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Transcript of Reinforced concrete
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University of Salahaddin Hawler
College of Engineering
Civil Engineering Department
Course Book
Design of Reinforced Concrete Structures
Dr. Omar Qarani Aziz
Assistant Professor
BSc. Civil Engineering, MSc. & PhD in Structural Engineering
(Fourth year)
Four hours per week
Six units
2013-2014
Phone No.: 066 226 0198; 455 82 55
E-mail: Dr_ [email protected]
www.OmarQarani.com
-
CE-401 Design of Reinforced Concrete Structures Course aims
This module is intended to give students a good understanding of the design and behavior of design of
reinforced concrete structures at the design ultimate limit state. We will look at the design of framed
building structures in some detail with particular emphasis on the design of torsion of beams, two way
slabs, shear walls, reinforced concrete tanks, reinforced concrete bridges and Prestressed concrete.
Learning outcomes
Emphasis is placed on understanding structural behavior and the background to the design methods in
ACI and other codes where appropriate. By the end of this module you will have a good understanding
of the design and behavior of reinforced concrete structures.
Course syllabus
The following topics are included:
Introduction
Torsion in beams
Two way slabs, Introduction, types, .. etc
Direct design method
Equivalent frame method
Yield line theory of slabs
Multi-story buildings, applied loads, Methods, Software application
Shear walls
Reinforced Concrete Water tanks
Reinforce Concrete Bridges
Prestress Concrete; Introduction, advantages and disadvantages, losses, check of stresses, design of prestressed girders.
Forms of teaching
Different forms of teaching will be used to reach the objects of the course. Notes to be written
on the board especially design equations, head titles, definitions and summary of conclusions,
classification of materials and any other illustration, there will be class room discussions and the lecture
will give enough background to solve examples. Power points presentation will be use when required;
besides work sheets will be designed to let the chance for practicing. Students should read the lectures
notes regularly and to participate the class room discussions.
Assessments (Grading)
Students are required to first semester exam on January, second semester exam on April, class room
activities, quizzes, home works and final exam on June. So that the final grade will be based upon the
following criteria:
- First semester exam -----------------17 %
- Second semester exam --------------17 %
- Activities ------------------------------06 %
- Final Exam ----------------------------60 %
-
Course program
Month Week No. Description
September Week 1 General introduction , objectives , Ref. Course program
October Week 2
Week 3
Week 4
Week 5
Torsion of R.C. Beams
Examples, Quiz 1
Analysis and design of two-way slabs , Introduction
Direct design method procedure.
November Week 6
Week 7
Week 8
Week 9
Continue design procedure of DDM , Design Example1
Continue design Example1 of DDM, Quiz 2
Design Example 2
Design Example 2 Continue
December Week 10
Week 11
Week 12
One Week
Design Example 3, Quiz 3
Equivalent frame method ( EFM )
Design Example
Holiday
January Week 13
Week 14
Week 15
Week 16
Yield line theory (YLT)
Continue Yield line theory (YLT), Examples, Quiz 4
First semester exam
First semester exam
February Week 17
Week 18
Week 19
Week 20
Design of multi-storey Building
Design of multi-storey Building Continue
Design of shear walls
Design of R. C. Tanks
March Week 21
Week 22
Two Weeks
Circular water tanks
Rectangular water tanks
Holiday
April Week 23
Week 24
Week 25
Week 26
Rectangular water tanks, Continue Quiz 5
Design of R.C. Bridges
Example 1
Example 2 Quiz 6
May Week 27
Week 28
Week 29
Second semester exam
Pre-stress Concrete, Introduction. Types of Pre-stress and Loading stages,
Losses, Design of Pre-stress Beams for bending.
-
References:
1. ACI 318M-11 Building code requirements for structural concrete Farmington Hills, 2011.
2. Arthur H. Nilson, David Darwin and Charles W. Dolan Design of Concrete Structure 13th
edition, 2004.
3. AASHTO Specifications , Standard Specifications for Highway Bridges 2005
4. W.F.Chen Hand book of Structural Engineering New York, 1997.
5. S. Unnikrikrishna, Pillai and Devdas Menon Reinforced Concrete Design New Delhi, 2004.
6. Other related to the Topics and published in 21st century the core materials of the course consists
of the above references ( articles from internet and notes, students should read all the material and
prepare well before going to the exams).
Overview
General Introduction and objectives
Design means presentation of the clients demand in an engineering manner, ready to be
executed by specialized company according to specifications and engineering laws. Design must
sure that an acceptable probability is achieved that the structure does not fail during the specified
life. Reinforced concrete structures must be capable of carrying any combination of loads that can
reasonably be expected to be applied during its intended life; it must be designed for the
combination that will produce the worst effects. Structures and structural members shall be
designed to have design strengths at all sections at least equal to the required strength calculated for
the factored loads and forces in such combinations. The course will give students understanding of
structural design of different subjects such as Torsion of Beams, Two-Way slabs, Shear walls,
Multi-Storey Building, R.C. Tanks, R.C. Bridges, Prestressed concrete. The followings are brief
description for each subject.
Torsion of R.C. Beams
In the ACI Code 2008 ( 11.6), the design for torsion is based on a thin walled tube space truss
analogy. A beam subjected to torsion is idealized as a thin - walled tube with the core concrete in a
solid beam cross-section neglected. Once a reinforced concrete beam has cracked in torsion its
torsional resistance is provided primarily by closed stirrups and longitudinal bars located near the
surface of the member. In the thin- walled tube analogy the resistance is assumed to be provided by the
outer of the cross-section roughly centered on the closed stirrups. Both hollow and solid sections are
idealized as thin walled tubes both before and after cracking.
-
Example : Check the cantilver beam shown below for torsion only.The uniform load is 20 KN/m and
concentrated load ( P= 20KN ) act (applied ) at 250mm distance from the centr of the cross-section.
Use f`c = 21MPa, fyl = 414 MPa for( 28mm ) for main rinforement for bending and fyv = 276 MPa
for closed stirrups ( 10mm ) and fyl= 414 MPa for longitudinal reinforcement of torsion ( 12mm ) if
required .
Solution:
Acp = 300 * 500 = 15 * 104mm
2
Pcp = 600 + 100 = 1600 mm
Tu = P * 0.25 = 20 * 0.25 = 5 Kn.m
Check equation 4 , 5 KN.m >/12 f c Acp2 /pcp =4 kN.m The effect of torsion considered
Analysis and design of two-way slabs
Common type of floor is the slab-beam girder construction, when length of the slab is two or
more times its width, almost all of the floor load goes to the beams, and very little, except some
near the edge of the girders, goes directly to the girders, thus slab may be designed as one way slab
with the main reinforcement parallel to the girder and the shrinkage and temperature reinforcement
parallel to the beams. The deflected surface of a one-way slab is primarily one of single curvature.
When the ratio of long-span L1, to the short span L2 is less than two, the slab is designed as
two-way slab with main reinforcement in both directions, the floor load is carried in both directions
to the four supporting beams around the panel and the deflected surface of the shaded area becomes
of double curvature. Both the flat-slab and flat-plate floors are characterized by the absence of the
P e =250 mm
h
b
P=20 KN
l =3.9 m
-
beams along the interior column lines, but edge beams may or may not be used at the exterior edges
of the floor. Flat-slab floors differ from flat-plate floors in that flat-slab floor provide adequate
shear strength by having either one or both of drop panels and column capitals In flat-plate floors a
uniform slab thickness is used and the shear strength is obtained by the embedment of multiple-U-
stirrups, flat slabs are more suitable for larger panel size or heavier loading than flat plates.
Methods of Analysis and Design of Two Way Slabs
First: Direct Design Method (DDM)
Design of slab system within the following limitations by the direct design method shall be
permitted:
1. Shall be a minimum of three continuous spans in each direction.
2. Panels shall be rectangular with a ratio of longer to shorter span c/c of supports within a panel
not greater than two.
3. Successive span length c/c of supports in each direction shall not differ by more than one-third
the long span.
4. Offset of columns by a maximum of 10% of the span (in direction of offset) from either axis
between center lines of successive columns shall be permitted.
5. All loads shall be due to gravity only and uniformly distributed of an entire panel. Live load
shall not exceed two times dead load.
6. For a panel with beams between supports on all sides, the relative stiffness of beams in two
perpendicular directions.
Design Example by DDM
Example 1: two-way slabs with beams
A two-way slab floor system shown below. It is divided into 25 panels with panel size of 7.62 *
6.1m. Concrete compressive strength, fc`=20.7MPa and steel yield strength, fy=272MPa. Service live
load is to be taken 5.09kN/m2, storey height is 3.65m. The preliminary sizes are as follows:
Slab thickness is 165mm, Long beams are 350*700mm overall, short beams are 300*600mm overall,
upper and tower columns are 375*375mm. The four kinds of panels (corner, long-sided edge, short
sided edge, and Interior) are numbered 1,2, 3 and 4.
Calculate:
1. Total factored static moment in a loaded span in frame A.
2. The negative and positive factored moments in frame A.
-
Solution:
Limitations of the DDM
1. more than 3 panels in each direction, ok satisfy
2. ysatisok21.256.1
7.62l
l
2
1 f
3. spans have the same length, ok satisfy
4. 0% ok satisfy
5. wl = 5.09 kN /m2
wd = wt of slab + wt of tile and mortar + wt of plastering or false ceiling
= 0.165 x 24 + 0.07 x 23 + 0.27 = 5.84 kN /m2
wl / wd = 5.09 / 5.84 = 0.87 < 2 ok satisfy
6. calculate 1 and 2
1 2
4 3
3 4
2 1
A
5 a
t 6.1
m =
30.5
m
5 at 7.62m = 38.1m
-
All limitations are satisfy and DDM can be apply.
2. Total factored static moment (Mo)
B1
B1
B3
B2
B2
B4
B8
B7
B6 B6
B5 B5 1
3 4
2
1
3 4
2
=7-77 =7-77
=7-77 =7-77 =
3.3
36
=
3.3
36
=
3.3
36
=
3.3
36
=13 =13
=
5.6
=5.6
7.62m
h=
700m
m
6.1
m 350
bE= 1420mm
-
Ext. eq. Frame 12. lorslo
Interior equivalent frame
Middle strip
.m606.367.2456.115.158
1MA,Frame
m15.151.w1.W
llW8
1M
2
o
2ldu
2
N2uo
kN
kNw
62
3. Negative and positive moments in frames are computed using case 2 for the exterior span as shown
below.
All the moments calculated (Negative and positive) and shown in the figures below.
Second: Equivalent frame method (EFM)
If the floor systems dont satisfy loading and geometric conditions required by the direct design method. The ACI code permits a building to be analyzed by subdividing the structure in to equivalent
frames, which are then analyzed elastically.
Design Procedure
1. Divide the structure to longitudinal and transverse frames centered on column centerline and bounded by panel center lines.
2. Each frame shall consist of a raw of columns and slab-beam strips bounded laterally by center line of panels.
3. Frames adjacent and parallel to an edge shall be bounded by that edge and the CL of adjacent panel.
0.16Mo
0.57Mo
0.7Mo 0.65Mo
0.35Mo
0.65Mo
97
345.6
424.5
212.2
394.1 394.1
212.2
Frame A
-
Col.
Col. Column a. b.
4. Columns shall be assumed to be attached to slab-beam strips by torsional members transverse to the direction of span for which moments being determined.
5. The slab-beam may be assumed to be fixed at any support to panels distance from the support of the span where critical moments are being obtained. Provide the slab is continues beyond that
edge.
6. Moment of inertia of slab-beam strip between the center of the column to face of the column or capital is to be assumed equal to that of slab-beam @ the face of the column or capital divided
by the quantity 2
2
2
lC
1
C2: size of rectangular or equivalent rectangular or column capital measured transverse to the
direction of the span for which moments are being determined.
7. Torsional members
A. Slabs without beams: A portion of slab having a width equal to that of the columns, or
capital in the direction of the span for which moments are being determined.
2l
B. Slab supported on beams
8. The stiffness Kt of the torsional member shall be calculated by the following eq.
3
2
22
S
lC
1l
C9EKt C
9. If the panel contains a beam parallel to the direction in which moments are being determined, the
value of Kt may lead to frame stiffness that is too law. In such cases, the value of Kt most be increased
by the ratio of the moment of inertia of the slab-beam to that of the slab a lone (Isb/Is).
10. Equivalent column stiffness Kec
Torsional
member
Torsional
member
Slab-beam strip
Torsional
member
-
Kc1
Kc2
For (Mneg.)max. For (Mpos.)max.
Kt
Kc1
KcKec
stiffness
1yflexibilitKcKcKc
Kt
1
Kc
1
Kec
1
21
or
11. Loads on the equivalent frame (13.7.6)
a. when dW4
3W l (un factored)
All panels will be loaded with lW
b. when dW4
3W l
Third: Yield Line Theory (YLT)
Although the yield line theory not included in the ACI code, slab analysis by yield line theory
may be useful in providing the needed information for understanding the behaviors of irregular or
single panel with various boundary conditions. The fundamentals concept of the yield line theory for
the ultimate load design of slabs has been expanded by K.W. Johansen (1948). In this theory the
strength of a slab is assumed to be governed by flexure alone, other effects such as shear & deflection
are to be separately considered. The steel reinforcement is assumed to be fully yielded along the yield
lines at collapse and the bending and twisting moments are assumed to be uniformly distributed along
the yield lines. Yield line theory for one-way slab is not much different from the limit analysis of
continues beams. Yield line theory for two-way slabs requires a different treatment from limit analysis
of continue beams, because in this case the yield lines will not in general be parallel to each other. The
entire slab area will be divided into several segments which can rotate along the yield lines as rigid
bodies at condition of collapse.
Example7: for two-way slab shown, subjected to a concentrated load (P).calculate the ultimate
moment (m) per unit length based on P.
Solution: Each segment carries P/4 @ center
-
P
8p
m
4p
2m
2m
2
1m
2m
2
pmMo
4
pWd
Or for all segments
8p
m
8m42mMo
pW
Design of multi storey buildings
A building frame may contain a number of bays, and may have several stories. A multi-storey,
multi-panel frame is a complicated statically indeterminate structure. It consists of a number of beams
and columns built monolithically, forming a net work. The floors and the walls are supported on beams
which transmit the loads to the columns. A building frame is subjected to both the vertical and as well
as horizontal loads. The vertical loads consist of the dead weight of structural components such as
beams, slabs, columns etc, and live load. The horizontal loads consist of the wind forces and the earth
quake forces. The ability of multi-storey-buildings to resist wind and other lateral forces depends upon
the rigidity of connecting between the beams and columns. When the connections of beams acting on
the structure.
In ordinary reinforced concrete skeleton buildings, a continuous beam is rigidly connected with
columns. Due to this, the moments in the beam depend not only upon the number and length of spans
composing the beam itself, but also upon the rigidity of columns with which it is connected. The
bending moment at the end of any one span of the continuous beam can not be transferred to the beam
in the next span without subjecting the columns to bending. Instead of transmitting the bending
moment in full of the beam in the next span, part of the moment is transferred to the columns above
span upon the other spans is much lower the beam. Due to this, the effect of loading on one span upon
the other spans is much lower than in ordinary continuous beams which are not connected to the
columns.
Shear Walls
Shear walls are deep relatively thin vertically cantilever reinforced concrete beams. They are
commonly used in structures to resist the effects of gravity loads and storey shears due to wind or earth
quick forces. In the multi-story buildings the loads can be resisted by Rigid frames (R.C. frames) or
Shear walls. Shear walls act as a cantilever fixed at their basis with the foundation to carry loads down
to the foundation, they are subjected to shear as well as bending and vertical compression due to
gravity of loading from the structure (weight of the shear wall and any additional weight), the shear
walls must be reinforced vertically as well as horizontally as shown.
Example:
Calculate the bending moment and shear force for the shear wall shown, the building consists of
18 floors, height of each floor is 3m, the wind load uniformly distributed along (over) all building
height and it value equal to 1.5 kN/m2, use fc`= 20.7 MPa and fy= 414 MPa. Each shear wall carry 21.9
kN/ story as vertical load (8.74 live load & 13.16 dead load) in addition to its own weight.
l
l
-
Solution:
54mh,0mm60,mm300h WW 0
Load calculation :
1. wind load - load per wall = m
KN2
4.534.51. 95
- Point load on the roof joint. 21.57KN2
31.
2P 96
- Point load on all other joints (p) =1.693= 43.15 kN
2. Vertical load
The weight of the shear wall = kN2333245460.3 The weight of the shear wall / story =233/18=129.6 kN Total factored d.l./story = 1.2(129.6+13.16)=171.4 kN Factored l.l./story = 1.68.74 = 14.00 kN Total factored load / storey = 171.4+14 =185.4 kN Total factored load (comp.) = 185.4 x 18 = 3337 kN
3*18=54m
4.5m3m4.5m
7.5m
6m
7.5m
hw=54m
lw=6m
h=30cm
win
d lo
ad
3m
She
ar w
all
She
ar w
all
-
Distribution of the wind load and its shear & bending moment diagram shown below. Use P=43.15kN,
h1=3m (heigh of each storey).
p/2
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p/2
18P
17.5P
16.5P
15.5P
14.5P
13.5P
12.5P
11.5P
10.5P
9.5P
8.5P
7.5P
6.5P
5.5P
4.5P
3.5P
2.5P
1.5P
0.5P
162Ph1
144.5Ph1
128Ph1
112.5Ph1
98Ph1
84.5Ph1
72Ph1
60.5Ph1
50Ph1
40.5Ph1
32Ph1
24.5Ph1
18Ph1
12.5Ph1
8Ph1
4.5Ph1
2Ph1
0.5Ph1
0
Bending moment diagram Shear force diagram
Reinforced Concrete Water Tanks
A water tank is used to store water to tied over the daily requirements in general, water tanks can be classified under three heads: (i) tanks resting on ground (ii) elevated tanks supported
on staging and (iii) under ground tanks. From the shape point of view, water tanks may be of several
types, such as (i) circular tanks (ii) rectangular tanks (iii) spherical tanks (iv) intze tanks and (v)
circular tanks with conical bottoms.
In the construction of concrete structures for the storage of water and other liquids, the
imperviousness of concrete is most essential. The permeability of any uniform and thoroughly
compacted concrete of given mix proportions is mainly dependent on the water cement ratio. The increase in water cement ratio results in increase in the permeability.
Design of liquid retaining structure has to be based on the avoidance of cracking in the concrete
having regard to its tensile strength. It has to be insured in its design that concrete does not crack on its
water face. Cracking may also result from the restrained to shrinkage free expansion and contraction of
concrete due to temperature and shrinkage and swelling due to moisture effects, correct placing of
reinforcement , use of small sized bars and use of deformed bars lead to a diffuse distribution of cracks.
The risk of cracking due to overall temperature and shrinkage effects may be minimized by limiting the
changes in moisture content and temperature to which the structure as a whole is subjected. Cracks can
be prevented by avoiding the use of thick timber shuttering which prevent the ease escape of heat of
hydration from the concrete mass. The risk of cracking can also be minimized by reducing the
-
restraints on the free expansion or contraction of the structure. For long walls or slabs founded at or
below the ground level, restraints can be minimized by founding the structure on a flat layer of concrete
with interposition of sliding layer of some material to break the bond and facilitate movement.
Whoever, it should be recognized that common and more serious cases of leakage in practice, other
than cracking, are defects such as segregation and honey combing and in particular all joints are
potential source of leakage
Example:
Design the wall of a circular water tank Restrained at base for the following design data:
h = 4.5 m
D = 9.0 m
1c
f0.56ftc
0.4ffc,20.7Mpac
f `
`
fy = 350 Mpa for all types of reinforcement , fs= 0.6fy
qall = 94 KN / m2
Solution:
Assume tw = 150mm
0.5
94.5
Dh
300.15
4.5tw
h
From the table get
K1 = 0.32
F = 0.009
h1 = K1 h = 0.32 4.5 = 1.44 m
m
KN137.70.32194.5102
1k1whD
2
1T 1max
Main reinforcement, m
mm655.7
mm
N3500.6
m
N10137.7
fs
TAs
2
2
3
max
Use 12mm @ 15cm c/c, (As = 753 > 655 )
Check thickness of the wall
15048mm7538.352.54
10137.7
10
1As1n
ft
T
10
1t
9.3520.74700
10200Ec
Esn
3
33W
3
assume 0.006F0.27,K40mmt
h11
W
,704As,m
KN147.8Tmax use 12mm @15cm c/c .
New tw = 52mm
Use tw = 113mm (to keep 40t
hW
) .
2818fc,82mm2
1225113d
mKN.m5.474.5100.006Fwh@baseM
33
max
-
mmm349
8291210
105.47
djfs
MAs
0.913
k1j
0.27
9.352108.28
8.28k
26
Use 12mm @ 30cm c/c , (As = 377 > 349 )
Negative binding moment @ point of m
KN.m1.8233
MS maxmax
mmm116
820.91210101.823As
26
Use 10mm @ 50cm , c/c , (As = 157 > 116 )
Secondary reinforcement m
mm2831131000100
0.25 2
Use 10mm @ 25 cm c/c , (As = 312 > 283 )
Reinforced concrete Bridges
The design of reinforced concrete bridges is based on the AASHTO specifications (American
Association of State Highway and Transportation Officials). Reinforced concrete in increasingly used
for highway and railway bridge construction due to its durability , rigidity , economy , ease of
construction and ease with which pleasing appearance can be made in it. Reinforced concrete bridges
may be of following types :
1- Solid Slab Bridge or Deck Bridge. 2- Deck girder bridge (T-beam Bridge). 3- Balanced cantilever bridge. 4- Rigid frame culvert or bridge (single span as well as multi-span). 5- Arch bridge. 6- Bowstring Girder Bridge. 7- Continuous girder or arch bridge.
A deck slab bridge or solid slab bridge is the simplest type of construction, used mostly
for culverts or small bridges with a span not existing 8.0 m. Though the thickness of deck slab
is considerable, its construction is much simpler and the cost of form work is also minimum.
Deck Girder Bridge or T-Beam bridge is another type of a simple R.C. bridge used for spans
between 10 to 20 meters. The monolithic with girders, so that T-beam effect is achieved.
Example: Design the slab of the bridge for the following given data and according to AASHTO specifications:
- Clear span 4.5m - Clear width 8m
- Live loading 4420Hs (Assume W=324KN weight of truck &load)
- Wearing surface 1.4 KN/m2 - fc=21MPa , fc=0.4fc - fy=276Mpa, fs=0.5fy - n=10
-
Solution: Asmin // Traffic
Assume thickness of the slab =32cm
S=4.5+0.45=4.95 > 4.5+0.32=4.82m
Use S=4.82m
- Wd.L.= 1.4+0.3224=9.08KN/m2
mKN.m26.374.829.08
8
1WS
8
1M
22
d.L.
For 4420Hs & W=324KN,
Load on each area , P=72KN
E=1.22+0.6S = 1.22+0.64.82=1.51m < 2.13m
The load on the unit width of the slab = m
KN47.71.51
72
ML.L. = 13.12S= 13.12 4.82 = 63.2 KN.m/m
Or m
KN.m57.54
4.8242.7
4
SPML.L.
.
Impact coff. , 30%0.356384.82
15.24
38L
15.24I
Use I=30%
mKN.m108.5318.9663.226.33M
mKN.m18.9663.20.3M
total
I
1000mmb
0.8743
k1j
0.378k
10n
138Mpa0.5FyFs
mmN8.40.4FFc,,
nFsFc
FcK 2
1
c
280mm0.8740.37810008.4
10108.532
FbF
Mt2d
6
jc
Assume using 25mm bar diam. And 25mm clear cover.
assumedh317.5mm2
2525280h
Use h=320mm
Dprovide =320-25-25/2=282.5mm
Main reinforcement required, m
mm31850.874282.5138
10108.53
jdfs
MAs
26
Use 25mm @ 15cm,
mmm3400
15
100510As
2
-
Transverse reinforcement, m
mm796or25%As31854.82
55As
5
55%
2
16mm@25cm c/c, Asv = 804mm2/m < 50%
Prestressed concrete
Prestress: means a stress that acts even though no dead or live load is acting. Prestress involves
the imposition of stresses opposite in sign to those which are caused by the subsequent application of
service loads. Concrete produce an axial compression as well as negative bending moment. Thus it is
possible to keep the entire section in compression when service loads are added. This is a great
advantage since concrete is weak in tension.
The general concepts of prestressed concrete were first formulated in 1885-1890 by Dochring in
Germany & Jakson in USA with low tensile stress.
Mandle 1896 in Germany produced a theory of prestressed concrete. Koenen 907 first recognition of
losses in pre stress force. Dill 928 in US produced prestressed planks and fence posts.
Circular prestressing of tanks began about 1935, but no significant liner prestressing (beams, slabs) was done untile 1950. The walnut lane bridge built in 949-950, was the first major of liner prestressing
in the United States. T.Y. lin has been a leading proponent and practitioner.
Example:
Determine the nominal moment strength Mn of the pretension bonded section shown below. The
concrete has 35Mpaf 1c and the stress relieved pre stressing strand fpu = 1750 Mpa. Assume 20% pre
stress losses and an average stress-section relationship for the steel as given in Fig. below
Solution:
For fully pre stressed member,
1
c
P
1 f
fpuP
p1fpufpu
As=1438
750
500
125
-
1551Mpa35
17500.0046
0.81
0.4011750fps
0.8130350.0080.85
0.40p0.850.891750
1560fpu
fpy
0.0046625500
1438
bd
ApsP
1
p
Check 0.290.360.20435
15510.0046
f
fpspWp 11
c
P
1227KN.m1206KN.m0.9
5051.71621.4
0.9
1.7M1.4Md
MuMnRequired
1227KN.m2
1506251023303382
adTuMn
185.2mm0.81
150
aC
150mma
TuCu
2230338N15511438fpsApsTu
14875aa500350.85ab0.85fCu
L.L
6
1
1
c