Project Truss Bridge Frequency Analysis

V V V

d

It’s straightforward to see that each sine has an offset of T = one another,

V

⇒

therefore the excitation happens every T seconds period T . If we take the

k2π 2kπV

the F F T of such force the armonics are = , which is like considering

T d

d meters of the beam.

From this we obtain the same results previously reported.

18

For completeness, it is shown for the first frequency what happens for V =

m

m

{51.22, which are resonance velocities and for V = 34 which is not

25.6} s s

(notice that damping was not considered, and that images are not scaled, so

check the amplitude):

(a) Resonance velocity V = 51m/s (b) Resonance velocity V = 25.6m/s

(a) Not resonant velocity V = 34m/s

19

0.2.6 Structural change

In this part of the project we have to either make a structural change of the

bridge in order to increase the first natural frequency by 20% or to make a

structural change on the bridge constraints to reduce by 15% the maximum

amplitude of vibration evaluated at point A when the bridge is subjected to the

seismic excitation described at point 4.

Structural change of the bridge

The maximum increase in mass admitted is 3%. Therefore, since the starting

mass is 10990.4209Kg the 3 % increase is 11320Kg. To do so it’s important

to observe that increasing the height of the bridge leads to an increase of the

first natural frequency of the system. This is due to the fact that the beams are

not parallel to the vertical axis, therefore they try to make the system oscillate

along the vertical direction, then increasing the frequency of the vibration.

Increasing the height by a factor of 1.3 leads to a change of mass that is under

the 3%, and the new mass is 11183.6377Kg. The new natural frequency is:

2.4987Hz. It’s trivial to observe though that from a certain point onwards the

horizontal oscillations prevail, leading to a reduction of the natural frequency

∞)

(just imagine increasing the height to

Figure 16: New natural frequency

20

Structural change of the constraints

To increase the natural frequency by simply changing the constraint, it’s trivial

to observe that hindering the bridge to elongate on the horizontal direction leads

to an automatic increase of the natural frequency (remember that the longer an

⇒

object, the lower the natural frequencies if the bridge doesn’t elongate, then

we increase the frequencies). This can be done by replacing the cart on O with

2

a pin.

This leads to an increase of the first frequency up to 2.5Hz, and the reduction

of the maximum amplitude of the quake is of about 35%.

In fact in point 4 the maximum amplitude was 0.24727m, whilst now it’s

0.16095m.

This is simple to understand since the quake has no frequency content in 2.5Hz.

The same reasoning can be applied to point 6.a.

(a) New natural frequency (b) y displacement of Node A after the

structural change

0.2.7 Previous version of the assignment

The first request for the project was to reduce the maximum amplitude of vibra-

tion of node A when subjected to an earthquake of about 20%, with a maximum

increase of the bridge’s mass of 5%, which ends up being 11540Kg. The dif-

ference is that it was modelled as a kart, with slider constraint and this meant

that the first natural frequency was in the range of frequencies of the earthquake

frequency content (see FRF of A in figure 18b). This led to the maximum vi-

bration of A being of A being 0.2867cm.

The first approach was to try changing the bridge’s beams.

From the natural frequencies formula there is a direct evidence that reducing J

and increasing m decreases the natural frequencies. Remember that m = ρA,

so we have a linear mass density.

To do so the bottom and top beams were replaced with a beam type IPE

220, which has larger section and lower inertia.

This leads to a bridge of mass 11657Kg but with a natural frequency of 1.9817Hz.This,

however, was not enough, since it reduced by 4cm the maximum amplitude

(about 15%). 21

To improve the bridge the diagonal beams were modified. Some beams were

replaced with the HEB 120, which have lower area and inertia (the inertia in

particular is much lower, about 1 order of magnitude, while the area changed by

10%). This led to a bridge with acceptable mass, but instead of decreasing, the

natural frequency rose. This is due to diagonal beams vibrating along the y axis.

Higher frequencies are thus preferred. Some HEB 140 beams were then added,

these have larger mass and about the same inertia as HEB 120. The bridge

mass then was 11519Kg and the first natural frequency was 1.968Hz. Again,

−

the maximum reduction of amplitude was of about 4.5 5cm, which again is

reasonable since it’s the same first frequency of the hinge,kart constraint.

(a) First natural frequency with (b) FFT node A when subjected to an

hinge,slider constraints earthquake

(a) First natural frequency after first (b) First natural frequency after all the

change changes

22

0.3 Appendix

In the following sections I’ll consider the analysis of the following function:

2

kω

0 ∈ ∈ ≤ ≤ ≥

s k 0 ξ 1, ω = 2πf 0

g(s) = C, R, 0 0

2

2

s + 2ξω s + ω

0 0

By setting s = j2πf we obtain: k

T (f ) = g(j2πf ) = 2

f 2ξf

−

1 + j

2 f

f 0

0

The modulus of T is given by: |k|

|T (f )| = q 2

f 2ξf 2

2

− )

(1 ) + (

2 f

f 0

0

. 23

0.3.1 Half-Power Rule 12

This rule determines when the resonance peak’s power drops by . For this

reason we consider a resonance peak with damping coefficient ξ equal to 0 and

k = 1.

Therefore the problem is to find the values of f such that:

√ 2 1

√

|T (f )| = =

2 2

The following steps are trivial: 2 2

⇒

2|T (f )| = 1 2T (f ) = 1

We obtain: 2 2

4

2

f f

f

− −

2= 1 2

=1+

2 4 2

f f f

0 0 0

2 4

By setting z = f and by multiplying for f :

0

2 2 4

− −

z 2zf f = 0

0 0

Then: √

p

2 4 4

± 4f + 4f

2f 0 0 0 2 ±

= f (1 2)

z =

1,2 0

2 √

q 2

⇒ ± ±

f 2)

f = (1

0

24

0.3.2 Derivative Rule

This method determines when the slope of the resonance peak is nearly the

value given by the user, which we will call m. For simplicity we’ll assume ξ = 0.

d |T (f )| = m

df |k|

d = m

q

df 2

f 2

−

(1 )

2

f

0

Therefore: 2

f

2f 1 1

−

− − 2 1

q 2

f

d 2 2 2 2

− − 2

(1 ) f f f 2

(1− )

d 0 0

2

df f 2

f

0

|T |k| |k| 0

(f )| = =

2 2

f f

df 2 2

− −

(1 (1

) )

2 2

f f

0 0

2 2

f f

− −

f (1 f (1

) )

2|k|

2|k| 2 2

f f

0 0

=

= = m

2 2

2 2

32

f f

f f

2 3

− |(1 −

((1 ) ) )|

0 0

2 2

f f

0 0

We can do the semplification as long as we remember that we need to con-

sider the sign, but T (f ) = T (−f ), therefore we can proceed and use this rule

thereafter: 2|k| f = m

2 2

f

f 2

− )

(1

0 2

f

0

2|k|

By letting = k̂:

2

mf

0 2

4 f

f − 2 = k̂f

1+ 4 2

f f

0 0

4 2 2 4 4

− −

f 2f f k̂f f + f = 0

0 0 0

From this point onward an approximation method is used, but matlab provides

function for finding roots of a polynomial.

Since we seek for a solution around f , a Taylor expansion in f might be

0 0

suitable: 4 2 2 4 4

− −

W (f ) = f 2f f k̂f f + f

0 0 0

5

−

W (f ) = k̂f

0 0

0 4

−

W (f ) = k̂f

0 0

00 2

W (f ) = 8f

0 0

000

W (f ) = 24f

0 0

So: 5 4 2 2 3

≈ − − − − −

W (f ) k̂f k̂f (f f ) + 4f (f f ) + 4f (f f )

0 0 0 0

0 0 0

5 4 2 2 3

− − − − −

0 = k̂f k̂f (f f ) + 4f (f f ) + 4f (f f )

0 0 0 0

0 0 0

3 2 2 3 3 2 2

− − − −

0 = k̂f f + 4f (f + f 2f f ) + 4(f f + 3f f 3f f )

0 0 0

0 0 0 0

3 2 2 3 2

− −

0 = f (− k̂f + 4f + 4f 8f f ) = f (−

k̂f + 4(f f ) )

0 0

0 0 0

25

Finally: 1 q

3 2 3

− − ⇒ ±

k̂f + 4(f f ) = 0 f = f k̂f

0 0

0 0

2

−f

Since T (−f ) = T (f ) also is solution.

26

0.3.3 BeamLength function

function [L] = beamLength(...)

1 %

2 % Parameters is a vector with parameters P,A,J,E

3 %−> parameters = [P,A,J,E];

4 % Frequency denotes the frequency range to be approximated

5 % Approximation is a scale factor that multiplied by the frequency

6 % gives the approximation, by default is 1 decade (10)

7 %

8 % The formula is 2*pi*f = (pi/L)ˆ2 sqrt(E*J/M);

*

9 % Therefore L = sqrt(pi/(2*f) sqrt(E*J/M));

*

10

11 % switch(nargin)

12 % case 2

13 % approximation =10 ;

14 % case 3 ;

15 % otherwise

16 % error('beamLength:TooManyInputs', ...

17 % 'requires at least 2 inputs');

18 % end

19

20 frequency = abs(frequency);

21 parameters = abs(parameters);

22 switch(approximationType)

23 case ApproxType.HalfPower

24 fmax = frequency*sqrt((1+sqrt(2)));

25 case ApproxType.FreqRange

26 fmax = frequency*approximationParam;

27 case ApproxType.DerivativeRule

28 a=frequency;

29 k=approximationParam;

30 r = roots([−k,0,3*aˆ2*k,0,−3*aˆ2*k,−4*aˆ2,k*aˆ6]);

31 r = round(r*10000)/10000.0;

32 fmax = r(real(r)>a & imag(r)==0);

33 otherwise

34 fmax = frequency*10;

35 end

36

37

38

39 M = parameters(1)*parameters(2);

40 sq1 = sqrt(parameters(4)*parameters(3)/M);

41 L = sqrt(pi*sq1/(2*fmax));

42

43

44 end

45 27

0.4 BuildStructure function

function [] = buildStructure(fileName, dampingCoefficients,

1 frequency, approximationType, approximationParam)

2 % Usage: buildstructure('file.txt',[a,b], ApproxType.DerivativeRule,

3 % 0.5);

4 %

5 %

6 %

7 | |

< >

if nargin 3 nargin 5

8 error('buildStructure:TooManyInputs', ...

9 'requires at least 3 inputs');

10 end

11

12 if (size(dampingCoefficients) ˜= 2)

13 error('buildStructure:WrongDampingCoeffi