𝒄𝒊
=
(1−0.85)50000+200000+300000
464.75−(0.85×2520)
=379.78 in3
1422in3
> 345.51in3
OK
1422in3
> 379.78in3
OK
STEP 4:-
INITIAL STRESSES
Y' =
𝑨𝟏𝒀𝟏+𝑨𝟐𝒀𝟐+𝑨𝟑𝒀𝟑
𝑨𝟏+𝑨𝟐+𝑨𝟑
=
(6 X 12)3+(6 X 16)14+(6 X 12)25
(6 X 12)+(6 X 16)+(6 X 12 )
= 14in
I’=
𝒃𝒅²
𝟏𝟐
+ 𝑨𝒚²
I1+I2+I3= 19904in4
S1=
𝑰
𝑪𝟏
= 1422in4
S2=
𝑰
𝑪𝟐
=1422in4
STEP 5:-
Concrete Centroid Stresses
fcci = 194.43-
14
28
(194.42+2520) = -1162.79Psi

36. REGISTER
36
STEP 6:-
Calculate of Initial pre – Stress forces
Pi = fcci×Ac = -1162.79X 240= 279069.6 N = 279.12 KN
Pe = Pi ×R= 279069.6 x .85 = 237209.16 N
STEP 7:-
Calculate of Eccentricity
e = (fti- fcci) ×(S1/Pi) + (MO/Pi)
= (194+ 1162.79) ×(1422/279069.6) + (50000/279069.6) = 9.07 in
STEP 8:-
Pre stress Steel calculation
Design of beam with variable Eccentricity
Use Table 3.2
fpy = 0.85× fpu
fpy = 0.85× 40
fpy = 204 ksi
.82 fpy = .85 x 204 = 167.28 ksi
.74 fpu =.77 x 240 = 177.6 ksi
STEP 9:-
Evolution of area of pre stress steel
Ap =𝑃𝑖 𝑓𝑝
⁄ = 279.12 167.28
⁄ = 1.66 𝐢𝐧𝟐
STEP 10:-
Number of pre stress steel wire Number of wires
Number of Pre-stress steel wire equal to =
Ap
A
= 1.66/0.0491
= 33.39
= 34 no
A Wire =
𝜋𝑑2
4
= 0.049 in2

37. REGISTER
37
STRESS CHECK
ƒ1 = -
𝑷𝒊
𝑨𝒄
+
𝑷𝒊 𝒆 𝑪𝟏
𝑰
ƒ1 = -
𝑷𝒊
𝑨𝒄
+
𝑷𝒊 𝒆 𝑪𝟏
𝑰
= -
279.11x1000
240
+
279.11X1000 (9.07) 12
19904
= 𝟑𝟔𝟑. 𝟎𝟓 psi
ƒ2 = -
𝑷𝒊
𝑨𝒄
-
𝑷𝒊 𝒆 𝑪𝟐
𝑰
ƒ2 = -
𝑷𝒊
𝑨𝒄
-
𝑷𝒊 𝒆 𝑪𝟐
𝑰
= -
279.11x1000
240
+
279.11X1000 (9.07) 12
19904
= - 2688.95 psi
ƒ1 = -
𝑷𝒆
𝑨𝒄
+
𝑷𝒆 𝒆 𝑪𝟏
𝑰
-
𝑴𝑶
𝑺𝟏
−
𝑴𝑫+𝑳
𝑺𝟏
= -
237209.16
240
+
237209.16 (9.07) 12
19904
-
𝟓𝟎𝟎𝟎𝟎
𝟏𝟗𝟗𝟎𝟒
−
𝟐𝟎𝟎𝟎𝟎𝟎+𝟒𝟎
𝟏𝟗𝟗𝟎𝟒
=- 414. 37 𝐩𝐬i
ƒ2 = -
𝑷𝒆
𝑨𝒄
+
𝑷𝒆 𝒆 𝑪𝟐
𝑰
+
𝑴𝑶
𝑺𝟐
+
𝑴𝑫+𝑳
𝑺𝟐
=-
237209.16
240
-
237209.16 (9.07) 12
19904
+
𝟓𝟎𝟎𝟎𝟎
𝟏𝟗𝟗𝟎𝟒
+
𝟐𝟎𝟎𝟎𝟎𝟎+𝟒𝟎
𝟏𝟗𝟗𝟎𝟒
= 995.91psi