PRESTRESSED CONCRETE
INTRODUCTION
Prestresssed concrete is the most recent of the major form
s
of construction to be introduced
into Structural Engineering. The introduction came in the early part of the twentieth century
following the acute shortage of steel that occurred in Europe after t
he Second World War.
The technique of presstressing has several advantages/ applications in Civil Engineering,
notably in prestress concrete where a prestress force is applied
to a concrete member and this
induces an axial compression that counteracts all or part of the tensile stresses set up in the
member by the applied loading.
OBJECTIVES OF PRESTRESSING
1.
Proper prestressing ordinar
i
ly prevent
s
the formation of tension crack under working
load
s
. This is a good reason for prestresssing tanks to avoid leakage or for
prestressing structural members subject to severe corrosion condition
s
.
2.
With lack of cracking, the entire cross

section r
emains effective for stress; as such a
small
section normally results. With lighter sections, precasting of longer members
becomes possible.
3.
S
ince prestressing put camber (curves)
into member
s
under dead loading, deflection at
working
load
is greatly redu
ced.
4.
Prestressing reduce
s
diagonal tension stresses at working loads.
ADVANTAGES OF PRESTRESSED CONCRETE
1.
Prestressed
concrete is a
crack
free
under service loads. When
exposed to wea
ther,
e
limination of cracks prevents
corrosion.
2.
A pre

stressed member
usually has greater stiffness than non

prestressed member
s
.
3.
Precompression of
concrete
reduces the tendency for inclined cracks to be formed.
4.
Shear strength is more co
nsistent than in ordinary reinforced concrete.
5.
Presstressed concrete
has a high abi
lity to absorb energy (
impact
resistance
etc
)
.
DISA
DVANTAGES OF PRESTRESSED CONCRETE
1.
Stronger material
s are used as such
higher unit cost.
2.
More complicated
formwork
s may be necessary
.
3.
Labor cost
s are usual
greater.
4.
More condition
s
must be checked in desi
gn and close
r
contr
ol of every phase is
required et
c
.
BASIC CONCEPTS
Consider a plain concrete beam
for rectangular
cross

section
W
L
+
f
(
compres
sion)
WB
h
L

f (tension)
b
The bending stress at m
i
d

span due to t
he loading condition,
f
=
Since concrete is a poor material in tension, usually
(

f) will exceed the tensile strength/stress
of concrete.
Option #1: A
dd steel in the tension zone to carry the tensile stresses as
in reinforced
concrete
Option #2: Pre
compress tension zone as in pre
s
tressed concrete.
Considering option #2
F or
P
Wa
F or
P
W
B
The
stresses at mid

span are:
+
=
0 An ideal situation
Stress distribution
The effect
of pre

compression her
e is to eliminate all tensile stresses in the concrete
.
However,
the
compressive stress (
+
) applied to the concrete at the top face may
exceed the
permis
sible compressive stress of concrete.
To avoid this
, the
position of F
is
relocate
d
.
F or P
F or P
Sum
med
up the stresses at mid
–
span
we have,
+
=
SUMMARY:
The objective of prestressing is to lim
it tension in concrete (precompress to
prevent cracks from forming)
. T
he design constraint
is the allowable stress of concrete.
DESIGN CONCEPT
In dealing with the stress in prestressed concrete under service load
conditions, ,three
concept may be applied.
1.
HOMOGENNOUS BEAM CONCEPT
As applied above,
the beam
is assumed
to be made up of one material. T
he
prestressing effectively eliminates cracking.
The combined stress formula is:
Or
this is used to investigate the section.
2.
THE INTERNAL FORCE CONCEPT
It
employ
s
the equilibrium of internal f
orces. Steel takes the tension
and
concrete
takes the compression.
This is analog
ous
to the internal couple method used for non

prestressed
reinforced concrete.
At service loads in reinforced concrete
the point
s
of
action of the force
s (C&
T)
, C=T
are independent of the magnitude
of applied bending
moment,
depending only on the cross

sectional
dimensions and the modular
ratio M.
This concept
is summarized thus:
i)
A know prestress force put into the steel define T.
ii)
An appl
ied moment M is put on the beam
.
iii)
Knowing the magnitu
de and point of action of the force
C, the stress in the concrete
may be computed as:
f =
3.
LOAD BALANCING
CONCEP
T
The concept visualizes prestressing primarily as a process of balancing loads on the
member. The prestresssing tendons are placed so that the eccentricity of the
prestressing force varies in the same manner as the moment from applied load
s. .I
f
this is done well, zero flexural stress
would
result. Thus only the axial stress
(F is
the horizontal component of the force in the tendon) would
act.
T
T
ℓ
max
parabolic tendon
L/2
ℓ
max
Parabolic tendon applied to beam
Force acting on
beam due to prestressing
T
T
The prestressing may be consider
ed as an upward uni
form loads
if the tension is parabolically
draped.
The max prestress moment o
f
*
at mid

span can be equated to an equivalent
uniformly loaded beam
mo
ment.
Thus,
*
Therefore
=
If
=
–
then,
=
And f =
ME
THOD
OF BEAM ANALYSIS
(
Stress analysis by superposition)
Consider a rectangular beam with uniform
load
M
ℓ
or e
F
F
Stress distribution due to an eccentric load is
f
t
=
f
b
=
ℓ
or e is +
, if it is below the member centroidal axis
A = cros
s

sectional area
And
are the stres
ses at the top and bottom fibers
and
are the section modulli at the top
and bottom
fibers
Compressive stresses
are +
as sagging
Bending Moments.
If an external bending moment is introduced, the additional stress dist
ribution of stresses is
introduced and the resultant stress distribution due to the prestress force and applied bending
moment is found by superposition.
=
…………………………………………………………………(1)
=
…………………………………………………………………(2)
In addition to the applied bending moment at the section there is also an axial load therefore,
the force (F) in equations 1 and 2 is the sum of the prestress force and the applied axial load.
EXAMPLE 1
A simply supported pretensioned concrete beam
has dimension
s
as shown in fig.1 and spans
15m. It has an initial prestress force
of 1100KN
applied to it and carries
a UDL (imposed) of
12KN/m.
Determine the extreme fiber stresses at mid

span
:
(1)
Under the self weight of the beam, if the short term losses are 10% and the eccentricity is
325mm below the beam centroid
(2)
Under
the service load,
when the prestress force has been reduced by a further 10%.
a
200
15m
750
a
150
325
Figure 1
0 0 0 200
400
Section A

A
SOLUTION
P= 1100KN
; e= 325mm (given)
A =
∑
areas =
2.13 *10
⁵
mm
²
I=
Also y
therefore, Z=
=
= 3
5. 12 *10
⁶
mm
ᵌ
w
=
24A
=5.1 KN/m
Mi =
143.4KNm (moment due to self weight)
q= w + imposed
load
M
S
=
= 4
80.9KNm (moment due to service load)
= 0.9P = 990KN (for 10% reduction)
= 0.8P= 88OKN (for 20% i.e. further 10% reduction)
(1)
=
=

0.43N/mm²
= 4.65 + 9.16
–
4.08 = 9.73N/
mm²
(ii)
=
= 9.68N/
mm²
= 4
.13 + 8.14
–
13. 69 =

1.42N/mm²

0.43 9.68
Transfer stress
service stress
9.73

1.42
STRESS DISTRIBUTION FOR EXAMPLE
1
The stress dis
tribution for example 1 exemplifies those in a prestressed concrete member
under max
imum
and min
imum
loads.
This demonstrates
the fact that in prestressed concrete
the minimum load condit
ion is always an important one.
The previous considerations; the
pres
tress force has provided by one layer of tendons,
so that
the resultant prestress force coincide with the physical locat
ion of tendons in prestresed
concrete members.
However
,
there are usually more than one layer of tendons
in prestressed
concrete members
,
in this case the resultant prestress force coincide with the location of the
resultant of all the individual prestressing tendons, even if it not physically possible to locate
a tendon at this position.
For post

tensioned members where the duct diameter
is not negligible in comparison with the
section dimension, due allowance for the duct must be made when determining the member
section properties. For pretensioned members, the transformed cross
–
section should be used.
In practice the section properties
are determined on the basis of the gross cross

section.
ADDITIONAL STEEL STRESS DUE TO BENDING
There is no bond between the presstressin
g steel and the surrounding concrete.
In pre

tensioned members and grou
ted post
–
tensioned member
s,
bond is
pres
ent
and bending of
the member induces stress in the steel as in
a reinforced concrete m
ember. It is the bond
between
the steel and concrete
that makes the ultimate loads behavior of pre

tensioned and
grou
ted post
–
tensioned members very similar to that
of reinforced concrete
members and
di
fferent from un

g
routed post
–
tensioned members.
EXAMPLE
Design a rectangular pre

tensioned beam to support a live load of 9kN/
m
on a span of 8m. Determine
the dimensions of the beam section, prestressing force and th
e eccentricity.
Final compres
sive strength
of concrete
(
')
=
3.5
/
cm²
Initial compressive strength
(
= 2.8
/
cm²
Tensile
strength of tendons (
) = 189
/
cm²
Solution
Given design data
are shown below;
W
L
=
9.0kN
/
m
L = 8.0m
'
= 3.5
/
cm²
= 2.8
/
cm²
= 189
/
cm²
TRIAL SECTION:
Let
=
20%
of W
L
= 0.2*9 = 1.8kN/m
Therefore (W) =
+
W
L
= 9+1.8 = 10.8
/m
=
= 10.8 *8² = 86.4kNm
Since б =
=
fc
'
=
=
2469cm
3
Assume b=
Z =
=
*
=
= 2469cm
ᶟ
h = 28cm
30 cm
Try 20cm x
30cm
Z =
=
3000cm
3
>
. (hence dimensions are adequate)
= {0.2*0.3*1} 24 = 1.44kN/m
1.8kN/m
w= 9+1.44
= 10.44
kN/m
Actual
=
= 83.52kNm
Actual
)
=
= 2386c
m
3
3
000cm
3
o.k.
ACI
DESIGN
SPECIFICATIONS
Permissible stress
:
a)
Initial co
ndition: compression =
= 0.6(2.8) = 1.68kN/cm
2
Tension
=
= 0.13kN/
cm²
b)
Final condition:
compression =
0.45
= 1
. 58kN/cm
2
Tension =
= 0.31kN/
cm²
c)
Steel stress:
Immediate stress after transfer (
)
= 0.7
f
pu
= 132. 3kN/
cm²
Assume 18% loss for pre

tensioning
Total loss
)
=
23.81kN/
cm²
Effective stress
(
)
=

= 108.49kN/
cm²
If immediate stress
loss (
) =
0.5
Immediate
stress loss
(
) = 11.9kN/
cm²
Immediate stress
(
)
=

= 120.4kN/
cm²
The ratio of
=
= 1.11F
Determine
force and eccentricity for maximum tension at the bottom
of the beam

=
2.474


(
equation
1)
Determine
force and eccentricity for minimum tension at the top (initial condition of self
weight)
Where
,
Mb
=
=
11.52kNm
But,
=
1.11F
=

0.463

(
equation
2)
Solving equation
s
1 & 2 together
=
2.474
–
0.463
F = 603.3KN
To obtain the value of the eccentricity (e), substitute the value of F = 603.3kN
into equation
(1)
e = [
2.474

]
=
7
.30cm
SELECTION OF STRANDS
Try 8
nos
ϴ
10mm strands.
G
iving
the following:
=
189kN/
cm²
F = 603.
3kN
e
= 7.30cm
Calculate area of strands =
8
2
= 6.28cm²
F = A
rea
of strands *
F = 681.32kN
>603.3
kN
Stran
ds
selected are
adequate
= 7.30cm
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