Training Course onTraining Course onCivil/Structural Codes and Inspectionp
BMA Engineering, Inc.
1BMA Engineering, Inc. – 5000
Overall OutlineOverall Outline
1000. Introduction
2000. Federal Regulations, Guides, and Reports
3000. Site Investigation
4000. Loads, Load Factors, and Load Combinations
5000. Concrete Structures and Construction
6000. Steel Structures and Construction
7000. General Construction Methods
8000. Exams and Course Evaluation
9000. References and Sources
2BMA Engineering, Inc. – 5000
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Objective and ScopeObjective and Scope– Provide introductory and intermediate level review of analysis and design of concretereview of analysis and design of concrete structures including concrete structures in nuclear power plants
– Present and discuss• Concrete Materials
• Reinforced Concrete: Beams, columns & Footings , g
• Strut & Tie Model
• Two‐way Slabsy
BMA Engineering, Inc. – 5000 3
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Objective and ScopeObjective and Scope
– Present and discuss (Cont’d)• Prestressed Concrete
• Reinforcing Bars, Reinforcing Details & Tolerances
• Concrete Containments and Modular Construction
• Mass Concrete
• Durability of NPP Concrete Structures
• Introduction to NDE of Concrete Structures
• Introduction to Masonry
BMA Engineering, Inc. – 5000 4
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Reinforced Concrete5100
Reinforced Concrete
• Prestressed Concrete5200
Prestressed Concrete
• Reinforcing Bars Reinforcing Details & Tolerances5300
• Reinforcing Bars, Reinforcing Details & Tolerances
• Concrete Containments, Modular Construction& 5400
,Mass Concrete
5500 • Durability, NDE & Masonry
BMA Engineering, Inc. – 5000 5
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, andApplicable ACI Codes and Specifications, and applicable NRC Publications – Applicable ACI Codes and Specifications– Applicable ACI Codes and Specifications
ACI 318 Building Code Requirements for Structural
ConcreteConcrete
ACI 349 Code Requirements for Nuclear Safety‐Related
Concrete Structures and CommentaryConcrete Structures and Commentary
ACI 359 (ASME B&PV Code, Section III, Division 2 for
C t R t V l d C t i t )Concrete Reactor Vessels and Containments)
Code for Concrete Containments
BMA Engineering, Inc. – 5000 6
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, andApplicable ACI Codes and Specifications, and applicable NRC Publications – Applicable ACI Codes and Specifications (Cont’d)– Applicable ACI Codes and Specifications (Cont d)
ACI 301 Specification for Structural Concrete
ACI 116R Cement and Concrete TerminologyACI 116R Cement and Concrete Terminology
ACI 121R Quality Management System for Concrete
ConstructionConstruction
ACI 207.1 Mass Concrete
ACI 207 2R Eff t f R t i t V l Ch dACI 207.2R Effect of Restraint, Volume Change, and
Reinforcement on Cracking of Mass Concrete
BMA Engineering, Inc. – 5000 7
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, andApplicable ACI Codes and Specifications, and applicable NRC Publications – Applicable ACI Codes and Specifications (Cont’d)– Applicable ACI Codes and Specifications (Cont d)
ACI 207.4R Cooling and Insulating Systems for Mass
ConcreteConcrete
ACI 304R Guide for Measuring, Mixing, Transporting, and
Placing ConcretePlacing Concrete
ACI 305R Hot Weather Concreting
ACI 306R C ld W th C tiACI 306R Cold Weather Concreting
ACI 309‐Guide for Consolidation of Concrete
BMA Engineering, Inc. – 5000 8
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, andApplicable ACI Codes and Specifications, and applicable NRC Publications – Applicable ACI Codes and Specifications (Cont’d)– Applicable ACI Codes and Specifications (Cont d)
ACI 311.5 Guide for Concrete Plant Inspection and Field
Testing of Ready Mixed ConcreteTesting of Ready‐Mixed Concrete
ACI 347 Guide to Formwork for Concrete
ACI 439 3R Types of Mechanical Splices for ReinforcingACI 439.3R Types of Mechanical Splices for Reinforcing
Bars
BMA Engineering, Inc. – 5000 9
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, andApplicable ACI Codes and Specifications, and applicable NRC Publications – Applicable NRC Publications– Applicable NRC Publications
NUREG/CR‐6927 ‐ Primer on Durability of Nuclear Power
Plant Reinforced Concrete StructuresPlant Reinforced Concrete Structures –
A Review of Pertinent Factors
NUREG/CR 6486 Assessment of Modular ConstructionNUREG/CR‐6486, Assessment of Modular Construction
for Safety‐Related Structures at
Ad d N l P Pl tAdvanced Nuclear Power Plants
BMA Engineering, Inc. – 5000 10
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Applicable ACI Codes and Specifications, and pp p ,applicable NRC Publications – Applicable NRC PublicationsApplicable NRC Publications
NUREG‐0800, Chapter 3 ‐ Standard Review Plan for the
Review of Safety Analysis Reports fory y p
Nuclear Power Plants: LWR Edition —
Design of Structures, Components,
Equipment, and Systems
RG 1.142 ‐ Safety‐Related Concrete Structures for
Nuclear Power Plants (Other Than Reactor Vessels andNuclear Power Plants (Other Than Reactor Vessels and
Containments) [Revision 2, November 2001];
RG 1.199 ‐ Anchoring Components and Structural Supports in g p pp
Concrete [November 2003]
BMA Engineering, Inc. – 5000 11
5000 Concrete Structures and Construction5000. Concrete Structures and Construction
• Reinforced Concrete5100
Reinforced Concrete
• Prestressed Concrete5200
Prestressed Concrete
• Reinforcing Bars Reinforcing Details & Tolerances5300
• Reinforcing Bars, Reinforcing Details & Tolerances
• Concrete Containments, Modular Construction& 5400
,Mass Concrete
5500 • Durability, NDE & Masonry
BMA Engineering, Inc. – 5000 12
5100 Reinforced Concrete5100. Reinforced Concrete
• Objective and ScopeObjective and Scope– Provide introductory and intermediate level review of analysis and design of reinforcedreview of analysis and design of reinforced concrete structures
– Present and discuss• Materials and Design Methods
• Moment & Shear Design of Beams
• Footing & Column Design
• Introduce Development of Reinforcement
• Strut & Tie Model, and Two‐way Slabsy
BMA Engineering, Inc. – 5000 13
5100 Reinforced Concrete5100. Reinforced Concrete
5110 • 5110 ‐ Introduction: Materials & Design Methods
5120 & 5130
• 5120 ‐Moment Design of Beams• 5130 ‐ Shear Design of Beams5130 Shear Design of Beams
5140 & 5150 • 5140 ‐ Footing Design
• 5150 ‐ Column Design
5160 & • 5160 ‐ Development & Splices of Reinforcement5170
p p• 5170 ‐ Strut and Tie Model
5180
BMA Engineering, Inc. – 5000 14
5180 • 5180 ‐ Two‐way Slabs
5100 Reinforced Concrete5100. Reinforced Concrete
5110 • 5110 ‐ Introduction: Materials & Design Methods
5120 & 5130
• 5120 ‐Moment Design of Beams• 5130 ‐ Shear Design of Beams5130 Shear Design of Beams
5140 & 5150 • 5140 ‐ Footing Design
• 5150 ‐ Column Design
5160 & • 5160 ‐ Development & Splices of Reinforcement5170
p p• 5170 ‐ Strut and Tie Model
5180
BMA Engineering, Inc. – 5000 15
5180 • 5180 ‐ Two‐way Slabs
5110 ‐ Introduction: Materials & Design Methods
• Concrete consists of primarily a proportioned mixture of fine and coarse aggregates (sand, f gg g ( ,gravel, crushed rock, and/or other materials), cement admixtures and watercement, admixtures and water
• Hydration, the reaction between cement and water, chemically binds the aggregatewater, chemically binds the aggregate particles into a solid mass
16BMA Engineering, Inc. – 5000
Design of Concrete Structures IntroductionDesign of Concrete Structures ‐ Introduction
• Historical Background– Concrete use dates back to ancient civilization, as ,far back as 12,000 B.C.
– Around the middle of the 19th century, steel bars were placed in concrete to overcome it’s poorwere placed in concrete to overcome it s poor tensile strength
17BMA Engineering, Inc. – 5000
Design of Concrete Structures IntroductionDesign of Concrete Structures ‐ Introduction
• Typical Reinforced Concrete Structure
BMA Engineering, Inc. – 5000 18
Design of Concrete Structures‐Advantages & Disadvantages
• Advantages– High compressive strengthg p g
– Good fire and moisture resistance
Low maintenance and long service life– Low maintenance and long service life • Strength increases with time in poorly designed structures
f– Easy to cast in a variety of shapes
– Local materials may be used
– Skill requirements for labor are low
19BMA Engineering, Inc. – 5000
Design of Concrete Structures‐Advantages & Disadvantages
• Disadvantages– Low tensile strength g
– Possible requirement of expensive forms
Low strength per unit weight disadvantage for– Low strength per unit weight, disadvantage for long span structures.
li l i diffi l d h– Quality control is difficult compared to other materials
20BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials
• Concrete– Plain concrete = cement + fine aggregate + coarse gg gaggregate + water + (admixtures)
• Aggregate• Aggregate– Composes 70 to 75% of concrete by volume.
– Aggregates are usually graded by size, a proper mix will have specified percentages of both fine and coarse aggregates
21BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials – Aggregate
• Aggregate (cont’d)– Fine Aggregate (sand)gg g ( )
• Any material passing through a number 4 sieve (four openings per linear inch. i.e. less than 3/16 inch
– Course Aggregate (sand)• Any material larger than 3/16 inchAny material larger than 3/16 inch
BMA Engineering, Inc. – 5000 22
Design of Concrete StructuresConcrete Materials ‐ Admixtures
• Admixtures– May be added to concrete mix immediately before y yor during the mixing to modify the properties of the concrete to make it better serve its intended use
• Fly ash (by product of exhaust fumes of coal fired Fly ash (by product of exhaust fumes of coal fired power stations): Like silica fume used to replace part of the Portland cement. Tends to increase the strength of concrete at ages over 28 days
23BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Admixtures
• Silica Fume– used in substitution for a part of the Portland pcement: very finely divided solid ‐micro‐silica comes off in fumes of electric furnaces that produce ferro‐silicon or silicon metal
• Provides a more impermeable concrete to resist Provides a more impermeable concrete to resist chloride intrusion into concrete exposed to deicing chemicals
• To achieve high strength and greatly improved durability in high performance concrete. Tends to i h l f 3 28 dincrease strength at early ages, from 3 to 28 days.
24BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Admixtures
• Accelerating admixtures– For example, calcium chloride is used to p ,accelerate the rate of strength development at early ages. (Restrictions see ACI 3.6.3). Advantages y g ( ) gare reduced times required for curing and protection of the concrete and earlier removal of pforms. Particularly useful in cold climates. Various soluble salts and organic compounds are available g pas accelerating admixtures
25BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Admixtures
• Retarding Admixtures– Are used to slow the setting and to mitigate g gtemperature increases. They may consist of various acids or sugars or sugar derivatives. g gParticularly useful for large pours. In addition to mitigating the temperature these admixtures may g g p yprolong the plasticity of the concrete, enabling better blending or bonding of successive poursg g p
26BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Admixtures
• Super plasticizers– Enable contractors to use less water to produce phigh strength workable concrete. A higher slump is obtained using superplasticizers while at the g p psame time reducing the water cement ratio
• Waterproofing admixtures• Waterproofing admixtures – Are typically applied to hardened concrete but
l b dd d h i Thmay also be added to the concrete mix. These may consist of some type of soap or petroleum
d f l h l l iproducts, for example asphalt emulsions
27BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Cement
• Cement– Material with adhesive and cohesive properties, p p ,enabling it to bond inert aggregates into a solid mass
• Hydraulic CementC h d h d i h f– Cements that can set and harden in the presence of water
– Most common hydraulic cement is Portland Cement
– ACI 318‐02 also recognizes expansive cement
28
g pconforming to ASTM Specification C845
BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Cement
• Portland Cement– Consist mainly of calcium and aluminum silicatesy
– Made from limestone (calcium carbonate) and clayMade from limestone (calcium carbonate) and clay (or shale) which are ground, blended and fused in a kiln and then crushed into powdered formkiln and then crushed into powdered form
29BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Cement
• Types of Portland Cement (ASTM C150)– Type I: Ordinary Portland Cement attains yp yadequate strength in about 14 days (forms can be removed), design strength reached in 28 days), g g y
– Type II: For precaution against moderate sulfate attack (ACI Table 4 3 1) Also when moderate heatattack. (ACI Table 4.3.1). Also when moderate heat of hydration is desired
Type III: When high early strength is required Has– Type III: When high early strength is required. Has considerably high heat of hydration
30BMA Engineering, Inc. – 5000
Design of Concrete StructuresConcrete Materials ‐ Cement
• Types of Portland Cement (Cont’d)– Type IV: When low heat of hydration is required yp y q(for mass concrete such as dams)
– Type V: For precaution against severe sulfateType V: For precaution against severe sulfate attack. (ACI Table 4.3.1)
There are also several categories of blended– There are also several categories of blended hydraulic cements (ASTM C595)
D i ti IA IIA IIA d f i t i i• Designations IA, IIA or IIA are used for air entraining Portland cement for improved durability against frost action as well as better workabilityaction, as well as better workability
31BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– The main mechanical strength property of concrete
– Usually measured by a compression test on a p6 in. diameter x 12 in.long cylinder at a specified g y ploading rate
– Two cylinder tests needed– Two cylinder tests needed for every 150 cubic yards placed
32BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– In ACI Specification, the acceptability of concrete is p , p ybased upon the average of two cylinder strengths
– (f ') ’d = average from two cylinders(fc )req’d = average from two cylinders
33BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength
– Compression of the concrete makesCompression of the concrete makes the concrete crack sideways, while pieces break from the side as it ispieces break from the side as it is more and more compressed.
34BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– Depending on water/cement ratio, concrete will p g / ,have different strengths
35BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– A specific age is specified (28 days = 4weeks) p g p ( y )because concrete gets stronger with age. In design, strength above 28 day fc’ is neglectedg , g y c g
36BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– A quick on site estimation of water/cement ratio q /for fresh concrete is given by the slump test.
– ASTM specifies the placing and tamping of freshASTM specifies the placing and tamping of fresh concrete in a truncated cone open at top.
37BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Compressive Strength– The height difference between fresh concrete with gand without metal cone removed is the slump. Slumps of 2" to 6" are usual, with slumps of 3" to p , p4" common for building construction.
38BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
BMA Engineering, Inc. – 5000 39
Design of Concrete StructuresMechanical Properties
• Tensile Strength– Tensile strength of concrete is more difficult to gmeasure and more variable than compressive strength. One method is the split cylinder test. g p y
40BMA Engineering, Inc. – 5000
Design of Concrete StructuresMechanical Properties
• Tensile Strength– Another method is the modulus of rupture test p(tensile strength in flexure)
41BMA Engineering, Inc. – 5000
Design of Concrete StructuresElastic Behavior
• At moderate loads (service loads) the stress in the concrete is linearly proportional to the strain.
• fc’ = 4 ksi
• fc ≤ 2 ksi
• εc ≤ 0.005 in/in
• εs ≤ 0.005 in/ins
• fs ≤ 15 ksi < fy
BMA Engineering, Inc. – 5000 42
Design of Concrete StructuresElastic Behavior
• For linear‐elastic concrete it is important to satisfy the requirements of y q
1) ilib i1) Equilibrium
2) Compatibility (εc= εs)) p y ( c s)
3) Constitutive (σ = Eε, Hooke’s Law)
BMA Engineering, Inc. – 5000 43
Design of Concrete StructuresStrength Design Method
• Members are designed to resist design load
• Service loads are increased by load factors toService loads are increased by load factors to obtain ultimate (design) loads.
D i St th R i d St th (U)– Design Strength > Required Strength (U)• Required Strength= Load factors x Service Load Effects
• Service Load= load specified by general building code
• Design Strength= (φ) x Nominal Strength– Where, φ = Strength Reduction Factor
– Nominal strength= strength based on code equations
44BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method
• Flow chart to obtain Required Strength, U
BMA Engineering, Inc. – 5000 45
Design of Concrete StructuresStrength Design Method
• Failure results when load exceeds strength– Load and strength are assumed to be random variables
• Members are designed to resist design load with an g gadequate margin of safety against failure
• Therefore, we modify the load and the strength using factorsfactors
BMA Engineering, Inc. – 5000 46
Design of Concrete StructuresStrength Design Method
• The Strength Reduction Factor, φ reflects– Variability in material properties and dimensionsy p p
– Imprecise construction • Reinforcement may not be in correct position• Reinforcement may not be in correct position
– Inaccuracies in the design equations
f– Importance of the member in the structure• Beam vs. column
47BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method
• Strength Reduction factors for Load Combinations
48BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method
• Figure provides φ’s for transition between compression‐ and tension‐controlled sectionsp
49BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method
• Compression‐Controlled Sections– Columns with small moments
– Failure due to concrete crushing
– εt ≤ 0.002
• Tension Controlled• Tension‐Controlled
Sections– Beams or columns with large moments
– Failure due to yielding of steel
50
– εt ≥ 0.005 (εt = net tensile strain in extreme tension steel)
BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method
• Required strength (U)– Expressed in terms of factored loadsp
• D = Dead Load
• L = Live Load
• W = Wind Load
• E = Earthquake Load
R R f L d• R = Roof Load
• Lr = Live Roof Load
• H = Soil Load
• T = Temperature Effect
• S = Snow Load
51BMA Engineering, Inc. – 5000
Design of Concrete StructuresStrength Design Method – Deflection Check
• Deflection is an essential consideration in designing beams by the Strength Design Method– Deflection Control Based On:
• Nonstructural Elements• Nonstructural Elements
• Equipment Alignment Considerations
P l R i A id “B ” B• Personnel Response, i.e., Avoid “Bouncy” Beams
• Both Immediate and Long‐Term Conditions Need to be Considered
BMA Engineering, Inc. – 5000 52
Design of Concrete StructuresStrength Design Method – Deflection Check (cont’d)
• Limitations Expressed as Fraction of Beam SpanACI Table 9.5b
Type of Member Deflection to be consideredDeflection Limitation
Flat roofs not supporting or attached to nonstructural elements likely to be damaged by large deflections.
Immediate deflection due to live load L.
l i h d
l*180
Floors not supporting or attached to nonstructural elements likely to be damaged by large deflections.
Immediate deflection due to live load L.
l360
• * Limit not intended to safeguard against ponding. Ponding should be checked by
suitable calculations of deflection…considering long‐time effects of all sustained
loads camber construction tolerances and reliability of provisions for drainageloads, camber, construction tolerances, and reliability of provisions for drainage.
BMA Engineering, Inc. – 5000 53
Design of Concrete StructuresStrength Design Method – Deflection Check (cont’d)
• ACI Table 9.5b (cont’d)Deflection
Type of Member Deflection to be consideredDeflection Limitation
Roof or floor construction supporting or attached to nonstructural elements likely to be damaged by large deflections
That part of the total deflection occurring after attachment of
l l ( f h l
l480be damaged by large deflections.
gnonstructural elements (sum of the long-time deflection due to all.sustained loads and the immediate deflection due to any additional live load).
Roof or floor construction supporting or attached to nonstructural elements not likely to be damaged by large deflections
480
l **
240 • **But not greater than the tolerance provided for non‐structural elements. Limits
may be exceeded if camber is provided so that total deflection minus camber does
)likely to be damaged by large deflections. 240
may be exceeded if camber is provided so that total deflection minus camber does not exceed limit.
BMA Engineering, Inc. – 5000 54
Design of Concrete StructuresAlternate Design Method – Service Load Design
• Design for service loads using the mechanics of elastic members so that stresses do not exceed some pre‐designated allowable valuesdesignated allowable values
• All relevant provisions of the code such as distribution of flexural reinforcement and slenderness effects apply also toflexural reinforcement and slenderness effects apply also to members designed by the Alternate Design Method
• Will generally result in designs that are more conservativeg y g
• Permitted for nonprestressed members (serviceability control provisions based on linear stress‐strain assumptions for prestressed members, given ACI Code, Ch.18)
55BMA Engineering, Inc. – 5000
Design of Concrete StructuresUniform Design Provisions
• An alternate method of design for reinforced and prestressed concrete flexural members, was given in appendix B but now is the main body of the code
• Similar to strength method (factored loads, strength reduction)
• Tries to unify design of beam & columns & reinforced & prestressed concrete
56BMA Engineering, Inc. – 5000
5100 Reinforced Concrete5100. Reinforced Concrete
5110 • 5110 ‐ Introduction: Materials & Design Methods
5120 & 5130
• 5120 ‐Moment Design of Beams• 5130 ‐ Shear Design of Beams5130 Shear Design of Beams
5140 & 5150 • 5140 ‐ Footing Design
• 5150 ‐ Column Design
5160 & • 5160 ‐ Development & Splices of Reinforcement5170
p p• 5170 ‐ Strut and Tie Model
5180
BMA Engineering, Inc. – 5000 57
5180 • 5180 ‐ Two‐way Slabs
5120 ‐Moment Design of BeamsBeam Flexure Assumptions & Principles
• Consider a simply supported R/C beam subject to pure moment in the midspan regionj p p g
58BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Examine behavior at cross section b‐b of beam
BMA Engineering, Inc. – 5000 59
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Behavior of Beams under Load– Concrete beams with tensile reinforcements under gradually increasing load go through three distinct stages before failure. The three stages g ginclude:
I. Uncracked (small loads)
II. Cracked with elastic stresses (moderate loads)
III. Ultimate strength stage
60BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Figure 1 – Uncracked Beam at Small Loads• fs = n times corresponding concrete stress; were modular ratio n = Es/Ec
61BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Figure 2 ‐ Cracked beam with Elastic Stresses at Moderate Loads
62BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Figure 3 – Ultimate Strength at Failure
63BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Cracking Moment– Approximate bending stresses can be obtained by pp g yneglecting the steel reinforcement (usually 2% or less) and based on gross properties of the cross ) g p psection.
• Therefore,Therefore,• f = bending stress;
• M = bending moment;Myf =• Y = distance from neutral axis; and
• Ig = gross moment of inertiag
fΙ
64BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Cracking Moment as given in ACI code,• ACI Eqn. 9‐9
• fr = bending stress (modulus of rupture of concrete)
• MCR = cracking moment;f grΙ=Μ • yt = distance from neutral axis to extreme tension fiber
• Ig= gross moment of inertiat
CR yΜ
– For normal weight concrete,'57 ff
• fc’ = specified compressive strength of concrete, psi
'5.7 cffr =
65BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Design Example:– Determine the cracking moment for the following g gsection assuming = 4000 psi and the modulus of rupture '5.7 cffr =p
66BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Design Example:– Determine the cracking moment for the following g gsection assuming = 4000 psi and the modulus of rupture '5.7 cffr =p
67BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Design Example:– Calculate the uniform load (in addition to the beam weight) which will
cause the section to begin to crack Assume 28 ft simple span f ’ 4000cause the section to begin to crack. Assume 28 ft simple span, fc = 4000 psi, fr = 7.5(fc’)1/2, beam weight = 150 lb/ft3
68BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Cracked Concrete – Elastic Stresses
– Even when concrete is cracked in tension zone, perfect bond is assumed between concrete and steel. Therefore, concrete and steel at similar distances from the neutral axis will undergo the same strain.
– The section can be handled by the usual methods for elastic h b f h l l d bhomogenous beams, if the steel area is replaced by a transformed area (an equivalent area of fictitious concrete)
69BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Transformed Area = nAs• n = Es/Ec
• n = modular ratio
• Es = modulus of elasticity of the steel
• E = modulus of elasticity of the concrete• Ec = modulus of elasticity of the concrete
Fi E l f T f d S iFigure ‐ Example of Transformed Section
70BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Stress Variation on a Transformed Section
– The dashed line indicates discontinuity in the tensile stress sinceThe dashed line indicates discontinuity in the tensile stress since the concrete is assumed to be cracked and unable to resist tension.
Th t f th ti b d t i d ft l ti th– The stresses for the section can be determined after locating the neutral axis and finding the moment of inertia of the transformed section.
71BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Design Example:– For the beam section shown assume that the section has cracked.
Using the transformed area method calculate the flexural stresses forUsing the transformed area method calculate the flexural stresses for the given moment.
72BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Flexural Strength of Rectangular Beams– To simplify analysis and design, ACI has adopted a p y y g pfictitious but equivalent stress distribution proposed by Whitney in 1942
– Other shapes may be used if they are shown to be in substantial agreement with comprehensive test resultssubstantial agreement with comprehensive test results (ACI 10.2.7)
73BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Flexural Strength of Rectangular Beams– For more general cross‐sections, a simpler g , papproach is to visualize the distribution of concrete stress at ultimate load as a uniform distribution. ACI 10.2.7.
74BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Flexural Strength of Rectangular Beams
– Depth of compression block determined from a = βc
Where,
Depth of compression block determined from a βc
,• a = depth of compression block
• c = distance from N.A. to extreme compression fiberp
• β1 = factor relating a to c
75BMA Engineering, Inc. – 5000
– Figure - Balanced Strain Condition in Flexure
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Flexural Strength of Rectangular Beams– β1 = factor relating a to cβ1 g
• a = β1c
2 00 i f’ 000 i β 0 8• For 2500 psi ≤ f’c ≤ 4000 psi, β1=0.85
• For f’ > 4000 psi
6501000
40000508501 .)'(.. ≥−
−=cfβ
• For f’c > 4000 psi,
76BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Flexural Strength of Rectangular Beams– Nominal or Theoretical Strength & Design Strengthg g g
• Usable flexural strength of a member, φMn, must be equal to or greater than the calculated factored moment, Mu, caused by the factored loads
φMn ≥ Mu
77BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• The following steps may be used for analysis and design of rectangular beams
1) Compute the Compression stress resultant
C = 0.85f’cab
2) Compute the Tension Stress resultant, T = Asfy3) Solving for “a” using equilibrium
C = T
0.85f’cab = Asfy → ,850850 ''
yys
fdf
bffA
aρ
==
Where, = percentage of Tensile Steel
85.085.0 cc fbf
bdAs=ρ
78BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
4) Determine the distance between C & T (for rectangular sections = d ‐ a/2)
5) Determine the nominal or theoretical moment Mn (C or T times distance between C & T)times distance between C & T)
⎟⎞
⎜⎛ −=⎟
⎞⎜⎛ −=
adfAadTM
6) For the analysis the usable flexural strength may be
⎟⎠
⎜⎝
=⎟⎠
⎜⎝
=22
dfAdTM ysn
determined from,
⎟⎠⎞
⎜⎝⎛ −φ=φ
2adfAM ysn
79BMA Engineering, Inc. – 5000
⎠⎝ 2y
Design of Concrete Structures Beam Flexure Assumptions & Principles
7) For design the reinforcement ratio ρ may be determined as follows: substitute into the above expression the value previously obtained for a (i e ) and equate to M andpreviously obtained for a (i.e., ) and equate to Mu, and obtain,
⎟⎞
⎜⎛ yf
dfAMMρ
φφ 11
Replacing A with and letting we can solve this expression
⎟⎟⎠
⎜⎜⎝−== '. c
yysun fdfAMM
ρφφ
711
Replacing As with and letting , we can solve this expression for (the percentage of steel required for a particular beam):
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=ρ '
'
85.021185.0
c
n
y
c
fR
ff
80BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Design Example:– For the beam section show, determine the nominal moment Mn
81BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Minimum Steel– Sometimes a beam with very small loads may be y ysized large for architectural reasons. Such beams will theoretically require very little reinforcementy q y
– For such beams the capacity of the plain concrete beam (cracking moment) can exceed the ultimatebeam (cracking moment) can exceed the ultimate moment capacity of the section
Failure of these type of beam will occur– Failure of these type of beam will occur immediately and without warning when a crack occursoccurs
BMA Engineering, Inc. – 5000 82
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Minimum Steel– Immediate failure is undesirable and therefore the code specifies minimum amount of tensile reinforcement to avoid these type of failures. yp
• ACI 10.5.1 specifies,
f '3 db200or less than
where b = web width of beam
dbffA wy
cs min,
3=
y
w
fdb200
where, bw = web width of beam
BMA Engineering, Inc. – 5000 83
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Minimum Steel
– The value was obtained by calculating the cracking moment of a plain concrete section and
⎟⎟⎠
⎞⎜⎜⎝
⎛
y
w
fdb200
cracking moment of a plain concrete section and equating it to the strength of a reinforced concrete section of the same size applying aconcrete section of the same size, applying a safety factor of 2.5 and solving for the steel requiredrequired.
BMA Engineering, Inc. – 5000 84
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Minimum Steel– For higher values fc’ the value from may not ⎟
⎟⎠
⎞⎜⎜⎝
⎛
y
w
fdb200
⎞⎛g c y
be sufficient. Thus, is also required to be met and will control when fc’ > 4440 psi
⎠⎝ y
dbff
wy
c
⎟⎟⎠
⎞⎜⎜⎝
⎛ '3
c p
ACI Eqn (10 3) for minimum amount of flexural– ACI Eqn. (10‐3) for minimum amount of flexural reinforcing can be written as a percentage as follows
BMA Engineering, Inc. – 5000 85
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Balanced Steel Reinforcement– At balanced condition, the steel simultaneously yields as the concrete reaches it’s theoretical failure strain of 0.003
– The percentage of steel required for a balanced design, ρbmay be determined as follows
Figure ‐ Balanced Condition
BMA Engineering, Inc. – 5000 86
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Balanced Steel Reinforcement– Strain compatibility: p y
• The neutral axis may be located using similar triangles and noting that Es = 29x106 psi
( )61029003.000300.0
)(00300.000300.0
×+=
+=
ysy fEfdc
– This may be simplified and rearranged as,
d00087 df
cy+
=000,87
000,87
BMA Engineering, Inc. – 5000 87
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Balanced Steel Reinforcement– An expression was derived for “a” by equating the values of C and T. This value can be converted to “c” by dividing it by β1
ydfρydfa ρ
Therefore, '85.0 c
y
fa
ρ=
'11 85.0 c
y
fdfacβ
ρβ
==
– Setting the two expressions for “c” equal
dfρ 00087 dff
dfyc
y
+=
β
ρ
000,87000,87
85.0 '1
BMA Engineering, Inc. – 5000 88
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Balanced Steel Reinforcement
– We obtain the balanced reinforcement ratio ρb
⎟⎞
⎜⎛
⎟⎞
⎜⎛ f 00087850 'β
N
⎟⎟⎠
⎞⎜⎜⎝
⎛
+⎟⎟⎠
⎞⎜⎜⎝
⎛=
yy
cb ff
f00087
00087850 1
,,*. βρ
Note,
BMA Engineering, Inc. – 5000 89
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Failure Modes– The location of the N.A. will vary based on the yamount of tension steel since the location of the stress block is just deep enough to satisfy equilibrium
– More steel than the balanced condition would mean– More steel than the balanced condition would mean brittle failure without steel yielding, therefore to avoid brittle failure ACI code has provisionsavoid brittle failure ACI code has provisions
90BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Failure Modes– Less steel than required for the balanced condition qleads to ductile failure
• (Refer to figure on next slide)( g )
– Code limits the amount of reinforcement to ensureCode limits the amount of reinforcement to ensure ductile failure by requiring that εt ≥ 0.004 (ACI 10.3.5)
91BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Flexure Assumptions & Principles
• Strain Distribution and Failure Modes
92BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Equations
• Singly Reinforced Rectangular Sections
1) Using force equilibrium1) Using force equilibrium
C = T'
yysc bdffAabf ρ=='85.0
dffA ρ,
85.085.0 ''c
y
c
ys
fdf
bffA
aρ
==
93BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Equations
• Singly Reinforced Rectangular Sections (cont’d)
2) Using moment equilibrium determine the nominal moment Mn (C or T times distance between C & T)n
or⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
22adfAadTM ysn
where,
We design flexural members so that,
94BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Equations
• Singly Reinforced Rectangular Sections (cont’d)
– Divide both sides of Eq. (2) by db2 to obtain, Rn, a nominal strength coefficient of resistance:
⎟⎟⎠
⎞⎜⎜⎝
⎛−==
c
yy
nn f
ff
bdMR
'85.05.0
12
ρρ
95BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Equations
• Singly Reinforced Rectangular Sections (cont’d)
⎟⎞
⎜⎛ yn f
fM
R5.0
1ρ
– Above equation can be used to determine the steel ratio, ρ given
⎟⎟⎠
⎞⎜⎜⎝
⎛−==
c
yy
nn f
ff
bdR
'85.012
ρρ
q ρ gMu or vice‐versa if section properties b & d are known
– Substituting Mn=Mu/φ and dividing each side by fc’
96BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Equations
• Singly Reinforced Rectangular Sections (cont’d)
– The above relation is plotted as vs. ρ for various material properties
Th f h i h b d l d id i l i– Therefore, the equation has been developed as an aid in analysis and design
97BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Steps
• Singly Reinforced Rectangular Sections1) Select an approximate value of tension reinforcement ratio ρt given by
y
ct f
f '. 13190 βρ =
where β1 = 0.85 for fc’ ≤ 4000 psi
f 4000 i f ’ 8000 ifor 4000 psi < fc’ < 8000 psi
= 0.65 for fc’ ≥ 8000 psi
– ρt is the reinforcement ratio at the strain limit of 0.005 for tension controlled sections and may be obtained from Tables.y
98BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Steps (cont’d)
• Singly Reinforced Rectangular Sections2) Using the approximate ρ (ρmin ≤ ρ ≤ ρt) compute bd2min trequired:
u
RMrequiredbdφ
=)(2
Where φ = 0 90 for flexure with ρ ≤ ρ
nRφ
⎟⎟⎞
⎜⎜⎛−= yf
fR5.0
1ρ
ρWhere, φ 0.90 for flexure with ρ ≤ ρt
and,
⎟⎟⎠
⎜⎜⎝ c
yn ffR
'85.01ρ
Mu = applied factored moment (required flexural strength)
99BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Steps (cont’d)
• Singly Reinforced Rectangular Sections3) Size the member so that the value of bd2 provided is greater than or equal to the value of bd2 required
100BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Steps (cont’d)
• Singly Reinforced Rectangular Sections4) Based on the provided bd2, may compute a revised value of ρ by one of the following methods
1 B h ( t th d)⎟⎞
⎜⎛ Rf 211'85.0 M
R u1. By where (exact method)
2. Using an ω table. Where ω = ρfy/fc’ and is given in terms of
⎟⎟⎠
⎜⎜⎝
−−=c
n
y
c
fR
ff
'85.021185.0ρ 2bd
R un φ=
g ρfy/fc gmoment strength
2'bdfM
c
u
φ)(revisedR
3. By approximate proportion
Plots represent a linear relationship between Rn and ρ
)()(
)(n
n
originalRrevisedRoriginalρρ ≈
101BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Design Steps
• Singly Reinforced Rectangular Sections5) Compute required Ass
– As = (revised ρ)(bd provided)
When b and d are preset, the required As is compute from:
– As = ρ(bd provided)s ρ( p )Where ρ is computed using one of the methods outlined in step 4
102BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design
• Sections with Multiple Layers of Steel– For sections with multiple layers, a conservative design would be to use d, rather than using dt which is allowed by code. If ρ2 stands for maximum ρ,
Cand
bdfC
y
=2ρty
t bdfC
=ρ
Therefore,
anddtρ2 ⎟
⎞⎜⎛=
dtρρandd
t
t
=ρρ2 ⎟
⎠⎜⎝
=dtρρ2
103BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design
• Sections with Multiple Layers of Steel– Below are strain and stress diagrams for a section with an extreme
steel layer at the tension controlled strain limit of 0.005
– Figure also shows Grade 60 steel with a yield strain of 0.002
104
– All steel within 0.366dt of the bottom layer will be at yieldBMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design ‐ Example
• Select a rectangular beam size and required reinforcement As to carry service load moments MD = 56 ft‐kips and ML = 35 ft‐kips. Select reinforcement to control flexural cracking.reinforcement to control flexural cracking.
fc’ = 4000 psi (normalweight)
fy = 60,000 psi
105BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Example (cont’d)
106BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Example (cont’d)
107BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Example (cont’d)
Figure Strength curves (R vs ρ) for Grade 60 reinforcementFigure – Strength curves (Rn vs. ρ) for Grade 60 reinforcement
2. By strength curves:
for Rn = 901 psi, ρ ≈ 0.0178
108BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Example (cont’d)
Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of
Rectangular Sections with Tension Reinforcement OnlyRectangular Sections with Tension Reinforcement Only
109BMA Engineering, Inc. – 5000
Design of Concrete Structures Beam Design – Example (cont’d)
BMA Engineering, Inc. – 5000 110
Design of Concrete Structures Beam Design – Example (cont’d)
BMA Engineering, Inc. – 5000 111
Design of Concrete Structures T‐Beams
• When a slab is cast monolithically with a beam, the slab participation will add strength to the beam
– The beam can then bedesigned as a flangedsection (T or L section)
112BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• The width of the slab effective as a T‐beam flange is limited by the ACI code (8.10)
• Depending on depth of neutral axis, c, two distinct cases can be identified
113
can be identified
BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• a) Depth of compression block ≤ flange thickness• In this case the section is analyzed and designed as a
l i i h fl id h i h irectangular section using the flange width in the compression side as the beam width
114BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• From Equilibrium• 0.85fc’ba = Asfy a= (Asfy /0.85fc’b) < hf
• The section is therefore treated, in design and analysis, as a standard rectangular section using the flange width as the beam width and the Ulti t t t th i
115
Ultimate moment strength is:
BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• b) Depth of compression block > flange thickness (True T‐ Beam)– From equilibrium: a = As fy/ 0.85 fc’ b
• For a real T‐ beam:For a real T beam:
• (1) a > hf or hf < (1.18 ω’ d = a )
• (2) Where ω’ (A /bd) (f /f ’)• (2) Where, ω = (As/bd) (fy/fc )
• (3) c > hf / β1
• If the section satisfies equation 1 or 3, it is treated as a true‐T section
116BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• b) Depth of compression block > flange thickness (True T‐ Beam)
117BMA Engineering, Inc. – 5000
Design of Concrete Structures
b d h f l b ff b ( )
T‐Beams
• b = width of slab effective as a T‐beam (ACI 8.10.2)
• bw = width of beam web
• hf = flange depthhf = flange depth
• c = depth of neutral axis
• a = β1c = depth of equivalent rectangular stress block (ACI 10.2.7.1)
• Asf = Area of reinforcement required to equilibrate compressive
strength of overhanging flange
f ’ S ifi d i h f• fc’= Specified compressive strength of concrete
• fy = Yield stress for steel
• ε & ε = Ultimate strain for concrete and yield strain of steel respectivelyεu & εy Ultimate strain for concrete and yield strain of steel respectively• T & C = Total tensile and compressive forces respectively
BMA Engineering, Inc. – 5000 118
Design of Concrete Structures T‐Beams
• b) Depth of compression block > flange thickness
(cont’d)
Tw = (As‐Asf) fy & Cw= 0.85fc’abwT = A f & C = 0 85 f ’ (b b ) hTf = Asf fy & Cf = 0.85 fc (b – bw) hf
• Tf = Cf Asf = 0.85 fc’ (b – bw) hf / fy (4a)
119
• Tw= Cw a = (As‐Asf) fy/0.85fc’bw (4b)
BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams
• b) Depth of compression block > flange thickness(cont’d)
– When the compression flange thickness hf < a, the nominal moment strength Mn is
⎞⎛ h
• Asf = Area of reinforcement required to equilibrate strength of overhanging flanges
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛=
2h
dfA2adf)AA(M f
ysfysfsn ---
sf
=
• A = wcysfs bffAA '85.0/)-(yfwC fhbbf /)(85.0 ' −
• b = width of effective flange (see ACI 8.10)
• bw = width of web
• h = thickness of flange
120
• hf = thickness of flange
BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams – Design Steps
• The following steps are summarized for the design of flanged sections with tension reinforcement only– Step 1: Determine effective flange width, b
(according to ACI 8.10)• Using the ω table, determine the depth of the equivalent stress block “a”
assuming rectangular section behavior with “b” equal to the flange width (i e a ≤ h )(i.e., a ≤ hf)
dω18.1'f85.0
dfρb'f85.0
fAa
c
y
c
ys ===
• Where ω is obtained from the Table for – Assume tension‐controlled section with φ = 0.9
2bdfM cu '/φ
121BMA Engineering, Inc. – 5000 BMA Engineering, Inc. – 5000 122
Design of Concrete Structures T‐Beams – Design Steps (cont’d)
• Step 2: If a ≤ hf , determine the reinforcement as for a rectangular section with tension reinforcement only– If a > hf , go to step 3
Step 3 If > h t th i d i f t A• Step 3: If a > hf, compute the required reinforcement Asf
and the moment strength corresponding to the overhanging beam flange in compression:overhanging beam flange in compression:
fwcf h)bb('f85.0CA
−⎥⎤
⎢⎡
⎟⎞
⎜⎛ fhdfAM φφ
y
fwc
y
fsf f
)(f
A == ⎥⎦
⎢⎣
⎟⎟⎠
⎜⎜⎝
−=2f
ysfnf dfAM φφ
123BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams – Design Steps (cont’d)
• Step 4: Compute the required moment strength to be carried by the beam web: Muw = Mu – φMnf
• Step 5: Using the ω table, compute the reinforcement A required to develop the moment strength to beAsw required to develop the moment strength to be carried by the web:
y
wwc'
sw fabf85.0A =
– Where aw = 1.18ωwd with ωw obtained from the table for
Mu/φfc’bd2
y
u/φ c
– Alternatively, obtain Asw from the followingy
wcwsw f
dbfA'ω
=
124BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams – Design Steps (cont’d)
• Step 6: Determine the total required reinforcementAs = Asf + Asw
• Step 7: Check to see if section if tension‐controlledφ = 0.9 c = aw/β1
• If c/d ≤ 0 375 section is tension controlled• If c/dt ≤ 0.375, section is tension‐controlled
• If c/dt > 0.375, add compression reinforcementIf c/dt 0.375, add compression reinforcement
125BMA Engineering, Inc. – 5000
Design of Concrete Structures T‐Beams – Design Steps (cont’d)
• Step 8: Check moment capacity
fhdfadf ⎤⎡ )()()( uf
ysfw
ysfsn MhdfAadfAAM ≥⎥⎦
⎤⎢⎣
⎡−+−−= )()()(
22φφ
Where,y
fwc'
sf fh)bb(f85.0
A−
=
and
y
fsfsw bf
fAAa '
)(850−
=wcbf.850
126BMA Engineering, Inc. – 5000
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only
S l i f f h T i h i d d d li l d• Select reinforcement for the T‐section shown, to carry a service dead and live load moments of MD = 72 ft‐kips and ML = 88 ft‐kips
fc’ = 4000 psi (normalweight)
f = 60 000 psify = 60,000 psi
1. Determine required flexural strength
Mu = (1.2 x 72) + (1.6 x 88) = 227 ft‐kips
2. Using Table 7‐1, determine depth of equivalent
stress block “a,” as for a rectangular section.
Assume φ = 0.9Assume φ 0.9
BMA Engineering, Inc. – 5000 127
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
St 2 (C t’d)Step 2. (Cont’d)
Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of
Rectangular Sections with Tension Reinforcement Only
Check φ:
ca1 = a/β1 = 1.64/0.85 = 1.93 in.
ca1/dt = 1.93/19 = 0.102 < 0.375
Section is tension‐controlled, and φ=0.9.
BMA Engineering, Inc. – 5000 128
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 129
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 130
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only
S l i f f h T i h f d f M 400 f ki• Select reinforcement for the T‐section shown, to carry a factored moment of Mu = 400 ft‐kipsfc’ = 4000 psi (normalweight)
fy = 60,000 psi
1 D t i i d i f t1. Determine required reinforcement.
Step 1. Determine depth of equivalent stress block “a,” as for a rectangular section
BMA Engineering, Inc. – 5000 131
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
St 1 (C t’d)Step 1. (Cont’d)
Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of
Rectangular Sections with Tension Reinforcement Only
Step 2. Since the value of “a” as a rectangular section exceeds the
flange thickness, the equivalent stress block extends in the web,
and the design must be based on T‐section behavior.
BMA Engineering, Inc. – 5000 132
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 133
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 134
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 135
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 136
Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)
BMA Engineering, Inc. – 5000 137
Design of Concrete Structures Doubly Reinforced Beams
• Doubly reinforced beams refer to flexural members with steel reinforcement in the compression regionof the beam as well as the tension region
• The most common reason for compression steel is
To reduce the long term deflection by inhibiting creep andTo reduce the long‐term deflection by inhibiting creep and shrinkage of concrete in compression
138BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams
• Compression Steel– Compression steel is also allows the development of additional moment strength beyond a singly‐reinforced section. This is especially beneficial in a cross‐section limited by architectural considerations.
139BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams
• For a doubly reinforced section with compression reinforcement As’, two possible situations can occur
• I. compression reinforcement As’ yields
• II compression reinforcement does no yield
140
II. compression reinforcement does no yield
BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams
• I. Compression reinforcement As’ yields
f ’ = f f)'AA(a yss -=fs fy
– The nominal moment strength is
b'f85.0a
c=
( )'ddf'A2adf)'AA(M ysyssn -+⎟⎠⎞
⎜⎝⎛= --
• Note that A’s yields when the following (for Grade 60 reinforcement, with Єy = 0.00207) is satisfiedreinforcement, with Єy 0.00207) is satisfied
d’/c ≤ 0.31 where, 1βac =
141BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams
• II. Compression reinforcement does not yield– The nominal moment strength isg
yussss fc
dcEEf <⎟⎠⎞
⎜⎝⎛ −
==''' εε
• The neutral axis depth c can be determined from
Wh f ’ d f h h i f k i( )
087872 − dAcAfAssys '''
Where fc’ and fy have the units of ksi
• The nominal moment strength is
( )0
850850 11
2 =−−bfbf
cc
s
c
sys
'.'. ββ
g
where a = β1c)'dd('f'A)2ad(ab'f85.0M sscn −+−=
142BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams – Design Steps
• The following steps summarize the design of rectangular beams (with b and d preset) requiring compression i freinforcement
– Step 1: Check to see if compression reinforcement is needed
Compute
• Compare this to the maximum R for tension controlled sections2bd
MR nn =
• Compare this to the maximum Rn for tension‐controlled sections given in Table 2. If Rn exceeds this, use compression reinforcement
• If compression reinforcement is needed, it is likely that two layers of tension reinforcement will be needed
Estimate d /d ratio
143
– Estimate dt/d ratio
BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams – Design Steps (cont’d)
• Step 2: Find the nominal moment strength resisted by a section without compression reinforcement, and the additional moment strength Mn’ to be resisted by the compression reinforcement and by added tension reinforcement. From Table 6‐1, find ρt
⎟⎞
⎜⎛d
⎟⎠
⎞⎜⎝
⎛=
ddρρ t
t
f'c
y
ff
ρω =
144BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams – Design Steps (cont’d)
• Step 2 (cont’d): Determine Mnt from ω table
• Compute moment strength to be resisted by compression reinforcement:
M ’ =M MMn = Mn –Mnt
Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of
Rectangular Sections with Tension Reinforcement Only
145BMA Engineering, Inc. – 5000
Rectangular Sections with Tension Reinforcement Only
Design of Concrete Structures Doubly Reinforced Beams – Design Steps (cont’d)
• Step 3: Check yielding of compression reinforcement
• If d’/c < 0.31, compressive reinforcement has yielded and f’s=fyf s fy
• When the compression reinforcement does not yield• When the compression reinforcement does not yield, determine fs’
yussss fdcEEf <⎟⎠⎞
⎜⎝⎛ −
==''' εε yussss f
cEEf ⎟
⎠⎜⎝
εε
146BMA Engineering, Inc. – 5000
Design of Concrete Structures Doubly Reinforced Beams – Design Steps (cont’d)
• Step 4. Determine the total required reinforcement, A’s and As '
, s s
s
ns 'f)'dd(
'M'A−
=
• Step 5. Check moment capacity⎤⎡
h
uysyssn MddfAadfAAM ≥⎥⎦⎤
⎢⎣⎡ −+−−= )'(')()'(
2φφ
where,bffAA
ac
ySs
'85.0)'( −
=
147BMA Engineering, Inc. – 5000
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement
fc’ = 4000 psi (normalweight)c p ( g )
fy = 60,000 psi
1. Determine required reinforcement
Step 1. Determine if compression reinforcement is needed
This exceeds the maximum Rn of 911 for tension‐controlled sections of 4000 psi concrete without compression reinforcement. It appears likely that two layers of tension reinforcement will becompression reinforcement. It appears likely that two layers of tension reinforcement will be necessary. Estimate d = dt – 1.2in. = 28.8in.
148BMA Engineering, Inc. – 5000
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
149BMA Engineering, Inc. – 5000
Design of Concrete Structures
St 2 (C t’d)
gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
Step 2. (Cont’d)
Required moment strength to be resisted by the compression
reinforcement
M’n = 884‐780 = 104 ft‐kips
Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of
Rectangular Sections with Tension Reinforcement Only
BMA Engineering, Inc. – 5000 150
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
BMA Engineering, Inc. – 5000 151
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
BMA Engineering, Inc. – 5000 152
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
BMA Engineering, Inc. – 5000 153
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
BMA Engineering, Inc. – 5000 154
Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)
BMA Engineering, Inc. – 5000 155
5100 Reinforced Concrete5100. Reinforced Concrete
5110 • 5110 ‐ Introduction: Materials & Design Methods
5120 & 5130
• 5120 ‐Moment Design of Beams• 5130 ‐ Shear Design of Beams5130 Shear Design of Beams
5140 & 5150 • 5140 ‐ Footing Design
• 5150 ‐ Column Design
5160 & • 5160 ‐ Development & Splices of Reinforcement5170
p p• 5170 ‐ Strut and Tie Model
5180
BMA Engineering, Inc. – 5000 156
5180 • 5180 ‐ Two‐way Slabs
5130 ‐ Shear Design of BeamsShear and Diagonal Tension
• Principal stresses in a homogeneous isotropic beam
Figure – Trajectories of principal stresses
BMA Engineering, Inc. – 5000 157
Shear and Diagonal TensionShear and Diagonal Tension
• Shear Stresses in BeamsFrom mechanics of materials, the shear stresses in elastic homogeneous b dbeams are expressed as:
v = VQ/IbWhere v = shear stress at point on cross section;
V = shear force on cross section;
I = moment of inertia of gross section about centroidal
axis;
b = thickness at level at which v is computed; and
Q = moment about centroidal axis of area lying betweenQ = moment about centroidal axis of area lying between
point at which the shear stress is being computed and
the outside surface.
BMA Engineering, Inc. – 5000 158
Shear Stresses in Reinforced ConcreteShear Stresses in Reinforced Concrete
• Precise analytical evaluation of shear stresses in reinforced concrete is difficult because– Not homogeneous
Tensile strength is about 10% of compressive– Tensile strength is about 10% of compressive strength
N t l ti– Not elastic • Creep, etc.
– Extent of cracking varies
BMA Engineering, Inc. – 5000 159
ACI Code ProcedureACI Code Procedure
• ACI code uses a simple procedure to establish the order of magnitude of the average shear g g
v = V/bwd
Where v = nominal shear stress;
V = shear force at section;
bw= width of beam web; and
d = distance between compression surface and
centroid of tension steel.
BMA Engineering, Inc. – 5000 160
Shear Stress Based on Linear Stress DistributionShear Stress Based on Linear Stress Distribution
From equilibrium & assuming linear compressive stresses,
⎤⎡⎟⎞
⎜⎛ − 2
12 11 yMM ⎤⎡ 2
1yV
⎥⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛=
212 11
)()( kdy
armbdzMMvy ⎥
⎦
⎤⎢⎣
⎡−=
21
)()( kdy
armbV
maximum shear stress is obtained at the neutral axis y=0:(arm = arm of internal couple) )(
varmbV
=
BMA Engineering, Inc. – 5000 161
(arm arm of internal couple) )(armb
Shear Stress at Neutral AxisShear Stress at Neutral Axis
BMA Engineering, Inc. – 5000 162
Diagonal Tension FailureDiagonal Tension Failure
BMA Engineering, Inc. – 5000 163
Cracking in Concrete BeamsCracking in Concrete Beams
Figure – Types of Cracking in Concrete Beams (R11.4.3)
BMA Engineering, Inc. – 5000 164
(R11.4.3)
Types of Shear ReinforcementTypes of Shear Reinforcement
BMA Engineering, Inc. – 5000 165
Vertical Shear ReinforcementVertical Shear Reinforcement
Figure – Isometric View
BMA Engineering, Inc. – 5000 166
The Effects of Shear & Moment on Beam CracksThe Effects of Shear & Moment on Beam Cracks
BMA Engineering, Inc. – 5000 167
Computation of Maximum Factored Shear Force
• Vu may be at a distance d from face of support when (ACI 11.1.3) ( )– Support reaction, in direction of applied shear force introduces compression into the end regionforce, introduces compression into the end region
– Loads are applied at or near the top of the member
– No concentrated load occurs within d
BMA Engineering, Inc. – 5000 168
Computation of Maximum Factored Shear Force (cont’d)
Figure – Typical Support Conditions for Locating Factored Shear Force Vu (R11.1.3.1)
BMA Engineering, Inc. – 5000 169
Types of Shear ReinforcementTypes of Shear Reinforcement
• shear reinforcements recognized by the ACI code (11.5.1.1, 11.5.1.2); vertical stirrups are the most common
Figure – Types and Arrangements of Shear Reinforcement
BMA Engineering, Inc. – 5000 170
Minimum Shear Reinforcement (ACI 11 5 6)Minimum Shear Reinforcement (ACI 11.5.6)
h h f d h f d h lf h h h• When the factored shear force Vu exceeds one‐half the shear strength provided by concrete (Vu > φVc /2), a minimum amount of shear reinforcement must be provided in concrete flexural members, except for slabs and footings, joists defined by 8.11, and wide, shallow beams. When required, the minimum shear reinforcement area (ACI Eq. 11‐13), Av,min , is:
but not less than
Where b = web width; s= spacing; and f t = yield strength
BMA Engineering, Inc. – 5000 171
Where, bw web width; s spacing; and fyt yield strength
Minimum Shear Reinforcement SpacingMinimum Shear Reinforcement ‐ Spacing
• ACI also requires that the spacing of minimum• ACI also requires that the spacing of minimum shear reinforcement must not exceed d/2 or 24 in.
BMA Engineering, Inc. – 2000 172
Shear reinforcement RequirementsShear reinforcement Requirements
BMA Engineering, Inc. – 5000 173
ACI Provisions for Shear ReinforcementACI Provisions for Shear Reinforcement
f d h f h d d b h fVu = factored shear force; Vs = strength provided by shear reinforcement;
Av = shear reinforcement area; fyt = yield strength; and s = spacing
> Eq. (11‐1)nVφuV
VVVwhere Eq. (11‐2)
E (11 15)
scn VVV +=
dfA Eq. (11‐15)
Substituting V into Eq (11‐2) and V into Eq (11‐1) & solving for As
dfAV ytv
s =
Substituting Vs into Eq. (11 2) and Vn into Eq. (11 1), & solving for Av
dfsVVA cu
v φφ )-(
=
BMA Engineering, Inc. – 5000 174
dfytφ
Design Steps for Shear ReinforcementDesign Steps for Shear Reinforcement
1. Determine maximum factored shear force Vu at critical
sections of the member per 11.1.3.
2. Determine shear strength provided by the concrete φV E (11 3)φVc per Eq. (11.3):
φVc = φ2(f’c)1/2bwdWhere, φ = 0.75 ACI 9.3.2.3
3 C t V φV t th iti l ti3. Compute Vu ‐ φVc at the critical section– If Vu ‐ φVc > φ8(f’c)1/2bwd, then increase the size of the
ti th t i t th
BMA Engineering, Inc. – 5000 175
section or the concrete compressive strength.
Design Steps for Shear Reinforcement (Cont’d)Design Steps for Shear Reinforcement (Cont d)
4. Compute the distance from the support beyond which the concrete can carry the total y yshear force (i.e., where ).2/cu VV φ=
5. Determine the required area of vertical stirrups Av or stirrup space s at a few controlling sections along the length of the g g gmember, which includes the critical sections.
BMA Engineering, Inc. – 5000 176
ACI Provisions for Shear DesignACI Provisions for Shear Design
BMA Engineering, Inc. – 5000 177
Shear Strength φVS for Given Bar Sizes and Spacings
BMA Engineering, Inc. – 5000 178
Example—Design for Shear ‐Members Subject to Shear and Flexure
Determine required size and spacing of vertical U‐stirrups for a 30‐foot span, simply supported p p , p y ppnormal weight reinforced concrete beam.
b = 13 inbw = 13 in.
d = 20 in.
fc′ = 3000 psif 40 000 ifyt = 40,000 psi
wu = 4.5 kips/ft (includes self weight)
BMA Engineering, Inc. – 5000 179
u p g
Shear Design Example (Cont’d)Shear Design Example (Cont d)
1. Determine factored shear forces
@ support:@ support:
Vu = 4.5 (15) = 67.5 kips
@ distance d from support:
Vu = 67 5 ‐ 4 5 (20/12) = 60 kipsVu = 67.5 4.5 (20/12) = 60 kips
(ACI 11.1.3.1)
BMA Engineering, Inc. – 5000 180
Shear Design Example (Cont’d)Shear Design Example (Cont d)
2. Determine shear strength provided by concrete
φVc = φ2λ(fc′)1/2 bwd ACI Eq. (11‐3)λ = 1 0 ACI 8 6 1λ = 1.0 ACI 8.6.1
φ = 0.75 ACI 9.3.2.31/2 /φVc = 0.75(2)(1.0) (3000)1/2 ×13 × 20 / 1000 = 21.4 kips
Vu = 60 kips > φVc = 21.4 kipsu c
Therefore shear reinforcement is required ACI 11 1 1
BMA Engineering, Inc. – 5000 181
Therefore, shear reinforcement is required ACI 11.1.1
Shear Design Example (Cont’d)Shear Design Example (Cont d)
3. Compute Vu – φVc at critical section.
Vu – φVc = 60 – 21.4
= 38.6 kips
< φ8 (f ′)1/2 b d = 85 4 kips< φ8 (fc ) / bwd = 85.4 kips O.K. ACI 11.4.7.9
BMA Engineering, Inc. – 5000 182
Shear Design Example (Cont’d)Shear Design Example (Cont d)
4. Determine distance xc from support beyond which minimum shear reinforcement is required (Vu = φVc):
xc = (Vu @ support – φVc ) /wu = (67.5 – 21.4)/4.5 = 10.2 ft
Determine distance xm from support beyond which concrete can carry total shear force (Vu φVc / 2):can carry total shear force (Vu = φVc / 2):
[V @ t (φV / 2)]/xm = [Vu @ support – (φVc / 2)]/wu
=[67.5 – (21.4/2)]/4.5 = 12.6 ft
BMA Engineering, Inc. – 5000 183
Shear Design Example (Cont’d)Shear Design Example (Cont d)
BMA Engineering, Inc. – 5000 184
Shear Design Example (Cont’d)Shear Design Example (Cont d)
5. Use Table above to determine required spacing of vertical U‐stirrups.
At critical section, Vu = 60 kips > φVc = 21.4 kips
s(req'd) = φAvfytd/(Vu – φVc) ACI Eq. (11‐15)
Assuming No. 4 U‐stirrups (Av = 0.40 in.2),
s(req'd) = 0.75 × 0.40 × 40 × 20/38.6 = 6.2 in.
BMA Engineering, Inc. – 5000 185
Shear Design Example (Cont’d)Shear Design Example (Cont d)
Check maximum permissible spacing of stirrups:
s (max) ≤ d/2 = 20/2 = 10 in. (governs) ACI 11.4.5.1
≤ 24 in. since Vu – φVc = 38.6 kips
< φ4 (fc′)1/2 bwd = 42.7 kipsMaximum stirrup spacing based on minimum shear reinforcement
(ACI 11.4.6.3 ):
( ) / ( ′)1/2s(max) ≤ Avfyt/0.75 (fc′)1/2 bw= 0.4 × 40,000/0.75 3000(13)= 30 in.
≤Avfyt/50bw= 0.4 × 40,000/50 ×13
= 24.6 in.
BMA Engineering, Inc. – 5000 186
Shear Design Example (Cont’d)Shear Design Example (Cont d)
Determine distance x from support beyond which 10 in. stirrup spacing may be used:
10 = 0.75 × 0.4 × 40 × 20/(Vu – 21.4)10 0.75 0.4 40 20/(Vu 21.4)
V 21 4 24 ki V 24 + 21 4 45 4 kiVu – 21.4 = 24 kips or Vu = 24 + 21.4 = 45.4 kips
x = (67.5 – 45.4)/4.5 = 4.9 ft
BMA Engineering, Inc. – 5000 187
Shear Design Example (Cont’d)Shear Design Example (Cont d)
Stirrup spacing using No. 4 U‐stirrups:Stirrup spacing using No. 4 U stirrups:
BMA Engineering, Inc. – 5000 188
Shear Design Example (Cont’d)Shear Design Example (Cont d)
6. As an alternate procedure, use simplified method presented in a Table above to determine stirrup size and spacing.
At iti l tiAt critical section,
φVs = Vu – φVc = 60 – 21.4 = 38.6 kips
F T bl b f G d 40 iFrom Table above for Grade 40 stirrups:
No. 4 U‐stirrups @ d/4 provides φVs = 48 kips
/ φ kNo. 4 U‐stirrups @ d/3 provides φVs = 36 kips
By interpolation, No. 4 U‐stirrups @ d/3.22 = 38.6 kips
Stirrup spacing = d/3.22 = 20/3.22 = 6.2 in.
Stirrup spacing is determined as shown previously
BMA Engineering, Inc. – 5000 189
5100 Reinforced Concrete5100. Reinforced Concrete
5110 • 5110 ‐ Introduction: Materials & Design Methods
5120 & 5130
• 5120 ‐Moment Design of Beams• 5130 ‐ Shear Design of Beams5130 Shear Design of Beams
5140 & 5150 • 5140 ‐ Footing Design
• 5150 ‐ Column Design
5160 & • 5160 ‐ Development & Splices of Reinforcement5170
p p• 5170 ‐ Strut and Tie Model
5180
BMA Engineering, Inc. – 5000 190
5180 • 5180 ‐ Two‐way Slabs
5140 Footing Design5140 ‐ Footing Design
• Structural elements used to reduce the pressure applied directly to the soil by p pp y yspreading the supported loads from walls or columns over a larger areacolumns over a larger area
• Used to prevent excessive settlement or ffrotation, to minimize differential settlement,
and to provide safety against sliding and overturning
BMA Engineering, Inc. – 5000 191
Design of Concrete Structures Footings
• Footing types selected may be influenced by:
‐ Types of soil strata (strength, compressibility, etc.)
‐ Load to be transmitted to soil
‐ Position of water table
‐ Depth of adjacent foundations‐ Depth of adjacent foundations
‐ Frost line
BMA Engineering, Inc. – 5000 192
Design of Concrete Structures Footings ‐ Types
• Wall footings– Under bearing walls. May be flat or stepped
– Long and narrow
– Bend in single curvature
P i i f t l d i th b tt i th h t di ti– Primary reinforcement placed in the bottom in the short direction
– Temperature & shrinkage steel in the long direction provide
h l i i i i i f• help in positioning transverse reinforcements;
• bending resistance to bridge over weak soil.
BMA Engineering, Inc. – 5000 193
Design of Concrete Structures Footings ‐ Types
• Isolated spread footing– Used to support individual columns. May be square, rectangular,
i lor circular
– Two way action
– reinforced in both directions with reinforcements placed in the– reinforced in both directions with reinforcements placed in the bottom
– If clearances permit, square sections are usually used
BMA Engineering, Inc. – 5000 194
Design of Concrete Structures Footings ‐ Types
• Combined Footing– Rectangular or trapezoidal
– Strap or cantilever (isolated footings joined by a beam)
– Usually rectangular, but trapezoidal more economical when l diff i t b t l l dlarge difference exists between column loads
– Useful when:• Columns are closely spacedColumns are closely spaced
• When there are property line restrictions
BMA Engineering, Inc. – 5000 195
Design of Concrete Structures Footings ‐ Types
• Raft, Mat or Floating Foundations– Usually more economical when combined area for individual
footings is greater than 50% the area within the perimeter of the building
– Useful in reducing differential settlement– Useful in reducing differential settlement
– A system of beams & slabs may be more economical instead of constant thickness slab
– Required, along with perimeter wall, if water table is above basement floor
BMA Engineering, Inc. – 5000 196
Design of Concrete Structures Footings ‐ Types
• Continuous Strip Footings/Grid Foundations
h l h l b– When soil has low bearing capacity
– Continuous strips in two directions result in a grid foundation
– Strips represent continuous beams and the p pmoments are much smaller than the cantilever moments from isolated footingsg
BMA Engineering, Inc. – 5000 197
Design of Concrete Structures Footings ‐ Types
• Pile Caps– Reinforced concrete slabs used to distribute loads from walls,
isolated columns, or groups of columns to groups of piles
BMA Engineering, Inc. – 5000 198
Design of Concrete Structures Footings
• ACI 318‐2008– Combined footings may be designed as beams in the longitudinal direction and as an isolated footing in the transverse direction over a defined width on each side of th t d lthe supported columns
• Loads– Plan dimensions are determined using service loads and permissible soil capacity (ACI 15.2.2)
– Depth & reinforcement are determined using factored loads (ACI 15.2.1)
BMA Engineering, Inc. – 5000 199
Design of Concrete Structures Footings ‐Moments
• External moment on any section of a footing shall be determined by passing a vertical plane through the f ti d ti th t f f tifooting, and computing the moment of forces acting over entire area of footing on one side of vertical plane
(ACI 15 4 1)(ACI 15.4.1)
BMA Engineering, Inc. – 5000 200
Design of Concrete Structures Footings ‐Moments
• Reinforcement– In one‐way footings & two‐way square footings, reinforcement shall be distributed uniformly across entire width of footing (ACI 15.4.3)
– In two‐way rectangular footings (ACI 15.4.4):
Uniform distribution for reinforcement in long– Uniform distribution for reinforcement in long direction
– Distribute reinforcement in short direction asDistribute reinforcement in short direction as shown in the figure below
BMA Engineering, Inc. – 5000 201
Design of Concrete Structures Footings – Moments, Reinforcements
BMA Engineering, Inc. – 5000 202
Design of Concrete Structures Footings ‐ Shear
• Check two conditions– Wide beam action (rarely controls)
– Two‐way action (punching shear)
• Shear strength around the critical section (perimeter bo) is calculated in accordance with ACI 11.12.2.1
b d l• Tributary Areas and Critical
Sections for Shear
BMA Engineering, Inc. – 5000 203
Design of Concrete Structures Footings ‐ Shear
• For footing design (without shear reinforcement), the shear strength equations ), g qmay be summarized as follows:– Wide beam action (rarely controls)Wide beam action (rarely controls)
– Vu ≤ φVnV ≤ φ[2(f’ )1/2b d]– Vu ≤ φ[2(f’c)1/2bwd]
– Where bw and Vu are computed for the critical section
(ACI 11.12.1.1)
BMA Engineering, Inc. – 5000 204
Design of Concrete Structures Footings ‐ Shear
C L Di i f C lCL – Long Dimension of Column
CS – Short Dimension of Column
βO – Critical Section Dimension βOβC – Shape Factor for Shear
BMA Engineering, Inc. – 5000 205
Design of Concrete Structures Footings ‐ Shear
• Two‐way action (ACI 11.11.2.1)– For footings, Vc shall be the smallest of (a), (b), and (c):
BMA Engineering, Inc. – 5000 206
Design of Concrete Structures Footings – Design for Base Area of Footing
• Determine the base area Af required for a square spread footing with the following design conditions:
• Service dead load = 350 kips• Service live load = 275 kips• Service surcharge = 100 psf
• Column dimensions = 30 x 12 inColumn dimensions = 30 x 12 in.
• Allowable soil pressure at bottom of footing = 4.5 ksf
• Assume average weight of soil and concrete above footing base = 130 g g gpcf
BMA Engineering, Inc. – 5000 207
Design of Concrete Structures Footings – Design for Base Area of Footing (cont’d)
BMA Engineering, Inc. – 5000 208
Design of Concrete Structures Footings – Design for Base Area of Footing (cont’d)
BMA Engineering, Inc. – 5000 209
Design of Concrete Structures Footings – Design for Depth of Footing
• Determine the overall thickness of footing required (Use the design conditions of the previous example)
fc’ = 3,000 psi
Pu = 860 kips
q = 5 10 ksfqs = 5.10 ksf
Determine depth based on shear strength
without shear reinforcement. Depth required
for shear usually controls the footing thickness.
Both Wide beam action and two‐way action for
Strength computation need to be investigated to determine the controlling shear
it i f d th
BMA Engineering, Inc. – 5000 210
criteria for depth
Design of Concrete Structures Footings – Design for Depth of Footing (cont’d)
BMA Engineering, Inc. – 5000 211
Design of Concrete Structures Footings – Design for Depth of Footing (cont’d)
BMA Engineering, Inc. – 5000 212
Design of Concrete Structures
( ’d)
Footings – Design for Depth of Footing (cont’d)
• 2. Two‐way Action (cont’d)
• Bo = 2 (30 + 28) + 2 (12 + 28) = 196 in.
/• Β = 30/12 = 2.5
• Bo/d = 196/28 = 7
0 f i i l• αs = 40 for interior columns
BMA Engineering, Inc. – 5000 213
Design of Concrete Structures Footings – Design for Footing Reinforcement
• Determine required footing reinforcement(Use the design conditions of the previous example)
fc’ = 3,000 psi Pu = 860 kips
fy = 60,000 psi qs = 5.10 ksf
BMA Engineering, Inc. – 5000 214
Design of Concrete Structures
l f f f l
Footings – Design for Footing Reinforcement (cont’d)
1. Critical Section for moment is at face of column
Mu = 5.10 x 13 x (62)/2
BMA Engineering, Inc. – 5000 215
Design of Concrete Structures
d ll d (φ ) ( ’d)
Footings – Design for Footing Reinforcement (cont’d)
2. Compute required As assuming tension‐controlled section (φ = 0.9) (cont’d)
Note that the lesser amount of reinforcement is required in the perpendicular direction
due to lesser M but for ease of placement the same uniformly distributed
BMA Engineering, Inc. – 5000 216
due to lesser Mu, but for ease of placement, the same uniformly distributed
reinforcement will be used each way. Also note that dt = 27 in for perpendicular direction
Design of Concrete Structures
h k l ( )
Footings – Design for Footing Reinforcement (cont’d)
3. Check net tensile strain (εt)
BMA Engineering, Inc. – 5000 217
Therefore, section is tension‐controlled and initial assumption is valid, O.K.
Thus, use 13‐No. 8 bars each way.
Design of Concrete Structures
h k d l f f
Footings – Design for Footing Reinforcement (cont’d)
3. Check development of reinforcementCritical section for development is the same as that for moment (at face of column)
ACI (12‐1)ACI (12 1)
BMA Engineering, Inc. – 5000 218
Design of Concrete Structures
h k d l f f ( ’d)
Footings – Design for Footing Reinforcement (cont’d)
3. Check development of reinforcement (cont’d)
BMA Engineering, Inc. – 5000 219
Design of Concrete Structures Footings – Plain Concrete Footings
• ACI Section 22.7– Design for factored loads & reactionsg
– Determine base area from unfactored forces & moments & permissible soil pressurep p
– Not to be used for footing on piles
– Thickness ≥ 8 in (ACI 22 7 4)Thickness ≥ 8 in. (ACI 22.7.4)– h shall be taken 2 in less than actual (ACI 22.4.8)
Critical sections for moment (ACI 22 7 5)– Critical sections for moment (ACI 22.7.5)
(same as reinforced footing)
BMA Engineering, Inc. – 5000 220
Design of Concrete Structures Footings – Plain Concrete Footings Cont’d
• For Flexure Design• φMn ≥ Muφ n u
• Mn = 5√[fc’]S
S = elastic section modulusS = elastic section modulus
∴ design for ft ≤ 5 φ√[fc’] (ACI 22.5.3)g t φ [ c ] ( )where φ = 0.60 (ACI 9.3.5)
= Strength reduction factor for flexure, compression, shear g , p ,and bearing of structural plain concrete
BMA Engineering, Inc. – 5000 221
Design of Concrete Structures Footings – Plain Concrete Footings
• Design for shear based on beam action (ACI 22.5.4)
BMA Engineering, Inc. – 5000 222
Design of Concrete Structures Footings – with Vertical and Horizontal Loads
h f d l b d l h l l d bl h• For the footing and column subjected to vertical & horizontal loads, Establish the footing depth and select the reinforcement
h h l d d f• Design the most heavily stressed side of the footing and use that reinforcement through out in both directions
fc’ = 3,500 psi Footing 7.5 ft2
f = 60 000 psi γ = 130 lb/FT3fy = 60,000 psi γs = 130 lb/FT
Vertical HorizontalVertical Horizontal
D = 165k L = 105k D = 6k L = 5k
BMA Engineering, Inc. – 5000 223
Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Factored Loads• Pu = 1.2 (165) + 1.6(105) = 366K
• Hu = 1.2(6) + 1.6(5) = 15.2K
• *WT of soil W1 = (7.5)22(.13)(1.2) = 17.55K
* Neglect weight of pier above grade and diff. between WT of concrete and soil
• WT of FTG W2 = (7.5)22(.15)(1.2) = 20.25K
BMA Engineering, Inc. – 5000 224
Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Factored Loads (cont’d)
• R = ΣF = P + W + W = 366 + 17 55 + 20 25 = 403 8k• Rv = ΣFy = Pu + W1 + W2 = 366 + 17.55 + 20.25 = 403.8k
• ΣMA = 15 2(6) + 366(7 5/2) + 20 25(7 5/2) + 17 55(7 5/2) ‐ R x = 0ΣMA 15.2(6) + 366(7.5/2) + 20.25(7.5/2) + 17.55(7.5/2) Rv x 0
χ = 3.98 ft
• Eccentricity of soil reaction from center line of FTG
e = 7.5/2 ‐ 3.98 = .23 ft
BMA Engineering, Inc. – 5000 225
Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Soil Pressure on Base
• p = P/A ± (Pe)c = 403 8 ± (403 8 x 0 23)7 5/2• p = P/A ± (Pe)c = 403.8 ± (403.8 x 0.23)7.5/2I (7.5)2 (7.5(7.5)3)/12
• = 7.18 K/ft2 ± 1.32 K/ft2;
• Pmax = 7.18 + 1.32 = 8.5 K/ft2
• P i = 7 18 ‐ 1 32 = 5 86 K/ft2Pmin 7.18 1.32 5.86 K/ft
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Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Check Punching Shear – Consider only column load. Neglect variation in soil pressure. Failure plane at d/2
• V = 366 ‐ [366/(7.5)2](30/12 x 30/12) = 325.4KVu 366 [366/(7.5) ](30/12 x 30/12) 325.4
• φVc = (0.75) 4 (fc’)0.5 bod = (0.75) 4 (3500/1000)0.5 (30 x 4)20 = 425.96Kφ c ( ) ( c ) o ( ) ( / ) ( )
• Since φVc > Vu, reduce depth. With h = 22” and d = 18”
Vu = 330.61K < φVc = 357.8K ∴ O.K. use h = 22”
BMA Engineering, Inc. – 5000 227
Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Check Beam Shear– at d from face of column. Consider WT of soil and FTG.
• Vu = 22(7.5)/12 [(8.5 + 7.85)/2] –
• [ 13(26/12) + 15 (22/12)]7 5 x 22/12[.13(26/12) + .15 (22/12)]7.5 x 22/12
• Vu = 112.4K ‐ 7.65K = 104.75KVu 112.4 7.65 104.75
• φVc = φ 2(fc’)0.5bwd φ c φ ( c ) w
= (0.75) 2((3500)0.5/1000) (7.5 x 12)18
= 143.76K > Vu
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Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Design of Flexural As
• Ws = 26/12(.13)(3.33) 7.5 = 7.03K
• Wc = 22/12(.15)(3.33) 7.5 = 6.87K
• P1 = 7.32/2(3.33 x 7.5) = 91.4K
• P2 = 8.5/2(3.33 x 7.5) = 106.14K
• Mu at face of columnMu at face of column
= P1 1.11 + P2 2.22 ‐Ws 3.33/2 ‐Wc 3.33/2
= 313.94 ft • K
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Design of Concrete Structures Footings – with Vertical and Horizontal Loads (cont’d)
• Compute As By Trial Method
• M = φA f (d ‐ a/2) Guess a = 2”• Mu = φAsfy (d ‐ a/2) Guess a = 212 x 313.94 = .9As60(18 ‐ 2/2); As = 4.1 in2
• Check a, C = T, ab .85fc’ = 3.84 x 60
a = 3.84 x 60/[(7.5 x 12).85(3.5)] = 0.86” < 2” O.K.a 3.84 x 60/[(7.5 x 12).85(3.5)] 0.86 < 2 O.K.
• min As = 0.0018(22)(7.5x12) (ACI 10.5.4)
= 3.56 in.2 Does not control. Use As = 4.1
BMA Engineering, Inc. – 5000 230