Question

A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observed to vibrate with a natural frequency of 9.3 HZ. What is the moment of inertia of the beam's cross section about its neutral axis?

Answer 1

Answer:

moment of inertia = 4.662 * 10^6 $mm^4$

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 $\pi f$ = 58.434 rad/secs

also Wm = $\sqrt{\frac{g}{t} }$  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = $\frac{wl^3}{48EI}$

insert the value of t into equation 1

W$m^2$ = $\frac{g*48*E*I}{WL^3}$   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = $\frac{WL^3* Wn^2}{48*g*E}$

I = $\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}$  = 4.662 * 10^6 $mm^4$