Solution to Question 1:

1) Natural period, s 6.1 nD TT

Structural stiffness kN/m 1750N/m 1750000101

105.172

3

Pk

Mass of structure k

mTn 2

tons5.113kg 1134804 2

kT

m n

2) Natural frequency rad/s 93.32

n

nT

3) Damping ratio %58.40458.0ln2

1

1

0 u

u

Damping coefficient Ns/m 6.408072 nmc

4) Amplitude after ten cycles

10

0ln2

1

u

u

j

10

1ln

102

105.0

u

cm 0563.010 u

Solution to Question 2:

1) Stead-state displacement is given as

)sin()( tRutu dst

where

m 0171.01750000

100030

k

pust ;

8.093.3

14.3

93.3

25.0

;

72.2)2()1(

1222

dR ;

5.11rad 2.01

2tan

2

1

Thus,

)2.014.3sin(0467.0)( ttu

2) Dynamic amplification factor Rd is given as

222 )2()1(

1

dR where

n

Following table summarised the value of Rd with different values of β

f (Hz) w (rad/s) beta Rd

0.1 0.63 0.16 1.03

0.2 1.26 0.32 1.11

0.4 2.51 0.64 1.69

0.5 3.14 0.80 2.72

0.625 3.93 1.00 10.92

0.7 4.40 1.12 3.65

0.8 5.03 1.28 1.54

1.0 6.28 1.60 0.64

3) Frequency-response curve

4) The total displacement of the building with zero initial conditions at resonant frequency can

be expressed as follows

𝑢(𝑡) = (𝑢𝑠𝑡)𝑜

1

2𝜉[𝑒−𝜉𝜔𝑛𝑡 (cos 𝜔𝐷𝑡 +

𝜉

√1 − 𝜉2sin 𝜔𝐷𝑡) − cos 𝜔𝑛𝑡]

The above function is plotted using Excel. The following figure shows the relationship

between total displacement and time. It can be seen that the total displacement increases

gradually until reaching its steady-state.

From the figure, the steady-state response vibration amplitude is approximately equal to

0.185 m. The result can be verified using the closed-form solution

𝑢𝑜 = (𝑢𝑠𝑡)𝑜

1

2𝜉= 0.187 m

which agrees with the graphical solution.

0.00

2.00

4.00

6.00

8.00

10.00

12.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80

Dyn

amic

Am

plif

icat

ion

Fac

tor

Frequency Ratio

5) About four cycles are required for the system to achieve 50% of the steady-state vibration

amplitude.

6) Total base shear

kN 6.327101872.01750000 3

o

sb

ku

fV

Over-turning moment

kNm 8.982

6.3273

bb hVM

7) Transmissibility

97.10

)2()1(

)2(12/1

222

2

TR

Total force

kN 329

97.1030

TRpf oT

-0.2500

-0.2000

-0.1500

-0.1000

-0.0500

0.0000

0.0500

0.1000

0.1500

0.2000

0.2500

0.0 5.0 10.0 15.0 20.0 25.0 30.0

Dis

pla

em

en

t (m

)

Time (s)

8) When the system reaches it vibration amplitude (says, at time t = τ) at steady-state, the

displacement and velocity are

m 1872.0)( u ; m/s 0)( u

If the applied force is suddenly removed at time t = τ, the system will undergo free vibration.

The displacement at time t = τ will become the new “initial conditions” of the free vibration

after the load removal.

The displacement history is then given as

)](sin)(cos[)()(

tBtAetu DD

tn for t

where

nD ;

m 1872.0)( uA ;

m 00857.0)()(

D

nuuB

Solution to Question 3:

Since

25.00625.06.1

1.0

n

d

T

t

The structural responses can be estimated using the approximation method.

Impulse magnitude

Ns 1000001.0101000 3 I

Maximum displacement

m 224.093.3113480

100000

n

om

Iu

Maximum base shear force on column

kN 2.1962

224.01750

2

o

b

kuV

Solution to Question 4:

The following two tables show some calculation steps for both Central Difference Method and

Newmark’s Method (up to 2 sec only).

Central Difference Method

1. initial condition

p0 25 kN

u0 0 m

v0 0 m/s

a0 5.00E-03 m/s2

u-1 2.50E-05 m

k^ 5.04E+08 N/m

a 4.96E+08

b -9.00E+08

2. Calculation

t pi ui-1 ui pi^ ui+1

0.0 25000 2.50E-05 0 1.26E+04 2.50E-05

0.1 25000 0 2.50E-05 4.75E+04 9.42E-05

0.2 25000 0.000025 9.42E-05 9.74E+04 1.93E-04

0.3 25000 9.4158E-05 1.93E-04 1.52E+05 3.01E-04

0.4 25000 0.00019298 3.01E-04 2.01E+05 3.98E-04

0.5 25000 0.00030136 3.98E-04 2.34E+05 4.63E-04

0.6 25000 0.00039763 4.63E-04 2.45E+05 4.85E-04

0.7 25000 0.00046293 4.85E-04 2.32E+05 4.60E-04

0.8 25000 0.00048487 4.60E-04 1.99E+05 3.94E-04

0.9 25000 0.00045986 3.94E-04 1.51E+05 3.00E-04

1.0 25000 0.00039369 3.00E-04 1.00E+05 1.98E-04

1.1 25000 0.00030021 1.98E-04 5.48E+04 1.09E-04

1.2 25000 0.00019844 1.09E-04 2.45E+04 4.85E-05

1.3 25000 0.00010869 4.85E-05 1.48E+04 2.94E-05

1.4 25000 4.8548E-05 2.94E-05 2.74E+04 5.43E-05

1.5 25000 2.9402E-05 5.43E-05 5.93E+04 1.18E-04

1.6 25000 5.4324E-05 1.18E-04 1.04E+05 2.06E-04

1.7 25000 0.00011759 2.06E-04 1.52E+05 3.02E-04

1.8 25000 0.00020599 3.02E-04 1.94E+05 3.85E-04

1.9 25000 0.00030154 3.85E-04 2.22E+05 4.41E-04

2.0 25000 0.00038518 4.41E-04 2.31E+05 4.57E-04

Newmark's Method

1. Initial condition

gamma 0.5

beta 0.25

u0 0 m

v0 0 m/s

a0 5.00E-03 m/s2

k^ 2.12E+09

a 2.02E+08

b 1.00E+07

2. Calculation

t pi ai dpi dp^ dui dvi dai vi ui

0.0 25000 5.00E-03 0 5.00E+04 2.36E-05 4.72E-04 -5.57E-04 0 0

0.1 25000 4.44E-03 0 1.40E+05 6.60E-05 3.75E-04 -1.39E-03 4.72E-04 2.36E-05

0.2 25000 3.06E-03 0 2.02E+05 9.52E-05 2.09E-04 -1.94E-03 8.47E-04 8.96E-05

0.3 25000 1.12E-03 0 2.24E+05 1.06E-04 5.73E-06 -2.12E-03 1.06E-03 1.85E-04

0.4 25000 -1.00E-03 0 2.04E+05 9.64E-05 -1.95E-04 -1.89E-03 1.06E-03 2.91E-04

0.5 25000 -2.90E-03 0 1.46E+05 6.89E-05 -3.55E-04 -1.31E-03 8.67E-04 3.87E-04

0.6 25000 -4.21E-03 0 6.11E+04 2.89E-05 -4.46E-04 -4.97E-04 5.11E-04 4.56E-04

0.7 25000 -4.71E-03 0 -3.38E+04 -1.60E-05 -4.51E-04 4.00E-04 6.56E-05 4.85E-04

0.8 25000 -4.31E-03 0 -1.21E+05 -5.70E-05 -3.70E-04 1.21E-03 -3.85E-04 4.69E-04

0.9 25000 -3.10E-03 0 -1.83E+05 -8.66E-05 -2.21E-04 1.77E-03 -7.55E-04 4.12E-04

1.0 25000 -1.33E-03 0 -2.10E+05 -9.94E-05 -3.32E-05 1.99E-03 -9.77E-04 3.25E-04

1.1 25000 6.65E-04 0 -1.97E+05 -9.31E-05 1.58E-04 1.83E-03 -1.01E-03 2.26E-04

1.2 25000 2.50E-03 0 -1.47E+05 -6.94E-05 3.16E-04 1.33E-03 -8.52E-04 1.33E-04

1.3 25000 3.83E-03 0 -6.98E+04 -3.29E-05 4.12E-04 5.85E-04 -5.36E-04 6.33E-05

1.4 25000 4.41E-03 0 1.93E+04 9.09E-06 4.29E-04 -2.58E-04 -1.23E-04 3.04E-05

1.5 25000 4.16E-03 0 1.03E+05 4.87E-05 3.64E-04 -1.04E-03 3.05E-04 3.95E-05

1.6 25000 3.12E-03 0 1.66E+05 7.84E-05 2.31E-04 -1.61E-03 6.69E-04 8.82E-05

1.7 25000 1.51E-03 0 1.97E+05 9.29E-05 5.73E-05 -1.87E-03 9.00E-04 1.67E-04

1.8 25000 -3.61E-04 0 1.90E+05 8.95E-05 -1.24E-04 -1.77E-03 9.57E-04 2.59E-04

1.9 25000 -2.13E-03 0 1.47E+05 6.93E-05 -2.80E-04 -1.34E-03 8.33E-04 3.49E-04

2.0 25000 -3.46E-03 0 7.70E+04 3.63E-05 -3.79E-04 -6.59E-04 5.53E-04 4.18E-04

The displacement histories of the structure obtained using (1) Central difference method; and (2)

Newmark’s method are shown as follows:

As indicated above, the structure undergoes forced vibration up to 5 sec, followed by free

vibration until all energy has been dissipated due to damping.

-0.0002

-0.0001

0

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

0.0007

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0

Dis

pla

cem

en

t (m

)

Time (s)

Central Difference Method

Newmark's Method

Forced vibration Free vibration

Solution to Question 5:

1) Natural period of building, Tn = 1.6 sec

From the response spectra,

Maximum displacement m 2.0D

Pseudo velocity m/s 8.0V

Pseudo acceleration g 3.0A

2) Total base shear

kN 0.3341081.93.0113480 3 mAVb

Overturning moment

kNm 10023334 bb hVM