lateral forces on building due to an earthquake using response spectrum method 20x20 matrix1

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?



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



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%



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


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


##### ###### 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

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


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


#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 =



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!



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


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



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



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

lateral forces on building due to an earthquake using response spectrum method 20x20 matrix1

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