Dhaval

Example on Multi storey building Analysis and Design

Prepare by Dhorajia Dhaval H. 06MCL003 1

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MODEL-1

MODEL-2

No. of stories=12

Keeping other data same only increase no of stories No. of stories=25

No. of shear wall=2 Thickness=0.365 m

Overcome additional quantity Increase thickness of shear wall by keeping same no. of shear wall as M1 & M-2

Of deflections, drift, B.M.,S.F. By two way

No. of stories=25

MODEL-3 Thickness of shear wall=0.5m

MODEL-4 Thickness of shear wall=0.6m

Increase nos. of shear wall by keeping same thickness of shear wall as M-1 & M-2

MODEL-5

MODEL-6

No. of shear wall=4

No. of shear wall=6 3

Problem solution Design

Analyses

Devendra

Dhaval •Basic fundamental calculation check of model under gravity loading •Comparison of base shear computation by seismic coefficient method by:1893(2002) & software 3D model & spring mass model • Comments on analyses under gravity load and lateral load

•Modeling in ETAB •Parametric study of diff. results of diff. models –Deflections, drift ,S.F., B.M., mode shapes •Conclusion related cost & reduction of actions Deepak 4

Analysis

Gravity Load

Lateral load

•Load transfer slab to beam as per One way & two way consideration

•Seismic coefficient method as per IS:1893(2002)

•Beam to column by reaction method

•Done by C++

•Done by STAAD modeling 5

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D.L.=0.2x24=4.8 kN/m2 in Y-range S.W. L.L.=0.25x3=0.75 kN/m2 in Y-range except on terrace

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Elevation: load case :L.L.

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Elevation: load case :L.L.

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Basic primary check of model under gravity load consideration

Model No.

Total gravity load (kN) by C++

Total gravity load (kN) by STAAD Pro.

Variation (%)

1

143110.9688

1460316.008

9.8

2

300854.6875

3539466.912

8.5

3

331215.1563

4246348.157

7.8

4

349615.4375

3928263.343

8.9

6

424136.5938

4560608.535

9.3 13

Lateral load Analysis Seismic coefficient method as per IS:1893 (Part-I) (2002)

Time period as per codal formula

Time period as per 3D ETAB model

Time period as per spring mass model

Ta=0.09xh/d1/2 Other values: Z=zone-III=0.16 I=1 R=5 Soil Type-II

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[K]-[Mw2]=0

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RESULTS

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Mod el No.

Sa/g value

Time period By By ETAB By spring codal 3D model mass formula model

By By ETAB By spring codal 3D mass formula model model

1

0.8792

1.203

2

1.83

2.32

1.54

1.13

3

1.83

2.2

0.7431

0.586

4

1.83

1.9

0.7431

0.618

5

1.83

1.87

0.7431

0.715

6

1.83

1.67

0.7431

0.7272

0.7431

0.7683 18

Ah value

Base shear Vb (kN)

By By ETAB By spring By By ETAB By spring codal 3D model mass codal 3D model mass formula model formula model 0.02464

0.01808

3526.25 2587.445

0.01189 0.009376

3577.04

3278.97

0.01189 0.009888

3938.14

3368.97

0.01189

0.01144

0.01189 0.011635 0.01189 0.012293

4156.78 3999.600 4309.92 4217.709 5042.81 5213.826 19

M 1 C

M 1 S

M 2 C

M 2 S

M 3 C

M 3 S

M 4 C

M 4 S

M 5 C

M 5 S

M 6 C

M 6 S

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Why base shear for same model we get diff. in comparison of analyses done by codal formula & by software solution?

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Time period •Based upon stiffness and mass of the system [K]-[Mw2]=0 •Solution can give w value and based upon that find T

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If time period increase

Sa/g value decrease

Ah decrease

Base shear decrease 23

If time period decrease

Sa/g value increase

Ah increase

Base shear increase 24

Time period Finding out by codal formula •Independent of stiffness and mass of structure. •Formula evaluate based on observation of seismic performance during seismic event of exciting structure and experiment base •Give lower value compare to software analyses for same structure and same condition.

Finding out by software •Dependent of stiffness and mass of structure. •Value find base on stiffness and mass of spring mass model & eigen-values and eigen-vectors. •Give higher value compare to codal formula for same structure and same condition. •If in any change in structural system is occur there will be change in time period value.

•If change in structural system is occur regarding to stiffness and mass other than storey height and base width there is no change in time period value. 25

Some observations and comparisons in model • M1 has lesser time period than the all the models because less height. • In M2 to M6 has constant time period as per codal formula but comparatively decrease in case of finding out by software analyses because stiffness increase.

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Compare to M3 & M4 , model M5 & M6 has lesser time period because they are more stiffer than M3 & M4. Conclusion: •Time period is very impoartant key factor in case of seismic analyses of structure. •In case of codal foumula which find based on experimental work and past observation not depended mass and stiffness of the structure. •Generally for one structure get diff. time period one by codal formula and one by software analyses and in comparison of both by coad get lesser time period.

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Other consideration to overcome lateral effects due to lateral loads on structure •By wall-frame structure •By framed structure •Infill wall •Effect of mass and stiffness variable and it’s effect on time period & comparisons with codal formula for time period. •For finding Equivalent stiffness by small C++ programme. •Then find time period by spring mass model. 28

Importance of studying time period • Time period find by codal formula and actual time period due to stiffness & mass of structure is also different when keeping geometrical data same. • Real time period find only by spring-mass model of structure.

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Thank You

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