Ch+10+HW+Solutions

Ch+10+HW+Solutions - 3 2 Solve for the string response...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
3 2. Solve for the string response, Equation 10.27, given the following initial conditions: (a) y ( x; 0) = 0 , _ y ( x; 0) = g ( x ) (b) y ( x; 0) = sin x , _ y ( x; 0) = 0 (c) y ( x; 0) = 0 , _ y ( x; 0) = sin x (d) y ( x; 0) = sin x , _ y ( x; 0) = sin x . In all cases, sketch the initial conditions and plot the response y ( x;t ) . Solution: Equation 10.27 is y ( x;t ) = 1 X r =1 r 2 mL sin L A r sin L t + B r cos L t ± : Given initial conditions, y ( x; 0) and _ y ( x; 0) ; the constant coe¢ cients are obtained using the or- thogonality condition, R L 0 mY r Y p dx = ± rp ; where Y r ( x ) is the normalized r th modal function, Y r ( x ) = r 2 mL sin L : Applying the initial displacement, y ( x; 0) = 1 X r =1 r 2 mL sin L B r : Multiplying both sides by mY p ( x ) and integrating over 0 to L; we have Z L 0 y ( x; 0) m r 2 mL sin L dx = 1 X r =1 B r Z L 0 m r 2 mL sin L ! r 2 mL sin L ! dx: The only non-zero term on the right-hand side occurs when r = p; B p = Z L 0 y ( x; 0) m r 2 mL sin L dx: (1) A p = L Z L 0 _ y ( x; 0) m r 2 mL sin L dx: (2) (a) y ( x; 0) = 0 , _ y ( x; 0) = g ( x ) Zero initial displacement gives B r = 0 for all r: A r : Substituting _ y ( x; 0) with g ( x ) in Equation (2), we have A p = L m r 2 mL Z L 0 g ( x )) sin L dx:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
9 6. For the uniform cantilever beam in longitudinal vibration shown in Figure 10.29 derive the Figure 10.29: Fixed beam in longitudinal vibration. Solution: The eigenvalue problem is given by (see Section 10.3) d dx EA dU dx ± = ! 2 mU; with boundary conditions, U (0) = 0 U ( L ) = 0 : For this uniform beam, the eigenvalue problem can be written as d 2 U dx 2 + 2 U = 0 ; 2 = ! 2 m EA : The solution to this harmonic equation is given by U ( x ) = C 1 sin + C 2 cos Applying the two boundary conditions yields the two equations for the constants of integration: U (0) = C 2 = 0 U ( L ) = C 1 sin = 0 ; from which sin = 0 since C 1 6 = 0 : Thus, r = L ; r = 1 ; 2 ; 3 ;::::; and 2 r = ! 2 r m EA or ! r = L r EA m = r EA mL 2 :
Background image of page 2
10 CHAPTER 10 CONTINUOUS MODELS Therefore, the modes are given by U r ( x ) = C r sin = C r sin L x; r = 1 ; 2 ; 3 ;:::: C r using the relation Z L 0 mU 2 r dx = 1 : For the graphs of the modes we assume that the coe¢ cients C r = 1 , and that L = 10 : The MATLAB code used for plotting the mode shapes are given below. clear L=10; for r=1:3 x=[0:0.02:L]± ; b=r*pi/L; plot(x, sin(b*x)) hold on end
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
27 13. Consider the uniform system shown in Figure 10.33. Solve for the lowest natural frequency of axial motion of the steel beam given the following parameter values: E = 210 GN/m 2 = 7850 kg/m 3 ; L = 1 : 2 m, k = 100 kN/m, and A = 3 cm 2 : Figure 10.33: Axially vibrating beam with spring boundary.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/26/2012 for the course 650 443 taught by Professor Benaroya during the Fall '11 term at Rutgers.

Page1 / 11

Ch+10+HW+Solutions - 3 2 Solve for the string response...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online