Question

3) (30 points) A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in the figure below. Use the principle of superposition and the convolution integral to determine the response of the system to each force. (Fo= 250 N, to=1 sec) (Hint: use table 5.1) F(1) 2F F 10 210

Answer 1

Solution:

Principle of superposition dictates that we can find the response of a systems which is being acted upon by a number of forces by vectorally adding their responses. Thus the force acting of the system may be assumed to be composed of two parts,

F =

1. 2F0 (0<t<to)
2. F0 (to<t<2t0)

Now we are given a SDOF undapmed system.

Respomse of such a system for a time period of a multi force system is given as:

maal e-In (3-7) sin wat - T) dt Fo mod e-w (1-7) (w, sin wat - 7) + We cos wat - 7) (Wx)2 + (wa) "}] -=0 Fo 1 = 1 .e K e-wyd cos wat _ ;2 where φ lan V1 - 42

For undapmed system,

Fo [1 k cos wat]

The response could be depicted as:

11) 2F0 k AAAAAT

Natural frequency fo the system is given as, wn = (k/m)^0.5 = (1500/30)^0.5 = 7.07 rad/s

Fo = 150, to = 1sec

Now for 0<t<t0

x(t) = (2x150/1500)(1-cos7.07(t-1)) = 0.2(1-cos7.07(t-1)) m

Now for t0<t<2t0

x(t) = (x150/1500)(1-cos7.07(t)) = 0.1(1-cos7.07(t)) m

From the principal of superposition,

x(t) = 0.1{ 2-cos7.07(t-1)-7.07cos(t) }

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