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 = M0 sin ωft , obtain the steady state response of this system.

Question

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 = [latex]M_{0}[/latex] sin [latex]ω_{f}[/latex]t , obtain the steady state response of this system.

Accepted Answer

Assuming small oscillations, one can show that the linear differential equation of motion is
[latex]I_{O} \ddot{θ}[/latex] + (kl + [latex]\frac{1}{2}[/latex]mg) lθ = [latex]M_{0}[/latex] sin [latex]ω_{f}[/latex]t
where [latex]I_{O}[/latex] 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 [latex]I_{O}[/latex] = ml²/3. The equation of motion can be written in a form similar to Eq. 4.2 as

[latex]m\ddot{x}+kx=F(t)[/latex] (4.2)

[latex]m_{e}\ddot{θ}[/latex] + [latex]k_{e}[/latex]θ = [latex]F_{e}[/latex] sin [latex]ω_{f}[/latex]t
where

[latex]m_{e} = I_{O} = \frac{ml²}{3}[/latex] , [latex]k_{e}[/latex] = (kl + [latex]\frac{1}{2}[/latex] mg) l, [latex]F_{e} = M_{0}[/latex]
The circular frequency ω is

ω = [latex]\sqrt{\frac{k_{e}}{m_{e}}}[/latex] = [latex]\sqrt{\frac{(kl + \frac{1}{2}mg) l}{I_{O}}}[/latex]
The steady state response is θ = [latex]Θ_{0}[/latex]β sin [latex]ω_{f}[/latex]t , where

[latex]Θ_{0} = \frac{F_{e}}{k_{e}} = \frac{M_{0}}{(kl + \frac{1}{2}mg) l}[/latex], β= [latex]\frac{1}{1 − r²}[/latex]
where the frequency ratio r is defined as r = [latex]ω_{f}[/latex]/ω.

Question 4.2: The uniform slender bar shown in Fig. 4.4, which has length ...

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