1 CHAPTER 1 Problem 1.1 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Fig. P1.1. Figure P1.1 Solution: If k e is the effective stiffness, f k S e u = f S f S k u 2 k u 1 k 1 k 2 u Equilibrium of forces: f k k u S = + ( 1 ) 2 Effective stiffness: + 2 k f u k k e S = = 1 Equation of motion: mu k u p t e  ( ) + = Problem 1.2 2 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Fig. P1.2. Figure P1.2 Solution: If k e is the effective stiffness, f k u = ( S e a) f S k 1 k 2 u If the elongations of the two springs are 1 u and , 2 u u u u = + (b 1 2 ) Because the force in each spring is f S , f k u S = 1 1 f k u S = 2 2 (c) Solving for u 1 and u 2 and substituting in Eq. (b) gives f 1 1 1 f f k k k S e S S = + 1 2 ⇒ 1 2 k k k e = + ⇒ k k k k k e = + 1 2 1 2 Equation of motion: mu  + k e u p ( t = ) . 3 Problem 1.3 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Fig. P1.3. Figure P1.3 Solution: k 1 k 2 k 3 m ⇓ Figure P1.3a k 1 k 3 k 2 + 1 m ⇓ Figure P1.3b u m k e Figure P1.3c This problem can be solved either by starting from the definition of stiffness or by using the results of Problems P1.1 and P1.2. We adopt the latter approach to illustrate the procedure of reducing a system with several springs to a single equivalent spring. First, using Problem 1.1, the parallel arrangement of k 1 and k 2 is replaced by a single spring, as shown in Fig. 1.3b. Second, using the result of Problem 1.2, the series arrangement of springs in Fig. 1.3b is replaced by a single spring, as shown in Fig. 1.3c: 1 1 1 1 2 3 k k k k e = + + Therefore the effective stiffness is k k k k k k k e = + + + ( ) 1 2 3 1 2 3 The equation of motion is mu  + = k e u p ( t ) . 4 Problem 1.4 Derive the equation governing the free motion of a simple pendulum that consists of a rigid massless rod pivoted at point O with a mass m attached at the tip (Fig. P1.4). Linearize the equation, for small oscillations, and determine the natural frequency of oscillation. Figure P1.4 Solution: 1. Draw a free body diagram of the mass. m O θ L T m g sin θ m g cos θ 2. Write equation of motion in tangential direction. Method 1: By Newton’s law. - = m m g sin θ a - = m m g s in θ L θ  (a) mL θ θ  + = m g sin 0 This nonlinear differential equation governs the motion for any rotation θ . Method 2: Equilibrium of moments about O yields 2 g sin mL m L θ θ = -  or g sin 0 mL m θ +  θ = 3. Linearize for small θ . For small θ , sin θ ≈ θ , and Eq. (a) becomes g mL m θ +  0 θ = g 0 L θ θ + = ( ) | │ ( )  (b) 4. Determine natural frequency. g ω n = L 5 Problem 1.5 Consider the free motion in the xy plane of a compound pendulum that consists of a rigid rod suspended from a point (Fig. P1.5). The length of the rod is L , and its mass m is uniformly distributed. The width of the uniform rod is b and the thickness is t . The angular displacement of the centerline of the pendulum measured from the y -axis is denoted by θ ( t ). (a) Derive the equation governing θ ( t ). (b) Linearize the equation for small θ . (c) Determine the natural frequency of small oscillations. Figure P1.5 Solution: 1. Find the moment of inertia about O. From Appendix 8, 2 1 2 2 0 1 12 2 3 L I mL m mL = + = ( ) | │ ( ) 2. Draw a free body diagram of the body in an arbitrary displaced position. mg y x θ L/2 y m g x L/2 3. Write the equation of motion using Newton’s second law of motion. 0 0 M I θ = ∑  2 1 g sin 2 3 L m θ θ - =  mL ) 2 g sin 0 3 2 mL m L θ θ + =  (a 4. Specialize for small θ . For small θ , sin θ ≅ θ and Eq. (a) becomes θ 2 g 0 3 2 mL m L θ + =  b) 3 g 0 2 L θ θ + =  ( 5. Determine natural frequency. L n g 2 3 = ω 6 Problem 1.6 Repeat Problem 1.5 for the system shown in Fig. P1.6, which differs in only one sense: its width varies from zero at O to b at the free end. Figure P1.6 Solution: 1. Find the moment of inertia about about O. I r L 0 2 0 = ∫ ρ d A α r ) = = = ∫ ρ ρ α r r d L mL L 2 0 4 2 4 1 2 ( x L α r L r 2. Draw a free body diagram of the body in an arbitrary displaced position. mg y x θ 2L/3 x y m g 2L/3 3. Write the equation of motion using Newton’s second law of motion. 0 0 2 2 g sin 3 2 M I L m θ θ θ = - ∑   1 m L = 0 sin 3 g 2 2 2 = + θ θ L m mL   (a) 4. Specialize for small θ . For small θ , sin θ ≅ θ , and Eq. (a) becomes 2 2 g 0 2 3 mL m L θ θ + =  or 4 g 0 3 L θ θ + =  (b) 5. Determine natural frequency. L n g 3 4 = ω In each case the system is equivalent to the spring- mass system shown for which the equation of motion is 0 g u ku + = ( ) | | ( )  w w u k w The spring stiffness is determined from the deflection u under a vertical force f S applied at the location of the lumped weight: Simply-supported beam: u f L EI k EI L S = ⇒ = 3 3 48 48 Cantilever beam: u f L EI k EI L S = ⇒ = 3 3 3 3 Clamped beam: u f L EI k EI L S = ⇒ = 3 3 192 192 7 Problem 1.7 Develop the equation governing the longitudinal motion of the system of Fig. P1.7. The rod is made of an elastic material with elastic modulus E ; its cross-sectional area is A and its length is L . Ignore the mass of the rod and measure u from the static equilibrium position. Figure P1.7 Solution: Draw a free body diagram of the mass: mü p ( t ) u f S Write equation of dynamic equilibrium: ( ) S mu f p t + =  (a) Write the force-displacement relation: b) S AE f u L = ( ) | │ ( ) ( Substitute Eq. (b) into Eq. (a) to obtain the equation of motion: = ( ) AE mu u p t L + ( ) | │ ( )  8 Problem 1.8 A rigid disk of mass m is mounted at the end of a flexible shaft (Fig. P1.8). Neglecting the weight of the shaft and neglecting damping, derive the equation of free torsional vibration of the disk. The shear modulus (of rigidity) of the shaft is G . Figure P1.8 Solution: Show forces on the disk: O θ f S R Write the equation of motion using Newton’s second law of motion: S O f I θ - =  where 2 2 O m R I = (a) Write the torque-twist relation: S GJ f L θ = ( ) | │ ( ) where J d = π 4 32 (b) Substitute Eq. (b) into Eq. (a): θ = 0 O GJ I L θ + ( ) | │ ( )  or, θ = 2 4 0 2 32 mR d G L π θ + ( ) ( ) | │ | │ ( ) ( )  9 Problems 1.9 through 1.11 Write the equation governing the free vibration of the systems shown in Figs. P1.9 to P1.11. Assuming the beam to be massless, each system has a single DOF defined as the vertical deflection under the weight w . The flexural rigidity of the beam is EI and the length is L . Solution: In each case the system is equivalent to the spring- mass system shown for which the equation of motion is 0 g u ku + = ( ) | │ ( )  w w u k w The spring stiffness is determined from the deflection u under a vertical force f S applied at the location of the lumped weight: Simply-supported beam: u f L k EI L S = ⇒ = 3 3 48 EI 48 Cantilever beam: I 3 3 3 3 S f L u k EI L = ⇒ = E Clamped beam: 3 3 192 192 S f L u k EI L = ⇒ = EI 10 Problem 1.12 Determine the natural frequency of a weight w suspended from a spring at the midpoint of a simply supported beam (Fig. P1.12). The length of the beam is L , and its flexural rigidity is EI . The spring stiffness is k . Assume the beam to be massless. Figure P1.12 Solution: L EI w k Simply supported w Figure 1.12a Static Equlibrium Deformed position u u δ st Undeformed position Figure 1.12b p(t) w mu .. f s f s p ( t ) w Figure 1.12c 1. Write the equation of motion. Equilibrium of forces in Fig. 1.12c gives a) ( ) S mu f p t + = +  w ( where b) S e f k u = ( The equation of motion is: c) ( ) e mu k u p t + = +  w ( 2. Determine the effective stiffness. d) u k f e s = ( where e) spring beam u δ δ = + ( spring beam beam S f k k δ δ = = (f) Substitute for the δ ’s from Eq. (f) and for u from Eq. (d): beam S S S e f f f k k k = + beam beam e kk k k k = + ( ) 3 3 48 48 e k EI L k EI k L = + 3. Determine the natural frequency. m k e n = ω 11 Problem 1.13 Derive the equation of motion for the frame shown in Fig. P1.13. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is m ; otherwise, assume the frame to be massless and neglect damping. By comparing the result with Eq. (1.3.2), comment on the effect of base fixity. Figure P1.13 Solution: Compute lateral stiffness: h 3 EI /h c 3 1 column 3 3 3 6 2 2 c c EI EI k k h h = × = × = Equation of motion: mu ku p t  ( ) + = Base fixity increases k by a factor of 4. 12 Problem 1.14 Write the equation of motion for the one-story, one-bay frame shown in Fig. P1.14. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is m ; otherwise, assume the frame to be massless and neglect damping. By comparing this equation of motion with the one for Example 1.1, comment on the effect of base fixity. Figure P1.14 Solution: 1. Define degrees-of-freedom (DOF). 1 2 3 2. Reduced stiffness coefficients. Since there are no external moments applied at the pinned supports, the following reduced stiffness coefficients are used for the columns. Joint rotation: L 1 3 2 EI L 3 2 EI L 3 EI L EI Joint translation: 1 3 2 EI L 3 3 EI L 3 3 EI L EI L 3. Form structural stiffness matrix. u u u 1 2 3 1 = = = , 0 k 21 k 31 2 1 3 1, 0 u u u = = = k 22 k 23 k c EI h EI h EI h c c 22 3 4 2 5 = + = ( ) k EI h EI h c c 32 2 2 = = ( ) k EI h c 12 2 3 = u u u 0 3 1 2 1 = = = , k 13 k 23 k 33 k EI h EI h EI h c c 33 3 4 2 5 = + = ( ) c k EI h EI h c c 23 2 2 = = ( ) k EI h c 13 2 3 = Hence k = ┌ └ │ │ │ ┐ ┘ │ │ │ EI h h h h h h h h h c 3 2 2 2 2 6 3 3 3 5 3 5 13 4. Determine lateral stiffness. The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3: ┌ 6 3 h h 3 ┐( u ) ( f EI 1 S ) c │ ││ │ │ │ 3 3 5 h h 2 2 h u } = │ │ { { 0 h 2 } │ 2 2 │ │ │ │ │ └ 3 5 h h h ┘( u 3 ) ( 0 ) First partition k as k k k k k = ┌ └ │ │ │ ┐ ┘ │ │ │ = ┌ └ │ ┐ ┘ │ EI h h h h h h h h h c tt t t 3 2 2 2 2 0 0 00 6 3 3 3 5 3 5 where [ ] = 3 6 h EI c tt k [ ] = 3 0 3 3 h h h EI c t k │ │ ┘ ┐ │ │ └ ┌ = 2 2 2 2 3 00 5 5 h h h h h EI c k Then compute the lateral stiffness k from T t t tt k 0 1 00 0 k k k k - - = Since k 00 1 24 5 1 1 5 - = - - ┌ └ │ ┐ ┘ │ h EI c we get [ ] k EI h EI h h h h EI EI h h h c c c c = -  - - ┌ └ │ ┐ ┘ │  ┌ └ │ ┐ ┘ │ 6 3 3 24 5 1 1 5 3 3 3 3 3 [ ] k EI h c = - 6 3 3 k EI h c = 3 3 5. Equation of motion. = m u E I h u p t c  ( ) + 3 3 14 Problem 1.15 Write the equation of motion of the one-story, one-bay frame shown in Fig. P1.15. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is m ; otherwise, assume the frame to be massless and neglect damping. Check your result from Problem 1.15 against Eq. (1.3.5). Comment on the effect of base fixity by comparing the two equations of motion. Figure P1.15 Solution: 1 k h I c I c I = I / b c 2 h 2 Define degrees-of-freedom (DOF): 3 1 2 Form structural stiffness matrix: u u u 1 2 3 1 = = = , 0 k 11 k 21 k 31 11 3 12 24 2 c EI EI k h = = 3 c h 21 31 2 6 c EI k k h = = u u u 2 1 3 1 = = = , 0 k 12 k 22 k 32 22 4 4 4 (2 ) c b c c EI EI EI EI EI k h h h h = + = + = 5 c h 32 2 (2 ) 2 b c EI EI k h h = = 12 2 6 c EI k h = u u u 3 1 1 = = = , 2 0 k 13 k 23 k 33 33 4 4 4 (2 ) c b c c EI EI EI EI EI k h h h h = + = + = 5 c h 23 2 (2 ) 2 b EI EI k h = = c h 13 2 6 c EI k h = Hence │ │ │ ┘ ┐ │ │ │ └ ┌ = 2 2 2 1 2 2 1 2 3 5 6 5 6 6 6 24 h h h h h h h h h EI c k The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3: │ ) │ } ) │ ( │ { ( = │ ) │ } ) │ ( │ { ( │ │ │ ┘ ┐ │ │ │ └ ┌ 0 0 5 6 5 6 6 6 24 3 2 1 2 2 2 1 2 2 1 2 3 S c f u u u h h h h h h h h h EI 15 First partition k as │ ┘ ┐ │ └ ┌ = │ │ │ ┘ ┐ │ │ │ └ ┌ = 00 0 0 2 2 2 1 2 2 1 2 3 5 6 5 6 6 6 24 k k k k k T t t tt c h h h h h h h h h EI where k tt c EI h = 3 24 k t c EI h h h 0 3 6 6 = │ │ ┘ ┐ │ │ └ ┌ = 2 2 2 1 2 2 1 2 3 00 5 5 h h h h h EI c k Then compute the lateral stiffness k from k tt t t T = - - k k k k 0 00 1 0 Since │ │ ┘ ┐ │ │ └ ┌ - - = - 5 5 99 4 2 1 2 1 1 00 c EI h k we get [ ] 3 3 3 2 1 2 1 3 3 11 120 ) 11 144 24 ( 6 6 5 5 99 4 6 6 24 h EI h EI h h h EI EI h h h h EI h EI k c c c c c c = - = │ ┘ ┐ │ └ ┌  │ │ ┘ ┐ │ │ └ ┌ - -  - = This result can be checked against Eq. 1.3.5: │ │ ) ) | | ( ( + + = 4 12 1 12 24 3 ρ ρ h EI k c Substituting ρ I b I = c 4 = 1 8 gives 3 3 8 1 8 1 3 11 120 11 5 24 4 12 1 12 24 h EI h EI h EI k c c c = │ ) ) | ( ( = │ │ ) ) | | ( ( + + = Equation of motion: ) ( 11 120 3 t p u h EI u m c = │ │ ) ) | | ( ( +  