Solutions Manual for Structural Dynamics: Concepts and Applications Henry R. Busby George H. Staab Chapter 1 1.1 Write the equations of motion for the one-degree-of-freedom systems shown in Figures1.72 (a) … (i). Assume that the loading is in the form of a force P(t), a given displacement a(t), or a given rotation the figure. Figure 1.72 One-degree-of-freedom systems t as indicated in Solutions (a) (b) spring force = 3EI / L3 u spring force = 48EI / L u 3 mu mu 48EI u P(t ) L3 (c) 3EI u P(t ) L3 (d) spring force = 3EI / L3 u 3EI / L2 (t ) spring force = 3EI / L3 u a mu 3EI u a 0 L3 3EI 3EI mu 3 u 3 a(t ) L L mu (e) 3EI 3EI u 2 (t ) 3 L L (f) spring force = EA / L u EA mu u P(t ) L spring force = 2 3EI / L3 u 6 EI / L3 u mu 6 EI u P(t ) L3 (g) (h) For m: u " spring force = 2 EI / L u 6 EI / L u 3 mu 6EI 6EI u 3 a (t ) 3 L L 3 P L L 3 L 6EI 2 2 muL3 3EI 5PL3 48EI 5PL3 muL3 L3 For P: u u ' u " 5P 16mu 48EI 3EI 48EI 2 For P: u ' 16mu 48EI u 5P(t ) L3 3 mu 3EI 5 u P( t ) 3 L 16 (i) spring force = 48EI / L3 u mu 48EI 48EI u 3 a(t ) 3 L L 1.2 Find the natural frequency 0 of each of the systems in Figures 1.72 (a) … (i) (write the general expression for 0 ) and calculate values using E = 30 x 106 psi, I = 80 in4, A = 10 in2, L = 100 in., and the weight of the mass 500 lbs. Solutions m 500 1.294 1.3 32.2(12) (a) o 48EI 48(30 106 )(80) 297.7 rad/sec mL3 1.3(100)3 (b) o 3EI 3(30 106 )(80) 74.4 rad/sec mL3 1.3(100)3 (c) o 3EI 74.4 rad/sec mL3 (d) o 3EI 74.4 rad/sec mL3 (e) o EA (30 106 )(10) 1519 rad/sec mL 1.3(100) (f) o 6 EI 6(30 106 )(80) 105.3 rad/sec 3 mL 1.3(100)3 (g) o 6 EI 105.3 rad/sec mL3 (h) o 3EI 74.4 rad/sec mL3 (i) o 48EI 297.7 rad/sec mL3 1.3 Find the frequency response functions U * i for the output u(t) and inputs as indicated in Figures 1.72 (a) … (i), assuming: (A) An elastic material (without damping) (B) An inelastic material with the complex modulus E* E iE For part (A), plot U * versus the frequency for 0 500 , except for case e, where 0 2000 . For part (B), plot the absolute values of the frequency functions U * versus the frequency for 0 500 . Assume the same numerical data as given in Problem 1.2, with E 0.03E and E 30 106 psi. Solutions (A) Elastic material – no damping (a) U * u* 1 1 2 3 P * m(i ) 48EI / L 115, 200 1.3 2 (b) U * u* 1 1 P * m(i ) 2 3EI / L3 7200 1.3 2 (c) U * u* 3EI / L3 1 2 3 2 P * m(i ) 3EI / L 7200 1.3 (d) U * 720,000 u* 3EI / L2 2 3 P * m(i ) 3EI / L 7200 1.3 2 (e) U * u* 1 1 2 2 P * m(i ) EA / L 3,000,000 1.3 (f) U * u* 1 1 P * m(i ) 2 6 EI / L3 14, 400 1.3 2 (g) U * u* 6 EI / L3 14, 400 P * m(i ) 2 6 EI / L3 14, 400 1.3 2 (h) U * u* 5 / 16 0.3125 2 3 2 P * m(i ) 3EI / L 72,000 1.3 (i) U * u* 48EI / L3 115,000 2 3 2 P * m(i ) 48EI / L 115,000 1.3 U* Plots for cases (a),(f),(g),(h) 0,005 0,004 0,003 0,002 0,001 0 -0,001 0 -0,002 -0,003 -0,004 -0,005 -0,006 a 100 200 300 400 500 f,g h Plots for cases (b), (c) b,c 0,015 U* 0,01 0,005 b,c 0 0 -0,005 100 200 300 400 500 Plot for case (e) e 0,00005 0 U* -0,00005 0 500 1000 1500 2000 -0,0001 e -0,00015 -0,0002 -0,00025 -0,0003 Plot for case (i) i 200 100 U* 0 0 100 200 300 400 i 500 -100 -200 -300 Plot for case (d) d 10000 8000 U* 6000 4000 d 2000 0 -2000 0 -4000 100 200 300 400 500 (B) For an inelastic material we can use the same solution if we replace E with E * E iE , where E 0.03E and E 30 106 psi. 48E ' I m 2 i 48E " I 3 L3 u* 1 L (a) U * 2 2 P * m(i ) 2 48 I ( E ' iE ") 48E ' I 48E " I 2 m 3 3 L L3 L 1 U* 2 2 48E ' I 48E " I 2 m 3 L3 L 1 115, 200 1.3 3456 2 2 2 48E " I 3456 1 L3 tan 1 tan 2 48E ' I 115, 200 1.3 m 2 L3 1 (b) U * 2 2 3E ' I 3E " I 2 m 3 L3 L 1 7200 1.3 216 2 2 2 3E " I L3 216 tan tan 1 2 3E ' I 7200 1.3 2 3 m L 1 3E ' I 3E " I i 3 3 u* L L (c) U * a * m(i ) 2 3E ' I i 3E " I L3 L3 3E ' I L3 3 E " I 3E ' I 3E " I m 2 i 3 L3 L3 L 2 2 3E ' I 3E '' I 2 3 m 3 L L i 2 3E ' I 3 E ' I 3E " I 3E " I 2 2 m 3 3 3 i 3 m L L L L U* 2 2 3E ' I 3E '' I 2 m 3 3 L L 2 2 2 3E ' I 3E ' I m 2 3E " I 3E " I m 2 3 3 L3 L3 L L U* 2 2 3E ' I 3E '' I 2 3 m 3 L L 2 2 2 2 2 7200 7200 1.3 216 216(1.3) U* 2 7200 1.3 2 2162 3E " I 3 m 2 216(1.3 2 ) 1 L tan 1 tan 2 2 2 2 2 3E ' I 3E ' I m 2 3E '' I 7200 7200 1.3 216 3 L3 L3 L 3E ' I 3E ' I 2 3 u * L L (d) U * * 2 2 3E " I 3E " I 3E " I m 2 2 3 2 m 2 L L L 2 2 3E ' I 3E '' I 2 3 m 3 L L 2 2 2 2 720,000 7200 1.3 21,600 216 21,600(1.3) U* 2 2 7200 1.3 2 216 3E " I 2 m 2 L tan 1 2 3E ' I 3E ' I m 2 3E '' I 3E '' I 2 3 L2 L3 L L 21,600(1.3 2 ) tan 1 2 2 720,000 7200 1.3 21,600 216 (e) u* 1 U* 2 2 P* E ' A E" A 2 m L L 1 3,000,000 1.3 90,000 2 2 90, 000 2 3, 000, 000 1.3 tan 1 1 (f) U * 2 2 6E ' I 6E " I 2 m 3 L3 L 1 14, 400 1.3 432 6E " I L3 432 tan tan 1 2 6E ' I 14, 400 1.3 2 3 m L 1 2 2 2 2 2 2 2 6 E ' I 6 E ' I m 2 6 E " I 6 E " I m 2 3 3 L3 L3 L L (g) U * 2 2 6E ' I 6 E '' I 2 3 m 3 L L 2 2 2 2 2 14, 400 14, 400 1.3 432 432(1.3) U* 2 14, 400 1.3 2 4322 tan 1 3E ' I L3 3E " I m 2 3 432(1.3 2 ) 1 L tan 2 2 2 2 2 3 E ' I 3 E '' I 2 14, 400 14, 400 1.3 432 3 m 3 L L 5 / 16 (h) U * 2 2 3E ' I 3E " I 2 m 3 L3 L 0.3125 7200 1.3 216 2 2 2 3E " I L3 216 1 tan tan 2 3E ' I 7200 1.3 2 3 m L 1 (i) U* 48E ' I 3 L 2 2 2 48E ' I 48 E " I 48 E " I 2 2 m m 3 3 3 L L L 2 2 48E ' I 48E '' I m 2 3 3 L L 2 2 2 2 2 115, 200 115, 200 1.3 3456 3456(1.3) U* 2 115, 200 1.3 2 3456 2 48E " I 2 m 3456(1.3 2 ) L3 tan 1 tan 1 2 2 2 2 2 48E ' I 48E ' I m 2 48E '' I 115, 200 115, 200 1.3 3456 3 L3 L3 L Plots for cases (a), (b), (e), (f), (h) 0,005 0,0045 U* 0,004 0,0035 a 0,003 b 0,0025 e 0,002 f 0,0015 h 0,001 0,0005 0 0 100 200 300 400 500 Plots for cases (c), (g, (i) 35 30 U* 25 20 c 15 g i 10 5 0 0 100 200 300 400 500 Plot for case (d) d 3500 3000 U* 2500 2000 d 1500 1000 500 0 0 100 200 300 400 500 1.4 Find the frequency response functions of the maximum bending moments M * i for systems (a), (b), (c), (d), (f), (g), (h), and (i) of problem 1.1; plot values of M * / P * or M * / a* or M * / Q* , as appropriate, versus the frequency . Find the frequency response of the axial force, N * i , in the system (e); plot N * / P * versus the frequency for 0 500 being careful to avoid the singularity. Assume EI 6.25 106 N-m2 , EA 1.25 106 N , L 2.5 m , and m 200 kg . Solutions (a) M Fd 1 48EI L 12 EI u u 2 L3 2 L2 12 EI 12 EI M * 2 u* 2 U * P * L L M * 12 EI 12 EI 1 75,000,000 2 U * 2 2 48 EI P* L L m 2 19, 200,000 200 L3 (a) 4000 3000 2000 (a) 1000 0 -1000 0 100 200 300 400 3EI 3EI 3EI uL 2 u M * 2 U * P * 3 L L L M * 3EI 1 3,000,000 3EI 2 U * 2 2 3 EI P* L L m 2 1, 200,000 200 L3 (b) M Fd 500 (b) 200 150 100 (b) 50 0 0 100 200 300 400 500 3EI 3EI 3EI 3EI u a L 2 u a M * 2 u * a * 2 U * a * a * 3 L L L L 3EI 2 3 3EI m 2 3,000,000 200 M * 3EI 3EI L 1 2 U * 1 2 3EI 2 3EI a * L2 L m 2 L 3 m 2 1, 200,000 200 3 L L (c) M Fd (c) 300000000 200000000 (c) 100000000 0 0 (d) M Fd 100 200 3EI 3EI u QL L 2 u QL 3 L L 300 400 500 3EI 3EI M * 2 u * Q * L 2 U * Q * Q * L L L 3EI 3EI mL 2 2 M * 3EI 3EI L L 2 U * L 2 3EI Q * L2 2 L L 3EI m 2 m 3 3 L L 3,000,000 2.5(3,000,000 200 2 ) 1, 200,000 200 2 (d) 100 80 60 (d) 40 20 0 0 (e) N F EA u L 100 N* 200 300 400 500 EA EA u* U * P * L L N * EA 1 500,000 EA U * EA 2 P* L L m 2 500,000 200 L (e) 14 12 10 8 6 4 2 0 (e) 0 100 200 300 400 3EI 3EI 3EI 3EI uL 2 u M * 2 u* 2 U * P * L3 L L L M * 3EI 1 3,000,000 3EI U * 2 6 EI 2 P * L2 2, 400,000 200 2 L m L3 (f) M Fd 500 (f) 200 150 100 (f) 50 0 0 100 200 300 400 500 3EI 3EI M * 2 u * a * 2 U * a * a * L L 6 EI 2 3 3EI m 2 3,000,000 200 M * 3EI 3EI L 2 U * 1* 2 1 2 2 a* L L 6 EI m 2 L 6 EI m 2 2, 400,000 200 L3 L3 (g) M Fd 3EI 3EI u a L 2 u a 3 L L (g) 400000000 300000000 200000000 (g) 100000000 0 0 200 300 400 3EI 3EI M * 2 u* 2 U * P * L L M * 3EI 3EI 5 / 16 0.3125 3,000,000 2 U * 2 2 P* L L 3EI m 2 1, 200,000 200 L3 (h) M Fd 3EI 3EI uL 2 u 3 L L 100 500 (h) 60 40 (h) 20 0 0 100 200 300 400 500 1 48EI L 12EI 12EI 12EI u a 2 u a M * 2 u * a * 2 U * a * a * 2 L3 2 L L L 48EI 2 3 12 EI 12,000,000 200 M * 12 EI 200 2 12 EI L 2 U * 1 2 1 2 2 a* L L 48EI m 2 L 48EI 200 2 19, 200,000 200 L3 L3 (i) M Fd (i) 1,5E+10 1E+10 5E+09 (i) 0 0 100 200 300 400 500 -5E+09 1.5 Using Laplace transformations, find the transfer functions for the systems given in Problem 1.1. Determine the transient responses. Assume: (A) Elastic material. (B) Kelvin model, with E = 30 x 106 psi, 60 106 psi.sec., (C) Maxwell model, with E = 30 x 106 psi, 18 108 psi.sec., (D) Standard solid model, with E0 30 106 psi., E1 60 106 psi. 60 106 psi.sec. Solutions (A) Elastic material – no damping (a) U ( s) u 1 2 . For the transient response 3 P ms 48EI / L U (t ) L U (s) L -1 -1 1/ m 2 3 s 48EI / mL 48EI sin 3 48EI mL m mL3 1 t Or U (t ) (b) U ( s) U (s) L -1 -1 1/ m 1 sin o t 2 3 s 3EI / mL mo o2 3EI mL3 U (s) L -1 -1 3EI / mL3 1 sin o t 2 3 s 3EI / mL mo o2 3EI mL3 3EI / L2 2 . For the transient response ms 3EI / L3 U (s) L -1 -1 3EI / mL2 EI 2 sin ot 2 3 s 3EI / mL mL o o2 3EI mL3 u 1 2 . For the transient response P ms EA / L U (t ) L (f) U ( s) mL3 u U (t ) L (e) U ( s) 48EI u 3EI / L3 . For the transient response P ms 2 3EI / L3 U (t ) L (d) U ( s) o u 1 2 . For the transient response 3 P ms 3EI / L U (t ) L (c) U ( s) 1 sin o t mo U (s) L -1 -1 1/ m 1 sin o t 2 s EA / mL mo u 1 . For the transient response P ms 2 6EI / L3 o2 EA mL (g) U ( s ) o2 6EI mL3 U (s) L -1 -1 6EI / mL3 sin ot 2 3 s 6EI / mL o2 6EI mL3 u 5 /16 2 . For the transient response 3 P ms 3EI / L U (t ) L (i) U ( s ) 1/ m 1 sin ot 2 3 m o s 6EI / mL -1 u 6 EI / L3 . For the transient response a ms 2 6 EI / L3 U (t ) L (h) U ( s) -1 U (s) L U (t ) L U (s) L -1 -1 5 /16m 5 sin ot 2 3 s 3EI / mL 16mo o2 3EI mL3 u 48EI / L3 . For the transient response a ms 2 48EI / L3 U (t ) L U (s) L -1 -1 48EI / mL3 sin ot 2 3 s 48EI / mL o2 48EI mL3 (B) For the Kelvin model, replace E by E ( s ) E s with E = 30 x 106 psi, 60 106 psi.sec., 1/ m 1/ m (a) U ( s ) 48EI 48 Is 2 48 I 48EI 2 s s s mL3 mL3 mL3 mL3 1/ m -1 U (t ) L 48 I 48EI 2 s s mL3 mL3 1 U (t ) m 48EI 24 I mL3 mL3 2 e L 24 I mL3 1/ m -1 2 2 24 I 48EI 24 I s mL3 mL3 mL3 2 48EI 24 I sin t mL3 mL3 (C) For the Maxwell model, replace E by E ( s ) Es with E = 30 x 106 psi, 18 108 psi.sec., E s 1 / m E s 1 / m E s 1/ m (a) U ( s ) 48 EI s 2 48EI ES s 3 Es 2 E 48EI s s2 s 3 s mL3 E s mL mL3 (b) U ( s) u 1 . For the transient response 2 P ms 3EI / L3 U (t ) L (c) U ( s) o2 3EI mL3 U (s) L -1 -1 3EI / mL3 1 sin o t 2 3 m o s 3 EI / mL o2 3EI mL3 3EI / L2 2 . For the transient response ms 3EI / L3 U (s) L -1 -1 3EI / mL2 EI 2 sin ot 2 3 s 3EI / mL mL o o2 3EI mL3 u 1 . For the transient response P ms 2 EA / L U (t ) L (f) U ( s) 1/ m 1 sin o t 2 s 3EI / mL3 mo u U (t ) L (e) U ( s) -1 u 3EI / L3 . For the transient response P ms 2 3EI / L3 U (t ) L (d) U ( s) U (s) L -1 U (s) L -1 -1 1/ m 1 sin o t 2 m o s EA / mL o2 EA mL u 1 2 . For the transient response 3 P ms 6EI / L U (t ) L U (s) L -1 -1 1/ m 1 sin ot 2 3 s 6EI / mL mo o2 6EI mL3 (g) U ( s ) u 6 EI / L3 . For the transient response a ms 2 6 EI / L3 U (t ) L (h) U ( s) -1 6EI / mL3 sin ot 2 3 s 6EI / mL o2 6EI mL3 u 5 /16 . For the transient response P ms 2 3EI / L3 U (t ) L (i) U ( s ) U (s) L -1 U (s) L -1 -1 5 /16m 5 sin ot 2 3 16 m o s 3EI / mL o2 3EI mL3 u 48EI / L3 . For the transient response a ms 2 48EI / L3 U (t ) L U (s) L -1 -1 48EI / mL3 sin ot 2 3 s 48 EI / mL o2 48EI mL3 1.6 Find the particular solution (steady state) for the displacement if the loading is P(t ) P0 sin t , P(t ) P0 cos t a (t ) a0 sin t , a (t ) a0 cos t (t ) 0 sin , (t ) 0 cos Consider the elastic and inelastic materials as in Problem 1.3, viscoelastic materials as in Problem 1.5. Solution (a) For P(t ) P0 sin t , u(t ) P0 Im U *eit Elastic material: U *(i ) eit cos t sin t i 48EI 48EI 2 2 3 m 3 m L L u (t ) Inelastic material: Po sin t 48EI 2 3 m L U *(i ) eit 48E ' I 48E ' I m 2 cos t m 2 sin t 3 3 L L 2 2 2 2 48E ' I 48E " I 48E ' I 48E " I 2 2 m m 3 3 3 3 L L L L 48E ' I 48E " I m 2 sin t cos t 3 3 L L i 2 2 48 E ' I 48 E " I 2 m 3 3 L L 48E ' I 48E " I m 2 sin t cos t 3 L3 L u(t ) Po 2 2 48E ' I 48E " I 2 m 3 3 L L Viscoelastic material: Kelvin model - E *(i ) E i 30 106 1 2i E ' iE 144 107 I 288 107 I m 2 sin t cos t 3 3 L L u(t ) Po 2 2 7 7 144 10 I 288 10 I 2 m 3 3 L L For P(t ) P0 cos t , u(t ) P0 Re U *eit Elastic material: u (t ) Po cos t 48EI 2 3 m L 48E ' I 48E " I m 2 cos t sin t 3 3 L L Inelastic material: u(t ) Po 2 2 48 E ' I 48 E " I 2 m 3 3 L L Viscoelastic material: Kelvin model - E *(i ) E i 30 106 1 2i E ' iE 144 107 I 288 107 I m 2 cos t sin t 3 3 L L u(t ) Po 2 2 7 7 144 10 I 288 10 I 2 m 3 3 L L 1.7 Find the solution for the displacement u(t) if the loading is as in Problem 1.6 for t > 0, and the initial conditions are u u 0 for t 0 Solution From Problem 1.6, the particular solution for an elastic material is up Po sin t 48EI 2 3 m L o2 The natural frequency is up 48EI , so mL3 Po / m sin t 2 o 2 The corresponding solution is u A sin o t B cos o t Po / m sin t 2 o 2 With initial conditions u(0) u(0) 0 u(0) A(0) B(1) 0 B 0 u(0) o A(1) o B (0) Po / m 2 o 2 0 A Po / mo 2 2 o Therefore u Po / m sin t sin o t o 2 o 2 For inelastic or viscoelastic materials the solution is the same with the new particular solution used. 1.8 Find the particular solution (steady state) for the displacement for the loading shown in Figure 1.74 Figure 1.73 Loading sequence Consider systems as in Problem 1.1, with various types of material. Calculate the first three terms of the Fourier series involved. Solution f ( x) ao n x n x an cos bn sin 2 n 1 p p 1 p 2 P T /2 f ( x )dx dt P p p T 0 Where ao T /2 n x 1 p 2 P T /2 P 2n t 2n t an f ( x ) cos cos sin dt 0 dx p 0 p T n T T 0 p T /2 n x 1 p 2 P T /2 2n t P 2n t f ( x ) sin sin cos dt dx p 0 p T n T T 0 p P 2P n 1,3,5,... cos n 1 n n bn Therefore f (t ) P 2P 2n t sin 2 n 1,3,5 n T P (t ) P 4 1 2n t 1 sin 2 n 1,3,5 n T Using the first three terms P( t ) Since 2 T P 4 2 t 4 6 t 1 sin sin 2 T 3 T P (t ) P 2P 2P sin t sin 3 t 2 3 As in Problem 6, the steady-state displacement is P / 2 2P 2P it 3it ImU *(i )e ImU *(3i )e K 3 u(t ) u (t ) PL3 96 EI 2 P sin t 2 P sin 3t 48EI 48EI 2 3 m 3 3 9m 2 L L For inelastic and viscoelastic materials proceed as in Problem 6 Chapter 2 2.1 What are the dimensions of the correlation function R x and the spectral density S x if the random function X (t ) is (A) Displacement (in.) (B) Acceleration ( in./sec2 ) (C) Force (lb.) (D) Stress (psi) Solution R x 1 T x(t ) x(t )d T T S x ( ) 2 1 * 1 xT i 2T 2T Part Function R x Sx ( ) A Displacement (in.) in.2 in.2 sec B Acceleration ( in./sec2 ) in.2 / sec4 in.2 / sec3 C Force (lb.) lb.2 lb.2 sec D Stress (psi) lb.2 / in.4 lb.2 sec/ in.4 xT (t )eit dt 2 2.2. What are the dimensions of the probability function Fx and the probability density function f x if the random function X (t ) is (A) Displacement (in.) (B) Acceleration ( in./sec2 ) (C) Force (lb.) (D) Stress (psi) Solution Fx f ( x)dx Part Function A fx 2 2 1 e ( x x ) /2 c 2 c Displacement (in.) Fx Dimensionless fx 1/ in B Acceleration ( in./sec2 ) Dimensionless sec2 / in. C Force (lb.) Dimensionless 1/lbs. D Stress (psi) Dimensionless in.2 /lbs 2.3. Define a stationary random process. Solution A stationary random process is defined by the correlations function of a random variable being constant regardless of the time at which it is evaluated, so long as the period over which each evaluation is taken is unchanged in duration. 2.4. Consider a stationary ergodic process X (t ) ; express E X , E X 2 , and R x as the time averages. Solution For a stationary ergodic process X (t ) we have in terms of time averages 1 T x(t )dt 2T T 1 T 2 E X 2 (t ) lim x (t )dt T 2T T E X (t ) lim T R x E X (t ) X (t ) lim T 1 2T T T x(t ) x(t )dt 2.5. Prove the relations between the spectral density S and the correlation function R : R 1 2 S e d i S R ei d Solution 1 2T R x E X (t ) X (t ) lim T T T x(t ) x(t )dt 1 i 1 e d x (t )x(t )dt d x (t )x(t )e i ( t )eit dt 2T 2T 1 it i ( t ) x( t) e dt x( t )e d 2T R ei d Since X *(i) x(t )eit dt R ei d and X *(i) x(t )e i ( t )d 1 X *(i) X *( i) 2T Take the limit of both sides as T lim R ei d lim T T 1 2 X *(i) X *( i) 2T lim R ei d S () T S () R ei d Thus, S () F R( ) R( ) and R( ) 1 2 F S () 1 S ei d t 2.6. Assuming that the input-output relation is y(t ) (t ) x( )d . Show that R y R x d d and S y *(i ) S x Solution Given y (t ) we know that the correlation function is 2 R y lim T 1 2T T T y(t ) y(t )dt lim T 1 2T T T dt x(t ) ( )d x(t ) ( )d where y t (t )x( )d x(t ) ( )d x(t ) ( )d t t y t t x(t ) ( )d x(t ) ( )d R y lim d d ( ) ( ) T lim T 1 2T T T 1 x(t ) x(t )dt 2T x(t )x(t )dt R x t t1 and t t2 so t2 t1 = therefore R y R x d d S ( ) R e i d R x e i d d d d d d e i R x d d d ei e i e i ( ) R x e i e i d ei d R x e i ( ) d *(i ) *(i )Sx ( ) therefore S y *(i ) S x 2 2.7. Derive the expression for the average number of crossings of the value x assuming that the joint distribution function f ( , ) of X (t ) and X ( t ) is (A) two-dimensional normal (B) arbitrary Solution (a) The joint distribution function f ( , ) of X (t ) and X ( t ) is given by 2 2 2 2 1 2C 2C f ( , ) e 1 2 2 C1C2 where C12 E X 2 and C22 E X 2 The probability that X d and X d is f ( , )d d . The duration of the passage from to d is d / . Thus the passages form to d at a velocity per unit time is f ( , )d d f ( , )d d f ( , )d d d The passages from to d at any rate is equal to the sum of the passages at each value of , or N 1 f ( , )d 2 C1C2 e 2 2 2 2 2 C1 2 C2 d 2.8. The spectral density of the loading (input) is given in the form shown in Figure 2.29 Figure 2.29Loading input Find the spectral density’s of the outputs (displacement, bending moment, etc.) of the systems described in problem 1(a) of Chapter 1. Solution From Problem 3(a) of Chapter 1 we know that 1 1 U* or *(i ) 2 115, 200 1.3 2 115, 200 1.3 S y *(i ) Sx 2 At 100 *(i100) 1 9.785 106 9.8 106 115, 200 1.3(100)2 *(i100) 9.8 106 9.6 1011 so S y 9.6 1011 S x 2 2 Also, the mean square value of u( x, t ) is E (u 2 ) 1 2 S x ( )d 1 2 100 101 (1)d 1 2 1001 100 (1)d 1 2.9. A random function X (t ) has its spectral density as in Problem 2.8. Find the mean square value of X (t ) . Find the average number of crossings of the values x 0 , x 1 , and x 10 . Solution From Problem 8 we have E (u 2 ) 1 For x 0 1/ 2 2 1 0 S x ( )d N0 S ( )d 0 x For x 1 1 10, 000 100 rad/sec 1 N N0 e 2 2C12 N1 100 N1 100 where C12 E (u 2 ) e (1) 2 /2 100 1 e / 2 rad/sec For x 10 e (10) 2 /2 100 e50 rad/sec Chapter 3 3.1 A 4 lb cart is attached to a 12 lb/in spring starts from rest and is subjected to an applied force F(t) as illustrated in Figure 3.15. The equation of motion for this system is mx kx F (t ) . Using Euler finite differences solve the differential equation above and note the position, velocity and acceleration at various time intervals up to 0.2 seconds. Assume a time interval of t T /10 . Figure 3.15 Spring-mass system with an applied force. Solution The mass of the cart is m 4 lb 0.124 lb-s2 /ft 0.01035 lb-s2 /in . The equation of motion is therefore 32.2 ft/s2 0.01035x 12x F (t ) or x 96.6F (t ) 1159.4 x .The time period is T 2 time interval is t m 0.01035 2 0.1845 s . The k 12 T 0.1845 0.01845 0.02 . Since the cart starts from rest, x1 x1 0 and the acceleration is 10 10 x1 96.6(10) 1159.4(0) 966 . Since x2 x1 x1t 1 2 x1 t and x2 x1t we have, at t 0.02 2 1 2 x2 0 0 (966) 0.02 0.1932 and x2 966(0.02) 19.32 2 x2 96.6( F2 ) 1159.4x2 96.6(12) 1159.4(0.1932) 935.2 At t 2t 0.04 : x3 x2 x2 t 1 1 2 2 x2 t 0.1932 19.32(0.02) (935.2) 0.02 0.7664 2 2 x3 2 x2 t x1 2(935.2)(0.02) 0 37.4 x3 96.6( F3 ) 1159.4x3 96.6(14) 1159.4(0.7664) 463.8 Following the same procedures as above we construct the table below t (s) F (t ) 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 10 12 14 16 18 20 18 16 15 15 15 Position 0 0.193 0.766 1.608 2.301 2..608 2.404 1.813 0.877 0.137 -0.084 Velocity 0 19.32 37.408 37.859 24.676 0.689 -19.035 -41.253 -41.322 -23.969 10.283 Acceleration 966 935.2 463.5 -318.3 -929.2 -1092.8 -1048.6 -577.2 432.1 1290.1 1546.8 3.2. Repeat problem 1 using t T 20 0.01 . Compare this solution to that of problem 3.1 by plotting the displacements, velocities, and accelerations for both cases. Solution Since the cart starts from rest, x1 x1 0 and the acceleration is x1 96.6(10) 1159.4(0) 966 . Since x2 x1 x1t 1 2 x1 t and x2 x1t we have, at t 0.01 2 1 2 x2 0 0 (966) 0.01 0.0483 and x2 966(0.01) 9.66 2 x2 96.6( F2 ) 1159.4x2 96.6(11) 1159.4(0.0483) 1006.6 At t 2t 0.02 : x3 x2 x2 t 1 1 2 2 x2 t 0.0483 9.66(0.01) (1006.6) 0.01 0.1952 2 2 x3 2 x2 t x1 2(1006.6)(0.01) 0 20.13 x3 96.6( F3 ) 1159.4x3 96.6(12) 1159.4(0.19523) 932.8 velocity displacement This continues in the same manner and the plots below can be generated using an appropriate computer code. 3 2,5 2 1,5 1 0,5 0 -0,5 0 50 40 30 20 10 0 -10 0 -20 -30 -40 -50 Dt=0.02 Dt=0.01 0,05 0,1 time 0,15 0,2 Dt=0.02 0,05 0,1 time 0,15 0,2 Dt=0.01 2000 1500 acceleration 1000 500 Dt=0.02 0 -500 0 0,05 0,1 0,15 0,2 Dt=0.01 -1000 -1500 time 3.3. A spring mass system is governed by the equation 0.2 x 32 x 5 100t , where 0 t 1 s . Assume the system starts from rest, the time interval is t T /10 , and determine the system response using the Euler finite difference method for 0 t 0.5 s. Solution T 0.5 m 0.2 2 0.4967 0.5 s .The time interval is t 0.05 . The equation k 32 10 10 of motion is written as x 25 500t 160 x . The time period is T 2 Since the system starts from rest, x1 x1 0 and the acceleration at t 0 is x1 25 . Since x2 x1 x1t 1 2 x1 t and x2 x1t we have, at t 0.05 2 1 2 x2 0 0 (25) 0.05 0.03125 and x2 25(0.05) 1.25 2 x2 25 500t 160 x2 25 500(0.05) 160(0.03125) 45 At t 2t 0.10 : x3 x2 x2 t 1 1 2 2 x2 t 0.03125 1.25(0.05) (45) 0.05 0.15 2 2 x3 2 x2 t x1 2(45)(0.05) 0 4.5 x3 25 500(0.1) 160 x3 25 500(0.1) 160(0.15) 51 Following the same procedures as above we construct the table below t (s) 0 0.05 0.1 0.15 Position 0 0.03125 0.15 0.43875 Velocity 0 1.25 4.5 6.35 Acceleration 25 45 51 29.8 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.7935 1.165 1.427 1.552 1.533 1.475 1.418 7.48 6.154 3.839 0.818 -1.001 -1.208 0.679 -1.96 -36.41 -53.36 -48.4 -20.27 16.79 48.1 3.4. Repeat problem 3.3 using t T 20 0.025 . Compare this solution to that of problem 3.3 by plotting the displacements, velocities, and accelerations for both cases. Solution Since the system starts from rest, x1 x1 0 and the acceleration at t 0 is x1 25 . Since x2 x1 x1t 1 2 x1 t and x2 x1t we have, at t 0.05 2 1 2 x2 0 0 (25) 0.025 0.007813 and x2 25(0.025) 0.625 2 x2 25 500t 160 x2 25 500(0.025) 160(0.007813) 36.25 At t 2t 0.05 : x3 x2 x2 t 1 1 2 2 x2 t 0.007813 0.625(0.025) (36.25) 0.025 0.03477 2 2 x3 2 x2 t x1 2(36.25)(0.025) 0 1.8125 x3 25 500(0.05) 160 x3 25 500(0.05) 160(0.03477) 44.437 This continues in the same manner and the plots below can be generated using an appropriate computer code. 8 7 6 velocity 5 4 3 Dt=0/05 2 Dt=0.25 1 0 -1 0 0,1 0,2 displacement -2 0,3 0,4 0,5 time 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 Dt=0/05 Dt=0.25 0 0,1 0,2 0,3 0,4 0,5 time 60 acceleration 40 20 Dt=0/05 0 -20 0 0,1 0,2 0,3 0,4 0,5 Dt=0.25 -40 -60 time 3.5. A dynamic system is governed by the equation u 10u 250u F (t ) , where F (t ) is shown in Figure 3.16. Assume the system starts from rest, the time interval is t T /10 , and plot the position and velocity using the Euler finite difference method for 0 t 0.5 s. Figure 3.16. Forcing function for a dynamic system. Solution From the given equation we can assume m 1 so the time period is T 2 interval is t m 1 2 0.397 0.4 s . The time k 250 T 0.4 0.04 . Within the first 2 seconds, the force is expressed as F (t ) 5 75t . Between 0.2 10 10 and 0.3 seconds F (t ) 20 100t then remains constant for 0.1 sec until the last segment where the force is F (t ) 10 50t . Since the system starts from rest, u1 u1 0 and the acceleration at t 0 is u1 5 . 1 2 Since u2 u1 u1t u1 t and u2 u1t we have, at t 0.04 , F (t ) 8 2 1 2 u2 0 0 (5) 0.04 0.004 and u2 5(0.04) 0.20 2 u2 8 10(0.2) 250(0.004) 5 At t 2t 0.08 : F (t ) 11 and 1 1 2 2 u3 u2 u2 t u2 t 0.004 0.2(0.04) (5) 0.04 0.016 2 2 u3 2u2 t u1 2(5)(0.04) 0 0.4 u2 11 10(0.4) 250(0.016) 3 Following the same procedure we end up with the plot below 1,5 1 0,5 0 -0,5 0 0,1 0,2 0,3 0,4 0,5 Position Velocity -1 -1,5 -2 -2,5 Time (s) 3.6. Solve problem 3.1 using the Runge-Kutta method with h 0.02 . Plot position, velocity, and acceleration as a function of time using the Runge-Kutta method and compare it to the solutions obtained in problem 1. Solution From problem 1 we have x 96.6F (t ) 1159.4 x , y x , y x 96.6F (t ) 1159.4x , h t 0.02 F (t ) 10 100t . For the Runge-Kutta method we follow the procedure outlined in the table below t x (position) y x (velocity) f y x (acceleration) T1 ti X 1 xi Y1 yi F1 f T1 , X1 ,Y1 h 2 h X 3 xi Y2 2 X 4 xi Y3h h 2 h Y3 yi F2 2 Y4 yi F3h F2 f T2 , X 2 ,Y2 h 2 h T3 ti 2 T4 ti h T2 ti X 2 xi Y1 Y2 yi F1 F3 f T3 , X 3 ,Y3 F4 f T4 , X 4 ,Y4 For subsequent steps we use xi 1 xi h Y1 2Y2 2Y3 Y4 6 and yi 1 yi h F1 2F2 2F3 F4 6 As a result, we can generate a simple computer code to establish the required parameters. The first several steps of the solution are as presented below step 1 t 0 0.01 0.01 0.02 F 10 11 11 12 x 0 0 0.0966 0.10626 y 0 9.66 10.626 9.50602 f 966 1062.6 950.602 1036.002 2 0.02 12 0.166927 20.09469 965.6651 0.03 0.03 0.04 0.04 3 13 13 14 14 0.367874 0.46444 0.734678 0.736306 29.75134 28.38756 34.44125 35.29306 829.2873 717.3281 500.6144 498.727 Following this procedure and incorporating the results form problem 1we can generate the curves shown displacement 3 2,5 2 1,5 Euler 1 Runge-Kutta 0,5 0 0 0,05 0,1 0,15 0,2 time 60 velocity 40 20 Euler 0 -20 0 0,05 0,1 0,15 0,2 Runge-Kutta -40 -60 time 2000 acceleration 1500 1000 500 Euler 0 -500 0 0,05 0,1 0,15 0,2 Runge-Kutta -1000 -1500 time 3.7. Solve problem 3.3 using the Runge-Kutta method with h 0.05 . Plot position, velocity, and acceleration as a function of time using the Runge-Kutta method and compare it to the solutions obtained in problem 3. Solution From problem 3 we have x 25 500t 160 x , y x , h t 0.05 . For the Runge-Kutta method we follow the procedure outlined in the table below t x (position) y x (velocity) f y x (acceleration) T1 ti X 1 xi Y1 yi F1 f T1 , X1 ,Y1 h 2 h X 3 xi Y2 2 X 4 xi Y3h h 2 h Y3 yi F2 2 Y4 yi F3h F2 f T2 , X 2 ,Y2 h 2 h T3 ti 2 T4 ti h T2 ti X 2 xi Y1 Y2 yi F1 F3 f T3 , X 3 ,Y3 F4 f T4 , X 4 ,Y4 For subsequent steps we use xi 1 xi h Y1 2Y2 2Y3 Y4 6 h F1 2F2 2F3 F4 6 and yi 1 yi As a result, we can generate a simple computer code to establish the required parameters. The first several steps of the solution are as presented in this table. step 1 t 0 0.025 0.025 0.05 0.05 0.075 0.075 0.1 0.1 2 Following this procedure and incorporating the results form problem 1we can generate the curves shown displacement 3 x 0 0 0.015625 0.046875 0.048438 0.092708 0.119115 0.196563 0.192545 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 y 0 0.625 0.9375 2.6875 1.770833 2.827083 2.9625 3.942917 4.004306 f 25.00 37.50 35.00 42.50 42.25 47.67 43.44 43.55 44.19 Euler Runge-Kutta 0 0,1 0,2 0,3 time 0,4 0,5 8 velocity 6 4 Euler 2 Runge-Kutta 0 0 0,1 0,2 -2 0,3 0,4 0,5 time 60,00 acceleration 40,00 20,00 Euler 0,00 0 0,1 0,2 0,3 -20,00 0,4 0,5 Runge-Kutta -40,00 -60,00 time 3.8. A beam of mass m is supported by two identical square steel columns of length L. The beam is acted on by a 6 EI time varying force P(t ) . The equation of motion for this system is mu 3 u P(t ) . The dimension of each L column is a a , and the overall dimensions of the system are such that the equation of motion can be written as 1800u 3.7 109 a4u 50t . The application of this system is such that the horizontal displacement, u, must not exceed 25 mm. We have available column sizes of a 25 mm , 38 mm , 50 mm and want to approximate the size of the column. Use the Newmark Beta method to determine the displacement of this system and approximate the required column size so that the allowable horizontal displacement is not exceeded. Assume h 0.2 , 1/ 2 , 1/ 6 , and that the system starts from rest with the applied load duration being 6 seconds. Solution Using the equation of motion, 1800u 3.7 109 a4u 50t , and following Table 3.6 we have 9 4 uo (0) uo (0) uo (0) 0 . With 1/ 2 , 1/ 6 , m 1800 , 0 , k 3.7 10 a , h 0.2, and p(t ) 50t we determine 1 kˆ m k 150m 15 k 150(1800) 15(0) 3.7 109 a 4 270,000 3.7 109 a 4 2 h h a 1 m 150m 15 270,000, 2 h h b 1 m 1 30m 2 54,000, h 1 c 1 m h 1 2m 0.1 3600 2 2 pˆ i 1 pi 1 aui bui cui pi 1 270,000ui 54,000ui 3600ui ui 1 pˆ i 1 pi 1 270,000ui 54,000ui 3600ui 270,000 3.7 109 a 4 kˆ ui 1 1 1 1 u ui ui 1 ui 150 ui 1 ui 30ui 2ui 2 i 1 h 2 h ui 1 ui 1 hui hui 1 ui 0.1 ui ui1 Horizontal Displacement (m) Developing a simple computer code we can generate a plot of the displacements as shown below 0,300 0,250 0,200 0,150 0,100 0,050 0,000 a = 25 mm a = 38 mm a = 50 mm 0 1 2 3 4 5 6 time Based on this we can see that a 50 mm is an appropriate solution. 3.9. Assume a single degree of freedom system starts from rest and is defined by 6u 3.6u 54u 360t . Use the Newmark Beta method to determine the response (displacement, velocity, and acceleration) of this system. Assume h t 0.1T , 1/ 2 and 1/ 6 . Plot the responses above 0 t 5 sec. Solution The natural angular frequency is k / m 3 and the period is T 2 / 2 / 3 2.094 . As a result, h t 0.1T 0.2 . Following Table 3.6 we have uo (0) uo (0) uo (0) 0 . Next, using 1/ 2 , 1/ 6 , m 6 , 3.6, k 54 , h 0.2, and p(t ) 360t we determine 1 kˆ m k 150m 15 k 150(6) 15(3.6) 54 1008 2 h h a 1 m h2 150m 15 954, h b 1 h m 1 30m 2 187.2, 1 c 1 m h 1 2m 0.1 12.36 2 2 pˆ i 1 pi 1 aui bui cui pi 1 954ui 187.2ui 12.36ui ui 1 ui 1 pˆ i 1 pi 1 954ui 187.2ui 12.36ui 1008 kˆ 1 h2 ui 1 ui 1 h 1 ui 1 ui 150 ui 1 ui 30ui 2ui 2 ui 1 ui 1 hui hui 1 ui 0.1 ui ui1 For i 0 , pˆ i 1 pi 1 p1 360(0.2) 72 , u1 72 0.0714 , u1 150 0.0714 0 2(0) 30(0) 10.714 , and 1008 u1 0 0.1 0 1.0714 0.10714 . Subsequently, the equations above can be programed into an appropriate computer code and the plot below can be generated. Position, Velocity, Accekleration 35,000 25,000 15,000 Position Acceleration 5,000 Velocity -5,000 0 -15,000 1 2 3 4 5 time 3.10. Work Problem 3.3 using the Newmark Beta method with 1/ 2 , 1/ 6 , and h t 0.1T . Compare the results by plotting the displacement for both cases. Solution From problem 3.3 we had 0.2 x 32 x 5 100t , T 0.5 , so h 0.05 . Following Table 3.6 we have uo (0) uo (0) 0 and uo (0) 5 / 0.2 25 . Next, using 1/ 2 , 1/ 6 , m 0.2 , 0, k 32 , h 0.05, and p(t) 100t 5 we determine 1 kˆ m k 2400m 60 k 2400(0.2) 15(0) 32 512 2 h h a 1 h2 m 1 2400m 60 480, b m 1 120m 2 24, h h 1 c 1 m h 1 2m 0.1 0.4 2 2 pˆ i 1 pi 1 aui bui cui pi 1 480ui 24ui 0.4ui ui 1 ui 1 pˆ i 1 pi 1 480ui 24ui 0.4ui 512 kˆ 1 h 2 ui 1 ui 1 1 ui 1 ui 2400 ui 1 ui 120ui 2ui h 2 ui 1 ui 1 hui hui 1 ui 0.025 ui ui1 Following the procedures of problem 8 and comparing those displacements with the ones from problem 3 we generate the plot below. displacement 2 1,5 1 Euler Newmark-Beta 0,5 0 0 0,1 0,2 0,3 0,4 0,5 time Chapter 4 4.1. Write the equations of motion for the systems shown in Figure 4.21 (a)...(e). Write the elements of the mass and stiffness (or flexibility) matrices (no damping). Write the frequency equation of each system and solve the equation. Figure 4.21 Multiple degree of freedom systems. Solution (a) The deformation will resemble that shown. Assume no axial deformation of the columns. The horizontal members will experience the forces shown (related to the spring rates k1 and k2 , defined as k1 k2 k 12 EI L3 The equations of motion are m2 x2 2k2 x2 x1 P(t ) m1 x1 2k1 x1 a(t ) 2k2 x2 x1 0 Since k1 k2 k 12 EI , L3 m1 x1 4kx1 2kx2 2ka(t ) m2 x2 2kx1 2kx2 P(t ) In matrix notation 2k x1 2ka (t ) 2k x2 P (t ) m1 0 x1 4k 0 m x 2k 2 2 The mass and stiffness matrices are 0 m M 1 0 m2 4k and K 2k 2k 2k Assume x1 A sin t , x2 B sin t , a(t ) P(t ) 0 so that we have x1 A 2 sin t and x2 B 2 sin t m1 A 2 sin t 4kA sin t 2kB sin t 0 m2 B 2 sin t 2 Ak sin t 2 Bk sin t 0 Thus; 2k 2k m 4k m12 2k 2 2 0 4k 2 4k m1 2 2k m2 2 Therefore the frequency equation is given by m1m24 2k2 m1 2m2 4k 2 0 Solving for 2 gives 2k m1 2m2 4k 2 m1 2m2 16m1m2 k 2 2 2 2 (b) 2m1m2 2k m1 2m2 4m12 k 2 16m1m2 k 2 16m22 k 2 16m1m2 k 2 2m1m2 2k m1 2m2 4k 2 m12 4m12 k m1m2 2m1m2 m 2m 1 2 m12 4m22 Because of the loading, the system will undergo a vertical displacement as well as a rotation. The two equations governing these are Y m d 2v Pa d 2 and M I z dt 2 2 dt 2 Where Y k11V k12 and M k21V k22 k11 k12 V Py k 21 k22 M Apply a unit displacement in the y direction of V 1 k11V Py and k21V M 0 k12 3EI k11V Py 2 3 V b k11 6EI b3 Apply a unit rotation of 1 so that k12 Py 0 and k22 M . Assuming small angles sin a/2 2 a Fb3 1 a 2 3EI a 3EI k22 M F 3 a 2 2 4b So Y k22 F 3EI a 2b 3 3EI 2 a 4b3 6EI 3EI v and M 3 a 2 and the equations of motion are 3 b 4b mv 6 EI v0 b3 m 0 M 0 Iz and I z 3EIa 2 Pa 3 4b 2 6 EI b3 and K 0 3EIa 2 4b3 0 Assume v Asin t and B sin t , so that v A2 sin t and B2 sin t results in 12 6 EI mb3 1 6 EI mb3 22 3EIa 2 4 Ib3 2 3Ea 2 4b 3 (c) This is a 5 degree of freedom system with 3 displacements (u, v, and w) and 2 rotations ( and ) as illustrated in the figure. We have d 2u dt 2 d 2v -Y m 2 dt -X m - M P (t ) - M I w - Z P (t ) m k11 k 21 k31 k41 k51 k12 k22 k32 k42 k52 L d 2 Iv 2 2 dt d 2 dt 2 d 2w dt 2 k13 k14 k23 k24 k33 k34 k43 k44 k53 k54 k15 u X k25 v Y k35 w Z k45 M k55 M Only the major diagonal terms are non-zero Spring pairs 2 and 3 are in parallel and spring pair 1 is in series so the equivalent springs are k1 eq K1 k1 / 2 , k2 eq K 2 2k2 , and k3 eq K 3 2k3 For rotation about the v and w axes we get k44 k3 L / 2 ( L) so that k44 k3 L2 / 2 and k55 k2 L / 2 ( L) so that k55 k2 L2 / 2 k1 / 2 0 0 2k 2 0 K 0 0 0 0 0 0 0 0 0 0 0 2k 3 0 0 0 k3 L2 / 2 0 0 0 k2 L2 / 2 mu (k1 / 2)u 0 mv 2k2 v 0 mw 2k3 w P(t ) I v k3 ( L2 / 2) P(t )(L/ 2) I w k2 ( L2 / 2) 0 m 0 0 0 0 m 0 0 m 0 0 m 0 0 0 0 Iv 0 0 0 0 0 0 0 0 I w Assume u Asin t , v B sin t , w C sin t , D sin t , E sin t mA12 sin t (k1 / 2) A sin t 0 1 k1 / 2m mB sin t 2k2 Bsin t 0 2 2k2 / m mC sin t 2k3C sin t 0 3 2k3 / m 2 2 2 3 I v D42 sin t (k3 L2 / 2) D sin t 0 4 k3 L2 / 2 I v I w E52 sin t ( k2 L2 / 2) E sin t 0 4 k2 L2 / 2 I w (d) - X P1 (t ) m k11 k12 k k 21 22 k31 k32 d 2u dt 2 , - Y P2 (t ) m , - Z P3 (t ) m d 2w dt 2 k13 u X k23 v Y k33 w Z For each rod we know k11 d 2v dt 2 EA1 L1 k22 , PL P AE k AE L EA2 L2 , k33 EA3 L3 k12 k21 k13 k31 k23 k 32 0 EA1 L1 K 0 0 mu 0 EA2 L2 0 0 0 EA3 L3 m 0 0 m 0 m 0 0 0 m EA EA1 EA P1 (t ) , mv 2 P2 (t ) , mw 3 P3 (t ) L1 L2 L3 Setting P1 (t ) P2 (t ) P3 (t ) 0 and assuming u Asin t , v B sin t , w C sin t EA1 A sin t 0 L1 mA12 sin t 12 EA1 mL1 1 EA1 mL1 mB12 sin t EA2 EA2 B sin t 0 2 L2 mL2 mC32 sin t EA3 EA3 Csin t 0 3 L3 mL3 (e) d 2u d 2v , - Y P2 (t ) m 2 2 dt dt d 2 - M P1 (t ) a I 2 dt - X P1 (t ) m k11 k12 k k 21 22 k31 k32 k13 u X k23 v Y k33 M Let u 1 , v 0 and get k11u X , k21u Y , k31u M Each structural member will carry an axial load and will have a stiffness. Looking at one member, we note its elongation to be PL1 AE P AE AE u cos L1 L1 Resolving this into components: Px Px' AE AE u cos2 , Py u cos sin . For the other member we find L1 L1 AE AE u cos2 , Py' u cos sin . L2 L2 cos2 cos2 k11u 2 Px 2 Px' 2 AEu L2 L1 cos2 cos2 k11 2 AE L2 L1 k21v Y Py Py' Py Py' 0 k21 0 2a cos2 2a cos2 b cos sin b cos sin k31u M 2 Px a 2 Px' a Py' b Py b AEu L1 L2 L2 L1 2a cos2 2a cos2 cos sin cos sin k31 AE 2a b L1 L2 L2 L1 Next, let v 1 , then k12 v X , k22v Y , k32v M . As before, k21 k12 0 . As before, we note PL1 AE AE AE v sin L1 L1 AE AE Resolving this into components: Py * v sin 2 , Px * v cos sin L1 L1 AE AE For the other member we find Py' * v sin 2 , Px' * v cos sin L2 L2 P sin 2 sin 2 k22 v Y 2 Py * 2 Py' * 2 AEv L2 L1 sin 2 sin 2 k22 2 AE L2 L1 k32 v M 0 k32 0 Next, let 1 , then k13 X , k23 Y , k33 M . From the rotation of the mass we see that for small rotations sin v / b v b u / a u a The forces due to horizontal ( u a ) and vertical ( v b ) displacements are as shown. k13 k31 , k23 Y 0 k23 k32 0 k33 a 2 Px Px' b Py' Py b Py Py' b 2 Py * Py' * a 2 Px' * Px * 2a Px Px' 2a Px' * Px * 2bPy' 2bPy 2b Py * Py' * AE AE k33 M 2a u cos2 u cos2 L1 L2 AE AE 2a v cos sin v cos sin L1 L2 AE AE 2b u cos sin 2b u cos sin L2 L1 AE AE 2b v sin 2 v sin 2 L2 L1 With u a and v b this becomes AE AE AE AE AE k33 2a a cos2 a cos2 2a b cos sin b cos sin 2b a cos sin L L L L L2 1 1 2 2 2b AE AE AE a cos sin 2b b sin 2 b sin 2 L1 L L 2 1 AE AE AE AE AE k33 2a 2 cos2 cos 2 2ab cos sin cos sin 2ab cos sin L1 L1 L2 L2 L2 2ab AE 2 AE AE 2 cos sin 2b2 sin sin L1 L L2 1 AE AE AE 2 AE AE AE 2 k33 2a 2 cos2 cos2 4ab cos sin cos sin 2b2 sin sin L L L L L L2 1 1 2 2 1 cos 2 cos 2 2 AE 0 L2 L1 sin 2 sin 2 K 0 2 AE L1 L2 2 2 2a cos 2a cos AE 2 a L1 L2 0 cos sin cos sin b L2 L1 2 2 2a cos 2a cos AE 2a L1 L2 cos sin cos sin b L2 L1 0 cos 2 cos 2 AE 2a 2 L1 L2 cos sin cos sin 4ab L2 L1 sin 2 sin 2 2b 2 L2 L1 The equations of motion are mu k11u k13 P1 (t) , mv k22 v P2 (t ) , I k13u k33 P1 (t )a Assume P1 (t ) P2 (t ) 0 and u A sin t , v B sin t , C sin t (1) mA 2 sin t k11 A sin t k13C sin t 0 (2) mB 2 sin t k22 B sin t 0 (3) IC 2 sin t k13 A sin t k33C sin t 0 From (1) and (3) k m 2 A k13C 0 and k33 I 2 C k13 0 11 For a non-trivial solution k 11 m 2 k13 k k13 33 I 2 0 k11 m 2 k33 I 2 k132 mI4 k33m k11I 2 k11k33 k132 0 2 k33m k11 I k33m k11I 2 4mI (k11k 33 k132 ) 2mI k33m k11 I k m 2k11k 33mI k112 I 2 4mIk132 k 33m k 11I 2 33 2 2 2mI m 0 0 m 0 m 0 0 0 I 4.2. Find two normalized natural modes for the system in problem 4.1(a). Assume m 1 . Write the elements i( r ) of the modal matrix . Solution From Problem 4.1 we have A1 2k m2 2 A2 2k For 1 we have 2k m2 2 (1) A1(1) A2 or 2k A1(1) XA2(1) For 1 we have A1(2) YA2(2) . In addition we have i( r ) Ai( r ) For 1 ; 1(1) A1(1) , 2(1) A2(1) For 2 ; 1(2) A1(2) , 2(2) A2(2) 1(1) XA2(1) X 2(1) ; 2(1) 1(1) / X 1(2) A2(2) YA2(2) Y 2(2) ; 2(2) 1(2) / Y For normalized modes, we have m j(s) rs m (r) ij i i j m mi 2i( r )2(s) rs m ( r ) (s) i1 i 1 i m111( r )1(s) m121( r )2(s) m212( r )1(s) m222(r )2(s) rs m Since m12 m21 0 , m11 m1 , m22 m2 , and m 1 m11(1)1(1) m22(1)2(1) 1 m11(2)1(2) m22(2)2(2) 1 Substituting and solving for 1(1) , 2(1) , 1(2) , and 2(2) gives m1 (1) 2 1 m2 (1) 2 1 X2 1 X 2 m1 m2 1 X2 (1) 2 1 X 1(1) 2(1) X m1 m2 2 m1 1(2) m2 2 (2) 2 1 (2) 2 1 Y2 1 X m1 m2 2 1 Y 2 m1 m2 1 Y2 Y 1(2) Y m1 m2 2 The modal matrix is 1(1) 1(2) (1) (2) 2 2 2(2) 1 Y m1 m2 2 4.3. Write the uncoupled equations of motion for forced vibrations of system of problem 4.1 (a) Solution Assume q1 g1 (t )1(1) g2 (t )1(2) and q2 g1 (t )2(1) g 2 (t )2(2) , then g1 12 g1 f1 f and g2 22 g2 2 , where m 1 and m m f1 p11(1) p22(1) and f 2 p11(2) p22(2) Where p1 p1 (t ) P(t ) and p2 p2 (t ) 2ka(t ) The solution is given by gr Ar cos r t Br sin r t gr (t ) Where t t 0 0 gr (t ) Gr (t t ') f r (t ')dt ' 1 sin r (t t ') f r (t ')dt ' mr Therefore qi Ari( r ) cos r t Bri( r ) sin r t gr (t )i( r ) r r r 4.4. Assume that the material of the structure in problem 4.1(a) is an arbitrary viscoelastic material (all of the elements of each structure are made from the same material). Write the uncoupled equations of motion for forced vibrations using: (A) Laplace transformations (B) Fourier transformations Solution (a) L Kij Kij ( s) , L pi (t ) pi (s) Assume q1 g1 (t )1(1) g2 (t )1(2) and q2 g1 (t )2(1) g 2 (t )2(2) Then q1 g11(1) g 21(2) and q2 g12(1) g 22(2) Then and g 2 22 g 2 f 2 And g1 s g2 s 2 1 f1 ( s ) 1 f 2 ( s) 12 ( s ) 22 ( s ) 2 Where f1 ( s ) P1 ( s )1(1) P2 ( s )2(1) f 2 ( s ) P1 ( s )1(2) P2 ( s )2(2) P(t ) And from the system P1( s) L ; P2 ( s) 2kL a(t) From Problem 4.2 we have 1(1) , 1(2) , 2(1) , 2(2) , 12 , and 22 . Substitute these values to obtain f1 ( s ) and f 2 ( s ) , then knowing f ( s ) , g (s) can be calculated and then qi ( s ) g( s )i( r ) gi( r ) ( s ) qi (s) 1 q(t ) L (b) m g k g ij i ij j i Pi (t ) j mij 2 g *j kij g *j Pi* j j For a viscoelastic material replace kij* with kij mij 2 g *j kij g *j Pi * j j * * (r) Assume q j gr j r mij 2 g r* j( r ) kij g r* j( r ) Pi * j r j mij g 2 i j * (s) r i r (r ) j kij g r*i( s ) j( r ) Pi *i( s ) r i j r i m rs g m rs g P 2 * r r r * * (s) Let f s Pi i i *2 s * r 2 g r* f s* * r * (s ) i i i k kij" kij* ' ij Pi (t ) Pi* Assume q1* g1*1(1) g 2*2(1) and q2* g1*1(2) g 2*2(2) . Thus g1* f1* f 2* * and g 2 1*2 2 2*2 2 Where f1* P1*1(1) P2*2(1) and f 2* P1*1(2) P2*2(2) . For the system P1* P1 (t ) and P2* 2kij* a (t ) Knowing 1(1) , 1(2) , 2(1) , and 2(2) we find f1*and f 2* to obtain g1* and g 2* 4.5. For the system shown in Figure 4.21(a), assume a structural rigidity of EI 117 103 N-m2 and that the masses are related through m1 nm2 , where 0.2 n 5 . If m2 225 kg and the distance between masses is L 3 m , plot the frequencies as a function of n. Solution From the solution of problem 4.1(a) we had 2 2k m1 2m2 4k 2 m12 4m12 2m1m2 or 12 With k k m1m2 12 117 103 (3)3 12 22 k nm22 k nm22 m 2m 1 2 m12 4m22 22 k m1m2 m 2m 1 2 m12 4m22 52,000 , and m1 nm2 n 2 m m 2 2 n 2 m m 2 2 n2 4 n2 4 k nm2 k nm2 n 2 n 2 n2 4 n2 4 231 n 2 n2 4 n 231 n 2 n2 4 n omega Plotting these two equations results in 6000 5000 4000 3000 2000 1000 0 Omega 1 Omega 2 0 1 2 3 4 5 n 4.6. For the system shown in Figure 4.21 (b) , assume the beam is made from aluminum ( E 10 106 psi ) and has dimensions such that I 2 in 4 . The weight of the rigid body is 300 lb. The beam segments are related by b na , where 1 n 5 and a 24" . Plot the frequencies as a function of n. Solution From the solution of problem 4.1 (b) 1 6 EI mb3 and 2 6 Ea 2 4b 3 With E 10 106 , I 2 , EI 20 106 , b na , a 24 , and m 300 / 32.2(12) 0.776 these become 6 20 106 6 EI 1 =105.7 1/ n3 3 mb3 0.776 24n 2 6 Ea 2 6(10 106 )(24) 2 =790.6 1/ n3 4b3 4(24n)3 1000 Omega 800 600 400 Omega 1 200 Omega 2 0 0 1 2 3 4 5 n 4.7. For the system shown in Figure 4.21 (c), assume the rigid bar is a cylinder made from an aluminum alloy ( E 70 GPa ) and has a weight of 30 N/m. The length of the bar is between 1 and 4 meters. The horizontal spring is relatively soft k1 880 N/m while springs 2 and 3 are stiffer with k2 k3 3520 N/m . Plot the frequencies as a function of L. Solution From the solution of problem 4.1(c) we have 5 equations: mu (k1 / 2)u 0 , mv 2k2 v 0 , mw 2k3w P(t ) I v k3 ( L / 2) P(t )(L/ 2) , I w k2 ( L / 2) 0 2 2 The mass matrix is m 0 0 0 0 m 0 0 m 0 0 m 0 0 0 0 Iv 0 0 0 0 0 0 0 0 I w Assuming u Asin t , v B sin t , w C sin t , D sin t , E sin t we get mA12 sin t (k1 / 2) A sin t 0 1 k1 / 2m mB22 sin t 2k2 Bsin t 0 2 2k2 / m mC32 sin t 2k3C sin t 0 3 2k3 / m I v D42 sin t (k3 L2 / 2) D sin t 0 4 k3 L2 / 2 I v I w E52 sin t ( k2 L2 / 2) E sin t 0 5 k2 L2 / 2 I w Using the information from the problem statement m 2 30L mL2 3.06L L 3.06L , I v I w 0.255L3 . Therefore 9.81 12 12 1 k1 / 2m 880 / 2(3.06L) 11.99 1/ L 12 1/ L 2 3 2(3520) / 3.06L 47.97 1/ L 48 1/ L Since I v I w 4 5 3520 L2 / 2(0.255L3 ) 83.1 1 / L 83 1 / L Plotting the results gives Omega 90 80 70 60 50 40 30 20 10 0 Omega 1 Omega 2,3 Omega 4,5 0 1 2 3 4 L (m) 4.8. For the system shown in Figure 4.21 (d) , assume m 180 kg and the three identical supporting rods are aluminum ( E 70 GPa ) . Each rod has a length which is related to its diameter by L nd , where 5 n 30 . Three possible rod diameters are being considered; d 12.5 , 25 , 37.5 mm . Plot the frequencies as a function of n. Solution From problem 1(d) we had 1 EA3 EA1 EA2 , 2 , and 3 . mL2 mL3 mL1 Since all rods are identical, we need only consider one frequency. With E 70 109 , L nd , and m 180 , the frequency become E d 2 / 4 m nd 70 10 d 9 2 4(180) nd 17, 477 d n Omega Plotting this equation using the data given yields the figure shown 1600 1400 1200 1000 800 600 400 200 0 d=12.5 mm d=25 mm d=37.5 mm 0 10 20 n 30 4.9. For the system shown in Figure 4.21 (e), assume the box weighs 200 lb and the three supporting rods are all 1” in diameter aluminum with AE 7.9 106 lb . For the crate we approximate I 0.863 in4 . The box is a rectangle with dimensions a 1 ft, b 2 ft The lengths of the rods are L1 5 ft and L2 10 ft . The box is suspended between 2 and 4 feet above the ground. Plot the frequencies as a function height above the ground. Solution From the solution of problem 1(e) we had 2 k33m k11I 2 k33 m 2 2k11k 33mI k112 I 2 4mIk132 k 33m k11I 2 2mI h L (1 cos ) Where sin 1 and sin 1 1 and m 200 / 32.2(12) 0.5175 L2 L 1 cos2 cos2 sin 2 sin 2 k11 2 AE , k22 2 AE L2 L2 L1 L1 cos2 cos2 cos sin cos sin k31 AE 2a b L2 L2 L1 L1 Omega AE AE AE 2 AE AE AE 2 k33 2a 2 cos2 cos2 4ab cos sin cos sin 2b2 sin sin L L L L L L2 1 1 2 2 1 18000 16000 14000 12000 10000 8000 6000 4000 2000 0 Omega 1 Omega 2 2 2,5 3 3,5 h 4 4,5 5 Chapter 5 5.1 A plate with the dimension shown in Figure 5.21 is placed between two rigid walls and subjected to an axial compressive force F. Determine the displacement equations (u, v, and w) of the plate in the x, y, and zdirections. Figure 5.21 Plate fixed between two rigid walls. Solution F and the rigid walls produce a compressive stress y o . ht xz yz 0 . From the axial displacement of the plate and the The applied force is modeled an axial stress x The remaining stress components are z xy fixed walls we note that x L and y 0 . In addition, the shear stresses are all zero ( xy xz yz 0 ). From Hook’s law x L y 0 1 1 F x y ( o ) E E ht 1 y x E y x o 1 2 z F ht FL Eht F F x y o o E E ht E ht 1 L Integrating the strains gives u L x 1 2 F x Eht , v0 , w 1 L z F z 1 Eht 5.2 An elastic material ( E 70 GPa, 0.33 ) fills a cavity in a rigid block. The dimensions of the cavity are a 75 mm, b 125 mm, L 300 mm . A rigid cap is placed on the material and a force Po is applied to the center of the cap as illustrated in Figure 5.22. The elastic material is compressed an amount . Determine general expressions for the applied force Po as well as the net forces that exist on the x and z faces of the material Px , Pz . In addition, determine the explicit force in each direction if the axial strain is 0.01%. Figure 5.22 Elastic material compressed by an applied force. Solution From the axial displacement of the cap and the fixed walls we note that x 0 , y L , and z 0 . In addition, Po and ab compressive normal stresses x and z will exist on other two faces of the material. We can express the strainthe shear stresses are all zero ( xy xz yz 0 ). The normal stress in the y-direction will be y stress relationship in matrix form as x 1 1 y E z 1 From this we find x z L L 0 x 11 or y L E 1 z 0 1 y 1 x P o ab 1 z Po . From the second row of the matrix we have 1 ab 1 Po Po Po 2 2 1 2 E ab 1 ab Eab 1 Po 1 x z E ab Po 1 2 2 Po 1 2 1 Eab 1 1 Eab L Po E 1 ab 1 2 1 L Px x a L and Pz z b L Px E 1 a L 1 1 2 1 L Pz E 1 bL 1 1 2 1 L From the information provided, E 70 109 , 0.33 , a 75 mm , b 125 mm , L 300 mm , / L 0.0001 0.03 mm 0.0003 m Po 70 109 1 0.33 (0.075)(0.125) 0.0001 97.2 kN 1 0.661 0.33 Px Pz 9 0.33 70 10 1 0.33 0.0001 0.075 0.3 0.0003 114.8 kN 1 0.33 1 0.661 0.33 9 0.33 70 10 1 0.33 0.0001 0.125 0.30 0.0003 191.3 kN 1 0.33 1 0.661 0.33 Po 97.2 kN Px 114.8 kN Pz 191.3 kN 5.3 Consider a cantilevered beam of length L subjected to a uniform load q applied over its entire length. Assume ( x) is approximated by the static deflection curve for the beam. In addition, assume the beam dimensions are such that I bh3 /12 and A bh . Use Rayleigh’s method to determine the fundamental frequency for this beam. Assume the beam is aluminum and the height can be 25, 50, or 75 mm and the length is between 1 and 3 meters. Plot the frequency as a function of beam length for each height. Solution The static deflection curve is w( x) q x 4 4 Lx3 6 L2 x 2 . 24 EI We therefore approximate ( x) a1 x 4 4Lx3 6L2 x 2 L L 144 5 2U () EI ('' )2 dx EIa12 (12 x 2 24Lx 12 L2 ) 2dx EIa12 L 0 0 5 L L 0 0 2K ' () A ()2 dx A a12 (x 4 4Lx3 6L2 x 2 ) 2 dx A a12 (2.31L9 ) 12 R U ( ) K ( ) ' EIa12 144 / 5L5 A a12 (2.31L9 ) 12.467 EI A L4 EI 3.53 bh3 I 1.019h E 2 A 12 bh L L2 3.53 2 L Frequency 1.019h E 450 400 350 300 250 200 150 100 50 0 L2 h=25 mm h=50 mm h=75 mm 0 0,5 1 1,5 2 2,5 3 Beam Length (m) 5.4 The static deflection curve of a uniform beam with a distributed weight q fixed to rigid walls at both ends is w( x ) q / 24 EI x 4 2 Lx 3 L2 x 2 . Assume ( x) is approximated by this deflection curve and use Rayleigh’s method to determine the fundamental frequency for this beam. Solution '( x ) a 4 x 2L x We approximate ( x) a1 x 4 2 Lx 3 L2 x 2 1 3 6Lx 2 "( x ) a1 12 x 2 12 Lx 2 L2 2 2 L L 4 2U () EI ('' )2 dx EIa12 12 x 2 12 Lx 2 L2 dx EIa12 L5 0 0 5 L L 0 0 2 K ' () A ()2 dx A a12 x 4 2 Lx3 L2 x 2 12 R U ( ) K ( ) ' EIa12 0.8 L5 A a12 (0.00158L9 ) 2 dx A a12 (0.00158L9 ) 506.3EI A L4 1 22.5 2 L EI A 5.5 The tapered circular shaft shown in Figure 5.23 is subjected to a longitudinal vibration, for which the EI term in the strain energy is replaced by EA and " is replaced by du / dx . Using Rayleigh’s method, approximate the x lowest natural frequency of this shaft. Assume the displacement is approximated by u( x) A sin . 2L Figure 5.23 Tapered circular shaft. Solution The radius of the shaft is a function of x, resulting in the area being a function of x. Therefore x r( x ) r 1 2L L x A( x ) r 2 1 2L 2U ( ) EA( x)(du / dx) dx 2 0 L 0 2 x x E r 1 cos dx 2L 2L 2L 2 2 E 3r 2 4 L2 E 3r 2 4 L2 2 L 0 x x x2 x 2x cos2 2 cos2 cos dx 2L L 2L 4L 2L x L x 1 x 2 Lx x L2 x sin sin 2 cos L L 4 2 L 2 L 2 2 L 1 x 3 x 2 L L3 xL2 x 2 3 sin 2 cos L L 4 L 6 2 0 L 1 L2 L2 1 L3 2 L3 2 2 2 2 L 4 4 L 6 E 3r 2 4 L2 E 3r 2 1 1 1 1 1 2 Er 2 2.6533 4 L 2 4 2 4 6 2 L L L x x 2 K ' () A (u)2 dx r 2 1 sin 2 dx 0 0 2L 2L 2 L x x 2 x x2 x r 2 sin 2 sin 2 sin2 dx 0 2L L 2L 4L 2L x L x 1 x 2 2 Lx x 4 L2 x r 2 sin sin 2 cos L L 4 4 L 8 L 2 1 x 3 x 2 L L3 xL2 x 2 3 sin 2 cos L 4 L 6 2 L L 0 L 1 L2 L2 1 L3 2 L3 r 2 2 2 2 2 L 4 4 L 6 1 1 1 1 1 2 L r 2 2 2 0.7567 Lr 2 2 4 4 6 12 R U ( ) 2.6533Er 2 / L E 3.506 2 ' 2 K ( ) 0.7567 Lr L 1 1.87 E L 5.6 Assume that ( x) a1x2 a2 x3 for a uniform cantilever beam. Determine the first two natural frequencies using the Ritz method. Solution d d 2 2a1 x 3a2 x 2 and 2a1 6a2 x dx dx 2 L L U11 EI (2)(2)dx 4EIL U12 U 21 EI (2)(6 x)dx 6EIL2 0 0 L U 22 EI (6 x)(6 x)dx 12EIL3 0 mL5 5 L K11 m( x 2 )( x 2 )dx 0 L K 22 m( x3 )( x3 )dx 0 L K12 K 21 m( x 2 )( x3 )dx 0 7 mL 7 The frequency equation is 4 EIL 6 EIL 2 2 mL5 5 2 mL6 6 6 EIL2 12 EIL 3 2 mL6 6 2 mL7 7 0 mL6 6 2 2 7 2 6 2 mL5 3 mL 2 mL 4 EIL 12 EIL 6 EIL 0 5 7 6 104 1 1 48E 2 I 2 L4 EI 2 mL8 4 m2 L12 36E 2 I 2 L4 2 EI 2 mL8 4 m2 L12 0 35 35 36 12 E 2 I 2 L4 0.971EI 2mL8 0.000794 4m 2 L12 0 2 EI 2 EI 15,120 4 0 4 mL mL 4 1223.46 EI mL4 1 3.53 2 34.8 EI mL4 5.7 For the fixed-fixed beam described in problem 5.4, we can assume the mode shape to be expressed as w( x) a11 ( x) a2 2 ( x) , where 1 ( x) x 4 2 Lx3 L2 x 2 and 2 ( x) x5 3L2 x3 2 L3 x 2 . Using these expressions for 1 ( x) and 2 ( x) , approximate the lowest two natural frequencies using Ritz’s method. Solution L Uij EI ("i )("j )dx 0 L Kij A(i )( j )dx 0 1' 4x3 6Lx2 2L2 x 1" 12x2 12Lx 2L2 '2 5x4 9L2 x2 4L3 x "2 20x3 18L2 x 4L3 L 2 U11 EI 12 x 2 12 Lx 2 L2 dx 0.8EIL5 0 L U12 U 21 EI 12 x2 12Lx 2L2 20 x3 18L2 x 4L3 dx 2EIL6 0 A x L U 22 EI 20 x3 18L2 x 4 L3 0 K11 L 4 0 L 2 dx 5.142EIL7 2 2Lx3 L2 x 2 dx A (0.001587 L9 ) K12 K21 A x4 2Lx3 L2 x2 0 L x 5 3L2 x3 2 L3 x2 dx A (0.003968L10 ) 2 K 22 A x5 3L2 x3 2L3 x 2 dx A (0.0099567 L11 ) 0 0.8EIL5 2 A (0.001587L9 ) 2EIL6 2 A (0.003968L10 ) 2EIL6 2 A (0.003968L10 ) 5.142EIL7 2 A (0.0099567L11 ) 0.8EI 2 A (0.001587 L4 ) 2EIL 2 A (0.003968L5 ) 2EIL 2 A (0.003968L5 ) 5.142EIL2 2 A (0.0099567 L6 ) 0 0.8EI 0.001587 2 A L4 5.142EIL 0.0099567 2 2 A L6 2EIL 0.003968 2 A L5 2 0 2 EI 2 EI 4612.7 2065454.5 0 4 4 A L A L 4 1 22.4 EI A 2 L 2 64.1 EI L2 A 5.8 Use the Ritz method to determine the two lowest frequencies for shaft in problem 5.5 assuming 1 sin and 2 sin x 2L 3 x . 2L Solution L L Uij EA(i' )('j )dx Kij A(i )( j )dx 0 x r( x ) r 1 2L 0 x A( x ) r 2 1 2L 2 1' 2L cos x 2L '2 3 3 x cos 2L 2L 2 2 3 2 L L E r x x x x2 E r 2 2 x 2 x 2 x U11 E r 2 1 cos dx cos cos cos dx 0.9017 0 4L2 2L L 0 2L 2 L 4 L2 2 L L 2L 2L L x x 3 3 x U12 U 21 E r 2 1 cos cos dx 0 2 L 2 L 2L 2L 2L 2 r x 3 x x x 3 x x 2 x 3 x E r 2 cos cos cos cos cos cos dx 0.6094 2L 2L L 2L 2 L 4 L2 2L 2 L L 4 L2 L 3E 0 3 2 3 2 L L 9 E r x 3 3 x x x2 3 x 2 3 x 2 3 x U 22 E r 2 1 cos dx cos cos cos 2 dx 2 2 0 0 2L 2L L 2L 4L 2 L 4L 2L 2L 2 2 6.664 E r 2 L L L x x x x 2 x x2 x 2 K11 r 2 1 sin dx r 2 sin 2 sin 2 sin 2 dx 0.2157 r L 0 0 2 L 2 L 2 L L 2 L 2 L 4L 2 2 L x x 3 x K12 K 21 r 2 1 sin sin dx 0 2 L 2 L 2L 2 x L 3 x x x 3 x x 2 x 3 x 2 r 2 sin sin sin sin 2 sin sin dx 0.0697 r L 0 2 L 2 L L 2 L 2 L 2 L 2 L 4L 2 2 L L x 3 x 3 x x 2 3 x x 2 3 x 2 K 22 r 2 1 sin dx r 2 sin 2 sin 2 sin 2 dx 0.2832 r L 0 0 2 L 2 L 2 L L 2 L 2 L 4L E r 2 0.2157 r 2 L 2 L E r 2 0.6094 0.0697 r 2 L 2 L 0.9017 E r 2 0.0697 r 2 L 2 L 0 E r 2 2 2 6.664 0.2832 r L L 0.6094 2 E r 2 E r 2 E r 2 2 2 0.2157 r 2 L 2 6.664 0.2832 r L 0.6094 0.0697 r 2 L 2 0 0.9017 L L L E 2 E 100.25 2 0 2 L L 4 28.59 1 2.02 E L 2 L L 0 0 4.95 E L 5.9 For a torsion bar Uij GJ (i / x)( j / x)dx and Kij J (i )( j )dx . The 1” diameter, 48” long solid aluminum bar in Figure 5.24 is fixed to a wall at one end and has a torsional spring with a spring rate of k is attached at one end. Assume 1 x3 3L2 x and 2 x 2 2Lx . Use the Ritz method to determine the two lowest frequency and plot the results as a function of k for 500 k 2000 . Figure 5.24 Torsion rod with a torsional spring at the end. Solution For this shaft J (0.5)4 2 0.0982 , G 3 106 , GJ 0.2946 106 , 0.10 , J 0.00982 L j Uij GJ i dx k i ( L) j ( L) 0 x x L Kij J ( i )( j )dx 0 1 3x 2 3L2 x 1 x3 3L2 x , L U11 GJ 3x 2 3L2 0 2 dx k 2 L3 2 2 x 2L x and 2 x 2 2Lx , 2 4.8GJL5 4k L6 4.8GJ 4k L L5 U12 U21 GJ 3x2 3L2 2 x 2 L dx k 2 L3 L2 2.5GJ 2k L L4 L 0 U 22 GJ 2 x 2 L dx k L2 L 2 0 2 1.333GJ k L L3 L K11 J ( x3 3L2 x)2 dx 1.9428 JL7 0 L K12 K21 J ( x3 3L2 x)( x2 2 Lx)dx 1.0167 JL6 0 L K22 J ( x 2 2Lx)2 dx 0.5333 JL5 0 4.8GJ 4k L L2 1.9428 JL4 2 2.5GJ 2k L L 1.016 JL3 2 3 L 0 2.5GJ 2k L L 1.016 JL3 2 1.333GJ k L 0.5333 JL2 2 4.8GJ 4k L L 1.9428 JL 1.333GJ k L 0.5333 JL 2.5GJ 2k L L 1.016 JL 0 2 4 2 2 2 3 2 2 2 2 G G kG k 3.1478 38.5 34.34 0 2 2 2 3 JL L L JL 4 17.87 Using an appropriate computer code, the graph below is produced. 350 lowest frequency 300 250 200 150 100 50 0 0 5000 10000 torsional spring rate 15000 20000 Chapter 6 6.1. A string is fixed to a mass and a spring at point A and fixed to a wall at point B as shown in Figure 6.21 (a) and a free body diagram of the mass is shown in Figure 6.21 (b). Use a free body diagram approach to develop an expression which allows you to determine the natural frequencies. Approximate the first two natural frequencies by assuming the cable length is l 1 , the spring rate is k A 25 , the mass is m 0.005 , and T / 10,000 Figure 6.21 String with a spring and mass at one end. Solution Using the free body diagram we sum forces vertically and get mvA (0, t ) k Ava (0, t ) T sin Assuming small angles so that sin dy / dx mvA (0, t ) k Av A (0, t ) T v A (0, t ) we can write this as x v A (0, t ) x Using vA X ( x) A cos t B sin t and X x C cos x c D sin x c we determine v A (0, t ) 2 X (0) A cos t B sin t 2C A cos t B sin t k Av A (0, t ) k A X (0) k AC v A X (0, t ) (0, t ) D x x c Therefore m2C A cos t B sin t k AC A cos t B sin t D or k A m C D 2 c k D C At end B we have vB (l , t ) 0 X (l ) C cos l c A m 2 c D sin l c or c A cost B sin t C cos Letting c l c k C A m 2 c sin l c 0 and 2 2c2 this becomes k A m 2 c 2 C cos l sin l 0 The non-trivial solution to this is tan l k A m 2 c 2 Since c T / this becomes tan l T k A m 2 Using the data from the problem statement, this becomes tan 25 50 2 . Solving results in 1 0.6059 , 2 2.5355 Since / c and c T / 100 we get 1 60.59 , 2 253.55 6.2. Assume a fixed-fixed string of length l is subjected to an applied load. Determine an expression for the steadystate forced vibration response if the applied load is; (A) A concentrated force p( x, t ) P(t ) x xo Po x xo (B) A constant velocity (v) moving load P(t ) x vt 0 vt l p ( x, t ) 0 vt l Solution (A) The steady-state forced vibration response is given in Equation (6.27) without the complementary solution as v ( x, t ) 1 n x l nc (t ) sin d 0 Pn ( )sin c n 1 n l l 2 for n 1, 2,3,... Where Pn (t ) is given by Equation (6.24) as Pn (t ) 0l p( x, t )sin n xo n x n x dx 0l P(t ) x xo sin dx P(t )sin l l l Since P(t ) Po , we can write P(t ) Po H (t ) , where H (t ) is the Heaviside function defined by 0 t 0 H (t ) 1 t 0 Therefore 2 Po 1 n xo n x l nc (t ) sin d sin 0 H (t ) sin c n 1 n l l l v ( x, t ) 2 Po l c 2 2 n 1 n xo 1 n x n ct sin sin 1 cos l l l n2 (B) The steady-state response is given by v ( x, t ) 1 n x l nc (t ) sin d 0 Pn ( )sin c n 1 n l l 2 for n 1, 2,3,... Where Pn (t ) is given by Equation (6.24) as Pn (t ) 0l p( x, t )sin n x n x n vt dx 0l P(t ) x vt sin dx P(t )sin l l l Therefore v ( x, t ) 2 c 1 n x l n vt nc (t ) sin sin d 0 P( )sin n l l l n 1 Using P(t ) Po H (t ) v ( x, t ) 2 Po 1 n x l n vt nc (t ) sin d sin 0 H ( ) sin c n 1 n l l l 2P o c n 1 n x sin l n c n c n v n v sin t sin t l l l l 2 n v n c n l l 2 6.3. For the bars shown in Figure 6.22, find the natural frequencies and modes of longitudinal vibrations and show the shapes of the first five modes. All bars have a length L . Figure 6.22 Bars with different end conditions Solution 2 2u E 2u Eg 2u 2 u a t 2 x 2 x 2 x 2 The equation of motion is Assume u( x, t ) X ( x)(t ) and substitute into the equation of motion, giving X ( x) (t ) a 2 (t ) X ( x ) Since separation of variables was assumed, each must be equal to a constant, i.e. (t ) X ( x) a2 2 (t ) X ( x) (t ) 2 (t ) 0 D 2 X (x ) 2 (t ) 0 2 a2 X (x ) 0 D 2 2 where D i (t ) C1eit C2eit C1 cos t i sin t C2 cost i sin t C1 C2 cost i C1 C2 sin t A cost B sint 2 2 D X ( x ) 0 a X ( x) C cos a D i x E sin (a) Free-free ends: @ x 0 : E a a x u X 0 and 0 x x x 0 X C sin x E cos x 0 E 0 so E 0 x a a a a a @x L: E u X 0 and 0 x x x L X L C sin x E cos x C sin (L) 0 C sin 0 x a a a a a a a a For a non-trivial solution sin L a 0 , so Therefore the natural frequencies are L a x or an L where n 1,2,3,... an n = L L n Eg where n 1, 2,3,... The modes are X n ( x ) Cn cos n x a Cn cos n x L u cos n 1,2,3,.. n x n at n at Bn sin An cos L L L The first 5 modes are shown below. 1 0,8 0,6 0,4 n=1 0,2 n=2 0 -0,2 n=3 0 0,2 0,4 0,6 0,8 1 n=4 n=5 -0,4 -0,6 -0,8 -1 (b) Cantilevered beam: u( x, t ) X ( x) (t ) , (t ) A cos t B sin t , X ( x ) C cos Boundary conditions: u x 0 0 X x 0 0 a E cos L a 0 L a n 2 n 1,3,5,... Therefore the natural frequencies are n The modes are an n Eg = 2L 2L a x E sin X (0) 0 C 0 L X 0 E cos a a x x L u 0 x x L where n 1,3,5,... E is arbitrary a x X n ( x ) En sin n x 2a En sin n x 2L u sin n 1,3,5,.. n x n at n at Bn sin An cos 2L 2L 2L The first 5 modes are shown below. 1 0,8 0,6 0,4 n=1 0,2 n=3 0 n=5 -0,2 0 0,2 0,4 0,6 0,8 1 n=7 n=9 -0,4 -0,6 -0,8 -1 (c) Fixed-fixed: u( x, t ) X ( x) (t ) , (t ) A cos t B sin t , X ( x ) C cos Boundary conditions: u x 0 0 X x 0 0 L a L 0 a a x E sin a x X (0) 0 C 0 u xL 0 u x L 0 E sin E sin L a n E is arbitrary n 1, 2,3,... Therefore the natural frequencies are n an n = L L Eg where n 1, 2,3,... The modes are X n ( x ) En sin n x a En sin n x L u n 1,2,3,.. sin n x n at n at Bn sin An cos L 2L 2L The first 5 modes are shown below. 1 0,8 0,6 0,4 n=1 0,2 n=2 0 -0,2 n=3 0 0,2 0,4 0,6 0,8 1 n=4 n=5 -0,4 -0,6 -0,8 -1 6.4. For the bars in Figure 6.23 find the general expressions for the displacement u caused by arbitrary forces P(t ) acting on the bars. Use normal mode expansions. Figure 6.23 End loaded bars with different end conditions Solution 2u 2 2u 1 a p ( x, t ) t 2 x 2 A The equation of motion is n 1 n 1 (a) Free-free with external loading. Assume u( x, t ) f n (t ) X n (t ) and p( x, t ) Pn (t ) X n (t ) where P (t ) L 0 n p( x, t ) X n dx L 0 2 X dx n Substituting into the equation of motion f n 1 n 1 Pn X n A n 1 X n a 2 f n X n" n 1 For free vibrations a 2 X n" 2 X n , resulting in f n 1 X n f nn2 X n" n n 1 1 Pn X n A n 1 Since the series on the left equals the series on the right we have f n n2 f n Pn (t ) A From Problem 1 for the free-free beam we had X n ( x ) Cn cos n x L where Cn 1, n n L Eg where n 1, 2,3,... Assume p( x, t ) P(t ) ( x L) Pn (t ) L 0 n x dx 2 P(t ) cos n L L L 2 n x 0 cos L dx P(t ) x L cos so f n n2 f n 2 P(t ) cos n A L Taking the Laplace transform gives s 2 f n n2 f n 2 P cos n A L s 2cos n P A L 2 n2 f n f n (t ) L 2 n f ( s) A L 0 sinn t P d n -1 2 n A n t sinn t P d Eg 0 or f n (t ) Thus 2 n An 2cos n P A L s 2 n2 fn g E 0 sinn t P d t t u( x, t ) 2 n g E An n 1 0 sinn t P d cos t n x L or 1 u( x, t ) A g E 2 n n x t cos 0 sinn t P d L n 1 n (b) From Problem 1; X n ( x) En sin n x n Eg with En 1, n where n 1,3,5,... 2L 2L Assume p( x, t ) P(t ) ( x L) Pn (t ) L 0 n x dx n 1 2 P (t ) n 2 P (t ) 2L 2 sin 1 L L 2 2 n x L 0 sin 2 L dx P(t ) x L sin so f n n2 f n 1 u( x, t ) 1 A g E n 1 2 2 P (t ) A L 1n n x t sin 0 sinn t P d n 2L n 1 6.5. Find the frequency response functions of the displacement u for the bar and loadings shown in Problem 6.4. Use natural mode expansions. Develop expressions for U *( x, ) if the material is elastic, Inelastic with E* E ' iE " , a Kelvin-type material, and a Maxwell-type material. Solution For a free-free elastic beam 2u 2 2u g a P (t ) t 2 x 2 A Assume P(t ) eit and u( x, t ) U * ( x, i )eit . Substituting into the equation of motion (i ) 2 U * ( x, i ) a 2 2U * ( x, i ) x 2 U * ( x, i) C sin a x D cos u Boundary conditions: E 0 x x 0 a x U * E 0 x x 0 U * C cos x D sin x 0 C x a a a a a x 0 P u E E x A xL 1 U * x EA xL U * 1 L D sin a EA a x x L D cos x a a U ( x, i ) EA sin L a For an inelastic beam, replace E with E *(i) and a with a *(i) so that cos a *(i ) x a *( i ) U * ( x, i ) E *(i ) A sin L a *(i ) For a Kelvin model, replace E *(i) with E i so that cos a *(i ) x a *( i ) U * ( x, i ) E i A sin L a *(i ) Ei so that i E / cos x a *(i ) a *(i ) U * ( x, i ) Ei sin L A a *(i ) i E / (b) Fixed-free beam: 2u 2 2u g a P (t ) t 2 x 2 A for P 1 a 1 EA sin L a * For a Maxwell model, replace E *(i) with C0 Assume P(t ) eit and u( x, t ) U * ( x, i )eit . Substituting into the equation of motion (i ) 2 U * ( x, i ) a 2 2U * ( x, i ) x 2 U * ( x, i) C sin Boundary conditions: a x D cos u x 0 0 a x U *x0 0 sin x D cos x 0 D a a a a a U *x0 C P u E E x A xL 1 U * x EA xL U * L 1 C cos a EA a x x L C sin x a a U ( x, i ) EA cos L a For an inelastic beam, replace E with E *(i) and a with a *(i) so that sin a *(i ) x a *( i ) U * ( x, i ) E *(i ) A cos L a *(i ) For a Kelvin model, replace E *(i) with E i so that sin x a *(i ) a *(i ) U ( x, i ) E i A cos L a *(i ) * For a Maxwell model, replace E *(i) with for P 1 a 1 EA cos L a * Ei so that i E / sin a *(i ) x a *( i ) U * ( x, i ) Ei cos L A a *(i ) i E / D0 6.6. A viscoelastic bar as shown in Figure 6.24 whose properties are specified by the complex modulus E* E i , where E and are constants, is subjected to an end load P(t ) Po cos t . Determine a particular solution for the axial displacement u( x, t ) and the axial stress ( x, t ) using (A) The natural mode method (B) The closed form solution. Figure 6.24 Viscoelastic bar with an end load. Solution (A) A 2u t 2 E* A 2u x 2 p ( x, t ) 2u For free vibrations and E being elastic t 2u t 2 2 E* 2u x 2 E 2u x 2 p ( x, t ) A 0 Assume u( x, t ) X ( x) A cos t B sin t and substitute into the equation of motion. The result is 2 X E d2X X C cos dX 2 E x D sin E x The boundary conditions are x 0: u 0 X 0 Substituting, C 0 and D cos cos E X n Dn sin E L 0 x L: E n x Dn sin x u / x 0 dX / dx 0 L 0 . Since D 0 is the trivial solution we explorer E n x 2L L n 2 d2 Xn dx 2 n 1 n 1 n n E 2L n2 X n ( x) E Assume u ( x, t ) X n ( x)n (t ) and p ( x, t ) X n ( x) f n (t ) , where n 1,3,5,.... l fn p( x, t ) X m dx (t ) 0 l 2 u( x, t ) X n ( x)n (t ) E* 0 X m dx Substituting results in n 1 Recalling that d2 Xn dx 2 n2 E n 1 d 2 Xn dx 2 1 X n f n (t ) A n1 X n ( x) E* 2 1 X ( x ) ( t ) n X nn n n X n fn (t ) E n 1 A n 1 or E* n (t ) E n2 n f n (t ) A For the loading condition given ( P(t ) Po cos t ) n x n 1 ( x L )dx 2 P cos t 2 Po cos t n 2L o sin (1) 2 n 1,3,5,... L L 2 L 2 n x sin dx 0 2L L f n (t ) Po cos t sin 0 E* n (t ) E n2 n (1) n 1 2 2 Po cos t n 1,3,5,.. A L Using the convolution integral gives for the particular solution n (1) n 1 2 2 Po A Ln E* E L 0 E* (t t ') cos t ' dt ' n 1,3,5,.... sin n E n 1 2P u ( x, t ) o A L n x (1) 2 sin E 2L E n 1 n * L 0 sin n E* (t t ') cos t ' dt ' E (B) Assume u( x, t ) U * ( x, i)eit and p( x, t ) Po eit . This results in (i ) 2 U *eit U * C sin E * d 2U * dx 2 E * eit 0 x D cos E* or x d 2U * dx 2 2 E * U* 0 The boundary conditions become C0 D E * cos E * x 0: L Po * E A U * 0 and D x L: Po 1 A E* P dU * *o . Substituting yields dx E A 1 cos E* L So that x cos * E U* * A E cos L * E Po x cos * E u ( x, L ) * A E cos L * E Po eit