reinforced concrete design i
TRANSCRIPT
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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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Reinforced Concrete Design
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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.
.
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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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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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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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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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
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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
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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
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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.
'
c
Mechanical Concrete Properties
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Compressive Strength,
• 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).
'
c
Mechanical Concrete Properties
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Compressive Strength,
• 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.
'
c
Mechanical Concrete Properties
Cylinder 0 8 Cubec c. f
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Mechanical Concrete PropertiesCompressive-Strength Test
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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
Mechanical Concrete Properties
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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
Mechanical Concrete Properties
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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
Mechanical Concrete Properties
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Tensile Strength – Splitting Tensile Strength, f ct
– Split Cylinder Test
PConcrete Cylinder
Poisson’s
Effect
2ct
P
Ld
Mechanical Concrete Properties
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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
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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, , #
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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)
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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.
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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.
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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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Load paths loads on (two-way) beams
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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 (two-way) beams
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Load paths loads on (two-way) beams
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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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Load paths loads on structural members
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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Load paths loads on structural members
Example
B1:
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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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Load paths loads on structural members
Example
B2:
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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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Load paths loads on structural members
Example
B1:
B2:
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w = 20 kN/m
w = 45 kN/m
L d th
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Load paths loads on structural members
Example
C1:
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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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Load paths loads on structural members
Example
C2:
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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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Building Codes, Standards, and Specifications:
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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Building Codes, Standards, and Specifications:
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3 Model Codes in the U.S.
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Building Codes, Standards, and Specifications:
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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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Building Codes, Standards, and Specifications:
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
12
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.
13
Elastic stresses – Cracked section
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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
Elastic stresses – Cracked section
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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
Elastic stresses – Cracked section
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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
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Lecture 5
Strength analysis of beams according to ACI Code
Dr. Nader Okasha
Strength requirement for flexure in beams
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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
The equivalent stress (Whitney) block
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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
The equivalent stress (Whitney) block
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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
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p
C = T
6
F x =0
Derivation of beam expressions
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p
7
Derivation of beam expressions
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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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Tension strain in flexural members
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Tension strain in flexural members
y
y
s
y ?t
f E
Strain Distribution
11
Types of flexural failure:
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Types of flexural failure:
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
Types of flexural failure:
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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
Types of flexural failure:
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[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
Types of flexural failure:
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[3] Tension Failure
εcu=0.003εcu=0.003
c
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[ ] 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
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Allowed strains for sections in bending ACI 10.3.5
17
Strength reduction factor Φ
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Strength reduction factor Φ
ct
d c
c
εcu=0.003εcu=0.003
c
d
εt
ACI R9.3.2.2 18
Balanced steel
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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
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= 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
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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
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22
Summary:
To calculate the moment capacity of a section:
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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:
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5-)
if t > 0.005: tension controlled f = 0.9
if 0.004 < t
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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
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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
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Example
Solut ion
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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
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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
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Lecture 6
Design of singly reinforced rectangular beams
Dr. Nader Okasha
Design of Beams For Flexure
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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
Design of Beams For Flexure
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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
Design of Beams For Flexure
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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
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5
Design of Beams For Flexure
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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
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l= span length measured center to center of support.
minh
7
minh h
b
h d
As
Beam cross section
Concrete Cover
Detailing issues:
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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:
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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
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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.
Estimation of applied moments M u
The magnitude of each moment is found from structural analysis of the
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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
+
+ +
–
Estimation of applied moments M u
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12
Continuous Beams
Moment Diagram
Simply Supported Beams
Moment Diagram
+
Section at midspan Section over support
+ + –
Approximate Structural AnalysisACI 8 3 3
Estimation of applied moments M u
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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
Estimation of applied moments M u
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More than two spans
14
ACI 8.3.3
ACI 8.3.3Approximate Structural Analysis
Estimation of applied moments M u
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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29
11Φ16
800
316 260
4- Sketch the cross section and its reinforcement
Reinforced Concrete Design I
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Lecture 7
Design of T and L beams
Dr. Nader Okasha
T Beams
Reinforced concrete systems may consist of slabs and dropped
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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
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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
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be is the width that is stressed uniformly to give the same compression
force actually developed in the compression zone of width b.
4
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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
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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
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Various Possible Geometries of T-Beams
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Single Tee
Double Tee
Box
8
Various Possible Geometries of T-Beams
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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
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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
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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
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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
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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
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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:
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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:
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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
Design of T-Beams --- Positive moment
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24
Step 1
Step 2
+
24
Design of T-Beams --- Positive moment
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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:
Design of T-Beams --- Positive moment
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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
Determine be according to ACI requirements
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
Solution (A) L = 8 m
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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
Solution (A) L = 8 m
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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
Determine be according to ACI requirements
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
Solution (B) L = 2 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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Solution (B) L = 2 m
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
Reinforced Concrete Design I
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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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Analysis of Doubly Reinforced Rectangular Sections
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
Analysis of Doubly Reinforced Rectangular Sections
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
Analysis of Doubly Reinforced Rectangular Sections
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 '
Analysis of Doubly Reinforced Rectangular Sections
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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