Ch+10+HW+Solutions

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

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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:

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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 :
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

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

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

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