Reinforced Concrete Design g I
Dr. Nader Okasha
Lecture 0 Syllabus
Reinforced Concrete Design
I Instructor
D N Dr. Nader d Ok Okasha. h
Email
[email protected]
Offi Hours Office H
A needed. As d d
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Reinforced Concrete Design
This course iis only Thi l offered ff d for f 2010 students d who h have passed strength of materials.
If you d don’t ’ meet this hi criteria i i you will ill not be b 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. 2008 Design of Reinforced Concrete. 7th edition, McCormac, J.C. and nd N Nelson, l n JJ.K., K 2006. 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. Design Kenneth Leet. Leet Reinforced Concrete: Mechanics and Design. James K. Wight, and James G G. MacGregor MacGregor. ٥
Reinforced Concrete Design
The art of design Design g is an analysis y of trial sections. The strength g of each trial section is compared with the expected load effect. The load effect on a section is determined using g structural analysis and mechanics of materials. The strength of a reinforced concrete section is g the concepts p taught g in this class. determined using
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Reinforced Concrete Design
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Reinforced Concrete Design Course outline Week
1
2 3,4 2, 34
Topic Introduction: Syllabus and course policies. policies -Syllabus -Introduction to reinforced concrete. -Load types, yp load p paths and tributaryy 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. beams -Design of T and L beams. -Design of doubly reinforced beams.
4
Design of beams for shear.
5
Midterm 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. 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 t exam
20%
Final exam
60%
.
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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. Markingg will consider pprimarilyy neatness of presentation, p , 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 cell-phone phone use
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Reinforced Concrete Design
Policy towards discipline during class Zero tolerance will be practiced. practiced No talking with other students is allowed. allowed Raise your hand before answering or asking questions questions. Leaving during class is not allowed (especially for answering the cell-phone) unless a previous permission is g granted. ed. Violation of discipline p rules mayy have you y dismissed from class and jeopardize your participation points.
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Reinforced Concrete Design
Policy towards missed classes Any collectively missed class MUST be made up. up p either on a A collectivelyy missed class will be made up 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 h llecture videos. id No late students will be allowed in class. class Anything mentioned in class is binding. binding No excuse for not being there or not paying attention.
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Reinforced Concrete Design
Units used in class In all equations, equations the input and output units are as follows: Distance (L,b,d,h L b d h): mm Area (Ac,Ag,As): mm2 Volume (V): mm3 Force (P,V,N): N Moment (M): N.mm N mm Stress (fy, fc’): N/mm2 = MPa = 106 N/m2 Pressure (qs): N/mm2 Distributed load per unit length (wu): N/mm Distributed load per unit area (qu): N/mm2 Weight per unit volume (γ): N/mm3
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Reinforced Concrete Design
Units used in class However these quantities may be presented as However, Distance (L,b,d,h L b d h): cm , m Area (Ac,Ag,As): cm2, m2 Volume (V): cm3, m3 Force (P,V,N): kN Moment (M): kN.m kN m Pressure (qs): kN/m2 Distributed load per unit length (wu): kN/m Distributed load per unit area (qu): kN/m2 Weight per unit volume (γ): 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 kN m = 106 N.mm N mm 1 kN/m2 = 10-3 N/mm2 1 kN/m3 = 10-66 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 throughout the slides by their section or equation number in the code provided in shading. Examples:
Ec = 4700 4 00 f c′ f r = 0.62 f c′
ACI 8.5.1 851 ACI E Eq. 9-10 9 10
Some of the original equations may have included the symbol λ = 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. class Study the lectures up to date. date Re-do Re do the lecture examples examples. Look at additional resources.
DO YOUR HOMEWORK!!!!! Check your solution with the HW solution uploaded to upinar. 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.
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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.
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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 members are very large and heavy because of the low strength per unit weight of concrete. 4. Properties of concrete vary due to variations in proportioning and mixing.
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 Admixtures
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.
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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
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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 admixtures 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
Admixtures – 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 ' f Compressive Strength, c
• Normally, 28-day strength is used as the design strength.
18
Mechanical Concrete Properties ' f Compressive Strength, 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).
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Mechanical Concrete Properties ' f Compressive Strength, 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.
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Mechanical Concrete Properties Compressive-Strength Test
21
Mechanical Concrete Properties Modulus of Elasticity, Ec ' f • Corresponds to the secant modulus at 0.45 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 fr 2 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 fs = ε Es ≤ fy Yield point
elastic
plastic
All steel grades have same modulus of elasticity Es= 2x105 MPa = 200 GPa
32
Steel Reinforcement Bar sizes, f, # 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 mm
N/m
1
2
3
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
Number of bars 4 5 6 7 113 201 314 452 616 804 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
8
9
10
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
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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 Supports Floor Loads Above
Girders Support Joists
Joists Support Floor Deck
Columns Support 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 members 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}
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Load paths loads on (one-way) beams This tributary load is determined by multiplying q by the tributary width for the beam.
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 panels and the centerlines of the adjacent panels parallel to the long sides. A panel is part of the slab formed by column centerlines.
9
Load paths loads on (two-way) beams An edge beam is bounded by panels from one side. An interior beam is bounded by panels from two sides. qD
For edge beams: D=S/2 qD
For interior beams:
D=S 10
Load paths loads on (two-way) beams
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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 members 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
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C2
C1 6m
5.5 m
Load paths loads on structural members 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 members Example B2:
w = 10 (4+5)/2 = 45 kN/m B1 4m B2 5m 4.5 m
16
C2
C1 6m
5.5 m
Load paths loads on structural members Example B1:
w = 20 kN/m
B2:
w = 45 kN/m
17
Load paths loads on structural members Example C1:
P = 10 (4.5/2 6/2) 5 = 337.5 kN B1 4m B2 5m 4.5 m
18
C2
C1 6m
5.5 m
Load paths loads on structural members Example C2:
P = 10 [(4.5+5)/2 (6+5.5)/2] 5 = 1366 kN B1 4m B2 5m 4.5 m
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C2
C1 6m
5.5 m
Load types Classification by direction
1- Gravity loads 2- Lateral loads
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Load types Classification by source and activity
1- Dead loads 2- Live loads 3- Environmental loads 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
Minimum live Load values on slabs Type of Use Uniform Live Load
Live Loads (L.L) The live load is a moving or movable type of load such as occupants, furniture, etc. Live loads used in designing 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.
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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
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.
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Structural Design Requirements: The design of a structure must satisfy three basic requirements: 1)Strength to resist safely the stresses induced by the loads in the various structural members. 2)Serviceability to ensure satisfactory performance under service load conditions, which implies providing adequate stiffness 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 Administrators 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.
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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.
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Design Methods (Philosophies) Two methods of design have long prevalent. Working Stress Method focuses on conditions at service loads. Strength Design Method focusing on conditions at loads greater than the service loads when failure may be imminent. The Strength Design Method is deemed conceptually more realistic to establish structural safety.
The Working-Stress Design Method This method is based on the condition that the stresses caused by service loads without load factors are not to exceed the allowable stresses which are taken as a fraction of the ultimate stresses of the materials, fc’ for concrete and fy for steel.
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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 sections, Members with spiral reinforcement Φ = 0.75 Other reinforced members Φ = 0.65 c- For shear and torsion Φ = 0.75
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 support 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: 1- Cracking moment. 2- Elastic stresses due to a given moment. 3- Moments at given (allowable) elastic stresses. 4- Ultimate strength moment (next lecture).
Note:
In calculating stresses and moments (Parts 1 and 2), you need to always check the maximum tensile stress with the modulus of rupture to determine if cracked or uncracked section analysis is appropriate. 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 bh 12 1 I g (350)(750) 3 1.2305 1010 mm4 12 f r 0.62 f c 0.62 30 3.4MPa
750 mm
1500 mm2
Ig
M cr
fr I g yt
f c 30MPa
3.4 1.2305 1010 1.1143 108 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 As d x 0 2
1 b x2 2
n As x n As d 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: f c 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 E s 2 105 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 nA s ( d x ) 2 3 1 I t ( 350 )( 185.16 ) 3 7.77 1500( 700 185.16 ) 2 3 750 mm I t 3.8295 109 mm 4
My 180 106 185.16 fc 8.7 MPa 9 It 3.8295 10 f c 8.7 MPa 0.45f c 0.45( 30 ) 13.5MPa
f c 30MPa
1500 mm2
OK
My 180 106 ( 700 185.16 ) fs n 7.77 188MPa It 3.8295 109
15
Elastic stresses – Cracked section Example 3: f c 30MPa
Calculate the allowable moment for the section shown, f s(allowable)= 180 MPa, f c(allowable)= 12 MPa f s It 180 3.8295 109 Ms ny ( 7.77 )( 700 185.16 )
750 mm
1500 mm2
M s 1.7234 108 N .mm 172.34kN .m f c I t 12 3.8295 109 Mc y 185.16 M c 2.4819 108 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=b1c. ACI 10.2.7.1 4
The equivalent stress (Whitney) block
The equivalent rectangular concrete stress distribution has what is known as the b1 coefficient. It relates the actual NA depth to the depth of the compression block by a=b1c. b1 0.85 for f c ' 28 MPa ACI 10.2.7.1 0.05( f c '28) b1 0.85 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
fMn=fRnbd2
Rn is given in tables and figures of design aids.
8
Design Aids
9
Design Aids
10
Tension strain in flexural members
y
fy 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>c c>cb b h
d
h
b
b
d
εt 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 mm 2 414 fy max 1.4 1.4 bw d (254)(457)=393 mm 2 fy 414
=412 mm 2 < A s,sup =2580 mm 2 OK
2580 414 2 a 143.4mm 0.85f c b 0.85 34.5 254 As f y
0.05( f c ' 28 ) 0.65 for f c ' 34.5MPa 28 MPa 7 0.05( 34.5 28 ) b1 0.85 0.804 0.65 7
3 b1 0.85
28
Example Solution b) f c 34.5MPa a
143.4 4 c 178.5mm b1 0.804 d c 457 178.5 5 t 0.003 0.003 0.00468 c 178.5 0.004 t 0.005 Section is in transision zone f =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
a 6 M d M n A s f y d 2 143.4 6 0.874 2850 414 457 360 10 N .mm 2 360 kN .m 7 M u 350kN .m ΦM n 360kN .m Section is adequate
30
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 414 fy max 1.4 1.4 bw d (254)(457)=393 mm 2 fy 414
=552 mm 2 < A s,sup =2580 mm 2 OK
2580 414 2 a 80mm 0.85f c b 0.85 62.1 254 As f y
31
Example Solution c) f c 62.1MPa 0.05( f c ' 28 ) 0.65 for f c ' 62.1MPa 28 MPa 7 0.05( 62.1 28 ) b1 0.85 0.61 0.65 7 b1 0.65
3 b1 0.85
a
80 4 c 123mm b1 0.65 d c 457 123 5 t 0.003 0.003 0.0081 c 123 t 0.005 Section is tension controlled ==> Satisfes ACI requirements ==> f =0.9
32
Example Solution c) f c 62.1MPa
a 6 M d M n A s f y d 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 r: 0.85 f c' ρ fy
2M u 1 1 2 0 . 85 f ' b d c
Remember: 1 kN.m = 106 N.mm
As = rbd 3
Design of Beams For Flexure Design aids can also be used:
0.85 f c' ρ fy
2M u 1 1 2 0 . 85 f ' b d c
Calculate: Then r 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 support.
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 Supported Beams
Continuous Beams
Determinate
Indeterminate
–
+ + 11
Moment Diagram
+
Moment Diagram
Estimation of applied moments Mu Simply Supported Beams
Continuous Beams
–
+ +
Moment Diagram Section at midspan
12
+
Moment Diagram Section over support
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. • Members are of similar section dimensions along their lengths (prismatic).
13
Estimation of applied moments Mu Approximate Structural Analysis More than two spans
14
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 supports. 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 1- Determine h (h>hmin from deflection control) and assume b. Estimate beam weight and include it with dead load. 2- Calculate the factored load wu and bending moment Mu. 3- Assume that Φ=0.9 and calculate the reinforcement (ρ and As). 4- Check solution: (a) (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 1- Calculate the factored load wu and bending moment Mu. 2- Assume that Φ=0.9 and calculate the reinforcement (ρ and As). 3- Check solution: (a) (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 supported. 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) 0.85f c ' ρ fy
2 Mu 1 1 2 Φ 0.85f ' b d c
0.85(25) ρ 420
2 227.3 106 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
5Φ20
a) Check spacing between bars sc
300 2 40 2 8 5 20 26 mm d b 20 mm 5 1 25 mm
300
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 b d (300)(442)=395 mm 2 w 420 fy max 1.4 1.4 bw d (300)(442)=442 mm 2 fy 420 =442 mm 2 < A s,sup =1571 mm 2 OK 21
Example 1 c) Check Φ =0.9 (tension controlled assumption) a
As f y 0.85f c ' b
1 0.85
1571 420 103.5 mm 0.85(25)300
for f c ' 25MPa 28 MPa c
a 103.5 121.7 mm β1 0.85
dc 442 121.7 εt 0.003 0.003 0.0079 0.005 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 Φ As f y d 2 103.5 6 0.90 1571 420 442 231.7 10 N.mm = 231.7 kN.m 2 M d 231.7 kN.m M u 227.3kN.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 support. 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 S1 = S2= S3 = 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)
0.85f c ' ρ fy
2 Mu 1 1 2 Φ 0.85f ' b d c
0.85(25) ρ 420
2 187.5 106 1 1 2 (0.9) 0.85(25)800 (260)
0.0102
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 2 8 1116 52.8 mm 11 1
d b 16 mm 25 mm
OK
b) Check minimum steel requirement
A s,min
0.25 f c 0.25 25 b d (800)(260)=620 mm 2 w 420 fy max 1.4 1.4 bw d (800)(260)=693 mm 2 fy 420 =693 mm 2 < A s,sup =2212 mm 2 OK 27
Example 2 c) Check Φ =0.9 a
As f y 0.85f c ' b
1 0.85
2212 420 55 mm 0.85(25)800
for f c ' 25MPa 28 MPa c
a 55 64 mm β1 0.85
dc 260 64 εt 0.003 0.003 0.0091 0.005 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 Φ As f y d 2 55 0.9 2212 420 260 194.5 106 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 L beams
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 supporting, 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.
beff
5
L /4 min b w +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 w L /12 beff min b w 6hf b 0.5b c w
6
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 support Negative moment
Analysis of T-beams Case 1: when a ≤ hf
[Same as rectangular section]
T C Asf y a 0.85 f c b e
12
a ΦM n Φ A s f y d 2
Analysis of T-beams Case 2: when a > hf C f 0.85 f c be bw hf C w 0.85 f c bw a T As f y
From equilibrium of forces T C f Cw
A s f y 0.85 f c be bw hf a 0.85 f c bw
a hf ΦM n Φ C w d Cf d 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: Asf y a 0.85 f c b e If a ≤ hf → a is correct If a > hf
As f y 0.85 f c be bw hf → a 0.85 f c bw
4- Calculate b1, c, and check εt 5- Calculate ΦMn, and check M u ΦM n 15
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 4 4 1370mm be min 16hf b w 16 150 300=2700mm b 2150 mm
16
Example 1 Check min. steel 1.4 0.25 f c ' As,min max bw d ; bw d max fy fy As,min 400 mm 2 As,sup 5000 mm 2 OK
0.25 25 1.4 300 400 ; 300 400 420 420
Calculate a (assuming a h f 140mm 0.85 28 500
The assumption is wrong T section design
36
14 55 As 30
Solution (B) L = 2 m 14
50
55
Calculate required reinforcement Asf
0.85 f c '( b bw ) hf fy
Asf
0.85 ( 28 )( 500 300 )140 1586mm 2 420
30
hf M uf As f y d 2 140 6 0.9 1586 420 550 288 10 N .m 2
M uw M u M uf 784 106 288 106 496 106 N .m 37
Solution (B) L = 2 m 0.85 f c ' fy
2M u 1 1 2 0 . 85 f ' b d c w
0.85 ( 28 ) ( 420 )
2( 496 ) 106 1 1 2 0 . 9 0 . 85 ( 28 ) ( 300 ) ( 550 )
0.017
As Asf Asw 1586 2808 4395mm
2
55
Asw bw d 0.017( 300 )( 550 ) 2808mm 2
14
50
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 C s As 2f y Asf s
Asf s As 2 fy 5
We need fs’ to find As2
As As 1 As 2 A s 1 A s A s 2
Analysis of Doubly Reinforced Rectangular Sections Find fs’: c d 0.003 c
s
c d f s sE s 0.003E s f y c E s 2 105 MPa
6
c
Analysis of Doubly Reinforced Rectangular Sections Find c:
T C c C s As f y 0.85f cab Asf s
c d As f y 0.85f c1cb As 0.003E s c 7
find c by solving the quadratic equation find fs’ from equation in slide 6
Analysis of Doubly Reinforced Rectangular Sections Find Md:
M d M n 8
As 1f y
a d - A s f s d - d ' 2
Analysis of Doubly Reinforced Rectangular Sections Procedure: 1) 2)
c d As f y 0.85f c1cb As 0.003E s c c d f s 0.003E s f y c
find c, a
3) As 2 Asf s
fy 4) As 1 As As 2
5) 6) 9
d c Check if f = 0.9 s c 0.003 0.005? a M d M n As 1f y d - Asf 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:0.05( f c ' 28 ) 0.65 for f c ' 35MPa 28 MPa 7 0.05( 35 28 ) 1 0.85 0.8 0.65 7 c d As f y 0.85f c 1cb A s 0.003E s c
1 0.85
10
c 50 5 4825(420) 0.85(35)(0.8)c (300) 982 0.003(2 10 ) c
30
Example 1 c 50 5 4825(420) 0.85(35)(0.8)c (300) 982 0.003(2 10 ) c 229.5c 2 1437300c 29460000 0 5.0 c 220mm
2Φ25
a 1c 0.8 220 176mm
60 6Φ32
c d fs 0.003E s f y c 220 50 5 f s 0.003(2 10 ) 463 f y 420MPa 220 f s f y 420 11
30
Example 1 Asf s 982(420) As 2 982mm 2 fy (420)
5.0 2Φ25 60 6Φ32
As 1 As As 2 4825 982 3843mm 2 d c s 0.003 0.005? c
30
600 220 0.003 0.0052 0.005 Tension Controlled , f 0.9 220
s
a M d M n A s 1f y d - A sf s d - d ' 2 176 M d 0.9 3843( 420 ) 600 982 ( 420 ) 600 50 2 12
M d 948 106 N .mm 948kN .m
Maximum allowed steel for a singly reinforced section
0.003 cmax d 0.003 0.005 3 cmax d 8
=1c 1cmax
3 0.85 1 f c ' max 8 fy 3 0.85 1f c ' As ,max bd 8 fy
3 d1 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 f 5) Mn2 = Mn – Mn1
c d 0.003E s f y c Asf s M n2 7) As As 2 fy f s(d d ) 6) f s sE s
As As 1 As 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: 0.85f c ' ρ fy
2 Mu 1 1 2 Φ 0.85f ' b d c
0.85(21) ρ 420
2 1225 106 1 1 2 (0.9) 0.85(21) 350 (700)
0.0284
As= ρ b d = 0.0284(350)(700) = 6947 mm2 15
Use 10 Φ32 mm in two rows (As,sup =7069 mm2)
Example 2 Check the ductility of the singly reinforced section: a
As f y 0.85f c ' b
7069 420 475 mm 0.85(21)350
c
a
1
475 559mm 0.85
dc 700 559 εt 0.003 0.003 0.00076 0.004 c 559 Section is brittle! can not be used. Use compression reinforcement. Mu
1225 Mn 1361kN .m f 0.9 As 1 As ,max 16
3 0.85 1f c ' 8 fy
As 1 3307mm 2
3 0.85 ( 0.85 )( 21) bd ( 350 )( 700 ) 8 ( 420 )
Example 2 As f y
3307( 420 ) a 222.3mm 0.85f cb 0.85( 21)( 350 ) a
222.3 c 261.55mm 1 0.85 222.3 a M n 1 A s f y d - ( 3307 )( 420 )( 700 ) 2 2 M n 1 818 106 N .mm 818kN .m Mn2 = Mn – Mn1 = 1361 – 818 = 543 kN.m
17
Example 2 c d f s 0.003E s f y c 261.55 70 5 fs 0.003(2 10 ) 439MPa f y 420MPa 261.55 f s f y 420 M n2 543 106 As 2052mm 2 f s(d d ) 420(700 70) Asf s (2052)(420) As 2 2052mm 2 fy (420)
As As 1 As 2 3307 2052 5359mm 2 Use 830 in two rows for tension steel (As,sup = 5655 mm 2 )
18
Use 4 26 for compression steel (As,sup = 2124 mm 2 )
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
Designing 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 shear reinforcement: Vn Vc Vs ACI Eq. 11-2 Vc = nominal shear force resisted by concrete Vs = nominal shear force resisted by shear reinforcement 12
Strength of Concrete in Shear For members 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 Vc
13
Vu d bw d 0.29 f c ' bw d 0.16 f c ' 17 w Mu
As where w b wd
ACI Eq. 11-5
Strength of Concrete in Shear For members subject to axial compression Nu plus shear Vu, ACI Code gives the following equation for calculating Vc N Vc 0.17 1 u 14 A g
f c' bw d
ACI Eq. 11-4
For members subject to axial tension Nu plus shear Vu, ACI Code gives the following equation for calculating Vc
14
0.29 N u f c' bw d Vc 0.17 1 A g
ACI Eq. 11-8
Designing to Resist Shear To find the force required to be resisted by shear reinforcement:
Vu ΦVn
Vn Vc Vs
Vu V s V c 15
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 is required
16
Design of Stirrups Shear reinforcement required when
Vu Vc
Vs
Vu V c
ACI 11.4.7.1 The bar size of the stirrups is established and the spacing is calculated:
Vs
A vf yd
s
Av f y d
s
For inclined stirrups (with angle a)
Vs
Av f y d sin α cos α s
s
ACI Eq. 11-15
Vs
Av f y d sin α cos α
ACI Eq. 11-16
Vs
where Av = the area of shear reinforcement within spacing s (for a 2-legged stirrup in a beam: Av = 2 times the area of the stirrup bar). 17
Minimum Amount of Shear Reinforcement
ACI 11.4.6.1
1 Minimum Shear Reinforcement (Av,min) required when Vu Vc 2 bw s bw s Av min 0.062 f c ' 0.35 ACI Eq. 11-13 f ys f ys Av f ys Av f ys s=min ; 0.062 f c ' bw 0.35 bw
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
d min ;600mm 2 d min ;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 support if: (a) Support 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 support 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 support as in cases (a) and (b), but must be taken at face of the support 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. • Members 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 supports.
24
Summary of ACI Shear Design Procedure for Beams 1- Draw 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 Vu V s V c 4- Check the code limit on Vs Vu V s 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 is required. * For Vu ≥ 0.50ΦVc , minimum shear reinforcement is required Av f ys Av f ys s=min ; 0.062 f c ' bw 0.35 bw
*For Vu ≥ ΦVc , shear reinforcement is required (in addition, check min shear) A v f yd Av f y d sin α cos α For vertical For inclined s s stirrups stirrups Vs Vs 6- Maximum spacing smax must be checked
26
d If V s 0.33 f c bw d s max min ;600mm 2 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 support 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 support.
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 28 300 542 146.3 103 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. V 247.1 V s u V c 146.3 183.2kN 0.75
4- Check the code limit on Vs : 0.66 f c ' bw d 0.66 28 300 542 567. 9 103 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 ; 400 mm 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
A v f yd Vs
2 50 420 542 125 mm 3 183.2 10
Example
Check maximum stirrup spacing: 0.33 f c ' bw d 0.33 28 300 542 284 kN 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 s max =min ; 0.062 f c ' bw 0.35 bw s max
2(50) 420 2(50) 420 min 427mm ; 400 mm 400mm 0.35 300 0.062 28 300
So, use 33
Φ8 mm vertical stirrups spaced at 12 cm.
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
Φ8@25
30 Section in zone C
34
Reinforced Concrete Design I
Dr. Nader Okasha
Lecture 10 Design of slabs
Regula (3
y
Introduction
Plate/Shell (2D) x z t 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.] 40Atr 2( 113 ) 15.1 mm sn ( 150 )( 4 )
C K tr 44 15 1.85 2.5 db 32 ld
50 cm
K tr
OK
αβ γ λ 1.1 f c C K tr db
4Φ32 Φ12@15
fy
d b 300 mm 420 ( 1.7 )( 1.0 )( 1.0 ) ld 32 2127 mm 300 mm OK 1 . 85 1 . 1 28
40 cm
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
l fc'
db
0.24 1.2 420 32 732 mm 1.0 28 150mm 8( 32 ) 256mm OK
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 Contact Splice T
T ls
Direct contact
2. Non-Contact 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
Class A: 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 K tr 83 0 C K tr 5.19 2.5 i .e., 2.5 db 16 db 420 ( 1.0 )( 1.0 )( 1.0 ) ld 16 446 mm 2 . 5 1.1 30 Required splice length ls 446( 1.3 ) 580 mm 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:
ACI 12.16.1
0.071 fy db ≥ 300 mm
for fy ≤ 420 MPa
(0.13 fy – 24) db ≥ 300 mm
for fy > 420 MPa
The computed splice length should be increase by 33% if fc’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 modification factors
0.24f y d b 0.24 420 18 331mm fc' 30 l dc max 333mm 0.043 f 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 supports 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 members and one-fourth the positive moment reinforcement in continuous members shall extend along the same face of member into the support. In beams, such reinforcement shall extend into the support at least 150 mm. ACI 12.11.1
At simple supports 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 support shall be the embedment length beyond the center of support; 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 supporting 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 support 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 members and one-fourth the positive moment reinforcement in continuous members shall extend along the same face of member into the support. In beams, such reinforcement shall extend into the support at least 150 mm. Negative moment:
At least one-third the total tension reinforcement provided for negative moment at a support 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. • Members 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 support 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 support single columns. This is one of the most economical types of footings and is used when columns are spaced at relatively long distances.
P kN
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 support structural walls that carry loads from other floors or to support 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 supporting soil. P1
P2
P2 kN
L
B
PP1 kN 1 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 terms 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 support 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 support 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 contact 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 supported 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 contact 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 0.17 1 f c 'bo d c
s d
C2
B
V C 0.083 2
C1
C2 + d
V C 0.33 f c 'bo d
C1 + d
f c 'bo 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 for moment
(L-C1)/2
Reinforcement in the long direction: 2
0.85f c 2M u 1 1 2 fy 0.85 f c B d As ,req B d
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 fy 0.85 f L d c As ,req L d 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 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. 2 s 1
long side of footing
B
where
Band width
short side of footing
ACI 15.4.4
B L
30
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 joint 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 supported column, the allowed bearing capacity ΦPn,c is
Pn ,c 0.85f cA1
ACI 10.14.1
For a supporting footing where the supporting surface is wider on all sides than the loaded area, the allowed bearing capacity ΦPn,f is
Pn ,f
A2 min 0.85f cA1 ; 2 0.85f cA1 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.
As ,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 support 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 1- Select a trial footing depth: Assume that the footing is 55 cm thick. 2- Evaluate 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/m2 40 40
245
3- Establish the required base area of the footing : P P 800 600 A req D L 7.839 m 2 q all (net) 178.6 7.84 Let L 3.20 m , B 2.45 m 3.20 Use 320x245x55 cm footing
320
4- Evaluate the net factored soil pressure
Pu 1.2PD 1.6PL 1.2 800 1.6 600 1920 kN Pu 1920 q u net 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 d avg 55-7.5-1.6 45.9cm bo 2[ 40 45.9 40 45.9 ] 343.6 cm
320
Vu 244.9 3.2 2.45 0.40 0.459 0.40 0.459 1740 kN Φ VC is the smallest of Φ 0.33 f c ' b o d 0.75 0.33 25 3436 459 1952 kN 2 2 Φ 0.17 f c ' 1 b o d 0.75 0.17 25 1 3436 459 3016 kN 0.4/0.4 βc αs d 40 459 Φ 0.083 f c ' 2 b d 0.75 0.083 25 2 o 3436 459 3605 kN b 3436 o Φ VC 1952 kN Vu 1740 kN 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 0.75 0.17 25 2450 459 717 kN
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 0.75 0.17 25 3200 459 936 kN 45.9
Vu is located at distance d from face of column
2.45 0.4 Vu 244.9 3.2 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 2
2 588 106 1- 12 0.9 0.85 25 2450 459 0.0031 A s 0.0031 459 2450 3500 mm 2
0.85 25 ρ 420
1.4
2
245
B L C1 2.45 3.2 0.4 M u q u net 244.9 2 2 2 2 588 kN .m
Critical section for moment
320 24.49 x 2.45
A s,min 0.0018 550 2450 2430 mm 2 A s,req 3500 mm 2 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
L B C2 3.2 2.45 0.4 M u q u net 244.9 2 2 2 2 Critical section for moment 412 kN .m
245 24.49 x 2.8
A s,min 0.0018 550 3200 3170 mm 2
320
1.025
2 412 106 1- 12 0.9 0.85 25 3200 459 0.0016 A s 0.0016 459 3200 2411mm 2
0.85 25 ρ 420
2
1.025
2
A s,req 317 0 mm 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
42.5
2Φ14 B
L 3.2 1.3 B 2.45 2 Central band reinft. As β 1 2 2 3170 2757 mm 1.3 1 Use 18 14 mm in the central band.
320
3170 2756 2 For each of the side bands, A s 207 mm 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 25 160000 ) 2210 103 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 6002500 mm 2
Pn ,f
A 2 min 0.85f cA1 ; 2 0.85f cA1 A1
Pn ,f
6002500 min 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 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 34 cm 49
Example 9- Check for anchorage of the reinforcement Dowel reinforcement (Φ16mm):
0.24f y d b 0.24 420 16 323mm fc' 25 l dc max 323mm 200mm 0.043 f d 0.043 420 16=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 y d b 0.071 420 16 478 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)
42.5
2.45 m
2Φ14 B
Width band =245
2Φ14 B
18Φ14 B
3.20 m
23Φ14 B
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 terms •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 terms •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 supported stair (transversely supported)
9
Simply supported stair (longitudinally supported)
Cantilever stair
Design of transversely supported 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 supported stairs Direction of bending Main reinforcement Shrinkage reinforcement
11
Direction of bending
Design of transversely supported 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 supported stairs Example 1 Design a straight flight stair in a residential building supported 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
have
0.075
0.30 0.34
0.16 0.165m 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 0.16
D.L (total) = 1.74 kN/m L.L =30.3 =0.9 kN/m
14
0.3
0.302 0.162 0.34
1.5 m
Shear diagram
Moment diagram 15
Design for moment M u 1kN .m d 165 20 6 139mm bw 300mm 0.85 f c ' fy
2M u 1 1 0.85 f c ' b d 2
0.85 25 2 1106 1 1 0.0005 2 420 0.9 0.85 25 300 139 A s 0.0005 300 139 20.9mm 2 A s ,min 0.0018 300 165 89.1mm 2 A s A s A s ,min 89.1mm 2 Use 112 for each step
16
Design for shear V C 0.75 0.17 25 139 300 /1000 26kN V u 2.65kN OK
17
Design of longitudinally supported stairs Direction of bending Shrinkage reinforcement
Main reinforcement
18
Design of longitudinally supported stairs
19
Design of longitudinally supported 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 supporting elements. n = number of goings X = Width of supporting landing slab at one end of the stairs slab
Y = Width of
20
supporting landing slab at the other end of the stairs slab.
Design of longitudinally supported 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 supporting elements. n = number of goings X = Width of supporting landing slab at one end of the stairs slab
Y = Width of
21
supporting landing slab at the other end of the stairs slab.
Design of longitudinally supported 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 supported 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 supported stairs Joint 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 joints.
24
Design of longitudinally supported 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 l 525 t 0.85 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
0.16
0.3 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 0.85 25 2 52.2 106 1 1 0.0037 2 420 0.85 0.9 25 1000 194 A s 0.0037 1000 194 718mm 2 A s ,min 0.0018 1000 220 396mm 2 A s OK Use 812
(22)=3.96 cm2/m
Design for shear 28
V C 0.75 0.17 25 194 1000 / 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
