CHAPTER 1 Problem 1.1 If k e is the effective stiffness, f k u S e = 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: k f u k k e S = = + 1 2 Equation of motion: mu k u p t e && ( ) + = Problem 1.2 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 u 1 and u 2 , u u u = + 1 2 (b) 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 k f k f k S e S S = + 1 2 ⇒ 1 1 1 1 2 k k k e = + ⇒ k k k k k e = + 1 2 1 2 Equation of motion: mu k u p t e && ( ) + = Problem 1.3 k 1 k 2 u k 1 k 3 k 2 + 1 k 3 m m m k e Fig. 1.3(a) Fig. 1.3(b) Fig. 1.3(c) ⇓ ⇓ 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. 1.3(b). Second, using the result of Problem 1.2, the series arrangement of springs in Fig. 1.3(b) is replaced by a single spring, as shown in Fig. 1.3(c): 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 u p t e && ( ) + = . Problem 1.4 1. Draw a free body diagram of the mass. O L m θ T mg sin θ mg cos θ 2. Write equation of motion in tangential direction. Method 1: By Newton’s law. 0 sin sin sin = + = − = − θ θ θ θ θ mg mL mL mg ma mg & & & & (a) This nonlinear differential equation governs the motion for any rotation θ . Method 2: Equilibrium of moments about O yields θ θ sin 2 mgL mL − = & & or 0 sin = + θ θ mg mL & & 3. Linearize for small θ . For small θ , θ θ ≈ sin , and Eq. (a) becomes 0 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = + θ θ θ θ L g mg mL & & & & (b) 4. Determine natural frequency. L g n = ω Problem 1.5 1. Find the moment of inertia about O. From Appendix 8, I mL m L mL 0 2 2 2 1 12 2 1 3 = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2. Draw a free body diagram of the body in an arbitrary displaced position. mg y x θ L/2 3. Write the equation of motion using Newton’s second law of motion. M I 0 0 = ∑ && θ − = mg L mL 2 1 3 2 sin && θ θ mL mgL 2 3 2 0 && sin θ θ + = (a) 4. Specialize for small θ . For small θ , sin θ ≅ θ and Eq. (a) becomes mL mgL 2 3 2 0 && θ θ + = && θ + = 3 2 0 g L (b) 5. Determine natural frequency. ω n g L = 3 2 Problem 1.6 1. Find the moment of inertia about about O . I r d A L 0 2 0 = ∫ ρ = = = ∫ ρ α ρ α r r dr L mL L 2 0 4 2 4 1 2 ( ) x L α r 2. Draw a free body diagram of the body in an arbitrary displaced position. mg y x θ 2L/3 3. Write the equation of motion using Newton’s second law of motion. M I 0 0 = ∑ && θ − = mg L mL 2 3 1 2 2 sin && θ θ mL mgL 2 2 2 3 0 && sin θ θ + = (a) 4. Specialize for small θ . For small θ , sin θ ≅ θ , and Eq. (a) becomes mL mgL 2 2 2 3 0 && θ θ + = or && θ + = 4 3 0 g L (b) 5. Determine natural frequency. ω n g L = 4 3 In each case the system is equivalent to the spring- mass system shown for which the equation of motion is 0 g = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ku u w & & w u k 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 E I k EI L S = ⇒ = 3 3 48 48 Cantilever beam: u f L E I k EI L S = ⇒ = 3 3 3 3 Clamped beam: u f L E I k EI L S = ⇒ = 3 3 192 192 Problem 1.7 Draw a free body diagram of the mass: mü p ( t ) u f S Write equation of dynamic equilibrium: mu f p t S && ( ) + = (a) Write the force-displacement relation: f AE L u S = F H G I K J (b) Substitute Eq. (b) into Eq. (a) to obtain the equation of motion: mu AE L u p t && ( ) + F H G I K J = Problem 1.8 Show forces on the disk: O θ f S R Write the equation of motion using Newton's second law of motion: − = f I S O && θ where I m R O = 2 2 (a) Write the torque-twist relation: f GJ L S = F H G I K J θ where J d = π 4 32 (b) Substitute Eq. (b) into Eq. (a): I GJ L O && θ θ + F H G I K J = 0 or, mR d G L 2 4 2 32 0 F H G I K J + F H G I K J = && θ π θ Problems 1.9 through 1.11 In each case the system is equivalent to the spring- mass system shown for which the equation of motion is w u ku g F H G I K J + = && 0 w u k 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 Problem 1.12 L EI w k Fig. 1.12a Static Equlibrium Deformed position u u δ st Undeformed position Fig. 1.12b p(t) w mu .. f s f s Fig. 1.12c 1. Write the equation of motion. Equilibrium of forces in Fig. 1.12c gives ) ( t p w f u m s + = + & & (a) where f s = k e u (b) The equation of motion is: ) ( t p w u k u m e + = + & & (c) 2. Determine the effective stiffness . u k f e s = (d) where beam spring u δ δ + = (e) beam beam spring s k k f δ δ = = (f) Substitute for the δ ’s from Eq. (f) and for u from Eq. (d): ( ) 3 3 48 / 48 L EI k L EI k k k k kk k k f k f k f e beam beam e beam s s e s + = + = + = 3. Determine the natural frequency. m k e n = ω simply supported Problem 1.13 Compute lateral stiffness: 1 h 3 EI /h c 3 k k EI h EI h column c c = × = × = 2 2 3 6 3 3 Equation of motion: mu ku p t && ( ) + = Base fixity increases k by a factor of 4. Problem 1.14 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 k EI h EI h k k EI h c c c 11 3 3 21 31 2 2 3 6 3 = = = = u u u 2 1 3 1 0 = = = , k 22 k 23 k EI h EI h EI h k EI h EI h k EI h c c c c c c 22 32 12 2 3 4 2 5 2 2 3 = + = = = = ( ) ( ) u u u 3 1 2 1 0 = = = , k 13 k 23 k 33 k EI h EI h EI h k EI h EI h k EI h c c c c c c 33 23 13 2 3 4 2 5 2 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 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: EI h h h h h h h h h u u u f c S 3 2 2 2 2 1 2 3 6 3 3 3 5 3 5 0 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 [ ] [ ] ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = = = 2 2 2 2 3 00 3 0 3 5 5 3 3 6 h h h h h EI h h h EI h EI c c t c tt k k 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 k EI h k EI h c c c c c c = − ⋅ − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − = 6 3 3 24 5 1 1 5 3 3 6 3 3 3 3 3 3 3 5. Equation of motion. m u E I h u p t c && ( ) + = 3 3 Problem 1.15 1 k h I c I c I = I / b c 2 h 2 Define degrees of freedom (DOF): 1 2 3 Form structural stiffness matrix: u u u 1 2 3 1 0 = = = , k 11 k 21 k 31 k EI h EI h c c 11 3 3 2 12 24 = = k k EI h c 21 31 2 6 = = u u u 2 1 3 1 0 = = = , k 12 k 22 k 32 k EI h EI h EI h EI h EI h c b c c c 22 4 4 2 4 5 = + = + = ( ) k EI h EI h b c 32 2 2 2 = = ( ) k EI h c 12 2 6 = u u u 3 1 2 1 0 = = = , k 13 k 23 k 33 k EI h EI h EI h EI h EI h c b c c c 33 4 4 2 4 5 = + = + = ( ) k EI h EI h b c 23 2 2 2 = = ( ) k EI h c 13 2 6 = 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 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 I b 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 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + & &