singularity functions

5 lb fy 0 r1 r2 200015 30000 r2 65625 lb slide

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Unformatted text preview: LARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 4 (cont’d) y 2000 lb/ft x 11 ft 4 ft 5 ft R1 + ∑M R2 2 Find the reactions: = 0; R1 (16 ) − 2000(15)(5 + 15 )=0 2 ∴ R1 = 23,437.5 lb + ↑ ∑ Fy = 0; R1 + R2 − 2000(15) = 30,000 ∴ R2 = 6,562.5 lb Slide No. 31 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 4 (cont’d) A single expression for the bending moment can be obtained using the singularity functions: y 2000 lb/ft x 4 ft R1 11 ft 5 ft M (x ) = − R2 2 2000( x ) 2000 x − 15 + 2 2 2 + 23,437.5 x − 4 1 (23a) 16 Slide No. 32 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 4 (cont’d) The elastic curve is found by integrating Eq. 23 twice 2000 x 2 2000 x − 15 EIy′′ = M (x ) = − + 2 2 2 + 23,437.5 x − 4 3 EIy′ = EIθ = −2000 23,437.5 x − 4 x 3 2000 x − 15 + + 6 6 2 x 4 2000 x − 15 EIy = −2000 + 24 24 4 + 1 2 + C1 (23b) 3 23,437.5 x − 4 + C1 x + C2 6 (23c) Sli...
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