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4 Pages

### SolSec1_8

Course: MECHANICAL MAE351, Spring 2009
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Word Count: 647

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and Problems Solutions Section 1.8 (1.90 through 1.93) 1.90 Consider the system of Figure 1.90 and (a) write the equations of motion in terms of the angle, , the bar makes with the vertical. Assume linear deflections of the springs and linearize the equations of motion. Then (b) discuss the stability of the linear systems solutions in terms of the physical constants, m, k, and ! . Assume the mass of the rod acts...

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and Problems Solutions Section 1.8 (1.90 through 1.93) 1.90 Consider the system of Figure 1.90 and (a) write the equations of motion in terms of the angle, , the bar makes with the vertical. Assume linear deflections of the springs and linearize the equations of motion. Then (b) discuss the stability of the linear systems solutions in terms of the physical constants, m, k, and ! . Assume the mass of the rod acts at the center as indicated in the figure. Figure P1.90 Solution: Note that from the geometry, the springs deflect a distance kx = k(!sin!) and the cg moves a distance ! 2 cos! . Thus the total potential energy is 1 mg! U = 2 ! k(!sin")2 # cos" 2 2 and the total kinetic energy is 1! 1 m"2 ! 2 T = J O! 2 = ! 2 23 The Lagrange equation (1.64) becomes d # !T & !U d # m"2 ! & 1 % !" ( + !" = dt % 3 " ( + 2k"sin" cos" ) 2 mg"sin" = 0 dt \$ ! ' \$ ' Using the linear, small angle approximations sin! " ! and cos! " 1 yields m!2 "" # mg! & a) ! + % 2k!2 " !=0 3 2( \$ ' Since the leading coefficient is positive the sign of the coefficient of determines the stability. mg mg if 2k! ! > 0 " 4k > " the system is stable 2 ! if 4k = mg " #(t) = at + b " the system is unstable b) if 2k! ! mg mg < 0 " 4k < " the system is unstable 2 ! Note that physically this results states that the systems response is stable as long as the spring stiffness is large enough to over come the force of gravity. 1.91 Consider the inverted pendulum of Figure 1.37 as discussed in Example 1.8.1. Assume that a dashpot (of damping rate c) also acts on the pendulum parallel to the two springs. How does this affect the stability properties of the pendulum? Solution: The equation of motion is found from the following FBD: m c l F dash mg 2Fsp k 0 k ! + Moment about O: !Mo = I" ml 2! = mgl sin! " 2 kl #l #l #l sin! \$ cos! % " c\$ ! % \$ cos! % & & 2 2 2&2 2 2 + cl ! + # kl " mgl& ! = 0 % ml ! 4 \$2 ' 2 When is small, sin and cos ml! 1 + cl # kl ! + \$ " mg& ! = 0 ' 4 2 For stability, kl > mg and c > 0. 2 The result of adding a dashpot is to make the system asymptotically stable. 1.92 Replace the massless rod of the inverted pendulum of Figure 1.37 with a solid object compound pendulum of Figure 1.20(b). stable. Solution: m2 m1 2Fsp k 0 ! Moment about O: "M o = I!! Calculate the equations of vibration and discuss values of the parameter relations for which the system is m2g + ! k l kl #l #1 m1 g sin! + m2 gl sin! " 2 sin! \$ cos! % = \$ m1l 2 + m2 l 2 % ! & & 2 2 2 3 When is small, sin and cos 1. 2 ! m1 + m # l 2% + ! kl & m1 gl & m gl# % = 0 ' 2 2 "3 \$ "2 2 \$ ! m1 + ( kl ! m + m2 # l% + * & " 1 + m2 # g-% = 0 "3 \$ \$, )2 2 For stability, kl ! m1 >" + m2 # g. \$ 2 2 1.93 A simple model of a control tab for an airplane is sketched in Figure P1.93. The equation of motion for the tab about the hinge point is written in terms of the angle from the centerline to be !! ! J! + (c " f d )! + k! = 0 . Here J is the moment of inertia of the tab, k is the rotational stiffness of the hinge, ! c is the rotational damping in the hinge and f d! is the negative damping provided by the aerodynamic forces (indicated by arrows in the figure). stability of the solution in terms of the parameters c and fd . Discuss the Figure P1.93 A simple model of an airplane control tab ! Solution: The stability of the system is determined by the coefficient of ! since the inertia and stiffness terms are both positive. There are three cases Case 1 c - fd > 0 and the systems solution is of the form !(t) = e" at sin(# nt + \$) and the solution is asymptotically stable. Case 2 c - fd < 0 and the systems solution is of the form !(t) = eat sin(" nt + #) and the solution is oscillates and grows without bound, and exhibits flutter instability as illustrated in Figure 1.36. Case 3 c = fd and the systems solution is of the form !(t) = Asin(" nt + #) and the solution is stable as illustrated in Figure 1.34.
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Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions Section 1.9 (1.94 through 1.101) 1.94* Reproduce Figure 1.38 for the various time steps indicated. Solution: The code is given here in Mathcad, which can be run repeatedly with different t to see the importance of step size. Ma
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions Section 1.10 (1.102 through 1.114) 1.102 A 2-kg mass connected to a spring of stiffness 103 N/m has a dry sliding friction force (Fc) of 3 N. As the mass oscillates, its amplitude decreases 20 cm. How long does this take? Solut
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.1 (4.1 through 4.16) 4.1 Consider the system of Figure P4.1. For c1 = c2 = c3 = 0, derive the equation of motion and calculate the mass and stiffness matrices. Note that setting k3 = 0 in your solution should resu
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.2 (4.19 through 4.33) 4.19 Calculate the square root of the matrix&quot; 13 !10 % M=\$ ' # !10 8 &amp;&quot; &quot; a !b % 1/ 2 1/ 2 \$ Hint: Let M = \$ ' ; calculate M !b c &amp; # #()2% and compare to M.' &amp;Solution: Given:&quot;
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.3 (4.34 through 4.43) 4.34 Solve Problem 4.11 by modal analysis for the case where the rods have equal stiffness! (i.e., k1 = k2 ), J1 = 3J 2 , and the initial conditions are x(0) = !0 1# and x 0 = 0. &quot; \$T()
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.4 (4.44 through 4.55) 4.44 A vibration model of the drive train of a vehicle is illustrated as the three-degreeof-freedom system of Figure P4.44. Calculate the undamped free response [i.e. ! M(t) = F(t) = 0, c1 =
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.5 (4.56 through 4.66) 4.56 Consider the example of the automobile drive train system discussed in Problem 4.44. Add 10% modal damping to each coordinate, calculate and plot the system response. Solution: Let k1 =
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.6 (4.67 through 4.76) 4.67 Calculate the response of the system of Figure 4.16 discussed in Example 4.6.1 if F1(t) = (t) and the initial conditions are set to zero. This might correspond to a two-degree-of-freedom
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.7 (4.76 through 4.79) 4.76 Use Lagrange's equation to derive the equations of motion of the lathe of Fig. 4.21 for the undamped case. Solution: Let the generalized coordinates be !1 ,! 2 and ! 3 . The kinetic ener
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions for Section 4.9 (4.80 through 4.90) 4.80 Consider the mass matrix &quot;10 !1% M=\$ ' # !1 1 &amp; and calculate M-1, M-1/2, and the Cholesky factor of M. Show that LLT = MM !1/ 2 M !1/ 2 = I M 1/ 2 M 1/ 2 = MSolution: Given&quot;10 !1% M
Korea Advanced Institute of Science and Technology - MECHANICAL - MAE351
Problems and Solutions Section 4.10 (4.91 through 4.98) 4.91* Solve the system of Example 1.7.3 for the vertical suspension system of a car with m = 1361 kg, k = 2.668 x 105 N/m, and c = 3.81 x 104 kg/s subject to the initial conditions of x(0) = 0 a
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N.C. State - MA - 242
N.C. State - MA - 242
N.C. State - MA - 242
N.C. State - MA - 242