Question 4.2: The uniform slender bar shown in Fig. 4.4, which has length ...
The uniform slender bar shown in Fig. 4.4, which has length l and mass m, is pinned at point O and connected to a linear spring which has a stiffness coefficient k. If the bar is subjected to the moment M = M_{0} sin ω_{f}t , obtain the steady state response of this system.
Assuming small oscillations, one can show that the linear differential equation of motion is
I_{O} \ddot{θ} + (kl + \frac{1}{2}mg) lθ = M_{0} sin ω_{f}t
where I_{O} is the mass moment of inertia about point O and g is the gravitational constant. For a uniform slender bar, the mass moment of inertia is I_{O} = ml²/3. The equation of motion can be written in a form similar to Eq. 4.2 as
m\ddot{x}+kx=F(t) (4.2)
m_{e}\ddot{θ} + k_{e}θ = F_{e} sin ω_{f}t
where
m_{e} = I_{O} = \frac{ml²}{3} , k_{e} = (kl + \frac{1}{2} mg) l, F_{e} = M_{0}
The circular frequency ω is
ω = \sqrt{\frac{k_{e}}{m_{e}}} = \sqrt{\frac{(kl + \frac{1}{2}mg) l}{I_{O}}}
The steady state response is θ = Θ_{0}β sin ω_{f}t , where
Θ_{0} = \frac{F_{e}}{k_{e}} = \frac{M_{0}}{(kl + \frac{1}{2}mg) l}, β= \frac{1}{1 − r²}
where the frequency ratio r is defined as r = ω_{f}/ω.