singularity functions

28 lecture 16 beams deformation by singularity

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Unformatted text preview: moment at section 1 due to distributed load alone is 1 x2 L y w0 w1 M w0 = − x x1 -w0 -w1 x2 L w0 w0 3 x − x1 + x − x2 6(x2 − x1 ) 6(x2 − x1 ) + w0 x − x2 3 2 2 Figure 15 14 Slide No. 28 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions ENES 220 ©Assakkaf Moment due to Distributed Loads Note that in Fig. 14, the linearly varying load at any point x ≥ x1 is w0 x1 x2 w w= w0 ( x − x1 ) x2 − x1 The moment of this load for Any point x ≥ x1 is 1 w ( x − x1 ) (x − x1 ) x − x1 = w0 (x − x1 )3 M =− 0 2 x2 − x1 3 6( x2 − x1 ) x From similar triangles : w x − x1 = w0 x2 − x1 LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Singularity Functions Slide No. 29 ENES 220 ©Assakkaf Example 4 Use singularity functions to write the moment equation for the beam shown in Figure 16. Employ this equation to obtain the elastic curve, and find the deflection at x = 10 ft. y 2000 lb/ft Figure 16 E = 29 × 106 psi I = 464 in x 2 4 ft 11 ft 5 ft 15 Slide No. 30 LECTURE 16. BEAMS: DEFORMATION BY SINGU...
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