Determine the response of the movable support for the system

Determine the response of the movable support for the system shown above if the
support is displaced 1and released from rest. The beam and moveable support have
the following properties; Mass (support) = 200lb, L = 10ft, Cross section of beam = 2\" x 4\"

Solution

solution:

1)for given rectangular beam with mass m attached to its free end(movable support restrict movement in x direction only) and its left end is fixed in wall,so it is type of cantilever beam with mass m at its free end.

2)here equation of motion for beam for small dispalcement x at its free end and K is stiffness if beam so we can write by equillibrium method as

inertia force(upward)+external force (downward)=0

mx\'\'+Kx=0

so we get euqtion similar to

x\'\'+(K/m)x=0

x\'\'+wn^2*x=0

so natural frequency of beam and movable support is given by

wn=(K/m)^.5

2)when initial beam is loaded with mass m of movable support then there is displacement of beam at its free end as dx

so we can write as

m*g=K*dx

K/m=g/dx

3)putting value in above equation we get that natural frequency or response of movable support as

wn=(K/m)^.5

wn=(g/dx)^.5

and equation of motion as

x\'\'+(K/m)x=0

x\'\'+(g/dx)x=0

4)for cantilever beam deflection under load of mass m and at length L is given as

dx=WL^3/3EI

but we are given it as

dx=1\'\' or 25.6 mm

5)putting value in above equation of motion we get that response as

x\'\'+(g/dx)x=0

x\'\'+(9.81/(.0256))x=0

x\'\'+383.203x=0

and natural frequency is given by

wn=383.203 rad/s