An object with mass = 500 kg is attached to a table with four steel legs of diam
ID: 1717476 • Letter: A
Question
An object with mass = 500 kg is attached to a table with four steel legs of diameter = 0.015 m and length = 0.1 m Derive the differential equation for vibration of the object in vertical direction. Determine the natural frequency of vibration. What is the maximum possible amplitude of vibration so that the maximum vibratory stress in each table leg is less than 70% of the yield point stress? For this maximum amplitude, plot the allowable region of initial displacement and the initial velocity of the object. Develop a MATLAB program to solve the governing differential equation. Compare the solution from your program to that obtained analytically for an allowable initial displacement and the initial velocity determined in part (c).Explanation / Answer
when weight W is moving downward and is at a disatace of x from equlibrium position
the net force acting on the weight = downward force - upward force
= W - (delta +x )
net force is equal to product of mass x acceleration
= m x d^2/dt^2
mx d^x2/dt^2 = -x
d^2/dt^2 +x/m = 0
the above equation is the equation of simple harmonic motion , but the fundamental equation of simple harmonic motion is d^2/ dt^2 + omega ^2 x = 0
where omega is angular velocity
w^2 = 1 /m
omega = sq root of 1/m
in the above equation omega is natural angular velocity and generaly denoted by omega n = sq root of 1/m
time period t 2 x pie / omega n
frequency fn = 1/t = 1/2xpie sq root of 1/m
fn= sq root of g / delta
=0.4985 /sq root of delta
delta = Wx L /Ax E
A = pie / 4 X d^2
3.14 /4 x 0.015^2= 0.00176m^2
E for steel = 200 x 10^9
delta = 500 x 9.81x 0.1 / 0.00176 x 200 x 10^4
= 4.05 Hz
= 0.015 m
natural frequency fn = 0.4985 / sq root of 0.015