singularity functions

5 4 4 6 2000 20 15 20 24 24 4 20c1 c2 2718750 3

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Unformatted text preview: de No. 33 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 4 (cont’d) Boundary conditions: y = 0 at x = 4 ft and at x = 20 ft EIy (4) = 0 = −2000 (4) 4 2000 4 − 15 + 24 24 + 23,437.5 4 − 4 6 2000 20 − 15 ( 20) + 24 24 4 ∴ 20C1 + C2 = −2,718,750 3 + C1 (4) + C2 0 0 ∴ 4C1 + C2 = 21,333.3 EIy (20) = 0 = −2000 4 4 + 23,437.5 20 − 4 6 (23d) 3 + C1 (20) + C2 (23e) 17 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions Slide No. 34 ENES 220 ©Assakkaf Example 4 (cont’d) From Eqs 23d and 23e, the constants of integrations are found to C1 = 168,588.54 and C2 = −653,020.88 Therefore, the elastic curve is given by y ( x) = 1 ( x) 4 2000 x − 15 + − 2000 24 24 EI 4 + 3 23,437.5 x − 4 6 + 168,588.5 x − 653,020.9 (23f) LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions Slide No. 35 ENES 220 ©Assakkaf Example 4 (cont’d) The def...
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This document was uploaded on 02/04/2014.

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