Question

Example 24 A beam of length L and mass m, is pivoted at one end and suspended from a spring of stiffness k distance L from the pivot. A dashpot is also positioned a distance R from the pivot. Show how the equation of motion below can be derived. de 3cR2 de dt 3kLi 0 = 0 mL2 mL dt where x is the displacement at the free end, and c is the damping coefficient (Ns/m)

Answer 1

A LI R 'L wwtt

drawing free body diagram by considering small deflection $\theta$ at free end,

kerLI LI R

now using D'Alemberts principle,

Inertial torque + resisting Torque = 0

Iek L1 c R=

As x = $\theta$ * L₁ and x^. = R*$\theta$^.

I0 kL1 0* L1+cR* 0* R 0

I = moment of inertia about pivot

d I = Iem+M

I_cm = moment of inertia about center of gravity

d = distance of pivot from center of gravity

ML2 I cm 12

$d = \frac{L}{12}$

Hence on solving we get

ML I 3

hence

M* L2 k*L0+c* R2 * e = 0

on solving we get

3k Li M2* MI2 0 3c R2

Hence Proved.