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_ Omar_Qarani@yahoo.com

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