lateral forces on building due to an earthquake using response spectrum method 20x20 matrix1
DESCRIPTION
gTRANSCRIPT
Lateral forces on building due to an Earthquake using Response Spectrum Method
S.No Description
1 Type of structure : Multi Storeyed RC Lateral Load Resisting System
2 Seismic zone : III Moderate (Z=0.16)
3 No of storeys : 10 G+9 No of Bays 10
4 Floor Height : 6 M Width of bay 6
5 Infill wall :
Longitudinal : 250 MM THK INCLUDING PLASTER
Transverse : 150 MM THK INCLUDING PLASTER
6 Imposed Load : 3.5
7 Materials : Concrete Grade 20 N/Sqmm Steel Grade 500 N/Sqmm
8 Size of Columns : Base in mm 250 Height in mm 450
9 Size of Beams :
Longitudinal Base in mm 250 Height in mm 400
Transverse Base in mm 250 Height in mm 350
10 Depth of slab : 100 mm Assumption
11 Specific weight of RCC : 25 KN/Cum
12 Specific weight of Infill wall : 20 KN/Cum
13 Type of soil : Soft Soil
14 Response Spectrum : As per IS1893 PART I 2002
15 Time History : As per IS1893 PART I 2002
Step:1 Calculation of Lumped Masses to various floor levels
Nos
Size in m
M10
Density
in
KN/m3
L B HWeight
in KN
Total Weigth
in KN
Mass of Infill 1 20 12 0.25 3 180
1 20 18 0.15 3 162
Mass of Columns 3 25 0.45 0.25 3 25.31
Mass of beams in longitudinal
direction of that floor 1 25 12.0 0.25 0.4 30
Mass of beams in Transverse
direction of that floor 1 25 18 0.15 0.35 23.63
Mass of slab 1 25 12.0 6 0.1 180 600.938
Imposed Load 50% as per IS 1893 PART I:2002
For Floors From M1 To M9
Mass of Infill 1 20 12 0.25 6 360
1 20 18 0.15 6 324
Mass of Columns 3 25 0.45 0.25 6 50.63
Mass of beams in longitudinal and
transverse direction of that floor 1 25 12 0.25 0.4 30
1 25 18 0.15 0.35 23.63
Mass of slab 1 25 12 6 0.1 18050 % of imposed load, if imposed load is
greater than 3KN/sqm50%
1 - 12 6 6 216 1184.25
Seismic weight of building M1 M2 M3
Seismic weight of all floors M1 + M2 + M3 120.720 120.720 61.26
3570.38 KN 363.953 TON
Masses of Roof and Floors
From M1 To M9 1086.47
M10 61.258
Roof
Nos
Step:2 Determination of Fundamental Natural Period
Ta = 0.075 X h0.75
Where h is the height of building in metres H= 60
1.617 Seconds
Step:3 Determination of Design Base Shear
Design of Seismic base shear Vb = AhW
Ah = Z/2 I/R Sa/g = #NAME?
Zone Factor Z = #NAME?
Importance factor = 1 From Table 6 of IS 1893(Part I:2002)
Response Reduction factor R = 5
From Table 7(Lateral Load Resisting system ) of IS 1893(Part I:2002) For RC
Lateral Load Resisting System
For Ta = 1.617 1/Ta for range 0.40 to 4.00
Sa/g = 1/Ta for rock site from Figure 2 of IS 1893(Part I:2002)
= 0.618
Design Seismic shear,Vb= AhW = #NAME?
Step:4 Vertical Distribution of Base shearThe design base shear(Vb) computed shall be distributed along the height of the building as oer the expression
Qi = Vb W1 x h1 x h1/ €W1 x h1 x h1Where
Qi = Design Lateral Forces at floor i
Wi = Seismic weights of floor i
hi = Height of the floor I, measured from base
n = Number of Stories
Using the above equation , base shear is distributd as follows
M1 1184.25 H1 6 W1 h12
42633
M2 1184.25 H2 12 W2 h22
170532
M3 600.94 H3 18 W3 h32
194703.75
Q1 = Vb ( W1 x h1 x h1 / W1 x h1 x h1 + W2 x h2 x h2 + W3 x h3 x h3 +W4 x h4 x h4 ) #NAME?
Q2 = Vb ( W2 x h2 x h2 / W1 x h1 x h1 + W2 x h2 x h2 + W3 x h3 x h3 +W4 x h4 x h4 ) #NAME?
Q3 = Vb ( W3 x h3 x h3 / W1 x h1 x h1 + W2 x h2 x h2 + W3 x h3 x h3 +W4 x h4 x h4 ) #NAME?
Lateral force distribution at various floor levels
Loading Diagram Shear Diagram
Q3 #NAME? #NAME?
Q2 #NAME? #NAME?
Q1 #NAME? #NAME?
0.0
0
0.0
0
0.0
0
1 2 3
0.0
0
0.0
0
1 2
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 Storey 6 1 1 1 1 1 1 1 1 1 19 Storey 6 1 1 1 1 1 1 1 1 1 18 Storey 6 1 1 1 1 1 1 1 1 1 17 Storey 6 1 1 1 1 1 1 1 1 1 16 Storey 6 1 1 1 1 1 1 1 1 1 15 Storey 6 1 1 1 1 1 1 1 1 1 1
4 Storey 61 1 1 1 1 1 1 1 1 1
3 Storey 61 1 1 1 1 1 1 1 1 1
2 Storey 61 1 1 1 1 1 1 1 1 1
1 Storey 61 1 1 1 1 1 1 1 1 1
6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0
BAY 1 BAY 2 BAY 3 BAY 4 BAY 5 BAY 6 BAY 7 BAY 8 BAY 9 BAY 10
3 3 3
3 3 3
6.0 6.0
0.0
0
0.0
0
2 3
0 0 0 0 0 0 0 0 0 0 0 0 25
0 0 0 0 0 0 0 0 0 0 0 0 24
0 0 0 0 0 0 0 0 0 0 0 0 23
0 0 0 0 0 0 0 0 0 0 0 0 22
0 0 0 0 0 0 0 0 0 0 0 0 21
0 0 0 0 0 0 0 0 0 0 0 0 20
0 0 0 0 0 0 0 0 0 0 0 0 19
0 0 0 0 0 0 0 0 0 0 0 0 18
0 0 0 0 0 0 0 0 0 0 0 0 17
0 0 0 0 0 0 0 0 0 0 0 0 16
0 0 0 0 0 0 0 0 0 0 0 0 15
0 0 0 0 0 0 0 0 0 0 0 0 14
0 0 0 0 0 0 0 0 0 0 0 0 13
0 0 0 0 0 0 0 0 0 0 0 0 12
0 0 0 0 0 0 0 0 0 0 0 0 11
0 0 0 0 0 0 0 0 0 0 3.5 1 1 10
0 0 0 0 0 0 0 0 0 0 3.5 1 1 9
0 0 0 0 0 0 0 0 0 0 3.5 1 1 8
0 0 0 0 0 0 0 0 0 0 3.5 1 1 7
0 0 0 0 0 0 0 0 0 0 3.5 1 1 6
0 0 0 0 0 0 0 0 0 0 3.5 1 1 5
0 0 0 0 0 0 0 0 0 03.5
1 1 4
0 0 0 0 0 0 0 0 0 03.5
1 1 3
0 0 0 0 0 0 0 0 0 03.5
1 1 2
0 0 0 0 0 0 0 0 0 03.5
1 1 1
Bays Floors
Storey 25 Storey
Storey 24 Storey
Storey 23 Storey
Storey 22 Storey
Storey 21 Storey
Storey 20 Storey
Storey 19 Storey
Storey 18 Storey
Storey 17 Storey
Storey 16 Storey
Storey 15 Storey
Storey 14 Storey
Storey 13 Storey
Storey 12 Storey
Storey 11 Storey
Storey 10 Storey
Storey 9 Storey
Storey 8 Storey
Storey 7 Storey
Storey 6 Storey
Storey 5 Storey
Storey 4 Storey
Storey 3 Storey
Storey 2 Storey
Storey 1 Storey
1 + 15 T 2.5 1/T 1 + 15 T 2.5 1.36/T 1 + 15 T 2.5 1.67/T
0 - 0.1 0. - 0.4 0.4 - 4 0 - 0.1 0. - 0.4 0.4 - 4 0 - 0.1 0. - 0.4 0.4 - 4
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.20
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
0.50
For Rocky Soils For Medium Soils For Soft Soils
Sa/g Values
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.60
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.70
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.80
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.90
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1.00
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
1.41
1.42
1.43
1.44
1.45
1.46
1.47
1.48
1.49
1.50
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.77
1.78
1.79
1.80
1.81
1.82
1.83
1.84
1.85
1.86
1.87
1.88
1.89
1.90
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2.00
2.01
2.02
2.03
2.04
2.05
2.06
2.07
2.08
2.09
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
2.23
2.24
2.25
2.26
2.27
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.38
2.39
2.40
2.41
2.42
2.43
2.44
2.45
2.46
2.47
2.48
2.49
2.50
2.51
2.52
2.53
2.54
2.55
2.56
2.57
2.58
2.59
2.60
2.61
2.62
2.63
2.64
2.65
2.66
2.67
2.68
2.69
2.70
2.71
2.72
2.73
2.74
2.75
2.76
2.77
2.78
2.79
2.80
2.81
2.82
2.83
2.84
2.85
2.86
2.87
2.88
2.89
2.90
2.91
2.92
2.93
2.94
2.95
2.96
2.97
2.98
2.99
3.00
3.01
3.02
3.03
3.04
3.05
3.06
3.07
3.08
3.09
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
3.26
3.27
3.28
3.29
3.30
3.31
3.32
3.33
3.34
3.35
3.36
3.37
3.38
3.39
3.40
3.41
3.42
3.43
3.44
3.45
3.46
3.47
3.48
3.49
3.50
3.51
3.52
3.53
3.54
3.55
3.56
3.57
3.58
3.59
3.60
3.61
3.62
3.63
3.64
3.65
3.66
3.67
3.68
3.69
3.70
3.71
3.72
3.73
3.74
3.75
3.76
3.77
3.78
3.79
3.80
3.81
3.82
3.83
3.84
3.85
3.86
3.87
3.88
3.89
3.90
3.91
3.92
3.93
3.94
3.95
3.96
3.97
3.98
3.99
4.00
Step 1 Determination of Eigen Values and Eigen vectors
Mass Matrix M and stifness Matrix K of the plane frame lumped mass model are:
M1 0 0 60.59 0 0
M= 0 M2 0 = 0 60.59 0
0 0 M3 0 0 30.59
Column Stiffness storey
K= = = 11881 KN/m
Total Stiffness of Each Storey
k1=k2=k3= 3 x11881.148 35643 KN/m
Stiffness of lumped mass modelled structure
M3= 30.59
k3 35643
k1=35643.44 X1 k2=35643.44 X2 k3=35643.44 X3 M2= 60.59
M1=60.59 M2=60.59 M3=30.59
k2 35643
Free Body Diagram M1= 60.59
m1x''1
M1=60.59 k2(X1-X2) k1 35643
k1X1
m2x''2
M2=60.59 k3(X2-X3)
k2(X2-X1)
m3x''3
M3=30.59
k3(X3-X2)
By writing the equations in the form we get
m1x''1 + k1X1 + k2(X1-X2) = 0
m2x''2 + k2(X2-X1)+ k3(X2-X3) = 0
m3x''3 + k3(X3-X2)= 0
m1x''1 + k1X1 + k2X1 - k2X2 = 0
m2x''2 + k2X2 - k2X1 + k3X2 - k3X3 = 0
m3x''3 + k3X3 - k3X2 = 0
k1+k2 -k2 0
K= -k2 k2+k3 -k30 -k3 k3
K1 Remains same for all three
storey because it is column
12𝐸𝐼/𝐿^3 (12∗5000√20∗〖10〗^3∗(0.25∗〖0.45〗^3)/12)/〖3.5〗^3
m𝑥 ̈+𝑘𝑥+𝑐=0
71286.9 -35643.44 0
K= -35643.4 71286.88 -35643
0 -35643.44 35643
Writing into Matrix Form
m1 0 0 X''1 k1+k2 -k2 0 X1
0 m2 0 X''2 -k2 k2+k3 -k3 X2
0 0 m3 X''3 0 -k3 k3 X3
60.59 0 0 X''1 71286.88 -35643 0 X1
0 60.59 0 X''2 -35643.4 71287 -35643 X2
0 0 30.59 X''3 0 -35643 35643 X3
By using Characteristic equation
2 -1 0 1.98 0 0
35643 -1 2 -1 - 30.59 0 1.98 0
0 -1 1 0 0 1
£ = 30.59 = 0.00086
35643
2 -1 0 1.981 0 0
-1 2 -1 £ 0 1.981 0
0 -1 1 0 0 1
2 -1 0 1.981£ 0 0
-1 2 -1 - 0 1.981£ 0
0 -1 1 0 0 1£
2-1.981£ -1 0
-1 2-1.981£ -1
0 -1 1-1£
By Finding DETERMINANT we get
£1 = 2.414
£2 = 0.126
£2 = 1.06
Frequency calculation
£ =
= 53.036 rad/sec
= 12.12 rad/sec
= 35.14 rad/sec
First Mode Shape
For £1 =2.414 2-1.981£ -1 0 -2.249 -1 0
-1 2-1.981£ -1 -1 -2.249 -1
0 -1 1-1£ 0 -1 -1.414
-1
-2.249 -1.000 x1 -1
- -1.000 -1.414 x2 = 0
0.000858
1235.111
2812.791
146.815
k-𝜔2m=0
𝜔^2
𝜔^2 𝜔2
𝜔^2
𝜔_1 √
𝜔_2 √
𝜔_3 √
1.500
Mode Shape 1 1.000
-0.648
0.458
For £2 =0.126 2-1.981£ -1 0 1.778 -1 0
-1 2-1.981£ -1 -1 1.778 -1
0 -1 1-1£ 0 -1 0.874
-1
1.778 -1.000 x1 -1
- -1.000 0.874 x2 = 0
Mode Shape 2 1.000
1.577
1.805
For £3 =1.06 2-1.981£ -1 0 0.134 -1 0
-1 2-1.981£ -1 -1 0.134 -1
0 -1 1-1£ 0 -1 -0.060
-1
0.134 -1.000 x1 -1
- -1.000 -0.060 x2 = 0
Mode Shape 3 1.000
0.059
-0.992
Eigen Vectors фT
1Mф 1.000 -0.648 0.458 60.59 0.00 0.00 1.000 = 124
0.00 60.59 0.00 -0.648
0.00 0.00 30.59 0.458
ф1 = 1/√ф1Mф = 0.090 1.000 = 0.090
-0.648 -0.058
0.458 0.041
Eigen Vectors фT
2Mф 1.000 1.577 1.805 60.59 0.00 0.00 1.000 = 429
0.00 60.59 0.00 1.577
0.00 0.00 30.59 1.805
ф2 = 1/√ф2Mф = 0.048 1.000 = 0.048
1.577 0.076
1.805 0.087
Eigen Vectors фT
3Mф 1.000 0.059 -0.992 60.59 0.00 0.00 1.000 = 126
0.00 60.59 0.00 0.059
0.00 0.00 30.59 -0.992
ф3 = 1/√ф3Mф = 0.089 1.000 = 0.089
0.059 0.005
-0.992 -0.089
1.000
-0.648
0.458
-1.000
-0.500
0.000
0.500
1.000
1.500
1 2 3
Mode Shape 1
1.000 1.577 1.805
0.000
0.500
1.000
1.500
2.000
1 2 3
Mode Shape 2
1.000
0.059
-0.992
-1.500
-1.000
-0.500
0.000
0.500
1.000
1.500
1 2 3
Mode Shape 3
Eigen Vector = ф = 0.090 0.048 0.089
-0.058 0.076 0.005
0.041 0.087 -0.089
Natural Time Period T1 = 2PI/Ѡn= 0.118
Natural Time Period T2 = 2PI/Ѡn= 0.519
Natural Time Period T3 = 2PI/Ѡn= 0.179
T= 0.118 0 0
0 0.519 0
0 0 0.179
Step 2 : Determination of Modal Participation Factors
The modal Participation factor ( Pk) of mode K is
Pk=
P1 P2 P3
2.376 7.399 2.185
Step3 : Determination of Modal Mass
The Modal Mass (Mk) of mode k is given by
Mk=
Where
g= Accleration due to gravity
ф= Mode shape coeficient at floor I in mode k and
Wi= Seismic weight of floor i
M1 M2 M3
13.51 100.116 29.609
M= 143.233
Modal contributions of various modes
For Mode 1 M1/M 9.4%
For Mode 2 M2/M 69.9%
For Mode 3 M3/M 20.7%
Step 4: Determination of Lateral Force at Each Floor in Each Mode
The design lateral force Qik at floor i in mode k is given by
Qik =
Where
The design horizontal seismic coefficeint Ah for various modes are
Design of Seismic base shear Vb = AhW
Ah = Z/2 I/R Sa/g =
Zone Factor Z = 0.16
= 1 From Table 6 of IS 1893(Part I:2002)
= 5
From
Table
Importance factor
Response
Ak = Design Horizontal accleleration spectrum value as per clause 6.4.2 of IS 1893 Part
I:2002 using natural period of vibration Tk of mode k
∑24_(𝑖=1)^𝑛▒𝑊𝑖𝜑𝑖𝑘/〖𝑊𝑖(𝜑𝑖𝑘)〗^2
∑24_(𝑖=1)^𝑛▒((𝑊𝑖𝜑𝑖𝑘)2)/(〖𝑔[𝑊𝑖(𝜑𝑖𝑘)〗^2])
(𝐴𝑘𝜑𝑖𝑘𝑃𝑘𝑊𝑖)
For Rocky Soil Sa/g Range Sa/g factor
Sag
Value
For T1 = 0.118 0.10 to 0.40 2.5 2.5
For T2 = 0.519 0.40 to 4.00 1/Ta 1.927
For T3 = 0.179 0.40 to 4.00 1/Ta 5.587
Ah1 = 0.040
Ah2 = 0.031
Ah3 = 0.089
Design Lateral force in each mode=
Qi1 Ah1 P1 фi1 wi
Ah1 P1 ф11 w1= 5.080
Qi1 Ah1 P1 ф21 w2= -3.292
Ah1 P1 ф31 w3= 1.175
Ah2 P2 ф11 w1= 6.546
Qi2 Ah2 P2 ф21 w2= 10.323
Ah2 P2 ф31 w3= 5.966
Ah3 P3 ф11 w1= 10.362
Qi3 Ah3 P3 ф21 w2= 0.611
Ah3 P3 ф31 w3= -5.190
Step 5 : Determination of Storey shear forces in Each Mode Vik =
The peak Shear force in given by,
V11 Q11 + Q12 + Q31 2.963
Vi1 = = V21 = Q12 + Q31 = -2.117
V31 Q31 1.175
V21 Q21 + Q22 + Q23 22.835
Vi2 = V22 = Q22 + Q23 = 16.289
V32 Q23 5.966
V13 Q31 + Q32 + Q33 5.784
Vi3 = V23 = Q32 + Q33 = -4.578
V33 Q33 -5.190
Step 6: Determination of Storey shear force due to All Modes
The Peak Storey Shear force Vi in storey I due to all modes considered is obtained by combining
those due to each mode in accordance with modal combination
i.e SRSS (Square Root of Sum of Squares) or (CQC (Complete Quadratic Combination) Methods.
SRSS (Square Root of Sum of Squares)
If the building does not have closely spaced modes, the peak response quantity (ʎ) due to obtained as
all modes considered shall be
ʎ =
Where,
ʎk = absolute value of quantity in mode "k", and r is the numbers of modes being considered.
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖𝑘)
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖1)
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖2)
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖3)
√(∑24_(𝑘=1)^𝑟▒〖(ʎ
Using the above method m the storey shears are:
V1 [(V11)2
+ (V12)2
+ (V13)2]1/2
= 3.83 KN
V2 [(V21)2
+ (V22)2
+ (V23)2]1/2
= 22.84 KN
V3 [(V31)2
+ (V32)2
+ (V33)2]1/2
= 5.97 KN
Step:7 Determination of Lateral Forces at Each Storey
The Design Lateral Forces Froof and Fi at roof and ith floor are calculated as,
Froof = Vroof and Fi = Vi - Vi+1
Square root of sum of squares ( SRSS )
Froof =F3 = V3 5.97
Ffloor2 =F2 = V2-V3 16.87
Ffloor1 =F1 = V1-V2 -19.01
3.8
3
22
.84
5.9
7
0.00
5.00
10.00
15.00
20.00
25.00
Shear Force Distribution
Shear Force
69 0 0 0 0 0 0 0 0
0 69 0 0 0 0 0 0 0
0 0 69 0 0 0 0 0 0
0 0 0 69 0 0 0 0 0
m = 0 0 0 0 69 0 0 0 0
0 0 0 0 0 69 0 0 0
0 0 0 0 0 0 69 0 0
0 0 0 0 0 0 0 69 0
0 0 0 0 0 0 0 0 69
0 0 0 0 0 0 0 0 0
By using Characteristic equation
##### ###### 0 0 0 0 0 0 0
##### ###### ##### 0 0 0 0 0 0
0 ###### ##### -2358.353 0 0 0 0 0
0 0 ##### 4716.706 -2358.353 0 0 0 0
0 0 0 -2358.353 4716.706 -2358.353 0 0 0
0 0 0 0 -2358.353 4716.706 ##### 0 0
0 0 0 0 0 -2358.353 ##### ##### 0
0 0 0 0 0 0 ##### ##### #####
0 0 0 0 0 0 0 ##### #####
0 0 0 0 0 0 0 0 #####
2 -1 0 0 0 0 0 0 0
-1 2 -1 0 0 0 0 0 0
0 -1 2 -1 0 0 0 0 0
0 0 -1 2 -1 0 0 0 0
0 0 0 -1 2 -1 0 0 0
2358.353 0 0 0 0 -1 2 -1 0 0
0 0 0 0 0 -1 2 -1 0
0 0 0 0 0 0 -1 2 -1
0 0 0 0 0 0 0 -1 2
0 0 0 0 0 0 0 0 -1
Let 0.015 W2
= λ
2-1.97λ -1-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ
-1-0λ 2-1.97λ -1-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ
0-0λ -1-0λ 2-1.97λ -1-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ
0-0λ 0-0λ -1-0λ 2-1.97λ -1-0λ 0-0λ 0-0λ 0-0λ 0-0λ
λ 0-0λ 0-0λ 0-0λ -1-0λ 2-1.97λ -1-0λ 0-0λ 0-0λ 0-0λ
0-0λ 0-0λ 0-0λ 0-0λ -1-0λ 2-1.97λ -1-0λ 0-0λ 0-0λ
0-0λ 0-0λ 0-0λ 0-0λ 0-0λ -1-0λ 2-1.97λ -1-0λ 0-0λ
0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ -1-0λ 2-1.97λ -1-0λ
0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ -1-0λ 2-1.97λ
0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ 0-0λ -1-0λ
k-𝜔2m=0
2-1.97λ -1 0 0 0 0 0 0 0
-1 2-1.97λ -1 0 0 0 0 0 0
0 -1 2-1.97λ -1 0 0 0 0 0
0 0 -1 2-1.97λ -1 0 0 0 0
2-1.97λ 0 0 0 -1 2-1.97λ -1 0 0 0
0 0 0 0 -1 2-1.97λ -1 0 0
0 0 0 0 0 -1 2-1.97λ -1 0
0 0 0 0 0 0 -1 2-1.97λ -1
0 0 0 0 0 0 0 -1 1-1λ
2-1.97λ -1 0 0 0 0 0 0
-1 2-1.97λ -1 0 0 0 0 0
0 -1 2-1.97λ -1 0 0 0 0
2-1.97λ 2-1.97λ 0 0 -1 2-1.97λ -1 0 0 0
0 0 0 -1 2-1.97λ -1 0 0
0 0 0 0 -1 2-1.97λ -1 0
0 0 0 0 0 -1 2-1.97λ -1
0 0 0 0 0 0 -1 1-1λ
2-1.97λ -1 0 0 0 0 0
-1 2-1.97λ -1 0 0 0 0
0 -1 2-1.97λ -1 0 0 0
2-1.97λ 2-1.97λ 2-1.97λ 0 0 -1 2-1.97λ -1 0 0
0 0 0 -1 2-1.97λ -1 0
0 0 0 0 -1 2-1.97λ -1
0 0 0 0 0 -1 1-1λ
2-1.97λ -1 0 0 0 0
-1 2-1.97λ -1 0 0 0
0 -1 2-1.97λ -1 0 0
2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 0 0 -1 2-1.97λ -1 0
0 0 0 -1 2-1.97λ -1
0 0 0 0 -1 1-1λ
2-1.97λ -1 0 0 0
-1 2-1.97λ -1 0 0
2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 0 -1 2-1.97λ -1 0
0 0 -1 2-1.97λ -1
0 0 0 -1 1-1λ
2-1.97λ -1 0 0
2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ -1 2-1.97λ -1 0
0 -1 2-1.97λ -1
0 0 -1 1-1λ
2-1.97λ -1 0
2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ -1 2-1.97λ -1
0 -1 1-1λ
2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ 2-1.97λ -1
-1 1-1λ
7 6 5 4 3 2 1
2-1.97A -1 2 1-1A - 1.97A 1-1A -
-1 1-1A
2-2A - -1.97A + 1.97A2
-
0
0
0
0
0
0
0
0
0
35
0 69 0 0 0 0 0 0 0
0 0 69 0 0 0 0 0 0
0 0 0 69 0 0 0 0 0
0 0 0 0 69 0 0 0 0
0 0 0 0 0 69 0 0 0
0 - w2 0 0 0 0 0 69 0 0
0 0 0 0 0 0 0 69 0
0 0 0 0 0 0 0 0 69
##### 0 0 0 0 0 0 0 0
##### 0 0 0 0 0 0 0 0
0 1.971 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0 0.000 1.971 0.000 0.000 0.000 0.000 0.000 0.000
0 0.000 0.000 1.971 0.000 0.000 0.000 0.000 0.000
0 0.000 0.000 0.000 1.971 0.000 0.000 0.000 0.000
0 0.000 0.000 0.000 0.000 1.971 0.000 0.000 0.000
0 - 35w2 0.000 0.000 0.000 0.000 0.000 1.971 0.000 0.000
0 0.000 0.000 0.000 0.000 0.000 0.000 1.971 0.000
0 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.971
-1 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
1 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
+ - + - + - + -
0-0λ 2-1.97λ -1 0 0 0 0 0 0
0-0λ -1 2-1.97λ -1 0 0 0 0 0
0-0λ 0 -1 2-1.97λ -1 0 0 0 0
0-0λ 0 0 -1 2-1.97λ -1 0 0 0
0-0λ 0 0 0 -1 2-1.97λ -1 0 0
0-0λ 0 0 0 0 -1 2-1.97λ -1 0
0-0λ 0 0 0 0 0 -1 2-1.97λ -1
0-0λ 0 0 0 0 0 0 -1 2-1.97λ
-1-0λ 0 0 0 0 0 0 0 -1
1-1λ 0 0 0 0 0 0 0 0
m=0
if λ = 1 0.03 -1 0 0 0 0 0 0
-1 0.03 -1 0 0 0 0 0
0 -1 0.03 -1 0 0 0 0
0 0 -1 0.03 -1 0 0 0
0 0 0 -1 0.03 -1 0 0
0 0 0 0 -1 0.03 -1 0
0 0 0 0 0 -1 0.03 -1
0 0 0 0 0 0 -1 0.03
0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0
if λ = 1.01 1 0.03 -1 0 0 0 0 0 0
-1 0.03 -1 0 0 0 0 0
0 -1 0.03 -1 0 0 0 0
0 0 -1 0.03 -1 0 0 0
0 0 0 -1 0.03 -1 0 0
0 0 0 0 -1 0.03 -1 0
0 0 0 0 0 -1 0.03 -1
0 0 0 0 0 0 -1 0.03
0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0
1
1
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
69 0
0 35
0.000 0.000
0.000 0.000
0.000 0.000
0.000 0.000
0.000 0.000
0.000 0.000
0.000 0.000
0.000 0.000
1.971 0.000
0.000 1.000
+ -
0 0
0 0
0 0
0 0
0 0
0 0
0 0
-1 0
2-1.97λ -1
-1 1-1λ
0 0
0 0
0 0
0 0
0 0 if λ = 1 -0.99101
0 0
0 0
-1 0
0.03 -1
-1 0
0 0
0 0
0 0
0 0
0 0 if λ = 1.01 -0.99101
0 0
0 0
-1 0
0.03 -1
-1 0
Step 1 Determination of Eigen Values and Eigen vectors
Mass Matrix M and stifness Matrix K of the plane frame lumped mass model are:
`
M1 0 0 0 0 0 0 0 0 0 0 0
0 M2 0 0 0 0 0 0 0 0 0 0
0 0 M3 0 0 0 0 0 0 0 0 0
0 0 0 M4 0 0 0 0 0 0 0 0
0 0 0 0 M5 0 0 0 0 0 0 0
0 0 0 0 0 M6 0 0 0 0 0 0
0 0 0 0 0 0 M7 0 0 0 0 0
0 0 0 0 0 0 0 M8 0 0 0 0
0 0 0 0 0 0 0 0 M9 0 0 0
M= 0 0 0 0 0 0 0 0 0 M10 0 0
0 0 0 0 0 0 0 0 0 0 M11 0
0 0 0 0 0 0 0 0 0 0 0 M12
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
120.72 0 0 0 0 0 0 0 0 0 0 0
0 120.72 0 0 0 0 0 0 0 0 0 0
0 0 120.72 0 0 0 0 0 0 0 0 0
0 0 0 120.72 0 0 0 0 0 0 0 0
0 0 0 0 120.72 0 0 0 0 0 0 0
0 0 0 0 0 120.72 0 0 0 0 0 0
0 0 0 0 0 0 120.72 0 0 0 0 0
0 0 0 0 0 0 0 120.72 0 0 0 0
0 0 0 0 0 0 0 0 120.72 0 0 0
0 0 0 0 0 0 0 0 0 120.72 0 0
0 0 0 0 0 0 0 0 0 0 120.72 0
0 0 0 0 0 0 0 0 0 0 0 120.72
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
Column Stiffness storey
K= = = 2358.4 KN/m
Total Stiffness of Each Storey
K= K1+Kn…… 23584 KN/m
Stiffness of lumped mass modelled structure
K1+K2 -K2
-K2 K2+K3
K3+K4
K4+K5
K5+K6
12𝐸𝐼/𝐿^3
(12∗5000√20∗〖10〗^3∗(0.25∗〖0.45〗^3)/12)/〖3.5〗^3
K6+K7
K7+K8
K8+K9
K9+K10
K= K10+K11
K11+K12
K12+K13
K13+K14
1 2 3 4 5 6 7 8 9 10 11 12 13
4716.70589 -2358.353 0 0 0 0 0 0 0 0 0 0 0
-2358.35295 4716.706 -2358.4 0 0 0 0 0 0 0 0 0 0
0 -2358.353 4716.7 -2358.4 0 0 0 0 0 0 0 0 0
0 0 -2358.4 4716.7 -2358.4 0 0 0 0 0 0 0 0
0 0 0 -2358.4 4716.7 -2358.4 0 0 0 0 0 0 0
0 0 0 0 -2358.4 4716.7 -2358.4 0 0 0 0 0 0
0 0 0 0 0 -2358.4 4716.7 -2358.4 0 0 0 0 0
0 0 0 0 0 0 -2358.4 4716.7 -2358.4 0 0 0 0
0 0 0 0 0 0 0 -2358.4 4716.7 -2358.4 0 0 0
0 0 0 0 0 0 0 0 -2358.4 4716.7 -2358.4 0 0
0 0 0 0 0 0 0 0 0 -2358.4 4716.7 -2358.4 0
K 0 0 0 0 0 0 0 0 0 0 -2358.4 4716.7 -2358.4
0 0 0 0 0 0 0 0 0 0 0 -2358.4 4716.7
0 0 0 0 0 0 0 0 0 0 0 0 -2358.4
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
By using Characteristic equation
1 2 3 4 5 6 7 8 9 10 11 12 13
2 10 10 10 10 10 10 10 10 10 10 10 10
3 10 10 10 10 10 10 10 10 10 10 10 10
4 10 10 10 10 10 10 10 10 10 10 10 10
5 10 10 10 10 10 10 10 10 10 10 10 10
6 10 10 10 10 10 10 10 10 10 10 10 10
7 10 10 10 10 10 80 10 10 10 10 10 10
8 10 10 10 10 10 2 10 10 10 10 10 10
9 10 10 10 10 10 10 10 10 10 10 10 10
10 10 10 10 10 10 10 10 10 10 10 10 10
11 10 10 10 10 10 10 10 10 10 10 10 10
12 10 10 10 10 10 10 10 10 10 10 10 10
13 10 10 10 10 10 10 10 10 10 10 10 10
14 10 10 10 10 10 10 10 10 10 10 10 10
15 10 10 10 10 10 10 10 10 10 10 10 10
16 10 10 10 10 10 10 10 10 10 10 10 10
17 10 10 10 10 10 10 10 10 10 10 10 10
18 10 10 10 10 10 10 10 10 10 10 10 10
19 10 10 10 10 10 10 10 10 10 10 10 10
20 10 10 10 10 10 10 10 10 10 10 10 10
0
1 2 3
2 10 10
3 10 10
k-𝜔2m=0
-10
Free Body Diagram
m1x''1
0
m2x''2
M2=0
k2(X2-)
m3x''3
M3=0
k3(X3-X2)
By writing the equations in the form
m1x''1 +
m2x''2 + k2(X2-)
m3x''3 + k3(X3-X2)
m1x''1 +
m2x''2 + k2X2
m3x''3 + k3X3
+k2 2k2 0
K= -k2 k2+k3 -k30 -k3 k3
0 0 0
K= 0 0 0
0 0 0
Writing into Matrix Form
m1 0 0
0 m2 0
0 0 m3
0.00 0 0
0 0.00 0
0 0 0.00
By using Characteristic equation
#DIV/0! #DIV/0! #DIV/0!
0 #DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
£ = 0.00
0
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
#DIV/0! #DIV/0! #DIV/0!
By Finding DETERMINANT we get
£1 = 2.414
£2 = 0.126
£2 = 1.06
Frequency calculation
£ =
=
=
=
First Mode Shape
For £1 = 2.414 #DIV/0! #DIV/0!
#DIV/0! #DIV/0!
#DIV/0! #DIV/0!
-1
-2.249 -1.000
- #DIV/0! -1.414
Mode Shape 1 1.000
-0.648
0.458
For £2 = 0.126 #DIV/0! #DIV/0!
#DIV/0! #DIV/0!
#DIV/0! #DIV/0!
-1
1.778 -1.000
#DIV/0!
#DIV/0!
#DIV/0!
#DIV/0!
𝜔^2
𝜔^2
𝜔_1 √
𝜔_2 √
𝜔_3 √
- #DIV/0! 0.874
Mode Shape 2 1.000
1.577
1.805
For £3 = 1.06 #DIV/0! #DIV/0!
#DIV/0! #DIV/0!
#DIV/0! #DIV/0!
-1
0.134 -1.000
- #DIV/0! -0.060
Mode Shape 3 1.000
0.059
-0.992
Eigen Vectors фT
1Mф 1.000
ф1 =
Eigen Vectors фT
2Mф 1.000
ф2 =
Eigen Vectors фT
3Mф 1.000
ф3 =
Eigen Vector = ф =
Natural Time Period T1 =
Natural Time Period T2 =
Natural Time Period T3 =
T= #DIV/0! 0
0 #DIV/0!
0 0
Step 2 : Determination of Modal Participation Factors
The modal Participation factor ( Pk) of mode K is
Pk=
P1 P2 P3
0.000 0.000 0.000
Step3 : Determination of Modal Mass
The Modal Mass (Mk) of mode k is given by
Mk=
Where
g= Accleration due to gravity
ф= Mode shape coeficient at floor I in mode k and
Wi= Seismic weight of floor i
M1 M2 M3
#DIV/0! 0.000 0.000
M= #DIV/0!
Modal contributions of various modes
For Mode 1 M1/M #DIV/0!
For Mode 2 M2/M #DIV/0!
For Mode 3 M3/M #DIV/0!
Step 4: Determination of Lateral Force at Each Floor in Each Mode
The design lateral force Qik at floor i in mode k is given by
Qik =
Where
The design horizontal seismic coefficeint Ah for various modes are
Design of Seismic base shear Vb = AhW
Ah = Z/2
Zone Factor Z = 0.16
= 1
= 5
Importance factor
Response Reduction
Ak = Design Horizontal accleleration spectrum value as per clause 6.4.2 of IS 1893 Part I:2002 using natural period of vibration Tk of mode k
∑24_(𝑖=1)^𝑛▒𝑊𝑖𝜑𝑖𝑘/〖𝑊𝑖(𝜑𝑖𝑘)〗^2
∑24_(𝑖=1)^𝑛▒((𝑊𝑖𝜑𝑖𝑘)2)/(〖𝑔[𝑊𝑖(𝜑𝑖𝑘)〗^2])
(𝐴𝑘𝜑𝑖𝑘𝑃𝑘𝑊𝑖)
For Rocky Soil
For T1 = #DIV/0!
For T2 = #DIV/0!
For T3 = #DIV/0!
Ah1 = 0.040
Ah2 = #DIV/0!
Ah3 = #DIV/0!
Design Lateral force in each mode
Qi1 Ah1 P1 фi1 wi
Ah1 P1 ф11 w1=
Qi1 Ah1 P1 ф21 w2=
Ah1 P1 ф31 w3=
Ah2 P2 ф11 w1=
Qi2 Ah2 P2 ф21 w2=
Ah2 P2 ф31 w3=
Ah3 P3 ф11 w1=
Qi3 Ah3 P3 ф21 w2=
Ah3 P3 ф31 w3=
Step 5 : Determination of Storey shear forces in Each Mode
The peak Shear force in given by,
Vi1 = =
Vi2 =
Vi3 =
Step 6: Determination of Storey shear force due to All Modes
The Peak Storey Shear force Vi in storey I due to all modes considered is obtained by combining
those due to each mode in accordance with modal combination
i.e SRSS (Square Root of Sum of Squares) or (CQC (Complete Quadratic Combination) Methods.
SRSS (Square Root of Sum of Squares)
If the building does not have closely spaced modes, the peak response quantity (ʎ) due to obtained as
all modes considered shall be
Where,
ʎk = absolute value of quantity in mode "k", and r is the numbers of modes being considered.
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖1)
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖2)
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖3)
Using the above method m the storey shears are:
V1 [(V11)2
+
V2 [(V21)2
+
V3 [(V31)2
+
Step:7 Determination of Lateral Forces at Each Storey
The Design Lateral Forces Froof and Fi at roof and ith floor are calculated as,
Froof = Vroof and Fi = Vi - Vi+1
Square root of sum of squares ( SRSS )
Froof = F3 = V3 #DIV/0!
Ffloor2 = F2 = V2-V3 #DIV/0!
Ffloor1 = F1 = V1-V2 #DIV/0!
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
M13 0 0 0 0 0 0 0
0 M14 0 0 0 0 0 0
0 0 M15 0 0 0 0 0
0 0 0 M16 0 0 0 0
0 0 0 0 M17 0 0 0
0 0 0 0 0 M18 0 0
0 0 0 0 0 0 M19 0
0 0 0 0 0 0 0 M20 20X20
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
120.72 0 0 0 0 0 0 0
0 120.72 0 0 0 0 0 0
0 0 120.72 0 0 0 0 0
0 0 0 120.72 0 0 0 0
0 0 0 0 120.72 0 0 0
0 0 0 0 0 120.72 0 0
0 0 0 0 0 0 120.72 0
0 0 0 0 0 0 0 120.72
= 11881 KN/m
3 x11881.148 35643 KN/m
K13+K14
K14+K15
K15+K16
K16+K17
K17+K18
K18+K19
K19+K20
14 15 16 17 18 19 20
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
-2358.4 0 0 0 0 0 0
4716.7 -2358.4 0 0 0 0 0
-2358.4 4716.7 -2358.4 0 0 0 0
0 -2358.4 4716.7 -2358.4 0 0 0
0 0 -2358.4 4716.7 -2358.4 0 0
0 0 0 -2358.4 4716.7 -2358.4 0
0 0 0 0 -2358.4 4716.7 -2358.4
0 0 0 0 0 -2358.4 2358.4
14 15 16 17 18 19 20
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
10 10 10 10 10 10 10
k2= X2 k3= X3
M2=0 M3=0
k2(-X2)
k3(X2-X3)
we get
+ k2(-X2) = 0
+ k3(X2-X3) = 0
= 0
+ k2-X 2 k2) = 0
- k2) + k3X2 - k3X3 = 0
- k3X2 = 0
X''1 +k2 2k2 0 X1
X''2 -k2 k2+k3 -k3 X2
X''3 0 -k3 k3 X3
K1 Remains same for all three
storey because it is column
m𝑥 ̈+𝑘𝑥+𝑐=0
X''1 0 0 0 X1
X''2 0 0 0 X2
X''3 0 0 0 X3
##### #### #####
- 0.00 ##### #### #####
##### #### #####
= #DIV/0!
##### #### #####
£ ##### #### #####
##### #### #####
##### #### #####
- ##### #### #####
##### #### #####
##### rad/sec
##### rad/sec
##### rad/sec
##### -2.249 ##### 0
##### -1 -2.249 -1
##### 0 ##### -1.414
x1 -1
x2 = 0
##### 1.778 ##### 0
##### -1 1.778 -1
##### 0 ##### 0.874
x1 -1
k-𝜔2m=0
𝜔^2
𝜔2
1.000
-0.648
0.458
-1.000
-0.500
0.000
0.500
1.000
1.500
1 2 3
Mode Shape 1
x2 = 0
##### 0.134 ##### 0
##### -1 0.134 -1
##### 0 ##### -0.060
x1 -1
x2 = 0
-0.648 0.458 0.00 0.00 0.00 1.000 =
0.00 0.00 0.00 -0.648
0.00 0.00 0.00 0.458
1/√ф1Mф = 0.090 1.000 = 0.090
-0.648 -0.058
0.458 0.041
1.577 1.805 0.00 0.00 0.00 1.000 =
0.00 0.00 0.00 1.577
0.00 0.00 0.00 1.805
1/√ф2Mф = 0.048 1.000 = 0.048
1.577 0.076
1.805 0.087
0.059 -0.992 0.00 0.00 0.00 1.000 =
0.00 0.00 0.00 0.059
0.00 0.00 0.00 -0.992
1/√ф3Mф = 0.089 1.000 = 0.089
0.059 0.005
-0.992 -0.089
1.000 1.577 1.805
0.000
0.500
1.000
1.500
2.000
1 2 3
Mode Shape 2
1.000
0.059
-0.992
-1.500
-1.000
-0.500
0.000
0.500
1.000
1.500
1 2 3
Mode Shape 3
0.090 0.048 0.089
-0.058 0.076 0.005
0.041 0.087 -0.089
2PI/Ѡn= #####
2PI/Ѡn= #####
2PI/Ѡn= #####
0
0
#####
I/R Sa/g =
From Table 6 of IS 1893(Part I:2002)From
Table
Ak = Design Horizontal accleleration spectrum value as per clause 6.4.2 of IS 1893 Part I:2002 using natural period of vibration Tk of mode k
Sa/g Range Sa/g factor
Sag
Value
0.10 to 0.40 2.5 2.5
0.40 to 4.00 1/Ta #####
0.40 to 4.00 1/Ta #####
=
0.000
0.000
0.000
#####
#####
#####
#####
#####
#DIV/0!
Vik =
V11 Q11 + Q12 + Q31 0.000
V21 = Q12 + Q31 = 0.000
V31 Q31 0.000
V21 Q21 + Q22 + Q23 #DIV/0!
V22 = Q22 + Q23 = #DIV/0!
V32 Q23 #DIV/0!
V13 Q31 + Q32 + Q33 #####
V23 = Q32 + Q33 = #####
V33 Q33 #DIV/0!
ʎ =
∑_(𝑗=𝑖+1)^𝑛▒(𝑄𝑖𝑘)
√(∑24_(𝑘=1)^𝑟▒〖(ʎ
(V12)2
+ (V13)2]1/2
= 0.00 KN
(V22)2
+ (V23)2]1/2
= #### KN
(V32)2
+ (V33)2]1/2
= #### KN
0.0
0
0.0
0
0.0
0
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
Shear Force Distribution
Shear Force
M3= 0.00
k3
M2= 0.00
k2
M1= 0.00
k1
K1 Remains same for all three
storey because it is column
124
429
126
Shear Force
-1
A B C D E F G H I
A a
B b
C c
D d
E e
F f
G g
H h
I i
J j
K k
L l
M m
N n
O o
P p
Q r
R q
S u
T v
U t
V v
W s
X x
Y y
Z z
J K L M N O P Q R
S T U V W X Y Z