A COMPARATIVE STUDY OF ANALYSIS OF A G+3 RESIDENTIAL BUILDING BY THE EQUIVALENT STATIC LOAD METHOD,

RESPONSE SPECTRA METHOD AND SAP2000

Under the guidance of Mr. Debaraj Bailung Sonowal

Presented By :- Susmit Boruah( CIB12021) Kamal Singh (CIB12046) Roshan Kumar (CIB12054) Kumar Aman (CIB12058) Ravindra Kumar Verma (CIB12060)

Objectives:-

To determine the storey shear forces and lateral forces acting at each storey of the considered frame under gravity and earthquake loads using Response Spectrum Method.

To carry out a comparative study between the lateral forces as well as shear forces acting on each storey obtained by Equivalent Static Load Analysis ,Dynamic Load Analysis and Software Analysis.

Introduction:-

• A (G+3) R.C. residential building is adopted for analysis & design.

• Ground floor is an open space for parking & floors 1st to 3rd are residential blocks.

• The location of the building is assumed to be at Guwahati (Zone V).

• Dynamic load analysis has been carried out by Response Spectrum Method.

D’ Alembert’s principle has been used to develop the mathematical model for the multiple degree of freedom system.

Seismic Analysis has been carried out as per IS1893 (Part 1): 2002.

A damping of 5 % of the critical damping has been considered during the analysis.

The building has been considered to be constructed on medium soil.

Plan of the Building:-

Fig 1. Plan of the Building

Elevation of the Building:-

Fig 2. Elevation of Building

Building detail:-

Building size: 20.4×22.60 square metre. Front setback =4.5mRear setback=4.5mSide setback=2.4mPlot size : 29.4x27.4 m2

Total plot area =805.56 sq. m.Percentage occupied space= 57.2%Percentage of free space=42.8%Tread of stairs is 0.23mRise of stairs is 0.16m

Specifications:-

Grade of concrete – M25 , grade of steel – Fe 450. Floor to floor height – 3.1 m Plinth height above GL – 0.9 m Depth of foundation below GL – 3.0 m Parapet Wall height – 1.0 m Slab thickness- 150 mm External wall thickness – 250mm , internal wall thickness- 150mm. Size of column – 500mm x 500mm . Size of beam – 300mm x 450mm. Live load on floor – 3 kN /m2 , Live load on roof – 3.0 kN/m2

Roof treatment & floor finish (F.F.) – 1.0 kN/m2

Site located on Seismic Zone V , Building resting on Medium Soil. Building frame type is Special Moment Resting Frame. Density of concrete -25 kN/m3 , Density of masonry wall – 20 kN/ m3

Bearing capacity of foundation soil= 100kN/m2

Dynamic Analysis:-

Calculation of lumped mass on each floor by considering dead loads, floor finish and imposed loads as per Cl 7.3.1 ,Table 8, IS 1893 ( Part 1): 2002

The lumped masses of each floor are worked out as follows: Lumped Mass, M = Mass of infill wall (longitudinal + transverse)+ Mass of columns + Mass of beams

(longitudinal + transverse) + Mass of slab + Imposed load of that floor if permissible + Floor Finish The elevation of the frame considered and plan showing the columns,

beams and slab at floor level is shown and the results have been tabulated.

Elevation of the frame considered for analysis:-

Fig 3. Elevation of the frame considered for analysis

Mass Roof 3rd Floor 2nd Floor 1st Floor Ground Floor

Mass of Infill (KN)

247.800 434.335 434.335 268.268 264.825

Mass of Columns

(KN)

58.125 116.250 116.250 129.844 132.187

Mass of beams (KN)

163.680 163.680 163.680 163.680 163.680

Mass of slabs(KN)

447.525 447.525 447.525 447.525 0

Imposed loads (KN)

0 119.340 119.340 119.340 0

Floor Finish(KN)

119.340 89.505 89.505 89.505 0

Total Lumped

Mass

1036.440 1370.630 1370.630 1218.160 560.690

Table 1. Lumped mass

Plan showing columns, beams and portion of the slab to be considered for calculating lumped mass at floor level:-

Fig 4. Plan showing columns, beams and portion of the slab

Calculation of Stiffness against lateral force on each storey

STIFFNESS

COLUMN

STIFFNESS

INFILL WALLSTIFFNESS

Modelling of the infill wall as equivalent diagonal strut

Fig 5. Modelling of the infill wall

2. Stiffness of infill which is determined by modelling the infill as an equivalent diagonal strut.

Width of strut is given by:- w=0.5×(αh

2+αl2)0.5

αh =(π/2)×[ Ef×Ic×h]1/4/[2×Em×t×sin2θ]1/4

αl =π×[ Ef× Ib×l]1/4/[Em×t×sin2θ]1/4

Where, Ef=Elastic modulus of frame material=5000×fck

0.5=25000N/m2 Em= Elastic modulus of masonry wall=13800N/m2 Ic & Ib= Moment of inertia of column & beam respectively h & l= height & length of infill wall; t= thickness of masonry wall θ=tan-1(h/l)

The lateral strength of the structure is contributed by two important elements.

1. Column stiffness- The column act as supporting member against horizontal force. The governing formula for calculating column stiffness, Kc = (12EI)/ L3

Stiffness of infill wall, Kw = (AEm cos2θ )/ ld

where A= Cross-sectional area of diagonal stiffness= W x t

ld = Diagonal length of strut

= (h2+l2)0.5

θ = tan-1(h/l) Total stiffness , K = Kc + Kw

Modified stiffness is given by:- K5=K5+2×Kw1×3×Kw2= modified stiffness of 3rd storey columns K4=K4+2×Kw1+3×Kw2=modified stiffness of 2nd storey columns K3=K3+2×Kw1+3×Kw2= modified stiffness of 1st storey columns K2=K2 ( Since plinth & ground floor falls under Soft Storey, so no modification) K1=K1

Table 2. Stiffness values at different floor levels

Stiffness Substructure

Ground Floor

1st Floor 2nd Floor 3rd Floor

K( KN/m)

279500.220

167523.960

1932580.169

1932580.169

1932580.169

Natural frequency Eigen vectors Time periods Modal participation factors Modal mass Design lateral forces at each floor in each mode Storey shear forces in each mode Storey shear forces due to all modes considered Lateral forces at each storey due to all modes considered

Determination of lateral forces at each storey:-

Using D’Alembert’s Principle,

Eqn of motion for the frame considered in X direction is

M(d2x/dt2 )+KX=0 => [K-Mω2]X=0

Where, K= M= 5x5 5x5

On solving the above eqn,

we have, ωi2= {194, 7277, 10725, 31150, 49000}

Calculation of natural frequency or Eigen values:-

K1+K2 -K2 0 0 0

-K2 K2+K3 -K3 0 0

0 -K3 K3+K4 -K4 0

0 0 -K4 K4+K5 -K5

0 0 0 -K5 K5

m1 0 0 0 0

0 m2 0 0 0

0 0 m3 0 0

0 0 0 m4 0

0 0 0 0 m5

Model of the building expressed as multiple degree of freedom system

Fig 6. Model of the Building

MATLAB FUNCTIONS

Corresponding to each eigen value, Eigen vector is calculated which is

φ1= φ2= φ3=

φ4= φ5=

Eigen vectors:-

-0.0162

-0.0420

-0.0437

-0.0449

-0.0453

0.1192

0.0219

0.0032

-0.0171

-0.0285

-0.0546

0.0542

0.0263

-0.0220

-0.0533

-0.0059

0.0471

-0.0426

-0.0364

0.0518

0.0018

-0.0254

0.0522

-0.0551

0.0328

Mode Shapes based on Eigen vectors:-

Height of floor in meters (m)

Amplitude of Displacement in meters(m)

Combined plot of mode shapes

Mode 1 Mode 2 Mode 3 Mode 4 Mode 5

Height of floor in meters (m)

Amplitude of Displacement in meters(m)

 

Time period, modal mass & modal participation factor:-

Design lateral force at floor i in mode k is given by :-

Qik=Ak×Φik×Pk×Wi

where,

Ak= Design horizontal seismic coefficient for mode k

= .

Qik =

Design lateral force:-

19.02 23.21 2.99 0.001 0.001

107.42 9.24 -6.46 -0.168 -0.016

125.82 1.52 -3.52 0.171 0.038

129.0 -8.15 2.95 0.146 -0.040

98.59 -10.23 5.40 -0.157 0.018

Response spectra for rock and soil sites for 5 % damping:-

Fig 7. Response spectra for rock and soil sites for 5 % damping

 

Storey shear force due to all modes considered:-

479.85460.83353.41227.5998.59

15.58-7.62

-16.86-18.38-10.23

1.36-1.624.838.365.39

0.002-0.0080.16-0.01

-0.16

0.0001-0.00050.0159-0.02200.0180

At roof, F5=V5=99.27kN

F4=V4-V5=129.23kN

F3=V3-V4=125.35kN

F2=V2-V3=107.04kN

F1=V1-V2=19.21kN

Determination of lateral force at each storey:-

Analysis by using software• Analysis of the frame has been done by using SAP 2000.• SAP 2000 is exclusively made for modelling, analysis and design of buildings. • A 3-D model is prepared by using the plan, elevation and sectional data of the

G+3 RC residential building.• Gravity loads have been assigned by defining the following load cases: a) Wall Ext(Load due to external wall) b) Wall Int(Load due to internal wall) c) Dead (Self weight of members) d) Dead Slab (Load due to self weight of slab ) e) Live (load due to live load acting over slab area) f) FF(Load due floor finish over slab)

• Earthquake load cases have been assigned by defining load cases as EQx and EQy

Software generated 3-D model of the building

Fig 8. Software generated 3-D model of the building

Calculation of distributed load on the beams due to Dead load, Live load and Floor finish acting on the slab

Fig 9. Load distribution pattern on slabs

Fig 10. Load distribution diagram(Due to self weight of slab)

Fig 11. Load distribution diagram(Due to Live loadon slab)

Fig 12 . Load distribution diagram(Due to Floor Finish on slab)

Fig 13. Load distribution diagram (Due to Plinth wall)

Fig 14. Load distribution diagram (Due to Internal wall)

Software generated undeformed and deformed 3-D model of the building after application of Gravity and Earthquake loads

Fig 15. Undeformed Shape Fig 16. Deformed Shape

Deformed shape of the considered frame under Gravity and Earthquake load

Fig. 17 Deformed Shape of frame

Deformation of Building

Results of analysis

Shear Force Diagram

Fig18. Shear Force diagram Load Case=1.5(DL+LL)

Bending Moment Diagram

Fig 19.Bending Moment diagram, Load Case=1.5(DL+LL)

Axial Force Diagram

Fig 20. Axial force diagram, Load Case=1.5(DL+LL)

Comparison of Design lateral forces acting on each floor

Table 3. Design lateral force acting on each floor

Graphical comparison of Design Lateral Loads

Design lateral forces acting on each floor has been calculated by using the Equivalent Static Load analysis, Dynamic load analysis and Software analysis.

The graphical representation is shown in Graph 1

Graph 1. Design lateral force acting on each floor

• Comparison of Shear forces acting on each floor

Table 4. Shear force acting on each floor

Graphical comparison of Shear Force

Shear forces acting on each floor has been calculated by using the Equivalent Static Load analysis, Dynamic load analysis and Software analysis.

The graphical representation is shown in Graph 2

Graph 2. Shear force acting on each floor