Geotechnical Engineering and the Structural Engineering

Engineering Mathematics 2
Civil Engineering Challenge

Contents
Abstract:................................................................................................................................................3
Research:...............................................................................................................................................3
Modelling:.............................................................................................................................................3
Testing and Verification:........................................................................................................................5
Discussion of Results :...........................................................................................................................9
Additional Work :...................................................................................................................................9
Conclusion :...........................................................................................................................................9
References :.........................................................................................................................................10
MATLAB Code :....................................................................................................................................11

Abstract:
The domain of earthquake engineering includes the concepts of Geotechnical engineering and
the structural engineering. It helps to protect the buildings and structures from natural
calamities like earthquake. The study of seismic waves is necessary for various structures to
design methods to save them. Their sub – parts need to be designed in such a way that they
can be protected from the seismic wave effect [1].
Research:
The second order equation of motion is considered with a natural frequency ( wn ) and a
damping ratio ‘ξ’. The excitation given to the system is Üg ( t ). The steps followed for sDOF
are as follows : The online tables are used to determine the total stiffness of the structure
( k ) , the dead ( self weight ) and live load on the structure. On the basis of DL and LL, the
total mass (m) is determined [2]. The natural frequency ( wn ) is found using the value of the
natural period ( Tn ). Initially, the excitation is defined ( Üg ( t ) ). The Fourier transform Üg
( w ) is found. The transfer function of the system is determined and the range of frequency is
also found [3]. H ( w ) is defined in MATLAB and the response is found in frequency domain
U(w). It is converted to u ( t ). Finally, the base shear is found for the structure along with the
peak absolute values.
Modelling:
Here, i = 11 and n = 2
Single Degree freedom :
Self weight concrete = 2400 kN / m3 x volume = DL
Self weight steel = 7850 kN / m3 x volume = DL
G = dead load
Q = Live load ( from online table )
Steel frame includes beam and column. Another part is concrete slab.
G, concrete slab :
L = ( 2 - 1 ) x 4 = 4

L x t x w = 4 x 0.25 x 4 = 4 m3 = volume
Volume x self weight = Dead load = 9600 kN
G, steel beam :
Area of beam = 0.5381 m2
Volume x self weight = Dead load = 2.1524 x 4 x self weight = 16896 kN
G, column
Area of column = 0.2 x 0.2 = 0.04
Volume = 0.04 x 3 = 0.12 m3
Volume x self weight = Dead load = 942 kg
For 2 columns, DL = 2 x 942 = 1884 kN
G = 9600 + 16896 + 1884 = 28380
Q = 2 kN / m2
MT = ( G + 0.3 Q ) / 9 = ( 28380 + 0.3 ( 2000 ) )/ 9.81 = 2954.128
KT = 12 E I / H = 12 x 2.15 x 108 x 5.696 x 10-6 / 33 = 4544.28
E = Young’s Modulus for steel = 215 x 1011
Ixx = Inertia for HEB200 = 56.96 m4
H = 3
Tn = 2 ∏ √ Mtotal / Ktotal = 2 ∏ √ 29.54 / 544.28 = 1.465
. wn = 2 ∏ / Tn = 4.3
. ξ = c / 2mwn = 0.05 / 2 x 29.54 x 4.3 = 0.0002
Equation :
Üg ( t ) = 0.05 g x i x sin ( wgt ) = 0.05 g x i x sin (2 ∏ t / Tn )

Here, i = 11, Tn = 1.46, t = 5s, g = 9.81 m/s2
Üg ( t ) = 0.05 x 9.81 x 11 x sin ( wgt ) = 0.05 x 9.81 x 11 x sin (2 ∏ t / 0.8x Tn ) = 5.3955
sin (2 ∏ t / 0.8 x 1.46 ) = 5.3955 x sin ( 5.38 t )
Testing and Verification:
Here, the value of ‘t’ is varied from 0 to 5 with an interval of 0.01. The data is stored in the
Excel file ‘abc.xlsx’. It consists of 2 columns, 1 for time ‘t’ and the other for the value of Üg
( t ). The number of rows is 501. This data is imported to Matlab and plotted with respect to
time ‘t’. The plot is obtained as shown in Figure 1. The Fourier transform is taken for the data
of second column which gives the plot as shown in Figure 2.
Figure 1

-400 -200 0 200 400 600 800 1000
-800
-600
-400
-200
0
200
400
600
800
Figure 2
The data given in the file ‘C2N.txt’ is plotted as shown in Figure 3. The Fourier transform
obtained is shown in the Figure 4.
0 2000 4000 6000 8000 10000 12000
-25
-20
-15
-10
-5
0
5
10
15
20
25
Figure 3

-1500 -1000 -500 0 500 1000 1500 2000
-1500
-1000
-500
0
500
1000
1500
Figure 4
If the Fourier Transform is found mathematically, then
Üg(t) = 5.3955 x sin ( 5.38 t )
The above equation represents the time domain signal.
Taking its Fourier transform,
Üg(w) = 5.38 x 5.3955 / ( 5.38 )2 – w2
Üg(w) = 29 / 29 - w2
The value of ‘w’ is varied from 0 to 10 with an interval of 0.01. The plot of the amplitude of
Fourier Transform with respect to ‘w’ is shown in the Figure 5.

0 1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
500
600
Figure 5
The Figure 6 shows the plot of U(w) with respect to frequency ‘w’.
0 1 2 3 4 5 6 7 8 9 10
0
5
10
15
20
25
Figure 6
Discussion of Results :
Here, a part of structural dynamics is studied to observe a structure’s dynamic behaviour. The major
part is the single-degree-of-freedom system ( SDoF ) [4]. The mathematical model can be
designed for the system to study its dynamic behaviour to a seismic wave [5]. The main
parameters involved are the mass of the system, the stiffness of the system and the damping
coefficient for the system [6]. The various parts of the structure show some amount of
restoring force [7]. It is referred to as stiffness and shown by a spring part using Hooke’s law
[8]. The damping coefficient ( viscous damping , force directly proportional to velocity )
takes into account any energy loss which occurs due to the excitation [9].

Additional Work :
The second order equation is given by :
. d2 u ( t ) / dt2 + 2 ξ wn d u ( t ) / dt + wn2 u ( t ) = - Üg ( t )
On taking the Fourier Transform of the 2nd order equation :
S2 U ( s ) + 2 ξ wn s U ( s ) + wn2 U ( s ) = - Üg ( s )
- ( w2 – 2 j ξ wnw – wn2 ) U ( w ) = - Üg ( w )
U (w) = Üg ( w ) / ( w2 – 2 j ξ wnw – wn2 )
Üg(w) = 29 / 29 - w2
U (w) = 29 / ( 29 - w2) ( w2 – 2 j ξ wnw – wn2 )
Conclusion :
Hence, the second order equation of motion has been studied with a natural frequency ( wn )
and a damping ratio ‘ξ’. The excitation given to the system was Üg ( t ). The various steps
were used for sDOF. The online tables were used to determine the total stiffness of the
structure ( k ) , the dead ( self weight ) and live load on the structure. On the basis of DL and
LL, the total mass (m) was determined. The natural frequency ( wn ) was found using the
value of the natural period ( Tn ). Initially, the excitation was defined as ( Üg ( t ) ). The
Fourier transform Üg ( w ) was found. The transfer function of the system was determined
and the range of frequency was also found. H ( w ) was defined in MATLAB and the
response was found in frequency domain U(w). It was converted to u ( t ). Finally, the base
shear was found for the structure along with the peak absolute values.

References :
[1] Li, H.N., Qu, C., Huo, L. and Nagarajaiah, S., 2016. Equivalent bilinear elastic single
degree of freedom system of multi-degree of freedom structure with negative
stiffness. Journal of Sound and Vibration, 365, pp.1-14.
[2] Cormie, D. and Arkinstall, M., 2012. SDOF Isn't Dead—The Role of Single Degree of
Freedom Analysis in the Design of Columns against Close-In Blast. In Structures Congress
2012 (pp. 114-125).
[3] Figuli, L. and Papán, D., 2014. Single degree of freedom analysis of steel beams under
blast loading. In Applied mechanics and materials (Vol. 617, pp. 92-95). Trans Tech
Publications Ltd.
[4] Gavin, H.P., 2014. Vibrations of single degree of freedom systems. CEE Structural
Dynamics.
[5] Ikago, K., Saito, K. and Inoue, N., 2012. Seismic control of single‐degree‐of‐freedom
structure using tuned viscous mass damper. Earthquake Engineering & Structural
Dynamics, 41(3), pp.453-474.
[6] Dragos, J. and Wu, C., 2015. Single-degree-of-freedom approach to incorporate axial load
effects on pressure impulse curves for steel columns. Journal of Engineering
Mechanics, 141(1), p.04014098.
[7] Lee, K. and Shin, J., 2016. Equivalent single-degree-of-freedom analysis for blast-
resistant design. International Journal of Steel Structures, 16(4), pp.1263-1271.
[8] Gao, C., Zhang, W., Liu, Y., Ye, Z. and Jiang, Y., 2015. Numerical study on the
correlation of transonic single-degree-of-freedom flutter and buffet. Science China Physics,
Mechanics & Astronomy, 58(8), p.84701.

[9] Hernández-Montes, E., Aschheim, M.A. and Gil-Martín, L.M., 2015. Energy components
in nonlinear dynamic response of SDOF systems. Nonlinear Dynamics, 82(1-2), pp.933-945.
[10] Wang, H.Y., Shan, X.B. and Xie, T., 2012. An energy harvester combining a
piezoelectric cantilever and a single degree of freedom elastic system. Journal of Zhejiang
University SCIENCE A, 13(7), pp.526-537.
MATLAB Code :
a=xlsread('abc.xlsx')
plot(a(:,1),a(:,2))
a1=fft(a(:,2))
figure(2)
plot(a1)
b=load('C2N.txt')
figure (1)
plot(b)
b1=fft(b)
figure(2)
plot(b1)
for w = 0:0.01:10
udw = 29/(29-w*w);
plot(w,abs(udw),'o');
hold on
end
for w = 0:0.01:10
udw = 29/(29-w*w);
udw1 = udw/sqrt((w*w-4.3*4.3)*(w*w-4.3*4.3)+(2*0.5*4.3*w)*(2*0.5*4.3*w));
plot(w,abs(udw1),'o');
hold on

end