L-07 EQ Resistant Design of Concrete_seismic for Print

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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures

Department of Civil Engineering, University of Engineering and Technology Peshawar

Lecture-07

Introduction to Earthquake Resistant Design of Reinforced

Concrete Structures

By: Prof Dr. Qaisar Ali

Civil Engineering Department

NWFP UET [email protected]

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Topics

Introduction

Effects of Earthquake

Deformation due to Earthquake

Horizontal and Vertical Earthquake Shaking

How Architectural Features Affect Buildings DuringEarthquakes

Earthquake Design Philosophy

Seismic Loading Criteria

Static & Dynamic Lateral Force Procedures

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Topics

Gravity vs. Earthquake Loading in RC Buildings

ACI Special Provisions for Seismic Design

ACI Provisions for SMRF

ACI Provisions for IMRF

Miscellaneous Considerations

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Introduction

Earth’s Interior

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Introduction

Earthquake results from the sudden movement ofthe tectonic plates in the earth’s crust.

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Introduction

Effect of Earthquake

The movement, taking place at the fault lines, causesenergy release which is transmitted through the earth inthe form of waves. These waves reach the structurecausing shaking.

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Introduction

Seismic Events around the globe

Mostly takes place at boundaries of Tectonic plates

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Dots represents an earthquake



Introduction

Types of Waves Generated Due to Earthquake

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Body Waves Surface Waves

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Displacement due to Earthquake

Introduction

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Horizontal and Vertical Shaking

Earthquake causes shaking of the ground in all three directions.

The structures designed for gravity loading (DL+LL) will benormally safe against vertical component of ground shaking.

The vertical acceleration during ground shaking either adds to orsubtracts from the acceleration due to gravity.

Introduction

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Horizontal and Vertical Shaking

The structures are normally designed for horizontal shakingto minimize the effect of damages due to earthquakes.

Introduction

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Earthquake Types with respect to Depth of Focus

Shallow

Depth of focus varies between 0 and 70 km.

Deep

Depth of focus varies between 70 and 700 km.

Introduction

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Earthquake characteristics with respect to distancefrom epicenter

Introduction

0.05 ≤ T ≤ 0.320 Hz ≥ f ≥ 3.33 Hz

Low period & high frequency field

0.3 ≤ T ≤ 1.0 sec3.33 Hz ≥ f ≥ 1 Hz

1.0 ≤ T ≤ 10 sec

25 km

Large period & low frequency field

Moderate period & low frequency field

Epicenter

1 Hz ≥ f ≥ 0.1 Hz

Near Field: 0 to 25 km

Intermediate Field: 25 to 50 km

Far Field: Beyond 50 km

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Resonance risk for structures w.r.t near, intermediateand far field earthquakes

The natural time period of a structure is its important characteristicto predict behavior during an earthquake of certain time period(Resonance phenomenon).

For a particular structure, the natural time period is a function ofmass and stiffness {T = 2π√(M/K)}

“T” can be roughly estimated from: T = 0.1 × number of stories

Introduction

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Resonance risk for structures w.r.t near, intermediateand far field earthquakes

Introduction

Low rise Structure

(upto 3 stories)

Epicenter

Medium rise Structure

(upto 5 stories)High rise Structure

(Above 5 stories)

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Earthquake Recording

Seismograph

Using multiple seismographsaround the world, accuratelocation of the epicenter of theearthquake, as well as itsmagnitude or size can bedetermined.

Working of seismograph shownin figure.

Introduction

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Richter Scale

In 1935, Charles Richter (US)developed this scale.

The Richter scale is logarithmic,So, a magnitude 5 Richtermeasurement is ten timesgreater than a magnitude 4;while it is 10 x 10, or 100 timesgreater than a magnitude 3measurement.

Introduction

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Some of the famous

earthquake records

Introduction

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Earthquake Occurrence

Introduction

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Seismic Zones

Introduction

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How Architectural Features Affect Buildings During Earthquakes?

Importance of Architectural Features

The behavior of a building duringearthquakes depend critically on its overallshape, size and geometry, in addition tohow the earthquake forces are carried to theground.

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Importance of Architectural Features

At the planning stage, architects and structural engineers mustwork together to ensure that the unfavorable features are avoidedand a good building configuration is chosen.

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Other Undesirable Scenarios

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Soft Storey


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Earthquake Design Philosophy

Performance level

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Building Code of Pakistan

In Pakistan, the design criteria for earthquake loading are basedon design procedures presented in chapter 5, division II ofBuilding Code of Pakistan, seismic provision 2007 (BCP, SP2007), which have been adopted from chapter 16, division II ofUBC-97 (Uniform Building Code), volume 2.


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Lateral Force Determination Procedures

The total design seismic force imposed by an earthquake onthe structure at its base is referred to as base shear “V” in theUBC-97.

The design seismic force can be determined based on:

Dynamic lateral force procedure [sec. 1631, UBC-97 or sec. 5.31, BCP-2007].

Static lateral force procedure [sec. 1630.2, UBC-97 or Sec. 5.30.2, BCP 2007], and/or




Dynamic Lateral Force Procedure

UBC-97 section 1631 include information on dynamic lateral forceprocedures that involve the use of:

Time history analysis.

Response spectrum analysis.

The details of these methods are presented in sections 1631.5and 1631.6 of the UBC-97.


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Time History Analysis (THA)

T

Ground acceleration T

LateralDisplacement





Response Spectrum Analysis (RSA)

T

a (ft/sec2)

T

Response

a (ft/sec2)

T

a (ft/sec2)

T

Response

Response

Ts1 = 0.3 sec

Ts2 = 1.0 sec

Ts3 = 2.0 sec

D1

D2

D3

(Ts1,D1)

(Ts2,D2)

(Ts3,D3)

Structural Time period

Peak Response

T

T


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UBC-97 Response Spectrum Curve(Acceleration vs. Time period)


Response Spectrum Analysis (RSA)




=


Static Lateral Force Procedure

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The total design base shear (V) in a given direction can bedetermined from the following formula:

V = (CνI/RT) W

Where,

Cν = Seismic coefficient (Table 16-R of UBC-97).

I = Seismic importance factor (Table 16-K of UBC-97 )

R = numerical coefficient representative of inherent over strength andglobal ductility capacity of lateral force-resisting systems (Table 16-Nor 16-P).

W = the total seismic dead load defined in Section 1630.1.1.





The total design base need not exceed [ V = (2.5CaI/R) W ]

Where, Ca = Seismic coefficient (Table 16-Q of UBC-97)

The total design base shear shall not be less than [ V = 0.11CaIW ]

In addition for seismic zone 4, the total base shear shall also notbe less [ V = (0.8ZNνI/R) W ]

Where, Nν = near source factor (Table 16-T of UBC-97);

Z = Seismic zone factor (Table 16-I of UBC-97)


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Steps for Calculation of “V”:

Step 1: Find Site Specific details.

Step 2: Determine Seismic Coefficients

Step 3: Determine Seismic Importance factor “I”

Step 4: Determine “R” factor

Step 5: Determine structure’s time period

Step 6:Determine base shear “V” and apply code maximum andminimum.

Step 7: Determine vertical distribution of “V”.







Following list of data needs to be obtained:

Seismic Zone

Soil type

Past earthquake magnitude (required only for highest seismic zone).

Closest distance to known seismic source (required only for highest seismiczone).


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Step 1: Site Specific details.

i. Seismic Zone Source: BCP SP-2007







ii. Soil Type

As per UBC code, if soil type is not known, type SD shall be taken.


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iii. Past Earthquake magnitude: This is required only for seismic zone 4to decide about seismic source type so that certain additional coefficientscan be determined.

iv. Distance to known seismic zone is also required to determineadditional coefficients for zone 4.






Step 2: Determination of Seismic Coefficients.

Cv:

Nv (required only for zone 4):


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Step 2: Determination of Seismic Coefficients.

Ca:

Na (required only for zone 4):






Step 3: Determination of Seismic Importance Factor.


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Step 4: Determination of “R” Factor.

R factor basically reduces base shear “V” to make the systemeconomical. However the structure will suffer some damage as explainedin the earthquake design philosophy.

R factor depends on overall structural response of the structure underlateral loading.

For structures exhibiting good performance, R factor will be high.






Step 4: Determination of “R” Factor.


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Step 5: Determination of structure’s time period.

Structural Period (By Method A, UBC 97): For all buildings, the value Tmay be approximated from the following formula:

T = Ct (hn)3/4

Where,

Ct = 0.035 (0.0853) for steel moment-resisting frames.

Ct = 0.030 (0.0731) for reinforced concrete moment-resisting frames andeccentrically braced frames.

Ct = 0.020 (0.0488) for all other buildings.

hn = Actual height (feet or meters) of the building above the base to the nth level.






Step 6: Determination of Base Shear (V).

Calculate base shear meeting the following criteria:

0.11CaIW ≤ V = (CνI/RT) W ≤ (2.5CaI/R) W

For zone 4, the total base shear shall also not be less than:

V = (0.8ZNνI/R) W


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Step 7: Vertical Distribution of V to storeys.

The joint force at a particular level x of the structure is given as:

Fx = (V – Ft)ωxhx/∑ωihi (UBC sec. 1630.5)

{ i ranges from 1 to n, where n = number of stories }

Ft = Additional force that is applied to the top level (i.e., the roof) inaddition to the Fx force at that level.

Ft = 0.07TV {for T > 0.7 sec}

Ft = 0 {for T ≤ 0.7 sec}




Example: Calculate the base shear and storey forces of a five storeybuilding given in the figure. The structure is constructed on stiff soilwhich comes under soil type SD of table 16-J of UBC-97. Thestructure is located in zone 3.


60′-0″

12′-0″

800 kip

800 kip

800 kip

800 kip

700 kip

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Solution: Selection of Ca and Cv.

Base Shear (V) = {CvI/RT}W


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Solution: Selection of I.


Therefore, I = 1.00


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Solution: Selection of R.


Therefore, R = 8.5 (for SMRF)


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Solution: Calculation of T and W


T = Ct (hn)3/4

= 0.030 × (60)3/4

= 0.646 sec.

W = w1 + w2 + w3 + w4 + w5

= 4 × 800 + 700

= 3900 kip


hn = 60′-0″

12′-0″

w1 = 800 kip

w2= 800 kip

w3 = 800 kip

w4 = 800 kip

w5 = 700 kip

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Solution: Therefore,

V = {CvI/RT}W

= {0.54 × 1.00/ (8.5 × 0.646)} × 3900 = 383 kips

The total design base need not exceed the following:

V = (2.5CaI/R) W

= {(2.5 × 0.36 × 1.00)/ (8.5)} × 3900 = 413 kips

The total design base shear shall not be less than the following:

V = 0.11CaIW

= 0.11 × 0.36 × 1.00 × 3900 = 154.44 kips

Therefore, V = 383 kip


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Solution: Vertical distribution of base shear to stories

The joint force at a level x of the structure is given as:

Fx = (V – Ft)ωxhx/∑ωihi

{ i ranges from 1 to n, where n = number of stories }

Ft = Additional force that is applied to the top level (i.e., the roof) in addition to theFx force at that level.

Ft = 0.07TV {for T > 0.7 sec}

Ft = 0 {for T ≤ 0.7 sec}


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Fx = (V – Ft)ωxhx/∑ωihi

∑ωihi = 800×12 + 800×24 + 800×36 + 800×48 + 700×60 = 138000 kip

Therefore for the case under consideration, Force for storey 1 is:

F1 = (383 – 0) × 800 × 12/ {(138000)} = 27 kip

Storey forces for other stories are given in table below.


Table Storey shears.Level

x hx (ft) wx(kip) wxhx (ft-kip) wxhx /(Σwihi) Fx (kip)

5 60 700 42000 0.304 1174 48 800 38400 0.278 1073 36 800 28800 0.209 802 24 800 19200 0.139 531 12 800 9600 0.070 27

Σwihi = 138000 Check ΣFx =V = 383 kip OK

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Table: Storey shears.Level

x hx (ft) wx(kip) wxhx (ft-kip) wxhx /(Σwihi) Fx (kip)

5 60 700 42000 0.304 1174 48 800 38400 0.278 1073 36 800 28800 0.209 802 24 800 19200 0.139 531 12 800 9600 0.070 27

Σwihi = 138000 Check ΣFx =V = 383 kip OK

800 kip

800 kip

800 kip

800 kip

700 kip

383 kips

117 kips

107 kips

80 kips

53 kips

27 kips

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Reversal of stresses takes placeduring earthquake shaking, fig.

Earthquake shaking reversestension and compression inmembers

Reinforcement is required onboth faces of members.

Gravity vs. Earthquake Loading in Reinforced Concrete Building

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The principal goal of the Special Provisions is to ensure adequatetoughness under inelastic displacement reversals brought on byearthquake loading.

The provisions accomplish this goal by requiring the designer toprovide for concrete confinement and inelastic rotation capacity.

No special requirements are placed on structures subjected to low orno seismic risk.

Structural systems designed for high and moderate seismic risk arereferred to as Special and Intermediate respectively.

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Based on moment resisting capacity, there are three types of RCframes,

SMRF (Special Moment Resisting Frame),

IMRF (Intermediate Moment Resisting Frame),

OMRF (Ordinary Moment Resisting Frame).

Some general requirements will be presented first, which arecommon to all frames. Specific requirements for each type of frameare presented later on.

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General Requirements

Concrete in members resisting earthquake induced forces

Min f’c = 3000 Psi (cylinder strength) for all types

No maximum limit on ordinary concrete

5000 psi is maximum limit for light weight

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Reinforcement in members resisting earthquake induced forces

Grade 60, conforming to ASTM A 706 (low alloy steel)

Grade 40 or 60, conforming to ASTM A 615 (billet steel) provided that

Fy (actual) – Fy (specified) ≤ +18 Ksi

Actual Ultimate / Actual Fy ≥ 1.25

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Hoops, Ties and Cross Ties

Confinement for concrete is provided by transverse reinforcementconsisting of stirrups. hoops, and crossties.

To ensure adequate anchorage, a seismic hook (shown in figure) is usedon stirrups, hoops and crossties .

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(Seismic Hook)

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Hoops, ties and Crossties: Advantages

Closely spaced horizontal closed ties in column help in three ways:

i. they carry the horizontal shear forces induced by earthquakes, and therebyresist diagonal shear cracks,

ii. they hold together the vertical bar and prevent them from excessivelybending outwards (in technical terms, this bending phenomenon is calledbuckling), and

iii. they contain the concrete in the column. The ends of the ties must be bentat 135° hooks. Such hook ends prevent opening of hoops andconsequently buckling of concrete and buckling of vertical bars.

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Provisions for Flexural Members

These provision applies to flexural members with:

Factored axial compressive force ≤ Agf’c/10.

Note: These provisions generally apply to beams because axial load on beamsis generally 0 or less than Agfc′/10.

However they are also applicable to columns subjected to axial load lessthan Agfc′/10.

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ACI Provisions for Special Moment Resisting Frames (SMRF)

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1. Size: The members must have:

a. clear span-to-effective-depth ratio of at least 4, (Ln/d ≥ 4)

e.g., for Ln = 15 ft, d = 16″, Ln/d = 15 × 12/16 = 11.25 > 4, O.K.

b. a width-to-depth ratio of at least 0.3, b/d ≥ 0.3

e.g., for width (b) = 12″ and depth (h) = 18″, b/h = 12/18 = 0.67 > 0.3, O.K.

c. A web width of not less 10 inches.

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2. Flexural Reinforcement

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Asl−

Asl+ ≥ (Asl−)/2

As− or As+ (at all section) ≥ (maximum of As at either joint)/4

Asr−

Asr+ ≥ (Asr−)/2

ρmin = 3√f’c/fy, 200/fy (at critical sections)

ρmax = 0.025 (at critical sections)

Min. 2 bars continuous at all sections

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3. Transverse Reinforcement

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s ≤

d/48 × smallest longitudinal bar diameter24 × hoop bar diameter12”≤ 2”

≥ 2h ≥ 2hs ≤ d/2

Column

Column




4. Lap Splice

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Lap splice length =1.3 ld = 1.3× 0.05 (fy/ √fc′)db

50 db for fc′ 3 and fy 40 ksi


Spacing of stirrups Least of d/4 or 4 inches

Lapping prohibited in regions where longitudinal bars can yield in tension

Lapping of Longitudinal bars

≥ 2h ≥ 2h

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Mechanical Splice of Longitudinal Reinforcement

Mechanical Splices shall conform to 21.2.6.

Section 21.2.6 says that welded splice shall conform to 12.14.3.2which states “A full mechanical splice shall develop in tension orcompression, as required, at least 125 % of the specified yieldstrength (fy) of the bar.




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Welded Splice of Longitudinal Reinforcement

Welded Splices shall conform to 21.2.7.

Section 7.3.6 says that welded splice shall conform to 12.14.3.4which states “ A full welded splice shall develop at least 125 % ofthe specified yield strength (fy) of the bar.


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Provision for Frame Members Subjected to Bending and Axial Load

The provision applies to members with:

Factored axial compressive force > Agf’c/10

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1. Size

a) Each side at least 12 inches

b) Shorter to longer side ratio ≥ 0.4.

i.e. 12/12, 12/18, 12/24 OK; but 12/36 not O.K

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Provision for Frame MembersSubjected to Bending and AxialLoad

2. Longitudinal Reinforcement

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0.01 ≤ ρg ≤ 0.06

Clear span, hc



Provision for FrameMembers Subjectedto Bending andAxial Load

3. Trans. Rein.

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h2

h1

s ≤ 0.25 × (smaller of h1 or h2)

6 × long. bar dia.so

s ≤ 6 × long. bar dia.6”

lo

lo ≥ Larger of h1 or h2

Clear span/618”

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3. Trans. Rein.

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4” ≤ so = 4 + [(14 – hx)/3] ≤ 6”

hx = max. value of hx on all column faces

hx ≤ 14”

hx hx hx

hx

hx

6db ≥ 3” 6db extension

Alternate90-deg hooks

Provide add.trans. reinf. if thickness > 4”



Provision for Frame Members Subjectedto Bending and Axial Load

4. Lap Splice

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Tension lap splicewithin center half ofmember length

Lap splice length =1.3 ld = 1.3 × 0.05 (fy/ √fc′)db



Spacing of ties in lap splice not more than smaller of d/4 or 4″

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Joints of Special Moment Frame


Beam

Column

Beam Column Joint

Column ties (with 135o) hook continued through joint(ACI 21.5.2)

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Successful seismic design of frames require that the structures beproportioned so that hinges occur at locations that leastcompromise strength. For this, “weak Beam-strong column”approach is used.

After design, the member capacities are calculated based ondesigned section.

At joint, Column flexural capacity ≥ 1.20 x Beam flexural capacity

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Minimum Flexural Strength of Column at Joint

M+nc,t

M-nc,b

M+nb,r

M-nb,l

M+nc,b + M-

nc,t ≥ 6(M+nb,l + M-

nb,r)/5

M-nc,t

M+nc,b

M+nb,l M-

nb,r

M-nc,b + M+

nc,t ≥ 6(M-nb,l + M+

nb,r)/5





To prevent beam column joint cracking, ACI Code 21.5.1requires that the column dimension parallel to the beamreinforcement must be at least 20 times the diameter of thelargest longitudinal bar.

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Beam longitudinal reinforcement with diameter (db)

20×db

Beam

Column

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Beam longitudinal reinforcement that is terminated within acolumn. must be extended to the far face of the column core.The development length (ldh) of bars with 90° hooks must be notless than 8db, 6 inch, Or fydb/ (65 √ fc′).

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Beam longitudinal reinforcement

Beam

Column

ldh



Provision for Flexural Members

1. Size: No special requirement (Just as ordinary beamrequirement).

2. Flexural steel: Less stringent requirement as discussed next.

3. Transverse steel: Same as for SMRF.

4. Lap: No special requirement (Just as ordinary beamrequirement).

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ACI Provisions for Intermediate Moment Resisting Frames (IMRF)

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2. Flexural Reinforcement

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Asl−

Asl+ ≥ (Asl−)/3

As− or As+ (at all section) ≥ (maximum of As at either joint)/5

Asr−

Asr+ ≥ (Asr−)/3

ρmin = 3√f’c/fy, 200/fy (at critical sections)

εt ≥ 0.004



Provision for Columns

1. Size: No special requirement (Just as ordinary columnrequirement)

2. Flexural steel: No special requirement (Just as ordinary columnrequirement)

3. Transverse steel: Less Stringent requirement as given next.

4. Lap: No special requirement (Just as ordinary columnrequirement)

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h2

h1

≤ so/2

Trans. reinf. based on Mn and factored tributary gravity load

so ≤

8 × smallest long. bar dia.24 × tie bar dia.0.5 × min. (h1 or h2)12”

lo

lo ≥ Larger of h1 or h2

Clear span/618”

s ≤ d/2 (As per ACI 11.5.4)



IMRF are not allowed in regions of high seismic risk, however,SMRF are allowed in regions of moderate seismic risk.

Unlike regions of high seismic risk, two way slab system withoutbeams are allowed in regions of moderate seismic risk.

In regions of low or no seismic risk ordinary moment resistingframes OMRF are allowed but IMRF and SMRF may also beprovided.

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Miscellaneous Considerations

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Example 1: Detail the selected frame of E-W interior frame of thegiven structure as per SMRF requirements. The structure isalready designed for the following seismic zone data.

Seismic zone: 4

Magnitude of earthquake ≥ 7.0

Slip rate ≥ 5.0

Closest distance to known seismic source > 15 km.

Soil type: SD (stiff).

Concrete compressive strength = 3 ksi,

Steel yield strength = 40 ksi

Modulus of elasticity of concrete = 3000 ksi.

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Design Example



Given 3D structure:

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Design Example

20 ft 20 ft 20 ft 20 ft

15 ft

15 ft

15 ft

10.5 ft

10.5 ft

10.5 ft (floor to floor)

SDL = 40 psfLL = 60 psf



fc′ = 3 ksify = 40 ksi

Slab-Beam Frame Structure

Beams: 15″ × 18″ (d =15 inch)Columns: 15″ square

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Design Example

Selected portion of E-W interior Frame:

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20 ft 20 ft 20 ft 20 ft

Portion of frame considered



Design Example

Reinforcement from Design:

90

For the beam in the given frame, we haveAsmin = 0.005 x15 x 15 = 1.125 in2

Asmax = 0.025x15 x 15 = 5.625 in2

Negative reinforcement at the ends and positive reinforcement at the mid span must be greater than Asmin and reinforcement at all locations must be less than the Asmax

This is OK for the values on the given beam.For the column, we haveAsmin = 0.01 x15 x 15 = 2.25 in2

This is also OK.20 ft 20 ft 20 ft 20 ft

1.61 1.611.02 1.021.21

2.25 2.25

Reinforcement in in2

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Design Example

Calculation of number of bars:

91

20 ft 20 ft 20 ft 20 ft

No. of bars = As/Ab

Use No. 5 bar,Negative reinforcement at joint:Left joint:1.61/0.31= 5.19 (take 6 bars in 2 layers)Right joint:1.61/0.31= 5.19 (take 6 bars in 2 layers)Positive bars (mid span):1.21/0.31 = 3.9 (take 4 bars in 1 layer)Positive bars (at joint):1.01/0.31 = 3.29 (take 4 bars in 1 layer)Column reinforcement:2.25/ 0.31 = 7.25 (take 8 bars for even distribution of bars at all faces of column)

6 bars 6 bars4 bars 4 bars4 bars

8 bars 8 bars

No. of #5 bars



SMRF Requirements Checklist

Provisions for Beams

Sizes

ln/d = 20 × 12/15 = 16 > 4 (ACI 21.3.1.2 satisfied)

Width/ depth = 15/18= 0.833 > 0.3 (ACI 21.3.1.3 satisfied)

Width = 15″ > 10″, O.K.

Therefore 12″ wide and 18″ beams are OK

Design Example

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Flexural Reinforcement

Design Example

As+ (at joints) ≥ ½ As− (at joints)4 #5 bars ≥ ½ (6 #5 bars)

As (any section) ≥ ¼ Max. As at joints2 #5 bars ≥ ¼ (6 #5 bars)OK OK

Asl− = 6 #5 Asr− = 6 #5

Asmid+ = 4 #5Asl+ = 4 #5 Asr+ = 4 #5

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Transverse Reinforcement

Design Example

s ≤

d/4 = /4 = 3.75″8 × smallest long. bar dia.= 8 × 5/8= 5″24 × hoop bar diameter = 24 × 3/8= 9″12”≤ 2”

2h = 36″s ≤ d/2 = 15/2 = 7.5″

Column

2h = 36″

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Lap Splice: If required then,

Not to be provided within joints. Not to be provided within 2h region from face ofthe support.

Spacing of hoops within lap = least of d/4 or 4″ c/c = 3.75″ c/c

Lap splice length =1.3 ld = 1.3×0.05 (fy/ √fc′)db ≈ 30″ = 2.5′

50 db = 50 × (5/8) = 31.25″ ≈ 2.5′ for fc′ 3 and fy 40 ksi

Design Example

2h=36″ 2h=36″

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Provisions for Columns

Size: All columns are 15″ square, therefore the size is more than the

least required for SMRF (i.e., 12″).

Flexural Reinforcement: All columns are reinforced with 8 #5 barswhich gives ρg = 0.011, within the specified range 0.01 ≤ ρg ≤ 0.06.

Design Example

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Design ExampleSMRF RequirementsChecklist


Transverse Reinforcement:

lo = max (larger columndimension, hc/6, 18″) = 18″

Spacing of ties in lo region isleast of = smaller columndimension/4, 6 × long bardia = 3.75″

Spacing in the remaining

region will be least of 6 ×

long bar dia or 6″ = 3.75″

hc = 8.5′

lo

lo

15” × 15” column

8, #5 bars

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#3 @ 3.75” c/cthroughout height



Design ExampleSMRF Requirements Checklist


Lap Splice:

Tension lap splice within centerhalf of member length.

Spacing of ties in lap splice notmore than smaller of d/4 or 4″.Continue 3.75 inch spacing.

Lap length = 1.3 × 0.05 (fy/ √fc′ )db=30″ ≈ 2.5′

And from 50db = 50×(5/8) =

31.25″

hc = 8.5′

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Provision for Joints

To prevent beam column joint cracking, ACI Code 21.5.1 requires thatthe column dimension parallel to the beam reinforcement must be atleast 20 times the diameter of the largest longitudinal bar.

20 × 5/8 = 12.5″

Column dimension parallel to beam long bar = 15″, OK

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6 #5 bars

6 #5 bars2″

ColumnBeam

Joint

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Example 2: A two dimensional frame of the building is shown infigure 01. All beams in the frame are 12″ wide and 18″ deep. Allcolumns are 12″ square. It is required only for frame BFGC to:

a. Provide suitable number of bars in beams and columns for which flexuralreinforcement is shown in the figure (Use only #5 bars as main reinforcementand #3 bars as transverse reinforcement).

b. Provide suitable spacing for stirrups and ties when shear reinforcement asper design is 0.23 in2/ft in all parts of the frame.

c. Satisfy all SMRF requirements for beams and columns.

d. Present appropriately proportioned structural details of beam and column.Also draw the beam to column joint detail at detail X as shown in fig. 01.Take fc′ = 3 ksi and fy = 40 ksi.

Design Example

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Given Building:

Design Example

Figure 01: RC Building Frame

3.2 in2

2.4 in2

1.2 in2

2.6 in2 2.6 in2 2.4 in2

1.2 in21.8 in2

3.0 in2 3.0 in2

Detail X

3.2 in2 3.2 in2 3.2 in2

F G

E H

B CA D

J K

1.44 in2 1.44 in2

1.0 in21.2 in2 1.2 in2

10 ft ln =15 ft 10 ft

10 ft

10 ft

1.2 in2 1.2 in2 1.0 in21.0 in2 1.0 in2 1.0 in2

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Solution of Part (a)

As given in problem, # of bars using #5 bars are as below:

Design Example

2.6 in2 2.6 in2

1.8 in2

3.0 in2 3.0 in2

3.2 in2 3.2 in2

F G

E H

B CA D

J K

10 ft 15 ft 10 ft

1.2 in2 1.2 in2

# of bars = As/Ab

Negative reinforcement at joint:2.6/0.31= 8.38 (take 9 bars in 2 layers)3.0/0.31= 9.67 (take 10 bars in 2 layers)Take 10 bars at joint for beam FG.Positive bars (mid span):1.8/0.31 = 5.80 (take 6 bars in 2 layers)Positive bars (at joint):1.2/0.31 = 3.87 (take 4 bars in 1 layer)Column reinforcement:3.2/ 0.31 = 10.32 (take 12 bars for even distribution of bars at all faces of column)

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Solution of Part (a)

Design Example

(4+2) #5(5+5) #5 (5+5) #5

12 #5 12 #5

F G

E H

B CA D

J K

10 ft 15 ft 10 ft

4 #5 4 #5

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Solution of Part (b)As given in problem, Shear reinforcement spacing with #3 bar is:

Design Example

0.23 in2/ft

F G

E H

B CA D

J K

10 ft 15 ft 10 ft

Spacing= (Ab/As) × 12With #3, 2 leg vertical stirrups (Ab = 0.22 in2)Stirrup spacing for beams:sb = (0.22/0.23) × 12 = 11.47″ c/cAs per ACI minimum reinforcement and maximum spacing requirement, spacing should not exceed:i. Avfy/50bw = 14″ c/cii. d/2 = 16.5/2 = 8.25″ c/ciii. 24″ c/civ. Avfy/{0.75√(fc′)bw = 17.85″ c/cFinally sb = 8.25″ c/cTie spacing for column:s = (0.22/0.23) × 12 = 11.47″ c/cSpacing for tie bars according to ACI 7.10.5.1 is minimum of:16 × dia of main bar =16 × 5/4 =10″ c/c48 × dia of tie bar = 48 × (3/8) =18″ c/cLeast column dimension =12″ c/cFinally sc = 10″ c/c

0.23 in2/ft

0.23 in2/ft

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Solution of Part (b)As given in problem, Shear reinforcement spacing with #3 bar is:

Design Example

F G

E H

B CA D

J K

10 ft 15 ft 10 ft

#3, 2 legged @ 8.25″ c/c

#3, ties @ 10″ c/c #3, ties @ 10″ c/c

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Solution of Part (c)



Sizes

ln/d = 15 × 12/16.5 = 10.9 > 4 (ACI 21.3.1.2 satisfied)

Width/ depth = 12/18 = 0.66 > 0.3 (ACI 21.3.1.3 satisfied)

Width = 12″ > 10″, O.K.

Therefore 12″ wide and 18″ beams are OK.

Design Example

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Solution of Part (c): SMRF Requirements Checklist


Flexural Reinforcement

Design Example

Asl− = (5+5) #5Asr− = (5+5) #5

Asmid+ = (4+2) #5Asl+ = 4 #5 Asr+ = 4 #5

As (at all critical sections) ≥ Asmin

As Asmin = 4 #5 bars

As (at any section) ≤ Asmax

Asmax = 0.025bd = 16 #5 barsOK

As+ (at joints) ≥ ½ As− (at joints)4 #5 bars ≥ ½ {(5+5) #5 bars}

As (any section) ≥ ¼ Max. As at jointsProvide at least 3 bar

N.G

Provide at least 5 bars at joint. O.K

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Recommended Flexural Reinforcement

Design Example

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Asl− = (5+5) #5 Asr− = (5+5) #5

Asmid+ = (4+2) #5Asl+ = (4+2) #5 Asr+ = (4+2) #5

3 #5





Transverse Reinforcement

Design Example

s ≤

d/4 = 16.5/4 = 4.125″ ≈ 4″8 × smallest long. bar dia.= 8 × 5/8= 5″24 × hoop bar diameter = 24 × 3/8= 9″12”≤ 2”

2h = 36″s ≤ d/2 = 16.5/2 = 8.25″

Column

2h = 36″

Note: Spacing calculated in part (b) (i.e., sb = 8.25″)of this problem does not govern for 2h regions.However it governs elsewhere.

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Lap Splice: If required then,

Not to be provided within joints. Not to be provided within 2h region from face ofthe support.

Spacing of hoops within lap = least of d/4 or 4″ c/c = 4″ c/c

Lap splice length =1.3 ld = 1.3×0.05 (fy/ √fc′)db ≈ 30″ = 2.5′

50 db = 50 × (5/8) = 31.25″ ≈ 2.5′ for fc′ 3 and fy 40 ksi

Design Example

2h=36″ 2h=36″

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Size: All columns are 12″ square, which is equal to least required for

SMRF (i.e., 12″).

Flexural Reinforcement: All columns are reinforced with 12 #5 barswhich gives ρg = 0.025, within the specified range 0.01 ≤ ρg ≤ 0.06.

Design Example

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Design ExampleSolution of Part (c): SMRF Requirements Checklist


Transverse Reinforcement:

lo = max (larger column dimension,hc/6, 18″) = 18.5″

Spacing of ties in lo region is least of =smaller column dimension/4, 6 × longbar dia = 3″

Spacing in the remaining region will be

least of 6 × long bar dia or 6″ = 3.75″

hc = 9.25′

lo

lo

12” × 12” column

12, #5 bars

G.L

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Lap Splice:

Tension lap splice within centerhalf of member length.

Spacing of ties in lap splice notmore than smaller of d/4 or 4″

Lap length = 1.3 × 0.05 (fy/ √fc′ )db=30″ ≈ 2.5′

And from 50db = 50×(5/8) =

31.25″

hc = 9.25′

lo

lo

12” × 12” column

G.L

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To prevent beam column joint cracking, ACI Code 21.5.1 requires thatthe column dimension parallel to the beam reinforcement must be atleast 20 times the diameter of the largest longitudinal bar.

20 × 5/8 = 12.5″

Column dimension parallel to beam long bar = 12″, Almost OK.

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(5+5) #5 bars

(4+2) #5 bars2″

ColumnBeam

Joint

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Design ExampleSolution of Part (d):

Presentation of appropriately proportioned structural details ofbeam, column and selected joint.

In this part, structural drawings showing all calculated SMRF detailshave been asked to draw. This has already been done in part (c) alongwith each member (beam and column) SMRF checks.

However in the examination, it is better to draw all drawings here in part(d) at the same place to avoid waste of time.

Part (d) is shown next.

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Structural Details for beam (Long section):

s = 4″

≤ 2”

2h = 36″s = 8.25″

Column

2h = 36″

(5+5) #5(5+5) #5

(4+2) #5(4+2) #5 (4+2) #5

Note: Lap splice is provided only if required.

A

A

B

B

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Structural Details for Beam (Cross sections):

12″

18″

(5+5) #5

(4+2) #5

Section A-A

12″

18″

3 #5

(4+2) #5

Section B-B

#3, 2 legged @ 4″ c/c

#3, 2 legged @ 8.25″ c/c

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Structural Details for Column:

lo = 18.5″

12” × 12” column

12, #5 bars

G.Llo = 18.5″

#3 ties @ 3″ c/c (throughout)

Lap length = 2.5′

12 #5

#3ties @ 3″ c/c

12″

12″

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Structural Details for Joints:

(5+5) #5 bars

(4+2) #5 bars2″

ColumnBeam

Joint

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ReferencesACI 318UBC-97BCP SP-2007Earthquake tips from IITK.

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The End

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L-07 EQ Resistant Design of Concrete_seismic for Print

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