lecture-09 introduction to earthquake resistant analysis

1

Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Lecture-09

Introduction to Earthquake Resistant Analysis & Design of

RC Structures (Part I)

1

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

www.drqaisarali.com



Topics

Introduction

Earthquake Design Philosophy

Seismic Loading Criteria

Analysis for Seismic Loads

Approximate Lateral Load Analysis

References

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2



Introduction

Earth’s Interior

3



Earthquake results from the sudden movement of

the tectonic plates in the earth’s crust.

4

Introduction

3



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.

5

Introduction



Seismic Events around the globe

Mostly takes place at boundaries of Tectonic plates

6

Dots represents an earthquake

Introduction

4



Types of Waves Generated Due to Earthquake

7

Body Waves Surface Waves

Introduction



Displacement due to Earthquake

8

Introduction

5



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 toor subtracts from the acceleration due to gravity.

9

Introduction



Horizontal and Vertical Shaking

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

10

Introduction

6



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.

11

Introduction



Earthquake characteristics with respect to distancefrom 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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Introduction

7



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 = 2p√(m/k)}

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

13

Introduction



Resonance risk for structures w.r.t near, intermediateand 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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Introduction

8



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.

15

Introduction




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.

16

Introduction

9




Some of the famous

earthquake records

17

Introduction



Earthquake Occurrence

18

Introduction

10



Seismic Zones

19

Introduction



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.

20

Introduction

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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.

21

Introduction



Other Undesirable Scenarios

22

Introduction

12



Soft Storey

23

Introduction



Earthquake Design Philosophy

Performance level

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13



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],


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14



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


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15




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)


30

16



=

Static Lateral Force Procedure


31




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

determined 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.


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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 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”.


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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 seismiczone).


35





Step 1: Site Specific details.

i. Seismic Zone

Source: BCP SP-2007


36

19






ii. Soil Type

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


37






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.


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20





Step 2: Determination of Seismic Coefficients.

Cv:

Nv (required only for zone 4):


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Ca:

Na (required only for zone 4):


40

21





Step 3: Determination of Seismic Importance Factor.


41





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.


42

22







43





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 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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23





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) 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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24




Example: Calculation of “V” for E-W interior frame of the given

structure. Structure is located in Peshawar. Soil type is stiff.

25 ft

25 ft

25 ft

25 ft

20 ft

20 ft

20 ft

10 ft

10 ft

10 ft (floor to floor)

SDL = NilLL = 144 psf



fc′ = 4 ksify = 60 ksi

Slab-Beam Frame

Structure


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Example:

E-W interior frame4 spans @ 25′-0″

3 spans @

20′-0″

l2 = 20′


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25



49


Example:

Step 1: Site specific details.

i. Seismic Zone:

From seismic zoning map of

Pakistan, Peshawar lies in

seismic zone 2B (Z = 0.20)





Example:


ii. Soil Type: Stiff soil is classified as SD (stiff soil).


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26




Example:


iii. Past earthquake magnitude: Not determined as it required for zone

4 only.

iv. Distance to known seismic zone: Not determined as it is required

for zone 4 only.


51




Example:


For seismic zone 2B, only Ca and Cv determination is required.


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53


Example:

Step 3: Determination of Seismic Importance Factor.

I = 1.00 (Standard Occupancy

Structures)





Example:


R = 8.5 (Concrete SMRF)


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Example:

Step 5: Determination of Structure’s time period.

By method A:

T = Ct (hn)3/4

Ct = 0.003; hn = 30 ft

T = 0.003 × (30)3/4 = 0.384 sec


55




Example:

Step 6: Determination of base shear (V).

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

W (self weight of E-W interior frame + super imposed dead load)

= 613 kips

25 % floor live load will also be added up (for warehouses, see

UBC sec.1630.1.1.)

W = 613 + 0.25 0.144 20×(4×25) = 685 kips


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Example:

Step 6: Determination of base shear (V).

V = {CvI/RT}W = {0.40 1.00/ (8.5 0.384)} 685 = 83.94 kips

The total design base need not exceed the following:

V = (2.5CaI/R) W = {(2.5 × 0.28 × 1.00)/ (8.5)} × 685 = 56.41 kips,

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

V = 0.11CaIW = 0.11 × 0.28 × 1.00 × 685= 21.098 kips, O.K.

Therefore, V = 56.41 kip (8 % of seismic weight W)


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Example:

Step 7: Vertical distribution of V to storeys.

Fx = (V – Ft)ωxhx/∑ωihi

∑ωihi = 228 ×10 + 228×20 + 228×30 = 13680 kip

F1 = (56.41 – 0) × 228 × 10/ {(13680)} = 9.402 kip

Storey forces for other stories are given in table below:

Table Storey shears.Level

xhx (ft) wx (kip) wxhx (ft-kip) wxhx /(Swihi) Fx (kip)

3 30 228 6840 0.5 28.212 20 228 4560 0.33 18.611 10 228 2280 0.166 9.36

Swihi = 13680 Check SFx =V = 56.18 kip OK


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Example (Storey Forces):

Same forces will be obtained for other E-W interior frame because it has

same dimensions and loading conditions as of E-W interior frame

considered.

Half values shall be applied to E-W exterior frames.


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Example (Storey Forces):

Note: Base shear can

also be computed for

complete structure and

then can be divided to

different frames.

25 ft25 ft25 ft25 ft

20 ft

20 ft

20 ft

28.21 kips

18.61 kips

9.36 kips

28.21 kips

18.61 kips

9.36 kips

14.1 kips

9.3 kips

4.68 kips

14.1 kips

9.3 kips

4.68 kips


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Base Shear using UBC Response Spectra

Example:


Period (sec)

Spectral Acceleration

(g’s)

Ca = 0.28

Cv = 0.4

Ts = Cv/2.5Ca = 0.57 sec

To = 0.2Ts = 0.114 sec

R = 8.5; W = 685 kips

Now, T of given structure = 0.384 sec

At T = 0.384 sec,

Spectral acceleration = 0.7g

V = W × (a/g)/R = 56 kips

Base shear computed here is same as

computed using the static lateral force

procedure.

Ca = 0.28

2.5Ca = 0.7

Cv/T

0.114 0.57

Line at T = 0.384 sec

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Automated lateral force procedure of SAP2000

Steps for the given 3D structure are shown next.

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SAP2000 3D Model (20ft × 15 ft) panelsSeismic Zone: 2B

Soil Type: SDMethod A used for time period calculation

Mass source: SDL only

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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures63


Static Lateral Force Procedure1. Automated Lateral Force Procedure of SAP2000

It is important to add SDL as Load for masssource with 3rd option selected to avoid load tobe taken two times.


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures64



Case Study 2: Base shear calculation for E-W direction using

SAP2000 automated lateral load feature and comparison with

results of manually applied lateral loads.

1. Automated Lateral Force Procedure of SAP2000

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Analysis for Seismic Loads

Methods of Seismic (lateral load) Analysis

Exact: FEM using SAP 2000, etc.

This method was demonstrated in previous example

Approximate lateral load analysis:

This will be discussed next

65



ACI Requirements on Lateral Load Analysis

Unlike ACI 6.4 which allows separate floor analysis for

gravity loads, ACI R6.4 states that for lateral load analysis, a

full frame from top to bottom must be considered.

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For Lateral LoadFor Gravity Load


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Portal Method

This is a method used to estimate the effects of side swaydue to lateral forces acting on multistory building frame.

This method is specialized form of point of inflection method.

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F3

F2

F1

Side sway (Δ)




Portal Method

Prepositions:

1. The total horizontal shear in all columns of a given storey is equal and

opposite to the sum of all horizontal loads acting above that storey.

This preposition follows from the requirement that horizontal

forces be in equilibrium at any level.

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F3

F2

F1

H31 H32

H21 H22

H11 H12

H31 + H32 = F3

H21 + H22 = F3 + F2

H11+ H12 = F3 + F2 + F1


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Portal Method

Prepositions:

2. The horizontal shear is the same in both exterior columns. The

horizontal shear in each interior column is twice that in exterior column.

This preposition is due to the fact that interior columns are generally more rigid than

exterior columns (interior column with larger axial load will require larger cross section).

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H3

H2

F3

F2

F1

H3 2H3 2H3

H2 2H2 2H2

H1 2H1 2H1 H1

6 H3 = F3

or H3 = F3 /6H3 = F3 / 2nWhere n= no. of baysAnd 2H3 = F3 / n




Portal Method

Prepositions:

3. The inflection points of all members (columns and beams) are located

midway between the joints except for bottom storey.

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F3

F2

F1

Point of Inflection

2h/3

h/3

Location of P.O.I depends on end restraints:

2h/3 (restraints with more resistance to rotation)

h/3 (restraints with less resistance to rotation)

At base (ideal hinge)


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Portal Method Analysis Steps

Step 1: Location of points of inflection on frame using preposition 3.

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F3

F2

F1

l1 l2 l3





Step 2: Determine column shears using proposition 1 and 2.

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F3

H3ext=F3/2n

F2

F1

l1 l2 l3

H3int=F3/n H3int=F3/n H3ext=F3/2n

H2ext=(F3 + F2)/2n H2int=(F3 + F2)/n H2int=(F3 + F2)/n H2ext=(F3 + F2)/2n

H1int=(F3 + F2 + F1)/n H1ext=(F3 + F2 + F1)/2n H1int=(F3 + F2 + F1)/n H1ext=(F3 + F2 + F1)/2n

n = number of bays


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Step 3a: Determine column moments from statics.

73

F3

F2

F1

l1 l2 l3

H3ext H3int H3extH3int



h

h

h

M3ext= H3exth/2

M3ext= H3exth/2

M3int= H3inth/2

M3int= H3inth/2

M2ext= H2exth/2

M2ext= H2exth/2

M2int= H2inth/2

M2int= H2inth/2

M1ext= H1exth/3

M1ext= H1ext2h/3

M1int= H1inth/3

M1int= H1int2h/3

M3int= H3inth/2

M3int= H3inth/2

M2int= H2inth/2

M2int= H2inth/2

M1int= H1inth/3

M1int= H1int2h/3

M3ext= H3exth/2

M3ext= H3exth/2

M2ext= H2exth/2

M2ext= H2exth/2

M1ext= H1exth/3

M1ext= H1ext2h/3





Step 3b: Determine beam moments from statics.

Beam moments at a joint can be determined from equilibrium. The beam

moments to the left (MBL) and right (MBR) of a joint can be determined

from the following formulae:

MBL= ∑Mcol/m

MBR= ∑Mcol/m

Where,

m = # of connecting beams at a joint.

∑Mcol = summation of column moments at a joint.

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Step 3b: Determine beam moments from statics.

75

F3

F2

F1

l1 l2 l3

M3ext

M3int

M2int

Note: The direction of beam moment shall be opposite to the direction of column moment.

M3int

MBL= M3ext/1

MBR= M3int/2

MBL= M3int/2

MBR= (M3int+M2int)/2 MBL= (M3int+M2int)/2





Step 3c: Determine beam shear from statics.

As the point of inflection is assumed to lie at mid span, the beam shear

equals beam end moment divided by ½ beam span.

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Step 3c: Determine beam shear from statics.

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F3

F2

F1

l1 l2 l3

MBLMBR

PL=MBL/0.5l1PR=MBR/0.5l1

MBLMBR

PL=MBL/0.5l2 PR=MBR/0.5l2PL PR

PL PR

PL PR

PL PR

PL PR

PL PR

PL PR





Step 3d: Determine column axial force from statics.

For a segment (abc for example), the axial force shall be arithmeticsum of beam shears within that segment, but in opposite direction.

Axial force in lower storey shall be the sum of axial force in storeyunder question plus the axial forces in all above stories.

78

F3

F2

F1

l1 l2 l3

PL PRPL PR PR

PL PRPL PR PL PR

PL PRPL PR PL PR

a

b

c


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Step 3d: Determine column axial force from statics.

79

F3

F2

F1

l1 l2 l3

PL3 PR3PL3 PR3 PR3

PL2 PR2PL2 PR2 PL2 PR2

PL1 PR1PL1 PR1 PL1 PR1

PL3 PL3

PL3+PL2

PR3+PL3

Similarly all other column axial forces can be determined

PL3+PL2 +PL1




Portal Method (Case Study 1)

Lateral load analysis for E-W Interior Frame of given 3D

structure by portal method and its comparison with SAP2000.

The objective of this study is to check the level of accuracy of portal

method.

80


41




Given 3D structure.

25 ft 25 ft 25 ft 25 ft

20 ft

20 ft

20 ft

10 ft

10 ft

10 ft (floor to floor)





Slab-Beam Frame

Structure

Note: Zone 2BSDL = NilLL = 144 psfSlab = 7″

81





E-W Interior FrameF3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft

h=10 ft

F2 =18.61 kip

F1 =9.36 kip

h=10 ft

h=10 ft

l4=25 ft

82


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Step 1: Locate points of inflection.F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

For Hinge

83





Step 2: Determine column shear.F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

H3ext=F3/2n H3int=F3/n

n = 4

7.05 7.05 3.5

5.85

3.5 7.05

5.8511.7 11.7 11.7

H2ext=(F3 + F2)/2n H2int=(F3 + F2)/n

H1int=(F3 + F2 + F1)/n H1ext=(F3 + F2 + F1)/2n

7.00 14.0 14.0 14.0 7.00

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Step 2: Determine column shear (comparison with SAP).

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

3.5(4)

5.85(7)

7.00(11)

7.05(7)

11.7(13)

14.0(13)

7.05(7)

11.7(12)

14.0(13)

Portal MethodSAP 3D

85





Step 3a: Determine column moments.F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

M = H ×h/2 (for all stories except bottom)M = H × h (for bottom storey)

7.05 7.05 3.5

5.85

3.5 7.05

5.8511.7 11.7 11.7

7.00 14.0 14.0 14.07.00

17.5

17.5

29.3

29.3

70 140

58.5

35.25

35.25

58.5

140

58.5

35.25

35.25

58.5

140

58.5

35.25

35.25

58.5

17.5

17.5

29.3

29.3

70

86


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Step 3a: Determine column moments (comparison with SAP).

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

17.5(28)

17.5(20)

35.25(39)

35.25(33)

35.25(37)

35.25(32)

29.3(45)

29.3(28)

58.5(69)

58.5(62)

58.5(65)

58.5(56)

70(111)

140(133)

140(129)

Portal MethodSAP 3D

87





Step 3b: Determine beam moments.F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

MBL= ∑Mcol/m

MBR= ∑Mcol/m

17.5

17.5

29.329.3

70 140

58.5

35.25

35.25

58.5

140

58.5

35.25

35.25

58.5

140

58.5

35.25

35.25

58.5

17.5

17.5

29.329.3

70

17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5

46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.8

99 99 99 99 99 99 99 99

88


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Step 3b: Determine beam moments (comparison with SAP).

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

17.5(19)

17.5(16)

17.5(14)

17.5(14)

46.8(46)

46.8(41)

46.8(37)

46.8(37)

99(99)

99(81)

99(70)

99(68)

Portal MethodSAP 3D

89





Step 3c: Determine beam shear.

F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

PL= MBL/0.5l

PR= MBR/0.5l

17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5

46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.8

99 99 99 99 99 99 99 99

1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4

3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74

7.92 7.92 7.92 7.92 7.92 7.92 7.92 7.92

90


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Step 3c: Determine beam shear (comparison with SAP).

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

1.4(1.8)

1.4(1.5)

3.74(4.4)

3.74(4)

7.92(9)

7.92(7)

Portal MethodSAP 3D





Step 3d: Determine column axial loads.F3 =28.21 kip

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

F2 =18.61 kip

F1 =9.36 kip

1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4

3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74

7.92 7.92 7.92 7.92 7.92 7.92 7.92 7.92

1.4 0 0 0

0 0 0

0 0 0

5.14

13.06

1.4

5.14

13.06

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47




Step 3d: Determine column axial loads (comparison with SAP).

l1=25 ft l2=25 ft l3=25 ft l4=25 ft

1.4(2)

13.06(17)

5.14(7)

0(-0.5)

0(-1.4)

0(-4.4)

0(0)

0(0)

0(0)

Portal MethodSAP 3D

93





l1=20 ft l2=20 ft l3=20 ft l4=20 ft

16(16)

16(13)

31(22)

31(20)

26(26)

26(19)

52(38)

52(36)

61(68)

123(77)

Portal MethodSAP 3D


Similar comparison for 20 × 15 ft structure is shown below:

94

48




l1=20 ft l2=20 ft l3=20 ft l4=20 ft

16(12)

16(9)

16(8.5)

16(8)

41(29)

41(25)

41(22)

41(22)

87(64)

87(50)

87(43)

87(40)

Portal MethodSAP 3D


Similar comparison for 20 × 15 ft structure is shown below:

95




Lateral Load Analysis of a frame corresponding to seismic

demand in seismic zones 1 to 4 using Portal Method.

96


49




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




Slab-Beam Frame

Structure

97





Zone 1 (Bending moments)

Ca = 0.12Cv = 0.18R = 8.5V = 20.04 kips

98


50




Zone 2A (Bending moments)

Ca = 0.22Cv = 0.32R = 8.5V = 36.74 kips

99





Zone 2B (Bending moments)

Ca = 0.28Cv = 0.40R = 8.5V = 46.76 kips

100


51





Ca = 0.36Cv = 0.54R = 8.5V = 60.11 kips

101






Ca = 0.44Cv = 0.64R = 8.5V = 73.47 kips

102


52




Comparison (Interior Negative Beam Moment)

Top

Intermediate

Bottom

103





Comparison (Column Moment)

Top

Intermediate

Bottom

104


53



References

ACI 318-14

UBC-97

BCP SP-2007

Earthquake tips from IITK.Intermediate

Bottom

105



The End

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lecture-09 introduction to earthquake resistant analysis

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