{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

singularity functions

# 40 lecture 16 beams deformation by singularity

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online