Seismic Analysis/Design of Multi-storied RC Buildings using STAAD.Pro & ETABS according to...

Seismic Analysis/Design of Multi-storied RC Buildings

using STAAD.Pro & ETABSaccording to IS:1893-2002

Presented by . Rahul Leslie

Assistant Director,Buildings Design,

DRIQ, Kerala PWDTrivandrum, India

Topics Covered:

• Computer modelling and analysis using STAAD.Pro & ETABS for– Seismic Coefficient method as per IS:1893

(Part 1)-2002

– Response Spectrum method as per IS:1893(Part 1)-2002

• Miscellaneous points

Seismic Analysis of Multi-storied RC Building Presented by Rahul Leslie

Aspects of Computer Model:• Modelling is done using analysis packages like

STAAD.Pro, STRAP, NISA Des. Studio, ETABS, GT STRUDL, RISA-3D, MIDAS-Gen, etc.

• Model contains • Beams• Columns• Shear walls

But not usually• Slabs (except in Flat slab - Shear wall construction)

• Masonry wall infills• Stair slabs

Foundation is represented by support points only

Model for ETABS: B+G+4=6 stories

ETABS Model

Model for STAAD: G+4 = 5 stories

STAAD Model

Aspects of Computer Model (Cont…)• A model must ideally represent the complete three

dimensional (3D) characteristics of the building, including – geometry– stiffness of various members– supports– load distribution– mass distribution

Beams and columns• Beams and columns are modelled by frame elements• Plinth beams should also be modelled as beams• Slabs are not usually modelled

Supports:• The type of support to be provided is decided by

considering the degree of fixity provided by the foundation.

• Fixed Supports:– Raft foundation: Support to be provided at the column

ends (located at top of the raft)– Pile cap for multiple piles: Support to be provided at the

column ends (located at top of the pile cap)– Isolated footing: When it is founded on hard rock, the

column end may be modelled as fixed (located at the top of the footing)

– Single pile: Fixed support of the column is recommended at a depth of five to ten times the diameter of pile, depending upon the type of soil, from the top of pile cap.

Supports (cont…):• Pinned supports:

– Isolated footing: Support to be provided at the column ends, (located at the bottom of the foundation).

• Spring supports: – Spring supports can be provided with spring constants ,

eg., as per ASCE/SEI 41(2006)

• In General– Engineering judgement must be exercised in modelling the

support

Slabs and Masonry walls• The weight of slabs are distributed, as 2-way load, on

the supporting beams. • The weight of masonry walls are applied as uniform

load on the supporting beam

Slabs and Masonry walls• Since the slabs are not modelled by plate elements, the

structural effect due to their in-plane stiffness (7.7.2.1, IS:1893(Part 1)-2002) can be taken into account by – using ‘Master/Slave’ option (STAAD.Pro)– assigning ‘Diaphragm’ action (ETABS , STAAD.Pro V8i

Sel.4)

Assign diaphragms: select all slabs in a storey and …

ETABS: Floor Diaphragm

STAAD: Floor Diaphragm

Shear walls• Structural shear walls and Shear core which are

integrally connected to the frame and floor slabs, can be modelled by plate elements – ‘Surface elements’ (STAAD.Pro)

– ‘Wall element’ (ETABS)

Analysis as per IS:1893-2002

Seismic Coefficient Method (Static Analysis)

Static analysis:

• The Design horizontal seismic coefficient Ah is calculated from (6.4.2, IS:1893(Part 1)-2002)

– Zone factor Z (Table 2, IS:1893(Part 1)-2002)– Importance factor I (Table 6, IS:1893(Part 1)-2002)– Response reduction coefficient R (Table 7, IS:1893(Part 1)-2002)

– Horizontal Acceleration coefficient Sa/g

Static analysis:• Where the Horizontal acceleration Sa/g is determined

from the Response spectrum curve (Fig.2, IS:1893(Part 1)-2002)

Static analysis (cont…):

Static analysis:• The time period of the structure is determined using

(7.6.1 & 7.6.2, IS:1893(Part 1)-2002)– RC frames without brick infills

h = height of building in m– RC frames with brick infills

d = base dimension in m(parallel to direction of earthquake)

Static analysis (cont…):• Where the type of soils are

– Type I (Rock or Hard soil): N > 30, among other descriptions– Type II (Medium soils): 10 ≤ N ≤ 30 for all soils

N >15 for poorly graded, among other descriptions

– Type III (Soft soils): N < 10(Table 1, IS:1893(Part 1)-2002)

Presented by Rahul LeslieSeismic Analysis of Multi-storied RC Building

Cl.6.4.4, IS:1893(Part 1)-2002

Static analysis:• Importance factor (Table 6, IS:1893(Part 1)-2002)

– I = 1.5 for special buildings– I = 1.0 for other buildings

• Response reduction factor (Table 7, IS:1893(Part 1)-2002)– R = 3 for ordinary detailing (with ordinary detailed shear

wall, if any)– R = 5 for ductile detailing (with ductile detailed shear wall, if

any) ie., as per IS:13920-1993– R = 4 for ductile detailing with ordinary detailed shear wall– R = 4.5 for ordinary detailing with ductile detailed shear wall

Static analysis (cont…):• The base shear is determined by (7.5.3, IS:1893(Part 1)-

• Design lateral force for each level is determined by (7.7.1, IS:1893(Part 1)-2002)

A Simple Example

A six storied structure

Lumped mass model

height 18 m

Period 0.075x(18)0.75 = 0.6554 sec

(Assumed to be open structure)

Levels W (kN) h (m) Wh2

ΣW*(Wh2/ΣWh2) Qi (kN)

1 0 0 0 0 0

2 23.57 3 212.13 1.5541 0.0632

3 23.57 6 848.52 6.2163 0.2529

4 23.57 9 1909.17 13.9866 0.5691

5 23.57 12 3394.08 24.8651 1.0117

6 23.57 15 5303.25 38.8516 1.5807

7 23.57 18 7636.68 55.9464 2.2763

ΣW 141.42 (kN) 19303.83 141.42 kN 5.7539 kN

Vb 5.7539 (kN)

height 18 m

Period 0.6554 sec

Sa/g 1.5258

Ah 0.0407

= (ZI)/(2R) x sa/g= (0.16*1)/(2*3) * 1.5258

The forces are applied …and analysed

The forces are applied

Seismic Coeff. method

ETABS: Define & Apply Seismic

parameters:• Direction• T

• Z, I, R

• Soil Type

STAAD: Define Seismic parameters:

• Z, I, R• Structure Type or

Tx & Tz

• Soil Type • Damping ratio ξ

• Direction (X, Y,Z), factor

41ETABS: Seismic coeff. method

42STAAD: Seismic coeff. method

43STAAD: Seismic coeff. method

Masses to be included:• For seismic analysis, the effective masses to be

included for analysis are :– Full dead load– 0.25 times Imposed Loads having intensity ≤ 3 kN/m2

– 0.5 times Imposed Loads having intensity > 3 kN/m2

– Imposed Load on roof need not be considered(7.3.1,7.3.2 & Table 8, IS:1893(Part 1)-2002)

• Live load reduction for upper floors (as per 3.2, IS:875(Part 2) - 1987) shall not be applied further for mass calculation (7.3.3, IS:1893(Part 1) - 2002)

Add Seismic Masses

EATBS• Select loads to

combine

STAAD• Add self wt., Joint

loads, Member loads, Floor loads

OR• Put lumped mass

Define Mass Source:-

ETABS: Seismic masses

47STAAD: Seismic masses

STAAD.Pro V8i SELECT 4

Results of Seismic Analysis – Bending Moment & Shear Force

• Gravity Loads – Bending Moment

• Gravity Loads – Shear Force

• Seismic Loads – Bending Moment

• Seismic Loads – Shear Force

55Load combinations: will be covered later

ETABS: Run Analysis

57ETABS: Gravity Loads Bending Moment

58ETABS: Gravity Loads Shear Force

59ETABS: Seismic Force in Z direction Bending Moment

60ETABS: Seismic Force in Z direction Shear Force

61STAAD: Run Analysis

62STAAD: Gravity Loads Bending Moment

63STAAD: Gravity Loads Shear Force

64STAAD: Seismic force in Z direction Bending Moment

65STAAD: Seismic force in Z direction Shear Force

Analysis as per IS:1893-2002

Response Spectrum Method (Dynamic Analysis)

Response spectrum analysis:• Response spectrum analysis is performed using multi-

mode responses, where the free vibration modes are computed using – Eigen vector analysis (STAAD.Pro)– Eigen vector or Ritz Vector analysis (ETABS)

• The modal parameters for a structure come as pairs of – Natural Frequency f (in Hz) – Mode shape φ

• The modal parameters for few of the lower frequencies are considered for further calculations (7.8.4.2, IS:1893(Part 1)-2002), based on the following;– Modes for frequencies > 33 Hz need not be considered– The no. of modes considered should be such that the total

mass participation factor should be at least 90%– Missing mass correction for modes beyond 33Hz

Response spectrum analysis (cont…):• For the modal parameters considered, the following

factors are determined for each mode– Mode participation factor of each mode Pk

– Mass participation factor for each mode Mk – Spectral Acceleration coefficient (Sa/g)

• The Design horizontal seismic coefficient Ah is calculated for each mode from (6.4.2, IS:1893(Part 1)-2002)

– Zone factor Z – Importance factor I– Response reduction coefficient R– Spectral Acceleration coefficient Sa/g

Response spectrum analysis (cont…):• Where the Horizontal acceleration Sa/g is determined

from the Response spectrum curve (Fig.2, IS:1893(Part 1)-2002)

Response spectrum analysis (cont…):• The lateral force due to the modal response

(considering the mode participation factor) is obtained for each mode of all the modes considered. The the force at each level for each mode is calculated as (7.8.4.5 (c), IS:1893(Part 1)-2002):

Where– Ak is the design horizontal seismic coefficient

– φik is the mode shape value for that floor level– Pk is mode participation factor– Wi is mass at that floor level

The Same Simple Example

A six storied structure

Lumped mass model

MODAL ANALSYSEIGEN VALUES

EIGEN VECTORS

Mode Shapes

Natural Frequencies

Different types of modes: 1.Translational mode in X direction

Different types of modes: 2.Translational mode in Y direction

Different types of modes: 3.Torsional mode

Mode 1

Freq. = 1.44119 Hz

Period = 0.69387 sec Mode shape = 0

0.15130.36870.57960.76310.90591

Mode 2

Freq. = 4.51039 Hz

Period = 0.22171 sec Mode shape = 0

-0.5293-1-0.9604-0.41100.366290.98395

Modes I II III IV V

Levels Mode Shapes

1 0 0 0 0 0

2 0.151 -0.53 0.824 -1 1

3 0.369 -1 0.861 -0.047 -0.836

4 0.58 -0.96 -0.24 0.999 0.027

5 0.763 -0.41 -1 -0.265 0.813

6 0.906 0.366 -0.38 -0.902 -0.989

7 1 0.984 0.804 0.606 0.415

Period (sec) 0.694 0.222 0.123 0.085 0.065

Freq(Hz) 1.441 4.51 8.098 11.78 15.31

For kth mode,

Where, n = no. of levels

m = no. of modes

Mode I II III IV V

Mk(kN) 115.5 16.31 5.424 2.686 1.237

Participating mass =115.5+16.31+5.424+2.686+1.237=141.183kN

Total mass = 23.57 x 6 = 141.42 kN

Mass participation = 100x141.183/141.42 = 99.8326% > 90% .:Okay

Check Mass Participation (7.8.4.5(a), IS:1893(Part 1)-2002)

For kth mode,

Where, n = no. of levels

Mode I II III IV V

Pk 1.30049 -0.44635 0.26529 -0.18718 0.12223

Calculate Mode Participation factors (7.8.4.5(b), IS:1893(Part 1)-2002)

x 1.30049

x 0.26529

x -0.44635

x -0.18718

x 0.12223+

Calculate Mode Participation factors

(Had all the mode shapes been utilized)

Calculate Mode Participation factors

Modes I II III IV V

Φ’s

1 0 0 0 0 0

2 0.1513 -0.5293 0.8237 -1 1

3 0.3688 -1 0.8614 -0.0466 -0.8364

4 0.5796 -0.9604 -0.239 0.999 0.0268

5 0.7631 -0.4111 -1 -0.2653 0.8128

6 0.906 0.3663 -0.382 -0.9023 -0.9892

7 1 0.984 0.8038 0.6064 0.4154

0 0 0 0 0

3.566 -12.48 19.415 -23.57 23.57

8.692 -23.57 20.303 -1.097 -19.71

13.66 -22.64 -5.6354 23.547 0.631

17.99 -9.689 -23.57 -6.253 19.16

21.35 8.6335 -9.0113 -21.27 -23.32

23.57 23.192 18.946 14.292 9.791

88.83 -36.55 20.447 -14.35 10.12

0 0 0 0 0

0.54 6.6043 15.992 23.57 23.57

3.206 23.57 17.488 0.0511 16.49

7.919 21.74 1.3474 23.524 0.017

13.73 3.9826 23.57 1.659 15.57

19.34 3.1623 3.4452 19.19 23.07

23.57 22.819 15.228 8.6658 4.067

68.3 81.879 77.072 76.66 82.78

Mk 115.5 16.313 5.4243 2.6859 1.2369

Pk 1.300 -0.4464 0.2653 -0.1872 0.1222

Period 0.693 0.221 0.123 0.084 0.065

Sa/g 1.4411 2.5 2.5 1.1273 1.0979

Calculate Design horizontal seismic coefficient

Z = Zone factor = 0.16 I = Importance factor = 1R = Response reduction coefficient = 3

Period 0.693 0.221 0.123 0.084 0.065

Sa/g 1.441 2.5 2.5 1.1273 1.0979

Ah 0.0384 0.0667 0.0667 0.0300 0.0292

Calculate horizontal force due to each mode

0.0384 x 1.30049 x

Eg: - for mode 1 :

Combining Modes:(a) Indirect method

(Supported by all codes –

IS, BS, EC, etc.)

Mode I II III IV V

Levels Qik (kN)

1 0 0 0 0 0

2 0.178 0.371 0.343 0.133 0.084

3 0.434 0.701 0.359 0.006 -0.071

4 0.683 0.674 -0.1 -0.133 0.002

5 0.899 0.288 -0.42 0.035 0.069

6 1.067 -0.26 -0.16 0.12 -0.083

7 1.178 -0.69 0.335 -0.08 0.035

Calculate horizontal force due to each mode…

From each mode characteristics…

Calculate horizontal force due to each mode… apply and…

Analyse for forces, and combine the forces to get final forces…

Combining Modes:(a) Direct method

(Supported by FEMA 356, etc.

for Non-linear Static Analysis*)

*Pushover Analysis

Mode I II III IV V

Levels Qik (kN)

1 0 0 0 0 0

2 0.178 0.371 0.343 0.133 0.084

3 0.434 0.701 0.359 0.006 -0.071

4 0.683 0.674 -0.1 -0.133 0.002

5 0.899 0.288 -0.42 0.035 0.069

6 1.067 -0.26 -0.16 0.12 -0.083

7 1.178 -0.69 0.335 -0.08 0.035

Calculate horizontal force due to each mode… and combined

Eg: - for level 7:

SRSS = √[(1.178)2 + (-0.69)2 + (0.335)2 + (-0.08)2 + (0.035)2] = 1.409

Calculate horizontal force due to each mode… and combined

The forces are applied… …and analysed

Indirect method Direct method

Combination of modal responses with…

(1) Dominant mode:• When activated, all modal combination results will

have the same sign as the dominant mode shape alone would have if it were excited and then the scaled results were used as a static displacements result.

(2) Signed value:• This option results in the creation of signed values for

all results. The sum of squares of positive values from the modes are compared to sum of squares of negative values from the modes. If the negative values are larger, the result is given a negative sign.

Methods of mode combinations

IS:1893 and other sources

I II III IV V VI VII VIII

1.441 4.51 8.098 11.782 15.309 17.908 69.357 103.083 … (Hz)

93.223 %

99.922 % 0.077 %

Out-of-phase modesOR

Periodic Response

In-phase modesOR

Rigid Response

Modal Combination(ABS, SRSS,CQC, etc.)

Missing Mass Correction

Combination of Periodic and Rigid modes

(Lindley-Yow, Hadjian, etc. )

Mode frequencies – an overview:

Combination of modal responses:

1. Absolute Combination (ABS)– The modal responses of all the individual modes are

summed up (to be used in modal combination as per IS:1893-1984) :

– Eg. for level 7,ABS = │1.178│ + │-0.69│ + │0.335│ + │-0.08│ + │0.035│ = 2.318

Combination of modal responses (cont…):

2. Square Root of Sum of Squares (SRSS)– The modal responses are squared, summed, and the root of

the sum taken:

– Eg. for level 7,SRSS = √[(1.178)2 + (-0.69)2 + (0.335)2 + (-0.08)2 + (0.035)2] = 1.409

104104

3. Combination of ABS and SRSS (ABS&SRSS)– The weighted sum of ABS and SRSS is taken– The method was prescribed in IS:1893-1984 (4.2.2.2, IS:1893-

1984), removed in IS:1893(Part 1)-2002

– Values for γ are given by table – Intermediate values can be

obtained by interpolation Note: γ1 + γ2 = 1

– Eg. for level 7,

ABS&SRSS = 0.6 x 2.318 + 0.4 x 1.408 = 1.954

H (m) γ1 γ2

20 0.6 0.4

40 0.4 0.6

60 0.2 0.8

90 0.0 1.0

105105

4. Out-of-phase response combination– This is a general category of methods, which includes CQC

and SRSS– The general equation is

where ρij is the cross modal coefficient

a) SRSS method• SRSS can be represented as ρij = 1 for i=j = 0 for i≠j

106106

5. Out-of-phase response combination (cont..)b) Complete Quadratic Combination (CQC)

• The CQC equation prescribed by IS:1893-2002 is (7.8.4.4, IS:1893(Part 1)-2002)

where ζ is the damping ratio ( 0.05 for RCC)and β = ωj/ωi

( )Seismic Analysis of Multi-storied RC Building Presented by Rahul Leslie

Response spectrum method

ETABS Define Seismic

parameters:• Zone • Soil Type

Apply Seismic parameters

• Damping ratio ξ• Method of comb (SRSS,

CQC, etc.)• Method of Dir. Comb• Direction (X, Y, Z)• I, R

STAAD Define& apply Seismic

parameters:• Damping ratio ξ• Method of comb (SRSS,

CQC, etc.)• Soil Type• Direction (X, Y, Z)• Scale (= Z.I / 2.R)• Normalization Scale (= 9.8)

108ETABS: Define Response SpectrumSeismic Analysis of Multi-storied RC Building Presented by Rahul Leslie

Scale factor = gI/(2R) = 9.81*1/(2*5) = 0.981

ETABS: Apply Response Spectrum

110110

111111

112112STAAD: Response SpectrumSeismic Analysis of Multi-storied RC Building Presented by Rahul Leslie

ZI / 2R = 0.024

113113STAAD: Response Spectrum

115115

Directional Combination of Modal Responses

Directional combination of modal responses:

• As per IS:1893, directional combination need to be done only when the lateral load resisting members are not oriented along horizontal orthogonal directions (6.3.2.2, IS:1893(Part 1)-2002)

Directional combination of modal responses:1. SRSS Method (6.3.4.2, IS:1893(Part 1)-2002)

– The method combines the directional components by SRSS method. The method is prescribed in IS:1893-2002 as an alternative to SAS method

2. Scaled Absolute Sum method (SAS)– The method combines the directional components by sum

of values in one direction, with 0.3 times the sum of values in the other directions (6.3.4.1, IS:1893(Part 1)-2002)

No. of Modes & Missing mass correction

1.441 4.51 8.098 11.782 15.309 17.908 69.357 103.083 … (Hz)

93.223 %

99.922 % 0.077 %

Out-of-phase modesOR

Periodic Response

In-phase modesOR

Rigid Response

Modal Combination(ABS, SRSS,CQC, etc.)

Missing Mass Correction

Mode frequencies – an overview:

Missing Mass Correction :

• Missing Mass = 0.077% of total mass• Zero Period Acceleration (3.11, IS:1893(Part 1)-2002) = 1.6

1.441 4.51 8.098 11.782 15.309 17.908 69.357 103.083 … (Hz)

99.922 % 0.077 %

Missing Mass Correction :

• Missing Mass = 0.077% of total mass• Zero Period Acceleration, (Sa/g)ZPA = 1.6 (3.11, IS:1893(Part 1)-

2002) • Horizontal acceleration coefficient, Ah

1.441 4.51 8.098 11.782 15.309 17.908 69.357 103.083 … (Hz)

99.922 % 0.077 %

Missing Mass Correction (cont…):

• Lateral force, QR is given by

where Mmiss.mass is the missing mass

Mk(ZPA) is sum of Participating masses up to ZPA

Storey wise Mass Participation for 1st mode

4.6285

11.310

17.778

23.387

27.771

30.652

For 1st mode, Mk = 0+4.6285+11.310+17.778+23.387+27.771+30.652 = 115.52

Thus, storey wise Mass Participation can be obtained for each mode, from which the total storey wise Mass Participation,

and consequently, the storey wise Missing Mass can be determined.

Storey wise Missing mass

0.0155

0.0243

0.0256

0.0234

0.0151

0.0047

X (1.6)0.16 x 1

2 x 3Ah = = 0.04266

:. 0.04266 x

0.0155

0.0243

0.0256

0.0234

0.0151

0.0047

0.0006

0.0010

0.0006

0.0002

0.5755

0.9114

0.9741

1.0422

1.4002

SRSS Final Results

Results

Set number of modes & Check mass participation

(2.8.4.2, IS:1893(Part 1)-2002) • Consider as many no. of modes so as to have a

total Participating Mass, Σ(Mk) > 90%

• One shouldn’t consider modes having frequencies > 33 Hz

• In case one has considered modes up to 33Hz, but still hasn’t obtained Σ(Mk) > 90%, the only option is a Missing Mass correction

Set number of Modes & check Mass Participation

ETABS & STAAD• Increase no. of modes

• Check output tables to see – Modes at which participating mass > 90%– Mode having freq. > 33Hz

• Set no. of modes to be considered, accordingly• If all modes are considered up to 33Hz, but still

hasn’t obtained Σ(Mk) > 90%, apply Missing Mass Correction

128ETABS: Set no. of modes

ETABS: Check mass participation

Set this to mode for 33 Hz cut-off freq.

135STAAD: Check mass participation

STAAD: Check mass participation

Base Shear Correction

When dynamic method is followed :• When dynamic analysis is done, a static analysis is

also to be done and base shears compared (7.8.2, IS:1893(Part 1)-2002) – If scale all forces in the ratio as

Base shear correction• Do analysis with seismic coeff. method

Base shear = Vb(Stat.Anal)

• Do analysis with Resp. Spec. method Base shear = Vb(Dyn.Anal)

• Is Vb(Dyn.Anal) >= Vb(Stat.Anal) ?

If yes (Vb(Dyn.Anal) >= Vb(Stat.Anal)) No problem

If no (Vb(Dyn.Anal) < Vb(Stat.Anal)) scale resp. spec. results by [Vb(Stat.Anal) / Vb(Dyn.Anal)] > 1

ETABS: Base shear correction

147147

VB(Dyn.Anal) = 1607kNVB(Stat.Anal) = 1769kN

VB(Dyn.Anal) < VB(Stat.Anal)

:.VB(Stat.Anal)/ VB(Dyn.Anal) = 1769/1607 = 1.1 (STAAD)

New Coeff. = Coeff.*1.1 = 0.981*1.1 = 1.08 (ETABS)

149149

STAAD: Base shear correction

Seismic Load Combinations

ETABS: Load combination

STAAD: Load combination

Design Combinations Service Combinations

STAAD: Load combination

Load Combinations:The analysis results are to be combined using the

following load combinations for RC structures (6.3.1.2, IS:1893(Part 1)-2002, Table 18, IS:456-2000) :

Where EL represents ELx and Ely

#Live load reduction for upper floors as per 3.2, IS:875(Part 2)-1987 not to be included (Note 6, 8.1, IS:875(Part 5)-1987), but reduction as per Table 8, IS:1893 shall be used (7.3.3, IS:1893 (Part 1)-2002)

*To be considered when stability against overturning or stress reversal is critical (foot note to Table 18, IS:456-2000)

Design Combinations:-•COMB-I = 1.5(DL+LL)#

•COMB-II = 1.2(DL+LL±EL)#

•COMB-III = 1.5(DL±EL)•COMB-IV = 0.9DL±1.5EL*

Service Combinations:-•COMB-I = 1.0(DL+LL)•COMB-II = 1.0DL+0.8(LL±EL)•COMB-III = 1.0(DL±EL)

Load Combinations (cont…):

Design Combinations: Service Combinations:

•COMB1 = 1.5(DL+LL)•COMB2 = 1.2(DL+LL+ELx)•COMB3 = 1.2(DL+LL − ELx)•COMB4 = 1.2(DL+LL+ELy)•COMB5 = 1.2(DL+LL − ELy)•COMB6 = 1.5(DL+ELx)•COMB7 = 1.5(DL − ELx)•COMB8 = 1.5(DL+ELy)•COMB9 = 1.5(DL − ELy)•COMB10 = 0.9DL+1.5ELx•COMB11 = 0.9DL − 1.5ELx•COMB13 = 0.9DL+1.5ELy•COMB14 = 0.9DL − 1.5ELy

•COMB1 = 1.0(DL+LL)•COMB2 = 1.0DL+0.8(LL+ELx)#

•COMB3 = 1.0DL+0.8(LL − ELx) #

•COMB4 = 1.0DL+0.8(LL+ELy) #

•COMB5 = 1.0DL+0.8(LL − ELy) #

•COMB6 = 1.0(DL+ELx) #

•COMB7 = 1.0(DL − ELx) #

•COMB8 = 1.0(DL+ELy) #

•COMB9 = 1.0(DL − ELy) #

# Use enhanced SBC for these cases (6.3.5.2 & Table 1, IS:1893(Part 1)-2002)

Type I :Hard Soil (N > 30)

Type II : Medium Soil(30 ≤ N ≤ 10)

Type III :Soft soil(N < 10)

Pile/Well 50% 25% 25%

Raft 50% 50% 50%

Isolated 50% 25% 0%

Permissible Increase in SBC of soil :-(6.3.5.2 & Table 1, IS:1893(Part 1)-2002)

Load Combinations (Cont.):The percentage of imposed loads given [0.25 LL ≤ 3 kN/m2, 0.5 LL

> 3kN/m2] shall also be used for ‘whole frame loaded’ condition in the load combinations specified [1.2(DL+LL±EL)] where the gravity loads are combined with the earth quake loads (7.3.3, IS:1893-2002)

Further reduction in imposed load as per 3.2, IS:875 (Part 2)-1987 need not be considered (7.3.3, IS:1893 (Part 1)-2002, similar to Note 6, 8.1, IS:875(Part 5)-1987).

• COMB-I = 1.5[DL+LL]• COMB-II = 1.2[DL+(0.25 LL≤3, 0.5 LL >3)±EL]• COMB-III = 1.5(DL±EL)• COMB-IV = 0.9DL±1.5EL

For Design:-• COMB1 = 1.5 ( DL + LL )• COMB2 = 1.2 ( DL + 0.25 LL≤3 + 0.5 LL >3 + ELx )• COMB3 = 1.2 ( DL + 0.25 LL≤3 + 0.5 LL >3 − ELx )• COMB4 = 1.2 ( DL + 0.25 LL≤3 + 0.5 LL >3 + ELy )• COMB5 = 1.2 ( DL + 0.25 LL≤3 + 0.5 LL >3 − ELy )• COMB6 = 1.5 ( DL + ELx )• COMB7 = 1.5 ( DL − ELx )• COMB8 = 1.5 ( DL + ELy )• COMB9 = 1.5 ( DL − ELy )• COMB10 = 0.9 DL + 1.5 ELx• COMB11 = 0.9 DL − 1.5 ELx• COMB13 = 0.9 DL + 1.5 ELy• COMB14 = 0.9 DL − 1.5 ELy

Force Envelopes

1.5x(DL + LL)

1.5x(DL + EQx)

1.5x(DL - EQx)

Envelope

1.5x(DL + LL)

1.5x(DL + EQx)

1.5x(DL - EQx)

Envelope

Torsion & Accidental Eccentricity

Accidental eccentricity:• Minimum Design eccentricity edi to be considered

during analysis (7.9.2, IS:1893(Part 1)-2002)

where esi = actual eccentricity bi = breadth of building

-- In case 3D dynamic analysis is carried out, the dynamic amplification factor of 1.5 be replaced by 1.0 (Note 2 to 7.9.2, IS:1893(Part 1)-2002 – Amendment No.1, Jan 2005)

Accidental eccentricity:

Specifying Accidental Eccentricity

ETABS• Give eccentricity = 0.05 (ie., 5% of respective dimension) OR (For Seismic Coeff. Method)• For each storey,• If Accidental eccentricity (0.05b) to be provided is in the

same direction as the Actual eccentricity (e), provide Accidental eccentricity as 0.5e+0.05b

• If Accidental eccentricity (0.05b) to be provided is in the opposite direction as the Actual eccentricity (e), provide Accidental eccentricity 0.05b

STAAD• Tick the Accidental Eccentricity / Include Torsion

checkbox

ETABS: Accidental eccentricity

STAAD: Accidental eccentricity

Specifying Accidental Eccentricity

STAAD 8i SELECT 4 and above

• If Accidental eccentricity (0.05b) to be given is in the same direction as the Actual eccentricity (e), provide Accidental eccentricity as (0.5e+0.05b)/b

• If Accidental eccentricity (0.05b) to be given is in the opposite direction as the Actual eccentricity (e), provide Accidental eccentricity 0.05

- Note: The above are only to demonstrate ECC and OCC options in STAAD. As per Note 2 to 7.9.2, IS:1893(Part 1)-2002 – Amendment No.1, Jan 2005, only 0.05 need to be given for eccentricity in Response Sectrum Analysis

STAAD: Accidental eccentricity

For Design:-• 1.5 ( DL + LL ) 1 Load Case• 1.2 ( DL + LL ± EL x/y ± e ) 8 Load Cases• 1.5 ( DL ± EL x/y ± e ) 8 Load Cases• 0.9 DL ± 1.5 (EL x/y ± e) 8 Load Cases

------------------- 25 Load Cases

For support reactions:-• 1.0 ( DL + LL ) 1 Load Case• 1.0 DL + 0.8 ( LL ± EL x/y ± e ) 8 Load Cases• 1.0 ( DL ± EL x/y ± e ) 8 Load Cases

------------------- 17 Load Cases 42 Load Cases

173173

Load Combinations (non-ortho beams…):For Design:-• 1.5 ( DL + LL ) 1 Load Case• 1.2 ( DL + LL ± (EL x/y ± e) ± 0.3 (EL y/x ± e))

32 Load Cases• 1.5 ( DL ± (EL x/y ± e) ± 0.3 (EL y/x ± e))

32 Load Cases• 0.9 DL ± 1.5((EL x/y ± e) ± 0.3 (EL y/x ± e))

32 Load Cases-------------------

97 Load Cases

For support reactions:-• 1.0 ( DL + LL ) 1 Load Case• 1.0 DL + 0.8 ((EL x/y ± e) ± 0.3 (EL y/x ± e))

32 Load Cases• 1.0 ( DL ± (EL x/y ± e) ± 0.3 (EL y/x ± e))

32 Load Cases-------------------

65 Load Cases 162 Load Cases

Reinforced Concrete Design

Concrete Design

ETABS• Load combinations • Design Code (Indian) • Material: ρ, E, μ, fck, fymain,

fyshear

• Ductile/ordinary• Col. Effective lengths• Etc…

STAAD• Load combinations• Design Code (Indian,

456/13920)• fck, fymain, fyshear

• Load case for beam shear design (DL+LL) (Cl. 6.3.3(b) , IS:13920)

• Col. Effective lengths• Etc…

Ref:- Fig.4, IS:13920

ETABS: Concrete Design

ETABS: Run Analysis

ETABS: Run Concrete Design

183STAAD: Concrete Design

184STAAD: Run Analysis

Miscellaneous points

Selection of method (7.8.1, IS:1893(Part 1)-2002) :• Dynamic analysis to be performed for

1. In case of regular buildings– h > 40 m in Zones IV & V– h > 90 m in Zones II & III

2. In case of irregular buildings– h > 12 m in Zones IV & V– h > 40 m in Zones II & III

Selection of method (7.8.1, IS:1893(Part 1)-2002) :

• When the structure is irregular as per Table 4, IS:1893(Part 1)-2002, the method of dynamic analysis with masses lumped at floor levels, as per 7.8.4.5, IS:1893(Part 1)-2002, cannot be done, rendering computer modelling the only option.

Soft Stories (7.10, IS:1893(Part 1)-2002) :• A soft storey is one in which the lateral stiffness is

less than 70 percent of that in the storey above or less than 80 percent of the average lateral stiffness of the three storeys above (4.20, IS:1893(Part 1)-2002)

(Fig.4, IS:1893(Part 1)-2002)

Soft Stories (7.10, IS:1893(Part 1)-2002) :• In case of buildings with soft stories,

– Dynamic analysis of building be carried out including the strength and stiffness effects of infills, and the members designed accordingly (7.10.2, IS:1893(Part 1)-2002)

– Alternatively, • Columns and beams of the soft storey be designed for 2.5 times

forces obtained by analysis • Shear walls of the soft storey be designed for 1.5 times forces

obtained by analysis

(7.10.3, IS:1893(Part 1)-2002)

Separation Between Adjacent Units (7.11.3, IS:1893(Part 1)-2002) :• To avoid damaging contact when the two units deflect towards

each other

• Separation equal to R/2 times the sum of the calculated storey displacements as per 7.11.1, IS:1893(Part 1)-2002 (ie., with

partial load factor of 1.0) of each of them.

• Separation equal to R times the sum of the calculated storey displacements, when floor levels of the adjacent units or buildings are not at the same elevation levels.

When to include ductile detailing :-

Ductile detailing provisions shall be adopted in RCC buildings for– More than 5 stories high in zone III– Industrial structure in zone III– Importance factor > 1.0 in zone III– In zone IV & V

(1.1.1, IS:13920-1993)

Ductile detailing provisions shall be adopted in RCC buildings for– All buildingd in zone III, IV & V

(1.1.1, IS:13920-1993, Edition 1.2)

Concluding Remarks

193193

Concluding remarks• The best way to implement advanced seismic analysis

techniques is through an analysis software package.• To use a software package, one has to know it• More importantly, one has to know its limitations,• Still more important, one has to know its pitfalls.• Software Demonstrators/Instructors may tell you the

limitations, but not the pitfalls. Mostly it can be learned only through experience.

• The user should have a good base in Seismic Analysis & Design, and Structural Dynamics. Also a basic understanding of FEM is desirable (but not necessary).

• Also one has to know the code provisions, and have them ready reference (IS:456, SP 34, IS:1893, IS:13920, IS:875 Part-I, II, III)

Thank you!

rahul.leslie@gmail.com

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