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HW3soln - CE 573 Structural Dynamics Solutions to HW#3...

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CE 573: Structural Dynamics Solutions to HW#3 Problem #1 R y (x) x kc g m v = constant Frame Center of Mass h y(t) Y(t) Wheel y o Given : A car can be (crudely) modeled as a single degree of freedom system having the entire car’s mass m lumped at the center of mass and connected to the frame (modeled as a rigid bar) by a linear spring (stiffness = k , unstretched length = l o ) and a linear damper (damping constant = c ). The frame, in turn, is a fixed height h above the wheel, which rides over a rough road whose vertical height above the fixed x - axis is given by a known function y R (x) . The car starts at the location x=0 at the initial time t=0 (assume that y R (x=0) = 0) and it moves to the right with a constant horizontal speed v . Let Y(t) be the height of the center of mass above the fixed x -axis, let y o be the static equilibrium height of the center of mass at time t=0 , and let y(t) be the height of the center of mass above this static equilibrium position. Gravity acts downward as shown. Required : (a) Determine the equation of motion for the car’s center of mass in terms of y(t) and the system and road parameters given above. Hint : you may find it helpful to express the equation of motion first in terms of Y(t) , then determine the relationship that exists between y o , l o , and h in order to simplify the result and apply it to y(t) . (b) Assume the following values for the parameters: mass 1430 m = kg, stiffness kN/m, and damping factor 140 k = 0.40 ξ = . Also, assume that the car travels over an idealized harmonic surface ( ) () s i n2 / R yx Y xL π = , with amplitude 80 Y = mm and wavelength 30 L = meters. Finally, assume the constant speed of the car is m/s. Determine the car’s steady-state vibration amplitude and the steady-state force it transmits to the road. 30 o v = (c) Suppose that the speed of the car is increased to a value above 30 m/s. What effect this would have on the answers to part (b)? Explain qualitatively why increasing the speed has this effect. Solution : (a) Start with a free body diagram of the center of mass –
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m F d Y(t) F s mg Notice that the displacement is not measured from static equilibrium, so the effect of the car’s weight must be included. Newton’s second law then tells us: () Yt , or 0. sd mF F m g m Y m Y F F m g =⇒ −−= + ++= Fa We need to evaluate s F and d F , which depend upon the deformation of the center of mass relative to the frame. Starting with the stiffness force, we note that the force is proportional to spring current o ll ∆= − l x , where ( ) . current R lY t h y = −− Since the force must be a function of time, we need to replace with an appropriate time- dependent function. To do this, we note that the constant horizontal velocity of the car implies that R yx o x tv tx =+ , with since the car is at the origin when 0 o x = 0 t = . Thus, we get { } .
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HW3soln - CE 573 Structural Dynamics Solutions to HW#3...

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