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

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

Step-by-step 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{(g)/(t) }` ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t =
`(wl^3)/(48EI)`

insert the value of t into equation 1

W
`m^2` =
`(g*48*E*I)/(WL^3)` make I the subject of the equation

I ( Moment of inertia about the neutral axis ) =
`(WL^3* Wn^2)/(48*g*E)`

I =
`(3924*3.2^3*58.434^2)/(48*9.81*200*10^9)` = 4.662 * 10^6
`mm^4`