Seismic Analysis using Dynamic Procedure

1.) Weight per Level:

a.) Weight of Slab:

Ws1 ＝ L ⋅ W ⋅ tslab ⋅ γc

kN
Ws1 ≔ 30 m ⋅ 10 m ⋅ 0.125 m ⋅ 24 ―― = 900 kN
m3

Ws1 ＝ Ws2 ＝ Ws3 ＝ 900 kN

b.) Weight of Beams: (Beam dimension = 300mm x 400mm)

Wb1 ＝ no.beams ⋅ L ⋅ W ⋅ tbeam ⋅ γc

⎛ kN ⎞ ⎛ kN ⎞
Wb1 ≔ ⎜4 ⋅ 30 m ⋅ 0.3 m ⋅ 0.4 m ⋅ 24 ―― + ⎜7 ⋅ 10 m ⋅ 0.3 m ⋅ 0.4 m ⋅ 24 ―― ⎟ = 547.2 kN
3 ⎟
⎝ m ⎠ ⎝ m3 ⎠

Wb1 ＝ Wb2 ＝ Wb3 ＝ 547.2 kN

c.) Weight of Wall: (Wall thickness = 150 mm)

Ww1 ＝ no.wall ⋅ L ⋅ twall ⋅ htributary ⋅ γc

⎛ kN ⎞ ⎛ kN ⎞
Ww1 ≔ ⎜2 ⋅ 30 m ⋅ 0.15 m ⋅ 3.4 m ⋅ 24 ―― ⎟ + ⎜2 ⋅ 10 m ⋅ 0.15 m ⋅ 3.4 m ⋅ 24 ―― ⎟ = 979.2 kN
⎝ m3 ⎠ ⎝ m3 ⎠

⎛ kN ⎞ ⎛ kN ⎞
Ww3 ≔ ⎜2 ⋅ 30 m ⋅ 0.15 m ⋅ 1.7 m ⋅ 24 ―― + ⎜2 ⋅ 10 m ⋅ 0.15 m ⋅ 1.7 m ⋅ 24 ―― ⎟ = 489.6 kN
3 ⎟
⎝ m ⎠ ⎝ m3 ⎠

Ww1 ＝ Ww2 ＝ 979.2 kN

Ww3 ＝ 489.6 kN

d.) Weight of Columns: (28 pcs - Column dimension = 400mm x 400mm)

Wc1 ＝ no.columns ⋅ L ⋅ W ⋅ htributary ⋅ γc

⎛ kN ⎞
Wc1 ≔ ⎜28 ⋅ 0.4 m ⋅ 0.4 m ⋅ 3.4 m ⋅ 24 ―― ⎟ = 365.568 kN
⎝ m3 ⎠

⎛ kN ⎞
Wc3 ≔ ⎜28 ⋅ 0.4 m ⋅ 0.4 m ⋅ 1.7 m ⋅ 24 ―― ⎟ = 182.784 kN
⎝ m3 ⎠

Wc1 ＝ Wc2 ＝ 365.568 kN

Wc3 ＝ 182.784 kN
 Level Hstory Wslab Wbeam Wwall Wcolumn Wi Hi
((m)) ((kN )) ((kN )) ((kN )) ((kN )) ((kN )) ((m))
3 3.4 900 547.2 489.6 182.784 2119.584 10.2
2 3.4 900 547.2 979.2 365.568 2791.968 6.8
1 3.4 900 547.2 979.2 365.568 2791.968 3.4

Wi1 ≔ 900 kN + 547.2 kN + 979.2 kN + 365.568 kN = 2791.968 kN

Wi2 ≔ 900 kN + 547.2 kN + 979.2 kN + 365.568 kN = 2791.968 kN

Wi3 ≔ 900 kN + 547.2 kN + 489.6 kN + 182.784 kN = 2119.584 kN

Wi ≔ Wi1 + Wi2 + Wi3

Wi ＝ 2 ((2791.968 kN)) + 2119.584 kN

Wi = 7703.52 kN

2.) Site Parameters:

I ≔ 1.0

Z ≔ 0.4

Soil_Type ＝ Sd

Na ≔ 1.0

Nv ≔ 1.0

Ca ≔ 0.44 Na = 0.44

Cv ≔ 0.64 Nv = 0.64

R ≔ 8.5

3.) Calculate Base Shear:

3
―
4
T ＝ Ct ⎛⎝Hi⎞⎠ ; Ct ≔ 0.0731
3
―
4
T ≔ 0.0731 ((10.2))

T = 0.417

T ＝ 0.417 sec T>7 ; Ft ≔ 0

 Cv ⋅ I 0.64 ⋅ 1
V0 ＝ ――⋅ W V0 ≔ ―――― ⋅ 7703.52 kN = 1390.959 kN Exceeds_V1
R⋅T 8.5 ⋅ 0.417

2.5 Ca ⋅ I 2.5 ⋅ 0.44 ⋅ 1

V1 ≤ ―――⋅ W V1 ≔ ――――⋅ 7703.52 kN = 996.926 kN
R 8.5

V2 ≥ 0.11 Ca ⋅ I ⋅ W V2 ≔ 0.11 ⋅ 0.44 ⋅ 1 ⋅ 7703.52 kN = 372.85 kN OK

0.8 Z ⋅ Nv ⋅ I 0.8 ⋅ 0.4 ⋅ 1 ⋅ 1

V3 ≥ ――――⋅ W V3 ≔ ――――― ⋅ 7703.52 kN = 290.015 kN OK
R 8.5

Static Base Shear:

V ≔ 996.926 kN

4.) Lateral Forces:

Level Wi Hi W iH i Fx
((kN )) ((m)) ((kN ⋅ m)) ((kN ))
3 2119.584 10.2 21619.757 430.224
2 2791.968 6.8 18985.382 377.801
1 2791.968 3.4 9492.691 188.901

V ＝ Fx + Ft ; Ft ≔ 0
WiHi ≔ 9492.691 + 18985.382 + 21619.757 = 50097.83
⎛⎝V - Ft⎞⎠ WiHi
Fx ＝ ――――― WiHi ≔ 50097.83 kN ⋅ m
3
∑ W iH i
i =1

((996.926 kN - 0)) ((9492.691 kN ⋅ m))

Fx1 ≔ ―――――――――――― = 188.901 kN
50097.83 kN ⋅ m

((996.926 kN - 0)) ((18985.382 kN ⋅ m))

Fx2 ≔ ――――――――――――= 377.801 kN
50097.83 kN ⋅ m

((996.926 kN - 0)) ((21619.757 kN ⋅ m))

Fx3 ≔ ――――――――――――= 430.224 kN
50097.83 kN ⋅ m

5.) Eigen Values

12 EI
Kx ＝ Ky ＝ K ＝ ――
Ec ＝ 2400 1.5 ⋅ 0.043 ⋅ ‾‾‾
f'c L3

⎛ ⎛ 0.4 ⋅ 0.4 3 ⎞ ⎞
Ec ≔ 2400 1.5 ⋅ 0.043 ⋅ ‾‾‾
21 ⎜ 12 ((23168.343)) ⎜―――⎟ ((1000)) ⎟
⎝ 12 ⎠
K ≔ 28 ⎜―――――――――――⎟ = 422528.706
⎜⎝ 3.4 3 ⎟⎠
Ec ＝ 23168.343 MPa

kN
K ＝ 422528.706 ――
m
Level Wi Mi Ki
⎛ kN ⎞
((kN )) ((tonne )) ⎜―― ⎟
⎝ m ⎠
3 2119.584 216.137 422528.706
2 2791.968 284.702 422528.706
1 2791.968 284.702 422528.706
S1 ＝ 2.5 Ca

S1 ≔ 2.5 ⋅ 0.44 = 1.1

T1 0.3618
―＝ ――― ＝ 0.6216
Ts 0.582
Cv
Ts ＝ ―――
2.5 Ca T2 0.1225
―＝ ――― ＝ 0.2105
Ts 0.582
0.64
Ts ≔ ―― = 0.582
1.1 T3 0.0853
―＝ ――― ＝ 0.1466
Ts 0.582

Ts ≔ 0.582 sec

T1 T2 T3 T1 T2 T3
―＝ ― > 0.2 > ― 1 > ―＝ ― > ―
Ts Ts Ts Ts Ts Ts

Sa1 ≔ 10.791
⎛ m⎞
Sa1 ＝ Sa2 ＝ 1.1 ⋅ g ＝ 1.1 ⎜9.81 ―⎟
⎝ s2 ⎠ Sa2 ≔ 10.791

1.1 - 0.44 = 0.66

0.66 y
―― ＝ ―――
0.2 0.1466

0.66 ((0.1466))
y ≔ ――――― = 0.484
0.2

Sa3 ≔ ((0.484 + 0.44)) ((9.81))

Sa3 = 9.064

2
⎛ Tk ⎞
Sdk ＝ Sak ⎜―― ⎟
⎝2 π⎠

⎛ 0.3618 ⎞
Sd1 ≔ 10.791 ⎜――― ⎟ = 0.6214 ; Sa1 ＝ 10.791
⎝ 2π ⎠

⎛ 0.1225 ⎞
Sd2 ≔ 10.791 ⎜――― ⎟ = 0.2104 ; Sa2 ＝ 10.791
⎝ 2π ⎠

⎛ 0.0853 ⎞
Sd3 ≔ 9.064 ⎜――― ⎟ = 0.1231 ; Sa3 ＝ 9.064
⎝ 2π ⎠
6.) Compute Mode Participating factor (mpf), Effective mass (emk) and Effective Participating
mass (epmk) for each mode k

@K ＝ 1 @K ＝ 2 @K ＝ 3
n＝3 n＝3 n＝3
∑ m1 ⋅ ϕ1 ＝ 1401.094 ∑ m2 ⋅ ϕ2 ＝ 160.7345 ∑ m3 ⋅ ϕ3 ＝ 77.7725
i =1 i =1 i =1

n＝3 2 n＝3 2 n＝3 2

∑ m1 ⎛⎝ϕ1⎞⎠ ＝ 2713.629 ∑ m2 ⎛⎝ϕ2⎞⎠ ＝ 544.0418 ∑ m3 ⎛⎝ϕ3⎞⎠ ＝ 413.2252
i =1 i =1 i =1

n＝3 n＝3 n＝3

∑ mi ＝ 785.5404 ∑ mi ＝ 785.5404 ∑ mi ＝ 785.5404
i =1 i =1 i =1

*Mode Participating Factor, mpf

n＝3
∑ mi ⋅ ϕi
i =1
mpfk ＝ ――――
n＝3 2
∑ mi ⎛⎝ϕi⎞⎠
i =1

@K ＝ 1 @K ＝ 2 @K ＝ 3

1401.094 160.7345 77.7725

mpf1 ≔ ――― mpf2 ≔ ――― mpf3 ≔ ―――
2713.629 544.0418 413.2252

mpf1 = 0.516 mpf2 = 0.295 mpf3 = 0.188

*Effective Mass, emk

⎛ n＝3 ⎞2
⎜ ∑ mi ⋅ ϕi⎟
⎝ i =1 ⎠
emk ＝ ―――――
n＝3 2
∑ mi ⎛⎝ϕi⎞⎠
i =1

@K ＝ 1 @K ＝ 2 @K ＝ 3

((1401.094)) 2 ((160.7345)) 2 ((77.7725)) 2

em1 ≔ ―――― em2 ≔ ―――― em3 ≔ ――――
2713.629 544.0418 413.2252

em1 = 723.409 em2 = 47.488 em3 = 14.637

 *Effective Participating Mass, empk

emk
empk ＝ ―――
n＝3
∑ mi
i =1

@K ＝ 1 @K ＝ 2 @K ＝ 3

723.409 47.488 14.637

emp1 ≔ ――― emp2 ≔ ――― emp3 ≔ ―――
785.5404 785.5404 785.5404

emp1 = 0.921 emp2 = 0.06 emp3 = 0.019

7.) Compute displacement at each level i, mode k

Uik ＝ mpfk ⋅ Sdk ⋅ ϕk

Ui ＝ 0.7 R ⋅ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝Ui1⎞⎠ 2 + ⎛⎝Ui2⎞⎠ 2 + ⎛⎝Ui3⎞⎠ 2 ; R ＝ 8.5
8.) Compute Story Drift at each level i, mode k

mpf1 ＝ 0.516 Sd1 ＝ 0.6214

mpf2 ＝ 0.295 Sd2 ＝ 0.2104

mpf3 ＝ 0.188 Sd3 ＝ 0.1231

Udi ＝ 0.7 R ⋅ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

⎛⎝Ui1⎞⎠ 2 + ⎛⎝Ui2⎞⎠ 2 + ⎛⎝Ui3⎞⎠ 2

*Revise Column Size from 0.6 x 0.6 to 0.7 x 0.7

9.) Compute Forces at each level, mode k

Fik ＝ mpfk ⋅ Sak ⋅ mi ⋅ ϕik

Sa1 ＝ 10.791 mpf1 ＝ 0.516

Sa2 ＝ 10.791 mpf2 ＝ 0.295

Sa3 ＝ 9.064 mpf3 ＝ 0.188

 Fi ＝ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝Fxi1⎞⎠ 2 + ⎛⎝Fxi2⎞⎠ 2 + ⎛⎝Fxi3⎞⎠ 2

I
F'i ＝ ―⋅ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝Fxi1⎞⎠ 2 + ⎛⎝Fxi2⎞⎠ 2 + ⎛⎝Fxi3⎞⎠ 2
R

10.) Compute Dynamic Base Shear

Vdynk ＝ emk ⋅ Sak

em1 ＝ 723.409 Sa1 ＝ 10.791

em2 ＝ 47.488 Sa2 ＝ 10.791

em3 ＝ 14.637 Sa3 ＝ 9.064

Vdesign1 ＝ em1 ⋅ Sd1 Vdesign3 ＝ em3 ⋅ Sd3

Vdesign1 ≔ 723.409 ⋅ 10.791 = 7806.307 Vdesign3 ≔ 14.637 ⋅ 9.064 = 132.67

Vdesign1 ＝ 7806.307 kN Vdesign3 ＝ 132.67 kN

Vdesign2 ＝ em2 ⋅ Sd2

Vdesign2 ≔ 47.488 ⋅ 10.791 = 512.443

Vdesign2 ＝ 512.443 kN

1
Vdesign ＝ ―⋅ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝Vdesign1⎞⎠ 2 + ⎛⎝Vdesign2⎞⎠ 2 + ⎛⎝Vdesign3⎞⎠ 2
R

1
Vdesign ≔ ―― ⋅ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
((7806.307)) 2 + ((512.443)) 2 + ((132.67)) 2 = 920.498
8.5

Vdesign ＝ 920.498 kN

Vdesign < 0.9 Vstatic Use V ＝ 0.9 Vstatic

Vdesign > 0.9 Vstatic Use V ＝ Vdesign

Vstatic ＝ 996.926 kN

0.9 Vstatic ＝ 897.233 kN < Vdesign

V ＝ 920.498 kN

*No Scaling Factor required