L-06-EQ-Resistant-Design-of RCC Structures

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Prof. Dr. Qaisar Ali Reinforced Concrete Design II

Department of Civil Engineering, University of Engineering and Technology Peshawar

Lecture - 06

Introduction to Earthquake Resistant Design of Reinforced

Concrete Structures

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

www.drqaisarali.com

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Topics

Introduction to Earthquakes and Its Effects on Buildings

Earthquake Design Philosophy

Seismic Loading Criteria

Example 1 (Static Lateral Load Procedure)

Gravity vs. Earthquake Loading in RC Buildings

ACI Special Provisions for Seismic Design

ACI Provisions for Special Moment Resisting Frames (SMRF)

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Topics

ACI Provisions for Intermediate Moment Resisting Frames

(IMRF)

Miscellaneous Considerations

Design Example 2 (SMRF)


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Earths Interior

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Earthquake results from the sudden movement of

the tectonic plates in the earths crust.

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Effect of Earthquake

The movement, taking place at the fault lines, causes

energy release which is transmitted through the earth in

the form of waves. These waves reach the structure

causing shaking.

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Seismic Events around the globe

Mostly takes place at boundaries of Tectonic plates

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Dots represents an earthquake




Types of Waves Generated Due to Earthquake

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Body Waves Surface Waves


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Displacement due to Earthquake

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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 be

normally safe against vertical component of ground shaking.

The vertical acceleration during ground shaking either adds to or

subtracts from the acceleration due to gravity.

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Horizontal and Vertical Shaking

The structures are normally designed for horizontal shaking to

minimize the effect of damages due to earthquakes.



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.

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Earthquake characteristics with respect to distance

from epicenter

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, intermediate

and far field earthquakes

The natural time period of a structure is its important characteristic

to predict behavior during an earthquake of certain time period

(Resonance phenomenon).

For a particular structure, the natural time period is a function of

mass and stiffness {T = 2p(m/k)}

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

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Resonance risk for structures w.r.t near, intermediate

and far field earthquakes

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 seismographs

around the world, accurate

location of the epicenter of the

earthquake, as well as its

magnitude or size can be

determined.

Working of seismograph shown

in figure.

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Richter Scale

In 1935, Charles Richter (US)

developed this scale.

The Richter scale is logarithmic,

So, a magnitude 5 Richter

measurement is ten times

greater than a magnitude 4;

while it is 10 x 10, or 100 times

greater than a magnitude 3

measurement.

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Some of the famous

earthquake records

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Earthquake Occurrence

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Seismic Zones

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Importance of Architectural Features

The behavior of a building during

earthquakes depend critically on its overall

shape, size and geometry, in addition to

how the earthquake forces are carried to the

ground.

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Importance of Architectural Features

At the planning stage, architects and structural engineers

must work together to ensure that the unfavorable features

are avoided and a good building configuration is chosen.

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Other Undesirable Scenarios



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 based

on design procedures presented in chapter 5, division II of

Building Code of Pakistan, seismic provision 2007 (BCP, SP

2007), which have been adopted from chapter 16, division II of

UBC-97 (Uniform Building Code), volume 2.


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Lateral Force Determination Procedures

The total design seismic force imposed by an earthquake on

the structure at its base is referred to as base shear V in the

UBC-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],


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Dynamic Lateral Force Procedure

UBC-97 section 1631 include information on dynamic lateral force

procedures that involve the use of:

Time history analysis.

Response spectrum analysis.

The details of these methods are presented in sections 1631.5

and 1631.6 of the UBC-97.


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Time History Analysis (THA)

T

Ground acceleration

T

LateralDisplacement


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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)


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=

Static Lateral Force Procedure


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The total design base shear (V) in a given direction can be

determined from the following formula:

V = (CI/RT) W

Where,

C = 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 and

global ductility capacity of lateral force-resisting systems (Table 16-N

or 16-P).

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


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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 not

be less [ V = (0.8ZNI/R) W ]

Where, N = 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 structures time period

Step 6:Determine base shear V and apply code maximum and

minimum.

Step 7: Determine vertical distribution of V.


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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 seismic

zone).


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Step 1: Site Specific details.

i. Seismic ZoneSource: BCP SP-2007


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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 4

to decide about seismic source type so that certain additional coefficients

can be determined.

iv. Distance to known seismic zone is also required to determine

additional coefficients for zone 4.


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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):


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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 system

economical. However the structure will suffer some damage as explained

in the earthquake design philosophy.

R factor depends on overall structural response of the structure under

lateral loading.

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


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Step 4: Determination of R Factor.


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Step 5: Determination of structures time period.

Structural Period (By Method A, UBC 97): For all buildings, the value T

may 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 and eccentrically

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.


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Step 6: Determination of Base Shear (V).

Calculate base shear meeting the following criteria:

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

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

V = (0.8ZNI/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)xhx/ihi (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) in

addition to the Fx force at that level.

Ft = 0.07TV {for T > 0.7 sec}

Ft = 0 {for T 0.7 sec}


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Example: Calculate the base shear and storey forces of a five storey

building given in the figure. The structure is constructed on stiff soil

which comes under soil type SD of table 16-J of UBC-97. The

structure is located in zone 3.

Example 1(Static Lateral Force Procedure)

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)xhx/ihi

{ 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 the

Fx 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)xhx/ihi

ihi = 80012 + 80024 + 80036 + 80048 + 70060 = 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

xhx (ft)

wx(kip)

wxhx (ft-kip)wxhx

/(wihi)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

wihi = 138000Check SFx =V = 383 kip OK

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Table: Storey shearsLevel

xhx (ft)

wx(kip)

wxhx (ft-kip)wxhx

/(wihi)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

wihi = 138000Check Fx =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 place

during earthquake shaking, fig.

Earthquake shaking reverses

tension and compression in

members

Reinforcement is required on

both 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 adequate

toughness under inelastic displacement reversals brought on by

earthquake loading.

The provisions accomplish this goal by requiring the designer to

provide for concrete confinement and inelastic rotation capacity.

No special requirements are placed on structures subjected to low or

no seismic risk.

Structural systems designed for high and moderate seismic risk are

referred to as Special and Intermediate respectively.


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

frames,

SMRF (Special Moment Resisting Frame),

IMRF (Intermediate Moment Resisting Frame),

OMRF (Ordinary Moment Resisting Frame).

Some general requirements will be presented first, which are

common to all frames. Specific requirements for each type of frame

are presented later on.


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General Requirements

Concrete in members resisting earthquake induced forces

Min fc = 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 reinforcement

consisting of stirrups. hoops, and crossties.

To ensure adequate anchorage, a seismic hook (shown in figure) is used

on 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 thereby

resist diagonal shear cracks,

ii. they hold together the vertical bar and prevent them from excessively

bending outwards (in technical terms, this bending phenomenon is called

buckling), and

iii. they contain the concrete in the column. The ends of the ties must be bent

at 135 hooks. Such hook ends prevent opening of hoops and

consequently 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 Agfc/10.

Note: These provisions generally apply to beams because axial load

on beams is generally 0 or less than Agfc/10.

However they are also applicable to columns subjected to axial load less

than Agfc/10.


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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/h 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

Asl

Asl+ (Asl)/2

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

Asr

Asr+ (Asr)/2

rmin = 3fc/fy, 200/fy (at critical sections)

rmax = 0.025 (at critical sections)

Min. 2 bars continuous at all sections


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3. Transverse Reinforcement

s

d/4

8 smallest longitudinal bar diameter

24 hoop bar diameter

12 2

2h 2hs d/2

Column

Column


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4. Lap Splice Lap splice length =1.3 ld = 1.3 0.05 (fy/ fc)db50 db for fc 3 and fy 40 ksi

70 db for fc 3 and fy 60 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 12.14.3.2

which states A full mechanical splice shall develop in tension or

compression, as required, at least 125 % of the specified yield

strength (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 12.14.3.4

which states A full welded splice shall develop at least 125 % of

the 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 > Agfc/10



Provision for Frame Members Subjected to Bending

and Axial Load

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 Members

Subjected to Bending and Axial

Load

2. Longitudinal Reinforcement

0.01 rg 0.06

Clear span, hc


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Provision for Frame

Members Subjected

to Bending and

Axial Load

3. Trans. Rein.

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/6

18


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Provision for Frame Members Subjected to Bending and Axial Load

3. Trans. Rein.

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


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Provision for Frame Members Subjected

to Bending and Axial Load

4. Lap Splice

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 be

proportioned so that hinges occur at locations that least

compromise strength. For this, weak-beam strong-column

approach is used.

After design, the member capacities are calculated based on

designed 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


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To prevent beam column joint cracking, ACI Code 21.5.1

requires that the column dimension parallel to the beam

reinforcement must be at least 20 times the diameter of the

largest longitudinal bar.

Beam longitudinal reinforcement with diameter (db)

20db

Beam

Column


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Beam longitudinal reinforcement that is terminated within a

column. must be extended to the far face of the column core.

The development length (ldh) of bars with 90 hooks must be not

less than 8db, 6 inch, Or fydb/ (65 fc).

Beam longitudinal reinforcement

Beam

Column

ldh


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Provision for Flexural Members

1. Size: No special requirement (Just as ordinary beam

requirement).

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

3. Transverse steel: Same as for SMRF.

4. Lap: No special requirement (Just as ordinary beam

requirement).

ACI Provisions for Intermediate Moment Resisting Frames (IMRF)

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2. Flexural Reinforcement

Asl

Asl+ (Asl)/3

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

Asr

Asr+ (Asr)/3

rmin = 3fc/fy, 200/fy (at critical sections)

t 0.004


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Provision for Columns

1. Size: No special requirement (Just as ordinary column

requirement)

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

requirement)

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

4. Lap: No special requirement (Just as ordinary column

requirement)


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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/6

18

s d/2 (As per ACI 11.5.4)


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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 without

beams are allowed in regions of moderate seismic risk.

In regions of low or no seismic risk ordinary moment resisting

frames OMRF are allowed but IMRF and SMRF may also be

provided.

Miscellaneous Considerations

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Example 2: Detail the selected frame of E-W interior frame of the

given structure as per SMRF requirements. The structure is

already 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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Given 3D structure:

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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Selected portion of E-W interior Frame:

20 ft 20 ft 20 ft 20 ft

Portion of frame considered


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Reinforcement from Design:

For the beam in the given frame, we have

Asmin = 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 have

Asmin = 0.01 x15 x 15 = 2.25 in2

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

1.61 1.61

1.02 1.021.21

2.25 2.25

Reinforcement in in2


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Calculation of number of bars:

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:


Positive bars (mid span):

1.21/0.31 = 3.9 (take 4 bars in 1 layer)

Positive bars (at joint):


Column reinforcement:

2.25/ 0.31 = 7.25 (take 8 bars for even distribution

of bars at all faces of column)

6 bars 6 bars

4 bars 4 bars4 bars

8 bars 8 bars

No. of #5 bars


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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


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Flexural Reinforcement

As+ (at joints) As (at joints)

4 #5 bars (6 #5 bars)

As (any section) Max. As at joints

2 #5 bars (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

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 = 36s 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 of

the support.

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

Lap splice length =1.3 ld = 1.30.05 (fy/ fc)db 30 = 2.5

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

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 bars

which gives g = 0.011, within the specified range 0.01 g 0.06.

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SMRF Requirements

Checklist


Transverse Reinforcement:

lo = max (larger column

dimension, hc/6, 18) = 18

Spacing of ties in lo region is

least of = smaller column

dimension/4, 6 long bar

dia = 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






Lap Splice:

Tension lap splice within center

half of member length.

Spacing of ties in lap splice not

more 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 that

the column dimension parallel to the beam reinforcement must be at

least 20 times the diameter of the largest longitudinal bar.

20 5/8 = 12.5

Column dimension parallel to beam long bar = 15, OK

99






6 #5 bars

6 #5 bars

2

ColumnBeam

Joint

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Example 3: A two dimensional frame of the building is shown in

figure 01. All beams in the frame are 12 wide and 18 deep. All

columns are 12 square. It is required only for frame BFGC to:

a. Provide suitable number of bars in beams and columns for which flexural

reinforcement is shown in the figure (Use only #5 bars as main reinforcement

and #3 bars as transverse reinforcement).

b. Provide suitable spacing for stirrups and ties when shear reinforcement as

per 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.

101




Given Building:

Figure 01: RC Building Frame

3.2 in2

2.4 in2

1.2 in22.6 in2 2.6 in2 2.4 in2

1.2 in21.8 in23.0 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:

2.6 in2 2.6 in2

1.8 in23.0 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:



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):


Column reinforcement:

3.2/ 0.31 = 10.32 (take 12 bars for even

distribution of bars at all faces of column)

103




Solution of Part (a)

(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:

0.23 in2/ft

F G

E H

B CA D

J K

10 ft 15 ft 10 ft

Spacing= (Ab/As) 12

With #3, 2 leg vertical stirrups (Ab = 0.22 in2)

Stirrup spacing for beams:

sb = (0.22/0.23) 12 = 11.47 c/c

As per ACI minimum reinforcement and maximum

spacing requirement, spacing should not exceed:

i. Avfy/50bw = 14 c/c

ii. d/2 = 16.5/2 = 8.25 c/c

iii. 24 c/c

iv. Avfy/{0.75(fc)bw = 17.85 c/c

Finally sb = 8.25 c/c

Tie spacing for column:

s = (0.22/0.23) 12 = 11.47 c/c

Spacing 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

105




Solution of Part (b)

As given in problem, Shear reinforcement spacing with #3 bar is:

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 12/15 = 15 > 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.

107




Solution of Part (c): SMRF Requirements Checklist


Flexural Reinforcement

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) As (at joints)

4 #5 bars {(5+5) #5 bars}

As (any section) Max. As at joints

Provide at least 3 bar

N.G

Provide at least 5 bars at joint.O.K

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Recommended Flexural Reinforcement

109

Asl = (5+5) #5 Asr = (5+5) #5

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

3 #5






Transverse Reinforcement

s

d/4 = 15/4 = 3.75

8 smallest long. bar dia.= 8 5/8= 5

24 hoop bar diameter = 24 3/8= 9

12 2

2h = 36s 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 of

the support.

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

Lap splice length =1.3 ld = 1.30.05 (fy/ fc)db 30 = 2.5

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

2h=36 2h=36

111






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 bars

which gives g = 0.025, within the specified range 0.01 g 0.06.

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Solution 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 long

bar 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

113




Solution of Part (c): SMRF

Requirements Checklist


Lap Splice:

Tension lap splice within center

half of member length.

Spacing of ties in lap splice not

more 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 that

the column dimension parallel to the beam reinforcement must be at

least 20 times the diameter of the largest longitudinal bar.

20 5/8 = 12.5

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

115






(5+5) #5 bars

(4+2) #5 bars

2

ColumnBeam

Joint

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Solution of Part (d):

Presentation of appropriately proportioned structural details of

beam, column and selected joint.

In this part, structural drawings showing all calculated SMRF details

have been asked to draw. This has already been done in part (c) along

with 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.

117





Structural Details for beam (Long section):

s = 3.75

2

2h = 36s = 7.5

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

118

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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 @ 3.75 c/c

#3, 2 legged @ 7.5 c/c

119





Structural Details for Column:

lo = 18.5

12 12 column

12, #5 bars

G.L

lo = 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 bars

2

ColumnBeam

Joint

121




References

ACI 318

UBC-97

BCP SP-2007

Earthquake tips from IITK.

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The End

123

Documents

L-06-EQ-Resistant-Design-of RCC Structures