Reinforced Concrete Design I

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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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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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


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


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 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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Grading

Course work:

20%

-Homework

4%

-Attendance

4%

-Project

12%

Mid-term t exam

20%

Final exam

60%

.

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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


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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Policy towards cell cell-phone phone use

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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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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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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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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


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


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


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


ENJOY THE COURSE!!

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Reinforced Concrete Design I

Dr. Nader Okasha

Lecture 1 Introduction to reinforced concrete

Contents 1.

2.

Concrete-producing materials Mechanical properties of concrete 3.

Steel reinforcement

2

Part 1: Concrete-Producing Materials

3

Advantages of reinforced concrete as a structural material 1. It has considerable compressive strength. 2. It has great resistance to the actions of fire and water. 3. Reinforced concrete structures are very rigid. 4. It is a low maintenance material. 5. It has very long service life.

4

Advantages of reinforced concrete as a structural material 6. It is usually the only economical material for footings, basement walls, etc. 7. It can be cast into many shapes. 8. It can be made from inexpensive local materials. 9. A lower grade of skilled labor is required for erecting.

5

Disadvantages of reinforced concrete as a structural material 1. It has a very low tensile strength. 2. Forms are required to hold the concrete in place until it hardens. 3. Concrete 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.

10

Types of Cement  Type I: General Purpose  Type II: Lower heat of hydration than Type I  Type III: High Early Strength • Quicker strength • Higher heat of hydration

11

Types of Cement  Type IV: Low Heat of Hydration • Slowly dissipates heat  less distortion (used for large structures).  Type V: Sulfate Resisting • For footings, basements, sewers, etc. exposed to soils with sulfates. If the desired type of cement is not available, different 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


• It is determined through testing standard cylinders 15 cm in diameter and 30 cm in height in uniaxial compression at 28 days (ASTM C470).

• Test cubes 10 cm × 10 cm × 10 cm are also tested in uniaxial compression at 28 days (BS 1881).

19


• The ACI Code is based on the concrete compressive strength as measured by a standard test cylinder. f c Cylinder  0.8f c Cube

• For ordinary applications, concrete compressive strengths from 20 MPa to 30 MPa are usually used.

20

Mechanical Concrete Properties Compressive-Strength Test

21

Mechanical Concrete Properties Modulus of Elasticity, Ec ' 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

38


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}

7

Load paths  loads on (one-way) beams This tributary load is determined by multiplying q by the tributary width for the beam.

w {kN/m} = q {kN/m2}  (S1+S2)/2 {m}

L

8

S1

S2

Load paths  loads on (two-way) beams The tributary areas for a beam in a two way system are areas which are bounded by 45-degree lines drawn from the corners of the 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

11

Load paths  loads on (two-way) beams

12

Load paths  loads on columns The tributary load for the column is concentrated. It has units of (force) e.g., kN. It is determined by multiplying q by the tributary area for the column.

P {kN} = q {kN/m2}  (x y){m2}

13

Load paths  loads on structural 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

14

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


w = 10  (4+5)/2 = 45 kN/m B1 4m B2 5m 4.5 m

16

C2

C1 6m

5.5 m


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

19

C2

C1 6m

5.5 m

Load types Classification by direction

1- Gravity loads 2- Lateral loads

20

Load types Classification by source and activity

1- Dead loads 2- Live loads 3- Environmental loads 21

Loads on Structures All structural elements must be designed for all loads anticipated to act during the life span of such elements. These loads should not cause the structural elements to fail or deflect excessively under working conditions.

Dead load (D.L) • Weight of all permanent construction • Constant magnitude and fixed location Examples: * Weight of the Structure (Walls, Floors, Roofs, Ceilings, Stairways, Partitions) * Fixed Service Equipment 22

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.

23

See IBC 2009 TABLE 1607.1 for more live loads. http://publicecodes.citation. com/icod/ibc/2009/index.ht m?bu=IC-P-2009000001&bu2=IC-P-2009000019

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


Dr. Nader Okasha

Lecture 3 Design philosophies and design codes

Design Versus Analysis Design involves the determination of the type of structural system to be used, the cross sectional dimensions, and the required reinforcement. The designed structure should be able to resist all forces expected to act during the life span of the structure safely and without excessive deformation or cracking. Analysis involves the determination of the capacity of a section of known dimensions, material properties and steel reinforcement, if any to external forces and moments.

2

Structural Design Requirements: The design of a structure must satisfy three basic requirements: 1)Strength to resist safely the stresses induced by the loads in the various structural 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.

7

Building Codes, Standards, and Specifications: Summary: The standards that will be used extensively throughout this course is Building Code Requirements for Reinforced Concrete and commentary, known as the ACI 318M-08 code. The building code that will be used for this course is the IBC 2009, in conjunction with the ANSI/ASCE7-02.

8

Design Methods (Philosophies) Two methods of design have long prevalent. Working Stress Method focuses on conditions at service loads. Strength Design Method focusing on conditions at loads greater than the service loads when failure may be imminent. The Strength Design Method is deemed conceptually more realistic to establish structural safety.

The Working-Stress Design Method This method is based on the condition that the stresses caused by service loads without load factors are not to exceed the allowable stresses which are taken as a fraction of the ultimate stresses of the materials, fc’ for concrete and fy for steel.

9

The Ultimate – Strength Design Method At the present time, the ultimate-strength design method is the method adopted by most prestigious design codes. In this method, elements are designed so that the internal forces produced by factored loads do not exceed the corresponding reduced strength capacities. reduced strength provided  factored loads 

The factored loads are obtained by multiplying the working loads (service loads) by factors usually greater than unity. 10

Safety Provisions (the strength requirement) Safety is required to insure that the structure can sustain all expected loads during its construction stage and its life span with an appropriate factor of safety. There are three main reasons why some sort of safety factor are necessary in structural design • Variability in resistance. *Variability of fc’ and fy, *assumptions are made during design and *differences between the as-built dimensions and those found in structural drawings.

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

• Consequences of failure. *Potential loss of life, *cost of clearing the debris and replacement of the structure and its contents and *cost to society. 11

Safety Provisions (the strength requirement) The strength design method, involves a two-way safety measure. The first of which involves using load factors, usually greater than unity to increase the service loads. The second safety measure specified by the ACI Code involves a strength reduction factor multiplied by the nominal strength to obtain design strength. The magnitude of such a reduction factor is usually smaller than unity Design strength ≥ Factored loads

 R    i Li i

ACI 9.3

ACI 9.2 12

Load factors ACI 9.2.1 Dead only U = 1.4D Dead and Live Loads U = 1.2D+1.6L Dead, Live, and Wind Loads U=1.2D+1.0L+1.6W Dead and Wind Loads U=1.2D+0.8W or U=0.9D+1.6W Dead, Live and Earthquake Loads U=1.2D+1.0L+1.0E Dead and Earthquake Loads U=0.9D+1.0E 13

Load factors ACI 9.2 Symbols

14

Strength Reduction Factors

ACI 9.3 According to ACI, strength reduction factors Φ are given as follows: a- For tension-controlled sections Φ = 0.90 b- For compression-controlled 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


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


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 <εεt y< εy

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

brittle,

thus sudden, and is not allowed by the ACI Strength Design Method. 14

Types of flexural failure: [3] Tension Failure The reinforcement yields before the concrete crushes. εcu=0.003 εcu=0.003

c
d h

b

d

b

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

ductile,

thus giving a sufficient amount of warning before collapse, and is required by the ACI Strength Design Method 15

Types of sections according to the ACI Code ACI 10.3.4 [1] Tension-controlled section The tensile strain in the tension steel is equal to or greater than 0.005 when the concrete in compression reaches its crushing strain of 0.003. This is a ductile section. [2] Compression-controlled section The tensile strain in the tension steel is equal to or less than εy (εy = fy/Es=0.002 for fy =420 MPa) when the concrete in compression reaches its crushing strain of 0.003. This is a brittle section. [3] Transition section The tensile strain in the tension steel is between 0.005 and εy (εy =fy/Es=0.002 for fy =420 MPa) when the concrete in compression reaches its crushing strain of 0.003. 16

Allowed strains for sections in bending

ACI 10.3.5

17

Strength reduction factor Φ εcu=0.003 c

d

y  c

t

y Es aεt

b1

d c     c

c

 y

ACI R9.3.2.2

18

Balanced steel 0.003 cb  d 0.003  f y E S Es  2105 MPa

600 cb  d 600  f y

=b1c  0.85 b1 f c '  600  b    600  f fy y  

   

19

Maximum allowed steel

0.003 cmax  d 0.003  0.005 3 cmax  d 8

=b1c  b1cmax 

 max

3 db1 8

3  0.85 b1 f c '    8  fy  20

Minimum steel allowed ACI 10.5.1

A s,min

 0.25 f c bw d   fy  max   1.4 b d w  f y 

bw = width of section d = effective depth of section

21

Design Aids

22

Summary: To calculate the moment capacity of a section:  0.25 f c bw d   fy  max   1.4 b d w  f y 

1-) As,min

if As,min > As,sup  reject section 2-) a 

3-)

As f y 0.85f c b

b1  0.85 b1  0.85 

or a 

df y 0.85f c 

for f c '  28 MPa 0.05( f c '28)  0.65 for f c '  28 MPa 7

a

4-) c  b 1

23

Summary:  d c 

5-) t   c  0.003 if t> 0.005: tension controlled f = 0.9 if 0.004 < t <0.005: in transition zone f =0.65+( t -0.002)  (250/3) if t < 0.004: compression controlled  reject section a  6-) M d  M n   As f y  d   2 

or fMn=fRnbd2 (find Rn from table)

7-) M u  ΦM n if not  reject section

24

Example A singly reinforced concrete beam has the cross-section shown in the figure below. Calculate the design moment strength. Can the section carry an Mu = 350 kN.m?

f y  414MPa a) f c   20.7MPa, b) f c   34.5MPa, c) f c   62.1M Pa

25

Example Solution a) f c  20.7MPa 1 A s,min

 0.25 f c 0.25 20.7 bw d  (254)(457)=319 mm 2  414  fy  max  1.4 1.4  bw d  (254)(457)=393 mm 2  fy 414 

=393 mm 2 < A s,sup =2580 mm 2 OK

2580  414 2 a    239mm 0.85f c b 0.85  20.7  254 As f y

3 b1  0.85

for f c '  20.7MPa  28 MPa

26

Example Solution a) f c  20.7MPa a

239 4 c    281mm b1 0.85  d c   457  281  5 t    0.003    0.003  0.00186  c   281   t  0.004 Section is compression controlled ==> Does not satisfy ACI requirements ==> Reject section

27

Example Solution b) f c  34.5MPa 1 A s,min

 0.25 f c 0.25 34.5 bw d  (254)(457)=412 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


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

 dc  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  1116  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

 dc  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


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
71.9 c   85 mm b1 0.85  d c 

17

 400  85   0.003  0.011  0.005  Tension controled 85 

es    0.003    c  

Example 1 Calculate Md a  M d  A s f y  d   2  71.8    0.9  5000  420   400   2    688 106 N.mm  688 kN.m

18

Example 2 Determine the ACI design moment strength Md (ΦMn) of the T-beam shown in the figure if fc’ =25 MPa and fy = 420 MPa. 10

90

1- Check min. steel

25 d  750-40-10-32-  655.5mm 8Φ32 2  0.25 f c '  1.4 A s,min  max  bw d ; bw d  fy  f y   0.25 25  1.4 A s,min  max  300  655.5 ; 300  655.5 420  420  A s,min  656 mm 2  A s,sup  6434 mm 2 19

Φ10

h= 75

Solution:-

OK

30

Example 2 2- Check if a < hf = 10cm Asf y 6434  420 a   141.3mm  0.85 f c b e 0.85  25  900

Section is T  NA is in the web

20

10 Φ10

h= 75

a= 141.3> hf = 100 mm i.e. assumption is wrong

90

8Φ32 30

Example 2 3- Calculate b1, c, and check εt A s f y - 0.85 f c be  bw  hf a 0.85 f c bw a

6434  420  0.85  25  900  300 100  224mm 0.85  25  300

c

a 224   264 mm β1 0.85

 dc  655.5  264   εt   0.003     0.003 264  c    ε t  0.00447  0.004  0.005  Transision zone

 =0.65+(e t -0.002)  (250/3) =0.65+(0.00447-0.002)  (250/3)=0.855 21

Example 2 4- Calculate Md Cf  0.85f c ' (be  b w ) h f  0.85  25  900  300 100  1275 103 N  1275 kN Cw  0.85f c ' a b w  0.85  25  224  300  1427.4 103 N  1427.4 kN  M d  Φ C w 

hf    d    2     224  100    3  0.855 1427.4 103  655.5   1275  10 655 . 5     2 2      a  d    C f 2 

 1323.4 106 N .mm  1323.4 kN .m

22

Design of T-Beams --- Positive moment

+

To analyze the section, the steel is divided in two portions: (1) Asf, which provides a tension force in equilibrium with the compression force of the overhanging flanges, and providing a section with capacity Muf and (2) Asw, the remaining of the steel, providing a section with capacity Muw.

M u  M uf  M uw

23

Mu : Ultimate moment applied, requiring steel As. Muf : Moment resisted by overhanging flange parts, requiring steel Asf. Muw : Moment resisted by web, requiring steel Asw.


+

Step 1

24

24

Step 2


+

M u  M uf  M uw

M uw  M u  M uf 0.85 f c '  fy

25

 2M uw 1  1   2  0 . 85 f ' b d c w 

Step 3

  

Step 4

Asw   bw d

Step 5

As  Asf  Asw

Step 6

Design of L-Beams --- Positive moment be

be

Same as bw

26

Design of T-Beams --- Negative moment be

bw

Design as a rectangular section with width bw

27

Flange Reinforcement When flanges of T-beams are in tension, part of the flexural reinforcement shall be distributed over effective flange width, or a width equal to one-tenth of the span, whichever is smaller

-ve moment

Additional Reinforcement

min (beff & l/10)

Additional Reinforcement

Main Reinforcement

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

Design of T-Beams --- Positive moment Design Procedure: 1- Establish h based on serviceability requirements of the slab and calculate d

2- Choose bw 3- Determine be according to ACI requirements. 4- Calculate As assuming that a < hf with beam width = be & Φ=0.90 b   2M u 1  1   hf 2   d Φ 0.85 f ' b d c e   As As f y As = ρ be d → a bw 0.85 f c ' b e 5- If a ≤ hf: the assumption is right  continue as rectangular section If a > hf: revise As using T-beam equations (steps 1-6). 6- Check the Φ=0.90 assumption (εt ≥0.005) and As,sup ≥ As,min

0.85 f c ' ρ fy

29

e

Example 3

fy = 420 MPa.

Lm

A floor system consists of a 14.0cm concrete slab supported by continuous T-beams with a span L. Given that bw=30cm and d=55cm, fc’ =28 MPa and Determine the steel required at midspan of an interior beam to resist a service dead load moment 320 kN.m and a service live load moment 250 kN.m in the following two cases: (A) L = 8 m Spandrel (B) L = 2 m beam

3.0 m

3.0 m

Slab

hf

bw

30

3.0 m

Solution (A) L = 8 m Determine be according to ACI requirements

784 kN.m

200

 L 8000  4  4  2000mm  be  min 16hf  b w  16 140   300=2540mm  b  3000 mm  

14 55 As 30

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

Calculate As assuming that a < hf with beam width = be & Φ=0.90 Mu = 1.2(320)+1.6(250)=784 kN.m 0.85 f c '  fy 31

 2M u 1- 12  0 . 85 f ' b d c e 

  

Solution (A) L = 8 m  2  784 10 1  1  2  0 . 9  0.85  28  2000  550  6

   

784 kN.m

0.85  28 ρ 420

200

14 55 As 30

 0.00354 As  ρ be d  0.00354  2000  550  3892 mm 2

Check a ≤ hf assumption a

As f y 0.85f c 'be



3892  420  34.3mm  h f  140mm 0.85  28  2000

The assumption is right  Rectangular section design

Use 8Φ25mm (As,sup= 3927 mm2) arranged in two layers. sc  32

300  2  40  2  8  4  25  34.5 mm  d b  25 mm  4  1  25 mm

OK

Solution (A) L = 8 m Check the Φ=0.90 assumption (εt ≥0.005) and As,sup ≥ As,min   1.4  0.25 f c '  As ,min  max  bw d ; bw d   max fy    fy  As,min  550 mm 2  As,sup  3927 mm 2 OK

1.4  0.25 28  300  550 ; 300  550  420  420   

3927  420 a   34.7 mm 0.85f c 'be 0.85  28  2000 As f y

55

14

200

c

33

a 34.7   40.8 mm β1 0.85

dc  550  40.8  εt    0.003    0.003  c   40.8   0.0374  0.005    0.9 OK

8Φ25 30

Solution (A) L = 8 m Check moment capacity a  M d   As f y  d   2  34.7    0.9  3927  420  550   2   M d  790.7 106 N.mm  790.7 kN.m  M u  784 kN.m

55

14

200

8Φ25 30

34

Solution (B) L = 2 m 784 kN.m

50

Determine be according to ACI requirements  L 2000  4  4  500mm  be  min 16hf  b w  16 140   300=2540mm  b  3000 mm  

14 55 As 30

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

Calculate As assuming that a < hf with beam width = be & Φ=0.90 Mu = 1.2(320)+1.6(250)=784 kN.m 0.85 f c '  fy 35

 2M u 1  1   2  0 . 85 f ' b d c e 

  

Solution (B) L = 2 m  2  784 10 1  1  2  0 . 9  0.85  28  500  550  6

   

784 kN.m

0.85  28 ρ 420

50

 0.0159 As  ρ be d  0.0159  500  550  4389 mm 2

Check a ≤ h assumption f

a

As f y 0.85 f c ' be



4389  420  155mm > 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


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  Asf s

Asf 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   sE 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 cab  Asf s

 c d   As f y  0.85f c1cb  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 c1cb  As   0.003E s  c  c d    f s    0.003E s  f y  c 

find c, a

3) As 2  Asf 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 -   Asf s d - d '   2   

Example 1 For the beam with double reinforcement shown in the figure, calculate the design moment Md. 5.0 2Φ25 fc’ =35MPa and fy = 420 MPa. 60 6Φ32

Solution: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 Asf 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 sf 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 d1 8

13

Design of Doubly Reinforced Rectangular Sections

1) Design the section as singly reinforced, and calculate t 2) If t < 0.004 Comp. steel is needed (or enlarge section if possible)

3) Design As1 for maximum reinforcement (slide 13) and find Mn1, a, c 4) M n  M u f 5) Mn2 = Mn – Mn1

c d    0.003E s  f y  c  Asf s M n2 7) As   As 2  fy  f s(d  d ) 6) f s   sE 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

 dc  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 cb 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) Asf s (2052)(420) As 2    2052mm 2 fy (420)

As  As 1  As 2  3307  2052  5359mm 2 Use 830 in two rows for tension steel (As,sup = 5655 mm 2 )

18

Use 4 26 for compression steel (As,sup = 2124 mm 2 )


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


Dr. Nader Okasha

Lecture 10 Design of slabs

Regula (3

y

Introduction

Plate/Shell (2D) x z t<<(x,z)

z

x A slab is a structural element whose thickness is small compared to its own length and width.

t  L , S

h t

zS

t

Lx

Slabs in buildings are usually used to transmit the loads on floors and Loads roofs to the supporting beams Dimensional Hierarchy of Structural Beam

Beam

Slab Column

Beam

Beam Footing

Slab Beam

Column

Beam

Beam Soil

2

Introduction Slabs are flexural members. Their flexure strength requirement may be expressed by

Mu  M n

Types of Slabs

Solid slabs :- which are divided into - One way solid slabs - Two way solid slabs Ribbed slabs :- which are divided into - One way ribbed slabs - Two way ribbed slabs 3

One-way slab

Two-way slab

Solid Slab

Two way slab

Ribbed Slab (joist construction)

4

Two way slab

L 2 S

One-way slab

One-way slab

L 2 S

Ribbed slab with hollow blocks

5

Ribbed slab with hollow blocks

6

One-way solid slabs

shrinkage Reinft.

A one-way solid slab curves under loads in one direction only. Accordingly, slabs supported on two opposite sides only and slabs supported on all four sides, but L/S ≥ 2 are classified as one-way slabs.

Main Reinft.

Main reinforcement is placed in the shorter direction, while the longer direction is provided with shrinkage reinforcement to limit cracking. 7

Two-way solid slabs

Main Reinft.

A two-way solid slab curves under loads in two directions. Accordingly slabs supported on all four sides, and L/S < 2 are classified as two-way slabs.

S Main Reinft.

L

Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement is required in the two directions. 8

One-way v.s two-way ribbed slabs If the ribs are provided in one direction only, the slab is classified as being one-way, regardless of the ratio of longer to shorter panel dimensions. It is classified as two-way if the ribs are provided in two directions .

9

Minimum thickness of one way slabs

Minimum Cover

10

ACI Table 9.5(a)

ACI 7.7.1

a - Concrete exposed to earth or weather for Φ<16mm------40 mm and for Φ>16mm----- 50 mm b - Concrete not exposed to earth or weather for Φ<32mm------20 mm, otherwise ------ 40 mm

Spacing of Reinforcement Bars a- Flexural Reinforcement Bars Flexural reinforcement is to be spaced not farther than three times the slab thickness (hs), nor farther apart than 45 cm, center-to-center.  3 hs Smax  smaller of  ACI 10.5.4 45cm b- Shrinkage Reinforcement Bars Shrinkage reinforcement is to be spaced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center.  5 hs Smax  smaller of  ACI 7.12.2.2 45cm

11

Loads Assigned to Slabs wu=1.2 D.L + 1.6 L.L

a- Dead Load (D.L) : 1- Weight of slab covering materials 2-Equivalent partition weight 3- Own weight of slab

b- Live Load (L.L)

12

a- Dead Load (D.L)

1- Weight of slab covering materials, total =2.315 kN/m2

tiles (2.5cm thick) =0.025×23 = 0.575 kN/m2 cement mortar (2.5cm thick) =0.025×21 = 0.525 kN/m2 sand (5.0cm thick) =0.05×18 = 0.9 kN/m2 plaster (1.5cm thick) =0.015×21 = 0.315 kN/m2

tiles cement mortar sand

2.5 cm 2.5 cm 5 cm

slab

plaster 13

1.5 cm

2-Equivalent partition weight

This load is usually taken as the weight of all walls (weight of 1m span of wall × total spans of all walls) carried by the slab divided by the floor area and treated as a dead load rather than a live load. To calculate the weight of 1m span of wall: Each 1m2 surface of wall contains 12.5 blocks A block with thickness 10cm weighs 10 kg A block with thickness 20cm weighs 20 kg Each face of 1m2 surface has 30kg plaster Load / 1m2 surface for 10 cm block = 12.5 × 10 +2×30=185 kg/m2 = 1.85 kN/m2 Load / 1m2 surface for 20 cm block = 12.5 × 20 +2×30=310 kg/m2 = 3.1 kN/m2

14

20 cm

Weight of 1m span of wall with height 3m: For 10 cm block wt. = 1.85 kN/m2 × 3 = 5.6 kN/m For 20 cm block wt. = 3.1 kN/m2 × 3 = 9.3 kN/m

3- Own weight of slab 1- Solid slab: Own weight of solid slab (per 1m2)=

gc h = 25 h

kN/m2

2- Ribbed slab:

Example Find the total ultimate load per rib for the ribbed slab shown: Assume depth of slab = 25 cm (20cm block +5cm toping slab) Hollow blocks are 40 cm × 25 cm × 20 cm in dimension Assume ribs have 10 cm width of web Assume equivalent partition load = 0.75 kN/m2 Consider live load = 2 kN/m2.

15

3- Own weight of slab

Solution •

Total volume (hatched) = 0.5 × 0.25 × 0.25 = 0.03125 m3

•

Volume of one hollow block = 0.4 × 0.20 × 0.25 = 0.02 m3

•

Net concrete volume = 0.03125 - 0.02 = 0.01125 m3

•

Weight of concrete = 0.01125 × 25= 0.28125 kN

•

Weight of concrete /m2 = 0.28125 /[(0.5)(0.25)] = 2.25 kN/ m2

•

Weight of hollow blocks /m2 = 0.2/[(0.5)(0.25)] = 1.6 kN/ m2

•

Total slab own weight= 2.25 + 1.6= 3.85kN/m2

Load per rib Total dead load= 3.85 + 2.315 + 0.75 = 6.915 kN/m2 Ultimate load = 1.2(6.915) + 1.6(2) = 11.5 kN/m2 16

Ultimate load per rib = 11.5 × 0.5 = 5.75 kN/m

Minimum live Load values on slabs Type of Use Uniform Live Load

kN/m2

b- Live Load (L.L)

It depends on the function for which the floor is constructed.

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 

17

2.5 5 2.5

Retail

4

 wholesale Exit facilities Manufacturing

5 5



Light

4



Heavy

6

Loads Assigned to Beams Beams are usually designed to carry the following loads: - Their own weight - Weights of partitions applied directly on them - Floor loads

L

S1 18

S2

Design of one way SOLID slabs

19

One-way solid slabs

S1

S2

1m

One-way solid slabs are designed as a number of independent 1 m wide strips which span in the short direction and are supported on crossing beams. These strips are designed as rectangular beams.

L

  0.85f c  2M u 1  1     2  fy   0.85  f c bd    S1

20

S2

shrinkage Reinft.

One-way solid slabs

Main Reinft.

21

Check on tension/compression control (maximum allowed steel)

Method 1: Check et

Method 2: Check max 0.003 cmax  d 0.003  0.005 3 cmax  d 8 =b1c  b1cmax

 max

3  0.85 b1 f c '    8  fy 

3  db1 8

22

Shrinkage Reinforcement Ratio According to ACI Code and for fy =420 MPa

ACI 7.12.2.1

 shrinkage  0.0018  As ,shrinkage  0.0018 b  h where, b = width of strip, and h = slab thickness

Minimum Reinforcement Ratio for Main Reinforcement

As ,min  As ,shrinkage  0.0018 b  h

ACI 10.5.4

Check shear capacity of the section

V u   V c  0.17 f c ' bwd Otherwise enlarge depth of slab 23


ACI 8.3.3

ACI Code permits the use of the following approximate moments and shears for design of continuous beams and one-way slabs, provided: • There are two or more spans. • Spans are approximately equal, with the larger of two adjacent spans not greater than the shorter by more than 20 percent. • Loads are uniformly distributed. • Unfactored live load does not exceed three times the unfactored dead load. • Members are of similar section dimensions along their lengths (prismatic).

24


Bending Moment More than two spans

25

ACI 8.3.3


Bending Moment 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.

26

ACI 8.3.3


Shear More than two spans

27

ACI 8.3.3


Shear Two spans

28

ACI 8.3.3

Summary of One-way Solid Slab Design Procedure 1- Select representative 1m wide design strip/strips to span in the short direction. 2- Choose a slab thickness to satisfy deflection control requirements. When several numbers of slab panels exist, select the largest calculated thickness. 3- Calculate the factored load wu 4- Draw the shear force and bending moment diagrams for each of the strips. 5- Check adequacy of slab thickness in terms of resisting shear by satisfying the following equation: V u  0.17  f c ' b d where b = 1000 mm

If the previous equation is not satisfied, go ahead and enlarge the thickness to do so. 29

Summary of One-way Solid Slab Design Procedure 6- Design flexural and shrinkage reinforcement: Flexural reinforcement ratio is calculated from the following equation   0.85f c  2M u 1  1     2  fy   0.85  f c bd    where b = 1000 mm

You need to check tension controlled requirement, minimum reinforcement requirement and spacing of selected bars. Compute the area of shrinkage reinforcement, where Ashrinkage=0.0018bh, where b = 1000 mm.

30

7- Draw a plan of the slab and representative cross sections showing the dimensions and the selected reinforcement.

Example 1 Using the ACI-Code approximate structural analysis, design for a warehouse, a continuous one-way solid slab supported on beams 4.0 m apart as shown in the figure. Assume that the beam webs are 30 cm wide.

The dead load is 3kN/m2 in addition to the own weight of the slab, and the live load is 3kN/m2.

8.0 m

Use fc’=28MPa, fy=420MPa

4.0 m

31

4.0 m

4.0 m

Solution: 1- Select a representative 1 m wide slab strip:

The selected representative strip is shown in the figure

2- Select slab thickness:

For one-end continuous spans, hmin = l/24 =4.0/24=0.167m Slab thickness is taken as 17 cm

8.0 m

1.0 m

17cm

4.0 m Wu

32

4.0 m

4.0 m

Solution: 3- Calculate the factored load wu per unit length of the selected strip:

Own weight of slab = 0.17× 25 = 4.25 kN/m2 wu= 1.20 (3+4.25) +1.60 (3)= 13.5 kN/m2 For a strip 1 m wide, wu=13.5 kN/m 4- Evaluate the maximum factored shear forces and bending moments in the strip:

The clear span length, ln = 4.0 – 0.30 = 3.70 m wu=13.5 kN/m

33

Solution:

18.5 7.7

16.8

16.8

16.8

Units of moment are in kN.m 34

18.5 7.7 16.8

Solution:

25

25

28.7

Units of shear are in kN 35

28.7

25

25

Solution: 5- Check slab thickness for beam shear:

Effective depth d = 17 – 2 – 0.60 = 14.40 cm, assuming Φ12 mm bars. Vu,max = 28.7 kN. V c   0.17 f c ' bd  0.75  0.17 28 1000 144  95.8 kN

i.e. , slab thickness is adequate in terms of resisting beam shear. 6- Design flexural and shrinkage reinforcement:

Assume that Φ=0.9

  0.85f c  2M u 1  1     2  fy   0.85 f c bd   

Where b = 1000 mm & d = 144mm

36

Solution: For max. negative moment, Mu = 18.5 kN.m   0.85  28  2 18.5 106 1  1     0.00241 ρ 2    420  0.85  0.9  28  1000 144      3  0.85 b1f c '  3  0.85  0.85  28  ρ max        0.01806  ρ    0.90  8 fy 420   8 As, ve  0.002411000 144  347 mm 2 As,min  0.0018 1000 170  306 mm 2  As, ve OK 79 347mm 2  φ10  S  227.5 mm 1000mmstrip S Smax  min(450 or 3 170)  450mm 37

 use 10@20cm

Solution: For max. positive moment, Mu = 16.8 kN.m   0.85  28  2 16.8 106 1  1     0.00219 ρ 2    420   0.85  0.9  28  1000 144    3  0.85 b1f c '  3  0.85  0.85  28  ρ max        0.01806  ρ    0.90 8  fy 8 420    As, ve  0.00219 1000 144  315 mm 2 As,min  0.0018 1000 170  306 mm 2  As, ve OK 79φ10 315mm 2   S  251mm 1000mmstrip S Smax  min(450 or 3 170)  450  use 10@25cm 38

Solution: Calculate the area of shrinkage reinforcement: Area of shrinkage reinforcement = 0.0018 (100) (17) = 306 mm2 For shrinkage reinforcement use Φ 10 mm @ 25 cm (from previous slides calculations) Shrinkage reinft. Φ10@25 Φ10@25

Φ10@20

Φ10@20

Φ10@25 17cm

Φ10@25

Φ10@25

Φ10@25

See Lecture 12 for information on detailing requirements

39

Solution:

8.0 m

Φ10@25 Φ10@25

4.0 m

40

Φ10@20

Φ10@20 Φ10@25

4.0 m

Φ10@25 Φ10@25

4.0 m

Design of one way RIBBED slabs

41

One-way ribbed slabs Ribbed slabs consist of regularly spaced ribs monolithically built with a toping slab. The voids between the ribs may be either light material such as hollow blocks [figure 1] or it may be left unfilled [figure 2]. Topping slab

Rib

Hollow block

Figure [1] Hollow block floor

Temporary form Figure [2] Moulded floor

The use of these blocks makes it possible to have smooth ceiling which is often required for architectural considerations and have good sound and temperature insulation properties besides reducing the dead load of the slab greatly. 42

Key components of one-way ribbed slabs ACI 8.13.6.1 Topping slab thickness (t) is not to be less than 1/12 the clear distance (lc) between ribs, nor less than 50 mm a. Topping slab:

 lc  t  12 50 mm

and should satisfy for a unit strip: t

lc Slab thickness (t)

w u l c2 1240 f c 

Shrinkage reinforcement is provided in the topping slab in both directions in a mesh form. 43

Key components of one-way ribbed slabs b. Regularly spaced ribs: Minimum dimensions:

Ribs are not to be less than 100 mm in width, and a depth of not more than 3.5 times the minimum web width and clear spacing between ribs is not to exceed 750 mm. ACI 8.13.2 ACI 8.13.3 l ≤ 750 mm c

h ≤ 3.5 bw

bw ≥ 100

44

Key components of one-way ribbed slabs

ACI 8.13.8 Shear strength provided by rib concrete Vc may be taken 10% greater than those for beams. Shear strength:

Flexural strength:

Ribs are designed as rectangular beams in the regions of negative moment at the supports and as T-shaped beams in the regions of positive moments between the supports. Effective flange width be is taken as half the distance between ribs, center-to-center. b e

45

Key components of one-way ribbed slabs Hollow blocks: Hollow blocks are made of lightweight concrete or other lightweight materials. The most common concrete hollow block sizes are 40 × 25 cm in plan and heights of 14, 17, 20, and 24 cm.

46

Summary of one-way ribbed slab design procedure 1. The direction of ribs is chosen. 2. Determine h, and select the hollow block size, bw and t 3. Provide shrinkage reinforcement for the topping slab in both directions. 4. The factored load on each of the ribs is computed. 5. The shear force and bending moment diagrams are drawn. 6. The strength of the web in shear is checked. 7. Design the ribs as T-section shaped beams in the positive moment regions and rectangular beams in the regions of negative moment. 8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared. 47

Example Design a one-way ribbed slab to cover a 3.8 m x 10 m panel, shown in the figure below. The covering materials weigh 2.25 kN/m2, equivalent partition load is equal to 0.75 kN/m2, and the live load is 2 kN/m2.

3.8 m

Use fc’=25 MPa, fy=420MPa

10 m

48

Solution 1. The direction of ribs is chosen:

3.8 m

Ribs are arranged in the short direction as shown in the figure

5.0 m

5.0 m

2. Determine h, and select the hollow block size, bw and t:

From ACI Table 9.5(a), hmin = 380/16 = 23.75cm  use h = 24 cm. Let width of web, bw =10 cm Use hollow blocks of size 40 cm × 25 cm × 17 cm (weight=0.17 kN) Topping slab thickness = 24 – 17 = 7cm > lc/12 =40/12= 3.3cm > 5cm OK For a unit strip of topping slab: wu=[1.2(0.07 × 25 + 0.75 + 2.25) + 1.6(2)] ×1m = 8.9 kN/m = 8.9 N/mm w u l c2

8.9( 400 ) 2 t   16mm OK ( 0.9 )1240 25 1240 f c 

49

Solution 3. Provide shrinkage reinforcement for the topping slab in both directions:

Area of shrinkage reinforcement, As=0.0018(1000)70=126 mm2 Use 5 Φ 6 mm/m in both directions. 4. The factored load on each of the ribs is to be computed:

50

1.0 m

0.4 m

0.1 m

0.4 m

7 cm

0.25 m

1.0 m

0.05 m

0.24 m

Total volume (in 1m2 surface) = 1.0 × 1.0 × 0.24 = 0.24 m3 Volume of hollow blocks in 1m2 = 8 × 0.4 × 0.25 × 0.17 = 0.136 m3 Net concrete volume in 1m2 = 0.24- 0.136 = 0.104 m3 Weight of concrete in 1m2 = 0.104 × 25 = 2.6 kN/m2 Weight of hollow blocks in 1m2 = 8 × 0.17= 1.36 kN/m2 Total dead load /m2 = 2.25 + 0.75 + 2.6 + 1.36 = 7.0 kN/m2

Solution wu=1.2(7)+1.6(2)=11.6 kN/m2 wu/m of rib =11.6x0.5= 5.8 kN/m of rib 5. Critical shear forces and bending moments are determined (simply supported beam):

Maximum factored shear force = wul/2 = 5.8 (3.8/2) = 11 kN Maximum factored bending moment = wul2/8 = 5.8 (3.8)2/8 = 10.5 kN.m 6. Check rib strength for beam shear:

Effective depth d = 24–2–0.6–0.6 =20.8 cm, assuming 12mm reinforcing bars and Φ 6 mm stirrups. 1.1ΦV c  1.1 0.75  0.17  25 100  208  14400 N = 14.4 kN  Vu,max  11kN

Though shear reinforcement is not required, 4  6 mm stirrups per meter run are to be used to carry the bottom flexural reinforcement.

51

Solution 7. Design flexural reinforcement for the ribs:

There is only positive moments over the simply supported beam, and the section of maximum positive moment is to be designed as a T-section Assume that a<70mm and Φ=0.90→Rectangular section with b = be =500mm

As  ρ be d  0.0013  500  208  135 mm

2

Use 210mm (As,sup= 157 mm2) 157  420  6.2 mm  70mm 0.85f c 'be 0.85  25  500 The assumption is right

a

52

As f y



   

50

105 kN.m

0.85  25  2 10.5 106 ρ 1  1  420  0.9  0.85  25  500  2082  0.0013

7 24

As 10

Solution Check As,min

  1.4  0.25 f c '  As,min  max  bw d ; bw d  fy  fy    As,min  70 mm 2  A s,sup  157 mm 2

OK

Check Φ=0.9

c

a 6.2   7.3 mm β1 0.85

 dc  208  7.3  εt   0.003     0.003  c   7.3  ε t  0.083  0.005    0.9 OK

53

Solution

A

1Φ10 m

A

1Φ10 m

1Φ10 m

1Φ10 m

3.8 m

8. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared

5.0 m

5.0 m Φ6mm mesh @20 cm

Φ6mm stirrups @25 cm

7cm 24cm 17cm

2Φ10mm

10

40 cm

10

2Φ10mm

Section A-A

54

See Lecture 12 for information on detailing requirements


Dr. Nader Okasha

Lecture 11 Design of short concentric columns

Columns Columns are vertical compression members of a structural frame intended to support the load-carrying beams. They transmit loads from the upper floors to the lower levels and then to the soil through the foundations.

Loads Beam

P

Column

h

Slab

b

Beam

l

h

Beam

Column

Beam

b

Beam

Slab Footing Beam

Beam Soil

2

Columns Usually columns carry bending moment as well, about one or both axes of the cross section, and the bending action may produce tensile forces over a part of the cross section.

The main reinforcement in a columns is longitudinal, parallel to the direction of the load and consists of bars arranged in a square, rectangular, or circular shape. 3

Length of the column in relation to its lateral dimensions Columns can be classified as

1- Short Columns, for which the strength is governed by the strength of the materials and the dimensions of the cross section

2- Slender Columns, for which the strength may be significantly reduced by lateral deflections.

Position of the load on the cross-section Columns can be classified as

1-Concentrically loaded columns, which are subjected to axial force only 2-Eccentrically loaded columns, which are subjected to moment in addition to the axial force. 4

Analysis and Design of Short Columns

Column Types: 1. Tied 2. Spiral 3. Composite

5

Behavior of Tied and Spirally-Reinforced Columns Axial load tests have proven that tied and spirally reinforced columns having the same cross-sectional areas of concrete and steel reinforcement behave in the same manner up to the ultimate load. At that load, tied columns fail suddenly due to excessive cracking in

the concrete section followed by buckling of the longitudinal reinforcement between ties within the failure region. For spirally reinforced columns, once the ultimate load is reached, the concrete shell covering the spiral starts to spall off but the core will continue to carry additional loads because the spiral provides a confining force to the concrete core, thus enabling the column to sustain large deformations before final collapse. 6

Behavior of Tied and Spirally-Reinforced Columns

Failure of a tied column

7

Failure of a spiral column

Nominal Capacity under Concentric Axial Loads

P0  0.85f c  Ag  Ast   f y Ast or

P0  Ag 0.85f c  Ast (f y  0.85f c) Ag = gross area = b*h Ast = area of longitudinal steel fc′ =concrete compressive strength 8

fy = steel yield stress

Maximum Nominal Capacity under Concentric Axial Loads

Pn  rP0 Pn  r Ag 0.85f c  Ast (f y  0.85f c)  r = Reduction factor to account for accidental eccentricity

r = 0.80 ( tied ) r = 0.85 ( spiral ) 9

ACI 10.3.6.3

Design Capacity under Concentric Axial Loads

 Pn  Pu  Pn   r Ag  0.85f c  Ast  f y  0.85f c  Pu or

 Pn   r A g 0.85f c  g  f y  0.85f c  Pu where g = Ast / Ag ACI 9.3.2.2

10

 = 0.65 for tied columns

ACI 10.3.6.3 r = 0.80 ( tied )

 = 0.75 for spiral columns

r = 0.85 ( spiral )

Design of Short Concentrically Loaded Columns

 Pn  Pu  Pn   r Ag 0.85f c  g  f y  0.85f c  Pu Both Ag and g are unknown in this equation. There are two options to design the column: 1- Select Ag and calculate g. The Ag may be selected from initial sizing (Ag = Pu / 0.5fc′ ).

11

2- Select g and calculate Ag. Usually g is assumed as 2% as a starting point.

Calculation of required cross section, if steel ratio is known

 Pn  Pu  Pn   r Ag 0.85f c  g  f y  0.85f c  Pu * when g is known or assumed:

Ag 

Pu

  r 0.85f c  g  f y  0.85f c 

12



Calculation of required steel ratio, if cross section is known

 Pn  Pu  Pn   r Ag 0.85f c  g  f y  0.85f c  Pu * when Ag is known or assumed:

  Pu 1 g    0.85f c       r  A g   f y  0.85f c 13

Design of spirals Spiral Reinforcement Ratio, s

Volume of Spiral 4 Asp s   Volume of Core Dc s

Asp  D c    from: s   2 [( / 4 ) D c ] s   Asp  cross-sectional area of spiral reinforcement D c  core diameter: outside edge to outside edge of spiral 14

s  spacing pitch of spiral steel (center to center)

Design of spirals Spiral Reinforcement Spacing, s

 A g   f c  s  0.45    1    ACI Eq. 10-5    Ac  fy  4Asp s  from previous slide D cs s 

4Asp  Ag  fc ' 0.45Dc   1     Ac  f y 

A c  core area  15

 D c2

A g  gross area 

4

 D2 4

Design Considerations

Longitudinal Steel

- Limits on reinforcement ratio: ACI 10.9.1

0.01Ag  Ast  0.08Ag or

0.01  g  0.08 16


Longitudinal Steel

- Minimum number of bars ACI 10.9.2

min. of 6 bars in spiral arrangement min. of 4 bars in rectangular or circular ties min. of 3 bars in triangular ties 17


Longitudinal Steel - Clear Distance between Reinforcing Bars (Longitudinal Steel) For tied or spirally reinforced columns, clear distance between bars, shown in the figure, is not to be less than the larger of 1.50 times bar diameter or 40 mm. This is done to ensure free flow of concrete among reinforcing bars.

ACI 7.6.3

S c  max 1.5  d b , 40mm  18


Lateral Ties

- Arrangement of ties and longitudinal bars: ACI 7.10.5.3 1.) At least every other longitudinal bar shall have lateral support from the corner of a tie with an included angle 135o. 2.) No longitudinal bar shall be more than 150 mm clear on either side from a laterally supported bar. 19


Lateral Ties

- Arrangement of ties and longitudinal bars: ACI 7.10.5.3

20

Ties shown dashed may be omitted if x < 150 mm


Lateral Ties

- Maximum vertical spacing: ACI 7.10.5.2

s s s

21

 16 db ( db = diameter for longitudinal bars )  48 dstirrup (dstirrup = diameter for stirrups)  least lateral dimension of column


Lateral Ties

- Minimum size of ties ACI 7.10.5.1

size

22

  8 bar if longitudinal bar  30 bar  12 bar if longitudinal bar   32 bar  12 bar if longitudinal bars are bundled


Spirals

- Size and spacing of spiral ACI 7.10.4.2

size  10 mm diameter ACI 7.10.4.3 25mm 23



clear spacing between spirals



75mm

Design Considerations Concrete Protection Cover

ACI 7.7.1

The clear concrete cover is not to be taken less than 4 cm for columns not exposed to weather or in contact with ground. Minimum Cross Sectional Dimensions The ACI Code does not specify minimum cross sectional dimensions for columns. Column cross sections 20 × 25 cm are considered as the smallest practicable sections. For practical considerations, column dimensions are taken as multiples of 5 cm. Lateral Reinforcement Ties are effective in restraining the longitudinal bars from buckling out through the surface of the column, holding the reinforcement cage together during the construction process, confining the concrete core and when columns are subjected to horizontal forces, they serve as shear reinforcement. 24

Design Procedure for Short Concentrically Loaded Columns 1. Evaluate the factored axial load Pu acting on the column. This can be done by: a- Tributary Area Method

b- Pu is the sum of the reactions of the beams supported by the column. 2. Assume a starting reinforcement ratio ρg that satisfies ACI Code limits. Usually a 2 % ratio is chosen for economic considerations. 3. Determine the gross sectional area Ag of the concrete section. 4. Choose the dimensions of the cross section based on its shape. 5. Readjust the reinforcement ratio by substituting the actual cross sectional area in the respective equation. This ratio has to fall within the specified code limits. 25

Design Procedure for Short Concentrically Loaded Columns 6.

Calculate the needed area of longitudinal reinforcement ratio based on the adjusted reinforced ratio and the chosen concrete dimensions.

7.

From reinforcement tables, choose the number and diameters of needed reinforcing bars. For rectangular sections, a minimum of four bars is needed, while a minimum of six bars is used for circular columns.

8.

Design the lateral reinforcement according to the type of column, either ties or spirals.

9.

Check whether the spacing between longitudinal reinforcing bars satisfies ACI Code requirements.

10. Draw the designed section showing concrete dimensions and with required longitudinal and lateral reinforcement.

26

Example 1 The cross section of a short axially loaded tied column is shown in the figure. It is reinforced with 616mm bars. Calculate the design load Ties Φ8@25cm capacity of the cross section. Use fc′ =28 MPa and fy = 420 MPa. 25

Solution: A 1206 ρg  st   0.012  1.21% A g 250  400 ρ min  1 %  ρ g  1.21%  ρ max  8%

6Φ16

40 Figure [1]

OK

Clear distance between bars Sc

40  2(4)  2(0.8)  3(1.6)  12.8cm 3 1 max 1.5  d b  2.4cm , 4cm 
Sc=12.8 cm

25

6Φ16

Sc 

Only, one tie is required for the cross section 27

40

Example 1 The spacing between ties 16 db =16(1.6) = 25.4 cm ≥ S = 25 cm 48 ds = 48(0.8) = 38.4 cm ≥ S = 25 cm smaller of b or d = 25 cm ≥ S = 25 cm Thus, ACI requirements regarding reinforcement ratio, clear distance between bars and tie spacing are all satisfied.

The design load capacity ΦPn

 Pn  0.65(0.8) A g 0.85f c  g  f y  0.85f c Φ Pn  0.52A g 0.85f c '  ρg  f y  0.85f c '   Φ Pn  0.65(0.8)  400  250  0.85  28  0.0121 420  0.85  28   Φ Pn  1487 kN

28

Example 2 Design a short tied column to support a factored concentric load of 1000 kN, with one side of the cross section equals to 25 cm. f c  30MPa

f y  420MPa

Solution Assume first that  g  2% Ag 

Pu





0.65  0.8 0.85f c  g f y  0.85f c   

1000 103 Ag  0.65  0.8 0.85  30   0.02  420  0.85  30   29

A g  57594mm 2

Example 2 A g  57594mm 2 b  250mm h  230mm use column 25cm  25cm Determine adjusted steel ratio

  Pu 1  g    0.85f c      r  A g   f y  0.85f c   1000 103 1 g =   0.85(30)   =0.0134  0.65  0.8  250  250   420  0.85(30)  0.01<g <0.08 OK A s   g bh  0.0134(250)(250)  835mm 2 30

Use 614 (A s,sup = 924 mm 2 )

Example 2 Check spacing s

h  (No. of bars/2)  d b  2  cover   2 d stirrup 

 (No. of bars/2)  1 250  (6 / 2) 14   2  40   2(8)   56mm 3 1

max 1.5  d b  21mm , 40mm   56mm < 150mm

OK

Stirrup design Use  8 mm (for longitudinal bars with  14 mm <  30 mm)

s max

31

 16d b  16 14cm   224mm   min  48d stirrup  48 8   384mm smaller of b or d  250mm 

Use  8 mm @ 200 mm

 governs

6  14 mm

Example 2

 8 mm @ 200 mm

250 mm

250 mm

32

Example 3 Design a short, spirally reinforced column to support a service dead load of 800 kN and a service live load of 400 kN. f c  28MPa

f y  420MPa

Use  g 1%

Solution

Pu  1.20 PD 1.60 PL  1.2  800  1.6  400  1600kN Ag 

Pu





0.75  0.85 0.85f c  g f y  0.85f c   

1600 103 Ag  0.75  0.85 0.85  28   0.01 420  0.85  28   33

A g  90405mm 2

Example 3 360/N

A g  90405mm 2 for circular column D=

Ag

 4

=339mm 0.5D’

use column with D = 350 mm   A s  0.01  (3502 )  962mm 2 4 use 714 (A s,sup =1078 mm 2 ) Check spacing between longitudinal bars D’ =350-2(40)-2(8)-14=240 mm,

N=7

 360/N   51.43  S  D'sin   240  sin     104.1mm  2   2  Sc  104.1  14  90.1mm  1.5(14)=21mm 34

 40mm

OK

 360/N     2  0.5D’

0.5Sc = 0.5D’  sin(360/N/2)

Example 3 Design the needed spiral, try  8 Dc  350  2(40)  270 mm S

4 A sp



4  50

  π/4  350 2   28  0.45  270   1   2   π/4  270    420  S  36.3mm, taken as 35 mm (center  to  center)  Ag   fc'  0.45Dc   1      Ac   f y 

Sc  35  8/2  8/2  27 mm,i.e within ACI code limit (  25mm &  75mm) Use Φ 8mm spiral with a pitch of 35mm center  to  center.

35


Dr. Nader Okasha

Lecture 12 Part I Bond, development length, and splicing

Bond

2

Concept of Bond Stress Bond stresses are existent whenever the tensile stress or force in a reinforcing bar changes from point to point along the length of the bar in order to maintain equilibrium. Without bond stresses, the reinforcement will pull out of the concrete.

Concrete Reinforcing bar

PL/4

M

M+dM

dx Moment diagram

3

µavg

Concept of Bond Stress

 F  0.0 T2  T1  Fbond If this equation is not true (bond force Fbond is not strong enough), the bar will pull out A bar  f s2  f s1    avg A surface d 4

2 b

db

µ=Bond stress

 fs2  fs1    avg ( d b ) l

 μ avg

f - f d = s2 s1 b 4l

µavg= average bond stress

4

T1=fs1Ab

l

T2=fs2Ab fs2=fs1+∆fs

Mechanism of Bond Transfer A smooth bar embedded in concrete develops bond by adhesion between concrete &

reinforcement, and a small amount of friction.

This is different in a deformed bar. Once adhesion is lost at high bar stress and some slight movement between the reinforcement and the concrete occurs, bond is then provided by friction and bearing on the deformations of the bar. At much higher bar stress, bearing on the deformations of the bar will be the only component contributing to bond strength.

(a) Forces on bar

5

(b) Forces on concrete

Splitting cracks The radial component of the bearing force will cause circumferential stress on the concrete that may cause splitting that creates cracks.

6

Splitting cracks Splitting of concrete may occur along the bars, either in vertical planes as in figure (a) or in horizontal plane as in figure (b).

7

Splitting cracks The load at which splitting failure develops is a function of : •

The minimum distance from the bar to the surface of the concrete or to the next bar. The smaller the distance, the smaller is the splitting load.

•

The tensile strength of the concrete. The higher the tensile strength, the higher is the splitting resistance.

•

The average bond stress. The higher the average bond stress, the higher is the splitting resistance.

If the concrete cover and bar spacing are large compared to the bar diameter, a pullout failure can occur, where the bar and the ring of concrete between successive deformations pullout along a cylindrical failure surface joining the tips of the deformations. 8

Development Length

9

Development Length The bars found to be needed at a section from design calculations have to be embedded a certain distance into the concrete. This distance has to be equal or larger than the development length (ld).

10

Development Length The development length ld is that length of embedment necessary to develop the full tensile strength of the bar (on both sides of sections where fy stress is required), controlled by either pullout or splitting.

μ avg

f s2  f s1  d b  = 4l

f s2  f s1  f y

 ld  11

f y db 4  avg,u

, where  avg,u is the value  avg at bond failure

Development Length of Deformed Bars in Tension

The development length for deformed bars in tension is given by

ld 

αβ γ λ 1.1 f c  C  K tr   db fy

  

d b  300 mm,

 C  K tr  where    2.5  db  where, ld = development length db = nominal diameter of bar fy = specified yield strength of reinforcement C = spacing or cover dimension (see next slide) Ktr = transverse reinforcement index (see slide 12) 12

abgl = see next slides

ACI Eq. 12-1

ACI 12.2.3

Development Length of Deformed Bars in Tension [contd.] C is the smaller of

ACI 12.2.4

(a) the smallest distance measured from the center of the bar to the nearest concrete surface (b) one-half the center-to-center spacing of bars being developed. α is a bar location factor (a) Horizontal reinforcement so placed that more than 30 cm of fresh concrete is cast in the member below the development length or splice………………………………………………………………. (b) Other reinforcement…………………………………………………………..

1.3 1.0

β is a coating factor that reflects the adverse effects of epoxy coating (a) Epoxy-coated bars or wires with cover less than 3db or clear spacing less than 6db …………………………………………………. 1.5 (b) All other epoxy-coated bars or wires………………………………............…. 1.2 (c) Uncoated reinforcement……………………………………………………… 1.0 13

However, the product β α is not to be greater than 1.7.

Development Length of Deformed Bars in Tension [contd.]

ACI 12.2.4 γ is a reinforcement size factor that reflects better performance of the smaller diameter reinforcement (a) Φ20mm and smaller bars.……………………………………………….. 0.8 (b) Φ22mm and larger bars.…………..…….………………………………. 1.0 λ is a lightweight aggregate concrete factor that reflects lower tensile strength of lightweight concrete, & resulting reduction in splitting resistance. (a) When lightweight aggregate concrete is used…….……..……………… 0.8 (b) When normal weight concrete is used…………………..………………. 1.0

14

Development Length of Deformed Bars in Tension [contd.] Ktr is a transverse reinforcement index that represents the contribution of confining reinforcement

40 Atr K tr  sn

ACI Eq. 12-2

where: Atr = total cross sectional area of all transverse reinforcement within the spacing s, which crosses the potential plane of splitting along the reinforcement being developed within the development length s = maximum center-to-center spacing of transverse reinforcement within development length ld n = number of LONGITUDINAL bars being developed along the plane of splitting. Note: It is permitted to use Ktr= 0.0 as design simplification even if transverse reinforcement is present. 15

Atr

Potential plane of splitting

n=4

Development Length of Deformed Bars in Tension [contd.]

ACI 12.2.5

Excessive Reinforcement

Reduction in development length is allowed where As provided > As required. the reduction is given by

Reduction factor 

As required As provided

-Except as required for seismic design -Good practice to ignore this factor, since the use of the structure may change over time.

Simplified Expression for Development Length See ACI 12.2.2. This will not be used in this class

16

Example 1

60 cm

Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown

Φ10@20 4Φ20

40 cm Cover is 4 cm on all sides

Section A-A

17

Example 1 Determine the development length in tension required for the uncoated bottom bars as shown in the figure. If (a) Ktr is calculated (b) Ktr is assumed = 0.0 Use fc’ = 25 MPa normal weight concrete and fy = 420 MPa (c) Check if space is available for bar development in the beam shown

Φ10@20

(a) Ktr is calculated

4Φ20

α=1.0 for bars over concrete < 30 cm thick β=1.0 for uncoated bars α β =1.0 <1.7 OK


γ=0.8 for Φ20mm, λ=1.0 for normal weight concrete, C the smallest of

40+10+(20/2)=60 mm {[400-2(40)-2(10)-2(20/2)]/(3)}/(2)=46.7 mm

18

60 cm

Solution:

i.e., C is taken as 46.7 mm

Example 1 [contd.] K tr 

40A tr 40(2  79)   7.9 mm sn (200)(4)

i.e., use ld 

C  K tr  2.5 db

60 cm

C  K tr 46.7  7.9   2.73  2.5 db 20 Φ10@20 4Φ20

αβ γ λ 1.1 f c  C  K tr   db fy

d b  300 mm 40 cm   Cover is 4 cm on all sides   420   (1.0)(1.0)(0.8)(1.0)  ld    20  489 mm  300 mm OK  2 .5   1.1 25   b) Assuming K tr  0.0 C  K tr 467  0   2.33  2.5 db 20 19

OK

 420   (1.0)(1.0)(0.8)(1.0)  ld    20  524 mm  300 mm OK  2.33   1.1 25  

Example 1 [contd.]

60 cm

(c) Check if space is available for bar development

Φ10@20 4Φ20


Section A-A

Available length for bar development = 2000+ 150– 40 = 2110 mm > ld = 524 mm

OK 20

Development Length of Deformed Bars in Compression

ACI 12.3

Shorter development lengths are required for compression than for tension since flexural tension cracks are not present for bars in compression. In addition, there is some bearing of the ends of the bars on concrete. The development length ld for deformed bars in compression is computed as the product of the basic development length ldc and applicable modification factors, but ld is not to be less than 200 mm.

ld = ldc x applicable modification factors ≥ 200 mm. The basic development length ldb for deformed bars in compression is given as

   0.24 f y d b  ldc  max  ;0.043 f y d b  fc '     21

Development Length of Deformed Bars in Compression [contd.]

ACI 12.3

Applicable Modification Factors 1. Excessive reinforcement factor =As required / As provided 2. Spirals or Ties: the modification factor for reinforcement, enclosed with spiral reinforcement ≥ 6mm in diameter and ≤ 10 cm pitch or within Φ12mm ties spaced at ≤ 10 cm on center is given as 0.75

Development Lengths for Bundled Bars

ACI 12.4

Development length of individual bars within a bundle, in tension or compression, is taken as that for individual bar, increased 20% for three-bar bundle, and 33% for fourbar bundle. For determining the appropriate modification factors, a unit of bundled bars is treated as a single bar of a diameter derived from the equivalent total area of bars. 22

ldh

Critical section

Development of Standard Hooks in Tension Hooks are used to provide additional anchorage when there is insufficient length available to develop a bar. Development length ldh for deformed bars in tension terminating in a standard hook is computed as the product of the basic development length lhb and applicable

modification factors, but ldh is not to be less than 8db, nor less than 150 mm.

ldh = lhb x applicable modification factors ≥ 15 cm or 8db. The basic development length lhb for hooked bars is given as

lhb 

0.24  e f y l fc '

For lightweight aggregate concrete, l = 0.75. For epoxy-coated reinforcement,  e= 1.2. 23

Otherwise, l = 1.0,  e= 1.0

ACI 12.5.1

db

ACI 12.5.2

Development of Standard Hooks in Tension [contd.]

ACI 12.5.3

Applicable Modification Factors 1. Concrete cover: for db ≤ Φ36mm, side cover (normal to plane of hook) ≥ 65 mm, and for 90 degree hook, cover on bar extension beyond hook ≥ 50 mm, the modification factor is taken as 0.7. not less than 50 mm 65 mm

65 mm

24


ACI 12.5.3

Applicable Modification Factors 2. Excessive reinforcement factor =As required / As provided 3. Spirals or Ties: for db ≤ Φ36mm, hooks enclosed vertically or horizontally within ties or stirrups spaced along the full development length ldh not greater than 3db , where db is the diameter of the hooked bar, and the first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, the modification factor is taken as 0.8.

25

Development of Standard Hooks in Tension [contd.] Development length ldh is measured

ACI 7.1

90-degree hook

from the critical section of the bar to the out-side end or edge of the hooks. Either a 90 or a 180-degree hook, shown in the figure, may be used ldh

Development of reinforcement- General * Hooks are not considered effective in compression and may not be used as anchorage.

Part (a) 180-degree hook

ACI 12.5.5 * The values of f c ' used in this lecture shall not exceed 8.3 MPa. 26

ACI 12.1.2

4db ≥ 65mm

Φ10 through Φ25 Φ28 through Φ36 Φ44 through Φ56

ldh

Part (b)


ACI 12.5.4

Confinement of hooks For bars being developed by a standard hook at discontinuous ends of members with both side cover and top (or bottom) cover over hook less than 65 mm, the hooked bar shall be enclosed within ties or stirrups perpendicular to the bar being developed, spaced not greater than 3db along ldh. The first tie or stirrup shall enclose the bent portion of the hook, within 2db of the outside of the bend, where db is the diameter of the hooked bar.

27

Example 2

50 cm

Determine the development length or anchorage required for the epoxy-coated top bars of the beam shown in the figure. The beam frames into an exterior 80cm x 30cm column (the bars extend parallel to the 80 cm side). Show the details if: (a) If a 180-degree hook is used (b) If a 90-degree hook is used Use fc’ = 28 MPa and fy = 420 MPa 4Φ32 Φ12@15

Solution: α=1.3 for bars over concrete > 30 cm thick β=1.5 for coated bars (take the larger of 1.2 and 1.5 conservatively) α β =1.3x1.5 = 1.95 > 1.7 use 1.7 γ=1.0 for Φ32mm, C the smallest of

λ=1.0 for normal weight concrete 40+12+16=68 mm

{[400-2(40)-2(12)-32]/(3)}/(2)=44 mm i.e., C is taken as 44 mm 28

40 cm

Example 2 [contd.] 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’<21 MPa When bars of different size are lap-spliced in compression, splice length shall be the larger of either development length of the larger bar, or splice length of the smaller bar.

ACI 12.16.2 ACI 12.15.3 37

Example 4 Design a compression lap splice for a tied column whose cross section is shown in the figure when: (a) Φ16 mm bars are used on both sides of the splice. (b) Φ 16 mm bars are lap spliced with Φ 18 mm bars. Use fc’ = 30 MPa and fy = 420 MPa

Solution: (a) For bars of same Φ16 mm diameter

Splice length in compression and for fy =420 MPa is equal to 0.071 fy db = 0.071 (420)(16) = 477 mm >300 mm taken as 480 mm

38

Example 4 [contd.] (b) For bars of different diameters

The development length of the larger bar ldc = ldb x applicable 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


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


4


5


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


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.


ACI 12.11.3

16


17


18


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


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


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


9


10

Typical details for one way solid slabs Bent-up bars

11

Typical details for beams Straight bars

12


13


14

Typical details for columns

15

Typical details for columns

16

17

18

19


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


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 cA1 

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 cA1  ; 2  0.85f cA1     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


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


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


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


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


320 24.49 x 2.45

A s,min  0.0018  550  2450  2430 mm 2 A s,req  3500 mm 2  2314mm 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 214 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 cA1   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 cA1  ; 2  0.85f cA1    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 416, As,sup = 804 mm2

47

Example 9- Check for anchorage of the reinforcement Bottom longitudinal reinforcement (Φ14mm) α=1.0 for bottom bars,

β=1.0 for uncoated bars 1.4

α β =1.0 <1.7 OK γ=0.8 for Φ14mm,

7.5+0.7=8.3 cm

245

C the smallest of

λ=1.0 for normal weight concrete

[245-2(7.5)-1.4]/(22)/(2)=5.2 cm i.e., C is taken as 5.2 cm

C  K tr 5.2  0 C  K tr   3.7  2.5  i.e.,use  2.5 db 1.4 db

320

 420   (1.0)(1.0)(0.8)(1.0)  ld   1.4  34 cm  2.5   1.1 25   Available length in long direction =140-7.5=132.5 > 34 cm 48

Example 9- Check for anchorage of the reinforcement Bottom reinforcement in short direction (Φ14mm) α=1.0 for bottom bars,

β=1.0 for uncoated bars

α β =1.0 <1.7 OK 7.5+0.7=8.3 cm [320-2(7.5)-1.4]/(19)/(2)=8 cm

1.025

C the smallest of

λ=1.0 for normal weight concrete 245

γ=0.8 for Φ14mm,

i.e., C is taken as 5.2 cm

320

C  K tr 8  0 C  K tr   5.7  2.5  i.e.,use  2.5 db 1.4 db  420   (1.0)(1.0)(0.8)(1.0)  ld   1.4  34 cm  2.5   1.1 25   Available length in short direction =102.5-7.5=95 > 34 cm 49

Example 9- Check for anchorage of the reinforcement Dowel reinforcement (Φ16mm):

 0.24f y d b 0.24  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


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) =0.340.075  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 =30.3 =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 1106 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 112 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) = 0.221.025/0.88 =6kN/m D.L (step) = (1/2)  0.161.0  25=2 kN/m D.L (covering material) = 21.0=2 kN/m D.L (flight) = 10 kN/m D.L (landing) = 8 kN/m L.L =3 1.0=3 kN/m

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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 812

(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

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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

Reinforced Concrete Design I

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