Solutions for Homework#7

Transcription

1 Solutions for Homewor#7 PROBEM (P. 5 on page 77) To a first approximation, a nuclear reactor building can be modeled as a rigid cylinder of radius R=m and height H=6m, with an average mass density of ρ = 65 g/m ( =5% of solid concrete ). This building is supported by a soil with shear wave velocity C s = m/s, mass density ρ s = g/m, and Poisson s ratio ν=/. Using approximate expressions for the foundation stiffnesses, determine the six natural frequencies and modal shapes of this system. SOUTION The nuclear reactor is modeled (to a first approximation!) as follows: z R H y x The mass of the cylinder is calculated as follows: ( ) m = V ρ = π R H ρ = π g {} 6 c c c c The shear modulus, G, of the soil is calculated as: G = ρ C = = 8 N / m {} 6 s s s The six natural frequencies and modal shapes of the system are: Vertical Oscillation Vertical Stiffness K v 6 4 Gs R 4 8 = = = v / 9.6 / N m {} Therefore, the natural frequency for vertical oscillation is: Its mode shape is a any scalar. v 9 Kv.6 = = = 6 m rad /sec {4} Torsional Oscillation

2 Torsional Stiffness K T 6 G R s = = = 7.68 Nm {5} The moment of inertia for the circular cross-section of the cylinder is defined as: J T m R c 9 = = = 9.4 gm {6} Therefore, the natural frequency for torsional oscillation is: Its mode shape is a any scalar. T KT 7.68 = = = 9 J 9.4 c 8.55 rad /sec {7} Horizontal and Rocing Oscillation (coupled) Recall the solution of P. in the lecture note. +z -z mu J θ H R Jc = m + 4 u z = u +θ z ( ) K x u K R θ u gb We formulate the equation of motion with respect to the center of mass: Horizontal Equuilibrium: mu + K u = {8} x b H Moment Equilibrium: Jc θ + KRθ Ku x b = {9} We substitute in equations {8} and {9}, the displacement of the base u b, as a function of the displacement of the center of mass u and the angle of rotation, θ, i.e.: H H ub u = = u θ {} Note that the frequencies are the same for x and y direction, due to symmetry of the structure.

3 The equation of motion for the -dof system is now formulated as follows: H mu K u K θ + x x = H H J K K u K 4 c θ + Rθ x + xθ = {} Equation {} can be expressed in matrix form as follows: H Kx Kx m u u J + = c θ H H θ Kx KR K + x 4 {} We now calculate: H R 6 Jc = m + = = 8.85 gm Gs R 88 9 Kx = Ky = = = 7.8 N / m ν / {} {4} 8 G R 88 / 6 x y s KR = KR = = = 5.76 Nm ( ν ) ( ) {5} The eigenvalue problem is formulated as follows: ( ) For the solution of {6} to be non-trivial, we demand: K M = K M f = {6} i = {7} =

4 The solution of the quadratic equation in {7}, has the following solutions: = 8.89 rad /sec = 7.66 rad /sec To define the modal shapes, we set φ n =., and solve for φ n, using the eigenvalues calculated above. We therefore have: n 7.8. = φn n 4 {8} From the first equation in {8}, we evaluate φ n as follows: φn = Substituting in {9} n=,, we evaluate: n {9} φ =.6 rad φ = -.96 rad The modal shapes for the coupled horizontal rocing motion are schematically shown below:.. f =, f =

5 PROBEM (P. 7 on page 77) A simply supported massless beam has two masses lumped, as shown. Determine a) The frequencies and modal shapes b) The dynamic response that the system would have if at t = gravity were to be suddenly switched on (physically, but not mentally impossible ). Basically, we are asing here to compute the modal contributions to the response, to write down the expressions, and perhaps setch the response of one of the masses. m m EI EI SOUTION Recall the deflection shape of the beam, from Problem, Homewor No.. P a b P b x b x u( x) = x a 6EI P a x a x = x a 6EI {} The structure under consideration is shown in the figure below. The system has dynamic dof s, namely u and u. m m / / / m = m m = m The mass matrix of the structure is simply: m M = m {}

6 To evaluate the stiffness matrix, we first construct the flexibility matrix as follows: We apply a unit load P =. at the rst degree of freedom, and calculate the deflection at the rst degree of freedom, f, and at the nd degree of freedom, f. P =. f f For a = / and b = /, we evaluate the flexibility coefficients by substituting in equation {}: f / / / / 4 = u( /) = = 6EI 4 EI / / / / 7 f = u( /) = 6EI = 486 EI We apply a unit load P =. at the nd degree of freedom, and calculate the deflection at the rst degree of freedom, f, and at the nd degree of freedom, f. P =. f f For a = / and b = /, we evaluate the flexibility coefficients by substituting in equation {}: f / / / / 7 = u( /) = = 6EI 486 EI / / / / 4 f = u( /) = 6EI = 4 EI

7 The flexibility matrix of the structure is formulated as follows: 4 7 f f F = f f = {} EI Therefore, the stiffness matrix is: 64.8 EI 4.5 F = K = =.5 4 {} The eigenvalue problem is formulated as follows: ( ) For the solution of {4} to be non-trivial, we demand: K M f = {4} i K M = 64.8 EI 4.5 m.5 4 m = {5} 4 λ.5 m = where λ =.5 4 λ 64.8 EI The solution of the quadratic equation in {5}, has the following solutions: λ λ = 5.67 = 9.7 =. = 4.6 EI m EI m {6} To define the modal shapes, we set φ n =., and solve for φ n, using the eigenvalues calculated above. We therefore have: 4 λ λ = φn {7} From the first equation in {7}, we evaluate φ n as follows: φ n 4 λn = {8}.5 Substituting in {8} n=,, we evaluate:

8 φ = m, φ =.49 m The modal shapes are schematically shown in the figure below:.. f = f = We now calculate the response of the system, it at t= a step load of amplitude P i = -m i g and infinite duration, is applied to the i th degree of freedom. We first calculate the modal masses (µ i ) and modal stiffnesses (κ i ) as follows: m. µ = {..477} =.455m m.477 µ κ m. = {..49} =.m m EI 4.5. EI = {..477} =.477 {9} κ 64.8 EI 4.5. = {..49} EI =.49 The modal forces are calculated as follows: π π mg = {..477} =.46mg mg mg = {..49} =.98mg mg {} The response of a -dof system with zero initial conditions, u = u =, due to a step load of infinite duration is: π j qj( t) = ( cos( j t) ) {} κ j Therefore, the response of the system, is calculated using modal superposition, as follows:

9 u u ( t) = = j qj( t) u f j=. mg. mg =.477 EI.49 EI cos.45 cos ( ( t) ) ( t) ( ) In what follows, the displacement time history of the masses is plotted. u (t), u (t) t As it can be readily observed, the second mode mainly contributes to the response as the oscillation of the two masses is essentially in phase.

10 PROBEM (P. 8 on page 78) In the structure shown, determine all frequencies and modal shapes (but thin before you proceed ) m m m m (You may verify your answers with MATAB, but you should also wor out this problem by hand.) SOUTION Recall the solution of Problem, Homewor No.. The system above can be analyzed using symmetry and antisymmetry concepts, as follows: Antisymmetric mode The two left masses are moving in phase with the two right masses, and due to the symmetry of the structure, the middle point is not moving. Therefore, the system can be analyzed as follows: m m For this system, the modal frequencies are calculated as follows: π n = sin ( n ) m 4N which is derived on pages to 4 in the lecture note for a discrete shear beam. {} Therefore, for N =, substituting in {} n =,, we calculate: π = sin ( ) =.54 m 8 m π = sin ( 4 ) =.7 m 8 m The modal shapes are now calculated as follows: For n =, we have: π φin = cos ( i )( n ) N {}

11 φ φ π = cos ( )( ) =. 4 π = cos ( )( ) =.77 4 and for n =, we have: φ φ π = cos ( )( 4 ) =. 4 π = cos ( )( 4 ) =.77 4 Symmetric mode The two left masses are moving to the opposite direction from the two right masses, and due to the symmetry of the structure, the system can be analyzed as follows: m m The mass and stiffness matrix of the structure are shown below: K = m M = The eigenvalue problem is formulated as follows: ( ) For the solution of {4} to be non-trivial, we demand: K M f = {} i K M = m m = {4} λ λ m = where λ = The solution of the quadratic equation in {4}, has the following solutions:

12 λ λ =. =.8 =.78 =.88 m m {5} To define the modal shapes, we set φ n =., and solve for φ n, using the eigenvalues calculated above. We therefore have: λ. λ = φn {6} From the first equation in {6}, we evaluate φ n as follows: φn = λ n {7} Substituting in {7} n=,, we evaluate: φ = -4.44, φ =.44 Therefore, the spectral and modal matrices of the structure are the following: ( ) Ω = diag = i m F = { f f f f 4} = Note-.The eigenvalues and eigenvectors could have been obtained directly from the following EVP: ( ) K M f = where: K = M = m yet the problem demands much more computational effort. i