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?

moment of inertia = 4.662 * 10^6

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 = 58.434 rad/secs

also Wm =  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t =

insert the value of t into equation 1

W =   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) =

I =  = 4.662 * 10^6

b. check your rearview mirror

explanation: