Reinforced Concrete Design I Nader Okasha 6u5w48
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Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 0 Syllabus
Reinforced Concrete Design
Instructor
Dr. Nader Okasha.
Email
[email protected]
Office Hours
As needed.
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Reinforced Concrete Design
This course is only offered for 2010 students who have ed strength of materials.
If you don’t meet this criteria you will not be allowed to continue this course.
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Reinforced Concrete Design
References: Building Code Requirements for Reinforced Concrete and commentary (ACI 318M-08). American Concrete Institute, 2008. Design of Reinforced Concrete. 7th edition, McCormac, J.C. and Nelson, J.K., 2006. Reinforced Concrete Design. By Dr. Sameer Shihada. ٤
Reinforced Concrete Design
Additional references (internationally recognized books in reinforced concrete design): Reinforced Concrete, A fundamental Approach. Edward Nawy. Design of Concrete Structure. Nilson A. et al.
Reinforced Concrete Design. Kenneth Leet. Reinforced Concrete: Mechanics and Design. James K. Wight, and James G. MacGregor. ٥
Reinforced Concrete Design
The art of design
Design is an analysis of trial sections. The strength of each trial section is compared with the expected load effect. The load effect on a section is determined using structural analysis and mechanics of materials. The strength of a reinforced concrete section is determined using the concepts taught in this class.
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Reinforced Concrete Design
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Reinforced Concrete Design Course outline Week
1
2, 3,4
Topic
Introduction: -Syllabus and course policies. -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 loads. -Strength analysis of beams according to ACI Code. -Design of singly reinforced rectangular beams. -Design of T and L beams. -Design of doubly reinforced beams.
4
Design of beams for shear.
5
Midterm. ٨
Reinforced Concrete Design Course outline Week
Topic
6
Design of slabs: One way solid slabs – One way ribbed slabs.
7
Design of short concentric columns.
7,8
Bond, development length, splicing and bar cutoff.
8,9
Design of isolated footings.
9
Staircase design.
10
Final
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Reinforced Concrete Design
Grading
Course work:
20%
-Homework
4%
-Attendance
4%
-Project
12%
Mid-term exam
20%
Final exam
60% ١٠
Reinforced Concrete Design
Exam Policy
Mid-term exam: Only one A4 cheat-sheet is allowed.
Necessary figures and tables will be provided with the exam forms.
Final exam: Open book. ١١
Reinforced Concrete Design
Homework Policy Show all your assumptions and work details. Prepare neat sketches showing the reinforcement and dimensions. Marking will consider primarily neatness of presentation, completeness and accuracy of results. You may get the HW points if you copy the solution from other students. However, you will have lost your chance in practicing the concepts through doing the HW. This will lead you to loosing points in the exams, which you could have gained if you did your HWs on your own. No late HWs will be accepted. Homework solutions will be posted on upinar immediately after the submission deadline.
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Reinforced Concrete Design
Policy towards cell-phone use
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Reinforced Concrete Design
Policy towards discipline during class Zero tolerance will be practiced.
No talking with other students is allowed. Raise your hand before answering or asking questions.
Leaving during class is not allowed (especially for answering the cell-phone) unless a previous permission is granted. Violation of discipline rules may have you dismissed from class and jeopardize your participation points.
١٤
Reinforced Concrete Design
Policy towards missed classes Any collectively missed class MUST be made up.
A collectively missed class will be made up either on a Thursday or during the discussion lecture. An absence from a lecture will loose you attendance points, and the lecture will not be repeated for you. You are on your own. You may use the lecture videos. No late students will be allowed in class.
Anything mentioned in class is binding. No excuse for not being there or not paying attention.
١٥
Reinforced Concrete Design
Units used in class In all equations, the input and output units are as follows: Distance (L,b,d, ): mm Area (Ac,Ag,As): mm2 Volume (V): mm3 Force (P,V,N): N Moment (M): N.mm Stress (fy, fc’): N/mm2 = MPa = 106 N/m2 Pressure ( s): N/mm2 Distributed load per unit length (wu): N/mm Distributed load per unit area ( u): N/mm2 Weight per unit volume (): N/mm3
١٦
Reinforced Concrete Design
Units used in class However, these quantities may be presented as Distance (L,b,d, ): cm , m Area (Ac,Ag,As): cm2, m2 Volume (V): cm3, m3
Force (P,V,N): kN Moment (M): kN.m Pressure (qs): kN/m2 Distributed load per unit length ( u): kN/m Distributed load per unit area (qu): kN/m2
Weight per unit volume ( ): kN/m3 ١٧
Reinforced Concrete Design
Unit conversions 1 m = 102 cm = 103 mm 1 m2 = 104 cm2 = 106 mm2 1 m3 = 106 cm3 = 109 mm3 1 kN = 103 N 1 kN.m = 106 N.mm 1 kN/m2 = 10-3 N/mm2 1 kN/m3 = 10-6 N/mm3
You MUST specify the unit of each result you obtain ١٨
Reinforced Concrete Design
ACI Equations The equations taken from the ACI code will be indicated throughoutthe slides by their section or equation number in the code provided in shading.
Examples:
Ec 4700
f c
ACI 8.5.1
f r 0.62
f c
ACI Eq. 9-10
Some of the original equations may have included the symbol = 1.0 for normal weight concrete and omitted in slides. ١٩
Reinforced Concrete Design
Advices for excelling in this course: Keep up with the teacher and pay attention in class.
Study the lectures up to date. Re-do the lecture examples.
Look at additional resources.
DO YOUR HOMEWORK!!!!! Check your solution with the HW solution ed to upinar. ٢٠
Reinforced Concrete Design
ENJOY THE COURSE!!
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Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 1 Introduction to reinforced concrete
Contents 1.
2.
Concrete-producing materials Mechanical properties of concrete 3.
Steel reinforcement
2
Part 1: Concrete-Producing Materials
3
Advantages of reinforced concrete as a structural material 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.
4
Advantages of reinforced concrete as a structural material 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.
5
Disadvantages of reinforced concrete as a structural material 1. It has a very low tensile strength. 2. Forms are required to hold the concrete in place until it hardens.
3. Concrete 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.
6
Compatibility of concrete and steel 1. Concrete is strong in compression, and steel is strong in tension. 2. The two materials bond very well together.
3. Concrete protects the steel from corrosive environments and high temperatures in fire. 4. The coefficients of thermal expansion for the two materials are quite close.
7
Concrete Concrete is a mixture of cement, fine and coarse aggregates, and water. This mixture creates a formable paste that hardens into a rocklike mass.
8
Concrete Producing Materials • • • •
Portland Cement Aggregates Water ixtures
9
Portland Cement The most common type of hydraulic cement used in the manufacture 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 ordinary cement. Concrete made with normal Portland cement require about two weeks to achieve a sufficient strength to permit the removal of forms and the application of moderate loads.
10
Types of Cement
Type I: General Purpose
Type II: Lower heat of hydration than Type I
Type III: High Early Strength • Quicker strength • Higher heat of hydration
11
Types of Cement 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. If the desired type of cement is not available, different ixtures may be used to modify the properties of Type 1 cement and produce the desired effect. 12
Aggregates Aggregates are particles that form about three-fourths of the volume of finished concrete. According to their particle size, aggregates are classified as fine or coarse.
Coarse Aggregates Coarse aggregates consist of gravel or crushed rock particles not less than 5 mm in size.
Fine Aggregates Fine aggregates consist of sand or pulverized rock particles usually less than 5 mm in size. 13
Water Mixing water should be clean and free of organic materials that react 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. 14
ixtures – Applications: • Improve workability (superplasticizers) • Accelerate or retard setting and hardening • Aid in curing • Improve durability
15
Concrete Mixing In the design of concrete mixes, three principal requirements for concrete are of importance: • Quality • Workability • Economy
16
Part 2: Mechanical Properties of Concrete
17
Mechanical Concrete Properties '
Compressive Strength, f c
• Normally, 28-day strength is used as the design strength.
18
Mechanical Concrete Properties '
Compressive Strength, f c
• It is determined through testing standard cylinders 15 cm in diameter and 30 cm in height in uniaxial compression at 28 days (ASTM C470). • Test cubes 10 cm × 10 cm × 10 cm are also tested in uniaxial compression at 28 days (BS 1881).
19
Mechanical Concrete Properties '
Compressive Strength, f c
• The ACI Code is based on the concrete compressive strength as measured by a standard test cylinder. f c Cylinder 0.8f c Cube
• For ordinary applications, concrete compressive strengths from 20 MPa to 30 MPa are usually used. 20
Mechanical Concrete Properties Compressive-Strength Test
21
Mechanical Concrete Properties Modulus of Elasticity, Ec ' • Corresponds to the secant modulus at 0.45 f c • For normal-weight concrete: Ec 4700 f c
22
0.002
ACI 8.5.1
0.003
Mechanical Concrete Properties Tensile Strength – Tensile strength ~ 8% to 15% of f c
'
– Tensile strength of concrete is quite difficult to measure with direct axial tension loads because of problems of gripping the specimen and due to the secondary stresses developing at the ends of the specimens. – Instead, two indirect tests are used to measure the tensile strength of concrete. These are given in the next two slides.
23
Mechanical Concrete Properties Tensile Strength – Modulus of Rupture, fr
f r 0.62 f c
ACI Eq. 9-10
– Modulus of Rupture Test (or flexural test): P
24
Mc 6M 2 fr I bh
unreinforced concrete beam
fr
Mechanical Concrete Properties Tensile Strength – Splitting Tensile Strength, fct
f ct 0.56 f c
ACI R8.6.1
– Split Cylinder Test Concrete Cylinder
P
Poisson’s Effect
f ct
2P Ld
25
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.
26
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
27
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 ordinary conditions about 90% of it occurs during the first year.
28
Part 3: Steel Reinforcement
29
Steel Reinforcement Tensile tests
30
Steel Reinforcement Tensile tests
31
Steel Reinforcement Stress-strain diagrams f s = ε Es ≤ f y Yield point
elastic
plastic
All steel grades have same modulus of elasticity Es= 2x105 MPa = 200 GPa
32
Steel Reinforcement Bar sizes, , # 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, fy ksi
MPa
40
276
60
414
80
552 33
Steel Reinforcement Bars are deformed to increase bonding with concrete
34
Steel Reinforcement Marks for ASTM Standard bars
35
Steel Reinforcement Bar sizes according to ASTM Standards U.S. customary units
36
Steel Reinforcement Bar sizes according to ASTM Standards SI Units
37
Steel Reinforcement Bar sizes according to European Standard (EN 10080) W
Number of bars
mm
N/m
1
2
3
4
5
6
7
8
9
10
6 8 10 12 14 16 18 20 22 24 25 26 28 30 32
2.2 3.9 6.2 8.9 12.1 15.8 19.9 24.7 29.8 35.5 38.5 41.7 45.4 55.4 63.1
28 50 79 113 154 201 254 314 380 452 491 531 616 707 804
57 101 157 226 308 402 509 628 760 905 982 1062 1232 1414 1608
85 151 236 339 462 603 763 942 1140 1357 1473 1593 1847 2121 2413
113 201 314 452 616 804 1018 1257 1521 1810 1963 2124 2463 2827 3217
141 251 393 565 770 1005 1272 1571 1901 2262 2454 2655 3079 3534 4021
170 302 471 679 924 1206 1527 1885 2281 2714 2945 3186 3695 4241 4825
198 352 550 792 1078 1407 1781 2199 2661 3167 3436 3717 4310 4948 5630
226 402 628 905 1232 1608 2036 2513 3041 3619 3927 4247 4926 5655 6434
254 452 707 1018 1385 1810 2290 2827 3421 4072 4418 4778 5542 6362 7238
283 503 785 1131 1539 2011 2545 3142 3801 4524 4909 5309 6158 7069 8042
Areas are in mm2
38
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 2 Load types, load paths and tributary areas
Load paths Structural systems transfer gravity loads from the floors and 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. 2
Metal Deck/Slab System s Floor Loads Above
Girders Joists
Joists Floor Deck
Columns Girders
The area tributary to a joist equals the length of the joist times the sum of half the distance to each adjacent joist.
The area tributary to a girder equals the length of the girder times the sum of half the distance to each adjacent girder.
Load paths loads on structural Load is distributed over the area of the floor. This distributed load
has units of (force/area), e.g. kN/m2. w {kN/m} q {kN/m2} Beam
Loads
Beam
Slab Column
Column
Beam
Beam
Beam Footing
Slab Beam
Beam Soil
6
P {kN}
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.
w {kN/m}
7
Load paths loads on (one-way) beams This tributary load is determined by multiplying q by the tributary
width for the beam.
w {kN/m} = q {kN/m2} (S1+S2)/2 {m}
L
8
S1
S2
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 s and the centerlines of the adjacent s parallel to the long sides. A is part of the slab formed by column centerlines.
9
Load paths loads on (two-way) beams An edge beam is bounded by s from one side. An interior beam is bounded by s from two sides. qD
For edge beams: D=S/2 qD
For interior beams: D=S 10
Load paths loads on (two-way) beams
11
Load paths loads on (two-way) beams
12
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.
P {kN} = q {kN/m2} (x y){m2}
13
Load paths loads on structural Example Determine 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. B1 4m B2
5m
4.5 m
14
C2
C1 6m
5.5 m
Load paths loads on structural Example B1:
w = 10 (4)/2 = 20 kN/m B1 4m B2 5m
4.5 m
15
C2
C1 6m
5.5 m
Load paths loads on structural Example B2:
w = 10 (4+5)/2 = 45 kN/m B1
4m B2 5m
4.5 m
C2 C1
16
6m
5.5 m
Load paths loads on structural Example B1:
w = 20 kN/m
B2:
w = 45 kN/m
17
Load paths loads on structural Example C1:
P = 10 (4.5/2 6/2) 5 = 337.5 kN B1
4m B2 5m
4.5 m
18
C1
C2 6m
5.5 m
Load paths loads on structural Example C2:
P = 10 [(4.5+5)/2 (6+5.5)/2] 5 = 1366 kN B1
4m B2 5m C2
4.5 m C1 19
6m
5.5 m
Load types Classification by direction
1 Gravity loads
2 Lateral loads
20
Load types Classification by source and activity
1 Dead loads 2 Live loads 3 Environmental loads 21
Loads on Structures All structural elements must be designed for all loads anticipated to act 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 22
Live Loads (L.L) The live load is a moving or movable type of load such as occupants, furniture, etc. Live loads used in deg buildings are usually specified by local building codes. Live loads depend on the intended use of the structure and the number of occupants at a particular time.
23
See IBC 2009 TABLE 1607.1 for more live loads. http://publicecodes.citation. com/icod/ibc/2009/index.ht m?bu=IC-P-2009000001&bu2=IC-P-2009000019
Minimum live Load values on slabs Type of Use Uniform Live Load kN/m2 Residential 2 Residential balconies Computer use Offices Warehouses
3 5 2
6
Light storage
Heavy Storage Schools
12
Classrooms Libraries
2
Rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
2.5 5 2.5
Retail
4
Wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
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 the type 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. 24
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 3 Design philosophies and design codes
Design Versus Analysis Design involves the determination of the type of structural system to be used, the cross sectional dimensions, and the required reinforcement. The designed structure should be able to resist all forces expected to act during the life span of the structure safely and without excessive deformation or cracking.
Analysis involves the determination of the capacity of a section of known dimensions, material properties and steel reinforcement, if any to external forces and moments.
2
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 . 2)Serviceability to ensure satisfactory performance under service load conditions, which implies providing adequate stiffness to contain 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. 3
Building Codes, Standards, and Specifications: 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
4
Building Codes, Standards, and Specifications: 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 International Conference of Building Officials (ICBO).
2.
BOCA National Building Code (NBC): published by Building Officials and Code s International (BOCA).
3.
Standard Building Code (SBC): published by Southern Building Code Congress International (SBCCI).
5
Building Codes, Standards, and Specifications: 3 Model Codes in the U.S.
6
Building Codes, Standards, and Specifications: International Building Code (IBC): published by International Code Council (2000 ,1st edition). To replace the 3 model codes for national and international use.
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.
7
Building Codes, Standards, and Specifications: Summary: The standards that will be used extensively throughout this course is Building Code Requirements for Reinforced Concrete and commentary, known as the ACI 318M-08 code. The building code that will be used for this course is the IBC 2009, in conjunction with the ANSI/ASCE7-02.
8
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, fc’ for concrete and f y for steel.
9
The Ultimate – Strength Design Method At the present time, the ultimate-strength design method is the method 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.
reduced strength provided factored loads
The factored loads are obtained by multiplying the working loads (service loads) by factors usually greater than unity. 10
Safety Provisions (the strength requirement) Safety is required to insure that the structure can sustain all expected loads 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 fc’ and fy, *assumptions are made during 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 the debris and replacement of the structure and its contents and *cost to society. 11
Safety Provisions (the strength requirement) The strength design method, involves a two-way safety measure. The first 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 a reduction factor is usually smaller than unity Design strength ≥ Factored loads
R i Li i
ACI 9.3
ACI 9.2 12
Load factors ACI 9.2.1 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 13
Load factors ACI 9.2 Symbols
14
Strength Reduction Factors
ACI 9.3 According to ACI, strength reduction factors Φ are given as follows: a. For tension-controlled sections Φ = 0.90 b. For compression-controlled Φ = 0.75 sections, with spiral Φ = 0.65 reinforcement Other reinforced Φ = 0.75 c. For shear and torsion
Tension-controlled section
compression-controlled section
15
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 4 Analysis of beams in bending at service loads
Introduction A beam is a structural member used to the internal moments and shears and in some cases torsion.
2
Basic Assumptions in Beam Theory •Plane sections remain plane after bending. This means that in an initially straight beam, strain varies linearly over the depth of the section after bending.
Unloaded beam
Beam after bending
Its cross section
Strain distribution
3
Basic Assumptions in Beam Theory •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.
4
Stages of flexural behavior w {kN/m}
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 5
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.
6
Stage II: Concrete cracked –Elastic Stress range Once the tensile stresses reach the modulus of rupture, the section cracks. The bending moment at which this transformation takes place is called the cracking moment Mcr.
7
Stage III: Beam failure –Ultimate Strength stage As the stresses in the concrete exceed the linear limit (0.45 fc’), the concrete stress distribution over the depth of the beam varies non-linearly.
8
0.002 0.003
Stages of flexural behavior w {kN/m}
9
Flexural properties to be determined: 1Cracking moment. 2 Elastic stresses due to a given moment. 3Moments 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. 10
Cracking moment Mcr 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 the uncracked section properties.
11
Cracking moment Mcr Example 1: Calculate the cracking moment for the section shown 1 3 I g bh 12 3 10 4 1 I g (350)(750) 1.230510 mm 12 fr 0.62 fc 0.62 30 3.4MPa M cr
frI g yt
750 mm 1500 mm2
fc 30MPa
3.41.23051010 1.1143108 N.mm 111.43kN.m (750 / 2) 12
Elastic stresses – Cracked section • After cracking, the steel bars carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load. • In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where
n = the modular ratio= Es/Ec • To determine the location of the neutral axis, bx x n Asd x 0 2
1 b x2 2
n As x n Asd 0
• The height of the concrete compression block is x.
• The normal stress in the concrete and steel fc
My It
fs n
My It
13
Elastic stresses – Cracked section Example 2: fc 30MPa
Calculate the bending stresses for the section shown, M= 180 kN.m Note: M > Mcr = 111 kN.m from previous example. Thus, section is cracked.
750 mm 1500 mm2
E c 4700 f c 4700 30 25743MPa 5 E 210 s n 7.77 E c 25743 x ( 350) x ( ) 1500( 7.77)( 700 x ) 2 x 185.16mm
14
Elastic stresses – Cracked section Example 2: 1 I t bx 3 nAs ( d x ) 2 3 1 I t ( 350)( 185.16) 3 7.771500( 700 185.16) 2 3 9 4 750 mm I t 3.829510 mm
My 180106 185.16 fc 8.7MPa 9 It 3.829510 f c 8.7MPa 0.45f c 0.45( 30) 13.5MPa
fc 30MPa
1500 mm2
OK
My 180106 ( 700 185.16) f s n 188MPa 7.77 9 It 3.829510
15
Elastic stresses – Cracked section Example 3: fc 30MPa
Calculate the allowable moment for the section shown, f s(allowable)= 180 MPa, f c(allowable)= 12 MPa f s It 1803.829510 9 Ms ny ( 7.77)( 700 185.16)
750 mm 1500 mm2
8
M s 1.723410 N .mm 172.34kN .m f c I t 12 3.8295109 Mc y 185.16 8
M c 2.481910 N .mm 248.19kN.m M allowable 172.34kN.m
16
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 5 Strength analysis of beams according to ACI Code
Strength requirement for flexure in beams
Md Mu M d Design moment strength (also known as moment resistance) M u Internal ultimate moment M u 1.2M D 1.6M L
Md Φ Mn M n Theoretical or nominal resisting moment. 2
The equivalent stress (Whitney) block
Strain Distribution
Actual Stress Distribution
Approximate Stress Distribution
3
The equivalent stress (Whitney) block •The shape of the stress block is not important. •However, the equivalent block must provide the same resultant (volume) acting at the same location (centroid). •The Whitney block has average stress 0.85fc’ and depth a=1c. ACI 10.2.7.1 4
The equivalent stress (Whitney) block
The equivalent rectangular concrete stress distribution has what is known as the 1 coefficient. It relates the actual NAdepth to the depth of the compression block by a=1c. for f c ' 28 MPa 1 0.85 ACI 10.2.7.1
1 0.85 0.05( fc '28) 0.65 for f c ' 28 MPa 7
5
Derivation of beam expressions
Fx=0
C =T
6
Derivation of beam expressions
7
Derivation of beam expressions Design aids can also be used: Assume Md = Mu = ΦMn
= Rn
Mn=R nbd2
Rn is given in tables and figures of design aids.
8
Design Aids
9
Design Aids
10
Tension strain in flexural
y
f y
Es
t y ?
Strain Distribution 11
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
Types of flexural failure: [1] Balanced Failure The concrete crushes and the steel yields simultaneously. εcu=0.003 cb d
h
b
εt = εy
Such a beam has a balanced reinforcement, its failure mode is
brittle, thus sudden, and is not allowed by the ACI Strength Design Method. 13
Types of flexural failure: [2] Compression Failure The concrete will crush before the steel yields. εcu=0.003 εcu=0.003
c>cb c>cb h
d
h
b
b
d
ε <ε εt < εt y y
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. 14
Types of flexural failure: [3] Tension Failure The reinforcement yields before the concrete crushes. εcu=0.003 εcu=0.003
c
h
d
d
bb
Such a beam is called under-reinforced beam, and its failure mode is
ductile,
thus giving a sufficient amount of warning before collapse, and is required by the ACI Strength Design Method 15
Types of sections according to the ACI Code ACI 10.3.4 [1] Tension-controlled section 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 εy (εy = fy/Es=0.002 for f y =420 MPa) when the concrete in compression reaches 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 εy (εy =fy/Es=0.002 for f y =420 MPa) when the concrete in compression reaches its crushing strain of 0.003. 16
Allowed strains for sections in bending
ACI 10.3.5
17
Strength reduction factor Φ εcu=0.003 c d
εt
t
d c c
c
ACI R9.3.2.2
18
Balanced steel cb
0.003 d 0.003 f y E S 5
Es 210 MPa
cb
600 d 600 f y =1c
0.85 1 f c ' 600 b 600 f y fy
19
Maximum allowed steel
0.003 d 0.003 0.005 3 d 8
c max c max
=1c 1c max
max
3 d 1 8
3 0.85 1 f c ' 8 fy 20
Minimum steel allowed ACI 10.5.1
A s,min
0.25 f c bw d fy max 1.4 b d w f y
bw = width of section d = effective depth of section
21
Design Aids
22
Summary: To calculate the moment capacity of a section: 0.25 f c bw d fy max 1.4 b d w f y
1-) As,min
if As,min > As,sup reject section
2-) a
3-)
As f y 0.85f c b
1 0.85
or a
df y 0.85f c
for f c ' 28 MPa
1 0.85 0.05( fc '28) 0.65 7
a c 4-) 1
for fc' 28 MPa 23
Summary: d c
5-) t c 0.003 if t> 0.005: tension controlled = 0.9 if 0.004 < t <0.005: in transition zone =0.65+( t -0.002) (250/3)
if t < 0.004: compression controlled reject section d a 6-) M d M n As f y 2 or Mn=Rnbd2 (find Rn from table)
7-) M u ΦM n
if not reject section
24
Example A singly reinforced concrete beam has the cross-section shown in the figure below. Calculate the design moment strength. Can the section carry an Mu = 350 kN.m?
f y 414MPa a) f c 20.7MPa, b) f c 34.5MPa, c) f c 62.1MPa
25
Example Solution a) f c 20.7MPa 1 A s,min
0.25 f c 0.25 20.7 bw d (254)(457)=319 mm2 fy 414 max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=393 mm
2 a
As f y 0.85f cb
3 1 0.85
2
2
< A s,sup =2580 mm OK
2580 414 239mm 0.85 20.7 254
for f c ' 20.7MPa 28 MPa 26
Example Solution a) f c 20.7MPa 4 c
a
1
239 281mm 0.85
d c 457 281 0.003 0.00186 5 t 0.003 c 281 t 0.004 Section is compression controlled ==> Does not satisfy ACI requirements ==> Reject section
27
Example Solution b) f c 34.5MPa 1 A s,min
0.25 f c 0.25 34.5 bw d (254)(457)=412 mm2 fy 414 max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=412 mm
2 a
As f y 0.85f c b
2
2
< A s,sup =2580 mm OK
2580 414 143.4mm 0.8534.5 254
0.05( f c ' 28) 0.65 for f c ' 34.5MPa 28 MPa 7 1 0.85 0.05( 34.5 28) 0.804 0.65 7
3 1 0.85
28
Example Solution b) f c 34.5MPa 4 c
a
1
143.4 178.5mm 0.804
d c 457 178.5 0.003 0.00468 5 t 0.003 c 178.5 0.004 t 0.005 Section is in transision zone
=0.65+(t -0.002)(250/3) =0.65+(0.00468-0.002)(250/3)=0.874
29
Example Solution b) f c 34.5MPa
d a 6 M d M n As f y 2 143.4 6 0.874 2850 414 457 360 10 N .mm 2 360kN .m 7 M u 350kN .m ΦM n 360kN .m Section is adequate
39
Example Solution c) f c 62.1MPa
1 A s,min
0.25 f c 0.25 62.1 2 b d (254)(457)=552 mm w f 414 y max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=552 mm
2 a
As f y
0.85f cb
2
2
< A s,sup =2580 mm OK
2580 414 80mm 0.85 62.1254
31
Example Solution c) f c 62.1MPa 0.05( f c ' 28) 3 1 0.85 0.65 for f c ' 62.1MPa 28 MPa 7 1 0.85 0.05( 62.1 28) 0.61 0.65 7 1 0.65
4 c
a
1
80 123mm 0.65
d c 457 123 0.003 0.0081 5 t 0.003 c 123 t 0.005 Section is tension controlled ==> Satisfes ACI requirements ==> =0.9
32
Example Solution c) f c 62.1MPa
d a 6 M d M n As f y 2 80 0.9 2850 414 457 520 106 N .mm 2 520kN .m 7 M u 350kN .m ΦM n 520kN .m Section is adequate
33
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 6 Design of singly reinforced rectangular beams
Design of Beams For Flexure The main two objectives of design is to satisfy the: 1) Strength and 2) Serviceability requirements 1) Strength M d Φ M n Mu M d Design moment strength (also known as moment resistance)
M u Internal ultimate moment Mn
Theoretical or nominal resisting moment.
M u 1.2M D 1.6M L 2
Design of Beams For Flexure Derivation of design expressions
h
d As
Assume ΦMn = Mu
b Beam cross section
Solve for 0.85 f c' ρ fy
2M u 1 1 2 0.85f ' b d c
: 1 kN.m = 106 N.mm
A s = bd 3
Design of Beams For Flexure Design aids can also be used:
0.85 f c' ρ fy
2M u 1 1 2 0.85f ' b d c
Calculate: Then is found from tables and figures of design aids.
4
Design Aids
5
Design of Beams For Flexure 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
hmin l = span length measured center to center of .
h hmin
h
d As b Beam cross section
7
Detailing issues: Concrete Cover 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.
Side cover
ACI 7.7.1
Bottom cove
8
Detailing issues: Spacing of Reinforcing Bars •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 Smin must be satisfied,where:
S min
bar diameter, d b ACI 7.6.1 max 25 mm 4/3 maximum size of coarse aggregate
ACI 3.3.2 •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. ACI 7.6.2
Clear distance
Clear spacing S
9
Estimation of applied moments Mu Beams are designed for maximum moments along the spans in both negative and positive directions.
10
Positive moment
Negative moment
Tension at bottom Needs bottom reinforcement
Tension at top Needs top reinforcement
Estimation of applied moments Mu 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.
Simply ed Beams
Continuous Beams
Determinate
Indeterminate
– + 11
Moment Diagram
+
Moment Diagram
Estimation of applied moments Mu Simply ed Beams
Continuous Beams
– +
Moment Diagram Section at midspan
12
+
Moment Diagram Section over
Estimation of applied moments Mu Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate moments for design 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. • are of similar section dimensions along their lengths (prismatic).
13
Estimation of applied moments Mu Approximate Structural Analysis More than two spans
1 4
ACI 8.3.3
Estimation of applied moments Mu Approximate Structural Analysis Two spans
l n = length of clear
span measured face-to-face of s. For calculating negative moments, l n is taken as the average of the adjacent clear span lengths.
15
ACI 8.3.3
Design procedures Method 1: When b and h are unknown 1Determine 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. 3Assume that Φ=0.9 and calculate the reinforcement (ρ and As).
4- Check solution: (a) (b) (c) (d)
Check spacing between bars Check minimum steel requirement Check Φ = 0.9 (tension controlled assumption) Check moment capacity (Md ≥ Mu ?)
5- Sketch the cross section and its reinforcement. 16
Design procedures Method 2: When b and h are known 1Calculate the factored load wu and bending moment Mu.
2Assume that Φ=0.9 and calculate the reinforcement (ρ and As). 3- Check solution: (a) (b) (c) (d)
Check spacing between bars Check minimum steel requirement Check Φ = 0.9 (tension controlled assumption) Check moment capacity (Md ≥ Mu ?)
4- Sketch the cross section and its reinforcement.
17
Example 1 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 a service live load of 10 kN/m are to be ed. wd=25 kN/m & wl =10 kN/m Use fc’ =25 MPa and fy = 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
6m
wu=50.5 kN/m 6m
2 Calculate wu and Mu wu = 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
227.3 kN.m
18
Example 1 3- Assume that Φ=0.9 and calculate ρ and As d = 500 – 40 – 8 – (20/2) = 442 mm (assuming one layer of Φ20mm reinforcement and Φ8mm stirrups) ρ
0.85fc ' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(25) ρ 420
2 227.3106 1 1 2 (0.9) 0.85(25) 300 (442)
0.0116
As = ρ b d = 0.0116(300)(442) =1536 mm2 Use 5 Φ 20 mm (As,sup=1571 mm2)
19
W
Number of bars
mm
N/m
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
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
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
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
20
Example 1 4- Check solution a) Check spacing between bars sc
5Φ20
300 2 40 28 5 20 26 mm d b 20 mm 51 25 mm
300
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 b d (300)(442)=395 mm2 w fy 420 max 1.4 1.4 bw d (300)(442)=442 mm2 fy 420 =442 mm
2
2
< A s,sup =1571 mm OK 21
Example 1 c) Check Φ =0.9 (tension controlled assumption) a
As f y 0.85 f c 'b
1 0.85
1571420 103.5mm 0.85(25)300
for f c ' 25MPa 28 MPa c
a 103.5 121.7 mm β1 0.85
dc 442 121.7 0.003 0.0079 0.005 ε t 0.003 c 121.7 for ε t 0.005 Φ 0.90, the assumption is true the section is tension controlled
d) Check moment capacity a M d ΦA sf y d 2
103.5 6 0.901571 420 442 231.710 N.mm = 231.7 kN.m 2 M d 231.7 kN.m M u 227.3 kN.m OK
22
Example 1 5- Sketch the cross section and its reinforcement
44.2
50 5Φ20
30 Beam cross section
23
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 . Columns have a cross section of 800x300 mm. The factored distributed load over the slab is qu =14.4 kN/m2. Use fc’ =25 MPa and fy = 420 MPa. L1 = L2 = L3 = 6 m S 1 = S 2= S 3 = 4 m B1
Solution: b & d are known 1- Calculate wu and Mu wu=4(14.4) = 57.6 kN/m ln = 6 – 0.3=5.7 m wu 24
Example 2 Moment diagram using the approximate ACI method:
Design for the maximum negative moment throughout the beam:
Mu = wu(ln )2/10 = 57.6 (5.7)2/10 Mu = 187.5 kN.m
25
Example 2 2- Assume Φ=0.9 and calculate ρ and As d = 316 – 40 – (16/2) – 8 = 260 mm (assuming one layer of Φ16 mm reinforcement and Φ8mm stirrups)
ρ
0.85fc' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(25) ρ 420
2187.5106 1 1 0.0102 2 (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) 26
Example 2 3- Check solution a) Check spacing between bars
sc
800 2 40 28 1116 52.8 mm 111
d b 16 mm 25 mm
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 bw d (800)(260)=620 mm2 fy 420 max 1.4 1.4 bw d (800)(260)=693 mm2 fy 420 2
=693 mm2 < A s,sup =2212 mm OK 27
Example 2 c) Check Φ =0.9 a
As f y
0.85 f c 'b
1 0.85
2212 420 55mm 0.85(25)800
for f c ' 25MPa 28 MPa c
55 a 64 mm β1 0.85
dc 260 64 0.003 0.0091 0.005 ε t 0.003 c 64 for ε t 0.005 Φ 0.90, the assumption is true the section is tension controlled
d) Check moment capacity a M d ΦA sf y d 2 55 0.9 2212 420 260 194.5106 N.mm=194.5 kN.m 2 M d 194.5 kN.m M u 187.5 kN.m OK
28
Example 2 4- Sketch the cross section and its reinforcement
11Φ16
316
260
800
29
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 7 Design of T and Lbeams
T Beams Reinforced concrete systems may consist of slabs and dropped 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 ing, and the part below the slab is called the web or stem.
Flange web
2
Flange Width b Parts of the slab near the webs are more highly stressed than areas away from the web.
effective flange width be
effective flange width be
hf
d
stirrup bw
bw
L-beam
3
T-beam
d: effective depth. hf : height of flange. bw : width of web. be : effective width. b: distance from center to center of adjacent web spacings
Effective Flange Width be be is the width that is stressed uniformly to give the same compression force actually developed in the compression zone of width b.
4
Effective Flange Width be ACI Code Provisions for Estimating be
ACI 8.12.2 According to the ACI code, the effective flange width of a T-beam, be is not to exceed the smallest of: 1. One-fourth the span length of the beam, L/4. 2. Width of web plus 16 times slab thickness, bw +16 hf . 3. Center-to-center spacing of beams, b.
b eff
5
L/4 min bw +16hf b
Effective Flange Width be ACI Code Provisions for Estimating be
ACI 8.12.3 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). b eff
6
bw L /12 min bw 6hf b 0.5b w c
A T-beam does not have to look like a T
7
Various Possible Geometries of T-Beams Single Tee
Double Tee
Box
8
Various Possible Geometries of T-Beams
Flange
Flange
web
web
Same as
9
T- versus Rectangular Sections 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 usually neglected in ultimate strength design, the sections are treated as 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 Tsection.
Compression zone
–
+ 11
Tension zone
+
Moment Diagram
Section at midspan Positive moment
Section at Negative moment
Analysis of T-beams Case 1: when a ≤ hf
[Same as rectangular section]
TC Asf y a 0.85f c b e
12
a ΦM n ΦA sf y d 2
Analysis of T-beams Case 2: when a > hf C f 0.85f c be bw hf Cw 0.85 f c bw a T As f y
From equilibrium of forces T C f C w a
As f y 0.85f c be bw hf
0.85f c bw
a h ΦM n Φ C w d C f d f 2 2 13
Minimum Reinforcement, As,min ACI 10.5.2
hf As
be
As
hf
bw
14
d
bw
-ve Moment
A s,min
0.25 f c bw d fy max 1.4 b d w f y
+ve Moment
be
d
Analysis procedure for calculating he ultimate strength of T-beams To calculate the moment capacity of a T-section: 1 Calculate be 2 Check As,sup> As,min 3 Assume a ≤ hf and calculate a using: a
Asf y 0.85fc b e
If a ≤ hf → a is correct If a > hf → a
A s f y 0.85f c be bw hf 0.85f c bw
4- Calculate 1, c, and check εt 5- Calculate ΦMn, and check 15
M u ΦM n
Example 1 Calculate Md for the T-Beam: hf = 150 mm
d = 400 mm
As = 5000mm2
fy = 420MPa fc’= 25MPa bw= 300mm
L = 5.5m
b=2.15m Determine be according to ACI requirements
L 5500 1370mm 4 4 be min 16hf b w 16 150 300=2700mm b 2150 mm
16
Example 1 Check min. steel 0.25 f ' 1.4 c b d ; b A s,min max w w d max fy f y 2 5000 mm 2 OK A s,min 400 mm A
0.25 25 300 400 ; 1.4 300400 420 420
s,sup
Calculate a (assuming a
c
Asf y 0.85f c b e a
1
5000420 71.9mm hf 0.85251370
150mm OK
71.9 85 mm 0.85
s d c 0.003 400 85 0.003 0.011 0.005 Tension controled c 85 17
Example 1 Calculate Md a M d As f y d 2 71.8 0.9 5000420 400 2 688106 N.mm 688 kN.m
18
Example 2 Determine the ACI design moment strength Md (ΦMn) of the T-beam shown in the figure if fc’ =25 MPa and fy = 420 MPa. 10
90
Φ10
h= 75
Solution:1- Check min. steel
25 655.5mm 8Φ32 2 0.25 f ' 1.4 c bw d A s,min max bw d ; fy f y 0.25 25 1.4 A s,min max 300 655.5 ; 300 655.5 420 420 2 2 656 mm A 6434 mm OK A s,min
d 750-40-10-32-
s,sup
19
30
Example 2 2- Check if a < hf = 10cm Asf y
6434 420 141.3mm 0.85fc b e 0.85 25 900
a= 141.3> hf = 100 mm i.e. assumption is wrong Section is T NA is in the web
20
10
90
Φ10
h= 75
a
8Φ32 30
Example 2 3- Calculate 1, c, and check εt a a
c
A s f y - 0.85f c be bw hf 0.85f c bw 6434 420 0.85 25900 300100 0.85 25 300
224mm
a 224 264 mm β1 0.85
d c 0.003 655.5 264 0.003 εt 264 c ε t 0.00447 0.004 0.005 Transision zone
=0.65+(t -0.002)(250/3) =0.65+(0.00447-0.002)(250/3)=0.855 21
Example 2 4- Calculate Md Cf 0.85fc' (be b w ) h f 0.85 25900 300100 127510 3N 1275 kN 3
Cw 0.85fc' a b w 0.85 25 224 300 1427.4 10 N 1427.4 kN M d Φ C w
a d C f 2
hf d 2
100 224 3 0.855 1427.4 10 3 655.5 127510 655.5 2 2 1323.4106 N .mm 1323.4 kN .m
22
Design of T-Beams --- Positive moment
+
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 Muf and (2) Asw, the remaining of the steel, providing a section with capacity Muw.
M u M uf M u w
23
Mu : Ultimate moment applied, requiring steel As. Muf : Moment resisted by overhanging flange parts, requiring steel Asf. Muw : Moment resisted by web, requiring steel Asw.
Design of T-Beams --- Positive moment
+
Step 1
24
24
Step 2
Design of T-Beams --- Positive moment
+
M u M uf M u w
Muw Mu Muf
0.85 f c ' 1 1 fy
25
2M uw 0.85f c ' bw d 2
Step 3
Step 4
Asw bw d
Step 5
As Asf Asw
Step 6
Design of L-Beams --- Positive moment be
be
Same as bw
26
Design of T-Beams --- Negative moment be
bw
Design as a rectangular section with width bw
27
Flange Reinforcement When flanges of T-beams are in tension, part of the flexural reinforcement shall be distributed over effective flange width, or a width equal to one-tenth of the span, whichever is smaller
-ve moment
Additional Reinforcement
min (beff & l/10)
Additional Reinforcement
Main Reinforcement
If beff > l/10, some longitudinal reinforcement shall be provided in outer portions of flange.
Design of T-Beams --- Positive moment Design Procedure: 1 Establish h based on serviceability requirements of the slab and calculate d
2 Choose bw 3 Determine be according to ACI requirements. 4 Calculate As assuming that a < hf with beam width = be & Φ=0.90 b 2M u 1 1 hf 2 d Φ 0.85f ' b d c e As As fy As = ρ be d → a bw 0.85f c ' b e 5 If a ≤ hf: the assumption is right continue as rectangular section If 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
0.85f c' ρ fy
29
e
Example 3
fy = 420 MPa.
Lm
A floor system consists of a 14.0cm concrete slab ed by continuous T-beams with a span L. Given that bw=30cm and d=55cm, fc’ =28 MPa and Determine the steel required at midspan 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 Spandrel (B) L = 2 m beam
3.0 m
3.0 m
Slab
hf
bw
39
3.0 m
Solution (A) L = 8 m Determine be according to ACI requirements
784 kN.m
200
L 8000 4 4 2000mm be min 16hf bw 16 140 300=2540mm b 3000 mm
14 As 30
be is taken as 2000 mm, as shown in the figure
Calculate As assuming that a < hf with beam width = be & Φ=0.90 Mu = 1.2(320)+1.6(250)=784 kN.m
0.85f c ' fy 31
2M u 11 0.85f ' b d 2 c e
55
Solution (A) L = 8 m 0.85 28 278410 ρ 1 1 2 420 0.90.85282000550 0.00354 6
As ρ be d 0.00354 2000 550 3892 mm
784 kN.m
200
14 As 30
2
Check a ≤ hf assumption a
As f y 0.85f c 'b e
3892420 34.3mm h f 140mm 0.85 28 2000
The assumption is right Rectangular section design
Use 8Φ25mm (As,sup= 3927 mm2) arranged in two layers. sc 32
300 2 40 28 4 25 34.5 mm d b 25 mm 41 25 mm
OK
55
Solution (A) L = 8 m Check the Φ=0.90 assumption (εt ≥0.005) and As,sup ≥ As,min A s ,min
A s,min
0.25 f ' 1.4 c b d ; b max w w d max fy f y 550 mm 2 A 3927 mm 2 OK
0.25 28 300 550 ; 1.4 300 550 420 420
s,sup
200
As f y
55
14
3927 420 a 34.7 mm 0.85 f c 'b e 0.85 28 2000
a 34.7 40.8 mm β1 0.85 550 40.8 dc εt 0.003 0.003 40.8 c 0.0374 0.005 0.9 OK c
33
8Φ25 30
Solution (A) L = 8 m Check moment capacity a M d As f y d 2 34.7 0.93927 420 550 2 6
Md 790.710 N.mm 790.7 kN.m M u 784kN.m
55
14
200
8Φ25 30
34
Solution (B) L = 2 m 784 kN.m
50
Determine be according to ACI requirements L 2000 500mm 4 4 be min 16hf bw 16 140 300=2540mm b 3000 mm
14 As 30
be is taken as 500 mm, as shown in the figure Calculate As assuming that a < hf with beam width = be & Φ=0.90
Mu = 1.2(320)+1.6(250)=784 kN.m
2M u 0.85f c ' 1 1 0.85f ' b d 2 fy
35
c
e
55
Solution (B) L = 2 m 0.85 28 278410 ρ 1 1 2 420 0.90.8528500550 0.0159 A ρ b d 0.0159500 550 4389 mm 2 6
s
784 kN.m
50
e
Check a ≤ h assumption f
a
As f y 0.85f c ' be
4389 420 155mm > h f 140mm 0.8528500
The assumption is wrong T section design
36
14 As 30
55
Solution (B) L = 2 m 14
50
55
Calculate required reinforcement A sf
0.85f c '( b bw ) hf fy
A sf
0.85( 28)( 500 300 )140 1586mm 2 420
30
h M uf A s f y d f 2 140 6 0.91586 420 550 28810 N .m 2 6
6
6
M uw M u M u f 78410 28810 49610 N .m 37
Solution (B) L = 2 m
0.85f c ' 1 1 fy
0.85( 28) ( 420)
sw
2( 496) 106 1 1 2 0.9 0.85( 28) ( 300) ( 550)
0.017
50
w
As Asf Asw 1586 2808 4395mm
14
b d 0.017( 300)( 550) 2808mm 2 55
A
2M u 0.85f c ' bw d 2
2
8Φ28 30
Use 8Φ28 mm (As,sup= 4926mm2) arranged in two layers. Check solution: (Do as in Example 2)
38
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 8 Design of doubly reinforced beams
Doubly Reinforced Rectangular Sections Beams having steel reinforcement on both the tension and compression sides are called doubly reinforced sections. Doubly reinforced 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. 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:
Mn
Mn2
Mn1
To analyze the section, the tension steel is divided in two portions: (1) As2, which is in equilibrium with the compression steel, and providing a section with capacity Mn2 and (2) As1, the remaining of the tension steel, providing a section with capacity Mn1. 4
Analysis of Doubly Reinforced Rectangular Sections Find As1 and As2:
T s 2 Cs As 2f y Asf s As 2 5
Asf s fy
We need f s’ to find A
As A s1 As 2 A s1 As As 2
s2
Analysis of Doubly Reinforced Rectangular Sections Find fs’: s c d 0.003
c
c d f s s Es 0.003Es f y c 5
E s 2 10 MPa
6
c
Analysis of Doubly Reinforced Rectangular Sections Find c:
T C c Cs As f y 0.85f cab Asf s
7
c d As f y 0.85f c1cb As 0.003Es c find c by solving the quadratic equation find fs’ from equation in slide 6
Analysis of Doubly Reinforced Rectangular Sections Find Md:
8
M d M n A s 1f y
d - a A f d - d ' s s 2
Analysis of Doubly Reinforced Rectangular Sections Procedure: 1) 2)
c d As f y 0.85f c1cb As 0.003Es c c d fs 0.003Es f y c
find c, a
3) A s 2 Asf s
fy 4) As 1 As As 2
5) 6) 9
d c Check if = 0.9 s c 0.003 0.005? a M d M n A s 1f y d - A s f s d - d ' 2
Example 1 For the beam with double reinforcement shown in the figure, calculate the design moment Md. 5.0 2Φ25 fc’ =35MPa and fy = 420 MPa. 60 6Φ32
Solution:1 0.85 0.05( f c ' 28) 0.65 for f c ' 35MPa 28 MPa 7 1 0.85 0.05( 35 28) 0.8 0.65 7 c d As f y 0.85f c 1cb As 0.003Es c 10
c 50 5 4825(420) 0.85(35)(0.8)c(300) 982 0.003(210 ) c
30
Example 1 c 50 5 4825(420) 0.85(35)(0.8)c(300) 982 0.003(210 ) c 229.5c 2 1437300c 29460000 0 5.0 c 220mm
2Φ25
a 1c 0.8 220 176mm
60 6Φ32
c d f s 0.003Es f y c 220 50 5 f s 0.003(210 ) 463 f 220 f s f y 420 11
30 y
420MPa
Example 1
5.0 2Φ25
Asf s 982(420) 982mm 2 fy (420)
As 2
A s1 As As 2 4825 982 3843mm
60 6Φ32
2
d c s 0.003 0.005? c
30
s 600 220 0.003 0.0052 0.005 Tension Controlled , 0.9
220
M d M
n
A s 1f y
d - a A f d - d ' s s 2
176 M d 0.9 3843( 420) 600 982( 420) 600 50 2 12
M
6
d
94810 N .mm 948kN .m
Maximum allowed steel for a singly reinforced section
0.003 d 0.003 0.005 3 d 8
c max c max
=1c 1c max
3 0.85 1 f c ' max 8 fy 3 0.85 1f c ' bd A s ,max 8 fy
3 d 1 8
13
Design of Doubly Reinforced Rectangular Sections
1) Design the section as singly reinforced, and calculate t 2) If t < 0.004 Comp. steel is needed (or enlarge section if possible)
3) Design As1 for maximum reinforcement (slide 13) and find Mn1, a, c 4) M n M u 5) Mn2 = Mn – Mn1 c d 6) f s sEs 0.003Es f y c Asf s Mn2 A s2 7) A s fy f s(d d )
As As 1 A s 2
14
Example 2 Design the beam shown in the figure to resist Mu=1225 kN.m. If compression steel is required, place it 70 mm from the compression face. fc’ =21 MPa and fy = 420 MPa. Solution: Try first to design the section as a singly reinforced section: ρ
0.85fc ' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(21) ρ 420
21225106 1 1 2 (0.9) 0.85(21)350 (700)
As= ρ b d = 0.0284(350)(700) = 6947 mm2 15
Use 10 Φ32 mm in two rows (As,sup =7069 mm2)
0.0284
Example 2 Check the ductility of the singly reinforced section: a
As f y 0.85f c ' b
7069 420 475mm 0.85(21)350
c
a
1
475 559mm 0.85
dc 700 559 0.003 0.00076 0.004 ε t 0.003 c 559 Section is brittle! can not be used. Use compression reinforcement. Mn
Mu
A s 1 A s ,max 16
1225 1361kN .m 0.9
3 0.85 1f c ' 8 fy
A s 1 3307mm 2
3 0.85( 0.85)( 21) ( 350)( 700) bd 8 ( 420)
Example 2 As f y
3307( 420) a 222.3mm 0.85f cb 0.85( 21)( 350)
c
a
1
222.3 261.55mm 0.85
a 222.3 M n1 A s f y d - ( 3307)( 420)( 700 ) 2 2 6
M n1 81810 N .mm 818kN .m Mn2 = Mn – Mn1 = 1361 – 818 = 543 kN.m
17
Example 2 c d f s 0.003Es f y c 261.55 70 5 f s 0.003(210 ) 439MPa f 261.55 f s f y 420
y
420MPa
Mn 2 543106 A s 2052mm 2 f s(d d ) 420(700 70) As 2
Asf s (2052)(420) 2052mm 2 fy (420)
A s A s 1 A s 2 3307 2052 5359mm 2 2
Use 830 in two rows for tension steel (A s,sup = 5655 mm ) 2
18
Use 426 for compression steel (A s,sup = 2124 mm )
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 9 Design of beams for shear
Shear Design vs Moment Design Beams are usually designed for bending moment first. Accordingly, cross sectional dimensions are determined along with the required amounts of longitudinal reinforcement.
Once this is done, sections are checked for shear to determine whether shear reinforcement is required or not. 2
Shear Design vs Moment Design This by no means indicates that shear is less important than bending. On the contrary, shear failure which is usually initiated by diagonal tension, is far more dangerous than flexural failure due to its brittle nature. It occurs without warning. Therefore, beams are designed to rather fail in bending. This is done by providing larger safety factor against shear failure than those provided for bending.
3
Shear and flexural stresses In linearly elastic beams, two types of stresses occur:
Flexural stresses:
Shear stresses: An element of a beam not on the NA or an extreme fiber is subjected to both stress types combined 4
Shear and flexural stresses The combined stress (called principal stresses) are calculated as:
which act on a direction inclined with respect to the beam axis by the angle:
5
Shear and cracks in beams
6
Shear and cracks in beams
7
7
Types of Shear Cracks Two types of inclined cracking occur in beams:
1- Web Shear Cracks Web shear cracking begins from an interior point in a member at the level of the centroid of the uncracked section and moves on a diagonal path to the tension face when the diagonal tensile stresses produced by shear exceed the tensile strength of concrete.
2- Flexure-Shear Cracks The most common type, develops from the tip of a flexural crack at the tension side of the beam and propagates towards mid depth until it reaches the compression side of the beam. 8
Shear and cracks in beams It is concluded that the shearing force acting on a vertical section in a reinforced concrete beam does not cause direct rupture of that section. Shear by itself or in combination with flexure may cause failure indirectly by producing tensile stresses on inclined planes. If these stresses exceed the relatively low tensile strength of concrete, diagonal cracks develop. If these cracks are not checked, splitting of the beam
or what is known as diagonal tension failure will take place.
9
Failure by shear in beams
10
Types of Shear Reinforcement The code allows the use of three types of Shear Reinforcement • Vertical stirrups • Inclined stirrups • Bent up bars Inclined Stirrups
Bent up bars
11
Vertical Stirrups
Deg to Resist Shear The strength requirement for shear that has to be satisfied is:
ΦVn Vu
ACI Eq. 11-1
Vu = factored shear force at section Vn = nominal shear strength Φ = strength reduction factor for shear = 0.75
The nominal shear force is generally resisted by concrete and reinforcement: Vn Vc Vs ACI Eq. 11-2 Vc = nominal shear force resisted by concrete Vs = nominal shear force resisted by shear reinforcement 12
shear
Strength of Concrete in Shear For subject to shear Vu and bending Mu only, ACI Code gives the following equation for calculating Vc
Simple formula
Vc 0.17 f c ' bw d
ACI Eq. 11-3
Detailed formula Vu d bw d 0.29 fc ' bw d Vc 0.16 f c ' 17 w Mu
13
As where w b wd
ACI Eq. 11-5
Strength of Concrete in Shear
For subject to axial compression Nu plus shear Vu, ACI Code gives the following equation for calculating Vc
N u f c' b wd Vc 0.17 1 14Ag
ACI Eq. 11-4
For subject to axial tension Nu plus shear Vu, ACI Code gives the following equation for calculating Vc
14
0.29N u fc' b w d Vc 0.17 1 A g
ACI Eq. 11-8
Deg to Resist Shear To find the force required to be resisted by shear reinforcement:
Vu ΦVn
Vn Vc Vs
Vs 15
Vu
V c
Three cases of shear requirement: Case 1: For Vu ≥ ΦVc shear reinforcement is required Case 2: For Vu ≥0.50ΦVc minimum shear reinforcement is required
Case 3: For Vu < 0.50ΦVc no shear reinforcement isrequired
16
Design of Stirrups Shear reinforcement required when V s V u V c
Vu Vc
ACI 11.4.7.1
The bar size of the stirrups is established and the spacing is calculated:
Vs
Avf yd
s
Afd v y
s
For inclined stirrups (with angle )
Vs
Av f y d sin α cos α
s
s
ACI Eq. 11-15
Vs
Av fy d sinαcos α
ACI Eq. 11-16
Vs
where A v = the area of shear reinforcement within spacing s (for a 2-legged stirrup in a beam: A v = 2 times the area of the stirrup bar). 17
ACI 11.4.6.1
Minimum Amount of Shear Reinforcement
1 Minimum Shear Reinforcement (Av,min) required when Vu Vc 2 bw s bw s Av min 0.062 f c ' 0.35 f ys f ys Av f ys Av f ys ; s=min 0.062 f c ' bw 0.35 bw
ACI Eq. 11-13
except in: (a)Footings and solid slabs (b) Concrete joist construction (c) Beams with h not greater than 250 mm (d)Beams integral with slabs with h not greater than 600 mm and not greater than the larger of 2.5 times the thickness of flange, and 0.5 times width of web.
18
Spacing limits for Shear Reinforcement
If V s 0.33 f c bw d s max If V s 0.33 f c bw d s max
min d ;600mm 2 min d ;300mm 4
ACI 11.4.5
Upper limit for Vs ACI Code requires that the maximum force resisted by shear reinforcement Vs is as follows
V s 0.66 f c ' bw d
ACI 11.4.7.9
If this condition is not satisfied
Section dimensions must be increased 19
Critical Section for Shear
ACI 11.1.3.1
Critical section for shear may be taken a distance d away from the face of the if: (a) reaction, introduces compression into the end regions of member; (b) Loads are applied at or near the top of the member; (c)No concentrated load occurs between face of and location of critical section.
20
Critical Section for Shear
ACI 11.1.3.1
Critical section for shear may be taken a distance d away from the face of the as in cases (a) and (b), but must be taken at face of the as in cases (c) and (d).
21
Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate shears for design of continuous beams, provided: • 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. • are of similar section dimensions along their lengths (prismatic).
22
Approximate Structural Analysis More than two spans
23
ACI 8.3.3
Approximate Structural Analysis ACI 8.3.3 Two spans
l n = length of clear
span measured face-to-face of s.
24
Summary of ACI Shear Design Procedure for Beams 1Draw the shearing force diagram and establish the critical section for shear Vu. 2 Calculate the nominal capacity of concrete in shear Vs. Vc 0.17 f c ' bw d 3- Calculate the force required to be resisted by shear reinforcement V V s u V c 4- Check the code limit on V s Vu Vs V c 0.66 f c ' bw d If this condition is not satisfied, the concrete dimensions should be increased. 25
Summary of ACI Shear Design Procedure for Beams 5- Classify the factored shearing forces acting on the beam according to the following * For Vu < 0.50ΦVc , no shear reinforcement isrequired. * For Vu ≥ 0.50ΦVc , minimum shear reinforcement is required Av f ys Av f ys ; s=min 0.35 b 0.062 f 'b w c w
*For Vu ≥ ΦVc , shear reinforcement is required (in addition, check min shear) Af d Av f y d sin α cos α For vertical v y For inclined s s stirrups stirrups Vs Vs 6- Maximum spacing smax must be checked If V s 0.33 f c bw d s max min d ;600mm 2
26
d If V s 0.33 f c bw d s max min ;300mm 4
Example
A rectangular beam has the dimensions shown in the figure and is loaded with a uniform service dead load of 40 kN/m (including own weight of beam) and a uniform service live load of 25 kN/m. Design the necessary shear reinforcement given that fc’ =28 MPa and fy=420 MPa. Width of is equal to 30 cm. wD=40 kNm & wL=25 kN/m
60
0.3m
0.3m 7.0 m
27
30
Example
Solution: Assuming Φ8 mm stirrups and Φ20 mm flexural steel, d=60-4-0.8-1.0=54.2 cm
wu=1.2(40)+1.6(25)=88 kN/m
0.3m 54.2
308 kN
1- Draw shearing force diagram:
Critical section for shear is located at a distance of d = 54.2 cm from the face of . Vu,critical is equal to 247.1 kN. 28
7.0 m 247.1 kN
308 kN
Example
2- Calculate the shear capacity of concrete: V c 0.17 f c ' bw d 0.17 28 300 542 146.3 10 3 N 146.3kN
V c 0.75 144.2kN 109.7kN V c 54.85 kN 2
3- Calculate the force required to be resisted by shear reinforcement Vs. Vu Vs V c 247.1 146.3 183.2kN 0.75
4- Check the code limit on V s : 0.66 f c ' bw d 0.66 28 300 542 567.9103 N 567.9kN V s 183.2kN 0.66 f c ' bw d 567.9kN 29
OK
Example
5- Classify the factored shear force: Vu= 247.1 kN > ΦVc = 109.7 kN, shear reinforcement is required. The beam can be designed to resist shear based on Vu= 247.1 kN over the entire span. However, to reduce reinforcement cost, the beam will not be designed for this shear over the entire span. The span will rather be divided into zones of different shear demands as shown below 308 kN
247.1 kN ΦVc=109.7 kN Zone C
Zone B
0.5ΦVc=54.85 kN Zone A 0.61 m
1.23 m
30
Example
Zone (A): [ Vu ≤ 0.5ΦVc ]
No shear reinforcement is required, but it is recommended to use minimum area of shear reinforcement. Try Φ8 mm vertical stirrups Av f ys Av f ys s=min ; 0.062 f c ' bw 0.35 bw 2(50) 420 2(50) 420 s min 427mm ; 400mm s 400mm 0.35 300 0.0062 28 300
Maximum stirrup spacing is not to exceed the smaller of d/2 = 271 mm or 600mm. So, use Φ8 mm vertical stirrups spaced at 250 mm. 31
Example
Zone (B): [ΦVc > Vu > 0.5ΦVc ] minimum shear reinforcement is required. use Φ8 mm vertical stirrups spaced at 25 cm (Calculated from Zone A). Zone (C): [Vu > ΦVc ] V s 183.2kN
Trying two-legged Φ8 mm vertical stirrups, s
32
Af d v y
Vs
2 50 420 542 183.2 10
3
125mm
Example
Check maximum stirrup spacing: 0.33 f c ' bw d 0.33 28 300 542 284kN V s 183.2kN Maximum stirrup spacing is not to exceed the smaller of d/2 = 271 mm or 600mm.
Check with minimum stirrup requirement:
Av f ys Av f ys ; smax =min 0.062 f c 'bw 0.35 bw 2(50) 420 2(50) 420 427mm ; 400mm 400mm s max min 0.35 300 0.062 28 300
So, use Φ8 mm vertical stirrups spaced at 12 cm. 33
Example 308 kN
247.1 kN
Zone C
Φ8@12
Zone B
Φ8@25
60
Φ8@25
ΦVc=109.7 kN 0.5ΦVc=54.85 kN Zone A
Φ8@25
0.61 m
30 Section in zones A&B
1.23 m 60
Φ8@12
Φ8@12
34
Φ8@25
30 Section in zone C
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 10 Design of slabs
Introduction A slab is a structural element whose thickness is small compared to its own length and width. t L ,S
t
zS
t
Lx
Slabs in buildings are usually used to transmit the loads on floors and Loads roofs to the ing beams Beam
Beam
Beam
Slab Column
Beam
Beam
Footing
Slab Beam
Column
Beam Soil
2
Introduction Slabs are flexural . Their flexure strength requirement may be expressed by
Mu M
n
Types of Slabs Solid slabs :- which are divided into - One way solid slabs - Two way solid slabs Ribbed slabs :- which are divided into - One way ribbed slabs - Two way ribbed slabs 3
One-way slab
Two-way slab
Solid Slab
Two way slab
Ribbed Slab (joist construction)
4
Two way slab
L 2 S
One-way slab
One-way slab
L 2 S
Ribbed slab with hollow blocks
5
Ribbed slab with hollow blocks
6
One-way solid slabs
shrinkage Reinft.
A one-way solid slab curves under loads in one direction only. Accordingly, slabs ed on two opposite sides only and slabs ed on all four sides, but L/S ≥ 2 are classified as one-way slabs.
Main Reinft.
Main reinforcement is placed in the shorter direction, while the longer direction is provided with shrinkage reinforcement to limit cracking. 7
Two-way solid slabs
Main Reinft.
A two-way solid slab curves under loads in two directions. Accordingly slabs ed on all four sides, and L/S < 2 are classified as two-way slabs.
S Main Reinft.
L
Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement is required in the two directions. 8
One-way v.s two-way ribbed slabs If the ribs are provided in one direction only, the slab is classified as being one-way, regardless of the ratio of longer to shorter dimensions. It is classified as two-way if the ribs are provided in two directions .
9
Minimum thickness of one way slabs
Minimum Cover
10
ACI Table 9.5(a)
ACI 7.7.1
a - Concrete exposed to earth or weather for Φ<16mm------40 mm and for Φ>16mm----- 50 mm b - Concrete not exposed to earth or weather for Φ<32mm------20 mm, otherwise ------ 40 mm
Spacing of Reinforcement Bars a- Flexural Reinforcement Bars Flexural reinforcement is to be spaced not farther than three times the slab thickness (hs), nor farther apart than 45 cm, center-to-center. 3 hs Smax smaller of ACI 10.5.4 45cm b- Shrinkage Reinforcement Bars Shrinkage reinforcement is to be spaced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center. 5 hs Smax smaller of ACI 7.12.2.2 45cm
11
Loads Assigned to Slabs wu=1.2 D.L + 1.6 L.L
a- Dead Load (D.L) : 1- Weight of slab covering materials 2-Equivalent partition weight 3- Own weight of slab
b- Live Load (L.L)
12
a- Dead Load (D.L)
1- Weight of slab covering materials, total =2.315 kN/m2
tiles (2.5cm thick) =0.025×23 = 0.575 kN/m2 cement mortar (2.5cm thick) =0.025×21 = 0.525 kN/m2 sand (5.0cm thick) =0.05×18 = 0.9 kN/m2 plaster (1.5cm thick) =0.015×21 = 0.315 kN/m2
tiles cement mortar sand
2.5 cm 2.5 cm 5 cm
slab
plaster 13
1.5 cm
2-Equivalent partition weight
This load is usually taken as the weight of all walls (weight of 1m span of wall × total spans of all walls) carried by the slab divided by the floor area and treated as a dead load rather than a live load. To calculate the weight of 1m span of wall: Each 1m2 surface of wall contains 12.5 blocks A block with thickness 10cm weighs 10 kg A block with thickness 20cm weighs 20 kg Each face of 1m2 surface has 30kg plaster Load / 1m2 surface for 10 cm block = 12.5 × 10 +2×30=185 kg/m2 = 1.85 kN/m2 Load / 1m2 surface for 20 cm block = 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2
14
20 cm
Weight of 1m span of wall with height 3m: For 10 cm block wt. = 1.85 kN/m2 × 3 = 5.6 kN/m For 20 cm block wt. = 3.1 kN/m2 × 3 = 9.3 kN/m
3- Own weight of slab
1- Solid slab: Own weight of solid slab (per 1m2)=
c h = 25 h kN/m2
2- Ribbed slab:
Example Find the total ultimate load per rib for the ribbed slab shown: Assume depth of slab = 25 cm (20cm block +5cm toping slab)
Hollow blocks are 40 cm × 25 cm × 20 cm in dimension Assume ribs have 10 cm width of web Assume equivalent partition load = 0.75 kN/m2 Consider live load = 2 kN/m2.
15
3- Own weight of slab
Solution •
Total volume (hatched) = 0.5 × 0.25 × 0.25 = 0.03125 m3
•
Volume of one hollow block = 0.4 × 0.20 × 0.25 = 0.02 m3
•
Net concrete volume = 0.03125 - 0.02 = 0.01125 m3
•
Weight of concrete = 0.01125 × 25= 0.28125 kN
•
Weight of concrete /m2 = 0.28125 /[(0.5)(0.25)] = 2.25 kN/ m2
•
Weight of hollow blocks /m2 = 0.2/[(0.5)(0.25)] = 1.6 kN/ m2
•
Total slab own weight= 2.25 + 1.6= 3.85kN/m2
Load per rib Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m2
16
Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m2 Ultimate load per rib = 11.5 × 0.5 = 5.75 kN/m
Minimum live Load values on slabs Type of Use Uniform Live Load kN/m2 Residential 2
b- Live Load (L.L)
It depends on the function for which the floor is constructed.
Residential balconies Computer use Offices Warehouses
3 5 2
6
Light storage
Heavy Storage Schools
12
Classrooms Libraries
2
rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
17
2.5 5 2.5
Retail
4
wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
Loads Assigned to Beams Beams are usually designed to carry the following loads: - Their own weight - Weights of partitions applied directly on them - Floor loads
L
S1
18
S2
Design of one way SOLID slabs
19
One-way solid slabs
S1
S2
1m
One-way solid slabs are designed as a number of independent 1 m wide strips which span in the short direction and are ed on crossing beams. These strips are designed as rectangular beams.
L
2M u 0.85f c 1 1 2 fy 0.85 f c bd S1
20
S2
shrinkage Reinft.
One-way solid slabs
Main Reinft.
21
Check on tension/compression control (maximum allowed steel) Method 1: Check t
Method 2: Check max 0.003 c max d 0.003 0.005 3 c max d 8 =1c 1c max
max
3 0.85 1 f c ' 8 fy
3 d 1 8
22
Shrinkage Reinforcement Ratio According to ACI Code and for fy =420 MPa
ACI 7.12.2.1
shrinkage 0.0018 As,shrinkage 0.0018 bh where, b = width of strip, and h = slab thickness
Minimum Reinforcement Ratio for Main Reinforcement
A s ,min A s ,shrinkage 0.0018 b h
ACI 10.5.4
Check shear capacity of the section
V u V c 0.17 f c ' b wd Otherwise enlarge depth of slab 23
Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate moments and shears for design of continuous beams and one-way slabs, provided: • 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. • are of similar section dimensions along their lengths (prismatic).
24
Approximate Structural Analysis Bending Moment More than two spans
25
ACI 8.3.3
Approximate Structural Analysis Bending Moment Two spans
l n = length of clear
span measured face-to-face of s. For calculating negative moments, l n is taken as the average of the adjacent clear span lengths.
26
ACI 8.3.3
Approximate Structural Analysis Shear More than two spans
27
ACI 8.3.3
Approximate Structural Analysis Shear Two spans
28
ACI 8.3.3
Summary of One-way Solid Slab Design Procedure 1 Select representative 1m wide design strip/strips to span in the short direction. 2 Choose a slab thickness to satisfy deflection control requirements. When several numbers of slab s exist, select the largest calculated thickness. 3 Calculate the factored load wu 4 Draw the shear force and bending moment diagrams for each of the strips. 5 Check adequacy of slab thickness in of resisting shear by satisfying the following equation: V u 0.17 f c ' b d where b = 1000 mm
If the previous equation is not satisfied, go ahead and enlarge the thickness to do so. 29
Summary of One-way Solid Slab Design Procedure 6- Design flexural and shrinkage reinforcement: Flexural reinforcement ratio is calculated from the following equation 2M u 0.85f c 1 1 2 fy 0.85 f c bd where b = 1000 mm
You need to check tension controlled requirement, minimum reinforcement requirement and spacing of selected bars. Compute the area of shrinkage reinforcement, where Ashrinkage=0.0018bh, where b = 1000 mm.
7- Draw a plan of the slab and representative cross sections showing the dimensions and the selected reinforcement. 30
Example 1 Using the ACI-Code approximate structural analysis, design for a warehouse, a continuous one-way solid slab ed on beams 4.0 m apart as shown in the figure. Assume that the beam webs are 30 cm wide. The dead load is 3kN/m2 in addition to the own weight of the slab, and the live load is 3kN/m2.
8.0 m
Use fc’=28MPa, fy=420MPa
4.0 m
31
4.0 m
4.0 m
Solution: 1- Select a representative 1 m wide slab strip:
The selected representative strip is shown in the figure
2- Select slab thickness:
For one-end continuous spans, hmin = l/24 =4.0/24=0.167m Slab thickness is taken as 17 cm
8.0 m
1.0 m
17cm
4.0 m Wu
32
4.0 m
4.0 m
Solution: 3Calculate the factored load wu per unit length of the selected strip:
Own weight of slab = 0.17× 25 = 4.25 kN/m2 wu= 1.20 (3+4.25) +1.60 (3)= 13.5 kN/m2 For a strip 1 m wide, wu=13.5 kN/m 4Evaluate the maximum factored shear forces and bending moments in the strip:
The clear span length, ln = 4.0 – 0.30 = 3.70 m wu=13.5 kN/m
33
Solution:
18.5 7.7
16.8
16.8
16.8
Units of moment are in kN.m 34
18.5 7.7 16.8
Solution:
25
25
28.7
Units of shear are in kN 35
28.7
25
25
Solution: 5- Check slab thickness for beam shear:
Effective depth d = 17 – 2 – 0.60 = 14.40 cm, assuming Φ12 mm bars. Vu,max = 28.7 kN. V c 0.17 f c ' bd 0.750.17 28 1000144 95.8 kN
i.e. , slab thickness is adequate in of resisting beam shear. 6- Design flexural and shrinkage reinforcement:
Assume that Φ=0.9
2M u 0.85f c 1 1 2 fy 0.85 f c bd
Where b = 1000 mm & d = 144mm
36
Solution: For max. negative moment, Mu = 18.5 kN.m 0.8528 1 1 218.5106 ρ 0.00241 2 420 0.85 0.9 28 1000 144 3 0.85 1f c ' 3 0.85 0.85 28 0.01806 ρ 0.90 ρmax 8 fy 420 8 2 As,ve 0.002411000144 347 mm 2
As,min 0.00181000170 306mm As,ve OK 79φ10 347mm2 S 227.5mm 1000mmstrip S Smax min(450 or 3170) 450mm
37
use 10@20cm
Solution: For max. positive moment, Mu = 16.8 kN.m 0.8528 1 1 216.8106 0.00219 ρ 2 0.85 0.9 28 1000 144 420 3 0.85 1f c ' 3 0.85 0.85 28 0.01806 ρ 0.90 ρmax 8 fy 420 8 2 As,ve 0.002191000144 315mm 2
As,min 0.00181000170 306mm As,ve OK 79φ10 315mm2 S 251mm 1000mmstrip S Smax min(450 or 3170) 450
38
use 10@25cm
Solution: Calculate the area of shrinkage reinforcement: Area of shrinkage reinforcement = 0.0018 (100) (17) = 306 mm2 For shrinkage reinforcement use Φ 10 mm @ 25 cm (from previous slides calculations) Shrinkage reinft. Φ10@25 Φ10@25
Φ10@20
Φ10@25
Φ10@20
17cm Φ10@25
Φ10@25
Φ10@25
See Lecture 12 for information on detailing requirements
39
Solution:
8.0 m
Φ10@25 Φ10@25
4.0 m
40
Φ10@20
Φ10@20 Φ10@25
4.0 m
Φ10@25 Φ10@25
4.0 m
Design of one way RIBBED slabs
41
One-way ribbed slabs Ribbed slabs consist of regularly spaced ribs monolithically built with a toping slab. The voids between the ribs may be either light material such as hollow blocks [figure 1] or it may be left unfilled [figure 2]. Topping slab
Rib
Hollow block
Figure [1] Hollow block floor
Temporary form Figure [2] Moulded floor
The use of these blocks makes it possible to have smooth ceiling which is often required for architectural considerations and have good sound and temperature insulation properties besides reducing the dead load of the slab greatly. 42
Key components of one-way ribbed slabs ACI 8.13.6.1 Topping slab thickness (t) is not to be less than 1/12 the clear distance (lc) between ribs, nor less than 50 mm a. Topping slab:
lc t 12 50 mm
and should satisfy for a unit strip: t
lc Slab thickness (t)
w u lc2
1240 f c
Shrinkage reinforcement is provided in the topping slab in both directions in a mesh form. 43
Key components of one-way ribbed slabs b. Regularly spaced ribs: Minimum dimensions:
Ribs are not to be less than 100 mm in width, and a depth of not more than 3.5 times the minimum web width and clear spacing between ribs is not to exceed 750 mm. ACI 8.13.2 ACI 8.13.3 l ≤ 750 mm c
h ≤ 3.5 bw
bw ≥ 100
44
Key components of one-way ribbed slabs ACI 8.13.8 Shear strength provided by rib concrete Vc may be taken 10% greater than those for beams. Shear strength:
Flexural strength:
Ribs are designed as rectangular beams in the regions of negative moment at the s and as T-shaped beams in the regions of positive moments between the s. Effective flange width be is taken as half the distance between ribs, center-to-center. b e
45
Key components of one-way ribbed slabs Hollow blocks: Hollow blocks are made of lightweight concrete or other lightweight materials. The most common concrete hollow block sizes are 40 × 25 cm in plan and heights of 14, 17, 20, and 24 cm.
46
Summary of one-way ribbed slab design procedure 1.The direction of ribs is chosen. 2. Determine h, and select the hollow block size, bw and t 3.Provide shrinkage reinforcement for the topping slab in both directions. 4. The factored load on each of the ribs is computed. 5. The shear force and bending moment diagrams are drawn. 6. The strength of the web in shear is checked. 7.Design the ribs as T-section shaped beams in the positive moment regions and rectangular beams in the regions of negative moment. 8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared. 47
Example Design a one-way ribbed slab to cover a 3.8 m x 10 m , shown in the figure below. The covering materials weigh 2.25 kN/m2, equivalent partition load is equal to 0.75 kN/m2, and the live load is 2 kN/m2.
3.8 m
Use fc’=25 MPa, fy=420MPa
10 m
48
Solution 1. The direction of ribs is chosen:
3.8 m
Ribs are arranged in the short direction as shown in the figure
5.0 m
5.0 m
2. Determine h, and select the hollow block size, bw and t:
From ACI Table 9.5(a), hmin = 380/16 = 23.75cm use h = 24 cm. Let width of web, bw =10 cm Use hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN) Topping slab thickness = 24 – 17 = 7cm > lc/12 =40/12= 3.3cm > 5cm OK For a unit strip of topping slab: wu=[1.2(0.07 × 25 + 0.75 + 2.25) + 1.6(2)] ×1m = 8.9 kN/m = 8.9 N/mm t
w u l c2 1240 f c
8.9( 400) 2 ( 0.9)1240 25
16mm OK 49
Solution 3. Provide shrinkage reinforcement for the topping slab in both directions:
Area of shrinkage reinforcement, As=0.0018(1000)70=126 mm2 Use 5 Φ 6 mm/m in both directions. 4. The factored load on each of the ribs is to be computed:
50
1.0 m
0.4 m
0.1 m
0.4 m
7 cm
0.25 m
1.0 m
0.05 m
0.24 m
Total volume (in 1m2 surface) = 1.0 × 1.0 × 0.24 = 0.24 m3 Volume of hollow blocks in 1m2 = 8 × 0.4 × 0.25 × 0.17 = 0.136 m3 Net concrete volume in 1m2 = 0.24- 0.136 = 0.104 m3 Weight of concrete in 1m2 = 0.104 × 25 = 2.6 kN/m2 Weight of hollow blocks in 1m2 = 8 × 0.17= 1.36 kN/m2 Total dead load /m2 = 2.25 + 0.75 + 2.6 + 1.36 = 7.0 kN/m2
Solution wu=1.2(7)+1.6(2)=11.6 kN/m2 wu/m of rib =11.6x0.5= 5.8 kN/m of rib 5. Critical shear forces and bending moments are determined (simply ed beam):
Maximum factored shear force = wul/2 = 5.8 (3.8/2) = 11 kN Maximum factored bending moment = wul2/8 = 5.8 (3.8)2/8 = 10.5 kN.m 6. Check rib strength for beam shear:
Effective depth d = 24–2–0.6–0.6 =20.8 cm, assuming 12mm reinforcing bars and Φ 6 mm stirrups. 1.1ΦVc 1.10.75 0.17 25 100 208 14400 N = 14.4 kN Vu,max 11kN
Though shear reinforcement is not required, 4 6 mm stirrups per meter run are to be used to carry the bottom flexural reinforcement.
51
Solution 7. Design flexural reinforcement for the ribs:
There is only positive moments over the simply ed beam, and the section of maximum positive moment is to be designed as a T-section Assume that a<70mm and Φ=0.90→Rectangular section with b = be =500mm
s
e
Use 210mm (As,sup= 157 mm2) As f y
157 420 6.2 mm 70mm 0.85f c 'b e 0.8525500 The assumption is right
a
52
50
105 kN.m
0.85 25 210.5106 ρ 1 1 420 0.90.85255002082 0.0013 A ρ b d 0.0013500 208 135 mm 2
7 24
As 10
Solution CheckAs,min
0.25 f c ' 1.4 bw d A s,min max bw d ; fy f y 2 2 A s,min 70 mm A s,sup 157 mm OK
Check Φ=0.9
c
a 6.2 7.3 mm β1 0.85
dc 208 7.3 0.003 ε t 0.003 7.3 c ε t 0.083 0.005 0.9 OK
53
Solution
A
reinforcement are to be
1Φ10 m
A
1Φ10 m
1Φ10 m
1Φ10 m
3.8 m
8. Neat sketches showing arrangement of ribs and details of the prepared
5.0 m
5.0 m Φ6mm mesh @20 cm
Φ6mm stirrups @25 cm
7cm 24cm 17cm
2Φ10mm
10
40 cm
10
2Φ10mm
SectionA-A
See Lecture 12 for information on detailing requirements 54
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 11 Design of short concentric columns
Columns Columns are vertical compression of a structural frame intended to the load-carrying beams. They transmit loads from the upper floors to the lower levels and then
to the soil through the foundations.
Loads Beam
P
Beam
Column
h
Slab
b
Beam
l
h
Column
Beam
b
Beam
Slab Footing Beam
Beam Soil
2
Columns Usually columns carry bending moment as well, about one or both axes of the cross section, and the bending action may produce tensile forces over a part of the cross section.
The main reinforcement in a columns is longitudinal, parallel to the direction of the load and consists of bars arranged in a square, rectangular, or circular shape.
3
Length of the column in relation to its lateral dimensions Columns can be classified as
1Short Columns, for which the strength is governed by the strength of the materials and the dimensions of the cross section
2Slender Columns, for which the strength may be significantly reduced by lateral deflections.
Position of the load on the cross-section Columns can be classified as
1Concentrically loaded columns, which are subjected to axial force only
2Eccentrically loaded columns, which are subjected to moment in addition to the axial force. 4
Analysis and Design of Short Columns
Column Types: 1. Tied
2. Spiral 3. Composite
5
Behavior of Tied and Spirally-Reinforced Columns Axial load tests have proven that tied and spirally reinforced columns having the same cross-sectional areas of concrete and steel reinforcement behave in the same manner up to the ultimate load. At that load, tied columns fail suddenly due to excessive cracking in the concrete section followed by buckling of the longitudinal reinforcement between ties within the failure region. For spirally reinforced columns, once the
ultimate load is reached, the concrete shell covering the spiral starts to spall off but the core will continue to carry additional loads because the spiral provides a confining force to the concrete core, thus enabling the column to sustain large deformations before final collapse. 6
Behavior of Tied and Spirally-Reinforced Columns
Failure of a tied column
7
Failure of a spiral column
Nominal Capacity under Concentric Axial Loads
P0 0.85f c A g A st f y A st or
P0 A g 0.85f c Ast (f y 0.85f c) A g = gross area = b*h Ast = area of longitudinal steel fc′ =concrete compressive strength 8
f y = steel yield stress
Maximum Nominal Capacity under Concentric Axial Loads
Pn rP0 Pn r Ag 0.85f c Ast (f y 0.85f c) r = Reduction factor to for accidental eccentricity
r = 0.80 ( tied )
r = 0.85 ( spiral ) 9
ACI 10.3.6.3
Design Capacity under Concentric Axial Loads
Pn Pu Pn r A g 0.85f c A st f y 0.85f c Pu or
Pn r A g 0.85f c g f y 0.85f c Pu where g = Ast / Ag ACI 9.3.2.2
10
= 0.65 for tied columns
ACI 10.3.6.3 r = 0.80 ( tied )
= 0.75 for spiral columns
r = 0.85 ( spiral )
Design of Short Concentrically Loaded Columns
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu Both Ag and g are unknown in this equation. There are two options to design the column: 1Select Ag and calculate g. The A g sizing (Ag = Pu / 0.5fc′ ).
may be selected from initial
2Select g and calculate Ag. Usually g is assumed as 2% as a starting point. 11
Calculation of required cross section, if steel ratio is known
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu * when g is known or assumed:
Ag
12
Pu
r 0.85f c g f y 0.85f c
Calculation of required steel ratio, if cross section is known
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu * when g is known or assumed:
1 Pu 0.85f c g r A g f y 0.85f c 13
Design of spirals Spiral Reinforcement Ratio, s
Volume of Spiral 4Asp s Volume of Core Dcs from:
A sp D c
s 2 [( / 4) D c ] s
Asp cross-sectional area of spiral reinforcement D c core diameter: outside edge to outside edge of spiral 14
s spacing pitch of spiral steel (center to center)
Design of spirals Spiral Reinforcement Spacing, s
Ag f c s 0.45 1 ACI Eq. 10-5 Ac f y 4A s sp from previous slide D cs s
4A sp f c ' Ag 0.45D c 1 Ac f y
A c core area 15
Dc2
Ag gross area
4
D2 4
Design Considerations
Longitudinal Steel
- Limits on reinforcement ratio: ACI 10.9.1
0.01Ag Ast 0.08Ag or
0.01 g 0.08 16
Design Considerations
Longitudinal Steel
- Minimum number of bars ACI 10.9.2
min. of 6 bars in spiral arrangement min. of 4 bars in rectangular or circular ties min. of 3 bars in triangular ties 17
Design Considerations
Longitudinal Steel - Clear Distance between Reinforcing Bars (Longitudinal Steel) For tied or spirally reinforced columns, clear distance between bars, shown in the figure, is not to
be less than the larger of 1.50 times bar diameter or 40 mm. This is done to ensure free flow of concrete among reinforcing bars.
ACI 7.6.3
S c max 1.5d b , 40mm 18
Design Considerations
Lateral Ties
- Arrangement of ties and longitudinal bars: ACI 7.10.5.3 1.) At least every other longitudinal bar shall have lateral from the corner of a tie with an included angle 135o. 2.) No longitudinal bar shall be more than 150 mm clear on either side from a laterally ed bar. 19
Design Considerations
Lateral Ties
- Arrangement of ties and longitudinal bars: ACI 7.10.5.3
20
Ties shown dashed may be omitted if x < 150 mm
Design Considerations
Lateral Ties
- Maximum vertical spacing: ACI 7.10.5.2
s 16 db ( db = diameter for longitudinal bars ) s 48 dstirrup (dstirrup = diameter for stirrups) s least lateral dimension of column
21
Design Considerations
Lateral Ties
- Minimum size of ties ACI 7.10.5.1
size 8 bar if longitudinal bar 30 bar 12 bar if longitudinal bar 32 bar 12 bar if longitudinal bars are bundled
22
Design Considerations
Spirals
- Size and spacing of spiral ACI 7.10.4.2
size 10 mm diameter ACI 7.10.4.3 25mm 23
clear spacing between spirals
75mm
Design Considerations ACI 7.7.1 Concrete Protection Cover The clear concrete cover is not to be taken less than 4 cm for columns not exposed to weather or in with ground.
Minimum Cross Sectional Dimensions The ACI Code does not specify minimum cross sectional dimensions for columns. Column cross sections 20 × 25 cm are considered as the smallest practicable sections. For practical considerations, column dimensions are taken as multiples of 5 cm. Lateral Reinforcement
Ties are effective in restraining the longitudinal bars from buckling out through the surface of the column, holding the reinforcement cage together during the construction process, confining the concrete core and when columns are subjected to horizontal forces, they serve as shear reinforcement. 24
Design Procedure for Short Concentrically Loaded Columns 1. Evaluate the factored axial load Pu acting on the column. This can be done by:
a- Tributary Area Method b- Pu is the sum of the reactions of the beams ed by the column. 2. Assume a starting reinforcement ratio ρg that satisfies ACI Code limits. Usually a 2
% ratio is chosen for economic considerations. 3. Determine the gross sectional area Ag of the concrete section. 4. Choose the dimensions of the cross section based on its shape. 5. Readjust the reinforcement ratio by substituting the actual cross sectional area in the respective equation. This ratio has to fall within the specified code limits. 25
Design Procedure for Short Concentrically Loaded Columns 6.
Calculate the needed area of longitudinal reinforcement ratio based on the adjusted
reinforced ratio and the chosen concrete dimensions. 7.
From reinforcement tables, choose the number and diameters of needed reinforcing bars. For rectangular sections, a minimum of four bars is needed, while a minimum of six bars is used for circular columns.
8.
Design the lateral reinforcement according to the type of column, either ties or spirals.
9.
Check whether the spacing between longitudinal reinforcing bars satisfies ACI Code requirements.
10. Draw the designed section showing concrete dimensions and with required longitudinal and lateral reinforcement.
26
Example 1 The cross section of a short axially loaded tied column is shown in the figure. It is reinforced with 616mm bars. Calculate the design load Ties Φ8@25cm capacity of the cross section. Use fc′ =28 MPa and fy = 420 MPa. 25
6Φ16
Solution: ρg
1206 Ast 0.012 1.21% A g 250 400
ρ min 1 % ρ g 1.21% ρ max 8%
40 Figure [1]
OK
Clear distance between bars Sc
40 2(4) 2(0.8) 3(1.6) 12.8cm 31 max 1.5d b 2.4cm , 4cm <Sc 12.8cm
Sc=12.8 cm
25
6Φ16
Sc
Only, one tie is required for the cross section 27
40
Example 1 The spacing between ties 16 db =16(1.6) = 25.4 cm ≥ S = 25 cm 48 ds = 48(0.8) = 38.4 cm ≥ S = 25 cm smaller of b or d = 25 cm ≥ S = 25 cm Thus, ACI requirements regarding reinforcement ratio, clear distance between bars and tie spacing are all satisfied. The design load capacity ΦPn Pn 0.65(0.8) A g 0.85f c g f y 0.85f c Φ Pn 0.52A g 0.85f c' ρ g f y 0.85fc'
Φ Pn 0.65(0.8) 400 250 0.8528 0.0121420 0.85 28 Φ Pn 1487 kN 28
Example 2 Design a short tied column to a factored concentric load of 1000 kN, with one side of the cross section equals to 25 cm. f c 30MPa
f y 420MPa
Solution
Assume first that g 2% Ag
Pu
0.65 0.8 0.85f c g f y 0.85f c
1000103 Ag 0.65 0.8 0.8530 0.02 420 0.85 30 29
A g 57594mm
2
Example 2 A
g
57594mm 2
b 250mm
h 230mm use column 25cm 25cm Determine adjusted steel ratio Pu 1 0.85f c g r A g f y 0.85f c 1 1000103 0.85(30) =0.0134 g = 0.65 0.8 250 250 420 0.85(30) 0.01<g <0.08 OK A bh 0.0134(250)(250) 835mm 2 s
g
2
30
Use 614 (As,sup = 924 mm )
Example 2 Check spacing s
h (No. of bars/2) d b 2 cover 2 dstirrup
(No. of bars/2) 1 250 (6 / 2) 14 2 40 2(8) 31
56mm
max 1.5d b 21mm , 40mm 56mm < 150mm
OK
Stirrup design Use 8 mm (for longitudinal bars with 14 mm < 30 mm)
smax
16d b 1614cm 224mm min 48d stirrup 488 384mm smaller of b or d 250mm
Use 8 mm @ 200 mm 31
governs
6 14 mm
Example 2
8 mm @ 200 mm
250 mm
250 mm
32
Example 3 Design a short, spirally reinforced column to a service dead load of 800 kN and a service live load of 400 kN. f c 28MPa
f y 420MPa
Use g 1%
Solution
Pu 1.20 PD 1.60 PL 1.2800 1.6 400 1600kN Ag
Pu
0.75 0.85 0.85f c g f y 0.85f c
1600103 Ag 0.75 0.85 0.8528 0.01420 0.85 28 33
A g 90405mm 2
Example 3 A g 90405mm
360/N
2
for circular column D=
Ag
4
=339mm 0.5D’
use column with D = 350 mm
A s 0.01 (3502 ) 962mm 2 4 2
use 714 (As,sup =1078 mm ) Check spacing between longitudinal bars D’ =350-2(40)-2(8)-14=240 mm,
N=7
360/N 51.43 104.1mm S D'sin 240sin 2 2 Sc 104.114 90.1mm 1.5(14)=21mm 34
40mm
OK
360/N 2 0.5D’
0.5Sc = 0.5D’ sin(360/N/2)
Example 3 Design the needed spiral, try 8 Dc 350 2(40) 270 mm S
4Asp
450
π/4 350 2 28 0.45 270 1 2 420 π/4270 S 36.3mm, taken as 35 mm (center to center) Sc 35 8/2 8/2 27 mm, i.e within ACIcode limit ( 25mm& 75mm) A g fc ' 1 0.45Dc Ac fy
Use Φ 8mmspiral with a pitch of 35mmcenter to center.
35
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 Part I Bond, development length, and splicing
Bond
2
Concept of Bond Stress Bond stresses are existent whenever the tensile stress or force in a reinforcing bar changes from point to point along the length of the bar in order to maintain equilibrium. Without bond stresses, the reinforcement will pull out of the concrete.
Concrete Reinforcing bar
PL/4
M
M+dM
dx Moment diagram
3
µavg
Concept of Bond Stress
F 0.0 T2 T1 Fbond If this equation is not true (bond force Fbond is not strong enough), the bar will pull out A bar fs2 fs1 avg A surface
d 4
2 b
db
µ=Bond stress
fs2 fs1 avg ( db ) l
μ avg
f - f d = s2 s1 b 4l
µavg= average bond stress
4
T1=fs1Ab
l
T2=fs2Ab fs2=fs1+∆fs
Mechanism of Bond Transfer A smooth bar embedded in concrete develops bond by adhesion between concrete & reinforcement, and a small amount of friction.
This is different in a deformed bar. Once adhesion is lost at high bar stress and some slight movement between the reinforcement and the concrete occurs, bond is then provided by friction and bearing on the deformations of the bar. At much higher bar stress, bearing on the deformations of the bar will be the only component contributing to
bond strength.
(a) Forces on bar
5
(b) Forces on concrete
Splitting cracks The radial component of the bearing
force will cause circumferential stress on the concrete that may cause splitting that creates cracks.
6
Splitting cracks Splitting of concrete may occur along the bars, either in vertical planes as in figure (a) or in horizontal plane as in figure (b).
7
Splitting cracks The load at which splitting failure develops is a function of :
•
The minimum distance from the bar to the surface of the concrete or to the next bar. The smaller the distance, the smaller is the splitting load.
•
The tensile strength of the concrete. The higher the tensile strength, the higher is the splitting resistance.
•
The average bond stress. The higher the average bond stress, the higher is the splitting resistance.
If the concrete cover and bar spacing are large compared to the bar diameter, a pullout failure can occur, where the bar and the ring of concrete between successive deformations pullout along a cylindrical failure surface ing the tips of the deformations. 8
Development Length
9
Development Length The bars found to be needed at a section from design calculations have to be embedded a certain distance into the concrete. This distance has to be equal or larger than the development length (ld).
10
Development Length The development length ld is that length of embedment necessary to develop the full tensile strength of the bar (on both sides of sections where fy stress is required),
controlled by either pullout or splitting.
μavg =
f s2 f s1 db 4l
f s2 f s1 f y
ld 11
f d y
b
4 avg,u
, where avg,u is the value avg at bondfailure
Development Length of Deformed Bars in Tension The development length for deformed bars in tension is given by
ld
fy
αβ γ λ
1.1 fc C K tr d b
d b 300 mm,
C K tr where 2.5 db where, ld = development length db = nominal diameter of bar fy = specified yield strength of reinforcement C = spacing or cover dimension (see next slide) Ktr = transverse reinforcement index (see slide 12) 12
= see next slides
ACI Eq. 12-1
ACI 12.2.3
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.4 C is the smaller of (a)the smallest distance measured from the center of the bar to the nearest concrete surface (b)one-half the center-to-center spacing of bars being developed. α is a bar location factor (a) Horizontal reinforcement so placed that more than 30 cm of fresh concrete is cast in the member below the development length or splice………………………………………………………………. (b) Other reinforcement…
1.3 1.0
β is a coating factor that reflects the adverse effects of epoxy coating (a)Epoxy-coated bars or wires with cover less than 3db or clear spacing less than 6db (b)All other epoxy-coated bars or wires… (c) Uncoated reinforcement……………………………………………………… 13
However, the product β α is not to be greater than 1.7.
1.5 1.2 1.0
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.4 γ is a reinforcement size factor that reflects better performance of the smaller diameter reinforcement (a) Φ20mm and smaller bars. (b) Φ22mm and larger bars.
0.8 1.0
λ is a lightweight aggregate concrete factor that reflects lower tensile strength of lightweight concrete, & resulting reduction in splitting resistance. (a) When lightweight aggregate concrete is used…….……..……………… 0.8 (b) When normal weight concrete is used… 1.0
14
Development Length of Deformed Bars in Tension [contd.] Ktr is a transverse reinforcement index that represents the contribution of confining reinforcement
K tr
40 Atr sn
ACI Eq. 12-2
where: Atr = total cross sectional area of all transverse reinforcement within the spacing s, which crosses the potential plane of splitting along the reinforcement being developed within the development length s = maximum center-to-center spacing of transverse reinforcement within development length ld n = number of LONGITUDINAL bars being developed along the plane of splitting. Note: It is permitted to use Ktr= 0.0 as design simplification even if transverse reinforcement is present. 15
Atr
Potential plane of splitting
n=4
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.5 Excessive Reinforcement Reduction in development length is allowed where As provided > As required. the reduction is given by Reduction factor
As required As provided
-Except as required for seismic design
-Good practice to ignore this factor, since the use of the structure may change over time.
Simplified Expression for Development Length See ACI 12.2.2. This will not be used in this class
16
Example 1
60 cm
Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown
Φ10@20 4Φ20
40 cm Cover is 4 cm on all sides
SectionA-A
17
Example 1 Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown
(a) Ktr is calculated α=1.0 for bars over concrete < 30 cm thick
Φ10@20 4Φ20
β=1.0 for uncoated bars
α β =1.0 <1.7 OK
40 cm Cover is 4 cm on all sides
γ=0.8 for Φ20mm, λ=1.0 for normal weight concrete, C the smallest of
40+10+(20/2)=60 mm {[400-2(40)-2(10)-2(20/2)]/(3)}/(2)=46.7 mm
18
60 cm
Solution:
i.e., C is taken as 46.7 mm
Example 1 [contd.] K tr
40A t r 40(2 79) 7.9 mm sn (200)(4)
i.e., use
ld
C Ktr 2.5 db
Φ10@20 4Φ20
αβ γ λ d b 300 mm 1.1 fc C K t r d b fy
40 cm Cover is 4 cm on all sides
420 (1.0)(1.0)(0.8)(1.0) 20 489 mm 300 mm OK ld 2.5 1.1 25 b) Assuming K t r 0.0 C Ktr 467 0 2.33 2.5 db 20 19
60 cm
C Ktr 46.7 7.9 2.73 2.5 db 20
OK
420 (1.0)(1.0)(0.8)(1.0) 20 524 mm 300 mm OK ld 2.33 1.1 25
Example 1 [contd.]
60 cm
(c) Check if space is available for bar development
Φ10@20 4Φ20
40 cm Cover is 4 cm on all sides
SectionA-A
Available length for bar development = 2000+ 150– 40 = 2110 mm > ld = 524 mm
OK 20
Development Length of Deformed Bars in Compression
ACI 12.3
Shorter development lengths are required for compression than for tension since flexural tension cracks are not present for bars in compression. In addition, there is some bearing of the ends of the bars on concrete.
The development length ld for deformed bars in compression is computed as the product of the basic development length ldc and applicable modification factors, but ld is not to be less than 200 mm.
ld = ldc x applicable modification factors ≥ 200 mm. The basic development length ldb for deformed bars in compression is given as
0.24 f y db ldc max ;0.043 f y db fc ' 21
Development Length of Deformed Bars in Compression [contd.]
ACI 12.3
Applicable Modification Factors 1.Excessive reinforcement factor =As required / As provided 2.Spirals or Ties: the modification factor for reinforcement, enclosed with spiral reinforcement ≥ 6mm in diameter and ≤ 10 cm pitch or within Φ12mm ties spaced at ≤ 10 cm on center is given as 0.75
Development Lengths for Bundled Bars
ACI 12.4
Development length of individual bars within a bundle, in tension or compression, is
taken as that for individual bar, increased 20% for three-bar bundle, and 33% for fourbar bundle. For determining the appropriate modification factors, a unit of bundled bars is treated as a single bar of a diameter derived from the equivalent total area of bars.
22
ldh
Critical section
Development of Standard Hooks in Tension Hooks are used to provide additional anchorage when there is insufficient length available
to develop a bar. Development length ldh for deformed bars in tension terminating in a standard hook is computed as the product of the basic development length lhb and applicable modification factors, but ldh is not to be less than 8db, nor less than 150 mm.
ldh = lhb x applicable modification factors ≥ 15 cm or 8db. The basic development length lhb for hooked bars is given as
lhb
0.24 f
e y
fc '
For lightweight aggregate concrete, = 0.75. For epoxy-coated reinforcement, e= 1.2. 23
Otherwise, = 1.0, e= 1.0
ACI 12.5.1
db
ACI 12.5.2
Development of Standard Hooks in Tension [contd.]
ACI 12.5.3
Applicable Modification Factors 1. Concrete cover: for db ≤ Φ36mm, side cover (normal to plane of hook) ≥ 65 mm, and for 90 degree hook, cover on bar extension beyond hook ≥ 50 mm, the modification factor is taken as 0.7. not less than 50 mm 65 mm
65 mm
24
Development of Standard Hooks in Tension [contd.]
ACI 12.5.3
Applicable Modification Factors 2.Excessive reinforcement factor =As required / As provided 3.Spirals or Ties: for db ≤ Φ36mm, hooks enclosed vertically or horizontally within ties or stirrups spaced along the full development length ldh not greater than 3db , where db is the diameter of the hooked bar, and the first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, the modification factor is taken as 0.8.
25
Development of Standard Hooks in Tension [contd.] Development length ldh is measured
ACI 7.1
90-degree hook
from the critical section of the bar to the out-side end or edge of the hooks. Either a 90 or a 180-degree hook, shown in the figure, may be used ldh
Development of reinforcement- General * Hooks are not considered effective in compression and may not be used as anchorage.
Part (a)
180-degree hook
ACI 12.5.5 * The values of fc ' used in this lecture shall not exceed 8.3 MPa. 26
ACI 12.1.2
4db ≥ 65mm
Φ10 through Φ25 Φ28 through Φ36 Φ44 through Φ56
ldh
Part (b)
Development of Standard Hooks in Tension [contd.]
ACI 12.5.4
Confinement of hooks For bars being developed by a standard hook at discontinuous ends of with both side cover and top (or bottom) cover over hook less than 65 mm, the hooked bar shall be enclosed within ties or stirrups perpendicular to the bar being developed, spaced not greater than 3db along ldh. The first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, where db is the diameter of the hooked bar.
27
Example 2
(a) If a 180-degree hook is used (b) If a 90-degree hook is used Use fc’ = 28 MPa and fy = 420 MPa
50 cm
Determine the development length or anchorage required for the epoxy-coated top bars of the beam shown in the figure. The beam frames into an exterior 80cm x 30cm column (the bars extend parallel to the 80 cm side). Show the details if:
4Φ32 Φ12@15
Solution: α=1.3 for bars over concrete > 30 cm thick β=1.5 for coated bars (take the larger of 1.2 and 1.5 conservatively)
α β =1.3x1.5 = 1.95 > 1.7 use 1.7 γ=1.0 for Φ32mm, C the smallest of
λ=1.0 for normal weight concrete 40+12+16=68 mm
{[400-2(40)-2(12)-32]/(3)}/(2)=44 mm i.e., C is taken as 44 mm 28
40 cm
Example 2 [contd.] 40A tr 2( 113) 15.1mm sn ( 150)( 4)
C Ktr 44 15 1.85 2.5 db 32
ld
fy
αβ γ λ
1.1 fc C K t r d b
50 cm
K tr
OK
4Φ32 Φ12@15
d b 300 mm 40 cm
420 ( 1.7 )( 1.0)( 1.0) ld 1.85 1.1 28
32 2127 m m 300 mm OK
Available length for bar development = 800 – 40 = 760 mm < ld = 2127 cm
Thus, a standard hook is required at column side
ldh = lhb x applicable modification factors ≥ 150 mm or 8db. (use a factor 1.2 for epoxy-coated hooks. Modification factors are inapplicable) l dh 29
0.24 e f y
fc '
db
0.24 1.2 420 32 732 m m 1.0 28
150m m 8( 32) 256m m O K
Example 2 [contd.] (b) If a 180-degree hook is used
ldh=732 mm
4db =128 mm Critical section 5db =160 mm 180o hook
12db=384 mm
(c) If a 90-degree hook is used
30
ldh=732 mm
Critical section
90o hook
Splicing
31
Splices of Reinforcement
ACI 12.14
Splicing of reinforcement bars is necessary, either because the available bars are not long enough, or to ease construction, in order to guarantee continuity of the reinforcement according to design requirements.
Types of Splices: (a) Welding (b) Mechanical connectors (c) Lap splices (simplest and most economical method) In a lapped splice, the force in one bar is transferred to the concrete, which transfers it to
the adjacent bar. Splice length is the distance over which the two bars overlap.
Forces on bar at splice
32
Splice length
Splices of Reinforcement Important note: Lap splices have a number of disadvantages, including congestion of reinforcement at the lap splice and development of transverse cracks due to stress concentrations. It is recommended to locate splices at sections where stresses are low. Types of Lap Splices: 1. Direct Splice T
T ls
Direct
2. Non- Splice (spaced) the distance between two bars cannot be greater than 1/5 of the splice length nor 15 cm
ACI 12.14.2.3 T
s
T ls
33
Bars are spaced
Splices of Deformed Bars in Tension
ACI 12.15
ACI code divides tension lap splices into two classes, A and B. The class of splice used is dependent on the level of stress in the reinforcing and on the percentage of steel that is spliced at particular location.
ACI 12.15.1
ClassA: A splice must satisfy the following two conditions to be in this class: (a) the area of reinforcement provided is at least twice that required by analysis over the entire length of the splice; and (b) one-half or less of the total reinforcement is spliced within the required lap length. Class B: If conditions above are not satisfied classify as Class B. The splice lengths for each class of splice are as follows: Class A splice: 1.0 ld 300 mm Class B splice: 1.3 ld 300 mm 34
ACI 12.15.2
Example 3 To facilitate construction of a cantilever retaining wall, the vertical reinforcement shown in the figure, is to be spliced with dowels extending from the foundation. Determine the required splice length when all reinforcement bars are spliced at the same location.
Use fc’ = 30 MPa and fy = 420 MPa Φ16 @ 250 Cover = 7.5 cm
Solution: Class B splice is required where ls = 1.3 ld α=1.0, β=1.0 → α β =1.0 < 1.7 OK ls
γ=1.0, λ=1.0 C the smallest of
75+8=83 mm 250/2=125 mm
i.e., C is taken as 83 mm 35
Ktr =0.0, since no stirrups are used
Φ16 @ 250 Cover = 7.5 cm
Example 3 [contd.] C Ktr 83 0 C Ktr 5.19 2.5 i .e., 2.5 db 16 db
420 ( 1.0)( 1.0)( 1.0) 16 446 m m ld 2.5 1.1 30 Required splice length l s 446( 1.3 ) 580 m m 300 OK Φ16 @ 25
ls=58 cm
Φ16 @ 25
36
Splices of Deformed Bars in Compression
ACI 12.16
Bond behavior of compression bars is not complicated by the problem of transverse tension cracking and thus compression splices do not require provisions as strict as those specified for tension Compression lap splice length shall be: 0.071 fy db ≥ 300 mm (0.13 fy – 24) db ≥ 300 mm
ACI 12.16.1 for fy ≤ 420 MPa for fy > 420 MPa
The computed splice length should be increase by 33% if fc’<21 MPa When bars of different size are lap-spliced in compression, splice length shall be the larger of either development length of the larger bar, or splice length of the smaller bar.
ACI 12.16.2 ACI 12.15.3 37
Example 4 Design a compression lap splice for a tied column whose cross section is shown in the figure when: (a)Φ16 mm bars are used on both sides of the splice. (b)Φ 16 mm bars are lap spliced with Φ 18 mm bars. Use fc’ = 30 MPa and fy = 420 MPa
Solution: (a) For bars of same Φ16 mm diameter
Splice length in compression and for fy =420 MPa is equal to 0.071 fy db = 0.071 (420)(16) = 477 mm >300 mm taken as 480 mm
38
Example 4 [contd.] (b) For bars of different diameters
The development length of the larger bar
ldc = ldb x applicable modificationfactors
0.24f y d b 0.24 420 18 331mm fc ' 30 l dc max 333mm 0.043f d 0.043 420 18=333mm y b Splice length of smaller diameter bar was calculated in part (a) as 477 mm. Thus, the splice length is taken as 480 mm.
39
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 PART II Bar cutoff
Bar cutoff It is economical to cut unnecessary bars as shown in the scenario below.
2
Bar cutoff: Theoretical points of cutoff or bent Example
3
Bar cutoff: Theoretical points of cutoff or bent Example
4
Bar cutoff: Theoretical points of cutoff or bent Example
5
Bar cutoff: Theoretical points of cutoff or bent Example
6
Bar cutoff: Theoretical points of cutoff or bent Using moment diagrams drawn to scale:
7
Bar cutoff: Theoretical points of cutoff or bent Using moment envelopes drawn to scale:
8
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/11, +1/16, -1/11)
9
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/16, +1/14, -1/10)
10
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/24, +1/14, -1/10)
11
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: 0, +1/11, -1/10)
12
Bar cutoff: Theoretical points of cutoff or bent Development length requirements
ACI 12.10.3 Reinforcement shall extend beyond the point at which it is no longer required to resist flexure for a distance equal to d or 12db, whichever is greater, except at s of simple spans and at free end of cantilevers.
ACI 12.10.4 Continuing reinforcement shall have an embedment length not less than ld beyond the point where bent or terminated tension reinforcement is no longer required to resist flexure. 13
Bar cutoff: Theoretical points of cutoff or bent Development length requirements
ACI 12.10.5
The ACI Code does not permit flexural reinforcement to be cutoff in a tension zone unless at least one of the special conditions, shown below, is satisfied: a.Factored shear force at the cutoff point does not exceed two-thirds of the design shear strength, ΦVn . b.Stirrup area exceeding that required for shear and torsion is provided along each cutoff bar over a distance from the termination point equal to three-fourths of the effective depth of the member. Excess stirrup area Av is not to be less than 0.41bwS /fy . Spacing S is not to exceed d/8βb where βb is the ratio of area of reinforcement cutoff to total area of tension reinforcement at the section. c.For φ 36 mm bars and smaller, continuing reinforcement provides double the area required for flexure at the cutoff point and factored shear does not exceed three-fourths of the design shear strength, ΦVn .
14
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
At least one-third the positive moment reinforcement in simple and one-fourth the positive moment reinforcement in continuous shall extend along the same face of member into the . In beams, such reinforcement shall extend into the at least 150 mm. ACI 12.11.1
At simple s and at points of inflection, positive moment tension reinforcement shall be limited to a diameter such that
ACI 12.11.3 Mn is calculated assuming all reinforcement at the section to be stressed to fy; Vu is calculated at the section; la at a shall be the embedment length beyond the center of ; or: la at a point of inflection shall be limited to d or 12db, whichever is greater.
15
An increase of 30 percent in the value of Mn /Vu shall be permitted when the ends of reinforcement are confined by a compressive reaction.
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
ACI 12.11.3
16
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
17
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
18
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
19
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Negative moment:
Negative moment reinforcement in a continuous, restrained cantilever member, or in any member of rigid frame, is to be anchored in or through the ing member by development length, hooks, or mechanical anchorage.
ACI 12.12.1
At least one-third the total tension reinforcement provided for negative moment at a shall have an embedment length beyond the point of inflection not less than d, 12db, or ln/16, whichever is greater
ACI 12.12.3
20
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 PART III Detailing of reinforcement
References for detailing ACI-318
2
References for detailing ACI-315 ACI Detailing Manual
3
References for detailing CRSI Design Handbook
4
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
At least one-third the positive moment reinforcement in simple and one-fourth the positive moment reinforcement in continuous shall extend along the same face of member into the . In beams, such reinforcement shall extend into the at least 150 mm. Negative moment:
At least one-third the total tension reinforcement provided for negative moment at a shall have an embedment length beyond the point of inflection not less than d, 12db, or ln/16, whichever is greater 5
Typical details for one way solid slabs
6
Requirements for using standard detailing for beams and one way slabs: ACI 8.3.3 • 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. • are of similar section dimensions along their lengths (prismatic).
7
Typical details for one way solid slabs Straight bars
8
Typical details for one way solid slabs Straight bars
9
Typical details for one way solid slabs Straight bars
10
Typical details for one way solid slabs Bent-up bars
11
Typical details for beams Straight bars
12
Typical details for beams Straight bars
13
Typical details for beams Straight bars
14
Typical details for columns
15
Typical details for columns
16
17
18
19
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 13 Design of isolated footings
Footing Introduction Footings are structural elements used to columns and walls and transmit their loads to the underlying soil without exceeding its safe bearing capacity below the structure. Loads
B
B
L
Column
L
P
Beam
P M Footing
Soil
2
Footing Introduction The design of footings calls for the combined efforts of geotechnical and structural engineers. The geotechnical engineer, on one hand, conducts the site investigation and on the light of his findings, recommends the most suitable type of foundation and the allowable
bearing capacity of the soil at the suggested foundation level.
The structural engineer, on the other hand, determines the concrete dimensions and reinforcement details of the approved foundation.
3
Types of Footing Isolated Footings Isolated or single footings are used to single columns. This is one of the most economical types of footings and is used when columns are spaced at relatively long distances.
PkN
B
C2 C1 L P
4
Types of Footing Isolated Footings
Shapes of isolated footings 5
Types of Footing Isolated Footings
Shapes of isolated footings 6
Types of Footing Wall Footings Wall footings are used to structural walls that carry loads from other floors or to nonstructural walls. W kN/m
Secondary reinft
Main reinft.
7
Types of Footing Combined Footings Combined footings are used when two columns are so close that single footings cannot be used. Or, when one column is located at or near a property line. In such a case, the load on the footing will be eccentric and hence this will result in an uneven distribution of load to the ing soil. P1
P2
P2 kN
L
B
PP11kN kN
C2
C2 C1
C1 L1
L2
L2
8
Types of Footing Combined Footings The shape of a combined footing in plan shall be such that the centroid of the foundation plan coincides with the centroid of the loads in the columns. Combined footings are either rectangular or trapezoidal. Rectangular footings are favored due to their simplicity in of design and construction. However, rectangular footings are not always practicable because of the limitations that may be imposed on their longitudinal projections beyond the two columns or the large difference that may exist between the magnitudes of the two column loads. Under these conditions, the provision of a trapezoidal footing is more economical.
9
Types of Footing Continuous Footings Continuous footings a row of three or more columns.
P1
P2
P3
P4 kN
P4 P3 kN
P2 kN L P1 kN
B
10
Types of Footing Strap (Cantilever) footings Strap footings consists of two separate footings, one under each column, connected together by a beam called “strap beam”. The purpose of the strap beam is to prevent overturning of the eccentrically loaded footing. It is also used when the distance between this column and the nearest internal column is long that a combined footing will be too narrow. P2 kN P2
property line
P1
Strap Beam P1 kN L1
L2
C2
B1 C1
C2
B2
C1
11
Types of Footing Mat (Raft) Footings Mat footings consist of one footing usually placed under the entire building area. They are used when soil bearing capacity is low, column loads are heavy and differential settlement for single footings are very large or must be reduced.
L
12
Types of Footing Pile caps Pile caps are thick slabs used to tie a group of piles together to and transmit column loads to the piles. P
B
L
13
Footing Loading Distribution of Soil Pressure The distribution of soil pressure under a footing is a function of the type of soil, the relative rigidity of the soil and the footing, and the depth of the foundation at the level of between footing and soil. P
P
P Centroidal axis
L
Footing on sand
L
Footing on clay
L
Equivalent uniform distribution
For design purposes, it is common to assume the soil pressure is uniformly distributed. The pressure distribution will be uniform if the centroid of the footing coincides with the resultant of the applied loads.
14
Footing Loading Pressure Distribution Below Footings The maximum intensity of loading at the base of a foundation which causes failure of soil is called ultimate bearing capacity of soil, denoted by qu. The allowable bearing capacity of soil is obtained by dividing the ultimate bearing capacity of soil by a factor of safety on the order of 2.50 to 3.0. The allowable soil pressure for soil may be either gross or net pressure permitted on the soil directly under the base of the footing. The gross pressure represents the total stress in the soil created by all the loads above the base of the footing. For design, the net soil pressure is used instead of the gross pressure value. P
Df hc
15
Footing Loading Concentrically Loaded Footings If the resultant of the loads acting at the base of the footing coincides with the centroid of the footing area, the footing is concentrically loaded and a uniform distribution of soil pressure is assumed in design. P Centroidal axis
L P/A L
B
16
Footing Loading Eccentrically Loaded Footings Footings are often designed for both axial load and moment. Moment may be caused by lateral forces due to wind or earthquake, and by lateral soil pressures. A footing is eccentrically loaded if the ed column is not concentric with the footing area or if the column transmits at its juncture with the footing not only a vertical load but also a bending moment. P
P
e M Centroidal axis
Centroidal axis
y
y
L
L P/A
P/A
Pey/I
My/I
17
Design of Isolated Footings Deformation of isolated footings
18
Design of Isolated Footings Deformation of isolated footings
19
Design of Isolated Footings The design of isolated rectangular footings is detailed in the following steps:
1- Select a trial footing depth. Depth of footing above reinforcement is not to be less than 15 cm.
ACI 15.7 Note that 7.5 cm of clear concrete cover is required if concrete is cast against soil.
ACI 7.7.1
20
Design of Isolated Footings 2- Evaluate the net allowable soil pressure:
qall (net) = qall (gross) - γc hc - γs (Df -hc) P
Df hc
where
qall(net)
hc is the assumed footing depth, df is the distance from ground surface to the surface between footing base and soil, γc is the weight density of concrete, and γs is the weight density of soil on top of footing. 21
Design of Isolated Footings 3- Establish the required base area of the footing Base area of footing is determined from unfactored forces transmitted by footing to soil and the allowable soil pressure evaluated through principles of soil mechanics.
Areq
PD PL qall (net)
ACI 15.2.2
where PD and PL are column service dead and live loads, respectively. Select appropriate L, and B values, if possible, use a square footing to achieve greatest economy.
4- Evaluate the net factored soil pressure: qu(net )
1.2PD 1.6PL LB
ACI 15.2.1 22
Design of Isolated Footings 5- Check footing thickness for punching shear. When loads are applied over small areas of slabs and footings with no beams, punching failure may occur. The sloping failure surface takes the shape of a truncated pyramid in case of rectangular columns, and a truncated cone in case of circular columns.
The ACI Code assumes that failure takes place on vertical planes located at distance d/2 from the faces of the column.
ACI 11.11.1.2
23
Design of Isolated Footings 5- Check footing thickness for punching shear [contd.] The depth of the footing must be checked so that the shear capacity of the concrete equals or exceeds the critical shear forces produced by factored loads
Vu Vc The critical punching shear forceVu can be evaluated as follows
Vu qu (net)L B C1 d C2 d
C1
B
C2
C2 + d
C1 + d
ACI 11.11.1.2
L
Since there are two layers of reinforcement, an average value of d may be used: d = h − 7.5cm− db , where db is the bar diameter.
24
Design of Isolated Footings 5- Check footing thickness for punching shear [contd.] Punching shear force resisted by concrete Vc is given as the smallest of
2 V C 0.171 f c 'bod c
C2
B
V C 0.083 2 sd f c 'bod
C1
C2 + d
V C 0.33 f c 'bod
C1 + d
b
L
βc = long side/short side of column, αs = 40 for interior, 30 for side, and 20 for corner columns, bo =length of critical perimeter around the column = 2[(C1+d)+(C2+d)]
Interior
ACI 11.11.2.1 Corner
Exterior
25
Design of Isolated Footings 6- Check footing thickness for beam shear in each direction.
If Vu ≤ ΦVc, thickness will be adequate for resisting beam shear. The critical section for beam shear is located at distance d from column faces.
The factored shear force is given by
Critical section for beam shear (short direction)
x
L C 1 Vu qu (net ) B x qu (net ) B d 2
V c 0.17 f c ' B d
C2
The factored beam shear capacity of the concrete is given as
C1
d
B
In the short direction:
L
ACI 11.2.1.1 26
Design of Isolated Footings 6- Check footing thickness for beam shear in each direction [contd.]
The factored beam shear capacity of the concrete is given as
V c 0.17 f c ' L d
B
C2
C1
d
B C 2 Vu qu (net ) L y qu (net ) L d 2
y
The factored shear force is given by
Critical section for beam Shear (long direction)
In the long direction:
L
ACI 11.2.1.1
Increase footing thickness if necessary until the condition Vu ≤ ΦVc is satisfied.
27
Design of Isolated Footings 7-Compute the area of flexural reinforcement in each direction. The footing is designed as rectangular-section beam in both directions. The critical section for bending is located at the face of the column.
ACI 15.4.2
Critical section formoment
(L-C1)/2
Reinforcement in the long direction: 2
0.85f c 2M u 1 1 2 0.85 f Bd fy c As ,req Bd
C1
B
C2
B L C1 M u qu (net) 2 2
L
As ,min 0.0018Bh As ,req
ACI 15.4.1 ACI 10.5.4 ACI 7.12.2.1
28
Design of Isolated Footings
0.85f c 2M u 1 1 2 0.85 f Ld fy c A s,req Ld
As ,min 0.0018Lh As ,req
C1
B
2
C2
L B C2 M u qu (net) 2 2
(B-C2)/2
Reinforcement in the short direction
Critical section for moment
7-Compute the area of flexural reinforcement in each direction [contd.]
L
ACI 15.4.1
ACI 10.5.4 ACI 7.12.2.1 29
Design of Isolated Footings
where
long side of footing
B
7-Compute the area of flexural reinforcement in each direction [contd.] For square footings, the reinforcement is identical in both directions. For rectangular footings, the reinforcement in the long direction is uniformly distributed. However, a portion of the total reinforcement in the short direction, γsAs is distributed uniformly over a band width (centered on centerline of column) as shown in the figure. Remainder of reinforcement required in the short direction, (1 – γs)As, shall be distributed uniformly outside the center band width of the footing. Band width 2 s 1
short side offooting
ACI 15.4.4
B L
39
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete
All forces applied at the base of a column or wall must be transferred to the footing by bearing on concrete and/or by reinforcement.
ACI 15.8.1
Bearing on concrete for column and footing must not exceed the concrete bearing strength.
ACI 15.8.1.1
Pn Pu Otherwise, the t would fail by crushing of the concrete at the bottom of the column where the column bars are no longer effective or by crushing the concrete in the footing under the column.
Pn min Pn ,c ; Pn ,f 31
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.] For a ed column, the allowed bearing capacity ΦPn,c is
Pn ,c 0.85f cA1
ACI 10.14.1
For a ing footing where the ing surface is wider on all sides than the loaded area, the allowed bearing capacity ΦPn,f is
Pn ,f
A2 0.85f cA1 ; 2 0.85f cA1 min A1
Φ = strength reduction factor for bearing = 0.65 A1= column cross-sectional area A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal (see next slide) 32
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal
33
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal
34
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
Dowel Reinforcement: •If
Pn Pu :
Reinforcement in the form of dowel bars must be provided to transfer the excess load.
A s ,req
Pu Pn f y
ACI 15.8.1.2
The dowel bars are usually extended into the footing, bent at the ends, and tied to the main footing reinforcement.
35
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
Minimum Dowel Reinforcement: •If Pn
Pu ::
Use minimum dowel reinforcement.
As ,min 0.005A1
ACI 15.8.2.1
36
Design of Isolated Footings 9- Check for anchorage of the reinforcement > ls (compn.)
10-Prepare neat design drawings showing footing dimensions and provided reinforcement.
37
Example Design an isolated rectangular footing to an interior column 40×40cm in cross
section and carry a dead load of 800 kN and a live load of 600 kN. One of the dimensions of the footing must not exceed 3.2 m. PD= 800 kN PL= 600 kN
Use fc’= 25 MPa and fy = 420 MPa, qall (gross) = 200 kN/m2, γsoil =17 kN/m3, γconc =25 kN/m3 Df=1.0
40 40
38
Example Solution 1Select a trial footing depth: Assume that the footing is 55 cm thick. 2Evaluate the net allowable soil pressure: qall (net) = qall (gross) - γs (Df - hc) - γc hc
qall net 200 ( 1 0.55) 17 0.55 25 178.6 kN/m 2 3 Establish the required base area of the footing :
Let L 3.20 m, B
7.84 2.45m 3.20
40
40
245
A req
P P D L 800 600 7.839 m2 qall (net) 178.6
320
Use 320x245x55 cm footing 4- Evaluate the net factored soil pressure
Pu 1.2PD 1.6P L 1.2800 1.6 600 1920 kN q u net
1920 Pu 244.9 kN/m 2 LB 3.2 2.45
39
245
40+45.9
Example
40+45.9
5- Check footing thickness for punching shear:
Average effective depth davg 55-7.5-1.6 45.9cm bo 2[ 40 45.9 40 45.9] 343.6 cm
320
Vu 244.9 3.22.45 0.40 0.4590.40 0.459 1740 kN Φ VC is the smallest of Φ0.33 fc' bod 0.75 0.33 253436 459 1952 kN 2 2 3436 459 3016 kN Φ0.17 fc' 1 bod 0.75 0.17 25 1 0.4/0.4 βc αd 40 459 Φ0.083 f c' 2 s bo d 0.750.083 25 2 3436 459 3605kN bo 3436 Φ VC 1952 kN Vu 1740kN OK i.e. footing thickness is adequate for resisting punching shear. 40
Example 6- Check footing thickness for beam shear in each direction: In short direction
ΦVc 0.750.17 25 2450459 717kN
245
45.9
Vu is located at distance d from face of column
3.2 0.4 Vu 244.9 2.45 0.459 565 kN 2 ΦVc= 717 kN > Vu= 565 kN
OK
320
ΦVc 0.750.17 25 3200459 936 kN 45.9
Vu is located at distance d from face of column
2.45 0.4 Vu 244.93.2 0.459 444 kN 2 ΦVc= 936 kN > Vu= 444 kN
245
In long direction
320
OK 41
Example 7- Calculate the area of flexural reinforcement in each direction: a- Reinforcement in the long direction: The critical section for bending is shown in the figure
Critical section for moment
2 2 B L C1 2.45 3.2 0.4 244.9 M u q u net 2 2 2 2
1.4
0.85 25 ρ 420
245
588 kN .m 2 58810 6 1- 10.9 0.85 25 2450 4592
0.0031 A s 0.0031 459 2450 3500 mm
2
320 24.49 x 2.45
A s,min 0.0018 550 2450 2430 mm2 2
As,req 3500 mm 2314mm in long direction
42
Example 7- Calculate the area of flexural reinforcement in each direction: b- Reinforcement in the short direction: The critical section for bending is shown in the figure
2 412106 1- 10.9 0.85 25 3200 4592
As,req 3170 mm
2
2 24.49 x 2.8
As,min 0.0018 550 3200 3170 mm
320
1.025
0.0016 A s 0.0016 459 3200 2411mm
1.025
0.85 25 ρ 420
245
2 2 L B C2 3.2 2.45 0.4 244.9 M u q u net 2 2 2 2 Critical section for moment 412 kN .m
2
43
Example 7- Calculate the area of flexural reinforcement in each direction: b- Reinforcement in the short direction: The distribution of the reinforcement is as follows:
245
42.5
2Φ14 B
Width band =245
18Φ14 B
B 2.45 2 As Central band reinft. β 1 2 2 3170 2757 mm 1.3 1 Use 18 14 mm in the central band.
42.5
2Φ14 B
L 3.2 1.3
320
3170 2756 207 mm 2 For each of the side bands, A s 2 Use 214 mm in each of the two side bands.
44
Example 8- Check for bearing strength of column and footing concrete For the column
A1 400 400 160000mm 2
Pn ,c 0.85f cA1 0.65( 0.85 25160000) 221010 3 N 2210kN For the footing
In short direcion: 1025mm 1100mm Use 1025 mm 1400
2
1
h= 550
1025
245
1100
320
45
Example 8- Check for bearing strength of column and footing concrete
A 2 400 2 1025 400 2 1025 6002500mm
Pn ,f Pn ,f
min min
2
A2 0.85f cA1 ;2 0.85f cA1 A1 6002500 2210 ; 2 2210 4420kN 160000
Pn min Pn ,c ; Pn ,f min 2210; 4420 2210kN Pu 1920 kN Use minimum dowel reinforcement
1025 + 400+ 1025
1025 + 400+ 1025
46
Example 8- Check for bearing strength of column and footing concrete Minimum dowel reinforcement
As ,min 0.005A1 0.005 400 400 800mm
2
Use 416, As,sup = 804 mm2
47
Example 9- Check for anchorage of the reinforcement Bottom longitudinal reinforcement (Φ14mm) α=1.0 for bottom bars,
β=1.0 for uncoated bars 1.4
α β =1.0 <1.7 OK γ=0.8 for Φ14mm,
7.5+0.7=8.3 cm
245
C the smallest of
λ=1.0 for normal weight concrete
[245-2(7.5)-1.4]/(22)/(2)=5.2 cm i.e., C is taken as 5.2 cm
C K tr 5.2 0 C K tr 3.7 2.5 i.e.,use 2.5 db 1.4 db
320
420 (1.0)(1.0)(0.8)(1.0) 1.4 34 cm ld 1.1 25 2.5 Available length in long direction =140-7.5=132.5 > 34 cm 48
Example 9- Check for anchorage of the reinforcement Bottom reinforcement in short direction (Φ14mm) α=1.0 for bottom bars,
β=1.0 for uncoated bars
α β =1.0 <1.7 OK γ=0.8 for Φ14mm,
245
7.5+0.7=8.3 cm
[320-2(7.5)-1.4]/(19)/(2)=8 cm
1.025
C the smallest of
λ=1.0 for normal weight concrete
i.e., C is taken as 5.2 cm
320
C K tr 8 0 C K tr 5.7 2.5 i.e.,use 2.5 db 1.4 db 420 (1.0)(1.0)(0.8)(1.0) 1.4 34 cm ld 1.1 25 2.5 Available length in short direction =102.5-7.5=95 > 34 cm 49
Example 9- Check for anchorage of the reinforcement Dowel reinforcement (Φ16mm):
0.24f y d b 0.24 42016 323mm fc ' 25 l dc max 323mm 200mm 0.043f d 0.04342016=289mm y b Available length = 550-75-14-14 = 447 mm > 323 mm OK Column reinforcement splices:
Considering that the column is reinforced with 16 bars ls 0.071f yd b 0.071 42016 478 mm 300 mm taken as 48cm
> ls (compn.)
50
Example
55 cm
48cm
10- Prepare neat design drawings showing footing dimensions and provided reinforcement
245 (18Φ14)
2.45 m
2Φ14 B
2Φ14 B
18Φ14 B
3.20 m
23Φ14 B 42.5
Width band =245
42.5
51
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 14 Staircase Design
Stair Types
2
Stair Types
3
Stair Types
4
Stair Types
5
Technical •Going: horizontal upper portion of a step. •Rise: vertical distance between two consecutive treads. •Flight: a series of steps provided between two landings.
•Landing: a horizontal slab provided between two flights. •Waist: the least thickness of a stair slab.
6
Technical •Winder: radiating or angular tapering steps. •Soffit: the bottom surface of a stair slab. •Nosing: the intersection of the going and the riser.
•Headroom: the vertical distance from a line connecting the nosings of all treads and the soffit above.
7
General Design Requirements
8
Stair type based on the structural loading type
Simply ed stair (transversely ed)
9
Simply ed stair (longitudinally ed)
Cantilever stair
Design of transversely ed stairs Loading: a. Dead load: The dead load includes own weight of the step, own weight of the waist slab, and surface finishes on the steps and on the soffit.
b. Live Load: Live load is taken as building design live load plus 1.5 kN/m2, with a maximum value of 5 kN/m2.
10
Design of transversely ed stairs Direction of bending Main reinforcement Shrinkage reinforcement
11
Direction of bending
Design of transversely ed stairs Design for Shear and Flexure: Each step is designed for shear and flexure as if it is a beam. Main reinforcement runs in the transverse direction at the bottom side of the steps while shrinkage reinforcement runs at the bottom side of the slab in the longitudinal direction. Since the step is not rectangular, the effective depth d is found by an equivalent rectangular section that can be used with an average height equal to:
t
R
havg
12
t
Design of transversely ed stairs Example 1 Design a straight flight stair in a residential building ed on reinforced concrete walls 1.5 m apart (center to center), given: L.L = 3 kN/m2; covering material = 0.5 kN/m; The risers are 16 cm and goings are 30 cm; fc’=25 MPa, fy= 420 MPa
13
Loads and Analysis
t
l 1.5 0.075m 20 20 t
0.16 0.165m have 0.30 0.34 2 D.L(O.W) =0.340.075 25 + (1/2) 0.16 0.3 25=1.24 kN/m D.L (covering material) = 0.5 kN/m
0.075
0.3
D.L (total) = 1.74 kN/m L.L =30.3 =0.9 kN/m
0.16
0.302 0.162 0.34
1.5 m 14
Shear diagram
Moment diagram 15
Design for moment M u 1kN .m d 165 20 6 139mm bw 300mm
0.85 f c ' 1 1 fy
2M u 0.85f c ' b d 2
0.8525 1 1 21106 0.0005 2 420 0.90.85 25 300 139 As 0.0005 300139 20.9mm
2 2
As ,min 0.0018 300165 89.1mm As As As ,min 89.1mm Use 112 for each step
16
2
Design for shear V C 0.75 0.17 25 139 300 /1000 26kN V u 2.65kN OK
17
Design of longitudinally ed stairs Direction of bending Shrinkage reinforcement
Main reinforcement
18
Design of longitudinally ed stairs
19
Design of longitudinally ed stairs Deflection Requirement: Since a flight of stairs is stiffer than a slab of thickness equal to the waist t, minimum required slab depth is reduced by 15 %.
Effective Span: The effective span is taken as the horizontal distance between centerlines of ing elements. n = number of goings X = Width of ing landing slab at one end of the stairs slab
20
Y = Width of ing landing slab at the other end of the stairs slab.
Design of longitudinally ed stairs Deflection Requirement: Since a flight of stairs is stiffer than a slab of thickness equal to the waist t, minimum required slab depth is reduced by 15 %.
Effective Span: The effective span is taken as the horizontal distance between centerlines of ing elements. n = number of goings X = Width of ing landing slab at one end of the stairs slab
21
Y = Width of ing landing slab at the other end of the stairs slab.
Design of longitudinally ed stairs Loading: a. Dead Load: The dead load, which can be calculated on horizontal plan, includes: •Own weight of the steps. •Own weight of the slab. •Surface finishes on the flight and on the landings. Note: For flight load calculations, the part of load acting on slope is to be increased by dividing it by cosα. This is because analysis for moment and shear is conducted on the horizontal span of the flight, but the load is that carried on the inclined span.
P P= wo.w.Linc .Linc
22
.L
w=P/L= wo.w.Linc/L= wo.w./cosα
Design of longitudinally ed stairs Loading: b. Live Load: Live load is taken as the building design live load plus 1.5 kN/m2, with a maximum value of 5 kN/m2. Live load is always given on the horizontal projection.
23
Design of longitudinally ed stairs t detail: The stairs slab is designed for maximum shear and flexure. Main reinforcement runs in the longitudinal direction, while shrinkage reinforcement runs in the transverse direction. Special attention has to be paid to reinforcement detail at opening ts.
24
Design of longitudinally ed stairs Example 2 Design the U- stair in a residential building shown in the figure, given: L.L = 3 kN/m2; covering material = 2 kN/m2; The rises are 16 cm and goings are 30 cm, fc’=25 MPa, fy= 420 MPa
25
Loads and Analysis t 0.85
l 525 22cm 20 20
cos() = 0.3/ 0.34 = 0.88 Take a unit strip along the span: D.L (slab) = 0.221.025/0.88 =6kN/m D.L (step) = (1/2) 0.161.0 25=2 kN/m D.L (covering material) = 21.0=2 kN/m D.L (flight) = 10 kN/m D.L (landing) = 8 kN/m L.L =3 1.0=3 kN/m
26
Wu (flight) = 1.2(10)+1.6(3)=16.8kN/m Wu (landing) = 1.2(8)+1.6(3)=14.4kN/m
0.3
0.16 0.34
Moment and shear diagram 14.4kN/m
27
16.8kN/m
14.4kN/m
Design for moment M u 52.2kN .m d 22 2 0.6 19.4cm 194mm bw 1000mm 0.8525 1 1 252.2106 0.0037 2 0.85 0.9 25 1000194 420 2
As 0.00371000194 718mm 2 0.00181000 220 396mm A As ,min
s
OK
Use 812
(22)=3.96 cm2/m
Design for shear 28
V C 0.75 0.17 25 1941000 / 1000 127.3kN V u 38.25kN OK
29
Design of quarter-turn stairs
A landing may be shared on two different stair slabs. The load of the shared landing can be assumed to be divided equally and each stair slab carries on 30
half.
Design of stair beams
Ls
P=wsLs/2
31
ws
P w=P/(L/2) L/2
Dr. Nader Okasha
Lecture 0 Syllabus
Reinforced Concrete Design
Instructor
Dr. Nader Okasha.
[email protected]
Office Hours
As needed.
٢
Reinforced Concrete Design
This course is only offered for 2010 students who have ed strength of materials.
If you don’t meet this criteria you will not be allowed to continue this course.
٣
Reinforced Concrete Design
References: Building Code Requirements for Reinforced Concrete and commentary (ACI 318M-08). American Concrete Institute, 2008. Design of Reinforced Concrete. 7th edition, McCormac, J.C. and Nelson, J.K., 2006. Reinforced Concrete Design. By Dr. Sameer Shihada. ٤
Reinforced Concrete Design
Additional references (internationally recognized books in reinforced concrete design): Reinforced Concrete, A fundamental Approach. Edward Nawy. Design of Concrete Structure. Nilson A. et al.
Reinforced Concrete Design. Kenneth Leet. Reinforced Concrete: Mechanics and Design. James K. Wight, and James G. MacGregor. ٥
Reinforced Concrete Design
The art of design
Design is an analysis of trial sections. The strength of each trial section is compared with the expected load effect. The load effect on a section is determined using structural analysis and mechanics of materials. The strength of a reinforced concrete section is determined using the concepts taught in this class.
٦
Reinforced Concrete Design
٧
Reinforced Concrete Design Course outline Week
1
2, 3,4
Topic
Introduction: -Syllabus and course policies. -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 loads. -Strength analysis of beams according to ACI Code. -Design of singly reinforced rectangular beams. -Design of T and L beams. -Design of doubly reinforced beams.
4
Design of beams for shear.
5
Midterm. ٨
Reinforced Concrete Design Course outline Week
Topic
6
Design of slabs: One way solid slabs – One way ribbed slabs.
7
Design of short concentric columns.
7,8
Bond, development length, splicing and bar cutoff.
8,9
Design of isolated footings.
9
Staircase design.
10
Final
٩
Reinforced Concrete Design
Grading
Course work:
20%
-Homework
4%
-Attendance
4%
-Project
12%
Mid-term exam
20%
Final exam
60% ١٠
Reinforced Concrete Design
Exam Policy
Mid-term exam: Only one A4 cheat-sheet is allowed.
Necessary figures and tables will be provided with the exam forms.
Final exam: Open book. ١١
Reinforced Concrete Design
Homework Policy Show all your assumptions and work details. Prepare neat sketches showing the reinforcement and dimensions. Marking will consider primarily neatness of presentation, completeness and accuracy of results. You may get the HW points if you copy the solution from other students. However, you will have lost your chance in practicing the concepts through doing the HW. This will lead you to loosing points in the exams, which you could have gained if you did your HWs on your own. No late HWs will be accepted. Homework solutions will be posted on upinar immediately after the submission deadline.
١٢
Reinforced Concrete Design
Policy towards cell-phone use
١٣
Reinforced Concrete Design
Policy towards discipline during class Zero tolerance will be practiced.
No talking with other students is allowed. Raise your hand before answering or asking questions.
Leaving during class is not allowed (especially for answering the cell-phone) unless a previous permission is granted. Violation of discipline rules may have you dismissed from class and jeopardize your participation points.
١٤
Reinforced Concrete Design
Policy towards missed classes Any collectively missed class MUST be made up.
A collectively missed class will be made up either on a Thursday or during the discussion lecture. An absence from a lecture will loose you attendance points, and the lecture will not be repeated for you. You are on your own. You may use the lecture videos. No late students will be allowed in class.
Anything mentioned in class is binding. No excuse for not being there or not paying attention.
١٥
Reinforced Concrete Design
Units used in class In all equations, the input and output units are as follows: Distance (L,b,d, ): mm Area (Ac,Ag,As): mm2 Volume (V): mm3 Force (P,V,N): N Moment (M): N.mm Stress (fy, fc’): N/mm2 = MPa = 106 N/m2 Pressure ( s): N/mm2 Distributed load per unit length (wu): N/mm Distributed load per unit area ( u): N/mm2 Weight per unit volume (): N/mm3
١٦
Reinforced Concrete Design
Units used in class However, these quantities may be presented as Distance (L,b,d, ): cm , m Area (Ac,Ag,As): cm2, m2 Volume (V): cm3, m3
Force (P,V,N): kN Moment (M): kN.m Pressure (qs): kN/m2 Distributed load per unit length ( u): kN/m Distributed load per unit area (qu): kN/m2
Weight per unit volume ( ): kN/m3 ١٧
Reinforced Concrete Design
Unit conversions 1 m = 102 cm = 103 mm 1 m2 = 104 cm2 = 106 mm2 1 m3 = 106 cm3 = 109 mm3 1 kN = 103 N 1 kN.m = 106 N.mm 1 kN/m2 = 10-3 N/mm2 1 kN/m3 = 10-6 N/mm3
You MUST specify the unit of each result you obtain ١٨
Reinforced Concrete Design
ACI Equations The equations taken from the ACI code will be indicated throughoutthe slides by their section or equation number in the code provided in shading.
Examples:
Ec 4700
f c
ACI 8.5.1
f r 0.62
f c
ACI Eq. 9-10
Some of the original equations may have included the symbol = 1.0 for normal weight concrete and omitted in slides. ١٩
Reinforced Concrete Design
Advices for excelling in this course: Keep up with the teacher and pay attention in class.
Study the lectures up to date. Re-do the lecture examples.
Look at additional resources.
DO YOUR HOMEWORK!!!!! Check your solution with the HW solution ed to upinar. ٢٠
Reinforced Concrete Design
ENJOY THE COURSE!!
٢١
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 1 Introduction to reinforced concrete
Contents 1.
2.
Concrete-producing materials Mechanical properties of concrete 3.
Steel reinforcement
2
Part 1: Concrete-Producing Materials
3
Advantages of reinforced concrete as a structural material 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.
4
Advantages of reinforced concrete as a structural material 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.
5
Disadvantages of reinforced concrete as a structural material 1. It has a very low tensile strength. 2. Forms are required to hold the concrete in place until it hardens.
3. Concrete 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.
6
Compatibility of concrete and steel 1. Concrete is strong in compression, and steel is strong in tension. 2. The two materials bond very well together.
3. Concrete protects the steel from corrosive environments and high temperatures in fire. 4. The coefficients of thermal expansion for the two materials are quite close.
7
Concrete Concrete is a mixture of cement, fine and coarse aggregates, and water. This mixture creates a formable paste that hardens into a rocklike mass.
8
Concrete Producing Materials • • • •
Portland Cement Aggregates Water ixtures
9
Portland Cement The most common type of hydraulic cement used in the manufacture 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 ordinary cement. Concrete made with normal Portland cement require about two weeks to achieve a sufficient strength to permit the removal of forms and the application of moderate loads.
10
Types of Cement
Type I: General Purpose
Type II: Lower heat of hydration than Type I
Type III: High Early Strength • Quicker strength • Higher heat of hydration
11
Types of Cement 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. If the desired type of cement is not available, different ixtures may be used to modify the properties of Type 1 cement and produce the desired effect. 12
Aggregates Aggregates are particles that form about three-fourths of the volume of finished concrete. According to their particle size, aggregates are classified as fine or coarse.
Coarse Aggregates Coarse aggregates consist of gravel or crushed rock particles not less than 5 mm in size.
Fine Aggregates Fine aggregates consist of sand or pulverized rock particles usually less than 5 mm in size. 13
Water Mixing water should be clean and free of organic materials that react 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. 14
ixtures – Applications: • Improve workability (superplasticizers) • Accelerate or retard setting and hardening • Aid in curing • Improve durability
15
Concrete Mixing In the design of concrete mixes, three principal requirements for concrete are of importance: • Quality • Workability • Economy
16
Part 2: Mechanical Properties of Concrete
17
Mechanical Concrete Properties '
Compressive Strength, f c
• Normally, 28-day strength is used as the design strength.
18
Mechanical Concrete Properties '
Compressive Strength, f c
• It is determined through testing standard cylinders 15 cm in diameter and 30 cm in height in uniaxial compression at 28 days (ASTM C470). • Test cubes 10 cm × 10 cm × 10 cm are also tested in uniaxial compression at 28 days (BS 1881).
19
Mechanical Concrete Properties '
Compressive Strength, f c
• The ACI Code is based on the concrete compressive strength as measured by a standard test cylinder. f c Cylinder 0.8f c Cube
• For ordinary applications, concrete compressive strengths from 20 MPa to 30 MPa are usually used. 20
Mechanical Concrete Properties Compressive-Strength Test
21
Mechanical Concrete Properties Modulus of Elasticity, Ec ' • Corresponds to the secant modulus at 0.45 f c • For normal-weight concrete: Ec 4700 f c
22
0.002
ACI 8.5.1
0.003
Mechanical Concrete Properties Tensile Strength – Tensile strength ~ 8% to 15% of f c
'
– Tensile strength of concrete is quite difficult to measure with direct axial tension loads because of problems of gripping the specimen and due to the secondary stresses developing at the ends of the specimens. – Instead, two indirect tests are used to measure the tensile strength of concrete. These are given in the next two slides.
23
Mechanical Concrete Properties Tensile Strength – Modulus of Rupture, fr
f r 0.62 f c
ACI Eq. 9-10
– Modulus of Rupture Test (or flexural test): P
24
Mc 6M 2 fr I bh
unreinforced concrete beam
fr
Mechanical Concrete Properties Tensile Strength – Splitting Tensile Strength, fct
f ct 0.56 f c
ACI R8.6.1
– Split Cylinder Test Concrete Cylinder
P
Poisson’s Effect
f ct
2P Ld
25
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.
26
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
27
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 ordinary conditions about 90% of it occurs during the first year.
28
Part 3: Steel Reinforcement
29
Steel Reinforcement Tensile tests
30
Steel Reinforcement Tensile tests
31
Steel Reinforcement Stress-strain diagrams f s = ε Es ≤ f y Yield point
elastic
plastic
All steel grades have same modulus of elasticity Es= 2x105 MPa = 200 GPa
32
Steel Reinforcement Bar sizes, , # 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, fy ksi
MPa
40
276
60
414
80
552 33
Steel Reinforcement Bars are deformed to increase bonding with concrete
34
Steel Reinforcement Marks for ASTM Standard bars
35
Steel Reinforcement Bar sizes according to ASTM Standards U.S. customary units
36
Steel Reinforcement Bar sizes according to ASTM Standards SI Units
37
Steel Reinforcement Bar sizes according to European Standard (EN 10080) W
Number of bars
mm
N/m
1
2
3
4
5
6
7
8
9
10
6 8 10 12 14 16 18 20 22 24 25 26 28 30 32
2.2 3.9 6.2 8.9 12.1 15.8 19.9 24.7 29.8 35.5 38.5 41.7 45.4 55.4 63.1
28 50 79 113 154 201 254 314 380 452 491 531 616 707 804
57 101 157 226 308 402 509 628 760 905 982 1062 1232 1414 1608
85 151 236 339 462 603 763 942 1140 1357 1473 1593 1847 2121 2413
113 201 314 452 616 804 1018 1257 1521 1810 1963 2124 2463 2827 3217
141 251 393 565 770 1005 1272 1571 1901 2262 2454 2655 3079 3534 4021
170 302 471 679 924 1206 1527 1885 2281 2714 2945 3186 3695 4241 4825
198 352 550 792 1078 1407 1781 2199 2661 3167 3436 3717 4310 4948 5630
226 402 628 905 1232 1608 2036 2513 3041 3619 3927 4247 4926 5655 6434
254 452 707 1018 1385 1810 2290 2827 3421 4072 4418 4778 5542 6362 7238
283 503 785 1131 1539 2011 2545 3142 3801 4524 4909 5309 6158 7069 8042
Areas are in mm2
38
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 2 Load types, load paths and tributary areas
Load paths Structural systems transfer gravity loads from the floors and 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. 2
Metal Deck/Slab System s Floor Loads Above
Girders Joists
Joists Floor Deck
Columns Girders
The area tributary to a joist equals the length of the joist times the sum of half the distance to each adjacent joist.
The area tributary to a girder equals the length of the girder times the sum of half the distance to each adjacent girder.
Load paths loads on structural Load is distributed over the area of the floor. This distributed load
has units of (force/area), e.g. kN/m2. w {kN/m} q {kN/m2} Beam
Loads
Beam
Slab Column
Column
Beam
Beam
Beam Footing
Slab Beam
Beam Soil
6
P {kN}
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.
w {kN/m}
7
Load paths loads on (one-way) beams This tributary load is determined by multiplying q by the tributary
width for the beam.
w {kN/m} = q {kN/m2} (S1+S2)/2 {m}
L
8
S1
S2
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 s and the centerlines of the adjacent s parallel to the long sides. A is part of the slab formed by column centerlines.
9
Load paths loads on (two-way) beams An edge beam is bounded by s from one side. An interior beam is bounded by s from two sides. qD
For edge beams: D=S/2 qD
For interior beams: D=S 10
Load paths loads on (two-way) beams
11
Load paths loads on (two-way) beams
12
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.
P {kN} = q {kN/m2} (x y){m2}
13
Load paths loads on structural Example Determine 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. B1 4m B2
5m
4.5 m
14
C2
C1 6m
5.5 m
Load paths loads on structural Example B1:
w = 10 (4)/2 = 20 kN/m B1 4m B2 5m
4.5 m
15
C2
C1 6m
5.5 m
Load paths loads on structural Example B2:
w = 10 (4+5)/2 = 45 kN/m B1
4m B2 5m
4.5 m
C2 C1
16
6m
5.5 m
Load paths loads on structural Example B1:
w = 20 kN/m
B2:
w = 45 kN/m
17
Load paths loads on structural Example C1:
P = 10 (4.5/2 6/2) 5 = 337.5 kN B1
4m B2 5m
4.5 m
18
C1
C2 6m
5.5 m
Load paths loads on structural Example C2:
P = 10 [(4.5+5)/2 (6+5.5)/2] 5 = 1366 kN B1
4m B2 5m C2
4.5 m C1 19
6m
5.5 m
Load types Classification by direction
1 Gravity loads
2 Lateral loads
20
Load types Classification by source and activity
1 Dead loads 2 Live loads 3 Environmental loads 21
Loads on Structures All structural elements must be designed for all loads anticipated to act 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 22
Live Loads (L.L) The live load is a moving or movable type of load such as occupants, furniture, etc. Live loads used in deg buildings are usually specified by local building codes. Live loads depend on the intended use of the structure and the number of occupants at a particular time.
23
See IBC 2009 TABLE 1607.1 for more live loads. http://publicecodes.citation. com/icod/ibc/2009/index.ht m?bu=IC-P-2009000001&bu2=IC-P-2009000019
Minimum live Load values on slabs Type of Use Uniform Live Load kN/m2 Residential 2 Residential balconies Computer use Offices Warehouses
3 5 2
6
Light storage
Heavy Storage Schools
12
Classrooms Libraries
2
Rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
2.5 5 2.5
Retail
4
Wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
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 the type 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. 24
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 3 Design philosophies and design codes
Design Versus Analysis Design involves the determination of the type of structural system to be used, the cross sectional dimensions, and the required reinforcement. The designed structure should be able to resist all forces expected to act during the life span of the structure safely and without excessive deformation or cracking.
Analysis involves the determination of the capacity of a section of known dimensions, material properties and steel reinforcement, if any to external forces and moments.
2
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 . 2)Serviceability to ensure satisfactory performance under service load conditions, which implies providing adequate stiffness to contain 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. 3
Building Codes, Standards, and Specifications: 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
4
Building Codes, Standards, and Specifications: 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 International Conference of Building Officials (ICBO).
2.
BOCA National Building Code (NBC): published by Building Officials and Code s International (BOCA).
3.
Standard Building Code (SBC): published by Southern Building Code Congress International (SBCCI).
5
Building Codes, Standards, and Specifications: 3 Model Codes in the U.S.
6
Building Codes, Standards, and Specifications: International Building Code (IBC): published by International Code Council (2000 ,1st edition). To replace the 3 model codes for national and international use.
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.
7
Building Codes, Standards, and Specifications: Summary: The standards that will be used extensively throughout this course is Building Code Requirements for Reinforced Concrete and commentary, known as the ACI 318M-08 code. The building code that will be used for this course is the IBC 2009, in conjunction with the ANSI/ASCE7-02.
8
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, fc’ for concrete and f y for steel.
9
The Ultimate – Strength Design Method At the present time, the ultimate-strength design method is the method 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.
reduced strength provided factored loads
The factored loads are obtained by multiplying the working loads (service loads) by factors usually greater than unity. 10
Safety Provisions (the strength requirement) Safety is required to insure that the structure can sustain all expected loads 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 fc’ and fy, *assumptions are made during 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 the debris and replacement of the structure and its contents and *cost to society. 11
Safety Provisions (the strength requirement) The strength design method, involves a two-way safety measure. The first 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 a reduction factor is usually smaller than unity Design strength ≥ Factored loads
R i Li i
ACI 9.3
ACI 9.2 12
Load factors ACI 9.2.1 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 13
Load factors ACI 9.2 Symbols
14
Strength Reduction Factors
ACI 9.3 According to ACI, strength reduction factors Φ are given as follows: a. For tension-controlled sections Φ = 0.90 b. For compression-controlled Φ = 0.75 sections, with spiral Φ = 0.65 reinforcement Other reinforced Φ = 0.75 c. For shear and torsion
Tension-controlled section
compression-controlled section
15
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 4 Analysis of beams in bending at service loads
Introduction A beam is a structural member used to the internal moments and shears and in some cases torsion.
2
Basic Assumptions in Beam Theory •Plane sections remain plane after bending. This means that in an initially straight beam, strain varies linearly over the depth of the section after bending.
Unloaded beam
Beam after bending
Its cross section
Strain distribution
3
Basic Assumptions in Beam Theory •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.
4
Stages of flexural behavior w {kN/m}
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 5
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.
6
Stage II: Concrete cracked –Elastic Stress range Once the tensile stresses reach the modulus of rupture, the section cracks. The bending moment at which this transformation takes place is called the cracking moment Mcr.
7
Stage III: Beam failure –Ultimate Strength stage As the stresses in the concrete exceed the linear limit (0.45 fc’), the concrete stress distribution over the depth of the beam varies non-linearly.
8
0.002 0.003
Stages of flexural behavior w {kN/m}
9
Flexural properties to be determined: 1Cracking moment. 2 Elastic stresses due to a given moment. 3Moments 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. 10
Cracking moment Mcr 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 the uncracked section properties.
11
Cracking moment Mcr Example 1: Calculate the cracking moment for the section shown 1 3 I g bh 12 3 10 4 1 I g (350)(750) 1.230510 mm 12 fr 0.62 fc 0.62 30 3.4MPa M cr
frI g yt
750 mm 1500 mm2
fc 30MPa
3.41.23051010 1.1143108 N.mm 111.43kN.m (750 / 2) 12
Elastic stresses – Cracked section • After cracking, the steel bars carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load. • In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where
n = the modular ratio= Es/Ec • To determine the location of the neutral axis, bx x n Asd x 0 2
1 b x2 2
n As x n Asd 0
• The height of the concrete compression block is x.
• The normal stress in the concrete and steel fc
My It
fs n
My It
13
Elastic stresses – Cracked section Example 2: fc 30MPa
Calculate the bending stresses for the section shown, M= 180 kN.m Note: M > Mcr = 111 kN.m from previous example. Thus, section is cracked.
750 mm 1500 mm2
E c 4700 f c 4700 30 25743MPa 5 E 210 s n 7.77 E c 25743 x ( 350) x ( ) 1500( 7.77)( 700 x ) 2 x 185.16mm
14
Elastic stresses – Cracked section Example 2: 1 I t bx 3 nAs ( d x ) 2 3 1 I t ( 350)( 185.16) 3 7.771500( 700 185.16) 2 3 9 4 750 mm I t 3.829510 mm
My 180106 185.16 fc 8.7MPa 9 It 3.829510 f c 8.7MPa 0.45f c 0.45( 30) 13.5MPa
fc 30MPa
1500 mm2
OK
My 180106 ( 700 185.16) f s n 188MPa 7.77 9 It 3.829510
15
Elastic stresses – Cracked section Example 3: fc 30MPa
Calculate the allowable moment for the section shown, f s(allowable)= 180 MPa, f c(allowable)= 12 MPa f s It 1803.829510 9 Ms ny ( 7.77)( 700 185.16)
750 mm 1500 mm2
8
M s 1.723410 N .mm 172.34kN .m f c I t 12 3.8295109 Mc y 185.16 8
M c 2.481910 N .mm 248.19kN.m M allowable 172.34kN.m
16
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 5 Strength analysis of beams according to ACI Code
Strength requirement for flexure in beams
Md Mu M d Design moment strength (also known as moment resistance) M u Internal ultimate moment M u 1.2M D 1.6M L
Md Φ Mn M n Theoretical or nominal resisting moment. 2
The equivalent stress (Whitney) block
Strain Distribution
Actual Stress Distribution
Approximate Stress Distribution
3
The equivalent stress (Whitney) block •The shape of the stress block is not important. •However, the equivalent block must provide the same resultant (volume) acting at the same location (centroid). •The Whitney block has average stress 0.85fc’ and depth a=1c. ACI 10.2.7.1 4
The equivalent stress (Whitney) block
The equivalent rectangular concrete stress distribution has what is known as the 1 coefficient. It relates the actual NAdepth to the depth of the compression block by a=1c. for f c ' 28 MPa 1 0.85 ACI 10.2.7.1
1 0.85 0.05( fc '28) 0.65 for f c ' 28 MPa 7
5
Derivation of beam expressions
Fx=0
C =T
6
Derivation of beam expressions
7
Derivation of beam expressions Design aids can also be used: Assume Md = Mu = ΦMn
= Rn
Mn=R nbd2
Rn is given in tables and figures of design aids.
8
Design Aids
9
Design Aids
10
Tension strain in flexural
y
f y
Es
t y ?
Strain Distribution 11
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
Types of flexural failure: [1] Balanced Failure The concrete crushes and the steel yields simultaneously. εcu=0.003 cb d
h
b
εt = εy
Such a beam has a balanced reinforcement, its failure mode is
brittle, thus sudden, and is not allowed by the ACI Strength Design Method. 13
Types of flexural failure: [2] Compression Failure The concrete will crush before the steel yields. εcu=0.003 εcu=0.003
c>cb c>cb h
d
h
b
b
d
ε <ε εt < εt y y
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. 14
Types of flexural failure: [3] Tension Failure The reinforcement yields before the concrete crushes. εcu=0.003 εcu=0.003
c
h
d
d
bb
Such a beam is called under-reinforced beam, and its failure mode is
ductile,
thus giving a sufficient amount of warning before collapse, and is required by the ACI Strength Design Method 15
Types of sections according to the ACI Code ACI 10.3.4 [1] Tension-controlled section 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 εy (εy = fy/Es=0.002 for f y =420 MPa) when the concrete in compression reaches 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 εy (εy =fy/Es=0.002 for f y =420 MPa) when the concrete in compression reaches its crushing strain of 0.003. 16
Allowed strains for sections in bending
ACI 10.3.5
17
Strength reduction factor Φ εcu=0.003 c d
εt
t
d c c
c
ACI R9.3.2.2
18
Balanced steel cb
0.003 d 0.003 f y E S 5
Es 210 MPa
cb
600 d 600 f y =1c
0.85 1 f c ' 600 b 600 f y fy
19
Maximum allowed steel
0.003 d 0.003 0.005 3 d 8
c max c max
=1c 1c max
max
3 d 1 8
3 0.85 1 f c ' 8 fy 20
Minimum steel allowed ACI 10.5.1
A s,min
0.25 f c bw d fy max 1.4 b d w f y
bw = width of section d = effective depth of section
21
Design Aids
22
Summary: To calculate the moment capacity of a section: 0.25 f c bw d fy max 1.4 b d w f y
1-) As,min
if As,min > As,sup reject section
2-) a
3-)
As f y 0.85f c b
1 0.85
or a
df y 0.85f c
for f c ' 28 MPa
1 0.85 0.05( fc '28) 0.65 7
a c 4-) 1
for fc' 28 MPa 23
Summary: d c
5-) t c 0.003 if t> 0.005: tension controlled = 0.9 if 0.004 < t <0.005: in transition zone =0.65+( t -0.002) (250/3)
if t < 0.004: compression controlled reject section d a 6-) M d M n As f y 2 or Mn=Rnbd2 (find Rn from table)
7-) M u ΦM n
if not reject section
24
Example A singly reinforced concrete beam has the cross-section shown in the figure below. Calculate the design moment strength. Can the section carry an Mu = 350 kN.m?
f y 414MPa a) f c 20.7MPa, b) f c 34.5MPa, c) f c 62.1MPa
25
Example Solution a) f c 20.7MPa 1 A s,min
0.25 f c 0.25 20.7 bw d (254)(457)=319 mm2 fy 414 max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=393 mm
2 a
As f y 0.85f cb
3 1 0.85
2
2
< A s,sup =2580 mm OK
2580 414 239mm 0.85 20.7 254
for f c ' 20.7MPa 28 MPa 26
Example Solution a) f c 20.7MPa 4 c
a
1
239 281mm 0.85
d c 457 281 0.003 0.00186 5 t 0.003 c 281 t 0.004 Section is compression controlled ==> Does not satisfy ACI requirements ==> Reject section
27
Example Solution b) f c 34.5MPa 1 A s,min
0.25 f c 0.25 34.5 bw d (254)(457)=412 mm2 fy 414 max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=412 mm
2 a
As f y 0.85f c b
2
2
< A s,sup =2580 mm OK
2580 414 143.4mm 0.8534.5 254
0.05( f c ' 28) 0.65 for f c ' 34.5MPa 28 MPa 7 1 0.85 0.05( 34.5 28) 0.804 0.65 7
3 1 0.85
28
Example Solution b) f c 34.5MPa 4 c
a
1
143.4 178.5mm 0.804
d c 457 178.5 0.003 0.00468 5 t 0.003 c 178.5 0.004 t 0.005 Section is in transision zone
=0.65+(t -0.002)(250/3) =0.65+(0.00468-0.002)(250/3)=0.874
29
Example Solution b) f c 34.5MPa
d a 6 M d M n As f y 2 143.4 6 0.874 2850 414 457 360 10 N .mm 2 360kN .m 7 M u 350kN .m ΦM n 360kN .m Section is adequate
39
Example Solution c) f c 62.1MPa
1 A s,min
0.25 f c 0.25 62.1 2 b d (254)(457)=552 mm w f 414 y max 1.4 1.4 bw d (254)(457)=393 mm2 fy 414
=552 mm
2 a
As f y
0.85f cb
2
2
< A s,sup =2580 mm OK
2580 414 80mm 0.85 62.1254
31
Example Solution c) f c 62.1MPa 0.05( f c ' 28) 3 1 0.85 0.65 for f c ' 62.1MPa 28 MPa 7 1 0.85 0.05( 62.1 28) 0.61 0.65 7 1 0.65
4 c
a
1
80 123mm 0.65
d c 457 123 0.003 0.0081 5 t 0.003 c 123 t 0.005 Section is tension controlled ==> Satisfes ACI requirements ==> =0.9
32
Example Solution c) f c 62.1MPa
d a 6 M d M n As f y 2 80 0.9 2850 414 457 520 106 N .mm 2 520kN .m 7 M u 350kN .m ΦM n 520kN .m Section is adequate
33
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 6 Design of singly reinforced rectangular beams
Design of Beams For Flexure The main two objectives of design is to satisfy the: 1) Strength and 2) Serviceability requirements 1) Strength M d Φ M n Mu M d Design moment strength (also known as moment resistance)
M u Internal ultimate moment Mn
Theoretical or nominal resisting moment.
M u 1.2M D 1.6M L 2
Design of Beams For Flexure Derivation of design expressions
h
d As
Assume ΦMn = Mu
b Beam cross section
Solve for 0.85 f c' ρ fy
2M u 1 1 2 0.85f ' b d c
: 1 kN.m = 106 N.mm
A s = bd 3
Design of Beams For Flexure Design aids can also be used:
0.85 f c' ρ fy
2M u 1 1 2 0.85f ' b d c
Calculate: Then is found from tables and figures of design aids.
4
Design Aids
5
Design of Beams For Flexure 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
hmin l = span length measured center to center of .
h hmin
h
d As b Beam cross section
7
Detailing issues: Concrete Cover 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.
Side cover
ACI 7.7.1
Bottom cove
8
Detailing issues: Spacing of Reinforcing Bars •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 Smin must be satisfied,where:
S min
bar diameter, d b ACI 7.6.1 max 25 mm 4/3 maximum size of coarse aggregate
ACI 3.3.2 •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. ACI 7.6.2
Clear distance
Clear spacing S
9
Estimation of applied moments Mu Beams are designed for maximum moments along the spans in both negative and positive directions.
10
Positive moment
Negative moment
Tension at bottom Needs bottom reinforcement
Tension at top Needs top reinforcement
Estimation of applied moments Mu 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.
Simply ed Beams
Continuous Beams
Determinate
Indeterminate
– + 11
Moment Diagram
+
Moment Diagram
Estimation of applied moments Mu Simply ed Beams
Continuous Beams
– +
Moment Diagram Section at midspan
12
+
Moment Diagram Section over
Estimation of applied moments Mu Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate moments for design 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. • are of similar section dimensions along their lengths (prismatic).
13
Estimation of applied moments Mu Approximate Structural Analysis More than two spans
1 4
ACI 8.3.3
Estimation of applied moments Mu Approximate Structural Analysis Two spans
l n = length of clear
span measured face-to-face of s. For calculating negative moments, l n is taken as the average of the adjacent clear span lengths.
15
ACI 8.3.3
Design procedures Method 1: When b and h are unknown 1Determine 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. 3Assume that Φ=0.9 and calculate the reinforcement (ρ and As).
4- Check solution: (a) (b) (c) (d)
Check spacing between bars Check minimum steel requirement Check Φ = 0.9 (tension controlled assumption) Check moment capacity (Md ≥ Mu ?)
5- Sketch the cross section and its reinforcement. 16
Design procedures Method 2: When b and h are known 1Calculate the factored load wu and bending moment Mu.
2Assume that Φ=0.9 and calculate the reinforcement (ρ and As). 3- Check solution: (a) (b) (c) (d)
Check spacing between bars Check minimum steel requirement Check Φ = 0.9 (tension controlled assumption) Check moment capacity (Md ≥ Mu ?)
4- Sketch the cross section and its reinforcement.
17
Example 1 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 a service live load of 10 kN/m are to be ed. wd=25 kN/m & wl =10 kN/m Use fc’ =25 MPa and fy = 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
6m
wu=50.5 kN/m 6m
2 Calculate wu and Mu wu = 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
227.3 kN.m
18
Example 1 3- Assume that Φ=0.9 and calculate ρ and As d = 500 – 40 – 8 – (20/2) = 442 mm (assuming one layer of Φ20mm reinforcement and Φ8mm stirrups) ρ
0.85fc ' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(25) ρ 420
2 227.3106 1 1 2 (0.9) 0.85(25) 300 (442)
0.0116
As = ρ b d = 0.0116(300)(442) =1536 mm2 Use 5 Φ 20 mm (As,sup=1571 mm2)
19
W
Number of bars
mm
N/m
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
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
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
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
20
Example 1 4- Check solution a) Check spacing between bars sc
5Φ20
300 2 40 28 5 20 26 mm d b 20 mm 51 25 mm
300
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 b d (300)(442)=395 mm2 w fy 420 max 1.4 1.4 bw d (300)(442)=442 mm2 fy 420 =442 mm
2
2
< A s,sup =1571 mm OK 21
Example 1 c) Check Φ =0.9 (tension controlled assumption) a
As f y 0.85 f c 'b
1 0.85
1571420 103.5mm 0.85(25)300
for f c ' 25MPa 28 MPa c
a 103.5 121.7 mm β1 0.85
dc 442 121.7 0.003 0.0079 0.005 ε t 0.003 c 121.7 for ε t 0.005 Φ 0.90, the assumption is true the section is tension controlled
d) Check moment capacity a M d ΦA sf y d 2
103.5 6 0.901571 420 442 231.710 N.mm = 231.7 kN.m 2 M d 231.7 kN.m M u 227.3 kN.m OK
22
Example 1 5- Sketch the cross section and its reinforcement
44.2
50 5Φ20
30 Beam cross section
23
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 . Columns have a cross section of 800x300 mm. The factored distributed load over the slab is qu =14.4 kN/m2. Use fc’ =25 MPa and fy = 420 MPa. L1 = L2 = L3 = 6 m S 1 = S 2= S 3 = 4 m B1
Solution: b & d are known 1- Calculate wu and Mu wu=4(14.4) = 57.6 kN/m ln = 6 – 0.3=5.7 m wu 24
Example 2 Moment diagram using the approximate ACI method:
Design for the maximum negative moment throughout the beam:
Mu = wu(ln )2/10 = 57.6 (5.7)2/10 Mu = 187.5 kN.m
25
Example 2 2- Assume Φ=0.9 and calculate ρ and As d = 316 – 40 – (16/2) – 8 = 260 mm (assuming one layer of Φ16 mm reinforcement and Φ8mm stirrups)
ρ
0.85fc' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(25) ρ 420
2187.5106 1 1 0.0102 2 (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) 26
Example 2 3- Check solution a) Check spacing between bars
sc
800 2 40 28 1116 52.8 mm 111
d b 16 mm 25 mm
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 bw d (800)(260)=620 mm2 fy 420 max 1.4 1.4 bw d (800)(260)=693 mm2 fy 420 2
=693 mm2 < A s,sup =2212 mm OK 27
Example 2 c) Check Φ =0.9 a
As f y
0.85 f c 'b
1 0.85
2212 420 55mm 0.85(25)800
for f c ' 25MPa 28 MPa c
55 a 64 mm β1 0.85
dc 260 64 0.003 0.0091 0.005 ε t 0.003 c 64 for ε t 0.005 Φ 0.90, the assumption is true the section is tension controlled
d) Check moment capacity a M d ΦA sf y d 2 55 0.9 2212 420 260 194.5106 N.mm=194.5 kN.m 2 M d 194.5 kN.m M u 187.5 kN.m OK
28
Example 2 4- Sketch the cross section and its reinforcement
11Φ16
316
260
800
29
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 7 Design of T and Lbeams
T Beams Reinforced concrete systems may consist of slabs and dropped 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 ing, and the part below the slab is called the web or stem.
Flange web
2
Flange Width b Parts of the slab near the webs are more highly stressed than areas away from the web.
effective flange width be
effective flange width be
hf
d
stirrup bw
bw
L-beam
3
T-beam
d: effective depth. hf : height of flange. bw : width of web. be : effective width. b: distance from center to center of adjacent web spacings
Effective Flange Width be be is the width that is stressed uniformly to give the same compression force actually developed in the compression zone of width b.
4
Effective Flange Width be ACI Code Provisions for Estimating be
ACI 8.12.2 According to the ACI code, the effective flange width of a T-beam, be is not to exceed the smallest of: 1. One-fourth the span length of the beam, L/4. 2. Width of web plus 16 times slab thickness, bw +16 hf . 3. Center-to-center spacing of beams, b.
b eff
5
L/4 min bw +16hf b
Effective Flange Width be ACI Code Provisions for Estimating be
ACI 8.12.3 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). b eff
6
bw L /12 min bw 6hf b 0.5b w c
A T-beam does not have to look like a T
7
Various Possible Geometries of T-Beams Single Tee
Double Tee
Box
8
Various Possible Geometries of T-Beams
Flange
Flange
web
web
Same as
9
T- versus Rectangular Sections 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 usually neglected in ultimate strength design, the sections are treated as 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 Tsection.
Compression zone
–
+ 11
Tension zone
+
Moment Diagram
Section at midspan Positive moment
Section at Negative moment
Analysis of T-beams Case 1: when a ≤ hf
[Same as rectangular section]
TC Asf y a 0.85f c b e
12
a ΦM n ΦA sf y d 2
Analysis of T-beams Case 2: when a > hf C f 0.85f c be bw hf Cw 0.85 f c bw a T As f y
From equilibrium of forces T C f C w a
As f y 0.85f c be bw hf
0.85f c bw
a h ΦM n Φ C w d C f d f 2 2 13
Minimum Reinforcement, As,min ACI 10.5.2
hf As
be
As
hf
bw
14
d
bw
-ve Moment
A s,min
0.25 f c bw d fy max 1.4 b d w f y
+ve Moment
be
d
Analysis procedure for calculating he ultimate strength of T-beams To calculate the moment capacity of a T-section: 1 Calculate be 2 Check As,sup> As,min 3 Assume a ≤ hf and calculate a using: a
Asf y 0.85fc b e
If a ≤ hf → a is correct If a > hf → a
A s f y 0.85f c be bw hf 0.85f c bw
4- Calculate 1, c, and check εt 5- Calculate ΦMn, and check 15
M u ΦM n
Example 1 Calculate Md for the T-Beam: hf = 150 mm
d = 400 mm
As = 5000mm2
fy = 420MPa fc’= 25MPa bw= 300mm
L = 5.5m
b=2.15m Determine be according to ACI requirements
L 5500 1370mm 4 4 be min 16hf b w 16 150 300=2700mm b 2150 mm
16
Example 1 Check min. steel 0.25 f ' 1.4 c b d ; b A s,min max w w d max fy f y 2 5000 mm 2 OK A s,min 400 mm A
0.25 25 300 400 ; 1.4 300400 420 420
s,sup
Calculate a (assuming a
c
Asf y 0.85f c b e a
1
5000420 71.9mm hf 0.85251370
150mm OK
71.9 85 mm 0.85
s d c 0.003 400 85 0.003 0.011 0.005 Tension controled c 85 17
Example 1 Calculate Md a M d As f y d 2 71.8 0.9 5000420 400 2 688106 N.mm 688 kN.m
18
Example 2 Determine the ACI design moment strength Md (ΦMn) of the T-beam shown in the figure if fc’ =25 MPa and fy = 420 MPa. 10
90
Φ10
h= 75
Solution:1- Check min. steel
25 655.5mm 8Φ32 2 0.25 f ' 1.4 c bw d A s,min max bw d ; fy f y 0.25 25 1.4 A s,min max 300 655.5 ; 300 655.5 420 420 2 2 656 mm A 6434 mm OK A s,min
d 750-40-10-32-
s,sup
19
30
Example 2 2- Check if a < hf = 10cm Asf y
6434 420 141.3mm 0.85fc b e 0.85 25 900
a= 141.3> hf = 100 mm i.e. assumption is wrong Section is T NA is in the web
20
10
90
Φ10
h= 75
a
8Φ32 30
Example 2 3- Calculate 1, c, and check εt a a
c
A s f y - 0.85f c be bw hf 0.85f c bw 6434 420 0.85 25900 300100 0.85 25 300
224mm
a 224 264 mm β1 0.85
d c 0.003 655.5 264 0.003 εt 264 c ε t 0.00447 0.004 0.005 Transision zone
=0.65+(t -0.002)(250/3) =0.65+(0.00447-0.002)(250/3)=0.855 21
Example 2 4- Calculate Md Cf 0.85fc' (be b w ) h f 0.85 25900 300100 127510 3N 1275 kN 3
Cw 0.85fc' a b w 0.85 25 224 300 1427.4 10 N 1427.4 kN M d Φ C w
a d C f 2
hf d 2
100 224 3 0.855 1427.4 10 3 655.5 127510 655.5 2 2 1323.4106 N .mm 1323.4 kN .m
22
Design of T-Beams --- Positive moment
+
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 Muf and (2) Asw, the remaining of the steel, providing a section with capacity Muw.
M u M uf M u w
23
Mu : Ultimate moment applied, requiring steel As. Muf : Moment resisted by overhanging flange parts, requiring steel Asf. Muw : Moment resisted by web, requiring steel Asw.
Design of T-Beams --- Positive moment
+
Step 1
24
24
Step 2
Design of T-Beams --- Positive moment
+
M u M uf M u w
Muw Mu Muf
0.85 f c ' 1 1 fy
25
2M uw 0.85f c ' bw d 2
Step 3
Step 4
Asw bw d
Step 5
As Asf Asw
Step 6
Design of L-Beams --- Positive moment be
be
Same as bw
26
Design of T-Beams --- Negative moment be
bw
Design as a rectangular section with width bw
27
Flange Reinforcement When flanges of T-beams are in tension, part of the flexural reinforcement shall be distributed over effective flange width, or a width equal to one-tenth of the span, whichever is smaller
-ve moment
Additional Reinforcement
min (beff & l/10)
Additional Reinforcement
Main Reinforcement
If beff > l/10, some longitudinal reinforcement shall be provided in outer portions of flange.
Design of T-Beams --- Positive moment Design Procedure: 1 Establish h based on serviceability requirements of the slab and calculate d
2 Choose bw 3 Determine be according to ACI requirements. 4 Calculate As assuming that a < hf with beam width = be & Φ=0.90 b 2M u 1 1 hf 2 d Φ 0.85f ' b d c e As As fy As = ρ be d → a bw 0.85f c ' b e 5 If a ≤ hf: the assumption is right continue as rectangular section If 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
0.85f c' ρ fy
29
e
Example 3
fy = 420 MPa.
Lm
A floor system consists of a 14.0cm concrete slab ed by continuous T-beams with a span L. Given that bw=30cm and d=55cm, fc’ =28 MPa and Determine the steel required at midspan 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 Spandrel (B) L = 2 m beam
3.0 m
3.0 m
Slab
hf
bw
39
3.0 m
Solution (A) L = 8 m Determine be according to ACI requirements
784 kN.m
200
L 8000 4 4 2000mm be min 16hf bw 16 140 300=2540mm b 3000 mm
14 As 30
be is taken as 2000 mm, as shown in the figure
Calculate As assuming that a < hf with beam width = be & Φ=0.90 Mu = 1.2(320)+1.6(250)=784 kN.m
0.85f c ' fy 31
2M u 11 0.85f ' b d 2 c e
55
Solution (A) L = 8 m 0.85 28 278410 ρ 1 1 2 420 0.90.85282000550 0.00354 6
As ρ be d 0.00354 2000 550 3892 mm
784 kN.m
200
14 As 30
2
Check a ≤ hf assumption a
As f y 0.85f c 'b e
3892420 34.3mm h f 140mm 0.85 28 2000
The assumption is right Rectangular section design
Use 8Φ25mm (As,sup= 3927 mm2) arranged in two layers. sc 32
300 2 40 28 4 25 34.5 mm d b 25 mm 41 25 mm
OK
55
Solution (A) L = 8 m Check the Φ=0.90 assumption (εt ≥0.005) and As,sup ≥ As,min A s ,min
A s,min
0.25 f ' 1.4 c b d ; b max w w d max fy f y 550 mm 2 A 3927 mm 2 OK
0.25 28 300 550 ; 1.4 300 550 420 420
s,sup
200
As f y
55
14
3927 420 a 34.7 mm 0.85 f c 'b e 0.85 28 2000
a 34.7 40.8 mm β1 0.85 550 40.8 dc εt 0.003 0.003 40.8 c 0.0374 0.005 0.9 OK c
33
8Φ25 30
Solution (A) L = 8 m Check moment capacity a M d As f y d 2 34.7 0.93927 420 550 2 6
Md 790.710 N.mm 790.7 kN.m M u 784kN.m
55
14
200
8Φ25 30
34
Solution (B) L = 2 m 784 kN.m
50
Determine be according to ACI requirements L 2000 500mm 4 4 be min 16hf bw 16 140 300=2540mm b 3000 mm
14 As 30
be is taken as 500 mm, as shown in the figure Calculate As assuming that a < hf with beam width = be & Φ=0.90
Mu = 1.2(320)+1.6(250)=784 kN.m
2M u 0.85f c ' 1 1 0.85f ' b d 2 fy
35
c
e
55
Solution (B) L = 2 m 0.85 28 278410 ρ 1 1 2 420 0.90.8528500550 0.0159 A ρ b d 0.0159500 550 4389 mm 2 6
s
784 kN.m
50
e
Check a ≤ h assumption f
a
As f y 0.85f c ' be
4389 420 155mm > h f 140mm 0.8528500
The assumption is wrong T section design
36
14 As 30
55
Solution (B) L = 2 m 14
50
55
Calculate required reinforcement A sf
0.85f c '( b bw ) hf fy
A sf
0.85( 28)( 500 300 )140 1586mm 2 420
30
h M uf A s f y d f 2 140 6 0.91586 420 550 28810 N .m 2 6
6
6
M uw M u M u f 78410 28810 49610 N .m 37
Solution (B) L = 2 m
0.85f c ' 1 1 fy
0.85( 28) ( 420)
sw
2( 496) 106 1 1 2 0.9 0.85( 28) ( 300) ( 550)
0.017
50
w
As Asf Asw 1586 2808 4395mm
14
b d 0.017( 300)( 550) 2808mm 2 55
A
2M u 0.85f c ' bw d 2
2
8Φ28 30
Use 8Φ28 mm (As,sup= 4926mm2) arranged in two layers. Check solution: (Do as in Example 2)
38
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 8 Design of doubly reinforced beams
Doubly Reinforced Rectangular Sections Beams having steel reinforcement on both the tension and compression sides are called doubly reinforced sections. Doubly reinforced 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. 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:
Mn
Mn2
Mn1
To analyze the section, the tension steel is divided in two portions: (1) As2, which is in equilibrium with the compression steel, and providing a section with capacity Mn2 and (2) As1, the remaining of the tension steel, providing a section with capacity Mn1. 4
Analysis of Doubly Reinforced Rectangular Sections Find As1 and As2:
T s 2 Cs As 2f y Asf s As 2 5
Asf s fy
We need f s’ to find A
As A s1 As 2 A s1 As As 2
s2
Analysis of Doubly Reinforced Rectangular Sections Find fs’: s c d 0.003
c
c d f s s Es 0.003Es f y c 5
E s 2 10 MPa
6
c
Analysis of Doubly Reinforced Rectangular Sections Find c:
T C c Cs As f y 0.85f cab Asf s
7
c d As f y 0.85f c1cb As 0.003Es c find c by solving the quadratic equation find fs’ from equation in slide 6
Analysis of Doubly Reinforced Rectangular Sections Find Md:
8
M d M n A s 1f y
d - a A f d - d ' s s 2
Analysis of Doubly Reinforced Rectangular Sections Procedure: 1) 2)
c d As f y 0.85f c1cb As 0.003Es c c d fs 0.003Es f y c
find c, a
3) A s 2 Asf s
fy 4) As 1 As As 2
5) 6) 9
d c Check if = 0.9 s c 0.003 0.005? a M d M n A s 1f y d - A s f s d - d ' 2
Example 1 For the beam with double reinforcement shown in the figure, calculate the design moment Md. 5.0 2Φ25 fc’ =35MPa and fy = 420 MPa. 60 6Φ32
Solution:1 0.85 0.05( f c ' 28) 0.65 for f c ' 35MPa 28 MPa 7 1 0.85 0.05( 35 28) 0.8 0.65 7 c d As f y 0.85f c 1cb As 0.003Es c 10
c 50 5 4825(420) 0.85(35)(0.8)c(300) 982 0.003(210 ) c
30
Example 1 c 50 5 4825(420) 0.85(35)(0.8)c(300) 982 0.003(210 ) c 229.5c 2 1437300c 29460000 0 5.0 c 220mm
2Φ25
a 1c 0.8 220 176mm
60 6Φ32
c d f s 0.003Es f y c 220 50 5 f s 0.003(210 ) 463 f 220 f s f y 420 11
30 y
420MPa
Example 1
5.0 2Φ25
Asf s 982(420) 982mm 2 fy (420)
As 2
A s1 As As 2 4825 982 3843mm
60 6Φ32
2
d c s 0.003 0.005? c
30
s 600 220 0.003 0.0052 0.005 Tension Controlled , 0.9
220
M d M
n
A s 1f y
d - a A f d - d ' s s 2
176 M d 0.9 3843( 420) 600 982( 420) 600 50 2 12
M
6
d
94810 N .mm 948kN .m
Maximum allowed steel for a singly reinforced section
0.003 d 0.003 0.005 3 d 8
c max c max
=1c 1c max
3 0.85 1 f c ' max 8 fy 3 0.85 1f c ' bd A s ,max 8 fy
3 d 1 8
13
Design of Doubly Reinforced Rectangular Sections
1) Design the section as singly reinforced, and calculate t 2) If t < 0.004 Comp. steel is needed (or enlarge section if possible)
3) Design As1 for maximum reinforcement (slide 13) and find Mn1, a, c 4) M n M u 5) Mn2 = Mn – Mn1 c d 6) f s sEs 0.003Es f y c Asf s Mn2 A s2 7) A s fy f s(d d )
As As 1 A s 2
14
Example 2 Design the beam shown in the figure to resist Mu=1225 kN.m. If compression steel is required, place it 70 mm from the compression face. fc’ =21 MPa and fy = 420 MPa. Solution: Try first to design the section as a singly reinforced section: ρ
0.85fc ' fy
2M u 1 1 2 Φ 0.85f ' bd c
0.85(21) ρ 420
21225106 1 1 2 (0.9) 0.85(21)350 (700)
As= ρ b d = 0.0284(350)(700) = 6947 mm2 15
Use 10 Φ32 mm in two rows (As,sup =7069 mm2)
0.0284
Example 2 Check the ductility of the singly reinforced section: a
As f y 0.85f c ' b
7069 420 475mm 0.85(21)350
c
a
1
475 559mm 0.85
dc 700 559 0.003 0.00076 0.004 ε t 0.003 c 559 Section is brittle! can not be used. Use compression reinforcement. Mn
Mu
A s 1 A s ,max 16
1225 1361kN .m 0.9
3 0.85 1f c ' 8 fy
A s 1 3307mm 2
3 0.85( 0.85)( 21) ( 350)( 700) bd 8 ( 420)
Example 2 As f y
3307( 420) a 222.3mm 0.85f cb 0.85( 21)( 350)
c
a
1
222.3 261.55mm 0.85
a 222.3 M n1 A s f y d - ( 3307)( 420)( 700 ) 2 2 6
M n1 81810 N .mm 818kN .m Mn2 = Mn – Mn1 = 1361 – 818 = 543 kN.m
17
Example 2 c d f s 0.003Es f y c 261.55 70 5 f s 0.003(210 ) 439MPa f 261.55 f s f y 420
y
420MPa
Mn 2 543106 A s 2052mm 2 f s(d d ) 420(700 70) As 2
Asf s (2052)(420) 2052mm 2 fy (420)
A s A s 1 A s 2 3307 2052 5359mm 2 2
Use 830 in two rows for tension steel (A s,sup = 5655 mm ) 2
18
Use 426 for compression steel (A s,sup = 2124 mm )
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 9 Design of beams for shear
Shear Design vs Moment Design Beams are usually designed for bending moment first. Accordingly, cross sectional dimensions are determined along with the required amounts of longitudinal reinforcement.
Once this is done, sections are checked for shear to determine whether shear reinforcement is required or not. 2
Shear Design vs Moment Design This by no means indicates that shear is less important than bending. On the contrary, shear failure which is usually initiated by diagonal tension, is far more dangerous than flexural failure due to its brittle nature. It occurs without warning. Therefore, beams are designed to rather fail in bending. This is done by providing larger safety factor against shear failure than those provided for bending.
3
Shear and flexural stresses In linearly elastic beams, two types of stresses occur:
Flexural stresses:
Shear stresses: An element of a beam not on the NA or an extreme fiber is subjected to both stress types combined 4
Shear and flexural stresses The combined stress (called principal stresses) are calculated as:
which act on a direction inclined with respect to the beam axis by the angle:
5
Shear and cracks in beams
6
Shear and cracks in beams
7
7
Types of Shear Cracks Two types of inclined cracking occur in beams:
1- Web Shear Cracks Web shear cracking begins from an interior point in a member at the level of the centroid of the uncracked section and moves on a diagonal path to the tension face when the diagonal tensile stresses produced by shear exceed the tensile strength of concrete.
2- Flexure-Shear Cracks The most common type, develops from the tip of a flexural crack at the tension side of the beam and propagates towards mid depth until it reaches the compression side of the beam. 8
Shear and cracks in beams It is concluded that the shearing force acting on a vertical section in a reinforced concrete beam does not cause direct rupture of that section. Shear by itself or in combination with flexure may cause failure indirectly by producing tensile stresses on inclined planes. If these stresses exceed the relatively low tensile strength of concrete, diagonal cracks develop. If these cracks are not checked, splitting of the beam
or what is known as diagonal tension failure will take place.
9
Failure by shear in beams
10
Types of Shear Reinforcement The code allows the use of three types of Shear Reinforcement • Vertical stirrups • Inclined stirrups • Bent up bars Inclined Stirrups
Bent up bars
11
Vertical Stirrups
Deg to Resist Shear The strength requirement for shear that has to be satisfied is:
ΦVn Vu
ACI Eq. 11-1
Vu = factored shear force at section Vn = nominal shear strength Φ = strength reduction factor for shear = 0.75
The nominal shear force is generally resisted by concrete and reinforcement: Vn Vc Vs ACI Eq. 11-2 Vc = nominal shear force resisted by concrete Vs = nominal shear force resisted by shear reinforcement 12
shear
Strength of Concrete in Shear For subject to shear Vu and bending Mu only, ACI Code gives the following equation for calculating Vc
Simple formula
Vc 0.17 f c ' bw d
ACI Eq. 11-3
Detailed formula Vu d bw d 0.29 fc ' bw d Vc 0.16 f c ' 17 w Mu
13
As where w b wd
ACI Eq. 11-5
Strength of Concrete in Shear
For subject to axial compression Nu plus shear Vu, ACI Code gives the following equation for calculating Vc
N u f c' b wd Vc 0.17 1 14Ag
ACI Eq. 11-4
For subject to axial tension Nu plus shear Vu, ACI Code gives the following equation for calculating Vc
14
0.29N u fc' b w d Vc 0.17 1 A g
ACI Eq. 11-8
Deg to Resist Shear To find the force required to be resisted by shear reinforcement:
Vu ΦVn
Vn Vc Vs
Vs 15
Vu
V c
Three cases of shear requirement: Case 1: For Vu ≥ ΦVc shear reinforcement is required Case 2: For Vu ≥0.50ΦVc minimum shear reinforcement is required
Case 3: For Vu < 0.50ΦVc no shear reinforcement isrequired
16
Design of Stirrups Shear reinforcement required when V s V u V c
Vu Vc
ACI 11.4.7.1
The bar size of the stirrups is established and the spacing is calculated:
Vs
Avf yd
s
Afd v y
s
For inclined stirrups (with angle )
Vs
Av f y d sin α cos α
s
s
ACI Eq. 11-15
Vs
Av fy d sinαcos α
ACI Eq. 11-16
Vs
where A v = the area of shear reinforcement within spacing s (for a 2-legged stirrup in a beam: A v = 2 times the area of the stirrup bar). 17
ACI 11.4.6.1
Minimum Amount of Shear Reinforcement
1 Minimum Shear Reinforcement (Av,min) required when Vu Vc 2 bw s bw s Av min 0.062 f c ' 0.35 f ys f ys Av f ys Av f ys ; s=min 0.062 f c ' bw 0.35 bw
ACI Eq. 11-13
except in: (a)Footings and solid slabs (b) Concrete joist construction (c) Beams with h not greater than 250 mm (d)Beams integral with slabs with h not greater than 600 mm and not greater than the larger of 2.5 times the thickness of flange, and 0.5 times width of web.
18
Spacing limits for Shear Reinforcement
If V s 0.33 f c bw d s max If V s 0.33 f c bw d s max
min d ;600mm 2 min d ;300mm 4
ACI 11.4.5
Upper limit for Vs ACI Code requires that the maximum force resisted by shear reinforcement Vs is as follows
V s 0.66 f c ' bw d
ACI 11.4.7.9
If this condition is not satisfied
Section dimensions must be increased 19
Critical Section for Shear
ACI 11.1.3.1
Critical section for shear may be taken a distance d away from the face of the if: (a) reaction, introduces compression into the end regions of member; (b) Loads are applied at or near the top of the member; (c)No concentrated load occurs between face of and location of critical section.
20
Critical Section for Shear
ACI 11.1.3.1
Critical section for shear may be taken a distance d away from the face of the as in cases (a) and (b), but must be taken at face of the as in cases (c) and (d).
21
Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate shears for design of continuous beams, provided: • 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. • are of similar section dimensions along their lengths (prismatic).
22
Approximate Structural Analysis More than two spans
23
ACI 8.3.3
Approximate Structural Analysis ACI 8.3.3 Two spans
l n = length of clear
span measured face-to-face of s.
24
Summary of ACI Shear Design Procedure for Beams 1Draw the shearing force diagram and establish the critical section for shear Vu. 2 Calculate the nominal capacity of concrete in shear Vs. Vc 0.17 f c ' bw d 3- Calculate the force required to be resisted by shear reinforcement V V s u V c 4- Check the code limit on V s Vu Vs V c 0.66 f c ' bw d If this condition is not satisfied, the concrete dimensions should be increased. 25
Summary of ACI Shear Design Procedure for Beams 5- Classify the factored shearing forces acting on the beam according to the following * For Vu < 0.50ΦVc , no shear reinforcement isrequired. * For Vu ≥ 0.50ΦVc , minimum shear reinforcement is required Av f ys Av f ys ; s=min 0.35 b 0.062 f 'b w c w
*For Vu ≥ ΦVc , shear reinforcement is required (in addition, check min shear) Af d Av f y d sin α cos α For vertical v y For inclined s s stirrups stirrups Vs Vs 6- Maximum spacing smax must be checked If V s 0.33 f c bw d s max min d ;600mm 2
26
d If V s 0.33 f c bw d s max min ;300mm 4
Example
A rectangular beam has the dimensions shown in the figure and is loaded with a uniform service dead load of 40 kN/m (including own weight of beam) and a uniform service live load of 25 kN/m. Design the necessary shear reinforcement given that fc’ =28 MPa and fy=420 MPa. Width of is equal to 30 cm. wD=40 kNm & wL=25 kN/m
60
0.3m
0.3m 7.0 m
27
30
Example
Solution: Assuming Φ8 mm stirrups and Φ20 mm flexural steel, d=60-4-0.8-1.0=54.2 cm
wu=1.2(40)+1.6(25)=88 kN/m
0.3m 54.2
308 kN
1- Draw shearing force diagram:
Critical section for shear is located at a distance of d = 54.2 cm from the face of . Vu,critical is equal to 247.1 kN. 28
7.0 m 247.1 kN
308 kN
Example
2- Calculate the shear capacity of concrete: V c 0.17 f c ' bw d 0.17 28 300 542 146.3 10 3 N 146.3kN
V c 0.75 144.2kN 109.7kN V c 54.85 kN 2
3- Calculate the force required to be resisted by shear reinforcement Vs. Vu Vs V c 247.1 146.3 183.2kN 0.75
4- Check the code limit on V s : 0.66 f c ' bw d 0.66 28 300 542 567.9103 N 567.9kN V s 183.2kN 0.66 f c ' bw d 567.9kN 29
OK
Example
5- Classify the factored shear force: Vu= 247.1 kN > ΦVc = 109.7 kN, shear reinforcement is required. The beam can be designed to resist shear based on Vu= 247.1 kN over the entire span. However, to reduce reinforcement cost, the beam will not be designed for this shear over the entire span. The span will rather be divided into zones of different shear demands as shown below 308 kN
247.1 kN ΦVc=109.7 kN Zone C
Zone B
0.5ΦVc=54.85 kN Zone A 0.61 m
1.23 m
30
Example
Zone (A): [ Vu ≤ 0.5ΦVc ]
No shear reinforcement is required, but it is recommended to use minimum area of shear reinforcement. Try Φ8 mm vertical stirrups Av f ys Av f ys s=min ; 0.062 f c ' bw 0.35 bw 2(50) 420 2(50) 420 s min 427mm ; 400mm s 400mm 0.35 300 0.0062 28 300
Maximum stirrup spacing is not to exceed the smaller of d/2 = 271 mm or 600mm. So, use Φ8 mm vertical stirrups spaced at 250 mm. 31
Example
Zone (B): [ΦVc > Vu > 0.5ΦVc ] minimum shear reinforcement is required. use Φ8 mm vertical stirrups spaced at 25 cm (Calculated from Zone A). Zone (C): [Vu > ΦVc ] V s 183.2kN
Trying two-legged Φ8 mm vertical stirrups, s
32
Af d v y
Vs
2 50 420 542 183.2 10
3
125mm
Example
Check maximum stirrup spacing: 0.33 f c ' bw d 0.33 28 300 542 284kN V s 183.2kN Maximum stirrup spacing is not to exceed the smaller of d/2 = 271 mm or 600mm.
Check with minimum stirrup requirement:
Av f ys Av f ys ; smax =min 0.062 f c 'bw 0.35 bw 2(50) 420 2(50) 420 427mm ; 400mm 400mm s max min 0.35 300 0.062 28 300
So, use Φ8 mm vertical stirrups spaced at 12 cm. 33
Example 308 kN
247.1 kN
Zone C
Φ8@12
Zone B
Φ8@25
60
Φ8@25
ΦVc=109.7 kN 0.5ΦVc=54.85 kN Zone A
Φ8@25
0.61 m
30 Section in zones A&B
1.23 m 60
Φ8@12
Φ8@12
34
Φ8@25
30 Section in zone C
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 10 Design of slabs
Introduction A slab is a structural element whose thickness is small compared to its own length and width. t L ,S
t
zS
t
Lx
Slabs in buildings are usually used to transmit the loads on floors and Loads roofs to the ing beams Beam
Beam
Beam
Slab Column
Beam
Beam
Footing
Slab Beam
Column
Beam Soil
2
Introduction Slabs are flexural . Their flexure strength requirement may be expressed by
Mu M
n
Types of Slabs Solid slabs :- which are divided into - One way solid slabs - Two way solid slabs Ribbed slabs :- which are divided into - One way ribbed slabs - Two way ribbed slabs 3
One-way slab
Two-way slab
Solid Slab
Two way slab
Ribbed Slab (joist construction)
4
Two way slab
L 2 S
One-way slab
One-way slab
L 2 S
Ribbed slab with hollow blocks
5
Ribbed slab with hollow blocks
6
One-way solid slabs
shrinkage Reinft.
A one-way solid slab curves under loads in one direction only. Accordingly, slabs ed on two opposite sides only and slabs ed on all four sides, but L/S ≥ 2 are classified as one-way slabs.
Main Reinft.
Main reinforcement is placed in the shorter direction, while the longer direction is provided with shrinkage reinforcement to limit cracking. 7
Two-way solid slabs
Main Reinft.
A two-way solid slab curves under loads in two directions. Accordingly slabs ed on all four sides, and L/S < 2 are classified as two-way slabs.
S Main Reinft.
L
Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement is required in the two directions. 8
One-way v.s two-way ribbed slabs If the ribs are provided in one direction only, the slab is classified as being one-way, regardless of the ratio of longer to shorter dimensions. It is classified as two-way if the ribs are provided in two directions .
9
Minimum thickness of one way slabs
Minimum Cover
10
ACI Table 9.5(a)
ACI 7.7.1
a - Concrete exposed to earth or weather for Φ<16mm------40 mm and for Φ>16mm----- 50 mm b - Concrete not exposed to earth or weather for Φ<32mm------20 mm, otherwise ------ 40 mm
Spacing of Reinforcement Bars a- Flexural Reinforcement Bars Flexural reinforcement is to be spaced not farther than three times the slab thickness (hs), nor farther apart than 45 cm, center-to-center. 3 hs Smax smaller of ACI 10.5.4 45cm b- Shrinkage Reinforcement Bars Shrinkage reinforcement is to be spaced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center. 5 hs Smax smaller of ACI 7.12.2.2 45cm
11
Loads Assigned to Slabs wu=1.2 D.L + 1.6 L.L
a- Dead Load (D.L) : 1- Weight of slab covering materials 2-Equivalent partition weight 3- Own weight of slab
b- Live Load (L.L)
12
a- Dead Load (D.L)
1- Weight of slab covering materials, total =2.315 kN/m2
tiles (2.5cm thick) =0.025×23 = 0.575 kN/m2 cement mortar (2.5cm thick) =0.025×21 = 0.525 kN/m2 sand (5.0cm thick) =0.05×18 = 0.9 kN/m2 plaster (1.5cm thick) =0.015×21 = 0.315 kN/m2
tiles cement mortar sand
2.5 cm 2.5 cm 5 cm
slab
plaster 13
1.5 cm
2-Equivalent partition weight
This load is usually taken as the weight of all walls (weight of 1m span of wall × total spans of all walls) carried by the slab divided by the floor area and treated as a dead load rather than a live load. To calculate the weight of 1m span of wall: Each 1m2 surface of wall contains 12.5 blocks A block with thickness 10cm weighs 10 kg A block with thickness 20cm weighs 20 kg Each face of 1m2 surface has 30kg plaster Load / 1m2 surface for 10 cm block = 12.5 × 10 +2×30=185 kg/m2 = 1.85 kN/m2 Load / 1m2 surface for 20 cm block = 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2
14
20 cm
Weight of 1m span of wall with height 3m: For 10 cm block wt. = 1.85 kN/m2 × 3 = 5.6 kN/m For 20 cm block wt. = 3.1 kN/m2 × 3 = 9.3 kN/m
3- Own weight of slab
1- Solid slab: Own weight of solid slab (per 1m2)=
c h = 25 h kN/m2
2- Ribbed slab:
Example Find the total ultimate load per rib for the ribbed slab shown: Assume depth of slab = 25 cm (20cm block +5cm toping slab)
Hollow blocks are 40 cm × 25 cm × 20 cm in dimension Assume ribs have 10 cm width of web Assume equivalent partition load = 0.75 kN/m2 Consider live load = 2 kN/m2.
15
3- Own weight of slab
Solution •
Total volume (hatched) = 0.5 × 0.25 × 0.25 = 0.03125 m3
•
Volume of one hollow block = 0.4 × 0.20 × 0.25 = 0.02 m3
•
Net concrete volume = 0.03125 - 0.02 = 0.01125 m3
•
Weight of concrete = 0.01125 × 25= 0.28125 kN
•
Weight of concrete /m2 = 0.28125 /[(0.5)(0.25)] = 2.25 kN/ m2
•
Weight of hollow blocks /m2 = 0.2/[(0.5)(0.25)] = 1.6 kN/ m2
•
Total slab own weight= 2.25 + 1.6= 3.85kN/m2
Load per rib Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m2
16
Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m2 Ultimate load per rib = 11.5 × 0.5 = 5.75 kN/m
Minimum live Load values on slabs Type of Use Uniform Live Load kN/m2 Residential 2
b- Live Load (L.L)
It depends on the function for which the floor is constructed.
Residential balconies Computer use Offices Warehouses
3 5 2
6
Light storage
Heavy Storage Schools
12
Classrooms Libraries
2
rooms
3
Stack rooms Hospitals Assembly Halls
6 2
Fixed seating
Movable seating Garages (cars) Stores
17
2.5 5 2.5
Retail
4
wholesale Exit facilities Manufacturing
5 5
Light
4
Heavy
6
Loads Assigned to Beams Beams are usually designed to carry the following loads: - Their own weight - Weights of partitions applied directly on them - Floor loads
L
S1
18
S2
Design of one way SOLID slabs
19
One-way solid slabs
S1
S2
1m
One-way solid slabs are designed as a number of independent 1 m wide strips which span in the short direction and are ed on crossing beams. These strips are designed as rectangular beams.
L
2M u 0.85f c 1 1 2 fy 0.85 f c bd S1
20
S2
shrinkage Reinft.
One-way solid slabs
Main Reinft.
21
Check on tension/compression control (maximum allowed steel) Method 1: Check t
Method 2: Check max 0.003 c max d 0.003 0.005 3 c max d 8 =1c 1c max
max
3 0.85 1 f c ' 8 fy
3 d 1 8
22
Shrinkage Reinforcement Ratio According to ACI Code and for fy =420 MPa
ACI 7.12.2.1
shrinkage 0.0018 As,shrinkage 0.0018 bh where, b = width of strip, and h = slab thickness
Minimum Reinforcement Ratio for Main Reinforcement
A s ,min A s ,shrinkage 0.0018 b h
ACI 10.5.4
Check shear capacity of the section
V u V c 0.17 f c ' b wd Otherwise enlarge depth of slab 23
Approximate Structural Analysis
ACI 8.3.3
ACI Code permits the use of the following approximate moments and shears for design of continuous beams and one-way slabs, provided: • 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. • are of similar section dimensions along their lengths (prismatic).
24
Approximate Structural Analysis Bending Moment More than two spans
25
ACI 8.3.3
Approximate Structural Analysis Bending Moment Two spans
l n = length of clear
span measured face-to-face of s. For calculating negative moments, l n is taken as the average of the adjacent clear span lengths.
26
ACI 8.3.3
Approximate Structural Analysis Shear More than two spans
27
ACI 8.3.3
Approximate Structural Analysis Shear Two spans
28
ACI 8.3.3
Summary of One-way Solid Slab Design Procedure 1 Select representative 1m wide design strip/strips to span in the short direction. 2 Choose a slab thickness to satisfy deflection control requirements. When several numbers of slab s exist, select the largest calculated thickness. 3 Calculate the factored load wu 4 Draw the shear force and bending moment diagrams for each of the strips. 5 Check adequacy of slab thickness in of resisting shear by satisfying the following equation: V u 0.17 f c ' b d where b = 1000 mm
If the previous equation is not satisfied, go ahead and enlarge the thickness to do so. 29
Summary of One-way Solid Slab Design Procedure 6- Design flexural and shrinkage reinforcement: Flexural reinforcement ratio is calculated from the following equation 2M u 0.85f c 1 1 2 fy 0.85 f c bd where b = 1000 mm
You need to check tension controlled requirement, minimum reinforcement requirement and spacing of selected bars. Compute the area of shrinkage reinforcement, where Ashrinkage=0.0018bh, where b = 1000 mm.
7- Draw a plan of the slab and representative cross sections showing the dimensions and the selected reinforcement. 30
Example 1 Using the ACI-Code approximate structural analysis, design for a warehouse, a continuous one-way solid slab ed on beams 4.0 m apart as shown in the figure. Assume that the beam webs are 30 cm wide. The dead load is 3kN/m2 in addition to the own weight of the slab, and the live load is 3kN/m2.
8.0 m
Use fc’=28MPa, fy=420MPa
4.0 m
31
4.0 m
4.0 m
Solution: 1- Select a representative 1 m wide slab strip:
The selected representative strip is shown in the figure
2- Select slab thickness:
For one-end continuous spans, hmin = l/24 =4.0/24=0.167m Slab thickness is taken as 17 cm
8.0 m
1.0 m
17cm
4.0 m Wu
32
4.0 m
4.0 m
Solution: 3Calculate the factored load wu per unit length of the selected strip:
Own weight of slab = 0.17× 25 = 4.25 kN/m2 wu= 1.20 (3+4.25) +1.60 (3)= 13.5 kN/m2 For a strip 1 m wide, wu=13.5 kN/m 4Evaluate the maximum factored shear forces and bending moments in the strip:
The clear span length, ln = 4.0 – 0.30 = 3.70 m wu=13.5 kN/m
33
Solution:
18.5 7.7
16.8
16.8
16.8
Units of moment are in kN.m 34
18.5 7.7 16.8
Solution:
25
25
28.7
Units of shear are in kN 35
28.7
25
25
Solution: 5- Check slab thickness for beam shear:
Effective depth d = 17 – 2 – 0.60 = 14.40 cm, assuming Φ12 mm bars. Vu,max = 28.7 kN. V c 0.17 f c ' bd 0.750.17 28 1000144 95.8 kN
i.e. , slab thickness is adequate in of resisting beam shear. 6- Design flexural and shrinkage reinforcement:
Assume that Φ=0.9
2M u 0.85f c 1 1 2 fy 0.85 f c bd
Where b = 1000 mm & d = 144mm
36
Solution: For max. negative moment, Mu = 18.5 kN.m 0.8528 1 1 218.5106 ρ 0.00241 2 420 0.85 0.9 28 1000 144 3 0.85 1f c ' 3 0.85 0.85 28 0.01806 ρ 0.90 ρmax 8 fy 420 8 2 As,ve 0.002411000144 347 mm 2
As,min 0.00181000170 306mm As,ve OK 79φ10 347mm2 S 227.5mm 1000mmstrip S Smax min(450 or 3170) 450mm
37
use 10@20cm
Solution: For max. positive moment, Mu = 16.8 kN.m 0.8528 1 1 216.8106 0.00219 ρ 2 0.85 0.9 28 1000 144 420 3 0.85 1f c ' 3 0.85 0.85 28 0.01806 ρ 0.90 ρmax 8 fy 420 8 2 As,ve 0.002191000144 315mm 2
As,min 0.00181000170 306mm As,ve OK 79φ10 315mm2 S 251mm 1000mmstrip S Smax min(450 or 3170) 450
38
use 10@25cm
Solution: Calculate the area of shrinkage reinforcement: Area of shrinkage reinforcement = 0.0018 (100) (17) = 306 mm2 For shrinkage reinforcement use Φ 10 mm @ 25 cm (from previous slides calculations) Shrinkage reinft. Φ10@25 Φ10@25
Φ10@20
Φ10@25
Φ10@20
17cm Φ10@25
Φ10@25
Φ10@25
See Lecture 12 for information on detailing requirements
39
Solution:
8.0 m
Φ10@25 Φ10@25
4.0 m
40
Φ10@20
Φ10@20 Φ10@25
4.0 m
Φ10@25 Φ10@25
4.0 m
Design of one way RIBBED slabs
41
One-way ribbed slabs Ribbed slabs consist of regularly spaced ribs monolithically built with a toping slab. The voids between the ribs may be either light material such as hollow blocks [figure 1] or it may be left unfilled [figure 2]. Topping slab
Rib
Hollow block
Figure [1] Hollow block floor
Temporary form Figure [2] Moulded floor
The use of these blocks makes it possible to have smooth ceiling which is often required for architectural considerations and have good sound and temperature insulation properties besides reducing the dead load of the slab greatly. 42
Key components of one-way ribbed slabs ACI 8.13.6.1 Topping slab thickness (t) is not to be less than 1/12 the clear distance (lc) between ribs, nor less than 50 mm a. Topping slab:
lc t 12 50 mm
and should satisfy for a unit strip: t
lc Slab thickness (t)
w u lc2
1240 f c
Shrinkage reinforcement is provided in the topping slab in both directions in a mesh form. 43
Key components of one-way ribbed slabs b. Regularly spaced ribs: Minimum dimensions:
Ribs are not to be less than 100 mm in width, and a depth of not more than 3.5 times the minimum web width and clear spacing between ribs is not to exceed 750 mm. ACI 8.13.2 ACI 8.13.3 l ≤ 750 mm c
h ≤ 3.5 bw
bw ≥ 100
44
Key components of one-way ribbed slabs ACI 8.13.8 Shear strength provided by rib concrete Vc may be taken 10% greater than those for beams. Shear strength:
Flexural strength:
Ribs are designed as rectangular beams in the regions of negative moment at the s and as T-shaped beams in the regions of positive moments between the s. Effective flange width be is taken as half the distance between ribs, center-to-center. b e
45
Key components of one-way ribbed slabs Hollow blocks: Hollow blocks are made of lightweight concrete or other lightweight materials. The most common concrete hollow block sizes are 40 × 25 cm in plan and heights of 14, 17, 20, and 24 cm.
46
Summary of one-way ribbed slab design procedure 1.The direction of ribs is chosen. 2. Determine h, and select the hollow block size, bw and t 3.Provide shrinkage reinforcement for the topping slab in both directions. 4. The factored load on each of the ribs is computed. 5. The shear force and bending moment diagrams are drawn. 6. The strength of the web in shear is checked. 7.Design the ribs as T-section shaped beams in the positive moment regions and rectangular beams in the regions of negative moment. 8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared. 47
Example Design a one-way ribbed slab to cover a 3.8 m x 10 m , shown in the figure below. The covering materials weigh 2.25 kN/m2, equivalent partition load is equal to 0.75 kN/m2, and the live load is 2 kN/m2.
3.8 m
Use fc’=25 MPa, fy=420MPa
10 m
48
Solution 1. The direction of ribs is chosen:
3.8 m
Ribs are arranged in the short direction as shown in the figure
5.0 m
5.0 m
2. Determine h, and select the hollow block size, bw and t:
From ACI Table 9.5(a), hmin = 380/16 = 23.75cm use h = 24 cm. Let width of web, bw =10 cm Use hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN) Topping slab thickness = 24 – 17 = 7cm > lc/12 =40/12= 3.3cm > 5cm OK For a unit strip of topping slab: wu=[1.2(0.07 × 25 + 0.75 + 2.25) + 1.6(2)] ×1m = 8.9 kN/m = 8.9 N/mm t
w u l c2 1240 f c
8.9( 400) 2 ( 0.9)1240 25
16mm OK 49
Solution 3. Provide shrinkage reinforcement for the topping slab in both directions:
Area of shrinkage reinforcement, As=0.0018(1000)70=126 mm2 Use 5 Φ 6 mm/m in both directions. 4. The factored load on each of the ribs is to be computed:
50
1.0 m
0.4 m
0.1 m
0.4 m
7 cm
0.25 m
1.0 m
0.05 m
0.24 m
Total volume (in 1m2 surface) = 1.0 × 1.0 × 0.24 = 0.24 m3 Volume of hollow blocks in 1m2 = 8 × 0.4 × 0.25 × 0.17 = 0.136 m3 Net concrete volume in 1m2 = 0.24- 0.136 = 0.104 m3 Weight of concrete in 1m2 = 0.104 × 25 = 2.6 kN/m2 Weight of hollow blocks in 1m2 = 8 × 0.17= 1.36 kN/m2 Total dead load /m2 = 2.25 + 0.75 + 2.6 + 1.36 = 7.0 kN/m2
Solution wu=1.2(7)+1.6(2)=11.6 kN/m2 wu/m of rib =11.6x0.5= 5.8 kN/m of rib 5. Critical shear forces and bending moments are determined (simply ed beam):
Maximum factored shear force = wul/2 = 5.8 (3.8/2) = 11 kN Maximum factored bending moment = wul2/8 = 5.8 (3.8)2/8 = 10.5 kN.m 6. Check rib strength for beam shear:
Effective depth d = 24–2–0.6–0.6 =20.8 cm, assuming 12mm reinforcing bars and Φ 6 mm stirrups. 1.1ΦVc 1.10.75 0.17 25 100 208 14400 N = 14.4 kN Vu,max 11kN
Though shear reinforcement is not required, 4 6 mm stirrups per meter run are to be used to carry the bottom flexural reinforcement.
51
Solution 7. Design flexural reinforcement for the ribs:
There is only positive moments over the simply ed beam, and the section of maximum positive moment is to be designed as a T-section Assume that a<70mm and Φ=0.90→Rectangular section with b = be =500mm
s
e
Use 210mm (As,sup= 157 mm2) As f y
157 420 6.2 mm 70mm 0.85f c 'b e 0.8525500 The assumption is right
a
52
50
105 kN.m
0.85 25 210.5106 ρ 1 1 420 0.90.85255002082 0.0013 A ρ b d 0.0013500 208 135 mm 2
7 24
As 10
Solution CheckAs,min
0.25 f c ' 1.4 bw d A s,min max bw d ; fy f y 2 2 A s,min 70 mm A s,sup 157 mm OK
Check Φ=0.9
c
a 6.2 7.3 mm β1 0.85
dc 208 7.3 0.003 ε t 0.003 7.3 c ε t 0.083 0.005 0.9 OK
53
Solution
A
reinforcement are to be
1Φ10 m
A
1Φ10 m
1Φ10 m
1Φ10 m
3.8 m
8. Neat sketches showing arrangement of ribs and details of the prepared
5.0 m
5.0 m Φ6mm mesh @20 cm
Φ6mm stirrups @25 cm
7cm 24cm 17cm
2Φ10mm
10
40 cm
10
2Φ10mm
SectionA-A
See Lecture 12 for information on detailing requirements 54
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 11 Design of short concentric columns
Columns Columns are vertical compression of a structural frame intended to the load-carrying beams. They transmit loads from the upper floors to the lower levels and then
to the soil through the foundations.
Loads Beam
P
Beam
Column
h
Slab
b
Beam
l
h
Column
Beam
b
Beam
Slab Footing Beam
Beam Soil
2
Columns Usually columns carry bending moment as well, about one or both axes of the cross section, and the bending action may produce tensile forces over a part of the cross section.
The main reinforcement in a columns is longitudinal, parallel to the direction of the load and consists of bars arranged in a square, rectangular, or circular shape.
3
Length of the column in relation to its lateral dimensions Columns can be classified as
1Short Columns, for which the strength is governed by the strength of the materials and the dimensions of the cross section
2Slender Columns, for which the strength may be significantly reduced by lateral deflections.
Position of the load on the cross-section Columns can be classified as
1Concentrically loaded columns, which are subjected to axial force only
2Eccentrically loaded columns, which are subjected to moment in addition to the axial force. 4
Analysis and Design of Short Columns
Column Types: 1. Tied
2. Spiral 3. Composite
5
Behavior of Tied and Spirally-Reinforced Columns Axial load tests have proven that tied and spirally reinforced columns having the same cross-sectional areas of concrete and steel reinforcement behave in the same manner up to the ultimate load. At that load, tied columns fail suddenly due to excessive cracking in the concrete section followed by buckling of the longitudinal reinforcement between ties within the failure region. For spirally reinforced columns, once the
ultimate load is reached, the concrete shell covering the spiral starts to spall off but the core will continue to carry additional loads because the spiral provides a confining force to the concrete core, thus enabling the column to sustain large deformations before final collapse. 6
Behavior of Tied and Spirally-Reinforced Columns
Failure of a tied column
7
Failure of a spiral column
Nominal Capacity under Concentric Axial Loads
P0 0.85f c A g A st f y A st or
P0 A g 0.85f c Ast (f y 0.85f c) A g = gross area = b*h Ast = area of longitudinal steel fc′ =concrete compressive strength 8
f y = steel yield stress
Maximum Nominal Capacity under Concentric Axial Loads
Pn rP0 Pn r Ag 0.85f c Ast (f y 0.85f c) r = Reduction factor to for accidental eccentricity
r = 0.80 ( tied )
r = 0.85 ( spiral ) 9
ACI 10.3.6.3
Design Capacity under Concentric Axial Loads
Pn Pu Pn r A g 0.85f c A st f y 0.85f c Pu or
Pn r A g 0.85f c g f y 0.85f c Pu where g = Ast / Ag ACI 9.3.2.2
10
= 0.65 for tied columns
ACI 10.3.6.3 r = 0.80 ( tied )
= 0.75 for spiral columns
r = 0.85 ( spiral )
Design of Short Concentrically Loaded Columns
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu Both Ag and g are unknown in this equation. There are two options to design the column: 1Select Ag and calculate g. The A g sizing (Ag = Pu / 0.5fc′ ).
may be selected from initial
2Select g and calculate Ag. Usually g is assumed as 2% as a starting point. 11
Calculation of required cross section, if steel ratio is known
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu * when g is known or assumed:
Ag
12
Pu
r 0.85f c g f y 0.85f c
Calculation of required steel ratio, if cross section is known
Pn Pu Pn r A g 0.85f c g f y 0.85f c Pu * when g is known or assumed:
1 Pu 0.85f c g r A g f y 0.85f c 13
Design of spirals Spiral Reinforcement Ratio, s
Volume of Spiral 4Asp s Volume of Core Dcs from:
A sp D c
s 2 [( / 4) D c ] s
Asp cross-sectional area of spiral reinforcement D c core diameter: outside edge to outside edge of spiral 14
s spacing pitch of spiral steel (center to center)
Design of spirals Spiral Reinforcement Spacing, s
Ag f c s 0.45 1 ACI Eq. 10-5 Ac f y 4A s sp from previous slide D cs s
4A sp f c ' Ag 0.45D c 1 Ac f y
A c core area 15
Dc2
Ag gross area
4
D2 4
Design Considerations
Longitudinal Steel
- Limits on reinforcement ratio: ACI 10.9.1
0.01Ag Ast 0.08Ag or
0.01 g 0.08 16
Design Considerations
Longitudinal Steel
- Minimum number of bars ACI 10.9.2
min. of 6 bars in spiral arrangement min. of 4 bars in rectangular or circular ties min. of 3 bars in triangular ties 17
Design Considerations
Longitudinal Steel - Clear Distance between Reinforcing Bars (Longitudinal Steel) For tied or spirally reinforced columns, clear distance between bars, shown in the figure, is not to
be less than the larger of 1.50 times bar diameter or 40 mm. This is done to ensure free flow of concrete among reinforcing bars.
ACI 7.6.3
S c max 1.5d b , 40mm 18
Design Considerations
Lateral Ties
- Arrangement of ties and longitudinal bars: ACI 7.10.5.3 1.) At least every other longitudinal bar shall have lateral from the corner of a tie with an included angle 135o. 2.) No longitudinal bar shall be more than 150 mm clear on either side from a laterally ed bar. 19
Design Considerations
Lateral Ties
- Arrangement of ties and longitudinal bars: ACI 7.10.5.3
20
Ties shown dashed may be omitted if x < 150 mm
Design Considerations
Lateral Ties
- Maximum vertical spacing: ACI 7.10.5.2
s 16 db ( db = diameter for longitudinal bars ) s 48 dstirrup (dstirrup = diameter for stirrups) s least lateral dimension of column
21
Design Considerations
Lateral Ties
- Minimum size of ties ACI 7.10.5.1
size 8 bar if longitudinal bar 30 bar 12 bar if longitudinal bar 32 bar 12 bar if longitudinal bars are bundled
22
Design Considerations
Spirals
- Size and spacing of spiral ACI 7.10.4.2
size 10 mm diameter ACI 7.10.4.3 25mm 23
clear spacing between spirals
75mm
Design Considerations ACI 7.7.1 Concrete Protection Cover The clear concrete cover is not to be taken less than 4 cm for columns not exposed to weather or in with ground.
Minimum Cross Sectional Dimensions The ACI Code does not specify minimum cross sectional dimensions for columns. Column cross sections 20 × 25 cm are considered as the smallest practicable sections. For practical considerations, column dimensions are taken as multiples of 5 cm. Lateral Reinforcement
Ties are effective in restraining the longitudinal bars from buckling out through the surface of the column, holding the reinforcement cage together during the construction process, confining the concrete core and when columns are subjected to horizontal forces, they serve as shear reinforcement. 24
Design Procedure for Short Concentrically Loaded Columns 1. Evaluate the factored axial load Pu acting on the column. This can be done by:
a- Tributary Area Method b- Pu is the sum of the reactions of the beams ed by the column. 2. Assume a starting reinforcement ratio ρg that satisfies ACI Code limits. Usually a 2
% ratio is chosen for economic considerations. 3. Determine the gross sectional area Ag of the concrete section. 4. Choose the dimensions of the cross section based on its shape. 5. Readjust the reinforcement ratio by substituting the actual cross sectional area in the respective equation. This ratio has to fall within the specified code limits. 25
Design Procedure for Short Concentrically Loaded Columns 6.
Calculate the needed area of longitudinal reinforcement ratio based on the adjusted
reinforced ratio and the chosen concrete dimensions. 7.
From reinforcement tables, choose the number and diameters of needed reinforcing bars. For rectangular sections, a minimum of four bars is needed, while a minimum of six bars is used for circular columns.
8.
Design the lateral reinforcement according to the type of column, either ties or spirals.
9.
Check whether the spacing between longitudinal reinforcing bars satisfies ACI Code requirements.
10. Draw the designed section showing concrete dimensions and with required longitudinal and lateral reinforcement.
26
Example 1 The cross section of a short axially loaded tied column is shown in the figure. It is reinforced with 616mm bars. Calculate the design load Ties Φ8@25cm capacity of the cross section. Use fc′ =28 MPa and fy = 420 MPa. 25
6Φ16
Solution: ρg
1206 Ast 0.012 1.21% A g 250 400
ρ min 1 % ρ g 1.21% ρ max 8%
40 Figure [1]
OK
Clear distance between bars Sc
40 2(4) 2(0.8) 3(1.6) 12.8cm 31 max 1.5d b 2.4cm , 4cm <Sc 12.8cm
Sc=12.8 cm
25
6Φ16
Sc
Only, one tie is required for the cross section 27
40
Example 1 The spacing between ties 16 db =16(1.6) = 25.4 cm ≥ S = 25 cm 48 ds = 48(0.8) = 38.4 cm ≥ S = 25 cm smaller of b or d = 25 cm ≥ S = 25 cm Thus, ACI requirements regarding reinforcement ratio, clear distance between bars and tie spacing are all satisfied. The design load capacity ΦPn Pn 0.65(0.8) A g 0.85f c g f y 0.85f c Φ Pn 0.52A g 0.85f c' ρ g f y 0.85fc'
Φ Pn 0.65(0.8) 400 250 0.8528 0.0121420 0.85 28 Φ Pn 1487 kN 28
Example 2 Design a short tied column to a factored concentric load of 1000 kN, with one side of the cross section equals to 25 cm. f c 30MPa
f y 420MPa
Solution
Assume first that g 2% Ag
Pu
0.65 0.8 0.85f c g f y 0.85f c
1000103 Ag 0.65 0.8 0.8530 0.02 420 0.85 30 29
A g 57594mm
2
Example 2 A
g
57594mm 2
b 250mm
h 230mm use column 25cm 25cm Determine adjusted steel ratio Pu 1 0.85f c g r A g f y 0.85f c 1 1000103 0.85(30) =0.0134 g = 0.65 0.8 250 250 420 0.85(30) 0.01<g <0.08 OK A bh 0.0134(250)(250) 835mm 2 s
g
2
30
Use 614 (As,sup = 924 mm )
Example 2 Check spacing s
h (No. of bars/2) d b 2 cover 2 dstirrup
(No. of bars/2) 1 250 (6 / 2) 14 2 40 2(8) 31
56mm
max 1.5d b 21mm , 40mm 56mm < 150mm
OK
Stirrup design Use 8 mm (for longitudinal bars with 14 mm < 30 mm)
smax
16d b 1614cm 224mm min 48d stirrup 488 384mm smaller of b or d 250mm
Use 8 mm @ 200 mm 31
governs
6 14 mm
Example 2
8 mm @ 200 mm
250 mm
250 mm
32
Example 3 Design a short, spirally reinforced column to a service dead load of 800 kN and a service live load of 400 kN. f c 28MPa
f y 420MPa
Use g 1%
Solution
Pu 1.20 PD 1.60 PL 1.2800 1.6 400 1600kN Ag
Pu
0.75 0.85 0.85f c g f y 0.85f c
1600103 Ag 0.75 0.85 0.8528 0.01420 0.85 28 33
A g 90405mm 2
Example 3 A g 90405mm
360/N
2
for circular column D=
Ag
4
=339mm 0.5D’
use column with D = 350 mm
A s 0.01 (3502 ) 962mm 2 4 2
use 714 (As,sup =1078 mm ) Check spacing between longitudinal bars D’ =350-2(40)-2(8)-14=240 mm,
N=7
360/N 51.43 104.1mm S D'sin 240sin 2 2 Sc 104.114 90.1mm 1.5(14)=21mm 34
40mm
OK
360/N 2 0.5D’
0.5Sc = 0.5D’ sin(360/N/2)
Example 3 Design the needed spiral, try 8 Dc 350 2(40) 270 mm S
4Asp
450
π/4 350 2 28 0.45 270 1 2 420 π/4270 S 36.3mm, taken as 35 mm (center to center) Sc 35 8/2 8/2 27 mm, i.e within ACIcode limit ( 25mm& 75mm) A g fc ' 1 0.45Dc Ac fy
Use Φ 8mmspiral with a pitch of 35mmcenter to center.
35
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 Part I Bond, development length, and splicing
Bond
2
Concept of Bond Stress Bond stresses are existent whenever the tensile stress or force in a reinforcing bar changes from point to point along the length of the bar in order to maintain equilibrium. Without bond stresses, the reinforcement will pull out of the concrete.
Concrete Reinforcing bar
PL/4
M
M+dM
dx Moment diagram
3
µavg
Concept of Bond Stress
F 0.0 T2 T1 Fbond If this equation is not true (bond force Fbond is not strong enough), the bar will pull out A bar fs2 fs1 avg A surface
d 4
2 b
db
µ=Bond stress
fs2 fs1 avg ( db ) l
μ avg
f - f d = s2 s1 b 4l
µavg= average bond stress
4
T1=fs1Ab
l
T2=fs2Ab fs2=fs1+∆fs
Mechanism of Bond Transfer A smooth bar embedded in concrete develops bond by adhesion between concrete & reinforcement, and a small amount of friction.
This is different in a deformed bar. Once adhesion is lost at high bar stress and some slight movement between the reinforcement and the concrete occurs, bond is then provided by friction and bearing on the deformations of the bar. At much higher bar stress, bearing on the deformations of the bar will be the only component contributing to
bond strength.
(a) Forces on bar
5
(b) Forces on concrete
Splitting cracks The radial component of the bearing
force will cause circumferential stress on the concrete that may cause splitting that creates cracks.
6
Splitting cracks Splitting of concrete may occur along the bars, either in vertical planes as in figure (a) or in horizontal plane as in figure (b).
7
Splitting cracks The load at which splitting failure develops is a function of :
•
The minimum distance from the bar to the surface of the concrete or to the next bar. The smaller the distance, the smaller is the splitting load.
•
The tensile strength of the concrete. The higher the tensile strength, the higher is the splitting resistance.
•
The average bond stress. The higher the average bond stress, the higher is the splitting resistance.
If the concrete cover and bar spacing are large compared to the bar diameter, a pullout failure can occur, where the bar and the ring of concrete between successive deformations pullout along a cylindrical failure surface ing the tips of the deformations. 8
Development Length
9
Development Length The bars found to be needed at a section from design calculations have to be embedded a certain distance into the concrete. This distance has to be equal or larger than the development length (ld).
10
Development Length The development length ld is that length of embedment necessary to develop the full tensile strength of the bar (on both sides of sections where fy stress is required),
controlled by either pullout or splitting.
μavg =
f s2 f s1 db 4l
f s2 f s1 f y
ld 11
f d y
b
4 avg,u
, where avg,u is the value avg at bondfailure
Development Length of Deformed Bars in Tension The development length for deformed bars in tension is given by
ld
fy
αβ γ λ
1.1 fc C K tr d b
d b 300 mm,
C K tr where 2.5 db where, ld = development length db = nominal diameter of bar fy = specified yield strength of reinforcement C = spacing or cover dimension (see next slide) Ktr = transverse reinforcement index (see slide 12) 12
= see next slides
ACI Eq. 12-1
ACI 12.2.3
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.4 C is the smaller of (a)the smallest distance measured from the center of the bar to the nearest concrete surface (b)one-half the center-to-center spacing of bars being developed. α is a bar location factor (a) Horizontal reinforcement so placed that more than 30 cm of fresh concrete is cast in the member below the development length or splice………………………………………………………………. (b) Other reinforcement…
1.3 1.0
β is a coating factor that reflects the adverse effects of epoxy coating (a)Epoxy-coated bars or wires with cover less than 3db or clear spacing less than 6db (b)All other epoxy-coated bars or wires… (c) Uncoated reinforcement……………………………………………………… 13
However, the product β α is not to be greater than 1.7.
1.5 1.2 1.0
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.4 γ is a reinforcement size factor that reflects better performance of the smaller diameter reinforcement (a) Φ20mm and smaller bars. (b) Φ22mm and larger bars.
0.8 1.0
λ is a lightweight aggregate concrete factor that reflects lower tensile strength of lightweight concrete, & resulting reduction in splitting resistance. (a) When lightweight aggregate concrete is used…….……..……………… 0.8 (b) When normal weight concrete is used… 1.0
14
Development Length of Deformed Bars in Tension [contd.] Ktr is a transverse reinforcement index that represents the contribution of confining reinforcement
K tr
40 Atr sn
ACI Eq. 12-2
where: Atr = total cross sectional area of all transverse reinforcement within the spacing s, which crosses the potential plane of splitting along the reinforcement being developed within the development length s = maximum center-to-center spacing of transverse reinforcement within development length ld n = number of LONGITUDINAL bars being developed along the plane of splitting. Note: It is permitted to use Ktr= 0.0 as design simplification even if transverse reinforcement is present. 15
Atr
Potential plane of splitting
n=4
Development Length of Deformed Bars in Tension [contd.]
ACI 12.2.5 Excessive Reinforcement Reduction in development length is allowed where As provided > As required. the reduction is given by Reduction factor
As required As provided
-Except as required for seismic design
-Good practice to ignore this factor, since the use of the structure may change over time.
Simplified Expression for Development Length See ACI 12.2.2. This will not be used in this class
16
Example 1
60 cm
Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown
Φ10@20 4Φ20
40 cm Cover is 4 cm on all sides
SectionA-A
17
Example 1 Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown
(a) Ktr is calculated α=1.0 for bars over concrete < 30 cm thick
Φ10@20 4Φ20
β=1.0 for uncoated bars
α β =1.0 <1.7 OK
40 cm Cover is 4 cm on all sides
γ=0.8 for Φ20mm, λ=1.0 for normal weight concrete, C the smallest of
40+10+(20/2)=60 mm {[400-2(40)-2(10)-2(20/2)]/(3)}/(2)=46.7 mm
18
60 cm
Solution:
i.e., C is taken as 46.7 mm
Example 1 [contd.] K tr
40A t r 40(2 79) 7.9 mm sn (200)(4)
i.e., use
ld
C Ktr 2.5 db
Φ10@20 4Φ20
αβ γ λ d b 300 mm 1.1 fc C K t r d b fy
40 cm Cover is 4 cm on all sides
420 (1.0)(1.0)(0.8)(1.0) 20 489 mm 300 mm OK ld 2.5 1.1 25 b) Assuming K t r 0.0 C Ktr 467 0 2.33 2.5 db 20 19
60 cm
C Ktr 46.7 7.9 2.73 2.5 db 20
OK
420 (1.0)(1.0)(0.8)(1.0) 20 524 mm 300 mm OK ld 2.33 1.1 25
Example 1 [contd.]
60 cm
(c) Check if space is available for bar development
Φ10@20 4Φ20
40 cm Cover is 4 cm on all sides
SectionA-A
Available length for bar development = 2000+ 150– 40 = 2110 mm > ld = 524 mm
OK 20
Development Length of Deformed Bars in Compression
ACI 12.3
Shorter development lengths are required for compression than for tension since flexural tension cracks are not present for bars in compression. In addition, there is some bearing of the ends of the bars on concrete.
The development length ld for deformed bars in compression is computed as the product of the basic development length ldc and applicable modification factors, but ld is not to be less than 200 mm.
ld = ldc x applicable modification factors ≥ 200 mm. The basic development length ldb for deformed bars in compression is given as
0.24 f y db ldc max ;0.043 f y db fc ' 21
Development Length of Deformed Bars in Compression [contd.]
ACI 12.3
Applicable Modification Factors 1.Excessive reinforcement factor =As required / As provided 2.Spirals or Ties: the modification factor for reinforcement, enclosed with spiral reinforcement ≥ 6mm in diameter and ≤ 10 cm pitch or within Φ12mm ties spaced at ≤ 10 cm on center is given as 0.75
Development Lengths for Bundled Bars
ACI 12.4
Development length of individual bars within a bundle, in tension or compression, is
taken as that for individual bar, increased 20% for three-bar bundle, and 33% for fourbar bundle. For determining the appropriate modification factors, a unit of bundled bars is treated as a single bar of a diameter derived from the equivalent total area of bars.
22
ldh
Critical section
Development of Standard Hooks in Tension Hooks are used to provide additional anchorage when there is insufficient length available
to develop a bar. Development length ldh for deformed bars in tension terminating in a standard hook is computed as the product of the basic development length lhb and applicable modification factors, but ldh is not to be less than 8db, nor less than 150 mm.
ldh = lhb x applicable modification factors ≥ 15 cm or 8db. The basic development length lhb for hooked bars is given as
lhb
0.24 f
e y
fc '
For lightweight aggregate concrete, = 0.75. For epoxy-coated reinforcement, e= 1.2. 23
Otherwise, = 1.0, e= 1.0
ACI 12.5.1
db
ACI 12.5.2
Development of Standard Hooks in Tension [contd.]
ACI 12.5.3
Applicable Modification Factors 1. Concrete cover: for db ≤ Φ36mm, side cover (normal to plane of hook) ≥ 65 mm, and for 90 degree hook, cover on bar extension beyond hook ≥ 50 mm, the modification factor is taken as 0.7. not less than 50 mm 65 mm
65 mm
24
Development of Standard Hooks in Tension [contd.]
ACI 12.5.3
Applicable Modification Factors 2.Excessive reinforcement factor =As required / As provided 3.Spirals or Ties: for db ≤ Φ36mm, hooks enclosed vertically or horizontally within ties or stirrups spaced along the full development length ldh not greater than 3db , where db is the diameter of the hooked bar, and the first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, the modification factor is taken as 0.8.
25
Development of Standard Hooks in Tension [contd.] Development length ldh is measured
ACI 7.1
90-degree hook
from the critical section of the bar to the out-side end or edge of the hooks. Either a 90 or a 180-degree hook, shown in the figure, may be used ldh
Development of reinforcement- General * Hooks are not considered effective in compression and may not be used as anchorage.
Part (a)
180-degree hook
ACI 12.5.5 * The values of fc ' used in this lecture shall not exceed 8.3 MPa. 26
ACI 12.1.2
4db ≥ 65mm
Φ10 through Φ25 Φ28 through Φ36 Φ44 through Φ56
ldh
Part (b)
Development of Standard Hooks in Tension [contd.]
ACI 12.5.4
Confinement of hooks For bars being developed by a standard hook at discontinuous ends of with both side cover and top (or bottom) cover over hook less than 65 mm, the hooked bar shall be enclosed within ties or stirrups perpendicular to the bar being developed, spaced not greater than 3db along ldh. The first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, where db is the diameter of the hooked bar.
27
Example 2
(a) If a 180-degree hook is used (b) If a 90-degree hook is used Use fc’ = 28 MPa and fy = 420 MPa
50 cm
Determine the development length or anchorage required for the epoxy-coated top bars of the beam shown in the figure. The beam frames into an exterior 80cm x 30cm column (the bars extend parallel to the 80 cm side). Show the details if:
4Φ32 Φ12@15
Solution: α=1.3 for bars over concrete > 30 cm thick β=1.5 for coated bars (take the larger of 1.2 and 1.5 conservatively)
α β =1.3x1.5 = 1.95 > 1.7 use 1.7 γ=1.0 for Φ32mm, C the smallest of
λ=1.0 for normal weight concrete 40+12+16=68 mm
{[400-2(40)-2(12)-32]/(3)}/(2)=44 mm i.e., C is taken as 44 mm 28
40 cm
Example 2 [contd.] 40A tr 2( 113) 15.1mm sn ( 150)( 4)
C Ktr 44 15 1.85 2.5 db 32
ld
fy
αβ γ λ
1.1 fc C K t r d b
50 cm
K tr
OK
4Φ32 Φ12@15
d b 300 mm 40 cm
420 ( 1.7 )( 1.0)( 1.0) ld 1.85 1.1 28
32 2127 m m 300 mm OK
Available length for bar development = 800 – 40 = 760 mm < ld = 2127 cm
Thus, a standard hook is required at column side
ldh = lhb x applicable modification factors ≥ 150 mm or 8db. (use a factor 1.2 for epoxy-coated hooks. Modification factors are inapplicable) l dh 29
0.24 e f y
fc '
db
0.24 1.2 420 32 732 m m 1.0 28
150m m 8( 32) 256m m O K
Example 2 [contd.] (b) If a 180-degree hook is used
ldh=732 mm
4db =128 mm Critical section 5db =160 mm 180o hook
12db=384 mm
(c) If a 90-degree hook is used
30
ldh=732 mm
Critical section
90o hook
Splicing
31
Splices of Reinforcement
ACI 12.14
Splicing of reinforcement bars is necessary, either because the available bars are not long enough, or to ease construction, in order to guarantee continuity of the reinforcement according to design requirements.
Types of Splices: (a) Welding (b) Mechanical connectors (c) Lap splices (simplest and most economical method) In a lapped splice, the force in one bar is transferred to the concrete, which transfers it to
the adjacent bar. Splice length is the distance over which the two bars overlap.
Forces on bar at splice
32
Splice length
Splices of Reinforcement Important note: Lap splices have a number of disadvantages, including congestion of reinforcement at the lap splice and development of transverse cracks due to stress concentrations. It is recommended to locate splices at sections where stresses are low. Types of Lap Splices: 1. Direct Splice T
T ls
Direct
2. Non- Splice (spaced) the distance between two bars cannot be greater than 1/5 of the splice length nor 15 cm
ACI 12.14.2.3 T
s
T ls
33
Bars are spaced
Splices of Deformed Bars in Tension
ACI 12.15
ACI code divides tension lap splices into two classes, A and B. The class of splice used is dependent on the level of stress in the reinforcing and on the percentage of steel that is spliced at particular location.
ACI 12.15.1
ClassA: A splice must satisfy the following two conditions to be in this class: (a) the area of reinforcement provided is at least twice that required by analysis over the entire length of the splice; and (b) one-half or less of the total reinforcement is spliced within the required lap length. Class B: If conditions above are not satisfied classify as Class B. The splice lengths for each class of splice are as follows: Class A splice: 1.0 ld 300 mm Class B splice: 1.3 ld 300 mm 34
ACI 12.15.2
Example 3 To facilitate construction of a cantilever retaining wall, the vertical reinforcement shown in the figure, is to be spliced with dowels extending from the foundation. Determine the required splice length when all reinforcement bars are spliced at the same location.
Use fc’ = 30 MPa and fy = 420 MPa Φ16 @ 250 Cover = 7.5 cm
Solution: Class B splice is required where ls = 1.3 ld α=1.0, β=1.0 → α β =1.0 < 1.7 OK ls
γ=1.0, λ=1.0 C the smallest of
75+8=83 mm 250/2=125 mm
i.e., C is taken as 83 mm 35
Ktr =0.0, since no stirrups are used
Φ16 @ 250 Cover = 7.5 cm
Example 3 [contd.] C Ktr 83 0 C Ktr 5.19 2.5 i .e., 2.5 db 16 db
420 ( 1.0)( 1.0)( 1.0) 16 446 m m ld 2.5 1.1 30 Required splice length l s 446( 1.3 ) 580 m m 300 OK Φ16 @ 25
ls=58 cm
Φ16 @ 25
36
Splices of Deformed Bars in Compression
ACI 12.16
Bond behavior of compression bars is not complicated by the problem of transverse tension cracking and thus compression splices do not require provisions as strict as those specified for tension Compression lap splice length shall be: 0.071 fy db ≥ 300 mm (0.13 fy – 24) db ≥ 300 mm
ACI 12.16.1 for fy ≤ 420 MPa for fy > 420 MPa
The computed splice length should be increase by 33% if fc’<21 MPa When bars of different size are lap-spliced in compression, splice length shall be the larger of either development length of the larger bar, or splice length of the smaller bar.
ACI 12.16.2 ACI 12.15.3 37
Example 4 Design a compression lap splice for a tied column whose cross section is shown in the figure when: (a)Φ16 mm bars are used on both sides of the splice. (b)Φ 16 mm bars are lap spliced with Φ 18 mm bars. Use fc’ = 30 MPa and fy = 420 MPa
Solution: (a) For bars of same Φ16 mm diameter
Splice length in compression and for fy =420 MPa is equal to 0.071 fy db = 0.071 (420)(16) = 477 mm >300 mm taken as 480 mm
38
Example 4 [contd.] (b) For bars of different diameters
The development length of the larger bar
ldc = ldb x applicable modificationfactors
0.24f y d b 0.24 420 18 331mm fc ' 30 l dc max 333mm 0.043f d 0.043 420 18=333mm y b Splice length of smaller diameter bar was calculated in part (a) as 477 mm. Thus, the splice length is taken as 480 mm.
39
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 PART II Bar cutoff
Bar cutoff It is economical to cut unnecessary bars as shown in the scenario below.
2
Bar cutoff: Theoretical points of cutoff or bent Example
3
Bar cutoff: Theoretical points of cutoff or bent Example
4
Bar cutoff: Theoretical points of cutoff or bent Example
5
Bar cutoff: Theoretical points of cutoff or bent Example
6
Bar cutoff: Theoretical points of cutoff or bent Using moment diagrams drawn to scale:
7
Bar cutoff: Theoretical points of cutoff or bent Using moment envelopes drawn to scale:
8
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/11, +1/16, -1/11)
9
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/16, +1/14, -1/10)
10
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: -1/24, +1/14, -1/10)
11
Bar cutoff: Theoretical points of cutoff or bent Bending moment envelope for typical span (moment coefficient: 0, +1/11, -1/10)
12
Bar cutoff: Theoretical points of cutoff or bent Development length requirements
ACI 12.10.3 Reinforcement shall extend beyond the point at which it is no longer required to resist flexure for a distance equal to d or 12db, whichever is greater, except at s of simple spans and at free end of cantilevers.
ACI 12.10.4 Continuing reinforcement shall have an embedment length not less than ld beyond the point where bent or terminated tension reinforcement is no longer required to resist flexure. 13
Bar cutoff: Theoretical points of cutoff or bent Development length requirements
ACI 12.10.5
The ACI Code does not permit flexural reinforcement to be cutoff in a tension zone unless at least one of the special conditions, shown below, is satisfied: a.Factored shear force at the cutoff point does not exceed two-thirds of the design shear strength, ΦVn . b.Stirrup area exceeding that required for shear and torsion is provided along each cutoff bar over a distance from the termination point equal to three-fourths of the effective depth of the member. Excess stirrup area Av is not to be less than 0.41bwS /fy . Spacing S is not to exceed d/8βb where βb is the ratio of area of reinforcement cutoff to total area of tension reinforcement at the section. c.For φ 36 mm bars and smaller, continuing reinforcement provides double the area required for flexure at the cutoff point and factored shear does not exceed three-fourths of the design shear strength, ΦVn .
14
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
At least one-third the positive moment reinforcement in simple and one-fourth the positive moment reinforcement in continuous shall extend along the same face of member into the . In beams, such reinforcement shall extend into the at least 150 mm. ACI 12.11.1
At simple s and at points of inflection, positive moment tension reinforcement shall be limited to a diameter such that
ACI 12.11.3 Mn is calculated assuming all reinforcement at the section to be stressed to fy; Vu is calculated at the section; la at a shall be the embedment length beyond the center of ; or: la at a point of inflection shall be limited to d or 12db, whichever is greater.
15
An increase of 30 percent in the value of Mn /Vu shall be permitted when the ends of reinforcement are confined by a compressive reaction.
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
ACI 12.11.3
16
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
17
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
18
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
19
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Negative moment:
Negative moment reinforcement in a continuous, restrained cantilever member, or in any member of rigid frame, is to be anchored in or through the ing member by development length, hooks, or mechanical anchorage.
ACI 12.12.1
At least one-third the total tension reinforcement provided for negative moment at a shall have an embedment length beyond the point of inflection not less than d, 12db, or ln/16, whichever is greater
ACI 12.12.3
20
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 12 PART III Detailing of reinforcement
References for detailing ACI-318
2
References for detailing ACI-315 ACI Detailing Manual
3
References for detailing CRSI Design Handbook
4
Bar cutoff: Theoretical points of cutoff or bent Development length requirements Positive moment:
At least one-third the positive moment reinforcement in simple and one-fourth the positive moment reinforcement in continuous shall extend along the same face of member into the . In beams, such reinforcement shall extend into the at least 150 mm. Negative moment:
At least one-third the total tension reinforcement provided for negative moment at a shall have an embedment length beyond the point of inflection not less than d, 12db, or ln/16, whichever is greater 5
Typical details for one way solid slabs
6
Requirements for using standard detailing for beams and one way slabs: ACI 8.3.3 • 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. • are of similar section dimensions along their lengths (prismatic).
7
Typical details for one way solid slabs Straight bars
8
Typical details for one way solid slabs Straight bars
9
Typical details for one way solid slabs Straight bars
10
Typical details for one way solid slabs Bent-up bars
11
Typical details for beams Straight bars
12
Typical details for beams Straight bars
13
Typical details for beams Straight bars
14
Typical details for columns
15
Typical details for columns
16
17
18
19
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 13 Design of isolated footings
Footing Introduction Footings are structural elements used to columns and walls and transmit their loads to the underlying soil without exceeding its safe bearing capacity below the structure. Loads
B
B
L
Column
L
P
Beam
P M Footing
Soil
2
Footing Introduction The design of footings calls for the combined efforts of geotechnical and structural engineers. The geotechnical engineer, on one hand, conducts the site investigation and on the light of his findings, recommends the most suitable type of foundation and the allowable
bearing capacity of the soil at the suggested foundation level.
The structural engineer, on the other hand, determines the concrete dimensions and reinforcement details of the approved foundation.
3
Types of Footing Isolated Footings Isolated or single footings are used to single columns. This is one of the most economical types of footings and is used when columns are spaced at relatively long distances.
PkN
B
C2 C1 L P
4
Types of Footing Isolated Footings
Shapes of isolated footings 5
Types of Footing Isolated Footings
Shapes of isolated footings 6
Types of Footing Wall Footings Wall footings are used to structural walls that carry loads from other floors or to nonstructural walls. W kN/m
Secondary reinft
Main reinft.
7
Types of Footing Combined Footings Combined footings are used when two columns are so close that single footings cannot be used. Or, when one column is located at or near a property line. In such a case, the load on the footing will be eccentric and hence this will result in an uneven distribution of load to the ing soil. P1
P2
P2 kN
L
B
PP11kN kN
C2
C2 C1
C1 L1
L2
L2
8
Types of Footing Combined Footings The shape of a combined footing in plan shall be such that the centroid of the foundation plan coincides with the centroid of the loads in the columns. Combined footings are either rectangular or trapezoidal. Rectangular footings are favored due to their simplicity in of design and construction. However, rectangular footings are not always practicable because of the limitations that may be imposed on their longitudinal projections beyond the two columns or the large difference that may exist between the magnitudes of the two column loads. Under these conditions, the provision of a trapezoidal footing is more economical.
9
Types of Footing Continuous Footings Continuous footings a row of three or more columns.
P1
P2
P3
P4 kN
P4 P3 kN
P2 kN L P1 kN
B
10
Types of Footing Strap (Cantilever) footings Strap footings consists of two separate footings, one under each column, connected together by a beam called “strap beam”. The purpose of the strap beam is to prevent overturning of the eccentrically loaded footing. It is also used when the distance between this column and the nearest internal column is long that a combined footing will be too narrow. P2 kN P2
property line
P1
Strap Beam P1 kN L1
L2
C2
B1 C1
C2
B2
C1
11
Types of Footing Mat (Raft) Footings Mat footings consist of one footing usually placed under the entire building area. They are used when soil bearing capacity is low, column loads are heavy and differential settlement for single footings are very large or must be reduced.
L
12
Types of Footing Pile caps Pile caps are thick slabs used to tie a group of piles together to and transmit column loads to the piles. P
B
L
13
Footing Loading Distribution of Soil Pressure The distribution of soil pressure under a footing is a function of the type of soil, the relative rigidity of the soil and the footing, and the depth of the foundation at the level of between footing and soil. P
P
P Centroidal axis
L
Footing on sand
L
Footing on clay
L
Equivalent uniform distribution
For design purposes, it is common to assume the soil pressure is uniformly distributed. The pressure distribution will be uniform if the centroid of the footing coincides with the resultant of the applied loads.
14
Footing Loading Pressure Distribution Below Footings The maximum intensity of loading at the base of a foundation which causes failure of soil is called ultimate bearing capacity of soil, denoted by qu. The allowable bearing capacity of soil is obtained by dividing the ultimate bearing capacity of soil by a factor of safety on the order of 2.50 to 3.0. The allowable soil pressure for soil may be either gross or net pressure permitted on the soil directly under the base of the footing. The gross pressure represents the total stress in the soil created by all the loads above the base of the footing. For design, the net soil pressure is used instead of the gross pressure value. P
Df hc
15
Footing Loading Concentrically Loaded Footings If the resultant of the loads acting at the base of the footing coincides with the centroid of the footing area, the footing is concentrically loaded and a uniform distribution of soil pressure is assumed in design. P Centroidal axis
L P/A L
B
16
Footing Loading Eccentrically Loaded Footings Footings are often designed for both axial load and moment. Moment may be caused by lateral forces due to wind or earthquake, and by lateral soil pressures. A footing is eccentrically loaded if the ed column is not concentric with the footing area or if the column transmits at its juncture with the footing not only a vertical load but also a bending moment. P
P
e M Centroidal axis
Centroidal axis
y
y
L
L P/A
P/A
Pey/I
My/I
17
Design of Isolated Footings Deformation of isolated footings
18
Design of Isolated Footings Deformation of isolated footings
19
Design of Isolated Footings The design of isolated rectangular footings is detailed in the following steps:
1- Select a trial footing depth. Depth of footing above reinforcement is not to be less than 15 cm.
ACI 15.7 Note that 7.5 cm of clear concrete cover is required if concrete is cast against soil.
ACI 7.7.1
20
Design of Isolated Footings 2- Evaluate the net allowable soil pressure:
qall (net) = qall (gross) - γc hc - γs (Df -hc) P
Df hc
where
qall(net)
hc is the assumed footing depth, df is the distance from ground surface to the surface between footing base and soil, γc is the weight density of concrete, and γs is the weight density of soil on top of footing. 21
Design of Isolated Footings 3- Establish the required base area of the footing Base area of footing is determined from unfactored forces transmitted by footing to soil and the allowable soil pressure evaluated through principles of soil mechanics.
Areq
PD PL qall (net)
ACI 15.2.2
where PD and PL are column service dead and live loads, respectively. Select appropriate L, and B values, if possible, use a square footing to achieve greatest economy.
4- Evaluate the net factored soil pressure: qu(net )
1.2PD 1.6PL LB
ACI 15.2.1 22
Design of Isolated Footings 5- Check footing thickness for punching shear. When loads are applied over small areas of slabs and footings with no beams, punching failure may occur. The sloping failure surface takes the shape of a truncated pyramid in case of rectangular columns, and a truncated cone in case of circular columns.
The ACI Code assumes that failure takes place on vertical planes located at distance d/2 from the faces of the column.
ACI 11.11.1.2
23
Design of Isolated Footings 5- Check footing thickness for punching shear [contd.] The depth of the footing must be checked so that the shear capacity of the concrete equals or exceeds the critical shear forces produced by factored loads
Vu Vc The critical punching shear forceVu can be evaluated as follows
Vu qu (net)L B C1 d C2 d
C1
B
C2
C2 + d
C1 + d
ACI 11.11.1.2
L
Since there are two layers of reinforcement, an average value of d may be used: d = h − 7.5cm− db , where db is the bar diameter.
24
Design of Isolated Footings 5- Check footing thickness for punching shear [contd.] Punching shear force resisted by concrete Vc is given as the smallest of
2 V C 0.171 f c 'bod c
C2
B
V C 0.083 2 sd f c 'bod
C1
C2 + d
V C 0.33 f c 'bod
C1 + d
b
L
βc = long side/short side of column, αs = 40 for interior, 30 for side, and 20 for corner columns, bo =length of critical perimeter around the column = 2[(C1+d)+(C2+d)]
Interior
ACI 11.11.2.1 Corner
Exterior
25
Design of Isolated Footings 6- Check footing thickness for beam shear in each direction.
If Vu ≤ ΦVc, thickness will be adequate for resisting beam shear. The critical section for beam shear is located at distance d from column faces.
The factored shear force is given by
Critical section for beam shear (short direction)
x
L C 1 Vu qu (net ) B x qu (net ) B d 2
V c 0.17 f c ' B d
C2
The factored beam shear capacity of the concrete is given as
C1
d
B
In the short direction:
L
ACI 11.2.1.1 26
Design of Isolated Footings 6- Check footing thickness for beam shear in each direction [contd.]
The factored beam shear capacity of the concrete is given as
V c 0.17 f c ' L d
B
C2
C1
d
B C 2 Vu qu (net ) L y qu (net ) L d 2
y
The factored shear force is given by
Critical section for beam Shear (long direction)
In the long direction:
L
ACI 11.2.1.1
Increase footing thickness if necessary until the condition Vu ≤ ΦVc is satisfied.
27
Design of Isolated Footings 7-Compute the area of flexural reinforcement in each direction. The footing is designed as rectangular-section beam in both directions. The critical section for bending is located at the face of the column.
ACI 15.4.2
Critical section formoment
(L-C1)/2
Reinforcement in the long direction: 2
0.85f c 2M u 1 1 2 0.85 f Bd fy c As ,req Bd
C1
B
C2
B L C1 M u qu (net) 2 2
L
As ,min 0.0018Bh As ,req
ACI 15.4.1 ACI 10.5.4 ACI 7.12.2.1
28
Design of Isolated Footings
0.85f c 2M u 1 1 2 0.85 f Ld fy c A s,req Ld
As ,min 0.0018Lh As ,req
C1
B
2
C2
L B C2 M u qu (net) 2 2
(B-C2)/2
Reinforcement in the short direction
Critical section for moment
7-Compute the area of flexural reinforcement in each direction [contd.]
L
ACI 15.4.1
ACI 10.5.4 ACI 7.12.2.1 29
Design of Isolated Footings
where
long side of footing
B
7-Compute the area of flexural reinforcement in each direction [contd.] For square footings, the reinforcement is identical in both directions. For rectangular footings, the reinforcement in the long direction is uniformly distributed. However, a portion of the total reinforcement in the short direction, γsAs is distributed uniformly over a band width (centered on centerline of column) as shown in the figure. Remainder of reinforcement required in the short direction, (1 – γs)As, shall be distributed uniformly outside the center band width of the footing. Band width 2 s 1
short side offooting
ACI 15.4.4
B L
39
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete
All forces applied at the base of a column or wall must be transferred to the footing by bearing on concrete and/or by reinforcement.
ACI 15.8.1
Bearing on concrete for column and footing must not exceed the concrete bearing strength.
ACI 15.8.1.1
Pn Pu Otherwise, the t would fail by crushing of the concrete at the bottom of the column where the column bars are no longer effective or by crushing the concrete in the footing under the column.
Pn min Pn ,c ; Pn ,f 31
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.] For a ed column, the allowed bearing capacity ΦPn,c is
Pn ,c 0.85f cA1
ACI 10.14.1
For a ing footing where the ing surface is wider on all sides than the loaded area, the allowed bearing capacity ΦPn,f is
Pn ,f
A2 0.85f cA1 ; 2 0.85f cA1 min A1
Φ = strength reduction factor for bearing = 0.65 A1= column cross-sectional area A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal (see next slide) 32
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal
33
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
A2= area of the lower base of the largest frustum of a pyramid, cone, or tapered wedge contained wholly within the footing and having for its upper base the loaded area, and having side slopes of 1 vertical to 2 horizontal
34
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
Dowel Reinforcement: •If
Pn Pu :
Reinforcement in the form of dowel bars must be provided to transfer the excess load.
A s ,req
Pu Pn f y
ACI 15.8.1.2
The dowel bars are usually extended into the footing, bent at the ends, and tied to the main footing reinforcement.
35
Design of Isolated Footings 8- Check for bearing strength of column and footing concrete [contd.]
Minimum Dowel Reinforcement: •If Pn
Pu ::
Use minimum dowel reinforcement.
As ,min 0.005A1
ACI 15.8.2.1
36
Design of Isolated Footings 9- Check for anchorage of the reinforcement > ls (compn.)
10-Prepare neat design drawings showing footing dimensions and provided reinforcement.
37
Example Design an isolated rectangular footing to an interior column 40×40cm in cross
section and carry a dead load of 800 kN and a live load of 600 kN. One of the dimensions of the footing must not exceed 3.2 m. PD= 800 kN PL= 600 kN
Use fc’= 25 MPa and fy = 420 MPa, qall (gross) = 200 kN/m2, γsoil =17 kN/m3, γconc =25 kN/m3 Df=1.0
40 40
38
Example Solution 1Select a trial footing depth: Assume that the footing is 55 cm thick. 2Evaluate the net allowable soil pressure: qall (net) = qall (gross) - γs (Df - hc) - γc hc
qall net 200 ( 1 0.55) 17 0.55 25 178.6 kN/m 2 3 Establish the required base area of the footing :
Let L 3.20 m, B
7.84 2.45m 3.20
40
40
245
A req
P P D L 800 600 7.839 m2 qall (net) 178.6
320
Use 320x245x55 cm footing 4- Evaluate the net factored soil pressure
Pu 1.2PD 1.6P L 1.2800 1.6 600 1920 kN q u net
1920 Pu 244.9 kN/m 2 LB 3.2 2.45
39
245
40+45.9
Example
40+45.9
5- Check footing thickness for punching shear:
Average effective depth davg 55-7.5-1.6 45.9cm bo 2[ 40 45.9 40 45.9] 343.6 cm
320
Vu 244.9 3.22.45 0.40 0.4590.40 0.459 1740 kN Φ VC is the smallest of Φ0.33 fc' bod 0.75 0.33 253436 459 1952 kN 2 2 3436 459 3016 kN Φ0.17 fc' 1 bod 0.75 0.17 25 1 0.4/0.4 βc αd 40 459 Φ0.083 f c' 2 s bo d 0.750.083 25 2 3436 459 3605kN bo 3436 Φ VC 1952 kN Vu 1740kN OK i.e. footing thickness is adequate for resisting punching shear. 40
Example 6- Check footing thickness for beam shear in each direction: In short direction
ΦVc 0.750.17 25 2450459 717kN
245
45.9
Vu is located at distance d from face of column
3.2 0.4 Vu 244.9 2.45 0.459 565 kN 2 ΦVc= 717 kN > Vu= 565 kN
OK
320
ΦVc 0.750.17 25 3200459 936 kN 45.9
Vu is located at distance d from face of column
2.45 0.4 Vu 244.93.2 0.459 444 kN 2 ΦVc= 936 kN > Vu= 444 kN
245
In long direction
320
OK 41
Example 7- Calculate the area of flexural reinforcement in each direction: a- Reinforcement in the long direction: The critical section for bending is shown in the figure
Critical section for moment
2 2 B L C1 2.45 3.2 0.4 244.9 M u q u net 2 2 2 2
1.4
0.85 25 ρ 420
245
588 kN .m 2 58810 6 1- 10.9 0.85 25 2450 4592
0.0031 A s 0.0031 459 2450 3500 mm
2
320 24.49 x 2.45
A s,min 0.0018 550 2450 2430 mm2 2
As,req 3500 mm 2314mm in long direction
42
Example 7- Calculate the area of flexural reinforcement in each direction: b- Reinforcement in the short direction: The critical section for bending is shown in the figure
2 412106 1- 10.9 0.85 25 3200 4592
As,req 3170 mm
2
2 24.49 x 2.8
As,min 0.0018 550 3200 3170 mm
320
1.025
0.0016 A s 0.0016 459 3200 2411mm
1.025
0.85 25 ρ 420
245
2 2 L B C2 3.2 2.45 0.4 244.9 M u q u net 2 2 2 2 Critical section for moment 412 kN .m
2
43
Example 7- Calculate the area of flexural reinforcement in each direction: b- Reinforcement in the short direction: The distribution of the reinforcement is as follows:
245
42.5
2Φ14 B
Width band =245
18Φ14 B
B 2.45 2 As Central band reinft. β 1 2 2 3170 2757 mm 1.3 1 Use 18 14 mm in the central band.
42.5
2Φ14 B
L 3.2 1.3
320
3170 2756 207 mm 2 For each of the side bands, A s 2 Use 214 mm in each of the two side bands.
44
Example 8- Check for bearing strength of column and footing concrete For the column
A1 400 400 160000mm 2
Pn ,c 0.85f cA1 0.65( 0.85 25160000) 221010 3 N 2210kN For the footing
In short direcion: 1025mm 1100mm Use 1025 mm 1400
2
1
h= 550
1025
245
1100
320
45
Example 8- Check for bearing strength of column and footing concrete
A 2 400 2 1025 400 2 1025 6002500mm
Pn ,f Pn ,f
min min
2
A2 0.85f cA1 ;2 0.85f cA1 A1 6002500 2210 ; 2 2210 4420kN 160000
Pn min Pn ,c ; Pn ,f min 2210; 4420 2210kN Pu 1920 kN Use minimum dowel reinforcement
1025 + 400+ 1025
1025 + 400+ 1025
46
Example 8- Check for bearing strength of column and footing concrete Minimum dowel reinforcement
As ,min 0.005A1 0.005 400 400 800mm
2
Use 416, As,sup = 804 mm2
47
Example 9- Check for anchorage of the reinforcement Bottom longitudinal reinforcement (Φ14mm) α=1.0 for bottom bars,
β=1.0 for uncoated bars 1.4
α β =1.0 <1.7 OK γ=0.8 for Φ14mm,
7.5+0.7=8.3 cm
245
C the smallest of
λ=1.0 for normal weight concrete
[245-2(7.5)-1.4]/(22)/(2)=5.2 cm i.e., C is taken as 5.2 cm
C K tr 5.2 0 C K tr 3.7 2.5 i.e.,use 2.5 db 1.4 db
320
420 (1.0)(1.0)(0.8)(1.0) 1.4 34 cm ld 1.1 25 2.5 Available length in long direction =140-7.5=132.5 > 34 cm 48
Example 9- Check for anchorage of the reinforcement Bottom reinforcement in short direction (Φ14mm) α=1.0 for bottom bars,
β=1.0 for uncoated bars
α β =1.0 <1.7 OK γ=0.8 for Φ14mm,
245
7.5+0.7=8.3 cm
[320-2(7.5)-1.4]/(19)/(2)=8 cm
1.025
C the smallest of
λ=1.0 for normal weight concrete
i.e., C is taken as 5.2 cm
320
C K tr 8 0 C K tr 5.7 2.5 i.e.,use 2.5 db 1.4 db 420 (1.0)(1.0)(0.8)(1.0) 1.4 34 cm ld 1.1 25 2.5 Available length in short direction =102.5-7.5=95 > 34 cm 49
Example 9- Check for anchorage of the reinforcement Dowel reinforcement (Φ16mm):
0.24f y d b 0.24 42016 323mm fc ' 25 l dc max 323mm 200mm 0.043f d 0.04342016=289mm y b Available length = 550-75-14-14 = 447 mm > 323 mm OK Column reinforcement splices:
Considering that the column is reinforced with 16 bars ls 0.071f yd b 0.071 42016 478 mm 300 mm taken as 48cm
> ls (compn.)
50
Example
55 cm
48cm
10- Prepare neat design drawings showing footing dimensions and provided reinforcement
245 (18Φ14)
2.45 m
2Φ14 B
2Φ14 B
18Φ14 B
3.20 m
23Φ14 B 42.5
Width band =245
42.5
51
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 14 Staircase Design
Stair Types
2
Stair Types
3
Stair Types
4
Stair Types
5
Technical •Going: horizontal upper portion of a step. •Rise: vertical distance between two consecutive treads. •Flight: a series of steps provided between two landings.
•Landing: a horizontal slab provided between two flights. •Waist: the least thickness of a stair slab.
6
Technical •Winder: radiating or angular tapering steps. •Soffit: the bottom surface of a stair slab. •Nosing: the intersection of the going and the riser.
•Headroom: the vertical distance from a line connecting the nosings of all treads and the soffit above.
7
General Design Requirements
8
Stair type based on the structural loading type
Simply ed stair (transversely ed)
9
Simply ed stair (longitudinally ed)
Cantilever stair
Design of transversely ed stairs Loading: a. Dead load: The dead load includes own weight of the step, own weight of the waist slab, and surface finishes on the steps and on the soffit.
b. Live Load: Live load is taken as building design live load plus 1.5 kN/m2, with a maximum value of 5 kN/m2.
10
Design of transversely ed stairs Direction of bending Main reinforcement Shrinkage reinforcement
11
Direction of bending
Design of transversely ed stairs Design for Shear and Flexure: Each step is designed for shear and flexure as if it is a beam. Main reinforcement runs in the transverse direction at the bottom side of the steps while shrinkage reinforcement runs at the bottom side of the slab in the longitudinal direction. Since the step is not rectangular, the effective depth d is found by an equivalent rectangular section that can be used with an average height equal to:
t
R
havg
12
t
Design of transversely ed stairs Example 1 Design a straight flight stair in a residential building ed on reinforced concrete walls 1.5 m apart (center to center), given: L.L = 3 kN/m2; covering material = 0.5 kN/m; The risers are 16 cm and goings are 30 cm; fc’=25 MPa, fy= 420 MPa
13
Loads and Analysis
t
l 1.5 0.075m 20 20 t
0.16 0.165m have 0.30 0.34 2 D.L(O.W) =0.340.075 25 + (1/2) 0.16 0.3 25=1.24 kN/m D.L (covering material) = 0.5 kN/m
0.075
0.3
D.L (total) = 1.74 kN/m L.L =30.3 =0.9 kN/m
0.16
0.302 0.162 0.34
1.5 m 14
Shear diagram
Moment diagram 15
Design for moment M u 1kN .m d 165 20 6 139mm bw 300mm
0.85 f c ' 1 1 fy
2M u 0.85f c ' b d 2
0.8525 1 1 21106 0.0005 2 420 0.90.85 25 300 139 As 0.0005 300139 20.9mm
2 2
As ,min 0.0018 300165 89.1mm As As As ,min 89.1mm Use 112 for each step
16
2
Design for shear V C 0.75 0.17 25 139 300 /1000 26kN V u 2.65kN OK
17
Design of longitudinally ed stairs Direction of bending Shrinkage reinforcement
Main reinforcement
18
Design of longitudinally ed stairs
19
Design of longitudinally ed stairs Deflection Requirement: Since a flight of stairs is stiffer than a slab of thickness equal to the waist t, minimum required slab depth is reduced by 15 %.
Effective Span: The effective span is taken as the horizontal distance between centerlines of ing elements. n = number of goings X = Width of ing landing slab at one end of the stairs slab
20
Y = Width of ing landing slab at the other end of the stairs slab.
Design of longitudinally ed stairs Deflection Requirement: Since a flight of stairs is stiffer than a slab of thickness equal to the waist t, minimum required slab depth is reduced by 15 %.
Effective Span: The effective span is taken as the horizontal distance between centerlines of ing elements. n = number of goings X = Width of ing landing slab at one end of the stairs slab
21
Y = Width of ing landing slab at the other end of the stairs slab.
Design of longitudinally ed stairs Loading: a. Dead Load: The dead load, which can be calculated on horizontal plan, includes: •Own weight of the steps. •Own weight of the slab. •Surface finishes on the flight and on the landings. Note: For flight load calculations, the part of load acting on slope is to be increased by dividing it by cosα. This is because analysis for moment and shear is conducted on the horizontal span of the flight, but the load is that carried on the inclined span.
P P= wo.w.Linc .Linc
22
.L
w=P/L= wo.w.Linc/L= wo.w./cosα
Design of longitudinally ed stairs Loading: b. Live Load: Live load is taken as the building design live load plus 1.5 kN/m2, with a maximum value of 5 kN/m2. Live load is always given on the horizontal projection.
23
Design of longitudinally ed stairs t detail: The stairs slab is designed for maximum shear and flexure. Main reinforcement runs in the longitudinal direction, while shrinkage reinforcement runs in the transverse direction. Special attention has to be paid to reinforcement detail at opening ts.
24
Design of longitudinally ed stairs Example 2 Design the U- stair in a residential building shown in the figure, given: L.L = 3 kN/m2; covering material = 2 kN/m2; The rises are 16 cm and goings are 30 cm, fc’=25 MPa, fy= 420 MPa
25
Loads and Analysis t 0.85
l 525 22cm 20 20
cos() = 0.3/ 0.34 = 0.88 Take a unit strip along the span: D.L (slab) = 0.221.025/0.88 =6kN/m D.L (step) = (1/2) 0.161.0 25=2 kN/m D.L (covering material) = 21.0=2 kN/m D.L (flight) = 10 kN/m D.L (landing) = 8 kN/m L.L =3 1.0=3 kN/m
26
Wu (flight) = 1.2(10)+1.6(3)=16.8kN/m Wu (landing) = 1.2(8)+1.6(3)=14.4kN/m
0.3
0.16 0.34
Moment and shear diagram 14.4kN/m
27
16.8kN/m
14.4kN/m
Design for moment M u 52.2kN .m d 22 2 0.6 19.4cm 194mm bw 1000mm 0.8525 1 1 252.2106 0.0037 2 0.85 0.9 25 1000194 420 2
As 0.00371000194 718mm 2 0.00181000 220 396mm A As ,min
s
OK
Use 812
(22)=3.96 cm2/m
Design for shear 28
V C 0.75 0.17 25 1941000 / 1000 127.3kN V u 38.25kN OK
29
Design of quarter-turn stairs
A landing may be shared on two different stair slabs. The load of the shared landing can be assumed to be divided equally and each stair slab carries on 30
half.
Design of stair beams
Ls
P=wsLs/2
31
ws
P w=P/(L/2) L/2
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