Ch 3-1 HW Solutions - Answers

Ch 3-1 HW Solutions - Answers - 6 CHAPTER 3 SDOF VIBRATION...

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Unformatted text preview: 6 CHAPTER 3 SDOF VIBRATION: WITH DAMPING 4. For a mass-spring-damper system in free vibration: (a) Derive the equation of motion and solve for the transient response x(t) that is driven by the initial conditions x(0) = x0 and x(0) = v0 . (b) Assume parameter values: k = 1 lb/in, weight W = 100 lb, x0 = 0, v0 = 10 in/s. _ Vary the damping constant c so that both underdamped and overdamped responses can be demonstrated. Try values of c such that the cases = 0:1 and = 0:9 are obtained. (c) For the case above with = 0:1, vary initial velocity v0 and study the variation of the …rst intercept (zero displacement) as a function of initial velocity. Solution: (a) The equation of motion for a free mass-spring-damper system is given by mx + cx + kx = 0 • _ or x + 2 ! n x + ! 2 x = 0: • _ n We assume a solution of the form x(t) = A exp(rt) or x(t) = A cos rt + B sin rt: Taking the …rst form, di¤erentiating to …nd x and x; and then substituting into the …rst governing equation of _ • motion, we …nd A exp(rt)[mr2 + cr + k ] = 0: Since neither A nor the exponential equal zero, the polynomial in the brackets must equal zero, mr2 + cr + k = 0: This second order polynomial is the characteristic equation for the system and it can be solved for the two values of r using the quadratic equation, p c2 4mk c : r 1 ;2 = 2m The form of the solution changes depending on the value of c2 Overdamped Response: For c2 solution takes the form 4mk > 0 (equivalently + 4mk: > 1); the system is overdamped, and the p 2 x (t) = Ae p 2 1 !n t + Be 1 !n t : Upon substituting the initial conditions, the constant coe¢ cients of integration are found to be ! ! 1 vo + x o ! n vo + x o ! n 1 p xo + p 2 xo A= and B = : 2 2 2 !n 1 !n 1 Critically Damped Response: For c2 4mk = 0 (equivalently and the solution takes the form = 1); the system is critically damped, x (t) = (A + Bt) exp ( ! n t) ; 7 where the constant coe¢ cients of integration are found to be A = x0 and B = v0 + x0 ! n : Underdamped Response: For c2 and the solution takes the form 4mk < 0 (equivalently 0 < < 1), the system is underdamped, x (t) = exp ( ! n t) (A cos ! d t + B sin ! d t) ; p 2 where ! d = ! n 1 . Upon substituting the initial conditions, the constant coe¢ cients of integration are found to be v0 + x0 ! n : A = x0 and B = !d An alternate form of the solution for the underdamped system is x (t) = exp ( ! n t) [C cos (! d t )] : Upon substituting the initial conditions, we …nd that x0 = C cos and v0 = C ( ! n cos + ! d sin ): The amplitude and the phase, C and ; are found to be s 2 v0 + ! n x0 C = x2 + and = tan 0 !d 1 v0 + ! n x0 ; x0 ! d where it should be checked that the correct phase angle is chosen. (b) It is given that k = 1 lb/in, W = 100 lb, x0 = 0 in, v0 = 10 in/s. The mass is m = W=g = 100=(32:2 12) = 0:259 lb-s2 /in: p The damping coe¢ cient is given by c = 2 mk and the damped natural frequency by ! d = p 2 : For = 0:1 and 0.9, we …nd !n 1 = 0:1 = 0:9 c=2 0:259 1 = 0:102 lb s/ft c = 2 0:9 0:259 1 = 0:916 lb s/ft p ! n = 1=0:259 = 1: 97 rad/s ! n = 1:97 rad/s p p ! d = 1:97 1 0:12 = 1: 96 rad/s ! d = 1:97 1 0:92 = 0: 857 rad/s 0:1 tan x =0:1 (t) 1 p p 1 = =2 rad 0:197t tan x =0:9 (t) 1 1 = =2 rad 1:77t = 5: 10e sin 1:96t = 11:7e sin 0:857t These two curves are plotted below. 8 CHAPTER 3 SDOF VIBRATION: WITH DAMPING 5 4 3 2 x (t) (in.) 1 0 -1 -2 -3 -4 0 2 4 6 8 10 t (s) x =0:1 (t) and x =0:9 (t): While the amplitude 11:7 would lead us to expect that x =0:9 (t) would have larger amplitudes than those of x =0:1 (t); the very large value for quickly dissipates the oscillation. (c) We use the general solution and vary the initial velocity, v0 x(t) = e 0:197t sin 1:96t: 1:96 The …gure below shows the solution when the initial velocities are 1, 5, 10, 20 in/s. The amplitudes approach zero exponentially as t ! 1: It is notable that all curves intersect the x = 0 axis at the same times regardless of the value of the damping. 10 8 6 4 x (t) (in.) 2 0 -2 -4 -6 -8 0 2 4 6 8 10 t (s) x(t) for various v0 : 9 5. A mass-spring-damper system is tested to determine the value of c. Assume k = 10 lb/in and m = 2 slug. (a) If the vibrational amplitude is observed to decrease to 33% of its initial value after 2 consecutive cycles, what is the value of c? (b) If k , rather than having the exact value given above, has a range of possible values of 8 to 12 lb/in with equal likelihood, then what is the range of possible values of c? Solution: (a) Vibration amplitude decreases to 33% of its initial value after 2 cycles: Let us keep three signi…cant …gures. We can use the logarithmic decrement to evaluate c: We know = We also can relate to 1 ln n 100 x = 1 ln 2 100 33 = 0:554: as follows: = ln = = = = x1 x2 ! n Td 2 !n !d !2 pn !n 1 2 p : 2 1 2 1 2 2 Therefore, and solve for to …nd = Note that 1 slug = 1 lb-s2 /ft: In order to calculate c; we …rst need to change the sti¤ness units to lb/ft (or change the mass units to lb-s2 =in): The sti¤ness is k = 120 lb/ft. Now we can evaluate c; p p c = 2 km = 2(0:0886) 120(2) = 2:74 lb-s/ft: (b) Possible range for c if k has a range of possible values of 8 to 12 lb/in with equal likelihood: In this case we can substitute the upper and obtain the range of values for c; as follows, p 2 km p min 2(0:0 886) 8(12)(2) 2: 45 lb-s/ft lower bounds into the equation for c and thereby c c c p kmax m p 2(0:0 886) 12(12)(2) 3:01 lb-s/ft. 2 q 0:554 = p 2 2 ; =(4 2 + ) = 0:0886: 9 5. A mass-spring-damper system is tested to determine the value of c. Assume k = 10 lb/in and m = 2 slug. (a) If the vibrational amplitude is observed to decrease to 33% of its initial value after 2 consecutive cycles, what is the value of c? (b) If k , rather than having the exact value given above, has a range of possible values of 8 to 12 lb/in with equal likelihood, then what is the range of possible values of c? Solution: (a) Vibration amplitude decreases to 33% of its initial value after 2 cycles: Let us keep three signi…cant …gures. We can use the logarithmic decrement to evaluate c: We know = We also can relate to 1 ln n 100 x = 1 ln 2 100 33 = 0:554: as follows: = ln = = = = x1 x2 ! n Td 2 !n !d !2 pn !n 1 2 p : 2 1 2 1 2 2 Therefore, and solve for to …nd = Note that 1 slug = 1 lb-s2 /ft: In order to calculate c; we …rst need to change the sti¤ness units to lb/ft (or change the mass units to lb-s2 =in): The sti¤ness is k = 120 lb/ft. Now we can evaluate c; p p c = 2 km = 2(0:0886) 120(2) = 2:74 lb-s/ft: (b) Possible range for c if k has a range of possible values of 8 to 12 lb/in with equal likelihood: In this case we can substitute the upper and obtain the range of values for c; as follows, p 2 km p min 2(0:0 886) 8(12)(2) 2: 45 lb-s/ft lower bounds into the equation for c and thereby c c c p kmax m p 2(0:0 886) 12(12)(2) 3:01 lb-s/ft. 2 q 0:554 = p 2 2 ; =(4 2 + ) = 0:0886: ...
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This note was uploaded on 09/24/2010 for the course MECHANICAL 14:650:443 taught by Professor Benaroya during the Spring '10 term at Rutgers.

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