0401 - e120 - civil codes and inspection - 05 - concrete … · accelerate the rate of strength...

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


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


• 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



• Reinforced Concrete5100

Reinforced Concrete

• Prestressed Concrete5200

Prestressed Concrete

• Reinforcing Bars Reinforcing Details & Tolerances5300


• Concrete Containments, Modular Construction& 5400

,Mass Concrete

5500 • Durability, NDE & Masonry



• 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



• 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




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




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



• 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



• 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]



• Reinforced Concrete5100

Reinforced Concrete

• Prestressed Concrete5200

Prestressed Concrete

• Reinforcing Bars Reinforcing Details & Tolerances5300


• Concrete Containments, Modular Construction& 5400

,Mass Concrete

5500 • Durability, NDE & Masonry


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



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


5180 • 5180 ‐ Two‐way Slabs



5120 & 5130


5140 & 5150 • 5140 ‐ Footing Design




5180



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


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


Design of Concrete Structures IntroductionDesign of Concrete Structures ‐ Introduction

• Typical Reinforced Concrete Structure


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


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


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


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


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



• 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.



• 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



• 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



• 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


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


• 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



• 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



• 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


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



• 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



• 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.



• Compressive Strength– Depending on water/cement ratio, concrete will p g / ,have different strengths



• 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



• 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.



• 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.





• 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



• Tensile Strength– Another method is the modulus of rupture test p(tensile strength in flexure)


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


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)


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



• Flow chart to obtain Required Strength, U



• 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



• 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



• Strength Reduction factors for Load Combinations



• Figure provides φ’s for transition between compression‐ and tension‐controlled sectionsp



• 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)



• 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


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


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.


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.


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)


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




5120 & 5130


5140 & 5150 • 5140 ‐ Footing Design




5180



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


Design of Concrete Structures Beam Flexure Assumptions & Principles

• Examine behavior at cross section b‐b of beam



• 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



• Figure 1 – Uncracked Beam at Small Loads• fs = n times corresponding concrete stress; were modular ratio n = Es/Ec



• Figure 2 ‐ Cracked beam with Elastic Stresses at Moderate Loads



• Figure 3 – Ultimate Strength at Failure



• 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



• 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 =



• Design Example:– Determine the cracking moment for the following g gsection assuming = 4000 psi and the modulus of rupture '5.7 cffr =p



• Design Example:– Determine the cracking moment for the following g gsection assuming = 4000 psi and the modulus of rupture '5.7 cffr =p



• 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



• 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)



• 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



• 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.



• 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.



• 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)



• 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.



• 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


– Figure - Balanced Strain Condition in Flexure


• 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,



• 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



• 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



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


⎠⎝ 2y


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



• Design Example:– For the beam section show, determine the nominal moment Mn



• 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



• 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



• 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.



• 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



• 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



• 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



• 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



• Balanced Steel Reinforcement

– We obtain the balanced reinforcement ratio ρb

⎟⎞

⎜⎛

⎟⎞

⎜⎛ f 00087850 'β

N

⎟⎟⎠

⎞⎜⎜⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

yy

cb ff

f00087

00087850 1

,,*. βρ

Note,



• 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



• 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)



• Strain Distribution and Failure Modes


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

==



• 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,




– 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




⎟⎞

⎜⎛ 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’




– 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


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


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)



• 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



• 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 ≈


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


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


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


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



Table 7‐1 Flexural Strength Mu/φfc’bd2 or Mn/φfc’bd2 of

Rectangular Sections with Tension Reinforcement OnlyRectangular Sections with Tension Reinforcement Only






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)



• 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



• 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



• 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:



• 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



• b) Depth of compression block > flange thickness (True T‐ Beam)


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



• 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)



• 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


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



• 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

=



• 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



• 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


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


Design of Concrete Structures gDesign of Flanged Section with Tension Reinforcement Only (cont’d)

St 2 (C t’d)Step 2. (Cont’d)


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.






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



St 1 (C t’d)Step 1. (Cont’d)



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.




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



• 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.



• 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



• 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 =



• 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 −+−=


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


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 =



• Step 2 (cont’d): Determine Mnt from ω table

• Compute moment strength to be resisted by compression reinforcement:

M ’ =M MMn = Mn –Mnt






• 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



• 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)'( −

=


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.


Design of Concrete Structures gDesign of Rectangular Beam with Compression Reinforcement (cont’d)



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
















5120 & 5130


5140 & 5150 • 5140 ‐ Footing Design




5180



5130 ‐ Shear Design of BeamsShear and Diagonal Tension

• Principal stresses in a homogeneous isotropic beam

Figure – Trajectories of principal stresses


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.


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


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.


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

=


(arm arm of internal couple) )(armb

Shear Stress at Neutral AxisShear Stress at Neutral Axis


Diagonal Tension FailureDiagonal Tension Failure


Cracking in Concrete BeamsCracking in Concrete Beams

Figure – Types of Cracking in Concrete Beams (R11.4.3)


(R11.4.3)

Types of Shear ReinforcementTypes of Shear Reinforcement


Vertical Shear ReinforcementVertical Shear Reinforcement

Figure – Isometric View


The Effects of Shear & Moment on Beam CracksThe Effects of Shear & Moment on Beam Cracks


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


Computation of Maximum Factored Shear Force (cont’d)

Figure – Typical Support Conditions for Locating Factored Shear Force Vu (R11.1.3.1)


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


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


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.


Shear reinforcement RequirementsShear reinforcement Requirements


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 )-(

=


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


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.


ACI Provisions for Shear DesignACI Provisions for Shear Design


Shear Strength φVS for Given Bar Sizes and Spacings


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)


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)



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


Therefore, shear reinforcement is required ACI 11.1.1


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



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





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.



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.



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



Stirrup spacing using No. 4 U‐stirrups:Stirrup spacing using No. 4 U stirrups:



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




5120 & 5130


5140 & 5150 • 5140 ‐ Footing Design




5180



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


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


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.



• 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



• 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



• 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



• 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



• Pile Caps– Reinforced concrete slabs used to distribute loads from walls,

isolated columns, or groups of columns to groups of piles


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)


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)


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


Design of Concrete Structures Footings – Moments, Reinforcements


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



• 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)



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



• Two‐way action (ACI 11.11.2.1)– For footings, Vc shall be the smallest of (a), (b), and (c):


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


Design of Concrete Structures Footings – Design for Base Area of Footing (cont’d)


Design of Concrete Structures Footings – Design for Base Area of Footing (cont’d)


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


criteria for depth

Design of Concrete Structures Footings – Design for Depth of Footing (cont’d)


Design of Concrete Structures Footings – Design for Depth of Footing (cont’d)



( ’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


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



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



d ll d (φ ) ( ’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


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


h k l ( )


3. Check net tensile strain (εt)


Therefore, section is tension‐controlled and initial assumption is valid, O.K.

Thus, use 13‐No. 8 bars each way.


h k d l f f


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)



h k d l f f ( ’d)


3. Check development of reinforcement (cont’d)


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)


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


Design of Concrete Structures Footings – Plain Concrete Footings

• Design for shear based on beam action (ACI 22.5.4)


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


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



• 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



• 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



• 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”



• 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



• 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



• 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


0401 - e120 - civil codes and inspection - 05 - concrete … · accelerate the rate of strength...

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