lecture 14 introduction to earthquake resistant design of rc structures (part i)_2011.pdf

7/27/2019 Lecture 14 Introduction to Earthquake Resistant Design of RC Structures (Part I)_2011.pdf

http://slidepdf.com/reader/full/lecture-14-introduction-to-earthquake-resistant-design-of-rc-structures-part 1/57

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

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011

Lecture-14

Introduction to Earthquake

Resistant Design of RC

Structures (Part I)

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]

1



Topics

Introduction

How Architectural Features Affect Buildings During

Earthquakes

Earthquake Design Philosophy

Seismic Loading Criteria

Analysis for Seismic Loads

Approximate Lateral Load Analysis (Portal method)

2





Introduction

Earth’s Interior

3



Introduction

Earthquake occurrence

Earthquake results from the sudden movement of the

tectonic plates in the earth’s crust.

4





Introduction

Seismic Events around the globe

Mostly takes place at boundaries of Tectonic plates

5

Dots

represents an

earthquake



Introduction

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.

6





Introduction

Types of Waves Generated Due to Earthquake

7

Body Waves Surface Waves



Displacement due to Earthquake

Introduction





Introduction

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.



Introduction

Horizontal and Vertical Shaking

The structures are normally designed for horizontal shaking

to minimize the effect of damages due to earthquakes.





Introduction

Earthquake characteristics with respect to distance

from epicenter

0.05 ≤ T ≤ 0.3

20 Hz ≥ f ≥ 3.33 Hz

Low period & high

frequency field

0.3 ≤ T ≤ 1.0 sec

3.33 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



Introduction

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 = 2π√(M/K)}

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





Introduction

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)



Introduction

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 bedetermined.

Working of seismograph shown

in figure.





Introduction


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 3measurement.



Introduction


Some of the famous

earthquake records





Introduction

Number of Earthquakes per year



Introduction

Seismic Zones





How Architectural Features Affect

Buildings During Earthquakes?

Other Undesirable Scenarios



How Architectural Features Affect

Buildings During Earthquakes?

Soft Storey





Earthquake Design Philosophy

Performance level



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.






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], and/or




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.







Time History Analysis (THA)

T

Ground

accelerationT

Lateral

Displacement





Response Spectrum Analysis (RSA)

a (ft/sec2)

T

Response

T

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






UBC-97 Response Spectrum Curve

(Acceleration vs. Time period)


Response Spectrum Analysis (RSA)




=


Static Lateral Force Procedure






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 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.





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 = (0.8ZNνI/R) W ]

Where, Nν

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

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







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 and

minimum.

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 seismic

zone).








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.









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.






Step 2: Determination of Seismic Coefficients.

Cv:

Nv (required only for zone 4):









Ca:

Na (required only for zone 4):






Step 3: Determination of Seismic Importance Factor.








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.














Step 5: Determination of structure’s 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.






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








Step 7: Vertical Distribution of V to storeys.

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

F x = (V – Ft)ω x h x /∑ωi hi (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 F x force at that level.

Ft = 0.07TV {for T > 0.7 sec}

Ft = 0 {for T ≤ 0.7 sec}





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

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

46

25 ft 25 ft 25 ft 25 ft

20 ft

20 ft

20 ft

10 ft

10 ft

10 ft (floor to floor)

SDL = Nil

LL = 144 psf

SDL = NilLL = 144 psf

SDL = Nil

LL = 144 psf

f c′ = 4 ksi

f y = 60 ksi

Slab-

Beam

Frame

Structure







Example:

E-W interior frame

4 spans @ 25′-0″

3 s p a n s

@ 2 0 ′ - 0 ″

l 2 = 20′





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).





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 requiredfor zone 4 only.







Example:


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





Example:

Step 3: Determination of Seismic Importance Factor.

I = 1.00 (Standard Occupancy

Structures)







Example:


R = 8.5 (Concrete SMRF)





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







Example:

Step 6: Determination of base shear (V).

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

W = 3 (w slab + w beams + w columns)E-W Frame

=3[(7/12)×20×(4×25)+{(14×13)/144}×(4×25)+5×{(14×14)/144}×10]× 0.15

= 3 × 204.33 = 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





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, N.G.

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)







Example:

Step 7: Vertical distribution of V to storeys.

F x = (V – Ft)ω x h x /∑ωi hi

∑ωi hi = 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

xh x (ft) w x (kip) w xh x (ft-kip) w xh x /(Swihi) F x (kip)

3 30 228 6840 0.5 28.21

2 20 228 4560 0.33 18.61

1 10 228 2280 0.166 9.36

Swihi = 13680 Check SF x =V = 56.18 kip OK





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.







Example (Storey Forces):

25 ft25 ft25 ft25 ft

20 ft

20 ft

20 ft

28.21 kips

18.61 kips

9.36 kips

Note: Base shear can

also be computed for

complete structure and

then can be divided to

different frames.

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





Class Activity: Calculate the base shear for the given frame.


200 kip

200 kipCv = 0.28I = 1.00

R = 8.5







Case Study 1: Comparison between the procedures of SAP2000 utilizing

automated lateral force feature and manually applied lateral loads for the

given frame.

The objectives of this study are:

To present the automated lateral force calculation feature of software

SAP2000.

To compare the manual lateral load application procedure with

automated load application feature of SAP2000.





Case Study 1:

3D structure under study

SAP2000 3D Model

(20ft × 15 ft) panels

Seismic Zone: 2B

Soil Type: SD

Method A used for time period calculation

Mass source: SDL only







Case Study 1:

1. Automated Lateral Force Procedure of SAP2000

Steps of this method are shown next.




Static Lateral Force Procedure1. Automated Lateral Force Procedure of SAP2000

It is important to add SDL as Load for mass source

with 3rd option selected to avoid load to be taken

two times.







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





Case Study 1:

2. Manual Lateral Force Procedure using SAP2000

Steps of this method are shown next.






Case Study 1:


Cv = 0.4; Ca = 0.28; I = 1.00; R = 8.5; T = 0.398 sec; W = 1703 kip

V = (CvI/RT)W = 201 kip; V = 2.5CaI/R)W/R = 140 kip (governs)


Table Storey shears.

Le

vel

x

h x

(ft)

w x

(kip)

w xh x (ft-

kip)

w xh x

/(Swihi)F x (kip)

Fx (kip)

Interior frame

Fx (kip)

Exterior frame

331.

5567 17860.5 0.5 70 70/3 = 23.33 23.33/2=11.65

2 21 567 11907 0.33 46.6 46.6/3=15.5 15.5/2=7.75

110.

5567 5953 0.166 23.24 23.34/3=7.74 7.74/2=3.87

Swihi = 35720.5Check SF x =V = 140

kip OK

Divided by 3 because of 3complete frames in E-W direction.


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 20116820 ft20 ft20 ft20 ft

15 ft

15 ft

15 ft

23.33 kips

15.5 kips

7.74 kips

23.33 kips

15.5 kips

7.74 kips

11.65 kips

7.75 kips

3.87 kips

11.65 kips

7.75 kips

3.87 kips



Case Study 1:





Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 201169l 1=20 ft l 2=20 ft l 3=20 ft l 4=20 ft

16

(16)

13

(13)

22

(22)

20

(20)

27

(26)

19

(19)

38

(38)

36

(36)

66

(68)77

(77)

Manual

SAP Automated

Seismic Loading Criteria Static Lateral Force Procedure

Case Study 1:

Results: Comparison of Column Bending Moment. In the case of “Manual” the forces are

applied at nodes as shown on previous slide. In “SAP automated” SAP does this automatically


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 201170l 1=20 ft l 2=20 ft l 3=20 ft l 4=20 ft

12

(12)

9.4

(9)8.5

(8.5)

8

(8)

29(29)

24

(25)22

(22)

21(22)

63

(64)

49

(50)

42

(43)

39

(40)

Manual

SAP Automated



Case Study 1:

Results: Comparison of Beam Bending Moment.





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

Ca = 0.28

2.5Ca = 0.7

Cv/T

0.114 0.57

Line at T = 0.384 sec



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

72





Approximate Lateral Load

Analysis ACI Requirements on Lateral Load Analysis

Unlike ACI 8.9 which allows separate floor analysis for

gravity loads, ACI R8.9 states that for lateral load analysis, a

full frame from top to bottom must be considered.

73

For Lateral LoadFor Gravity Load




Analysis Portal Method

This is a method used to estimate the effects of side sway

due to lateral forces acting on multistory building frame.

This method is specialized form of point of inflection method.

74

E3

E2

E1

Side sway (Δ)







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.

75

F3

F2

F1

H31 H32

H21 H22

H11 H12

H31 + H32 = F3

H21 + H22 = F3 + F2

H11+ H12 = F3 + F2 + F1





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).

76

H

H

F3

F2

F1

H 2H 2H

H 2H 2H

H 2H 2H H

6 H = F3

or H = F3 /6

H = F3 / 2n

Where n= no. of bays







Prepositions:

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

midway between the joints except for bottom storey.

77

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)





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

78

F3

F2

F1

l 1 l 2 l 3







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

79

F3

H3ext=F3/2n

F2

F1

l 1 l 2 l 3

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)/nH1ext=(F3 + F2 + F1)/2n H1int=(F3 + F2 + F1)/n H1ext=(F3 + F2 + F1)/2n

n = number of bays





Step 3a: Determine column moments from statics.

80

F3

F2

F1

l 1 l 2 l 3

H3ext H3int H3ext H3int



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.

81





Step 3b: Determine beam moments from statics.

82

F3

F2

F1

l 1 l 2 l 3

M3ext

M3int

M2int

Note: The direction

of beam moment

shall be opposite tothe direction of

column moment.

M3int

MBL= M3ext/1

MBR = M3int/2

MBL= M3int/2

MBR = (M3in3+M2int)/2 MBL= (M3in3+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.

83





Step 3c: Determine beam shear from statics.

84

F3

F2

F1

l 1 l 2 l 3

MBLMBR

PL=MBL/0.5l 1 PR =MBR /0.5l 1

MBLMBR

PL=MBL/0.5l 2 PR =MBR /0.5l 2 PL 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 arithmetic

sum of beam shears within that segment, but in opposite direction.

Axial force in lower storey shall be the sum of axial force in storey

under question plus the axial forces in all above stories.

85

F3

F2

F1

l 1 l 2 l 3

PL PR PL PR PR

PL PR PL PR PL PR

PL PR PL PR PL PR

a

b

c





Step 3d: Determine column axial force from statics.

86

F3

F2

F1

l 1 l 2 l 3

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







Case Study 2: 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.

87





Case Study 2: Given 3D structure.

88

25 ft 25 ft 25 ft 25 ft

20 ft

20 ft

20 ft

10 ft

10 ft

10 ft (floor to floor)

SDL = Nil

LL = 144 psf

SDL = NilLL = 144 psf

SDL = Nil

LL = 144 psf

f c′ = 4 ksi

f y = 60 ksi

Slab-

Beam

Frame

Structure

Note: Zone 2B

SDL = Nil

LL = 144 psf

Slab = 7″







Case Study 2: E-W Interior Frame

89

F3 =28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft

h=10 ft

F2 =18.61 kip

F1 =9.36 kip

h=10 ft

h=10 ft

l 4=25 ft




Analysis Portal Method (Case Study 2)

Step 1: Locate points of inflection.

90

F3 =28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft l 4=25 ft

F2 =18.61 kip

F1 =9.36 kip

For

Hinge







Step 2: Determine column shear.

91

F3 =28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft l 4=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

5.85 11.7 11.7 11.7

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

H1int=(F3 + F2 + F1)/nH1ext=(F3 + F2 + F1)/2n

7.00 14.0 14.0 14.0 7.00





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

92l 1=25 ft l 2=25 ft l 3=25 ft l 4=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 Method

SAP 3D







Step 3a: Determine column moments.

93

F3 =

28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft l 4=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

5.85 11.7 11.7 11.7

7.00 14.0 14.0 14.0 7.00

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





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

94l 1=25 ft l 2=25 ft l 3=25 ft l 4=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 Method

SAP 3D







Step 3c: Determine beam shear.

97

F3 =

28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft l 4=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





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

98l 1=25 ft l 2=25 ft l 3=25 ft l 4=25 ft

1.4

(1.8)

1.4

(1.5)

3.74(4.4) 3.74(4)

7.92

(9)

7.92

(7)

Portal Method

SAP 3D







Step 3d: Determine column axial loads.

99

F3 =

28.21 kip

l 1=25 ft l 2=25 ft l 3=25 ft l 4=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





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

100l 1=25 ft l 2=25 ft l 3=25 ft l 4=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 Method

SAP 3D





Approximate Lateral Load Analysis

101l 1=20 ft l 2=20 ft l 3=20 ft l 4=20 ft

16

(16)

16

(13)

31

(22)

31

(20)

26

(26)

26

(19)

52

(38)

52

(36)

61

(68)123

(77)

Portal Method

SAP 3D

Portal Method (Case Study 2) Similar comparison for 20 × 15 ft structure is shown below:



Approximate Lateral Load Analysis

102l 1=20 ft l 2=20 ft l 3=20 ft l 4=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 Method

SAP 3D

Portal Method (Case Study 2)

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







Case Study 3: Lateral Load Analysis of a frame

corresponding to seismic demand in seismic zones 1 to 4

using Portal Method.

103





Case Study 3

104

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 psf

LL = 60 psf

SDL = 40 psf LL = 60 psf

SDL = 40 psf

LL = 60 psf

f c′ = 3 ksi

f y = 40 ksi

Slab-

Beam

Frame

Structure







Case Study 3: Zone 1 (Bending moments)

105

Ca = 0.12

Cv = 0.18

R = 8.5

V = 20.04 kips





Case Study 3: Zone 2A (Bending moments)

106

Ca = 0.22

Cv = 0.32

R = 8.5

V = 36.74 kips







Case Study 3: Zone 2B (Bending moments)

107

Ca = 0.28

Cv = 0.40

R = 8.5

V = 46.76 kips






108

Ca = 0.36

Cv = 0.54

R = 8.5

V = 60.11 kips








109

Ca = 0.44

Cv = 0.64

R = 8.5

V = 73.47 kips





Case Study 3: Comparison (Interior Negative Beam Moment)

110

Top

Intermediate

Bottom







Case Study 3: Comparison (Column Moment)

111

Top

Intermediate

Bottom



References

ACI 318

UBC-97

BCP SP-2007

Earthquake tips from IITK.

112

Intermediate

Bottom





The End

113

lecture 14 introduction to earthquake resistant design of rc structures (part i)_2011.pdf

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