singularity functions

40 lecture 16 beams deformation by singularity

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Unformatted text preview: ORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 5 (cont’d) Singularity functions to describe the bending moment: y M A R A x L Mx +M x−L 2L or M ( x) = − E, I C B L R B EIy′′ = − Mx +M x−L 2L 0 0 (24a) (24b) 20 Slide No. 40 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 5 (cont’d) Integrating Eq. 24.b twice, we get Mx 0 EIy′′ = − +M x−L 2L Mx 2 1 (24c) EIy′ = − + M x − L + C1 4L 2 Mx 3 M x − L (24d) EIy = − + + C1 x + C2 12 L 2 Slide No. 41 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Example 5 (cont’d) Boundary conditions: At x = 0, y = 0 At x = 2L, y = 0 Thus, Therefore: C2 = 0 Therefore: C1= ML/12 [ M 2 − x 3 + 6 L x − L + L2 x 48EIL M y ( L) = − L3 + L3 = 0 48EIL y= [ (24e) (24f) 21 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions Slide No. 42 ENES 220 ©Assakkaf Example 5 (cont’d) (b) Finding the maximum deflection: y′ = [ M 1 − 3 x 2 + 12 L x − L + L2 12 EIL ymax when y′ = 0 1 Therefore : − 3 x 3 + 12 x − L + L2 = 0 ⇒ x = L / 3 ( ) ymax = y L / 3 = = M L3 L3 2 ML2 + − = 12 EIL 3 3 3 36 3EI 3ML2 54 EI 22...
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This document was uploaded on 02/04/2014.

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