The homogeneous slender bar of mass m and length L in Fig. (a) is supported by a pin at O. The bar is also connected to an ideal spring and viscous damper at points A and B, respectively. The bar is initially in equilibrium in the position shown with the spring undeformed. (1) Derive the

Question

The homogeneous slender bar of mass m and length L in Fig. (a) is supported by a pin at O. The bar is also connected to an ideal spring and viscous damper at points A and B, respectively. The bar is initially in equilibrium in the position shown with the spring undeformed. (1) Derive the differential equation of motion for small angular displacements of the bar. (2) Determine whether the bar is overdamped or underdamped, given that m = 12 kg, L = 800 mm, a = 400 mm, k = 80 N/m, and c = 20 N · s/m.

Accepted Answer

Part 1

Figure (b) shows the free-body diagram (FBD) of the bar when it is displaced a small counterclockwise angle θ from the vertical. The horizontal displacements [latex]x_G[/latex] (G is the mass center), [latex]x_A[/latex], and [latex]x_B[/latex] shown in the figure were obtained from the small angle approximation sin θ ≈ θ. The forces acting on the bar are its weight mg, the spring force k(aθ ) (recall that the spring is undeformed when [latex]x_A[/latex] = 0), and the damping force c(L[latex]\dot θ[/latex]) due to the dashpot.
We derive the differential equation of motion by summing moments about point O, thereby eliminating the pin reaction. Because O is a fixed point, a valid equation of motion is [latex]∑M_O = I_O\ddot θ[/latex]. From the FBD in Fig. (b) and the approximation cos θ ≈ 1, we obtain

[latex]∑M_O = I_O\ddot θ \overset{\curvearrowleft}{+} − mg\frac{L}{2}θ − (kaθ)a − (cL\dot θ) L = I_O\ddot θ[/latex]

which, on rearranging the terms, becomes

[latex]I_O\ddot θ + cL\dot θ +(ka^2 + \frac{mgL}{2})θ = 0[/latex] (a)

Part 2

Comparing Eq. (a) with Eq. (20.33), we conclude that

[latex]M\ddot q + C\dot q + K q = F_0 sin ωt[/latex] (20.33)

[latex]M = I_O[/latex] [latex]C = cL^2[/latex] [latex]K = ka^2 + \frac{mgL}{2}[/latex]

Therefore, the undamped circular frequency is [see Eq. (20.34d)]

[latex]\left.\begin{aligned}Undamped circular frequency p = \sqrt{\frac{K}{M}}\Critical damping constant C_{cr} = 2Mp\Damping factor ζ = \frac{C}{C_{cr}}=\frac{C}{2Mp}\end{aligned} ight\} [/latex] (20.34d)

[latex]p = \sqrt{\frac{K}{M}} = \sqrt{\frac{ka^2 + (mgL/2)}{I_O}}[/latex]

With [latex]I_O[/latex] = mL²/3 = 12(0.8)²/3 = 2.560 kg · m², this becomes

[latex]p = \sqrt{\frac{80(0.4)^2 + [12(9.81)(0.8)/2]}{2.560}}[/latex] = 4.837 rad/s

From Eq. (20.34d), we also obtain [latex]C_{cr} = 2Mp[/latex], which in our case becomes [latex]c_{cr}L^2 = 2I_O p[/latex], yielding

[latex]C_{cr} = \frac{2I_O p}{L^2} = \frac{2(2.560)(4.837)}{(0.8)^2}[/latex] = 38.70 N · s/m

Because the given damping constant c = 20 N · s/m is less than [latex]C_{cr}[/latex], we conclude that the bar is underdamped.

Question 20.7: The homogeneous slender bar of mass m and length L in Fig. (...

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