Question

A mass of 50kg is hung from a spring of stiffness ?=?.?×10⁵ (?/? ) and damper ?=100 ?.?/? , which is attached to two aluminum beams with ?=71×10⁹ (Pa), moment of inertia ?=?.?×10^−? (?^?), and length of 255 (mm ). The beam is supported at its free end. Determine:
(a) Equivalent stiffness and equation of motion of the system.
(b) Damped natural frequency of the system (?_?) in (Hz).
(c) Free vibration response of the system in time domain,?(?), when ?(? )=10mm and ?̇(? )=100 mm/s.
(d) Simulation and plot of the response of system in MatLab/Simulink

255 mm Aluminum I= 3,5 x 10-8m+ 32.5 x 10 N/m 50 kg 100 N. s/m -

Answer 1

a)

k_A = stiffness of one aluminium beam in shown orientation = 3*E*I/L^3

= 3*71*3.5*10/0.255^3

= 449600.8323N/m

(here L= length of one aluminium beam)

The equivalent system can be shown as:

here,

k₁ = spring stiffness = 2.5*10^5 N/m and

C = damping coefficient = 100N.s/m

The two springs with stiffness k_A are in parallel and can be combined to get one spring (net stiffness = k_A + k_A) as shown:

Now, the springs with stiffness 2kA and k1 are in series with each other and can be combined to give one spring (of net stiffness
2kA *k1 ki + 2kA

2 * 449600.8323* 250000 = 195614.4192N/m = k) 250000 + 2 * 449600.8323
as shown below:

Therefore, equivalent stiffness of the system is, k = 195614.4192N/m ____________ eq.1

Consider the free body diagram as shown below:

к их та Сх

equation of motion for the system is:

n* 7 +*i k *I=0

__________________ Answer to part a

b)

Ccr = critical damping coefficient of the system = 2*(k*n)^(0.5)

= 2*(195614.4192*50)^(0.5)

= 6254.8288N.s/m

therefore, $\xi$ = damping ratio = C/Ccr = 100/6254.8288 = 0.01599

since $\xi$ < 1 => the system is under damped.

w_n = natural frequency of the system = (k/n)^(0.5) = 62.5483rad/s,

therefore, w_d = damped natural frequency of the system = w_n*(1-$\xi^2$)^0.5 = 62.54030331rad/s

____________________________ Answer to part b

c)

free vibration response of such a system is given by:   

A* e-*W* * sin(wd*t + o)

where, A and $\phi$ are constants that can be found using boundary conditions.

applying boundary conditions:

i) at t=0s, x = 10mm,

=> A*sin($\phi$) = 10 ____________________ eq. 2

ii) at t=0s, $\dot{x}$ = 100mm/s and using A*sin($\phi$) = 10 we get:

=> A*cos($\phi$)*w_d - 10*$\xi$*wn = 100

______________ eq.3

or A * cos(0)=- 100 + 10 * 0.01599 * 62.5483 62.54030331 = 1.75889

performing (eq.2)^2 + (eq.3)^2 and using (sin($\phi$))^2 + (cos($\phi$))^2 = 1, we get:

or A = (1.75889) + 100

____________ eq. 4

=> A= 10.15350649mm

putting this in eq.2, we get:

$\phi$ = 1.396688rad_________ eq.5

therefore, system response is:

_________eq. 6

(t) = 10.15350649 * e-1.0001473*+ * sin(62.54030331 *t + 1.396688)

___________________ Answer to part c

d)

In MATLAB, the code is written as follows:

ou WN IIIIII t = 0:0.01:7; $7s analysis time x = 10.15350649*exp(-1.0001473*t).*sin (62.54030331*t + 1.396688); $displacement of the mass plot(t, x); xlabel('time (3)'); ylabel('displacement response'); title('displacement response curve of the system');

Graph plotted for first 7s is as shown below:

displacement response curve of the system displacement response Wwwwwwwwwwww 1 2 3 5 67 4 time(s)

It can be seen that, the system nearly reaches equilibrium at t = 7s.

_________________ Answer to part d