This document presents a comparative study of the analysis of a G+3 residential building using the equivalent static load method, response spectra method, and SAP2000 software. A 3D model of the building was created in SAP2000 and analyzed under gravity and earthquake loads. Storey shear forces and lateral forces obtained from each method were compared. The response spectra method calculated natural frequencies, mode shapes, time periods, and design lateral forces considering multiple modes of vibration. SAP2000 provided shear force, bending moment, and axial force diagrams as well as deformed shape under loading. Lateral forces and shear forces from each analysis method were plotted for comparison.

1. 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)

2. 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.

3. 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.

4.  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.

5. Plan of the Building:-
Fig 1. Plan of the Building

6. Elevation of the Building:-
Fig 2. Elevation of Building

7. Building detail:-
 Building size: 20.4×22.60 square metre.
 Front setback =4.5m
 Rear setback=4.5m
 Side setback=2.4m
 Plot 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 0.23m
 Rise of stairs is 0.16m

8. 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

9. 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.

10. Elevation of the frame considered for analysis:-
Fig 3. Elevation of the frame considered for analysis

11. 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

12. 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

13.  Calculation of Stiffness against lateral force on each storey
STIFFNESS
COLUMN
STIFFNE
SS
INFILL
WALL
STIFFNE
SS

14. Modelling of the infill wall as equivalent diagonal strut
Fig 5. Modelling of the infill wall

15. 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+αl
2)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

16. 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 167523.96
0
1932580.16
9
1932580.16
9
1932580.16
9

17.  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:-

18.  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, ωi
2= {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

19. Model of the building expressed as multiple degree
of freedom system
Fig 6. Model of the Building

20. MATLAB FUNCTIONS

21.  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

22. Mode Shapes based on Eigen vectors:-
Height of floor in
meters (m)
Amplitude of Displacement in
meters(m)

23. 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)

24. Time period, modal mass & modal participation
factor:-

25.  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

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

27. Storey shear force due to all modes considered:-
479.85
460.83
353.41
227.59
98.59
15.58
-7.62
-16.86
-18.38
-10.23
1.36
-1.62
4.83
8.36
5.39
0.002
-0.008
0.16
-0.01
-0.16
0.0001-
0.0005
0.0159
-0.0220
0.0180

28.  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:-

29. 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

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

31. 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

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

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

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

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

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

37. 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

38. Deformed shape of the considered frame under
Gravity and Earthquake load
Fig. 17 Deformed Shape of frame

39. Deformation of Building

40. Results of analysis
 Shear Force Diagram
Fig18. Shear Force diagram Load Case=1.5(DL+LL)

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

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

43.  Comparison of Design lateral forces acting on each floor
Table 3. Design lateral force acting on each floor

44. 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

45. • Comparison of Shear forces acting on each floor
Table 4. Shear force acting on each floor

46. 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