Question

A precision miling machince wighing 1000lb is supported on a rubber mount. The force deflection relationship of the rubber mount I'd given by F=3500x+55x^3. Where the force F and the deflection x are measured in pounds and inches, respectively. Determine the equivalent linerized spring constant of he rubber mount at its static equilibrium position.

Answer 1

Answer:

0.2846 in

Explanation:

The static equilibrium position of the rubber mount ( $x^*$), under the weight of the milling machine,  can be determined from:

$1000=3500(x^*)+ 55(x^*)^3\\\\55(x^*)^3+ 3500x^*-1000=0\\\\Solving\ for\ the\ roots\ using \ a \ calculator\ or\ matlab\ gives:\\ \\x^*_1=0.28535\\x^*_2=-0.14267+7.89107i\\x^*_3=-0.14267-7.89107i\\\\We\ are\ using\ the \ real\ root\ which\ is\ x^*_1=0.28535$

$k=\frac{dF}{dx}|_{x^*}\\ \\k=\frac{d}{dx}(3500x+55x^3)|_{x^*}\\\\k=3500+165x^2|_{x^*}\\\\k=3500+165(x^*)^2\\\\k=3500+165(0.28535)^2=3513.435\ lb/in$

The static equilibrium position is at:

$x=\frac{F}{k}=\frac{1000}{3513.435} =0.2846\ in$