Forcing Interpolation

Forcing Interpolation - CE 573 Structural Dynamics...

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CE 573: Structural Dynamics Numerical Evaluation of Dynamic Response – Forcing Interpolation Example Given: Mass of tank , stiffness k = 10 kips/in, undamped natural period ( 2 0.2533 kip-sec /in m = 1.0 sec T = 6.283 rad/sec ω = ), and 0.05 ξ = . Tank starts at rest and is subjected to half-cycle sine pulse F(t) as shown. Required: Determine the response x(t) from 0.0 sec t = to 1.0 sec t = using linear interpolation of F(t) with . 0.1 sec t ∆= Solution: 1. Preliminary Calculations – () ( ) ( ) () () ( ) () () ( ) 0.05 6.283 0.1 2 0.9691; 1 6.275 rad/sec. sin sin 6.275 0.1 0.5871; cos sin 6.275 0.1 0.8095. t d dd ee tt ξω ωω −∆ == = = ∆= ⎡ = = ⎣⎦ 2. Recurrence Formulas: ( ) ( ) ( ) ( ) 2 0.05 6.283 rad/s 6.275 rad/s 3 0.5871 6.275 rad/s 1 2 0.05 0.05 6.283 rad/s 2 0.05 1 10 kip/in 6.275 rad/s 6.275 rad/s 0.1 s 6.283 rad/s 0.1 s 0.9691* *0.5871 0.8095 0.8130 0.9691 90.67 10 s. 0.9691* *0.5871 1 A B C =+ = =∗ =− + ( ) ( ) ( ) ( ) {} 2 20 .05 6.283 rad/s 0.1 s 3 in kip 2 0.05 1 2 0.05 2 0.05 3 1 in 10 kip/in kip 6.275 rad/s 0.1 s 6.283 rad/s 0.1 s 6.283 rad/s 0.1 s *0.8095 = 12.350 10 0.9691* *0.5871 *0.8095 1 6.352 10 D ⎡⎤ + ⎢⎥ × + = ×
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() ( ) ( ) ( ) 2 2 6.283 rad/s 6.275 rad/s 0.05 6.283 rad/s 6.275 rad/s 6.283 rad/s 6.283 rad/s 0.05 1 1 6.275 rad/s 0.1 10 k/in 0.1 s 6.275 rad/s 0.9691* *0.5871 3.579 rad/s 0.9691* 0.8095 *0.5871 0.7560 * 0.9691* *0.5871 A B C ′′=− =− ′′ = =+ + ( ) ( ) {} 1 s 0.1 s in kip-s 0.05 6.283 rad/s 1 in 6.275 rad/s kip-s 10 k/in 0.1 s *0.8095 0.1709 * 0.9691* *0.5871 0.8095 1 0.1870 D ⎧⎫ ⎡⎤ ⎨⎬ ⎢⎥ ⎣⎦ ⎩⎭ = + + = 3. Numerical solution (via spreadsheet) CE 573: Example on Forcing Interpolation A' B' (sec) C' (in/kip) D' (in/kip) 8.130E-01 9.067E-02 1.2350E-02
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Forcing Interpolation - CE 573 Structural Dynamics...

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