Reinforced Concrete Design I

8/20/2019 Reinforced Concrete Design I

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Reinforced Concrete D

Dr. Nader Okasha


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Reinforced Concrete Design

nstructor r. a er as a.

Email [email protected]

ce ours s nee e .


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s course s on y o ere or stu enhave passed strength of materials.

you on t meet t s cr ter a you w not

allowed to continue this course.


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References:

Building Code Requirements for Reinforced Concrete

commentary (ACI 318M-08). American Concr

.

Design of Reinforced Concrete . 7th edition, McC

an e son, . ., .


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Additional references (internationally recogreinforced concrete design):

Reinforced Concrete, A fundamental Approach . Ed

Design of Concrete Structure. Nilson A. et al.

. .


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Desi n is an anal sis of trial sections. Th

of each trial section is compared with the

load effect.

The load effect on a section is determinestructural analysis and mechanics of mat

The strength of a reinforced concrete secdetermined usin the conce ts tau ht in t


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Course outline

ee op c

Introduction:

1- .-Introduction to reinforced concrete.

-Load types, load paths and tributary areas.

-Design philosophies and design codes.

Analysis and design of beams for bending:-Analysis of beams in bending at service loa

-Strength analysis of beams according to AC

-, ,

-Design of T and L beams.Design of doubly reinforced beams


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Course outline

Week Topic

– 6

ribbed slabs.

.

7,8 Bond, development length, splicing and b

8,9 Design of isolated footings.

9 Staircase design.


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Course work: 20%

-Homework 4%-Attendance 4%

-Project 12% .


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Mid-term exam:

Only one A4 cheat-sheet is allowed. Necessary figures and tables will be provided with

Final exam:


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Show all our assum tions and work details. Pr

sketches showing the reinforcement and dimens

Markin will consider rimaril neatness of re

completeness and accuracy of results.

You may get the HW points if you copy the solu

other students. However, you will have lost you

practicing the concepts through doing the HW. Tyou to loosing points in the exams which you c


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-


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.

.

answering the cell-phone) unless a previous permt d


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A collectivel missed class will be made u eith

Thursday or during the discussion lecture.

An absence from a lecture will loose you attend

and the lecture will not be repeated for you. You

own. You may use the lecture videos.


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,

, , ,

Area (Ac ,A g ,As ): mm2

3

Force (P,V,N ): N

.

Stress (f y , f c ’): N/mm2 = MPa = 106 N/m2


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,

, , , ,

Area (Ac ,A g ,As ): cm2, m2

3 3 ,

Force (P,V,N ): kN

.

Pressure (q s ): kN/m2


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slides by their section or equation number in the code pro

shading.

Examples:

cc f E = 4700

ACI Eq. cr f f

′=62.0


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.

- .

.


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Reinforced Concrete Design I

Lecture 1

Introduction to reinforced concrete

Dr. Nader Okasha


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Contents

1. Concrete-producing materials

2. Mechanical properties of concrete

3. Steel reinforcement

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Part 1:

Concrete-ProducingMaterials

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4

1. It has considerable compressive strength.

2. It has great resistance to the actions of fire and water.

3. Reinforced concrete structures are very rigid.

4. It is a low maintenance material.

5. It has very long service life.

Advantages of reinforced concrete

as a structural material


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5

6. It is usually the only economical material for footings,

basement walls, etc.

7. It can be cast into many shapes.

8. It can be made from inexpensive local materials.

9. A lower grade of skilled labor is required for erecting.

Advantages of reinforced concrete

as a structural material


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6

1. It has a very low tensile strength.

2. Forms are required to hold the concrete in place until it

hardens.

3. Concrete members are very large and heavy because of the

low strength per unit weight of concrete.

4. Properties of concrete vary due to variations in

proportioning and mixing.

Disadvantages of reinforced

concrete as a structural material


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Concrete

Concrete is a mixture of cement, fine and coarseaggregates, and water. This mixture creates a formable

paste that hardens into a rocklike mass.

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Concrete Producing Materials

• Portland Cement

• Aggregates

• Water

• Admixtures

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Portland Cement

10

The most common type of hydraulic cement used in themanufacture of concrete is known as Portland cement, which is

available in various types.

Although there are several types of ordinary Portland cements,

most concrete for buildings is made from Type I ordinarycement.

Concrete made with normal Portland cement require about two

weeks to achieve a sufficient strength to permit the removal offorms and the application of moderate loads.


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Type I: General Purpose

Type II: Lower heat of hydration than

Type I

Type III: High Early Strength

• Quicker strength

• Higher heat of hydration

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Types of Cement


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Type IV: Low Heat of Hydration

• Slowly dissipates heat less distortion (used for

large structures).

Type V: Sulfate Resisting

• For footings, basements, sewers, etc. exposed to

soils with sulfates.

Types of Cement

12

If the desired type of cement is not available, different

admixtures may be used to modify the properties of Type 1

cement and produce the desired effect.


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Aggregates

Coarse Aggregates

Fine Aggregates

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Aggregates are particles that form about three-fourths of thevolume of finished concrete. According to their particle size,

aggregates are classified as fine or coarse.

Coarse aggregates consist of gravel or crushed rock particles

not less than 5 mm in size.

Fine aggregates consist of sand or pulverized rock particles

usually less than 5 mm in size.


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Water

14

Mixing water should be clean and free of organic materials thatreact with the cement or the reinforcing bars.

The quantity of water relative to that of the cement, called

water-cement ratio, is the most important item in determining

concrete strength.

An increase in this ratio leads to a reduction in the compressive

strength of concrete.

It is important that concrete has adequate workability to assure

its consolidation in the forms without excessive voids.


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– Applications:

• Improve workability (superplasticizers)

• Accelerate or retard setting and hardening

• Aid in curing

• Improve durability

Admixtures

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Concrete Mixing

• Quality

• Workability

• Economy

In the design of concrete mixes, three principal

requirements for concrete are of importance:

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Part 2:

Mechanical Properties ofConcrete

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Compressive Strength,

• Normally, 28-day strength is used as the design

strength.

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c

Mechanical Concrete Properties


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• It is determined through testing standard cylinders 15

cm in diameter and 30 cm in height in uniaxialcompression at 28 days (ASTM C470).

• Test cubes 10 cm × 10 cm × 10 cm are also tested in

uniaxial compression at 28 days (BS 1881).

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c



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• The ACI Code is based on the concrete compressive

strength as measured by a standard test cylinder.

• For ordinary applications, concrete compressive

strengths from 20 MPa to 30 MPa are usually used.

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c


Cylinder 0 8 Cubec c. f


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Mechanical Concrete PropertiesCompressive-Strength Test


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22

Modulus of Elasticity, Ec• Corresponds to the secant modulus at 0.45

• For normal-weight concrete:

'

c

0.0030.002

cc f E 4700 ACI 8.5.1



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Tensile Strength – Tensile strength ~ 8% to 15% of

– Tensile strength of concrete is quite difficult to measure

with direct axial tension loads because of problems ofgripping the specimen and due to the secondary stresses

developing at the ends of the specimens.

– Instead, two indirect tests are used to measure the tensilestrength of concrete. These are given in the next two slides.

'

c



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24

Tensile Strength – Modulus of Rupture, f

r

– Modulus of Rupture Test (or flexural test):

2

6

bh

M

I

Mc f r

ACI Eq. 9-10

P

f r

unreinforced

concrete beam

cr f f 62.0



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Tensile Strength – Splitting Tensile Strength, f ct

– Split Cylinder Test

PConcrete Cylinder

Poisson’s

Effect

2ct

P

Ld


25

cct f f 56.0 ACI R8.6.1


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Creep

• Creep is defined as the long-term deformation caused

by the application of loads for long periods of time,

usually years.

• Creep strain occurs due to sustaining the same load

over time.

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Creep

The total deformation is divided into two parts; the first

is called elastic deformation occurring right after the

application of loads, and the second which is time

dependent, is called creep

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Shrinkage

Shrinkage of concrete is defined as the reduction in

volume of concrete due to loss of moisture. As a

result, shrinkage cracks develop.

Shrinkage continues for many years, but under ordinaryconditions about 90% of it occurs during the first

year.

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Part 3:

Steel Reinforcement

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Steel Reinforcement

Tensile tests

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Steel Reinforcement

Tensile tests

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Steel Reinforcement

Stress-strain diagrams

32

Yield point

elastic plastic

All steel grades have same modulus of elasticity E s = 2x105 MPa

= 200 GPa

f s = ε Es ≤ f y


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Steel Reinforcement

Bar sizes, , #

33

Bars are available in nominal diameters ranging from 5mm

to 50mm, and may be plain or deformed. When bars have

smooth surfaces, they are called plain, and when they have

projections on their surfaces, they are called deformed.

Steel grades, f y

MPaksi

27640

41460

55280


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Steel Reinforcement

Bars are deformed to increase bonding with concrete

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Steel Reinforcement

Marks for ASTM Standard bars

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Steel Reinforcement

Bar sizes according to ASTM StandardsU.S. customary units

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Steel Reinforcement

Bar sizes according to European Standard (EN 10080)

38

W

N/m

Number of bars

mm 1 2 3 4 5 6 7 8 9 10

6 2.2 28 57 85 113 141 170 198 226 254 283

8 3.9 50 101 151 201 251 302 352 402 452 503

10 6.2 79 157 236 314 393 471 550 628 707 78512 8.9 113 226 339 452 565 679 792 905 1018 1131

14 12.1 154 308 462 616 770 924 1078 1232 1385 1539

16 15.8 201 402 603 804 1005 1206 1407 1608 1810 2011

18 19.9 254 509 763 1018 1272 1527 1781 2036 2290 2545

20 24.7 314 628 942 1257 1571 1885 2199 2513 2827 3142

22 29.8 380 760 1140 1521 1901 2281 2661 3041 3421 380124 35.5 452 905 1357 1810 2262 2714 3167 3619 4072 4524

25 38.5 491 982 1473 1963 2454 2945 3436 3927 4418 4909

26 41.7 531 1062 1593 2124 2655 3186 3717 4247 4778 5309

28 45.4 616 1232 1847 2463 3079 3695 4310 4926 5542 6158

30 55.4 707 1414 2121 2827 3534 4241 4948 5655 6362 7069

32 63.1 804 1608 2413 3217 4021 4825 5630 6434 7238 8042

Areas

are in

mm2



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Lecture 2

Load types, load paths and tributary areas

Dr. Nader Okasha


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Load paths

Structural systems transfer gravity loads from the floorsand roof to the ground through load paths that need to

be clearly identified in the design process.

Identifying the correct path is important for determining

the load carried by each structural member.

The tributary area concept is used to determine the load

that each structural component is subjected to.

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Metal Deck/Slab System

Supports Floor Loads Above

Girders Support Joists

Columns Support Girders

Joists Support Floor Deck


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The area tributary to a

joist equals the length of

the joist times the sum of

half the distance to eachadjacent joist.


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The area tributary to a girder

equals the length of the

girder times the sum of halfthe distance to each adjacent

girder.


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Load paths loads on structural members

Load is distributed over the area of the floor. This distributed load

has units of (force/area), e.g. kN/m2.

6

Column

Beam

Loads

Footing

Soil

Slab

Slab

Beam Beam

Beam Beam

Beam Beam

Column

q {kN/m2}

w {kN/m}

P {kN}


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Load paths loads on (one-way) beams

In order to design a beam, the tributary load from the floor carried

by the beam and distributed over its span is determined. This load

has units of (force/distance), e.g. kN/m.

Notes:

-In some cases, there may be concentrated loads carried by the beams as well.

-All spans of the beam must be considered together (as a continuous beam) for design.

7

w {kN/m}


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Load paths loads on (one-way) beams

This tributary load is determined by multiplying q by the tributary

width for the beam.

8 S 2 S 1

L

w {kN/m} = q {kN/m2

}

(S 1+S 2)/2 {m}


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Load paths loads on (two-way) beams

The tributary areas for a beam in a two way system are areas which

are bounded by 45-degree lines drawn from the corners of the

panels and the centerlines of the adjacent panels parallel to the long

sides.

A panel is part of the slab formed by column centerlines.

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For edge beams:D=S /2

For interior beams:

D=S

An edge beam is bounded by panels from one side.

An interior beam is

bounded by panels fromtwo sides.

qD

qD


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Load paths loads on columns

The tributary load for the column is concentrated. It has units of

(force) e.g., kN. It is determined by multiplying q by the tributary

area for the column.

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P {kN} = q {kN/m2} (x y){m2}


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ExampleDetermine the loads acting on beams B1 and B2 and columns C1

and C2. Distributed load over the slab is q = 10 kN/m2. This is a 5

story structure.

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B1

B2

C1C2

4 m

5 m

4.5 m

6 m 5.5 m


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Example

B1:

15

B1

B2

C1C2

4 m

5 m

4.5 m

6 m 5.5 m

w = 10 (4)/2 = 20 kN/m

L d h


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Example

B2:

16

B1

B2

C1C2

4 m

5 m

4.5 m

6 m 5.5 m

w = 10 (4+5)/2 = 45 kN/m

L d h


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Example

B1:

B2:

17

w = 20 kN/m

w = 45 kN/m

L d th


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Example

C1:

18

B1

B2

C1C2

4 m

5 m

4.5 m

6 m 5.5 m

P = 10 (4.5/2 6/2) 5 = 337.5 kN

L d th


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Example

C2:

19

B1

B2

C1C2

4 m

5 m

4.5 m

6 m 5.5 m

P = 10 [(4.5+5)/2 (6+5.5)/2] 5 = 1366 kN


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Load types

Classification by direction

1- Gravity loads

2- Lateral loads

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Load types

Classification by source and activity

1- Dead loads

2- Live loads

3- Environmental loads

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Loads on Structures

All structural elements must be designed for all loads anticipated toact during the life span of such elements. These loads should not

cause the structural elements to fail or deflect excessively under

working conditions.

Dead load (D.L)• Weight of all permanent construction

• Constant magnitude and fixed location

Examples: * Weight of the Structure(Walls, Floors, Roofs, Ceilings, Stairways, Partitions)

* Fixed Service Equipment

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Minimum live Load values on slabs

Type of Use Uniform Live Load

kN/m2 Live Loads (L.L)


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kN/m

Residential

Residential balconies

2

3

Computer use 5

Offices 2

Warehouses

Light storage

Heavy Storage

6

12

Schools

Classrooms 2

Libraries

Rooms

Stack rooms

3

6

Hospitals 2

Assembly Halls

Fixed seating

Movable seating

2.5

5

Garages (cars) 2.5Stores

Retail

Wholesale

4

5

Exit facilities 5

Manufacturing

Light

Heavy

4

6

( )

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The live load is a moving ormovable type of load such

as occupants, furniture, etc.Live loads used in designing

buildings are usuallyspecified by local buildingcodes. Live loads depend on

the intended use of thestructure and the number of

occupants at a particulartime.

See IBC 2009 TABLE

1607.1 for more live loads.http://publicecodes.citation.com/icod/ibc/2009/index.htm?bu=IC-P-2009-000001&bu2=IC-P-2009-

000019


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Environmental loads

Wind load (W.L)The wind load is a lateral load produced by wind pressure and

gusts. It is a type of dynamic load that is considered static to

simplify analysis. The magnitude of this force depends on the

shape of the building, its height, the velocity of the wind and thetype of terrain in which the building exists.

Earthquake load (E.L) or seismic load

The earthquake load is a lateral load caused by ground motions

resulting from earthquakes. The magnitude of such a load depends

on the mass of the structure and the acceleration caused by the

earthquake.

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e o ced Co c e e es g

Dr. Nader Okasha

Lecture 3

Design philosophies and design codes


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Structural Design Requirements:

The design of a structure must satisfy three basic requirements:1)Strength to resist safely the stresses induced by the loads in the

various structural members.

2)Serviceability to ensure satisfactory performance under service

load conditions, which implies providing adequate stiffness tocontain deflections, crack widths and vibrations within acceptable

limits.

3)Stability to prevent overturning, sliding or buckling of the

structure, or part of it under the action of loads.

There are two other considerations that a sensible designer should

keep in mind: Economy and aesthetics.

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Building Codes, Standards, and Specifications:

4

Standards and Specifications: Detailed statement of

procedures for design (i.e., AISC Structural Steel Spec;

ACI 318 Standards, ANSI/ASCE7-05). Not legally

binding. Think of as Recommended Practice.

Code: Systematically arranged and comprehensive

collection of laws and regulations


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5

Model Codes: Consensus documents that can be adopted

by government agencies as legal documents.

3 Model Codes in the U.S.

1. Uniform Building Code (UBC): published by InternationalConference of Building Officials (ICBO).

2. BOCA National Building Code (NBC): published by Building

Officials and Code Administrators International (BOCA).

3. Standard Building Code (SBC): published by Southern BuildingCode Congress International (SBCCI).


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3 Model Codes in the U.S.


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Building Code: covers all aspects related to structural safety -loads, structural design using various kinds of materials (e.g., structural

steel, reinforced concrete, timber), architectural details, fire protection,

plumbing, HVAC. Is a legal document. Purpose of building codes: to

establish minimum acceptable requirements considered necessary for

preserving public health, safety, and welfare in the built environment.

International Building Code (IBC): published by InternationalCode Council (2000 ,1st edition). To replace the 3 model codes for

national and international use.


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8

Summary:

The standards that will be used extensively throughoutthis course is Building Code Requirements for ReinforcedConcrete and commentary, known as the ACI 318M-08 code.

The building code that will be used for this course isthe IBC 2009, in conjunction with the ANSI/ASCE7-02.


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Design Methods (Philosophies)

Two methods of design have long prevalent.

Working Stress Method focuses on conditions at service

loads.

Strength Design Method focusing on conditions at loads

greater than the service loads when failure may be imminent.

The Strength Design Method is deemed conceptually more realistic

to establish structural safety.

The Working-Stress Design Method

This method is based on the condition that the stresses caused by

service loads without load factors are not to exceed the allowable

stresses which are taken as a fraction of the ultimate stresses of the

materials, f c’ for concrete and f y for steel.9


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The Ultimate – Strength Design Method

reduced strength provided factored loads

At the present time, the ultimate-strength design method is themethod adopted by most prestigious design codes.

In this method, elements are designed so that the internal forces

produced by factored loads do not exceed the corresponding

reduced strength capacities.

The factored loads are obtained by multiplying the working loads

(service loads) by factors usually greater than unity.

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Safety Provisions (the strength requirement)

Safety is required to insure that the structure can sustain all expectedloads during its construction stage and its life span with an

appropriate factor of safety.

There are three main reasons why some sort of safety factor are

necessary in structural design• Variability in resistance. *Variability of f c’ and f y, *assumptions are madeduring design and *differences between the as-built dimensions and those found in

structural drawings.

•Variability in loading. Real loads may differ from assumed design loads,or distributed differently.

• Consequences of failure. *Potential loss of life, *cost of clearing thedebris and replacement of the structure and its contents and *cost to society.

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The strength design method, involves a two-way safety measure. Thefirst of which involves using load factors, usually greater than unity

to increase the service loads. The second safety measure specified by

the ACI Code involves a strength reduction factor multiplied by the

nominal strength to obtain design strength. The magnitude of such areduction factor is usually smaller than unity

Design strength ≥ Factored loads

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i i

i

R L ACI 9.3 ACI 9.2

Safety Provisions (the strength requirement)


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Dead only

U = 1.4D

Dead and Live Loads

U = 1.2D+1.6L

Dead, Live, and Wind Loads

U=1.2D+1.0L+1.6W

Dead and Wind Loads

U=1.2D+0.8W or U=0.9D+1.6W

Dead, Live and Earthquake Loads

U=1.2D+1.0L+1.0E

Dead and Earthquake Loads

U=0.9D+1.0E

Load factorsACI 9.2.1

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Load factorsACI 9.2

14

Symbols


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Strength Reduction Factors

According to ACI, strength reduction factors Φ are given as follows:

a- For tension-controlled sections Φ = 0.90b- For compression-controlled sections,

Members with spiral reinforcement Φ = 0.75Other reinforced members Φ = 0.65

c- For shear and torsion Φ = 0.75

Tension-controlled section compression-controlled section

15

ACI 9.3



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Lecture 4

Analysis of beams in bending at service loads

Dr. Nader Okasha


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Introduction

A beam is a structural member used to support the internal momentsand shears and in some cases torsion.

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Basic Assumptions in Beam Theory

4

•The strain in the reinforcement is equal to the strain in the concrete at the same

level, i.e. εs = εc at same level.

• Concrete is assumed to fail in compression, when εc = 0.003.

•Tensile strength of concrete is neglected in flexural strength.

•Perfect bond is assumed between concrete and steel.


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Stages of flexural behavior

If load w varies from zero to until the beam fails, the beam will

go through three stages of behavior:

1. Uncracked concrete stage

2. Concrete cracked – Elastic Stress stage

3. Beam failure – Ultimate Strength stage

w {kN/m}

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Stage I: Uncracked concrete stage

At small loads, when the tensile stresses are less than the

modulus of rupture, the beam behaves like a solid rectangular

beam made completely of concrete.

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Stage II: Concrete cracked – Elastic Stress range

Once the tensile stresses reach the modulus of rupture, thesection cracks. The bending moment at which this

transformation takes place is called the cracking moment M cr .

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Stage III: Beam failure – Ultimate Strength stage

As the stresses in the concrete exceed the linear limit (0.45f c’), the concrete stress distribution over the depth of the beam

varies non-linearly.

0.0030.0028


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Stages of flexural behaviorw {kN/m}

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Flexural properties to be determined:

1- Cracking moment.

2- Elastic stresses due to a given moment.

3- Moments at given (allowable) elastic stresses.

4- Ultimate strength moment (next lecture).

Note: In calculating stresses and moments (Parts 1 and 2),you need to always check the maximum tensile stress with the

modulus of rupture to determine if cracked or uncracked

section analysis is appropriate.

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Cracking moment M cr

When the section is still uncracked, the contribution of the

steel to the strength is negligible because it is a very small

percentage of the gross area of the concrete.

Therefore, the cracking moment can be calculated using theuncracked section properties.

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Cracking moment M cr

Example 1:

Calculate the cracking moment

for the section shown

mkN mm N y

I f M

MPa f f

mm I

bh I

t

g r

cr

cr

g

g

.43.111.101143.1)2/750(

102305.14.3

4.33062.062.0

102305.1)750)(350(12

1

121

810

4103

3

MPa f c 30

1500 mm2750 mm

12

Elastic stresses – Cracked section


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• In the transformed section, the cross sectional area

of the steel, As , is replaced by the equivalent area

nAs where

n = the modular ratio= E s /E c

• To determine the location of the neutral axis,

0

022

2

1

d An x An xb

xd An x

bx

s s

s

• The normal stress in the concrete and steel

c s

t t

My My f n

I I

• After cracking, the steel bars carry the entire

tensile load below the neutral surface. The

upper part of the concrete beam carries thecompressive load.

• The height of the concrete compression block is x.

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Example 2:

Calculate the bending stresses for thesection shown, M= 180 kN.m

5

4700 4700 30 25743

2 107 77

25743

350 1500 7 77 7002

185 16

c c

s

c

E f MPa

E n .

E

x( ) x ( ) ( . )( x )

x . mm

750 mmNote: M > Mcr = 111 kN.m from previous

example. Thus, section is cracked.

14

1500 mm2

MPa f c 30



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3 2

3 2

9 4

6

9

6

9

1

3

1350 185 16 7 77 1500 700 185 16

3

3 8295 10

180 10 185 168 7

3 8295 10

8 7 0 45 0 45 30 13 5 OK

180 10 700 185 167 77 18

3 8295 10

t s

t

t

c

t

c c

s

t

I bx nA ( d x )

I ( )( . ) . ( . )

I . mm

My .. MPa

I .

. MPa . f . ( ) . MPa

My ( . )n .

I .

8 MPa

1500 mm2750 mm

15

Example 2:

MPa f c 30



114/514

Example 3:

Calculate the allowable moment for the

section shown, f s(allowable) = 180 MPa,

f c(allowable) = 12 MPa

750 mm

16

9

8

9

8

180 3 8295 10

7 77 700 185 16

1 7234 10 172 34

12 3 8295 10

185 16

2 4819 10 248 19

172 34

s t s

s

c t c

c

allowable

f I . M

ny ( . )( . ) M . N .mm . kN .m

f I . M

y .

M . N .mm . kN .m

M . kN .m

1500 mm2

MPa f c 30



115/514

Lecture 5

Strength analysis of beams according to ACI Code

Dr. Nader Okasha

Strength requirement for flexure in beams


116/514

Strength requirement for flexure in beams

Design moment strength (also known as moment resistance)

Internal ultimate momentu M

L Du M M M 6.12.1

nd M Φ M

Theoretical or nominal resisting moment.

ud M M

d M

n M

2

The equivalent stress (Whitney) block


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3

Strain

Distribution

Actual

Stress Distribution

Approximate

Stress Distribution



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•The shape of the

stress block is notimportant.

•However, the

equivalent block must

provide the sameresultant (volume)

acting at the same

location (centroid).

•The Whitney blockhas average stress

0.85f c ’ and depth

a= b 1c.

4

ACI 10.2.7.1



119/514

q ( y)

The equivalent rectangular concrete stress distribution has what

is known as the b 1 coefficient. It relates the actual NA depth tothe depth of the compression block by a= b 1c.

MPa f for f

MPa f for

cc

c

28'65.0

7

)28'(05.085.0

28'85.0

1

1

b

b ACI 10.2.7.1

5

Derivation of beam expressions


120/514

p

C = T

6

F x =0



121/514

p

7



122/514

p

= R n

8

Design aids can also be used:

Assume

M d = M u = Φ M n

R n

is given in tables and figures of design aids.

f M n =f R n bd2

Design Aids


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9


124/514

Tension strain in flexural members


125/514

Tension strain in flexural members

y

y

s

y ?t

f E

Strain Distribution

11

Types of flexural failure:


126/514


Flexural failure may occur in three different ways

[1] Balanced Failure –

(balanced reinforcement)

[2] Compression Failure –

(over-reinforced beam)

[3] Tension Failure -

(under-reinforced beam)

12

[1] Balanced Failure



127/514

The concrete crushes and the steel yields simultaneously.

Such a beam has a balanced reinforcement, its failure mode is

brittle , thus sudden, and is not allowed by the ACI Strength Design

Method.

[1] Balanced Failure

εcu=0.003εcu=0.003

c b

b

h

d

εt = εy

13

[2] C i F il



128/514

[2] Compression Failure

εcu=0.003εcu=0.003

c>cb

b

h

d

εt< εy

The concrete will crush before the steel yields.

Such a beam is called over-reinforced beam, and its failure mode is

brittle , thus sudden, and is not allowed by the ACI Strength

Design Method.

εcu=0.003εcu=0.003

c>cb

b

hd

εt< εy

14

[3] T i F il



129/514

[3] Tension Failure

εcu=0.003εcu=0.003

c


130/514

[ ] e s o co o ed sec o

The tensile strain in the tension steel is equal to or greater than 0.005

when the concrete in compression reaches its crushing strain of 0.003. This is a ductile section.

[2] Compression-controlled section

The tensile strain in the tension steel is equal to or less than ε y (ε y =

f y /E s =0.002 for f y =420 MPa) when the concrete in compressionreaches its crushing strain of 0.003. This is a brittle section.

[3] Transition section

The tensile strain in the tension steel is between 0.005 and ε y (ε y =f y /E s =0.002 for f y =420 MPa) when the concrete in compression

reaches its crushing strain of 0.003.

16

All d t i f ti i b di


131/514

Allowed strains for sections in bending ACI 10.3.5

17

Strength reduction factor Φ


132/514

Strength reduction factor Φ

ct

d c

c

εcu=0.003εcu=0.003

c

d

εt

ACI R9.3.2.2 18

Balanced steel


133/514

600

600b

y

c d f

0.003

0.003b

y S

c d f E

MPa E s5

102

y y

cb

f f

f

600

600'85.0 1 b

= b 1c

19

Maximum allowed steel


134/514

= b 1c

y

c

f

f '85.0

8

3 1max

b

d c

d c

8

3

005.0003.0

003.0

max

max

1max1

8

3 b b d c

20

Minimum steel allowed


135/514

c

y

s,min

y

0 25

1 4

w

w

. f b d

f A max

.

b d f

bw = width of section

d = effective depth of section

21

ACI 10.5.1

Design Aids


136/514

22

Summary:

To calculate the moment capacity of a section:


137/514

p y

1-)

if As,min > As,sup reject section

2-) or

3-)

4-)

0.85

y

c

df a

f

c

y

s,min

y

0 25

1 4

w

w

. f b d

f A max.

b d f

MPa f for f

MPa f for

cc

c

28'65.07

)28'(05.085.0

28'85.0

1

1

b

b

0.85

s y

c

A f a

f b

1

ac

b

23

Summary:


138/514

5-)

if t > 0.005: tension controlled f = 0.9

if 0.004 < t


139/514

A singly reinforced concrete beam has the cross-section shown in the figure

below. Calculate the design moment strength. Can the section carry an

M u

= 350 kN.m?

a) 20.7 , b) 34.

41

5 , c) 62.1

4

c c c

y

f MPa f MPa f P

MP

a

a

M

25

Solut ion

Example


140/514

a) 7.20c f MPa

c 2

y

s,min

2

y

2 2

s,sup

0 25 0 25 20.7(254)(457)=319 mm

4141

14 14(254)(457)=393 mm

414

=393 mm < A =2580 mm OK

w

w

. f .b d

f A max

. .b d

f

2580 4142 239

0.85 20.7 2540.85

s y

c

A f a mm

f b

13 0 85 20 7 28c. for f ' . MPa MPa b 26


141/514


142/514

Example

Solut ion


143/514

457 178 55 0 003 0 003 0 00468178 5

0 004 0 005 Section is in transision zone

t

t

d c .. . .c .

. .

1

143.44 178.5

0.804

ac mm

b

t=0.65+( -0.002) (250/3) =0.65+(0.00468-0.002) (250/3)=0.874

f

29

b) 5.34c f MPa

Example

Solut ion


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6

62

143 4 0 874 2850 414 457 360 102

360

7 350 360

Section is adequate

d n s y

u n

a M M A f d

.. N .mm

kN .m

M kN .m ΦM kN .m

30

b) 5.34c f MPa

Example

Solut ion


145/514

2580 414

2 800.85 62.1 2540.85

s y

c

A f a mm

f b

c) 1.62c f MPa

31

c 2

y

s,min

2

y

2 2

s,sup

0 25 0 25 62.1(254)(457)=552 mm

4141

14 14(254)(457)=393 mm

414

=552 mm < A =2580 mm OK

w

w

. f .b d

f A max

. .b d

f

Example

Solut ion


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1

1

1

0 05 283 0 85 0 65 62 1 28

7

0 05 62 1 28 0 85 0 61 0 65

7 0 65

cc

. ( f ' ). . for f ' . MPa MPa

. ( . ). . .

.

b

b

b

1

804 123

0.65

ac mm

b

457 123

5 0 003 0 003 0 0081123

0 005 Section is tension controlled

==> Satisfes ACI requirements ==> =0 9

t

t

d c. . .

c

.

f 32

c) 1.62c f MPa

Example

Solut ion


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6

62

80 0 9 2850 414 457 520 102

520

7 350 520

Section is adequate

d n s y

u n

a M M A f d

. N .mm

kN .m

M kN .m ΦM kN .m

33

c) 1.62c f MPa



148/514

Lecture 6

Design of singly reinforced rectangular beams

Dr. Nader Okasha

Design of Beams For Flexure


149/514

The main two objectives of design is to satisfy the:1) Strength and 2) Serviceability requirements

d n u M Φ M M

Design moment strength (also known as moment resistance)

Internal ultimate moment

d M

u M

L Du M M M 6.12.1

Theoretical or nominal resisting moment.n M

1) Strength

2



150/514

Derivation of design expressions

3

Assume

Φ M n = M u

Solve for r :

2

0 85 21 1

0 85

c u

y c

. f ' M ρ

f . f ' b d

Remember: 1 kN.m = 106

N.mm

As = r bd

b

h d

As

Beam cross section



151/514

Design aids can also be used:

4

2

0 85 21 1

0 85

c u

y c

. f ' M ρ

f . f ' b d

Then r is found from tables and figures of design aids.

Calculate:

Design Aids


152/514

5



153/514

2) Serviceability

The serviceability requirement ensures adequate performance

at service load without excessive deflection and cracking.

Two methods are given by the ACI for controlling deflections:

1) by calculating the deflection and comparing it with code

specified maximum values.

2) by using member thickness equal to the minimum values provided in by the code as shown in the next slide.

6

Minimum Beam Thickness

ACI 9.5.2.2


154/514

l= span length measured center to center of support.

minh

7

minh h

b

h d

As

Beam cross section

Concrete Cover

Detailing issues:


155/514

Concrete cover is necessary for protecting the reinforcement from

fire, corrosion, and other effects. Concrete cover is measured from

the concrete surface to the closest surface of steel reinforcement.

8

ACI 7 7 1

Side

cover

Bottom

cove

Spacing of Reinforcing Bars

Detailing issues:


156/514

9

p g g

•The ACI Code specifies limits for bar spacing to permit concrete to

flow smoothly into spaces between bars without honeycombing.

According to the ACI code, S S min must be satisfied, where:

•When two or more layers are used, bars in

the upper layers are placed directly above

the bars in the bottom layer with clear distance

between layers not less than 25 mm.Clear

distance

Clear spacing S

ACI 3.3.2

b

min

bar diameter, d

25 mm4/3 maximum size of coarse aggregate

S max

ACI 7.6.1

ACI 7.6.2

Estimation of applied moments M u

Beams are designed for maximum moments along the spans in both


157/514

10

Positive moment

Tension at bottom

Needs bottom reinforcement

Negative moment

Tension at top

Needs top reinforcement

Beams are designed for maximum moments along the spans in both

negative and positive directions.


The magnitude of each moment is found from structural analysis of the


158/514

11

The magnitude of each moment is found from structural analysis of the

beam. To find the moments in a continuous (indeterminate) beam, one

can use: (1) indeterminate structural analysis (2) structural analysis

software (3) ACI approximate method for the analysis.

Continuous Beams

Indeterminate

Moment Diagram

Simply Supported Beams

Determinate

Moment Diagram

+

+ +

–



159/514

12

Continuous Beams

Moment Diagram

Simply Supported Beams

Moment Diagram

+

Section at midspan Section over support

+ + –

Approximate Structural AnalysisACI 8 3 3



160/514

13

ACI Code permits the use of the following approximate moments fordesign of continuous beams, provided that:

• There are two or more spans.

• Spans are approximately equal, with the larger of two adjacent spans

not greater than the shorter by more than 20 percent.• Loads are uniformly distributed.

• Unfactored live load does not exceed three times the unfactored dead

load.

• Members are of similar section dimensions along their lengths(prismatic).

ACI 8.3.3

ACI 8 3 3Approximate Structural Analysis



161/514

More than two spans

14

ACI 8.3.3

ACI 8.3.3Approximate Structural Analysis



162/514

15

Two spans

l n = length of clear

span measured

face-to-face of

supports.

For calculating

negative moments, l n is taken as the

average of the

adjacent clear spanlengths.

Method 1: When b and h are unknown

Design procedures


163/514

1- Determine h (h>hmin from deflection control) and assume b. Estimate beam weight and include it with dead load.

2- Calculate the factored load wu and bending moment Mu.

3- Assume that Φ=0.9 and calculate the reinforcement (ρ and As).4- Check solution:

(a) Check spacing between bars

(b) Check minimum steel requirement

(c) Check Φ = 0.9 (tension controlled assumption)(d) Check moment capacity (Md ≥ Mu ?)

5- Sketch the cross section and its reinforcement.

16

Method 2: When b and h are known

Design procedures


164/514

1- Calculate the factored load wu and bending moment Mu.2- Assume that Φ=0.9 and calculate the reinforcement (ρ and As).

3- Check solution:

(a) Check spacing between bars

(b) Check minimum steel requirement

(c) Check Φ = 0.9 (tension controlled assumption)

(d) Check moment capacity (Md ≥ Mu ?)

4- Sketch the cross section and its reinforcement.

17

Example 1


165/514

Design a rectangular reinforced concrete beam having a 6 m simple span. A

service dead load of 25 kN/m (not including the beam weight) and aservice live load of 10 kN/m are to be supported.

Use f c’ =25 MPa and f y = 420 MPa.

Solution:-

b & d are unknown

1- Estimate beam dimensions and weight

hmin = l /16 =6000/16 = 375 mm

Assume that h = 500mm and b = 300mm

Beam wt. = 0.5x0.3x25 = 3.75 kN/m

2- Calculate wu and Muwu = 1.2 D+1.6 L =1.2(25+3.75)+1.6(10)

=50.5 kN/m

Mu = wul2/8 = 50.5(6)2/8 =227.3 kN.m

6 m

w d=25 kN/m & w l =10 kN/m

6 m

w u=50.5 kN/m

227.3 kN.m

18

Example 1


166/514

c u

2

y c

6

2

0.85f ' 2Mρ 1 1

f Φ0.85f ' bd

0.85(25) 2 227.3 10ρ 1 1 0.0116

420 (0.9)0.85(25)300(442)

3- Assume that Φ=0.9 and calculate ρ and Asd = 500 – 40 – 8 – (20/2) = 442 mm

(assuming one layer of Φ20mm reinforcement and Φ8mm stirrups)

As = ρ b d = 0.0116(300)(442) =1536 mm2

Use 5 Φ 20 mm (As,sup=1571 mm2)

19

WN/m

Number of barsmm 1 2 3 4 5 6 7 8 9 10

6 2.2 28 57 85 113 141 170 198 226 254 283


167/514

20

8 3.9 50 101 151 201 251 302 352 402 452 503

10 6.2 79 157 236 314 393 471 550 628 707 785

12 8.9 113 226 339 452 565 679 792 905 1018 1131

14 12.1 154 308 462 616 770 924 1078 1232 1385 1539

16 15.8 201 402 603 804 1005 1206 1407 1608 1810 2011

18 19.9 254 509 763 1018 1272 1527 1781 2036 2290 254520 24.7 314 628 942 1257 1571 1885 2199 2513 2827 3142

22 29.8 380 760 1140 1521 1901 2281 2661 3041 3421 3801

24 35.5 452 905 1357 1810 2262 2714 3167 3619 4072 4524

25 38.5 491 982 1473 1963 2454 2945 3436 3927 4418 4909

26 41.7 531 1062 1593 2124 2655 3186 3717 4247 4778 5309

28 45.4 616 1232 1847 2463 3079 3695 4310 4926 5542 6158

30 55.4 707 1414 2121 2827 3534 4241 4948 5655 6362 7069

32 63.1 804 1608 2413 3217 4021 4825 5630 6434 7238 8042

Example 1


168/514

4- Check solution

a) Check spacing between bars

b) Check minimum steel requirement

300 2 40 2 8 5 20

26 205 1

25

c b s mm d mm

mm OK

21

c 2

y

s,min

2

y

2 2

s,sup

0 25 0 25 25(300)(442)=395 mm

420

1 4 1 4 (300)(442)=442 mm420

=442 mm < A =1571 mm OK

w

w

. f .b d

f A max

. .b d f

300

5Φ

20

Example 1

c) Check Φ =0.9 (tension controlled assumption)

A f 1571 420


169/514

d) Check moment capacity

s y

c

1

1

t

t

A f 1571 420a 103.5mm

0.85f 'b 0.85(25)300a 103.5

0 85 25 28 c 121.7mmβ 0.85

d c 442 121.7ε 0.003 0.003 0.0079 0.005

c 121.7

for ε 0.005 Φ 0.90, the assumption is true

c. for f ' MPa MPa

d s y

6

d u

M Φ A f d2

103 50.90 1571 420 442 231.7 10 N.mm = 231.7 kN.m

2

M 231.7 kN.m M 227.3kN.m OK

a

.

22

the section is tension controlled

Example 1


170/514

23

30

50 44.2

5Φ20

Beam cross section

5- Sketch the cross section and its reinforcement

Example 2

The rectangular beam B1 shown in the figure has b = 800mm and h =

316mm Design the section of the beam over an interior support Columns


171/514

316mm. Design the section of the beam over an interior support. Columns

have a cross section of 800x300 mm. The factored distributed load over theslab is q u =14.4 kN/m2.

Use f c’ =25 MPa and f y = 420 MPa.

Solution:

24

b & d are known

1- Calculate wu and Mu

w u =4(14.4) = 57.6 kN/ml n = 6 – 0.3=5.7 m

w u

B1

L 1 = L 2 = L 3 = 6 m

S 1 = S 2= S 3 = 4 m

Example 2


172/514

25

M u = w u (l n )2/10 = 57.6 (5.7)2/10

M u = 187.5 kN.m

Moment diagram using the approximate ACI method:

Design for the maximum negative moment throughout the beam:

Example 2


173/514

26

2- Assume Φ=0.9 and calculate ρ and Asd = 316 – 40 – (16/2) – 8 = 260 mm

(assuming one layer of Φ16 mm reinforcement and Φ8mm stirrups)

c u

2y c

6

2

0.85f ' 2Mρ 1 1

f Φ0.85f ' bd

0.85(25) 2 187 5 10ρ 1 1 0 0102

420 (0.9)0.85(25)800(260)

..

As= ρ b d = 0.0102(800)(260) = 2120 mm2

Use 11 Φ16 mm (As,sup =11[(16)2/4]=2212 mm2)

Example 2


174/514

3- Check solutiona) Check spacing between bars

b) Check minimum steel requirement

27

800 2 40 2 8 11 16

52 8 1611 1

25

c b s . mm d mm

mm OK

c 2

y

s,min

2

y

2 2

s,sup

0 25 0 25 25(800)(260)=620 mm

420

14 14(800)(260)=693 mm

420

=693 mm < A =2212 mm OK

w

w

. f .b d

f

A max . .b d

f

c) Check Φ =0.9

Example 2


175/514

d) Check moment capacity

s y

c

1

1

t

t

A f 2212 420a 55mm

0.85f 'b 0.85(25)800

a 550 85 25 28 c 64mm

β 0.85

d c 260 64ε 0.003 0.003 0.0091 0.005

c 64for ε 0.005 Φ 0.90, the assumption is true

c. for f ' MPa MPa

d s y

6

d u

M Φ A f d2

550.9 2212 420 260 194 5 10 N.mm=194 5 kN.m

2

M 194 5 kN.m M 187 5kN.m OK

a

. .

. .

28

the section is tension controlled

Example 2


176/514

29

11Φ16

800

316 260

4- Sketch the cross section and its reinforcement



177/514

Lecture 7

Design of T and L beams

Dr. Nader Okasha

T Beams

Reinforced concrete systems may consist of slabs and dropped


178/514

2

Flange

web

beams that are placed monolithically. As a result, the two parts act

together to resist loads. The beams have extra widths at their tops

called flanges, which are parts of the slabs they are supporting, and

the part below the slab is called the web or stem.

Parts of the slab near the webs are more highly stressed than areas

Flange Width b


179/514

away from the web.

b w b w

effective flange

width be

effective flange

width be

dhf

stirrup

L-beam T-beam

3

d : effective depth. h f: height of flange.

b w

: width of web. b e: effective width.

b : distance from center to center of adjacent web spacings

Effective Flange Width be

b is the width that is stressed uniformly to give the same compression


180/514

be is the width that is stressed uniformly to give the same compression

force actually developed in the compression zone of width b.

4


181/514

Effective Flange Width be

ACI Code Provisions for Estimating be

A di h ACI d h ff i fl id h f L b

ACI 8.12.3


182/514

According to the ACI code, the effective flange width of an L-beam,

be is not to exceed the smallest of:1. bw + L/12.

2. bw

+ 6 hf .

3. bw

+ 0.5(clear distance to next web).

w

eff w f

/12

min 6

0.5

w c

b L

b b h

b b

6


183/514

Various Possible Geometries of T-Beams


184/514

Single Tee

Double Tee

Box

8

Various Possible Geometries of T-Beams


185/514

Same as

Flange

web

Flange

web

9

T- versus Rectangular Sections

If the neutral axis falls within the slab depth: analyze the beam as a


186/514

If the neutral axis falls within the slab depth: analyze the beam as a

rectangular beam, otherwise as a T-beam.

10

T- versus Rectangular Sections

When T-beams are subjected to negative moments, the flange is

located in the tension zone Since concrete strength in tension is


187/514

11

Compression zone Tension

zone

located in the tension zone. Since concrete strength in tension is

usually neglected in ultimate strength design, the sections are treatedas rectangular sections of width bw.

When sections are subjected to positive moments, the flange is

located in the compression zone and the section is treated as a T-section.

Moment Diagram

+ +

–

Section at midspan

Positive moment

Section at support

Negative moment

C 1 h ≤ h [S l i ]

Analysis of T-beams


188/514

Case 1: when a ≤ hf [Same as rectangular section]

ec

ys

bf 0.85

f A

C

a

T

12

2

adf AΦΦM ysn

C 2 h > h

Analysis of T-beams


189/514

0.85

0.85

0.85

0.85

f c e w f

w c w

s y

f w

s y c e w f

c w

C f b b h

C f b a

T A f

T C C

A f f b b h

a f b

Case 2: when a > hf

From equilibrium of forces

13

2

hdC

2

adCΦΦM f f wn

Minimum Reinforcement, As,minACI 10.5.2


190/514

b w

dhf

As

be

b w

dhf

As

be

+ v e M o m e n t

- v e M

o m e n t

14

c

ys,min

y

0 25

1 4

w

w

. f b d

f A max

.b d

f

Analysis procedure for calculating he ultimate strength of T-beams

To calculate the moment capacity of a T-section:


191/514

1- Calculate be

2- Check As,sup> As,min

3- Assume a ≤ h f and calculate a using:

If a ≤ h f

→ a is correct

If a > h f

→

4- Calculate b 1, c, and check εt

5- Calculate Φ M n , and check

ec

ys

bf 0.85

f A

a

0.85

0.85

s y c e w f

c w

A f f b b ha

f b

15

u n M ΦM

Example 1

Calculate Md for the T-Beam:


192/514

d

hf = 150 mm

d = 400 mm As = 5000mm2

f y = 420MPa f c’= 25MPa

bw= 300mm L = 5.5m

b=2.15m

e f w

5500 13704 4

min 16 16 150 300=2700mm

2150 mm

L mm

b h b

b

16

Determine be according to ACI requirements

Example 1

Check min. steel


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17

c

s,min w w

y y

2 2

s,min s,sup

0.25 f ' 1.4 0.25 25 1.4A b d b d 300 400 300 400

f f 420 420

A 400 mm A 5000mm OK

max ; max ;

Calculate a (assuming a


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Determine the ACI design moment strength M (ΦM ) of the T beam

Example 2


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Determine the ACI design moment strength Md (ΦMn) of the T-beam

shown in the figure if f c’ =25 MPa and f y = 420 MPa.

Solution:-1- Check min. steel

30

8Φ32

Φ10

h = 7 5

1 0

90

19

c

s,min w w

y y

s,min

2 2

s,min s,sup

25d 750-40-10-32- 655.5mm

2

0.25 f ' 1.4A b d b d

f f

0.25 25 1.4A 300 655.5 300 655 5

420 420

A 656mm A 6434mm OK

max ;

max ; .

2 Check if a < h = 10cm

Example 2


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2- Check if a < h f

= 10cm

a= 141.3> h f

= 100 mm

i.e. assumption is wrong

Section is T NA is in the web

20

mma 3.1419002585.0

4206434

bf 0.85

f A

ec

ys

30

8Φ32

Φ10

h = 7 5

1 0

90


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Example 2

4- Calculate Md


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d

3 3

6

M Φ2 2

224 1000 855 1427 4 10 655 5 1275 10 655 5

2 2

1323 4 10 1323 4

f w f haC d C d

. . . .

. N .mm . kN .m

3f c e w f 3

w c w

C 0.85f ' (b b ) h 0.85 25 900 300 100 1275 10 N 1275 kN

C 0.85f ' a b 0.85 25 224 300 1427.4 10 N 1427.4 kN

4 Calculate Md

22

Design of T-Beams --- Positive moment


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uwuf u M M M

+

To analyze the section, the steel is divided in two portions: (1) Asf , which provides a

tension force in equilibrium with the compression force of the overhanging flanges, and

providing a section with capacity M uf

and (2) Asw

, the remaining of the steel, providing

a section with capacity M uw

.

M u: Ultimate moment applied, requiring steelA

s .

M uf

: Moment resisted by overhanging flange parts, requiring steelAsf .

M uw : Moment resisted by web requiring steel

Asw

23



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24

Step 1

Step 2

+

24



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uf uuw M M M uwuf u M M M

2

0 85 2

1 1 0 85

c uw

y c w

. f ' M

f . f ' b d

Step 3

Step 4

Step 5

Step 6

+

sw w

s sf sw

A b d

A A A

25

Design of L-Beams --- Positive moment


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b w

be

Same as

be

26


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Flange Reinforcement

When flanges of T-beams are in tension, part of the flexural


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Main Reinforcement

min (beff & l /10) Additional

Reinforcement Additional

Reinforcement

- v e m o m e n t

When flanges of T beams are in tension, part of the flexural

reinforcement shall be distributed over effective flange width, or awidth equal to one-tenth of the span, whichever is smaller

If beff > l /10, some longitudinal reinforcement shall be provided inouter portions of flange.

Design Procedure:



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Design Procedure:

1- Establish h based on serviceability requirements of the slab and calculate d

2- Choose bw3- Determine be according to ACI requirements.

4- Calculate As assuming that a < hf with beam width = be & Φ=0.90

As = ρ be d →

5- If a ≤ hf : the assumption is right continue as rectangular sectionIf a > hf : revise As using T-beam equations (steps 1-6).

6- Check the Φ=0.90 assumption (εt ≥0.005) and As,sup ≥ As,min

b w

dhf

As

be

2

ec

u

y

c

d b'f 0.85Φ

2M11

f

'f 0.85ρ

ec

ys

b'f 0.85

f Aa

29

Example 3

A floor system consists of a 14.0cm


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Spandrelbeam

L m

3.0 m 3.0 m 3.0 m

hf Slab

b w

y

concrete slab supported by continuousT-beams with a span L. Given that

bw=30cm and d=55cm, f c’ =28 MPa andf y = 420 MPa.

Determine the steel required atmidspan of an interior beam to resist

a service dead load moment 320

kN.m and a service live load moment

250 kN.m in the following two cases:

(A) L = 8 m(B) L = 2 m

30

5514

200

N . m

Solution (A) L = 8 m

Determine b according to ACI requirements


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be is taken as 2000 mm, as shown in the figure

Mu = 1.2(320)+1.6(250)=784 kN.m

30

55

As

2

0 85 21 1

0 85

c u

y c e

. f ' M - -

f . f ' b d

7 8 4

k N

31


Calculate As assuming that a < hf with beam width = be & Φ=0.90

e f w

80002000

4 4

min 16 16 140 300=2540mm

3000 mm

Lmm

b h b

b


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Ch k th Φ=0 90 ti ( t ≥0 005) d A ≥ A i



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OK 9.0005.00374.0

0.00340.8

8.405500.003

c

cdε

mm8.400.85

34.7

β

a

c

t

1

5 5

308Φ25

1 4

200

Check the Φ=0.90 assumption (εt ≥0.005 ) and As,sup ≥ As,min

2 2

s,min s,sup

0 25 1 4 0 25 28 1 4300 550 300 550

420 420

A 550mm A 3927 mm OK

c

s , min w w

y y

. f ' . . . A max b d ; b d max ;

f f

33

s y

c e

A f 3927 420a 34.7 mm

0.85f 'b 0.85 28 2000

Check moment capacity



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6

d u

2

34 70 9 3927 420 550

2

M 790.7 10 N.mm 790.7 kN.m M 784kN.m

d s y

a M A f d

..

5 5

30

8Φ25

1 4

200

p y

34

5514

50

N . m

Determine b according to ACI requirements

Solution (B) L = 2 m


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30

55

As

2

0 85 21 1

0 85

c u

y c e

. f ' M

f . f ' b d

7 8 4

k N

be is taken as 500 mm, as shown in the figure

Mu = 1.2(320)+1.6(250)=784 kN.m

35


Calculate As assuming that a < hf with beam width = be & Φ=0.90

e f w

2000 5004 4

min 16 16 140 300=2540mm

3000 mm

L mm

b h b

b

60 85 28 2 784 10 55

14

50

k N . m



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Check a ≤ hf

assumption

The assumption is wrong T section design

2

2

0.85 28 2 784 10

ρ 1 1420 0 9 0.85 28 500 550

0 0159

ρ 0 0159 500 550 4389 s e

.

.

A b d . mm

30

55

As 7 8 4 k

36

f

4389 420155mm > h 140mm

0 85 0.85 28 500

s y

c e

A f a

. f ' b


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0 85 2f ' M


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Use 8Φ28 mm (As,sup= 4926mm2 ) arranged in two layers.

Check solution: (Do as in Example 2) 38

5 5

30

8Φ28

1 4

50

6

2

0 85 28 2 496 101 1 0 017

420 0 9 0 85 28 300 550

. ( ) ( ).

( ) . . ( ) ( ) ( )

2

2

0 017 300 550 2808

1586 2808 4395

sw w

s sf sw

A b d . ( )( ) mm

A A A mm

2

0 85 21 1

0 85c u

y c w

. f ' M

f . f ' b d



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Lecture 8

Design of doubly reinforced beams

Dr. Nader Okasha

Doubly Reinforced Rectangular Sections

Beams having steel reinforcement on both the tension and


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compression sides are called doubly reinforced sections. Doublyreinforced sections are useful in the case of limited cross sectional

dimensions being unable to provide the required bending strength.

Increasing the area of reinforcement makes the section brittle.

2

Reasons for Providing Compression Reinforcement

1- Increased strength


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1- Increased strength.

2- Increased ductility.

3- Reduced sustained load deflections due to shrinkage and creep.

4- Ease of fabrication. Use corner bars to hold & anchor stirrups.

3

Analysis of Doubly Reinforced Rectangular Sections

Divide the section:


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4

To analyze the section, the tension steel is divided in two portions: (1) As2 , which is inequilibrium with the compression steel, and providing a section with capacity M n2 and

(2) As1 , the remaining of the tension steel, providing a section with capacity M n1 .

M n2 M n1 M n


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Find f s ’:


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003.0s

c

d c

s 0.003 s s s yc d

E E f c

52 10 MPa s E

6

c


Find c:


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7

c sT C C

0.85 s y c s s A f f ab A f

10.85 0.003 s y c s sc d

A f f cb A E c

find c by solving the quadratic equation find f s ’ from equation in slide 6


Find M d :


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8

12

d n s y s s

a M M A f d - A f d - d '



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Procedure:1)

2)

3)

4)

5) Check if f = 0.9

6)

10.85 0.003 s y c s sc d

A f f cb A E c

find c, a

0.003 s s yc d

E f c

s 0.003 0.005?d c

c

9

2 s s

s

y

A f A f

1 2 s s s A A A

12

d n s y s s

a M M A f d - A f d - d '

Example 1

For the beam with double reinforcement shown in the figure,


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30

6Φ32

2Φ25

60

5.0calculate the design moment Md.f c’ =35MPa and f y = 420 MPa.

Solution:-

1

5

0.85 0.003

504825(420) 0.85(35)(0.8) (300) 982 0.003(2 10 )

s y c s s

c d A f f cb A E

c

cc

c

1

1

0 05 28 0 85 0 65 35 28

7

0 05 35 28 0 85 0 8 0 65

7

cc

. ( f ' ). . for f ' MPa MPa

. ( ). . .

10

Example 1


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30

6Φ32

2Φ25

60

5.0

5

2

1

504825(420) 0.85(35)(0.8) (300) 982 0.003(2 10 )

229.5 1437300 29460000 0

220

0.8 220 176

ccc

c c

c mm

a c mm

5

0.003

220 500.003(2 10 ) 463 420

220

420

s s y

s y

s y

c d E f

c

f MPa

f f

11

Example 12Φ25

5.0

2982(420)A f


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30

6Φ32

60

2

1 2 4825 982 3843

Documents

Reinforced Concrete Design I